PREFACE BACKGROUND OBJECTIVES PHILOSOPHY AND GOAL. xvii

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1 PREFACE BACKGROUND Thermodynamics is an exciting and fascinating subject that deals with energy, which is essential for sustenance of life, and thermodynamics has long been an essential part of engineering curricula all over the world. It has a broad application area ranging from microscopic organisms to common household appliances, transportation vehicles, power generation systems, and even philosophy. This introductory book contains sufficient material for two sequential courses in thermodynamics. Students are assumed to have an adequate background in calculus and physics. OBJECTIVES This book is intended for use as a textbook by undergraduate engineering students in their sophomore or junior year, and as a reference book for practicing engineers. The objectives of this text are To cover the basic principles of thermodynamics. To present a wealth of real-world engineering examples to give students a feel for how thermodynamics is applied in engineering practice. To develop an intuitive understanding of thermodynamics by emphasizing the physics and physical arguments. It is our hope that this book, through its careful explanations of concepts and its use of numerous practical examples and figures, helps students develop the necessary skills to bridge the gap between knowledge and the confidence to properly apply knowledge. PHILOSOPHY AND GOAL The philosophy that contributed to the overwhelming popularity of the prior editions of this book has remained unchanged in this edition. Namely, our goal has been to offer an engineering textbook that Communicates directly to the minds of tomorrow s engineers in a simple yet precise manner. Leads students toward a clear understanding and firm grasp of the basic principles of thermodynamics. Encourages creative thinking and development of a deeper understanding and intuitive feel for thermodynamics. Is read by students with interest and enthusiasm rather than being used as an aid to solve problems. xvii

2 xviii Preface Special effort has been made to appeal to students natural curiosity and to help them explore the various facets of the exciting subject area of thermodynamics. The enthusiastic responses we have received from users of prior editions from small colleges to large universities all over the world indicate that our objectives have largely been achieved. It is our philosophy that the best way to learn is by practice. Therefore, special effort is made throughout the book to reinforce material that was presented earlier. Yesterday s engineer spent a major portion of his or her time substituting values into the formulas and obtaining numerical results. However, formula manipulations and number crunching are now being left mainly to computers. Tomorrow s engineer will need a clear understanding and a firm grasp of the basic principles so that he or she can understand even the most complex problems, formulate them, and interpret the results. A conscious effort is made to emphasize these basic principles while also providing students with a perspective of how computational tools are used in engineering practice. The traditional classical, or macroscopic, approach is used throughout the text, with microscopic arguments serving in a supporting role as appropriate. This approach is more in line with students intuition and makes learning the subject matter much easier. NEW IN THIS EDITION All the popular features of the previous editions are retained while new ones are added. With the exception of reorganizing the first law coverage and updating the steam and refrigerant properties, the main body of the text remains largely unchanged. The most significant changes in this fifth edition are highlighted below. EARLY INTRODUCTION OF THE FIRST LAW OF THERMODYNAMICS The first law of thermodynamics is now introduced early in the new Chapter, Energy, Energy Transfer, and General Energy Analysis. This introductory chapter sets the framework of establishing a general understanding of various forms of energy, mechanisms of energy transfer, the concept of energy balance, thermo-economics, energy conversion, and conversion efficiency using familiar settings that involve mostly electrical and mechanical forms of energy. It also exposes students to some exciting real-world applications of thermodynamics early in the course, and helps them establish a sense of the monetary value of energy. SEPARATE COVERAGE OF CLOSED SYSTEMS AND CONTROL VOLUME ENERGY ANALYSES The energy analysis of closed systems is now presented in a separate chapter, Chapter 4, together with the boundary work and the discussion of specific heats for both ideal gases and incompressible substances. The conservation of mass is now covered together with conservation of energy in new Chapter 5. A formal derivation of the general energy equation is also given in this chapter as the Topic of Special Interest. REVISED COVERAGE OF COMPRESSIBLE FLOW The chapter on compressible flow that deals with compressibility effects (now Chapter 17) is greatly revised and expanded. This chapter now includes

3 Preface xix coverage of oblique shocks and flow with heat transfer (Rayleigh flow) with some exciting photographs and extended discussions of shock waves. UPDATED STEAM AND REFRIGERANT-134A TABLES The steam and refrigerant-134a tables are updated using the most current property data from EES. Tables A-4 through A-8 and A-11 through A-13, as well as their counterparts in English units, have all been revised. All the examples and homework problems in the text that involve steam or refrigerant- 134a are also revised to reflect the small changes in steam and refrigerant properties. An added advantage of this update is that students will get the same result when solving problems whether they use steam or refrigerant properties from EES or property tables in the appendices. OVER 300 NEW COMPREHENSIVE PROBLEMS This edition includes over 300 new comprehensive problems that come mostly from industrial applications. Problems whose solutions require parametric investigations, and thus the use of a computer, are identified by a computer-ees icon, as before. CONTENT CHANGES AND REORGANIZATION The noteworthy changes in various chapters are summarized below for those who are familiar with the previous edition. Chapter 1 is greatly revised, and its title is changed to Introduction and Basic Concepts. A new section Density and Specific Gravity and a new subsection The International Temperature Scale of 1990 are added. The sections Forms of Energy and Energy and the Environment are moved to new Chapter, and the Topic of Special Interest Thermodynamic Aspects of Biological Systems is moved to new Chapter 4. The new Chapter Energy, Energy Transfer, and General Energy Analysis mostly consists of the sections Forms of Energy and Energy and the Environment moved from Chapter 1, Energy Transfer by Heat and Energy Transfer by Work, and Mechanical Forms of Energy from Chapter 3, The First Law of Thermodynamics from Chapter 4, and Energy Conversion Efficiencies from Chapter 5. The Topic of Special Interest in this chapter is Mechanisms of Heat Transfer moved from Chapter 3. Chapter 3 Properties of Pure Substance is essentially the previous edition Chapter, except that the last three sections on specific heats are moved to new Chapter 4. Chapter 4 Energy Analysis of Closed Systems consists of Moving Boundary Work from Chapter 3, sections on Specific Heats from Chapter, and Energy Balance for Closed Systems from Chapter 4. Also, the Topic of Special Interest Thermodynamic Aspects of Biological Systems is moved here from Chapter 1. Chapter 5 Mass and Energy Analysis of Control Volumes consists of Mass Balance for Control Volumes and Flow Work and the Energy of a Flowing Fluid from Chapter 3 and the sections on Energy Balance for Steady- and Unsteady-Flow Systems from Chapter 4. The

4 xx Preface Topic of Special Interest Refrigeration and Freezing of Foods is deleted and is replaced by a formal derivation of the General Energy Equation. Chapter 6 The Second Law of Thermodynamics is identical to the previous edition Chapter 5, except the section Energy Conversion Efficiencies is moved to Chapter. Chapters 7 through 15 are essentially identical to the previous edition Chapters 6 through 14, respectively. Chapter 17 Compressible Flow is an updated version of the previous edition Chapter 16. The entire chapter is greatly revised, the section Flow Through Actual Nozzles and Diffusers is deleted, and a new section Duct Flow with Heat Transfer and Negligible Friction (Rayleigh Flow) is added. In Appendices 1 and, the steam and refrigerant-134a tables (Tables 4 through 8 and 11 through 13) are entirely revised, but the table numbers are kept the same. The tables for isentropic compressible flow functions and the normal shock functions (Tables A-3 and A-33) are updated and plots of functions are now included. Also, Rayleigh flow functions are added as Table A-34. Appendix 3 Introduction to EES is moved to the Student Resources DVD that comes packaged free with the text. The conversion factors on the inner cover pages and the physical constants are updated, and some nomenclature symbols are revised. LEARNING TOOLS EMPHASIS ON PHYSICS A distinctive feature of this book is its emphasis on the physical aspects of the subject matter in addition to mathematical representations and manipulations. The authors believe that the emphasis in undergraduate education should remain on developing a sense of underlying physical mechanisms and a mastery of solving practical problems that an engineer is likely to face in the real world. Developing an intuitive understanding should also make the course a more motivating and worthwhile experience for students. EFFECTIVE USE OF ASSOCIATION An observant mind should have no difficulty understanding engineering sciences. After all, the principles of engineering sciences are based on our everyday experiences and experimental observations. Therefore, a physical, intuitive approach is used throughout this text. Frequently, parallels are drawn between the subject matter and students everyday experiences so that they can relate the subject matter to what they already know. The process of cooking, for example, serves as an excellent vehicle to demonstrate the basic principles of thermodynamics. SELF-INSTRUCTING The material in the text is introduced at a level that an average student can follow comfortably. It speaks to students, not over students. In fact, it is self-instructive. The order of coverage is from simple to general. That is, it

5 starts with the simplest case and adds complexities gradually. In this way, the basic principles are repeatedly applied to different systems, and students master how to apply the principles instead of how to simplify a general formula. Noting that the principles of sciences are based on experimental observations, all the derivations in this text are based on physical arguments, and thus they are easy to follow and understand. EXTENSIVE USE OF ARTWORK Figures are important learning tools that help students get the picture, and the text makes very effective use of graphics. The fifth edition of Thermodynamics: An Engineering Approach contains more figures and illustrations than any other book in this category. This edition incorporates an expanded photo program and updated art style. Figures attract attention and stimulate curiosity and interest. Most of the figures in this text are intended to serve as a means of emphasizing some key concepts that would otherwise go unnoticed; some serve as page summaries. The popular cartoon feature Blondie is used to make some important points in a humorous way and also to break the ice and ease the nerves. Who says studying thermodynamics can t be fun? LEARNING OBJECTIVES AND SUMMARIES Each chapter begins with an overview of the material to be covered and chapter-specific learning objectives. A summary is included at the end of each chapter, providing a quick review of basic concepts and important relations, and pointing out the relevance of the material. NUMEROUS WORKED-OUT EXAMPLES WITH A SYSTEMATIC SOLUTIONS PROCEDURE Each chapter contains several worked-out examples that clarify the material and illustrate the use of the basic principles. An intuitive and systematic approach is used in the solution of the example problems, while maintaining an informal conversational style. The problem is first stated, and the objectives are identified. The assumptions are then stated, together with their justifications. The properties needed to solve the problem are listed separately, if appropriate. Numerical values are used together with their units to emphasize that numbers without units are meaningless, and that unit manipulations are as important as manipulating the numerical values with a calculator. The significance of the findings is discussed following the solutions. This approach is also used consistently in the solutions presented in the instructor s solutions manual. A WEALTH OF REAL-WORLD END-OF-CHAPTER PROBLEMS The end-of-chapter problems are grouped under specific topics to make problem selection easier for both instructors and students. Within each group of problems are Concept Questions, indicated by C, to check the students level of understanding of basic concepts. The problems under Review Problems are more comprehensive in nature and are not directly tied to any specific section of a chapter in some cases they require review of material learned in previous chapters. Problems designated as Design and Essay are intended to encourage students to make engineering judgments, to conduct independent exploration of topics of interest, and to communicate Preface xxi

6 xxii Preface their findings in a professional manner. Problems designated by an E are in English units, and SI users can ignore them. Problems with the are solved using EES, and complete solutions together with parametric studies are included on the enclosed DVD. Problems with the are comprehensive in nature and are intended to be solved with a computer, preferably using the EES software that accompanies this text. Several economics- and safety-related problems are incorporated throughout to enhance cost and safety awareness among engineering students. Answers to selected problems are listed immediately following the problem for convenience to students. In addition, to prepare students for the Fundamentals of Engineering Exam (that is becoming more important for the outcome-based ABET 000 criteria) and to facilitate multiple-choice tests, over 00 multiple-choice problems are included in the end-of-chapter problem sets. They are placed under the title Fundamentals of Engineering (FE) Exam Problems for easy recognition. These problems are intended to check the understanding of fundamentals and to help readers avoid common pitfalls. RELAXED SIGN CONVENTION The use of a formal sign convention for heat and work is abandoned as it often becomes counterproductive. A physically meaningful and engaging approach is adopted for interactions instead of a mechanical approach. Subscripts in and out, rather than the plus and minus signs, are used to indicate the directions of interactions. PHYSICALLY MEANINGFUL FORMULAS The physically meaningful forms of the balance equations rather than formulas are used to foster deeper understanding and to avoid a cookbook approach. The mass, energy, entropy, and exergy balances for any system undergoing any process are expressed as Mass balance: Energy balance: Entropy balance: m in m out m system E in E out E system Net energy transfer by heat, work, and mass Change in internal, kinetic, potential, etc., energies S in S out S gen S system Net entropy transfer Entropy Change by heat and mass generation in entropy Exergy balance: X in X out X destroyed X system Net exergy transfer Exergy Change by heat, work, and mass destruction in exergy These relations reinforce the fundamental principles that during an actual process mass and energy are conserved, entropy is generated, and exergy is destroyed. Students are encouraged to use these forms of balances in early chapters after they specify the system, and to simplify them for the particular problem. A more relaxed approach is used in later chapters as students gain mastery.

7 A CHOICE OF SI ALONE OR SI/ENGLISH UNITS In recognition of the fact that English units are still widely used in some industries, both SI and English units are used in this text, with an emphasis on SI. The material in this text can be covered using combined SI/English units or SI units alone, depending on the preference of the instructor. The property tables and charts in the appendices are presented in both units, except the ones that involve dimensionless quantities. Problems, tables, and charts in English units are designated by E after the number for easy recognition, and they can be ignored by SI users. Preface xxiii TOPICS OF SPECIAL INTEREST Most chapters contain a section called Topic of Special Interest where interesting aspects of thermodynamics are discussed. Examples include Thermodynamic Aspects of Biological Systems in Chapter 4, Household Refrigerators in Chapter 6, Second-Law Aspects of Daily Life in Chapter 8, and Saving Fuel and Money by Driving Sensibly in Chapter 9. The topics selected for these sections provide intriguing extensions to thermodynamics, but they can be ignored if desired without a loss in continuity. GLOSSARY OF THERMODYNAMIC TERMS Throughout the chapters, when an important key term or concept is introduced and defined, it appears in boldface type. Fundamental thermodynamic terms and concepts also appear in a glossary located on our accompanying website ( This unique glossary helps to reinforce key terminology and is an excellent learning and review tool for students as they move forward in their study of thermodynamics. In addition, students can test their knowledge of these fundamental terms by using the flash cards and other interactive resources. CONVERSION FACTORS Frequently used conversion factors and physical constants are listed on the inner cover pages of the text for easy reference. SUPPLEMENTS The following supplements are available to the adopters of the book. STUDENT RESOURCES DVD Packaged free with every new copy of the text, this DVD provides a wealth of resources for students including Physical Experiments in Thermodynamics, an Interactive Thermodynamics Tutorial, and EES Software. Physical Experiments in Thermodynamics: A new feature of this book is the addition of Physical Experiments in Thermodynamics created by Ronald Mullisen of the Mechanical Engineering Department at California Polytechnic State University (Cal Poly), San Luis Obispo. At appropriate places in the margins of Chapters 1, 3, and 4, photos with captions show physical experiments that directly relate to material covered on that page. The captions

8 xxiv Preface refer the reader to end-of-chapter problems that give a brief description of the experiments. These experiments cover thermodynamic properties, thermodynamic processes, and thermodynamic laws. The Student Resources DVD contains complete coverage of the nine experiments. Each experiment contains a video clip, a complete write-up including historical background, and actual data (usually in an Excel file). The results are also provided on the website that accompanies the text, and they are password protected for instructor use. After viewing the video and reading the write-up, the student will be ready to reduce the data and obtain results that directly connect with material presented in the chapters. For all of the experiments the final results are compared against published information. Most of the experiments give final results that come within 10 percent or closer to these published values. Interactive Thermodynamics Tutorial: Also included on the Student Resources DVD is the Interactive Thermodynamics Tutorial developed by Ed Anderson of Texas Tech University. The revised tutorial is now tied directly to the text with an icon to indicate when students should refer to the tutorial to further explore specific topics such as energy balance and isentropic processes. Engineering Equation Solver (EES): Developed by Sanford Klein and William Beckman from the University of Wisconsin Madison, this software combines equation-solving capability and engineering property data. EES can do optimization, parametric analysis, and linear and nonlinear regression, and provides publication-quality plotting capabilities. Thermodynamics and transport properties for air, water, and many other fluids are built in, and EES allows the user to enter property data or functional relationships. ONLINE LEARNING CENTER (OLC) Web support is provided for the book on our Online Learning Center at Visit this robust site for book and supplement information, errata, author information, and further resources for instructors and students. INSTRUCTOR S RESOURCE CD-ROM (AVAILABLE TO INSTRUCTORS ONLY) This CD, available to instructors only, offers a wide range of classroom preparation and presentation resources including the solutions manual in PDF files by chapter, all text chapters and appendices as downloadable PDF files, and all text figures in JPEG format. COSMOS CD-ROM (COMPLETE ONLINE SOLUTIONS MANUAL ORGANIZATION SYSTEM) (AVAILABLE TO INSTRUCTORS ONLY) This CD, available to instructors only, provides electronic solutions delivered via our database management tool. McGraw-Hill s COSMOS allows instructors to streamline the creation of assignments, quizzes, and tests by using problems and solutions from the textbook as well as their own custom material.

9 Chapter 9 GAS POWER CYCLES Two important areas of application for thermodynamics are power generation and refrigeration. Both are usually accomplished by systems that operate on a thermodynamic cycle. Thermodynamic cycles can be divided into two general categories: power cycles, which are discussed in this chapter and Chap. 10, and refrigeration cycles, which are discussed in Chap. 11. The devices or systems used to produce a net power output are often called engines, and the thermodynamic cycles they operate on are called power cycles. The devices or systems used to produce a refrigeration effect are called refrigerators, air conditioners, or heat pumps, and the cycles they operate on are called refrigeration cycles. Thermodynamic cycles can also be categorized as gas cycles and vapor cycles, depending on the phase of the working fluid. In gas cycles, the working fluid remains in the gaseous phase throughout the entire cycle, whereas in vapor cycles the working fluid exists in the vapor phase during one part of the cycle and in the liquid phase during another part. Thermodynamic cycles can be categorized yet another way: closed and open cycles. In closed cycles, the working fluid is returned to the initial state at the end of the cycle and is recirculated. In open cycles, the working fluid is renewed at the end of each cycle instead of being recirculated. In automobile engines, the combustion gases are exhausted and replaced by fresh air fuel mixture at the end of each cycle. The engine operates on a mechanical cycle, but the working fluid does not go through a complete thermodynamic cycle. Heat engines are categorized as internal combustion and external combustion engines, depending on how the heat is supplied to the working fluid. In external combustion engines (such as steam power plants), heat is supplied to the working fluid from an external source such as a furnace, a geothermal well, a nuclear reactor, or even the sun. In internal combustion engines (such as automobile engines), this is done by burning the fuel within the system boundaries. In this chapter, various gas power cycles are analyzed under some simplifying assumptions. Objectives The objectives of Chapter 9 are to: Evaluate the performance of gas power cycles for which the working fluid remains a gas throughout the entire cycle. Develop simplifying assumptions applicable to gas power cycles. Review the operation of reciprocating engines. Analyze both closed and open gas power cycles. Solve problems based on the Otto, Diesel, Stirling, and Ericsson cycles. Solve problems based on the Brayton cycle; the Brayton cycle with regeneration; and the Brayton cycle with intercooling, reheating, and regeneration. Analyze jet-propulsion cycles. Identify simplifying assumptions for second-law analysis of gas power cycles. Perform second-law analysis of gas power cycles. 487

10 488 Thermodynamics Potato WATER 175ºC OVEN ACTUAL IDEAL FIGURE 9 1 Modeling is a powerful engineering tool that provides great insight and simplicity at the expense of some loss in accuracy. P Actual cycle Ideal cycle FIGURE 9 The analysis of many complex processes can be reduced to a manageable level by utilizing some idealizations. FIGURE 9 3 Care should be exercised in the interpretation of the results from ideal cycles. Reprinted with special permission of King Features Syndicate. v 9 1 BASIC CONSIDERATIONS IN THE ANALYSIS OF POWER CYCLES Most power-producing devices operate on cycles, and the study of power cycles is an exciting and important part of thermodynamics. The cycles encountered in actual devices are difficult to analyze because of the presence of complicating effects, such as friction, and the absence of sufficient time for establishment of the equilibrium conditions during the cycle. To make an analytical study of a cycle feasible, we have to keep the complexities at a manageable level and utilize some idealizations (Fig. 9 1). When the actual cycle is stripped of all the internal irreversibilities and complexities, we end up with a cycle that resembles the actual cycle closely but is made up totally of internally reversible processes. Such a cycle is called an ideal cycle (Fig. 9 ). A simple idealized model enables engineers to study the effects of the major parameters that dominate the cycle without getting bogged down in the details. The cycles discussed in this chapter are somewhat idealized, but they still retain the general characteristics of the actual cycles they represent. The conclusions reached from the analysis of ideal cycles are also applicable to actual cycles. The thermal efficiency of the Otto cycle, the ideal cycle for spark-ignition automobile engines, for example, increases with the compression ratio. This is also the case for actual automobile engines. The numerical values obtained from the analysis of an ideal cycle, however, are not necessarily representative of the actual cycles, and care should be exercised in their interpretation (Fig. 9 3). The simplified analysis presented in this chapter for various power cycles of practical interest may also serve as the starting point for a more in-depth study. Heat engines are designed for the purpose of converting thermal energy to work, and their performance is expressed in terms of the thermal efficiency h th, which is the ratio of the net work produced by the engine to the total heat input: h th W net Q in or h th w net q in (9 1) Recall that heat engines that operate on a totally reversible cycle, such as the Carnot cycle, have the highest thermal efficiency of all heat engines operating between the same temperature levels. That is, nobody can develop a cycle more efficient than the Carnot cycle. Then the following question arises naturally: If the Carnot cycle is the best possible cycle, why do we not use it as the model cycle for all the heat engines instead of bothering with several so-called ideal cycles? The answer to this question is hardwarerelated. Most cycles encountered in practice differ significantly from the Carnot cycle, which makes it unsuitable as a realistic model. Each ideal cycle discussed in this chapter is related to a specific work-producing device and is an idealized version of the actual cycle. The ideal cycles are internally reversible, but, unlike the Carnot cycle, they are not necessarily externally reversible. That is, they may involve irreversibilities external to the system such as heat transfer through a finite temperature difference. Therefore, the thermal efficiency of an ideal cycle, in general, is less than that of a totally reversible cycle operating between the

11 Chapter FIGURE 9 4 An automotive engine with the combustion chamber exposed. Courtesy of General Motors same temperature limits. However, it is still considerably higher than the thermal efficiency of an actual cycle because of the idealizations utilized (Fig. 9 4). The idealizations and simplifications commonly employed in the analysis of power cycles can be summarized as follows: 1. The cycle does not involve any friction. Therefore, the working fluid does not experience any pressure drop as it flows in pipes or devices such as heat exchangers.. All expansion and compression processes take place in a quasiequilibrium manner. 3. The pipes connecting the various components of a system are well insulated, and heat transfer through them is negligible. Neglecting the changes in kinetic and potential energies of the working fluid is another commonly utilized simplification in the analysis of power cycles. This is a reasonable assumption since in devices that involve shaft work, such as turbines, compressors, and pumps, the kinetic and potential energy terms are usually very small relative to the other terms in the energy equation. Fluid velocities encountered in devices such as condensers, boilers, and mixing chambers are typically low, and the fluid streams experience little change in their velocities, again making kinetic energy changes negligible. The only devices where the changes in kinetic energy are significant are the nozzles and diffusers, which are specifically designed to create large changes in velocity. In the preceding chapters, property diagrams such as the P-v and T-s diagrams have served as valuable aids in the analysis of thermodynamic processes. On both the P-v and T-s diagrams, the area enclosed by the process curves of a cycle represents the net work produced during the cycle (Fig. 9 5), which is also equivalent to the net heat transfer for that cycle.

12 490 Thermodynamics P T 3 3 FIGURE 9 5 On both P-v and T-s diagrams, the area enclosed by the process curve represents the net work of the cycle. 1 w net 4 v 1 w net 4 s P T T H T L 1 Isentropic Isentropic 4 q in 1 4 q in TH = const. q out q out Isentropic T L = const. 3 Isentropic FIGURE 9 6 P-v and T-s diagrams of a Carnot cycle. 3 v s The T-s diagram is particularly useful as a visual aid in the analysis of ideal power cycles. An ideal power cycle does not involve any internal irreversibilities, and so the only effect that can change the entropy of the working fluid during a process is heat transfer. On a T-s diagram, a heat-addition process proceeds in the direction of increasing entropy, a heat-rejection process proceeds in the direction of decreasing entropy, and an isentropic (internally reversible, adiabatic) process proceeds at constant entropy. The area under the process curve on a T-s diagram represents the heat transfer for that process. The area under the heat addition process on a T-s diagram is a geometric measure of the total heat supplied during the cycle q in, and the area under the heat rejection process is a measure of the total heat rejected q out. The difference between these two (the area enclosed by the cyclic curve) is the net heat transfer, which is also the net work produced during the cycle. Therefore, on a T-s diagram, the ratio of the area enclosed by the cyclic curve to the area under the heat-addition process curve represents the thermal efficiency of the cycle. Any modification that increases the ratio of these two areas will also increase the thermal efficiency of the cycle. Although the working fluid in an ideal power cycle operates on a closed loop, the type of individual processes that comprises the cycle depends on the individual devices used to execute the cycle. In the Rankine cycle, which is the ideal cycle for steam power plants, the working fluid flows through a series of steady-flow devices such as the turbine and condenser, whereas in the Otto cycle, which is the ideal cycle for the spark-ignition automobile engine, the working fluid is alternately expanded and compressed in a piston cylinder device. Therefore, equations pertaining to steady-flow systems should be used in the analysis of the Rankine cycle, and equations pertaining to closed systems should be used in the analysis of the Otto cycle. 9 THE CARNOT CYCLE AND ITS VALUE IN ENGINEERING The Carnot cycle is composed of four totally reversible processes: isothermal heat addition, isentropic expansion, isothermal heat rejection, and isentropic compression. The P-v and T-s diagrams of a Carnot cycle are replotted in Fig The Carnot cycle can be executed in a closed system (a piston cylinder device) or a steady-flow system (utilizing two turbines and two compressors, as shown in Fig. 9 7), and either a gas or a vapor can

13 Chapter Isothermal compressor Isentropic compressor Isothermal turbine Isentropic turbine w net q out 4 q in FIGURE 9 7 A steady-flow Carnot engine. be utilized as the working fluid. The Carnot cycle is the most efficient cycle that can be executed between a heat source at temperature T H and a sink at temperature T L, and its thermal efficiency is expressed as h th,carnot 1 T L T H (9 ) Reversible isothermal heat transfer is very difficult to achieve in reality because it would require very large heat exchangers and it would take a very long time (a power cycle in a typical engine is completed in a fraction of a second). Therefore, it is not practical to build an engine that would operate on a cycle that closely approximates the Carnot cycle. The real value of the Carnot cycle comes from its being a standard against which the actual or the ideal cycles can be compared. The thermal efficiency of the Carnot cycle is a function of the sink and source temperatures only, and the thermal efficiency relation for the Carnot cycle (Eq. 9 ) conveys an important message that is equally applicable to both ideal and actual cycles: Thermal efficiency increases with an increase in the average temperature at which heat is supplied to the system or with a decrease in the average temperature at which heat is rejected from the system. The source and sink temperatures that can be used in practice are not without limits, however. The highest temperature in the cycle is limited by the maximum temperature that the components of the heat engine, such as the piston or the turbine blades, can withstand. The lowest temperature is limited by the temperature of the cooling medium utilized in the cycle such as a lake, a river, or the atmospheric air. EXAMPLE 9 1 Derivation of the Efficiency of the Carnot Cycle Show that the thermal efficiency of a Carnot cycle operating between the temperature limits of T H and T L is solely a function of these two temperatures and is given by Eq. 9. Solution It is to be shown that the efficiency of a Carnot cycle depends on the source and sink temperatures alone.

14 49 Thermodynamics T T H T L q 1 in 4 3 q out s 1 = s 4 s = s 3 FIGURE 9 8 T-s diagram for Example 9 1. s Analysis The T-s diagram of a Carnot cycle is redrawn in Fig All four processes that comprise the Carnot cycle are reversible, and thus the area under each process curve represents the heat transfer for that process. Heat is transferred to the system during process 1- and rejected during process 3-4. Therefore, the amount of heat input and heat output for the cycle can be expressed as q in T H 1s s 1 and q out T L 1s 3 s 4 T L 1s s 1 since processes -3 and 4-1 are isentropic, and thus s s 3 and s 4 s 1. Substituting these into Eq. 9 1, we see that the thermal efficiency of a Carnot cycle is h th w net q in 1 q out q in 1 T L 1s s 1 T H 1s s 1 1 T L T H Discussion Notice that the thermal efficiency of a Carnot cycle is independent of the type of the working fluid used (an ideal gas, steam, etc.) or whether the cycle is executed in a closed or steady-flow system. AIR FUEL AIR Combustion chamber (a) Actual HEAT Heating section (b) Ideal COMBUSTION PRODUCTS AIR FIGURE 9 9 The combustion process is replaced by a heat-addition process in ideal cycles. 9 3 AIR-STANDARD ASSUMPTIONS In gas power cycles, the working fluid remains a gas throughout the entire cycle. Spark-ignition engines, diesel engines, and conventional gas turbines are familiar examples of devices that operate on gas cycles. In all these engines, energy is provided by burning a fuel within the system boundaries. That is, they are internal combustion engines. Because of this combustion process, the composition of the working fluid changes from air and fuel to combustion products during the course of the cycle. However, considering that air is predominantly nitrogen that undergoes hardly any chemical reactions in the combustion chamber, the working fluid closely resembles air at all times. Even though internal combustion engines operate on a mechanical cycle (the piston returns to its starting position at the end of each revolution), the working fluid does not undergo a complete thermodynamic cycle. It is thrown out of the engine at some point in the cycle (as exhaust gases) instead of being returned to the initial state. Working on an open cycle is the characteristic of all internal combustion engines. The actual gas power cycles are rather complex. To reduce the analysis to a manageable level, we utilize the following approximations, commonly known as the air-standard assumptions: 1. The working fluid is air, which continuously circulates in a closed loop and always behaves as an ideal gas.. All the processes that make up the cycle are internally reversible. 3. The combustion process is replaced by a heat-addition process from an external source (Fig. 9 9). 4. The exhaust process is replaced by a heat-rejection process that restores the working fluid to its initial state. Another assumption that is often utilized to simplify the analysis even more is that air has constant specific heats whose values are determined at

15 room temperature (5 C, or 77 F). When this assumption is utilized, the air-standard assumptions are called the cold-air-standard assumptions. A cycle for which the air-standard assumptions are applicable is frequently referred to as an air-standard cycle. The air-standard assumptions previously stated provide considerable simplification in the analysis without significantly deviating from the actual cycles. This simplified model enables us to study qualitatively the influence of major parameters on the performance of the actual engines. 9 4 AN OVERVIEW OF RECIPROCATING ENGINES Despite its simplicity, the reciprocating engine (basically a piston cylinder device) is one of the rare inventions that has proved to be very versatile and to have a wide range of applications. It is the powerhouse of the vast majority of automobiles, trucks, light aircraft, ships, and electric power generators, as well as many other devices. The basic components of a reciprocating engine are shown in Fig The piston reciprocates in the cylinder between two fixed positions called the top dead center (TDC) the position of the piston when it forms the smallest volume in the cylinder and the bottom dead center (BDC) the position of the piston when it forms the largest volume in the cylinder. The distance between the TDC and the BDC is the largest distance that the piston can travel in one direction, and it is called the stroke of the engine. The diameter of the piston is called the bore. The air or air fuel mixture is drawn into the cylinder through the intake valve, and the combustion products are expelled from the cylinder through the exhaust valve. The minimum volume formed in the cylinder when the piston is at TDC is called the clearance volume (Fig. 9 11). The volume displaced by the piston as it moves between TDC and BDC is called the displacement volume. The ratio of the maximum volume formed in the cylinder to the minimum (clearance) volume is called the compression ratio r of the engine: (9 3) Notice that the compression ratio is a volume ratio and should not be confused with the pressure ratio. Another term frequently used in conjunction with reciprocating engines is the mean effective pressure (MEP). It is a fictitious pressure that, if it acted on the piston during the entire power stroke, would produce the same amount of net work as that produced during the actual cycle (Fig. 9 1). That is, or W net MEP Piston area Stroke MEP Displacement volume MEP r V max V min V BDC V TDC W net w net 1kPa V max V min v max v min (9 4) The mean effective pressure can be used as a parameter to compare the performances of reciprocating engines of equal size. The engine with a larger value of MEP delivers more net work per cycle and thus performs better. Intake valve Bore Chapter Exhaust valve Stroke TDC BDC FIGURE 9 10 Nomenclature for reciprocating engines. TDC BDC (a) Displacement (b) Clearance volume volume FIGURE 9 11 Displacement and clearance volumes of a reciprocating engine.

16 494 Thermodynamics P MEP W net = MEP(V max V min ) V min V max V TDC W net BDC FIGURE 9 1 The net work output of a cycle is equivalent to the product of the mean effective pressure and the displacement volume. Reciprocating engines are classified as spark-ignition (SI) engines or compression-ignition (CI) engines, depending on how the combustion process in the cylinder is initiated. In SI engines, the combustion of the air fuel mixture is initiated by a spark plug. In CI engines, the air fuel mixture is self-ignited as a result of compressing the mixture above its selfignition temperature. In the next two sections, we discuss the Otto and Diesel cycles, which are the ideal cycles for the SI and CI reciprocating engines, respectively. 9 5 OTTO CYCLE: THE IDEAL CYCLE FOR SPARK-IGNITION ENGINES The Otto cycle is the ideal cycle for spark-ignition reciprocating engines. It is named after Nikolaus A. Otto, who built a successful four-stroke engine in 1876 in Germany using the cycle proposed by Frenchman Beau de Rochas in 186. In most spark-ignition engines, the piston executes four complete strokes (two mechanical cycles) within the cylinder, and the crankshaft completes two revolutions for each thermodynamic cycle. These engines are called four-stroke internal combustion engines. A schematic of each stroke as well as a P-v diagram for an actual four-stroke spark-ignition engine is given in Fig. 9 13(a). End of combustion P Exhaust gases Air fuel mixture Expansion Ignition Exhaust valve opens Intake valve opens Compression Exhaust Air fuel mixture P atm P TDC 3 Intake BDC v Compression stroke Power (expansion) stroke (a) Actual four-stroke spark-ignition engine q in Exhaust stroke Intake stroke q out q in AIR () AIR () (3) AIR (3) AIR Isentropic Isentropic 4 1 q out TDC BDC v Isentropic compression (1) (b) Ideal Otto cycle v = const. heat addition Isentropic expansion (4) v = const. heat rejection (4) (1) FIGURE 9 13 Actual and ideal cycles in spark-ignition engines and their P-v diagrams.

17 Initially, both the intake and the exhaust valves are closed, and the piston is at its lowest position (BDC). During the compression stroke, the piston moves upward, compressing the air fuel mixture. Shortly before the piston reaches its highest position (TDC), the spark plug fires and the mixture ignites, increasing the pressure and temperature of the system. The high-pressure gases force the piston down, which in turn forces the crankshaft to rotate, producing a useful work output during the expansion or power stroke. At the end of this stroke, the piston is at its lowest position (the completion of the first mechanical cycle), and the cylinder is filled with combustion products. Now the piston moves upward one more time, purging the exhaust gases through the exhaust valve (the exhaust stroke), and down a second time, drawing in fresh air fuel mixture through the intake valve (the intake stroke). Notice that the pressure in the cylinder is slightly above the atmospheric value during the exhaust stroke and slightly below during the intake stroke. In two-stroke engines, all four functions described above are executed in just two strokes: the power stroke and the compression stroke. In these engines, the crankcase is sealed, and the outward motion of the piston is used to slightly pressurize the air fuel mixture in the crankcase, as shown in Fig Also, the intake and exhaust valves are replaced by openings in the lower portion of the cylinder wall. During the latter part of the power stroke, the piston uncovers first the exhaust port, allowing the exhaust gases to be partially expelled, and then the intake port, allowing the fresh air fuel mixture to rush in and drive most of the remaining exhaust gases out of the cylinder. This mixture is then compressed as the piston moves upward during the compression stroke and is subsequently ignited by a spark plug. The two-stroke engines are generally less efficient than their four-stroke counterparts because of the incomplete expulsion of the exhaust gases and the partial expulsion of the fresh air fuel mixture with the exhaust gases. However, they are relatively simple and inexpensive, and they have high power-to-weight and power-to-volume ratios, which make them suitable for applications requiring small size and weight such as for motorcycles, chain saws, and lawn mowers (Fig. 9 15). Advances in several technologies such as direct fuel injection, stratified charge combustion, and electronic controls brought about a renewed interest in two-stroke engines that can offer high performance and fuel economy while satisfying the stringent emission requirements. For a given weight and displacement, a well-designed two-stroke engine can provide significantly more power than its four-stroke counterpart because two-stroke engines produce power on every engine revolution instead of every other one. In the new two-stroke engines, the highly atomized fuel spray that is injected into the combustion chamber toward the end of the compression stroke burns much more completely. The fuel is sprayed after the exhaust valve is closed, which prevents unburned fuel from being ejected into the atmosphere. With stratified combustion, the flame that is initiated by igniting a small amount of the rich fuel air mixture near the spark plug propagates through the combustion chamber filled with a much leaner mixture, and this results in much cleaner combustion. Also, the advances in electronics have made it possible to ensure the optimum operation under varying engine load and speed conditions. Exhaust port Chapter SEE TUTORIAL CH. 9, SEC. ON THE DVD. Crankcase Spark plug FIGURE 9 14 Schematic of a two-stroke reciprocating engine. Intake port Fuel air mixture FIGURE 9 15 Two-stroke engines are commonly used in motorcycles and lawn mowers. Vol. 6/PhotoDisc INTERACTIVE TUTORIAL

18 496 Thermodynamics T 1 v = const. q in v = const. 3 4 q out FIGURE 9 16 T-s diagram of the ideal Otto cycle. s Major car companies have research programs underway on two-stroke engines which are expected to make a comeback in the future. The thermodynamic analysis of the actual four-stroke or two-stroke cycles described is not a simple task. However, the analysis can be simplified significantly if the air-standard assumptions are utilized. The resulting cycle, which closely resembles the actual operating conditions, is the ideal Otto cycle. It consists of four internally reversible processes: 1- Isentropic compression -3 Constant-volume heat addition 3-4 Isentropic expansion 4-1 Constant-volume heat rejection The execution of the Otto cycle in a piston cylinder device together with a P-v diagram is illustrated in Fig. 9 13b. The T-s diagram of the Otto cycle is given in Fig The Otto cycle is executed in a closed system, and disregarding the changes in kinetic and potential energies, the energy balance for any of the processes is expressed, on a unit-mass basis, as (9 5) No work is involved during the two heat transfer processes since both take place at constant volume. Therefore, heat transfer to and from the working fluid can be expressed as and (9 6a) (9 6b) Then the thermal efficiency of the ideal Otto cycle under the cold air standard assumptions becomes Processes 1- and 3-4 are isentropic, and v v 3 and v 4 v 1. Thus, (9 7) Substituting these equations into the thermal efficiency relation and simplifying give where h th,otto w net q in 1q in q out 1w in w out u 1kJ>kg q in u 3 u c v 1T 3 T q out u 4 u 1 c v 1T 4 T 1 1 q out 1 T 4 T 1 1 T 1 1T 4 >T 1 1 q in T 3 T T 1T 3 >T 1 T 1 a v k 1 b a v k 1 3 b T 4 T v 1 v 4 T 3 h th,otto 1 1 r k 1 (9 8) r V max V 1 v 1 (9 9) V min V v is the compression ratio and k is the specific heat ratio c p /c v. Equation 9 8 shows that under the cold-air-standard assumptions, the thermal efficiency of an ideal Otto cycle depends on the compression ratio of the engine and the specific heat ratio of the working fluid. The thermal efficiency of the ideal Otto cycle increases with both the compression ratio

19 and the specific heat ratio. This is also true for actual spark-ignition internal combustion engines. A plot of thermal efficiency versus the compression ratio is given in Fig for k 1.4, which is the specific heat ratio value of air at room temperature. For a given compression ratio, the thermal efficiency of an actual spark-ignition engine is less than that of an ideal Otto cycle because of the irreversibilities, such as friction, and other factors such as incomplete combustion. We can observe from Fig that the thermal efficiency curve is rather steep at low compression ratios but flattens out starting with a compression ratio value of about 8. Therefore, the increase in thermal efficiency with the compression ratio is not as pronounced at high compression ratios. Also, when high compression ratios are used, the temperature of the air fuel mixture rises above the autoignition temperature of the fuel (the temperature at which the fuel ignites without the help of a spark) during the combustion process, causing an early and rapid burn of the fuel at some point or points ahead of the flame front, followed by almost instantaneous inflammation of the end gas. This premature ignition of the fuel, called autoignition, produces an audible noise, which is called engine knock. Autoignition in spark-ignition engines cannot be tolerated because it hurts performance and can cause engine damage. The requirement that autoignition not be allowed places an upper limit on the compression ratios that can be used in sparkignition internal combustion engines. Improvement of the thermal efficiency of gasoline engines by utilizing higher compression ratios (up to about 1) without facing the autoignition problem has been made possible by using gasoline blends that have good antiknock characteristics, such as gasoline mixed with tetraethyl lead. Tetraethyl lead had been added to gasoline since the 190s because it is an inexpensive method of raising the octane rating, which is a measure of the engine knock resistance of a fuel. Leaded gasoline, however, has a very undesirable side effect: it forms compounds during the combustion process that are hazardous to health and pollute the environment. In an effort to combat air pollution, the government adopted a policy in the mid-1970s that resulted in the eventual phase-out of leaded gasoline. Unable to use lead, the refiners developed other techniques to improve the antiknock characteristics of gasoline. Most cars made since 1975 have been designed to use unleaded gasoline, and the compression ratios had to be lowered to avoid engine knock. The ready availability of high octane fuels made it possible to raise the compression ratios again in recent years. Also, owing to the improvements in other areas (reduction in overall automobile weight, improved aerodynamic design, etc.), today s cars have better fuel economy and consequently get more miles per gallon of fuel. This is an example of how engineering decisions involve compromises, and efficiency is only one of the considerations in final design. The second parameter affecting the thermal efficiency of an ideal Otto cycle is the specific heat ratio k. For a given compression ratio, an ideal Otto cycle using a monatomic gas (such as argon or helium, k 1.667) as the working fluid will have the highest thermal efficiency. The specific heat ratio k, and thus the thermal efficiency of the ideal Otto cycle, decreases as the molecules of the working fluid get larger (Fig. 9 18). At room temperature it is 1.4 for air, 1.3 for carbon dioxide, and 1. for ethane. The working η th,otto FIGURE 9 17 Chapter Typical compression ratios for gasoline engines Compression ratio, r Thermal efficiency of the ideal Otto cycle as a function of compression ratio (k 1.4). η th,otto k = k = 1.4 k = Compression ratio, r FIGURE 9 18 The thermal efficiency of the Otto cycle increases with the specific heat ratio k of the working fluid.

20 498 Thermodynamics fluid in actual engines contains larger molecules such as carbon dioxide, and the specific heat ratio decreases with temperature, which is one of the reasons that the actual cycles have lower thermal efficiencies than the ideal Otto cycle. The thermal efficiencies of actual spark-ignition engines range from about 5 to 30 percent. EXAMPLE 9 The Ideal Otto Cycle P, kpa q in Isentropic 1 v = v 3 = v 8 1 Isentropic q out FIGURE 9 19 P-v diagram for the Otto cycle discussed in Example v 1 = v 4 v An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 100 kpa and 17 C, and 800 kj/kg of heat is transferred to air during the constant-volume heat-addition process. Accounting for the variation of specific heats of air with temperature, determine (a) the maximum temperature and pressure that occur during the cycle, (b) the net work output, (c) the thermal efficiency, and (d ) the mean effective pressure for the cycle. Solution An ideal Otto cycle is considered. The maximum temperature and pressure, the net work output, the thermal efficiency, and the mean effective pressure are to be determined. Assumptions 1 The air-standard assumptions are applicable. Kinetic and potential energy changes are negligible. 3 The variation of specific heats with temperature is to be accounted for. Analysis The P-v diagram of the ideal Otto cycle described is shown in Fig We note that the air contained in the cylinder forms a closed system. (a) The maximum temperature and pressure in an Otto cycle occur at the end of the constant-volume heat-addition process (state 3). But first we need to determine the temperature and pressure of air at the end of the isentropic compression process (state ), using data from Table A 17: Process 1- (isentropic compression of an ideal gas): v r v r1 v v 1 1 r T 1 90 K S u kj>kg v r S v r v r1 r S T 65.4 K u kj>kg P v T P 1v 1 T 1 S P P 1 a T T 1 ba v 1 v b Process -3 (constant-volume heat addition): 1100 kpa a 65.4 K b kpa 90 K q in u 3 u 800 kj>kg u kj>kg u kj>kg S T K v r

21 Chapter P 3 v 3 T 3 P v T S P 3 P a T 3 T ba v v 3 b (b) The net work output for the cycle is determined either by finding the boundary (PdV) work involved in each process by integration and adding them or by finding the net heat transfer that is equivalent to the net work done during the cycle. We take the latter approach. However, first we need to find the internal energy of the air at state 4: Process 3-4 (isentropic expansion of an ideal gas): Process 4-1 (constant-volume heat rejection): MPa a K b MPa 65.4 K v r4 v r3 v 4 v 3 r S v r4 rv r S T K u kj>kg Thus, q out u 1 u 4 S q out u 4 u 1 q out kj>kg w net q net q in q out kj/kg (c) The thermal efficiency of the cycle is determined from its definition: Under the cold-air-standard assumptions (constant specific heat values at room temperature), the thermal efficiency would be (Eq. 9 8) which is considerably different from the value obtained above. Therefore, care should be exercised in utilizing the cold-air-standard assumptions. (d ) The mean effective pressure is determined from its definition, Eq. 9 4: where Thus, v 1 RT 1 P 1 MEP h th w net q in h th,otto 1 1 r k 1 1 r 1 k or 56.5% MEP kj>kg 0.53 or 5.3% 800 kj>kg w net v 1 v kpa # m 3 >kg # K190 K 100 kpa kj>kg w net v 1 v 1 >r m 3 >kg a 1 kpa # m3 1 kj w net v >r 0.83 m 3 >kg b 574 kpa Discussion Note that a constant pressure of 574 kpa during the power stroke would produce the same net work output as the entire cycle.

22 500 Thermodynamics Spark plug Air fuel mixture Gasoline engine Spark Fuel injector AIR Fuel spray Diesel engine FIGURE 9 0 In diesel engines, the spark plug is replaced by a fuel injector, and only air is compressed during the compression process. P T q in 3 1 v Isentropic Isentropic (a) P- v diagram P = constant v = constant (b) T-s diagram q in q out FIGURE 9 1 T-s and P-v diagrams for the ideal Diesel cycle. q out v s 9 6 DIESEL CYCLE: THE IDEAL CYCLE FOR COMPRESSION-IGNITION ENGINES The Diesel cycle is the ideal cycle for CI reciprocating engines. The CI engine, first proposed by Rudolph Diesel in the 1890s, is very similar to the SI engine discussed in the last section, differing mainly in the method of initiating combustion. In spark-ignition engines (also known as gasoline engines), the air fuel mixture is compressed to a temperature that is below the autoignition temperature of the fuel, and the combustion process is initiated by firing a spark plug. In CI engines (also known as diesel engines), the air is compressed to a temperature that is above the autoignition temperature of the fuel, and combustion starts on contact as the fuel is injected into this hot air. Therefore, the spark plug and carburetor are replaced by a fuel injector in diesel engines (Fig. 9 0). In gasoline engines, a mixture of air and fuel is compressed during the compression stroke, and the compression ratios are limited by the onset of autoignition or engine knock. In diesel engines, only air is compressed during the compression stroke, eliminating the possibility of autoignition. Therefore, diesel engines can be designed to operate at much higher compression ratios, typically between 1 and 4. Not having to deal with the problem of autoignition has another benefit: many of the stringent requirements placed on the gasoline can now be removed, and fuels that are less refined (thus less expensive) can be used in diesel engines. The fuel injection process in diesel engines starts when the piston approaches TDC and continues during the first part of the power stroke. Therefore, the combustion process in these engines takes place over a longer interval. Because of this longer duration, the combustion process in the ideal Diesel cycle is approximated as a constant-pressure heat-addition process. In fact, this is the only process where the Otto and the Diesel cycles differ. The remaining three processes are the same for both ideal cycles. That is, process 1- is isentropic compression, 3-4 is isentropic expansion, and 4-1 is constant-volume heat rejection. The similarity between the two cycles is also apparent from the P-v and T-s diagrams of the Diesel cycle, shown in Fig Noting that the Diesel cycle is executed in a piston cylinder device, which forms a closed system, the amount of heat transferred to the working fluid at constant pressure and rejected from it at constant volume can be expressed as and q in w b,out u 3 u S q in P 1v 3 v 1u 3 u (9 10a) (9 10b) Then the thermal efficiency of the ideal Diesel cycle under the cold-airstandard assumptions becomes h th,diesel w net q in h 3 h c p 1T 3 T q out u 1 u 4 S q out u 4 u 1 c v 1T 4 T 1 1 q out q in 1 T 4 T 1 k 1T 3 T 1 T 1 1T 4 >T 1 1 kt 1T 3 >T 1

23 We now define a new quantity, the cutoff ratio r c, as the ratio of the cylinder volumes after and before the combustion process: r c V 3 v 3 (9 11) V v Utilizing this definition and the isentropic ideal-gas relations for processes 1- and 3-4, we see that the thermal efficiency relation reduces to h th,diesel 1 1 r c r k c 1 k 1 k 1r c 1 d (9 1) where r is the compression ratio defined by Eq Looking at Eq. 9 1 carefully, one would notice that under the cold-air-standard assumptions, the efficiency of a Diesel cycle differs from the efficiency of an Otto cycle by the quantity in the brackets. This quantity is always greater than 1. Therefore, h th,otto 7 h th,diesel (9 13) when both cycles operate on the same compression ratio. Also, as the cutoff ratio decreases, the efficiency of the Diesel cycle increases (Fig. 9 ). For the limiting case of r c 1, the quantity in the brackets becomes unity (can you prove it?), and the efficiencies of the Otto and Diesel cycles become identical. Remember, though, that diesel engines operate at much higher compression ratios and thus are usually more efficient than the spark-ignition (gasoline) engines. The diesel engines also burn the fuel more completely since they usually operate at lower revolutions per minute and the air fuel mass ratio is much higher than spark-ignition engines. Thermal efficiencies of large diesel engines range from about 35 to 40 percent. The higher efficiency and lower fuel costs of diesel engines make them attractive in applications requiring relatively large amounts of power, such as in locomotive engines, emergency power generation units, large ships, and heavy trucks. As an example of how large a diesel engine can be, a 1- cylinder diesel engine built in 1964 by the Fiat Corporation of Italy had a normal power output of 5,00 hp (18.8 MW) at 1 rpm, a cylinder bore of 90 cm, and a stroke of 91 cm. Approximating the combustion process in internal combustion engines as a constant-volume or a constant-pressure heat-addition process is overly simplistic and not quite realistic. Probably a better (but slightly more complex) approach would be to model the combustion process in both gasoline and diesel engines as a combination of two heat-transfer processes, one at constant volume and the other at constant pressure. The ideal cycle based on this concept is called the dual cycle, and a P-v diagram for it is given in Fig The relative amounts of heat transferred during each process can be adjusted to approximate the actual cycle more closely. Note that both the Otto and the Diesel cycles can be obtained as special cases of the dual cycle. η th,diesel Chapter r c = 1 (Otto) 3 Typical compression ratios for diesel engines Compression ratio, r FIGURE 9 Thermal efficiency of the ideal Diesel cycle as a function of compression and cutoff ratios (k 1.4). P X q in FIGURE Isentropic INTERACTIVE TUTORIAL SEE TUTORIAL CH. 9, SEC. 3 ON THE DVD. Isentropic P-v diagram of an ideal dual cycle q out v EXAMPLE 9 3 The Ideal Diesel Cycle An ideal Diesel cycle with air as the working fluid has a compression ratio of 18 and a cutoff ratio of. At the beginning of the compression process, the working fluid is at 14.7 psia, 80 F, and 117 in 3. Utilizing the cold-airstandard assumptions, determine (a) the temperature and pressure of air at

24 50 Thermodynamics P, psia 14.7 q in 3 Isentropic Isentropic 4 1 q out V = V 1 /18 V 3 = V V 1 = V 4 V FIGURE 9 4 P-V diagram for the ideal Diesel cycle discussed in Example 9 3. the end of each process, (b) the net work output and the thermal efficiency, and (c) the mean effective pressure. Solution An ideal Diesel cycle is considered. The temperature and pressure at the end of each process, the net work output, the thermal efficiency, and the mean effective pressure are to be determined. Assumptions 1 The cold-air-standard assumptions are applicable and thus air can be assumed to have constant specific heats at room temperature. Kinetic and potential energy changes are negligible. Properties The gas constant of air is R psia ft 3 /lbm R and its other properties at room temperature are c p 0.40 Btu/lbm R, c v Btu/lbm R, and k 1.4 (Table A Ea). Analysis The P-V diagram of the ideal Diesel cycle described is shown in Fig We note that the air contained in the cylinder forms a closed system. (a) The temperature and pressure values at the end of each process can be determined by utilizing the ideal-gas isentropic relations for processes 1- and 3-4. But first we determine the volumes at the end of each process from the definitions of the compression ratio and the cutoff ratio: Process 1- (isentropic compression of an ideal gas, constant specific heats): Process -3 (constant-pressure heat addition to an ideal gas): Process 3-4 (isentropic expansion of an ideal gas, constant specific heats): (b) The net work for a cycle is equivalent to the net heat transfer. But first we find the mass of air: m P 1V 1 RT 1 P 3 P 841 psia P V T P 3V 3 T 3 S T 3 T a V 3 V b R R T 4 T 3 a V k in3 b 1343 Ra V in b 145 R 3 P 4 P 3 a V k in3 b 1841 psiaa V in b 38.8 psia 3 V V 1 r 117 in3 18 V 3 r c V in 3 13 in 3 V 4 V in 3 T T 1 a V k 1 1 b 1540 R R V P P 1 a V k 1 b psia psia V psia 1117 in in psia # ft 3 >lbm # R1540 R a 1 ft 3 b lbm in

25 Chapter Process -3 is a constant-pressure heat-addition process, for which the boundary work and u terms can be combined into h. Thus, Process 4-1 is a constant-volume heat-rejection process (it involves no work interactions), and the amount of heat rejected is Thus, Then the thermal efficiency becomes The thermal efficiency of this Diesel cycle under the cold-air-standard assumptions could also be determined from Eq (c) The mean effective pressure is determined from its definition, Eq. 9 4: MEP Q in m 1h 3 h mc p 1T 3 T W net W net V max V min V 1 V 110 psia lbm10.40 Btu>lbm # R R4.051 Btu Q out m 1u 4 u 1 mc v 1T 4 T lbm Btu>lbm # R R Btu W net Q in Q out Btu h th W net 1.97 Btu 0.63 or 63.% Q in.051 Btu 1.97 Btu in a lbf # ft 3 1 Btu Discussion Note that a constant pressure of 110 psia during the power stroke would produce the same net work output as the entire Diesel cycle. ba 1 in. b 1 ft 9 7 STIRLING AND ERICSSON CYCLES The ideal Otto and Diesel cycles discussed in the preceding sections are composed entirely of internally reversible processes and thus are internally reversible cycles. These cycles are not totally reversible, however, since they involve heat transfer through a finite temperature difference during the nonisothermal heat-addition and heat-rejection processes, which are irreversible. Therefore, the thermal efficiency of an Otto or Diesel engine will be less than that of a Carnot engine operating between the same temperature limits. Consider a heat engine operating between a heat source at T H and a heat sink at T L. For the heat-engine cycle to be totally reversible, the temperature difference between the working fluid and the heat source (or sink) should never exceed a differential amount dt during any heat-transfer process. That is, both the heat-addition and heat-rejection processes during the cycle must take place isothermally, one at a temperature of T H and the other at a temperature of T L. This is precisely what happens in a Carnot cycle.

26 504 Thermodynamics Working fluid Energy REGENERATOR Energy FIGURE 9 5 A regenerator is a device that borrows energy from the working fluid during one part of the cycle and pays it back (without interest) during another part. There are two other cycles that involve an isothermal heat-addition process at T H and an isothermal heat-rejection process at T L : the Stirling cycle and the Ericsson cycle. They differ from the Carnot cycle in that the two isentropic processes are replaced by two constant-volume regeneration processes in the Stirling cycle and by two constant-pressure regeneration processes in the Ericsson cycle. Both cycles utilize regeneration, a process during which heat is transferred to a thermal energy storage device (called a regenerator) during one part of the cycle and is transferred back to the working fluid during another part of the cycle (Fig. 9 5). Figure 9 6(b) shows the T-s and P-v diagrams of the Stirling cycle, which is made up of four totally reversible processes: 1- T constant expansion (heat addition from the external source) -3 v constant regeneration (internal heat transfer from the working fluid to the regenerator) 3-4 T constant compression (heat rejection to the external sink) 4-1 v constant regeneration (internal heat transfer from the regenerator back to the working fluid) The execution of the Stirling cycle requires rather innovative hardware. The actual Stirling engines, including the original one patented by Robert Stirling, are heavy and complicated. To spare the reader the complexities, the execution of the Stirling cycle in a closed system is explained with the help of the hypothetical engine shown in Fig This system consists of a cylinder with two pistons on each side and a regenerator in the middle. The regenerator can be a wire or a ceramic mesh T T T T H q 1 in T H q in 1 T H q in 1 s = const. s = const. v = const. Regeneration v = const. P = const. Regeneration P = const. T L 4 q out 3 T L 4 q out 3 T L 4 q out 3 s s s FIGURE 9 6 T-s and P-v diagrams of Carnot, Stirling, and Ericsson cycles. P 1 T H = const. q in 4 4 q out q out T L = const. 3 P P 1 q in 4 1 TH = const. Regeneration T L = const. T L = const. Regeneration TH = const. 3 q out 3 v v v (a) Carnot cycle (b) Stirling cycle (c) Ericsson cycle q in

27 or any kind of porous plug with a high thermal mass (mass times specific heat). It is used for the temporary storage of thermal energy. The mass of the working fluid contained within the regenerator at any instant is considered negligible. Initially, the left chamber houses the entire working fluid (a gas), which is at a high temperature and pressure. During process 1-, heat is transferred to the gas at T H from a source at T H. As the gas expands isothermally, the left piston moves outward, doing work, and the gas pressure drops. During process -3, both pistons are moved to the right at the same rate (to keep the volume constant) until the entire gas is forced into the right chamber. As the gas passes through the regenerator, heat is transferred to the regenerator and the gas temperature drops from T H to T L. For this heat transfer process to be reversible, the temperature difference between the gas and the regenerator should not exceed a differential amount dt at any point. Thus, the temperature of the regenerator will be T H at the left end and T L at the right end of the regenerator when state 3 is reached. During process 3-4, the right piston is moved inward, compressing the gas. Heat is transferred from the gas to a sink at temperature T L so that the gas temperature remains constant at T L while the pressure rises. Finally, during process 4-1, both pistons are moved to the left at the same rate (to keep the volume constant), forcing the entire gas into the left chamber. The gas temperature rises from T L to T H as it passes through the regenerator and picks up the thermal energy stored there during process -3. This completes the cycle. Notice that the second constant-volume process takes place at a smaller volume than the first one, and the net heat transfer to the regenerator during a cycle is zero. That is, the amount of energy stored in the regenerator during process -3 is equal to the amount picked up by the gas during process 4-1. The T-s and P-v diagrams of the Ericsson cycle are shown in Fig. 9 6c. The Ericsson cycle is very much like the Stirling cycle, except that the two constant-volume processes are replaced by two constant-pressure processes. A steady-flow system operating on an Ericsson cycle is shown in Fig Here the isothermal expansion and compression processes are executed in a compressor and a turbine, respectively, and a counter-flow heat exchanger serves as a regenerator. Hot and cold fluid streams enter the heat exchanger from opposite ends, and heat transfer takes place between the two streams. In the ideal case, the temperature difference between the two fluid streams does not exceed a differential amount at any point, and the cold fluid stream leaves the heat exchanger at the inlet temperature of the hot stream. T H T H Chapter q in Regenerator T L T L q out FIGURE 9 7 The execution of the Stirling cycle. State 1 State State 3 State 4 Regenerator Heat T L = const. Compressor T H = const. Turbine w net q in q out FIGURE 9 8 A steady-flow Ericsson engine.

28 506 Thermodynamics Both the Stirling and Ericsson cycles are totally reversible, as is the Carnot cycle, and thus according to the Carnot principle, all three cycles must have the same thermal efficiency when operating between the same temperature limits: h th,stirling h th,ericsson h th,carnot 1 T L (9 14) T H This is proved for the Carnot cycle in Example 9 1 and can be proved in a similar manner for both the Stirling and Ericsson cycles. EXAMPLE 9 4 Thermal Efficiency of the Ericsson Cycle Using an ideal gas as the working fluid, show that the thermal efficiency of an Ericsson cycle is identical to the efficiency of a Carnot cycle operating between the same temperature limits. Solution It is to be shown that the thermal efficiencies of Carnot and Ericsson cycles are identical. Analysis Heat is transferred to the working fluid isothermally from an external source at temperature T H during process 1-, and it is rejected again isothermally to an external sink at temperature T L during process 3-4. For a reversible isothermal process, heat transfer is related to the entropy change by q T s The entropy change of an ideal gas during an isothermal process is s c p ln T e 0 R ln P e R ln P e T i P i The heat input and heat output can be expressed as P i and q in T H 1s s 1 T H a R ln P P 1 b RT H ln P 1 P q out T L 1s 4 s 3 T L a R ln P 4 P 3 b RT L ln P 4 P 3 Then the thermal efficiency of the Ericsson cycle becomes h th,ericsson 1 q out q in 1 RT L ln 1P 4 >P 3 RT H ln 1P 1 >P 1 T L T H since P 1 P 4 and P 3 P. Notice that this result is independent of whether the cycle is executed in a closed or steady-flow system. Stirling and Ericsson cycles are difficult to achieve in practice because they involve heat transfer through a differential temperature difference in all components including the regenerator. This would require providing infinitely large surface areas for heat transfer or allowing an infinitely long time for the process. Neither is practical. In reality, all heat transfer processes take place through a finite temperature difference, the regenerator does not have an efficiency of 100 percent, and the pressure losses in the regenerator are considerable. Because of these limitations, both Stirling and Ericsson cycles

29 have long been of only theoretical interest. However, there is renewed interest in engines that operate on these cycles because of their potential for higher efficiency and better emission control. The Ford Motor Company, General Motors Corporation, and the Phillips Research Laboratories of the Netherlands have successfully developed Stirling engines suitable for trucks, buses, and even automobiles. More research and development are needed before these engines can compete with the gasoline or diesel engines. Both the Stirling and the Ericsson engines are external combustion engines. That is, the fuel in these engines is burned outside the cylinder, as opposed to gasoline or diesel engines, where the fuel is burned inside the cylinder. External combustion offers several advantages. First, a variety of fuels can be used as a source of thermal energy. Second, there is more time for combustion, and thus the combustion process is more complete, which means less air pollution and more energy extraction from the fuel. Third, these engines operate on closed cycles, and thus a working fluid that has the most desirable characteristics (stable, chemically inert, high thermal conductivity) can be utilized as the working fluid. Hydrogen and helium are two gases commonly employed in these engines. Despite the physical limitations and impracticalities associated with them, both the Stirling and Ericsson cycles give a strong message to design engineers: Regeneration can increase efficiency. It is no coincidence that modern gas-turbine and steam power plants make extensive use of regeneration. In fact, the Brayton cycle with intercooling, reheating, and regeneration, which is utilized in large gas-turbine power plants and discussed later in this chapter, closely resembles the Ericsson cycle. 9 8 BRAYTON CYCLE: THE IDEAL CYCLE FOR GAS-TURBINE ENGINES The Brayton cycle was first proposed by George Brayton for use in the reciprocating oil-burning engine that he developed around Today, it is used for gas turbines only where both the compression and expansion processes take place in rotating machinery. Gas turbines usually operate on an open cycle, as shown in Fig Fresh air at ambient conditions is drawn into the compressor, where its temperature and pressure are raised. The highpressure air proceeds into the combustion chamber, where the fuel is burned at constant pressure. The resulting high-temperature gases then enter the turbine, where they expand to the atmospheric pressure while producing power. The exhaust gases leaving the turbine are thrown out (not recirculated), causing the cycle to be classified as an open cycle. The open gas-turbine cycle described above can be modeled as a closed cycle, as shown in Fig. 9 30, by utilizing the air-standard assumptions. Here the compression and expansion processes remain the same, but the combustion process is replaced by a constant-pressure heat-addition process from an external source, and the exhaust process is replaced by a constantpressure heat-rejection process to the ambient air. The ideal cycle that the working fluid undergoes in this closed loop is the Brayton cycle, which is made up of four internally reversible processes: 1- Isentropic compression (in a compressor) -3 Constant-pressure heat addition Chapter INTERACTIVE TUTORIAL SEE TUTORIAL CH. 9, SEC. 4 ON THE DVD.

30 508 Thermodynamics Fuel Combustion chamber q in 3 Heat exchanger Compressor Turbine w net 3 1 Fresh air Exhaust gases 4 Compressor Turbine w net FIGURE 9 9 An open-cycle gas-turbine engine. 1 Heat exchanger 4 q out FIGURE 9 30 A closed-cycle gas-turbine engine. T 1 q in P = const. P = const. 3 4 q out 3-4 Isentropic expansion (in a turbine) 4-1 Constant-pressure heat rejection The T-s and P-v diagrams of an ideal Brayton cycle are shown in Fig Notice that all four processes of the Brayton cycle are executed in steadyflow devices; thus, they should be analyzed as steady-flow processes. When the changes in kinetic and potential energies are neglected, the energy balance for a steady-flow process can be expressed, on a unit mass basis, as 1q in q out 1w in w out h exit h inlet (9 15) Therefore, heat transfers to and from the working fluid are (a) T-s diagram s q in h 3 h c p 1T 3 T (9 16a) and P s = const. q out h 4 h 1 c p 1T 4 T 1 (9 16b) q in 3 Then the thermal efficiency of the ideal Brayton cycle under the cold-airstandard assumptions becomes s = const. h th,brayton w net q in 1 q out q 1 c p 1T 4 T 1 in c p 1T 3 T 1 T 1 1T 4 >T 1 1 T 1T 3 >T q out (b) P-v diagram FIGURE 9 31 T-s and P-v diagrams for the ideal Brayton cycle. v Processes 1- and 3-4 are isentropic, and P P 3 and P 4 P 1. Thus, T a P 1k 1>k b a P 1k 1>k 3 b T 3 T 1 P 1 P 4 T 4 Substituting these equations into the thermal efficiency relation and simplifying give h th,brayton 1 1 r 1k 1>k p (9 17)

31 where r p P P 1 (9 18) is the pressure ratio and k is the specific heat ratio. Equation 9 17 shows that under the cold-air-standard assumptions, the thermal efficiency of an ideal Brayton cycle depends on the pressure ratio of the gas turbine and the specific heat ratio of the working fluid. The thermal efficiency increases with both of these parameters, which is also the case for actual gas turbines. A plot of thermal efficiency versus the pressure ratio is given in Fig. 9 3 for k 1.4, which is the specific-heat-ratio value of air at room temperature. The highest temperature in the cycle occurs at the end of the combustion process (state 3), and it is limited by the maximum temperature that the turbine blades can withstand. This also limits the pressure ratios that can be used in the cycle. For a fixed turbine inlet temperature T 3, the net work output per cycle increases with the pressure ratio, reaches a maximum, and then starts to decrease, as shown in Fig Therefore, there should be a compromise between the pressure ratio (thus the thermal efficiency) and the net work output. With less work output per cycle, a larger mass flow rate (thus a larger system) is needed to maintain the same power output, which may not be economical. In most common designs, the pressure ratio of gas turbines ranges from about 11 to 16. The air in gas turbines performs two important functions: It supplies the necessary oxidant for the combustion of the fuel, and it serves as a coolant to keep the temperature of various components within safe limits. The second function is accomplished by drawing in more air than is needed for the complete combustion of the fuel. In gas turbines, an air fuel mass ratio of 50 or above is not uncommon. Therefore, in a cycle analysis, treating the combustion gases as air does not cause any appreciable error. Also, the mass flow rate through the turbine is greater than that through the compressor, the difference being equal to the mass flow rate of the fuel. Thus, assuming a constant mass flow rate throughout the cycle yields conservative results for open-loop gas-turbine engines. The two major application areas of gas-turbine engines are aircraft propulsion and electric power generation. When it is used for aircraft propulsion, the gas turbine produces just enough power to drive the compressor and a small generator to power the auxiliary equipment. The high-velocity exhaust gases are responsible for producing the necessary thrust to propel the aircraft. Gas turbines are also used as stationary power plants to generate electricity as stand-alone units or in conjunction with steam power plants on the high-temperature side. In these plants, the exhaust gases of the gas turbine serve as the heat source for the steam. The gas-turbine cycle can also be executed as a closed cycle for use in nuclear power plants. This time the working fluid is not limited to air, and a gas with more desirable characteristics (such as helium) can be used. The majority of the Western world s naval fleets already use gas-turbine engines for propulsion and electric power generation. The General Electric LM500 gas turbines used to power ships have a simple-cycle thermal efficiency of 37 percent. The General Electric WR-1 gas turbines equipped with intercooling and regeneration have a thermal efficiency of 43 percent and η th,brayton FIGURE 9 3 Chapter Typical pressure ratios for gasturbine engines Pressure ratio, r p Thermal efficiency of the ideal Brayton cycle as a function of the pressure ratio. T T max 1000 K T min 300 K 1 r p = 15 w net,max r p = 8. 3 r p = FIGURE 9 33 For fixed values of T min and T max, the net work of the Brayton cycle first increases with the pressure ratio, then reaches a maximum at r p (T max /T min ) k/[(k 1)], and finally decreases. 4 s

32 510 Thermodynamics w turbine w compressor Back work w net FIGURE 9 34 The fraction of the turbine work used to drive the compressor is called the back work ratio. produce 1.6 MW (9,040 hp). The regeneration also reduces the exhaust temperature from 600 C (1100 F) to 350 C (650 F). Air is compressed to 3 atm before it enters the intercooler. Compared to steam-turbine and dieselpropulsion systems, the gas turbine offers greater power for a given size and weight, high reliability, long life, and more convenient operation. The engine start-up time has been reduced from 4 h required for a typical steampropulsion system to less than min for a gas turbine. Many modern marine propulsion systems use gas turbines together with diesel engines because of the high fuel consumption of simple-cycle gas-turbine engines. In combined diesel and gas-turbine systems, diesel is used to provide for efficient low-power and cruise operation, and gas turbine is used when high speeds are needed. In gas-turbine power plants, the ratio of the compressor work to the turbine work, called the back work ratio, is very high (Fig. 9 34). Usually more than one-half of the turbine work output is used to drive the compressor. The situation is even worse when the isentropic efficiencies of the compressor and the turbine are low. This is quite in contrast to steam power plants, where the back work ratio is only a few percent. This is not surprising, however, since a liquid is compressed in steam power plants instead of a gas, and the steady-flow work is proportional to the specific volume of the working fluid. A power plant with a high back work ratio requires a larger turbine to provide the additional power requirements of the compressor. Therefore, the turbines used in gas-turbine power plants are larger than those used in steam power plants of the same net power output. Development of Gas Turbines The gas turbine has experienced phenomenal progress and growth since its first successful development in the 1930s. The early gas turbines built in the 1940s and even 1950s had simple-cycle efficiencies of about 17 percent because of the low compressor and turbine efficiencies and low turbine inlet temperatures due to metallurgical limitations of those times. Therefore, gas turbines found only limited use despite their versatility and their ability to burn a variety of fuels. The efforts to improve the cycle efficiency concentrated in three areas: 1. Increasing the turbine inlet (or firing) temperatures This has been the primary approach taken to improve gas-turbine efficiency. The turbine inlet temperatures have increased steadily from about 540 C (1000 F) in the 1940s to 145 C (600 F) and even higher today. These increases were made possible by the development of new materials and the innovative cooling techniques for the critical components such as coating the turbine blades with ceramic layers and cooling the blades with the discharge air from the compressor. Maintaining high turbine inlet temperatures with an air-cooling technique requires the combustion temperature to be higher to compensate for the cooling effect of the cooling air. However, higher combustion temperatures increase the amount of nitrogen oxides (NO x ), which are responsible for the formation of ozone at ground level and smog. Using steam as the coolant allowed an increase in the turbine inlet temperatures by 00 F without an increase in the combustion temperature. Steam is also a much more effective heat transfer medium than air.

33 . Increasing the efficiencies of turbomachinery components The performance of early turbines suffered greatly from the inefficiencies of turbines and compressors. However, the advent of computers and advanced techniques for computer-aided design made it possible to design these components aerodynamically with minimal losses. The increased efficiencies of the turbines and compressors resulted in a significant increase in the cycle efficiency. 3. Adding modifications to the basic cycle The simple-cycle efficiencies of early gas turbines were practically doubled by incorporating intercooling, regeneration (or recuperation), and reheating, discussed in the next two sections. These improvements, of course, come at the expense of increased initial and operation costs, and they cannot be justified unless the decrease in fuel costs offsets the increase in other costs. The relatively low fuel prices, the general desire in the industry to minimize installation costs, and the tremendous increase in the simple-cycle efficiency to about 40 percent left little desire for opting for these modifications. The first gas turbine for an electric utility was installed in 1949 in Oklahoma as part of a combined-cycle power plant. It was built by General Electric and produced 3.5 MW of power. Gas turbines installed until the mid-1970s suffered from low efficiency and poor reliability. In the past, the base-load electric power generation was dominated by large coal and nuclear power plants. However, there has been an historic shift toward natural gas fired gas turbines because of their higher efficiencies, lower capital costs, shorter installation times, and better emission characteristics, and the abundance of natural gas supplies, and more and more electric utilities are using gas turbines for base-load power production as well as for peaking. The construction costs for gas-turbine power plants are roughly half that of comparable conventional fossil-fuel steam power plants, which were the primary base-load power plants until the early 1980s. More than half of all power plants to be installed in the foreseeable future are forecast to be gasturbine or combined gas steam turbine types. A gas turbine manufactured by General Electric in the early 1990s had a pressure ratio of 13.5 and generated MW of net power at a thermal efficiency of 33 percent in simple-cycle operation. A more recent gas turbine manufactured by General Electric uses a turbine inlet temperature of 145 C (600 F) and produces up to 8 MW while achieving a thermal efficiency of 39.5 percent in the simple-cycle mode. A 1.3-ton small-scale gas turbine labeled OP-16, built by the Dutch firm Opra Optimal Radial Turbine, can run on gas or liquid fuel and can replace a 16-ton diesel engine. It has a pressure ratio of 6.5 and produces up to MW of power. Its efficiency is 6 percent in the simple-cycle operation, which rises to 37 percent when equipped with a regenerator. Chapter EXAMPLE 9 5 The Simple Ideal Brayton Cycle A gas-turbine power plant operating on an ideal Brayton cycle has a pressure ratio of 8. The gas temperature is 300 K at the compressor inlet and 1300 K at the turbine inlet. Utilizing the air-standard assumptions, determine (a) the

34 51 Thermodynamics T, K q in P = const. w comp r p = 8 P = const. FIGURE 9 35 T-s diagram for the Brayton cycle discussed in Example w turb 4 q out s gas temperature at the exits of the compressor and the turbine, (b) the back work ratio, and (c) the thermal efficiency. Solution A power plant operating on the ideal Brayton cycle is considered. The compressor and turbine exit temperatures, back work ratio, and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 The variation of specific heats with temperature is to be considered. Analysis The T-s diagram of the ideal Brayton cycle described is shown in Fig We note that the components involved in the Brayton cycle are steady-flow devices. (a) The air temperatures at the compressor and turbine exits are determined from isentropic relations: Process 1- (isentropic compression of an ideal gas): T K S h kj>kg P r P r P P 1 P r S T 540 K 1at compressor exit Process 3-4 (isentropic expansion of an ideal gas): T K S h kj>kg P r4 P 4 P 3 P r3 a 1 8 b S T K 1at turbine exit (b) To find the back work ratio, we need to find the work input to the compressor and the work output of the turbine: Thus, That is, 40.3 percent of the turbine work output is used just to drive the compressor. (c) The thermal efficiency of the cycle is the ratio of the net power output to the total heat input: Thus, P r w comp,in h h kj>kg w turb,out h 3 h kj>kg r bw w comp,in kj>kg w turb,out kj>kg q in h 3 h kj>kg w net w out w in kj>kg h th w net q in h kj>kg h kj>kg 36.4 kj>kg 0.46 or 4.6% kj>kg

35 Chapter The thermal efficiency could also be determined from where h th 1 q out q in q out h 4 h kj>kg Discussion Under the cold-air-standard assumptions (constant specific heat values at room temperature), the thermal efficiency would be, from Eq. 9 17, h th,brayton r 1k 1>k p >1.4 8 which is sufficiently close to the value obtained by accounting for the variation of specific heats with temperature. Deviation of Actual Gas-Turbine Cycles from Idealized Ones The actual gas-turbine cycle differs from the ideal Brayton cycle on several accounts. For one thing, some pressure drop during the heat-addition and heatrejection processes is inevitable. More importantly, the actual work input to the compressor is more, and the actual work output from the turbine is less because of irreversibilities. The deviation of actual compressor and turbine behavior from the idealized isentropic behavior can be accurately accounted for by utilizing the isentropic efficiencies of the turbine and compressor as and h C w s w a h s h 1 h a h 1 (9 19) h T w a h 3 h 4a (9 0) w s h 3 h 4s where states a and 4a are the actual exit states of the compressor and the turbine, respectively, and s and 4s are the corresponding states for the isentropic case, as illustrated in Fig The effect of the turbine and compressor efficiencies on the thermal efficiency of the gas-turbine engines is illustrated below with an example. T a s 1 Pressure drop during heat addition 4s 3 4a Pressure drop during heat rejection FIGURE 9 36 The deviation of an actual gas-turbine cycle from the ideal Brayton cycle as a result of irreversibilities. s EXAMPLE 9 6 An Actual Gas-Turbine Cycle Assuming a compressor efficiency of 80 percent and a turbine efficiency of 85 percent, determine (a) the back work ratio, (b) the thermal efficiency, and (c) the turbine exit temperature of the gas-turbine cycle discussed in Example 9 5. Solution The Brayton cycle discussed in Example 9 5 is reconsidered. For specified turbine and compressor efficiencies, the back work ratio, the thermal efficiency, and the turbine exit temperature are to be determined.

36 514 Thermodynamics T, K s 1 a q in q out 3 4a FIGURE 9 37 T-s diagram of the gas-turbine cycle discussed in Example s s Analysis (a) The T-s diagram of the cycle is shown in Fig The actual compressor work and turbine work are determined by using the definitions of compressor and turbine efficiencies, Eqs and 9 0: Compressor: Turbine: Thus, w comp,in w s kj>kg kj>kg h C 0.80 w turb,out h T w s kj>kg kj>kg r bw w comp,in kj>kg 0.59 w turb,out kj>kg That is, the compressor is now consuming 59. percent of the work produced by the turbine (up from 40.3 percent). This increase is due to the irreversibilities that occur within the compressor and the turbine. (b) In this case, air leaves the compressor at a higher temperature and enthalpy, which are determined to be Thus, and w comp,in h a h 1 S h a h 1 w comp,in h th w net q in That is, the irreversibilities occurring within the turbine and compressor caused the thermal efficiency of the gas turbine cycle to drop from 4.6 to 6.6 percent. This example shows how sensitive the performance of a gas-turbine power plant is to the efficiencies of the compressor and the turbine. In fact, gas-turbine efficiencies did not reach competitive values until significant improvements were made in the design of gas turbines and compressors. (c) The air temperature at the turbine exit is determined from an energy balance on the turbine: Then, from Table A 17, kj>kg 1and T a 598 K q in h 3 h a kj>kg w net w out w in kj>kg kj>kg 0.66 or 6.6% kj>kg w turb,out h 3 h 4a S h 4a h 3 w turb,out T 4a 853 K kj>kg Discussion The temperature at turbine exit is considerably higher than that at the compressor exit (T a 598 K), which suggests the use of regeneration to reduce fuel cost.

37 Chapter Regenerator 1 Heat 5 Combustion chamber 3 4 Compressor Turbine w net FIGURE 9 38 A gas-turbine engine with regenerator. 9 9 THE BRAYTON CYCLE WITH REGENERATION In gas-turbine engines, the temperature of the exhaust gas leaving the turbine is often considerably higher than the temperature of the air leaving the compressor. Therefore, the high-pressure air leaving the compressor can be heated by transferring heat to it from the hot exhaust gases in a counter-flow heat exchanger, which is also known as a regenerator or a recuperator. A sketch of the gas-turbine engine utilizing a regenerator and the T-s diagram of the new cycle are shown in Figs and 9 39, respectively. The thermal efficiency of the Brayton cycle increases as a result of regeneration since the portion of energy of the exhaust gases that is normally rejected to the surroundings is now used to preheat the air entering the combustion chamber. This, in turn, decreases the heat input (thus fuel) requirements for the same net work output. Note, however, that the use of a regenerator is recommended only when the turbine exhaust temperature is higher than the compressor exit temperature. Otherwise, heat will flow in the reverse direction (to the exhaust gases), decreasing the efficiency. This situation is encountered in gas-turbine engines operating at very high pressure ratios. The highest temperature occurring within the regenerator is T 4, the temperature of the exhaust gases leaving the turbine and entering the regenerator. Under no conditions can the air be preheated in the regenerator to a temperature above this value. Air normally leaves the regenerator at a lower temperature, T 5. In the limiting (ideal) case, the air exits the regenerator at the inlet temperature of the exhaust gases T 4. Assuming the regenerator to be well insulated and any changes in kinetic and potential energies to be negligible, the actual and maximum heat transfers from the exhaust gases to the air can be expressed as and q regen,act h 5 h q regen,max h 5 h h 4 h (9 1) (9 ) The extent to which a regenerator approaches an ideal regenerator is called the effectiveness ` and is defined as T q regen 1 q in 5' 5 Regeneration 6 q out 3 4 q saved = q regen FIGURE 9 39 T-s diagram of a Brayton cycle with regeneration. s P q regen,act q regen,max h 5 h h 4 h (9 3)

38 516 Thermodynamics η th,brayton With regeneration Without regeneration T 1 /T 3 = 0.33 T 1 /T 3 = 0. T 1 /T 3 = Pressure ratio, r p FIGURE 9 40 Thermal efficiency of the ideal Brayton cycle with and without regeneration. When the cold-air-standard assumptions are utilized, it reduces to P T 5 T T 4 T (9 4) A regenerator with a higher effectiveness obviously saves a greater amount of fuel since it preheats the air to a higher temperature prior to combustion. However, achieving a higher effectiveness requires the use of a larger regenerator, which carries a higher price tag and causes a larger pressure drop. Therefore, the use of a regenerator with a very high effectiveness cannot be justified economically unless the savings from the fuel costs exceed the additional expenses involved. The effectiveness of most regenerators used in practice is below Under the cold-air-standard assumptions, the thermal efficiency of an ideal Brayton cycle with regeneration is h th,regen 1 a T 1 T 3 b1r p 1k 1>k (9 5) Therefore, the thermal efficiency of an ideal Brayton cycle with regeneration depends on the ratio of the minimum to maximum temperatures as well as the pressure ratio. The thermal efficiency is plotted in Fig for various pressure ratios and minimum-to-maximum temperature ratios. This figure shows that regeneration is most effective at lower pressure ratios and low minimum-to-maximum temperature ratios. EXAMPLE 9 7 Actual Gas-Turbine Cycle with Regeneration T, K a 5 q in 3 4a q regen = q saved FIGURE 9 41 T-s diagram of the regenerative Brayton cycle described in Example 9 7. s Determine the thermal efficiency of the gas-turbine described in Example 9 6 if a regenerator having an effectiveness of 80 percent is installed. Solution The gas-turbine discussed in Example 9 6 is equipped with a regenerator. For a specified effectiveness, the thermal efficiency is to be determined. Analysis The T-s diagram of the cycle is shown in Fig We first determine the enthalpy of the air at the exit of the regenerator, using the definition of effectiveness: Thus, P h 5 h a h 4a h a h kj>kg kj>kg S h kj>kg q in h 3 h kj>kg kj>kg This represents a savings of 0.0 kj/kg from the heat input requirements. The addition of a regenerator (assumed to be frictionless) does not affect the net work output. Thus, h th w net q in kj>kg or 36.9% kj>kg

39 Chapter Discussion Note that the thermal efficiency of the gas turbine has gone up from 6.6 to 36.9 percent as a result of installing a regenerator that helps to recuperate some of the thermal energy of the exhaust gases THE BRAYTON CYCLE WITH INTERCOOLING, REHEATING, AND REGENERATION The net work of a gas-turbine cycle is the difference between the turbine work output and the compressor work input, and it can be increased by either decreasing the compressor work or increasing the turbine work, or both. It was shown in Chap. 7 that the work required to compress a gas between two specified pressures can be decreased by carrying out the compression process in stages and cooling the gas in between (Fig. 9 4) that is, using multistage compression with intercooling. As the number of stages is increased, the compression process becomes nearly isothermal at the compressor inlet temperature, and the compression work decreases. Likewise, the work output of a turbine operating between two pressure levels can be increased by expanding the gas in stages and reheating it in between that is, utilizing multistage expansion with reheating. This is accomplished without raising the maximum temperature in the cycle. As the number of stages is increased, the expansion process becomes nearly isothermal. The foregoing argument is based on a simple principle: The steady-flow compression or expansion work is proportional to the specific volume of the fluid. Therefore, the specific volume of the working fluid should be as low as possible during a compression process and as high as possible during an expansion process. This is precisely what intercooling and reheating accomplish. Combustion in gas turbines typically occurs at four times the amount of air needed for complete combustion to avoid excessive temperatures. Therefore, the exhaust gases are rich in oxygen, and reheating can be accomplished by simply spraying additional fuel into the exhaust gases between two expansion states. The working fluid leaves the compressor at a lower temperature, and the turbine at a higher temperature, when intercooling and reheating are utilized. This makes regeneration more attractive since a greater potential for regeneration exists. Also, the gases leaving the compressor can be heated to a higher temperature before they enter the combustion chamber because of the higher temperature of the turbine exhaust. A schematic of the physical arrangement and the T-s diagram of an ideal two-stage gas-turbine cycle with intercooling, reheating, and regeneration are shown in Figs and The gas enters the first stage of the compressor at state 1, is compressed isentropically to an intermediate pressure P,is cooled at constant pressure to state 3 (T 3 T 1 ), and is compressed in the second stage isentropically to the final pressure P 4. At state 4 the gas enters the regenerator, where it is heated to T 5 at constant pressure. In an ideal regenerator, the gas leaves the regenerator at the temperature of the turbine exhaust, that is, T 5 T 9. The primary heat addition (or combustion) process takes P P P 1 D B Isothermal process paths Polytropic process paths INTERACTIVE TUTORIAL SEE TUTORIAL CH. 9, SEC. 5 ON THE DVD. C Work saved as a result of intercooling A Intercooling FIGURE 9 4 Comparison of work inputs to a single-stage compressor (1AC) and a two-stage compressor with intercooling (1ABD). 1 v

40 518 Thermodynamics 10 Regenerator 5 1 Combustion chamber Reheater Compressor I Compressor II Turbine I Turbine II w net 3 Intercooler FIGURE 9 43 A gas-turbine engine with two-stage compression with intercooling, two-stage expansion with reheating, and regeneration. T 4 3 q regen q in q regen = q saved q out place between states 5 and 6. The gas enters the first stage of the turbine at state 6 and expands isentropically to state 7, where it enters the reheater. It is reheated at constant pressure to state 8 (T 8 T 6 ), where it enters the second stage of the turbine. The gas exits the turbine at state 9 and enters the regenerator, where it is cooled to state 10 at constant pressure. The cycle is completed by cooling the gas to the initial state (or purging the exhaust gases). It was shown in Chap. 7 that the work input to a two-stage compressor is minimized when equal pressure ratios are maintained across each stage. It can be shown that this procedure also maximizes the turbine work output. Thus, for best performance we have s P P 1 P 4 P 3 and P 6 P 7 P 8 P 9 (9 6) FIGURE 9 44 T-s diagram of an ideal gas-turbine cycle with intercooling, reheating, and regeneration. In the analysis of the actual gas-turbine cycles, the irreversibilities that are present within the compressor, the turbine, and the regenerator as well as the pressure drops in the heat exchangers should be taken into consideration. The back work ratio of a gas-turbine cycle improves as a result of intercooling and reheating. However, this does not mean that the thermal efficiency also improves. The fact is, intercooling and reheating always decreases the thermal efficiency unless they are accompanied by regeneration. This is because intercooling decreases the average temperature at which heat is added, and reheating increases the average temperature at which heat is rejected. This is also apparent from Fig Therefore, in gasturbine power plants, intercooling and reheating are always used in conjunction with regeneration.

41 If the number of compression and expansion stages is increased, the ideal gas-turbine cycle with intercooling, reheating, and regeneration approaches the Ericsson cycle, as illustrated in Fig. 9 45, and the thermal efficiency approaches the theoretical limit (the Carnot efficiency). However, the contribution of each additional stage to the thermal efficiency is less and less, and the use of more than two or three stages cannot be justified economically. T T H,avg Chapter P = const. P = const. EXAMPLE 9 8 A Gas Turbine with Reheating and Intercooling T L,avg An ideal gas-turbine cycle with two stages of compression and two stages of expansion has an overall pressure ratio of 8. Air enters each stage of the compressor at 300 K and each stage of the turbine at 1300 K. Determine the back work ratio and the thermal efficiency of this gas-turbine cycle, assuming (a) no regenerators and (b) an ideal regenerator with 100 percent effectiveness. Compare the results with those obtained in Example 9 5. Solution An ideal gas-turbine cycle with two stages of compression and two stages of expansion is considered. The back work ratio and the thermal efficiency of the cycle are to be determined for the cases of no regeneration and maximum regeneration. Assumptions 1 Steady operating conditions exist. The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. Analysis The T-s diagram of the ideal gas-turbine cycle described is shown in Fig We note that the cycle involves two stages of expansion, two stages of compression, and regeneration. For two-stage compression and expansion, the work input is minimized and the work output is maximized when both stages of the compressor and the turbine have the same pressure ratio. Thus, P P 1 P 4 P and P 6 P 7 P 8 P Air enters each stage of the compressor at the same temperature, and each stage has the same isentropic efficiency (100 percent in this case). Therefore, the temperature (and enthalpy) of the air at the exit of each compression stage will be the same. A similar argument can be given for the turbine. Thus, At inlets: T 1 T 3, h 1 h 3 and T 6 T 8, h 6 h 8 At exits: T T 4, h h 4 and T 7 T 9, h 7 h 9 Under these conditions, the work input to each stage of the compressor will be the same, and so will the work output from each stage of the turbine. (a) In the absence of any regeneration, the back work ratio and the thermal efficiency are determined by using data from Table A 17 as follows: T K S h kj>kg P r P r P P 1 P r S T K h kj>kg FIGURE 9 45 As the number of compression and expansion stages increases, the gasturbine cycle with intercooling, reheating, and regeneration approaches the Ericsson cycle. T, K q primary q out q reheat FIGURE 9 46 T-s diagram of the gas-turbine cycle discussed in Example s s

42 50 Thermodynamics T K S h kj>kg P r P r7 P 7 P 6 P r S T K Then w comp,in 1w comp,in,i 1h h kj>kg w turb,out 1w turb,out,i 1h 6 h kj>kg Thus, and A comparison of these results with those obtained in Example 9 5 (singlestage compression and expansion) reveals that multistage compression with intercooling and multistage expansion with reheating improve the back work ratio (it drops from 40.3 to 30.4 percent) but hurt the thermal efficiency (it drops from 4.6 to 35.8 percent). Therefore, intercooling and reheating are not recommended in gas-turbine power plants unless they are accompanied by regeneration. (b) The addition of an ideal regenerator (no pressure drops, 100 percent effectiveness) does not affect the compressor work and the turbine work. Therefore, the net work output and the back work ratio of an ideal gas-turbine cycle are identical whether there is a regenerator or not. A regenerator, however, reduces the heat input requirements by preheating the air leaving the compressor, using the hot exhaust gases. In an ideal regenerator, the compressed air is heated to the turbine exit temperature T 9 before it enters the combustion chamber. Thus, under the air-standard assumptions, h 5 h 7 h 9. The heat input and the thermal efficiency in this case are and w net w turb,out w comp,in kj>kg q in q primary q reheat 1h 6 h 4 1h 8 h kj>kg r bw w comp,in 08.4 kj>kg or 30.4% w turb,out kj>kg h th w net q in q in q primary q reheat 1h 6 h 5 1h 8 h kj>kg Discussion Note that the thermal efficiency almost doubles as a result of regeneration compared to the no-regeneration case. The overall effect of twostage compression and expansion with intercooling, reheating, and regenerah th w net q in h kj>kg kj>kg or 35.8% kj>kg kj>kg or 69.6% kj>kg

43 Chapter 9 51 tion on the thermal efficiency is an increase of 63 percent. As the number of compression and expansion stages is increased, the cycle will approach the Ericsson cycle, and the thermal efficiency will approach h th,ericsson h th,carnot 1 T L K T H 1300 K Adding a second stage increases the thermal efficiency from 4.6 to 69.6 percent, an increase of 7 percentage points. This is a significant increase in efficiency, and usually it is well worth the extra cost associated with the second stage. Adding more stages, however (no matter how many), can increase the efficiency an additional 7.3 percentage points at most, and usually cannot be justified economically IDEAL JET-PROPULSION CYCLES Gas-turbine engines are widely used to power aircraft because they are light and compact and have a high power-to-weight ratio. Aircraft gas turbines operate on an open cycle called a jet-propulsion cycle. The ideal jetpropulsion cycle differs from the simple ideal Brayton cycle in that the gases are not expanded to the ambient pressure in the turbine. Instead, they are expanded to a pressure such that the power produced by the turbine is just sufficient to drive the compressor and the auxiliary equipment, such as a small generator and hydraulic pumps. That is, the net work output of a jetpropulsion cycle is zero. The gases that exit the turbine at a relatively high pressure are subsequently accelerated in a nozzle to provide the thrust to propel the aircraft (Fig. 9 47). Also, aircraft gas turbines operate at higher pressure ratios (typically between 10 and 5), and the fluid passes through a diffuser first, where it is decelerated and its pressure is increased before it enters the compressor. Aircraft are propelled by accelerating a fluid in the opposite direction to motion. This is accomplished by either slightly accelerating a large mass of fluid ( propeller-driven engine) or greatly accelerating a small mass of fluid ( jet or turbojet engine) or both (turboprop engine). A schematic of a turbojet engine and the T-s diagram of the ideal turbojet cycle are shown in Fig The pressure of air rises slightly as it is decelerated in the diffuser. Air is compressed by the compressor. It is mixed with fuel in the combustion chamber, where the mixture is burned at constant pressure. The high-pressure and high-temperature combustion gases partially expand in the turbine, producing enough power to drive the compressor and other equipment. Finally, the gases expand in a nozzle to the ambient pressure and leave the engine at a high velocity. In the ideal case, the turbine work is assumed to equal the compressor work. Also, the processes in the diffuser, the compressor, the turbine, and the nozzle are assumed to be isentropic. In the analysis of actual cycles, however, the irreversibilities associated with these devices should be considered. The effect of the irreversibilities is to reduce the thrust that can be obtained from a turbojet engine. The thrust developed in a turbojet engine is the unbalanced force that is caused by the difference in the momentum of the low-velocity air entering the engine and the high-velocity exhaust gases leaving the engine, and it is Turbine Nozzle High T and P V exit FIGURE 9 47 In jet engines, the high-temperature and high-pressure gases leaving the turbine are accelerated in a nozzle to provide thrust.

44 5 Thermodynamics T 3 q in P = const P = const. q out s Diffuser Compressor Burner section Turbine Nozzle FIGURE 9 48 Basic components of a turbojet engine and the T-s diagram for the ideal turbojet cycle. Source: The Aircraft Gas Turbine Engine and Its Operation. United Aircraft Corporation (now United Technologies Corp.), 1951, F V, m/s W P = F V FIGURE 9 49 Propulsive power is the thrust acting on the aircraft through a distance per unit time. F determined from Newton s second law. The pressures at the inlet and the exit of a turbojet engine are identical (the ambient pressure); thus, the net thrust developed by the engine is F 1m # V exit 1m # V inlet m # 1V exit V inlet 1N (9 7) where V exit is the exit velocity of the exhaust gases and V inlet is the inlet velocity of the air, both relative to the aircraft. Thus, for an aircraft cruising in still air, V inlet is the aircraft velocity. In reality, the mass flow rates of the gases at the engine exit and the inlet are different, the difference being equal to the combustion rate of the fuel. However, the air fuel mass ratio used in jetpropulsion engines is usually very high, making this difference very small. Thus, ṁ in Eq. 9 7 is taken as the mass flow rate of air through the engine. For an aircraft cruising at a constant speed, the thrust is used to overcome air drag, and the net force acting on the body of the aircraft is zero. Commercial airplanes save fuel by flying at higher altitudes during long trips since air at higher altitudes is thinner and exerts a smaller drag force on aircraft. The power developed from the thrust of the engine is called the propulsive power W. P, which is the propulsive force (thrust) times the distance this force acts on the aircraft per unit time, that is, the thrust times the aircraft velocity (Fig. 9 49): W # P FV aircraft m # 1V exit V inlet V aircraft 1kW (9 8) The net work developed by a turbojet engine is zero. Thus, we cannot define the efficiency of a turbojet engine in the same way as stationary gasturbine engines. Instead, we should use the general definition of efficiency, which is the ratio of the desired output to the required input. The desired output in a turbojet engine is the power produced to propel the aircraft W. P, and the required input is the heating value of the fuel Q. in. The ratio of these two quantities is called the propulsive efficiency and is given by Propulsive power h P (9 9) Energy input rate W # P Q # in Propulsive efficiency is a measure of how efficiently the thermal energy released during the combustion process is converted to propulsive energy. The

45 remaining part of the energy released shows up as the kinetic energy of the exhaust gases relative to a fixed point on the ground and as an increase in the enthalpy of the gases leaving the engine. Chapter 9 53 EXAMPLE 9 9 The Ideal Jet-Propulsion Cycle A turbojet aircraft flies with a velocity of 850 ft/s at an altitude where the air is at 5 psia and 40 F. The compressor has a pressure ratio of 10, and the temperature of the gases at the turbine inlet is 000 F. Air enters the compressor at a rate of 100 lbm/s. Utilizing the cold-air-standard assumptions, determine (a) the temperature and pressure of the gases at the turbine exit, (b) the velocity of the gases at the nozzle exit, and (c) the propulsive efficiency of the cycle. Solution The operating conditions of a turbojet aircraft are specified. The temperature and pressure at the turbine exit, the velocity of gases at the nozzle exit, and the propulsive efficiency are to be determined. Assumptions 1 Steady operating conditions exist. The cold-air-standard assumptions are applicable and thus air can be assumed to have constant specific heats at room temperature (c p 0.40 Btu/lbm F and k 1.4). 3 Kinetic and potential energies are negligible, except at the diffuser inlet and the nozzle exit. 4 The turbine work output is equal to the compressor work input. Analysis The T-s diagram of the ideal jet propulsion cycle described is shown in Fig We note that the components involved in the jet-propulsion cycle are steady-flow devices. (a) Before we can determine the temperature and pressure at the turbine exit, we need to find the temperatures and pressures at other states: Process 1- (isentropic compression of an ideal gas in a diffuser): For convenience, we can assume that the aircraft is stationary and the air is moving toward the aircraft at a velocity of V ft/s. Ideally, the air exits the diffuser with a negligible velocity (V 0): h V 0 h 1 V 1 0 c p 1T T 1 V 1 T T 1 V 1 c p 40 R 480 R 1850 ft>s Btu>lbm # a 1 Btu>lbm R 5,037 ft >s b P P 1 a T k>1k 1 b 15 psia a 480 R 1.4> T 1 40 R b 8.0 psia Process -3 (isentropic compression of an ideal gas in a compressor): P 3 1r p 1P psia 80 psia 1 P 4 T 3 T a P 1k 1>k 3 b 1480 R > R P T, F q in P = const. P = const. q out FIGURE 9 50 T-s diagram for the turbojet cycle described in Example s

46 54 Thermodynamics Process 4-5 (isentropic expansion of an ideal gas in a turbine): Neglecting the kinetic energy changes across the compressor and the turbine and assuming the turbine work to be equal to the compressor work, we find the temperature and pressure at the turbine exit to be w comp,in w turb,out h 3 h h 4 h 5 c p 1T 3 T c p 1T 4 T 5 T 5 T 4 T 3 T R P 5 P 4 a T k>1k 1 5 b 180 psiaa 013 R 1.4> T R b 39.7 psia (b) To find the air velocity at the nozzle exit, we need to first determine the nozzle exit temperature and then apply the steady-flow energy equation. Process 5-6 (isentropic expansion of an ideal gas in a nozzle): T 6 T 5 a P 1k 1>k 6 b 1013 Ra 5 psia >1.4 P psia b 1114 R 0 h 6 V 6 h 5 V 5 0 c p 1T 6 T 5 V 6 V 6 c p 1T 5 T 6 B Btu>lbm # R R4 a 5,037 ft >s 1 Btu>lbm b 388 ft/s (c) The propulsive efficiency of a turbojet engine is the ratio of the propulsive power developed W. P to the total heat transfer rate to the working fluid: W # P m # 1V exit V inlet V aircraft 1100 lbm>s ft>s ft>sa 1 Btu>lbm 5,037 ft >s b 876 Btu>s 1or 11,707 hp Q # in m # 1h 4 h 3 m # c p 1T 4 T lbm>s Btu>lbm # R R4 36,794 Btu>s h P W# P Q # in 876 Btu>s.5% 36,794 Btu>s That is,.5 percent of the energy input is used to propel the aircraft and to overcome the drag force exerted by the atmospheric air.

47 Chapter 9 55 Discussion For those who are wondering what happened to the rest of the energy, here is a brief account: KE # out # V g m 850ft>s lbm>se31388 fa 1 Btu>lbm 5,037 ft >s b 11,867 Btu>s 13.% Q # out m # 1h 6 h 1 m # c p 1T 6 T lbm>s 10.4 Btu>lbm # R R4 16,651 Btu>s 145.3% Thus, 3. percent of the energy shows up as excess kinetic energy (kinetic energy of the gases relative to a fixed point on the ground). Notice that for the highest propulsion efficiency, the velocity of the exhaust gases relative to the ground V g should be zero. That is, the exhaust gases should leave the nozzle at the velocity of the aircraft. The remaining 45.3 percent of the energy shows up as an increase in enthalpy of the gases leaving the engine. These last two forms of energy eventually become part of the internal energy of the atmospheric air (Fig. 9 51). Q in AIRCRAFT (propulsive power) W P KE out (excess kinetic energy) Q out (excess thermal energy) FIGURE 9 51 Energy supplied to an aircraft (from the burning of a fuel) manifests itself in various forms. Modifications to Turbojet Engines The first airplanes built were all propeller-driven, with propellers powered by engines essentially identical to automobile engines. The major breakthrough in commercial aviation occurred with the introduction of the turbojet engine in 195. Both propeller-driven engines and jet-propulsion-driven engines have their own strengths and limitations, and several attempts have been made to combine the desirable characteristics of both in one engine. Two such modifications are the propjet engine and the turbofan engine. The most widely used engine in aircraft propulsion is the turbofan (or fanjet) engine wherein a large fan driven by the turbine forces a considerable amount of air through a duct (cowl) surrounding the engine, as shown in Figs. 9 5 and The fan exhaust leaves the duct at a higher velocity, enhancing the total thrust of the engine significantly. A turbofan engine is based on the principle that for the same power, a large volume of slowermoving air produces more thrust than a small volume of fast-moving air. The first commercial turbofan engine was successfully tested in Low-pressure compressor Fan Duct Burners Low-pressure turbine Fan exhaust Turbine exhaust FIGURE 9 5 A turbofan engine. Fan High-pressure compressor High-pressure turbine Source: The Aircraft Gas Turbine and Its Operation. United Aircraft Corporation (now United Technologies Corp.), 1951, 1974.

48 56 Thermodynamics Fan Low pressure compressor Fan air bypassing the jet engine -stage high pressure turbine to turn outer shaft Combustors Low pressure turbine to turn inner shaft High pressure compressor Thrust FIGURE 9 53 A modern jet engine used to power Boeing 777 aircraft. This is a Pratt & Whitney PW4084 turbofan capable of producing 84,000 pounds of thrust. It is 4.87 m (19 in.) long, has a.84 m (11 in.) diameter fan, and it weighs 6800 kg (15,000 lbm). Courtesy of Pratt & Whitney Corp. Air inlet Twin spool shaft to turn the fan and the compressors Thrust Propeller Compressor Burners Turbine FIGURE 9 54 A turboprop engine. Source: The Aircraft Gas Turbine Engine and Its Operation. United Aircraft Corporation (now United Technologies Corp.), 1951, Gear reduction The turbofan engine on an airplane can be distinguished from the lessefficient turbojet engine by its fat cowling covering the large fan. All the thrust of a turbojet engine is due to the exhaust gases leaving the engine at about twice the speed of sound. In a turbofan engine, the high-speed exhaust gases are mixed with the lower-speed air, which results in a considerable reduction in noise. New cooling techniques have resulted in considerable increases in efficiencies by allowing gas temperatures at the burner exit to reach over 1500 C, which is more than 100 C above the melting point of the turbine blade materials. Turbofan engines deserve most of the credit for the success of jumbo jets that weigh almost 400,000 kg and are capable of carrying over 400 passengers for up to a distance of 10,000 km at speeds over 950 km/h with less fuel per passenger mile. The ratio of the mass flow rate of air bypassing the combustion chamber to that of air flowing through it is called the bypass ratio. The first commercial high-bypass-ratio engines had a bypass ratio of 5. Increasing the bypass ratio of a turbofan engine increases thrust. Thus, it makes sense to remove the cowl from the fan. The result is a propjet engine, as shown in Fig Turbofan and propjet engines differ primarily in their bypass ratios: 5 or 6 for turbofans and as high as 100 for propjets. As a general rule, propellers

49 Chapter 9 57 Fuel nozzles or spray bars Air inlet Flame holders Jet nozzle FIGURE 9 55 A ramjet engine. Source: The Aircraft Gas Turbine Engine and Its Operation. United Aircraft Corporation (now United Technologies Corp.), 1951, are more efficient than jet engines, but they are limited to low-speed and low-altitude operation since their efficiency decreases at high speeds and altitudes. The old propjet engines (turboprops) were limited to speeds of about Mach 0.6 and to altitudes of around 9100 m. The new propjet engines ( propfans) are expected to achieve speeds of about Mach 0.8 and altitudes of about 1,00 m. Commercial airplanes of medium size and range propelled by propfans are expected to fly as high and as fast as the planes propelled by turbofans, and to do so on less fuel. Another modification that is popular in military aircraft is the addition of an afterburner section between the turbine and the nozzle. Whenever a need for extra thrust arises, such as for short takeoffs or combat conditions, additional fuel is injected into the oxygen-rich combustion gases leaving the turbine. As a result of this added energy, the exhaust gases leave at a higher velocity, providing a greater thrust. A ramjet engine is a properly shaped duct with no compressor or turbine, as shown in Fig. 9 55, and is sometimes used for high-speed propulsion of missiles and aircraft. The pressure rise in the engine is provided by the ram effect of the incoming high-speed air being rammed against a barrier. Therefore, a ramjet engine needs to be brought to a sufficiently high speed by an external source before it can be fired. The ramjet performs best in aircraft flying above Mach or 3 (two or three times the speed of sound). In a ramjet, the air is slowed down to about Mach 0., fuel is added to the air and burned at this low velocity, and the combustion gases are expended and accelerated in a nozzle. A scramjet engine is essentially a ramjet in which air flows through at supersonic speeds (above the speed of sound). Ramjets that convert to scramjet configurations at speeds above Mach 6 are successfully tested at speeds of about Mach 8. Finally, a rocket is a device where a solid or liquid fuel and an oxidizer react in the combustion chamber. The high-pressure combustion gases are then expanded in a nozzle. The gases leave the rocket at very high velocities, producing the thrust to propel the rocket. 9 1 SECOND-LAW ANALYSIS OF GAS POWER CYCLES The ideal Carnot, Ericsson, and Stirling cycles are totally reversible; thus they do not involve any irreversibilities. The ideal Otto, Diesel, and Brayton cycles, however, are only internally reversible, and they may involve irreversibilities

50 58 Thermodynamics external to the system. A second-law analysis of these cycles reveals where the largest irreversibilities occur and where to start improvements. Relations for exergy and exergy destruction for both closed and steadyflow systems are developed in Chap. 8. The exergy destruction for a closed system can be expressed as X dest T 0 S gen T 0 1 S sys S in S out (9 30) where T b,in and T b,out are the temperatures of the system boundary where heat is transferred into and out of the system, respectively. A similar relation for steady-flow systems can be expressed, in rate form, as X # dest T 0 S # gen T 0 1S # out S # in T 0 a aout m # s ain m # s Q# in Q # out b 1kW T b,in T b,out or, on a unit mass basis for a one-inlet, one-exit steady-flow device, as (9 31) (9 3) where subscripts i and e denote the inlet and exit states, respectively. The exergy destruction of a cycle is the sum of the exergy destructions of the processes that compose that cycle. The exergy destruction of a cycle can also be determined without tracing the individual processes by considering the entire cycle as a single process and using one of the relations above. Entropy is a property, and its value depends on the state only. For a cycle, reversible or actual, the initial and the final states are identical; thus s e s i. Therefore, the exergy destruction of a cycle depends on the magnitude of the heat transfer with the high- and low-temperature reservoirs involved and on their temperatures. It can be expressed on a unit mass basis as (9 33) For a cycle that involves heat transfer only with a source at T H and a sink at T L, the exergy destruction becomes (9 34) The exergies of a closed system f and a fluid stream c at any state can be determined from and T 0 c1s S 1 sys Q in T b,in Q out T b,out d 1kJ X dest T 0 s gen T 0 a s e s i q in T b,in q out T b,out b 1kJ>kg x dest T 0 a a q out T b,out a q in T b,in b 1kJ>kg x dest T 0 a q out T L q in T H b 1kJ>kg f 1u u 0 T 0 1s s 0 P 0 1v v 0 V gz 1kJ>kg c 1h h 0 T 0 1s s 0 V gz 1kJ>kg where subscript 0 denotes the state of the surroundings. (9 35) (9 36)

51 Chapter 9 59 EXAMPLE 9 10 Second-Law Analysis of an Otto Cycle Determine the exergy destruction associated with the Otto cycle (all four processes as well as the cycle) discussed in Example 9, assuming that heat is transferred to the working fluid from a source at 1700 K and heat is rejected to the surroundings at 90 K. Also, determine the exergy of the exhaust gases when they are purged. Solution The Otto cycle analyzed in Example 9 is reconsidered. For specified source and sink temperatures, the exergy destruction associated with the cycle and the exergy purged with the exhaust gases are to be determined. Analysis In Example 9, various quantities of interest were given or determined to be Processes 1- and 3-4 are isentropic (s 1 s, s 3 s 4 ) and therefore do not involve any internal or external irreversibilities; that is, X dest,1 0 and X dest,34 0. Processes -3 and 4-1 are constant-volume heat-addition and heat-rejection processes, respectively, and are internally reversible. However, the heat transfer between the working fluid and the source or the sink takes place through a finite temperature difference, rendering both processes irreversible. The exergy destruction associated with each process is determined from Eq However, first we need to determine the entropy change of air during these processes: s 3 s s 3 s R ln P 3 P Also, Thus, kj>kg # K r 8 P MPa T 0 90 K P MPa T 1 90 K q in 800 kj>kg T 65.4 K q out kj>kg T K w net kj>kg kj>kg # K kj>kg # K ln MPa q in 800 kj>kg and T source 1700 K x dest,3 T 0 c1s 3 s sys 190 Kc kj>kg # K 800 kj>kg 1700 K d 8. kj>kg q in T source d For process 4-1, s 1 s 4 s s kj/kg K, q R,41 q out kj/kg, and T sink 90 K. Thus, x dest,41 T 0 c1s 1 s 4 sys q out T sink d MPa

52 530 Thermodynamics Therefore, the irreversibility of the cycle is The exergy destruction of the cycle could also be determined from Eq Notice that the largest exergy destruction in the cycle occurs during the heat-rejection process. Therefore, any attempt to reduce the exergy destruction should start with this process. Disregarding any kinetic and potential energies, the exergy (work potential) of the working fluid before it is purged (state 4) is determined from Eq. 9 35: where Thus, kj>kg 190 K c kj>kg # K d 90 K 163. kj>kg x dest,cycle x dest,1 x dest,3 x dest,34 x dest, kj>kg kj>kg 45.4 kj/kg f 4 1u 4 u 0 T 0 1s 4 s 0 P 0 1v 4 v 0 s 4 s 0 s 4 s kj>kg # K u 4 u 0 u 4 u 1 q out kj>kg v 4 v 0 v 4 v 1 0 f kj>kg 190 K kj>kg # K kj /kg which is equivalent to the exergy destruction for process 4-1. (Why?) Discussion Note that 163. kj/kg of work could be obtained from the exhaust gases if they were brought to the state of the surroundings in a reversible manner. TOPIC OF SPECIAL INTEREST* Saving Fuel and Money by Driving Sensibly Two-thirds of the oil used in the United States is used for transportation. Half of this oil is consumed by passenger cars and light trucks that are used to commute to and from work (38 percent), run a family business (35 percent), and for recreational, social, and religious activities (7 percent). The overall fuel efficiency of the vehicles has increased considerably over the years due to improvements primarily in aerodynamics, materials, and electronic controls. However, the average fuel consumption of new vehicles has not changed much from about 0 miles per gallon (mpg) because of the increasing consumer trend toward purchasing larger and less fuel-efficient cars, trucks, and sport utility vehicles. Motorists also continue to drive more each year: 11,75 miles in 1999 compared to 10,77 miles in Consequently, the annual gasoline *This section can be skipped without a loss in continuity. Information in this section is based largely on the publications of the U.S. Department of Energy, Environmental Protection Agency, and the American Automotive Association.

53 Chapter use per vehicle in the United States has increased to 603 gallons in 1999 (worth $106 at $.00/gal) from 506 gallons in 1990 (Fig. 9 56). Saving fuel is not limited to good driving habits. It also involves purchasing the right car, using it responsibly, and maintaining it properly. A car does not burn any fuel when it is not running, and thus a sure way to save fuel is not to drive the car at all but this is not the reason we buy a car. We can reduce driving and thus fuel consumption by considering viable alternatives such as living close to work and shopping areas, working at home, working longer hours in fewer days, joining a car pool or starting one, using public transportation, combining errands into a single trip and planning ahead, avoiding rush hours and roads with heavy traffic and many traffic lights, and simply walking or bicycling instead of driving to nearby places, with the added benefit of good health and physical fitness. Driving only when necessary is the best way to save fuel, money, and the environment too. Driving efficiently starts before buying a car, just like raising good children starts before getting married. The buying decision made now will affect the fuel consumption for many years. Under average driving conditions, the owner of a 30-mpg vehicle will spend $400 less each year on fuel than the owner of a 0-mpg vehicle (assuming a fuel cost of $.00 per gallon and 1,000 miles of driving per year). If the vehicle is owned for 5 years, the 30-mpg vehicle will save $000 during this period (Fig. 9 57). The fuel consumption of a car depends on many factors such as the type of the vehicle, the weight, the transmission type, the size and efficiency of the engine, and the accessories and the options installed. The most fuelefficient cars are aerodynamically designed compact cars with a small engine, manual transmission, low frontal area (the height times the width of the car), and bare essentials. At highway speeds, most fuel is used to overcome aerodynamic drag or air resistance to motion, which is the force needed to move the vehicle through the air. This resistance force is proportional to the drag coefficient and the frontal area. Therefore, for a given frontal area, a sleek-looking aerodynamically designed vehicle with contoured lines that coincide with the streamlines of air flow has a smaller drag coefficient and thus better fuel economy than a boxlike vehicle with sharp corners (Fig. 9 58). For the same overall shape, a compact car has a smaller frontal area and thus better fuel economy compared to a large car. Moving around the extra weight requires more fuel, and thus it hurts fuel economy. Therefore, the lighter the vehicle, the more fuel-efficient it is. Also as a general rule, the larger the engine is, the greater its rate of fuel consumption is. So you can expect a car with a 1.8 L engine to be more fuel efficient than one with a 3.0 L engine. For a given engine size, diesel engines operate on much higher compression ratios than the gasoline engines, and thus they are inherently more fuel-efficient. Manual transmissions are usually more efficient than the automatic ones, but this is not always the case. A car with automatic transmission generally uses 10 percent more fuel than a car with manual transmission because of the losses associated with the hydraulic connection between the engine and the transmission, and the added weight. Transmissions with an overdrive gear (found in four-speed automatic transmissions and five-speed manual transmissions) save fuel and reduce FIGURE 9 56 The average car in the United States is driven about 1,000 miles a year, uses about 600 gallons of gasoline, worth $100 at $.00/gal. 30 MPG 0 MPG $800/yr $100/yr FIGURE 9 57 Under average driving conditions, the owner of a 30-mpg vehicle spends $400 less each year on gasoline than the owner of a 0-mpg vehicle (assuming $.00/gal and 1,000 miles/yr).

