分离过程习题答案

1、第一章绪论1 .列出5种使用ESA和5种使用MSA的别离操作.答：属于ESA别离操作的有精储、萃取精储、吸收蒸出、再沸蒸出、共沸精储.属于MSA别离操作的有萃取精储、液-液萃取、液-液萃取(双溶剂)、吸收、吸附.2 .比拟使用ESA与MSA别离方法的优缺点.答：当被别离组分间相对挥发度很小,必须采用具有大量塔板数的精储塔才能分离时,就要考虑采用萃取精储(MSA),但萃取精储需要参加大量萃取剂,萃取剂的别离比拟困难,需要消耗较多能量,因此,别离混合物优先选择能量媒介(ESA)方法.3 .气体别离与渗透蒸发这两种膜别离过程有何区别？答：气体别离与渗透蒸发式两种正在开发应用中的膜技术.气体别离更成熟

2、些,渗透蒸发是有相变的膜别离过程,利用混合液体中不同组分在膜中溶解与扩散性能的差异而实现别离.5.海水的渗透压由下式近似计算：兀=RTC/M式中C为溶解盐的浓度,g/cm3;M为离子状态的各种溶剂的平均分子量.假设从含盐0.035g/cm3的海水中制取纯水,M=31.5,操作温度为298K.问反渗透膜两侧的最小压差应为多少kPa?答：渗透压tt=RTC/W8.314X298X0.035/31.5=2.753kPa所以反渗透膜两侧的最小压差应为2.753kPa.9.假定有一绝热平衡闪蒸过程,所有变量表示在所附简图中.求:(1)总变更量数Nv;(2)有关变更量的独立方程数Nc；(3) 设计变量数N

3、i;(4)固定和可调设计变量数Nx,Na；(5)对典型的绝热闪蒸过程,你将推荐规定哪些变量？思路1:3股物流均视为单相物流,总变量数Nv=3(C+2)=3c+6独立方程数Nc物料衡算式C个热量衡算式1个相平衡组成关系式C个1个平衡温度等式1个平衡压力等式共2C+3个故设计变量Ni=Nv-Ni=3C+6-(2C+3)=C+3固定设计变量Nx=C+2,加上节流后的压力,共C+3个可调设计变量Na=0解：(4) Nv=3(c+2)(5) Nc物c能1相c内在(P.T)2Nc=2c+3(6) Ni=Nv-Nc=c+3(7) Nxu=(c+2)+1=c+3(8) Nau=c+3-(c+3)=0思路2:输

4、出的两股物流看成是相平衡物流,所以总变量数Nv=2(C+2)独立方程数Nc：物料衡算式C个,热量衡算式1个,共C+1个设计变量数Ni=Nv-Ni=2C+4-(C+1)=C+3固定设计变量Nx:有C+2个加上节流后的压力共C+3个可调设计变量Na：有0塔顶产物进料,227K,2068kPa/力2123456fiNCCCCCCKmol/h1.054.467.6141.154.756.033.3211.满足以下要求而设计再沸汽提塔见附图,求:(1)设计变更量数是多少？(2)如果有,请指出哪些附加变量需要规定？解：Nxu进料c+2压力9c+11=7+11=18Nau小级单元传热合计2Nvu=Nxu+N

5、au=20附加变量：总理论板数.第二章单级平衡过程2.计算在0.1013MPa和378.47K下苯1-甲苯2-对二甲苯3三元系,当xi=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值比拟.三个二元系的Wilson方程参数.12111035.33;1222977.832322442.15;233313111510.14;1333在T=378.47K时液相摩尔体积为：v1L100.91103m3kmol安托尼公式为：460.051642.81(单位：J/mol)vL117.55103;v：136.6910苯:lnP1s甲苯：ln

6、P2s20.79362788.51/T52.36；20.90653096.52/T53.67；对二甲苯：lnP3s20.98913346.65T57.84(Ps:Pa;T:K)LVj解1:由Wilson参数万程ij4expj./RTViLv12expVi1211RT117.55100.911033101035.33.8.314378.47=1.619L2122RTVi-expV2977.838.314378.47100.91103exp117.5510=0.629kixk231.010;320.995kkjXj30.9930由Wilson方程Ini1Inijxj10.9184;20.9718根

