Mass Transfer Principles for Vapor-Liquid Unit Operations (3 of 3)

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 NOTE:
 k-type mass-transfer coefficients are not constant, but usually depend on the concentration
driving force
 In dilute cases:
1 2 1 2
1 2 1 2
( ) (c )
(c ) (c )
A G A A c A A
A L A A c A A
N k p p k c
N k c k c
   
   
1 2 1 2
1 2
A
A
N ( ) k (c )
N (c )
G A A c A A
L A A
k p p c
k c
   
 
, ,
,
, ,
B LM B LM
G G B LM c y
L L B LM x B LM
p p
F k p k k
RT P
F k x c k x
  
 
, , 1.0B LM B LMy x 

 Blood oxygenators are used to replace the human lungs during open-heart surgery.
 To improve oxygenator design, you are studying mass transfer of oxygen into water
at 310 K in one specific blood oxygenator.
 Correlated values of the MTC are
 A) Calculate the corresponding mass-transfer coefficient based on the mole
fraction of oxygen in the liquid.
Example 2.1 Mass-Transfer Coefficients in a
Blood Oxygenator. Principles and Modern
Applications of Mass Transfer Operations,
Jaime Benitez, 2nd Edition
5
3.3 10 /MTC x m s

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 A) Calculate the corresponding mass-transfer coefficient based on the mole
fraction of oxygen in the liquid.
 Interpretation:
 First, verify which type of units / type of MTC is given
 Verify the type of process (UMD/EMD/OTHER)
 Calculate the required MTC in mol fraction of Oxygen.
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 Assumptions:
 Neglect evaporation of water
 Diffusion of A (oxygen) takes place
 Stagnant B (water).
 Assume Dilute solution, since the solubility of oxygen in water at 310 K is extremely low,
 If this is true (dilute solution):
 k-type coefficients is appropriate.

 Solution
 The MTC value must be diluted, and valid for UMD, as it is the only species being diffused.
 This must be k-type value
 This is specified for the liquid so, kL
 We need to change form kL to kx. From previous work (actually for any UMD Case)
5
5
3.3 10 /
3.3 10 /L
MTC x m s
k x m s




x Lk k c
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 For this, we need concentration of the mixture.
 The concentration for any mixture:
 Since this is very dilute, assume:
 We know the density of water in mass per volume…
 Change to mol per volume
x Lk k c
3
993
993
18
55.2 55.2 /
mix
mix
g
Lwater
g
mix mol mol
Lwater
mix
mol
c
vol
mass
mol
MW
gmass
mol MWc c kmol m
vol L L




     
mix
mix
mol
c
vol

mix solvent
mix solvent
mol mol
vol vol


 Now,
3 2
2
5 m
2
(3.3 10 )(55.2 ) 0.018216
1.8 10
kmol
m s
x L
kmol
kmol
m s
x s m
x
k k c
k x
k x



 

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 A gas absorber is used to remove ammonia from air by scrubbing the gas mixture
with water at 300 K and 1 atm.
 T=300K, P = 1 atm
 At a certain point in the absorber:
 the ammonia mole fraction in the bulk of the gas phase is 0.80
 while the corresponding interfacial ammonia gas-phase concentration is 0.732.
 The ammonia flux at that point is measured as
 Evaporation of water can be neglected
 A) Calculate the mass-transfer coefficient in the gas phase at that point in the
equipment.
Example 2.2 Mass-Transfer Coefficient in a Gas
Absorber. Principles and Modern Applications of Mass
Transfer Operations, Jaime Benitez, 2nd Edition
2
4
4.3 10 kmol
m s
x 

 Analysis:
 We are given concentrations in mol fraction of bulk-interphases
 The rate is given (that’s not common)
 Assumptions:
 Diffusion of ammonia through water
 Water is stagnant
 Non-dilute case!
 Do not use k-type coefficients!
 Use:
2
1
1
ln
1
A
A G
A
y
N F
y
 
  
  
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 The ammonia flux is from the bulk of the gas phase to the interface
 yA1 = 0.800 and yA2 = 0.732
2
1
2
2
1
2
4
4
1
ln
1
4.3 10
1 0.7321 lnln 1 0.801
1.47 10
A
A G
A
kmol
m sA
G
A
A
kmol
G m s
y
N F
y
xN
F
y
y
F x 
 
  
  
 
   
      


 Only valid for:
 EMD
 UMD
Relations between Mass Transfer Coefficients (MTC)
Phase EMD UMD Units
Gas
Gas
Gas
Liquid
Liquid
Conversion
'A G AN k p 
'A y AN k y 
'A c AN k c 
A G AN k p 
A y AN k y 
A c AN k c 
2
mol
m s Pa 
2
.
mol
m s mol frac 
2
/ .
mol
m s mol vol 
'A L AN k c 
'A x AN k x 
A L AN k c 
A x AN k x  2
.
mol
m s mol frac 
2
/ .
mol
m s mol vol 
' ' 'LM LMx B L B L L xMWF k x k x k c k k
    

 So far, Mass Transfer coefficients discussed are all based on:
 unit surface of contact between the phases
 They are expressed as
 But in most of the widely used industrial mass transfer equipment like the packed
or plate columns:
 the interfacial area of contact between the phases cannot be measured.
     
moles
unit time unit area unit driving force
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 In order to overcome this difficulty, volumetric mass transfer coefficients are
generally used for calculating the rate of mass transfer.
 Both mass transfer coefficient and specific surface (m2/m3) of packing depend on:
 Type of Packing (configuration)
 Size of packing
 Flow rates of the concerned fluids

 They can be combined into a single product to provide mass transfer coefficient on
volumetric basis.
 Thus, for transfer of a component within the gas phase, the mass flux and the rate
equation may be written as:
 Where,
 NAa = mass flux per unit volume of the equipment
 kya = volumetric mass transfer coefficient in the gas phase
 moles/(unit time) (unit volume) (unit mole fraction)
 kxa = volumetric mass transfer coefficient in the liquid phase
 moles/(unit time) (unit volume) (unit mole fraction)
( )
(x )
G i
L
A y A A
A x Ai A
N a k a y y
N a k a x
 
 

 Recall the Heat Transfer – Mass Transfer Analogy
 These are useful in understanding the underlying transport phenomena and as a satisfactory
means of predicting behavior of systems for which limited quantitative data exist.
 Many of these coefficients have been derived by
analogy with heat transfer
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 In order to get these:
 The dimensionless groups of heat transfer are replaced by the corresponding groups of mass transfer.

 Reynolds
 Schmidt
 Prandtl
 Sherwood
 Nusselt
 Stanton
 Peclet
 Fourier
 Grashof
 Colburn
 Lewis
Heat Trasfer Mass Transfer
Related Dimensionless Numbers & Anologies
Pr pC
k


Re
vl

 Re
vl


Sc
ABD



'
, , ,LM LMG B c B y
AB AB AB AB
k p RTl k p l k RTlFl
Sh
cD PD PD PD

hl
N
k

3
2D
gl
Gr
 



ReD
AB
lv
Pe Sc
D
  RePr p
H
C lv
Pe
k

 
3 2
2H
gl T
Gr
 



Re
D
D
Sh Sh F
St
Sc Pe cv
  
RePr
H
H p
Nu Nu h
St
Pe C v
  
2/3
D Dj St Sc 2/3
PrH Hj St
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 Conditions:
 1. The flow conditions and geometry must be the same.
 2. Most heat-transfer data are based on situations involving no mass transfer.
 Use of the analogy would then produce mass-transfer coefficients corresponding to no net mass transfer, in turn
corresponding most closely to:
 Sherwood numbers are commonly written in terms of any of the coefficients
 BUT:
 when derived by replacement of Nusselt numbers for use where the net mass transfer is not zero
 they should be taken as and the “F” used in:
2
1
ln
A
A
c
A c
A A c
A c
N F
  
   
   
 ' ’ ’G c yk k or k F
AB
Fl
cDSh 

 Common analogies:
 Reynolds Analogy
 Chilton Colburn Analogy
 As in other systems, Chilton–Colburn analogy holds very well in single-phase flow
through packed beds:
 jD and jH have been found to be essentially equal for the same bed geometry and flux
conditions.
 In many cases:
 mass transfer coefficient can be estimated from heat transfer data.
 See  Mass Transfer Coefficient correlation for packed beds

 Check out this vid:
 https://www.youtube.com/watch?v=7YlQ_4jL_gs

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 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.
 Experimental data are usually obtained by blowing gases
over various shapes wet with evaporating liquids, or causing
liquids to flow past solids which dissolve.

