Queens Land

Transport Phenomena Section 3: Mass Transfer D. D. Do University of Queensland

Mode of Mass Transfer †

Diffusion: „ „

Molecular diffusion Knudsen 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 †

Concentration „

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

Mass concentration

ρj =

Molar concentration Mass fraction

ρj

Cj =

ωj =

Mj

ρj ρ

mass of j volume of solution

=

=

moleof j volumeof solution

ρj n

∑ρ

=

mass of j total mass

=

mole of j total moles

k

k =1

Mole fraction xj =

Cj C

=

Cj n

∑C k =1

k

Velocities and Average Velocities (1) †

Different components move at different velocities 2 1

†

v1

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) n

∑ρ v k =1

†

v2

k

k

The molar flux is (molar rate per unit area) n

∑C v k =1

k

k

Velocities and Average Velocities (2) †

Different components move at different velocities 2 1

†

v2 v1

The local mass velocity is defined as a velocity „

If we multiply it by the total mass concentration, we get the total mass flux n n ρk vk ∑ ρk vk ∑ n n n ρk k =1 k =1 v= n = = ∑ vk = ∑ ω k vk ρk v k ρ k =1 ρ k =1 ρ k =1 ∑ k

ρv = ∑

k =1

Velocities and Average Velocities (3) †

Different components move at different velocities v2

2 1

†

v1

The local molar velocity is defined as a velocity „

If we multiply it by the total molar concentration, we get the total molar flux n

n

n

Cv = ∑ C k v k k =1

v = *

∑C k =1 n

k

vk

∑ Ck k =1

=

∑C k =1

k

C

vk

n

n Ck =∑ vk = ∑ x k vk k =1 C k =1

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 2 1

„

V

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

Diffusion Velocities (2) †

This gives rise to the need to define the so-called diffusion velocity „

There are two diffusion velocities † †

Diffusion velocity relative to the mass average velocity Diffusion velocity relative to the molar average velocity

⎛ diffusion velocity of species j⎞ vj − v ≡ ⎜ ⎟ ⎠ ⎝ with respect to v

⎛ diffusion velocity of species j⎞ v j − v* ≡ ⎜ ⎟ ⎝ with respect to v* ⎠

2 1

V

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

Mass Fluxes „

The mass flux of the component j is the product of the mass concentration and the velocity of that component ⎛ mass of species j⎞ ⎛ dis tan ce ⎞ n j = ρ jv j = ⎜ ⎟ ⎟ ×⎜ ⎝ ⎠ ⎝ time ⎠ volume

„

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

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

Molar Fluxes „

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

⎛ mole of species N j = C jv j = ⎜ ⎝ volume „

j ⎞ ⎛ dis tan ce ⎞ ⎟ ⎟ ×⎜ ⎠ ⎝ time ⎠

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

Diffusion Mass and Molar Fluxes (1) †

The diffusion mass flux is defined the flux relative to the bulk motion of the solution

(

jj = ρ j v j − v „

)

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

Diffusion Mass and Molar Fluxes (2) †

The diffusion molar flux is defined the flux relative to the bulk motion of the solution

(

J j = C j v j − v* „

)

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

Properties of Diffusion Fluxes (1) †

The diffusion fluxes satisfy the following equation n

∑j k =1

k

= 0;

n

∑J

k

=0

k =1

2

„

For example, for a system of two components

∑J k =1

„

k

=0

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?

Properties of Diffusion Fluxes (2) †

The diffusion fluxes satisfy the following equation n

∑j

k

n

∑J

= 0;

k =1

†

=0

k

k =1

Proof: „

By definition n

∑J

=

k

k =1

∑ C (v n

k

k

− v*

k =1

n

∑J

)

k

=

k =1

n

∑C

k

vk − v

k =1

n

*

∑C

k

k =1

n

n

∑J k =1

k

=

n

∑C

k

n

∑J

v k − v *C

k =1

k =1

k

=

n

∑C k =1

k

vk −

∑C k =1

k

vk

C

n

v = *

∑C

k

k =1

C

vk

n

∑J k =1

k

=0

C

Diffusion Flux vs Flux (1) †

Recall the definitions of molar flux and molar diffusion flux

N j = C jv j

J j = C j (v j − v *)

This is necessary for the diffusion analysis This is useful for engineering calculation of mass transfer †

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 n

Jk = Nk − xk ∑ N j j= 1

Diffusion Flux vs Flux (2) †

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

Jk = Nk − xk ∑ N j j= 1

†

Proof: Always go back to definition = Ckvk − Ckv * J k = C k v k − v*

(

Ck Jk = Ckvk − C

)

j =1

j

= Ckvk − Ck

∑C v

j

j

j =1

C n

n

∑C v

n

Jk = Nk − xk ∑ N j j= 1

j

Diffusion Flux vs Flux (3) †

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

Jk = Nk − xk ∑ N j j= 1

†

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

Diffusion Flux vs Flux (4) †

Similarly, the relationship between the mass flux and the mass diffusive flux is n

jk = n k − ω k ∑ n j j= 1

†

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

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

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

†

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

The Fick’s law for the first component is dx J 1 = − cD 12 1 dz „ „ „

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 above The coefficient D12 is called the binary diffusivity

Fick’s law of Diffusion for Binary Mixtures (2) †

Similarly, we can write the same equation for the second component by simply interchanging the subscripts 1 and 2 dx 2 dx 1 J cD = − J 1 = − cD 12 2 21 dz dz

†

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

J1 + J 2 = 0 „

Thus adding the two molar diffusive fluxes, we have J1 + J 2 = −cD12

„

„

dx dx1 − cD 21 2 = 0 dz dz

But x 1 + x 2 = 1 The above equation will become J1 + J 2 = −c(D12 − D 21 )

dx1 =0 dz

J1 + J 2 = −cD12

dx1 dx + cD 21 1 = 0 dz dz

(D12 − D 21 ) = 0

Fick’s law of Diffusion for Binary Mixtures (3) †

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

†

What it means is that the two diffusion equations

„

†

J 1 = − cD12

dx1 dz

J 2 = −cD 21

dx 2 dz

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

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

J 1 = − cD 12 ∇ x 1

Fick’s law of Diffusion for Binary Mixtures (4) †

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 dx1 J 1 = − cD12 dz

†

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

J k = Nk − xk ∑ N j †

which for binary mixtures, it is 2

J1 = N1 − x1 ∑ N j j=1

j=1

N 1 = J 1 + x1 ( N 1 + N 2 )

Fick’s law of Diffusion for Binary Mixtures (5) †

So the equation for the molar flux is

J 1 = − cD12

dx1 dz

N 1 = J 1 + x1 ( N 1 + N 2 ) dx 1 N 1 = − cD12 + x1 ( N 1 + N 2 ) dz †

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

dx 2 N 2 = − cD 21 + x 2 (N 2 + N1 ) dz †

But only one of them is independent!

Fick’s law of Diffusion for Binary Mixtures (6) †

Summary

J 1 = − cD12

N 1 = − cD12 †

dx 1 + x1 ( N 1 + N 2 ) dz

The above equation involves two unknown variables: „ „

†

dx1 dz

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.

Diffusion Coefficient †

†

Units

D12

⎡ m2 ⎤ ≡⎢ ⎥ sec ⎣ ⎦

Order of magnitude

State

Order of magnitude (cm2/s)

Gas

0.1 - 1

Liquid 1 × 10-7 – 1 × 10-5 Solid

†

Pressure and Temperature Dependence

1 × 10-12 – 1 × 10-7

State

P

T

Gas

decrease

increase

Liquid

-

increase

Solid

-

increase

Stefan-Maxwell’s law for mixtures † †

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

n

− c ∇x i = ∑ j=1

„

†

j

i

− xi N j D ij

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

Another form of the Stefan-Maxwell equation is n

− c ∇x i = ∑ j=1

†

(x N

(x J j

i

− xi J j

)

D ij

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

)

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 directions Set up the mass balance equation around the shell ⎛ Rate of ⎞ ⎛ Rate of ⎞ ⎛ Rate of mass⎞ ⎟ = ( Accummulation) ⎟ +⎜ ⎟ −⎜ ⎜ ⎝ mass in ⎠ ⎝ mass out⎠ ⎝ production ⎠

„

„

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 flux Apply the Fick’s law, and we will get a second-order differential equation in terms of concentration

First Principles for Binary Mixtures (2) †

The procedure (continued) „ „ „

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

Boundary Conditions †

There are five boundary conditions that you will regularly encounter in mass transfer „

