mass transfer 2. diffusion in dilute solutions transfer –diffusion in dilute solutions_...

Mass Transfer – Diffusion in Dilute Solutions_ Fick‘s Laws 2-1

2. Diffusion in Dilute Solutions

2.1 Diffusion across thin films and membranes

2.2 Diffusion into a semi-infinite slab (strength of weld, tooth decay)

2.3 Examples

2.4 Dilute diffusion and convection

Graham (1850) monitored the diffusion

of salt (NaCl)

solutions in a larger jar containing

water. Every so often he removed the

bottle and analyzed it.

Mass Transfer


Initial salt

concentration,

Weight-% of NaCl

Relative Flux

1 1.00

2 1.99

3 3.01

4 4.00

He postulated that

a) The quantities diffused appear to be proportional to the salt

concentration.

b) Diffusion must follow diminishing progression.

Fick (1855) analyzed these data and wrote

“The diffusion of the dissolved material ... is left completely to the influence

of the molecular forces basic to the same law ... for the spreading of

warmth in a conductor and which has already been applied with such great

success to the spreading of electricity.”


Fick’s first law:dc

j Ddz

This is analogous to Newton’s lawdy

dvxyx

This is analogous to Fourier’s lawdx

dTqx

Tq cDj or

These equations imply no convection (dilute solutions !).


2.1 Diffusion across thin films and membranes

Example 2.1.1: Diffusion across a thin film

Dz

z z 0

1C

C10

Goal: concentration profile in the

film, and the flux across it at steady

state.

Mass balance across arbitrary thin layer Dz:

D

zz at

layer of out

diffusion of rate

z at layer the into

diffusion of rate

onaccumulati

solute

0

Steady state

)jj(A0 zz1z1 D


)jj(A0 zz1z1 D

Divide this equation by the film volume A‧Dz

D

D

z)zz(

jj0

z1zz1

21

2

1dz

cdD0j

dz

d0 0z DAs

Fick’s

first law(1)


If we solve this equation we have the concentration profile of c in

and then we can calculate the flux

from Fick’s first law 11

dcj D

dz (2)

by estimating the dz

dc10z zorat

The boundary conditions are 0z

z

10cc

1cc

Then the solution to eq.1 is bzac1


and using the boundary condition gives:

z)cc(cc 101101

1011010z1

1 ccD1

cc0Ddz

dcDj

101101z1 cc

D1cc0D

dz

dcD

or


Example 2.1.2: Membrane diffusion

Derive the concentration profile and the flux for a single

solute diffusing across a thin membrane.

The analysis is the same as before leading to

21

2

zz1z1dz

cdDjjA0 D

but the boundary conditions differ:

11

101

Hcc,z

Hcc,0z

where H is a partition coefficient (the concentration in the

membrane divided by that in the adjacent solution e.g. Henry’s or

Raoult’s law).


Then the concentration profile becomes:

z

ccHHcc 101101

10c

1c

The solute is more soluble in

the membrane than in the

adjacent solution

10c

1c

The solute is less soluble in

the membrane than in the

adjacent solution


Example 2.1.3: Concentration–dependent diffusion coefficient

The diffusion coefficient D can vary with concentration c.

(water across films and in detergent solutions)

Assumption:

cc1

ccc 1

ccc 1

slow diffusion (small D), DS

fast diffusion (large D), D

10c

sD

1CcD

1c

l-ZcZc

Consider two-films in series.

At steady state j1 =const in

both films.


In film 1: Large sD small dz

dc

c cz c

css dcDdzj

dz

dcDj

011

11

1

10

)cc(z

Dj c

c

s1101 (1)

In film 2: Smalldc

large Ddz

c cz

c

c

dcDdzjdz

dcDj

1

1

111

1 )cc(z

Dj 1c1

c1

(2)

)cc(D)cc(D

D)cc(z

1c1c110s

sc110c

)cc(D

)cc(D1

z

c110s

1c1c

(1) = (2)

The flux becomes then:

)cc(D)cc(Dj ccs 111101

2

1101

ccc c

2

DDD s then If


In the following film two compounds A and B diffuse from 1 to 2

through the film Dz.

Which one diffuses faster or which one has the largest

Diffusivity?

1

2

BD

ADAc1

Bc1Ac2

Bc 2zD


A compound diffuses through two films in series. When it

diffuses faster in film A than in film B, which concentration profile

best describes this process, 1,2 or 3 and why?

