CHAPTER 19 HEAT ENGINES AND REFRIGERATORS

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1 CHAPTER 19 HEAT ENGINES AND REFRIGERATORS In this chapter, we will examine devices that turn heat into work. I swear, I haven t confused you people for Mechanical Engineering majors! Biological processes (such as muscle action) rely on the same physics. Slide 1

2 19.1 Turning Heat Into Work To start, recall that we ve always been calculating W as the work done on a system. Since we re now looking at what a system can do for us, we re more interested in the work done by a system. These quantities are equal in magnitude and opposite in sign. We ll use the new notation W s = W, where W s is the work done by the system. We can immediately make two claims based on earlier chapters: 1. On a pv diagram, W s is the area under the curve 2. The first law of thermodynamics can be written E th = W s + Q or Q = W s + E th The latter equation can be summed up nicely: Any energy transferred into a system as heat is either used to do work or it is stored within the system as an increased thermal energy. Slide 2

3 19.1 Turning Heat Into Work Energy Reservoirs A couple of chapters ago, we went through a problem in which a hot pan was plunged into a sink of water. The pan cooled down and the water warmed up. What if the pan was instead plunged into the ocean? Due to the incredibly large mass of water, we would find that T water would be essentially zero. An energy reservoir is an object or a portion of the environment that is so large, its temperature does not change when heat is transferred into it or out of it. Slide 3

19.1 Turning Heat Into Work Some more notation: Q H (Q C ) is the amount of heat transferred to or from a hot (cold) reservoir. By definition, both Q H and Q C are positive quantities.

4 19.1 Turning Heat Into Work Some more notation: Q H (Q C ) is the amount of heat transferred to or from a hot (cold) reservoir. By definition, both Q H and Q C are positive quantities. The direction of heat transfer will always be clear for any particular device. The top figure shows a copper bar, with its two ends immersed in hot and cold reservoirs. The bottom figure is an energy-transfer diagram for this situation, showing the directions of both Q s. Try not to confuse the direction of heat transfer terminology with direction in a vector sense. It simply indicates whether we are transferring heat into or out of a system or reservoir. Slide 4

5 19.1 Turning Heat Into Work How does the first law relate to the bottom figure from the previous page? Q = W s + E th refers to the system, so Q is the net heat into the system; according to the figure, this is Q = Q H Q C. The copper bar does no work, so W s = 0. When the bar is initially placed between the reservoirs, it will start to heat up. Thus, Q H > Q C, Q > 0, and E th > 0; this means that the temperature will increase. Eventually, the bar will heat up to a point where its temperature stops changing. Then, E th = 0, Q = 0, and Q H = Q C. This means that all of the heat transferred into the bar from the hot reservoir is transferred out of the bar into the cold reservoir Slide 5

19.1 Turning Heat Into Work The process shown in this figure is allowable according to the first law, if Q H = Q C. However, it is forbidden by the 2 nd law, which says that for an isolated system (i.

6 19.1 Turning Heat Into Work The process shown in this figure is allowable according to the first law, if Q H = Q C. However, it is forbidden by the 2 nd law, which says that for an isolated system (i.e. with no external work), the transfer of heat from a colder object to a hotter object is not possible. This is a problem for those of us who enjoy a cold beer on a hot day. We d like to have a device that makes the beer colder by transferring heat into the hotter environment. Damn you, thermodynamics! Slide 6

19.1 Turning Heat Into Work Turning work into heat is quite simple. Friction accomplishes this well rub your hands together, and you re converting mechanical energy into thermal energy.

7 19.1 Turning Heat Into Work Turning work into heat is quite simple. Friction accomplishes this well rub your hands together, and you re converting mechanical energy into thermal energy. When you stop rubbing, your hands will cool down, as heat is transferred into the colder environment. The 2 nd law is quite happy to permit both of these processes, which are illustrated in the energy-transfer diagram shown below (note that work is shown entering or leaving from the side it does not originate or terminate at a reservoir). Slide 7

8 19.1 Turning Heat Into Work Turning heat into work isn t so easy. It s not impossible we ve already examined isothermal expansion - but the problem is that this process results in a change in the state of the system (it s expanded). If we wanted to turn more heat into work, we d need to find a way to compress the system to its initial volume and we can t do this by running the original isothermal process in reverse, since that would exactly undo the original process, turning work into heat! What we seek, then, is a device that transforms heat into work, while returning to its initial state at the end of the process, ready for continued use. This describes a perfect engine (which, unfortunately, doesn t exist). Slide 8

9 19.2 Heat Engines and Refrigerators A closed-cycle device is one that cyclically returns to its initial state (that is, all of the state variables pressure, volume, temperature, etc. are returned to) in order to repeat the same process over and over. A heat engine is a closed-cycle device that extracts heat Q H from a hot reservoir, does some sort of useful work, and then exhausts heat Q C to a cold reservoir. Examples will be given in class. The energy-transfer diagram of a heat engine is shown here. Slide 9

19.2 Heat Engines and Refrigerators Since the temperature of the heat engine returns to its original value at the end of the cycle, its net change in thermal energy ( E th ) net is zero.