54 53 Thermodynamics FIGURE 9 58 Aerodynamically designed vehicles have a smaller drag coefficient and thus better fuel economy than boxlike vehicles with sharp corners. noise and engine wear during highway driving by decreasing the engine rpm while maintaining the same vehicle speed. Front wheel drive offers better traction (because of the engine weight on top of the front wheels), reduced vehicle weight and thus better fuel economy, with an added benefit of increased space in the passenger compartment. Four-wheel drive mechanisms provide better traction and braking thus safer driving on slippery roads and loose gravel by transmitting torque to all four wheels. However, the added safety comes with increased weight, noise, and cost, and decreased fuel economy. Radial tires usually reduce the fuel consumption by 5 to 10 percent by reducing the rolling resistance, but their pressure should be checked regularly since they can look normal and still be underinflated. Cruise control saves fuel during long trips on open roads by maintaining steady speed. Tinted windows and light interior and exterior colors reduce solar heat gain, and thus the need for air-conditioning. BEFORE DRIVING Certain things done before driving can make a significant difference on the fuel cost of the vehicle while driving. Below we discuss some measures such as using the right kind of fuel, minimizing idling, removing extra weight, and keeping the tires properly inflated. Use Fuel with the Minimum Octane Number Recommended by the Vehicle Manufacturer Many motorists buy higher-priced premium fuel, thinking that it is better for the engine. Most of today s cars are designed to operate on regular unleaded fuel. If the owner s manual does not call for premium fuel, using anything other than regular gas is simply a waste of money. Octane number is not a measure of the power or quality of the fuel, it is simply a measure of fuel s resistance to engine knock caused by premature ignition. Despite the implications of flashy names like premium, super, or power plus, a fuel with a higher octane number is not a better fuel; it is simply more expensive because of the extra processing involved to raise the octane number (Fig. 9 59). Older cars may need to go up one grade level from the recommended new car octane number if they start knocking. FIGURE 9 59 Despite the implications of flashy names, a fuel with a higher octane number is not a better fuel; it is simply more expensive. Vol. 1/PhotoDisc Do Not Overfill the Gas Tank Topping off the gas tank may cause the fuel to backflow during pumping. In hot weather, an overfilled tank may also cause the fuel to overflow due to thermal expansion. This wastes fuel, pollutes the environment, and may damage the car s paint. Also, fuel tank caps that do not close tightly allow some gasoline to be lost by evaporation. Buying fuel in cool weather such as early in the mornings minimizes evaporative losses. Each gallon of spilled or evaporated fuel emits as much hydrocarbon to the air as 7500 miles of driving.

55 Chapter Park in the Garage The engine of a car parked in a garage overnight is warmer the next morning. This reduces the problems associated with the warming-up period such as starting, excessive fuel consumption, and environmental pollution. In hot weather, a garage blocks the direct sunlight and reduces the need for airconditioning. Start the Car Properly and Avoid Extended Idling With today s cars, it is not necessary to prime the engine first by pumping the accelerator pedal repeatedly before starting. This only wastes fuel. Warming up the engine isn t necessary either. Keep in mind that an idling engine wastes fuel and pollutes the environment. Don t race a cold engine to warm it up. An engine warms up faster on the road under a light load, and the catalytic converter begins to function sooner. Start driving as soon as the engine is started, but avoid rapid acceleration and highway driving before the engine and thus the oil fully warms up to prevent engine wear. In cold weather, the warm-up period is much longer, the fuel consumption during warm-up is much higher, and the exhaust emissions are much larger. At 0 C, for example, a car needs to be driven at least 3 miles to warm up fully. A gasoline engine uses up to 50 percent more fuel during warm-up than it does after it is warmed up. Exhaust emissions from a cold engine during warm-up are much higher since the catalytic converters do not function properly before reaching their normal operating temperature of about 390 C. Don t Carry Unnecessary Weight in or on the Vehicle Remove any snow or ice from the vehicle, and avoid carrying unneeded items, especially heavy ones (such as snow chains, old tires, books) in the passenger compartment, trunk, or the cargo area of the vehicle (Fig. 9 60). This wastes fuel since it requires extra fuel to carry around the extra weight. An extra 100 lbm decreases fuel economy of a car by about 1 percent. Some people find it convenient to use a roof rack or carrier for additional cargo space. However, if you must carry some extra items, place them inside the vehicle rather than on roof racks to reduce drag. Any snow that accumulates on a vehicle and distorts its shape must be removed for the same reason. A loaded roof rack can increase fuel consumption by up to 5 percent in highway driving. Even the most streamlined empty rack increases aerodynamic drag and thus fuel consumption. Therefore, the roof rack should be removed when it is no longer needed. FIGURE 9 60 A loaded roof rack can increase fuel consumption by up to 5 percent in highway driving. Keep Tires Inflated to the Recommended Maximum Pressure Keeping the tires inflated properly is one of the easiest and most important things one can do to improve fuel economy. If a range is recommended by the manufacturer, the higher pressure should be used to maximize fuel efficiency. Tire pressure should be checked when the tire is cold since tire pressure changes with temperature (it increases by 1 psi for every 10 F rise in temperature due to a rise in ambient temperature or just road friction). Underinflated tires run hot and jeopardize safety, cause the tires to wear prematurely, affect

56 534 Thermodynamics the vehicle s handling adversely, and hurt the fuel economy by increasing the rolling resistance. Overinflated tires cause unpleasant bumpy rides, and cause the tires to wear unevenly. Tires lose about 1 psi pressure per month due to air loss caused by the tire hitting holes, bumps, and curbs. Therefore, the tire pressure should be checked at least once a month. Just one tire underinflated by psi results in a 1 percent increase in fuel consumption (Fig. 9 61). Underinflated tires often cause fuel consumption of vehicles to increase by 5 or 6 percent. It is also important to keep the wheels aligned. Driving a vehicle with the front wheels out of alignment increases rolling resistance and thus fuel consumption while causing handling problems and uneven tire wear. Therefore, the wheels should be aligned properly whenever necessary. FIGURE 9 61 Underinflated tires often cause fuel consumption of vehicles to increase by 5 or 6 percent. The McGraw-Hill Companies/Jill Braaten, photographer MPG Speed (mph) FIGURE 9 6 Aerodynamic drag increases and thus fuel economy decreases rapidly at speeds above 55 mph. Source: EPA and U.S. Dept. of Energy. WHILE DRIVING The driving habits can make a significant difference in the amount of fuel used. Driving sensibly and practicing some fuel-efficient driving techniques such as those discussed below can improve fuel economy easily by more than 10 percent. Avoid Quick Starts and Sudden Stops Despite the attention they may get, the abrupt, aggressive jackrabbit starts waste fuel, wear the tires, jeopardize safety, and are harder on vehicle components and connectors. The squealing stops wear the brake pads prematurely, and may cause the driver to lose control of the vehicle. Easy starts and stops save fuel, reduce wear and tear, reduce pollution, and are safer and more courteous to other drivers. Drive at Moderate Speeds Avoiding high speeds on open roads results in safer driving and better fuel economy. In highway driving, over 50 percent of the power produced by the engine is used to overcome aerodynamic drag (i.e., to push air out of the way). Aerodynamic drag and thus fuel consumption increase rapidly at speeds above 55 mph, as shown in Fig On average, a car uses about 15 percent more fuel at 65 mph and 5 percent more fuel at 70 mph than it does at 55 mph. (A car uses about 10 percent more fuel at 100 km/h and 0 percent more fuel at 110 km/h than it does at 90 km/h.) The discussion above should not lead one to conclude that the lower the speed, the better the fuel economy because it is not. The number of miles that can be driven per gallon of fuel drops sharply at speeds below 30 mph (or 50 km/h), as shown in the chart. Besides, speeds slower than the flow of traffic can create a traffic hazard. Therefore, a car should be driven at moderate speeds for safety and best fuel economy. Maintain a Constant Speed The fuel consumption remains at a minimum during steady driving at a moderate speed. Keep in mind that every time the accelerator is hard pressed, more fuel is pumped into the engine. The vehicle should be accelerated gradually and smoothly since extra fuel is squirted into the engine during quick

57 Chapter acceleration. Using cruise control on highway trips can help maintain a constant speed and reduce fuel consumption. Steady driving is also safer, easier on the nerves, and better for the heart. Anticipate Traffic Ahead and Avoid Tailgating A driver can reduce fuel consumption by up to 10 percent by anticipating traffic conditions ahead and adjusting the speed accordingly, and avoiding tailgating and thus unnecessary braking and acceleration (Fig. 9 63). Accelerations and decelerations waste fuel. Braking and abrupt stops can be minimized, for example, by not following too closely, and slowing down gradually by releasing the gas pedal when approaching a red light, a stop sign, or slow traffic. This relaxed driving style is safer, saves fuel and money, reduces pollution, reduces wear on the tires and brakes, and is appreciated by other drivers. Allowing sufficient time to reach the destination makes it easier to resist the urge to tailgate. Avoid Sudden Acceleration and Sudden Braking (Except in Emergencies) Accelerate gradually and smoothly when passing other vehicles or merging with faster traffic. Pumping or hard pressing the accelerator pedal while driving causes the engine to switch to a fuel enrichment mode of operation that wastes fuel. In city driving, nearly half of the engine power is used for acceleration. When accelerating with stick-shifts, the RPM of the engine should be kept to a minimum. Braking wastes the mechanical energy produced by the engine and wears the brake pads. FIGURE 9 63 Fuel consumption can be decreased by up to 10 percent by anticipating traffic conditions ahead and adjusting accordingly. Vol. 3/PhotoDisc Avoid Resting Feet on the Clutch or Brake Pedal while Driving Resting the left foot on the brake pedal increases the temperature of the brake components, and thus reduces their effectiveness and service life while wasting fuel. Similarly, resting the left foot on the clutch pedal lessens the pressure on the clutch pads, causing them to slip and wear prematurely, wasting fuel. Use Highest Gear (Overdrive) During Highway Driving Overdrive improves fuel economy during highway driving by decreasing the vehicle s engine speed (or RPM). The lower engine speed reduces fuel consumption per unit time as well as engine wear. Therefore, overdrive (the fifth gear in cars with overdrive manual transmission) should be used as soon as the vehicle s speed is high enough. Turn the Engine Off Rather Than Letting It Idle Unnecessary idling during lengthy waits (such as waiting for someone or for service at a drive-up window, being stuck in traffic, etc.) wastes fuel, pollutes the air, and causes engine wear (more wear than driving) (Fig. 9 64). Therefore, the engine should be turned off rather than letting it idle. Idling for more than a minute consumes much more fuel than restarting the engine. Fuel consumption in the lines of drive-up windows and the pollution emitted can be avoided altogether by simply parking the car and going inside. FIGURE 9 64 Unnecessary idling during lengthy waits wastes fuel, costs money, and pollutes the air.

58 536 Thermodynamics FIGURE 9 65 Air conditioning increases fuel consumption by 3 to 4 percent during highway driving, and by as much as 10 percent during city driving. FIGURE 9 66 Proper maintenance maximizes fuel efficiency and extends engine life. Use the Air Conditioner Sparingly Air-conditioning consumes considerable power and thus increases fuel consumption by 3 to 4 percent during highway driving, and by as much as 10 percent during city driving (Fig. 9 65). The best alternative to air-conditioning is to supply fresh outdoor air to the car through the vents by turning on the flowthrough ventilation system (usually by running the air conditioner in the economy mode) while keeping the windows and the sunroof closed. This measure is adequate to achieve comfort in pleasant weather, and it saves the most fuel since the compressor of the air conditioner is off. In warmer weather, however, ventilation cannot provide adequate cooling effect. In that case we can attempt to achieve comfort by rolling down the windows or opening the sunroof. This is certainly a viable alternative for city driving, but not so on highways since the aerodynamic drag caused by wide-open windows or sunroof at highway speeds consumes more fuel than does the air conditioner. Therefore, at highway speeds, the windows or the sunroof should be closed and the air conditioner should be turned on instead to save fuel. This is especially the case for the newer, aerodynamically designed cars. Most air conditioners have a maximum or recirculation setting that reduces the amount of hot outside air that must be cooled, and thus the fuel consumption for air-conditioning. A passive measure to reduce the need for air conditioning is to park the vehicle in the shade, and to leave the windows slightly open to allow for air circulation. AFTER DRIVING You cannot be an efficient person and accomplish much unless you take good care of yourself (eating right, maintaining physical fitness, having checkups, etc.), and the cars are no exception. Regular maintenance improves performance, increases gas mileage, reduces pollution, lowers repair costs, and extends engine life. A little time and money saved now may cost a lot later in increased fuel, repair, and replacement costs. Proper maintenance such as checking the levels of fluids (engine oil, coolant, transmission, brake, power steering, windshield washer, etc.), the tightness of all belts, and formation of cracks or frays on hoses, belts, and wires, keeping tires properly inflated, lubricating the moving components, and replacing clogged air, fuel, or oil filters maximizes fuel efficiency (Fig. 9 66). Clogged air filters increase fuel consumption (by up to 10 percent) and pollution by restricting airflow to the engine, and thus they should be replaced. The car should be tuned up regularly unless it has electronic controls and a fuelinjection system. High temperatures (which may be due to a malfunction of the cooling fan) should be avoided as they may cause the break down of the engine oil and thus excessive wear of the engine, and low temperatures (which may be due to a malfunction of the thermostat) may extend the engine s warm-up period and may prevent the engine from reaching the optimum operating conditions. Both effects reduce fuel economy. Clean oil extends engine life by reducing engine wear caused by friction, removes acids, sludge, and other harmful substances from the engine, improves performance, reduces fuel consumption, and decreases air pollution. Oil also helps to cool the engine, provides a seal between the cylinder walls and the

59 Chapter pistons, and prevents the engine from rusting. Therefore, oil and oil filter should be changed as recommended by the vehicle manufacturer. Fuel-efficient oils (indicated by Energy Efficient API label) contain certain additives that reduce friction and increase a vehicle s fuel economy by 3 percent or more. In summary, a person can save fuel, money, and the environment by purchasing an energy-efficient vehicle, minimizing the amount of driving, being fuel-conscious while driving, and maintaining the car properly. These measures have the added benefits of enhanced safety, reduced maintenance costs, and extended vehicle life. SUMMARY A cycle during which a net amount of work is produced is called a power cycle, and a power cycle during which the working fluid remains a gas throughout is called a gas power cycle. The most efficient cycle operating between a heat source at temperature T H and a sink at temperature T L is the Carnot cycle, and its thermal efficiency is given by h th,carnot 1 T L T H The actual gas cycles are rather complex. The approximations used to simplify the analysis are known as the airstandard assumptions. Under these assumptions, all the processes are assumed to be internally reversible; the working fluid is assumed to be air, which behaves as an ideal gas; and the combustion and exhaust processes are replaced by heat-addition and heat-rejection processes, respectively. The air-standard assumptions are called cold-air-standard assumptions if air is also assumed to have constant specific heats at room temperature. In reciprocating engines, the compression ratio r and the mean effective pressure MEP are defined as r V max V min V BDC V TDC w net MEP v max v min The Otto cycle is the ideal cycle for the spark-ignition reciprocating engines, and it consists of four internally reversible processes: isentropic compression, constant-volume heat addition, isentropic expansion, and constant-volume heat rejection. Under cold-air-standard assumptions, the thermal efficiency of the ideal Otto cycle is h th,otto 1 1 r k 1 where r is the compression ratio and k is the specific heat ratio c p /c v. The Diesel cycle is the ideal cycle for the compressionignition reciprocating engines. It is very similar to the Otto cycle, except that the constant-volume heat-addition process is replaced by a constant-pressure heat-addition process. Its thermal efficiency under cold-air-standard assumptions is h th,diesel 1 1 r k 1c r k c 1 k 1r c 1 d where r c is the cutoff ratio, defined as the ratio of the cylinder volumes after and before the combustion process. Stirling and Ericsson cycles are two totally reversible cycles that involve an isothermal heat-addition process at T H and an isothermal heat-rejection process at T L. They differ from the Carnot cycle in that the two isentropic processes are replaced by two constant-volume regeneration processes in the Stirling cycle and by two constant-pressure regeneration processes in the Ericsson cycle. Both cycles utilize regeneration, a process during which heat is transferred to a thermal energy storage device (called a regenerator) during one part of the cycle that is then transferred back to the working fluid during another part of the cycle. The ideal cycle for modern gas-turbine engines is the Brayton cycle, which is made up of four internally reversible processes: isentropic compression, constant-pressure heat addition, isentropic expansion, and constant-pressure heat rejection. Under cold-air-standard assumptions, its thermal efficiency is h th,brayton 1 1 r 1k 1>k p where r p P max /P min is the pressure ratio and k is the specific heat ratio. The thermal efficiency of the simple Brayton cycle increases with the pressure ratio. The deviation of the actual compressor and the turbine from the idealized isentropic ones can be accurately accounted for by utilizing their isentropic efficiencies, defined as h C w s w a h s h 1 h a h 1

60 538 Thermodynamics and h T w a w s h 3 h 4a h 3 h 4s where states 1 and 3 are the inlet states, a and 4a are the actual exit states, and s and 4s are the isentropic exit states. In gas-turbine engines, the temperature of the exhaust gas leaving the turbine is often considerably higher than the temperature of the air leaving the compressor. Therefore, the high-pressure air leaving the compressor can be heated by transferring heat to it from the hot exhaust gases in a counterflow heat exchanger, which is also known as a regenerator. The extent to which a regenerator approaches an ideal regenerator is called the effectiveness P and is defined as P q regen,act q regen,max Under cold-air-standard assumptions, the thermal efficiency of an ideal Brayton cycle with regeneration becomes h th,regen 1 a T 1 T 3 b1r p 1k 1>k where T 1 and T 3 are the minimum and maximum temperatures, respectively, in the cycle. The thermal efficiency of the Brayton cycle can also be increased by utilizing multistage compression with intercooling, regeneration, and multistage expansion with reheating. The work input to the compressor is minimized when equal pressure ratios are maintained across each stage. This procedure also maximizes the turbine work output. Gas-turbine engines are widely used to power aircraft because they are light and compact and have a high powerto-weight ratio. The ideal jet-propulsion cycle differs from the simple ideal Brayton cycle in that the gases are partially expanded in the turbine. The gases that exit the turbine at a relatively high pressure are subsequently accelerated in a nozzle to provide the thrust needed to propel the aircraft. The net thrust developed by the engine is F m # 1V exit V inlet where m. is the mass flow rate of gases, V exit is the exit velocity of the exhaust gases, and V inlet is the inlet velocity of the air, both relative to the aircraft. The power developed from the thrust of the engine is called the propulsive power W. P, and it is given by W # P m # 1V exit V inlet V aircraft Propulsive efficiency is a measure of how efficiently the energy released during the combustion process is converted to propulsive energy, and it is defined as Propulsive power h P Energy input rate W Q # in For an ideal cycle that involves heat transfer only with a source at T H and a sink at T L, the exergy destruction is x dest T 0 a q out T L q in T H b # P REFERENCES AND SUGGESTED READINGS 1. W. Z. Black and J. G. Hartley. Thermodynamics. New York: Harper & Row, V. D. Chase. Propfans: A New Twist for the Propeller. Mechanical Engineering, November 1986, pp C. R. Ferguson and A. T. Kirkpatrick, Internal Combustion Engines: Applied Thermosciences, nd ed., New York: Wiley, R. A. Harmon. The Keys to Cogeneration and Combined Cycles. Mechanical Engineering, February 1988, pp J. Heywood, Internal Combustion Engine Fundamentals, New York: McGraw-Hill, L. C. Lichty. Combustion Engine Processes. New York: McGraw-Hill, H. McIntosh. Jumbo Jet. 10 Outstanding Achievements Washington, D.C.: National Academy of Engineering, 1989, pp W. Pulkrabek, Engineering Fundamentals of the Internal Combustion Engine, nd ed., Upper Saddle River, NJ: Prentice-Hall, W. Siuru. Two-stroke Engines: Cleaner and Meaner. Mechanical Engineering. June 1990, pp C. F. Taylor. The Internal Combustion Engine in Theory and Practice. Cambridge, MA: M.I.T. Press, K. Wark and D. E. Richards. Thermodynamics. 6th ed. New York: McGraw-Hill, 1999.

61 Chapter PROBLEMS* Actual and Ideal Cycles, Carnot Cycle, Air-Standard Assumptions, Reciprocating Engines 9 1C Why is the Carnot cycle not suitable as an ideal cycle for all power-producing cyclic devices? 9 C How does the thermal efficiency of an ideal cycle, in general, compare to that of a Carnot cycle operating between the same temperature limits? 9 3C What does the area enclosed by the cycle represent on a P-v diagram? How about on a T-s diagram? 9 4C What is the difference between air-standard assumptions and the cold-air-standard assumptions? 9 5C How are the combustion and exhaust processes modeled under the air-standard assumptions? 9 6C What are the air-standard assumptions? 9 7C What is the difference between the clearance volume and the displacement volume of reciprocating engines? 9 8C Define the compression ratio for reciprocating engines. 9 9C How is the mean effective pressure for reciprocating engines defined? 9 10C Can the mean effective pressure of an automobile engine in operation be less than the atmospheric pressure? 9 11C As a car gets older, will its compression ratio change? How about the mean effective pressure? 9 1C What is the difference between spark-ignition and compression-ignition engines? 9 13C Define the following terms related to reciprocating engines: stroke, bore, top dead center, and clearance volume An air-standard cycle with variable specific heats is executed in a closed system and is composed of the following four processes: 1- Isentropic compression from 100 kpa and 7 C to 800 kpa -3 v constant heat addition to 1800 K 3-4 Isentropic expansion to 100 kpa 4-1 P constant heat rejection to initial state (a) Show the cycle on P-v and T-s diagrams. (b) Calculate the net work output per unit mass. (c) Determine the thermal efficiency. * Problems designated by a C are concept questions, and students are encouraged to answer them all. Problems designated by an E are in English units, and the SI users can ignore them. Problems with a CD-EES icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed DVD. Problems with a computer-ees icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text Reconsider Problem Using EES (or other) software, study the effect of varying the temperature after the constant-volume heat addition from 1500 K to 500 K. Plot the net work output and thermal efficiency as a function of the maximum temperature of the cycle. Plot the T-s and P-v diagrams for the cycle when the maximum temperature of the cycle is 1800 K An air-standard cycle is executed in a closed system and is composed of the following four processes: 1- Isentropic compression from 100 kpa and 7 C to 1 MPa -3 P constant heat addition in amount of 800 kj/kg 3-4 v constant heat rejection to 100 kpa 4-1 P constant heat rejection to initial state (a) Show the cycle on P-v and T-s diagrams. (b) Calculate the maximum temperature in the cycle. (c) Determine the thermal efficiency. Assume constant specific heats at room temperature. Answers: (b) 3360 K, (c) 1.0 percent 9 17E An air-standard cycle with variable specific heats is executed in a closed system and is composed of the following four processes: 1- v constant heat addition from 14.7 psia and 80 F in the amount of 300 Btu/lbm -3 P constant heat addition to 300 R 3-4 Isentropic expansion to 14.7 psia 4-1 P constant heat rejection to initial state (a) Show the cycle on P-v and T-s diagrams. (b) Calculate the total heat input per unit mass. (c) Determine the thermal efficiency. Answers: (b) 61.4 Btu/lbm, (c) 4. percent 9 18E Repeat Problem 9 17E using constant specific heats at room temperature An air-standard cycle is executed in a closed system with kg of air and consists of the following three processes: 1- Isentropic compression from 100 kpa and 7 C to 1 MPa -3 P constant heat addition in the amount of.76 kj 3-1 P c 1 v + c heat rejection to initial state (c 1 and c are constants) (a) Show the cycle on P-v and T-s diagrams. (b) Calculate the heat rejected. (c) Determine the thermal efficiency. Assume constant specific heats at room temperature. Answers: (b) kj, (c) 39. percent

62 540 Thermodynamics 9 0 An air-standard cycle with variable specific heats is executed in a closed system with kg of air and consists of the following three processes: 1- v constant heat addition from 95 kpa and 17 C to 380 kpa -3 Isentropic expansion to 95 kpa 3-1 P constant heat rejection to initial state (a) Show the cycle on P-v and T-s diagrams. (b) Calculate the net work per cycle, in kj. (c) Determine the thermal efficiency. 9 1 Repeat Problem 9 0 using constant specific heats at room temperature. 9 Consider a Carnot cycle executed in a closed system with kg of air. The temperature limits of the cycle are 300 and 900 K, and the minimum and maximum pressures that occur during the cycle are 0 and 000 kpa. Assuming constant specific heats, determine the net work output per cycle. 9 3 An air-standard Carnot cycle is executed in a closed system between the temperature limits of 350 and 100 K. The pressures before and after the isothermal compression are 150 and 300 kpa, respectively. If the net work output per cycle is 0.5 kj, determine (a) the maximum pressure in the cycle, (b) the heat transfer to air, and (c) the mass of air. Assume variable specific heats for air. Answers: (a) 30,013 kpa, (b) kj, (c) kg 9 4 Repeat Problem 9 3 using helium as the working fluid. 9 5 Consider a Carnot cycle executed in a closed system with air as the working fluid. The maximum pressure in the cycle is 800 kpa while the maximum temperature is 750 K. If the entropy increase during the isothermal heat rejection process is 0.5 kj/kg K and the net work output is 100 kj/kg, determine (a) the minimum pressure in the cycle, (b) the heat rejection from the cycle, and (c) the thermal efficiency of the cycle. (d) If an actual heat engine cycle operates between the same temperature limits and produces 500 kw of power for an air flow rate of 90 kg/s, determine the second law efficiency of this cycle. Otto Cycle 9 6C What four processes make up the ideal Otto cycle? 9 7C How do the efficiencies of the ideal Otto cycle and the Carnot cycle compare for the same temperature limits? Explain. 9 8C How is the rpm (revolutions per minute) of an actual four-stroke gasoline engine related to the number of thermodynamic cycles? What would your answer be for a two-stroke engine? 9 9C Are the processes that make up the Otto cycle analyzed as closed-system or steady-flow processes? Why? 9 30C How does the thermal efficiency of an ideal Otto cycle change with the compression ratio of the engine and the specific heat ratio of the working fluid? 9 31C Why are high compression ratios not used in sparkignition engines? 9 3C An ideal Otto cycle with a specified compression ratio is executed using (a) air, (b) argon, and (c) ethane as the working fluid. For which case will the thermal efficiency be the highest? Why? 9 33C What is the difference between fuel-injected gasoline engines and diesel engines? 9 34 An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kpa and 7 C, and 750 kj/kg of heat is transferred to air during the constant-volume heat-addition process. Taking into account the variation of specific heats with temperature, determine (a) the pressure and temperature at the end of the heataddition process, (b) the net work output, (c) the thermal efficiency, and (d) the mean effective pressure for the cycle. Answers: (a) 3898 kpa, 1539 K, (b) 39.4 kj/kg, (c) 5.3 percent, (d ) 495 kpa 9 35 Reconsider Problem Using EES (or other) software, study the effect of varying the compression ratio from 5 to 10. Plot the net work output and thermal efficiency as a function of the compression ratio. Plot the T-s and P-v diagrams for the cycle when the compression ratio is Repeat Problem 9 34 using constant specific heats at room temperature The compression ratio of an air-standard Otto cycle is 9.5. Prior to the isentropic compression process, the air is at 100 kpa, 35 C, and 600 cm 3. The temperature at the end of the isentropic expansion process is 800 K. Using specific heat values at room temperature, determine (a) the highest temperature and pressure in the cycle; (b) the amount of heat transferred in, in kj; (c) the thermal efficiency; and (d) the mean effective pressure. Answers: (a) 1969 K, 607 kpa, (b) 0.59 kj, (c) 59.4 percent, (d) 65 kpa 9 38 Repeat Problem 9 37, but replace the isentropic expansion process by a polytropic expansion process with the polytropic exponent n E An ideal Otto cycle with air as the working fluid has a compression ratio of 8. The minimum and maximum temperatures in the cycle are 540 and 400 R. Accounting for the variation of specific heats with temperature, determine (a) the amount of heat transferred to the air during the heat-addition process, (b) the thermal efficiency, and (c) the thermal efficiency of a Carnot cycle operating between the same temperature limits. 9 40E Repeat Problem 9 39E using argon as the working fluid A four-cylinder, four-stroke,.-l gasoline engine operates on the Otto cycle with a compression ratio of 10. The air is at 100 kpa and 60 C at the beginning of the compression process, and the maximum pressure in the cycle is 8 MPa. The compression and expansion processes may be

63 modeled as polytropic with a polytropic constant of 1.3. Using constant specific heats at 850 K, determine (a) the temperature at the end of the expansion process, (b) the net work output and the thermal efficiency, (c) the mean effective pressure, (d ) the engine speed for a net power output of 70 kw, and (e) the specific fuel consumption, in g/kwh, defined as the ratio of the mass of the fuel consumed to the net work produced. The air fuel ratio, defined as the amount of air divided by the amount of fuel intake, is 16. DIESEL CYCLE 9 4C How does a diesel engine differ from a gasoline engine? 9 43C How does the ideal Diesel cycle differ from the ideal Otto cycle? 9 44C For a specified compression ratio, is a diesel or gasoline engine more efficient? 9 45C Do diesel or gasoline engines operate at higher compression ratios? Why? 9 46C What is the cutoff ratio? How does it affect the thermal efficiency of a Diesel cycle? 9 47 An air-standard Diesel cycle has a compression ratio of 16 and a cutoff ratio of. At the beginning of the compression process, air is at 95 kpa and 7 C. Accounting for the variation of specific heats with temperature, determine (a) the temperature after the heat-addition process, (b) the thermal efficiency, and (c) the mean effective pressure. Answers: (a) K, (b) 56.3 percent, (c) kpa 9 48 Repeat Problem 9 47 using constant specific heats at room temperature. 9 49E An air-standard Diesel cycle has a compression ratio of 18.. Air is at 80 F and 14.7 psia at the beginning of the compression process and at 3000 R at the end of the heataddition process. Accounting for the variation of specific heats with temperature, determine (a) the cutoff ratio, (b) the heat rejection per unit mass, and (c) the thermal efficiency. 9 50E Repeat Problem 9 49E using constant specific heats at room temperature An ideal diesel engine has a compression ratio of 0 and uses air as the working fluid. The state of air at the beginning of the compression process is 95 kpa and 0 C. If the maximum temperature in the cycle is not to exceed 00 K, determine (a) the thermal efficiency and (b) the mean effective pressure. Assume constant specific heats for air at room temperature. Answers: (a) 63.5 percent, (b) 933 kpa 9 5 Repeat Problem 9 51, but replace the isentropic expansion process by polytropic expansion process with the polytropic exponent n Reconsider Problem 9 5. Using EES (or other) software, study the effect of varying the compression ratio from 14 to 4. Plot the net work output, mean Chapter effective pressure, and thermal efficiency as a function of the compression ratio. Plot the T-s and P-v diagrams for the cycle when the compression ratio is A four-cylinder two-stroke.4-l diesel engine that operates on an ideal Diesel cycle has a compression ratio of 17 and a cutoff ratio of.. Air is at 55 C and 97 kpa at the beginning of the compression process. Using the cold-airstandard assumptions, determine how much power the engine will deliver at 1500 rpm Repeat Problem 9 54 using nitrogen as the working fluid The compression ratio of an ideal dual cycle is 14. Air is at 100 kpa and 300 K at the beginning of the compression process and at 00 K at the end of the heat-addition process. Heat transfer to air takes place partly at constant volume and partly at constant pressure, and it amounts to kj/kg. Assuming variable specific heats for air, determine (a) the fraction of heat transferred at constant volume and (b) the thermal efficiency of the cycle Reconsider Problem Using EES (or other) software, study the effect of varying the compression ratio from 10 to 18. For the compression ratio equal to 14, plot the T-s and P-v diagrams for the cycle Repeat Problem 9 56 using constant specific heats at room temperature. Is the constant specific heat assumption reasonable in this case? 9 59 A six-cylinder, four-stroke, 4.5-L compression-ignition engine operates on the ideal diesel cycle with a compression ratio of 17. The air is at 95 kpa and 55 C at the beginning of the compression process and the engine speed is 000 rpm. The engine uses light diesel fuel with a heating value of 4,500 kj/kg, an air fuel ratio of 4, and a combustion efficiency of 98 percent. Using constant specific heats at 850 K, determine (a) the maximum temperature in the cycle and the cutoff ratio (b) the net work output per cycle and the thermal efficiency, (c) the mean effective pressure, (d ) the net power output, and (e) the specific fuel consumption, in g/kwh, defined as the ratio of the mass of the fuel consumed to the net work produced. Answers: (a) 383 K,.7 (b) 4.36 kj, 0.543, (c) 969 kpa, (d ) 7.7 kw, (e) 159 g/kwh Stirling and Ericsson Cycles 9 60C Consider the ideal Otto, Stirling, and Carnot cycles operating between the same temperature limits. How would you compare the thermal efficiencies of these three cycles? 9 61C Consider the ideal Diesel, Ericsson, and Carnot cycles operating between the same temperature limits. How would you compare the thermal efficiencies of these three cycles? 9 6C What cycle is composed of two isothermal and two constant-volume processes? 9 63C How does the ideal Ericsson cycle differ from the Carnot cycle?

64 54 Thermodynamics 9 64E An ideal Ericsson engine using helium as the working fluid operates between temperature limits of 550 and 3000 R and pressure limits of 5 and 00 psia. Assuming a mass flow rate of 14 lbm/s, determine (a) the thermal efficiency of the cycle, (b) the heat transfer rate in the regenerator, and (c) the power delivered Consider an ideal Ericsson cycle with air as the working fluid executed in a steady-flow system. Air is at 7 C and 10 kpa at the beginning of the isothermal compression process, during which 150 kj/kg of heat is rejected. Heat transfer to air occurs at 100 K. Determine (a) the maximum pressure in the cycle, (b) the net work output per unit mass of air, and (c) the thermal efficiency of the cycle. Answers: (a) 685 kpa, (b) 450 kj/kg, (c) 75 percent 9 66 An ideal Stirling engine using helium as the working fluid operates between temperature limits of 300 and 000 K and pressure limits of 150 kpa and 3 MPa. Assuming the mass of the helium used in the cycle is 0.1 kg, determine (a) the thermal efficiency of the cycle, (b) the amount of heat transfer in the regenerator, and (c) the work output per cycle. Ideal and Actual Gas-Turbine (Brayton) Cycles 9 67C Why are the back work ratios relatively high in gasturbine engines? 9 68C What four processes make up the simple ideal Brayton cycle? 9 69C For fixed maximum and minimum temperatures, what is the effect of the pressure ratio on (a) the thermal efficiency and (b) the net work output of a simple ideal Brayton cycle? 9 70C What is the back work ratio? What are typical back work ratio values for gas-turbine engines? 9 71C How do the inefficiencies of the turbine and the compressor affect (a) the back work ratio and (b) the thermal efficiency of a gas-turbine engine? 9 7E A simple ideal Brayton cycle with air as the working fluid has a pressure ratio of 10. The air enters the compressor at 50 R and the turbine at 000 R. Accounting for the variation of specific heats with temperature, determine (a) the air temperature at the compressor exit, (b) the back work ratio, and (c) the thermal efficiency A simple Brayton cycle using air as the working fluid has a pressure ratio of 8. The minimum and maximum temperatures in the cycle are 310 and 1160 K. Assuming an isentropic efficiency of 75 percent for the compressor and 8 percent for the turbine, determine (a) the air temperature at the turbine exit, (b) the net work output, and (c) the thermal efficiency Reconsider Problem Using EES (or other) software, allow the mass flow rate, pressure ratio, turbine inlet temperature, and the isentropic efficiencies of the turbine and compressor to vary. Assume the compressor inlet pressure is 100 kpa. Develop a general solution for the problem by taking advantage of the diagram window method for supplying data to EES software Repeat Problem 9 73 using constant specific heats at room temperature Air is used as the working fluid in a simple ideal Brayton cycle that has a pressure ratio of 1, a compressor inlet temperature of 300 K, and a turbine inlet temperature of 1000 K. Determine the required mass flow rate of air for a net power output of 70 MW, assuming both the compressor and the turbine have an isentropic efficiency of (a) 100 percent and (b) 85 percent. Assume constant specific heats at room temperature. Answers: (a) 35 kg/s, (b) 1037 kg/s 9 77 A stationary gas-turbine power plant operates on a simple ideal Brayton cycle with air as the working fluid. The air enters the compressor at 95 kpa and 90 K and the turbine at 760 kpa and 1100 K. Heat is transferred to air at a rate of 35,000 kj/s. Determine the power delivered by this plant (a) assuming constant specific heats at room temperature and (b) accounting for the variation of specific heats with temperature Air enters the compressor of a gas-turbine engine at 300 K and 100 kpa, where it is compressed to 700 kpa and 580 K. Heat is transferred to air in the amount of 950 kj/kg before it enters the turbine. For a turbine efficiency of 86 percent, determine (a) the fraction of the turbine work output used to drive the compressor and (b) the thermal efficiency. Assume variable specific heats for air Repeat Problem 9 78 using constant specific heats at room temperature. 9 80E A gas-turbine power plant operates on a simple Brayton cycle with air as the working fluid. The air enters the turbine at 10 psia and 000 R and leaves at 15 psia and 100 R. Heat is rejected to the surroundings at a rate of 6400 Btu/s, and air flows through the cycle at a rate of 40 lbm/s. Assuming the turbine to be isentropic and the compresssor to have an isentropic efficiency of 80 percent, determine the net power output of the plant. Account for the variation of specific heats with temperature. Answer: 3373 kw 9 81E For what compressor efficiency will the gas-turbine power plant in Problem 9 80E produce zero net work? 9 8 A gas-turbine power plant operates on the simple Brayton cycle with air as the working fluid and delivers 3 MW of power. The minimum and maximum temperatures in the cycle are 310 and 900 K, and the pressure of air at the compressor exit is 8 times the value at the compressor inlet. Assuming an isentropic efficiency of 80 percent for the compressor and 86 percent for the turbine, determine the mass flow rate of air through the cycle. Account for the variation of specific heats with temperature Repeat Problem 9 8 using constant specific heats at room temperature.

65 9 84 A gas-turbine power plant operates on the simple Brayton cycle between the pressure limits of 100 and 100 kpa. The working fluid is air, which enters the compressor at 30 C at a rate of 150 m 3 /min and leaves the turbine at 500 C. Using variable specific heats for air and assuming a compressor isentropic efficiency of 8 percent and a turbine isentropic efficiency of 88 percent, determine (a) the net power output, (b) the back work ratio, and (c) the thermal efficiency. Answers: (a) 659 kw, (b) 0.65, (c) Compressor 100 kpa 30 C 1. MPa Combustion chamber FIGURE P Turbine 500 C Brayton Cycle with Regeneration 9 85C How does regeneration affect the efficiency of a Brayton cycle, and how does it accomplish it? 9 86C Somebody claims that at very high pressure ratios, the use of regeneration actually decreases the thermal efficiency of a gas-turbine engine. Is there any truth in this claim? Explain. 9 87C Define the effectiveness of a regenerator used in gas-turbine cycles. 9 88C In an ideal regenerator, is the air leaving the compressor heated to the temperature at (a) turbine inlet, (b) turbine exit, (c) slightly above turbine exit? 9 89C In 1903, Aegidius Elling of Norway designed and built an 11-hp gas turbine that used steam injection between the combustion chamber and the turbine to cool the combustion gases to a safe temperature for the materials available at the time. Currently there are several gas-turbine power plants that use steam injection to augment power and improve thermal efficiency. For example, the thermal efficiency of the General Electric LM5000 gas turbine is reported to increase from 35.8 percent in simple-cycle operation to 43 percent when steam injection is used. Explain why steam injection increases the power output and the efficiency of gas turbines. Also, explain how you would obtain the steam. 9 90E The idea of using gas turbines to power automobiles was conceived in the 1930s, and considerable research was done in the 1940s and 1950s to develop automotive gas turbines by major automobile manufacturers such as the Chrysler and Ford corporations in the United States and 4 Chapter Rover in the United Kingdom. The world s first gas-turbinepowered automobile, the 00-hp Rover Jet 1, was built in 1950 in the United Kingdom. This was followed by the production of the Plymouth Sport Coupe by Chrysler in 1954 under the leadership of G. J. Huebner. Several hundred gasturbine-powered Plymouth cars were built in the early 1960s for demonstration purposes and were loaned to a select group of people to gather field experience. The users had no complaints other than slow acceleration. But the cars were never mass-produced because of the high production (especially material) costs and the failure to satisfy the provisions of the 1966 Clean Air Act. A gas-turbine-powered Plymouth car built in 1960 had a turbine inlet temperature of 1700 F, a pressure ratio of 4, and a regenerator effectiveness of 0.9. Using isentropic efficiencies of 80 percent for both the compressor and the turbine, determine the thermal efficiency of this car. Also, determine the mass flow rate of air for a net power output of 95 hp. Assume the ambient air to be at 540 R and 14.5 psia The 7FA gas turbine manufactured by General Electric is reported to have an efficiency of 35.9 percent in the simple-cycle mode and to produce 159 MW of net power. The pressure ratio is 14.7 and the turbine inlet temperature is 188 C. The mass flow rate through the turbine is 1,536,000 kg/h. Taking the ambient conditions to be 0 C and 100 kpa, determine the isentropic efficiency of the turbine and the compressor. Also, determine the thermal efficiency of this gas turbine if a regenerator with an effectiveness of 80 percent is added. 9 9 Reconsider Problem Using EES (or other) software, develop a solution that allows different isentropic efficiencies for the compressor and turbine and study the effect of the isentropic efficiencies on net work done and the heat supplied to the cycle. Plot the T-s diagram for the cycle An ideal Brayton cycle with regeneration has a pressure ratio of 10. Air enters the compressor at 300 K and the turbine at 100 K. If the effectiveness of the regenerator is 100 percent, determine the net work output and the thermal efficiency of the cycle. Account for the variation of specific heats with temperature Reconsider Problem Using EES (or other) software, study the effects of varying the isentropic efficiencies for the compressor and turbine and regenerator effectiveness on net work done and the heat supplied to the cycle for the variable specific heat case. Plot the T-s diagram for the cycle Repeat Problem 9 93 using constant specific heats at room temperature A Brayton cycle with regeneration using air as the working fluid has a pressure ratio of 7. The minimum and maximum temperatures in the cycle are 310 and 1150 K. Assuming an isentropic efficiency of 75 percent for the compressor and

66 544 Thermodynamics 8 percent for the turbine and an effectiveness of 65 percent for the regenerator, determine (a) the air temperature at the turbine exit, (b) the net work output, and (c) the thermal efficiency. Answers: (a) 783 K, (b) kj/kg, (c).5 percent 9 97 A stationary gas-turbine power plant operates on an ideal regenerative Brayton cycle (P 100 percent) with air as the working fluid. Air enters the compressor at 95 kpa and 90 K and the turbine at 760 kpa and 1100 K. Heat is transferred to air from an external source at a rate of 75,000 kj/s. Determine the power delivered by this plant (a) assuming constant specific heats for air at room temperature and (b) accounting for the variation of specific heats with temperature Air enters the compressor of a regenerative gas-turbine engine at 300 K and 100 kpa, where it is compressed to 800 kpa and 580 K. The regenerator has an effectiveness of 7 percent, and the air enters the turbine at 100 K. For a turbine efficiency of 86 percent, determine (a) the amount of heat transfer in the regenerator and (b) the thermal efficiency. Assume variable specific heats for air. Answers: (a) 15.5 kj/kg, (b) 36.0 percent 9 99 Repeat Problem 9 98 using constant specific heats at room temperature Repeat Problem 9 98 for a regenerator effectiveness of 70 percent. Brayton Cycle with Intercooling, Reheating, and Regeneration 9 101C Under what modifications will the ideal simple gas-turbine cycle approach the Ericsson cycle? 9 10C The single-stage compression process of an ideal Brayton cycle without regeneration is replaced by a multistage compression process with intercooling between the same pressure limits. As a result of this modification, (a) Does the compressor work increase, decrease, or remain the same? (b) Does the back work ratio increase, decrease, or remain the same? (c) Does the thermal efficiency increase, decrease, or remain the same? 9 103C The single-stage expansion process of an ideal Brayton cycle without regeneration is replaced by a multistage expansion process with reheating between the same pressure limits. As a result of this modification, (a) Does the turbine work increase, decrease, or remain the same? (b) Does the back work ratio increase, decrease, or remain the same? (c) Does the thermal efficiency increase, decrease, or remain the same? 9 104C A simple ideal Brayton cycle without regeneration is modified to incorporate multistage compression with inter- cooling and multistage expansion with reheating, without changing the pressure or temperature limits of the cycle. As a result of these two modifications, (a) Does the net work output increase, decrease, or remain the same? (b) Does the back work ratio increase, decrease, or remain the same? (c) Does the thermal efficiency increase, decrease, or remain the same? (d) Does the heat rejected increase, decrease, or remain the same? 9 105C A simple ideal Brayton cycle is modified to incorporate multistage compression with intercooling, multistage expansion with reheating, and regeneration without changing the pressure limits of the cycle. As a result of these modifications, (a) Does the net work output increase, decrease, or remain the same? (b) Does the back work ratio increase, decrease, or remain the same? (c) Does the thermal efficiency increase, decrease, or remain the same? (d) Does the heat rejected increase, decrease, or remain the same? 9 106C For a specified pressure ratio, why does multistage compression with intercooling decrease the compressor work, and multistage expansion with reheating increase the turbine work? 9 107C In an ideal gas-turbine cycle with intercooling, reheating, and regeneration, as the number of compression and expansion stages is increased, the cycle thermal efficiency approaches (a) 100 percent, (b) the Otto cycle efficiency, or (c) the Carnot cycle efficiency Consider an ideal gas-turbine cycle with two stages of compression and two stages of expansion. The pressure ratio across each stage of the compressor and turbine is 3. The air enters each stage of the compressor at 300 K and each stage of the turbine at 100 K. Determine the back work ratio and the thermal efficiency of the cycle, assuming (a) no regenerator is used and (b) a regenerator with 75 percent effectiveness is used. Use variable specific heats Repeat Problem 9 108, assuming an efficiency of 80 percent for each compressor stage and an efficiency of 85 percent for each turbine stage Consider a regenerative gas-turbine power plant with two stages of compression and two stages of expansion. The overall pressure ratio of the cycle is 9. The air enters each stage of the compressor at 300 K and each stage of the turbine at 100 K. Accounting for the variation of specific heats with temperature, determine the minimum mass flow rate of air needed to develop a net power output of 110 MW. Answer: 50 kg/s

67 9 111 Repeat Problem using argon as the working fluid. Jet-Propulsion Cycles 9 11C What is propulsive power? How is it related to thrust? 9 113C What is propulsive efficiency? How is it determined? 9 114C Is the effect of turbine and compressor irreversibilities of a turbojet engine to reduce (a) the net work, (b) the thrust, or (c) the fuel consumption rate? 9 115E A turbojet is flying with a velocity of 900 ft/s at an altitude of 0,000 ft, where the ambient conditions are 7 psia and 10 F. The pressure ratio across the compressor is 13, and the temperature at the turbine inlet is 400 R. Assuming ideal operation for all components and constant specific heats for air at room temperature, determine (a) the pressure at the turbine exit, (b) the velocity of the exhaust gases, and (c) the propulsive efficiency E Repeat Problem 9 115E accounting for the variation of specific heats with temperature A turbojet aircraft is flying with a velocity of 30 m/s at an altitude of 9150 m, where the ambient conditions are 3 kpa and 3 C. The pressure ratio across the compressor is 1, and the temperature at the turbine inlet is 1400 K. Air enters the compressor at a rate of 60 kg/s, and the jet fuel has a heating value of 4,700 kj/kg. Assuming ideal operation for all components and constant specific heats for air at room temperature, determine (a) the velocity of the exhaust gases, (b) the propulsive power developed, and (c) the rate of fuel consumption Repeat Problem using a compressor efficiency of 80 percent and a turbine efficiency of 85 percent Consider an aircraft powered by a turbojet engine that has a pressure ratio of 1. The aircraft is stationary on the ground, held in position by its brakes. The ambient air is at 7 C and 95 kpa and enters the engine at a rate of 10 kg/s. The jet fuel has a heating value of 4,700 kj/kg, and it is burned completely at a rate of 0. kg/s. Neglecting the effect of the diffuser and disregarding the slight increase in mass at the engine exit as well as the inefficiencies of engine components, determine the force that must be applied on the brakes to hold the plane stationary. Answer: 9089 N 9 10 Reconsider Problem In the problem statement, replace the inlet mass flow rate by an inlet volume flow rate of m 3 /s. Using EES (or other) software, investigate the effect of compressor inlet temperature in the range of 0 to 30 C on the force that must be applied to the brakes to hold the plane stationary. Plot this force as a function in compressor inlet temperature Air at 7 C enters a turbojet engine at a rate of 16 kg/s and at a velocity of 300 m/s (relative to the engine). Chapter Air is heated in the combustion chamber at a rate 15,000 kj/s and it leaves the engine at 47 C. Determine the thrust produced by this turbojet engine. (Hint: Choose the entire engine as your control volume.) Second-Law Analysis of Gas Power Cycles 9 1 Determine the total exergy destruction associated with the Otto cycle described in Problem 9 34, assuming a source temperature of 000 K and a sink temperature of 300 K. Also, determine the exergy at the end of the power stroke. Answers: 45.1 kj/kg, 145. kj/kg 9 13 Determine the total exergy destruction associated with the Diesel cycle described in Problem 9 47, assuming a source temperature of 000 K and a sink temperature of 300 K. Also, determine the exergy at the end of the isentropic compression process. Answers: 9.7 kj/kg, kj/kg 9 14E Determine the exergy destruction associated with the heat rejection process of the Diesel cycle described in Problem 9 49E, assuming a source temperature of 3500 R and a sink temperature of 540 R. Also, determine the exergy at the end of the isentropic expansion process Calculate the exergy destruction associated with each of the processes of the Brayton cycle described in Problem 9 73, assuming a source temperature of 1600 K and a sink temperature of 90 K Determine the total exergy destruction associated with the Brayton cycle described in Problem 9 93, assuming a source temperature of 1800 K and a sink temperature of 300 K. Also, determine the exergy of the exhaust gases at the exit of the regenerator Reconsider Problem Using EES (or other) software, investigate the effect of varying the cycle pressure ratio from 6 to 14 on the total exergy destruction for the cycle and the exergy of the exhaust gas leaving the regenerator. Plot these results as functions of pressure ratio. Discuss the results Determine the exergy destruction associated with each of the processes of the Brayton cycle described in Problem 9 98, assuming a source temperature of 160 K and a sink temperature of 300 K. Also, determine the exergy of the exhaust gases at the exit of the regenerator. Take P exhaust P kpa A gas-turbine power plant operates on the simple Brayton cycle between the pressure limits of 100 and 700 kpa. Air enters the compressor at 30 C at a rate of 1.6 kg/s and leaves at 60 C. A diesel fuel with a heating value of 4,000 kj/kg is burned in the combustion chamber with an air fuel ratio of 60 and a combustion efficiency of 97 percent. Combustion gases leave the combustion chamber and enter the turbine whose isentropic efficiency is 85 percent. Treating the combustion gases as air and using constant specific heats at 500 C, determine (a) the isentropic efficiency

68 546 Thermodynamics Diesel fuel Compressor 700 kpa 60 C Combustion chamber 3 Turbine 6 Regenerator 100 kpa 30 C kpa 60 C C Combustion chamber 871 C kpa 30 C 4 Compressor Turbine FIGURE P9 19 of the compressor, (b) the net power output and the back work ratio, (c) the thermal efficiency, and (d) the second-law efficiency A four-cylinder, four-stroke,.8-liter modern, highspeed compression-ignition engine operates on the ideal dual cycle with a compression ratio of 14. The air is at 95 kpa and 55 C at the beginning of the compression process and the engine speed is 3500 rpm. Equal amounts of fuel are burned at constant volume and at constant pressure. The maximum allowable pressure in the cycle is 9 MPa due to material strength limitations. Using constant specific heats at 850 K, determine (a) the maximum temperature in the cycle, (b) the net work output and the thermal efficiency, (c) the mean effective pressure, and (d) the net power output. Also, determine (e) the second-law efficiency of the cycle and the rate of exergy output with the exhaust gases when they are purged. Answers: (a) 354 K, (b) 1349 kj/kg, 0.587, (c) 1466 kpa, (d) 10 kw, (e) 0.646, 50.4 kw A gas-turbine power plant operates on the regenerative Brayton cycle between the pressure limits of 100 and 700 kpa. Air enters the compressor at 30 C at a rate of 1.6 kg/s and leaves at 60 C. It is then heated in a regenerator to 400 C by the hot combustion gases leaving the turbine. A diesel fuel with a heating value of 4,000 kj/kg is burned in the combustion chamber with a combustion efficiency of 97 percent. The combustion gases leave the combustion chamber at 871 C and enter the turbine whose isentropic efficiency is 85 percent. Treating combustion gases as air and using constant specific heats at 500 C, determine (a) the isentropic efficiency of the compressor, (b) the effectiveness of the regenerator, (c) the air fuel ratio in the combustion chamber, (d) the net power output and the back work ratio, (e) the thermal efficiency, and ( f ) the second-law efficiency of the plant. Also determine (g) the second-law (exergetic) efficiencies of the compressor, the turbine, and the regenerator, and (h) the rate of the exergy flow with the combustion gases at the regenerator exit. Answers: (a) 0.881, (b) 0.63, (c) 78.1, (d) 67 kw, 0.583, (e) 0.345, (f ) 0.469, (g) 0.99, 0.93, 0.890, (h) 1351 kw Review Problems FIGURE P A four-stroke turbocharged V-16 diesel engine built by GE Transportation Systems to power fast trains produces 3500 hp at 100 rpm. Determine the amount of power produced per cylinder per (a) mechanical cycle and (b) thermodynamic cycle Consider a simple ideal Brayton cycle operating between the temperature limits of 300 and 1500 K. Using constant specific heats at room temperature, determine the pressure ratio for which the compressor and the turbine exit temperatures of air are equal An air-standard cycle with variable coefficients is executed in a closed system and is composed of the following four processes: 1- v constant heat addition from 100 kpa and 7 C to 300 kpa -3 P constant heat addition to 107 C 3-4 Isentropic expansion to 100 kpa 4-1 P constant heat rejection to initial state (a) Show the cycle on P-v and T-s diagrams. (b) Calculate the net work output per unit mass. (c) Determine the thermal efficiency Repeat Problem using constant specific heats at room temperature An air-standard cycle with variable specific heats is executed in a closed system with kg of air, and it consists of the following three processes: 1- Isentropic compression from 100 kpa and 7 C to 700 kpa -3 P constant heat addition to initial specific volume 3-1 v constant heat rejection to initial state (a) Show the cycle on P-v and T-s diagrams. (b) Calculate the maximum temperature in the cycle. (c) Determine the thermal efficiency. Answers: (b) 100 K, (c) 15.8 percent

69 9 137 Repeat Problem using constant specific heats at room temperature A Carnot cycle is executed in a closed system and uses kg of air as the working fluid. The cycle efficiency is 60 percent, and the lowest temperature in the cycle is 300 K. The pressure at the beginning of the isentropic expansion is 700 kpa, and at the end of the isentropic compression it is 1 MPa. Determine the net work output per cycle A four-cylinder spark-ignition engine has a compression ratio of 8, and each cylinder has a maximum volume of 0.6 L. At the beginning of the compression process, the air is at 98 kpa and 17 C, and the maximum temperature in the cycle is 1800 K. Assuming the engine to operate on the ideal Otto cycle, determine (a) the amount of heat supplied per cylinder, (b) the thermal efficiency, and (c) the number of revolutions per minute required for a net power output of 60 kw. Assume variable specific heats for air Reconsider Problem Using EES (or other) software, study the effect of varying the compression ratio from 5 to 11 on the net work done and the efficiency of the cycle. Plot the P-v and T-s diagrams for the cycle, and discuss the results An ideal Otto cycle has a compression ratio of 9. and uses air as the working fluid. At the beginning of the compression process, air is at 98 kpa and 7 C. The pressure is doubled during the constant-volume heat-addition process. Accounting for the variation of specific heats with temperature, determine (a) the amount of heat transferred to the air, (b) the net work output, (c) the thermal efficiency, and (d) the mean effective pressure for the cycle Repeat Problem using constant specific heats at room temperature Consider an engine operating on the ideal Diesel cycle with air as the working fluid. The volume of the cylinder is 100 cm 3 at the beginning of the compression process, 75 cm 3 at the end, and 150 cm 3 after the heat-addition process. Air is at 17 C and 100 kpa at the beginning of the compression process. Determine (a) the pressure at the beginning of the heat-rejection process, (b) the net work per cycle, in kj, and (c) the mean effective pressure Repeat Problem using argon as the working fluid E An ideal dual cycle has a compression ratio of 1 and uses air as the working fluid. At the beginning of the compression process, air is at 14.7 psia and 90 F, and occupies a volume of 75 in 3. During the heat-addition process, 0.3 Btu of heat is transferred to air at constant volume and 1.1 Btu at constant pressure. Using constant specific heats evaluated at room temperature, determine the thermal efficiency of the cycle Consider an ideal Stirling cycle using air as the working fluid. Air is at 350 K and 00 kpa at the beginning of the Chapter isothermal compression process, and heat is supplied to air from a source at 1800 K in the amount of 900 kj/kg. Determine (a) the maximum pressure in the cycle and (b) the net work output per unit mass of air. Answers: (a) 5873 kpa, (b) 75 kj/kg Consider a simple ideal Brayton cycle with air as the working fluid. The pressure ratio of the cycle is 6, and the minimum and maximum temperatures are 300 and 1300 K, respectively. Now the pressure ratio is doubled without changing the minimum and maximum temperatures in the cycle. Determine the change in (a) the net work output per unit mass and (b) the thermal efficiency of the cycle as a result of this modification. Assume variable specific heats for air. Answers: (a) 41.5 kj/kg, (b) 10.6 percent Repeat Problem using constant specific heats at room temperature Helium is used as the working fluid in a Brayton cycle with regeneration. The pressure ratio of the cycle is 8, the compressor inlet temperature is 300 K, and the turbine inlet temperature is 1800 K. The effectiveness of the regenerator is 75 percent. Determine the thermal efficiency and the required mass flow rate of helium for a net power output of 60 MW, assuming both the compressor and the turbine have an isentropic efficiency of (a) 100 percent and (b) 80 percent A gas-turbine engine with regeneration operates with two stages of compression and two stages of expansion. The pressure ratio across each stage of the compressor and turbine is 3.5. The air enters each stage of the compressor at 300 K and each stage of the turbine at 100 K. The compressor and turbine efficiencies are 78 and 86 percent, respectively, and the effectiveness of the regenerator is 7 percent. Determine the back work ratio and the thermal efficiency of the cycle, assuming constant specific heats for air at room temperature. Answers: 53. percent, 39. percent Reconsider Problem Using EES (or other) software, study the effects of varying the isentropic efficiencies for the compressor and turbine and regenerator effectiveness on net work done and the heat supplied to the cycle for the variable specific heat case. Let the isentropic efficiencies and the effectiveness vary from 70 percent to 90 percent. Plot the T-s diagram for the cycle Repeat Problem using helium as the working fluid Consider the ideal regenerative Brayton cycle. Determine the pressure ratio that maximizes the thermal efficiency of the cycle and compare this value with the pressure ratio that maximizes the cycle net work. For the same maximumto-minimum temperature ratios, explain why the pressure ratio for maximum efficiency is less than the pressure ratio for maximum work Consider an ideal gas-turbine cycle with one stage of compression and two stages of expansion and regeneration.

70 548 Thermodynamics The pressure ratio across each turbine stage is the same. The high-pressure turbine exhaust gas enters the regenerator and then enters the low-pressure turbine for expansion to the compressor inlet pressure. Determine the thermal efficiency of this cycle as a function of the compressor pressure ratio and the high-pressure turbine to compressor inlet temperature ratio. Compare your result with the efficiency of the standard regenerative cycle A four-cylinder, four-stroke spark-ignition engine operates on the ideal Otto cycle with a compression ratio of 11 and a total displacement volume of 1.8 liter. The air is at 90 kpa and 50 C at the beginning of the compression process. The heat input is 1.5 kj per cycle per cylinder. Accounting for the variation of specific heats of air with temperature, determine (a) the maximum temperature and pressure that occur during the cycle, (b) the net work per cycle per cyclinder and the thermal efficiency of the cycle, (c) the mean effective pressure, and (d) the power output for an engine speed of 3000 rpm A gas-turbine plant operates on the regenerative Brayton cycle with two stages of reheating and two-stages of intercooling between the pressure limits of 100 and 100 kpa. The working fluid is air. The air enters the first and the second stages of the compressor at 300 K and 350 K, respectively, and the first and the second stages of the turbine at 1400 K and 1300 K, respectively. Assuming both the compressor and the turbine have an isentropic efficiency of 80 percent and the regenerator has an effectiveness of 75 percent and using variable specific heats, determine (a) the back work ratio and the net work output, (b) the thermal efficiency, and (c) the second-law efficiency of the cycle. Also determine (d) the exergies at the exits of the combustion chamber (state 6) and the regenerator (state 10) (See Figure 9 43 in the text). Answers: (a) 0.53, 317 kj/kg, (b) 0.553, (c) 0.704, (d) 931 kj/kg, 19 kj/kg Electricity and process heat requirements of a manufacturing facility are to be met by a cogeneration plant consisting of a gas turbine and a heat exchanger for steam production. 1 Compressor 100 kpa 30 C 350 C 5 C Heat Combustion exchanger chamber 3 1. MPa C Sat. vapor 00 C FIGURE P9 157 Turbine The plant operates on the simple Brayton cycle between the pressure limits of 100 and 100 kpa with air as the working fluid. Air enters the compressor at 30 C. Combustion gases leave the turbine and enter the heat exchanger at 500 C, and leave the heat exchanger of 350 C, while the liquid water enters the heat exchanger at 5 C and leaves at 00 C as a saturated vapor. The net power produced by the gas-turbine cycle is 800 kw. Assuming a compressor isentropic efficiency of 8 percent and a turbine isentropic efficiency of 88 percent and using variable specific heats, determine (a) the mass flow rate of air, (b) the back work ratio and the thermal efficiency, and (c) the rate at which steam is produced in the heat exchanger. Also determine (d) the utilization efficiency of the cogeneration plant, defined as the ratio of the total energy utilized to the energy supplied to the plant A turbojet aircraft flies with a velocity of 900 km/h at an altitude where the air temperature and pressure are 35 C and 40 kpa. Air leaves the diffuser at 50 kpa with a velocity of 15 m/s, and combustion gases enter the turbine at 450 kpa and 950 C. The turbine produces 500 kw of power, all of which is used to drive the compressor. Assuming an isentropic efficiency of 83 percent for the compressor, turbine, and nozzle, and using variable specific heats, determine (a) the pressure of combustion gases at the turbine exit, (b) the mass flow rate of air through the compressor, (c) the velocity of the gases at the nozzle exit, and (d) the propulsive power and the propulsive efficiency for this engine. Answers: (a) 147 kpa, (b) 1.76 kg/s, (c) 719 m/s, (d) 06 kw, Using EES (or other) software, study the effect of variable specific heats on the thermal efficiency of the ideal Otto cycle using air as the working fluid. At the beginning of the compression process, air is at 100 kpa and 300 K. Determine the percentage of error involved in using constant specific heat values at room temperature for the following combinations of compression ratios and maximum cycle temperatures: r 6, 8, 10, 1, and T max 1000, 1500, 000, 500 K Using EES (or other) software, determine the effects of compression ratio on the net work output and the thermal efficiency of the Otto cycle for a maximum cycle temperature of 000 K. Take the working fluid to be air that is at 100 kpa and 300 K at the beginning of the compression process, and assume variable specific heats. Vary the compression ratio from 6 to 15 with an increment of 1. Tabulate and plot your results against the compression ratio Using EES (or other) software, determine the effects of pressure ratio on the net work output and the thermal efficiency of a simple Brayton cycle for a maximum cycle temperature of 1800 K. Take the working fluid to be air that is at 100 kpa and 300 K at the beginning of the compression process, and assume variable specific heats. Vary the pressure ratio from 5 to 4 with an increment of 1. Tabulate and plot your results against the pressure ratio. At what pressure ratio does the net work output become a

71 maximum? At what pressure ratio does the thermal efficiency become a maximum? 9 16 Repeat Problem assuming isentropic efficiencies of 85 percent for both the turbine and the compressor Using EES (or other) software, determine the effects of pressure ratio, maximum cycle temperature, and compressor and turbine efficiencies on the net work output per unit mass and the thermal efficiency of a simple Brayton cycle with air as the working fluid. Air is at 100 kpa and 300 K at the compressor inlet. Also, assume constant specific heats for air at room temperature. Determine the net work output and the thermal efficiency for all combinations of the following parameters, and draw conclusions from the results. Pressure ratio: 5, 8, 14 Maximum cycle temperature: 800, 100, 1600 K Compressor isentropic efficiency: 80, 100 percent Turbine isentropic efficiency: 80, 100 percent Repeat Problem by considering the variation of specific heats of air with temperature Repeat Problem using helium as the working fluid Using EES (or other) software, determine the effects of pressure ratio, maximum cycle temperature, regenerator effectiveness, and compressor and turbine efficiencies on the net work output per unit mass and on the thermal efficiency of a regenerative Brayton cycle with air as the working fluid. Air is at 100 kpa and 300 K at the compressor inlet. Also, assume constant specific heats for air at room temperature. Determine the net work output and the thermal efficiency for all combinations of the following parameters. Pressure ratio: 6, 10 Maximum cycle temperature: 1500, 000 K Compressor isentropic efficiency: 80, 100 percent Turbine isentropic efficiency: 80, 100 percent Regenerator effectiveness: 70, 90 percent Repeat Problem by considering the variation of specific heats of air with temperature Repeat Problem using helium as the working fluid Using EES (or other) software, determine the effect of the number of compression and expansion stages on the thermal efficiency of an ideal regenerative Brayton cycle with multistage compression and expansion. Assume that the overall pressure ratio of the cycle is 1, and the air enters each stage of the compressor at 300 K and each stage of the turbine at 100 K. Using constant specific heats for air at room temperature, determine the thermal efficiency of the cycle by varying the number of stages from 1 to in increments of 3. Plot the thermal efficiency versus the number Chapter of stages. Compare your results to the efficiency of an Ericsson cycle operating between the same temperature limits Repeat Problem using helium as the working fluid. Fundamentals of Engineering (FE) Exam Problems An Otto cycle with air as the working fluid has a compression ratio of 8.. Under cold-air-standard conditions, the thermal efficiency of this cycle is (a) 4 percent (b) 43 percent (c) 5 percent (d) 57 percent (e) 75 percent 9 17 For specified limits for the maximum and minimum temperatures, the ideal cycle with the lowest thermal efficiency is (a) Carnot (b) Stirling (c) Ericsson (d ) Otto (e) All are the same A Carnot cycle operates between the temperature limits of 300 and 000 K, and produces 600 kw of net power. The rate of entropy change of the working fluid during the heat addition process is (a) 0 (b) kw/k (c) kw/k (d ) 0.61 kw/k (e).0 kw/k Air in an ideal Diesel cycle is compressed from 3 to 0.15 L, and then it expands during the constant pressure heat addition process to 0.30 L. Under cold air standard conditions, the thermal efficiency of this cycle is (a) 35 percent (b) 44 percent (c) 65 percent (d) 70 percent (e) 8 percent Helium gas in an ideal Otto cycle is compressed from 0 C and.5 to 0.5 L, and its temperature increases by an additional 700 C during the heat addition process. The temperature of helium before the expansion process is (a) 1790 C (b) 060 C (c) 140 C (d) 60 C (e) 80 C In an ideal Otto cycle, air is compressed from 1.0 kg/m 3 and. to 0.6 L, and the net work output of the cycle is 440 kj/kg. The mean effective pressure (MEP) for this cycle is (a) 61 kpa (b) 599 kpa (c) 58 kpa (d) 416 kpa (e) 367 kpa In an ideal Brayton cycle, air is compressed from 95 kpa and 5 C to 800 kpa. Under cold-air-standard conditions, the thermal efficiency of this cycle is (a) 46 percent (b) 54 percent (c) 57 percent (d) 39 percent (e) 61 percent Consider an ideal Brayton cycle executed between the pressure limits of 100 and 100 kpa and temperature limits of 0 and 1000 C with argon as the working fluid. The net work output of the cycle is (a) 68 kj/kg (b) 93 kj/kg (c) 158 kj/kg (d) 186 kj/kg (e) 310 kj/kg

72 550 Thermodynamics An ideal Brayton cycle has a net work output of 150 kj/kg and a back work ratio of 0.4. If both the turbine and the compressor had an isentropic efficiency of 85 percent, the net work output of the cycle would be (a) 74 kj/kg (b) 95 kj/kg (c) 109 kj/kg (d) 18 kj/kg (e) 177 kj/kg In an ideal Brayton cycle, air is compressed from 100 kpa and 5 C to 1 MPa, and then heated to 100 C before entering the turbine. Under cold-air-standard conditions, the air temperature at the turbine exit is (a) 490 C (b) 515 C (c) 6 C (d) 763 C (e) 895 C In an ideal Brayton cycle with regeneration, argon gas is compressed from 100 kpa and 5 C to 400 kpa, and then heated to 100 C before entering the turbine. The highest temperature that argon can be heated in the regenerator is (a) 46 C (b) 846 C (c) 689 C (d) 368 C (e) 573 C 9 18 In an ideal Brayton cycle with regeneration, air is compressed from 80 kpa and 10 C to 400 kpa and 175 C, is heated to 450 C in the regenerator, and then further heated to 1000 C before entering the turbine. Under cold-air-standard conditions, the effectiveness of the regenerator is (a) 33 percent (b) 44 percent (c) 6 percent (d) 77 percent (e) 89 percent Consider a gas turbine that has a pressure ratio of 6 and operates on the Brayton cycle with regeneration between the temperature limits of 0 and 900 C. If the specific heat ratio of the working fluid is 1.3, the highest thermal efficiency this gas turbine can have is (a) 38 percent (b) 46 percent (c) 6 percent (d) 58 percent (e) 97 percent An ideal gas turbine cycle with many stages of compression and expansion and a regenerator of 100 percent effectiveness has an overall pressure ratio of 10. Air enters every stage of compressor at 90 K, and every stage of turbine at 100 K. The thermal efficiency of this gas-turbine cycle is (a) 36 percent (b) 40 percent (c) 5 percent (d) 64 percent (e) 76 percent Air enters a turbojet engine at 60 m/s at a rate of 30 kg/s, and exits at 800 m/s relative to the aircraft. The thrust developed by the engine is (a) 8 kn (b) 16 kn (c) 4 kn (d) 0 kn (e) 3 kn Design and Essay Problems Design a closed-system air-standard gas power cycle composed of three processes and having a minimum thermal efficiency of 0 percent. The processes may be isothermal, isobaric, isochoric, isentropic, polytropic, or pressure as a linear function of volume. Prepare an engineering report describ- ing your design, showing the system, P-v and T-s diagrams, and sample calculations Design a closed-system air-standard gas power cycle composed of three processes and having a minimum thermal efficiency of 0 percent. The processes may be isothermal, isobaric, isochoric, isentropic, polytropic, or pressure as a linear function of volume; however, the Otto, Diesel, Ericsson, and Stirling cycles may not be used. Prepare an engineering report describing your design, showing the system, P-v and T-s diagrams, and sample calculations Write an essay on the most recent developments on the two-stroke engines, and find out when we might be seeing cars powered by two-stroke engines in the market. Why do the major car manufacturers have a renewed interest in two-stroke engines? In response to concerns about the environment, some major car manufacturers are currently marketing electric cars. Write an essay on the advantages and disadvantages of electric cars, and discuss when it is advisable to purchase an electric car instead of a traditional internal combustion car Intense research is underway to develop adiabatic engines that require no cooling of the engine block. Such engines are based on ceramic materials because of the ability of such materials to withstand high temperatures. Write an essay on the current status of adiabatic engine development. Also determine the highest possible efficiencies with these engines, and compare them to the highest possible efficiencies of current engines Since its introduction in 1903 by Aegidius Elling of Norway, steam injection between the combustion chamber and the turbine is used even in some modern gas turbines currently in operation to cool the combustion gases to a metallurgical-safe temperature while increasing the mass flow rate through the turbine. Currently there are several gasturbine power plants that use steam injection to augment power and improve thermal efficiency. Consider a gas-turbine power plant whose pressure ratio is 8. The isentropic efficiencies of the compressor and the turbine are 80 percent, and there is a regenerator with an effectiveness of 70 percent. When the mass flow rate of air through the compressor is 40 kg/s, the turbine inlet temperature becomes 1700 K. But the turbine inlet temperature is limited to 1500 K, and thus steam injection into the combustion gases is being considered. However, to avoid the complexities associated with steam injection, it is proposed to use excess air (that is, to take in much more air than needed for complete combustion) to lower the combustion and thus turbine inlet temperature while increasing the mass flow rate and thus power output of the turbine. Evaluate this proposal, and compare the thermodynamic performance of high air flow to that of a steam-injection gas-turbine power plant under the following design conditions: the ambient air is at 100 kpa and 5 C, adequate water supply is available at 0 C, and the amount of fuel supplied to the combustion chamber remains constant.

73 Chapter 10 VAPOR AND COMBINED POWER CYCLES In Chap. 9 we discussed gas power cycles for which the working fluid remains a gas throughout the entire cycle. In this chapter, we consider vapor power cycles in which the working fluid is alternatively vaporized and condensed. We also consider power generation coupled with process heating called cogeneration. The continued quest for higher thermal efficiencies has resulted in some innovative modifications to the basic vapor power cycle. Among these, we discuss the reheat and regenerative cycles, as well as combined gas vapor power cycles. Steam is the most common working fluid used in vapor power cycles because of its many desirable characteristics, such as low cost, availability, and high enthalpy of vaporization. Therefore, this chapter is mostly devoted to the discussion of steam power plants. Steam power plants are commonly referred to as coal plants, nuclear plants, or natural gas plants, depending on the type of fuel used to supply heat to the steam. However, the steam goes through the same basic cycle in all of them. Therefore, all can be analyzed in the same manner. Objectives The objectives of Chapter 10 are to: Analyze vapor power cycles in which the working fluid is alternately vaporized and condensed. Analyze power generation coupled with process heating called cogeneration. Investigate ways to modify the basic Rankine vapor power cycle to increase the cycle thermal efficiency. Analyze the reheat and regenerative vapor power cycles. Analyze power cycles that consist of two separate cycles known as combined cycles and binary cycles. 551

74 55 Thermodynamics 10 1 THE CARNOT VAPOR CYCLE We have mentioned repeatedly that the Carnot cycle is the most efficient cycle operating between two specified temperature limits. Thus it is natural to look at the Carnot cycle first as a prospective ideal cycle for vapor power plants. If we could, we would certainly adopt it as the ideal cycle. As explained below, however, the Carnot cycle is not a suitable model for power cycles. Throughout the discussions, we assume steam to be the working fluid since it is the working fluid predominantly used in vapor power cycles. Consider a steady-flow Carnot cycle executed within the saturation dome of a pure substance, as shown in Fig. 10-1a. The fluid is heated reversibly and isothermally in a boiler (process 1-), expanded isentropically in a turbine (process -3), condensed reversibly and isothermally in a condenser (process 3-4), and compressed isentropically by a compressor to the initial state (process 4-1). Several impracticalities are associated with this cycle: 1. Isothermal heat transfer to or from a two-phase system is not difficult to achieve in practice since maintaining a constant pressure in the device automatically fixes the temperature at the saturation value. Therefore, processes 1- and 3-4 can be approached closely in actual boilers and condensers. Limiting the heat transfer processes to two-phase systems, however, severely limits the maximum temperature that can be used in the cycle (it has to remain under the critical-point value, which is 374 C for water). Limiting the maximum temperature in the cycle also limits the thermal efficiency. Any attempt to raise the maximum temperature in the cycle involves heat transfer to the working fluid in a single phase, which is not easy to accomplish isothermally.. The isentropic expansion process (process -3) can be approximated closely by a well-designed turbine. However, the quality of the steam decreases during this process, as shown on the T-s diagram in Fig. 10 1a. Thus the turbine has to handle steam with low quality, that is, steam with a high moisture content. The impingement of liquid droplets on the turbine blades causes erosion and is a major source of wear. Thus steam with qualities less than about 90 percent cannot be tolerated in the operation of power plants. T T FIGURE 10 1 T-s diagram of two Carnot vapor cycles. (a) s (b) s

75 This problem could be eliminated by using a working fluid with a very steep saturated vapor line. 3. The isentropic compression process (process 4-1) involves the compression of a liquid vapor mixture to a saturated liquid. There are two difficulties associated with this process. First, it is not easy to control the condensation process so precisely as to end up with the desired quality at state 4. Second, it is not practical to design a compressor that handles two phases. Some of these problems could be eliminated by executing the Carnot cycle in a different way, as shown in Fig. 10 1b. This cycle, however, presents other problems such as isentropic compression to extremely high pressures and isothermal heat transfer at variable pressures. Thus we conclude that the Carnot cycle cannot be approximated in actual devices and is not a realistic model for vapor power cycles. 10 RANKINE CYCLE: THE IDEAL CYCLE FOR VAPOR POWER CYCLES Many of the impracticalities associated with the Carnot cycle can be eliminated by superheating the steam in the boiler and condensing it completely in the condenser, as shown schematically on a T-s diagram in Fig. 10. The cycle that results is the Rankine cycle, which is the ideal cycle for vapor power plants. The ideal Rankine cycle does not involve any internal irreversibilities and consists of the following four processes: 1- Isentropic compression in a pump -3 Constant pressure heat addition in a boiler 3-4 Isentropic expansion in a turbine 4-1 Constant pressure heat rejection in a condenser Chapter INTERACTIVE TUTORIAL SEE TUTORIAL CH. 10, SEC. 1 ON THE DVD. q in Boiler 3 w turb,out T w pump,in Pump Turbine 4 q in 3 w turb,out 1 Condenser q out 1 q out w pump,in 4 s FIGURE 10 The simple ideal Rankine cycle.

76 554 Thermodynamics Water enters the pump at state 1 as saturated liquid and is compressed isentropically to the operating pressure of the boiler. The water temperature increases somewhat during this isentropic compression process due to a slight decrease in the specific volume of water. The vertical distance between states 1 and on the T-s diagram is greatly exaggerated for clarity. (If water were truly incompressible, would there be a temperature change at all during this process?) Water enters the boiler as a compressed liquid at state and leaves as a superheated vapor at state 3. The boiler is basically a large heat exchanger where the heat originating from combustion gases, nuclear reactors, or other sources is transferred to the water essentially at constant pressure. The boiler, together with the section where the steam is superheated (the superheater), is often called the steam generator. The superheated vapor at state 3 enters the turbine, where it expands isentropically and produces work by rotating the shaft connected to an electric generator. The pressure and the temperature of steam drop during this process to the values at state 4, where steam enters the condenser. At this state, steam is usually a saturated liquid vapor mixture with a high quality. Steam is condensed at constant pressure in the condenser, which is basically a large heat exchanger, by rejecting heat to a cooling medium such as a lake, a river, or the atmosphere. Steam leaves the condenser as saturated liquid and enters the pump, completing the cycle. In areas where water is precious, the power plants are cooled by air instead of water. This method of cooling, which is also used in car engines, is called dry cooling. Several power plants in the world, including some in the United States, use dry cooling to conserve water. Remembering that the area under the process curve on a T-s diagram represents the heat transfer for internally reversible processes, we see that the area under process curve -3 represents the heat transferred to the water in the boiler and the area under the process curve 4-1 represents the heat rejected in the condenser. The difference between these two (the area enclosed by the cycle curve) is the net work produced during the cycle. Energy Analysis of the Ideal Rankine Cycle All four components associated with the Rankine cycle (the pump, boiler, turbine, and condenser) are steady-flow devices, and thus all four processes that make up the Rankine cycle can be analyzed as steady-flow processes. The kinetic and potential energy changes of the steam are usually small relative to the work and heat transfer terms and are therefore usually neglected. Then the steady-flow energy equation per unit mass of steam reduces to 1q in q out 1w in w out h e h i 1kJ>kg (10 1) The boiler and the condenser do not involve any work, and the pump and the turbine are assumed to be isentropic. Then the conservation of energy relation for each device can be expressed as follows: Pump (q 0): w pump,in h h 1 (10 )

77 or, where w pump,in v 1P P 1 h 1 h P1 and v v 1 v P1 (10 3) (10 4) Chapter Boiler (w 0): q in h 3 h (10 5) Turbine (q 0): w turb,out h 3 h 4 (10 6) Condenser (w 0): q out h 4 h 1 (10 7) The thermal efficiency of the Rankine cycle is determined from where h th w net q in 1 q out q in w net q in q out w turb,out w pump,in (10 8) The conversion efficiency of power plants in the United States is often expressed in terms of heat rate, which is the amount of heat supplied, in Btu s, to generate 1 kwh of electricity. The smaller the heat rate, the greater the efficiency. Considering that 1 kwh 341 Btu and disregarding the losses associated with the conversion of shaft power to electric power, the relation between the heat rate and the thermal efficiency can be expressed as 341 1Btu>kWh h th Heat rate 1Btu>kWh (10 9) For example, a heat rate of 11,363 Btu/kWh is equivalent to 30 percent efficiency. The thermal efficiency can also be interpreted as the ratio of the area enclosed by the cycle on a T-s diagram to the area under the heat-addition process. The use of these relations is illustrated in the following example. EXAMPLE 10 1 The Simple Ideal Rankine Cycle Consider a steam power plant operating on the simple ideal Rankine cycle. Steam enters the turbine at 3 MPa and 350 C and is condensed in the condenser at a pressure of 75 kpa. Determine the thermal efficiency of this cycle. Solution A steam power plant operating on the simple ideal Rankine cycle is considered. The thermal efficiency of the cycle is to be determined. Assumptions 1 Steady operating conditions exist. Kinetic and potential energy changes are negligible. Analysis The schematic of the power plant and the T-s diagram of the cycle are shown in Fig We note that the power plant operates on the ideal Rankine cycle. Therefore, the pump and the turbine are isentropic, there are no pressure drops in the boiler and condenser, and steam leaves the condenser and enters the pump as saturated liquid at the condenser pressure.

78

79 Chapter q in w pump,in 3 MPa Boiler 3 3 MPa 350 C Turbine w turb,out T, C MPa Pump 75 kpa 4 q out 1 75 kpa Condenser kpa 75 kpa FIGURE 10 3 Schematic and T-s diagram for Example s 1 = s s 3 = s 4 s It is also interesting to note the thermal efficiency of a Carnot cycle operating between the same temperature limits h th,carnot 1 T min T max K K The difference between the two efficiencies is due to the large external irreversibility in Rankine cycle caused by the large temperature difference between steam and combustion gases in the furnace DEVIATION OF ACTUAL VAPOR POWER CYCLES FROM IDEALIZED ONES The actual vapor power cycle differs from the ideal Rankine cycle, as illustrated in Fig. 10 4a, as a result of irreversibilities in various components. Fluid friction and heat loss to the surroundings are the two common sources of irreversibilities. Fluid friction causes pressure drops in the boiler, the condenser, and the piping between various components. As a result, steam leaves the boiler at a somewhat lower pressure. Also, the pressure at the turbine inlet is somewhat lower than that at the boiler exit due to the pressure drop in the connecting pipes. The pressure drop in the condenser is usually very small. To compensate for these pressure drops, the water must be pumped to a sufficiently higher pressure than the ideal cycle calls for. This requires a larger pump and larger work input to the pump. The other major source of irreversibility is the heat loss from the steam to the surroundings as the steam flows through various components. To maintain the same level of net work output, more heat needs to be transferred to INTERACTIVE TUTORIAL SEE TUTORIAL CH. 10, SEC. ON THE DVD.

80 558 Thermodynamics T IDEAL CYCLE T Irreversibility in the pump 1 ACTUAL CYCLE Pressure drop in the boiler Pressure drop in the condenser 3 Irreversibility in the turbine 4 a s 1 4s 3 4a (a) s (b) s FIGURE 10 4 (a) Deviation of actual vapor power cycle from the ideal Rankine cycle. (b) The effect of pump and turbine irreversibilities on the ideal Rankine cycle. the steam in the boiler to compensate for these undesired heat losses. As a result, cycle efficiency decreases. Of particular importance are the irreversibilities occurring within the pump and the turbine. A pump requires a greater work input, and a turbine produces a smaller work output as a result of irreversibilities. Under ideal conditions, the flow through these devices is isentropic. The deviation of actual pumps and turbines from the isentropic ones can be accounted for by utilizing isentropic efficiencies, defined as and h P w s w a h s h 1 h a h 1 (10 10) h T w a w s h 3 h 4a h 3 h 4s (10 11) where states a and 4a are the actual exit states of the pump and the turbine, respectively, and s and 4s are the corresponding states for the isentropic case (Fig. 10 4b). Other factors also need to be considered in the analysis of actual vapor power cycles. In actual condensers, for example, the liquid is usually subcooled to prevent the onset of cavitation, the rapid vaporization and condensation of the fluid at the low-pressure side of the pump impeller, which may damage it. Additional losses occur at the bearings between the moving parts as a result of friction. Steam that leaks out during the cycle and air that leaks into the condenser represent two other sources of loss. Finally, the power consumed by the auxiliary equipment such as fans that supply air to the furnace should also be considered in evaluating the overall performance of power plants. The effect of irreversibilities on the thermal efficiency of a steam power cycle is illustrated below with an example.

81 Chapter EXAMPLE 10 An Actual Steam Power Cycle A steam power plant operates on the cycle shown in Fig If the isentropic efficiency of the turbine is 87 percent and the isentropic efficiency of the pump is 85 percent, determine (a) the thermal efficiency of the cycle and (b) the net power output of the plant for a mass flow rate of 15 kg/s. Solution A steam power cycle with specified turbine and pump efficiencies is considered. The thermal efficiency and the net power output are to be determined. Assumptions 1 Steady operating conditions exist. Kinetic and potential energy changes are negligible. Analysis The schematic of the power plant and the T-s diagram of the cycle are shown in Fig The temperatures and pressures of steam at various points are also indicated on the figure. We note that the power plant involves steady-flow components and operates on the Rankine cycle, but the imperfections at various components are accounted for. (a) The thermal efficiency of a cycle is the ratio of the net work output to the heat input, and it is determined as follows: Pump work input: w pump,in w s,pump,in h p v 1 1P P 1 h p m3 >kg3116,000 9 kpa kj>kg 1 kj a 1 kpa # b m MPa 35 C 15. MPa 3 65 C Boiler 4 15 MPa 5 16 MPa 600 C w turb,out T 4 w pump,in Turbine η T = Pump η P = kpa 1 9 kpa 38 C Condenser 3 s 1 6s 6 s FIGURE 10 5 Schematic and T-s diagram for Example 10.

82 560 Thermodynamics Turbine work output: w turb,out h T w s,turb,out h T 1h 5 h 6s kj>kg kj>kg Boiler heat input: q in h 4 h kj>kg kj>kg Thus, w net w turb,out w pump,in kj>kg kj>kg h th w net q in kj>kg or 36.1% kj>kg (b) The power produced by this power plant is W # net m # 1w net 115 kg>s kj>kg 18.9 MW Discussion Without the irreversibilities, the thermal efficiency of this cycle would be 43.0 percent (see Example 10 3c). T ' 1 1' Increase in w net 4 3 4' P'4 < P 4 FIGURE 10 6 The effect of lowering the condenser pressure on the ideal Rankine cycle. s 10 4 HOW CAN WE INCREASE THE EFFICIENCY OF THE RANKINE CYCLE? Steam power plants are responsible for the production of most electric power in the world, and even small increases in thermal efficiency can mean large savings from the fuel requirements. Therefore, every effort is made to improve the efficiency of the cycle on which steam power plants operate. The basic idea behind all the modifications to increase the thermal efficiency of a power cycle is the same: Increase the average temperature at which heat is transferred to the working fluid in the boiler, or decrease the average temperature at which heat is rejected from the working fluid in the condenser. That is, the average fluid temperature should be as high as possible during heat addition and as low as possible during heat rejection. Next we discuss three ways of accomplishing this for the simple ideal Rankine cycle. Lowering the Condenser Pressure (Lowers T low,avg ) Steam exists as a saturated mixture in the condenser at the saturation temperature corresponding to the pressure inside the condenser. Therefore, lowering the operating pressure of the condenser automatically lowers the temperature of the steam, and thus the temperature at which heat is rejected. The effect of lowering the condenser pressure on the Rankine cycle efficiency is illustrated on a T-s diagram in Fig For comparison purposes, the turbine inlet state is maintained the same. The colored area on this diagram represents the increase in net work output as a result of lowering the condenser pressure from P 4 to P 4. The heat input requirements also increase (represented by the area under curve -), but this increase is very small. Thus the overall effect of lowering the condenser pressure is an increase in the thermal efficiency of the cycle.

83 To take advantage of the increased efficiencies at low pressures, the condensers of steam power plants usually operate well below the atmospheric pressure. This does not present a major problem since the vapor power cycles operate in a closed loop. However, there is a lower limit on the condenser pressure that can be used. It cannot be lower than the saturation pressure corresponding to the temperature of the cooling medium. Consider, for example, a condenser that is to be cooled by a nearby river at 15 C. Allowing a temperature difference of 10 C for effective heat transfer, the steam temperature in the condenser must be above 5 C; thus the condenser pressure must be above 3. kpa, which is the saturation pressure at 5 C. Lowering the condenser pressure is not without any side effects, however. For one thing, it creates the possibility of air leakage into the condenser. More importantly, it increases the moisture content of the steam at the final stages of the turbine, as can be seen from Fig The presence of large quantities of moisture is highly undesirable in turbines because it decreases the turbine efficiency and erodes the turbine blades. Fortunately, this problem can be corrected, as discussed next. Superheating the Steam to High Temperatures (Increases T high,avg ) The average temperature at which heat is transferred to steam can be increased without increasing the boiler pressure by superheating the steam to high temperatures. The effect of superheating on the performance of vapor power cycles is illustrated on a T-s diagram in Fig The colored area on this diagram represents the increase in the net work. The total area under the process curve 3-3 represents the increase in the heat input. Thus both the net work and heat input increase as a result of superheating the steam to a higher temperature. The overall effect is an increase in thermal efficiency, however, since the average temperature at which heat is added increases. Superheating the steam to higher temperatures has another very desirable effect: It decreases the moisture content of the steam at the turbine exit, as can be seen from the T-s diagram (the quality at state 4 is higher than that at state 4). The temperature to which steam can be superheated is limited, however, by metallurgical considerations. Presently the highest steam temperature allowed at the turbine inlet is about 60 C (1150 F). Any increase in this value depends on improving the present materials or finding new ones that can withstand higher temperatures. Ceramics are very promising in this regard. Increasing the Boiler Pressure (Increases T high,avg ) Another way of increasing the average temperature during the heat-addition process is to increase the operating pressure of the boiler, which automatically raises the temperature at which boiling takes place. This, in turn, raises the average temperature at which heat is transferred to the steam and thus raises the thermal efficiency of the cycle. The effect of increasing the boiler pressure on the performance of vapor power cycles is illustrated on a T-s diagram in Fig Notice that for a fixed turbine inlet temperature, the cycle shifts to the left and the moisture content of steam at the turbine exit increases. This undesirable side effect can be corrected, however, by reheating the steam, as discussed in the next section. T 1 Chapter Increase in w net 3' FIGURE 10 7 The effect of superheating the steam to higher temperatures on the ideal Rankine cycle. T Increase in w net 3' ' ' 4' 3 4 T max s s Decrease in w net FIGURE 10 8 The effect of increasing the boiler pressure on the ideal Rankine cycle.

84 56 Thermodynamics T 1 Critical point 4 3 Operating pressures of boilers have gradually increased over the years from about.7 MPa (400 psia) in 19 to over 30 MPa (4500 psia) today, generating enough steam to produce a net power output of 1000 MW or more in a large power plant. Today many modern steam power plants operate at supercritical pressures (P.06 MPa) and have thermal efficiencies of about 40 percent for fossil-fuel plants and 34 percent for nuclear plants. There are over 150 supercritical-pressure steam power plants in operation in the United States. The lower efficiencies of nuclear power plants are due to the lower maximum temperatures used in those plants for safety reasons. The T-s diagram of a supercritical Rankine cycle is shown in Fig The effects of lowering the condenser pressure, superheating to a higher temperature, and increasing the boiler pressure on the thermal efficiency of the Rankine cycle are illustrated below with an example. FIGURE 10 9 A supercritical Rankine cycle. s EXAMPLE 10 3 Effect of Boiler Pressure and Temperature on Efficiency Consider a steam power plant operating on the ideal Rankine cycle. Steam enters the turbine at 3 MPa and 350 C and is condensed in the condenser at a pressure of 10 kpa. Determine (a) the thermal efficiency of this power plant, (b) the thermal efficiency if steam is superheated to 600 C instead of 350 C, and (c) the thermal efficiency if the boiler pressure is raised to 15 MPa while the turbine inlet temperature is maintained at 600 C. Solution A steam power plant operating on the ideal Rankine cycle is considered. The effects of superheating the steam to a higher temperature and raising the boiler pressure on thermal efficiency are to be investigated. Analysis The T-s diagrams of the cycle for all three cases are given in Fig T T T T 3 = 600 C 3 3 T 3 = 600 C 3 MPa 3 T 3 = 350 C 3 MPa 15 MPa 1 10 kpa kpa kpa 4 (a) s (b) s (c) s FIGURE T-s diagrams of the three cycles discussed in Example 10 3.

85 Chapter (a) This is the steam power plant discussed in Example 10 1, except that the condenser pressure is lowered to 10 kpa. The thermal efficiency is determined in a similar manner: State 1: State : P 1 10 kpa f h 1 h 10 kpa kj>kg Sat. liquid v 1 v 10 kpa m 3 >kg P 3 MPa s s 1 1 kj w pump,in v 1 1P P m 3 >kg kpa4a 1 kpa # b m kj>kg h h 1 w pump,in kj>kg kj>kg State 3: P 3 3 MPa T C f h kj>kg s kj>kg # K State 4: Thus, and Therefore, the thermal efficiency increases from 6.0 to 33.4 percent as a result of lowering the condenser pressure from 75 to 10 kpa. At the same time, however, the quality of the steam decreases from 88.6 to 81.3 percent (in other words, the moisture content increases from 11.4 to 18.7 percent). (b) States 1 and remain the same in this case, and the enthalpies at state 3 (3 MPa and 600 C) and state 4 (10 kpa and s 4 s 3 ) are determined to be Thus, and P 4 10 kpa 1sat. mixture s 4 s 3 x 4 s 4 s f s fg h 4 h f x 4 h fg kj>kg q in h 3 h kj>kg 91.3 kj>kg q out h 4 h kj>kg kj>kg h th 1 q out q in h kj>kg h kj>kg 1x q in h 3 h kj>kg q out h 4 h kj>kg h th 1 q out q in kj>kg or 33.4% 91.3 kj>kg kj>kg or 37.3% kj>kg

86 564 Thermodynamics Therefore, the thermal efficiency increases from 33.4 to 37.3 percent as a result of superheating the steam from 350 to 600 C. At the same time, the quality of the steam increases from 81.3 to 91.5 percent (in other words, the moisture content decreases from 18.7 to 8.5 percent). (c) State 1 remains the same in this case, but the other states change. The enthalpies at state (15 MPa and s s 1 ), state 3 (15 MPa and 600 C), and state 4 (10 kpa and s 4 s 3 ) are determined in a similar manner to be Thus, and h kj>kg h kj>kg h kj>kg 1x q in h 3 h kj>kg q out h 4 h kj>kg h th 1 q out q in kj>kg or 43.0% kj>kg Discussion The thermal efficiency increases from 37.3 to 43.0 percent as a result of raising the boiler pressure from 3 to 15 MPa while maintaining the turbine inlet temperature at 600 C. At the same time, however, the quality of the steam decreases from 91.5 to 80.4 percent (in other words, the moisture content increases from 8.5 to 19.6 percent). INTERACTIVE TUTORIAL SEE TUTORIAL CH. 10, SEC. 3 ON THE DVD THE IDEAL REHEAT RANKINE CYCLE We noted in the last section that increasing the boiler pressure increases the thermal efficiency of the Rankine cycle, but it also increases the moisture content of the steam to unacceptable levels. Then it is natural to ask the following question: How can we take advantage of the increased efficiencies at higher boiler pressures without facing the problem of excessive moisture at the final stages of the turbine? Two possibilities come to mind: 1. Superheat the steam to very high temperatures before it enters the turbine. This would be the desirable solution since the average temperature at which heat is added would also increase, thus increasing the cycle efficiency. This is not a viable solution, however, since it requires raising the steam temperature to metallurgically unsafe levels.. Expand the steam in the turbine in two stages, and reheat it in between. In other words, modify the simple ideal Rankine cycle with a reheat process. Reheating is a practical solution to the excessive moisture problem in turbines, and it is commonly used in modern steam power plants. The T-s diagram of the ideal reheat Rankine cycle and the schematic of the power plant operating on this cycle are shown in Fig The ideal reheat Rankine cycle differs from the simple ideal Rankine cycle in that the

87 expansion process takes place in two stages. In the first stage (the highpressure turbine), steam is expanded isentropically to an intermediate pressure and sent back to the boiler where it is reheated at constant pressure, usually to the inlet temperature of the first turbine stage. Steam then expands isentropically in the second stage (low-pressure turbine) to the condenser pressure. Thus the total heat input and the total turbine work output for a reheat cycle become and q in q primary q reheat 1h 3 h 1h 5 h 4 w turb,out w turb,i w turb,ii 1h 3 h 4 1h 5 h 6 (10 1) (10 13) The incorporation of the single reheat in a modern power plant improves the cycle efficiency by 4 to 5 percent by increasing the average temperature at which heat is transferred to the steam. The average temperature during the reheat process can be increased by increasing the number of expansion and reheat stages. As the number of stages is increased, the expansion and reheat processes approach an isothermal process at the maximum temperature, as shown in Fig The use of more than two reheat stages, however, is not practical. The theoretical improvement in efficiency from the second reheat is about half of that which results from a single reheat. If the turbine inlet pressure is not high enough, double reheat would result in superheated exhaust. This is undesirable as it would cause the average temperature for heat rejection to increase and thus the cycle efficiency to decrease. Therefore, double reheat is used only on supercritical-pressure (P.06 MPa) power plants. A third reheat stage would increase the cycle efficiency by about half of the improvement attained by the second reheat. This gain is too small to justify the added cost and complexity. Chapter T 3 High-pressure turbine 3 Reheating 5 Boiler Reheater 4 High-P turbine Low-P turbine 4 Low-pressure turbine P 4 = P 5 = P reheat 5 Condenser Pump s FIGURE The ideal reheat Rankine cycle.

88 566 Thermodynamics T FIGURE 10 1 T avg,reheat The average temperature at which heat is transferred during reheating increases as the number of reheat stages is increased. s The reheat cycle was introduced in the mid-190s, but it was abandoned in the 1930s because of the operational difficulties. The steady increase in boiler pressures over the years made it necessary to reintroduce single reheat in the late 1940s and double reheat in the early 1950s. The reheat temperatures are very close or equal to the turbine inlet temperature. The optimum reheat pressure is about one-fourth of the maximum cycle pressure. For example, the optimum reheat pressure for a cycle with a boiler pressure of 1 MPa is about 3 MPa. Remember that the sole purpose of the reheat cycle is to reduce the moisture content of the steam at the final stages of the expansion process. If we had materials that could withstand sufficiently high temperatures, there would be no need for the reheat cycle. EXAMPLE 10 4 The Ideal Reheat Rankine Cycle Consider a steam power plant operating on the ideal reheat Rankine cycle. Steam enters the high-pressure turbine at 15 MPa and 600 C and is condensed in the condenser at a pressure of 10 kpa. If the moisture content of the steam at the exit of the low-pressure turbine is not to exceed 10.4 percent, determine (a) the pressure at which the steam should be reheated and (b) the thermal efficiency of the cycle. Assume the steam is reheated to the inlet temperature of the high-pressure turbine. Solution A steam power plant operating on the ideal reheat Rankine cycle is considered. For a specified moisture content at the turbine exit, the reheat pressure and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. Kinetic and potential energy changes are negligible. Analysis The schematic of the power plant and the T-s diagram of the cycle are shown in Fig We note that the power plant operates on the ideal reheat Rankine cycle. Therefore, the pump and the turbines are isentropic, there are no pressure drops in the boiler and condenser, and steam leaves the condenser and enters the pump as saturated liquid at the condenser pressure. (a) The reheat pressure is determined from the requirement that the entropies at states 5 and 6 be the same: State 6: Also, Thus, State 5: P 6 10 kpa x sat. mixture s 6 s f x 6 s fg kj>kg # K h 6 h f x 6 h fg kj>kg T C f P MPa s 5 s 6 h kj>kg Therefore, steam should be reheated at a pressure of 4 MPa or lower to prevent a moisture content above 10.4 percent.

89 Chapter (b) To determine the thermal efficiency, we need to know the enthalpies at all other states: State 1: P 1 10 kpa f h 1 h 10 kpa kj>kg Sat. liquid v 1 v 10 kpa m 3 >kg State : P 15 MPa s s 1 w pump,in v 1 1P P m 3 >kg 1 kj 3115,000 10kPa4 a 1 kpa # b m kj>kg h h 1 w pump,in kj>kg kj>kg State 3: State 4: Thus P 3 15 MPa T C f h kj>kg s kj>kg # K P 4 4 MPa f h kj>kg s 4 s 3 1T C q in 1h 3 h 1h 5 h kj>kg kj>kg kj>kg q out h 6 h kj>kg kj>kg T, C 15 MPa MPa 3 Reheating 5 Boiler 4 High-P turbine Low-P turbine 15 MPa 4 Reheater P 4 = P 5 = P reheat 6 10 kpa 10 kpa 5 15 MPa Condenser 1 6 Pump 10 kpa s 1 FIGURE Schematic and T-s diagram for Example 10 4.

90 568 Thermodynamics and Open Feedwater Heaters An open (or direct-contact) feedwater heater is basically a mixing chamber, where the steam extracted from the turbine mixes with the feedwater exiting the pump. Ideally, the mixture leaves the heater as a saturated liquid at the heater pressure. The schematic of a steam power plant with one open feedwater heater (also called single-stage regenerative cycle) and the T-s diagram of the cycle are shown in Fig In an ideal regenerative Rankine cycle, steam enters the turbine at the boiler pressure (state 5) and expands isentropically to an intermediate presh th 1 q out q in kj>kg or 45.0% kj>kg Discussion This problem was solved in Example 10 3c for the same pressure and temperature limits but without the reheat process. A comparison of the two results reveals that reheating reduces the moisture content from 19.6 to 10.4 percent while increasing the thermal efficiency from 43.0 to 45.0 percent. T Low-temperature heat addition 1 ' INTERACTIVE TUTORIAL SEE TUTORIAL CH. 10, SEC. 4 ON THE DVD. Steam entering boiler Steam exiting boiler FIGURE The first part of the heat-addition process in the boiler takes place at relatively low temperatures. 4 3 s 10 6 THE IDEAL REGENERATIVE RANKINE CYCLE A careful examination of the T-s diagram of the Rankine cycle redrawn in Fig reveals that heat is transferred to the working fluid during process - at a relatively low temperature. This lowers the average heataddition temperature and thus the cycle efficiency. To remedy this shortcoming, we look for ways to raise the temperature of the liquid leaving the pump (called the feedwater) before it enters the boiler. One such possibility is to transfer heat to the feedwater from the expanding steam in a counterflow heat exchanger built into the turbine, that is, to use regeneration. This solution is also impractical because it is difficult to design such a heat exchanger and because it would increase the moisture content of the steam at the final stages of the turbine. A practical regeneration process in steam power plants is accomplished by extracting, or bleeding, steam from the turbine at various points. This steam, which could have produced more work by expanding further in the turbine, is used to heat the feedwater instead. The device where the feedwater is heated by regeneration is called a regenerator, or a feedwater heater (FWH). Regeneration not only improves cycle efficiency, but also provides a convenient means of deaerating the feedwater (removing the air that leaks in at the condenser) to prevent corrosion in the boiler. It also helps control the large volume flow rate of the steam at the final stages of the turbine (due to the large specific volumes at low pressures). Therefore, regeneration has been used in all modern steam power plants since its introduction in the early 190s. A feedwater heater is basically a heat exchanger where heat is transferred from the steam to the feedwater either by mixing the two fluid streams (open feedwater heaters) or without mixing them (closed feedwater heaters). Regeneration with both types of feedwater heaters is discussed below.

91 sure (state 6). Some steam is extracted at this state and routed to the feedwater heater, while the remaining steam continues to expand isentropically to the condenser pressure (state 7). This steam leaves the condenser as a saturated liquid at the condenser pressure (state 1). The condensed water, which is also called the feedwater, then enters an isentropic pump, where it is compressed to the feedwater heater pressure (state ) and is routed to the feedwater heater, where it mixes with the steam extracted from the turbine. The fraction of the steam extracted is such that the mixture leaves the heater as a saturated liquid at the heater pressure (state 3). A second pump raises the pressure of the water to the boiler pressure (state 4). The cycle is completed by heating the water in the boiler to the turbine inlet state (state 5). In the analysis of steam power plants, it is more convenient to work with quantities expressed per unit mass of the steam flowing through the boiler. For each 1 kg of steam leaving the boiler, y kg expands partially in the turbine and is extracted at state 6. The remaining (1 y) kg expands completely to the condenser pressure. Therefore, the mass flow rates are different in different components. If the mass flow rate through the boiler is m., for example, it is (1 y)m. through the condenser. This aspect of the regenerative Rankine cycle should be considered in the analysis of the cycle as well as in the interpretation of the areas on the T-s diagram. In light of Fig , the heat and work interactions of a regenerative Rankine cycle with one feedwater heater can be expressed per unit mass of steam flowing through the boiler as follows: q in h 5 h 4 q out 11 y1h 7 h 1 w turb,out 1h 5 h 6 11 y 1h 6 h 7 (10 14) (10 15) (10 16) Chapter w pump,in 11 yw pump I,in w pump II,in (10 17) 5 Boiler Turbine T 5 4 Open FWH y 1 y Pump II Condenser 1 7 Pump I 1 s FIGURE The ideal regenerative Rankine cycle with an open feedwater heater.

92 570 Thermodynamics where y m # 6>m # 5 1fraction of steam extracted w pump I,in v 1 1P P 1 Closed Feedwater Heaters Another type of feedwater heater frequently used in steam power plants is the closed feedwater heater, in which heat is transferred from the extracted steam to the feedwater without any mixing taking place. The two streams now can be at different pressures, since they do not mix. The schematic of a steam power plant with one closed feedwater heater and the T-s diagram of the cycle are shown in Fig In an ideal closed feedwater heater, the feedwater is heated to the exit temperature of the extracted steam, which ideally leaves the heater as a saturated liquid at the extraction pressure. In actual power plants, the feedwater leaves the heater below the exit temperaw pump II,in v 3 1P 4 P 3 The thermal efficiency of the Rankine cycle increases as a result of regeneration. This is because regeneration raises the average temperature at which heat is transferred to the steam in the boiler by raising the temperature of the water before it enters the boiler. The cycle efficiency increases further as the number of feedwater heaters is increased. Many large plants in operation today use as many as eight feedwater heaters. The optimum number of feedwater heaters is determined from economical considerations. The use of an additional feedwater heater cannot be justified unless it saves more from the fuel costs than its own cost. T 6 6 Boiler 5 Mixing chamber 4 9 Pump II 3 Closed FWH Turbine 7 Pump I 1 8 Condenser s FIGURE The ideal regenerative Rankine cycle with a closed feedwater heater.

93 Chapter Turbine Boiler Condenser Deaerating Closed FWH Closed FWH Open FWH Closed FWH Pump Pump Trap Trap Trap FIGURE A steam power plant with one open and three closed feedwater heaters. ture of the extracted steam because a temperature difference of at least a few degrees is required for any effective heat transfer to take place. The condensed steam is then either pumped to the feedwater line or routed to another heater or to the condenser through a device called a trap. A trap allows the liquid to be throttled to a lower pressure region but traps the vapor. The enthalpy of steam remains constant during this throttling process. The open and closed feedwater heaters can be compared as follows. Open feedwater heaters are simple and inexpensive and have good heat transfer characteristics. They also bring the feedwater to the saturation state. For each heater, however, a pump is required to handle the feedwater. The closed feedwater heaters are more complex because of the internal tubing network, and thus they are more expensive. Heat transfer in closed feedwater heaters is also less effective since the two streams are not allowed to be in direct contact. However, closed feedwater heaters do not require a separate pump for each heater since the extracted steam and the feedwater can be at different pressures. Most steam power plants use a combination of open and closed feedwater heaters, as shown in Fig EXAMPLE 10 5 The Ideal Regenerative Rankine Cycle Consider a steam power plant operating on the ideal regenerative Rankine cycle with one open feedwater heater. Steam enters the turbine at 15 MPa and 600 C and is condensed in the condenser at a pressure of 10 kpa.

94 57 Thermodynamics Some steam leaves the turbine at a pressure of 1. MPa and enters the open feedwater heater. Determine the fraction of steam extracted from the turbine and the thermal efficiency of the cycle. Solution A steam power plant operates on the ideal regenerative Rankine cycle with one open feedwater heater. The fraction of steam extracted from the turbine and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. Kinetic and potential energy changes are negligible. Analysis The schematic of the power plant and the T-s diagram of the cycle are shown in Fig We note that the power plant operates on the ideal regenerative Rankine cycle. Therefore, the pumps and the turbines are isentropic; there are no pressure drops in the boiler, condenser, and feedwater heater; and steam leaves the condenser and the feedwater heater as saturated liquid. First, we determine the enthalpies at various states: State 1: P 1 10 kpa f h 1 h 10 kpa kj>kg Sat. liquid v 1 v 10 kpa m 3 >kg State : P 1. MPa s s 1 w pump I,in v 1 1P P m 3 1 kj >kg kpa4a 1 kpa # b m kj>kg h h 1 w pump I,in kj>kg kj>kg T 5 15 MPa 600 C w turb,out 5 q in Boiler Open FWH Turbine 1. MPa kpa MPa 3 Pump II 1. MPa 1. MPa Pump I 1 10 kpa q out Condenser 1 7 s FIGURE Schematic and T-s diagram for Example 10 5.

95 Chapter State 3: State 4: P 3 1. MPa f v 3 v 1. MPa m 3 >kg Sat. liquid h 3 h 1. MPa kj>kg P 4 15 MPa s 4 s 3 w pump II,in v 3 1P 4 P m 3 1 kj >kg3115, kpa4a 1 kpa # b m kj>kg h 4 h 3 w pump II,in kj>kg kj>kg State 5: State 6: P 5 15 MPa T C f h kj>kg s kj>kg # K P 6 1. MPa f h kj>kg s 6 s 5 1T C State 7: P 7 10 kpa s 7 s 5 x 7 s 7 s f s fg h 7 h f x 7 h fg kj>kg The energy analysis of open feedwater heaters is identical to the energy analysis of mixing chambers. The feedwater heaters are generally well insulated (Q. 0), and they do not involve any work interactions (W. 0). By neglecting the kinetic and potential energies of the streams, the energy balance reduces for a feedwater heater to E # in E # out S ain m # h aout m # h or where y is the fraction of steam extracted from the turbine ( m. 6 /m. 5 ). Solving for y and substituting the enthalpy values, we find Thus, and y h 3 h h 6 h q in h 5 h kj>kg kj>kg q out 11 y 1h 7 h kj>kg kj>kg h th 1 q out q in yh 6 11 yh 1 1h kj>kg or 46.3% kj>kg

96 574 Thermodynamics Discussion This problem was worked out in Example 10 3c for the same pressure and temperature limits but without the regeneration process. A comparison of the two results reveals that the thermal efficiency of the cycle has increased from 43.0 to 46.3 percent as a result of regeneration. The net work output decreased by 171 kj/kg, but the heat input decreased by 607 kj/kg, which results in a net increase in the thermal efficiency. EXAMPLE 10 6 The Ideal Reheat Regenerative Rankine Cycle Consider a steam power plant that operates on an ideal reheat regenerative Rankine cycle with one open feedwater heater, one closed feedwater heater, and one reheater. Steam enters the turbine at 15 MPa and 600 C and is condensed in the condenser at a pressure of 10 kpa. Some steam is extracted from the turbine at 4 MPa for the closed feedwater heater, and the remaining steam is reheated at the same pressure to 600 C. The extracted steam is completely condensed in the heater and is pumped to 15 MPa before it mixes with the feedwater at the same pressure. Steam for the open feedwater heater is extracted from the low-pressure turbine at a pressure of 0.5 MPa. Determine the fractions of steam extracted from the turbine as well as the thermal efficiency of the cycle. Solution A steam power plant operates on the ideal reheat regenerative Rankine cycle with one open feedwater heater, one closed feedwater heater, and one reheater. The fractions of steam extracted from the turbine and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. Kinetic and potential energy changes are negligible. 3 In both open and closed feedwater heaters, feedwater is heated to the saturation temperature at the feedwater heater pressure. (Note that this is a conservative assumption since extracted steam enters the closed feedwater heater at 376 C and the saturation temperature at the closed feedwater pressure of 4 MPa is 50 C). Analysis The schematic of the power plant and the T-s diagram of the cycle are shown in Fig The power plant operates on the ideal reheat regenerative Rankine cycle and thus the pumps and the turbines are isentropic; there are no pressure drops in the boiler, reheater, condenser, and feedwater heaters; and steam leaves the condenser and the feedwater heaters as saturated liquid. The enthalpies at the various states and the pump work per unit mass of fluid flowing through them are h kj>kg h kj>kg h kj>kg h kj>kg h kj>kg h kj>kg h kj>kg h kj>kg h kj>kg h kj>kg h kj>kg w pump I,in 0.49 kj>kg h kj>kg w pump II,in 3.83 kj>kg h kj>kg w pump III,in kj>kg

97 Chapter The fractions of steam extracted are determined from the mass and energy balances of the feedwater heaters: Closed feedwater heater: y Open feedwater heater: The enthalpy at state 8 is determined by applying the mass and energy equations to the mixing chamber, which is assumed to be insulated: E # in E # out E # in E # out z 11 y1h 3 h h 1 h 11h 8 11 yh 5 yh kj>kg E # in E # out yh yh 4 11 yh 5 yh 6 h 5 h 4 1h 10 h 6 1h 5 h zh 1 11 y zh 11 yh h kj>kg kj>kg 1 kg 9 15 MPa 600 C T Boiler Reheater 10 High-P turbine Low-P turbine 15 MPa 1 kg y Mixing chamber Closed FWH y P 10 = P 11 = 4 MPa 600 C z 4 MPa y Open FWH MPa 1 1 y z kpa Condenser y 4 MPa MPa z 10 kpa 1 y 1 1 y z 13 s Pump III Pump II Pump I FIGURE Schematic and T-s diagram for Example 10 6.