BC of the first kind: †

„

BC of the second kind: †

„

Molar flux into a medium is equal to the molar flux through the stagnant film surrounding the medium

BC of the fourth kind: †

„

Molar flux is specified at the boundary

BC of the third kind: †

„

Concentration is specified at the boundary

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 tube Example 2: Dissolution of a spherical objects Example 3: Diffusion and heterogeneous reaction at surface Example 4: Diffusion and homogeneous reaction Example 5: Diffusion into a falling film Example 6: Gas absorption in a rising bubble Example 7: Diffusion and reaction in a porous catalyst Example 8: Transient diffusion through a polymer film Example 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 tube Gas 2 is flowing across the tube mouth Gas 2 is non-soluble in liquid 1

Diffusion in a Stefan tube (2) †

Shell balance ⎛ Rate of ⎞ ⎛ Rate of ⎞ ⎛ Rate of mass⎞ ⎟ = ( Accummulation) ⎟ +⎜ ⎟ −⎜ ⎜ ⎝ mass in ⎠ ⎝ mass out⎠ ⎝ production ⎠

S⋅ N1 z

− S ⋅ N1 z + ∆z

+0 = 0

S ∆z

†

Taking the shell as thin as possible − „

dN 1 =0 dz

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

Diffusion in a Stefan tube (3) †

So the shell balance equation is dN 1 − =0 dz

†

Apply the Fick’s law for binary system

N 1 = − cD 12 „

dx 1 + x1(N 1 + N 2 ) dz

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:

N 1 = − cD 12

dx 1 + x 1 (N 1 ) dz

N1 = −

cD 12 dx 1 1 − x 1 dz

Diffusion in a Stefan tube (4) †

The mass balance equation is dN 1 − =0 dz „

The Fick’s law

N1 = − „

cD 12 dx 1 1 − x 1 dz

So the mass balance equation in terms of concentration is

d ⎛ cD 12 dx 1 ⎞ ⎜ ⎟=0 dz ⎝ 1 − x 1 dz ⎠

cD 12 For constant P and D12

d dz

⎛ 1 dx 1 ⎞ ⎜⎜ ⎟⎟ = 0 ⎝ 1 − x 1 dz ⎠

Diffusion in a Stefan tube (5) †

The mass balance equation in terms of concentration is a second-order differential equation ⎛ 1 dx 1 ⎞ ⎜⎜ ⎟⎟ = 0 cD 12 ⎝ 1 − x 1 dz ⎠ Physical constraints (Boundary conditions) d dz

†

z = z1 ;

x 1 = x 1, 0 =

z = z2 ;

x 1 = x 1, L

p10 P

z2

z1

0

Diffusion in a Stefan tube (6) †

Solution procedure cD 12

d dz

⎛ 1 dx 1 ⎞ ⎜⎜ ⎟⎟ = 0 ⎝ 1 − x 1 dz ⎠

„

Integrate it once

„

cD 12 dx 1 = K 1 ( cons tan t ) 1 − x 1 dz Integrate it one more time ⎛ 1 ⎞ cD 12 ln ⎜ ⎟ = K1z + K 2 ⎝ 1 − x1 ⎠ †

†

z2

z1

0

which involves two constants of integration

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

Diffusion in a Stefan tube (7) †

Solution ⎛ 1 ⎞ cD 12 ln ⎜ ⎟ = K1z + K 2 ⎝ 1 − x1 ⎠

†

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

@ z = z1 ; @ z = z2 ;

†

⎛ 1 ⎞ ⎟ = K1z1 + K 2 cD 12 ln ⎜ ⎝ 1 − x 1, 0 ⎠ ⎛ 1 ⎞ ⎟ = K1z 2 + K 2 cD 12 ln ⎜ ⎝ 1 − x 1, L ⎠

So you can solve for K1 and K2

z2

z1

0

Diffusion in a Stefan tube (8) †

Solution cD ln ⎛⎜ 1 ⎞⎟ = K z + K 12 1 2 ⎝ 1 − x1 ⎠

†

z2

K1 and K2 K1 =

⎛ 1 − x 1, 0 ⎞ cD 12 ln ⎜ ⎟ z 2 − z 1 ⎝ 1 − x 1, L ⎠

⎛ 1 ⎞ cD 12 z 1 ⎛ 1 − x 1, 0 ⎞ ⎟− ⎟ K 2 = cD 12 ln ⎜ ln ⎜ ⎝ 1 − x 1, L ⎠ z 2 − z 1 ⎝ 1 − x 1, L ⎠

z1

0 †

So the concentration distribution is ⎛ 1 − x 1 ⎞ ⎛ 1 − x 1,L ⎞ ⎜ ⎟ =⎜ ⎟ ⎝ 1 − x 1,0 ⎠ ⎝ 1 − x 1,0 ⎠

( z − z1 )/ ( z 2 − z1 )

Diffusion in a Stefan tube (9) †

So the final solution for the concentration distribution is z ⎛ 1 − x 1 ⎞ ⎛ 1 − x 1,L ⎞ ⎜ ⎟ =⎜ ⎟ ⎝ 1 − x 1,0 ⎠ ⎝ 1 − x 1,0 ⎠

2

( z − z1 )/ ( z 2 − z1 )

z1 †

To find the evaporation flux, we simply apply the Fick’s law cD 12 dx 1 N1 = − 1 − x 1 dz „

0

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

Diffusion in a Stefan tube (10) †

Let us find the flux formally

⎛ 1 − x 1 ⎞ ⎛ 1 − x 1,L ⎞ ⎜ ⎟ =⎜ ⎟ ⎝ 1 − x 1, 0 ⎠ ⎝ 1 − x 1, 0 ⎠ †

( z − z1 )/ ( z 2 − z1 )

Taking logarithm both side ⎧⎪⎛ 1 − x ⎞ (z − z1 )/ (z 2 − z1 ) ⎫⎪ ⎛ 1 − x1 ⎞ 1, L ⎟ = ln ⎨⎜ ⎟ ln⎜⎜ ⎬ ⎜ 1− x ⎟ ⎟ 1 x − 1, 0 ⎠ 1, 0 ⎠ ⎪⎩⎝ ⎪⎭ ⎝

†

z2

z1

0 z − z1 ⎛⎜ 1 − x1,L ⎞⎟ ln (1 − x1 ) − ln (1 − x1, 0 ) = ln⎜ z 2 − z1 ⎝ 1 − x1, 0 ⎟⎠ Now taking differentiation: ⎛ 1 − x1,L ⎞ ⎛ 1 − x1,L ⎞ dx1 dz 1 dx1 1 ⎜ ⎟ ⎜ ⎟ − = − = ln⎜ ln 1 − x1 z 2 − z1 ⎝ 1 − x1, 0 ⎟⎠ 1 − x1 dz z 2 − z1 ⎜⎝ 1 − x1, 0 ⎟⎠

Diffusion in a Stefan tube (11) †

So we have ⎛ 1 − x1,L ⎞ 1 dx1 1 ⎟ − = ln⎜⎜ 1 − x1 dz z 2 − z1 ⎝ 1 − x1, 0 ⎟⎠

†

Substitute this into the Fick’s law

cD 12 dx 1 N1 = − 1 − x 1 dz †

z2

The evaporation flux is

z1

0

cD12 ⎛⎜ 1 − x1,L ⎞⎟ N1 = ln z 2 − z1 ⎜⎝ 1 − x1, 0 ⎟⎠ †

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

Diffusion in a Stefan tube (12) †

†

However there is a short cut. Remember that during the integration of the mass balance we got: cD 12 dx 1 = K1 1 − x 1 dz

And K1 was found from the application of the z 1 boundary condition as

⎛ 1 − x 1,0 ⎞ cD12 ⎟ K1 = ln⎜ z 2 − z1 ⎝ 1 − x 1,L ⎠ †

z2

Therefore the evaporation flux is cD12 ⎛⎜ 1 − x1,L ⎞⎟ N1 = ln z 2 − z1 ⎜⎝ 1 − x1, 0 ⎟⎠

0

Diffusion in a Stefan tube (13) †

So the solution for the evaporation flux is

cD12 ⎛⎜ 1 − x1,L ⎞⎟ N1 = ln⎜ z 2 − z1 ⎝ 1 − x1, 0 ⎟⎠ †

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 ⎛ 1 ⎞ cD 12 ⎟ ln ⎜ N1 = z 2 − z 1 ⎝ 1 − x 1, 0 ⎠ N1 =

