reactionchemicalby

destroyedorproduced

1speciesofmass

outthatusmin

in1species

offluxmass

zyx

inngaccumulati

1speciesofmass

zyxr

yxnyxn

zxnzxn

zynzyn

zyxct

zzz1zz1

yyy1yy1

xxx1xx1

1

Divide by the volume x y z:

1z1y1x11 r

z

n

y

n

x

n

t

c

GENERALIZED MASS BALANCESEquations that describe both steady and non-steady state

Mass balance for species 1 in volume Δx Δy Δz

1

5‐2

Mass balance for species “1” for different coordinate systems:

2

5‐3

111 rnt

c

011

0111 vccDvcjn

where v0 is the volume average velocity.

10

1121 r)vc(cD

t

c

(1)

Express contributions of diffusion and convection to the speciesmass flux:

Mass balance for species “1” in vector notation:

3

5‐4

Dividing by x y z yields the continuity equation:

Overall (total) mass balance, considering all species

zzzzz

yyyyy

xxxxx

yxvyxv

zxvzxv

zyvzyv

zyxt

where vx, vy, vz are the components of the mass average velocity.

Note: The overall mass balance has no reaction term since no total mass is generated or destroyed.

zyx vz

vy

vxt

v

t

or (2)

4

5‐5

Overall (total) mass balances for different coordinate systems:

5

5‐6

Simplify the species mass balance (1) with the help of the continuity equation (2)!Problem: Different reference velocities used.

For constant density, the mass and volume average velocities are the same and

00 v)v()v(0t

Divide by ρ, multiply by c1 and note that 011

001 vccv)vc(

100

1100

10

1 cvvcsocvvcvc0 →

so eqn. (1) becomes 112

101 rcDcv

t

c

6

5‐7

Mass balance for species “1” with diffusive and convective terms for constant density systems :

1

01

010

r1 c

sinr

vc

r

v

r

cv

t

c

121

2

221

212

2r

c

sinr

1csin

sinr

1

r

cr

rr

1D

7

5‐8

ExamplesFast diffusion through a stagnant film

1. Step:Select appropriate mass balance

1z11

r11 r

z

nn

r

1nr

rr

1

t

c

or for const. density:

12

12

2

12

2

1

10z

10

10r

1

rz

cc

r

1

r

cr

rr

1D

z

cv

c

r

v

r

cv

t

c

Remember few weeks back….

8

y1l

5‐9

2. Step: Simplify mass balance

1z11

r11 r

z

nn

r1

nrrr

1t

c

=0

steady state

=0

no reaction

=0

no flow inr-direction

=0

symmetry

gives: 0z

n z1

or:

21

210

zz

cD

zc

v

Then solve as before:

9

5‐10

Fast diffusion into a semi-infinite slab

Remember again:1. Step:Select appropriate mass balance

Here: capillary 1z11

r11 r

z

nn

r1

nrrr

1t

c

10

2. Step: Simplify mass balance

• No flow in x-direction• No flow in y-direction• No chemical reaction

0z

n

t

c z,11

Or: 21

210

z1

z

cD

z

cv

t

c

Then solve as before (Ex. 3.2.4).

5‐11

The flux near a spinning disk

A solvent flow approaches a spinning disk made out of a sparingly soluble solute, as shown below. Calculate the diffusion-controlled rate at which the disk slowly dissolves at steady state.

1. Step:Select appropriate mass balance

12

12

2

12

2

1

10z

10

10r

1

rz

cc

r

1

r

cr

rr

1D

z

cv

c

r

v

r

cv

t

c

11

5‐12

2. Step: Simplify mass balance

12

12

2

12

2

110z

10

10r

1 rz

cc

r

1

r

cr

rr

1D

z

cv

c

r

v

r

cv

t

c

=0

steadystate

=0

noreaction

=0 =0

angularsymmetry

angularsymmetry

seeflow pattern

=0 =0

seeflow pattern

21

210

z z

cD

z

cv

12

5‐13

Velocity over a spinning disk From: Levich, “Physiochemical Hydrodynamics”, Prentice-Hall, 1962.

ϴ

The velocity profile suggests that there is no gradient in radial direction! This is true for sufficiently large disks where edge effects are negligible.

This special flow behavior is taken advantage of in the application of homogeneous coatings, e.g. on wafers.

13

5‐14

3. Step: Boundary conditions

B.C.: z=0 c1=c1(sat)z= c1=0

bdrds)s(vD/1expacz

0

r

0 z1

Where a and b are integration constants calculated from the boundary conditions, so

The solution is:

drds)s(vD/1exp

drds)s(vD/1exp1

)sat(cc

0

r

0 z

z

0

r

0 z

1

1

14

5‐15

If we know the velocity vz we can get c1. From the literature (Levich, 1962) we know

22123z s51.0v

and inserting in the above eqns. gives:

duuexp

duuexp

)sat(cc

03

03

1

1

where

1

2/1

6/13/1D82.1z

is the kinematic viscosity of the liquid and is the angular velocity of the disk.

