1、Advanced Chemical Engineering Thermodynamics:Solution ManualDue on Oct. 10, 2016Associate Professor Diannan Lu , Seminar 1stDiannan LuDigital Version Created byXiaoyu Hu1Diannan Lu Seminar 1st: Solution Manual Homework 3-3Homework 3-3SolutionaB = 1:29 10 3 m2; eB = 1=2 0:508 = 0:254 mA, B, Cylinder!
2、nonconducting, no heat capacity, having some friction.What we know:Thus PEaA = PDaA + (mA +mB +mC)g)PE = PD + (mA +mB +mC)gaA= 1:013 105 + (9:07 + 4:53 + 18:14) 9:816:45 10 3= 1:50 105 PaIf we consider air pressure,P0E = PE +Pair aBaA= 1:72 105 Pa2Diannan Lu Seminar 1st: Solution Manual Homework 3-3
3、for aB=aA, in present study, we omit thisVD;i = eB (aA aB) = 0:254 (6:45 1:29) 10 3 = 1:31 10 3 m3PD;iVD;i = NDR TD;i)ND = 1:013 105 1:31 10 38:314 311 = 5:13 102 molVE;i = eE aA = 0:254 6:45 10 3 = 1:64 10 3 m3PE;iVE;i = NER TD;i)NE = 1:50 105 1:64 10 38:314 311 = 9:51 102 molAll initial conditions
4、:PD;i = 1:013 105 Pa; TD;i = 311 K; ND = 5:13 10 2 mol; VD;i = 1:31 10 3 m3PE;i = 1:50 105 Pa; TE;i = 311 K; NE = 9:51 10 2 mol; VE;i = 1:64 10 3 m3(a) Expansion diathermalConstraints:1 diathermal TD;f = TE;f = Tf2 Force balance (omit air pressure)Piston aB +PD;f(aA aB) + (mA +mB)g = PE;faA (1)3 Ide
5、al Gas:D : PD;f(aA aB)(eB x) = NDR Tf (2)E : PE;f aA(eE +x) = NER Tf (3)4 1st Law: System(D+E)+A+BE = Q+WU = NDCV (Tf TD;i) +NECV (Tf TE;i)E = U + (mA +mB)g xQ = 0W = Patm aB x3Diannan Lu Seminar 1st: Solution Manual Homework 3-3) (ND +NE)CV (Tf Ti) + (mA +mB)g x = PatmaBx (4)Combine (1)(4) Four var
6、iables, Four Equations, Done.Lets work it out!1:013 105 1:29 10 2 +PD;f (6:45 1:29) 10 3 + (9:07 + 4:53) 9:81 = PE;f 6:45 10 3) PE;f = 0:8PD;f + 0:409 105(2) PD;f(6:45 1:29) 10 3(0:254 x) = 5:13 10 2 8:314 Tf) PD;f(0:254 x) = 82:66Tf(3) PE;f 6:45 10 3(0:254 +x) = 9:51 10 2 8:314 Tf) PE;f(0:254 +x) =
7、 122:58Tf(b) Piston is adiabaticTD;f 6= TE;f1 First EquationPD;f(aA aB) + (mA +mB)g +PatmaB = PE;faAPD;f(6:45 1:29) 10 3 + (9:07 + 4:53) 9:81 + 1:013 105 1:09 10 3 = PE;f 6:45 10 3)0:8PD;f + 0:4094 105 = PE;f2 Second EquationPD;f(aA aB)(eB x) = RNDTD;f)PD;f(6:45 1:29) 10 3(0:254 x) = 5:13 10 2 8:314
8、 Tf)PD;f(0:254 x) = 82:66 TD;f3 Third EquationPE;f(aA)(eB +x) = NERTE;f)PE;f 6:45 10 3 (0:254 +x) = 9:51 10 2 8:314 TE;f)PE;f(0:254 +x) = 122:58TE;f4 Fourth EquationNDCV (TD;f Ti) +NECV (TE;f Ti) + (mA +mB)g x = PatmaB x)5:13 10 2 12:6(TD;f Ti) + 9:51 10 2 12:6(TE;f Ti) + (9:07 + 4:53) 9:81 x= 1:013