( P / RT ) D 12 ln ⎛ z 2 − z1

⎞ 1 ⎜ ⎟ ⎝ 1 − p i0 / P ⎠

z2

z1

0

Diffusion in a Stefan tube (14) †

Given the flux of species 1 as N1 =

†

z 2 − z1

⎞ 1 ⎜ ⎟ ⎝ 1 − p i0 / P ⎠

z2

The total flux is

( P / RT ) D 12 ln ⎛ =

z1

The flux of the component 2 is

0

N1 + N2 †

( P / RT ) D 12 ln ⎛

z 2 − z1

N 2 = − cD 12

⎞ 1 ⎜ ⎟ 0 p P 1 − / ⎝ ⎠ i

dx 2 + x 2 (N 1 + N 2 ) = 0 dz

Diffusion in a Stefan tube (15) †

Because of the evaporation, the liquid level will drop with time N1

†

P / RT ) D 12 ⎛ ( = ln z 2 − z1

⎞ 1 ⎜ ⎟ 0 p P 1 − / ⎝ ⎠ i

Therefore you need to write a mass balance around the liquid d (Sz 1ρ L ) = − SM 1 N 1 dt

d (Sz 1ρ L ) ( P / RT ) D 12 ln ⎛ 1 ⎞ = − SM 1 ⎜ ⎟ dt z 2 − z1 ⎝ 1 − p 10 / P ⎠ †

This is a differential equation with respect to t

z2

z1

0

Diffusion in a Stefan tube (16) †

Differential equation for the liquid level with respect to t d (Sz 1ρ L ) ( P / RT ) D 12 ln ⎛ 1 ⎞ = − SM 1 ⎜ ⎟ dt z 2 − z1 ⎝ 1 − p 10 / P ⎠

†

z1

The initial condition is

t = 0; †

z2

z 1 = z 1, 0

The solution is

[ z 2 − z 1 ( t ) ]2 − ( z 2 − z 10 ) 2

0 The time taken to empty the liquid is:

=

⎡ M 1 ( P / RT ) D 12 ⎛ ⎞⎤ 1 ln ⎜ ⎟ ⎥t ⎢2 0 ρ / 1 − p P ⎝ ⎠⎦ L 1 ⎣

t* =

z 22 − (z 2 − z 10 )

2

⎡ M 1 (P / RT )D 12 ⎛ ⎞⎤ 1 ⎜ ⎟⎥ ln ⎜ ⎢2 0 ⎟ ρL ⎝ 1 − p 1 / P ⎠ ⎦⎥ ⎣⎢

Diffusion in a Stefan tube (17) †

The time taken to empty the liquid is: t = *

z 22 − (z 2 − z 10 )

z2

2

⎡ M 1 (P / RT )D 12 ⎛ ⎞⎤ 1 ⎜ ⎟ 2 ln ⎢ ⎜ 1 − p 0 / P ⎟⎥ ρ ⎢⎣ L 1 ⎝ ⎠ ⎥⎦

z1 †

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

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 ∆r

2R

r

Dissolution of a Sphere (2) †

Physical system: Sparingly soluble sphere in a surrounding fluid of infinite extent „

The shell balance

( 4πr

2

N1

) − ( 4πr

2

r

N1

)

r + ∆r

+0= 0

4 π ∆r

lim

∆r → 0

(

r2 N1

) ( r

− r2 N1 ∆r

)

r + ∆r

(

)

d 2 =− r N1 = 0 dr

Dissolution of a Sphere (3) †

The shell balance

†

The Fick’s law „

†

−

(

r

)

d 2 r N1 = 0 dr

N 1 = − cD 12

dx 1 + x1 (N 1 + N 2 ) dr

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 dx N1 ≈ −cD12 1 dr

Substitute the Fick’s law into the mass balance equation dx 1 ⎞ d ⎛ 2 ⎜ r cD 12 ⎟=0 ⎝ dr dr ⎠

Dissolution of a Sphere (4) †

So the final form of the mass balance equation is d⎛ 2 dC1 ⎞ r D ⎜ ⎟=0 12 dr ⎝ dr ⎠

dx 1 ⎞ d ⎛ 2 r cD ⎜ ⎟=0 12 dr ⎝ dr ⎠ †

The boundary conditions are: r = R;

†

C 1 = C 1, 0

C1 = 0

The concentration distribution C 1 = C 10

†

r = ∞;

The dissolution rate „ „

R r

dx1 dC1 = − D 12 Back to the Fick’s law: dr dr The dissolution rate is simply the flux at the surface of the object D C dC N1 r = R ≈ − D12 1 = 12 10 dr r = R R N1 ≈ −cD12

r

Dissolution of a Sphere (5) †

Very often the dissolution is calculated by using the concept of mass transfer coefficient:

N1 †

r

r= R

= k m ( C 10 − 0)

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

N1 r = R

D12 C10 = R

„

You will get

„

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

k m (2 R ) =2 D 12

Dissolution of a Sphere (6) †

So the Sherwood number for a stagnant environment is k m (2 R ) =2 D12

†

Recall the equivalent number in heat transfer is Nu h ( 2R ) =2 kf

†

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

Nu = 2 + 1.1 Re 0.6 Pr1/ 3 Sh = 2 + 1.1 Re 0.6 Sc1/ 3

Pr =

C pµ k

; Sc =

µ ρD12

Diffusion with Heterogeneous Reaction (1) †

Physical system: „ „ „ „ „

Chemical reaction occurs on a catalytic surface Catalytic surface is surrounded by a thin stagnant film Diffusion occurs in the stagnant film Isothermal conditions Reaction is n A → A n

A An

Diffusion with Heterogeneous Reaction (2) †

A

An

Physical system: „ „ „

†

n A → An

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

The chemical reaction demands that

N 1 = − nN 2 „

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

„

This equation basically provides the additional equation to the Fick’s law equation

Diffusion with Heterogeneous Reaction (3) †

n A → An

A

An

The shell balance equation: „

Setting a shell at the distance z in the stagnant film

dN 1 − =0 dz †

The Fick’s law N 1 = − cD 12 „

dx 1 + x1 (N 1 + N 2 ) dz

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

N 1 = − nN 2

N2 = −

1 N1 n

Diffusion with Heterogeneous Reaction (4) †

†

The mass balance equation: dN 1 − =0 dz The Fick’s law dx 1 N 1 = − cD 12 + x1 (N 1 + N 2 ) dz „

n A → An

A

An

1 N 2 = − N1 n

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

cD 12 dx 1 N1 = − ⎡ ( n − 1) ⎤ dz x1 ⎥ ⎢1 − n ⎣ ⎦

Diffusion with Heterogeneous Reaction (5) †

†

The mass balance equation: dN 1 − =0 dz The Fick’s law cD 12 dx 1 N1 = − ⎡ ( n − 1) ⎤ dz x1 ⎥ ⎢1 − n ⎣ ⎦

†

The mass balance equation, then, gives: cD 12 dx 1 N1 = − ≡ −K1 ⎡ (n − 1) ⎤ dz ⎢⎣1 − n x 1 ⎥⎦

n A → An

A

An

Diffusion with Heterogeneous Reaction (6) †

The mass balance equation:

†

cD 12 dx 1 N1 = − ≡ −K1 ⎡ (n − 1) ⎤ dz ⎢⎣1 − n x 1 ⎥⎦ Now we pose the boundary conditions

z = 0;

x1 = x10

z = L;

x1 = 0

n A→An

A An

Diffusion with Heterogeneous n A→A Reaction (7) n

†

†

The mass balance equation: cD 12 dx 1 N1 = − ≡ −K1 ⎡ (n − 1) ⎤ dz ⎢⎣1 − n x 1 ⎥⎦

A An

Integrating the mass balance equation one more time cD 12 „

⎡ ⎤ ( n − 1) 1 ln ⎢ K1z + K 2 ⎥= n ⎣ 1 − ( n − 1) x 1 / n ⎦

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

Diffusion with Heterogeneous n A→A Reaction (8) n

†

The solution:

⎡ ⎤ ( n − 1) 1 cD 12 ln ⎢ K1z + K 2 ⎥= n ⎣ 1 − ( n − 1) x 1 / n ⎦ †

A An

Applying the two boundary conditions

z = 0;

x1 = x10

z = L;

x1 = 0

⎤ ( n − 1) ⎡ 1 cD 12 ln ⎢ K 1 (0) + K 2 ⎥= n ⎣⎢ 1 − (n − 1)x 1, 0 / n ⎥⎦

⎡ ⎤ ( n − 1) 1 cD 12 ln ⎢ = K 1 (L) + K 2 ⎥ n ⎣ 1 − (n − 1)( 0 ) / n ⎦

Diffusion with Heterogeneous n A→A Reaction (9) n

†

The solution:

⎡ ⎤ ( n − 1) 1 cD 12 ln ⎢ K1z + K 2 ⎥= n ⎣ 1 − ( n − 1) x 1 / n ⎦ †

A An

Solving for K1 and K2, we get ⎤ ⎡ 1 K 2 = cD 12 ln ⎢ ⎥ ( ) 1 − n − 1 x / n ⎥⎦ 1, 0 ⎣⎢ ⎤ 1 ⎛ n ⎞ cD 12 ⎡ K 1 = −⎜ ln ⎢ ⎟ ⎥ ( ) n − 1 L 1 − n − 1 x / n ⎝ ⎠ ⎢⎣ ⎥⎦ 1, 0

Remember that N1 = - K1; So the negative of the constant K1 is simply the flux

Diffusion with Heterogeneous n A→A Reaction (10) n

†

The solution for the concentration distribution:

n − 1) ⎤ ⎡ n − 1) ⎡ ⎤ ( ( x 1 ⎥ = ⎢1 − x 10 ⎥ ⎢1 − n n ⎣ ⎦ ⎣ ⎦ †

( 1− z / L )

A An

The molar flux across the stagnant film is the negative of the constant K1: ⎤ 1 ⎛ n ⎞ cD12 ⎡ ln ⎢ N1 = ⎜ ⎟ ⎥ ( ) n − 1 L 1 − n − 1 x / n ⎝ ⎠ ⎢⎣ ⎥⎦ 1, 0 „

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

SO THIS IS THE REACTION RATE PER UNIT AREA OF THE CATALYST

Diffusion with Heterogeneous Reaction (11) †

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 chemical reaction. 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

Diffusion with Heterogeneous Reaction (12) †

The mass balance equation:

†

cD 12 dx 1 N1 = − ≡ −K1 ⎡ (n − 1) ⎤ dz ⎢⎣1 − n x 1 ⎥⎦ Now we pose the boundary conditions

z = 0;

z = L;

x1 = x1,0

N1 z =L = kcx1 z =L

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

n A→An

A An

Diffusion with Heterogeneous n A→A Reaction (13) n

†

†

The mass balance equation: cD 12 dx 1 N1 = − ≡ −K1 ⎡ (n − 1) ⎤ dz ⎢⎣1 − n x 1 ⎥⎦

A An

Integrating the mass balance equation one more time cD 12 „

⎡ ⎤ ( n − 1) 1 ln ⎢ K1z + K 2 ⎥= n ⎣ 1 − ( n − 1) x 1 / n ⎦

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

Diffusion with Heterogeneous n A→A Reaction (14) n

†

The solution:

⎡ ⎤ ( n − 1) 1 cD 12 ln ⎢ K1z + K 2 ⎥= n ⎣ 1 − ( n − 1) x 1 / n ⎦ †

An

Applying the two boundary conditions

z = 0;

z = L;

A

x1 = x10

N1 z =L = kcx1 z =L

⎡ ⎤ ( n − 1) 1 cD 12 ln ⎢ K 1 (0) + K 2 ⎥= n ⎣⎢ 1 − (n − 1)x 1, 0 / n ⎥⎦

− K 1 = kcx 1, L

Diffusion with Heterogeneous n A→A Reaction (15) n

†

The solution:

⎡ ⎤ ( n − 1) 1 cD 12 ln ⎢ K1z + K 2 ⎥= n ⎣ 1 − ( n − 1) x 1 / n ⎦ †

Solving for K1 and K2, we get ⎤ ⎡ 1 K 2 = cD 12 ln ⎢ ⎥ ( ) 1 − n − 1 x / n ⎥⎦ 1, 0 ⎣⎢

K 1 = − kcx 1, L

A An

Diffusion with Heterogeneous n A→A Reaction (16) n

†

The solution for the concentration distribution:

⎡ 1 − ( n − 1) x 1, 0 / n ⎤ ( n − 1) kx 1, L = − ⋅ z ln ⎢ ⎥ n D 12 ⎣ 1 − ( n − 1) x 1 / n ⎦ †

A

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

⎡ 1 − ( n − 1) x 1, L / n ⎤ ( n − 1) ⎛ kL ⎞ ⋅⎜ ln ⎢ ⎟ x 1, L ⎥= n ⎝ D 12 ⎠ ⎣ 1 − ( n − 1) x 1, 0 / n ⎦ „

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

An

Diffusion with Heterogeneous Reaction (17) †

How to get x1,L?

⎡1 − ( n − 1) x 1,L / n ⎤ ( n − 1) ⎛ kL ⎞ ⋅⎜ ln ⎢ ⎟ x 1,L ⎥= n ⎝ D 12 ⎠ ⎣ 1 − ( n − 1) x 1,0 / n ⎦ †

Knowing this concentration at the catalytic surface, the reaction rate is N1 „

z=L

= kcx 1, L

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

Diffusion with Heterogeneous Reaction (18) †

Equation for x1,L

⎡1 − ( n − 1) x 1,L / n ⎤ ( n − 1) ⎛ kL ⎞ ⋅⎜ ln ⎢ ⎟ x 1,L ⎥= n ⎝ D 12 ⎠ ⎣ 1 − ( n − 1) x 1,0 / n ⎦ †

This equation involves a non-dimensional group ⎛ kL ⎞ chemical reaction rate ⎜ ⎟≡ diffusion rate ⎝ D 12 ⎠ „

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

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

Diffusion & Homogeneous Reaction (3) †

The system: Absorption and reaction „ „

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

⎛ moles of A reacted ⎞ ⎜ ⎟ = kC 1 ⎝ volume − time ⎠ „

†

Isothermal system

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

Diffusion & Homogeneous Reaction (4) †

The system: Absorption and reaction „ „

„

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) Gas

Mole fraction

Sulfur dioxide Chlorine Hydrogen sulfide Carbon dioxide Ethylene Carbon monoxide

0.03 0.0017 0.002 0.00062 0.0001 0.000018

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

Diffusion & Homogeneous Reaction (5) †

The system: Absorption and reaction „

Shell mass balance

⎛ Rate of ⎞ ⎛ Rate of ⎞ ⎛ Rate of mass⎞ ⎜ ⎟ −⎜ ⎟ +⎜ ⎟ = ( Accummulation) ⎝ mass in ⎠ ⎝ mass out⎠ ⎝ production ⎠

(SN1 ) z − (SN1 ) z+ ∆z − (S∆z)kC1 = 0 S ∆z dN 1 − − kC1 = 0 dz

Diffusion & Homogeneous Reaction (6) †

†

The Fick’s law N 1 = − cD 12

dx 1 + x1(N 1 + N 2 ) dz

N 1 = − D 12

dC 1 dz

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

„

d 2C1 D 12 − kC 1 = 0 2 dz This is one of the classic equation in diffusion and reaction theory!

dN 1 − − kC1 = 0 dz

Diffusion & Homogeneous Reaction (7) †

The mass balance equation D 12

†

d 2C1 − kC 1 = 0 2 dz

Physical constraints (Boundary conditions) „

„

At the gas-liquid interface, the concentration is always equal to the solubility

z = 0; C1 = C1, 0 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.

z = L;

dC1 / dz = 0

dN 1 − − kC1 = 0 dz

Diffusion & Homogeneous Reaction (8) †

So we have the mass balance equation & the two necessary boundary conditions d 2C1 D 12 − kC 1 = 0 2 dz z = 0; C1 = C1, 0 z = L; dC1 / dz = 0

