Mass Transfer Principles for Vapor-Liquid Unit Operations (2 of 3)

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

 We will analyze mostly Gas-Liquid Interaction
 If a certain quantity of a gaseous mixture and a non-
volatile liquid are brought into contact, some components
of the gas may dissolve into the liquid.
 The resulting concentration of the dissolved gas in the
liquid is said to be the gas solubility at the prevailing
temperature and pressure.

 Example:
 Consider the system of:
 ammonia-air gas mixture and liquid water
 Where a fixed amount of water is in contact with
gaseous ammonia and air in a closed container.

 Ammonia is very soluble in water
 Some ammonia molecules will immediately travel from the gas phase into the liquid phase
 Ammonia crosses the interface separating the 2 phases.
 As the molecules of dissolved ammonia increases:
 some of the molecules will start to escape back to the gas phase.

 As more ammonia enters the liquid:
 The rate of ammonia returning to the gas increases
 Eventually the rate at which it enters the liquid
exactly equals that at which it leaves.
 An equilibrium condition now exist between the
gas and liquid.
 Although the molecules of ammonia still travel
back and forth from one phase to another
 The net transfer is zero.

 If additional ammonia is injected into the container,
the existing equilibrium is disturbed.
 *Le Chatelier 
 The ammonia molecules will re-distribute themselves
until a new equilibrium is re-established
 higher concentrations of ammonia in both gas phase and
liquid phase.
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 Such equilibrium relationship between the concentrations is known as the equilibrium
distribution curve or simply
 the solubility curve
 Most operations will involve interaction between:
 Equilibrium Line
 Operation Line
 Mass Transfer involves the:
 Operation Point reaching the Equilibrium Point

 Recall from Phase Diagrams that:
 X-axis  mol fraction of Solute (gas) in Liquid Mixture
 Y-axis  mol fraction of Solute (gas) in Gas Mixture
 Typically:
 Set for a Given Pressure and Temperature
 WHY?

 The gas absorption process involves the re-distribution of
solute between
 the gas phase and the liquid phase
 the 2 phases must come into close contact and achieves
equilibrium condition.
 The equilibrium distribution curve is the relationship
between
 solute concentration in the gas phase and in the liquid phase
 constant temperature and pressure
 NOTE: Not same for Distillation & Absorption!
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 In Distillation:
 we have seen that the equilibrium curve simply shows the equilibrium relationship
between 2 components
 the more volatile and the less volatile
 e.g. Water-ethanol
 the points represent vapour-liquid equilibrium at different temperatures.
 the solubility curve can lie anywhere in the x-y plot (or p-y plot, etc).
 In gas absorption
 as noted at the beginning, in the simplest case:
 there will be 3 components
 e.g. NH3 , air and water
 The equilibrium solubility curve is plotted for a particular constant temperature
 Any point on the same curve represent gas solubility at the same temperature.

Not the one we need!This one is great!

 Before we actually go and do a proper analysis…
 Draw curves depicting the changes that will happen if:
 Pressure of system increases
 Pressure of system decreases
 Temperature increases
 Temperature decreases
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 Before we actually go and do a proper analysis…
 Draw curves depicting the changes that will happen if:

 Draw curves depicting the changes that will happen if;:
SOLUTION
If P increases 
• Gas is more soluble in Liquid
• Expect Equilibrium Line to Shift to the Right
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 Compare:
 HCl, NH3 and SO2
 Temperature Effect
 Compare
 Diff. Species
 Vapor Pressure
 Relatively insoluble gas is high in concentration in the gas phase
 i.e. high partial pressure at equilibrium.

 High soluble gas has low partial pressure.
 In many practical cases, only one component in the gas
mixture is relatively soluble in the liquid
 e.g. in the NH3-Air-H2O system
 since NH3 is relatively more soluble than air in water
 pAir >> pNH3

 When the gas mixture in equilibrium with an ideal liquid solution follows the ideal
gas behavior
 When the solution is non-ideal
 Raoult's Law cannot be applied
 Most cases, it is more convenient to apply Henry’s Law
 Models for ideal solution, real gas
 Dilute solution
 “ideal-dilute” case
1 1 1 Tx P y P 
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 The equation for describing non-ideal solutions is the Henry's Law, which states
that (for component-A in a mixture of 2 components):
 Where,
 pA is the partial pressure of component-A
 xA is the mole fraction of component-A in liquid
 H is the Henry's Law Constant
 Henry's Law is often used for describing gas solubility relationships.

 Solubility of Gases in Liquids
 Typically, as Pressure increases, solubility of gases increases
 Henry’s Constant is dependent directly proportional to Temperature
 Henry's Law:
 Used to represent equilibrium solubility curves.
 predicts a linear equilibrium relationship.
 Still, most equilibrium relationships are actually non-linear.
 Valid at low concentrations  dilute.
E Line  Henry’s Law

 Typical Henry’s Law Constants for gases
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 Gas Absorption is further explored in the following lectures
 Application to two-film theory
 We will use this  Equilibrium Line!

https://demonstrations.wolfram.com/HenrysLawForGasesDissolvedInWater/

https://demonstrations.wolfram.com/TemperatureDependenceOfHenrysLawConstant/
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 Given that typically  0.348 g CO2 / 100 mL water at 0°C, P = 1.00 atm
 A) Calculate Henry’s constant in (atm/M; atm/xa)
 B) Calculate the solubility of CO2 in water at 0°C P = 3atm
 First… Calculate molar quantities.
2
2
44 /
18 /
CO
H O
MW g mol
MW g mol


2
2
0.007909
0.348
44 /
100
5.556
18 /
CO
H O
mass g
mol
MW g mol
mass g
mol mol
MW g mol
  
  

 Now, total mol:
 Get molar content of CO2
2
2
0.007909
0.348
44 /
100
5.556
18 /
CO
H O
mass g
mol
MW g mol
mass g
mol mol
MW g mol
  
  
2 2
2
2
5.5630.0079 5.556 9
0.0014
.
0.0
5639
079
5
CO H O
CO
CO
mol mol mol
mol
x
mol
    
  
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 Get Molarity (mol of solute per liter of solution)
 Henry’s Constant (atm/M)
0.0079
0.079
0.1
A
solvent
mol mol
M M
L L
  
1
0.079
A A
A
A
P HxM
P atm
H
M M

 
12.658atm
MH
Henry’s Constant (atm/xa)
1
0.0014
714.28 A
A
A
atm
x
P atm
H
x
  

 B) Calculate the solubility of CO2 in water at 0°C P = 3atm
3.00
12.65
0.2371
8
A A
A
A atm
M
A
P HxM
P atm
M
M
H
M

 

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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) 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
Molecular Diffusion is STRONG
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1. Introduction to Mass Transfer
 Fluxes & Velocities
 Diffusion Coefficient

 When a system contains two or more components whose concentrations vary from
point to point, there is a natural tendency for mass to be transferred, minimizing
the concentration differences within the system and moving it towards equilibrium.

 The transport of one component from a region of higher concentration to that of a
lower concentration is called mass transfer.
 In this example, flow goes from:
 High Concentration to Low Concentration
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 Mass transfer plays an important role in many industrial processes.
 A group of operations for separating the components of mixtures is based on the
transfer of material from one homogeneous phase to another.
 These methods-covered by the term “mass transfer operations” include such
techniques as:
 distillation, gas absorption, humidification, liquid extraction, adsorption, membrane
separations, and others.

 The driving force for transfer in these operations is a concentration gradient.
 Similar to a temperature gradient provides the driving force for heat transfer.
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 Do NOT confuse Mass Transfer with Movement of Mass!
 There must be a change in concentration in order to have a Mass Transfer Phenomena
• If color is T
• If color if V
• If color if C

 Typically occurs in stagnant conditions or laminar flow
 Focuses on layer study
 Studied in this Chapter
 Convection Mass Transfer
 Typically occurs in Turbulent flow
 Aka Eddy Diffusion
 Focuses in Bulk Studies
 Studied in Next Chapter
 Interphase Mass Transfer
 Mass Transfer between two phases at least… Turbulent Flow typically
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 As early as 1815 it was observed qualitatively that whenever a gas mixture contains
two or more molecular species, whose relative concentrations vary from point to
point, an apparently natural process results which tends to diminish any inequalities
in composition.
 This macroscopic transport of mass, independent of any convection effects* within
the system, is defined as molecular diffusion
Convection Effects 
Due to movement or changes in
velocity/viscosities
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 Check out this video…
 https://www.youtube.com/watch?v=aCtzlQL3GaM
 Try to ask this question:
 Why does the dye moves?

https://demonstrations.wolfram.com/DiffusionOfGasesInATube/

 Let us consider:
 Non-uniform multi-component fluid mixture
 Having bulk motion owing to pressure difference
 The different components are moving:
 At different molecular velocities as a result of diffusion.
 Two types of average velocities with respect to a
stationary observer have been defined for such cases…
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 Two types of average velocities
 For an n-component system, the mass average velocity in the x-direction is defined:
 Another form of average velocity for the mixture is the molar average velocity defined as:
1
1
u
n
i i
i
u
 
 
1
1 n
i i
i
U c u
c 
 

 The velocities u and U:
 Are approximately equal at low solute concentrations in binary systems & in nonuniform
mixtures of components having the same molecular weight.
 The velocities u and U:
 Are also equal in the bulk flow of a mixture with uniform concentration throughout regardless
of the relative molecular weights of the components.

