Moran, energetyka, mgr 2, termodynamika procesów nieodwracalnych
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//-->6.6Entropy Rate Balance for Control Volumes237Combining the mass and entropy rate balances#####1m2m32s1m2s2m3s3scv###m21s1s22m31s1s32scv###0.4m11s1s220.6m11s1s32scvp3dp1##Solving forscvm1and using Eq. 6.23 to evaluate changes in specific entropy#scvT3p2T20.4ccplnRlnd0.6ccpln#p1m1T1T1Rln❸0.4c a1.0kJ352#Kbln 294kga18.314 kJ#°Kbln 5.0d28.97 kga8.314 kJ1#°Kbln 5.0d28.97 kg❹❺kJ255#°Kbln 294kg0.454 kJ/kg#°K0.6c a1.0Thus, the second law of thermodynamics is also satisfied.On the basis of this evaluation, the inventor’s claim does not violate principles of thermodynamics.❶❷❸❹❺Since the specific heatcpof air varies little over the temperature interval from 0 to 79 C,cpcan be taken as constant.From Table A-20,cp1.0 kJ /kg#K.Since temperaturedifferencesare involved in this calculation, the temperatures can be either in C or K.In this calculation involving temperatureratios,the temperatures must be in K.If the value of the rate of entropy production had been negative or zero, the claim would be rejected. A negative value isimpossible by the second law and a zero value would indicate operation without irreversibilities.Such devicesdoexist. They are known asvortex tubesand are used in industry forspot cooling.In Example 6.8, we evaluate and compare the rates of entropy production for three com-ponents of a heat pump system. Heat pumps are considered in detail in Chap. 10.EXAMPLE6.8Entropy Production in Heat Pump ComponentsComponents of a heat pump for supplying heated air to a dwelling are shown in the schematic below. At steady state,Refrigerant 22 enters the compressor at 5 C, 3.5 bar and is compressed adiabatically to 75 C, 14 bar. From the com-pressor, the refrigerant passes through the condenser, where it condenses to liquid at 28 C, 14 bar. The refrigerant thenexpands through a throttling valve to 3.5 bar. The states of the refrigerant are shown on the accompanyingT–sdiagram.Return air from the dwelling enters the condenser at 20 C, 1 bar with a volumetric flow rate of 0.42 m3/s and exits at50 C with a negligible change in pressure. Using the ideal gas model for the air and neglecting kinetic and potential en-ergy effects,(a)determine the rates of entropy production, in kW/K, for control volumes enclosing the condenser,compressor, and expansion valve, respectively.(b)Discuss the sources of irreversibility in the components considered inpart (a).SOLUTIONKnown:Refrigerant 22 is compressed adiabatically, condensed by heat transfer to air passing through a heat exchanger, andthen expanded through a throttling valve. Steady-state operating data are known.Find:Determine the entropy production rates for control volumes enclosing the condenser, compressor, and expansion valve,respectively, and discuss the sources of irreversibility in these components.238Chapter6Using EntropySchematic and Given Data:Indoor return airT5= 20°C5p5= 1 bar3/s(AV)5= 0.42 m3p3= 14 barT3= 28°CExpansionvalvep4= 3.5 bar4CondenserSupply airT= 50°C6p6= 1 bar6T2p2= 14 barT2= 75°C14 barCompressorT1= –5°Cp1= 3.5 bar4328°C275°C13.5 bar1–5°COutdoor airEvaporatorsFigure E6.8Assumptions:1.Each component is analyzed as a control volume at steady state.2.The compressor operates adiabatically, and the expansion across the valve is athrottling process.##3.For the control volume enclosing the condenser,Wcv0 andQcv0.4.Kinetic and potential energy effects can be neglected.❶5.The air is modeled as an ideal gas with constantcpAnalysis:1.005 kJ/kg#K.(a)Let us begin by obtaining property data at each of the principal refrigerant states located on the accompanying schematic andT–sdiagram. At the inlet to the compressor, the refrigerant is a superheated vapor at 5 C, 3.5 bar, so from Table A-9,s10.9572 kJ/kg#K. Similarly, at state 2, the refrigerant is a superheated vapor at 75 C, 14 bar, so interpolating in Table A-9givess20.98225 kJ/kg#K andh2294.17 kJ/kg.State 3 is compressed liquid at 28 C, 14 bar. From Table A-7,s3sf(28 C) 0.2936 kJ/kg#K andh3hf(28 C) 79.05 kJ/kg.The expansion through the valve is athrottling process,soh3h4. Using data from Table A-8, the quality at state 4 isx4and the specific entropy iss4Condenser.Consider the control volume enclosing the condenser. With assumptions 1 and 3, the entropy rate balance reduces to###mref1s2s32mair1s5s62scond###To