化工原理第五章-精馏-答案

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1、五 蒸馏习题解答1解: (1)作x-y图及t-x(y)图,作图依据如下: xA=(p-pB0)/(pA0-pB0); yA=pA0xA/p 以t=90为例,xA=(760-208.4)/(1008-208.4)=0.6898 yA=10080.6898/760=0.9150 计算结果汇总: t80.0290100110120130131.8x10.68980.4483 0.2672 0.12870.01950 y10.91500.7875 0.61180.37770.07240 4.612x/(1+3.612x)10.91120.7894 0.6271 0.4052 0.08400(2)用相对

2、挥发度计算x-y值: y=x/1+(-1)x 式中=M=1/2(1+2) =pA0/pB0 1=760/144.8=5.249 ;2=3020/760=3.974M=1/2(1+2)=1/2(5.249+3.974)=4.612 y=4.612x/(1+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 2解: (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+pB=pA0xA+xB0xB=p求泡点要用试差法,先设泡点为87 lgpA0=6.89740-1206.350/(87+220.237)=2.971请预览后下载！ pA0=1

3、02.971=935.41mmHg lgpB0=6.95334-1343.943/(87+219.337)=2.566 pB0=102.566=368.13mmHg 935.410.4+368.130.6=595600mmHg泡点为87,气相平衡组成为 y=pA/p=pA0xA/P=935.410.4/600=0.624 (2)求露点: 露点时,液滴中参与甲苯组成应符合下列关系: xA+xB=1或pA/pA0+pB/pB0=1式中 pA=0.4760=304mmHg; pB=0.6760=456mmHg求露点亦要用试差法,先设露点为103,则:lgpA0=6.8974-120.635/ (10

4、3+220.237)=3.165 pA0=1462.2mmHg lgpB0=6.95334-1343.943/(103+219.337)=2.784 pB0=608.14mmHg于是 : 304/1462.2+456/608.14=0.961再设露点为102,同时求得pA0=1380.4; pB0=588.84 304/1380.4+456/588.84=0.9951故露点为102,平衡液相组成为 xA=pA/pA0=304/1380.4=0.223解: (1)xA=(p总-pB0)/(pA0-pB0) 0.4=(p总-40)/(106.7-40) p总=66.7KPa yA=xApA0/p=

5、0.4106.7/66.7=0.64 (2)=pA0/pB0=106.7/40=2.674解: (1) yD=? D =(y/x)A/(y/x)B =(yD /0.95)/(1-yD )/0.05)=2 yD =0.974 (2) L/VD =? V=VD +L (V/VD )=1+(L/VD ) V0.96=VD 0.974+L0.95 (V/VD )0.96=0.974+(L/VD )0.95 (1+L/VD )0.96=0.974+(L/VD )0.95 (L/VD )=1.45解: 简单蒸馏计算:请预览后下载！ lnW1/W2= W2=(1-1/3)W1=2/3W1;y=0.46x+0

6、.549,x1=0.6,代入上式积分解得: 釜液组成:x2=0.498, 馏出液组成:WD xD =W1x1 -W2x2 (1/3W1)xD =W10.6-(2/3W1)0.498 xD =0.804 6解: FxF=Vy+Lx 0.4=0.5y+0.5x -(1) y=x/(1+(-1)x)=3x/(1+2x) -(2) (1),(2)联立求解,得y=0.528,x=0.272 回收率=(Vy)/(FxF )=0.50.528/0.4=66% 7.解: F=D+W FxF =DxD +WxW已知xF =0.24,xD =0.95,xW =0.03,解得: D/F=(xF -xW )/(xD

7、-xW )=(0.24-0.03)/(0.95-0.03)=0.228回收率 DxD /FxF =0.2280.95/0.24=90.4%残液量求取: W/D=F/D-1=1/0.228-1=3.38 W=3.38D=3.38(V-L)=3.38(850-670)=608.6kmol/h8解: (1) 求D及W,全凝量V F=D+W FxF =DxD +WxW xF =0.1,xD =0.95,xW =0.01(均为质量分率) F=100Kg/h,代入上两式解得: D=9.57Kg/h; W=90.43Kg/h由恒摩尔流得知: F(0.1/78+0.9/92)=V(0.95/78+0.05/9