7、据安托尼方程：坪0.2075MPa;P2s8.693104Pa;P3s3.823104Pa由式(2-38)计算得：K11.88;K20.834;K30.375如视为完全理想系,根据式(2-36)计算得：K12.048解2:在T=378.47K下K20.858;K30.377甲苯:对二甲苯:ln坪lnP2s20.793620.90652788.5/(378.473096.52/(378.4752.36)；53.67)；lnP3s20.98913346.65/(378.4757.84)；Rs=207.48Kpas_F2=86.93KpaBs=38.23KpaWilson方程参数求取12212332

8、1331LV3.-exp(V11211100.91103RT)117.551031222117.55103RT)100.911032322117.55103RT)136.691032333、136.69103RT)117.551031311、136.6910:3RT)100.9110：3,1333、136.69103exp(exp(exp(exp(exp(RTLV2/exp(V1LV2.exp(V3LV1.exp(V2LV1.-exp(V2v3rexp(V3exp(100.91103lnr11ln(x12x213x3)(x1035.338.314378.47977.83)8.314378.47

9、,442.158.314378.47460.058.314378.471510.148.314378.471642.818.314378.4721x21.1930.8540.74721.3460.4572.28331x312x213x321x1x223x331xi32%x3ln(0.31251.1930.29780.4570.3897)(0.31250.8540.29780.31251.1930.29780.4570.38972.2830.38970.8540.31250.29780.7夕20.38972.2830.31251.3460.29780.3890.09076r1=0.9132Inr

10、21ln(x121x2X323)(一X1xl1212X213X3X221xlX223x332X331X132X2一)X31ln(0.31250.8540.29780.74720.3897)(0.21251.1930.31251.1930.29780.4570.38970.29780.38971.3460.8540.31250.29780.74720.38972.2830.31251.3460.29780.3890.0188r2=1.019Inr31ln(x131X2、,X23X3)(X12X21313X323X221X1X223X331X1ln(0.31250.4570.29871.3460.

11、3897)/0.31250.457(0.31251.1930.29780.4570.38970.74720.29780.3897)0.8540.31250.29780.74720.38972.2830.31251.3460.29780.3897)0.2431r3=1.2752K1rP/0.9132207.4801.31.87K2r2P*1.01986.9%01.30.8744K3/1.275238.2%01.30.4813而完全理想系：K1P/P207.4%1.32.048K2%86咪1J0.8581K3P%3&%1.30.37743.在361K和4136.8kPa下,甲烷和正丁烷二元

12、系呈汽液平衡,汽相含甲烷0.603873%(mol与其平衡的液相含甲烷0.1304%.用R-K方程计算可,穹,和Ki值.0.42748R2Tc2.5c1解：an=pc1=3.222MPa?dm6?$5?mol-222.50.42748RTc2a22=pc2=28.9926MPa?dm6?k0.5?mo20.08664R2TC1_2_250.42748RT&C2b2=pc2=0.0806dm3mol-1其中Tci=190.6K,Pci=4.60MpaTc2=425.5K,Pc2=3.80Mpa均为查表所得.ai2=Vii?a22=9.6651MPa?dm6?k0.5?mo2液相：a=ai

13、ixi2+2ai2XiX2+a22X22=3.220.13042+2X9.66510.13040.8696+28.99260潟6962=24.1711b=bixi+b2X2=0.02980.1304+0.08060.8696=0.0740由RK方程：P=RT/(V-b)-a/T0.5V(V+b)0.008314536124.1711V；0.0740361%.：0.0740)4.1368解得Vml=0.l349?In0.074020.00831453611.550.1349)0.07404.13680.13490.13470.0740-In0.0083145361=1.3297?=3.7780?