10 19.2 Heat Engines and Refrigerators Since the temperature of the heat engine returns to its original value at the end of the cycle, its net change in thermal energy ( E th ) net is zero. The first law then tells us that for a full cycle, Q W s = 0. As we re concerned about the work output from the engine, we ll rewrite W s as W out. Thus, W out = Q = Q H Q C. The figure illustrates this nicely. The thermal efficiency of a heat engine is the ratio of the output work to the heat required: η = W out Q H = 1 Q C Q H. A perfect heat engine would have η = 1. Such an engine would produce no waste heat. An elegant proof that no such heat engine can exist is provided on p. 533 of the text. Slide 10

11 19.2 Heat Engines and Refrigerators A heat-engine example is shown here, and will be discussed thoroughly in class. It s analyzed mathematically in Example 19.1 of the text. Slide 11

19.2 Heat Engines and Refrigerators At first glance, the 2 nd law seems to suggest that a refrigerator (or an conditioner) is physically impossible, since both are used to transfer heat energy from a

12 19.2 Heat Engines and Refrigerators At first glance, the 2 nd law seems to suggest that a refrigerator (or an conditioner) is physically impossible, since both are used to transfer heat energy from a colder region (the interior of the fridge or of your apartment) to a warmer region (the air of your kitchen or outside). But remember, neither device is isolated they both have a source of external work. Hence the electrical cables. Physically speaking, a refrigerator is a closed-cycle device that uses external work W in to remove heat Q C from a cold reservoir and exhaust heat Q H to a hot reservoir. This is illustrated in the energy-transfer diagram to the right. Slide 12

13 19.2 Heat Engines and Refrigerators As with the heat engine, the cyclical nature of the process requires that E th = 0. Thus, by the first law, Q H = Q C + W in This equation makes it clear that a refrigerator must exhaust more heat to the outside than it removes from the inside. So no, you can not cool down an insulated room by running a refrigerator in it with the door open! The coefficient of performance of a refrigerator is K = Q C W in. That is, a better refrigerator (one with a high value of K) requires less work to remove a given amount of heat. A perfect refrigerator requires no work, and would have K =. However, the resulting energytransfer diagram would look like the one from slide 22, which we know is not possible with no input, we would be expecting spontaneous heat flow from cold to hot, which violates the 2 nd law. Slide 13

14 Problem #1: Heat Engines Rank in order, from largest to smallest, the thermal efficiencies of the four heat engines illustrated in the figure. Solution: in class RDK CQ Slide 14

15 Problem #2: Heat Engines A heat engine extracts 55 kj of heat from the hot reservoir each cycle and exhausts 40 kj of heat. What are the thermal efficiency and the work done per cycle? Solution: in class RDK EX 19.3 Slide 15

19.3 Ideal-Gas Heat Engines Let s take a look at an arbitrary closed-cycle heat engine. As shown below, a full cycle consists of an expansion and a subsequent compression.

16 19.3 Ideal-Gas Heat Engines Let s take a look at an arbitrary closed-cycle heat engine. As shown below, a full cycle consists of an expansion and a subsequent compression. We know that the work done BY the gas is equal to the area under the curve. In the cycle shown here, the work done by the gas during expansion is greater than the work done on the gas during compression. There is a net positive work done, which is determined by the area enclosed within the pv curve. Clearly, the trajectory of the cycle must be clockwise for a heat engine. Slide 16

17 19.3 Ideal-Gas Heat Engines Ideal-Gas Summary Our upcoming discussion of heat engines will involve the thermodynamics of various idealgas processes. It s a good idea to tabulate them here. You ve seen them all before, except for the ones that are circled. We ll omit the proofs. Slide 17

18 19.3 Ideal-Gas Heat Engines Solving heat-engine problems The general procedure for solving heat-engine problems is as follows: Draw a pv diagram that includes the trajectory of the closed cycle. Use the ideal-gas law to compute p, V, n, and T at one point in the cycle. Use the ideal-gas law and equations from the previous slide to calculate p, V, and T at the beginning and end of each process (assume that n is constant, although technically speaking this isn t exactly true for most engines). Calculate Q, W s, and ΔE th for each process. Calculate W out (the output from one full cycle) by summing the W s from each cycle. For very simple cycles, this can be done by finding the area inside the pv trajectory. Add each of the positive values of Q to find Q H. Verify that ( E th ) net = 0. Since this is a requirement for a closed cycle, if you do not get this answer, you ve done something wrong. Calculate the thermal efficiency η based on W out and Q H. A good example can be found on pp of the text. Slide 18

19 Problem #3: Simple Heat Engine RDK STT 19.3 What is the thermal efficiency of this heat engine? A 0.10 B 0.20 C 0.25 D 4 Slide 19

compression stroke (2 3) at point 2, the spark plug fires.