98 576 Thermodynamics Thus, and q in 1h 9 h 8 11 y 1h 11 h 10 Discussion This problem was worked out in Example 10 4 for the same pressure and temperature limits with reheat but without the regeneration process. A comparison of the two results reveals that the thermal efficiency of the cycle has increased from 45.0 to 49. percent as a result of regeneration. The thermal efficiency of this cycle could also be determined from where kj>kg kj>kg 91.4 kj>kg q out 11 y z1h 13 h kj>kg kj>kg h th 1 q out q in kj>kg or 49.% 91.4 kj>kg h th w net q in w turb,out w pump,in q in w turb,out 1h 9 h y1h 11 h 1 11 y z 1h 1 h 13 w pump,in 11 y zw pump I,in 11 yw pump II,in 1yw pump III,in Also, if we assume that the feedwater leaves the closed FWH as a saturated liquid at 15 MPa (and thus at T 5 34 C and h kj/kg), it can be shown that the thermal efficiency would be SECOND-LAW ANALYSIS OF VAPOR POWER CYCLES The ideal Carnot cycle is a totally reversible cycle, and thus it does not involve any irreversibilities. The ideal Rankine cycles (simple, reheat, or regenerative), however, are only internally reversible, and they may involve irreversibilities external to the system, such as heat transfer through a finite temperature difference. A second-law analysis of these cycles reveals where the largest irreversibilities occur and what their magnitudes are. Relations for exergy and exergy destruction for steady-flow systems are developed in Chap. 8. The exergy destruction for a steady-flow system can be expressed, in the rate form, as X # dest T 0 S # gen T 0 1S # out S # in T 0 a aout m # s Q# out T ain m # s Q # in b 1kW b,out T b,in or on a unit mass basis for a one-inlet, one-exit, steady-flow device as (10 18) x dest T 0 s gen T 0 a s e s i q out T b,out q in T b,in b 1kJ>kg (10 19)

99 where T b,in and T b,out are the temperatures of the system boundary where heat is transferred into and out of the system, respectively. The exergy destruction associated with a cycle depends on the magnitude of the heat transfer with the high- and low-temperature reservoirs involved, and their temperatures. It can be expressed on a unit mass basis as Chapter q out q in x dest T 0 a a (10 0) T a b 1kJ>kg b,out T b,in For a cycle that involves heat transfer only with a source at T H and a sink at T L, the exergy destruction becomes x dest T 0 a q out T L q in T H b 1kJ>kg The exergy of a fluid stream c at any state can be determined from c 1h h 0 T 0 1s s 0 V gz 1kJ>kg where the subscript 0 denotes the state of the surroundings. (10 1) (10 ) EXAMPLE 10 7 Second-Law Analysis of an Ideal Rankine Cycle Determine the exergy destruction associated with the Rankine cycle (all four processes as well as the cycle) discussed in Example 10 1, assuming that heat is transferred to the steam in a furnace at 1600 K and heat is rejected to a cooling medium at 90 K and 100 kpa. Also, determine the exergy of the steam leaving the turbine. Solution The Rankine cycle analyzed in Example 10 1 is reconsidered. For specified source and sink temperatures, the exergy destruction associated with the cycle and exergy of the steam at turbine exit are to be determined. Analysis In Example 10 1, the heat input was determined to be 78.6 kj/kg, and the heat rejected to be kj/kg. Processes 1- and 3-4 are isentropic (s 1 s, s 3 s 4 ) and therefore do not involve any internal or external irreversibilities, that is, Processes -3 and 4-1 are constant-pressure heat-addition and heatrejection processes, respectively, and they are internally reversible. But the heat transfer between the working fluid and the source or the sink takes place through a finite temperature difference, rendering both processes irreversible. The irreversibility associated with each process is determined from Eq The entropy of the steam at each state is determined from the steam tables: Thus, s s 1 s 75 kpa 1.13 kj>kg # K s 4 s kj>kg # K 1at 3 MPa, 350 C x dest,3 T 0 a s 3 s q in,3 T source b 190 K c kj>kg # K 78.6 kj>kg 1600 K d 1110 kj/kg x dest,1 0 and x dest,34 0

100 578 Thermodynamics Therefore, the irreversibility of the cycle is The total exergy destroyed during the cycle could also be determined from Eq Notice that the largest exergy destruction in the cycle occurs during the heat-addition process. Therefore, any attempt to reduce the exergy destruction should start with this process. Raising the turbine inlet temperature of the steam, for example, would reduce the temperature difference and thus the exergy destruction. The exergy (work potential) of the steam leaving the turbine is determined from Eq. 10. Disregarding the kinetic and potential energies, it reduces to where Thus, x dest,41 T 0 a s 1 s 4 q out,41 b T sink c kj>kg 190 K kj>kg # K4 449 kj/kg kj>kg 190 Kc kj>kg # K d 90 K 414 kj/kg x dest,cycle x dest,1 x dest,3 x dest,34 x dest, kj>kg kj>kg 154 kj/kg c 4 1h 4 h 0 T 0 1s 4 s 0 V 4 Q 0 gz Q 0 4 1h 4 h 0 T 0 1s 4 s 0 h 0 90 K,100 kpa h 90 K kj>kg s 0 90 K,100 kpa s 90 K kj>kg # K Discussion Note that 449 kj/kg of work could be obtained from the steam leaving the turbine if it is brought to the state of the surroundings in a reversible manner COGENERATION In all the cycles discussed so far, the sole purpose was to convert a portion of the heat transferred to the working fluid to work, which is the most valuable form of energy. The remaining portion of the heat is rejected to rivers, lakes, oceans, or the atmosphere as waste heat, because its quality (or grade) is too low to be of any practical use. Wasting a large amount of heat is a price we have to pay to produce work, because electrical or mechanical work is the only form of energy on which many engineering devices (such as a fan) can operate. Many systems or devices, however, require energy input in the form of heat, called process heat. Some industries that rely heavily on process heat are chemical, pulp and paper, oil production and refining, steel making,

101 food processing, and textile industries. Process heat in these industries is usually supplied by steam at 5 to 7 atm and 150 to 00 C (300 to 400 F). Energy is usually transferred to the steam by burning coal, oil, natural gas, or another fuel in a furnace. Now let us examine the operation of a process-heating plant closely. Disregarding any heat losses in the piping, all the heat transferred to the steam in the boiler is used in the process-heating units, as shown in Fig Therefore, process heating seems like a perfect operation with practically no waste of energy. From the second-law point of view, however, things do not look so perfect. The temperature in furnaces is typically very high (around 1400 C), and thus the energy in the furnace is of very high quality. This high-quality energy is transferred to water to produce steam at about 00 C or below (a highly irreversible process). Associated with this irreversibility is, of course, a loss in exergy or work potential. It is simply not wise to use high-quality energy to accomplish a task that could be accomplished with low-quality energy. Industries that use large amounts of process heat also consume a large amount of electric power. Therefore, it makes economical as well as engineering sense to use the already-existing work potential to produce power instead of letting it go to waste. The result is a plant that produces electricity while meeting the process-heat requirements of certain industrial processes. Such a plant is called a cogeneration plant. In general, cogeneration is the production of more than one useful form of energy (such as process heat and electric power) from the same energy source. Either a steam-turbine (Rankine) cycle or a gas-turbine (Brayton) cycle or even a combined cycle (discussed later) can be used as the power cycle in a cogeneration plant. The schematic of an ideal steam-turbine cogeneration plant is shown in Fig Let us say this plant is to supply process heat Q. p at 500 kpa at a rate of 100 kw. To meet this demand, steam is expanded in the turbine to a pressure of 500 kpa, producing power at a rate of, say, 0 kw. The flow rate of the steam can be adjusted such that steam leaves the processheating section as a saturated liquid at 500 kpa. Steam is then pumped to the boiler pressure and is heated in the boiler to state 3. The pump work is usually very small and can be neglected. Disregarding any heat losses, the rate of heat input in the boiler is determined from an energy balance to be 10 kw. Probably the most striking feature of the ideal steam-turbine cogeneration plant shown in Fig is the absence of a condenser. Thus no heat is rejected from this plant as waste heat. In other words, all the energy transferred to the steam in the boiler is utilized as either process heat or electric power. Thus it is appropriate to define a utilization factor u for a cogeneration plant as or Net work output Process heat delivered u Total heat input W# net Q # p Q # in (10 3) Q in Boiler Chapter Pump Process heater FIGURE 10 0 A simple process-heating plant. 10 kw Boiler 3 Pump Turbine Process heater W ~ pump = 0 FIGURE 10 1 An ideal cogeneration plant. 0 kw kw Q p u 1 Q# out (10 4) where Q. Q # in out represents the heat rejected in the condenser. Strictly speaking, Q. out also includes all the undesirable heat losses from the piping and other components, but they are usually small and thus neglected. It also includes combustion inefficiencies such as incomplete combustion and stack losses

102 580 Thermodynamics 3 Boiler Expansion valve Pump II 4 5 Pump I Process heater Turbine 7 Condenser FIGURE 10 A cogeneration plant with adjustable loads when the utilization factor is defined on the basis of the heating value of the fuel. The utilization factor of the ideal steam-turbine cogeneration plant is obviously 100 percent. Actual cogeneration plants have utilization factors as high as 80 percent. Some recent cogeneration plants have even higher utilization factors. Notice that without the turbine, we would need to supply heat to the steam in the boiler at a rate of only 100 kw instead of at 10 kw. The additional 0 kw of heat supplied is converted to work. Therefore, a cogeneration power plant is equivalent to a process-heating plant combined with a power plant that has a thermal efficiency of 100 percent. The ideal steam-turbine cogeneration plant described above is not practical because it cannot adjust to the variations in power and process-heat loads. The schematic of a more practical (but more complex) cogeneration plant is shown in Fig. 10. Under normal operation, some steam is extracted from the turbine at some predetermined intermediate pressure P 6. The rest of the steam expands to the condenser pressure P 7 and is then cooled at constant pressure. The heat rejected from the condenser represents the waste heat for the cycle. At times of high demand for process heat, all the steam is routed to the process-heating units and none to the condenser (ṁ 7 0). The waste heat is zero in this mode. If this is not sufficient, some steam leaving the boiler is throttled by an expansion or pressure-reducing valve (PRV) to the extraction pressure P 6 and is directed to the process-heating unit. Maximum process heating is realized when all the steam leaving the boiler passes through the PRV (ṁ 5 ṁ 4 ). No power is produced in this mode. When there is no demand for process heat, all the steam passes through the turbine and the condenser (ṁ 5 ṁ 6 0), and the cogeneration plant operates as an ordinary steam power plant. The rates of heat input, heat rejected, and process heat supply as well as the power produced for this cogeneration plant can be expressed as follows: Q # in m # 3 1h 4 h 3 Q # out m # 7 1h 7 h 1 (10 5) (10 6) Q # (10 7) W # p m # 5h 5 m # 6h 6 m # 8h 8 turb 1m # 4 m # 51h 4 h 6 m # 7 1h 6 h 7 (10 8) Under optimum conditions, a cogeneration plant simulates the ideal cogeneration plant discussed earlier. That is, all the steam expands in the turbine to the extraction pressure and continues to the process-heating unit. No steam passes through the PRV or the condenser; thus, no waste heat is rejected (ṁ 4 ṁ 6 and ṁ 5 ṁ 7 0). This condition may be difficult to achieve in practice because of the constant variations in the process-heat and power loads. But the plant should be designed so that the optimum operating conditions are approximated most of the time. The use of cogeneration dates to the beginning of this century when power plants were integrated to a community to provide district heating, that is, space, hot water, and process heating for residential and commercial buildings. The district heating systems lost their popularity in the 1940s owing to low fuel prices. However, the rapid rise in fuel prices in the 1970s brought about renewed interest in district heating.

103 Cogeneration plants have proved to be economically very attractive. Consequently, more and more such plants have been installed in recent years, and more are being installed. Chapter EXAMPLE 10 8 An Ideal Cogeneration Plant Consider the cogeneration plant shown in Fig Steam enters the turbine at 7 MPa and 500 C. Some steam is extracted from the turbine at 500 kpa for process heating. The remaining steam continues to expand to 5 kpa. Steam is then condensed at constant pressure and pumped to the boiler pressure of 7 MPa. At times of high demand for process heat, some steam leaving the boiler is throttled to 500 kpa and is routed to the process heater. The extraction fractions are adjusted so that steam leaves the process heater as a saturated liquid at 500 kpa. It is subsequently pumped to 7 MPa. The mass flow rate of steam through the boiler is 15 kg/s. Disregarding any pressure drops and heat losses in the piping and assuming the turbine and the pump to be isentropic, determine (a) the maximum rate at which process heat can be supplied, (b) the power produced and the utilization factor when no process heat is supplied, and (c) the rate of process heat supply when 10 percent of the steam is extracted before it enters the turbine and 70 percent of the steam is extracted from the turbine at 500 kpa for process heating. Solution A cogeneration plant is considered. The maximum rate of process heat supply, the power produced and the utilization factor when no process heat is supplied, and the rate of process heat supply when steam is extracted from the steam line and turbine at specified ratios are to be determined. Assumptions 1 Steady operating conditions exist. Pressure drops and heat losses in piping are negligible. 3 Kinetic and potential energy changes are negligible. Analysis The schematic of the cogeneration plant and the T-s diagram of the cycle are shown in Fig The power plant operates on an ideal 7 MPa 500 C T 1,, Mixing chamber 11 9 Boiler 10 7 MPa Expansion valve kpa Pump II Process heater Pump I 5 7 Turbine 500 kpa 6 5 kpa Condenser 8 5 kpa s FIGURE 10 3 Schematic and T-s diagram for Example 10 8.

104 58 Thermodynamics cycle and thus the pumps and the turbines are isentropic; there are no pressure drops in the boiler, process heater, and condenser; and steam leaves the condenser and the process heater as saturated liquid. The work inputs to the pumps and the enthalpies at various states are as follows: w pump I,in v 8 1P 9 P m 3 1 kj >kg kPa4a 1 kpa # b m 3 (a) The maximum rate of process heat is achieved when all the steam leaving the boiler is throttled and sent to the process heater and none is sent to the turbine (that is, m. 4 m. 7 m kg/s and m. 3 m. 5 m. 6 0). Thus, Q # p,max m # 1 1h 4 h kg>s kj>kg4 41,570 kw The utilization factor is 100 percent in this case since no heat is rejected in the condenser, heat losses from the piping and other components are assumed to be negligible, and combustion losses are not considered. (b) When no process heat is supplied, all the steam leaving the boiler passes through the turbine and expands to the condenser pressure of 5 kpa (that is, m. 3 m. 6 m kg/s and m. m. 5 0). Maximum power is produced in this mode, which is determined to be W # turb,out m # 1h 3 h kg>s kj>kg4 0,076 kw W # pump,in 115 kg>s17.03 kj>kg 105 kw W # net,out W # turb,out W # pump,in 10, kw 19,971 kw 0.0 MW Thus, 7.03 kj>kg 1 kj w pump II,in v 7 1P 10 P m 3 >kg kpa4 a 1 kpa # b m kj>kg h 1 h h 3 h kj>kg h kj>kg h kj>kg h 7 h 500 kpa kj>kg h 8 h 5 kpa kj>kg h 9 h 8 w pump I,in kj>kg kj>kg h 10 h 7 w pump II,in kj>kg kj>kg Q # in m # 1 1h 1 h kg>s kj>kg4 48,999 kw u W# net Q # p Q # in 119,971 0 kw or 40.8% 48,999 kw That is, 40.8 percent of the energy is utilized for a useful purpose. Notice that the utilization factor is equivalent to the thermal efficiency in this case. (c) Neglecting any kinetic and potential energy changes, an energy balance on the process heater yields

105 Chapter E # in E # out or m # 4h 4 m # 5h 5 Q # p,out m # 7h 7 where Thus 6. MW Q # p,out m # 4h 4 m # 5h 5 m # 7h 7 m # kg>s 1.5 kg>s m # kg>s 10.5 kg>s m # 7 m # 4 m # kg>s Q # p,out 11.5 kg>s kj>kg kg>s kj>kg 11 kg>s kj>kg Discussion Note that 6. MW of the heat transferred will be utilized in the process heater. We could also show that 11.0 MW of power is produced in this case, and the rate of heat input in the boiler is 43.0 MW. Thus the utilization factor is 86.5 percent COMBINED GAS VAPOR POWER CYCLES The continued quest for higher thermal efficiencies has resulted in rather innovative modifications to conventional power plants. The binary vapor cycle discussed later is one such modification. A more popular modification involves a gas power cycle topping a vapor power cycle, which is called the combined gas vapor cycle, or just the combined cycle. The combined cycle of greatest interest is the gas-turbine (Brayton) cycle topping a steamturbine (Rankine) cycle, which has a higher thermal efficiency than either of the cycles executed individually. Gas-turbine cycles typically operate at considerably higher temperatures than steam cycles. The maximum fluid temperature at the turbine inlet is about 60 C (1150 F) for modern steam power plants, but over 145 C (600 F) for gas-turbine power plants. It is over 1500 C at the burner exit of turbojet engines. The use of higher temperatures in gas turbines is made possible by recent developments in cooling the turbine blades and coating the blades with high-temperature-resistant materials such as ceramics. Because of the higher average temperature at which heat is supplied, gas-turbine cycles have a greater potential for higher thermal efficiencies. However, the gas-turbine cycles have one inherent disadvantage: The gas leaves the gas turbine at very high temperatures (usually above 500 C), which erases any potential gains in the thermal efficiency. The situation can be improved somewhat by using regeneration, but the improvement is limited. It makes engineering sense to take advantage of the very desirable characteristics of the gas-turbine cycle at high temperatures and to use the hightemperature exhaust gases as the energy source for the bottoming cycle such as a steam power cycle. The result is a combined gas steam cycle, as shown

106 584 Thermodynamics in Fig In this cycle, energy is recovered from the exhaust gases by transferring it to the steam in a heat exchanger that serves as the boiler. In general, more than one gas turbine is needed to supply sufficient heat to the steam. Also, the steam cycle may involve regeneration as well as reheating. Energy for the reheating process can be supplied by burning some additional fuel in the oxygen-rich exhaust gases. Recent developments in gas-turbine technology have made the combined gas steam cycle economically very attractive. The combined cycle increases the efficiency without increasing the initial cost greatly. Consequently, many new power plants operate on combined cycles, and many more existing steam- or gas-turbine plants are being converted to combined-cycle power plants. Thermal efficiencies well over 40 percent are reported as a result of conversion. A 1090-MW Tohoku combined plant that was put in commercial operation in 1985 in Niigata, Japan, is reported to operate at a thermal efficiency of 44 percent. This plant has two 191-MW steam turbines and six 118-MW gas turbines. Hot combustion gases enter the gas turbines at 1154 C, and steam enters the steam turbines at 500 C. Steam is cooled in the condenser by cooling water at an average temperature of 15 C. The compressors have a pressure ratio of 14, and the mass flow rate of air through the compressors is 443 kg/s. Q in Compressor Combustion chamber 6 7 GAS CYCLE Gas turbine T 7 Air in 5 Heat exchanger 9 8 Exhaust gases Q in GAS CYCLE 8 3 Pump Condenser STEAM CYCLE 6 3 Steam turbine STEAM CYCLE 4 1 Q out 4 Q out s FIGURE 10 4 Combined gas steam power plant.

107 A 1350-MW combined-cycle power plant built in Ambarli, Turkey, in 1988 by Siemens of Germany is the first commercially operating thermal plant in the world to attain an efficiency level as high as 5.5 percent at design operating conditions. This plant has six 150-MW gas turbines and three 173-MW steam turbines. Some recent combined-cycle power plants have achieved efficiencies above 60 percent. Chapter EXAMPLE 10 9 A Combined Gas Steam Power Cycle Consider the combined gas steam power cycle shown in Fig The topping cycle is a gas-turbine cycle that has a pressure ratio of 8. Air enters the compressor at 300 K and the turbine at 1300 K. The isentropic efficiency of the compressor is 80 percent, and that of the gas turbine is 85 percent. The bottoming cycle is a simple ideal Rankine cycle operating between the pressure limits of 7 MPa and 5 kpa. Steam is heated in a heat exchanger by the exhaust gases to a temperature of 500 C. The exhaust gases leave the heat exchanger at 450 K. Determine (a) the ratio of the mass flow rates of the steam and the combustion gases and (b) the thermal efficiency of the combined cycle. Solution A combined gas steam cycle is considered. The ratio of the mass flow rates of the steam and the combustion gases and the thermal efficiency are to be determined. Analysis The T-s diagrams of both cycles are given in Fig The gasturbine cycle alone was analyzed in Example 9 6, and the steam cycle in Example 10 8b, with the following results: Gas cycle: h kj>kg 1T K q in kj>kg w net kj>kg h th 6.6% h K kj>kg T, K ' 4' ' C 7 MPa 450 5' 7 MPa 300 1' 1 5 kpa 4 s FIGURE 10 5 T-s diagram of the gas steam combined cycle described in Example 10 9.

108 586 Thermodynamics Steam cycle: (a) The ratio of mass flow rates is determined from an energy balance on the heat exchanger: Thus, That is, 1 kg of exhaust gases can heat only kg of steam from 33 to 500 C as they are cooled from 853 to 450 K. Then the total net work output per kilogram of combustion gases becomes w net w net,gas yw net,steam kj>kg gas kg steam>kg gas kj>kg steam kj>kg gas h kj>kg 1T 33 C h kj>kg 1T C w net kj>kg h th 40.8% m # s m # g Therefore, for each kg of combustion gases produced, the combined plant will deliver kj of work. The net power output of the plant is determined by multiplying this value by the mass flow rate of the working fluid in the gas-turbine cycle. (b) The thermal efficiency of the combined cycle is determined from h th w net q in E # in E # out m # g h 5 m # s h 3 m # g h 4 m # s h m # s 1h 3 h m # g 1h 4 h 5 m # s m # y g kj>kg gas or 48.7% kj>kg gas Discussion Note that this combined cycle converts to useful work 48.7 percent of the energy supplied to the gas in the combustion chamber. This value is considerably higher than the thermal efficiency of the gas-turbine cycle (6.6 percent) or the steam-turbine cycle (40.8 percent) operating alone. TOPIC OF SPECIAL INTEREST* Binary Vapor Cycles With the exception of a few specialized applications, the working fluid predominantly used in vapor power cycles is water. Water is the best working fluid presently available, but it is far from being the ideal one. The binary cycle is an attempt to overcome some of the shortcomings of water and to approach the ideal working fluid by using two fluids. Before we discuss the binary cycle, let us list the characteristics of a working fluid most suitable for vapor power cycles: *This section can be skipped without a loss in continuity.

109 Chapter A high critical temperature and a safe maximum pressure. A critical temperature above the metallurgically allowed maximum temperature (about 60 C) makes it possible to transfer a considerable portion of the heat isothermally at the maximum temperature as the fluid changes phase. This makes the cycle approach the Carnot cycle. Very high pressures at the maximum temperature are undesirable because they create material-strength problems.. Low triple-point temperature. A triple-point temperature below the temperature of the cooling medium prevents any solidification problems. 3. A condenser pressure that is not too low. Condensers usually operate below atmospheric pressure. Pressures well below the atmospheric pressure create air-leakage problems. Therefore, a substance whose saturation pressure at the ambient temperature is too low is not a good candidate. 4. A high enthalpy of vaporization (h fg ) so that heat transfer to the working fluid is nearly isothermal and large mass flow rates are not needed. 5. A saturation dome that resembles an inverted U. This eliminates the formation of excessive moisture in the turbine and the need for reheating. 6. Good heat transfer characteristics (high thermal conductivity). 7. Other properties such as being inert, inexpensive, readily available, and nontoxic. Not surprisingly, no fluid possesses all these characteristics. Water comes the closest, although it does not fare well with respect to characteristics 1, 3, and 5. We can cope with its subatmospheric condenser pressure by careful sealing, and with the inverted V-shaped saturation dome by reheating, but there is not much we can do about item 1. Water has a low critical temperature (374 C, well below the metallurgical limit) and very high saturation pressures at high temperatures (16.5 MPa at 350 C). Well, we cannot change the way water behaves during the high-temperature part of the cycle, but we certainly can replace it with a more suitable fluid. The result is a power cycle that is actually a combination of two cycles, one in the high-temperature region and the other in the low-temperature region. Such a cycle is called a binary vapor cycle. In binary vapor cycles, the condenser of the high-temperature cycle (also called the topping cycle) serves as the boiler of the low-temperature cycle (also called the bottoming cycle). That is, the heat output of the high-temperature cycle is used as the heat input to the low-temperature one. Some working fluids found suitable for the high-temperature cycle are mercury, sodium, potassium, and sodium potassium mixtures. The schematic and T-s diagram for a mercury water binary vapor cycle are shown in Fig The critical temperature of mercury is 898 C (well above the current metallurgical limit), and its critical pressure is only about 18 MPa. This makes mercury a very suitable working fluid for the topping cycle. Mercury is not suitable as the sole working fluid for the entire cycle, however, since at a condenser temperature of 3 C its saturation pressure is 0.07 Pa. A power plant cannot operate at this vacuum because of air-leakage problems. At an acceptable condenser pressure of 7 kpa, the saturation temperature of mercury is

110 588 Thermodynamics T 3 Boiler Mercury pump MERCURY CYCLE Mercury turbine Saturation dome (mercury) 3 1 Heat exchanger 4 MERCURY CYCLE Q 4 7 Steam pump 6 STEAM CYCLE 7 Steam turbine Superheater 6 1 STEAM CYCLE 5 8 Saturation dome (steam) 5 Condenser 8 s FIGURE 10 6 Mercury water binary vapor cycle. 37 C, which is too high as the minimum temperature in the cycle. Therefore, the use of mercury as a working fluid is limited to the high-temperature cycles. Other disadvantages of mercury are its toxicity and high cost. The mass flow rate of mercury in binary vapor cycles is several times that of water because of its low enthalpy of vaporization. It is evident from the T-s diagram in Fig that the binary vapor cycle approximates the Carnot cycle more closely than the steam cycle for the same temperature limits. Therefore, the thermal efficiency of a power plant can be increased by switching to binary cycles. The use of mercury water binary cycles in the United States dates back to 198. Several such plants have been built since then in the New England area, where fuel costs are typically higher. A small (40-MW) mercury steam power plant that was in service in New Hampshire in 1950 had a higher thermal efficiency than most of the large modern power plants in use at that time. Studies show that thermal efficiencies of 50 percent or higher are possible with binary vapor cycles. However, binary vapor cycles are not economically attractive because of their high initial cost and the competition offered by the combined gas steam power plants.

111 Chapter SUMMARY The Carnot cycle is not a suitable model for vapor power cycles because it cannot be approximated in practice. The model cycle for vapor power cycles is the Rankine cycle, which is composed of four internally reversible processes: constant-pressure heat addition in a boiler, isentropic expansion in a turbine, constant-pressure heat rejection in a condenser, and isentropic compression in a pump. Steam leaves the condenser as a saturated liquid at the condenser pressure. The thermal efficiency of the Rankine cycle can be increased by increasing the average temperature at which heat is transferred to the working fluid and/or by decreasing the average temperature at which heat is rejected to the cooling medium. The average temperature during heat rejection can be decreased by lowering the turbine exit pressure. Consequently, the condenser pressure of most vapor power plants is well below the atmospheric pressure. The average temperature during heat addition can be increased by raising the boiler pressure or by superheating the fluid to high temperatures. There is a limit to the degree of superheating, however, since the fluid temperature is not allowed to exceed a metallurgically safe value. Superheating has the added advantage of decreasing the moisture content of the steam at the turbine exit. Lowering the exhaust pressure or raising the boiler pressure, however, increases the moisture content. To take advantage of the improved efficiencies at higher boiler pressures and lower condenser pressures, steam is usually reheated after expanding partially in the high-pressure turbine. This is done by extracting the steam after partial expansion in the high-pressure turbine, sending it back to the boiler where it is reheated at constant pressure, and returning it to the low-pressure turbine for complete expansion to the condenser pressure. The average temperature during the reheat process, and thus the thermal efficiency of the cycle, can be increased by increasing the number of expansion and reheat stages. As the number of stages is increased, the expansion and reheat processes approach an isothermal process at maximum temperature. Reheating also decreases the moisture content at the turbine exit. Another way of increasing the thermal efficiency of the Rankine cycle is regeneration. During a regeneration process, liquid water (feedwater) leaving the pump is heated by steam bled off the turbine at some intermediate pressure in devices called feedwater heaters. The two streams are mixed in open feedwater heaters, and the mixture leaves as a saturated liquid at the heater pressure. In closed feedwater heaters, heat is transferred from the steam to the feedwater without mixing. The production of more than one useful form of energy (such as process heat and electric power) from the same energy source is called cogeneration. Cogeneration plants produce electric power while meeting the process heat requirements of certain industrial processes. This way, more of the energy transferred to the fluid in the boiler is utilized for a useful purpose. The fraction of energy that is used for either process heat or power generation is called the utilization factor of the cogeneration plant. The overall thermal efficiency of a power plant can be increased by using a combined cycle. The most common combined cycle is the gas steam combined cycle where a gas-turbine cycle operates at the high-temperature range and a steam-turbine cycle at the low-temperature range. Steam is heated by the high-temperature exhaust gases leaving the gas turbine. Combined cycles have a higher thermal efficiency than the steam- or gas-turbine cycles operating alone. REFERENCES AND SUGGESTED READINGS 1. R. L. Bannister and G. J. Silvestri. The Evolution of Central Station Steam Turbines. Mechanical Engineering, February 1989, pp R. L. Bannister, G. J. Silvestri, A. Hizume, and T. Fujikawa. High Temperature Supercritical Steam Turbines. Mechanical Engineering, February 1987, pp M. M. El-Wakil. Powerplant Technology. New York: McGraw-Hill, K. W. Li and A. P. Priddy. Power Plant System Design. New York: John Wiley & Sons, H. Sorensen. Energy Conversion Systems. New York: John Wiley & Sons, Steam, Its Generation and Use. 39th ed. New York: Babcock and Wilcox Co., Turbomachinery 8, no. (March/April 1987). Norwalk, CT: Business Journals, Inc. 8. K. Wark and D. E. Richards. Thermodynamics. 6th ed. New York: McGraw-Hill, J. Weisman and R. Eckart. Modern Power Plant Engineering. Englewood Cliffs, NJ: Prentice-Hall, 1985.

112 590 Thermodynamics PROBLEMS* Carnot Vapor Cycle 10 1C Why is excessive moisture in steam undesirable in steam turbines? What is the highest moisture content allowed? 10 C Why is the Carnot cycle not a realistic model for steam power plants? 10 3E Water enters the boiler of a steady-flow Carnot engine as a saturated liquid at 180 psia and leaves with a quality of Steam leaves the turbine at a pressure of 14.7 psia. Show the cycle on a T-s diagram relative to the saturation lines, and determine (a) the thermal efficiency, (b) the quality at the end of the isothermal heat-rejection process, and (c) the net work output. Answers: (a) 19.3 percent, (b) 0.153, (c) 148 Btu/lbm 10 4 A steady-flow Carnot cycle uses water as the working fluid. Water changes from saturated liquid to saturated vapor as heat is transferred to it from a source at 50 C. Heat rejection takes place at a pressure of 0 kpa. Show the cycle on a T-s diagram relative to the saturation lines, and determine (a) the thermal efficiency, (b) the amount of heat rejected, in kj/kg, and (c) the net work output Repeat Prob for a heat rejection pressure of 10 kpa Consider a steady-flow Carnot cycle with water as the working fluid. The maximum and minimum temperatures in the cycle are 350 and 60 C. The quality of water is at the beginning of the heat-rejection process and 0.1 at the end. Show the cycle on a T-s diagram relative to the saturation lines, and determine (a) the thermal efficiency, (b) the pressure at the turbine inlet, and (c) the net work output. Answers: (a) 0.465, (b) 1.40 MPa, (c) 163 kj/kg The Simple Rankine Cycle 10 7C What four processes make up the simple ideal Rankine cycle? 10 8C Consider a simple ideal Rankine cycle with fixed turbine inlet conditions. What is the effect of lowering the condenser pressure on *Problems designated by a C are concept questions, and students are encouraged to answer them all. Problems designated by an E are in English units, and the SI users can ignore them. Problems with a CD-EES icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed DVD. Problems with a computer-ees icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text. Pump work input: (a) increases, (b) decreases, (c) remains the same Turbine work (a) increases, (b) decreases, output: (c) remains the same Heat supplied: (a) increases, (b) decreases, (c) remains the same Heat rejected: (a) increases, (b) decreases, (c) remains the same Cycle efficiency: (a) increases, (b) decreases, (c) remains the same Moisture content (a) increases, (b) decreases, at turbine exit: (c) remains the same 10 9C Consider a simple ideal Rankine cycle with fixed turbine inlet temperature and condenser pressure. What is the effect of increasing the boiler pressure on Pump work input: (a) increases, (b) decreases, (c) remains the same Turbine work (a) increases, (b) decreases, output: (c) remains the same Heat supplied: (a) increases, (b) decreases, (c) remains the same Heat rejected: (a) increases, (b) decreases, (c) remains the same Cycle efficiency: (a) increases, (b) decreases, (c) remains the same Moisture content (a) increases, (b) decreases, at turbine exit: (c) remains the same 10 10C Consider a simple ideal Rankine cycle with fixed boiler and condenser pressures. What is the effect of superheating the steam to a higher temperature on Pump work input: (a) increases, (b) decreases, (c) remains the same Turbine work (a) increases, (b) decreases, output: (c) remains the same Heat supplied: (a) increases, (b) decreases, (c) remains the same Heat rejected: (a) increases, (b) decreases, (c) remains the same Cycle efficiency: (a) increases, (b) decreases, (c) remains the same Moisture content (a) increases, (b) decreases, at turbine exit: (c) remains the same 10 11C How do actual vapor power cycles differ from idealized ones?

113 10 1C Compare the pressures at the inlet and the exit of the boiler for (a) actual and (b) ideal cycles C The entropy of steam increases in actual steam turbines as a result of irreversibilities. In an effort to control entropy increase, it is proposed to cool the steam in the turbine by running cooling water around the turbine casing. It is argued that this will reduce the entropy and the enthalpy of the steam at the turbine exit and thus increase the work output. How would you evaluate this proposal? 10 14C Is it possible to maintain a pressure of 10 kpa in a condenser that is being cooled by river water entering at 0 C? A steam power plant operates on a simple ideal Rankine cycle between the pressure limits of 3 MPa and 50 kpa. The temperature of the steam at the turbine inlet is 300 C, and the mass flow rate of steam through the cycle is 35 kg/s. Show the cycle on a T-s diagram with respect to saturation lines, and determine (a) the thermal efficiency of the cycle and (b) the net power output of the power plant Consider a 10-MW steam power plant that operates on a simple ideal Rankine cycle. Steam enters the turbine at 10 MPa and 500 C and is cooled in the condenser at a pressure of 10 kpa. Show the cycle on a T-s diagram with respect to saturation lines, and determine (a) the quality of the steam at the turbine exit, (b) the thermal efficiency of the cycle, and (c) the mass flow rate of the steam. Answers: (a) 0.793, (b) 40. percent, (c) 165 kg/s Repeat Prob assuming an isentropic efficiency of 85 percent for both the turbine and the pump. Answers: (a) 0.874, (b) 34.1 percent, (c) 194 kg/s 10 18E A steam power plant operates on a simple ideal Rankine cycle between the pressure limits of 150 and psia. The mass flow rate of steam through the cycle is 75 lbm/s. The moisture content of the steam at the turbine exit is not to exceed 10 percent. Show the cycle on a T-s diagram with respect to saturation lines, and determine (a) the minimum turbine inlet temperature, (b) the rate of heat input in the boiler, and (c) the thermal efficiency of the cycle E Repeat Prob E assuming an isentropic efficiency of 85 percent for both the turbine and the pump Consider a coal-fired steam power plant that produces 300 MW of electric power. The power plant operates on a simple ideal Rankine cycle with turbine inlet conditions of 5 MPa and 450 C and a condenser pressure of 5 kpa. The coal has a heating value (energy released when the fuel is burned) of 9,300 kj/kg. Assuming that 75 percent of this energy is transferred to the steam in the boiler and that the electric generator has an efficiency of 96 percent, determine (a) the overall plant efficiency (the ratio of net electric power output to the energy input as fuel) and (b) the required rate of coal supply. Answers: (a) 4.5 percent, (b) 150 t/h Chapter Consider a solar-pond power plant that operates on a simple ideal Rankine cycle with refrigerant-134a as the working fluid. The refrigerant enters the turbine as a saturated vapor at 1.4 MPa and leaves at 0.7 MPa. The mass flow rate of the refrigerant is 3 kg/s. Show the cycle on a T-s diagram with respect to saturation lines, and determine (a) the thermal efficiency of the cycle and (b) the power output of this plant. 10 Consider a steam power plant that operates on a simple ideal Rankine cycle and has a net power output of 45 MW. Steam enters the turbine at 7 MPa and 500 C and is cooled in the condenser at a pressure of 10 kpa by running cooling water from a lake through the tubes of the condenser at a rate of 000 kg/s. Show the cycle on a T-s diagram with respect to saturation lines, and determine (a) the thermal efficiency of the cycle, (b) the mass flow rate of the steam, and (c) the temperature rise of the cooling water. Answers: (a) 38.9 percent, (b) 36 kg/s, (c) 8.4 C 10 3 Repeat Prob. 10 assuming an isentropic efficiency of 87 percent for both the turbine and the pump. Answers: (a) 33.8 percent, (b) 41.4 kg/s, (c) 10.5 C 10 4 The net work output and the thermal efficiency for the Carnot and the simple ideal Rankine cycles with steam as the working fluid are to be calculated and compared. Steam enters the turbine in both cases at 10 MPa as a saturated vapor, and the condenser pressure is 0 kpa. In the Rankine cycle, the condenser exit state is saturated liquid and in the Carnot cycle, the boiler inlet state is saturated liquid. Draw the T-s diagrams for both cycles A binary geothermal power plant uses geothermal water at 160 C as the heat source. The cycle operates on the simple Rankine cycle with isobutane as the working fluid. Heat is transferred to the cycle by a heat exchanger in which geothermal liquid water enters at 160 C at a rate of kg/s and leaves at 90 C. Isobutane enters the turbine at 3.5 MPa and 147 C at a rate of kg/s, and leaves at 79.5 C and 4 Turbine 3 Geothermal water in Air-cooled condenser Heat exchanger FIGURE P10 5 Pump 1 Geothermal water out

114 59 Thermodynamics 410 kpa. Isobutane is condensed in an air-cooled condenser and pumped to the heat exchanger pressure. Assuming the pump to have an isentropic efficiency of 90 percent, determine (a) the isentropic efficiency of the turbine, (b) the net power output of the plant, and (c) the thermal efficiency of the cycle The schematic of a single-flash geothermal power plant with state numbers is given in Fig. P10 6. Geothermal resource exists as saturated liquid at 30 C. The geothermal liquid is withdrawn from the production well at a rate of 30 kg/s, and is flashed to a pressure of 500 kpa by an essentially isenthalpic flashing process where the resulting vapor is separated from the liquid in a separator and directed to the turbine. The steam leaves the turbine at 10 kpa with a moisture content of 10 percent and enters the condenser where it is condensed and routed to a reinjection well along with the liquid coming off the separator. Determine (a) the mass flow rate of steam through the turbine, (b) the isentropic efficiency of the turbine, (c) the power output of the turbine, and (d) the thermal efficiency of the plant (the ratio of the turbine work output to the energy of the geothermal fluid relative to standard ambient conditions). Answers: (a) 38. kg/s, (b) 0.686, (c) 15.4 MW, (d) 7.6 percent Flash chamber Separator 6 3 Steam turbine 4 Condenser 5 6 Flash chamber 1 Production well Separator I 7 Flash chamber Separator II Steam turbine Condenser Reinjection well 10 8 Reconsider Prob Now, it is proposed that the liquid water coming out of the separator be used as the heat source in a binary cycle with isobutane as the working fluid. Geothermal liquid water leaves the heat exchanger at 90 C while isobutane enters the turbine at 3.5 MPa and 145 C and leaves at 80 C and 400 kpa. Isobutane is condensed in an air-cooled condenser and then pumped to the heat exchanger pressure. Assuming an isentropic efficiency of 90 percent for the pump, determine (a) the mass flow rate of isobutane in the binary cycle, (b) the net power outputs of both the flashing and the binary sections of the plant, and (c) the thermal efficiencies of the binary cycle and the combined plant. Answers: (a) kg/s, (b) 15.4 MW, 6.14 MW, (c) 1. percent, 10.6 percent 3 8 FIGURE P Production well FIGURE P10 6 Reinjection well 10 7 Reconsider Prob Now, it is proposed that the liquid water coming out of the separator be routed through another flash chamber maintained at 150 kpa, and the steam produced be directed to a lower stage of the same turbine. Both streams of steam leave the turbine at the same state of 10 kpa and 90 percent quality. Determine (a) the temperature of steam at the outlet of the second flash chamber, (b) the power produced by the lower stage of the turbine, and (c) the thermal efficiency of the plant. Production well Flash chamber 1 Separator 6 3 Isobutane turbine 8 Heat exchanger Steam turbine Air-cooled condenser BINARY CYCLE 11 FIGURE P10 8 Condenser Pump 5 10 Reinjection well

115 The Reheat Rankine Cycle 10 9C How do the following quantities change when a simple ideal Rankine cycle is modified with reheating? Assume the mass flow rate is maintained the same. Pump work input: (a) increases, (b) decreases, (c) remains the same Turbine work (a) increases, (b) decreases, output: (c) remains the same Heat supplied: (a) increases, (b) decreases, (c) remains the same Heat rejected: (a) increases, (b) decreases, (c) remains the same Moisture content (a) increases, (b) decreases, at turbine exit: (c) remains the same 10 30C Show the ideal Rankine cycle with three stages of reheating on a T-s diagram. Assume the turbine inlet temperature is the same for all stages. How does the cycle efficiency vary with the number of reheat stages? 10 31C Consider a simple Rankine cycle and an ideal Rankine cycle with three reheat stages. Both cycles operate between the same pressure limits. The maximum temperature is 700 C in the simple cycle and 450 C in the reheat cycle. Which cycle do you think will have a higher thermal efficiency? 10 3 A steam power plant operates on the ideal reheat Rankine cycle. Steam enters the highpressure turbine at 8 MPa and 500 C and leaves at 3 MPa. Steam is then reheated at constant pressure to 500 C before it expands to 0 kpa in the low-pressure turbine. Determine the turbine work output, in kj/kg, and the thermal efficiency of the cycle. Also, show the cycle on a T-s diagram with respect to saturation lines Reconsider Prob Using EES (or other) software, solve this problem by the diagram window data entry feature of EES. Include the effects of the turbine and pump efficiencies and also show the effects of reheat on the steam quality at the low-pressure turbine exit. Plot the cycle on a T-s diagram with respect to the saturation lines. Discuss the results of your parametric studies Consider a steam power plant that operates on a reheat Rankine cycle and has a net power output of 80 MW. Steam enters the high-pressure turbine at 10 MPa and 500 C and the low-pressure turbine at 1 MPa and 500 C. Steam leaves the condenser as a saturated liquid at a pressure of 10 kpa. The isentropic efficiency of the turbine is 80 percent, and that of the pump is 95 percent. Show the cycle on a T-s diagram with respect to saturation lines, and determine (a) the quality (or temperature, if superheated) of the steam at the turbine exit, (b) the thermal efficiency of the cycle, and (c) the mass flow rate of the steam. Answers: (a) 88.1 C, (b) 34.1 percent, (c) 6.7 kg/s Chapter Repeat Prob assuming both the pump and the turbine are isentropic. Answers: (a) 0.949, (b) 41.3 percent, (c) 50.0 kg/s 10 36E Steam enters the high-pressure turbine of a steam power plant that operates on the ideal reheat Rankine cycle at 800 psia and 900 F and leaves as saturated vapor. Steam is then reheated to 800 F before it expands to a pressure of 1 psia. Heat is transferred to the steam in the boiler at a rate of Btu/s. Steam is cooled in the condenser by the cooling water from a nearby river, which enters the condenser at 45 F. Show the cycle on a T-s diagram with respect to saturation lines, and determine (a) the pressure at which reheating takes place, (b) the net power output and thermal efficiency, and (c) the minimum mass flow rate of the cooling water required A steam power plant operates on an ideal reheat Rankine cycle between the pressure limits of 15 MPa and 10 kpa. The mass flow rate of steam through the cycle is 1 kg/s. Steam enters both stages of the turbine at 500 C. If the moisture content of the steam at the exit of the low-pressure turbine is not to exceed 10 percent, determine (a) the pressure at which reheating takes place, (b) the total rate of heat input in the boiler, and (c) the thermal efficiency of the cycle. Also, show the cycle on a T-s diagram with respect to saturation lines A steam power plant operates on the reheat Rankine cycle. Steam enters the high-pressure turbine at 1.5 MPa and 550 C at a rate of 7.7 kg/s and leaves at MPa. Steam is then reheated at constant pressure to 450 C before it expands in the low-pressure turbine. The isentropic efficiencies of the turbine and the pump are 85 percent and 90 percent, respectively. Steam leaves the condenser as a saturated liquid. If the moisture content of the steam at the exit of the turbine is not to exceed 5 percent, determine (a) the condenser pressure, (b) the net power output, and (c) the thermal efficiency. Answers: (a) 9.73 kpa, (b) 10. MW, (c) 36.9 percent Boiler 4 5 Pump 3 FIGURE P10 38 Turbine 1 6 Condenser

116 594 Thermodynamics Regenerative Rankine Cycle 10 39C How do the following quantities change when the simple ideal Rankine cycle is modified with regeneration? Assume the mass flow rate through the boiler is the same. Turbine work (a) increases, (b) decreases, output: (c) remains the same Heat supplied: (a) increases, (b) decreases, (c) remains the same Heat rejected: (a) increases, (b) decreases, (c) remains the same Moisture content (a) increases, (b) decreases, at turbine exit: (c) remains the same 10 40C During a regeneration process, some steam is extracted from the turbine and is used to heat the liquid water leaving the pump. This does not seem like a smart thing to do since the extracted steam could produce some more work in the turbine. How do you justify this action? 10 41C How do open feedwater heaters differ from closed feedwater heaters? 10 4C Consider a simple ideal Rankine cycle and an ideal regenerative Rankine cycle with one open feedwater heater. The two cycles are very much alike, except the feedwater in the regenerative cycle is heated by extracting some steam just before it enters the turbine. How would you compare the efficiencies of these two cycles? 10 43C Devise an ideal regenerative Rankine cycle that has the same thermal efficiency as the Carnot cycle. Show the cycle on a T-s diagram A steam power plant operates on an ideal regenerative Rankine cycle. Steam enters the turbine at 6 MPa and 450 C and is condensed in the condenser at 0 kpa. Steam is extracted from the turbine at 0.4 MPa to heat the feedwater in an open feedwater heater. Water leaves the feedwater heater as a saturated liquid. Show the cycle on a T-s diagram, and determine (a) the net work output per kilogram of steam flowing through the boiler and (b) the thermal efficiency of the cycle. Answers: (a) 1017 kj/kg, (b) 37.8 percent Repeat Prob by replacing the open feedwater heater with a closed feedwater heater. Assume that the feedwater leaves the heater at the condensation temperature of the extracted steam and that the extracted steam leaves the heater as a saturated liquid and is pumped to the line carrying the feedwater A steam power plant operates on an ideal regenerative Rankine cycle with two open feedwater heaters. Steam enters the turbine at 10 MPa and 600 C and exhausts to the condenser at 5 kpa. Steam is extracted from the turbine at 0.6 and 0. MPa. Water leaves both feedwater heaters as a saturated liquid. The mass flow rate of steam through the boiler is kg/s. Show the cycle on a T-s diagram, and determine (a) the net power output of the power plant and (b) the thermal efficiency of the cycle. Answers: (a) 30.5 MW, (b) 47.1 percent Consider an ideal steam regenerative Rankine cycle with two feedwater heaters, one closed and one open. Steam enters the turbine at 1.5 MPa and 550 C and exhausts to the condenser at 10 kpa. Steam is extracted from the turbine at 0.8 MPa for the closed feedwater heater and at 0.3 MPa for the open one. The feedwater is heated to the condensation temperature of the extracted steam in the closed feedwater heater. The extracted steam leaves the closed feedwater heater as a saturated liquid, which is subsequently throttled to the open feedwater heater. Show the cycle on a T-s diagram with respect to saturation lines, and determine (a) the mass flow rate of steam through the boiler for a net power output of 50 MW and (b) the thermal efficiency of the cycle. Boiler 5 Closed FWH 6 4 P II Open FWH y 10 Turbine z P I 11 Condenser 1 y z FIGURE P Reconsider Prob Using EES (or other) software, investigate the effects of turbine and pump efficiencies as they are varied from 70 percent to 100 percent on the mass flow rate and thermal efficiency. Plot the mass flow rate and the thermal efficiency as a function of turbine efficiency for pump efficiencies of 70, 85, and 100 percent, and discuss the results. Also plot the T-s diagram for turbine and pump efficiencies of 85 percent A steam power plant operates on an ideal reheat regenerative Rankine cycle and has a net power output of 80 MW. Steam enters the high-pressure turbine at 10 MPa and 550 C and leaves at 0.8 MPa. Some steam is extracted at this pressure to heat the feedwater in an open feedwater heater. The rest of the steam is reheated to 500 C and is expanded in the low-pressure turbine to the condenser pressure of 10 kpa. Show the cycle on a T-s diagram with respect to saturation lines, and determine (a) the mass flow rate of steam through the boiler and (b) the thermal efficiency of the cycle. Answers: (a) 54.5 kg/s, (b) 44.4 percent Repeat Prob , but replace the open feedwater heater with a closed feedwater heater. Assume that the feed- 1

117 water leaves the heater at the condensation temperature of the extracted steam and that the extracted steam leaves the heater as a saturated liquid and is pumped to the line carrying the feedwater. Boiler Mixing chamber P II 1 y 7 y Closed FWH 8 Condenser 10 51E A steam power plant operates on an ideal reheat regenerative Rankine cycle with one reheater and two open feedwater heaters. Steam enters the high-pressure turbine at 1500 psia and 1100 F and leaves the low-pressure turbine at 1 psia. Steam is extracted from the turbine at 50 and 40 psia, and it is reheated to 1000 F at a pressure of 140 psia. Water leaves both feedwater heaters as a saturated liquid. Heat is transferred to the steam in the boiler at a rate of Btu/s. Show the cycle on a T-s diagram with respect to saturation lines, and determine (a) the mass flow rate of steam through the boiler, (b) the net power output of the plant, and (c) the thermal efficiency of the cycle. 6 Boiler P III Reheater y 7 Open FWH II FIGURE P P II High-P turbine 8 y Open FWH I 1 y 3 FIGURE P10 51E High-P turbine z P I z P I Low-P turbine 1 Low-P turbine 11 1 y z 1 Condenser 1 Chapter A steam power plant operates on the reheatregenerative Rankine cycle with a closed feedwater heater. Steam enters the turbine at 1.5 MPa and 550 C at a rate of 4 kg/s and is condensed in the condenser at a pressure of 0 kpa. Steam is reheated at 5 MPa to 550 C. Some steam is extracted from the low-pressure turbine at 1.0 MPa, is completely condensed in the closed feedwater heater, and pumped to 1.5 MPa before it mixes with the feedwater at the same pressure. Assuming an isentropic efficiency of 88 percent for both the turbine and the pump, determine (a) the temperature of the steam at the inlet of the closed feedwater heater, (b) the mass flow rate of the steam extracted from the turbine for the closed feedwater heater, (c) the net power output, and (d) the thermal efficiency. Answers: (a) 38 C, (b) 4.9 kg/s, (c) 8.6 MW, (d) 39.3 percent 4 Boiler Mixing chamber Closed FWH 10 3 P II 7 FIGURE P10 5 High-P turbine 8 P I y Low-P turbine y Condenser Second-Law Analysis of Vapor Power Cycles 10 53C How can the second-law efficiency of a simple ideal Rankine cycle be improved? Determine the exergy destruction associated with each of the processes of the Rankine cycle described in Prob , assuming a source temperature of 1500 K and a sink temperature of 90 K Determine the exergy destruction associated with each of the processes of the Rankine cycle described in Prob , assuming a source temperature of 1500 K and a sink temperature of 90 K. Answers: 0, 111 kj/kg, 0, 17.3 kj/kg Determine the exergy destruction associated with the heat rejection process in Prob. 10. Assume a source temperature of 1500 K and a sink temperature of 90 K. Also, determine the exergy of the steam at the boiler exit. Take P kpa Determine the exergy destruction associated with each of the processes of the reheat Rankine cycle described in Prob Assume a source temperature of 1800 K and a sink temperature of 300 K.

118 596 Thermodynamics Reconsider Prob Using EES (or other) software, solve this problem by the diagram window data entry feature of EES. Include the effects of the turbine and pump efficiencies to evaluate the irreversibilities associated with each of the processes. Plot the cycle on a T-s diagram with respect to the saturation lines. Discuss the results of your parametric studies Determine the exergy destruction associated with the heat addition process and the expansion process in Prob Assume a source temperature of 1600 K and a sink temperature of 85 K. Also, determine the exergy of the steam at the boiler exit. Take P kpa. Answers: 189 kj/kg, 47.9 kj/kg, 1495 kj/kg Determine the exergy destruction associated with the regenerative cycle described in Prob Assume a source temperature of 1500 K and a sink temperature of 90 K. Answer: 1155 kj/kg Determine the exergy destruction associated with the reheating and regeneration processes described in Prob Assume a source temperature of 1800 K and a sink temperature of 90 K The schematic of a single-flash geothermal power plant with state numbers is given in Fig. P10 6. Geothermal resource exists as saturated liquid at 30 C. The geothermal liquid is withdrawn from the production well at a rate of 30 kg/s and is flashed to a pressure of 500 kpa by an essentially isenthalpic flashing process where the resulting vapor is separated from the liquid in a separator and is directed to the turbine. The steam leaves the turbine at 10 kpa with a moisture content of 5 percent and enters the condenser where it is condensed; it is routed to a reinjection well along with the liquid coming off the separator. Determine (a) the power output of the turbine and the thermal efficiency of the plant, (b) the exergy of the geothermal liquid at the exit of the flash chamber, and the exergy destructions and the second-law (exergetic) efficiencies for (c) the flash chamber, (d) the turbine, and (e) the entire plant. Answers: (a) 10.8 MW, 0.053, (b) 17.3 MW, (c) 5.1 MW, 0.898, (d) 10.9 MW, 0.500, (e) 39.0 MW, 0.18 Cogeneration 10 63C How is the utilization factor P u for cogeneration plants defined? Could P u be unity for a cogeneration plant that does not produce any power? 10 64C Consider a cogeneration plant for which the utilization factor is 1. Is the irreversibility associated with this cycle necessarily zero? Explain C Consider a cogeneration plant for which the utilization factor is 0.5. Can the exergy destruction associated with this plant be zero? If yes, under what conditions? 10 66C What is the difference between cogeneration and regeneration? Steam enters the turbine of a cogeneration plant at 7 MPa and 500 C. One-fourth of the steam is extracted from the turbine at 600-kPa pressure for process heating. The remaining steam continues to expand to 10 kpa. The extracted steam is then condensed and mixed with feedwater at constant pressure and the mixture is pumped to the boiler pressure of 7 MPa. The mass flow rate of steam through the boiler is 30 kg/s. Disregarding any pressure drops and heat losses in the piping, and assuming the turbine and the pump to be isentropic, determine the net power produced and the utilization factor of the plant. 6 Flash chamber Separator 6 3 Steam turbine 4 Condenser 5 Boiler 5 P II Q process Process heater Turbine P I FIGURE P Condenser 1 1 Production well FIGURE P10 6 Reinjection well 10 68E A large food-processing plant requires lbm/s of saturated or slightly superheated steam at 80 psia, which is extracted from the turbine of a cogeneration plant. The boiler generates steam at 1000 psia and 1000 F at a rate of 5 lbm/s,

119 and the condenser pressure is psia. Steam leaves the process heater as a saturated liquid. It is then mixed with the feedwater at the same pressure and this mixture is pumped to the boiler pressure. Assuming both the pumps and the turbine have isentropic efficiencies of 86 percent, determine (a) the rate of heat transfer to the boiler and (b) the power output of the cogeneration plant. Answers: (a) 6667 Btu/s, (b) 06 kw Steam is generated in the boiler of a cogeneration plant at 10 MPa and 450 C at a steady rate of 5 kg/s. In normal operation, steam expands in a turbine to a pressure of 0.5 MPa and is then routed to the process heater, where it supplies the process heat. Steam leaves the process heater as a saturated liquid and is pumped to the boiler pressure. In this mode, no steam passes through the condenser, which operates at 0 kpa. (a) Determine the power produced and the rate at which process heat is supplied in this mode. (b) Determine the power produced and the rate of process heat supplied if only 60 percent of the steam is routed to the process heater and the remainder is expanded to the condenser pressure Consider a cogeneration power plant modified with regeneration. Steam enters the turbine at 6 MPa and 450 C and expands to a pressure of 0.4 MPa. At this pressure, 60 percent of the steam is extracted from the turbine, and the remainder expands to 10 kpa. Part of the extracted steam is used to heat the feedwater in an open feedwater heater. The rest of the extracted steam is used for process heating and leaves the process heater as a saturated liquid at 0.4 MPa. It is subsequently mixed with the feedwater leaving the feedwater heater, and the mixture is pumped to the boiler pressure. 6 Chapter Assuming the turbines and the pumps to be isentropic, show the cycle on a T-s diagram with respect to saturation lines, and determine the mass flow rate of steam through the boiler for a net power output of 15 MW. Answer: 17.7 kg/s Reconsider Prob Using EES (or other) software, investigate the effect of the extraction pressure for removing steam from the turbine to be used for the process heater and open feedwater heater on the required mass flow rate. Plot the mass flow rate through the boiler as a function of the extraction pressure, and discuss the results. 10 7E Steam is generated in the boiler of a cogeneration plant at 600 psia and 800 F at a rate of 18 lbm/s. The plant is to produce power while meeting the process steam requirements for a certain industrial application. One-third of the steam leaving the boiler is throttled to a pressure of 10 psia and is routed to the process heater. The rest of the steam is expanded in an isentropic turbine to a pressure of 10 psia and is also routed to the process heater. Steam leaves the process heater at 40 F. Neglecting the pump work, determine (a) the net power produced, (b) the rate of process heat supply, and (c) the utilization factor of this plant A cogeneration plant is to generate power and 8600 kj/s of process heat. Consider an ideal cogeneration steam plant. Steam enters the turbine from the boiler at 7 MPa and 500 C. One-fourth of the steam is extracted from the turbine at 600-kPa pressure for process heating. The remainder of the steam continues to expand and exhausts to the condenser at 10 kpa. The steam extracted for the process heater is condensed in the heater and mixed with the feedwater at 600 kpa. The mixture is pumped to the boiler pressure of 7 MPa. Show the cycle on a T-s diagram with respect to saturation lines, and determine (a) the mass flow rate of steam that must be supplied by the boiler, (b) the net power produced by the plant, and (c) the utilization factor. Boiler Turbine Process heater 8 5 Boiler Process heater 7 Turbine 8 P I 4 3 FWH Condenser 3 Condenser P I 1 4 P II 1 P I FIGURE P10 70 FIGURE P10 73

120 598 Thermodynamics Combined Gas Vapor Power Cycles 10 74C In combined gas steam cycles, what is the energy source for the steam? 10 75C Why is the combined gas steam cycle more efficient than either of the cycles operated alone? The gas-turbine portion of a combined gas steam power plant has a pressure ratio of 16. Air enters the compressor at 300 K at a rate of 14 kg/s and is heated to 1500 K in the combustion chamber. The combustion gases leaving the gas turbine are used to heat the steam to 400 C at 10 MPa in a heat exchanger. The combustion gases leave the heat exchanger at 40 K. The steam leaving the turbine is condensed at 15 kpa. Assuming all the compression and expansion processes to be isentropic, determine (a) the mass flow rate of the steam, (b) the net power output, and (c) the thermal efficiency of the combined cycle. For air, assume constant specific heats at room temperature. Answers: (a) 1.75 kg/s, (b) 7819 kw, (c) 66.4 percent Consider a combined gas steam power plant that has a net power output of 450 MW. The pressure ratio of the gas-turbine cycle is 14. Air enters the compressor at 300 K and the turbine at 1400 K. The combustion gases leaving the gas turbine are used to heat the steam at 8 MPa to 400 C in a heat exchanger. The combustion gases leave the heat exchanger at 460 K. An open feedwater heater incorporated with the steam cycle operates at a pressure of 0.6 MPa. The condenser pressure is 0 kpa. Assuming all the compression and expansion processes to be isentropic, determine (a) the mass flow rate ratio of air to steam, (b) the required rate of heat input in the combustion chamber, and (c) the thermal efficiency of the combined cycle Reconsider Prob Using EES (or other) software, study the effects of the gas cycle pressure ratio as it is varied from 10 to 0 on the ratio of gas flow rate to steam flow rate and cycle thermal efficiency. Plot your results as functions of gas cycle pressure ratio, and discuss the results Repeat Prob assuming isentropic efficiencies of 100 percent for the pump, 8 percent for the compressor, and 86 percent for the gas and steam turbines Reconsider Prob Using EES (or other) software, study the effects of the gas cycle pressure ratio as it is varied from 10 to 0 on the ratio of gas flow rate to steam flow rate and cycle thermal efficiency. Plot your results as functions of gas cycle pressure ratio, and discuss the results Consider a combined gas steam power cycle. The topping cycle is a simple Brayton cycle that has a pressure ratio of 7. Air enters the compressor at 15 C at a rate of 10 kg/s and the gas turbine at 950 C. The bottoming cycle is a reheat Rankine cycle between the pressure limits of 6 MPa and 10 kpa. Steam is heated in a heat exchanger at a rate of 1.15 kg/s by the exhaust gases leaving the gas turbine and the exhaust gases leave the heat exchanger at 00 C. Steam leaves the high-pressure turbine at 1.0 MPa and is reheated to 400 C in the heat exchanger before it expands in the lowpressure turbine. Assuming 80 percent isentropic efficiency for all pumps and turbine, determine (a) the moisture content at the exit of the low-pressure turbine, (b) the steam temperature at the inlet of the high-pressure turbine, (c) the net power output and the thermal efficiency of the combined plant. Compressor Combustion chamber Heat exchanger 5 Pump 3 FIGURE P Gas turbine 1 10 Steam turbine 6 Condenser Special Topic: Binary Vapor Cycles 10 8C What is a binary power cycle? What is its purpose? 10 83C By writing an energy balance on the heat exchanger of a binary vapor power cycle, obtain a relation for the ratio of mass flow rates of two fluids in terms of their enthalpies C Why is steam not an ideal working fluid for vapor power cycles? 10 85C Why is mercury a suitable working fluid for the topping portion of a binary vapor cycle but not for the bottoming cycle? 10 86C What is the difference between the binary vapor power cycle and the combined gas steam power cycle?

121 Review Problems Show that the thermal efficiency of a combined gas steam power plant h cc can be expressed as h cc h g h s h g h s where h g W g /Q in and h s W s /Q g,out are the thermal efficiencies of the gas and steam cycles, respectively. Using this relation, determine the thermal efficiency of a combined power cycle that consists of a topping gas-turbine cycle with an efficiency of 40 percent and a bottoming steam-turbine cycle with an efficiency of 30 percent It can be shown that the thermal efficiency of a combined gas steam power plant h cc can be expressed in terms of the thermal efficiencies of the gas- and the steam-turbine cycles as h cc h g h s h g h s Prove that the value of h cc is greater than either of h g or h s. That is, the combined cycle is more efficient than either of the gas-turbine or steam-turbine cycles alone Consider a steam power plant operating on the ideal Rankine cycle with reheat between the pressure limits of 5 MPa and 10 kpa with a maximum cycle temperature of 600 C and a moisture content of 8 percent at the turbine exit. For a reheat temperature of 600 C, determine the reheat pressures of the cycle for the cases of (a) single and (b) double reheat E The Stillwater geothermal power plant in Nevada, which started full commercial operation in 1986, is designed to operate with seven identical units. Each of these seven units consists of a pair of power cycles, labeled Level I and Level II, operating on the simple Rankine cycle using an organic fluid as the working fluid. The heat source for the plant is geothermal water (brine) entering the vaporizer (boiler) of Level I of each unit at 35 F at a rate of 384,86 lbm/h and delivering.79 MBtu/h ( M stands for million ). The organic fluid that enters the vaporizer at 0. F at a rate of 157,895 lbm/h leaves it at 8.4 F and 5.8 psia as saturated vapor. This saturated vapor expands in the turbine to 95.8 F and 19.0 psia and produces 171 kw of electric power. About 00 kw of this power is used by the pumps, the auxiliaries, and the six fans of the condenser. Subsequently, the organic working fluid is condensed in an air-cooled condenser by air that enters the condenser at 55 F at a rate of 4,195,100 lbm/h and leaves at 84.5 F. The working fluid is pumped and then preheated in a preheater to 0. F by absorbing MBtu/h of heat from the geothermal water (coming from the vaporizer of Level II) entering the preheater at 11.8 F and leaving at F. Taking the average specific heat of the geothermal water to be 1.03 Btu/lbm F, determine (a) the exit temperature of the geothermal water from the vaporizer, (b) the rate of heat Chapter rejection from the working fluid to the air in the condenser, (c) the mass flow rate of the geothermal water at the preheater, and (d) the thermal efficiency of the Level I cycle of this geothermal power plant. Answers: (a) 67.4 F, (b) 9.7 MBtu/h, (c) 187,10 lbm/h, (d) 10.8 percent Electricity Generator 1 4 Hot geothermal brine Vapor 9 Vaporizer Vapor Turbine Working fluid Land surface 3 Production pump Condenser Fluid pump Injection pump Cooler geothermal brine 8 Air Preheater m geo FIGURE P10 90E Schematic of a binary geothermal power plant. Courtesy of ORMAT Energy Systems, Inc Steam enters the turbine of a steam power plant that operates on a simple ideal Rankine cycle at a pressure of 6 MPa, and it leaves as a saturated vapor at 7.5 kpa. Heat is transferred to the steam in the boiler at a rate of 40,000 kj/s. Steam is cooled in the condenser by the cooling water from a nearby river, which enters the condenser at 15 C. Show the cycle on a T-s diagram with respect to saturation lines, and determine (a) the turbine inlet temperature, (b) the net power output and thermal efficiency, and (c) the minimum mass flow rate of the cooling water required A steam power plant operates on an ideal Rankine cycle with two stages of reheat and has a net power output of

122 600 Thermodynamics 10 MW. Steam enters all three stages of the turbine at 500 C. The maximum pressure in the cycle is 15 MPa, and the minimum pressure is 5 kpa. Steam is reheated at 5 MPa the first time and at 1 MPa the second time. Show the cycle on a T-s diagram with respect to saturation lines, and determine (a) the thermal efficiency of the cycle and (b) the mass flow rate of the steam. Answers: (a) 45.5 percent, (b) 64.4 kg/s Consider a steam power plant that operates on a regenerative Rankine cycle and has a net power output of 150 MW. Steam enters the turbine at 10 MPa and 500 C and the condenser at 10 kpa. The isentropic efficiency of the turbine is 80 percent, and that of the pumps is 95 percent. Steam is extracted from the turbine at 0.5 MPa to heat the feedwater in an open feedwater heater. Water leaves the feedwater heater as a saturated liquid. Show the cycle on a T-s diagram, and determine (a) the mass flow rate of steam through the boiler and (b) the thermal efficiency of the cycle. Also, determine the exergy destruction associated with the regeneration process. Assume a source temperature of 1300 K and a sink temperature of 303 K. Boiler 4 P II 5 Open FWH 3 FIGURE P y P I Turbine 7 Condenser 1 1 y Repeat Prob assuming both the pump and the turbine are isentropic Consider an ideal reheat regenerative Rankine cycle with one open feedwater heater. The boiler pressure is 10 MPa, the condenser pressure is 15 kpa, the reheater pressure is 1 MPa, and the feedwater pressure is 0.6 MPa. Steam enters both the high- and low-pressure turbines at 500 C. Show the cycle on a T-s diagram with respect to saturation lines, and determine (a) the fraction of steam extracted for regeneration and (b) the thermal efficiency of the cycle. Answers: (a) 0.144, (b) 4.1 percent Repeat Prob assuming an isentropic efficiency of 84 percent for the turbines and 100 percent for the pumps A steam power plant operates on an ideal reheat regenerative Rankine cycle with one reheater and two feedwater heaters, one open and one closed. Steam enters the high-pressure turbine at 15 MPa and 600 C and the lowpressure turbine at 1 MPa and 500 C. The condenser pressure is 5 kpa. Steam is extracted from the turbine at 0.6 MPa for the closed feedwater heater and at 0. MPa for the open feedwater heater. In the closed feedwater heater, the feedwater is heated to the condensation temperature of the extracted steam. The extracted steam leaves the closed feedwater heater as a saturated liquid, which is subsequently throttled to the open feedwater heater. Show the cycle on a T-s diagram with respect to saturation lines. Determine (a) the fraction of steam extracted from the turbine for the open feedwater heater, (b) the thermal efficiency of the cycle, and (c) the net power output for a mass flow rate of 4 kg/s through the boiler. 5 Boiler Reheater Closed FWH Pump II 3 7 High-P Turbine 11 Open FWH FIGURE P y 1 Low-P Turbine z Pump I Condenser 1 y z Consider a cogeneration power plant that is modified with reheat and that produces 3 MW of power and supplies 7 MW of process heat. Steam enters the high-pressure turbine at 8 MPa and 500 C and expands to a pressure of 1 MPa. At this pressure, part of the steam is extracted from the turbine and routed to the process heater, while the remainder is reheated to 500 C and expanded in the low-pressure turbine to the condenser pressure of 15 kpa. The condensate from the condenser is pumped to 1 MPa and is mixed with the extracted steam, which leaves the process heater as a com- 13 1

123 pressed liquid at 10 C. The mixture is then pumped to the boiler pressure. Assuming the turbine to be isentropic, show the cycle on a T-s diagram with respect to saturation lines, and disregarding pump work, determine (a) the rate of heat input in the boiler and (b) the fraction of steam extracted for process heating. 5 Boiler 6 Process heater 3 7 High-P Turbine 7 MW Low-P Turbine 8 9 Condenser 3 MW Chapter Steam is to be supplied from a boiler to a highpressure turbine whose isentropic efficiency is 75 percent at conditions to be determined. The steam is to leave the high-pressure turbine as a saturated vapor at 1.4 MPa, and the turbine is to produce 1 MW of power. Steam at the turbine exit is extracted at a rate of 1000 kg/min and routed to a process heater while the rest of the steam is supplied to a low-pressure turbine whose isentropic efficiency is 60 percent. The low-pressure turbine allows the steam to expand to 10 kpa pressure and produces 0.8 MW of power. Determine the temperature, pressure, and the flow rate of steam at the inlet of the high-pressure turbine A textile plant requires 4 kg/s of saturated steam at MPa, which is extracted from the turbine of a cogeneration plant. Steam enters the turbine at 8 MPa and 500 C at a rate of 11 kg/s and leaves at 0 kpa. The extracted steam leaves the process heater as a saturated liquid and mixes with the feedwater at constant pressure. The mixture is pumped to the boiler pressure. Assuming an isentropic efficiency of 88 percent for both the turbine and the pumps, determine (a) the rate of process heat supply, (b) the net power output, and (c) the utilization factor of the plant. Answers: (a) 8.56 MW, (b) 8.60 MW, (c) 53.8 percent P II 4 P I 1 6 FIGURE P The gas-turbine cycle of a combined gas steam power plant has a pressure ratio of 8. Air enters the compressor at 90 K and the turbine at 1400 K. The combustion gases leaving the gas turbine are used to heat the steam at 15 MPa to 450 C in a heat exchanger. The combustion gases leave the heat exchanger at 47 C. Steam expands in a highpressure turbine to a pressure of 3 MPa and is reheated in the combustion chamber to 500 C before it expands in a lowpressure turbine to 10 kpa. The mass flow rate of steam is 30 kg/s. Assuming all the compression and expansion processes to be isentropic, determine (a) the mass flow rate of air in the gas-turbine cycle, (b) the rate of total heat input, and (c) the thermal efficiency of the combined cycle. Answers: (a) 63 kg/s, (b) kj/s, (c) 55.6 percent Repeat Prob assuming isentropic efficiencies of 100 percent for the pump, 80 percent for the compressor, and 85 percent for the gas and steam turbines Starting with Eq. 10 0, show that the exergy destruction associated with a simple ideal Rankine cycle can be expressed as i q in (h th,carnot h th ), where h th is efficiency of the Rankine cycle and h th,carnot is the efficiency of the Carnot cycle operating between the same temperature limits. Boiler 5 P II Process heater 3 4 FIGURE P P I Turbine 1 8 Condenser Using EES (or other) software, investigate the effect of the condenser pressure on the performance of a simple ideal Rankine cycle. Turbine inlet conditions of steam are maintained constant at 5 MPa and 500 C while the condenser pressure is varied from 5 to 100 kpa. Determine the thermal efficiency of the cycle and plot it against the condenser pressure, and discuss the results Using EES (or other) software, investigate the effect of the boiler pressure on the performance of a simple ideal Rankine cycle. Steam enters the turbine at 500 C and exits at 10 kpa. The boiler pressure is

124 60 Thermodynamics varied from 0.5 to 0 MPa. Determine the thermal efficiency of the cycle and plot it against the boiler pressure, and discuss the results Using EES (or other) software, investigate the effect of superheating the steam on the performance of a simple ideal Rankine cycle. Steam enters the turbine at 3 MPa and exits at 10 kpa. The turbine inlet temperature is varied from 50 to 1100 C. Determine the thermal efficiency of the cycle and plot it against the turbine inlet temperature, and discuss the results Using EES (or other) software, investigate the effect of reheat pressure on the performance of an ideal Rankine cycle. The maximum and minimum pressures in the cycle are 15 MPa and 10 kpa, respectively, and steam enters both stages of the turbine at 500 C. The reheat pressure is varied from 1.5 to 0.5 MPa. Determine the thermal efficiency of the cycle and plot it against the reheat pressure, and discuss the results Using EES (or other) software, investigate the effect of number of reheat stages on the performance of an ideal Rankine cycle. The maximum and minimum pressures in the cycle are 15 MPa and 10 kpa, respectively, and steam enters all stages of the turbine at 500 C. For each case, maintain roughly the same pressure ratio across each turbine stage. Determine the thermal efficiency of the cycle and plot it against the number of reheat stages 1,, 4, and 8, and discuss the results Using EES (or other) software, investigate the effect of extraction pressure on the performance of an ideal regenerative Rankine cycle with one open feedwater heater. Steam enters the turbine at 15 MPa and 600 C and the condenser at 10 kpa. Determine the thermal efficiency of the cycle, and plot it against extraction pressures of 1.5, 10, 7, 5,, 1, 0.5, 0.1, and 0.05 MPa, and discuss the results Using EES (or other) software, investigate the effect of the number of regeneration stages on the performance of an ideal regenerative Rankine cycle. Steam enters the turbine at 15 MPa and 600 C and the condenser at 5 kpa. For each case, maintain about the same temperature difference between any two regeneration stages. Determine the thermal efficiency of the cycle, and plot it against the number of regeneration stages for 1,, 3, 4, 5, 6, 8, and 10 regeneration stages. Fundamentals of Engineering (FE) Exam Problems Consider a steady-flow Carnot cycle with water as the working fluid executed under the saturation dome between the pressure limits of 8 MPa and 0 kpa. Water changes from saturated liquid to saturated vapor during the heat addition process. The net work output of this cycle is (a) 494 kj/kg (b) 975 kj/kg (c) 596 kj/kg (d) 845 kj/kg (e) 1148 kj/kg A simple ideal Rankine cycle operates between the pressure limits of 10 kpa and 3 MPa, with a turbine inlet temperature of 600 C. Disregarding the pump work, the cycle efficiency is (a) 4 percent (b) 37 percent (c) 5 percent (d) 63 percent (e) 71 percent A simple ideal Rankine cycle operates between the pressure limits of 10 kpa and 5 MPa, with a turbine inlet temperature of 600 C. The mass fraction of steam that condenses at the turbine exit is (a) 6 percent (b) 9 percent (c) 1 percent (d) 15 percent (e) 18 percent A steam power plant operates on the simple ideal Rankine cycle between the pressure limits of 10 kpa and 10 MPa, with a turbine inlet temperature of 600 C. The rate of heat transfer in the boiler is 800 kj/s. Disregarding the pump work, the power output of this plant is (a) 43 kw (b) 84 kw (c) 508 kw (d) 335 kw (e) 800 kw Consider a combined gas-steam power plant. Water for the steam cycle is heated in a well-insulated heat exchanger by the exhaust gases that enter at 800 K at a rate of 60 kg/s and leave at 400 K. Water enters the heat exchanger at 00 C and 8 MPa and leaves at 350 C and 8 MPa. If the exhaust gases are treated as air with constant specific heats at room temperature, the mass flow rate of water through the heat exchanger becomes (a) 11 kg/s (b) 4 kg/s (c) 46 kg/s (d) 53 kg/s (e) 60 kg/s An ideal reheat Rankine cycle operates between the pressure limits of 10 kpa and 8 MPa, with reheat occurring at 4 MPa. The temperature of steam at the inlets of both turbines is 500 C, and the enthalpy of steam is 3185 kj/kg at the exit of the high-pressure turbine, and 47 kj/kg at the exit of the low-pressure turbine. Disregarding the pump work, the cycle efficiency is (a) 9 percent (b) 3 percent (c) 36 percent (d) 41 percent (e) 49 percent Pressurized feedwater in a steam power plant is to be heated in an ideal open feedwater heater that operates at a pressure of 0.5 MPa with steam extracted from the turbine. If the enthalpy of feedwater is 5 kj/kg and the enthalpy of extracted steam is 665 kj/kg, the mass fraction of steam extracted from the turbine is (a) 4 percent (b) 10 percent (c) 16 percent (d) 7 percent (e) 1 percent Consider a steam power plant that operates on the regenerative Rankine cycle with one open feedwater heater. The enthalpy of the steam is 3374 kj/kg at the turbine inlet, 797 kj/kg at the location of bleeding, and 346 kj/kg at the

125 turbine exit. The net power output of the plant is 10 MW, and the fraction of steam bled off the turbine for regeneration is If the pump work is negligible, the mass flow rate of steam at the turbine inlet is (a) 117 kg/s (b) 16 kg/s (c) 19 kg/s (d) 68 kg/s (e) 679 kg/s Consider a simple ideal Rankine cycle. If the condenser pressure is lowered while keeping turbine inlet state the same, (a) the turbine work output will decrease. (b) the amount of heat rejected will decrease. (c) the cycle efficiency will decrease. (d) the moisture content at turbine exit will decrease. (e) the pump work input will decrease Consider a simple ideal Rankine cycle with fixed boiler and condenser pressures. If the steam is superheated to a higher temperature, (a) the turbine work output will decrease. (b) the amount of heat rejected will decrease. (c) the cycle efficiency will decrease. (d) the moisture content at turbine exit will decrease. (e) the amount of heat input will decrease Consider a simple ideal Rankine cycle with fixed boiler and condenser pressures. If the cycle is modified with reheating, (a) the turbine work output will decrease. (b) the amount of heat rejected will decrease. (c) the pump work input will decrease. (d) the moisture content at turbine exit will decrease. (e) the amount of heat input will decrease Consider a simple ideal Rankine cycle with fixed boiler and condenser pressures. If the cycle is modified with regeneration that involves one open feedwater heater (select the correct statement per unit mass of steam flowing through the boiler), (a) the turbine work output will decrease. (b) the amount of heat rejected will increase. (c) the cycle thermal efficiency will decrease. (d) the quality of steam at turbine exit will decrease. (e) the amount of heat input will increase Consider a cogeneration power plant modified with regeneration. Steam enters the turbine at 6 MPa and 450 C at a rate of 0 kg/s and expands to a pressure of 0.4 MPa. At this pressure, 60 percent of the steam is extracted from the turbine, and the remainder expands to a pressure of 10 kpa. Part of the extracted steam is used to heat feedwater in an open feedwater heater. The rest of the extracted steam is used for process heating and leaves the process heater as a saturated liquid at 0.4 MPa. It is subsequently mixed with the feedwater leaving the feedwater heater, and the mixture is pumped to the boiler pressure. The steam in the condenser Chapter is cooled and condensed by the cooling water from a nearby river, which enters the adiabatic condenser at a rate of 463 kg/s. 1. The total power output of the turbine is (a) 17.0 MW (b) 8.4 MW (c) 1. MW (d) 0.0 MW (e) 3.4 MW. The temperature rise of the cooling water from the river in the condenser is (a) 8.0 C (b) 5. C (c) 9.6 C (d) 1.9 C (e) 16. C 3. The mass flow rate of steam through the process heater is (a) 1.6 kg/s (b) 3.8 kg/s (c) 5. kg/s (d) 7.6 kg/s (e) 10.4 kg/s 4. The rate of heat supply from the process heater per unit mass of steam passing through it is (a) 46 kj/kg (b) 893 kj/kg (c) 1344 kj/kg (d) 1891 kj/kg (e) 060 kj/kg 5. The rate of heat transfer to the steam in the boiler is (a) 6.0 MJ/s (b) 53.8 MJ/s (c) 39.5 MJ/s (d) 6.8 MJ/s (e) 15.4 MJ/s Boiler 5 9 h P II 6 h 7 h 8 h kj/kg 4 3 h 3 h 4 h Design and Essay Problems h kj/kg Process heater FWH 8 FIGURE P Turbine Condenser P I h kg/s h T 463 kg/s 1 h Design a steam power cycle that can achieve a cycle thermal efficiency of at least 40 percent under the conditions that all turbines have isentropic efficiencies of 85 percent and all pumps have isentropic efficiencies of 60 percent. Prepare

126 604 Thermodynamics an engineering report describing your design. Your design report must include, but is not limited to, the following: (a) Discussion of various cycles attempted to meet the goal as well as the positive and negative aspects of your design. (b) System figures and T-s diagrams with labeled states and temperature, pressure, enthalpy, and entropy information for your design. (c) Sample calculations Contact your power company and obtain information on the thermodynamic aspects of their most recently built power plant. If it is a conventional power plant, find out why it is preferred over a highly efficient combined power plant Several geothermal power plants are in operation in the United States and more are being built since the heat source of a geothermal plant is hot geothermal water, which is free energy. An 8-MW geothermal power plant is being considered at a location where geothermal water at 160 C is available. Geothermal water is to serve as the heat source for a closed Rankine power cycle with refrigerant-134a as the working fluid. Specify suitable temperatures and pressures for the cycle, and determine the thermal efficiency of the cycle. Justify your selections A 10-MW geothermal power plant is being considered at a site where geothermal water at 30 C is available. Geothermal water is to be flashed into a chamber to a lower pressure where part of the water evaporates. The liquid is returned to the ground while the vapor is used to drive the steam turbine. The pressures at the turbine inlet and the turbine exit are to remain above 00 kpa and 8 kpa, respectively. High-pressure flash chambers yield a small amount of steam with high exergy whereas lower-pressure flash chambers yield considerably more steam but at a lower exergy. By trying several pressures, determine the optimum pressure of 30 C Geothermal water Turbine Flash chamber FIGURE P10 17 the flash chamber to maximize the power production per unit mass of geothermal water withdrawn. Also, determine the thermal efficiency for each case assuming 10 percent of the power produced is used to drive the pumps and other auxiliary equipment A natural gas fired furnace in a textile plant is used to provide steam at 130 C. At times of high demand, the furnace supplies heat to the steam at a rate of 30 MJ/s. The plant also uses up to 6 MW of electrical power purchased from the local power company. The plant management is considering converting the existing process plant into a cogeneration plant to meet both their process-heat and power requirements. Your job is to come up with some designs. Designs based on a gas turbine or a steam turbine are to be considered. First decide whether a system based on a gas turbine or a steam turbine will best serve the purpose, considering the cost and the complexity. Then propose your design for the cogeneration plant complete with pressures and temperatures and the mass flow rates. Show that the proposed design meets the power and process-heat requirements of the plant E A photographic equipment manufacturer uses a flow of 64,500 lbm/h of steam in its manufacturing process. Presently the spent steam at 3.8 psig and 4 F is exhausted to the atmosphere. Do the preliminary design of a system to use the energy in the waste steam economically. If electricity is produced, it can be generated about 8000 h/yr and its value is $0.05/kWh. If the energy is used for space heating, the value is also $0.05/kWh, but it can only be used about 3000 h/yr (only during the heating season ). If the steam is condensed and the liquid H O is recycled through the process, its value is $0.50/100 gal. Make all assumptions as realistic as possible. Sketch the system you propose. Make a separate list of required components and their specifications (capacity, efficiency, etc.). The final result will be the calculated annual dollar value of the energy use plan (actually a saving because it will replace electricity or heat and/or water that would otherwise have to be purchased) Design the condenser of a steam power plant that has a thermal efficiency of 40 percent and generates 10 MW of net electric power. Steam enters the condenser as saturated vapor at 10 kpa, and it is to be condensed outside horizontal tubes through which cooling water from a nearby river flows. The temperature rise of the cooling water is limited to 8 C, and the velocity of the cooling water in the pipes is limited to 6 m/s to keep the pressure drop at an acceptable level. From prior experience, the average heat flux based on the outer surface of the tubes can be taken to be 1,000 W/m. Specify the pipe diameter, total pipe length, and the arrangement of the pipes to minimize the condenser volume Water-cooled steam condensers are commonly used in steam power plants. Obtain information about water-cooled steam condensers by doing a literature search on the topic and

127 also by contacting some condenser manufacturers. In a report, describe the various types, the way they are designed, the limitation on each type, and the selection criteria Steam boilers have long been used to provide process heat as well as to generate power. Write an essay on the history of steam boilers and the evolution of modern supercritical steam power plants. What was the role of the American Society of Mechanical Engineers in this development? The technology for power generation using geothermal energy is well established, and numerous geothermal power plants throughout the world are currently generating electricity economically. Binary geothermal plants utilize a volatile secondary fluid such as isobutane, n-pentane, and R-114 in a closed loop. Consider a binary geothermal plant Chapter with R-114 as the working fluid that is flowing at a rate of 600 kg/s. The R-114 is vaporized in a boiler at 115 C by the geothermal fluid that enters at 165 C, and is condensed at 30 C outside the tubes by cooling water that enters the tubes at 18 C. Based on prior experience, the average heat flux based on the outer surface of the tubes can be taken to be 4600 W/m. The enthalpy of vaporization of R-114 at 30 C is h fg 11.5 kj/kg. Specify (a) the length, diameter, and number of tubes and their arrangement in the condenser to minimize overall volume of the condenser; (b) the mass flow rate of cooling water; and (c) the flow rate of make-up water needed if a cooling tower is used to reject the waste heat from the cooling water. The liquid velocity is to remain under 6 m/s and the length of the tubes is limited to 8 m.

128

129 Chapter 11 REFRIGERATION CYCLES Amajor application area of thermodynamics is refrigeration, which is the transfer of heat from a lower temperature region to a higher temperature one. Devices that produce refrigeration are called refrigerators, and the cycles on which they operate are called refrigeration cycles. The most frequently used refrigeration cycle is the vapor-compression refrigeration cycle in which the refrigerant is vaporized and condensed alternately and is compressed in the vapor phase. Another well-known refrigeration cycle is the gas refrigeration cycle in which the refrigerant remains in the gaseous phase throughout. Other refrigeration cycles discussed in this chapter are cascade refrigeration, where more than one refrigeration cycle is used; absorption refrigeration, where the refrigerant is dissolved in a liquid before it is compressed; and, as a Topic of Special Interest, thermoelectric refrigeration, where refrigeration is produced by the passage of electric current through two dissimilar materials. Objectives The objectives of Chapter 11 are to: Introduce the concepts of refrigerators and heat pumps and the measure of their performance. Analyze the ideal vapor-compression refrigeration cycle. Analyze the actual vapor-compression refrigeration cycle. Review the factors involved in selecting the right refrigerant for an application. Discuss the operation of refrigeration and heat pump systems. Evaluate the performance of innovative vapor-compression refrigeration systems. Analyze gas refrigeration systems. Introduce the concepts of absorption-refrigeration systems. Review the concepts of thermoelectric power generation and refrigeration. 607

130 608 Thermodynamics WARM environment R Q H COLD refrigerated space Q L (desired output) (a) Refrigerator INTERACTIVE TUTORIAL SEE TUTORIAL CH. 11, SEC. 1 ON THE DVD. W net,in (required input) WARM house HP COLD environment (b) Heat pump Q H (desired output) Q L W net,in (required input) FIGURE 11 1 The objective of a refrigerator is to remove heat (Q L ) from the cold medium; the objective of a heat pump is to supply heat (Q H ) to a warm medium REFRIGERATORS AND HEAT PUMPS We all know from experience that heat flows in the direction of decreasing temperature, that is, from high-temperature regions to low-temperature ones. This heat-transfer process occurs in nature without requiring any devices. The reverse process, however, cannot occur by itself. The transfer of heat from a low-temperature region to a high-temperature one requires special devices called refrigerators. Refrigerators are cyclic devices, and the working fluids used in the refrigeration cycles are called refrigerants. A refrigerator is shown schematically in Fig. 11 1a. Here Q L is the magnitude of the heat removed from the refrigerated space at temperature T L,Q H is the magnitude of the heat rejected to the warm space at temperature T H, and W net,in is the net work input to the refrigerator. As discussed in Chap. 6, Q L and Q H represent magnitudes and thus are positive quantities. Another device that transfers heat from a low-temperature medium to a high-temperature one is the heat pump. Refrigerators and heat pumps are essentially the same devices; they differ in their objectives only. The objective of a refrigerator is to maintain the refrigerated space at a low temperature by removing heat from it. Discharging this heat to a higher-temperature medium is merely a necessary part of the operation, not the purpose. The objective of a heat pump, however, is to maintain a heated space at a high temperature. This is accomplished by absorbing heat from a low-temperature source, such as well water or cold outside air in winter, and supplying this heat to a warmer medium such as a house (Fig. 11 1b). The performance of refrigerators and heat pumps is expressed in terms of the coefficient of performance (COP), defined as Desired output COP R Required input Desired output COP HP Required input Cooling effect Work input Heating effect Work input (11 1) (11 ) These relations can also be expressed in the rate form by replacing the quantities Q L, Q H, and W net,in by Q. L, Q. H, and Ẇ net,in, respectively. Notice that both COP R and COP HP can be greater than 1. A comparison of Eqs and 11 reveals that COP HP COP R 1 (11 3) for fixed values of Q L and Q H. This relation implies that COP HP 1 since COP R is a positive quantity. That is, a heat pump functions, at worst, as a resistance heater, supplying as much energy to the house as it consumes. In reality, however, part of Q H is lost to the outside air through piping and other devices, and COP HP may drop below unity when the outside air temperature is too low. When this happens, the system normally switches to the fuel (natural gas, propane, oil, etc.) or resistance-heating mode. The cooling capacity of a refrigeration system that is, the rate of heat removal from the refrigerated space is often expressed in terms of tons of refrigeration. The capacity of a refrigeration system that can freeze 1 ton (000 lbm) of liquid water at 0 C (3 F) into ice at 0 C in 4 h is said to be Q L W net,in Q H W net,in

131 1 ton. One ton of refrigeration is equivalent to 11 kj/min or 00 Btu/min. The cooling load of a typical 00-m residence is in the 3-ton (10-kW) range. Chapter THE REVERSED CARNOT CYCLE Recall from Chap. 6 that the Carnot cycle is a totally reversible cycle that consists of two reversible isothermal and two isentropic processes. It has the maximum thermal efficiency for given temperature limits, and it serves as a standard against which actual power cycles can be compared. Since it is a reversible cycle, all four processes that comprise the Carnot cycle can be reversed. Reversing the cycle does also reverse the directions of any heat and work interactions. The result is a cycle that operates in the counterclockwise direction on a T-s diagram, which is called the reversed Carnot cycle. A refrigerator or heat pump that operates on the reversed Carnot cycle is called a Carnot refrigerator or a Carnot heat pump. Consider a reversed Carnot cycle executed within the saturation dome of a refrigerant, as shown in Fig. 11. The refrigerant absorbs heat isothermally from a low-temperature source at T L in the amount of Q L (process 1-), is compressed isentropically to state 3 (temperature rises to T H ), rejects heat isothermally to a high-temperature sink at T H in the amount of Q H (process 3-4), and expands isentropically to state 1 (temperature drops to T L ). The refrigerant changes from a saturated vapor state to a saturated liquid state in the condenser during process 3-4. WARM medium at T H T Q H 4 T H Condenser 3 Q H 4 3 Turbine Compressor 1 Evaporator T L 1 Q L Q L COLD medium at T L s FIGURE 11 Schematic of a Carnot refrigerator and T-s diagram of the reversed Carnot cycle.

132 610 Thermodynamics INTERACTIVE TUTORIAL SEE TUTORIAL CH. 11, SEC. ON THE DVD. The coefficients of performance of Carnot refrigerators and heat pumps are expressed in terms of temperatures as and COP R,Carnot 1 T H >T L 1 1 COP HP,Carnot 1 T L >T H (11 4) (11 5) Notice that both COPs increase as the difference between the two temperatures decreases, that is, as T L rises or T H falls. The reversed Carnot cycle is the most efficient refrigeration cycle operating between two specified temperature levels. Therefore, it is natural to look at it first as a prospective ideal cycle for refrigerators and heat pumps. If we could, we certainly would adapt it as the ideal cycle. As explained below, however, the reversed Carnot cycle is not a suitable model for refrigeration cycles. The two isothermal heat transfer processes are not difficult to achieve in practice since maintaining a constant pressure automatically fixes the temperature of a two-phase mixture at the saturation value. Therefore, processes 1- and 3-4 can be approached closely in actual evaporators and condensers. However, processes -3 and 4-1 cannot be approximated closely in practice. This is because process -3 involves the compression of a liquid vapor mixture, which requires a compressor that will handle two phases, and process 4-1 involves the expansion of high-moisture-content refrigerant in a turbine. It seems as if these problems could be eliminated by executing the reversed Carnot cycle outside the saturation region. But in this case we have difficulty in maintaining isothermal conditions during the heat-absorption and heat-rejection processes. Therefore, we conclude that the reversed Carnot cycle cannot be approximated in actual devices and is not a realistic model for refrigeration cycles. However, the reversed Carnot cycle can serve as a standard against which actual refrigeration cycles are compared THE IDEAL VAPOR-COMPRESSION REFRIGERATION CYCLE Many of the impracticalities associated with the reversed Carnot cycle can be eliminated by vaporizing the refrigerant completely before it is compressed and by replacing the turbine with a throttling device, such as an expansion valve or capillary tube. The cycle that results is called the ideal vapor-compression refrigeration cycle, and it is shown schematically and on a T-s diagram in Fig The vapor-compression refrigeration cycle is the most widely used cycle for refrigerators, air-conditioning systems, and heat pumps. It consists of four processes: 1- Isentropic compression in a compressor -3 Constant-pressure heat rejection in a condenser 3-4 Throttling in an expansion device 4-1 Constant-pressure heat absorption in an evaporator In an ideal vapor-compression refrigeration cycle, the refrigerant enters the compressor at state 1 as saturated vapor and is compressed isentropically to the condenser pressure. The temperature of the refrigerant increases during

133 Chapter WARM environment T Q H 3 Condenser Saturated liquid Q H Expansion valve Compressor W in 3 W in 4 1 Evaporator Q L 4' 4 Q L Saturated vapor 1 COLD refrigerated space s FIGURE 11 3 Schematic and T-s diagram for the ideal vapor-compression refrigeration cycle. this isentropic compression process to well above the temperature of the surrounding medium. The refrigerant then enters the condenser as superheated vapor at state and leaves as saturated liquid at state 3 as a result of heat rejection to the surroundings. The temperature of the refrigerant at this state is still above the temperature of the surroundings. The saturated liquid refrigerant at state 3 is throttled to the evaporator pressure by passing it through an expansion valve or capillary tube. The temperature of the refrigerant drops below the temperature of the refrigerated space during this process. The refrigerant enters the evaporator at state 4 as a low-quality saturated mixture, and it completely evaporates by absorbing heat from the refrigerated space. The refrigerant leaves the evaporator as saturated vapor and reenters the compressor, completing the cycle. In a household refrigerator, the tubes in the freezer compartment where heat is absorbed by the refrigerant serves as the evaporator. The coils behind the refrigerator, where heat is dissipated to the kitchen air, serve as the condenser (Fig. 11 4). Remember that the area under the process curve on a T-s diagram represents the heat transfer for internally reversible processes. The area under the process curve 4-1 represents the heat absorbed by the refrigerant in the evaporator, and the area under the process curve -3 represents the heat rejected in the condenser. A rule of thumb is that the COP improves by to 4 percent for each C the evaporating temperature is raised or the condensing temperature is lowered. Freezer compartment 18 C 3 C Q L Kitchen air 5 C Evaporator coils Capillary tube Q H Condenser coils Compressor FIGURE 11 4 An ordinary household refrigerator.

134 61 Thermodynamics P 3 4 Q L Q H 1 Win FIGURE 11 5 The P-h diagram of an ideal vapor-compression refrigeration cycle. h Another diagram frequently used in the analysis of vapor-compression refrigeration cycles is the P-h diagram, as shown in Fig On this diagram, three of the four processes appear as straight lines, and the heat transfer in the condenser and the evaporator is proportional to the lengths of the corresponding process curves. Notice that unlike the ideal cycles discussed before, the ideal vaporcompression refrigeration cycle is not an internally reversible cycle since it involves an irreversible (throttling) process. This process is maintained in the cycle to make it a more realistic model for the actual vapor-compression refrigeration cycle. If the throttling device were replaced by an isentropic turbine, the refrigerant would enter the evaporator at state 4 instead of state 4. As a result, the refrigeration capacity would increase (by the area under process curve 4-4 in Fig. 11 3) and the net work input would decrease (by the amount of work output of the turbine). Replacing the expansion valve by a turbine is not practical, however, since the added benefits cannot justify the added cost and complexity. All four components associated with the vapor-compression refrigeration cycle are steady-flow devices, and thus all four processes that make up the cycle can be analyzed as steady-flow processes. The kinetic and potential energy changes of the refrigerant are usually small relative to the work and heat transfer terms, and therefore they can be neglected. Then the steadyflow energy equation on a unit mass basis reduces to 1q in q out 1w in w out h e h i (11 6) The condenser and the evaporator do not involve any work, and the compressor can be approximated as adiabatic. Then the COPs of refrigerators and heat pumps operating on the vapor-compression refrigeration cycle can be expressed as and COP R q L w net,in h 1 h 4 h h 1 (11 7) COP HP q H h h 3 (11 8) w net,in h h 1 where h 1 h P1 and h 3 h P3 for the ideal case. Vapor-compression refrigeration dates back to 1834 when the Englishman Jacob Perkins received a patent for a closed-cycle ice machine using ether or other volatile fluids as refrigerants. A working model of this machine was built, but it was never produced commercially. In 1850, Alexander Twining began to design and build vapor-compression ice machines using ethyl ether, which is a commercially used refrigerant in vapor-compression systems. Initially, vapor-compression refrigeration systems were large and were mainly used for ice making, brewing, and cold storage. They lacked automatic controls and were steam-engine driven. In the 1890s, electric motordriven smaller machines equipped with automatic controls started to replace the older units, and refrigeration systems began to appear in butcher shops and households. By 1930, the continued improvements made it possible to have vapor-compression refrigeration systems that were relatively efficient, reliable, small, and inexpensive.

135 Chapter EXAMPLE 11 1 The Ideal Vapor-Compression Refrigeration Cycle A refrigerator uses refrigerant-134a as the working fluid and operates on an ideal vapor-compression refrigeration cycle between 0.14 and 0.8 MPa. If the mass flow rate of the refrigerant is 0.05 kg/s, determine (a) the rate of heat removal from the refrigerated space and the power input to the compressor, (b) the rate of heat rejection to the environment, and (c) the COP of the refrigerator. Solution A refrigerator operates on an ideal vapor-compression refrigeration cycle between two specified pressure limits. The rate of refrigeration, the power input, the rate of heat rejection, and the COP are to be determined. Assumptions 1 Steady operating conditions exist. Kinetic and potential energy changes are negligible. Analysis The T-s diagram of the refrigeration cycle is shown in Fig We note that this is an ideal vapor-compression refrigeration cycle, and thus the compressor is isentropic and the refrigerant leaves the condenser as a saturated liquid and enters the compressor as saturated vapor. From the refrigerant-134a tables, the enthalpies of the refrigerant at all four states are determined as follows: P MPa h 1 h 0.14 MPa kj>kg s 1 s 0.14 MPa kj>kg # K T P 0.8 MPa f h s s kj>kg 1 P MPa h 3 h 0.8 MPa kj>kg h 4 h 3 1throttling h kj>kg (a) The rate of heat removal from the refrigerated space and the power input to the compressor are determined from their definitions: Q H 0.8 MPa 3 W in Q # L m # 1h 1 h kg>s kj>kg kw and W # in m # 1h h kg>s kj>kg kw 0.14 MPa 4s 4 Q L 1 (b) The rate of heat rejection from the refrigerant to the environment is Q # H m # 1h h kg>s kj>kg4 9.0 kw It could also be determined from Q # H Q # L W # in kw (c) The coefficient of performance of the refrigerator is COP R Q# L W # 7.18 kw 3.97 in 1.81 kw That is, this refrigerator removes about 4 units of thermal energy from the refrigerated space for each unit of electric energy it consumes. Discussion It would be interesting to see what happens if the throttling valve were replaced by an isentropic turbine. The enthalpy at state 4s (the turbine exit with P 4s 0.14 MPa, and s 4s s kj/kg K) is kj/kg, FIGURE 11 6 T-s diagram of the ideal vapor-compression refrigeration cycle described in Example s

136 614 Thermodynamics and the turbine would produce 0.33 kw of power. This would decrease the power input to the refrigerator from 1.81 to 1.48 kw and increase the rate of heat removal from the refrigerated space from 7.18 to 7.51 kw. As a result, the COP of the refrigerator would increase from 3.97 to 5.07, an increase of 8 percent. INTERACTIVE TUTORIAL SEE TUTORIAL CH. 11, SEC. 3 ON THE DVD ACTUAL VAPOR-COMPRESSION REFRIGERATION CYCLE An actual vapor-compression refrigeration cycle differs from the ideal one in several ways, owing mostly to the irreversibilities that occur in various components. Two common sources of irreversibilities are fluid friction (causes pressure drops) and heat transfer to or from the surroundings. The T-s diagram of an actual vapor-compression refrigeration cycle is shown in Fig In the ideal cycle, the refrigerant leaves the evaporator and enters the compressor as saturated vapor. In practice, however, it may not be possible to control the state of the refrigerant so precisely. Instead, it is easier to design the system so that the refrigerant is slightly superheated at the compressor inlet. This slight overdesign ensures that the refrigerant is completely vaporized when it enters the compressor. Also, the line connecting WARM environment T Q H 4 3 Condenser 3 5 ' 6 Expansion valve Compressor 1 W in Evaporator 7 8 Q L COLD refrigerated space s FIGURE 11 7 Schematic and T-s diagram for the actual vapor-compression refrigeration cycle.

137 the evaporator to the compressor is usually very long; thus the pressure drop caused by fluid friction and heat transfer from the surroundings to the refrigerant can be very significant. The result of superheating, heat gain in the connecting line, and pressure drops in the evaporator and the connecting line is an increase in the specific volume, thus an increase in the power input requirements to the compressor since steady-flow work is proportional to the specific volume. The compression process in the ideal cycle is internally reversible and adiabatic, and thus isentropic. The actual compression process, however, involves frictional effects, which increase the entropy, and heat transfer, which may increase or decrease the entropy, depending on the direction. Therefore, the entropy of the refrigerant may increase (process 1-) or decrease (process 1- ) during an actual compression process, depending on which effects dominate. The compression process 1- may be even more desirable than the isentropic compression process since the specific volume of the refrigerant and thus the work input requirement are smaller in this case. Therefore, the refrigerant should be cooled during the compression process whenever it is practical and economical to do so. In the ideal case, the refrigerant is assumed to leave the condenser as saturated liquid at the compressor exit pressure. In reality, however, it is unavoidable to have some pressure drop in the condenser as well as in the lines connecting the condenser to the compressor and to the throttling valve. Also, it is not easy to execute the condensation process with such precision that the refrigerant is a saturated liquid at the end, and it is undesirable to route the refrigerant to the throttling valve before the refrigerant is completely condensed. Therefore, the refrigerant is subcooled somewhat before it enters the throttling valve. We do not mind this at all, however, since the refrigerant in this case enters the evaporator with a lower enthalpy and thus can absorb more heat from the refrigerated space. The throttling valve and the evaporator are usually located very close to each other, so the pressure drop in the connecting line is small. Chapter EXAMPLE 11 The Actual Vapor-Compression Refrigeration Cycle Refrigerant-134a enters the compressor of a refrigerator as superheated vapor at 0.14 MPa and 10 C at a rate of 0.05 kg/s and leaves at 0.8 MPa and 50 C. The refrigerant is cooled in the condenser to 6 C and 0.7 MPa and is throttled to 0.15 MPa. Disregarding any heat transfer and pressure drops in the connecting lines between the components, determine (a) the rate of heat removal from the refrigerated space and the power input to the compressor, (b) the isentropic efficiency of the compressor, and (c) the coefficient of performance of the refrigerator. Solution A refrigerator operating on a vapor-compression cycle is considered. The rate of refrigeration, the power input, the compressor efficiency, and the COP are to be determined. Assumptions 1 Steady operating conditions exist. Kinetic and potential energy changes are negligible.

138 616 Thermodynamics Analysis The T-s diagram of the refrigeration cycle is shown in Fig We note that the refrigerant leaves the condenser as a compressed liquid and enters the compressor as superheated vapor. The enthalpies of the refrigerant at various states are determined from the refrigerant tables to be P MPa T 1 10 C f P 0.8 MPa f T 50 C h kj/kg h kj/kg P MPa f h 3 h 6 C kj/kg T 3 6 C h 4 h 3 (throttling) h kj/kg T 0.7 MPa 6 C MPa Q L Q H s FIGURE 11 8 T-s diagram for Example MPa 50 C W in 0.14 MPa 10 C s (a) The rate of heat removal from the refrigerated space and the power input to the compressor are determined from their definitions: and Q # L m # 1h 1 h kg>s kj>kg kw W # in m # 1h h kg>s kj>kg4.0 kw (b) The isentropic efficiency of the compressor is determined from where the enthalpy at state s (P s 0.8 MPa and s s s kj/kg K) is 84.1 kj/kg. Thus, h C h C h s h 1 h h or 93.9% (c) The coefficient of performance of the refrigerator is COP R Q# L W # 7.93 kw 3.93 in.0 kw Discussion This problem is identical to the one worked out in Example 11 1, except that the refrigerant is slightly superheated at the compressor inlet and subcooled at the condenser exit. Also, the compressor is not isentropic. As a result, the heat removal rate from the refrigerated space increases (by 10.4 percent), but the power input to the compressor increases even more (by 11.6 percent). Consequently, the COP of the refrigerator decreases from 3.97 to SELECTING THE RIGHT REFRIGERANT When designing a refrigeration system, there are several refrigerants from which to choose, such as chlorofluorocarbons (CFCs), ammonia, hydrocarbons (propane, ethane, ethylene, etc.), carbon dioxide, air (in the air-conditioning of aircraft), and even water (in applications above the freezing point). The right

139 choice of refrigerant depends on the situation at hand. Of these, refrigerants such as R-11, R-1, R-, R-134a, and R-50 account for over 90 percent of the market in the United States. Ethyl ether was the first commercially used refrigerant in vapor-compression systems in 1850, followed by ammonia, carbon dioxide, methyl chloride, sulphur dioxide, butane, ethane, propane, isobutane, gasoline, and chlorofluorocarbons, among others. The industrial and heavy-commercial sectors were very satisfied with ammonia, and still are, although ammonia is toxic. The advantages of ammonia over other refrigerants are its low cost, higher COPs (and thus lower energy cost), more favorable thermodynamic and transport properties and thus higher heat transfer coefficients (requires smaller and lower-cost heat exchangers), greater detectability in the event of a leak, and no effect on the ozone layer. The major drawback of ammonia is its toxicity, which makes it unsuitable for domestic use. Ammonia is predominantly used in food refrigeration facilities such as the cooling of fresh fruits, vegetables, meat, and fish; refrigeration of beverages and dairy products such as beer, wine, milk, and cheese; freezing of ice cream and other foods; ice production; and low-temperature refrigeration in the pharmaceutical and other process industries. It is remarkable that the early refrigerants used in the light-commercial and household sectors such as sulfur dioxide, ethyl chloride, and methyl chloride were highly toxic. The widespread publicity of a few instances of leaks that resulted in serious illnesses and death in the 190s caused a public cry to ban or limit the use of these refrigerants, creating a need for the development of a safe refrigerant for household use. At the request of Frigidaire Corporation, General Motors research laboratory developed R-1, the first member of the CFC family of refrigerants, within three days in 198. Of several CFCs developed, the research team settled on R-1 as the refrigerant most suitable for commercial use and gave the CFC family the trade name Freon. Commercial production of R-11 and R-1 was started in 1931 by a company jointly formed by General Motors and E. I. du Pont de Nemours and Co., Inc. The versatility and low cost of CFCs made them the refrigerants of choice. CFCs were also widely used in aerosols, foam insulations, and the electronic industry as solvents to clean computer chips. R-11 is used primarily in large-capacity water chillers serving airconditioning systems in buildings. R-1 is used in domestic refrigerators and freezers, as well as automotive air conditioners. R- is used in window air conditioners, heat pumps, air conditioners of commercial buildings, and large industrial refrigeration systems, and offers strong competition to ammonia. R-50 (a blend of R-115 and R-) is the dominant refrigerant used in commercial refrigeration systems such as those in supermarkets because it allows low temperatures at evaporators while operating at singlestage compression. The ozone crisis has caused a major stir in the refrigeration and airconditioning industry and has triggered a critical look at the refrigerants in use. It was realized in the mid-1970s that CFCs allow more ultraviolet radiation into the earth s atmosphere by destroying the protective ozone layer and thus contributing to the greenhouse effect that causes global warming. As a result, the use of some CFCs is banned by international treaties. Fully Chapter

140 618 Thermodynamics halogenated CFCs (such as R-11, R-1, and R-115) do the most damage to the ozone layer. The nonfully halogenated refrigerants such as R- have about 5 percent of the ozone-depleting capability of R-1. Refrigerants that are friendly to the ozone layer that protects the earth from harmful ultraviolet rays have been developed. The once popular refrigerant R-1 has largely been replaced by the recently developed chlorine-free R-134a. Two important parameters that need to be considered in the selection of a refrigerant are the temperatures of the two media (the refrigerated space and the environment) with which the refrigerant exchanges heat. To have heat transfer at a reasonable rate, a temperature difference of 5 to 10 C should be maintained between the refrigerant and the medium with which it is exchanging heat. If a refrigerated space is to be maintained at 10 C, for example, the temperature of the refrigerant should remain at about 0 C while it absorbs heat in the evaporator. The lowest pressure in a refrigeration cycle occurs in the evaporator, and this pressure should be above atmospheric pressure to prevent any air leakage into the refrigeration system. Therefore, a refrigerant should have a saturation pressure of 1 atm or higher at 0 C in this particular case. Ammonia and R-134a are two such substances. The temperature (and thus the pressure) of the refrigerant on the condenser side depends on the medium to which heat is rejected. Lower temperatures in the condenser (thus higher COPs) can be maintained if the refrigerant is cooled by liquid water instead of air. The use of water cooling cannot be justified economically, however, except in large industrial refrigeration systems. The temperature of the refrigerant in the condenser cannot fall below the temperature of the cooling medium (about 0 C for a household refrigerator), and the saturation pressure of the refrigerant at this temperature should be well below its critical pressure if the heat rejection process is to be approximately isothermal. If no single refrigerant can meet the temperature requirements, then two or more refrigeration cycles with different refrigerants can be used in series. Such a refrigeration system is called a cascade system and is discussed later in this chapter. Other desirable characteristics of a refrigerant include being nontoxic, noncorrosive, nonflammable, and chemically stable; having a high enthalpy of vaporization (minimizes the mass flow rate); and, of course, being available at low cost. In the case of heat pumps, the minimum temperature (and pressure) for the refrigerant may be considerably higher since heat is usually extracted from media that are well above the temperatures encountered in refrigeration systems HEAT PUMP SYSTEMS Heat pumps are generally more expensive to purchase and install than other heating systems, but they save money in the long run in some areas because they lower the heating bills. Despite their relatively higher initial costs, the popularity of heat pumps is increasing. About one-third of all single-family homes built in the United States in the last decade are heated by heat pumps. The most common energy source for heat pumps is atmospheric air (airto-air systems), although water and soil are also used. The major problem with air-source systems is frosting, which occurs in humid climates when the temperature falls below to 5 C. The frost accumulation on the evapo-

141 rator coils is highly undesirable since it seriously disrupts heat transfer. The coils can be defrosted, however, by reversing the heat pump cycle (running it as an air conditioner). This results in a reduction in the efficiency of the system. Water-source systems usually use well water from depths of up to 80 m in the temperature range of 5 to 18 C, and they do not have a frosting problem. They typically have higher COPs but are more complex and require easy access to a large body of water such as underground water. Ground-source systems are also rather involved since they require long tubing placed deep in the ground where the soil temperature is relatively constant. The COP of heat pumps usually ranges between 1.5 and 4, depending on the particular system used and the temperature of the source. A new class of recently developed heat pumps that use variable-speed electric motor drives are at least twice as energy efficient as their predecessors. Both the capacity and the efficiency of a heat pump fall significantly at low temperatures. Therefore, most air-source heat pumps require a supplementary heating system such as electric resistance heaters or an oil or gas furnace. Since water and soil temperatures do not fluctuate much, supplementary heating may not be required for water-source or ground-source systems. However, the heat pump system must be large enough to meet the maximum heating load. Heat pumps and air conditioners have the same mechanical components. Therefore, it is not economical to have two separate systems to meet the heating and cooling requirements of a building. One system can be used as a heat pump in winter and an air conditioner in summer. This is accomplished by adding a reversing valve to the cycle, as shown in Fig As Chapter HEAT PUMP OPERATION HEATING MODE Outdoor coil Reversing valve Indoor coil Fan Fan Expansion valve Compressor High-pressure liquid Low-pressure liquid vapor Low-pressure vapor High-pressure vapor HEAT PUMP OPERATION COOLING MODE Outdoor coil Reversing valve Fan Indoor coil Compressor Fan Expansion valve FIGURE 11 9 A heat pump can be used to heat a house in winter and to cool it in summer.

142 60 Thermodynamics a result of this modification, the condenser of the heat pump (located indoors) functions as the evaporator of the air conditioner in summer. Also, the evaporator of the heat pump (located outdoors) serves as the condenser of the air conditioner. This feature increases the competitiveness of the heat pump. Such dual-purpose units are commonly used in motels. Heat pumps are most competitive in areas that have a large cooling load during the cooling season and a relatively small heating load during the heating season, such as in the southern parts of the United States. In these areas, the heat pump can meet the entire cooling and heating needs of residential or commercial buildings. The heat pump is least competitive in areas where the heating load is very large and the cooling load is small, such as in the northern parts of the United States INNOVATIVE VAPOR-COMPRESSION REFRIGERATION SYSTEMS The simple vapor-compression refrigeration cycle discussed above is the most widely used refrigeration cycle, and it is adequate for most refrigeration applications. The ordinary vapor-compression refrigeration systems are simple, inexpensive, reliable, and practically maintenance-free (when was the last time you serviced your household refrigerator?). However, for large industrial applications efficiency, not simplicity, is the major concern. Also, for some applications the simple vapor-compression refrigeration cycle is inadequate and needs to be modified. We now discuss a few such modifications and refinements. Cascade Refrigeration Systems Some industrial applications require moderately low temperatures, and the temperature range they involve may be too large for a single vaporcompression refrigeration cycle to be practical. A large temperature range also means a large pressure range in the cycle and a poor performance for a reciprocating compressor. One way of dealing with such situations is to perform the refrigeration process in stages, that is, to have two or more refrigeration cycles that operate in series. Such refrigeration cycles are called cascade refrigeration cycles. A two-stage cascade refrigeration cycle is shown in Fig The two cycles are connected through the heat exchanger in the middle, which serves as the evaporator for the topping cycle (cycle A) and the condenser for the bottoming cycle (cycle B). Assuming the heat exchanger is well insulated and the kinetic and potential energies are negligible, the heat transfer from the fluid in the bottoming cycle should be equal to the heat transfer to the fluid in the topping cycle. Thus, the ratio of mass flow rates through each cycle should be Also, m # A 1h 5 h 8 m # B 1h h 3 m# A m # h h 3 B h 5 h 8 COP R,cascade Q# L W # m # B 1h 1 h 4 net,in m # A 1h 6 h 5 m # B 1h h 1 (11 9) (11 10)

143 Chapter WARM environment 7 Q H 6 Condenser Expansion A valve Compressor Heat exchanger Evaporator T Decrease in compressor work Q H 6 8 Heat 5 3 Condenser A 5 Expansion valve 4 B Q L Evaporator Q L Compressor 1 4 B 1 Q L Increase in refrigeration capacity s COLD refrigerated space FIGURE A two-stage cascade refrigeration system with the same refrigerant in both stages. In the cascade system shown in the figure, the refrigerants in both cycles are assumed to be the same. This is not necessary, however, since there is no mixing taking place in the heat exchanger. Therefore, refrigerants with more desirable characteristics can be used in each cycle. In this case, there would be a separate saturation dome for each fluid, and the T-s diagram for one of the cycles would be different. Also, in actual cascade refrigeration systems, the two cycles would overlap somewhat since a temperature difference between the two fluids is needed for any heat transfer to take place. It is evident from the T-s diagram in Fig that the compressor work decreases and the amount of heat absorbed from the refrigerated space increases as a result of cascading. Therefore, cascading improves the COP of a refrigeration system. Some refrigeration systems use three or four stages of cascading. EXAMPLE 11 3 A Two-Stage Cascade Refrigeration Cycle Consider a two-stage cascade refrigeration system operating between the pressure limits of 0.8 and 0.14 MPa. Each stage operates on an ideal vaporcompression refrigeration cycle with refrigerant-134a as the working fluid. Heat rejection from the lower cycle to the upper cycle takes place in an adiabatic counterflow heat exchanger where both streams enter at about 0.3 MPa.

144 6 Thermodynamics (In practice, the working fluid of the lower cycle is at a higher pressure and temperature in the heat exchanger for effective heat transfer.) If the mass flow rate of the refrigerant through the upper cycle is 0.05 kg/s, determine (a) the mass flow rate of the refrigerant through the lower cycle, (b) the rate of heat removal from the refrigerated space and the power input to the compressor, and (c) the coefficient of performance of this cascade refrigerator. Solution A cascade refrigeration system operating between the specified pressure limits is considered. The mass flow rate of the refrigerant through the lower cycle, the rate of refrigeration, the power input, and the COP are to be determined. Assumptions 1 Steady operating conditions exist. Kinetic and potential energy changes are negligible. 3 The heat exchanger is adiabatic. Properties The enthalpies of the refrigerant at all eight states are determined from the refrigerant tables and are indicated on the T-s diagram. Analysis The T-s diagram of the refrigeration cycle is shown in Fig The topping cycle is labeled cycle A and the bottoming one, cycle B. For both cycles, the refrigerant leaves the condenser as a saturated liquid and enters the compressor as saturated vapor. (a) The mass flow rate of the refrigerant through the lower cycle is determined from the steady-flow energy balance on the adiabatic heat exchanger, E # out E # in m # Ah 5 m # Bh 3 m # Ah 8 m # Bh m # A 1h 5 h 8 m # B 1h h kg>s kj>kg4 m # B kj>kg4 m # B kg/s (b) The rate of heat removal by a cascade cycle is the rate of heat absorption in the evaporator of the lowest stage. The power input to a cascade cycle is the sum of the power inputs to all of the compressors: Q # L m # B 1h 1 h kg>s kj>kg kw W # in W # comp I,in W # comp II,in m # A 1h 6 h 5 m # B 1h h 1 T FIGURE T-s diagram of the cascade refrigeration cycle described in Example h 3 = h 7 = h 6 = 70.9 kj/kg 7 h = MPa A MPa 5 h 8 = h 5 = B 0.14 MPa h 1 = h 4 = s

145 Chapter kg>s kj>kg kg>s kj>kg kw (c) The COP of a refrigeration system is the ratio of the refrigeration rate to the net power input: COP R Q# L W # 7.18 kw 4.46 net,in 1.61 kw Discussion This problem was worked out in Example 11 1 for a single-stage refrigeration system. Notice that the COP of the refrigeration system increases from 3.97 to 4.46 as a result of cascading. The COP of the system can be increased even more by increasing the number of cascade stages. Multistage Compression Refrigeration Systems When the fluid used throughout the cascade refrigeration system is the same, the heat exchanger between the stages can be replaced by a mixing chamber (called a flash chamber) since it has better heat transfer characteristics. Such systems are called multistage compression refrigeration systems. A twostage compression refrigeration system is shown in Fig WARM environment Q H 5 Condenser 4 T 4 Expansion valve High-pressure compressor 6 Flash chamber Expansion valve Low-pressure compressor Evaporator 1 Q L s COLD refrigerated space FIGURE 11 1 A two-stage compression refrigeration system with a flash chamber.

