Correlations of Mass Transfer Coefficients
Mass transfer coefficients (MTCs) are not physical properties like the
diffusion coefficient. They differ from case to case and even within a
system, depending on their definition.
With the help of experimental observations, correlations for mass
transfer coefficients have been developed for standard cases
(e.g. fluid flow through a packed bed of particles, gas bubbles rising
in a tank, falling films, flow over surfaces and within tubes, …).
Such MTC correlations are typically expressed with dimensionless
numbers, frequently in the following form:
X
kd
 dv    
 C   
D
   D
Y
or
Sh  C  Re   Sc 
X
Y
7.2-1
Mass Transfer – Mass Transfer Coefficients II
Sh  Nu 
Convective heat transfer h  l

Conductive heat transfer

Equivalent in heat transfer
h: convective heat transfer coefficient
λ: thermal conductivity
diffusivity of momentum
thermal diffusivity
Pr is the equivalent to Sc
in heat transfer
α: thermal diffusivity
Dimensionless numbers
Table 8.3-1 from Cussler, 3rd ed.
Mass Transfer – Mass Transfer Coefficients II
7.2-2
Selected mass transfer correlations for fluid-fluid interfacesa
Table 8.3.-2 from Cussler
7.2-3
Mass Transfer – Mass Transfer Coefficients II
Selected mass transfer correlations for fluid-solid interfacesa
Mass Transfer – Mass Transfer Coefficients II
Table 8.3.-3 from Cussler 7.2-4
MTC Correlations
Excellent for preliminary design of small pilot plants. For design of full scale
equipment you must supplement them with data of the SPECIFIC chemical system.
Fluid-Fluid interface
Fluid-Solid interface
MTC error at best 30%
The MTC is expressed mostly as
The error is about 10% and as low as
1% especially when dealing with laminar
flow in a single tube. This high accuracy
is attributed to the heat transfer origin of
these correlations and the fact that
SIMPLER geometries are involved. Also
heat transfer is an older subject than
mass transfer. For example, laminar
flow of one fluid in a tube is much better
understood than turbulent flow of gas
and liquid in a packed tower!
Again the MTC is written in the Sh- or
St-notation.
kl
Sherwood-# 
D
or a Stanton-# 
k

The Sh is typically expressed by powers of
Re and Sc corresponding to “convection”
and “diffusion”, respectively.
When the convection is not a typical
”forced” one but one generated by density
gradients, it is “free convection” and the
Re is replaced by the Grashof-#.
7.2-5
Mass Transfer – Mass Transfer Coefficients II
Example: Dissolution rate of a spinning disk
Remember from Chapter 4, “Generalized Mass Balances”, Ex. 4.2.3:
“A solvent flow approaches a spinning
disk made out of a sparingly soluble
solute. Calculate the diffusioncontrolled rate at which the disk
slowly dissolves at steady state.”
→ The diffusion flux is:
c
j1 z0  D 1
z
z 0
 D 2 / 3 1/ 2 
 c (sat)
 0.62 
 1/ 6  1


1/ 2
D  d2  

 j1  0.62 
d   
1/ 3

 
D
c1(sat)
D
 j1  0.62 Re1/ 2 Sc1/ 3c1(sat)
d
Mass Transfer – Mass Transfer Coefficients II
7.2-6
Now:
A solid disc of benzoic acid (BA) 2.5 cm in diameter is spinning at
20 rpm and 25°C. How fast will it dissolve in a large volume of a)
water and b) air?
DBA/W =10-5cm2 /s
DBA/A =0.233 cm2 /s
Solubility of benzoic acid in water is 0.003 g/cm3.
Equilibrium vapor pressure of benzoic acid in air is 0.3 mmHg at 25°C.
The molecular weight of BA is 122 g/mol.
Will the mass transfer be faster in air or in water?
7.2-7
Mass Transfer – Mass Transfer Coefficients II
1/2
1/3
    
