C0 COP Cooling Mode2 5 (22.3) Aswarphysics.weebly

22.4 The Carnot Engine661Hot reservoirat ThHeat WengineQ hCWC CarnotrnheatpumpQcFigure 22.9 A Carnot engineoperated as a heat pump andanother engine with a proposedhigher efficiency operate betweentwo energy reservoirs. The workoutput and input are matched.QcComCold reservoirat Tc INTERFOTO/AlamyQhSadi CarnotFrench engineer (1796–1832)0W 00Qh 0.0 WC 00 Q hC 0ice . eC Ss.weebly.cmaximum possible efficiency for real engines. Let us confirm that the Carnotengine is the most efficient. We imagine a hypothetical engine with an efficiencygreater than that of the Carnot engine. Consider Figure 22.9, which shows thehypothetical engine with e 7 e C on the left connected between hot and cold reservoirs. In addition, let us attach a Carnot engine between the same reservoirs.Because the Carnot cycle is reversible, the Carnot engine can be run in reverse asa Carnot heat pump as shown on the right in Figure 22.9. We match the outputwork of the engine to the input work of the heat pump, W 5 W C , so there is noexchange of energy by work between the surroundings and the engine–heat pumpcombination.Because of the proposed relation between the efficiencies, we must haveysThe numerators of these two fractions cancel because the works have beenmatched. This expression requires thatarph0 Q hC 0 . 0 Q h 0(22.5)From Equation 22.1, the equality of the works gives us0 W 0 5 0 W C 0 S 0 Q h 0 2 0 Q c 0 5 0 Q hC 0 2 0 Q c C 0swwhich can be rewritten to put the energies exchanged with the cold reservoir onthe left and those with the hot reservoir on the right:.a0 Q hC 0 2 0 Q h 0 5 0 Q c C 0 2 0 Q c 0(22.6)wwwNote that the left side of Equation 22.6 is positive, so the right side must be positivealso. We see that the net energy exchange with the hot reservoir is equal to the netenergy exchange with the cold reservoir. As a result, for the combination of theheat engine and the heat pump, energy is transferring from the cold reservoir tothe hot reservoir by heat with no input of energy by work.This result is in violation of the Clausius statement of the second law. Therefore,our original assumption that e 7 e C must be incorrect, and we must conclude thatthe Carnot engine represents the highest possible efficiency for an engine. The keyfeature of the Carnot engine that makes it the most efficient is its reversibility; it canbe run in reverse as a heat pump. All real engines are less efficient than the Carnotengine because they do not operate through a reversible cycle. The efficiency of areal engine is further reduced by such practical difficulties as friction and energylosses by conduction.To describe the Carnot cycle taking place between temperatures Tc and Th , let’sassume the working substance is an ideal gas contained in a cylinder fitted with amovable piston at one end. The cylinder’s walls and the piston are thermally nonconducting. Four stages of the Carnot cycle are shown in Figure 22.10(page 662),Carnot was the first to show the quantitative relationship between work andheat. In 1824, he published his onlywork, Reflections on the Motive Powerof Heat, which reviewed the industrial,political, and economic importance ofthe steam engine. In it, he defined workas “weight lifted through a height.”

662 Chapter 22 Heat Engines, Entropy, and the Second Law of ThermodynamicsFigure 22.10The Carnot cycle.The letters A, B, C, and D referto the states of the gas shown inFigure 22.11. The arrows on thepiston indicate the direction of itsmotion during each process.ASBThe gas undergoes anisothermal expansion.QhEnergy reservoir at ThBSCThe gas undergoesan adiabaticexpansion.ysiclys.wCSDThe gas undergoesan isothermalcompression.arphQ 0eebCycleThermal insulationd.cDSAThe gas undergoesan adiabaticcompression.Q 0omaThermal insulationbQcEnergy reservoir at Tccswand the PV diagram for the cycle is shown in Figure 22.11. The Carnot cycle consistsof two adiabatic processes and two isothermal processes, all reversible:1. Process A S B (Fig. 22.10a) is an isothermal expansion at temperature Th .The gas is placed in thermal contact with an energy reservoir at temperature Th . During the expansion, the gas absorbs energy Q h from the reservoir through the base of the cylinder and does work WAB in raising thepiston.2. In process B S C (Fig. 22.10b), the base of the cylinder is replaced by athermally nonconducting wall and the gas expands adiabatically; that is, noenergy enters or leaves the system by heat. During the expansion, the temperature of the gas decreases from Th to Tc and the gas does work W BC inraising the piston.3. In process C S D (Fig. 22.10c), the gas is placed in thermal contact with anenergy reservoir at temperature Tc and is compressed isothermally at temperature Tc . During this time, the gas expels energy Q c to the reservoirand the work done by the piston on the gas is WCD .4. In the final process D S A (Fig. 22.10d), the base of the cylinder is replacedby a nonconducting wall and the gas is compressed adiabatically. The temperature of the gas increases to Th , and the work done by the piston on thegas is W DA .awwAThe work doneduring the cycleequals the areaenclosed by the pathon the PV diagram.wPQhBWengDThQcCTcVFigure 22.11 PV diagram for theCarnot cycle. The net work doneWeng equals the net energy transferred into the Carnot engine inone cycle, Q h 2 Q c .

