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CHAPTER 3 HEAT ENGINES AND THE SECOND LAW OF THERMODYNAMICS INTRODUCTION In the last chapter we utilized an atomic kinetic model to develop an equation for the internal energy of an ideal polyatomic gas and to develop the equation of state of such an ideal gas (the ideal gas law). We then discussed the different modes of energy transfer across the boundary of a system, in particular heat and work. Utilizing these concepts and the basic assumption of conservation of energy, we developed the first law of thermodynamics. Using the first law, we were able to derive some useful relationships between the various heat capacities of ideal gases. In this chapter we wish to investigate the second law of thermodynamics: heat is always observed to flow from a hotter system to a colder system, never the other way around. Likewise, if we use a paddle wheel to stir a container filled with water, we find that the water will heat up, but heating up the water will never cause the paddle wheel to rotate. This later example would seem to indicate that the kinetic energy associated with the organized motion of the paddle wheel can be changed into a random form of internal energy characterized by molecular motion, whereas the random form of internal energy can never be “reorganized” to produce the organized motion of the paddle wheel. The connection of the second law to the microscopic motion of particles is a fundamental one. However, the early study of thermodynamics was undertaken before the atomic theory was well established. Early workers in this field found that a macroscopic property called entropy could be used to predict the direction in which energy would flow in a given process. Although the second law in some way determines the direction of energy transfer, the first law puts some limits on the amount of energy that can be exchanged in a certain process. A real process cannot occur, however, unless both the first and second laws are satisfied. As mentioned earlier, much of the present theory of thermodynamics was developed during the age of industrialization in an attempt to develop more efficient engines. In this chapter we will discuss heat engines, refrigerators, and heat pumps, and the techniques for calculating the efficiency of these devices. We will find that the first and second laws taken together place some limitations upon the efficiencies of these devices. HEAT ENGINES As mentioned in the introduction, a paddle wheel in a water bath can be used to convert organized mechanical motion into heat. In fact all the mechanical motion of the paddle wheel can be converted into heat. It is not such an easy task, however, to convert heat energy into mechanical energy (organized energy which can be used to do work). A device which accomplishes this task is called a heat engine. In the last chapter, we actually introduced a process which acts like a heat engine. Let's look again at this process (Fig. 3.1). Process E p F is an isobaric expansion, where the work done by the gas is given by [EF œ TE ZF ZE (3.1) During this process, the temperature of the gas also increases (as can be seen from the isothermal curves), increasing the internal energy of the gas. We can find the heat added to the system during this process using the first law UEF œ ?IEF [EF (3.2) Since the work done by the gas and the change in internal energy of the gas are both positive, the heat added to the system is positive, i.e., UEF !. Process F p G is isochoric, with no increase in the volume of the system. In this process, there is no work done by (or on) the gas. However, the internal energy decreases due to the drop in temperature of the gas. Thus, in process F p G heat is removed from the system, so that UFG !. Process G p E is an isothermal compression, where work is done on the system as the volume decreases. But, since this process is isothermal, the internal energy remains constant. Thus, the heat added to the system in this process must be equal to the work done by the system. Since the work done by the system is negative, heat must be removed from the system during this process, so we find that UGE !. Chapter 3: The Second Law of Thermodynamics 2 P TB > TA A PA B TB C PB TA VA VB V Fig. 3.1 A T -Z diagram for an ideal gas as it is carried through the cyclic process EpFpG pE. Process EpF is isobaric, process FpG is isochoric, and process G pE is isothermal. The work done in each process is the area under the T -Z curve. The work is positive if the volume increases and negative if the volume decreases during a process (it is zero if the volume does not change). When this cycle is completed, the positive work done by the gas is greater than the negative work done on the gas, so that the net work done in the complete cycle is positive. This process, then, acts like a heat engine. The work done in the process G p E is negative, since the volume decreases, but is still just the area under the T -Z curve for that process. Obviously, when this cycle is completed, the positive work done by the gas in process E p F is greater than the negative work done on the gas in process G p E, so that the net work done in the complete cycle is positive. This cyclic process, therefore, provides us a mechanism whereby we can supply an amount of heat energy to a gas and accomplish a net positive mechanical work. Any cyclic process which absorbs heat energy and accomplishes a net positive mechanical work is known as a heat engine. But you will notice that we were not able to convert all the heat energy supplied to the system into useful mechanical work - some heat was removed from the system. Heat is added to the system during process E p F, but removed from the system during processes F p G and G p E. Since the total change in the internal energy of the system must be zero in this cyclic process, the net positive work done by the system must be equal to the net positive heat added to the system. This means that |Uin | |Uout |, or lUEF l lUFG l lUGE l, where |Uin | and |Uout | represent the magnitude of the heat flow into and out of the system. As you might imagine, there are a large number of different processes which could be used as heat engines. It would obviously be desirable to devise a system whereby all the heat energy supplied to the system could be used to do useful mechanical work. In fact, one of the major tasks of classical thermodynamics was to determine which processes were the most efficient. In order to determine this, we need to examine some of the more fundamental aspects of heat engines. No matter what the actual process, we find that the following characteristics are true of all heat engines: 1. 2. 3. 