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=pA0×xA/p
以t=90℃为例,xA=(760-208.4)/(1008-208.4)=0.6898 yA=1008×0.6898/760=0.9150 计算结果汇总: t℃ x y 4.612x/(1+3.612x) 80.02 1 1 1 90 0.6898 0.9150 0.9112 100 110 120 0.3777 130 0.0195 0.0724 131.8 0 0 0 0.4483 0.2672 0.1287 0.7875 0.6118 0.7894 0.6271 0.4052 0.0840 (2)用相对挥发度计算x-y值: y=αx/[1+(α-1)x]
式中α=αM=1/2(α1+α2) ∵α=pA0/pB0
α1=760/144.8=5.249 ;α2=3020/760=3.974 ∴αM=1/2(α1+α2)=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
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pA0=102.971=935.41[mmHg]
lgpB0=6.95334-1343.943/(87+219.337)=2.566 pB0=102.566=368.13[mmHg]
935.41×0.4+368.13×0.6=595≈600mmHg ∴泡点为87℃,气相平衡组成为
y=pA/p=pA0xA/P=935.41×0.4/600=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=0.4×760=304[mmHg]; pB=0.6×760=456[mmHg] 求露点亦要用试差法,先设露点为103℃,则:lgpA0=6.8974-120.635/ (103+220.237)=3.165 ∴pA0=1462.2[mmHg]
lgpB0=6.95334-1343.943/(103+219.337)=2.784 ∴pB0=608.14[mmHg] 于是 :
304/1462.2+456/608.14=0.96<1
再设露点为102℃,同时求得pA0=1380.4; pB0=588.84 304/1380.4+456/588.84=0.995≈1 故露点为102℃,平衡液相组成为 xA=pA/pA0=304/1380.4=0.22 3解:
(1)xA=(p总-pB0)/(pA0-pB0)
0.4=(p总-40)/(106.7-40) ∴p总=66.7KPa yA=xA·pA0/p=0.4×106.7/66.7=0.64
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(2)α=pA/pB=106.7/40=2.67 4解:
(1) yD=?
αD =(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 )=1.4 5解:
简单蒸馏计算:
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lnW1/W2=
?x1x2dx y?x W2=(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:x2=0.498,
馏出液组成:WD xD =W1x1 -W2x2 (1/3W1)xD =W1×0.6-(2/3W1)×0.498 ∴xD =0.804 6解:
FxF=Vy+Lx ∴0.4=0.5y+0.5x --------(1) y=αx/(1+(α-1)x)=3x/(1+2x) --------(2) (1),(2)联立求解,得y=0.528,x=0.272 回收率=(V·y)/(FxF )=0.5×0.528/0.4=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 )=(0.24-0.03)/(0.95-0.03)=0.228 回收率 DxD /FxF =0.228×0.95/0.24=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)=608.6[kmol/h] 8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=100[Kg/h],代入上两式解得: D=9.57[Kg/h]; W=90.43[Kg/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=87[Kg/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=77.43[Kg/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
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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+0.961/9.09=0.89xn +0.106 操作线方程为:yn+1 =0.89xn +0.106 9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 '+1.5×0.44 1.25xq '=1.5×0.44-0.2075=0.4425 xq '=0.362 (5)0 (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=0.4×78+0.6×92=86.4 F=1000/86.4=11.6[Kmol/h] FxF =DxD +WxW F=D+W 11.6×0.4=D×0.97+(11.6-D)0.02 ∴D=4.64[Kmol/h] W=6.96[Kmol/h] R=L/D, ∴L=3.7×4.64=17.17[Kmol/h] V=(R+1)D=4.7×4.64=21.8[Kmol/h] 平均气化潜热r=30807×0.4+33320×0.6=32313.6[KJ/Kmol] 从手册中查得xF =0.4时泡点为95℃,则: q=[r+cp(95-20)]/r=(32313.6+159.2×75)/32313.6=1.37 ∴L'=L+qF=17.17+1.37×11.6=33.1[Kmol/h] V'=V-(1-q)F=21.8+0.37×11.6=26.1[Kmol/h] (2) 求塔顶全凝器热负荷及每小时耗水量. Qc=Vr ∴r=0.97×30804+33320×0.03=30879.5[KJ/Kmol] ∴Qc=21.8×30879.5=673172.7[KJ/h] 耗水量 Gc=673172.7/4.18(50-20)=5368.2[Kg/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: QB≈QV=V·Iv Iv=(r+Cpt)=30879.5+159.2×8.2=43933.9[KJ/Kmol] 119 ∴QB=21.8×43933.9=957759.02[KJ/h] 2.5[KgF/cm2]下蒸汽潜热r=522Kcal/Kg=522×4.18×18=39275.3[KJ/Kmol] ∴蒸汽需量为Gv Gv =QB/r=957759.02/39275.3=24.4Kmol/h =24.4×18=39.04[Kg/h] (4) 提馏段方程 y=L'x/(L'-W)-WxW /(L'-W)=1.26x-0.005 11解: 提馏段: ym+1’=1.25xM’-0.0187---------(1) =L'xM'/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=1.25×0.5-0.0187=0.606,代入(2) 0.606=0.75×0.5+0.25xD , xD =0.924 12解: (1) y1=xD =0.84, 0.84=0.45x1+0.55 x1=0.64, yW =3×0.64/(3+1)+0.84/(3+1)=0.69, 0.69=0.45×xW +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=L'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=xW xW =1.25 xW -0.0187 ∴xW =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=0.65 120
化工原理第五章 精馏 答案
