化工基本知识第五章精馏答案解析

1、五蒸馏习题解答1解: 作x-y图及t-x(y)图,作图依据如下：yA= p A0 XXA/pXA=(p-p B0)/(p A0-p B0)；以 t=90 C为例，xa=(760-208.4)/(1008-208.4)=0.6898yA=1008 XO.6898/76O=O.915O计算结果汇总：t C8O.O29O1OO11O12O13O131.8x1O.689O.448O.267O.128O.O19O83275y1O.915O.787O.611O.377O.O72OO58744.612x/(1O.911O.789O.627O.4O5O.O84O1+3.6122412Ox) 用相对挥发度计算x

2、-y值:y= ax/1+(a-1)x式中a = aM =1/2( a1 + a2)a=p A0/p B0a1 =760/144.8=5.249 ;a2=3020/760=3.974 om=1/2( a1+ a2)=1/2(5.249+3.974)=4.612y=4.612x心+3.612X)由此计算x-y值亦列于计算表中，y-x图,t-x(y)图如下：1题附图(1)求泡点：在泡点下两组分的蒸汽分压之和等于总压P即:Pa+p b=p a0xa+x b0xb=p 求泡点要用试差法,先设泡点为87 CIgp a0=6.8974O-12O6.35O/(87+22O.237)=2.971pa0=10 2

3、.971 =935.41mmHgIgp bO=6.95334-1343.943/(87+219.337)=2.566pB0=10 2.566 =368.13mmHg935.41 XO.4+368.13XO.6=595 600mmHg泡点为87 C ,气相平衡组成为y=p A/p=p aOxa/P=935.41XO.4/6OO=O.624304/1380.4+456/588.84=0.9951求露点：露点时，液滴中参与甲苯组成应符合下列关系:x A+X B=1 或 p A/p A0+p B/p B0 = 1pb=0.6 X760=456mmHg式中 pa=0.4 X760=304mmHg;求露点

4、亦要用试差法，先设露点为103 C，则:lgP a0=6.8974-120.635/(103+220.237)=3.165.p A0=1462.2mmHglgp B0 =6.95334-1343.943/(103+219.337)=2.784p B0=608.14mmHg于是：3O4/1462.2+456/6O8.14=O.96<1pb0=588.84再设露点为1O2 C，同时求得pa0=1380.4;故露点为102 C,平衡液相组成为XA= p A/p a0=304/1380.4=0.223解:(1) X A=( p 总-p B0)/( p A0-pB0)0.4=( p总-40&quo

5、t;(106.7-40)p 总=66.7KPayA=x A A0/p=0.4 X106.7/66.7=0.64(2) a=p A0/p b0=106.7/40=2.67(1) yD=?aD =(y/x) A/(y/x)B=(y D /0.95)/(1-y d )/0.05)=2y D =0.974L/V D =?V=V D +L(V/V D )=1+(L/V D )V0.96=V D 0.974+L0.95(V/V D )0.96=0.974+(L/V d )0.95(1+L/V D )0.96=0.974+(L/V d )0.95(L/V D )=1.45解:简单蒸馏计算：InW 1/W 2

6、 =x1 dx x2 y xW2=(1-1/3)Wi =2/3W i ;y=0.46x+0.549,x i=0.6,代入上式积分解得：釜液组成：x2=0.498.馏出液组成：Wd XD =W 1X1 -W 2X2(1/3W i)xd =W i XO.6-(2/3W i) XO.498/Xd =O.8O4Fx F=Vy+Lx0.4=0.5y+0.5x(1)(2)y= ax/(1+(a-1)x)=3x/(1+2x)(1),(2)联立求解，得 y=0.528,x=0.272回收率=(V )/(Fx f )=O.5 XO.528/O.4=66%7.解:f=d+wFxf =Dx d +Wx w已知 xf