From Table 8.3.-3: k  0.62D    
   D
a) For water
1/2
and
1/3
cm2  20 / 60  2 / s   0.01cm2 / s 
k  0.62  10
s  0.01cm2 / s   10 5 cm2 / s 
 0.9  10 3 cm / s
5
N  0.9  10 cm / s  0.003g / cm
-3
 2.7  10-6 g / cm2 s
3
1
b) For air
N1  kc1(sat)
1/2
1/3
cm2  20 / 60  2 / s   0.15cm2 / s 
k A  0.62  0.233


s  0.15cm2 / s   0.233cm2 / s 
 0.47cm / s
 0.3mmHg
1mol
273 122g 
6
2

0.9

10
g
/
cm
s
N1  0.47cm / s  

3
3
760mmHg
22.4

10
cm
298
mol


The flux in air is 1/3 of that in water even though the k in air is 500
times that in water.
Mass Transfer – Mass Transfer Coefficients II
7.2-8
7.3 Mass Transfer across Interfaces (very important)
Often we encounter the following scenario: Bulk  interface  Bulk
Basic equation:
N  K c
1
1
where K is the overall MTC and c1 an APPROPRIATE
concentration difference and most notably the equilibrium or
asymptotic concentration.
Case a: Heat transfer
Hot benzene on cold water.
The benzene cools while the water
warms until they reach the same T.
Equal T is the criterion for equilibrium.
The amount of energy transferred is
always equal to the T.
No problem!
7.2-9
Mass Transfer – Mass Transfer Coefficients II
Case b: Bromine extraction
A benzene solution of bromine is placed
on top of a water solution containing the
SAME concentration of bromine.
Later the initially equal concentrations
have CHANGED and the Br
concentration in C6H6 is higher than in
H2O. Why?
Bromine concentration:
→ Bromine is more soluble in C6H6 than in H2O.
Now the c1 here should be: c1= c1(in benzene) - Hc1(in water)
Otherwise the initial c1 is zero and we still have flux.
 concentration of Br in benzene 
The partition coefficient H is: H= 

 concentration of Br in water at equilibrium
Mass Transfer – Mass Transfer Coefficients II
7.2-10
Case c: Bromine vaporization
Initially Br evaporates from water
into air. Initially the Br concentration
in water is higher than that in air; at
the end it is lower.
This might be a problem of units: Concentrations in the liquid are
expressed in mol/L and those in air by the partial pressure ??
Mass transfer should be described in terms of the more
fundamental chemical potentials. If this was done, the
concentration difference would disappear.
7.2-11
Mass Transfer – Mass Transfer Coefficients II
7.4 The Overall MTC
The flux in the gas is:
N  k (p  p )
1
p
10
1i
(7.4)
Because the interfacial region
is thin, it is at steady state.
Thus, the flux will be equal to
that in the liquid.
N  k (c  c )
1
L
1i
10
(7.5)
where kP and kL are the gas and
liquid MTC’s!
So, kP p10  p1i   kL c1i  c10 
Mass Transfer – Mass Transfer Coefficients II
7.2-12
Always we must remove the dependency on the interfacial
concentration or partial pressures, as these are difficult to determine.
Usually there is equilibrium at the interface:
p1i
kP  p10  kL  c10 H is Henry’s constant or
 c1i 
the partition coefficient (7.6)
c1i 
kP  H  kL
H
in the simplest case
So the flux N1 from equation (7.5) should be derived as:
1
N 
(p  Hc )
1/ k  H / k
1
10
p
KP 
1
1 kp  H kL
(7.7)
10
L
is the “overall gas-side MTC”
7.2-13
Mass Transfer – Mass Transfer Coefficients II
Analogy with electric circuits:
“Voltage difference”
“Current”
N1