14、i?同理In'2=-1.16696,2=0.3113汽相：a=3.222>0.603872+2X9.66510.603870(39613+28.99260.396132=10.3484b=0.02980.60387+0.08060.39613=0.04990.008314536110.3484由4.1368=v；v0.0499_36i0.5vmv(vmv0.0499)=lnV/(V-b)+bi/(V-b)-22yiaij/bmRT1.5*In(V+b)/V+abi/b2RT1.5In(V+b)/V-b/(V+b)-In(PV/RT),0.1349、0.0298?()lni=ln0

15、.13490.0740+0.13490.07402(0.13043.2220.86969.6651)0.13490.0740_15_0.07400.0083145361/(0.1340)+24.17110.02980.13470.07400.58610.0298v1ni=ln(0.58610.0499)+0.58610.04992(0.603873.2220.396139.66510.58610.0499、ri1n()0.04990.008314536150.586110.34840.02980.049920.00831453611.5xln(0.58610.0499)0.58610.0499

16、4.13680.58610.58610.0499_1n(0.0083145361)=0.0334942v故1=1.0341同理,In2=-0.522819,2=0.5928故K1=y1/x1=0.60387/0.1304=4.63110.60387K2=y2/x2=10.1304=0.4555?v(K1=1/1)4.(1)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)平衡常数法由于汽相、液相均为完全理想物系,故符合乌拉尔定律pyi=pisxisyiPi而

17、Ki=xi=p设T为80c时,由安托尼公式(见习题1)求出格组分的饱和蒸汽压.p1=101.29kPa,P2=38.82kPa,sp3=15.63kPa故y1y2y3=K1X1+K2X2+K3X3sssP1P2P3一X1X2一X3=PPP101.29=1000.538.821000.2515.631000.25=0.64<1故所设温度偏低,重设T为95c时sssP1=176.00kPa,P2=63.47kPa,P3=27.01kPay1y2y3=1.11>1故所设温度偏高,重设T为91.19C,sssp1=160.02kPa,p2=56.34kPa,p3=23.625kPay1y2

18、y3=1.00001251故用平衡常数法计算该物系在100kPa时的平衡温度为91.19C汽相组成：y1=K1X1=sp1X1p160.021000.5=0.8001sP2一X2y2=K2X2=P56.341000.25=0.1409sP3y3=K3X3=PX323.6251000.25=0.059(2)相对挥发度法1i由于是理想混合物,所以/y1、X1、()/()yiyiXi,得y11i(X1/Xi)对于理想混合物,得设T为80c时,p1S1i=P2Ssp3=15.63kPay1/5.22,y3=y1/12.964.(2)p1=101.29kPa,p2=38.82kPa,故12=2.61,1

19、3=6.48,y2=由于y1y2y3=1,故y1=0.788s又由于py1=100X0.788=78.8kPa,而p1X1=101.29>0.5=50.645kPa<py1故所设温度偏低；sSs重设T=92C时p1=163.31kPa,p2=57.82kPa,p3=24.31kPa得故12=2.824,13=6.718,y2=y1/5.648,y3=y1/13.436由于y1y2y3=1,故y1=0.799,y2=0.141,y3=0.0595s且py1=100X0.799=79.9kPa,而p1x1=163.31>0.5=81.655kPa,根本相等因此,由相对挥发度计算该

20、物系平衡温度为92,此时y1=0.799,y2=0.141,y3=0.0595液体混合物的组成为：苯0.50;甲苯0.25;对二甲苯0.25(摩尔分率)分别用平衡常数法和相对挥发度法计算该物系在100kPa式的平衡温度和汽相组成.假设为完全理想系.解1:平衡常数法：设T=368K用安托尼公式得:P1s156.24kPa由式(2-36)得：P2s63.28kPa;P3s26.88kPaK11.562;K20.633;K30.269yi0.781;y0.158;y0.067由于y>1.001,说明所设温度偏高.y1.006由题意知液相中含量最大的是苯,由式(2-62)得:K1上二1.553可

21、得T367.78K重复上述步谑Ki1.553;K20.6284y10.7765;y20.1511K30.2667,_y30.066675V1.0003在温度为367.78K时,存在与之平衡的汽相,组成为：苯0.7765、甲苯0.1511、对二甲苯0.066675(2)用相对挥发度法:设温度为368K,取对二甲苯为相对组分.计算相对挥发度的:组分i苯甲苯对二甲苯Xi0.500.250.251.0005.8072.3531.000ijxi2.90350.58830.25003.7418ijXiijXi0.77600.15720.06681.0000135.807;232.353;331.000解2