7、 =0.24,x d =0.95,x w =O.O3,解得:D/F=(x f -x w )/(x d -x w )=(O.24-O.O3)/(O.95-O.O3)=O.228回收率 Dx d /Fx f =O.228 XO.95/O.24=9O.4%残液量求取：W/D=F/D-1=1/0.228-1=3.38W=3.38D=3.38(V-L)=3.38(850-670)=608.6kmol/h(1)求D及W,全凝量VF=D+WFXF =Dx D +Wx WXF =0.1,x D =0.95,x W =0.01(均为质量分率)F=100Kg/h,代入上两式解得：D=9.57Kg/h;W=90.4

8、3Kg/h由恒摩尔流得知：F(0.1/78+0.9/92)=V(0.95/78+0.05/92)注意:如用质量百分数表示组成，平均分子量 Mm=1/(a a/M A+a b/M b)解得 V=87Kg/h由于塔顶为全凝器，故上升蒸汽量V即为冷凝量，求回流比RV=D+LL=V-D=87-9.57=77.43Kg/hR=L/D=77.43/9.57=8.09(因为L与D的组成相同，故8.09亦即为摩尔比)(3) 操作线方程.因塔只有精馏段，故精馏段操作线方程为yn+1 =Rx n /(R+1)+x D /(R+1)式中XD应为摩尔分率XD =( X D /M A)/X D /M A + (1-X

9、D )/M=(0.95/78)/(0.95/78+0.05/92)=0.961操作线方程为:y n+1 =0.89x+0.106Bn +0.106yn+1 =8.09x n/9.09+0.961/9.09=0.89x9解:y=R/(R+1)x+xD /(R+1)(1) R/(R+1)=0.75R=0.75R+0.75R=0.75/0.25=3 X D /(R+1)=0.2075XD /(3+1)=0.2079XD =0.83(3) q/(q-1)=-0.5q=-0.5q+0.5q=0.5/1.5=0.333 0.75x+0.2075=-0.5x+1.5xF 0.75X q'+0.207

10、5=-0.5x q '+1.5 X0.441.25X q '=1.5 X0.44-0.2075=0.4425xq '=0.3620<q<1原料为汽液混合物10解:(1)求精馏段上升蒸汽量V和下降的液体量L,提馏段上升蒸汽量V'和下降的液体量L'.进料平均分子量：Mm=0.4 X78+0.6 X92=86.4F=1000/86.4=11.6Kmol/hFxf =Dx d +Wx wF=D+W11.6 X0.4=D X0.97+(11.6-D)0.02D=4.64Kmol/hW=6.96Kmol/hR=L/D,L=3.7 X4.64=17.17K

11、mol/hV=(R+1)D=4.7X4.64=21.8Kmol/h平均气化潜热r=30807X0.4+33320 X0.6=32313.6KJ/Kmol从手册中查得XF =0.4时泡点为95 C，则:q=r+cp(95-20)/r=(32313.6+159.2X75)/32313.6=1.37L'=L+qF=17.17+1.37X11.6=33.1Kmol/hV'=V-(1-q)F=21.8+0.37X11.6=26.1Kmol/h(2)求塔顶全凝器热负荷及每小时耗水量Qc=Vrr=0.97 X30804+33320X0.03=30879.5KJ/KmolQc=21.8 X30

12、879.5=673172.7KJ/h耗水量 Gc=673172.7/4.18(50-20)=5368.2Kg/h(3)求再沸器热负荷及蒸汽耗量塔的热量衡算Qb+Q F +Q r=Q v+Q W +Q LQb=Q v+Q W +Q l-Q f -Q r该式右边第一项是主要的，其它四项之总和通常只占很小比例，故通常有:Iv=(r+C pt)=30879.5+159.2X8.2=43933.9KJ/KmolQ b=21.8 X43933.9=957759.02KJ/h2.5KgF/cm2下蒸汽潜热 r=522Kcal/Kg=522X4.18 X18=39275.3KJ/Kmol蒸汽需量为GvGv =

13、Q B/r=957759.02/39275.3=24.4Kmol/h=24.4 X18=39.04Kg/h(4) 提馏段方程 y=L'x/(L'-W)-Wx w /(L'-W)=1.26x-0.00511解:提馏段：ym+1 ' =1.25x M ' -0.0187(1)=L'x m'/V'-Wx w /V',L'=L+qF=RD+FV'=(R+1)DW=F-D,精馏段：yn+1 =Rx n /(R+1)+X D /(R+1)=0.75x n +0.25X D(2)(3)q 线:XF =0.50将代入（1）

