《化工原理二》复习题及答案

内容发布更新时间 : 2025/6/27 9:28:45星期一 下面是文章的全部内容请认真阅读。

x1?y10.96??0.906

2.5?1.5y12.5?1.5?0.96y2?0.580x1?0.403?0.508?0.906?0.403?0.928

6. 解:(1)由全塔物料衡算: D?xF?xW0.4?0.05F??150?58.33kmol/h

xD?xW0.95?0.05 ∴ W?F?D?150?58.33?91.67kmol/h

(2)精馏段操作线方程:

yn?1?xR40.95xn?D?xn??0.8xn?0.19 R?1R?155 提馏段操作线方程:

WxWRDxn?(R?1)D?(1?q)F(R?1)D?(1?q)F

4?58.3391.67?0.05?xn??1.647xn?0.03245?58.33?1505?58.33?150(3)饱和蒸汽进料 ye?yF?0.4

ye0.4??0.2125 xe???(??1)ye2.47?1.47?0.4x?ye0.95?0.4??2.93 Rmin?Dye?xe0.4?0.2125?4?1.36 RRmin2.93yn?1?

7. 解:(1)由于为理想干燥过程,所以为等焓干燥。有:I1?I2

I1??1.01?1.88H1?t1?2490H1??1.01?1.88?0.005??90?2490?0.005?104.2kJ/kg干空气 I2??1.01?1.88H2?t2?2490H2??1.01?1.88H2??55?2490H2?55.55?2593.4H2

得:H2?0.0189kg/kg绝干气

(2)GC?G(1?w1)?0.8?(1?0.05)?0.76kg/s

X1?w10.05??0.05261?w11?0.05X2?0.01?0.01011?0.01

W?G(H ) C(X1?X2)?V2?H1W?GC(X1?X2)?0.76?(0.0526?0.0101)?0.0323kg/s

∴ V?W0.0323??2.324kg/s

H2?H10.0189?0.005QP?V(1.01?1.88H1)(t1?t0)?2.324?(1.01?1.88?0.005)(90?20)?165.8kW

8. 解:(1)40℃下的空气湿度为:

Hw,40℃?0.622ps,40℃p?ps,40℃?0.622?7.3766?0.0488kg水/kg干气

101.325?7.3766H1?Hw,40℃??0.0488?1.09t1?tw,1??rw,40℃1.09??65?40??0.0375kg水/kg干气

2401.1ps,25℃p?ps,25℃H2?Hw,25℃?0.622?0.622?3.1684?0.0201kg水/kg干气

101.325?3.1684W水?H1?H2?0.0375?0.0201?0.0174kg水/kg干气

(2)确定放热量

I1?(1.01?1.88H1)t1?2500H1?(1.01?1.88?0.0375)?65?2500?0.0375?164.0kJ/kg干气 I2?(1.01?1.88H2)t2?2500H2?(1.01?1.88?0.0201)?25?2500?0.0201?76.4kJ/kg干气

Q?I1?I2?164.0?76.4?87.6kJ/kg干气

9. 解:(1)由题给条件可知干物料的流率:

Gc?G1(1?w1)?500?(1?0

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