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Thermo solutions chapter07

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• An insulated rigid vessel is divided into two parts by a membrane. One part of the vessel contains air at 10 MPa and other part is fully evacuated. The membrane ruptures and the air fill the entire vessel. Is there any heat and/or work transfer during this process? Justify your answer

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Thermo solutions chapter07

1. 1. 7-1 Chapter 7 ENTROPYEntropy and the Increase of Entropy Principle7-1C Yes. Because we used the relation (QH/TH) = (QL/TL) in the proof, which is the defining relation ofabsolute temperature.7-2C No. The Q represents the net heat transfer during a cycle, which could be positive.7-3C Yes.7-4C No. A system may reject more (or less) heat than it receives during a cycle. The steam in a steampower plant, for example, receives more heat than it rejects during a cycle.7-5C No. A system may produce more (or less) work than it receives during a cycle. A steam power plant,for example, produces more work than it receives during a cycle, the difference being the net work output.7-6C The entropy change will be the same for both cases since entropy is a property and it has a fixedvalue at a fixed state.7-7C No. In general, that integral will have a different value for different processes. However, it will havethe same value for all reversible processes.7-8C Yes.7-9C That integral should be performed along a reversible path to determine the entropy change.7-10C No. An isothermal process can be irreversible. Example: A system that involves paddle-wheel workwhile losing an equivalent amount of heat.7-11C The value of this integral is always larger for reversible processes.7-12C No. Because the entropy of the surrounding air increases even more during that process, makingthe total entropy change positive.7-13C It is possible to create entropy, but it is not possible to destroy it.
2. 2. 7-27-14C Sometimes.7-15C Never.7-16C Always.7-17C Increase.7-18C Increases.7-19C Decreases.7-20C Sometimes.7-21C Yes. This will happen when the system is losing heat, and the decrease in entropy as a result of thisheat loss is equal to the increase in entropy as a result of irreversibilities.7-22C They are heat transfer, irreversibilities, and entropy transport with mass.7-23C Greater than.7-24 A rigid tank contains an ideal gas that is being stirred by a paddle wheel. The temperature of the gasremains constant as a result of heat transfer out. The entropy change of the gas is to be determined.Assumptions The gas in the tank is given to be an ideal gas.Analysis The temperature and the specific volume of the gas remain IDEAL GASconstant during this process. Therefore, the initial and the final states 40 C Heatof the gas are the same. Then s2 = s1 since entropy is a property. 200 kJTherefore, 30 C S sys 0
3. 3. 7-37-25 Air is compressed steadily by a compressor. The air temperature is maintained constant by heatrejection to the surroundings. The rate of entropy change of air is to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 Air is an ideal gas. 4 The process involves no internal irreversibilitiessuch as friction, and thus it is an isothermal, internally reversible process.Properties Noting that h = h(T) for ideal gases, we have h1 = h2 since T1 = T2 = 25 C.Analysis We take the compressor as the system. Noting that the enthalpy of air remains constant, theenergy balance for this steady-flow system can be expressed in the rate form as 0 (steady) P2 Ein Eout Esystem 0 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies · Q Ein Eout AIR T = const. 12 kW Win QoutTherefore, Qout Win 12 kWNoting that the process is assumed to be an isothermal and internally P1reversible process, the rate of entropy change of air is determined tobe Qout,air 12 kW Sair 0.0403 kW/K Tsys 298 K7-26 Heat is transferred isothermally from a source to the working fluid of a Carnot engine. The entropychange of the working fluid, the entropy change of the source, and the total entropy change during thisprocess are to be determined.Analysis (a) This is a reversible isothermal process, and the entropy change during such a process is givenby Q S TNoting that heat transferred from the source is equal to the heat transferred to the working fluid, theentropy changes of the fluid and of the source become Qfluid Qin,fluid 900 kJ Sfluid 1.337 kJ/K Tfluid Tfluid 673 K Source Qsource Qout, source 900 kJ(b) Ssource 1.337 kJ/K 400 C Tsource Tsource 673 K 900 kJ(c) Thus the total entropy change of the process is Sgen S total Sfluid Ssource 1.337 1.337 0
4. 4. 7-47-27 EES Problem 7-26 is reconsidered. The effects of the varying the heat transferred to the working fluidand the source temperature on the entropy change of the working fluid, the entropy change of the source,and the total entropy change for the process as the source temperature varies from 100°C to 1000°C are tobe investigated. The entropy changes of the source and of the working fluid are to be plotted against thesource temperature for heat transfer amounts of 500 kJ, 900 kJ, and1300 kJ.Analysis The problem is solved using EES, and the results are tabulated and plotted below."Knowns:"{T_H = 400 [C]}Q_H = 1300 [kJ]T_Sys = T_H"Analysis: (a) & (b) This is a reversible isothermal process, and the entropy change during such a processis given byDELTAS = Q/T""Noting that heat transferred from the source is equal to the heat transferred to the working fluid,the entropy changes of the fluid and of the source become "DELTAS_source = -Q_H/(T_H+273)DELTAS_fluid = +Q_H/(T_Sys+273)"(c) entropy generation for the process:"S_gen = DELTAS_source + DELTAS_fluid Sfluid Ssource Sgen TH [kJ/K] [kJ/K] [kJ/K] [C] 3.485 -3.485 0 100 2.748 -2.748 0 200 2.269 -2.269 0 300 1.932 -1.932 0 400 1.682 -1.682 0 500 1.489 -1.489 0 600 1.336 -1.336 0 700 1.212 -1.212 0 800 1.108 -1.108 0 900 1.021 -1.021 0 1000 4 3.5 Ssource = - Sfluid ] 3 K / J 2.5 K [ d 2 QH = 1300 kJ i u l f 1.5 S QH = 900 kJ 1 0.5 QH = 500 kJ 0 100 200 300 400 500 600 700 800 900 1000 TH [C]