 Frames of reference:
 are the co-ordinates on the basis of which the measurements are made.
 Three frames of reference or co-ordinates are commonly used for measuring the
flux of a diffusing component
 It is assumed that in a frame of reference
there is an observer who observes or
measures the velocity or flux of a component
in a mixture.
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 If the frame of reference or the observer is stationary with respect to the earth:
 He notes a velocity ui of the ith component.
That guy is at
u = 1m/s
u = 1 m/s

 If the observer is located in a frame of reference that moves with a mass average
velocity u of component i:
 He will note a velocity (ui – u0) of the component i
 This is the relative average velocity of the component with respect to the observer who
himself is moving with the velocity u in the same direction.
Now, I’m at
u = 0 m/s
u = 1 m/s

 The flux is defined as the rate of transport of species i per unit area in a direction
normal to the transport.
 The flux is calculated with respect to a fixed reference frame.
 Another convenient expression is:
 Relates Velocity + Concentration instead of Mass Flow per unit area (SAME UNITS)
 ui  fixed reference!
i mol i iN C u 
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 But…
 In a system, since several molecular species move with different average velocities,
a frame of moving reference must be chosen.
 The important moving references are mass average, molar average and volume
average velocities.
 Imagine moving a bottle inside a car…
 The car is moving at 60 km/h
 You move the bottle inside at 1m/s
 We care the velocity of the bottle inside the car!

 That’s why it is convenient to interpret the total flux of species “i” with respect to
an arbitrary reference frame rather than a fixed set of reference frame (typically,
the velocity of material in a pipe or equipment)
 The molar flux of species i based on arbitrary reference velocity u0 is denoted by Ji-
mol which can be defined as
0( )i mol i iJ C u u  

 A gas mixture containing 65% NH3, 8% N2, 24% H2 and 3% Ar is flowing through a pipe
25 mm in diameter at a total pressure of 4.0 atm.
 The velocities of the components are as follows:
 Topic: Estimation of mass average, molar average velocity and volume average
velocity
 A) Calculate the mass average velocity
 B) the molar average velocity
 C) the volume average velocity of the gas mixture.
3
2
2
0.030
0.030
0.035
0.020
m
s
m
s
m
s
m
s
NH
N
H
Ar




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 A) Calculate the mass average velocity
 From the definition of velocity:
 We can know that:
 ti = density of ith component
 t = total density of gas mix
 MWi = molar weight of ith component
 MW = average molar weight of the mix.
1
1 1 2 2 3 3 4 4
1
1
( )
n
i i
i
u u
u t u t u t u t u





   

;i
i i
p p
t MW t MW
RT RT
 

 Substitute in the expression for mass average velocity (u)
 Now, calculate the average Molecular Weight (MW)=
 Calculating the “sum”
 Substitute in previous equation:
4
1
1
i i i
i
u y u MW
MW 
 
1 1 2 2 3 3 4 4
(0.65 17) (0.08 28) (0.24 2) (0.03 40) 14.97
MW y MW y MW y MW y MW
MW x x x x
   
    
4
1
(0.65 17 0.03) (0.08 28 0.03) (0.24 2 0.035) (0.03 40 0.02 0.43 5) 9i i i
i
y u MW x x x x x x x x

    
4
1
0.4395
1
0.02935
4.
8
97
i i i
m
i
s
y u MW
u
MW
u


 

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 B) Molar average velocity
 From the equation “U”:
 Now, substitute the values given:
1 1 2 2 3 3 4 4
1 1 2 2 3 3 4 4
1 1 2 2 3 3 4 4
1
( )
1 1 1 1
( ) ( ) ( ) ( )
U
U c u c u c u c u
c
U c u c u c u c u
c c c c
y u y u y u y u
   
   
   
1 1 2 2 3 3 4 4U (0.65 0.03) (0.08x 0.03) (0.24x 0.035) (0.03 0.02)
U 0.0309m
s
y u y u y u y u x x       


 C) the volume average velocity of the gas mixture.
 In this specific case, since we are talking about ideal gases:
. . 0.0309m
sVolumeavg Velocity Molar avg Velocity 
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 Molecular diffusion or molecular transport can be defined as:
 transfer or movement of individual molecules through a fluid by means of random,
individual movements of the molecules.
 The molecules travel only in straight lines and in the process, may collide with
other molecules in their path.
 The molecules then change direction (still in a straight line) after the collision.
 This is sometimes referred to as a random-walk process as shown below:

 Movement due to thermal molecules
 Kinetic theory  Erratic Movement of A and B
 rate of diffusion is defined the total distance travelled per unit time
 Rate of Diffusion = Total Distance / total time
 Increasing Temperature favors increase in thermal energy, therefore, increase in
Rate of Diffusion
 Decrease in Pressure favors decrease in collisions.
 If this is true, the total distance or free mean path will be higher, hence, faster rate.
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 Molecular Diffusion can happen anywhere!
 Solids
 Liquids
 Gases
 In this Case, we will focus mainly in
Diffusion of Gases!

 Molecular diffusion in gases is caused by random
movement of molecules due to their thermal energy
and hence the kinetic theory of gases helps to
understand the mechanism of molecular diffusion in
gases.
 According to this theory:
 molecules of gases move with very high speed.
 For instance, at 273 K temperature and 101.3 kN/m2
pressure, the mean speed of oxygen molecules is
about 462 m/s.
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 It may, therefore, be expected that the rate of molecular diffusion should be very
high.
 But actually molecular diffusion is an extremely slow process since the molecules
undergo several billion collisions per second
 Therefore, their velocities frequently change both in magnitude and direction, thus
making the effective velocity very low.
 From now on:
 Assume convective MT is faster in rate!
 Convective MT is aka Eddys Diffusion

 Molecular  Due to thermal/kinetic energies
 Eddys  Due to movement

 Identify:
 Molecular Diffusion Phenomena
 Convective MT
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 Identify:
 Molecular Diffusion Phenomena
 Convective MT

 Diffusivity or Diffusion Coefficient is a measure of the capability of a substance or
energy to be diffused or to allow something to pass by diffusion.
 The constant value: DAB
 As you can imagine, the higher the coefficient/diffusivity, the faster the rate of
mass transfer
 Diffusivity is a rate of diffusion, a measure of the rate at which particles or fluids
can spread.

 Typically we would cover:
 Diffusivities
 Gas
 Liquid
 Solids
 Estimation
 Fullers Equations
 Chapman-Enskog Equation
 Wilke-Chang Equation
 Experiment
 Twin Bulb
 Stefan Tube
 Diaphragm Method
We will only cover this topic for
GAS/Vapor-Liquid Operations

 In general, you will obtain diffusivity of A through B:
 Reference/Bibliography/Literature or previous experimentation reported data
 Estimations via correlation equations
 Experimentation
 Units
 m2/s
 cm2/s
 Typical ranges
 Gases 
 Liquids
 Solids 
 Clearly, solids have much lower rates of diffusion… Why?
25
10 m
s

210
10 m
s

2 210 15
10 10m m
s sto 
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https://demonstrations.wolfram.com/BinaryDiffusionCoefficientsForGases/

https://demonstrations.wolfram.com/DiffusionCoefficientsForMulticomponentGases/

1. Fick’s Law
 Fick’s Model
 Case (A) Equimolar Counter-Diffusion (EMD)
 Case (B) Unimolecular Diffusion (UMD)
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 Fick’s Law
 commonly used in basic sciences to describe diffusion
 Typically under no turbulence
 Mass Transfer Coefficient
 The coefficient, k takes care of several parameters which cannot be directly measured
 Requires correlations commonly used in engineering
 Cylinders, Heaters, Spheres, Tubes, etc.
 Both the models have striking similarity with Ohm’s law.