†

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

C1 = e λz †

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

dN 1 − − kC1 = 0 dz

Diffusion & Homogeneous Reaction (9) †

The mass balance equation

†

d 2C1 D 12 − kC 1 = 0 2 dz If the assumed form C = e λz 1

„

is the solution to the mass balance equation, it must satisfy 2 λz

( )

d e λz D12 − ke =0 dz 2

dN 1 − − kC1 = 0 dz

D12 λ2 e λz − ke λz = 0

(D λ − k ) e = 0 (D λ − k ) = 0 λz

2

12

2

12

k λ = D12 2

λ=±

k D12

Diffusion & Homogeneous Reaction (10) †

†

Solution (continued) d 2C1 D 12 − kC 1 = 0 2 dz The assumed form C1 = e λz „

where

λ=± †

k D12

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

C1 = e λ1 z „

dN 1 − − kC1 = 0 dz

C1 = e λ 2 z

That is

C1 = A e

λ1 z

+Be

λ2 z

So our solution has two constants of integration, A and B

Diffusion & Homogeneous Reaction (11) †

The equation

D 12 †

d 2C1 − kC 1 = 0 2 dz

The solution

C1 = A e †

λ1 z

+Be

λ2 z

λ=±

k D12

Apply the two boundary conditions

z = 0; z = L;

C1 = C1, 0 dC1 / dz = 0

C1, 0 = A e λ1 ( 0 ) + B e λ 2 ( 0 ) = A + B

dC1 = A λ 1 e λ1 ( L ) + B λ 2 e λ 2 ( L ) = 0 dz

Diffusion & Homogeneous Reaction (12) †

The equation

D 12 †

d 2C1 − kC 1 = 0 2 dz

The solution

C1 = A e λ1 z + B e λ 2 z †

λ=±

Solution for A and B

A = −C1,0

B = C1, 0

λ 2eλ 2L λ1e λ1L − λ 2 e λ 2 L

λ1e λ1L λ1e λ1L − λ 2 e λ 2 L

k D12

A = C1, 0

e − λL e λL + e − λ L

B = C1,0

e λL e λL + e − λL

Diffusion & Homogeneous Reaction (13) †

So after a long and tedious (but fun) calculus and algebra, the solution for the concentration distribution is − λL λL

e C1 e = λL − λ L e λ z + λ L − λ L e − λ z e +e C1, 0 e + e

†

Simplify it

[e C1 = C1, 0

C1 e −λ ( L−z ) eλ ( L−z ) = λL − λL + λ L − λL C1, 0 e + e e +e

C1 e − λ ( L − z ) + e λ ( L − z ) = C1, 0 e λL + e − λL

]

−λ ( L−z )

+ eλ ( L−z ) cosh[λ(L − z )] cosh[λL(1 − z / L)] 2 = = λL − λL e +e cosh (λL ) cosh(λL) 2

(

)

Diffusion & Homogeneous Reaction (14) †

So the final solution is C1 cosh[λL(1 − z / L)] = C1, 0 cosh(λL)

†

Let us have a good look at the group λL k k L2 λL = L = = D12 D12

†

L2 Diffusion Time def D12 = ≡ 1 Reaction Time k

So we can write solution as

[

C1 cosh Da (1 − z / L) = C1, 0 cosh Da

(

)

]

Da

To credit German scientist Damkohler

Diffusion & Homogeneous Reaction (15) †

So the final solution is „

†

With

[

C1 cosh Da (1 − z / L) = C1, 0 cosh Da

(

)

]

k L2 Diffusion Time = Da = D12 Reaction Time

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 uniform The 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 interface The absorption rate is controlled by diffusion and reaction

Diffusion & Homogeneous Reaction (16) †

So the final solution is „

†

with

[

C1 cosh Da (1 − z / L) = C1, 0 cosh Da

(

)

]

k L2 Diffusion Time = Da = D12 Reaction Time

The absorption rate per unit interfacial area is just simply the molar flux at the gas liquid interface dC 1 N 1 z = 0 = − D 12 dz z = 0 C D N 1 z = 0 = 10 12 Da ⋅ tanh Da L ⎛ kL2 ⎞ C 10 D 12 kL2 ⎟ N 1 z= 0 = ⋅ tanh ⎜⎜ ⎟ L D 12 ⎝ D 12 ⎠

(

)

Diffusion & Homogeneous Reaction (17) †

Summary

[

C1 cosh Da (1 − z / L) = C1, 0 cosh Da

N1 z=0 =

(

C 10 D 12 L

)

]

Da ⋅ tanh

k L2 Diffusion Time = Da = D12 Reaction Time

(

Da

)

Diffusion & Homogeneous Reaction (18) †

Special case: Da << 1 C1 ≈1 C1, 0

N 1 z = 0 ≈ k L C 1, 0 †

The rate of absorption is controlled purely by chemical reaction

Diffusion & Homogeneous Reaction (19) †

Special case: Da >> 1

C1 −z ≈e C1, 0 lim N 1

Da >>1

†

z= 0

C 10 D 12 = L

k / D12

Da = C 10 k ⋅ D 12

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

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

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 film „ „ „

Flow of liquid film is laminar No end effects Solubility is constant

Diffusion into a Falling Film (2) †

Gas absorption in a falling film „

Shell balance equation

⎛ Rate of ⎞ ⎛ Rate of ⎞ ⎛ Rate of mass⎞ ⎟ = ( Accummulation) ⎟ +⎜ ⎟ −⎜ ⎜ ⎝ mass in ⎠ ⎝ mass out⎠ ⎝ production ⎠

⎡(W∆zN x ,1 ) ⎤ x ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ( ) + ∆ W xN z ,1 z ⎥ ⎢⎣ ⎦

⎤ ⎡(W∆zN x ,1 ) x + ∆x ⎥ ⎢ −⎢ ⎥ ⎥ ⎢ ( ) W xN + ∆ z ,1 z + ∆z ⎥ ⎢⎣ ⎦

+

0 = 0

Diffusion into a Falling Film (3) †

The mass balance equation ⎤ ⎡(W∆zN x ,1 ) ⎤ ⎡(W∆zN x ,1 ) x + ∆x x ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ −⎢ ⎥ ⎢ ⎥ ⎢ ( ) + ∆ W xN ( ) + ∆ W xN z ,1 z + ∆z ⎥ z ,1 z ⎥ ⎢⎣ ⎦ ⎦ ⎢⎣

+

0 = 0

W ∆x ∆z

∂N x ,1 ∂x †

+

∂N z ,1 ∂z

=0

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

Diffusion into a Falling Film (4) †

†

The mass balance equation

∂x

+

∂N z ,1 ∂z

=0

The molar flux in the x-direction is given by the Fick’s law ∂ C1 N x ,1 = − D 12 + x 1 ( N x ,1 + N x , 2 ) ∂x „

†

∂N x ,1

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 N z ,1 = − D 12

∂C 1 + x 1 ( N z ,1 + N z ,2 ) ∂z

Diffusion into a Falling Film (5) †

The molar flux in the z-direction is also by the Fick’s law ∂C 1 N z ,1 = − D 12 + x 1 ( N z ,1 + N z ,2 ) ∂z

†

By definition of flux

†

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

N z ,2 = C 2 v z ,2

The Fick’s law can now be written as ∂C 1 N z ,1 = − D 12 + x 1 ( C 1 v z ,1 + C 2 v z ,2 ) ∂z N z ,1 = − D 12

⎛ C 1 v z ,1 + C 2 v z ,2 ⎞ ∂C 1 + x1 ⎜ ⎟ (C 1 + C 2 ) ∂z C1 + C 2 ⎝ ⎠

N z ,1 = − D 12

∂C 1 + v z ( x)C 1 ∂z

But ⎛ C 1 v z ,1 + C 2 v z ,2 ⎞ ⎜ ⎟ = v z (x) C1 + C 2 ⎝ ⎠

Diffusion into a Falling Film (6) †

The mass balance equation ∂N x ,1 ∂x „

+

∂N z ,1 ∂z

where the fluxes are N x ,1 = −D12 N z ,1 = − D12

†

=0

∂C1 ∂x

∂C1 + v z ( x ) C1 ≈ v z ( x ) C1 ∂z

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

D 12

∂ 2 C1 ∂C 1 = ( ) v x z ∂x 2 ∂z

Diffusion into a Falling Film (7) †

The mass balance equation

D 12

∂ 2 C1 ∂C 1 = v z (x) ∂x 2 ∂z

†

The heat balance equation α

1 ∂ ⎛ ∂T ⎞ ∂T r = ( ) v x ⎜ ⎟ z r ∂r ⎝ ∂r ⎠ ∂z

∂ 2T ∂T α 2 = vz (x) ∂x ∂z

Diffusion into a Falling Film (8) †

†

The mass balance equation ∂ 2 C1 ∂C 1 = v z (x) D 12 2 ∂x ∂z The boundary conditions: z = 0;

C1 = 0

x = 0;

C1 = C1, 0

x = δ;