5. 5. 7-57-28E Heat is transferred isothermally from the working fluid of a Carnot engine to a heat sink. Theentropy change of the working fluid is given. The amount of heat transfer, the entropy change of the sink,and the total entropy change during the process are to be determined.Analysis (a) This is a reversible isothermal process, and the Heatentropy change during such a process is given by Q SINK 95 F S T 95 F Carnot heat engineNoting that heat transferred from the working fluid is equal tothe heat transferred to the sink, the heat transfer become Qfluid Tfluid Sfluid 555 R 0.7 Btu/R 388.5 Btu Qfluid,out 388.5 Btu(b) The entropy change of the sink is determined from Qsink,in 388.5 Btu Ssink 0.7 Btu/R Tsink 555 R(c) Thus the total entropy change of the process is Sgen S total Sfluid Ssink 0.7 0.7 0This is expected since all processes of the Carnot cycle are reversible processes, and no entropy isgenerated during a reversible process.7-29 R-134a enters an evaporator as a saturated liquid-vapor at a specified pressure. Heat is transferred tothe refrigerant from the cooled space, and the liquid is vaporized. The entropy change of the refrigerant, theentropy change of the cooled space, and the total entropy change for this process are to be determined.Assumptions 1 Both the refrigerant and the cooled space involve no internal irreversibilities such asfriction. 2 Any temperature change occurs within the wall of the tube, and thus both the refrigerant and thecooled space remain isothermal during this process. Thus it is an isothermal, internally reversible process.Analysis Noting that both the refrigerant and the cooled space undergo reversible isothermal processes, theentropy change for them can be determined from Q S T(a) The pressure of the refrigerant is maintained constant. Therefore, the temperature of the refrigerantalso remains constant at the saturation value, T Tsat @160 kPa 15.6 C 257.4 K (Table A-12)Then, R-134a 160 kPa Qrefrigerant,in 180 kJ S refrigerant 0.699 kJ/K 180 kJ Trefrigerant 257.4 K -5 C(b) Similarly, Qspace,out 180 kJ S space 0.672 kJ/K Tspace 268 K(c) The total entropy change of the process is Sgen S total S refrigerant Sspace 0.699 0.672 0.027 kJ/K
6. 6. 7-6Entropy Changes of Pure Substances7-30C Yes, because an internally reversible, adiabatic process involves no irreversibilities or heat transfer.7-31 The radiator of a steam heating system is initially filled with superheated steam. The valves areclosed, and steam is allowed to cool until the temperature drops to a specified value by transferring heat tothe room. The entropy change of the steam during this process is to be determined.Analysis From the steam tables (Tables A-4 through A-6), P1 200 kPa v 1 0.95986 m 3 /kg T1 150 C s1 7.2810 kJ/kg K H2O 200 kPa T2 40 C v2 v f 0.95986 0.001008 150 C x2 0.04914 Q v2 v1 v fg 19.515 0.001008 s2 sf x 2 s fg 0.5724 0.04914 7.6832 0.9499 kJ/kg KThe mass of the steam is V 0.020 m 3 m 0.02084 kg v1 0.95986 m 3 /kgThen the entropy change of the steam during this process becomes S m s2 s1 0.02084 kg 0.9499 7.2810 kJ/kg K 0.132 kJ/K
7. 7. 7-77-32 A rigid tank is initially filled with a saturated mixture of R-134a. Heat is transferred to the tank from asource until the pressure inside rises to a specified value. The entropy change of the refrigerant, entropychange of the source, and the total entropy change for this process are to be determined.Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero. 2 Thereare no work interactions.Analysis (a) From the refrigerant tables (Tables A-11 through A-13), u1 uf x1u fg 38.28 0.4 186.21 112.76 kJ/kg P1 200 kPa s1 sf x1 s fg 0.15457 0.4 0.78316 0.4678 kJ/kg K x1 0.4 v1 v f x1v fg 0.0007533 0.4 0.099867 0.0007533 0.04040 m 3 /kg v2 v f 0.04040 0.0007907 x2 0.7857 v fg 0.051201 0.0007907 P2 400 kPa u2 uf x 2 u fg 63.62 0.7857 171.45 198.34 kJ/kg v2 v1 s2 sf x 2 s fg 0.24761 0.7857 0.67929 0.7813 kJ/kg KThe mass of the refrigerant is V 0.5 m 3 m 12.38 kg v1 0.04040 m 3 /kgThen the entropy change of the refrigerant becomes Ssystem m s2 s1 12.38 kg 0.7813 0.4678 kJ/kg K 3.880 kJ/K(b) We take the tank as the system. This is a closed system since no mass enters or leaves. Noting that thevolume of the system is constant and thus there is no boundary work, the energy balance for this stationaryclosed system can be expressed as Ein Eout Esystem Net energy transfer Change in internal, kinetic, R-134a by heat, work, and mass potential, etc. energies 200 kPa Source Q Qin U m(u2 u1 ) 35 CSubstituting, Qin m u2 u1 12.38 kg 198.34 112.76 1059 kJThe heat transfer for the source is equal in magnitude but opposite in direction. Therefore, Qsource, out = - Qtank, in = - 1059 kJand Qsource,out 1059 kJ Ssource 3.439 kJ/K Tsource 308 K(c) The total entropy change for this process is S total S system S source 3.880 3.439 0.442 kJ/K
8. 8. 7-87-33 EES Problem 7-32 is reconsidered. The effects of the source temperature and final pressure on thetotal entropy change for the process as the source temperature varies from 30°C to 210°C, and the finalpressure varies from 250 kPa to 500 kPa are to be investigated. The total entropy change for the process isto be plotted as a function of the source temperature for final pressures of 250 kPa, 400 kPa, and 500 kPa.Analysis The problem is solved using EES, and the results are tabulated and plotted below."Knowns:"P_1 = 200 [kPa]x_1 = 0.4V_sys = 0.5 [m^3]P_2 = 400 [kPa]{T_source = 35 [C]}"Analysis: "" Treat the rigid tank as a closed system, with no work in, neglect changes in KE and PE of theR134a."E_in - E_out = DELTAE_sysE_out = 0 [kJ]E_in = QDELTAE_sys = m_sys*(u_2 - u_1)u_1 = INTENERGY(R134a,P=P_1,x=x_1)v_1 = volume(R134a,P=P_1,x=x_1)V_sys = m_sys*v_1"Rigid Tank: The process is constant volume. Then P_2 and v_2 specify state 2."v_2 = v_1u_2 = INTENERGY(R134a,P=P_2,v=v_2)"Entropy calculations:"s_1 = entropy(R134a,P=P_1,x=x_1)s_2 = entropY(R134a,P=P_2,v=v_2)DELTAS_sys = m_sys*(s_2 - s_1)"Heat is leaving the source, thus:"DELTAS_source = -Q/(T_source + 273)"Total Entropy Change:"DELTAS_total = DELTAS_source + DELTAS_sys Stotal Tsource 3 [kJ/K] [C] 0.3848 30 2.5 P2 = 250 kPa 0.6997 60 = 400 kPa 0.9626 90 ] 2 = 500 kPa 1.185 120 K / 1.376 150 J 1.5 k [ 1.542 180 l 1.687 210 a t o 1 t S 0.5 0 25 65 105 145 185 225 Tsource [C]
9. 9. 7-97-34 An insulated rigid tank contains a saturated liquid-vapor mixture of water at a specified pressure. Anelectric heater inside is turned on and kept on until all the liquid vaporized. The entropy change of thewater during this process is to be determined.Analysis From the steam tables (Tables A-4 through A-6) P1 100 kPa v1 v f x1v fg 0.001 0.25 1.6941 0.001 0.4243 m3/kg x1 0.25 s1 s f x1s fg 1.3028 0.25 6.0562 2.8168 kJ/kg K H 2O 2 kg v2 v1 100 kPa s2 6.8649 kJ/kg K sat. vapor WeThen the entropy change of the steam becomes S m s2 s1 (2 kg )(6.8649 2.8168) kJ/kg K 8.10 kJ/K7-35 [Also solved by EES on enclosed CD] A rigid tank is divided into two equal parts by a partition. Onepart is filled with compressed liquid water while the other side is evacuated. The partition is removed andwater expands into the entire tank. The entropy change of the water during this process is to be determined.Analysis The properties of the water are (Table A-4) P1 300 kPa v1 v f @ 60 C 0.001017 m3/kg 1.5 kg compressed T1 60 C s1 s f @ 60 C 0.8313 kJ/kg K liquid Vacuum 300 kPaNoting that v 2 2v1 2 0.001017 0.002034 m3/kg 60 C v2 v f 0.002034 0.001014 P2 15 kPa x2 0.0001018 v fg 10.02 0.001014 v2 0.002034 m3/kg s2 sf x2 s fg 0.7549 0.0001018 7.2522 0.7556 kJ/kg KThen the entropy change of the water becomes S m s2 s1 1.5 kg 0.7556 0.8313 kJ/kg K 0.114 kJ/K
