Queens Land

Transport PhenomenaSection 3: Mass TransferD. D. DoUniversity of Queensland

Mode of Mass TransferDiffusion:

Molecular diffusionKnudsen diffusion

Convection

Diffusion is more complicated than viscous flow and heat conduction because we have to deal with mixture (more than one component)

But this is what chemical engineers like!

Definitions of Concentration, Velocity and Fluxes

ConcentrationSolution of one phase whether it be gas, liquid or solid phase

Mole fraction

Mass fraction

Molar concentration

Mass concentration ρ jmass of j

volume of solution=

solutionofvolumejofmole

MC

j

jj =

ρ=

ωρρ

ρ

ρj

j j

kk

n

mass of jtotal mass

= = =

=∑

1

xCC

C

C

mole of jtotal molesj

j j

kk

n= = =

=∑

1

Velocities and Average Velocities (1)Different components move at different velocities

Let vj be the velocity of the component j, relative to the fixed frame of coordinate. Thus the mass flux is (mass rate per unit cross-sectional area perpendicular to the flow)

The molar flux is (molar rate per unit area)

1

2

v1

v2

∑=

ρn

1kkkv

∑=

n

1kkkvC


The local mass velocity is defined as a velocityIf we multiply it by the total mass concentration, we get the total mass flux

1

2

v1

v2

vv v

v vk k

k

n

kk

n

k kk

n

kk

k

n

k kk

n

= = = ==

=

=

= =

∑

∑

∑∑ ∑

ρ

ρ

ρ

ρρρ

ω1

1

1

1 1∑=

ρ=ρn

1kkkvv


The local molar velocity is defined as a velocityIf we multiply it by the total molar concentration, we get the total molar flux

1

2

v1

v2

∑=

=n

1kkkvCCv ∑∑

∑

∑

∑==

=

=

= ====n

1kkk

n

1kk

k

n

1kkk

n

1kk

n

1kkk

* vxvC

CC

vC

C

vCv

Diffusion Velocities (1)The velocity relative to the fixed frame of coordinate is sufficient to calculate the flux

But in flow systems, it is extremely useful to know the diffusion rate of different components relative to the flow of the bulk solution

In this example, both components move to the right according to the person standing on a fixed position. However, according to the person riding with the flow, component 1 moves to the right while component 2 moves to the left

V1

2

Diffusion Velocities (2)This gives rise to the need to define the so-called diffusion velocity

There are two diffusion velocitiesDiffusion velocity relative to the mass average velocityDiffusion velocity relative to the molar average velocity

V1

2

v vdiffusion velocity of species jwith respect to vj − ≡⎛⎝⎜

⎞⎠⎟ v v

diffusion velocity of species jwith respect to vj − ≡⎛

⎝⎜

⎞

⎠⎟*

*

Mass Fluxes and Molar Fluxes relative to the Fixed Frame (1)

Mass FluxesThe mass flux of the component j is the product of the mass concentration and the velocity of that component

which is the mass of the component j transported per unit time and per unit cross sectional area perpendicular to the velocity vj

n vmass of species j

volumedis ce

timej j j= = ⎛⎝⎜

⎞⎠⎟× ⎛⎝⎜

⎞⎠⎟

ρtan

Mass Fluxes and Molar Fluxes relative to the Fixed Frame (2)

Molar FluxesThe molar flux of the component j is the product of the molar concentration and the velocity of that component

which is the number of moles of the component j transported per unit time and per unit cross sectional area perpendicular to the velocity vj

N C vmole of species j

volumedis ce

timej j j= = ⎛⎝⎜

⎞⎠⎟× ⎛⎝⎜

⎞⎠⎟

tan

Diffusion Mass and Molar Fluxes (1)The diffusion mass flux is defined the flux relative to the bulkmotion of the solution

which is the mass transferred per unit time per unit area, relative to the bulk flow

( )j v vj j j= −ρ

Diffusion Mass and Molar Fluxes (2)The diffusion molar flux is defined the flux relative to the bulk motion of the solution

which is the moles transferred per unit time per unit area, relative to the bulk flow

( )J C v vj j j= − *

Properties of Diffusion Fluxes (1)The diffusion fluxes satisfy the following equation

For example, for a system of two components

which simply states that the diffusion flux of component 1 (to the right, say) MUST be the same to the diffusion of component 2 (to the left).

If this is NOT satisfied, what would be the consequence?

j Jkk

n

kk

n

= =∑ ∑= =

1 1

0 0;

0J2

1kk =∑

=

Properties of Diffusion Fluxes (2)The diffusion fluxes satisfy the following equation

Proof:By definition

j Jkk

n

kk

n

= =∑ ∑= =

1 1

0 0;

( )J C v vkk

n

k kk

n

= =∑ ∑= −

1 1

* J C v v Ckk

n

k kk

n

kk

n

= = =∑ ∑ ∑= −

1 1 1

*

J C v v Ckk

n

k kk

n

= =∑ ∑= −

1 1

*

vC v

C

k kk

n

* = =∑

1

CC

vCvCJ

n

1kkkn

1kkk

n

1kk

∑∑∑ =

==

−=

J kk

n

=∑ =

1

0

Diffusion Flux vs Flux (1)Recall the definitions of molar flux and molar diffusion flux

So we need to relate these two most important quantities in mass transfer.

The relationship between the molar flux and the molar diffusive flux is

jjj vCN = ( )*vvCJ jjj −=

J N x Nk k k jj

n

= −=∑

1

This is useful for engineering calculation of mass transfer

This is necessary for the diffusion analysis

Diffusion Flux vs Flux (2)The relationship between the molar flux and the molar diffusive flux is

Proof: Always go back to definition

J N x Nk k k jj

n

= −=∑

1

( )*kkk vvCJ −= *vCvC kkk −=

C

vCCvC

n

1jjj

kkk

∑=−=

∑=

−=n

1jjj

kkkk vC

CCvCJ J N x Nk k k j

j

n

= −=∑

1

Diffusion Flux vs Flux (3)The relationship between the molar flux and the molar diffusive flux is

This equation simply states that the diffusive flux of the component k is equal to the flux of that component minus the fraction of that component in the bulk flow

J N x Nk k k jj

n

= −=∑

1

Diffusion Flux vs Flux (4)Similarly, the relationship between the mass flux and the mass diffusive flux is

This equation again states that the diffusive flux of the component k is equal to the flux of that component minus the fraction of that component in the bulk flow

j n nk k k jj

n

= −=∑ω

1

Fick’s law of Diffusion for Binary Mixtures (1)

The basic law for diffusion study is the widely known Fick’slaw.

It is only applicable for BINARY mixtureFor mixtures of three or more components, the proper law is the Maxwell-Stefan law

The Fick’s law for the first component is

which is only correct for isobaric and isothermal system.This equation states that if there is a gradient in the mole fraction of the component 1, the molar diffusive flux is calculated as aboveThe coefficient D12 is called the binary diffusivity

J cD dxdz1 12

1= −


Similarly, we can write the same equation for the second component by simply interchanging the subscripts 1 and 2

But we know that the molar diffusive fluxes satisfy the following equation

Thus adding the two molar diffusive fluxes, we have

But

The above equation will become

dzdxcDJ 2

212 −=J cD dxdz1 12

1= −

0JJ 21 =+

0dz

dxcDdzdxcDJJ 2

211

1221 =−−=+

1xx 21 =+

0dzdxcD

dzdxcDJJ 1

211

1221 =+−=+

( ) 0dzdxDDcJJ 1

211221 =−−=+ ( ) 0DD 2112 =−


Thus the diffusivity (diffusion coefficient) D12 is equal to the diffusivity D21

What it means is that the two diffusion equations

are not independent. So only one is used. Either one will do.

For three-dimensional coordinates, the Fick’s law equation is

J cD dxdz1 12

1= −

dzdxcDJ 2

212 −=

J cD x1 12 1= − ∇


So the Fick’s law gives us the molar diffusive flux. What we need for mass transfer calculation is the molar flux relative to the fixed frame of coordinates

Here is the place where we need the relationship between the molar diffusive flux and the molar flux. This relationship is

which for binary mixtures, it is

J cD dxdz1 12

1= −

J N x Nk k k jj

n

= −=∑

1

∑=

−=2

1jj111 NxNJ ( )N J x N N1 1 1 1 2= + +


So the equation for the molar flux is

This is the equation for the molar flux for the component 1. You can also easily write an equation for the component 2 by interchanging the subscripts 1 and 2

But only one of them is independent!

