Asummaryofthermodynamicprocesses：热力学过程的总结.doc

Thermodynamic ProcessesCalorimetryChange of phase?Type of heatEquationTemperatureCalculationChange?Nospecific heatdQ = mCdT 1YesYeslatent heatdQ = mL No1 Caution! Be careful of molar versus mass based specific heat constants.Ideal Gas LawPV = nRT,P = pressure in Pascals (N/m2)V = volume in m3n = number of moles (dimensionless)R = gas constantT = temperature in Kelvin (not Celsius!)Other Key EquationsdU = dQ dW(first law of thermodynamics)dQ = nCVDT(ideal gas, specific heat at constant volume)dQ = nCPDT(ideal gas, specific heat at constant pressure)dU = nCVDT(ideal gas, derivation attached)CP CV = R(statistical mechanics)Internal Energy of an Ideal GasThe internal energy depends only on the endpoints. Pick a constant volume and constant pressure line segments to connect the endpoints. Using the first law:DU = nCV(T T0) + nCP(Tf T) 0 Pf(Vf-V0) = nCV(Tf T0), sincePfV0 = nRTLaws of Thermodynamics for Ideal Gasesprocessmeaningwork (DW)heat (DQ)entropy (DS)isobaricconstant pressureP0(VF V0)nCP(TF T0)nCP ln(TF/T0)isochoricconstant volume0nCV(TF T0)nCV ln(TF/T0)isothermalconstant temperature(nRT0)ln(VF/V0)1(nR)ln(VF/V0)adiabatic 2no heat exchange(PFVF- P0V0) /(1 - g) 0031From the first law of thermodynamics, dU = dQ dW; dU=0 for an isothermal process, so dQ = dW, or DQ = DW2From the first law of thermodynamics, dU = dQ dW; dQ=0 for an adiabatic process, so dU = -dW =nCVdT = -PdV;from the ideal gas law, dT = d(PV/nR), son(CV/nR) d(PV) = -PdVn(CV/nR) PdV + VdP = -PdV,VdP = -PdV 1 + (CV/R) / (CV/R);since R = CP - CV, g CP / CVVdP = -gPdV , orP/P0 = (V/V0)-g,orPVg = P0V0g.3dW = PdV, so DW = (P0V0g) V-gdV = (P0V0g) V1-g/(1 - g), V V0,VFDW = (P0V0g) V1-g/(1 - g) = (PFVF- P0V0) /(1 - g)Examples follow(1) a simple example(2) Carnot cycle(3) Otto cycle(4) Diesel cycle(5) Stirling cycleExample 1: A Simple ExamplePV4P0P03V0V01243Heat calculations:Work calculations:DQ12 = 8(CP/R)P0V0DW12 = 8P0V0DQ23 = -9(CV/R)P0V0DW23 = DW41 = 0DQ34 = -2(CP/R)P0V0DW34 = -2P0V0DQ41 = 3(CV/R)P0V0Entropy calculations:Sums:DS12 = nCPln(3)DQ = DW = 6P0V0DS23 = -nCVln(4)DU = DQ - DW = 0 (expected, closed cycle)DS34 = -nCPln(3)DS = 0 (reversible process)DS41 = nCVln(4)efficiency:QH = DQ12 + DQ41 = (8CP + 3CV)P0V0/R(sum of positive heat results)QC = |DQ23 + DQ34| = (2CP + 9CV)P0V0/R(sum