14、得出:ym+1 =1.25 X0.5-0.0187=0.606, 代入0.606=0.75 X0.5+0.25X d ,XD =0.92412解: yi=XD =0.84,0.84=0.45x1+0.55xi =0.64,yw =3 X0.64/(3+1)+0.84/(3+1)=0.69,0.69=0.45 XXW +0.55,X W =0.311, D=100(0.4-0.311)/(0.84-0.311)=16.8(Kmol/h),W=100-16.8=83.2(Kmol/h)13解:(1)求 R,X D,XW精馏段操作线斜率为R/(R+1)=0.723R=2.61提馏段方程y=L'

15、;x/(L'-W)-Wxw/(L'-W)=1.25x-0.0187精馏段操作线截距为xd/(R+1)=0.263/XD =0.95提馏段操作线与对角线交点坐标为y=x=x WX W =1.25 X W -0.0187xw =0.0748(2)饱和蒸汽进料时，求取进料组成y=0.723x+0.263 y=1.25x-0.0187联立求解，得x=0.535,y=0.6514i=x D =0.9,x i=0.9/(4-3XO.9)=O.692,2=1 XO.692心+1)+0.9/2=0.796d =x F =O.5, y d =O.5/2+O.9/2=O.715(1)FXF=Vy

16、q + Lx qO.45=(1 y q+(2/3)x qy q =2.5x q /(1 + 1.5xq)Xq =0.375yq=0.6Rmin=(x D-y q)/(y q-xq)=(0.95-0.6)/(0.6-0.375)=1.56R=1.5Rmi n=2.34D=O.95 XO.45/O.95=O.45W=1-O.45=O.55xw=(Fx f-Dx d)/W=(0.45-0.45XO.95)/O.55=O.O41L=RD=2.34 XO.45=1.O53;V=(R+1)D=1.5O3L'=L+qF=1.053+(2X1=1.72;V'=V-(1-q)F=1.503-1/

17、3=1.17y'=(L7V')x'-Wxw/V'=1.72/1.17x'-0.55XO.041/1.17=1.47x'-0.0193因饱和蒸汽进料,q线为水平线，可得原料组成y=x f=0.65#解:精馏段操作线方程yn+1 =3/4x n +0.24平衡线方程y= ax/1+(a-1)x=2.5x/(1 + 1.5x)提馏段操作线方程y=1.256x-0.01278其计算结果如下：No0.9060.96100.8210.7070.5730.4620.3440.2240.1280.0650.0290.920.860.770.700.5670.41

18、90.2680.1480.069由计算结果得知：理论板为10块（包括釜）,加料板位置在第五块;17解:D/F=(x F -x w )/(x D -x w )=(0.52-x w )/(0.8-x w )=0.5解得:x w =0.24精馏段操作线方程：yn+1 =(R/(R+1)x n +x D /(R+1)=0.75xn +0.2(1)平衡线方程：y= ax/(1+( a-1)x)=3x/(1+2x)(2)或:x=y/( a-( a-1)y)=y/(3-2y)交替运用式（1）,（2）逐板计算：x D =y 1=0.8 .x 1=0.571;y 2=0.628,x 2=0.360;=0.241

19、8解:y 3=0.470,x 3=0.228<x共需Nt=3块（包括釜）.q=0,x D =0.9,x F =0.5,x w =0.1,R=5,精馏段操作线方程：yn+1 =Rx n/(R+1)+X d/(R+1)=5x n/(5+1)+0.9/(5+1)=0.833x n+0.15图解:得理论板数为11块（不包括釜）,包括釜为12块由图读得:y W =0.06, xW-1 =0.03#解:18题附图(1) F=D+WFxf =Dx D +Wx w19解:D=F(x F -x w )/(x d -x w ) =100(0.3-0.015)/(0.95-0.015) =30.48 Kmol