10. 10. 7-107-36 EES Problem 7-35 is reconsidered. The entropy generated is to be evaluated and plotted as a functionof surroundings temperature, and the values of the surroundings temperatures that are valid for thisproblem are to be determined. The surrounding temperature is to vary from 0°C to 100°C.Analysis The problem is solved using EES, and the results are tabulated and plotted below."Input Data"P[1]=300 [kPa]T[1]=60 [C]m=1.5 [kg]P[2]=15 [kPa]Fluid\$=Steam_IAPWSV[1]=m*spv[1]spv[1]=volume(Fluid\$,T=T[1], P=P[1]) "specific volume of steam at state 1, m^3/kg"s[1]=entropy(Fluid\$,T=T[1],P=P[1]) "entropy of steam at state 1, kJ/kgK"V[2]=2*V[1] "Steam expands to fill entire volume at state 2""State 2 is identified by P[2] and spv[2]"spv[2]=V[2]/m "specific volume of steam at state 2, m^3/kg"s[2]=entropy(Fluid\$,P=P[2],v=spv[2]) "entropy of steam at state 2, kJ/kgK"T[2]=temperature(Fluid\$,P=P[2],v=spv[2])DELTAS_sys=m*(s[2]-s[1]) "Total entopy change of steam, kJ/K""What does the first law tell us about this problem?""Conservation of Energy for the entire, closed system"E_in - E_out = DELTAE_sys"neglecting changes in KE and PE for the system:"DELTAE_sys=m*(intenergy(Fluid\$, P=P[2], v=spv[2]) - intenergy(Fluid\$,T=T[1],P=P[1]))E_in = 0"How do you interpert the energy leaving the system, E_out? Recall this is a constant volumesystem."Q_out = E_out"What is the maximum value of the Surroundings temperature?""The maximum possible value for the surroundings temperature occurs when we setS_gen = 0=Delta S_sys+sum(DeltaS_surr)"Q_net_surr=Q_outS_gen = 0S_gen = DELTAS_sys+Q_net_surr/Tsurr"Establish a parametric table for the variables S_gen, Q_net_surr, T_surr, and DELTAS_sys. Inthe Parametric Table window select T_surr and insert a range of values. Then place { and }about the S_gen = 0 line; press F3 to solve the table. The results are shown in Plot Window 1.What values of T_surr are valid for this problem?" Sgen Qnet,surr Tsurr Ssys [kJ/K] [kJ] [K] [kJ/K] 0.02533 37.44 270 -0.1133 0.01146 37.44 300 -0.1133 0.0001205 37.44 330 -0.1133 -0.009333 37.44 360 -0.1133 -0.01733 37.44 390 -0.1133
11. 11. 7-11 0.030 0.020S gen [kJ/K] 0.010 -0.000 -0.010 -0.020 260 280 300 320 340 360 380 400 Tsurr [K]
12. 12. 7-127-37E A cylinder is initially filled with R-134a at a specified state. The refrigerant is cooled and condensedat constant pressure. The entropy change of refrigerant during this process is to be determinedAnalysis From the refrigerant tables (Tables A-11E through A-13E), P1 120 psia s1 0.22361 Btu/lbm R T1 100 F T2 50 F s2 s f @90 F 0.06039 Btu/lbm R R-134a P2 120 psia 120 psia 100 F QThen the entropy change of the refrigerant becomes S m s2 s1 2 lbm 0.06039 0.22361 Btu/lbm R 0.3264 Btu/R7-38 An insulated cylinder is initially filled with saturated liquid water at a specified pressure. The water isheated electrically at constant pressure. The entropy change of the water during this process is to bedetermined.Assumptions 1 The kinetic and potential energy changes are negligible. 2 The cylinder is well-insulatedand thus heat transfer is negligible. 3 The thermal energy stored in the cylinder itself is negligible. 4 Thecompression or expansion process is quasi-equilibrium.Analysis From the steam tables (Tables A-4 through A-6), v 1 v f @150 kPa 0.001053 m 3 /kg H2O P1 150 kPa h1 h f @150 kPa 467.13 kJ/kg 150 kPa sat.liquid 2200 kJ Sat. liquid s1 s f @150 kPa 1.4337 kJ/kg K V 0.005 m 3Also, m 4.75 kg v1 0.001053 m 3 /kg We take the contents of the cylinder as the system. This is a closed system since no mass enters orleaves. The energy balance for this stationary closed system can be expressed as Ein Eout Esystem Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies We,in Wb,out U We,in m(h2 h1 )since U + Wb = H during a constant pressure quasi-equilibrium process. Solving for h2, We,in 2200 kJ h2 h1 467.13 930.33 kJ/kg m 4.75 kgThus, h2 hf 930.33 467.13 P2 150 kPa x2 0.2081 h fg 2226.0 h2 930.33 kJ/kg s2 sf x 2 s fg 1.4337 0.2081 5.7894 2.6384 kJ/kg KThen the entropy change of the water becomes S m s2 s1 4.75 kg 2.6384 1.4337 kJ/kg K 5.72 kJ/K
13. 13. 7-137-39 An insulated cylinder is initially filled with saturated R-134a vapor at a specified pressure. Therefrigerant expands in a reversible manner until the pressure drops to a specified value. The finaltemperature in the cylinder and the work done by the refrigerant are to be determined.Assumptions 1 The kinetic and potential energy changes are negligible. 2 The cylinder is well-insulatedand thus heat transfer is negligible. 3 The thermal energy stored in the cylinder itself is negligible. 4 Theprocess is stated to be reversible.Analysis (a) This is a reversible adiabatic (i.e., isentropic) process, and thus s2 = s1. From the refrigeranttables (Tables A-11 through A-13), v1 v g @ 0.8 MPa 0.025621 m3/kg P1 0.8 MPa u1 u g @ 0.8 MPa 246.79 kJ/kg sat. vapor s1 s g @ 0.8 MPa 0.91835 kJ/kg KAlso, V 0.05 m 3 R-134a m 1.952 kg v1 0.025621 m 3 /kg 0.8 MPa 0.05 m3and s2 sf 0.91835 0.24761 P2 0.4 MPa x2 0.9874 s fg 0.67929 s2 s1 u2 uf x 2 u fg 63.62 0.9874 171.45 232.91 kJ/kg T2 Tsat @ 0.4 MPa 8.91 C(b) We take the contents of the cylinder as the system. This is a closed system since no mass enters orleaves. The energy balance for this adiabatic closed system can be expressed as Ein Eout Esystem Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies Wb,out U Wb,out m(u1 u2 )Substituting, the work done during this isentropic process is determined to be Wb, out m u1 u2 (1.952 kg )(246.79 232.91) kJ/kg 27.09 kJ
