Chapter 6. The Second Law of Thermodynamics PDF

cen84959_ch06.qxd 3/31/05 3:51 PM Page 279 Chapter 6 THE SECOND LAW OF THERMODYNAMICS Objectives...

cen84959_ch06.qxd 3/31/05 3:51 PM Page 279 Chapter 6 THE SECOND LAW OF THERMODYNAMICS Objectives T o this point, we have focused our attention on the first law of thermodynamics, which requires that energy be conserved during a process. In this chapter, we intro- duce the second law of thermodynamics, which asserts that processes occur in a certain direction and that energy has The objectives of Chapter 6 are to: Introduce the second law of thermodynamics. quality as well as quantity. A process cannot take place Identify valid processes as those that satisfy both the first unless it satisfies both the first and second laws of thermody- and second laws of thermodynamics. namics. In this chapter, the thermal energy reservoirs, Discuss thermal energy reservoirs, reversible and reversible and irreversible processes, heat engines, refrigera- irreversible processes, heat engines, refrigerators, and tors, and heat pumps are introduced first. Various statements heat pumps. of the second law are followed by a discussion of perpetual- Describe the Kelvin–Planck and Clausius statements of the motion machines and the thermodynamic temperature scale. second law of thermodynamics. The Carnot cycle is introduced next, and the Carnot princi- Discuss the concepts of perpetual-motion machines. ples are discussed. Finally, the idealized Carnot heat engines, refrigerators, and heat pumps are examined. Apply the second law of thermodynamics to cycles and cyclic devices. Apply the second law to develop the absolute thermodynamic temperature scale. Describe the Carnot cycle. Examine the Carnot principles, idealized Carnot heat engines, refrigerators, and heat pumps. Determine the expressions for the thermal efficiencies and coefficients of performance for reversible heat engines, heat pumps, and refrigerators. | 279 cen84959_ch06.qxd 4/25/05 3:10 PM Page 280 280 | Thermodynamics INTERACTIVE 6–1 INTRODUCTION TO THE SECOND LAW TUTORIAL In Chaps. 4 and 5, we applied the first law of thermodynamics, or the conser- SEE TUTORIAL CH. 6, SEC. 1 ON THE DVD. vation of energy principle, to processes involving closed and open systems. As pointed out repeatedly in those chapters, energy is a conserved property, and no process is known to have taken place in violation of the first law of thermodynamics. Therefore, it is reasonable to conclude that a process must satisfy the first law to occur. However, as explained here, satisfying the first law alone does not ensure that the process will actually take place. It is common experience that a cup of hot coffee left in a cooler room eventually cools off (Fig. 6–1). This process satisfies the first law of thermo- dynamics since the amount of energy lost by the coffee is equal to the amount gained by the surrounding air. Now let us consider the reverse process—the hot coffee getting even hotter in a cooler room as a result of HOT heat transfer from the room air. We all know that this process never takes COFFEE Heat place. Yet, doing so would not violate the first law as long as the amount of energy lost by the air is equal to the amount gained by the coffee. As another familiar example, consider the heating of a room by the passage of electric current through a resistor (Fig. 6–2). Again, the first law dictates that the amount of electric energy supplied to the resistance wires be equal to FIGURE 6–1 the amount of energy transferred to the room air as heat. Now let us attempt A cup of hot coffee does not get hotter to reverse this process. It will come as no surprise that transferring some heat in a cooler room. to the wires does not cause an equivalent amount of electric energy to be generated in the wires. Finally, consider a paddle-wheel mechanism that is operated by the fall of Heat a mass (Fig. 6–3). The paddle wheel rotates as the mass falls and stirs a fluid within an insulated container. As a result, the potential energy of the I=0 mass decreases, and the internal energy of the fluid increases in accordance with the conservation of energy principle. However, the reverse process, raising the mass by transferring heat from the fluid to the paddle wheel, FIGURE 6–2 does not occur in nature, although doing so would not violate the first law of thermodynamics. Transferring heat to a wire will not It is clear from these arguments that processes proceed in a certain direc- generate electricity. tion and not in the reverse direction (Fig. 6–4). The first law places no restriction on the direction of a process, but satisfying the first law does not ensure that the process can actually occur. This inadequacy of the first law to identify whether a process can take place is remedied by introducing another general principle, the second law of thermodynamics. We show later in this chapter that the reverse processes discussed above violate the second law of thermodynamics. This violation is easily detected with the help of a property, Heat called entropy, defined in Chap. 7. A process cannot occur unless it satisfies both the first and the second laws of thermodynamics (Fig. 6–5). There are numerous valid statements of the second law of thermodynam- ics. Two such statements are presented and discussed later in this chapter in FIGURE 6–3 relation to some engineering devices that operate on cycles. Transferring heat to a paddle wheel The use of the second law of thermodynamics is not limited to identifying will not cause it to rotate. the direction of processes, however. The second law also asserts that energy has quality as well as quantity. The first law is concerned with the quantity of energy and the transformations of energy from one form to another with no regard to its quality. Preserving the quality of energy is a major concern cen84959_ch06.qxd 4/25/05 3:10 PM Page 281 Chapter 6 | 281 to engineers, and the second law provides the necessary means to determine the quality as well as the degree of degradation of energy during a process. ONE WAY As discussed later in this chapter, more of high-temperature energy can be converted to work, and thus it has a higher quality than the same amount of energy at a lower temperature. FIGURE 6–4 The second law of thermodynamics is also used in determining the Processes occur in a certain direction, theoretical limits for the performance of commonly used engineering sys- and not in the reverse direction. tems, such as heat engines and refrigerators, as well as predicting the degree of completion of chemical reactions. PROCESS 1st law 2nd law 6–2 THERMAL ENERGY RESERVOIRS In the development of the second law of thermodynamics, it is very conve- FIGURE 6–5 nient to have a hypothetical body with a relatively large thermal energy A process must satisfy both the first capacity (mass
