Appendices, Thermodynamics, Phases of Water, Laws of Thermodynamics and Thermodynamic Processes By

Transcription

1 Appendices, Thermodynamics, Phases of Water, Laws of Thermodynamics and Thermodynamic Processes By S. Bobby Rauf, P.E., CEM, MBA Credit: 11 PDH Thermodynamics Fundamentals Series

2 Appendix A This appendix includes the solutions and answers to end of segment self-assessment problems and questions Segment 1 Solutions - Self-assessment Problems & Questions 1. A boiler is relocated from sea level to a location that is at an elevation of 10,000 ft MSL. Using the table below and Satrurated Steam Tables, determine the temperature at which the water will boil if the boiler is assumed to be open to atmosphere. Altitude With Mean Sea Level as Ref. Feet Meters Absolute Pressure in Hg Column Inches Hg Column mm Hg Column Absolute Atmospheric Pressure psia kg/cm 2 kpa

3 Solution/Answer: In order to determine the boiling point of the water, at an altitude of 10,000 ft, we need to assess the pressure at that altitude. The pressures at a range of altitudes are available through the given table. In US units, the pressure at 10,000 ft is 10.1 psia, or simply, 10 psia. The next step is to retrieve the saturation temperature at 10 psia, from the saturated steam tables. An excerpt of the Saturated Steam Tables, for the range of pressures surrounding 10 psia, is shown below in form of Table A-1.2. Abs. Temp in Press. psia F Properties of Saturated Steam By Pressure US/Imperial Units Specific Volume Enthalpy Entropy Abs. Sat. Liquid ft 3 /lbm Sat. Vapor Sat. Liquid BTU/lbm BTU/(lbm. R) Press. Evap. Sat. Sat. Sat. Vapor Liquid Vapor psia L V h L h fg h V s L s V Table A-1.2, Properties of Saturated Steam By Pressure, US Units Reading directly across, to the right, from the 10 psia saturation pressure field, we see the listed saturation temperature of 193 F. Answer: Therefore, the boiling point for the water at the listed altitude of 10,000 ft would be 193 F. 2. In problem (1), if the objective is just to heat the water close to the boiling point, will the boiler consume more or less fuel than it did when it was located at the sea level? Solution/Answer: In order to assess if it would take more heat or less heat to bring the water to a boil at the higher altitude of 10,000 ft, we need to compare the difference between the enthalpy values of water, in saturated liquid phase, at the two different altitudes and pressures.

4 In other words, we need to compare the enthalpy, hl, at 14 psia (pressure at sea level) and 10 psia (pressure at 10,000 ft altitude). Based on the saturated steam table excerpt in Table A-1.2: hl, at 10 psia = BTU/lbm, and hl, at 14 psia = BTU/lbm Answer: As apparent from the heat content values or enthalpy values above, the water that comes to a boil at the higher altitude of 10,000 ft will do so with less heat content, or would require less heat. Conversely, water that comes to a boil at higher pressure would do so at higher enthalpy value, or greater heat content. NOTE: This question pertains to escalating the water to saturated liquid state and not fully evaporating it. 3. Answer the following questions for water at a temperature of 193 F and pressure of 10 psia: a) Heat content for saturated water. b) Specific heat (BTU/lbm) required to evaporate the water. c) If the water were evaporated, what would the saturated vapor heat content be? d) What state or phase would the water be in at the stated temperature and pressure? e) What would the entropy of the water be while it is in saturated liquid phase? f) What would the specific volume of the water be while it is in saturated vapor phase? g) What would the phase of the water if the pressure is increased to 20 psia while keeping the temperature constant at 193 F? Solution/Answer: a) At193 F and pressure of 10 psia, the water is in saturated liquid form. According to saturated steam table excerpt in Table A-1.2, the saturated water enthalpy at 193 F and pressure of 10 psia is BTU/lbm. This value is listed under column labeled hl, in Table A-1.2, in the row representing temperature of193 F and pressure of 10 psia. Therefore, the enthalpy or heat content for saturated water, at the given temperature and pressure, is BTU/lbm. b) The specific heat, in BTU/lbm, required to evaporate the water from saturated liquid phase to saturated vapor phase, is represented by the term hfg. The value of hfg, for saturated water at 193 F and a pressure of 10 psia, as read from Table A-1.2, is 986 BTU/lbm. See circled values in Table A-1.2. Answer: hfg at193 F and 10 psia = 986 BTU/lbm

