发布时间 : 星期一 文章2011复习化工原理下计算题及答案更新完毕开始阅读
?Y1 =Y1-Y1*=8.7×10-2-6.10×10-2=2.60×10-2 ?Y2 =Y2-Y2*=4.4×10-3
?Y1??Y22.60?10?2?4.4?10?3-2
?Ym ===1.22×10 ?2?Y12.60?10lnln?Y24.4?10?3 N0G=
Y1?Y2=6.67 ?Ym H0G=
FB推出Ω=4.51 m2 KYaS推出D得2.4 m
2. D4944 (1)
y=6x-1.5 xq==0.333
3解得 y= Rmin=
12x yq=0.5 1+xxd?yqyq?xq?0.94?050.=2.64
050.?0.333 R=1.3Rmin=1.3×2.64=3.432 yn +1=
x3.4320.94Rxn?xn +d=
4.432R?1R?13.432?1 yn +1=0.774xn +0.212
(2)
y=6x-1.5 x=0.30
解得 y=x(对角线) y=0.30 故xf=0.30 Ff=Fd+Fw
Ff xf=Fdxd+Fwxw=(Ff-Fw)xd+Fwxw
Ff=
Fw(xd?xw)150(0.94?0.04)?=211 kmol·h-1
xd?xf0.94?0.30 Fd=Ff-Fw=61 kmol·h-1。
3.D4511
(1) 求水的用量qv（水）,m3·h-1: M(CH3OH)=32
32=0.07527 1000100?22.432 Y2= Y1 (1-?) =0.07527×(1-0.98)=1.5054×10-3
Y0.07527 X1*=1?==0.06545
m115. Y1=
X1=0.67X1*=0.67×0.06545=0.04385 FB=
1001000100?=41.52 kmol·h-1 22.432FCY1?Y20.07527?15054.?103 = ==1.682
0.04385?0FBX1?X2 FC=1.682FB=1.682×41.52≈69.84 kmol·h-1
水的密度?=1000 kg·m-3 则qv （水）= (2)求塔径D,m: 塔截面S= D=
69.84?18=1.3 m3·h-1;
1000qv1000?=0.556 m2 u0.5?3600S??0.5564?=0.841 m
4 (3)求填料层高H,m: H0G=
FB41.52==0.786 m KYaS0.5?190?0.556Y?mX2mFBmFB1ln[(1)(1-)+] mFBFCFCY2?mX21?FC N0G =
=
1115.0.07527?0115.ln[(1?)?]=8.86 ?3115.1682..15054.?10?016821?1682. H=H0G·N0G=0.786×8.86≈7.0 m。
4.D4501
(1) 进入吸收塔的惰性气体摩尔流量为: FB=
273108.24000×××(1-0.022)=168.07 kmol·h-1
.22.4273?301013Y1=
y10.022==0.0225 1?0.0221?y1 Y2=
y20.001==0.001 1?0.0011?y2Y1?Y20.0225?0.001==95.56%
0.0225Y1x20.006==0.00604 1?0.0061?x2FB(Y1-Y2) FC168.07(0.0225-0.001) 312. ?=
(2) X2=
则X1=X2+
=0.00604+ =0.1219 x1=
5.D4970
X101219.==0.1087。
.1?X11?01219 (1) 因为是气液混合物进料，所以q等于原料中液相所占的摩尔分率。即 q=1-
12? 33 (2) yq =
2.5xq1?(2.5?1)xq
20.44 yq =3xq?
22?1?133 联立解得：xq =0.37
yq =0.59 (3) Rmin=
xd?yqyq?xq?0.957?059.=1.67
059.?0.376.D4528
已知:溶解度常数H=1.53 kmol·m-3·kPa-1 (1) KG=
111 + kgHkl?111?345.?10?6153.?2.02?10?4
=3.41×10-6 kmol·m-2·s-1·kPa-1
KY =pKG=101.3×3.41×10-6=3.46 ×10-4 kmol·m-2·s-1
(2) KL=
1 H1+ kgkl?1153.1?345.?10?62.02?10?41000=55.56 kmol·m-3 18=2.23×10-6 m·s-1
由于溶液浓度甚低,故cl≈
KX =KL·cl=2.23×10-6 ×55.56=1.24×10-6 kmol·m-2·s-1。
7.D4931
Rmin=
xd-yqyq-xq=
0.950-yqyq-xq=1.137
?xq2.50xq yq==
1+(?-1)xq1+1.50xq xq =0.4873 yq =0.7038 yq =
xqxq -f q?1q?10.400q×0.4873- q?1q?1 0.7038=
q=1.403
故为冷液进料。
8.D1654
SI制密度?=960 kg·m-3
在SI制中，每立方米的重量=960×9.81=9418 N
9.D4552
pA*2.499?103E===8.33×104 Pa ?23.00?10xAH=
1
MAxA?MA(1?xA)E??L1000?1031? = 417?0.03?18?0.978.33?10 =6.88×10-1 mol·m-3·Pa-1
E8.33?104m===0.276。 5p3.022?1010. D1153
设蒸发水的质量为m1吨，剩下煤的质量为m2吨，
?m?m{1000 1000(1?0.20)?m(1?0.08)122m2=870吨，m1=130吨