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# Thermo solutions chapter07

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## Thermo solutions chapter07Document Transcript

• 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.
• 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
• 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
• 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]
• 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
• 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
• 7-77-32 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
• 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]
• 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
• 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
• 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]
• 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
• 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
• 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)
• 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]
• 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
• 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
• 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]
• 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
• 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
• 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
• 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]
• 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]
• 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]
• 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
• 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
• 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.
• 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
• 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
• 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
• 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.
• 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.
• 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
• 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
• 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
• 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
• 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
• 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.
• 7-397-75 Air is compressed in a piston-cylinder device in a reversible and adiabatic manner. The finaltemperature and the work are to be determined for the cases of constant and variable specific heats.Assumptions 1 At specified conditions, air can be treated as an ideal gas. 2 The process is given to bereversible and adiabatic, and thus isentropic. Therefore, isentropic relations of ideal gases apply.Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1). The specific heat ratio of air at lowto moderately high temperatures is k = 1.4 (Table A-2).Analysis (a) Assuming constant specific heats, the ideal gas isentropic relations give k 1 k 0.4 1.4 P 800 kPa T2 T1 2 290 K 525.3 K P1 100 kPaThen, Tavg 290 525.3 /2 407.7 K cv , avg 0.727 kJ/kg KWe take the air in the cylinder as the system. The energy balance for AIRthis stationary closed system can be expressed as Reversible Ein Eout Esystem Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies Win U m(u2 u1 ) mcv (T2 T1 )Thus, win cv ,avg T2 T1 0.727 kJ/kg K 525.3 290 K 171.1 kJ/kg(b) Assuming variable specific heats, the final temperature can be determined using the relative pressuredata (Table A-17), Pr1 1.2311 T1 290 K u1 206.91 kJ/kgand P2 800 kPa T2 522.4 K Pr2 Pr 1.2311 9.849 P 1 1 100 kPa u2 376.16 kJ/kgThen the work input becomes win u2 u1 376.16 206.91 kJ/kg 169.25 kJ/kg
• 7-407-76 EES Problem 7-75 is reconsidered. The work done and final temperature during the compressionprocess are to be calculated and plotted as functions of the final pressure for the two cases as the finalpressure varies from 100 kPa to 800 kPa.Analysis The problem is solved using EES, and the results are tabulated and plotted below.Procedure ConstPropSol(P_1,T_1,P_2,Gas\$:Work_in_ConstProp,T2_ConstProp)C_P=SPECHEAT(Gas\$,T=27)MM=MOLARMASS(Gas\$)R_u=8.314 [kJ/kmol-K]R=R_u/MMC_V = C_P - Rk = C_P/C_VT2= (T_1+273)*(P_2/P_1)^((k-1)/k)T2_ConstProp=T2-273 "[C]"DELTAu = C_v*(T2-(T_1+273))Work_in_ConstProp = DELTAuEnd"Knowns:"P_1 = 100 [kPa]T_1 = 17 [C]P_2 = 800 [kPa]"Analysis: "" Treat the piston-cylinder as a closed system, with no heat transfer in, neglectchanges in KE and PE of the air. The process is reversible and adiabatic thus isentropic.""The isentropic work is determined from:"e_in - e_out = DELTAe_syse_out = 0 [kJ/kg]e_in = Work_inDELTAE_sys = (u_2 - u_1)u_1 = INTENERGY(air,T=T_1)v_1 = volume(air,P=P_1,T=T_1)s_1 = entropy(air,P=P_1,T=T_1)" The process is reversible andadiabatic or isentropic.Then P_2 and s_2 specify state 2."s_2 = s_1u_2 = INTENERGY(air,P=P_2,s=s_2)T_2_isen=temperature(air,P=P_2,s=s_2) 180Gas\$ = airCall ConstPropSol(P_1,T_1,P_2,Gas\$: 160 Work_in_ConstProp,T2_ConstProp) 140 ] g 120 P2 Workin Workin,ConstProp k / [kPa] [kJ/kg] [kJ/kg] J 100 k [ 100 0 0 80 n i 200 45.63 45.6 k 60 300 76.84 76.77 r o 40 400 101.3 101.2 W 500 121.7 121.5 20 600 139.4 139.1 0 700 155.2 154.8 100 200 300 400 500 600 700 800 800 169.3 168.9 P2 [kPa]
• 7-417-77 Helium gas is compressed in a piston-cylinder device in a reversible and adiabatic manner. The finaltemperature and the work are to be determined for the cases of the process taking place in a piston-cylinderdevice and a steady-flow compressor.Assumptions 1 Helium is an ideal gas with constant specific heats. 2 The process is given to be reversibleand adiabatic, and thus isentropic. Therefore, isentropic relations of ideal gases apply.Properties The specific heats and the specific heat ratio of helium are cv = 3.1156 kJ/kg.K, cp = 5.1926kJ/kg.K, and k = 1.667 (Table A-2).Analysis (a) From the ideal gas isentropic relations, k 1 k 0.667 1.667 2 P 450 kPaT2 T1 2 303 K 576.9 K P1 90 kPa(a) We take the air in the cylinder as the system. The He He Rev.energy balance for this stationary closed system can beexpressed as Rev. Ein Eout Esystem Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies 1 Win U m(u2 u1 ) mcv (T2 T1 )Thus, win cv T2 T1 (3.1156 kJ/kg K )(576.9 303)K 853.4 kJ/kg(b) If the process takes place in a steady-flow device, the final temperature will remain the same but thework done should be determined from an energy balance on this steady-flow device, 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 Win m(h2 h1 ) mc p (T2 T1 )Thus, win c p T2 T1 (5.1926 kJ/kg K )(576.9 303)K 1422.3 kJ/kg7-78 An insulated rigid tank contains argon gas at a specified pressure and temperature. A valve is opened,and argon escapes until the pressure drops to a specified value. The final mass in the tank is to bedetermined.Assumptions 1 At specified conditions, argon can be treated as an ideal gas. 2 The process is given to bereversible and adiabatic, and thus isentropic. Therefore, isentropic relations of ideal gases apply.Properties The specific heat ratio of argon is k = 1.667 (Table A-2).Analysis From the ideal gas isentropic relations, k 1 k 0.667 1.667 P 200 kPa ARGON T2 T1 2 303 K 219.0 K 4 kg P1 450 kPa 450 kPaThe final mass in the tank is determined from the ideal gas relation, 30 C P1V m1 RT1 P2 T1 200 kPa 303 K m2 m1 4 kg 2.46 kg P2V m 2 RT2 P1T2 450 kPa 219 K
• 7-427-79 EES Problem 7-78 is reconsidered. The effect of the final pressure on the final mass in the tank is tobe investigated as the pressure varies from 450 kPa to 150 kPa, and the results are to be plotted.Analysis The problem is solved using EES, and the results are tabulated and plotted below."UNIFORM_FLOW SOLUTION:""Knowns:"C_P = 0.5203"[kJ/kg-K ]"C_V = 0.3122 "[kJ/kg-K ]"R=0.2081 "[kPa-m^3/kg-K]"P_1= 450"[kPa]"T_1 = 30"[C]"m_1 = 4"[kg]"P_2= 150"[kPa]""Analysis:We assume the mass that stays in the tank undergoes an isentropic expansionprocess. This allows us to determine the final temperature of that gas at the finalpressure in the tank by using the isentropic relation:"k = C_P/C_VT_2 = ((T_1+273)*(P_2/P_1)^((k-1)/k)-273)"[C]"V_2 = V_1P_1*V_1=m_1*R*(T_1+273)P_2*V_2=m_2*R*(T_2+273) m2 P2 [kg] [kPa] 2.069 150 2.459 200 2.811 250 3.136 300 3.44 350 3.727 400 4 450 4 3.6 3.2 m 2 [kg] 2.8 2.4 2 150 200 250 300 350 400 450 P [kPa] 2
• 7-437-80E Air is accelerated in an adiabatic nozzle. Disregarding irreversibilities, the exit velocity of air is tobe determined.Assumptions 1 Air is an ideal gas with variable specific heats. 2 The process is given to be reversible andadiabatic, and thus isentropic. Therefore, isentropic relations of ideal gases apply. 2 The nozzle operatessteadily.Analysis Assuming variable specific heats, the inlet and exit properties are determined to be Pr1 12.30 T1 1000 R h1 240.98 Btu/lbmand P2 12 psia T2 635.9 R 1 AIR 2 Pr2 Pr 12.30 2.46 P 1 1 60 psia h2 152.11 Btu/lbmWe take the nozzle as the system, which is a control volume. The energybalance for this steady-flow system can 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 m(h1 V12 / 2) 2 m(h2 + V2 /2) V22 V12 h2 h1 0 2Therefore, 25,037 ft 2 /s 2 2 V2 2 h1 h2 V12 2 240.98 152.11 Btu/lbm 200 ft/s 1 Btu/lbm 2119 ft/s
• 7-447-81 Air is accelerated in an nozzle, and some heat is lost in the process. The exit temperature of air andthe total entropy change during the process are to be determined.Assumptions 1 Air is an ideal gas with variable specific heats. 2 The nozzle operates steadily.Analysis (a) Assuming variable specific heats, the inlet properties are determined to be, h1 350.49 kJ / kg T1 350 K (Table A-17) 3.2 kJ/s s1 1.85708 / kJ / kg KWe take the nozzle as the system, which is a control volume. The energy AIR 1 2balance for this steady-flow system can 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 m(h1 V12 / 2) m(h2 + V22 /2) + Qout V22 V12 0 qout + h2 h1 2Therefore, 2 2 V22 V12 320 m/s 50 m/s 1 kJ/kg h2 h1 qout 350.49 3.2 2 2 1000 m 2 /s 2 297.34 kJ/kgAt this h2 value we read, from Table A-17, T2 297.2 K, s2 1.6924 kJ / kg K(b) The total entropy change is the sum of the entropy changes of the air and of the surroundings, and isdetermined from stotal sair ssurrwhere P2 85 kPa sair s2 s1 R ln 1.6924 1.85708 0.287 kJ/kg K ln 0.1775 kJ/kg K P1 280 kPaand qsurr,in 3.2 kJ/kg ssurr 0.0109 kJ/kg K Tsurr 293 KThus, stotal 0.1775 0.0109 0.1884 kJ/kg K
• 7-457-82 EES Problem 7-76 is reconsidered. The effect of varying the surrounding medium temperature from10°C to 40°C on the exit temperature and the total entropy change for this process is to be studied, and theresults are to be plotted.Analysis The problem is solved using EES, and the results are tabulated and plotted below.Function HCal(WorkFluid\$, Tx, Px)"Function to calculate the enthalpy of an ideal gas or real gas" If Air = WorkFluid\$ then HCal:=ENTHALPY(Air,T=Tx) "Ideal gas equ." else HCal:=ENTHALPY(WorkFluid\$,T=Tx, P=Px)"Real gas equ." endifend HCal"System: control volume for the nozzle""Property relation: Air is an ideal gas""Process: Steady state, steady flow, adiabatic, no work""Knowns - obtain from the input diagram"WorkFluid\$ = AirT[1] = 77 [C]P[1] = 280 [kPa]Vel[1] = 50 [m/s]P[2] = 85 [kPa]Vel[2] = 320 [m/s]q_out = 3.2 [kJ/kg]"T_surr = 20 [C]""Property Data - since the Enthalpy function has different parametersfor ideal gas and real fluids, a function was used to determine h."h[1]=HCal(WorkFluid\$,T[1],P[1])h[2]=HCal(WorkFluid\$,T[2],P[2])"The Volume function has the same form for an ideal gas as for a real fluid."v[1]=volume(workFluid\$,T=T[1],p=P[1])v[2]=volume(WorkFluid\$,T=T[2],p=P[2])"If we knew the inlet or exit area, we could calculate the mass flow rate. Since we dont knowthese areas, we write the conservation of energy per unit mass.""Conservation of mass: m_dot[1]= m_dot[2]""Conservation of Energy - SSSF energy balance for neglecting the change in potential energy, nowork, but heat transfer out is:"h[1]+Vel[1]^2/2*Convert(m^2/s^2, kJ/kg) = h[2]+Vel[2]^2/2*Convert(m^2/s^2, kJ/kg)+q_outs[1]=entropy(workFluid\$,T=T[1],p=P[1])s[2]=entropy(WorkFluid\$,T=T[2],p=P[2])"Entropy change of the air and the surroundings are:"DELTAs_air = s[2] - s[1]q_in_surr = q_outDELTAs_surr = q_in_surr/(T_surr+273)DELTAs_total = DELTAs_air + DELTAs_surr
• 7-46 stotal Tsurr T2 [kJ/kg-K] [C] [C] 0.189 10 24.22 0.1888 15 24.22 0.1886 20 24.22 0.1884 25 24.22 0.1882 30 24.22 0.188 35 24.22 0.1879 40 24.22 0.189 0.1888 0.1886s total [kJ/kg-K] 0.1884 0.1882 0.188 0.1878 10 15 20 25 30 35 40 T [C] surr
• 7-477-83 A container is filled with liquid water is placed in a room and heat transfer takes place between thecontainer and the air in the room until the thermal equilibrium is established. The final temperature, theamount of heat transfer between the water and the air, and the entropy generation are to be determined.Assumptions 1 Kinetic and potential energy changes are negligible. 2 Air is an ideal gas with constantspecific heats. 3 The room is well-sealed and there is no heat transfer from the room to the surroundings. 4Sea level atmospheric pressure is assumed. P = 101.3 kPa.Properties The properties of air at room temperature are R = 0.287 kPa.m3/kg.K, cp = 1.005 kJ/kg.K, cv =0.718 kJ/kg.K. The specific heat of water at room temperature is cw = 4.18 kJ/kg.K (Tables A-2, A-3).Analysis (a) The mass of the air in the room is PV (101.3 kPa)(90 m 3 ) Room ma 111.5 kg RTa1 3 (0.287 kPa m /kg K)(12 273 K) 90 m3 12 CAn energy balance on the system that consists of the water in the Watercontainer and the air in the room gives the final equilibrium 45 kgtemperature 95 C 0 m w c w (T2 Tw1 ) m a cv (T2 Ta1 ) 0 (45 kg)(4.18 kJ/kg.K)(T2 95) (111.5 kg)(0.718 kJ/kg.K)(T2 12) T2 70.2 C(b) The heat transfer to the air is Q m a cv (T2 Ta1 ) (111.5 kg)(0.718 kJ/kg.K)(70.2 12) 4660 kJ(c) The entropy generation associated with this heat transfer process may be obtained by calculating totalentropy change, which is the sum of the entropy changes of water and the air. T2 (70.2 273) K Sw mwcw ln (45 kg)(4.18 kJ/kg.K)ln 13.11 kJ/K Tw1 (95 273) K ma RT2 (111.5 kg)(0.287 kPa m3/kg K)(70.2 273 K) P2 122 kPa V (90 m3 ) T2 P2 Sa m a c p ln R ln Ta1 P1 (70.2 273) K 122 kPa (111.5 kg) (1.005 kJ/kg.K)ln (0.287 kJ/kg.K)ln 14.88 kJ/K (12 273) K 101.3 kPa S gen S total Sw Sa 13.11 14.88 1.77 kJ/K
• 7-487-84 Air is accelerated in an isentropic nozzle. The maximum velocity at the exit is to be determined.Assumptions 1 Air is an ideal gas with constant specific heats. 2 The nozzle operates steadily.Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K, k = 1.4 (Table A-2a).Analysis The exit temperature is determined from ideal gas isentropic relation to be, ( k 1) / k 0.4/1.4 P2 100 kPa T2 T1 400 273 K 371.5 K P1 800 kPaWe take the nozzle as the system, which is a control volume. The energy balance for this steady-flowsystem can 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 1 AIR 2 m(h1 V12 / 2) m(h2 + V22 /2) V22 0 0 h2 h1 2 V22 0 c p (T2 T1 ) 2Therefore, V2 2c p (T2 T1 ) 2(1.005 kJ/kg.K)(673 - 371.5)K 778.5 m/s
• 7-497-85 An ideal gas is compressed in an isentropic compressor. 10% of gas is compressed to 400 kPa and90% is compressed to 600 kPa. The compression process is to be sketched, and the exit temperatures at thetwo exits, and the mass flow rate into the compressor are to be determined.Assumptions 1 The compressor operates steadily. 2 The process is reversible-adiabatic (isentropic)Properties The properties of ideal gas are given to be cp = 1.1 kJ/kg.K and cv = 0.8 kJ/kg.K.Analysis (b) The specific heat ratio of the gas is cp 1.1 k 1.375 P3 = 600 kPa cv 0.8The exit temperatures are determined from ideal gas isentropicrelations to be, ( k 1) / k 0.375/1.375 COMPRESSOR P2 400 kPa T2 T1 27 273 K 437.8 K 32 kW P1 100 kPa ( k 1) / k 0.375/1.375 P3 600 kPa P2 = 400 kPa T3 T1 27 273 K 489.0 K P1 100 kPa P1 = 100 kPa T1 = 300 K(c) A mass balance on the control volume gives m1 m2 m3 m2 0.1m1 P3where T m3 0.9m1 P2We take the compressor as the system, which is a controlvolume. The energy balance for this steady-flow system canbe expressed in the rate form as P1 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 s m1h1 Win m2 h2 m3h3 m1c pT1 Win 0.1m1c pT2 0.9m1c pT3Solving for the inlet mass flow rate, we obtain Win m1 c p 0.1(T2 T1 ) 0.9(T3 T1 ) 32 kW (1.1 kJ/kg K) 0.1(437.8 - 300) 0.9(489.0 - 300) 0.158 kg/s
• 7-507-86 Air contained in a constant-volume tank s cooled to ambient temperature. The entropy changes of theair and the universe due to this process are to be determined and the process is to be sketched on a T-sdiagram.Assumptions 1 Air is an ideal gas with constant specific heats.Properties The specific heat of air at room temperature iscv = 0.718 kJ/kg.K (Table A-2a). Air 5 kgAnalysis (a) The entropy change of air is determined from 327 C T2 100 kPa Sair mcv ln T1 (27 273) K (5 kg)(0.718 kJ/kg.K)ln (327 273) K 2.488 kJ/K T 1(b) An energy balance on the system gives 327ºC air Qout mcv (T2 T1 ) 2 (5 kg)(0.718 kJ/kg.K)(327 27) 1077 kJ 27ºC 1 surr 2The entropy change of the surroundings is Qout 1077 kJ s ssurr 3.59 kJ/K Tsurr 300 KThe entropy change of universe due to this process is Sgen S total Sair Ssurr 2.488 3.59 1.10 kJ/K
• 7-51Reversible Steady-Flow Work7-87C The work associated with steady-flow devices is proportional to the specific volume of the gas.Cooling a gas during compression will reduce its specific volume, and thus the power consumed by thecompressor.7-88C Cooling the steam as it expands in a turbine will reduce its specific volume, and thus the workoutput of the turbine. Therefore, this is not a good proposal.7-89C We would not support this proposal since the steady-flow work input to the pump is proportional tothe specific volume of the liquid, and cooling will not affect the specific volume of a liquid significantly.7-90 Liquid water is pumped reversibly to a specified pressure at a specified rate. The power input to thepump is to be determined.Assumptions 1 Liquid water is an incompressible substance. 2 Kinetic and potential energy changes arenegligible. 3 The process is reversible.Properties The specific volume of saturated liquid water at 20 kPa is v1 = vf @ 20 kPa = 0.001017 m3/kg(Table A-5).Analysis The power input to the pump can be determined directly from the 2steady-flow work relation for a liquid, 45 kg/s 2 Win m vdP ke 0 pe 0 mv1 P2 P1 1 H2OSubstituting, 1 kJ Win (45 kg/s)(0.001017 m3/kg )(6000 20) kPa 274 kW 1 kPa m3 17-91 Liquid water is to be pumped by a 25-kW pump at a specified rate. The highest pressure the watercan be pumped to is to be determined.Assumptions 1 Liquid water is an incompressible substance. 2 Kinetic and potential energy changes arenegligible. 3 The process is assumed to be reversible since we will determine the limiting case.Properties The specific volume of liquid water is given to be v1 = 0.001 m3/kg.Analysis The highest pressure the liquid can have at the pump exit can be determined from the reversiblesteady-flow work relation for a liquid, 2 P2 Win m vdP ke 0 pe 0 mv1 P2 P1 1Thus, 25 kW 3 1 kJ PUMP 25 kJ/s (5 kg/s)(0.001 m /kg )( P2 100) k Pa 1 kPa m3It yields P2 5100 kPa 100 kPa
• 7-527-92E Saturated refrigerant-134a vapor is to be compressed reversibly to a specified pressure. The powerinput to the compressor is to be determined, and it is also to be compared to the work input for the liquidcase.Assumptions 1 Liquid refrigerant is an incompressible substance. 2 Kinetic and potential energy changesare negligible. 3 The process is reversible. 4 The compressor is adiabatic.Analysis The compression process is reversible and adiabatic, and thus isentropic, s1 = s2. Then theproperties of the refrigerant are (Tables A-11E through A-13E) P1 15 psia h1 100.99 Btu/lbm sat. vapor s1 0.22715 Btu/lbm R 2 2 P1 80 psia h2 115.80 Btu/lbm s2 s1 R-134a R-134aThe work input to this isentropic compressor isdetermined from the steady-flow energy balance to be 0 (steady) Ein Eout Esystem 0 Rate of net energy transfer by heat, work, and mass Rate of change in internal, kinetic, potential, etc. energies 1 1 Ein Eout Win mh1 mh2 Win m(h2 h1 )Thus, win h2 h1 115.80 100.99 14.8 Btu/lbmIf the refrigerant were first condensed at constant pressure before it was compressed, we would use a pumpto compress the liquid. In this case, the pump work input could be determined from the steady-flow workrelation to be 2 win v dP ke 0 pe 0 v1 P2 P1 1where v3 = vf @ 15 psia = 0.01165 ft3/lbm. Substituting, 1 Btu win (0.01165 ft 3/lbm) 80 15 psia 0.14 Btu/lbm 5.4039 psia ft 3
• 7-537-93 A steam power plant operates between the pressure limits of 10 MPa and 20 kPa. The ratio of theturbine work to the pump work is to be determined.Assumptions 1 Liquid water is an incompressible substance. 2 Kinetic and potential energy changes arenegligible. 3 The process is reversible. 4 The pump and the turbine are adiabatic.Properties The specific volume of saturated liquid water at 20 kPa is v1 = vf @ 20 kPa = 0.001017 m3/kg(Table A-5).Analysis Both the compression and expansion processes are reversible and adiabatic, and thus isentropic,s1 = s2 and s3 = s4. Then the properties of the steam are P4 20 kPa h4 h g @ 20 kPa 2608.9 kJ/kg 2 s4 s g @ 20 kPa 7.9073 kJ/kg K 3 sat.vapor P3 10 MPa h3 4707.2 kJ/kg H2O s3 s4 H2OAlso, v1 = vf @ 20 kPa = 0.001017 m3/kg.The work output to this isentropic turbine is determinedfrom the steady-flow energy balance to be 1 4 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 mh3 mh4 Wout Wout m(h3 h4 )Substituting, wturb,out h3 h4 4707.2 2608.9 2098.3 kJ/kgThe pump work input is determined from the steady-flow work relation to be 2 wpump,in v dP ke 0 pe 0 v1 P2 P1 1 1 kJ (0.001017 m3/kg )(10,000 20) kPa 1 kPa m3 10.15 kJ/kgThus, wturb,out 2098.3 206.7 wpump,in 10.15
• 7-547-94 EES Problem 7-93 is reconsidered. The effect of the quality of the steam at the turbine exit on the network output is to be investigated as the quality is varied from 0.5 to 1.0, and the net work output us to beplotted as a function of this quality.Analysis The problem is solved using EES, and the results are tabulated and plotted below."System: control volume for the pump and turbine""Property relation: Steam functions""Process: For Pump and Turbine: Steady state, steady flow, adiabatic, reversible orisentropic""Since we dont know the mass, we write the conservation of energy per unit mass.""Conservation of mass: m_dot[1]= m_dot[2]""Knowns:"WorkFluid\$ = Steam_IAPWSP[1] = 20 [kPa]x[1] = 0P[2] = 10000 [kPa]x[4] = 1.0"Pump Analysis:"T[1]=temperature(WorkFluid\$,P=P[1],x=0)v[1]=volume(workFluid\$,P=P[1],x=0)h[1]=enthalpy(WorkFluid\$,P=P[1],x=0)s[1]=entropy(WorkFluid\$,P=P[1],x=0)s[2] = s[1]h[2]=enthalpy(WorkFluid\$,P=P[2],s=s[2])T[2]=temperature(WorkFluid\$,P=P[2],s=s[2])"The Volume function has the same form for an ideal gas as for a real fluid."v[2]=volume(WorkFluid\$,T=T[2],p=P[2])"Conservation of Energy - SSSF energy balance for pump"" -- neglect the change in potential energy, no heat transfer:"h[1]+W_pump = h[2]"Also the work of pump can be obtained from the incompressible fluid, steady-flow result:"W_pump_incomp = v[1]*(P[2] - P[1])"Conservation of Energy - SSSF energy balance for turbine -- neglecting the change in potentialenergy, no heat transfer:"P[4] = P[1]P[3] = P[2]h[4]=enthalpy(WorkFluid\$,P=P[4],x=x[4])s[4]=entropy(WorkFluid\$,P=P[4],x=x[4])T[4]=temperature(WorkFluid\$,P=P[4],x=x[4])s[3] = s[4]h[3]=enthalpy(WorkFluid\$,P=P[3],s=s[3])T[3]=temperature(WorkFluid\$,P=P[3],s=s[3])h[3] = h[4] + W_turbW_net_out = W_turb - W_pump
• 7-55 Wnet,out Wpump Wpump,incomp Wturb x4 [kJ/kg] [kJ/kg] [kJ/kg] [kJ/kg] 557.1 10.13 10.15 567.3 0.5 734.7 10.13 10.15 744.8 0.6 913.6 10.13 10.15 923.7 0.7 1146 10.13 10.15 1156 0.8 1516 10.13 10.15 1527 0.9 2088 10.13 10.15 2098 1 Steam 1100 1000 3 900 800 700 x = 1.0 4T [°C] 600 = 0.5 500 10000 kPa 400 300 200 1, 2 100 4 0.2 20 kPa 0.6 0.8 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] 2250 1900 1550W net,out [kJ/kg] 1200 850 500 0.5 0.6 0.7 0.8 0.9 1 x[4]
• 7-567-95 Liquid water is pumped by a 70-kW pump to a specified pressure at a specified level. The highestpossible mass flow rate of water is to be determined.Assumptions 1 Liquid water is an incompressible substance. 2 Kinetic P2 = 5 MPaenergy changes are negligible, but potential energy changes may besignificant. 3 The process is assumed to be reversible since we willdetermine the limiting case.Properties The specific volume of liquid water is given to be v1 = 0.001m3/kg. PUMPAnalysis The highest mass flow rate will be realized when the entireprocess is reversible. Thus it is determined from the reversible steady-flow work relation for a liquid, 2 Water Win m v dP ke 0 pe m v P2 P1 g z2 z1 P1 = 120 kPa 1Thus, 1 kJ 1 kJ/kg 7 kJ/s m (0.001 m3/kg )(5000 120)kPa 3 (9.8 m/s 2 )(10 m) 1 kPa m 1000 m 2 /s 2It yields m 1.41 kg/s
• 7-577-96E Helium gas is compressed from a specified state to a specified pressure at a specified rate. Thepower input to the compressor is to be determined for the cases of isentropic, polytropic, isothermal, andtwo-stage compression.Assumptions 1 Helium is an ideal gas with constant specific heats. 2 The process is reversible. 3 Kineticand potential energy changes are negligible.Properties The gas constant of helium is R = 2.6805 psia.ft3/lbm.R = 20.4961 Btu/lbm.R. The specific heat ratio of helium is k = 1.667(Table A-2E).Analysis The mass flow rate of helium is · W He P1V 1 14 psia 5 ft 3 /s m 0.0493 lbm/s RT1 2.6805 psia ft 3 /lbm R 530 R(a) Isentropic compression with k = 1.667: 5 ft3/s k 1 /k 1 kRT1 P2 Wcomp,in m 1 k 1 P1 0.667/1.667 1.667 0.4961 Btu/lbm R 530 R 120 psia 0.0493 lbm/s 1 1.667 1 14 psia 44.11 Btu/s 62.4 hp since 1 hp = 0.7068 Btu/s(b) Polytropic compression with n = 1.2: n 1 /n nRT1 P2 Wcomp,in m 1 n 1 P1 0.2/1.2 1.2 0.4961 Btu/lbm R 530 R 120 psia 0.0493 lbm/s 1 1.2 1 14 psia 33.47 Btu/s 47.3 hp since 1 hp = 0.7068 Btu/s(c) Isothermal compression: P2 120 psia Wcomp,in mRT ln 0.0493 lbm/s 0.4961 Btu/lbm R 530 R ln 27.83 Btu/s 39.4 hp P1 14 psia(d) Ideal two-stage compression with intercooling (n = 1.2): In this case, the pressure ratio across eachstage is the same, and its value is determined from Px P P2 1 14 psia 120 psia 41.0 psiaThe compressor work across each stage is also the same, thus total compressor work is twice thecompression work for a single stage: n 1 /n nRT1 Px Wcomp,in 2mwcomp,I 2m 1 n 1 P1 0.2/1.2 1.2 0.4961 Btu/lbm R 530 R 41 psia 2 0.0493 lbm/s 1 1.2 1 14 psia 30.52 Btu/s 43.2 hp since 1 hp = 0.7068 Btu/s
• 7-587-97E EES Problem 7-96E is reconsidered. The work of compression and entropy change of the helium isto be evaluated and plotted as functions of the polytropic exponent as it varies from 1 to 1.667.Analysis The problem is solved using EES, and the results are tabulated and plotted below.Procedure FuncPoly(m_dot,k, R,T1,P2,P1,n:W_dot_comp_polytropic,W_dot_comp_2stagePoly,Q_dot_Out_polytropic,Q_dot_Out_2stagePoly)If n =1 thenT2=T1W_dot_comp_polytropic= m_dot*R*(T1+460)*ln(P2/P1)*convert(Btu/s,hp) "[hp]"W_dot_comp_2stagePoly = W_dot_comp_polytropic "[hp]"Q_dot_Out_polytropic=W_dot_comp_polytropic*convert(hp,Btu/s) "[Btu/s]"Q_dot_Out_2stagePoly = Q_dot_Out_polytropic*convert(hp,Btu/s) "[Btu/s]"ElseC_P = k*R/(k-1) "[Btu/lbm-R]"T2=(T1+460)*((P2/P1)^((n+1)/n)-460)"[F]"W_dot_comp_polytropic = m_dot*n*R*(T1+460)/(n-1)*((P2/P1)^((n-1)/n) -1)*convert(Btu/s,hp)"[hp]"Q_dot_Out_polytropic=W_dot_comp_polytropic*convert(hp,Btu/s)+m_dot*C_P*(T1-T2)"[Btu/s]"Px=(P1*P2)^0.5T2x=(T1+460)*((Px/P1)^((n+1)/n)-460)"[F]"W_dot_comp_2stagePoly = 2*m_dot*n*R*(T1+460)/(n-1)*((Px/P1)^((n-1)/n) -1)*convert(Btu/s,hp)"[hp]"Q_dot_Out_2stagePoly=W_dot_comp_2stagePoly*convert(hp,Btu/s)+2*m_dot*C_P*(T1-T2x)"[Btu/s]"endifENDR=0.4961[Btu/lbm-R]k=1.667n=1.2P1=14 [psia]T1=70 [F]P2=120 [psia]V_dot = 5 [ft^3/s]P1*V_dot=m_dot*R*(T1+460)*convert(Btu,psia-ft^3)W_dot_comp_isentropic = m_dot*k*R*(T1+460)/(k-1)*((P2/P1)^((k-1)/k) -1)*convert(Btu/s,hp)"[hp]"Q_dot_Out_isentropic = 0"[Btu/s]"Call FuncPoly(m_dot,k, R,T1,P2,P1,n:W_dot_comp_polytropic,W_dot_comp_2stagePoly,Q_dot_Out_polytropic,Q_dot_Out_2stagePoly)W_dot_comp_isothermal= m_dot*R*(T1+460)*ln(P2/P1)*convert(Btu/s,hp)"[hp]"Q_dot_Out_isothermal = W_dot_comp_isothermal*convert(hp,Btu/s)"[Btu/s]"
• 7-59 n Wcomp2StagePoly Wcompisentropic Wcompisothermal Wcomppolytropic [hp] [hp] [hp] [hp] 1 39.37 62.4 39.37 39.37 1.1 41.36 62.4 39.37 43.48 1.2 43.12 62.4 39.37 47.35 1.3 44.68 62.4 39.37 50.97 1.4 46.09 62.4 39.37 54.36 1.5 47.35 62.4 39.37 57.541.667 49.19 62.4 39.37 62.4 65 60 Polytropic Isothermal 55 Isentropic W comp,polytropic [hp] 2StagePoly 50 45 40 35 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 n
• 7-607-98 Nitrogen gas is compressed by a 10-kW compressor from a specified state to a specified pressure. Themass flow rate of nitrogen through the compressor is to be determined for the cases of isentropic,polytropic, isothermal, and two-stage compression.Assumptions 1 Nitrogen is an ideal gas with constant specific heats. 2 The process is reversible. 3 Kineticand potential energy changes are negligible.Properties The gas constant of nitrogen is R = 0.297 kJ/kg.K (Table A-1). The specific heat ratio ofnitrogen is k = 1.4 (Table A-2). 2Analysis (a) Isentropic compression: kRT1 k 1 /k Wcomp,in m P2 P1 1 k 1 N2or, · 10 kW m 1.4 0.297 kJ/kg K 300 K 0.4/1.4 10 kJ/s m 480 kPa 80 kPa 1 1.4 1It yields m 0.048 kg / s 1(b) Polytropic compression with n = 1.3: nRT1 n 1 /n Wcomp,in m P2 P1 1 n 1or, 1.3 0.297 kJ/kg K 300 K 0.3/1.3 10 kJ/s m 480 kPa 80 kPa 1 1.3 1It yields m 0.051 kg / s(c) Isothermal compression: P1 480 kPa Wcomp,in mRT ln 10 kJ/s m 0.297 kJ/kg K 300 K ln P2 80 kPaIt yields m 0.063 kg / s(d) Ideal two-stage compression with intercooling (n = 1.3): In this case, the pressure ratio across eachstage is the same, and its value is determined to be Px P P2 1 80 kPa 480 kPa 196 kPaThe compressor work across each stage is also the same, thus total compressor work is twice thecompression work for a single stage: nRT1 n 1 /n Wcomp,in 2mwcomp,I 2m Px P1 1 n 1or, 1.3 0.297 kJ/kg K 300 K 0.3/1.3 10 kJ/s 2m 196 kPa 80 kPa 1 1.3 1It yields m 0.056 kg/s
• 7-617-99 Water mist is to be sprayed into the air stream in the compressor to cool the air as the waterevaporates and to reduce the compression power. The reduction in the exit temperature of the compressedair and the compressor power saved are to be determined.Assumptions 1 Air is an ideal gas with variable specific heats. 2 The process is reversible. 3 Kinetic andpotential energy changes are negligible. 3 Air is compressed isentropically. 4 Water vaporizes completelybefore leaving the compressor. 4 Air properties can be used for the air-vapor mixture.Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1). The specific heat ratio of air is k =1.4. The inlet enthalpies of water and air are (Tables A-4 and A-17) hw1 = hf@20 C = 83.29 kJ/kg , hfg@20 C = 2453.9 kJ/kg and ha1 = h@300 K =300.19 kJ/kgAnalysis In the case of isentropic operation (thus no cooling or water spray), the exit temperature and thepower input to the compressor are ( k 1) / k (1.4 1) / 1.4 T2 P2 1200 kPa T2 (300 K) = 610.2 K T1 P1 100 kPa kRT1 Wcomp,in m P2 P k 1 / k 1 1 k 1 1.4 0.287 kJ/kg K 300 K 0.4/1.4 (2.1 kg/s) 1200 kPa/100 kPa 1 654.3 kW 1.4 1When water is sprayed, we first need to check the accuracy of theassumption that the water vaporizes completely in the compressor. In the 2limiting case, the compression will be isothermal at the compressor inlet 1200 kPatemperature, and the water will be a saturated vapor. To avoid thecomplexity of dealing with two fluid streams and a gas mixture, we · Wdisregard water in the air stream (other than the mass flow rate), and Heassume air is cooled by an amount equal to the enthalpy change of water. Water 20 C The rate of heat absorption of water as it evaporates at the inlettemperature completely is 100 kPa 300 K Qcooling,max mw h fg @ 20 C (0.2 kg/s)(2453.9 kJ/kg) = 490.8 kW 1The minimum power input to the compressor is P2 1200 kPa Wcomp,in, min mRT ln (2.1 kg/s)(0.287 kJ/kg K)(300 K) ln 449.3 kW P1 100 kPaThis corresponds to maximum cooling from the air since, at constant temperature, h = 0 and thus Qout Win 449.3 kW , which is close to 490.8 kW. Therefore, the assumption that all the water vaporizesis approximately valid. Then the reduction in required power input due to water spray becomes Wcomp,in Wcomp, isentropic Wcomp, isothermal 654.3 449.3 205 kWDiscussion (can be ignored): At constant temperature, h = 0 and thus Qout Win 449.3 kW correspondsto maximum cooling from the air, which is less than 490.8 kW. Therefore, the assumption that all the watervaporizes is only roughly valid. As an alternative, we can assume the compression process to be polytropicand the water to be a saturated vapor at the compressor exit temperature, and disregard the remainingliquid. But in this case there is not a unique solution, and we will have to select either the amount of wateror the exit temperature or the polytropic exponent to obtain a solution. Of course we can also tabulate theresults for different cases, and then make a selection.
