CP302 MassTransfer 04 OK

Prof. R. Shanthini 06 Oct 2011

1

Course content of Mass transfer section

L T A

Diffusion Theory of interface mass transfer Mass transfer coefficients, overall coefficients and transfer units

04 01 03

Application of absorption, extraction and adsorptionConcept of continuous contacting equipment

04 01 04

Simultaneous heat and mass transfer in gas-liquid contacting, and solids drying

04 01 03

CP302 Separation Process PrinciplesMass Transfer - Set 4


2

CP302 Separation Process Principles

Reference books used for ppts

1. C.J. Geankoplis Transport Processes and Separation Process Principles4th edition, Prentice-Hall India

2. J.D. Seader and E.J. HenleySeparation Process Principles2nd edition, John Wiley & Sons, Inc.

3. J.M. Coulson and J.F. RichardsonChemical Engineering, Volume 15th edition, Butterworth-Heinemann


3

Microscopic (or Fick’s Law) approach:

Macroscopic (or mass transfer coefficient) approach:

NA = - k ΔCA

where k is known as the mass transfer coefficient

(1)JA = - DAB

dCA

dz

(50)

good for diffusion dominated problems

good for convection dominated problems


4

NA

Mass Transfer Coefficient Approach

kc ΔCA=

NA

CA1

CA2

A & B

kc (CA1 – CA2 )= (51)

kc is the liquid-phase mass-transfer coefficient based on a concentration driving force.

What is the unit of kc?


5

Mass Transfer Coefficient Approach

CA1 = pA1 / RT; CA2 = pA2 / RT

Using the following relationships between concentrations and partial pressures:

Equation (51) can be written as

NA kc(pA1 – pA2) / RT= (52)kp

(pA1 – pA2)=

where kp = kc / RT (53)

kp is a gas-phase mass-transfer coefficient based on a partial-pressure driving force.

What is the unit of kp?

NA kc ΔCA= kc (CA1 – CA2 )= (51)


6

Mass transfer between phases across the following interfaces are of great interest in separation processes:

- gas/liquid interface

- liquid/liquid interface

Such interfaces are found in the following separation processes:

- absorption

- distillation

- extraction

- stripping

Models for mass transfer between phases


7

Models for mass transfer at a fluid-fluid interface

Theoretical models used to describe mass transfer between a fluid and such an interface:

- Film Theory

- Penetration Theory

- Surface-Renewal Theory

- Film Penetration Theory


8

Film Theory

Entire resistance to mass transfer in a given turbulent phase is in a thin, stagnant region of that phase at the interface, called a film.

Bulk liquid

Liquid film

Gas

pA

z=0 z=δL

CAi

CAb

For the system shown, gas is taken as pure component A, which diffuses into nonvolatile liquid B.

Mass transport

In reality, there may be mass transfer resistances in both liquid and gas phases. So we need to add a gas film in which gas is stagnant.


9

Two Film Theory

Liquid phase

Liquid film

Gas phase

pAb

CAi

CAb

pAi

Gas film

Mass transport

There are two stagnant films (on either side of the fluid-fluid interface).

Each film presents a resistance to mass transfer.

Concentrations in the two fluid at the interface are assumed to be in phase equilibrium.


10

Liquid phase

Gas phase

pAb

CAi

CAb

pAi

Mass transport

Concentration gradients for the film theory

More realistic concentration gradients

Liquid phase

Liquid film

Gas phase

pAb

CAi

CAb

pAi

Gas film

Mass transport

Interface Interface

Two Film Theory


11

Liquid phase

Liquid film

Gas phase

pAb

CAi

CAb

pAi

Gas film

Mass transport, NA

Two Film Theory applied at steady-state

NA kc (CAi – CAb )= (51)

NA (52)kp(pAb – pAi)=

Mass transfer in the gas phase:

Mass transfer in the liquid phase:

Phase equilibrium is assumed at the gas-liquid interface.

Applying Henry’s law,

pAi = HA CAi (53)


12

Liquid film

pAb

CAb

Gas film

Henry’s Law

pAi = HA CAi at equilibrium,

where HA is Henry’s constant for A

Unit of H:

[Pressure]/[concentration] = [ bar / (kg.m3) ]

CAi

pAi

Note that pAi is the gas phase pressure and CAi is the liquid phase concentration.


13

Liquid phase

Liquid film

Gas phase

pAb

CAi

CAb

pAi

Gas film

Mass transport, NA

Two Film Theory applied at steady-state

We know the bulk concentration and partial pressure.

We do not know the interface concentration and partial pressure.

Therefore, we eliminate pAi and CAi from (51), (52) and (53) by combining them appropriately.


14

Two Film Theory applied at steady-stateNA (54)From (52):

Substituting the above in (53) and rearranging:

(56)

pAi = pAb - kp

NA (55)From (51): CAi = CAb + kc

NA =pAb - HA CAb

HA / kc + 1 / kp

The above expression is based on gas-phase and liquid-phase mass transfer coefficients.