146 64 Thermodynamics In this system, the liquid refrigerant expands in the first expansion valve to the flash chamber pressure, which is the same as the compressor interstage pressure. Part of the liquid vaporizes during this process. This saturated vapor (state 3) is mixed with the superheated vapor from the low-pressure compressor (state ), and the mixture enters the high-pressure compressor at state 9. This is, in essence, a regeneration process. The saturated liquid (state 7) expands through the second expansion valve into the evaporator, where it picks up heat from the refrigerated space. The compression process in this system resembles a two-stage compression with intercooling, and the compressor work decreases. Care should be exercised in the interpretations of the areas on the T-s diagram in this case since the mass flow rates are different in different parts of the cycle. EXAMPLE 11 4 A Two-Stage Refrigeration Cycle with a Flash Chamber Consider a two-stage compression refrigeration system operating between the pressure limits of 0.8 and 0.14 MPa. The working fluid is refrigerant-134a. The refrigerant leaves the condenser as a saturated liquid and is throttled to a flash chamber operating at 0.3 MPa. Part of the refrigerant evaporates during this flashing process, and this vapor is mixed with the refrigerant leaving the low-pressure compressor. The mixture is then compressed to the condenser pressure by the high-pressure compressor. The liquid in the flash chamber is throttled to the evaporator pressure and cools the refrigerated space as it vaporizes in the evaporator. Assuming the refrigerant leaves the evaporator as a saturated vapor and both compressors are isentropic, determine (a) the fraction of the refrigerant that evaporates as it is throttled to the flash chamber, (b) the amount of heat removed from the refrigerated space and the compressor work per unit mass of refrigerant flowing through the condenser, and (c) the coefficient of performance. Solution A two-stage compression refrigeration system operating between specified pressure limits is considered. The fraction of the refrigerant that evaporates in the flash chamber, the refrigeration and work input per unit mass, and the COP are to be determined. Assumptions 1 Steady operating conditions exist. Kinetic and potential energy changes are negligible. 3 The flash chamber is adiabatic. Properties The enthalpies of the refrigerant at various states are determined from the refrigerant tables and are indicated on the T-s diagram. Analysis The T-s diagram of the refrigeration cycle is shown in Fig We note that the refrigerant leaves the condenser as saturated liquid and enters the low-pressure compressor as saturated vapor. (a) The fraction of the refrigerant that evaporates as it is throttled to the flash chamber is simply the quality at state 6, which is x 6 h 6 h f h fg (b) The amount of heat removed from the refrigerated space and the compressor work input per unit mass of refrigerant flowing through the condenser are q L 11 x 6 1h 1 h kj>kg kj/kg

147 Chapter T 4 h 4 = kj/kg h 5 = h = h 7 = h 6 = h 3 = h 9 = h 8 = h 1 = s FIGURE T-s diagram of the two-stage compression refrigeration cycle described in Example and w in w comp I,in w comp II,in 11 x 6 1h h h 4 h 9 The enthalpy at state 9 is determined from an energy balance on the mixing chamber, E # out E # in 11h 9 x 6 h 3 11 x 6 h h kj>kg Also, s kj/kg K. Thus the enthalpy at state 4 (0.8 MPa, s 4 s 9 ) is h kj/kg. Substituting, w in kj>kg kj>kg 3.71 kj/kg (c) The coefficient of performance is COP R q L kj>kg 4.47 w in 3.71 kj>kg Discussion This problem was worked out in Example 11 1 for a single-stage refrigeration system (COP 3.97) and in Example 11 3 for a two-stage cascade refrigeration system (COP 4.46). Notice that the COP of the refrigeration system increased considerably relative to the single-stage compression but did not change much relative to the two-stage cascade compression. Multipurpose Refrigeration Systems with a Single Compressor Some applications require refrigeration at more than one temperature. This could be accomplished by using a separate throttling valve and a separate compressor for each evaporator operating at different temperatures. However, such a system is bulky and probably uneconomical. A more practical and

148 66 Thermodynamics Kitchen air T Q H 3 Condenser Expansion valve 4 Refrigerator Compressor 3 Q H A Q L,R Expansion 1 valve 6 (Alternative path) 5 Freezer 4 Q L,R 5 A 6 Q L,F 1 Q L,F s FIGURE Schematic and T-s diagram for a refrigerator freezer unit with one compressor. economical approach would be to route all the exit streams from the evaporators to a single compressor and let it handle the compression process for the entire system. Consider, for example, an ordinary refrigerator freezer unit. A simplified schematic of the unit and the T-s diagram of the cycle are shown in Fig Most refrigerated goods have a high water content, and the refrigerated space must be maintained above the ice point to prevent freezing. The freezer compartment, however, is maintained at about 18 C. Therefore, the refrigerant should enter the freezer at about 5 C to have heat transfer at a reasonable rate in the freezer. If a single expansion valve and evaporator were used, the refrigerant would have to circulate in both compartments at about 5 C, which would cause ice formation in the neighborhood of the evaporator coils and dehydration of the produce. This problem can be eliminated by throttling the refrigerant to a higher pressure (hence temperature) for use in the refrigerated space and then throttling it to the minimum pressure for use in the freezer. The entire refrigerant leaving the freezer compartment is subsequently compressed by a single compressor to the condenser pressure. Liquefaction of Gases The liquefaction of gases has always been an important area of refrigeration since many important scientific and engineering processes at cryogenic temperatures (temperatures below about 100 C) depend on liquefied gases. Some examples of such processes are the separation of oxygen and nitrogen from air, preparation of liquid propellants for rockets, the study of material properties at low temperatures, and the study of some exciting phenomena such as superconductivity.

149 At temperatures above the critical-point value, a substance exists in the gas phase only. The critical temperatures of helium, hydrogen, and nitrogen (three commonly used liquefied gases) are 68, 40, and 147 C, respectively. Therefore, none of these substances exist in liquid form at atmospheric conditions. Furthermore, low temperatures of this magnitude cannot be obtained by ordinary refrigeration techniques. Then the question that needs to be answered in the liquefaction of gases is this: How can we lower the temperature of a gas below its critical-point value? Several cycles, some complex and others simple, are used successfully for the liquefaction of gases. Below we discuss the Linde-Hampson cycle, which is shown schematically and on a T-s diagram in Fig Makeup gas is mixed with the uncondensed portion of the gas from the previous cycle, and the mixture at state is compressed by a multistage compressor to state 3. The compression process approaches an isothermal process due to intercooling. The high-pressure gas is cooled in an aftercooler by a cooling medium or by a separate external refrigeration system to state 4. The gas is further cooled in a regenerative counter-flow heat exchanger by the uncondensed portion of gas from the previous cycle to state 5, and it is throttled to state 6, which is a saturated liquid vapor mixture state. The liquid (state 7) is collected as the desired product, and the vapor (state 8) is routed through the regenerator to cool the high-pressure gas approaching the throttling valve. Finally, the gas is mixed with fresh makeup gas, and the cycle is repeated. Chapter Heat exchanger 3 Multistage compressor T 5 Q 9 Regenerator 1 Makeup gas Vapor recirculated s 7 Liquid removed FIGURE Linde-Hampson system for liquefying gases.

150 68 Thermodynamics This and other refrigeration cycles used for the liquefaction of gases can also be used for the solidification of gases GAS REFRIGERATION CYCLES As explained in Sec. 11, the Carnot cycle (the standard of comparison for power cycles) and the reversed Carnot cycle (the standard of comparison for refrigeration cycles) are identical, except that the reversed Carnot cycle operates in the reverse direction. This suggests that the power cycles discussed in earlier chapters can be used as refrigeration cycles by simply reversing them. In fact, the vapor-compression refrigeration cycle is essentially a modified Rankine cycle operating in reverse. Another example is the reversed Stirling cycle, which is the cycle on which Stirling refrigerators operate. In this section, we discuss the reversed Brayton cycle, better known as the gas refrigeration cycle. Consider the gas refrigeration cycle shown in Fig The surroundings are at T 0, and the refrigerated space is to be maintained at T L. The gas is compressed during process 1-. The high-pressure, high-temperature gas at state is then cooled at constant pressure to T 0 by rejecting heat to the surroundings. This is followed by an expansion process in a turbine, during which the gas temperature drops to T 4. (Can we achieve the cooling effect by using a throttling valve instead of a turbine?) Finally, the cool gas absorbs heat from the refrigerated space until its temperature rises to T 1. WARM environment Q H T Heat exchanger 3 Q H Turbine Compressor W net,in Heat exchanger 1 4 Q L Q L COLD refrigerated space s FIGURE Simple gas refrigeration cycle.

151 All the processes described are internally reversible, and the cycle executed is the ideal gas refrigeration cycle. In actual gas refrigeration cycles, the compression and expansion processes deviate from the isentropic ones, and T 3 is higher than T 0 unless the heat exchanger is infinitely large. On a T-s diagram, the area under process curve 4-1 represents the heat removed from the refrigerated space, and the enclosed area represents the net work input. The ratio of these areas is the COP for the cycle, which may be expressed as where COP R q L w net,in q L h 1 h 4 w turb,out h 3 h 4 w comp,in h h 1 (11 11) The gas refrigeration cycle deviates from the reversed Carnot cycle because the heat transfer processes are not isothermal. In fact, the gas temperature varies considerably during heat transfer processes. Consequently, the gas refrigeration cycles have lower COPs relative to the vapor-compression refrigeration cycles or the reversed Carnot cycle. This is also evident from the T-s diagram in Fig The reversed Carnot cycle consumes a fraction of the net work (rectangular area 1A3B) but produces a greater amount of refrigeration (triangular area under B1). Despite their relatively low COPs, the gas refrigeration cycles have two desirable characteristics: They involve simple, lighter components, which make them suitable for aircraft cooling, and they can incorporate regeneration, which makes them suitable for liquefaction of gases and cryogenic applications. An open-cycle aircraft cooling system is shown in Fig Atmospheric air is compressed by a compressor, cooled by the surrounding air, and expanded in a turbine. The cool air leaving the turbine is then directly routed to the cabin. q L w comp,in w turb,out T Gas refrigeration cycle 3 B 4 Chapter A 1 Reversed Carnot cycle FIGURE A reserved Carnot cycle produces more refrigeration (area under B1) with less work input (area 1A3B). s Q Heat exchanger 3 W net,in Turbine Compressor 4 Cool air out 1 Warm air in FIGURE An open-cycle aircraft cooling system.

152 630 Thermodynamics COLD refrigerated space Heat exchanger 6 Regenerator 5 4 WARM environment Q 3 Heat exchanger 1 T Q H Turbine Compressor 5 6 Q L W net,in s FIGURE Gas refrigeration cycle with regeneration. T, F The regenerative gas cycle is shown in Fig Regenerative cooling is achieved by inserting a counter-flow heat exchanger into the cycle. Without regeneration, the lowest turbine inlet temperature is T 0, the temperature of the surroundings or any other cooling medium. With regeneration, the high-pressure gas is further cooled to T 4 before expanding in the turbine. Lowering the turbine inlet temperature automatically lowers the turbine exit temperature, which is the minimum temperature in the cycle. Extremely low temperatures can be achieved by repeating this process. T max 80 0 T min Q H Q L EXAMPLE 11 5 The Simple Ideal Gas Refrigeration Cycle An ideal gas refrigeration cycle using air as the working medium is to maintain a refrigerated space at 0 F while rejecting heat to the surrounding medium at 80 F. The pressure ratio of the compressor is 4. Determine (a) the maximum and minimum temperatures in the cycle, (b) the coefficient of performance, and (c) the rate of refrigeration for a mass flow rate of 0.1 lbm/s. FIGURE 11 0 T-s diagram of the ideal-gas refrigeration cycle described in Example s Solution An ideal gas refrigeration cycle using air as the working fluid is considered. The maximum and minimum temperatures, the COP, and the rate of refrigeration are to be determined. Assumptions 1 Steady operating conditions exist. Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Analysis The T-s diagram of the gas refrigeration cycle is shown in Fig We note that this is an ideal gas-compression refrigeration cycle, and thus, both the compressor and the turbine are isentropic, and the air is cooled to the environment temperature before it enters the turbine.

153 Chapter (a) The maximum and minimum temperatures in the cycle are determined from the isentropic relations of ideal gases for the compression and expansion processes. From Table A 17E, T R h Btu/lbm and P r P P r P r1 (4)(0.7913) e h Btu/lbm T 683 R (or 3 F) P 1 T R h Btu/lbm and P r P 4 P 3 P r4 P r3 (0.5)(1.386) Therefore, the highest and the lowest temperatures in the cycle are 3 and 97 F, respectively. (b) The COP of this ideal gas refrigeration cycle is COP R q L w net,in q L w comp,in W turb,out e h Btu/lbm T R (or 97 F) where Thus, q L h 1 h Btu>lbm W turb,out h 3 h Btu>lbm W comp,in h h Btu>lbm COP R (c) The rate of refrigeration is Q # refrig m # 1q L 10.1 lbm>s 13. Btu>lbm.3 Btu/s Discussion It is worth noting that an ideal vapor-compression cycle working under similar conditions would have a COP greater than ABSORPTION REFRIGERATION SYSTEMS Another form of refrigeration that becomes economically attractive when there is a source of inexpensive thermal energy at a temperature of 100 to 00 C is absorption refrigeration. Some examples of inexpensive thermal energy sources include geothermal energy, solar energy, and waste heat from cogeneration or process steam plants, and even natural gas when it is available at a relatively low price. As the name implies, absorption refrigeration systems involve the absorption of a refrigerant by a transport medium. The most widely used absorption refrigeration system is the ammonia water system, where ammonia (NH 3 ) serves as the refrigerant and water (H O) as the transport medium. Other absorption refrigeration systems include water lithium bromide and water lithium chloride systems, where water serves as the refrigerant. The latter two systems are limited to applications such as air-conditioning where the minimum temperature is above the freezing point of water.

154 63 Thermodynamics To understand the basic principles involved in absorption refrigeration, we examine the NH 3 H O system shown in Fig The ammonia water refrigeration machine was patented by the Frenchman Ferdinand Carre in Within a few years, the machines based on this principle were being built in the United States primarily to make ice and store food. You will immediately notice from the figure that this system looks very much like the vapor-compression system, except that the compressor has been replaced by a complex absorption mechanism consisting of an absorber, a pump, a generator, a regenerator, a valve, and a rectifier. Once the pressure of NH 3 is raised by the components in the box (this is the only thing they are set up to do), it is cooled and condensed in the condenser by rejecting heat to the surroundings, is throttled to the evaporator pressure, and absorbs heat from the refrigerated space as it flows through the evaporator. So, there is nothing new there. Here is what happens in the box: Ammonia vapor leaves the evaporator and enters the absorber, where it dissolves and reacts with water to form NH 3 H O. This is an exothermic reaction; thus heat is released during this process. The amount of NH 3 that can be dissolved in H O is inversely proportional to the temperature. Therefore, it is necessary to cool the absorber to maintain its temperature as low as possible, hence to maximize the amount of NH 3 dissolved in water. The liquid NH 3 H O solution, which is rich in NH 3, is then pumped to the generator. Heat is transferred to the solution from a source to vaporize some of the solution. The vapor, which is rich in NH 3, passes through a rectifier, which separates the water and returns it to the generator. The high-pressure pure NH 3 vapor then continues its journey through the rest of the cycle. The WARM environment Q H Condenser Pure NH 3 Rectifier Generator Q gen NH 3 + H O Solar energy Expansion valve H O Q Regenerator Absorber Expansion valve NH 3 + H O Evaporator Pure NH 3 W pump FIGURE 11 1 Ammonia absorption refrigeration cycle. Q L COLD refrigerated space Q cool Cooling water Pump

155 hot NH 3 H O solution, which is weak in NH 3, then passes through a regenerator, where it transfers some heat to the rich solution leaving the pump, and is throttled to the absorber pressure. Compared with vapor-compression systems, absorption refrigeration systems have one major advantage: A liquid is compressed instead of a vapor. The steady-flow work is proportional to the specific volume, and thus the work input for absorption refrigeration systems is very small (on the order of one percent of the heat supplied to the generator) and often neglected in the cycle analysis. The operation of these systems is based on heat transfer from an external source. Therefore, absorption refrigeration systems are often classified as heat-driven systems. The absorption refrigeration systems are much more expensive than the vapor-compression refrigeration systems. They are more complex and occupy more space, they are much less efficient thus requiring much larger cooling towers to reject the waste heat, and they are more difficult to service since they are less common. Therefore, absorption refrigeration systems should be considered only when the unit cost of thermal energy is low and is projected to remain low relative to electricity. Absorption refrigeration systems are primarily used in large commercial and industrial installations. The COP of absorption refrigeration systems is defined as COP absorption Desired output Required input Q L Q L Q gen W pump,in Q gen (11 1) The maximum COP of an absorption refrigeration system is determined by assuming that the entire cycle is totally reversible (i.e., the cycle involves no irreversibilities and any heat transfer is through a differential temperature difference). The refrigeration system would be reversible if the heat from the source (Q gen ) were transferred to a Carnot heat engine, and the work output of this heat engine (W h th,rev Q gen ) is supplied to a Carnot refrigerator to remove heat from the refrigerated space. Note that Q L W COP R,rev h th,rev Q gen COP R,rev. Then the overall COP of an absorption refrigeration system under reversible conditions becomes (Fig. 11 ) COP rev,absorption Q L h th,rev COP R,rev a 1 T 0 ba b Q gen T s T 0 T L (11 13) where T L, T 0, and T s are the thermodynamic temperatures of the refrigerated space, the environment, and the heat source, respectively. Any absorption refrigeration system that receives heat from a source at T s and removes heat from the refrigerated space at T L while operating in an environment at T 0 has a lower COP than the one determined from Eq For example, when the source is at 10 C, the refrigerated space is at 10 C, and the environment is at 5 C, the maximum COP that an absorption refrigeration system can have is 1.8. The COP of actual absorption refrigeration systems is usually less than 1. Air-conditioning systems based on absorption refrigeration, called absorption chillers, perform best when the heat source can supply heat at a high temperature with little temperature drop. The absorption chillers are typically rated at an input temperature of 116 C (40 F). The chillers perform at lower temperatures, but their cooling capacity decreases sharply with T L Source T s Reversible heat engine T 0 environment Chapter Q gen W = h rev Q gen Reversible refrigerator Q L = COP R,rev W W = h rev Q gen = ( ) 1 T 0 Q gen T s TL Q L = COP R,rev W = ( T ) W 0 T L Environment T 0 T L Refrigerated space Q COP rev,absorption = L = ( )( ) 1 T 0 T L Q gen T s T 0 T L FIGURE 11 Determining the maximum COP of an absorption refrigeration system.

156 634 Thermodynamics decreasing source temperature, about 1.5 percent for each 6 C (10 F) drop in the source temperature. For example, the capacity goes down to 50 percent when the supply water temperature drops to 93 C (00 F). In that case, one needs to double the size (and thus the cost) of the chiller to achieve the same cooling. The COP of the chiller is affected less by the decline of the source temperature. The COP drops by.5 percent for each 6 C (10 F) drop in the source temperature. The nominal COP of single-stage absorption chillers at 116 C (40 F) is 0.65 to Therefore, for each ton of refrigeration, a heat input of (1,000 Btu/h)/ ,460 Btu/h is required. At 88 C (190 F), the COP drops by 1.5 percent and thus the heat input increases by 1.5 percent for the same cooling effect. Therefore, the economic aspects must be evaluated carefully before any absorption refrigeration system is considered, especially when the source temperature is below 93 C (00 F). Another absorption refrigeration system that is quite popular with campers is a propane-fired system invented by two Swedish undergraduate students. In this system, the pump is replaced by a third fluid (hydrogen), which makes it a truly portable unit. TOPIC OF SPECIAL INTEREST* Metal B I Metal A FIGURE 11 3 When one of the junctions of two dissimilar metals is heated, a current I flows through the closed circuit. Metal B I = 0 + V FIGURE 11 4 When a thermoelectric circuit is broken, a potential difference is generated. I Metal A Thermoelectric Power Generation and Refrigeration Systems All the refrigeration systems discussed above involve many moving parts and bulky, complex components. Then this question comes to mind: Is it really necessary for a refrigeration system to be so complex? Can we not achieve the same effect in a more direct way? The answer to this question is yes. It is possible to use electric energy more directly to produce cooling without involving any refrigerants and moving parts. Below we discuss one such system, called thermoelectric refrigerator. Consider two wires made from different metals joined at both ends (junctions), forming a closed circuit. Ordinarily, nothing will happen. However, when one of the ends is heated, something interesting happens: A current flows continuously in the circuit, as shown in Fig This is called the Seebeck effect, in honor of Thomas Seebeck, who made this discovery in 181. The circuit that incorporates both thermal and electrical effects is called a thermoelectric circuit, and a device that operates on this circuit is called a thermoelectric device. The Seebeck effect has two major applications: temperature measurement and power generation. When the thermoelectric circuit is broken, as shown in Fig. 11 4, the current ceases to flow, and we can measure the driving force (the electromotive force) or the voltage generated in the circuit by a voltmeter. The voltage generated is a function of the temperature difference and the materials of the two wires used. Therefore, temperature can be measured by simply measuring voltages. The two wires used to measure the temperature in *This section can be skipped without a loss in continuity.

157 Chapter this manner form a thermocouple, which is the most versatile and most widely used temperature measurement device. A common T-type thermocouple, for example, consists of copper and constantan wires, and it produces about 40 mv per C difference. The Seebeck effect also forms the basis for thermoelectric power generation. The schematic diagram of a thermoelectric generator is shown in Fig Heat is transferred from a high-temperature source to the hot junction in the amount of Q H, and it is rejected to a low-temperature sink from the cold junction in the amount of Q L. The difference between these two quantities is the net electrical work produced, that is, W e Q H Q L. It is evident from Fig that the thermoelectric power cycle closely resembles an ordinary heat engine cycle, with electrons serving as the working fluid. Therefore, the thermal efficiency of a thermoelectric generator operating between the temperature limits of T H and T L is limited by the efficiency of a Carnot cycle operating between the same temperature limits. Thus, in the absence of any irreversibilities (such as I R heating, where R is the total electrical resistance of the wires), the thermoelectric generator will have the Carnot efficiency. The major drawback of thermoelectric generators is their low efficiency. The future success of these devices depends on finding materials with more desirable characteristics. For example, the voltage output of thermoelectric devices has been increased several times by switching from metal pairs to semiconductors. A practical thermoelectric generator using n-type (heavily doped to create excess electrons) and p-type (heavily doped to create a deficiency of electrons) materials connected in series is shown in Fig Despite their low efficiencies, thermoelectric generators have definite weight and reliability advantages and are presently used in rural areas and in space applications. For example, silicon germanium-based thermoelectric generators have been powering Voyager spacecraft since 1980 and are expected to continue generating power for many more years. If Seebeck had been fluent in thermodynamics, he would probably have tried reversing the direction of flow of electrons in the thermoelectric circuit (by externally applying a potential difference in the reverse direction) to create a refrigeration effect. But this honor belongs to Jean Charles Athanase Peltier, who discovered this phenomenon in He noticed during his experiments that when a small current was passed through the junction of two dissimilar wires, the junction was cooled, as shown in Fig This is called the Peltier effect, and it forms the basis for thermoelectric refrigeration. A practical thermoelectric refrigeration circuit using semiconductor materials is shown in Fig Heat is absorbed from the refrigerated space in the amount of Q L and rejected to the warmer environment in the amount of Q H. The difference between these two quantities is the net electrical work that needs to be supplied; that is, W e Q H Q L. Thermoelectric refrigerators presently cannot compete with vapor-compression refrigeration systems because of their low coefficient of performance. They are available in the market, however, and are preferred in some applications because of their small size, simplicity, quietness, and reliability. Hot junction Cold junction High-temperature source T H Q H I Q L Low-temperature sink T L I W net FIGURE 11 5 Schematic of a simple thermoelectric power generator. Hot plate Cold plate SOURCE SINK Q H p n p n p n + I W net Q L FIGURE 11 6 A thermoelectric power generator.

158 636 Thermodynamics Heat rejected Heat absorbed EXAMPLE 11 6 Cooling of a Canned Drink by a Thermoelectric Refrigerator FIGURE 11 7 When a current is passed through the junction of two dissimilar materials, the junction is cooled. Hot plate Cold plate + WARM environment Q H p n p n p n Q L Refrigerated space FIGURE 11 8 A thermoelectric refrigerator. + I A thermoelectric refrigerator that resembles a small ice chest is powered by a car battery and has a COP of 0.1. If the refrigerator cools a L canned drink from 0 to 4 C in 30 min, determine the average electric power consumed by the thermoelectric refrigerator. Solution A thermoelectric refrigerator with a specified COP is used to cool canned drinks. The power consumption of the refrigerator is to be determined. Assumptions Heat transfer through the walls of the refrigerator is negligible during operation. Properties The properties of canned drinks are the same as those of water at room temperature, r 1 kg/l and c 4.18 kj/kg C (Table A 3). Analysis The cooling rate of the refrigerator is simply the rate of decrease of the energy of the canned drinks, m rv 11 kg>l L kg Q cooling mc T kg kj>kg # C 10 4 C 3.4 kj Q # cooling Q cooling t Then the average power consumed by the refrigerator becomes W # in Q# cooling COP R 3.4 kj kw 13 W s 13 W W Discussion In reality, the power consumption will be larger because of the heat gain through the walls of the refrigerator. SUMMARY The transfer of heat from lower temperature regions to higher temperature ones is called refrigeration. Devices that produce refrigeration are called refrigerators, and the cycles on which they operate are called refrigeration cycles. The working fluids used in refrigerators are called refrigerants. Refrigerators used for the purpose of heating a space by transferring heat from a cooler medium are called heat pumps. The performance of refrigerators and heat pumps is expressed in terms of coefficient of performance (COP), defined as Desired output COP R Required output Desired output COP HP Required input Cooling effect Work input Heating effect Work input Q L W net,in Q H W net,in The standard of comparison for refrigeration cycles is the reversed Carnot cycle. A refrigerator or heat pump that operates on the reversed Carnot cycle is called a Carnot refrigerator or a Carnot heat pump, and their COPs are COP R,Carnot 1 T H >T L 1 1 COP HP,Carnot 1 T L >T H The most widely used refrigeration cycle is the vaporcompression refrigeration cycle. In an ideal vapor-compression refrigeration cycle, the refrigerant enters the compressor as a saturated vapor and is cooled to the saturated liquid state in the condenser. It is then throttled to the evaporator pressure and vaporizes as it absorbs heat from the refrigerated space.

159 Very low temperatures can be achieved by operating two or more vapor-compression systems in series, called cascading. The COP of a refrigeration system also increases as a result of cascading. Another way of improving the performance of a vapor-compression refrigeration system is by using multistage compression with regenerative cooling. A refrigerator with a single compressor can provide refrigeration at several temperatures by throttling the refrigerant in stages. The vapor-compression refrigeration cycle can also be used to liquefy gases after some modifications. The power cycles can be used as refrigeration cycles by simply reversing them. Of these, the reversed Brayton cycle, which is also known as the gas refrigeration cycle, is used to cool aircraft and to obtain very low (cryogenic) temperatures after it is modified with regeneration. The work output of the turbine can be used to reduce the work input requirements to the compressor. Thus the COP of a gas refrigeration cycle is COP absorption q L w net,in q L w comp,in w turb,out Chapter Another form of refrigeration that becomes economically attractive when there is a source of inexpensive thermal energy at a temperature of 100 to 00 C is absorption refrigeration, where the refrigerant is absorbed by a transport medium and compressed in liquid form. The most widely used absorption refrigeration system is the ammonia water system, where ammonia serves as the refrigerant and water as the transport medium. The work input to the pump is usually very small, and the COP of absorption refrigeration systems is defined as COP absorption Desired output Required input Q L Q L Q gen W pump,in Q gen The maximum COP an absorption refrigeration system can have is determined by assuming totally reversible conditions, which yields COP rev,absorption h th,rev COP R,rev a1 T 0 ba b T s T 0 T L where T 0, T L, and T s are the thermodynamic temperatures of the environment, the refrigerated space, and the heat source, respectively. T L REFERENCES AND SUGGESTED READINGS 1. ASHRAE, Handbook of Fundamentals. Atlanta: American Society of Heating, Refrigerating, and Air-Conditioning Engineers, Heat Pump Systems A Technology Review. OECD Report, Paris, B. Nagengast. A Historical Look at CFC Refrigerants. ASHRAE Journal 30, no. 11 (November 1988), pp W. F. Stoecker. Growing Opportunities for Ammonia Refrigeration. Proceedings of the Meeting of the International Institute of Ammonia Refrigeration, Austin, Texas, W. F. Stoecker and J. W. Jones. Refrigeration and Air Conditioning. nd ed. New York: McGraw-Hill, K. Wark and D. E. Richards. Thermodynamics. 6th ed. New York: McGraw-Hill, PROBLEMS* The Reversed Carnot Cycle 11 1C Why is the reversed Carnot cycle executed within the saturation dome not a realistic model for refrigeration cycles? *Problems designated by a C are concept questions, and students are encouraged to answer them all. Problems designated by an E are in English units, and the SI users can ignore them. Problems with a CD-EES icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed DVD. Problems with a computer-ees icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text. 11 A steady-flow Carnot refrigeration cycle uses refrigerant-134a as the working fluid. The refrigerant changes from saturated vapor to saturated liquid at 30 C in the condenser as it rejects heat. The evaporator pressure is 160 kpa. Show the cycle on a T-s diagram relative to saturation lines, and determine (a) the coefficient of performance, (b) the amount of heat absorbed from the refrigerated space, and (c) the net work input. Answers: (a) 5.64, (b) 147 kj/kg, (c) 6.1 kj/kg 11 3E Refrigerant-134a enters the condenser of a steadyflow Carnot refrigerator as a saturated vapor at 90 psia, and it leaves with a quality of The heat absorption from the refrigerated space takes place at a pressure of 30 psia. Show

160 638 Thermodynamics the cycle on a T-s diagram relative to saturation lines, and determine (a) the coefficient of performance, (b) the quality at the beginning of the heat-absorption process, and (c) the net work input. Ideal and Actual Vapor-Compression Refrigeration Cycles 11 4C Does the ideal vapor-compression refrigeration cycle involve any internal irreversibilities? 11 5C Why is the throttling valve not replaced by an isentropic turbine in the ideal vapor-compression refrigeration cycle? 11 6C It is proposed to use water instead of refrigerant- 134a as the working fluid in air-conditioning applications where the minimum temperature never falls below the freezing point. Would you support this proposal? Explain. 11 7C In a refrigeration system, would you recommend condensing the refrigerant-134a at a pressure of 0.7 or 1.0 MPa if heat is to be rejected to a cooling medium at 15 C? Why? 11 8C Does the area enclosed by the cycle on a T-s diagram represent the net work input for the reversed Carnot cycle? How about for the ideal vapor-compression refrigeration cycle? 11 9C Consider two vapor-compression refrigeration cycles. The refrigerant enters the throttling valve as a saturated liquid at 30 C in one cycle and as subcooled liquid at 30 C in the other one. The evaporator pressure for both cycles is the same. Which cycle do you think will have a higher COP? 11 10C The COP of vapor-compression refrigeration cycles improves when the refrigerant is subcooled before it enters the throttling valve. Can the refrigerant be subcooled indefinitely to maximize this effect, or is there a lower limit? Explain A commercial refrigerator with refrigerant-134a as the working fluid is used to keep the refrigerated space at 30 C by rejecting its waste heat to cooling water that enters 6 C 4 C 3 Expansion valve Q H Condenser Water 18 C Evaporator 4 1 Q L 1. MPa 65 C Compressor Q in 60 kpa 34 C FIGURE P11 11 W in the condenser at 18 C at a rate of 0.5 kg/s and leaves at 6 C. The refrigerant enters the condenser at 1. MPa and 65 C and leaves at 4 C. The inlet state of the compressor is 60 kpa and 34 C and the compressor is estimated to gain a net heat of 450 W from the surroundings. Determine (a) the quality of the refrigerant at the evaporator inlet, (b) the refrigeration load, (c) the COP of the refrigerator, and (d) the theoretical maximum refrigeration load for the same power input to the compressor A refrigerator uses refrigerant-134a as the working fluid and operates on an ideal vapor-compression refrigeration cycle between 0.1 and 0.7 MPa. The mass flow rate of the refrigerant is 0.05 kg/s. Show the cycle on a T-s diagram with respect to saturation lines. Determine (a) the rate of heat removal from the refrigerated space and the power input to the compressor, (b) the rate of heat rejection to the environment, and (c) the coefficient of performance. Answers: (a) 7.41 kw, 1.83 kw, (b) 9.3 kw, (c) Repeat Prob for a condenser pressure of 0.9 MPa If the throttling valve in Prob is replaced by an isentropic turbine, determine the percentage increase in the COP and in the rate of heat removal from the refrigerated space. Answers: 4. percent, 4. percent Consider a 300 kj/min refrigeration system that operates on an ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid. The refrigerant enters the compressor as saturated vapor at 140 kpa and is compressed to 800 kpa. Show the cycle on a T-s diagram with respect to saturation lines, and determine (a) the quality of the refrigerant at the end of the throttling process, (b) the coefficient of performance, and (c) the power input to the compressor Reconsider Prob Using EES (or other) software, investigate the effect of evaporator pressure on the COP and the power input. Let the evaporator pressure vary from 100 to 400 kpa. Plot the COP and the power input as functions of evaporator pressure, and discuss the results Repeat Prob assuming an isentropic efficiency of 85 percent for the compressor. Also, determine the rate of exergy destruction associated with the compression process in this case. Take T 0 98 K Refrigerant-134a enters the compressor of a refrigerator as superheated vapor at 0.14 MPa and 10 C at a rate of 0.1 kg/s, and it leaves at 0.7 MPa and 50 C. The refrigerant is cooled in the condenser to 4 C and 0.65 MPa, and it is throttled to 0.15 MPa. Disregarding any heat transfer and pressure drops in the connecting lines between the components, show the cycle on a T-s diagram with respect to saturation lines, and determine (a) the rate of heat removal from the refrigerated

161 space and the power input to the compressor, (b) the isentropic efficiency of the compressor, and (c) the COP of the refrigerator. Answers: (a) 19.4 kw, 5.06 kw, (b) 8.5 percent, (c) E An ice-making machine operates on the ideal vapor-compression cycle, using refrigerant-134a. The refrigerant enters the compressor as saturated vapor at 0 psia and leaves the condenser as saturated liquid at 80 psia. Water enters the ice machine at 55 F and leaves as ice at 5 F. For an ice production rate of 15 lbm/h, determine the power input to the ice machine (169 Btu of heat needs to be removed from each lbm of water at 55 F to turn it into ice at 5 F) Refrigerant-134a enters the compressor of a refrigerator at 140 kpa and 10 C at a rate of 0.3 m 3 /min and leaves at 1 MPa. The isentropic efficiency of the compressor is 78 percent. The refrigerant enters the throttling valve at 0.95 MPa and 30 C and leaves the evaporator as saturated vapor at 18.5 C. Show the cycle on a T-s diagram with respect to saturation lines, and determine (a) the power input to the compressor, (b) the rate of heat removal from the refrigerated space, and (c) the pressure drop and rate of heat gain in the line between the evaporator and the compressor. Answers: (a) 1.88 kw, (b) 4.99 kw, (c) 1.65 kpa, 0.41 kw 11 1 Reconsider Prob Using EES (or other) software, investigate the effects of varying the compressor isentropic efficiency over the range 60 to 100 percent and the compressor inlet volume flow rate from 0.1 to 1.0 m 3 /min on the power input and the rate of refrigeration. Plot the rate of refrigeration and the power input to the compressor as functions of compressor efficiency for compressor inlet volume flow rates of 0.1, 0.5, and 1.0 m 3 /min, and discuss the results. 11 A refrigerator uses refrigerant-134a as the working fluid and operates on the ideal vapor-compression refrigeration cycle. The refrigerant enters the evaporator at 10 kpa with a quality of 30 percent and leaves the compressor at 3 Expansion valve Q H Condenser Evaporator kpa Q L x = C Compressor FIGURE P11 W in Chapter C. If the compressor consumes 450 W of power, determine (a) the mass flow rate of the refrigerant, (b) the condenser pressure, and (c) the COP of the refrigerator. Answers: (a) kg/s, (b) 67 kpa, (c).43 Selecting the Right Refrigerant 11 3C When selecting a refrigerant for a certain application, what qualities would you look for in the refrigerant? 11 4C Consider a refrigeration system using refrigerant- 134a as the working fluid. If this refrigerator is to operate in an environment at 30 C, what is the minimum pressure to which the refrigerant should be compressed? Why? 11 5C A refrigerant-134a refrigerator is to maintain the refrigerated space at 10 C. Would you recommend an evaporator pressure of 0.1 or 0.14 MPa for this system? Why? 11 6 A refrigerator that operates on the ideal vaporcompression cycle with refrigerant-134a is to maintain the refrigerated space at 10 C while rejecting heat to the environment at 5 C. Select reasonable pressures for the evaporator and the condenser, and explain why you chose those values A heat pump that operates on the ideal vaporcompression cycle with refrigerant-134a is used to heat a house and maintain it at C by using underground water at 10 C as the heat source. Select reasonable pressures for the evaporator and the condenser, and explain why you chose those values. Heat Pump Systems 11 8C Do you think a heat pump system will be more cost-effective in New York or in Miami? Why? 11 9C What is a water-source heat pump? How does the COP of a water-source heat pump system compare to that of an air-source system? 11 30E A heat pump that operates on the ideal vaporcompression cycle with refrigerant-134a is used to heat a house and maintain it at 75 F by using underground water at 50 F as the heat source. The house is losing heat at a rate of 60,000 Btu/h. The evaporator and condenser pressures are 50 and 10 psia, respectively. Determine the power input to the heat pump and the electric power saved by using a heat pump instead of a resistance heater. Answers:.46 hp, 1.1 hp A heat pump that operates on the ideal vaporcompression cycle with refrigerant-134a is used to heat water from 15 to 45 C at a rate of 0.1 kg/s. The condenser and evaporator pressures are 1.4 and 0.3 MPa, respectively. Determine the power input to the heat pump A heat pump using refrigerant-134a heats a house by using underground water at 8 C as the heat source. The house is losing heat at a rate of 60,000 kj/h. The refrigerant enters the compressor at 80 kpa and 0 C, and it leaves at 1 MPa

162 640 Thermodynamics and 60 C. The refrigerant exits the condenser at 30 C. Determine (a) the power input to the heat pump, (b) the rate of heat absorption from the water, and (c) the increase in electric power input if an electric resistance heater is used instead of a heat pump. Answers: (a) 3.55 kw, (b) 13.1 kw, (c) 13.1 kw Reconsider Prob Using EES (or other) software, investigate the effect of varying the compressor isentropic efficiency over the range 60 to 100 percent. Plot the power input to the compressor and the electric power saved by using a heat pump rather than electric resistance heating as functions of compressor efficiency, and discuss the results Refrigerant-134a enters the condenser of a residential heat pump at 800 kpa and 55 C at a rate of kg/s and leaves at 750 kpa subcooled by 3 C. The refrigerant enters the compressor at 00 kpa superheated by 4 C. Determine (a) the isentropic efficiency of the compressor, (b) the rate of heat supplied to the heated room, and (c) the COP of the heat pump. Also, determine (d) the COP and the rate of heat supplied to the heated room if this heat pump operated on the ideal vapor-compression cycle between the pressure limits of 00 and 800 kpa. Q 750 kpa H 800 kpa 55 C Condenser 3 Expansion valve Evaporator 4 1 Q L Compressor FIGURE P A heat pump with refrigerant-134a as the working fluid is used to keep a space at 5 C by absorbing heat from geothermal water that enters the evaporator at 50 C at a rate of kg/s and leaves at 40 C. The refrigerant enters the evaporator at 0 C with a quality of 3 percent and leaves at the inlet pressure as saturated vapor. The refrigerant loses 300 W of heat to the surroundings as it flows through the compressor and the refrigerant leaves the compressor at 1.4 MPa at the same entropy as the inlet. Determine (a) the degrees of W in 3 Expansion valve Evaporator C Sat. Q L x = C Water 50 C subcooling of the refrigerant in the condenser, (b) the mass flow rate of the refrigerant, (c) the heating load and the COP of the heat pump, and (d) the theoretical minimum power input to the compressor for the same heating load. Answers: (a) 3.8 C, (b) kg/s, (c) 3.07 kw, 4.68, (d) 0.38 kw Innovative Refrigeration Systems Q H Condenser 1.4 MPa s = s 1 Compressor 11 36C What is cascade refrigeration? What are the advantages and disadvantages of cascade refrigeration? 11 37C How does the COP of a cascade refrigeration system compare to the COP of a simple vapor-compression cycle operating between the same pressure limits? 11 38C A certain application requires maintaining the refrigerated space at 3 C. Would you recommend a simple refrigeration cycle with refrigerant-134a or a two-stage cascade refrigeration cycle with a different refrigerant at the bottoming cycle? Why? 11 39C Consider a two-stage cascade refrigeration cycle and a two-stage compression refrigeration cycle with a flash chamber. Both cycles operate between the same pressure limits and use the same refrigerant. Which system would you favor? Why? 11 40C Can a vapor-compression refrigeration system with a single compressor handle several evaporators operating at different pressures? How? 11 41C In the liquefaction process, why are gases compressed to very high pressures? 11 4 Consider a two-stage cascade refrigeration system operating between the pressure limits of 0.8 and 0.14 MPa. FIGURE P11 35 W in

163 Each stage operates on the ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid. Heat rejection from the lower cycle to the upper cycle takes place in an adiabatic counterflow heat exchanger where both streams enter at about 0.4 MPa. If the mass flow rate of the refrigerant through the upper cycle is 0.4 kg/s, determine (a) the mass flow rate of the refrigerant through the lower cycle, (b) the rate of heat removal from the refrigerated space and the power input to the compressor, and (c) the coefficient of performance of this cascade refrigerator. Answers: (a) kg/s, (b) 34. kw, 7.63 kw, (c) Repeat Prob for a heat exchanger pressure of 0.55 MPa A two-stage compression refrigeration system operates with refrigerant-134a between the pressure limits of 1 and 0.14 MPa. The refrigerant leaves the condenser as a saturated liquid and is throttled to a flash chamber operating at 0.5 MPa. The refrigerant leaving the low-pressure compressor at 0.5 MPa is also routed to the flash chamber. The vapor in the flash chamber is then compressed to the condenser pressure by the high-pressure compressor, and the liquid is throttled to the evaporator pressure. Assuming the refrigerant leaves the evaporator as saturated vapor and both compressors are isentropic, determine (a) the fraction of the refrigerant that evaporates as it is throttled to the flash chamber, (b) the rate of heat removed from the refrigerated space for a mass flow rate of 0.5 kg/s through the condenser, and (c) the coefficient of performance Reconsider Prob Using EES (or other) software, investigate the effect of the various refrigerants for compressor efficiencies of 80, 90, and 100 percent. Compare the performance of the refrigeration system with different refrigerants Repeat Prob for a flash chamber pressure of 0.3 MPa Consider a two-stage cascade refrigeration system operating between the pressure limits of 1. MPa and 00 kpa with refrigerant-134a as the working fluid. Heat rejection from the lower cycle to the upper cycle takes place in an adiabatic counterflow heat exchanger where the pressure in the upper and lower cycles are 0.4 and 0.5 MPa, respectively. In both cycles, the refrigerant is a saturated liquid at the condenser exit and a saturated vapor at the compressor inlet, and the isentropic efficiency of the compressor is 80 percent. If the mass flow rate of the refrigerant through the lower cycle is 0.15 kg/s, determine (a) the mass flow rate of the refrigerant through the upper cycle, (b) the rate of heat removal from the refrigerated space, and (c) the COP of this refrigerator. Answers: (a) 0.1 kg/s, (b) 5.7 kw, (c).68 7 Expansion valve Evaporator 4 1 Q H Condenser 8 Evaporator 5 3 Expansion valve Heat Condenser Q L Chapter Compressor Consider a two-stage cascade refrigeration system operating between the pressure limits of 1. MPa and 00 kpa with refrigerant-134a as the working fluid. The refrigerant leaves the condenser as a saturated liquid and is throttled to a flash chamber operating at 0.45 MPa. Part of the refrigerant evaporates during this flashing process, and this vapor is mixed with the refrigerant leaving the low-pressure compressor. The mixture is then compressed to the condenser pressure by the high-pressure compressor. The liquid in the flash chamber is throttled to the evaporator pressure and cools the refrigerated space as it vaporizes in the evaporator. The mass flow rate of the refrigerant through the low-pressure compressor is 0.15 kg/s. Assuming the refrigerant leaves the evaporator as a saturated vapor and the isentropic efficiency is 80 percent for both compressors, determine (a) the mass flow rate of the refrigerant through the high-pressure compressor, (b) the rate of heat removal from the refrigerated space, and (c) the COP of this refrigerator. Also, determine (d) the rate of heat removal and the COP if this refrigerator operated on a single-stage cycle between the same pressure limits with the same compressor efficiency and the same flow rate as in part (a). 6 Compressor FIGURE P11 47 W in W in

164 64 Thermodynamics 6 7 Gas Refrigeration Cycle 5 Flash chamber Expansion valve Expansion valve Q H Condenser 3 Evaporator 8 1 Q L High-pressure compressor Low-pressure compressor FIGURE P C How does the ideal-gas refrigeration cycle differ from the Brayton cycle? 11 50C Devise a refrigeration cycle that works on the reversed Stirling cycle. Also, determine the COP for this cycle C How does the ideal-gas refrigeration cycle differ from the Carnot refrigeration cycle? 11 5C How is the ideal-gas refrigeration cycle modified for aircraft cooling? 11 53C In gas refrigeration cycles, can we replace the turbine by an expansion valve as we did in vapor-compression refrigeration cycles? Why? 11 54C How do we achieve very low temperatures with gas refrigeration cycles? An ideal gas refrigeration cycle using air as the working fluid is to maintain a refrigerated space at 3 C while rejecting heat to the surrounding medium at 7 C. If the pressure ratio of the compressor is 3, determine (a) the maximum and minimum temperatures in the cycle, (b) the coefficient of performance, and (c) the rate of refrigeration for a mass flow rate of 0.08 kg/s Air enters the compressor of an ideal gas refrigeration cycle at 1 C and 50 kpa and the turbine at 47 C and 50 kpa. The mass flow rate of air through the cycle is 0.08 kg/s. Assuming variable specific 4 9 heats for air, determine (a) the rate of refrigeration, (b) the net power input, and (c) the coefficient of performance. Answers: (a) 6.67 kw, (b) 3.88 kw, (c) Reconsider Prob Using EES (or other) software, study the effects of compressor and turbine isentropic efficiencies as they are varied from 70 to 100 percent on the rate of refrigeration, the net power input, and the COP. Plot the T-s diagram of the cycle for the isentropic case E Air enters the compressor of an ideal gas refrigeration cycle at 40 F and 10 psia and the turbine at 10 F and 30 psia. The mass flow rate of air through the cycle is 0.5 lbm/s. Determine (a) the rate of refrigeration, (b) the net power input, and (c) the coefficient of performance Repeat Prob for a compressor isentropic efficiency of 80 percent and a turbine isentropic efficiency of 85 percent A gas refrigeration cycle with a pressure ratio of 3 uses helium as the working fluid. The temperature of the helium is 10 C at the compressor inlet and 50 C at the turbine inlet. Assuming adiabatic efficiencies of 80 percent for both the turbine and the compressor, determine (a) the minimum temperature in the cycle, (b) the coefficient of performance, and (c) the mass flow rate of the helium for a refrigeration rate of 18 kw A gas refrigeration system using air as the working fluid has a pressure ratio of 4. Air enters the compressor at 7 C. The high-pressure air is cooled to 7 C by rejecting heat to the surroundings. It is further cooled to 15 C by regenerative cooling before it enters the turbine. Assuming both the turbine and the compressor to be isentropic and using constant specific heats at room temperature, determine (a) the lowest temperature that can be obtained by this cycle, (b) the coefficient of performance of the cycle, and (c) the mass flow rate of air for a refrigeration rate of 1 kw. Answers: (a) 99.4 C, (b) 1.1, (c) 0.37 kg/s 11 6 Repeat Prob assuming isentropic efficiencies of 75 percent for the compressor and 80 percent for the turbine A gas refrigeration system using air as the working fluid has a pressure ratio of 5. Air enters the compressor at 0 C. The high-pressure air is cooled to 35 C by rejecting heat to the surroundings. The refrigerant leaves the turbine at 80 C and then it absorbs heat from the refrigerated space before entering the regenerator. The mass flow rate of air is 0.4 kg/s. Assuming isentropic efficiencies of 80 percent for the compressor and 85 percent for the turbine and using constant specific heats at room temperature, determine (a) the effectiveness of the regenerator, (b) the rate of heat removal from the refrigerated space, and (c) the COP of the cycle. Also, determine (d) the refrigeration load and the COP if this system operated on the simple gas refrigeration cycle. Use the

165 5 Q L 6 Heat exch. 4 Turbine Regenerator same compressor inlet temperature as given, the same turbine inlet temperature as calculated, and the same compressor and turbine efficiencies. Answers: (a) 0.434, (b) 1.4 kw, (c) 0.478, (d) 4.7 kw, Absorption Refrigeration Systems Heat exch. Compressor 11 64C What is absorption refrigeration? How does an absorption refrigeration system differ from a vapor-compression refrigeration system? 11 65C What are the advantages and disadvantages of absorption refrigeration? 11 66C Can water be used as a refrigerant in air-conditioning applications? Explain C In absorption refrigeration cycles, why is the fluid in the absorber cooled and the fluid in the generator heated? 11 68C How is the coefficient of performance of an absorption refrigeration system defined? 11 69C What are the functions of the rectifier and the regenerator in an absorption refrigeration system? An absorption refrigeration system that receives heat from a source at 130 C and maintains the refrigerated space at 5 C is claimed to have a COP of. If the environment temperature is 7 C, can this claim be valid? Justify your answer An absorption refrigeration system receives heat from a source at 10 C and maintains the refrigerated space at 0 C. If the temperature of the environment is 5 C, what is the maximum COP this absorption refrigeration system can have? 11 7 Heat is supplied to an absorption refrigeration system from a geothermal well at 130 C at a rate of kj/h. The environment is at 5 C, and the refrigerated space is maintained at 30 C. Determine the maximum rate at which this system can remove heat from the refrigerated space. Answer: kj/h 3 FIGURE P11 63 Q H 1 Chapter E Heat is supplied to an absorption refrigeration system from a geothermal well at 50 F at a rate of 10 5 Btu/h. The environment is at 80 F, and the refrigerated space is maintained at 0 F. If the COP of the system is 0.55, determine the rate at which this system can remove heat from the refrigerated space A reversible absorption refrigerator consists of a reversible heat engine and a reversible refrigerator. The system removes heat from a cooled space at 10 C at a rate of kw. The refrigerator operates in an environment at 5 C. If the heat is supplied to the cycle by condensing saturated steam at 00 C, determine (a) the rate at which the steam condenses and (b) the power input to the reversible refrigerator. (c) If the COP of an actual absorption chiller at the same temperature limits has a COP of 0.7, determine the second law efficiency of this chiller. Answers: (a) kg/s, (b).93 kw, (c) 0.5 T s Rev. HE T 0 FIGURE P11 74 T 0 Rev. Ref. Special Topic: Thermoelectric Power Generation and Refrigeration Systems 11 75C What is a thermoelectric circuit? 11 76C Describe the Seebeck and the Peltier effects C Consider a circular copper wire formed by connecting the two ends of a copper wire. The connection point is now heated by a burning candle. Do you expect any current to flow through the wire? 11 78C An iron and a constantan wire are formed into a closed circuit by connecting the ends. Now both junctions are heated and are maintained at the same temperature. Do you expect any electric current to flow through this circuit? T L

166 644 Thermodynamics 11 79C A copper and a constantan wire are formed into a closed circuit by connecting the ends. Now one junction is heated by a burning candle while the other is maintained at room temperature. Do you expect any electric current to flow through this circuit? 11 80C How does a thermocouple work as a temperature measurement device? 11 81C Why are semiconductor materials preferable to metals in thermoelectric refrigerators? 11 8C Is the efficiency of a thermoelectric generator limited by the Carnot efficiency? Why? 11 83E A thermoelectric generator receives heat from a source at 340 F and rejects the waste heat to the environment at 90 F. What is the maximum thermal efficiency this thermoelectric generator can have? Answer: 31.3 percent A thermoelectric refrigerator removes heat from a refrigerated space at 5 C at a rate of 130 W and rejects it to an environment at 0 C. Determine the maximum coefficient of performance this thermoelectric refrigerator can have and the minimum required power input. Answers: 10.7, 1.1 W A thermoelectric cooler has a COP of 0.15 and removes heat from a refrigerated space at a rate of 180 W. Determine the required power input to the thermoelectric cooler, in W E A thermoelectric cooler has a COP of 0.15 and removes heat from a refrigerated space at a rate of 0 Btu/min. Determine the required power input to the thermoelectric cooler, in hp A thermoelectric refrigerator is powered by a 1-V car battery that draws 3 A of current when running. The refrigerator resembles a small ice chest and is claimed to cool nine canned drinks, L each, from 5 to 3 C in 1 h. Determine the average COP of this refrigerator. FIGURE P E Thermoelectric coolers that plug into the cigarette lighter of a car are commonly available. One such cooler is claimed to cool a 1-oz (0.771-lbm) drink from 78 to 38 F or to heat a cup of coffee from 75 to 130 F in about 15 min in a well-insulated cup holder. Assuming an average COP of 0. in the cooling mode, determine (a) the average rate of heat removal from the drink, (b) the average rate of heat supply to the coffee, and (c) the electric power drawn from the battery of the car, all in W It is proposed to run a thermoelectric generator in conjunction with a solar pond that can supply heat at a rate of 10 6 kj/h at 80 C. The waste heat is to be rejected to the environment at 30 C. What is the maximum power this thermoelectric generator can produce? Review Problems Consider a steady-flow Carnot refrigeration cycle that uses refrigerant-134a as the working fluid. The maximum and minimum temperatures in the cycle are 30 and 0 C, respectively. The quality of the refrigerant is 0.15 at the beginning of the heat absorption process and 0.80 at the end. Show the cycle on a T-s diagram relative to saturation lines, and determine (a) the coefficient of performance, (b) the condenser and evaporator pressures, and (c) the net work input A large refrigeration plant is to be maintained at 15 C, and it requires refrigeration at a rate of 100 kw. The condenser of the plant is to be cooled by liquid water, which experiences a temperature rise of 8 C as it flows over the coils of the condenser. Assuming the plant operates on the ideal vapor-compression cycle using refrigerant-134a between the pressure limits of 10 and 700 kpa, determine (a) the mass flow rate of the refrigerant, (b) the power input to the compressor, and (c) the mass flow rate of the cooling water Reconsider Prob Using EES (or other) software, investigate the effect of evaporator pressure on the COP and the power input. Let the evaporator pressure vary from 10 to 380 kpa. Plot the COP and the power input as functions of evaporator pressure, and discuss the results Repeat Prob assuming the compressor has an isentropic efficiency of 75 percent. Also, determine the rate of exergy destruction associated with the compression process in this case. Take T 0 5 C A heat pump that operates on the ideal vaporcompression cycle with refrigerant-134a is used to heat a house. The mass flow rate of the refrigerant is 0.3 kg/s. The condenser and evaporator pressures are 900 and 00 kpa, respectively. Show the cycle on a T-s diagram with respect to saturation lines, and determine (a) the rate of heat supply to the house, (b) the volume flow rate of the refrigerant at the compressor inlet, and (c) the COP of this heat pump Derive a relation for the COP of the two-stage refrigeration system with a flash chamber as shown in Fig in terms of the enthalpies and the quality at state 6. Consider a unit mass in the condenser Consider a two-stage compression refrigeration system operating between the pressure limits of 0.8 and

167 0.14 MPa. The working fluid is refrigerant-134a. The refrigerant leaves the condenser as a saturated liquid and is throttled to a flash chamber operating at 0.4 MPa. Part of the refrigerant evaporates during this flashing process, and this vapor is mixed with the refrigerant leaving the low-pressure compressor. The mixture is then compressed to the condenser pressure by the high-pressure compressor. The liquid in the flash chamber is throttled to the evaporator pressure, and it cools the refrigerated space as it vaporizes in the evaporator. Assuming the refrigerant leaves the evaporator as saturated vapor and both compressors are isentropic, determine (a) the fraction of the refrigerant that evaporates as it is throttled to the flash chamber, (b) the amount of heat removed from the refrigerated space and the compressor work per unit mass of refrigerant flowing through the condenser, and (c) the coefficient of performance. Answers: (a) 0.165, (b) kj/kg, 3.6 kj/kg, (c) An aircraft on the ground is to be cooled by a gas refrigeration cycle operating with air on an open cycle. Air enters the compressor at 30 C and 100 kpa and is compressed to 50 kpa. Air is cooled to 70 C before it enters the turbine. Assuming both the turbine and the compressor to be isentropic, determine the temperature of the air leaving the turbine and entering the cabin. Answer: 9 C Consider a regenerative gas refrigeration cycle using helium as the working fluid. Helium enters the compressor at 100 kpa and 10 C and is compressed to 300 kpa. Helium is then cooled to 0 C by water. It then enters the regenerator where it is cooled further before it enters the turbine. Helium leaves the refrigerated space at 5 C and enters the regenerator. Assuming both the turbine and the compressor to be isentropic, determine (a) the temperature of the helium at the turbine inlet, (b) the coefficient of performance of the cycle, and (c) the net power input required for a mass flow rate of 0.45 kg/s An absorption refrigeration system is to remove heat from the refrigerated space at 10 C at a rate of 1 kw while operating in an environment at 5 C. Heat is to be supplied from a solar pond at 85 C. What is the minimum rate of heat supply required? Answer: 9.53 kw Reconsider Prob Using EES (or other) software, investigate the effect of the source temperature on the minimum rate of heat supply. Let the source temperature vary from 50 to 50 C. Plot the minimum rate of heat supply as a function of source temperature, and discuss the results A typical 00-m house can be cooled adequately by a 3.5-ton air conditioner whose COP is 4.0. Determine the rate of heat gain of the house when the air conditioner is running continuously to maintain a constant temperature in the house Rooms with floor areas of up to 15-m are cooled adequately by window air conditioners whose cooling capacity Chapter is 5000 Btu/h. Assuming the COP of the air conditioner to be 3.5, determine the rate of heat gain of the room, in Btu/h, when the air conditioner is running continuously to maintain a constant room temperature A gas refrigeration system using air as the working fluid has a pressure ratio of 5. Air enters the compressor at 0 C. The high-pressure air is cooled to 35 C by rejecting heat to the surroundings. The refrigerant leaves the turbine at 80 C and enters the refrigerated space where it absorbs heat before entering the regenerator. The mass flow rate of air is 0.4 kg/s. Assuming isentropic efficiencies of 80 percent for the compressor and 85 percent for the turbine and using variable specific heats, determine (a) the effectiveness of the regenerator, (b) the rate of heat removal from the refrigerated space, and (c) the COP of the cycle. Also, determine (d) the refrigeration load and the COP if this system operated on the simple gas refrigeration cycle. Use the same compressor inlet temperature as given, the same turbine inlet temperature as calculated, and the same compressor and turbine efficiencies. 5 Q L 6 Heat exch. 4 Turbine Regenerator 3 FIGURE P Heat exch. Q H Compressor An air conditioner with refrigerant-134a as the working fluid is used to keep a room at 6 C by rejecting the waste heat to the outside air at 34 C. The room is gaining heat through the walls and the windows at a rate of 50 kj/min while the heat generated by the computer, TV, and lights amounts to 900 W. An unknown amount of heat is also generated by the people in the room. The condenser and evaporator pressures are 100 and 500 kpa, respectively. The refrigerant is saturated liquid at the condenser exit and saturated vapor at the compressor inlet. If the refrigerant enters the compressor at a rate of 100 L/min and the isentropic efficiency of the compressor is 75 percent, determine (a) the temperature of the refrigerant at the compressor exit, (b) the rate of heat generation by the people in the room, (c) the COP of the air 1

168 646 Thermodynamics 3 34 C Q H Condenser Expansion valve Compressor 100 kpa Evaporator 4 1 Q 500 kpa L 6 C conditioner, and (d) the minimum volume flow rate of the refrigerant at the compressor inlet for the same compressor inlet and exit conditions. Answers: (a) 54.5 C, (b) 670 W, (c) 5.87, (d) 15.7 L/min A heat pump water heater (HPWH) heats water by absorbing heat from the ambient air and transferring it to water. The heat pump has a COP of. and consumes kw of electricity when running. Determine if this heat pump can be used to meet the cooling needs of a room most of the time for free by absorbing heat from the air in the room. The rate of heat gain of a room is usually less than 5000 kj/h. Warm air from the room FIGURE P Cool air to the room FIGURE P W in Cold water in Water heater Hot water out The vortex tube (also known as a Ranque or Hirsch tube) is a device that produces a refrigeration effect by expanding pressurized gas such as air in a tube (instead of a turbine as in the reversed Brayton cycle). It was invented and patented by Ranque in 1931 and improved by Hirsch in 1945, and is commercially available in various sizes. The vortex tube is simply a straight circular tube equipped with a nozzle, as shown in the figure. The compressed gas at temperature T 1 and pressure P 1 is accelerated in the nozzle by expanding it to nearly atmospheric pressure and is introduced into the tube tangentially at a very high (typically supersonic) velocity to produce a swirling motion (vortex) within the tube. The rotating gas is allowed to exit through the full-size tube that extends to the right, and the mass flow rate is controlled by a valve located about 30 diameters downstream. A smaller amount of air at the core region is allowed to escape to the left through a small aperture at the center. It is observed that the gas that is in the core region and escapes through the central aperture is cold while the gas that is in the peripheral region and escapes through the full-size tube is hot. If the temperature and the mass flow rate of the cold stream are T c and m. c, respectively, the rate of refrigeration in the vortex tube can be expressed as Q # refrig,vortex tube m # c 1h 1 h c m # cc p 1T 1 T c where c p is the specific heat of the gas and T 1 T c is the temperature drop of the gas in the vortex tube (the cooling effect). Temperature drops as high as 60 C (or 108 F) are obtained at high pressure ratios of about 10. The coefficient of performance of a vortex tube can be defined as the ratio of the refrigeration rate as given above to the power used to compress the gas. It ranges from about 0.1 to 0.15, which is well below the COPs of ordinary vapor compression refrigerators. This interesting phenomenon can be explained as follows: the centrifugal force creates a radial pressure gradient in the vortex, and thus the gas at the periphery is pressurized and heated by the gas at the core region, which is cooled as a result. Also, energy is transferred from the inner layers toward the outer layers as the outer layers slow down the inner layers because of fluid viscosity that tends to produce a solid vortex. Both of these effects cause the energy and thus the temperature of the gas in the core region to decline. The conservation of energy requires the energy of the fluid at the outer layers to increase by an equivalent amount. The vortex tube has no moving parts, and thus it is inherently reliable and durable. The ready availability of the compressed air at pressures up to 10 atm in most industrial facilities makes the vortex tube particularly attractive in such settings. Despite its low efficiency, the vortex tube has found application in small-scale industrial spot-cooling operations such as cooling of soldered parts or critical electronic components, cooling drinking water, and cooling the suits of workers in hot environments.

169 Consider a vortex tube that receives compressed air at 500 kpa and 300 K and supplies 5 percent of it as cold air at 100 kpa and 78 K. The ambient air is at 300 K and 100 kpa, and the compressor has an isentropic efficiency of 80 percent. The air suffers a pressure drop of 35 kpa in the aftercooler and the compressed air lines between the compressor and the vortex tube. (a) Without performing any calculations, explain how the COP of the vortex tube would compare to the COP of an actual air refrigeration system based on the reversed Brayton cycle for the same pressure ratio. Also, compare the minimum temperatures that can be obtained by the two systems for the same inlet temperature and pressure. (b) Assuming the vortex tube to be adiabatic and using specific heats at room temperature, determine the exit temperature of the hot fluid stream. (c) Show, with calculations, that this process does not violate the second law of thermodynamics. (d) Determine the coefficient of performance of this refrigeration system, and compare it to the COP of a Carnot refrigerator. Cold air Compressed air FIGURE P Warm air Repeat Prob for a pressure of 600 kpa at the vortex tube intake Using EES (or other) software, investigate the effect of the evaporator pressure on the COP of an ideal vapor-compression refrigeration cycle with R-134a as the working fluid. Assume the condenser pressure is kept constant at 1 MPa while the evaporator pressure is varied from 100 kpa to 500 kpa. Plot the COP of the refrigeration cycle against the evaporator pressure, and discuss the results Using EES (or other) software, investigate the effect of the condenser pressure on the COP of an ideal vapor-compression refrigeration cycle with R-134a as the working fluid. Assume the evaporator pressure is kept constant at 10 kpa while the condenser pressure is varied from 400 to 1400 kpa. Plot the COP of the refrigeration cycle against the condenser pressure, and discuss the results. Chapter Fundamentals of Engineering (FE) Exam Problems Consider a heat pump that operates on the reversed Carnot cycle with R-134a as the working fluid executed under the saturation dome between the pressure limits of 140 and 800 kpa. R-134a changes from saturated vapor to saturated liquid during the heat rejection process. The net work input for this cycle is (a) 8 kj/kg (b) 34 kj/kg (c) 49 kj/kg (d) 144 kj/kg (e) 75 kj/kg A refrigerator removes heat from a refrigerated space at 5 C at a rate of 0.35 kj/s and rejects it to an environment at 0 C. The minimum required power input is (a) 30 W (b) 33 W (c) 56 W (d) 14 W (e) 350 W A refrigerator operates on the ideal vapor compression refrigeration cycle with R-134a as the working fluid between the pressure limits of 10 and 800 kpa. If the rate of heat removal from the refrigerated space is 3 kj/s, the mass flow rate of the refrigerant is (a) 0.19 kg/s (b) 0.15 kg/s (c) 0.3 kg/s (d) 0.8 kg/s (e) 0.81 kg/s A heat pump operates on the ideal vapor compression refrigeration cycle with R-134a as the working fluid between the pressure limits of 0.3 and 1. MPa. If the mass flow rate of the refrigerant is kg/s, the rate of heat supply by the heat pump to the heated space is (a) 3.3 kw (b) 3 kw (c) 6 kw (d) 31 kw (e) 45 kw An ideal vapor compression refrigeration cycle with R-134a as the working fluid operates between the pressure limits of 10 kpa and 1000 kpa. The mass fraction of the refrigerant that is in the liquid phase at the inlet of the evaporator is (a) 0.65 (b) 0.60 (c) 0.40 (d) 0.55 (e) Consider a heat pump that operates on the ideal vapor compression refrigeration cycle with R-134a as the working fluid between the pressure limits of 0.3 and 1. MPa. The coefficient of performance of this heat pump is (a) 0.17 (b) 1. (c) 3.1 (d) 4.9 (e) An ideal gas refrigeration cycle using air as the working fluid operates between the pressure limits of 80 and 80 kpa. Air is cooled to 35 C before entering the turbine. The lowest temperature of this cycle is (a) 58 C (b) 6 C (c) 5 C (d) 11 C (e) 4 C Consider an ideal gas refrigeration cycle using helium as the working fluid. Helium enters the compressor at 100 kpa and 10 C and compressed to 50 kpa. Helium is

170 648 Thermodynamics then cooled to 0 C before it enters the turbine. For a mass flow rate of 0. kg/s, the net power input required is (a) 9.3 kw (b) 7.6 kw (c) 48.8 kw (d) 93.5 kw (e) 119 kw An absorption air-conditioning system is to remove heat from the conditioned space at 0 C at a rate of 150 kj/s while operating in an environment at 35 C. Heat is to be supplied from a geothermal source at 140 C. The minimum rate of heat supply is (a) 86 kj/s (b) 1 kj/s (c) 30 kj/s (d) 61 kj/s (e) 150 kj/s Consider a refrigerator that operates on the vapor compression refrigeration cycle with R-134a as the working fluid. The refrigerant enters the compressor as saturated vapor at 160 kpa, and exits at 800 kpa and 50 C, and leaves the condenser as saturated liquid at 800 kpa. The coefficient of performance of this refrigerator is (a).6 (b) 1.0 (c) 4. (d) 3. (e) 4.4 Design and Essay Problems Design a vapor-compression refrigeration system that will maintain the refrigerated space at 15 C while operating in an environment at 0 C using refrigerant-134a as the working fluid Write an essay on air-, water-, and soil-based heat pumps. Discuss the advantages and the disadvantages of each system. For each system identify the conditions under which that system is preferable over the other two. In what situations would you not recommend a heat pump heating system? 11 1 Consider a solar pond power plant operating on a closed Rankine cycle. Using refrigerant-134a as the working fluid, specify the operating temperatures and pressures in the cycle, and estimate the required mass flow rate of refrigerant- 134a for a net power output of 50 kw. Also, estimate the surface area of the pond for this level of continuous power production. Assume that the solar energy is incident on the pond at a rate of 500 W per m of pond area at noontime, and that the pond is capable of storing 15 percent of the incident solar energy in the storage zone Design a thermoelectric refrigerator that is capable of cooling a canned drink in a car. The refrigerator is to be powered by the cigarette lighter of the car. Draw a sketch of your design. Semiconductor components for building thermoelectric power generators or refrigerators are available from several manufacturers. Using data from one of these manufacturers, determine how many of these components you need in your design, and estimate the coefficient of performance of your system. A critical problem in the design of thermoelectric refrigerators is the effective rejection of waste heat. Discuss how you can enhance the rate of heat rejection without using any devices with moving parts such as a fan It is proposed to use a solar-powered thermoelectric system installed on the roof to cool residential buildings. The system consists of a thermoelectric refrigerator that is powered by a thermoelectric power generator whose top surface is a solar collector. Discuss the feasibility and the cost of such a system, and determine if the proposed system installed on one side of the roof can meet a significant portion of the cooling requirements of a typical house in your area. Waste heat Thermoelectric generator Thermoelectric refrigerator Solar energy FIGURE P11 14 SUN Electric current A refrigerator using R-1 as the working fluid keeps the refrigerated space at 15 C in an environment at 30 C. You are asked to redesign this refrigerator by replacing R-1 with the ozone-friendly R-134a. What changes in the pressure levels would you suggest in the new system? How do you think the COP of the new system will compare to the COP of the old system? In the 1800s, before the development of modern air-conditioning, it was proposed to cool air for buildings with the following procedure using a large piston cylinder device [ John Gorrie: Pioneer of Cooling and Ice Making, ASHRAE Journal 33, no. 1 (Jan. 1991)]: 1. Pull in a charge of outdoor air.. Compress it to a high pressure. 3. Cool the charge of air using outdoor air. 4. Expand it back to atmospheric pressure. 5. Discharge the charge of air into the space to be cooled. Suppose the goal is to cool a room 6 m 10 m.5 m. Outdoor air is at 30 C, and it has been determined that 10 air changes per hour supplied to the room at 10 C could provide adequate cooling. Do a preliminary design of the system and

171 Chapter do calculations to see if it would be feasible. (You may make optimistic assumptions for the analysis.) (a) Sketch the system showing how you will drive it and how step 3 will be accomplished. (b) Determine what pressure will be required (step ). (c) Estimate (guess) how long step 3 will take and what size will be needed for the piston cylinder to provide the required air changes and temperature. (d) Determine the work required in step for one cycle and per hour. (e) Discuss any problems you see with the concept of your design. (Include discussion of any changes that may be required to offset optimistic assumptions.) Solar or photovoltaic (PV) cells convert sunlight to electricity and are commonly used to power calculators, satellites, remote communication systems, and even pumps. The conversion of light to electricity is called the photoelectric effect. It was first discovered in 1839 by Frenchman Edmond Becquerel, and the first PV module, which consisted of several cells connected to each other, was built in 1954 by Bell Laboratories. The PV modules today have conversion efficiencies of about 1 to 15 percent. Noting that the solar energy incident on a normal surface on earth at noontime is about 1000 W/m during a clear day, PV modules on a 1-m surface can provide as much as 150 W of electricity. The annual average daily solar energy incident on a horizontal surface in the United States ranges from about to 6 kwh/m. A PV-powered pump is to be used in Arizona to pump water for wildlife from a depth of 180 m at an average rate of 400 L/day. Assuming a reasonable efficiency for the pumping system, which can be defined as the ratio of the increase in the potential energy of the water to the electrical energy consumed by the pump, and taking the conversion efficiency of the PV cells to be 0.13 to be on the conservative side, determine the size of the PV module that needs to be installed, in m The temperature in a car parked in the sun can approach 100 C when the outside air temperature is just 5 C, and it is desirable to ventilate the parked car to avoid such high temperatures. However, the ventilating fans may run down the battery if they are powered by it. To avoid that happening, it is proposed to use the PV cells discussed in the preceding problem to power the fans. It is determined that the air in the car should be replaced once every minute to avoid excessive rise in the interior temperature. Determine if this can be accomplished by installing PV cells on part of the roof of the car. Also, find out if any car is currently ventilated this way. Solar panels Solar energy Sun Solar-powered exhaust fan PV panel Water FIGURE P11 17 PV-powered pump FIGURE P A company owns a refrigeration system whose refrigeration capacity is 00 tons (1 ton of refrigeration 11 kj/min), and you are to design a forced-air cooling system for fruits whose diameters do not exceed 7 cm under the following conditions: The fruits are to be cooled from 8 C to an average temperature of 8 C. The air temperature is to remain above C and below 10 C at all times, and the velocity of air approaching the fruits must remain under m/s. The cooling section can be as wide as 3.5 m and as high as m. Assuming reasonable values for the average fruit density, specific heat, and porosity (the fraction of air volume in a box), recommend reasonable values for (a) the air velocity approaching the cooling section, (b) the product-cooling capacity of the system, in kg fruit/h, and (c) the volume flow rate of air.

172

173 Chapter 1 THERMODYNAMIC PROPERTY RELATIONS In the preceding chapters we made extensive use of the property tables. We tend to take the property tables for granted, but thermodynamic laws and principles are of little use to engineers without them. In this chapter, we focus our attention on how the property tables are prepared and how some unknown properties can be determined from limited available data. It will come as no surprise that some properties such as temperature, pressure, volume, and mass can be measured directly. Other properties such as density and specific volume can be determined from these using some simple relations. However, properties such as internal energy, enthalpy, and entropy are not so easy to determine because they cannot be measured directly or related to easily measurable properties through some simple relations. Therefore, it is essential that we develop some fundamental relations between commonly encountered thermodynamic properties and express the properties that cannot be measured directly in terms of easily measurable properties. By the nature of the material, this chapter makes extensive use of partial derivatives. Therefore, we start by reviewing them. Then we develop the Maxwell relations, which form the basis for many thermodynamic relations. Next we discuss the Clapeyron equation, which enables us to determine the enthalpy of vaporization from P, v, and T measurements alone, and we develop general relations for c v, c p, du, dh, and ds that are valid for all pure substances under all conditions. Then we discuss the Joule-Thomson coefficient, which is a measure of the temperature change with pressure during a throttling process. Finally, we develop a method of evaluating the h, u, and s of real gases through the use of generalized enthalpy and entropy departure charts. Objectives The objectives of Chapter 1 are to: Develop fundamental relations between commonly encountered thermodynamic properties and express the properties that cannot be measured directly in terms of easily measurable properties. Develop the Maxwell relations, which form the basis for many thermodynamic relations. Develop the Clapeyron equation and determine the enthalpy of vaporization from P, v, and T measurements alone. Develop general relations for c v, c p, du, dh, and ds that are valid for all pure substances. Discuss the Joule-Thomson coefficient. Develop a method of evaluating the h, u, and s of real gases through the use of generalized enthalpy and entropy departure charts. 651

174 65 Thermodynamics f(x) (x+ x) f(x) Slope x x + x f FIGURE 1 1 The derivative of a function at a specified point represents the slope of the function at that point. x x 1 1 A LITTLE MATH PARTIAL DERIVATIVES AND ASSOCIATED RELATIONS Many of the expressions developed in this chapter are based on the state postulate, which expresses that the state of a simple, compressible substance is completely specified by any two independent, intensive properties. All other properties at that state can be expressed in terms of those two properties. Mathematically speaking, where x and y are the two independent properties that fix the state and z represents any other property. Most basic thermodynamic relations involve differentials. Therefore, we start by reviewing the derivatives and various relations among derivatives to the extent necessary in this chapter. Consider a function f that depends on a single variable x, that is, f f (x). Figure 1 1 shows such a function that starts out flat but gets rather steep as x increases. The steepness of the curve is a measure of the degree of dependence of f on x. In our case, the function f depends on x more strongly at larger x values. The steepness of a curve at a point is measured by the slope of a line tangent to the curve at that point, and it is equivalent to the derivative of the function at that point defined as df dx z z 1x, y f f 1x x f 1x lim lim xs0 x xs0 x (1 1) Therefore, the derivative of a function f(x) with respect to x represents the rate of change of f with x. EXAMPLE 1 1 Approximating Differential Quantities by Differences h(t ), kj/kg Slope = c p (T ) FIGURE 1 Schematic for Example 1 1. T, K The c p of ideal gases depends on temperature only, and it is expressed as c p (T ) dh(t )/dt. Determine the c p of air at 300 K, using the enthalpy data from Table A 17, and compare it to the value listed in Table A b. Solution The c p value of air at a specified temperature is to be determined using enthalpy data. Analysis The c p value of air at 300 K is listed in Table A b to be kj/kg K. This value could also be determined by differentiating the function h(t ) with respect to T and evaluating the result at T 300 K. However, the function h(t ) is not available. But, we can still determine the c p value approximately by replacing the differentials in the c p (T ) relation by differences in the neighborhood of the specified point (Fig. 1 ): dh 1T c p 1300 K c dt d T 300 K h 1T c T d h 1305 K h 195 K T 300 K K kj>kg kj/kg # K K Discussion Note that the calculated c p value is identical to the listed value. Therefore, differential quantities can be viewed as differences. They can

175 even be replaced by differences, whenever necessary, to obtain approximate results. The widely used finite difference numerical method is based on this simple principle. z Chapter z ( x ) y Partial Differentials Now consider a function that depends on two (or more) variables, such as z z(x, y). This time the value of z depends on both x and y. It is sometimes desirable to examine the dependence of z on only one of the variables. This is done by allowing one variable to change while holding the others constant and observing the change in the function. The variation of z(x, y) with x when y is held constant is called the partial derivative of z with respect to x, and it is expressed as (1 ) This is illustrated in Fig The symbol represents differential changes, just like the symbol d. They differ in that the symbol d represents the total differential change of a function and reflects the influence of all variables, whereas represents the partial differential change due to the variation of a single variable. Note that the changes indicated by d and are identical for independent variables, but not for dependent variables. For example, ( x) y dx but ( z) y dz. [In our case, dz ( z) x ( z) y.] Also note that the value of the partial derivative ( z/ x) y, in general, is different at different y values. To obtain a relation for the total differential change in z(x, y) for simultaneous changes in x and y, consider a small portion of the surface z(x, y) shown in Fig When the independent variables x and y change by x and y, respectively, the dependent variable z changes by z, which can be expressed as Adding and subtracting z(x, y y), we get or a 0z 0x b y z lim a xs0 x b z 1x x, y z 1x, y lim y xs0 x z z 1x x, y y z 1x, y z z 1x x, y y z 1x, y y z 1x, y y z 1x, y z 1x x, y y z 1x, y y z 1x, y y z 1x, y z x y x y Taking the limits as x 0 and y 0 and using the definitions of partial derivatives, we obtain dz a 0z 0x b dx a 0z y 0y b dy x (1 3) Equation 1 3 is the fundamental relation for the total differential of a dependent variable in terms of its partial derivatives with respect to the independent variables. This relation can easily be extended to include more independent variables. x x FIGURE 1 3 Geometric representation of partial derivative ( z/ x) y. z x + x, y z(x, y) z(x + x, y + y) x + x, y + y y x, y + y FIGURE 1 4 Geometric representation of total derivative dz for a function z(x, y). y

176 654 Thermodynamics EXAMPLE 1 Total Differential versus Partial Differential Consider air at 300 K and 0.86 m 3 /kg. The state of air changes to 30 K and 0.87 m 3 /kg as a result of some disturbance. Using Eq. 1 3, estimate the change in the pressure of air. P, kpa ( P) T = T, K 300 ( P) v = dp = FIGURE 1 5 Geometric representation of the disturbance discussed in Example 1. v, m 3 /kg Solution The temperature and specific volume of air changes slightly during a process. The resulting change in pressure is to be determined. Assumptions Air is an ideal gas. Analysis Strictly speaking, Eq. 1 3 is valid for differential changes in variables. However, it can also be used with reasonable accuracy if these changes are small. The changes in T and v, respectively, can be expressed as and An ideal gas obeys the relation Pv RT. Solving for P yields Note that R is a constant and P P(T, v). Applying Eq. 1 3 and using average values for T and v, Therefore, the pressure will decrease by kpa as a result of this disturbance. Notice that if the temperature had remained constant (dt 0), the pressure would decrease by kpa as a result of the 0.01 m 3 /kg increase in specific volume. However, if the specific volume had remained constant (dv 0), the pressure would increase by kpa as a result of the -K rise in temperature (Fig. 1 5). That is, and dp a 0P 0T b dt a 0P v 0v b dv R dt RT dv T v v kpa # m 3 >kg # Kc kpa kpa kpa dt T K K dv v m 3 >kg 0.01 m 3 >kg P RT v K m 3 >kg 1301 K10.01 m3 >kg d m 3 >kg a 0P 0T b dt 10P v kpa v a 0P 0v b dv 10P T kpa T dp 10P v 10P T kpa Discussion Of course, we could have solved this problem easily (and exactly) by evaluating the pressure from the ideal-gas relation P RT/v at the final state (30 K and 0.87 m 3 /kg) and the initial state (300 K and 0.86 m 3 /kg) and taking their difference. This yields kpa, which is exactly the value obtained above. Thus the small finite quantities ( K, 0.01 m 3 /kg) can be approximated as differential quantities with reasonable accuracy.

177 Partial Differential Relations Now let us rewrite Eq. 1 3 as where (1 4) Taking the partial derivative of M with respect to y and of N with respect to x yields The order of differentiation is immaterial for properties since they are continuous point functions and have exact differentials. Therefore, the two relations above are identical: a 0M (1 5) 0y b a 0N x 0x b y This is an important relation for partial derivatives, and it is used in calculus to test whether a differential dz is exact or inexact. In thermodynamics, this relation forms the basis for the development of the Maxwell relations discussed in the next section. Finally, we develop two important relations for partial derivatives the reciprocity and the cyclic relations. The function z z(x, y) can also be expressed as x x(y, z) if y and z are taken to be the independent variables. Then the total differential of x becomes, from Eq. 1 3, Eliminating dx by combining Eqs. 1 3 and 1 6, we have Rearranging, a 0M 0y b x dz M dx N dy M a 0z 0x b and N a 0z y 0y b x 0 z 0N and a 0x 0y 0x b y dx a 0x 0y b dy a 0x z 0z b dz y 0 z 0y 0x dz ca 0z 0x b a 0x y 0y b a 0z z 0y b d dy a 0x x 0z b a 0z y 0x b dz y ca 0z 0x b a 0x y 0y b a 0z z 0y b d dy c 1 a 0x x 0z b a 0z y 0x b d dz y (1 6) (1 7) Chapter The variables y and z are independent of each other and thus can be varied independently. For example, y can be held constant (dy 0), and z can be varied over a range of values (dz 0). Therefore, for this equation to be valid at all times, the terms in the brackets must equal zero, regardless of the values of y and z. Setting the terms in each bracket equal to zero gives a 0x 0z b a 0z y 0x b 1 S a 0x y 0z b y a 0z 0x b a 0x y 0y b a 0x z 0y b S a 0x x 0y b a 0y z 0z b a 0z x 0x b 1 y 1 10z>0x y (1 8) (1 9)

178 656 Thermodynamics Function: z + xy 3y z = 0 ( ) y xy 1) z = z = 3y 1 x ( ) y ) x = 3y z z x y z = Thus, z 1 ( = x) y x z ( ) y FIGURE 1 6 Demonstration of the reciprocity relation for the function z xy 3y z 0. y 3y 1 3y 1 y FIGURE 1 7 Partial differentials are powerful tools that are supposed to make life easier, not harder. Reprinted with special permission of King Features Syndicate. The first relation is called the reciprocity relation, and it shows that the inverse of a partial derivative is equal to its reciprocal (Fig. 1 6). The second relation is called the cyclic relation, and it is frequently used in thermodynamics (Fig. 1 7). EXAMPLE 1 3 Verification of Cyclic and Reciprocity Relations Using the ideal-gas equation of state, verify (a) the cyclic relation and (b) the reciprocity relation at constant P. Solution The cyclic and reciprocity relations are to be verified for an ideal gas. Analysis The ideal-gas equation of state Pv RT involves the three variables P, v, and T. Any two of these can be taken as the independent variables, with the remaining one being the dependent variable. (a) Replacing x, y, and z in Eq. 1 9 by P, v, and T, respectively, we can express the cyclic relation for an ideal gas as where Substituting yields which is the desired result. (b) The reciprocity rule for an ideal gas at P constant can be expressed as Performing the differentiations and substituting, we have Thus the proof is complete. a 0P 0v b a 0v T 0T b a 0T P 0P b 1 v P P 1v, T RT v v v 1P, T RT P T T 1P, v Pv R a RT v bar P bav R b RT Pv 1 a 0v 0T b P S a 0P 0v b RT T v S a 0v 0T b P R P S a 0T 0P b v v R 1 10T>0v P R P 1 P>R S R P R P 1 THE MAXWELL RELATIONS The equations that relate the partial derivatives of properties P, v, T, and s of a simple compressible system to each other are called the Maxwell relations. They are obtained from the four Gibbs equations by exploiting the exactness of the differentials of thermodynamic properties.

179 Two of the Gibbs relations were derived in Chap. 7 and expressed as du T ds P dv (1 10) dh T ds v dp (1 11) The other two Gibbs relations are based on two new combination properties the Helmholtz function a and the Gibbs function g, defined as a u Ts (1 1) g h Ts (1 13) Differentiating, we get da du T ds s dt dg dh T ds s dt Simplifying the above relations by using Eqs and 1 11, we obtain the other two Gibbs relations for simple compressible systems: da s dt P dv (1 14) dg s dt v dp (1 15) A careful examination of the four Gibbs relations reveals that they are of the form with dz M dx N dy a 0M 0y b a 0N x 0x b y (1 4) (1 5) since u, h, a, and g are properties and thus have exact differentials. Applying Eq. 1 5 to each of them, we obtain a 0T 0v b a 0P s 0s b v a 0T 0P b a 0v s 0s b P a 0s 0v b a 0P T 0T b v a 0s 0P b a 0v T 0T b P (1 16) (1 17) (1 18) (1 19) These are called the Maxwell relations (Fig. 1 8). They are extremely valuable in thermodynamics because they provide a means of determining the change in entropy, which cannot be measured directly, by simply measuring the changes in properties P, v, and T. Note that the Maxwell relations given above are limited to simple compressible systems. However, other similar relations can be written just as easily for nonsimple systems such as those involving electrical, magnetic, and other effects. Chapter T ( = ( P v ) s ) s v T ( ) = v P ( ) s s P ( ) T ( ) P s = v T v s v ( P) = ( ) T T P FIGURE 1 8 Maxwell relations are extremely valuable in thermodynamic analysis.

180 658 Thermodynamics EXAMPLE 1 4 Verification of the Maxwell Relations Verify the validity of the last Maxwell relation (Eq. 1 19) for steam at 50 C and 300 kpa. Solution The validity of the last Maxwell relation is to be verified for steam at a specified state. Analysis The last Maxwell relation states that for a simple compressible substance, the change in entropy with pressure at constant temperature is equal to the negative of the change in specific volume with temperature at constant pressure. If we had explicit analytical relations for the entropy and specific volume of steam in terms of other properties, we could easily verify this by performing the indicated derivations. However, all we have for steam are tables of properties listed at certain intervals. Therefore, the only course we can take to solve this problem is to replace the differential quantities in Eq with corresponding finite quantities, using property values from the tables (Table A 6 in this case) at or about the specified state. P s s 400 kpa s 00 (400 00) kpa P s? T? T 50 C? T 50 C ( ) kj kg # K (400 00) kpa? a 0v 0T b P a 0v 0T b P 300 kpa c v 300 C v 00 C (300 00) C d m 3 /kg K m 3 /kg K P 300 kpa ( ) m 3 kg (300 00) C since kj kpa m 3 and K C for temperature differences. The two values are within 4 percent of each other. This difference is due to replacing the differential quantities by relatively large finite quantities. Based on the close agreement between the two values, the steam seems to satisfy Eq at the specified state. Discussion This example shows that the entropy change of a simple compressible system during an isothermal process can be determined from a knowledge of the easily measurable properties P, v, and T alone. 1 3 THE CLAPEYRON EQUATION The Maxwell relations have far-reaching implications in thermodynamics and are frequently used to derive useful thermodynamic relations. The Clapeyron equation is one such relation, and it enables us to determine the enthalpy change associated with a phase change (such as the enthalpy of vaporization h fg ) from a knowledge of P, v, and T data alone. Consider the third Maxwell relation, Eq. 1 18: a 0P 0T b a 0s v 0v b T During a phase-change process, the pressure is the saturation pressure, which depends on the temperature only and is independent of the specific

181 volume. That is, P sat f (T sat ). Therefore, the partial derivative ( P/ T ) v can be expressed as a total derivative (dp/dt ) sat, which is the slope of the saturation curve on a P-T diagram at a specified saturation state (Fig. 1 9). This slope is independent of the specific volume, and thus it can be treated as a constant during the integration of Eq between two saturation states at the same temperature. For an isothermal liquid vapor phase-change process, for example, the integration yields or s g s f a dp dt b 1v g v f sat a dp dt b s fg sat v fg (1 0) (1 1) During this process the pressure also remains constant. Therefore, from Eq. 1 11, 0 g g dh T ds v dp S dh T ds S h fg Ts fg Substituting this result into Eq. 1 1, we obtain a dp dt b h fg sat Tv fg (1 ) which is called the Clapeyron equation after the French engineer and physicist E. Clapeyron ( ). This is an important thermodynamic relation since it enables us to determine the enthalpy of vaporization h fg at a given temperature by simply measuring the slope of the saturation curve on a P-T diagram and the specific volume of saturated liquid and saturated vapor at the given temperature. The Clapeyron equation is applicable to any phase-change process that occurs at constant temperature and pressure. It can be expressed in a general form as a dp dt b h 1 sat Tv 1 where the subscripts 1 and indicate the two phases. f f (1 3) P SOLID Chapter LIQUID VAPOR T P ( ) = const. T sat FIGURE 1 9 The slope of the saturation curve on a P-T diagram is constant at a constant T or P. T EXAMPLE 1 5 Evaluating the h fg of a Substance from the P-v-T Data Using the Clapeyron equation, estimate the value of the enthalpy of vaporization of refrigerant-134a at 0 C, and compare it with the tabulated value. Solution The h fg of refrigerant-134a is to be determined using the Clapeyron equation. Analysis From Eq. 1, h fg Tv fg a dp dt b sat

182 660 Thermodynamics where, from Table A 11, v fg 1v g v 0 C m 3 >kg a dp d T b a P sat,0 C T b P 4 C P 16 C sat,0 C 4 C 16 C kpa 8 C since T( C) T(K). Substituting, we get 1 kj h fg K m 3 >kg kpa>k a 1 kpa # b m kj/kg kpa>k The tabulated value of h fg at 0 C is 18.7 kj/kg. The small difference between the two values is due to the approximation used in determining the slope of the saturation curve at 0 C. The Clapeyron equation can be simplified for liquid vapor and solid vapor phase changes by utilizing some approximations. At low pressures v g v f, and thus v fg v g. By treating the vapor as an ideal gas, we have v g RT/P. Substituting these approximations into Eq. 1, we find or a dp dt b Ph fg sat RT a dp P b sat h fg R a dt T b sat For small temperature intervals h fg can be treated as a constant at some average value. Then integrating this equation between two saturation states yields ln a P b h fg P 1 sat R a 1 1 b T 1 T sat (1 4) This equation is called the Clapeyron Clausius equation, and it can be used to determine the variation of saturation pressure with temperature. It can also be used in the solid vapor region by replacing h fg by h ig (the enthalpy of sublimation) of the substance. EXAMPLE 1 6 Extrapolating Tabular Data with the Clapeyron Equation Estimate the saturation pressure of refrigerant-134a at 50 F, using the data available in the refrigerant tables. Solution The saturation pressure of refrigerant-134a is to be determined using other tabulated data. Analysis Table A 11E lists saturation data at temperatures 40 F and above. Therefore, we should either resort to other sources or use extrapolation

183 Chapter to obtain saturation data at lower temperatures. Equation 1 4 provides an intelligent way to extrapolate: ln a P b h fg P 1 sat R a 1 1 b T 1 T sat In our case T 1 40 F and T 50 F. For refrigerant-134a, R Btu/lbm R. Also from Table A 11E at 40 F, we read h fg Btu/lbm and P 1 P 40 F 7.43 psia. Substituting these values into Eq. 1 4 gives P Btu>lbm ln a b 7.43 psia Btu>lbm # R a 1 40 R R b P 5.56 psia Therefore, according to Eq. 1 4, the saturation pressure of refrigerant-134a at 50 F is 5.56 psia. The actual value, obtained from another source, is psia. Thus the value predicted by Eq. 1 4 is in error by about 1 percent, which is quite acceptable for most purposes. (If we had used linear extrapolation instead, we would have obtained psia, which is in error by 7 percent.) 1 4 GENERAL RELATIONS FOR du, dh, ds, c v, AND c p The state postulate established that the state of a simple compressible system is completely specified by two independent, intensive properties. Therefore, at least theoretically, we should be able to calculate all the properties of a system at any state once two independent, intensive properties are available. This is certainly good news for properties that cannot be measured directly such as internal energy, enthalpy, and entropy. However, the calculation of these properties from measurable ones depends on the availability of simple and accurate relations between the two groups. In this section we develop general relations for changes in internal energy, enthalpy, and entropy in terms of pressure, specific volume, temperature, and specific heats alone. We also develop some general relations involving specific heats. The relations developed will enable us to determine the changes in these properties. The property values at specified states can be determined only after the selection of a reference state, the choice of which is quite arbitrary. Internal Energy Changes We choose the internal energy to be a function of T and v; that is, u u(t, v) and take its total differential (Eq. 1 3): Using the definition of c v, we have du a 0u 0T b dt a 0u v 0v b dv T du c v dt a 0u 0v b dv T (1 5)

184 66 Thermodynamics Now we choose the entropy to be a function of T and v; that is, s s(t, v) and take its total differential, ds a 0s 0T b dt a 0s v 0v b dv T Substituting this into the Tdsrelation du Tds Pdv yields du T a 0s 0T b dt c T a 0s v 0v b P d dv T Equating the coefficients of dt and dv in Eqs. 1 5 and 1 7 gives a 0s 0T b c v v T a 0u 0v b T a 0s T 0v b P T Using the third Maxwell relation (Eq. 1 18), we get a 0u 0v b T a 0P T 0T b P v Substituting this into Eq. 1 5, we obtain the desired relation for du: du c v dt c T a 0P 0T b P d dv v (1 6) (1 7) (1 8) (1 9) The change in internal energy of a simple compressible system associated with a change of state from (T 1, v 1 ) to (T, v ) is determined by integration: T v u u 1 c v dt T 1 v 1 c T a 0P 0T b v P d dv (1 30) Enthalpy Changes The general relation for dh is determined in exactly the same manner. This time we choose the enthalpy to be a function of T and P, that is, h h(t, P), and take its total differential, Using the definition of c p, we have dh a 0h 0T b dt a 0h P 0P b dp T dh c p dt a 0h 0P b dp T (1 31) Now we choose the entropy to be a function of T and P; that is, we take s s(t, P) and take its total differential, ds a 0s 0T b dt a 0s P 0P b dp T Substituting this into the T ds relation dh T ds v dp gives dh T a 0s 0T b dt c v T a 0s P 0P b d dp T (1 3) (1 33)

185 Equating the coefficients of dt and dp in Eqs and 1 33, we obtain a 0s 0T b c p P T a 0h 0P b v T a 0s T 0P b T Using the fourth Maxwell relation (Eq. 1 19), we have a 0h 0P b v T a 0v T 0T b P Substituting this into Eq. 1 31, we obtain the desired relation for dh: dh c p d T c v T a 0v 0T b d dp P (1 34) (1 35) The change in enthalpy of a simple compressible system associated with a change of state from (T 1, P 1 ) to (T, P ) is determined by integration: T P h h 1 c p dt T 1 (1 36) In reality, one needs only to determine either u u 1 from Eq or h h 1 from Eq. 1 36, depending on which is more suitable to the data at hand. The other can easily be determined by using the definition of enthalpy h u Pv: h h 1 u u 1 1P v P 1 v 1 P 1 c v T a 0v 0T b d dp P (1 37) Chapter Entropy Changes Below we develop two general relations for the entropy change of a simple compressible system. The first relation is obtained by replacing the first partial derivative in the total differential ds (Eq. 1 6) by Eq. 1 8 and the second partial derivative by the third Maxwell relation (Eq. 1 18), yielding and (1 38) (1 39) The second relation is obtained by replacing the first partial derivative in the total differential of ds (Eq. 1 3) by Eq. 1 34, and the second partial derivative by the fourth Maxwell relation (Eq. 1 19), yielding and ds c v T s s 1 T T 1 ds c P T c v T dt v 0v dt a 0T b dp P c p 0P dt a 0T b dv v (1 40) T s s 1 (1 41) T dt T 1 P a 0v 0T b dp P P 1 Either relation can be used to determine the entropy change. The proper choice depends on the available data. v 1 a 0P 0T b dv v

186 664 Thermodynamics Specific Heats c v and c p Recall that the specific heats of an ideal gas depend on temperature only. For a general pure substance, however, the specific heats depend on specific volume or pressure as well as the temperature. Below we develop some general relations to relate the specific heats of a substance to pressure, specific volume, and temperature. At low pressures gases behave as ideal gases, and their specific heats essentially depend on temperature only. These specific heats are called zero pressure, or ideal-gas, specific heats (denoted c v 0 and c p0 ), and they are relatively easier to determine. Thus it is desirable to have some general relations that enable us to calculate the specific heats at higher pressures (or lower specific volumes) from a knowledge of c v 0 or c p 0 and the P-v-T behavior of the substance. Such relations are obtained by applying the test of exactness (Eq. 1 5) on Eqs and 1 40, which yields and a 0c v 0v b T a 0 P T 0T b v (1 4) a 0c p (1 43) 0P b Ta 0 v T 0T b P The deviation of c p from c p 0 with increasing pressure, for example, is determined by integrating Eq from zero pressure to any pressure P along an isothermal path: 1c p c p0 T T P (1 44) The integration on the right-hand side requires a knowledge of the P-v-T behavior of the substance alone. The notation indicates that v should be differentiated twice with respect to T while P is held constant. The resulting expression should be integrated with respect to P while T is held constant. Another desirable general relation involving specific heats is one that relates the two specific heats c p and c v. The advantage of such a relation is obvious: We will need to determine only one specific heat (usually c p ) and calculate the other one using that relation and the P-v-T data of the substance. We start the development of such a relation by equating the two ds relations (Eqs and 1 40) and solving for dt: dt T 10P>0T v c p c v dv T 10v>0T P c p c v Choosing T T(v, P) and differentiating, we get dt a 0T 0v b dv a 0T P 0P b dp v Equating the coefficient of either dv or dp of the above two equations gives the desired result: c p c v T a 0v 0T b a 0P P 0T b v 0 a 0 v 0T b dp P dp (1 45)

187 An alternative form of this relation is obtained by using the cyclic relation: Substituting the result into Eq gives (1 46) This relation can be expressed in terms of two other thermodynamic properties called the volume expansivity b and the isothermal compressibility a, which are defined as (Fig. 1 10) and a 0P 0T b a 0T v 0v b a 0v P 0P b 1S a 0P T 0T b a 0v v 0T b a 0P P 0v b T c p c v T a 0v 0T b b 1 v a 0v 0T b P (1 47) a 1 (1 48) v a 0v 0P b T Substituting these two relations into Eq. 1 46, we obtain a third general relation for c p c v : c p c v vtb a (1 49) It is called the Mayer relation in honor of the German physician and physicist J. R. Mayer ( ). We can draw several conclusions from this equation: 1. The isothermal compressibility a is a positive quantity for all substances in all phases. The volume expansivity could be negative for some substances (such as liquid water below 4 C), but its square is always positive or zero. The temperature T in this relation is thermodynamic temperature, which is also positive. Therefore we conclude that the constant-pressure specific heat is always greater than or equal to the constant-volume specific heat: c p c v (1 50). The difference between c p and c v approaches zero as the absolute temperature approaches zero. 3. The two specific heats are identical for truly incompressible substances since v constant. The difference between the two specific heats is very small and is usually disregarded for substances that are nearly incompressible, such as liquids and solids. P a 0P 0v b T 0 C 100 kpa 1 kg 0 C 100 kpa 1 kg Chapter v ( T ) P 1 C 100 kpa 1 kg (a) A substance with a large β ( v T ) P 1 C 100 kpa 1 kg (b) A substance with a small β FIGURE 1 10 The volume expansivity (also called the coefficient of volumetric expansion) is a measure of the change in volume with temperature at constant pressure. EXAMPLE 1 7 Internal Energy Change of a van der Waals Gas Derive a relation for the internal energy change as a gas that obeys the van der Waals equation of state. Assume that in the range of interest c v varies according to the relation c v c 1 c T, where c 1 and c are constants. Solution A relation is to be obtained for the internal energy change of a van der Waals gas.

188 666 Thermodynamics Analysis The change in internal energy of any simple compressible system in any phase during any process can be determined from Eq. 1 30: The van der Waals equation of state is Then Thus, Substituting gives Integrating yields T v u u 1 c v dt T a 0P 0T b P v u u 1 c 1 1T T 1 c 1T T 1 a a 1 v 1 1 v b which is the desired relation. T 1 P a 0P 0T b R v v b T v a u u 1 1c 1 c T dt v dv T 1 v 1 RT v b a v RT v b c T a 0P 0T b P d dv v RT v b a v a v v 1 EXAMPLE 1 8 Internal Energy as a Function of Temperature Alone Show that the internal energy of (a) an ideal gas and (b) an incompressible substance is a function of temperature only, u u(t). Solution It is to be shown that u u(t) for ideal gases and incompressible substances. Analysis The differential change in the internal energy of a general simple compressible system is given by Eq. 1 9 as (a) For an ideal gas Pv RT. Then Thus, du c v dt c T a 0P 0T b P d dv v T a 0P 0T b P T a R b P P P 0 v v du c v dt To complete the proof, we need to show that c v is not a function of v either. This is done with the help of Eq. 1 4: a 0c v 0v b T a 0 P T 0T b v

189 Chapter For an ideal gas P RT/v. Then Thus, a 0P 0T b R v v and a 0 P 0T b c 0 1R>v d 0 v 0T v a 0c v 0v b 0 T which states that c v does not change with specific volume. That is, c v is not a function of specific volume either. Therefore we conclude that the internal energy of an ideal gas is a function of temperature only (Fig. 1 11). (b) For an incompressible substance, v constant and thus dv 0. Also from Eq. 1 49, c p c v c since a b 0 for incompressible substances. Then Eq. 1 9 reduces to du c dt Again we need to show that the specific heat c depends on temperature only and not on pressure or specific volume. This is done with the help of Eq. 1 43: u = u(t) c v = c v (T) c p = c p (T) u = u(t) c = c(t) AIR LAKE a 0c p 0P b Ta 0 v T 0T b 0 P since v constant. Therefore, we conclude that the internal energy of a truly incompressible substance depends on temperature only. FIGURE 1 11 The internal energies and specific heats of ideal gases and incompressible substances depend on temperature only. EXAMPLE 1 9 The Specific Heat Difference of an Ideal Gas Show that c p c v R for an ideal gas. Solution It is to be shown that the specific heat difference for an ideal gas is equal to its gas constant. Analysis This relation is easily proved by showing that the right-hand side of Eq is equivalent to the gas constant R of the ideal gas: Substituting, Therefore, P RT v v RT P T a 0v 0T b S a 0P 0v b RT T v P v S a 0v 0T b P c p c v Ta 0v 0T b P a R P b a 0P 0v b Ta R T P b a P v b R c p c v R P a 0P 0v b T

190 668 Thermodynamics > T 1 = 0 C T = 0 C < P 1 = 800 kpa P = 00 kpa FIGURE 1 1 The temperature of a fluid may increase, decrease, or remain constant during a throttling process. T P, T (varied) h = constant line Exit states P 1, T 1 (fixed) Inlet state 1 P 1 FIGURE 1 13 The development of an h constant line on a P-T diagram. T µ JT > 0 µ JT < 0 Maximum inversion temperature h = const. Inversion line FIGURE 1 14 Constant-enthalpy lines of a substance on a T-P diagram. P P 1 5 THE JOULE-THOMSON COEFFICIENT When a fluid passes through a restriction such as a porous plug, a capillary tube, or an ordinary valve, its pressure decreases. As we have shown in Chap. 5, the enthalpy of the fluid remains approximately constant during such a throttling process. You will remember that a fluid may experience a large drop in its temperature as a result of throttling, which forms the basis of operation for refrigerators and air conditioners. This is not always the case, however. The temperature of the fluid may remain unchanged, or it may even increase during a throttling process (Fig. 1 1). The temperature behavior of a fluid during a throttling (h constant) process is described by the Joule-Thomson coefficient, defined as (1 51) Thus the Joule-Thomson coefficient is a measure of the change in temperature with pressure during a constant-enthalpy process. Notice that if m JT m a 0T 0P b h 6 0 temperature increases 0 temperature remains constant 7 0 temperature decreases during a throttling process. A careful look at its defining equation reveals that the Joule-Thomson coefficient represents the slope of h constant lines on a T-P diagram. Such diagrams can be easily constructed from temperature and pressure measurements alone during throttling processes. A fluid at a fixed temperature and pressure T 1 and P 1 (thus fixed enthalpy) is forced to flow through a porous plug, and its temperature and pressure downstream (T and P ) are measured. The experiment is repeated for different sizes of porous plugs, each giving a different set of T and P. Plotting the temperatures against the pressures gives us an h constant line on a T-P diagram, as shown in Fig Repeating the experiment for different sets of inlet pressure and temperature and plotting the results, we can construct a T-P diagram for a substance with several h constant lines, as shown in Fig Some constant-enthalpy lines on the T-P diagram pass through a point of zero slope or zero Joule-Thomson coefficient. The line that passes through these points is called the inversion line, and the temperature at a point where a constant-enthalpy line intersects the inversion line is called the inversion temperature. The temperature at the intersection of the P 0 line (ordinate) and the upper part of the inversion line is called the maximum inversion temperature. Notice that the slopes of the h constant lines are negative (m JT 0) at states to the right of the inversion line and positive (m JT 0) to the left of the inversion line. A throttling process proceeds along a constant-enthalpy line in the direction of decreasing pressure, that is, from right to left. Therefore, the temperature of a fluid increases during a throttling process that takes place on the right-hand side of the inversion line. However, the fluid temperature decreases during a throttling process that takes place on the left-hand side of the inversion line. It is clear from this diagram that a cooling effect cannot be achieved by throttling unless the fluid is below its maximum inversion

191 temperature. This presents a problem for substances whose maximum inversion temperature is well below room temperature. For hydrogen, for example, the maximum inversion temperature is 68 C. Thus hydrogen must be cooled below this temperature if any further cooling is to be achieved by throttling. Next we would like to develop a general relation for the Joule-Thomson coefficient in terms of the specific heats, pressure, specific volume, and temperature. This is easily accomplished by modifying the generalized relation for enthalpy change (Eq. 1 35) dh c p dt c v T a 0v 0T b d dp P For an h constant process we have dh 0. Then this equation can be rearranged to give 1 c v T a 0v (1 5) c p 0T b d a 0T P 0P b m JT h which is the desired relation. Thus, the Joule-Thomson coefficient can be determined from a knowledge of the constant-pressure specific heat and the P-v-T behavior of the substance. Of course, it is also possible to predict the constant-pressure specific heat of a substance by using the Joule-Thomson coefficient, which is relatively easy to determine, together with the P-v-T data for the substance. Chapter EXAMPLE 1 10 Joule-Thomson Coefficient of an Ideal Gas Show that the Joule-Thomson coefficient of an ideal gas is zero. Solution Analysis It is to be shown that m JT 0 for an ideal gas. For an ideal gas v RT/P, and thus a 0v 0T b P R P Substituting this into Eq. 1 5 yields m JT 1 c v T a 0v c p 0T b d 1 c v T R P c p P d 1 1v v 0 c p Discussion This result is not surprising since the enthalpy of an ideal gas is a function of temperature only, h h(t), which requires that the temperature remain constant when the enthalpy remains constant. Therefore, a throttling process cannot be used to lower the temperature of an ideal gas (Fig. 1 15). T h = constant line P 1 P P FIGURE 1 15 The temperature of an ideal gas remains constant during a throttling process since h constant and T constant lines on a T-P diagram coincide. 1 6 THE h, u, AND s OF REAL GASES We have mentioned many times that gases at low pressures behave as ideal gases and obey the relation Pv RT. The properties of ideal gases are relatively easy to evaluate since the properties u, h, c v, and c p depend on temperature only. At high pressures, however, gases deviate considerably from ideal-gas behavior, and it becomes necessary to account for this deviation.

192 670 Thermodynamics T T Actual process path T 1 1 P 1 P P 0 = 0 1* * Alternative process path FIGURE 1 16 An alternative process path to evaluate the enthalpy changes of real gases. s In Chap. 3 we accounted for the deviation in properties P, v, and T by either using more complex equations of state or evaluating the compressibility factor Z from the compressibility charts. Now we extend the analysis to evaluate the changes in the enthalpy, internal energy, and entropy of nonideal (real) gases, using the general relations for du, dh, and ds developed earlier. Enthalpy Changes of Real Gases The enthalpy of a real gas, in general, depends on the pressure as well as on the temperature. Thus the enthalpy change of a real gas during a process can be evaluated from the general relation for dh (Eq. 1 36) where P 1, T 1 and P, T are the pressures and temperatures of the gas at the initial and the final states, respectively. For an isothermal process dt 0, and the first term vanishes. For a constant-pressure process, dp 0, and the second term vanishes. Properties are point functions, and thus the change in a property between two specified states is the same no matter which process path is followed. This fact can be exploited to greatly simplify the integration of Eq Consider, for example, the process shown on a T-s diagram in Fig The enthalpy change during this process h h 1 can be determined by performing the integrations in Eq along a path that consists of two isothermal (T 1 constant and T constant) lines and one isobaric (P 0 constant) line instead of the actual process path, as shown in Fig Although this approach increases the number of integrations, it also simplifies them since one property remains constant now during each part of the process. The pressure P 0 can be chosen to be very low or zero, so that the gas can be treated as an ideal gas during the P 0 constant process. Using a superscript asterisk (*) to denote an ideal-gas state, we can express the enthalpy change of a real gas during process 1- as where, from Eq. 1 36, T P h h 1 c p dt T 1 h h 1 1h h* 1h* h* 1 1h* 1 h 1 P h h* 0 c v T a 0v P 0T b d dp P* P T T c v T a 0v 0T b d dp P P T T 0 P 1 c v T a 0v 0T b d dp P (1 53) (1 54) T T h* h* 1 c p dt 0 c p0 1T dt (1 55) T 1 h* 1 h 1 0 P 1 * P 1 T 1 cv T a 0v P1 0T b d dp P T T 1 cv T a 0v 0T b d dp P P T T 0 1 (1 56) The difference between h and h* is called the enthalpy departure, and it represents the variation of the enthalpy of a gas with pressure at a fixed temperature. The calculation of enthalpy departure requires a knowledge of the P-v-T behavior of the gas. In the absence of such data, we can use the relation Pv ZRT, where Z is the compressibility factor. Substituting

193 v ZRT/P and simplifying Eq. 1 56, we can write the enthalpy departure at any temperature T and pressure P as 1h* h T RT P The above equation can be generalized by expressing it in terms of the reduced coordinates, using T T cr T R and P P cr P R. After some manipulations, the enthalpy departure can be expressed in a nondimensionalized form as 0 a 0Z 0T b dp P P Chapter Z h 1h* h T R u T cr T R PR (1 57) where Z h is called the enthalpy departure factor. The integral in the above equation can be performed graphically or numerically by employing data from the compressibility charts for various values of P R and T R. The values of Z h are presented in graphical form as a function of P R and T R in Fig. A 9. This graph is called the generalized enthalpy departure chart, and it is used to determine the deviation of the enthalpy of a gas at a given P and T from the enthalpy of an ideal gas at the same T. By replacing h* by h ideal for clarity, Eq for the enthalpy change of a gas during a process 1- can be rewritten as 0 a 0Z 0T R b P R d 1ln P R h h 1 1h h 1 ideal R u T cr 1Z h Z h1 (1 58) or h h 1 1h h 1 ideal RT cr 1Z h Z h1 (1 59) where the values of Z h are determined from the generalized enthalpy departure chart and (h h 1 ) ideal is determined from the ideal-gas tables. Notice that the last terms on the right-hand side are zero for an ideal gas. Internal Energy Changes of Real Gases The internal energy change of a real gas is determined by relating it to the enthalpy change through the definition h u Pv u ZR u T: u u 1 1h h 1 R u 1Z T Z 1 T 1 (1 60) Entropy Changes of Real Gases The entropy change of a real gas is determined by following an approach similar to that used above for the enthalpy change. There is some difference in derivation, however, owing to the dependence of the ideal-gas entropy on pressure as well as the temperature. The general relation for ds was expressed as (Eq. 1 41) s s 1 T T 1 c p T dt P where P 1, T 1 and P, T are the pressures and temperatures of the gas at the initial and the final states, respectively. The thought that comes to mind at this point is to perform the integrations in the previous equation first along a T 1 constant line to zero pressure, then along the P 0 line to T, and P 1 a 0v 0T b dp P

194 67 Thermodynamics finally along the T constant line to P, as we did for the enthalpy. This approach is not suitable for entropy-change calculations, however, since it involves the value of entropy at zero pressure, which is infinity. We can avoid this difficulty by choosing a different (but more complex) path between the two states, as shown in Fig Then the entropy change can be expressed as T T T 1 Actual process path P * P 1 Alternative process path 1 1* P 0 a* b* FIGURE 1 17 An alternative process path to evaluate the entropy changes of real gases during process 1-. s s s 1 1s s b * 1s b * s * 1s * s 1 * 1s 1 * s a * 1s a * s 1 (1 61) States 1 and 1* are identical (T 1 T 1 * and P 1 P 1 *) and so are states and *. The gas is assumed to behave as an ideal gas at the imaginary states 1* and * as well as at the states between the two. Therefore, the entropy change during process 1*-* can be determined from the entropy-change relations for ideal gases. The calculation of entropy change between an actual state and the corresponding imaginary ideal-gas state is more involved, however, and requires the use of generalized entropy departure charts, as explained below. Consider a gas at a pressure P and temperature T. To determine how much different the entropy of this gas would be if it were an ideal gas at the same temperature and pressure, we consider an isothermal process from the actual state P, T to zero (or close to zero) pressure and back to the imaginary idealgas state P*, T* (denoted by superscript *), as shown in Fig The entropy change during this isothermal process can be expressed as 1s P s P * T 1s P s 0 * T 1s 0 * s P * T where v ZRT/P and v* v ideal RT/P. Performing the differentiations and rearranging, we obtain P 11 ZR 1s P s P * T c P By substituting T T cr T R and P P cr P R and rearranging, the entropy departure can be expressed in a nondimensionalized form as Z s 1 s * s T,P R u P 0 0 PR 0 a 0v 0 0T b dp P a 0v* 0T b dp P P RT P a 0Zr 0T b d dp P c Z 1 T R a 0Z 0T R b P R d d 1ln P R (1 6) The difference (s * s ) T,P is called the entropy departure and Z s is called the entropy departure factor. The integral in the above equation can be performed by using data from the compressibility charts. The values of Z s are presented in graphical form as a function of P R and T R in Fig. A 30. This graph is called the generalized entropy departure chart, and it is used to determine the deviation of the entropy of a gas at a given P and T from the entropy of an ideal gas at the same P and T. Replacing s* by s ideal for clarity, we can rewrite Eq for the entropy change of a gas during a process 1- as s s 1 1 s s 1 ideal R u 1Z s Z s1 (1 63)

195 or s s 1 1s s 1 ideal R 1Z s Z s1 (1 64) where the values of Z s are determined from the generalized entropy departure chart and the entropy change (s s 1 ) ideal is determined from the idealgas relations for entropy change. Notice that the last terms on the right-hand side are zero for an ideal gas. Chapter EXAMPLE 1 11 The h and s of Oxygen at High Pressures Determine the enthalpy change and the entropy change of oxygen per unit mole as it undergoes a change of state from 0 K and 5 MPa to 300 K and 10 MPa (a) by assuming ideal-gas behavior and (b) by accounting for the deviation from ideal-gas behavior. Solution Oxygen undergoes a process between two specified states. The enthalpy and entropy changes are to be determined by assuming ideal-gas behavior and by accounting for the deviation from ideal-gas behavior. Analysis The critical temperature and pressure of oxygen are T cr K and P cr 5.08 MPa (Table A 1), respectively. The oxygen remains above its critical temperature; therefore, it is in the gas phase, but its pressure is quite high. Therefore, the oxygen will deviate from ideal-gas behavior and should be treated as a real gas. (a) If the O is assumed to behave as an ideal gas, its enthalpy will depend on temperature only, and the enthalpy values at the initial and the final temperatures can be determined from the ideal-gas table of O (Table A 19) at the specified temperatures: 1h h 1 ideal h,ideal h 1,ideal kj>kmol 33 kj/kmol The entropy depends on both temperature and pressure even for ideal gases. Under the ideal-gas assumption, the entropy change of oxygen is determined from 1 s s 1 ideal s s 1 R u ln P P kj>kmol # K kj>kmol # 10 MPa K ln 5 MPa 3.8 kj/kmol # K (b) The deviation from the ideal-gas behavior can be accounted for by determining the enthalpy and entropy departures from the generalized charts at each state: T R1 T 1 0 K 1.4 T cr K P R1 P 1 5 MPa Z h1 0.53, Z s P cr 5.08 MPa

196 674 Thermodynamics and Then the enthalpy and entropy changes of oxygen during this process are determined by substituting the values above into Eqs and 1 63, and T R T 300 K 1.94 T cr K P R P Z 10 MPa h 0.48, Z s P cr 5.08 MPa h h 1 1h h 1 ideal R u T cr 1Z h Z h1 33 kj>kmol kj>kmol # K K kj/kmol s s 1 1 s s 1 ideal R u 1Z s Z s1 3.8 kj>kmol # K kj>kmol # K kj/kmol # K Discussion Note that the ideal-gas assumption would underestimate the enthalpy change of the oxygen by.7 percent and the entropy change by 11.4 percent. SUMMARY Some thermodynamic properties can be measured directly, but many others cannot. Therefore, it is necessary to develop some relations between these two groups so that the properties that cannot be measured directly can be evaluated. The derivations are based on the fact that properties are point functions, and the state of a simple, compressible system is completely specified by any two independent, intensive properties. The equations that relate the partial derivatives of properties P, v, T, and s of a simple compressible substance to each other are called the Maxwell relations. They are obtained from the four Gibbs equations, expressed as du T ds P dv dh T ds v dp da s dt P dv dg s dt v dp The Maxwell relations are a 0T 0v b a 0P s 0s b v a 0T 0P b a 0v s 0s b P a 0s 0v b a 0P T 0T b v a 0s 0P b a 0v T 0T b P The Clapeyron equation enables us to determine the enthalpy change associated with a phase change from a knowledge of P, v, and T data alone. It is expressed as a dp dt b h fg sat T v fg

197 For liquid vapor and solid vapor phase-change processes at low pressures, it can be approximated as The changes in internal energy, enthalpy, and entropy of a simple compressible substance can be expressed in terms of pressure, specific volume, temperature, and specific heats alone as du c v dt c T a 0P 0T b P d dv v dh c p dt c v T a 0v 0T b d dp P ds c v 0P dt a T 0T b dv v or ln a P b h fg P 1 sat ds c p 0v dt a T 0T b dp P For specific heats, we have the following general relations: a 0c v 0v b T a 0 P T 0T b v a 0c p 0P b Ta 0 v T 0T b P c p,t c p0,t T P R a T T 1 T 1 T c p c v Ta 0v 0T b 0 a 0 v 0T b dp P P b sat a 0P 0v b T c p c v vtb a Chapter where b is the volume expansivity and a is the isothermal compressibility, defined as b 1 v a 0v 0T b and a 1 P v a 0v 0P b T The difference c p c v is equal to R for ideal gases and to zero for incompressible substances. The temperature behavior of a fluid during a throttling (h constant) process is described by the Joule-Thomson coefficient, defined as m JT a 0T 0P b h The Joule-Thomson coefficient is a measure of the change in temperature of a substance with pressure during a constantenthalpy process, and it can also be expressed as m JT 1 c v T a 0v c p 0T b d P The enthalpy, internal energy, and entropy changes of real gases can be determined accurately by utilizing generalized enthalpy or entropy departure charts to account for the deviation from the ideal-gas behavior by using the following relations: h h 1 1h h 1 ideal R u T cr 1Z h Z h1 u u 1 1h h 1 R u 1Z T Z 1 T 1 s s 1 1 s s 1 ideal R u 1Z s Z s1 where the values of Z h and Z s are determined from the generalized charts. REFERENCES AND SUGGESTED READINGS 1. G. J. Van Wylen and R. E. Sonntag. Fundamentals of Classical Thermodynamics. 3rd ed. New York: John Wiley & Sons, K. Wark and D. E. Richards. Thermodynamics. 6th ed. New York: McGraw-Hill, PROBLEMS* Partial Derivatives and Associated Relations 1 1C Consider the function z(x, y). Plot a differential surface on x-y-z coordinates and indicate x, dx, y, dy, ( z) x, ( z) y, and dz. 1 C What is the difference between partial differentials and ordinary differentials? *Problems designated by a C are concept questions, and students are encouraged to answer them all. Problems designated by an E are in English units, and the SI users can ignore them. Problems with a CD-EES icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed DVD. Problems with a computer-ees icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text.

