内容发布更新时间 : 2024/12/23 18:07:32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?0.2)?400kg干料/h
X1?w10.2??0.25kg水/kg干料 1?w11?0.2w20.02??0.020408kg水/kg干料 1?w21?0.02X2?干燥器入口（预热器出口H1?H0）的焓：
I1?(1.01?1.88H1)t1?2490H1?(1.01?1.88?0.01)?100?2490?0.01?127.78kJ/kg干气
假设此干燥过程为理想干燥过程：
I2?I1?(1.01?1.88H2)t2?2490H2 ?(1.01?1.88H2)?60?2490H2?60.6?2602.8H2?127.78kJ/kg干气
81水)/kg(干气) 解得： H2?0.025kg(
由干燥器的水分平衡式： L?H2?H1??Gc?X1?X2?
Gc?X1?X2?400??0.25?0.020408?得： L???5807.0kg干气/h
H2?H10.02581?0.01所需湿空气量： L1?L?1?H0??5807.0??1?0.01??5865.1kg湿气/h （2）空气在预热器进口处的焓值为：
I0?(1.01?1.88H0)t0?2490H0?(1.01?1.88?0.01)?20?2490?0.01?45.476kJ/kg干气
则空气经预热器获得的热量为：
Qp?L?I1?I0??5807.0?(127.78?45.476)?4.78?105kJ/h