p10  Hc10 

1 kP  H kL
“2 resistances in series”
Many times the engineer’s job is to determine which is the rate
limiting resistance: in the gas or the liquid ??
Mass Transfer – Mass Transfer Coefficients II
7.2-14
Now we can write the flux equation in two ways:
A) N  K (p  p *) where K p 
1
p
10
1
1
1/ k  H / k
p
and
p *  Hc
1
10
L
KP is the “overall gas-side mass transfer coefficient” and p1* is the
hypothetical gas-phase concentration that would be in equilibrium
with the bulk liquid concentration.
1
B) N  K (c * c ) where K L 
and c *  p / H
1/ k L  1/ k pH
1
L
1
1
10
10
KL is called the “overall liquid-side mass transfer coefficient” and
c1* is the hypothetical liquid concentration in equilibrium with the
bulk gas concentration.
7.2-15
Mass Transfer – Mass Transfer Coefficients II
Example 7.4.1: Oxygen Mass Transfer
Estimate the overall liquid-side MTC for O2 transfer from water into
air assuming that each MTC is k = D/0.01 cm, Henry‘s law constant
is H = 4.4×104 atm, Dair = 0.23 cm2/s and Dwater = 2.1×10-5 cm2/s.
Goal: Calculate kL and kP and substitute in the appropriate equation.
D
2.1 10 cm / s

 2.1 10 cm / s
k 
0.01cm
0.01cm
5
2
3
L
L
Finding kP and H is more difficult for unit conversion
k 
p
k
D

RT (0.01cm)RT
G
G
0.23cm / s

 9.4  10 mol / (cm  s  atm)
(0.01cm)(82cm  atm / (mol  K))(298K)
2
4
2
3
Mass Transfer – Mass Transfer Coefficients II
7.2-16
From the way H’ is given (unit consistency)
H` 4.4  104 atm
5
3
H


7
.
9

10
atm

cm
/ mol
3
c 1 mol / 18 cm
Insert these values into the equation for KL
KL 

1
1/ k L  1/ k p H
1
1/ (2.1 10 cm / s)  1/ (9.4  10 mol / cm s  atm  7.9  10 cm  atm / mol)
3
4
2
5
3
 2.1 10 cm / s
3
The mass transfer is dominated by the liquid-side resistance!!
7.2-17
Mass Transfer – Mass Transfer Coefficients II
Example 7.4.2: Perfume Extraction
Jasmone (C11H16O) is a valuable aroma from jasmine flowers that
is used in soaps and cosmetics. We are recovering this from its
water solution (jasmine flowers in water) with benzene drops the kB
=3.0 x 10-4 cm/s while kW = 2.4x10-3 cm/s
However, jasmone is 170 times more soluble in C6H6 than in H2O.
What is the overall MTC?
Assuming steady state
N1 = k W (c 10 W - c 1iW ) = k B (c 1iB - c 10B )
(7.8)
The interfacial concentration is in equilibrium, so:
c1iB  H  c1iW
(7.9)
Mass Transfer – Mass Transfer Coefficients II
7.2-18
Eliminate the interfacial concentration using equation (7.8) and (7.9):
k c
W
10 W
k c
W
10 W
k c
W
1 iW
k c
B
10 B
 k Hc  k c
B
1 iW
B
 ( k  k H )c
W
B
10 B
→ c 1iW
1 iW
k c k c
c


H
k k H
W
1 iB
10 W
W
B
(7.10)
10 B
B
Replace this in the N1 for benzene
 k W c10 W  k Bc10B

- c10B 
N1  k B  H
k

k
H


W
B
 Hk c
k 

W
B
10 W
 Hk c  k c
k k H
B
10 B
W


1

 (Hc

1/
k
H
/
k


B
W
10 B
B
10 W
 Hk c
B
10 B

kk
(Hc


 k k H
B
W
W
10 W
c )
10 B
B
c )
10 B
W
Mass Transfer – Mass Transfer Coefficients II
7.2-19
The overall MTC, K´ is
K' 
1
1/ (3.0  10 4 cm / s)  170 / (2.4  10 3 cm / s)
 1.3  10 5 cm / s
Again the mass transfer in water controls the process because
jasmone is more soluble in C6H6.
Mass Transfer – Mass Transfer Coefficients II
7.2-20