 First described the molecular diffusion in:
 Isothermal, isobaric binary system of components A and B
 According to his idea of molecular diffusion:
 the molar flux of a species relative to an observer moving with molar average velocity is
proportional to the concentration gradient in a certain direction.
 Proportionality Constant  Diffusivity or Diffusivity Constant
 Typically Done per Axis (Z)
Minus (-) Sign
Drop in concentration toward
direction of Diffusion
AdC
J
dZ

A
AB
dC
J D
dZ
 
B
B BA
dC
J D
dZ
 
A
A AB
dC
J D
dZ
 
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 Assumptions:
 Isobaric
 Isothermal
 No chemical reaction
 DAB is constant
 One Dimension
 Dilute solutions
 Steady State
Minus (-) Sign
Drop in concentration toward
direction of Diffusion
A
AB
dC
J D
dZ
 

https://demonstrations.wolfram.com/SteadyStateBinaryFickianDiffusion/
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 Fick’s Law must be applied to each individual case.
 Some common applications:
 Steady-State Diffusion:
 Through Non-Diffusing Component in Steady-State
 Through a Constant Area in Steady-State
 Through Variable Area in Steady-State
 Through Figures (Plane Wall, Hollow Cylinder, Spheres, Disks, etc...)
 Unsteady-State Diffusion
 Diffusion through Liquids & Solids
 It is not our scope to account for all of the mass transfer cases
 We use the most common ones of interest in engineering (Diffusion of Gases)

 Examples:
 Diffusion of component A through a stagnant layer of component B
 Equimolar counter-diffusion of two components
 Non-equimolal Counter-Diffusion of Two Components
 Diffusion Through Moving Bulk Fluid
 Case (A) - Equimolar Counter-Diffusion (EMD) in Steady-State
 Case (B) - Unimolecular Diffusion (UMD) in Steady State

 For any given case, we know that the molar flux of A is given by:
 Total Molar Flux of A = Molar Flux of A due to Molecular Diffusion + Molar Flux of A due to Bulk Velocity
 For a binary case:
A A A
A A A
N J c V
N J y N
 
 
( )A A A A BN J y N N  
A BN N N 
PA2
PB2A
B
Total P
Distance, z
PA1
PB1

 In equimolar counter-diffusion, the molar fluxes of A and B are equal, but opposite
in direction, and the total pressure is constant throughout.
 Hence we can write:
 We know that Fick’s Law:
(0)A A A
A A
N J y
N J
 

PA2
PB2A
B
Total P
Distance, z
PA1
PB1
0
( )A A A B
A B
AN J
f N N
N
i
y N
N   
  
A
A AB
dC
J D
dZ
 
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 For this specific Case:
PA2
PB2A
B
Total P
Distance, z
PA1
PB1
A
A A AB
A
AA B
dC
N J D
dZ
dC
D
dZ
N
  

 Under equimolar counter-diffusion, the diffusivity
of A in B is the same as the diffusivity of B in A:
 i.e.
 Also: AB BAD D
A B
A B
N N
J J
 
 

 Now, for Ideal Gas Case
 We know that:
A A
A
A
A A
A
A
PV nRT
P V n RT
n
C
V
n P
V RT
P
C
RT





PA2
PB2A
B
Total P
Distance, z
PA1
PB1
A
ABA
d
d
N
C
D
z
 
A
ABA
d
d
N
C
D
z
 
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 The differentials
 Which can be simplified to:
 Solving now for the 1st Order Differential Equation
PA2
PB2A
B
Total P
Distance, z
PA1
PB1
A
AB B
A
A AA
P
RTN N
d
dC
D D
dZ dz
  
AB A
A A
P
N J
R
D
dT
d
z
  
AB A
AB
A
A A
P
J
RT
J
D d
dz
D
P
R
dz d
T
 
 

 Integrating
 Limits:
 Z1 to Z2
 PA1 to PA2
 Further Simplification for JA:
2 2
1 1
A
A
A A
Pz
AB
A A
z P
AB
J P
RT
D
J dz dP
D
RT
dz d 
  
PA2
PB2A
B
Total P
Distance, z
PA1
PB1
2 1
2 1
2 1
2 1
( ) ( )
( )
( )
AB
A A A
A AAB
A
D
J z z P P
RT
P PD
J
RT z z
   

 

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 For B is the same case, as
PA2
PB2A
B
Total P
Distance, z
PA1
PB1
2 1
1 2
2 1
2 1
( )
( )
( )
( )
( )
0
B B B A B
B B
B BA B
B B BA
B BBA
B
A B
B BAB
B
A B
N J y N N
N J
dC D dP
N J D
dz RT dz
P PD
J
RT z z
J J
if N N N
P PD
J
RT z z
  

    

 

 
   

  

 Referencing to A

 Our Equations:
1 2
2 1
2 1
2 1
( )
( )
( )
( )
B BAB
B
B BAB
B
P PD
J
RT z z
P PD
J
RT z z

 


  

2 1
2 1
( )
( )
A AAB
A
P PD
J
RT z z

 


 Ammonia gas (A) and nitrogen (B) are stored in 2 large storage tanks respectively.
 A uniform tube 0.10 m long connects the 2 tanks
 The partial pressure of A:
A B
L=0.1m
5
1.013 10
298
tanks
tanks
P x Pa
T K


1
2
4
4
1 1.013 10
2, 0.507 10
A
A
point is p x Pa
point p x Pa


2
3
4
0.230 10
8.314
m
sAB
m Pa
molK
D x
R



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 (a) Calculate the flux JA at steady-state.
 (b) Repeat for JB.
A B
L=0.1m

 We can assume this is equimolar since:
 Moles flow from A to  and B to 
 Total Mol flux = 0, since P, T, V are constant
 Recall that:
 Since these are gases, then… Assume Ideal Gas Law (Why?)
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 From our previous equations… Get JA
 Note that we have all!
PA1
PA2
2 1
2 1
( )
( )
A AAB
A
P PD
J
RT z z

 

2 1 0.10L z z m   2
3
4
0.230 10
8.314
m
sAB
m Pa
molK
D x
R



A B
L=0.1m

 Substitute all in equation
 NOTE: Use only SI-units
 
2 1
2
3
2
2
2 1
4 4 4
4
( )
( )
(2.3 10 )(0.507 10 1.1013 10 )
8.314 (298 )(0.10 )
0.0004932
4.932 10
A AAB
A
m
s
A m Pa
molK
mol
A m s
mol
A m s
P PD
J
RT z z
x x x Pa
J
K m
J
J x



 


 


A B
L=0.1m
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 (b) Repeat for JB.
 Use Dalton law of Partial Pressure to get PB1 and PB2
1 2
2 2
1 1
2 2
1 1
4 4
5 4 4
5 4 4
(1.013 10 ); (1.013 10 );P (0.507 10 );
(1.0132x10 0.507 10 ) 9.625 10
(1.013x10 1.1013 10 ) 9.119 10
T A B
B T A
T A A
B T A
B T A
B T A
B T A
P P P
P P P
P x Pa P x Pa x Pa
P P P
P P P
P P P x Pa x Pa
P P P x Pa x Pa

 
 
  
 
 
    
    
2
1
4
4
9.625 10
9.119 10
B
B
P x Pa
P x Pa


A BL=0.1m
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 
2 1
2
3
2
2
2 1
4 4 4
4
( )
( )
(0.23 10 )(9.625 10 9.119x10 )
8.314 (298 )(0.10 )
0.00049322
4.9322 10
B BBA
B
BA AB
m
s
B m Pa
molK
mol
B m s
mol
B m s
P PD
J
RT z z
D D
x x Pa
J
K m
J
J x



 



 



 Compare JA and JB
 As expected… (in different directions)
 Assumption of equimolar condition is correct!
2
2
4
4
4.9322 10
4.9322 10
mol
A m s
mol
B m s
A B
J x
J x
J J





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 Two bulbs are connected by a straight tube, 0.001m (0.1cm) in diameter and 0.15m
(15cm) in length.
 Initially the bulb at End 1 contains N2 and the bulb at End 2 contains H2
 Pressure and temperature are constant at 25°C and 1 atm.
 At a time after diffusion starts, tf, the nitrogen content of the gas at:
 End 1 of the tube is 80 mol%
 End 2 is 25 mol%
Example 3.1 Separation Process
Principles, J. D. Seader, 3rd Edition
L=0.15m 21
N2 H2

 If the binary diffusion coefficient is 0.784 cm2/s, determine:
 (a) The rates and directions of mass transfer in mol/s
 (b) The species velocities relative to stationary coordinates, in cm/s
Example 3.1 Separation Process
Principles, J. D. Seader, 3rd Edition
L=0.15m 21
N2 H2
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 (A) The rates and directions of mass transfer in mol/s
 Because the gas system is closed and at constant pressure and temperature
 no bulk flow occurs
 mass transfer in the connecting tube is EMD 
 The area for mass transfer through the tube, in cm2 is given as follows:
A BN N
   
22 0.1
2 2
3 2
A
7.85 10
D
A x cm
 

 

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 By the ideal gas law:
 Take as the reference plane End 1 of the connecting tube.
 Apply the equation:
3
51
(82.06c)(298 ) 4.09 10
P
RT
atm molP
RT K cm
PV nRT C
C x 
  
