mass transfer 2. diffusion indilute solutions · mass transfer –diffusion in dilutesolutions_...

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

2. Diffusion in Dilute Solutions2.1 Diffusion across thin films and membranes2.2 Diffusion into a semi-infinite slab (strength of weld, tooth decay)2.3 Examples2.4 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 1.002 1.993 3.014 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: dcj Ddz

This is analogous to Newton’s law dydvx

yx

This is analogous to Fourier’s lawdxdTqx

Tq cDjor

These equations imply no convection (dilute solutions !).


2.1 Diffusion across thin films and membranes

Example 2.1.1: Diffusion across a thin filmz

zz 0

1C

C10

Goal: concentration profile in the film, and the flux across it at steady state.

Mass balance across arbitrary thin layer z:

zzatlayerofout

diffusionofrate

zatlayertheintodiffusionofrate

onaccumulatisolute

0

Steady state

)jj(A0 zz1z1


)jj(A0 zz1z1

Divide this equation by the film volume A z

z)zz(jj

0 z1zz1

21

2

1 dzcdD0j

dzd00zAs

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 Ddz

(2)

by estimating the dzdc1 0z zorat

The boundary conditions are 0z

z10cc

1cc

Then the solution to eq.1 is bzac1


and using the boundary condition gives:

z)cc(cc 101101

1011010z1

1 ccD1cc0DdzdcDj

101101z1 ccD1cc0D

dzdcD

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

2zz1z1 dz

cdDjjA0

but the boundary conditions differ:11

101Hcc,zHcc,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:zccHHcc 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

10csD

1CcD

1c

-ZcZc

Consider two-films in series.

At steady state j1 =const in both films.


In film 1: Large sDsmalldzdc

c cz c

css dcDdzj

dzdcDj

011

11

1

10

)cc(zDj c

c

s1101 (1)

In film 2: Small dc large Ddz

c cz

c

cdcDdzj

dzdcDj

1

111

11 )cc(

zDj 1c1

c1 (2)

)cc(D)cc(DD)cc(z

1c1c110s

sc110c

)cc(D)cc(D1

z

c110s

1c1c(1) = (2)

The flux becomes then: )cc(D)cc(Dj ccs 111101

2110

1ccc c 2

DDD sthen If


Two compounds A and B diffuse from 1 to 2 through the film z.

Which one diffuses faster or which one has the largest Diffusivity?

12

BD

ADAc1Bc1 Ac2

Bc 2z

12

3

BA1c

1c

When a compound diffuses faster in film A than in B, which conc’nprofile best describes this process, 1,2 or 3 and why?


2.2 Diffusion in a Semi-infinite SlabFick’s Second LawDiffusion 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

x

yz

j jx

xx

x2 j j

xx

xx

2


Rate at which particles enter the elemental volume x y z across the left side of that volume

zy2x

xjj X

XIN

IN - OUT =

zy2x

xjj x

x

gradient of jx at the center of x y z

Net rate of transport into that element

xjzyx x

A

B

C

D

x

yz

jjx

xx

x2 j j

xx

xx

2

Similarly for the x z face:yj

zyx y

zjzyx zand for x y face:

OUT


The rate of change of the number of particles per unit volume (& size), n, in the elemental volume x y z is:

c x y zt

jx

jy

jz

x y zx y z

jzj

yj

xj

tc zyx

From experimental observations: xcDjx

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

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

cDzc

yc

xcD

tc 2

2

2

2

2

2

2


Unsteady diffusion in a semi-infinite slabz

zz 0

1C

C10

Steady-state diffusion

10c

1c


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 (or more accurately, complimentary) to steady-diffusion through films. Everything else in the course is in between.


At 11 cc but at t > 0: 101 cc

Mass balance:zzat

layertheofoutdiffusionofrate

zatlayertheinto

diffusionofrate

z Avolumeinonaccumulatisolute

)jj(A)czA(t zzz 111

Divide by A z: zjj

tc zzz 111 0z

zj

tc 11

Combine this with Fick’s first law gives:

21

21

zcD

tc (1)

dzdcDj


21

21

zcD

tc (1)

Boundary Conditions: t = 0 all z: 11 cc

101 cc

11 cct > 0 z=0:

z= :How to solve Fick’s 2nd law?