specific heat) that can supply or absorb finite amounts of and second laws of thermodynamics to heat without undergoing any change in temperature. Such a body is called a proceed. thermal energy reservoir, or just a reservoir. In practice, large bodies of water such as oceans, lakes, and rivers as well as the atmospheric air can be modeled accurately as thermal energy reservoirs because of their large ther- INTERACTIVE mal energy storage capabilities or thermal masses (Fig. 6–6). The atmo- TUTORIAL sphere, for example, does not warm up as a result of heat losses from SEE TUTORIAL CH. 6, SEC. 2 ON THE DVD. residential buildings in winter. Likewise, megajoules of waste energy dumped in large rivers by power plants do not cause any significant change in water temperature. ATMOSPHERE A two-phase system can be modeled as a reservoir also since it can absorb RIVER and release large quantities of heat while remaining at constant temperature. Another familiar example of a thermal energy reservoir is the industrial fur- LAKE nace. The temperatures of most furnaces are carefully controlled, and they are capable of supplying large quantities of thermal energy as heat in an essentially isothermal manner. Therefore, they can be modeled as reservoirs. OCEAN A body does not actually have to be very large to be considered a reser- voir. Any physical body whose thermal energy capacity is large relative to FIGURE 6–6 the amount of energy it supplies or absorbs can be modeled as one. The air in a room, for example, can be treated as a reservoir in the analysis of the Bodies with relatively large thermal heat dissipation from a TV set in the room, since the amount of heat transfer masses can be modeled as thermal from the TV set to the room air is not large enough to have a noticeable energy reservoirs. effect on the room air temperature. A reservoir that supplies energy in the form of heat is called a source, and Thermal energy one that absorbs energy in the form of heat is called a sink (Fig. 6–7). Ther- SOURCE mal energy reservoirs are often referred to as heat reservoirs since they supply or absorb energy in the form of heat. HEAT Heat transfer from industrial sources to the environment is of major con- cern to environmentalists as well as to engineers. Irresponsible manage- HEAT ment of waste energy can significantly increase the temperature of portions of the environment, causing what is called thermal pollution. If it is not Thermal energy carefully controlled, thermal pollution can seriously disrupt marine life in SINK lakes and rivers. However, by careful design and management, the waste energy dumped into large bodies of water can be used to improve the qual- FIGURE 6–7 ity of marine life by keeping the local temperature increases within safe A source supplies energy in the form and desirable levels. of heat, and a sink absorbs it. cen84959_ch06.qxd 4/25/05 3:10 PM Page 282 282 | Thermodynamics INTERACTIVE 6–3 HEAT ENGINES TUTORIAL As pointed out earlier, work can easily be converted to other forms of energy, SEE TUTORIAL CH. 6, SEC. 3 ON THE DVD. but converting other forms of energy to work is not that easy. The mechani- cal work done by the shaft shown in Fig. 6–8, for example, is first converted to the internal energy of the water. This energy may then leave the water as Work No work heat. We know from experience that any attempt to reverse this process will fail. That is, transferring heat to the water does not cause the shaft to rotate. From this and other observations, we conclude that work can be converted to Heat heat directly and completely, but converting heat to work requires the use of some special devices. These devices are called heat engines. Heat engines differ considerably from one another, but all can be charac- Heat terized by the following (Fig. 6–9): WATER WATER 1. They receive heat from a high-temperature source (solar energy, oil fur- nace, nuclear reactor, etc.). FIGURE 6–8 2. They convert part of this heat to work (usually in the form of a rotating Work can always be converted to heat shaft). directly and completely, but the 3. They reject the remaining waste heat to a low-temperature sink (the reverse is not true. atmosphere, rivers, etc.). 4. They operate on a cycle. Heat engines and other cyclic devices usually involve a fluid to and from High-temperature which heat is transferred while undergoing a cycle. This fluid is called the SOURCE working fluid. The term heat engine is often used in a broader sense to include work- producing devices that do not operate in a thermodynamic cycle. Engines Qin that involve internal combustion such as gas turbines and car engines fall into this category. These devices operate in a mechanical cycle but not in a thermodynamic cycle since the working fluid (the combustion gases) does HEAT not undergo a complete cycle. Instead of being cooled to the initial tempera- ENGINE Wnet,out ture, the exhaust gases are purged and replaced by fresh air-and-fuel mixture at the end of the cycle. The work-producing device that best fits into the definition of a heat Qout engine is the steam power plant, which is an external-combustion engine. That is, combustion takes place outside the engine, and the thermal energy released during this process is transferred to the steam as heat. The Low-temperature SINK schematic of a basic steam power plant is shown in Fig. 6–10. This is a rather simplified diagram, and the discussion of actual steam power plants FIGURE 6–9 is given in later chapters. The various quantities shown on this figure are as follows: Part of the heat received by a heat engine is converted to work, while the Qin  amount of heat supplied to steam in boiler from a high-temperature rest is rejected to a sink. source (furnace) Qout  amount of heat rejected from steam in condenser to a low- temperature sink (the atmosphere, a river, etc.) Wout  amount of work delivered by steam as it expands in turbine Win  amount of work required to compress water to boiler pressure Notice that the directions of the heat and work interactions are indicated by the subscripts in and out. Therefore, all four of the described quantities are always positive. cen84959_ch06.qxd 3/31/05 3:51 PM Page 283 Chapter 6 | 283 Energy source (such as a furnace) Qin System boundary Boiler Wout Win Pump Turbine Condenser Qout Energy sink (such as the atmosphere) Wout Wnet,out FIGURE 6–10 HEAT ENGINE Schematic of a steam power plant. The net work output of this power plant is simply the difference between Win the total work output of the plant and the total work input (Fig. 6–11): Wnet,out  Wout  Win¬¬1kJ2 (6–1) FIGURE 6–11 A portion of the work output of a heat The net work can also be determined from the heat transfer data alone. The engine is consumed internally to four components of the steam power plant involve mass flow in and out, and maintain continuous operation. therefore they should be treated as open systems. These components, together with the connecting pipes, however, always contain the same fluid (not count- ing the steam that may leak out, of course). No mass enters or leaves this com- SOURCE bination system, which is indicated by the shaded area on Fig. 6–10; thus, it can be analyzed as a closed system. Recall that for a closed system undergoing Heat input a cycle, the change in internal energy U is zero, and therefore the net work 100 kJ 100 kJ output of the system is also equal to the net heat transfer to the system: Wnet,out  Qin  Qout¬¬1kJ2 (6–2) 1 2 Thermal Efficiency Net Net work work In Eq. 6–2, Qout represents the magnitude of the energy wasted in order to output output complete the cycle. But Qout is never zero; thus, the net work output of a heat 20 kJ 30 kJ engine is always less than the amount of heat input. That is, only part of the heat transferred to the heat engine is converted to work. The fraction of the Waste heat SINK Waste heat heat input that is converted to net work output is a measure of the perfor- 80 kJ 70 kJ mance of a heat engine and is called the thermal efficiency hth (Fig. 6–12). ηth,1 = 20% ηth,2 = 30% For heat engines, the desired output is the net work output, and the required input is the amount of heat supplied to the working fluid. Then the FIGURE 6–12 thermal efficiency of a heat engine can be expressed as Some heat engines perform better than Net work output others (convert more of the heat they Thermal efficiency  (6–3) Total heat input receive to work). cen84959_ch06.qxd 3/31/05 3:51 PM Page 284 284 | Thermodynamics or Wnet,out hth  (6–4) Qin It can also be expressed as Q out h th  1  (6–5) Q in since Wnet,out  Qin  Qout. Cyclic devices of practical interest such as heat engines, refrigerators, and heat pumps operate between a high-temperature medium (or reservoir) at temperature TH and a low-temperature medium (or reservoir) at temperature TL. To bring uniformity to the treatment of heat engines, refrigerators, and High-temperature reservoir heat pumps, we define these two quantities: at TH QH  magnitude of heat transfer between the cyclic device and the high- QH temperature medium at temperature TH QL  magnitude of heat transfer between the cyclic device and the low- temperature medium at temperature TL Wnet,out Notice that both QL and QH are defined as magnitudes and therefore are HE positive quantities. The direction of QH and QL is easily determined by inspection. Then the net work output and thermal efficiency relations for any heat engine (shown in Fig. 6–13) can also be expressed as QL Wnet,out  QH  QL Low-temperature reservoir and at TL Wnet,out hth  FIGURE 6–13 QH Schematic of a heat engine. or QL h th  1  (6–6) Furnace QH QH = 100 MJ The thermal efficiency of a heat engine is always less than unity since both QL and QH are defined as positive quantities. Thermal efficiency is a measure of how efficiently a heat engine converts Wnet,out = 55 MJ the heat that it receives to work. Heat engines are built for the purpose of HE converting heat to work, and engineers are constantly trying to improve the efficiencies of these devices since increased efficiency means less fuel con- sumption and thus lower fuel bills and less pollution. QL = 45 MJ The thermal efficiencies of work-producing devices are relatively low. Ordinary spark-ignition automobile engines have a thermal efficiency of about 25 percent. That is, an automobile engine converts about 25 percent The atmosphere of the chemical energy of the gasoline to mechanical work. This number is as high as 40 percent for diesel engines and large gas-turbine plants and as FIGURE 6–14 high as 60 percent for large combined gas-steam power plants. Thus, even with the most efficient heat engines available today, almost one-half of the Even the most efficient heat engines energy supplied ends up in the rivers, lakes, or the atmosphere as waste or reject almost one-half of the energy useless energy (Fig. 6–14). they receive as waste heat. cen84959_ch06.qxd 3/31/05 3:51 PM Page 285 Chapter 6 | 285 Can We Save Qout? In a steam power plant, the condenser is the device where large quantities of waste heat is rejected to rivers, lakes, or the atmosphere. Then one may ask, can we not just take the condenser out of the plant and save all that waste energy? The answer to this question is, unfortunately, a firm no for the sim- ple reason that without a heat rejection process in a condenser, the cycle cannot be completed. (Cyclic devices such as steam power plants cannot run continuously unless the cycle is completed.) This is demonstrated next with the help of a simple heat engine. Consider the simple heat engine shown in Fig. 6–15 that is used to lift weights. It consists of a piston–cylinder device with two sets of stops. The working fluid is the gas contained within the cylinder. Initially, the gas tem- perature is 30°C. The piston, which is loaded with the weights, is resting on top of the lower stops. Now 100 kJ of heat is transferred to the gas in the cylinder from a source at 100°C, causing it to expand and to raise the loaded piston until the piston reaches the upper stops, as shown in the figure. At this point, the load is removed, and the gas temperature is observed to be 90°C. The work done on the load during this expansion process is equal to the increase in its potential energy, say 15 kJ. Even under ideal conditions (weightless piston, no friction, no heat losses, and quasi-equilibrium expan- sion), the amount of heat supplied to the gas is greater than the work done since part of the heat supplied is used to raise the temperature of the gas. Now let us try to answer this question: Is it possible to transfer the 85 kJ of excess heat at 90°C back to the reservoir at 100°C for later use? If it is, then we will have a heat engine that can have a thermal efficiency of 100 percent under ideal conditions. The answer to this question is again no, for the very simple reason that heat is always transferred from a high- temperature medium to a low-temperature one, and never the other way around. Therefore, we cannot cool this gas from 90 to 30°C by transferring heat to a reservoir at 100°C. Instead, we have to bring the system into con- tact with a low-temperature reservoir, say at 20°C, so that the gas can return to its initial state by rejecting its 85 kJ of excess energy as heat to this reser- voir. This energy cannot be recycled, and it is properly called waste energy. We conclude from this discussion that every heat engine must waste some energy by transferring it to a low-temperature reservoir in order to complete (15 kJ) LOAD LOAD GAS 90°C GAS GAS 30°C 30°C FIGURE 6–15 Heat in Heat out Reservoir at (100 kJ) (85 kJ) A heat-engine cycle cannot be 100°C Reservoir at completed without rejecting some heat 20°C to a low-temperature sink. cen84959_ch06.qxd 3/31/05 3:51 PM Page 286 286 | Thermodynamics the cycle, even under idealized conditions. The requirement that a heat engine exchange heat with at least two reservoirs for continuous operation forms the basis for the Kelvin–Planck expression of the second law of ther- modynamics discussed later in this section. EXAMPLE 6–1 Net Power Production of a Heat Engine Heat is transferred to a heat engine from a furnace at a rate of 80 MW. If the rate of waste heat rejection to a nearby river is 50 MW, determine the FURNACE net power output and the thermal efficiency for this heat engine. · Solution The rates of heat transfer to and from a heat engine are given. QH = 80 MW The net power output and the thermal efficiency are to be determined. · Assumptions Heat losses through the pipes and other components are Wnet,out negligible. HE Analysis A schematic of the heat engine is given in Fig. 