( )N J x N N1 1 1 1 2= + +

J cD dxdz1 12

1= −

( )N cD dxdz

x N N1 121

1 1 2= − + +

( )N cD dxdz

x N N2 212

2 2 1= − + +


Summary

The above equation involves two unknown variables:The molar flux of component 1, N1

The molar flux of component 2, N2

So we need to find another equation. This equation is specific to a given system. More about this later when we deal with a number of examples.

J cD dxdz1 12

1= −

( )N cD dxdz

x N N1 121

1 1 2= − + +

Diffusion CoefficientUnits

Order of magnitude

Pressure and Temperature Dependence

1 × 10-12 – 1 × 10-7Solid

1 × 10-7 – 1 × 10-5Liquid

0.1 - 1Gas

Order of magnitude (cm2/s)StateDm

12

2

≡⎡

⎣⎢

⎤

⎦⎥sec

-

-

decrease

P

increaseSolid

increaseLiquid

increaseGas

TState

Stefan-Maxwell’s law for mixturesA bit excursion into the Stefan-Maxwell law.The equation for mixtures of n components is

for i = 1, 2, …, (n-1). Only (n-1) equations are independent as the n-thequation can be derived from the other equations

Another form of the Stefan-Maxwell equation is

Details of this exact Stefan Maxwell equation can be found in D. D. Do, “Adsorption Analysis”, Imperial College Press, New Jersey, 1998, Chapter 8

( )− ∇ =

−

=∑c x

x N x NDi

j i i j

ijj

n

1

( )− ∇ =

−

=∑c x

x J x JDi

j i i j

ijj

n

1

First Principles for Binary Mixtures (1)The procedure is identical to what you have learnt before in momentum and heat transfers

Draw physical diagram and then a thin shell whose surfaces are perpendicular to the transport directionsSet up the mass balance equation around the shell

Take the limit of the mass balance equation as the shell shrinks to zero. This will lead to a first-order differential equation with respect to fluxApply the Fick’s law, and we will get a second-order differential equation in terms of concentration

( )Rate ofmass in

Rate ofmass out

Rate of massproduction

Accummulation⎛⎝⎜

⎞⎠⎟ −

⎛⎝⎜

⎞⎠⎟ +

⎛⎝⎜

⎞⎠⎟ =

First Principles for Binary Mixtures (2)The procedure (continued)

Impose the constraints on the system (boundary conditions)Solve for the concentration distributionKnowing the concentration distribution, derive the average concentration, molar fluxes, etc.

Boundary ConditionsThere are five boundary conditions that you will regularly encounter in mass transfer

BC of the first kind: Concentration is specified at the boundary

BC of the second kind:Molar flux is specified at the boundary

BC of the third kind:Molar flux into a medium is equal to the molar flux through the stagnant film surrounding the medium

BC of the fourth kind:Concentrations and fluxes are continuous across the interface

BC of the fifth kind:Molar flux to a surface is equal to the chemical reaction occurring on that surface

What is next?That is all about mass transfer!

What come next are simple examples to illustrate the mass transfer principles.Example 1: Diffusion in a Stefan tubeExample 2: Dissolution of a spherical objectsExample 3: Diffusion and heterogeneous reaction at surfaceExample 4: Diffusion and homogeneous reactionExample 5: Diffusion into a falling filmExample 6: Gas absorption in a rising bubbleExample 7: Diffusion and reaction in a porous catalystExample 8: Transient diffusion through a polymer filmExample 9: Transient diffusion in a finite reservoir

Diffusion in a Stefan tube (1)Stefan tube is simply a test tube with liquid in it.

Physical system:Liquid 1 is in the tubeGas 2 is flowing across the tube mouthGas 2 is non-soluble in liquid 1

Diffusion in a Stefan tube (2)Shell balance

Taking the shell as thin as possible

Which simply state that the evaporation flux of the component 1 is constant along the tube

( )Rate ofmass in

Rate ofmass out



⎞⎠⎟ −

⎛⎝⎜

⎞⎠⎟ +

⎛⎝⎜

⎞⎠⎟ =

z1NS ⋅zz1NS

∆+⋅− 00 =+

zS ∆

− =dNdz

1 0

Diffusion in a Stefan tube (3)So the shell balance equation is

Apply the Fick’s law for binary system

Since the gas 2 is insoluble in liquid 1, the molar flux of component 2 (with respect to a person standing outside the tube) is simply ZERO. So the Fick’s law is reduced to:

− =dNdz

1 0

( )N cD dxdz

x N N1 121

1 1 2= − + +

( )111

121 NxdzdxcDN +−= dz

dxx1

cDN 1

1

121 −

−=

Diffusion in a Stefan tube (4)The mass balance equation is

The Fick’s law

So the mass balance equation in terms of concentration is

− =dNdz

1 0

dzdx

x1cDN 1

1

121 −

−=

ddz

cDx

dxdz

12

1

1

10

−⎛⎝⎜

⎞⎠⎟ = 0

dzdx

x11

dzdcD 1

112 =⎟⎟

⎠

⎞⎜⎜⎝

⎛−

For constant P and D12

Diffusion in a Stefan tube (5)The mass balance equation in terms of concentration is a second-order differential equation

Physical constraints (Boundary conditions)

0dzdx

x11

dzdcD 1

112 =⎟⎟

⎠

⎞⎜⎜⎝

⎛−

0

z1

z2

z z x xpP

= = =1 1 1 010

; ,

z z x x L= =2 1 1; ,

Diffusion in a Stefan tube (6)Solution procedure

Integrate it once

Integrate it one more time

which involves two constants of integration

Did you note that K1 is simply the negative of the molar flux N1?

0dzdx

x11

dzdcD 1

112 =⎟⎟

⎠

⎞⎜⎜⎝

⎛−

0

z1

z2

cDx

dxdz

K cons t12

1

111−

= ( tan )

cDx

K z K121

1 21

1ln

−⎛⎝⎜

⎞⎠⎟ = +

Diffusion in a Stefan tube (7)Solution

To find K1 and K2, you apply the two boundary conditions

So you can solve for K1 and K2

0

z1

z2cDx

K z K121

1 21

1ln

−⎛⎝⎜

⎞⎠⎟ = +

@ ; ln,

z z cDx

K z K=−

⎛

⎝⎜

⎞

⎠⎟ = +1 12

1 01 1 2

11

@ ; ln,

z z cDx

K z KL

=−

⎛

⎝⎜

⎞

⎠⎟ = +2 12

11 2 2

11

Diffusion in a Stefan tube (8)Solution

K1 and K2

So the concentration distribution is0

z1

z2

cDx

K z K121

1 21

1ln

−⎛⎝⎜

⎞⎠⎟ = +

K cDz z

xx L

112

2 1

1 0

1

11

=−

−−

⎛

⎝⎜

⎞

⎠⎟ln ,

,

K cDx

cD zz z

xxL L

2 121

12 1

2 1

1 0

1

11

11

=−

⎛

⎝⎜

⎞

⎠⎟ −

−−−

⎛

⎝⎜

⎞

⎠⎟ln ln

,

,

,

( ) ( )1

111

1

1 0

1

1 0

1 2 1−−

⎛

⎝⎜

⎞

⎠⎟ =

−−

⎛

⎝⎜

⎞

⎠⎟

− −x

xxx

L

z z z z

,

,

,

/

Diffusion in a Stefan tube (9)So the final solution for the concentration distribution is

To find the evaporation flux, we simply apply the Fick’s law

This means that we have to take the derivative of the mole fraction x1 with respect to distance z!!

