传质分离过程课后习题答案

1第一章绪论略第二章习题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时液相摩尔体积（m3/kmol）为：=100.91×10-3；=177.55×10-3；=136.69×10-3安托尼公式为（ps：Pa；T：K）：苯：1n=20.7936-2788.51/（T-52.36）；甲苯：1n=20.9065-3096.52/（T-53.67）；对-二甲苯：1n=20.9891-3346.65/（T-57.84）；解：由Wilson方程得：Λ12=llVV12exp[-(λ12-λ11)/RT]=331091.1001055.177×exp[-(1035.33)/(8.314×378.47)]=2.4450Λ21=0.4165Λ13=0.8382Λ31=1.24432Λ23=0.6689Λ32=1.5034lnγ1=1-ln(Λ12X2+Λ13X3)-[3322311313233221122131321211XXXXXXXXXXXX]=0.054488γ1=1.056同理,γ2=1.029;γ3=1.007lnP1S=20.7936-2788.51/(378.47-52.36)=12.2428,P1S=0.2075MpalnP2S=20.9062-3096.52/(378.47-53.67)=11.3729,P2S=0.0869MpalnP3S=20.9891-3346.65/(378.47-57.84)=10.5514,P3S=0.0382Mpa作为理想气体实际溶液,K1=PPS11=2.16,K2=0.88,K3=0.38003若完全为理想系,K1=PPS1=2.0484K2=0.8578K3=0.37712.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方程计算和Ki值。解：a11=115.2242748.0ccpTR=3.222MPa•dm6•k0.5•mol-2a22=225.2242748.0ccpTR=28.9926MPa•dm6•k0.5•mol-2b1=11208664.0ccpTR=0.0298dm3mol-13b2=225.2242748.0ccpTR=0.0806dm3mol-1其中Tc1=190.6K,Pc1=4.60MpaTc2=425.5K,Pc2=3.80Mpa均为查表所得。a12=√a11•a22=9.6651MPa•dm6•k0.5•mol-2液相：a＝a11x12+2a12x1x2+a22x22=3.22×0.13042+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：P=RT/(V-b)-a/[T0.5V(V+b)]4.1368＝0740.03610083145.0lmV－)0740.0(3611711.245.0lmlmVV解得Vml=0.1349lnl1ˆ=ln[V/(V-b)]+[bi/(V-b)]-2Σyiaij/bmRT1.5*ln[(V+b)/V]+abi/b2RT1.5{[ln[(V+b)/V]-[b/(V+b)]}-ln(PV/RT)lnl1ˆ=ln)0740.01349.01349.0(+0740.01349.00298.0－5.13610083145.00740.0)6651.98696.0222.31304.0(2×ln(1340.00740.01349.0)+5.123610083145.00740.00298.01711.24×[ln(1349.00740.01347.0)－0740.01347.00740.0]-ln3610083145.01349.01368.4=1.32974l1ˆ=3.7780同理lnl2ˆ＝-1.16696，l2ˆ=0.3113汽相：a＝3.222×0.603872+2×9.6651×0.60387×0.39613+28.9926×0.396132=10.3484b=0.0298×0.60387+0.0806×0.39613=0.0499由4.1368=0499.03610083145.0vmV－)0499.0(3613484.105.0vmvmVV得vmV=0.5861lnΦv1=ln(0499.05861.05861.0)+0499.05861.00298.0－5.125.13610083145.00499.00298.03484.10)5861.00499.05861.0ln(3610083145.00499.06651.939613.0222.360387.0(2×[ln0499.05861.00499.0)5861.00499.05861.0(]－ln(3610083145.05861.01368.4)=0.0334942故Φv1＝1.0341同理，lnl2ˆ＝-0.522819，l2ˆ=0.5928故K1=y1/x1=0.60387/0.1304=4.631(K1=l1ˆ/Φv1)K2＝y2/x2＝1304.0160387.01＝0.45553.乙酸甲酯（1）-丙酮（2）-甲醇（3）三组分蒸汽混合物的组成为y1＝0.33，y2＝0.34，y3=0.33(摩尔分率)。汽相假定为理想气体，液相活度系数用Wilson方程表示，试求50℃时该蒸汽混合物之露点压力。