 Extensive data have been obtained for the transfer of mass between a moving fluid
and certain shapes, such as:
 flat plates, spheres and cylinders.
 fluid flow through a packed bed of particles, gas bubbles rising in a tank, falling films, flow
over surfaces and within tubes
 The techniques include:
 sublimation of a solid
 vapourization of a liquid into a moving stream of air
 the dissolution of a solid into water
 Dissolution of gases in liquids
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 These data have been correlated in terms of dimensionless parameters and the
equations obtained are used to estimate the mass transfer coefficients in other
moving fluids and geometrically similar surfaces.
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 Average, rather than local, mass-transfer coefficients are usually obtained
 In most cases, the data are reported in terms of the k-type coefficients applicable to the
binary systems used with NB = 0.
 Typically, there are no details concerning the actual concentrations of solute during the experiments.
 Fortunately, the experimental concentrations are usually fairly low, so that if necessary,
conversion of the data to the corresponding F is possible by taking:


1LMBp
P

1LMBx 
Typically valid when solvent >> solute
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www.ChemicalEngineeringGuy.com Typical Mass Transfer Correlations
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Typical Mass Transfer Correlations
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 Mass Transfer in Flow Past Flat Plates
 Mass Transfer in Fluids Flowing Through Pipes*
 Mass Transfer Through Packed Bed*
 Mass Transfer Coefficient for Flow Past a Solid Sphere
 Mass Transfer in Flow Normal to a Single Cylinder
 Mass Transfer in the Neighbourhood of a Rotating Disk
 Mass Transfer to Drops and Bubbles*

 Fluids moving through pipes are used extensively in the industry.
 Many materials will mix inside a pipe regardless of the operation.
 Correlations for mass transfer inside pipes are similar in nature to those in heat
transfer.
 For laminar flow:
 Sherwood number shows the same trend:
 As Nusselt number with a limiting value of 3.66 for very long pipes.
 For short pipes:
 Sherwood number varies as one-third power of flow rate.

 In view of inadequate data however, it has not been possible to develop working
equations for mass transfer coefficient in laminar flow through pipes.
 Several authors have studied gas side coefficients for mass transfer between liquids
and gases in wetted wall columns.
 Most of their data agree approximately with the following equation suggested by
Gilliland and Sherwood (1934)
 Its based on 400 tests
 Test:
 Evaporation of water + eight other organic liquids into air flowing
 Through a wetted wall column of 2.54 cm internal diameter and 117 cm long
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 We get:
 kC = gas-side mass transfer coefficient based on log-mean driving forces at
two ends of the column, cm/s
 Re = air stream Reynold’s number based on velocity of air relative to the
pipe wall.
 For engineering calculations, the following equation of Johnstone
and Pigford (1942) may be used
 Where,
 Re′ = Reynolds number based on gas velocity relative to the liquid surface
 3,000 < Re′ < 40,000
 Sc lies between 0.5 and 3.0.
0.83 0.44
'
0.023ReLMBc
AB
pk d
Sc
D P

0.23
0.0328(Re')Dj 

Pr pC
k


Re
vl

 Re
vl


Sc
ABD



'
, , ,LM LMG B c B y
AB AB AB AB
Sh
cD PD PD PD

hl
N
k

3
2D
gl
Gr
 



ReD
AB
lv
Pe Sc
D
  RePr p
H
C lv
Pe
k

 
3 2
2H
gl T
Gr
 



Re
D
D
Sh Sh F
St
Sc Pe cv
  
RePr
H
H p
Nu Nu h
St
Pe C v
  
2/3
D Dj St Sc 2/3
PrH Hj St

 Estimation of mass transfer coefficient of fluid flowing through a pipe from
Gilliland’s equation
 Estimate the value of the mass transfer coefficient kG for:
 The absorption of ammonia by an acid from a turbulent air-NH3 stream
 Its moving velocity is 3 m/s through
 Operation counter-current wetted wall tube of 25 mm id.
 The temperature and pressure are 38°C and 101.3 kN/m2, respectively.
 The inlet gas contains 10% NH3 by volume
 The exit gas contains 1% NH3 by volume.
EXAMPLE 3.5
Mass-Transfer-Principles-
and-Operations-de-Sinha

 From databases:
 Analysis:
 Clearly, we can use Gilliland equation.
EXAMPLE 3.5
3
3
5
5 2
1.15
1.85 10
2.4 10 /
kg
gas m
kg
msgas
NH Air
x
D x m s

 





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 Solution
 Verify that the operation is within the ranges:
EXAMPLE 3.5
3
2
3
5
5
5 2
(1.14 )(3m/ s)(0.025m)
Re 4620
1.85 10
1.85 10
(1.14 )(2.4 1
0.6761
0 / )
6
kg
i m
kg
m s
kg
ms
kg
AB m
vDvl
x
x
Sc
D x m s

 





   
  
Pr pC
k


Re
vl

 Re
vl


Sc
ABD



'
, , ,LM LMG B c B y
AB AB AB AB
Sh
cD PD PD PD

hl
N
k

3
2D
gl
Gr
 



ReD
AB
lv
Pe Sc
D
  RePr p
H
C lv
Pe
k

 
3 2
2H
gl T
Gr
 



Re
D
D
Sh Sh F
St
Sc Pe cv
  
RePr
H
H p
Nu Nu h
St
Pe C v
  
2/3
D Dj St Sc 2/3
PrH Hj St
3,000 < Re′ < 40,000
0.5 < Sc < 3.0.
0.83 0.44
'
0.023ReLMBc
AB
pk d
Sc
D P

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 Now, we have Re, Sc, and all other data BUT pressures:
 Calculate Pressure of A:
 Calculate Pressure of B:
0.83 0.44
'
0.023ReLMBc
AB
pk d
Sc
D P

1 1
2 2
' ' 101.3 10.13 91.17
' ' 101.3 10.13 100.29
B T A
B T A
p P p kPa
p P p kPa
    
    
1 1
2 2
' (0.10)(101.3 ) 10.13
' (0.01)(101.3 ) 1.013
A A T
A A T
p y P kPa kPa
p y P kPa kPa
  
  
EXAMPLE 3.5

 Calculate log mean. pressure of B:
 Now, solve for kc:
EXAMPLE 3.5
2 1
2
1
( ' ' ) (100.29 91.17)
p' 95.6
100.29' lnln 91.17'
LM
B B
B
B
B
p p
kPa
p
p
 
  
   
     
 
0.83 0.44
0.83 0.44
'
0.023Re
'
0.023Re
LM
LM
Bc
AB
B AB
c
pk d
Sc
D P
p D
k Sc
P d


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 Now, substitute data:
 Now, for kG
0.83 0.44
5 2
0.83 0.44
'
0.023Re
95.67 2.4 10 /
0.023(4621) (0.676)
101.3 0.025
0.02167
LMB AB
c
c
c
p D
k Sc
P d
kPa x m s
k
kPa m
k


  
   
  

23
60.02167
8.38 10
8.314 (311 )
c
G
m
s kmol
G s m kPam kPa
kmolK
k
k
RT
k x
K



 