198 676 Thermodynamics 1 3C Consider the function z(x, y), its partial derivatives ( z/ x) y and ( z/ y) x, and the total derivative dz/dx. (a) How do the magnitudes ( x) y and dx compare? (b) How do the magnitudes ( z) y and dz compare? (c) Is there any relation among dz, ( z) x, and ( z) y? 1 4C Consider a function z(x, y) and its partial derivative ( z/ y) x. Under what conditions is this partial derivative equal to the total derivative dz/dy? 1 5C Consider a function z(x, y) and its partial derivative ( z/ y) x. If this partial derivative is equal to zero for all values of x, what does it indicate? 1 6C Consider a function z(x, y) and its partial derivative ( z/ y) x. Can this partial derivative still be a function of x? 1 7C Consider a function f(x) and its derivative df/dx. Can this derivative be determined by evaluating dx/df and taking its inverse? 1 8 Consider air at 400 K and 0.90 m 3 /kg. Using Eq. 1 3, determine the change in pressure corresponding to an increase of (a) 1 percent in temperature at constant specific volume, (b) 1 percent in specific volume at constant temperature, and (c) 1 percent in both the temperature and specific volume. 1 9 Repeat Problem 1 8 for helium Prove for an ideal gas that (a) the P constant lines on a T-v diagram are straight lines and (b) the high-pressure lines are steeper than the low-pressure lines Derive a relation for the slope of the v constant lines on a T-P diagram for a gas that obeys the van der Waals equation of state. Answer: (v b)/r 1 1 Nitrogen gas at 400 K and 300 kpa behaves as an ideal gas. Estimate the c p and c v of the nitrogen at this state, using enthalpy and internal energy data from Table A 18, and compare them to the values listed in Table A b. 1 13E Nitrogen gas at 600 R and 30 psia behaves as an ideal gas. Estimate the c p and c v of the nitrogen at this state, using enthalpy and internal energy data from Table A 18E, and compare them to the values listed in Table A Eb. Answers: 0.49 Btu/lbm R, Btu/lbm R 1 14 Consider an ideal gas at 400 K and 100 kpa. As a result of some disturbance, the conditions of the gas change to 404 K and 96 kpa. Estimate the change in the specific volume of the gas using (a) Eq. 1 3 and (b) the ideal-gas relation at each state Using the equation of state P(v a) RT, verify (a) the cyclic relation and (b) the reciprocity relation at constant v. The Maxwell Relations 1 16 Verify the validity of the last Maxwell relation (Eq. 1 19) for refrigerant-134a at 80 C and 1. MPa Reconsider Prob Using EES (or other) software, verify the validity of the last Maxwell relation for refrigerant-134a at the specified state. 1 18E Verify the validity of the last Maxwell relation (Eq. 1 19) for steam at 800 F and 400 psia Using the Maxwell relations, determine a relation for ( s/ P) T for a gas whose equation of state is P(v b) RT. Answer: R/P 1 0 Using the Maxwell relations, determine a relation for ( s/ v) T for a gas whose equation of state is (P a/v ) (v b) RT. 1 1 Using the Maxwell relations and the ideal-gas equation of state, determine a relation for ( s/ v) T for an ideal gas. Answer: R/v The Clapeyron Equation 1 C What is the value of the Clapeyron equation in thermodynamics? 1 3C Does the Clapeyron equation involve any approximations, or is it exact? 1 4C What approximations are involved in the Clapeyron- Clausius equation? 1 5 Using the Clapeyron equation, estimate the enthalpy of vaporization of refrigerant-134a at 40 C, and compare it to the tabulated value. 1 6 Reconsider Prob Using EES (or other) software, plot the enthalpy of vaporization of refrigerant-134a as a function of temperature over the temperature range 0 to 80 C by using the Clapeyron equation and the refrigerant-134a data in EES. Discuss your results. 1 7 Using the Clapeyron equation, estimate the enthalpy of vaporization of steam at 300 kpa, and compare it to the tabulated value. 1 8 Calculate the h fg and s fg of steam at 10 C from the Clapeyron equation, and compare them to the tabulated values. 1 9E Determine the h fg of refrigerant-134a at 50 F on the basis of (a) the Clapeyron equation and (b) the Clapeyron-Clausius equation. Compare your results to the tabulated h fg value Plot the enthalpy of vaporization of steam as a function of temperature over the temperature range 10 to 00 C by using the Clapeyron equation and steam data in EES Using the Clapeyron-Clausius equation and the triplepoint data of water, estimate the sublimation pressure of water at 30 C and compare to the value in Table A 8.

199 General Relations for du, dh, ds, c v, and c p 1 3C Can the variation of specific heat c p with pressure at a given temperature be determined from a knowledge of P- v-t data alone? 1 33 Show that the enthalpy of an ideal gas is a function of temperature only and that for an incompressible substance it also depends on pressure Derive expressions for (a) u, (b) h, and (c) s for a gas that obeys the van der Waals equation of state for an isothermal process Derive expressions for (a) u, (b) h, and (c) s for a gas whose equation of state is P(v a) RT for an isothermal process. Answers: (a) 0, (b) a(p P 1 ), (c) R ln (P /P 1 ) 1 36 Derive expressions for ( u/ P) T and ( h/ v) T in terms of P, v, and T only Derive an expression for the specific-heat difference c p c v for (a) an ideal gas, (b) a van der Waals gas, and (c) an incompressible substance Estimate the specific-heat difference c p c v for liquid water at 15 MPa and 80 C. Answer: 0.3 kj/kg K 1 39E Estimate the specific-heat difference c p c v for liquid water at 1000 psia and 150 F. Answer: Btu/lbm R 1 40 Derive a relation for the volume expansivity b and the isothermal compressibility a (a) for an ideal gas and (b) for a gas whose equation of state is P(v a) RT Estimate the volume expansivity b and the isothermal compressibility a of refrigerant-134a at 00 kpa and 30 C. The Joule-Thomson Coefficient 1 4C What does the Joule-Thomson coefficient represent? 1 43C Describe the inversion line and the maximum inversion temperature. 1 44C The pressure of a fluid always decreases during an adiabatic throttling process. Is this also the case for the temperature? 1 45C Does the Joule-Thomson coefficient of a substance change with temperature at a fixed pressure? 1 46C Will the temperature of helium change if it is throttled adiabatically from 300 K and 600 kpa to 150 kpa? 1 47 Consider a gas whose equation of state is P(v a) RT, where a is a positive constant. Is it possible to cool this gas by throttling? 1 48 Derive a relation for the Joule-Thomson coefficient and the inversion temperature for a gas whose equation of state is (P a/v )v RT Estimate the Joule-Thomson coefficient of steam at (a) 3 MPa and 300 C and (b) 6 MPa and 500 C. Chapter E Estimate the Joule-Thomson coefficient of nitrogen at (a) 00 psia and 500 R and (b) 000 psia and 400 R. Use nitrogen properties from EES or other source. 1 51E Reconsider Prob. 1 50E. Using EES (or other) software, plot the Joule-Thomson coefficient for nitrogen over the pressure range 100 to 1500 psia at the enthalpy values 100, 175, and 5 Btu/lbm. Discuss the results. 1 5 Estimate the Joule-Thomson coefficient of refrigerant-134a at 0.7 MPa and 50 C Steam is throttled slightly from 1 MPa and 300 C. Will the temperature of the steam increase, decrease, or remain the same during this process? The dh, du, and ds of Real Gases 1 54C What is the enthalpy departure? 1 55C On the generalized enthalpy departure chart, the normalized enthalpy departure values seem to approach zero as the reduced pressure P R approaches zero. How do you explain this behavior? 1 56C Why is the generalized enthalpy departure chart prepared by using P R and T R as the parameters instead of P and T? 1 57 Determine the enthalpy of nitrogen, in kj/kg, at 175 K and 8 MPa using (a) data from the ideal-gas nitrogen table and (b) the generalized enthalpy departure chart. Compare your results to the actual value of 15.5 kj/kg. Answers: (a) kj/kg, (b) 11.6 kj/kg 1 58E Determine the enthalpy of nitrogen, in Btu/lbm, at 400 R and 000 psia using (a) data from the ideal-gas nitrogen table and (b) the generalized enthalpy chart. Compare your results to the actual value of Btu/lbm What is the error involved in the (a) enthalpy and (b) internal energy of CO at 350 K and 10 MPa if it is assumed to be an ideal gas? Answers: (a) 50%, (b) 49% 1 60 Determine the enthalpy change and the entropy change of nitrogen per unit mole as it undergoes a change of state from 5 K and 6 MPa to 30 K and 1 MPa, (a) by assuming ideal-gas behavior and (b) by accounting for the deviation from ideal-gas behavior through the use of generalized charts Determine the enthalpy change and the entropy change of CO per unit mass as it undergoes a change of state from 50 K and 7 MPa to 80 K and 1 MPa, (a) by assuming ideal-gas behavior and (b) by accounting for the deviation from ideal-gas behavior. 1 6 Methane is compressed adiabatically by a steady-flow compressor from MPa and 10 C to 10 MPa and 110 C at a rate of 0.55 kg/s. Using the generalized charts, determine the required power input to the compressor. Answer: 133 kw

200 678 Thermodynamics 1 63 Propane is compressed isothermally by a piston cylinder device from 100 C and 1 MPa to 4 MPa. Using the generalized charts, determine the work done and the heat transfer per unit mass of propane Reconsider Prob Using EES (or other) software, extend the problem to compare the solutions based on the ideal-gas assumption, generalized chart data, and real fluid data. Also extend the solution to methane. 1 65E Propane is compressed isothermally by a piston cylinder device from 00 F and 00 psia to 800 psia. Using the generalized charts, determine the work done and the heat transfer per unit mass of the propane. Answers: 45.3 Btu/lbm, 141 Btu/lbm 1 66 Determine the exergy destruction associated with the process described in Prob Assume T 0 30 C Carbon dioxide enters an adiabatic nozzle at 8 MPa and 450 K with a low velocity and leaves at MPa and 350 K. Using the generalized enthalpy departure chart, determine the exit velocity of the carbon dioxide. Answer: 384 m/s 1 68 Reconsider Prob Using EES (or other) software, compare the exit velocity to the nozzle assuming ideal-gas behavior, the generalized chart data, and EES data for carbon dioxide A 0.08-m 3 well-insulated rigid tank contains oxygen at 0 K and 10 MPa. A paddle wheel placed in the tank is turned on, and the temperature of the oxygen rises to 50 K. Using the generalized charts, determine (a) the final pressure in the tank and (b) the paddle-wheel work done during this process. Answers: (a) 1,190 kpa, (b) 393 kj 1 70 Carbon dioxide is contained in a constant-volume tank and is heated from 100 C and 1 MPa to 8 MPa. Determine the heat transfer and entropy change per unit mass of the carbon dioxide using (a) the ideal-gas assumption, (b) the generalized charts, and (c) real fluid data from EES or other sources. Review Problems 10 MPa 110 C CH 4 m = 0.55 kg/s MPa 10 C FIGURE P For b 0, prove that at every point of a singlephase region of an h-s diagram, the slope of a constantpressure (P constant) line is greater than the slope of a constant-temperature (T constant) line, but less than the slope of a constant-volume (v constant) line. 1 7 Using the cyclic relation and the first Maxwell relation, derive the other three Maxwell relations Starting with the relation dh T ds + v dp, show that the slope of a constant-pressure line on an h-s diagram (a) is constant in the saturation region and (b) increases with temperature in the superheated region Derive relations for (a) u, (b) h, and (c) s of a gas that obeys the equation of state (P + a/v )v RT for an isothermal process Show that c v T a 0v 0T b a 0P s 0T b and c p T a 0P v 0T b a 0v s 0T b P 1 76 Estimate the c p of nitrogen at 300 kpa and 400 K, using (a) the relation in the above problem and (b) its definition. Compare your results to the value listed in Table A b Steam is throttled from 4.5 MPa and 300 C to.5 MPa. Estimate the temperature change of the steam during this process and the average Joule-Thomson coefficient. Answers: 6.3 C, 13.1 C/MPa 1 78 A rigid tank contains 1. m 3 of argon at 100 C and 1 MPa. Heat is now transferred to argon until the temperature in the tank rises to 0 C. Using the generalized charts, determine (a) the mass of the argon in the tank, (b) the final pressure, and (c) the heat transfer. Answers: (a) 35.1 kg, (b) 1531 kpa, (c) 151 kj 1 79 Argon gas enters a turbine at 7 MPa and 600 K with a velocity of 100 m/s and leaves at 1 MPa and 80 K with a velocity of 150 m/s at a rate of 5 kg/s. Heat is being lost to the surroundings at 5 C at a rate of 60 kw. Using the generalized charts, determine (a) the power output of the turbine and (b) the exergy destruction associated with the process. 7 MPa 600 K 100 m/s T 0 = 5 C 60 kw Ar m = 5 kg/s 1 MPa 80 K 150 m/s FIGURE P1 79 W

201 1 80 Reconsider Prob Using EES (or other) software, solve the problem assuming steam is the working fluid by using the generalized chart method and EES data for steam. Plot the power output and the exergy destruction rate for these two calculation methods against the turbine exit pressure as it varies over the range 0.1 to 1 MPa when the turbine exit temperature is 455 K. 1 81E Argon gas enters a turbine at 1000 psia and 1000 R with a velocity of 300 ft/s and leaves at 150 psia and 500 R with a velocity of 450 ft/s at a rate of 1 lbm/s. Heat is being lost to the surroundings at 75 F at a rate of 80 Btu/s. Using the generalized charts, determine (a) the power output of the turbine and (b) the exergy destruction associated with the process. Answers: (a) 9 hp, (b) 11.5 Btu/s 1 8 An adiabatic 0.-m 3 storage tank that is initially evacuated is connected to a supply line that carries nitrogen at 5 K and 10 MPa. A valve is opened, and nitrogen flows into the tank from the supply line. The valve is closed when the pressure in the tank reaches 10 MPa. Determine the final temperature in the tank (a) treating nitrogen as an ideal gas and (b) using generalized charts. Compare your results to the actual value of 93 K. N 0. m 3 Initially evacuated 5 K 10 MPa FIGURE P For a homogeneous (single-phase) simple pure substance, the pressure and temperature are independent properties, and any property can be expressed as a function of these two properties. Taking v v(p, T), show that the change in specific volume can be expressed in terms of the volume expansivity b and isothermal compressibility a as dv v b dt a dp Also, assuming constant average values for b and a, obtain a relation for the ratio of the specific volumes v /v 1 as a homogeneous system undergoes a process from state 1 to state Repeat Prob for an isobaric process. Chapter The volume expansivity of water at 0 C is b K 1. Treating this value as a constant, determine the change in volume of 1 m 3 of water as it is heated from 10 C to 30 C at constant pressure The volume expansivity b values of copper at 300 K and 500 K are K 1 and K 1, respectively, and b varies almost linearly in this temperature range. Determine the percent change in the volume of a copper block as it is heated from 300 K to 500 K at atmospheric pressure Starting with m JT (1/c p ) [T( v/ T ) p v] and noting that Pv ZRT, where Z Z(P, T ) is the compressibility factor, show that the position of the Joule-Thomson coefficient inversion curve on the T-P plane is given by the equation ( Z/ T) P Consider an infinitesimal reversible adiabatic compression or expansion process. By taking s s(p, v) and using the Maxwell relations, show that for this process Pv k constant, where k is the isentropic expansion exponent defined as k v P a 0P 0v b s Also, show that the isentropic expansion exponent k reduces to the specific heat ratio c p /c v for an ideal gas Refrigerant-134a undergoes an isothermal process at 60 C from 3 to 0.1 MPa in a closed system. Determine the work done by the refrigerant-134a by using the tabular (EES) data and the generalized charts, in kj/kg Methane is contained in a piston cylinder device and is heated at constant pressure of 4 MPa from 100 to 350 C. Determine the heat transfer, work and entropy change per unit mass of the methane using (a) the ideal-gas assumption, (b) the generalized charts, and (c) real fluid data from EES or other sources. Fundamentals of Engineering (FE) Exam Problems 1 91 A substance whose Joule-Thomson coefficient is negative is throttled to a lower pressure. During this process, (select the correct statement) (a) the temperature of the substance will increase. (b) the temperature of the substance will decrease. (c) the entropy of the substance will remain constant. (d) the entropy of the substance will decrease. (e) the enthalpy of the substance will decrease. 1 9 Consider the liquid vapor saturation curve of a pure substance on the P-T diagram. The magnitude of the slope of the tangent line to this curve at a temperature T (in Kelvin) is

202 680 Thermodynamics (a) proportional to the enthalpy of vaporization h fg at that temperature. (b) proportional to the temperature T. (c) proportional to the square of the temperature T. (d) proportional to the volume change v fg at that temperature. (e) inversely proportional to the entropy change s fg at that temperature Based on the generalized charts, the error involved in the enthalpy of CO at 350 K and 8 MPa if it is assumed to be an ideal gas is (a) 0 (b) 0% (c) 35% (d) 6% (e) 65% 1 94 Based on data from the refrigerant-134a tables, the Joule-Thompson coefficient of refrigerant-134a at 0.8 MPa and 100 C is approximately (a) 0 (b) 5 C/MPa (c) 11 C/MPa (d) 8 C/MPa (e) 6 C/MPa 1 95 For a gas whose equation of state is P(v b) RT, the specified heat difference c p c v is equal to (a) R (b) R b (c) R b (d) 0 (e) R(1 + v/b) Design and Essay Problems 1 96 Consider the function z z(x, y). Write an essay on the physical interpretation of the ordinary derivative dz/dx and the partial derivative ( z/ x) y. Explain how these two derivatives are related to each other and when they become equivalent There have been several attempts to represent the thermodynamic relations geometrically, the best known of these v u s a h FIGURE P1 97 being Koenig s thermodynamic square shown in the figure. There is a systematic way of obtaining the four Maxwell relations as well as the four relations for du, dh, dg, and da from this figure. By comparing these relations to Koenig s diagram, come up with the rules to obtain these eight thermodynamic relations from this diagram Several attempts have been made to express the partial derivatives of the most common thermodynamic properties in a compact and systematic manner in terms of measurable properties. The work of P. W. Bridgman is perhaps the most fruitful of all, and it resulted in the well-known Bridgman s table. The 8 entries in that table are sufficient to express the partial derivatives of the eight common properties P, T, v, s, u, h, f, and g in terms of the six properties P, v, T, c p, b, and a, which can be measured directly or indirectly with relative ease. Obtain a copy of Bridgman s table and explain, with examples, how it is used. T g P

203 Chapter 13 GAS MIXTURES Up to this point, we have limited our consideration to thermodynamic systems that involve a single pure substance such as water. Many important thermodynamic applications, however, involve mixtures of several pure substances rather than a single pure substance. Therefore, it is important to develop an understanding of mixtures and learn how to handle them. In this chapter, we deal with nonreacting gas mixtures. A nonreacting gas mixture can be treated as a pure substance since it is usually a homogeneous mixture of different gases. The properties of a gas mixture obviously depend on the properties of the individual gases (called components or constituents) as well as on the amount of each gas in the mixture. Therefore, it is possible to prepare tables of properties for mixtures. This has been done for common mixtures such as air. It is not practical to prepare property tables for every conceivable mixture composition, however, since the number of possible compositions is endless. Therefore, we need to develop rules for determining mixture properties from a knowledge of mixture composition and the properties of the individual components. We do this first for ideal-gas mixtures and then for real-gas mixtures. The basic principles involved are also applicable to liquid or solid mixtures, called solutions. Objectives The objectives of Chapter 13 are to: Develop rules for determining nonreacting gas mixture properties from knowledge of mixture composition and the properties of the individual components. Define the quantities used to describe the composition of a mixture, such as mass fraction, mole fraction, and volume fraction. Apply the rules for determining mixture properties to idealgas mixtures and real-gas mixtures. Predict the P-v-T behavior of gas mixtures based on Dalton s law of additive pressures and Amagat s law of additive volumes. Perform energy and exergy analysis of mixing processes. 681

204 68 Thermodynamics H 6 kg + O 3 kg H + O 38 kg FIGURE 13 1 The mass of a mixture is equal to the sum of the masses of its components. H 3 kmol + FIGURE 13 O 1 kmol H + O 4 kmol The number of moles of a nonreacting mixture is equal to the sum of the number of moles of its components. H + O y H = 0.75 y O = FIGURE 13 3 The sum of the mole fractions of a mixture is equal to COMPOSITION OF A GAS MIXTURE: MASS AND MOLE FRACTIONS To determine the properties of a mixture, we need to know the composition of the mixture as well as the properties of the individual components. There are two ways to describe the composition of a mixture: either by specifying the number of moles of each component, called molar analysis, or by specifying the mass of each component, called gravimetric analysis. Consider a gas mixture composed of k components. The mass of the mixture m m is the sum of the masses of the individual components, and the mole number of the mixture N m is the sum of the mole numbers of the individual components* (Figs and 13 ). That is, m m a k (13 1a, b) The ratio of the mass of a component to the mass of the mixture is called the mass fraction mf, and the ratio of the mole number of a component to the mole number of the mixture is called the mole fraction y: mf i m i m m and y i N i N m (13 a, b) Dividing Eq. 13 1a by m m or Eq. 13 1b by N m, we can easily show that the sum of the mass fractions or mole fractions for a mixture is equal to 1 (Fig. 13 3): k k a mf i 1 and a y i 1 i 1 i 1 m i and N m a k The mass of a substance can be expressed in terms of the mole number N and molar mass M of the substance as m NM. Then the apparent (or average) molar mass and the gas constant of a mixture can be expressed as i 1 N i i 1 M m m m a m i a N i M k i N m N m N a y i M i and R m R u m i 1 M m The molar mass of a mixture can also be expressed as M m m m m m 1 N m a m i >M i a m i >1m m M i 1 k mf i a Mass and mole fractions of a mixture are related by mf i m i m m N i M i N m M m y i M i M m i 1 M i (13 3a, b) (13 4) (13 5) *Throughout this chapter, the subscript m denotes the gas mixture and the subscript i denotes any single component of the mixture.

205 Chapter EXAMPLE 13 1 Mass and Mole Fractions of a Gas Mixture Consider a gas mixture that consists of 3 kg of O, 5 kg of N, and 1 kg of CH 4, as shown in Fig Determine (a) the mass fraction of each component, (b) the mole fraction of each component, and (c) the average molar mass and gas constant of the mixture. Solution The masses of components of a gas mixture are given. The mass fractions, the mole fractions, the molar mass, and the gas constant of the mixture are to be determined. Analysis (a) The total mass of the mixture is m m m O m N m CH kg Then the mass fraction of each component becomes mf O m O 3 kg 0.15 m m 0 kg mf N m N 5 kg 0.5 m m 0 kg mf CH4 m CH 4 m m 1 kg kg 3 kg O 5 kg N 1 kg CH 4 FIGURE 13 4 Schematic for Example (b) To find the mole fractions, we need to determine the mole numbers of each component first: Thus, and N O m O M O N N m N M N N CH4 m CH 4 M CH4 N m N O N N N CH kmol y O N O kmol 0.09 N m 1.03 kmol y N N N kmol N m 1.03 kmol y CH4 N CH 4 N m 3 kg kmol 3 kg>kmol 5 kg kmol 8 kg>kmol 1 kg kmol 16 kg>kmol kmol kmol (c) The average molar mass and gas constant of the mixture are determined from their definitions, M m m m 0 kg N m 1.03 kmol 19.6 kg /kmol

206 684 Thermodynamics or Also, M m a y i M i y O M O y N M N y CH4 M CH kg>kmol R m R u kj>1kmol # K 0.44 kj/kg # K M m 19.6 kg>kmol Discussion When mass fractions are available, the molar mass and mole fractions could also be determined directly from Eqs and Gas A V, T P A + Gas B V, T P B Gas mixture A + B V, T P A + P B FIGURE 13 5 Dalton s law of additive pressures for a mixture of two ideal gases. Gas A P, T V A + Gas B P, T V B Gas mixture A + B P, T V A + V B FIGURE 13 6 Amagat s law of additive volumes for a mixture of two ideal gases. 13 P-v-T BEHAVIOR OF GAS MIXTURES: IDEAL AND REAL GASES An ideal gas is defined as a gas whose molecules are spaced far apart so that the behavior of a molecule is not influenced by the presence of other molecules a situation encountered at low densities. We also mentioned that real gases approximate this behavior closely when they are at a low pressure or high temperature relative to their critical-point values. The P-v-T behavior of an ideal gas is expressed by the simple relation Pv RT, which is called the ideal-gas equation of state. The P-v-T behavior of real gases is expressed by more complex equations of state or by Pv ZRT, where Z is the compressibility factor. When two or more ideal gases are mixed, the behavior of a molecule normally is not influenced by the presence of other similar or dissimilar molecules, and therefore a nonreacting mixture of ideal gases also behaves as an ideal gas. Air, for example, is conveniently treated as an ideal gas in the range where nitrogen and oxygen behave as ideal gases. When a gas mixture consists of real (nonideal) gases, however, the prediction of the P-v-T behavior of the mixture becomes rather involved. The prediction of the P-v-T behavior of gas mixtures is usually based on two models: Dalton s law of additive pressures and Amagat s law of additive volumes. Both models are described and discussed below. Dalton s law of additive pressures: The pressure of a gas mixture is equal to the sum of the pressures each gas would exert if it existed alone at the mixture temperature and volume (Fig. 13 5). Amagat s law of additive volumes: The volume of a gas mixture is equal to the sum of the volumes each gas would occupy if it existed alone at the mixture temperature and pressure (Fig. 13 6). Dalton s and Amagat s laws hold exactly for ideal-gas mixtures, but only approximately for real-gas mixtures. This is due to intermolecular forces that may be significant for real gases at high densities. For ideal gases, these two laws are identical and give identical results.

207 Dalton s and Amagat s laws can be expressed as follows: k Dalton s law: P m a P i 1T m, V m exact for ideal gases, (13 6) i 1 approximate k for real gases Amagat s law: V m a V i 1T m, P m (13 7) i 1 In these relations, P i is called the component pressure and V i is called the component volume (Fig. 13 7). Note that V i is the volume a component would occupy if it existed alone at T m and P m, not the actual volume occupied by the component in the mixture. (In a vessel that holds a gas mixture, each component fills the entire volume of the vessel. Therefore, the volume of each component is equal to the volume of the vessel.) Also, the ratio P i /P m is called the pressure fraction and the ratio V i /V m is called the volume fraction of component i. O + N 100 kpa 400 K 1 m 3 Chapter O 100 kpa 400 K 0.3 m 3 N 100 kpa 400 K 0.7 m 3 FIGURE 13 7 The volume a component would occupy if it existed alone at the mixture T and P is called the component volume (for ideal gases, it is equal to the partial volume y i V m ). Ideal-Gas Mixtures For ideal gases, P i and V i can be related to y i by using the ideal-gas relation for both the components and the gas mixture: Therefore, P i 1T m, V m P m V i 1T m, P m V m N ir u T m >V m N m R u T m >V m N i N m y i N ir u T m >P m N m R u T m >P m N i N m y i P i P m V i V m N i N m y i (13 8) Equation 13 8 is strictly valid for ideal-gas mixtures since it is derived by assuming ideal-gas behavior for the gas mixture and each of its components. The quantity y i P m is called the partial pressure (identical to the component pressure for ideal gases), and the quantity y i V m is called the partial volume (identical to the component volume for ideal gases). Note that for an ideal-gas mixture, the mole fraction, the pressure fraction, and the volume fraction of a component are identical. The composition of an ideal-gas mixture (such as the exhaust gases leaving a combustion chamber) is frequently determined by a volumetric analysis (called the Orsat Analysis) and Eq A sample gas at a known volume, pressure, and temperature is passed into a vessel containing reagents that absorb one of the gases. The volume of the remaining gas is then measured at the original pressure and temperature. The ratio of the reduction in volume to the original volume (volume fraction) represents the mole fraction of that particular gas. Real-Gas Mixtures Dalton s law of additive pressures and Amagat s law of additive volumes can also be used for real gases, often with reasonable accuracy. This time, however, the component pressures or component volumes should be evaluated from relations that take into account the deviation of each component

208 686 Thermodynamics P m V m = Z m N m R u T m Z m = y i Z i i = 1 FIGURE 13 8 One way of predicting the P-v-T behavior of a real-gas mixture is to use compressibility factor. k Pseudopure substance k P ' cr,m = y i P cr,i i = 1 k T ' cr,m = y i T cr,i i = 1 FIGURE 13 9 Another way of predicting the P-v-T behavior of a real-gas mixture is to treat it as a pseudopure substance with critical properties P cr and T cr. from ideal-gas behavior. One way of doing that is to use more exact equations of state (van der Waals, Beattie Bridgeman, Benedict Webb Rubin, etc.) instead of the ideal-gas equation of state. Another way is to use the compressibility factor (Fig. 13 8) as (13 9) The compressibility factor of the mixture Z m can be expressed in terms of the compressibility factors of the individual gases Z i by applying Eq to both sides of Dalton s law or Amagat s law expression and simplifying. We obtain (13 10) where Z i is determined either at T m and V m (Dalton s law) or at T m and P m (Amagat s law) for each individual gas. It may seem that using either law gives the same result, but it does not. The compressibility-factor approach, in general, gives more accurate results when the Z i s in Eq are evaluated by using Amagat s law instead of Dalton s law. This is because Amagat s law involves the use of mixture pressure P m, which accounts for the influence of intermolecular forces between the molecules of different gases. Dalton s law disregards the influence of dissimilar molecules in a mixture on each other. As a result, it tends to underpredict the pressure of a gas mixture for a given V m and T m. Therefore, Dalton s law is more appropriate for gas mixtures at low pressures. Amagat s law is more appropriate at high pressures. Note that there is a significant difference between using the compressibility factor for a single gas and for a mixture of gases. The compressibility factor predicts the P-v-T behavior of single gases rather accurately, as discussed in Chapter 3, but not for mixtures of gases. When we use compressibility factors for the components of a gas mixture, we account for the influence of like molecules on each other; the influence of dissimilar molecules remains largely unaccounted for. Consequently, a property value predicted by this approach may be considerably different from the experimentally determined value. Another approach for predicting the P-v-T behavior of a gas mixture is to treat the gas mixture as a pseudopure substance (Fig. 13 9). One such method, proposed by W. B. Kay in 1936 and called Kay s rule, involves the use of a pseudocritical pressure P cr,m and pseudocritical temperature T cr,m for the mixture, defined in terms of the critical pressures and temperatures of the mixture components as P cr,m a k i 1 PV ZNR u T Z m a k y i P cr,i and T cr,m a k (13 11a, b) The compressibility factor of the mixture Z m is then easily determined by using these pseudocritical properties. The result obtained by using Kay s rule is accurate to within about 10 percent over a wide range of temperatures and pressures, which is acceptable for most engineering purposes. Another way of treating a gas mixture as a pseudopure substance is to use a more accurate equation of state such as the van der Waals, Beattie Bridgeman, or Benedict Webb Rubin equation for the mixture, and to determine the constant coefficients in terms of the coefficients of the components. i 1 y i Z i i 1 y i T cr,i

209 In the van der Waals equation, for example, the two constants for the mixture are determined from k k a m a a y i a 1> i b and b m a y i b i (13 1a, b) i 1 i 1 where expressions for a i and b i are given in Chapter 3. Chapter EXAMPLE 13 P-v-T Behavior of Nonideal Gas Mixtures A rigid tank contains kmol of N and 6 kmol of CO gases at 300 K and 15 MPa (Fig ). Estimate the volume of the tank on the basis of (a) the ideal-gas equation of state, (b) Kay s rule, (c) compressibility factors and Amagat s law, and (d ) compressibility factors and Dalton s law. Solution The composition of a mixture in a rigid tank is given. The volume of the tank is to be determined using four different approaches. Assumptions Stated in each section. Analysis (a) When the mixture is assumed to behave as an ideal gas, the volume of the mixture is easily determined from the ideal-gas relation for the mixture: V m N mr u T m P m since 18 kmol kpa # m 3 >kmol # K1300 K 15,000 kpa N m N N N CO 6 8 kmol m 3 (b) To use Kay s rule, we need to determine the pseudocritical temperature and pseudocritical pressure of the mixture by using the critical-point properties of N and CO from Table A 1. However, first we need to determine the mole fraction of each component: y N N N kmol N m 8 kmol 0.5 and y CO N CO 6 kmol 0.75 N m 8 kmol kmol N 6 kmol CO 300 K 15 MPa V m =? FIGURE Schematic for Example 13. T cr,m a y i T cr,i y N T cr,n y CO T cr,co K K 59.7 K Then, Thus, P cr,m a y i P cr,i y N P cr,n y CO P cr,co MPa MPa 6.39 MPa T R T m 300 K 1.16 T cr,m 59.7 K P R P Z m Fig. A 15b m 15 MPa.35 P cr,m 6.39 MPa V m Z m N m R u T m P m Z m V ideal m m 3

210 688 Thermodynamics (c) When Amagat s law is used in conjunction with compressibility factors, Z m is determined from Eq But first we need to determine the Z of each component on the basis of Amagat s law: N : T R,N T m P R,N T cr,n P m P cr,n 300 K K 15 MPa MPa Z N 1.0 1Fig. A 15b CO : T R,CO T m P R,CO T cr,co P m P cr,co 300 K K 15 MPa MPa Z CO Fig. A 15b Mixture: Thus, Z m a y i Z i y N Z N y CO Z CO The compressibility factor in this case turned out to be almost the same as the one determined by using Kay s rule. (d) When Dalton s law is used in conjunction with compressibility factors, Z m is again determined from Eq However, this time the Z of each component is to be determined at the mixture temperature and volume, which is not known. Therefore, an iterative solution is required. We start the calculations by assuming that the volume of the gas mixture is m 3, the value determined by assuming ideal-gas behavior. The T R values in this case are identical to those obtained in part (c) and remain constant. The pseudoreduced volume is determined from its definition in Chap. 3: Similarly, From Fig. A 15, we read Z N 0.99 and Z CO Thus, and V m Z m N m R u T m P m Z m V ideal m m 3 v R,N v R,CO v N R u T cr,n >P cr,n V m >N N R u T cr,n >P cr,n m 3 >1 kmol kpa # m 3 >kmol # K116. K>13390 kpa m 3 >16 kmol kpa # m 3 >kmol # K1304. K>17390 kpa Z m y N Z N y CO Z CO V m Z m N m RT m P m Z m V ideal m m 3

211 Chapter This is 33 percent lower than the assumed value. Therefore, we should repeat the calculations, using the new value of V m. When the calculations are repeated we obtain m 3 after the second iteration, m 3 after the third iteration, and m 3 after the fourth iteration. This value does not change with more iterations. Therefore, V m m 3 Discussion Notice that the results obtained in parts (b), (c), and (d ) are very close. But they are very different from the ideal-gas values. Therefore, treating a mixture of gases as an ideal gas may yield unacceptable errors at high pressures PROPERTIES OF GAS MIXTURES: IDEAL AND REAL GASES Consider a gas mixture that consists of kg of N and 3 kg of CO. The total mass (an extensive property) of this mixture is 5 kg. How did we do it? Well, we simply added the mass of each component. This example suggests a simple way of evaluating the extensive properties of a nonreacting idealor real-gas mixture: Just add the contributions of each component of the mixture (Fig ). Then the total internal energy, enthalpy, and entropy of a gas mixture can be expressed, respectively, as U m a k S m a k (13 13) (13 14) (13 15) By following a similar logic, the changes in internal energy, enthalpy, and entropy of a gas mixture during a process can be expressed, respectively, as U m a k i 1 H m a k i 1 S m a k i 1 i 1 H m a k i 1 i 1 U i a k S i a k U i a k H i a k S i a k i 1 i 1 i 1 i 1 i 1 H i a k i 1 m i u i a k m i h i a k m i s i a k m i u i a k m i h i a k m i s i a k (13 16) (13 17) (13 18) Now reconsider the same mixture, and assume that both N and CO are at 5 C. The temperature (an intensive property) of the mixture is, as you would expect, also 5 C. Notice that we did not add the component temperatures to determine the mixture temperature. Instead, we used some kind of averaging scheme, a characteristic approach for determining the intensive properties of a mixture. The internal energy, enthalpy, and entropy of a mixture per unit mass or per unit mole of the mixture can be determined by dividing the equations above by the mass or the mole number of the mixture (m m or N m ). We obtain (Fig. 13 1) i 1 i 1 i 1 N i s i 1kJ>K i 1 i 1 i 1 N i u i 1kJ N i h i 1kJ N i u i 1kJ N i h i 1kJ N i s i 1kJ>K kmol A 6 kmol B U A =1000 kj U B =1800 kj U m =800 kj FIGURE The extensive properties of a mixture are determined by simply adding the properties of the components. kmol A 3 kmol B ū A =500 kj/kmol ū B =600 kj/kmol ū m =560 kj/kmol FIGURE 13 1 The intensive properties of a mixture are determined by weighted averaging.

212 690 Thermodynamics Similarly, the specific heats of a gas mixture can be expressed as c v,m a k c p,m a k u m a k i 1 i 1 i 1 h m a k i 1 s m a k i 1 mf i u i mf i h i 1kJ>kg 1kJ>kg and u m a k mf i s i 1kJ>kg # K and s m a k mf i c v,i 1kJ>kg # K and c v,m a k mf i c p,i 1kJ>kg # K and c p,m a k (13 19) (13 0) (13 1) (13 ) (13 3) Notice that properties per unit mass involve mass fractions (mf i ) and properties per unit mole involve mole fractions (y i ). The relations given above are exact for ideal-gas mixtures, and approximate for real-gas mixtures. (In fact, they are also applicable to nonreacting liquid and solid solutions especially when they form an ideal solution. ) The only major difficulty associated with these relations is the determination of properties for each individual gas in the mixture. The analysis can be simplified greatly, however, by treating the individual gases as ideal gases, if doing so does not introduce a significant error. Ideal-Gas Mixtures The gases that comprise a mixture are often at a high temperature and low pressure relative to the critical-point values of individual gases. In such cases, the gas mixture and its components can be treated as ideal gases with negligible error. Under the ideal-gas approximation, the properties of a gas are not influenced by the presence of other gases, and each gas component in the mixture behaves as if it exists alone at the mixture temperature T m and mixture volume V m. This principle is known as the Gibbs Dalton law, which is an extension of Dalton s law of additive pressures. Also, the h, u, c v, and c p of an ideal gas depend on temperature only and are independent of the pressure or the volume of the ideal-gas mixture. The partial pressure of a component in an ideal-gas mixture is simply P i y i P m, where P m is the mixture pressure. Evaluation of u or h of the components of an ideal-gas mixture during a process is relatively easy since it requires only a knowledge of the initial and final temperatures. Care should be exercised, however, in evaluating the s of the components since the entropy of an ideal gas depends on the pressure or volume of the component as well as on its temperature. The entropy change of individual gases in an ideal-gas mixture during a process can be determined from i 1 i 1 i 1 and h m a k i 1 i 1 y i u i 1kJ>kmol y i h i 1kJ>kmol y i s i 1kJ>kmol # K y i c v,i 1kJ>kmol # K y i c p,i 1kJ>kmol # K or s i s i, s i,1 R i ln P i, P i,1 c p,i ln T i, T i,1 R i ln P i, P i,1 s i s i, s i,1 R u ln P i, P i,1 c p,i ln T i, T i,1 R u ln P i, P i,1 (13 4) (13 5)

213 where P i, y i, P m, and P i,1 y i,1 P m,1. Notice that the partial pressure P i of each component is used in the evaluation of the entropy change, not the mixture pressure P m (Fig ). EXAMPLE 13 3 Mixing Two Ideal Gases in a Tank An insulated rigid tank is divided into two compartments by a partition, as shown in Fig One compartment contains 7 kg of oxygen gas at 40 C and 100 kpa, and the other compartment contains 4 kg of nitrogen gas at 0 C and 150 kpa. Now the partition is removed, and the two gases are allowed to mix. Determine (a) the mixture temperature and (b) the mixture pressure after equilibrium has been established. Solution A rigid tank contains two gases separated by a partition. The pressure and temperature of the mixture are to be determined after the partition is removed. Assumptions 1 We assume both gases to be ideal gases, and their mixture to be an ideal-gas mixture. This assumption is reasonable since both the oxygen and nitrogen are well above their critical temperatures and well below their critical pressures. The tank is insulated and thus there is no heat transfer. 3 There are no other forms of work involved. Properties The constant-volume specific heats of N and O at room temperature are c v,n kj/kg K and c v,o kj/kg K (Table A a). Analysis We take the entire contents of the tank (both compartments) as the system. This is a closed system since no mass crosses the boundary during the process. We note that the volume of a rigid tank is constant and thus there is no boundary work done. (a) Noting that there is no energy transfer to or from the tank, the energy balance for the system can be expressed as E in E out E system Chapter Partial pressure of component i at state P i, s i = s i, s i,1 R i ln Partial pressure of component i at state 1 i, P i,1 FIGURE Partial pressures (not the mixture pressure) are used in the evaluation of entropy changes of ideal-gas mixtures. O 7 kg 40 C 100 kpa N 4 kg 0 C 150 kpa Partition FIGURE Schematic for Example U U N U O 3mc v 1T m T 1 4 N 3mc v 1T m T 1 4 O 0 By using c v values at room temperature, the final temperature of the mixture is determined to be 14 kg kj>kg # K1Tm 0 C 17 kg kj>kg # K1Tm 40 C 0 T m 3. C (b) The final pressure of the mixture is determined from the ideal-gas relation where N O m O M O N N m N M N P m V m N m R u T m 7 kg 0.19 kmol 3 kg>kmol 4 kg kmol 8 kg>kmol N m N O N N kmol

214 69 Thermodynamics and V O a NR ut 1 b P 1 V N a NR ut 1 b P 1 V m V O V N m 3 Thus, P m N m R u T m V m kmol kpa # m 3 >kmol # K1313 K 5.70 m 3 O 100 kpa kmol kpa # m 3 >kmol # K193 K.3 m 3 N 150 kpa kmol kpa # m 3 >kmol # K1305. K 8.0 m kpa Discussion We could also determine the mixture pressure by using P m V m m m R m T m, where R m is the apparent gas constant of the mixture. This would require a knowledge of mixture composition in terms of mass or mole fractions. EXAMPLE 13 4 Exergy Destruction during Mixing of Ideal Gases O 5 C 00 kpa CO 5 C 00 kpa FIGURE Schematic for Example An insulated rigid tank is divided into two compartments by a partition, as shown in Fig One compartment contains 3 kmol of O, and the other compartment contains 5 kmol of CO. Both gases are initially at 5 C and 00 kpa. Now the partition is removed, and the two gases are allowed to mix. Assuming the surroundings are at 5 C and both gases behave as ideal gases, determine the entropy change and exergy destruction associated with this process. Solution A rigid tank contains two gases separated by a partition. The entropy change and exergy destroyed after the partition is removed are to be determined. Assumptions Both gases and their mixture are ideal gases. Analysis We take the entire contents of the tank (both compartments) as the system. This is a closed system since no mass crosses the boundary during the process. We note that the volume of a rigid tank is constant, and there is no energy transfer as heat or work. Also, both gases are initially at the same temperature and pressure. When two ideal gases initially at the same temperature and pressure are mixed by removing a partition between them, the mixture will also be at the same temperature and pressure. (Can you prove it? Will this be true for nonideal gases?) Therefore, the temperature and pressure in the tank will still be 5 C and 00 kpa, respectively, after the mixing. The entropy change of each component gas can be determined from Eqs and 13 5: 0 S m a S i a N i s i a N i a c p,i ln T i, R T u ln P i, b i,1 P i,1 R u a N i ln y i,p m, P i,1 R u a N i ln y i, since P m, P i,1 00 kpa. It is obvious that the entropy change is independent of the composition of the mixture in this case and depends on only

215 Chapter the mole fraction of the gases in the mixture. What is not so obvious is that if the same gas in two different chambers is mixed at constant temperature and pressure, the entropy change is zero. Substituting the known values, the entropy change becomes N m N O N CO 13 5 kmol 8 kmol y O N O 3 kmol N m 8 kmol y CO N CO N m 44.0 kj/k 5 kmol kmol S m R u 1N O ln y O N CO ln y CO kj>kmol # K313 kmol1ln kmol 1ln The exergy destruction associated with this mixing process is determined from X destroyed T 0 S gen T 0 S sys 198 K144.0 kj>k 13.1 MJ Discussion This large value of exergy destruction shows that mixing processes are highly irreversible. Real-Gas Mixtures When the components of a gas mixture do not behave as ideal gases, the analysis becomes more complex because the properties of real (nonideal) gases such as u, h, c v, and c p depend on the pressure (or specific volume) as well as on the temperature. In such cases, the effects of deviation from ideal-gas behavior on the mixture properties should be accounted for. Consider two nonideal gases contained in two separate compartments of an adiabatic rigid tank at 100 kpa and 5 C. The partition separating the two gases is removed, and the two gases are allowed to mix. What do you think the final pressure in the tank will be? You are probably tempted to say 100 kpa, which would be true for ideal gases. However, this is not true for nonideal gases because of the influence of the molecules of different gases on each other (deviation from Dalton s law, Fig ). When real-gas mixtures are involved, it may be necessary to account for the effect of nonideal behavior on the mixture properties such as enthalpy and entropy. One way of doing that is to use compressibility factors in conjunction with generalized equations and charts developed in Chapter 1 for real gases. Consider the following T ds relation for a gas mixture: It can also be expressed as dh m T m ds m v m dp m Real gas A 5 C 0.4 m kpa Real gas mixture A + B 5 C 1 m 3 10 kpa? Real gas B 5 C 0.6 m kpa FIGURE It is difficult to predict the behavior of nonideal-gas mixtures because of the influence of dissimilar molecules on each other. d a a mf i h i b T m d a a mf i s i b a a mf i v i b dp m

216 694 Thermodynamics or which yields a mf i 1dh i T m ds i v i dp m 0 dh i T m ds i v i dp m (13 6) This is an important result because Eq is the starting equation in the development of the generalized relations and charts for enthalpy and entropy. It suggests that the generalized property relations and charts for real gases developed in Chapter 1 can also be used for the components of real-gas mixtures. But the reduced temperature T R and reduced pressure P R for each component should be evaluated by using the mixture temperature T m and mixture pressure P m. This is because Eq involves the mixture pressure P m, not the component pressure P i. The approach described above is somewhat analogous to Amagat s law of additive volumes (evaluating mixture properties at the mixture pressure and temperature), which holds exactly for ideal-gas mixtures and approximately for real-gas mixtures. Therefore, the mixture properties determined with this approach are not exact, but they are sufficiently accurate. What if the mixture volume and temperature are specified instead of the mixture pressure and temperature? Well, there is no need to panic. Just evaluate the mixture pressure, using Dalton s law of additive pressures, and then use this value (which is only approximate) as the mixture pressure. Another way of evaluating the properties of a real-gas mixture is to treat the mixture as a pseudopure substance having pseudocritical properties, determined in terms of the critical properties of the component gases by using Kay s rule. The approach is quite simple, and the accuracy is usually acceptable. T 1 = 0 K P 1 = 10 MPa AIR 79% N 1% O Heat FIGURE Schematic for Example T = 160 K P = 10 MPa EXAMPLE 13 5 Cooling of a Nonideal Gas Mixture Air is a mixture of N, O, and small amounts of other gases, and it can be approximated as 79 percent N and 1 percent O on mole basis. During a steady-flow process, air is cooled from 0 to 160 K at a constant pressure of 10 MPa (Fig ). Determine the heat transfer during this process per kmol of air, using (a) the ideal-gas approximation, (b) Kay s rule, and (c) Amagat s law. Solution Air at a low temperature and high pressure is cooled at constant pressure. The heat transfer is to be determined using three different approaches. Assumptions 1 This is a steady-flow process since there is no change with time at any point and thus m CV 0 and E CV 0. The kinetic and potential energy changes are negligible. Analysis We take the cooling section as the system. This is a control volume since mass crosses the system boundary during the process. We note that heat is transferred out of the system. The critical properties are T cr 16. K and P cr 3.39 MPa for N and T cr K and P cr 5.08 MPa for O. Both gases remain above their

217 Chapter critical temperatures, but they are also above their critical pressures. Therefore, air will probably deviate from ideal-gas behavior, and thus it should be treated as a real-gas mixture. The energy balance for this steady-flow system can be expressed on a unit mole basis as e in e out e system 0 0 S e in e out S h 1 h qout q out h 1 h y N 1h 1 h N y O 1h 1 h O where the enthalpy change for either component can be determined from the generalized enthalpy departure chart (Fig. A 9) and Eq. 1 58: h 1 h h 1,ideal h,ideal R u T cr 1Z h1 Z h The first two terms on the right-hand side of this equation represent the ideal-gas enthalpy change of the component. The terms in parentheses represent the deviation from the ideal-gas behavior, and their evaluation requires a knowledge of reduced pressure P R and reduced temperature T R, which are calculated at the mixture temperature T m and mixture pressure P m. (a) If the N and O mixture is assumed to behave as an ideal gas, the enthalpy of the mixture will depend on temperature only, and the enthalpy values at the initial and the final temperatures can be determined from the ideal-gas tables of N and O (Tables A 18 and A 19): T 1 0 K S h 1,ideal,N 6391 kj>kmol h 1,ideal,O 6404 kj>kmol T 160 K S h,ideal,n 4648 kj>kmol h,ideal,o 4657 kj>kmol q out y N 1h 1 h N y O 1h 1 h O kj>kmol kj>kmol 1744 kj/kmol (b) Kay s rule is based on treating a gas mixture as a pseudopure substance whose critical temperature and pressure are and T cr,m a y i T cr,i y N T cr,n y O T cr,o K K 13. K Then, P cr,m a y i P cr,i y N P cr,n y O P cr,o MPa MPa 3.74 MPa T R,1 T m,1 0 K 1.66 T cr,m 13. K P R P m 10 MPa f Z h1,m P cr,m 3.74 MPa T R, T m, 160 K 1.1 T cr,m 13. K f Z h,m.6

218 696 Thermodynamics Also, Therefore, (c) The reduced temperatures and pressures for both N and O at the initial and final states and the corresponding enthalpy departure factors are, from Fig. A 9, N : h m1,ideal y N h 1,ideal,N y O h 1,ideal,O 3503 kj/kmol kj>kmol kj>kmol 6394 kj>kmol h m,ideal y N h,ideal,n y O h,ideal,o kj>kmol kj>kmol 4650 kj>kmol q out 1h m1,ideal h m,ideal R u T cr 1Z h1 Z h m kj>kmol kj>kmol # K113. K T R1,N T m,1 0 K 1.74 T cr,n 16. K P R,N P m P cr,n 10 MPa MPa T R,N T m, 160 K 1.7 T cr,n 16. K f Z h1,n 0.9 f Z h,n.4 O : From Eq. 1 58, T R1,O T m,1 0 K 1.4 T cr,o K P R,O P m P cr,o 10 MPa MPa T R1,O T m, 160 K 1.03 T cr,o K f Z h1,o 1.3 f Z h,o 4.0 1h 1 h N 1h 1,ideal h,ideal N R u T cr 1Z h1 Z h N kj>kmol kj>kmol # K116. K kj>kmol 1h 1 h O 1h 1,ideal h,ideal O R u T cr 1Z h1 Z h O kj>kmol kj>kmol # K K kj>kmol Therefore, q out y N 1h 1 h N y O 1h 1 h O kj>kmol kj>kmol 3717 kj/kmol

219 Chapter Discussion This result is about 6 percent greater than the result obtained in part (b) by using Kay s rule. But it is more than twice the result obtained by assuming the mixture to be an ideal gas. TOPIC OF SPECIAL INTEREST* Chemical Potential and the Separation Work of Mixtures When two gases or two miscible liquids are brought into contact, they mix and form a homogeneous mixture or solution without requiring any work input. That is, the natural tendency of miscible substances brought into contact is to mix with each other. As such, these are irreversible processes, and thus it is impossible for the reverse process of separation to occur spontaneously. For example, pure nitrogen and oxygen gases readily mix when brought into contact, but a mixture of nitrogen and oxygen (such as air) never separates into pure nitrogen and oxygen when left unattended. Mixing and separation processes are commonly used in practice. Separation processes require a work (or, more generally, exergy) input, and minimizing this required work input is an important part of the design process of separation plants. The presence of dissimilar molecules in a mixture affect each other, and therefore the influence of composition on the properties must be taken into consideration in any thermodynamic analysis. In this section we analyze the general mixing processes, with particular emphasis on ideal solutions, and determine the entropy generation and exergy destruction. We then consider the reverse process of separation, and determine the minimum (or reversible) work input needed for separation. The specific Gibbs function (or Gibbs free energy) g is defined as the combination property g h Ts. Using the relation dh v dp Tds, the differential change of the Gibbs function of a pure substance is obtained by differentiation to be dg v dp s dt or dg V dp S dt 1pure substance (13 7) For a mixture, the total Gibbs function is a function of two independent intensive properties as well as the composition, and thus it can be expressed as G G(P, T, N 1, N,..., N i ). Its differential is dg a 0G 0P b dp a 0G T,N 0T b dt ai P,N a 0G 0N i b P,T,N j dn i 1mixture (13 8) where the subscript N j indicates that the mole numbers of all components in the mixture other than component i are to be held constant during differentiation. For a pure substance, the last term drops out since the composition is fixed, and the equation above must reduce to the one for a pure substance. Comparing Eqs and 13 8 gives dg V dp S dt ai m i dn i or dg v dp s dt ai m i dy i (13 9) *This section can be skipped without a loss in continuity.

220 698 Thermodynamics Mixture: i h i,mixture Ts i,mixture g i,mixture Pure substance: h Ts g FIGURE For a pure substance, the chemical potential is equivalent to the Gibbs function. where y i N i /N m is the mole fraction of component i (N m is the total number of moles of the mixture) and m i a 0G 0N i b P,T,N j g i h i Ts i 1for component i of a mixture (13 30) is the chemical potential, which is the change in the Gibbs function of the mixture in a specified phase when a unit amount of component i in the same phase is added as pressure, temperature, and the amounts of all other components are held constant. The symbol tilde (as in v, h, and s ) is used to denote the partial molar properties of the components. Note that the summation term in Eq is zero for a single component system and thus the chemical potential of a pure system in a given phase is equivalent to the molar Gibbs function (Fig ) since G Ng Nm, where m a 0G 0N b g h Ts 1pure substance P,T (13 31) Therefore, the difference between the chemical potential and the Gibbs function is due to the effect of dissimilar molecules in a mixture on each other. It is because of this molecular effect that the volume of the mixture of two miscible liquids may be more or less than the sum of the initial volumes of the individual liquids. Likewise, the total enthalpy of the mixture of two components at the same pressure and temperature, in general, is not equal to the sum of the total enthalpies of the individual components before mixing, the difference being the enthalpy (or heat) of mixing, which is the heat released or absorbed as two or more components are mixed isothermally. For example, the volume of an ethyl alcohol water mixture is a few percent less than the sum of the volumes of the individual liquids before mixing. Also, when water and flour are mixed to make dough, the temperature of the dough rises noticeably due to the enthalpy of mixing released. For reasons explained above, the partial molar properties of the components (denoted by an tilde) should be used in the evaluation of the extensive properties of a mixture instead of the specific properties of the pure components. For example, the total volume, enthalpy, and entropy of a mixture should be determined from, respectively, V ai N i v i H ai N i h i and S ai N i s i 1mixture (13 3) A y A B Mixing chamber A + B mixture instead of V * ai N i vi H* ai N i h i and S* ai N i s i (13 33) y B Then the changes in these extensive properties during mixing become FIGURE The amount of heat released or absorbed during a mixing process is called the enthalpy (or heat) of mixing, which is zero for ideal solutions. V mixing ai N i 1v i v i, H mixing ai N i 1h i h i, S mixing ai N i 1 s i s i (13 34) where H mixing is the enthalpy of mixing and S mixing is the entropy of mixing (Fig ). The enthalpy of mixing is positive for exothermic mix-

221 Chapter ing processes, negative for endothermic mixing processes, and zero for isothermal mixing processes during which no heat is absorbed or released. Note that mixing is an irreversible process, and thus the entropy of mixing must be a positive quantity during an adiabatic process. The specific volume, enthalpy, and entropy of a mixture are determined from v ai y i v i h ai y i h i and s ai y i s i (13 35) where y i is the mole fraction of component i in the mixture. Reconsider Eq for dg. Recall that properties are point functions, and they have exact differentials. Therefore, the test of exactness can be applied to the right-hand side of Eq to obtain some important relations. For the differential dz Mdx Ndy of a function z(x, y), the test of exactness is expressed as ( M/ y) x ( N/ x) y. When the amount of component i in a mixture is varied at constant pressure or temperature while other components (indicated by j ) are held constant, Eq simplifies to dg S dt m i dn i 1for P constant and N j constant (13 36) dg V dp m i dn i 1for T constant and N j constant (13 37) Applying the test of exactness to both of these relations gives a 0 m i 0T b a 0S b s P,N 0N i and a 0 m i i T,P,N j 0P b a 0V b vi T,N 0N i T,P,N j (13 38) where the subscript N indicates that the mole numbers of all components (and thus the composition of the mixture) is to remain constant. Taking the chemical potential of a component to be a function of temperature, pressure, and composition and thus m i m i (P, T, y 1, y,..., y j...), its total differential can be expressed as dm i dg i a 0 m i 0P b dp a 0 m i T,y 0T b dt ai P,y (13 39) where the subscript y indicates that the mole fractions of all components (and thus the composition of the mixture) is to remain constant. Substituting Eqs into the above relation gives dm i v i dp s i dt a i a 0 m i 0y i b P,T,y j dy i a 0 m i 0y i b P,T,y j dy i (13 40) For a mixture of fixed composition undergoing an isothermal process, it simplifies to dm i vi dp 1T constant, y i constant (13 41) Ideal-Gas Mixtures and Ideal Solutions When the effect of dissimilar molecules in a mixture on each other is negligible, the mixture is said to be an ideal mixture or ideal solution and the chemical potential of a component in such a mixture equals the Gibbs function

222 700 Thermodynamics of the pure component. Many liquid solutions encountered in practice, especially dilute ones, satisfy this condition very closely and can be considered to be ideal solutions with negligible error. As expected, the ideal solution approximation greatly simplifies the thermodynamic analysis of mixtures. In an ideal solution, a molecule treats the molecules of all components in the mixture the same way no extra attraction or repulsion for the molecules of other components. This is usually the case for mixtures of similar substances such as those of petroleum products. Very dissimilar substances such as water and oil won t even mix at all to form a solution. For an ideal-gas mixture at temperature T and total pressure P, the partial molar volume of a component i is v i v i R ut/p. Substituting this relation into Eq gives dm i R ut P dp R utd ln P R u Td ln P i 1T constant, y i constant, ideal gas (13 4) since, from Dalton s law of additive pressures, P i y i P for an ideal gas mixture and d ln P i d ln 1y i P d 1ln y i ln P d ln P 1y i constant (13 43) for constant y i. Integrating Eq at constant temperature from the total mixture pressure P to the component pressure P i of component i gives m i 1T, P i m i 1T, P R u T ln P i P m i 1T, P R u T ln y i 1ideal gas (13 44) For y i 1 (i.e., a pure substance of component i alone), the last term in the above equation drops out and we end up with m i (T, P i ) m i (T, P), which is the value for the pure substance i. Therefore, the term m i (T, P) is simply the chemical potential of the pure substance i when it exists alone at total mixture pressure and temperature, which is equivalent to the Gibbs function since the chemical potential and the Gibbs function are identical for pure substances. The term m i (T, P) is independent of mixture composition and mole fractions, and its value can be determined from the property tables of pure substances. Then Eq can be rewritten more explicitly as m i,mixture,ideal 1T, P i m i,pure 1T, P R u T ln y i (13 45) Note that the chemical potential of a component of an ideal gas mixture depends on the mole fraction of the components as well as the mixture temperature and pressure, and is independent of the identity of the other constituent gases. This is not surprising since the molecules of an ideal gas behave like they exist alone and are not influenced by the presence of other molecules. Eq is developed for an ideal-gas mixture, but it is also applicable to mixtures or solutions that behave the same way that is, mixtures or solutions in which the effects of molecules of different components on each other are negligible. The class of such mixtures is called ideal solutions (or ideal mixtures), as discussed before. The ideal-gas mixture described is just one cate-

223 Chapter gory of ideal solutions. Another major category of ideal solutions is the dilute liquid solutions, such as the saline water. It can be shown that the enthalpy of mixing and the volume change due to mixing are zero for ideal solutions (see Wark, 1995). That is, V mixing,ideal ai N i 1v i v i 0 and H mixing,ideal ai N i 1h i h i 0 (13 46) Then it follows that v i v i and h i h i. That is, the partial molar volume and the partial molar enthalpy of a component in a solution equal the specific volume and enthalpy of that component when it existed alone as a pure substance at the mixture temperature and pressure. Therefore, the specific volume and enthalpy of individual components do not change during mixing if they form an ideal solution. Then the specific volume and enthalpy of an ideal solution can be expressed as (Fig. 13 0) V mixing,ideal 0 v i,mixture v i,pure v mixture y i v i,pure i H mixing,ideal 0 h i,mixing h i,pure h mixture y i h i,pure i v mixing,ideal ai y i v i ai y i v i,pure and h mixture,ideal ai y i h i ai y i h i,pure (13 47) Note that this is not the case for entropy and the properties that involve entropy such as the Gibbs function, even for ideal solutions. To obtain a relation for the entropy of a mixture, we differentiate Eq with respect to temperature at constant pressure and mole fraction, a 0 m i,mixing 1T, P i 0T b a 0 m i,pure 1T, P b R P,y 0T u ln y i P,y (13 48) We note from Eq that the two partial derivatives above are simply the negative of the partial molar entropies. Substituting, s i,mixture,ideal 1T, P i s i,pure 1T, P R u ln y 1 1ideal solution (13 49) Note that ln y i is a negative quantity since y i 1, and thus R u ln y i is always positive. Therefore, the entropy of a component in a mixture is always greater than the entropy of that component when it exists alone at the mixture temperature and pressure. Then the entropy of mixing of an ideal solution is determined by substituting Eq into Eq to be FIGURE 13 0 The specific volume and enthalpy of individual components do not change during mixing if they form an ideal solution (this is not the case for entropy). S mixing,ideal ai N i 1 s i s i R u ai N i ln y i 1ideal solution (13 50a) or, dividing by the total number of moles of the mixture N m, s mixing,ideal ai y i 1 s i s i R u ai y i ln y i 1per unit mole of mixture (13 50b) Minimum Work of Separation of Mixtures The entropy balance for a steady-flow system simplifies to S in S out S gen 0. Noting that entropy can be transferred by heat and mass only, the

224 70 Thermodynamics entropy generation during an adiabatic mixing process that forms an ideal solution becomes S gen S out S in S mixing R u ai N i ln y i 1ideal solution (13 51a) or s gen s out s in s mixing R u ai y i ln y i 1per unit mole of mixture (13 51b) Also noting that X destroyed T 0 S gen, the exergy destroyed during this (and any other) process is obtained by multiplying the entropy generation by the temperature of the environment T 0. It gives X destroyed T 0 S gen R u T 0 ai N i ln y i 1ideal soluton (13 5a) or x destroyed T 0 s gen R u T 0 ai y i ln y i 1per unit mole of mixture (13 5b) A B Mixing chamber T 0 W rev X destruction T 0 S gen A + B mixture FIGURE 13 1 For a naturally occurring process during which no work is produced or consumed, the reversible work is equal to the exergy destruction. Exergy destroyed represents the wasted work potential the work that would be produced if the mixing process occurred reversibly. For a reversible or thermodynamically perfect process, the entropy generation and thus the exergy destroyed is zero. Also, for reversible processes, the work output is a maximum (or, the work input is a minimum if the process does not occur naturally and requires input). The difference between the reversible work and the actual useful work is due to irreversibilities and is equal to the exergy destruction. Therefore, X destroyed W rev W actual. Then it follows that for a naturally occurring process during which no work is produced, the reversible work is equal to the exergy destruction (Fig. 13 1). Therefore, for the adiabatic mixing process that forms an ideal solution, the reversible work (total and per unit mole of mixture) is, from Eq. 13 5, W rev R u T 0 ai N i ln y i and w rev R u T 0 ai y i ln y i (13 53) A reversible process, by definition, is a process that can be reversed without leaving a net effect on the surroundings. This requires that the direction of all interactions be reversed while their magnitudes remain the same when the process is reversed. Therefore, the work input during a reversible separation process must be equal to the work output during the reverse process of mixing. A violation of this requirement will be a violation of the second law of thermodynamics. The required work input for a reversible separation process is the minimum work input required to accomplish that separation since the work input for reversible processes is always less than the work input of corresponding irreversible processes. Then the minimum work input required for the separation process can be expressed as W min,in R u T 0 ai N i ln y i and w min,in R u T 0 ai y i ln y i (13 54)

225 Chapter It can also be expressed in the rate form as W # min,in R u T 0 ai N # i ln y i N # mr u T 0 ai y i ln y i 1kW (13 55) where Ẇ min,in is the minimum power input required to separate a solution that approaches at a rate of N # (or m # m N # m kmol>s mm m kg>s) into its components. The work of separation per unit mass of mixture can be determined from w min,in w min,in>m m, where M m is the apparent molar mass of the mixture. The minimum work relations above are for complete separation of the components in the mixture. The required work input will be less if the exiting streams are not pure. The reversible work for incomplete separation can be determined by calculating the minimum separation work for the incoming mixture and the minimum separation works for the outgoing mixtures, and then taking their difference. Reversible Mixing Processes The mixing processes that occur naturally are irreversible, and all the work potential is wasted during such processes. For example, when the fresh water from a river mixes with the saline water in an ocean, an opportunity to produce work is lost. If this mixing is done reversibly (through the use of semipermeable membranes, for example) some work can be produced. The maximum amount of work that can be produced during a mixing process is equal to the minimum amount of work input needed for the corresponding separation process (Fig. 13 ). That is, A y A B y B (a) Mixing W max,out = 5 kj/kg mixture Mixing chamber A + B mixture W max,out,mixing W min,in,separation (13 56) Therefore, the minimum work input relations given above for separation can also be used to determine the maximum work output for mixing. The minimum work input relations are independent of any hardware or process. Therefore, the relations developed above are applicable to any separation process regardless of actual hardware, system, or process, and can be used for a wide range of separation processes including the desalination of sea or brackish water. Second-Law Efficiency The second-law efficiency is a measure of how closely a process approximates a corresponding reversible process, and it indicates the range available for potential improvements. Noting that the second-law efficiency ranges from 0 for a totally irreversible process to 100 percent for a totally reversible process, the second-law efficiency for separation and mixing processes can be defined as h II,separation W# min,in W # act,in w min,in and h w II,mixing W # act,out act,in W # max,out w act,out w max,out (13 57) where Ẇ act,in is the actual power input (or exergy consumption) of the separation plant and Ẇ act,out is the actual power produced during mixing. Note that A y A B y B (b) Separation W max,in = 5 kj/kg mixture Separation unit A + B mixture FIGURE 13 Under reversible conditions, the work consumed during separation is equal to the work produced during the reverse process of mixing.

226 704 Thermodynamics the second-law efficiency is always less than 1 since the actual separation process requires a greater amount of work input because of irreversibilities. Therefore, the minimum work input and the second-law efficiency provide a basis for comparison of actual separation processes to the idealized ones and for assessing the thermodynamic performance of separation plants. A second-law efficiency for mixing processes can also be defined as the actual work produced during mixing divided by the maximum work potential available. This definition does not have much practical value, however, since no effort is done to produce work during most mixing processes and thus the second-law efficiency is zero. Special Case: Separation of a Two Component Mixture Consider a mixture of two components A and B whose mole fractions are y A and y B. Noting that y B 1 y A, the minimum work input required to separate 1 kmol of this mixture at temperature T 0 completely into pure A and pure B is, from Eq , w min,in = R u T 0 ln y A (kj/kmol A) A + B y A, y B A + B 1 kmol y A, y B Separation unit Separation unit pure A (1 kmol) A + B (a) Separating 1 kmol of A from a large body of mixture wmin,in = R u T 0 ( y A ln y A + y B ln y B ) (kj/kmol mixture) (b) Complete separation of 1 kmol mixture into its components A and B pure A pure B FIGURE 13 3 The minimum work required to separate a two-component mixture for the two limiting cases. or w min,in R u T 0 1y A ln y A y B ln y B 1kJ>kmol mixture or, from Eq , W min,in R u T 0 1N A ln y A N B ln y B 1kJ W # min,in N # mr u T 0 1y A ln y A y B ln y B m # mr m T 0 1y A ln y A y B ln y B 1kW (13 58a) (13 58b) (13 58c) Some separation processes involve the extraction of just one of the components from a large amount of mixture so that the composition of the remaining mixture remains practically the same. Consider a mixture of two components A and B whose mole fractions are y A and y B, respectively. The minimum work required to separate 1 kmol of pure component A from the mixture of N m N A N B kmol (with N A 1) is determined by subtracting the minimum work required to separate the remaining mixture R u T 0 [(N A 1)ln y A N B ln y B ] from the minimum work required to separate the initial mixture W min,in R u T 0 (N A ln y A N B ln y B ). It gives (Fig. 13 3) w min,in R u T 0 ln y A R u T 0 ln 11>y A 1kJ>kmol A (13 59) The minimum work needed to separate a unit mass (1 kg) of component A is determined from the above relation by replacing R u by R A (or by dividing the relation above by the molar mass of component A) since R A R u /M A. Eq also gives the maximum amount of work that can be done as one unit of pure component A mixes with a large amount of A B mixture. An Application: Desalination Processes The potable water needs of the world is increasing steadily due to population growth, rising living standards, industrialization, and irrigation in agriculture. There are over 10,000 desalination plants in the world, with a total desalted

227 Chapter water capacity of over 5 billion gallons a day. Saudi Arabia is the largest user of desalination with about 5 percent of the world capacity, and the United States is the second largest user with 10 percent. The major desalination methods are distillation and reverse osmosis. The relations can be used directly for desalination processes, by taking the water (the solvent) to be component A and the dissolved salts (the solute) to be component B. Then the minimum work needed to produce 1 kg of pure water from a large reservoir of brackish or seawater at temperature T 0 in an environment at T 0 is, from Eq , Desalination: w min,in R w T 0 ln 11>y w 1kJ>kg pure water (13 60) where R w kj/kg K is the gas constant of water and y w is the mole fraction of water in brackish or seawater. The relation above also gives the maximum amount of work that can be produced as 1 kg of fresh water (from a river, for example) mixes with seawater whose water mole fraction is y w. The reversible work associated with liquid flow can also be expressed in terms of pressure difference P and elevation difference z (potential energy) as w min,in P/r g z where r is the density of the liquid. Combining these relations with Eq gives and P min rw min,in rr w T 0 ln 11>y w 1kPa z min w min,in >g R w T 0 ln 11>y w >g 1m (13 61) (13 6) where P min is the osmotic pressure, which represents the pressure difference across a semipermeable membrane that separates fresh water from the saline water under equilibrium conditions, r is the density of saline water, and z min is the osmotic rise, which represents the vertical distance the saline water would rise when separated from the fresh water by a membrane that is permeable to water molecules alone (again at equilibrium). For desalination processes, P min represents the minimum pressure that the saline water must be compressed in order to force the water molecules in saline water through the membrane to the fresh water side during a reverse osmosis desalination process. Alternately, z min represents the minimum height above the fresh water level that the saline water must be raised to produce the required osmotic pressure difference across the membrane to produce fresh water. The z min also represents the height that the water with dissolved organic matter inside the roots will rise through a tree when the roots are surrounded by fresh water with the roots acting as semipermeable membranes. The reverse osmosis process with semipermeable membranes is also used in dialysis machines to purify the blood of patients with failed kidneys. EXAMPLE 13 6 Obtaining Fresh Water from Seawater Fresh water is to be obtained from seawater at 15 C with a salinity of 3.48 percent on mass basis (or TDS 34,800 ppm). Determine (a) the mole fractions of the water and the salts in the seawater, (b) the minimum work input required to separate 1 kg of seawater completely into pure water and pure salts, (c) the minimum work input required to obtain 1 kg of fresh

228 706 Thermodynamics water from the sea, and (d ) the minimum gauge pressure that the seawater must be raised if fresh water is to be obtained by reverse osmosis using semipermeable membranes. Solution Fresh water is to be obtained from seawater. The mole fractions of seawater, the minimum works of separation needed for two limiting cases, and the required pressurization of seawater for reverse osmosis are to be determined. Assumptions 1 The seawater is an ideal solution since it is dilute. The total dissolved solids in water can be treated as table salt (NaCl). 3 The environment temperature is also 15 C. Properties The molar masses of water and salt are M w 18.0 kg/kmol and M s kg/kmol. The gas constant of pure water is R w kj/kg K (Table A 1). The density of seawater is 108 kg/m 3. Analysis (a) Noting that the mass fractions of salts and water in seawater are mf s and mf w 1 mf s 0.965, the mole fractions are determined from Eqs and 13 5 to be M m 1 mf i a (b) The minimum work input required to separate 1 kg of seawater completely into pure water and pure salts is w min,in w min,in M m M i 147. kj>kmol 1 mf s M s mf w M w M m kg>kmol y w mf w M w 18.0 kg>kmol y s 1 y w % w min,in R u T 0 1y A ln y A y B ln y B R u T 0 1y w ln y w y s ln y s kj>kmol # K K ln ln kj>kmol kg>kmol 7.98 kj /kg seawater Therefore, it takes a minimum of 7.98 kj of work input to separate 1 kg of seawater into kg of salt and kg (nearly 1 kg) of fresh water. (c) The minimum work input required to produce 1 kg of fresh water from seawater is w min,in R w T 0 ln 11>y w kj>kg # K Kln 11> kj/kg fresh water Note that it takes about 5 times more work to separate 1 kg of seawater completely into fresh water and salt than it does to produce 1 kg of fresh water from a large amount of seawater kg>kmol

229 Chapter (d ) The osmotic pressure in this case is P min r m R w T 0 ln 11>y w 1108 kg>m kpa # m 3 >kg # K Kln 11> kpa which is equal to the minimum gauge pressure to which seawater must be compressed if the fresh water is to be discharged at the local atmospheric pressure. As an alternative to pressurizing, the minimum height above the fresh water level that the seawater must be raised to produce fresh water is (Fig. 13 4) z min w min,in g 1.50 kj>kg Discussion The minimum separation works determined above also represent the maximum works that can be produced during the reverse process of mixing. Therefore, 7.98 kj of work can be produced when kg of salt is mixed with kg of water reversibly to produce 1 kg of saline water, and 1.50 kj of work can be produced as 1 kg of fresh water is mixed with seawater reversibly. Therefore, the power that can be generated as a river with a flow rate of 10 6 m 3 /s mixes reversibly with seawater through semipermeable membranes is (Fig. 13 5) W # max,out rv # w max,out kg>m m 3 >s kj>kga 1 MW 10 3 kj>s b MW 9.81 m>s a 1 kg # m>s 1 N ba 1000 N # m b 153 m 1 kj which shows the tremendous amount of power potential wasted as the rivers discharge into the seas. Saline water z Membrane P P 1 P P P 1 Pure water FIGURE 13 4 The osmotic pressure and the osmotic rise of saline water. Fresh and saline water mixing irreversibly Fresh river water Sea water salinity 3.48% z 153 m Fresh and saline water mixing reversibly through semi-permeable membranes, and producing power FIGURE 13 5 Power can be produced by mixing solutions of different concentrations reversibly.

230 708 Thermodynamics SUMMARY A mixture of two or more gases of fixed chemical composition is called a nonreacting gas mixture The composition of a gas mixture is described by specifying either the mole fraction or the mass fraction of each component, defined as where The apparent (or average) molar mass and gas constant of a mixture are expressed as Also, Dalton s law of additive pressures states that the pressure of a gas mixture is equal to the sum of the pressures each gas would exert if it existed alone at the mixture temperature and volume. Amagat s law of additive volumes states that the volume of a gas mixture is equal to the sum of the volumes each gas would occupy if it existed alone at the mixture temperature and pressure. Dalton s and Amagat s laws hold exactly for ideal-gas mixtures, but only approximately for real-gas mixtures. They can be expressed as Dalton s law: Amagat s law: mf i m i m m and y i N i N m m m a k i 1 M m m m N m a k m i and N m a k i 1 y i M i and R m R u M m mf i y M i i and M M m 1 k m mf i a P m a k V m a k Here P i is called the component pressure and V i is called the component volume. Also, the ratio P i /P m is called the pressure fraction and the ratio V i /V m is called the volume fraction of component i. For ideal gases, P i and V i can be related to y i by P i V i N i y P m V m N i m The quantity y i P m is called the partial pressure and the quantity y i V m is called the partial volume. The P-v-T behavior of real-gas mixtures can be predicted by using generalized i 1 i 1 N i i 1 i 1 M i P i 1T m, V m V i 1T m, P m compressibility charts. The compressibility factor of the mixture can be expressed in terms of the compressibility factors of the individual gases as where Z i is determined either at T m and V m (Dalton s law) or at T m and P m (Amagat s law) for each individual gas. The P-v-T behavior of a gas mixture can also be predicted approximately by Kay s rule, which involves treating a gas mixture as a pure substance with pseudocritical properties determined from The extensive properties of a gas mixture, in general, can be determined by summing the contributions of each component of the mixture. The evaluation of intensive properties of a gas mixture, however, involves averaging in terms of mass or mole fractions: and P cr,m a k i 1 U m a k u m a k i 1 h m a k i 1 s m a k i 1 c v,m a k i 1 c p,m a k i 1 i 1 H m a k i 1 S m a k i 1 k Z m a y i Z i U i a k H i a k S i a k i 1 i 1 y i P cr,i and T cr,m a k i 1 i 1 m i u i a k m i s i a k mf i u i and u m a k mf i h i and h m a k mf i s i and s m a k mf i c v,i and c v,m a k mf i c p,i and c p,m a k These relations are exact for ideal-gas mixtures and approximate for real-gas mixtures. The properties or property changes of individual components can be determined by using idealgas or real-gas relations developed in earlier chapters. i 1 m i h i a k i 1 i 1 i 1 i 1 i 1 N i u i N i h i N i s i y i u i y i h i y i s i i 1 i 1 i 1 y i T cr,i y i c v,i y i c p,i

231 Chapter REFERENCES AND SUGGESTED READING 1. A. Bejan. Advanced Engineering Thermodynamics. nd ed. New York: Wiley Interscience, Y. A. Çengel, Y. Cerci, and B. Wood, Second Law Analysis of Separation Processes of Mixtures, ASME International Mechanical Engineering Congress and Exposition, Nashville, Tennessee, Y. Cerci, Y. A. Çengel, and B. Wood, The Minimum Separation Work for Desalination Processes, ASME International Mechanical Engineering Congress and Exposition, Nashville, Tennessee, J. P. Holman. Thermodynamics. 3rd ed. New York: McGraw-Hill, K. Wark, Jr. Advanced Thermodynamics for Engineers. New York: McGraw-Hill, PROBLEMS* Composition of Gas Mixtures 13 1C What is the apparent gas constant for a gas mixture? Can it be larger than the largest gas constant in the mixture? 13 C Consider a mixture of two gases. Can the apparent molar mass of this mixture be determined by simply taking the arithmetic average of the molar masses of the individual gases? When will this be the case? 13 3C What is the apparent molar mass for a gas mixture? Does the mass of every molecule in the mixture equal the apparent molar mass? 13 4C Consider a mixture of several gases of identical masses. Will all the mass fractions be identical? How about the mole fractions? 13 5C The sum of the mole fractions for an ideal-gas mixture is equal to 1. Is this also true for a real-gas mixture? 13 6C What are mass and mole fractions? 13 7C Using the definitions of mass and mole fractions, derive a relation between them. 13 8C Somebody claims that the mass and mole fractions for a mixture of CO and N O gases are identical. Is this true? Why? 13 9C Consider a mixture of two gases A and B. Show that when the mass fractions mf A and mf B are known, the mole fractions can be determined from y A M B M A 11>mf A 1 M B and y B 1 y A where M A and M B are the molar masses of A and B. *Problems designated by a C are concept questions, and students are encouraged to answer them all. Problems designated by an E are in English units, and the SI users can ignore them. Problems with a CD-EES icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed DVD. Problems with a computer-ees icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text The composition of moist air is given on a molar basis to be 78 percent N, 0 percent O, and percent water vapor. Determine the mass fractions of the constituents of air A gas mixture has the following composition on a mole basis: 60 percent N and 40 percent CO. Determine the gravimetric analysis of the mixture, its molar mass, and gas constant Repeat Prob by replacing N by O A gas mixture consists of 5 kg of O, 8 kg of N, and 10 kg of CO. Determine (a) the mass fraction of each component, (b) the mole fraction of each component, and (c) the average molar mass and gas constant of the mixture Determine the mole fractions of a gas mixture that consists of 75 percent CH 4 and 5 percent CO by mass. Also, determine the gas constant of the mixture A gas mixture consists of 8 kmol of H and kmol of N. Determine the mass of each gas and the apparent gas constant of the mixture. Answers: 16 kg, 56 kg, kj/kg K 13 16E A gas mixture consists of 5 lbmol of H and 4 lbmol of N. Determine the mass of each gas and the apparent gas constant of the mixture A gas mixture consists of 0 percent O, 30 percent N, and 50 percent CO on mass basis. Determine the volumetric analysis of the mixture and the apparent gas constant. P-v-T Behavior of Gas Mixtures 13 18C Is a mixture of ideal gases also an ideal gas? Give an example C Express Dalton s law of additive pressures. Does this law hold exactly for ideal-gas mixtures? How about nonidealgas mixtures? 13 0C Express Amagat s law of additive volumes. Does this law hold exactly for ideal-gas mixtures? How about nonidealgas mixtures?

232 710 Thermodynamics 13 1C How is the P-v-T behavior of a component in an ideal-gas mixture expressed? How is the P-v-T behavior of a component in a real-gas mixture expressed? 13 C What is the difference between the component pressure and the partial pressure? When are these two equivalent? 13 3C What is the difference between the component volume and the partial volume? When are these two equivalent? 13 4C In a gas mixture, which component will have the higher partial pressure the one with the higher mole number or the one with the larger molar mass? 13 5C Consider a rigid tank that contains a mixture of two ideal gases. A valve is opened and some gas escapes. As a result, the pressure in the tank drops. Will the partial pressure of each component change? How about the pressure fraction of each component? 13 6C Consider a rigid tank that contains a mixture of two ideal gases. The gas mixture is heated, and the pressure and temperature in the tank rise. Will the partial pressure of each component change? How about the pressure fraction of each component? 13 7C Is this statement correct? The volume of an idealgas mixture is equal to the sum of the volumes of each individual gas in the mixture. If not, how would you correct it? 13 8C Is this statement correct? The temperature of an ideal-gas mixture is equal to the sum of the temperatures of each individual gas in the mixture. If not, how would you correct it? 13 9C Is this statement correct? The pressure of an idealgas mixture is equal to the sum of the partial pressures of each individual gas in the mixture. If not, how would you correct it? 13 30C Explain how a real-gas mixture can be treated as a pseudopure substance using Kay s rule A rigid tank contains 8 kmol of O and 10 kmol of CO gases at 90 K and 150 kpa. Estimate the volume of the tank. Answer: 89 m Repeat Prob for a temperature of 400 K A rigid tank contains 0.5 kmol of Ar and kmol of N at 50 kpa and 80 K. The mixture is now heated to 400 K. Determine the volume of the tank and the final pressure of the mixture A gas mixture at 300 K and 00 kpa consists of 1 kg of CO and 3 kg of CH 4. Determine the partial pressure of each gas and the apparent molar mass of the gas mixture E A gas mixture at 600 R and 0 psia consists of 1 lbm of CO and 3 lbm of CH 4. Determine the partial pressure of each gas and the apparent molar mass of the gas mixture A 0.3-m 3 rigid tank contains 0.6 kg of N and 0.4 kg of O at 300 K. Determine the partial pressure of each gas and the total pressure of the mixture kpa, 8.0 kpa Answers: kpa, A gas mixture at 350 K and 300 kpa has the following volumetric analysis: 65 percent N, 0 percent O, and 15 percent CO. Determine the mass fraction and partial pressure of each gas A rigid tank that contains 1 kg of N at 5 C and 300 kpa is connected to another rigid tank that contains 3 kg of O at 5 C and 500 kpa. The valve connecting the two tanks is opened, and the two gases are allowed to mix. If the final mixture temperature is 5 C, determine the volume of each tank and the final mixture pressure. Answers: 0.95 m 3, m 3, 4 kpa N 1 kg 5 C 300 kpa Ar 1 kmol 0 K 5 MPa FIGURE P13 38 O 3 kg 5 C 500 kpa A volume of 0.3 m 3 of O at 00 K and 8 MPa is mixed with 0.5 m 3 of N at the same temperature and pressure, forming a mixture at 00 K and 8 MPa. Determine the volume of the mixture, using (a) the ideal-gas equation of state, (b) Kay s rule, and (c) the compressibility chart and Amagat s law. Answers: (a) 0.8 m 3, (b) 0.79 m 3, (c) 0.80 m A rigid tank contains 1 kmol of Ar gas at 0 K and 5 MPa. A valve is now opened, and 3 kmol of N gas is allowed to enter the tank at 190 K and 8 MPa. The final mixture temperature is 00 K. Determine the pressure of the mixture, using (a) the ideal-gas equation of state and (b) the compressibility chart and Dalton s law. FIGURE P13 40 N 3 kmol 190 K 8 MPa Reconsider Prob Using EES (or other) software, study the effect of varying the moles of nitrogen supplied to the tank over the range of 1 to 10 kmol of N. Plot the final pressure of the mixture as a function of the amount of nitrogen supplied using the ideal-gas equation of state and the compressibility chart with Dalton s law.

233 13 4E A rigid tank contains 1 lbmol of argon gas at 400 R and 750 psia. A valve is now opened, and 3 lbmol of N gas is allowed to enter the tank at 340 R and 100 psia. The final mixture temperature is 360 R. Determine the pressure of the mixture, using (a) the ideal-gas equation of state and (b) the compressibility chart and Dalton s law. Answers: (a) 700 psia, (b) 507 psia Properties of Gas Mixtures 13 43C Is the total internal energy of an ideal-gas mixture equal to the sum of the internal energies of each individual gas in the mixture? Answer the same question for a real-gas mixture C Is the specific internal energy of a gas mixture equal to the sum of the specific internal energies of each individual gas in the mixture? 13 45C Answer Prob C and 13 44C for entropy C Is the total internal energy change of an ideal-gas mixture equal to the sum of the internal energy changes of each individual gas in the mixture? Answer the same question for a real-gas mixture C When evaluating the entropy change of the components of an ideal-gas mixture, do we have to use the partial pressure of each component or the total pressure of the mixture? 13 48C Suppose we want to determine the enthalpy change of a real-gas mixture undergoing a process. The enthalpy change of each individual gas is determined by using the generalized enthalpy chart, and the enthalpy change of the mixture is determined by summing them. Is this an exact approach? Explain A process requires a mixture that is 1 percent oxygen, 78 percent nitrogen, and 1 percent argon by volume. All three gases are supplied from separate tanks to an adiabatic, constant-pressure mixing chamber at 00 kpa but at different temperatures. The oxygen enters at 10 C, the nitrogen at 60 C, and the argon at 00 C. Determine the total entropy change for the mixing process per unit mass of mixture A mixture that is 15 percent carbon dioxide, 5 percent carbon monoxide, 10 percent oxygen, and 70 percent nitrogen by volume undergoes an adiabatic compression process having a compression ratio of 8:1. If the initial state of the mixture is 300 K and 100 kpa, determine the makeup of the mixture on a mass basis and the internal energy change per unit mass of mixture Propane and air are supplied to an internal combustion engine such that the air-fuel ratio is 16:1 when the pressure is 95 kpa and the temperature is 30 C. The compression ratio of the engine is 9.5:1. If the compression process is isentropic, determine the required work input for this compression process, in kj/kg of mixture An insulated rigid tank is divided into two compartments by a partition. One compartment contains.5 kmol of Chapter CO at 7 C and 00 kpa, and the other compartment contains 7.5 kmol of H gas at 40 C and 400 kpa. Now the partition is removed, and the two gases are allowed to mix. Determine (a) the mixture temperature and (b) the mixture pressure after equilibrium has been established. Assume constant specific heats at room temperature for both gases. CO.5 kmol 7 C 00 kpa H 7.5 kmol 40 C 400 kpa FIGURE P A 0.9-m 3 rigid tank is divided into two equal compartments by a partition. One compartment contains Ne at 0 C and 100 kpa, and the other compartment contains Ar at 50 C and 00 kpa. Now the partition is removed, and the two gases are allowed to mix. Heat is lost to the surrounding air during this process in the amount of 15 kj. Determine (a) the final mixture temperature and (b) the final mixture pressure. Answers: (a) 16. C, (b) kpa Repeat Prob for a heat loss of 8 kj Ethane (C H 6 ) at 0 C and 00 kpa and methane (CH 4 ) at 45 C and 00 kpa enter an adiabatic mixing chamber. The mass flow rate of ethane is 9 kg/s, which is twice the mass flow rate of methane. Determine (a) the mixture temperature and (b) the rate of entropy generation during this process, in kw/k. Take T 0 5 C Reconsider Prob Using EES (or other) software, determine the effect of the mass fraction of methane in the mixture on the mixture temperature and the rate of exergy destruction. The total mass flow rate is maintained constant at 13.5 kg/s, and the mass fraction of methane is varied from 0 to 1. Plot the mixture temperature and the rate of exergy destruction against the mass fraction, and discuss the results An equimolar mixture of helium and argon gases is to be used as the working fluid in a closed-loop gas-turbine cycle..5 MPa 1300 K He - Ar turbine 00 kpa FIGURE P13 57

234 71 Thermodynamics The mixture enters the turbine at.5 MPa and 1300 K and expands isentropically to a pressure of 00 kpa. Determine the work output of the turbine per unit mass of the mixture E A mixture of 80 percent N and 0 percent CO gases (on a mass basis) enters the nozzle of a turbojet engine at 90 psia and 1800 R with a low velocity, and it expands to a pressure of 1 psia. If the isentropic efficiency of the nozzle is 9 percent, determine (a) the exit temperature and (b) the exit velocity of the mixture. Assume constant specific heats at room temperature E Reconsider Prob E. Using EES (or other) software, first solve the stated problem and then, for all other conditions being the same, resolve the problem to determine the composition of the nitrogen and carbon dioxide that is required to have an exit velocity of 600 ft /s at the nozzle exit A piston cylinder device contains a mixture of 0.5 kg of H and 1.6 kg of N at 100 kpa and 300 K. Heat is now transferred to the mixture at constant pressure until the volume is doubled. Assuming constant specific heats at the average temperature, determine (a) the heat transfer and (b) the entropy change of the mixture An insulated tank that contains 1 kg of O at 15 C and 300 kpa is connected to a -m 3 uninsulated tank that contains N at 50 C and 500 kpa. The valve connecting the two tanks is opened, and the two gases form a homogeneous mixture at 5 C. Determine (a) the final pressure in the tank, (b) the heat transfer, and (c) the entropy generated during this process. Assume T 0 5 C. Answers: (a) kpa, (b) 187. kj, (c) 0.96 kj/k during this process by treating the mixture (a) as an ideal gas and (b) as a nonideal gas and using Amagat s law. Answers: (a) 473 kj, (b) 4745 kj Heat 6 kg H 1 kg N 160 K 5 MPa FIGURE P Determine the total entropy change and exergy destruction associated with the process described in Prob by treating the mixture (a) as an ideal gas and (b) as a nonideal gas and using Amagat s law. Assume constant specific heats at room temperature and take T 0 30 C Air, which may be considered as a mixture of 79 percent N and 1 percent O by mole numbers, is compressed isothermally at 00 K from 4 to 8 MPa in a steady-flow device. The compression process is internally reversible, and the mass flow rate of air is.9 kg/s. Determine the power input to the compressor and the rate of heat rejection by treating the mixture (a) as an ideal gas and (b) as a nonideal gas and using Amagat s law. Answers: (a) kw, kw, (b) kw, 94. kw O 1 kg 15 C 300 kpa N m 3 50 C 500 kpa 79% N 1% O 00 K 8 MPa W FIGURE P Reconsider Prob Using EES (or other) software, compare the results obtained assuming ideal-gas behavior with constant specific heats at the average temperature, and using real-gas data obtained from EES by assuming variable specific heats over the temperature range A piston cylinder device contains 6 kg of H and 1 kg of N at 160 K and 5 MPa. Heat is now transferred to the device, and the mixture expands at constant pressure until the temperature rises to 00 K. Determine the heat transfer 00 K 4 MPa FIGURE P Reconsider Prob Using EES (or other) software, compare the results obtained by assuming ideal behavior, real gas behavior with Amagat s law, and real gas behavior with EES data The combustion of a hydrocarbon fuel with air results in a mixture of products of combustion having the composition on a volume basis as follows: 4.89 percent carbon dioxide,

235 6.50 percent water vapor, 1.0 percent oxygen, and percent nitrogen. Determine the average molar mass of the mixture, the average specific heat at constant pressure of the mixture at 600 K, in kj/kmol K, and the partial pressure of the water vapor in the mixture for a mixture pressure of 00 kpa A mixture that is 0 percent carbon dioxide, 10 percent oxygen, and 70 percent nitrogen by volume undergoes a process from 300 K and 100 kpa to 500 K and 400 kpa. Determine the makeup of the mixture on a mass basis and the enthalpy change per unit mass of mixture. Special Topic: Chemical Potential and the Separation Work of Mixtures 13 69C It is common experience that two gases brought into contact mix by themselves. In the future, could it be possible to invent a process that will enable a mixture to separate into its components by itself without any work (or exergy) input? 13 70C A -L liquid is mixed with 3 L of another liquid, forming a homogeneous liquid solution at the same temperature and pressure. Can the volume of the solution be more or less than the 5 L? Explain C A -L liquid at 0 C is mixed with 3 L of another liquid at the same temperature and pressure in an adiabatic container, forming a homogeneous liquid solution. Someone claims that the temperature of the mixture rose to C after mixing. Another person refutes the claim, saying that this would be a violation of the first law of thermodynamics. Who do you think is right? 13 7C What is an ideal solution? Comment on the volume change, enthalpy change, entropy change, and chemical potential change during the formation of ideal and nonideal solutions Brackish water at 1 C with total dissolved solid content of TDS 780 ppm (a salinity of percent on mass basis) is to be used to produce fresh water with negligible salt content at a rate of 80 L/s. Determine the minimum power input required. Also, determine the minimum height to which the brackish water must be pumped if fresh water is to be obtained by reverse osmosis using semipermeable membranes A river is discharging into the ocean at a rate of 400,000 m 3 /s. Determine the amount of power that can be generated if the river water mixes with the ocean water reversibly. Take the salinity of the ocean to be 3.5 percent on mass basis, and assume both the river and the ocean are at 15 C Reconsider Prob Using EES (or other) software, investigate the effect of the salinity of the ocean on the maximum power generated. Let the salinity vary from 0 to 5 percent. Plot the power produced versus the salinity of the ocean, and discuss the results. Chapter E Fresh water is to be obtained from brackish water at 65 F with a salinity of 0.1 percent on mass basis (or TDS 100 ppm). Determine (a) the mole fractions of the water and the salts in the brackish water, (b) the minimum work input required to separate 1 lbm of brackish water completely into pure water and pure salts, and (c) the minimum work input required to obtain 1 lbm of fresh water A desalination plant produces fresh water from seawater at 10 C with a salinity of 3. percent on mass basis at a rate of 1.4 m 3 /s while consuming 8.5 MW of power. The salt content of the fresh water is negligible, and the amount of fresh water produced is a small fraction of the seawater used. Determine the second-law efficiency of this plant Fresh water is obtained from seawater at a rate of 0.5 m 3 /s by a desalination plant that consumes 3.3 MW of power and has a second-law efficiency of 18 percent. Determine the power that can be produced if the fresh water produced is mixed with the seawater reversibly. Review Problems Air has the following composition on a mole basis: 1 percent O, 78 percent N, and 1 percent Ar. Determine the gravimetric analysis of air and its molar mass. Answers: 3. percent O, 75.4 percent N, 1.4 percent Ar, 8.96 kg/kmol Using Amagat s law, show that k Z m a y i Z i for a real-gas mixture of k gases, where Z is the compressibility factor Using Dalton s law, show that i 1 Z m a k i 1 y i Z i for a real-gas mixture of k gases, where Z is the compressibility factor A mixture of carbon dioxide and nitrogen flows through a converging nozzle. The mixture leaves the nozzle at a temperature of 500 K with a velocity of 360 m/s. If the velocity is equal to the speed of sound at the exit temperature, determine the required makeup of the mixture on a mass basis A piston cylinder device contains products of combustion from the combustion of a hydrocarbon fuel with air. The combustion process results in a mixture that has the composition on a volume basis as follows: 4.89 percent carbon dioxide, 6.50 percent water vapor, 1.0 percent oxygen, and percent nitrogen. This mixture is initially at 1800 K and 1 MPa and expands in an adiabatic, reversible process to 00 kpa. Determine the work done on the piston by the gas, in kj/kg of mixture. Treat the water vapor as an ideal gas.

236 714 Thermodynamics A rigid tank contains kmol of N and 6 kmol of CH 4 gases at 00 K and 1 MPa. Estimate the volume of the tank, using (a) the ideal-gas equation of state, (b) Kay s rule, and (c) the compressibility chart and Amagat s law A steady stream of equimolar N and CO mixture at 100 kpa and 18 C is to be separated into N and CO gases at 100 kpa and 18 C. Determine the minimum work required per unit mass of mixture to accomplish this separation process. Assume T 0 18 C A gas mixture consists of O and N. The ratio of the mole numbers of N to O is 3:1. This mixture is heated during a steady-flow process from 180 to 10 K at a constant pressure of 8 MPa. Determine the heat transfer during this process per mole of the mixture, using (a) the ideal-gas approximation and (b) Kay s rule Reconsider Prob Using EES (or other) software, investigate the effect of the mole fraction of oxygen in the mixture on heat transfer using realgas behavior with EES data. Let the mole fraction of oxygen vary from 0 to 1. Plot the heat transfer against the mole fraction, and discuss the results Determine the total entropy change and exergy destruction associated with the process described in Prob , using (a) the ideal-gas approximation and (b) Kay s rule. Assume constant specific heats and T 0 30 C A rigid tank contains a mixture of 4 kg of He and 8 kg of O at 170 K and 7 MPa. Heat is now transferred to the tank, and the mixture temperature rises to 0 K. Treating the He as an ideal gas and the O as a nonideal gas, determine (a) the final pressure of the mixture and (b) the heat transfer A mixture of 60 percent carbon dioxide and 40 percent methane on a mole basis expands through a turbine from 1600 K and 800 kpa to 100 kpa. The volume flow rate at the turbine entrance is 10 L/s. Determine the rate of work done by the mixture using (a) ideal-gas approximation and (b) Kay s rule A pipe fitted with a closed valve connects two tanks. One tank contains a 5-kg mixture of 6.5 percent CO and 37.5 percent O on a mole basis at 30 C and 15 kpa. The second tank contains 10 kg of N at 15 C and 00 kpa. The valve in the pipe is opened and the gases are allowed to mix. During the mixing process 100 kj of heat energy is supplied to the combined tanks. Determine the final pressure and temperature of the mixture and the total volume of the mixture Using EES (or other) software, write a program to determine the mole fractions of the components of a mixture of three gases with known molar masses when the mass fractions are given, and to determine the mass fractions of the components when the mole fractions are given. Run the program for a sample case, and give the results Using EES (or other) software, write a program to determine the apparent gas constant, constant volume specific heat, and internal energy of a mixture of three ideal gases when the mass fractions and other properties of the constituent gases are given. Run the program for a sample case, and give the results Using EES (or other) software, write a program to determine the entropy change of a mixture of three ideal gases when the mass fractions and other properties of the constituent gases are given. Run the program for a sample case, and give the results. Fundamentals of Engineering (FE) Exam Problems An ideal-gas mixture whose apparent molar mass is 36 kg/kmol consists of N and three other gases. If the mole fraction of nitrogen is 0.30, its mass fraction is (a) 0.15 (b) 0.3 (c) 0.30 (d ) 0.39 (e) An ideal-gas mixture consists of kmol of N and 6 kmol of CO. The mass fraction of CO in the mixture is (a) (b) 0.50 (c) (d ) (e) An ideal-gas mixture consists of kmol of N and 4 kmol of CO. The apparent gas constant of the mixture is (a) 0.15 kj/kg K (b) 0.5 kj/kg K (c) 0.43 kj/kg K (d) kj/kg K (e) 1.4 kj/kg K A rigid tank is divided into two compartments by a partition. One compartment contains 3 kmol of N at 600 kpa and the other compartment contains 7 kmol of CO at 00 kpa. Now the partition is removed, and the two gases form a homogeneous mixture at 300 kpa. The partial pressure of N in the mixture is (a) 75 kpa (b) 90 kpa (c) 150 kpa (d ) 175 kpa (e) 5 kpa An 80-L rigid tank contains an ideal-gas mixture of 5 g of N and 5 g of CO at a specified pressure and temperature. If N were separated from the mixture and stored at mixture temperature and pressure, its volume would be (a) 3 L (b) 36 L (c) 40 L (d) 49 L (e) 80 L An ideal-gas mixture consists of 3 kg of Ar and 6 kg of CO gases. The mixture is now heated at constant volume from 50 K to 350 K. The amount of heat transfer is (a) 374 kj (b) 436 kj (c) 488 kj (d ) 55 kj (e) 664 kj An ideal-gas mixture consists of 30 percent helium and 70 percent argon gases by mass. The mixture is now expanded isentropically in a turbine from 400 C and 1. MPa to a pressure of 00 kpa. The mixture temperature at turbine exit is (a) 195 C (b) 56 C (c) 11 C (d ) 130 C (e) 400 C

237 13 10 One compartment of an insulated rigid tank contains kmol of CO at 0 C and 150 kpa while the other compartment contains 5 kmol of H gas at 35 C and 300 kpa. Now the partition between the two gases is removed, and the two gases form a homogeneous ideal-gas mixture. The temperature of the mixture is (a) 5 C (b) 9 C (c) C (d ) 3 C (e) 34 C A piston cylinder device contains an ideal-gas mixture of 3 kmol of He gas and 7 kmol of Ar gas at 50 C and 400 kpa. Now the gas expands at constant pressure until its volume doubles. The amount of heat transfer to the gas mixture is (a) 6. MJ (b) 4. MJ (c) 7 MJ (d ) 10 MJ (e) 67 MJ An ideal-gas mixture of helium and argon gases with identical mass fractions enters a turbine at 100 K and 1 MPa at a rate of 0.3 kg/s, and expands isentropically to 100 kpa. The power output of the turbine is (a) 478 kw (b) 619 kw (c) 96 KW (d ) 79 kw (e) 564 kw Chapter Design and Essay Problem Prolonged exposure to mercury even at relatively low but toxic concentrations in the air is known to cause permanent mental disorders, insomnia, and pain and numbness in the hands and the feet, among other things. Therefore, the maximum allowable concentration of mercury vapor in the air at work places is regulated by federal agencies. These regulations require that the average level of mercury concentration in the air does not exceed 0.1 mg/m 3. Consider a mercury spill that occurs in an airtight storage room at 0 C in San Francisco during an earthquake. Calculate the highest level of mercury concentration in the air that can occur in the storage room, in mg/m 3, and determine if it is within the safe level. The vapor pressure of mercury at 0 C is Pa. Propose some guidelines to safeguard against the formation of toxic concentrations of mercury vapor in air in storage rooms and laboratories.

238

239 Chapter 14 GAS VAPOR MIXTURES AND AIR-CONDITIONING At temperatures below the critical temperature, the gas phase of a substance is frequently referred to as a vapor. The term vapor implies a gaseous state that is close to the saturation region of the substance, raising the possibility of condensation during a process. In Chap. 13, we discussed mixtures of gases that are usually above their critical temperatures. Therefore, we were not concerned about any of the gases condensing during a process. Not having to deal with two phases greatly simplified the analysis. When we are dealing with a gas vapor mixture, however, the vapor may condense out of the mixture during a process, forming a two-phase mixture. This may complicate the analysis considerably. Therefore, a gas vapor mixture needs to be treated differently from an ordinary gas mixture. Several gas vapor mixtures are encountered in engineering. In this chapter, we consider the air water-vapor mixture, which is the most commonly encountered gas vapor mixture in practice. We also discuss air-conditioning, which is the primary application area of air water-vapor mixtures. Objectives The objectives of Chapter 14 are to: Differentiate between dry air and atmospheric air. Define and calculate the specific and relative humidity of atmospheric air. Calculate the dew-point temperature of atmospheric air. Relate the adiabatic saturation temperature and wet-bulb temperatures of atmospheric air. Use the psychrometric chart as a tool to determine the properties of atmospheric air. Apply the principles of the conservation of mass and energy to various air-conditioning processes. 717

240 718 Thermodynamics T, C 50 T, C DRY AIR c p,kj/kg C FIGURE 14 1 The c p of air can be assumed to be constant at kj/kg C in the temperature range 10 to 50 C with an error under 0. percent. h = const. FIGURE 14 At temperatures below 50 C, the h constant lines coincide with the T constant lines in the superheated vapor region of water. s 14 1 DRY AND ATMOSPHERIC AIR Air is a mixture of nitrogen, oxygen, and small amounts of some other gases. Air in the atmosphere normally contains some water vapor (or moisture) and is referred to as atmospheric air. By contrast, air that contains no water vapor is called dry air. It is often convenient to treat air as a mixture of water vapor and dry air since the composition of dry air remains relatively constant, but the amount of water vapor changes as a result of condensation and evaporation from oceans, lakes, rivers, showers, and even the human body. Although the amount of water vapor in the air is small, it plays a major role in human comfort. Therefore, it is an important consideration in air-conditioning applications. The temperature of air in air-conditioning applications ranges from about 10 to about 50 C. In this range, dry air can be treated as an ideal gas with a constant c p value of kj/kg K [0.40 Btu/lbm R] with negligible error (under 0. percent), as illustrated in Fig Taking 0 C as the reference temperature, the enthalpy and enthalpy change of dry air can be determined from and h dry air c p T kj>kg # CT 1kJ>kg h dry air c p T kj>kg # C T 1kJ>kg (14 1a) (14 1b) where T is the air temperature in C and T is the change in temperature. In air-conditioning processes we are concerned with the changes in enthalpy h, which is independent of the reference point selected. It certainly would be very convenient to also treat the water vapor in the air as an ideal gas and you would probably be willing to sacrifice some accuracy for such convenience. Well, it turns out that we can have the convenience without much sacrifice. At 50 C, the saturation pressure of water is 1.3 kpa. At pressures below this value, water vapor can be treated as an ideal gas with negligible error (under 0. percent), even when it is a saturated vapor. Therefore, water vapor in air behaves as if it existed alone and obeys the ideal-gas relation Pv RT. Then the atmospheric air can be treated as an ideal-gas mixture whose pressure is the sum of the partial pressure of dry air* P a and that of water vapor P v : P P a P v 1kPa (14 ) The partial pressure of water vapor is usually referred to as the vapor pressure. It is the pressure water vapor would exert if it existed alone at the temperature and volume of atmospheric air. Since water vapor is an ideal gas, the enthalpy of water vapor is a function of temperature only, that is, h h(t ). This can also be observed from the T-s diagram of water given in Fig. A 9 and Fig. 14 where the constantenthalpy lines coincide with constant-temperature lines at temperatures *Throughout this chapter, the subscript a denotes dry air and the subscript v denotes water vapor.

241 below 50 C. Therefore, the enthalpy of water vapor in air can be taken to be equal to the enthalpy of saturated vapor at the same temperature. That is, (14 3) The enthalpy of water vapor at 0 C is kj/kg. The average c p value of water vapor in the temperature range 10 to 50 C can be taken to be 1.8 kj/kg C. Then the enthalpy of water vapor can be determined approximately from or (14 4) (14 5) in the temperature range 10 to 50 C (or 15 to 10 F), with negligible error, as shown in Fig SPECIFIC AND RELATIVE HUMIDITY OF AIR The amount of water vapor in the air can be specified in various ways. Probably the most logical way is to specify directly the mass of water vapor present in a unit mass of dry air. This is called absolute or specific humidity (also called humidity ratio) and is denoted by v: The specific humidity can also be expressed as or h v 1T, low P h g 1T h g 1T T 1kJ>kg T in C h g 1T T 1Btu>lbm T in F v m v m a 1kg water vapor>kg dry air v m v m a P vv>r v T P a V>R a T P v >R v P a >R a 0.6 P v P a v 0.6P v P P v 1kg water vapor>kg dry air (14 6) (14 7) (14 8) where P is the total pressure. Consider 1 kg of dry air. By definition, dry air contains no water vapor, and thus its specific humidity is zero. Now let us add some water vapor to this dry air. The specific humidity will increase. As more vapor or moisture is added, the specific humidity will keep increasing until the air can hold no more moisture. At this point, the air is said to be saturated with moisture, and it is called saturated air. Any moisture introduced into saturated air will condense. The amount of water vapor in saturated air at a specified temperature and pressure can be determined from Eq by replacing P v by P g, the saturation pressure of water at that temperature (Fig. 14 4). The amount of moisture in the air has a definite effect on how comfortable we feel in an environment. However, the comfort level depends more on the amount of moisture the air holds (m v ) relative to the maximum amount of moisture the air can hold at the same temperature (m g ). The ratio of these two quantities is called the relative humidity f (Fig. 14 5) f m v m g P vv>r v T P g V>R v T P v P g (14 9) Chapter WATER VAPOR h g,kj/kg T, C Table A-4 Eq Difference, kj/kg FIGURE 14 3 In the temperature range 10 to 50 C, the h g of water can be determined from Eq with negligible error. AIR 5 C,100 kpa (P sat,h 5 C = kpa) P v = 0 P v < kpa P v = kpa dry air unsaturated air saturated air FIGURE 14 4 For saturated air, the vapor pressure is equal to the saturation pressure of water.

242 70 Thermodynamics AIR 5 C,1 atm m a = 1 kg m v = 0.01 kg m v, max = 0.0 kg Specific humidity: ω = 0.01 kg H O kg dry air Relative humidity: φ = 50% FIGURE 14 5 Specific humidity is the actual amount of water vapor in 1 kg of dry air, whereas relative humidity is the ratio of the actual amount of moisture in the air at a given temperature to the maximum amount of moisture air can hold at the same temperature. Dry air 1 kg h a moisture ω kg h g h = h a + ωh g,kj/kg dry air (1 + ω) kg of moist air where (14 10) Combining Eqs and 14 9, we can also express the relative humidity as (14 11a, b) The relative humidity ranges from 0 for dry air to 1 for saturated air. Note that the amount of moisture air can hold depends on its temperature. Therefore, the relative humidity of air changes with temperature even when its specific humidity remains constant. Atmospheric air is a mixture of dry air and water vapor, and thus the enthalpy of air is expressed in terms of the enthalpies of the dry air and the water vapor. In most practical applications, the amount of dry air in the air water-vapor mixture remains constant, but the amount of water vapor changes. Therefore, the enthalpy of atmospheric air is expressed per unit mass of dry air instead of per unit mass of the air water vapor mixture. The total enthalpy (an extensive property) of atmospheric air is the sum of the enthalpies of dry air and the water vapor: Dividing by m a gives or f P g P T vp 10.6 vp g and v 0.6fP g P fp g H H a H v m a h a m v h v h H m a h a m v m a h v h a vh v h h a vh g 1kJ>kg dry air (14 1) since h v h g (Fig. 14 6). Also note that the ordinary temperature of atmospheric air is frequently referred to as the dry-bulb temperature to differentiate it from other forms of temperatures that shall be discussed. FIGURE 14 6 The enthalpy of moist (atmospheric) air is expressed per unit mass of dry air, not per unit mass of moist air. EXAMPLE 14 1 The Amount of Water Vapor in Room Air ROOM 5 m 5 m 3 m T = 5 C P = 100 kpa φ = 75% FIGURE 14 7 Schematic for Example A 5-m 5-m 3-m room shown in Fig contains air at 5 C and 100 kpa at a relative humidity of 75 percent. Determine (a) the partial pressure of dry air, (b) the specific humidity, (c) the enthalpy per unit mass of the dry air, and (d ) the masses of the dry air and water vapor in the room. Solution The relative humidity of air in a room is given. The dry air pressure, specific humidity, enthalpy, and the masses of dry air and water vapor in the room are to be determined. Assumptions The dry air and the water vapor in the room are ideal gases. Properties The constant-pressure specific heat of air at room temperature is c p kj/kg K (Table A a). For water at 5 C, we have T sat kpa and h g kj/kg (Table A 4).