Define a new variable (Boltzmann): Dt

z4 (2)

(It requires the wild imagination of mathematicians)

So eqn. (1) becomes:2

21

21

zdcdD

tddc

or using eqn. 2: 02 121

2

ddc

dcd (3)


The B.C. become: 1010 cc

11 cc

Set yddc1 so eqn.(3) becomes: 02 y

ddy

22 alnylndy

dyor:integrate

)exp(ay 2

Resubstitution: )exp(addc 21

0

21 kds)sexp(ac (4)


at 0

0

2100 kds)sexp(ac

kc10

21

0

2101 ds)sexp(accat

so 2101

/cca

in (4):0

2101101 ds)sexp(

2/)cc(cc

0

2

101

101 2 erfds)sexp(cccc

Dtz4


So the flux can be obtained as :2z-1 4Dt

1 10 1cj -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 mass (moles) transferred per unit area after time t t

z dtj0

01


2.3 ExamplesExample 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:

shelltheofoutdiffusion

shelltheintodiffusion

shellthiswithinonaccumulatisolute

rrr )jr()jr()crr(t 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 0r

12

210 jr

drd

r(2)

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

2 120 dcD d r

r dr dr (3)


Boundary Conditions: )sat(ccRr 110

01cr(4)(5)

Integrating eqn. 3 gives:2

1

ra

drdc (6)

where a is an integration constant.

Integrating eqn. 6 again gives:rabc1

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

rR)sat(cc 0

11 (8)


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

2010

11

1 rR)sat(Dc

rR)sat(c

drdD

drdcDj

which at the sphere’s surface is )sat(cRDj 10

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 x z) (w = width of film wall)

xxatoutflowingsolute-xatinflowingsolute

zzatoutdiffusingsolute-zatindiffusingsolute

zx w inonaccumulatisolute

0xxx1xx1

zz1z11

zwvczwvc

xwjxwj)zxwc(t

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 dzdj0 1

Combining it with Fick’s law gives: 21

2

dzcdD0

11

101

cczcc0zBoundary conditions:

The solution is: z)cc(cc 101101

)cc(Dj 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 dilute2. 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:

xxatoutflowingmass-xatinflowingassm

zzatoutdiffusingmass-zatindiffusingmass

zx w inonaccumulatimass


xxx1xx1

zz1z11

vczwvczw

jxwjxw)zxwc(t

At steady state and after dividing by the volume (w x z) and taking the limit as this volume goes to zero:

x11 vc

xzj0

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

1zcD

)v/x(c

)5(0cz)4()sat(cc0z0x)3(0cz0x

1

11

1Boundary 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

1v/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 solutions2.1 Across thin film and steady-state2.2 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 slab1 / Fo << 1 => steady-state1 / Fo ~ 1 => detailed analysis

if

Fo1

timetcoefficien

diffusionlength 2


Example: Membrane for industrial separation:Thickness = 0.01 cmD = 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

cm10Fo1

27

24

1.0s10scm10

cm10Fo1

427

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.

cm8length1s600sm10

lengthFo1

25-

2

days11s10t1tsm10

m10Fo1 6

212-

2-6


Another important difference of the two limiting cases stems from the interfacial fluxes.

1 1Dj c

1 1Dj c

t

(thin film)

(thick slab)

Note that both and have velocity units (dimensions), Dt

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, 0upper,1c

Lower compartment = solution, 0lower,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,upperDHj (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(dtd

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

where the geometric constant is )V

1V

1(AHupperlower

Boundary condition: 0upper,1

0lower,1upper,1lower,1 cccct=0: (5)

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

Dt0

upper,10

lower,1

upper,1lower,1 ecc

cc

upper,1lower,1

0upper,1

0lower,1

cc

ccln

t1Dor (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

0upper,1

0lower,1

cc

ccln

t1D

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

D1tC (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(e1cc 0

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

And compare with eq. (6): 0upper,1

0lower,1

Dtupper,1lower,1 ccecc

So set:

1Fo1


So eqn. (6) can be written as: Dtee1 Dt1 ee

CR tD1t

or

Now the above analysis is accurate when DC tt

upperlowervolumediaphragm

2

V1

V1HA

D1D1

upperlowervoidsdiaphragm V1

V1

V1

or

so

This is engineering MAGIC !!!

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