• 7-62Sample Analysis: We take the compressor exit temperature to be T2 = 200 C = 473 K. Then, hw2 = hg@200 C = 2792.0 kJ/kg and ha2 = h@473 K = 475.3 kJ/kgThen, ( n 1) / n ( n 1) / n T2 P2 473 K 1200 kPa n 1.224 T1 P1 300 K 100 kPa nRT1 nR Wcomp,in m P2 P1 n 1 / n 1 m (T2 T1 ) n 1 n 1 1.224 0.287 kJ/kg K (2.1 kg/s) (473 300)K 570 kW 1.224 1Energy balance: Wcomp,in Qout m(h2 h1 ) Qout Wcomp,in m(h2 h1 ) 569.7 kW (2.1 kg/s)(475.3 300.19) 202.0 kWNoting that this heat is absorbed by water, the rate at which water evaporates in the compressor becomes Qin, water 202.0 kJ/s Qout,air Qin, water mw (hw2 hw1 ) mw 0.0746 kg/s hw2 hw1 (2792.0 83.29) kJ/kgThen the reductions in the exit temperature and compressor power input become T2 T2,isentropic T2,water cooled 610.2 473 137.2 C Wcomp,in Wcomp,isentropic Wcomp,water cooled 654.3 570 84.3 kWNote that selecting a different compressor exit temperature T2 will result in different values.7-100 A water-injected compressor is used in a gas turbine power plant. It is claimed that the power outputof a gas turbine will increase when water is injected into the compressor because of the increase in themass flow rate of the gas (air + water vapor) through the turbine. This, however, is not necessarily rightsince the compressed air in this case enters the combustor at a low temperature, and thus it absorbs muchmore heat. In fact, the cooling effect will most likely dominate and cause the cyclic efficiency to drop.
• 7-63Isentropic Efficiencies of Steady-Flow Devices7-101C The ideal process for all three devices is the reversible adiabatic (i.e., isentropic) process. Theadiabatic efficiencies of these devices are defined as actual work output insentropic work input actual exit kineticenergy T , C , and N insentropic work output actual work input insentropic exit kinetic energy7-102C No, because the isentropic process is not the model or ideal process for compressors that arecooled intentionally.7-103C Yes. Because the entropy of the fluid must increase during an actual adiabatic process as a resultof irreversibilities. Therefore, the actual exit state has to be on the right-hand side of the isentropic exitstate7-104 Steam enters an adiabatic turbine with an isentropic efficiency of 0.90 at a specified state with aspecified mass flow rate, and leaves at a specified pressure. The turbine exit temperature and power outputof the turbine are to be determined.Assumptions 1 This is a steady-flow process since there is no changewith time. 2 Kinetic and potential energy changes are negligible. 3 P1 = 8 MPaThe device is adiabatic and thus heat transfer is negligible. T1 = 500 CAnalysis (a) From the steam tables (Tables A-4 through A-6), P1 8 MPa h1 3399.5 kJ/kg STEAM T1 500 C s1 6.7266 kJ/kg K TURBINE T = 90% s2s sf 6.7266 0.9441 P2 s 30 kPa x2s 0.8475 s fg 6.8234 s 2s s1 h2 s hf x 2 s h fg 289.27 0.8475 2335.3 2268.3 kJ/kg P2 = 30 kPaFrom the isentropic efficiency relation, h1 h2 a T h2 a h1 T h1 h2 s 3399.5 0.9 3399.5 2268.3 2381.4 kJ/kg h1 h2 sThus, P2 a 30 kPa T2 a Tsat @ 30 kPa 69.09 C h2 a 2381.4 kJ/kg(b) There is only one inlet and one exit, and thus m1 m2 m . We take the actual turbine 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 mh1 Wa,out mh2 (since Q ke pe 0) Wa,out m(h1 h2 )Substituting, Wa,out 3kg/s 3399.5 2381.4 kJ/kg 3054 kW
• 7-647-105 EES Problem 7-104 is reconsidered. The effect of varying the turbine isentropic efficiency from0.75 to 1.0 on both the work done and the exit temperature of the steam are to be investigated, and theresults are to be plotted.Analysis The problem is solved using EES, and the results are tabulated and plotted below."System: control volume for turbine""Property relation: Steam functions""Process: Turbine: Steady state, steady flow, adiabatic, reversible or isentropic""Since we dont know the mass, we write the conservation of energy per unit mass.""Conservation of mass: m_dot[1]= m_dot[2]=m_dot" Steam 700"Knowns:"WorkFluid\$ = Steam_iapws 600m_dot = 3 [kg/s]P[1] = 8000 [kPa] 500 1T[1] = 500 [C]P[2] = 30 [kPa] T [°C] 400"eta_turb = 0.9" 300 8000 kPa"Conservation of Energy - SSSF energy 200balance for turbine -- neglecting the 0.4 0.6 2 0.8change in potential energy, no heat 100 0.2 2transfer:" 30 kPa sh[1]=enthalpy(WorkFluid\$,P=P[1],T=T[1]) 0s[1]=entropy(WorkFluid\$,P=P[1],T=T[1]) 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0T_s[1] = T[1] s [kJ/kg-K]s[2] =s[1]s_s[2] = s[1]h_s[2]=enthalpy(WorkFluid\$,P=P[2],s=s_s[2])T_s[2]=temperature(WorkFluid\$,P=P[2],s=s_s[2])eta_turb = w_turb/w_turb_sh[1] = h[2] + w_turbh[1] = h_s[2] + w_turb_sT[2]=temperature(WorkFluid\$,P=P[2],h=h[2])W_dot_turb = m_dot*w_turb 3400 turb Wturb [kW] 3300 0.75 2545 0.8 2715 3200 0.85 2885 3100 0.9 3054 W turb [kW ] 0.95 3224 3000 1 3394 2900 2800 2700 2600 2500 0.75 0.8 0.85 0.9 0.95 1 turb
• 7-657-106 Steam enters an adiabatic turbine at a specified state, and leaves at a specified state. The mass flowrate of the steam and the isentropic efficiency are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changesare negligible. 3 The device is adiabatic and thus heat transfer is negligible.Analysis (a) From the steam tables (Tables A-4 and A-6), P1 7 MPa h1 3650.6 kJ/kg T1 600 C s1 7.0910 kJ/kg K P2 50 kPa h2 a 2780.2 kJ/kg T2 150 CThere is only one inlet and one exit, and thus m1 m2 m . We take the actual turbine as the system, whichis a control volume since mass crosses the boundary. The energy balance for this steady-flow system canbe expressed in the rate form as 0 (steady) Ein Eout Esystem 0 1 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies Ein Eout 6 MW H2O m(h1 V12 / 2) Wa,out m(h2 + V12 /2) (since Q pe 0) V22 V12 Wa,out m h2 h1 2Substituting, the mass flow rate of the steam is determined to be 2 (140 m/s) 2 (80 m/s) 2 1 kJ/kg 6000 kJ/s m 2780.2 3650.6 2 1000 m 2 /s 2 m 6.95 kg/s(b) The isentropic exit enthalpy of the steam and the power output of the isentropic turbine are s 2s sf7.0910 1.0912 P2 s 50 kPa x 2s 0.9228 s fg 6.5019 s 2s s1 h2 s hf x 2 s h fg 340.54 0.9228 2304.7 2467.3 kJ/kgand Ws, out m h2 s h1 V22 V12 / 2 (140 m/s) 2 (80 m/s) 2 1 kJ/kg Ws, out 6.95 kg/s 2467.3 3650.6 2 1000 m 2 /s 2 8174 kWThen the isentropic efficiency of the turbine becomes Wa 6000 kW T 0.734 73.4% Ws 8174 kW
• 7-667-107 Argon enters an adiabatic turbine at a specified state with a specified mass flow rate, and leaves at aspecified pressure. The isentropic efficiency of the turbine 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. 4 Argon is anideal gas with constant specific heats.Properties The specific heat ratio of argon is k = 1.667. The constant pressure specific heat of argon is cp =0.5203 kJ/kg.K (Table A-2).Analysis There is only one inlet and one exit, and thus m1 m2 m . We take the isentropic turbine as thesystem, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as Ein Eout 1 mh1 Ws ,out mh2 s (since Q ke pe 0) Ws, out m(h1 h2 s ) 370 kW ArFrom the isentropic relations, T k 1 /k 0.667/1.667 P2 s 200 kPa T2 s T1 1073 K 479 K P1 1500 kPaThen the power output of the isentropic turbine becomes 2 Ws, out mc p T1 T2 s 80/60 kg/min 0.5203 kJ/kg K 1073 479 412.1 kWThen the isentropic efficiency of the turbine is determined from Wa 370 kW T 0.898 89.8% Ws 412.1 kW7-108E Combustion gases enter an adiabatic gas turbine with an isentropic efficiency of 82% at a specifiedstate, and leave at a specified pressure. The work output of the turbine 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. 4 Combustiongases can be treated as air that is an ideal gas with variable specific heats.Analysis From the air table and isentropic relations, 1 h1 = 504.71 Btu / lbm T1 2000 R Pr1 174.0 P2 60 psia AIR Pr2 Pr 174.0 87.0 h2s 417.3 Btu/lbm T = 82% P 1 1 120 psiaThere is only one inlet and one exit, and thus m1 m2 m . We take theactual turbine as the system, which is a control volume since mass crosses the 2boundary. The energy balance for this steady-flow system can be expressedas Ein Eout mh1 Wa,out mh2 (since Q ke pe 0) Wa,out m(h1 h2 )Noting that wa = Tws, the work output of the turbine per unit mass is determined from wa 0.82 504.71 417.3 Btu/lbm 71.7 Btu/lbm
• 7-677-109 [Also solved by EES on enclosed CD] Refrigerant-134a enters an adiabatic compressor with anisentropic efficiency of 0.80 at a specified state with a specified volume flow rate, and leaves at a specifiedpressure. The compressor exit temperature and power input to the compressor are 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 (a) From the refrigerant tables (Tables A-11E through A-13E), 2 h1 hg @120 kPa 236.97 kJ/kg P1 120 kPa s1 s g @120 kPa 0.94779 kJ/kg K sat. vapor R-134a v1 v g @120 kPa 0.16212 m3/kg C = 80% P2 1 MPa h2 s 281.21 kJ/kg s2 s s1 0.3 m3/minFrom the isentropic efficiency relation, 1 h2 s h1 C h2 a h1 h2 s h1 / C 236.97 281.21 236.97 /0.80 292.26 kJ/kg h2 a h1Thus, P2 a 1 MPa T2 a 58.9 C h2 a 292.26 kJ/kg(b) The mass flow rate of the refrigerant is determined from V1 0.3/60 m 3 /s m 0.0308 kg/s v1 0.16212 m 3 /kgThere is only one inlet and one exit, and thus m1 m2 m . We take the actual compressor as the system,which is a control volume since mass crosses the boundary. The energy balance for this steady-flow systemcan be expressed 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 Wa,in mh1 mh2 (since Q ke pe 0) Wa,in m(h2 h1 )Substituting, the power input to the compressor becomes, Wa,in 0.0308 kg/s 292.26 236.97 kJ/kg 1.70 kW
• 7-687-110 EES Problem 7-109 is reconsidered. The problem is to be solved by considering the kinetic energyand by assuming an inlet-to-exit area ratio of 1.5 for the compressor when the compressor exit pipe insidediameter is 2 cm.Analysis The problem is solved using EES, and the solution is given below."Input Data from diagram window"{P[1] = 120 "kPa"P[2] = 1000 "kPa"Vol_dot_1 = 0.3 "m^3/min"Eta_c = 0.80 "Compressor adiabatic efficiency"A_ratio = 1.5d_2 = 2/100 "m"}"System: Control volume containing the compressor, see the diagram window.Property Relation: Use the real fluid properties for R134a.Process: Steady-state, steady-flow, adiabatic process."Fluid\$=R134a"Property Data for state 1"T[1]=temperature(Fluid\$,P=P[1],x=1)"Real fluid equ. at the sat. vapor state"h[1]=enthalpy(Fluid\$, P=P[1], x=1)"Real fluid equ. at the sat. vapor state"s[1]=entropy(Fluid\$, P=P[1], x=1)"Real fluid equ. at the sat. vapor state"v[1]=volume(Fluid\$, P=P[1], x=1)"Real fluid equ. at the sat. vapor state""Property Data for state 2"s_s[1]=s[1]; T_s[1]=T[1] "needed for plot"s_s[2]=s[1] "for the ideal, isentropic process across the compressor"h_s[2]=ENTHALPY(Fluid\$, P=P[2], s=s_s[2])"Enthalpy 2 at the isentropic state 2s andpressure P[2]"T_s[2]=Temperature(Fluid\$, P=P[2], s=s_s[2])"Temperature of ideal state - needed only forplot.""Steady-state, steady-flow conservation of mass" m_dot_1 = m_dot_2m_dot_1 = Vol_dot_1/(v[1]*60)Vol_dot_1/v[1]=Vol_dot_2/v[2]Vel[2]=Vol_dot_2/(A[2]*60)A[2] = pi*(d_2)^2/4A_ratio*Vel[1]/v[1] = Vel[2]/v[2] "Mass flow rate: = A*Vel/v, A_ratio = A[1]/A[2]"A_ratio=A[1]/A[2]"Steady-state, steady-flow conservation of energy, adiabatic compressor, see diagramwindow"m_dot_1*(h[1]+(Vel[1])^2/(2*1000)) + W_dot_c= m_dot_2*(h[2]+(Vel[2])^2/(2*1000))"Definition of the compressor adiabatic efficiency, Eta_c=W_isen/W_act"Eta_c = (h_s[2]-h[1])/(h[2]-h[1])"Knowing h[2], the other properties at state 2 can be found."v[2]=volume(Fluid\$, P=P[2], h=h[2])"v[2] is found at the actual state 2, knowing P and h."T[2]=temperature(Fluid\$, P=P[2],h=h[2])"Real fluid equ. for T at the known outlet h and P."s[2]=entropy(Fluid\$, P=P[2], h=h[2]) "Real fluid equ. at the known outlet h and P."T_exit=T[2]"Neglecting the kinetic energies, the work is:"m_dot_1*h[1] + W_dot_c_noke= m_dot_2*h[2]
• 7-69SOLUTIONA[1]=0.0004712 [m^2] s_s[1]=0.9478 [kJ/kg-K]A[2]=0.0003142 [m^2] s_s[2]=0.9478 [kJ/kg-K]A_ratio=1.5 T[1]=-22.32 [C]d_2=0.02 [m] T[2]=58.94 [C]Eta_c=0.8 T_exit=58.94 [C]Fluid\$=R134a T_s[1]=-22.32 [C]h[1]=237 [kJ/kg] T_s[2]=48.58 [C]h[2]=292.3 [kJ/kg] Vol_dot_1=0.3 [m^3 /min]h_s[2]=281.2 [kJ/kg] Vol_dot_2=0.04244 [m^3 /min]m_dot_1=0.03084 [kg/s] v[1]=0.1621 [m^3/kg]m_dot_2=0.03084 [kg/s] v[2]=0.02294 [m^3/kg]P[1]=120.0 [kPa] Vel[1]=10.61 [m/s]P[2]=1000.0 [kPa] Vel[2]=2.252 [m/s]s[1]=0.9478 [kJ/kg-K] W_dot_c=1.704 [kW]s[2]=0.9816 [kJ/kg-K] W_dot_c_noke=1.706 [kW] R134a 125 T-s diagram for real and ideal com pressor 100 Ideal Compressor Real Com pressor 75 Tem perature [C] 50 1000 kPa 25 0 -25 120 kPa -50 0.0 0.2 0.4 0.6 0.8 1.0 1.2 Entropy [kJ/kg-K]
• 7-707-111 Air enters an adiabatic compressor with an isentropic efficiency of 84% at a specified state, andleaves at a specified temperature. The exit pressure of air and the power input to the compressor are to bedetermined.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. 4 Air is anideal gas with variable specific heats.Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1) 2Analysis (a) From the air table (Table A-17), T1 290 K h1 290.16 kJ/kg, Pr1 1.2311 AIR T2 530 K h2 a 533.98 kJ/kg C = 84% h2 s h1From the isentropic efficiency relation , C h2 a h1 2.4 m3/s h2 s h1 C h2 a h1 1 290.16 0.84 533.98 290.16 495.0 kJ/kg Pr2 7.951Then from the isentropic relation , P2 Pr2 Pr2 7.951 P2 P1 100 kPa 646 kPa P1 Pr1 Pr1 1.2311(b) There is only one inlet and one exit, and thus m1 m2 m . We take the actual compressor as thesystem, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed 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 Wa,in mh1 mh2 (since Q ke pe 0) Wa,in m(h2 h1 ) P1V 1 (100 kPa )(2.4 m 3 /s)where m 2.884 kg/s RT1 (0.287 kPa m 3 /kg K )(290 K )Then the power input to the compressor is determined to be Wa,in (2.884 kg/s)(533.98 290.16) kJ/kg 703 kW
• 7-717-112 Air is compressed by an adiabatic compressor from a specified state to another specified state. Theisentropic efficiency of the compressor and the exit temperature of air for the isentropic case are to bedetermined.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. 4 Air is anideal gas with variable specific heats.Analysis (a) From the air table (Table A-17), 2 T1 300 K h1 300.19 kJ / kg, Pr1 1.386 T2 550 K h2 a 554.74 kJ / kgFrom the isentropic relation, AIR P2 600 kPa Pr2 Pr 1.386 8.754 h2 s 508.72 kJ/kg P 1 1 95 kPaThen the isentropic efficiency becomes 1 h2 s h1 508.72 30019 . 0.819 81.9% C h2a h1 554.74 30019 .(b) If the process were isentropic, the exit temperature would be h2 s 508.72 kJ / kg T2 s 505.5 K
• 7-727-113E Argon enters an adiabatic compressor with an isentropic efficiency of 80% at a specified state, andleaves at a specified pressure. The exit temperature of argon and the work input to the compressor are to bedetermined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changesare negligible. 3 The device is adiabatic and thus heat transfer is negligible. 4 Argon is an ideal gas withconstant specific heats.Properties The specific heat ratio of argon is k = 1.667. The constant 2pressure specific heat of argon is cp = 0.1253 Btu/lbm.R (Table A-2E).Analysis (a) The isentropic exit temperature T2s is determined from k 1 /k 0.667/1.667 Ar P2 s 200 psia = 80% T2 s T1 550 R 1381.9 R C P1 20 psiaThe actual kinetic energy change during this process is 2 2 V 22 V12 240 ft/s 60 ft/s 1 Btu/lbm ke a 1.08 Btu/lbm 1 2 2 25,037 ft 2 /s 2The effect of kinetic energy on isentropic efficiency is very small. Therefore, we can take the kineticenergy changes for the actual and isentropic cases to be same in efficiency calculations. From theisentropic efficiency relation, including the effect of kinetic energy, ws (h2 s h1 ) ke c p T2 s T1 ke s 0.1253 1381.9 550 1.08 C 0.8 wa ( h2 a h1 ) ke c p T2 a T1 ke a 0.1253 T2 a 550 1.08It yields T2a = 1592 R(b) There is only one inlet and one exit, and thus m1 m2 m . We take the actual compressor as thesystem, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed 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 Wa,in m(h1 V12 / 2) m(h2 + V22 /2) (since Q pe 0) V22 V12 Wa,in m h2 h1 wa,in h2 h1 ke 2Substituting, the work input to the compressor is determined to be wa,in 0.1253 Btu/lbm R 1592 550 R 1.08 Btu/lbm 131.6 Btu/lbm
• 7-737-114 CO2 gas is compressed by an adiabatic compressor from a specified state to another specified state.The isentropic efficiency of 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. 4 CO2 is anideal gas with constant specific heats.Properties At the average temperature of (300 + 450)/2 = 375 K, the constant pressure specificheat and the specific heat ratio of CO2 are k = 1.260 and cp = 0.917 kJ/kg.K (Table A-2). 2Analysis The isentropic exit temperature T2s is k 1 /k 0.260/1.260 P2 s 600 kPa T2 s T1 300 K 434.2 K CO2 P1 100 kPa 1.8 kg/sFrom the isentropic efficiency relation, w s h2 s h1 c p T2 s T1 T2 s T1 434.2 300 C 0.895 89.5% wa h2 a h1 c p T2 a T1 T2 a T1 450 300 17-115E Air is accelerated in a 90% efficient adiabatic nozzle from low velocity to a specified velocity. Theexit temperature and pressure of the air are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changesare negligible. 3 The device is adiabatic and thus heat transfer is negligible. 4 Air is an ideal gas withvariable specific heats.Analysis From the air table (Table A-17E), T1 1480 R h1 363.89 Btu / lbm, Pr1 53.04There is only one inlet and one exit, and thus m1 m2 m . We take the nozzle as the system, which is acontrol volume since mass crosses the boundary. The energy balance for this steady-flow system can beexpressed 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 AIR Ein Eout 1 2 N = 90% m(h1 V12 / 2) m(h2 + V22 /2) (since W Q pe 0) 0 V22 V12 h2 h1 2Substituting, the exit temperature of air is determined to be 2 800 ft/s 0 1 Btu/lbm h2 363.89 kJ/kg 351.11 Btu/lbm 2 25,037 ft 2 /s 2From the air table we read T2a = 1431.3 R h2 a h1From the isentropic efficiency relation N , h2 s h1 h2 s h1 h2 a h1 / N 363.89 351.11 363.89 / 0.90 349.69 Btu/lbm Pr2 46.04Then the exit pressure is determined from the isentropic relation to be P2 Pr2 Pr2 46.04 P2 P1 60 psia 52.1 psia P1 Pr1 Pr1 53.04
• 7-747-116E EES Problem 7-115E is reconsidered. The effect of varying the nozzle isentropic efficiency from0.8 to 1.0 on the exit temperature and pressure of the air is to be investigated, and the results are to beplotted.Analysis The problem is solved using EES, and the results are tabulated and plotted below."Knowns:"WorkFluid\$ = AirP[1] = 60 [psia]T[1] = 1020 [F]Vel[2] = 800 [ft/s]Vel[1] = 0 [ft/s]eta_nozzle = 0.9"Conservation of Energy - SSSF energy balance for turbine -- neglecting the change in potentialenergy, no heat transfer:"h[1]=enthalpy(WorkFluid\$,T=T[1])s[1]=entropy(WorkFluid\$,P=P[1],T=T[1])T_s[1] = T[1]s[2] =s[1]s_s[2] = s[1]h_s[2]=enthalpy(WorkFluid\$,T=T_s[2])T_s[2]=temperature(WorkFluid\$,P=P[2],s=s_s[2])eta_nozzle = ke[2]/ke_s[2]ke[1] = Vel[1]^2/2ke[2]=Vel[2]^2/2h[1]+ke[1]*convert(ft^2/s^2,Btu/lbm) = h[2] + ke[2]*convert(ft^2/s^2,Btu/lbm)h[1] +ke[1]*convert(ft^2/s^2,Btu/lbm) = h_s[2] + ke_s[2]*convert(ft^2/s^2,Btu/lbm)T[2]=temperature(WorkFluid\$,h=h[2])P_2_answer = P[2]T_2_answer = T[2] P2 T2 Ts,2 980 nozzle [psia [F [F] 0.8 51.09 971.4 959.2 0.85 51.58 971.4 962.8 0.9 52.03 971.4 966 970 0.95 52.42 971.4 968.8 1 52.79 971.4 971.4 T s[2] 960 T 2 T s[2] 950 0.8 0.84 0.88 0.92 0.96 1 nozzle