14. 14. 7-147-40 EES Problem 7-39 is reconsidered. The work done by the refrigerant is to be calculated and plotted asa function of final pressure as the pressure varies from 0.8 MPa to 0.4 MPa. The work done for this processis to be compared to one for which the temperature is constant over the same pressure range.Analysis The problem is solved using EES, and the results are tabulated and plotted below.ProcedureIsothermWork(P_1,x_1,m_sys,P_2:Work_out_Isotherm,Q_isotherm,DELTAE_isotherm,T_isotherm)T_isotherm=Temperature(R134a,P=P_1,x=x_1)T=T_isothermu_1 = INTENERGY(R134a,P=P_1,x=x_1)v_1 = volume(R134a,P=P_1,x=x_1)s_1 = entropy(R134a,P=P_1,x=x_1)u_2 = INTENERGY(R134a,P=P_2,T=T)s_2 = entropy(R134a,P=P_2,T=T)"The process is reversible and Isothermal thus the heat transfer is determined by:"Q_isotherm = (T+273)*m_sys*(s_2 - s_1)DELTAE_isotherm = m_sys*(u_2 - u_1)E_in = Q_isothermE_out = DELTAE_isotherm+E_inWork_out_isotherm=E_outEND"Knowns:"P_1 = 800 [kPa]x_1 = 1.0V_sys = 0.05[m^3]"P_2 = 400 [kPa]""Analysis: "" Treat the rigid tank as a closed system, with no heat transfer in, neglectchanges in KE and PE of the R134a.""The isentropic work is determined from:"E_in - E_out = DELTAE_sysE_out = Work_out_isenE_in = 0DELTAE_sys = m_sys*(u_2 - u_1)u_1 = INTENERGY(R134a,P=P_1,x=x_1)v_1 = volume(R134a,P=P_1,x=x_1)s_1 = entropy(R134a,P=P_1,x=x_1)V_sys = m_sys*v_1"Rigid Tank: The process is reversible and adiabatic or isentropic.Then P_2 and s_2 specify state 2."s_2 = s_1u_2 = INTENERGY(R134a,P=P_2,s=s_2)T_2_isen = temperature(R134a,P=P_2,s=s_2)CallIsothermWork(P_1,x_1,m_sys,P_2:Work_out_Isotherm,Q_isotherm,DELTAE_isotherm,T_isotherm)
15. 15. 7-15 P2 Workout,isen Workout,isotherm Qisotherm [kPa] [kJ] [kJ] [kJ] 400 27.09 60.02 47.08 500 18.55 43.33 33.29 600 11.44 28.2 21.25 700 5.347 13.93 10.3 800 0 0 0 60 50] 40J Isentropick[t 30uokr 20oW 10 Isothermal 0 400 450 500 550 600 650 700 750 800 P2 [kPa] 50 40 Qisentropic = 0 kJ]J 30k[mre 20htosiQ 10 0 400 450 500 550 600 650 700 750 800 P2 [kPa]
16. 16. 7-167-41 Saturated Refrigerant-134a vapor at 160 kPa is compressed steadily by an adiabatic compressor. Theminimum power input to the compressor is to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible.Analysis The power input to an adiabatic compressor will be a minimum when the compression process isreversible. For the reversible adiabatic process we have s2 = s1. From the refrigerant tables (Tables A-11through A-13), v1 v g @160 kPa 0.12348 m3/kg 2 P1 160 kPa h1 hg @160 kPa 241.11 kJ/kg sat. vapor s1 s g @160 kPa 0.9419 kJ/kg K P2 900 kPa R-134a h2 277.06 kJ/kg s2 s1Also, V1 2 m 3 /min m 16.20 kg/min 0.27 kg/s 1 v1 0.12348 m 3 /kg There is only one inlet and one exit, and thus m1 m2 m . We take the compressor as the system,which is a control volume since mass crosses the boundary. The energy balance for this steady-flow systemcan be expressed in the rate form as 0 (steady) Ein Eout Esystem 0 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies Ein Eout Win mh1 mh2 (since Q ke pe 0) Win m(h2 h1 )Substituting, the minimum power supplied to the compressor is determined to be Win 0.27 kg/s 277.06 241.11 kJ/kg 9.71 kW
17. 17. 7-177-42E Steam expands in an adiabatic turbine. The maximum amount of work that can be done by theturbine is to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible.Analysis The work output of an adiabatic turbine is maximum when the expansion process is reversible.For the reversible adiabatic process we have s2 = s1. From the steam tables (Tables A-4E through A-6E), P1 800 psia h1 1456.0 Btu/lbm T1 900 F s1 1.6413 Btu/lbm R 1 s2 sf 1.6413 0.39213 P2 40 psia x2 0.9725 s fg 1.28448 s2 s1 h2 hf x 2 h fg 236.14 0.9725 933.69 1144.2 Btu/lbm H2OThere is only one inlet and one exit, and thus m1 m2 m . We take theturbine as the system, which is a control volume since mass crosses theboundary. The energy balance for this steady-flow system can beexpressed in the rate form as 0 (steady) 2 Ein Eout Esystem 0 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies Ein Eout mh1 Wout mh2 Wout m(h1 h2 )Dividing by mass flow rate and substituting, wout h1 h2 1456.0 1144.2 311.8 Btu/lbm
18. 18. 7-187-43E EES Problem 7-42E is reconsidered. The work done by the steam is to be calculated and plotted as afunction of final pressure as the pressure varies from 800 psia to 40 psia. Also the effect of varying theturbine inlet temperature from the saturation temperature at 800 psia to 900°F on the turbine work is to beinvestigated.Analysis The problem is solved using EES, and the results are tabulated and plotted below."Knowns:"P_1 = 800 [psia]T_1 = 900 [F]P_2 = 40 [psia]T_sat_P_1= temperature(Fluid\$,P=P_1,x=1.0)Fluid\$=Steam_IAPWS"Analysis: "" Treat theturbine as a steady-flow control volume, with no heat transfer in, neglectchanges in KE and PE of the Steam.""The isentropic work is determined from the steady-flow energy equation written per unit mass:"e_in - e_out = DELTAe_sysE_out = Work_out+h_2 "[Btu/lbm]" 350e_in = h_1 "[Btu/lbm]" 300 Isentropic ProcessDELTAe_sys = 0 "[Btu/lbm]" ] P1 = 800 psiah_1 = enthalpy(Fluid\$,P=P_1,T=T_1) m 250s_1 = entropy(Fluid\$,P=P_1,T=T_1) b l T1 = 900 F / 200 u t"The process is reversible B [ 150and adiabatic or isentropic. t u o 100Then P_2 and s_2 specify state 2." k rs_2 = s_1 "[Btu/lbm-R]" o 50h_2 = enthalpy(Fluid\$,P=P_2,s=s_2) WT_2_isen=temperature(Fluid\$,P=P_2,s=s_2) 0 0 100 200 300 400 500 600 700 800 P2 [psia] T1 Workout [F] [Btu/lbm] 320 520 219.3 Isentropic Process 560 229.6 600 239.1 298 P1 = 800 psia ] 650 250.7 m P2 = 40 psia b l 276 690 260 / u t 730 269.4 B [ 770 279 t 254 820 291.3 u o 860 301.5 k r 900 311.9 o 232 W 210 500 550 600 650 700 750 800 850 900 T1 [F]
19. 19. 7-197-44 An insulated cylinder is initially filled with superheated steam at a specified state. The steam iscompressed in a reversible manner until the pressure drops to a specified value. The work input during thisprocess is to be determined.Assumptions 1 The kinetic and potential energy changes are negligible. 2 The cylinder is well-insulatedand thus heat transfer is negligible. 3 The thermal energy stored in the cylinder itself is negligible. 4 Theprocess is stated to be reversible.Analysis This is a reversible adiabatic (i.e., isentropic) process, and thus s2 = s1. From the steam tables(Tables A-4 through A-6), v1 0.63402 m 3 /kg P1 300 kPa u1 2571.0 kJ/kg T1 150 C s1 7.0792 kJ/kg K P2 1 MPa u2 2773.8 kJ/kg s2 s1 H2OAlso, 300 kPa 150 C V 0.05 m 3 m 0.0789 kg v1 0.63402 m 3 /kgWe take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves.The energy balance for this adiabatic closed system can be expressed as Ein Eout Esystem Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies Wb,in U m(u2 u1 )Substituting, the work input during this adiabatic process is determined to be Wb,in m u2 u1 0.0789 kg 2773.8 2571.0 kJ/kg 16.0 kJ