5 c) Saturated vapor heat content if the water were evaporated would be the value for hv at193 F and 10 psia, and from Table A-1.2 this value is 1143 BTU/lbm. d) The water would be in saturated liquid phase at the stated temperature and pressure. All stated saturation temperatures and pressures, in the saturated steam tables, represent the current state of water in saturated liquid phase. e) The entropy of water at 193 F and a pressure of 10 psia, in saturated liquid phase, as read from Table A-1.2 would be sl= BTU/(lbm. R). Note that the sl value is retrieved form the table and not the sv value. This is because the problem statement specifies the liquid phase. f) The specific volume of water at 193 F and a pressure of 10 psia, in saturated vapor phase, as read from Table A-1.2 would be V = ft 3 /lbm. Note that the V value is retrieved form the table and not the L value. This is because the problem statement specifies the vapor phase. g) The phase of the water if the pressure is increased to 20 psia while keeping the temperature constant at 193 F can be determined through the saturated team table excerpt in Table A-1.2. As established earlier in part (d), the current phase of the water is saturated liquid. If, however, the pressure is doubled to 20 psia, according to Table A-1.2, it would take well over 209 F - and additional heat - to reinstate the water into the saturated water phase. Therefore, the water at 20 psia and 193 F would be in subcooled phase or state.

6 Segment 2 Solutions - Self-assessment Problems & Questions 1. Why is the efficiency of this power plant in Case Study 2.1 rather low (17%)? Solution/Answer: The efficiency of the power station is low (17%) because of the fact that the working fluid is introduced into the system as -10 C ice. If the working fluid were room temperature water or return condensate from the discharge side of the turbine, as is the case in Self-Assessment problem number 4 the overall system efficiency would be substantially higher. 2. Using the steam tables in Appendix B and the Double Interpolation Method described in Case Study 2.1, US Unit Version, determine the exact enthalpy of a superheated steam at a pressure of 400 psia and temperature of 950 F. Solution: As apparent from the superheated steam tables in Appendix B, the enthalpy value for 400 psia and 950 F is not readily available and, therefore, double interpolation must be conducted between the enthalpy values given for 360 psia, 900 F, and 450 psia, 1000 F, to derive h 400 psia and 950 F. The double interpolation approach, as applied here, will entail three steps. First step involves determination of h400 psia and 900 F, the enthalpy value at 400 psia and 900 F. The enthalpy values available and used in this interpolation step are circled in Table A-2.1. The following formula sums up the mathematical approach to this first step: h400 psia and 900 F = ((h360 psia and 900 F - h450 psia and 900 F)/(450 psia 360 psia)).(450 psia-400 psia) + h450 psia and 900 F Substituting enthalpy values and other given data from superheated steam table excerpt, shown in Table A-2.1 below: h400 psia and 900 F = (( )/(450 psia psia)).(450 psia-400 psia) = BTU/lbm

7 Properties of Superheated Steam US/Imperial Units Abs. Temp. Note: is in ft 3 /lbm, h is in BTU/lbm and s is in BTU/(lbm- Press. F R) psia (Sat. T, F ) (404.45) h s (434.43) h s (456.32) h s (486.25) h s Table A-2.1, Superheated Steam Table Excerpt, US/Imperial Units Second step involves determination of h400 psia and 1000 F, the enthalpy value at 400 psia and 1000 F. The enthalpy values available and used in this interpolation step are circled in Table A The following formula sums up the mathematical approach to this first step: h400 psia and 1000 F = ((h360 psia and 1000 F - h450 psia and 1000 F)/(450 psia 360 psia)). (450 psia-400 psia) + h450 psia and 1000 F