0

z1

z2( ) ( )

11

11

1

1 0

1

1 0

1 2 1−−

⎛

⎝⎜

⎞

⎠⎟ =

−−

⎛

⎝⎜

⎞

⎠⎟

− −x

xxx

L

z z z z

,

,

,

/

N cDx

dxdz1

12

1

1

1= −

−

Diffusion in a Stefan tube (10)Let us find the flux formally

Taking logarithm both side

Now taking differentiation:

( ) ( )

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎟⎟⎠

⎞⎜⎜⎝

⎛

−−

=⎟⎟⎠

⎞⎜⎜⎝

⎛

−−

−− 121 zz/zz

0,1

L,1

0,1

1

x1x1

lnx1x1ln

0

z1

z2

( ) ( )1

111

1

1 0

1

1 0

1 2 1−−

⎛

⎝⎜

⎞

⎠⎟ =

−−

⎛

⎝⎜

⎞

⎠⎟

− −x

xxx

L

z z z z

,

,

,

/

( ) ( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛

−−

−−

=−−−0,1

L,1

12

10,11 x1

x1ln

zzzzx1lnx1ln

⎟⎟⎠

⎞⎜⎜⎝

⎛

−−

−=

−−

0,1

L,1

121

1

x1x1

lnzz

dzx1

dx⎟⎟⎠

⎞⎜⎜⎝

⎛

−−

−=

−−

0,1

L,1

12

1

1 x1x1

lnzz

1dzdx

x11

Diffusion in a Stefan tube (11)So we have

Substitute this into the Fick’s law

The evaporation flux is

This is a long, but formal, way of getting the flux

0

z1

z2⎟⎟⎠

⎞⎜⎜⎝

⎛

−−

−=

−−

0,1

L,1

12

1

1 x1x1

lnzz

1dzdx

x11

N cDx

dxdz1

12

1

1

1= −

−

⎟⎟⎠

⎞⎜⎜⎝

⎛

−−

−=

0,1

L,1

12

121 x1

x1ln

zzcDN

Diffusion in a Stefan tube (12)However there is a short cut. Remember that during the integration of the mass balance we got:

And K1 was found from the application of the boundary condition as

Therefore the evaporation flux is

⎟⎟⎠

⎞⎜⎜⎝

⎛

−−

−=

0,1

L,1

12

121 x1

x1ln

zzcDN

0

z1

z2cD

xdxdz

K12

1

111−

=

K cDz z

xx L

112

2 1

1 0

1

11

=−

−−

⎛

⎝⎜

⎞

⎠⎟ln ,

,

Diffusion in a Stefan tube (13)So the solution for the evaporation flux is

If the gas 2 is sweeping past the tube fast enough, the concentration of the species 1 at the mouth is effectively zero. Hence the evaporation rate is

⎟⎟⎠

⎞⎜⎜⎝

⎛

−−

−=

0,1

L,1

12

121 x1

x1ln

zzcDN

0

z1

z2

N cDz z x1

12

2 1 1 0

11

=− −

⎛

⎝⎜

⎞

⎠⎟ln

,

( )NP RT D

z z p Pi1

12

2 10

11

=− −

⎛⎝⎜

⎞⎠⎟

/ln

/

Diffusion in a Stefan tube (14)Given the flux of species 1 as

The total flux is

The flux of the component 2 is 0

z1

z2( )NP RT D

z z p Pi1

12

2 10

11

=− −

⎛⎝⎜

⎞⎠⎟

/ln

/

( )N NP RT D

z z p Pi1 2

12

2 10

11

+ =− −

⎛⎝⎜

⎞⎠⎟

/ln

/

( ) 0NNxdz

dxcDN 2122

122 =++−=

Diffusion in a Stefan tube (15)Because of the evaporation, the liquid level will drop with time

Therefore you need to write a mass balance around the liquid

This is a differential equation with respect to t

0

z1

z2( )NP RT D

z z p Pi1

12

2 10

11

=− −

⎛⎝⎜

⎞⎠⎟

/ln

/

( ) ( )d Szdt

SMP RT D

z z p PL1

112

2 1 10

11

ρ= −

− −⎛⎝⎜

⎞⎠⎟

/ln

/

( )11

L1 NSMdt

Szd−=

ρ

Diffusion in a Stefan tube (16)Differential equation for the liquid level with respect to t

The initial condition is

The solution is

0

z1

z2

( ) ( )d Szdt

SMP RT D

z z p PL1

112

2 1 10

11

ρ= −

− −⎛⎝⎜

⎞⎠⎟

/ln

/

0,11 zz;0t ==

[ ] ( )( )

z z t z z

M P RT Dp P

tL

2 12

2 102

1 12

102 1

1

− − − =

−⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

( )

/ln

/ρ

The time taken to empty the liquid is:

( )( )

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛

−ρ

−−=

P/p11ln

DRT/PM2

zzzt

01L

121

2102

22*

Diffusion in a Stefan tube (17)The time taken to empty the liquid is:

Example:Carbon tetrachloride (1) in air (2) in a Stefan tube of length 40 cm and the initial level of the liquid is 25 cm.The time taken to empty the liquid is 651 days!!

Conclusion: DIFFUSION IS A SLOW PROCESS. Get rid of it IF YOU CAN

0

z1

z2

( )( )

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛

−ρ

−−=

P/p11ln

DRT/PM2

zzzt

01L

121

2102

22*

Dissolution of a Sphere (1)Physical system: Sparingly soluble sphere in a surrounding fluid of infinite extent

The object (1) dissolves into a liquid (2)The solubility is C10

r

2R

∆r

Dissolution of a Sphere (2)Physical system: Sparingly soluble sphere in a surrounding fluid of infinite extent

The shell balance

r

( ) ( )4 4 0 021

21π πr N r N

r r r− + =

+ ∆

r4 ∆π

( ) ( ) ( )lim∆

∆

∆r

r r rr N r N

rddr

r N→

+−

= − =0

21

21 2

1 0

Dissolution of a Sphere (3)The shell balance

The Fick’s law

The second term is the contribution of the species 1 in the bulk flow. Since the object is only sparingly soluble, x1 is expected very small. Therefore we can ignore the bulk flow contribution in the Fick’s law equation

Substitute the Fick’s law into the mass balance equation

r

( ) 0Nrdrd

12 =−

( )N cD dxdr

x N N1 121

1 1 2= − + +

drdxcDN 1

121 −≈

ddr

r cD dxdr

212

1 0⎛⎝⎜

⎞⎠⎟=

Dissolution of a Sphere (4)So the final form of the mass balance equation is

The boundary conditions are:

The concentration distribution

The dissolution rateBack to the Fick’s law:The dissolution rate is simply the flux at the surface of the object

0dr

dCDrdrd 1

122 =⎟

⎠⎞

⎜⎝⎛

r

ddr

r cD dxdr

212

1 0⎛⎝⎜

⎞⎠⎟=

0,11 CC;Rr == 0C;r 1 =∞=

C C Rr1 10=

drdCD

drdxcDN 1

121

121 −=−≈

RCD

drdCDN 1012

Rr

112Rr1 =−≈

==

Dissolution of a Sphere (5)Very often the dissolution is calculated by using the concept of mass transfer coefficient:

Comparing this equation with the dissolution rate obtained from the first principles

You will get

This group is known as the Sherwood number to describe the mass transfer coefficient

r

RCDN 1012

Rr1 ==

( )N k Cr R m1 10 0=

= −

k RD

m ( )2 212

=

Dissolution of a Sphere (6)So the Sherwood number for a stagnant environment is

Recall the equivalent number in heat transfer is Nu

It, therefore, comes as no surprise that some correlations for heat and mass transfer take the following form:

k RD

m ( )2 212

=

2k

)R2(h

f

=

3/16.0 PrRe1.12Nu +=3/16.0 ScRe1.12Sh += 12

p

DSc;

kC

Prρµ

=µ

=

Diffusion with Heterogeneous Reaction (1)

Physical system:Chemical reaction occurs on a catalytic surfaceCatalytic surface is surrounded by a thin stagnant filmDiffusion occurs in the stagnant filmIsothermal conditionsReaction is nAAn →

AAn


Physical system:Let the reactant A be the species 1Let the product An be the species 2To restrict ourselves to only binary mixture, we shall assume that there will be no other species

The chemical reaction demands that

The above equation simply states that “n” moles of reactant (A) coming down towards the catalytic surface is equal to “1” mole of product Angoing out of the surface

This equation basically provides the additional equation to the Fick’slaw equation

nAAn →

A An

21 nNN −=


The shell balance equation:Setting a shell at the distance z in the stagnant film

