化工热力学国际班ch

Chapter7ApplicationsofThermodynamicstoFlowProcess

流动过程热力学ConservationofMassConservationEqnMassbalanceConservationofenergyTheentropybalancesSgenerationSflowwithmassSflowwithheatTheSbalancefortheopensystem：S-balanceEx.Saturatedsteam150℃Saturatedwater150℃.Tsur=20℃,min=mout=100kg/s.Sg=?Solution:Massbalance:Heatbalance:byconsultingsteamtable:(p684)Entropybalance:Sm1=6837.9J/(kg.K),Sm2=1841.8J/(kg.K)reTheprocessisirreversible.Ifweassembleaheatengineoperatedbetween150℃and20℃,wecanrecoveramountofworkof：Ex.Air,T1=500K,P1=0.1MPa,m1=10kg/s;T2=300K,P2=0.1MPa,m2=5kg/s,mixedadiabaticlly,Sg=?Solution:Idealgas:Massbalance:Energybalance:Setreferencestate:T0=300K,P0=0.1MPa,H0=0,S0=0.Entropybalance:Steadyadiabaticprocess稳态绝热过程：Mixing----irreversible!7.4.1DuctFlowTheBernouliiEqn7.4.2Compressionprocess

TS123T2P1P2T3Ex.NH3,-8℃,0.304MPa,compressedadiabaticallyto1.419MPa.s=0.8.Ws(R)=?,Ws=?Sg=?Solution:a.reversiblecompressionFromT-Sgraph,H1=1443.5kJ/kg,S1=S2=5.5438kJ/(kg.K),H2=1665.2kJ/kg,then,Ws(R)=H2-H1=1665.2-1443.5=221.7kJ/kgb.Irreversiblewiths=0.8FromT-Sgraph,S2=5.6484kJ/(kg.K)Thefundamentalthermodynamicrelation:Forisothermalcompressionofideagas:ReversibleshaftworkReversibleprocess:Then:The1stLaw:Foradiabaticcompressionofideagas:Forpolytropiccompressionofideagas:Forrealgas:IsothermalcompressionForideagas:AdiabaticcompressionPolytropiccompressionEx.Air1kg,0.10814MPa,288.75K1.8424MPaWs(R)=?Solution:takeairasidealgasa.Isothermalcompression:b.Adiabaticcompression:K=1.4c.Polytropiccompression:m=1.25Theterminaltemperature:Polytropiccompression:Adiabaticcompression:Thecalculationresults：CompressionprocessTerminalTWorkconsumedisothermal15.6234.6polytropic235.83315.7adiabatic375.79361.44PumpFormostliquid:7.4.3Turbines(expanders)SteamturbineAdiabaticReversibleIsentropicIsentropicefficiencyTurbineefficiency:TS123T2P1P2T3Isotropicexpansion12Ifitchangesirreversibly:13：Theturbineefficiency(isentropicefficiency)：Ex.Ethylene(C2H4),573K,4.5MPa,expendsadiabaticallyandreversiblyto0.2Mpa,ws=?Solution:Basedon1molofethylene:a)Byidealgaslaw:Firstly,solveentropybalanceforT2withNewtonmethod:Till:T2=370.79KThen:b)BygeneralizedVirialcoefficientcorrelation:realgasForethylene:Atinitialstate:WithgeneralizedVirialcorrelation60Theexhaustedgas，P=0.2MPa,realgas(ifwereferitasidealgas,theresultdefersalittle).SelectinitialvalueofT2accordinga)：T2=371KThenSolveforT2byNewtoniteration,andfinallywehave:T2=365.79KWiththenewT2：7.4.4Throttling节流TheprocessoccursatconstantenthalpyTS1234P1P256s1s2Ex.7.8C3H8(propane),①400K,2MPathrottledto②0.1MPa.T2=?S=?forthrottlingH=0Solution:Forpropane:Atinitialstate:With