8 Substituting enthalpy values and other given data from superheated steam table excerpt, shown in Table A-2.1: h400 psia and 1000 F = ((1525 BTU/lbm BTU/lbm)/(450 psia psia)). (450 psia-400 psia) BTU/lbm = BTU/lbm The final step in the double interpolation process, as applied in this case, involves interpolating between h400 psia and 1000 F and h400 psia and 900 F, the enthalpy values derived in the first two steps above, to obtain the desired final enthalpy h400 psia and 950 F. The formula for this final step is as follows: h400 psia and 950 F = ((h400 psia and 1000 F - h400 psia and 900 F)/( 1000 F F)).( 950 F F) + h400 psia and 900 F Substituting enthalpy values derived in the first two steps above: h400 psia and 950 F = (( BTU/lbm BTU/lbm)/(1000 F F)).( 950 F F) BTU/lbm = BTU/lbm 3. In Case Study 2.1, as an energy engineer you have been retained by Station Zebra to explore or develop an alternative integrated steam turbine and electric power generating system that is capable of generating 10 MW of power with only 60 truck loads, or 54,432 kg, of ice per hour. With all other parameters the same as in the original Case Study 2.1 scenario, determine the total heat flow rate, in kj/hr, needed to produce 10 MW of electrical power. Solution: This problem requires accounting for heat added during each of the five (5) stages of the overall process, with a working fluid mass flow rate of 60 truck loads, or 54,432 kg, per hour. Therefore, the solution is divided into five sub-parts, each involving either a sensible or a latent heat calculation, based on the entry and exit temperature and phase status. Table 2.2 lists specific heat for water and ice. These heat values will be used in the sensible heat calculations. Table 2.3 lists latent heat values for water. These values will be used to compute the latent heats associated with stages that involve phase transformation.

9 (i) Calculate the heat required to heat the ice from -10 C to 0 C. Since there is no change in phase involved, the entire heat absorbed by the ice (working substance) in this stage would be sensible heat. First stage of the overall power generating system is illustrated in Figure A-2.4, below. Stage 1, Sensible Heat Harvested Ice Staging Tank & Heater -10 C Ice 0 C Heat Given: Ti = -10 C Tf = 0 C Figure A-2.4, Case Study 2.1 Stage 1 Sensible Heat Calculation cice = 2.1 kj/kg. K {Table 2.2} Utilizing the given information: ΔT = Tf - Ti ΔT = 0 (-10 C) = +10 C Since ΔT represents the change in temperature and not a specific absolute temperature, ΔT = +10 K Mathematical relationship between sensible heat, mass of the working substance, specific heat of the working substance and change in temperature can be stated as: And, Qs(heat ice) = m. cice. ΔT Eq s(heat ice) = ṁ. cice. ΔT Eq. 2.15

10 Where, Qs(heat ice) = Sensible heat required to heat the ice over ΔT s(heat ice) = Sensible heat flow rate required to heat the ice over ΔT m = Mass of ice being heated cice = Specific heat of ice = 2.1 kj/kg. K ΔT = Change in temperature, in C or K ṁ = Mass flow rate of water/ice = 60 tons/hr = (60 tons/hr). (907.2 kg/ton) = 54,432 kg/hr Then, by application of Eq. 2.15: Or, s(heat ice) = (54,432 kg/hr). (2.1 kj/kg. K). (10 K) s(heat ice) = 1,143,072 kj/hr Since there are kj per BTU, Or, s(heat ice) = (1,143,072 kj/hr)/(1.055 kj/btu) s(heat ice) = 1,083,481 BTU/hr (ii) Calculate the heat required to melt the ice at 0 C. Since change in phase is involved in this case, the heat absorbed by the ice (working substance) in this stage would be latent heat. The 2 nd stage of the overall power generating system is illustrated in Figure A-2.5, below. Stage 2, Latent Heat Water Heater 0 C Ice Water 0 C Heat Potable Water Discharge Figure A-2.5, Case Study 2.1 Stage 2 Latent Heat Calculation