  
   
 
2 2
2
1 2
2 1
5
3
9
( )
(4.09 10 )(0.784)
0.80 0.25 (7.85 10 )
(15)
9.23 10 /
AB
A N N
A
A N
cD
J x x A
z z
x
J x
J J x mol s



   
 
  
 
1 2A A
2 1
(x x )
( )
AB
A
cD
J
z z
 


 This case is valid for EMD:
 the molar-average velocity of the mixture,
 Therefore, assume that:
 the species velocities are equal to species diffusion velocities.
  2 2
2 2
2 2
N N
N N D
N N
J n
v v
C Acx
  
0 /U mol s
  2 2
2 2
2 2
H H
H H D
H H
J n
v v
C Acx
  
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 Solving:
 When End 1, xN2 = 0.80
 When End 1, xN2 = 0.80
2
2
2 2 2
2 2
2
2 2
9
3 5
9.23 10 0.0287
(7.85 10 )(4.09 10 )
0.0287 0.0287
N
N
N N N
N H
H
H N
n x
v
Acx x x x x
v v
v
x x

 
  
 
  
2
2
0.0287 0.028
0
7
0.8
.035875 /
0
N
N
cm sv
x
  
When End 2, xN2 = 0.25
2
2
0.0287 0.0287
0.25
0.1148 /N
N
v
x
cm s  
When End 2, xH2 = 0.75
2
2
0.0287 0.0287
0.75
0.03826 /H
N
v
x
cm s  
2
2
0.0287 0.0287
0.20
0.1435 /H
N
v
x
cm s  
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 So… when is v = 0m/s?
 CASE: End 2:
 Then:
When End 2, xN2 = 0.25
2
2
0.0287 0.0287
0.25
0.1148 /N
N
v
x
cm s  
When End 2, xH2 = 0.75
2
2
0.0287 0.0287
0.75
0.03826 /H
N
v
x
cm s  
2 2 2 2
0.1148 / ) 0.03826 / ) 0cm/ s(0.25)( (0.75)(
M N N H H
M cm s cm
v x v x
s
v
v  
 

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 Note that we are in fact using steady state
 Many questions I get:
 Why if Concentration changing?
 Why is Pressure changing?
 Why velocities changes?
 My main Answer:
 changes with respect to LOCATION that is, the distance
 Time is irrelevant
PA2
PB2A
B
Total P
Distance, z
PA1
PB1
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 A binary gaseous mixture of components A and B at a pressure of 1 bar and
temperature of 300 K undergoes steady-state equimolar counter-diffusion along a 1-
mm-thick diffusion path
 At one end of the path the mole fraction of component A is 70%, while at the other
end it is 20%.
 Under these conditions, DAB = 0.1 cm2/s.
 A) Calculate the molar flux of component A.

 A) Calculate the molar flux of component A.
 Direct substitution into equation 
 
 
  
1 2 1 2
1 2 1 2
3
2
A A A A
2 1 2 1
A A A A
2 1 2 1
5 2 5
3
mol
m s
(x x ) (x x )
( ) ( )
( ) ( )
( ) ( )
(1.0x10 m / s)(1.0x10 )(0.7 0.2)
8.314 300 (1.0 10 )
0.20
AB
P
ABRTAB
A
D P
RT AB
A
A m Pa
molK
A
DcD
J
z z z z
D P
J x x x x
z z RT z z
Pa
J
K x m
J


   
 
   
 



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 A mixture of He and N2 gas is contained in a pipe at:
 T= 298K
 P = 1 atm
 Flow = constant flow throughout
 Conditions:
 At one end, point 1, PA1 of He = 0.6 atm
 At the other end, point 2, PA2 of He = 0.2 atm
 Note that the distance is 20 cm between each other.
 A) Calculate the flux of He at steady state
Example 6.1-1 From Transport Processes
and Unit Operations, Geankopolis 3rd.
Molecular diffusion of he in N2
24
0.687 10 m
AB sD x 


 Solution
 Total Pressure is constant trough time and pipe!
 C must be constant as well.
 Use either:
 Since P is given, use “P”
; / /V nRT n V P RT C  
2 1 2 1 2 1
2 1 2 1 2 1
( ) ( ) (P )
( ) ( ) ( )
A A A A A AAB
A AB AB
C C x x PD
J D cD
z z z z RT z z
  
     
  

 Substitute in:
2 1
2
2
3
2
2 1
4
4
8
(P )
( )
(0.687 10 ) (0.6 0.2 )
(0.082 )(298K) (0.2m)
(0.687 10 ) (0.6 0.2 )x(101325Pa/ atm)
(0.2m)(8.314 )(298K)
(2.772 10 )(151987.5)
0.0
A AAB
A
m
s
A Latm
molK
m
s
A Pa m
molK
mol
A m s
A
PD
J
RT z z
x atm atm
J
x atm atm
J
J x
J





 


 

 

 2
2
3
04214
4.2 10
mol
m s
mol
A m s
J x 


Air
 As the name implies, there is only a species diffusing through a stagnant medium
 Typical Examples:
 Evaporation of Species A through B
 Quasi-steady State will be required
B
e
n
z
e
n
e
Air Flow
B
e
n
z
e
n
e
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 The general 1-Dimensional, steady state equation
 And we know…
 Substituting Fick’s Law and Total mol flux
;A
A A B
C
y N N N
C
  
( )
( )
A A
A AB A B
A A
A AB A
dC C
N D N N
dz C
dC C
N D N
dz C
   
  
A A A
A A A
N J c V
N J y N
 
 
Total Molar Flux of A = Molar Flux of A due to Molecular Diffusion + Molar Flux of A due to Bulk Velocity

 This is our “general equation”
 In a stagnant medium:
 We will work these cases:
 Concentrations
 Partial Pressures
 Molar Fractions
( )A A
A AB A B
dC C
N D N N
dz C
   
A A
A AB A
dC C
N D N
dz C
  
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 UMD Case – Partial Pressures
 We will get from:
 To:
Working:
Concentrations
A A
A AB A
dC C
N D N
dz C
  
2
1
A
2 1 A
C
ln
( ) C
AB
A
ccD
N
z z c
 
     
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 Since there is no diffusion of B:
 Substitute in
 We get
BN 0
( )A A
A AB A B
dC C
N D N N
dz C
   
( 0)
( )
A A
A AB A
A A
A AB A
dC C
N D N
dz C
dC C
N D N
dz C
   
  
Working:
Concentrations

( )
1
1
A A
A AB A
A A
A A AB
A A
A AB
A A
A AB
A
A AB
A
A
A AB
A
dC C
N D N
dz C
C dC
N N D
C dz
C dC
N D
C dz
c C dC
N D
c dz
dCc
N D
c C dz
dC
N cD
c C dz
  
  
 
   
 
 
  
 
 

 

Working:
Concentrations
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2 2
1 1
2 2
1 1
1
1
1
1
A
A
A
A
A
A AB
A
A AB A
A
Cz
A AB A
Az C
Cz
A AB A
Az C
dC
N cD
c C dz
N dz cD dC
c C
N dz cD dC
c C
N dz cD dC
c C
 

 

 

 

 
 
Working:
Concentrations
2 1
1
2
1
2
2
1
2 1 A A
A
2 1
A
A
2 1 A
A
2 1 A
( ) ln( C ) ( ln( C ))
C
( ) ln
C
C
ln
( ) C
C
ln
( ) C
A AB
A AB
AB
A
AB
A
N z z cD c c
c
N z z cD
c
ccD
N
z z c
ccD
N
z z c
         
 
      
 
      
 
     

 Our main equation:
Working:
Concentrations2
1
A
2 1 A
C
ln
( ) C
AB
A
ccD
N
z z c
 
     
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 UMD Case – Partial Pressures
 To:
Working:
Partial Pressures
2
12 1
ln
( )
AAB
A
A
P PD P
N
RT z z P P
 
     
( )A A
A AB A B
dC C
N D N N
dz C
   

 Since there is no diffusion of B:
 Substitute in
 We get
Working:
Partial PressuresBN 0
( )A A
A AB A B
dC C
N D N N
dz C
   
( 0)
( )
A A
A AB A
A A
A AB A
dC C
N D N
dz C
dC C
N D N
dz C
   
  
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 From ideal gas:
 Substitute in all CA 
Working:
Partial Pressures
AC AP
RT

1
A
A A A
P
d
dC dP dPRT
dz dz RTdz RT dz
 
 
   
( )
1
( )
1
( )
A A
A AB A
A A
A AB A
A A
A AB A
dC C
N D N
dz C
dP C
N D N
RT dz C
dP C
N D N
RT dz C
 
    
 
    
  