• 7-757-117 Hot combustion gases are accelerated in a 92% efficient adiabatic nozzle from low velocity to aspecified velocity. The exit velocity and the exit temperature are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changesare negligible. 3 The device is adiabatic and thus heat transfer is negligible. 4 Combustion gases can betreated as air that is an ideal gas with variable specific heats.Analysis From the air table (Table A-17),T1 1020 K h1 1068.89 kJ/kg, Pr1 123.4 P1 = 260 kPa AIR T1 = 747 C P2 = 85 kPa N = 92%From the isentropic relation , V1 = 80 m/s P2 85 kPaPr2 Pr 123.4 40.34 h2 s 783.92 kJ/kg P 1 1 260 kPaThere is only one inlet and one exit, and thus m1 m2 m . We take the nozzle as the system, which is acontrol volume since mass crosses the boundary. The energy balance for this steady-flow system for theisentropic process can be expressed 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 m(h1 V12 / 2) m(h2 s + V 22s /2) (since W Q pe 0) V 22s V12 h2 s h1 2Then the isentropic exit velocity becomes 2 1000 m 2 /s 2 V2s V12 2 h1 h2 s 80 m/s 2 1068.89 783.92 kJ/kg 759.2 m/s 1 kJ/kgTherefore, V2 a N V2 s 0.92 759.2 m/s 728.2 m/sThe exit temperature of air is determined from the steady-flow energy equation, 2 2 728.2 m/s 80 m/s 1 kJ/kg h2 a 1068.89 kJ/kg 806.95 kJ/kg 2 1000 m 2 /s 2From the air table we read T2a = 786.3 K
• 7-51Reversible Steady-Flow Work7-87C The work associated with steady-flow devices is proportional to the specific volume of the gas.Cooling a gas during compression will reduce its specific volume, and thus the power consumed by thecompressor.7-88C Cooling the steam as it expands in a turbine will reduce its specific volume, and thus the workoutput of the turbine. Therefore, this is not a good proposal.7-89C We would not support this proposal since the steady-flow work input to the pump is proportional tothe specific volume of the liquid, and cooling will not affect the specific volume of a liquid significantly.7-90 Liquid water is pumped reversibly to a specified pressure at a specified rate. The power input to thepump is to be determined.Assumptions 1 Liquid water is an incompressible substance. 2 Kinetic and potential energy changes arenegligible. 3 The process is reversible.Properties The specific volume of saturated liquid water at 20 kPa is v1 = vf @ 20 kPa = 0.001017 m3/kg(Table A-5).Analysis The power input to the pump can be determined directly from the 2steady-flow work relation for a liquid, 45 kg/s 2 Win m vdP ke 0 pe 0 mv1 P2 P1 1 H2OSubstituting, 1 kJ Win (45 kg/s)(0.001017 m3/kg )(6000 20) kPa 274 kW 1 kPa m3 17-91 Liquid water is to be pumped by a 25-kW pump at a specified rate. The highest pressure the watercan be pumped to is to be determined.Assumptions 1 Liquid water is an incompressible substance. 2 Kinetic and potential energy changes arenegligible. 3 The process is assumed to be reversible since we will determine the limiting case.Properties The specific volume of liquid water is given to be v1 = 0.001 m3/kg.Analysis The highest pressure the liquid can have at the pump exit can be determined from the reversiblesteady-flow work relation for a liquid, 2 P2 Win m vdP ke 0 pe 0 mv1 P2 P1 1Thus, 25 kW 3 1 kJ PUMP 25 kJ/s (5 kg/s)(0.001 m /kg )( P2 100) k Pa 1 kPa m3It yields P2 5100 kPa 100 kPa
• 7-527-92E Saturated refrigerant-134a vapor is to be compressed reversibly to a specified pressure. The powerinput to the compressor is to be determined, and it is also to be compared to the work input for the liquidcase.Assumptions 1 Liquid refrigerant is an incompressible substance. 2 Kinetic and potential energy changesare negligible. 3 The process is reversible. 4 The compressor is adiabatic.Analysis The compression process is reversible and adiabatic, and thus isentropic, s1 = s2. Then theproperties of the refrigerant are (Tables A-11E through A-13E) P1 15 psia h1 100.99 Btu/lbm sat. vapor s1 0.22715 Btu/lbm R 2 2 P1 80 psia h2 115.80 Btu/lbm s2 s1 R-134a R-134aThe work input to this isentropic compressor isdetermined from the steady-flow energy balance to be 0 (steady) Ein Eout Esystem 0 Rate of net energy transfer by heat, work, and mass Rate of change in internal, kinetic, potential, etc. energies 1 1 Ein Eout Win mh1 mh2 Win m(h2 h1 )Thus, win h2 h1 115.80 100.99 14.8 Btu/lbmIf the refrigerant were first condensed at constant pressure before it was compressed, we would use a pumpto compress the liquid. In this case, the pump work input could be determined from the steady-flow workrelation to be 2 win v dP ke 0 pe 0 v1 P2 P1 1where v3 = vf @ 15 psia = 0.01165 ft3/lbm. Substituting, 1 Btu win (0.01165 ft 3/lbm) 80 15 psia 0.14 Btu/lbm 5.4039 psia ft 3
• 7-537-93 A steam power plant operates between the pressure limits of 10 MPa and 20 kPa. The ratio of theturbine work to the pump work is to be determined.Assumptions 1 Liquid water is an incompressible substance. 2 Kinetic and potential energy changes arenegligible. 3 The process is reversible. 4 The pump and the turbine are adiabatic.Properties The specific volume of saturated liquid water at 20 kPa is v1 = vf @ 20 kPa = 0.001017 m3/kg(Table A-5).Analysis Both the compression and expansion processes are reversible and adiabatic, and thus isentropic,s1 = s2 and s3 = s4. Then the properties of the steam are P4 20 kPa h4 h g @ 20 kPa 2608.9 kJ/kg 2 s4 s g @ 20 kPa 7.9073 kJ/kg K 3 sat.vapor P3 10 MPa h3 4707.2 kJ/kg H2O s3 s4 H2OAlso, v1 = vf @ 20 kPa = 0.001017 m3/kg.The work output to this isentropic turbine is determinedfrom the steady-flow energy balance to be 1 4 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 mh3 mh4 Wout Wout m(h3 h4 )Substituting, wturb,out h3 h4 4707.2 2608.9 2098.3 kJ/kgThe pump work input is determined from the steady-flow work relation to be 2 wpump,in v dP ke 0 pe 0 v1 P2 P1 1 1 kJ (0.001017 m3/kg )(10,000 20) kPa 1 kPa m3 10.15 kJ/kgThus, wturb,out 2098.3 206.7 wpump,in 10.15
• 7-547-94 EES Problem 7-93 is reconsidered. The effect of the quality of the steam at the turbine exit on the network output is to be investigated as the quality is varied from 0.5 to 1.0, and the net work output us to beplotted as a function of this quality.Analysis The problem is solved using EES, and the results are tabulated and plotted below."System: control volume for the pump and turbine""Property relation: Steam functions""Process: For Pump and Turbine: Steady state, steady flow, adiabatic, reversible orisentropic""Since we dont know the mass, we write the conservation of energy per unit mass.""Conservation of mass: m_dot[1]= m_dot[2]""Knowns:"WorkFluid\$ = Steam_IAPWSP[1] = 20 [kPa]x[1] = 0P[2] = 10000 [kPa]x[4] = 1.0"Pump Analysis:"T[1]=temperature(WorkFluid\$,P=P[1],x=0)v[1]=volume(workFluid\$,P=P[1],x=0)h[1]=enthalpy(WorkFluid\$,P=P[1],x=0)s[1]=entropy(WorkFluid\$,P=P[1],x=0)s[2] = s[1]h[2]=enthalpy(WorkFluid\$,P=P[2],s=s[2])T[2]=temperature(WorkFluid\$,P=P[2],s=s[2])"The Volume function has the same form for an ideal gas as for a real fluid."v[2]=volume(WorkFluid\$,T=T[2],p=P[2])"Conservation of Energy - SSSF energy balance for pump"" -- neglect the change in potential energy, no heat transfer:"h[1]+W_pump = h[2]"Also the work of pump can be obtained from the incompressible fluid, steady-flow result:"W_pump_incomp = v[1]*(P[2] - P[1])"Conservation of Energy - SSSF energy balance for turbine -- neglecting the change in potentialenergy, no heat transfer:"P[4] = P[1]P[3] = P[2]h[4]=enthalpy(WorkFluid\$,P=P[4],x=x[4])s[4]=entropy(WorkFluid\$,P=P[4],x=x[4])T[4]=temperature(WorkFluid\$,P=P[4],x=x[4])s[3] = s[4]h[3]=enthalpy(WorkFluid\$,P=P[3],s=s[3])T[3]=temperature(WorkFluid\$,P=P[3],s=s[3])h[3] = h[4] + W_turbW_net_out = W_turb - W_pump
• 7-55 Wnet,out Wpump Wpump,incomp Wturb x4 [kJ/kg] [kJ/kg] [kJ/kg] [kJ/kg] 557.1 10.13 10.15 567.3 0.5 734.7 10.13 10.15 744.8 0.6 913.6 10.13 10.15 923.7 0.7 1146 10.13 10.15 1156 0.8 1516 10.13 10.15 1527 0.9 2088 10.13 10.15 2098 1 Steam 1100 1000 3 900 800 700 x = 1.0 4T [°C] 600 = 0.5 500 10000 kPa 400 300 200 1, 2 100 4 0.2 20 kPa 0.6 0.8 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] 2250 1900 1550W net,out [kJ/kg] 1200 850 500 0.5 0.6 0.7 0.8 0.9 1 x[4]
• 7-567-95 Liquid water is pumped by a 70-kW pump to a specified pressure at a specified level. The highestpossible mass flow rate of water is to be determined.Assumptions 1 Liquid water is an incompressible substance. 2 Kinetic P2 = 5 MPaenergy changes are negligible, but potential energy changes may besignificant. 3 The process is assumed to be reversible since we willdetermine the limiting case.Properties The specific volume of liquid water is given to be v1 = 0.001m3/kg. PUMPAnalysis The highest mass flow rate will be realized when the entireprocess is reversible. Thus it is determined from the reversible steady-flow work relation for a liquid, 2 Water Win m v dP ke 0 pe m v P2 P1 g z2 z1 P1 = 120 kPa 1Thus, 1 kJ 1 kJ/kg 7 kJ/s m (0.001 m3/kg )(5000 120)kPa 3 (9.8 m/s 2 )(10 m) 1 kPa m 1000 m 2 /s 2It yields m 1.41 kg/s
• 7-577-96E Helium gas is compressed from a specified state to a specified pressure at a specified rate. Thepower input to the compressor is to be determined for the cases of isentropic, polytropic, isothermal, andtwo-stage compression.Assumptions 1 Helium is an ideal gas with constant specific heats. 2 The process is reversible. 3 Kineticand potential energy changes are negligible.Properties The gas constant of helium is R = 2.6805 psia.ft3/lbm.R = 20.4961 Btu/lbm.R. The specific heat ratio of helium is k = 1.667(Table A-2E).Analysis The mass flow rate of helium is · W He P1V 1 14 psia 5 ft 3 /s m 0.0493 lbm/s RT1 2.6805 psia ft 3 /lbm R 530 R(a) Isentropic compression with k = 1.667: 5 ft3/s k 1 /k 1 kRT1 P2 Wcomp,in m 1 k 1 P1 0.667/1.667 1.667 0.4961 Btu/lbm R 530 R 120 psia 0.0493 lbm/s 1 1.667 1 14 psia 44.11 Btu/s 62.4 hp since 1 hp = 0.7068 Btu/s(b) Polytropic compression with n = 1.2: n 1 /n nRT1 P2 Wcomp,in m 1 n 1 P1 0.2/1.2 1.2 0.4961 Btu/lbm R 530 R 120 psia 0.0493 lbm/s 1 1.2 1 14 psia 33.47 Btu/s 47.3 hp since 1 hp = 0.7068 Btu/s(c) Isothermal compression: P2 120 psia Wcomp,in mRT ln 0.0493 lbm/s 0.4961 Btu/lbm R 530 R ln 27.83 Btu/s 39.4 hp P1 14 psia(d) Ideal two-stage compression with intercooling (n = 1.2): In this case, the pressure ratio across eachstage is the same, and its value is determined from Px P P2 1 14 psia 120 psia 41.0 psiaThe compressor work across each stage is also the same, thus total compressor work is twice thecompression work for a single stage: n 1 /n nRT1 Px Wcomp,in 2mwcomp,I 2m 1 n 1 P1 0.2/1.2 1.2 0.4961 Btu/lbm R 530 R 41 psia 2 0.0493 lbm/s 1 1.2 1 14 psia 30.52 Btu/s 43.2 hp since 1 hp = 0.7068 Btu/s
• 7-587-97E EES Problem 7-96E is reconsidered. The work of compression and entropy change of the helium isto be evaluated and plotted as functions of the polytropic exponent as it varies from 1 to 1.667.Analysis The problem is solved using EES, and the results are tabulated and plotted below.Procedure FuncPoly(m_dot,k, R,T1,P2,P1,n:W_dot_comp_polytropic,W_dot_comp_2stagePoly,Q_dot_Out_polytropic,Q_dot_Out_2stagePoly)If n =1 thenT2=T1W_dot_comp_polytropic= m_dot*R*(T1+460)*ln(P2/P1)*convert(Btu/s,hp) "[hp]"W_dot_comp_2stagePoly = W_dot_comp_polytropic "[hp]"Q_dot_Out_polytropic=W_dot_comp_polytropic*convert(hp,Btu/s) "[Btu/s]"Q_dot_Out_2stagePoly = Q_dot_Out_polytropic*convert(hp,Btu/s) "[Btu/s]"ElseC_P = k*R/(k-1) "[Btu/lbm-R]"T2=(T1+460)*((P2/P1)^((n+1)/n)-460)"[F]"W_dot_comp_polytropic = m_dot*n*R*(T1+460)/(n-1)*((P2/P1)^((n-1)/n) -1)*convert(Btu/s,hp)"[hp]"Q_dot_Out_polytropic=W_dot_comp_polytropic*convert(hp,Btu/s)+m_dot*C_P*(T1-T2)"[Btu/s]"Px=(P1*P2)^0.5T2x=(T1+460)*((Px/P1)^((n+1)/n)-460)"[F]"W_dot_comp_2stagePoly = 2*m_dot*n*R*(T1+460)/(n-1)*((Px/P1)^((n-1)/n) -1)*convert(Btu/s,hp)"[hp]"Q_dot_Out_2stagePoly=W_dot_comp_2stagePoly*convert(hp,Btu/s)+2*m_dot*C_P*(T1-T2x)"[Btu/s]"endifENDR=0.4961[Btu/lbm-R]k=1.667n=1.2P1=14 [psia]T1=70 [F]P2=120 [psia]V_dot = 5 [ft^3/s]P1*V_dot=m_dot*R*(T1+460)*convert(Btu,psia-ft^3)W_dot_comp_isentropic = m_dot*k*R*(T1+460)/(k-1)*((P2/P1)^((k-1)/k) -1)*convert(Btu/s,hp)"[hp]"Q_dot_Out_isentropic = 0"[Btu/s]"Call FuncPoly(m_dot,k, R,T1,P2,P1,n:W_dot_comp_polytropic,W_dot_comp_2stagePoly,Q_dot_Out_polytropic,Q_dot_Out_2stagePoly)W_dot_comp_isothermal= m_dot*R*(T1+460)*ln(P2/P1)*convert(Btu/s,hp)"[hp]"Q_dot_Out_isothermal = W_dot_comp_isothermal*convert(hp,Btu/s)"[Btu/s]"
• 7-59 n Wcomp2StagePoly Wcompisentropic Wcompisothermal Wcomppolytropic [hp] [hp] [hp] [hp] 1 39.37 62.4 39.37 39.37 1.1 41.36 62.4 39.37 43.48 1.2 43.12 62.4 39.37 47.35 1.3 44.68 62.4 39.37 50.97 1.4 46.09 62.4 39.37 54.36 1.5 47.35 62.4 39.37 57.541.667 49.19 62.4 39.37 62.4 65 60 Polytropic Isothermal 55 Isentropic W comp,polytropic [hp] 2StagePoly 50 45 40 35 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 n
• 7-607-98 Nitrogen gas is compressed by a 10-kW compressor from a specified state to a specified pressure. Themass flow rate of nitrogen through the compressor is to be determined for the cases of isentropic,polytropic, isothermal, and two-stage compression.Assumptions 1 Nitrogen is an ideal gas with constant specific heats. 2 The process is reversible. 3 Kineticand potential energy changes are negligible.Properties The gas constant of nitrogen is R = 0.297 kJ/kg.K (Table A-1). The specific heat ratio ofnitrogen is k = 1.4 (Table A-2). 2Analysis (a) Isentropic compression: kRT1 k 1 /k Wcomp,in m P2 P1 1 k 1 N2or, · 10 kW m 1.4 0.297 kJ/kg K 300 K 0.4/1.4 10 kJ/s m 480 kPa 80 kPa 1 1.4 1It yields m 0.048 kg / s 1(b) Polytropic compression with n = 1.3: nRT1 n 1 /n Wcomp,in m P2 P1 1 n 1or, 1.3 0.297 kJ/kg K 300 K 0.3/1.3 10 kJ/s m 480 kPa 80 kPa 1 1.3 1It yields m 0.051 kg / s(c) Isothermal compression: P1 480 kPa Wcomp,in mRT ln 10 kJ/s m 0.297 kJ/kg K 300 K ln P2 80 kPaIt yields m 0.063 kg / s(d) Ideal two-stage compression with intercooling (n = 1.3): In this case, the pressure ratio across eachstage is the same, and its value is determined to be Px P P2 1 80 kPa 480 kPa 196 kPaThe compressor work across each stage is also the same, thus total compressor work is twice thecompression work for a single stage: nRT1 n 1 /n Wcomp,in 2mwcomp,I 2m Px P1 1 n 1or, 1.3 0.297 kJ/kg K 300 K 0.3/1.3 10 kJ/s 2m 196 kPa 80 kPa 1 1.3 1It yields m 0.056 kg/s
• 7-617-99 Water mist is to be sprayed into the air stream in the compressor to cool the air as the waterevaporates and to reduce the compression power. The reduction in the exit temperature of the compressedair and the compressor power saved are to be determined.Assumptions 1 Air is an ideal gas with variable specific heats. 2 The process is reversible. 3 Kinetic andpotential energy changes are negligible. 3 Air is compressed isentropically. 4 Water vaporizes completelybefore leaving the compressor. 4 Air properties can be used for the air-vapor mixture.Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1). The specific heat ratio of air is k =1.4. The inlet enthalpies of water and air are (Tables A-4 and A-17) hw1 = hf@20 C = 83.29 kJ/kg , hfg@20 C = 2453.9 kJ/kg and ha1 = h@300 K =300.19 kJ/kgAnalysis In the case of isentropic operation (thus no cooling or water spray), the exit temperature and thepower input to the compressor are ( k 1) / k (1.4 1) / 1.4 T2 P2 1200 kPa T2 (300 K) = 610.2 K T1 P1 100 kPa kRT1 Wcomp,in m P2 P k 1 / k 1 1 k 1 1.4 0.287 kJ/kg K 300 K 0.4/1.4 (2.1 kg/s) 1200 kPa/100 kPa 1 654.3 kW 1.4 1When water is sprayed, we first need to check the accuracy of theassumption that the water vaporizes completely in the compressor. In the 2limiting case, the compression will be isothermal at the compressor inlet 1200 kPatemperature, and the water will be a saturated vapor. To avoid thecomplexity of dealing with two fluid streams and a gas mixture, we · Wdisregard water in the air stream (other than the mass flow rate), and Heassume air is cooled by an amount equal to the enthalpy change of water. Water 20 C The rate of heat absorption of water as it evaporates at the inlettemperature completely is 100 kPa 300 K Qcooling,max mw h fg @ 20 C (0.2 kg/s)(2453.9 kJ/kg) = 490.8 kW 1The minimum power input to the compressor is P2 1200 kPa Wcomp,in, min mRT ln (2.1 kg/s)(0.287 kJ/kg K)(300 K) ln 449.3 kW P1 100 kPaThis corresponds to maximum cooling from the air since, at constant temperature, h = 0 and thus Qout Win 449.3 kW , which is close to 490.8 kW. Therefore, the assumption that all the water vaporizesis approximately valid. Then the reduction in required power input due to water spray becomes Wcomp,in Wcomp, isentropic Wcomp, isothermal 654.3 449.3 205 kWDiscussion (can be ignored): At constant temperature, h = 0 and thus Qout Win 449.3 kW correspondsto maximum cooling from the air, which is less than 490.8 kW. Therefore, the assumption that all the watervaporizes is only roughly valid. As an alternative, we can assume the compression process to be polytropicand the water to be a saturated vapor at the compressor exit temperature, and disregard the remainingliquid. But in this case there is not a unique solution, and we will have to select either the amount of wateror the exit temperature or the polytropic exponent to obtain a solution. Of course we can also tabulate theresults for different cases, and then make a selection.
• 7-62Sample Analysis: We take the compressor exit temperature to be T2 = 200 C = 473 K. Then, hw2 = hg@200 C = 2792.0 kJ/kg and ha2 = h@473 K = 475.3 kJ/kgThen, ( n 1) / n ( n 1) / n T2 P2 473 K 1200 kPa n 1.224 T1 P1 300 K 100 kPa nRT1 nR Wcomp,in m P2 P1 n 1 / n 1 m (T2 T1 ) n 1 n 1 1.224 0.287 kJ/kg K (2.1 kg/s) (473 300)K 570 kW 1.224 1Energy balance: Wcomp,in Qout m(h2 h1 ) Qout Wcomp,in m(h2 h1 ) 569.7 kW (2.1 kg/s)(475.3 300.19) 202.0 kWNoting that this heat is absorbed by water, the rate at which water evaporates in the compressor becomes Qin, water 202.0 kJ/s Qout,air Qin, water mw (hw2 hw1 ) mw 0.0746 kg/s hw2 hw1 (2792.0 83.29) kJ/kgThen the reductions in the exit temperature and compressor power input become T2 T2,isentropic T2,water cooled 610.2 473 137.2 C Wcomp,in Wcomp,isentropic Wcomp,water cooled 654.3 570 84.3 kWNote that selecting a different compressor exit temperature T2 will result in different values.7-100 A water-injected compressor is used in a gas turbine power plant. It is claimed that the power outputof a gas turbine will increase when water is injected into the compressor because of the increase in themass flow rate of the gas (air + water vapor) through the turbine. This, however, is not necessarily rightsince the compressed air in this case enters the combustor at a low temperature, and thus it absorbs muchmore heat. In fact, the cooling effect will most likely dominate and cause the cyclic efficiency to drop.