6–16. The furnace serves as the high-temperature reservoir for this heat engine and the river as the low-temperature reservoir. The given quantities can be expressed as · # # QL = 50 MW QH  80 MW¬and¬QL  50 MW The net power output of this heat engine is Wnet,out  QH  QL  180  502 MW  30 MW RIVER # # # Then the thermal efficiency is easily determined to be FIGURE 6–16 #  0.375 1or 37.5% 2 Schematic for Example 6–1. Wnet,out 30 MW hth  #  QH 80 MW Discussion Note that the heat engine converts 37.5 percent of the heat it receives to work. m· fuel EXAMPLE 6–2 Fuel Consumption Rate of a Car Combustion chamber A car engine with a power output of 65 hp has a thermal efficiency of 24 · percent. Determine the fuel consumption rate of this car if the fuel has a QH heating value of 19,000 Btu/lbm (that is, 19,000 Btu of energy is released for each lbm of fuel burned). · Wnet,out = 65 hp CAR Solution The power output and the efficiency of a car engine are given. ENGINE The rate of fuel consumption of the car is to be determined. (idealized) Assumptions The power output of the car is constant. Analysis A schematic of the car engine is given in Fig. 6–17. The car · engine is powered by converting 24 percent of the chemical energy released QL during the combustion process to work. The amount of energy input required to produce a power output of 65 hp is determined from the definition of Atmosphere thermal efficiency to be a b  689,270 Btu>h # Wnet,out 65 hp 2545 Btu>h QH   FIGURE 6–17 h th 0.24 1 hp Schematic for Example 6–2. cen84959_ch06.qxd 4/25/05 3:10 PM Page 287 Chapter 6 | 287 To supply energy at this rate, the engine must burn fuel at a rate of # 689,270 Btu>h m  36.3 lbm/h 19,000 Btu>lbm since 19,000 Btu of thermal energy is released for each lbm of fuel burned. Discussion Note that if the thermal efficiency of the car could be doubled, the rate of fuel consumption would be reduced by half. The Second Law of Thermodynamics: Kelvin–Planck Statement We have demonstrated earlier with reference to the heat engine shown in Fig. 6–15 that, even under ideal conditions, a heat engine must reject some heat to a low-temperature reservoir in order to complete the cycle. That is, no heat engine can convert all the heat it receives to useful work. This limi- tation on the thermal efficiency of heat engines forms the basis for the Kelvin–Planck statement of the second law of thermodynamics, which is Thermal energy reservoir expressed as follows: · QH = 100 kW It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce a net amount of work. · Wnet,out = 100 kW That is, a heat engine must exchange heat with a low-temperature sink as well HEAT as a high-temperature source to keep operating. The Kelvin–Planck statement ENGINE can also be expressed as no heat engine can have a thermal efficiency of · 100 percent (Fig. 6–18), or as for a power plant to operate, the working fluid QL = 0 must exchange heat with the environment as well as the furnace. Note that the impossibility of having a 100 percent efficient heat engine is FIGURE 6–18 not due to friction or other dissipative effects. It is a limitation that applies A heat engine that violates the to both the idealized and the actual heat engines. Later in this chapter, we Kelvin–Planck statement of the develop a relation for the maximum thermal efficiency of a heat engine. We second law. also demonstrate that this maximum value depends on the reservoir temper- atures only. 6–4 REFRIGERATORS AND HEAT PUMPS INTERACTIVE TUTORIAL We all know from experience that heat is transferred in the direction of decreasing temperature, that is, from high-temperature mediums to low- SEE TUTORIAL CH. 6, SEC. 4 ON THE DVD. temperature ones. This heat transfer process occurs in nature without requir- ing any devices. The reverse process, however, cannot occur by itself. The transfer of heat from a low-temperature medium to a high-temperature one requires special devices called refrigerators. Refrigerators, like heat engines, are cyclic devices. The working fluid used in the refrigeration cycle is called a refrigerant. The most frequently used refrigeration cycle is the vapor-compression refrigeration cycle, which involves four main components: a compressor, a condenser, an expansion valve, and an evaporator, as shown in Fig. 6–19. cen84959_ch06.qxd 3/31/05 3:51 PM Page 288 288 | Thermodynamics Surrounding medium such as the kitchen air QH CONDENSER 800 kPa 800 kPa 30°C 60°C Wnet,in EXPANSION VALVE COMPRESSOR 120 kPa 120 kPa –25°C –20°C EVAPORATOR FIGURE 6–19 QL Basic components of a refrigeration system and typical operating Refrigerated space conditions. The refrigerant enters the compressor as a vapor and is compressed to the condenser pressure. It leaves the compressor at a relatively high temperature and cools down and condenses as it flows through the coils of the condenser by rejecting heat to the surrounding medium. It then enters a capillary tube Warm environment where its pressure and temperature drop drastically due to the throttling effect. at TH > TL The low-temperature refrigerant then enters the evaporator, where it evapo- rates by absorbing heat from the refrigerated space. The cycle is completed as Required the refrigerant leaves the evaporator and reenters the compressor. QH input In a household refrigerator, the freezer compartment where heat is absorbed Wnet,in by the refrigerant serves as the evaporator, and the coils usually behind the R refrigerator where heat is dissipated to the kitchen air serve as the condenser. A refrigerator is shown schematically in Fig. 6–20. Here QL is the magni- tude of the heat removed from the refrigerated space at temperature TL, QH Desired output is the magnitude of the heat rejected to the warm environment at tempera- QL ture TH, and Wnet,in is the net work input to the refrigerator. As discussed before, QL and QH represent magnitudes and thus are positive quantities. Cold refrigerated space at TL Coefficient of Performance The efficiency of a refrigerator is expressed in terms of the coefficient of FIGURE 6–20 performance (COP), denoted by COPR. The objective of a refrigerator is to The objective of a refrigerator is to remove heat (QL) from the refrigerated space. To accomplish this objective, remove QL from the cooled space. it requires a work input of Wnet,in. Then the COP of a refrigerator can be expressed as Desired output QL COPR   (6–7) Required input Wnet,in. This relation. can also be expressed in rate form by replacing QL by QL and Wnet,in by Wnet,in. The conservation of energy principle for a cyclic device requires that Wnet,in  QH  QL¬¬1kJ2 (6–8) cen84959_ch06.qxd 3/31/05 5:53 PM Page 289 Chapter 6 | 289 Then the COP relation becomes Warm heated space QL at TH > TL 1 COPR

