传质分离过程答案(精简)

《传质分离过程答案(精简)》由会员上传分享，免费在线阅读，更多相关内容在行业资料-天天文库。

1、分离工程答案第2章1.计算在0.1013MPa和378.47K下苯（1）-甲苯（2）-对二甲苯（3）三元系，当x1=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。已知三个二元系的wilson方程参数（单位：J/mol）：λ12－λ11=－1035.33；λ12－λ22=977.83λ23－λ22=442.15；λ23－λ33=－460.05λ13－λ11=1510.14；λ13－λ33=－1642.81在T=378.4K时液相摩尔体

2、积（m3/kmol）为：ν1L=100.91×10-3；ν2L=177.55×10-3；ν3L=136.69×10-3安托尼公式为（ps：Pa；T：K）：苯：1np1s=20.7936-2788.51/（T-52.36）；甲苯：1np2s=20.9065-3096.52/（T-53.67）；对-二甲苯：1np3s=20.9891-3346.65/（T-57.84）；解：由Wilson方程得：Λ12=ν2Lν1Leλ11-λ12RT=177.55×103100.91×103e-1035.338.314×378

3、.47=2.4450Λ21=0.4165Λ13=0.8382Λ31=1.2443Λ23=0.6689Λ32=1.5034lnγ1=1-lnx1+Λ12x2+Λ13x3-x1x1+Λ12x2+Λ13x3+Λ21x2Λ21x1+x2+Λ23x3+Λ31x3Λ31x1+Λ32x2+x3求得γ1=0.8897，同理γ2=，γ3=lnp1s=20.7936-2788.51/(378.47-52.36)=12.2428,p1s=0.2075Mpalnp2s=20.9062-3096.52/(378.47-53.67)=

4、11.3729,p2s=0.0869Mpalnp3s=20.9891-3346.65/(378.47-57.84)=10.5514,p3s=0.0382Mpa作为理想气体实际溶液K1=γ1P1SP=,同理K2=,K3=若完全为理想系K1=P1SP=,同理K2=,K3=2.在361K和4136.8kPa下，甲烷和正丁烷二元系呈汽液平衡，汽相含甲烷0.60387%（mol）,与其平衡的液相含甲烷0.1304%。用R-K方程计算ФiV,ФiL和Ki值。解：a11=0.42748R2×Tc12.5pc1=3.222

5、MPa•dm6K0.5mol-2a22=0.42748R2×Tc22.5pc2=28.9926MPa•dm6K0.5mol-2b1=0.08664R2×Tc12.5pc1=0.0298dm3mol-1b2=0.08664R2×Tc22.5pc2=0.0806dm3mol-1其中Tc1=190.6K,Pc1=4.60Mpa,Tc2=425.5K,Pc2=3.80Mpa均为查表所得a12=a11•a22=9.6651MPa•dm6K0.5mol-2液相：a＝a11x12+2a12x1x2+a22x22=3.22

6、×0.13042+2×9.6651×0.1304×0.8696+28.9926×0.86962=24.1711b=b1x1+b2x2=0.0298×0.1304+0.0806×0.8696=0.0740由R－K方程：4.1368=0.0083145×361Vm1-0.0740-24.1711361Vm1(Vm1+0.0740)解得Vm1=0.1349lnФ1L=ln0.13490.1349-0.0740+0.02980.1349-0.0740-2×0.1304×3.222+0.8696×9.66510.07

7、40×0.0083145×3611.5ln0.1347+0.07400.1340+24.1711×0.02980.07402×0.0083145*×ln0.1347+0.07400.1349-0.07400.1347+0.0740-ln4.1368×0.13490.0083145×361=1.3297Ф1L=3.7780,同理Ф2L=0.3113汽相：a＝3.222×0.603872+2×9.6651×0.60387×0.39613+28.9926×0.396132=10.3484b=0.0298×0.60

8、387+0.0806×0.39613=0.0499由R－K方程：4.1368=0.0083145×361Vmv-0.0499-10.3484361Vmv(Vmv+0.0499)解得Vmv=0.5861lnФ1v=ln0.58610.5861-0.0499+0.02980.5861-0.0499-2×0.60387×3.222+0.39613×9.66510.0499×0.0083145×3611.5ln0.586