• 7-63Isentropic Efficiencies of Steady-Flow Devices7-101C The ideal process for all three devices is the reversible adiabatic (i.e., isentropic) process. Theadiabatic efficiencies of these devices are defined as actual work output insentropic work input actual exit kineticenergy T , C , and N insentropic work output actual work input insentropic exit kinetic energy7-102C No, because the isentropic process is not the model or ideal process for compressors that arecooled intentionally.7-103C Yes. Because the entropy of the fluid must increase during an actual adiabatic process as a resultof irreversibilities. Therefore, the actual exit state has to be on the right-hand side of the isentropic exitstate7-104 Steam enters an adiabatic turbine with an isentropic efficiency of 0.90 at a specified state with aspecified mass flow rate, and leaves at a specified pressure. The turbine exit temperature and power outputof the turbine are to be determined.Assumptions 1 This is a steady-flow process since there is no changewith time. 2 Kinetic and potential energy changes are negligible. 3 P1 = 8 MPaThe device is adiabatic and thus heat transfer is negligible. T1 = 500 CAnalysis (a) From the steam tables (Tables A-4 through A-6), P1 8 MPa h1 3399.5 kJ/kg STEAM T1 500 C s1 6.7266 kJ/kg K TURBINE T = 90% s2s sf 6.7266 0.9441 P2 s 30 kPa x2s 0.8475 s fg 6.8234 s 2s s1 h2 s hf x 2 s h fg 289.27 0.8475 2335.3 2268.3 kJ/kg P2 = 30 kPaFrom the isentropic efficiency relation, h1 h2 a T h2 a h1 T h1 h2 s 3399.5 0.9 3399.5 2268.3 2381.4 kJ/kg h1 h2 sThus, P2 a 30 kPa T2 a Tsat @ 30 kPa 69.09 C h2 a 2381.4 kJ/kg(b) There is only one inlet and one exit, and thus m1 m2 m . We take the actual turbine 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 mh1 Wa,out mh2 (since Q ke pe 0) Wa,out m(h1 h2 )Substituting, Wa,out 3kg/s 3399.5 2381.4 kJ/kg 3054 kW
• 7-647-105 EES Problem 7-104 is reconsidered. The effect of varying the turbine isentropic efficiency from0.75 to 1.0 on both the work done and the exit temperature of the steam are to be investigated, and theresults are to be plotted.Analysis The problem is solved using EES, and the results are tabulated and plotted below."System: control volume for turbine""Property relation: Steam functions""Process: Turbine: Steady state, steady flow, adiabatic, reversible or isentropic""Since we dont know the mass, we write the conservation of energy per unit mass.""Conservation of mass: m_dot[1]= m_dot[2]=m_dot" Steam 700"Knowns:"WorkFluid\$ = Steam_iapws 600m_dot = 3 [kg/s]P[1] = 8000 [kPa] 500 1T[1] = 500 [C]P[2] = 30 [kPa] T [°C] 400"eta_turb = 0.9" 300 8000 kPa"Conservation of Energy - SSSF energy 200balance for turbine -- neglecting the 0.4 0.6 2 0.8change in potential energy, no heat 100 0.2 2transfer:" 30 kPa sh[1]=enthalpy(WorkFluid\$,P=P[1],T=T[1]) 0s[1]=entropy(WorkFluid\$,P=P[1],T=T[1]) 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0T_s[1] = T[1] s [kJ/kg-K]s[2] =s[1]s_s[2] = s[1]h_s[2]=enthalpy(WorkFluid\$,P=P[2],s=s_s[2])T_s[2]=temperature(WorkFluid\$,P=P[2],s=s_s[2])eta_turb = w_turb/w_turb_sh[1] = h[2] + w_turbh[1] = h_s[2] + w_turb_sT[2]=temperature(WorkFluid\$,P=P[2],h=h[2])W_dot_turb = m_dot*w_turb 3400 turb Wturb [kW] 3300 0.75 2545 0.8 2715 3200 0.85 2885 3100 0.9 3054 W turb [kW ] 0.95 3224 3000 1 3394 2900 2800 2700 2600 2500 0.75 0.8 0.85 0.9 0.95 1 turb
• 7-657-106 Steam enters an adiabatic turbine at a specified state, and leaves at a specified state. The mass flowrate of the steam and the isentropic efficiency are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changesare negligible. 3 The device is adiabatic and thus heat transfer is negligible.Analysis (a) From the steam tables (Tables A-4 and A-6), P1 7 MPa h1 3650.6 kJ/kg T1 600 C s1 7.0910 kJ/kg K P2 50 kPa h2 a 2780.2 kJ/kg T2 150 CThere is only one inlet and one exit, and thus m1 m2 m . We take the actual turbine as the system, whichis a control volume since mass crosses the boundary. The energy balance for this steady-flow system canbe expressed in the rate form as 0 (steady) Ein Eout Esystem 0 1 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies Ein Eout 6 MW H2O m(h1 V12 / 2) Wa,out m(h2 + V12 /2) (since Q pe 0) V22 V12 Wa,out m h2 h1 2Substituting, the mass flow rate of the steam is determined to be 2 (140 m/s) 2 (80 m/s) 2 1 kJ/kg 6000 kJ/s m 2780.2 3650.6 2 1000 m 2 /s 2 m 6.95 kg/s(b) The isentropic exit enthalpy of the steam and the power output of the isentropic turbine are s 2s sf7.0910 1.0912 P2 s 50 kPa x 2s 0.9228 s fg 6.5019 s 2s s1 h2 s hf x 2 s h fg 340.54 0.9228 2304.7 2467.3 kJ/kgand Ws, out m h2 s h1 V22 V12 / 2 (140 m/s) 2 (80 m/s) 2 1 kJ/kg Ws, out 6.95 kg/s 2467.3 3650.6 2 1000 m 2 /s 2 8174 kWThen the isentropic efficiency of the turbine becomes Wa 6000 kW T 0.734 73.4% Ws 8174 kW
• 7-667-107 Argon enters an adiabatic turbine at a specified state with a specified mass flow rate, and leaves at aspecified pressure. The isentropic efficiency of the turbine 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. 4 Argon is anideal gas with constant specific heats.Properties The specific heat ratio of argon is k = 1.667. The constant pressure specific heat of argon is cp =0.5203 kJ/kg.K (Table A-2).Analysis There is only one inlet and one exit, and thus m1 m2 m . We take the isentropic turbine as thesystem, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as Ein Eout 1 mh1 Ws ,out mh2 s (since Q ke pe 0) Ws, out m(h1 h2 s ) 370 kW ArFrom the isentropic relations, T k 1 /k 0.667/1.667 P2 s 200 kPa T2 s T1 1073 K 479 K P1 1500 kPaThen the power output of the isentropic turbine becomes 2 Ws, out mc p T1 T2 s 80/60 kg/min 0.5203 kJ/kg K 1073 479 412.1 kWThen the isentropic efficiency of the turbine is determined from Wa 370 kW T 0.898 89.8% Ws 412.1 kW7-108E Combustion gases enter an adiabatic gas turbine with an isentropic efficiency of 82% at a specifiedstate, and leave at a specified pressure. The work output of the turbine 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. 4 Combustiongases can be treated as air that is an ideal gas with variable specific heats.Analysis From the air table and isentropic relations, 1 h1 = 504.71 Btu / lbm T1 2000 R Pr1 174.0 P2 60 psia AIR Pr2 Pr 174.0 87.0 h2s 417.3 Btu/lbm T = 82% P 1 1 120 psiaThere is only one inlet and one exit, and thus m1 m2 m . We take theactual turbine as the system, which is a control volume since mass crosses the 2boundary. The energy balance for this steady-flow system can be expressedas Ein Eout mh1 Wa,out mh2 (since Q ke pe 0) Wa,out m(h1 h2 )Noting that wa = Tws, the work output of the turbine per unit mass is determined from wa 0.82 504.71 417.3 Btu/lbm 71.7 Btu/lbm
• 7-677-109 [Also solved by EES on enclosed CD] Refrigerant-134a enters an adiabatic compressor with anisentropic efficiency of 0.80 at a specified state with a specified volume flow rate, and leaves at a specifiedpressure. The compressor exit temperature and power input to the compressor are 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 (a) From the refrigerant tables (Tables A-11E through A-13E), 2 h1 hg @120 kPa 236.97 kJ/kg P1 120 kPa s1 s g @120 kPa 0.94779 kJ/kg K sat. vapor R-134a v1 v g @120 kPa 0.16212 m3/kg C = 80% P2 1 MPa h2 s 281.21 kJ/kg s2 s s1 0.3 m3/minFrom the isentropic efficiency relation, 1 h2 s h1 C h2 a h1 h2 s h1 / C 236.97 281.21 236.97 /0.80 292.26 kJ/kg h2 a h1Thus, P2 a 1 MPa T2 a 58.9 C h2 a 292.26 kJ/kg(b) The mass flow rate of the refrigerant is determined from V1 0.3/60 m 3 /s m 0.0308 kg/s v1 0.16212 m 3 /kgThere is only one inlet and one exit, and thus m1 m2 m . We take the actual compressor as the system,which is a control volume since mass crosses the boundary. The energy balance for this steady-flow systemcan be expressed 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 Wa,in mh1 mh2 (since Q ke pe 0) Wa,in m(h2 h1 )Substituting, the power input to the compressor becomes, Wa,in 0.0308 kg/s 292.26 236.97 kJ/kg 1.70 kW
• 7-687-110 EES Problem 7-109 is reconsidered. The problem is to be solved by considering the kinetic energyand by assuming an inlet-to-exit area ratio of 1.5 for the compressor when the compressor exit pipe insidediameter is 2 cm.Analysis The problem is solved using EES, and the solution is given below."Input Data from diagram window"{P[1] = 120 "kPa"P[2] = 1000 "kPa"Vol_dot_1 = 0.3 "m^3/min"Eta_c = 0.80 "Compressor adiabatic efficiency"A_ratio = 1.5d_2 = 2/100 "m"}"System: Control volume containing the compressor, see the diagram window.Property Relation: Use the real fluid properties for R134a.Process: Steady-state, steady-flow, adiabatic process."Fluid\$=R134a"Property Data for state 1"T[1]=temperature(Fluid\$,P=P[1],x=1)"Real fluid equ. at the sat. vapor state"h[1]=enthalpy(Fluid\$, P=P[1], x=1)"Real fluid equ. at the sat. vapor state"s[1]=entropy(Fluid\$, P=P[1], x=1)"Real fluid equ. at the sat. vapor state"v[1]=volume(Fluid\$, P=P[1], x=1)"Real fluid equ. at the sat. vapor state""Property Data for state 2"s_s[1]=s[1]; T_s[1]=T[1] "needed for plot"s_s[2]=s[1] "for the ideal, isentropic process across the compressor"h_s[2]=ENTHALPY(Fluid\$, P=P[2], s=s_s[2])"Enthalpy 2 at the isentropic state 2s andpressure P[2]"T_s[2]=Temperature(Fluid\$, P=P[2], s=s_s[2])"Temperature of ideal state - needed only forplot.""Steady-state, steady-flow conservation of mass" m_dot_1 = m_dot_2m_dot_1 = Vol_dot_1/(v[1]*60)Vol_dot_1/v[1]=Vol_dot_2/v[2]Vel[2]=Vol_dot_2/(A[2]*60)A[2] = pi*(d_2)^2/4A_ratio*Vel[1]/v[1] = Vel[2]/v[2] "Mass flow rate: = A*Vel/v, A_ratio = A[1]/A[2]"A_ratio=A[1]/A[2]"Steady-state, steady-flow conservation of energy, adiabatic compressor, see diagramwindow"m_dot_1*(h[1]+(Vel[1])^2/(2*1000)) + W_dot_c= m_dot_2*(h[2]+(Vel[2])^2/(2*1000))"Definition of the compressor adiabatic efficiency, Eta_c=W_isen/W_act"Eta_c = (h_s[2]-h[1])/(h[2]-h[1])"Knowing h[2], the other properties at state 2 can be found."v[2]=volume(Fluid\$, P=P[2], h=h[2])"v[2] is found at the actual state 2, knowing P and h."T[2]=temperature(Fluid\$, P=P[2],h=h[2])"Real fluid equ. for T at the known outlet h and P."s[2]=entropy(Fluid\$, P=P[2], h=h[2]) "Real fluid equ. at the known outlet h and P."T_exit=T[2]"Neglecting the kinetic energies, the work is:"m_dot_1*h[1] + W_dot_c_noke= m_dot_2*h[2]
• 7-69SOLUTIONA[1]=0.0004712 [m^2] s_s[1]=0.9478 [kJ/kg-K]A[2]=0.0003142 [m^2] s_s[2]=0.9478 [kJ/kg-K]A_ratio=1.5 T[1]=-22.32 [C]d_2=0.02 [m] T[2]=58.94 [C]Eta_c=0.8 T_exit=58.94 [C]Fluid\$=R134a T_s[1]=-22.32 [C]h[1]=237 [kJ/kg] T_s[2]=48.58 [C]h[2]=292.3 [kJ/kg] Vol_dot_1=0.3 [m^3 /min]h_s[2]=281.2 [kJ/kg] Vol_dot_2=0.04244 [m^3 /min]m_dot_1=0.03084 [kg/s] v[1]=0.1621 [m^3/kg]m_dot_2=0.03084 [kg/s] v[2]=0.02294 [m^3/kg]P[1]=120.0 [kPa] Vel[1]=10.61 [m/s]P[2]=1000.0 [kPa] Vel[2]=2.252 [m/s]s[1]=0.9478 [kJ/kg-K] W_dot_c=1.704 [kW]s[2]=0.9816 [kJ/kg-K] W_dot_c_noke=1.706 [kW] R134a 125 T-s diagram for real and ideal com pressor 100 Ideal Compressor Real Com pressor 75 Tem perature [C] 50 1000 kPa 25 0 -25 120 kPa -50 0.0 0.2 0.4 0.6 0.8 1.0 1.2 Entropy [kJ/kg-K]
• 7-707-111 Air enters an adiabatic compressor with an isentropic efficiency of 84% at a specified state, andleaves at a specified temperature. The exit pressure of air and the power input to the compressor are to bedetermined.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. 4 Air is anideal gas with variable specific heats.Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1) 2Analysis (a) From the air table (Table A-17), T1 290 K h1 290.16 kJ/kg, Pr1 1.2311 AIR T2 530 K h2 a 533.98 kJ/kg C = 84% h2 s h1From the isentropic efficiency relation , C h2 a h1 2.4 m3/s h2 s h1 C h2 a h1 1 290.16 0.84 533.98 290.16 495.0 kJ/kg Pr2 7.951Then from the isentropic relation , P2 Pr2 Pr2 7.951 P2 P1 100 kPa 646 kPa P1 Pr1 Pr1 1.2311(b) There is only one inlet and one exit, and thus m1 m2 m . We take the actual compressor as thesystem, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed 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 Wa,in mh1 mh2 (since Q ke pe 0) Wa,in m(h2 h1 ) P1V 1 (100 kPa )(2.4 m 3 /s)where m 2.884 kg/s RT1 (0.287 kPa m 3 /kg K )(290 K )Then the power input to the compressor is determined to be Wa,in (2.884 kg/s)(533.98 290.16) kJ/kg 703 kW
• 7-717-112 Air is compressed by an adiabatic compressor from a specified state to another specified state. Theisentropic efficiency of the compressor and the exit temperature of air for the isentropic case are to bedetermined.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. 4 Air is anideal gas with variable specific heats.Analysis (a) From the air table (Table A-17), 2 T1 300 K h1 300.19 kJ / kg, Pr1 1.386 T2 550 K h2 a 554.74 kJ / kgFrom the isentropic relation, AIR P2 600 kPa Pr2 Pr 1.386 8.754 h2 s 508.72 kJ/kg P 1 1 95 kPaThen the isentropic efficiency becomes 1 h2 s h1 508.72 30019 . 0.819 81.9% C h2a h1 554.74 30019 .(b) If the process were isentropic, the exit temperature would be h2 s 508.72 kJ / kg T2 s 505.5 K
• 7-727-113E Argon enters an adiabatic compressor with an isentropic efficiency of 80% at a specified state, andleaves at a specified pressure. The exit temperature of argon and the work input to the compressor are to bedetermined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changesare negligible. 3 The device is adiabatic and thus heat transfer is negligible. 4 Argon is an ideal gas withconstant specific heats.Properties The specific heat ratio of argon is k = 1.667. The constant 2pressure specific heat of argon is cp = 0.1253 Btu/lbm.R (Table A-2E).Analysis (a) The isentropic exit temperature T2s is determined from k 1 /k 0.667/1.667 Ar P2 s 200 psia = 80% T2 s T1 550 R 1381.9 R C P1 20 psiaThe actual kinetic energy change during this process is 2 2 V 22 V12 240 ft/s 60 ft/s 1 Btu/lbm ke a 1.08 Btu/lbm 1 2 2 25,037 ft 2 /s 2The effect of kinetic energy on isentropic efficiency is very small. Therefore, we can take the kineticenergy changes for the actual and isentropic cases to be same in efficiency calculations. From theisentropic efficiency relation, including the effect of kinetic energy, ws (h2 s h1 ) ke c p T2 s T1 ke s 0.1253 1381.9 550 1.08 C 0.8 wa ( h2 a h1 ) ke c p T2 a T1 ke a 0.1253 T2 a 550 1.08It yields T2a = 1592 R(b) There is only one inlet and one exit, and thus m1 m2 m . We take the actual compressor as thesystem, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed 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 Wa,in m(h1 V12 / 2) m(h2 + V22 /2) (since Q pe 0) V22 V12 Wa,in m h2 h1 wa,in h2 h1 ke 2Substituting, the work input to the compressor is determined to be wa,in 0.1253 Btu/lbm R 1592 550 R 1.08 Btu/lbm 131.6 Btu/lbm
• 7-737-114 CO2 gas is compressed by an adiabatic compressor from a specified state to another specified state.The isentropic efficiency of 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. 4 CO2 is anideal gas with constant specific heats.Properties At the average temperature of (300 + 450)/2 = 375 K, the constant pressure specificheat and the specific heat ratio of CO2 are k = 1.260 and cp = 0.917 kJ/kg.K (Table A-2). 2Analysis The isentropic exit temperature T2s is k 1 /k 0.260/1.260 P2 s 600 kPa T2 s T1 300 K 434.2 K CO2 P1 100 kPa 1.8 kg/sFrom the isentropic efficiency relation, w s h2 s h1 c p T2 s T1 T2 s T1 434.2 300 C 0.895 89.5% wa h2 a h1 c p T2 a T1 T2 a T1 450 300 17-115E Air is accelerated in a 90% efficient adiabatic nozzle from low velocity to a specified velocity. Theexit temperature and pressure of the air are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changesare negligible. 3 The device is adiabatic and thus heat transfer is negligible. 4 Air is an ideal gas withvariable specific heats.Analysis From the air table (Table A-17E), T1 1480 R h1 363.89 Btu / lbm, Pr1 53.04There is only one inlet and one exit, and thus m1 m2 m . We take the nozzle as the system, which is acontrol volume since mass crosses the boundary. The energy balance for this steady-flow system can beexpressed 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 AIR Ein Eout 1 2 N = 90% m(h1 V12 / 2) m(h2 + V22 /2) (since W Q pe 0) 0 V22 V12 h2 h1 2Substituting, the exit temperature of air is determined to be 2 800 ft/s 0 1 Btu/lbm h2 363.89 kJ/kg 351.11 Btu/lbm 2 25,037 ft 2 /s 2From the air table we read T2a = 1431.3 R h2 a h1From the isentropic efficiency relation N , h2 s h1 h2 s h1 h2 a h1 / N 363.89 351.11 363.89 / 0.90 349.69 Btu/lbm Pr2 46.04Then the exit pressure is determined from the isentropic relation to be P2 Pr2 Pr2 46.04 P2 P1 60 psia 52.1 psia P1 Pr1 Pr1 53.04
• 7-747-116E EES Problem 7-115E is reconsidered. The effect of varying the nozzle isentropic efficiency from0.8 to 1.0 on the exit temperature and pressure of the air is to be investigated, and the results are to beplotted.Analysis The problem is solved using EES, and the results are tabulated and plotted below."Knowns:"WorkFluid\$ = AirP[1] = 60 [psia]T[1] = 1020 [F]Vel[2] = 800 [ft/s]Vel[1] = 0 [ft/s]eta_nozzle = 0.9"Conservation of Energy - SSSF energy balance for turbine -- neglecting the change in potentialenergy, no heat transfer:"h[1]=enthalpy(WorkFluid\$,T=T[1])s[1]=entropy(WorkFluid\$,P=P[1],T=T[1])T_s[1] = T[1]s[2] =s[1]s_s[2] = s[1]h_s[2]=enthalpy(WorkFluid\$,T=T_s[2])T_s[2]=temperature(WorkFluid\$,P=P[2],s=s_s[2])eta_nozzle = ke[2]/ke_s[2]ke[1] = Vel[1]^2/2ke[2]=Vel[2]^2/2h[1]+ke[1]*convert(ft^2/s^2,Btu/lbm) = h[2] + ke[2]*convert(ft^2/s^2,Btu/lbm)h[1] +ke[1]*convert(ft^2/s^2,Btu/lbm) = h_s[2] + ke_s[2]*convert(ft^2/s^2,Btu/lbm)T[2]=temperature(WorkFluid\$,h=h[2])P_2_answer = P[2]T_2_answer = T[2] P2 T2 Ts,2 980 nozzle [psia [F [F] 0.8 51.09 971.4 959.2 0.85 51.58 971.4 962.8 0.9 52.03 971.4 966 970 0.95 52.42 971.4 968.8 1 52.79 971.4 971.4 T s[2] 960 T 2 T s[2] 950 0.8 0.84 0.88 0.92 0.96 1 nozzle
• 7-757-117 Hot combustion gases are accelerated in a 92% efficient adiabatic nozzle from low velocity to aspecified velocity. The exit velocity and the exit temperature are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changesare negligible. 3 The device is adiabatic and thus heat transfer is negligible. 4 Combustion gases can betreated as air that is an ideal gas with variable specific heats.Analysis From the air table (Table A-17),T1 1020 K h1 1068.89 kJ/kg, Pr1 123.4 P1 = 260 kPa AIR T1 = 747 C P2 = 85 kPa N = 92%From the isentropic relation , V1 = 80 m/s P2 85 kPaPr2 Pr 123.4 40.34 h2 s 783.92 kJ/kg P 1 1 260 kPaThere is only one inlet and one exit, and thus m1 m2 m . We take the nozzle as the system, which is acontrol volume since mass crosses the boundary. The energy balance for this steady-flow system for theisentropic process can be expressed 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 m(h1 V12 / 2) m(h2 s + V 22s /2) (since W Q pe 0) V 22s V12 h2 s h1 2Then the isentropic exit velocity becomes 2 1000 m 2 /s 2 V2s V12 2 h1 h2 s 80 m/s 2 1068.89 783.92 kJ/kg 759.2 m/s 1 kJ/kgTherefore, V2 a N V2 s 0.92 759.2 m/s 728.2 m/sThe exit temperature of air is determined from the steady-flow energy equation, 2 2 728.2 m/s 80 m/s 1 kJ/kg h2 a 1068.89 kJ/kg 806.95 kJ/kg 2 1000 m 2 /s 2From the air table we read T2a = 786.3 K
• 7-107Special Topic: Reducing the Cost of Compressed Air7-150 The total installed power of compressed air systems in the US is estimated to be about 20 millionhorsepower. The amount of energy and money that will be saved per year if the energy consumed bycompressors is reduced by 5 percent is to be determined.Assumptions 1 The compressors operate at full load during one-third of the time on average, and are shutdown the rest of the time. 2 The average motor efficiency is 85 percent.Analysis The electrical energy consumed by compressors per year is Energy consumed = (Power rating)(Load factor)(Annual Operating Hours)/Motor efficiency = (20 106 hp)(0.746 kW/hp)(1/3)(365 24 hours/year)/0.85 = 5.125 1010 kWh/year 2Then the energy and cost savings corresponding to a 5% reduction inenergy use for compressed air become Energy Savings = (Energy consumed)(Fraction saved) W=20 106 hp Air = (5.125 1010 kWh)(0.05) Compressor = 2.563 109 kWh/year Cost Savings = (Energy savings)(Unit cost of energy) = (2.563 109 kWh/year)(\$0.07/kWh) 1 = \$0.179 109 /yearTherefore, reducing the energy usage of compressors by 5% will save \$179 million a year.7-151 The total energy used to compress air in the US is estimated to be 0.5 1015 kJ per year. About 20%of the compressed air is estimated to be lost by air leaks. The amount and cost of electricity wasted peryear due to air leaks is to be determined. 2Assumptions About 20% of the compressed air is lost by air leaks.Analysis The electrical energy and money wasted by air leaks are Energy wasted = (Energy consumed)(Fraction wasted) W=0.5 1015 kJ Air = (0.5 1015 kJ)(1 kWh/3600 kJ)(0.20) Compressor = 27.78 109 kWh/year Money wasted = (Energy wasted)(Unit cost of energy) = (27.78 109 kWh/year)(\$0.07/kWh) 1
• 7-108 = \$1.945 109 /yearTherefore, air leaks are costing almost \$2 billion a year in electricity costs. The environment also suffersfrom this because of the pollution associated with the generation of this much electricity.
• 7-1097-152 The compressed air requirements of a plant is being met by a 125 hp compressor that compresses airfrom 101.3 kPa to 900 kPa. The amount of energy and money saved by reducing the pressure setting ofcompressed air to 750 kPa is to be determined.Assumptions 1 Air is an ideal gas with constant specific heats. 2 Kinetic and potential energy changes arenegligible. 3 The load factor of the compressor is given to be 0.75. 4 The pressures given are absolutepressure rather than gage pressure.Properties The specific heat ratio of air is k = 1.4 (Table A-2).Analysis The electrical energy consumed by this compressor per year is Energy consumed = (Power rating)(Load factor)(Annual Operating Hours)/Motor efficiency = (125 hp)(0.746 kW/hp)(0.75)(3500 hours/year)/0.88 900 kPa 2 = 278,160 kWh/yearThe fraction of energy saved as a result of reducing the W=125 hppressure setting of the compressor is Air Compressor ( P2, reduced / P )( k 1 1) / k 1 Power Reduction Factor 1 ( k 1) / k ( P2 / P ) 1 1 (1.4 1) / 1, 4 (750 / 101.3) 1 1 101 kPa (900 / 101.3)(1.4 1) / 1, 4 1 1 15 C 0.1093That is, reducing the pressure setting will result in about 11 percent savings from the energy consumed bythe compressor and the associated cost. Therefore, the energy and cost savings in this case become Energy Savings = (Energy consumed)(Power reduction factor) = (278,160 kWh/year)(0.1093) = 30,410 kWh/year Cost Savings = (Energy savings)(Unit cost of energy) = (30,410 kWh/year)(\$0.085/kWh) = \$2585/yearTherefore, reducing the pressure setting by 150 kPa will result in annual savings of 30.410 kWh that isworth \$2585 in this case.Discussion Some applications require very low pressure compressed air. In such cases the need can be metby a blower instead of a compressor. Considerable energy can be saved in this manner, since a blowerrequires a small fraction of the power needed by a compressor for a specified mass flow rate.
• 7-1107-153 A 150 hp compressor in an industrial facility is housed inside the production area where the averagetemperature during operating hours is 25 C. The amounts of energy and money saved as a result ofdrawing cooler outside air to the compressor instead of using the inside air are to be determined.Assumptions 1 Air is an ideal gas with constant specific heats. 2 Kinetic and potential energy changes arenegligible.Analysis The electrical energy consumed by this compressor per year is 2Energy consumed = (Power rating)(Load factor)(Annual Operating Hours)/Motor efficiency = (150 hp)(0.746 kW/hp)(0.85)(4500 hours/year)/0.9 = 475,384 kWh/year W=150 hp T0 = 10 C AirAlso, Compressor Cost of Energy = (Energy consumed)(Unit cost of energy) = (475,384 kWh/year)(\$0.07/kWh) = \$33,277/year 101 kPa 25 CThe fraction of energy saved as a result of drawing in cooler outside air is 1 Toutside 10 273 Power Reduction Factor 1 1 0.0503 Tinside 25 273That is, drawing in air which is 15 C cooler will result in 5.03 percent savings from the energy consumedby the compressor and the associated cost. Therefore, the energy and cost savings in this case become Energy Savings = (Energy consumed)(Power reduction factor) = (475,384 kWh/year)(0.0503) = 23,929 kWh/year Cost Savings = (Energy savings)(Unit cost of energy) = (23,929 kWh/year)(\$0.07/kWh) = \$1675/yearTherefore, drawing air in from the outside will result in annual savings of 23,929 kWh, which is worth\$1675 in this case.Discussion The price of a typical 150 hp compressor is much lower than \$50,000. Therefore, it isinteresting to note that the cost of energy a compressor uses a year may be more than the cost of thecompressor itself. The implementation of this measure requires the installation of an ordinary sheet metal or PVCduct from the compressor intake to the outside. The installation cost associated with this measure isrelatively low, and the pressure drop in the duct in most cases is negligible. About half of themanufacturing facilities we have visited, especially the newer ones, have the duct from the compressorintake to the outside in place, and they are already taking advantage of the savings associated with thismeasure.
• 7-1117-154 The compressed air requirements of the facility during 60 percent of the time can be met by a 25 hpreciprocating compressor instead of the existing 100 hp compressor. The amounts of energy and moneysaved as a result of switching to the 25 hp compressor during 60 percent of the time are to be determined.Analysis Noting that 1 hp = 0.746 kW, the electrical energy consumed by each compressor per year isdetermined from(Energy consumed)Large = (Power)(Hours)[(LFxTF/ motor)Unloaded + (LFxTF/ motor)Loaded] = (100 hp)(0.746 kW/hp)(3800 hours/year)[0.35 0.6/0.82+0.90 0.4/0.9] = 185,990 kWh/year(Energy consumed)Small =(Power)(Hours)[(LFxTF/ motor)Unloaded + (LFxTF/ motor)Loaded] = (25 hp)(0.746 kW/hp)(3800 hours/year)[0.0 0.15+0.95 0.85]/0.88 2 = 65,031 kWh/yearTherefore, the energy and cost savings in this case become W=100 hp Energy Savings = (Energy consumed)Large- (Energy consumed)Small Air Compressor = 185,990 - 65,031 kWh/year = 120,959 kWh/year Cost Savings = (Energy savings)(Unit cost of energy) = (120,959 kWh/year)(\$0.075/kWh) 1 = \$9,072/yearDiscussion Note that utilizing a small compressor during the times of reduced compressed air requirementsand shutting down the large compressor will result in annual savings of 120,959 kWh, which is worth\$9,072 in this case.7-155 A facility stops production for one hour every day, including weekends, for lunch break, but the 125hp compressor is kept operating. If the compressor consumes 35 percent of the rated power when idling,the amounts of energy and money saved per year as a result of turning the compressor off during lunchbreak are to be determined.Analysis It seems like the compressor in this facility is kept on unnecessarily for one hour a day and thus365 hours a year, and the idle factor is 0.35. Then the energy and cost savings associated with turning thecompressor off during lunch break are determined to be Energy Savings = (Power Rating)(Turned Off Hours)(Idle Factor)/ motor 2 = (125 hp)(0.746 kW/hp)(365 hours/year)(0.35)/0.84 = 14,182 kWh/year Cost Savings = (Energy savings)(Unit cost of energy) W=125 hp Air = (14,182 kWh/year)(\$0.09/kWh) Compressor = \$1,276/year
• 7-112Discussion Note that the simple practice of turning the compressor offduring lunch break will save this facility \$1,276 a year in energy costs.There are also side benefits such as extending the life of the motor andthe compressor, and reducing the maintenance costs.
• 7-1137-156 It is determined that 40 percent of the energy input to the compressor is removed from thecompressed air as heat in the aftercooler with a refrigeration unit whose COP is 3.5. The amounts of theenergy and money saved per year as a result of cooling the compressed air before it enters the refrigerateddryer are to be determined.Assumptions The compressor operates at full load when operating.Analysis Noting that 40 percent of the energy input to the compressor is removed by the aftercooler, therate of heat removal from the compressed air in the aftercooler under full load conditions is Qaftercooling (Rated Power of Compressor)(Load Factor)(Aftercooling Fraction) = (150 hp)(0.746 kW/hp)(1.0)(0.4) = 44.76 kW After- QaftercoolingThe compressor is said to operate at full load for 1600 hours a cooleryear, and the COP of the refrigeration unit is 3.5. Then the energyand cost savings associated with this measure become Energy Savings = ( Qaftercooling )(Annual Operating Hours)/COP W=150 hp Air = (44.76 kW)(1600 hours/year)/3.5 Compressor = 20,462 kWh/year Cost Savings = (Energy savings)(Unit cost of energy saved) = (20,462 kWh/year)(\$0.06/kWh) = \$1227/yearDiscussion Note that the aftercooler will save this facility 20,462 kWh of electrical energy worth \$1227per year. The actual savings will be less than indicated above since we have not considered the powerconsumed by the fans and/or pumps of the aftercooler. However, if the heat removed by the aftercooler isutilized for some useful purpose such as space heating or process heating, then the actual savings will bemuch more.7-157 The motor of a 150 hp compressor is burned out and is to be replaced by either a 93% efficientstandard motor or a 96.2% efficient high efficiency motor. It is to be determined if the savings from thehigh efficiency motor justify the price differential.Assumptions 1 The compressor operates at full load when operating. 2 The life of the motors is 10 years. 3There are no rebates involved. 4 The price of electricity remains constant.Analysis The energy and cost savings associated with the installation of the high efficiency motor in thiscase are determined to beEnergy Savings = (Power Rating)(Operating Hours)(Load Factor)(1/ standard - 1/ efficient) = (150 hp)(0.746 kW/hp)(4,368 hours/year)(1.0)(1/0.930 - 1/0.962) = 17,483 kWh/year 150 hp AirCost Savings = (Energy savings)(Unit cost of energy) Compressor = (17,483 kWh/year)(\$0.075/kWh) = \$1311/yearThe additional cost of the energy efficient motor is
• 7-114 Cost Differential = \$10,942 - \$9,031 = \$1,911Discussion The money saved by the high efficiency motor will pay for this cost difference in\$1,911/\$1311 = 1.5 years, and will continue saving the facility money for the rest of the 10 years of itslifetime. Therefore, the use of the high efficiency motor is recommended in this case even in the absence ofany incentives from the local utility company.
• 7-1157-158 The compressor of a facility is being cooled by air in a heat-exchanger. This air is to be used to heatthe facility in winter. The amount of money that will be saved by diverting the compressor waste heat intothe facility during the heating season is to be determined.Assumptions The compressor operates at full load when operating.Analysis Assuming operation at sea level and taking the density of air to be 1.2 kg/m3, the mass flow rateof air through the liquid-to-air heat exchanger is determined to be Mass flow rate of air = (Density of air)(Average velocity)(Flow area) = (1.2 kg/m3)(3 m/s)(1.0 m2) = 3.6 kg/s = 12,960 kg/hNoting that the temperature rise of air is 32 C, the rate at which heat can berecovered (or the rate at which heat is transferred to air) isRate of Heat Recovery = (Mass flow rate of air)(Specific heat of air)(Temperature rise) Hot Compressed = (12,960 kg/h)(1.0 kJ/kg. C)(32 C) air = 414,720 kJ/h AirThe number of operating hours of this compressor during the 20 C 52 Cheating season is 3 m/sOperating hours = (20 hours/day)(5 days/week)(26 weeks/year) = 2600 hours/yearThen the annual energy and cost savings becomeEnergy Savings = (Rate of Heat Recovery)(Annual Operating Hours)/Efficiency = (414,720 kJ/h)(2600 hours/year)/0.8 = 1,347,840,000 kJ/year = 12,776 therms/year Cost Savings = (Energy savings)(Unit cost of energy saved) = (12,776 therms/year)(\$1.0/therm) = \$12,776/yearTherefore, utilizing the waste heat from the compressor will save \$12,776 per year from the heating costs.Discussion The implementation of this measure requires the installation of an ordinary sheet metal ductfrom the outlet of the heat exchanger into the building. The installation cost associated with this measure isrelatively low. A few of the manufacturing facilities we have visited already have this conservation systemin place. A damper is used to direct the air into the building in winter and to the ambient in summer. Combined compressor/heat-recovery systems are available in the market for both air-cooled(greater than 50 hp) and water cooled (greater than 125 hp) systems.