22.4 The Carnot Engine663The thermal efficiency of the engine is given by Equation 22.2:e5120Qc00Qh0In Example 22.3, we show that for a Carnot cycle,0Qc00Qh05TcTh(22.7)Hence, the thermal efficiency of a Carnot engine isTcTh(22.8)WWEfficiency of a Carnot engineomeC 5 1 2W0Qh0ys0Qh0arphCOPC 1 heating mode 2 5ics.weebly.cThis result indicates that all Carnot engines operating between the same two temperatures have the same efficiency.5Equation 22.8 can be applied to any working substance operating in a Carnotcycle between two energy reservoirs. According to this equation, the efficiency iszero if Tc 5 Th , as one would expect. The efficiency increases as Tc is lowered andTh is raised. The efficiency can be unity (100%), however, only if Tc 5 0 K. Suchreservoirs are not available; therefore, the maximum efficiency is always less than100%. In most practical cases, Tc is near room temperature, which is about 300 K.Therefore, one usually strives to increase the efficiency by raising Th .Theoretically, a Carnot-cycle heat engine run in reverse constitutes the mosteffective heat pump possible, and it determines the maximum COP for a given combination of hot and cold reservoir temperatures. Using Equations 22.1 and 22.4, wesee that the maximum COP for a heat pump in its heating mode is50Qh0 2 0Qc0512150Qc00Qh0112TcTh5ThTh 2 TcswThe Carnot COP for a heat pump in the cooling mode isCOPC 1 cooling mode 2 5TcTh 2 Tcww.aAs the difference between the temperatures of the two reservoirs approaches zeroin this expression, the theoretical COP approaches infinity. In practice, the lowtemperature of the cooling coils and the high temperature at the compressor limitthe COP to values below 10.wQ uick Quiz 22.3  Three engines operate between reservoirs separated in temperature by 300 K. The reservoir temperatures are as follows: Engine A: Th 51 000 K, Tc 5 700 K; Engine B: Th 5 800 K, Tc 5 500 K; Engine C: Th 5 600 K,Tc 5 300 K. Rank the engines in order of theoretically possible efficiency fromhighest to lowest.5Forthe processes in the Carnot cycle to be reversible, they must be carried out infinitesimally slowly. Therefore,although the Carnot engine is the most efficient engine possible, it has zero power output because it takes an infinitetime interval to complete one cycle! For a real engine, the short time interval for each cycle results in the workingsubstance reaching a high temperature lower than that of the hot reservoir and a low temperature higher than thatof the cold reservoir. An engine undergoing a Carnot cycle between this narrower temperature range was analyzedby F. L. Curzon and B. Ahlborn (“Efficiency of a Carnot engine at maximum power output,” Am. J. Phys. 43(1), 22,1975), who found that the efficiency at maximum power output depends only on the reservoir temperatures Tc andTh and is given by e C-A 5 1 2 (Tc /Th )1/2. The Curzon–Ahlborn efficiency e C-A provides a closer approximation to theefficiencies of real engines than does the Carnot efficiency.