4. They receive energy from one or more high-temperature sources (such as solar heat, burning coal, or a nuclear reaction). They convert part of this energy into useful (organized) mechanical work (or energy). The remaining energy is released to one or more low-temperature sinks (such as a river or cooling tower). They all work in a cycle. A schematic of a simple heat engine with only one source and one sink is shown in Fig. 3.2. The best practical example of a heat engine is a steam power plant (see Fig. 3.3). Chapter 3: The Second Law of Thermodynamics 3 High-Temperature Source Qin Worknet,out Heat Engine Qout Low-Temperature Sink Fig. 3.2 A schematic of a ideal heat engine. Energy in the form of heat is added to the heat engine from a high-temperature source. Part of this added energy is then converted into useful mechanical work, while the remaining amount is dumped into a low-temperature sink. The circle which designates the actual heat engine represents the fact that the engine operates in a continual cycle. Energy Source (coal fire or nuclear reactor) System Boundary Qin Boiler Wout Win Turbine Pump Condenser Qout Energy Sink (river or cooling tower) Fig. 3.3 Schematic of a steam power plant. The net work out is the shaft work done by the turbine less the work necessary to drive the pump. The net heat added to the system is the heat added at the high temperature source less the heat removed at the low temperature sink. This might represent either a coal-fired steam plant or a nuclear steam plant. The only real difference is the operating temperatures. The maximum allowed temperatures for coal-fired steam plants is approximately 500 ‰ C, whereas the maximum allowed temperatures for nuclear power plants is approximately 300 ‰ C. Chapter 3: The Second Law of Thermodynamics 4 A steam power plant is an external combusion engine, as opposed to the internal combution gasoline engine. In this external combustion engine, the fuel is burned outside the engine and the heat of this combustion is transfered to the boiler in the form of heat. In the boiler, the working fluid (water and steam) is heated to produce high temperature steam that is directed by a nozzle onto a turbine. This turbine produces shaft work which turns a generator and produces electricity. The steam from the turbine is then cooled in a condensing unit and then pumped back into the boiler, where it is again heated. You might wonder why one would want a condenser unit since some of the energy is lost from the system at this point. If there were no condenser unit to cool the steam, the temperature of the steam leaving the turbine would rise until it becomes nearly equal to the temperature of the entering steam. This would produce a large back-pressure and reduce the difference in pressure across the turbine. If this occurred the efficienty of the nozzle-turbine subsystem would drop. Although each component of the steam power plant is actually an open system, with mass flowing into and out of it, the system as a whole is a closed system, with a constant amount of steam contained within the subsystems and the connecting pipes. If we take the whole system as our thermodynamic system, we are dealing with a closed system for which the total internal energy change in a complete cycle is zero! This means, according to the first law, * .I œ * $ U * $ [ œ ! (3.3) or * $U œ * $[ Ê |U38 | |U9?> | œ |[9?> | |[38 | Ê |U8/>ß38 | œ |[8/>ß 9?> | (3.4) where, again, the symbols |Uin | and |Uout |, as well as |[38 | and |[9?> | represent magnitudes. This equation simply states that the net work done by the system is equal to the net heat added to the system. That is, the heat added to the system at the high-temperature source less the heat given up at the low-temperature sink is equal to the shaft work done by the turbine less the shaft work done on the pump. The amount of heat that is removed from the system at the low-temperature sink represents the amount of heat “wasted” in order to complete the cycle. For those situations where there is wasted heat, the net work done by the system is obviously less than the energy put into the system at the high-temperature source. Thus, only part of the heat transfered to the system is converted to useful work. We define the fraction of the heat energy which is converted into useful work in a heat engine as the thermal efficiency ( of the heat engine, or (œ R /> [ 9<5 S?> |[8/>ß9?> | œ L/+> M 8 |U38 | (3.5) Notice that this definition can also be expressed in the form (œ H/=3</. S?>:?> V/;?3</. M 8:?> (3.6) In our discussion so far, we have implied that the temperature of the high-temperature source and the low-temperature sink do not change during the operation of the system. At first thought, this might seem difficult to accomplish in practice. However, a large enough body of water (say the ocean) can be treated as a constanttemperature source or sink, since the energy removed from, or added to, the ocean would not appreciably increase or decrease the average temperature of the ocean. The same is true of the atmosphere, although local heating or cooling would occur. In practice, we don't really need an ocean to maintain a reasonably constant temperature source. All we need is a body which has a large enough “thermal mass” (the product of mass and specific heat capacity) that the temperature of the body can be maintained at an approximately constant temperature as heat energy is added to or removed from the body, or a thermistatically controlled heater. Such bodies are called heat reservoirs. Thus, we will often assume that the sources and sinks of our heat engine are heat reservoirs which are maintained at constant temperature. We will designate the temperature of the high-temperature reservoir as XL and the temperature of the low-temperature reservoir as XP . Likewise, we will use |UL | to designate the magnitude of the heat added to or removed from the system at temperature XL and |UP | as the magnitude of the heat added to or removed from the Chapter 3: The Second Law of Thermodynamics 5 system at temperature XP . Using this notation, we can express the net work done by a heat engine as it moves through a complete cycle as |[8/>ß 9?> | œ |UL | |UP | (3.7) and the thermal efficiency as (œ |[8/>ß 9?