Interphase Mass Transfer

Interphase Mass Transfer

Interphase Mass Transfer

the transfer of solute A from one fluid phase by convection and then through a second fluid phase by convection

the two phases are in direct contact with each other

the interfacial area is not well defined mass transfer is caused by a concentration

gradient in each phase equilibrium exist at the interface

Equilibrium Relations

Gas-liquid equilibrium data (A.3-19-25) Henry’s law

where H = Henry’s law constant (atm/mole frac)H’ = Henry’s law constant (mole frac gas/mole frac

liquid) = H/P

AA Hxp AA xHy '

Concentration Profile

NA

yAG

yAi

xAL

xAi

Gas-phase mixture of A in gas G

Liquids-phase solution of A in liquid L

Distance from the interface

Concentration Profile

)( AiAi xfy

At the interface:

• No resistance to mass transfer is present

• yAi is at equilibrium with xAi

Film Mass-Transfer Coefficients and Interface Concentrations

)()( ''

ALAixAiAGyA xxkyykN

Equimolar counterdiffusion

For A diffusingfrom gas to liquid and B from liquid to gas:

Rearranging:

)(

)('

'

AiAL

AiAG

y

x

xx

yy

k

k

Driving forces


xAL xAi

yAG

yAi

Slope = -kx’/ky’

Equilibrium line


)()( ALAixAiAGyA xxkyykN

Diffusion of A through stagnant or non-diffusing B

For A diffusing through a stagnant gas phase then through a stagnant liquid phase:

iMA

y

y y

kk

)1(

'

iMA

xx x

kk

)1(

'


Equilibrium line

xAL xAi

yAG

yAi

Slope = iMAy

iMAx

yk

xk

)1/(

)1/('

'

Overall Mass-Transfer Coefficients

)()( *'*'

ALAxAAGyA xxKyyKN

where

Ky’ = mass transfer coefficient based on the overall gas-phase

driving force

Kx’ = mass transfer coefficient based on the overall liquid-phase

driving force

y*A = mole fraction of a in equilibrium with xAL

x*A = mole fraction of a in equilibrium with yAG


)()( ''

ALAixAiAGyA xxkyykN

Equimolar counterdiffusion and/or diffusion in dilute solutions

)(

)( *'

ALAi

AAi

xx

yym

'

'

''

11

xyy k

m

kK

where


xAL xAi

yAG

yAi

Slope = -kx’/ky’

Equilibrium line

x*A

y*A

Slope=m’

Slope=m’’


Also:

)(

)(*

''

AiA

AiAi

xx

yym

'''''

111

xxx kkmK

where


Special cases:

''

11

yy kK

Gas-phase controlling

m’ is very small

Liquid -phase controlling

m’’ is very large

''

11

xx kK

AiAGAAG yyyy * *

AAi xx


)()1(

)()1(

''

ALAi

iMA

xAiAG

iMA

y

A xxx

kyy

y

kN

Unimolar diffusion

)()1(

)()1(

*

*

'*

*

'

ALA

MA

xAAG

MA

y

A xxx

Kyy

y

KN


Unimolar diffusion

iMAxiMAyMAy xk

m

ykyK )1/()1/(

1

)1/(

1'

'

'

*

'

xyy k

m

kK

'11

Ky


Unimolar diffusion

iMAxiMAyMAx xkykmxK )1/(

1

)1/(

1

)1/(

1''''

*

'

xyx kmkK

111

Kx


where

)1()1(

ln

)1()1()1( *

*

*

AG

A

AGAMA

yyyy

y

)1()1(

ln

)1()1()1(

*

*

*

A

AL

AALMA

xx

xxx


xAL xAi

yAG

yAi

Slope =

Equilibrium line

x*A

y*A

Slope=m’

Slope=m’’

iMAy

iMAx

yk

xk

)1/(

)1/('

'

Sample ProblemThe solute A is being absorbed from a gas mixture of A and B in a wetted-wall

tower with the liquid flowing as a film downward along the wall. At a certain point in the tower the bulk gas concentration yAG

= 0.380 mol fraction and the bulk liquid concentration is xAL=0.100. The tower is operating at 298K and 1.1013 x 105Pa and the equilibrium data are as follows:

xA yA xA yA

0 0 0.20 0.131

0.05 0.022 0.25 0.187

0.10 0.052 0.30 0.265

0.15 0.087 0.35 0.385

The solute A diffuses through stagnant B in the gas phase and then through a nondiffusing liquid.

Using correlations for the dilute solutions in wetted-wall towers, the film mass-transfer coefficient for A in the gas phase is predicted as ky=1.465 x 10-3 kgmol A/s m2mol frac and for the liquid phase as kx=1.967 x 10-3kg mol A/s m2

mol frac. Calculate the interface concentrations yAi and xAi and flux NA. Calculate the overall mass transfer coefficient K’y, the flux, and the percent resistance in the gas and liquid films. Do this for the case of A diffusing through stagnant B.

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Interphase Mass Transfer