资源描述
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 Lu Seminar 1st: Solution Manual Homework 3-3 Homework 3-3 Solution aB = 1:29 10 3 m2; eB = 1=2 0:508 = 0:254 m A, B, Cylinder!nonconducting, no heat capacity, having some friction. What we know: Thus PEaA = PDaA + (mA +mB +mC)g )PE = PD + (mA +mB +mC)ga A = 1:013 105 + (9:07 + 4:53 + 18:14) 9:816:45 10 3 = 1:50 105 Pa If we consider air pressure, P0E = PE +Pair aBa A = 1:72 105 Pa 2 Diannan Lu Seminar 1st: Solution Manual Homework 3-3 for aB=aA, in present study, we omit this VD;i = eB (aA aB) = 0:254 (6:45 1:29) 10 3 = 1:31 10 3 m3 PD;iVD;i = NDR TD;i)ND = 1:013 10 5 1:31 10 3 8:314 311 = 5:13 10 2 mol VE;i = eE aA = 0:254 6:45 10 3 = 1:64 10 3 m3 PE;iVE;i = NER TD;i)NE = 1:50 10 5 1:64 10 3 8:314 311 = 9:51 10 2 mol All initial conditions: PD;i = 1:013 105 Pa; TD;i = 311 K; ND = 5:13 10 2 mol; VD;i = 1:31 10 3 m3 PE;i = 1:50 105 Pa; TE;i = 311 K; NE = 9:51 10 2 mol; VE;i = 1:64 10 3 m3 (a) Expansion diathermal Constraints: 1 diathermal TD;f = TE;f = Tf 2 Force balance (omit air pressure) Piston aB +PD;f(aA aB) + (mA +mB)g = PE;faA (1) 3 Ideal 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+B E = Q+W U = NDCV (Tf TD;i) +NECV (Tf TE;i) E = U + (mA +mB)g x Q = 0 W = Patm aB x 3 Diannan Lu Seminar 1st: Solution Manual Homework 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 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) = 122:58Tf (b) Piston is adiabatic TD;f 6= TE;f 1 First Equation PD;f(aA aB) + (mA +mB)g +PatmaB = PE;faA PD;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;f 2 Second Equation PD;f(aA aB)(eB x) = RNDTD;f )PD;f(6:45 1:29) 10 3(0:254 x) = 5:13 10 2 8:314 Tf )PD;f(0:254 x) = 82:66 TD;f 3 Third Equation PE;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;f 4 Fourth Equation NDCV (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 = 1:013 105 1:29 10 3 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 NECV dTE = PEdVE; PEVE = NER TE 4 Diannan Lu Seminar 1st: Solution Manual Homework 3-3 Therefore, )CV dTE = RdTE + (RTEP E )dPE ) (CV +R)dTE = RT dlnPE ) TE;fT i = P E;f PE;i R=Cp+R = P E;f PE;i 0:3975 )TE;f = Ti PE;f=1:013 105 0:3975 (5) We get the ANSWER! 8 : x = 0:0187 m PD;f = 1:16 105 Pa PE;f = 1:34 105 Pa 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 PD;f=(1:013 105) 0:3975 (5) Then we get the ANSWER! 8 : x = 0:0195 m PD;f = 1:16 105 Pa PE;f = 1:33 105 Pa TE;f = 297:5 K TD;f = 327:7 K END of SOLUTION 5 Diannan Lu Seminar 1st: Solution Manual Homework 3-8 Homework 3-8 Solution Data: PCylinder = 15:17 MPa TCylinder = 311:0 K VR;i =? PR;i = 0:101 MPa TR;i = 311 K CP = 29:3 J/mol K CV = 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 VR originally. ) dUR = dUR Nin +URdNin = HindNin CV dT Nin +URdNin = HindNin )NinCV dT = (Hin UR)dNin; where dHin = CpdT, so Hin = Hin(T0) +CP(Te T0), where T0 is the reference temperature, and Te is 311 K. 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. Thus we have, NinCV dT = (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 = 1RT (VRdP RNindT) 6 Diannan Lu Seminar 1st: Solution Manual Homework 3-8 Replace dNin in equation (1), and we get NinCV dT = (CPTe CVT) 1RT ( VRdP R NindT) CPTe T dT = (CPTe CVT) dP P : Let = CPC V = 29:320:9 = 1:4; so we get 1 T T e Te T dT = dPP : 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 NCV dT = PdV From PV = NRT, we have d(PV) = d(NRT), )PdV +VdP = NRdT ) NCV dT +VdP = NRdT ) NCV dT + NRTP dP = NRdT ) C V +R RT dT = 1P dP ) TfT i = P f Pi RC V+R )T = 46711 K 800 K 7 Diannan Lu Seminar 1st: Solution Manual Homework 4-2 END of SOLUTION Homework 4-2 Solution Birds or co 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 TH and TL. For reversible engine, rev = WQ H = TH TLT H What is TH, heat ows from the environment to the engine. What is TL, heat ows to the environment from the engine. Vapour to liquid at TL For L it is easy. All water is condensed! so QL = Hvap at 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 ( WQH = TH TLTH 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 TLT H = 2485 300 280280 = 177:5 kJ/kg 8 Diannan Lu Seminar 1st: Solution Manual Homework 4-11 Using W = gh, we have h = 177:5 10 3 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 = d _HH = _nHCPHdTH _QC = d _HC = _nCCPCdTC ) Z T TH _nHCPHdTH TH + Z T TC _nCCPC TC dTC = 0 ) _nHCPH ln TT H + _nCCPC ln TT C = 0 We de ne _nHCPH_n CCPC = 20 10 6 16 106 = 1:25 9 Diannan Lu Seminar 1st: Solution Manual Homework 4-11 Therefore, ln TT H + ln TT C = 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 = d _HH = _nHCPHdTH _QC = d _HC = _nCCPCdTC Therefore, Z TH;f TH _nHCPHdTH TH Z TC TC;f _nCCPC TC dTC = 0 Let _nHCPH_n CCPC 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 Lu Seminar 1st: Solution Manual Homework 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 T H;f = 0 and we get, TH;f = (TC(TH) )1=1+ and TC;f = TC T H;f TH = TC T1=1+ C T =1+ H TH ! = (TC(TH) )1=1+ END of SOLUTION 11 Diannan Lu Seminar 1st: Solution Manual The Speed of Sound The Speed of Sound Solution 1st Law dU = Q+ W + dmin Hin + 12u2in dmout Hout + 12u2out with mass balance dmin = dmout, we get Hin + 12u2in = Hout + 12u2out: Therefore, dH + 12du2 = 0: (1) With mass balance, we get d u A V = 0) udV +VduV2 = 0) duu = dVV : Using dH = TdS +VdP, and for the adiabatic, reversible process, dS = 0, we have dH = VdP: With equation (1), we have VdP + 12du2 = 0)VdP = u 2 V dV )u 2 = V2dP dV : Thus, we get uc = s P S : How to calculate ( P)S Assuming Ideal Gas, and is adiabatic dU = Q+ W: With 12 Diannan Lu Seminar 1st: Solution Manual The Speed of Sound NCV dT = PdV PV = NRT we have PV = const where = Cp=CV = 1:4. Thus we have lnP V S = V ) P V S = PV : So the answer is uc = p RT=MW = r1:4 8:314 298:15 29 10 3 = 346 m/s END of SOLUTION 13
展开阅读全文