8、2) 注意:如用质量百分数表示组成,平均分子量Mm=1/(aA/MA+aB/MB) 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, (2) 求回流比R V=D+L L=V-D=87-9.57=77.43Kg/h R=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 =Rxn /(R+1)+xD /(R+1)式中xD 应为摩尔分率请预览后下载！ xD =( xD /MA)/xD /MA+(1-xD )/MB =(0.95/78)/(0.95/78+0.05/92)=0

9、.961 yn+1=8.09xn/9.09+0.961/9.09=0.89xn +0.106操作线方程为:yn+1 =0.89xn +0.1069解: y=R/(R+1)x+xD /(R+1) (1) R/(R+1)=0.75 R=0.75R+0.75 R=0.75/0.25=3 (2) xD /(R+1)=0.2075 xD /(3+1)=0.2079 xD =0.83 (3) q/(q-1)=-0.5 q=-0.5q+0.5 q=0.5/1.5=0.333 (4) 0.75x+0.2075=-0.5x+1.5xF 0.75xq+0.2075=-0.5xq +1.50.44 1.25xq =

10、1.50.44-0.2075=0.4425 xq =0.362 (5)0q1 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=0.478+0.692=86.4 F=1000/86.4=11.6Kmol/h FxF =DxD +WxW F=D+W 11.60.4=D0.97+(11.6-D)0.02 D=4.64Kmol/h W=6.96Kmol/h R=L/D, L=3.74.64=17.17Kmol/h V=(R+1)D=4.74.64=21.8Kmol/h平均气化潜热r=308070.4+333200.6=

11、32313.6KJ/Kmol从手册中查得xF =0.4时泡点为95,则: q=r+cp(95-20)/r=(32313.6+159.275)/32313.6=1.37 L=L+qF=17.17+1.3711.6=33.1Kmol/h V=V-(1-q)F=21.8+0.3711.6=26.1Kmol/h (2) 求塔顶全凝器热负荷及每小时耗水量. Qc=Vr r=0.9730804+333200.03=30879.5KJ/Kmol Qc=21.830879.5=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

12、量. 塔的热量衡算 QB+QF +QR=Qv+QW +QL QB=Qv+QW +QL-QF -QR 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: QBQV=VIv Iv=(r+Cpt)=30879.5+159.28.2=43933.9KJ/Kmol请预览后下载！ QB=21.843933.9=957759.02KJ/h2.5KgF/cm2下蒸汽潜热r=522Kcal/Kg=5224.1818=39275.3KJ/Kmol 蒸汽需量为Gv Gv =QB/r=957759.02/39275.3=24.4Kmol/h =24.418=39.04Kg/h (4) 提馏段方程 y=

13、Lx/(L-W)-WxW /(L-W)=1.26x-0.00511解: 提馏段: ym+1=1.25xM-0.0187-(1) =LxM/V-WxW /V, L=L+qF=RD+F V=(R+1)D W=F-D, 精馏段: yn+1 =Rxn /(R+1)+xD /(R+1) =0.75xn +0.25xD -(2) q线:xF =0.50 -(3) 将(3)代入(1)得出: ym+1=1.250.5-0.0187=0.606,代入(2) 0.606=0.750.5+0.25xD , xD =0.92412解: (1) y1=xD =0.84, 0.84=0.45x1+0.55 x1=0.64

14、, yW =30.64/(3+1)+0.84/(3+1)=0.69, 0.69=0.45xW +0.55,xW =0.311, (2) 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,xD,xW精馏段操作线斜率为R/(R+1)=0.723 R=2.61提馏段方程 y=Lx/(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=xW xW =1.25 xW -0.0187 xW =