evaluatescondrequires the two mass flow rates,mairandmref, and the change in specific entropy for the air. These are ob-tained next.Evaluating the mass flow rate of air using the ideal gas model (assumption 5)#mair1AV25v51AV25p5RT511bar2sf4x41sg4sf421h41hfg24hf42179.051212.91233.0920.2160.13280.21610.94310.132820.3078 kJ/kg#Ka0.42m3105N/m21 kJ`b` `3#`s1 bar10 N m8.314 kJb 1293K2a28.97 kg#K0.5 kg/s6.6Entropy Rate Balance for Control Volumes239The refrigerant mass flow rate is determined using an energy balance for the control volume enclosing the condenser togetherwith assumptions 1, 3, and 4 to obtain#mair1h6h52#mref1h2h32With assumption 5,h6h5cp(T6T5). Inserting valuesa0.5kgkJb a1.005#b 13232932Kskg K1294.17 79 .052 kJ/kgp6p5❷#mref0.07 kg/sUsing Eq. 6.23, the change in specific entropy of the air iss6s5cplnT6T5Rln3231.0kJa1.005#blnabRlna bkg K2931.0#Finally, solving the entropy balance forscondand inserting values###scondmref1s3s22mair1s6s52c a0.077.95Compressor.kgb 10.2936s1040.098 kJ/kg#K0.982252kJkg#K10.52 10.0982 d `1 kW`1 kJ/skWKFor the control volume enclosing the compressor, the entropy rate balance reduces with assumptions 1 and 3 to##mref1s1s22scompor#scomp#mref1s2s12kga0.07 b 10.98225s17.5Valve.Finally, for the control volume enclosing the throttling valve, the entropy rate balance reduces to##mref1s3s42svalve#Solving forsvalveand inserting values#svalve#mref1s49.94s321040.95722a1 kWkJ#Kb `1 kJ/s`kg104kW/Ka0.07kW/Kkgb 10.3078s0.29362akJ1 kW#Kb `1 kJ/s`kg(b)The following table summarizes, in rank order, the calculated entropy production rates:Component.cv(kW/K)101010444❸compressorvalvecondenser17.59.947.95240Chapter6Using EntropyEntropy production in the compressor is due to fluid friction, mechanical friction of the moving parts, and internal heat trans-fer. For the valve, the irreversibility is primarily due to fluid friction accompanying the expansion across the valve. The prin-cipal source of irreversibility in the condenser is the temperature difference between the air and refrigerant streams. In thisexample, there are no pressure drops for either stream passing through the condenser, but slight pressure drops due to fluidfriction would normally contribute to the irreversibility of condensers. The evaporator lightly shown in Fig. E6.8 has not beenanalyzed.❶❷❸Due to the relatively small temperature change of the air, the specific heatcpcan be taken as constant at the average ofthe inlet and exit air temperatures.#Temperatures in K are used to evaluatemref, but since a temperaturedifferenceis involved the same result would be ob-tained if temperatures in C were used. Temperatures in Kmustbe used when a temperatureratiois involved, as in Eq. 6.23used to evaluates6s5.By focusing attention on reducing irreversibilities at the sites with the highest entropy production rates,thermodynamicimprovements may be possible. However, costs and other constraints must be considered, and can be overriding.6.7Isentropic ProcessesThe termisentropicmeans constant entropy. Isentropic processes are encountered in manysubsequent discussions. The object of the present section is to explain how properties are re-lated at any two states of a process in which there is no change in specific entropy.6.7.1General ConsiderationsThe properties at states having the same specific entropy can be related using the graphicaland tabular property data discussed in Sec. 6.3.1. For example, as illustrated by Fig. 6.9, tem-perature–entropy and enthalpy–entropy diagrams are particularly convenient for determiningproperties at states having the same value of specific entropy. All states on a vertical linepassing through a given state have the same entropy. If state 1 on Fig. 6.9 is fixed by pres-surep1and temperatureT1, states 2 and 3 are readily located once one additional property,such as pressure or temperature, is specified. The values of several other properties at states2 and 3 can then be read directly from the figures.Tabular data also can be used to relate two states having the same specific entropy. Forthe case shown in Fig. 6.9, the specific entropy at state 1 could be determined from thesuperheated vapor table. Then, withs2s1and one other property value, such asp2orT2,T1p1T1h1p1T1p2223p2T2p3T3sFigure 6.9s3T2T3p3T–sandh–sdiagrams showing states having the same value ofspecific entropy.
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