20、/h=30.5 Kmol/hW=F-D=69.50 Kmol/h Nt及N F =?xd =0.95 、xW =0.015、q=1 、R=1.5 ; xd /(R+1)=0.38作图得:N t =9-1=8（ 不含釜） 进料位置：N F =619题附图L ' ,V ' ,y W 及 XW-1 q=1,V'=V=(R+1)DV'=30.5(1.5+1)=76.25Kmol/hL'=L+qF=RD+F=1.5 X30.5+100=145.8Kmol/h平衡线方程 y= ax/1+(a-1)x=4.6x/(1+3.6x)q线方程(q=2/(1+2)=2/3)

21、则y=q/(q-1)x-x f /(q-1)=-2x+1.35联解(1),(2)两式，经整理得：-2x+1.35=4.6x心+3.6X) 7.2x 2 +1.740x-1.35=0解知,x=0.329y=0.693(2) Rmi n=(x d -ye)/(y e-x e)=(0.95-0.693)/(0.693-0.329)=0.70621解:因为饱和液体进料,q=iye= aXe/1+(a-1)x e=2.47 X0.6/(1+1.47 X0.6)=0.788Rmin =(x D -y e)/(ye-x e)=(0.98-0.788)/(0.788-0.6)=1.02R=1.5 XRmin

22、=1.53N min =lg(x D /(1-x D )(1-x W )/x W )/lg a=lg(0.98/0.02)(0. 95/0. 05)/lg2.47= 7.56x=(R-R min )/(R+1)=(1.53-1.02)/(1.53+1)=0.202Y=(N-N min )/(N+1)Y=0.75(1-x0.567 )(N-7.56)/(N+1)=0.75(1-0.2020.567 )解得N=14.5 取15块理论板(包括釜)实际板数：N=(15-1)/0.7+1=21(包括釜）求加料板位置，先求最小精馏板数(N min )精=lgx D /(1-x D ) X(1-X F )/

23、X F /lg a=lg0.98/0.02.4/0.6/lg2.47=3.85N 精/N=(Nmin )精/N minN 精=N(N min )精/N min =14.5 X3.85/7.56=7.4则精馏段实际板数为 7.4/0.7=10.6取11块 故实际加料板位置为第12块板上.22解:作y-x图(1)由 y= ax/1+(a-1)x=2.4x/(1 + 1.4x)由于精馏段有侧线产品抽出，故精馏段被分为上，下两段，抽出侧线以上的操作线方程式(1)y n+1 =Rx n /(R+1)+x D /(R+1)=2/3x n +0.3侧线下操作线方程推导如下以虚线范围作物料衡算V=L+D 1+

24、D 2Vys+1 =Lx s+D 1 xDi +D 2xd2 ；ys+1 =Lx s/V +(D 1 XDi +D 2Xd2 )/V=Lxs/(L+D 1+D 2)+(D 1XD 1+D 2xd2)/(L+D 1+D 2);L=L 0-D2,则:(R=L 0/D 1)00.£0.40.60.8 LO x将已知条件代入上式，得到:y S+1 =0.5x+0.416(2)用图解法，求得理论塔板数为(5-1 )块,见附图.23解:根据所给平衡数据作x-y图.精馏段操作线yn+1 =Rx n/(R+1)+x D /(R+1)=1.5xn /(1.5+1)+0.95/(1.5+1)=0.6xn

25、 +0.3822题附图Xq线方程与q线:料液平均分子量：Mm=0.35 X+0.65 X18=22.9甲醇分子汽化潜热：r=252 X32 X4.2=33868.8KJ/Kmol水的分子汽化潜热：r=552 X18 X4.2=41731.2KL/Kmol23题附图料液的平均分子汽化潜热r=0.35 X33868.8+0.65 X41731.2=38979.4KL/Kmol料液的平均分子比热Cp=0.88 X22.9 X4.2=84.6KL/Kmolq=r+C p(ts-t f )/r=38979.4+84.6(78-20)/38979.4=1.13q 线斜率 q/(q-1)=1/13/0.13