> | |UL | |UP | |UP | œ œ" |U L | |UL | |UL | (3.8) This equation for the thermal efficiency of a heat engine is very interesting. It tells us that the maximum thermal efficiency will occur when |UP | p ! or when |UL | p ∞. To obtain the latter condition might be somewhat difficult, but perhaps we might be able to achieve the former, or at least approach it. In the next section we will consider whether or not this is possible. THE NECESSITY OF EXHAUST HEAT AND THE KELVIN-PLANCK STATEMENT OF THE SECOND LAW Consider an ideal gas confined in a cylinder with a movable lid (see Fig. 3.4). The system is originally at an equilibrium temperature of 30‰ C. At this point, the lid of the system is at the lower stops, and we place a load on the lid. If we now take the cylinder and put it in contact with a heat reservoir at 100‰ C, heat will enter the system and the pressure will increase. As the volume of the gas increases, the system does work lifting the lid and load. As long as the process proceeds slowly enough that it can be considered quasi-static, the pressure of the gas inside will remain constant (equal to the outside pressure, and the pressure due to the lid and load). Under these circumstances, the system is always in a state of thermodynamic equilibrium and the ideal gas law requires that the temperature of the gas must rise. Thus, the energy added to the gas while it is in contact with the heat reservoir is equal to the work done as the gas expands plus the increase in internal energy of the gas as the temperature rises. We wait until the lid reaches the upper stops, and find that the gas temperature is 90‰ C. At this point, we remove the cylinder from the heat reservoir. At this point we also remove the load, having accomplished the work objective. Let's assume that 100 Joules of energy is added to the system from the hightemperature reservoir during this process, and that 25 Joules were required to lift the lid and load to the upper stops, while the internal energy has been increased by 75 Joules. Load Load Gas o 30 C Gas o 30 C Gas o 90 C Gas o 30 C Qin Qout o 20 C 100 C o Fig 3.4 A gas is contained in a cylinder with a movable lid. The lid rests on the lower stops when the gas is at 30‰ C. A load is placed on the lid and the cylinder is placed in contact with a high-temperature reservoir. Heat flows into the cylinder causing the pressure to rise and lift the lid. The internal energy also increases. When the lid reaches the upper stops, the gas has reached 90‰ C and is removed from the high-temperature reservoir. The load is now removed (useful work has been accomplished in lifting this load). We now wish to return the system to its original state so that the lid rests on the lower stops and we can again place a load on the lid to be lifted. To accomplish this, we must remove some of the internal energy from the gas. The only way this can be done is to dump the heat into a low-temperature reservoir. This amounts to “wasting” heat in this cycle. For this system to act as a cyclic heat engine, we must restore the system to its original configuration in order to repeat the cycle. To accomplish this, we must lower the temperature and pressure of the system (remove Chapter 3: The Second Law of Thermodynamics 6 the 75 Joules of extra internal energy from this system now as 90‰ C) and restore the lid to its original position. If we could remove the heat from our system and somehow put it back into the high-temperature reservoir, we would be saving the heat energy we now have in our system. The second law, however, reminds us that it is impossible to remove heat from a 90‰ C system and deposit that heat into a 100‰ C system. In order to remove the necessary heat from our gas and return it to its original state (original temperature), we will need to bring it into contact with a low-temperature reservoir which has a temperature less than or equal to 30‰ C. The heat removed from our system in this process is exhaust heat and is essentially wasted. Or is it? Can't we use this heat to do more work? After all, you have just added some extra energy to the lower temperature reservoir. But that is just the point! In the process of moving through the cycle, we have removed heat from a higher temperature reservoir and deposited heat to a lower temperature reservoir. To utilize the excess heat deposited in the lower temperature (30°C) reservoir, we would have to utilize this reservoir as a high-temperature reservoir is some other process. This you might be able to do - you might even be able to continue the process, transferring exhausted heat to a lower temperature reservoir further and further down the temperature ladder. However, you will eventually reach a limit beyond which you cannot practically go. Thus, we conclude that it is impossible to devise a system operating in a cycle that doesn't exhaust heat. This is the foundation of the Kelvin-Planck statement of the second law. This statement of the second law can be expressed as follows: It is impossible for a heat engine, which operates in a cycle, to remove an amount of heat energy from a single heat reservoir and to accomplish that same amount of mechanical work - there must be some exhaust heat. Another way of saying this is that no heat engine can be 100% efficient. REFRIGERATORS AND HEAT PUMPS If you look back at the process described in Fig. 3.1, you will notice that this heat engine passes from state E to F to G and back to E along a clockwise path enclosing a specific area on the T -Z graph. Since the area under a T -Z curve is the work done by the system, it should be obvious that the net work done by the system in the cylic process is just the area enclosed by this loop. Furthermore, it should be obvious that the enclosed area corresponds to work done by the system if it is traversed in a clockwise direction, and corresponds to work done on the system if it is traversed in a counter-clockwised direction. If the system is operated in a counter-clockwise direction work is done on the system by the surroundings and there is a net amount of heat removed from the system, i.e., more heat is removed from the system in the form of heat than is added to the system in the form of heat. Schematically, we can represent a simple form of this process by the drawing in Fig. 3.5 High-Temperature Reservoir QH Refrigerator Worknet,in QL Low-Temperature Reservoir Fig. 3.5 A schematic representation of a cyclic process working as a refrigerator. Chapter 3: The Second Law of Thermodynamics 7 From Fig. 3.5 it should be clear that the net effect of this “refrigeration” process is to remove an amount of heat from a low-temperature reservoir and deposit a somewhat larger amount of heat at a high-temperature reservoir. This process is driven by the mechanical work done on the system. From the first law principle of conservation of energy and the knowledge that the internal energy cannot change in a cyclic process we can see from the schematic that |UP | |[8/>ß38 | œ |UL | (remember that these quantities all express magnitudes). The amount of heat |UP | which can be removed from the low temperature reservoir in a cycle is obviously related to the amount of mechanical work done on the system. Now if we are interested in operating a refrigerator to keep something cool, the desired output of our device is the amount of heat that can be removed from the low-temperature reservoir. The required input for this system is the mechanical work which must be done on the system. We define the coefficient of performance of a refrigerator in terms of our previous definition of thermal efficiency, i.e., GST</0 œ H/=3</. S?