1

2

3

BA

1c

1c


2.2 Diffusion in a Semi-infinite Slab

Fick’s Second Law

Diffusion is the net migration (mass transfer-transport) of

molecules from regions of HIGH to LOW concentration.

jX: flux of particles in the x-direction

A

B

C

D

dx

dy

dz

j jx

xx

x

d

2 jjx

xx

x

d

2


Rate at which particles enter the

elemental volume dxdydz across the

left side of that volume

zy

2

x

x

jj XX dd

d

IN

IN - OUT =

zy2

x

x

jj xx dd

d

gradient of jx at the

center of dxdydz

Net rate of transport

into that element

x

jzyx x

ddd

A

B

C

D

dx

dy

dz

jj

x

xx

x

d

2j

j

x

xx

x

d

2

Similarly for the dxdz face:y

jzyx

y

ddd

z

jzyx z

dddand for dxdy face:

OUT


The rate of change of the number of particles per unit volume

(& size), n, in the elemental volume dxdydz is:

d d d

d d d c x y z

t

j

x

j

y

j

z

x y z x y z

jz

j

y

j

x

j

t

c zyx

From experimental observations:x

cDjx

(Fick’s first law without convection, dilute solutions).

Substituting it in the above gives Fick’s second law:

cDz

c

y

c

x

cD

t

c 2

2

2

2

2

2

2


Example 2.2.1:

Unsteady diffusion in a semi-infinite slab

Consider that suddenly the

concentration at the interface

changes.

Goal: To find how the

concentration and flux

varies with time.

Very important in diffusion in solids (tooth decay, corrosion of

metals). This is the opposite to diffusion through films. Everything

else in the course is in between.


At t ≤ 0: 11 cc but at t > 0: 101 cc

Mass balance:

D

Dzz at

layer the of out

diffusion of rate

z at

layer the into

diffusion of rate

z Avolume in

onaccumulati solute

)jj(A)czA(t

zzz DD

111

Divide by ADz:

D

D

z

jj

t

c zzz 111 0Dzz

j

t

c

11

Combine this with Fick’s first law gives:

21

21

z

cD

t

c

(1)

dz

dcDj


21

21

z

cD

t

c

(1)

Boundary Conditions: t = 0 all z: 11 cc

101 cc

11 cc

t > 0 z=0:

z=:How to solve Fick’s 2nd law?

Define a new variable (Boltzmann): Dt

z

4 (2)

(It requires the wild imagination of

mathematicians)

So eqn. (1) becomes:2

21

21

zd

cdD

td

dc

or using eqn. 2: 02 121

2

d

dc

d

cd (3)


The B.C. become: 1010 cc

11 cc

Set yd

dc

1 so eqn.(3) becomes: 02 y

d

dy

22 alnylndy

dyor:

integrate

)exp(ay 2

Resubstitution: )exp(ad

dc 21

0

21 kds)sexp(ac (4)


at 0

0

2100 kds)sexp(ac

kc 10

2

1

0

2101 ds)sexp(accat

so2

101

/

cca

in (4):

0

2101101 ds)sexp(

2/

)cc(cc

0

2

101

101 2erfds)sexp(

cc

cc


So the flux can be obtained as :

2z-

1 4Dt1 10 1

cj -D D / t e (c - c )

z

and the flux across the interface becomes (z=0) :

)cc(t

Dj z 11001

This is the flux at time t.

Total flux at time t

t

z dtj0

01


2.3 Examples

Example 2.3.1: Steady dissolution of a sphere

Consider a sphere that dissolves

slowly in a large tank. The sphere

volume does not change.

Find the dissolution rate and the

concentration c1(r) profile away

from the sphere at steady-state.

www.sciencebasedmedicine.org www.what-when-how.com


Mass balance on a spherical shell:

shell the of out

diffusion

shell the into

diffusion

shell this within

onaccumulati solute

rrr )jr()jr()crr(t

DD

1

21

21

2 444 (1)

Divide both sides by the spherical shell’s volume, note that

LHS=0 at steady-state and take the limit as 0Dr

12

2

10 jr

dr

d

r (2)

Combine this with Fick’s law at spherical coordinates and D = const:

2 1

20

dcD dr

r dr dr (3)


Boundary Conditions: )sat(ccRr 110

01 cr

(4)

(5)

Integrating eqn. 3 gives:2

1

r

a

dr

dc (6)

where a is an integration constant.

Integrating eqn. 6 again gives:r

abc 1

where b is another integration constant.

(7)

Using the B.C. gives b=0 from eqn. 5 and a =c1(sat)R0 from

eqn. 4 so eqn. 7 becomes

r

R)sat(cc 0

11 (8)


The dissolution flux can be found from Fick’s law:

2010

11

1r

R)sat(Dc

r

R)sat(c

dr

dD

dr

dcDj

which at the sphere’s surface is )sat(cR

Dj 1

0

1

If you double the sphere (particle) size, the dissolution rate per unit

area is only half as large even though the total dissolution rate

over the entire surface is doubled.