15、0.0748 (2)饱和蒸汽进料时,求取进料组成 将 y=0.723x+0.263 y=1.25x-0.0187 联立求解,得x=0.535,y=0.65因饱和蒸汽进料,q线为水平线,可得原料组成y=xF=0.65请预览后下载！14解: (1) y1=xD =0.9,x1=0.9/(4-30.9)=0.692, (2) y2=10.692/(1+1)+0.9/2=0.796 (3) xD =xF =0.5, yD =0.5/2+0.9/2=0.715解: (1) FxF=Vyq+Lxq 0.45=(1/3)yq+(2/3)xq y q =2.5xq /(1+1.5xq) xq=0.375 yq

16、=0.6 (2) Rmin=(xD-yq)/(yq-xq) =(0.95-0.6)/(0.6-0.375)=1.56 R=1.5Rmin=2.34 D=0.950.45/0.95=0.45 W=1-0.45=0.55 xW=(FxF-DxD)/W=(0.45-0.450.95)/0.55=0.041 L=RD=2.340.45=1.053; V=(R+1)D=1.503 L=L+qF=1.053+(2/3)1=1.72; V=V-(1-q)F=1.503-1/3=1.17 y=(L/V)x-WxW/V=1.72/1.17x-0.550.041/1.17 =1.47x-0.019316解: 精馏

17、段操作线方程 yn+1 =3/4xn +0.24平衡线方程 y=x/1+(-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: N0 x y 1 0.906 0.96 2 0.821 0.92 3 0.707 0.86 4 0.573 0.77 5 0.462 0.70 6 0.344 0.567 7 0.224 0.419 8 0.128 0.268 9 0.065 0.148 10 0.029 0.069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=(xF -xW )/(xD -xW )=(0.52-x

18、W )/(0.8-xW )=0.5请预览后下载！ 解得:xW =0.24精馏段操作线方程: yn+1 =(R/(R+1)xn +xD /(R+1)=0.75xn +0.2 -(1)平衡线方程:y=x/(1+(-1)x)=3x/(1+2x) 或:x=y/(-(-1)y)=y/(3-2y) -(2)交替运用式(1),(2)逐板计算: xD =y1=0.8 .x1=0.571; y2=0.628,x2=0.360; y3=0.470,x3=0.228xW =0.24 共需NT=3块(包括釜). 18解:q=0,xD =0.9,xF =0.5,xW =0.1,R=5,精馏段操作线方程: yn+1=Rx

19、n/(R+1)+xD/(R+1)=5xn/(5+1)+0.9/(5+1)=0.833xn+0.15图解: 得理论板数为11块(不包括釜),包括釜为12块18题附图 19解: (1) F=D+W FxF =DxD +WxW D=F(xF -xW )/(xD -xW )=100(0.3-0.015)/(0.95-0.015) =30.48 Kmol/h=30.5 Kmol/hW=F-D=69.50 Kmol/h (2) NT及NF =? xD =0.95、xW =0.015、q=1、R=1.5；xD /(R+1)=0.38作图得:NT =9-1=8(不含釜)进料位置: NF =6(3)L,V,yW

20、 及xW-1 19题附图q=1,V=V=(R+1)D V=30.5(1.5+1)=76.25Kmol/h L=L+qF=RD+F=1.530.5+100=145.8Kmol/h由图读得:yW =0.06, xW-1=0.0320解:请预览后下载！ (1) 原料为汽液混合物,成平衡的汽液相组成为x ,y平衡线方程 y=x/1+(-1)x=4.6x/(1+3.6x) - (1)q线方程 (q=2/(1+2)=2/3)则 y=q/(q-1)x-xF /(q-1)=-2x+1.35 - (2) 联解(1),(2)两式,经整理得: -2x+1.35=4.6x/(1+3.6x) 7.2x2 +1.740x