11 Mathematical relationship between latent heat, mass of the working substance, and the heat of fusion of ice can be stated as: And, Ql(latent ice) = hsl (ice). m Eq l(latent ice) = hsl (ice). ṁ Eq Where, Ql(latent ice) = Latent heat required to melt a specific mass of ice, isothermally l(latent ice) = Latent heat flow rate required to melt a specific mass of ice, isothermally, over a period of time m = Mass of ice being melted ṁ = Mass flow rate of water/ice = 60 tons/hr = (60 tons/hr). (907.2 kg/ton) = 54,432 kg/hr, same as part (a) (i) hsl (ice) = Heat of fusion for Ice = kj/kg {Table 2.3} Then, by application of Eq. 2.17: l(latent ice) = hsl (ice). ṁ l(latent ice) = (333.5 kj/kg). (54,432 kg/hr) l(latent ice) = 18,153,072 kj/hr Since there are kj per BTU, Or, l(latent ice) = (18,153,072 kj/hr)/(1.055 kj/btu) l(latent ice) = 17,206,703 BTU/hr Note that the specific heat required to melt ice is called heat of fusion because of the fact that the water molecules come closer together as heat is added in the melting process. The water molecules are held apart at specific distances in the crystallographic structure of solid ice. The heat of fusion allows the molecules to overcome the crystallographic forces and fuse to form liquid water. This also explains why the density of water is higher than the density of ice. (iii) Calculate the heat reqd. to heat the water from 0 C to 100 C

12 The 3 rd stage of the overall power generating system is illustrated in Figure A-2.6, below. Since no phase change is involved in this stage, the heat absorbed by the water in this stage would be sensible heat. Stage 3, Sensible Heat Heat Heater/Boiler 0 C Hot Water 100 C, 100 C Water Given: Ti = 0 C Tf = 100 C Figure A-2.6, Case Study 2.1 Stage 3 Sensible Heat Calculation cp-water = 4.19 kj/kg. K {Table 2.3} Utilizing the given information: ΔT = Tf - Ti ΔT = 100 C 0 C = 100 C Since ΔT represents the change in temperature and not a specific absolute temperature, ΔT = 100 K Mathematical relationship between sensible heat, mass of the working substance, specific heat of water (the working substance), and change in temperature can be stated as: And, Qs(water) = m. cp-water. ΔT Eq s(water) = ṁ. cp-water. ΔT Eq Where, Qs(water) = Sensible heat required to heat the water over ΔT s(water) = Sensible heat flow rate required to heat the water over ΔT

13 m = Mass of water being heated cp-water = Specific heat of water = 4.19 kj/kg. K ṁ = Mass flow rate of water = 54,432 kg/hr, as calculated in part (a) ΔT = Change in temperature, in C or K Then, by applying Eq. 2.19: s(water) = ṁ. cp-water. ΔT s(water) = (54,432 kg/hr). (4.19 kj/kg. K). (100 K) s(water) = 22,807,008 kj/hr Since there are kj per BTU, Or, s(water) = (22,807,008 kj/hr)/(1.055 kj/btu) s(water) = 21,618,017 BTU/hr (iv) Calculate the heat required to convert 100 C water to 100 C steam The 4 th stage of the overall power generating system is illustrated in Figure A-2.7, below. Since change in phase is involved in this case, the heat absorbed by the water in this stage would be latent heat. Stage 4, Latent Heat Heat Boiler / Steam Generator 100 C Water Steam 100 C, Steam 100 C Steam Figure A-2.7, Case Study 2.1 Stage 4 Latent Heat Calculation Mathematical relationship between latent heat of vaporization for water, hfg(water), mass of the water, and the total heat of vaporization of water, Ql(latent water), can be stated as: And, Ql(latent water) = hfg (water). m Eq. 2.20

