化工原理第五章-精馏-答案

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-p B0)/(p A0-p B0); y A=p A0×x A/p

以t=90℃为例,x A=(760-208.4)/(1008-208.4)=0.6898

y A=1008×0.6898/760=0.9150

t℃80.02 90 100 110 120 130 131.8

x 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

1 0.911

2 0.7894 0.6271 0.4052 0.0840 0

4.612x/(1+

3.612x)

(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=p A0/p B0

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:p A+p B=p A0x A+x B0x B=p求泡点要用试差法,先设泡点为87℃

lgp A0=6.89740-1206.350/(87+220.237)=2.971

p A0=102.971=935.41[mmHg]

lgp B0=6.95334-1343.943/(87+219.337)=2.566

p B0=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=p A/p=p A0x A/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: x A+x B=1或p A/p A0+p B/p B0=1 式中p A=0.4×760=304[mmHg]; p B=0.6×760=456[mmHg]

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)=3.165

∴p A0=1462.2[mmHg]

lgp B0=6.95334-1343.943/(103+219.337)=2.784

∴p B0=608.14[mmHg]

于是:

304/1462.2+456/608.14=0.96<1

再设露点为102℃,同时求得p A0=1380.4; p B0=588.84

304/1380.4+456/588.84=0.995≈1

故露点为102℃,平衡液相组成为

x A=p A/p A0=304/1380.4=0.22

3解:

(1)x A=(p总-p B0)/(p A0-p B0)

0.4=(p总-40)/(106.7-40)

∴p总=66.7KPa

y A=x A·p A0/p=0.4×106.7/66.7=0.64

(2)α=p A0/p B0=106.7/40=2.67

4解:

(1) y D=?

αD =(y/x)A/(y/x)B

=(y D /0.95)/((1-y D )/0.05)=2

y D =0.974

(2) L/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 )=1.4

5解:

简单蒸馏计算:

lnW 1/W 2=

⎰

-1

2

x x x

y dx

W 2=(1-1/3)W 1=2/3W 1;y=0.46x+0.549,x 1=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 x D =W 1x 1 -W 2x 2 (1/3W 1)x D =W 1×0.6-(2/3W 1)×0.498 ∴x D =0.804 6解:

Fx F =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)/(Fx F )=0.5×0.528/0.4=66% 7.解:

F=D+W

Fx F =Dx D +Wx W

已知x F =0.24,x D =0.95,x W =0.03,解得:

D/F=(x F -x W )/(x D -x W )=(0.24-0.03)/(0.95-0.03)=0.228 回收率 Dx D /Fx F =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

Fx F =Dx D +Wx W

x F =0.1,x D =0.95,x W =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M m =1/(a A /M A +a B /M B )]

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 y n+1 =Rx n /(R+1)+x D /(R+1) 式中x D 应为摩尔分率