21、-1.35=0 解知,x=0.329 y=0.693 (2) Rmin=(xD -ye)/(ye-xe)=(0.95-0.693)/(0.693-0.329)=0.70621解: 因为饱和液体进料,q=1 ye=xe/1+(-1)xe=2.470.6/(1+1.470.6)=0.788 Rmin=(xD -ye)/(ye-xe)=(0.98-0.788)/(0.788-0.6)=1.02 R=1.5Rmin=1.53 Nmin=lg(xD /(1-xD )(1-xW )/xW )/lg =lg(0.98/0.02)(0. 95/0. 05)/lg2.47= 7.56 x=(R-Rmin)/(R

22、+1)=(1.53-1.02)/(1.53+1)=0.202 Y=(N-Nmin)/(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 (Nmin)精=lgxD /(1-xD )(1-xF )/xF /lg =lg0.98/0.020.4/0.6/lg2.47=3.85 N精/N=(Nmin)精/Nmin N精=N(Nmin)精/Nmin=14.53.85/7.56=7.4则精馏段实际板数为 7.4/0

23、.7=10.6 取11块 故实际加料板位置为第12块板上.22解: (1) 由 y=x/1+(-1)x=2.4x/(1+1.4x) 作y-x图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: yn+1 =Rxn /(R+1)+xD /(R+1)=2/3xn +0.3 - (1)侧线下操作线方程推导如下: 以虚线范围作物料衡算 V=L+D1+D2Vys+1=Lxs+D1xD1+D2xD2 ;ys+1=Lxs/V +(D1xD1+D2xD2)/V=Lxs/(L+D1+D2)+(D1xD 1+D2xD2)/(L+D1+D2); L=L0-D2, 则:请预览后下载！y

24、s+1=(L0-D2)xs/(L0-D2+D1+D2)+(D1xD 1+D2xD 2)/(L0-D2+D1+D2)=(R-D2/D1)xs/(R+1)+(xD1+D2xD2/D1)/(R+1)(R=L0/D1)将已知条件代入上式,得到: yS+1=0.5x+0.416 (2) 用图解法,求得理论塔板数为（5-1）块,见附图. 22题附图23解: 根据所给平衡数据作x-y图.精馏段操作线 yn+1 =Rxn /(R+1)+xD /(R+1)=1.5xn /(1.5+1)+0.95/(1.5+1) =0.6xn +0.38q线方程与q线: 料液平均分子量:Mm=0.35+0.6518=22.9甲醇

25、分子汽化潜热:r=252324.2=33868.8KJ/Kmol 水的分子汽化潜热:r=552184.2=41731.2KL/Kmol 23题附图料液的平均分子汽化潜热: r=0.3533868.8+0.6541731.2=38979.4KL/Kmol 料液的平均分子比热 Cp=0.8822.94.2=84.6KL/Kmolq=r+Cp(ts-tF )/r=38979.4+84.6(78-20)/38979.4=1.13 q线斜率 q/(q-1)=1/13/0.13=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)一点,于是在x-y图上,作出三条线,

26、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) F1xF1+F2xF2=DxD +WxW 1000.6+2000.2=D0.8+W0.02 100=0.8D+0.02W -(2)请预览后下载！ 由式(1) W=F1+F2-D=100+200-D=300-D 代入式(2)得:D=120.5Kmol/h L=RD=2120.5=241kmol/h V=L+D=241+120.5=361.5Kmol/h 在两进料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+F1=D+L Vys+1+F1xF1=Lxs+DxD

27、ys+1=(L/V)xs+(DxD -F1xF1)/V L=L+q1F1=241+1100=341Kmol/h V=V=361.5 ys+1=(341/361.5)xs+(120.50.8-1000.6)/361.5 ys+1=0.943xs+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xD 0.98, 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: 130.5=6.5取苯-甲苯溶液相对挥发度为=2.54 用捷算法求精馏段最小理论板数 (Nmin)精=ln0.98/0.02-0.5

28、/0.5/ln2.54=4.175 y=N精馏段-(Nmin)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)(1/0.567)=0.392=(R-Rmin)/(R+1) R=(0.392+Rmin)/(1-0.392) Rmin=(xD -ye)/(ye-xe) 对泡点进料xe=xF =0.5 ye=x/1+(-1)x =2.540.5/(1+1.540.5)=1.27/1.77=0.72 Rmin=(0.98-0.72)/(0.72-0.5)=0.26/0.22=1.18 R=(0.392+1.18)/(