9、 105 1:29 10 3 x)0:6464(TD;f Ti) + 1:198(TE;f Ti) = 264:093xFive variables, four equations.1. assuming E expands reversibly and adiabaticallyFor E,dUE = Q+ WNECV dTE = PEdVE; PEVE = NER TE4Diannan Lu Seminar 1st: Solution Manual Homework 3-3Therefore,)CV dTE = RdTE + (RTEPE)dPE) (CV +R)dTE = RT dlnP
10、E) TE;fTi=PE;fPE;iR=Cp+R=PE;fPE;i0:3975)TE;f = Ti PE;f=1:013 105 0:3975 (5)We get the ANSWER!8:x = 0:0187 mPD;f = 1:16 105 PaPE;f = 1:34 105 PaTE;f = 296:9 KTD;f = 329:4 K2. assuming D expands reversibly and adiabaticallyTD;fTi =PD;fPD;i0:3975)TD;f = Ti PD;f=(1:013 105) 0:3975 (5)Then we get the ANS
11、WER!8:x = 0:0195 mPD;f = 1:16 105 PaPE;f = 1:33 105 PaTE;f = 297:5 KTD;f = 327:7 KEND of SOLUTION5Diannan Lu Seminar 1st: Solution Manual Homework 3-8Homework 3-8SolutionData:PCylinder = 15:17 MPa TCylinder = 311:0 K VR;i =? PR;i = 0:101 MPa TR;i = 311 KCP = 29:3 J/mol K CV = 20:9 J/mol Ka) mixing c
12、ompletelyour system, open, adiabatic, rigiddUR = QR + W +HindNinW= 0, rigid, no movement of bounding andUR = UR Nin;we omit the gas in the VR originally.) dUR = dUR Nin +URdNin = HindNinCV dT Nin +URdNin = HindNin)NinCV dT = (Hin UR)dNin;where dHin = CpdT, so Hin = Hin(T0) +CP(Te T0), where T0 is th
13、e reference temperature, and Te is 311K.Besides, we know U = U(T0) +CV (T T0), where T is unknown.So we get,NinCV dT = (Hin(T0) U(T0) (CP CV )T0 +CPTe CVTdNin;and the reason why we can cancel these is that for ideal gas, Hin(T0) U(T0) = (CP CV )T0 = RT0. Thuswe have,NinCV dT = (CPTe CVT)dNin: (1)Nex
14、t question: Nin?Another Equation is from PV = NinRT: Use d(P(V) = d(NinRT), we getVRdP = NinRdT +RTdNin;where weve known change of P, so we want the relation between T and Nin, and we get.dNin = 1RT (VRdP RNindT)6Diannan Lu Seminar 1st: Solution Manual Homework 3-8Replace dNin in equation (1), and w
15、e getNinCV dT = (CPTe CVT) 1RT ( VRdP R NindT)CPTeT dT = (CPTe CVT)dPP :Let= CPCV= 29:320:9 = 1:4;so we get1TTeTe TdT = dPP :Integrate this we get ZTfTi1Te T +1TdT =Z PfPi1P dPhere Ti = Te = 311 K; Pi = 0:101 MPa; Pf = 15:17 MPa, and we get Tf = 434 Kb) no mixingPf = 15:17 MPa; Pi = 0:101 MPa; Ti =
16、311 K; N = constTf =? Vi =?; Vf =?Use 1st law,dU = Q+ W;and we getNCV dT = PdVFrom PV = NRT, we have d(PV) = d(NRT),)PdV +VdP = NRdT) NCV dT +VdP = NRdT) NCV dT + NRTP dP = NRdT)CV +RRTdT = 1P dP) TfTi=PfPiRCV+R)T = 46711 K 800 K7Diannan Lu Seminar 1st: Solution Manual Homework 4-2END of SOLUTIONHom
17、ework 4-2Solution“Birds or co ee percolator“ is regarded as “Heat machine“Assumption: no heat losses, totally reversible, normal ambient conditions, working substance is waterWe know the heat engine works between TH and TL.For reversible engine,rev = WQH= TH TLTHWhat is TH, heat ows from the environ