 Recall from the “Mass Transfer” Analogies with Heat Transfer that:
 As in other systems, Chilton–Colburn analogy holds very well in single-
phase flow through packed beds:
 jD and jH have been found to be essentially equal for the same bed
geometry and flux conditions.
 In many cases:
 mass transfer coefficient can be estimated from heat transfer data.
 This is a specific case in which jD = jH
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 Mass transfer to and form packed beds occur often in processing
operations:
 Drying, adsorption, desorption of gas
 Mass transfer from gases and liquids to catalyst particles.
 Using a packed bed a large amount of mass transfer area can be
obtained in relatively small volume.
 The void fraction, , is the volume of void space divided by the total
volume of the void space & solid.
 Typical values for void fraction 

0.3 0.5 

 For a Packed Bed having:
 Gas
 Spheres
 And within the following range
 A correlation can be given for gases
0.40690.4545
ReD Hj j


 
10 Re 10,000 
jD = jH = Mass-transfer and heat-transfer factors (dimensionless)
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 And within the following range:


 A correlation can be given for Liquid Mass Transfer Coefficients:
2
3
1.09
ReDj



0.0016 Re 55 
165 70,600Sc 
 And within the following range:


 A correlation can be given for Liquid Mass Transfer Coefficients:
0.310.250
ReDj



55 Re 1,500 
165 10,690Sc 
Pr pC
k


Re
vl

 Re
vl


Sc
ABD



'
, , ,LM LMG B c B y
AB AB AB AB
Sh
cD PD PD PD

hl
N
k

3
2D
gl
Gr
 



ReD
AB
lv
Pe Sc
D
  RePr p
H
C lv
Pe
k

 
3 2
2H
gl T
Gr
 



Re
D
D
Sh Sh F
St
Sc Pe cv
  
RePr
H
H p
Nu Nu h
St
Pe C v
  
2/3
D Dj St Sc 2/3
PrH Hj St
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 Pure water at 26.1°C flows at a rate of through a packed bed of
benzoic acid spheres.
 The Diameter of the spheres .
 The total surface area of the spheres in the bed 
 The void fraction 
 The Tower Diameter is approx.
 The solubility of Benzoic Acid in water =
 For this process:
 A) Predict the mass-transfer coefficient kc.
 B) Compare with experimental value
7 3
5.515 10 /x m s
6.375 mm
2
0.01198sphereA m
0.436 
0.0677towerD m
3
2
. . 2.948 10 kgmol
benz acid m
S x 

6
4.66 10 /ck x m s

Ex. 7.3-4 Mass Transfer of a Liquid in
Packed Bed. Transport Processes and Unit
Operations, C. J. Geankoplis, 3rd Edition

 Analysis:
 Use Mass Transfer Rate (Coefficient) Approach
 Benzoic acid is acting as the particle
 Water is the solvent
 Expect dissolution of benzoic acid in water
 Mass Transfer will be assumed as non-laminar
 Case can be assumed to be dilute
Ex. 7.3-4 Mass Transfer of a Liquid in
Packed Bed. Transport Processes and Unit
Operations, C. J. Geankoplis, 3rd Edition
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 For A)
 Solution
 Get all the data required for water & benzoic acid:
 For Water:

3
2
3
9
26. C
0.8718x10
996.7
1.25x10
kg
m
m
sAB
T
Pa s
D




 
 


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 Now, we need to verify the ranges:
 We need Reynolds & Schmidt numbers
 We will need to get the velocity
 Calculate the properties of the equipment:
 Calculate the superficial average velocity of the material
2 2
3 2
(0.0677 )
4 4
3.49 10
tower
tower
Area D m
Area x m
 

 

3
3 2
4
5.51 10 7
'
3.49 10
' 1.578 10
m
sQ x
v
A x m
v x



 

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 Calculate ranges:
 Reynolds Number:
 Schmidt numbers
3
4
3
' (996.7 )(1.578 10 )(0.00638
0.872 10
Re 1.15
)kg m
sp m
kg
m s
v D x m
Re
x





 

 
  2
3
3
9 m
0.872 10
996.7 (1.25 10 )
2.670
kg
m s
kg
sAB m
x
Sc
D x x
Sc





 


 Assume that in dilute solution,
 Since Re and Sc are within the required range, we must use:
 Now, substitute in our original equation
’c ck k
2/3 2/31.09 1.09
(1.15)
0.436
2.277
D
D
j Re
j

 
 
 2
3
2
3
2
3
2
3
4
4
6
’
’
’
2.277
’
’
2.277 (702.6 )
1.578 10 /
(2.277)(1.578 10 / )
’
(702.6 )
’ 4.45 10 /
c
D
c
c
c
c
k
j Sc
v
k
Sc
v
k
x m s
x m s
k
k x m s








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 For now, flow in pipes & packed beds are good
correlations to understand!
 If you need more MTC Correlations:
 Check out the Mass Transfer – Diffusion & Convection Course
 The important thing:
 Know that there are lots of correlations for given ranges &
processes
 Understand how to use Heat transfer correlations & Mass
Transfer ones
 The scope of the course does not limits to MTC Correlations
but:
 Interphase Mass Trasfer!

1. Introduction to Mass Transfer within Interphases
2. Theories for Diffusion between Phases
 Original Film Theory
 The Penetration Theory
 Surface Renewal Theory
 Film Penetration Model
 Surface-Stretch Theory
 The Two-Film (aka Two Resistance Theory)
3. Application to Industry
 Gas Absorption & Distillation

 (1) diffusion in a quiescent medium
 Concentration Gradient from Point A to B
 (2) mass transfer in laminar flow
 Flow in pipes and Concentration Distribution
 (3) mass transfer in the turbulent flow
 Mixing in an Agitation Vessel
 (4) mass exchange between phases
 Gas-Liquid Absorption
 Vapor-Liquid Distillation
Convective MT is STRONG between phases
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 In most Mass Transfer Operations:
 two insoluble phases are brought into contact to permit transfer of constituent substances
between them.
 We are now concerned with simultaneous application of the diffusional mechanism
for each phase to the combined system.
 Example of Operations:
 Gas absorption
 Liquid extraction
 Distillation
 Drying
 Crystallization
 Leaching

 We have seen that:
 the rate of diffusion within each phase is dependent
upon the concentration gradient existing within it.
 the concentration gradients of the two-phase system
are indicative of the departure from equilibrium which
exists between the phases.
 If equilibrium be established:
 the concentration gradients will fall to zero
 Hence  the rate of diffusion will fall to zero.
 We need to consider:
 Diffusional phenomena
 Equilibria

 Is a boundary between two spatial regions occupied by different matter, or by
matter in different physical states.
 In thermal equilibrium, the regions in contact are called phases, and the interface is
called a phase boundary.
 The limit between two immiscible phases
 Vapor-Liquid
 Liquid 1 – Liquid 2
 Solid-Liquid
 Solid-Vapor
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 Common Interphase Interaction/Processes:
 Gas absorption
 Distillation

 We need to revisit Equilibrium:
 Understand the concept of equilibrium between two insoluble phases
and its importance from the point of view of mass transfer.
 Represent interphase equilibrium data in the form of an equilibrium
distribution curve
 Review and use in phase equilibria calculations the concepts:
 Raoult’s law
 Modified Raoult’s law
 Henry’s law.
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 Let us first consider the equilibrium characteristics of a particular operation
 Distillation of 2 components
 Gas-Liquid Absorption
 This way, we can then use this model to generalize the result for others.
 For now, we will consider:
 Gas-absorption operation

 Suppose that:
 An amount of liquid water is placed in a closed container
 It also has a gaseous mixture of ammonia and air
 The system will be maintained at:
 constant temperature and pressure.
 Since ammonia is very soluble in water:
 Expect some ammonia molecules to transfer:
 from the gas  into the liquid crossing the interphase.
NH3Interphase

