《化工原理第五章-精馏-答案》由会员分享，可在线阅读，更多相关《化工原理第五章-精馏-答案（21页珍藏版）》请在金锄头文库上搜索。
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 pA0=102.971=
3、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/ (103+220.2
4、37)=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=0.4106.
5、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: lnW1/W2= W2=(1-1/3)W1=2/3W1;y=0.46x+0.549,x1=0.6,代入
6、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 -xW )=(0.24-0.
7、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+0.05/92) 注意:如用质量百分数表
8、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 xD =( xD /MA)/xD /MA+(1-xD )/MB =(0.95/78)/(0.95/78+0.05/92)=0.961 yn+1=8.09xn/9.09
9、+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 =1.50.44-0.2075=0.4425
10、 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=32313.6KJ/Kmol从手册中查得x
11、F =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 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
