1. PM3125: Lectures 1 to 5 Content: Mass transfer: concept and theory
uploaded at
Prof. R. Shanthini Feb & 05 Mar 2012

2. Reference books used for ppts
C.J. Geankoplis
Transport Processes and Separation Process Principles
4th edition, Prentice-Hall India
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 Feb & 05 Mar 2012

3. Modes of mass transfer
Mass transfer could occur by the following three ways:
Diffusion is caused by concentration gradient.
Advection is caused by moving fluid. (It cannot therefore happen in solids.)
Convection is the net transport caused by both diffusion and advection. (It occurs only in fluids.)
Diffusion
Advection
Convection
Prof. R. Shanthini Feb & 05 Mar 2012

4. Stirring the water with a spoon creates forced convection.
That helps the sugar molecules to transfer to the bulk water much faster.
Diffusion
(slower)
Convection
(faster)
Prof. R. Shanthini Feb & 05 Mar 2012

5. Diffusion Solvent B Solute A concentration of A concentration of A
is high
concentration of A
is low
Mass transfer by diffusion occurs when a component in a stationary solid or fluid goes from one point to another driven by a concentration gradient of the component.
Prof. R. Shanthini Feb & 05 Mar 2012

6. Air Blood Example of diffusion mass transfer
At the surface of the lung:
Air
Blood
Oxygen
High oxygen concentration
Low carbon dioxide concentration
Low oxygen concentration
High carbon dioxide concentration
Carbon dioxide
Prof. R. Shanthini Feb & 05 Mar 2012

7. dCA JA = - DAB dz JA (1) CA CA + dCA dz Fick’s First Law of Diffusion
for mass transfer in z-direction only CA
A & B JA
CA + dCA dz
Prof. R. Shanthini Feb & 05 Mar 2012

8. dCA JA = - DAB dz What is the unit of diffusivity?
concentration gradient of A
in z-direction
(mass/moles per volume per distance)
diffusion coefficient
(or diffusivity) of A in B
diffusion flux of A in relation to the bulk motion in z-direction
(mass/moles per area per time)
What is the unit of diffusivity?
Prof. R. Shanthini Feb & 05 Mar 2012

9. Unit and Scale of Diffusivity
For dissolved matter in water:
D ≈ 10-5 cm2/s
For gases in air at 1 atm and at room temperature:
D ≈ 0.1 to 0.01 cm2/s
Diffusivity depends on the type of solute, type of solvent, temperature, pressure, solution phase (gas, liquid or solid) and other characteristics.
Prof. R. Shanthini Feb & 05 Mar 2012

10. Example 6.1.1 from Ref. 1 dCA JA = - DAB dz
Molecular diffusion of Helium in Nitrogen: A mixture of He and N2 gas is contained in a pipe (0.2 m long) at 298 K and 1 atm total pressure which is constant throughout. The partial pressure of He is 0.60 atm at one end of the pipe, and it is 0.20 atm at the other end. Calculate the flux of He at steady state if DAB of He-N2 mixture is x 10-4 m2/s.
Solution:
Use Fick’s law of diffusion given by equation (1) as
JA = - DAB dz
dCA
Rearranging equation (1) and integrating gives the following:
Prof. R. Shanthini Feb & 05 Mar 2012

11. ⌠ ⌡ ⌠ ⌡ JA = - DAB dz dCA (z2 – z1) (CA2 – CA1) JA = - DAB z1 z2 CA1
(2)
At steady state, diffusion flux is constant.
Diffusivity is taken as constant.
Therefore, equation (2) gives
(z2 – z1)
(CA2 – CA1)
JA = - DAB
(3)
DAB is given as x 10-4 m2/s
(z2 – z1) is given as 0.2 m
(CA2 – CA1) = ?
Prof. R. Shanthini Feb & 05 Mar 2012

12. Combining the above we get CA = RT
Even though CA is not given at points 1 and 2, partial pressures are given. We could relate partial pressure to concentration as follows: nA
Number of moles of A
CA = V
Total volume
Absolute temperature
pA V = nA RT
Gas constant
Partial pressure of A pA
Combining the above we get
CA = RT
Prof. R. Shanthini Feb & 05 Mar 2012

