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Solution:C.V.: Stage 1 air, Steady flowProcess:adibatic: q = 0,reversible:sgen= 0Energy Eq.6.13:-wC1= h2²h1,Entropy Eq.9.8:s2= s1Assume constant CP0= 1.004 from A.5and isentropic leads to Eq.8.32T2= T1(P2/P1)k-1k= 300(2000/100)0.286= 706.7 KwC1= h1- h2= CP0(T1- T2) = 1.004(300 – 706.7) =-408.3 kJ/kgC.V. Intercooler, no work and no changes in kinetic or potential energy.q23= h3- h2= CP0(T3- T2) = 1.004(340 – 706.7) =-368.2 kJ/kgC.V. Stage 2. Analysis the same as stage 1. So from Eq.8.32T4= T3(P4/P3) = 340(15.74/2)0.286= 613.4 KwC2= h3- h4= CP0(T3- T4) = 1.004(340 – 613.4) =-274.5 kJ/kgSame flow rate through both stages so the total work is the sum of the twowcomp= wC1+ wC2= –408.3 – 274.5 = –682.8 kJ/kgFor no intercooler (P2= 15.74 MPa) same analysis as stage 1. So Eq.8.32T2= 300(15740/100)0.286= 1274.9 Kwcomp= 1.004(300 – 1274.9) = –978.8 kJ/kgC1C212342-W·Q·1-W·coolerinter-
9-249.22A heat-powered portable air compressor consists of three components: (a) anadiabatic compressor; (b) a constant pressure heater (heat supplied from an outsidesource); and (c) an adiabatic turbine. Ambient air enters the compressor at 100 kPa,300 K, and is compressed to 600 kPa. All of the power from the turbine goes intothe compressor, and the turbine exhaust is the supply of compressed air. If thispressure is required to be 200 kPa, what must the temperature be at the exit of theheater?Solution:HeaterCTqH1234P2= 600 kPa,P4= 200 kPaAdiabatic and reversible compressor:Process:q = 0andsgen= 0Energy Eq.6.13:h²wc= h2Entropy Eq.9.8:s2= s1For constant specific heat the isentropic relation becomes Eq.8.32T2= T1©¨§¹¸·P2P1k-1k= 300(6)0.286= 500.8 K²wc= CP0(T2- T1) = 1.004(500.8²300) = 201.5 kJ/kgAdiabatic and reversible turbine:q = 0andsgen= 0Energy Eq.6.13:h3= wT+ h4;Entropy Eq.9.8:s4= s3For constant specific heat the isentropic relation becomes Eq.8.32T4= T3(P4/P3)k-1k= T3(200/600)0.286= 0.7304 T3Energy Eq. for shaft:²wc= wT= CP0(T3²T4)201.5 = 1.004 T3(1²0.7304)=>T3=744.4 K21vTs1300100 kPa3P234200 kPa600 kPa4
9-259.23A certain industrial process requires a steady 0.5 kg/s supply of compressed air at500 kPa, at a maximum temperature of 30qC. This air is to be supplied byinstalling a compressor and aftercooler. Local ambient conditions are 100 kPa,20qC. Using an reversible compressor, determine the power required to drive thecompressor and the rate of heat rejection in the aftercooler.Solution:Air Table A.5:R = 0.287 kJ/kg-K,Cp= 1.004 kJ/kg K,k = 1.4State 1:T1= To= 20oC,P1= Po= 100 kPa,m.= 0.5 kg/sState 2: P2= P3= 500 kPaState 3: T3= 30oC, P3= 500 kPaCompressor: Assume Isentropic (adiabaticq = 0and reversiblesgen= 0 )From entropy equation Eq.9.8 this gives constant s which is expressed for anideal gas in Eq.8.32T2= T1(P2/P1)k-1k,T2= 464.6 K1stLaw Eq.6.13:qc+ h1= h2+ wc;qc= 0,assume constant specific heat from Table A.5wc= Cp(T1- T2) = -172.0 kJ/kgW.C= m.wC=-86 kWAftercooler:1stLaw Eq.6.13:q + h2= h3+ w;w = 0,assume constant specific heatq = Cp(T3- T2) = -205 kJ/kg,Q.= m.q =-102.5 kWCompressorcoolerQ-WC123..coolerafter-
9-26Steady state irreversible processes9.24Analyze the steam turbine described in Problem 6.44. Is it possible?

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