Also in the growth of fog droplets and spraying, as well as in

growth of particles by condensation or by surface reaction limited

by transport.

Very important in pharmaceutics!!


Challenging Mathematics:

Text: 2.4.1 Decay of a Pulse

2.4.3 Unsteady Diffusion into Cylinders

Decay of a Pulse Unsteady Diffusion into Cylinders


2.4 Dilute Diffusion and Convection

Till now we did not consider any flow.Convection

Diffusion

Here we address a special

case where diffusion and

convection occur normal to

each other:


2.4.1 Steady Diffusion across a falling film

Assumptions:

1. The solution is dilute (no diffusion-driven flow)

2. The liquid is the only resistance to mass transfer.

3. Mass transfer by diffusion in z-direction and flow in x-direction


Mass balance on volume ( w Dx Dz) (w = width of film wall)

D

D

DD

xx at out flowing solute

- x at in flowing solute

zz at out diffusing solute

- z at in diffusing solute

z x w in

onaccumulati solute

0

xxx1xx1

zz1z11

zwvczwvc

xwjxwj)zxwc(t

D

D

DD

DDDD

as c1 and vx are constant in x

c1 varies in z but not in x ! (The film is long)

vx varies in z but not in x ! (Couette flow, no pressure drop in x)


Now we can write dz

dj0 1

Combining it with Fick’s law gives:21

2

dz

cdD0

11

101

ccz

cc0z

Boundary conditions:

The solution is:

z)cc(cc 101101

)cc(D

j

1101

Unbelievable ! The flow has no effect.

That’s right! When solutions are dilute this is correct.


2.4.2 Diffusion into a falling film

A thin liquid film flows slowly without

ripples (waves) down a flat surface.

One side of the film wets the surface

while the other is in contact with the

gas which is slightly (sparingly)

soluble in the liquid.

Goal: Find how much gas dissolves

in the liquid.

(important to

“Penetration Theory”)


Assumptions:

1. The solution is dilute

2. Mass transfer in z-direction and flow

(convection) in x-direction

3. The gas over the film is pure (no resistance to

diffusion)

4. Short contact between liquid and gas (for

convenience)

Mass balance:

D

D

DD

xx at out flowing mass

- x at in flowing assm

zz at out diffusing mass

- z at in diffusing mass

z x w in

onaccumulati mass


xxx1xx1

zz1z11

vczwvczw

jxwjxw)zxwc(t

D

D

DD

DDDD

At steady state and after dividing by the volume (w Dx Dz) and taking

the limit as this volume goes to zero:

x11 vc

xz

j0

We combine this with Fick’s law and set vx= vmax (fluid velocity at

the interface) as the gas-liquid contact time is short (based on our

bold (too strong) assumption #4)

(1)


The implication here is that the solute barely has a

chance to cross the interface so slightly diffuses into the

fluid.

So equation (1) becomes:

(2)21

2

max

1

z

cD

)v/x(

c

)5(0cz

)4()sat(cc0z0x

)3(0cz0x

1

11

1

Boundary conditions:


Now revoking (recalling) again assumption #4 the last B.C. is

replaced by

)6(0cz0x 1

meaning that the solute diffuses only shortly into the liquid. As a

result, the solute does not “see” the wall.

In this case this problem reduces to that of diffusion in a semi-

infinite slab with maxv/xt

and the solution is the same: (slide 2-23)

max1

1

v/xD4

zerf1

)sat(c

c

and the flux at the interface is: )sat(cx/vDj 1max0z1


What did we learn so far ?

2. Diffusion of dilute solutions

2.1 Across thin film and steady-state

2.2 Across a thick slab and no steady-state

How to choose between these two ?

This is the variable in the

error function of the semi-

infinite slab problem.

Fo: Fourier number

1 / Fo >> 1 => semi-infinite slab

1 / Fo << 1 => steady-state

1 / Fo ~ 1 => detailed analysis

if

Fo

1

timetcoefficien

diffusion

length2


Example: Membrane for industrial separation:

Thickness = 0.01 cm

D = 10-11 m2/s

If the duration of the experiment is

a) t=10 s

This is a semi-infinite slab problem!

b) t=3 hrs 104 s

This is a thin film, steady-state problem.

The value of Fo = 1 indicates that mass transfer is significantly

advanced in a given process. As a result it can be used to estimate

the EXTENT (or DEGREE) of advancement (or progress) for

unsteady-state processes.