26、=8.7提馏段操作线方程与操作线:由于塔釜用直接蒸汽加热，故提馏段操作线过横轴上（XW ,0） 点,于是在x-y图上，作出三条线,用图解法所得理论板数为7.6块，可取8块（包括釜）.24 解 :对全塔进行物料衡算：F1+F2=D+W(1)Fixfi+F 2xf2=Dx d +Wx w100 X0.6+200 X0.2=D X0.8+W X0.02100=0.8D+0.02W(2)由式(1) W=F 1+F 2-D=100+200-D=300-D代入式(2)得:D=120.5Kmol/hL=RD=2 X 120.5=241kmol/hV=L+D=241+120.5=361.5Kmol/h在两进料

27、间和塔顶进行物料衡算，并设其间液汽流率为L",V",塔板序号为s.V”+F 1=D+L”V”y s+1 "+F 1XF1=L”xs”+Dxdys+1 =(L'7V”)xs”+(DxD -F1XF1)/V”L”=L+q 1F1 =241 + 1X100=341Kmol/hV”=V=361.5ys+1 "=(341/361.5)xs”+(120.5X0.8-100 X0.6)/361.5ys+1 "=0.943x s”+0.125解:对于给定的最大 V',V=(R+1)D, 回流比R愈小，塔顶产品量 D愈大，但R需满足产品的质量要求

28、xD0.98,故此题的关键是求得回流比R.由题已知加料板为第 14层,故精馏段实际板数为 13层,精馏段板数为：13 X0.5=6.5取苯-甲苯溶液相对挥发度为a =2.54用捷算法求精馏段最小理论板数(N min )精=ln0.98/0.02-0.5心5/ln2.54=4.175y=N 精馏段-(N min )精/(N 精馏段 +1)=(6.5-4.175)/(6.5+1)=1.31由 y=0.75(1-x0.567 )x=(1-Y/0.75)Cl/O.567) =0.392=(R-R min )/(R+1)R=(0.392+R min )/(1-0.392)Rmin =(X D -y e)

29、/(y e-X e)对泡点进料 Xe=X F =0.5ye= ax/1+(a-1)x=2.54 X0.5/(1 + 1.54X0.5)=1.27/1.77=0.72Rmin =(0.98-0.72)/(0.72-0.5)=0.26/0.22=1.18 .R=(0.392+1.18)/(1-0.392)=1.572/0.608=2.59 D=V/(R+L)=2.5/(2.59+1)=0.696Kmol/h故最大馏出量为 0.696Kmol/h求n板效率:26解:Emv =(y n -y n+1 )/(y n -y n+1 ),因全回流操作，故有yn+1 =X n ,y n =X n-1与 xn

30、成平衡的 yn *= axn /1+( a-1)x n =2.43 X0.285/(1 + 1.43 X0.285)=0.492于是：Emv=(x n-1 -X n )/(y n -X n )=(0.43-0.285)/(0.492-0.285)=0.7Emv=(y n+1-y n+2 )/(y n+1Xf求n+1板板效率：由图可知：该板的板效率为Emv=(y1 -y )/(y 1 -y w)从图中看出,y 1=x D =0.28,关键要求y 1* 与 y W ./> IXw<-y n+2 )=(x n -x n+ )/(y n+1 -X n+1 )y n+1 =2.43 X0.1

31、73/(1 + 1.43X0.173)=0.337 Emv=(0.285-0.173)/(0.337-0.173)=0.683 27解:由已知条件Dx D /Fx F =0.8D/F=0.8 X0.2/0.28=0.57作系统的物料衡算：Fx F =Dx D +WxF=D+W联立求解：xf =Dx D /F+(1-D/F)x0.2=0.57X0.28+(1-0.57)xw习题27附图X0.093)=0.204解得 XW =0.093因塔釜溶液处于平衡状态，故yW = aXW /1+(a-1)x W =2.5 X0.093/(1+1.5yW与X1是操作线关系.yn+1 =L'x n /V

32、'-Wx W N'=Fx n /D-Wx w/D=Fx n /D-(F-D)x w /D=Fx n /D-(F/D-1)x w当 yn+1 =y W 时,xn =x 1xi=(y w +0.07)/1.75=(0.204+0.07)/1.75=0.157与X1成平衡气相组成为y1X0.157)=0.318yi = aX1/1+(a-1)x 1=2.5 X0.157/(1 + 1.5Emv=(0.28-0.204)/(0.318-0.204)=66.8%28解:(1)精馏段有两层理论板，XD =0.85,x F =0.5,用试差法得精馏段操作线ac,与x=x F =0.5线交于d