 A portion of the ammonia molecules escape back into the gas
 They do it at a rate proportional to their concentration in the
liquid.
 As more ammonia moves into the liquid, with the consequent
increase in concentration within the liquid:
 the rate at which ammonia returns to the gas increases
 This is true until eventually :
 the rate at which it enters the liquid = rate at which it leaves.
NH3Interphase
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 At the same time, through the mechanism of diffusion, the
concentrations throughout each phase become uniform.
 A dynamic equilibrium develops
 The ammonia molecules continue to transfer back and forth from one
phase to the other, the net transfer falls to zero.
 The concentrations within each phase no longer change.
 To the observer who cannot see the individual molecules:
 the diffusion has apparently stopped.
NH3Interphase

 If we now inject additional ammonia into the container:
 a new set of equilibrium concentrations will eventually be
established
 It will now contain higher concentrations in each phase than
were initially obtained.
NH3Interphase

 We can eventually obtain the complete relationship between
the equilibrium concentrations in both phases
 Ammonia in aqueous phase
 Ammonia in gaseous phase
 If the ammonia is designated as substance A
 We can then get an equilibrium-distribution curve shown 
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 This curve results irrespective of the amounts of air and
water that we start with.
 It will be influenced only by:
 the temperature and/or pressure of the system.
 It is important to note that at equilibrium the
concentrations in the two phases are not equal!
 Instead, the chemical potential of the ammonia is the
same in both phases.
Concentrationof Ammoniain Liquid Concentration of Ammonia in Gas
Chemical Potential of Ammoniain Liquid Chemical Potential of Ammonia in Gas

 In general, Gibbs’ phase rule stipulates that a set of
equilibrium relations exists which may be shown in the form
of an equilibrium-distribution curve.
 Typically when fixing:
 Pressure
 Temperature
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 When a system is in equilibrium
 there is no net mass transfer between the phases.

 When a system is not in equilibrium:
 diffusion of the components between the phases will occur in such
a manner as to cause the system composition to shift toward
equilibrium.

 If sufficient time is allowed, the system will eventually reach
equilibrium.
0MassTrasfer Rate
0MassTrasfer Rate

 Equations relating the equilibrium concentrations in the two phases have been developed
and are presented in textbooks on thermodynamics.
 In cases involving ideal gas and liquid phases, the fairly simple, yet useful relation known as
Raoult's law applies,
 Where,
 PA is the vapor pressure of pure A at the equilibrium temperature and P is the equilibrium pressure.
A A Ay P x P 

 If the liquid phase does not behave ideally, a modified form of
 Modified Raoult’s Law:
 Where,
is the activity coeficient of species A in solution.
A A A Ay P x P 
A
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 Raoult’s law may be used to determine phase compositions for:
 the binary system benzene-toluene at low temperatures and pressures.
 (A) Estimate the vapor pressure of benzene and toluene at 300 K from the Antoine
equation
 (B) Calculate the total equilibrium pressure.
 (C) Determine the composition of the vapor in equilibrium with a liquid containing:
 0.4 mole fraction of benzene at 300 K
Example 3.1 Principles and Modern
Applications of Mass Transfer Operations,
Jaime Benitez, 2nd Edition

 Analysis:
 Benzene is more volatile than toluene, therefore, let it be component A.
 Benzene is nonpolar as Toluene, expect “ideal” case
 No azeotrope formation
 Vapor pressures are within ranges of ideal cases, no Equation of State required

 (A) Estimate the vapor pressure of benzene and toluene at 300 K from the Antoine
equation
 From Antoine Model:
 Where,
 Pi is the vapor pressure of component i, in mmHg
 T is the temperature, in K.
 Ai, Bi, Ci, are the Antoine's Constant
 For Benzene (A,B,C)  15.9008; 2788.51; -52.36
 For Toluene (A,B,C)  16.0137; 3096.52; - 53.67
ln i
i i
i
B
P A
C T
 

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 From Antoine Model, for Benzene
ln
exp
2788.51
exp 15.9008
52.33 300
exp(4.6419)
103.7
benzene
benzene benzene
benzene
benzene
benzene benzene
benzene
benzene
benzene
benzene
B
P A
C T
B
P A
C T
P
P
P mmHg
 

 
  
 
 
  
  



 From Antoine Model, for Toluene
ln
exp
3096.52
exp 16.0137
53.67 300
exp(3.4437)
31.30
toluene
toluene toluene
toluene
toluene
toluene toluene
toluene
toluene
toluene
toluene
B
P A
C T
B
P A
C T
P
P
P mmHg
 

 
  
 
 
  
  


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 From Antoine Model:
31.30tolueneP mmHg
103.7benzeneP mmHg

 (B) Calculate the total equilibrium pressure.
 We need to calculate individual pressures first:
 The total pressure is:
(0.4)(103.7 ) 41.48
(0.6)(31.30 ) 18.78
benzene benzene
toluene toluene
p x P mmHg
p
mmH
x
g
mmHP gmmHg
  
  
2
1
41.48 18.78
60.26
benzene tolue
i
nei pP P mmHg
P mmHg
p

    



 (C) Determine the composition of the vapor in equilibrium with a liquid containing:
 0.4 mole fraction of benzene at 300 K
 Now, substitute in Raoult’s Law
0.68
0.40 103.
3
.26
8
7
60
benzene benzene benzene
benzene benzene
benzene
benzene
x p y P
x p x mmHg
y
P mmHg
y
 

 

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 Another equilibrium relation which is found to be true for dilute solutions is Henry’s
law, expressed by:
 Where,
 is the equilibrium partial pressure of component A in the vapor phase
 H is the Henry’s law constant.
A A Ap y P Hx 
Ap

 An equation similar to Henry’s law relation describes the distribution of a solute
between two immiscible liquids
 Typical for Liquid-Liquid Extractions
 This equation, the distribution-law equation, is:
 Where,
 K is the distribution coefficient.
1 2, ,A liq A liqc Kc
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 The Henry's law constant for oxygen dissolved in water at 298 K is:
 A) Determine the saturation concentration of oxygen in water exposed to dry air at
298 K and 1 atm (units must be mg/L)
 Note:
 Dry air contains 21% oxygen; therefore, Poxygen = 0.21 atm.
 If using Molarity, recall that basis is 1L
4
4.5 10
.
atm
H x
mol frac

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 Solution
 The equilibrium liquid concentration is, from Henry's law:
 Assume that:
4
6
(0.21 ) 4.5 10
.
4.67 10
A
A A A
A
x atm
p x p H atm x
H mol frac
x x 
 
     
 

1 1
1 1 1000
1000
55.6
18 /
L solution kg solution
kg solution kg solvent g solvent
mass solvent
mol solvent
MW solvent
g solvent
mol solvent mol solvent
g mol

 

 
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 Calculating moles of solute
 Since basis was 1 Liter of solution…
6
4
2
4
2
55.6 55.6
( )( ) (4.67 10 )
0.0083
(55.6 )
2.596 10
( ) x(MW) ( 8.32.596 10 )(32 / )
solute
mol solvent mol solution
mol water mol solution
mol solute x total mol x mol
mol solute x mol O
mass mol x mol O g mol g mg






 

  
2
2 8.3mgO
LSaturationO in water 

 Main Goals:
 Understand several theories and concepts related to Interphase Mass Transfer
 Define local vs. overall mass transfer coefficients
 Calculate interfacial mass-transfer rates in terms of the local mass-transfer coefficients
for each phase.
 Use and model, where appropriate, overall mass-transfer coefficients.
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 Recall that the equilibrium provides the driving force for
diffusion
 We can now study the rates of diffusion in terms of the
driving forces.
 Since we have defined the rate of mass transfer as such:
( )(d )coefficientRateof MT MT riving force
( )A c f iN k C C 
mol
time area driving force 
If you have no idea go back to
Mass Transfer Coefficient

 Many of the mass-transfer operations will be carried out:
 Steady-flow fashion
 Continuous flow
 Invariant flow of the contacted phases
 Concentrations won’t change with time.
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 For this purpose, let us consider the absorption of a soluble gas.
 Pretty similar as the previous case…
 Note that it is now an Unit Operation / Equipment
 Similarly:
 Gas being absorbed  ammonia (substance A)
 from a mixture with air
 Liquid absorbing  liquid water
 Equipment  in a wetted-wall tower.

