清华大学高等化工热力学习题答案.pdf

Advanced Chemical Engineering Thermodynamics: Solution Manual Due on Oct. 10, 2016 Associate Professor Diannan Lu , Seminar 1st Diannan Lu Digital Version Created by Xiaoyu Hu 1 Diannan LuSeminar 1st: Solution ManualHomework 3-3 Homework 3-3 Solution aB= 1.29 103m2, eB= 1/2 0.508 = 0.254 m A, B, Cylindernonconducting, no heat capacity, having some friction. What we know: Thus PEaA= PDaA+ (mA+ mB+ mC)g PE= PD+ (mA+ mB+ mC)g aA = 1.013 105+ (9.07 + 4.53 + 18.14) 9.81 6.45 103 = 1.50 105Pa If we consider air pressure, P0 E = PE+ Pair aB aA = 1.72 105Pa 2 Diannan LuSeminar 1st: Solution ManualHomework 3-3 for aB/aA, in present study, we omit this VD,i= eB (aA aB) = 0.254 (6.45 1.29) 103= 1.31 103m3 PD,iVD,i= NDR TD,i ND= 1.013 105 1.31 103 8.314 311 = 5.13 102mol VE,i= eE aA= 0.254 6.45 103= 1.64 103m3 PE,iVE,i= NER TD,i NE= 1.50 105 1.64 103 8.314 311 = 9.51 102mol All initial conditions: PD,i= 1.013 105Pa,TD,i= 311 K,ND= 5.13 102mol,V D,i = 1.31 103m3 PE,i= 1.50 105Pa,TE,i= 311 K,NE= 9.51 102mol,VE,i= 1.64 103m3 (a) Expansion diathermal Constraints: 1diathermalTD,f= TE,f= Tf 2Force balance (omit air pressure) PistonaB+ PD,f(aA aB) + (mA+ mB)g = PE,faA(1) 3Ideal Gas: D : PD,f(aA aB)(eB x) = NDR Tf(2) E : PE,faA(eE+ x) = NER Tf(3) 41st Law: System(D+E)+A+B E = Q + W U = NDCV(Tf TD,i) + NECV(Tf TE,i) E = U + (mA+ mB)g x Q = 0 W = PatmaBx 3 Diannan LuSeminar 1st: Solution ManualHomework 3-3 (ND+ NE)CV(Tf Ti) + (mA+ mB)g x = PatmaBx(4) Combine (1)(4) Four variables, Four Equations, Done. Lets work it out! 1.013 105 1.29 102+ PD,f (6.45 1.29) 103+ (9.07 + 4.53) 9.81 = PE,f 6.45 103 PE,f= 0.8PD,f+ 0.409 105 (2)PD,f(6.45 1.29) 103(0.254 x) = 5.13 102 8.314 Tf PD,f(0.254 x) = 82.66Tf (3)PE,f 6.45 103(0.254 + x) = 9.51 102 8.314 Tf PE,f(0.254 + x) = 122.58Tf (b) Piston is adiabatic TD,f6= TE,f 1First Equation PD,f(aA aB) + (mA+ mB)g + PatmaB= PE,faA PD,f(6.45 1.29) 103+ (9.07 + 4.53) 9.81 + 1.013 105 1.09 103= PE,f 6.45 103 0.8PD,f+ 0.4094 105= PE,f 2Second Equation PD,f(aA aB)(eB x) = RNDTD,f PD,f(6.45 1.29) 103(0.254 x) = 5.13 102 8.314 Tf PD,f(0.254 x) = 82.66 TD,f 3Third Equation PE,f(aA)(eB+ x) = NERTE,f PE,f 6.45 103 (0.254 + x) = 9.51 102 8.314 TE,f PE,f(0.254 + x) = 122.58TE,f 4Fourth Equation NDCV(TD,f Ti) + NECV(TE,f Ti) + (mA+ mB)g x = PatmaB x 5.13 102 12.6(TD,f Ti) + 9.51 102 12.6(TE,f Ti) + (9.07 + 4.53) 9.81 x = 1.013 105 1.29 103 x 0.6464(TD,f Ti) + 1.198(TE,f Ti) = 264.093x Five variables, four equations. 1. assuming E expands reversibly and adiabatically For E, dUE=Q + W NECVdTE= PEdVE,PEVE= NER TE 4 Diannan LuSeminar 1st: Solution ManualHomework 3-3 Therefore, CVdTE= RdTE+ (RTE PE )dPE (CV+ R)dTE= RT dlnPE TE,f Ti = ?P E,f PE,i ?R/Cp+R = ?P E,f PE,i ?0.3975 TE,f= Ti ?P E,f/1.013 105 ?0.3975 (5) We get the ANSWER! x= 0.0187 m PD,f= 1.16 105Pa PE,f= 1.34 105Pa TE,f= 296.9 K TD,f= 329.4 K 2. assuming D expands reversibly and adiabatically TD,f Ti = ?P D,f PD,i ?0.3975 TD,f= Ti ?P D,f/(1.013 105) ?0.3975 (5) Then we get the ANSWER! x= 0.0195 m PD,f= 1.16 105Pa PE,f= 1.33 105Pa TE,f= 297.5 K TD,f= 327.7 K END of SOLUTION 5 Diannan LuSeminar 1st: Solution ManualHomework 3-8 Homework 3-8 Solution Data: PCylinder= 15.17 MPaTCylinder= 311.0 KVR,i=?PR,i= 0.101 MPaTR,i= 311 K CP= 29.3 J/mol KCV= 20.9 J/mol K a) mixing completely our system, open, adiabatic, rigid dUR=? ? QR +? ? W + HindNin W= 0, rigid, no movement of bounding and UR= UR Nin, we omit the gas in the VRoriginally. dUR= dUR Nin+ URdNin= HindNin CVdT Nin+ URdNin= HindNin NinCVdT = (Hin UR)dNin, where dHin= CpdT, so Hin= Hin(T0)+CP(TeT0), where T0is the reference temperature, and Teis 311 K. Besides, we know U = U(T0) + CV(T T0), where T is unknown. So we get, NinCVdT = ( ( ( ( ( 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) = (CPCV)T0= RT0. Thus we have, NinCVdT = (CPTe CVT)dNin.