化工原理课后习题答案第七章吸收习题解答 下载本文

LL?1.3()min?1.3?1.08?1.404 VV吸收剂用量:L?1.404?V?1.404?144.196?202.452kmol/h

实际液气比:

(2)求塔底排出水中丙酮含量

V(Y1?Y2)0.0526?0.00526??0.0337 L1.404π2π22塔横截面积:??D??1.3?1.327m

44X1?传质单元高度:HOG?VKYa??144.196?0.581m

187?1.327(Y1?Y1*)?(Y2?Y2*)(0.0526?1.2?0.0337)?0.00526?Ym???0.00823 *0.0526?1.2?0.0337Y?Ylnln11*0.00526Y2?Y2传质单元数:NOG?Y1?Y20.0526?0.00526??5.75 ?Ym0.00823填料层高度为:Z?HOGNOG?0.581?5.75?3.34m (3)Y2??Y1?1?????0.0526??1?0.95??0.00263

V(Y1?Y2?)0.0526?0.00263?X1???0.0356

L1.404HOG?VKYa??144.196?0.581m

187?1.327(Y1?Y1*?)?(Y2??Y2*)(0.0526?1.2?0.0356)?0.00263?Ym???0.00548

*?0.0526?1.2?0.0356Y?Ylnln110.00263Y2??Y2*Y1?Y2?0.0526?0.00263??NOG??9.12

0.00548?Ym???0.581?9.12?5.30m Z??HOGNOG

7-14 由矿石焙烧炉送出来的气体含有SO2 6%(体积分数),其余可视为空气,冷却送入填料塔用水

吸收。该填料层高度为6米,可以将混合气中的SO2 回收95%,气体速率为600kg惰气/(m2?h),

0.7液体速率为900 kg/(m2?h)。在操作范围内,SO2的气液平衡关系为Y?2X,KYa?W气,受液

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体速率影响很小,而W气是单位时间内通过塔截面的气体质量,试计算将操作条件作下列变动,所需填料层有何增减(假设气、液体速率变动后,塔内不会发生液泛)?(1)气体速度增加20%; (2)液体速度增加20%。

解:Y1?0.06/0.94?0.0638,Y2?0.0638??1?0.95??0.00319,X2?0

单位塔截面积空气流量:WV?V/??600/29?20.69kmol/m2?h?0.0057kmol/m2?s 单位塔截面积水的流量:WL?L/Q?900/18?50kmol/m2?h?0.0139kmol/m2?s

????????1/A?mV/L?2?20.69/50?0.8276

NOG??Y?mX21?110.0638??ln??1?1/A?1???ln??1?0.8276??0.8276??8.4271?1/A?Y2?mX2A?1?0.8276?0.00319?HOG?Z/NOG?6/8.427?0.712m

(1)当气体流量增加20%,1/A1?mV1/L?1.2?0.8276?0.993

NOG??Y?mX21?110.0638??ln??1?1/A1?1???ln??1?0.993??0.993??17.838

1?1/A1?Y2?mX2A1?1?0.993?0.00319?0.7?KYa?1HOG1??1.2MV????KYaMV???1.13?6?KYa?1?1.1KY3a6

MV11.2MV??1.056HOG?1.056?0.712?0.752m

?KYa?11.136KYa所以:Z?HOG1NOG1?17.838?0.752?13.412m

(2)当液体流量增加20%,1/A2?mV/1.2L?A/1.2?0.8276/1.2?0.6897

NOG??Y?mX21?110.0638??ln??1?1/A2?1???ln??1?0.6897??0.6897??6.221?1/A2?Y2?mX2A2?1?0.6897?0.00319? 又:气体流量不变,HOG也不变,HOG2?HOG?0.712m

所以:Z?HOG2NOG2?6.22?0.712?4.43m

7-15 某制药厂现有一直径为1米,填料层的高度为4米的吸收塔。用纯溶剂逆流吸收气体混合物中

的某可溶组分,该组分进口浓度为0.08(摩尔分率),混合气流率为40kmol/h,要求回收率不低

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于95%。操作液气比为最小液气比的1.5倍,相平衡关系为Y* = 2X,试计算:(1)操作条件下的液气比为多少;(2)塔高为3米处气相浓度;(3)若塔高不受限制,最大吸收率为多少? 解:(1)Y1?8?0.0870,Y2?0.0870?0.05?0.0043 92?L/V?min?Y1?Y2Y?Y?M12?M??2?0.95?1.9

X1,max?X2Y1L/V?1.5?1.9?2.85,1/A?MV/L?2/2.85?0.702

(2)NOGY1?Y2*111?ln[(1?1/A)?1/A]?ln[(1?1/A)?1/A] 1?1/AY2?Y2*1?1/A1???11ln[(1?0.702)?0.702]?6.364

1?0.7021?0.950.62 85m4 ?HOG?Z/NOG?4/6.36? 当Z'?3m时,设其气液相浓度分别为X'、Y',

HOG不变,则:Z'/N'OG?Z/NO G又由V(Y1?Y2)?L(X1?X2)得:

