化工热力学习题参考答案第三版1

1、化工热力学（第二版）习题参考答案58页第2章2- 1求温度673.15K、压力4.053MPa的甲烷气体摩尔体积。 解：（a）理想气体方程pV RTRTP8.314 673.1564.053 10331.381 10 m mol（b ）用R-K方程查表求Tc、计算a、b；利用迭代法计算RTV bVi 1RT ba Vi bp. T Vi Vi b33V01.381 10 m molVi 11.3896 10 3 m3 mol（c）用PR方程步骤同（b），计算结果：Vi11.389310 33 mmol 1。（d）利用维里截断式Zpv1空1BPcpr 1B0PrB1 PrRTRTRTcTrTrT

2、r00.0830.422J0.1390.172B1 6B4 2Tr.r、Tr、B0、B1Tr查表可计算 p和Z由ZPVv ZRT1.391 10 33 mmol1RTp2-2 V=0.5 m 3,耐压2.7 MPa容器。规定丙烷在 T=400.15K时，p<1.35MPa。求可充丙烷多少千克？解：（a）用理想气体方程6mMpV 0.0441 1.35 100.53.1pV nRT pV RT m8.948 m molMRT8.314 400.15（b ）用R-K方程查表求Tc、Pc :计算a、b;利用迭代法计算 V。pV bT V V bV “RTba Vi bV i 1pu、T Vi

3、VibV。2.4641033.1m molRTa33V 12.241 10 m molmn M1(c)利用维里截断式：ZpVm1 Bp 1RTRTJ0.422B0.0831 6Tr则可充丙烷质量计算如下：查表可计算pr、Tr、B0、B1和Z0.04410.59.838kg2.24110 3BPcPr1 B0PrRTcTrTrB10.1390.172T421 rB1PrTr,pVm*0.916 8.314 400.15田 ZVm6RT1.35 106则可充丙烷质量计算如下：2.257 10 3 m3 molVi 10.0441 0.52.25710 39.77 kg2- 4 V=1.213 m3

4、,乙醇 45.40 kg ,T=500.15K，求压力。解：（a）理想气体状态方程nRT m RT pV M V(b )用R-K方程45.40 8.314 500.15461.2133.383 MPa22 50.42748R TcPC二 28.0390.08664R $PC二 0.058RTap Vm b T Vm Vm bV1.229 m3 kmoln 45.40/46p 2.759 MPa(c)用SRK方程计算345-400-母ti:=ihilieacr莎 mT := (227 + 273 l KR = £.314?&-301_1< 1 Pc -6.3S 10 Pa

5、 TC =516.2 Ko =0635R T gg(T) R T 就巩片-b)V -fc V<7 十 b)1+1 p十 b)cl(T> ：= 0.2743PC0J04R-T.b：=7-(7 4-功a(I) := 0457240 07780.R-TC6clO) - 1 372 - Pt 翻2+ W，勢北4十 1 J422fl-co -V- b出V (V十可(d)用PR方程计算345-400= IN-母ti:=ihil"act45.01T := (227 + 273 l K3- 1- 16R = Sr314Pa-m mol Kpc -<5.38-10 Pa TC =51

6、.2 Ko :=Dd35RT工Tl V-RT+b_一 y - bv<?十 ® 十 b(7 - b)1+1 ppf 卧十 q 十 b <V - b)J(e)用三参数普遍化关联m 45.40.987 kmolM46V1.213,3.- 1Vm1.229m kmoln0.987ZpVm彳Bp1pRTrt rt p vm_BBPcB0B1B B0 B1 R Tc pcRTCc 匕B00.361, B10.057,0.635B 0.267RTVm B2.779 MPa2-6 （本题以计算机编程计算提供参考，考试时不能使用）(a) R-K方程Tr 0.858Pr0.3290.176Z