†

N1, x = −D12

∂C1 =0 ∂x

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

Diffusion into a Falling Film (8) †

Short contact time problem

D 12 „

∂ 2 C1 ∂C 1 = ( ) v x z ∂x 2 ∂z

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

D 12

∂ 2C1 ∂C 1 v ≈ max ∂z ∂x 2

z = 0;

C1 = 0

x = 0;

C1 = C1, 0

x = ∞;

C1 = 0

Diffusion into a Falling Film (9) †

Short contact time problem for gas absorption in falling film D 12

†

Transient heat conduction in a semi-infinite object

∂ 2 T ∂T α 2 = ∂x ∂t

∂ 2C1 ∂C 1 v ≈ max ∂z ∂x 2

z = 0;

C1 = 0

t = 0;

T = T0

x = 0;

C1 = C1, 0

x = 0;

T = Ts

C1 = 0

x = ∞;

x = ∞;

T = T0

Ts x

Diffusion into a Falling Film (10) †

†

Short contact time problem for gas absorption in falling film

∂ 2 T ∂T α 2 = ∂x ∂t

∂ 2C1 ∂C 1 v D 12 ≈ max ∂z ∂x 2 z = 0; C1 = 0 x = 0; x = ∞;

C1 ( x , t ) − C1, 0 0 − C1, 0

Transient heat conduction in a semi-infinite object

t = 0;

T = T0

C1 = C1, 0

x = 0;

T = Ts

C1 = 0

x = ∞;

⎛ x = erf ⎜ ⎜ 4D z / v 12 max ⎝

⎡ ⎛ ⎞⎤ x ⎟⎟ ⎥ C 1 = C 10 ⎢1 − erf ⎜⎜ ⎝ 4 D 12 z / v max ⎠ ⎥⎦ ⎢⎣

⎞ ⎟ ⎟ ⎠

T = T0

T( x, t ) − Ts ⎛ x ⎞ = erf ⎜ ⎟ ⎝ 4αt ⎠ T0 − Ts

Diffusion into a Falling Film (11) †

Short contact time problem for gas absorption in falling film ⎡ ⎛ ⎞⎤ x ⎟⎟ ⎥ C 1 = C 10 ⎢1 − erf ⎜⎜ ⎝ 4 D 12 z / v max ⎠ ⎥⎦ ⎢⎣

†

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

N x ,1

x =0

= − D12

∂C1 ∂x

x =0

∂ 2C1 ∂C 1 v D 12 ≈ max ∂z ∂x 2 z = 0; C1 = 0 x = 0; x = ∞;

C1 = C1, 0 C1 = 0

Diffusion into a Falling Film (12) †

Find the derivative ⎡ ⎛ x C1 = C10 ⎢1 − erf ⎜ ⎜ 4D z / v ⎢⎣ 12 max ⎝

1 ∂C1 ⎛ 2 − η2 ⎞ ⎡ e ⎟⎢ = −C10 ⎜ ∂x ⎠ ⎢⎣ 4D12 z / v max ⎝ π

⎞⎤ ⎟⎥ ⎟⎥ ⎠⎦

2 C10 ∂C1 e −η =− ∂x πD12 z / v max

⎛ ∂C1 ∂ ⎡ x = C10 ⎢1 − erf ⎜ ⎜ 4D z / v ∂x ∂x ⎢⎣ 12 max ⎝

⎞⎤ ⎟⎥ ⎟⎥ ⎠⎦

∂C1 ∂ ⎡ ⎛⎜ x = −C10 ⎢erf ∂x ∂x ⎢⎣ ⎜⎝ 4D12 z / v max

⎞⎤ ⎟⎥ ⎟⎥ ⎠⎦

∂C1 ∂ = −C10 [erf (η)] ∂x ∂x

∂C1 ∂ [erf (η)] ∂η = −C10 ∂x ∂η ∂x

⎤ ⎥ ⎥⎦

η=

⎡ ⎤ C10 x2 ∂C1 =− exp ⎢− ⎥ ∂x πD12 z / v max ⎣ 4D12 z / v max ⎦ x 4D12 z / v max

Diffusion into a Falling Film (13) †

The partial derivative of C1 with respect to x ⎤ ⎡ C10 x2 ∂C1 exp ⎢− =− ⎥ 4 D z / v ∂x πD12 z / v max 12 max ⎦ ⎣

†

Therefore the molar flux at the gas-liquid interface is N1 x = 0

∂C = − D12 1 ∂x

x =0

⎡ ⎤ C10 D12 x2 exp ⎢− = ⎥ 4 D z / v πD12 z / v max 12 max ⎦ x = 0 ⎣

N1 x = 0 = „

C10 D12 πD12 z / v max

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

Diffusion into a Falling Film (14) †

Therefore the molar flux at the gas-liquid interface is C10 D12 N1 x = 0 = πD12 z / v max

†

The mass transfer rate that occurs over the whole interfacial area is L L D 12 v max M 1 = W ∫ N x ,1 x= 0 dz = W ∫ C10 dz πz 0 0

M 1 = (WL) C10 †

4 D 12 v max πL

Hence the average molar flux into the falling film is N ave =

M1 4D12 v max = C10 WL πL

END

Gas Absorption from a Rising Bubble (1) †

The physical system: „ „ „

†

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

Given a bubble of initial size R0 injected into the column, what is the distance that the bubble can travel before it completely dissolves?

Gas Absorption from a Rising Bubble (2) †

The physical system: „ „ „

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

D

Species 1

Species 2

Gas Absorption from a Rising Bubble (3) †

The physical system: „

†

The physical system: „

Rising bubble

Falling film

vt L D

⎛ Average ⎞ 4D12 v t ⎜⎜ ⎟⎟ = C10 πD ⎝ molarflux ⎠

⎛ Average ⎞ 4D12 v max ⎟⎟ = C10 ⎜⎜ πL ⎝ molarflux ⎠

Gas Absorption from a Rising Bubble (4) †

The physical system: „

†

Rising bubble vt

„

D †

⎛ Average ⎞ 4D12 v t ⎜⎜ ⎟⎟ = C10 πD ⎝ molarflux ⎠

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 4D12 v t d ⎛ πD3 p 0 ⎞ 2 = − π D C ⎜ ⎟ 10 dt ⎝ 6 RT⎠ πD

(

)

Gas Absorption from a Rising Bubble (5) †

The mass balance equation around the bubble: vt

4D12 v t d ⎛ πD3 p 0 ⎞ 2 ⎜ ⎟ = − πD C10 dt ⎝ 6 RT⎠ πD

(

)

†

2 g ∆ρ D The terminal velocity is v = t 18µ

†

Simplification:

D

3πD 2 p 0 d 4D12 g∆ρD 2 2 (D ) = − πD C10 πD 18µ 6RT dt

(

)

p0 d (D ) = −C10 2D12 g∆ρD 2RT dt π 9µ dD C RT = −2 D 10 dt p0

2D12 g∆ρ π 9µ

dD = −2γ D dt

Gas Absorption from a Rising Bubble (6) †

SO the final mass balance of the bubble is

dD = −2 γ D dt † †

t = 0;

The initial condition is The solution:

dD = − γdt 2 D γ= „

D = D0

D = D0 − γt C10 RT p0

2D12 g∆ρ π 9µ

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

1 t = D0 γ *

Gas Absorption from a Rising Bubble (7) †

Knowing the bubble size as a function of time

D = D0 − γt „

γ=

2D12 g∆ρ π 9µ

we can obtain the terminal velocity as a function of time

g∆ρD 2 g∆ρ vt = = 18µ 18µ †

C10 RT p0

(

D 0 − γt

)

4

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

dx g∆ρ = vt = 18µ dt

(

D 0 − γt

)

4

Gas Absorption from a Rising Bubble (8) †

Thus the distance traveled by the bubble is 4 dx g∆ρ = vt = D 0 − γt dt 18µ

(

g∆ρ x= 18µ ∫0 t

1 g∆ρ x=− γ 18µ ∫0 t

1 g∆ρ x=− 5γ 18µ

(

(

(

)

)

4

D 0 − γt dt

) ( 4

D 0 − γt d D 0 − γt

D 0 − γt

)

5 t 0

(

1 g∆ρ ⎡ x= D0 ⎢ 5γ 18µ ⎣

) −( 5

)

D 0 − γt ⎤ ⎥⎦ 5

)

Gas Absorption from a Rising Bubble (9) †

At the time when the bubble completely dissolve

t* = †

1 D0 γ

the distance that the bubble has traveled is

H=

1 g∆ρ 5 / 2 D0 5γ 18µ

⎛ π ⎞ ⎛ g∆ρD5o ⎟⋅⎜ H = ⎜⎜ ⎟ ⎜ ⎝ 30 2 ⎠ ⎝ µD12

⎞ ⎛ p / RT ⎞ ⎟⋅⎜ 0 ⎟ ⎟ ⎜⎝ C10 ⎟⎠ ⎠ γ=

C10 RT p0

2D12 g∆ρ π 9µ

Diffusion and Reaction in a Porous Catalyst (1) †

Porous Catalysts: very complex because „ „ „

†

Nonporous catalyst: Simple „ „

†

Pores of different size and shape Pores are tortuous Resistance to mass transfer

Nice simple geometry No problem with mass transfer

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

Diffusion and Reaction in a Porous Catalyst (2) †

Porous Catalysts vs non-porous catalysts Porous Catalysts

Non-porous Catalysts

High surface area

Very low surface area per unit mass Low pressure drop High pressure drop Easy to handle †

Difficult to handle

This is why most catalysts used in industries are porous!