29、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/h26解:求n板效率: Emv =(yn -yn+1 )/(yn*-yn+1 ),因全回流操作,故有yn+1 =xn ,yn =xn-1与xn 成平衡的yn *=xn /1+(-1)xn =2.430.285/(1+1.430.285)=0.492于是: Emv=(xn-1 -xn )/(yn *-xn )=(0.43-0.285)/(0.492-0.285)=0.7请预览后下载！求n+1板板效率: Emv=(yn+1 -yn+2)/(yn

30、+1* -yn+2)=(xn-xn+)/(yn+1*-xn+1 ) yn+1 =2.430.173/(1+1.430.173)=0.337Emv=(0.285-0.173)/(0.337-0.173)=0.68327解:由图可知:该板的板效率为 Emv=(y1-y )/(y1*-yW)从图中看出,y1=xD =0.28,关键要求y1* 与yW .由已知条件 DxD /FxF =0.8 D/F=0.80.2/0.28=0.57作系统的物料衡算: FxF =DxD +WxW F=D+W联立求解: xF =DxD /F+(1-D/F)xW 0.2=0.570.28+(1-0.57)xW 解得xW =

31、0.093 习题27附图因塔釜溶液处于平衡状态,故 yW =xW /1+(-1)xW =2.50.093/(1+1.50.093)=0.204yW 与x1是操作线关系. yn+1 =Lxn /V-WxW /V =Fxn /D-WxW/D =Fxn /D-(F-D)xW /D=Fxn /D-(F/D-1)xW yn+1 =xn /0.57-(1/0.57-1)0.093=1.75xn -0.07当 yn+1 =yW 时,xn =x1 x1=(yW +0.07)/1.75=(0.204+0.07)/1.75=0.157与x1成平衡气相组成为y1* y1*=x1/1+(-1)x1=2.50.157/

32、(1+1.50.157)=0.318 Emv=(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,xF =0.5,用试差法得精馏段操作线ac,与x=xF =0.5线交于d.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：xW =0.17 xD/(R+1)=0.103R=0.85/0.103-1=7.25 F=D+W FxF =DxD +WxW 100=D+W 1000.5=D0.85+W0.17 得 D=48.5Kmol/h V=V=(R+1)D=8.2548.5=400Kmol/h 28题附图 (2)此时

33、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=,FxF /xD =0.91000.4/0.92=39.13Kmol/h,W=60.9Kmol/h xW =0.1FxF /W=0.11000.4/60.9=0.0656请预览后下载！ q=1 xq =0.4 查图得yq =0.61 Rmin=(xD -yq )/(yq -xq )=(0.92-0.61)/(0.61-0.4)=1.48 R=1.51.48=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/

34、0.7=20块(不包括釜) Np精=5/0.7=7.14,取8块,第九块为实际加料板 (2) 可用措施:(1)加大回流比,xD ,xW ,= (2)改为冷液进料,NT NT q=1, NT =const xD q约为const,下移加料点,xD . 29题附图30解:(1) DxD /FxF =0.922; DxD =0.9221500.4=55.32 DxD =FxF -WxW =FxF -(F-D)xW =55.32 1500.4-(150-D)0.05=55.32 D=56.4Kmol/h W=F-D=93.6Kmol/h xD =55.32/56.4=0.981 (2) NT 及NF

35、(进料位置) xD =0.981,xW =0.05,q=1, xD /(R+1)=0.981/(2.43+1)=0.286 a(0.981,0.981), b(0.05,0.05) q线: xF=0.4、q=1, q线为垂线。 作图得:NT =12-1=11,不含釜,NF =7(3) 液气比精馏段:L/V=R/(R+1) =2.43/(2.43+1)=0.708请预览后下载！ 提馏段: L/V=(L+qF)/(L+qF-W)或V=V ,L=RD L/V=(RD+F)/(R+1)D)=(2.4356.4+150)/(3.43 56.4)=1.484 (4)由于再沸器结垢,则QB,V,RxD 若要