13. (pA2 – pA1) (z2 – z1) JA = - DAB RT (pA2 – pA1) JA = - DAB RT(z2 – z1)
Equation (3) can therefore be written as
(pA2 – pA1)
(z2 – z1)
JA = - DAB RT
which gives the flux as
(pA2 – pA1)
JA = - DAB
RT(z2 – z1)
(0.6 – 0.2) x x 105 Pa
JA = - (0.687x10-4 m2/s)
(8314 J/kmol.K) x (298 K) x (0.20–0) m
JA = 5.63 x 10-6 kmol/m2.s
Prof. R. Shanthini Feb & 05 Mar 2012

14. Diffusion of gases A & B plus convection
Diffusion is the net transport of substances in a stationary solid or fluid under a concentration gradient.
Advection is the net transport of substances by the moving fluid, and so cannot happen in solids. It does not include transport of substances by simple diffusion.
Convection is the net transport of substances caused by both advective transport and diffusive transport in fluids.
JA is the diffusive flux described by Fick’s law, and we have already studied about it.
Let us use NA to denote the total flux by convection (which is diffusion plus advection.
Prof. R. Shanthini Feb & 05 Mar 2012

15. dCA JA = - DAB (1) dz Molar diffusive flux of A in B:
The velocity of the above diffusive flux of A in B can be given by
JA (mol/m2.s)
(4)
vA,diffusion (m/s) =
CA (mol/m3)
The velocity of the net flux of A in B can be given by
NA (mol/m2.s)
(5)
vA,convection (m/s) =
CA (mol/m3)
The velocity of the bulk motion can be given by
(NA + NB) (mol/m2.s)
(6)
vbulk (m/s) =
(CT) (mol/m3)
Prof. R. Shanthini Feb & 05 Mar 2012
Total concentration
Ignore the derivation, if you wish

16. Multiplying the above by CA, we get
vA,convection =
vA,diffusion +
vbulk
Multiplying the above by CA, we get
CA vA,convection =
CA vA,diffusion +
CA vbulk
Using equations (4) to (6) in the above, we get
(NA + NB)
NA =
JA + CA
(7) CT
Substituting JA from equation (1) in (7), we get
dCA
(NA + NB)
NA =
-DAB
(8)
+ CA dz CT
Prof. R. Shanthini Feb & 05 Mar 2012
Ignore the derivation, if you wish

17. Let us introduce partial pressure pA into (8) as follows:nA pA
(9a)
CA = = V RT nT P
(9b)
CT = = V RT
Total pressure
Total number of moles
Using (9a) and (9b), equation (8) can be written as
DAB
dpA pA
(NA + NB)
(10)
NA = RT dz P
Prof. R. Shanthini Feb & 05 Mar 2012
Ignore the derivation, if you wish

18. Let us introduce molar fractions xA into (8) as follows:NA CA
xA = =
(11)
(NA + NB) CT
Using (11), equation (8) can be written as
dxA
NA =
-CT DAB
(12)
+ xA
(NA + NB) dz
Prof. R. Shanthini Feb & 05 Mar 2012
Ignore the derivation, if you wish

19. Diffusion of gases A & B plus convection:
Summary equations for (one dimensional) flow in z direction
In terms of concentration of A:
dCA CA
NA =
-DAB
(NA + NB)
(8) + dz CT
Total concentration
In terms of partial pressures (using pA = CART and P = CTRT):
DAB
dpA pA
(NA + NB)
(10)
NA = dz P RT
Total pressure
In terms of molar fraction of A (using xA = CA /CT):
dxA
NA =
-CT DAB xA
(12) +
(NA + NB) dz NA CA
xA = =
Molar fraction
Prof. R. Shanthini Feb & 05 Mar 2012
(NA + NB) CT

20. A diffusing through stagnant, non-diffusing B
Evaporation of a pure liquid (A) is at the bottom of a narrow tube.
Large amount of inert or non-diffusing air (B) is passed over the top.
Vapour A diffuses through B in the tube.
The boundary at the liquid surface (at point 1) is impermeable to B, since B is insoluble in liquid A.
Hence, B cannot diffuse into or away from the surface.
Therefore, NB = 0
Air (B) 2
z2 – z1 1
Liquid
Benzene
(A)
Prof. R. Shanthini Feb & 05 Mar 2012

