化工原理杨祖荣1-7章习题答案(完美排版)

Y1?0.0561.2/58?0.0526 X2?0 X1??0.0202

1?0.05938.8/18?Y1?Y1?Y1*?0.0526?2?0.0202?0.0122 ?Y1?Y1?Y1*?0.0526?2?0.0202?0.0122 ?Y2?Y2?Y2*?0.00263

?Ym??Y1??Y20.0122?0.00263??0.00624 ?Y10.0122lnln0.00263?Y2Y1?Y20.0526?0.00263??8.008 ?Ym0.00624VZ?

KYa?NOGNOG?HOG?KYa?V77.7NOG??8.008?170kmol/(m3?h) Z?6?0.61W?V(Y1?Y2)?77.7?(0.0526?0.00263)?3.883kmol/h?225.19kg/h

(2) HOG?N'OG?ZVZ6???0.7493m

KYa?NOG8.008'HOG?6?3?12.01 0.7493Y?Y2L0.0526?0.00263?1??2.474 VX1?X20.0202?0S'?S?2/2.474?0.8084

N'OG???10.0526ln?(1?0.8084)?0.8084??12.01 '1?0.8084?Y??2?解得Y2?0.001096

'W'?V(Y1?Y'2)?77.7?(0.0526?0.001096)?4.002kmol/h?232.11kg/h ?W?W?W'?225.19?232.11?6.918kg/h

22.用纯溶剂在一填料吸收塔内,逆流吸收某混合气体中的可溶组分。混合气体处理量为1.25Nm/s,要求溶质的回收率为99.2%。操作液气比为1.71,吸收过程为气膜控制。已知10℃下,相平衡关系

3

Y*?0.5X,气相总传质单元高度为0.8m。试求:

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(1)吸收温度升为30℃时,溶质的吸收率降低到多少? (30℃时,相平衡关系Y?1.2X) (2)若维持原吸收率,应采取什么措施(定量计算其中的2个措施)。 解:

(1)原工况:S? NOG?*mV0.5??0.292 L1.71??Y?mX2??111ln?(1?S)1?S??ln?(1?S)?S? 1?S?Y2?mX21????1?S?11??ln?(1?0.292)?0.292??6.34

1?0.292?1?0.992?NOG?Z?NOG?HOG?6.34?0.8?5.072m

新工况: H'OG?HOG Z'?Z

m'1.2S??0.292?0.7008 S?m0.5' NOG?N'OG? 解得??0.95

'??11ln?(1?0.7008)?0.7008?6.34 ?'1?0.7008?1???(2)温度升高,平衡线上移,推动力减小,保持吸收率不变,可采取措施:

1)L/V增加,即增加溶剂量;

S?S'mVm'V ?'LLm'1.2L?L?L?2.4L

m0.5'2)增加填料层高度

m'1.2S??0.292?0.7008,推动力减小靠增加塔高弥补。 L/V不变,温度升高,S?m0.5' N'OG?11??ln?(1?0.7008)?0.7008??12.17

1?0.7008?1?0.992?'OG温度改变,对气膜控制吸收过程,传质单元高度不变,H?HOG?0.8

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