7、00.08664PrZi TrZi4.9340 hT 1.51 hTrZ10.835mZ1nZiwhilem n|10 3nmH0.08664Prn Trm14.9340H1 Ht1.51 HTrV(m n)m nV(1 0.8350.7950.795S-R-KTr 0.858Pr0.3290.176i 0Z012S 0.4801.5740.176丄1 S 1Tr0.08664PrT 0.51 rZiTrZi4.9340FZ10.828V(m n)ZiZiwhile10 30.08664Prn Tr4.9340FHV(1 0.8280.7840.7852-7 计算 T=523.15K , p=

8、2 MPa 解：（a）用维里截断式的水蒸气的Z和V必1旦£RT V Vrt BRT crtV ppVpV2采用迭代法计算V=2.006之后求得Z=0.923Z史1RTBP 1RTBpcRTcB00.0830.422B1f.61 r(d )利用维里截断式查表可计算pr、Pr彳0 Pr1 Pr-1B-B-TrTrTr0.1720.1394 2TrTr、B0、B1 可得到 Z=0.932 ;由 z 空 V ZRT 2.025 10 3 m3 mol 1 RTp(c)水蒸气表V 0.11144 m3 kg 10.11144 18 2.00592 m3 kmolpV 2000 2.00592R

9、T 8.314 523.1592页第三章3- 4R 8.314丁二烯 13T1127273.15t2227273.15P12.53610 PaP212.6710 6PaTc425Pc4.32610 6Pa0.181Cp(T)22.738 222.79610 3T73.879106t2利用三参数压缩因子计算方法，查图表，得到压缩因子:T1 T2Tr1ToT"ToTr10.942J1.177ccPr1岂PcPr2戈PoPr10.585Pr22.929ccZ10.677Z20.535Z2 R T2 Z R T43,1VV7.14610m molP2 P1H2RRTcP20.083Pc0.1

10、391.0971.60.8944.2H2R8.475103H1RRTcP10.083Pc0.1391.097T1TcT2Cp(T) dTH2RH1R1.60.894T1Tc4.2H1R2.704103T15.028103J molS2RR P20.6752.6T20.722 -Tc5.2S2R12.128S1RRP10.6752.60.722 25.2S1R4.7083-7: 解:T2T15.261注意:m3或者P2P1R InP2可S2RS1R3.212J mol10 3kPaP2P1Vdp2.095101.551310 2000381kPakg5.261 Jkg 1 K1031T VLdp

11、2705.2611.55110 31.6196101.091 kJ kg2.095 10 3 270 1.55110 3 2 106 3.81 1051090.6J kg 13- 9解：乙腈的An to nie方程为3271 24In ps 14.7258kPat/ c 241.85(1) 60C时，乙腈的蒸气压3271 24ln ps 14.7258.3.88860241.85ps 48.813 kPa(2 )乙腈的标准沸点ln 100 14.72584.605t/ c 241.85t 81.375 c(3) 20C、40C和标准沸点时的汽化焓d ln ps H3271.24HdTRT7t

12、241.85 2RT13271.24 8.314 T2241.85H 20 c34.09kJ /mol;H 40 c33.57kJ /mol;H 81.375 c 32.72kJ / mol117页第四章4- 11 2h - u gz_q w23 J104 kgz 3 mh (23003230) 10 q 0kg3600skgm 2.778 -410 kg2.778kgs3600ssh2.5836 J10 -s101 24 JJ-mu1.6510 -g z m 81.729-2ss1 26 J6 J6h-mug z m2.56710w =2.567102.56710 W2ss2.5832.56

13、7wc100%2.567wc 0.623%4- 2方法u13R 8.3141 2h 一 u = w2 u220.075u1u20.270.252h - u gz=q w T2353.15T1593.15HR2R 647.3Trr1dB01B010.344 dB1B11dPrHR2 56.91H = Cpmh T2T1hr2 hr1Pr0B00HR1 R 647.3 Tr1r0dB00.3440 Tr00B10dB1dPr HR1 576.7710 tr0经计算得Cpmh 35.03 Jmol 1体积流速为:Uid/2 223.14 3 凹50.0132 m3摩尔流速为:0.0132VmRT/

