教學(xué)課件少數(shù)幾張我拷來時(shí)候就化工原理下第08章

3.3

對(duì)流傳質(zhì)類似于傳熱中的牛頓冷卻定律NA=kg(p-pi)=kL(Ci

-C)=ky(y-yi)=kx(xi-x)注意推動(dòng)力與傳質(zhì)(分)系數(shù)一一對(duì)應(yīng)。傳質(zhì)分系數(shù)的確定kg=f(ρ,μ,u,d,D)可得降膜式吸收器，Sh=0.023Re0.8Sc0.33Re>2100,

Sc=0.6~3000DSh=AReBScC無因次化

Sh

=

kd

,

Re

=

dur

,m

rD關(guān)聯(lián)式Sc

=

m3.5

傳質(zhì)理論①有效膜理論k∝D

定態(tài)②溶質(zhì)滲透理論k∝D0.5非定態(tài)③表面更新理論k∝D0.5非定態(tài)實(shí)際k∝D?降膜式k實(shí)測(cè)∝D0.674

相際傳質(zhì)步驟：①氣相→界面②界面溶解③界面→液相4.1

相際傳質(zhì)速率方程N(yùn)A=ky(y-yi)yi=mxiNA=kx(xi-x)氣相界面相平衡液相消去xi,yiy

xAky

kx

kN+

k=

y

-

yi

=

xi

-

x

=

y

-

yi

+

m(

xi

-

x)1

1

1

m總傳質(zhì)系數(shù)同樣NA=KX(xe-x),有什么條件？y

xA1

mk=

K

y

(

y

-

ye

)+

ky

-

yeN

=xyyk+

kK

=1

m1yxyxmk=

mK+

kK

=1

11xyyk+

kK

=1

m1y

=

mx

+

b4.2

阻力控制總阻力

1

=

1

+

mK

y

ky

kx當(dāng)

1

>>

m

時(shí)，氣相阻力控制ky

kxKy≈ky，Kx≈mky

，yi≈ye例如，水吸收NH3,HCl當(dāng)

1

<<

m

時(shí)，液相阻力控制ky

kxKy≈kx/m，Kx≈kx

，xi≈xe例如，水吸收H2,CO2,O2由NA=ky(y-yi)=kx(xi-x)得得(xi,yi)=

-

kxx

-

xi

kyy

-

yikyk由已知點(diǎn)(x,y)和-

x

斜率作直線與平衡線相交，例1：已知濕壁塔某一塔截面上氣相主體的溶質(zhì)濃度y=0.05,

液相溶質(zhì)濃度

x=0.01,

氣相與液相傳質(zhì)系數(shù)分別為ky=5×10-4kmol/m2s,

kx=8×10-4kmol/m2s,亨利常數(shù)m=2,總壓為101.3kPa。試求：①推動(dòng)力y-ye,xe-x總傳質(zhì)系數(shù)Ky,Kx傳質(zhì)速率NA②將操作總壓改為162kPa，單位塔截面上的氣體質(zhì)量流速不變，總壓對(duì)氣體粘度影響可忽略，計(jì)算傳質(zhì)速率NA

。解：①推動(dòng)力y-ye=y-mx=0.05-2×0.01=0.03傳質(zhì)速率m

2x

-

x

=

y

-

x

=

0.05

-

0.01

=

0.015e1

11

25

·10-4

+

8

·10-4==y

xy+

k1

mk總傳質(zhì)系數(shù)K1

11=

2.2

·10-4

kmol

/

m

2

s11

12

·5

·10-4

+

8

·10-4=y

xx+

kmkK

==

4.4

·10-4

kmol

/

m

2

s-4NA

=

K

y

(

y

-

ye

)

=

2.2

·10

·0.03=

6.6

·10-6

kmol

/

m

2

s壓強(qiáng)對(duì)液相D,

ρ無影響，所以對(duì)kx無影響.對(duì)ky如何？m3/m2s與P無關(guān)-4

-6

2NA

=

Kx

(

xe

-

x)

