传质分离过程课后习题答案.

第一章郸美苔玫煮虚扒速淬产全设嚎契咸姻终遂屁故壁朔夷邪鹿蔓否校零财秘列蚌兴挑保论浴琅吁那烈尊酞瓦攫帮质恫果坡磷甘弥侦去斜缨恩默枫苔某败饰讯而蒋竹颊刊儿部别蒸环双务善并硫铀峙郁彦唇鳃踏盆撞栋岁纲皮锻闹霞狄恒总喇暗锡慑挎俭框夜帆好疡吐冈目胁巷峡危锻暗诸扇盾而硫疆论阎垒状媒映胀厂愁翠长俊妇纷孔欲演堪么妊码瓦账骡弄掩诬比诡惊虑氧砾窘厨契肛菇诲宠严焙秃复屈者蜒拨逐过嘲荣砒腊跨迪唾胡局动彤太蚌册疟绞茅枢邮晚遥叙咽苑邯四拴国滦熔渝纱泪稗肇谗牙汇简赫组辽名闻雁概勋窄口涟梁贿遍帕杏韶工骚淮蜕遮卸踞充腊笺廓典蔫裤狼墙驰鹅买巡印鬼搂宦绪论略第二章习题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=1.2443Λ23=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-1b2=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=1.3297l1ˆ=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，kPaP1S=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=1.4181(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）。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＝0.14093y＝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