1

Interface Mass Transfer

• Most of the mass transfer processes involve the transfer of solute A between two or more immiscible phases.

• The rate of mass transfer depends on a number of factors: solubility of solute A and its displacement form equilibrium.

• Design of mass transfer equipment should take into account not only the temperature and pressure, but other factors such as interfacial contact area and distribution of both phases affect the rate of mass transfer between the different phases.

2

Two resistance model of interface mass transferLewis and Whitman (1924)

• Transfer between 2 contacting phases.

• Three steps of interface transfer.

• The theory:– Rate of transfer is controlled

by diffusion through phases on each side of interface.

– No resistance across the interface mole fraction in the liquid and in the gas are at equilibrium.

3

Mass transfer using local film MTC.

1- Equimolar counter diffusion

/ /A y i x iN k y y k x x

x

y

P

M

yA

yAi

xA xAi

2

kgmol/ /

s . m . mol frac, are local liquid and gas MTC x yk k

/ /y i x ik y y k x x

At steady state:

/

/ the slope of the line P-M ix

y i

y yk

k x x

4

2- Diffusin of A through a stagnant or non diffusing B

Mass transfer using local film MTC.

x

y

P

M

yA

yAi

xA xAi

A y i x iN k y y k x x

/ /

1x x

xBLM LM

k kk

x x

/ /

1y y

yBLM LM

k kk

y y

1 11

1ln

1

i

LMi

y yy

y

y

1 11

1ln

1

i

LMi

x xx

x

x

/ /

1 1y x

A i i

LM LM

k kN y y x x

y x

/

/

1 slope of the line P-M

1x iLM

y iLM

k x y y

k y x x

• For dilute solutions k/ and k are equal.

• For concentrated solutions use trial and error method to find the slope

5

Overall MTC

x

y

P

M

yA

yAi

xA xAi

y*A

x*A

m”

m’

The previous equations for local MTC’s require the determination of interfacial concentrations which is impractical and usually difficult to measure experimentally, therefore it is more convenient to use the overall MTC’s instead. The overall MTC is based on the bulk concentrations.

/ * / *A y xN K y y K x x

2

kgmol/ /

s . m . mol frac, are overall liquid and gas MTC x yK K

y* mole fraction of solute A in the gas phase in equilibrium with the concentration of A in the bulk liquid phase

x* mole fraction of solute A in the liquid phase in equilibrium with the concentration of A in the bulk gas phase

6

Overall MTC1- Equimolar counter diffusion

/ /A y i x iN k y y k x x

* *i iy y y y y y

/A

iy

Ny y

k *

/A

y

Ny y

K

x

y

P

M

yA

yAi

xA xAi

y*A

x*A

m”

m’

* / //A

i ix

Ny y m x x m

k

// / /A A A

y y x

N N Nm

K k k

Liquid phaseOverall Gas phaseresistanceresistance resistance

/

/ / /

1 1

y y x

m

K k k

Similar analysis can be used to obtain the overall liquid phase resistance as

Gas phaseOverall Liquid phaseresistanceresistance resistance

/ / / / /

1 1 1

x y xK m k k

7

Overall MTC2- Diffusion of A through a stagnant or non diffusing B

/ /

1 1y x

A i i

LM LM

k kN y y x x

y x

x

y

P

M

yA

yAi

xA xAi

y*A

x*A

m”

m’

* *

/ /* *

1 1y x

A

LM LM

K KN y y x x

y x

* *i iy y y y y y

*

Liquid phaseOverall Gas phaseresistanceresistance resistance

/

/ / /

1 1

1 1 1y y xLM LM LM

m

K y k y k x

*

Overall Gas phase Liquid phaseresistance resistance resistance

/ / / / /

1 1 1

1 1 1x y xLM LM LMK x m k y k x

Overall MTC based on the gas phase

Overall MTC based on the liquid phase

*

*

*

1 11

1ln

1

LM

y yy

y

y

*

*

*

1 11

1ln

1

LM

x xx

x

x

8

Importance of the slope of the equilibrium curve

Case I: Gas phase controlled system ( resistance in the liquid phase is very small)

In this situation the equilibrium curve is almost a horizontal line and the slope of the line

m’ is quite small. This means that the a small concentration of A in the gas phase yA will provide a large value of x*. The gas solute is very soluble in the liquid phase and

hence the term m’/k’x is very small. Under such circumstances, in order to increase the mass transfer rate efforts should be directed toward decreasing the gas-phase resistance .

Case II: Liquid phase controlled system (resistance in the gas phase is very small)

In this situation the slope of the line m” is very large and the solute is very insoluble in the liquid phase. The major resistance to mass transfer resides within the liquid phase. In this situation efforts to increase mass transfer rate should be focused on conditions that

increase the liquid mass transfer coefficient kx.