14、p4.0158.314 593.15/1500000mol s 1根据热力学第一定律，绝热时Ws = -H，所以H 二 n C pmh T2 T 1 n HR? HR 1W s 4.0158.408 108.314593.153.28810( 56.91576.771)3.167101500000 W方法二：h2=2645.6 kJ kg-1 ； h1=3081.5 kJ kg-1根据过热蒸汽表，内插法应用可查得35kPa、80C的乏汽处在过热蒸汽区，其焓值1500 kPa、320C的水蒸汽在过热蒸汽区，其焓值4.464 10 3 二 435.904kJ kg1 2 2w - h U2 出 2

15、645.63081.5按理想气体体积计算的体积m3 mol 1molN 4.015s0.0132m3 s 14.015柬3.28810 3molw = 435.90418=3.151044-6解:二氧化碳8.314303.15P11.5 106 PaP20.1013310 6PaTc304.2Pc7.357 106Pa0.225Cp(T)45.3698.688 10 ' T 9.619 105TRTcpiH 1R 0.0830.139PcT11.097 Tc1.60.894T1Tc4.2RTcP2H 2R T 20.0830.139P cT21.097Tc1.60.894T24.2通过

16、t2H 1RCpmhT1迭代计算温度,T2=287.75 KT2CPms8 1 H - C p(T) dT H 2r T 2 Hr 1.82210 J molTcT2 Cp(T)TT1T2InT1dTS 1RR P 12.6T10.675 -Tc5.2T10.722 -TcPcT15.2R P2T2.6t2S2RPc0.675 -TcT2T20.722 -TcCp(T)P2S .dT R ln TP1T1S2R S1R 21.801J mol 1 K4-7解:T1473.15P12.5 106 PaTc305.40.098Cp2.6R P 1TiS 1R_0.675 -1 1 xpPcTcR8

17、.314P20.20106PaPc4.88106Pa9.403159.837103t6 246.23410T0.722T1Tc5.2S2R T22.65.2R P2T2T20.675 0.722 PcTcTc经迭代计算（参考101页例题4-3）得到T2=340.71K。RTcP1H 1R 0.0830.139PcT11.097 一Tc1.60.894T1Tc4.2103 J molH1R=8.32725RTcP2H2rT 20.0830.139P cT2H 二 Cp(T)cT H2R T2T 11.64.2T2T21.0970.894TcTC1146页第五章5- 1: b 5-2: c 5-4

18、: a 5-5: a5- 1:5解：可逆过程熵产为零，即SgSsysSfSsys0Ssys0。T05-2:解：不可逆过程熵产大于零，即SgSsysSfSsys50Ssys5。即T0T0系统熵变可小于零也可大于零。5-4：解：不可逆绝热过程熵产大于零，即SgSsysSfssys0。所以流体熵变大于零。5-5：解：不可逆过程熵产大于零，即SgSsysSfSsys100Ssys10。T0T05-3:解：电阻器作为系统，温度维持100 C,即373.15K，属于放热；环境温度298.15K，属于吸热，根据孤立体系的熵变为系统熵变加环境熵变，可计算如下：2 850(20 A) 2 3600 s 1.44

19、10 J8 81.4410 J 1.4410 J4 19.70710 J373.15K298.15KK5-6:解：理想气体节流过程即是等焓变化，温度不变，而且过程绝热，所以系统的熵变等于熵产,计算如下：8.314 J mol 1 K0.09807 In - 1.9624.901 JK mol1所以过程不可逆。5-7:M 二 m m2 H 二 0页 4-解:绝热稳流过程所以T320 kg s1 (90273.15) K 30 kg1 (50273.15) KT3339.15KSg =50 kg s 1339.15273.15mi* * * Si = m1CpmsIn66T3T1t3m2 Cpms

20、ln 0.336。解:水泵功与透平功之比(2)H2=3562.38 kJ kg-1,WnWS,TurW4 1Qh H2H15-12H4 VP H1(1)循环的热效率H3=2409.3 kJ kg-1,也=162.60 kJ 3162.60140.007 100.001kg-1, H5=2572.14 kJ kg-1,176.6 kJ kgWS,Tur |V ph2 H3。皿(14 呵 1030.0123562.38 2409.3查表可得 h二 376.92 h2 二 209.33376.92209.33119050H2TH3 比H2 H1H±0.345C pms - C pmh4.1