14 l(latent water) = hfg (water). ṁ Eq Where, Q l(latent water) = Latent heat of vaporization of water required to evaporate a specific mass of water, isothermally l(latent water) = Latent heat of vaporization flow rate required to evaporate a specific mass of water, isothermally, over a given period of time m = Mass of water being evaporated ṁ = Mass flow rate of water = 60 tons/hr = (60 tons/hr). (907.2 kg/ton) = 54,432 kg/hr, same as part (a) (i) hfg (water) = latent heat of vaporization for water = 2257 kj/kg {From the steam tables and Table 2.3} Then, by application of Eq. 2.21: l(latent water) = hfg (water). ṁ l(latent water) = (2257 kj/kg). (54,432 kg/hr) l(latent water) = 122,853,024 kj/hr Since there are kj per BTU, Or, l(latent water) = (122,853,024 kj/hr)/(1.055 kj/btu) l(latent water) = 116,448,364 BTU/hr (v) Calculate the heat reqd. to heat the steam from 100 C, 1-atm (102 KPa, or 1-bar) to 500 C, 2.5 MPa superheated steam The 5 th stage of the overall power generating system is illustrated in Figure A-2.8, below. Since this stage involves no phase change, the heat absorbed by the steam is sensible heat. In superheated steam phase, the heat required to raise the temperature and pressure of the steam can be determined using the enthalpy difference between the initial and final conditions.

15 Stage 5, Sensible Heat He at Boiler 100 C Super Heated Steam 500 C 500 C, 2.5 MPa Steam Turbine Figure A-2.8, Case Study 2.1 Stage 5 Sensible Heat Calculation Given: Ti = 100 C Pi = 1 - Atm. Note: At 100 C, the saturation pressure is 1- Atm, 1-Bar, or 102 kpa Tf = 500 C Pf = 2.5 MPa For the initial and final temperature and pressure conditions stated above, the enthalpy values, as read from saturated steam table excerpt in Table 2.4a and the superheated steam table excerpt in Table 2.4, are as follows: hi = 2676 kj/kg at 100 C, 1-Atm hf = 3462 kj/kg at 500 C, 2.5 MPa

16 Table 2.4a. Excerpt, Saturated Steam Table, SI Units Equations for determining the heat required to boost the steam from 100 C, 1-Atm to 500 C, 2.5 MPa are as follows:

17 ΔQ steam = (hf - hi ). m Eq steam= (hf - hi ). ṁ Eq Where, ΔQ steam = Addition of heat required for a specific change in enthalpy steam = Rate of addition of heat for a specific change in enthalpy hi = Initial enthalpy hf = Final enthalpy m = Mass of steam being heated ṁ = Mass flow rate of steam as calculated in part (a) of this case study = 54,432 kg/hr {From Part (a)} Then, by applying Eq. 2.23: steam= (hf - hi ). ṁ steam = (3462 kj/kg kj/kg). (54,432 kg/hr) steam = 42,783,552 kj/hr Since there are kj per BTU, Or, s(water) = (42,783,552 kj/hr)/(1.055 kj/btu) s(water) = 40,553,130 BTU/hr After assessing the heat added, per hour, during each of the five (5) stages of the steam generation process, add all of the heat addition rates to compile the total heat addition rate for the power generating station. The tallying of total heat is performed in BTU s/hr as well as kj/hr. Total Heat Addition Rate in kj/hr: Total Heat Required to Generate 500 C, 2.5 MPa steam from -10 C Ice, at 54,432 kg/hr = 1,143,072 kj/hr + 18,153,072 kj/hr + 22,807,008 kj/hr + 122,853,024 kj/hr + 42,783,552 kj/hr = 207,739,728 kj/hr