36、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,xD ,xW 各为多少?由题知,当塔板为无穷时： R=Rmin =0.8, 30题附图对泡点进料, Rmin =(xD -ye)/(ye-xe) xe=xF =0.5, ye=xe/1+(-1)xe=xF/1+(-1)xF=20.5/(1+0.5)=0.667 于是: (xD -0.667)/(0.667-0.5)=0.8 解得: xD =0.8 FxF =DxD +WxW xF =DxD /F+(1-D/F)xW 由题知D/F=0.6代入上式,解得xW =0.05, (2)R=1.5时,求xD ,xW

37、 . 由题知,当塔板为无穷多时,R=Rmin =1.5 Rmin =(xD-ye)/(ye-xe) 同理求得xD =0.917,代入物料衡算式 xF =DxD /F+(1-D/F)xW 0.5=0.60.917=(1-0.6)xW xW =-0.125,不成立. 31题附图 故操作线与平衡线应取xW =0处相交,即: xW =0； FxF =DxD +WxW xD =FxF /D=0.51/0.6=0.83此时精馏段与提馏段操作线示意图如上: 32解: (1) xF =yq =0.5,; xq =yq/(-(-1)yq)=0.5/(3-20.5)=0.25 Rmin =(xD-yq)/(yq-

38、xq)=(0.9-0.5)/(0.5-0.25)=1.6请预览后下载！ R=21.6=3.2 F=D+W FxF =DxD +WxW 0.5=0.9D+0.05W D=0.529 W=0.471 L=RD=3.20.529=1.693 V=2.222 L=L=1.693 V=V-F=1.222 y=1.385x-0.0193 (2)精馏段操作线 y=(L/V)x+DxD /V=(1.693/2.222)x+0.5290.9/2.222 y=0.762x+0.214 或y=Rx/(R+1)+xD /(R+1)=3.2x/4.2+0.9/4.2=0.762x+0.214 y1=xD =0.9 x1

39、=y1/(3-2y1)=0.9/(3-20.9)=0.75 y2=0.7620.75+0.214=0.7855 (3)应维持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0.238 xW=(FxF-DxD)/W=(0.5-0.2380.9)/(1-0.238)=0.375 33解: q=(r+(80-20)Cp)/r=(40000+60100)/40000=1.15 W=L+qF=1.15100=115 D=F+S-W=100+50-115=35 FxF =DxD +WxW y=(L/s)x-(W/S)xW =2.3x-2.3xW y2与xW 成

40、平衡 y2=3xW x1=y2/2.3+xW =2.304xW y1=3x1=6.913xW =xD 1000 2=356.913xW +115xW xW =0.056 xD =0.387 =350.387/(10000.2) =0.678 34解: 作精馏段物料衡算，得精馏段操作线方程：yn+1=(R/(R+1)xn+xD/(R+1) 将 x0=0.5、y1=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请预览后下载！再就q=1,xF=0.4作

41、进料线。从y1、xo开始作梯级，共得6块理论板。 35解：对第n块板:EmL=(xn-1-xn)/(xn-1-xn*)=0.5;xn=0.4-0.5(0.4-xn*) yn =xn*/1+(-1)xn*=2xn*/(1+xn*)对第n板作物料衡算：1000.3+1000.4=100(2xn*/（1+xn*）)+1000.4-0.5(0.4- xn*)解得：xn*=0.263 xn=0.4-0.5(0.4- 0.263)=0.318 yn=20.263/(1+0.236)=0.38236解：作全塔总物料衡算： F=D+W (1)作全塔易挥发组分物料衡算： FxF=DxD+WxW (2)作分凝器易