21. ⌠ ⌡ ⌠ ⌡ Substituting NB = 0 in equation (10), we get DAB dpA pA
(NA + 0)
NA = RT dz P
Rearranging and integrating
DAB
dpA
NA (1 - pA/P) = - RT dz ⌠ ⌡ z1 z2 ⌠ ⌡
pA1
pA2
DAB
dpA
NA dz = - RT
(1 - pA/P)
DAB P
P - pA2
NA = ln
(13)
RT(z2 – z1)
P – pA1
Prof. R. Shanthini Feb & 05 Mar 2012

22. Introduce the log mean value of inert B as follows:
(pB2 – pB1 )
(P – pA2 ) – (P – pA1 )
pB,LM = =
ln(pB2 /pB1 )
ln[(P - pA2 )/ (P - pA1 )]
(pA1 – pA2 ) =
ln[(P - pA2 )/ (P - pA1 )]
Equation (13) is therefore written as follows:
DAB P
NA =
(14)
(pA1 - pA2 )
RT(z2 – z1) pB,LM
Equation (14) is the most used form.
Prof. R. Shanthini Feb & 05 Mar 2012

23. Using xA = CA /CT, pA = CART and P = CTRT,
equation (13) can be converted to the following:
NA = ln
DAB CT
1 - xA2
(15)
(z2 – z1)
1 – xA1
Introduce the log mean value of inert B as follows:
(xB2 – xB1 )
(1 – xA2 ) – (1 – xA1 )
xB,LM = =
ln(xB2 /xB1 )
ln[(1 - xA2 )/ (1 - xA1 )]
(xA1 – xA2 ) =
ln[(1 - xA2 )/ (1 - xA1 )]
Therefore, equation (15) becomes the following:
DAB CT
NA = -
(xA1 - xA2 )
(16)
(z2 – z1) xB,LM
Prof. R. Shanthini Feb & 05 Mar 2012

24. Example from Ref. 1
Diffusion of water through stagnant, non-diffusing air:
Water in the bottom of a narrow metal tube is held at a constant temperature of 293 K. The total pressure of air (assumed to be dry) is 1 atm and the temperature is 293 K. Water evaporates and diffuses through the air in the tube, and the diffusion path is m long. Calculate the rate of evaporation at steady state. The diffusivity of water vapour at 1 atm and 293 K is x 10-4 m2/s. Assume that the vapour pressure of water at 293 K is atm.
Answer: x 10-7 kmol/m2.s
Prof. R. Shanthini Feb & 05 Mar 2012

25. Data provided are the following: DAB = 0.250 x 10-4 m2/s;
Solution: The set-up of Example is shown in the figure. Assuming steady state, equation (14) applies.
Air (B)
DAB P
NA =
(pA1 - pA2 )
RT(z2 – z1) pB,LM 2
(14)
where
z2 – z1
(pA1 – pA2 )
pB,LM =
ln[(P - pA2 )/ (P - pA1 )] 1
Data provided are the following:
DAB = x 10-4 m2/s;
P = 1 atm; T = 293 K;
z2 – z1 = m;
pA1 = atm (saturated vapour pressure);
pA2 = 0 atm (water vapour is carried away by air at point 2)
Water (A)
Prof. R. Shanthini Feb & 05 Mar 2012

26. (8314 J/kmol.K) (293 K) (0.1524 m) (0.988 atm)
Substituting the data provided in the equations given, we get the following:
( – 0 )
pB,LM =
= atm
ln[(1 - 0 )/ (1 – )]
(0.250x10-4 m2/s)(1x x105 Pa)
( ) atm
NA =
(8314 J/kmol.K) (293 K) ( m) (0.988 atm)
= x 10-7 kmol/m2.s
Prof. R. Shanthini Feb & 05 Mar 2012

27. Estimating Diffusivity
Diffusivities for different systems could be estimated using the empirical equations provided in the following slides as well as those provided in other reference texts available in the library and other sources.
Prof. R. Shanthini Feb & 05 Mar 2012