The Fick’s law

This equation involves two unknown variables N1 and N2. We need one more equation, and that equation is specific to this problem, which is the reaction stoichiometry

nAAn →

A An

21 nNN −=

− =dNdz

1 0

( )N cD dxdz

x N N1 121

1 1 2= − + +

Nn

N2 11

= −


The mass balance equation:

The Fick’s law

Substitute the stoichiometry relation into the Fick’s law, you can now solve for the molar flux N1 in terms of the concentration gradient

nAAn →

A An

− =dNdz

1 0

( )N cD dxdz

x N N1 121

1 1 2= − + +

Nn

N2 11

= −

( )N cD

nn

x

dxdz1

12

1

1

11

= −−

−⎡

⎣⎢

⎤

⎦⎥



The Fick’s law

The mass balance equation, then, gives:

nAAn →

A An

− =dNdz

1 0

( )N cD

nn

x

dxdz1

12

1

1

11

= −−

−⎡

⎣⎢

⎤

⎦⎥

( ) 11

1

121 K

dzdx

xn

1n1

cDN −≡

⎥⎦⎤

⎢⎣⎡ −−

−=



Now we pose the boundary conditionsnAAn →

AAn

( ) 11

1

121 K

dzdx

xn

1n1

cDN −≡

⎥⎦⎤

⎢⎣⎡ −−

−=

0x;Lzxx;0z

1

101

====



Integrating the mass balance equation one more time

We have two constants of integration, and we also have two boundary conditions

nAAn →

AAn

( ) 11

1

121 K

dzdx

xn

1n1

cDN −≡

⎥⎦⎤

⎢⎣⎡ −−

−=

( )cD

n x nn

nK z K12

11 2

11 1

1ln/

( )− −

⎡

⎣⎢

⎤

⎦⎥ =

−+


The solution:

Applying the two boundary conditions

( )cD

n x nn

nK z K12

11 2

11 1

1ln/

( )− −

⎡

⎣⎢

⎤

⎦⎥ =

−+

0x;Lzxx;0z

1

101

====

nAAn →

AAn

( ) 210,1

12 K)0(Kn

)1n(n/x1n1

1lncD +−

=⎥⎥⎦

⎤

⎢⎢⎣

⎡

−−

( ) 2112 K)L(Kn

)1n(n/)0(1n1

1lncD +−

=⎥⎦

⎤⎢⎣

⎡−−


The solution:

Solving for K1 and K2, we get

( )cD

n x nn

nK z K12

11 2

11 1

1ln/

( )− −

⎡

⎣⎢

⎤

⎦⎥ =

−+

nAAn →

AAn

( ) ⎥⎥⎦

⎤

⎢⎢⎣

⎡

−−=

n/x1n11lncDK

0,1122

( ) ⎥⎥⎦

⎤

⎢⎢⎣

⎡

−−⎟⎠⎞

⎜⎝⎛

−−=

n/x1n11ln

LcD

1nnK

0,1

121

Remember that N1 = - K1; Sothe negative of the constantK1 is simply the flux


The solution for the concentration distribution:

The molar flux across the stagnant film is the negative of the constant K1:

This molar flux tells you how many moles of reactant passing through the stagnant film per unit area and unit time. This is EXACTLY equal to the rate of reaction at the surface

nAAn →

AAn

( ) ( ) ( )

11

11

1 10

1

−−⎡

⎣⎢

⎤

⎦⎥ = −

−⎡

⎣⎢

⎤

⎦⎥

−n

nx

nn

xz L/

( ) ⎥⎥⎦

⎤

⎢⎢⎣

⎡

−−⎟⎠⎞

⎜⎝⎛

−=

n/x1n11ln

LcD

1nnN

0,1

121

SO THIS IS THEREACTION RATEPER UNIT AREAOF THE CATALYST


We just completed the analysis of diffusion and surface reaction for the case of extremely fast reaction, and you probably have noted that

The reaction rate does not involve the rate constant of chemicalreaction.This is so because reaction is so fast and since it is in series with a diffusion process, the slow step is the controlling step, which is the diffusion process.Therefore the rate of reaction only involves the diffusion characteristics, D12, and the reaction stoichiometry

Let’s have a look at the general case of finite reaction rate at the surface



Now we pose the boundary conditionsnAAn →

AAn

( ) 11

1

121 K

dzdx

xn

1n1

cDN −≡

⎥⎦⎤

⎢⎣⎡ −−

−=

0,11 xx;0z ==

Lz1Lz1 xkcN;Lz==

==This is the boundary condition of the fifth kind.Here k is the chemical reaction rate constant



Integrating the mass balance equation one more time

We have two constants of integration, and we also have two boundary conditions

nAAn →

AAn

( ) 11

1

121 K

dzdx

xn

1n1

cDN −≡

⎥⎦⎤

⎢⎣⎡ −−

−=

( )cD

n x nn

nK z K12

11 2

11 1

1ln/

( )− −

⎡

⎣⎢

⎤

⎦⎥ =

−+


The solution:

Applying the two boundary conditions

( )cD

n x nn

nK z K12

11 2

11 1

1ln/

( )− −

⎡

⎣⎢

⎤

⎦⎥ =

−+

101 xx;0z ==

nAAn →

AAn

( ) 210,1

12 K)0(Kn

)1n(n/x1n1

1lncD +−

=⎥⎥⎦

⎤

⎢⎢⎣

⎡

−−

L,11 kcxK =−Lz1Lz1 xkcN;Lz

====


The solution:

Solving for K1 and K2, we get

( )cD

n x nn

nK z K12

11 2

11 1

1ln/

( )− −

⎡

⎣⎢

⎤

⎦⎥ =

−+

nAAn →

AAn

( ) ⎥⎥⎦

⎤

⎢⎢⎣

⎡

−−=

n/x1n11lncDK

0,1122

L,11 kcxK −=


The solution for the concentration distribution:

Remember that x1,L is still unknown at this stage. But it can be found easily by putting z = L into the above equation

This is a nonlinear equation in terms of x1,L.

nAAn →

AAn

( )ln

( ) //

( ), ,1 11 1

11 0

1

1

12

− −− −

⎡

⎣⎢

⎤

⎦⎥ = −

−⋅

n x nn x n

nn

kxD

zL

( )ln

( ) //

( ),

,,

1 11 1

11

1 0 121

− −− −

⎡

⎣⎢

⎤

⎦⎥ =

−⋅⎛⎝⎜

⎞⎠⎟

n x nn x n

nn

kLD

xLL


How to get x1,L?

Knowing this concentration at the catalytic surface, the reaction rate is

Thus the reaction rate in the case of finite reaction is a function of the reaction rate constant and the diffusion coefficient

( )ln

( ) //

( ),

,,

1 11 1

11

1 0 121

− −− −

⎡

⎣⎢

⎤

⎦⎥ =

−⋅⎛⎝⎜

⎞⎠⎟

n x nn x n

nn

kLD

xLL

L,1Lz1 kcxN ==


Equation for x1,L

This equation involves a non-dimensional group

Which describes the interplay between the reaction rate and the diffusion rate

( )ln

( ) //

( ),

,,

1 11 1

11

1 0 121

− −− −

⎡

⎣⎢

⎤

⎦⎥ =

−⋅⎛⎝⎜

⎞⎠⎟

n x nn x n

nn

kLD

xLL

kLD

chemical reaction ratediffusion rate12

⎛⎝⎜

⎞⎠⎟ ≡

Diffusion & Homogeneous Reaction (1)Let us now consider a new example where we have diffusion and reaction occurring simultaneously, rather than reaction at the boundary.

Diffusion & Homogeneous Reaction (2)The system: Absorption and reaction


Gas A dissolves sparingly in liquid BDissolved A reacts with B, according to a first order chemical reaction

Isothermal system

Let the species A be 1; and the species B be 2

moles of A reactedvolume time

kC−

⎛⎝⎜

⎞⎠⎟= 1


Let us check the assumption “Gas A dissolves sparingly in liquid B”At 1 atm and ambient temperature (King, “Separation Processes, McGraw Hill, pg 273)

Solubility, in general, increases with pressure and decreases with temperature

0.030.00170.0020.000620.00010.000018

Sulfur dioxideChlorineHydrogen sulfideCarbon dioxideEthyleneCarbon monoxide

Mole fractionGas


Shell mass balance

( )Rate ofmass in

Rate ofmass out



⎞⎠⎟ −

⎛⎝⎜

⎞⎠⎟ +

⎛⎝⎜

⎞⎠⎟ =

( ) ( ) ( )SN SN S z kCz z z1 1 1 0− − =

+∆∆

zS ∆

− − =dNdz

kC11 0

Diffusion & Homogeneous Reaction (6)The Fick’s law

Substitute the Fick’s law into the mass balance equation, we get:

This is one of the classic equation in diffusion and reaction theory!