33、.提馏段有两层理论板,从点d开始再用试差法作图，得提馏段操作线bd，得:x W =0.17F=D+WFxf =Dx D +Wx W100=D+Wx D /(R+1)=0.103R=0.85/0.103-1=7.25100 X0.5=D X0.85+W X0.1728题附图得 D=48.5Kmol/hV'=V=(R+1)D=8.25 X48.5=400Kmol/h(2)此时加入的料液全被气化而从塔顶排出，其组成与原料组成相同，相当于一个提馏塔.29解:(1) D= n,FxF /x D =0.9 X100 X0.4/0.92=39.13Kmol/h,W=60.9Kmol/hXW =0.1

34、Fx f /W=0.1X100 X0.4/60.9=0.0656q=1.xq =0.4 查图得 yq =0.61Rmin =(x D -y q )/(y q -X q )=(0.92-0.61)/(0.61-0.4)=1.48R=1.5 X1.48=2.2 XD /(R+1)=0.92/3.2=0.29在y-x图中绘图得Nt =15-1=14块(未包括釜),N加料=第6块理论板Np=14/0.7=20块(不包括釜)N P 精=5/0.7=7.14,取8块,.第九块为实际加料板 可用措施： 加大回流比,XD f,XWJ, n= f点,X Df.改为冷液进料,N T <Nt' q=1

35、, N T =const.XDfq约为con st,下移加料30解:29题附图ye= aXe/1+( a-1)x e= aXF/1+( a-1)x f=2 X0.5/(1+0.5)=0.667DxD /Fx F =0.922; Dx d =0.922 X150 X0.4=55.32Dx D =Fx F -Wx W =Fx F -(F-D)x w =55.32150 X0.4-(150-D)X0.05=55.32D=56.4Kmol/hW=F-D=93.6Kmol/hXD =55.32/56.4=0.981及N F (进料位置)XD=0.981,x W =0.05,q=1.XD/(R+1)=0.

36、981/(2.43+1)=0.286a(0.981,0.981), b(0.05,0.05)q线:X F=0.4、q=1, q 线为垂线。(3)液气比精馏段：L/V=R/(R+1) =2.43/(2.43+1)=0.708提馏段：L'/V'=(L+qF)/(L+qF-W)或 V'=V ,L=RDL7V'=(RD+F)/(R+1)D)=(2.43 X56.4+150)/(3.43X56.4)=1.484(4)由于再沸器结垢，贝U Qb J,V' J,R J.XD J若要求维持XD不变,应提高再沸器加热蒸汽的压力Ps,及时清除污垢31解:(1)R=0.8 时

37、,xd ,xw 各为多少？30题附图由题知，当塔板为无穷时：R=R min =0.8,对泡点进料,Rmin =(x D -y e)/(y e-X e)Xe=X F =0.5,于 是(XD-0.667)/(0.667-0.5)=0.8解得：X D =0.8Fxf =Dx d +Wx wXF =Dx D /F+(1-D/F)x W由题知D/F=0.6 代入上式,T解得 xw =0.05,(2)R=1.5 时,求 XD ,xw .由题知，当塔板为无穷多时，R=R min =1.5Rmin =(X D-ye)/(y e-X e)同理求得XD =0.917,代入物料衡算式XF =Dx D /F+(1-D

38、/F)x w0.5=0.6 X0.917=(1-0.6)x w31题附图XW =-0.125,不成立.故操作线与平衡线应取 XW =0处相交，即:X w =0 ;Fxf =Dx d +Wx wXD =Fx F /D=0.5 X1/0.6=0.83此时精馏段与提馏段操作线示意图如上32解:(1) XF =y q =0.5,; X q =y q/( a-( a-1)y q)=0.5/(3-2 X0.5)=0.25Rmin =(x D-yq)/(y q-X q )=(0.9-0.5)/(0.5-0.25)=1.6R=2 X1.6=3.2F=D+WFxf =Dx d +Wx w0.5=0.9D+0.0