 The ammonia-air mixture will enter at:
 the bottom of the tower and flow upward
 the water flows downward around the inside of the pipe.
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 Notes:
 The ammonia concentration in the gas mixture
diminishes as it flows upward
 The water absorbs the ammonia as it flows downward
 It leaves at the bottom as an aqueous ammonia solution.
 Assume that the concentrations at any point in the
apparatus:
 Won’t change with respect to time
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 Several theories have been proposed with the objective of explaining the mechanism
of interphase mass transfer and develop quantitative relations for the same.
 Original Film Theory
 Penetration Theory
 Film Penetration Model
 Surface Stretch Model
 Two-Resistance Theory / Two-Film Theory***

 Personal note
 If you are in a hurry, you might want to skip directly to:
 Two-Resistance Theory / Two-Film Theory***
 These are just theoretical concepts
 They might help you understand two-film theory
 Yet, you do NOT need it 100%
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 Or simply “film theory”
 It is a model for turbulent mass transfer to or from a fluid-phase
boundary
 It was suggested in 1904 by Nernst
 He postulated that:
 The resistance to mass transfer in a given turbulent fluid phase is in a
thin, relatively stagnant region at the interface, called a film.

 This is similar to the laminar sublayer that forms when a fluid
flows in the turbulent regime parallel to a flat plate
 Analogy to Momentum!
 Arrows show the “sub-layer” or “film”
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 A process of absorption of A into liquid B takes place
 Note that:
 there is no vaporization of B
 there is no resistance to mass transfer of A in the gas phase, because it is pure A
 At the interface:
 phase equilibrium is assumed
 the concentration of A at the interface, , is related to the partial pressure of A at the
interface,
 This can be modeled by a solubility relation like Henry’s law, iA A Ac H p
iAc
Ap

 In the liquid film of thickness
 molecular diffusion occurs with a driving force of
 where  is the bulk-average concentration of A in the liquid
 Since the film is assumed to be very thin:
 all of the diffusing A is assumed to pass through the film and into the bulk liquid.
 Accordingly, integration of Fick’s first law for this case:
i iA A A A(c c ) (x )b b
AB AB
A
D cD
J x
 
   
( )i bulkA AdC c c 
bulkAc

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 If the liquid phase is dilute in A:
 the bulk-flow effect can be neglected so that
 Applies to the total flux, and the concentration gradient is linear
i
i
A A
A
A
x
(1 )
1
ln
1 x
b
b
A LM
x
x
x

 
 
 
  
i iA A A A(c c ) (x )b b
AB AB
A
D cD
J x
 
   
i iA A A(c c ) (x )bulk bulk
AB AB
A A
D cD
N x
 
   
iA A
1
ln (x )
1 (1 )
bulk
bulk
i
AAB AB
A
A A LM
xcD cD
N x
x x 
 
   
   

 In practice:
 the ratios and are replaced by empirical mass-transfer coefficients
and respectively
 Recall that “kc”subscript:
 c refers to the mass-transfer coefficient refers to a concentration driving force
 ‘c denotes that kc includes both:
 diffusion mechanisms
 bulk-flow effect.
ABD
 (1 )
AB
A LM
cD
x  ck
'ck
iA A
1
ln (x )
1 (1 )
bulk
bulk
i
AAB AB
A
A A LM
xcD cD
N x
x x 
 
   
   
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 The film theory is typically criticized because it predicts
that:
 The rate of mass transfer is proportional to molecular
diffusivity
 Which we know it is not 100% the Case!
 This is a very ideal case!
 Regardless of the criticism:
 the film theory continues to be widely used in design of
mass-transfer separation equipment.

https://demonstrations.wolfram.com/MembraneConcentrationProfile/
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 SO2 is absorbed from air into water in a packed absorption tower.
 At a location in the tower:
 the mass-transfer flux is
 At the two-phase interface and in the bulk liquid:
 the liquid-phase mole fractions are 0.0025 and 0.0003, respectively
 The accepted diffusivity is given as:
 A) Determine the mass-transfer coefficient, kc
 B) Calculate the corresponding film thickness (neglecting the bulk flow effect)
Ex 3.17 Mass-Transfer Flux in a
Packed Absorption Tower. Separation
Process Principles, J. D. Seader, 3rd
Edition
2
20.0270 kmol SO
m h
2
2 2
5
1.7 10 cm
sSO H OD x 

 Analysis:
 Assume This can be modeled with Original Film Theory
 The resistance to mass transfer in a given turbulent
fluid phase is in a thin, relatively stagnant region at
the interface, called a film.
 Assume Steady State
 This is Dilute case!
 Bulk properties can be ignored

 Solution of A – The MTC
 Since we are given mol fractions, we need concentration
  1 3 3 3
18 2 3
:
1 / 1000 /
1000 55.5 55.5 10 /
5.55 10 /
1 1 1
mol
g
for liquid water
mass
D g mL g L
vol
gmol mol mol cm dm
c x x mol cm
L L L L


  
    
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 Continue with:
    2
2 2 2
3
1 1 1000
3600 1100
2
3
( )( )
0.027
( )( ) 5.55 10 (0.0025 0.0003)
6.14 10 /
i bulk
i bulk
A c A A
kmol h m mol
s kmolm h cmA
c mol
A A cm
c
N k c x x
N
k
c x x x
k x cm s


 
 
 


 Now, b: The film thickness
5 2
3
1.7 10 /
6.14 10 /
0.0028
0.028
AB
c
AB
c
D
k
D x cm s
k x cm s
cm
mm







 



 Ralph Higbie (1935) proposed the Penetration Theory.
 Theory goes towards Eddies
 The principal mechanism of interphase mass transfer involves:
 Motion of turbulent eddies from the core of the fluid to the
interface
 They are followed by a short interval of unsteady-state molecular
diffusion into the other fluid before these eddies are displaced
from the surface by subsequent eddies.
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 The main assumption:
 All eddies that reach the interface have the same
exposure time
 The diffusing molecules cannot reach the depth ze of the
eddies due to slow diffusion and short exposure.
 According to this theory:
 in most cases the time of exposure of fluid elements to
mass transfer is too short for steady-state concentration
gradient to develop, which is the characteristic of the
two-film theory.

 (1) move from the bulk liquid to the interface;
 (2) stay at the interface for a short, fixed period of time
during which they remain static, allowing molecular
diffusion to take place in a direction normal to the
interface
 (3) leave the interface to mix with the bulk stream.
 When an eddy moves to the interface, it replaces a static
eddy.
 Thus, eddies are alternately static and moving.
 Turbulence extends to the interface.

 In the penetration theory, unsteady-state diffusion takes place at the interface
during the time the eddy is static.
 This process is governed by Fick’s second law with boundary conditions
 When solved:
2
2
A A
AB
C C
D
t z
 

 
2 AB
c
c
D
k
t
 tc = contact time
2 ( ) 2i b
AB AB
A A A c
c c
D D
N C C k
t t 
   
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 The model, which predicts that kc is proportional to the square root of the
diffusivity, which is at the lower limit of experimental data
 Penetration theory is most useful for bubble, droplet, or random-packing interfaces
 For bubbles, the contact time, tc, of the liquid surrounding the bubble is
approximated by the ratio of bubble diameter to its rise velocity.