− − =dNdz

kC11 0

( )N cD dxdz

x N N1 121

1 1 2= − + +

N D dCdz1 12

1= −

D d Cdz

kC12

21

2 1 0− =

Diffusion & Homogeneous Reaction (7)The mass balance equation

Physical constraints (Boundary conditions)At the gas-liquid interface, the concentration is always equal to the solubility

At the bottom of the liquid pool, mass can not penetrate through it. So the molar flux at the bottom of the pool is zero.

0,11 CC;0z ==− − =

dNdz

kC11 0

D d Cdz

kC12

21

2 1 0− =

0dz/dC;Lz 1 ==

Diffusion & Homogeneous Reaction (8)So we have the mass balance equation & the two necessary boundary conditions

Solution by the method of characteristics. In this method, the solution is assumed to take the form:

More about this useful method of characteristics, see the book Rice and Do “Applied Mathematics and Modeling for chemical engineers”, Wiley, 1995, page 63

0,11 CC;0z ==

− − =dNdz

kC11 0

D d Cdz

kC12

21

2 1 0− =

0dz/dC;Lz 1 ==

z1 eC λ=

Diffusion & Homogeneous Reaction (9)The mass balance equation

If the assumed form

is the solution to the mass balance equation, it must satisfy ( ) 0ke

dzedD z

2

z2

12 =− λλ

− − =dNdz

kC11 0

D d Cdz

kC12

21

2 1 0− =

z1 eC λ=

0keeD zz212 =−λ λλ

( ) 0ekD z212 =−λ λ

( ) 0kD 212 =−λ

12

2

Dk

=λ12D

k±=λ

Diffusion & Homogeneous Reaction (10)Solution (continued)

The assumed formwhere

Since we have two values for the eigenvalue λ, the general solution is a linear combination of the following two solutions

That is

z1

1eC λ=

− − =dNdz

kC11 0

D d Cdz

kC12

21

2 1 0− =

z1 eC λ=

12Dk

±=λ

z1

2eC λ=

zz1

21 eBeAC λλ +=So our solution has two constants ofintegration, A and B

Diffusion & Homogeneous Reaction (11)The equation

The solution

Apply the two boundary conditions

D d Cdz

kC12

21

2 1 0− =

zz1

21 eBeAC λλ +=

0,11 CC;0z ==

0dz/dC;Lz 1 ==

BAeBeAC )0()0(0,1

21 +=+= λλ

0eBeAdz

dC )L(2

)L(1

1 21 =λ+λ= λλ

12Dk

±=λ

Diffusion & Homogeneous Reaction (12)The equation

The solution

Solution for A and B

L2

L1

L1

0,1 21

1

eeeCB λλ

λ

λ−λλ

=

D d Cdz

kC12

21

2 1 0− =

zz1

21 eBeAC λλ +=

L2

L1

L2

0,1 21

2

eeeCA λλ

λ

λ−λλ

−=

12Dk

±=λ

LL

L

0,1 eeeCA λ−λ

λ−

+=

LL

L

0,1 eeeCB λ−λ

λ

+=

Diffusion & Homogeneous Reaction (13)So after a long and tedious (but fun) calculus and algebra, the solution for the concentration distribution is

Simplify it

zLL

Lz

LL

L

0,1

1 eee

eeee

eCC λ−

λ−λ

λλ

λ−λ

λ−

++

+=

LL

)zL(

LL

)zL(

0,1

1

eee

eee

CC

λ−λ

−λ

λ−λ

−λ−

++

+=

LL

)zL()zL(

0,1

1

eeee

CC

λ−λ

−λ−λ−

++

=

[ ]( )

( )[ ]( )

[ ])Lcosh(

)L/z1(LcoshLcosh

zLcosh

2ee

2ee

CC

LL

)zL()zL(

0,1

1

λ−λ

=λ−λ

=+

+

= λ−λ

−λ−λ−

Diffusion & Homogeneous Reaction (14)So the final solution is

Let us have a good look at the group λL

So we can write solution as

DaTimeReaction TimeDiffusion

k1

DL

DLk

DkLL

def12

2

12

2

12

≡====λ

[ ])Lcosh(

)L/z1(LcoshCC

0,1

1

λ−λ

=

[ ]( )Dacosh

)L/z1(DacoshCC

0,1

1 −=

To credit German scientist Damkohler


With

What does this group tell us?When Da << 1: the diffusion time is much smaller than the reaction time, we would expect

The concentration distribution is uniformThe absorption rate is controlled by reaction

When Da >> 1: the diffusion time is much greater than the reaction time, we would expect

The concentration distribution is very sharp and localized near the gas-liquid interfaceThe absorption rate is controlled by diffusion and reaction

TimeReaction TimeDiffusion

DLkDa12

2

==

[ ]( )Dacosh

)L/z1(DacoshCC

0,1

1 −=


with

The absorption rate per unit interfacial area is just simply the molar flux at the gas liquid interface


DLkDa12

2

==

[ ]( )Dacosh

)L/z1(DacoshCC

0,1

1 −=

( )

N DdCdz

NC D

LDa Da

NC D

LkLD

kLD

zz

z

z

1 0 121

0

1 010 12

1 010 12

2

12

2

12

==

=

=

= −

= ⋅

= ⋅⎛

⎝⎜⎜

⎞

⎠⎟⎟

tanh

tanh

Diffusion & Homogeneous Reaction (17)Summary


DLkDa12

2

==

[ ]( )Dacosh

)L/z1(DacoshCC

0,1

1 −=

( )DatanhDaLDCN 1210

0z1 ⋅==

Diffusion & Homogeneous Reaction (18)Special case: Da << 1

The rate of absorption is controlled purely by chemical reaction

1CC

0,1

1 ≈

0,10z1 CLkN ≈=

Diffusion & Homogeneous Reaction (19)Special case: Da >> 1

When the reaction is very fast, we see that the absorption rate is

proportional to the square root of reaction rate constant and diffusivityIndependent of the size of the pool

12D/kz

0,1

1 eCC −≈

limDa z

N C DL

Da C k D>> =

= = ⋅1 1

0

10 1210 12

Diffusion into a Falling Film (1)Now we shall deal with combination of diffusion and convection

The physical system is the gas absorption in a falling filmFlow of liquid film is laminarNo end effectsSolubility is constant

Diffusion into a Falling Film (2)Gas absorption in a falling film

Shell balance equation

( )

( ) ⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

∆+

∆

−

∆+

∆+

zz1,z

xx1,x

xNW

zNW

( )Rate ofmass in

Rate ofmass out



⎞⎠⎟ −

⎛⎝⎜

⎞⎠⎟ +

⎛⎝⎜

⎞⎠⎟ =

( )

( ) ⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

∆+

∆

z1,z

x1,x

xNW

zNW00 =+

Diffusion into a Falling Film (3)The mass balance equation

This is the mass balance equation in terms of fluxes. What needs to done is to relate these fluxes in terms of concentration

( )

( ) ⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

∆+

∆

−

∆+

∆+

zz1,z

xx1,x

xNW

zNW( )

( ) ⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

∆+

∆

z1,z

x1,x

xNW

zNW00 =+

∂

∂

∂

∂

Nx

Nz

x z, ,1 1 0+ =

zxW ∆∆


The molar flux in the x-direction is given by the Fick’s law

Since most gases dissolve sparingly in liquid, the bulk flow (second term in the above equation) is negligible compared to the diffusive term

The molar flux in the z-direction is also by the Fick’s law

∂

∂

∂

∂

Nx

Nz

x z, ,1 1 0+ =

( )N D Cx

x N Nx x x, , ,1 121

1 1 2= − + +∂∂

( )N DCz

x N Nz z z, , ,1 121

1 1 2= − + +∂∂

Diffusion into a Falling Film (5)The molar flux in the z-direction is also by the Fick’s law