>:?> V/;?3</. M 8:?> (3.9) orß GST</0 œ |UP | |UP | œ œ |[8/>ß38 | |UL | |UP | " " (3.10) |UL | |UP | Typically |UP | is larger that |[8/>ß38 |, so that the coefficient of performance of a refrigerator is greater than one. (That is the reason this quantity is not called an “efficiency”, but a coefficient of performance - efficiencies are typically less than one.) This means that the energy required to extract a certain amount of heat energy from the low-temperature reservoir is less than the amount of heat extracted. Energy Sink (kitchen room air or heated space) System Boundary QH Condenser 800 kPa 30 C Expansion Valve 120 kPa -25 C 800 kPa 60 C Wnet,in Compressor Evaporator 120 kPa -20 C QL Cool Energy Source (icebox or outdoors) Fig. 3.6 Schematic of a typical refrigeration system indicating typcial operating conditions. Notice also that the amount of heat added to the high-temperature reservoir must be larger than the mechanical work done on the system. This means that we could use this “refrigeration” process to efficiently add heat to the high-temperature reservoir. This is the principle behind the heat pump. A heat pump is simply a refrigeration cycle in which the “exhausted” heat is what is utilized rather than the removal of heat from the lowtemperature reservior. As an example, if you take a conventional air-conditioning unit and mount it in the window in reverse you will heat up your house and cool down the exterior. But the amount of heat supplied to the house will typically be greater than the amount of mechanical energy required to run the air-conditioner! The Chapter 3: The Second Law of Thermodynamics 8 coefficient of preformance for a heat pump is defined as GSTL/+> T ?7: œ H/=3</. S?>:?> |UL | |UL | " œ œ œ UP | |[8/>ß38 | |UL | |UP | V/;?3</ M 8:?> " ||U L| (3.11) A typical refrigeration system is presented schematically in Fig. 3.6. Here the individual subsystems are illustrated along with typical operating conditions. The refrigerant enters the compressor in the form of a vapor and is compressed to a relatively high pressure and temperature. This heated refrigerant then enters a condenser where it looses some of its heat energy to the surroundings. This cooled refrigerent then passes through a throttling (or expansion) valve allowing the gas to expand adiabatically - thus cooling significantly. This cooled gas then enters an evaporator where it absorbs heat from the space being cooled. The refrigerant is then drawn back into the compressor and the cycle begins again. As in the case of the steam power plant, each subsystem is actually an open system, but considered as a whole, we can treat the entire system as a closed system and apply the first law conservation of energy principle. It is this principle that Fig. 3.5 is based upon and upon which the definitions for the coefficient of performance are based. CLAUSIUS' STATEMENT OF THE SECOND LAW The simplest statement of the second law of thermodynamics is to say that it is impossible for heat to flow spontaneously from a cooler body to a warmer body. This is something we never observe. When an ice cube is placed in a glass of water, the heat energy from the hotter water enters the ice and melts it - heat is never extracted from the ice to heat up the hotter water. The refrigeration cycle that we mentioned above does not violate this principle, even though heat is removed from a colder object and deposited into a hotter object. This is because the cycle requires an input of mechanical work for the process to operate. However, since more energy is added to the high-temperature reservoir than is added to the system in the form of mechanical work, we might try to use some of this additional heat energy to actualy “run” the refrigerator. One way to do this might be to operate a heat engine between the same two heat reservoirs and let the mechanical output of the heat engine drive the refrigerator. If we could do something like that we might really have something! To test this idea, lets assume that the heat engine we operate between these two reservoirs has an efficiency of 100%, i.e., no heat is exhausted to the low-temperature reservoir. This is clearly in violation of the Kelvin-Planck statement of the second law. However, if this could be accomplished, then the refrigerator cycle we have developed could be driven by this special heat engine and effectively remove heat from the lowtemperature reservoir without any additional input of mechanical energy (see Fig. 3.7). This leads us to anotherß /;?3@+6/8> statement of the second law, the Claussius statement: No device can be constructed which operates in a cycle and produces no net effect other than the transfer of heat from a low-temperature reservoir to a hightemperature reservoir. High-Temperature Reservoir QH Refrigerator 100% Efficient Heat Engine QL Low-Temperature Reservoir Fig. 3.7 A cyclic refrigerator driven by a 100% efficient heat engine effectively transfers heat from a lowtemperature reservoir to a high-temperature reservoir with no other net effect - a clear violation of the second law. Chapter 3: The Second Law of Thermodynamics 9 You can see that this system would effectively act as if you could extract heat from a cold object, and deposit that same amount of heat at a higher temperature, violating the simple statement of the second law. THE CARNOT CYCLE To try and gain even better insight into the concepts of heat engines and refrigerators, as well as the implications of the second law, we will now consider a very special cyclic process that was first proposed by Sadi Carnot, a French engineer, in 1824. The Carnot cycle consists of an ideal, monatomic gas carried through four reversible processes as shown in Fig. 3.8: 1) an isothermal expansion, 2) an adiabatic expansion, 3) an isothermal contraction, and 4) an adiabatic contraction which returns the system to the original state. In this section we will examine each of the four segments of this cycle and calculate, for each segment, the net change in the internal energy of the gas, the work done, and the heat added to the system. We begin the cycle at state E where the pressure, volume and temperature are designated TE , ZE , and XE , respectively. Pressure A Qin B TH D Qout C TL Volume Fig. 3.8 T -Z diagram of a Carnot Cycle. E Ä F: First we carry out an isothermal expansion from state E to state F. During this process, the temperature of the system remains the same (i.e., TF œ XE œ XL ) so that the change in internal energy of the gas for this process is given by ?I œ $ 8VÐXF XE Ñ œ ! # (3.12) Now from the first law, the