 The Penetration Theory was modified by Danckwerts (1951) to form:
 According to this theory the main drawback of the penetration theory is the assumption
that:
 The assumption of a constant contact time for all eddies that reach the surface is not reasonable
 All the liquid elements or eddies are not really exposed to the gas for the same length of time.
 In a turbulent fluid:
 It is highly probable that some of the eddies are swept away while still young
 i.e. time of exposure is not the same!
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 Proposal:
 The replacement of constant eddy contact time with the assumption of a residence-time
distribution
 This assumes that the probability of an eddy at the surface being replaced by a fresh eddy is
independent of the age (time) of the surface eddy.
 As a result:
 there is a distribution of the eddies present at the interface into different ‘age groups’
depending upon their contact time with the gas.
 replacement of constant eddy contact time with the assumption of a residence-time
distribution

 At any time:
 each of the eddies at the interface has equal chance of being replaced by fresh eddies.
 He further assumed that unsteady-state mass transfer occurs to the eddies during
their stay at the interface.
 He introduced a fractional rate of surface renewal(s)
 where “s” is the fraction of the surface area renewed in unit time.
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 The functions of the theory
tA
0
{ }N dt
{ } se
1/
avgA
st
N t
t
s t








• The fractional rate of surface renewal
• For steady-state flow into and out of a well-mixed vessel.
• The equation models t with respect to molar flux
•  where “t” is the average residence time
• f{t}dt = the probability that a given surface eddy will
have a residence time t.
• The sum of probabilities must be equal to 1

 When solving the previous 3 equations & the mass transfer coefficient
c ABk D s
AB
c
D
k


Original Film Theory
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 An objection to the penetration theory and the surface renewal theory is:
 that both the theories have assumed the depth of the liquid element (depth of
penetration) to be infinite.
 In reality:
 it should have a finite value and the thickness should decrease as turbulence increases.
 Toor and Marchelo (1958) proposed the film penetration model:
 It is according to which the transfer to young elements at the interface with short
exposure follows the penetration theory, and transfer to old elements with long exposure
follows the film theory.
 They showed that the Film and Penetration theories are but limiting cases of their
more general film penetration model.
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 Toor and Marchello combined features of:
 The film
 Penetration
 Surface Renewal
 Note that if In the limit for a high rate of surface renewal 
2 20
2
( )
1
( ) ( )x 1 2
1
avg i b
i b i b
A c A A
AB
c A A A A
ABn
N k C C
D
k C C C C
D
n
s
 



 
 
 
    
 
 

2
ABs D 
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 Lightfoot and co-workers (Stewart et al. 1970) in their surface-stretch theory:
 Combined the penetration and surface renewal theories
 They modeled that the interfacial area through which mass transfer takes place, to change
periodically with time.
 Examples of such periodic changes are:
 transfer to bubbles rising through a liquid
 transfer to drops and bubbles being formed at a nozzle or to wavy or rippled liquid surfaces.

 Film theory
 Penetration Theory
 Film Penetration Theory 2 20
2
1
x 1 2
1
AB
c
ABn
D
k
D
n
s
 



 
 
  
 
 

2 AB
c
c
D
k
t

c ABk D s
AB
c
D
k


Advantages:
• Simple
• Base
• Better Physical
Approach
• It includes
Flow and time
• Accounts for both
• Penetration
• Film
Disadvantages:
• Film thickness
unkown
• Surface
Renewal time
unknown
• Contact time
unkown
• Complex model
• Unknown times
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 The credit of being the pioneers in attempting to
visualize the mechanism of interphase mass
transfer and offer some theoretical basis for the
same goes to W.K. Lewis and W.G. Whitman (1924)
 They proposed the two-film or two-resistance
theory of interphase mass transfer.
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 They assumed the existence of an imaginary film on each side of the interface
(phase boundary).
 Film 1  Vapor-Liquid
 Film 2  Liquid-Vapor
 Mass was visualized as being transferred by steady-state molecular diffusion through
these films

 Each film having its characteristic thickness.
 Also had its resistance to molecular diffusion being
equal to the total resistance due to both:
 molecular diffusion
 eddy diffusion actually
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 Gas Absorption is the process/operation of removing an undesired gas
from a Gas Stream.
 The gas of interest goes from Gas Stream to Liquid Solvent.
 This is done exploiting solubility properties.
 A Gas Absorber is an Equipment designed to perform this operation
 Typically, this is done counter-current:
 Liquid flows from top to bottom
 Gas flows from bottom to top
 The film concept is very powerful here!

 As previously noted, gas absorption operation involves mass
transfer from the gas phase to the liquid phase.
 That means that:
 gas molecules must diffuse from the main body of the gas phase to
the gas-liquid interface
 then cross this interface into the liquid side
 finally diffuses from the interface into the main body of the liquid.
 Typical gas-liquid interface is experienced in these type of
operations
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 In the gas phase, 3 flow regimes can be visualized :
 Fully developed turbulent region where most of the
mass transfer takes place by eddy diffusion
 A transition zone with some turbulence
 A laminar film with molecular diffusion
 Such phenomena are difficult to analyze.
 TWO-FILM THEORY is a simplified theory:
 BASIS for analysis
 Will allow us to get correlations of mass transfer
phenomena.

 Film-Film interaction
 Gas vs. Liquid
 Consider the interface between the gas
phase and the liquid phase.
 This interface can represent any point in
the gas absorption equipment where the
gas contacts the liquid.

 Analysis:
 the diffusion of solute A
 from the gas phase
 into the liquid phase
 Example:
 NH3 that is diffusing
 from an gaseous air-NH3 mixture
 into liquid phase water.
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 Assumptions
 Steady-state
 concentrations at any position in the tower do not change with time.
 Sharp boundary
 Interface between the gas phase and the liquid phase
 Laminar
 film exist at the interface on both sides of the interface
 Equilibrium
 at the interface
 there is negligible resistance to mass transfer across the interface:
 (xi, yi) is the equilibrium concentration.
 No chemical reaction:
 rate of diffusion across the gas-phase film must equal the rate of diffusion across the liquid-phase film.

 In the analysis of gas absorption we are interested in the transfer of materials
throughout the entire gas absorption equipment, not just a single location in the
equipment.
 Therefore the two-film theory can be analyzed more effectively by using the
equilibrium solubility curve.
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 The concentrations at the interface:
 in the gas ( yAi )
 in the liquid ( xAi )
 are represented as a point M on the equilibrium
solubility curve.
 Point M thus has the coordinates ( yAi, xAi ).
 As we move along the column along the continuous
interface, we can trace out an equilibrium curve.
 The “best case scenario” will be equilibrium
 From P to M

 Very often, the subscript "Ai" is dropped
 The equilibrium curve is simply a relationship between y
and x
 y= f(x).
 Bulk phase (P)
 bulk gas phase ( yAG )
 bulk liquid phase ( xAL )
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 Point P thus has the coordinates ( xAL, yAG ).
 Point P is located above the equilibrium curve.
(ABSORPTION)
 Point P is located below the equilibrium curve
(STRIPPING)
 Departure from equilibrium:
 provides the driving force for mass transfer

 In the gas-phase, the concentration falls from yAG in the
bulk gas to yAi at the interface.
 Thus, there is a concentration driving force for mass transfer
from the bulk gas to the gas film to the interface.
 At the interface, the component A crossed the interface and
enters the liquid side.
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 In the liquid-phase, the concentration falls from xAi at the
interface to xAL in the bulk liquid.
 Thus, there is a concentration driving force for mass transfer
from the interface to the liquid film to the bulk liquid.
 The mass transfer process can be represented by the line PM.

 Mass transfer can be described by a set of mass transfer
equations.
 NOTE that
 The bulk concentrations yAG, xAL are not equilibrium values
 If so, there would be no diffusion of A.

