7.2‐1

Correlations of Mass Transfer CoefficientsMass 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:

YX

DdvC

Dkd

YX ScReCSh or1

7.2‐2

Dimensionless numbers Table 8.3-1 from Cussler, 3rd ed.

Equivalent in heat transfer

lh

transfer heat Conductivetransfer heat ConvectiveNuSh

Pr is the equivalent to Scin heat transfer

h: convective heat transfer coefficientλ: thermal conductivity

α: thermal diffusivity

ydiffusivit thermalmomentum of ydiffusivit

2

7.2‐3

Table 8.3.-2 from Cussler

Selected mass transfer correlations for fluid-fluid interfacesa

3

7.2‐4Table 8.3.-3 from Cussler

Selected mass transfer coefficient correlations for fluid-solid interfacesa

4

7.2‐5

MTC CorrelationsExcellent 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

MTC error at best 30%The MTC is expressed mostly as

Sherwood-# or a Stanton-# Dkl

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-#.

Fluid-Solid interface

The error is about 10% and as low as 1% e.g., 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. 5

7.2‐6

Example: Dissolution rate of a spinning disk

Remember from “Generalized Mass Balances” : “A solvent flow approaches a spinning disk made out of a sparingly soluble solute. Calculate the diffusion-controlled rate at which the disk slowly dissolves at steady state.”

)sat(cD62.0zcDj 16/1

2/13/20z

10z1

→ The diffusion flux is:

)sat(cScRedD62.0j

)sat(cD

ddD62.0j

13/12/1

1

1

3/12/12

1

6

7.2‐7

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?

-5 2BA/W

2BA/A

D =10 cm /sD =0.233cm /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

7.2‐8

1 1N kc (sat)From Table 8.3.-3:1/2 1/3

k 0.62DD

a) For water

1/31/22 25

2 5 2

cm 20 / 60 2 / s 0.01cm / sk 0.62 10s 0.01cm / s 10 cm / s

30.9 10 cm / s

-3 3

1N 0.9 10 cm / s 0.003g / cm

b) For air1/31/22 2

A 2 2cm 20 / 60 2 / s 0.15cm / sk 0.62 0.233

s 0.15cm / s 0.233cm / s

0.47cm / s

1 3 30.3mmHg 1mol 273 122gN 0.47cm / s760mmHg 22.4 10 cm 298 mol

6 20.9 10 g / cm s

The flux in air is 1/3 of that in water even though the k in air is 500 times that in water.

-6 22.7 10 g / cm s

and

8

7.2‐9

Mass Transfer across Interfaces (very important) 

Bulk interface Bulk

Basic equation:1 1

N K c where K is the overall MTC and c1 an APPROPRIATE concentration difference and most notably the equilibrium or asymptotic concentration.Case a: Heat transferHot 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.

Often we encounter the following scenario:

No problem!9

7.2‐10

Case b: Bromine extraction

Now the c1 here should be: c1= c1(in benzene) - Hc1(in water) Otherwise the initial c1 is zero and we still have flux.

The partition coefficient H is:

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 Brconcentration in C6H6 is higher than in H2O. Why?

at equilibrium

concentration of Br in benzeneH=concentration of Br in water

Bromine concentration:

→ Bromine is more soluble in C6H6 than in H2O.

10

7.2‐11

Case c: Bromine vaporizationInitially 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.

11

7.2‐12

7.4 The Overall MTCThe flux in the gas is:

1 p 10 1iN k (p p )

Because the interfacial region is thin, it is at steady state. Thus, the flux will be equal to that in the liquid.

1 L 1i 10N k (c c ) (2)

(1)

where kP and kL are the gas and liquid MTC’s!

So, 10i1Li110P cckppk 12

7.2‐13

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:

So the flux N1 from equation (5) should be derived as:

1 10 10

p L

1N (p Hc )1/ k H / k

LpP kHk1

1K

is the “overall gas-side MTC”

(3)H is Henry’s constant or the partition coefficient in the simplest case

(4)

Hpc i1

i1 LP

10L10Pi1 kHk

ckpkc

13

7.2‐14

Analogy with electric circuits:

Many times the engineer’s job is to determine which is the rate limiting resistance: in the gas or the liquid ??

LP

10101 kHk1

HcpN

“Current”“Voltage difference”

“2 resistances in series”

14

7.2‐15

Now we can write the flux equation in two ways:

A)

1 L 1 10N K (c * c )

where

LL p

1K1/ k 1/ k H

and

1 10c * p / H

KL is called the “overall liquid-side mass transfer coefficient” and c1* is the hypothetical liquid concentration in equilibrium with the bulk gas concentration.

B)

1 p 10 1N K (p p *)

where

p

p L

1K1/ k H / k

1 10p * Hc

and

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.

15

7.2‐16

Example : Oxygen Mass TransferEstimate 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.5 2

3L

L

D 2.1 10 cm / sk 2.1 10 cm / s0.01cm 0.01cm

Finding kP and H is more difficult for unit conversion

G G

p

k DkRT (0.01cm)RT

2

4 2

3

0.23cm / s 9.4 10 mol / (cm s atm)(0.01cm)(82cm atm / (mol K))(298K)

16

7.2‐17

From the way H’ is given (unit consistency)

Insert these values into the equation for KL

LL p

1K1/ k 1/ k H

3 4 2 5 3

11/ (2.1 10 cm / s) 1/ (9.4 10 mol / cm s atm 7.9 10 cm atm / mol)

32.1 10 cm / s

The mass transfer is dominated by the liquid-side resistance!!

mol/cmatm 109.7cm 18 /mol 1atm 104.4

c`HH 35

3

4

17

7.2‐18

Example : Perfume ExtractionJasmone (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

1 W 10 W 1iW B 1iB 10BN = k (c - c ) = k (c - c ) (5)

The interfacial concentration is in equilibrium, so:

(6)iW1iB1 cHc

18

7.2‐19

Eliminate the interfacial concentration using equations 5 and 6:

10 1 1 10W W W iW B iW B Bk c k c k Hc k c

10 10 1W W B B W B iWk c k c k k H c ( ) 10 101

1

W W B BiB

iW

W B

k c k cccH k k H

(7)

Replace this in the N1 for benzene

10 101 10

W W B BB B

W B

k c k cN k H ck k H

-

W 10 W B 10B W 10B B 10B B W

B 10 W 10B

W B W B

10 W 10B

B W

Hk c Hk c k c Hk c k kk (Hc c )k k H k k H

1 (Hc c )1/ k H / k

→

19

7.2‐20

4 3

1K '1/ (3.0 10 cm / s) 170 / (2.4 10 cm / s)

51.3 10 cm / s

Again the mass transfer in water controls the process because jasmone is more soluble in C6H6.

The overall MTC, K´ is

20