Q H>Q L  1 (6–9) QH  QL Desired QH output Notice that the value of COPR can be greater than unity. That is, the amount of heat removed from the refrigerated space can be greater than the Wnet,in amount of work input. This is in contrast to the thermal efficiency, which can never be greater than 1. In fact, one reason for expressing the efficiency HP of a refrigerator by another term—the coefficient of performance—is the Required desire to avoid the oddity of having efficiencies greater than unity. input QL Heat Pumps Another device that transfers heat from a low-temperature medium to a Cold environment at TL high-temperature one is the heat pump, shown schematically in Fig. 6–21. Refrigerators and heat pumps operate on the same cycle but differ in their FIGURE 6–21 objectives. The objective of a refrigerator is to maintain the refrigerated space at a low temperature by removing heat from it. Discharging this heat The objective of a heat pump is to to a higher-temperature medium is merely a necessary part of the operation, supply heat QH into the warmer space. not the purpose. The objective of a heat pump, however, is to maintain a heated space at a high temperature. This is accomplished by absorbing heat from a low-temperature source, such as well water or cold outside air in winter, and supplying this heat to the high-temperature medium such as a house (Fig. 6–22). Warm An ordinary refrigerator that is placed in the window of a house with its indoors door open to the cold outside air in winter will function as a heat pump at 20°C since it will try to cool the outside by absorbing heat from it and rejecting this heat into the house through the coils behind it (Fig. 6–23). The measure of performance of a heat pump is also expressed in terms of QH = 7 kJ the coefficient of performance COPHP, defined as Wnet,in = 2 kJ Desired output QH COPHP

(6–10) COP = 3.5 HP Required input Wnet,in which can also be expressed as QH 1 QL = 5 kJ COPHP

1  Q L >Q H (6–11) QH  QL Cold outdoors A comparison of Eqs. 6–7 and 6–10 reveals that at 4°C COPHP
COPR  1 (6–12) FIGURE 6–22 for fixed values of QL and QH. This relation implies that the coefficient of The work supplied to a heat pump is performance of a heat pump is always greater than unity since COPR is a used to extract energy from the cold positive quantity. That is, a heat pump will function, at worst, as a resistance outdoors and carry it into the warm heater, supplying as much energy to the house as it consumes. In reality, indoors. however, part of QH is lost to the outside air through piping and other devices, and COPHP may drop below unity when the outside air temperature is too low. When this happens, the system usually switches to a resistance heating mode. Most heat pumps in operation today have a seasonally aver- aged COP of 2 to 3. cen84959_ch06.qxd 3/31/05 3:51 PM Page 290 290 | Thermodynamics Most existing heat pumps use the cold outside air as the heat source in winter, and they are referred to as air-source heat pumps. The COP of such heat pumps is about 3.0 at design conditions. Air-source heat pumps are not appropriate for cold climates since their efficiency drops consider- ably when temperatures are below the freezing point. In such cases, geo- thermal (also called ground-source) heat pumps that use the ground as the heat source can be used. Geothermal heat pumps require the burial of pipes in the ground 1 to 2 m deep. Such heat pumps are more expensive to install, but they are also more efficient (up to 45 percent more efficient than air-source heat pumps). The COP of ground-source heat pumps is about 4.0. Air conditioners are basically refrigerators whose refrigerated space is a room or a building instead of the food compartment. A window air- conditioning unit cools a room by absorbing heat from the room air and discharging it to the outside. The same air-conditioning unit can be used as a heat pump in winter by installing it backwards as shown in Fig. 6–23. In this mode, the unit absorbs heat from the cold outside and delivers it to the room. Air-conditioning systems that are equipped with proper controls FIGURE 6–23 and a reversing valve operate as air conditioners in summer and as heat When installed backward, an air pumps in winter. conditioner functions as a heat pump. The performance of refrigerators and air conditioners in the United States © Reprinted with special permission of King is often expressed in terms of the energy efficiency rating (EER), which is Features Syndicate. the amount of heat removed from the cooled space in Btu’s for 1 Wh (watt- hour) of electricity consumed. Considering that 1 kWh  3412 Btu and thus 1 Wh  3.412 Btu, a unit that removes 1 kWh of heat from the cooled space for each kWh of electricity it consumes (COP  1) will have an EER of 3.412. Therefore, the relation between EER and COP is EER  3.412 COPR Most air conditioners have an EER between 8 and 12 (a COP of 2.3 to 3.5). A high-efficiency heat pump manufactured by the Trane Company using a reciprocating variable-speed compressor is reported to have a COP of 3.3 in the heating mode and an EER of 16.9 (COP of 5.0) in the air- conditioning mode. Variable-speed compressors and fans allow the unit to operate at maximum efficiency for varying heating/cooling needs and weather conditions as determined by a microprocessor. In the air-conditioning mode, for example, they operate at higher speeds on hot days and at lower speeds on cooler days, enhancing both efficiency and comfort. The EER or COP of a refrigerator decreases with decreasing refrigera- tion temperature. Therefore, it is not economical to refrigerate to a lower temperature than needed. The COPs of refrigerators are in the range of 2.6–3.0 for cutting and preparation rooms; 2.3–2.6 for meat, deli, dairy, and produce; 1.2–1.5 for frozen foods; and 1.0–1.2 for ice cream units. Note that the COP of freezers is about half of the COP of meat refrigera- tors, and thus it costs twice as much to cool the meat products with refrig- erated air that is cold enough to cool frozen foods. It is good energy conservation practice to use separate refrigeration systems to meet different refrigeration needs. cen84959_ch06.qxd 3/31/05 3:51 PM Page 291 Chapter 6 | 291 EXAMPLE 6–3 Heat Rejection by a Refrigerator Kitchen The food compartment of a refrigerator, shown in Fig. 6–24, is maintained at 4°C by removing heat from it at a rate of 360 kJ/min. If the required power · QH input to the refrigerator is 2 kW, determine (a) the coefficient of perfor- mance of the refrigerator and (b) the rate of heat rejection to the room that · Wnet,in = 2 kW houses the refrigerator. R Solution The power consumption of a refrigerator is given. The COP and the rate of heat rejection are to be determined. Assumptions Steady operating conditions exist. · QL = 360 kJ/min Analysis (a) The coefficient of performance of the refrigerator is # ¬a b 3 QL 360 kJ>min 1 kW COPR  #  Food Wnet,in 2 kW 60 kJ>min compartment 4°C That is, 3 kJ of heat is removed from the refrigerated space for each kJ of work supplied. (b) The rate at which heat is rejected to the room that houses the refrigerator FIGURE 6–24 is determined from the conservation of energy relation for cyclic devices, Schematic for Example 6–3. QH  QL  Wnet,in  360 kJ>min  12 kW2 a b  480 kJ/min # # # 60 kJ>min 1 kW Discussion Notice that both the energy removed from the refrigerated space as heat and the energy supplied to the refrigerator as electrical work eventu- ally show up in the room air and become part of the internal energy of the air. This demonstrates that energy can change