 The two-film theory and equilibrium curve can be
expressed in other ways; e.g. in terms of
 partial pressure (for the gas phase)
 concentration (for the liquid phase)
 mole fractions ( x and y).
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 In commercial absorption equipment
 both the liquid and the gas are usually in turbulent flow
 the film thickness is not easy to determine.
 Therefore instead of analysis of mass transfer using Fick's Law, it is more convenient
to write the molar flux of A using 
 It can be written for liquid and gas phases
   , AMolar Flux N Mass Transfer Coefficient x Driving Force
 
 
L
G
x
y
=k
=k
i b
G i
A A A
A A A
N x x
N y y


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 At a point A (xAL, yAG), we can write the mass transfer equations for each of the
phases:
 where :
 NA molar flux of component A, mole/(area.time)
 ky mass transfer coefficients in the gas phase
 ( yAG - yAi ) concentration driving force in the gas phase (mole fraction)
 kx mass transfer coefficients in the liquid phase
 ( xAi - xAL ) concentration driving force in the liquid phase (mole fraction)
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 We need to name MTC as follows:
 MTC will depend on:
 Type of reference values
 i.e. Interphase compositions
 i.e. Equilibrium of interphase compositions

 At Point P; the Mass Transfer is given by:
 
 
L
G
x
y
=k
=k
i L
G i
A A A
A A A
N x x
N y y


   
 
 
 
 
G L
y x
x
y
x
y
=
k k
k
k
k
k
G i i L
G i
i L
G i
L i
A A
A A A A
A A
A A
A A
A A
N N
y y x x
y y
x x
y y
x x


 
 




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 At Point P; the Mass Transfer is given by:
 From Point P to Point M:
 The ratio of mass transfer coefficients is equal to the slope of line PM.
 The gradient of the line determines the relative resistances of the 2 phases.
 The above equation is useful if one does not know the interface equilibrium
concentrations.
 
 
L
G
x
y
=k
=k
i L
G i
A A A
A A A
N x x
N y y



 We can use the above equation to determine the equilibrium concentration at the
interface ( xAi, yAi )
 i.e. to locate point M, provided that kx and ky are known (or can be calculated using
appropriate correlations).
 We do so by plotting:
 A straight line originating from point P ( xAL, yAG )
 slope
 The point of intersection of this line with the equilibrium curve gives point M
 which yield the values of xAi and yAi .
 That way we can calculate the flux NA at that particular point.
x
y
k
k

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 As stated, we use correlations in order to derive the Mass Transfer Coefficients
 Most of the correlations are based on dimensionless groups
 Generally, the mass transfer correlations are more complex and difficult to use.
 In addition, they are very specific in applications and are limited to some simple
situations.

 In the above analysis of mass transfer across an interface, note that since the interface
concentrations varies throughout the gas absorption equipment (e.g. a tray column).
 It is worthwhile highlighting that NA depends on the conditions at the particular point in
the column.
 In other words, NA may vary throughout the entire length of the column.
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 The previous definitions for molar flux NA require:
 the knowledge of the interface concentrations.
 Since experimental sampling of the concentrations at the interface is very difficult or
virtually impossible:
 it is more useful to define the mass transfer equation using overall mass transfer coefficients
KX and KY:
 xA* is the concentration (mole fraction) in liquid phase that is in equilibrium with yAG.
yA* is the concentration (mole fraction) in vapor phase that is in equilibrium with xAL.
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 Driving force for mass transfer:
 ( yAG - yA* ) in the gas phase (as indicated by line PC)
 ( xA* - xAL ) in the liquid phase (line PD)

 Watch this video:
 https://www.youtube.com/watch?v=kKsxZy2mAeM
 See how the “K” Value is derived
 It requires:
 K1
 D/d
 K2
 The “overall” mass transfer coefficient
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 Local:
 ky mass transfer coefficients in the gas phase
 yAi; is the concentration (mole fraction) in vapor phase at the interface
 kx mass transfer coefficients in the liquid phase
 xAi; is the concentration (mole fraction) in liquid phase at the interface
 Overall:
 KY mass transfer coefficients in the gas phase
 yA* is the concentration (mole fraction) in vapor phase that is in equilibrium with xAL.
 KX mass transfer coefficients in the liquid phase
 xA* is the concentration (mole fraction) in liquid phase that is in equilibrium with yAG.

 Units for kx , ky , KX , and KY varies with the way the mass
transfer equation is written:
 By phase:
 Vapor phase
 Liquid phase
 Driving forces used:
 mole fractions ( y or x )
 mole ratios ( X or Y )
 weight fraction (not common/recommended)
 partial pressures (p)
 concentrations (c)
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 The resistance to mass transfer is defined as the reciprocal of the mass transfer
coefficient:
 represents the resistance to mass transfer in the liquid phase
 represents the resistance to mass transfer in the gas phase
 Refer to:
 Electric Resistance  More resistance, less flow
 Heat Resistance  is the convective resistance to heat transfer
 It is important to know if one of the 2 resistances is controlling the mass transfer.
1
xk
1
yk
1
h
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 It can be shown that kx, ky, KX, and KY are related through the following equations:
 where m" is the slope of line segment DM, and m' is the slope of line segment MC as
shown.
 If the equilibrium line is straight, then
 Dilute cases! ( Henry’s Law)
 kx , ky , KX , and KY all change with positions in the tower.
1 1 1
''X y xK m k k
 
1 1 '
Y y x
m
K k k
 
' ''m m

 Recall:
 the relationship between overall and film/local mass transfer coefficients
 The 1/k represents the mass transfer resistance.
 If m' is small
 i.e. the equilibrium curve is very flat, the term m'/kx is not significant, therefore:
 and the major resistance to diffusional mass transfer lies in:
 the gas phase  the mass transfer is said to be gas-phase controlled.
 In this case, solute A can be interpreted as being very soluble in the liquid:
 at equilibrium, a small concentration of A in the gas will bring about a very large
concentration in the liquid.
' 1 1
0
x Y x
m
k K k
  
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 Recall:
 the relationship between overall and film mass transfer coefficients
 The 1/k represents the mass transfer resistance.
 If m" is large
 (i.e. the equilibrium curve is very steep), then the term 1/m"ky is insignificant, therefore,
 and the majority of resistance to mass transfer lies
 In the liquid  The mass transfer is said to be liquid-phase controlled.
 Solute A is relatively insoluble in the liquid:
 a very large concentration of A in the gas phase is required to provide even a small change
of concentration in the liquid.
1 1 1
0
'' y X xm k K k
  
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 Determination of the packed height of a column (see later) most commonly involves
the overall gas-phase coefficients
 the liquid usually have a strong affinity for the solute so that resistance to mass transfer is
mostly in the gas.
 The following must be true in absorption (in order to favor gas dissolution)
1 1
Y yK k


 In an experimental study of the absorption of ammonia by water in a wetted-wall
column, the value of KG was found to be 2.75 x 10-6 kmol/m2-s-kPa.
 At one point in the column, the composition of the gas and liquid phases were:
 8.0 and 0.115 mol% NH3 respectively.
 The temperature was 300 K and the total pressure was 1 atm. Eighty five percent
(85%) of the total resistance to mass transfer was found to be in the gas phase.
 At 300 K, ammonia-water solutions follow Henry's law up to 5 mol% ammonia in the
liquid, with m = 1.64 when the total pressure is 1 atm.
 (A) Calculate the individual film coefficients
 (B) Calculate the interfacial concentrations.
Example 3.4 Mass-Transfer Resistances During
Absorption of Ammonia by Water. Principles
and Modern Applications of Mass Transfer
Operations, Jaime Benitez, 2nd Edition
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 Solution of (A)
 Let us first verify for the coefficient given.
 Since this is related to the gas, then local-overall MTC must be related as well
 Therefore, we must relate KY and KG:
Example 3.4 Mass-Transfer Resistances During
Absorption of Ammonia by Water. Principles
and Modern Applications of Mass Transfer
Operations, Jaime Benitez, 2nd Edition
 2 2
6 4
2.75 10 (101.325 ) 2.786 10
Y G
kmol kmol
Y m skPa m s
K K P
K x kPa x 

 
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 Now, since we know that there is 85% resistance, then:
 
2
2
4
4
1/tan
tan 2 1/
1/
0.85
1/
(0.85) 1/ 1/
0.85 1
2.786 101
0.85 0.85
3.28 10
y
Y
y
Y
Y y
Y y
kmol
m s
y y
kmol
y m s
kresis ce gas phase
total resis ce phases K
k
K
K k
K k
x
k K
k x






 