20. 20. 7-207-45 EES Problem 7-44 is reconsidered. The work done on the steam is to be determined and plotted as afunction of final pressure as the pressure varies from 300 kPa to 1 MPa.Analysis The problem is solved using EES, and the results are tabulated and plotted below."Knowns:"P_1 = 300 [kPa]T_1 = 150 [C]V_sys = 0.05 [m^3]"P_2 = 1000 [kPa]""Analysis: "Fluid\$=Steam_IAPWS" Treat the piston-cylinder as a closed system, with no heat transfer in, neglectchanges in KE and PE of the Steam. The process is reversible and adiabatic thus isentropic.""The isentropic work is determined from:"E_in - E_out = DELTAE_sysE_out = 0 [kJ]E_in = Work_inDELTAE_sys = m_sys*(u_2 - u_1)u_1 = INTENERGY(Fluid\$,P=P_1,T=T_1)v_1 = volume(Fluid\$,P=P_1,T=T_1)s_1 = entropy(Fluid\$,P=P_1,T=T_1)V_sys = m_sys*v_1" The process is reversible and adiabatic or isentropic.Then P_2 and s_2 specify state 2."s_2 = s_1u_2 = INTENERGY(Fluid\$,P=P_2,s=s_2)T_2_isen = temperature(Fluid\$,P=P_2,s=s_2) P2 Workin 16 [kPa] [kJ] W ork on Steam 14 300 0 P = 300 kPa 400 3.411 12 1 500 6.224 T = 150 C 600 8.638 10 1 W ork in [kJ] 700 10.76 800 12.67 8 900 14.4 6 1000 16 4 2 0 300 400 500 600 700 800 900 1000 P [kPa] 2
21. 21. 7-217-46 A cylinder is initially filled with saturated water vapor at a specified temperature. Heat is transferredto the steam, and it expands in a reversible and isothermal manner until the pressure drops to a specifiedvalue. The heat transfer and the work output for this process are to be determined.Assumptions 1 The kinetic and potential energy changes are negligible. 2 The cylinder is well-insulatedand thus heat transfer is negligible. 3 The thermal energy stored in the cylinder itself is negligible. 4 Theprocess is stated to be reversible and isothermal.Analysis From the steam tables (Tables A-4 through A-6), T1 200 C u1 u g @ 200 C 2594.2 kJ/kg sat.vapor s1 s g @ 200 C 6.4302 kJ/kg K H2O 200 C P2 800 kPa u2 2631.1 kJ/kg sat. vapor T = const T2 T1 s2 6.8177 kJ/kg K QThe heat transfer for this reversible isothermal process can be determined from Q T S Tm s 2 s1 (473 K )(1.2 kg )(6.8177 6.4302)kJ/kg K 219.9 kJ We take the contents of the cylinder as the system. This is a closed system since no mass enters orleaves. The energy balance for this closed system can be expressed as Ein Eout Esystem Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies Qin Wb,out U m(u2 u1 ) Wb,out Qin m(u2 u1 )Substituting, the work done during this process is determined to be Wb, out 219.9 kJ (1.2 kg )(2631.1 2594.2) kJ/kg 175.6 kJ
22. 22. 7-227-47 EES Problem 7-46 is reconsidered. The heat transferred to the steam and the work done are to bedetermined and plotted as a function of final pressure as the pressure varies from the initial value to thefinal value of 800 kPa.Analysis The problem is solved using EES, and the results are tabulated and plotted below."Knowns:"T_1 = 200 [C]x_1 = 1.0m_sys = 1.2 [kg]{P_2 = 800"[kPa]"}"Analysis: "Fluid\$=Steam_IAPWS" Treat the piston-cylinder as a closed system, neglect changes in KE and PE of the Steam. Theprocess is reversible and isothermal ."T_2 = T_1E_in - E_out = DELTAE_sys 200E_in = Q_inE_out = Work_out 160DELTAE_sys = m_sys*(u_2 - u_1)P_1 = pressure(Fluid\$,T=T_1,x=1.0) ] Ju_1 = INTENERGY(Fluid\$,T=T_1,x=1.0) K 120 [v_1 = volume(Fluid\$,T=T_1,x=1.0) t us_1 = entropy(Fluid\$,T=T_1,x=1.0) o 80V_sys = m_sys*v_1 k r o W 40" The process is reversible and isothermal.Then P_2 and T_2 specify state 2."u_2 = INTENERGY(Fluid\$,P=P_2,T=T_2) 0s_2 = entropy(Fluid\$,P=P_2,T=T_2) 800 1000 1200 1400 1600Q_in= (T_1+273)*m_sys*(s_2-s_1) P2 [kPa] P2 Qin Workout [kPa] [kJ] [kJ] 200 800 219.9 175.7 900 183.7 144.7 160 1000 150.6 117 1100 120 91.84 ] 120 1200 91.23 68.85 J k [ 1300 64.08 47.65 80 n i 1400 38.2 27.98 Q 1500 13.32 9.605 1553 219.9 175.7 40 0 800 1000 1200 1400 1600 P2 [kPa]
23. 23. 7-237-48 A cylinder is initially filled with saturated water vapor mixture at a specified temperature. Steamundergoes a reversible heat addition and an isentropic process. The processes are to be sketched and heattransfer for the first process and work done during the second process are to be determined.Assumptions 1 The kinetic and potential energy changes are negligible. 2 The thermal energy stored in thecylinder itself is negligible. 3 Both processes are reversible.Analysis (b) From the steam tables (Tables A-4 through A-6), T1 100 C h1 hf xh fg 419.17 (0.5)(2256.4) 1547.4 kJ/kg x 0.5 H2O h2 hg 2675.6 kJ/kg 100 C T2 100 C x=0.5 u 2 u g 2506.0 kJ/kg x2 1 s 2 7.3542 kJ/kg K Q P3 15 kPa u3 2247.9 kJ/kg s3 s2We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves.The energy balance for this closed system can be expressed as Ein Eout Esystem Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies Qin Wb,out U m(u2 u1 )For process 1-2, it reduces to Q12,in m(h2 h1 ) (5 kg)(2675.6 - 1547.4)kJ/kg 5641 kJ(c) For process 2-3, it reduces to W23, b,out m(u2 u3 ) (5 kg)(2506.0 - 2247.9)kJ/kg 1291 kJ SteamIAPWS 700 600 500 400 ] C ° [ T 300 101.42 kPa 200 15 kPa 1 2 100 3 0 0.0 1.1 2.2 3.3 4.4 5.5 6.6 7.7 8.8 9.9 11.0 s [kJ/kg-K]
24. 24. 7-247-49 A rigid tank contains saturated water vapor at a specified temperature. Steam is cooled to ambienttemperature. The process is to be sketched and entropy changes for the steam and for the process are to bedetermined.Assumptions 1 The kinetic and potential energy changes are negligible.Analysis (b) From the steam tables (Tables A-4 through A-6), v1 vg 1.6720 kJ/kg T1 100 C u1 u g 2506.0 kJ/kg H2O x 1 s1 7.3542 kJ/kg K 100 C x=1 x2 0.0386 T2 25 C Q u2 193.78 kJ/kg v2 v1 s2 1.0715 kJ/kg KThe entropy change of steam is determined from Sw m( s 2 s1 ) (5 kg)(1.0715 - 7.3542)kJ/kg K -31.41 kJ/K(c) We take the contents of the tank as the system. This is a closed system since no mass enters or leaves.The energy balance for this closed system can be expressed as Ein Eout Esystem Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies Qout U m(u2 u1 )That is, Qout m(u1 u2 ) (5 kg)(2506.0 - 193.78)kJ/kg 11,511 kJThe total entropy change for the process is Qout 11,511 kJ Sgen Sw 31.41 kJ/K 7.39 kJ/K Tsurr 298 K SteamIAPWS 700 600 500 400 ] C ° [ 300 T 200 1 100 101.4 kPa 3.17 kPa 2 0 10-3 10-2 10-1 100 101 102 103 3 v [m /kg]
