吸收习题(答案) 下载本文

?Y2?Y2?mX2?Y2?2.64?10?3

?Y1?Y1?mX1?0.0526?2?0.0202?0.0122

0.0122?2.64?10ln0.0122?3

?Ym??Y1??Y2??Y1?ln??Y2???Y1?Y2?Ym??2.64?10?3?3??6.25?10?3

?NOG??0.0526?2.64?106.25?10?3?7.99

HOG?zNOG?67.99?0.751m

而V?

H2000?101.3??1?0.05?8.314?298?0.785?0.8?VKYa2?154.74kmol/(m2?h)

OG

154.740.7513

KYa?VHOG??206.05kmol/(m?h)

5-36.体积流量为200m3/h(18℃、101.3kPa)的空气-氨混合物,用清水逆流吸收其中的氨,欲使氨含量由5%下降到0.04%(均为体积百分数)。出塔氨水组成为其最大组成的80%。今有一填料塔,塔径为0.3m,填料层高5m,操作条件下的相平衡关系为Y*=1.44X,问该塔是否合用?KGa可用下式计算:

KGa=0.0027m0.35W0.36 kmol/(m3.h.kPa) m---气体质量流速,kg/(m2.h); W---液体质量流速,kg/(m2.h)。 解:

y1?5%,Y1?y11?y1?0.051?0.05?0.0526

y2?0.04%,Y2?0.04%?4?10??4 0.05261.44

X2?0,X1?0.8X1?0.8Y1m?4?0.8??0.0292

?Y2?Y2?mX2?Y2?4?10

?Y1?Y1?mX1?0.0526?1.44?0.0292?0.0106

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?Ym??Y1??Y2??Y1?ln??Y2???Y1?Y2?Ym?0.0106?4?10ln0.0106?4?4?10?4?4??3.11?10?3

?NOG??0.0526?4?103.11?10?3?16.78

混合气体摩尔流率:

G?200?101.38.314??273?18??0.785?0.32?118.62kmol/(m2?h)

混合气体平均分子量:

M?0.05?17?0.95?29?28.4kg/kmol

混合气体质量流速:

m?118.62?28.4?3368.8kg/(m2?h)

惰性组分摩尔流率:

V?G?1?y1??118.62??1?0.05??112.69kmol/(m2?h)

又 L?X1?X2??V?Y1?Y2?

?L?VY1?Y2X1?X2?112.690.0526?4?100.0292?4?201.45kmol/(m2?h)

液体质量流速:

w?201.45?18?3626.1kg/(m?h)0.350.362

0.35KGa?0.0027mw?0.0027?3368.833626.10.36?0.886kmol/(m3?h?kPa)

?KYa?KGa?P?89.75kmol/(mHOG?VKYa??112.6989.75?h)

?1.26m

?z?HOGNOG?1.26?16.78?21.14mz需?z实?5m

所以,该塔不合适。

5-37.混合气中含0.1(摩尔分率,下同)CO2,其余为空气,于20℃及2026kPa下在填料塔中用清水逆流吸收,使CO2的浓度降到0.5%。已知混合气的处理量为2240m/h(标准状态下),溶液出口浓度为0.0006,亨利系数E为200MPa,液相总体积传质系数KLa为50 m3/h,塔径为1.5m。试求每小时的用水量(kg/h)及所需填料层的高度。

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解:

Y1?y11?y1?0.11?0.1?0.111,X1?0.5%1?0.5%x11?x1?0.00061?0.0006?3?0.0006

y2?0.5%,Y2?LVY1?Y2X1?X2?0.503%?5.03?10?3

??0.111?5.03?100.0006?176.6

而 V?

2240?1?0.1?0.785?1.5?22.42?50.96kmol/(m22?h)

L?176.6?50.96?8999.54kmol/(m?h)

液体流量

L?8999.54?0.785?1.52?15895.4kmol/h=15895.4?18=286117.2kg/h=286.117t/h

Ep200?1020263 相平衡常数 m???98.7

?X1?X1?X1??Y1mY2m??X1?0.11198.7?0.0006?5.247?10?4

?X2?X2?X2??X1??X??X1ln??X?X1?X?Xm2??X2?5.247?105.03?1098.7?4?3?5.097?10?5?5

?Xm?2?5.097?102????4??ln?5.247?10?5?5.097?10???2.032?10?4

NOL??0.0006?02.032?10?4?2.95

又 KXa?C?KLa?