By definition of flux

The Fick’s law can now be written as

( )N DCz

x N Nz z z, , ,1 121

1 1 2= − + +∂∂

N C v N C vz z z z, , , ,;1 1 1 2 2 2= =

( )N D Cz

x C v C vz z z, , ,1 121

1 1 1 2 2= − + +∂∂

( )N DCz

xC v C v

C CC Cz

z z,

, ,1 12

11

1 1 2 2

1 21 2= − +

++

⎛⎝⎜

⎞⎠⎟ +

∂∂

C v C vC C

v xz zz

1 1 2 2

1 2

, , ( )++

⎛⎝⎜

⎞⎠⎟ =

But

N DCz

v x Cz z, ( )1 121

1= − +∂∂


where the fluxes are

Substitute these flux equations into the mass balance equation, we get

∂

∂

∂

∂

Nx

Nz

x z, ,1 1 0+ =

xCDN 1

121,x ∂∂

−=

1z1z1

121,z C)x(vC)x(vz

CDN ≈+∂∂

−=

D Cx

v x Czz12

21

21∂

∂∂∂

= ( )


D Cx

v x Czz12

21

21∂

∂∂∂

= ( )

The heat balance equation

α∂∂

∂∂

∂∂

1r r

r Tr

v x Tzz

⎛⎝⎜

⎞⎠⎟= ( )

zT)x(v

xT

z2

2

∂∂

=∂∂

α


The boundary conditions:

Solution of this set of equations is possible, but let us consider the case of short contact time.

The dissolved species only travels a short distance into the bulk liquid

0C;0z 1 ==

D Cx

v x Czz12

21

21∂

∂∂∂

= ( )

0,11 CC;0x ==

0xCDN;x 1

12x,1 =∂∂

−=δ=

Diffusion into a Falling Film (8)Short contact time problem

Dissolved species only feel a convective motion of vmaxvelocity, and the plate seems to be far away. So the mass balance equation and three boundary conditions are replaced by

0C;0z 1 ==

D Cx

v x Czz12

21

21∂

∂∂∂

= ( )

0,11 CC;0x ==

0C;x 1 =∞=

DCx

vCz12

21

21∂

∂∂∂

≈ max

Diffusion into a Falling Film (9)Short contact time problem for gas absorption in falling film

0C;0z 1 ==

0,11 CC;0x ==

0C;x 1 =∞=

DCx

vCz12

21

21∂

∂∂∂

≈ max

Transient heat conduction in a semi-infinite object

x

Ts

tT

xT2

2

∂∂

=∂∂

α

0TT;0t ==

sTT;0x ==

0TT;x =∞=


0C;0z 1 ==

0,11 CC;0x ==

0C;x 1 =∞=

DCx

vCz12

21

21∂

∂∂∂

≈ max

Transient heat conduction in a semi-infinite object

tT

xT2

2

∂∂

=∂∂

α

0TT;0t ==

sTT;0x ==

0TT;x =∞=

T x t TT T

erf xt

s

s

( , ) −−

= ⎛⎝⎜

⎞⎠⎟0 4α⎟

⎟⎠

⎞⎜⎜⎝

⎛=

−−

max120,1

0,11

v/zD4xerf

C0C)t,x(C

C C erf xD z v1 10

12

14

= −⎛

⎝⎜⎜

⎞

⎠⎟⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥/ max


The quantity of interest is the gas absorption rate into the thin film. It simply is

0C;0z 1 ==

0,11 CC;0x ==

0C;x 1 =∞=

DCx

vCz12

21

21∂

∂∂∂

≈ max

C C erf xD z v1 10

12

14

= −⎛

⎝⎜⎜

⎞

⎠⎟⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥/ max

0x

1120x1,x x

CDN=

= ∂∂

−=

Diffusion into a Falling Film (12)Find the derivative

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−=

max12101 v/zD4

xerf1CC

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−

∂∂

=∂∂

max1210

1

v/zD4xerf1

xC

xC

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂

−=∂∂

max1210

1

v/zD4xerf

xC

xC

( )[ ]η∂∂

−=∂∂ erf

xC

xC

101

max12 v/zD4x

=η

( )[ ]x

erfCxC

101

∂η∂

ηη∂∂

−=∂∂

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛π

−=∂∂ η−

max1210

1

v/zD41e2C

xC 2

2

ev/zD

CxC

max12

101 η−

π−=

∂∂

⎥⎦

⎤⎢⎣

⎡−

π−=

∂∂

max12

2

max12

101

v/zD4xexp

v/zDC

xC

Diffusion into a Falling Film (13)The partial derivative of C1 with respect to x

Therefore the molar flux at the gas-liquid interface is

This is the gas absorption flux which is a function of z. To calculate the mass transfer rate that occurs over the whole interfacial area, we simply perform an integration with respect to area

0xmax12

2

max12

1210

0x

1120x1 v/zD4

xexpv/zD

DCxCDN

=== ⎥

⎦

⎤⎢⎣

⎡−

π=

∂∂

−=

⎥⎦

⎤⎢⎣

⎡−

π−=

∂∂

max12

2

max12

101

v/zD4xexp

v/zDC

xC

max12

12100x1 v/zD

DCNπ

==

Diffusion into a Falling Film (14)Therefore the molar flux at the gas-liquid interface is

The mass transfer rate that occurs over the whole interfacial area is

Hence the average molar flux into the falling film is

max12

12100x1 v/zD

DCNπ

==

M W N dz W CD v

zdz

M WL CD v

L

x x

L L

1 1 00

1012

0

1 10124

= =

=

=∫ ∫,max

max( )

π

π

LvD4C

WLMN max12

101

ave π==

END

Gas Absorption from a Rising Bubble (1)The physical system:

A column of liquid (species 2)A rising bubble containing pure species 1Mass transfer is by absorption of component 1 into the liquid 2

Given a bubble of initial size R0injected into the column, what is the distance that the bubble can travel before it completely dissolves?


A column of liquid (species 2)A rising bubble containing pure species 1Mass transfer is by absorption of component 1 into the liquid 2

Species 1 Species 2D

vt


Rising bubble

D

vt

The physical system:Falling film

L

LvD4C

fluxmolarAverage max12

10 π=⎟⎟

⎠

⎞⎜⎜⎝

⎛

DvD4C

fluxmolarAverage t12

10 π=⎟⎟

⎠

⎞⎜⎜⎝

⎛


Rising bubble

D

vt

So we know the average flux of mass transfer from the bubble to the surrounding

The remaining task is to carry out the mass balance around the bubble to find out how fast the bubble will shrink

The mass balance around the bubble

DvD4C

fluxmolarAverage t12

10 π=⎟⎟

⎠

⎞⎜⎜⎝

⎛( )d

dtD p

RTD C D v

Dtπ

ππ

30 2

1012

64⎛

⎝⎜

⎞⎠⎟ = −

Gas Absorption from a Rising Bubble (5)The mass balance equation around the bubble:

The terminal velocity is

Simplification:

( ) ( )µρ∆

ππ−=

π18

DgD

D4CDDdtd

RT6pD3 2

1210

202

D

vt

( )ddt

D pRT

D C D vD

tππ

π

30 2

1012

64⎛

⎝⎜

⎞⎠⎟ = −

v g Dt =

∆ρ 2

18µ

( )µρ∆

π−=

9DgD2CD

dtd

RT2p 12

100

µρ∆

π−=

9gD2

pRTCD2

dtdD 12

0

10 D2dtdD

γ−=

Gas Absorption from a Rising Bubble (6)SO the final mass balance of the bubble is

The initial condition isThe solution:

This solution tells us how the bubble size changes with time. So we can solve for the time when the bubble completely dissolves

0DD;0t ==

D2dtdD

γ−=

dtD2

dDγ−= D D t= −0 γ

µρ∆

π=γ

9gD2

pRTC 12

0

10

t D* =1

0γ

Gas Absorption from a Rising Bubble (7)Knowing the bubble size as a function of time

we can obtain the terminal velocity as a function of time

Thus given the velocity as a function of time, we can calculate the distance traveled by the bubble by applying the classical mechanics