18 Total Heat Addition Rate in BTU s/hr: Total Heat Required to Generate 500 C, 2.5 MPa steam from -10 C Ice, at 54,432 kg/hr = 1,083,481 BTU/hr +17,206,703 BTU/hr +21,618,017 BTU/hr +116,448,364 BTU/hr +40,553,130 BTU/hr = 196,909,695 BTU/hr 4. If all of the working fluid, or steam, discharged from the turbine in Case Study 2.1 is reclaimed, reheated and returned to the turbine, what would be the overall system efficiency? Assume that the mass flow rate is 58,860 kg/hr, or 65 tons per hour. Solution Strategy: Energy and process flow pertaining to this special case scenario of Case Study 1 is depicted in Figure A-2.9. In order to derive the efficiency for this scenario where all of the steam discharged from the turbine is reclaimed and used as working fluid fed into the boiler in the last, superheating, stage, we need to determine the rate of heat addition required to raise the temperature from 150 C to 500 C and the pressure from 50 kpa to 2.5 MPa. This rate of addition of heat can be determined using the following formula: steam= (hf - hi ). ṁ Eq Once steam is determined, we can convert it into equivalent power (kw or MW) units for computation of efficiency through the following formula: Total Station Energy Efficiency, in Percent = Power Output / Power Input * 100 Or, Total Station Energy Efficiency, in Percent = (Power Output in MW / steam in MW) * 100 Note: With the exception of the provisions stipulated in the problem statement above, all of the pertinent given data from Case Study 1 remains the same, as stated below: Solution: Given: Pshaft = (10 MW) / g = (10 MW) / (0.9) = 11.11x10 6 W = 11.11x10 6 J/s

19 ṁ = 58,860 kg/hr hi = 2780 kj/kg = 2780x10 3 J/kg {See Table 2.4} hf = 3462 kj/kg = 3462x10 3 J/kg {See Table 2.4} Then, by applying Eq. 2.23: Or, steam= (hf - hi ). ṁ = (3462 kj/kg kj/kg ). 58,860 kg/hr = 40,142,520 kj/hr steam in kw = (40,142,520 kj/hr) / (3600 sec/hr) = 11,151 kj/sec Or, since 1J/sec = 1W, and 1kJ/sec = 1kW, steam in kw = = 11,151 kw Or, Then, steam in MW = (11,151 kw)/(1000kw/mw) = MW Total Station Energy Efficiency, in Percent = Power Output / Power Input * 100 = 10 MW/11.15MW = 89.68%

20 Boiler / Steam Generator 150 C, 50 kpa Steam 150 C, 50 kpa Low Pressure Steam. h i = 2780 kj/kg Steam 500 C, 2.5 MPa 500 C, 2.5 MPa Steam to Turbine. Turbine Efficiency, t = 30% h f = 3462 kj/kg Generator Efficiency, g = 90% Electrical Output = 10 MW Figure A-2.9, Case Study 2.1, Mass Flow Rate Analysis

21 Properties of Superheated Steam Metric/SI Units Abs. Temp. Press. C MPa (Sat. T, C ) n (81.33) h s n (99.61) h s n (179.89) h s n (223.99) h s n (233.86) h s n (250.36) h s Table 2.4. Excerpt, Superheated Steam Table, SI.

22 Segment 3 Solutions for self-assessment problems and questions: 1. An ideal heat engine always includes a boiler with superheating function. Answer: B. False. Heat engines equipped with superheating function, albeit common, is just one type of heat engine. 2. In the heat engine represented by the enthalpy vs. entropy graph in Figure 3.22, the heat is added to working fluid in: Answer: (iv) Steam generation stage and the Steam superheating stage This is evident from Figure A-3.1. Enthalpy is converted to the kinetic mechanical energy of the turbine in the path 4 to 5 3. In the heat engine represented by the enthalpy vs. entropy graph in Figure 3.22, the energy contained in the superheated working fluid is converted into the rotational kinetic energy in: Answer: (ii) Process transition from point 4 to 5. Figure A-3.1 is a duplicate of Figure 3.22, referred to in the problem statement. As evident from Figure A-3.1 below, path 4 to 5 represents the conversion of enthalpy contained in the superheated steam into mechanical kinetic energy of the turbine. h Liquid Vapor Liquid &Vapor 5 s