(1) Next question: Nin? Another Equation is from PV = NinRT: Use d(P(V ) = d(NinRT), we get VRdP = NinRdT + RTdNin, where weve known change of P, so we want the relation between T and Nin, and we get. dNin= 1 RT (VRdP RNindT) 6 Diannan LuSeminar 1st: Solution ManualHomework 3-8 Replace dNinin equation (1), and we get ? NinCVdT = (CPTe CVT) 1 RT (VRdP R?NindT) CPTe T dT = (CPTe CVT)dP P . Let = CP CV = 29.3 20.9 = 1.4, so we get 1 T ? Te Te T ? dT = dP P . Integrate this we get Z Tf Ti ? 1 Te T + 1 T ? dT = Z Pf Pi 1 P dP here Ti= Te= 311 K, Pi= 0.101 MPa, Pf= 15.17 MPa, and we get Tf= 434 K b) no mixing Pf= 15.17 MPa,Pi= 0.101 MPa,Ti= 311 K,N = const Tf=?Vi=?,Vf=? Use 1st law, dU =Q + W, and we get NCVdT = PdV From PV = NRT, we have d(PV ) = d(NRT), PdV + V dP = NRdT NCVdT + V dP = NRdT NCVdT + NRT P dP =NRdT ?C V+ R RT ? dT = 1 P dP Tf Ti = ?P f Pi ? R CV+R T = 46711 K ? 800 K 7 Diannan LuSeminar 1st: Solution ManualHomework 4-2 END of SOLUTION Homework 4-2 Solution ”Birds or coff ee percolator” is regarded as ”Heat machine” Assumption: no heat losses, totally reversible, normal ambient conditions, working substance is water We know the heat engine works between THand TL. For reversible engine, rev= W QH = TH TL TH What is TH, heat fl ows from the environment to the engine. What is TL, heat fl ows to the environment from the engine. Vapour to liquid at TL For L it is easy. All water is condensed! so QL= Hvapat TL. Liquid to vapour at TH For H it is not easy to determine. We dont know how much water is vapoured! Thus we get ( W QH = THTL TH QH+QC+ W = 0 From these equations, we get W QC+ W = TH TL TH Therefore, W QC = TH TL TL Assuming: TH= 300 K,TL= 280 K,QL= 2485 kJ/kg then we get W = QL TH TL TH = 2485 300 280 280 = 177.5 kJ/kg 8 Diannan LuSeminar 1st: Solution ManualHomework 4-11 Using W = gh, we have h = 177.5 103 9.81 = 18.1 km END of SOLUTION Homework 4-11 Solution (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 = 0 QH TH + QC TC = 0 For steady and constant pressure, QH= dHH= nHCPHdTH QC= dHC= nCCPCdTC Z T TH nHCPHdTH TH + Z T TC nCCPC TC dTC= 0 nHCPHln T TH + nCCPCln T TC = 0 We defi ne nHCPH nCCPC = 20 106 16 106 = 1.25 9 Diannan LuSeminar 1st: Solution ManualHomework 4-11 Therefore, ln T TH + ln T TC = 0 and then T+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) = nHCPH h TH (T HTC) 1/1+i nCCPC h (T HTC) 1/1+ TC i (b) pinch temperature? (c) countercurrent case QH TH + QH TC = 0 For steady and constant pressure QH= dHH= nHCPHdTH QC= dHC= nCCPCdTC Therefore, Z TH,f TH nHCPHdTH TH Z TC TC,f nCCPC TC dTC= 0 Let nHCPH nCCPC then we get Z TH,f TH dTH TH Z TC TC,f dTC TC = 0 thus TC,f= TC ?T H,f TH ? 10 Diannan LuSeminar 1st: Solution ManualHomework 4-11 and Wmax= QH QC = nHCPHTH TH,f nCCPCTC,f TC For , we have = W nCCPC = TH T f H TC ? TH,f TH ? 1 # Let TH,f = 0 and we get, TH,f= (TC(TH)1/1+ and TC,f= TC ?T H,f TH ? = TC T 1/1+ C T /1+ H TH ! = (TC(TH)1/1+ END of SOLUTION 11 Diannan LuSeminar 1st: Solution ManualThe Speed of Sound The Speed of Sound So