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化工课程设计 - 图文 

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2

??????323600????2×0.0123)=0.0442()=0.54????23

1.0223

2

?????=0.0442(

????

???=2.5???=2.5×(???+?????)=2.5×(0.059+0.54????2

?6

)=0.15+1.35????2

3.2

23

5.7×10????=(1.11×

55.9×0.001????2

(0.45?(0.15+1.35????2

232))3)=0.1

整理得:????2=2.33?(10.47????2

)

(2)液相负荷下限

对平直堰,取堰上液层高度how=0.006m作为最小液体负荷标准。

22.84

=??()=0.54????23=0.006 1000????

2????3?????

取E=1,则:

?????=0.0442(

???????

)=0.006 ????

23??????????=1.7???07??3╱??

(3)液相负荷上限

液体的最大流量应保证降液管中停留时间不低于3-5s。 液体降液管内停留时间

??????????==3?5??

????

以θ=5s作为液体在降液管内停留时间的下限,则:

??????????

????????0.2261×0.45===0.02035??3╱??

??528

(4)漏液线

由??0??????=4.4??0√(0.0056+0.13×(???+?????)????)×(

2.84=??() 1000????

2????3??????????????)= ??????0

?????

⑴精馏段:

????12.84

????1??????=4.4??0??0√(0.0056+0.13×(???1+??())????1)×()

1000????????1

2

2

????32.84????13816.66

=4.4×0.5×0.164√(0.0056+0.13×(0.064+×1(3600×))?0.0044)×()

10001.021.02即????1??????

⑵提馏段:

=0.36×√7.62+68.52??

2

??13

????2??????=4.4??0??0√(0.0056+0.13×(???2+

??2.84

??())????2)×(??2) 1000????????2

2

32????3=4.4×0.5×0.164√(0.0056+0.13×(0.059+

2.84????2931.46

×1(3600×))?0.0061)×() 10001.020.752

??23

即????2??????

=0.36×√8.9+106.29??

(5)液泛线

????=??(????+???)=???+???+???=???+?1+???+???+???

?1=????? ????=???+?????

29

联立解得:??????+(??????1)×(???)=(??+1)×(?????)+???+???+???

忽略hσ,将how与Ls,hd与Ls,hc与Vs的关系式代入,并整理得:

??????=??????????

⑴精馏段:

22

2

??????3

0.051????10.0511.02??1===0.0095

(??0??0)2????1(0.164×0.5)2816.66??1=??????+(??????1)×(???1)=0.5×0.45+(0.5?0.564?1)×0.064=0.157

0.1530.153??1===1815.541

(?????0)2(1.02×0.009)2??1=2.84×10?3??(1+??)(

36003600

)=2.84×10?3×1(1+0.564)()=1.03 ????1.022

323将有关数据代入整理得:????12=16.53?191109.58????12?108.42????1?

⑵提馏段:

??2=

0.051????20.0510.75

==0.0061

(??0??0)2????2(0.164×0.5)2931.46??2=??????+(??????1)×(???2)=0.5×0.45+(0.5?0.573?1)×0.059=0.162

??2=

0.1530.153

==277.994

(?????0)2(1.02×0.023)2232336003600?3)=2.84×10×1(1+0.573)()=1.036 ??2=2.84×10??(1+??)(????1.02

?3

整理得:????22=26.56?45572.79????22?169.84????2?

在操作范围内任意取两个Ls值算出Vs。

30

由以上1-5作出塔板负荷性能

图4-5精馏段塔板负荷性能图

图4-6提馏段塔板负荷性能图

由塔板负荷性能图可以看出:

31

①在任务规定的气、液负荷下的操作点P(设计点)处在适宜操作区内的适中位置; ②塔板的气相负荷上限完全由雾沫夹带控制,操作下限由漏液控制; ③按固定的液气比,由图可查出塔板的气相负荷上限

Vs1max=3.54(m3╱s),Vs2max=3.57(m3╱s),气相负荷下限Vs1min=0.9939(m3╱s),Vs2min=1.0742(m3╱s)

精馏段操作弹性:

提馏段操作弹性:

故设计基本合理。

????1=??1??????????1??????=3.54

0.9939=3.56

????2=??2??????????2??????=3.571.0742=3.32

32

化工课程设计 - 图文 

2??????323600????2×0.0123)=0.0442()=0.54????231.02232?????=0.0442(???????=2.5???=2.5×(???+?????)=2.5×(0.059+0.54????2?6)=0.15+1.35??
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