243 Chapter Analysis (a) The partial pressure of dry air can be determined from Eq. 14 : where Thus, P a P P v P v fp g fp 5 C kpa.38 kpa P a kpa 97.6 kpa (b) The specific humidity of air is determined from Eq. 14 8: v 0.6P v P P v (c) The enthalpy of air per unit mass of dry air is determined from Eq. 14 1: 63.8 kj/kg dry air kpa kpa kg H O/kg dry air h h a vh v c p T vh g kj>kg # C15 C kj>kg The enthalpy of water vapor (546.5 kj/kg) could also be determined from the approximation given by Eq. 14 4: h 5 C kj>kg which is almost identical to the value obtained from Table A 4. (d ) Both the dry air and the water vapor fill the entire room completely. Therefore, the volume of each gas is equal to the volume of the room: V a V v V room 15 m 15 m13 m 75 m 3 The masses of the dry air and the water vapor are determined from the idealgas relation applied to each gas separately: m a P av a R a T kpa 175 m kpa # m 3 >kg # K198 K kg m v P vv v R v T 1.38 kpa175 m kpa # m 3 >kg # K198 K 1.30 kg The mass of the water vapor in the air could also be determined from Eq. 14 6: m v vm a kg 1.30 kg 14 3 DEW-POINT TEMPERATURE If you live in a humid area, you are probably used to waking up most summer mornings and finding the grass wet. You know it did not rain the night before. So what happened? Well, the excess moisture in the air simply condensed on the cool surfaces, forming what we call dew. In summer, a considerable amount of water vaporizes during the day. As the temperature falls during the

244 7 Thermodynamics T T 1 T dp P v = const. FIGURE 14 8 Constant-presssure cooling of moist air and the dew-point temperature on the T-s diagram of water. T < T dp 1 MOIST AIR Liquid water droplets (dew) FIGURE 14 9 When the temperature of a cold drink is below the dew-point temperature of the surrounding air, it sweats. AIR 0 C, 75% Typical temperature distribution s COLD OUTDOORS 10 C 18 C 0 C 0 C 0 C 18 C 16 C 16 C FIGURE Schematic for Example 14. night, so does the moisture capacity of air, which is the maximum amount of moisture air can hold. (What happens to the relative humidity during this process?) After a while, the moisture capacity of air equals its moisture content. At this point, air is saturated, and its relative humidity is 100 percent. Any further drop in temperature results in the condensation of some of the moisture, and this is the beginning of dew formation. The dew-point temperature T dp is defined as the temperature at which condensation begins when the air is cooled at constant pressure. In other words, T dp is the saturation temperature of water corresponding to the vapor pressure: T dp T Pv (14 13) This is also illustrated in Fig As the air cools at constant pressure, the vapor pressure P v remains constant. Therefore, the vapor in the air (state 1) undergoes a constant-pressure cooling process until it strikes the saturated vapor line (state ). The temperature at this point is T dp, and if the temperature drops any further, some vapor condenses out. As a result, the amount of vapor in the air decreases, which results in a decrease in P v. The air remains saturated during the condensation process and thus follows a path of 100 percent relative humidity (the saturated vapor line). The ordinary temperature and the dew-point temperature of saturated air are identical. You have probably noticed that when you buy a cold canned drink from a vending machine on a hot and humid day, dew forms on the can. The formation of dew on the can indicates that the temperature of the drink is below the dew-point temperature of the surrounding air (Fig. 14 9). The dew-point temperature of room air can be determined easily by cooling some water in a metal cup by adding small amounts of ice and stirring. The temperature of the outer surface of the cup when dew starts to form on the surface is the dew-point temperature of the air. EXAMPLE 14 Fogging of the Windows in a House In cold weather, condensation frequently occurs on the inner surfaces of the windows due to the lower air temperatures near the window surface. Consider a house, shown in Fig , that contains air at 0 C and 75 percent relative humidity. At what window temperature will the moisture in the air start condensing on the inner surfaces of the windows? Solution The interior of a house is maintained at a specified temperature and humidity. The window temperature at which fogging starts is to be determined. Properties The saturation pressure of water at 0 C is P sat.339 kpa (Table A 4). Analysis The temperature distribution in a house, in general, is not uniform. When the outdoor temperature drops in winter, so does the indoor temperature near the walls and the windows. Therefore, the air near the walls and the windows remains at a lower temperature than at the inner parts of a house even though the total pressure and the vapor pressure remain constant throughout the house. As a result, the air near the walls and the windows undergoes a P v constant cooling process until the moisture in the air

245 Chapter starts condensing. This happens when the air reaches its dew-point temperature T dp, which is determined from Eq to be T dp T Pv where Thus, P v fp 0 C kpa kpa T dp T kpa 15.4 C Discussion Note that the inner surface of the window should be maintained above 15.4 C if condensation on the window surfaces is to be avoided ADIABATIC SATURATION AND WET-BULB TEMPERATURES Relative humidity and specific humidity are frequently used in engineering and atmospheric sciences, and it is desirable to relate them to easily measurable quantities such as temperature and pressure. One way of determining the relative humidity is to determine the dew-point temperature of air, as discussed in the last section. Knowing the dew-point temperature, we can determine the vapor pressure P v and thus the relative humidity. This approach is simple, but not quite practical. Another way of determining the absolute or relative humidity is related to an adiabatic saturation process, shown schematically and on a T-s diagram in Fig The system consists of a long insulated channel that contains a pool of water. A steady stream of unsaturated air that has a specific humidity of v 1 (unknown) and a temperature of T 1 is passed through this channel. As the air flows over the water, some water evaporates and mixes with the airstream. The moisture content of air increases during this process, and its temperature decreases, since part of the latent heat of vaporization of the water that evaporates comes from the air. If the channel is long enough, the airstream exits as saturated air (f 100 percent) at temperature T, which is called the adiabatic saturation temperature. If makeup water is supplied to the channel at the rate of evaporation at temperature T, the adiabatic saturation process described above can be analyzed as a steady-flow process. The process involves no heat or work interactions, and the kinetic and potential energy changes can be neglected. Then the conservation of mass and conservation of energy relations for this twoinlet, one-exit steady-flow system reduces to the following: Mass balance: or m # a 1 m # a m # a m # w 1 m # f m # w (The mass flow rate of dry air remains constant) (The mass flow rate of vapor in the air increases by an amount equal to the rate of evaporation ṁ f ) Unsaturated air T 1, ω 1 f 1 T 1 Liquid water Adiabatic saturation temperature P v1 1 Saturated air T, ω f 100% Liquid water at T Dew-point temperature FIGURE The adiabatic saturation process and its representation on a T-s diagram of water. s m # av 1 m # f m # av

246 74 Thermodynamics Thus, Energy balance: m # f m # a 1v v 1 E # in E # out 1since Q # 0 and W # 0 or m # a h 1 m # f h f m # a h Dividing by ṁ a gives m # a h 1 m # a 1v v 1 h f m # a h or which yields h 1 1v v 1 h f h 1c p T 1 v 1 h g1 1v v 1 h f 1c p T v h g where, from Eq b, v 1 c p 1T T 1 v h fg h g1 h f v 0.6P g P P g (14 14) (14 15) Ordinary thermometer Wet-bulb thermometer Air flow Wick Liquid water FIGURE 14 1 A simple arrangement to measure the wet-bulb temperature. since f 100 percent. Thus we conclude that the specific humidity (and relative humidity) of air can be determined from Eqs and by measuring the pressure and temperature of air at the inlet and the exit of an adiabatic saturator. If the air entering the channel is already saturated, then the adiabatic saturation temperature T will be identical to the inlet temperature T 1, in which case Eq yields v 1 v. In general, the adiabatic saturation temperature is between the inlet and dew-point temperatures. The adiabatic saturation process discussed above provides a means of determining the absolute or relative humidity of air, but it requires a long channel or a spray mechanism to achieve saturation conditions at the exit. A more practical approach is to use a thermometer whose bulb is covered with a cotton wick saturated with water and to blow air over the wick, as shown in Fig The temperature measured in this manner is called the wet-bulb temperature T wb, and it is commonly used in air-conditioning applications. The basic principle involved is similar to that in adiabatic saturation. When unsaturated air passes over the wet wick, some of the water in the wick evaporates. As a result, the temperature of the water drops, creating a temperature difference (which is the driving force for heat transfer) between the air and the water. After a while, the heat loss from the water by evaporation equals the heat gain from the air, and the water temperature stabilizes. The thermometer reading at this point is the wet-bulb temperature. The wetbulb temperature can also be measured by placing the wet-wicked thermometer in a holder attached to a handle and rotating the holder rapidly, that is, by moving the thermometer instead of the air. A device that works

247 on this principle is called a sling psychrometer and is shown in Fig Usually a dry-bulb thermometer is also mounted on the frame of this device so that both the wet- and dry-bulb temperatures can be read simultaneously. Advances in electronics made it possible to measure humidity directly in a fast and reliable way. It appears that sling psychrometers and wet-wicked thermometers are about to become things of the past. Today, hand-held electronic humidity measurement devices based on the capacitance change in a thin polymer film as it absorbs water vapor are capable of sensing and digitally displaying the relative humidity within 1 percent accuracy in a matter of seconds. In general, the adiabatic saturation temperature and the wet-bulb temperature are not the same. However, for air water vapor mixtures at atmospheric pressure, the wet-bulb temperature happens to be approximately equal to the adiabatic saturation temperature. Therefore, the wet-bulb temperature T wb can be used in Eq in place of T to determine the specific humidity of air. Wet-bulb thermometer Chapter Dry-bulb thermometer EXAMPLE 14 3 The Specific and Relative Humidity of Air The dry- and the wet-bulb temperatures of atmospheric air at 1 atm ( kpa) pressure are measured with a sling psychrometer and determined to be 5 and 15 C, respectively. Determine (a) the specific humidity, (b) the relative humidity, and (c) the enthalpy of the air. Solution Dry- and wet-bulb temperatures are given. The specific humidity, relative humidity, and enthalpy are to be determined. Properties The saturation pressure of water is kpa at 15 C, and kpa at 5 C (Table A 4). The constant-pressure specific heat of air at room temperature is c p kj/kg K (Table A a). Analysis (a) The specific humidity v 1 is determined from Eq , where T is the wet-bulb temperature and v is Thus, (b) The relative humidity f 1 is determined from Eq a to be f kg H O/kg dry air v 1 P 10.6 v 1 P g1 v 1 c p 1T T 1 v h fg h g1 h f v 0.6P g P P g kpa kpa kg H O>kg dry air v kj>kg # C C kj>kg kj>kg kpa 0.33 or 33.% kpa FIGURE Sling psychrometer. Wet-bulb thermometer wick

248 76 Thermodynamics (c) The enthalpy of air per unit mass of dry air is determined from Eq. 14 1: h 1 h a1 v 1 h v1 c p T 1 v 1 h g kj>kg # C15 C kj>kg 41.8 kj/kg dry air Discussion The previous property calculations can be performed easily using EES or other programs with built-in psychrometric functions. Saturation line, φ = 100% v = const. φ = const. Twb = const. h = const. Dry-bulb temperature Specific humidity, ω FIGURE Schematic for a psychrometric chart. Saturation line 15 C T db = 15 C 15 C T dp = 15 C T wb = 15 C FIGURE For saturated air, the dry-bulb, wet-bulb, and dew-point temperatures are identical THE PSYCHROMETRIC CHART The state of the atmospheric air at a specified pressure is completely specified by two independent intensive properties. The rest of the properties can be calculated easily from the previous relations. The sizing of a typical airconditioning system involves numerous such calculations, which may eventually get on the nerves of even the most patient engineers. Therefore, there is clear motivation to computerize calculations or to do these calculations once and to present the data in the form of easily readable charts. Such charts are called psychrometric charts, and they are used extensively in air-conditioning applications. A psychrometric chart for a pressure of 1 atm ( kpa or psia) is given in Fig. A 31 in SI units and in Fig. A 31E in English units. Psychrometric charts at other pressures (for use at considerably higher elevations than sea level) are also available. The basic features of the psychrometric chart are illustrated in Fig The dry-bulb temperatures are shown on the horizontal axis, and the specific humidity is shown on the vertical axis. (Some charts also show the vapor pressure on the vertical axis since at a fixed total pressure P there is a one-to-one correspondence between the specific humidity v and the vapor pressure P v, as can be seen from Eq ) On the left end of the chart, there is a curve (called the saturation line) instead of a straight line. All the saturated air states are located on this curve. Therefore, it is also the curve of 100 percent relative humidity. Other constant relative-humidity curves have the same general shape. Lines of constant wet-bulb temperature have a downhill appearance to the right. Lines of constant specific volume (in m 3 /kg dry air) look similar, except they are steeper. Lines of constant enthalpy (in kj/kg dry air) lie very nearly parallel to the lines of constant wet-bulb temperature. Therefore, the constantwet-bulb-temperature lines are used as constant-enthalpy lines in some charts. For saturated air, the dry-bulb, wet-bulb, and dew-point temperatures are identical (Fig ). Therefore, the dew-point temperature of atmospheric air at any point on the chart can be determined by drawing a horizontal line (a line of v constant or P v constant) from the point to the saturated curve. The temperature value at the intersection point is the dew-point temperature. The psychrometric chart also serves as a valuable aid in visualizing the airconditioning processes. An ordinary heating or cooling process, for example, appears as a horizontal line on this chart if no humidification or dehumidification is involved (that is, v constant). Any deviation from a horizontal line indicates that moisture is added or removed from the air during the process.

249 Chapter EXAMPLE 14 4 The Use of the Psychrometric Chart Consider a room that contains air at 1 atm, 35 C, and 40 percent relative humidity. Using the psychrometric chart, determine (a) the specific humidity, (b) the enthalpy, (c) the wet-bulb temperature, (d ) the dew-point temperature, and (e) the specific volume of the air. Solution The relative humidity of air in a room is given. The specific humidity, enthalpy, wet-bulb temperature, dew-point temperature, and specific volume of the air are to be determined using the psychrometric chart. Analysis At a given total pressure, the state of atmospheric air is completely specified by two independent properties such as the dry-bulb temperature and the relative humidity. Other properties are determined by directly reading their values at the specified state. (a) The specific humidity is determined by drawing a horizontal line from the specified state to the right until it intersects with the v axis, as shown in Fig At the intersection point we read v kg H O/kg dry air (b) The enthalpy of air per unit mass of dry air is determined by drawing a line parallel to the h constant lines from the specific state until it intersects the enthalpy scale, giving h 71.5 kj/kg dry air (c) The wet-bulb temperature is determined by drawing a line parallel to the T wb constant lines from the specified state until it intersects the saturation line, giving T wb 4 C (d ) The dew-point temperature is determined by drawing a horizontal line from the specified state to the left until it intersects the saturation line, giving T dp 19.4 C (e) The specific volume per unit mass of dry air is determined by noting the distances between the specified state and the v constant lines on both sides of the point. The specific volume is determined by visual interpolation to be v m 3 /kg dry air Discussion Values read from the psychrometric chart inevitably involve reading errors, and thus are of limited accuracy. T dp h T wb T = 35 C φ = 40% FIGURE Schematic for Example v ω 14 6 HUMAN COMFORT AND AIR-CONDITIONING Human beings have an inherent weakness they want to feel comfortable. They want to live in an environment that is neither hot nor cold, neither humid nor dry. However, comfort does not come easily since the desires of the human body and the weather usually are not quite compatible. Achieving comfort requires a constant struggle against the factors that cause discomfort, such as high or low temperatures and high or low humidity. As engineers, it is our duty to help people feel comfortable. (Besides, it keeps us employed.)

250 78 Thermodynamics FIGURE We cannot change the weather, but we can change the climate in a confined space by air-conditioning. Vol. 77/PhotoDisc 37 C Waste heat 3 C FIGURE A body feels comfortable when it can freely dissipate its waste heat, and no more. It did not take long for people to realize that they could not change the weather in an area. All they can do is change it in a confined space such as a house or a workplace (Fig ). In the past, this was partially accomplished by fire and simple indoor heating systems. Today, modern air-conditioning systems can heat, cool, humidify, dehumidify, clean, and even deodorize the air in other words, condition the air to peoples desires. Air-conditioning systems are designed to satisfy the needs of the human body; therefore, it is essential that we understand the thermodynamic aspects of the body. The human body can be viewed as a heat engine whose energy input is food. As with any other heat engine, the human body generates waste heat that must be rejected to the environment if the body is to continue operating. The rate of heat generation depends on the level of the activity. For an average adult male, it is about 87 W when sleeping, 115 W when resting or doing office work, 30 W when bowling, and 440 W when doing heavy physical work. The corresponding numbers for an adult female are about 15 percent less. (This difference is due to the body size, not the body temperature. The deep-body temperature of a healthy person is maintained constant at about 37 C.) A body will feel comfortable in environments in which it can dissipate this waste heat comfortably (Fig ). Heat transfer is proportional to the temperature difference. Therefore in cold environments, a body loses more heat than it normally generates, which results in a feeling of discomfort. The body tries to minimize the energy deficit by cutting down the blood circulation near the skin (causing a pale look). This lowers the skin temperature, which is about 34 C for an average person, and thus the heat transfer rate. A low skin temperature causes discomfort. The hands, for example, feel painfully cold when the skin temperature reaches 10 C (50 F). We can also reduce the heat loss from the body either by putting barriers (additional clothes, blankets, etc.) in the path of heat or by increasing the rate of heat generation within the body by exercising. For example, the comfort level of a resting person dressed in warm winter clothing in a room at 10 C (50 F) is roughly equal to the comfort level of an identical person doing moderate work in a room at about 3 C ( 10 F). Or we can just cuddle up and put our hands between our legs to reduce the surface area through which heat flows. In hot environments, we have the opposite problem we do not seem to be dissipating enough heat from our bodies, and we feel as if we are going to burst. We dress lightly to make it easier for heat to get away from our bodies, and we reduce the level of activity to minimize the rate of waste heat generation in the body. We also turn on the fan to continuously replace the warmer air layer that forms around our bodies as a result of body heat by the cooler air in other parts of the room. When doing light work or walking slowly, about half of the rejected body heat is dissipated through perspiration as latent heat while the other half is dissipated through convection and radiation as sensible heat. When resting or doing office work, most of the heat (about 70 percent) is dissipated in the form of sensible heat whereas when doing heavy physical work, most of the heat (about 60 percent) is dissipated in the form of latent heat. The body helps out by perspiring or sweating more. As this sweat evaporates, it absorbs latent heat from the body and cools it. Perspiration is not much help, however, if the relative humidity of

251 the environment is close to 100 percent. Prolonged sweating without any fluid intake causes dehydration and reduced sweating, which may lead to a rise in body temperature and a heat stroke. Another important factor that affects human comfort is heat transfer by radiation between the body and the surrounding surfaces such as walls and windows. The sun s rays travel through space by radiation. You warm up in front of a fire even if the air between you and the fire is quite cold. Likewise, in a warm room you feel chilly if the ceiling or the wall surfaces are at a considerably lower temperature. This is due to direct heat transfer between your body and the surrounding surfaces by radiation. Radiant heaters are commonly used for heating hard-to-heat places such as car repair shops. The comfort of the human body depends primarily on three factors: the (dry-bulb) temperature, relative humidity, and air motion (Fig ). The temperature of the environment is the single most important index of comfort. Most people feel comfortable when the environment temperature is between and 7 C (7 and 80 F). The relative humidity also has a considerable effect on comfort since it affects the amount of heat a body can dissipate through evaporation. Relative humidity is a measure of air s ability to absorb more moisture. High relative humidity slows down heat rejection by evaporation, and low relative humidity speeds it up. Most people prefer a relative humidity of 40 to 60 percent. Air motion also plays an important role in human comfort. It removes the warm, moist air that builds up around the body and replaces it with fresh air. Therefore, air motion improves heat rejection by both convection and evaporation. Air motion should be strong enough to remove heat and moisture from the vicinity of the body, but gentle enough to be unnoticed. Most people feel comfortable at an airspeed of about 15 m/min. Very-high-speed air motion causes discomfort instead of comfort. For example, an environment at 10 C (50 F) with 48 km/h winds feels as cold as an environment at 7 C (0 F) with 3 km/h winds as a result of the body-chilling effect of the air motion (the wind-chill factor). Other factors that affect comfort are air cleanliness, odor, noise, and radiation effect. Chapter C f = 50% Air motion 15 m/min FIGURE A comfortable environment. Reprinted with special permission of King Features Syndicate AIR-CONDITIONING PROCESSES Maintaining a living space or an industrial facility at the desired temperature and humidity requires some processes called air-conditioning processes. These processes include simple heating (raising the temperature), simple cooling (lowering the temperature), humidifying (adding moisture), and dehumidifying (removing moisture). Sometimes two or more of these processes are needed to bring the air to a desired temperature and humidity level. Various air-conditioning processes are illustrated on the psychrometric chart in Fig Notice that simple heating and cooling processes appear as horizontal lines on this chart since the moisture content of the air remains constant (v constant) during these processes. Air is commonly heated and humidified in winter and cooled and dehumidified in summer. Notice how these processes appear on the psychrometric chart. Cooling Humidifying Cooling and dehumidifying Dehumidifying Heating and humidifying Heating FIGURE 14 0 Various air-conditioning processes.

252 730 Thermodynamics Most air-conditioning processes can be modeled as steady-flow processes, and thus the mass balance relation ṁ in ṁ out can be expressed for dry air and water as Mass balance for dry air: a m # a aout m # a 1kg>s (14 16) in Mass balance for water: a m # w aout m # w or ain m # a v aout m # a v (14 17) in Disregarding the kinetic and potential energy changes, the steady-flow energy balance relation E. in E. out can be expressed in this case as Q # in W # in ain m # h Q # out W # out aout m # h (14 18) The work term usually consists of the fan work input, which is small relative to the other terms in the energy balance relation. Next we examine some commonly encountered processes in air-conditioning. Heating coils Air T T 1, ω 1, f 1 Heat ω = ω 1 f < f 1 FIGURE 14 1 During simple heating, specific humidity remains constant, but relative humidity decreases. v = constant Cooling f = 80% f1 = 30% 1 1 C 30 C FIGURE 14 During simple cooling, specific humidity remains constant, but relative humidity increases. Simple Heating and Cooling (V constant) Many residential heating systems consist of a stove, a heat pump, or an electric resistance heater. The air in these systems is heated by circulating it through a duct that contains the tubing for the hot gases or the electric resistance wires, as shown in Fig The amount of moisture in the air remains constant during this process since no moisture is added to or removed from the air. That is, the specific humidity of the air remains constant (v constant) during a heating (or cooling) process with no humidification or dehumidification. Such a heating process proceeds in the direction of increasing dry-bulb temperature following a line of constant specific humidity on the psychrometric chart, which appears as a horizontal line. Notice that the relative humidity of air decreases during a heating process even if the specific humidity v remains constant. This is because the relative humidity is the ratio of the moisture content to the moisture capacity of air at the same temperature, and moisture capacity increases with temperature. Therefore, the relative humidity of heated air may be well below comfortable levels, causing dry skin, respiratory difficulties, and an increase in static electricity. A cooling process at constant specific humidity is similar to the heating process discussed above, except the dry-bulb temperature decreases and the relative humidity increases during such a process, as shown in Fig. 14. Cooling can be accomplished by passing the air over some coils through which a refrigerant or chilled water flows. The conservation of mass equations for a heating or cooling process that involves no humidification or dehumidification reduce to ṁ a1 ṁ a ṁa for dry air and v 1 v for water. Neglecting any fan work that may be present, the conservation of energy equation in this case reduces to Q # m # a 1h h 1 or q h h 1 where h 1 and h are enthalpies per unit mass of dry air at the inlet and the exit of the heating or cooling section, respectively.

253 Heating with Humidification Problems associated with the low relative humidity resulting from simple heating can be eliminated by humidifying the heated air. This is accomplished by passing the air first through a heating section (process 1-) and then through a humidifying section (process -3), as shown in Fig The location of state 3 depends on how the humidification is accomplished. If steam is introduced in the humidification section, this will result in humidification with additional heating (T 3 T ). If humidification is accomplished by spraying water into the airstream instead, part of the latent heat of vaporization comes from the air, which results in the cooling of the heated airstream (T 3 T ). Air should be heated to a higher temperature in the heating section in this case to make up for the cooling effect during the humidification process. Air Heating coils ω = ω 1 Chapter Heating section Humidifier ω 3 > ω Humidifying section FIGURE 14 3 Heating with humidification. EXAMPLE 14 5 Heating and Humidification of Air An air-conditioning system is to take in outdoor air at 10 C and 30 percent relative humidity at a steady rate of 45 m 3 /min and to condition it to 5 C and 60 percent relative humidity. The outdoor air is first heated to C in the heating section and then humidified by the injection of hot steam in the humidifying section. Assuming the entire process takes place at a pressure of 100 kpa, determine (a) the rate of heat supply in the heating section and (b) the mass flow rate of the steam required in the humidifying section. Solution Outdoor air is first heated and then humidified by steam injection. The rate of heat transfer and the mass flow rate of steam are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process. Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties The constant-pressure specific heat of air at room temperature is c p kj/kg K, and its gas constant is R a 0.87 kj/kg K (Table A a). The saturation pressure of water is 1.81 kpa at 10 C, and kpa at 5 C. The enthalpy of saturated water vapor is 519. kj/kg at 10 C, and kj/kg at C (Table A 4). Analysis We take the system to be the heating or the humidifying section, as appropriate. The schematic of the system and the psychrometric chart of the process are shown in Fig We note that the amount of water vapor in the air remains constant in the heating section (v 1 v ) but increases in the humidifying section (v 3 v ). (a) Applying the mass and energy balances on the heating section gives Dry air mass balance: m # a Water mass balance: m # 1 m # a Energy balance: Q # a in m # 1 v 1 m # m # a a a h 1 m # v S v 1 v a h S Q # in m # a 1h h 1 The psychrometric chart offers great convenience in determining the properties of moist air. However, its use is limited to a specified pressure only, which is 1 atm ( kpa) for the one given in the appendix. At pressures other than f 1 1 = 30% 10 C C 5 C Heating coils T 1 = 10 C T 3 = 5 C f Air 1 = 30% f 3 = 60% V 1 = 45 m 3 /min T = C f = 60% 3 Humidifier FIGURE 14 4 Schematic and psychrometric chart for Example 14 5.

254 73 Thermodynamics 1 atm, either other charts for that pressure or the relations developed earlier should be used. In our case, the choice is clear: since v v 1. Then the rate of heat transfer to air in the heating section becomes (b) The mass balance for water in the humidifying section can be expressed as or P v1 f 1 P g1 fp 10 C kpa kpa P a1 P 1 P v kpa kpa v 1 R at 1 P a v 1 0.6P v 1 P 1 P v1 where Thus, kpa # m 3 >kg # K183 K kpa m # a V# 1 45 m3 >min 55. kg>min v m 3 >kg 15.8 kj>kg dry air 8.0 kj>kg dry air kpa kpa kg H O>kg dry air h 1 c p T 1 v 1 h g kj>kg # C110 C kj>kg h c p T v h g kj>kg # C1 C kj>kg Q # in m # a 1h h kg>min kj>kg4 673 kj/min v 3 0.6f 3P g3 P 3 f 3 P g3 m # a v m # w m # a 3 v 3 m # w m # a 1v 3 v kg H O>kg dry air m # w 155. kg>min kg/min m 3 >kg dry air kpa kpa Discussion The result kg/min corresponds to a water requirement of close to one ton a day, which is significant. Cooling with Dehumidification The specific humidity of air remains constant during a simple cooling process, but its relative humidity increases. If the relative humidity reaches undesirably high levels, it may be necessary to remove some moisture from the air, that is, to dehumidify it. This requires cooling the air below its dewpoint temperature.

255 The cooling process with dehumidifying is illustrated schematically and on the psychrometric chart in Fig in conjunction with Example Hot, moist air enters the cooling section at state 1. As it passes through the cooling coils, its temperature decreases and its relative humidity increases at constant specific humidity. If the cooling section is sufficiently long, air reaches its dew point (state x, saturated air). Further cooling of air results in the condensation of part of the moisture in the air. Air remains saturated during the entire condensation process, which follows a line of 100 percent relative humidity until the final state (state ) is reached. The water vapor that condenses out of the air during this process is removed from the cooling section through a separate channel. The condensate is usually assumed to leave the cooling section at T. The cool, saturated air at state is usually routed directly to the room, where it mixes with the room air. In some cases, however, the air at state may be at the right specific humidity but at a very low temperature. In such cases, air is passed through a heating section where its temperature is raised to a more comfortable level before it is routed to the room. Chapter f 1 = 80% 1 x f = 100% 14 C 30 C Cooling coils EXAMPLE 14 6 Cooling and Dehumidification of Air Air enters a window air conditioner at 1 atm, 30 C, and 80 percent relative humidity at a rate of 10 m 3 /min, and it leaves as saturated air at 14 C. Part of the moisture in the air that condenses during the process is also removed at 14 C. Determine the rates of heat and moisture removal from the air. Solution Air is cooled and dehumidified by a window air conditioner. The rates of heat and moisture removal are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process. Dry air and the water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties The enthalpy of saturated liquid water at 14 C is 58.8 kj/kg (Table A 4). Also, the inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. Therefore, we can determine the properties of the air at both states from the psychrometric chart to be h kj/kg dry air h 39.3 kj/kg dry air v kg H O/kg dry air and v kg H O/kg dry air v m 3 /kg dry air Analysis We take the cooling section to be the system. The schematic of the system and the psychrometric chart of the process are shown in Fig We note that the amount of water vapor in the air decreases during the process (v v 1 ) due to dehumidification. Applying the mass and energy balances on the cooling and dehumidification section gives Dry air mass balance: Water mass balance: m # a 1 m # a m # a m # a 1 v 1 m # a v m # w S m # w m # a 1v 1 v Air Condensate 1 T = 14 C T 1 = 30 C f = 100% 14 C f 1 = 80% Condensate V 1 = 10 m 3 /min removal FIGURE 14 5 Schematic and psychrometric chart for Example Energy balance: ain m # h Q # out aout m # h S Q # out m # 1h 1 h m # w h w

256 734 Thermodynamics Then, Water that leaks out Hot, dry air FIGURE 14 6 Water in a porous jug left in an open, breezy area cools as a result of evaporative cooling. COOL, MOIST AIR ' 1 FIGURE 14 7 Evaporative cooling. T wb = ~ const. h = ~ const. Liquid water 1 HOT, DRY AIR Therefore, this air-conditioning unit removes moisture and heat from the air at rates of kg/min and 511 kj/min, respectively. Evaporative Cooling Conventional cooling systems operate on a refrigeration cycle, and they can be used in any part of the world. But they have a high initial and operating cost. In desert (hot and dry) climates, we can avoid the high cost of cooling by using evaporative coolers, also known as swamp coolers. Evaporative cooling is based on a simple principle: As water evaporates, the latent heat of vaporization is absorbed from the water body and the surrounding air. As a result, both the water and the air are cooled during the process. This approach has been used for thousands of years to cool water. A porous jug or pitcher filled with water is left in an open, shaded area. A small amount of water leaks out through the porous holes, and the pitcher sweats. In a dry environment, this water evaporates and cools the remaining water in the pitcher (Fig. 14 6). You have probably noticed that on a hot, dry day the air feels a lot cooler when the yard is watered. This is because water absorbs heat from the air as it evaporates. An evaporative cooler works on the same principle. The evaporative cooling process is shown schematically and on a psychrometric chart in Fig Hot, dry air at state 1 enters the evaporative cooler, where it is sprayed with liquid water. Part of the water evaporates during this process by absorbing heat from the airstream. As a result, the temperature of the airstream decreases and its humidity increases (state ). In the limiting case, the air leaves the evaporative cooler saturated at state. This is the lowest temperature that can be achieved by this process. The evaporative cooling process is essentially identical to the adiabatic saturation process since the heat transfer between the airstream and the surroundings is usually negligible. Therefore, the evaporative cooling process follows a line of constant wet-bulb temperature on the psychrometric chart. (Note that this will not exactly be the case if the liquid water is supplied at a temperature different from the exit temperature of the airstream.) Since the constant-wetbulb-temperature lines almost coincide with the constant-enthalpy lines, the enthalpy of the airstream can also be assumed to remain constant. That is, and m # a V# 1 v 1 m # w kg>min kg/min Q # out kg>min kj>kg kg>min158.8 kj>kg 511 kj/min 10 m 3 >min 11.5 kg>min m 3 >kg dry air T wb constant h constant (14 19) (14 0) during an evaporative cooling process. This is a reasonably accurate approximation, and it is commonly used in air-conditioning calculations.

257 Chapter EXAMPLE 14 7 Evaporative Cooling of Air by a Swamp Cooler Air enters an evaporative (or swamp) cooler at 14.7 psi, 95 F, and 0 percent relative humidity, and it exits at 80 percent relative humidity. Determine (a) the exit temperature of the air and (b) the lowest temperature to which the air can be cooled by this evaporative cooler. Solution Air is cooled steadily by an evaporative cooler. The temperature of discharged air and the lowest temperature to which the air can be cooled are to be determined. Analysis The schematic of the evaporative cooler and the psychrometric chart of the process are shown in Fig (a) If we assume the liquid water is supplied at a temperature not much different from the exit temperature of the airstream, the evaporative cooling process follows a line of constant wet-bulb temperature on the psychrometric chart. That is, T wb constant The wet-bulb temperature at 95 F and 0 percent relative humidity is determined from the psychrometric chart to be 66.0 F. The intersection point of the T wb 66.0 F and the f 80 percent lines is the exit state of the air. The temperature at this point is the exit temperature of the air, and it is determined from the psychrometric chart to be T 70.4 F (b) In the limiting case, air leaves the evaporative cooler saturated (f 100 percent), and the exit state of the air in this case is the state where the T wb 66.0 F line intersects the saturation line. For saturated air, the dry- and the wet-bulb temperatures are identical. Therefore, the lowest temperature to which air can be cooled is the wet-bulb temperature, which is T min T 66.0 F Discussion Note that the temperature of air drops by as much as 30 F in this case by evaporative cooling. AIR ' f = 80% ' 1 1 T min T 95 F f 1 = 0% T 1 = 95 F f = 0% P = 14.7 psia FIGURE 14 8 Schematic and psychrometric chart for Example Adiabatic Mixing of Airstreams Many air-conditioning applications require the mixing of two airstreams. This is particularly true for large buildings, most production and process plants, and hospitals, which require that the conditioned air be mixed with a certain fraction of fresh outside air before it is routed into the living space. The mixing is accomplished by simply merging the two airstreams, as shown in Fig The heat transfer with the surroundings is usually small, and thus the mixing processes can be assumed to be adiabatic. Mixing processes normally involve no work interactions, and the changes in kinetic and potential energies, if any, are negligible. Then the mass and energy balances for the adiabatic mixing of two airstreams reduce to Mass of dry air: m # a 1 m # a m # a 3 (14 1) Mass of water vapor: v 1 m # a 1 v m # a v 3 m # a 3 (14 ) Energy: m # a 1 h 1 m # a h m # a 3 h 3 (14 3)

258 736 Thermodynamics Eliminating ṁ a3 from the relations above, we obtain ω 1 h 1 1 m # a 1 m # a v v 3 v 3 v 1 h h 3 h 3 h 1 (14 4) h 1 h 3 h 1 ω h h 3 C h A Mixing section h h D 3 ω 3 h 3 ω ω 3 ω 3 ω B 1 ω ω 3 ω 1 This equation has an instructive geometric interpretation on the psychrometric chart. It shows that the ratio of v v 3 to v 3 v 1 is equal to the ratio of ṁ a1 to ṁ a. The states that satisfy this condition are indicated by the dashed line AB. The ratio of h h 3 to h 3 h 1 is also equal to the ratio of ṁ a1 to ṁ a, and the states that satisfy this condition are indicated by the dashed line CD. The only state that satisfies both conditions is the intersection point of these two dashed lines, which is located on the straight line connecting states 1 and. Thus we conclude that when two airstreams at two different states (states 1 and ) are mixed adiabatically, the state of the mixture (state 3) lies on the straight line connecting states 1 and on the psychrometric chart, and the ratio of the distances -3 and 3-1 is equal to the ratio of mass flow rates ṁ a1 and ṁ a. The concave nature of the saturation curve and the conclusion above lead to an interesting possibility. When states 1 and are located close to the saturation curve, the straight line connecting the two states will cross the saturation curve, and state 3 may lie to the left of the saturation curve. In this case, some water will inevitably condense during the mixing process. FIGURE 14 9 When two airstreams at states 1 and are mixed adiabatically, the state of the mixture lies on the straight line connecting the two states. EXAMPLE 14 8 Mixing of Conditioned Air with Outdoor Air Saturated air leaving the cooling section of an air-conditioning system at 14 C at a rate of 50 m 3 /min is mixed adiabatically with the outside air at 3 C and 60 percent relative humidity at a rate of 0 m 3 /min. Assuming that the mixing process occurs at a pressure of 1 atm, determine the specific humidity, the relative humidity, the dry-bulb temperature, and the volume flow rate of the mixture. Solution Conditioned air is mixed with outside air at specified rates. The specific and relative humidities, dry-bulb temperature, and the flow rate of the mixture are to be determined. Assumptions 1 Steady operating conditions exist. Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The mixing section is adiabatic. Properties The properties of each inlet stream are determined from the psychrometric chart to be h kj>kg dry air v kg H O>kg dry air v m 3 >kg dry air and h 79.0 kj>kg dry air v kg H O>kg dry air v m 3 >kg dry air

259 Chapter Analysis We take the mixing section of the streams as the system. The schematic of the system and the psychrometric chart of the process are shown in Fig We note that this is a steady-flow mixing process. The mass flow rates of dry air in each stream are m # a 1 V# 1 v 1 m # a V# v From the mass balance of dry air, 50 m 3 >min 60.5 kg>min 0.86 m 3 >kg dry air 0 m 3 >min.5 kg>min m 3 >kg dry air m # a 3 m # a 1 m # a kg>min 83 kg>min Saturated air T 1 = 14 C V 1 = 50 m 3 /min 1 T = 3 C f = 60% V = 0 m 3 /min Mixing section P = 1 atm 3 V 3 v 3 f 3 T 3 The specific humidity and the enthalpy of the mixture can be determined from Eq. 14 4, which yield m # a 1 m # a v v 3 v 3 v 1 h h 3 h 3 h v 3 v h 3 h v kg H O/kg dry air h kj>kg dry air These two properties fix the state of the mixture. Other properties of the mixture are determined from the psychrometric chart: T C f 3 89% v m 3 >kg dry air Finally, the volume flow rate of the mixture is determined from V # 3 m # a 3 v kg>min m 3 >kg 70.1 m 3 /min 1 f 1 = 100% 14 C 3 C 3 f = 60% FIGURE Schematic and psychrometric chart for Example Discussion Notice that the volume flow rate of the mixture is approximately equal to the sum of the volume flow rates of the two incoming streams. This is typical in air-conditioning applications. Wet Cooling Towers Power plants, large air-conditioning systems, and some industries generate large quantities of waste heat that is often rejected to cooling water from nearby lakes or rivers. In some cases, however, the cooling water supply is limited or thermal pollution is a serious concern. In such cases, the waste heat must be rejected to the atmosphere, with cooling water recirculating and serving as a transport medium for heat transfer between the source and the sink (the atmosphere). One way of achieving this is through the use of wet cooling towers. A wet cooling tower is essentially a semienclosed evaporative cooler. An induced-draft counterflow wet cooling tower is shown schematically in

260 738 Thermodynamics WARM WATER COOL WATER WARM WATER COOL WATER AIR EXIT FAN FIGURE 14 3 A natural-draft cooling tower. AIR INLET FIGURE An induced-draft counterflow cooling tower. AIR INLET Fig Air is drawn into the tower from the bottom and leaves through the top. Warm water from the condenser is pumped to the top of the tower and is sprayed into this airstream. The purpose of spraying is to expose a large surface area of water to the air. As the water droplets fall under the influence of gravity, a small fraction of water (usually a few percent) evaporates and cools the remaining water. The temperature and the moisture content of the air increase during this process. The cooled water collects at the bottom of the tower and is pumped back to the condenser to absorb additional waste heat. Makeup water must be added to the cycle to replace the water lost by evaporation and air draft. To minimize water carried away by the air, drift eliminators are installed in the wet cooling towers above the spray section. The air circulation in the cooling tower described is provided by a fan, and therefore it is classified as a forced-draft cooling tower. Another popular type of cooling tower is the natural-draft cooling tower, which looks like a large chimney and works like an ordinary chimney. The air in the tower has a high water-vapor content, and thus it is lighter than the outside air. Consequently, the light air in the tower rises, and the heavier outside air fills the vacant space, creating an airflow from the bottom of the tower to the top. The flow rate of air is controlled by the conditions of the atmospheric air. Natural-draft cooling towers do not require any external power to induce the air, but they cost a lot more to build than forced-draft cooling towers. The natural-draft cooling towers are hyperbolic in profile, as shown in Fig. 14 3, and some are over 100 m high. The hyperbolic profile is for greater structural strength, not for any thermodynamic reason. The idea of a cooling tower started with the spray pond, where the warm water is sprayed into the air and is cooled by the air as it falls into the pond, as shown in Fig Some spray ponds are still in use today. However, they require 5 to 50 times the area of a cooling tower, water loss due to air drift is high, and they are unprotected against dust and dirt. We could also dump the waste heat into a still cooling pond, which is basically a large artificial lake open to the atmosphere. Heat transfer from the pond surface to the atmosphere is very slow, however, and we would need about 0 times the area of a spray pond in this case to achieve the same cooling. EXAMPLE 14 9 Cooling of a Power Plant by a Cooling Tower FIGURE A spray pond. Photo by Yunus Çengel. Cooling water leaves the condenser of a power plant and enters a wet cooling tower at 35 C at a rate of 100 kg/s. Water is cooled to C in the cooling tower by air that enters the tower at 1 atm, 0 C, and 60 percent relative humidity and leaves saturated at 30 C. Neglecting the power input to the fan, determine (a) the volume flow rate of air into the cooling tower and (b) the mass flow rate of the required makeup water. Solution Warm cooling water from a power plant is cooled in a wet cooling tower. The flow rates of makeup water and air are to be determined. Assumptions 1 Steady operating conditions exist and thus the mass flow rate of dry air remains constant during the entire process. Dry air and the water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The cooling tower is adiabatic.

261 Chapter Properties The enthalpy of saturated liquid water is 9.8 kj/kg at C and kj/kg at 35 C (Table A 4). From the psychrometric chart, h 1 4. kj/kg dry air v kg H O/kg dry air v m 3 /kg dry air h kj/kg dry air v kg H O/kg dry air Analysis We take the entire cooling tower to be the system, which is shown schematically in Fig We note that the mass flow rate of liquid water decreases by an amount equal to the amount of water that vaporizes in the tower during the cooling process. The water lost through evaporation must be made up later in the cycle to maintain steady operation. (a) Applying the mass and energy balances on the cooling tower gives Dry air mass balance: Water mass balance: or Energy balance: or m # a 1 m # a m # a m # 3 m # a 1 v 1 m # 4 m # a v m # 3 m # 4 m # a 1v v 1 m # makeup a in m # h aout m # h S m # a 1 h 1 m # 3h 3 m # a h m # 4h 4 m # 3h 3 m # a 1h h 1 1m # 3 m # makeuph 4 WARM WATER 35 C 100 kg/s System boundary 100 kg/s COOL WATER 3 4 Makeup water C 30 C f = 100% FIGURE Schematic for Example AIR 1 atm 0 C f 1 = 60% V 1 Solving for m a gives m # a Substituting, m # 3 1h 3 h 4 1h h 1 1v v 1 h 4 m # 1100 kg>s kj>kg4 a 96.9 kg>s kj>kg kj>kg4 Then the volume flow rate of air into the cooling tower becomes V # 1 m # av kg>s m 3 >kg 81.6 m 3 /s (b) The mass flow rate of the required makeup water is determined from m # makeup m # a 1v v kg>s kg/s Discussion Note that over 98 percent of the cooling water is saved and recirculated in this case. SUMMARY In this chapter we discussed the air water-vapor mixture, which is the most commonly encountered gas vapor mixture in practice. The air in the atmosphere normally contains some water vapor, and it is referred to as atmospheric air. By contrast, air that contains no water vapor is called dry air. In the temperature range encountered in air-conditioning applications, both the dry air and the water vapor can be treated as ideal gases. The enthalpy change of dry air during a process can be determined from h dry air c p T kj>kg # C T The atmospheric air can be treated as an ideal-gas mixture whose pressure is the sum of the partial pressure of dry air P a and that of the water vapor P v, P P a P v

262 740 Thermodynamics The enthalpy of water vapor in the air can be taken to be equal to the enthalpy of the saturated vapor at the same temperature: h v (T, low P) h g (T) T (kj/kg) T in C T (Btu/lbm) T in F in the temperature range 10 to 50 C (15 to 10 F). The mass of water vapor present per unit mass of dry air is called the specific or absolute humidity v, v m v m a 0.6P v P P v 1kg H O>kg dry air where P is the total pressure of air and P v is the vapor pressure. There is a limit on the amount of vapor the air can hold at a given temperature. Air that is holding as much moisture as it can at a given temperature is called saturated air. The ratio of the amount of moisture air holds (m v ) to the maximum amount of moisture air can hold at the same temperature (m g ) is called the relative humidity f, where P g P T. The relative and specific humidities can also be expressed as f f m v m g P v P g vp 10.6 vp g and v 0.6fP g P fp g Relative humidity ranges from 0 for dry air to 1 for saturated air. The enthalpy of atmospheric air is expressed per unit mass of dry air, instead of per unit mass of the air water-vapor mixture, as h h a vh g 1kJ>kg dry air The ordinary temperature of atmospheric air is referred to as the dry-bulb temperature to differentiate it from other forms of temperatures. The temperature at which condensation begins if the air is cooled at constant pressure is called the dew-point temperature T dp : T dp T Pv Relative humidity and specific humidity of air can be determined by measuring the adiabatic saturation temperature of air, which is the temperature air attains after flowing over water in a long adiabatic channel until it is saturated, v 1 c p 1T T 1 v h fg h g1 h f where and T is the adiabatic saturation temperature. A more practical approach in air-conditioning applications is to use a thermometer whose bulb is covered with a cotton wick saturated with water and to blow air over the wick. The temperature measured in this manner is called the wet-bulb temperature T wb, and it is used in place of the adiabatic saturation temperature. The properties of atmospheric air at a specified total pressure are presented in the form of easily readable charts, called psychrometric charts. The lines of constant enthalpy and the lines of constant wet-bulb temperature are very nearly parallel on these charts. The needs of the human body and the conditions of the environment are not quite compatible. Therefore, it often becomes necessary to change the conditions of a living space to make it more comfortable. Maintaining a living space or an industrial facility at the desired temperature and humidity may require simple heating (raising the temperature), simple cooling (lowering the temperature), humidifying (adding moisture), or dehumidifying (removing moisture). Sometimes two or more of these processes are needed to bring the air to the desired temperature and humidity level. Most air-conditioning processes can be modeled as steadyflow processes, and therefore they can be analyzed by applying the steady-flow mass (for both dry air and water) and energy balances, Dry air mass: Water mass: Energy: v 0.6P g P P g a in m # a aout m # a a in m # w aout m # w or a in m # av aout m # av Q # in W # in ain m # h Q # out W # out aout m # h The changes in kinetic and potential energies are assumed to be negligible. During a simple heating or cooling process, the specific humidity remains constant, but the temperature and the relative humidity change. Sometimes air is humidified after it is heated, and some cooling processes include dehumidification. In dry climates, air can be cooled via evaporative cooling by passing it through a section where it is sprayed with water. In locations with limited cooling water supply, large amounts of waste heat can be rejected to the atmosphere with minimum water loss through the use of cooling towers.

263 Chapter REFERENCES AND SUGGESTED READINGS 1. ASHRAE Handbook of Fundamentals. Atlanta, GA: American Society of Heating, Refrigerating, and Air- Conditioning Engineers, S. M. Elonka. Cooling Towers. Power, March W. F. Stoecker and J. W. Jones. Refrigeration and Air Conditioning. nd ed. New York: McGraw-Hill, K. Wark and D. E. Richards. Thermodynamics. 6th ed. New York: McGraw-Hill, L. D. Winiarski and B. A. Tichenor. Model of Natural Draft Cooling Tower Performance. Journal of the Sanitary Engineering Division, Proceedings of the American Society of Civil Engineers, August PROBLEMS* Dry and Atmospheric Air: Specific and Relative Humidity 14 1C Is it possible to obtain saturated air from unsaturated air without adding any moisture? Explain. 14 C Is the relative humidity of saturated air necessarily 100 percent? 14 3C Moist air is passed through a cooling section where it is cooled and dehumidified. How do (a) the specific humidity and (b) the relative humidity of air change during this process? 14 4C What is the difference between dry air and atmospheric air? 14 5C Can the water vapor in air be treated as an ideal gas? Explain. 14 6C What is vapor pressure? 14 7C How would you compare the enthalpy of water vapor at 0 C and kpa with the enthalpy of water vapor at 0 C and 0.5 kpa? 14 8C What is the difference between the specific humidity and the relative humidity? 14 9C How will (a) the specific humidity and (b) the relative humidity of the air contained in a well-sealed room change as it is heated? *Problems designated by a C are concept questions, and students are encouraged to answer them all. Problems designated by an E are in English units, and the SI users can ignore them. Problems with a CD-EES icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed DVD. Problems with a computer-ees icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text C How will (a) the specific humidity and (b) the relative humidity of the air contained in a well-sealed room change as it is cooled? 14 11C Consider a tank that contains moist air at 3 atm and whose walls are permeable to water vapor. The surrounding air at 1 atm pressure also contains some moisture. Is it possible for the water vapor to flow into the tank from surroundings? Explain. 14 1C Why are the chilled water lines always wrapped with vapor barrier jackets? 14 13C Explain how vapor pressure of the ambient air is determined when the temperature, total pressure, and the relative humidity of air are given An 8 m 3 -tank contains saturated air at 30 C, 105 kpa. Determine (a) the mass of dry air, (b) the specific humidity, and (c) the enthalpy of the air per unit mass of the dry air A tank contains 1 kg of dry air and 0.3 kg of water vapor at 30 C and 100 kpa total pressure. Determine (a) the specific humidity, (b) the relative humidity, and (c) the volume of the tank Repeat Prob for a temperature of 4 C A room contains air at 0 C and 98 kpa at a relative humidity of 85 percent. Determine (a) the partial pressure of dry air, (b) the specific humidity of the air, and (c) the enthalpy per unit mass of dry air Repeat Prob for a pressure of 85 kpa E A room contains air at 70 F and 14.6 psia at a relative humidity of 85 percent. Determine (a) the partial pressure of dry air, (b) the specific humidity, and (c) the enthalpy per unit mass of dry air. Answers: (a) psia, (b) lbm H O/lbm dry air, (c) Btu/lbm dry air 14 0 Determine the masses of dry air and the water vapor contained in a 40-m 3 room at 98 kpa, 3 C, and 50 percent relative humidity. Answers: 73 kg,.5 kg

264 74 Thermodynamics Dew-Point, Adiabatic Saturation, and Wet-Bulb Temperatures 14 1C What is the dew-point temperature? 14 C Andy and Wendy both wear glasses. On a cold winter day, Andy comes from the cold outside and enters the warm house while Wendy leaves the house and goes outside. Whose glasses are more likely to be fogged? Explain. 14 3C In summer, the outer surface of a glass filled with iced water frequently sweats. How can you explain this sweating? 14 4C In some climates, cleaning the ice off the windshield of a car is a common chore on winter mornings. Explain how ice forms on the windshield during some nights even when there is no rain or snow. 14 5C When are the dry-bulb and dew-point temperatures identical? 14 6C When are the adiabatic saturation and wet-bulb temperatures equivalent for atmospheric air? 14 7 A house contains air at 5 C and 65 percent relative humidity. Will any moisture condense on the inner surfaces of the windows when the temperature of the window drops to 10 C? 14 8 After a long walk in the 8 C outdoors, a person wearing glasses enters a room at 5 C and 40 percent relative humidity. Determine whether the glasses will become fogged Repeat Prob for a relative humidity of 30 percent E A thirsty woman opens the refrigerator and picks up a cool canned drink at 40 F. Do you think the can will sweat as she enjoys the drink in a room at 80 F and 50 percent relative humidity? The dry- and wet-bulb temperatures of atmospheric air at 95 kpa are 5 and 17 C, respectively. Determine (a) the specific humidity, (b) the relative humidity, and (c) the enthalpy of the air, in kj/kg dry air The air in a room has a dry-bulb temperature of C and a wet-bulb temperature of 16 C. Assuming a pressure of 100 kpa, determine (a) the specific humidity, (b) the relative humidity, and (c) the dew-point temperature. Answers: (a) kg H O/kg dry air, (b) 54.1 percent, (c) 1.3 C Reconsider Prob Determine the required properties using EES (or other) software. What would the property values be at a pressure of 300 kpa? 14 34E The air in a room has a dry-bulb temperature of 80 F and a wet-bulb temperature of 65 F. Assuming a pressure of 14.7 psia, determine (a) the specific humidity, (b) the relative humidity, and (c) the dew-point temperature. Answers: (a) lbm H O/lbm dry air, (b) 44.7 percent, (c) 56.6 F Atmospheric air at 35 C flows steadily into an adiabatic saturation device and leaves as a saturated mixture at 5 C. Makeup water is supplied to the device at 5 C. Atmospheric pressure is 98 kpa. Determine the relative humidity and specific humidity of the air. Psychrometric Chart 14 36C How do constant-enthalpy and constant-wet-bulbtemperature lines compare on the psychrometric chart? 14 37C At what states on the psychrometric chart are the dry-bulb, wet-bulb, and dew-point temperatures identical? 14 38C How is the dew-point temperature at a specified state determined on the psychrometric chart? 14 39C Can the enthalpy values determined from a psychrometric chart at sea level be used at higher elevations? The air in a room is at 1 atm, 3 C, and 60 percent relative humidity. Determine (a) the specific humidity, (b) the enthalpy (in kj/kg dry air), (c) the wet-bulb temperature, (d ) the dew-point temperature, and (e) the specific volume of the air (in m 3 /kg dry air). Use the psychrometric chart or available software Reconsider Prob Determine the required properties using EES (or other) software instead of the psychrometric chart. What would the property values be at a location at 1500 m altitude? 14 4 A room contains air at 1 atm, 6 C, and 70 percent relative humidity. Using the psychrometric chart, determine (a) the specific humidity, (b) the enthalpy (in kj/kg dry air), (c) the wet-bulb temperature, (d ) the dew-point temperature, and (e) the specific volume of the air (in m 3 /kg dry air) Reconsider Prob Determine the required properties using EES (or other) software instead of the psychrometric chart. What would the property values be at a location at 000 m altitude? 14 44E A room contains air at 1 atm, 8 F, and 70 percent relative humidity. Using the psychrometric chart, determine (a) the specific humidity, (b) the enthalpy (in Btu/lbm dry air), (c) the wet-bulb temperature, (d ) the dew-point temperature, and (e) the specific volume of the air (in ft 3 /lbm dry air) E Reconsider Prob E. Determine the required properties using EES (or other) software instead of the psychrometric chart. What would the property values be at a location at 5000 ft altitude? The air in a room has a pressure of 1 atm, a dry-bulb temperature of 4 C, and a wet-bulb temperature of 17 C. Using the psychrometric chart, determine (a) the specific humidity, (b) the enthalpy (in kj/kg dry air), (c) the relative humidity, (d ) the dew-point temperature, and (e) the specific volume of the air (in m 3 /kg dry air).

265 14 47 Reconsider Prob Determine the required properties using EES (or other) software instead of the psychrometric chart. What would the property values be at a location at 3000 m altitude? Human Comfort and Air-Conditioning 14 48C What does a modern air-conditioning system do besides heating or cooling the air? 14 49C How does the human body respond to (a) hot weather, (b) cold weather, and (c) hot and humid weather? 14 50C What is the radiation effect? How does it affect human comfort? 14 51C How does the air motion in the vicinity of the human body affect human comfort? 14 5C Consider a tennis match in cold weather where both players and spectators wear the same clothes. Which group of people will feel colder? Why? 14 53C Why do you think little babies are more susceptible to cold? 14 54C How does humidity affect human comfort? 14 55C What are humidification and dehumidification? 14 56C What is metabolism? What is the range of metabolic rate for an average man? Why are we interested in the metabolic rate of the occupants of a building when we deal with heating and air-conditioning? 14 57C Why is the metabolic rate of women, in general, lower than that of men? What is the effect of clothing on the environmental temperature that feels comfortable? 14 58C What is sensible heat? How is the sensible heat loss from a human body affected by the (a) skin temperature, (b) environment temperature, and (c) air motion? 14 59C What is latent heat? How is the latent heat loss from the human body affected by the (a) skin wettedness and (b) relative humidity of the environment? How is the rate of evaporation from the body related to the rate of latent heat loss? An average person produces 0.5 kg of moisture while taking a shower and 0.05 kg while bathing in a tub. Consider a family of four who each shower once a day in a bathroom that is not ventilated. Taking the heat of vaporization of water to be 450 kj/kg, determine the contribution of showers to the latent heat load of the air conditioner per day in summer An average (1.8 kg or 4.0 lbm) chicken has a basal metabolic rate of 5.47 W and an average metabolic rate of 10. W (3.78 W sensible and 6.4 W latent) during normal activity. If there are 100 chickens in a breeding room, determine the rate of total heat generation and the rate of moisture production in the room. Take the heat of vaporization of water to be 430 kj/kg. Chapter A department store expects to have 10 customers and 15 employees at peak times in summer. Determine the contribution of people to the total cooling load of the store E In a movie theater in winter, 500 people, each generating sensible heat at a rate of 70 W, are watching a movie. The heat losses through the walls, windows, and the roof are estimated to be 130,000 Btu/h. Determine if the theater needs to be heated or cooled For an infiltration rate of 1. air changes per hour (ACH), determine sensible, latent, and total infiltration heat load of a building at sea level, in kw, that is 0 m long, 13 m wide, and 3 m high when the outdoor air is at 3 C and 50 percent relative humidity. The building is maintained at 4 C and 50 percent relative humidity at all times Repeat Prob for an infiltration rate of 1.8 ACH. Simple Heating and Cooling 14 66C How do relative and specific humidities change during a simple heating process? Answer the same question for a simple cooling process C Why does a simple heating or cooling process appear as a horizontal line on the psychrometric chart? Air enters a heating section at 95 kpa, 1 C, and 30 percent relative humidity at a rate of 6 m 3 /min, and it leaves at 5 C. Determine (a) the rate of heat transfer in the heating section and (b) the relative humidity of the air at the exit. Answers: (a) 91.1 kj/min, (b) 13.3 percent 14 69E A heating section consists of a 15-in.-diameter duct that houses a 4-kW electric resistance heater. Air enters the heating section at 14.7 psia, 50 F, and 40 percent relative humidity at a velocity of 5 ft/s. Determine (a) the exit temperature, (b) the exit relative humidity of the air, and (c) the exit velocity. Answers: (a) 56.6 F, (b) 31.4 percent, (c) 5.4 ft/s Air enters a 40-cm-diameter cooling section at 1 atm, 3 C, and 30 percent relative humidity at 18 m/s. Heat is removed from the air at a rate of 100 kj/min. Determine (a) the exit temperature, (b) the exit relative humidity of the air, and (c) the exit velocity. Answers: (a) 4.4 C, (b) 46.6 percent, (c) 17.6 m/s 3 C, 30% AIR 18 m/s 100 kj/min 1 atm FIGURE P14 70

266 744 Thermodynamics Repeat Prob for a heat removal rate of 800 kj/min. Heating with Humidification 14 7C Why is heated air sometimes humidified? Air at 1 atm, 15 C, and 60 percent relative humidity is first heated to 0 C in a heating section and then humidified by introducing water vapor. The air leaves the humidifying section at 5 C and 65 percent relative humidity. Determine (a) the amount of steam added to the air, and (b) the amount of heat transfer to the air in the heating section. Answers: (a) kg H O/kg dry air, (b) 5.1 kj/kg dry air 14 74E Air at 14.7 psia, 50 F, and 60 percent relative humidity is first heated to 7 F in a heating section and then humidified by introducing water vapor. The air leaves the humidifying section at 75 F and 55 percent relative humidity. Determine (a) the amount of steam added to the air, in lbm H O/lbm dry air, and (b) the amount of heat transfer to the air in the heating section, in Btu/lbm dry air An air-conditioning system operates at a total pressure of 1 atm and consists of a heating section and a humidifier that supplies wet steam (saturated water vapor) at 100 C. Air enters the heating section at 10 C and 70 percent relative humidity at a rate of 35 m 3 /min, and it leaves the humidifying section at 0 C and 60 percent relative humidity. Determine (a) the temperature and relative humidity of air when it leaves the heating section, (b) the rate of heat transfer in the heating section, and (c) the rate at which water is added to the air in the humidifying section. Heating coils 10 C 70% AIR 35 m 3 /min P = 1 atm FIGURE P14 75 Sat. vapor 100 C Humidifier 0 C 60% Repeat Prob for a total pressure of 95 kpa for the airstream. Answers: (a) 19.5 C, 37.7 percent, (b) 391 kj/min, (c) kg/min Cooling with Dehumidification 14 77C Why is cooled air sometimes reheated in summer before it is discharged to a room? Air enters a window air conditioner at 1 atm, 3 C, and 70 percent relative humidity at a rate of m 3 /min, and it leaves as saturated air at 15 C. Part of the moisture in the air that condenses during the process is also removed at 15 C. Determine the rates of heat and moisture removal from the air. Answers: 97.7 kj/min, 0.03 kg/min An air-conditioning system is to take in air at 1 atm, 34 C, and 70 percent relative humidity and deliver it at C and 50 percent relative humidity. The air flows first over the cooling coils, where it is cooled and dehumidified, and then over the resistance heating wires, where it is heated to the desired temperature. Assuming that the condensate is removed from the cooling section at 10 C, determine (a) the temperature of air before it enters the heating section, (b) the amount of heat removed in the cooling section, and (c) the amount of heat transferred in the heating section, both in kj/kg dry air Air enters a 30-cm-diameter cooling section at 1 atm, 35 C, and 60 percent relative humidity at 10 m/min. The air is cooled by passing it over a cooling coil through which cold water flows. The water experiences a temperature rise of 8 C. The air leaves the cooling section saturated at 0 C. Determine (a) the rate of heat transfer, (b) the mass flow rate of the water, and (c) the exit velocity of the airstream. 35 C 60% 10 m/min Water T Cooling coils AIR FIGURE P14 80 T + 8 C 0 C Saturated Reconsider Prob Using EES (or other) software, develop a general solution of the problem in which the input variables may be supplied and parametric studies performed. For each set of input variables for which the pressure is atmospheric, show the process on the psychrometric chart Repeat Prob for a total pressure of 95 kpa for air. Answers: (a) 93. kj/min, (b) 8.77 kg/min, (c) 113 m/min 14 83E Air enters a 1-ft-diameter cooling section at 14.7 psia, 90 F, and 60 percent relative humidity at 600 ft/min. The air is cooled by passing it over a cooling coil through which cold water flows. The water experiences a temperature rise of 14 F. The air leaves the cooling section saturated at 70 F. Determine (a) the rate of heat transfer, (b) the mass flow rate of the water, and (c) the exit velocity of the airstream E Reconsider Prob E. Using EES (or other) software, study the effect of the total pressure of the air over the range 14.3 to 15. psia on the

267 required results. Plot the required results as functions of air total pressure E Repeat Prob E for a total pressure of 14.4 psia for air Atmospheric air from the inside of an automobile enters the evaporator section of the air conditioner at 1 atm, 7 C and 50 percent relative humidity. The air returns to the automobile at 10 C and 90 percent relative humidity. The passenger compartment has a volume of m 3 and 5 air changes per minute are required to maintain the inside of the automobile at the desired comfort level. Sketch the psychrometric diagram for the atmospheric air flowing through the air conditioning process. Determine the dew point and wet bulb temperatures at the inlet to the evaporator section, in C. Determine the required heat transfer rate from the atmospheric air to the evaporator fluid, in kw. Determine the rate of condensation of water vapor in the evaporator section, in kg/min. AIR Cooling coils Chapter Air from a workspace enters an air conditioner unit at 30 C dry bulb and 5 C wet bulb. The air leaves the air conditioner and returns to the space at 5 C dry-bulb and 6.5 C dew-point temperature. If there is any, the condensate leaves the air conditioner at the temperature of the air leaving the cooling coils. The volume flow rate of the air returned to the workspace is 1000 m 3 /min. Atmospheric pressure is 98 kpa. Determine the heat transfer rate from the air, in kw, and the mass flow rate of condensate water, if any, in kg/h. Evaporative Cooling 14 90C Does an evaporation process have to involve heat transfer? Describe a process that involves both heat and mass transfer C During evaporation from a water body to air, under what conditions will the latent heat of vaporization be equal to the heat transfer from the air? 14 9C What is evaporative cooling? Will it work in humid climates? Air enters an evaporative cooler at 1 atm, 36 C, and 0 percent relative humidity at a rate of 4 m 3 /min, and it leaves with a relative humidity of 90 percent. Determine (a) the exit temperature of the air and (b) the required rate of water supply to the evaporative cooler. Condensate FIGURE P Two thousand cubic meters per hour of atmospheric air at 8 C with a dew point temperature of 5 C flows into an air conditioner that uses chilled water as the cooling fluid. The atmospheric air is to be cooled to 18 C. Sketch the system hardware and the psychrometric diagram for the process. Determine the mass flow rate of the condensate water, if any, leaving the air conditioner, in kg/h. If the cooling water has a 10 C temperature rise while flowing through the air conditioner, determine the volume flow rate of chilled water supplied to the air conditioner heat exchanger, in m 3 /min. The air conditioning process takes place at 100 kpa An automobile air conditioner uses refrigerant-134a as the cooling fluid. The evaporator operates at 75 kpa gauge and the condenser operates at 1.7 MPa gage. The compressor requires a power input of 6 kw and has an isentropic efficiency of 85 percent. Atmospheric air at C and 50 percent relative humidity enters the evaporator and leaves at 8 C and 90 percent relative humidity. Determine the volume flow rate of the atmospheric air entering the evaporator of the air conditioner, in m 3 /min. 1 atm 36 C f 1 = 0% AIR Water m ω FIGURE P14 93 Humidifier f = 90% 14 94E Air enters an evaporative cooler at 14.7 psia, 90 F, and 0 percent relative humidity at a rate of 150 ft 3 /min, and it leaves with a relative humidity of 90 percent. Determine (a) the exit temperature of air and (b) the required rate of water supply to the evaporative cooler. Answers: (a) 65 F, (b) 0.06 lbm/min Air enters an evaporative cooler at 95 kpa, 40 C, and 5 percent relative humidity and exits saturated. Determine the exit temperature of air. Answer: 3.1 C 14 96E Air enters an evaporative cooler at 14.5 psia, 93 F, and 30 percent relative humidity and exits saturated. Determine the exit temperature of air Air enters an evaporative cooler at 1 atm, 3 C, and 30 percent relative humidity at a rate of 5 m 3 /min and leaves

268 746 Thermodynamics at C. Determine (a) the final relative humidity and (b) the amount of water added to air What is the lowest temperature that air can attain in an evaporative cooler if it enters at 1 atm, 9 C, and 40 percent relative humidity? Answer: 19.3 C Air at 1 atm, 15 C, and 60 percent relative humidity is first heated to 30 C in a heating section and then passed through an evaporative cooler where its temperature drops to 5 C. Determine (a) the exit relative humidity and (b) the amount of water added to air, in kg H O/kg dry air. Adiabatic Mixing of Airstreams C Two unsaturated airstreams are mixed adiabatically. It is observed that some moisture condenses during the mixing process. Under what conditions will this be the case? C Consider the adiabatic mixing of two airstreams. Does the state of the mixture on the psychrometric chart have to be on the straight line connecting the two states? Two airstreams are mixed steadily and adiabatically. The first stream enters at 3 C and 40 percent relative humidity at a rate of 0 m 3 /min, while the second stream enters at 1 C and 90 percent relative humidity at a rate of 5 m 3 /min. Assuming that the mixing process occurs at a pressure of 1 atm, determine the specific humidity, the relative humidity, the dry-bulb temperature, and the volume flow rate of the mixture. Answers: kg H O/kg dry air, 63.4 percent, 0.6 C, 45.0 m 3 /min 1 3 C 40% 1 C 90% P = 1 atm AIR 3 FIGURE P14 10 v 3 f 3 T Repeat Prob for a total mixing-chamber pressure of 90 kpa E During an air-conditioning process, 900 ft 3 /min of conditioned air at 65 F and 30 percent relative humidity is mixed adiabatically with 300 ft 3 /min of outside air at 80 F and 90 percent relative humidity at a pressure of 1 atm. Determine (a) the temperature, (b) the specific humidity, and (c) the relative humidity of the mixture. Answers: (a) 68.7 F, (b) lbm H O/lbm dry air, (c) 5.1 percent E Reconsider Prob E. Using EES (or other) software, develop a general solution of the problem in which the input variables may be supplied and parametric studies performed. For each set of input variables for which the pressure is atmospheric, show the process on the psychrometric chart A stream of warm air with a dry-bulb temperature of 40 C and a wet-bulb temperature of 3 C is mixed adiabatically with a stream of saturated cool air at 18 C. The dry air mass flow rates of the warm and cool airstreams are 8 and 6 kg/s, respectively. Assuming a total pressure of 1 atm, determine (a) the temperature, (b) the specific humidity, and (c) the relative humidity of the mixture Reconsider Prob Using EES (or other) software, determine the effect of the mass flow rate of saturated cool air stream on the mixture temperature, specific humidity, and relative humidity. Vary the mass flow rate of saturated cool air from 0 to 16 kg/s while maintaining the mass flow rate of warm air constant at 8 kg/s. Plot the mixture temperature, specific humidity, and relative humidity as functions of the mass flow rate of cool air, and discuss the results. Wet Cooling Towers C How does a natural-draft wet cooling tower work? C What is a spray pond? How does its performance compare to the performance of a wet cooling tower? The cooling water from the condenser of a power plant enters a wet cooling tower at 40 C at a rate of 90 kg/s. The water is cooled to 5 C in the cooling tower by air that enters the tower at 1 atm, 3 C, and 60 percent relative humidity and leaves saturated at 3 C. Neglecting the power input to the fan, determine (a) the volume flow rate of air into the cooling tower and (b) the mass flow rate of the required makeup water E The cooling water from the condenser of a power plant enters a wet cooling tower at 110 F at a rate of 100 lbm/s. Water is cooled to 80 F in the cooling tower by air that enters the tower at 1 atm, 76 F, and 60 percent relative humidity and leaves saturated at 95 F. Neglecting the power input to the fan, determine (a) the volume flow rate of air into the cooling tower and (b) the mass flow rate of the required makeup water. Answers: (a) 135 ft 3 /s, (b).4 lbm/s A wet cooling tower is to cool 60 kg/s of water from 40 to 6 C. Atmospheric air enters the tower at 1 atm with dry- and wet-bulb temperatures of and 16 C, respectively, and leaves at 34 C with a relative humidity of 90 percent. Using the psychrometric chart, determine (a) the volume flow rate of air into the cooling tower and (b) the mass flow rate of the required makeup water. Answers: (a) 44.9 m 3 /s, (b) 1.16 kg/s

269 WARM WATER 60 kg/s 40 C AIR INLET 1 atm T db = C T wb = 16 C 6 C COOL WATER Makeup water AIR EXIT FIGURE P C 90% A wet cooling tower is to cool 5 kg/s of cooling water from 40 to 30 C at a location where the atmospheric pressure is 96 kpa. Atmospheric air enters the tower at 0 C and 70 percent relative humidity and leaves saturated at 35 C. Neglecting the power input to the fan, determine (a) the volume flow rate of air into the cooling tower and (b) the mass flow rate of the required makeup water. Answers: (a) 11. m 3 /s, (b) 0.35 kg/s A natural-draft cooling tower is to remove waste heat from the cooling water flowing through the condenser of a steam power plant. The turbine in the steam power plant receives 4 kg/s of steam from the steam generator. Eighteen percent of the steam entering the turbine is extracted for various feedwater heaters. The condensate of the higher pressure feedwater heaters is trapped to the next lowest pressure feedwater heater. The last feedwater heater operates at 0. MPa and all of the steam extracted for the feedwater heaters is throttled from the last feedwater heater exit to the condenser operating at a pressure of 10 kpa. The remainder of the steam produces work in the turbine and leaves the lowest pressure stage of the turbine at 10 kpa with an entropy of 7.96 kj/kg K. The cooling tower supplies the cooling water at 6 C to the condenser, and cooling water returns from the condenser to the cooling tower at 40 C. Atmospheric air enters the tower at 1 atm with dry- and wet-bulb temperatures of 3 and 18 C, respectively, and leaves saturated at 37 C. Determine (a) the mass flow rate of the cooling water, (b) the volume flow rate of air into the cooling tower, and (c) the mass flow rate of the required makeup water. Chapter Review Problems The condensation of the water vapor in compressedair lines is a major concern in industrial facilities, and the compressed air is often dehumidified to avoid the problems associated with condensation. Consider a compressor that compresses ambient air from the local atmospheric pressure of 9 kpa to a pressure of 800 kpa (absolute). The compressed air is then cooled to the ambient temperature as it flows through the compressed-air lines. Disregarding any pressure losses, determine if there will be any condensation in the compressed-air lines on a day when the ambient air is at 0 C and 50 percent relative humidity E The relative humidity of air at 80 F and 14.7 psia is increased from 5 to 75 percent during a humidification process at constant temperature and pressure. Determine the percent error involved in assuming the density of air to have remained constant Dry air whose molar analysis is 78.1 percent N, 0.9 percent O, and 1 percent Ar flows over a water body until it is saturated. If the pressure and temperature of air remain constant at 1 atm and 5 C during the process, determine (a) the molar analysis of the saturated air and (b) the density of air before and after the process. What do you conclude from your results? E Determine the mole fraction of the water vapor at the surface of a lake whose surface temperature is 60 F, and compare it to the mole fraction of water in the lake, which is very nearly 1.0. The air at the lake surface is saturated, and the atmospheric pressure at lake level can be taken to be 13.8 psia Determine the mole fraction of dry air at the surface of a lake whose temperature is 18 C. The air at the lake surface is saturated, and the atmospheric pressure at lake level can be taken to be 100 kpa E Consider a room that is cooled adequately by an air conditioner whose cooling capacity is 7500 Btu/h. If the room is to be cooled by an evaporative cooler that removes heat at the same rate by evaporation, determine how much water needs to be supplied to the cooler per hour at design conditions E The capacity of evaporative coolers is usually expressed in terms of the flow rate of air in ft 3 /min (or cfm), and a practical way of determining the required size of an evaporative cooler for an 8-ft-high house is to multiply the floor area of the house by 4 (by 3 in dry climates and by 5 in humid climates). For example, the capacity of an evaporative cooler for a 30-ft-long, 40-ft-wide house is cfm. Develop an equivalent rule of thumb for the selection of an evaporative cooler in SI units for.4-m-high houses whose floor areas are given in m A cooling tower with a cooling capacity of 100 tons (440 kw) is claimed to evaporate 15,800 kg of water per day. Is this a reasonable claim?

270 748 Thermodynamics 14 13E The U.S. Department of Energy estimates that 190,000 barrels of oil would be saved per day if every household in the United States raised the thermostat setting in summer by 6 F (3.3 C). Assuming the average cooling season to be 10 days and the cost of oil to be $0/barrel, determine how much money would be saved per year E The thermostat setting of a house can be lowered by F by wearing a light long-sleeved sweater, or by 4 F by wearing a heavy long-sleeved sweater for the same level of comfort. If each F reduction in thermostat setting reduces the heating cost of a house by 4 percent at a particular location, determine how much the heating costs of a house can be reduced by wearing heavy sweaters if the annual heating cost of the house is $ The air-conditioning costs of a house can be reduced by up to 10 percent by installing the outdoor unit (the condenser) of the air conditioner at a location shaded by trees and shrubs. If the air-conditioning costs of a house are $500 a year, determine how much the trees will save the home owner in the 0-year life of the system A 3-m 3 tank contains saturated air at 5 C and 97 kpa. Determine (a) the mass of the dry air, (b) the specific humidity, and (c) the enthalpy of the air per unit mass of the dry air. Answers: (a) 3.9 kg, (b) kg H O/kg dry air, (c) 78.6 kj/kg dry air Reconsider Prob Using EES (or other) software, determine the properties of the air at the initial state. Study the effect of heating the air at constant volume until the pressure is 110 kpa. Plot the required heat transfer, in kj, as a function of pressure E Air at 15 psia, 60 F, and 50 percent relative humidity flows in an 8-in.-diameter duct at a velocity of 50 ft/s. Determine (a) the dew-point temperature, (b) the volume flow rate of air, and (c) the mass flow rate of dry air Air enters a cooling section at 97 kpa, 35 C, and 30 percent relative humidity at a rate of 6 m 3 /min, where it is cooled until the moisture in the air starts condensing. Determine (a) the temperature of the air at the exit and (b) the rate of heat transfer in the cooling section Outdoor air enters an air-conditioning system at 10 C and 40 percent relative humidity at a steady rate of m 3 /min, and it leaves at 5 C and 55 percent relative humidity. The outdoor air is first heated to C in the heating section and then humidified by the injection of hot steam in the humidifying section. Assuming the entire process takes place at a pressure of 1 atm, determine (a) the rate of heat supply in the heating section and (b) the mass flow rate of steam required in the humidifying section Air enters an air-conditioning system that uses refrigerant-134a at 30 C and 70 percent relative humidity at a rate of 4 m 3 /min. The refrigerant enters the cooling section at 700 kpa with a quality of 0 percent and leaves as saturated vapor. The air is cooled to 0 C at a pressure of 1 atm. Determine (a) the rate of dehumidification, (b) the rate of heat transfer, and (c) the mass flow rate of the refrigerant Repeat Prob for a total pressure of 95 kpa for air An air-conditioning system operates at a total pressure of 1 atm and consists of a heating section and an evaporative cooler. Air enters the heating section at 10 C and 70 percent relative humidity at a rate of 30 m 3 /min, and it leaves the evaporative cooler at 0 C and 60 percent relatively humidity. Determine (a) the temperature and relative humidity of the air when it leaves the heating section, (b) the rate of heat transfer in the heating section, and (c) the rate of water added to air in the evaporative cooler. Answers: (a) 8.3 C,.3 percent, (b) 696 kj/min, (c) 0.13 kg/min Reconsider Prob Using EES (or other) software, study the effect of total pressure in the range 94 to 100 kpa on the results required in the problem. Plot the results as functions of total pressure Repeat Prob for a total pressure of 96 kpa Conditioned air at 13 C and 90 percent relative humidity is to be mixed with outside air at 34 C and 40 percent relative humidity at 1 atm. If it is desired that the mixture have a relative humidity of 60 percent, determine (a) the ratio of the dry air mass flow rates of the conditioned air to the outside air and (b) the temperature of the mixture Reconsider Prob Determine the desired quantities using EES (or other) software instead of the psychrometric chart. What would the answers be at a location at an atmospheric pressure of 80 kpa? A natural-draft cooling tower is to remove 50 MW of waste heat from the cooling water that enters the tower at 4 C and leaves at 7 C. Atmospheric air enters the tower at 1 atm with dry- and wet-bulb temperatures of 3 and 18 C, respectively, and leaves saturated at 37 C. Determine (a) the mass flow rate of the cooling water, (b) the volume flow rate of air into the cooling tower, and (c) the mass flow rate of the required makeup water Reconsider Prob Using EES (or other) software, investigate the effect of air inlet wet-bulb temperature on the required air volume flow rate and the makeup water flow rate when the other input data are the stated values. Plot the results as functions of wetbulb temperature Atmospheric air enters an air-conditioning system at 30 C and 70 percent relative humidity with a volume flow rate of 4 m 3 /min and is cooled to 0 C and 0 percent relative humidity at a pressure of 1 atm. The system uses refrigerant- 134a as the cooling fluid that enters the cooling section at

271 AIR Cooling coils Condensate FIGURE P kpa with a quality of 0 percent and leaves as a saturated vapor. Draw a schematic and show the process on the psychrometric chart. What is the heat transfer from the air to the cooling coils, in kw? If any water is condensed from the air, how much water will be condensed from the atmospheric air per min? Determine the mass flow rate of the refrigerant, in kg/min An uninsulated tank having a volume of 0.5 m 3 contains air at 35 C, 130 kpa, and 0 percent relative humidity. The tank is connected to a water supply line in which water flows at 50 C. Water is sprayed into the tank until the relative humidity of the air vapor mixture is 90 percent. Determine the amount of water supplied to the tank, in kg, the final pressure of the air vapor mixture in the tank, in kpa, and the heat transfer required during the process to maintain the air vapor mixture in the tank at 35 C Air flows steadily through an isentropic nozzle. The air enters the nozzle at 35 C, 00 kpa and 50 percent relative humidity. If no condensation is to occur during the expansion process, determine the pressure, temperature, and velocity of the air at the nozzle exit. Fundamentals of Engineering (FE) Exam Problems A room is filled with saturated moist air at 5 C and a total pressure of 100 kpa. If the mass of dry air in the room is 100 kg, the mass of water vapor is (a) 0.5 kg (b) 1.97 kg (c).96 kg (d ).04 kg (e) 3.17 kg A room contains 50 kg of dry air and 0.6 kg of water vapor at 5 C and 95 kpa total pressure. The relative humidity of air in the room is (a) 1.% (b) 18.4% (c) 56.7% (d ) 65.% (e) 78.0% A 40-m 3 room contains air at 30 C and a total pressure of 90 kpa with a relative humidity of 75 percent. The mass of dry air in the room is (a) 4.7 kg (b) 9.9 kg (c) 39.9 kg (d ) 41.4 kg (e) 5.3 kg Chapter A room contains air at 30 C and a total pressure of 96.0 kpa with a relative humidity of 75 percent. The partial pressure of dry air is (a) 8.0 kpa (b) 85.8 kpa (c) 9.8 kpa (d ) 90.6 kpa (e) 7.0 kpa The air in a house is at 0 C and 50 percent relative humidity. Now the air is cooled at constant pressure. The temperature at which the moisture in the air will start condensing is (a) 8.7 C (b) 11.3 C (c) 13.8 C (d ) 9.3 C (e) 10.0 C On the psychrometric chart, a cooling and dehumidification process appears as a line that is (a) horizontal to the left (b) vertical downward (c) diagonal upwards to the right (NE direction) (d ) diagonal upwards to the left (NW direction) (e) diagonal downwards to the left (SW direction) On the psychrometric chart, a heating and humidification process appears as a line that is (a) horizontal to the right (b) vertical upward (c) diagonal upwards to the right (NE direction) (d ) diagonal upwards to the left (NW direction) (e) diagonal downwards to the right (SE direction) An air stream at a specified temperature and relative humidity undergoes evaporative cooling by spraying water into it at about the same temperature. The lowest temperature the air stream can be cooled to is (a) the dry bulb temperature at the given state (b) the wet bulb temperature at the given state (c) the dew point temperature at the given state (d ) the saturation temperature corresponding to the humidity ratio at the given state (e) the triple point temperature of water Air is cooled and dehumidified as it flows over the coils of a refrigeration system at 85 kpa from 30 C and a humidity ratio of 0.03 kg/kg dry air to 15 C and a humidity ratio of kg/kg dry air. If the mass flow rate of dry air is 0.7 kg/s, the rate of heat removal from the air is (a) 5 kj/s (b) 10 kj/s (c) 15 kj/s (d ) 0 kj/s (e) 5 kj/s Air at a total pressure of 90 kpa, 15 C, and 75 percent relative humidity is heated and humidified to 5 C and 75 percent relative humidity by introducing water vapor. If the mass flow rate of dry air is 4 kg/s, the rate at which steam is added to the air is (a) 0.03 kg/s (b) kg/s (c) kg/s (d ) kg/s (e) 0 kg/s

272 750 Thermodynamics Design and Essay Problems The condensation and even freezing of moisture in building walls without effective vapor retarders are of real concern in cold climates as they undermine the effectiveness of the insulation. Investigate how the builders in your area are coping with this problem, whether they are using vapor retarders or vapor barriers in the walls, and where they are located in the walls. Prepare a report on your findings, and explain the reasoning for the current practice The air-conditioning needs of a large building can be met by a single central system or by several individual window units. Considering that both approaches are commonly used in practice, the right choice depends on the situation on hand. Identify the important factors that need to be considered in decision making, and discuss the conditions under which an air-conditioning system that consists of several window units is preferable over a large single central system, and vice versa Identify the major sources of heat gain in your house in summer, and propose ways of minimizing them and thus reducing the cooling load Write an essay on different humidity measurement devices, including electronic ones, and discuss the advantages and disadvantages of each device Design an inexpensive evaporative cooling system suitable for use in your house. Show how you would obtain a water spray, how you would provide airflow, and how you would prevent water droplets from drifting into the living space.

273 Chapter 15 CHEMICAL REACTIONS In the preceding chapters we limited our consideration to nonreacting systems systems whose chemical composition remains unchanged during a process. This was the case even with mixing processes during which a homogeneous mixture is formed from two or more fluids without the occurrence of any chemical reactions. In this chapter, we specifically deal with systems whose chemical composition changes during a process, that is, systems that involve chemical reactions. When dealing with nonreacting systems, we need to consider only the sensible internal energy (associated with temperature and pressure changes) and the latent internal energy (associated with phase changes). When dealing with reacting systems, however, we also need to consider the chemical internal energy, which is the energy associated with the destruction and formation of chemical bonds between the atoms. The energy balance relations developed for nonreacting systems are equally applicable to reacting systems, but the energy terms in the latter case should include the chemical energy of the system. In this chapter we focus on a particular type of chemical reaction, known as combustion, because of its importance in engineering. But the reader should keep in mind, however, that the principles developed are equally applicable to other chemical reactions. We start this chapter with a general discussion of fuels and combustion. Then we apply the mass and energy balances to reacting systems. In this regard we discuss the adiabatic flame temperature, which is the highest temperature a reacting mixture can attain. Finally, we examine the second-law aspects of chemical reactions. Objectives The objectives of Chapter 15 are to: Give an overview of fuels and combustion. Apply the conservation of mass to reacting systems to determine balanced reaction equations. Define the parameters used in combustion analysis, such as air fuel ratio, percent theoretical air, and dew-point temperature. Apply energy balances to reacting systems for both steadyflow control volumes and fixed mass systems. Calculate the enthalpy of reaction, enthalpy of combustion, and the heating values of fuels. Determine the adiabatic flame temperature for reacting mixtures. Evaluate the entropy change of reacting systems. Analyze reacting systems from the second-law perspective. 751

274 75 Thermodynamics CRUDE OIL Gasoline Kerosene Diesel fuel Fuel oil FIGURE 15 1 Most liquid hydrocarbon fuels are obtained from crude oil by distillation FUELS AND COMBUSTION Any material that can be burned to release thermal energy is called a fuel. Most familiar fuels consist primarily of hydrogen and carbon. They are called hydrocarbon fuels and are denoted by the general formula C n H m. Hydrocarbon fuels exist in all phases, some examples being coal, gasoline, and natural gas. The main constituent of coal is carbon. Coal also contains varying amounts of oxygen, hydrogen, nitrogen, sulfur, moisture, and ash. It is difficult to give an exact mass analysis for coal since its composition varies considerably from one geographical area to the next and even within the same geographical location. Most liquid hydrocarbon fuels are a mixture of numerous hydrocarbons and are obtained from crude oil by distillation (Fig. 15 1). The most volatile hydrocarbons vaporize first, forming what we know as gasoline. The less volatile fuels obtained during distillation are kerosene, diesel fuel, and fuel oil. The composition of a particular fuel depends on the source of the crude oil as well as on the refinery. Although liquid hydrocarbon fuels are mixtures of many different hydrocarbons, they are usually considered to be a single hydrocarbon for convenience. For example, gasoline is treated as octane, C 8 H 18, and the diesel fuel as dodecane, C 1 H 6. Another common liquid hydrocarbon fuel is methyl alcohol, CH 3 OH, which is also called methanol and is used in some gasoline blends. The gaseous hydrocarbon fuel natural gas, which is a mixture of methane and smaller amounts of other gases, is often treated as methane, CH 4, for simplicity. Natural gas is produced from gas wells or oil wells rich in natural gas. It is composed mainly of methane, but it also contains small amounts of ethane, propane, hydrogen, helium, carbon dioxide, nitrogen, hydrogen sulfate, and water vapor. On vehicles, it is stored either in the gas phase at pressures of 150 to 50 atm as CNG (compressed natural gas), or in the liquid phase at 16 C as LNG (liquefied natural gas). Over a million vehicles in the world, mostly buses, run on natural gas. Liquefied petroleum gas (LPG) is a byproduct of natural gas processing or the crude oil refining. It consists mainly of propane and thus LPG is usually referred to as propane. However, it also contains varying amounts of butane, propylene, and butylenes. Propane is commonly used in fleet vehicles, taxis, school buses, and private cars. Ethanol is obtained from corn, grains, and organic waste. Methonal is produced mostly from natural gas, but it can also be obtained from coal and biomass. Both alcohols are commonly used as additives in oxygenated gasoline and reformulated fuels to reduce air pollution. Vehicles are a major source of air pollutants such as nitric oxides, carbon monoxide, and hydrocarbons, as well as the greenhouse gas carbon dioxide, and thus there is a growing shift in the transportation industry from the traditional petroleum-based fuels such as gaoline and diesel fuel to the cleaner burning alternative fuels friendlier to the environment such as natural gas, alcohols (ethanol and methanol), liquefied petroleum gas (LPG), and hydrogen. The use of electric and hybrid cars is also on the rise. A comparison of some alternative fuels for transportation to gasoline is given in Table Note that the energy contents of alternative fuels per unit volume are lower than that of gasoline or diesel fuel, and thus the driving range of a

275 Chapter TABLE 15 1 A comparison of some alternative fuels to the traditional petroleum-based fuels used in transportation Energy content Gasoline equivalence,* Fuel kj/l L/L-gasoline Gasoline 31,850 1 Light diesel 33, Heavy diesel 35, LPG (Liquefied petroleum gas, primarily propane) 3, Ethanol (or ethyl alcohol) 9, Methanol (or methyl alcohol) 18, CNG (Compressed natural gas, primarily methane, at 00 atm) 8, LNG (Liquefied natural gas, primarily methane) 0, *Amount of fuel whose energy content is equal to the energy content of 1-L gasoline. vehicle on a full tank is lower when running on an alternative fuel. Also, when comparing cost, a realistic measure is the cost per unit energy rather than cost per unit volume. For example, methanol at a unit cost of $1.0/L may appear cheaper than gasoline at $1.80/L, but this is not the case since the cost of 10,000 kj of energy is $0.57 for gasoline and $0.66 for methanol. A chemical reaction during which a fuel is oxidized and a large quantity of energy is released is called combustion (Fig. 15 ). The oxidizer most often used in combustion processes is air, for obvious reasons it is free and readily available. Pure oxygen O is used as an oxidizer only in some specialized applications, such as cutting and welding, where air cannot be used. Therefore, a few words about the composition of air are in order. On a mole or a volume basis, dry air is composed of 0.9 percent oxygen, 78.1 percent nitrogen, 0.9 percent argon, and small amounts of carbon dioxide, helium, neon, and hydrogen. In the analysis of combustion processes, the argon in the air is treated as nitrogen, and the gases that exist in trace amounts are disregarded. Then dry air can be approximated as 1 percent oxygen and 79 percent nitrogen by mole numbers. Therefore, each mole of oxygen entering a combustion chamber is accompanied by 0.79/ mol of nitrogen (Fig. 15 3). That is, 1 kmol O 3.76 kmol N 4.76 kmol air (15 1) During combustion, nitrogen behaves as an inert gas and does not react with other elements, other than forming a very small amount of nitric oxides. However, even then the presence of nitrogen greatly affects the outcome of a combustion process since nitrogen usually enters a combustion chamber in large quantities at low temperatures and exits at considerably higher temperatures, absorbing a large proportion of the chemical energy released during combustion. Throughout this chapter, nitrogen is assumed to remain perfectly FIGURE 15 Combustion is a chemical reaction during which a fuel is oxidized and a large quantity of energy is released. Reprinted with special permission of King Features Syndicate. AIR ( 1% O 79% N ) 1 kmol O 3.76 kmol N FIGURE 15 3 Each kmol of O in air is accompanied by 3.76 kmol of N.

276 754 Thermodynamics Reactants FIGURE 15 4 Reaction chamber Products In a steady-flow combustion process, the components that enter the reaction chamber are called reactants and the components that exit are called products. kg hydrogen 1 H O H O 16 kg oxygen kg hydrogen 16 kg oxygen FIGURE 15 5 The mass (and number of atoms) of each element is conserved during a chemical reaction. inert. Keep in mind, however, that at very high temperatures, such as those encountered in internal combustion engines, a small fraction of nitrogen reacts with oxygen, forming hazardous gases such as nitric oxide. Air that enters a combustion chamber normally contains some water vapor (or moisture), which also deserves consideration. For most combustion processes, the moisture in the air and the H O that forms during combustion can also be treated as an inert gas, like nitrogen. At very high temperatures, however, some water vapor dissociates into H and O as well as into H, O, and OH. When the combustion gases are cooled below the dew-point temperature of the water vapor, some moisture condenses. It is important to be able to predict the dew-point temperature since the water droplets often combine with the sulfur dioxide that may be present in the combustion gases, forming sulfuric acid, which is highly corrosive. During a combustion process, the components that exist before the reaction are called reactants and the components that exist after the reaction are called products (Fig. 15 4). Consider, for example, the combustion of 1 kmol of carbon with 1 kmol of pure oxygen, forming carbon dioxide, C O S CO (15 ) Here C and O are the reactants since they exist before combustion, and CO is the product since it exists after combustion. Note that a reactant does not have to react chemically in the combustion chamber. For example, if carbon is burned with air instead of pure oxygen, both sides of the combustion equation will include N. That is, the N will appear both as a reactant and as a product. We should also mention that bringing a fuel into intimate contact with oxygen is not sufficient to start a combustion process. (Thank goodness it is not. Otherwise, the whole world would be on fire now.) The fuel must be brought above its ignition temperature to start the combustion. The minimum ignition temperatures of various substances in atmospheric air are approximately 60 C for gasoline, 400 C for carbon, 580 C for hydrogen, 610 C for carbon monoxide, and 630 C for methane. Moreover, the proportions of the fuel and air must be in the proper range for combustion to begin. For example, natural gas does not burn in air in concentrations less than 5 percent or greater than about 15 percent. As you may recall from your chemistry courses, chemical equations are balanced on the basis of the conservation of mass principle (or the mass balance), which can be stated as follows: The total mass of each element is conserved during a chemical reaction (Fig. 15 5). That is, the total mass of each element on the right-hand side of the reaction equation (the products) must be equal to the total mass of that element on the left-hand side (the reactants) even though the elements exist in different chemical compounds in the reactants and products. Also, the total number of atoms of each element is conserved during a chemical reaction since the total number of atoms is equal to the total mass of the element divided by its atomic mass. For example, both sides of Eq. 15 contain 1 kg of carbon and 3 kg of oxygen, even though the carbon and the oxygen exist as elements in the reactants and as a compound in the product. Also, the total mass of reactants is equal to the total mass of products, each being 44 kg. (It is common practice to round the molar masses to the nearest integer if great accuracy is

277 not required.) However, notice that the total mole number of the reactants ( kmol) is not equal to the total mole number of the products (1 kmol). That is, the total number of moles is not conserved during a chemical reaction. A frequently used quantity in the analysis of combustion processes to quantify the amounts of fuel and air is the air fuel ratio AF. It is usually expressed on a mass basis and is defined as the ratio of the mass of air to the mass of fuel for a combustion process (Fig. 15 6). That is, AF m air m fuel (15 3) The mass m of a substance is related to the number of moles N through the relation m NM, where M is the molar mass. The air fuel ratio can also be expressed on a mole basis as the ratio of the mole numbers of air to the mole numbers of fuel. But we will use the former definition. The reciprocal of air fuel ratio is called the fuel air ratio. Fuel 1 kg Air 17 kg Chapter Combustion chamber AF = 17 Products 18 kg FIGURE 15 6 The air fuel ratio (AF) represents the amount of air used per unit mass of fuel during a combustion process. EXAMPLE 15 1 Balancing the Combustion Equation One kmol of octane (C 8 H 18 ) is burned with air that contains 0 kmol of O, as shown in Fig Assuming the products contain only CO, H O, O, and N, determine the mole number of each gas in the products and the air fuel ratio for this combustion process. Solution The amount of fuel and the amount of oxygen in the air are given. The amount of the products and the AF are to be determined. Assumptions The combustion products contain CO, H O, O, and N only. Properties The molar mass of air is M air 8.97 kg/kmol 9.0 kg/kmol (Table A 1). Analysis The chemical equation for this combustion process can be written as C 8 H 18 1 kmol AIR Combustion chamber FIGURE 15 7 Schematic for Example x CO y H O z O w N where the terms in the parentheses represent the composition of dry air that contains 1 kmol of O and x, y, z, and w represent the unknown mole numbers of the gases in the products. These unknowns are determined by applying the mass balance to each of the elements that is, by requiring that the total mass or mole number of each element in the reactants be equal to that in the products: C: H: O: N : Substituting yields C 8 H O 3.76N S xco yh O zo wn 8 x S x 8 18 y S y 9 0 x y z S z w S w 75. C 8 H O 3.76N S 8CO 9H O 7.5O 75.N Note that the coefficient 0 in the balanced equation above represents the number of moles of oxygen, not the number of moles of air. The latter is obtained by adding moles of nitrogen to the 0 moles of

278 756 Thermodynamics oxygen, giving a total of 95. moles of air. The air fuel ratio (AF) is determined from Eq by taking the ratio of the mass of the air and the mass of the fuel, AF m air m fuel 1NM air 1NM C 1NM H 4. kg air/kg fuel kmol 19 kg>kmol 18 kmol 11 kg>kmol 19 kmol 1 kg>kmol That is, 4. kg of air is used to burn each kilogram of fuel during this combustion process. Fuel n CO m C n H m H O AIR Combustion chamber CH 4 + (O N ) CO + H O + 7.5N no unburned fuel no free oxygen in products Excess O N FIGURE 15 8 A combustion process is complete if all the combustible components of the fuel are burned to completion. FIGURE 15 9 The complete combustion process with no free oxygen in the products is called theoretical combustion. 15 THEORETICAL AND ACTUAL COMBUSTION PROCESSES It is often instructive to study the combustion of a fuel by assuming that the combustion is complete. A combustion process is complete if all the carbon in the fuel burns to CO, all the hydrogen burns to H O, and all the sulfur (if any) burns to SO. That is, all the combustible components of a fuel are burned to completion during a complete combustion process (Fig. 15 8). Conversely, the combustion process is incomplete if the combustion products contain any unburned fuel or components such as C, H, CO, or OH. Insufficient oxygen is an obvious reason for incomplete combustion, but it is not the only one. Incomplete combustion occurs even when more oxygen is present in the combustion chamber than is needed for complete combustion. This may be attributed to insufficient mixing in the combustion chamber during the limited time that the fuel and the oxygen are in contact. Another cause of incomplete combustion is dissociation, which becomes important at high temperatures. Oxygen has a much greater tendency to combine with hydrogen than it does with carbon. Therefore, the hydrogen in the fuel normally burns to completion, forming H O, even when there is less oxygen than needed for complete combustion. Some of the carbon, however, ends up as CO or just as plain C particles (soot) in the products. The minimum amount of air needed for the complete combustion of a fuel is called the stoichiometric or theoretical air. Thus, when a fuel is completely burned with theoretical air, no uncombined oxygen is present in the product gases. The theoretical air is also referred to as the chemically correct amount of air, or 100 percent theoretical air. A combustion process with less than the theoretical air is bound to be incomplete. The ideal combustion process during which a fuel is burned completely with theoretical air is called the stoichiometric or theoretical combustion of that fuel (Fig. 15 9). For example, the theoretical combustion of methane is CH 4 1O 3.76N S CO H O 7.5N Notice that the products of the theoretical combustion contain no unburned methane and no C, H, CO, OH, or free O.

279 In actual combustion processes, it is common practice to use more air than the stoichiometric amount to increase the chances of complete combustion or to control the temperature of the combustion chamber. The amount of air in excess of the stoichiometric amount is called excess air. The amount of excess air is usually expressed in terms of the stoichiometric air as percent excess air or percent theoretical air. For example, 50 percent excess air is equivalent to 150 percent theoretical air, and 00 percent excess air is equivalent to 300 percent theoretical air. Of course, the stoichiometric air can be expressed as 0 percent excess air or 100 percent theoretical air. Amounts of air less than the stoichiometric amount are called deficiency of air and are often expressed as percent deficiency of air. For example, 90 percent theoretical air is equivalent to 10 percent deficiency of air. The amount of air used in combustion processes is also expressed in terms of the equivalence ratio, which is the ratio of the actual fuel air ratio to the stoichiometric fuel air ratio. Predicting the composition of the products is relatively easy when the combustion process is assumed to be complete and the exact amounts of the fuel and air used are known. All one needs to do in this case is simply apply the mass balance to each element that appears in the combustion equation, without needing to take any measurements. Things are not so simple, however, when one is dealing with actual combustion processes. For one thing, actual combustion processes are hardly ever complete, even in the presence of excess air. Therefore, it is impossible to predict the composition of the products on the basis of the mass balance alone. Then the only alternative we have is to measure the amount of each component in the products directly. A commonly used device to analyze the composition of combustion gases is the Orsat gas analyzer. In this device, a sample of the combustion gases is collected and cooled to room temperature and pressure, at which point its volume is measured. The sample is then brought into contact with a chemical that absorbs the CO. The remaining gases are returned to the room temperature and pressure, and the new volume they occupy is measured. The ratio of the reduction in volume to the original volume is the volume fraction of the CO, which is equivalent to the mole fraction if ideal-gas behavior is assumed (Fig ). The volume fractions of the other gases are determined by repeating this procedure. In Orsat analysis the gas sample is collected over water and is maintained saturated at all times. Therefore, the vapor pressure of water remains constant during the entire test. For this reason the presence of water vapor in the test chamber is ignored and data are reported on a dry basis. However, the amount of H O formed during combustion is easily determined by balancing the combustion equation. BEFORE 100 kpa 5 C Gas sample including CO 1 liter Chapter V CO y CO = V AFTER 100 kpa 5 C Gas sample without CO 0.9 liter 0.1 = = FIGURE Determining the mole fraction of the CO in combustion gases by using the Orsat gas analyzer. EXAMPLE 15 Dew-Point Temperature of Combustion Products Ethane (C H 6 ) is burned with 0 percent excess air during a combustion process, as shown in Fig Assuming complete combustion and a total pressure of 100 kpa, determine (a) the air fuel ratio and (b) the dew-point temperature of the products. Solution The fuel is burned completely with excess air. The AF and the dew point of the products are to be determined. C H 6 Combustion chamber CO H O AIR O 100 kpa N (0% excess) FIGURE Schematic for Example 15.

280 758 Thermodynamics Assumptions 1 Combustion is complete. Combustion gases are ideal gases. Analysis The combustion products contain CO, H O, N, and some excess O only. Then the combustion equation can be written as C H 6 1.a th 1O 3.76N S CO 3H O 0.a th O a th N where a th is the stoichiometric coefficient for air. We have automatically accounted for the 0 percent excess air by using the factor 1.a th instead of a th for air. The stoichiometric amount of oxygen (a th O ) is used to oxidize the fuel, and the remaining excess amount (0.a th O ) appears in the products as unused oxygen. Notice that the coefficient of N is the same on both sides of the equation, and that we wrote the C and H balances directly since they are so obvious. The coefficient a th is determined from the O balance to be O : 1.a th a th S a th 3.5 Substituting, C H O 3.76N S CO 3H O 0.7O 15.79N (a) The air fuel ratio is determined from Eq by taking the ratio of the mass of the air to the mass of the fuel, AF m air m fuel 19.3 kg air/kg fuel kmol19 kg>kmol 1 kmol 11 kg>kmol 13 kmol 1 kg>kmol That is, 19.3 kg of air is supplied for each kilogram of fuel during this combustion process. (b) The dew-point temperature of the products is the temperature at which the water vapor in the products starts to condense as the products are cooled at constant pressure. Recall from Chap. 14 that the dew-point temperature of a gas vapor mixture is the saturation temperature of the water vapor corresponding to its partial pressure. Therefore, we need to determine the partial pressure of the water vapor P v in the products first. Assuming ideal-gas behavior for the combustion gases, we have P v a N v 3 kmol b1p N prod a b1100 kpa kpa prod 1.49 kmol Thus, T dp T kpa 5.3 C (Table A 5) EXAMPLE 15 3 Combustion of a Gaseous Fuel with Moist Air FUEL CH 4, O, H, N, CO AIR 0 C, φ = 80% Combustion chamber 1 atm FIGURE 15 1 Schematic for Example CO H O N A certain natural gas has the following volumetric analysis: 7 percent CH 4, 9 percent H, 14 percent N, percent O, and 3 percent CO. This gas is now burned with the stoichiometric amount of air that enters the combustion chamber at 0 C, 1 atm, and 80 percent relative humidity, as shown in Fig Assuming complete combustion and a total pressure of 1 atm, determine the dew-point temperature of the products. Solution A gaseous fuel is burned with the stoichiometric amount of moist air. The dew point temperature of the products is to be determined.

281 Chapter Assumptions 1 The fuel is burned completely and thus all the carbon in the fuel burns to CO and all the hydrogen to H O. The fuel is burned with the stoichiometric amount of air and thus there is no free O in the product gases. 3 Combustion gases are ideal gases. Properties The saturation pressure of water at 0 C is.339 kpa (Table A 4). Analysis We note that the moisture in the air does not react with anything; it simply shows up as additional H O in the products. Therefore, for simplicity, we balance the combustion equation by using dry air and then add the moisture later to both sides of the equation. Considering 1 kmol of fuel, fuel dry air 10.7CH H 0.14N 0.0O 0.03CO a th 1O 3.76N S xco yh O zn The unknown coefficients in the above equation are determined from mass balances on various elements, C: H: x S x y S y O : a th x y S a th N : a th z S z Next we determine the amount of moisture that accompanies 4.76a th (4.76)(1.465) 6.97 kmol of dry air. The partial pressure of the moisture in the air is Assuming ideal-gas behavior, the number of moles of the moisture in the air is which yields P v,air f air P 0 C kpa kpa N v,air a P v,air kpa b N P total a total kpa b16.97 N v,air N v,air kmol The balanced combustion equation is obtained by substituting the coefficients determined earlier and adding kmol of H O to both sides of the equation: fuel dry air CH H 0.14N 0.0O 0.03CO O 3.76N moisture includes moisture H O S 0.75CO 1.661H O 5.648N The dew-point temperature of the products is the temperature at which the water vapor in the products starts to condense as the products are cooled. Again, assuming ideal-gas behavior, the partial pressure of the water vapor in the combustion gases is P v,prod a N v,prod kmol b P N prod a b kpa 0.88 kpa prod kmol

282 760 Thermodynamics Thus, T dp T 0.88 kpa 60.9 C Discussion If the combustion process were achieved with dry air instead of moist air, the products would contain less moisture, and the dew-point temperature in this case would be 59.5 C. EXAMPLE 15 4 Reverse Combustion Analysis C 8 H % CO AIR Combustion chamber FIGURE Schematic for Example % O 0.88% CO 83.48% N Octane (C 8 H 18 ) is burned with dry air. The volumetric analysis of the products on a dry basis is (Fig ) CO : 10.0 percent O : 5.6 percent CO: 0.88 percent N : percent Determine (a) the air fuel ratio, (b) the percentage of theoretical air used, and (c) the amount of H O that condenses as the products are cooled to 5 C at 100 kpa. Solution Combustion products whose composition is given are cooled to 5 C. The AF, the percent theoretical air used, and the fraction of water vapor that condenses are to be determined. Assumptions Combustion gases are ideal gases. Properties The saturation pressure of water at 5 C is kpa (Table A 4). Analysis Note that we know the relative composition of the products, but we do not know how much fuel or air is used during the combustion process. However, they can be determined from mass balances. The H O in the combustion gases will start condensing when the temperature drops to the dewpoint temperature. For ideal gases, the volume fractions are equivalent to the mole fractions. Considering 100 kmol of dry products for convenience, the combustion equation can be written as xc 8 H 18 a 1O 3.76N S 10.0CO 0.88CO 5.6O 83.48N bh O The unknown coefficients x, a, and b are determined from mass balances, N : C: H: 3.76a S a.0 8x S x x b S b 1.4 O : a b S.0.0 The O balance is not necessary, but it can be used to check the values obtained from the other mass balances, as we did previously. Substituting, we get 1.36C 8 H 18. 1O 3.76N S 10.0CO 0.88CO 5.6O 83.48N 1.4H O

283 Chapter The combustion equation for 1 kmol of fuel is obtained by dividing the above equation by 1.36, C 8 H O 3.76N S (a) The air fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel (Eq. 15 3), AF m air m fuel kg air/kg fuel 7.37CO 0.65CO 4.13O 61.38N 9H O kmol19 kg>kmol 18 kmol 11 kg>kmol 19 kmol 1 kg>kmol (b) To find the percentage of theoretical air used, we need to know the theoretical amount of air, which is determined from the theoretical combustion equation of the fuel, O : Then, C 8 H 18 a th 1O 3.76N S 8CO 9H O 3.76a th N a th S a th 1.5 Percentage of theoretical air m air,act m air,th N air,act N air,th That is, 31 percent excess air was used during this combustion process. Notice that some carbon formed carbon monoxide even though there was considerably more oxygen than needed for complete combustion. (c) For each kmol of fuel burned, kmol of products are formed, including 9 kmol of H O. Assuming that the dew-point temperature of the products is above 5 C, some of the water vapor will condense as the products are cooled to 5 C. If N w kmol of H O condenses, there will be (9 N w ) kmol of water vapor left in the products. The mole number of the products in the gas phase will also decrease to 8.53 N w as a result. By treating the product gases (including the remaining water vapor) as ideal gases, N w is determined by equating the mole fraction of the water vapor to its pressure fraction, N v N prod,gas 131% 9 N w kpa 8.53 N w 100 kpa N w 6.59 kmol Therefore, the majority of the water vapor in the products (73 percent of it) condenses as the product gases are cooled to 5 C. P v P prod kmol kmol

284 76 Thermodynamics ATOM MOLECULE Nuclear energy Chemical energy Latent energy ATOM Sensible energy MOLECULE FIGURE The microscopic form of energy of a substance consists of sensible, latent, chemical, and nuclear energies ENTHALPY OF FORMATION AND ENTHALPY OF COMBUSTION We mentioned in Chap. that the molecules of a system possess energy in various forms such as sensible and latent energy (associated with a change of state), chemical energy (associated with the molecular structure), and nuclear energy (associated with the atomic structure), as illustrated in Fig In this text we do not intend to deal with nuclear energy. We also ignored chemical energy until now since the systems considered in previous chapters involved no changes in their chemical structure, and thus no changes in chemical energy. Consequently, all we needed to deal with were the sensible and latent energies. During a chemical reaction, some chemical bonds that bind the atoms into molecules are broken, and new ones are formed. The chemical energy associated with these bonds, in general, is different for the reactants and the products. Therefore, a process that involves chemical reactions involves changes in chemical energies, which must be accounted for in an energy balance (Fig ). Assuming the atoms of each reactant remain intact (no nuclear reactions) and disregarding any changes in kinetic and potential energies, the energy change of a system during a chemical reaction is due to a change in state and a change in chemical composition. That is, E sys E state E chem (15 4) 1 kmol C 5 C, 1 atm 1 kmol O 5 C, 1 atm Sensible energy ATOM ATOM ATOM FIGURE ,50 kj Combustion chamber Broken chemical bond FIGURE When the existing chemical bonds are destroyed and new ones are formed during a combustion process, usually a large amount of sensible energy is absorbed or released. CO 5 C, 1 atm The formation of CO during a steadyflow combustion process at 5 C and 1 atm. Therefore, when the products formed during a chemical reaction exit the reaction chamber at the inlet state of the reactants, we have E state 0 and the energy change of the system in this case is due to the changes in its chemical composition only. In thermodynamics we are concerned with the changes in the energy of a system during a process, and not the energy values at the particular states. Therefore, we can choose any state as the reference state and assign a value of zero to the internal energy or enthalpy of a substance at that state. When a process involves no changes in chemical composition, the reference state chosen has no effect on the results. When the process involves chemical reactions, however, the composition of the system at the end of a process is no longer the same as that at the beginning of the process. In this case it becomes necessary to have a common reference state for all substances. The chosen reference state is 5 C (77 F) and 1 atm, which is known as the standard reference state. Property values at the standard reference state are indicated by a superscript ( ) (such as h and u ). When analyzing reacting systems, we must use property values relative to the standard reference state. However, it is not necessary to prepare a new set of property tables for this purpose. We can use the existing tables by subtracting the property values at the standard reference state from the values at the specified state. The ideal-gas enthalpy of N at 500 K relative to the standard reference state, for example, is h 500 K h 14, kj/kmol. Consider the formation of CO from its elements, carbon and oxygen, during a steady-flow combustion process (Fig ). Both the carbon and the oxygen enter the combustion chamber at 5 C and 1 atm. The CO formed during this process also leaves the combustion chamber at 5 C and 1 atm. The combustion of carbon is an exothermic reaction (a reaction dur-

285 ing which chemical energy is released in the form of heat). Therefore, some heat is transferred from the combustion chamber to the surroundings during this process, which is 393,50 kj/kmol CO formed. (When one is dealing with chemical reactions, it is more convenient to work with quantities per unit mole than per unit time, even for steady-flow processes.) The process described above involves no work interactions. Therefore, from the steady-flow energy balance relation, the heat transfer during this process must be equal to the difference between the enthalpy of the products and the enthalpy of the reactants. That is, Q H prod H react 393,50 kj>kmol (15 5) Since both the reactants and the products are at the same state, the enthalpy change during this process is solely due to the changes in the chemical composition of the system. This enthalpy change is different for different reactions, and it is very desirable to have a property to represent the changes in chemical energy during a reaction. This property is the enthalpy of reaction h R, which is defined as the difference between the enthalpy of the products at a specified state and the enthalpy of the reactants at the same state for a complete reaction. For combustion processes, the enthalpy of reaction is usually referred to as the enthalpy of combustion h C, which represents the amount of heat released during a steady-flow combustion process when 1 kmol (or 1 kg) of fuel is burned completely at a specified temperature and pressure (Fig ). It is expressed as h R h C H prod H react (15 6) which is 393,50 kj/kmol for carbon at the standard reference state. The enthalpy of combustion of a particular fuel is different at different temperatures and pressures. The enthalpy of combustion is obviously a very useful property for analyzing the combustion processes of fuels. However, there are so many different fuels and fuel mixtures that it is not practical to list h C values for all possible cases. Besides, the enthalpy of combustion is not of much use when the combustion is incomplete. Therefore a more practical approach would be to have a more fundamental property to represent the chemical energy of an element or a compound at some reference state. This property is the enthalpy of formation h f, which can be viewed as the enthalpy of a substance at a specified state due to its chemical composition. To establish a starting point, we assign the enthalpy of formation of all stable elements (such as O,N,H, and C) a value of zero at the standard reference state of 5 C and 1 atm. That is, h f 0 for all stable elements. (This is no different from assigning the internal energy of saturated liquid water a value of zero at 0.01 C.) Perhaps we should clarify what we mean by stable. The stable form of an element is simply the chemically stable form of that element at 5 C and 1 atm. Nitrogen, for example, exists in diatomic form (N ) at 5 C and 1 atm. Therefore, the stable form of nitrogen at the standard reference state is diatomic nitrogen N, not monatomic nitrogen N. If an element exists in more than one stable form at 5 C and 1 atm, one of the forms should be specified as the stable form. For carbon, for example, the stable form is assumed to be graphite, not diamond. 1 kmol C 5 C, 1 atm 1 kmol O 5 C, 1 atm Chapter h C = Q = 393,50 kj/kmol C Combustion process 1 kmol CO 5 C, 1 atm FIGURE The enthalpy of combustion represents the amount of energy released as a fuel is burned during a steady-flow process at a specified state.

286 764 Thermodynamics 1 kmol C 5 C, 1 atm 1 kmol O 5 C, 1 atm h f = Q = 393,50 kj/kmol CO Combustion chamber 1 kmol CO 5 C, 1 atm FIGURE The enthalpy of formation of a compound represents the amount of energy absorbed or released as the component is formed from its stable elements during a steady-flow process at a specified state. Fuel 1 kg Air LHV = Q out Combustion chamber HHV = LHV + (mh fg ) H O (mh fg ) H O Products (vapor H O) Products (liquid H O) FIGURE The higher heating value of a fuel is equal to the sum of the lower heating value of the fuel and the latent heat of vaporization of the H O in the products. Now reconsider the formation of CO (a compound) from its elements C and O at 5 C and 1 atm during a steady-flow process. The enthalpy change during this process was determined to be 393,50 kj/kmol. However, H react 0 since both reactants are elements at the standard reference state, and the products consist of 1 kmol of CO at the same state. Therefore, the enthalpy of formation of CO at the standard reference state is 393,50 kj/kmol (Fig ). That is, h f,co 393,50 kj>kmol The negative sign is due to the fact that the enthalpy of 1 kmol of CO at 5 C and 1 atm is 393,50 kj less than the enthalpy of 1 kmol of C and 1 kmol of O at the same state. In other words, 393,50 kj of chemical energy is released (leaving the system as heat) when C and O combine to form 1 kmol of CO. Therefore, a negative enthalpy of formation for a compound indicates that heat is released during the formation of that compound from its stable elements. A positive value indicates heat is absorbed. You will notice that two h f values are given for H O in Table A 6, one for liquid water and the other for water vapor. This is because both phases of H O are encountered at 5 C, and the effect of pressure on the enthalpy of formation is small. (Note that under equilibrium conditions, water exists only as a liquid at 5 C and 1 atm.) The difference between the two enthalpies of formation is equal to the h fg of water at 5 C, which is kj/kg or 44,000 kj/kmol. Another term commonly used in conjunction with the combustion of fuels is the heating value of the fuel, which is defined as the amount of heat released when a fuel is burned completely in a steady-flow process and the products are returned to the state of the reactants. In other words, the heating value of a fuel is equal to the absolute value of the enthalpy of combustion of the fuel. That is, Heating value 0h C 0 1kJ>kg fuel The heating value depends on the phase of the H O in the products. The heating value is called the higher heating value (HHV) when the H O in the products is in the liquid form, and it is called the lower heating value (LHV) when the H O in the products is in the vapor form (Fig ). The two heating values are related by HHV LHV 1mh fg H O 1kJ>kg fuel (15 7) where m is the mass of H O in the products per unit mass of fuel and h fg is the enthalpy of vaporization of water at the specified temperature. Higher and lower heating values of common fuels are given in Table A 7. The heating value or enthalpy of combustion of a fuel can be determined from a knowledge of the enthalpy of formation for the compounds involved. This is illustrated with the following example. EXAMPLE 15 5 Evaluation of the Enthalpy of Combustion Determine the enthalpy of combustion of liquid octane (C 8 H 18 ) at 5 C and 1 atm, using enthalpy-of-formation data from Table A 6. Assume the water in the products is in the liquid form.