• 7-1167-159 The compressed air lines in a facility are maintained at a gage pressure of 850 kPa at a locationwhere the atmospheric pressure is 85.6 kPa. There is a 5-mm diameter hole on the compressed air line.The energy and money saved per year by sealing the hole on the compressed air line.Assumptions 1 Air is an ideal gas with constant specific heats. 2 Kinetic and potential energy changes arenegligible.Properties The gas constant of air is R = 0.287 kJ/kg.K. The specific heat ratio of air is k = 1.4 (Table A-2).Analysis Disregarding any pressure losses and noting that the absolute pressure is the sum of the gagepressure and the atmospheric pressure, the work needed to compress a unit mass of air at 15 C from theatmospheric pressure of 85.6 kPa to 850+85.6 = 935.6 kPa is determined to be ( k 1) / k kRT1 P2 wcomp, in 1 comp ( k 1) P1 (1.4 1) / 1.4 (1.4)(0.287 kJ/kg.K)(288 K) 935.6 kPa 1 (0.8)(1.4 1) 85.6 kPa 354.5 kJ/kg Patm = 85.6 kPa, 15 CThe cross-sectional area of the 5-mm diameter hole is Air leak A D2 / 4 (5 10 3 m) 2 / 4 19.63 10 6 m2Noting that the line conditions are T0 = 298 Kand P0 = 935.6 kPa, the mass flow rate of the airleaking through the hole is determined to be Compressed air line 1 /( k 1) 850 kPa, 25 C 2 P0 2 m air C loss A kR T0 k 1 RT0 k 1 1 /(1.4 1) 2 935.6 kPa 6 (0.65) (19.63 10 m2 ) 1.4 1 (0.287 kPa.m 3 / kg.K)(298 K) 1000 m 2 / s 2 2 (1.4)(0.287 kJ/kg.K) (298 K) 1 kJ/kg 1.4 1 0.02795 kg/sThen the power wasted by the leaking compressed air becomes Power wasted mair wcomp,in (0.02795 kg / s)(354.5 kJ / kg) 9.91 kWNoting that the compressor operates 4200 hours a year and the motor efficiency is 0.93, the annual energyand cost savings resulting from repairing this leak are determined to be Energy Savings = (Power wasted)(Annual operating hours)/Motor efficiency = (9.91 kW)(4200 hours/year)/0.93 = 44,755 kWh/year Cost Savings = (Energy savings)(Unit cost of energy) = (44,755 kWh/year)(\$0.07/kWh) = \$3133/year
• 7-117Therefore, the facility will save 44,755 kWh of electricity that is worth \$3133 a year when this air leak issealed.
• 7-118Review Problems7-160 A piston-cylinder device contains steam that undergoes a reversible thermodynamic cycle composedof three processes. The work and heat transfer for each process and for the entire cycle are to bedetermined.Assumptions 1 All processes are reversible. 2 Kinetic and potential energy changes are negligible.Analysis The properties of the steam at various statesare (Tables A-4 through A-6) P = const. u1 2884.5 kJ/kg P1 400 kPa 3 v1 0.71396 m 3 /kg 1 T1 350 C s = const. s1 7.7399 kJ/kg.K P2 150 kPa u 2 2888.0 kJ/kg T2 350 C s2 8.1983 kJ/kg.K T = const. 2 P3 400 kPa u3 3132.9 kJ/kg s3 s2 8.1983 kJ/kg.K v 3 0.89148 m 3 /kgThe mass of the steam in the cylinder and the volume at state 3 are V1 0.3 m 3 m 0.4202 kg v1 0.71396 m 3 /kg V3 mV 3 (0.4202 kg)(0.89148 m 3 /kg) 0.3746 m 3Process 1-2: Isothermal expansion (T2 = T1) S1 2 m ( s2 s1 ) (0.4202 kg)(8.1983 7.7399)kJ/kg.K 0.1926 kJ/kg.K Qin,1 2 T1 S1 2 (350 273 K)(0.1926 kJ/K) 120 kJ Wout,1 2 Qin,1 2 m(u 2 u1 ) 120 kJ (0.4202 kg)(2888.0 2884.5)kJ/kg 118.5 kJProcess 2-3: Isentropic (reversible-adiabatic) compression (s3 = s2) Win,2 3 m(u3 u2 ) (0.4202 kg)(3132.9 - 2888.0)kJ/kg 102.9 kJ Q2-3 = 0 kJProcess 3-1: Constant pressure compression process (P1 = P3) Win,3 1 P3 (V3 V1 ) (400 kPa)(0.3746 - 0.3) 29.8 kJ Qout,3 1 Win,3 1 m(u1 u 3 ) 29.8 kJ - (0.4202 kg)(2884.5 - 3132.9) 134.2 kJThe net work and net heat transfer are Wnet,in Win,3 1 Win,2 3 Wout,1 2 29.8 102.9 118.5 14.2 kJ Qnet,in Qin,1 2 Qout,3 1 120 134.2 14.2 kJ Qnet,out 14.2 kJDiscussion The results are not surprising since for a cycle, the net work and heat transfers must be equal toeach other.
• 7-1197-161 The work input and the entropy generation are to be determined for the compression of saturatedliquid water in a pump and that of saturated vapor in a compressor.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3Heat transfer to or from the fluid is zero.Analysis Pump Analysis: (Properties are obtained from EES) P1 100 kPa h1 417.51 kJ/kg P2 1 MPa 1 MPa h2 s 418.45 kJ/kg x1 0 (sat. liq.) s1 1.3028 kJ/kg.K s2 s1 h2 s h1 418.45 417.51 h2 h1 417.51 418.61 kJ/kg 100 kPa pump P 0.85 P2 1 MPa s2 1.3032 kJ/kg.K h2 418.61 kJ/kg wP h2 h1 418.61 417.51 1.10 kJ/kg s gen,P s2 s1 1.3032 1.3028 0.0004 kJ/kg.KCompressor Analysis: P 100 kPa 1 h1 2675.0 kJ/kg P2 1 MPa h2 s 3193.6 kJ/kg x1 1 (sat. vap.) s1 7.3589 kJ/kg.K s2 s1 1 MPa h2 s h1 3193.6 2675.0 h2 h1 2675.0 3285.1 kJ/kg C 0.85 Compresso P2 1 MPa s2 7.4974 kJ/kg.K h2 3285.1 kJ/kg 100 kPa wC h2 h1 3285.1 2675.0 610.1 kJ/kg s gen,C s2 s1 7.4974 7.3589 0.1384 kJ/kg.K7-162 A paddle wheel does work on the water contained in a rigid tank. For a zero entropy change ofwater, the final pressure in the tank, the amount of heat transfer between the tank and the surroundings, andthe entropy generation during the process are to be determined.Assumptions The tank is stationary and the kinetic and potential energy changes are negligible.Analysis (a) Using saturated liquid properties for the compressed liquid at the initial state (Table A-4) T1 120 C u1 503.60 kJ/kg x1 0 (sat. liq.) s1 1.5279 kJ/kg.K Water WpwThe entropy change of water is zero, and thus at the final state we have 120 C T2 95 C P2 84.6 kPa 500 kPa s2 s1 1.5279 kJ/kg.K u 2 492.63 kJ/kg(b) The heat transfer can be determined from an energy balance on the tank Qout W Pw,in m(u 2 u1 ) 22 kJ (1.5 kg)(492.63 503.60)kJ/kg 38.5 kJ(c) Since the entropy change of water is zero, the entropy generation is only due to the entropy increase ofthe surroundings, which is determined from
• 7-120 Qout 38.5 kJS gen S surr 0.134 kJ/K Tsurr (15 273) K
• 7-1217-163 A horizontal cylinder is separated into two compartments by a piston, one side containing nitrogenand the other side containing helium. Heat is added to the nitrogen side. The final temperature of thehelium, the final volume of the nitrogen, the heat transferred to the nitrogen, and the entropy generationduring this process are to be determined.Assumptions 1 Kinetic and potential energy changes are negligible. 2 Nitrogen and helium are ideal gaseswith constant specific heats at room temperature. 3 The piston is adiabatic and frictionless.Properties The properties of nitrogen at room temperature are R = 0.2968 kPa.m3/kg.K, cp = 1.039kJ/kg.K, cv = 0.743 kJ/kg.K, k = 1.4. The properties for helium are R = 2.0769 kPa.m3/kg.K, cp = 5.1926kJ/kg.K, cv = 3.1156 kJ/kg.K, k = 1.667 (Table A-2).Analysis (a) Helium undergoes an isentropic compressionprocess, and thus the final helium temperature isdetermined from N2 He P2 ( k 1) / k 120 kPa (1.667 1) / 1.667 0.2 m3 0.1 kgTHe,2 T1 (20 273)K P1 95 kPa 321.7 K(b) The initial and final volumes of the helium are mRT1 (0.1 kg)(2.0769 kPa m 3 /kg K)(20 273 K) V He,1 0.6406 m 3 P1 95 kPa mRT2 (0.1 kg)(2.0769 kPa m 3 /kg K)(321.7 K) V He,2 0.5568 m 3 P2 120 kPaThen, the final volume of nitrogen becomes V N2,2 V N2,1 V He,1 V He,2 0.2 0.6406 0.5568 0.2838 m 3(c) The mass and final temperature of nitrogen are P1V 1 (95 kPa)(0.2 m 3 ) m N2 0.2185 kg RT1 (0.2968 kPa m 3 /kg K)(20 273 K) P2V 2 (120 kPa)(0.2838 m 3 ) T N2,2 525.1 K mR (0.2185 kg)(0.2968 kPa m 3 /kg K)The heat transferred to the nitrogen is determined from an energy balance Qin U N2 U He mcv (T2 T1 ) N2 mcv (T2 T1 ) He (0.2185 kg)(0.743 kJ/kg.K)(525.1 293) (0.1 kg)(3.1156 kJ/kg.K)(321.7 293) 46.6 kJ(d) Noting that helium undergoes an isentropic process, the entropy generation is determined to be T2 P2 Qin S gen S N2 S surr m N2 c p ln R ln T1 P1 TR 525.1 K 120 kPa 46.6 kJ (0.2185 kg) (1.039 kJ/kg.K)ln (0.2968 kJ/kg.K)ln 293 K 95 kPa (500 273) K 0.057 kJ/K
• 7-1227-164 An electric resistance heater is doing work on carbon dioxide contained an a rigid tank. The finaltemperature in the tank, the amount of heat transfer, and the entropy generation are to be determined.Assumptions 1 Kinetic and potential energy changes are negligible. 2 Carbon dioxide is ideal gas withconstant specific heats at room temperature.Properties The properties of CO2 at an anticipated average temperature of 350 K are R = 0.1889kPa.m3/kg.K, cp = 0.895 kJ/kg.K, cv = 0.706 kJ/kg.K (Table A-2b).Analysis (a) The mass and the final temperature of CO2may be determined from ideal gas equation CO2 We P1V (100 kPa)(0.8 m 3 ) 250 K m 1.694 kg RT1 (0.1889 kPa m 3 /kg K)(250 K) 100 kPa P2V (175 kPa)(0.8 m 3 ) T2 437.5 K mR (1.694 kg)(0.1889 kPa m 3 /kg K)(b) The amount of heat transfer may be determined from an energy balance on the system Qout Ee,in t mcv (T2 T1 ) (0.5 kW)(40 60 s) - (1.694 kg)(0.706 kJ/kg.K)(437.5 - 250)K 975.8 kJ(c) The entropy generation associated with this process may be obtained by calculating total entropychange, which is the sum of the entropy changes of CO2 and the surroundings T2 P2 Qout S gen S CO2 S surr m c p ln R ln T1 P1 Tsurr 437.5 K 175 kPa 975.8 kJ (1.694 kg) (0.895 kJ/kg.K)ln (0.1889 kJ/kg.K)ln 250 K 100 kPa 300 K 3.92 kJ/K7-165 Heat is lost from the helium as it is throttled in a throttling valve. The exit pressure and temperatureof helium and the entropy generation are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3Helium is an ideal gas with constant specific heats. qProperties The properties of helium are R = 2.0769 HeliumkPa.m3/kg.K, cp = 5.1926 kJ/kg.K (Table A-2a). 500 kPaAnalysis (a) The final temperature of helium may be 70 Cdetermined from an energy balance on the control volume q out 2.5 kJ/kg q out c p (T1 T2 ) T2 T1 70 C 342.5 K 69.5 C cp 5.1926 kJ/kg. CThe final pressure may be determined from the relation for the entropy change of helium T2 P2 sHe c p ln R ln T1 P1 342.5 K P2 0.25 kJ/kg.K (5.1926 kJ/kg.K)ln (2.0769 kJ/kg.K)ln 343 K 500 kPa P2 441.7 kPa(b) The entropy generation associated with this process may be obtained by adding the entropy change ofhelium as it flows in the valve and the entropy change of the surroundings
• 7-123 qout 2.5 kJ/kgsgen sHe ssurr sHe 0.25 kJ/kg.K 0.258 kJ/kg.K Tsurr (25 273) K
• 7-1247-166 Refrigerant-134a is compressed in a compressor. The rate of heat loss from the compressor, the exittemperature of R-134a, and the rate of entropy generation are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) The properties of R-134a at the inlet of the compressor are (Table A-12) P1 200 kPa v 1 0.09987 m 3 /kg h1 244.46 kJ/kg Q 700 kPa x1 1 s1 0.93773 kJ/kg.KThe mass flow rate of the refrigerant is Compressor V1 0.03 m 3 /s m 0.3004 kg/s v1 0.09987 m 3 /kg R-134a 200 kPaGiven the entropy increase of the surroundings, the heat sat. vap.lost from the compressor is Qout S surr Qout Tsurr S surr (20 273 K)(0.008 kW/K) 2.344 kW Tsurr(b) An energy balance on the compressor gives Win Qout m( h2 h1 ) 10 kW - 2.344 kW (0.3004 kg/s)(h2 - 244.46) kJ/kg h2 269.94 kJ/kgThe exit state is now fixed. Then, P2 700 kPa T2 31.5 C h2 269.94 kJ/kg s 2 0.93620 kJ/kg.K(c) The entropy generation associated with this process may be obtained by adding the entropy change ofR-134a as it flows in the compressor and the entropy change of the surroundings S gen SR S surr m( s 2 s1 ) S surr (0.3004 kg/s)(0.93620 - 0.93773) kJ/kg.K 0.008 kW/K 0.00754 kJ/K
• 7-1257-167 Air flows in an adiabatic nozzle. The isentropic efficiency, the exit velocity, and the entropygeneration are to be determined.Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1).Assumptions 1 Steady operating conditions exist. 2 Potential energy changes are negligible.Analysis (a) (b) Using variable specific heats, the properties can be determined from air table as follows h1 400.98 kJ/kg 0 T1 400 K s1 1.99194 kJ/kg.K Air Pr1 3.806 500 kPa 300 kPa h2 350.49 kJ/kg 400 K 350 K T2 350 K 0 30 m/s s2 1.85708 kJ/kg.K P2 300 kPa Pr 2 Pr1 3.806 2.2836 h2 s 346.31 kJ/kg P1 500 kPaEnergy balances on the control volume for the actual and isentropic processes give V12 V22 h1 h2 2 2 (30 m/s)2 1 kJ/kg V22 1 kJ/kg 400.98 kJ/kg 350.49 kJ/kg 2 1000 m 2 /s 2 2 1000 m 2 /s 2 V2 319.1 m/s V12 2 V2s h1 h2 s 2 2 (30 m/s)2 1 kJ/kg 2 V2s 1 kJ/kg 400.98 kJ/kg 346.31 kJ/kg 2 1000 m 2 /s 2 2 1000 m 2 /s 2 V2s 331.8 m/sThe isentropic efficiency is determined from its definition, V 22 (319.1 m/s) 2 N 2 0.925 V 2s (331.8 m/s) 2(c) Since the nozzle is adiabatic, the entropy generation is equal to the entropy increase of the air as itflows in the nozzle 0 0 P2 sgen sair s2 s1 R ln P1 300 kPa (1.85708 1.99194)kJ/kg.K (0.287 kJ/kg.K)ln 0.0118 kJ/kg.K 500 kPa7-168 It is to be shown that the difference between the steady-flow and boundary works is the flow energy.Analysis The total differential of flow energy Pv can be expressed as d Pv P dv v dP wb w flow wb w flow
• 7-126Therefore, the difference between the reversible steady-flow work and the reversible boundary work is theflow energy.
• 7-1277-169 An insulated rigid tank is connected to a piston-cylinder device with zero clearance that ismaintained at constant pressure. A valve is opened, and some steam in the tank is allowed to flow into thecylinder. The final temperatures in the tank and the cylinder are to be determined. Assumptions 1 Both the tank and cylinder are well-insulated and thus heat transfer is negligible. 2 Thewater that remains in the tank underwent a reversible adiabatic process. 3 The thermal energy stored in thetank and cylinder themselves is negligible. 4 The system is stationary and thus kinetic and potential energychanges are negligible.Analysis (a) The steam in tank A undergoes a reversible, adiabatic process, and thus s2 = s1. From thesteam tables (Tables A-4 through A-6), v1 v g @500 kPa 0.37483 m 3 /kg P1 500 kPa u1 u g @500 kPa 2560.7 kJ/kg sat.vapor s1 s g @500 kPa 6.8207 kJ/kg K T 2, A Tsat @150 kPa 111.35 C P2 150 kPa s 2, A sf 6.8207 1.4337 x 2, A 0.9305 s2 s1 s fg 5.7894 sat.mixture v 2, A vf x 2, Av fg 0.001053 (0.9305)(1.1594 0.001053) 1.0789 m 3 /kg u 2, A uf x 2, A u fg 466.97 (0.9305)(2052.3 kJ/kg ) 2376.6 kJ/kgThe initial and the final masses in tank A are VA 0.4 m3 VA 0.4 m3 m1, A 1.067 kg and m2, A 0.371 kg v1, A 0.37483 m3/kg v 2, A 1.0789 m3/kgThus, m2, B m1, A m2, A 1.067 0.371 0.696 kg(b) The boundary work done during this process is 2 Wb,out P dV PB V 2, B 0 PB m 2, Bv 2, B 1Taking the contents of both the tank and the cylinder Sat.to be the system, the energy balance for this closed vapor 500 kPa 150 kPasystem can be expressed as 0 4 m3 Ein Eout Esystem Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies Wb, out U U A U B Wb, out U A U B 0or, PB m2, Bv 2, B m2u2 m1u1 A m2u2 B 0 m2, B h2, B m2u2 m1u1 A 0Thus, m1u1 m 2 u 2 A 1.067 2560.7 0.371 2376.6 h 2, B 2658.8 kJ/kg m 2, B 0.696At 150 kPa, hf = 467.13 and hg = 2693.1 kJ/kg. Thus at the final state, the cylinder will contain a saturatedliquid-vapor mixture since hf < h2 < hg. Therefore,
• 7-128T2, B Tsat @150 kPa 111.35 C
• 7-1297-170 One ton of liquid water at 80°C is brought into a room. The final equilibrium temperature in theroom and the entropy change during this process are to be determined.Assumptions 1 The room is well insulated and well sealed. 2 The thermal properties of water and air areconstant at room temperature. 3 The system is stationary and thus the kinetic and potential energy changesare zero. 4 There are no work interactions involved.Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1). The specific heat of water atroom temperature is c = 4.18 kJ/kg C (Table A-3). For air is cv = 0.718 kJ/kg C at room temperature.Analysis (a) The volume and the mass of the air in the room are V = 4x5x7 = 140 m³ PV1 100 kPa 140 m3 4m 5m 7m 1 mair 165.4 kg RT1 0.2870 kPa m3/kg K 295 K ROOM 22 CTaking the contents of the room, including the water, as our 100 kPasystem, the energy balance can be written as Heat Ein Eout Esystem 0 U U water U air Water Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies 80 Cor mc T2 T1 water mcv T2 T1 air 0Substituting, 1000 kg 4.18 kJ/kg C T f 80 C 165.4 kg 0.718 kJ/kg C T f 22 C 0It gives the final equilibrium temperature in the room to be Tf = 78.4 C(b) Considering that the system is well-insulated and no mass is entering and leaving, the total entropychange during this process is the sum of the entropy changes of water and the room air, S total S gen S air S waterwhere T2 V2 0 351.4 K Sair mcv ln mR ln 165.4 kg 0.718 kJ/kg K ln 20.78 kJ/K T1 V1 295 K T2 351.4 K S water mc ln 1000 kg 4.18 kJ/kg K ln 18.99 kJ/K T1 353 KSubstituting, the total entropy change is determined to be Stotal = 20.78 - 18.99 = 1.79 kJ/K
• 7-1307-171E A cylinder initially filled with helium gas at a specified state is compressed polytropically to aspecified temperature and pressure. The entropy changes of the helium and the surroundings are to bedetermined, and it is to be assessed if the process is reversible, irreversible, or impossible.Assumptions 1 Helium is an ideal gas with constant specific heats. 2 The cylinder is stationary and thus thekinetic and potential energy changes are negligible. 3 The thermal energy stored in the cylinder itself isnegligible. 4 The compression or expansion process is quasi-equilibrium.Properties The gas constant of helium is R = 2.6805 psia.ft3/lbm.R = 0.4961 Btu/lbm.R. The specific heatsof helium are cv = 0.753 and cp = 1.25 Btu/lbm.R (Table A-2E).Analysis (a) The mass of helium is PV1 1 25 psia 15 ft 3 m 0.264 lbm RT1 2.6805 psia ft 3/lbm R 530 R HELIUMThen the entropy change of helium becomes 15 ft3 n Q PV = const T2 P2 Ssys S helium m c p ,avg ln R ln T1 P1 760 R 70 psia (0.264 lbm) (1.25 Btu/lbm R ) ln (0.4961 Btu/lbm R ) ln 0.016 Btu/R 530 R 25 psia(b) The exponent n and the boundary work for this polytropic process are determined to be P1V1 P2V 2 T2 P1 760 R 25 psia V2 V1 15 ft 3 7.682 ft 3 T1 T2 T1 P2 530 R 70 psia n n P2 V1 70 15 P2V 2n P1V1n n 1.539 P1 V2 25 7.682Then the boundary work for this polytropic process can be determined from 2 P2V 2 PV1 mR T2 T1 Wb,in P dV 1 1 1 n 1 n 0.264 lbm 0.4961 Btu/lbm R 760 530 R 55.9 Btu 1 1.539We take the helium in the cylinder as the system, which is a closed system. Taking the direction of heattransfer to be from the cylinder, 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 Qout Wb,in U m(u2 u1 ) Qout m(u2 u1 ) Wb,in Qout Wb,in mcv (T2 T1 )Substituting, Qout 55.9 Btu 0.264 lbm 0.753 Btu/lbm R 760 530 R 10.2 BtuNoting that the surroundings undergo a reversible isothermal process, its entropy change becomes Qsurr,in 10.2 Btu Ssurr 0.019 Btu/R Tsurr 530 R(c) Noting that the system+surroundings combination can be treated as an isolated system,
• 7-131 S total Ssys Ssurr 0.016 0.019 0.003 Btu/R 0Therefore, the process is irreversible.
• 7-1327-172 Air is compressed steadily by a compressor from a specified state to a specified pressure. Theminimum power input required is to be determined for the cases of adiabatic and isothermal operation.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 with variable specific heats. 4 The process is reversiblesince the work input to the compressor will be minimum when the compression process is reversible.Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1).Analysis (a) For the adiabatic case, the process will be reversible and adiabatic (i.e., isentropic),thus the isentropic relations are applicable. T1 290 K Pr1 1.2311 and h1 290.16 kJ/kgand P2 700 kPa T2 503.3 K Pr2 Pr (1.2311) 8.6177 P 1 1 100 kPa h2 506.45 kJ/kgThe energy balance for the compressor, which is a steady-flow 2 2system, can be expressed in the rate form as 0 (steady) Ein Eout Esystem 0 AIR T = const Rate of net energy transfer Rate of change in internal, kinetic, Rev. by heat, work, and mass potential, etc. energies Ein Eout Win mh1 mh2 Win m( h 2 h1 ) 1 1Substituting, the power input to the compressor is determined to be Win (5/60 kg/s)(506.45 290.16) kJ/kg 18.0 kW(b) In the case of the reversible isothermal process, the steady-flow energy balance becomes 0 Ein Eout Win mh1 Qout mh2 Win Qout m(h2 h1 ) Qoutsince h = h(T) for ideal gases, and thus the enthalpy change in this case is zero. Also, for a reversibleisothermal process, Qout mT s1 s2 mT s2 s1where 0 P2 P2 700 kPa s2 s1 s2 s1 R ln R ln 0.287 kJ/kg K ln 0.5585 kJ/kg K P1 P1 100 kPaSubstituting, the power input for the reversible isothermal case becomes Win (5/60 kg/s)(290 K )( 0.5585 kJ/kg K ) 13.5 kW
• 7-1337-173 Air is compressed in a two-stage ideal compressor with intercooling. For a specified mass flow rateof air, the power input to the compressor is to be determined, and it is to be compared to the power input toa single-stage compressor.Assumptions 1 The compressor operates steadily. 2 Kinetic and potential energies are negligible. 3 Thecompression process is reversible adiabatic, and thus isentropic. 4 Air is an ideal gas with constant specificheats at room temperature.Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1). The specific heat ratio of air is k= 1.4 (Table A-2).Analysis The intermediate pressure between the two stages is 100 kPa 900 kPa Px P P2 1 100 kPa 900 kPa 300 kPa 27 CThe compressor work across each stage is thesame, thus total compressor work is twice thecompression work for a single stage: W Stage I Stage II kRT1 k 1 /kwcomp,in 2 wcomp,in, I 2 Px P1 1 k 1 27 C 0.4/1.4 1.4 0.287 kJ/kg K 300 K 300 kPa 2 1 1.4 1 100 kPa 222.2 kJ/kg Heatand Win mwcomp,in 0.02 kg/s 222.2 kJ/kg 4.44 kWThe work input to a single-stage compressor operating between the same pressure limits would be 0.4/1.4 kRT1 k 1 /k 1.4 0.287 kJ/kg K 300 K 900 kPa wcomp,in P2 P1 1 1 263.2 kJ/kg k 1 1.4 1 100 kPaand Win mwcomp,in 0.02 kg/s 263.2 kJ/kg 5.26 kWDiscussion Note that the power consumption of the compressor decreases significantly by using 2-stagecompression with intercooling.