28. Diffusivity of gases An example at 1 atm and 298 K: System
Diffusivity (cm2/s)
H2-NH3
0.783
H2-CH4
0.726
Ar-CH4
0.202
He-CH4
0.675
He-N2
0.687
Air-H2O
0.260
Air-C2H5OH
0.135
Air-benzene
0.0962
Prof. R. Shanthini Feb & 05 Mar 2012

29. DAB is proportional to 1/P and T1.75
Binary Gas Diffusivity
DAB - diffusivity in cm2/s
P - absolute pressure in atm
Mi - molecular weight
T - temperature in K
Vi - sum of the diffusion volume for component i
DAB is proportional to 1/P and T1.75
Prof. R. Shanthini Feb & 05 Mar 2012

30. Binary Gas Diffusivity
Prof. R. Shanthini Feb & 05 Mar 2012

31. Diffusivity in Liquids
For very large spherical molecules (A) of 1000 molecular weight or greater diffusing in a liquid solvent (B) of small molecules:
9.96 x T
DAB =
applicable for biological solutes such as proteins
μ VA1/3
DAB - diffusivity in cm2/s
T - temperature in K
μ - viscosity of solution in kg/m s
VA - solute molar volume at its normal boiling point
in m3/kmol
DAB is proportional to 1/μ and T
Prof. R. Shanthini Feb & 05 Mar 2012

32. Diffusivity in Liquids
For smaller molecules (A) diffusing in a dilute liquid solution of solvent (B):
1.173 x (Φ MB)1/2 T
DAB =
μB VA0.6
applicable for biological solutes
DAB - diffusivity in cm2/s
MB - molecular weight of solvent B
T - temperature in K
μ - viscosity of solvent B in kg/m s
VA - solute molar volume at its normal boiling point in m3/kmol
Φ - association parameter of the solvent, which 2.6 for water,
1.9 for methanol, 1.5 for ethanol, and so on
DAB is proportional to 1/μB and T
Prof. R. Shanthini Feb & 05 Mar 2012

33. DAB is proportional to T
Diffusivity of Electrolytes in Liquids
For smaller molecules (A) diffusing in a dilute liquid solution of solvent (B):
8.928 x T (1/n+ + 1/n-)
DoAB =
(1/λ+ + 1/ λ-)
DoAB is diffusivity in cm2/s
n+ is the valence of cation
n- is the valence of anion
λ+ and λ- are the limiting ionic conductances in very dilute solutions
T is when using the above at 25oC
DAB is proportional to T
Prof. R. Shanthini Feb & 05 Mar 2012

34. Diffusion in solids
Diffusion in solids are occurring at a very slow rate.
In gas: DAB = 0.1 cm2/s
Time taken 2.09 h
In liquid: DAB = 10-5 cm2/s
Time taken year
In solid: DAB = 10-9 cm2/s
Time taken 239 centuries
Prof. R. Shanthini Feb & 05 Mar 2012

35. Diffusion in solids
Diffusion in solids are occurring at a very slow rate.
However, mass transfer in solids are very important.
Examples:
Leaching of metal ores
Drying of timber, and foods
Diffusion and catalytic reaction in solid catalysts
Separation of fluids by membranes
Treatment of metal at high temperature by gases.
Prof. R. Shanthini Feb & 05 Mar 2012

36. Diffusion in solids Diffusion in solids occur in two different ways:
- Diffusion following Fick’s law (does not depend on the structure of the solid)
- Diffusion in porous solids where the actual structure and void channels are important
Prof. R. Shanthini Feb & 05 Mar 2012

37. Diffusion in solids following Fick’s Law
Start with equation (8):
dCA CA
NA =
-DAB
(NA + NB)
(8) + dz CT
Bulk term is set to zero in solids
Therefore, the following equation will be used to describe the process:
dCA
NA =
-DAB
(17) dz
Prof. R. Shanthini Feb & 05 Mar 2012

38. Diffusion through a slab
Applying equation (17) for steady-state diffusion through a solid slab, we get
DAB
(CA1 - CA2)
NA =
(18)
z2 - z1
where NA and DAB are taken as constants.
CA1
CA2
z2-z1
Similar to heat conduction.
Prof. R. Shanthini Feb & 05 Mar 2012

