分离工程（黄国文）分离工程习题及答案及思考题

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1、2章习题2.11.计算在0.1013MPa和378.47K下苯（1）■甲苯（2）■对二甲苯（3）三元系，当刘二0.3125、x2=0.2978.x3=0.3897时的K值。汽相为理想气体，液相为非理想溶液。并与完全理想系的K值比较。己知三个二元系的wilson方程参数（单位：J/mol）:人2一儿尸一1035.33；人2一人22=977.83人23—人22=442.15；人23—人33二—460.05^13-/111=1510.14：人3一人33=一1642.81在7=378.4K时液相摩尔体积（n/kmol

2、）为:^=100.91x10-3；^*=177.55x10-3；=136.69x10-3安托尼公式为（p：Pa；7■:K）：苯：1n*<=20.7936-2788.51/（7-52.36）；甲苯:inA=20.9065-3096.52/（7-53.67）；对■二甲苯：1n^=20.9891-3346.65/（7-57.84）；答案解:由Wilson方程得：V1A12=1exp卜(入12■入1J/RT]177.55X103A2i=0.4165Ai3=0.8382A3i=1.2443A23=0.6689A32=1

3、.5034XilnYi=l-ln(x1+A12x2+Ai3x3)-[+A12x2+A13x3+A31兀3A?"+A31X1+A32X2+兀3]=0.04095人2內+兀2+人23兀3Yi=1.0418同理,lnY2=0.05148,丫2=1・6732InY3=0.4190zY3=1.52031计=20.7936-2788.51/(378.47・52.36)=12.2428,P1s=0.2075MpalnP2s=20.9062-3096.52/(378.47-53.67)=11.3729,P2s=0.0869M

4、palnP3s=20.9891・3346.65/(378.47・57.84)=10.5514,P3s=0.0382Mpa作为理想气体实际溶液，r^sKi=P=2.134,©=1.4354,©=0.5733若完全为理想系，K]=P=2.0484©=0.8578K3=0.37711.在361K和4136.8kPaF,甲烷和正丁烷二元系呈汽液平衡，汽相含甲烷0.60387%(mol)，耳英平衡的液相含甲烷0.1304%。用R・K方程计算和Kj值。答案解:0.427482?2XT25^

5、_an=Pc、=3.222MP

6、a•dm6•k0-5•mol'20.42748/?2XT25a22=Pci=28.9926MPa*dm6»k°-5*mor20.086647?2x7；bi=P®=0.0298dm-^mor10.427482?2XT25b2=Pc?=0.0806dm3mor1Jt+Tcl=190.6KzPci=4.60MpaTc2=425.5K,Pc2=3.80Mpa均为查表所得。ai2=Vai1*a22=9.6651MPa*dm6*k0,5«mor2液和：8=61X12+262X1X2+822X22=3.22x0.13042

7、+2x9.6651x0.1304x0.8696+28.9926x0.86962=24.1711b=b1x1+b2x2=0.0298x0.1304+0.0806x0.8696=0.0740由R—K方程：0.0083145x3614.1368=^-0-0740_解得Vm'=0.1349~0.13490.0298In狄=片0.1349-0.0740+0.1349-0.0740-2x(0」304x3.222+0.8696x9.6651)0.1349+0.07400.0740x0.0083145x361'5X

8、n(0.

9、134024.1711x0.02980.1347+0.07400.07402x0.0083145x361'5X[ln(0.13490.07404.1368x0.1349-0.1347+0.0740j_

10、n0.0083145x361=1.3297处=3.7780同理In必=-1.16696,孑2=0.3113汽相：a=3.222x0.603872+2x9.6651x0.60387x0.39613+28.9926x0.396132=10.3484b=0.0298x0.60387+0.0806x0.39613=0.

11、0499由4.1368=0.0083145x36110.3484匕;_0.0499_361°“匕；(匕;+0.0499)得=0.5861’0.58610.0298In$=

12、0(0.5861-0.0499)+0.5861-0.0499_2x(0.60387x3.222+0.39613x9.66510.5861-0.04990.0499x0.0083145x361'5*°0.5861+10.3484x