22、:平衡常数法.假设为完全理想系.设t=95C苯：lnP1s20.79362788.5/(95273.1552.36)11.96；s5R1.56910Pa甲苯：lnF2s20.90653096.52/(95273.1553.67)11.06；s_4P2s6.358104Pa对二甲苯：lnP3s20.98913346.65/(95273.1557.84)10.204；P3sKiK31.569;K22.702P1sPP/P0.6358KiXi1.5960.50.27020.250.63580.251.011选苯为参考组分：K121.56%0111.552；解得T2=94.61CInP2s11.05;

23、P2s6.281104PaInP3s10.19；P3s2.6654104PaK2=0.6281K3=0.2665Kixi1.5520.50.62810.250.26650.250.99971故泡点温度为94.61C,且y11.5520.50.776；y20.62810.250.157;y0.26650.250.067(2)相对挥发度法设t=95C,同上求得K1=1.569,K2=0.6358,K3=0.2702135.807232.353331ixi5.8070.52.3530.2510.253.74yiiXi5.8070.5iXi3.742.3530.2510.253.743.741.0故泡

24、点温度为95C,且y15.8070.53.740.776；丫22.3530.250.1573.74y310.253.740.0672.513.台中羊30%、乙革40%、水30%均为庠尔%为淞体在总用为5O.kPd下通行施埃因羔,暇世平革和乙笨的充台恢米从世4尔定律,烧与水完全不互部.计算泡点温度R布应的汽相组成,2513.iff：改中K为1泡分,乙弟为2级分,火为灶和钓介马担尔定牛ttff.小吓P；Verr或沟点Si度为7.'已由安抵其方和裨/：-27.1644.：-11.3006在友得p,m-3L176此旨p=31.1764-27.L&44x(3/7)4-11.3QO6x(4

25、/7)=49.275kPa<50.6kPar效所设31度略低,重设泡力浊度为70.5X2此时依卜万法求得p=50.S50.SkRh成河岸温底M7O.5C,>>-23.3m,y,T3.01Mm0-63.68汁茸姑集rtl*7O.73oC：汽物组成：甲七23.31%>、乙*13.01%、水53.68%均为摩尔FT分立口口147(175妆薄木交档文榭5那么触25含30%甲苯1,40%乙苯2和30%水3的液体混合物,在总压为50.66kPa下进行连续闪蒸蒸镭,假设乙苯和甲苯混合物服从拉乌尔定律,烧和水完全不互溶.计算泡点温度和气相组成.解：压力较低,气体看为理想气体：烧与水完全

26、互不溶,那么液相按两相考虑,乙茶与甲苯混合物可看为理想溶液.在泡点P=6+匕+匕=夕X+px2+°查?化工热力学?附录得安托尼公式：T：K；P：105帕.h=11.6834-3816.44/7-46.13文in在线uiiBiiiiiaimi/44in"=9.3635-3096.52/(T-53.67)InP；=9.3993-3279.47/(T-79.47)设T为70,343.15KIn=11.68343816.44/(7-46.13)=11.6834-3816.44/(343.15-46.13)=-1.1657=31170PaInP；=9.3635-3096.52/(7-

27、53.67)=9.3635-3096.52/(343.15-53.67)=-1.3033P；=27163Pa文桂在线InP5O=9.3993-3279.47/(7-79.47)3I=9.3993-3279.47/(343.15-79.47)=-2.18075巴°=11296?尸=十尸：其I34=31170+27163x4-11296x=49266P77I得P小于所给P,前设温度偏低,再设T为71,344.15K,In尸=11.6834-3816.44/(7-46.13)I=11.6834-3816.44/(344.15-46.13)=-1.12259I占.=32544Pa|文档在我I

28、InP；=9.3635-3096.52/(7-53.67)=9.3635-3096.52/(344.15-53.67)=-1.26651P；=28181PaIn/°=9.3993-3279.47/(1,-79.47)=9.3993-3279.47/(344.15-79.47)=-2.1400P：=11766PaP=P.O4-尸;X；+32544+28181x+11766x=51345Va设T为70.65C,343.8K文柳在蜷In6.=11.6834-3816.44/(T-46.13)=11.6834-3816.44/(343.8-46.13)=-1.13764P：=32057.5P