Fractional gas resistance y

y

K

k Fractional liquid resistance x

x

K

k

9

Example 1

A solute A is being absorbed from a gas mixture of A and B n 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 is 38% and the bulk liquid concentration 10%. The tower is operating at 298 K

and 1 atm. If the k’y and k’x are 1.465 e-3, 1.967 e-3 kgmol of A/ (s.m2.m.f) respectively. Calculate the interface concentration and the molar flux.

x 0 0.05 0.10 0.15 0.2 0.25 0.3 0.35

y 0 0.022 0.052 0.087 0.131 0.187 0.265 0.385

Solution

Absorption process → diffusion through stagnant B

Graphical trail and error solution is needed

10

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.40

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

f(x) = 0.164613502354635 ln(x) + 0.453944171697482

x

y

P

Since the slope is almost equal to the one in the second trail then stop iteration

First trial x y(1-y)LM = 1

(1-x)LM = 1

slope = -k'x/k'y =

-1.34

Drawing the linesxi,yi = 0.247 0.183Second trial:(1-y)LM = 0.714

(1-x)LM = 0.824

slope = -k'x/k'y =

-1.163

xi,yi = 0.258 0.196Third trial:(1-y)LM = 0.708

(1-x)LM = 0.818

slope = -k'x/k'y =

-1.161

Example 1

/

42

1

0.001465 kgmol 0.38 0.196 3.78 10

0.708 s.m

yA i

LM

kN y y

y

11

Example 1

How to calculate the Overall MTC for the same example

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.40

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

f(x) = 0.164613502354635 ln(x) + 0.453944171697482

x

yy* = 0.052 x* =

0.347

P

*

/

/ / /

1 1

1 1 1y y xLM LM LM

m

K y k y k x

(1-y)LM = 0.708

(1-x)LM = 0.818

(1-y)LM* = 0.772

2

kgmol/ 4

s.m .(m.f)8.95 10 yK

*

/

/

4

3

1 % gas phase resistance

1

8.95 10 0.708 100

1.47 10 0.772

55.8%

y LM

y LM

K y

k y

*

/ 0.196 0.052

0.258 0.1

0.91

i

i

y ym

x x

12

Example 2

A wetted-wall absorption tower 2.54 cm of inside diameter is fed with water as the wall liquid and NH3-air mixture as the central core gas. At a particular level in the tower the NH3 concentration in the bulk gas is 0.8 (mole fraction) and in the bulk liquid is 0.05 (mole fraction). The temperature is 26.7 C and the pressure is 1 atm. The rates of flow are such that the local mass transfer coefficient in the liquid from a correlation obtained with dilute

solutions is 2.87 e-5 m/s and the local Sherwood number for the gas is 40. DNH3-air is 2.297

e-5 m2/s. Calculate y*, x*, xi, yi, Ky, Kx and the % gas phase resistance.

x 0 0.05 0.10 0.25 0.30

y 0 0.0707 0.1347 0.591 0.920

Solution

2

2

/ / 5

kgmol5 3

m . s . (m.f)

2.87 10.

1000 2.87 10 1.59 10

18

H Ox ck Ck

M W

Local MTC for the liquid phase

13

Example 2

Local MTC for the gas phase

5

/

2

40 2.297 100.0361

2.54 10Sh AB

c

N Dk

D

/ / / 101325

0.0361 0.001468314 26.7 273.15y c c

Pk Ck k

RT

14

Example 2

-0.05 0.00 0.05 0.10 0.15 0.20 0.25 0.300.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

f(x) = 0.447835749692882 ln(x) + 1.31229754197529

x

y

yi = 0.73

xi = 0.277

15

Example 2

-0.05 0.00 0.05 0.10 0.15 0.20 0.25 0.300.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

f(x) = 0.447835749692882 ln(x) + 1.31229754197529

x

y

y* = 0.07 x* =

0.286

P

16

*

/

/ / /

1 1

1 1 1y y xLM LM LM

m

K y k y k x

Example 2

*

/ / / / /

1 1 1

1 1 1x y xLM LM LMK x m k y k x

m' = 2.865

K'y = 2.85E-04

m'' = 7.73

K'x = 1.52E-03

(1-y)LM* = 0.475

(1-x)LM* = 0.826

Fourth trial:(1-y)LM = 0.234

(1-x)LM = 0.831slope =

-k'x/k'y =-0.306

xi,yi = 0.277 0.730

Third trial:(1-y)LM = 0.240

(1-x)LM = 0.832slope =

-k'x/k'y =-0.314

xi,yi = 0.277 0.729

% gas resistance = 9.62% liquid resistance = 96.19

Summary of the results

17

Example 3

In a dilute concentration region, equilibrium data for SO2 distributed between air and water

can be approximated by PA = 25 xA, where the partial pressure of SO2 in the vapor is expressed in atmospheres. For as absorption column operating at 10 atm, the bulk vapor and

liquid concentrations at one point in the column are yA = 0.01 and xA = 0.0. The mass

transfer coefficients for this process are, k’x = 10 and k’y = 8 [kgmol/(s.m2.mf)].

Assuming equimolar counter diffusion, (a) find K’x. (b) determine the interfacial

compositions, xAi and yAi, and (c) calculate the molar flux, NA.

Solution

2525 2.5

10A A A A AP x y x x

Since the diffusion is Equimolar counter diffusion, therefore no need for a trial and error method

18

Example 3

0.000 0.002 0.004 0.0060.000

0.002

0.004

0.006

0.008

0.010

0.012

0.014

f(x) = 0.00319763262539134 ln(x) + 0.0274909442175848

x

y

P

/ / / / /

1 1 1

x y xK m k k

/

/ / /

1 1

y y x

m

K k k

m' = m'' =

2.5

K'y = 2.67

K'x = 6.667

NA = 0.0267 kgmol/(s. m2)

/A y iN k y y

0.00667

0.00267Summary of the results