20 19.3 Ideal-Gas Heat Engines The Otto Cycle The engine in your (non-diesel) car can be modeled as an Otto Cycle. (5 1) fuel and air is input to the cylinder in an isobaric process (the intake ) (1 2) the mixture is compressed as the piston moves toward the spark plug during the adiabatic compression stroke (2 3) at point 2, the spark plug fires. This rapidly heats the fuel in an isochoric process (the piston doesn t have time to move) (3 4) the hot, high-pressure gas pushes the piston outward during the adiabatic power stroke (4 1) an exhaust valve opens to allow the gas pressure to quickly drop back to its initial value (1 5) isobaric cooling and compression complete the cycle (the exhaust ) 5 Slide 20

21 Problem #4: Otto Cycle RDK EX For the Otto cycle a) Show that the work done per cycle is W out = nr 1 γ T 2 T 1 + T 4 T 3 b) Use the adiabatic connection between T 1 and T 2 and also between T 3 and T 4 to show that the thermal efficiency of the Otto cycle is η = 1 1 r (γ 1) where r = V max / V min is the engine s compression ratio c) Graph η vs. r out to r = 30 for a diatomic gas. 5 Solution: I will run through this in class Slide 21

19.3 Ideal-Gas Heat Engines The Brayton Cycle A gas turbine engine undergoes the Brayton Cycle. We ll discuss it briefly in class, then I ll let you read the analysis on pp. 536-538.

22 19.3 Ideal-Gas Heat Engines The Brayton Cycle A gas turbine engine undergoes the Brayton Cycle. We ll discuss it briefly in class, then I ll let you read the analysis on pp Don t worry about memorizing the results just try to find the similarities and differences between this cycle and the Otto cycle. You may have to use this information in an upcoming assignment. Slide 22

23 19.4 Ideal-Gas Refrigerators Picture a heat engine operating in reverse. That is, the trajectory in the pv diagram is now counterclockwise. Starting from point 4 The gas is adiabatically compressed (4 3), increasing T and p. It flows through a high-temperature heat exchanger, where the gas is isobarically cooled (3 2). It then expands adiabatically (2 1), leaving it significantly colder than when it started. Finally, the gas flows through a low-temperature heat exchanger, where it warms up to the starting temperature (1 4). Slide 23

24 19.4 Ideal-Gas Refrigerators If the low-t heat exchanger (1 4) is a chamber of air which surrounds the pipe through which the cool gas is flowing, this process must cool the air as it warms the gas in the pipe. This air can represent the inside of a refrigerator. Since we re reversed the trajectory in the pv diagram, we have reversed the sign of the work done by the system. It s now negative. External work must be done ON the system (your fridge won t work if it s not plugged in). This work is used to extract heat Q C from the cold reservoir and exhaust a larger amount of heat Q H = Q C + W in into the hot reservoir. Slide 24

25 19.4 Ideal-Gas Refrigerators Let s clarify something heat is always transferred from a hotter object to a colder object, even though this seems to be the opposite of what the refrigerator is accomplishing. This requires that: The gas temperature must be lower than the coldreservoir temperature (i.e. the air inside the fridge) The gas temperature must be higher than that of the hot-reservoir (i.e. the air outside of the fridge) These two statements seem incompatible, since a refrigerator is only useful if it s colder inside than outside. However, the gas temperature is different during these two heat-exchange processes; this is accomplished by the adiabatic compression and expansion processes. Slide 25

26 Problem #5: Air Conditioner Parameters An air conditioner removes 5.0 x 10 5 J/min of heat from a house and exhausts 8.0 x 10 5 J/min of heat to the outdoors. a) How much power does the air conditioner s compressor require? b) What is the air conditioner s coefficient of performance? Solution: In class RDK EX Slide 26

19.5 The Limits of Efficiency Our study of heat engines has shown that the thermal efficiency η can be calculated by examining the trajectory in the pv diagram.