Diffusion and Reaction in a Porous Catalyst (3) †

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.

Diffusion and Reaction in a Porous Catalyst (4) †

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 size Effective diffusivity Particle Porosity Particle tortuosity factor Specific surface area Shape of the particle

Diffusion and Reaction in a Porous Catalyst (5) †

Pore size: „

†

Effective diffusivity „

† †

†

So much has been done in 60 and 70.

Particle Porosity Particle tortuosity factor „

†

Pore size distribution: Choose the appropriate average pore size

Pore is not straight

Specific surface area; quite high 100-1000 m2/g Shape of the particle: „

Make it simple

Diffusion and Reaction in a Porous Catalyst (6) †

Two ingredients required to model diffusion and reaction in a catalyst „ „

†

Diffusion flux Reaction 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

J = −D eff „ „

∂C ∂r

moles transported ⎡ ⎤ ⎢⎣ total cross sectional area - time ⎥⎦

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

Diffusion and Reaction in a Porous Catalyst (7) †

Equation for diffusion flux

J = −D eff „

†

Equation for heat flux

∂C ∂r

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

D eff = f (D12 , D K , ε, τ )

q = −k eff „

∂T ∂r

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

k eff = f (k f , k s , ε )

Diffusion and Reaction in a Porous Catalyst (8) †

Equation for diffusion flux

J = −D eff „

∂C ∂r

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

D eff = f (D12 , D K , ε, τ ) „

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

ε D eff = D c τ

Diffusion and Reaction in a Porous Catalyst (9) †

The effective diffusivity

„ „ „

ε D eff = D c τ

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 transfer The 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

Diffusion and Reaction in a Porous Catalyst (9) ε = Dc τ

†

The effective diffusivity

†

The combined diffusivity is given by: 1 1 1 = + D eff ,1 D12 D K ,1 „ „

†

†

D eff

Where D12 is the usual binary diffusivity And 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 equation The Knudsen diffusivity is calculated from DK =

2r 8R g T 3 πM

r = Pore radius (not particle radius) M = Molecular weight

Diffusion and Reaction in a Porous Catalyst (10) ε = Dc τ

†

The effective diffusivity

†

The combined diffusivity is given by: 1 1 1 = + D eff ,1 D12 D K ,1

†

Order of magnitude ε τ

D eff

0.2 – 0.7 2-6

D12 0.1 - 2

cm2/s

DK 0.1 - 100

cm2/s

Diffusion and Reaction in a Porous Catalyst (11) ε = Dc τ

†

The effective diffusivity

†

The combined diffusivity is given by: 1 1 1 = + D eff ,1 D12 D K ,1

†

The pore size, pressure and temperature dependence Diffusivity Binary diffusivity

D eff

P T r M P-1 T1.75 r0 1 M1

Knudsen diffusivity P0 T0.5

r1

+

1 M1

1 M2

Diffusion and Reaction in a Porous Catalyst (12) ε = Dc τ

†

The effective diffusivity

†

The combined diffusivity is given by: 1 1 1 = + D eff ,1 D12 D K ,1

†

Controlling mechanism „ „

D eff

Molecular diffusion dominates in Diffusivity P T r M large pores and high pressure P-1 T1.75 r0 1 1 Knudsen diffusion dominates in small Binary + diffusivity M1 M 2 pores and very low pressure

Knudsen P0 T0.5 diffusivity

r1

1 M1

Diffusion and Reaction in a Porous Catalyst (13) †

Reaction rate „

Since 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

ℜ( c )

„

⎡ moles reacted ⎤ ⎢⎣ surface area - time ⎥⎦

Let us deal with first-order reaction: ℜ(C) = kC

†

⎡ moles reacted ⎤ ⎢⎣ surface area - time ⎥⎦

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

Diffusion and Reaction in a Porous Catalyst (14) †

Summary „

Diffusion rate J = − D eff

„

∂C ∂r

1 D eff ,1

=

1 1 + D12 D K ,1

Reaction rate ℜ(C) = kC

†

moles transported ⎡ ⎤ ⎢ total cross section area - time ⎥ ⎣ ⎦

ε D eff = D c τ

⎡ moles reacted ⎤ ⎢⎣ surface area - time ⎥⎦

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

Diffusion and Reaction in a Porous Catalyst (15) †

Shell balance ∆r

⎛ Rate of ⎞ ⎛ Rate of ⎞ ⎛ Rate of mass⎞ ⎜ ⎟ −⎜ ⎟ +⎜ ⎟ = ( Accummulation) ⎝ mass in ⎠ ⎝ mass out⎠ ⎝ production ⎠

[(

)

]

4πr 2 N r = r − 4πr 2 N r = r + ∆r − 4πr 2 ∆r ρ pSg kC = 0

4π ∆r −

r 2 N r = r + ∆r − r 2 N r = r ∆r

−

− r 2ρ pSg kC = 0

∂ 2 r N − r 2ρ pSg kC = 0 ∂r

( )

r

Diffusion and Reaction in a Porous Catalyst (16) †

†

†

The mass balance equation ∂ 2 − r N − r 2ρ pSg kC = 0 ∂r But the diffusion flux is ∂C N = − D eff ∂r The final mass balance equation in terms of concentration is then

( )

D eff ∂ ⎛ 2 ∂C ⎞ ⎜r ⎟ − ρ pSg kC = 0 2 r ∂r ⎝ ∂r ⎠ †

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

∆r

r

Diffusion and Reaction in a Porous Catalyst (17) †

†

So, the mass balance equation D eff ∂ ⎛ 2 ∂C ⎞ ⎜r ⎟ − ρ pSg kC = 0 2 r ∂r ⎝ ∂r ⎠ The boundary conditions are:

∆r

r

∂C =0 ∂r C = C 0 (known bulk concentration )

r = 0; r = R; „

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

r = R;

− D eff

∂C = k m C r =R − C0 ∂r r = R

(

)

Diffusion and Reaction in a Porous Catalyst (18) †

So, the mass balance equation D eff ∂ ⎛ 2 ∂C ⎞ ⎜r ⎟ − ρ pSg kC = 0 2 r ∂r ⎝ ∂r ⎠

∆r

r

∂C =0 ∂r C = C 0 (known bulk concentration )

r = 0; r = R; †

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

r x= ; R

C y= C0

Diffusion and Reaction in a Porous Catalyst (19) †

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

∆r

r

D eff ∂ ⎛ 2 ∂C ⎞ ⎜r ⎟ − ρ pSg kC = 0 2 r ∂r ⎝ ∂r ⎠ r C y= x= ; R C0 D eff ∂ ⎡ 2 ∂ (C 0 y ) ⎤ ( ) Rx − ρ pSg k (C 0 y ) = 0 ⎢ ⎥ 2 ∂ (Rx ) ⎦ (Rx ) ∂Rx ⎣

D eff 1 ∂ ⎡ 2 ∂y ⎤ x − ρ pSg ky = 0 2 2 ⎢ ⎥ R x ∂x ⎣ ∂x ⎦

2 1 ∂ ⎡ 2 ∂y ⎤ ⎛⎜ ρ pSg k R ⎞⎟ x y=0 − 2 ⎢ ⎥ ⎜ ⎟ x ∂x ⎣ ∂x ⎦ ⎝ D eff ⎠

Diffusion and Reaction in a Porous Catalyst (20) †

The mass balance equation is ….. 2 1 ∂ ⎡ 2 ∂y ⎤ ⎛⎜ ρ pSg k R ⎞⎟ x y=0 − 2 ⎢ ⎥ ⎜ ⎟ x ∂x ⎣ ∂x ⎦ ⎝ D eff ⎠