work done by the system in expanding against its surroundings can only be accomplished as a result of heat being added to the system during this process. Thus, we have ?IEF œ UEF [EF œ 0 Ê UEF œ [EF (3.13) and since positive work is done by the system in this process, the heat UEF added to the system must also be positive. We will let the magnitude of the heat added to the system at the temperature XL be |UL |Þ We can calculate the work done by the ideal gas for this process from the equation F [EF œ ( T .Z œ 8VXL ( E This equation can also be expressed as E F .Z ZF œ 8VXL lnÐ Ñ Z ZE (3.14) Chapter 3: The Second Law of Thermodynamics 10 F [EF œ ( T .Z œ 8VXL ( E E F .Z TE œ 8VXL lnÐ Ñ Z TF (3.15) since TE ZE œ TF ZF Ê ZF ÎZE œ TE ÎTF (3.16) for an ideal gas during an isothermal process. Since TE TF this equation demonstrates that the net work is positive, as we know it should be. Notice that this process is a good example of a case where heat can be added to the system even though the temperature may not change. Again, since the heat added to the system during this process must be equal to the work done during this process, we have UEF œ 8VXL lnÐ TE ZF Ñ œ 8VXL lnÐ Ñ ! TF ZE (3.17) F Ä G: The system is now allowed to expand adiabatically from state F to state G . In this process, no heat is added to or removed from the system (i.e., UFG œ !). From the first law, we have ?IFG œ UFG [FG (3.18) ?IFG œ [FG (3.19) But since UFG œ !, this reduces to and we only need to find [FG or ?IFG . It is easier to find ?IFG in this case, since it is only a function of the temperatures. Thus, ?IFG œ [FG œ $ $ 8VÐXG XF Ñ œ 8VÐXP XL Ñ # # (3.20) Since XP XL the internal energy of the system must decrease, but the work done during this process must be positive, as the gas expands! G Ä H: Since this process is an isothermal process, we can find the changes in internal energy, work, and heat added by using the equations developed in the first isothermal process, E Ä F. We obtain: ?IGH œ ! [GH œ UGH œ 8VXP lnÐ TG ZH Ñ œ 8VXP lnÐ Ñ TH ZG (3.21) (3.22) which is less than zero, showing that heat is removed from the system as work is done compressing the system from ZG Ä ZH . Thus, UGH (which is less than zero) is the heat removed from the system at temperature XP and is given by UGH œ 8VXP lnÐ TG ZH Ñ œ 8VXP lnÐ Ñ ! TH ZG (3.23) H Ä E: Similarly, for the final adiabatic process, we can use the equations developed for the previous adiabatic process: UHE œ ! ?IHE œ [HE œ $ $ 8VÐXH XE Ñ œ 8VÐXL XP Ñ # # (3.24) (3.25) Since XL XP , this last equation indicates that the internal energy of the system increases while the work done by the gas is negative (i.e., work is done on the gas by the surroundings as the gas is further compressed)! Net Changes in the Carnot Cycle. To find the net changes in internal energy, the net work done by the system, and the net heat added to the system, we simply add the changes in each process together. This gives Chapter 3: The Second Law of Thermodynamics 11 ?I-C-6/ œ ! (3.26) [-C-6/ œ U-C-6/ œ UEF UGH œ 8VXL lnÐ TE TG Ñ 8VXP lnÐ Ñ TF TH (3.27) This last equation can be simplified, using the relationships we derived earlier for adiabatic processes in ideal gases. We showed that for an adiabatic process, T "-# X # œ -98=> (3.28) which means that "-# # "-# # (3.29) "-# # "-# # (3.30) TF XF œ TG XG and TE XE œ TH XH Now, since XE œ XF œ XL and XG œ XH œ XP (3.31) we can divide one of these last two equations by the other and obtain: TF TG œ TE TH (3.32) Thus, the net work done, which is equal to the net heat added to the system in one complete cycle is given by [-C-6/ œ U-C-6/ œ UEF UGH œ 8VXL lnÐ TE TG Ñ 8VXP lnÐ Ñ TF TH (3.33) [-C-6/ œ U-C-6/ œ UEF UGH œ 8VXL lnÐ TE TF Ñ 8VXP lnÐ Ñ TF TE (3.34) TE TE Ñ 8VXP lnÐ Ñ œ |UL | |UP | TF TF (3.35) [-C-6/ œ U-C-6/ œ UEF UGH œ 8VXL lnÐ where we let |UL | and |UP | be the magnitude of the heat added to or removed from the heat reservoirs at temperatures XL and XP , respectively. Finally, we can express the work done in the Carnot cycle by the equation [-C-6/ œ U-C-6/ œ UEF UGH œ |UL | |UP | œ 8V lnÐ TE Ñ ÐXL XP Ñ TF (3.36) which demonstrates that the net work done by the system is positive, as it should be since this cyclic process acts like a heat engine producing a net output of useful mechanical energy. If this process were operated in reverse, the system would be a refrigerator or a heat pump and all the quantities we have calculated would be negative. THE EFFICIENCY OF A CARNOT CYCLE As we mentioned earlier, the thermal efficiency ( of a heat engine is defined to be the net work done by the engine in a complete cycle divided by the amount of heat added to the system during that cycle. (Note that the net heat added to the system is equal to the net amount of work done by the system.) For the Carnot cycle, we know that the net work done in the cycle is equal to the net heat added during the cycle, or [-C-6/ œ |UL | |UP | (see Fig. 3.2). The efficiency of this cycle is given by (œ |[-C-6/ | |UL | |UP | |UP | œ œ" |UL | |UL | |UL | (3.37) Chapter 3: The Second Law of Thermodynamics 12 Using the calculated values for |[-C-6/ |, and |UL |, we also find the efficiency to be (œ 8V lnÐ TTFE Ñ ÐXL XP ) |[-C-6/ | XL XP XP œ œ œ" T E |UL | XL XL 8VXL lnÐ TF Ñ (3.38) From these last two equations we discover an extremely important relationship between the magnitude of the heat added to or taken from a system and the temperature of the system during the isothermal processes of a Carnot cycle. These equations imply that for an ideal gas carried through a Carnot cycle |UP | XP œ |UL | XL (3.39) where |UP | and |UL | are the magnitudes of heat flow and XP and XL are the temperatures of the low- and hightemperature reservoirs, respectively. Since the ratio of the temperatures can be determined by simply measuring the amount of heat transfered during the isothermal processes of a Carnot cycle, we can use the Carnot cycle to define a temperature scale. This temperature scale, known as the thermodynamic temperature scale, has been shown to be equivalent to the Kelvin scale. Thus, the ratio of the heat added to or removed from an ideal gas system during the isothermal processes of a Carnot cycle is exactly equal to the ratio of the absolute temperatures of the two heat reservoirs. This means that the equations developed earlier for the thermal efficiency of a heat engine and the coefficient of performance of a refrigerator and of a heat pump can now be evaluated for a Carnot cycle simply by knowing the temperatures of the heat reservoirs. One can show that now heat engine, operating between two heat reservoirs, can be more efficient than a Carnot heat engine operating between these same two heat reservoirs. Thus a Carnot heat engine is the most efficient heat engine operating between any two heat reservoirs of fixed temperature. As an example, let's apply what we have learned to the case of an ideal (Carnot) steam plant. The efficiency of the steam plant is given by (œ |[-C-6/ | |UL | |UP | |UP | XP œ œ" œ" |UL | |UL | |UL | XL (3.40) We can achieve the largest possible efficiency by letting XP Ä !, of by letting XL Ä ∞, or both. However, there is usually a practical limit to these temperatures. If we consider the efficiency of a coal fired steam plant, for example, the pipes which carry the heated steam to drive the turbine would melt if XL were too high (greater than about 500°G ). In addition, the heat exchangers which allow us to get rid of the excess heat do not function well when XP gets too low (at temperatures lower than !