25. 25. 7-257-50 Steam expands in an adiabatic turbine. Steam leaves the turbine at two different pressures. Theprocess is to be sketched on a T-s diagram and the work done by the steam per unit mass of the steam at theinlet are to be determined.Assumptions 1 The kinetic and potential energy changes are negligible. P1 = 6 MPaAnalysis (b) From the steam tables (Tables A-4 through A-6), T1 = 500 CT1 500 C h1 3423.1 kJ/kg P2 1 MPa h2 s 2921.3 kJ/kg P 6 MPa s1 6.8826 kJ/kg K s2 s1 1 TurbineP3 10 kPa h3s 2179.6 kJ/kgs3 s1 x3s 0.831A mass balance on the control volume gives P2 = 1 MPa m 2 0.1m1m1 m 2 m3 where P3 = 10 kPa m3 0.9m1 SteamIAPWSWe take the turbine as the system, which is a 700control volume. The energy balance for this 600steady-flow system can be expressed in the 1rate form as 500 Ein Eout 400 ] C ° m1h1 Ws ,out m2 h2 m3h3 [ T 300 6000 kPa 2 m1h1 Ws ,out 0.1m1h2 0.9m1h3 200 1000 kPaor 100 10 kPa 3 h1 ws ,out 0.1h2 0.9h3 0 0.0 1.1 2.2 3.3 4.4 5.5 6.6 7.7 8.8 9.9 11.0 ws ,out h1 0.1h2 0.9h3 s [kJ/kg-K] 3423.1 (0.1)(2921.3) (0.9)(2179.6) 1169.3 kJ/kgThe actual work output per unit mass of steam at the inlet is wout T ws , out (0.85)(1169.3 kJ/kg ) 993.9 kJ/kg7-51E An insulated rigid can initially contains R-134a at a specified state. A crack develops, andrefrigerant escapes slowly. The final mass in the can is to be determined when the pressure inside drops toa specified value. Assumptions 1 The can is well-insulated and thus heat transfer is negligible. 2 The refrigerant thatremains in the can underwent a reversible adiabatic process.Analysis Noting that for a reversible adiabatic (i.e., isentropic)process, s1 = s2, the properties of the refrigerant in the can are(Tables A-11E through A-13E) R-134 P1 140 psia s1 s f @70 F 0.07306 Btu/lbm R 140 psiaT1 70 F 70 F Leak s2 sf 0.07306 0.02605P2 20 psia x2 0.2355 s fg 0.19962s2 s1 v2 vf x 2v fg 0.01182 0.2355 2.2772 0.01182 0.5453 ft 3 /lbmThus the final mass of the refrigerant in the can is V 1.2 ft 3 m 2.201 lbm v2 0.5453 ft 3 /lbm
26. 26. 7-26Entropy Change of Incompressible Substances7-52C No, because entropy is not a conserved property.7-53 A hot copper block is dropped into water in an insulated tank. The final equilibrium temperature ofthe tank and the total entropy change are to be determined.Assumptions 1 Both the water and the copper block are incompressible substances with constant specificheats at room temperature. 2 The system is stationary and thus the kinetic and potential energies arenegligible. 3 The tank is well-insulated and thus there is no heat transfer.Properties The density and specific heat of water at 25 C are = 997 kg/m3 and cp = 4.18 kJ/kg. C. Thespecific heat of copper at 27 C is cp = 0.386 kJ/kg. C (Table A-3).Analysis We take the entire contents of the tank, water + copper block, as the system. This is a closedsystem since no mass crosses the system boundary during the process. The energy balance for this systemcan be expressed as Ein Eout Esystem WATER Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies 0 U Copper 50 kgor, U Cu U water 0 120 L [mc(T2 T1 )]Cu [mc(T2 T1 )]water 0where mwater V (997 kg/m3 )(0.120 m3 ) 119.6 kgUsing specific heat values for copper and liquid water at room temperature and substituting, (50 kg)(0.386 kJ/kg C)(T2 80) C (119.6 kg)(4.18 kJ/kg C)(T2 25) C 0 T2 = 27.0 CThe entropy generated during this process is determined from T2 300.0 K Scopper mcavg ln 50 kg 0.386 kJ/kg K ln 3.140 kJ/K T1 353 K T2 300.0 K S water mcavg ln 119.6 kg 4.18 kJ/kg K ln 3.344 kJ/K T1 298 KThus, S total Scopper S water 3.140 3.344 0.204 kJ/K
27. 27. 7-277-54 A hot iron block is dropped into water in an insulated tank. The total entropy change during thisprocess is to be determined.Assumptions 1 Both the water and the iron block are incompressible substances with constant specificheats at room temperature. 2 The system is stationary and thus the kinetic and potential energies arenegligible. 3 The tank is well-insulated and thus there is no heat transfer. 4 The water that evaporates,condenses back.Properties The specific heat of water at 25 C is cp = 4.18 kJ/kg. C. The specific heat of iron at roomtemperature is cp = 0.45 kJ/kg. C (Table A-3).Analysis We take the entire contents of the tank, water + iron block, as the system. This is a closed systemsince no mass crosses the system boundary during the process. The energy balance for this system can beexpressed as Ein Eout Esystem Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies WATER 0 U 18 Cor, Iron 350 C U iron U water 0 [mc(T2 T1 )]iron [mc(T2 T1 )]water 0Substituting, (25 kg )(0.45 kJ/kg K )(T2 350 C) (100 kg )(4.18 kJ/kg K )(T2 18 C) 0 T2 26.7 CThe entropy generated during this process is determined from T2 299.7 K Siron mcavg ln 25 kg 0.45 kJ/kg K ln 8.232 kJ/K T1 623 K T2 299.7 K S water mcavg ln 100 kg 4.18 kJ/kg K ln 12.314 kJ/K T1 291 KThus, Sgen S total Siron S water 8.232 12.314 4.08 kJ/KDiscussion The results can be improved somewhat by using specific heats at average temperature.
28. 28. 7-287-55 An aluminum block is brought into contact with an iron block in an insulated enclosure. The finalequilibrium temperature and the total entropy change for this process are to be determined.Assumptions 1 Both the aluminum and the iron block are incompressible substances with constant specificheats. 2 The system is stationary and thus the kinetic and potential energies are negligible. 3 The system iswell-insulated and thus there is no heat transfer.Properties The specific heat of aluminum at the anticipated average temperature of 450 K is cp = 0.973kJ/kg. C. The specific heat of iron at room temperature (the only value available in the tables) is cp = 0.45kJ/kg. C (Table A-3).Analysis We take the iron+aluminum blocks as the system, which is a closed system. The energy balancefor this system can be expressed as Ein Eout Esystem Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies Iron Aluminum 20 kg 20 kg 0 U 100 C 200 Cor, U alum U iron 0 [mc(T2 T1 )]alum [mc(T2 T1 )]iron 0Substituting, (20 kg )(0.45 kJ/kg K )(T2 100 C) (20 kg)(0.973 kJ/kg K )(T2 200 C) 0 T2 168.4 C 441.4 KThe total entropy change for this process is determined from T2 441.4 K Siron mcavg ln 20 kg 0.45 kJ/kg K ln 1.515 kJ/K T1 373 K T2 441.4 K Salum mcavg ln 20 kg 0.973 kJ/kg K ln 1.346 kJ/K T1 473 KThus, S total Siron Salum 1.515 1.346 0.169 kJ/K
29. 29. 7-297-56 EES Problem 7-55 is reconsidered. The effect of the mass of the iron block on the final equilibriumtemperature and the total entropy change for the process is to be studied. The mass of the iron is to varyfrom 1 to 10 kg. The equilibrium temperature and the total entropy change are to be plotted as a functionof iron mass.Analysis The problem is solved using EES, and the results are tabulated and plotted below."Knowns:"T_1_iron = 100 [C]{m_iron = 20 [kg]}T_1_al = 200 [C]m_al = 20 [kg] 198C_al = 0.973 [kJ/kg-K] "FromTable A-3 196at the anticipated average temperature of 194450 K."C_iron= 0.45 [kJ/kg-K] "FromTable A-3 192at room temperature, the only value 190available." 188 T2"Analysis: " 186" Treat the iron plus aluminum as a 184closed system, with no heat transfer in,no work out, neglect changes in KE and 182PE of the system. " 180"The final temperature is found from the 1 2 3 4 5 6 7 8 9 10energy balance." m iron [kg]E_in - E_out = DELTAE_sysE_out = 0E_in = 0DELTAE_sys = m_iron*DELTAu_iron + m_al*DELTAu_alDELTAu_iron = C_iron*(T_2_iron - T_1_iron)DELTAu_al = C_al*(T_2_al - T_1_al)"the iron and aluminum reach thermal equilibrium:"T_2_iron = T_2T_2_al = T_2DELTAS_iron = m_iron*C_iron*ln((T_2_iron+273) / (T_1_iron+273))DELTAS_al = m_al*C_al*ln((T_2_al+273) / (T_1_al+273))DELTAS_total = DELTAS_iron + DELTAS_al 0.1 Stotal miron T2 [kJ/kg] [kg] [C] 0.090.01152 1 197.7 0.080.0226 2 195.6 0.070.03326 3 193.5 S total [kJ/kg] 0.060.04353 4 191.50.05344 5 189.6 0.050.06299 6 187.8 0.040.07221 7 186.1 0.030.08112 8 184.40.08973 9 182.8 0.020.09805 10 181.2 0.01 1 2 3 4 5 6 7 8 9 10 m [kg] iron