• 7-1347-174 A three-stage compressor with two stages of intercooling is considered. The two intermediatepressures that will minimize the work input are to be determined in terms of the inlet and exit pressures.Analysis The work input to this three-stage compressor with intermediate pressures Px and Py and twointercoolers can be expressed as wcomp wcomp,I wcomp,II wcomp,III nRT1 n 1 /n nRT1 n 1 /n nRT1 n 1 /n 1 Px P1 1 Py Px 1 Px P1 n 1 n 1 n 1 nRT1 n 1 /n n 1 /n n 1 /n 1 Px P1 1 Py Px 1 Px P1 n 1 nRT1 n 1 /n n 1 /n n 1 /n 3 Px P1 Py Px Px P1 n 1The Px and Py values that will minimize the work input are obtained by taking the partial differential of wwith respect to Px and Py, and setting them equal to zero: n 1 n 1 1 1 w n 1 1 Px n n 1 1 Px n 0 0 Px n P1 P1 n Py Py n 1 n 1 1 1 w n 1 1 Py n n 1 1 Py n 0 0 Py n Px Px n P2 P2Simplifying, 1 2n 1 1 2n 1 Px n 1 Px n 1 P1 1 Px n 1 n Px2 1 P1 Py P1 P1 Py Py P1n Px Pyn Py 1 2n 1 1 2n 1 Py n 1 Py n 1 Px 1 Py n 1 n Py2 1 Px P2 Px Px P2 P2 Pxn Py P2n P2which yield 13 Px2 P1 Px P2 Px P12 P2 13 Py2 P2 P1 Py Py P1 P22
• 7-1357-175 Steam expands in a two-stage adiabatic turbine from a specified state to specified pressure. Somesteam is extracted at the end of the first stage. The power output of the turbine is to be determined for thecases of 100% and 88% isentropic efficiencies.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 The turbine is adiabatic and thus heat transfer is negligible.Properties From the steam tables (Tables A-4 through 6) P1 6 MPa h1 3423.1 kJ/kg T1 500 C s1 6.8826 kJ/kg K P2 1.2 MPa h2 2962.8 kJ / kg s2 s1 s3s sf 6.8826 0.8320 P3 20 kPa x3s 0.8552 s fg 7.0752 s3 s1 h3s hf x3s h fg 251.42 0.8552 2357.5 2267.5 kJ/kgAnalysis (a) The mass flow rate through the second stage is m3 0.9m1 0.9 15 kg/s 13.5 kg/sWe take the entire turbine, including the connection part 6 MPabetween the two stages, as the system, which is a 500 Ccontrol volume since mass crosses the boundary. Notingthat one fluid stream enters the turbine and two fluid STEAM STEAMstreams leave, the energy balance for this steady-flow 15 kg/s 13.5 kg/ssystem can be expressed in the rate form as I II 0 (steady) 1.2 MPa Ein Eout Esystem 0 20 kPa Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies 90% 10% Ein Eout m1h1 (m1 m3 )h2 Wout m3h3 Wout m1h1 (m1 m3 )h2 m3h3 m1 (h1 h2 ) m3 (h2 h3 )Substituting, the power output of the turbine is Wout 15 kg/s 3423.1 2962.8 kJ/kg 13.5 kg 2962.8 2267.5 kJ/kg 16,291 kW(b) If the turbine has an adiabatic efficiency of 88%, then the power output becomes Wa T Ws 0.88 16,291 kW 14,336 kW
• 7-1367-176 Steam expands in an 84% efficient two-stage adiabatic turbine from a specified state to a specifiedpressure. Steam is reheated between the stages. For a given power output, the mass flow rate of steamthrough the turbine 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 turbine is adiabatic and thus heat transfer is negligible.Properties From the steam tables (Tables A-4 through 6) Heat P1 8 MPa h1 3521.8 kJ/kg 2 MPa T1 550 C s1 6.8800 kJ/kg K 2 MPa 550 C P2 s 2 MPa h2 s 3089.7 kJ/kg s 2s s1 Stage I Stage II 80 MW P3 2 MPa h3 3579.0 kJ/kg T3 550 C s3 7.5725 kJ/kg K P4 s 200 kPa 8 MPa 200 kPa h4 s 2901.7 kJ/kg 550 C s 4s s3Analysis The power output of the actual turbine is given to be 80 MW. Then the power output for theisentropic operation becomes Ws,out Wa,out / T (80,000 kW) / 0.84 95,240 kWWe take the entire turbine, excluding the reheat section, as the system, which is a control volume sincemass crosses the boundary. The energy balance for this steady-flow system in isentropic operation can beexpressed 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 mh1 mh3 mh2 s mh4 s Ws,out Ws,out m[(h1 h2 s ) (h3 h4 s )]Substituting, 95,240 kJ/s m (3521.8 3089.7) kJ/kg (3579.0 2901.7) kJ/kgwhich gives m 85.8 kg/s
• 7-1377-177 Refrigerant-134a is compressed by a 0.7-kW adiabatic compressor from a specified state to anotherspecified state. The isentropic efficiency, the volume flow rate at the inlet, and the maximum flow rate atthe compressor inlet are 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.Properties From the R-134a tables (Tables A-11 through A-13) v1 0.14605 m 3 /kg 2 P1 140 kPa h1 246.36 kJ/kg T1 10 C s1 0.9724 kJ/kg K 0.7 kW R-134a P2 700 kPa h2 288.53 kJ/kg T2 50 C P2 700 kPa · V1 h2 s 281.16 kJ/kg s 2s s1 1Analysis (a) The isentropic efficiency is determined from its definition, h2 s h1 281.16 246.36 C 0.825 82.5% h2 a h1 288.53 246.36(b) There is only one inlet and one exit, and thus m1 m2 m . We take the actual compressor as thesystem, which is a control volume. The energy balance for this steady-flow system can be expressed 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 Wa,in mh1 mh2 (since Q ke pe 0) Wa,in m(h2 h1 )Then the mass and volume flow rates of the refrigerant are determined to be Wa,in 0.7 kJ/s m 0.0166 kg/s h2 a h1 288.53 246.36 kJ/kg V1 mv1 0.0166 kg/s 0.14605 m3/kg 0.00242 m3/s 145 L/min(c) The volume flow rate will be a maximum when the process is isentropic, and it is determined similarlyfrom the steady-flow energy equation applied to the isentropic process. It gives Ws,in 0.7 kJ/s mmax 0.0201 kg/s h2 s h1 281.16 246.36 kJ/kg V1, max mmaxv1 0.0201 kg/s 0.14605 m3/kg 0.00294 m3/s 176 L/minDiscussion Note that the raising the isentropic efficiency of the compressor to 100% would increase thevolumetric flow rate by more than 20%.
• 7-1387-178E Helium is accelerated by a 94% efficient nozzle from a low velocity to 1000 ft/s. The pressure andtemperature at the nozzle inlet are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Helium is an ideal gaswith constant specific heats. 3 Potential energy changes are negligible. 4 The device is adiabatic and thusheat transfer is negligible.Properties The specific heat ratio of helium is k = 1.667. The constant pressure specific heat of helium is1.25 Btu/lbm.R (Table A-2E).Analysis We take nozzle as the system, which is a control volume since mass crosses the boundary. Theenergy balance for this steady-flow system can 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 HELIUM Ein Eout 1 2 N = 94% m(h1 V12 / 2) m(h2 + V22 /2) (since Q W pe 0) V22 V12 V22 V12 0 h2 h1 0 c p , avg T2 T1 2 2Solving for T1 and substituting, 0 V 22s V12 1000 ft/s 2 1 Btu/lbm T1 T2 a 180 F 196.0 F 656 R 2C p 2 1.25 Btu/lbm R 25,037 ft 2 /s 2From the isentropic efficiency relation, h2 a h1 cP T2 a T1 N h2 s h1 cP T2 s T1or, T2 s T1 T2 a T1 / N 656 640 656 / 0.94 639 R k 1 /k T P2From the isentropic relation, 2 s T1 P1 k/ k 1 1.667/0.667 T 656 R P1 P2 1 14 psia 14.9 psia T2 s 639 R
• 7-1397-179 [Also solved by EES on enclosed CD] An adiabatic compressor is powered by a direct-coupled steamturbine, which also drives a generator. The net power delivered to the generator and the rate of entropygeneration are 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 devices are adiabatic and thus heat transfer is negligible. 4 Air is anideal gas with variable specific heats.Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1). From the steam tables (Tables A-4through 6) and air table (Table A-17), T1 295 K h1 295.17 kJ/kg, s1 1.68515 kJ/kg K T2 620 K h1 628.07 kJ/kg, s2 2.44356 kJ/kg K P3 12.5 MPa h3 3343.6 kJ/kg T3 500 C s3 6.4651 kJ/kg K P4 10 kPa h4 hf x4 h fg 191.81 0.92 2392.1 2392.5 kJ/kg x4 0.92 s4 sf x4 s fg 0.6492 0.92 7.4996 7.5489 kJ/kg KAnalysis There is only one inlet and one exit for either device, and thus min mout m . We take eitherthe turbine or the compressor as the system, which is a control volume since mass crosses the boundary.The energy balance for either steady-flow system can be expressed in the rate form as 0 (steady) Ein Eout Esystem 0 Ein Eout 12.5 MPa Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies 1 MPa 500 C 620 KFor the turbine and the compressor it becomes Air SteamCompressor: Wcomp, in mair h1 mair h2 Wcomp, in mair (h2 h1 ) comp turbineTurbine: msteam h3 Wturb, out msteam h4 Wturb, out msteam (h3 h4 )Substituting, 98 kPa 295 K Wcomp,in 10 kg/s 628.07 295.17 kJ/kg 3329 kW 10 kPa Wturb,out 25 kg/s 3343.6 2392.5 kJ/kg 23,777 kWTherefore, Wnet,out W turb,out Wcomp,in 23,777 3329 20,448 kWNoting that the system is adiabatic, the total rate of entropy change (or generation) during this process isthe sum of the entropy changes of both fluids, Sgen mair ( s2 s1 ) msteam ( s4 s3 )where P2 mair s2 s1 m s2 s1 R ln P1 1000 kPa 10 kg/s 2.44356 1.68515 0.287 ln kJ/kg K 0.92 kW/K 98 kPa msteam s4 s3 25 kg/s 7.5489 6.4651 kJ/kg K 27.1 kW/K
• 7-140Substituting, the total rate of entropy generation is determined to be Sgen, total Sgen, comp Sgen, turb 0.92 27.1 28.02 kW/K
• 7-1417-180 EES Problem 7-179 is reconsidered. The isentropic efficiencies for the compressor and turbine areto be determined, and then the effect of varying the compressor efficiency over the range 0.6 to 0.8 and theturbine efficiency over the range 0.7 to 0.95 on the net work for the cycle and the entropy generated for theprocess is to be investigated. The net work is to be plotted as a function of the compressor efficiency forturbine efficiencies of 0.7, 0.8, and 0.9.Analysis The problem is solved using EES, and the results are tabulated and plotted below."Input Data"m_dot_air = 10 [kg/s] "air compressor (air) data"T_air[1]=(295-273) "[C]" "We will input temperature in C"P_air[1]=98 [kPa]T_air[2]=(700-273) "[C]"P_air[2]=1000 [kPa]m_dot_st=25 [kg/s] "steam turbine (st) data"T_st[1]=500 [C]P_st[1]=12500 [kPa]P_st[2]=10 [kPa]x_st[2]=0.92 "quality""Compressor Analysis:""Conservation of mass for the compressor m_dot_air_in = m_dot_air_out =m_dot_air""Conservation of energy for the compressor is:"E_dot_comp_in - E_dot_comp_out = DELTAE_dot_compDELTAE_dot_comp = 0 "Steady flow requirement"E_dot_comp_in=m_dot_air*(enthalpy(air,T=T_air[1])) + W_dot_comp_inE_dot_comp_out=m_dot_air*(enthalpy(air,T=T_air[2]))"Compressor adiabatic efficiency:"Eta_comp=W_dot_comp_in_isen/W_dot_comp_inW_dot_comp_in_isen=m_dot_air*(enthalpy(air,T=T_air_isen[2])-enthalpy(air,T=T_air[1]))s_air[1]=entropy(air,T=T_air[1],P=P_air[1])s_air[2]=entropy(air,T=T_air[2],P=P_air[2])s_air_isen[2]=entropy(air, T=T_air_isen[2],P=P_air[2])s_air_isen[2]=s_air[1]"Turbine Analysis:""Conservation of mass for the turbine m_dot_st_in = m_dot_st_out =m_dot_st""Conservation of energy for the turbine is:"E_dot_turb_in - E_dot_turb_out = DELTAE_dot_turbDELTAE_dot_turb = 0 "Steady flow requirement"E_dot_turb_in=m_dot_st*h_st[1]h_st[1]=enthalpy(steam,T=T_st[1], P=P_st[1])E_dot_turb_out=m_dot_st*h_st[2]+W_dot_turb_outh_st[2]=enthalpy(steam,P=P_st[2], x=x_st[2])"Turbine adiabatic efficiency:"Eta_turb=W_dot_turb_out/W_dot_turb_out_isenW_dot_turb_out_isen=m_dot_st*(h_st[1]-h_st_isen[2])s_st[1]=entropy(steam,T=T_st[1],P=P_st[1])h_st_isen[2]=enthalpy(steam, P=P_st[2],s=s_st[1])"Note: When Eta_turb is specified as an independent variable inthe Parametric Table, the iteration process may put the steam state 2 in thesuperheat region, where the quality is undefined. Thus, s_st[2], T_st[2] arecalculated at P_st[2], h_st[2] and not P_st[2] and x_st[2]"s_st[2]=entropy(steam,P=P_st[2],h=h_st[2])T_st[2]=temperature(steam,P=P_st[2], h=h_st[2])s_st_isen[2]=s_st[1]"Net work done by the process:"W_dot_net=W_dot_turb_out-W_dot_comp_in
• 7-142"Entropy generation:""Since both the compressor and turbine are adiabatic, and thus there is no heat transferto the surroundings, the entropy generation for the two steady flow devices becomes:"S_dot_gen_comp=m_dot_air*( s_air[2]-s_air[1])S_dot_gen_turb=m_dot_st*(s_st[2]-s_st[1])S_dot_gen_total=S_dot_gen_comp+S_dot_gen_turb"To generate the data for Plot Window 1, Comment out the line T_air[2]=(700-273) Cand select values for Eta_comp in the Parmetric Table, then press F3 to solve the table.EES then solves for the unknown value of T_air[2] for each Eta_comp.""To generate the data for Plot Window 2, Comment out the two lines x_st[2]=0.92 quality and h_st[2]=enthalpy(steam,P=P_st[2], x=x_st[2]) and select values for Eta_turb in theParmetric Table, then press F3 to solve the table. EES then solves for the h_st[2] for eachEta_turb." Wnet Sgentotal turb comp Wnet Sgentotal turb comp [kW] [kW/K] [kW] [kW/K] 20124 27.59 0.75 0.6665 19105 30 0.7327 0.6 21745 22.51 0.8 0.6665 19462 29.51 0.7327 0.65 23365 17.44 0.85 0.6665 19768 29.07 0.7327 0.7 24985 12.36 0.9 0.6665 20033 28.67 0.7327 0.75 26606 7.281 0.95 0.6665 20265 28.32 0.7327 0.8 Effect of Compressor Efficiency on Net Work and Entropy Generated 20400 30.0 turb =0.7333 29.6 20000 ] ] K / W 29.2 W k [ k [ t 19600 l e a t n 28.8 o t , W n e g 19200 28.4 S 0.60 0.65 0.70 0.75 0.80 compb Effect of Turbine Efficiency on Net Work and Entropy Generated 30 = 0.6665 26000 comp 25 ] K / ] 20 W W 24000 k k [ [ l t e 15 a t n o t , W 22000 n e 10 g S 20000 5 0.75 0.78 0.81 0.84 0.87 0.90 0.93 turbb
• 7-1437-181 Two identical bodies at different temperatures are connected to each other through a heat engine. Itis to be shown that the final common temperature of the two bodies will be T f T1T2 when the workoutput of the heat engine is maximum.Analysis For maximum power production, the entropy generation must be zero. Taking the source, thesink, and the heat engine as our system, which is adiabatic, and noting that the entropy change for cyclicdevices is zero, the entropy generation for this system can be expressed as m, c Sgen S source S engine 0 S sink 0 T1 Tf Tf mcln 0 mcln 0 QH T1 T2 HE W Tf Tf Tf Tf ln ln 0 ln 0 T f2 T1T2 T1 T2 T1 T2 QLand thus m, c Tf T1T2 T2for maximum power production.7-182 The pressure in a hot water tank rises to 2 MPa, and the tank explodes. The explosion energy of thewater is to be determined, and expressed in terms of its TNT equivalence.Assumptions 1 The expansion process during explosion is isentropic. 2 Kinetic and potential energychanges are negligible. 3 Heat transfer with the surroundings during explosion is negligible.Properties The explosion energy of TNT is 3250 kJ/kg. From the steam tables (Tables A-4 through 6) v1 v f @ 2 MPa 0.001177 m3/kg P1 2 MPa u1 u f @ 2 MPa 906.12 kJ/kg sat. liquid s1 s f @ 2 MPa 2.4467 kJ/kg K P2 100 kPa u f 417.40, u fg 2088.2 kJ/kg Water Tank s2 s1 sf 1.3028, s fg 6.0562 kJ/kg K 2 MPa s2 sf 2.4467 1.3028 x2 0.1889 s fg 6.0562 u2 uf x2u fg 417.40 0.1889 2088.2 811.83 kJ/kgAnalysis We idealize the water tank as a closed system that undergoes a reversible adiabatic process withnegligible changes in kinetic and potential energies. The work done during this idealized process representsthe explosive energy of the tank, and is determined from the closed system energy balance to be Ein Eout Esystem Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies Wb,out U m(u2 u1 ) Eexp Wb,out m u1 u2 V 0.080 m3where m 67.99 kg v1 0.001177 m3/kgSubstituting, E exp 67.99 kg 906.12 811.83 kJ/kg 6410 kJ
• 7-144 6410 kJwhich is equivalent to mTNT 1.972 kg TNT 3250 kJ/kg
• 7-1457-183 A 0.35-L canned drink explodes at a pressure of 1.2 MPa. The explosive energy of the drink is to bedetermined, and expressed in terms of its TNT equivalence.Assumptions 1 The expansion process during explosion is isentropic. 2 Kinetic and potential energychanges are negligible. 3 Heat transfer with the surroundings during explosion is negligible. 4 The drinkcan be treated as pure water.Properties The explosion energy of TNT is 3250 kJ/kg. From the steam tables (Tables A-4 through 6) v1 v f @1.2 MPa 0.001138 m3/kg P1 1.2 MPa u1 u f @1.2 MPa 796.96 kJ/kg Comp. liquid s1 s f @1.2 MPa 2.2159 kJ/kg K P2 100 kPa u f 417.40, u fg 2088.2 kJ/kg s2 s1 sf 1.3028, s fg 6.0562 kJ/kg K COLA 1.2 MPa s2 sf 2.2159 1.3028 x2 0.1508 s fg 6.0562 u2 uf x2u fg 417.40 0.1508 2088.2 732.26 kJ/kgAnalysis We idealize the canned drink as a closed system that undergoes a reversible adiabatic processwith negligible changes in kinetic and potential energies. The work done during this idealized processrepresents the explosive energy of the can, and is determined from the closed system energy balance to be Ein Eout Esystem Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies Wb, out U m(u2 u1 ) Eexp Wb,out m u1 u2where V 0.00035 m 3 m 0.3074 kg v1 0.001138 m 3 /kgSubstituting, E exp 0.3074 kg 796.96 732.26 kJ/kg 19.9 kJwhich is equivalent to 19.9 kJ mTNT 0.00612 kg TNT 3250 kJ/kg
• 7-1357-184 The validity of the Clausius inequality is to be demonstrated using a reversible and an irreversibleheat engine operating between the same temperature limits.Analysis Consider two heat engines, one reversible and one irreversible, both operating between a high-temperature reservoir at TH and a low-temperature reservoir at TL. Both heat engines receive the sameamount of heat, QH. The reversible heat engine rejects heat in the amount of QL, and the irreversible one inthe amount of QL, irrev = QL + Qdiff, where Qdiff is a positive quantity since the irreversible heat engineproduces less work. Noting that QH and QL are transferred at constant temperatures of TH and TL,respectively, the cyclic integral of Q/T for the reversible and irreversible heat engine cycles become Q QH QL 1 1 QH QL QH QL 0 T rev TH TL TH TL TH TLsince (QH/TH) = (QL/TL) for reversible cycles. Also, Q QH Q L ,irrev QH QL Qdiff Qdiff 0 T irrev TH TL TH TL TL TL Qsince Qdiff is a positive quantity. Thus, 0. TH T · QH QH · · Wnet, rev Wnet, irrev Rev Irrev HE HE · · QL QL, irrev TL7-185 The inner and outer surfaces of a window glass are maintained at specified temperatures. Theamount of heat transfer through the glass and the amount of entropy generation within the glass in 5 h areto be determinedAssumptions 1 Steady operating conditions exist since the surface temperatures of the glass remainconstant at the specified values. 2 Thermal properties of the glass are constant.Analysis The amount of heat transfer over a period of 5 h is Glass Q Qcond t (3.2 kJ/s)(5 3600 s) 57,600 kJWe take the glass to be the system, which is a closed system. Under steadyconditions, the rate form of the entropy balance for the glass simplifies to 0 S in S out S gen S system 0 Rate of net entropy transfer Rate of entropy Rate of change by heat and mass generation of entropy 10 C 3 C Qin Qout S gen,glass 0 Tb,in Tb,out 3200 W 3200 W S gen, wall 0 S gen,glass 0.287 W/K 283 K 276 K
• 7-1367-186 Two rigid tanks that contain water at different states are connected by a valve. The valve is openedand steam flows from tank A to tank B until the pressure in tank A drops to a specified value. Tank B losesheat to the surroundings. The final temperature in each tank and the entropy generated during this processare to be determined.Assumptions 1 Tank A is insulated, and thus heat transfer is negligible. 2 The water that remains in tank Aundergoes a reversible adiabatic process. 3 The thermal energy stored in the tanks themselves is negligible.4 The system is stationary and thus kinetic and potential energy changes are negligible. 5 There are nowork interactions.Analysis (a) The steam in tank A undergoes a reversible, adiabatic process, and thus s2 = s1. From thesteam tables (Tables A-4 through A-6), Tank A : v 1, A vf x1v fg 0.001084 0.8 0.46242 0.001084 0.37015 m 3 /kg P1 400 kPa u1, A uf x1u fg 604.22 0.8 1948.9 2163.3 kJ/kg x1 0.8 s1, A sf x1 s fg 1.7765 0.8 5.1191 5.8717 kJ/kg K T 2, A Tsat @300 kPa 133.52 C P1 300 kPa s 2, A sf 5.8717 1.6717 x 2, A 0.7895 s 2 s1 s fg 5.3200 sat. mixture v 2, A vf x 2, Av fg 0.001073 0.7895 0.60582 0.001073 0.47850 m 3 /kg u 2, A uf x 2, A u fg 561.11 0.7895 1982.1 kJ/kg 2125.9 kJ/kg Tank B : 600 kJ v 1, B 1.1989 m 3 /kg P1 200 kPa u1, B 2731.4 kJ/kg T1 250 C A B s1, B 7.7100 kJ/kg K V = 0.2 m3 m = 3 kgThe initial and the final masses in tank A are steam steam P = 400 kPa T = 250 C VA 0.2 m 3 x = 0.8 P = 200 kPa m1, A 0.5403 kg v 1, A 0.37015 m 3 /kgand VA 0.2 m 3 m 2, A 0.4180 kg v 2, A 0.47850 m 3 /kgThus, 0.5403 - 0.4180 = 0.1223 kg of mass flows into tank B. Then, m 2, B m1, B 0.1223 3 0.1223 3.1223 kgThe final specific volume of steam in tank B is determined from VB m1v 1 3 kg 1.1989 m 3 /kg v 2, B B 1.1519 m 3 /kg m 2, B m 2, B 3.1223 kgWe take the entire contents of both tanks as the system, which is a closed system. The energy balance forthis stationary closed system can be expressed as E in E out E system Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies Qout U ( U)A ( U)B (since W KE = PE = 0) Qout m2 u 2 m1u1 A m2 u 2 m1u1 B
• 7-137Substituting, 600 0.418 2125.9 0.5403 2163.3 3.1223 u 2, B 3 2731.4 u 2, B 2522.0 kJ/kgThus, v 2, B 1.1519 m 3 /kg T2, B 113.2 C u 2, B 2522.0 kJ/kg s 2, B 7.2274 kJ/kg K(b) The total entropy generation during this process is determined by applying the entropy balance on anextended system that includes both tanks and their immediate surroundings so that the boundarytemperature of the extended system is the temperature of the surroundings at all times. It gives S in S out S gen S system Net entropy transfer Entropy Change by heat and mass generation in entropy Qout S gen SA SB Tb,surrRearranging and substituting, the total entropy generated during this process is determined to be Qout Qout S gen SA SB m2 s 2 m1 s1 A m2 s 2 m1 s1 B Tb,surr Tb,surr 600 kJ 0.418 5.8717 0.5403 5.8717 3.1223 7.2274 3 7.7100 273 K 0.916 kJ/K7-187 Heat is transferred steadily to boiling water in a pan through its bottom. The rate of entropygeneration within the bottom plate is to be determined.Assumptions Steady operating conditions exist since the surface temperatures of the pan remain constant atthe specified values.Analysis We take the bottom of the pan to be the system,which is a closed system. Under steady conditions, the rateform of the entropy balance for this system can beexpressed as 0 104 C S in S out S gen S system 0 Rate of net entropy transfer Rate of entropy Rate of change by heat and mass generation of entropy 500 W Qin Qout 105 C S gen,system 0 Tb,in Tb,out 500 W 500 W S gen,system 0 S gen,system 0.00351 W/K 378 K 377 KDiscussion Note that there is a small temperature drop across the bottom of the pan, and thus a smallamount of entropy generation.