( )40

2

t tD18g

18Dgv γ−

µρ∆

=µρ∆

=

D D t= −0 γµρ∆

π=γ

9gD2

pRTC 12

0

10

( )40t tD18gv

dtdx

γ−µρ∆

==

Gas Absorption from a Rising Bubble (8)Thus the distance traveled by the bubble is

( )40t tD18gv

dtdx

γ−µρ∆

==

( )∫ γ−µρ∆

=t

0

4

0 dttD18gx

( ) ( )∫ γ−γ−µρ∆

γ−=

t

00

4

0 tDdtD18g1x

( ) t

0

5

0 tD18g

51x γ−

µρ∆

γ−=

( ) ( ) ⎥⎦⎤⎢⎣⎡ γ−−

µρ∆

γ=

5

0

5

0 tDD18g

51x

Gas Absorption from a Rising Bubble (9)At the time when the bubble completely dissolve

the distance that the bubble has traveled is

t D* =1

0γ

2/50D

18g

51H

µρ∆

γ=

µρ∆

π=γ

9gD2

pRTC 12

0

10

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅⎟⎟⎠

⎞⎜⎜⎝

⎛

µρ∆

⋅⎟⎟⎠

⎞⎜⎜⎝

⎛ π=

10

0

12

5o

CRT/p

DDg

230H

Diffusion and Reaction in a Porous Catalyst (1)

Porous Catalysts: very complex becausePores of different size and shapePores are tortuousResistance to mass transfer

Nonporous catalyst: SimpleNice simple geometryNo problem with mass transfer

So why don’t we use non-porous catalysts?


Porous Catalysts vs non-porous catalysts

This is why most catalysts used in industries are porous!

Very low surface area per unit massHigh pressure drop

Difficult to handle

High surface area

Low pressure drop

Easy to handle

Non-porous CatalystsPorous Catalysts


So how do we model diffusion and reaction in a catalyst?Should we model as detailed as possible? For example, we consider every possible pores within a solid, assuming of course that we know in details the connectivity between all pores.


So how do we model diffusion and reaction in a catalyst?Or we shall model the catalyst particle as if it is homogeneous, and take advantage of all that we know about the macroscopic properties of the solid, for example

Pore sizeEffective diffusivityParticle PorosityParticle tortuosity factorSpecific surface areaShape of the particle


Pore size:Pore size distribution: Choose the appropriate average pore size

Effective diffusivitySo much has been done in 60 and 70.

Particle PorosityParticle tortuosity factor

Pore is not straight

Specific surface area; quite high 100-1000 m2/gShape of the particle:

Make it simple


Two ingredients required to model diffusion and reaction in a catalystDiffusion fluxReaction rate

The diffusion flux done so far is for homogeneous media. For heterogeneous media, like a porous catalyst, the molar flux could be defined in the way as the Fick’s law

where the effective diffusivity is a function of system parameters as well as the concentrationThis form is identical to the heat flux equation for heterogeneous media dealt with earlier

⎥⎦⎤

⎢⎣⎡

∂∂

−= time- area sectional cross total

ed transportmolesrCDJ eff


Equation for diffusion flux

The effective diffusivity is a function of binary diffusivity, the Knudsen diffusivity, the porosity and the tortuosity

rCDJ eff ∂∂

−=

Equation for heat flux

The effective thermal conductivity is a function of thermal conductivities of the phases constituting the media, and the porosity.

rTkq eff ∂∂

−=

( )ε= ,k,kfk sfeff( )τε= ,,D,DfD K12eff


Equation for diffusion flux

The effective diffusivity is a function of binary diffusivity, the Knudsen diffusivity, the porosity and the tortuosity

Extensive research was conducted in the 60 and 70 by engineers, and one of the simple formulas is given below for the effective diffusivity

rCDJ eff ∂∂

−=

( )τε= ,,D,DfD K12eff

ceff DDτε

=


The effective diffusivity

Where ε is the particle porosity. This is to account for the fact that only a fraction of ε of the cross sectional area is available for mass transferThe parameter τ is the tortuosity factor. It accounts for the zig-zag pattern of the pore.The parameter Dc is called the combined diffusivity. It accounts for two mechanisms for diffusion in pore, namely molecular diffusion and Knudsen diffusion

ceff DDτε

=



The combined diffusivity is given by:

Where D12 is the usual binary diffusivityAnd DK is the Knudsen diffusivity

The molecular diffusivity can be found in any books or handbook (such as Perry) or it can be calculated from Chapman-Enskog equationThe Knudsen diffusivity is calculated from

ceff DDτε

=

1,K121,eff D1

D1

D1

+=

MTR8

3r2D g

K π= r = Pore radius (not particle radius)

M = Molecular weight




Order of magnitude

ceff DDτε

=

1,K121,eff D1

D1

D1

+=

cm2/s0.1 - 100DK

cm2/s0.1 - 2D12

2 - 6τ0.2 – 0.7ε




The pore size, pressure and temperature dependence

ceff DDτε

=

1,K121,eff D1

D1

D1

+=

r1T0.5P0Knudsen diffusivity

r0T1.75P-1Binary diffusivityMrTPDiffusivity

21 M1

M1

+

1M1




Controlling mechanismMolecular diffusion dominates in large pores and high pressureKnudsen diffusion dominates in small pores and very low pressure

ceff DDτε

=

1,K121,eff D1

D1

D1

+=

r1T0.5P0Knudsen diffusivity

r0T1.75P-1Binary diffusivity

MrTPDiffusivity

21 M1

M1

+

1M1


Reaction rateSince reaction in a porous catalyst occurs on surface, the reaction rate is usually (but not always) defined as mole reacted per unit surface area and per unit time

Let us deal with first-order reaction:

With this definition of the reaction rate, the reaction rate constant of the first-order reaction has units of m/s

⎥⎦⎤

⎢⎣⎡ℜ

time- area surfacereacted moles)c(

⎥⎦⎤

⎢⎣⎡=ℜ

time- area surfacereacted moleskC)C(


SummaryDiffusion rate

Reaction rate

These are all we need to model diffusion and reaction in a porous catalysts

⎥⎦⎤

⎢⎣⎡=ℜ

time- area surfacereacted moleskC)C(

⎥⎦⎤

⎢⎣⎡

∂∂

−= time- areasection cross total

ed transportmolesrCDJ eff

ceff DDτε

=

1,K121,eff D1

D1

D1

+=


Shell balance

( )Rate ofmass in

Rate ofmass out



⎞⎠⎟ −

⎛⎝⎜

⎞⎠⎟ +

⎛⎝⎜

⎞⎠⎟ =

r

∆r

( )[ ] 0kCSrr4Nr4Nr4 gp2

rrr2

rr2 =ρ∆π−π−π

∆+==

r4 ∆π

0kCSrr

NrNrgp

2rr2

rrr2

=ρ−∆

−− =∆+=

( ) 0kCSrNrr gp

22 =ρ−∂∂

−


The mass balance equation

But the diffusion flux is

The final mass balance equation in terms of concentration is then

This is now known as the classical equation for diffusion and reaction in a porous catalyst

rCDN eff ∂∂

−=r

∆r( ) 0kCSrNrr gp

22 =ρ−∂∂

−

0kCSrCr

rrD

gp2

2eff =ρ−⎟

⎠⎞

⎜⎝⎛

∂∂

∂∂


So, the mass balance equation

The boundary conditions are:

For simplicity, we shall assume the boundary condition of the first kind at the catalyst surface. In general, you should use the boundary condition of the third kind

0kCSrCr

rrD

gp2

2eff =ρ−⎟

⎠⎞

⎜⎝⎛

∂∂

∂∂

r

∆r

0rC;0r =∂∂

=

( )ionconcentratbulk known CC;Rr 0==

( )0RrmRr

eff CCkrCD;Rr −=∂∂

−==

=


So, the mass balance equation

It is much more convenient and compact if we cast the above mass balance equation into non-dimensional form. To do this we scale the concentration against the bulk concentration C0 and the radial distance against the particle radius

0kCSrCr

rrD

gp2

2eff =ρ−⎟

⎠⎞

⎜⎝⎛

∂∂

∂∂ r

∆r

0rC;0r =∂∂

=

( )ionconcentratbulk known CC;Rr 0==

0CCy;

Rrx ==


Let us dimensionalize the mass balance equation, and something will evolve naturally out of this process

0kCSrCr

rrD

gp2

2eff =ρ−⎟

⎠⎞

⎜⎝⎛

∂∂

∂∂

r

∆r

0CCy;

Rrx ==

( )( ) ( )

( ) ( ) 0yCkSRx

yCRxRxRx

D0gp

022

eff =ρ−⎥⎦

⎤⎢⎣

⎡∂∂

∂∂

0kySxyx

xx1

RD

gp2

22eff =ρ−⎥⎦

⎤⎢⎣⎡

∂∂

∂∂ 0y

DRkS

xyx

xx1

eff

2gp2

2 =⎟⎟⎠

⎞⎜⎜⎝

⎛ ρ−⎥⎦

⎤⎢⎣⎡

∂∂

∂∂


The mass balance equation is …..