30. 30. 7-307-57 An iron block and a copper block are dropped into a large lake. The total amount of entropy changewhen both blocks cool to the lake temperature is to be determined.Assumptions 1 Both the water and the iron block are incompressible substances with constant specificheats at room temperature. 2 Kinetic and potential energies are negligible.Properties The specific heats of iron and copper at room temperature are ciron = 0.45 kJ/kg. C and ccopper =0.386 kJ/kg. C (Table A-3).Analysis The thermal-energy capacity of the lake is very large, and thus the temperatures of both the ironand the copper blocks will drop to the lake temperature (15 C) when the thermal equilibrium is established.Then the entropy changes of the blocks become T2 288 K Siron mcavg ln 50 kg 0.45 kJ/kg K ln 4.579 kJ/K T1 353 K T2 288 K Scopper mcavg ln 20 kg 0.386 kJ/kg K ln 1.571 kJ/K T1 353 KWe take both the iron and the copper blocks, as thesystem. This is a closed system since no mass crossesthe system boundary during the process. The energybalance for this system can be expressed as Iron Ein Eout Esystem 50 kg Net energy transfer 80 C Change in internal, kinetic, by heat, work, and mass potential, etc. energies Copper Qout U U iron U copper Lake 20 kg 15 C 80 Cor, Qout [mc(T1 T2 )]iron [mc(T1 T2 )]copperSubstituting, Qout 50 kg 0.45 kJ/kg K 353 288 K 20 kg 0.386 kJ/kg K 353 288 K 1964 kJThus, Qlake,in 1964 kJ Slake 6.820 kJ/K Tlake 288 KThen the total entropy change for this process is S total Siron Scopper Slake 4.579 1.571 6.820 0.670 kJ/K
31. 31. 7-317-58 An adiabatic pump is used to compress saturated liquid water in a reversible manner. The work inputis to be determined by different approaches.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3Heat transfer to or from the fluid is negligible.Analysis The properties of water at the inlet and exit of the pump are (Tables A-4 through A-6) h1 191.81 kJ/kg P1 10 kPa s1 0.6492 kJ/kg x1 0 v1 0.001010 m 3 /kg 15 MPa P2 15 MPa h2 206.90 kJ/kg s2 s1 v2 0.001004 m 3 /kg 10 kPa pump(a) Using the entropy data from the compressed liquid water table wP h2 h1 206.90 191.81 15.10 kJ/kg(b) Using inlet specific volume and pressure values wP v 1 ( P2 P1 ) (0.001010 m 3 /kg)(15,000 10)kPa 15.14 kJ/kg Error = 0.3%(b) Using average specific volume and pressure values wP v avg ( P2 P) 1 1 / 2(0.001010 0.001004) m3/kg (15,000 10)kPa 15.10 kJ/kg Error = 0%Discussion The results show that any of the method may be used to calculate reversible pump work.
32. 32. 7-32Entropy Changes of Ideal Gases7-59C For ideal gases, cp = cv + R and P2V 2 PV11 V 2 T2 P 1 T2 T1 V1 T1P2Thus, T V s2 s1 cv ln 2 R ln 2 T1 V1 T2 T2 P1 cv ln R ln T1 T1P2 T2 T2 P2 cv ln R ln R ln T1 T1 P1 T2 P2 c p ln R ln T1 P17-60C For an ideal gas, dh = cp dT and v = RT/P. From the second Tds relation, dh v dP c p dP RT dP dT dP ds cp R T T T P T T PIntegrating, T2 P2 s2 s1 c p ln R ln T1 P1Since cp is assumed to be constant.7-61C No. The entropy of an ideal gas depends on the pressure as well as the temperature.7-62C Setting s = 0 gives R Cp T P T R P T2 P2 c p ln 2 R ln 2 0 ln 2 ln 2 T1 P1 T1 cp P1 T1 P1But k 1 k R cp cv 1 k 1 T2 P2 1 since k c p / cv . Thus, cp cp k k T1 P17-63C The Pr and vr are called relative pressure and relative specific volume, respectively. They arederived for isentropic processes of ideal gases, and thus their use is limited to isentropic processes only.7-64C The entropy of a gas can change during an isothermal process since entropy of an ideal gas dependson the pressure as well as the temperature.7-65C The entropy change relations of an ideal gas simplify to s = cp ln(T2/T1) for a constant pressure processand s = cv ln(T2/T1) for a constant volume process.Noting that cp > cv, the entropy change will be larger for a constant pressure process.
33. 33. 7-337-66 Oxygen gas is compressed from a specified initial state to a specified final state. The entropy changeof oxygen during this process is to be determined for the case of constant specific heats.Assumptions At specified conditions, oxygen can be treated as an ideal gas.Properties The gas constant and molar mass of oxygen are R = 0.2598 kJ/kg.K and M = 32 kg/kmol (TableA-1).Analysis The constant volume specific heat of oxygen at the average temperature is (Table A-2) 298 560 Tavg 429 K cv ,avg 0.690 kJ/kg K 2Thus, T2V2 s2 s1 cv ,avg ln R ln O2 T1V1 0.8 m3/kg 560 K 0.1 m3/kg 25 C 0.690 kJ/kg K ln 0.2598 kJ/kg K ln 298 K 0.8 m3/kg 0.105 kJ/kg K7-67 An insulated tank contains CO2 gas at a specified pressure and volume. A paddle-wheel in the tankstirs the gas, and the pressure and temperature of CO2 rises. The entropy change of CO2 during this processis to be determined using constant specific heats.Assumptions At specified conditions, CO2 can be treated as an ideal gas with constant specific heats atroom temperature.Properties The specific heat of CO2 is cv = 0.657 kJ/kg.K (Table A-2).Analysis Using the ideal gas relation, the entropy change is determined to be CO2 P2V PV 1 T2 P2 150 kPa 1.5 m3 1.5 100 kPa T2 T1 T1 P1 100 kPa 2.7 kgThus, T2 V 0 T2 S m s2 s1 R ln 2 m cv ,avg ln mcv ,avg ln T1 V1 T1 2.7 kg 0.657 kJ/kg K ln 1.5 0.719 kJ/K