664 Chapter 22 Heat Engines, Entropy, and the Second Law of ThermodynamicsExample 22.3    Efficiency of the Carnot EngineShow that the ratio of energy transfers by heat in a Carnot engine is equal to the ratio of reservoir temperatures, asgiven by Equation 22.7.SolutionConceptualize Make use of Figures 22.10 and 22.11 to help you visualize the processes in the Carnot cycle.omCategorize Because of our understanding of the Carnot cycle, we can categorize the processes in the cycle as isothermal and adiabatic.Analyze For the isothermal expansion (process A S B0 Q h 0 5 0 DE int 2 WAB 0 5 0 0 2 WAB 0 5 nRTh lnIn a similar manner, find the energy transfer to the coldreservoir during the isothermal compression C S D :0 Q c 0 5 0 DE int 2 WCD 0 5 0 0 2 WCD 0 5 nRTc lnDivide the second expression by the first:(1)VCVDs.w0Qc0Tc ln 1 VC /VD 250Qh0Th ln 1 VB /VA 2ThV Bg21 5 TcVCg21Apply Equation 21.39 to the adiabatic processes B S Cand D S A:icThVAg21 5 TcV D g21aVC g21VB g215a bbVAVDysarphDivide the first equation by the second:eebly.cin Fig. 22.10), find the energy transfer by heat from thehot reservoir using Equation 20.14 and the first law ofthermodynamics:VBVAswSubstitute Equation (2) into Equation (1):(2)0Qc00Qh0VCVB5VAVD5Tc ln 1 VC /VD 2Tc ln 1 VC /VD 2Tc55Th ln 1 VB /VA 2Th ln 1 VC /VD 2Thw.aFinalize This last equation is Equation 22.7, the one we set out to prove.wExample 22.4    The Steam EnginewA steam engine has a boiler that operates at 500 K. The energy from the burning fuel changes water to steam, and thissteam then drives a piston. The cold reservoir’s temperature is that of the outside air, approximately 300 K. What is themaximum thermal efficiency of this steam engine?SolutionConceptualize In a steam engine, the gas pushing on the piston in Figure 22.10 is steam. A real steam engine does notoperate in a Carnot cycle, but, to find the maximum possible efficiency, imagine a Carnot steam engine.Categorize We calculate an efficiency using Equation 22.8, so we categorize this example as a substitution problem.Substitute the reservoir temperatures into Equation 22.8:eC 5 1 2Tc300 K5125 0.400  or  40.0%Th500 KThis result is the highest theoretical efficiency of the engine. In practice, the efficiency is considerably lower.

66522.5 Gasoline and Diesel Engines 22.4 c o n t i n u e dSuppose we wished to increase the theoretical efficiency of this engine. This increase can be achieved byraising Th by DT or by decreasing Tc by the same DT. Which would be more effective?W h at If ?Answer A given DT would have a larger fractional effect on a smaller temperature, so you would expect a largerchange in efficiency if you alter Tc by DT. Let’s test that numerically. Raising Th by 50 K, corresponding to Th 5 550 K,would give a maximum efficiency ofeC 5 1 2Tc300 K5125 0.455Th550 KeC 5 1 2omDecreasing Tc by 50 K, corresponding to Tc 5 250 K, would give a maximum efficiency ofTc250 K5125 0.500Th500 Kly.cAlthough changing Tc is mathematically more effective, often changing Th is practically more feasible.eeb22.5 Gasoline and Diesel Enginesarphysics.wIn a gasoline engine, six processes occur in each cycle; they are illustrated in Figure22.12. In this discussion, let’s consider the interior of the cylinder above the pistonto be the system that is taken through repeated cycles in the engine’s operation. Fora given cycle, the piston moves up and down twice, which represents a four-strokecycle consisting of two upstrokes and two downstrokes. The processes in the cyclecan be approximated by the Otto cycle shown in the PV diagram in Figure 22.13(page 666). In the following discussion, refer to Figure 22.12 for the pictorial representation of the strokes and Figure 22.13 for the significance on the PV diagram ofthe letter designations below:The piston movesup and compressesthe mixture.aThe intake valveopens, and the air–fuel mixture entersas the piston movesdown.sw1. During the intake stroke (Fig. 22.12a and O S A in Figure 22.13), the pistonmoves downward and a gaseous mixture of air and fuel is drawn into theThe spark plugfires and ignitesthe mixture.The hot gaspushes the pistondownward.The exhaust valveopens, and theresidual gas escapes.The piston movesup and pushes theremaining gas out.ExhaustwwAirandfuelwSpark austdefFigure 22.12 The four-stroke cycle of a conventional gasoline engine. The arrows on the pistonindicate the direction of its motion during each process.