39. Relating the concentration and solubility
The solubility of a solute gas in a solid is usually expressed by the notation S.
Unit used in general is the following:
m3 solute at STP
m3 solid . atm partial pressure of solute
Relationship between concentration and solubility:
S pA
CA = kmol solute /m3 solid
22.414
where pA is in atm
STP of 0oC and 1 atm
Prof. R. Shanthini Feb & 05 Mar 2012

40. Relating the concentration and permeability
The permeability of a solute gas (A) in a solid is usually expressed by the notation PM. in m3 solute at STP (0oC and 1 atm) diffusing per second per m2 cross-sectional area through a solid 1 m thick under a pressure difference of 1 atm.
Unit used in general is the following:
m3 solute at STP . 1 m thick solid
s . m2 cross-sectional area . atm pressure difference
Relationship between concentration and permeability:
PM = DAB S
where DAB is in m2/s and S is in m3/m3.atm
Prof. R. Shanthini Feb & 05 Mar 2012

41. Example 6.5.1 from Ref. 1 Diffusion of H2 through Neoprene membrane:
The gas hydrogen at 17oC and atm partial pressure is diffusing through a membrane on vulcanized neoprene rubber 0.5 mm thick. The pressure of H2 on the other side of neoprene is zero. Calculate the steady-state flux, assuming that the only resistance to diffusion is in the membrane. The solubility S of H2 gas in neoprene at 17oC is m3 (at STP of 0oC and 1 atm)/m3 solid. atm and the diffusivity DAB is 1.03 x m2/s at 17oC.
Answer: 4.69 x kmol H2/m2.s
Prof. R. Shanthini Feb & 05 Mar 2012

42. Example from Ref. 1
Diffusion through a packging film using permeability:
A polythene film m (0.15 mm) thick is being considered for use in packaging a pharmaceutical product at 30oC. If the partial pressure of O2 outside the package is 0.21 atm and inside it is 0.01 atm, calculate the diffusion flux of O2 at steady state. Assume that the resistances to diffusion outside the film and inside are negligible compared to the resistance of the film. Permeability of O2 in polythene at 303 K is 4.17 x m3 solute (STP)/(s.m2.atm.m).
Answer: x kmol O2/m2.s
Would you prefer nylon to polythene? Permeability of O2 in nylon at 303 K is x m3 solute (STP)/(s.m2.atm.m). Support your answer.
Prof. R. Shanthini Feb & 05 Mar 2012

43. Diffusion through a cylinder wall
Applying equation (17) for steady-state diffusion through a cylinder wall of inner radius r1 and outer radius r2 and length L in the radial direction outward, we get
Mass transfer per time
Mass transfer per area per time nA
dCA
(19)
NA = =
-DAB
2 π r L dr
Area of mass transfer
2πL DAB(CA1 - CA2)
CA2
nA =
(20)
CA1 r
ln(r2 / r1) r2 r1
Similar to heat conduction.
Prof. R. Shanthini Feb & 05 Mar 2012

44. Diffusion through a spherical shell
Applying equation (17) for steady-state diffusion through a spherical shell of inner radius r1 and outer radius r2 in the radial direction outward, we get
Mass transfer per time
Mass transfer per area per time nA
dCA
(21)
NA = =
-DAB
4 π r2 dr
Area of mass transfer
4πr1r2 DAB(CA1 - CA2)
nA =
(22)
CA2
CA1
(r2 - r1) r r2 r1
Similar to heat conduction.
Prof. R. Shanthini Feb & 05 Mar 2012

45. Microscopic (or Fick’s Law) approach:
dCA
JA = - DAB
(1) dz
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 Feb & 05 Mar 2012

46. Mass Transfer Coefficient ApproachNA = 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
CA2
Prof. R. Shanthini Feb & 05 Mar 2012

47. Mass Transfer Coefficient ApproachNA = kc
ΔCA = kc
(CA1 – CA2 )
(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 = kp
(pA1 – pA2)
(52)
where kp = kc / RT
(53)
kp is a gas-phase mass-transfer coefficient based on a partial-pressure driving force.
Prof. R. Shanthini Feb & 05 Mar 2012
What is the unit of kp?