34. 34. 7-347-68 An insulated cylinder initially contains air at a specified state. A resistance heater inside the cylinderis turned on, and air is heated for 15 min at constant pressure. The entropy change of air during this processis to be determined for the cases of constant and variable specific heats.Assumptions At specified conditions, air can be treated as an ideal gas.Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1).Analysis The mass of the air and the electrical work done during this process are PV1 1 120 kPa 0.3 m3 m 0.4325 kg RT1 0.287 kPa m3/kg K 290 K We,in We,in t 0.2 kJ/s 15 60 s 180 kJ AIRThe energy balance for this stationary closed system can be expressed as 0.3 m3 120 kPa Ein Eout Esystem 17 C Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies We We,in Wb,out U We,in m(h2 h1 ) c p (T2 T1 )since U + Wb = H during a constant pressure quasi-equilibrium process.(a) Using a constant cp value at the anticipated average temperature of 450 K, the final temperaturebecomes We,in 180 kJThus, T2 T1 290 K 698 K mc p 0.4325 kg 1.02 kJ/kg KThen the entropy change becomes 0 T2 P2 T2 Ssys m s2 s1 m c p ,avg ln R ln mc p ,avg ln T1 P1 T1 698 K 0.4325 kg 1.020 kJ/kg K ln 0.387 kJ/K 290 K(b) Assuming variable specific heats, We,in 180 kJ We,in m h2 h1 h2 h1 290.16 kJ/kg 706.34 kJ/kg m 0.4325 kgFrom the air table (Table A-17, we read s2 = 2.5628 kJ/kg·K corresponding to this h2 value. Then, 0 P2 Ssys m s2 s1 R ln m s2 s1 0.4325 kg 2.5628 1.66802 kJ/kg K 0.387 kJ/K P17-69 A cylinder contains N2 gas at a specified pressure and temperature. It is compressed polytropicallyuntil the volume is reduced by half. The entropy change of nitrogen during this process is to be determined.Assumptions 1 At specified conditions, N2 can be treated as an ideal gas. 2 Nitrogen has constant specificheats at room temperature.Properties The gas constant of nitrogen is R = 0.297 kJ/kg.K (Table A-1). The constant volume specificheat of nitrogen at room temperature is cv = 0.743 kJ/kg.K (Table A-2).Analysis From the polytropic relation, n 1 n 1 T2 v1 v T2 T1 1 300 K 2 1.3 1 369.3 K T1 v2 v2 N2Then the entropy change of nitrogen becomes T V PV1.3 = C S N 2 m cv ,avg ln 2 R ln 2 T1 V1 369.3 K 1.2 kg 0.743 kJ/kg K ln 0.297 kJ/kg K ln 0.5 0.0617 kJ/K 300 K
35. 35. 7-357-70 EES Problem 7-69 is reconsidered. The effect of varying the polytropic exponent from 1 to 1.4 on theentropy change of the nitrogen is to be investigated, and the processes are to be shown on a common P-vdiagram.Analysis The problem is solved using EES, and the results are tabulated and plotted below.Function BoundWork(P[1],V[1],P[2],V[2],n)"This function returns the Boundary Work for the polytropic process. This function is requiredsince the expression for boundary work depens on whether n=1 or n<>1" If n<>1 then BoundWork:=(P[2]*V[2]-P[1]*V[1])/(1-n)"Use Equation 3-22 when n=1" else BoundWork:= P[1]*V[1]*ln(V[2]/V[1]) "Use Equation 3-20 when n=1" endifendn=1P[1] = 120 [kPa]T[1] = 27 [C]m = 1.2 [kg]V[2]=V[1]/2Gas\$=N2MM=molarmass(Gas\$)R=R_u/MMR_u=8.314 [kJ/kmol-K]"System: The gas enclosed in the piston-cylinder device.""Process: Polytropic expansion or compression, P*V^n = C"P[1]*V[1]=m*R*(T[1]+273)P[2]*V[2]^n=P[1]*V[1]^nW_b = BoundWork(P[1],V[1],P[2],V[2],n)"Find the temperature at state 2 from the pressure and specific volume."T[2]=temperature(gas\$,P=P[2],v=V[2]/m)"The entropy at states 1 and 2 is:"s[1]=entropy(gas\$,P=P[1],v=V[1]/m)s[2]=entropy(gas\$,P=P[2],v=V[2]/m)DELTAS=m*(s[2] - s[1])"Remove the {} to generate the P-v plot data"{Nsteps = 10VP[1]=V[1]PP[1]=P[1]Duplicate i=2,Nsteps VP[i]=V[1]-i*(V[1]-V[2])/Nsteps PP[i]=P[1]*(V[1]/VP[i])^nEND } S [kJ/kg] n Wb [kJ] -0.2469 1 -74.06 -0.2159 1.05 -75.36 -0.1849 1.1 -76.69 -0.1539 1.15 -78.05 -0.1229 1.2 -79.44 -0.09191 1.25 -80.86 -0.06095 1.3 -82.32 -0.02999 1.35 -83.82 0.0009849 1.4 -85.34
36. 36. 7-36 350 300 n=1 = 1.4 250 PP[i] 200 150 100 0.4 0.5 0.6 0.7 0.8 0.9 1 VP[i] 0.05 0 S = 0 kJ/k -0.05 S [kJ/K] -0.1 -0.15 -0.2 -0.25 1 1.05 1.1 1.15 1.2 1.25 1.3 1.35 1.4 n -72.5 -75.5 -78.5W b [kJ] -81.5 -84.5 -87.5 1 1.05 1.1 1.15 1.2 1.25 1.3 1.35 1.4 n
37. 37. 7-377-71E A fixed mass of helium undergoes a process from one specified state to another specified state. Theentropy change of helium is to be determined for the cases of reversible and irreversible processes.Assumptions 1 At specified conditions, helium can be treated as an ideal gas. 2 Helium has constantspecific heats at room temperature.Properties The gas constant of helium is R = 0.4961 Btu/lbm.R (Table A-1E). The constant volumespecific heat of helium is cv = 0.753 Btu/lbm.R (Table A-2E).Analysis From the ideal-gas entropy change relation, T2 v2 S He m cv ,ave ln R ln T1 v1 He 660 R 10 ft 3/lbm T1 = 540 R (15 lbm) (0.753 Btu/lbm R) ln 0.4961 Btu/lbm R ln T2 = 660 R 540 R 50 ft 3/lbm 9.71 Btu/RThe entropy change will be the same for both cases.7-72 Air is compressed in a piston-cylinder device in a reversible and isothermal manner. The entropychange of air and the work done are to be determined.Assumptions 1 At specified conditions, air can be treated as an ideal gas. 2 The process is specified to bereversible.Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1).Analysis (a) Noting that the temperature remains constant, the entropy change of air is determined from 0 T2 P2 P2 Sair c p ,avg ln R ln R ln T1 P1 P1 400 kPa 0.287 kJ/kg K ln 0.428 kJ/kg K 90 kPa AIR QAlso, for a reversible isothermal process, T = const q T s 293 K 0.428 kJ/kg K 125.4 kJ/kg qout 125.4 kJ/kg(b) The work done during this process is determined from the closed system energy balance, Ein Eout Esystem Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies Win Qout U mcv (T2 T1 ) 0 win qout 125.4 kJ/kg
38. 38. 7-387-73 Air is compressed steadily by a 5-kW compressor from one specified state to another specified state.The rate of entropy change of air is to be determined.Assumptions At specified conditions, air can be treated as an ideal gas. 2 Air has variable specific heats.Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1).Analysis From the air table (Table A-17), P2 = 600 kPa T2 = 440 K T1 290 K s1 1.66802 kJ/kg K P1 100 kPa AIR T2 440 K COMPRESSOR s2 2.0887 kJ/kg K 5 kW P2 600 kPaThen the rate of entropy change of air becomes P2 P1 = 100 kPa Ssys m s2 s1 R ln P1 T1 = 290 K 600 kPa 1.6/60 kg/s 2.0887 1.66802 0.287 kJ/kg K ln 100 kPa 0.00250 kW/K7-74 One side of a partitioned insulated rigid tank contains an ideal gas at a specified temperature andpressure while the other side is evacuated. The partition is removed, and the gas fills the entire tank. Thetotal entropy change during this process is to be determined.Assumptions The gas in the tank is given to be an ideal gas, and thus ideal gas relations apply.Analysis Taking the entire rigid tank as the system, the energy balance can be expressed as Ein Eout Esystem Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies IDEAL 0 U m(u2 u1 ) GAS 5 kmol u2 u1 40 C T2 T1since u = u(T) for an ideal gas. Then the entropy change of the gas becomes T2 0 V V S N cv ,avg ln Ru ln 2 NRu ln 2 T1 V1 V1 5 kmol 8.314 kJ/kmol K ln 2 28.81 kJ/KThis also represents the total entropy change since the tank does not contain anything else, and there areno interactions with the surroundings.