29、aIn仆9.3635-3096.52/(r-53.67)=9.3635-3096.52/(343.8-53.67)=-1.27937P：=278213PaIn尸30=9.3993-3279.47/(T-79.47)=9.3993-3279.47/(343.8-79.47)=-2.15423尸30=11599.2PaBiff线iihP=+尸,?+内同3=32057.5+27821.3x-+11599.2x-=50610Pp讣E,巧'一1.0卜0.0010H=0.01所以泡点温度为343.8K.y1=0.633fy2=0,236,y3=O,131&p)假设用fj_1-(叫po(&#

30、187;)p.®)1+-*2+PPPTil*E格底til)C求与“(2L再由其求下一次所设温度值可否.重例在叫IIBill6.一烧类混合物含甲烷5%(mol),乙烷10%,丙烷30%及异丁烷55%,试求混合物在25c时的泡点压力和露点压力.解1:由于各组分都是烷姓,所以汽、液相均可以看成理想溶液,Ki值只取决于温度和压力.可使用姓类的P-T-K图.泡点压力的计算：75348假设P=2.0MPa,因T=25C,查图求Ki组分i甲烷乙烷(2)丙烷异丁烷(4)Xi0.050.100.300.551.00Ki8.51.80.570.26yiKiXi0.4250.180.1710.1430.9

31、19KiX=0.919<1,说明所设压力偏高,重设P=1.8MPa组分i甲烷乙烷(2)丙烷异丁烷(4)Xi0.050.100.300.551.00Ki9.41.950.620.28yiKiXi0.470.1950.1860.1541.005KiXi=1.005=,1故泡点压力为1.8MPa.露点压力的计算：假设P=0.6MPa,因T=25C,查图求Ki组分i甲烷乙烷(2)丙烷(3)异丁烷(4)Yi0.050.100.300.551.00Ki26.05.01.60.64xiy/K.0.00190.020.18750.85941.0688yi=1.0688>1.00,说明压力偏高,重设

32、P=0.56MPa/-i组分i甲烷(1)乙烷丙烷异丁烷小0.050.100.300.551.00Ki27.85.381.690.68xiyi/K0.00180.01860.17750.80881.006J故露点压力为0.56MPa.yiZ=1.006=Ki解2:(1)求泡点压力:设Pi=1000KPa,由25C,1000KPa,查P-T-K列线图得KiK1=16.5K2=3.2K3=1.0K4=0.43所以yi0.0516.50.13.20.31.00.550.431.681选异丁烷为参考组分0.2560.9070.282,查得P=1771KPa在此条件下求得yi=1.021,继续调整K44K

33、43/Z0.280.279,查得P=1800KPay1.02求得：yi1.0011,故混合物在25c的泡点压力为1800KPa厅P组分xi1000KPa2000KPa1770KPa1800KPaKiyiKiyiKiyiKiyi1甲烷0.0516.50.8258.40.429.60.489.40.472乙烷0.103.20.321.750.1751.950.1951.920.1923丙烷0.301.00.300.570.1710.630.1890.620.1864异丁烷0.550.430.240.2560.1410.2850.1570.2790.1531.001.680.9071.001(2)求

34、露点压力设P1=1000KPa,由25C,1000KPa,查P-T-K列线图得Ki所以XiyiKiK1=16.5K2=3.2K3=1.0K4=0.430.050.100.55161416.53.20.431.614选异丁烷为参考组分K42K41Xi0.431.6140.694由25C,K42=0.694查得P=560KPa,查得各组分的Ki值求得Xi0.9901故混合物在25c时的露点压力为560KPa厅p组成组成1000KPa560KPaKiXiKiXi1甲烷0.0516.50.00327.50.0022乙烷0.103.20.0315.200.0193丙烷0.301.00.301.700.1

35、764异丁烷0.550.431.280.6940.7931.6140.9907.含有80%(mol)醋酸乙酯(A)和20%(mol)乙醇(E)的二元物系.液相活度系数用VanLaar方程计算,Aae=0.144,Aea=0.170.试计算在101.3kPa压力下的泡点温度和露点温解：由Vanlaar方程得：lna(1AaeXAEAAE)2xeaAea(10.1440.1440.80.170.2)2,得a=1.0075lnrE(1AeaXEAAEA)2xaeAae(10.1700.1700.20.1440.8)2r-,彳导rB=1.1067由于低压气体可视为理想气体,故siPiXisyipyir