48. 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 Feb & 05 Mar 2012

49. 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 Feb & 05 Mar 2012

50. 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.
Liquid film
For the system shown, gas is taken as pure component A, which diffuses into nonvolatile liquid B. pA
Bulk liquid
CAi
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.
Gas
CAb
z=0
z=δL
Prof. R. Shanthini Feb & 05 Mar 2012
Mass transport

51. 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
Liquid phase
pAb
pAi
CAi
CAb
Prof. R. Shanthini Feb & 05 Mar 2012
Mass transport

52. Two Film Theory Interface Interface Liquid phase Liquid film Gas phase
pAb
CAi
CAb
pAi
Gas film
Mass transport
Gas phase
Liquid phase
pAb
pAi
CAi
CAb
Mass transport
Concentration gradients for the film theory
More realistic concentration gradients
Prof. R. Shanthini Feb & 05 Mar 2012

53. Two Film Theory applied at steady-state
Mass transfer in the gas phase: NA = kp
(pAb – pAi)
(52)
Mass transfer in the liquid phase:
Liquid film
Gas film NA = kc
(CAi – CAb )
(51)
Gas phase
Liquid phase
Phase equilibrium is assumed at the gas-liquid interface.
Applying Henry’s law,
pAb
pAi
CAi
CAb
pAi = HA CAi
(53)
Prof. R. Shanthini Feb & 05 Mar 2012
Mass transport, NA

54. 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
pAi
Unit of H:
[Pressure]/[concentration] = [ bar / (kg.m3) ]
CAi
CAb
Prof. R. Shanthini Feb & 05 Mar 2012

55. 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
Liquid phase
pAb
pAi
CAi
CAb
Prof. R. Shanthini Feb & 05 Mar 2012
Mass transport, NA

56. Two Film Theory applied at steady-stateNA
From (52):
pAi = pAb -
(54) kp NA
From (51):
CAi = CAb +
(55) kc
Substituting the above in (53) and rearranging:
pAb - HA CAb
NA =
(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 Feb & 05 Mar 2012

57. 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 1 HA ≡
(58) KG kp kc
Combining (56), (57) and (58):
NA = KG (pAb - pA* )
(59)
Prof. R. Shanthini Feb & 05 Mar 2012

58. 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 1 1 ≡
(61) KL
HAkp kc
Combining (56), (60) and (61):
NA = KL (CA* - CAb)
(62)
Prof. R. Shanthini Feb & 05 Mar 2012

59. CAb CAi CA* CA Gas-Liquid Equilibrium Partitioning Curve showing the
locations of p*A and C*A pA
pAb
pAb = HACA*
pAi
pAi = HA CAi
pA*
pA* = HA CAb
CAb
CAi
CA* CA
Prof. R. Shanthini Feb & 05 Mar 2012

60. Summary: NA = KL (CA* - CAb) (62) = KG (pAb - pA*) (59) where
CA* = pAb / HA
(60)
pA* = HA CAb
(57) 1 HA 1 HA
= = +
(58 and 61) KG KL kp kc
Prof. R. Shanthini Feb & 05 Mar 2012

61. 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 = 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)
Experimental values of the mass transfer coefficients are kc = 0.18 m/h and kp = 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 Feb & 05 Mar 2012

62. Solution: Data provided: T = 273oC + 50oC = 323 K; PT = 2 atm;
yAb = 0.085; xAb = 0.001;
kc = 0.18 m/h; kp = kmol/h.m2.kPa
HA = atm.m3/kmol
slope of the curve
HA = kPa.m3/kmol
Prof. R. Shanthini Feb & 05 Mar 2012

63. Equations to be used: NA = KL (CA* - CAb) (62) = KG (pAb - pA*) (59)
where
CA* = pAb / HA
(60)
pA* = HA CAb
(57) 1 HA 1 HA
= = +
(58 and 61) KG KL kp kc
Prof. R. Shanthini Feb & 05 Mar 2012

64. Calculation of overall mass transfer coefficients:1 HA 1 HA
= = +
(58 and 61) KG KL kp kc 1 1 =
h.m2.kPa/kmol
= 25 h.m2.kPa/kmol kp
0.040 HA
kPa.m3/kmol =
= 897 h.m2.kPa/kmol kc
0.18 m/h KG
= 1/( ) = 1/922 = kmol/h.m2.kPa KL
= HA KG = /922 = m/h
Prof. R. Shanthini Feb & 05 Mar 2012