21、9 kJ kg Kql tlW = T0 tl二 20%qlqhT01thtlT0 tl0.6 0.2 二 0.555QlQhir irc 60%20%1壬天°6 O21 273.151 -118 273.156 273.1521 60.12 0.555194页第六章6-1:解：水蒸气的摩尔流量为:1680 1000M 360018 3600二 25.926mol s(3)提供1 kw电功的蒸气循环量1000 1000m W聞08 0.857 g s5-15 题:4 15T0W1C thqhT 二 C 60%T3 52H(T) n 8.314 3.471.45 10 T 0.121

22、10 T dT703T3528.314 3.471.45 10 T 0.121 10 TpS(T p) n dT n 8.314 ln T3.727703(a)通过内插法求出 0.1049MPa时对应的温度，如下100 t100950.298101.325104.9101.32584.55t 二 100.964W id 二 H(374.114) T0 S(374.114 0.1049)二 3.607 105Js 1WsH(374.114)3.046105 J s(b)Ws 0.844idWid 二W s H(333)3.408105Js 1WsWid0.69251H(333) To S(333

23、 0.0147)二 4.92610 Js6- 3H = h2% = 292.98376.92 =83.94HSSULT0Sg 二 SsysSsur 二 0.95491.1925-83.94二 0.044 kJ1 1 kg K298根据热力学第-疋律热损失为131Q 二 H二 83.94 kJ kg或Q 二1.51110 J mol功损失为WL = T01Sg = 13.1 kJ kg或wl二 235.81 J mol6- 6:解：理想气体经一锐孔降压过程为节流过程,0,且 Q过程恒温。则绝热膨胀过程的理想功和损耗功计算如下:Wid 二 H T0S = T0S=298 8.314 ln0.098

24、071.96二 7.423110 J molWL-Wid_T0 Sa _298 8.314 l n 0.09807_ 7.42103 J mol 1L id 0 g1.966-8：解：(1 )产品是纯氮和纯氧时,WidRT0(y1 l n y1 y21 n y2) 8.314 298.15 (0.21l n 0.21 0.79l n0.79)1.274kJ mol 1(2)产品是98% N2和50% O2的空气时，设计计算流程如下:0.21 O2分离亠纯O2混合亠0.98 N2Wid纯N2W2W1498% N250% O2总的功WWidW1W2WT0R 0.981 n0.980.02l n0.

25、02W>T0R0.5l n0.50.5ln 0.5298.15 8.3140.098298.15 8.3140.693242.924 J mol 3N2 H2NH31.718 kJ mol 1W 1.274 242.924 10 3 1.71810.687kJ molEXCT 0 S 285.84298 169.7851000235.244kJ mol6-12:解:查表得 H2H2ONH3O2CH3OHCO2N2Hf0285.8446.190238.64393.510S130.5969.94192.51205.03126.8213.64191.49h21 HoOn2 2 2H2O(l)H

26、 = 285.84 kJ mol10.02061051 1S 二 69.94130.590.5 205.038.314 ln0.10133二 169.785 J molKNH3EXC (NH3)二 3 117.611 0.3351 16.63 二 336.535kJ molCH3OHEXC(CH30H)二 4 117.611 1.966410.54166.31 二 716.636kJ mol6-13解：m 1h3 h1m2 h3h2二 072000, 1108000, 1mkg sm2kg s36003600h1376.92kJ kg 1S11 11.1925 kJ kg Kh2209.33 kJ kg 1S20.7038kJ kg1K 1m 1h3 h1m2 h3h2二 0h 3 =276.366kJ kg 1由m1 C pmht311m2 Cpmh t 3t2 =°,可得 t3= 66.03 CS3°.89350.95490.893566.03657065S3 二 0.906kJ kg