36、iPiXi,得P(1)泡点温度时,设T=348.15K,由安托尼方程得SsPa=94.377kPa,Pe=88.651kPassaPaXarEPeXe1.007594.3770.81.106788.6510.2yiyAyE故P+P=101.3101.3=0.945<1,可知所设温度偏低,重设T=349.82K:Ss此时Pa=99.685kPa,Pe=94.819kPa1.106794.8190.2101.3ssyiaPaXarePeXe1.007599.6850.8P+P=101.31.000331故泡点温度为349.82KpyisXi(2)求露点温度,此体系可视为理想气体,由pyiri

37、PiXi,得piri设T=349.8KSs由安托尼方程得pA=99.620kPa,pE=94.743kPa,101.30.8101.30.2故xiXAxb=99.6201.007594.7431.1067=1.4>1,故所设温度偏低重设T=350.1K时XiXAXb=0.992=1故露点温度为350.1K11.组成为60%苯,25%甲苯和15%又-二甲苯(均为mol百分数)的液体混合物100kmol,在101.3kPa和100c下闪蒸.试计算液体和气体产物的数量和组成.假设该物系为理想溶液.用安托尼方程计算蒸汽压.解：设苯为组分1,甲苯为组分2,对二甲苯为组分3.100c时,【P33例2

38、-71p0.1(0.3161)=198.929kPa,p2=74.165kPa,p3=32.039kPa对于低压气体,气相可视为理想气体,液相可视为理想溶液,sp3K3=p=0.316ssspip1p2故Ki=p,得K1=p=1.964,K2=p=0.732,(1)核实闪蒸温度假设100c为进料的泡点温度,那么(Kizi)=1.964>0.6+0.732>0.25+0.316W15=1.41>1假设100c为进料的露点温度,那么(zi/Ki)=1.21>1说明实际的进料泡点温度和露点温度分别低于和高于规定的闪蒸温度,闪蒸问题成立.(2)求明令少=0.1(0.3161)0

39、.151(0.3161)(0.9641)0.6(0.7321)0.25f()=1(1.9641)+1(0.7321)(0.9641)0.6f(0.1)=10.1(1.9641)十(0.7321)0.2510.1(0.7321)(0.3161)0.15=0.366f(0.1)>0,应增大小值.计算R-K方程导数公式为:f'()=-1(Ki1)2zi2(K11)2+1(K21)2Z22(K21)2+12(K31)2Z32(K31)20.5580.018(10.964)2+(10.2684)2+(10.070.6844)2=0.1f(i)idf(i)d为初值进行迭代,得下表迭代次数f(

40、)df(10.10.3660.56420.750.04360.51130.840.0092一)/d可知f(3)数值已到达P-T-K图的精确度(3)计算xi,yiZ10.6x11(K11)=10.84(0.9641)=0.332K1Z11.9640.6y1-1(K11)=10.84(0.9641)=0.651同理,x2=0.323,y2=0.236x3=0.353,y3=0.112(4)计算V,LV=F=0.84M00=84kmolL=FV=100-84=16kmol(5)核实yiXi14.xi=0.999,yi=1.008,结果以满意98在101.3kPa下,对组成为45%(摩尔百分数,卜同)

41、正已烷,25%正庚烷及30%正辛烷的混合物计算.(1)泡点和露点温度(2)将此混合物在101.3kPa下进行闪蒸,使进料的50%汽化.求闪蒸温度,两相的组成.解：由于各组分都是烷烧,所得的汽、液相均可看成理想溶液,Ki只取决于温度和压力,假设计算精度不要求非常图可使用位关日勺P-T-K图,见图2-1假设T=82C,由P=101.3kPa得下表：组分XiKiyKiXi正己烷45%1.50.675正庚烷25%0.630.158正辛烷30%0.280.084KiXi=0.917<1,说明所设温度偏低,重设T=85.8C,得组分XiKiyKiXi正己烷45%1.60.72正庚烷25%0.70.1

42、75正辛烷30%0.310.093KiXi=1.008=故泡点温度为85.8Co同理,可迭代求出露点温度设T=95C,此时组分yiKiXi=yi/Ki正己烷45%2.00.225正庚烷25%0.90.278正辛烷30%0.4250.705yi/K；i=1.2068>1,所设温度偏低,重设T=102.4C,得组分yiKixi_yi/Ki/正己烷45%2.350.1915正庚烷25%1.080.2315正辛烷30%0.5200.5769yi/Ki=0.9999满足精度要求,故露点温度为102.4C.TTb(1)进料50%气化,那么由公式TDTB得T=94.1C为闪蒸温度,查表2-1得:正己烷