Let us now introduce overall gas-phase and overall liquid-phase mass transfer coefficients.


15

Introducing overall gas-phase mass transfer coefficient:Let’s start from (56).

Introduce the following imaginary gas-phase partial pressure:

pA* ≡ HA CAb

where pA* is a partial pressure that would have been in

equilibrium with the concentration of A in the bulk liquid.

Introduce an overall gas-phase mass-transfer coefficient (KG) as

(57)

(58)1

kp

HA

kc

≡ +1

KG

Combining (56), (57) and (58):

NA = KG (pAb - pA* ) (59)


16

Introducing overall liquid-phase mass transfer coefficient:Once again, let’s start from (56).

Introduce the following imaginary liquid-phase concentration:

pAb ≡ HA CA

*

where CA* is a concentration that would have been in

equilibrium with the partial pressure of A in the bulk gas.

Introduce an overall liquid-phase mass-transfer coefficient (KL) as

(60)

(61)1

HAkp

1

kc

≡ +1

KL

Combining (56), (60) and (61):

NA = KL (CA* - CAb) (62)


17CAi

pAi

pAb

CAb

pA*

CA* CA

pA

pAi = HA CAi

pA* = HA CAb

pAb = HACA*

Gas-Liquid Equilibrium Partitioning Curve showing the locations of p*

A and C*A


18

Summary:

where

NA = KL (CA* - CAb)

= KG (pAb - pA*)

1

kp

HA

kc

+1

KG

pA* = HA CAb

CA* = pAb

/ HA

KL

= =HA

(62)

(59)

(60)

(57)

(58 and 61)


19

Example 3.20 from Ref. 2 (modified)

Sulfur dioxide (A) is absorbed into water in a packed column. At a certain location, the bulk conditions are 50oC, 2 atm, yAb = 0.085, and xAb = 0.001. Equilibrium data for SO2 between air and water at 50oC are the following:

pA (atm) 0.0382 0.0606 0.1092 0.1700

CA (kmol/m3) 0.03126 0.04697 0.07823 0.10949

Experimental values of the mass transfer coefficients are kc = 0.18 m/h and kp = 0.040 kmol/h.m2.kPa.

Compute the mass-transfer flux by assuming an average Henry’s law constant and a negligible bulk flow.


20

Solution:

T = 273oC + 50oC = 323 K; PT = 2 atm;

yAb = 0.085; xAb = 0.001;

kc = 0.18 m/h;kp = 0.040 kmol/h.m2.kPa

y = 1.4652x

R2 = 0.9759

0

0.04

0.08

0.12

0.16

0.2

0.02 0.04 0.06 0.08 0.1 0.12

CA (kmol/m3)

pA (

atm

)

HA = 1.4652 atm.m3/kmol

Data provided:

slope of the curve

HA = 161.61 kPa.m3/kmol


21

where

NA = KL (CA* - CAb)

= KG (pAb - pA*)

1

kp

HA

kc

+1

KG

pA* = HA CAb

CA* = pAb

/ HA

KL

= =HA

(62)

(59)

(60)

(57)

(58 and 61)

Equations to be used:


22

1

kp

HA

kc

+1

KG KL

= =HA (58 and 61)

1

kp

=0.040

1h.m2.kPa/kmol = 25 h.m2.kPa/kmol

HA

kc

=0.18 m/h

161.61 kPa.m3/kmol= 897 h.m2.kPa/kmol

KG = 1/(25 + 897) = 1/922 = 0.001085 kmol/h.m2.kPa

KL = HA KG = 161.61/922 = 0.175 m/h

Calculation of overall mass transfer coefficients:


23

NA = KL (CA* - CAb)

CA* = pAb

/ HA = yAb PT / HA

(62) is used to calculate NA

= 0.085 x 2 atm / 1.4652 atm.m3/kmol = 0.1160 kmol/m3

CAb = xAb

CT = 0.001 CT

CT = concentration of water (assumed) = 1000 kg/m3 = 1000/18 kmol/m3 = 55.56 kmol/m3

CAb = 0.001 x 55.56 kmol/m3 = 0.05556 kmol/m3

NA = (0.175 m/h) (0.1160 - 0.05556) kmol/m3

= 0.01058 kmol/m2.h


24

NA = KG (pAb - pA*)

pA* = CAb

HA = xAb CT HA

(59) is used to calculate NA

= 0.001 x 55.56 x 161.61 kPa = 8.978 kPa

pAb = yAb

PT = 0.085 x 2 x 1.013 x 100 kPa = 17.221 kPa

NA = (1/922 h.m2.kPa/kmol) (17.221 - 8.978) kPa

= 0.00894 kmol/m2.h

Alternatively,

CP302 MassTransfer 04 OK

Documents

Transcript of CP302 MassTransfer 04 OK