27 19.5 The Limits of Efficiency Our study of heat engines has shown that the thermal efficiency η can be calculated by examining the trajectory in the pv diagram. A logical question to ask is, given the temperatures T H and T C of hot and cold reservoirs, what is the highest possible efficiency of a heat engine? Equivalently, we could seek the highest possible coefficient of performance K for a refrigerator operating between T H and T C. It turns out that such a maximum η and K does indeed exist. Showing this requires some slick qualitative reasoning. To start, this figure illustrates a heat engine and the corresponding refrigerator (that is, they follow the exact same cycle, but in opposite directions). These are referred to as a perfectly reversible heat engine. Slide 27

19.5 The Limits of Efficiency Now, we re going to join these two devices together. Since the devices follow identical cycles, we can use the heat engine s W out as the refrigerator s W in.

28 19.5 The Limits of Efficiency Now, we re going to join these two devices together. Since the devices follow identical cycles, we can use the heat engine s W out as the refrigerator s W in. For this combined device, there is now no external input or output of work, and no heat transferred. In practice, this is a completely useless device! However, we can squeeze some helpful information out of it. Let s say that this perfectly reversible engine has an efficiency η. I m going to claim that I have a heat engine that operates between the same T H and T C with an even higher efficiency. Remember, η = W out Q H and W out = Q H Q C. Therefore, my super-efficient device requires less heat Q H to produce the same W out, resulting in a lower Q C as well. Slide 28

19.5 The Limits of Efficiency When I connect my super-efficient heat engine to the original refrigerator, as shown in the figure, the heat it exhausts to the cold reservoir is less than the heat

29 19.5 The Limits of Efficiency When I connect my super-efficient heat engine to the original refrigerator, as shown in the figure, the heat it exhausts to the cold reservoir is less than the heat extracted from the cold reservoir by the refrigerator. Also, the heat extracted from the hot reservoir is less than the heat exhausted to the hot reservoir by the refrigerator. The result would be a device that transfers heat from a cold reservoir to a hot reservoir with no external assistance and as we know, this situation is prohibited by the 2 nd law of thermodynamics. Our conclusion therefore is that no heat engine operating between reservoir temperatures T H and T C can have an efficiency greater than that of a perfectly reversible engine. This is often considered to be another informal statement of the 2 nd law. Slide 29

30 19.6 The Carnot Cycle Since the highest possible efficiency is found in a perfectly reversible engine, let s try to design one. Here are the requirements: 1. The mechanical interactions (i.e. motion of a piston) must be frictionless, with no heat transfer 2. The thermal interactions in which heat is transferred must be isothermal ( E th = 0). These processes must occur very slowly in order to be reversible (see p. 542 for an argument as to why this is the case). An engine that obeys these two very strict requirements is called a Carnot engine. Based on our discussion on the last slide, we can make the claim that all Carnot engines operating between the same two reservoir temperatures T H and T C have the same efficiency. Slide 30

31 19.6 The Carnot Cycle Design and Analysis of a Carnot Engine To adhere to the requirements stated on the previous slide, our Carnot cycle must proceed as follows. Starting from point 1: 1. The gas is isothermally compressed while in thermal contact with the cold reservoir. To keep temperature constant, heat energy Q C is removed from the gas. 2. The gas is adiabatically compressed while thermally isolated from its environment. This increases the gas temperature until it reaches T H. 3. The gas expands isothermally while in contact with the hot reservoir. heat energy Q H is removed from the gas. 4. The gas expands adiabatically until the temperature decreases back to T C. Slide 31

32 19.6 The Carnot Cycle As with all engine efficiencies, the efficiency of a Carnot engine is η = W out = 1 Q C Q H Q H With the help of the chart on slide 17 of these notes, it can be shown (pp ) that η Carnot = 1 T C T H That is, it depends only on the reservoir temperatures. A similar analysis can be performed for a Carnot refrigerator, resulting in K Carnot = T H T C These equations show that device performance is maximized for T H T C. However, for practical reasons, the temperature difference between reservoirs often isn t very large. This limits η Carnot and K Carnot. T C Slide 32

33 Problem #6: Fun with Refrigerators The coefficient of performance of a refrigerator is 5.0 The compressor uses 10 J of energy per cycle. a) How much heat energy is exhausted per cycle? b) If the hot-reservoir temperature is 27 C, what is the lowest possible temperature of the cold reservoir, in C? Solution: In class RDK EX Slide 33

A) W B) - W C) 0 D) cannot be determined

A) W B) - W C) 0 D) cannot be determined Chapter 19. Heat Engines and Refrigerators That s not smoke. It s clouds of water vapor rising from the cooling towers around a large power plant. The power plant is generating electricity by turning heat