The diffusion flux is:

)sat(cD

62.0zc

Dj 16/1

2/13/2

0z1

0z1

15

5‐16

This equation can also be written as:

)sat(cScRed

D62.0j

)sat(cD

d

d

D62.0j

1

chaptersgminupcotheinMore)tcoefficientransfermass(k

3/12/11

1

3/12/12

1

There is an excellent agreement when plotting j1 d / D c1(sat) vs. Re1/2 justifying the strong assumption made in the analysis.

The fact that this device gives uniform mass transfer over the entire surface area makes it popular in CVD and electrochemistry.

16

5‐17

Example : Dissolving Pill

In pharmacy it is important to know HOW FAST a medicine can permeate the body to act. Estimate the time it takes to start a steady-state dissolution of a drug pill. Is dissolution diffusion controlled? This can be checked by gentle stirring. If it dissolves faster than in the absence of stirring then it is diffusion-controlled!!

1. Step:Select appropriate mass balance

1

21

2

22

12

122

10

10

10r

1

rc

sinr

1

csin

sinr

1

r

cr

rr

1

D

c

sinr

vc

r

v

r

cv

t

c

17

5‐18

2. Step: Simplify mass balance

121

2

221

212

2

10

10

10r

1

rc

sinr

1csin

sinr

1rc

rrr

1D

csinr

vcr

vrc

vt

c

=0

noreaction

=0

symmetry

=0

symmetry

=0=0symmetry

=0

“Sparingly” soluble→ stagnant

surroundings

r

cr

rr

D

t

c12

2

1

18

5‐19

3. Step: Boundary conditions

t=0 r c1=0t>0 r=R0 c1=c1(sat)

r= c1=0 (meaning that the drug is consumed at thegut wall)

This is easily solved by introducing a variable =c1r and the equation reduces to that describing unsteady diffusion through a semi-infinite slab, so

tD4

Rrerf1

r

R

)sat(cc 00

1

1

19

5‐20

Thus the flux for a sparingly soluble pill:

stateunsteady

0

0

1111

tD

R1

R

)sat(cD

r

cDjn

0Rr

Now you can calculate how long you need to reach steady-state by calculating when the term

11.0tD

R0

Say that the pill is 3 mm and the D=10-5 cm2/s (typical for D in liquids) then t80 hours.

This is clearly wrong. We know that usually a pill acts within 10-20 minutes. So where is the mistake? Revisit the approximations and think looking at slide 2-24…..

20

5‐21

Example: Diffusion through a Polymer Membrane

A diaphragm-cell separates olephins (ethylene) from aliphatic hydrocarbons. Upper compartment: vacuum, lower: ethylene.

Measure the ethylene concentration as a function of time. Find the diffusivity.

1. Step:Select appropriate mass balance

12

12

2

12

2

1

10z

10

10r

1

rz

cc

r

1

r

cr

rr

1D

z

cv

c

r

v

r

cv

t

c

21

5‐22

Under which other assumptions this eqn. becomes this?

2. Step: Simplify mass balance

21

21

zc

Dt

c

3. Step: Boundary conditions for the polymer film:

t=0 z c1=0t>0 z=0 c1=H p0

z=L c1=H pL0

where L is the film (membrane) thickness and H the partition coefficient relating ethylene conc’n(partial pressure) in the gas to ethylene conc’n in the membrane.

22

Now the solution to this equation is (J. Crank “The mathematics of diffusion”, 2nd ed., 1975, p.50 eqn. 4.22):

5‐23

2

22

1n0

1

L

tnDexp

n

L/znsin2

L

z1

pH

c

This is the concentration profile c1(z,t). We need to bring this result into a convenient form for experimental measurements.

Make a mass balance or better, a mole balance as we are dealing with gases. Start from the ideal gas law:

dtdp

TRV

dtdN1

(1)

moles enteringupper compartment at 

z=LLz

1Lz1 z

cDAjA

(2)

23

Substitute (1) into (2) and integrate for p assuming that at t=0 p=0 at upper compartment.

2

22

1n22

20

L

tnDexp1

n

ncosLH2tDH

LV

pTRAp

5‐24

At long times the exponential term 0

6

LHtDH

LV

pTRAp

2

known

0

Thus the intercept is related to Henry’s constant while the slope is related to DH, the permeability of ethylene through the membrane.

24

From another point of view now: As you can obtain H from the intercept, you can estimate D from the slope!

This is an elegant problem combining verification (extraction) of both equilibrium (H) and transport (D) properties!

The challenge of an engineer is to QUANTITATIVELY describe the phenomenon, e.g. connect the math to the physical problem.