42、挥发组分物料衡算： Vy1=DxD+LxL (3)因为：V=2D L=D，（3）式：2y1=xD+xL （3）相平衡方程：xD=xL /1+（-1）xL 即：0.8=2.46xL /1+（2.46-1）xL解得：xL=0.619; 代人（3）式：2y1=0.8+0.619,得y1=0.71y1=yW=0.71,代人平衡方程：0.71=2.46xW /1+（2.46-1）xW解得：xW=0.5代人（2）得：D=F(xF-xW)/(xD-xW)=66.7 Kmol/h, W=33.3Kmol/h汽化量：V=266.7=133.4 Kmol/h37解：(1) 精馏段操作线方程：yn+1=(R/(R+

43、1)xn+xD/(R+1) =（4/（4+1）x+0.7/(4+1)=0.8x+0.14 提馏段操作线方程：y=(L/V)x-(W/V)xWD/F=(xF-xW)/(xD-xW)=(0.3-xW)/(0.7-xW)=0.4 xW=0.0333因为 q=1,所以：L/V=(L+F)/(R+1)D=R(D/F)+1/(R+1)D/F =(40.4+1)/(4+1)0.4=1.3(W/V)xW=(F-D)/(R+1)DxW=(1-D/F)/(R+1)D/FxW =(1-0.4)/(50.4)0.0333=0.01所以：y=1.3x-0.01(2) yq=xF/1+(-1) xF =20.3/(1+(

44、2-1)0.3)=0.4615 若平衡点在进料口处： Rm/(Rm*1)=4/(4+1)=(xD-yq)/(xD-xF) =(xD-0.4615)/(xD-0.3) xD=1.11不可能在进料口平衡。请预览后下载！在塔顶平衡：即 xD=1 D/F=（xF-xW）/(xD-xW)=0.4; (0.3-xW)/(1-xW)=0.4 解得 xW=-0.167 故不可能。在塔底平衡：即xW=0 xDmax=FxF/D=0.3/0.4=0.7538解：（1）饱和水蒸气用量S=V=V=（R+1）D=2.5D，（q=1）y1=xD=0.95Emv=(y1-y2)/(y1*-y2)=(0.95-y2)/(0.

45、5x1+0.5-y2)=0.5整理得： 0.5y2=0.7-0.25x1 （1） Vy2=Lx1+DxD 2.5Dy2=1.5Dx1+DxD整理得： 2.5y2=1.5x1+0.95 （2）联解（1）、（2）式得：x1=0.927 (2) F+S=D+W; S=V=2.5D; F+2.5D=D+W即 F+1.5D=W (3) FxF=DxD+WxW (4) 式（3）代人（4）消去W得： D/F=（xF-xW）/(xD+1.5xW) =(0.5-0.1)/(0.95+1.50.1)=0.36439解：(1) =DxD /(FxF)= xD(xF-xW)/( xF(xD-xW )=xD (0.4-

46、0.05)/(0.4(xD-0.05)=0.955 xD=0.6 D/F=xF/xD=0.9550.4/0.6=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)xn-(W/D)xW=(100/64)xn-(36/64)0.05=1.56xn-0.028(3) 当NT时，可获得xDmax q=1. q线是垂线交平衡线上ye点ye=xF/(1+(-1)xF)=(30.4)/(1+20.4)=0.667,此值是否最大值必须校验，由于F，V不变，D，W不变xW=(xF-(D/F)xD)

47、/ (W/F)=(0.4-0.640.667)/0.36=-0.07470 当xW=0,夹点在塔底xDmax=（F/D）xF=0.4/0.64=0.62540解：(1) F1xF1+F2xF2=DxD+W xW10.6+0.50.4=0.99D+0.02WF1+F2=D+W 1+0.5=D+W 请预览后下载！D=0.794Kmol/s W=0.706Kmol/sL=RD=0.794Kmol/s V=L+D=1.588Kmol/sL=L+q1F1=1.794Kmol/s V=V=1.588Kmol/sy=(L/V)x+(DxD-F1xF1)/V=(1.794/1.588)x+(0.7940.99-10.6)/1.588 y=1.13x+0.117(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）：xq1=0.6 yq1=30.6/(1+20.6)