 Now, given that the ratios of pressure-concentration are the same 
 Substitute in:
 And we get:
Working:
Partial Pressures
1
( )A A
A AB A
dP C
N D N
RT dz C
  
A AP C
P C

1
( )
1
( )
1
1
A A
A AB A
A A
A A AB
A A
A AB
dP P
N D N
RT dz P
P dP
N N D
P RT dz
P dP
N D
P RT dz
  
  
 
   
 

 Solve Diff. Equation
Working:
Partial Pressures
2 2
1 1
2 2
1 1
1
1
1
1
1
1
1
A
A
A
A
A A
A AB
A
A AB
A
AB A
A
A
Pz
AB A
A
Az P
Pz
AB A
A
Az P
P dP
N D
P RT dz
dP
N dz D
PRT
P
D dP
N dz
PRT
P
D dP
N dz
PRT
P
D dP
N dz
PRT
P
 
   
 
 
 
 
 


 
 
 


 
 
 


 
 
 
 
 
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Working:
Partial Pressures2 2
1 1
2 2
1 1
2 2
1 1
2
1
2 1
1
1
1
( )
A
A
A
A
A
A
A
A
Pz
AB A
A
Az P
Pz
AB
A A
Az P
Pz
AB
A A
Az P
P
AB
A A
AP
D dP
N dz
PRT
P
D P
N dz dP
RT P P
D P
N dz dP
RT P P
D P
N z z dP
RT P P


 
 
 







 

 
 
 


Working:
Partial Pressures
2
12 1
ln
( )
AAB
A
A
P PD P
N
RT z z P P
 
     
1
2
2
1
2
1
2 1
2 1
2 1
( ) ln
( ) ln
ln
( )
AAB
A
A
AAB
A
A
AAB
A
A
P PD P
N z z
RT P P
P PD P
N z z
RT P P
P PD P
N
RT z z P P
 
     
 
     
 
     
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 We get our equation:
Working:
Partial Pressures
2
12 1
ln
( )
AAB
A
A
P PD P
N
RT z z P P
 
     

 UMD Case – Molar Fractions
 To:
( )A A
A AB A B
dC C
N D N N
dz C
    Working:
Molar Fractions
2
1
A
2 1
1
N ln
( ) 1
AAB
A
yCD
z z y
 
     
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 From our initial equation
 Force yA in Fick’s Law
 Get Diff. Equation
A
A AB A
dC
N D y N
dz
  
A
A AB A
dy
N cD y N
dz
  
A
A
N (1 )
N
(1 )
A
A A AB
A
A AB
AB A
A
dy
N y N cD
dz
dy
y cD
dz
cD dy
y dz
  
  



Working:
Molar Fractions

2 2
1 1
2 2
1 1
A
A
A
A
N
(1 )
N
(1 )
N
(1 )
N
(1 )
A
A
A
A
AB A
A
A
AB
A
z y
A
AB
Az y
z y
A
AB
Az y
cD dy
y dz
dy
dz cD
y
dy
dz cD
y
dy
dz cD
y



 

 

 

 
 
Working:
Molar Fractions

2 2
1 1
1
2
1
2
2
1
A
A 2 1
A
2 1
A
2 1
N
(1 )
1
N ( ) ln
1
1
N ln
( ) 1
1
N ln
( ) 1
A
A
z y
A
AB
Az y
A
AB
A
AAB
A
AAB
A
dy
dz cD
y
y
z z cD
y
ycD
z z y
ycD
z z y
 

 
      
 
     
 
     
 
2
1
A
2 1
1
N ln
( ) 1
AAB
A
yCD
z z y
 
     
Working:
Molar Fractions

 Compare!
2
1
A
2 1
1
N ln
( ) 1
AAB
A
ycD
z z y
 
     
2
12 1
ln
( )
AAB
A
A
P PPD
N
RT z z P P
 
     
Working:
Molar Fractions
Working:
Partial Pressures
Working:
Concentrations
2
1
A
2 1 A
C
ln
( ) C
AB
A
ccD
N
z z c
 
     
2
12 1
ln
( )
AAB
A
A
P PcD
N
z z P P
 
     
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 Oxygen (A) is diffusing through carbon dioxide (B) under steady-state conditions,
with the CO2 non-diffusing.
Path
Non-Diffusion
1
2
2
5
3
4
4
5
10
273
2.0 2.0 10
1.30 10
0.65 10
1.87 10
A
A
m
AB s
P Pa
T K
L mm x m
P x Pa
P x Pa
D x




 




 (a) Calculate the molar flux of O2 in the mixture.
 Assume Unimolecular Diffusion of Oxygen
 CO2 is not diffusing…
L=2.0mm
Non-Diffusion
5
10
273
P Pa
T K


1
4
1.30 10AP x Pa
2
4
0.65 10AP x Pa
25
1.87 10 m
AB sD x 


 Substitute all data:
L=2.0mm
Non-Diffusion
5
10
273
P Pa
T K


1
4
1.30 10AP x Pa
2
4
0.65 10AP x Pa
25
1.87 10 m
AB sD x 

 
 
2
1
3
2
2
2 1
5 2 5 4 4
4 43
ln
( )
(1.87 10 / )(10 ) 10 10 0.65 10
ln
10 10 1.30 108.314 (273 )(2 10 )
0.4119 ln(1.0747)
0.02967
AAB
A
A
A Pa m
molK
mol
A m s
mol
A m s
P PD P
N
RT z z P P
x m s Pa x x
N
x xK x m
N
N


 
     
 
  
 


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 (a) Calculate the molar flux of O2 in the mixture.
2 2 2
0.02967 / 2.967 10 /AN mol m s x mol m s
 
L=2.0mm
Non-Diffusion
5
10
273
P Pa
T K


1
4
1.30 10AP x Pa
2
4
0.65 10AP x Pa
25
1.87 10 m
AB sD x 


 (b) Calculate the total moles in diffusion per unit area in 5 seconds
 (c) if the spill is 0.45 m2; calculate molar flow
2
2
(2.967 10 / 2 )(5s)
0.14835 /
A A
A
mol N t x mol m s
mol mol m

 

2 2 2
(2.967 x10 mol/ m s)(0.45m )
0.01335 /
A A
A
F N A
F mol s

  

L=2.0mm
Non-Diffusion
5
10
273
P Pa
T K


1
4
1.30 10AP x Pa
2
4
0.65 10AP x Pa
25
1.87 10 m
AB sD x 


 An open beaker, 6 cm high, is filled with liquid benzene (A) at 25°C to within 0.5 cm of the top.
EXAMPLE 3.2 Evaporation from an Open
Beaker. Separation Process Principles, J.
D. Seader, 3rd Edition
 Dry air (B) at 25°C and 1 atm is blown across the mouth of the
beaker so that evaporated benzene is carried away by convection
after it transfers through a stagnant air layer in the beaker.
 The vapor pressure of benzene at 25°C is 0.131 atm.
Air
Liquid Benzene
6.0 cm
0.5 cm
1 atm; 25°C

 The mole fraction of benzene in the air at the top of the beaker is 0.0
 It can be determined by Raoult’s law at the gas–liquid interface.
 Diffusion coefficient (benzene in air):
 at 25°C and 1 atm
 DAB = 0.0905 cm2/s.
 Neglect the accumulation of benzene and air in the stagnant layer
with time as it increases in height
 This is te so called  Quasi-steady-state assumption
Air
Liquid Benzene
6.0
cm
0.5
cm
1 atm;
25°C
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 Calculate:
 (a) initial rate of evaporation of benzene as a molar flux in mol/cm2-s;
 (b) initial mole-fraction profiles in the stagnant air layer;
 (c) initial fractions of the mass-transfer fluxes due to molecular diffusion;
 (d) initial diffusion velocities

 Before we start…
 Since Dalton’s Law is valid
 And stated that Raoult’s Law can be applied:
1 1
1 1
1 1
1 1
1 1
0.131
1
0.131
1.0; 0.131
A sat A
sat
A A
A A
A A
A A
x P y P
P
y x
P
y x
y x
x y






 
We use x and y interchangeably in this case
Air
Liquid Benzene
6.0
cm
0.5
cm
1 atm;
25°C

 (a) initial rate of evaporation of benzene as a molar flux in mol/cm2-s;
 Use UMD equation for molar fractions (WHY?)
 The total vapor concentration by the ideal-gas law is:
5 3
1
82.06 298
4.09 10 /
P
C
RT
C
x
C x mol cm



2
1
A
2 1
1
N ln
( ) 1
AAB
A
yCD
z z y
 
     
First, let us calculate
C; DAB; Z2-Z1
2
1
A
2 1
1
N ln
( ) 1
AAB
A
yCD
z z y
 
     
Working:
Molar Fractions

 Now, assume that z equal to the distance down from the top of the beaker:
 z1 = 0.000m; at the top of beaker
 z2 = 0.005m; the distance from the top of the beaker to gas–liquid interface.
 Therefore, the layer will be:
 We can then state the compositions of Benzene in z1 and z2
2 1 0.005-0.000=0.005m=0.50z z dz cm  
1 1
2 2
( 0.00)
( 0.50)
0.131
0.000
A z
A z
x
x