65. (62) is used to calculate NA
NA = KL (CA* - CAb)
(62) is used to calculate NA
CA* = pAb / HA = yAb PT / HA
= x 2 atm / atm.m3/kmol = kmol/m3
CAb = xAb CT = CT
CT = concentration of water (assumed)
= 1000 kg/m3 = 1000/18 kmol/m3 = kmol/m3
CAb = x kmol/m3 = kmol/m3
NA = (0.175 m/h) ( ) kmol/m3
= kmol/m2.h
Prof. R. Shanthini Feb & 05 Mar 2012

66. (59) is used to calculate NA
Alternatively,
NA = KG (pAb - pA*)
(59) is used to calculate NA
pA* = CAb HA = xAb CT HA
= x x kPa = kPa
pAb = yAb PT = x 2 x x 100 kPa = kPa
NA = (1/922 h.m2.kPa/kmol) ( ) kPa
= kmol/m2.h
Prof. R. Shanthini Feb & 05 Mar 2012

67. Summary: Two Film Theory applied at steady-state
= KG (pAb - pA*) = KL (CA* - CAb)
NA = kp (pAb – pAi) = kc (CAi – CAb )
(52)
(51)
(59)
(62)
pAi = HA CAi
(53)
Liquid film
Gas film
pAb = HA CA*
(60)
pA* = HA CAb
(57)
Liquid phase
pAb 1 HA 1 HA
Gas phase
pAi
= = +
CAi KG KL kp kc
CAb
(58 and 61)
Prof. R. Shanthini Feb & 05 Mar 2012
Mass transport, NA

68. Summary equations with mole fractions
= Ky (yAb - yA*) = Kx (xA* - xAb)
NA = ky (yAb – yAi) = kx (xAi – xAb )
(63)
yAi = KA xAi
(64)
Liquid film
Gas film
yAb = KA xA*
(65)
yA* = KA xAb
(66)
Liquid phase
yAb 1 KA 1 KA
Gas phase
yAi
= = +
(67)
xAi Ky Kx ky kx
xAb
Prof. R. Shanthini Feb & 05 Mar 2012
Mass transport, NA

69. Notations used:
xAb : liquid-phase mole fraction of A in the bulk liquid
yAb : gas-phase mole fraction of A in the bulk gas
xAi : liquid-phase mole fraction of A at the interface
yAi : gas-phase mole fraction of A at the interface
xA* : liquid-phase mole fraction of A which would have been in equilibrium with yAb
yA* : gas-phase mole fraction of A which would have been in equilibrium with xAb
kx : liquid-phase mass-transfer coefficient
ky : gas-phase mass-transfer coefficient
Kx : overall liquid-phase mass-transfer coefficient
Ky : overall gas-phase mass-transfer coefficient
KA : vapour-liquid equilibrium ratio (or equilibrium distribution coefficient)
Prof. R. Shanthini Feb & 05 Mar 2012

70. xAb xAi xA* xA Gas-liquid equilibrium ratio (KA) curve yA yAb
yAb = KAxA*
yAi
yAi = KA xAi
yA*
yA* = KA xAb
xAb
xAi
xA* xA
Prof. R. Shanthini Feb & 05 Mar 2012

71. Gas & Liquid-side Resistances in Interfacial Mass Transfer1 KG kp = + H kc
fG = fraction of gas-side resistance =
1/KG
1/kp
1/kp =
+ H/kc kc =
+ H kp 1 KL
H kp = + kc
fL = fraction of liquid-side resistance =
1/KL
1/kc
1/Hkp
1/kc =
+ 1/kc
+ kc/H kp =
Prof. R. Shanthini Feb & 05 Mar 2012

72. Gas & Liquid-side Resistances in Interfacial Mass Transfer
If fG > fL, use the overall gas-side mass transfer coefficient and the overall gas-side driving force.
If fL > fG use the overall liquid-side mass transfer coefficient and the overall liquid-side driving force.
Prof. R. Shanthini Feb & 05 Mar 2012

73. 1 H 1 H = = + KOG KOL KG KL 1 HA 1 HA = = + (58 and 61) KG KL kp kc
= = +
(58 and 61) KG KL kp kc
The above is also written with the following notations: 1 H 1 H = = +
KOG
KOL KG KL
Prof. R. Shanthini Feb & 05 Mar 2012