43、正庚烷正辛烷58.90%22.45%17.14%组分为小31.0%27.0%42.85%结果(1)泡点：85.8oC,露点：102.4oC;(2)闪蒸温度94.1oC;气相组成：正已烷一0.31,正庚烷一0.27,正辛烷一0.43;液相组成：正已烷一0.59,正庚烷一0.23,正辛烷一0.17.(均为摩尔分数)第三章多组分精储和特殊精储1.附图为脱丁烷塔的物料平衡图.全塔平均操作压力为522kPa.求最小理论塔板数；估计非分配组分的分配.【P75例3-41/G(11)匚*(ni匚*i阳沿习题6附图解：(1)根据题意可列表如下:编号组分Xi,FXi,FFXi,DXi,DDXi,BXi,BB1C3

44、0.0139120.025612002iC40.5114480.9444420.014763nC40.0411360.0278130.0563234iC50.0171152.137*10-310.0343145C60.026223000.0563236C70.044639.1000.095839.17C80.311272.1000.667272.28C90.035431.0000.07593121.000F=876.21.0004681.000408.3(2)塔顶温度和塔釜温度的计算：塔顶温度的计算：第一次试差：假设TD=37.5C,查Ki(37.5C,522kPa)K1=2.91,K2=0.

45、93,K3=0.68,K4=0.30,A1=3.22,a2=1.368,a3=1.0a4=0.441,再由yi,D=xi,Dac=1ai>yi/aAy/aKc=2yi/a=0.727查彳3Kc值相应于37.5C.故假设正确,露点温度Td=37.5C塔釜温度的计算：第一次试差温度：假设TB=155C编号组分Xi,DDXi,BKiaiaixi1C31209.801.84202iC44420.01476.141.1540.01693nC4130.05635.321.00.05634iC510.03433.00.5640.01935C600.05631.420.2670.01506C700.09

46、580.790.1480.014187C800.6670.470.08830.05898C900.07590.280.05263.99*10-3所以：Kc=5.414;再有P-T-K图查塔釜温度为TB=155C,因与假设值符合,结束试差.(3)最小理论塔板数和组分分配：首先计算a值,aLH,D=l.368,aLH,B=l.l54aLH,w=(aLH,DaLH,B)0.5=1.265;XiD,L/x旧,Llg代入芬斯克方程NmX"H/婀H/有1gLH,avNm=lg(0.944/0.0147)*(0.0563/0.0278)/lg1.256=21.37其他组分的分配：a13,w=(3.

47、22*1.842)0.5=2.435Xi,dD/Xi,bB=(a13,w)Nm*X3,dD/X3,bB=(2.435)Nm*(13/23)=1.0273*108又Xi,fF=12=Xi,dD/+Xi,bB由以上两式得：d1=12同理可得：d2=447,d3=13,d4=2.138*10-7d5=d6=d7=d8=03.估计习题6脱丁烷塔的最小回流比.进料的液相分率q=0.8666.【P77例3-51解：由恩特伍德公式计算最小回流比iXi,F2.421612876.2(2.4216)1.2395448876.2(1.2395)136876.2(1)0.696515876.2(1)0.177923

48、876.2(0.1779)解得1.01339.10.0804876.2(0.0804)272.10.0418876.2(0.0418)310.02261876.2(0.0226)0.8668代入式X26、i(Xi,D)m44212442132.42160.02572.42161.013Rm2.0854Rm1,其中0.00230.9465X3X11212442130.00230.02571312442130.00231.23950.946510.02781.23951.01311.013Rm10.0278P,6用芬斯克方程计算附图精馋塔的最少理论板数与非关犍组分的分配.解题思路：:假设清楚分割-d,w巧计算塔顶、塔釜温度!计算各组分平均相对挥发度Nm=?验证清楚分割匕63,6解：假设清楚分割用清楚分割方法对各组进行分预分配,如下表:进科组分进料4kmol/h饰出液1釜液Wg12500250024004003nCs(LK)600600-664i-C5(HK)100151(M)-155n-Cg20020