• 7-1387-188 An electric resistance heater is immersed in water. The time it will take for the electric heater to raisethe water temperature to a specified temperature and the entropy generated during this process are to bedetermined.Assumptions 1 Water is an incompressible substance with constant specific heats. 2 The energy stored inthe container itself and the heater is negligible. 3 Heat loss from the container is negligible.Properties The specific heat of water at room temperature is c = 4.18 kJ/kg·°C (Table A-3).Analysis Taking the water in the container as the system, which is a closed system, the energy balance canbe expressed as E in E out E system Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies Water We,in ( U ) water 40 kg We,in t mc(T2 T1 ) water HeaterSubstituting, (1200 J/s) t = (40 kg)(4180 J/kg· C)(50 - 20) CSolving for t gives t = 4180 s = 69.7 min = 1.16 hAgain we take the water in the tank to be the system. Noting that no heat or mass crosses the boundaries ofthis system and the energy and entropy contents of the heater are negligible, the entropy balance for it canbe expressed as S in S out S gen S system Net entropy transfer Entropy Change by heat and mass generation in entropy 0 S gen S waterTherefore, the entropy generated during this process is T2 323 K S gen S water mc ln 40 kg 4.18 kJ/kg K ln 16.3 kJ/K T1 293 K7-189 A hot water pipe at a specified temperature is losing heat to the surrounding air at a specified rate.The rate of entropy generation in the surrounding air due to this heat transfer are to be determined.Assumptions Steady operating conditions exist.Analysis We take the air in the vicinity of the pipe (excluding the pipe) as our system, which is a closedsystem.. The system extends from the outer surface of the pipe to a distance at which the temperature dropsto the surroundings temperature. In steady operation, the rate form of the entropy balance for this systemcan be expressed as 0 S in S out S gen S system 0 80 C Rate of net entropy transfer Rate of entropy Rate of change by heat and mass generation of entropy Qin Qout S gen,system 0 Tb,in Tb,out Q Air, 5 C 2200 W 2200 W S gen,system 0 S gen,system 1.68 W/K 353 K 278 K
• 7-1397-190 The feedwater of a steam power plant is preheated using steam extracted from the turbine. The ratioof the mass flow rates of the extracted steam to the feedwater and entropy generation per unit mass offeedwater are 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 Heat loss from the device to the surroundings is negligible.Properties The properties of steam and feedwater are (Tables A-4 through A-6)P1 1 MPa h1 2828.3 kJ/kg 1T1 200 C s1 6.6956 kJ/kg K Steam 1 MPa from 200 C h2 h f @1 MPa 762.51 kJ/kg turbineP2 1 MPa Feedwater s2 s f @1 MPa 2.1381 kJ/kg K 3sat. liquid T2 179.88 C 2.5 MPaP3 2.5 MPa h3 h f @50 C 209.34 kJ/kgT3 50 C s3 s f @50 C 0.7038 kJ/kg K 4P4 2.5 MPa h4 h f @170 C 719.08 kJ/kgT4 T2 10 C 170 C s 4 s f @170 C 2.0417 kJ/kg KAnalysis (a) We take the heat exchanger as the system, which is a control 2volume. The mass and energy balances for this steady-flow system can be sat. liquidexpressed in the rate form as follows:Mass balance (for each fluid stream): 0 (steady) min m out msystem 0 min m out m1 m2 m s and m3 m4 m fwEnergy balance (for the heat exchanger): 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 m1h1 m3h3 m2h2 m4h4 (since Q W ke pe 0)Combining the two, m s h2 h1 m fw h3 h4 ms h 4 h3 719.08 209.34 kJ/kgDividing by m fw and substituting, 0.247 m fw h1 h2 2828.3 762.51 kJ/kg(b) The total entropy change (or entropy generation) during this process per unit mass of feedwater can bedetermined from an entropy balance expressed in the rate form as 0 S in S out S gen S system 0 Rate of net entropy transfer Rate of entropy Rate of change by heat and mass generation of entropy m1 s1 m 2 s 2 m3 s 3 m4 s 4 S gen 0 m s ( s1 s 2 ) m fw ( s 3 s 4 ) S gen 0 S gen ms s2 s1 s4 s3 0.247 2.1381 6.6956 2.0417 0.7038 m fw m fw 0.213 kJ/K per kg of feedwater
• 7-1407-191 EES Problem 7-190 is reconsidered. The effect of the state of the steam at the inlet to the feedwaterheater is to be investigated. The entropy of the extraction steam is assumed to be constant at the value for 1MPa, 200°C, and the extraction steam pressure is to be varied from 1 MPa to 100 kPa. Both the ratio of themass flow rates of the extracted steam and the feedwater heater and the total entropy change for thisprocess per unit mass of the feedwater are to be plotted as functions of the extraction pressure.Analysis The problem is solved using EES, and the results are tabulated and plotted below."Knowns:"WorkFluid\$ = Steam_iapws"P[3] = 1000 [kPa]" "place {} around P[3] and T[3] eqations to solve the table"T[3] = 200 [C]P[4] = P[3]x[4]=0T[4]=temperature(WorkFluid\$,P=P[4],x=x[4])P[1] = 2500 [kPa]T[1] = 50 [C]P[2] = 2500 [kPa]T[2] = T[4] - 10"[C]""Since we dont know the mass flow rates and we want to determine the ratio of mass flow rate ofthe extracted steam and the feedwater, we can assume the mass flow rate of the feedwater is 1kg/s without loss of generality. We write the conservation of energy.""Conservation of mass for the steam extracted from the turbine: "m_dot_steam[3]= m_dot_steam[4]"Conservation of mass for the condensate flowing through the feedwater heater:"m_dot_fw[1] = 1m_dot_fw[2]= m_dot_fw[1]"Conservation of Energy - SSSF energy balance for the feedwater heater -- neglecting thechange in potential energy, no heat transfer, no work:"h[3]=enthalpy(WorkFluid\$,P=P[3],T=T[3])"To solve the table, place {} around s[3] and remove them from the 2nd and 3rd equations"s[3]=entropy(WorkFluid\$,P=P[3],T=T[3]){s[3] =6.693 [kJ/kg-K] "This s[3] is for the initial T[3], P[3]"T[3]=temperature(WorkFluid\$,P=P[3],s=s[3]) "Use this equation for T[3] only when s[3] is given."}h[4]=enthalpy(WorkFluid\$,P=P[4],x=x[4])s[4]=entropy(WorkFluid\$,P=P[4],x=x[4])h[1]=enthalpy(WorkFluid\$,P=P[1],T=T[1])s[1]=entropy(WorkFluid\$,P=P[1],T=T[1])h[2]=enthalpy(WorkFluid\$,P=P[2],T=T[2])s[2]=entropy(WorkFluid\$,P=P[2],T=T[2])"For the feedwater heater:"E_dot_in = E_dot_outE_dot_in = m_dot_steam[3]*h[3] +m_dot_fw[1]*h[1]E_dot_out= m_dot_steam[4]*h[4] + m_dot_fw[2]*h[2]m_ratio = m_dot_steam[3]/ m_dot_fw[1]"Second Law analysis:"S_dot_in - S_dot_out + S_dot_gen = DELTAS_dot_sysDELTAS_dot_sys = 0 "[KW/K]" "steady-flow result"S_dot_in = m_dot_steam[3]*s[3] +m_dot_fw[1]*s[1]S_dot_out= m_dot_steam[4]*s[4] + m_dot_fw[2]*s[2]S_gen_PerUnitMassFWH = S_dot_gen/m_dot_fw[1]"[kJ/kg_fw-K]"
• 7-141 mratio Sgen,PerUnitMass P3 [kJ/kg-K] [kPa] 0.2109 0.1811 732 0.2148 0.185 760 0.219 0.189 790 0.223 0.1929 820 0.227 0.1968 850 0.2309 0.2005 880 0.2347 0.2042 910 0.2385 0.2078 940 0.2422 0.2114 970 0.2459 0.2149 1000] 0.22K-wfg 0.21 P3 = 1000 kPak/Jk[H 0.2W For P3 < 732 kPaFss 0.19a Sgen < 0MtinUr P3 = 732 kPae 0.18P,neg 0.21 0.22 0.23 0.24 0.25S mratio
• 7-1427-192E A rigid tank initially contains saturated R-134a vapor. The tank is connected to a supply line, andis charged until the tank contains saturated liquid at a specified pressure. The mass of R-134a that enteredthe tank, the heat transfer with the surroundings at 110 F, and the entropy generated during this process areto be determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remainsconstant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 Thedirection of heat transfer is to the tank (will be verified).Properties The properties of R-134a are (Tables A-11 through A-13) v 1 v g @100 psia 0.47760 ft 3 /lbm P1 100 psia u1 u g @100 psia 104.99 Btu/lbm R-134a 140 psia sat. vapor 80 F s1 s g @100 psia 0.2198 Btu/lbm R v2 v f @120 psia 0.01360 ft 3 /lbm P2 120 psia R-134a u2 u f @120 psia 41.49 Btu/lbm 110 F sat. liquid 3 ft3 s2 s f @120 psia 0.08589 Btu/lbm R Q Pi 140 psia hi h f @ 80 F 38.17 Btu/lbm Ti 80 F si s f @ 80 F 0.07934 Btu/lbm RAnalysis (a) We take the tank as the system, which is a control volume since mass crosses the boundary.Noting that the energies of flowing and nonflowing fluids are represented by enthalpy h and internalenergy u, respectively, the mass and energy balances for this uniform-flow system can be expressed asMass balance: min m out msystem mi m 2 m1Energy balance: E in E out E system Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies Qin mi hi m 2 u 2 m1u1 (since W ke pe 0)The initial and the final masses in the tank are V 3 ft 3 V 3 ft 3 m1 6.28 lbm m2 220.55 lbm v1 0.4776 ft 3/lbm v 2 0.01360 ft 3/lbmThen from the mass balance, mi m 2 m1 220.55 6.28 214.3 lbm(b) The heat transfer during this process is determined from the energy balance to beQin mi hi m 2 u 2 m1u1 214.3 lbm 38.17 Btu/lbm 220.55 lbm 41.49 Btu/lbm 6.28 lbm 104.99 Btu/lbm 312 Btu(c) The entropy generated during this process is determined by applying the entropy balance on anextended system that includes the tank and its immediate surroundings so that the boundary temperature ofthe extended system is the temperature of the surroundings at all times. The entropy balance for it can beexpressed as Qin S in S out S gen S system mi s i S gen S tank m 2 s 2 m1 s1 Tb,in Net entropy transfer Entropy Change by heat and mass generation in entropyTherefore, the total entropy generated during this process is Qin S gen mi s i (m 2 s 2 m1 s1 ) Tb,in 312 Btu 214.3 0.07934 220.55 0.08589 6.28 0.2198 0.0169 Btu/R 570 R
• 7-1437-193 It is to be shown that for thermal energy reservoirs, the entropy change relation S mc ln(T2 / T1 )reduces to S Q / T as T2 T1 .Analysis Consider a thermal energy reservoir of mass m, specific heat c, and initial temperature T1. Nowheat, in the amount of Q, is transferred to this reservoir. The first law and the entropy change relations forthis reservoir can be written as Q Q mc T2 T1 mc T2 T1and Q T ln T2 / T1 Thermal energy S mc ln 2 Q reservoir T1 T2 T1 m, c, TTaking the limit as T2 T1 by applying the LHospitals rule, 1 / T1 Q S Q 1 T1which is the desired result.7-194 The inner and outer glasses of a double pane window are at specified temperatures. The rates ofentropy transfer through both sides of the window and the rate of entropy generation within the windoware to be determined.Assumptions Steady operating conditions exist since the surface temperatures of the glass remain constantat the specified values.Analysis The entropy flows associated with heat transfer through the left and right glasses are Qleft 110 W S left 0.378 W/K Tleft 291 K Q right 110 W S right 0.394 W/K Air Tright 279 K 18 C 6 C · QWe take the double pane window as the system, which is aclosed system. In steady operation, the rate form of theentropy balance for this system can be expressed as 0 S in S out S gen S system 0 Rate of net entropy transfer Rate of entropy Rate of change by heat and mass generation of entropy Qin Qout S gen,system 0 Tb,in Tb,out 110 W 110 W S gen,system 0 S gen,system 0.016 W/K 291 K 279 K
• 7-1447-195 A well-insulated room is heated by a steam radiator, and the warm air is distributed by a fan. Theaverage temperature in the room after 30 min, the entropy changes of steam and air, and the entropygenerated during this process are to be determined.Assumptions 1 Air is an ideal gas with constant specific heats at room temperature. 2 The kinetic andpotential energy changes are negligible. 3 The air pressure in the room remains constant and thus the airexpands as it is heated, and some warm air escapes.Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1). Also, cp = 1.005 kJ/kg.K for airat room temperature (Table A-2).Analysis We first take the radiator 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 E in E out E system Net energy transfer Change in internal, kinetic,by heat, work, and mass potential, etc. energies 10 C Qout U m(u 2 u1 ) (since W KE = PE = 0) 4m 4m 5m Qout m(u1 u 2 )Using data from the steam tables (Tables A-4 through A-6), some Steamproperties are determined to be radiator P1 200 kPa v 1 1.0805 m 3 /kg u1 2654.6 kJ/kg T1 200 C s1 7.5081 kJ/kg.K vf 0.001043, v g 1.6941 m 3 /kg P2 100 kPa uf 417.40, u fg 2088.2 kJ/kg v2 v1 sf 1.3028, s fg 6.0562 kJ/kg.K v2 v f 1.0805 0.001043 x2 0.6376 v fg 1.6941 0.001043 u2 uf x 2 u fg 417.40 0.6376 2088.2 1748.7 kJ/kg s2 sf x 2 s fg 1.3028 0.6376 6.0562 5.1642 kJ/kg.K V1 0.015 m 3 m 0.01388 kg v1 1.0805 m 3 /kgSubstituting, Qout = (0.01388 kg)( 2654.6 - 1748.7)kJ/kg = 12.6 kJThe volume and the mass of the air in the room are V = 4 4 5 = 80 m³ and P1V 1 100 kPa (80 m 3 ) m air 98.5 kg RT1 (0.2870 kPa m 3 /kg K ) 283 KThe amount of fan work done in 30 min is Wfan,in Wfan,in t (0.120 kJ/s)(30 60 s) 216kJWe now take the air in the room as the system. The energy balance for this closed system is expressed as E in E out E system Qin Wfan,in W b,out U Qin Wfan,in H mc p (T2 T1 )
• 7-145since the boundary work and U combine into H for a constant pressure expansion or compressionprocess.Substituting, (12.6 kJ) + (216 kJ) = (98.5 kg)(1.005 kJ/kg C)(T2 - 10) Cwhich yields T2 = 12.3 CTherefore, the air temperature in the room rises from 10 C to 12.3 C in 30 min.(b) The entropy change of the steam is S steam m s2 s1 0.01388 kg 5.1642 7.5081 kJ/kg K 0.0325 kJ/K(c) Noting that air expands at constant pressure, the entropy change of the air in the room is 0 T2 P2 285.3 K S air mc p ln mR ln 98.5 kg 1.005 kJ/kg K ln 0.8013 kJ/K T1 P1 283 K(d) We take the air in the room (including the steam radiator) as our system, which is a closed system.Noting that no heat or mass crosses the boundaries of this system, the entropy balance for it can beexpressed as S in S out S gen S system Net entropy transfer Entropy Change by heat and mass generation in entropy 0 S gen S steam S airSubstituting, the entropy generated during this process is determined to be S gen S steam S air 0.0325 0.8013 0.7688 kJ/K
• 7-1467-196 The heating of a passive solar house at night is to be assisted by solar heated water. The length oftime that the electric heating system would run that night and the amount of entropy generated that nightare to be determined.Assumptions 1 Water is an incompressible substance with constant specific heats. 2 The energy stored inthe glass containers themselves is negligible relative to the energy stored in water. 3 The house ismaintained at 22°C at all times.Properties The density and specific heat of water at room temperature are = 1 kg/L and c = 4.18kJ/kg·°C (Table A-3).Analysis The total mass of water is mw V 1 kg/L 50 20 L 1000 kgTaking the contents of the house, including the water as our system, the energy balance relation can bewritten as E in E out E system 50,000 kJ/h Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies We,in Qout U ( U ) water ( U ) air ( U ) water 22 C mc(T2 T1 ) wateror, water We,in t Qout [mc(T2 T1 )] water 80 CSubstituting, (15 kJ/s) t - (50,000 kJ/h)(10 h) = (1000 kg)(4.18 kJ/kg· C)(22 - 80) CIt gives t = 17,170 s = 4.77 hWe take the house as the system, which is a closed system. The entropy generated during this process isdetermined by applying the entropy balance on an extended system that includes the house and itsimmediate surroundings so that the boundary temperature of the extended system is the temperature of thesurroundings at all times. The entropy balance for the extended system can be expressed as S in S out S gen S system Net entropy transfer Entropy Change by heat and mass generation in entropy Qout 0 S gen S water S air S water Tb,outsince the state of air in the house remains unchanged. Then the entropy generated during the 10-h periodthat night is Qout T2 Qout S gen S water mc ln Tb,out T1 water Tsurr 295 K 500,000 kJ 1000 kg 4.18 kJ/kg K ln 353 K 276 K 750 1811 1061 kJ/K
• 7-1477-197E A steel container that is filled with hot water is allowed to cool to the ambient temperature. Thetotal entropy generated during this process is to be determined.Assumptions 1 Both the water and the steel tank are incompressible substances with constant specific heatsat room temperature. 2 The system is stationary and thus the kinetic and potential energy changes are zero.3 Specific heat of iron can be used for steel. 4 There are no work interactions involved.Properties The specific heats of water and the iron at room temperature are cp, water = 1.00 Btu/lbm. F andCp, iron = 0.107 Btu/lbm. C. The density of water at room temperature is 62.1 lbm/ft³ (Table A-3E).Analysis The mass of the water is m water VV (62.1 lbm/ft 3 )(15 ft 3 ) 931.5 lbmWe take the steel container and the water in it as the system, which is a closed system. The energy balanceon the system can be expressed as E in E out E system Steel Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies WATER Qout U U container U water 120 F Q [mc(T2 T1 )] container [mc(T2 T1 )] water 70 FSubstituting, the heat loss to the surrounding air is determined to be Qout [mc(T1 T2 )] container [mc(T1 T2 )] water (75 lbm)(0.107 Btu/lbm F)(120 70) F (931.5 lbm)(1.00 Btu/lbm F)(120 70) F 46,976 BtuWe again take the container and the water In it as the system. The entropy generated during this process isdetermined by applying the entropy balance on an extended system that includes the container and itsimmediate surroundings so that the boundary temperature of the extended system is the temperature of thesurrounding air at all times. The entropy balance for the extended system can be expressed as S in S out S gen S system Net entropy transfer Entropy Change by heat and mass generation in entropy Qout S gen S container S water Tb,outwhere T2 530 R S container mc avg ln 75 lbm 0.107 Btu/lbm R ln 0.72 Btu/R T1 580 R T2 530 R S water mc avg ln 931.5 lbm 1.00 Btu/lbm R ln 83.98 Btu/R T1 580 RTherefore, the total entropy generated during this process is Qout 46,976 Btu S gen S container S water 0.72 83.98 3.93 Btu/R Tb,out 70 460 R
• 7-1487-198 Refrigerant-134a is vaporized by air in the evaporator of an air-conditioner. For specified flow rates,the exit temperature of air and the rate of entropy generation are to be determined for the cases of aninsulated and uninsulated evaporator.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 There are no work interactions. 4 Air is an ideal gas with constant specificheats at room temperature.Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A-1). The constant pressure specific heat ofair at room temperature is cp = 1.005 kJ/kg.K (Table A-2). The properties of R-134a at the inlet and the exitstates are (Tables A-11 through A-13) P1 120 kPa h1 hf x1 h fg 22.49 0.3 214.48 86.83 kJ/kg x1 0.3 s1 sf x1 s fg 0.09275 0.3(0.85503) 0.3493 kJ/kg K T2 120 kPa h2 hg @ 120 kPa 236.97 kJ/kg sat. vapor s2 hg @ 120 kPa 0.9478 kJ/kg K 6 m3/min AIRAnalysis (a) The mass flow rate of air is 3 R-134a P3V3 100 kPa 6 m3/min 1mair 3 6.97 kg/min RT3 0.287 kPa m /kg K 300 K 2 kg/min 2We take the entire heat exchanger as the system, which is acontrol volume. The mass and energy balances for this steady- 4 sat. vaporflow system can be expressed in the rate form asMass balance ( for each fluid stream): 0 (steady) min mout msystem 0 min mout m1 m2 mair and m3 m4 mREnergy balance (for the entire heat exchanger): 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 m1h1 m3h3 m2 h2 m4 h4 (since Q W ke pe 0)Combining the two, mR h2 h1 mair h3 h4 mair c p T3 T4 m R h2 h1Solving for T4, T4 T3 m air c p (2 kg/min)(236.97 86.83) kJ/kgSubstituting, T4 27 C 15.9 C = 257.1 K (6.97 kg/min)(1.005 kJ/kg K)Noting that the condenser is well-insulated and thus heat transfer is negligible, the entropy balance for thissteady-flow system can be expressed as 0 (steady) Sin Sout Sgen Ssystem Rate of net entropy transfer Rate of entropy Rate of change by heat and mass generation of entropy m1s1 m3 s3 m2 s2 m4 s4 Sgen 0 (since Q 0) mR s1 mair s3 m R s2 mair s4 Sgen 0or,
• 7-149 Sgen mR s 2 s1 mair s4 s3where 0 T P T4 257.1 K s4 s3 c p ln 4 R ln 4 c p ln (1.005 kJ/kg K) ln 0.1551 kJ/kg K T3 P3 T3 300 KSubstituting, Sgen 2 kg/min 0.9478 - 0.3493 kJ/kg K (6.97 kg/min)( 0.1551 kJ/kg K) 0.116 kJ/min K 0.00193 kW/K(b) When there is a heat gain from the surroundings at a rate of 30 kJ/min, the steady-flow energy equationreduces to Qin mR h2 h1 mair c p T4 T3 Qin mR h2 h1Solving for T4, T4 T3 mair c p (30 kJ/min) (2 kg/min)(236.97 86.83) kJ/kgSubstituting, T4 27 C + 11.6 C 261.4 K (6.97 kg/min)(1.005 kJ/kg K)The entropy generation in this case is determined by applying the entropy balance on an extended systemthat includes the evaporator and its immediate surroundings so that the boundary temperature of theextended system is the temperature of the surrounding air at all times. The entropy balance for the extendedsystem can be expressed as 0 (steady) Sin Sout Sgen Ssystem Rate of net entropy transfer Rate of entropy Rate of change by heat and mass generation of entropy Qin m1s1 m3 s3 m2 s2 m4 s4 Sgen 0 Tb,out Qin mR s1 mair s3 m R s2 mair s4 Sgen 0 Tsurr Qinor Sgen m R s2 s1 mair s4 s3 T0 0 T P 261.4 Kwhere s4 s3 c p ln 4 R ln 4 (1.005 kJ/kg K) ln 0.1384 kJ/kg K T3 P3 300 KSubstituting, 30 kJ/min Sgen 2 kg/min 0.9478 0.3493 kJ/kg K 6.97 kg/min 0.1384 kJ/kg K 305 K 0.1340 kJ/min K 0.00223 kW/KDiscussion Note that the rate of entropy generation in the second case is greater because of theirreversibility associated with heat transfer between the evaporator and the surrounding air.
• 7-1507-199 A room is to be heated by hot water contained in a tank placed in the room. The minimum initialtemperature of the water needed to meet the heating requirements of this room for a 24-h period and theentropy generated are to be determined.Assumptions 1 Water is an incompressible substance with constant specific heats. 2 Air is an ideal gas withconstant specific heats. 3 The energy stored in the container itself is negligible relative to the energy storedin water. 4 The room is maintained at 20°C at all times. 5 The hot water is to meet the heatingrequirements of this room for a 24-h period.Properties The specific heat of water at room temperature is c = 4.18 kJ/kg·°C (Table A-3).Analysis Heat loss from the room during a 24-h period is Qloss = (10,000 kJ/h)(24 h) = 240,000 kJTaking the contents of the room, including the water, as our system, the energy balance can be written as 0 Ein Eout Esystem Qout U U water U air Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies 10,000 kJ/hor -Qout = [mc(T2 - T1)]waterSubstituting, 20 C -240,000 kJ = (1500 kg)(4.18 kJ/kg· C)(20 - T1)It gives water T1 = 58.3 Cwhere T1 is the temperature of the water when it is first brought into the room.(b) We take the house as the system, which is a closed system. The entropy generated during this process isdetermined by applying the entropy balance on an extended system that includes the house and itsimmediate surroundings so that the boundary temperature of the extended system is the temperature of thesurroundings at all times. The entropy balance for the extended system can be expressed as Sin Sout Sgen Ssystem Net entropy transfer Entropy Change by heat and mass generation in entropy Qout 0 Sgen S water Sair S water Tb,outsince the state of air in the house (and thus its entropy) remains unchanged. Then the entropy generatedduring the 24 h period becomes Qout T2 Qout Sgen S water mc ln Tb,out T1 water Tsurr 293 K 240,000 kJ 1500 kg 4.18 kJ/kg K ln 331.3 K 278 K 770.3 863.3 93.0 kJ/K
• 7-1517-200 An insulated cylinder is divided into two parts. One side of the cylinder contains N2 gas and theother side contains He gas at different states. The final equilibrium temperature in the cylinder and theentropy generated are to be determined for the cases of the piston being fixed and moving freely.Assumptions 1 Both N2 and He are ideal gases with constant specific heats. 2 The energy stored in thecontainer itself is negligible. 3 The cylinder is well-insulated and thus heat transfer is negligible.Properties The gas constants and the constant volume specific heats are R = 0.2968 kPa.m3/kg.K, cv =0.743 kJ/kg·°C and cp =1.039 kJ/kg·°C for N2, and R = 2.0769 kPa.m3/kg.K, cv = 3.1156 kJ/kg·°C, and cp= 5.1926 kJ/kg·°C for He (Tables A-1 and A-2)Analysis The mass of each gas in the cylinder is PV1 1 500 kPa 1 m3 N2 He mN 2 3 4.77 kg 1 m3 1 m3 RT1 N2 0.2968 kPa m /kg K 353 K 500 kPa 500 kPa PV1 1 500 kPa 1 m3 80 C 25 C mHe 0.808 kg RT1 He 2.0769 kPa m3/kg K 298 KTaking the entire contents of the cylinder as our system, the 1st law relation can be written as Ein Eout Esystem Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies 0 U U N2 U He 0 [mcv (T2 T1 )]N 2 [mcv (T2 T1 )]HeSubstituting, 4.77 kg 0.743 kJ/kg C T f 80 C 0.808 kg 3.1156 kJ/kg C T f 25 C 0It gives Tf = 57.2 Cwhere Tf is the final equilibrium temperature in the cylinder. The answer would be the same if the piston were not free to move since it would effect onlypressure, and not the specific heats.(b) We take the entire cylinder as our system, which is a closed system. Noting that the cylinder is well-insulated and thus there is no heat transfer, the entropy balance for this closed system can be expressed as Sin Sout Sgen Ssystem Net entropy transfer Entropy Change by heat and mass generation in entropy 0 Sgen SN2 S HeBut first we determine the final pressure in the cylinder: m m 4.77 kg 0.808 kg N total NN2 N He 0.372 kmol M N2 M He 28 kg/kmol 4 kg/kmol N total RuT 0.372 kmol 8.314 kPa m3/kmol K 330.2 K P2 510.6 kPa V total 2 m3Then, T2 P2 SN2 m c p ln R ln T1 P1 N2 330.2 K 510.6 kPa 4.77 kg 1.039 kJ/kg K ln 0.2968 kJ/kg K ln 353 K 500 kPa 0.361 kJ/K
• 7-152 T2 P2 S He m c p ln R ln T1 P1 He 330.2 K 510.6 kPa 0.808 kg 5.1926 kJ/kg K ln 2.0769 kJ/kg K ln 298 K 500 kPa 0.395 kJ/K Sgen SN2 S He 0.361 0.395 0.034 kJ/KIf the piston were not free to move, we would still have T2 = 330.2 K but the volume of each gas wouldremain constant in this case: 0 T2 V2 330.2 K SN2 m cv ln R ln 4.77 kg 0.743 kJ/kg K ln 0.237 kJ/K T1 V1 353 K N2 0 T2 V2 330.2 K S He m cv ln R ln 0.808 kg 3.1156 kJ/kg K ln 0.258 kJ/K T1 V1 298 K He Sgen SN2 S He 0.237 0.258 0.021 kJ/K
• 7-1537-201 EES Problem 7-200 is reconsidered. The results for constant specific heats to those obtained usingvariable specific heats are to be compared using built-in EES or other functions.Analysis The problem is solved using EES, and the results are given below."Knowns:"R_u=8.314 [kJ/kmol-K]V_N2[1]=1 [m^3]Cv_N2=0.743 [kJ/kg-K] "From Table A-2(a) at 27C"R_N2=0.2968 [kJ/kg-K] "From Table A-2(a)"T_N2[1]=80 [C]P_N2[1]=500 [kPa]Cp_N2=R_N2+Cv_N2V_He[1]=1 [m^3]Cv_He=3.1156 [kJ/kg-K] "From Table A-2(a) at 27C"T_He[1]=25 [C]P_He[1]=500 [kPa]R_He=2.0769 [kJ/kg-K] "From Table A-2(a)"Cp_He=R_He+Cv_He"Solution:""mass calculations:"P_N2[1]*V_N2[1]=m_N2*R_N2*(T_N2[1]+273)P_He[1]*V_He[1]=m_He*R_He*(T_He[1]+273)"The entire cylinder is considered to be a closed system, allowing the piston to move.""Conservation of Energy for the closed system:""E_in - E_out = DELTAE, we neglect DELTA KE and DELTA PE for the cylinder."E_in - E_out = DELTAEE_in =0 [kJ]E_out = 0 [kJ]"At the final equilibrium state, N2 and He will have a common temperature."DELTAE= m_N2*Cv_N2*(T_2-T_N2[1])+m_He*Cv_He*(T_2-T_He[1])"Total volume of gases:"V_total=V_N2[1]+V_He[1]MM_He = 4 [kg/kmol]MM_N2 = 28 [kg/kmol]N_total = m_He/MM_He+m_N2/MM_N2"Final pressure at equilibrium:""Allowing the piston to move, the pressure on both sides is the same, P_2 is:"P_2*V_total=N_total*R_u*(T_2+273)S_gen_PistonMoving = DELTAS_He_PM+DELTAS_N2_PMDELTAS_He_PM=m_He*(Cp_He*ln((T_2+273)/(T_He[1]+273))-R_He*ln(P_2/P_He[1]))DELTAS_N2_PM=m_N2*(Cp_N2*ln((T_2+273)/(T_N2[1]+273))-R_N2*ln(P_2/P_N2[1]))"The final temperature of the system when the piston does not move will be the same as when itdoes move. The volume of the gases remain constant and the entropy changes are given by:"S_gen_PistNotMoving = DELTAS_He_PNM+DELTAS_N2_PNMDELTAS_He_PNM=m_He*(Cv_He*ln((T_2+273)/(T_He[1]+273)))DELTAS_N2_PNM=m_N2*(Cv_N2*ln((T_2+273)/(T_N2[1]+273)))