†

∆r

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

φ2 = „

ρ p Sg k R 2 D eff

This non-dimensional group is known as † † †

Thiele modulus in Western literature Damkohler number in German literature Zel’dowitch number in the Russian literature

r

Diffusion and Reaction in a Porous Catalyst (21) †

†

†

†

∆r

So the mass balance equation is ….. 1 ∂ ⎡ 2 ∂y ⎤ 2 r x − φ y = 0 2 x ∂x ⎢⎣ ∂x ⎥⎦ The physical significance of the Thiele modulus ⎛ R2 ⎞ ⎜⎜ ⎟⎟ 2 ρ p Sg k R D Diffusion time φ2 = = ⎝ eff ⎠ ≡ D eff ⎛ 1 ⎞ Reaction time ⎜ ⎟ ⎜ρ S k⎟ ⎝ p g ⎠ If φ << 1 (diffusion time is small compared to reaction time), we would expect uniform concentration profile and the reaction is kinetically-controlled If φ >> 1 (diffusion time is greater than reaction time), we would expect a very sharp concentration profile and the reaction is called difuusion-controlled

Diffusion and Reaction in a Porous Catalyst (22) †

†

So the mass balance equation is ….. 1 ∂ ⎡ 2 ∂y ⎤ 2 x −φ y = 0 2 ⎢ ⎥ x ∂x ⎣ ∂x ⎦ Solution? „

„

r

Introduce a new variable u y= x Substitute this into the above mass balance equation 1 ∂ ⎡ 2 ∂ (u / x ) ⎤ 2 u x −φ =0 x 2 ∂x ⎢⎣ ∂x ⎥⎦ x

1 ∂ ⎡ 2 ⎛ 1 ∂u u ⎞⎤ 2 u x ⎜ =0 − ⎟ −φ x 2 ∂x ⎢⎣ ⎝ x ∂x x 2 ⎠⎥⎦ x

∆r

1 ∂ ⎛ ∂u ⎞ 2u x − u =0 ⎜ ⎟−φ 2 x ∂x ⎝ ∂x x ⎠

Diffusion and Reaction in a Porous Catalyst (23) †

Solution (continued) 1 ∂ ⎛ ∂u ⎞ 2u − u =0 x ⎜ ⎟−φ x 2 ∂x ⎝ ∂x x ⎠

r

1 ⎛ ∂u ∂ 2 u ∂u ⎞ 2 u ⎜ + x 2 − ⎟⎟ − φ =0 x 2 ⎜⎝ ∂x ∂x ∂x ⎠ x

1 ∂ 2u 2 u − φ =0 2 x ∂x x ∂ 2u 2 − φ u=0 2 ∂x †

∆r

Solution of this equation is

u = Aeφx + Be − φx

Diffusion and Reaction in a Porous Catalyst (24) †

Solution (continued) „

†

u = Aeφx + Be − φx

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

r = 0;

∂C =0 ∂r

x = 0;

∂y =0 ∂x

x = 0;

u=0

r = R;

C = C0

x = 1;

y =1

x = 1;

u =1

Application of the two boundary conditions

1 e φ − e −φ

x = 0;

u=0

0=A+B

A=

x = 1;

u =1

1 = Aeφ + Be − φ

B=−

1 e φ − e −φ

Diffusion and Reaction in a Porous Catalyst (25) †

∆r

So the solution for u is

e φx − e − φx sinh (φx ) u = φ −φ = e −e sinh (φ)

u y= x

†

Since

†

The solution for the non-dimensional concentration is

y=

1 sinh (φx ) x sinh (φ)

r

Diffusion and Reaction in a Porous Catalyst (26) †

The solution for the non-dimensional concentration is

1 sinh (φx ) y= x sinh (φ) †

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

∂C N R = −D eff ∂r r = R †

⎡ ⎤ moles reacted ⎢ area of catalyst - time ⎥ ⎣ ⎦

Therefore the reaction rate per particle is

⎛ ∂C ⎞ ⎟⎟ W = 4πR 2 ⎜⎜ − D eff ∂r r = R ⎠ ⎝

(

)

⎡ moles reacted ⎤ ⎢ particle - time ⎥ ⎣ ⎦

∆r

r

Diffusion and Reaction in a Porous Catalyst (27) †

r

Reaction rate per particle

⎛ ∂C ⎞ ⎜ ⎟⎟ W = 4πR ⎜ − D eff ∂r r = R ⎠ ⎝

(

†

2

)

⎡ moles reacted ⎤ ⎢ particle - time ⎥ ⎣ ⎦

Let put the RHS into non-dimensional form: ⎛ ∂ (C 0 y ) ⎞⎟ 2 ⎜ W = 4πR ⎜ − D eff ∂ (Rx ) x =1 ⎟⎠ ⎝

(

)

⎛ ∂y W = −(4πRD eff C 0 )⎜⎜ ⎝ ∂x

∆r

⎞ ⎟⎟ x =1 ⎠

W = −(4πRD eff C 0 )(φ coth φ − 1)

y=

1 sinh (φx ) x sinh (φ)

dy 1 ⎡ φ cosh (φx ) sinh (φx )⎤ = − ⎢ dx sinh (φ) ⎣ x x 2 ⎥⎦ This is your reaction rate per particle written in terms of the bulk concentration and the system parameters

Diffusion and Reaction in a Porous Catalyst (28) †

The reaction rate per particle is

W = −(4πRD eff C 0 )(φ coth φ − 1) †

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

coth φ ≈

1 φ + +L φ 3

⎛ 4πR 3 ⎞ ⎟⎟ρ pSg k C 0 W = −⎜⎜ ⎝ 3 ⎠ „

The reaction is kinetically-controlled

∆r

r

Diffusion and Reaction in a Porous Catalyst (29) †

r

The reaction rate per particle is

W = −(4πRD eff C 0 )(φ coth φ − 1) †

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

coth φ ≈ 1

(

W = − 4πR 2

„

)(

)

ρ pSg D eff k C 0

The reaction is diffusion-controlled

∆r

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 membrane †

†

The 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 dC J = −D dx „

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

D = f (C, ∂C / ∂x , T )

Diffusion through a Polymer Film (2) †

This last example will show you how to determine the intramembrane 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 membrane The 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

∂C ∂ 2C =D 2 ∂t ∂x „

The initial condition and boundary conditions are:

t = 0;

†

C=0

x = 0;

C = HP0

x = L;

C≈0

The solution is: ⎛ n 2 π2 D ⎞ x 2 ∞ sin (nπx / L ) C = 1− − ∑ t ⎟⎟ exp⎜⎜ − 2 L L π n =1 n H P0 ⎝ ⎠

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 d ⎛ PV ⎞ ∂C ⎜ ⎟ = −A D dt ⎝ RT ⎠ ∂x x = L t = 0; P=0

†

Given

†

the pressure in the bottom chamber is

⎛ n 2 π2 D ⎞ x 2 ∞ sin (nπx / L ) C = 1− − ∑ t ⎟⎟ exp⎜⎜ − 2 L L π n =1 n H P0 ⎝ ⎠

A R T P0 P= VL

⎧⎪ 2H L2 ⎨H D t + π2 ⎪⎩

⎛ n 2 π 2 D ⎞⎤ ⎫⎪ cos(nπ) ⎡ t ⎟⎟⎥ ⎬ ⎢ 1 − exp⎜⎜ − ∑ 2 2 n L n =1 ⎝ ⎠⎦ ⎪⎭ ⎣ ∞

Diffusion through a Polymer Film (6) †

So the solution for the pressure of the bottom chamber is

A R T P0 P= VL †

⎧⎪ 2H L2 ⎨H D t + π2 ⎪⎩

⎛ n 2 π 2 D ⎞⎤ ⎫⎪ cos(nπ) ⎡ t ⎟⎟⎥ ⎬ ⎢ 1 − exp⎜⎜ − ∑ 2 2 L n n =1 ⎝ ⎠⎦ ⎪⎭ ⎣ ∞

At sufficiently large time, this solution has an asymptote A R T P0 H D ⎛ L2 ⎞ ⎜⎜ t − ⎟⎟ P= VL 6 D ⎝ ⎠ P time