39、5WD=0.529W=0.471L=RD=3.2 X0.529=1.693V=2.222L'=L=1.693V'=V-F=1.222y'=1.385x'-0.0193(2)精馏段操作线y=(L/V)x+Dx D /V=(1.693/2.222)x+0.529X0.9/2.222y=0.762x+0.214或 y=Rx/(R+1)+x D /(R+1)=3.2x/4.2+0.9/4.2=0.762x+0.214y1=x D =0.9X1=y 1/(3-2 Xy1)=0.9/(3-2X0.9)=0.75y2=0.762 X0.75+0.214=0.7855(3)应维

40、持 R不变，此时V=F=1此时 D=V/(R+1)=1/(3.2+1)=0.238即D/F应改为0.238xw=(Fx f-Dx d)/W=(0.5-0.238 X0.9)/(1-0.238)=0.37533解:q=(r+(80-20)C p)/r=(40000+60X100)/40000=1.15W=L+qF=1.15X100=115D=F+S-W=100+50-115=35FXF =Dx D +Wx Wy=(L/s)x-(W/S)x w=2.3x-2.3x Wy2与 Xw 成平衡.y2=3x wX1=y 2/2.3+x w =2.304x wyi=3x i=6.913x w =x d100

41、 X0 2=35 X6.913X w +115xXW =0.056XD =0.387n=35 x0.387/(1000 X0.2)=0.67834解:作精馏段物料衡算，得精馏段操作线方程:yn+1 =(R/(R+1)X n+x D/(R+1)将 X0=0.5、y 1=0.63、xd=0.9代人上述方程:0.63=(R/(R+1)0.5+0.9/(R+1)解得： R=2.08操作线：截距 XD/(R+1)=0.9/(2.08*1)=0.292作精馏段操作线ac 再就q=1,x f=0.4作进料线。从yi、xo开始作梯级，共得 6块理论板。35 解:对第 n 块板：EmL=(X n-1 -x n)

42、/(x n-1 -X n*)=0.5；Xn=0.4-0.5(0.4-X n )yn = aXn71+( a-1)X n*=2x n*/(1+X n*)对第n板作物料衡算:X0.4-0.5(0.4- X n*)100 X0.3+100 X0.4=100 X(2xn7 (1+x n*) )+100解得：xn*=0.263xn=0.4-0.5(0.4- 0.263)=0.318yn=2 X0.263心+0.236)=0.38236 解:作全塔总物料衡算:F=D+W作全塔易挥发组分物料衡算:Fxf=Dx D+Wx W作分凝器易挥发组分物料衡算:Vy 1=Dx D+Lx L因为：V=2D L=D , (

43、3)式:2y1=x D+x L(3)(1)联解(1 )、(2)式得：x1=0.927解得：xl=0.619;代人(3)式：2y 1=0.8+0.619, 得 y1=0.71y1=y w=0.71,代人平衡方程：0.71=2.46x w /1+(2.46-1 ) xw解得：xw=0.5代人(2)得：D=F(x F-xw)/(x d-xw)=66.7 Kmol/h, W=33.3Kmol/h汽化量：V=2 X66.7=133.4 Kmol/h37 解:(1)精馏段操作线方程:yn+1 =(R/(R+1)x n+x d/(R+1)(4/ (4+1 ) x+0.7/(4+1)=0.8x+0.14提馏段

44、操作线方程:y' =(L ' /V ' )x-(W/V' )xwD/F=(x F-x W)/(x D-x W)=(0.3-x w)/(0.7-x w)=0.4 f xw=0.0333因为q=1,所以:L' /V ' =(L+F)/(R+1)D=RX(D/F)+1/(R+1)D/F=(4 X0.4+1)/(4+1)X0.4=1.3(W/V ' )xw=(F-D)/(R+1)DXxW=(1-D/F)/(R+1)D/FXxw=(1-0.4)/(5 X0.4) X0.0333=0.01所以：y =1.3x ' -0.01(2) yq= a