666 Chapter 22 Heat Engines, Entropy, and the Second Law of 3.Figure 22.13ic4.s.weebly.cPV diagram forthe Otto cycle, which approximately represents the processesoccurring in an internal combustion engine.cylinder at atmospheric pressure. That is the energy input part of the cycle:energy enters the system (the interior of the cylinder) by matter transferas potential energy stored in the fuel. In this process, the volume increasesfrom V2 to V1. This apparent backward numbering is based on the compression stroke (process 2 below), in which the air–fuel mixture is compressedfrom V1 to V2 .During the compression stroke (Fig. 22.12b and A S B in Fig. 22.13), the piston moves upward, the air–fuel mixture is compressed adiabatically fromvolume V1 to volume V2 , and the temperature increases from TA to TB . Thework done on the gas is positive, and its value is equal to the negative of thearea under the curve AB in Figure 22.13.Combustion occurs when the spark plug fires (Fig. 22.12c and B S C in Fig.22.13). That is not one of the strokes of the cycle because it occurs in a veryshort time interval while the piston is at its highest position. The combustion represents a rapid energy transformation from potential energy storedin chemical bonds in the fuel to internal energy associated with molecularmotion, which is related to temperature. During this time interval, themixture’s pressure and temperature increase rapidly, with the temperaturerising from TB to TC . The volume, however, remains approximately constantbecause of the short time interval. As a result, approximately no work isdone on or by the gas. We can model this process in the PV diagram (Fig.22.13) as that process in which the energy Q h enters the system. (In reality, however, this process is a transformation of energy already in the cylinderfrom process O S A.)In the power stroke (Fig. 22.12d and C S D in Fig. 22.13), the gas expandsadiabatically from V2 to V1 . This expansion causes the temperature to dropfrom TC to TD . Work is done by the gas in pushing the piston downward,and the value of this work is equal to the area under the curve CD.Release of the residual gases occurs when an exhaust valve is opened (Fig.22.12e and D S A in Fig. 22.13). The pressure suddenly drops for a shorttime interval. During this time interval, the piston is almost stationary andthe volume is approximately constant. Energy is expelled from the interiorof the cylinder and continues to be expelled during the next process.In the final process, the exhaust stroke (Fig. 22.12e and A S O in Fig. 22.13),the piston moves upward while the exhaust valve remains open. Residualgases are exhausted at atmospheric pressure, and the volume decreasesfrom V1 to V2 . The cycle then repeats.arph5.ysTAomPwww.asw6.If the air–fuel mixture is assumed to be an ideal gas, the efficiency of the Ottocycle ise51211 V1 /V2 2 g211 Otto cycle 2(22.9)where V1 /V2 is the compression ratio and g is the ratio of the molar specific heatsCP /CV for the air–fuel mixture. Equation 22.9, which is derived in Example 22.5,shows that the efficiency increases as the compression ratio increases. For a typical compression ratio of 8 and with g 5 1.4, Equation 22.9 predicts a theoreticalefficiency of 56% for an engine operating in the idealized Otto cycle. This valueis much greater than that achieved in real engines (15% to 20%) because of sucheffects as friction, energy transfer by conduction through the cylinder walls, andincomplete combustion of the air–fuel mixture.Diesel engines operate on a cycle similar to the Otto cycle, but they do not employa spark plug. The compression ratio for a diesel engine is much greater than thatfor a gasoline engine. Air in the cylinder is compressed to a very small volume, and,as a consequence, the cylinder temperature at the end of the compression stroke is

22.6 Entropy667very high. At this point, fuel is injected into the cylinder. The temperature is highenough for the air–fuel mixture to ignite without the assistance of a spark plug.Diesel engines are more efficient than gasoline engines because of their greatercompression ratios and resulting higher combustion temperatures.Example 22.5    Efficiency of the Otto CycleShow that the thermal efficiency of an engine operating in an idealized Otto cycle (see Figs. 22.12 and 22.13) is givenby Equation 22.9. Treat the working substance as an ideal gas.omSolutionConceptualize Study Figures 22.12 and 22.13 to make sure you understand the working of the Otto cycle. Q h 5 nC V (TC 2 TB )DSA Q c 5 nC V (TD 2 TA )lyBSCs.wring by heat in processes B S C and D S A. (In reality,most of the energy enters and leaves by matter transferas the air–fuel mixture enters and leaves the cylinder.)Use Equation 21.23 to find the energy transfers by heatfor these processes, which take place at constant volume:eebAnalyze Model the energy input and output as occur-.cCategorize A s seen in Figure 22.13, we categorize the processes in the Otto cycle as isovolumetric and adiabatic.(1) e 5 1 2Substitute these expressions into Equation 22.2:ASB0Qh0512TD 2 TATC 2 T BTAVAg21 5 TBV Bg21ysicApply Equation 21.39 to the adiabatic processes A S Band C S D:0Qc 0arphSolve these equations for the temperatures TA and TD ,noting that VA 5 V D 5 V1 and V B 5 VC 5 V2:C S D TCVCg21 5 TDV D g21(2) TA 5 TB a(3) TD 5 TC aSubstitute Equation (4) into Equation (1):e512.asw(4)wV2 g21VC g215 TC a bbVDV1V2 g21TD 2 TA5a bTC 2 TBV1Subtract Equation (2) from Equation (3) and rearrange:11 V1 /V2 2 g21wwFinalize This final expression is Equation 22.9.V2 g21VB g215 TB a bbVAV122.6 EntropyThe zeroth law of thermodynamics involves the concept of temperature, and thefirst law involves the concept of internal energy. Temperature and internal energyare both state variables; that is, the value of each depends only on the thermodynamic state of a system, not on the process that brought it to that state. Anotherstate variable—this one related to the second law of thermodynam

22.3 Reversible and Irreversible Processes In the next section, we will discuss a theoretical heat engine that is the most effi-cient possible. To understand its nature, we must first examine the meaning of reversible and irreversible processes. In a process, the system undergoreversible -