of negative heat results)e = 1 - (2CP + 9CV)/ (8CP + 3CV) = 1 - (2g + 9)/ (8g + 3);for a monatomic gas, g = 5/3 and e = 0.24Carnot efficiency:TC = T4 = P0V0/(nR)TH = T2 = 12P0V0/(nR)e = 1 - TC/TH = 0.92 (notice that the actual efficiency is much lower)Example 2: Carnot CycleSTATETaTHbTHcTCdTC_STEPTYPEDQDWDUDSa-bisothermalnRTHln(Vb/Va)DQ0nRln(Vb/Va)b-cadiabatic0DUnCV(TC - TH)0c-disothermalnRTCln(Vd/Vc) DQ0nRln(Vd/Vc)d-aadiabatic0DUnCV (TH - TC)0_efficiency:DQab = nRTHln(Vb/Va) 0DQcd = -nRTCln(Vd/Vc) Vb/Va = Vc/Vd = |QC| / | QH | = (TC/TH)TdVdg-1 = TaVag-1|e = 1 - (TC/TH)entropy:DS = 0, see efficiency calculation. Reversible process.Example 3: Otto CycleSTATEPVTaPaVa = rVbTabPb = Pa rgVbTb = Ta rg-1cPc = Pb(Tc/Tb)VbTc = Td rg-1dPd = Pc(1/r)gVa = rVbTd_STEPTYPEDQDWDUDSa-badiabatic0nCV(Tb Ta)-DW0b-cisochoricnCV(Tc Tb)0DQnCVln(Tc/Tb)c-dadiabatic0nCV(Td Tc)-DW0d-aisochoric nCV(Ta Td)0DQnCVln(Ta/Td)_efficiency:DQbc = nCV(Tc Tb) 0DQcd = nCV(Ta Td) Tb Ta (since PcPb);Td/Ta = Tc/Tb 1 = Td Ta;so that Tc = TH and Ta = TCOLD; using these temperatures, the Carnot efficiency is e = 1 (1/ rg-1)( Ta/Td) Otto efficiencyentropy:DS = nCVln(Tc/Tb) + nCVln(Ta/Td) = nCVln(Tc/Tb)(Ta/Td)= nCVln(Td/Ta)(Ta/Td)= nCVln (1) = 0DS = 0. Reversible process.Example 4: Diesel CycleSTATEPVTaPaVa = rVbTabPb = Pa rgVbTb = Ta rg-1cPbVc= rcVbTc = (Vc/Vb)Tb = (Vc/Vb)Ta rg-1dPd = Pa (Vc/Vb)gVa = rVbTd = (Vc/Vb)g (1/r)g-1Tb = (Vc/Vb)g Ta_STEPTYPEDQDWDUDSa-badiabatic0nCV(Tb Ta)-DW0b-cisobaricnCP(Tc Tb)0DQnCPln(Tc/Tb) c-dadiabatic0nCV(Td Tc)-DW0d-aisochoric nCV(Ta Td)0DQnCVln(Ta/Td)_efficiency:DQbc = nCP(Tc Tb) 0DQcd = nCV(Ta Td) 1 2/ (5 r2/3) for monatomicNOTE: for rc 1, e - 1 (1/ g)(rc/r) g-1 = 1 3(r/rc)-2/3 for monatomicentropy:DS = nCPln(Tc/Tb) + nCVln(Ta/Td)= nCV gln(Tc/Tb) + ln(Ta/Td)= nCV ln(Tc/Tb)g(Ta/Td)= nCV ln(Vc/Vb)g (Vb/Vc)g= nCV ln(1) = 0DS = 0. Reversible process.Example 5: Stirling CycleSTATEPVTaPaVa = rVbTCbPbVbTCcPcVbTHdPdVa = rVbTH_STEPTYPEDQDWDUDSa-bisothermalDW-nRTC ln(r)0nRln(Vb/Va)b-cisochoricnCV(TH TC)0DQnCVln(Tc/Tb) c-disothermalDWnRTH ln(r)0nRln(Vd/Vc)d-aisochoric -nCV(TH TC)0DQnCVln(Ta/Td)_efficiency:DQcd = nRTH ln(r) + nCV(TH TC) 0DQab = -nRTC ln(r) - nCV(TH TC) 0 |QC| / | QH | = TH ln(r) + (CV/R)(TH TC) / TC ln(r) + (CV/R)(TH TC)= TH (ln(r) + CV/R) (CV/R)TC) / TC (ln(r) - CV/R) + (CV/R)THe = 1 -