 For the local MTC of liquid phase:
 Recall that we can further relate via the equation:
1 1
1 1
1 0.85
0.15
y y x
x y y
x y y
x y
m
K k k
m
k K k
m
k K K
m
k K
 
 
 

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 Substituting values:
 Answer for (A):
2
2
4
3
1.64 (2.786 10 )
0.15 0.15
3.05 10
kmol
m sY
x
kmol
x m s
x xmK
k
k x


 

2
3
3.05 10 kmol
x m s
k x 

2
4
3.28 10 kmol
y m s
k x 


 Solution of B:
 We can now calculate equilibrium concentration for the vapor:
 Now, substitute for Flux
* 3
,
* 3
(1.64)(1.15 10 )
1.887 10
A A L
A
y mx x
y x


 

2
*
4
5
( )
(2.768 10 )(0.08000 0.001866)
2.18 10
GA Y A A
A
kmol
A m s
N K y y
N x
N x


 
 

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 Since we now have equilibrium & bulk conditions for the vapor, calculate the
interphase for vapor.
5
3
4
( )
2.18 10
1.887 10
3.28 10
0.01362
G i
i G
i
G
A Y A A
A
A A
Y
A
A
N K y y
N
y y
K
x
y x
x
y



 
 
 


 Now for the liquids:
 Since the flux of liquid equals the flux of gas:
 Now, substitute for Flux
L GA AN N
5 3 3
5
3
3
. 3
(x )
2.18x10 (3.05 10 )(x 1.15 10 )
2.18x10
x 1.15 10
3.05 10
x 8.305 10
i L
i
i
i
A x A A
A
A
A
N k x
x x
x
x
x
  




 
 
 

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 Now, we have all the required data:
 Fluxes
 Interphase conditions
 Note for the student  Verify fluxes with the new conditions.

 Sulfur dioxide is absorbed into water in a packed column.
 The bulk conditions:
 50°C, 2atm
 yAG=0.085; xAL=0.001
 The mass transfer coefficients:
 kc=0.18m/s
 kp=0.040 kmol/m2-h-kPa
 For mole-fraction driving forces, compute the mass-transfer flux assuming an
average Henry’s law constant and a negligible bulk-flow effect
EXAMPLE 3.20 Absorption of SO2 into
Water. Separation Process Principles, J.
D. Seader, 3rd Edition
Equilibrium data
c(SO2) lbmol/ft3 p(SO2), atm
0.00193 0.0382
0.00290 0.0606
0.00483 0.1092
0.00676 0.1700
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 Solution of A
 The very first thing we need to do is change everything to mole fraction (vapor and liquid
conditions) as well as the mass transfer coefficients
 Recall that for pressure, verify partial pressure
 Recall that for liquid, verify concentration
 Pressure is constant
 Concentration is constant
; ;A A
A A
p c
y x
P c
 
c(SO2) lbmol/ft3 p(SO2), atm
0.00193 0.0382
0.00290 0.0606
0.00483 0.1092
0.00676 0.1700
x(SO2) y(SO2)
0.00056 0.00097
0.00085 0.00145
0.00141 0.00242
0.00197 0.00338

 Now that we have the data, we must fit it to a STRAIGHT line
 This is Henry’s Law!
 Equation:
 y=1.715x
 m=1.715
y = 1.715x
R² = 1
0.00000
0.00050
0.00100
0.00150
0.00200
0.00250
0.00300
0.00350
0.00400
0.00000 0.00050 0.00100 0.00150 0.00200 0.00250
y(SO2)
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 Converting the coefficients:
3 2
2 2
(0.18 )(55 ) 9.9
(0.040 )(202.6 ) 8.1
x c
m kmol kmol
hx m h m
y p
kmol kmol
y h m kPa h m
k k c
k
k k P
k kPa

  

 

 

 Now, verify which is the controlling (mass resistance) phase:
2 2
2
0.12345 0.1732 3.37154
1 1 1 1.715
8.1 9.9
1 1 1 1 1
9.9
41
0.10101 0
1.7
.07198 5.780
5
6
1 (8.1)
x
kmol kmol
y y x h m h m
kmol
x y h m
x
x
y
y
m
K
k k
K k m k
K
K  

  

    
     
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 Analysis is mostly into liquid phase (kx concept is always larger)
 Now, calculate Flux:
2
0.08
*
5
1.715
5.7806
( )
5.7806 ( 0.001
0.280
)
7
L
L L
x
A
A x A A x A
A
kmol
A h m
K
y
N K x x K x
m
N x
N 

 
    
 



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1. Introduction
2. Equilibrium Fundamentals
3. Molecular Diffusion
4. Convective Mass Transfer & MT Coefficients
5. Interphase Mass Transfer
6. Conclusion

1. Introduction
2. Equilibrium Fundamentals
1. Basics
2. Vapor-Liquid Equilibrium
3. Equilibrium Diagrams
4. VLE Cases
5. Deviations of VLE
6. Bonus – Getting VLE Data
7. Gas solubility in Liquids
3. Molecular Diffusion
1. Introduction to Mass Transfer
2. Fick’s Law & Cases
4. Convective Mass Transfer & MT Coefficients
1. Convective MT
2. MT Coefficients
3. Correlations
5. Interphase Mass Transfer
1. Introduction to MT in Interphases
2. Theories for Diffusion between Phases
3. Two Film Theory Applied to Industrial Processes
6. Conclusion

1. Basic Topics
 Ideal Gas and Solution, Real Gas/Solution, Phase Equilibrium, Solubility
2. Vapor-Liquid Equilibrium
 VLE Pure, VLE Binary, Volatility
3. Equilibrium Diagrams (Txy, Pxy, XY)
4. Vapor-Liquid Equilibrium Thermodynamics
 Raoult’s Law for IG-IS, Models for Real Cases
5. Deviations of VLE: Azeotropes
6. BONUS - Getting VLE Data on Aspen Plus ®
7. Gas Solubility in Liquids
 Overview, Equilibrium Diagrams, Analysis and Henry’s Law

 Ideal Solution & Gas, Equilibrium
 Concept of Ideality vs. Reality
 Non-ideal solution
 Vapor & Partial Pressures
 Phases & Equilibrium between Phases
 Solubility

1. VLE Pure Phase Equilibrium
2. VLE Binary Mixture Equilibrium
3. Volatility & Relative Volatility
4. Introduction to the K-Value

1. Phase Diagrams:
 Phase Rules
 Binary Diagrams
 Txy, Pxy, XY

1. Fundamentals
 Equilibrium Models (EOS & Activity Models)
2. Cases
 Case 1: IS – IG
 Case 2: IS - RG
 Case 3: RS - IG
 Case 4: RS - RG

1. Deviations
2. Azeotropes
 Positive Boiling Point Azeotrope
 Negative Boiling Point Azeotrope

 Getting Data From Aspen Plus – Existing Components
 Getting Data from NIIST Database

 Overview
 Equilibrium curve
 Analysis
 Henry's Law

1. Introduction to Mass Transfer
 Fluxes & Velocities
 Molecular Diffusion
 Diffusion Coefficient
2. Fick’s Law
 Fick’s Model
 Case (A) Equimolar Counter-Diffusion (EMD)
 Case (B) Unimolecular Diffusion (UMD)

1. Convective Mass Transfer
2. Mass Transfer Coefficients
 The MT Coefficient
 Analogies
3. MT Coefficient Correlations
 Correlation for Disks, Plates & Sphere
 Correlation for Cylinders & Pipes
 Correlation for Packed Beds & Fluidized Beds
 Correlation for Tray Columns
 Correlation Hollow-Fiber Membrane Modules (Shell & Tube)

 More Engineering Courses
 Process Design & Simulation
 Aspen Plus
 Aspen HYSYS
 Unit Operations
 Gas Absorption & Stripping
 Flash Distillation
 Binary Distillation
 Oil & Gas
 Petrochemical Industries
 Petroleum Refining

Mass Transfer Principles for Vapor-Liquid Unit Operations (3 of 3)

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