100s10scm10

cm10

Fo

127

24

1.0s10scm10

cm10

Fo

1427

24


For example:

a) Guess how far gasoline has evaporated into the stagnant air in

a regular glass-fiber filter. Say that evaporation is going on for

10 min and D = 10-5 m2/s.

b) Consider H2 diffusion in nickel making it rather brittle. If

D = 10-12 m2/s estimate how long it will take for H2 to diffuse

1 mm through the Ni specimen.

cm8length1

s600sm10

length

Fo

125-

2

days 11s10t1tsm10

m10

Fo

1 6

212-

2-6


Another important difference of the two limiting cases stems from

the interfacial fluxes.

1 1

Dj c D

1 1

Dj c

t D

(thin film)

(thick slab)

Note that both and have velocity units (dimensions),

D

t

D

some people even call them “the velocity of diffusion”. In fact

these are equivalent to the mass transfer coefficients we

talked earlier on !!


Example: Diaphragm-cell diffusion

Goal: To measure the diffusion coefficient

Cell: Two well-stirred volumes and a thin barrier (or diaphragm,

e.g. sintered glass frit or even a piece of paper).

Combination of a steady-state (inside diaphragm) and a transient

problem (in liquid reservoirs).


Upper compartment = solvent,0

upper,1c

Lower compartment = solution,0

lower,1c

After time t, measure new c1 at the upper and lower

compartment

Procedure:

Assumptions: Rapid attainment of steady state flux across the

diaphragm.

Note that this says the flux is steady even through the

concentrations are changing! Can we get away with that?


At this pseudo steady-state the flux across the diaphragm

(membrane) is:

1 1,lower 1,upper

DHj (c c )

(H can also be regarded as the fraction of the diaphragm area

available for diffusion)

Mass balance on each compartment

(1)

Lower: 1lower,1

lower jAdt

dcV

Upper: 1upper,1

upper jAdt

dcV

(2)

(3)

where A is the area of the diaphragm.


Dividing eqs. (2) and (3) by Vlower and Vupper, respectively, followed

by subtracting eqn. (3) from (2) and substituting eqn. (1), gives:

)cc(D)cc(dt

dlower,1upper,1upper,1lower,1 (4)

where the geometric constant is )V

1

V

1(

AH

upperlower

Boundary condition:0upper,1

0lower,1upper,1lower,1 cccc t=0: (5)

Integrating eqn. (4) subject to (5) gives

Dt

0

upper,1

0

lower,1

upper,1lower,1 ecc

cc

upper,1lower,1

0

upper,1

0

lower,1

cc

ccln

t

1Dor (7)(6)

is obtained by calibration with solute of known D.

Now as we can measure t and the solute concentration at the two

compartments, D can be obtained.


Let´s go back to our assumptions:

a) D is affected by the diaphragm and its tortuosity (internal

channel-like structure)

This can be accounted for by rewriting eqn. (7) as:

upper,1lower,1

0

upper,1

0

lower,1

cc

cc

lnt

1D

Where ’ is a new calibration constant that includes tortuosity.

Surprisingly this works well as D agrees with that measured by

other techniques.


b) Pseudo steady-state (steady-state flux across a diaphragm with

unsteady-state concentrations in the compartments)

Compare the volume of material (solvent and solute) in the

diaphragm voids (empty space) with that of each compartment.

The solute concentrations in the compartments changes

slooooowly because they are very large compared to the

diaphragm.

The solute concentration in the diaphragm changes much faster

as it has little volume.

Thus the concentration profile in the diaphragm will reach a

(pseudo) steady-state before the corresponding concentrations

change much. Thus the flux will reach steady-state!


Now more quantitatively and professionally: We can compare the

characteristic (or relaxation) times of the two units:

Diaphragm:

Compartment:

Dt

2

D

D

1tC (1/e)

(8)

(9)

Definition: The relaxation time is the time at which the distance to

equilibrium has been reduced to the fraction 1/e of its initial value.

)cc(e

1cc 0

upper,10lower,1upper,1lower,1

And compare with eq. (6): 0

upper,1

0

lower,1

Dt

upper,1lower,1 ccecc

So set:

1

Fo

1


So eqn. (6) can be written as: Dtee

1 Dt1 ee

CR tD

1t

or

Now the above analysis is accurate when DC tt

upperlower

volumediaphragm

2

V

1

V

1HA

D

1D1

upperlowervoidsdiaphragm V

1

V

1

V

1

or

so

This is engineering MAGIC !!!

mass transfer 2. diffusion in dilute solutions transfer –diffusion in dilute solutions_...

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