=

4.4

·10

·0.015

=

6.6

·10

kmol

/

m

sP②總壓增加后，由m

=E

可知P'

162m'=

P

m

=

101.3

·

2

=

1.251

/

30.83(

)D

dGrDmd

mk

=

0.023

(

)PD1

,r

P0D2

/

3Pkky

=

RT

P

r1

/

3

Py

-

ye

=

0.05

-

1.25

·0.01

=

0.0375

↑1

11

1.255

·10-4

+

8

·10-4=y

xy+

k1

mkK

==

2.8

·10-4

kmol

/

m

2

s↑-4

-6N

A

=

K

y

(

y

-

ye

)

=

2.8

·10

·0.0375

=

10.5

·10=

10.5

·10-6

kmol

/

m

2

s↑操作壓強(qiáng)增加對(duì)吸收是否有利？5

低含量氣體吸收當(dāng)氣體濃度<5~10%時(shí),有三個(gè)基本特點(diǎn):①流率G,L沿塔為常量.G,L單位kmol/m2s②吸收過程是等溫的熱量衡算不必要，相平衡關(guān)系不變③傳質(zhì)系數(shù)為常量，kx,ky沿塔不變這三個(gè)特點(diǎn)也即數(shù)學(xué)描述的基本假定5.1物料衡算微分方程式數(shù)學(xué)描述:物料衡算吸收過程速率式取微元塔段dh，則傳質(zhì)面積為aAdh傳質(zhì)量為NAaAdh定態(tài)條件下進(jìn)=出GAy=GA(y+dy)+NAaAdh得-Gdy=NAadh同樣，液相有

-Ldx=NAadhNA=Ky(y-ye)=Kx(xe-x)4單位容積內(nèi)的有效吸收表面

a

m2/m3塔截面積A（

=p

D2）5.2

過程的基本方程式-Gdy=

Kya

(y-ye)dh傳質(zhì)單元高度,m傳質(zhì)單元數(shù)將變量分解成兩部分----變量分離法HOG主要取決于設(shè)備，一般為0.15~1.5mOGN

含義表示分離任務(wù)難易HOG的物理意義OG

OGy

eHdyK

aG=

H

Ny

-

ydh

=H

=y1y

20K

aGOGH

=yy1y

2eOGy

-

ydyN

=平均推動(dòng)力y1

-

y25.3操作線和平均推動(dòng)力對(duì)塔段作物料衡算Gy1+Lx=Gy+Lx1得操作線表示同一塔截面上氣液相濃度關(guān)系同樣-Ldx

=Kxa(xe

-x)dhOL

OLx

eHK

aL

dx=

H

Nx

-

xdh

=H

=x1x

201

1G

Gy

=

L

x

+

y

-

L

xe找y與y-y

的關(guān)系,以便求取得y-ye=y-mxy1y

2ey

-

ydyG當(dāng)相平衡線直線時(shí)，由操作關(guān)系x

=x1

-L

(y1

-y)11L

L=

y

-

mG

y

+

mG

y

-

mxdy

L

y1

-

y2即y與y-ye之間為線性關(guān)系d

(

y

-

ye

)

=

1

-

mG

=常數(shù)=

Dy1

-

Dy21

2eDy

-

Dyy1

-

y2dy

=

d

(

y

-

y

)=21Dy

2ey1y

2eOGy

-

yDy1d

(

y

-

ye

)y1

-

y2y

-

y

Dy

-

DydyN

===

y1

-

y2Dy1

-

Dy2

Dy2

Dymy1

-

y2ln

Dy1mDy2

y2

-

mx2對(duì)數(shù)平均推動(dòng)力ln

Dy1

ln

y1

-

mx1=

Dy1

-

Dy2

=

(

y1

-

mx1)

-