from one form to another, can move from one place to another, but is never destroyed during a process. EXAMPLE 6–4 Heating a House by a Heat Pump House 80,000 kJ/h A heat pump is used to meet the heating requirements of a house and main- 20°C tain it at 20°C. On a day when the outdoor air temperature drops to 2°C, Heat loss the house is estimated to lose heat at a rate of 80,000 kJ/h. If the heat pump under these conditions has a COP of 2.5, determine (a) the power consumed by the heat pump and (b) the rate at which heat is absorbed from · QH the cold outdoor air. · Wnet,in = ? Solution The COP of a heat pump is given. The power consumption and the rate of heat absorption are to be determined. COP = 2.5 HP Assumptions Steady operating conditions exist. Analysis (a) The power consumed by this heat pump, shown in Fig. 6–25, is determined from the definition of the coefficient of performance to be Q· L = ? #  32,000 kJ/h 1or 8.9 kW2 # QH 80,000 kJ>h Wnet,in   Outdoor air at – 2°C COPHP 2.5 (b) The house is losing heat at a rate of 80,000 kJ/h. If the house is to be FIGURE 6–25 maintained at a constant temperature of 20°C, the heat pump must deliver Schematic for Example 6–4. cen84959_ch06.qxd 3/31/05 3:51 PM Page 292 292 | Thermodynamics heat to the house at the same rate, that is, at a rate of 80,000 kJ/h. Then the rate of heat transfer from the outdoor becomes QL  QH  Wnet,in  180,000  32,000 2 kJ>h  48,000 kJ/h # # # Discussion Note that 48,000 of the 80,000 kJ/h heat delivered to the house is actually extracted from the cold outdoor air. Therefore, we are pay- ing only for the 32,000-kJ/h energy that is supplied as electrical work to the heat pump. If we were to use an electric resistance heater instead, we would have to supply the entire 80,000 kJ/h to the resistance heater as electric energy. This would mean a heating bill that is 2.5 times higher. This explains the popularity of heat pumps as heating systems and why they are preferred to simple electric resistance heaters despite their considerably higher initial cost. The Second Law of Thermodynamics: Clausius Statement There are two classical statements of the second law—the Kelvin–Planck statement, which is related to heat engines and discussed in the preceding Warm environment section, and the Clausius statement, which is related to refrigerators or heat pumps. The Clausius statement is expressed as follows: It is impossible to construct a device that operates in a cycle and produces QH = 5 kJ no effect other than the transfer of heat from a lower-temperature body to a higher-temperature body. Wnet,in = 0 It is common knowledge that heat does not, of its own volition, transfer R from a cold medium to a warmer one. The Clausius statement does not imply that a cyclic device that transfers heat from a cold medium to a warmer one is impossible to construct. In fact, this is precisely what a com- QL = 5 kJ mon household refrigerator does. It simply states that a refrigerator cannot operate unless its compressor is driven by an external power source, such as an electric motor (Fig. 6–26). This way, the net effect on the surroundings Cold refrigerated space involves the consumption of some energy in the form of work, in addition to the transfer of heat from a colder body to a warmer one. That is, it leaves a FIGURE 6–26 trace in the surroundings. Therefore, a household refrigerator is in complete A refrigerator that violates the compliance with the Clausius statement of the second law. Clausius statement of the second law. Both the Kelvin–Planck and the Clausius statements of the second law are negative statements, and a negative statement cannot be proved. Like any other physical law, the second law of thermodynamics is based on experimen- tal observations. To date, no experiment has been conducted that contradicts the second law, and this should be taken as sufficient proof of its validity. Equivalence of the Two Statements The Kelvin–Planck and the Clausius statements are equivalent in their conse- quences, and either statement can be used as the expression of the second law of thermodynamics. Any device that violates the Kelvin–Planck statement also violates the Clausius statement, and vice versa. This can be demonstrated as follows. cen84959_ch06.qxd 4/25/05 3:10 PM Page 293 Chapter 6 | 293 High-temperature reservoir High-temperature reservoir at TH at TH QH QH + QL QL HEAT Wnet ENGINE REFRIG- REFRIG- η th = 100% ERATOR ERATOR = QH QL QL Low-temperature reservoir Low-temperature reservoir FIGURE 6–27 at TL at TL Proof that the violation of the (a) A refrigerator that is powered by (b) The equivalent refrigerator Kelvin–Planck statement leads to the a 100 percent efficient heat engine violation of the Clausius statement. Consider the heat-engine-refrigerator combination shown in Fig. 6–27a, operating between the same two reservoirs. The heat engine is assumed to have, in violation of the Kelvin–Planck statement, a thermal efficiency of 100 percent, and therefore it converts all the heat QH it receives to work W. This work is now supplied to a refrigerator that removes heat in the amount of QL from the low-temperature reservoir and rejects heat in the amount of QL  QH to the high-temperature reservoir. During this process, the high- temperature reservoir receives a net amount of heat QL (the difference between QL  QH and QH). Thus, the combination of these two devices can be viewed as a refrigerator, as shown in Fig. 6–27b, that transfers heat in an amount of QL from a cooler body to a warmer one without requiring any input from outside. This is clearly a violation of the Clausius statement. Therefore, a violation of the Kelvin–Planck statement results in the viola- tion of the Clausius statement. It can also be shown in a similar manner that a violation of the Clausius statement leads to the violation of the Kelvin–Planck statement. Therefore, the Clausius and the Kelvin–Planck statements are two equivalent expres- sions of the second law of thermodynamics. 6–5 PERPETUAL-MOTION MACHINES INTERACTIVE We have repeatedly stated that a process cannot take place unless it satisfies TUTORIAL both the first and second laws of thermodynamics. Any device that violates SEE TUTORIAL CH. 6, SEC. 5 ON THE DVD. either law is called a perpetual-motion machine, and despite numerous attempts, no perpetual-motion machine is known to have worked. But this has not stopped inventors from trying to create new ones. A device that violates the first law of thermodynamics (by creating energy) is called a perpetual-motion machine of the first kind (PMM1), and a device that violates the second law of thermodynamics is called a perpetual-motion machine of the second kind (PMM2). cen84959_ch06.qxd 3/31/05 3:51 PM Page 294 294 | Thermodynamics · Wnet,out System boundary BOILER Resistance heater PUMP TURBINE GENERATOR FIGURE 6–28 A perpetual-motion machine that CONDENSER violates the first law of · thermodynamics (PMM1). Qout Consider the steam power plant shown in Fig. 6–28. It is proposed to heat the steam by resistance heaters placed inside the boiler, instead of by the energy supplied from fossil or nuclear fuels. Part of the electricity