化工原理第五章精馏答案

116五 蒸馏习题解答1 解:(1)作 x-y 图及 t-x(y)图,作图依据如下:x A=(p-pB0)/(pA0-pB0); yA=pA0xA/p以 t=90为例,x A=(760-208.4)/(1008-208.4)=0.6898yA=10080.6898/760=0.9150计算结果汇总: t 80.02 90 100 110 120 130 131.8x 1 0.6898 0.4483 0.2672 0.1287 0.0195 0 y 1 0.9150 0.7875 0.6118 0.3777 0.0724 0 4.612x/(1+3.612x)1 0.9112 0.7894 0.6271 0.4052 0.0840 0(2)用相对挥发度计算 x-y 值:y=x/1+(-1)x式中 =M=1/2(1+2)=p A0/pB01=760/144.8=5.249 ;2=3020/760=3.974 M=1/2(1+2)=1/2(5.249+3.974)=4.612y=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:p A+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)=2.971117pA0=102.971=935.41mmHglgpB0=6.95334-1343.943/(87+219.337)=2.566pB0=102.566=368.13mmHg935.410.4+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: x A+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:lgp A0=6.8974-120.635/(103+220.237)=3.165p A0=1462.2mmHglgpB0=6.95334-1343.943/(103+219.337)=2.784p B0=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=588.84304/1380.4+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 =66.7KPayA=xApA0/p=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)=2yD =0.974(2) L/VD =?V=V D +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:118lnW1/W2=xydW2=(1-1/3)W1=2/3W1;y=0.46x+0.549,x1=0.6,代入上式积分解得:釜液组成:x 2=0.498,馏出液组成:W D xD =W1x1 -W2x2(1/3W1)xD =W10.6-(2/3W1)0.498x D =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)/(Fx F )=0.50.528/0.4=66% 7.解:F=D+WFxF =DxD +WxW已知 xF =0.24,xD =0.95,xW =0.03,解得:D/F=(xF -xW )/(xD -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=3.38W=3.38D=3.38(V-L)=3.38(850-670)=608.6kmol/h8 解:(1) 求 D 及 W,全凝量 VF=D+WFxF =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+0.05/92)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 RV=D+L L=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 =Rxn /(R+1)+xD /(R+1)式中 xD 应为摩尔分率119xD =( xD /MA)/xD /MA+(1-xD )/MB=(0.95/78)/(0.95/78+0.05/92)=0.961y n+1=8.09xn/9.09+0.961/9.09=0.89xn +0.106操作线方程为:y n+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 +1.50.441.25xq =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=86.4F=1000/86.4=11.6Kmol/hFxF =DxD +WxW F=D+W11.60.4=D0.97+(11.6-D)0.02D=4.64Kmol/hW=6.96Kmol/hR=L/D, L=3.74.64=17.17Kmol/hV=(R+1)D=4.74.64=21.8Kmol/h平均气化潜热 r=308070.4+333200.6=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=1.37L=L+qF=17.17+1.3711.6=33.1Kmol/hV=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=Vrr=0.9730804+333200.03=30879.5KJ/KmolQc=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量.塔的热量衡算QB+QF +QR=Qv+QW +QLQB=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=VIvIv=(r+Cpt)=30879.5+159.28.2=43933.9KJ/Kmol120Q B=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 GvGv =QB/r=957759.02/39275.3=24.4Kmol/h=24.418=39.04Kg/h(4) 提馏段方程 y=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+FV=(R+1)DW=F-D,精馏段: yn+1 =Rxn /(R+1)+xD /(R+1)=0.75xn +0.25xD -(2)q 线:x F =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+0.55x1=0.64,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 x D =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 x W =0.0748(2)饱和蒸汽进料时,求取进料组成将 y=0.723x+0.263y=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=0.6512114 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+Lxq0.45=(1/3)yq+(2/3)xq y q =2.5xq /(1+1.5xq)x q=0.375 yq=0.6(2) Rmin=(xD-yq)/(yq-xq)=(0.95-0.6)/(0.6-0.375)=1.56R=1.5Rmin=2.34D=0.950.45/0.95=0.45 W=1-0.45=0.55xW=(FxF-DxD)/W=(0.45-0.450.95)/0.55=0.041L=RD=2.340.45=1.053; V=(R+1)D=1.503L=L+qF=1.053+(2/3)1=1.72; V=V-(1-q)F=1.503-1/3=1.17y=(L/V)x-WxW/V=1.72/1.17x-0.550.041/1.17=1.47x-0.019316 解:精馏段操作线方程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 y1 0.906 0.962 0.821 0.923 0.707 0.864 0.573 0.775 0.462 0.706 0.344 0.5677 0.224 0.4198 0.128 0.2689 0.065 0.14810 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-xW )/(0.8-xW )=0.5122解得:x W =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=Rxn/(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+WFxF =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、x W =0.015、q=1、R=1.5；x D /(R+1)=0.38作图得:N T =9-1=8(不含釜)进料位置: NF =6(3)L,V,yW 及 xW-1 19 题附图q=1,V=V=(R+1)D V=30.5(1.5+1)=76.25Kmol/hL=L+qF=RD+F=1.530.5+100=145.8Kmol/h由图读得:y W =0.06, xW-1=0.0312320 解:(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-1.35=0解知,x=0.329y=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=1ye=xe/1+(-1)xe=2.470.6/(1+1.470.6)=0.788Rmin=(xD -ye)/(ye-xe)=(0.98-0.788)/(0.788-0.6)=1.02R=1.5Rmin=1.53Nmin=lg(xD /(1-xD )(1-xW )/xW )/lg=lg(0.98/0.02)(0. 95/0. 05)/lg2.47= 7.56x=(R-Rmin)/(R+1)=(1.53-1.02)/(1.53+1)=0.202Y=(N-Nmin)/(N+1) Y=0.75(1-x0.567)(N-7.56)/(N+1)=0.75(1-0.202 0.567) 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=3.85N 精 /N=(Nmin)精 /NminN 精 =N(Nmin)精 /Nmin=14.53.85/7.56=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:(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:124ys+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=1.13q 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(x W ,0)一点,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)F1xF1+F2xF2=DxD +WxW 1000.6+2000.2=D0.8+W0.02100=0.8D+0.02W -(2)125由式(1) W=F1+F2-D=100+200-D=300-D代入式(2)得:D=120.5Kmol/hL=RD=2120.5=241kmol/hV=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+LVys+1“+F1xF1=Lxs+DxD ys+1=(L/V)xs+(DxD