45、XF/1+( a-1) X F =2 XO.3/(1+(2-1)XO.3)=O.4615若平衡点在进料口处:Rm/(Rm*1)=4/(4+1)=(xD-y q)/(X D-X F)=(x d-0.4615)/(xd-0.3)7 XD=1.11不可能在进料口平衡。在塔顶平衡：即 XD=1(0.3-x w)/(1-x w)=0.4D/F=(XF-x w) /(x D-X w)=0.4;解得xw=-0.167故不可能。在塔底平衡：即xw=OXDmax =F Xx f/D=0.3/0.4=0.7538 解:(1 )饱和水蒸气用量 S=V'=V=( R+1y 1 =x d=0.95Emv=(y1

46、-y 2)/(y 1*-y 2)=(0.95-y2)/(0.5x1+0.5-y2)=O.5整理得:O.5y 2=0.7-0.25x1(1)Vy 2=Lx 1 +Dx D2.5D Xy2=1.5D Xx1+Dx d整理得:2.5y 2=1.5x 1+O.95F+S=D+W;(2)S=V ' =2.5D;F+2.5D=D+WF+1.5D=WF Xxf=D Xxd+W Xxw(3)代人(4)消去W得:D/F= ( xf-xW ) /(X d+1.5xW)XW=(X f-(D/F)x d)/ (W/F)=(0.4-0.64X0.667)/0.36=-0.0747<0=(0.5-0.1)/

47、(0.95+1.5X0.1)=0.36439 解:(1)7 xd=0.6n=Dx D /(Fx F)= X D(X F-X W)/( X F(X D-X W )=x D (0.4-0.05)/(0.4(x d-0.05)=0.955D/F= nXF/x d=0.955 X0.4/0.6=0.64D=0.64F=64Kmol/h,W=36Kmol/h(2)该塔只有提馏段，又q=1,L=F,V=D,故(L/V ) =F/D操作线方程：yn+1 =(F/D)X n -(W/D)X W=(100/64)X n-(36/64) X0.05=1.56x n-0.028(3)当Nt78时，可获得 XDmax

48、 q=1. q 线是垂线交平衡线上ye点ye= aXF/(1+( a-1)X f)=(3 X0.4)/(1+2X0.4)=0.667,此值是否最大值必须校验，由于F，V不变， D，W不变当XW=0,夹点在塔底XDmax = (F/D ) xf=0.4/0.64=0.62540解:(1) F 1XF1+F 2XF2=Dx D +W x W1 XO.6+0.5 XO.4=O.99D+O.O2WFi+F2=D+W1+O.5=D+W=(0.4-0.02)/(0.182-0.02)=2.35D=0.794Kmol/sW=0.706Kmol/sL=RD=0.794Kmol/sV=L+D=1.588Kmol

49、/sL" =L+q iFi=1.794Kmol/sV ''=V=1.588Kmol/sy ''=(L "N '')x ''+(Dx d-F1xf1)/V "=(1.794/1.588)x''+(O.794 XO.99-1 XO.6)/1.588 y"=1.13x "+O.117(2)若夹紧点在第一进料口处(第一段操作线与q线交点落在平衡线上)xqi =O.6 yqi =3 XO.6/(1+2XO.6)=O.82R' m=(x D-y qi )/(y qi -

50、x qi )=(O.99-O.82)/(O.82-O.6)=O.773若夹紧点在第二进料口处:yq2=0.4 xq2=y q2/( a-( a-1)y q2)=0.4/(3-2XO.4)=O.182提馏段操作线斜率:(yq2 -y W)/(x q2 -X W)L' =2.35V '，代人 L' -V ' =W=O.7O6 得：V' =0.523而 V ' =V ''-F2=V-F 2= ( Rm+1 ) D-F 2= ( Rm+1 )X 0.794-0.5=0.523解得：Rm =0.288取 Rmin =R m=0.773.41解:(1 ) D/F=(x F-XW)/(X D-X w)=(0.5-0.2)/(0.8-0.2)=0.5,令 F=1, D=0.5 W=0.5R=L/D=(2V/3)/(V/3)=2L=RD=2D=1L'V' =V=3D=1.5L' X' =V ' y+Wx W X '=