287 Chapter Solution The enthalpy of combustion of a fuel is to be determined using enthalpy of formation data. Properties The enthalpy of formation at 5 C and 1 atm is 393,50 kj/kmol for CO, 85,830 kj/kmol for H O( ), and 49,950 kj/kmol for C 8 H 18 ( ) (Table A 6). Analysis The combustion of C 8 H 18 is illustrated in Fig The stoichiometric equation for this reaction is C 8 H 18 a th 1O 3.76N S 8CO 9H O a th N Both the reactants and the products are at the standard reference state of 5 C and 1 atm. Also, N and O are stable elements, and thus their enthalpy of formation is zero. Then the enthalpy of combustion of C 8 H 18 becomes (Eq. 15 6) C 8 H 18 ( ) 5 C, 1 atm AIR 5 C, 1 atm h C = H prod H react FIGURE 15 0 Schematic for Example CO 5 C 1 atm H O( ) N h C H prod H react Substituting, a N p h f,p a N r h f,r 1Nh f CO 1Nh f H O 1Nh f C8 H 18 h C 18 kmol 1 393,50 kj>kmol 19 kmol 1 85,830 kj>kmol 11 kmol1 49,950 kj>kmol 5,471,000 kj>kmol C 8 H 18 47,891 kj>kg C 8 H 18 which is practially identical to the listed value of 47,890 kj/kg in Table A 7. Since the water in the products is assumed to be in the liquid phase, this h C value corresponds to the HHV of liquid C 8 H 18. Discussion It can be shown that the result for gaseous octane is 5,51,00 kj/kmol or 48,55 kj/kg. When the exact composition of the fuel is known, the enthalpy of combustion of that fuel can be determined using enthalpy of formation data as shown above. However, for fuels that exhibit considerable variations in composition depending on the source, such as coal, natural gas, and fuel oil, it is more practical to determine their enthalpy of combustion experimentally by burning them directly in a bomb calorimeter at constant volume or in a steady-flow device FIRST-LAW ANALYSIS OF REACTING SYSTEMS The energy balance (or the first-law) relations developed in Chaps. 4 and 5 are applicable to both reacting and nonreacting systems. However, chemically reacting systems involve changes in their chemical energy, and thus it is more convenient to rewrite the energy balance relations so that the changes in chemical energies are explicitly expressed. We do this first for steady-flow systems and then for closed systems. Steady-Flow Systems Before writing the energy balance relation, we need to express the enthalpy of a component in a form suitable for use for reacting systems. That is, we need to express the enthalpy such that it is relative to the standard reference

288 766 Thermodynamics Enthalpy at 5 C, 1 atm H = N (h f + h h ) Sensible enthalpy relative to 5 C, 1 atm FIGURE 15 1 The enthalpy of a chemical component at a specified state is the sum of the enthalpy of the component at 5 C, 1 atm (h f ), and the sensible enthalpy of the component relative to 5 C, 1 atm. state and the chemical energy term appears explicitly. When expressed properly, the enthalpy term should reduce to the enthalpy of formation h f at the standard reference state. With this in mind, we express the enthalpy of a component on a unit mole basis as (Fig. 15 1) where the term in the parentheses represents the sensible enthalpy relative to the standard reference state, which is the difference between h the sensible enthalpy at the specified state) and h (the sensible enthalpy at the standard reference state of 5 C and 1 atm). This definition enables us to use enthalpy values from tables regardless of the reference state used in their construction. When the changes in kinetic and potential energies are negligible, the steady-flow energy balance relation E. in E. out can be expressed for a chemically reacting steady-flow system more explicitly as Rate of net energy transfer in by heat, work, and mass Enthalpy h f 1h h 1kJ>kmol Q # in W # in a n # r 1h f h h r Q # out W # out a n # p 1h f h h p Rate of net energy transfer out by heat, work, and mass (15 8) where ṅ p and ṅ r represent the molal flow rates of the product p and the reactant r, respectively. In combustion analysis, it is more convenient to work with quantities expressed per mole of fuel. Such a relation is obtained by dividing each term of the equation above by the molal flow rate of the fuel, yielding Q in W in a N r 1h f h h r Q out W out a N p 1h f h h p Energy transfer in per mole of fuel Energy transfer out per mole of fuel by heat, work, and mass by heat, work, and mass (15 9) where N r and N p represent the number of moles of the reactant r and the product p, respectively, per mole of fuel. Note that N r 1 for the fuel, and the other N r and N p values can be picked directly from the balanced combustion equation. Taking heat transfer to the system and work done by the system to be positive quantities, the energy balance relation just discussed can be expressed more compactly as or as where Q W a N p 1h f h h p a N r 1h f h h r Q W H prod H react 1kJ>kmol fuel H prod a N p 1h f h h p 1kJ>kmol fuel (15 10) (15 11) H react a N r 1h f h h r 1kJ>kmol fuel If the enthalpy of combustion h C for a particular reaction is available, the steady-flow energy equation per mole of fuel can be expressed as Q W h C a N p 1h h p a N r 1h h r 1kJ>kmol (15 1) The energy balance relations above are sometimes written without the work term since most steady-flow combustion processes do not involve any work interactions.

289 A combustion chamber normally involves heat output but no heat input. Then the energy balance for a typical steady-flow combustion process becomes Chapter Q out a N r 1h f h h r a N p 1h f h h p Energy in by mass Energy out by mass per mole of fuel per mole of fuel (15 13) It expresses that the heat output during a combustion process is simply the difference between the energy of the reactants entering and the energy of the products leaving the combustion chamber. Closed Systems The general closed-system energy balance relation E in E out E system can be expressed for a stationary chemically reacting closed system as 1Q in Q out 1W in W out U prod U react 1kJ>kmol fuel (15 14) where U prod represents the internal energy of the products and U react represents the internal energy of the reactants. To avoid using another property the internal energy of formation u we f utilize the definition of enthalpy (u h Pv or u f u u h f h h Pv) and express the above equation as (Fig. 15 ) Q W a N p 1h f h h Pv p a N r 1h f h h Pv r (15 15) where we have taken heat transfer to the system and work done by the system to be positive quantities. The Pv terms are negligible for solids and liquids, and can be replaced by R u T for gases that behave as an ideal gas. Also, if desired, the h Pv terms in Eq can be replaced by u. The work term in Eq represents all forms of work, including the boundary work. It was shown in Chap. 4 that U W b H for nonreacting closed systems undergoing a quasi-equilibrium P constant expansion or compression process. This is also the case for chemically reacting systems. There are several important considerations in the analysis of reacting systems. For example, we need to know whether the fuel is a solid, a liquid, or a gas since the enthalpy of formation h f of a fuel depends on the phase of the fuel. We also need to know the state of the fuel when it enters the combustion chamber in order to determine its enthalpy. For entropy calculations it is especially important to know if the fuel and air enter the combustion chamber premixed or separately. When the combustion products are cooled to low temperatures, we need to consider the possibility of condensation of some of the water vapor in the product gases. EXAMPLE 15 6 First-Law Analysis of Steady-Flow Combustion Liquid propane (C 3 H 8 ) enters a combustion chamber at 5 C at a rate of 0.05 kg/min where it is mixed and burned with 50 percent excess air that enters the combustion chamber at 7 C, as shown in Fig An analysis of the combustion gases reveals that all the hydrogen in the fuel burns to H O but only 90 percent of the carbon burns to CO, with the remaining 10 percent forming CO. If the exit temperature of the combustion gases is U = H PV = N(h f + h h ) PV = N(h f + h h Pv) FIGURE 15 An expression for the internal energy of a chemical component in terms of the enthalpy. C 3 H 8 ( ) 5 C, 0.05 kg/min AIR 7 C Q =? Combustion chamber FIGURE 15 3 Schematic for Example H O CO 1500 K CO O N

290 768 Thermodynamics 1500 K, determine (a) the mass flow rate of air and (b) the rate of heat transfer from the combustion chamber. Solution Liquid propane is burned steadily with excess air. The mass flow rate of air and the rate of heat transfer are to be determined. Assumptions 1 Steady operating conditions exist. Air and the combustion gases are ideal gases. 3 Kinetic and potential energies are negligible. Analysis We note that all the hydrogen in the fuel burns to H O but 10 percent of the carbon burns incompletely and forms CO. Also, the fuel is burned with excess air and thus there is some free O in the product gases. The theoretical amount of air is determined from the stoichiometric reaction to be O balance: Then the balanced equation for the actual combustion process with 50 percent excess air and some CO in the products becomes (a) The air fuel ratio for this combustion process is Thus, C 3 H 8 1 a th 1O 3.76N S 3CO 4H O 3.76a th N AF m air m fuel 5.53 kg air>kg fuel m # air 1AF 1m # fuel a th 3 5 C 3 H O 3.76N S.7CO 0.3CO 4H O.65O 8.N kg air>kg fuel kg fuel>min 1.18 kg air/min kmol19 kg>kmol 13 kmol 11 kg>kmol 14 kmol 1 kg>kmol (b) The heat transfer for this steady-flow combustion process is determined from the steady-flow energy balance E out E in applied on the combustion chamber per unit mole of the fuel, or Q out a N p 1h f h h p a N r 1h f h h r Q out a N r 1h f h h r a N p 1h f h h p Assuming the air and the combustion products to be ideal gases, we have h h(t ), and we form the following minitable using data from the property tables: h f h 80 K h 98 K h 1500 K Substance kj/kmol kj/kmol kj/kmol kj/kmol C 3 H 8 ( ) 118,910 O ,9 N ,073 H O(g) 41, ,999 CO 393, ,078 CO 110, ,517

291 Chapter The h f of liquid propane is obtained by subtracting the h fg of propane at 5 C from the h f of gas propane. Substituting gives Q out 11 kmol C 3 H ,910 h 98 h 98 kj>kmol C 3 H kmol O kj>kmol O kmol N kj>kmol N kmol CO ,50 71, kj>kmol CO kmol CO31 110,530 47, kj>kmol CO4 14 kmol H O31 41,80 57, kj>kmol H O kmol O ,9 868 kj>kmol O kmol N , kj>kmol N 4 363,880 kj>kmol of C 3 H 8 Thus 363,880 kj of heat is transferred from the combustion chamber for each kmol (44 kg) of propane. This corresponds to 363,880/ kj of heat loss per kilogram of propane. Then the rate of heat transfer for a mass flow rate of 0.05 kg/min for the propane becomes Q # out m # q out kg>min1870 kj>kg kj>min 6.89 kw EXAMPLE 15 7 First-Law Analysis of Combustion in a Bomb The constant-volume tank shown in Fig contains 1 lbmol of methane (CH 4 ) gas and 3 lbmol of O at 77 F and 1 atm. The contents of the tank are ignited, and the methane gas burns completely. If the final temperature is 1800 R, determine (a) the final pressure in the tank and (b) the heat transfer during this process. Solution Methane is burned in a rigid tank. The final pressure in the tank and the heat transfer are to be determined. Assumptions 1 The fuel is burned completely and thus all the carbon in the fuel burns to CO and all the hydrogen to H O. The fuel, the air, and the combustion gases are ideal gases. 3 Kinetic and potential energies are negligible. 4 There are no work interactions involved. Analysis The balanced combustion equation is BEFORE REACTION AFTER REACTION CO 1 lbmol CH 4 H O 3 lbmol O O 77 F 1800 R 1 atm P FIGURE 15 4 Schematic for Example (a) At 1800 R, water exists in the gas phase. Using the ideal-gas relation for both the reactants and the products, the final pressure in the tank is determined to be Substituting, we get CH 4 1g 3O S CO H O O P react V N react R u T react P prod V N prod R u T prod f P prod P react a N prod N react ba T prod T react b P prod 11 atma 4 lbmol 4 lbmol ba1800 R b 3.35 atm 537 R

292 770 Thermodynamics (b) Noting that the process involves no work interactions, the heat transfer during this constant-volume combustion process can be determined from the energy balance E in E out E system applied to the tank, Q out a N p 1h f h h Pv p a N r 1h f h h Pv r Since both the reactants and the products are assumed to be ideal gases, all the internal energy and enthalpies depend on temperature only, and the Pv terms in this equation can be replaced by R u T. It yields Q out a N r 1h f R u T r a N p 1h f h 1800 R h 537 R R u T p since the reactants are at the standard reference temperature of 537 R. From h f and ideal-gas tables in the Appendix, h f h 537 R h 1800 R Substance Btu/lbmol Btu/lbmol Btu/lbmol CH 4 3,10 O ,485.8 CO 169, ,391.5 H O(g) 104, ,433.0 Substituting, we have Q out 11 lbmol CH , Btu>lbmol CH lbmol O Btu>lbmol O 4 11 lbmol CO ,300 18, Btu>lbmol CO 4 1 lbmol H O31 104,040 15, Btu>lbmol H O4 11 lbmol O , Btu>lbmol O 4 308,730 Btu/lbmol CH 4 Discussion On a mass basis, the heat transfer from the tank would be 308,730/16 19,300 Btu/lbm of methane. Insulation Fuel Air FIGURE 15 5 Combustion chamber Products T max The temperature of a combustion chamber becomes maximum when combustion is complete and no heat is lost to the surroundings (Q 0) ADIABATIC FLAME TEMPERATURE In the absence of any work interactions and any changes in kinetic or potential energies, the chemical energy released during a combustion process either is lost as heat to the surroundings or is used internally to raise the temperature of the combustion products. The smaller the heat loss, the larger the temperature rise. In the limiting case of no heat loss to the surroundings (Q 0), the temperature of the products reaches a maximum, which is called the adiabatic flame or adiabatic combustion temperature of the reaction (Fig. 15 5).

293 The adiabatic flame temperature of a steady-flow combustion process is determined from Eq by setting Q 0 and W 0. It yields or H prod H react (15 16) Chapter a N p 1h f h h p a N r 1h f h h r (15 17) Once the reactants and their states are specified, the enthalpy of the reactants H react can be easily determined. The calculation of the enthalpy of the products H prod is not so straightforward, however, because the temperature of the products is not known prior to the calculations. Therefore, the determination of the adiabatic flame temperature requires the use of an iterative technique unless equations for the sensible enthalpy changes of the combustion products are available. A temperature is assumed for the product gases, and the H prod is determined for this temperature. If it is not equal to H react, calculations are repeated with another temperature. The adiabatic flame temperature is then determined from these two results by interpolation. When the oxidant is air, the product gases mostly consist of N, and a good first guess for the adiabatic flame temperature is obtained by treating the entire product gases as N. In combustion chambers, the highest temperature to which a material can be exposed is limited by metallurgical considerations. Therefore, the adiabatic flame temperature is an important consideration in the design of combustion chambers, gas turbines, and nozzles. The maximum temperatures that occur in these devices are considerably lower than the adiabatic flame temperature, however, since the combustion is usually incomplete, some heat loss takes place, and some combustion gases dissociate at high temperatures (Fig. 15 6). The maximum temperature in a combustion chamber can be controlled by adjusting the amount of excess air, which serves as a coolant. Note that the adiabatic flame temperature of a fuel is not unique. Its value depends on (1) the state of the reactants, () the degree of completion of the reaction, and (3) the amount of air used. For a specified fuel at a specified state burned with air at a specified state, the adiabatic flame temperature attains its maximum value when complete combustion occurs with the theoretical amount of air. Fuel Air Heat loss Incomplete combustion Dissociation Products T prod < T max FIGURE 15 6 The maximum temperature encountered in a combustion chamber is lower than the theoretical adiabatic flame temperature. EXAMPLE 15 8 Adiabatic Flame Temperature in Steady Combustion Liquid octane (C 8 H 18 ) enters the combustion chamber of a gas turbine steadily at 1 atm and 5 C, and it is burned with air that enters the combustion chamber at the same state, as shown in Fig Determine the adiabatic flame temperature for (a) complete combustion with 100 percent theoretical air, (b) complete combustion with 400 percent theoretical air, and (c) incomplete combustion (some CO in the products) with 90 percent theoretical air. Solution Liquid octane is burned steadily. The adiabatic flame temperature is to be determined for different cases. C 8 H 18 5 C, 1 atm Air 5 C, 1 atm Combustion chamber T P 1 atm FIGURE 15 7 Schematic for Example CO H O N O

294 77 Thermodynamics Assumptions 1 This is a steady-flow combustion process. The combustion chamber is adiabatic. 3 There are no work interactions. 4 Air and the combustion gases are ideal gases. 5 Changes in kinetic and potential energies are negligible. Analysis (a) The balanced equation for the combustion process with the theoretical amount of air is C 8 H O 3.76N S 8CO 9H O 47N The adiabatic flame temperature relation H prod H react in this case reduces to since all the reactants are at the standard reference state and h f 0 for O and N. The h f and h values of various components at 98 K are h f h 98 K Substance kj/kmol kj/kmol C 8 H 18 ( ) 49,950 O N H O(g) 41, CO 393, Substituting, we have which yields a N p 1h f h h p a N r h f,r 1Nh f C8 H kmol CO ,50 h CO 9364 kj>kmol CO 4 19 kmol H O31 41,80 h H O 9904 kj>kmol H O4 147 kmol N 310 h N 8669 kj>kmol N 4 11 kmol C 8 H ,950 kj>kmol C 8 H 18 8h CO 9h H O 47h N 5,646,081 kj It appears that we have one equation with three unknowns. Actually we have only one unknown the temperature of the products T prod since h h(t ) for ideal gases. Therefore, we have to use an equation solver such as EES or a trial-and-error approach to determine the temperature of the products. A first guess is obtained by dividing the right-hand side of the equation by the total number of moles, which yields 5,646,081/( ) 88,0 kj/kmol. This enthalpy value corresponds to about 650 K for N, 100 K for H O, and 1800 K for CO. Noting that the majority of the moles are N, we see that T prod should be close to 650 K, but somewhat under it. Therefore, a good first guess is 400 K. At this temperature, 8h CO 9h H O 47h N 8 15, , ,30 5,660,88 kj This value is higher than 5,646,081 kj. Therefore, the actual temperature is slightly under 400 K. Next we choose 350 K. It yields 8 1, , ,496 5,56,654

295 Chapter which is lower than 5,646,081 kj. Therefore, the actual temperature of the products is between 350 and 400 K. By interpolation, it is found to be T prod 395 K. (b) The balanced equation for the complete combustion process with 400 percent theoretical air is C 8 H O 3.76N S 8CO 9H O 37.5O 188N By following the procedure used in (a), the adiabatic flame temperature in this case is determined to be T prod 96 K. Notice that the temperature of the products decreases significantly as a result of using excess air. (c) The balanced equation for the incomplete combustion process with 90 percent theoretical air is C 8 H O 3.76N S 5.5CO.5CO 9H O 4.3N Following the procedure used in (a), we find the adiabatic flame temperature in this case to be T prod 36 K. Discussion Notice that the adiabatic flame temperature decreases as a result of incomplete combustion or using excess air. Also, the maximum adiabatic flame temperature is achieved when complete combustion occurs with the theoretical amount of air ENTROPY CHANGE OF REACTING SYSTEMS So far we have analyzed combustion processes from the conservation of mass and the conservation of energy points of view. The thermodynamic analysis of a process is not complete, however, without the examination of the second-law aspects. Of particular interest are the exergy and exergy destruction, both of which are related to entropy. The entropy balance relations developed in Chap. 7 are equally applicable to both reacting and nonreacting systems provided that the entropies of individual constituents are evaluated properly using a common basis. The entropy balance for any system (including reacting systems) undergoing any process can be expressed as S in S out S gen S system 1kJ>K Net entropy transfer Entropy Change by heat and mass generation in entropy (15 18) Using quantities per unit mole of fuel and taking the positive direction of heat transfer to be to the system, the entropy balance relation can be expressed more explicitly for a closed or steady-flow reacting system as (Fig. 15 8) a Q k T k S gen S prod S react 1kJ>K (15 19) where T k is temperature at the boundary where Q k crosses it. For an adiabatic process (Q 0), the entropy transfer term drops out and Eq reduces to S gen,adiabatic S prod S react 0 (15 0) Reactants S react Surroundings Reaction chamber S sys Products S prod FIGURE 15 8 The entropy change associated with a chemical relation.

296 774 Thermodynamics T T s(t,p) s = R u ln P P0 P s (T,P 0 ) (Tabulated) P 0 = 1 atm FIGURE 15 9 At a specified temperature, the absolute entropy of an ideal gas at pressures other than P 0 1 atm can be determined by subtracting R u ln (P/P 0 ) from the tabulated value at 1 atm. s The total entropy generated during a process can be determined by applying the entropy balance to an extended system that includes the system itself and its immediate surroundings where external irreversibilities might be occurring. When evaluating the entropy transfer between an extended system and the surroundings, the boundary temperature of the extended system is simply taken to be the environment temperature, as explained in Chap. 7. The determination of the entropy change associated with a chemical reaction seems to be straightforward, except for one thing: The entropy relations for the reactants and the products involve the entropies of the components, not entropy changes, which was the case for nonreacting systems. Thus we are faced with the problem of finding a common base for the entropy of all substances, as we did with enthalpy. The search for such a common base led to the establishment of the third law of thermodynamics in the early part of this century. The third law was expressed in Chap. 7 as follows: The entropy of a pure crystalline substance at absolute zero temperature is zero. Therefore, the third law of thermodynamics provides an absolute base for the entropy values for all substances. Entropy values relative to this base are called the absolute entropy. The s values listed in Tables A 18 through A 5 for various gases such as N,O,CO,CO,H,H O, OH, and O are the ideal-gas absolute entropy values at the specified temperature and at a pressure of 1 atm. The absolute entropy values for various fuels are listed in Table A 6 together with the h f values at the standard reference state of 5 C and 1 atm. Equation 15 0 is a general relation for the entropy change of a reacting system. It requires the determination of the entropy of each individual component of the reactants and the products, which in general is not very easy to do. The entropy calculations can be simplified somewhat if the gaseous components of the reactants and the products are approximated as ideal gases. However, entropy calculations are never as easy as enthalpy or internal energy calculations, since entropy is a function of both temperature and pressure even for ideal gases. When evaluating the entropy of a component of an ideal-gas mixture, we should use the temperature and the partial pressure of the component. Note that the temperature of a component is the same as the temperature of the mixture, and the partial pressure of a component is equal to the mixture pressure multiplied by the mole fraction of the component. Absolute entropy values at pressures other than P 0 1 atm for any temperature T can be obtained from the ideal-gas entropy change relation written for an imaginary isothermal process between states (T,P 0 ) and (T,P), as illustrated in Fig. 15 9: s 1T,P s 1T,P0 R u ln P P 0 For the component i of an ideal-gas mixture, this relation can be written as s i 1T,P i s i 1T,P 0 R u ln y ip m 1kJ>kmol # K P 0 (15 1) (15 ) where P 0 1 atm, P i is the partial pressure, y i is the mole fraction of the component, and P m is the total pressure of the mixture.

297 If a gas mixture is at a relatively high pressure or low temperature, the deviation from the ideal-gas behavior should be accounted for by incorporating more accurate equations of state or the generalized entropy charts SECOND-LAW ANALYSIS OF REACTING SYSTEMS Once the total entropy change or the entropy generation is evaluated, the exergy destroyed X destroyed associated with a chemical reaction can be determined from (15 3) where T 0 is the thermodynamic temperature of the surroundings. When analyzing reacting systems, we are more concerned with the changes in the exergy of reacting systems than with the values of exergy at various states (Fig ). Recall from Chap. 8 that the reversible work W rev represents the maximum work that can be done during a process. In the absence of any changes in kinetic and potential energies, the reversible work relation for a steady-flow combustion process that involves heat transfer with only the surroundings at T 0 can be obtained by replacing the enthalpy terms by h f h h, yielding (15 4) An interesting situation arises when both the reactants and the products are at the temperature of the surroundings T 0. In that case, h T 0 s (h T 0 s ) T0 g 0, which is, by definition, the Gibbs function of a unit mole of a substance at temperature T 0. The W rev relation in this case can be written as or X destroyed T 0 S gen 1kJ W rev a N r 1h f h h T 0 s r a N p 1h f h h T 0 s p W rev a N r g 0,r a N p g 0,p W rev a N r 1g f g T 0 g r a N p 1g f g T 0 g p (15 5) (15 6) where g f is the Gibbs function of formation (g f 0 for stable elements like N and O at the standard reference state of 5 C and 1 atm, just like the enthalpy of formation) and g T 0 g represents the value of the sensible Gibbs function of a substance at temperature T 0 relative to the standard reference state. For the very special case of T react T prod T 0 5 C (i.e., the reactants, the products, and the surroundings are at 5 C) and the partial pressure P i 1 atm for each component of the reactants and the products, Eq reduces to W rev a N r g f,r a n p g f,p 1kJ (15 7) We can conclude from the above equation that the g f value (the negative of the Gibbs function of formation at 5 C and 1 atm) of a compound represents the reversible work associated with the formation of that compound from its stable elements at 5 C and 1 atm in an environment at 5 C and 1 atm (Fig ). The g f values of several substances are listed in Table A 6. Exergy Chapter Reactants Reversible work Products T, P State FIGURE The difference between the exergy of the reactants and of the products during a chemical reaction is the reversible work associated with that reaction. T 0 = 5 C Stable elements 5 C, 1 atm C + O CO 5 C, 1 atm Compound 5 C, 1 atm W rev = g f, CO = 394,360 kj/ J/kmol FIGURE The negative of the Gibbs function of formation of a compound at 5 C, 1 atm represents the reversible work associated with the formation of that compound from its stable elements at 5 C, 1 atm in an environment that is at 5 C, 1 atm.

298 776 Thermodynamics C 77 F, 1 atm O 77 F, 1 atm T 0 = 77 F P 0 = 1 atm Combustion chamber FIGURE 15 3 Schematic for Example CO 77 F, 1 atm EXAMPLE 15 9 Reversible Work Associated with a Combustion Process One lbmol of carbon at 77 F and 1 atm is burned steadily with 1 lbmol of oxygen at the same state as shown in Fig The CO formed during the process is then brought to 77 F and 1 atm, the conditions of the surroundings. Assuming the combustion is complete, determine the reversible work for this process. Solution Carbon is burned steadily with pure oxygen. The reversible work associated with this process is to be determined. Assumptions 1 Combustion is complete. Steady-flow conditions exist during combustion. 3 Oxygen and the combustion gases are ideal gases. 4 Changes in kinetic and potential energies are negligible. Properties The Gibbs function of formation at 77 F and 1 atm is 0 for C and O, and 169,680 Btu/lbmol for CO. The enthalpy of formation is 0 for C and O, and 169,300 Btu/lbmol for CO. The absolute entropy is 1.36 Btu/lbmol R for C, Btu/lbmol R for O, and Btu/lbmol R for CO (Table A 6E). Analysis The combustion equation is The C, O, and CO are at 77 F and 1 atm, which is the standard reference state and also the state of the surroundings. Therefore, the reversible work in this case is simply the difference between the Gibbs function of formation of the reactants and that of the products (Eq. 15 7): since the g f of stable elements at 77 F and 1 atm is zero. Therefore, 169,680 Btu of work could be done as 1 lbmol of C is burned with 1 lbmol of O at 77 F and 1 atm in an environment at the same state. The reversible work in this case represents the exergy of the reactants since the product (the CO ) is at the state of the surroundings. Discussion We could also determine the reversible work without involving the Gibbs function by using Eq. 15 4: W rev a N r 1h f h h T 0 s r a N p 1h f h h T 0 s p a N r 1h f T 0 s r a N p 1h f T 0 s p N C 1h f T 0 s C N O 1h f T 0 s O N CO 1h f T 0 s CO Substituting the enthalpy of formation and absolute entropy values, we obtain W rev 11 lbmol C R11.36 Btu>lbmol # R4 11 lbmol O R Btu>lbmol # R4 11 lbmol CO 3 169,300 Btu>lbmol 1537 R Btu>lbmol # R4 169,680 Btu W rev a N r g f,r a N p g f,p N C g f,c 0 N O g f,o 0 N CO g f,co N CO g f,co 1 1 lbmol1 169,680 Btu>lbmol 169,680 Btu C O S CO which is identical to the result obtained before.

299 Chapter EXAMPLE Second-Law Analysis of Adiabatic Combustion Methane (CH 4 ) gas enters a steady-flow adiabatic combustion chamber at 5 C and 1 atm. It is burned with 50 percent excess air that also enters at 5 C and 1 atm, as shown in Fig Assuming complete combustion, determine (a) the temperature of the products, (b) the entropy generation, and (c) the reversible work and exergy destruction. Assume that T 0 98 K and the products leave the combustion chamber at 1 atm pressure. Solution Methane is burned with excess air in a steady-flow combustion chamber. The product temperature, entropy generated, reversible work, and exergy destroyed are to be determined. Assumptions 1 Steady-flow conditions exist during combustion. Air and the combustion gases are ideal gases. 3 Changes in kinetic and potential energies are negligible. 4 The combustion chamber is adiabatic and thus there is no heat transfer. 5 Combustion is complete. Analysis (a) The balanced equation for the complete combustion process with 50 percent excess air is CH 4 1g 3 1O 3.76N S CO H O O 11.8N Under steady-flow conditions, the adiabatic flame temperature is determined from H prod H react, which reduces to CH 4 5 C, 1 atm AIR 5 C, 1 atm T 0 = 5 C Adiabatic combustion chamber FIGURE Schematic for Example CO H O O N since all the reactants are at the standard reference state and h f O for O and N. Assuming ideal-gas behavior for air and for the products, the h f and h values of various components at 98 K can be listed as Substituting, we have which yields a N p 1h f h h p a N r h f,r 1Nh f CH4 h f h 98 K Substance kj/kmol kj/kmol CH 4 (g) 74,850 O N H O(g) 41, CO 393, kmol CO ,50 h CO 9364 kj>kmol CO 4 1 kmol H O31 41,80 h H O 9904 kj>kmol H O kmol N 310 h N 8669 kj>kmol N 4 11 kmol O 310 h O 868 kj>kmol O 4 11 kmol CH ,850 kj>kmol CH 4 h CO h H O h O 11.8h N 937,950 kj By trial and error, the temperature of the products is found to be T prod 1789 K

300 778 Thermodynamics (b) Noting that combustion is adiabatic, the entropy generation during this process is determined from Eq. 15 0: S gen S prod S react a N p s p a N r s r The CH 4 is at 5 C and 1 atm, and thus its absolute entropy is s CH kj/kmol K (Table A 6). The entropy values listed in the ideal-gas tables are for 1 atm pressure. Both the air and the product gases are at a total pressure of 1 atm, but the entropies are to be calculated at the partial pressure of the components, which is equal to P i y i P total, where y i is the mole fraction of component i. From Eq. 15 : S i N i s i 1T, P i N i 3 s i 1T, P 0 R u ln y i P m 4 The entropy calculations can be represented in tabular form as follows: CH 4 5 C, 1 atm AIR 5 C, 1 atm T 0 = 5 C Combustion chamber 5 C, 1 atm FIGURE Schematic for Example CO H O O N N i y i s i (T, 1 atm) R u ln y i P m N i s i CH O N S react CO H O O N S prod Thus, S gen S prod S react kj>kmol # K CH 4 (c) The exergy destruction or irreversibility associated with this process is determined from Eq. 15 3, That is, 88 MJ of work potential is wasted during this combustion process for each kmol of methane burned. This example shows that even complete combustion processes are highly irreversible. This process involves no actual work. Therefore, the reversible work and exergy destroyed are identical: That is, 88 MJ of work could be done during this process but is not. Instead, the entire work potential is wasted. EXAMPLE kj/kmol # K X destroyed T 0 S gen 198 K kj>kmol # K 88 MJ/kmol CH 4 W rev 88 MJ/kmol CH 4 Second-Law Analysis of Isothermal Combustion Methane (CH 4 ) gas enters a steady-flow combustion chamber at 5 C and 1 atm and is burned with 50 percent excess air, which also enters at 5 C and 1 atm, as shown in Fig After combustion, the products are allowed to cool to 5 C. Assuming complete combustion, determine

301 Chapter (a) the heat transfer per kmol of CH 4, (b) the entropy generation, and (c) the reversible work and exergy destruction. Assume that T 0 98 K and the products leave the combustion chamber at 1 atm pressure. Solution This is the same combustion process we discussed in Example 15 10, except that the combustion products are brought to the state of the surroundings by transferring heat from them. Thus the combustion equation remains the same: At 5 C, part of the water will condense. The amount of water vapor that remains in the products is determined from (see Example 15 3) and CH 4 1g 3 1O 3.76N S CO H O O 11.8N N v N gas P v P total Therefore, 1.57 kmol of the H O formed is in the liquid form, which is removed at 5 C and 1 atm. When one is evaluating the partial pressures of the components in the product gases, the only water molecules that need to be considered are those that are in the vapor phase. As before, all the gaseous reactants and products are treated as ideal gases. (a) Heat transfer during this steady-flow combustion process is determined from the steady-flow energy balance E out E in on the combustion chamber, since all the reactants and products are at the standard reference of 5 C and the enthalpy of ideal gases depends on temperature only. Solving for Q out and substituting the h f values, we have 871,400 kj/kmol CH kpa kpa N v a P v P total b N gas N v S N v 0.43 kmol Q out a N p h f,p a N r h f,r Q out 11 kmol CH ,850 kj>kmol CH 4 11 kmol CO 1 393,50 kj>kmol CO kmol H O 1g43 41,80 kj>kmol H O 1g kmol H O kj>kmol H O 1 4 (b) The entropy of the reactants was evaluated in Example and was determined to be S react kj/kmol K CH 4. By following a similar approach, the entropy of the products is determined to be N i y i s i (T, 1 atm) R u ln y i P m N i s i H O( ) H O CO O N S prod

302 780 Thermodynamics Then the total entropy generation during this process is determined from an entropy balance applied on an extended system that includes the immediate surroundings of the combustion chamber S gen S prod S react Q out T surr 871,400 kj>kmol kj>kmol 98 K 746 kj/kmol # K CH4 (c) The exergy destruction and reversible work associated with this process are determined from X destroyed T 0 S gen 198 K 1746 kj>kmol # K and 818 MJ/kmol CH 4 W rev X destroyed 818 MJ/kmol CH 4 since this process involves no actual work. Therefore, 818 MJ of work could be done during this process but is not. Instead, the entire work potential is wasted. The reversible work in this case represents the exergy of the reactants before the reaction starts since the products are in equilibrium with the surroundings, that is, they are at the dead state. Discussion Note that, for simplicity, we calculated the entropy of the product gases before they actually entered the atmosphere and mixed with the atmospheric gases. A more complete analysis would consider the composition of the atmosphere and the mixing of the product gases with the gases in the atmosphere, forming a homogeneous mixture. There is additional entropy generation during this mixing process, and thus additional wasted work potential. TOPIC OF SPECIAL INTEREST* Fuel Cells Fuels like methane are commonly burned to provide thermal energy at high temperatures for use in heat engines. However, a comparison of the reversible works obtained in the last two examples reveals that the exergy of the reactants (818 MJ/kmol CH 4 ) decreases by 88 MJ/kmol as a result of the irreversible adiabatic combustion process alone. That is, the exergy of the hot combustion gases at the end of the adiabatic combustion process is MJ/kmol CH 4. In other words, the work potential of the hot combustion gases is about 65 percent of the work potential of the reactants. It seems that when methane is burned, 35 percent of the work potential is lost before we even start using the thermal energy (Fig ). Thus, the second law of thermodynamics suggests that there should be a better way of converting the chemical energy to work. The better way is, of course, the less irreversible way, the best being the reversible case. In chemi- *This section can be skipped without a loss in continuity.

303 Chapter cal reactions, the irreversibility is due to uncontrolled electron exchange between the reacting components. The electron exchange can be controlled by replacing the combustion chamber by electrolytic cells, like car batteries. (This is analogous to replacing unrestrained expansion of a gas in mechanical systems by restrained expansion.) In the electrolytic cells, the electrons are exchanged through conductor wires connected to a load, and the chemical energy is directly converted to electric energy. The energy conversion devices that work on this principle are called fuel cells. Fuel cells are not heat engines, and thus their efficiencies are not limited by the Carnot efficiency. They convert chemical energy to electric energy essentially in an isothermal manner. A fuel cell functions like a battery, except that it produces its own electricity by combining a fuel with oxygen in a cell electrochemically without combustion, and discards the waste heat. A fuel cell consists of two electrodes separated by an electrolyte such as a solid oxide, phosphoric acid, or molten carbonate. The electric power generated by a single fuel cell is usually too small to be of any practical use. Therefore, fuel cells are usually stacked in practical applications. This modularity gives the fuel cells considerable flexibility in applications: The same design can be used to generate a small amount of power for a remote switching station or a large amount of power to supply electricity to an entire town. Therefore, fuel cells are termed the microchip of the energy industry. The operation of a hydrogen oxygen fuel cell is illustrated in Fig Hydrogen is ionized at the surface of the anode, and hydrogen ions flow through the electrolyte to the cathode. There is a potential difference between the anode and the cathode, and free electrons flow from the anode to the cathode through an external circuit (such as a motor or a generator). Hydrogen ions combine with oxygen and the free electrons at the surface of the cathode, forming water. Therefore, the fuel cell operates like an electrolysis system working in reverse. In steady operation, hydrogen and oxygen continuously enter the fuel cell as reactants, and water leaves as the product. Therefore, the exhaust of the fuel cell is drinkable quality water. The fuel cell was invented by William Groves in 1839, but it did not receive serious attention until the 1960s, when they were used to produce electricity and water for the Gemini and Apollo spacecraft during their missions to the moon. Today they are used for the same purpose in the space shuttle missions. Despite the irreversible effects such as internal resistance to electron flow, fuel cells have a great potential for much higher conversion efficiencies. Currently fuel cells are available commercially, but they are competitive only in some niche markets because of their higher cost. Fuel cells produce high-quality electric power efficiently and quietly while generating low emissions using a variety of fuels such as hydrogen, natural gas, propane, and biogas. Recently many fuel cells have been installed to generate electricity. For example, a remote police station in Central Park in New York City is powered by a 00-kW phosphoric acid fuel cell that has an efficiency of 40 percent with negligible emissions (it emits 1 ppm NO x and 5 ppm CO). 5 C REACTANTS (CH 4, air) Exergy = 818 MJ (100%) Adiabatic combustion chamber 1789 K PRODUCTS Exergy = 530 MJ (65%) FIGURE The availability of methane decreases by 35 percent as a result of irreversible combustion process. H Porous anode e Load e e H H + H O O Porous cathode FIGURE The operation of a hydrogen oxygen fuel cell. O

304 78 Thermodynamics Hybrid power systems (HPS) that combine high-temperature fuel cells and gas turbines have the potential for very high efficiency in converting natural gas (or even coal) to electricity. Also, some car manufacturers are planning to introduce cars powered by fuel-cell engines, thus more than doubling the efficiency from less than 30 percent for the gasoline engines to up to 60 percent for fuel cells. In 1999, DaimlerChrysler unveiled its hydrogen fuel-cell powered car called NECAR IV that has a refueling range of 80 miles and can carry 4 passengers at 90 mph. Some research programs to develop such hybrid systems with an efficiency of at least 70 percent by 010 are under way. SUMMARY Any material that can be burned to release energy is called a fuel, and a chemical reaction during which a fuel is oxidized and a large quantity of energy is released is called combustion. The oxidizer most often used in combustion processes is air. The dry air can be approximated as 1 percent oxygen and 79 percent nitrogen by mole numbers. Therefore, 1 kmol O 3.76 kmol N 4.76 kmol air During a combustion process, the components that exist before the reaction are called reactants and the components that exist after the reaction are called products. Chemical equations are balanced on the basis of the conservation of mass principle, which states that the total mass of each element is conserved during a chemical reaction. The ratio of the mass of air to the mass of fuel during a combustion process is called the air fuel ratio AF: AF m air m fuel where m air (NM) air and m fuel (N i M i ) fuel. A combustion process is complete if all the carbon in the fuel burns to CO, all the hydrogen burns to H O, and all the sulfur (if any) burns to SO. The minimum amount of air needed for the complete combustion of a fuel is called the stoichiometric or theoretical air. The theoretical air is also referred to as the chemically correct amount of air or 100 percent theoretical air. The ideal combustion process during which a fuel is burned completely with theoretical air is called the stoichiometric or theoretical combustion of that fuel. The air in excess of the stoichiometric amount is called the excess air. The amount of excess air is usually expressed in terms of the stoichiometric air as percent excess air or percent theoretical air. During a chemical reaction, some chemical bonds are broken and others are formed. Therefore, a process that involves chemical reactions involves changes in chemical energies. Because of the changed composition, it is necessary to have a standard reference state for all substances, which is chosen to be 5 C (77 F) and 1 atm. The difference between the enthalpy of the products at a specified state and the enthalpy of the reactants at the same state for a complete reaction is called the enthalpy of reaction h R. For combustion processes, the enthalpy of reaction is usually referred to as the enthalpy of combustion h C, which represents the amount of heat released during a steadyflow combustion process when 1 kmol (or 1 kg) of fuel is burned completely at a specified temperature and pressure. The enthalpy of a substance at a specified state due to its chemical composition is called the enthalpy of formation h f. The enthalpy of formation of all stable elements is assigned a value of zero at the standard reference state of 5 C and 1 atm. The heating value of a fuel is defined as the amount of heat released when a fuel is burned completely in a steadyflow process and the products are returned to the state of the reactants. The heating value of a fuel is equal to the absolute value of the enthalpy of combustion of the fuel, Heating value 0h C 0 1kJ>kg fuel Taking heat transfer to the system and work done by the system to be positive quantities, the conservation of energy relation for chemically reacting steady-flow systems can be expressed per unit mole of fuel as Q W a N p 1h f h h p a N r 1h f h h r where the superscript represents properties at the standard reference state of 5 C and 1 atm. For a closed system, it becomes Q W a N p 1h f h h Pv p a N r 1h f h h Pv r The Pv terms are negligible for solids and liquids and can be replaced by R u T for gases that behave as ideal gases.

305 Chapter In the absence of any heat loss to the surroundings (Q 0), the temperature of the products will reach a maximum, which is called the adiabatic flame temperature of the reaction. The adiabatic flame temperature of a steady-flow combustion process is determined from H prod H react or a N p 1h f h h p a N r 1h f h h r Taking the positive direction of heat transfer to be to the system, the entropy balance relation can be expressed for a closed system or steady-flow combustion chamber as a Q k T k S gen S prod S react For an adiabatic process it reduces to S gen,adiabatic S prod S react 0 The third law of thermodynamics states that the entropy of a pure crystalline substance at absolute zero temperature is zero. The third law provides a common base for the entropy of all substances, and the entropy values relative to this base are called the absolute entropy. The ideal-gas tables list the absolute entropy values over a wide range of temperatures but at a fixed pressure of P 0 1 atm. Absolute entropy values at other pressures P for any temperature T are determined from For component i of an ideal-gas mixture, this relation can be written as where P i is the partial pressure, y i is the mole fraction of the component, and P m is the total pressure of the mixture in atmospheres. The exergy destruction and the reversible work associated with a chemical reaction are determined from and s 1T, P s 1T, P0 R u ln P P 0 s i 1T, P i s i 1T, P 0 R u ln y i P m P 0 X destroyed W rev W act T 0 S gen W rev a N r 1h f h h T 0 s r a N p 1h f h h T 0 s p When both the reactants and the products are at the temperature of the surroundings T 0, the reversible work can be expressed in terms of the Gibbs functions as W rev a N r 1g f g T 0 g r a N p 1g f g T 0 g p REFERENCES AND SUGGESTED READINGS 1. S. W. Angrist. Direct Energy Conversion. 4th ed. Boston: Allyn and Bacon, W. Z. Black and J. G. Hartley. Thermodynamics. New York: Harper & Row, I. Glassman. Combustion. New York: Academic Press, R. Strehlow. Fundamentals of Combustion. Scranton, PA: International Textbook Co., K. Wark and D. E. Richards. Thermodynamics. 6th ed. New York: McGraw-Hill, PROBLEMS* Fuels and Combustion 15 1C What are the approximate chemical compositions of gasoline, diesel fuel, and natural gas? *Problems designated by a C are concept questions, and students are encouraged to answer them all. Problems designated by an E are in English units, and the SI users can ignore them. Problems with a CD-EES icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed DVD. Problems with a computer-ees icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text. 15 C How does the presence of N in air affect the outcome of a combustion process? 15 3C How does the presence of moisture in air affect the outcome of a combustion process? 15 4C What does the dew-point temperature of the product gases represent? How is it determined? 15 5C Is the number of atoms of each element conserved during a chemical reaction? How about the total number of moles?

306 784 Thermodynamics 15 6C What is the air fuel ratio? How is it related to the fuel air ratio? 15 7C Is the air fuel ratio expressed on a mole basis identical to the air fuel ratio expressed on a mass basis? Theoretical and Actual Combustion Processes 15 8C What are the causes of incomplete combustion? 15 9C Which is more likely to be found in the products of an incomplete combustion of a hydrocarbon fuel, CO or OH? Why? 15 10C What does 100 percent theoretical air represent? 15 11C Are complete combustion and theoretical combustion identical? If not, how do they differ? 15 1C Consider a fuel that is burned with (a) 130 percent theoretical air and (b) 70 percent excess air. In which case is the fuel burned with more air? Methane (CH 4 ) is burned with stoichiometric amount of air during a combustion process. Assuming complete combustion, determine the air fuel and fuel air ratios Propane (C 3 H 8 ) is burned with 75 percent excess air during a combustion process. Assuming complete combustion, determine the air fuel ratio. Answer: 7.5 kg air/kg fuel Acetylene (C H ) is burned with stoichiometric amount of air during a combustion process. Assuming complete combustion, determine the air fuel ratio on a mass and on a mole basis One kmol of ethane (C H 6 ) is burned with an unknown amount of air during a combustion process. An analysis of the combustion products reveals that the combustion is complete, and there are 3 kmol of free O in the products. Determine (a) the air fuel ratio and (b) the percentage of theoretical air used during this process E Ethylene (C H 4 ) is burned with 00 percent theoretical air during a combustion process. Assuming complete combustion and a total pressure of 14.5 psia, determine (a) the air fuel ratio and (b) the dew-point temperature of the products. Answers: (a) 9.6 lbm air/lbm fuel, (b) 101 F Propylene (C 3 H 6 ) is burned with 50 percent excess air during a combustion process. Assuming complete combustion and a total pressure of 105 kpa, determine (a) the air fuel ratio and (b) the temperature at which the water vapor in the products will start condensing Propal alcohol (C 3 H 7 OH) is burned with 50 percent excess air. Write the balanced reaction equation for complete combustion and determine the air-to-fuel ratio. Answer: 15.5 kg air/kg fuel 15 0 Butane (C 4 H 10 ) is burned in 00 percent theoretical air. For complete combustion, how many kmol of water must be sprayed into the combustion chamber per kmol of fuel if the products of combustion are to have a dew-point temperature of 60 C when the product pressure is 100 kpa? 15 1 A fuel mixture of 0 percent by mass methane (CH 4 ) and 80 percent by mass ethanol (C H 6 O), is burned completely with theoretical air. If the total flow rate of the fuel is 31 kg/s, determine the required flow rate of air. Answer: 330 kg/s 15 Octane (C 8 H 18 ) is burned with 50 percent theoretical air, which enters the combustion chamber at 5 C. Assuming complete combustion and a total pressure of 1 atm, determine (a) the air fuel ratio and (b) the dew-point temperature of the products. C 8 H 18 AIR 5 C Combustion chamber P = 1 atm FIGURE P15 Products 15 3 Gasoline (assumed C 8 H 18 ) is burned steadily with air in a jet engine. If the air fuel ratio is 18 kg air/kg fuel, determine the percentage of theoretical air used during this process In a combustion chamber, ethane (C H 6 ) is burned at a rate of 8 kg/h with air that enters the combustion chamber at a rate of 176 kg/h. Determine the percentage of excess air used during this process. Answer: 37 percent 15 5 One kilogram of butane (C 4 H 10 ) is burned with 5 kg of air that is at 30 C and 90 kpa. Assuming that the combustion is complete and the pressure of the products is 90 kpa, determine (a) the percentage of theoretical air used and (b) the dew-point temperature of the products. 15 6E One lbm of butane (C 4 H 10 ) is burned with 5 lbm of air that is at 90 F and 14.7 psia. Assuming that the combustion is complete and the pressure of the products is 14.7 psia, determine (a) the percentage of theoretical air used and (b) the dew-point temperature of the products. Answers: (a) 161 percent, (b) 113 F 15 7 A certain natural gas has the following volumetric analysis: 65 percent CH 4, 8 percent H, 18 percent N, 3 percent O, and 6 percent CO. This gas is now burned completely with the stoichiometric amount of dry air. What is the air fuel ratio for this combustion process? 15 8 Repeat Prob by replacing the dry air by moist air that enters the combustion chamber at 5 C, 1 atm, and 85 percent relative humidity A gaseous fuel with a volumetric analysis of 60 percent CH 4, 30 percent H, and 10 percent N is burned to completion with 130 percent theoretical air. Determine (a) the

307 air fuel ratio and (b) the fraction of water vapor that would condense if the product gases were cooled to 0 C at 1 atm. Answers: (a) 18.6 kg air/kg fuel, (b) 88 percent Reconsider Prob Using EES (or other) software, study the effects of varying the percentages of CH 4,H, and N making up the fuel and the product gas temperature in the range 5 to 150 C A certain coal has the following analysis on a mass basis: 8 percent C, 5 percent H O, percent H, 1 percent O, and 10 percent ash. The coal is burned with 50 percent excess air. Determine the air fuel ratio. Answer: 15.1 kg air/kg coal 15 3 Octane (C 8 H 18 ) is burned with dry air. The volumetric analysis of the products on a dry basis is 9.1 percent CO, 0.61 percent CO, 7.06 percent O, and 83.1 percent N. Determine (a) the air fuel ratio and (b) the percentage of theoretical air used Carbon (C) is burned with dry air. The volumetric analysis of the products is percent CO, 0.4 percent CO, percent O, and percent N. Determine (a) the air fuel ratio and (b) the percentage of theoretical air used Methane (CH 4 ) is burned with dry air. The volumetric analysis of the products on a dry basis is 5.0 percent CO, 0.33 percent CO, 11.4 percent O, and 83.3 percent N. Determine (a) the air fuel ratio and (b) the percentage of theoretical air used. Answers: (a) 34.5 kg air/kg fuel, (b) 00 percent Enthalpy of Formation and Enthalpy of Combustion 15 35C What is enthalpy of combustion? How does it differ from the enthalpy of reaction? 15 36C What is enthalpy of formation? How does it differ from the enthalpy of combustion? 15 37C What are the higher and the lower heating values of a fuel? How do they differ? How is the heating value of a fuel related to the enthalpy of combustion of that fuel? 15 38C When are the enthalpy of formation and the enthalpy of combustion identical? 15 39C Does the enthalpy of formation of a substance change with temperature? 15 40C The h f of N is listed as zero. Does this mean that N contains no chemical energy at the standard reference state? 15 41C Which contains more chemical energy, 1 kmol of H or 1 kmol of H O? 15 4 Determine the enthalpy of combustion of methane (CH 4 ) at 5 C and 1 atm, using the enthalpy of formation data from Table A 6. Assume that the water in the products is in the liquid form. Compare your result to the value listed in Table A 7. Answer: 890,330 kj/kmol Chapter Reconsider Prob Using EES (or other) software, study the effect of temperature on the enthalpy of combustion. Plot the enthalpy of combustion as a function of temperature over the range 5 to 600 C Repeat Prob for gaseous ethane (C H 6 ) Repeat Prob for liquid octane (C 8 H 18 ). First-Law Analysis of Reacting Systems 15 46C Derive an energy balance relation for a reacting closed system undergoing a quasi-equilibrium constant pressure expansion or compression process C Consider a complete combustion process during which both the reactants and the products are maintained at the same state. Combustion is achieved with (a) 100 percent theoretical air, (b) 00 percent theoretical air, and (c) the chemically correct amount of pure oxygen. For which case will the amount of heat transfer be the highest? Explain C Consider a complete combustion process during which the reactants enter the combustion chamber at 0 C and the products leave at 700 C. Combustion is achieved with (a) 100 percent theoretical air, (b) 00 percent theoretical air, and (c) the chemically correct amount of pure oxygen. For which case will the amount of heat transfer be the lowest? Explain Methane (CH 4 ) is burned completely with the stoichiometric amount of air during a steady-flow combustion process. If both the reactants and the products are maintained at 5 C and 1 atm and the water in the products exists in the liquid form, determine the heat transfer from the combustion chamber during this process. What would your answer be if combustion were achieved with 100 percent excess air? Answer: 890,330 kj/kmol Hydrogen (H ) is burned completely with the stoichiometric amount of air during a steady-flow combustion process. If both the reactants and the products are maintained at 5 C and 1 atm and the water in the products exists in the liquid form, determine the heat transfer from the combustion chamber during this process. What would your answer be if combustion were achieved with 50 percent excess air? Liquid propane (C 3 H 8 ) enters a combustion chamber at 5 C at a rate of 1. kg/min where it is mixed and burned with 150 percent excess air that enters the combustion chamber at 1 C. If the combustion is complete and the exit C 3 H 8 5 C AIR 1 C Q out Combustion chamber FIGURE P15 51 Products 100 K

308 786 Thermodynamics temperature of the combustion gases is 100 K, determine (a) the mass flow rate of air and (b) the rate of heat transfer from the combustion chamber. Answers: (a) 47.1 kg/min, (b) 5194 kj/min 15 5E Liquid propane (C 3 H 8 ) enters a combustion chamber at 77 F at a rate of 0.75 lbm/min where it is mixed and burned with 150 percent excess air that enters the combustion chamber at 40 F. if the combustion is complete and the exit temperature of the combustion gases is 1800 R, determine (a) the mass flow rate of air and (b) the rate of heat transfer from the combustion chamber. Answers: (a) 9.4 lbm/min, (b) 4479 Btu/min Acetylene gas (C H ) is burned completely with 0 percent excess air during a steady-flow combustion process. The fuel and air enter the combustion chamber at 5 C, and the products leave at 1500 K. Determine (a) the air fuel ratio and (b) the heat transfer for this process E Liquid octane (C 8 H 18 ) at 77 F is burned completely during a steady-flow combustion process with 180 percent theoretical air that enters the combustion chamber at 77 F. If the products leave at 500 R, determine (a) the air fuel ratio and (b) the heat transfer from the combustion chamber during this process Benzene gas (C 6 H 6 ) at 5 C is burned during a steady-flow combustion process with 95 percent theoretical air that enters the combustion chamber at 5 C. All the hydrogen in the fuel burns to H O, but part of the carbon burns to CO. If the products leave at 1000 K, determine (a) the mole fraction of the CO in the products and (b) the heat transfer from the combustion chamber during this process. Answers: (a).1 percent, (b),11,800 kj/kmol C 6 H Diesel fuel (C 1 H 6 ) at 5 C is burned in a steadyflow combustion chamber with 0 percent excess air that also enters at 5 C. The products leave the combustion chamber at 500 K. Assuming combustion is complete, determine the required mass flow rate of the diesel fuel to supply heat at a rate of 000 kj/s. Answer: 49.5 g/s 15 57E Diesel fuel (C 1 H 6 ) at 77 F is burned in a steadyflow combustion chamber with 0 percent excess air that also enters at 77 F. The products leave the combustion chamber at 800 R. Assuming combustion is complete, determine the required mass flow rate of the diesel fuel to supply heat at a rate of 1800 Btu/s. Answer: 0.1 lbm/s Octane gas (C 8 H 18 ) at 5 C is burned steadily with 30 percent excess air at 5 C, 1 atm, and 60 percent relative humidity. Assuming combustion is complete and the products leave the combustion chamber at 600 K, determine the heat transfer for this process per unit mass of octane Reconsider Prob Using EES (or other) software, investigate the effect of the amount of excess air on the heat transfer for the combustion process. Let the excess air vary from 0 to 00 percent. Plot the heat transfer against excess air, and discuss the results Ethane gas (C H 6 ) at 5 C is burned in a steady-flow combustion chamber at a rate of 5 kg/h with the stoichiometric amount of air, which is preheated to 500 K before entering the combustion chamber. An analysis of the combustion gases reveals that all the hydrogen in the fuel burns to H O but only 95 percent of the carbon burns to CO, the remaining 5 percent forming CO. If the products leave the combustion chamber at 800 K, determine the rate of heat transfer from the combustion chamber. Answer: 00,170 kj/h C H 6 5 C AIR 500 K Q out Combustion chamber FIGURE P K H O CO CO O N A constant-volume tank contains a mixture of 10 g of methane (CH 4 ) gas and 600 g of O at 5 C and 00 kpa. The contents of the tank are now ignited, and the methane gas burns completely. If the final temperature is 100 K, determine (a) the final pressure in the tank and (b) the heat transfer during this process Reconsider Prob Using EES (or other) software, investigate the effect of the final temperature on the final pressure and the heat transfer for the combustion process. Let the final temperature vary from 500 to 1500 K. Plot the final pressure and heat transfer against the final temperature, and discuss the results A closed combustion chamber is designed so that it maintains a constant pressure of 300 kpa during a combustion process. The combustion chamber has an initial volume of 0.5 m 3 and contains a stoichiometric mixture of octane (C 8 H 18 ) gas and air at 5 C. The mixture is now ignited, and the product gases are observed to be at 1000 K at the end of the combustion process. Assuming complete combustion, and treating both the reactants and the products as ideal gases, determine the heat transfer from the combustion chamber during this process. Answer: 3610 kj A constant-volume tank contains a mixture of 1 kmol of benzene (C 6 H 6 ) gas and 30 percent excess air at 5 C and 1 atm. The contents of the tank are now ignited, and all the hydrogen in the fuel burns to H O but only 9 percent of the carbon burns to CO, the remaining 8 percent forming CO. If the final temperature in the tank is 1000 K, determine the heat transfer from the combustion chamber during this process.

309 Chapter E A constant-volume tank contains a mixture of 1 lbmol of benzene (C 6 H 6 ) gas and 30 percent excess air at 77 F and 1 atm. The contents of the tank are now ignited, and all the hydrogen in the fuel burns to H O but only 9 percent of the carbon burns to CO, the remaining 8 percent forming CO. If the final temperature in the tank is 1800 R, determine the heat transfer from the combustion chamber during this process. Answer: 946,870 Btu To supply heated air to a house, a high-efficiency gas furnace burns gaseous propane (C 3 H 8 ) with a combustion efficiency of 96 percent. Both the fuel and 140 percent theoretical air are supplied to the combustion chamber at 5 C and 100 kpa, and the combustion is complete. Because this is a high-efficiency furnace, the product gases are cooled to 5 C and 100 kpa before leaving the furnace. To maintain the house at the desired temperature, a heat transfer rate of 31,650 kj/h is required from the furnace. Determine the volume of water condensed from the product gases per day. Answer: 8.7 L/day Liquid ethyl alcohol (C H 5 OH( )) at 5 C is burned in a steady-flow combustion chamber with 40 percent excess air that also enters at 5 C. The products leave the combustion chamber at 600 K. Assuming combustion is complete, determine the required volume flow rate of the liquid ethyl alcohol, to supply heat at a rate of 000 kj/s. At 5 C the density of liquid ethyl alcohol is 790 kg/m 3, the specific heat at a constant pressure is kj/kmol K, and the enthalpy of vaporization is 4,340 kj/kmol. Answer: 6.81 L/min Adiabatic Flame Temperature Q out C 6 H 6 30% excess air 5 C 1 atm FIGURE P C A fuel is completely burned first with the stoichiometric amount of air and then with the stoichiometric amount of pure oxygen. For which case will the adiabatic flame temperature be higher? 15 69C A fuel at 5 C is burned in a well-insulated steady-flow combustion chamber with air that is also at 5 C. Under what conditions will the adiabatic flame temperature of the combustion process be a maximum? Hydrogen (H ) at 7 C is burned with 0 percent excess air that is also at 7 C during an adiabatic steady-flow combustion process. Assuming complete combustion, determine the exit temperature of the product gases. Answer: 51.4 K H 7 C Combustion Products AIR chamber T prod 7 C FIGURE P Reconsider Prob Using EES (or other) software, modify this problem to include the fuels butane, ethane, methane, and propane as well as H ; to include the effects of inlet air and fuel temperatures; and the percent theoretical air supplied. Select a range of input parameters and discuss the results for your choices. 15 7E Hydrogen (H ) at 40 F is burned with 0 percent excess air that is also at 40 F during an adiabatic steady-flow combustion process. Assuming complete combustion, find the exit temperature of the product gases Acetylene gas (C H ) at 5 C is burned during a steady-flow combustion process with 30 percent excess air at 7 C. It is observed that 75,000 kj of heat is being lost from the combustion chamber to the surroundings per kmol of acetylene. Assuming combustion is complete, determine the exit temperature of the product gases. Answer: 301 K An adiabatic constant-volume tank contains a mixture of 1 kmol of hydrogen (H ) gas and the stoichiometric amount of air at 5 C and 1 atm. The contents of the tank are now ignited. Assuming complete combustion, determine the final temperature in the tank Octane gas (C 8 H 18 ) at 5 C is burned steadily with 30 percent excess air at 5 C, 1 atm, and 60 percent relative humidity. Assuming combustion is complete and adiabatic, calculate the exit temperature of the product gases Reconsider Prob Using EES (or other) software, investigate the effect of the relative humidity on the exit temperature of the product gases. Plot the exit temperature of the product gases as a function of relative humidity for 0 f 100 percent. Entropy Change and Second-Law Analysis of Reacting Systems 15 77C Express the increase of entropy principle for chemically reacting systems C How are the absolute entropy values of ideal gases at pressures different from 1 atm determined? 15 79C What does the Gibbs function of formation g f of a compound represent? One kmol of H at 5 C and 1 atm is burned steadily with 0.5 kmol of O at the same state. The H O formed during the process is then brought to 5 C and 1 atm, the conditions

310 788 Thermodynamics of the surroundings. Assuming combustion is complete, determine the reversible work and exergy destruction for this process Ethylene (C H 4 ) gas enters an adiabatic combustion chamber at 5 C and 1 atm and is burned with 0 percent excess air that enters at 5 C and 1 atm. The combustion is complete, and the products leave the combustion chamber at 1 atm pressure. Assuming T 0 5 C, determine (a) the temperature of the products, (b) the entropy generation, and (c) the exergy destruction. Answers: (a) 69.6 K, (b) kj/kmol K, (c) 390,760 kj/kmol 15 8 Liquid octane (C 8 H 18 ) enters a steady-flow combustion chamber at 5 C and 1 atm at a rate of 0.5 kg/min. It is burned with 50 percent excess air that also enters at 5 C and 1 atm. After combustion, the products are allowed to cool to 5 C. Assuming complete combustion and that all the H O in the products is in liquid form, determine (a) the heat transfer rate from the combustion chamber, (b) the entropy generation rate, and (c) the exergy destruction rate. Assume that T 0 98 K and the products leave the combustion chamber at 1 atm pressure. C 8 H 18 ( ) T 0 = 98 K Q out 5 C Combustion Products AIR chamber 1 atm 5 C 5 C FIGURE P Acetylene gas (C H ) is burned completely with 0 percent excess air during a steady-flow combustion process. The fuel and the air enter the combustion chamber separately at 5 C and 1 atm, and heat is being lost from the combustion chamber to the surroundings at 5 C at a rate of 300,000 kj/kmol C H. The combustion products leave the combustion chamber at 1 atm pressure. Determine (a) the temperature of the products, (b) the total entropy change per kmol of C H, and (c) the exergy destruction during this process A steady-flow combustion chamber is supplied with CO gas at 37 C and 110 kpa at a rate of 0.4 m 3 /min and air at 5 C and 110 kpa at a rate of 1.5 kg/min. Heat is transferred to a medium at 800 K, and the combustion products leave the combustion chamber at 900 K. Assuming the combustion is complete and T 0 5 C, determine (a) the rate of heat transfer from the combustion chamber and (b) the rate of exergy destruction. Answers: (a) 3567 kj/min, (b) 1610 kj/min 15 85E Benzene gas (C 6 H 6 ) at 1 atm and 77 F is burned during a steady-flow combustion process with 95 percent theoretical air that enters the combustion chamber at 77 F and 1 atm. All the hydrogen in the fuel burns to H O, but part of the carbon burns to CO. Heat is lost to the surroundings at 77 F, and the products leave the combustion chamber at 1 atm and 1500 R. Determine (a) the heat transfer from the combustion chamber and (b) the exergy destruction Liquid propane (C 3 H 8 ) enters a steady-flow combustion chamber at 5 C and 1 atm at a rate of 0.4 kg/min where it is mixed and burned with 150 percent excess air that enters the combustion chamber at 1 C. If the combustion products leave at 100 K and 1 atm, determine (a) the mass flow rate of air, (b) the rate of heat transfer from the combustion chamber, and (c) the rate of entropy generation during this process. Assume T 0 5 C. Answers: (a) 15.7 kg/min, (b) 173 kj/min, (c) 34. kj/min K Reconsider Prob Using EES (or other) software, study the effect of varying the surroundings temperature from 0 to 38 C on the rate of exergy destruction, and plot it as a function of surroundings temperature. Review Problems A 1-g sample of a certain fuel is burned in a bomb calorimeter that contains kg of water in the presence of 100 g of air in the reaction chamber. If the water temperature rises by.5 C when equilibrium is established, determine the heating value of the fuel, in kj/kg E Hydrogen (H ) is burned with 100 percent excess air that enters the combustion chamber at 90 F, 14.5 psia, and 60 percent relative humidity. Assuming complete combustion, determine (a) the air fuel ratio and (b) the volume flow rate of air required to burn the hydrogen at a rate of 5 lbm/h A gaseous fuel with 80 percent CH 4, 15 percent N, and 5 percent O (on a mole basis) is burned to completion with 10 percent theoretical air that enters the combustion chamber at 30 C, 100 kpa, and 60 percent relative humidity. Determine (a) the air fuel ratio and (b) the volume flow rate of air required to burn fuel at a rate of kg/min A gaseous fuel with 80 percent CH 4, 15 percent N, and 5 percent O (on a mole basis) is burned with dry air that enters the combustion chamber at 5 C and 100 kpa. The volumetric analysis of the products on a dry basis is 3.36 percent CO, 0.09 percent CO, percent O, and percent N. Determine (a) the air fuel ratio, (b) the percent theoretical 15% N 80% CH % CO 5% O Combustion chamber 0.09% CO 14.91% O AIR 81.64% N FIGURE P15 91

311 air used, and (c) the volume flow rate of air used to burn fuel at a rate of 1.4 kg/min A steady-flow combustion chamber is supplied with CO gas at 37 C and 110 kpa at a rate of 0.4 m 3 /min and air at 5 C and 110 kpa at a rate of 1.5 kg/min. The combustion products leave the combustion chamber at 900 K. Assuming combustion is complete, determine the rate of heat transfer from the combustion chamber Methane gas (CH 4 ) at 5 C is burned steadily with dry air that enters the combustion chamber at 17 C. The volumetric analysis of the products on a dry basis is 5.0 percent CO, 0.33 percent CO, 11.4 percent O, and 83.3 percent N. Determine (a) the percentage of theoretical air used and (b) the heat transfer from the combustion chamber per kmol of CH 4 if the combustion products leave at 700 K A 6-m 3 rigid tank initially contains a mixture of 1 kmol of hydrogen (H ) gas and the stoichiometric amount of air at 5 C. The contents of the tank are ignited, and all the hydrogen in the fuel burns to H O. If the combustion products are cooled to 5 C, determine (a) the fraction of the H O that condenses and (b) the heat transfer from the combustion chamber during this process Propane gas (C 3 H 8 ) enters a steady-flow combustion chamber at 1 atm and 5 C and is burned with air that enters the combustion chamber at the same state. Determine the adiabatic flame temperature for (a) complete combustion with 100 percent theoretical air, (b) complete combustion with 300 percent theoretical air, and (c) incomplete combustion (some CO in the products) with 95 percent theoretical air Determine the highest possible temperature that can be obtained when liquid gasoline (assumed C 8 H 18 ) at 5 C is burned steadily with air at 5 C and 1 atm. What would your answer be if pure oxygen at 5 C were used to burn the fuel instead of air? 15 97E Determine the work potential of 1 lbmol of diesel fuel (C 1 H 6 ) at 77 F and 1 atm in an environment at the same state. Answer: 3,375,000 Btu Liquid octane (C 8 H 18 ) enters a steady-flow combustion chamber at 5 C and 8 atm at a rate of 0.8 kg/min. It is burned with 00 percent excess air that is compressed and preheated to 500 K and 8 atm before entering the combustion chamber. After combustion, the products enter an adiabatic turbine at 1300 K and 8 atm and leave at 950 K and atm. Assuming complete combustion and T 0 5 C, determine (a) the heat transfer rate from the combustion chamber, (b) the power output of the turbine, and (c) the reversible work and exergy destruction for the entire process. Answers: (a) 770 kj/min, (b) 63 kw, (c) 514 kw, 51 kw The combustion of a fuel usually results in an increase in pressure when the volume is held constant, or an increase in volume when the pressure is held constant, Chapter because of the increase in the number of moles and the temperature. The increase in pressure or volume will be maximum when the combustion is complete and when it occurs adiabatically with the theoretical amount of air. Consider the combustion of methyl alcohol vapor (CH 3 OH(g)) with the stoichiometric amount of air in an 0.8-L combustion chamber. Initially, the mixture is at 5 C and 98 kpa. Determine (a) the maximum pressure that can occur in the combustion chamber if the combustion takes place at constant volume and (b) the maximum volume of the combustion chamber if the combustion occurs at constant pressure Reconsider Prob Using EES (or other) software, investigate the effect of the initial volume of the combustion chamber over the range 0.1 to.0 liters on the results. Plot the maximum pressure of the chamber for constant volume combustion or the maximum volume of the chamber for constant pressure combustion as functions of the initial volume Repeat Prob using methane (CH 4 (g)) as the fuel instead of methyl alcohol A mixture of 40 percent by volume methane (CH 4 ), and 60 percent by volume propane (C 3 H 8 ), is burned completely with theoretical air and leaves the combustion chamber at 100 C. The products have a pressure of 100 kpa and are cooled at constant pressure to 39 C. Sketch the T-s diagram for the water vapor that does not condense, if any. How much of the water formed during the combustion process will be condensed, in kmol H O/kmol fuel? Answer: Liquid propane (C 3 H 8 ( )) enters a combustion chamber at 5 C and 1 atm at a rate of 0.4 kg/min where it is mixed and burned with 150 percent excess air that enters the combustion chamber at 5 C. The heat transfer from the combustion process is 53 kw. Write the balanced combustion equation and determine (a) the mass flow rate of air; (b) the average molar mass (molecular weight) of the product gases; (c) the average specific heat at constant pressure of the product gases; and (d ) the temperature of the products of combustion. Answers: (a) kg/min, (b) 8.63 kg/kmol, (c) kj/kmol K, (d ) 18 K A gaseous fuel mixture of 30 percent propane (C 3 H 8 ), and 70 percent butane (C 4 H 10 ), on a volume basis is burned in air such that the air fuel ratio is 0 kg air/kg fuel when the combustion process is complete. Determine (a) the moles of nitrogen in the air supplied to the combustion process, in kmol/kmol fuel; (b) the moles of water formed in the combustion process, in kmol/kmol fuel; and (c) the moles of oxygen in the product gases. Answers: (a) 9.41, (b) 4.7, (c) A liquid gas fuel mixture consists of 90 percent octane (C 8 H 18 ), and 10 percent alcohol (C H 5 OH), by moles. This fuel is burned with 00 percent theoretical dry air. Write the balanced reaction equation for complete combustion of

312 790 Thermodynamics this fuel mixture. Determine (a) the theoretical air fuel ratio for this reaction; (b) the product fuel ratio for this reaction; (c) the air-flow rate for a fuel mixture flow rate of 5 kg/s; and (d) the lower heating value of the fuel mixture with 00 percent theoretical air at 5 C. Answers: (a) kg air/kg fuel, (b) kg product/kg fuel, (c) kg/s, (d) 43,67 kj/kg fuel The furnace of a particular power plant can be considered to consist of two chambers: an adiabatic combustion chamber where the fuel is burned completely and adiabatically, and a heat exchanger where heat is transferred to a Carnot heat engine isothermally. The combustion gases in the heat exchanger are well-mixed so that the heat exchanger is at a uniform temperature at all times that is equal to the temperature of the exiting product gases, T p. The work output of the Carnot heat engine can be expressed as W Qh C Q a 1 T 0 T p b where Q is the magnitude of the heat transfer to the heat engine and T 0 is the temperature of the environment. The work output of the Carnot engine will be zero either when T p T af (which means the product gases will enter and exit the heat exchanger at the adiabatic flame temperature T af, and thus Q 0) or when T p T 0 (which means the temperature T 0 Fuel Air Adiabatic combustion chamber of the product gases in the heat exchanger will be T 0, and thus h C 0), and will reach a maximum somewhere in between. Treating the combustion products as ideal gases with constant specific heats and assuming no change in their composition in the heat exchanger, show that the work output of the Carnot heat engine will be maximum when T p T af T 0 Also, show that the maximum work output of the Carnot engine in this case becomes T 0 W max CT af a 1 b B T af where C is a constant whose value depends on the composition of the product gases and their specific heats The furnace of a particular power plant can be considered to consist of two chambers: an adiabatic combustion chamber where the fuel is burned completely and adiabatically and a counterflow heat exchanger where heat is transferred to a reversible heat engine. The mass flow rate of the working fluid of the heat engine is such that the working fluid is heated from T 0 (the temperature of the environment) to T af (the adiabatic flame temperature) while the combustion products are cooled from T af to T 0. Treating the combustion products as ideal gases with constant specific heats and assuming no change in their composition in the heat exchanger, show that the work output of this reversible heat engine is W CT 0 a T af T 0 1 ln T af T 0 b Heat exchanger T p = const. Q T p Fuel T 0 Air Adiabatic combustion chamber Heat exchanger T 0 T af T 0 W W Surroundings T 0 Surroundings T 0 FIGURE P FIGURE P15 107

313 where C is a constant whose value depends on the composition of the product gases and their specific heats. Also, show that the effective flame temperature T e of this furnace is T e T af T 0 ln 1T af >T 0 That is, the work output of the reversible engine would be the same if the furnace above is considered to be an isothermal furnace at a constant temperature T e Using EES (or other) software, determine the effect of the amount of air on the adiabatic flame temperature of liquid octane (C 8 H 18 ). Assume both the air and the octane are initially at 5 C. Determine the adiabatic flame temperature for 75, 90, 100, 10, 150, 00, 300, 500, and 800 percent theoretical air. Assume the hydrogen in the fuel always burns H O and the carbon CO, except when there is a deficiency of air. In the latter case, assume that part of the carbon forms CO. Plot the adiabatic flame temperature against the percent theoretical air, and discuss the results Using EES (or other) software, write a general program to determine the heat transfer during the complete combustion of a hydrocarbon fuel (C n H m ) at 5 C in a steady-flow combustion chamber when the percent of excess air and the temperatures of air and the products are specified. As a sample case, determine the heat transfer per unit mass of fuel as liquid propane (C 3 H 8 ) is burned steadily with 50 percent excess air at 5 C and the combustion products leave the combustion chamber at 1800 K Using EES (or other) software, write a general program to determine the adiabatic flame temperature during the complete combustion of a hydrocarbon fuel (C n H m ) at 5 C in a steady-flow combustion chamber when the percent of excess air and its temperature are specified. As a sample case, determine the adiabatic flame temperature of liquid propane (C 3 H 8 ) as it is burned steadily with 50 percent excess air at 5 C Using EES (or other) software, determine the adiabatic flame temperature of the fuels CH 4 (g), C H (g), CH 3 OH(g), C 3 H 8 (g), C 8 H 18 ( ). Assume both the fuel and the air enter the steady-flow combustion chamber at 5 C Using EES (or other) software, determine the minimum percent of excess air that needs to be used for the fuels CH 4 (g), C H (g), CH 3 OH(g), C 3 H 8 (g), C 8 H 18 ( ) if the adiabatic flame temperature is not to exceed 1500 K. Assume both the fuel and the air enter the steadyflow combustion chamber at 5 C Using EES (or other) software, repeat Prob for adiabatic flame temperatures of (a) 100 K, (b) 1750 K, and (c) 000 K. Chapter Using EES (or other) software, determine the adiabatic flame temperature of CH 4 (g) when both the fuel and the air enter the combustion chamber at 5 C for the cases of 0, 0, 40, 60, 80, 100, 00, 500, and 1000 percent excess air Using EES (or other) software, determine the rate of heat transfer for the fuels CH 4 (g), C H (g), CH 3 OH(g), C 3 H 8 (g), and C 8 H 18 ( ) when they are burned completely in a steady-flow combustion chamber with the theoretical amount of air. Assume the reactants enter the combustion chamber at 98 K and the products leave at 100 K Using EES (or other) software, repeat Prob for (a) 50, (b) 100, and (c) 00 percent excess air Using EES (or other) software, determine the fuel among CH 4 (g), C H (g), C H 6 (g), C 3 H 8 (g), C 8 H 18 ( ) that gives the highest temperature when burned completely in an adiabatic constant-volume chamber with the theoretical amount of air. Assume the reactants are at the standard reference state. Fundamentals of Engineering (FE) Exam Problems A fuel is burned with 90 percent theoretical air. This is equivalent to (a) 10% excess air (b) 90% excess air (c) 10% deficiency of air (d) 90% deficiency of air (e) stoichiometric amount of air Propane (C 3 H 8 ) is burned with 150 percent theoretical air. The air fuel mass ratio for this combustion process is (a) 5.3 (b) 10.5 (c) 15.7 (d) 3.4 (e) One kmol of methane (CH 4 ) is burned with an unknown amount of air during a combustion process. If the combustion is complete and there are kmol of free O in the products, the air fuel mass ratio is (a) 34.3 (b) 17. (c) 19.0 (d) 14.9 (e) A fuel is burned steadily in a combustion chamber. The combustion temperature will be the highest except when (a) the fuel is preheated. (b) the fuel is burned with a deficiency of air. (c) the air is dry. (d) the combustion chamber is well insulated. (e) the combustion is complete An equimolar mixture of carbon dioxide and water vapor at 1 atm and 60 C enter a dehumidifying section where the entire water vapor is condensed and removed from the mixture, and the carbon dioxide leaves at 1 atm and 60 C.

314 79 Thermodynamics The entropy change of carbon dioxide in the dehumidifying section is (a).8 kj/kg K (b) 0.13 kj/kg K (c) 0 (d) 0.13 kj/kg K (e).8 kj/kg K Methane (CH 4 ) is burned completely with 80 percent excess air during a steady-flow combustion process. If both the reactants and the products are maintained at 5 C and 1 atm and the water in the products exists in the liquid form, the heat transfer from the combustion chamber per unit mass of methane is (a) 890 MJ/kg (b) 80 MJ/kg (c) 75 MJ/kg (d) 56 MJ/kg (e) 50 MJ/kg The higher heating value of a hydrocarbon fuel C n H m with m 8 is given to be 1560 MJ/kmol of fuel. Then its lower heating value is (a) 1384 MJ/kmol (b) 108 MJ/kmol (c) 140 MJ/kmol (d) 1514 MJ/kmol (e) 1551 MJ/kmol Acetylene gas (C H ) is burned completely during a steady-flow combustion process. The fuel and the air enter the combustion chamber at 5 C, and the products leave at 1500 K. If the enthalpy of the products relative to the standard reference state is 404 MJ/kmol of fuel, the heat transfer from the combustion chamber is (a) 177 MJ/kmol (b) 7 MJ/kmol (c) 404 MJ/kmol (d) 631 MJ/kmol (e) 751 MJ/kmol Benzene gas (C 6 H 6 ) is burned with 90 percent theoretical air during a steady-flow combustion process. The mole fraction of the CO in the products is (a) 1.6% (b) 4.4% (c).5% (d) 10% (e) 16.7% A fuel is burned during a steady-flow combustion process. Heat is lost to the surroundings at 300 K at a rate of 110 kw. The entropy of the reactants entering per unit time is 17 kw/k and that of the products is 15 kw/k. The total rate of exergy destruction during this combustion process is (a) 50 kw (b) 600 kw (c) 110 kw (d) 340 kw (e) 739 kw Design and Essay Problems Design a combustion process suitable for use in a gas-turbine engine. Discuss possible fuel selections for the several applications of the engine Constant-volume vessels that contain flammable mixtures of hydrocarbon vapors and air at low pressures are frequently used. Although the ignition of such mixtures is very unlikely as there is no source of ignition in the tank, the Safety and Design Codes require that the tank withstand four times the pressure that may occur should an explosion take place in the tank. For operating gauge pressures under 5 kpa, determine the pressure for which these vessels must be designed in order to meet the requirements of the codes for (a) acetylene C H (g), (b) propane C 3 H 8 (g), and (c) n-octane C 8 H 18 (g). Justify any assumptions that you make The safe disposal of hazardous waste material is a major environmental concern for industrialized societies and creates challenging problems for engineers. The disposal methods commonly used include landfilling, burying in the ground, recycling, and incineration or burning. Incineration is frequently used as a practical means for the disposal of combustible waste such as organic materials. The EPA regulations require that the waste material be burned almost completely above a specified temperature without polluting the environment. Maintaining the temperature above a certain level, typically about 1100 C, necessitates the use of a fuel when the combustion of the waste material alone is not sufficient to obtain the minimum specified temperature. A certain industrial process generates a liquid solution of ethanol and water as the waste product at a rate of 10 kg/s. The mass fraction of ethanol in the solution is 0.. This solution is to be burned using methane (CH 4 ) in a steady-flow combustion chamber. Propose a combustion process that will accomplish this task with a minimal amount of methane. State your assumptions Obtain the following information about a power plant that is closest to your town: the net power output; the type and amount of fuel; the power consumed by the pumps, fans, and other auxiliary equipment; stack gas losses; and the rate of heat rejection at the condenser. Using these data, determine the rate of heat loss from the pipes and other components, and calculate the thermal efficiency of the plant What is oxygenated fuel? How would the heating value of oxygenated fuels compare to those of comparable hydrocarbon fuels on a unit-mass basis? Why is the use of oxygenated fuels mandated in some major cities in winter months? A promising method of power generation by direct energy conversion is through the use of magnetohydrodynamic (MHD) generators. Write an essay on the current status of MHD generators. Explain their operation principles and how they differ from conventional power plants. Discuss the problems that need to be overcome before MHD generators can become economical.

315 Chapter 16 CHEMICAL AND PHASE EQUILIBRIUM In Chapter 15 we analyzed combustion processes under the assumption that combustion is complete when there is sufficient time and oxygen. Often this is not the case, however. A chemical reaction may reach a state of equilibrium before reaching completion even when there is sufficient time and oxygen. A system is said to be in equilibrium if no changes occur within the system when it is isolated from its surroundings. An isolated system is in mechanical equilibrium if no changes occur in pressure, in thermal equilibrium if no changes occur in temperature, in phase equilibrium if no transformations occur from one phase to another, and in chemical equilibrium if no changes occur in the chemical composition of the system. The conditions of mechanical and thermal equilibrium are straightforward, but the conditions of chemical and phase equilibrium can be rather involved. The equilibrium criterion for reacting systems is based on the second law of thermodynamics; more specifically, the increase of entropy principle. For adiabatic systems, chemical equilibrium is established when the entropy of the reacting system reaches a maximum. Most reacting systems encountered in practice are not adiabatic, however. Therefore, we need to develop an equilibrium criterion applicable to any reacting system. In this chapter, we develop a general criterion for chemical equilibrium and apply it to reacting ideal-gas mixtures. We then extend the analysis to simultaneous reactions. Finally, we discuss phase equilibrium for nonreacting systems. Objectives The objectives of Chapter 16 are to: Develop the equilibrium criterion for reacting systems based on the second law of thermodynamics. Develop a general criterion for chemical equilibrium applicable to any reacting system based on minimizing the Gibbs function for the system. Define and evaluate the chemical equilibrium constant. Apply the general criterion for chemical equilibrium analysis to reacting ideal-gas mixtures. Apply the general criterion for chemical equilibrium analysis to simultaneous reactions. Relate the chemical equilibrium constant to the enthalpy of reaction. Establish the phase equilibrium for nonreacting systems in terms of the specific Gibbs function of the phases of a pure substance. Apply the Gibbs phase rule to determine the number of independent variables associated with a multicomponent, multiphase system. Apply Henry s law and Raoult s law for gases dissolved in liquids. 793

316 794 Thermodynamics CO CO CO O O CO CO CO O FIGURE 16 1 A reaction chamber that contains a mixture of CO, CO, and O at a specified temperature and pressure. S ds > 0 ds = 0 Violation of second law ds < CRITERION FOR CHEMICAL EQUILIBRIUM Consider a reaction chamber that contains a mixture of CO, O, and CO at a specified temperature and pressure. Let us try to predict what will happen in this chamber (Fig. 16 1). Probably the first thing that comes to mind is a chemical reaction between CO and O to form more CO : CO 1 O S CO This reaction is certainly a possibility, but it is not the only possibility. It is also possible that some CO in the combustion chamber dissociated into CO and O. Yet a third possibility would be to have no reactions among the three components at all, that is, for the system to be in chemical equilibrium. It appears that although we know the temperature, pressure, and composition (thus the state) of the system, we are unable to predict whether the system is in chemical equilibrium. In this chapter we develop the necessary tools to correct this. Assume that the CO, O, and CO mixture mentioned above is in chemical equilibrium at the specified temperature and pressure. The chemical composition of this mixture does not change unless the temperature or the pressure of the mixture is changed. That is, a reacting mixture, in general, has different equilibrium compositions at different pressures and temperatures. Therefore, when developing a general criterion for chemical equilibrium, we consider a reacting system at a fixed temperature and pressure. Taking the positive direction of heat transfer to be to the system, the increase of entropy principle for a reacting or nonreacting system was expressed in Chapter 7 as 100% reactants Equilibrium composition 100% products FIGURE 16 Equilibrium criteria for a chemical reaction that takes place adiabatically. REACTION CHAMBER δ Q Control mass T, P W b FIGURE 16 3 A control mass undergoing a chemical reaction at a specified temperature and pressure. δ (16 1) A system and its surroundings form an adiabatic system, and for such systems Eq reduces to ds sys 0. That is, a chemical reaction in an adiabatic chamber proceeds in the direction of increasing entropy. When the entropy reaches a maximum, the reaction stops (Fig. 16 ). Therefore, entropy is a very useful property in the analysis of reacting adiabatic systems. When a reacting system involves heat transfer, the increase of entropy principle relation (Eq. 16 1) becomes impractical to use, however, since it requires a knowledge of heat transfer between the system and its surroundings. A more practical approach would be to develop a relation for the equilibrium criterion in terms of the properties of the reacting system only. Such a relation is developed below. Consider a reacting (or nonreacting) simple compressible system of fixed mass with only quasi-equilibrium work modes at a specified temperature T and pressure P (Fig. 16 3). Combining the first- and the second-law relations for this system gives dq P dv du ds dq T ds sys dq T du P dv T ds 0 (16 )

317 The differential of the Gibbs function (G H TS) at constant temperature and pressure is G Chapter (16 3) From Eqs. 16 and 16 3, we have (dg) T,P 0. Therefore, a chemical reaction at a specified temperature and pressure proceeds in the direction of a decreasing Gibbs function. The reaction stops and chemical equilibrium is established when the Gibbs function attains a minimum value (Fig. 16 4). Therefore, the criterion for chemical equilibrium can be expressed as (16 4) A chemical reaction at a specified temperature and pressure cannot proceed in the direction of the increasing Gibbs function since this will be a violation of the second law of thermodynamics. Notice that if the temperature or the pressure is changed, the reacting system will assume a different equilibrium state, which is the state of the minimum Gibbs function at the new temperature or pressure. To obtain a relation for chemical equilibrium in terms of the properties of the individual components, consider a mixture of four chemical components A, B, C, and D that exist in equilibrium at a specified temperature and pressure. Let the number of moles of the respective components be N A, N B, N C, and N D. Now consider a reaction that occurs to an infinitesimal extent during which differential amounts of A and B (reactants) are converted to C and D (products) while the temperature and the pressure remain constant (Fig. 16 5): The equilibrium criterion (Eq. 16 4) requires that the change in the Gibbs function of the mixture during this process be equal to zero. That is, or 1dG T,P dh T ds S dt 0 1dU P dv V dp T ds S dt 0 du P dv T ds 1dG T,P 0 dn A A dn B B dn C C dn D D 1dG T,P a 1dG i T,P a 1g i dn i T,P 0 g C dn C g D dn D g A dn A g B dn B 0 (16 5) (16 6) where the g s are the molar Gibbs functions (also called the chemical potentials) at the specified temperature and pressure and the dn s are the differential changes in the number of moles of the components. To find a relation between the dn s, we write the corresponding stoichiometric (theoretical) reaction 100% reactants dg < 0 dg > 0 dg = 0 Violation of second law Equilibrium composition 100% products FIGURE 16 4 Criteria for chemical equilibrium for a fixed mass at a specified temperature and pressure. REACTION CHAMBER T, P N A moles of A N B moles of B N C moles of C N D moles of D dn A A + dn B B dn C C + dn D D FIGURE 16 5 An infinitesimal reaction in a chamber at constant temperature and pressure. n A A n B B n C C n D D (16 7) where the n s are the stoichiometric coefficients, which are evaluated easily once the reaction is specified. The stoichiometric reaction plays an important role in the determination of the equilibrium composition of the reacting

318 796 Thermodynamics H H 0.1H 0.H 0.01H 0.0H 0.001H 0.00H n = 1 H nh = FIGURE 16 6 The changes in the number of moles of the components during a chemical reaction are proportional to the stoichiometric coefficients regardless of the extent of the reaction. mixtures because the changes in the number of moles of the components are proportional to the stoichiometric coefficients (Fig. 16 6). That is, dn A en A dn C en C (16 8) dn B en B dn D en D where e is the proportionality constant and represents the extent of a reaction. A minus sign is added to the first two terms because the number of moles of the reactants A and B decreases as the reaction progresses. For example, if the reactants are C H 6 and O and the products are CO and H O, the reaction of 1 mmol (10 6 mol) of C H 6 results in a -mmol increase in CO,a 3-mmol increase in H O, and a 3.5-mmol decrease in O in accordance with the stoichiometric equation C H 6 3.5O S CO 3H O That is, the change in the number of moles of a component is one-millionth (e 10 6 ) of the stoichiometric coefficient of that component in this case. Substituting the relations in Eq into Eq and canceling e, we obtain n C g C n D g D n A g A n B g B 0 (16 9) This equation involves the stoichiometric coefficients and the molar Gibbs functions of the reactants and the products, and it is known as the criterion for chemical equilibrium. It is valid for any chemical reaction regardless of the phases involved. Equation 16 9 is developed for a chemical reaction that involves two reactants and two products for simplicity, but it can easily be modified to handle chemical reactions with any number of reactants and products. Next we analyze the equilibrium criterion for ideal-gas mixtures. 16 THE EQUILIBRIUM CONSTANT FOR IDEAL-GAS MIXTURES Consider a mixture of ideal gases that exists in equilibrium at a specified temperature and pressure. Like entropy, the Gibbs function of an ideal gas depends on both the temperature and the pressure. The Gibbs function values are usually listed versus temperature at a fixed reference pressure P 0, which is taken to be 1 atm. The variation of the Gibbs function of an ideal gas with pressure at a fixed temperature is determined by using the definition of the Gibbs function and the entropy-change relation for isothermal processes 3 1g h Ts s R u ln 1P >P 1 4. It yields 1 g T h 0 T 1 s T T1 s T R u T ln P Thus the Gibbs function of component i of an ideal-gas mixture at its partial pressure P i and mixture temperature T can be expressed as P 1 g i 1T, P i g * i 1T R u T ln P i (16 10)

319 where g * (T) represents the Gibbs function of component i at 1 atm pressure and temperature T, and P i represents the partial pressure of component i i in atmospheres. Substituting the Gibbs function expression for each component into Eq. 16 9, we obtain Chapter n C 3 g * C 1T R u T ln P C 4 n D 3 g D * 1T R u T ln P D 4 n A 3 g * A 1T R u T ln P A 4 n B 3 g * B 1T R u T ln P B 4 0 For convenience, we define the standard-state Gibbs function change as G* 1T n C g * C 1T n D g * D 1T n A g * A 1T n B g * B 1T (16 11) Substituting, we get G* 1T R u T 1n C ln P C n D ln P D n A ln P A n B ln P B R u T ln P C n C P D n D n (16 1) P A n A P B B Now we define the equilibrium constant K P for the chemical equilibrium of ideal-gas mixtures as K P P C n C P D nd P A n A P B n B (16 13) Substituting into Eq and rearranging, we obtain (16 14) Therefore, the equilibrium constant K P of an ideal-gas mixture at a specified temperature can be determined from a knowledge of the standard-state Gibbs function change at the same temperature. The K P values for several reactions are given in Table A 8. Once the equilibrium constant is available, it can be used to determine the equilibrium composition of reacting ideal-gas mixtures. This is accomplished by expressing the partial pressures of the components in terms of their mole fractions: where P is the total pressure and N total is the total number of moles present in the reaction chamber, including any inert gases. Replacing the partial pressures in Eq by the above relation and rearranging, we obtain (Fig. 16 7) where K P e G*1T>R ut P i y i P N i N total P K P N C n C N D n n N A n A N B a P b B N total n n C n D n A n B (16 15) Equation is written for a reaction involving two reactants and two products, but it can be extended to reactions involving any number of reactants and products. n (1) In terms of partial pressures K P = P nc nd P C D na nb P A P B () In terms of G*(T) K P = e G*(T)/R u T (3) In terms of the equilibrium composition K P = N nc nd C N D P n na nb N A N B ( N total FIGURE 16 7 Three equivalent K P relations for reacting ideal-gas mixtures. (

320 798 Thermodynamics EXAMPLE 16 1 Equilibrium Constant of a Dissociation Process Using Eq and the Gibbs function data, determine the equilibrium constant K P for the dissociation process N N at 5 C. Compare your result to the K P value listed in Table A 8. Solution The equilibrium constant of the reaction N N is listed in Table A 8 at different temperatures. It is to be verified using Gibbs function data. Assumptions 1 The constituents of the mixture are ideal gases. The equilibrium mixture consists of N and N only. Properties The equilibrium constant of this reaction at 98 K is ln K P (Table A 8). The Gibbs function of formation at 5 C and 1 atm is O for N and 455,510 kj/kmol for N (Table A 6). Analysis In the absence of K P tables, K P can be determined from the Gibbs function data and Eq , where, from Eq , Substituting, we find or G* 1T n N g * N 1T n N g * N 1T ,510 kj>kmol 0 911,00 kj>kmol 911,00 kj>kmol ln K P kj>kmol # K K K P e G*1T>R ut K P The calculated K P value is in agreement with the value listed in Table A 8. The K P value for this reaction is practically zero, indicating that this reaction will not occur at this temperature. Discussion Note that this reaction involves one product (N) and one reactant (N ), and the stoichiometric coefficients for this reaction are n N and n N 1. Also note that the Gibbs function of all stable elements (such as N ) is assigned a value of zero at the standard reference state of 5 C and 1 atm. The Gibbs function values at other temperatures can be calculated from the enthalpy and absolute entropy data by using the definition of the Gibbs function, g, where h 1T h*f ht * 1T h 1T Ts * 1T h98 K. EXAMPLE 16 Dissociation Temperature of Hydrogen Determine the temperature at which 10 percent of diatomic hydrogen (H ) dissociates into monatomic hydrogen (H) at a pressure of 10 atm. Solution The temperature at which 10 percent of H dissociates into H is to be determined. Assumptions 1 The constituents of the mixture are ideal gases. The equilibrium mixture consists of H and H only.

321 Chapter Analysis This is a dissociation process that is significant at very high temperatures only. For simplicity we consider 1 kmol of H, as shown in Fig The stoichiometric and actual reactions in this case are as follows: Stoichiometric: Actual: H H 1thus n H 1 and n H H 0.9H 0.H reactants products (leftover) A double-headed arrow is used for the stoichiometric reaction to differentiate it from the actual reaction. This reaction involves one reactant (H ) and one product (H). The equilibrium composition consists of 0.9 kmol of H (the leftover reactant) and 0. kmol of H (the newly formed product). Therefore, N H 0.9 and N H 0. and the equilibrium constant K P is determined from Eq to be Initial Equilibrium composition composition 10 atm 1 kmol H 0.9H 0.H FIGURE 16 8 Schematic for Example 16. K P N H n n N H a P n H n H b 10. H N total 0.9 a 10 b From Table A 8, the temperature corresponding to this K P value is T 3535 K Discussion We conclude that 10 percent of H dissociates into H when the temperature is raised to 3535 K. If the temperature is increased further, the percentage of H that dissociates into H will also increase. A double arrow is used in equilibrium equations as an indication that a chemical reaction does not stop when chemical equilibrium is established; rather, it proceeds in both directions at the same rate. That is, at equilibrium, the reactants are depleted at exactly the same rate as they are replenished from the products by the reverse reaction SOME REMARKS ABOUT THE K P OF IDEAL-GAS MIXTURES In the last section we developed three equivalent expressions for the equilibrium constant K P of reacting ideal-gas mixtures: Eq , which expresses K P in terms of partial pressures; Eq , which expresses K P in terms of the standard-state Gibbs function change G*(T); and Eq , which expresses K P in terms of the number of moles of the components. All three relations are equivalent, but sometimes one is more convenient to use than the others. For example, Eq is best suited for determining the equilibrium composition of a reacting ideal-gas mixture at a specified temperature and pressure. On the basis of these relations, we may draw the following conclusions about the equilibrium constant K P of ideal-gas mixtures: 1. The K P of a reaction depends on temperature only. It is independent of the pressure of the equilibrium mixture and is not affected by the presence of inert gases. This is because K P depends on G*(T), which depends on

322 800 Thermodynamics T, K H H P = 1 atm K P % mol H FIGURE 16 9 The larger the K P, the more complete the reaction. (a) (b) Initial composition 1 mol H 1 mol H 1 mol N Equilibrium composition at 3000 K, 1 atm 0.91 mol H mol H K P = mol H 1.40 mol H 1 mol N K P = FIGURE The presence of inert gases does not affect the equilibrium constant, but it does affect the equilibrium composition. temperature only, and the G*(T) of inert gases is zero (see Eq ). Thus, at a specified temperature the following four reactions have the same K P value: H 1 O H O at 1 atm H 1 O H O at 5 atm H 1 O 3N H O 3N at 3 atm H O 5N H O 1.5O 5N at atm. The K P of the reverse reaction is 1/K P. This is easily seen from Eq For reverse reactions, the products and reactants switch places, and thus the terms in the numerator move to the denominator and vice versa. Consequently, the equilibrium constant of the reverse reaction becomes 1/K P. For example, from Table A 8, K P for H 1 O H O at 1000 K K P for H O H 1 O at 1000 K 3. The larger the K P, the more complete the reaction. This is also apparent from Fig and Eq If the equilibrium composition consists largely of product gases, the partial pressures of the products (P C and P D ) are considerably larger than the partial pressures of the reactants (P A and P B ), which results in a large value of K P. In the limiting case of a complete reaction (no leftover reactants in the equilibrium mixture), K P approaches infinity. Conversely, very small values of K P indicate that a reaction does not proceed to any appreciable degree. Thus reactions with very small K P values at a specified temperature can be neglected. A reaction with K P 1000 (or ln K P 7) is usually assumed to proceed to completion, and a reaction with K P (or ln K P 7) is assumed not to occur at all. For example, ln K P 6.8 for the reaction N N at 5000 K. Therefore, the dissociation of N into monatomic nitrogen (N) can be disregarded at temperatures below 5000 K. 4. The mixture pressure affects the equilibrium composition (although it does not affect the equilibrium constant K P ). This can be seen from Eq , which involves the term P n, where n n P n R (the difference between the number of moles of products and the number of moles of reactants in the stoichiometric reaction). At a specified temperature, the K P value of the reaction, and thus the right-hand side of Eq , remains constant. Therefore, the mole numbers of the reactants and the products must change to counteract any changes in the pressure term. The direction of the change depends on the sign of n. An increase in pressure at a specified temperature increases the number of moles of the reactants and decreases the number of moles of products if n is positive, have the opposite effect if n is negative, and have no effect if n is zero. 5. The presence of inert gases affects the equilibrium composition (although it does not affect the equilibrium constant K P ). This can be seen from Eq , which involves the term (1/N total ) n, where N total is the total number of moles of the ideal-gas mixture at equilibrium, including inert gases. The sign of n determines how the presence of inert gases influences the equilibrium composition (Fig ). An increase in the number of moles of inert gases at a specified temperature and pressure decreases the number of

323 moles of the reactants and increases the number of moles of products if n is positive, have the opposite effect if n is negative, and have no effect if n is zero. 6. When the stoichiometric coefficients are doubled, the value of K P is squared. Therefore, when one is using K P values from a table, the stoichiometric coefficients (the n s) used in a reaction must be exactly the same ones appearing in the table from which the K P values are selected. Multiplying all the coefficients of a stoichiometric equation does not affect the mass balance, but it does affect the equilibrium constant calculations since the stoichiometric coefficients appear as exponents of partial pressures in Eq For example, For But for H 1 O H O K P1 P H O P H P 1> O H O H O K P 7. Free electrons in the equilibrium composition can be treated as an ideal gas. At high temperatures (usually above 500 K), gas molecules start to dissociate into unattached atoms (such as H H), and at even higher temperatures atoms start to lose electrons and ionize, for example, H H e P H O P H P O 1K P1 (16 16) The dissociation and ionization effects are more pronounced at low pressures. Ionization occurs to an appreciable extent only at very high temperatures, and the mixture of electrons, ions, and neutral atoms can be treated as an ideal gas. Therefore, the equilibrium composition of ionized gas mixtures can be determined from Eq (Fig ). This treatment may not be adequate in the presence of strong electric fields, however, since the electrons may be at a different temperature than the ions in this case. 8. Equilibrium calculations provide information on the equilibrium composition of a reaction, not on the reaction rate. Sometimes it may even take years to achieve the indicated equilibrium composition. For example, the equilibrium constant of the reaction H 1 O H O at 98 K is about 10 40, which suggests that a stoichiometric mixture of H and O at room temperature should react to form H O, and the reaction should go to completion. However, the rate of this reaction is so slow that it practically does not occur. But when the right catalyst is used, the reaction goes to completion rather quickly to the predicted value. Chapter H H + + e n H N + n e H + N e P n K P = n H ( N N total H where N total = N H + N H + + N e n= n H + + n e nh = = 1 FIGURE Equilibrium-constant relation for the ionization reaction of hydrogen. ( EXAMPLE 16 3 Equilibrium Composition at a Specified Temperature A mixture of kmol of CO and 3 kmol of O is heated to 600 K at a pressure of 304 kpa. Determine the equilibrium composition, assuming the mixture consists of CO, CO, and O (Fig. 16 1). Initial composition kmol CO 3 kmol O Equilibrium composition at 600 K, 304 kpa x CO y CO z O Solution A reactive gas mixture is heated to a high temperature. The equilibrium composition at that temperature is to be determined. FIGURE 16 1 Schematic for Example 16 3.

324 80 Thermodynamics Assumptions 1 The equilibrium composition consists of CO, CO, and O. The constituents of the mixture are ideal gases. Analysis The stoichiometric and actual reactions in this case are as follows: Stoichiometric: Actual: C balance: CO 1 O CO 1thus n CO 1, n CO 1, and n O 1 CO 3O xco yco zo products reactants (leftover) x y or y x O balance: 8 x y z or z 3 x Total number of moles: Pressure: N total x y z 5 x P 304 kpa 3.0 atm The closest reaction listed in Table A 8 is CO CO 1 O, for which ln K P.801 at 600 K. The reaction we have is the inverse of this, and thus ln K P.801, or K P in our case. Assuming ideal-gas behavior for all components, the equilibrium constant relation (Eq ) becomes Substituting, we get n CO K P N CO n O a P n CO n CO n O b N total n N CO CO N O > x 1 x13 x> a 3 1> 5 x> b Solving for x yields Then x y x z 3 x.047 Therefore, the equilibrium composition of the mixture at 600 K and 304 kpa is 1.906CO 0.094CO.074O Discussion In solving this problem, we disregarded the dissociation of O into O according to the reaction O O, which is a real possibility at high temperatures. This is because ln K P 7.51 at 600 K for this reaction, which indicates that the amount of O that dissociates into O is negligible. (Besides, we have not learned how to deal with simultaneous reactions yet. We will do so in the next section.)

325 Chapter EXAMPLE 16 4 Effect of Inert Gases on Equilibrium Composition A mixture of 3 kmol of CO,.5 kmol of O, and 8 kmol of N is heated to 600 K at a pressure of 5 atm. Determine the equilibrium composition of the mixture (Fig ). Solution A gas mixture is heated to a high temperature. The equilibrium composition at the specified temperature is to be determined. Assumptions 1 The equilibrium composition consists of CO, CO, O, and N. The constituents of the mixture are ideal gases. Analysis This problem is similar to Example 16 3, except that it involves an inert gas N. At 600 K, some possible reactions are O O (ln K P 7.51), N N (ln K P 8.304), 1 O 1 N NO (ln K P.671), and CO 1 O CO (ln K P.801 or K P ). Based on these K P values, we conclude that the O and N will not dissociate to any appreciable degree, but a small amount will combine to form some oxides of nitrogen. (We disregard the oxides of nitrogen in this example, but they should be considered in a more refined analysis.) We also conclude that most of the CO will combine with O to form CO. Notice that despite the changes in pressure, the number of moles of CO and O and the presence of an inert gas, the K P value of the reaction is the same as that used in Example The stoichiometric and actual reactions in this case are Initial composition 3 kmol CO.5 kmol O 8 kmol N Equilibrium composition at 600 K, 5 atm x CO y CO z O 8 N FIGURE Schematic for Example Stoichiometric: Actual: C balance: O balance: CO 1 O CO 1thus n CO 1, n CO 1, and n O 1 3CO.5O 8N xco yco zo 8N products reactants inert (leftover) 3 x y or y 3 x 8 x y z or z.5 x Total number of moles: N total x y z x Assuming ideal-gas behavior for all components, the equilibrium constant relation (Eq ) becomes Substituting, we get Solving for x yields n CO K P N CO n O a P n CO n CO n O b N total n N CO CO N O x 13 x1.5 x> 1> a x> b 1> x.754

326 804 Thermodynamics Then y 3 x 0.46 z.5 x 1.13 Therefore, the equilibrium composition of the mixture at 600 K and 5 atm is.754co 0.46CO 1.13O 8N Discussion Note that the inert gases do not affect the K P value or the K P relation for a reaction, but they do affect the equilibrium composition. Composition: CO, CO, O, O No. of components: 4 No. of elements: No. of K p relations needed: 4 = FIGURE The number of K P relations needed to determine the equilibrium composition of a reacting mixture is the difference between the number of species and the number of elements CHEMICAL EQUILIBRIUM FOR SIMULTANEOUS REACTIONS The reacting mixtures we have considered so far involved only one reaction, and writing a K P relation for that reaction was sufficient to determine the equilibrium composition of the mixture. However, most practical chemical reactions involve two or more reactions that occur simultaneously, which makes them more difficult to deal with. In such cases, it becomes necessary to apply the equilibrium criterion to all possible reactions that may occur in the reaction chamber. When a chemical species appears in more than one reaction, the application of the equilibrium criterion, together with the mass balance for each chemical species, results in a system of simultaneous equations from which the equilibrium composition can be determined. We have shown earlier that a reacting system at a specified temperature and pressure achieves chemical equilibrium when its Gibbs function reaches a minimum value, that is, (dg) T,P 0. This is true regardless of the number of reactions that may be occurring. When two or more reactions are involved, this condition is satisfied only when (dg) T,P 0 for each reaction. Assuming ideal-gas behavior, the K P of each reaction can be determined from Eq , with N total being the total number of moles present in the equilibrium mixture. The determination of the equilibrium composition of a reacting mixture requires that we have as many equations as unknowns, where the unknowns are the number of moles of each chemical species present in the equilibrium mixture. The mass balance of each element involved provides one equation. The rest of the equations must come from the K P relations written for each reaction. Thus we conclude that the number of K P relations needed to determine the equilibrium composition of a reacting mixture is equal to the number of chemical species minus the number of elements present in equilibrium. For an equilibrium mixture that consists of CO, CO, O, and O, for example, two K P relations are needed to determine the equilibrium composition since it involves four chemical species and two elements (Fig ). The determination of the equilibrium composition of a reacting mixture in the presence of two simultaneous reactions is here with an example.

327 Chapter EXAMPLE 16 5 Equilibrium Composition for Simultaneous Reactions A mixture of 1 kmol of H O and kmol of O is heated to 4000 K at a pressure of 1 atm. Determine the equilibrium composition of this mixture, assuming that only H O, OH, O, and H are present (Fig ). Solution A gas mixture is heated to a specified temperature at a specified pressure. The equilibrium composition is to be determined. Assumptions 1 The equilibrium composition consists of H O, OH, O, and H. The constituents of the mixture are ideal gases. Analysis The chemical reaction during this process can be expressed as H O O xh O yh zo woh Mass balances for hydrogen and oxygen yield H balance: x y w (1) O balance: 5 x z w () The mass balances provide us with only two equations with four unknowns, and thus we need to have two more equations (to be obtained from the K P relations) to determine the equilibrium composition of the mixture. It appears that part of the H O in the products is dissociated into H and OH during this process, according to the stoichiometric reactions H O H 1 O 1reaction 1 H O 1 H OH 1reaction The equilibrium constants for these two reactions at 4000 K are determined from Table A 8 to be ln K P K P ln K P K P The K P relations for these two simultaneous reactions are K P1 N n H n H N O O n N HO a P n H n O n HO b H O N total Initial composition 1 kmol H O kmol O Equilibrium composition at 4000 K, 1 atm x H O y H z O w OH FIGURE Schematic for Example where Substituting yields K P N n H n H N OH OH n N HO a P n H n OH n HO b H O N total N total N H O N H N O N OH x y z w y 1z1> x 1w 1y1> x 1> 1 a x y z w b 1> 1 a x y z w b (3) (4)

328 806 Thermodynamics Solving Eqs. (1), (), (3), and (4) simultaneously for the four unknowns x, y, z, and w yields x 0.71 y 0.13 z w 1.03 Therefore, the equilibrium composition of 1 kmol H O and kmol O at 1 atm and 4000 K is 0.71H O 0.13H 1.849O 1.03OH Discussion We could also solve this problem by using the K P relation for the stoichiometric reaction O O as one of the two equations. Solving a system of simultaneous nonlinear equations is extremely tedious and time-consuming if it is done by hand. Thus it is often necessary to solve these kinds of problems by using an equation solver such as EES VARIATION OF K P WITH TEMPERATURE It was shown in Section 16 that the equilibrium constant K P of an ideal gas depends on temperature only, and it is related to the standard-state Gibbs function change G*(T) through the relation (Eq ) G* 1T ln K P R u T In this section we develop a relation for the variation of K P with temperature in terms of other properties. Substituting G*(T) H*(T) T S*(T) into the above relation and differentiating with respect to temperature, we get d 1ln K p H* 1T d 3 H* 1T4 d 3 S* 1T4 dt R u T R u T dt R u dt At constant pressure, the second Tdsrelation, Tds dh v dp, reduces to Tds dh. Also, T d( S*) d( H*) since S* and H* consist of entropy and enthalpy terms of the reactants and the products. Therefore, the last two terms in the above relation cancel, and it reduces to d 1ln K p H* 1T h R 1T (16 17) dt R u T R u T where h R 1T is the enthalpy of reaction at temperature T. Notice that we dropped the superscript * (which indicates a constant pressure of 1 atm) from H(T), since the enthalpy of an ideal gas depends on temperature only and is independent of pressure. Equation is an expression of the variation of K P with temperature in terms of h R 1T, and it is known as the van t Hoff equation. To integrate it, we need to know how h varies with T. For R small temperature intervals, h R can be treated as a constant and Eq can be integrated to yield ln K P h R a 1 1 b K P1 R u T 1 T (16 18)

329 This equation has two important implications. First, it provides a means of calculating the h R of a reaction from a knowledge of K P, which is easier to determine. Second, it shows that exothermic reactions 1h R 6 0 such as combustion processes are less complete at higher temperatures since K P decreases with temperature for such reactions (Fig ). EXAMPLE 16 6 The Enthalpy of Reaction of a Combustion Process Estimate the enthalpy of reaction h R for the combustion process of hydrogen H + 0.5O H O at 000 K, using (a) enthalpy data and (b) K P data. Solution The h R at a specified temperature is to be determined using the enthalpy and K p data. Assumptions Both the reactants and the products are ideal gases. Analysis (a) The h R of the combustion process of H at 000 K is the amount of energy released as 1 kmol of H is burned in a steady-flow combustion chamber at a temperature of 000 K. It can be determined from Eq. 15 6, Chapter Reaction: C + O CO T, K K P FIGURE Exothermic reactions are less complete at higher temperatures. h R a N p 1h f h h p a N r 1h f h h r N H O 1h f h 000 K h 98 K H O N H 1h f h 000 K h 98 K H N O 1h f h 000 K h 98 K O Substituting yields h R 11 kmol H O31 41,80 8, kj>kmol H O4 (b) The h R value at 000 K can be estimated by using K P values at 1800 and 00 K (the closest two temperatures to 000 K for which K P data are available) from Table A 8. They are K P1 18,509 at T K and K P at T 00 K. By substituting these values into Eq , the value is determined to be h R 11 kmol H , kj>kmol H kmol O , kj>kmol O 4 51,663 kj/kmol ln ln K P K P1 h R R u a 1 T 1 1 T b ,509 h R kj>kmol # K a K 1 00 K b h R 51,698 kj/kmol Discussion Despite the large temperature difference between T 1 and T (400 K), the two results are almost identical. The agreement between the two results would be even better if a smaller temperature interval were used.

330 808 Thermodynamics FIGURE Wet clothes hung in an open area eventually dry as a result of mass transfer from the liquid phase to the vapor phase. Vol. OS36/PhotoDisc T, P VAPOR m g m f LIQUID FIGURE A liquid vapor mixture in equilibrium at a constant temperature and pressure PHASE EQUILIBRIUM We showed at the beginning of this chapter that the equilibrium state of a system at a specified temperature and pressure is the state of the minimum Gibbs function, and the equilibrium criterion for a reacting or nonreacting system was expressed as (Eq. 16 4) 1dG T,P 0 In the preceding sections we applied the equilibrium criterion to reacting systems. In this section, we apply it to nonreacting multiphase systems. We know from experience that a wet T-shirt hanging in an open area eventually dries, a small amount of water left in a glass evaporates, and the aftershave in an open bottle quickly disappears (Fig ). These examples suggest that there is a driving force between the two phases of a substance that forces the mass to transform from one phase to another. The magnitude of this force depends, among other things, on the relative concentrations of the two phases. A wet T-shirt dries much quicker in dry air than it does in humid air. In fact, it does not dry at all if the relative humidity of the environment is 100 percent. In this case, there is no transformation from the liquid phase to the vapor phase, and the two phases are in phase equilibrium. The conditions of phase equilibrium change, however, if the temperature or the pressure is changed. Therefore, we examine phase equilibrium at a specified temperature and pressure. Phase Equilibrium for a Single-Component System The equilibrium criterion for two phases of a pure substance such as water is easily developed by considering a mixture of saturated liquid and saturated vapor in equilibrium at a specified temperature and pressure, such as that shown in Fig The total Gibbs function of this mixture is G m f g f m g g g where g f and g g are the Gibbs functions of the liquid and vapor phases per unit mass, respectively. Now imagine a disturbance during which a differential amount of liquid dm f evaporates at constant temperature and pressure. The change in the total Gibbs function during this disturbance is 1dG T,P g f dm f g g dm g since g f and g g remain constant at constant temperature and pressure. At equilibrium, (dg) T,P 0. Also from the conservation of mass, dm g dm f. Substituting, we obtain 1dG T,P 1g f g g dm f which must be equal to zero at equilibrium. It yields g f g g (16 19) Therefore, the two phases of a pure substance are in equilibrium when each phase has the same value of specific Gibbs function. Also, at the triple point (the state at which all three phases coexist in equilibrium), the specific Gibbs functions of all three phases are equal to each other.

331 What happens if g f g g? Obviously the two phases are not in equilibrium at that moment. The second law requires that (dg) T, P (g f g g ) dm f 0. Thus, dm f must be negative, which means that some liquid must vaporize until g f g g. Therefore, the Gibbs function difference is the driving force for phase change, just as the temperature difference is the driving force for heat transfer. Chapter EXAMPLE 16 7 Phase Equilibrium for a Saturated Mixture Show that a mixture of saturated liquid water and saturated water vapor at 10 C satisfies the criterion for phase equilibrium. Solution It is to be shown that a saturated mixture satisfies the criterion for phase equilibrium. Properties The properties of saturated water at 10 C are h f kj/kg, s f kj/kg K, h g kj/kg, and s g 7.19 kj/kg K (Table A 4). Analysis Using the definition of Gibbs function together with the enthalpy and entropy data, we have and g f h f Ts f kj>kg K kj>kg # K 96.9 kj>kg g g h g Ts g kj>kg K kj>kg # K 96.8 kj>kg Discussion The two results are in close agreement. They would match exactly if more accurate property data were used. Therefore, the criterion for phase equilibrium is satisfied. The Phase Rule Notice that a single-component two-phase system may exist in equilibrium at different temperatures (or pressures). However, once the temperature is fixed, the system is locked into an equilibrium state and all intensive properties of each phase (except their relative amounts) are fixed. Therefore, a single-component two-phase system has one independent property, which may be taken to be the temperature or the pressure. In general, the number of independent variables associated with a multicomponent, multiphase system is given by the Gibbs phase rule, expressed as IV C PH (16 0) where IV the number of independent variables, C the number of components, and PH the number of phases present in equilibrium. For the single-component (C 1) two-phase (PH ) system discussed above, for example, one independent intensive property needs to be specified (IV 1, Fig ). At the triple point, however, PH 3 and thus IV 0. That is, none of the properties of a pure substance at the triple point can be varied. Also, based on this rule, a pure substance that exists in a single phase WATER VAPOR T 100 C 150 C 00 C. LIQUID WATER FIGURE According to the Gibbs phase rule, a single-component, two-phase system can have only one independent variable.