LKX?SMS?KLa?100018?50?2778kmol/(m3?h)

HOL?a??8999.542778?3.24m

?z?HOLNOL?3.24?2.95?9.56m5-38. 有一常压吸收塔,塔截面为0.5m2,填料层高为3m,用清水逆流吸收混合气中的丙酮(丙酮的分子量为58kg/kmol)。丙酮含量为0.05(摩尔比,下同),混合气中惰性气体的流量为1120m/h(标准状态)。已知在液气比为3的条件下,出塔气体中丙酮含量为0.005,操作条件下的平衡关系为Y=2X。试求:

(1) 出塔液中丙酮的质量分率;

(2) 气相总体积传质系数KYa(kmol/m3·s)

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(3) 若填料塔填料层增高3m,其它操作条件不变,问此吸收塔的吸收率为多大? 解: (1)

LV?Y1?Y2X1?X2?Y1?Y2X1

?0.015?X1?Y1?Y2L/V?0.05?0.0053

0.015?58w?x1M1x1M1?(1?x1)M2?58?0.015?0.985?18?0.0468

2(2) V?1120/(22.4?3600?0.5)?0.027k8mo/lms

S= m/(L/V)=2/3=0.667

(Y1-mX2)/(Y2-mX2)=Y1/Y2=0.05/0.005=10

NOG?11?Sln[(1?S)Y1?mXY2?mX22?S]?10.333ln[0.333?10?0.667]?4.16

HOG?z/NOG?3/4.16?0.721m KYa?V/HOG?0.0278/0.72?0.0386 kmol/(m·s)

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(3) z=3+3=6m

S、HOG不变, ?N?NOG?'’

'OG?Y1'Y2z'HOG?S]

?60.721?8.32

11?Sln[(1?S)解得: Y2'=0.00109

?'?Y1?Y2Y1''?0.05?0.001090.05?0.978

5-39.在逆流操作的填料吸收塔中,用清水吸收含氨0.05(摩尔比)的空气—氨混合气中的氨。已知混合气中空气的流量为2000m3/h(标准状态),气体空塔气速为1m/s(标准状态),操作条件下,平衡关系为Y*?1.2X,气相总体积传质系数KYa?180kmol/m3h,采用吸收剂用量为最小用量的1.5倍,要求吸收率为98%。试求: (1)溶液出口浓度x1;

(2)气相总传质单元高度HOG和气相总传质单元数NOG;

(3)若吸收剂改为含氨0.0015(摩尔比)的水溶液,问能否达到吸收率98%的要求,为

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什么?

解:(1)Y1?0.05 Y2?0.05(1?0.98)?0.001 X2?0 ?Y1?Y2X1?1.5Y1?Y2Y1mY1?X1?0.05/1.2m??0.02781.51.5

x1?X11?X1?0.02781?0.0278?0.0270

(2) ?Y1?Y1?mX1?0.05?1.2?0.0278?0.0166

?Y2?0.001?Ym?

?0.00560.0166?0.001ln0.01660.001

NOG?0.05?0.0010.0056?8.75

HOG?2000/22.4180?2000?0.89m

3600?1L0.05?0.001??1.76?m (3) V0.0278 ??max?Y1?Y2Y1??0.05?1.2?0.00150.05?96.4%

?? 不可能达到98%

5-40. 在一充有25mm阶梯环的填料塔中,用清水吸收混合气体中的NH3。吸收塔在20℃及101.3kPa(绝压)的条件下逆流操作,气液相平衡关系为Y??0.752X。已知混合气流率为0.045kmol/m2·s, NH3入塔浓度为0.05(摩尔分率),吸收率为99%,操作液气比为最小液气比的1.5倍,填料层高度为8.75m,试求: (1)气相总体积传质系数KYa;

(2)塔底截面处NH3吸收的体积传质速率NAa。 解: (1)

(LV)min?Y1?Y2X1?X?2

?Y1?Y2Y1/m

LV?m??0.752?0.99?0.744

=1.5×0.744=1.116

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