Air
Liquid Benzene
6.0
cm
0.5
cm
1 atm;
25°C
z2
z1
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 Substitute in Equation
2
1
A
2 1
1
N ln
( ) 1
AAB
A
yCD
z z y
 
     
Working:
Molar Fractions
2
1
A
2 1
5
A
A
6
A
6 2
A
1
N ln
( ) 1
(4.09 10 )(0.0905) 1 0.00
N ln
(0.5 0) 1 0.131
N (0.000007402)(0.14075)
N 0.00000104195 1.04 10
N 1.04 10 /
AAB
A
yCD
z z y
x
x
x mol cm s



 
     
 
  
  

 

1 1 1
2 2 2
( 0.00)
( 0.50)
0.131
0.000
A A z
A A z
y x
y x


 
 
We use x and y
interchangeably in
this case

 (b) initial mole-fraction profiles in the stagnant air layer;
 Assume quasi-steady-state
 No accumulation of species in time
 Constant molar density
Ctotal is constant, as:
• P,T are constant
NA is constant, as:
• Quasi state
Air
B
e
n
z
e
n
e
Z2=0.50cm
Z1=0.00cm x1=0.131
x2=0.000
P
r
o
f
i
l
e
Using zx and xAx for “x any
point between z2 and z1

 The integral form, solving for “xAx” the concentration
at any point “Zx”
1 1
1
1
1
6
5
2 1
( )
(1.04 10 )( 0)
(4.09 10 )(0.0905)
0.281
1
1
ln
( ) 1
1 (1 )
1 (1 0.131)
1 0.869
Ax x
A
x
A x
AB
x
x
x
x
x
xz
AB A
A Az x
AAB
A
A
N z z
cD
A A
x z
x
A
z
A
cD dx
dz
N x
xcD
N
z z x
x x e
x e
x e


 
 
 
 
 
  
 

 
     
  
  
 
 
Air
B
e
n
z
e
n
e
Z2=0.50cm
Z1=0.00cm x1=0.131
x2=0.000
P
r
o
f
i
l
e
Zx=x-cm xAx=?
We want to 
• propose “Z” value and obtain
“X” value

 The table:
 These profiles are only slightly curved
 In this case, you could model as straight line
0.281
1 0.869 ; 1x
x x x
z
A B Ax e x x   
z (cm) xA xB
0.0 0.1310 0.8690
0.05 0.1187 0.8813
0.1 0.1062 0.8938
0.15 0.0936 0.9064
0.2 0.0808 0.9192
0.25 0.0678 0.9322
0.3 0.0546 0.9454
0.35 0.0412 0.9588
0.4 0.0276 0.9724
0.45 0.0139 0.9861
0.5 0.0000 1.0000
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 (c) initial fractions of the mass-transfer fluxes due to molecular diffusion
 Bulk-Flow Flux
 Molecular Diffusion Flux
 For this specific case:
 For the stagnant component, (Air or B) becomes 
0;
0
B
B B A BA
B
B
B A BA
B
B A BA
B
B B A BA
B A
dx
N x N cD
dz
N stagnant
dx
x N cD
dz
dx
x N cD
dz
dx
J x N cD
dz
J J
 

 

 
 

 (c) initial fractions of the mass-transfer fluxes due to molecular diffusion
z (cm) xA xB xA*NA xB*NA JB JA
0.00 0.1310 0.8690 1.3624E-07 9.0376E-07 9.0376E-07 -9.0376E-07
0.05 0.1187 0.8813 1.2345E-07 9.1655E-07 9.1655E-07 -9.16547E-07
0.10 0.1062 0.8938 1.1048E-07 9.2952E-07 9.2952E-07 -9.29516E-07
0.15 0.0936 0.9064 9.7332E-08 9.4267E-07 9.4267E-07 -9.42668E-07
0.20 0.0808 0.9192 8.3994E-08 9.5601E-07 9.5601E-07 -9.56006E-07
0.25 0.0678 0.9322 7.0468E-08 9.6953E-07 9.6953E-07 -9.69532E-07
0.30 0.0546 0.9454 5.675E-08 9.8325E-07 9.8325E-07 -9.8325E-07
0.35 0.0412 0.9588 4.2837E-08 9.9716E-07 9.9716E-07 -9.97163E-07
0.40 0.0276 0.9724 2.8728E-08 1.0113E-06 1.0113E-06 -1.01127E-06
0.45 0.0139 0.9861 1.442E-08 1.0256E-06 1.0256E-06 -1.02558E-06
0.50 0.0000 1.0000 0 0.00000104 0.00000104 -0.00000104
B
B B A BA
dx
J x N cD
dz
 
B AJ J 

 (d)
 initial diffusion velocities
 From:
;
A B
M
i i A B
iD AD BD
i i A B
i iD M
N NN
v
C C
J J J J
v v v
C x C x C x C
v v v

 
    
 
vM= molar average mixture velocity
viD= diffusion velocity of species “i”
Vi= velocity of species “i”
viD= diffusion velocity of species “i”

 Accordingly:
z (cm) xA xB JB JA vAD vBD
0.00 0.1310 0.8690 9.0376E-07 -9.0376E-07 0.168678027 -0.025427873
0.05 0.1187 0.8813 9.16547E-07 -9.16547E-07 0.188783884 -0.025427873
0.10 0.1062 0.8938 9.29516E-07 -9.29516E-07 0.213927575 -0.025427873
0.15 0.0936 0.9064 9.42668E-07 -9.42668E-07 0.246270102 -0.025427873
0.20 0.0808 0.9192 9.56006E-07 -9.56006E-07 0.289414642 -0.025427873
0.25 0.0678 0.9322 9.69532E-07 -9.69532E-07 0.349850521 -0.025427873
0.30 0.0546 0.9454 9.8325E-07 -9.8325E-07 0.44056655 -0.025427873
0.35 0.0412 0.9588 9.97163E-07 -9.97163E-07 0.591906119 -0.025427873
0.40 0.0276 0.9724 1.01127E-06 -1.01127E-06 0.89508972 -0.025427873
0.45 0.0139 0.9861 1.02558E-06 -1.02558E-06 1.808515543 -0.025427873
0.50 0.0000 1.0000 0.00000104 -0.00000104 #DIV/0! -0.025427873
;A B
AD BD
A B
J J
v v
x C x C
 
C=Constant
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 Water in the bottom of a narrow metal tube is held at
constant
 Temperature = 293K
 The total pressure of air is P = 1 atm.
 Temperature of air is also T = 293K
 P°vap (20°C)= 17.54 mm Hg = 0.023 atm
 Water evaporates and diffuses through air in the tube which
length = 0.1524 m
 D = 0.25x10-4 m2/s
 A) Calculate the rate of evaporation (kgmol/s-m-2)
EXAMPLE 6.2-2 Diffusion of Water through
stagnant, non-diffusing Air. Transport Process
& Unit Operations. Genkopolis. 3rd Ed.

 Water in the bottom of a narrow metal tube is held at
constant
 Temperature = 293K
 The total pressure of air is P = 1 atm.
 Temperature of air is also T = 293K
 P°vap (20°C)= 17.54 mm Hg = 0.023 atm
 Water evaporates and diffuses through air in the tube which
length = 0.1524 m
 D = 0.25x10-4 m2/s
 A) Calculate the rate of evaporation (kgmol/s-m-2)
EXAMPLE 6.2-2 Diffusion of Water through
stagnant, non-diffusing Air. Transport Process
& Unit Operations. Genkopolis. 3rd Ed.
Point 2
Point 1
Air
Water
Length of tube

 Solution:
 Since we are working mostly with pressure:
 We must first calculate Partial Pressures in both ends:

2
12 1
ln
( )
AAB
A
A
P PD P
N
RT z z P P
 
     
2 1
; ;A A totalP P P
Point 2
Point 1
Air
Water
Length of tube
PT=1atm
PT=1atm
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 Note that:
 Only 2 gases, 2 points
 Point 1
 Point 2
Point 2
Point 1
Air
Water
Length of tube
PT=1atm
1
0
1
T
water
Air
P atm
P atm
P atm



0.023
0.023 0.977
1
1
T
water vap
Air
P atm
P P atm
atm a matm tP 

  
 
A B totalP P P 
A B totalP P P 

 Substitute all data:
Point 2
Point 1
Air
Water
Length of tube
PT=1atm
1
0
1
T
water
Air
P atm
P atm
P atm