generated by the plant is to be used to power the resistors as well as the pump. The rest of the electric energy is to be supplied to the electric network as the net work output. The inventor claims that once the system is started, this power plant will produce electricity indefinitely without requiring any energy input from the outside. Well, here is an invention that could solve the world’s energy problem—if it works, of course. A careful examination of this invention reveals that the system enclosed by the. shaded. area is continuously supplying energy to the outside at a rate of Qout  Wnet,out without. receiving. any energy. That is, this system is creating energy at a rate of Qout  Wnet,out, which is clearly a vio- lation of the first law. Therefore, this wonderful device is nothing more than a PMM1 and does not warrant any further consideration. Now let us consider another novel idea by the same inventor. Convinced that energy cannot be created, the inventor suggests the following modifica- tion that will greatly improve the thermal efficiency of that power plant without violating the first law. Aware that more than one-half of the heat transferred to the steam in the furnace is discarded in the condenser to the environment, the inventor suggests getting rid of this wasteful component and sending the steam to the pump as soon as it leaves the turbine, as shown in Fig. 6–29. This way, all the heat transferred to the steam in the boiler will be converted to work, and thus the power plant will have a theoretical effi- ciency of 100 percent. The inventor realizes that some heat losses and fric- tion between the moving components are unavoidable and that these effects will hurt the efficiency somewhat, but still expects the efficiency to be no less than 80 percent (as opposed to 40 percent in most actual power plants) for a carefully designed system. Well, the possibility of doubling the efficiency would certainly be very tempting to plant managers and, if not properly trained, they would proba- bly give this idea a chance, since intuitively they see nothing wrong with it. A student of thermodynamics, however, will immediately label this cen84959_ch06.qxd 3/31/05 3:51 PM Page 295 Chapter 6 | 295 System · boundary Qin BOILER · Wnet,out PUMP TURBINE FIGURE 6–29 A perpetual-motion machine that violates the second law of thermodynamics (PMM2). device as a PMM2, since it works on a cycle and does a net amount of work while exchanging heat with a single reservoir (the furnace) only. It satisfies the first law but violates the second law, and therefore it will not work. Countless perpetual-motion machines have been proposed throughout his- tory, and many more are being proposed. Some proposers have even gone so far as to patent their inventions, only to find out that what they actually have in their hands is a worthless piece of paper. Some perpetual-motion machine inventors were very successful in fund- raising. For example, a Philadelphia carpenter named J. W. Kelly col- lected millions of dollars between 1874 and 1898 from investors in his hydropneumatic-pulsating-vacu-engine, which supposedly could push a rail- road train 3000 miles on 1 L of water. Of course, it never did. After his death in 1898, the investigators discovered that the demonstration machine was powered by a hidden motor. Recently a group of investors was set to invest $2.5 million into a mysterious energy augmentor, which multiplied whatever power it took in, but their lawyer wanted an expert opinion first. Confronted by the scientists, the “inventor” fled the scene without even attempting to run his demo machine. Tired of applications for perpetual-motion machines, the U.S. Patent Office decreed in 1918 that it would no longer consider any perpetual- motion machine applications. However, several such patent applications were still filed, and some made it through the patent office undetected. Some applicants whose patent applications were denied sought legal action. For example, in 1982 the U.S. Patent Office dismissed as just another perpetual- motion machine a huge device that involves several hundred kilograms of rotating magnets and kilometers of copper wire that is supposed to be gen- erating more electricity than it is consuming from a battery pack. However, the inventor challenged the decision, and in 1985 the National Bureau of Standards finally tested the machine just to certify that it is battery-operated. However, it did not convince the inventor that his machine will not work. The proposers of perpetual-motion machines generally have innovative minds, but they usually lack formal engineering training, which is very unfor- tunate. No one is immune from being deceived by an innovative perpetual- motion machine. As the saying goes, however, if something sounds too good to be true, it probably is. cen84959_ch06.qxd 4/25/05 3:10 PM Page 296 296 | Thermodynamics INTERACTIVE 6–6 REVERSIBLE AND IRREVERSIBLE PROCESSES TUTORIAL The second law of thermodynamics states that no heat engine can have an SEE TUTORIAL CH. 6, SEC. 6 ON THE DVD. efficiency of 100 percent. Then one may ask, What is the highest efficiency that a heat engine can possibly have? Before we can answer this question, we need to define an idealized process first, which is called the reversible process. The processes that were discussed at the beginning of this chapter occurred in a certain direction. Once having taken place, these processes cannot reverse themselves spontaneously and restore the system to its initial state. For this reason, they are classified as irreversible processes. Once a cup of hot coffee cools, it will not heat up by retrieving the heat it lost from the sur- roundings. If it could, the surroundings, as well as the system (coffee), would be restored to their original condition, and this would be a reversible process. A reversible process is defined as a process that can be reversed without (a) Frictionless pendulum leaving any trace on the surroundings (Fig. 6–30). That is, both the system and the surroundings are returned to their initial states at the end of the reverse process. This is possible only if the net heat and net work exchange between the system and the surroundings is zero for the combined (original and reverse) process. Processes that are not reversible are called irreversible processes. It should be pointed out that a system can be restored to its initial state (b) Quasi-equilibrium expansion following a process, regardless of whether the process is reversible or irre- and compression of a gas versible. But for reversible processes, this restoration is made without leav- ing any net change on the surroundings, whereas for irreversible processes, FIGURE 6–30 the surroundings usually do some work on the system and therefore does Two familiar reversible processes. not return to their original state. Reversible processes actually do not occur in nature. They are merely ide- alizations of actual processes. Reversible processes can be approximated by actual devices, but they can never be achieved. That is, all the processes occurring in nature are irreversible. You may be wondering, then, why we are bothering