填料吸收塔设计说明(8)

=0.0095(

3.21

90.5 0.0181 10

3

)2/3(

1.005 10

3

)-1/2(1.005

9

9.811/3

998.2

)

998.2 1.334 10

=1.17×10-2 m/s

错误！未找到引用源。依式

kG= 0.237(

VG`0.7at G

)(

G GDG

)1\3(

atDGRT

)

3

= 0.237 (

1.30

132.5 0.0181 10

3

0.0181 10132.5 1.98 10

5

5

)0.7(

1.206 1.98 10

)1/3(

8.314 293

)

=8.1×10-6kmol/(m2·S·kPa) 故kGa kGaw

kLa kLaw

0.4

1.1

=8.1×10 6×90.5×1.451.1=1.1×10-3 (m/s)

=1.17×10-2×90.5×1.450.4=1.23m/s

(2)计算KYa

KYa＝KGaP，而

1KGa

1aKG

1akG1

1aHkL

，H=0.725kmol/(m3.kpa)

=

11.1 10

3

＋

1.23 0.725

-3

求得: KGa=1.09×10kmol/(m3.S.kPa)

KYa

＝KGaP=1.09×10-3×101.3=0.11kmol/(m3.S)

(3)计算HOG

HOG

=

VBKYa

=

19.671/36000.11 0.785 0.4

2

=0.40m

(4)．传质单元数NOG计算

Y1=Y1-mX1=0.0574-2.737×0.014=0.019