2
12 1
ln
( )
AAB
A
A
P PD P
N
RT z z P P
 
     
2
1
1 0
ln ln 0.02273
1 0.023
A
A
P P atm atm
P P atm atm
   
        
4 2
2 1
(0.25x10 / )(101325 )
0.0068232
( ) (8.314 / )(293 )(0.1524 )
ABD P m s Pa
RT z z J molK K m

 

4 2
7 2
0.02273 0.0068232 0.00015509
1.55 10 / m
1.55 10 / m
A
A
A
N x
N x mol s
N x kmol s


 


0.023
0.023 0.977
1
1
T
water vap
Air
P atm
P P atm
atm a matm tP 

  
 

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)
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 (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
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 Molecular Diffusion vs. Convective Mass Transfer
 Types of Models:
 Detailed physical description based on Fick‘s laws and the diffusion coefficient.
 Engineering approach based on the Mass Transfer Coefficient, MTC or k

 Check out:
 https://www.youtube.com/watch?v=BaBMXgVBQKk
 Try to imagine how to model:
 Fick’s Law
 Eddys Diffusion
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 Convective mass transfer occurs due to:
 the bulk motion of the fluid the mass transfer is faster compared to the molecular diffusion.
 Convective mass transfer is of two types
 forced convection mass transfer
 free convection mass transfer.
 Another analogy:
 mass diffusion + bulk fluid motion
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 Concentration Profiles
 Convective MT

 In some cases of convective mass transfer:
 may be possible to calculate the rate of mass transfer by solving the
differential equations obtained from mass and momentum balance provided
the nature of the flow is properly defined.
 Typical examples are:
 absorption of a gas in a laminar liquid film falling down along a wall
 dissolution of a solid coated on a flat plate in a liquid flowing over the plate.

 BUT!
 In more complex situations such as:
 gas-liquid contact in a packed or plate column
 dissolution of a solid in a mechanically stirred vessel
 theoretical calculations for rate of mass transfer
 It becomes quite difficult and may even be impossible
for the complex nature of the flow.
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 Recall Heat Transfer:
 Rate of Heat transfer = (HT Coefficient)(Area)(Change in Temperature)
 The coefficient, “h” depends on:
 Fluid Conditions:
 Temperature, Pressure, viscosity, density
 Figure being used:
 In this case a Pipe
 Characteristic Length: Diameter
 We would want to do this for MT as well!
  Q h A dT
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 When a fluid if flowing outside a solid surface, such as a cylinder, sphere, plate, etc.
in forced convection (i.e. Convective MT):
 We can express the rate of convective mass transfer from the surface to the fluid,
fluid to surface as follows:
 An analogy:
 Rate of Transfer  (Q) Heat transfer vs. Mass Transfer (NA)
 Driving Force  Change in Temperature vs. Change in Concentration
 Constant  h might be the thermal equivalent to kc for mass transfer
 As with Heat Transfer, kc has correlations for several figures
We will explore Kc in this section!
( )A c f iN k C C 
  Q h A dT

 REMEMBER
 Mass-transfer problems involving flowing fluids are often solved using mass-transfer coefficients,
which are analogous to heat-transfer coefficients.
 For mass transfer, a composition-driving force replaces DT.
 Because composition can be expressed in a number of ways:
 Expect different mass-transfer coefficients
 Mol fraction, mass/volume, mass/mass, volume/volume, mol/mol, etc…
 partial pressure difference (only for gases), concentration difference and mole fraction difference.
 If concentration (Mol per liter) is used, dCA is selected as the driving force:
 Units:

 the inverse of resistance to mass transfer.
( )A c f iN k C C 
Note:
The gradient for heat
was Temperature
change.
There is only one way
to “show”
temperature.
mol
time area driving force 
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 Let’s apply our new concept of “mass transfer coefficient” to our old friends!
 EMD – Equimolar diffusion
 UMD – Unimolecular diffusion
 Equations are already available for predicting the rates of molecular diffusion.
 Therefore, mass transfer coefficients as such are not required for cases where only
molecular diffusion is involved.
 But in order to have uniformity with eddy diffusion and to develop design
equations for some complex situations:
 Mass Transfer Coefficients are also used in molecular diffusion!
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 Let us consider the following transfer:
 From: component A from the bulk of a gas
 To: the bulk of a liquid through the gas-liquid interface.
 The transfer in the bulk of gas which is in turbulent motion
 is by eddy diffusion
 The near we go to the interface:
 is by molecular diffusion.
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 The rate of transfer for the gas section
 Similarly:
 the rate of transfer of A from the interface to the bulk of the
liquid may be expressed as
G iA A(p' p' )GA GN k 
i LA A(c c )LA LN k 
pAi
In this diagram:
L bulkA AC C

 Where;
= partial pressures of component A in the bulk of the gas
= partial pressures of component A at the interface
= concentrations of component A at the interface
= concentrations of component A at the interface in the bulk of the liquid
G iA A(p' p' )GA GN k 
i LA A(c c )LA LN k 
G
L
mol
A time area
mol
A time area
N rateof transferof A from gas phase
N rateof transferof A fromliquid phase


 
 
Pr
G
moles
k gas phaseMT coefficient
time area d essureof i
 
 
L
moles
k liquid phaseMT coefficient
time area dConcentration
 
 
' GAp
' iAp
iAc
LAc
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 Now we have two MTC (kG, kL) values
G iA A(p' p' )GA GN k  i LA A(c c )LA LN k 

 We will now try to “force” mass transfer coefficients in previous EMD & UMD Case.
 Recall that here, molecular diffusion is stronger than eddy’s diffusion
 The Main Goal:
 Obtain an equation with:
 Rate of Mass Transfer
 Mass Transfer Coefficient
 Driving Force

 We want this for:
 Bulk, interphase and liquid, gas phases
( )(d )coefficientRateof MT MT riving force
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 Try to imagine:
 EMD Case:
2 1 2 1 2 1
2 1 2 1 2 1
2 1 2 1 2 1
2 1 2 1 2 1
2 1 2 1 2 1
( ) ( ) (P )
( ) ( ) ( )
( ) ( ) (P )
( ) ( ) ( )
( ) ( ) (P )
A A A A A AAB
A AB AB
AB AB AB
A A A A A A A
A C A A x A A P A A
C C x x PD
J D cD
z z z z RT z z
D cD D
J C C x x P
z z z z z z RT
N F C C F x x F P
  
     
  
        
  
        

 Try to imagine:
 UMD Case:
1 2
1 2
A
,
A
,
N (y y )
N ( )
G
A A
B LM
L
A A
B LM
F
y
F
x x
x
 
 
2
1
2
1
2 1
2 1
1
ln
( ) 1
1
ln
( ) 1
G
L
AAB
A
A
AAB
A
A
yD
N
RT z z y
xD
N
RT z z x
 
     
 
     
Harder to get directly the:
Rate of MT = MTC x Driving Force
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 In Equimolar Diffusion:
 NA = NB (opposite directions)
 No bulk-motion contribution to the flux, which facilitates the equation
 Flux can be related almost linearly
A B
A A
A
A B A A
A
N N
N N
N N N N
undeffinined
 
  
  
 
2 1 2 1 2 1
2 1 2 1 2 1
( ) ( ) (P )
( ) ( ) ( )
A A A A A AAB
A A AB AB
C C x x PD
N J D cD
z z z z RT z z
  
      
  
A AN J

 Then, if applied to a Liquid – Gas Interphase:
 Where x  liquid phase, y vapor phase
2 1 2 1 2 1
2 1 2 1 2 1
( ) ( ) (P )
( ) ( ) ( )
A A A A A AAB
A A AB AB
C C x x PD
N J D cD
z z z z RT z z
  
      
  
2 1 2 1
2 1 2 1 2 1
2 1 2 1
2 1 2 1 2 1
( ) ( )
( ) ( )
(P ) (y ) ( )
( ) ( ) ( )
L
G
A A A A
A AB AB
A A A A A AAB
A AB AB
C C x x
N D cD
z z z z
P y C CD
N cD D
RT z z z z z z
 
   
 
  
     
  
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 Then, forcing the MTC:
 For the liquid phase:
 For the gas phase
1 2 1 2 1 2
' ( ) ' ( ) ' ( )
G
A G A A y A A c A AN k p p k y y k c c     
1 2 1 2
' ( ) ' ( )LA L A A x A AN k c c k x x   
2 1 2 1
2 1 2 1 2 1
2 1 2 1
2 1 2 1 2 1
( ) ( )
( ) ( )
(P ) (y ) ( )
( ) ( ) ( )
L
G
AB AB
A A A A A
AB AB AB
A A A A A A A
D cD
N C C x x
z z z z
D cD D
N P y C C
RT z z z z z z
     
 
        