74. Other Driving Forces
Mass transfer is driven by concentration gradient as well as by pressure gradient as we have just seen.
In pharmaceutical sciences, we also must consider mass transfer driven by electric potential gradient (as in the transport of ions) and temperature gradient.
Transport Processes in Pharmaceutical Systems (Drugs and the Pharmaceutical Sciences, vol. 102), edited by G.L. Amidon, P.I. Lee, and E.M. Topp (Nov 1999)
Encyclopedia of Pharmaceutical Technology (Hardcover)
by James Swarbrick (Author)
Prof. R. Shanthini Feb & 05 Mar 2012

75. Example 1 Example 1 worked out
An exhaust stream from a containing 3 mole% acetone and 90 mole% air is fed to a mass transfer column in which the acetone is stripped by a countercurrent, falling 293 K water stream. The tower is operated at a total pressure of 1.013x105 Pa. If a combination of Raoult-Dalton equilibrium relation may be used to determine the distribution of acetone between the air and aqueous phases, determine
(a) The mole fraction of acetone within the aqueous phase which would be in equilibrium with the 3 mole% acetone gas mixture, and
(b) The mole fraction of acetone in the gas phase which would be in equilibrium with 20 ppm acetone in the aqueous phase.
Example 1 worked out
Prof. R. Shanthini Feb & 05 Mar 2012

76. Prof. R. Shanthini 27 Feb & 05 Mar 2012

77. Example 2
The Henry’s law constant for oxygen dissolved in water is 4.06x109 Pa/(mole of oxygen per total mole of solution) at 293 K. Determine the solution concentration of oxygen in water which is exposed to dry air at 1.013x105 Pa and 293 K.
Prof. R. Shanthini Feb & 05 Mar 2012

78. Example 2 worked out
Henry’s law can be expressed in terms of the mole fraction units by
pA = H’ xA
where H’ is 4.06x109 Pa/(mol of oxygen/total mol of solution).
Dry air contains 21 mole percent oxygen. By Dalton’s law
pA = yA P = (0.21)(1.013x105 Pa)
= 2.13 x 104 Pa
Prof. R. Shanthini Feb & 05 Mar 2012

79. Prof. R. Shanthini 27 Feb & 05 Mar 2012

80. Example 3
In an experiment study of the absorption of ammonia by water in a wetted-wall column, the overall mass-transfer coefficient, KG, was found to be 2.74 x 10-9 kgmol/m2.s.Pa.
At one point in the column, the gas phase contained 8 mole% ammonia and the liquid-phase concentration was kgmole ammonia/m3 of solution. The tower operated at 293 K and 1.013x105 Pa. At that temperautre, the Henry’s law constant is 1.358x103 Pa/(kgmol/m3).
If 85% of the total resistance to mass transfer is encountered in the gas phase, determine the individual film mass-transfer coefficients and the interfacial compositions.
Prof. R. Shanthini Feb & 05 Mar 2012

81. Example 3 worked out
Prof. R. Shanthini Feb & 05 Mar 2012

82. Prof. R. Shanthini 27 Feb & 05 Mar 2012

83. Prof. R. Shanthini 27 Feb & 05 Mar 2012

84. Prof. R. Shanthini 27 Feb & 05 Mar 2012

85. Example 4 Example 4 worked out
A wastewater stream is introduced to the top of a mass-transfer tower where it flows countercurrent to an air stream. At one point in the tower, the wastewater stream contains 10-3 mole A/m3 and the air is essentially free of any A. At the operation conditions within the tower, the film mass-transfer coefficients are KL = 5x10-4 kmole/m2.s.(kmole/m3) and KG = 0.01 kmole/m2.s.atm. The concentrations are in the henry’s law region where pA,i = H CA,i with H =1 0 atm/(kmole/m3). Determine the following:
(a) The overall mass flux of A
(b) The overall mass-transfer coefficients, KOL and KOG.
Example 4 worked out
Prof. R. Shanthini Feb & 05 Mar 2012

86. Prof. R. Shanthini 27 Feb & 05 Mar 2012

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89. Prof. R. Shanthini 27 Feb & 05 Mar 2012