Since every term in the above equation is non-dimensional, the group in the bracket MUST be non-dimensional as well. We define that group as

This non-dimensional group is known asThiele modulus in Western literatureDamkohler number in German literatureZel’dowitch number in the Russian literature

r

∆r

0yD

RkSxyx

xx1

eff

2gp2

2 =⎟⎟⎠

⎞⎜⎜⎝

⎛ ρ−⎥⎦

⎤⎢⎣⎡

∂∂

∂∂

eff

2gp2

DRkSρ

=φ


So the mass balance equation is …..

The physical significance of the Thiele modulus

If φ << 1 (diffusion time is small compared to reaction time), we would expect uniform concentration profile and the reaction is kinetically-controlledIf φ >> 1 (diffusion time is greater than reaction time), we would expect a very sharp concentration profile and the reaction is called difuusion-controlled

r

∆r

0yxyx

xx1 22

2 =φ−⎥⎦⎤

⎢⎣⎡

∂∂

∂∂

imeReaction ttimeDiffusion

kS1

DR

DRkS

gp

eff

2

eff

2gp2 ≡

⎟⎟⎠

⎞⎜⎜⎝

⎛

ρ

⎟⎟⎠

⎞⎜⎜⎝

⎛

=ρ

=φ


So the mass balance equation is …..

Solution?Introduce a new variable

Substitute this into the above mass balance equationxuy =

r

∆r

0yxyx

xx1 22

2 =φ−⎥⎦⎤

⎢⎣⎡

∂∂

∂∂

( ) 0xu

xx/ux

xx1 22

2 =φ−⎥⎦⎤

⎢⎣⎡

∂∂

∂∂

0xu

xu

xu

x1x

xx1 2

22

2 =φ−⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −

∂∂

∂∂ 0

xuu

xux

xx1 2

2 =φ−⎟⎠⎞

⎜⎝⎛ −

∂∂

∂∂


Solution (continued)

Solution of this equation is

0xu

xu

xux

xu

x1 2

2

2

2 =φ−⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

−∂∂

+∂∂

r

∆r

0xuu

xux

xx1 2

2 =φ−⎟⎠⎞

⎜⎝⎛ −

∂∂

∂∂

0xu

xu

x1 2

2

2

=φ−∂∂

0uxu 22

2

=φ−∂∂

xx BeAeu φ−φ +=


Solution (continued)

The constants A and B can be found from the two boundary conditions

Application of the two boundary conditions

xx BeAeu φ−φ +=

0rC;0r =∂∂

=

0CC;Rr ==

0xy;0x =∂∂

=

1y;1x ==

0u;0x ==

1u;1x ==

0u;0x ==

1u;1x ==

BA0 +=

φ−φ += BeAe1

φ−φ −=

ee1A

φ−φ −−=

ee1B


So the solution for u is

Since

The solution for the non-dimensional concentration is

r

∆r

( )( )φφ

=−−

= φ−φ

φ−φ

sinhxsinh

eeeeu

xx

xuy =

( )( )φφ

=sinh

xsinhx1y


The solution for the non-dimensional concentration is

The reaction rate in the catalyst particle is simply to molar flux at the surface of the catalyst

Therefore the reaction rate per particle is

( )( )φφ

=sinh

xsinhx1y

r

∆r

⎥⎦

⎤⎢⎣

⎡∂∂

−== time-catalyst of area

reacted molesrCDN

RreffR

( ) ⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

−π== time-particle

reacted molesrCDR4W

Rreff

2


Reaction rate per particle

Let put the RHS into non-dimensional form:( )( )φφ

=sinh

xsinhx1y

r

∆r

( ) ⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

−π== time-particle

reacted molesrCDR4W

Rreff

2

( ) ( )( ) ⎟

⎟⎠

⎞⎜⎜⎝

⎛

∂∂

−π==1x

0eff

2

RxyCDR4W

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

π−==1x

0eff xyCRD4W ( )

( ) ( )⎥⎦⎤

⎢⎣⎡ φ

−φφ

φ= 2x

xsinhx

xcoshsinh

1dxdy

( )( )1cothCRD4W 0eff −φφπ−=This is your reaction rate perparticle written in terms of the bulk concentration and the systemparameters


The reaction rate per particle is

For very slow reaction compared to diffusion, φ << 1

The reaction is kinetically-controlled

L+φ

+φ

≈φ3

1coth

r

∆r

( )( )1cothCRD4W 0eff −φφπ−=

0gp

3

CkS3R4W ρ⎟⎟

⎠

⎞⎜⎜⎝

⎛ π−=


The reaction rate per particle is

For very fast reaction compared to diffusion, φ >> 1

The reaction is diffusion-controlled

1coth ≈φ

r

∆r

( )( )1cothCRD4W 0eff −φφπ−=

( )( ) 0effgp2 CkDSR4W ρπ−=

Diffusion through a Polymer Film (1)Type of membranes

Porous membrane: Pore sizes usually between 200 to 3000 nm. Transport through this type is by viscous flow. Separation is due to size

Non-porous membraneThe transport through this type of membrane is controlled by diffusion of adsorbed molecules inside the membrane. The diffusivity is called the intra-membrane diffusivity.The flux equation mimics the Fick’s law

This intra-membrane diffusivity must be determined from experiments. With the exception of very low concentration, this diffusivity is generally a complex function of concentration and its gradient

dxdCDJ −=

( )T,x/C,CfD ∂∂=

Diffusion through a Polymer Film (2)This last example will show you how to determine the intra-membrane diffusivity.

The method is the time-lag method, developed by Daynes in 1920 who studied the permeation of gases through rubbery membranes used in balloons

Diffusion through a Polymer Film (3)The time lag method is quite simple.

It consists of two chambers separated by the tested membraneThe top chamber is maintained at constant pressure P0

The pressure of the bottom chamber is monitored with respect to time

Maintained at constant pressure, P0

Diffusion through a Polymer Film (4)The time lag method

P

time

Diffusion through a Polymer Film (4)The shell is drawn inside the membrane

The mass balance equation is

The initial condition and boundary conditions are:

The solution is:

2

2

xCD

tC

∂∂

=∂∂

0C;0t ==

0HPC;0x ==

0C;Lx ≈=

( )⎟⎟⎠

⎞⎜⎜⎝

⎛ π−

ππ

−−= ∑∞

=

tL

Dnexpn

L/xnsin2Lx1

PHC

2

22

1n0

Diffusion through a Polymer Film (5)To determine the pressure change in the bottom chamber, we have to set up the mass balance around that chamber

Given

the pressure in the bottom chamber is

LxxCDA

RTPV

dtd

=∂∂

−=⎟⎠⎞

⎜⎝⎛

0P;0t ==

( )⎟⎟⎠

⎞⎜⎜⎝

⎛ π−

ππ

−−= ∑∞

=

tL

Dnexpn

L/xnsin2Lx1

PHC

2

22

1n0

( )⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛ π−−

ππ

+= ∑∞

=

tL

Dnexp1n

ncosLH2tDHLV

PTRAP 2

22

1n22

20

Diffusion through a Polymer Film (6)So the solution for the pressure of the bottom chamber is

At sufficiently large time, this solution has an asymptote

( )⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛ π−−

ππ

+= ∑∞

=

tL

Dnexp1n

ncosLH2tDHLV

PTRAP 2

22

1n22

20

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

D6Lt

LVDHPTRAP

20

P

time

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