‰ C, the water which passes over the heat exchanger freezes and the ice acts as a thermal insulator decreasing the efficiency of the heat exchanger). For a typical coal fired steam plant, then, the high temperature reservoir is at approximately 500‰ C (773 K) and the low temperature reservoir is at approximately 27°C (300K), so that the efficiency of this system can be no better than ( œ" 300 œ 0.6119 or 61.19% 773 (3.41) The total efficiency of a real coal-fired steam plant is much less than this, since we must take into account many other factors: the efficiency of heating the water to produce steam, the loss of heat through the pipes, the friction in the turbine, etc. In addition, a coal-fired steam plant is not exactly equivalent to a Carnot cycle. [Note: Nuclear powered steam plants are operated at lower temperatures than steam plants - for safety reasons - with the high temperature of the boiler restricted to about 300‰ C. This means that the efficiency of a nuclear powered steam plant must necessarily be less than a comparable coal or gas fired steam plant.] As a side note, notice that since no heat engine can ever by 100% efficient, we can never have |UP | œ !, or XP œ !. Thus, absolute zero Kelvin is unattainable. Experiments have been performed in which temperatures as low as several milliKelvin have been attained, but one can never reach absolute zero! This is often given as an alternate way of stating the second law of thermodynamics. We next look at the example of an icebox which is maintained at a temperature of &°C in a room which is maintained at a temperature of 23°C. The coefficient of performance of a Carnot refrigerator operating between Chapter 3: The Second Law of Thermodynamics 13 these two temperatures is given by GST</0 œ |UP | XP 268 œ œ œ 9.57 |[ | 296 268 XL XP (3.42) Notice that this number is greater than unity just as we said it would be! This means that the amount of heat which can be extracted from the icebox is 9.57 times greater than the amount of energy input to the system in the form of mechanical energy. This means that if energy were put into the compressor at a rate of 1 Watt, the rate at which heat energy would be removed from the icebox is 9.57 Watts, provided that the compressor is 100% efficient, which it isn't. However, you can see that even a 30% conversion of electrical energy to mechanical energy in this case would yield a gain of .30 ‚ 9.57 œ 2.87 over the amount of energy supplied by the electrical company. Now consider a heat pump which maintains a home at 23°C when the outside temperature is 5°C (these are the same temperatures we used for the refrigerator). The coefficient of performance of the heat pump is given by GST2: œ XL |UL | 296 œ œ œ 10.57 |[ | XL XP 296 268 (3.43) Notice that this value is different from the coefficient of performance of the refrigerator operating between the same two reservoirs. In fact, notice that GST2: GST</0 œ |UL | |UP | |UL | |UP | œ œ" |UL | |UP | |UL | |UP | |UL | |UP | (3.44) This equation is true for all systems, not just a Carnot system. The actual efficiency of a heat pump is less than what we calculated for a Carnot system because of the inefficiency of heat exchangers, and the fact that no mechanical motor is 100% efficient. This fact actually limits the utility of heat pumps is some areas of the country. However, we can use the ideal system to determine under what circumstances a heat pump might be most useful. For example, you should be able to see that the efficiency of the heat pump goes up when the average outside temperature is close to the desired temperature inside the house. The efficiency decreases when the average outside temperature drops. Thus, real heat pumps have been more popular in the south, than in the north. EFFICIENCIES OF REVERSIBLE AND IRREVERSIBLE PROCESSES Carnot discovered two very important principles while studying the efficiencies of thermodynamic systems: 1. The efficiency of an irreversible heat engine is always less that the efficiency of a reversible process when operating between the same two heat reservoirs. 2. The efficiencies of all reversible heat engines must be the same when operating between the same two heat reservoirs. A proof of these two statements is left as an exercise for the student. However, the method of proof is very similar to our discussion of the equivalence of the Claussius and Kelvin-Planck statements of the second law. You can show that the assumption that there is an irreversible heat engine which is more efficient than a reversible one leads to a contradiction of the second law. Once you show this, the proof of the second statement follows easily. The consequencies of these two principles are fairly straight forward. Since all reversible heat engines operating between the same two heat reservoirs must have the same efficiency, the special nature of the Carnot cycle is unimportant. Likewise, since no irreversible heat engine can be more efficient than a reversible one, the Carnot cycle can be used as a theoretical limit on any real heat engine operating between two heat reservoirs. Other types of heat engines, and their corresponding idealized models are discussed in the appendix. There you will find descriptions of the Otto cycle (internal combustion engine with spark ignition [the gasoline Chapter 3: The Second Law of Thermodynamics 14 engine]) as well as the Diesel cycle (internal combustion engines with compression ignition) and calculations of their respective efficiencies. THE DEFINITION OF ENTROPY We discovered that when an ideal gas is carried through a Carnot cycle the ratio of the heat added to the system at the high-temperature reservoir to the heat removed from the system at the low-temperature reservoir is just equal to the ratio of the absolute temperatures of the two reservoirs, or |UP | XP œ |UL | XL (3.45) where |UP | and |UL | are the magnitudes of heat flow and XP and XL are the temperatures of the low- and hightemperature reservoirs, respectively. Rearranging this last equation, we find that |UP | |UL | œ XP XL (3.46) for the Carnot cycle. This equation serves to define a new quantity which we will call the entropy change of a system during a reversible, isothermal process. More precisely, we define the change in entropy of a system during a reversible, isothermal process by the equation WEF œ Œ UEF X </@ (3.47) where WEF will be positive if heat is added to the system, but negative if heat is removed from the system. Notice that from the viewpoint of the thermal reservoirs, the heat removed from a reservoir is considered positive, while the heat depositied in a reservoir is considered negative. For the Carnot cycle, we have the total change in entropy of the ideal gas system for the complete cycle given by W-C-6/ œ Œ UEF UGH |UL | |UP | Œ œŒ Œ œ0 XL </@ XP </@ XL </@ XP </@ (3.48) where |UL | and |UP | are magnitudes, whereas UEF and UGH are either positive or negative depending upon the direction of energy flow. This last equation indicates the special nature of the quantity which we defined as the entropy. Since the change in entropy of the system is zero as the system is carried through a complete cycle, the entropy of a system must be a state variable! Because the entropy is actually a state variable, it is more appropriate to express the differential change in entropy by the following equation .W œ Œ $U X </@ (3.49) where $U is the differential amount of heat added to the system during a reversible, isothermal process at temperature X . If the system temperature in a process is changing, but the process is still a reversible one, we can integrate to determine the total change in entropy of the system ?WEF œ ( F E Œ $U X </@ (3.50) You might wonder how we can equate the change in a state variable with a change in a process variable. Remember that we did the same thing for the differentical work done in a volume expansion. As long as the process occurs very slowly (quasi-statically) this definition holds true. In a real, irreversible process, however, the change in entropy is not equal to the amount of heat added to the system divided by the temperature. The processes involved in the Carnot cycle from which we derived this expression are processes occurring for an ideal gas as that gas moves through certain reversible, or quasi-static, processes on the T ß @ß X surface of that gas. For this reason, all the equations we have developed for the Carnot cycle are valid only for reversible processes (including the equation relating the entropy to the heat transfer of an isothermal process)! However, if we change Chapter 3: The Second Law of Thermodynamics 15 a system from state E to state F by some arbitrary process, the state variables must change by the same amount, no matter if the process is reversible or not! This means that since the entropy of the system is a state variable, changes in entropy will be the same between any two states whether or not the process which connects the two states is reversible. However, we can actually calculate the change in entropy (or temperature, pressure, volume, etc.) only if we choose an appropriate reversible process for which we know the equation of state of the system (or the appropriate partials) and for which the equation .W œ Œ $U X </@ (3.51) is valid! THE SECOND LAW IN TERMS OF ENTROPY We now want to consider how the change in entropy of a system is related to the second law of thermodynamics. We want to demonstrate that the change in entropy can be used to indicate the direction of heat (or energy) flow in a complex system. To see this, we will consider an arbitrary cyclic heat engine (not necessarily reversible), in which a certain amount of heat, |UL |, flows from a high temperature reservoir at temperature XL into the system, and an equal amount of heat, |UP |, flows out of the system and into a colder reservoir at temperature XP . In this case the net work done by the system, [ , is zero, since |UL | œ |UP | œ |Uo |. (This is not a very useful heat engine, since it will perform no useful work - but is helps in understanding the relationship between entropy an heat flow.) This reversible engine is, therefore, equivalent to the case where heat simply flows from a hotter reservoir to a cooler one. In this case, the ratio |UL |ÎXL is not equal to the entropy change, since the process is not necessarily reversible. However, we can still calculate this ratio for our arbitrary system. Remembering that we take heat flowing out of a reservoir as positive and into a reservoir as negative, we can write the sum of these ratios as |UL | |UP | " " œ |Uo |œ XL XP XL XP (3.52) This equation tells us that if XL XP , as we assumed, this sum is less than zero. If, however, we were to assume that XP XL , (in direct violation of second law) this sum would be greater than zero! Thus, we conclude that the sum of the amount of heat removed from the heat reservoirs divided by the absolute temperature of those reservoirs must be less than zero, or U ! X (3.53) to be consistent with the second law. We, therefore, postulate that for any arbitrary process, carried through a complete cycle, the sum of the heats added to the system divided by the corresponding absolute temperatures must be less than or equal to zero (the equality occuring only in the case of a reversible system passing through a cyclic process). This same principle can be extended to a system removing heat from any number of heat reservoirs, and gives us the so-called Clausius inequality: U Ÿ! X (3.54) In the limit as we remove a small amount of heat at one temperature, and then a small amount of heat at another adjacent temperature, the sum becomes an integral, and we write * X Ÿ! $U (3.55) where the equal sign is true for a reversible process only. Note that this equation explicitly indicates that a cyclic process is being considered by the inclusion of a circle over the integral sign. This integral is zero only for the case of a reversible process and it is only for reversible processes that the ratio $UÎX is equal to the entropy, a state variable. To see how the Clausius inequality leads to a general statement of the law of entropy, consider a cyclic process in which a system is carried from state E to state F and then back to E, as shown in Fig 3.9: Chapter 3: The Second Law of Thermodynamics 16 I B A II Fig. 3.9 A closed-loop path for which we calculate the entropy change. Path I is an irreversible process, while path II is a reversible one. In this particular case, we consider a system which is allowed to go from state E to state F through some isolated, adiabatic process, which is irreversible. It is then returned to its original state by means of some reversible (quasi-static) process. The Clausius inequality for this cycle states that * X Ÿ! $U (3.56) Since the cyclic process as a whole is irreversible, we