23 Figure A-3.1.; Heat Cycle in a Rankin Engine with Superheat, h vs. s. 4. A thermodynamic system consists of a Rankin engine with superheat function. The enthalpy vs. entropy graph for this system is shown in Figure 3.33 (Figure A-3.2) Vapor a 15 h (BTU/lbm) 00 Liquid c Liquid &Vapor s {BTU/(lbm- R)} Figure A-3.2 (Same as Figure 3.33) Heat Cycle in a Rankin Engine with Superheat, h vs.. The mass flow rate of the system or working fluid is 100 lbm/sec. Answer the following questions based on the data provided: a) If the enthalpy of the fluid is approx BTU/lbm and the entropy of the system is approximately BTU/ (lbm- R), what phase or stage is the system in: Note: Figure 3.33, referred to in the problem statement is duplicated in form of Figure A- 3.2, with pertinent annotations, for solution illustration purposes. Solution/Answer: (i) Work producing turbine stage The point representing an enthalpy of approximately 1850 BTU/lbm and the entropy of approximately BTU/(lbm- R) is labeled as a in the Figure A-3.2 enthalpy entropy diagram. This point is clearly on thermodynamic transition path 4 to 5. This path represents the work producing turbine stage.

24 b) If the enthalpy of the subcooled water entering the boiler is 900 BTU/lbm and the enthalpy of the water at a downstream point in the boiler is 1080 BTU/lbm, what is approximate amount of heat added in MMBTU per hour? Assume that there is no heat loss. Answer: (ii) 1.08 MMBTU/hr Solution: Since we are allowed to assume that there is not heat loss, the difference between the enthalpy of the subcooled water entering the boiler is and the enthalpy of the water at a downstream point in the boiler would be equal to the heat energy added on per lbm basis. Also, the mass flow rate, ṁ, of the working fluid is given. Heat energy delivered to the water, in BTU s/sec Where, Q = (hf hi). (ṁ) ṁ = Mass flow rate of the working fluid or water = 100 lbm/sec {Given} hi = Enthalpy of steam entering the turbine = 900 BTU/lbm 1860 {Given} hf = Enthalpy of steam discharged = 1080 BTU/lbm {Given} Then, application of Eq. 3.2 would yield: Heat energy delivered to the water, in BTU s/sec = (1080 BTU/lbm 900 BTU/lbm). (100 lbm/sec) = 18,000 BTU/sec Heat energy delivered to the water, in BTU s/hr = (18,000 BTU/sec). (60 sec/hr.) = 1,080,000 BTU s/hr Since there are 1,000,000 BTU s per MMBTU,

25 Heat energy delivered to the water, in MMBTU s/hr 1, 080, 000 BTU ' s / hr 1,000,000 BTU ' s = MMBTU/hr Therefore, the answer is: (ii) 1.08 MMBTU/hr Ancillary to part (b): As apparent upon examination of Figure A-3.2, the rise in enthalpy stated in part (b) - from 900 BTU/lbm to1080 BTU/lbm - occurs while the working fluid remains in subcooled realm. If the final enthalpy, hf, were in the vapor-liquid mixture region, the solution for this part of the problem would have required more steps. This is because the enthalpy of the working fluid at final point would be a sum of the saturated vapor and liquid enthalpies, added in proportion determined by the humidity ratio, ω. c) If the enthalpy of the working fluid is 1440 BTU/lbm and the entropy is BTU/(lbm- R), what is the phase of the working fluid? Answer: (iii) A mixture of vapor and liquid To address this question, we must locate the point on the graph where h = 1440 BTU/lbm and s = BTU/(lbm- R). This point is labeled as point c on the enthalpy vs. entropy graph in Figure A-3.2. Point c lies in the region that falls between saturated liquid line and the saturated vapor line. In other words, at point c, the working fluid is in a phase that consists of a mixture of liquid and vapor.

26 Appendix B Steam Tables These steam tables, copyright ASME, published with ASME permission, do not include the heat of evaporation value, hfg, values for the saturation temperature and pressures. The saturated steam tables presented in this text are the compact version. However, the hfg values can be derived by simply subtracting the available values of hl from hv, for the respective saturation pressures and temperatures. In other words: hfg = hv - hl Appendix C Common Units and Unit Conversion Factors Included with Appendix B

27 Appendix D Common Symbols