• 7-154"The following uses the EES functions for the nitrogen. Since helium is monatomic, we use theconstant specific heat approach to find its property changes."E_in - E_out = DELTAE_VPDELTAE_VP= m_N2*(INTENERGY(N2,T=T_2_VP)-INTENERGY(N2,T=T_N2[1]))+m_He*Cv_He*(T_2_VP-T_He[1])"Final Pressure for moving piston:"P_2_VP*V_total=N_total*R_u*(T_2_VP+273)S_gen_PistMoving_VP = DELTAS_He_PM_VP+DELTAS_N2_PM_VPDELTAS_N2_PM_VP=m_N2*(ENTROPY(N2,T=T_2_VP,P=P_2_VP)-ENTROPY(N2,T=T_N2[1],P=P_N2[1]))DELTAS_He_PM_VP=m_He*(Cp_He*ln((T_2+273)/(T_He[1]+273))-R_He*ln(P_2/P_He[1]))"Fianl N2 Pressure for piston not moving."P_2_N2_VP*V_N2[1]=m_N2*R_N2*(T_2_VP+273)S_gen_PistNotMoving_VP = DELTAS_He_PNM_VP+DELTAS_N2_PNM_VPDELTAS_N2_PNM_VP = m_N2*(ENTROPY(N2,T=T_2_VP,P=P_2_N2_VP)-ENTROPY(N2,T=T_N2[1],P=P_N2[1]))DELTAS_He_PNM_VP=m_He*(Cv_He*ln((T_2_VP+273)/(T_He[1]+273)))SOLUTIONCp_He=5.193 [kJ/kg-K] P_2=511.1 [kPa]Cp_N2=1.04 [kJ/kg-K] P_2_N2_VP=467.7Cv_He=3.116 [kJ/kg-K] P_2_VP=511.2Cv_N2=0.743 [kJ/kg-K] P_He[1]=500 [kPa]DELTAE=0 [kJ] P_N2[1]=500 [kPa]DELTAE_VP=0 [kJ] R_He=2.077 [kJ/kg-K]DELTAS_He_PM=0.3931 [kJ/K] R_N2=0.2968 [kJ/kg-K]DELTAS_He_PM_VP=0.3931 [kJ/K] R_u=8.314 [kJ/kmol-K]DELTAS_He_PNM=0.258 [kJ/K] S_gen_PistMoving_VP=0.02993 [kJ/K]DELTAS_He_PNM_VP=0.2583 [kJ/K] S_gen_PistNotMoving=0.02089 [kJ/K]DELTAS_N2_PM=-0.363 [kJ/K] S_gen_PistNotMoving_VP=0.02106 [kJ/K]DELTAS_N2_PM_VP=-0.3631 [kJ/K] S_gen_PistonMoving=0.03004 [kJ/K]DELTAS_N2_PNM=-0.2371 [kJ/K] T_2=57.17 [C]DELTAS_N2_PNM_VP=-0.2372 [kJ/K] T_2_VP=57.2 [C]E_in=0 [kJ] T_He[1]=25 [C]E_out=0 [kJ] T_N2[1]=80 [C]MM_He=4 [kg/kmol] V_He[1]=1 [m^3]MM_N2=28 [kg/kmol] V_N2[1]=1 [m^3]m_He=0.8079 [kg] V_total=2 [m^3]m_N2=4.772 [kg]N_total=0.3724 [kmol]
• 7-1557-202 An insulated cylinder is divided into two parts. One side of the cylinder contains N2 gas and theother side contains He gas at different states. The final equilibrium temperature in the cylinder and theentropy generated are to be determined for the cases of the piston being fixed and moving freely.Assumptions 1 Both N2 and He are ideal gases with constant specific heats. 2 The energy stored in thecontainer itself, except the piston, is negligible. 3 The cylinder is well-insulated and thus heat transfer isnegligible. 4 Initially, the piston is at the average temperature of the two gases.Properties The gas constants and the constant volume specific heats are R = 0.2968 kPa.m3/kg.K, cv =0.743 kJ/kg·°C and cp =1.039 kJ/kg·°C for N2, and R = 2.0769 kPa.m3/kg.K, cv = 3.1156 kJ/kg·°C, and cp= 5.1926 kJ/kg·°C for He (Tables A-1 and A-2). The specific heat of the copper at room temperature is c= 0.386 kJ/kg·°C (Table A-3).Analysis The mass of each gas in the cylinder is N2 He PV1 1 500 kPa 1 m3 1 m3 1 m3 mN 2 4.77 kg RT1 N2 0.2968 kPa m3/kg K 353 K 500 kPa 500 kPa 80 C 25 C PV1 1 500 kPa 1m3 mHe 0.808 kg RT1 He 2.0769 kPa m3/kg K 298 K CopperTaking the entire contents of the cylinder as our system, the 1st law relation can be written as Ein Eout Esystem Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies 0 U U N2 U He U Cu 0 [mcv (T2 T1 )]N 2 [mcv (T2 T1 )]He [mc(T2 T1 )]Cuwhere T1, Cu = (80 + 25) / 2 = 52.5 CSubstituting, 4.77 kg 0.743 kJ/kg C T f 80 C 0.808 kg 3.1156 kJ/kg C T f 25 C 5.0 kg 0.386 kJ/kg C T f 52.5 C 0It gives Tf = 56.0 Cwhere Tf is the final equilibrium temperature in the cylinder. The answer would be the same if the piston were not free to move since it would effect onlypressure, and not the specific heats.(b) We take the entire cylinder as our system, which is a closed system. Noting that the cylinder is well-insulated and thus there is no heat transfer, the entropy balance for this closed system can be expressed as Sin Sout Sgen Ssystem Net entropy transfer Entropy Change by heat and mass generation in entropy 0 Sgen SN2 S He S pistonBut first we determine the final pressure in the cylinder: m m 4.77 kg 0.808 kg N total N N2 N He 0.372 kmol M N2 M He 28 kg/kmol 4 kg/kmol N total RuT 0.372 kmol 8.314 kPa m3/kmol K 329 K P2 508.8 kPa V total 2 m3
• 7-156Then, T2 P2 SN2 m c p ln R ln T1 P1 N2 329 K 508.8 kPa 4.77 kg 1.039 kJ/kg K ln 0.2968 kJ/kg K ln 353 K 500 kPa 0.374 kJ/K T P S He m c p ln 2 R ln 2 T1 P1 He 329 K 508.8 kPa 0.808 kg 5.1926 kJ/kg K ln 2.0769 kJ/kg K ln 298 K 500 kPa 0.386 kJ/K T 329 K S piston mc ln 2 5 kg 0.386 kJ/kg K ln 0.021 kJ/K T1 piston 325.5 K Sgen SN2 S He S piston 0.374 0.386 0.021 0.033 kJ/KIf the piston were not free to move, we would still have T2 = 329 K but the volume of each gas wouldremain constant in this case: T2 V2 0 329 K SN2 m cv ln R ln 4.77 kg 0.743 kJ/kg K ln 0.250 kJ/K T1 V1 353 K N2 T2 V2 0 329 K S He m cv ln R ln 0.808 kg 3.1156 kJ/kg K ln 0.249 kJ/K T1 V1 298 K He Sgen SN2 S He S piston 0.250 0.249 0.021 0.020 kJ/K
• 7-1577-203 An insulated rigid tank equipped with an electric heater initially contains pressurized air. A valve isopened, and air is allowed to escape at constant temperature until the pressure inside drops to a specifiedvalue. The amount of electrical work done during this process and the total entropy change are to bedetermined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process since the exit temperature (and enthalpy) of airremains constant. 2 Kinetic and potential energies are negligible. 3 The tank is insulated and thus heattransfer is negligible. 4 Air is an ideal gas with variable specific heats.Properties The gas constant is R = 0.287 kPa.m3/kg.K (Table A-1). The properties of air are (Table A-17) Te 330 K he 330.34 kJ/kg T1 330 K u1 235.61 kJ/kg T2 330 K u2 235.61 kJ/kgAnalysis We take the tank as the system, which is a control volume since mass crosses the boundary.Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h andinternal energy u, respectively, the mass and energy balances for this uniform-flow system can beexpressed asMass balance: min mout msystem me m1 m2Energy balance: Ein Eout Esystem Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies We,in me he m2u2 m1u1 (since Q ke pe 0)The initial and the final masses of air in the tank are PV 1 500 kPa 5 m3 m1 26.40 kg RT1 0.287 kPa m3/kg K 330 K AIR P2V 200 kPa 5 m3 5 m3 m2 10.56 kg 500 kPa We 0.287 kPa m3/kg K 330 K RT2 57 CThen from the mass and energy balances, me m1 m2 26.40 10.56 15.84 kg We,in me he m2u2 m1u1 15.84 kg 330.34 kJ/kg 10.56 kg 235.61 kJ/kg 26.40 kg 235.61 kJ/kg 1501 kJ(b) The total entropy change, or the total entropy generation within the tank boundaries is determined froman entropy balance on the tank expressed as Sin Sout Sgen Ssystem Net entropy transfer Entropy Change by heat and mass generation in entropy me se Sgen S tank S gen me s e S tank me s e (m 2 s 2 m1 s1 )or, m1 m 2 s e m 2 s 2 m1 s1 m 2 s 2 s e m1 s1 s eAssuming a constant average pressure of (500 + 200)/2 = 350 kPa for the exit stream, the entropy changesare determined to be T 0 P P 200 kPa s 2 s e c p ln 2 R ln 2 R ln 2 0.287 kJ/kg K ln 0.1606 kJ/kg K Te Pe Pe 350 kPa T1 0 P P 500 kPa s1 se c p ln R ln 2 R ln 1 0.287 kJ/kg K ln 0.1024 kJ/kg K Te Pe Pe 350 kPaTherefore, the total entropy generated within the tank during this process is Sgen 10.56 kg 0.1606 kJ/kg K 26.40 kg 0.1024 kJ/kg K 4.40 kJ/K
• 7-1587-204 A 1- ton (1000 kg) of water is to be cooled in a tank by pouring ice into it. The final equilibriumtemperature in the tank and the entropy generation are to be determined.Assumptions 1 Thermal properties of the ice and water are constant. 2 Heat transfer to the water tank isnegligible. 3 There is no stirring by hand or a mechanical device (it will add energy).Properties The specific heat of water at room temperature is c = 4.18 kJ/kg·°C, and the specific heat of iceat about 0 C is c = 2.11 kJ/kg·°C (Table A-3). The melting temperature and the heat of fusion of ice at 1atm are 0 C and 333.7 kJ/kg..Analysis (a) We take the ice and the water as the system, and disregard iceany heat transfer between the system and the surroundings. Then the -5 Cenergy balance for this process can be written as 80 kg Ein Eout Esystem Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies 0 U WATER 1 ton 0 U ice U water [mc(0 C T1 ) solid mhif mc(T2 0 C) liquid ] ice [mc(T2 T1 )] water 0Substituting, (80 kg){(2.11 kJ/kg C)[0 (-5)] C + 333.7 kJ/kg + (4.18 kJ/kg C)(T2 0) C} (1000 kg )(4.18 kJ/kg C)(T2 20) C 0It gives T2 = 12.42 Cwhich is the final equilibrium temperature in the tank.(b) We take the ice and the water as our system, which is a closed system. Considering that the tank iswell-insulated and thus there is no heat transfer, the entropy balance for this closed system can beexpressed as Sin Sout Sgen Ssystem Net entropy transfer Entropy Change by heat and mass generation in entropy 0 Sgen Sice S waterwhere T2 285.42 K S water mc ln 1000 kg 4.18 kJ/kg K ln 109.6 kJ/K T1 water 293 K Sice Ssolid S melting Sliquid ice Tmelting mhif T2 mc ln mc ln T1 solid Tmelting T1 liquid ice 273 K 333.7 kJ/kg 285.42 K 80 kg 2.11 kJ/kg K ln 4.18 kJ/kg K ln 268 K 273 K 273 K 115.8 kJ/KThen, Sgen S water Sice 109.6 115.8 6.2 kJ/K
• 7-1597-205 An insulated cylinder initially contains a saturated liquid-vapor mixture of water at a specifiedtemperature. The entire vapor in the cylinder is to be condensed isothermally by adding ice inside thecylinder. The amount of ice added and the entropy generation are to be determined.Assumptions 1 Thermal properties of the ice are constant. 2 The cylinder is well-insulated and thus heattransfer is negligible. 3 There is no stirring by hand or a mechanical device (it will add energy).Properties The specific heat of ice at about 0 C is c = 2.11 kJ/kg·°C (Table A-3). The melting temperatureand the heat of fusion of ice at 1 atm are 0 C and 333.7 kJ/kg.Analysis (a) We take the contents of the cylinder (ice and saturated water) as our system, which is a closedsystem. Noting that the temperature and thus the pressure remains constant during this phase changeprocess and thus Wb + U = H, the energy balance for this system can be written as Ein Eout Esystem Wb,in U H 0 H ice H water 0 Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energiesor [mc(0 C T1 ) solid mhif mc(T2 0 C) liquid ] ice [m(h2 h1 )] water 0The properties of water at 100 C are (Table A-4) v f 0.001043, v g 1.6720 m 3 /kg ice h f 419.17, h fg 2256.4 kJ.kg -18 C WATER sf 1.3072 s fg 6.0490 kJ/kg.K 0.02 m3 100 C v1 v f x1v fg 0.001043 0.1 1.6720 0.001043 0.16814 m3/kg h1 hf x1h fg 419.17 0.1 2256.4 644.81 kJ/kg s1 sf x1s fg 1.3072 0.1 6.0470 1.9119 kJ/kg K h2 h f @100 C 419.17 kJ/kg s2 s f @100 C 1.3072 kJ/kg K V1 0.02 m3 msteam 0.119 kg v10.16814 m3/kg Noting that T1, ice = -18 C and T2 = 100 C and substituting gives m{(2.11 kJ/kg.K)[0-(-18)] + 333.7 kJ/kg + (4.18 kJ/kg· C)(100-0) C} +(0.119 kg)(419.17 – 644.81) kJ/kg = 0 m = 0.034 kg = 34.0 g ice(b) We take the ice and the steam as our system, which is a closed system. Considering that the tank iswell-insulated and thus there is no heat transfer, the entropy balance for this closed system can beexpressed as Sin Sout Sgen Ssystem Net entropy transfer Entropy Change by heat and mass generation in entropy 0 Sgen Sice Ssteam Ssteam m s2 s1 0.119 kg 1.3072 1.9119 kJ/kg K 0.0719 kJ/K Tmelting mhif T2 Sice Ssolid S melting Sliquid ice mc ln mc ln T1 solid Tmelting T1 liquid ice 273.15 K 333.7 kJ/kg 373.15 K 0.034 kg 2.11 kJ/kg K ln 4.18 kJ/kg K ln 0.0907 kJ/K 255.15 K 273.15 K 273.15 KThen, Sgen Ssteam Sice 0.0719 0.0907 0.0188 kJ/K
• 7-1607-206 An evacuated bottle is surrounded by atmospheric air. A valve is opened, and air is allowed to fillthe bottle. The amount of heat transfer through the wall of the bottle when thermal and mechanicalequilibrium is established and the amount of entropy generated are to be determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remainsconstant. 2 Air is an ideal gas. 3 Kinetic and potential energies are negligible. 4 There are no workinteractions involved. 5 The direction of heat transfer is to the air in the bottle (will be verified).Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A-1).Analysis We take the bottle as the system, which is a control volume since mass crosses the boundary.Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h andinternal energy u, respectively, the mass and energy balances for this uniform-flow system can beexpressed asMass balance: min mout msystem mi m2 (since mout minitial 0)Energy balance: Ein Eout Esystem Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies Qin mi hi m2u2 (since W Eout Einitial ke pe 0)Combining the two balances: Qin m2 u2 hi 10 kPawhere 17 C P2V 100 kPa 0.005 m 3 m2 0.0060 kg RT2 0.287 kPa m 3 /kg K 290 K 5L Evacuated Table A -17 hi 290.16 kJ/kg Ti T2 290 K u2 206.91 kJ/kgSubstituting, Qin = (0.0060 kg)(206.91 - 290.16) kJ/kg = - 0.5 kJ Qout = 0.5 kJNote that the negative sign for heat transfer indicates that the assumed direction is wrong. Therefore, wereverse the direction. The entropy generated during this process is determined by applying the entropy balance on anextended system that includes the bottle and its immediate surroundings so that the boundary temperatureof the extended system is the temperature of the surroundings at all times. The entropy balance for it can beexpressed as Sin Sout Sgen Ssystem Net entropy transfer Entropy Change by heat and mass generation in entropy Qout 0 mi si Sgen S tank m2 s2 m1s1 m2 s2 Tb,inTherefore, the total entropy generated during this process is Qout 0 Qout Qout 0.5 kJ Sgen mi si m2 s2 m2 s2 si 0.0017 kJ/K Tb,out Tb, out Tsurr 290 K
• 7-1617-207 Water is heated from 16°C to 43°C by an electric resistance heater placed in the water pipe as itflows through a showerhead steadily at a rate of 10 L/min. The electric power input to the heater and therate of entropy generation are to be determined. The reduction in power input and entropy generation as aresult of installing a 50% efficient regenerator are also to be determined.Assumptions 1 This is a steady-flow process since there is no change with time at any point within thesystem and thus mCV 0 and E CV 0 . 2 Water is an incompressible substance with constant specificheats. 3 The kinetic and potential energy changes are negligible, ke pe 0 . 4 Heat losses from thepipe are negligible.Properties The density of water is given to be = 1 kg/L. The specific heat of water at room temperature isc = 4.18 kJ/kg·°C (Table A-3).Analysis (a) We take the pipe as the system. This is a control volume since mass crosses the systemboundary during the process. We observe that there is only one inlet and one exit and thus m1 m2 m .Then the energy balance for this steady-flow system can be expressed in the rate form as 0 (steady) Ein Eout Esystem 0 Ein Eout Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies WATER We,in mh1 mh2 (since ke pe 0) 16 C 43 C We,in m(h2 h1 ) mc(T2 T1 )where m V 1 kg/L 10 L/min 10 kg/minSubstituting, We,in 10/60 kg/s 4.18 kJ/kg C 43 16 C 18.8 kWThe rate of entropy generation in the heating section during this process is determined by applying theentropy balance on the heating section. Noting that this is a steady-flow process and heat transfer from theheating section is negligible, 0 Sin Sout Sgen Ssystem 0 Rate of net entropy transfer Rate of entropy Rate of change by heat and mass generation of entropy ms1 ms2 Sgen 0 Sgen m ( s2 s1 )Noting that water is an incompressible substance and substituting, T 316 K Sgen mc ln 2 10/60 kg/s 4.18 kJ/kg K ln 0.0622 kJ/K T1 289 K(b) The energy recovered by the heat exchanger is Qsaved Qmax mC Tmax Tmin 0.5 10/60 kg/s 4.18 kJ/kg C 39 16 C 8.0 kJ/s 8.0 kWTherefore, 8.0 kW less energy is needed in this case, and the required electric power in this case reduces to Win, new Win,old Qsaved 18.8 8.0 10.8 kWTaking the cold water stream in the heat exchanger as our control volume (a steady-flow system), thetemperature at which the cold water leaves the heat exchanger and enters the electric resistance heatingsection is determined from Q mc(Tc, out Tc,in )Substituting, 8 kJ/s (10/60 kg/s )( 4.18 kJ/kg C)(Tc, out 16 C)It yields Tc, out 27.5 C 300.5 KThe rate of entropy generation in the heating section in this case is determined similarly to be T 316 K Sgen mcln 2 10/60 kg/s 4.18 kJ/kg K ln 0.0350 kJ/K T1 300.5 KThus the reduction in the rate of entropy generation within the heating section is S reduction 0.0622 0.0350 0.0272 kW/K
• 7-1627-208 EES Using EES (or other) software, the work input to a multistage compressor is to be determinedfor a given set of inlet and exit pressures for any number of stages. The pressure ratio across each stage isassumed to be identical and the compression process to be polytropic. The compressor work is to betabulated and plotted against the number of stages for P1 = 100 kPa, T1 = 17°C, P2 = 800 kPa, and n = 1.35for air.Analysis The problem is solved using EES, and the results are tabulated and plotted below.GAS\$ = AirNstage = 2 "number of stages of compression with intercooling, each having same pressureratio."n=1.35MM=MOLARMASS(GAS\$)R_u = 8.314 [kJ/kmol-K]R=R_u/MMk=1.4P1=100 [kPa]T1=17 [C]P2=800 [kPa]R_p = (P2/P1)^(1/Nstage)W_dot_comp= Nstage*n*R*(T1+273)/(n-1)*((R_p)^((n-1)/n) - 1) Nstage Wcomp [kJ/kg] 1 229.4 2 198.7 3 189.6 4 185.3 5 182.8 6 181.1 7 179.9 8 179 9 178.4 10 177.8 230 220 210 W comp [kJ/kg] 200 190 180 170 1 2 3 4 5 6 7 8 9 10 Nstage
• 7-1637-209 A piston-cylinder device contains air that undergoes a reversible thermodynamic cycle composed ofthree processes. The work and heat transfer for each process are to be determined.Assumptions 1 All processes are reversible. 2 Kineticand potential energy changes are negligible. 3 Air is anideal gas with variable specific heats. P = const.Properties The gas constant of air is R = 0.287 3kPa.m3/kg.K (Table A-1). 1 s = const.Analysis Using variable specific heats, the propertiescan be determined using the air table as follows u1 u2 214.07 kJ/kg T = const. 2 0 0 T1 T2 300 K s1 s2 1.70203 kJ/kg.K Pr1 Pr 2 1.3860 P3 400 kPa u3 283.71 kJ/kg Pr 3 Pr 2 1.3860 3.696 P2 150 kPa T3 396.6 KThe mass of the air and the volumes at the various states are P1V1 (400 kPa)(0.3 m 3 ) m 1.394 kg RT1 (0.287 kPa m 3 /kg K)(300 K) mRT2 (1.394 kg)(0.287 kPa m 3 /kg K)(300 K) V2 0.8 m 3 P2 150 kPa mRT3 (1.394 kg)(0.287 kPa m 3 /kg K)(396.6 K) V3 0.3967 m 3 P3 400 kPaProcess 1-2: Isothermal expansion (T2 = T1) P2 150 kPa S1 2 mR ln (1.394 kg)(0.287 kJ/kg.K)ln 0.3924 kJ/kg.K P1 400 kPa Qin,1 2 T1 S1 2 (300 K)(0.3924 kJ/K) 117.7 kJ Wout,1 2 Qin,1 2 117.7 kJProcess 2-3: Isentropic (reversible-adiabatic) compression (s2 = s1) Win,2 3 m(u3 u2 ) (1.394 kg)(283.71 - 214.07) kJ/kg 97.1 kJ Q2-3 = 0 kJProcess 3-1: Constant pressure compression process (P1 = P3) Win,3 1 P3 (V3 V1 ) (400 kg)(0.3924 - 0.3) kJ/kg 37.0 kJ Qout,3 1 Win,3 1 m(u1 u3 ) 37.0 kJ - (1.394 kg)(214.07 - 283.71) kJ/kg 135.8 kJ
• 7-1647-210 The turbocharger of an internal combustion engine consisting of a turbine driven by hot exhaustgases and a compressor driven by the turbine is considered. The air temperature at the compressor exit andthe isentropic efficiency of the compressor are to be determined.Assumptions 1 Steady operating conditions exist. 2 Air, 70 CKinetic and potential energy changes are negligible. 3 95 kPaExhaust gases have air properties and air is an ideal gas 400 C 0.018 kg/swith constant specific heats.Properties The specific heat of exhaust gases at the Compressor Turbineaverage temperature of 425ºC is cp = 1.075 kJ/kg.K andproperties of air at an anticipated average temperature of100ºC are cp = 1.011 kJ/kg.K and k =1.397 (Table A-2). Exh. gasAnalysis (a) The turbine power output is determined from 135 kPa 450 C WT m exh c p (T1 T2 ) 0.02 kg/s (0.02 kg/s)(1.075 kJ/kg. C)(450 - 400) C 1.075 kWFor a mechanical efficiency of 95% between the turbine and the compressor, WC mWT (0.95)(1.075 kW) 1.021 kWThen, the air temperature at the compressor exit becomes WC mair c p (T2 T1 ) 1.021 kW (0.018 kg/s)(1.011 kJ/kg. C)(T2 - 70) C T2 126.1 C(b) The air temperature at the compressor exit for the case of isentropic process is ( k 1) / k (1.397 -1)/1.397 P2 135 kPa T2 s T1 (70 273 K) 379 K 106 C P1 95 kPaThe isentropic efficiency of the compressor is determined to be T2 s T1 106 70 C 0.642 T2 T1 126.1 70
• 7-1657-211 Air is compressed in a compressor that is intentionally cooled. The work input, the isothermalefficiency, and the entropy generation are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic andpotential energy changes are negligible. 3 Air is an ideal gas with Q 300 Cconstant specific heats. 1.2 MPaProperties The gas constant of air is R = 0.287 kJ/kg.K and thespecific heat of air at an average temperature of (20+300)/2 = 160ºC Compressor= 433 K is cp = 1.018 kJ/kg.K (Table A-2).Analysis (a) The power input is determined from an energy balanceon the control volume Air WC mc p (T2 T1 ) Qout 20 C, 100 kPa (0.4 kg/s)(1.018 kJ/kg. C)(300 20) C 15 kW 129.0 kW(b) The power input for a reversible-isothermal process is given by P2 1200 kPa WT const. mRT1 ln (0.4 kg/s)(0.287 kJ/kg.K)(20 273 K)ln 83.6 kW P1 100 kPaThen, the isothermal efficiency of the compressor becomes WT const. 83.6 kW T 0.648 WC 129.0 kW(c) The rate of entropy generation associated with this process may be obtained by adding the rate ofentropy change of air as it flows in the compressor and the rate of entropy change of the surroundings T2 P2 Qout S gen S air S surr c p ln R ln T1 P1 Tsurr 300 273 K 1200 kPa 15 kW (1.018 kJ/kg.K)ln (0.287 kJ/kg.K)ln 20 273 K 100 kPa (20 273) K 0.0390 kW/K
• 7-1667-212 Air is allowed to enter an insulated piston-cylinder device until the volume of the air increases by50%. The final temperature in the cylinder, the amount of mass that has entered, the work done, and theentropy generation are to be determined.Assumptions 1 Kinetic and potential energy changesare negligible. 2 Air is an ideal gas with constantspecific heats.Properties The gas constant of air is R = 0.287 AirkJ/kg.K and the specific heats of air at room 0.25 m3temperature are cp = 1.005 kJ/kg.K, cv = 0.718 0.7 kg AirkJ/kg.K (Table A-2). 20 C 500 kPaAnalysis The initial pressure in the cylinder is 70 C m1 RT1 (0.7 kg)(0.287 kPa m 3 /kg K)(20 273 K)P1 235.5 kPa V1 0.25 m 3 P2V 2 (235.5 kPa)(1.5 0.25 m 3 ) 307.71m2 3 RT2 (0.287 kPa m /kg K)T2 T2A mass balance on the system gives the expression for the mass entering the cylinder 307.71 mi m2 m1 0.7 T2(c) Noting that the pressure remains constant, the boundary work is determined to be W b,out P1 (V 2 V1 ) (235.5 kPa)(1.5 0.25 - 0.5)m 3 29.43 kJ(a) An energy balance on the system may be used to determine the final temperature mi hi Wb, out m2u2 m1u1 mi c pTi Wb, out m2cv T2 m1cv T1 307.71 307.71 0.7 (1.005)(70 273) 29.43 (0.718)T2 (0.7)(0.718)(20 273) T2 T2There is only one unknown, which is the final temperature. By a trial-error approach or using EES, we find T2 = 308.0 K(b) The final mass and the amount of mass that has entered are 307.71 m2 0.999 kg 308.0 mi m2 m1 0.999 0.7 0.299 kg(d) The rate of entropy generation is determined from S gen m2 s 2 m1 s1 mi s i m2 s2 m1 s1 (m 2 m1 ) s i m 2 (s 2 s i ) m1 ( s1 si ) T2 P2 T1 P1 m 2 c p ln R ln m1 c p ln R ln Ti Pi Ti Pi 308 K 235.5 kPa (0.999 kg) (1.005 kJ/kg.K)ln (0.287 kJ/kg.K)ln 343 K 500 kPa 293 K 235.5 kPa (0.7 kg) (1.005 kJ/kg.K)ln (0.287 kJ/kg.K)ln 343 K 500 kPa 0.0673 kJ/K
• 7-1677-213 A cryogenic turbine in a natural gas liquefaction plant produces 350 kW of power. The efficiency ofthe turbine is to be determined.Assumptions 1 The turbine operates steadily. 2 The properties of methane is used for natural gas.Properties The density of natural gas is given to be 423.8 kg/m3.Analysis The maximum possible power that can be obtained from thisturbine for the given inlet and exit pressures can be determined from 3 bar m (55 kg/s) Wmax ( Pin Pout ) (4000 300)kPa 480.2 kW Cryogenic 423.8 kg/m3 turbineGiven the actual power, the efficiency of this cryogenic turbine becomes W 350 kW 0.729 72.9% LNG, 40 bar Wmax 480.2 kW -160 C, 55 kg/sThis efficiency is also known as hydraulic efficiency since the cryogenicturbine handles natural gas in liquid state as the hydraulic turbine handlesliquid water.
• 7-168Fundamentals of Engineering (FE) Exam Problems7-214 Steam is condensed at a constant temperature of 30 C as it flows through the condenser of a powerplant by rejecting heat at a rate of 55 MW. The rate of entropy change of steam as it flows through thecondenser is(a) –1.83 MW/K (b) –0.18 MW/K (c) 0 MW/K (d) 0.56 MW/K (e) 1.22 MW/KAnswer (b) –0.18 MW/KSolution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines ona blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).T1=30 "C"Q_out=55 "MW"S_change=-Q_out/(T1+273) "MW/K""Some Wrong Solutions with Common Mistakes:"W1_S_change=0 "Assuming no change"W2_S_change=Q_out/T1 "Using temperature in C"W3_S_change=Q_out/(T1+273) "Wrong sign"W4_S_change=-s_fg "Taking entropy of vaporization"s_fg=(ENTROPY(Steam_IAPWS,T=T1,x=1)-ENTROPY(Steam_IAPWS,T=T1,x=0))7-215 Steam is compressed from 6 MPa and 300 C to 10 MPa isentropically. The final temperature of thesteam is(a) 290 C (b) 300 C (c) 311 C (d) 371 C (e) 422 CAnswer (d) 371 CSolution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines ona blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).P1=6000 "kPa"T1=300 "C"P2=10000 "kPa"s2=s1s1=ENTROPY(Steam_IAPWS,T=T1,P=P1)T2=TEMPERATURE(Steam_IAPWS,s=s2,P=P2)"Some Wrong Solutions with Common Mistakes:"W1_T2=T1 "Assuming temperature remains constant"W2_T2=TEMPERATURE(Steam_IAPWS,x=0,P=P2) "Saturation temperature at P2"W3_T2=TEMPERATURE(Steam_IAPWS,x=0,P=P2) "Saturation temperature at P1"
• 7-1697-216 An apple with an average mass of 0.15 kg and average specific heat of 3.65 kJ/kg. C is cooled from20 C to 5 C. The entropy change of the apple is(a) –0.0288 kJ/K (b) –0.192 kJ/K (c) -0.526 kJ/K (d) 0 kJ/K (e) 0.657 kJ/KAnswer (a) –0.0288 kJ/KSolution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines ona blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).C=3.65 "kJ/kg.K"m=0.15 "kg"T1=20 "C"T2=5 "C"S_change=m*C*ln((T2+273)/(T1+273))"Some Wrong Solutions with Common Mistakes:"W1_S_change=C*ln((T2+273)/(T1+273)) "Not using mass"W2_S_change=m*C*ln(T2/T1) "Using C"W3_S_change=m*C*(T2-T1) "Using Wrong relation"7-217 A piston-cylinder device contains 5 kg of saturated water vapor at 3 MPa. Now heat is rejected fromthe cylinder at constant pressure until the water vapor completely condenses so that the cylinder containssaturated liquid at 3 MPa at the end of the process. The entropy change of the system during this process is(a) 0 kJ/K (b) -3.5 kJ/K (c) -12.5 kJ/K (d) -17.7 kJ/K (e) -19.5 kJ/KAnswer (d) -17.7 kJ/KSolution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines ona blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).P1=3000 "kPa"m=5 "kg"s_fg=(ENTROPY(Steam_IAPWS,P=P1,x=1)-ENTROPY(Steam_IAPWS,P=P1,x=0))S_change=-m*s_fg "kJ/K"7-218 Helium gas is compressed from 1 atm and 25 C to a pressure of 10 atm adiabatically. The lowesttemperature of helium after compression is(a) 25 C (b) 63 C (c) 250 C (d) 384 C (e) 476 CAnswer (e) 476 CSolution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines ona blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).k=1.667P1=101.325 "kPa"T1=25 "C"P2=10*101.325 "kPa"
• 7-170"s2=s1""The exit temperature will be lowest for isentropic compression,"T2=(T1+273)*(P2/P1)^((k-1)/k) "K"T2_C= T2-273 "C""Some Wrong Solutions with Common Mistakes:"W1_T2=T1 "Assuming temperature remains constant"W2_T2=T1*(P2/P1)^((k-1)/k) "Using C instead of K"W3_T2=(T1+273)*(P2/P1)-273 "Assuming T is proportional to P"W4_T2=T1*(P2/P1) "Assuming T is proportional to P, using C"7-219 Steam expands in an adiabatic turbine from 8 MPa and 500 C to 0.1 MPa at a rate of 3 kg/s. If steamleaves the turbine as saturated vapor, the power output of the turbine is(a) 2174 kW (b) 698 kW (c) 2881 kW (d) 1674 kW (e) 3240 kWAnswer (a) 2174 kWSolution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines ona blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).P1=8000 "kPa"T1=500 "C"P2=100 "kPa"x2=1m=3 "kg/s"h1=ENTHALPY(Steam_IAPWS,T=T1,P=P1)h2=ENTHALPY(Steam_IAPWS,x=x2,P=P2)W_out=m*(h1-h2)"Some Wrong Solutions with Common Mistakes:"s1=ENTROPY(Steam_IAPWS,T=T1,P=P1)h2s=ENTHALPY(Steam_IAPWS, s=s1,P=P2)W1_Wout=m*(h1-h2s) "Assuming isentropic expansion"7-220 Argon gas expands in an adiabatic turbine from 3 MPa and 750 C to 0.2 MPa at a rate of 5 kg/s. Themaximum power output of the turbine is(a) 1.06 MW (b) 1.29 MW (c) 1.43 MW (d) 1.76 MW (e) 2.08 MWAnswer (d) 1.76 MWSolution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines ona blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).Cp=0.5203k=1.667P1=3000 "kPa"T1=750 "C"m=5 "kg/s"P2=200 "kPa"