must use the inequality, which gives E $U $U ( Œ ! X </@ E X F M MM ( F (3.57) Since path II is a reversible process, the ratio $UÎX is equal to the entropy change, .W , so that the integral over path II will give us the entropy change from state F to state E, ?WFE . But the process along path I was assumed to be isolated (adiabatic), so that $U must be zero for all points along that path. This means that the integral along path I (which is not equal to the entropy change since the process is irreversible) must be zero. Therefore, we have E E $U ( Œ œ ( .W œ WE WF ! X </@ F F MM MM (3.58) or WE WF (3.59) This equation asserts that the entropy of any isolated, irreversible process must increase! Notice that we can not say that the entropy will always increase for every process. In fact, this is not generally true, as we will see in the examples that follow. It is true only for isolated processes. Thus, if we can consider our galaxy as sufficiently isolated from the rest of the universe, we can claim that the entropy of our isolated galaxy must be increasing if there is anything, whatsoever, going on (any processes occuring)! CALCULATION OF ENTROPY CHANGES IN VARIOUS PROCESSES Since the entropy is defines by the equation .W œ Œ $U X </@ (3.60) we can calculate the entropy change for any process E p F, so long as we can devise a reversible process which will take us from E p F. In this section we will consider a process where the entropy of the system must increase Chapter 3: The Second Law of Thermodynamics 17 because the process is inherently irreversible, and we will demonstrate how to calculate the entropy change for this process. We will consider the irreversible heat flow that occurs when we mix water and ice. As a specific example, consider the case where 1 kg of ice at !‰ C is placed in an adiabatic container filled with 1 kg of water at 80‰ C. Since the temperature difference between the ice and water is fairly large, the cooling of the water and the melting of the ice can not be considered a quasi-static (reversible) process. We can, however, determine the amount of heat transferred from the water to the ice as the ice is melted, and we can determine the final temperature of the system when it reaches equilibrium. (This can be done whether or not all the ice is melted.) We know that melting 1 kg of ice at 0‰ C requires U7/6> œ 7P œ " kg )! kcal/kg œ )! kcal of energy. The amount of heat energy available in the water (i.e., the amount of energy that can be removed from the water as it is cooled to the point of freezing) is equal to U+@+36 œ 7-?X œ " kg " kcal/kg † ‰ C 8!‰ C œ 8! kcal so that for this particular example, all the ice is melted and the water's temperature is lowered to the freezing point. We have transferred 80 kcal of thermal energy from the water to the ice, melting the entire 1 kg of ice, and the water has been cooled to 0‰ C, the equilibrium temperature of the system. Although the process we have described is not a reversible one, we can take 1 kg of ice and melt it in a reversible process by bringing the ice in contact with a heat reservoir with a temperature infinitesimally above 0°C so that heat is allowed to slowly enter the ice and melt it. Since this reversible process takes place at a temperature of 273K, the heat added to the ice to melt it, divided by the absolute temperature of the heat reservoir is just the change in entropy of the ice, given by ?W3-/ œ )! 5-+6 œ "Þ#$ ‚ "!$ N ÎO #($ O The entropy change of the heat reservoir in contact with the ice is the same magnitude as the entropy change of the ice, but with an opposite sign, since heat is being removed from the reservoir. Thus to total change in entropy of the universe for this reversible process is zero. Likewise, we can take 1 kg of water at 80°C and lower its temperature infinitesimally by bringing it into contact with a heat reservoir infinitesimally cooler than 80°C. We can continue this process, by moving the water from one reservoir to another among an infinite number of reservoirs which are slightly cooler than the previous one, until we remove all the heat in the water and the water reaches the equilibrium temperature of 0°C. Since this process is reversible, we can calculate the entropy change of the water from the equation ?WA+>/< œ ( X œ#($ X œ#($)! Œ X œ#($ X œ#($ $U 7A -A .X .X œ 7 A -A ( œ( X </@ X X œ#($)! X œ#($)! X assuming that the heat capacity of water is a constant over this temperature range. For our particular problem, which gives ?WA+>/< œ 1 ‚ 10$ g ‚ %Þ")'N #($ $ 68Œ œ "Þ!) ‚ "! N ÎO 1O $&$ This is negative since heat is being removed from the water. The change in entropy of all the heat reservoirs that we had to use is just the negative of this quantity (the heat was added to the reservoirs), so that the total entropy change in the universe for this reversible process is zero. Now, the total entropy change of the water plus the ice is a state variable - independent of the process by which you reach the end points. Thus, the total change in entropy of the isolated system (water plus ice) must be given by ?W73B>?</ œ ?W3-/ ?WA+>/< œ "Þ#$ ‚ "!$ N ÎO "Þ!) ‚ "!$ N ÎO ?W73B>?</ œ !Þ"& ‚ "!$ N ÎO which is greater than unity as it must be for an isolated system! Chapter 3: The Second Law of Thermodynamics 18 A Molecular View of Entropy As we stated earlier, we can demonstrate several processes whereby mechanical energy can be converted into heat. One such example is a block of wood sliding across the floor. We have also demonstrated that it is impossible to convert a given amount of heat energy into an equivalent amount of mechanical energy. This “oneway” aspect of nature seems to be tied into the concept of entropy. We might expect, therefore, that the entropy of a block of wood which is sliding across the floor is increasing as it comes to a stop. When the block of wood come to a stop, we find that the block will be somewhat hotter than if there were no friction to slow the block down. This increase in the thermal energy of the block with a subsequent loss of mechanical energy of motion must be tied to the entropy change of the system. This would seem to indicate that the organized motion of all the molecules that make up the block of wood has less entropy than the increased random motion of the molecules within the block after it is heated by friction. Thus, we might conclude that the entropy of a system may be related somehow to the random nature of the motion of the molecules within a material. We will attempt to determine how the entropy of a system is related to the molecular structure of matter later in our study Chapter 3: The Second Law of Thermodynamics 19 Appendix 3.1 EFFICIENCIES OF OTHER CYCLIC PROCESS COMPARED WITH THE CARNOT CYCLE Diesel, Sterling, Joule, Otto, compared with Carnot