CP302 Separation Process Principles
Mass Transfer - Set 4
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
adsorption
Concept of continuous contacting equipment
04
01
04
Simultaneous heat and mass transfer in gasliquid contacting, and solids drying
04
01
03
Prof. R. Shanthini
06 Oct 2011
1
CP302 Separation Process Principles
Reference books used for ppts
1. C.J. Geankoplis
Transport Processes and Separation Process Principles
4th edition, Prentice-Hall India
2. J.D. Seader and E.J. Henley
Separation Process Principles
2nd edition, John Wiley & Sons, Inc.
3. J.M. Coulson and J.F. Richardson
Chemical Engineering, Volume 1
5th edition, Butterworth-Heinemann
Prof. R. Shanthini
06 Oct 2011
2
Microscopic (or Fick’s Law) approach:
JA = - DAB
dCA
dz
(1)
good for diffusion dominated problems
Macroscopic (or mass transfer coefficient) approach:
NA = - k ΔCA
(50)
where k is known as the mass transfer coefficient
good for convection dominated problems
Prof. R. Shanthini
06 Oct 2011
3
Mass Transfer Coefficient Approach
NA = kc ΔCA
= kc (CA1 – CA2 )
(51)
kc is the liquid-phase mass-transfer coefficient
based on a concentration driving force.
CA1
A&B
What is the unit of kc?
NA
Prof. R. Shanthini
06 Oct 2011
CA2
4
Mass Transfer Coefficient Approach
= kc (CA1 – CA2 )
NA = kc ΔCA
(51)
Using the following relationships between concentrations and
partial pressures:
CA1 = pA1 / RT;
CA2 = pA2 / RT
Equation (51) can be written as
NA = kc (pA1 – pA2) / RT
where kp = kc / RT
= kp (pA1 – pA2)
(52)
(53)
kp is a gas-phase mass-transfer coefficient
based on a partial-pressure driving force.
Prof. R. Shanthini
06 Oct 2011
What is the unit of kp?
5
Models for mass transfer between phases
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
Prof. R. Shanthini
06 Oct 2011
6
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
Prof. R. Shanthini
06 Oct 2011
7
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.
For the system shown, gas is
taken as pure component A,
which diffuses into nonvolatile
liquid B.
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.
Prof. R. Shanthini
06 Oct 2011
Liquid
film
pA
Bulk
liquid
CAi
Gas
CAb
z=0
z=δL
Mass transport
8
Two Film Theory
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.
Liquid
film
Gas
film
Gas
phase
pAb
Liquid
phase
pAi
CAi
CAb
Prof. R. Shanthini
06 Oct 2011
Mass transport
9
Two Film Theory
Gas
film
Interface
Gas
phase
pAb
Interface
Liquid
film
Liquid
phase
Gas
phase
pAb
Liquid
phase
pAi
pAi
CAi
CAi
CAb
Mass transport
Concentration gradients for
the film theory
Prof. R. Shanthini
06 Oct 2011
CAb
Mass transport
More realistic concentration
gradients
10
Two Film Theory applied at steady-state
Mass transfer in the gas phase:
NA = kp (pAb – pAi)
(52)
Mass transfer in the liquid phase:
NA = kc (CAi – CAb )
(51)
Phase equilibrium is assumed at
the gas-liquid interface.
Applying Henry’s law,
pAi = HA CAi
Prof. R. Shanthini
06 Oct 2011
Liquid
film
Gas
film
Gas
phase
pAb
Liquid
phase
pAi
CAi
CAb
(53)
Mass transport, NA 11
Henry’s Law
pAi = HA CAi at equilibrium,
where HA is Henry’s constant for A
Note that pAi is the gas phase pressure
and CAi is the liquid phase concentration.
Liquid
film
Gas
film
pAb
Unit of H:
pAi
CAi
[Pressure]/[concentration] = [ bar / (kg.m3) ]
Prof. R. Shanthini
06 Oct 2011
CAb
12
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.
Liquid
film
Gas
film
Gas
phase
pAb
Liquid
phase
pAi
CAi
CAb
Prof. R. Shanthini
06 Oct 2011
Mass transport, NA 13
Two Film Theory applied at steady-state
From (52):
pAi = pAb - NA
kp
(54)
From (51):
CAi = CAb + NA
kc
(55)
Substituting the above in (53) and rearranging:
NA =
pAb - HA CAb
(56)
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.
Prof. R. Shanthini
06 Oct 2011
14
Introducing overall gas-phase mass transfer
coefficient:
Let’s start from (56).
Introduce the following imaginary gas-phase partial pressure:
pA* ≡ HA CAb
(57)
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
1
KG
≡
1
kp
+
HA
kc
(58)
Combining (56), (57) and (58):
Prof. R. Shanthini
06 Oct 2011
NA = KG (pAb - pA* )
(59)
15
Introducing overall liquid-phase mass transfer
coefficient:
Once again, let’s start from (56).
Introduce the following imaginary liquid-phase concentration:
pAb ≡ HA CA*
(60)
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
1
KL
≡
1
+
HAkp
1
kc
(61)
Combining (56), (60) and (61):
Prof. R. Shanthini
06 Oct 2011
NA = KL (CA* - CAb)
(62)
16
Gas-Liquid Equilibrium Partitioning Curve showing the
locations of p*A and C*A
pA
pAb
pAb = HACA*
pAi
p A*
Prof. R. Shanthini
06 Oct 2011
pAi = HA CAi
pA* = HA CAb
CAb
CAi
CA*
CA
17
Summary:
NA = KL (CA* - CAb)
(62)
= KG (pAb - pA*)
(59)
CA* = pAb / HA
(60)
pA* = HA CAb
(57)
where
1
KG
=
HA
KL
Prof. R. Shanthini
06 Oct 2011
=
1
kp
+
HA
kc
(58 and 61)
18
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
CA (kmol/m3)
0.03126
0.04697 0.07823
0.1700
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.
Prof. R. Shanthini
06 Oct 2011
19
Solution:
Data provided:
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
0.2
HA = 1.4652 atm.m3/kmol
slope of
the curve
HA = 161.61
kPa.m3/kmol
pA (atm)
0.16
0.12
0.08
0.04
0
0.02
Prof. R. Shanthini
06 Oct 2011
y = 1.4652x
R2 = 0.9759
0.04
0.06
0.08
CA (kmol/m3)
0.1
20
0.12
Equations to be used:
NA = KL (CA* - CAb)
(62)
= KG (pAb - pA*)
(59)
CA* = pAb / HA
(60)
pA* = HA CAb
(57)
where
1
KG
=
HA
KL
Prof. R. Shanthini
06 Oct 2011
=
1
kp
+
HA
kc
(58 and 61)
21
Calculation of overall mass transfer coefficients:
1
KG
=
1
kp
1
=
h.m2.kPa/kmol
0.040
HA
161.61 kPa.m3/kmol
kc
=
HA
=
KL
1
+
kp
0.18 m/h
HA
kc
(58 and 61)
= 25 h.m2.kPa/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
Prof. R. Shanthini
06 Oct 2011
22
NA = KL (CA* - CAb)
(62) is used to calculate NA
CA* = pAb / HA = yAb PT / HA
= 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
Prof. R. Shanthini
06 Oct 2011
23
Alternatively,
NA = KG (pAb - pA*)
(59) is used to calculate NA
pA* = CAb HA = xAb CT HA
= 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
Prof. R. Shanthini
06 Oct 2011
24