X1?(V/L)(Y1?Y2)?(1/2.85)?(0.0870?0.0043)?0.0290

则:NOG??Y?MX1?11?Y1?MX1?'N?ln; ln?1OG???1?1/A?Y'?MX'?1?1/A?Y2?0??Y?MX1??Y1?MX1? 1?ln?1?4?ln????Y'?MX'??Y2?Y'?mX'?(Y2?Y2*)1/4(Y1?Y1*)3/4?0.00431/4(0.0870?2?0.0290)3/4?0.0180

即 Y'?0.018 (1) ?0X2又由操作线方程 Y'?LL??X'??Y2?X2??2.85X'?0.0043 (2) VV??? (1)、(2)式联解得:Y'?0.0502,X'?0.0161

(3)若塔高不受限制,因为2.85>2,所以,在塔顶达到平衡,最大吸收率为100%。

7-16 在装填有50mm拉西环的填料塔中用清水吸收空气中的甲醇,直径为880mm,填料层高6m,,

每小时处理2000m3甲醇-空气混合气,其中含甲醇5%(体积分数);操作条件为298.15 K,101.3

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kPa。塔顶放出废气中含甲醇0.263%(体积分数),塔底排出的溶液每kg含甲醇61.2g;在此操作条件下,平衡关系Y = 2.0X。根据上述测得数据试计算:(1)气相体积总传质系数KYa;(2)每小时回收多少甲醇;(3)若保持气液流量V、L不变,将填料层高度加高3m,可以多回收多少甲醇。

解:(1)Y1?0.05/0.95?0.0526;Y2?y2?0.00263;X2?0

V?2000/22.4?(273.15/298.15)?(1?0.05)?77.71kmol/h X1?61.2/581.0552??0.0202

(1000?61.2)/1852.16?Y1?Y1?2X1?0.0526?2?0.0202?0.0121;?Y2?Y2?2X2?0.00263

?Ym??Y1??Y20.0121?0.00263??0.0062

ln(?Y1/?Y2)ln(0.0121/0.00263)Y1?Y20.0526?0.00263??8.05 ?Ym0.0062则 NOG?HOG?V/(KYa?)?Z/NOG?6/8.05?0.745m

KYa?V/(HOG?)?77.71/(0.745?0.785?0.882)?171.59kmol/(m3h)

(2)G?V(Y1?Y2)?77.71?(0.0526?0.00263)?3.88kmol/h (3)

V、L不变,则

L?G/(X1?X2)?3.88/0.0202?192.24kmol/h

1/A?(2?77.71)/192.24?0.8085

即HOG不变,则NOG'?Z'/HOG?9/0.745?12.08 又 NOG'???Y?mX21ln?(1?1/A)1'?1/A?

1?1/A?Y2?mX2???10.0526ln?(1?0.8085)?0.8085??12.08 '1?0.8085?Y2????0.0526'2.31?ln?0.1915?0.8085? ? 0.01/Y2?10.0744?0.8085 'Y2?? 14

Y2'?0.01/9.266?0.0011

G'?V(Y1?Y2')?77.71?(0.0526?0.0011)?4.0kmol/h

??G?G'?G?4.0?3.88?0.12kmol/h

7-17 现需用纯溶剂除去混合气中的某组分,所用的填料塔的高度为6m,在操作条件下的平衡关系为

Y = 0.8X ,当L / V = 1.2时, 溶质回收率可达90%。在相同条件下,若改用另一种性能较好的填料,则其吸收率可提高到95%,请问第二种填料的体积传质系数是第一种填料的多少倍? 解:

''在同一吸收塔中操作,则有:Z?NOGHOG?NOGHOG

1/A?mV/L?0.8/1.2?2/3 NOG???Y111ln?(1?1/A)1?1/A??ln[(1?1/A)?1/A]

1?1/A?Y21?1/A1??????11ln??2/3??4.16

1?2/3?3?(1?0.9)???Y111ln?(1?1/A)1?1/A??ln[(1?1/A)?1/A]

1?1/A?Y21?1/A1??'???NOG???11ln??2/3??5.98

1?2/3?3?(1?0.95)?NOG'HOGV/KYa(Ka)'5.98???Y??1.44 NOGHOG'V/(KYa)'KYa4.16?

7-18 用填料塔解吸某含有二氧化碳的碳酸丙烯酯吸收液,已知进、出解吸塔的液相组成分别为0.0085

和0.0016(摩尔分数)。解吸所用载气为含二氧化碳0.0005(摩尔分数)的空气,解吸的操作条件为35℃、101.3kPa,此时平衡关系为Y = 106.03X。操作气液比为最小气液比的1.45倍。若取HOL=0.82m,求所需填料层的高度? 解:X1?(KYa)'?1.44 KYa0.00850.0016?0.0086,X2??0.0016

1?0.00851?0.00160.0005Y2??0.0005

1?0.000515