with such fictitious processes. There are two reasons. First, they are easy to analyze, since a system passes through a series of equilibrium states during a reversible process; second, they serve as idealized models to which actual processes can be compared. In daily life, the concepts of Mr. Right and Ms. Right are also idealiza- tions, just like the concept of a reversible (perfect) process. People who insist on finding Mr. or Ms. Right to settle down are bound to remain Mr. or Ms. Single for the rest of their lives. The possibility of finding the perfect prospective mate is no higher than the possibility of finding a perfect (reversible) process. Likewise, a person who insists on perfection in friends is bound to have no friends. Engineers are interested in reversible processes because work-producing devices such as car engines and gas or steam turbines deliver the most work, and work-consuming devices such as compressors, fans, and pumps consume the least work when reversible processes are used instead of irreversible ones (Fig. 6–31). Reversible processes can be viewed as theoretical limits for the corre- sponding irreversible ones. Some processes are more irreversible than others. We may never be able to have a reversible process, but we can certainly cen84959_ch06.qxd 3/31/05 3:51 PM Page 297 Chapter 6 | 297 Expansion Compression Expansion Compression Pressure distribution Water Water Water FIGURE 6–31 Water Reversible processes deliver the most (a) Slow (reversible) process (b) Fast (irreversible) process and consume the least work. approach it. The more closely we approximate a reversible process, the more work delivered by a work-producing device or the less work required by a work-consuming device. The concept of reversible processes leads to the definition of the second- law efficiency for actual processes, which is the degree of approximation to the corresponding reversible processes. This enables us to compare the per- formance of different devices that are designed to do the same task on the basis of their efficiencies. The better the design, the lower the irreversibili- ties and the higher the second-law efficiency. Irreversibilities Friction The factors that cause a process to be irreversible are called irreversibilities. They include friction, unrestrained expansion, mixing of two fluids, heat transfer across a finite temperature difference, electric resistance, inelastic deformation of solids, and chemical reactions. The presence of any of these effects renders a process irreversible. A reversible process involves none of GAS these. Some of the frequently encountered irreversibilities are discussed briefly below. Friction is a familiar form of irreversibility associated with bodies in FIGURE 6–32 motion. When two bodies in contact are forced to move relative to each other (a piston in a cylinder, for example, as shown in Fig. 6–32), a friction Friction renders a process irreversible. force that opposes the motion develops at the interface of these two bodies, and some work is needed to overcome this friction force. The energy sup- plied as work is eventually converted to heat during the process and is trans- ferred to the bodies in contact, as evidenced by a temperature rise at the interface. When the direction of the motion is reversed, the bodies are restored to their original position, but the interface does not cool, and heat is not converted back to work. Instead, more of the work is converted to heat while overcoming the friction forces that also oppose the reverse motion. Since the system (the moving bodies) and the surroundings cannot be returned to their original states, this process is irreversible. Therefore, any process that involves friction is irreversible. The larger the friction forces involved, the more irreversible the process is. Friction does not always involve two solid bodies in contact. It is also encountered between a fluid and solid and even between the layers of a fluid moving at different velocities. A considerable fraction of the power produced by a car engine is used to overcome the friction (the drag force) between the air and the external surfaces of the car, and it eventually becomes part of the internal energy of the air. It is not possible to reverse cen84959_ch06.qxd 3/31/05 3:51 PM Page 298 298 | Thermodynamics this process and recover that lost power, even though doing so would not violate the conservation of energy principle. Another example of irreversibility is the unrestrained expansion of a gas separated from a vacuum by a membrane, as shown in Fig. 6–33. When (a) Fast compression the membrane is ruptured, the gas fills the entire tank. The only way to restore the system to its original state is to compress it to its initial volume, while transferring heat from the gas until it reaches its initial temperature. From the conservation of energy considerations, it can easily be shown that the amount of heat transferred from the gas equals the amount of work done on the gas by the surroundings. The restoration of the surroundings involves (b) Fast expansion conversion of this heat completely to work, which would violate the second law. Therefore, unrestrained expansion of a gas is an irreversible process. A third form of irreversibility familiar to us all is heat transfer through a 700 kPa 50 kPa finite temperature difference. Consider a can of cold soda left in a warm room (Fig. 6–34). Heat is transferred from the warmer room air to the cooler soda. The only way this process can be reversed and the soda (c) Unrestrained expansion restored to its original temperature is to provide refrigeration, which requires some work input. At the end of the reverse process, the soda will be FIGURE 6–33 restored to its initial state, but the surroundings will not be. The internal Irreversible compression and energy of the surroundings will increase by an amount equal in magnitude expansion processes. to the work supplied to the refrigerator. The restoration of the surroundings to the initial state can be done only by converting this excess internal energy completely to work, which is impossible to do without violating the second law. Since only the system, not both the system and the surroundings, can be restored to its initial condition, heat transfer through a finite temperature difference is an irreversible process. 20°C Heat transfer can occur only when there is a temperature difference between a system and its surroundings. Therefore, it is physically impossi- Heat 20°C ble to have a reversible heat transfer process. But a heat transfer process becomes less and less irreversible as the temperature difference between the 5°C two bodies approaches zero. Then heat transfer through a differential tem- perature difference dT can be considered to be reversible. As dT approaches (a) An irreversible heat transfer process zero, the process can be reversed in direction (at least theoretically) without requiring any refrigeration. Notice that reversible heat transfer is a concep- tual process and cannot be duplicated in the real world. The smaller the temperature difference between two bodies, the smaller 20°C the heat transfer ra