18、ment to the engine.What is TL, heat ows to the environment from the engine.Vapour to liquid at TLFor L it is easy. All water is condensed! so QL = Hvap at TL.Liquid to vapour at THFor H it is not easy to determine. We dont know how much water is vapoured!Thus we get (WQH = TH TLTHQH+ QC +W = 0From t
19、hese equations, we getWQC +W =TH TLTHTherefore,WQC =TH TLTLAssuming:TH = 300 K; TL = 280 K; QL = 2485 kJ/kgthen we getW = QL TH TLTH= 2485 300 280280= 177:5 kJ/kg8Diannan Lu Seminar 1st: Solution Manual Homework 4-11Using W = gh, we haveh = 177:5 1039:81 = 18:1 kmEND of SOLUTIONHomework 4-11Solution
20、(a) cocurrent_Wmax = W(T; _nH; _nC;CPH;CPC)For Carnot Engine,_W = _QH _QC: pay attention to “sign“Clausius theorem. for reversible process, dS = 0QHTH +QCTC = 0For steady and constant pressure,_QH = d _HH = _nHCPHdTH_QC = d _HC = _nCCPCdTC)Z TTH_nHCPHdTHTH +Z TTC_nCCPCTC dTC = 0) _nHCPH ln TTH+ _nCC
21、PC ln TTC= 0We de ne_nHCPH_nCCPC= 20 10616 106 = 1:259Diannan Lu Seminar 1st: Solution Manual Homework 4-11Therefore,ln TTH+ ln TTC= 0and thenT +1 = T HTC ) T = (T HTC)1=1+ = (3001:25 278)1=2:25 = 290 K:Thus, we have_Wmax = _QH _QC= _nHCPH(TH T) _nCCPC(T TC)= _nHCPHhTH (T HTC)1=1+ i_nCCPCh(T HTC)1=1
22、+ TCi(b) pinch temperature?(c) countercurrent caseQHTH +QHTC = 0For steady and constant pressure_QH = d _HH = _nHCPHdTH_QC = d _HC = _nCCPCdTCTherefore, ZTH;fTH_nHCPHdTHTH Z TCTC;f_nCCPCTC dTC = 0Let_nHCPH_nCCPCthen we getZ TH;fTHdTHTH Z TCTC;fdTCTC = 0thusTC;f = TCTH;fTH10Diannan Lu Seminar 1st: So
23、lution Manual Homework 4-11and_Wmax = _QH _QC= _nHCPHTH TH;f _nCCPCTC;f TCFor , we have=_W_nCCPC = TH TfH TC“ TH;fTH1#Let TH;f= 0 and we get,TH;f = (TC(TH) )1=1+ andTC;f = TCTH;fTH= TCT1=1+ C T =1+ HTH! = (TC(TH) )1=1+ END of SOLUTION11Diannan Lu Seminar 1st: Solution Manual The Speed of SoundThe Sp
24、eed of SoundSolution1st LawdU = Q+ W + dminHin + 12u2indmoutHout + 12u2outwith mass balance dmin = dmout, we getHin + 12u2in = Hout + 12u2out:Therefore,dH + 12du2 = 0: (1)With mass balance, we getdu AV= 0) udV +VduV2 = 0) duu = dVV :Using dH = TdS +VdP, and for the adiabatic, reversible process, dS
25、= 0, we havedH = VdP:With equation (1), we haveVdP + 12du2 = 0)VdP = u2V dV )u2 = V2dPdV :Thus, we getuc =s P S:How to calculate ( P)SAssuming Ideal Gas, and is adiabaticdU = Q+ W:With12Diannan Lu Seminar 1st: Solution Manual The Speed of SoundNCV dT = PdVPV = NRTwe havePV = constwhere = Cp=CV = 1:4.Thus we havelnPVS= V )PVS= PV :So the answer isuc =pRT=MW=r1:4 8:314 298:1529 10 3= 346 m/sEND of SOLUTION13