  
NOTE!
Differ form “k” and
“k’”
k’ = EMD, k = UMD

 For the liquid phase:
 For the gas phase
 Relationship:
' ' '
' '
G G c y
L L x
P
F k p k k
RT
F k c k
  
 
1 2 1 2 1 2
' ( ) ' ( ) ' ( )
G
1 2 1 2
' ( ) ' ( )LA L A A x A AN k c c k x x   
FG and FL will be defined
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 Special notes:
 Valid at low and high concentrations (i.e. dilute and non-dilute cases)
 Must be low Mass Transfer Rates
 We ignored the bulk flow

 A packed-bed distillation column is used to adiabatically separate a mixture of
methanol and water at a total pressure of 1 atm.
 Methanol-the more volatile of the two component diffuses from the liquid phase
toward the vapor phase, while water diffuses in the opposite direction.
 Assuming that the molar latent heat of vaporization is similar for the two
components, this process is usually modeled as one of equimolar counter diffusion.
 At a point in the column, the mass-transfer coefficient is estimated as
 Mole fractions of methanol:
 The gas-phase methanol mole fraction at the interface is 0.707
 While at the bulk of the gas it is 0.656
 A) Estimate the methanol flux at that point.
Example 2.3 Mass-Transfer Coefficient in a
Packed-Bed Distillation Column. Principles
and Modern Applications of Mass Transfer
Operations, Jaime Benitez, 2nd Edition
2
5
1.62 10 kmol
kPa m s
x 


 Analysis:
 Assume EMD is true
 From the units given, it can be inferred that the coefficient given in the problem
statement is k’G
 The MTC is given with Pressure units as well, therefore, gas phase
 Flux will be calculate almost directly with the equation.
 Non-Diluted case! No worries!
' ' '
' '
G G c y
L L x
P
F k p k k
RT
F k c k
  
 
1 2 1 2 1 2
' ( ) ' ( ) ' ( )
G

 Solution:
 Equimolar counter diffusion can be assumed in this case
 It will be shown in a later chapter, this is the basis of the McCabe-Thiele method of analysis of
distillation columns
 Methanol diffuses from the interface towards the bulk of the gas phase; therefore
 yA1 = 0.707
 yA2 = 0.656.
 Since they are not limited to dilute solutions:
 k'-type mass-transfer coefficients may be used to estimate the methanol flux.

 Recall that we have related the equations for MTC in EMD
 We can either:
 Use 
 Requires Partial Pressure calculation
 Or Use 
 Requires k’y calculation
1 2 1 2 1 2 1 2
c
A A c A A A A G A A
' k' k'
' (p p ) k' ( ) k' ( ) F ( )
G y
y
G
A G
P
k P
RT
N k c c y y y
F
y
 
     

 
1 2A Ak' ( )A yN y y 
1 2A A' (p p )A GN k 
k' 'y Gk P
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 Let us calculate k’y
2 2
2 2
1 2
5 3
3 5
A A
k' ' (1.62 10 )(101.3 ) 1.64 10
' (y ) (1.64 10 )(0.707 0.656) 8.36 10
kmol kmol
y G kPa m s m s
kmol kmol
A y kPa m s m s
k P x kPa x
N k y x x
 

 

  
    

 Recall that we have related the equations for MTC in EMD
 We get
1 2 1 2 1 2 1 2
c
A A c A A A A G A A
' k' k'
' (p p ) k' ( ) k' ( ) F ( )
G y
y
G
A G
P
k P
RT
N k c c y y y
F
y
 
     

 
2 2
2 2
1 2
5 3
3 5
A A
k' ' (1.62 10 )(101.3 ) 1.64 10
' (y ) (1.64 10 )[101.3 (0.707 0.656)] 8.36 10
kmol kmol
y G kPa m s m s
kmol kmol
A G kPa m s m s
k P x kPa x
N k y x kPa x
 

 

  
    
2 2
2 2
1 2
5 3
3 5
A A
k' ' (1.62 10 )(101.3 ) 1.64 10
' (p p ) (1.64 10 )(0.707 0.656 ) 8.36 10
kmol kmol
y G kPa m s m s
kmol kmol
A G kPa m s m s
k P x kPa x
N k x kPa kPa x
 

 

  
    
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 In Molecular diffusion, we require “Ji” from Fick’s Law + Bulk Velocity:
 Which we then obtained:
 Developed:
i i i i iN J cV J y N   
( )A A
A AB A AB A A B
dy dy
N cD y N cD y N N
dz dz
      
2
1
A
2 1
1
N ln
( ) 1
AAB
A
ycD
z z y
 
     
2
12 1
ln
( )
AAB
A
A
P PcD
N
z z P P
 
     
Working:
Molar Fractions
Working:
Partial Pressures
Working:
Concentrations
2
1
A
2 1 A
C
ln
( ) C
AB
A
ccD
N
z z c
 
     

 If 
 Which can be translated to concentrations as well:
2 1( )
ABcD
F
z z


• Note that for any case (UMD, EMD) this expression:
• Repeats a lot…
• This is the so called “characteristic” property of molecular diffusion.
• We will replace it by F, a mass-transfer coefficient
• Technically, a “Local MTC”
2 1( )
ABcD
z z
2 2
1 1
A
2 1
1 1
N ln ln
( ) 1 1
A AAB
A A
y ycD
F
z z y y
    
            
2 2
1 1
A A
2 1 A A
C C
ln ln
( ) C C
AB
A
c ccD
N F
z z c c
    
            
2 2
1 12 1
ln ln
( )
A AAB
A
A A
P P P PcD
N F
z z P P P P
    
            

 Since the surface through which the transfer takes place may not be plane
 This will make that the diffusion path in the fluid may be of variable cross section
 NA is defined as the flux at the phase interface or boundary where:
 substance A enters or leaves the phase for which F has been defined.
2
G
1
A
1
N ln
1
A
G
A
y
F
y
 
    
2
1
A
A
C
ln
CGA G
c
N F
c
 
    
2
1
lnG
A
A G
A
P P
N F
P P
 
    
2
1
A
1
N ln
1L
A
L
A
x
F
x
 
    
2
1
A
A
C
ln
CLA L
c
N F
c
 
    
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 We can then assign coefficients to their respective phase:
 Liquid Phase:
 Gas Phase:
Force dY and dX
2 1
2 1
A A
A A
(x x )
(c )
II
A L
II
A L
N F
N F c
 
 
2 1
2 1
2 1
A A
A A
A A
( )
( )
(p )
II
A G
II
A G
II
A G
N F y y
N F c c
N F p
 
 
 
2
G
1
A
1
N ln
1
A
G
A
y
F
y
 
    
2
1
A
A
C
ln
CGA G
c
N F
c
 
    
2
1
lnG
A
A G
A
P P
N F
P P
 
    
2
1
A
1
N ln
1L
A
L
A
x
F
x
 
    
2
1
A
A
C
ln
CLA L
c
N F
c
 
    
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 In reality, this can’t be simply done, as we must algebraically manipulate the
natural logarithm differences.
 But… We could use a new coefficient:
2
1
1
ln
1
A
A G
A
y
N F
y
 
  
 
2
1
1
ln
1
A
A L
A
x
N F
x
 
  
 
2 1A A
,LM
(x x )L
A
B
F
N
x
 
2 1A A
,LM
( )G
A
B
F
N y y
y
 
,LM ,LM
; GL
x y
B B
FF
k k
x y
 
2 1A A(x x )A xN k 
2 1A A( )A yN k y y 
2 1 2 2 1
12
1
2 2 1 2 1 2
1 1
,LM
,LM
1 ,LM 1 ,LM
ln
ln
1 1 (1 ) 1
ln ln
1 1
B B B B B
B
B BB
B
B B B B B B
B B B B
x x x x x
x
x xx
x
x x x x x x
x x x x 
  
       
  
 
        
            
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 Note that we have many ways to model “concentrations”:
 Partial Pressure
 Molarity
 Molar fractions
 Each one will have then, its equivalent “Mass Transfer Coefficient”
2 1 2 1
2 1 2 1 2 1
A A A A
A A A A A A
(x x ) (c )
(y ) (c ) (p )
A x L
A y c G
N k k c
N k y k c k p
   
     
Note that any way you
calculate, the result must
be the same… WHY?
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 Proof:
Note that any way you
calculate, the result must
be the same… WHY?
,LM
L
x
B
F
k
x

,LM
G
y
B
F
k
y

2 1 2 1
2 1 2 1 2 1
A A A A
A A A A A A
(x x ) (c )
(y ) (c ) (p )
x L
y c G
k k c
k y k c k p
  
    
, ,L L B LM y B LMF k x c k x 
, ,
,
B LM B LM
G G B LM c y
p p
F k p k k
RT P
  

 Special notes:
 Valid mostly for low concentration, i.e. dilute cases
 Must be low mass transfer rates due to ignoring the bulk flow velocity
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