332 810 Thermodynamics T, P NH 3 + H O VAPOR g f,nh3 = g g,nh3 g f,h O = g g,h O LIQUID NH 3 + H O FIGURE 16 0 A multicomponent multiphase system is in phase equilibrium when the specific Gibbs function of each component is the same in all phases. 94 T, K VAPOR LIQUID + VAPOR LIQUID % O % N FIGURE 16 1 Equilibrium diagram for the two-phase mixture of oxygen and nitrogen at 0.1 MPa. (PH 1) has two independent variables. In other words, two independent intensive properties need to be specified to fix the equilibrium state of a pure substance in a single phase. Phase Equilibrium for a Multicomponent System Many multiphase systems encountered in practice involve two or more components. A multicomponent multiphase system at a specified temperature and pressure is in phase equilibrium when there is no driving force between the different phases of each component. Thus, for phase equilibrium, the specific Gibbs function of each component must be the same in all phases (Fig. 16 0). That is, g f,1 g g,1 g s,1 for component 1 g f, g g, g s, for component ppppp p g f,n g g,n g s,n for component N We could also derive these relations by using mathematical vigor instead of physical arguments. Some components may exist in more than one solid phase at the specified temperature and pressure. In this case, the specific Gibbs function of each solid phase of a component must also be the same for phase equilibrium. In this section we examine the phase equilibrium of two-component systems that involve two phases (liquid and vapor) in equilibrium. For such systems, C, PH, and thus IV. That is, a two-component, twophase system has two independent variables, and such a system will not be in equilibrium unless two independent intensive properties are fixed. In general, the two phases of a two-component system do not have the same composition in each phase. That is, the mole fraction of a component is different in different phases. This is illustrated in Fig for the twophase mixture of oxygen and nitrogen at a pressure of 0.1 MPa. On this diagram, the vapor line represents the equilibrium composition of the vapor phase at various temperatures, and the liquid line does the same for the liquid phase. At 84 K, for example, the mole fractions are 30 percent nitrogen and 70 percent oxygen in the liquid phase and 66 percent nitrogen and 34 percent oxygen in the vapor phase. Notice that y f,n y f,o y g,n y g,o (16 1a) (16 1b) Therefore, once the temperature and pressure (two independent variables) of a two-component, two-phase mixture are specified, the equilibrium composition of each phase can be determined from the phase diagram, which is based on experimental measurements. It is interesting to note that temperature is a continuous function, but mole fraction (which is a dimensionless concentration), in general, is not. The water and air temperatures at the free surface of a lake, for example, are always the same. The mole fractions of air on the two sides of a water air interface, however, are obviously very different (in fact, the mole fraction of air in water is close to zero). Likewise, the mole fractions of water on the

333 two sides of a water air interface are also different even when air is saturated (Fig. 16 ). Therefore, when specifying mole fractions in two-phase mixtures, we need to clearly specify the intended phase. In most practical applications, the two phases of a mixture are not in phase equilibrium since the establishment of phase equilibrium requires the diffusion of species from higher concentration regions to lower concentration regions, which may take a long time. However, phase equilibrium always exists at the interface of two phases of a species. In the case of air water interface, the mole fraction of water vapor in the air is easily determined from saturation data, as shown in Example The situation is similar at solid liquid interfaces. Again, at a given temperature, only a certain amount of solid can be dissolved in a liquid, and the solubility of the solid in the liquid is determined from the requirement that thermodynamic equilibrium exists between the solid and the solution at the interface. The solubility represents the maximum amount of solid that can be dissolved in a liquid at a specified temperature and is widely available in chemistry handbooks. In Table 16 1 we present sample solubility data for sodium chloride (NaCl) and calcium bicarbonate [Ca(HO 3 ) ] at various temperatures. For example, the solubility of salt (NaCl) in water at 310 K is 36.5 kg per 100 kg of water. Therefore, the mass fraction of salt in the saturated brine is simply mf salt,liquid side m salt m 36.5 kg or 6.7 percent kg whereas the mass fraction of salt in the pure solid salt is mf 1.0. Many processes involve the absorption of a gas into a liquid. Most gases are weakly soluble in liquids (such as air in water), and for such dilute solutions the mole fractions of a species i in the gas and liquid phases at the interface are observed to be proportional to each other. That is, y i,gas side y i,liquid side or P i,gas side Py i,liquid side since y i P i /P for ideal-gas mixtures. This is known as the Henry s law and is expressed as y i,liquid side P i,gas side H (16 ) where H is the Henry s constant, which is the product of the total pressure of the gas mixture and the proportionality constant. For a given species, it is a function of temperature only and is practically independent of pressure for pressures under about 5 atm. Values of the Henry s constant for a number of aqueous solutions are given in Table 16 for various temperatures. From this table and the equation above we make the following observations: 1. The concentration of a gas dissolved in a liquid is inversely proportional to Henry s constant. Therefore, the larger the Henry s constant, the smaller the concentration of dissolved gases in the liquid.. The Henry s constant increases (and thus the fraction of a dissolved gas in the liquid decreases) with increasing temperature. Therefore, the dissolved gases in a liquid can be driven off by heating the liquid (Fig. 16 3). 3. The concentration of a gas dissolved in a liquid is proportional to the partial pressure of the gas. Therefore, the amount of gas dissolved in a liquid can be increased by increasing the pressure of the gas. This can be used to advantage in the carbonation of soft drinks with CO gas. x Jump in concentration Water Chapter Air Concentration profile y H O,gas side y H O,liquid side 1 FIGURE 16 Unlike temperature, the mole fraction of species on the two sides of a liquid gas (or solid gas or solid liquid) interface are usually not the same. TABLE 16 1 Solubility of two inorganic compounds in water at various temperatures, in kg (in 100 kg of water) (from Handbook of Chemistry, McGraw-Hill, 1961) Solute Calcium Tempera- Salt bicarbonate ture, K NaCl Ca(HCO 3 )

334 81 Thermodynamics or or y A,gas side y A,liquid side P A,gas side P Gas A Gas: A Liquid: B y A,liquid side y A,gas side y A,liquid side TABLE 16 Henry s constant H (in bars) for selected gases in water at low to moderate pressures (for gas i, H P i,gas side /y i,water side ) (from Mills, Table A.1, p. 874) Solute 90 K 300 K 310 K 30 K 330 K 340 K H S CO 1,80 1,710,170,70 3,0 O 38,000 45,000 5,000 57,000 61,000 65,000 H 67,000 7,000 75,000 76,000 77,000 76,000 CO 51,000 60,000 67,000 74,000 80,000 84,000 Air 6,000 74,000 84,000 9,000 99, ,000 N 76,000 89, , , ,000 14,000 P A,gas side = Hy A,liquid side FIGURE 16 3 Dissolved gases in a liquid can be driven off by heating the liquid. Strictly speaking, the result obtained from Eq. 16 for the mole fraction of dissolved gas is valid for the liquid layer just beneath the interface, but not necessarily the entire liquid. The latter will be the case only when thermodynamic phase equilibrium is established throughout the entire liquid body. We mentioned earlier that the use of Henry s law is limited to dilute gas liquid solutions, that is, liquids with a small amount of gas dissolved in them. Then the question that arises naturally is, what do we do when the gas is highly soluble in the liquid (or solid), such as ammonia in water? In this case, the linear relationship of Henry s law does not apply, and the mole fraction of a gas dissolved in the liquid (or solid) is usually expressed as a function of the partial pressure of the gas in the gas phase and the temperature. An approximate relation in this case for the mole fractions of a species on the liquid and gas sides of the interface is given by Raoult s law as P i,gas side y i,gas side P total y i,liquid side P i,sat 1T (16 3) where P i,sat (T) is the saturation pressure of the species i at the interface temperature and P total is the total pressure on the gas phase side. Tabular data are available in chemical handbooks for common solutions such as the ammonia water solution that is widely used in absorption-refrigeration systems. Gases may also dissolve in solids, but the diffusion process in this case can be very complicated. The dissolution of a gas may be independent of the structure of the solid, or it may depend strongly on its porosity. Some dissolution processes (such as the dissolution of hydrogen in titanium, similar to the dissolution of CO in water) are reversible, and thus maintaining the gas content in the solid requires constant contact of the solid with a reservoir of that gas. Some other dissolution processes are irreversible. For example, oxygen gas dissolving in titanium forms TiO on the surface, and the process does not reverse itself. The molar density of the gas species i in the solid at the interface r i,solid side is proportional to the partial pressure of the species i in the gas P i,gas side on the gas side of the interface and is expressed as r i,solid side P i,gas side 1kmol>m 3 (16 4)

335 where is the solubility. Expressing the pressure in bars and noting that the unit of molar concentration is kmol of species i per m 3, the unit of solubility is kmol/m 3 bar. Solubility data for selected gas solid combinations are given in Table The product of solubility of a gas and the diffusion coefficient of the gas in a solid is referred to as the permeability, which is a measure of the ability of the gas to penetrate a solid. Permeability is inversely proportional to thickness and has the unit kmol/s m bar. Finally, if a process involves the sublimation of a pure solid such as ice or the evaporation of a pure liquid such as water in a different medium such as air, the mole (or mass) fraction of the substance in the liquid or solid phase is simply taken to be 1.0, and the partial pressure and thus the mole fraction of the substance in the gas phase can readily be determined from the saturation data of the substance at the specified temperature. Also, the assumption of thermodynamic equilibrium at the interface is very reasonable for pure solids, pure liquids, and solutions except when chemical reactions are occurring at the interface. TABLE 16 3 Chapter Solubility of selected gases and solids (from Barrer) (for gas i, r i,solid side /P i,gas side) Gas Solid T K kmol/m 3 bar O Rubber N Rubber CO Rubber He SiO H Ni EXAMPLE 16 8 Mole Fraction of Water Vapor Just over a Lake Determine the mole fraction of the water vapor at the surface of a lake whose temperature is 15 C, and compare it to the mole fraction of water in the lake (Fig. 16 4). Take the atmospheric pressure at lake level to be 9 kpa. Solution The mole fraction of water vapor at the surface of a lake is to be determined and to be compared to the mole fraction of water in the lake. Assumptions 1 Both the air and water vapor are ideal gases. The amount of air dissolved in water is negligible. Properties The saturation pressure of water at 15 C is kpa (Table A 4). Analysis There exists phase equilibrium at the free surface of the lake, and thus the air at the lake surface is always saturated at the interface temperature. The air at the water surface is saturated. Therefore, the partial pressure of water vapor in the air at the lake surface will simply be the saturation pressure of water at 15 C, The mole fraction of water vapor in the air at the surface of the lake is determined from Eq. 16 to be y v P v P P v P 15 C kpa kpa 9 kpa or 1.85 percent Water contains some dissolved air, but the amount is negligible. Therefore, we can assume the entire lake to be liquid water. Then its mole fraction becomes y water,liquid side 1.0 or 100 percent Lake 15 C Air 9 kpa Saturated air y H O,air side = y H O,liquid side 1.0 FIGURE 16 4 Schematic for Example Discussion Note that the concentration of water on a molar basis is 100 percent just beneath the air water interface and less than percent just above it even though the air is assumed to be saturated (so this is the highest value at 15 C). Therefore, large discontinuities can occur in the concentrations of a species across phase boundaries.

336 814 Thermodynamics Air EXAMPLE 16 9 The Amount of Dissolved Air in Water Saturated air P dry air,gas side Determine the mole fraction of air at the surface of a lake whose temperature is 17 C (Fig. 16 5). Take the atmospheric pressure at lake level to be 9 kpa. Lake 17 C y dry air,liquid side FIGURE 16 5 Schematic for Example Solution The mole fraction of air in lake water is to be determined. Assumptions Both the air and vapor are ideal gases. Properties The saturation pressure of water at 17 C is 1.96 kpa (Table A 4). The Henry s constant for air dissolved in water at 90 K is H 6,000 bar (Table 16 ). Analysis This example is similar to the previous example. Again the air at the water surface is saturated, and thus the partial pressure of water vapor in the air at the lake surface is the saturation pressure of water at 17 C, The partial pressure of dry air is P v P 17 C 1.96 kpa P dry air P P v kpa bar Note that we could have ignored the vapor pressure since the amount of vapor in air is so small with little loss in accuracy (an error of about percent). The mole fraction of air in the water is, from Henry s law, y dry air,liquid side P dry air,gas side H bar ,000 bar Discussion This value is very small, as expected. Therefore, the concentration of air in water just below the air water interface is 1.45 moles per 100,000 moles. But obviously this is enough oxygen for fish and other creatures in the lake. Note that the amount of air dissolved in water will decrease with increasing depth unless phase equilibrium exists throughout the entire lake. Hydrogen gas 358 K, 300 kpa H Nickel plate H FIGURE 16 6 Schematic for Example EXAMPLE Diffusion of Hydrogen Gas into a Nickel Plate Consider a nickel plate that is placed into a tank filled with hydrogen gas at 358 K and 300 kpa. Determine the molar and mass density of hydrogen in the nickel plate when phase equilibrium is established (Fig. 16 6). Solution A nickel plate is exposed to hydrogen gas. The density of hydrogen in the plate is to be determined. Properties The molar mass of hydrogen H is M kg/kmol, and the solubility of hydrogen in nickel at the specified temperature is given in Table 16 3 to be kmol/m 3 bar. Analysis Noting that 300 kpa 3 bar, the molar density of hydrogen in the nickel plate is determined from Eq to be r H,solid side P H,gas side It corresponds to a mass density of kmol>m 3 # bar13 bar 0.07 kmol/m 3 r H,solid side r H,solid side M H kmol>m 3 1 kg>kmol kg/m 3 That is, there will be 0.07 kmol (or kg) of H gas in each m 3 volume of nickel plate when phase equilibrium is established.

337 Chapter EXAMPLE Composition of Different Phases of a Mixture In absorption refrigeration systems, a two-phase equilibrium mixture of liquid ammonia (NH 3 ) and water (H O) is frequently used. Consider one such mixture at 40 C, shown in Fig If the composition of the liquid phase is 70 percent NH 3 and 30 percent H O by mole numbers, determine the composition of the vapor phase of this mixture. Solution A two-phase mixture of ammonia and water at a specified temperature is considered. The composition of the liquid phase is given, and the composition of the vapor phase is to be determined. Assumptions The mixture is ideal and thus Raoult s law is applicable. Properties The saturation pressures of H O and NH 3 at 40 C are P H O,sat kpa and P NH3,sat kpa. Analysis The vapor pressures are determined from P H O,gas side y H O,liquid side P H O,sat 1T kpa. kpa P NH3,gas side y NH3,liquid side P NH3,sat 1T kpa kpa VAPOR LIQUID H O + NH 3 40 C y g,h O =? y g,nh3 =? y f,h O = 0.30 y f,nh3 = 0.70 FIGURE 16 7 Schematic for Example The total pressure of the mixture is P total P H O P NH kpa Then the mole fractions in the gas phase are y H O,gas side P H O,gas side. kpa P total kpa y NH3,gas side P NH 3,gas side P total kpa kpa Discussion Note that the gas phase consists almost entirely of ammonia, making this mixture very suitable for absorption refrigeration. SUMMARY An isolated system is said to be in chemical equilibrium if no changes occur in the chemical composition of the system. The criterion for chemical equilibrium is based on the second law of thermodynamics, and for a system at a specified temperature and pressure it can be expressed as For the reaction 1dG T, P 0 n A A n B B n C C n D D where the n s are the stoichiometric coefficients, the equilibrium criterion can be expressed in terms of the Gibbs functions as n C g C n D g D n A g A n B g B 0 which is valid for any chemical reaction regardless of the phases involved. For reacting systems that consist of ideal gases only, the equilibrium constant K P can be expressed as where the standard-state Gibbs function change G*(T) and the equilibrium constant K P are defined as and K P e G*1T >R ut G* 1T n C g * C 1T n D g * D 1T n A g * A 1T n B g * B 1T K P P n C C P n D D P n A A P n B B Here, P i s are the partial pressures of the components in atm. The K P of ideal-gas mixtures can also be expressed in terms of the mole numbers of the components as K P N n C C N n D D N n A A N n B B a P b N total n

338 816 Thermodynamics where n n C n D n A n B, P is the total pressure in atm, and N total is the total number of moles present in the reaction chamber, including any inert gases. The equation above is written for a reaction involving two reactants and two products, but it can be extended to reactions involving any number of reactants and products. The equilibrium constant K P of ideal-gas mixtures depends on temperature only. It is independent of the pressure of the equilibrium mixture, and it is not affected by the presence of inert gases. The larger the K P, the more complete the reaction. Very small values of K P indicate that a reaction does not proceed to any appreciable degree. A reaction with K P 1000 is usually assumed to proceed to completion, and a reaction with K P is assumed not to occur at all. The mixture pressure affects the equilibrium composition, although it does not affect the equilibrium constant K P. The variation of K P with temperature is expressed in terms of other thermochemical properties through the van t Hoff equation d 1ln K P dt h R 1T R u T where h R 1T is the enthalpy of reaction at temperature T. For small temperature intervals, it can be integrated to yield ln K P h R a 1 1 b K P1 R u T 1 T This equation shows that combustion processes are less complete at higher temperatures since K P decreases with temperature for exothermic reactions. Two phases are said to be in phase equilibrium when there is no transformation from one phase to the other. Two phases of a pure substance are in equilibrium when each phase has the same value of specific Gibbs function. That is, g f g g In general, the number of independent variables associated with a multicomponent, multiphase system is given by the Gibbs phase rule, expressed as IV C PH where IV the number of independent variables, C the number of components, and PH the number of phases present in equilibrium. A multicomponent, multiphase system at a specified temperature and pressure is in phase equilibrium when the specific Gibbs function of each component is the same in all phases. For a gas i that is weakly soluble in a liquid (such as air in water), the mole fraction of the gas in the liquid y i,liquid side is related to the partial pressure of the gas P i,gas side by Henry s law y i,liquid side P i,gas side H where H is Henry s constant. When a gas is highly soluble in a liquid (such as ammonia in water), the mole fractions of the species of a two-phase mixture in the liquid and gas phases are given approximately by Raoult s law, expressed as P i,gas side y i,gas side P total y i,liquid side P i,sat 1T where P total is the total pressure of the mixture, P i,sat (T) is the saturation pressure of species i at the mixture temperature, and y i,liquid side and y i,gas side are the mole fractions of species i in the liquid and vapor phases, respectively. REFERENCES AND SUGGESTED READINGS 1. R. M. Barrer. Diffusion in and through Solids. New York: Macmillan, I. Glassman. Combustion. New York: Academic Press, A. M. Kanury. Introduction to Combustion Phenomena. New York: Gordon and Breach, A. F. Mills. Basic Heat and Mass Transfer. Burr Ridge, IL: Richard D. Irwin, J. M. Smith and H. C. Van Ness. Introduction to Chemical Engineering Thermodynamics. 3rd ed. New York: John Wiley & Sons, K. Wark and D. E. Richards. Thermodynamics. 6th ed. New York: McGraw-Hill, 1999.

339 Chapter PROBLEMS* K P and the Equilibrium Composition of Ideal Gases 16 1C Why is the criterion for chemical equilibrium expressed in terms of the Gibbs function instead of entropy? 16 C Is a wooden table in chemical equilibrium with the air? 16 3C Write three different K P relations for reacting idealgas mixtures, and state when each relation should be used. 16 4C 1 The equilibrium constant of the reaction CO O CO at 1000 K and 1 atm is K P1. Express the equilibrium constant of the following reactions at 1000 K in terms of K P1 : (a) CO 1 O CO at 3 atm (b) CO CO 1 O at 1 atm (c) CO O CO 1 O at 1 atm (d) CO O 5N CO 1.5O 5N at 4 atm (e) CO O CO at 1 atm 16 5C The equilibrium constant of the dissociation reaction H H at 3000 K and 1 atm is K P1. Express the equilibrium constants of the following reactions at 3000 K in terms of K P1 : (a) H H at atm (b) H H at 1 atm (c) H 4H at 1 atm (d) H N H N at atm (e) 6H 3H at 4 atm 16 6C Consider a mixture of CO, CO, and O in equilibrium at a specified temperature and pressure. Now the pressure is doubled. (a) (b) Will the equilibrium constant K P change? Will the number of moles of CO, CO, and O change? How? 16 7C Consider a mixture of NO, O, and N in equilibrium at a specified temperature and pressure. Now the pressure is tripled. (a) (b) Will the equilibrium constant K P change? Will the number of moles of NO, O, and N change? How? 16 8C A reaction chamber contains a mixture of CO,CO, and O in equilibrium at a specified temperature and pres- *Problems designated by a C are concept questions, and students are encouraged to answer them all. Problems designated by an E are in English units, and the SI users can ignore them. Problems with a CD-EES icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed DVD. Problems with a computer-ees icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text. sure. How will (a) increasing the temperature at constant pressure and (b) increasing the pressure at constant temperature affect the number of moles of CO? 16 9C A reaction chamber contains a mixture of N and N in equilibrium at a specified temperature and pressure. How will (a) increasing the temperature at constant pressure and (b) increasing the pressure at constant temperature affect the number of moles of N? 16 10C A reaction chamber contains a mixture of CO, CO, and O in equilibrium at a specified temperature and pressure. Now some N is added to the mixture while the mixture temperature and pressure are kept constant. Will this affect the number of moles of O? How? 16 11C Which element is more likely to dissociate into its monatomic form at 3000 K, H or N? Why? 16 1 Using the Gibbs function data, determine the equilibrium constant K P for the reaction H 1 O H O at (a) 98 K and (b) 000 K. Compare your results with the K P values listed in Table A E Using Gibbs function data, determine the equilibrium constant K P for the reaction H 1 O H O at (a) 537 R and (b) 340 R. Compare your results with the K P values listed in Table A 8. Answers: (a) , (b) Determine the equilibrium constant K P for the process CO O CO at (a) 98 K and (b) 000 K. Com- 1 pare your results with the values for K P listed in Table A Study the effect of varying the percent excess air during the steady-flow combustion of hydrogen at a pressure of 1 atm. At what temperature will 97 percent of H burn into H O? Assume the equilibrium mixture consists of H O, H,O, and N Determine the equilibrium constant K P for the reaction CH 4 O CO H O at 5 C. Answer: Using the Gibbs function data, determine the equilibrium constant K P for the dissociation process CO CO 1 O at (a) 98 K and (b) 1800 K. Compare your results with the K P values listed in Table A Using the Gibbs function data, determine the equilibrium constant K P for the reaction H O 1 H OH at 5 C. Compare your result with the K P value listed in Table A Determine the temperature at which 5 percent of diatomic oxygen (O ) dissociates into monatomic oxygen (O) at a pressure of 3 atm. Answer: 3133 K 16 0 Repeat Prob for a pressure of 6 atm.

340 818 Thermodynamics 16 1 Carbon monoxide is burned with 100 percent excess air during a steady-flow process at a pressure of 1 atm. At what temperature will 97 percent of CO burn to CO? Assume the equilibrium mixture consists of CO, CO, O, and N. Answer: 76 K 16 Reconsider Prob Using EES (or other) software, study the effect of varying the percent excess air during the steady-flow process from 0 to 00 percent on the temperature at which 97 percent of CO burns into CO. Plot the temperature against the percent excess air, and discuss the results. 16 3E Repeat Prob using data in English units Hydrogen is burned with 150 percent theoretical air during a steady-flow process at a pressure of 1 atm. At what temperature will 98 percent of H burn to H O? Assume the equilibrium mixture consists of H O, H,O, and N Air (79 percent N and 1 percent O ) is heated to 000 K at a constant pressure of atm. Assuming the equilibrium mixture consists of N,O, and NO, determine the equilibrium composition at this state. Is it realistic to assume that no monatomic oxygen or nitrogen will be present in the equilibrium mixture? Will the equilibrium composition change if the pressure is doubled at constant temperature? 16 6 Hydrogen (H ) is heated to 300 K at a constant pressure of 8 atm. Determine the percentage of H that will dissociate into H during this process. Answer: 5.0 percent 16 7 Carbon dioxide (CO ) is heated to 400 K at a constant pressure of 3 atm. Determine the percentage of CO that will dissociate into CO and O during this process A mixture of 1 mol of CO and 3 mol of O is heated to 00 K at a pressure of atm. Determine the equilibrium composition, assuming the mixture consists of CO, CO, and O. Answers: 0.995CO, 0.005CO,.505O 16 9E A mixture of mol of CO, mol of O, and 6 mol of N is heated to 430 R at a pressure of 3 atm. Determine the equilibrium composition of the mixture. Answers: 1.93CO, 0.07CO, 1.035O, 6N A mixture of 3 mol of N, 1 mol of O, and 0.1 mol of Ar is heated to 400 K at a constant pressure of 10 atm. Assuming the equilibrium mixture consists of N,O, Ar, and NO, determine the equilibrium composition. Answers: 0.083NO,.9589N, O, 0.1Ar Determine the mole fraction of sodium that ionizes according to the reaction Na Na e at 000 K and 0.8 atm (K P for this reaction). Answer: 67.5 percent 16 3 Liquid propane (C 3 H 8 ) enters a combustion chamber at 5 C at a rate of 1. kg/min where it is mixed and burned with 150 percent excess air that enters the combustion chamber at 1 C. If the combustion gases consist of CO,H O, CO, O, and N that exit at 100 K and atm, determine (a) the equilibrium composition of the product gases and (b) the rate of heat transfer from the combustion chamber. Is it realistic to disregard the presence of NO in the product gases? Answers: (a) 3CO, 7.5O, 4H O, 47N, (b) 5066 kj/min C 3 H 8 CO 5 C Combustion chamber 100 K CO H O AIR 1 C atm O N FIGURE P Reconsider Prob Using EES (or other) software, investigate if it is realistic to disregard the presence of NO in the product gases? 16 34E A steady-flow combustion chamber is supplied with CO gas at 560 R and 16 psia at a rate of 1.5 ft 3 /min and with oxygen (O ) at 537 R and 16 psia at a rate of 0.7 lbm/min. The combustion products leave the combustion chamber at 3600 R and 16 psia. If the combustion gases consist of CO, CO, and O, determine (a) the equilibrium composition of the product gases and (b) the rate of heat transfer from the combustion chamber Oxygen (O ) is heated during a steady-flow process at 1 atm from 98 to 3000 K at a rate of 0.5 kg/min. Determine the rate of heat supply needed during this process, assuming (a) some O dissociates into O and (b) no dissociation takes place. O 98 K Q in FIGURE P16 35 O, O 3000 K Estimate K P for the following equilibrium reaction at 500 K: CO H O CO H At 000 K it is known that the enthalpy of reaction is 6176 kj/kmol and K P is Compare your result with the value obtained from the definition of the equilibrium constant A constant-volume tank contains a mixture of 1 kmol H and 1 kmol O at 5 C and 1 atm. The contents are ignited. Determine the final temperature and pressure in the tank when the combustion gases are H O, H, and O.

341 Simultaneous Reactions 16 38C What is the equilibrium criterion for systems that involve two or more simultaneous chemical reactions? 16 39C When determining the equilibrium composition of a mixture involving simultaneous reactions, how would you determine the number of K P relations needed? One mole of H O is heated to 3400 K at a pressure of 1 atm. Determine the equilibrium composition, assuming that only H O, OH, O, and H are present. Answers: 0.574H O, 0.308H, 0.095O, 0.36OH A mixture of mol of CO and 1 mol of O is heated to 300 K at a pressure of atm. Determine the equilibrium composition of the mixture, assuming that only CO, CO, O, and O are present Air (1 percent O, 79 percent N ) is heated to 3000 K at a pressure of atm. Determine the equilibrium composition, assuming that only O,N, O, and NO are present. Is it realistic to assume that no N will be present in the final equilibrium mixture? AIR Q in Reaction chamber FIGURE P16 4 O, N, O, NO 3000 K 16 43E Air (1 percent O, 79 percent N ) is heated to 5400 R at a pressure of 1 atm. Determine the equilibrium composition, assuming that only O,N,O, and NO are present. Is it realistic to assume that no N will be present in the final equilibrium mixture? 16 44E Reconsider Prob E. Use EES (or other) software to obtain the equilibrium solution. Compare your solution technique with that used in Prob E Water vapor (H O) is heated during a steady-flow process at 1 atm from 98 to 3000 K at a rate of 0. kg/min. Determine the rate of heat supply needed during this process, assuming (a) some H O dissociates into H,O, and OH and (b) no dissociation takes place. Answers: (a) 055 kj/min, (b) 1404 kj/min Reconsider Prob Using EES (or other) software, study the effect of the final temperature on the rate of heat supplied for the two cases. Let the final temperature vary from 500 to 3500 K. For each of the two cases, plot the rate of heat supplied as a function final temperature. Chapter Ethyl alcohol (C H 5 OH(g)) at 5 C is burned in a steady-flow adiabatic combustion chamber with 40 percent excess air that also enters at 5 C. Determine the adiabatic flame temperature of the products at 1 atm assuming the significant equilibrium reactions are CO CO O and N O NO. Plot the adiabatic flame temperature and kmoles of CO, CO, and NO at equilibrium for values of percent excess air between 10 and 100 percent. Variations of K P with Temperature 16 48C What is the importance of the van t Hoff equation? 16 49C Will a fuel burn more completely at 000 or 500 K? Estimate the enthalpy of reaction h R for the combustion process of carbon monoxide at 00 K, using (a) enthalpy data and (b) K P data E Estimate the enthalpy of reaction h R for the combustion process of carbon monoxide at 3960 R, using (a) enthalpy data and (b) K P data. Answers: (a) 119,030 Btu/lbmol, (b) 119,041 Btu/lbmol 16 5 Using the enthalpy of reaction h R data and the K P value at 400 K, estimate the K P value of the combustion process H 1 O H O at 600 K. Answer: Estimate the enthalpy of reaction h R for the dissociation process CO CO 1 O at 00 K, using (a) enthalpy data and (b) K P data Estimate the enthalpy of reaction h R for the dissociation process O O at 3100 K, using (a) enthalpy data and (b) K P data. Answers: (a) 513,614 kj/kmol, (b) 51,808 kj/kmol Estimate the enthalpy of reaction for the equilibrium reaction CH 4 O CO H O at 500 K, using (a) enthalpy data and (b) K P data. Phase Equilibrium 16 56C Consider a tank that contains a saturated liquid vapor mixture of water in equilibrium. Some vapor is now allowed to escape the tank at constant temperature and pressure. Will this disturb the phase equilibrium and cause some of the liquid to evaporate? 16 57C Consider a two-phase mixture of ammonia and water in equilibrium. Can this mixture exist in two phases at the same temperature but at a different pressure? 16 58C Using the solubility data of a solid in a specified liquid, explain how you would determine the mole fraction of the solid in the liquid at the interface at a specified temperature C Using solubility data of a gas in a solid, explain how you would determine the molar concentration of the gas in the solid at the solid gas interface at a specified temperature.

342 80 Thermodynamics 16 60C Using the Henry s constant data for a gas dissolved in a liquid, explain how you would determine the mole fraction of the gas dissolved in the liquid at the interface at a specified temperature Show that a mixture of saturated liquid water and saturated water vapor at 100 C satisfies the criterion for phase equilibrium Show that a mixture of saturated liquid water and saturated water vapor at 300 kpa satisfies the criterion for phase equilibrium Show that a saturated liquid vapor mixture of refrigerant-134a at 10 C satisfies the criterion for phase equilibrium Consider a mixture of oxygen and nitrogen in the gas phase. How many independent properties are needed to fix the state of the system? Answer: In absorption refrigeration systems, a two-phase equilibrium mixture of liquid ammonia (NH 3 ) and water (H O) is frequently used. Consider a liquid vapor mixture of ammonia and water in equilibrium at 30 C. If the composition of the liquid phase is 60 percent NH 3 and 40 percent H O by mole numbers, determine the composition of the vapor phase of this mixture. Saturation pressure of NH 3 at 30 C is kpa Consider a liquid vapor mixture of ammonia and water in equilibrium at 5 C. If the composition of the liquid phase is 50 percent NH 3 and 50 percent H O by mole numbers, determine the composition of the vapor phase of this mixture. Saturation pressure of NH 3 at 5 C is kpa. Answers: 0.31 percent, percent A two-phase mixture of ammonia and water is in equilibrium at 50 C. If the composition of the vapor phase is 99 percent NH 3 and 1 percent H O by mole numbers, determine the composition of the liquid phase of this mixture. Saturation pressure of NH 3 at 50 C is kpa Using the liquid vapor equilibrium diagram of an oxygen nitrogen mixture, determine the composition of each phase at 80 K and 100 kpa Using the liquid vapor equilibrium diagram of an oxygen nitrogen mixture, determine the composition of each phase at 84 K and 100 kpa Using the liquid vapor equilibrium diagram of an oxygen nitrogen mixture at 100 kpa, determine the temperature at which the composition of the vapor phase is 79 percent N and 1 percent O. Answer: 8 K Using the liquid vapor equilibrium diagram of an oxygen nitrogen mixture at 100 kpa, determine the temperature at which the composition of the liquid phase is 30 percent N and 70 percent O Consider a rubber plate that is in contact with nitrogen gas at 98 K and 50 kpa. Determine the molar and mass density of nitrogen in the rubber at the interface A wall made of natural rubber separates O and N gases at 5 C and 500 kpa. Determine the molar concentrations of O and N in the wall Consider a glass of water in a room at 7 C and 97 kpa. If the relative humidity in the room is 100 percent and the water and the air are in thermal and phase equilibrium, determine (a) the mole fraction of the water vapor in the air and (b) the mole fraction of air in the water E Water is sprayed into air at 80 F and 14.3 psia, and the falling water droplets are collected in a container on the floor. Determine the mass and mole fractions of air dissolved in the water Consider a carbonated drink in a bottle at 7 C and 130 kpa. Assuming the gas space above the liquid consists of a saturated mixture of CO and water vapor and treating the drink as water, determine (a) the mole fraction of the water vapor in the CO gas and (b) the mass of dissolved CO in a 300-ml drink. Review Problems Using the Gibbs function data, determine the equilibrium constant K P for the dissociation process O O at 000 K. Compare your result with the K P value listed in Table A 8. Answer: A mixture of 1 mol of H and 1 mol of Ar is heated at a constant pressure of 1 atm until 15 percent of H dissociates into monatomic hydrogen (H). Determine the final temperature of the mixture A mixture of 1 mol of H O, mol of O, and 5 mol of N is heated to 00 K at a pressure of 5 atm. Assuming the equilibrium mixture consists of H O, O,N, and H, determine the equilibrium composition at this state. Is it realistic to assume that no OH will be present in the equilibrium mixture? Determine the mole fraction of argon that ionizes according to the reaction Ar Ar e at 10,000 K and 0.35 atm (K P for this reaction) Methane gas (CH 4 ) at 5 C is burned with the stoichiometric amount of air at 5 C during an adiabatic steady-flow combustion process at 1 atm. Assuming the product gases consist of CO, H O, CO, N, and O, determine (a) the equilibrium composition of the product gases and (b) the exit temperature Reconsider Prob Using EES (or other) software, study the effect of excess air on the equilibrium composition and the exit temperature by varying the percent excess air from 0 to 00 percent. Plot the exit temperature against the percent excess air, and discuss the results.

343 16 83 A constant-volume tank contains a mixture of 1 mol of H and 0.5 mol of O at 5 C and 1 atm. The contents of the tank are ignited, and the final temperature and pressure in the tank are 800 K and 5 atm, respectively. If the combustion gases consist of H O, H, and O, determine (a) the equilibrium composition of the product gases and (b) the amount of heat transfer from the combustion chamber. Is it realistic to assume that no OH will be present in the equilibrium mixture? Answers: (a) 0.944H O, 0.056H, 0.08O, (b) 13,574 J/mol H A mixture of mol of H O and 3 mol of O is heated to 3600 K at a pressure of 8 atm. Determine the equilibrium composition of the mixture, assuming that only H O, OH, O, and H are present A mixture of 3 mol of CO and 3 mol of O is heated to 3400 K at a pressure of atm. Determine the equilibrium composition of the mixture, assuming that only CO, CO, O, and O are present. Answers: 1.313CO, 1.687CO, 3.187O, 1.314O Reconsider Prob Using EES (or other) software, study the effect of pressure on the equilibrium composition by varying pressure from 1 atm to 10 atm. Plot the amount of CO present at equilibrium as a function of pressure Estimate the enthalpy of reaction h R for the combustion process of hydrogen at 400 K, using (a) enthalpy data and (b) K P data. Answers: (a) 5,377 kj/kmol, (b) 5,047 kj/kmol Reconsider Prob Using EES (or other) software, investigate the effect of temperature on the enthalpy of reaction using both methods by varying the temperature from 000 to 3000 K Using the enthalpy of reaction h R data and the K P value at 800 K, estimate the K P value of the dissociation process O O at 3000 K Show that when the three phases of a pure substance are in equilibrium, the specific Gibbs function of each phase is the same Show that when the two phases of a two-component system are in equilibrium, the specific Gibbs function of each phase of each component is the same A constant-volume tank initially contains 1 kmol of carbon monoxide CO and 3 kmol of oxygen O (no nitrogen) at 5 C and atm. Now the mixture is ignited and the CO burns completely to carbon dioxide CO. If the final temperature in the tank is 500 K, determine the final pressure in the tank and the amount of heat transfer. Is it realistic to assume that there will be no CO in the tank when chemical equilibrium is reached? Using Henry s law, show that the dissolved gases in a liquid can be driven off by heating the liquid. Chapter Consider a glass of water in a room at 5 C and 100 kpa. If the relative humidity in the room is 70 percent and the water and the air are in thermal equilibrium, determine (a) the mole fraction of the water vapor in the room air, (b) the mole fraction of the water vapor in the air adjacent to the water surface, and (c) the mole fraction of air in the water near the surface Repeat Prob for a relative humidity of 5 percent A carbonated drink is fully charged with CO gas at 17 C and 600 kpa such that the entire bulk of the drink is in thermodynamic equilibrium with the CO water vapor mixture. Now consider a -L soda bottle. If the CO gas in that bottle were to be released and stored in a container at 0 C and 100 kpa, determine the volume of the container Ethyl alcohol (C H 5 OH(g)) at 5 C is burned in a steady-flow adiabatic combustion chamber with 40 percent excess air that also enters at 5 C. Determine the adiabatic flame temperature of the products at 1 atm assuming the only significant equilibrium reaction is CO 1 CO O. Plot the adiabatic flame temperature as the percent excess air varies from 10 to 100 percent Tabulate the natural log of the equilibrium constant as a function of temperature between 98 to 3000 K for the equilibrium reaction CO H O CO H. Compare your results to those obtained by combining the ln K P values for the two equilibrium reactions CO CO 1 1 O and H O H O given in Table A It is desired to control the amount of CO in the products of combustion of octane C 8 H 18 so that the volume fraction of CO in the products is less than 0.1 percent. Determine the percent theoretical air required for the combustion of octane at 5 atm such that the reactant and product temperatures are 98 K and 000 K, respectively. Determine the heat transfer per kmol of octane for this process if the combustion occurs in a steady-flow combustion chamber. Plot the percent theoretical air required for 0.1 percent CO in the products as a function of product pressures between 100 and 300 kpa. Fundamentals of Engineering (FE) Exam Problems If the equilibrium constant for the reaction H 1 O S H O is K, the equilibrium constant for the reaction H O H O at the same temperature is (a) 1/K (b) 1/(K) (c) K (d) K (e) 1/K If the equilibrium constant for the reaction CO 1 O S CO is K, the equilibrium constant for the reaction CO 3N S CO 1 O 3N at the same temperature is (a) 1/K (b) 1/(K 3) (c) 4K (d) K (e) 1/K

344 8 Thermodynamics The equilibrium constant for the reaction H 1 O S H O at 1 atm and 1500 C is given to be K. Of the reactions given below, all at 1500 C, the reaction that has a different equilibrium constant is (a) H 1 O S H O at 5 atm (b) H O H O at 1 atm (c) H O S H O 1 O at atm (d) H 1 O 3N S H O 3N at 5 atm (e) H 1 O 3N S H O 3N at 1 atm Of the reactions given below, the reaction whose equilibrium composition at a specified temperature is not affected by pressure is (a) H 1 O S H O (b) CO 1 O S CO (c) N O NO (d) N N (e) all of the above Of the reactions given below, the reaction whose number of moles of products increases by the addition of inert gases into the reaction chamber at constant pressure and temperature is (a) H 1 O S H O (b) CO 1 O S CO (c) N O NO (d) N N (e) all of the above Moist air is heated to a very high temperature. If the equilibrium composition consists of H O, O,N, OH, H, and NO, the number of equilibrium constant relations needed to determine the equilibrium composition of the mixture is (a) 1 (b) (c) 3 (d) 4 (e) Propane C 3 H 8 is burned with air, and the combustion products consist of CO, CO, H O, O,N, OH, H, and NO. The number of equilibrium constant relations needed to determine the equilibrium composition of the mixture is (a) 1 (b) (c) 3 (d) 4 (e) Consider a gas mixture that consists of three components. The number of independent variables that need to be specified to fix the state of the mixture is (a) 1 (b) (c) 3 (d) 4 (e) The value of Henry s constant for CO gas dissolved in water at 90 K is 1.8 MPa. Consider water exposed to atmospheric air at 100 kpa that contains 3 percent CO by volume. Under phase equilibrium conditions, the mole fraction of CO gas dissolved in water at 90 K is (a) (b) (c) (d) (e) The solubility of nitrogen gas in rubber at 5 C is kmol/m 3 bar. When phase equilibrium is established, the density of nitrogen in a rubber piece placed in a nitrogen gas chamber at 800 kpa is (a) 0.01 kg/m 3 (b) 0.35 kg/m 3 (c) 0.4 kg/m 3 (d) 0.56 kg/m 3 (e) kg/m 3 Design and Essay Problems An article that appeared in the Reno Gazette- Journal on May 18, 199, quotes an inventor as saying that he has turned water into motor vehicle fuel in a breakthrough that would increase engine efficiency, save gasoline, and reduce smog. There is also a picture of a car that the inventor has modified to run on half water and half gasoline. The inventor claims that sparks from catalytic poles in the converted engine break down the water into oxygen and hydrogen, which is burned with the gasoline. He adds that hydrogen has a higher energy density than carbon and the high-energy density enables one to get more power. The inventor states that the fuel efficiency of his car increased from 0 mpg (miles per gallon) to more than 50 mpg of gasoline as a result of conversion and notes that the conversion has sharply reduced emissions of hydrocarbons, carbon monoxide, and other exhaust pollutants. Evaluate the claims made by the inventor, and write a report that is to be submitted to a group of investors who are considering financing this invention Automobiles are major emitters of air pollutants such as NO x, CO, and hydrocarbons HC. Find out the legal limits of these pollutants in your area, and estimate the total amount of each pollutant, in kg, that would be produced in your town if all the cars were emitting pollutants at the legal limit. State your assumptions.

345 Chapter 17 COMPRESSIBLE FLOW For the most part, we have limited our consideration so far to flows for which density variations and thus compressibility effects are negligible. In this chapter we lift this limitation and consider flows that involve significant changes in density. Such flows are called compressible flows, and they are frequently encountered in devices that involve the flow of gases at very high velocities. Compressible flow combines fluid dynamics and thermodynamics in that both are necessary to the development of the required theoretical background. In this chapter, we develop the general relations associated with one-dimensional compressible flows for an ideal gas with constant specific heats. We start this chapter by introducing the concepts of stagnation state, speed of sound, and Mach number for compressible flows. The relationships between the static and stagnation fluid properties are developed for isentropic flows of ideal gases, and they are expressed as functions of specificheat ratios and the Mach number. The effects of area changes for one-dimensional isentropic subsonic and supersonic flows are discussed. These effects are illustrated by considering the isentropic flow through converging and converging diverging nozzles. The concept of shock waves and the variation of flow properties across normal and oblique shocks are discussed. Finally, we consider the effects of heat transfer on compressible flows and examine steam nozzles. Objectives The objectives of Chapter 17 are to: Develop the general relations for compressible flows encountered when gases flow at high speeds. Introduce the concepts of stagnation state, speed of sound, and Mach number for a compressible fluid. Develop the relationships between the static and stagnation fluid properties for isentropic flows of ideal gases. Derive the relationships between the static and stagnation fluid properties as functions of specific-heat ratios and Mach number. Derive the effects of area changes for one-dimensional isentropic subsonic and supersonic flows. Solve problems of isentropic flow through converging and converging diverging nozzles. Discuss the shock wave and the variation of flow properties across the shock wave. Develop the concept of duct flow with heat transfer and negligible friction known as Rayleigh flow. Examine the operation of steam nozzles commonly used in steam turbines. 83

346 84 Thermodynamics FIGURE 17 1 Aircraft and jet engines involve high speeds, and thus the kinetic energy term should always be considered when analyzing them. (a) Photo courtesy of NASA, jpeg, and (b) Figure courtesy of Pratt and Whitney. Used by permission. h 1 h V Control 1 V volume h 01 h 0 = h 01 FIGURE 17 Steady flow of a fluid through an adiabatic duct STAGNATION PROPERTIES When analyzing control volumes, we find it very convenient to combine the internal energy and the flow energy of a fluid into a single term, enthalpy, defined per unit mass as h u Pv. Whenever the kinetic and potential energies of the fluid are negligible, as is often the case, the enthalpy represents the total energy of a fluid. For high-speed flows, such as those encountered in jet engines (Fig. 17 1), the potential energy of the fluid is still negligible, but the kinetic energy is not. In such cases, it is convenient to combine the enthalpy and the kinetic energy of the fluid into a single term called stagnation (or total) enthalpy h 0, defined per unit mass as (17 1) When the potential energy of the fluid is negligible, the stagnation enthalpy represents the total energy of a flowing fluid stream per unit mass. Thus it simplifies the thermodynamic analysis of high-speed flows. Throughout this chapter the ordinary enthalpy h is referred to as the static enthalpy, whenever necessary, to distinguish it from the stagnation enthalpy. Notice that the stagnation enthalpy is a combination property of a fluid, just like the static enthalpy, and these two enthalpies become identical when the kinetic energy of the fluid is negligible. Consider the steady flow of a fluid through a duct such as a nozzle, diffuser, or some other flow passage where the flow takes place adiabatically and with no shaft or electrical work, as shown in Fig. 17. Assuming the fluid experiences little or no change in its elevation and its potential energy, the energy balance relation (E. in E. out ) for this single-stream steady-flow system reduces to or h 0 h V 1kJ>kg h 1 V 1 h V (17 ) h 01 h 0 (17 3) That is, in the absence of any heat and work interactions and any changes in potential energy, the stagnation enthalpy of a fluid remains constant during a steady-flow process. Flows through nozzles and diffusers usually satisfy these conditions, and any increase in fluid velocity in these devices creates an equivalent decrease in the static enthalpy of the fluid. If the fluid were brought to a complete stop, then the velocity at state would be zero and Eq. 17 would become h 1 V 1 h h 0 Thus the stagnation enthalpy represents the enthalpy of a fluid when it is brought to rest adiabatically. During a stagnation process, the kinetic energy of a fluid is converted to enthalpy (internal energy flow energy), which results in an increase in the fluid temperature and pressure (Fig. 17 3). The properties of a fluid at the stagnation state are called stagnation properties (stagnation temperature,

347 stagnation pressure, stagnation density, etc.). The stagnation state and the stagnation properties are indicated by the subscript 0. The stagnation state is called the isentropic stagnation state when the stagnation process is reversible as well as adiabatic (i.e., isentropic). The entropy of a fluid remains constant during an isentropic stagnation process. The actual (irreversible) and isentropic stagnation processes are shown on the h-s diagram in Fig Notice that the stagnation enthalpy of the fluid (and the stagnation temperature if the fluid is an ideal gas) is the same for both cases. However, the actual stagnation pressure is lower than the isentropic stagnation pressure since entropy increases during the actual stagnation process as a result of fluid friction. The stagnation processes are often approximated to be isentropic, and the isentropic stagnation properties are simply referred to as stagnation properties. When the fluid is approximated as an ideal gas with constant specific heats, its enthalpy can be replaced by c p T and Eq can be expressed as or c p T 0 c p T V T 0 T V (17 4) Here T 0 is called the stagnation (or total) temperature, and it represents the temperature an ideal gas attains when it is brought to rest adiabatically. The term V /c p corresponds to the temperature rise during such a process and is called the dynamic temperature. For example, the dynamic temperature of air flowing at 100 m/s is (100 m/s) /( kj/kg K) 5.0 K. Therefore, when air at 300 K and 100 m/s is brought to rest adiabatically (at the tip of a temperature probe, for example), its temperature rises to the stagnation value of 305 K (Fig. 17 5). Note that for low-speed flows, the stagnation and static (or ordinary) temperatures are practically the same. But for high-speed flows, the temperature measured by a stationary probe placed in the fluid (the stagnation temperature) may be significantly higher than the static temperature of the fluid. The pressure a fluid attains when brought to rest isentropically is called the stagnation pressure P 0. For ideal gases with constant specific heats, P 0 is related to the static pressure of the fluid by P 0 P a T 0 T b k>1k 1 (17 5) By noting that r 1/v and using the isentropic relation Pv k P 0 v 0 k, the ratio of the stagnation density to static density can be expressed as r 0 r a T 0 T b 1>1k 1 (17 6) When stagnation enthalpies are used, there is no need to refer explicitly to kinetic energy. Then the energy balance E # in E # out for a single-stream, steady-flow device can be expressed as c p q in w in 1h 01 gz 1 q out w out 1h 0 gz (17 7) Chapter FIGURE 17 3 Kinetic energy is converted to enthalpy during a stagnation process. Reprinted with special permission of King Features Syndicate. h h0 h Isentropic stagnation state V P 0 P 0,act Actual stagnation state P Actual state FIGURE 17 4 The actual state, actual stagnation state, and isentropic stagnation state of a fluid on an h-s diagram. s

348 86 Thermodynamics Temperature rise during stagnation AIR 100 m/s 305 K 300 K FIGURE 17 5 The temperature of an ideal gas flowing at a velocity V rises by V /c p when it is brought to a complete stop. T 1 = 55.7 K P 1 = kpa V 1 = 50 m/s Diffuser FIGURE 17 6 Schematic for Example Compressor Aircraft engine where h 01 and h 0 are the stagnation enthalpies at states 1 and, respectively. When the fluid is an ideal gas with constant specific heats, Eq becomes 1q in q out 1w in w out c p 1T 0 T 01 g 1z z 1 (17 8) where T 01 and T 0 are the stagnation temperatures. Notice that kinetic energy terms do not explicitly appear in Eqs and 17 8, but the stagnation enthalpy terms account for their contribution. EXAMPLE 17 1 Compression of High-Speed Air in an Aircraft An aircraft is flying at a cruising speed of 50 m/s at an altitude of 5000 m where the atmospheric pressure is kpa and the ambient air temperature is 55.7 K. The ambient air is first decelerated in a diffuser before it enters the compressor (Fig. 17 6). Assuming both the diffuser and the compressor to be isentropic, determine (a) the stagnation pressure at the compressor inlet and (b) the required compressor work per unit mass if the stagnation pressure ratio of the compressor is 8. Solution High-speed air enters the diffuser and the compressor of an aircraft. The stagnation pressure of air and the compressor work input are to be determined. Assumptions 1 Both the diffuser and the compressor are isentropic. Air is an ideal gas with constant specific heats at room temperature. Properties The constant-pressure specific heat c p and the specific heat ratio k of air at room temperature are (Table A a) Analysis (a) Under isentropic conditions, the stagnation pressure at the compressor inlet (diffuser exit) can be determined from Eq However, first we need to find the stagnation temperature T 01 at the compressor inlet. Under the stated assumptions, T 01 can be determined from Eq to be 86.8 K Then from Eq. 17 5, c p kj>kg # K and k 1.4 T 01 T 1 V 1 c p 55.7 K P 01 P 1 a T k>1k 1 01 b kpa a 86.8 K 1.4> T K b kpa 150 m>s kj>kg # K a 1 kj>kg 1000 m >s b That is, the temperature of air would increase by 31.1 C and the pressure by 6.7 kpa as air is decelerated from 50 m/s to zero velocity. These increases in the temperature and pressure of air are due to the conversion of the kinetic energy into enthalpy. (b) To determine the compressor work, we need to know the stagnation temperature of air at the compressor exit T 0. The stagnation pressure ratio across the compressor P 0 /P 01 is specified to be 8. Since the compression process is assumed to be isentropic, T 0 can be determined from the idealgas isentropic relation (Eq. 17 5): T 0 T 01 a P 1k 1>k 0 b K > K P 01

349 Chapter Disregarding potential energy changes and heat transfer, the compressor work per unit mass of air is determined from Eq. 17 8: w in c p 1T 0 T kj>kg # K K 86.8 K 33.9 kj/kg Thus the work supplied to the compressor is 33.9 kj/kg. Discussion Notice that using stagnation properties automatically accounts for any changes in the kinetic energy of a fluid stream. 17 SPEED OF SOUND AND MACH NUMBER An important parameter in the study of compressible flow is the speed of sound (or the sonic speed), which is the speed at which an infinitesimally small pressure wave travels through a medium. The pressure wave may be caused by a small disturbance, which creates a slight rise in local pressure. To obtain a relation for the speed of sound in a medium, consider a pipe that is filled with a fluid at rest, as shown in Fig A piston fitted in the pipe is now moved to the right with a constant incremental velocity dv, creating a sonic wave. The wave front moves to the right through the fluid at the speed of sound c and separates the moving fluid adjacent to the piston from the fluid still at rest. The fluid to the left of the wave front experiences an incremental change in its thermodynamic properties, while the fluid on the right of the wave front maintains its original thermodynamic properties, as shown in Fig To simplify the analysis, consider a control volume that encloses the wave front and moves with it, as shown in Fig To an observer traveling with the wave front, the fluid to the right will appear to be moving toward the wave front with a speed of c and the fluid to the left to be moving away from the wave front with a speed of c dv. Of course, the observer will think the control volume that encloses the wave front (and herself or himself) is stationary, and the observer will be witnessing a steady-flow process. The mass balance for this single-stream, steady-flow process can be expressed as or rac 1r dra 1c dv By canceling the cross-sectional (or flow) area A and neglecting the higherorder terms, this equation reduces to No heat or work crosses the boundaries of the control volume during this steady-flow process, and the potential energy change, if any, can be neglected. Then the steady-flow energy balance e in e out becomes h c m # right m # left c dr r dv 0 h dh 1c dv (a) V 0 P Piston dv h + dh P + dp c h P r + dr r dv P + dp Moving wave front Stationary fluid FIGURE 17 7 Propagation of a small pressure wave along a duct. Control volume traveling with the wave front h + dh c dv c h P + dp P r + dr r FIGURE 17 8 Control volume moving with the small pressure wave along a duct. P x x

350 88 Thermodynamics which yields dh c dv 0 where we have neglected the second-order term dv. The amplitude of the ordinary sonic wave is very small and does not cause any appreciable change in the pressure and temperature of the fluid. Therefore, the propagation of a sonic wave is not only adiabatic but also very nearly isentropic. Then the second Tdsrelation developed in Chapter 7 reduces to T ds 0 dh dp r (b) or dh dp r (c) Combining Eqs. a, b, and c yields the desired expression for the speed of sound as c dp at s constant dr or c a 0P (17 9) 0r b s It is left as an exercise for the reader to show, by using thermodynamic property relations (see Chap. 1) that Eq can also be written as c k a 0P 0r b T (17 10) where k is the specific heat ratio of the fluid. Note that the speed of sound in a fluid is a function of the thermodynamic properties of that fluid. When the fluid is an ideal gas (P rrt), the differentiation in Eq can easily be performed to yield AIR HELIUM or c k a 0P 0r b k c 0 1rRT d krt T 0r T 84 m/s 347 m/s 634 m/s 00 K 300 K 1000 K 83 m/s 1019 m/s 1861 m/s c krt (17 11) Noting that the gas constant R has a fixed value for a specified ideal gas and the specific heat ratio k of an ideal gas is, at most, a function of temperature, we see that the speed of sound in a specified ideal gas is a function of temperature alone (Fig. 17 9). A second important parameter in the analysis of compressible fluid flow is the Mach number Ma, named after the Austrian physicist Ernst Mach ( ). It is the ratio of the actual velocity of the fluid (or an object in still air) to the speed of sound in the same fluid at the same state: FIGURE 17 9 The speed of sound changes with temperature and varies with the fluid. Ma V c (17 1) Note that the Mach number depends on the speed of sound, which depends on the state of the fluid. Therefore, the Mach number of an aircraft cruising

351 at constant velocity in still air may be different at different locations (Fig ). Fluid flow regimes are often described in terms of the flow Mach number. The flow is called sonic when Ma 1, subsonic when Ma 1, supersonic when Ma 1, hypersonic when Ma 1, and transonic when Ma 1. EXAMPLE 17 Mach Number of Air Entering a Diffuser Air enters a diffuser shown in Fig with a velocity of 00 m/s. Determine (a) the speed of sound and (b) the Mach number at the diffuser inlet when the air temperature is 30 C. Solution Air enters a diffuser with a high velocity. The speed of sound and the Mach number are to be determined at the diffuser inlet. Assumptions Air at specified conditions behaves as an ideal gas. Properties The gas constant of air is R 0.87 kj/kg K, and its specific heat ratio at 30 C is 1.4 (Table A a). Analysis We note that the speed of sound in a gas varies with temperature, which is given to be 30 C. (a) The speed of sound in air at 30 C is determined from Eq to be c krt B kj>kg # K1303 Ka 1000 m >s (b) Then the Mach number becomes Ma V c 00 m>s m>s 1 kj>kg b 349 m /s Discussion The flow at the diffuser inlet is subsonic since Ma 1. AIR 00 K AIR 300 K Chapter V = 30 m/s Ma = 1.13 V = 30 m/s Ma = 0.9 FIGURE The Mach number can be different at different temperatures even if the velocity is the same. AIR V = 00 m/s T = 30 C Diffuser FIGURE Schematic for Example ONE-DIMENSIONAL ISENTROPIC FLOW During fluid flow through many devices such as nozzles, diffusers, and turbine blade passages, flow quantities vary primarily in the flow direction only, and the flow can be approximated as one-dimensional isentropic flow with good accuracy. Therefore, it merits special consideration. Before presenting a formal discussion of one-dimensional isentropic flow, we illustrate some important aspects of it with an example. EXAMPLE 17 3 Gas Flow through a Converging Diverging Duct Carbon dioxide flows steadily through a varying cross-sectional-area duct such as a nozzle shown in Fig at a mass flow rate of 3 kg/s. The carbon dioxide enters the duct at a pressure of 1400 kpa and 00 C with a low velocity, and it expands in the nozzle to a pressure of 00 kpa. The duct is designed so that the flow can be approximated as isentropic. Determine the density, velocity, flow area, and Mach number at each location along the duct that corresponds to a pressure drop of 00 kpa. Solution Carbon dioxide enters a varying cross-sectional-area duct at specified conditions. The flow properties are to be determined along the duct. Stagnation region: 1400 kpa 00 C CO m kg/s P, kpa FIGURE 17 1 Schematic for Example 17 3.

352 830 Thermodynamics Assumptions 1 Carbon dioxide is an ideal gas with constant specific heats at room temperature. Flow through the duct is steady, one-dimensional, and isentropic. Properties For simplicity we use c p kj/kg K and k 1.89 throughout the calculations, which are the constant-pressure specific heat and specific heat ratio values of carbon dioxide at room temperature. The gas constant of carbon dioxide is R kj/kg K (Table A a). Analysis We note that the inlet temperature is nearly equal to the stagnation temperature since the inlet velocity is small. The flow is isentropic, and thus the stagnation temperature and pressure throughout the duct remain constant. Therefore, and To illustrate the solution procedure, we calculate the desired properties at the location where the pressure is 100 kpa, the first location that corresponds to a pressure drop of 00 kpa. From Eq. 17 5, From Eq. 17 4, From the ideal-gas relation, From the mass flow rate relation, From Eqs and 17 1, 1000 m c krt kj>kg # >s K1457 K a b m>s B 1 kj>kg Ma V c T T 0 a P 1k 1>k b P 0 V c p 1T 0 T A m# rv B kj>kg # K1473 K 457 K a 1000 m >s 3 1 kj>kg b m/s r P RT 3 kg>s kg>m m>s m 13.1 cm m>s m>s T 0 T 1 00 C 473 K P 0 P kpa > kpa 1473 Ka 1400 kpa b 457 K 100 kpa kpa # m 3 >kg # K1457 K 13.9 kg/m 3 The results for the other pressure steps are summarized in Table 17 1 and are plotted in Fig Discussion Note that as the pressure decreases, the temperature and speed of sound decrease while the fluid velocity and Mach number increase in the flow direction. The density decreases slowly at first and rapidly later as the fluid velocity increases.

353 Chapter TABLE 17 1 Variation of fluid properties in flow direction in duct described in Example 17 3 for m. 3 kg/s constant P, kpa T, K V, m/s r, kg/m 3 c, m/s A, cm Ma * * 767 kpa is the critical pressure where the local Mach number is unity. Flow direction A A, Ma, r, T, V T r Ma 1400 V P, kpa FIGURE Variation of normalized fluid properties and cross-sectional area along a duct as the pressure drops from 1400 to 00 kpa. We note from Example 17 3 that the flow area decreases with decreasing pressure up to a critical-pressure value where the Mach number is unity, and then it begins to increase with further reductions in pressure. The Mach number is unity at the location of smallest flow area, called the throat (Fig ). Note that the velocity of the fluid keeps increasing after passing the throat although the flow area increases rapidly in that region. This increase in velocity past the throat is due to the rapid decrease in the fluid density. The flow area of the duct considered in this example first decreases and then increases. Such ducts are called converging diverging nozzles. These nozzles are used to accelerate gases to supersonic speeds and should not be confused with Venturi nozzles, which are used strictly for incompressible flow. The first use of such a nozzle occurred in 1893 in a steam turbine

354 83 Thermodynamics Fluid Fluid Throat Converging nozzle Throat Converging diverging nozzle FIGURE The cross section of a nozzle at the smallest flow area is called the throat. designed by a Swedish engineer, Carl G. B. de Laval ( ), and therefore converging diverging nozzles are often called Laval nozzles. Variation of Fluid Velocity with Flow Area It is clear from Example 17 3 that the couplings among the velocity, density, and flow areas for isentropic duct flow are rather complex. In the remainder of this section we investigate these couplings more thoroughly, and we develop relations for the variation of static-to-stagnation property ratios with the Mach number for pressure, temperature, and density. We begin our investigation by seeking relationships among the pressure, temperature, density, velocity, flow area, and Mach number for onedimensional isentropic flow. Consider the mass balance for a steady-flow process: m # rav constant Differentiating and dividing the resultant equation by the mass flow rate, we obtain dr r da A dv V 0 (17 13) CONSERVATION OF ENERGY (steady flow, w = 0, q = 0, pe = 0) h 1 + V 1 = h + V or h + V = constant Differentiate, dh + V dv = 0 Also, 0 (isentropic) T ds = dh v dp dh = v dp = r 1 dp Substitute, dp r + V dv = 0 FIGURE Derivation of the differential form of the energy equation for steady isentropic flow. Neglecting the potential energy, the energy balance for an isentropic flow with no work interactions can be expressed in the differential form as (Fig ) (17 14) This relation is also the differential form of Bernoulli s equation when changes in potential energy are negligible, which is a form of the conservation of momentum principle for steady-flow control volumes. Combining Eqs and gives da A dp r a 1 dr V dp b (17 15) Rearranging Eq as ( r/ P) s 1/c and substituting into Eq yield da A dp r V dv 0 dp rv 11 Ma (17 16) This is an important relation for isentropic flow in ducts since it describes the variation of pressure with flow area. We note that A, r, and V are positive quantities. For subsonic flow (Ma 1), the term 1 Ma is positive; and thus da and dp must have the same sign. That is, the pressure of the fluid must increase as the flow area of the duct increases and must decrease as the flow area of the duct decreases. Thus, at subsonic velocities, the pressure decreases in converging ducts (subsonic nozzles) and increases in diverging ducts (subsonic diffusers). In supersonic flow (Ma 1), the term 1 Ma is negative, and thus da and dp must have opposite signs. That is, the pressure of the fluid must

355 increase as the flow area of the duct decreases and must decrease as the flow area of the duct increases. Thus, at supersonic velocities, the pressure decreases in diverging ducts (supersonic nozzles) and increases in converging ducts (supersonic diffusers). Another important relation for the isentropic flow of a fluid is obtained by substituting rv dp/dv from Eq into Eq : Chapter da A dv V 11 Ma (17 17) This equation governs the shape of a nozzle or a diffuser in subsonic or supersonic isentropic flow. Noting that A and V are positive quantities, we conclude the following: For subsonic flow 1Ma 6 1, For supersonic flow 1Ma 7 1, For sonic flow 1Ma 1, da dv 6 0 da dv 7 0 da dv 0 Thus the proper shape of a nozzle depends on the highest velocity desired relative to the sonic velocity. To accelerate a fluid, we must use a converging nozzle at subsonic velocities and a diverging nozzle at supersonic velocities. The velocities encountered in most familiar applications are well below the sonic velocity, and thus it is natural that we visualize a nozzle as a converging duct. However, the highest velocity we can achieve by a converging nozzle is the sonic velocity, which occurs at the exit of the nozzle. If we extend the converging nozzle by further decreasing the flow area, in hopes of accelerating the fluid to supersonic velocities, as shown in Fig , we are up for disappointment. Now the sonic velocity will occur at the exit of the converging extension, instead of the exit of the original nozzle, and the mass flow rate through the nozzle will decrease because of the reduced exit area. Based on Eq , which is an expression of the conservation of mass and energy principles, we must add a diverging section to a converging nozzle to accelerate a fluid to supersonic velocities. The result is a converging diverging nozzle. The fluid first passes through a subsonic (converging) section, where the Mach number increases as the flow area of the nozzle decreases, and then reaches the value of unity at the nozzle throat. The fluid continues to accelerate as it passes through a supersonic (diverging) section. Noting that ṁ rav for steady flow, we see that the large decrease in density makes acceleration in the diverging section possible. An example of this type of flow is the flow of hot combustion gases through a nozzle in a gas turbine. The opposite process occurs in the engine inlet of a supersonic aircraft. The fluid is decelerated by passing it first through a supersonic diffuser, which has a flow area that decreases in the flow direction. Ideally, the flow reaches a Mach number of unity at the diffuser throat. The fluid is further P 0, T 0 Converging MaA = 1 nozzle (sonic) P 0, T 0 Converging nozzle A A Ma A < 1 B Attachment MaB = 1 (sonic) FIGURE We cannot obtain supersonic velocities by attaching a converging section to a converging nozzle. Doing so will only move the sonic cross section farther downstream and decrease the mass flow rate.

356 834 Thermodynamics Ma < 1 P decreases V increases Ma increases T decreases r decreases Ma < 1 P increases V decreases Ma decreases T increases r increases Subsonic nozzle Subsonic diffuser (a) Subsonic flow FIGURE Variation of flow properties in subsonic and supersonic nozzles and diffusers. Ma > 1 P decreases V increases Ma increases T decreases r decreases Supersonic nozzle Ma > 1 (b) Supersonic flow P increases V decreases Ma decreases T increases r increases Supersonic diffuser decelerated in a subsonic diffuser, which has a flow area that increases in the flow direction, as shown in Fig Property Relations for Isentropic Flow of Ideal Gases Next we develop relations between the static properties and stagnation properties of an ideal gas in terms of the specific heat ratio k and the Mach number Ma. We assume the flow is isentropic and the gas has constant specific heats. The temperature T of an ideal gas anywhere in the flow is related to the stagnation temperature T 0 through Eq. 17 4: or Noting that c p kr/(k 1), c krt, and Ma V/c, we see that Substituting yields V c p T V 3kR>1k 14T T 0 T V c p T 0 V 1 T c p T a k 1 b V c k 1 a b Ma which is the desired relation between T 0 and T. T 0 T k 1 1 a b Ma (17 18)

357 The ratio of the stagnation to static pressure is obtained by substituting Eq into Eq. 17 5: P 0 P k>1k 1 k 1 c 1 a b Ma d (17 19) The ratio of the stagnation to static density is obtained by substituting Eq into Eq. 17 6: T t Pt rt Chapter Subsonic nozzle Throat T * P * r * (if Ma t = 1) r 0 r 1>1k 1 k 1 c 1 a b Ma d (17 0) Numerical values of T/T 0, P/P 0, and r/r 0 are listed versus the Mach number in Table A 3 for k 1.4, which are very useful for practical compressible flow calculations involving air. The properties of a fluid at a location where the Mach number is unity (the throat) are called critical properties, and the ratios in Eqs. (17 18) through (17 0) are called critical ratios (Fig ). It is common practice in the analysis of compressible flow to let the superscript asterisk (*) represent the critical values. Setting Ma 1 in Eqs through 17 0 yields T* T 0 k 1 k>1k 1 P* a P 0 k 1 b 1>1k 1 r* a r 0 k 1 b (17 1) (17 ) (17 3) These ratios are evaluated for various values of k and are listed in Table 17. The critical properties of compressible flow should not be confused with the properties of substances at the critical point (such as the critical temperature T c and critical pressure P c ). T o P ro o Throat T *, P *, r * (Ma t = 1) Supersonic nozzle FIGURE When Ma t 1, the properties at the nozzle throat become the critical properties. TABLE 17 The critical-pressure, critical-temperature, and critical-density ratios for isentropic flow of some ideal gases P* P 0 T* T 0 r* r 0 Superheated Hot products Monatomic steam, of combustion, Air, gases, k 1.3 k 1.33 k 1.4 k

358 836 Thermodynamics EXAMPLE 17 4 Critical Temperature and Pressure in Gas Flow P 0 = 1.4 MPa T 0 = 473 K P * T * CO FIGURE Schematic for Example Calculate the critical pressure and temperature of carbon dioxide for the flow conditions described in Example 17 3 (Fig ). Solution For the flow discussed in Example 17 3, the critical pressure and temperature are to be calculated. Assumptions 1 The flow is steady, adiabatic, and one-dimensional. Carbon dioxide is an ideal gas with constant specific heats. Properties The specific heat ratio of carbon dioxide at room temperature is k 1.89 (Table A a). Analysis The ratios of critical to stagnation temperature and pressure are determined to be T* T 0 k k>1k > P* a P 0 k 1 b a b Noting that the stagnation temperature and pressure are, from Example 17 3, T K and P kpa, we see that the critical temperature and pressure in this case are T* T K 413 K P* P kpa 767 kpa Discussion Note that these values agree with those listed in Table 17 1, as expected. Also, property values other than these at the throat would indicate that the flow is not critical, and the Mach number is not unity. Reservoir P r = P 0 P e T r = T 0 V r = 0 P/P 0 1 x 1 P b (Back pressure) P b = P 0 P b > P* 17 4 ISENTROPIC FLOW THROUGH NOZZLES Converging or converging diverging nozzles are found in many engineering applications including steam and gas turbines, aircraft and spacecraft propulsion systems, and even industrial blasting nozzles and torch nozzles. In this section we consider the effects of back pressure (i.e., the pressure applied at the nozzle discharge region) on the exit velocity, the mass flow rate, and the pressure distribution along the nozzle. P* P 0 0 Lowest exit pressure FIGURE 17 0 The effect of back pressure on the pressure distribution along a converging nozzle P b =P* P b < P* P b =0 x Converging Nozzles Consider the subsonic flow through a converging nozzle as shown in Fig The nozzle inlet is attached to a reservoir at pressure P r and temperature T r. The reservoir is sufficiently large so that the nozzle inlet velocity is negligible. Since the fluid velocity in the reservoir is zero and the flow through the nozzle is approximated as isentropic, the stagnation pressure and stagnation temperature of the fluid at any cross section through the nozzle are equal to the reservoir pressure and temperature, respectively.

359 Now we begin to reduce the back pressure and observe the resulting effects on the pressure distribution along the length of the nozzle, as shown in Fig If the back pressure P b is equal to P 1, which is equal to P r, there is no flow and the pressure distribution is uniform along the nozzle. When the back pressure is reduced to P, the exit plane pressure P e also drops to P. This causes the pressure along the nozzle to decrease in the flow direction. When the back pressure is reduced to P 3 ( P*, which is the pressure required to increase the fluid velocity to the speed of sound at the exit plane or throat), the mass flow reaches a maximum value and the flow is said to be choked. Further reduction of the back pressure to level P 4 or below does not result in additional changes in the pressure distribution, or anything else along the nozzle length. Under steady-flow conditions, the mass flow rate through the nozzle is constant and can be expressed as Chapter m # rav a P RT b A 1MakRT PAMa k B RT Solving for T from Eq and for P from Eq and substituting, m # AMaP 0 k>1rt k 1Ma >4 1k 1>31k 14 (17 4) Thus the mass flow rate of a particular fluid through a nozzle is a function of the stagnation properties of the fluid, the flow area, and the Mach number. Equation 17 4 is valid at any cross section, and thus ṁ can be evaluated at any location along the length of the nozzle. For a specified flow area A and stagnation properties T 0 and P 0, the maximum mass flow rate can be determined by differentiating Eq with respect to Ma and setting the result equal to zero. It yields Ma 1. Since the only location in a nozzle where the Mach number can be unity is the location of minimum flow area (the throat), the mass flow rate through a nozzle is a maximum when Ma 1 at the throat. Denoting this area by A*, we obtain an expression for the maximum mass flow rate by substituting Ma 1 in Eq. 17 4: m m max P e /P P* P P b P 0 m # 1k 1> 31k 14 k max A*P 0 a B RT 0 k 1 b (17 5) Thus, for a particular ideal gas, the maximum mass flow rate through a nozzle with a given throat area is fixed by the stagnation pressure and temperature of the inlet flow. The flow rate can be controlled by changing the stagnation pressure or temperature, and thus a converging nozzle can be used as a flowmeter. The flow rate can also be controlled, of course, by varying the throat area. This principle is vitally important for chemical processes, medical devices, flowmeters, and anywhere the mass flux of a gas must be known and controlled. A plot of ṁ versus P b /P 0 for a converging nozzle is shown in Fig Notice that the mass flow rate increases with decreasing P b /P 0, reaches a maximum at P b P*, and remains constant for P b /P 0 values less than this P* P 0 0 P* P P b P 0 FIGURE 17 1 The effect of back pressure P b on the mass flow rate m # and the exit pressure P e of a converging nozzle.

360 838 Thermodynamics critical ratio. Also illustrated on this figure is the effect of back pressure on the nozzle exit pressure P e. We observe that P e e P b for P b P* P* for P b 6 P* To summarize, for all back pressures lower than the critical pressure P*, the pressure at the exit plane of the converging nozzle P e is equal to P*, the Mach number at the exit plane is unity, and the mass flow rate is the maximum (or choked) flow rate. Because the velocity of the flow is sonic at the throat for the maximum flow rate, a back pressure lower than the critical pressure cannot be sensed in the nozzle upstream flow and does not affect the flow rate. The effects of the stagnation temperature T 0 and stagnation pressure P 0 on the mass flow rate through a converging nozzle are illustrated in Fig. 17 where the mass flow rate is plotted against the static-to-stagnation pressure ratio at the throat P t /P 0. An increase in P 0 (or a decrease in T 0 ) will increase the mass flow rate through the converging nozzle; a decrease in P 0 (or an increase in T 0 ) will decrease it. We could also conclude this by carefully observing Eqs and A relation for the variation of flow area A through the nozzle relative to throat area A* can be obtained by combining Eqs and 17 5 for the same mass flow rate and stagnation properties of a particular fluid. This yields A A* 1 1k 1> 31k 14 Ma ca k 1 ba1 Ma bd k 1 (17 6) Table A 3 gives values of A/A* as a function of the Mach number for air (k 1.4). There is one value of A/A* for each value of the Mach number, but there are two possible values of the Mach number for each value of A/A* one for subsonic flow and another for supersonic flow. m Ma t = 1 Ma t < 1 Increase in P 0, decrease in T 0, or both P 0, T 0 Decrease in P 0, increase in T 0, or both FIGURE 17 The variation of the mass flow rate through a nozzle with inlet stagnation properties. 0 P* P P t P 0

361 Another parameter sometimes used in the analysis of one-dimensional isentropic flow of ideal gases is Ma*, which is the ratio of the local velocity to the speed of sound at the throat: Chapter Ma* V c* (17 7) It can also be expressed as Ma* V c Mac MakRT c c* c* krt* Ma T B T* where Ma is the local Mach number, T is the local temperature, and T* is the critical temperature. Solving for T from Eq and for T* from Eq and substituting, we get Ma* Ma B k 1 1k 1Ma (17 8) Values of Ma* are also listed in Table A 3 versus the Mach number for k 1.4 (Fig. 17 3). Note that the parameter Ma* differs from the Mach number Ma in that Ma* is the local velocity nondimensionalized with respect to the sonic velocity at the throat, whereas Ma is the local velocity nondimensionalized with respect to the local sonic velocity. (Recall that the sonic velocity in a nozzle varies with temperature and thus with location.) Ma Ma * A P r T A * P0 r0 T FIGURE 17 3 Various property ratios for isentropic flow through nozzles and diffusers are listed in Table A 3 for k 1.4 for convenience. EXAMPLE 17 5 Effect of Back Pressure on Mass Flow Rate Air at 1 MPa and 600 C enters a converging nozzle, shown in Fig. 17 4, with a velocity of 150 m/s. Determine the mass flow rate through the nozzle for a nozzle throat area of 50 cm when the back pressure is (a) 0.7 MPa and (b) 0.4 MPa. Solution Air enters a converging nozzle. The mass flow rate of air through the nozzle is to be determined for different back pressures. Assumptions 1 Air is an ideal gas with constant specific heats at room temperature. Flow through the nozzle is steady, one-dimensional, and isentropic. Properties The constant-pressure specific heat and the specific heat ratio of air are c p kj/kg K and k 1.4, respectively (Table A a). Analysis We use the subscripts i and t to represent the properties at the nozzle inlet and the throat, respectively. The stagnation temperature and pressure at the nozzle inlet are determined from Eqs and 17 5: T 0i T i V i 1150 m>s 873 K c p kj>kg # a 1 kj>kg b 884 K K 1000 m >s P 0i P i a T k>1k 1 0i b 11 MPa a 884 K 1.4> T i 873 K b MPa These stagnation temperature and pressure values remain constant throughout the nozzle since the flow is assumed to be isentropic. That is, T 0 T 0i 884 K and P 0 P 0i MPa AIR P i = 1 MPa T i = 600 C V i = 150 m/s Converging nozzle FIGURE 17 4 Schematic for Example P b A t = 50 cm

362 840 Thermodynamics The critical-pressure ratio is determined from Table 17 (or Eq. 17 ) to be P*/P (a) The back pressure ratio for this case is which is greater than the critical-pressure ratio, Thus the exit plane pressure (or throat pressure P t ) is equal to the back pressure in this case. That is, P t P b 0.7 MPa, and P t /P Therefore, the flow is not choked. From Table A 3 at P t /P , we read Ma t and T t /T The mass flow rate through the nozzle can be calculated from Eq But it can also be determined in a step-by-step manner as follows: Thus, r t P t RT t V t Ma t c t Ma t krt t m # r t A t V t kg>m m m>s 6.77 kg/s (b) The back pressure ratio for this case is which is less than the critical-pressure ratio, Therefore, sonic conditions exist at the exit plane (throat) of the nozzle, and Ma 1. The flow is choked in this case, and the mass flow rate through the nozzle can be calculated from Eq. 17 5: m # 1k 1>31k 14 k A*P 0 a B RT 0 k 1 b.4> m kpa B kj>kg # a K1884 K b 7.10 kg/s B kj>kg # K Ka 1000 m >s 1 kj>kg b m>s P b 0.7 MPa P MPa T t 0.89T K K 700 kpa kpa # m 3 >kg # K K kg>m 3 P b 0.4 MPa P MPa since kpa # m > kj>kg 1000 kg>s. Discussion This is the maximum mass flow rate through the nozzle for the specified inlet conditions and nozzle throat area. EXAMPLE 17 6 Gas Flow through a Converging Nozzle Nitrogen enters a duct with varying flow area at T K, P kpa, and Ma Assuming steady isentropic flow, determine T, P, and Ma at a location where the flow area has been reduced by 0 percent.

363 Chapter Solution Nitrogen gas enters a converging nozzle. The properties at the nozzle exit are to be determined. Assumptions 1 Nitrogen is an ideal gas with k 1.4. Flow through the nozzle is steady, one-dimensional, and isentropic. Analysis The schematic of the duct is shown in Fig For isentropic flow through a duct, the area ratio A/A* (the flow area over the area of the throat where Ma 1) is also listed in Table A 3. At the initial Mach number of Ma 1 0.3, we read A 1 A*.0351 T 1 T P 1 P With a 0 percent reduction in flow area, A 0.8A 1, and T 1 = 400 K P 1 = 100 kpa Ma 1 = 0.3 A 1 N Nozzle T P Ma A = 0.8A 1 FIGURE 17 5 Schematic for Example 17 6 (not to scale). A A* A A A 1 A* For this value of A /A* from Table A 3, we read T T P P Ma Here we chose the subsonic Mach number for the calculated A /A* instead of the supersonic one because the duct is converging in the flow direction and the initial flow is subsonic. Since the stagnation properties are constant for isentropic flow, we can write T T >T 0 S T T 1 T 1 >T T 1 a T >T 0 b 1400 K a b 395 K 0 T 1 >T P P >P 0 S P P 1 P 1 >P P 1 a P >P 0 b 1100 kpaa b 95.7 kpa 0 P 1 >P which are the temperature and pressure at the desired location. Discussion Note that the temperature and pressure drop as the fluid accelerates in a converging nozzle. Converging Diverging Nozzles When we think of nozzles, we ordinarily think of flow passages whose cross-sectional area decreases in the flow direction. However, the highest velocity to which a fluid can be accelerated in a converging nozzle is limited to the sonic velocity (Ma 1), which occurs at the exit plane (throat) of the nozzle. Accelerating a fluid to supersonic velocities (Ma 1) can be accomplished only by attaching a diverging flow section to the subsonic nozzle at the throat. The resulting combined flow section is a converging diverging nozzle, which is standard equipment in supersonic aircraft and rocket propulsion (Fig. 17 6). Forcing a fluid through a converging diverging nozzle is no guarantee that the fluid will be accelerated to a supersonic velocity. In fact, the fluid may find itself decelerating in the diverging section instead of accelerating if the back pressure is not in the right range. The state of the nozzle flow is determined by the overall pressure ratio P b /P 0. Therefore, for given inlet conditions, the flow through a converging diverging nozzle is governed by the back pressure P b, as will be explained.

364 84 Thermodynamics FIGURE 17 6 Converging diverging nozzles are commonly used in rocket engines to provide high thrust. Courtesy of Pratt and Whitney, Used by permission. Consider the converging diverging nozzle shown in Fig A fluid enters the nozzle with a low velocity at stagnation pressure P 0. When P b P 0 (case A), there will be no flow through the nozzle. This is expected since the flow in a nozzle is driven by the pressure difference between the nozzle inlet and the exit. Now let us examine what happens as the back pressure is lowered. 1. When P 0 P b P C, the flow remains subsonic throughout the nozzle, and the mass flow is less than that for choked flow. The fluid velocity increases in the first (converging) section and reaches a maximum at the throat (but Ma 1). However, most of the gain in velocity is lost in the second (diverging) section of the nozzle, which acts as a diffuser. The pressure decreases in the converging section, reaches a minimum at the throat, and increases at the expense of velocity in the diverging section.. When P b P C, the throat pressure becomes P* and the fluid achieves sonic velocity at the throat. But the diverging section of the nozzle still acts as a diffuser, slowing the fluid to subsonic velocities. The mass flow rate that was increasing with decreasing P b also reaches its maximum value. Recall that P* is the lowest pressure that can be obtained at the throat, and the sonic velocity is the highest velocity that can be achieved with a converging nozzle. Thus, lowering P b further has no influence on the fluid flow in the converging part of the nozzle or the

365 Chapter P 0 V i 0 Throat P e P b x P P 0 P* Ma 0 Inlet 1 0 Inlet Sonic flow at throat Sonic flow at throat Throat Throat Shock in nozzle Shock in nozzle E, F, G A B C D E, F, G Exit D C P b B A Exit P A P B P C P D P E P F P G } } } } } } Subsonic flow at nozzle exit (no shock) Subsonic flow at nozzle exit (shock in nozzle) Supersonic flow at nozzle exit (no shock in nozzle) x Supersonic flow at nozzle exit (no shock in nozzle) Subsonic flow at nozzle exit (shock in nozzle) Subsonic flow at nozzle exit (no shock) x FIGURE 17 7 The effects of back pressure on the flow through a converging diverging nozzle. mass flow rate through the nozzle. However, it does influence the character of the flow in the diverging section. 3. When P C P b P E, the fluid that achieved a sonic velocity at the throat continues accelerating to supersonic velocities in the diverging section as the pressure decreases. This acceleration comes to a sudden stop, however, as a normal shock develops at a section between the throat and the exit plane, which causes a sudden drop in velocity to subsonic levels and a sudden increase in pressure. The fluid then continues to decelerate further in the remaining part of the converging diverging nozzle. Flow through the shock is highly irreversible, and thus it cannot be approximated as isentropic. The normal shock moves downstream away from the throat as P b is decreased, and it approaches the nozzle exit plane as P b approaches P E. When P b P E, the normal shock forms at the exit plane of the nozzle. The flow is supersonic through the entire diverging section in this case, and it can be approximated as isentropic. However, the fluid velocity drops to subsonic levels just before leaving the nozzle as it

366 844 Thermodynamics crosses the normal shock. Normal shock waves are discussed in Section When P E P b 0, the flow in the diverging section is supersonic, and the fluid expands to P F at the nozzle exit with no normal shock forming within the nozzle. Thus, the flow through the nozzle can be approximated as isentropic. When P b P F, no shocks occur within or outside the nozzle. When P b P F, irreversible mixing and expansion waves occur downstream of the exit plane of the nozzle. When P b P F, however, the pressure of the fluid increases from P F to P b irreversibly in the wake of the nozzle exit, creating what are called oblique shocks. T 0 = 800 K P 0 = 1.0 MPa V i 0 A t = 0 cm Ma e = EXAMPLE 17 7 Airflow through a Converging Diverging Nozzle Air enters a converging diverging nozzle, shown in Fig. 17 8, at 1.0 MPa and 800 K with a negligible velocity. The flow is steady, one-dimensional, and isentropic with k 1.4. For an exit Mach number of Ma and a throat area of 0 cm, determine (a) the throat conditions, (b) the exit plane conditions, including the exit area, and (c) the mass flow rate through the nozzle. FIGURE 17 8 Schematic for Example Solution Air flows through a converging diverging nozzle. The throat and the exit conditions and the mass flow rate are to be determined. Assumptions 1 Air is an ideal gas with constant specific heats at room temperature. Flow through the nozzle is steady, one-dimensional, and isentropic. Properties The specific heat ratio of air is given to be k 1.4. The gas constant of air is 0.87 kj/kg K. Analysis The exit Mach number is given to be. Therefore, the flow must be sonic at the throat and supersonic in the diverging section of the nozzle. Since the inlet velocity is negligible, the stagnation pressure and stagnation temperature are the same as the inlet temperature and pressure, P MPa and T K. The stagnation density is r 0 P 0 RT kpa kpa # m 3 >kg # K1800 K kg>m 3 (a) At the throat of the nozzle Ma 1, and from Table A 3 we read Thus, Also, P* 0.583P MPa MPa T* T K K r* r kg>m kg/m 3 V* c* krt* B kj>kg # K K a 1000 m >s 1 kj>kg b m/s P* T* r* P 0 T 0 r 0

367 Chapter (b) Since the flow is isentropic, the properties at the exit plane can also be calculated by using data from Table A 3. For Ma we read P e P Thus, and The nozzle exit velocity could also be determined from V e Ma e c e, where c e is the speed of sound at the exit conditions: 845. m>s T e T r e r Ma t * P e 0.178P MPa MPa T e T K K r e 0.300r kg>m kg/m 3 A e A* cm cm V e Ma e *c* m>s m/s A e A* V e Ma e c e Ma e krt e B kj>kg # K K a 1000 m >s 1 kj>kg b (c) Since the flow is steady, the mass flow rate of the fluid is the same at all sections of the nozzle. Thus it may be calculated by using properties at any cross section of the nozzle. Using the properties at the throat, we find that the mass flow rate is m # r*a*v* kg>m m m>s.86 kg/s Discussion Note that this is the highest possible mass flow rate that can flow through this nozzle for the specified inlet conditions SHOCK WAVES AND EXPANSION WAVES We have seen that sound waves are caused by infinitesimally small pressure disturbances, and they travel through a medium at the speed of sound. We have also seen that for some back pressure values, abrupt changes in fluid properties occur in a very thin section of a converging diverging nozzle under supersonic flow conditions, creating a shock wave. It is of interest to study the conditions under which shock waves develop and how they affect the flow. Normal Shocks First we consider shock waves that occur in a plane normal to the direction of flow, called normal shock waves. The flow process through the shock wave is highly irreversible and cannot be approximated as being isentropic. Next we follow the footsteps of Pierre Lapace ( ), G. F. Bernhard Riemann ( ), William Rankine ( ), Pierre Henry Hugoniot ( ), Lord Rayleigh ( ), and G. I. Taylor

368 846 Thermodynamics Ma 1 > 1 Flow V 1 V P1 P h1 h r1 r s s 1 Control volume Shock wave Ma < 1 FIGURE 17 9 Control volume for flow across a normal shock wave. ( ) and develop relationships for the flow properties before and after the shock. We do this by applying the conservation of mass, momentum, and energy relations as well as some property relations to a stationary control volume that contains the shock, as shown in Fig The normal shock waves are extremely thin, so the entrance and exit flow areas for the control volume are approximately equal (Fig 17 30). We assume steady flow with no heat and work interactions and no potential energy changes. Denoting the properties upstream of the shock by the subscript 1 and those downstream of the shock by, we have the following: Conservation of mass: r 1 AV 1 r AV (17 9) or r 1 V 1 r V Conservation of energy: h 1 V 1 h V (17 30) or h 01 h 0 (17 31) Conservation of momentum: Rearranging Eq and integrating yield A 1P 1 P m # 1V V 1 (17 3) Increase of entropy: s s 1 0 (17 33) FIGURE Schlieren image of a normal shock in a Laval nozzle. The Mach number in the nozzle just upstream (to the left) of the shock wave is about 1.3. Boundary layers distort the shape of the normal shock near the walls and lead to flow separation beneath the shock. Photo by G. S. Settles, Penn State University. Used by permission. We can combine the conservation of mass and energy relations into a single equation and plot it on an h-s diagram, using property relations. The resultant curve is called the Fanno line, and it is the locus of states that have the same value of stagnation enthalpy and mass flux (mass flow per unit flow area). Likewise, combining the conservation of mass and momentum equations into a single equation and plotting it on the h-s diagram yield a curve called the Rayleigh line. Both these lines are shown on the h-s diagram in Fig As proved later in Example 17 8, the points of maximum entropy on these lines (points a and b) correspond to Ma 1. The state on the upper part of each curve is subsonic and on the lower part supersonic. The Fanno and Rayleigh lines intersect at two points (points 1 and ), which represent the two states at which all three conservation equations are satisfied. One of these (state 1) corresponds to the state before the shock, and the other (state ) corresponds to the state after the shock. Note that the flow is supersonic before the shock and subsonic afterward. Therefore the flow must change from supersonic to subsonic if a shock is to occur. The larger the Mach number before the shock, the stronger the shock will be. In the limiting case of Ma 1, the shock wave simply becomes a sound wave. Notice from Fig that s s 1. This is expected since the flow through the shock is adiabatic but irreversible.

369 Chapter h 01 h h 01 h 0 = h 0 h h1 0 V 1 1 s1 P 01 Fanno line SHOCK WAVE Rayleigh line s P 0 V Subsonic flow (Ma < 1) b Ma = 1 a Ma = 1 Supersonic flow (Ma > 1) s FIGURE The h-s diagram for flow across a normal shock. The conservation of energy principle (Eq ) requires that the stagnation enthalpy remain constant across the shock; h 01 h 0. For ideal gases h h(t ), and thus (17 34) That is, the stagnation temperature of an ideal gas also remains constant across the shock. Note, however, that the stagnation pressure decreases across the shock because of the irreversibilities, while the thermodynamic temperature rises drastically because of the conversion of kinetic energy into enthalpy due to a large drop in fluid velocity (see Fig. 17 3). We now develop relations between various properties before and after the shock for an ideal gas with constant specific heats. A relation for the ratio of the thermodynamic temperatures T /T 1 is obtained by applying Eq twice: T 01 1 a k 1 T 1 T 01 T 0 b Ma 1 and T 0 1 a k 1 b Ma T Dividing the first equation by the second one and noting that T 01 T 0,we have T 1 Ma1 1k 1> T 1 1 Ma 1k 1> (17 35) Normal shock P increases P 0 decreases V decreases Ma decreases T increases T 0 remains constant r increases s increases FIGURE 17 3 Variation of flow properties across a normal shock. From the ideal-gas equation of state, r 1 P 1 RT 1 and r P RT Substituting these into the conservation of mass relation r 1 V 1 r V and noting that Ma V/c and c 1kRT, we have T P V P Ma c P Ma T a P b a Ma b T 1 P 1 V 1 P 1 Ma 1 c 1 P 1 Ma 1 T 1 P 1 Ma 1 (17 36)

370 848 Thermodynamics Combining Eqs and gives the pressure ratio across the shock: P Ma 1 1 Ma 1 1k 1> P 1 Ma 1 Ma 1k 1> (17 37) Equation is a combination of the conservation of mass and energy equations; thus, it is also the equation of the Fanno line for an ideal gas with constant specific heats. A similar relation for the Rayleigh line can be obtained by combining the conservation of mass and momentum equations. From Eq. 17 3, However, P 1 P m# A 1V V 1 r V r 1 V 1 rv a P RT b1mac a P RT b1makrt PkMa Thus, or P 1 11 kma 1 P 11 kma P P 1 1 kma 1 1 kma Combining Eqs and yields Ma Ma 1 >1k 1 Ma 1k>1k 1 1 (17 38) (17 39) This represents the intersections of the Fanno and Rayleigh lines and relates the Mach number upstream of the shock to that downstream of the shock. The occurrence of shock waves is not limited to supersonic nozzles only. This phenomenon is also observed at the engine inlet of a supersonic aircraft, where the air passes through a shock and decelerates to subsonic velocities before entering the diffuser of the engine. Explosions also produce powerful expanding spherical normal shocks, which can be very destructive (Fig ). Various flow property ratios across the shock are listed in Table A 33 for an ideal gas with k 1.4. Inspection of this table reveals that Ma (the Mach number after the shock) is always less than 1 and that the larger the supersonic Mach number before the shock, the smaller the subsonic Mach number after the shock. Also, we see that the static pressure, temperature, and density all increase after the shock while the stagnation pressure decreases. The entropy change across the shock is obtained by applying the entropychange equation for an ideal gas across the shock: s s 1 c p ln T R ln P (17 40) T 1 P 1 which can be expressed in terms of k, R, and Ma 1 by using the relations developed earlier in this section. A plot of nondimensional entropy change

371 Chapter FIGURE Schlieren image of the blast wave (expanding spherical normal shock) produced by the explosion of a firecracker detonated inside a metal can that sat on a stool. The shock expanded radially outward in all directions at a supersonic speed that decreased with radius from the center of the explosion. The microphone at the lower right sensed the sudden change in pressure of the passing shock wave and triggered the microsecond flashlamp that exposed the photograph. Photo by G. S. Settles, Penn State University. Used by permission. across the normal shock (s s 1 )/R versus Ma 1 is shown in Fig Since the flow across the shock is adiabatic and irreversible, the second law requires that the entropy increase across the shock wave. Thus, a shock wave cannot exist for values of Ma 1 less than unity where the entropy change would be negative. For adiabatic flows, shock waves can exist only for supersonic flows, Ma 1 1. EXAMPLE 17 8 The Point of Maximum Entropy on the Fanno Line Show that the point of maximum entropy on the Fanno line (point b of Fig ) for the adiabatic steady flow of a fluid in a duct corresponds to the sonic velocity, Ma 1. Solution It is to be shown that the point of maximum entropy on the Fanno line for steady adiabatic flow corresponds to sonic velocity. Assumptions The flow is steady, adiabatic, and one-dimensional. Analysis In the absence of any heat and work interactions and potential energy changes, the steady-flow energy equation reduces to (s s 1 )/R 0 s s 1 > 0 s s 1 < 0 Impossible Subsonic flow before shock Ma 1 = 1 Supersonic flow Ma 1 before shock FIGURE Entropy change across the normal shock. Differentiating yields h V constant dh V dv 0 For a very thin shock with negligible change of duct area across the shock, the steady-flow continuity (conservation of mass) equation can be expressed as rv constant

372 850 Thermodynamics Differentiating, we have Solving for dv gives r dv V dr 0 Combining this with the energy equation, we have which is the equation for the Fanno line in differential form. At point a (the point of maximum entropy) ds 0. Then from the second T ds relation (T ds dh v dp) we have dh v dp dp/r. Substituting yields Solving for V, we have dp r V dr r dv V dr r dh V dr r 0 0 at s constant V a 0P 0r b 1> which is the relation for the speed of sound, Eq Thus the proof is complete. s m =.86 kg/s Shock wave Ma 1 = 1 P 01 = 1.0 MPa P 1 = MPa T 1 = K r1 = 1.00 kg/m3 FIGURE Schematic for Example EXAMPLE 17 9 Shock Wave in a Converging Diverging Nozzle If the air flowing through the converging diverging nozzle of Example 17 7 experiences a normal shock wave at the nozzle exit plane (Fig ), determine the following after the shock: (a) the stagnation pressure, static pressure, static temperature, and static density; (b) the entropy change across the shock; (c) the exit velocity; and (d ) the mass flow rate through the nozzle. Assume steady, one-dimensional, and isentropic flow with k 1.4 from the nozzle inlet to the shock location. Solution Air flowing through a converging diverging nozzle experiences a normal shock at the exit. The effect of the shock wave on various properties is to be determined. Assumptions 1 Air is an ideal gas with constant specific heats at room temperature. Flow through the nozzle is steady, one-dimensional, and isentropic before the shock occurs. 3 The shock wave occurs at the exit plane. Properties The constant-pressure specific heat and the specific heat ratio of air are c p kj/kg K and k 1.4. The gas constant of air is 0.87 kj/kg K (Table A a). Analysis (a) The fluid properties at the exit of the nozzle just before the shock (denoted by subscript 1) are those evaluated in Example 17 7 at the nozzle exit to be P MPa P MPa T K r kg>m 3

373 Chapter The fluid properties after the shock (denoted by subscript ) are related to those before the shock through the functions listed in Table A 33. For Ma 1.0, we read Ma P 0 P P P T T r r Then the stagnation pressure P 0, static pressure P, static temperature T, and static density r after the shock are P P MPa 0.71 MPa P P MPa MPa T T K 750 K r.6667r kg>m 3.67 kg/m 3 (b) The entropy change across the shock is s s 1 c p ln T T 1 R ln P P kj>kg # Kln kj>kg # Kln kj/kg # K Thus, the entropy of the air increases as it experiences a normal shock, which is highly irreversible. (c) The air velocity after the shock can be determined from V Ma c, where c is the speed of sound at the exit conditions after the shock: V Ma c Ma krt B kj>kg # K1750 K a 1000 m >s 1 kj>kg b 317 m/s (d) The mass flow rate through a converging diverging nozzle with sonic conditions at the throat is not affected by the presence of shock waves in the nozzle. Therefore, the mass flow rate in this case is the same as that determined in Example 17 7: m #.86 kg/s Discussion This result can easily be verified by using property values at the nozzle exit after the shock at all Mach numbers significantly greater than unity. Example 17 9 illustrates that the stagnation pressure and velocity decrease while the static pressure, temperature, density, and entropy increase across the shock. The rise in the temperature of the fluid downstream of a shock wave is of major concern to the aerospace engineer because it creates heat transfer problems on the leading edges of wings and nose cones of space reentry vehicles and the recently proposed hypersonic space planes. Overheating, in fact, led to the tragic loss of the space shuttle Columbia in February of 003 as it was reentering earth s atmosphere.

374 85 Thermodynamics FIGURE Schlieren image of a small model of the space shuttle Orbiter being tested at Mach 3 in the supersonic wind tunnel of the Penn State Gas Dynamics Lab. Several oblique shocks are seen in the air surrounding the spacecraft. Photo by G. S. Settles, Penn State University. Used by permission. Ma 1 b Oblique shock Ma 1 Ma FIGURE An oblique shock of shock angle b formed by a slender, two-dimensional wedge of half-angle d. The flow is turned by deflection angle u downstream of the shock, and the Mach number decreases. d u Oblique Shocks Not all shock waves are normal shocks (perpendicular to the flow direction). For example, when the space shuttle travels at supersonic speeds through the atmosphere, it produces a complicated shock pattern consisting of inclined shock waves called oblique shocks (Fig ). As you can see, some portions of an oblique shock are curved, while other portions are straight. First, we consider straight oblique shocks, like that produced when a uniform supersonic flow (Ma 1 1) impinges on a slender, two-dimensional wedge of half-angle d (Fig ). Since information about the wedge cannot travel upstream in a supersonic flow, the fluid knows nothing about the wedge until it hits the nose. At that point, since the fluid cannot flow through the wedge, it turns suddenly through an angle called the turning angle or deflection angle u. The result is a straight oblique shock wave, aligned at shock angle or wave angle b, measured relative to the oncoming flow (Fig ). To conserve mass, b must obviously be greater than d. Since the Reynolds number of supersonic flows is typically large, the boundary layer growing along the wedge is very thin, and we ignore its effects. The flow therefore turns by the same angle as the wedge; namely, deflection angle u is equal to wedge half-angle d. If we take into account the displacement thickness effect of the boundary layer, the deflection angle u of the oblique shock turns out to be slightly greater than wedge half-angle d. Like normal shocks, the Mach number decreases across an oblique shock, and oblique shocks are possible only if the upstream flow is supersonic. However, unlike normal shocks, in which the downstream Mach number is always subsonic, Ma downstream of an oblique shock can be subsonic, sonic, or supersonic, depending on the upstream Mach number Ma 1 and the turning angle.

375 We analyze a straight oblique shock in Fig by decomposing the velocity vectors upstream and downstream of the shock into normal and tangential components, and considering a small control volume around the shock. Upstream of the shock, all fluid properties (velocity, density, pressure, etc.) along the lower left face of the control volume are identical to those along the upper right face. The same is true downstream of the shock. Therefore, the mass flow rates entering and leaving those two faces cancel each other out, and conservation of mass reduces to r 1 V 1,n A r V,n A S r 1 V 1,n r V,n (17 41) where A is the area of the control surface that is parallel to the shock. Since A is the same on either side of the shock, it has dropped out of Eq As you might expect, the tangential component of velocity (parallel to the oblique shock) does not change across the shock (i.e., V 1,t V,t ). This is easily proven by applying the tangential momentum equation to the control volume. When we apply conservation of momentum in the direction normal to the oblique shock, the only forces are pressure forces, and we get Chapter V,n Oblique shock P V 1 1,t P V 1,n V V 1 Control volume b V,t FIGURE Velocity vectors through an oblique shock of shock angle b and deflection angle u. u P 1 A P A rv,n AV,n rv 1,n AV 1,n S P 1 P r V,n r 1 V 1,n (17 4) Finally, since there is no work done by the control volume and no heat transfer into or out of the control volume, stagnation enthalpy does not change across an oblique shock, and conservation of energy yields h 01 h 0 h 0 S h 1 1 V 1,n 1 V 1,t h 1 V,n 1 V,t But since V 1,t V,t, this equation reduces to h 1 1 V 1,n h 1 V,n (17 43) Careful comparison reveals that the equations for conservation of mass, momentum, and energy (Eqs through 17 43) across an oblique shock are identical to those across a normal shock, except that they are written in terms of the normal velocity component only. Therefore, the normal shock relations derived previously apply to oblique shocks as well, but must be written in terms of Mach numbers Ma 1,n and Ma,n normal to the oblique shock. This is most easily visualized by rotating the velocity vectors in Fig by angle p/ b, so that the oblique shock appears to be vertical (Fig ). Trigonometry yields Ma 1,n Ma 1 sin b and Ma,n Ma sin 1b u (17 44) where Ma 1,n V 1,n /c 1 and Ma,n V,n /c. From the point of view shown in Fig , we see what looks like a normal shock, but with some superposed tangential flow coming along for the ride. Thus, All the equations, shock tables, etc., for normal shocks apply to oblique shocks as well, provided that we use only the normal components of the Mach number. In fact, you may think of normal shocks as special oblique shocks in which shock angle b p/, or 90. We recognize immediately that an oblique shock can exist only if Ma 1,n 1, and Ma,n 1. The normal shock Ma 1,n 1 V 1,t Oblique shock b V 1,n P 1 P V 1 V P 1 P V,n b u V,t Ma,n 1 FIGURE The same velocity vectors of Fig , but rotated by angle p/ b, so that the oblique shock is vertical. Normal Mach numbers Ma 1,n and Ma,n are also defined. u

376 854 Thermodynamics T [ (k 1)Ma 1, T 1 P 0 c P 01 r h 01 h 0 T 01 T 0 (k 1)Ma 1, n Ma, n B k Ma k 1 r 1 V 1, 1, n V, n (k 1)Ma 1, n (k 1)Ma 1, n Ma 1, n P k Ma 1, n k 1 P 1 k 1 (k 1)Ma 1, 1, n k/( /(k 1) d k ] Ma 1, n k 1 (k 1) Ma c 1, n (k 1)Ma 1, 1, n (k 1) Ma 1, n k Ma 1, n k 1 d 1/(k 1) 1) FIGURE Relationships across an oblique shock for an ideal gas in terms of the normal component of upstream Mach number Ma 1,n. equations appropriate for oblique shocks in an ideal gas are summarized in Fig in terms of Ma 1,n. For known shock angle b and known upstream Mach number Ma 1, we use the first part of Eq to calculate Ma 1,n, and then use the normal shock tables (or their corresponding equations) to obtain Ma,n. If we also knew the deflection angle u, we could calculate Ma from the second part of Eq But, in a typical application, we know either b or u, but not both. Fortunately, a bit more algebra provides us with a relationship between u, b, and Ma 1. We begin by noting that tan b V 1,n /V 1,t and tan(b u) V,n /V,t (Fig ). But since V 1,t V,t, we combine these two expressions to yield V,n tan 1b u 1k 1Ma 1,n 1k 1Ma 1 sin b V 1,n tan b 1k 1Ma 1,n 1k 1Ma 1 sin b (17 45) where we have also used Eq and the fourth equation of Fig We apply trigonometric identities for cos b and tan(b u), namely, tan b tan u cos b cos b sin b and tan 1b u 1 tan b tan u After some algebra, Eq reduces to The u-b-ma relationship: tan u cot b 1Ma 1 sin b 1 (17 46) Ma 1 1k cos b Equation provides deflection angle u as a unique function of shock angle b, specific heat ratio k, and upstream Mach number Ma 1. For air (k 1.4), we plot u versus b for several values of Ma 1 in Fig We note that this plot is often presented with the axes reversed (b versus u) in compressible flow textbooks, since, physically, shock angle b is determined by deflection angle u. Much can be learned by studying Fig , and we list some observations here: Figure displays the full range of possible shock waves at a given free-stream Mach number, from the weakest to the strongest. For any value of Mach number Ma 1 greater than 1, the possible values of u range from u 0 at some value of b between 0 and 90, to a maximum value u u max at an intermediate value of b, and then back to u 0 at b 90. Straight oblique shocks for u or b outside of this range cannot and do not exist. At Ma 1 1.5, for example, straight oblique shocks cannot exist in air with shock angle b less than about 4, nor with deflection angle u greater than about 1. If the wedge half-angle is greater than u max, the shock becomes curved and detaches from the nose of the wedge, forming what is called a detached oblique shock or a bow wave (Fig. 17 4). The shock angle b of the detached shock is 90 at the nose, but b decreases as the shock curves downstream. Detached shocks are much more complicated than simple straight oblique shocks to analyze. In fact, no simple solutions exist, and prediction of detached shocks requires computational methods. Similar oblique shock behavior is observed in axisymmetric flow over cones, as in Fig , although the u-b-ma relationship for axisymmetric flows differs from that of Eq

377 u, degrees Ma 1 3 Ma 1 Ma 1 Weak u u max When supersonic flow impinges on a blunt body a body without a sharply pointed nose, the wedge half-angle d at the nose is 90, and an attached oblique shock cannot exist, regardless of Mach number. In fact, a detached oblique shock occurs in front of all such blunt-nosed bodies, whether two-dimensional, axisymmetric, or fully three-dimensional. For example, a detached oblique shock is seen in front of the space shuttle model in Fig and in front of a sphere in Fig While u is a unique function of Ma 1 and b for a given value of k, there are two possible values of b for u u max. The dashed black line in Fig passes through the locus of u max values, dividing the shocks into weak oblique shocks (the smaller value of b) and strong oblique shocks (the larger value of b). At a given value of u, the weak shock is more common and is preferred by the flow unless the downstream pressure conditions are high enough for the formation of a strong shock. For a given upstream Mach number Ma 1, there is a unique value of u for which the downstream Mach number Ma is exactly 1. The dashed gray line in Fig passes through the locus of values where Ma 1. To the left of this line, Ma 1, and to the right of this line, Ma 1. Downstream sonic conditions occur on the weak shock side of the plot, with u very close to u max. Thus, the flow downstream of a strong oblique shock is always subsonic (Ma 1). The flow downstream of a weak oblique shock remains supersonic, except for a narrow range of u just below u max, where it is subsonic, although it is still called a weak oblique shock. As the upstream Mach number approaches infinity, straight oblique shocks become possible for any b between 0 and 90, but the maximum possible turning angle for k 1.4 (air) is u max 45.6, which occurs at b Straight oblique shocks with turning angles above this value of u max are not possible, regardless of the Mach number. For a given value of upstream Mach number, there are two shock angles where there is no turning of the flow (u 0 ): the strong case, b 90, b, degrees 1. Ma 1 Strong Chapter FIGURE The dependence of straight oblique shock deflection angle u on shock angle b for several values of upstream Mach number Ma 1. Calculations are for an ideal gas with k 1.4. The dashed black line connects points of maximum deflection angle (u u max ). Weak oblique shocks are to the left of this line, while strong oblique shocks are to the right of this line. The dashed gray line connects points where the downstream Mach number is sonic (Ma 1). Supersonic downstream flow (Ma 1) is to the left of this line, while subsonic downstream flow (Ma 1) is to the right of this line. Ma 1 Detached oblique shock d u max FIGURE 17 4 A detached oblique shock occurs upstream of a two-dimensional wedge of half-angle d when d is greater than the maximum possible deflection angle u. A shock of this kind is called a bow wave because of its resemblance to the water wave that forms at the bow of a ship.

378 856 Thermodynamics FIGURE Still frames from schlieren videography illustrating the detachment of an oblique shock from a cone with increasing cone half-angle d in air at Mach 3. At (a) d 0 and (b) d 40, the oblique shock remains attached, but by (c) d 60, the oblique shock has detached, forming a bow wave. Photos by G. S. Settles, Penn State University. Used by permission. Ma 1 d (a) (b) (c) corresponds to a normal shock, and the weak case, b b min, represents the weakest possible oblique shock at that Mach number, which is called a Mach wave. Mach waves are caused, for example, by very small nonuniformities on the walls of a supersonic wind tunnel (several can be seen in Figs and 17 43). Mach waves have no effect on the flow, since the shock is vanishingly weak. In fact, in the limit, Mach waves are isentropic. The shock angle for Mach waves is a unique function of the Mach number and is given the symbol m, not to be confused with the coefficient of viscosity. Angle m is called the Mach angle and is found by setting u equal to zero in Eq , solving for b m, and taking the smaller root. We get Mach angle: m sin 1 11>Ma 1 (17 47) Since the specific heat ratio appears only in the denominator of Eq , m is independent of k. Thus, we can estimate the Mach number of any supersonic flow simply by measuring the Mach angle and applying Eq FIGURE Shadowgram of a one-half-in diameter sphere in free flight through air at Ma The flow is subsonic behind the part of the bow wave that is ahead of the sphere and over its surface back to about 45. At about 90 the laminar boundary layer separates through an oblique shock wave and quickly becomes turbulent. The fluctuating wake generates a system of weak disturbances that merge into the second recompression shock wave. Photo by A. C. Charters, Army Ballistic Research Laboratory. Prandtl Meyer Expansion Waves We now address situations where supersonic flow is turned in the opposite direction, such as in the upper portion of a two-dimensional wedge at an angle of attack greater than its half-angle d (Fig ). We refer to this type of flow as an expanding flow, whereas a flow that produces an oblique shock may be called a compressing flow. As previously, the flow changes direction to conserve mass. However, unlike a compressing flow, an expanding flow does not result in a shock wave. Rather, a continuous expanding region called an expansion fan appears, composed of an infinite number of Mach waves called Prandtl Meyer expansion waves. In other words, the flow does not turn suddenly, as through a shock, but gradually each successive Mach wave turns the flow by an infinitesimal amount. Since each individual expansion wave is isentropic, the flow across the entire expansion fan is also isentropic. The Mach number downstream of the expansion increases (Ma Ma 1 ), while pressure, density, and temperature decrease, just as they do in the supersonic (expanding) portion of a converging diverging nozzle. Prandtl Meyer expansion waves are inclined at the local Mach angle m, as sketched in Fig The Mach angle of the first expansion wave is easily determined as m 1 sin 1 (1/Ma 1 ). Similarly, m sin 1 (1/Ma ),

379 where we must be careful to measure the angle relative to the new direction of flow downstream of the expansion, namely, parallel to the upper wall of the wedge in Fig if we neglect the influence of the boundary layer along the wall. But how do we determine Ma? It turns out that the turning angle u across the expansion fan can be calculated by integration, making use of the isentropic flow relationships. For an ideal gas, the result is (Anderson, 003), Turning angle across an expansion fan: u n 1Ma n 1Ma 1 (17 48) where n(ma) is an angle called the Prandtl Meyer function (not to be confused with the kinematic viscosity), n 1Ma B k 1 k 1 tan 1 c B k 1 k 1 1Ma 1 d tan 1 a Ma 1 b (17 49) Note that n(ma) is an angle, and can be calculated in either degrees or radians. Physically, n(ma) is the angle through which the flow must expand, starting with n 0 at Ma 1, in order to reach a supersonic Mach number, Ma 1. To find Ma for known values of Ma 1, k, and u, we calculate n(ma 1 ) from Eq , n(ma ) from Eq , and then Ma from Eq , noting that the last step involves solving an implicit equation for Ma. Since there is no heat transfer or work, and the flow is isentropic through the expansion, T 0 and P 0 remain constant, and we use the isentropic flow relations derived previously to calculate other flow properties downstream of the expansion, such as T, r, and P. Prandtl Meyer expansion fans also occur in axisymmetric supersonic flows, as in the corners and trailing edges of a cone-cylinder (Fig ). Some very complex and, to some of us, beautiful interactions involving both shock waves and expansion waves occur in the supersonic jet produced by an overexpanded nozzle, as in Fig Analysis of such flows is beyond the scope of the present text; interested readers are referred to compressible flow textbooks such as Thompson (197) and Anderson (003). Ma 1 1 Chapter Oblique shock Expansion waves m 1 m Ma FIGURE An expansion fan in the upper portion of the flow formed by a twodimensional wedge at the angle of attack in a supersonic flow. The flow is turned by angle u, and the Mach number increases across the expansion fan. Mach angles upstream and downstream of the expansion fan are indicated. Only three expansion waves are shown for simplicity, but in fact, there are an infinite number of them. (An oblique shock is present in the bottom portion of this flow.) d u FIGURE A cone-cylinder of 1.5 half-angle in a Mach number 1.84 flow. The boundary layer becomes turbulent shortly downstream of the nose, generating Mach waves that are visible in this shadowgraph. Expansion waves are seen at the corners and at the trailing edge of the cone. Photo by A. C. Charters, Army Ballistic Research Laboratory.

380 858 Thermodynamics FIGURE The complex interactions between shock waves and expansion waves in an overexpanded supersonic jet. The flow is visualized by a schlierenlike differential interferogram. Photo by H. Oertel sen. Reproduced by courtesy of the French-German Research Institute of Saint- Louis, ISL. Used with permission. EXAMPLE Estimation of the Mach Number from Mach Lines Estimate the Mach number of the free-stream flow upstream of the space shuttle in Fig from the figure alone. Compare with the known value of Mach number provided in the figure caption. Ma 1 Weak shock Solution We are to estimate the Mach number from a figure and compare it to the known value. Analysis Using a protractor, we measure the angle of the Mach lines in the free-stream flow: m 19. The Mach number is obtained from Eq , b weak d 10 m sin 1 a 1 Ma 1 b S Ma 1 1 sin 19 S Ma Our estimated Mach number agrees with the experimental value of Discussion The result is independent of the fluid properties. (a) Ma 1 Strong shock EXAMPLE Oblique Shock Calculations b strong d 10 Supersonic air at Ma 1.0 and 75.0 kpa impinges on a two-dimensional wedge of half-angle d 10 (Fig ). Calculate the two possible oblique shock angles, b weak and b strong, that could be formed by this wedge. For each case, calculate the pressure and Mach number downstream of the oblique shock, compare, and discuss. (b) FIGURE Two possible oblique shock angles, (a) b weak and (b) b strong, formed by a two-dimensional wedge of half-angle d 10. Solution We are to calculate the shock angle, Mach number, and pressure downstream of the weak and strong oblique shocks formed by a twodimensional wedge. Assumptions 1 The flow is steady. The boundary layer on the wedge is very thin. Properties The fluid is air with k 1.4.