解：由有关文献查得和回归的所需数据为：【P24例2-5，2-6】50℃时各纯组分的饱和蒸气压，kPa5P1S=78.049P2S=81.848P3S=55.58150℃时各组分的气体摩尔体积，cm3/molV1l=83.77V2l=76.81V3l=42.05由50℃时各组分溶液的无限稀释活度系数回归得到的Wilson常数：Λ11＝1.0Λ21＝0.71891Λ31＝0.57939Λ12＝1.18160Λ22＝1.0Λ32＝0.97513Λ13＝0.52297Λ23＝0.50878Λ33＝1.0（1）假定x值，取x1＝0.33，x2＝0.34，x3＝0.33。按理想溶液确定初值p＝78.049×0.33+81.8418×0.34+55.581×0.33=71.916kPa（2）由x和Λij求γi从多组分Wilson方程lnγi=1-ln∑cjijjx1)(-ckcjkjjkjkxx11得lnγ1=1-ln(x1+Λ12x2+Λ13x3)-[31321211xxxx[phone]xxxx[phone]xxxx=0.1834故γ1=1.2013同理，γ2=1.0298γ3=1.41816(3)求KiKi=RTppVppsiLisii)(expK1=916.71049.782013.1exp16.323314.810)049.7896.71(77.833=1.3035同理K2＝1.1913K3=1.0963(4)求∑xi∑xi=3035.133.0+1713.134.0+0963.133.0=0.8445整理得x1＝0.2998x2=0.3437x3=0.3565在p＝71.916kPa内层经7次迭代得到：x1＝0.28964,x2=0.33891,x3=0.37145(5)调整pp＝RTppVxpsiLiisii)(exp=piixK=71.916(1.3479×0.28964+1.18675×0.33891+1.05085×0.37145)=85.072kPa在新的p下重复上述计算，迭代至p达到所需精度。最终结果：露点压力85.101kPa平衡液相组成：x1＝0.28958x2＝0.33889x3＝0.371534.一液体混合物的组分为：苯0.50；甲苯0.25；对-二甲苯0.25（摩尔分数）。7分别用平衡常数法和相对挥发度法计算该物系在100kPa时的平衡温度和汽相组成。假设为完全理想物系。解：(1)平衡常数法因为汽相、液相均为完全理想物系，故符合乌拉尔定律pyi=pisxi而Ki=iixy=ppsi设T为80℃时,由安托尼公式（见习题1）求出格组分的饱和蒸汽压。sp1＝101.29kPa，sp2=38.82kPa,sp3=15.63kPa故321yyy=K1x1+K2x2+K3x3=332211xppxppxppsss=25.010063.1525.010082.385.010029.101=0.641故所设温度偏低，重设T为95℃时sp1＝176.00kPa,sp2=63.47kPa,sp3=27.01kPa321yyy=1.111故所设温度偏高，重设T为91.19℃，sp1＝160.02kPa,sp2=56.34kPa,sp3=23.625kPa321yyy＝1.0000125≈1故用平衡常数法计算该物系在100kPa时的平衡温度为91.19℃汽相组成：1y＝11xK＝11xpps＝5.010002.160＝0.80012y＝22xK＝22xpps＝25.010034.56＝0.140983y＝33xK＝33xpps＝25.0100625.23＝0.059(2)相对挥发度法由于是理想混合物，所以)/()(111iiixxyy，得)/(111iiixxyy对于理想混合物，得i1＝SSPp21设T为80℃时，sp1＝101.29kPa,Sp2=38.82kPa,sp3=15.63kPa故12＝2.61,13＝6.48,2y＝1y/5.22,3y=1y/12.96因为321yyy＝1，故1y＝0.788又因为1py＝100×0.788＝78.8kPa，而11xps＝101.29×0.5＝50.645kPa1py故所设温度偏低；重设T＝92℃时sp1＝163.31kPa,Sp2=57.82kPa,sp3=24.31kPa得故12＝2.824,13＝6.718,2y＝1y/5.648,3y=1y/13.436因为321yyy＝1，故1y＝0.799，2y＝0.141,3y=0.0595且1py＝100×0.799＝79.9kPa，而11xps＝163.31×0.5＝81.655kPa，基本相等因此，由相对挥发度计算该物系平衡温度为92℃，此时1y＝0.799，2y＝0.141,3y=0.05955.一烃类混合物含有甲烷5%、乙烷10%、丙烷30%及异丁烷55%（mol），试求混合物在25℃时的泡点压力和露点压力。解：设甲烷为1组分，乙烷为2组分，丙烷为3组分因为各组分都是烷烃，汽液相均可视为理想溶液，故符合乌拉尔定律。925℃时，sp1=30768.14kPa,sp2=4118.81kPa,sp3=347.59kPa(1)泡点压力iisiixppyp＝30768.14×5％＋4118.81×10％＋950.31×30％＋347.59×55％＝2426.56kPa(2)露点压力时由乌拉尔定律得isiixppy，isiiyppx代入4321xxxx＝1，并化简得sssspypypypyp433322111＝519.77kPa故露点压力为519.77kPa。