lecture 5: generalized mass balances

5-1

LECTURE 5: GENERALIZED MASS BALANCES

Equations that describe both steady and non-steady state

Figure shows:

• the arbitrary volume

• the flux in the x-direction but we should

also include the fluxes in the other two

directions

1

5-2

1 1

1 1

mass of species mass flux of mass flux of mass of species

accumulating in species species produced or destroyed

x y z in out by chemical reaction

1 1

1 1 1 1

1 1

x xx x x

y yy y y

z zz z z

n y z n y z

c x y z n x z n x z r x y zt

n x y n x y

Divide by the volume x y z:

1 1 1 1

1

x y zc n n nr

t x y z

• Mass balance for species 1 in volume Δx Δy Δz

Or, in vector form:1

1 1

cr

t

n

2

5-3

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

3

5-4

1

1 1

cr

t

n

0 0

1 1 1 1 1j c D c c n v v where v0 is the volume average velocity.

1 2 0

1 1 1( )c

D c c rt

v (1)

Express contributions of diffusion and convection to the species mass flux:

Mass balance for species “1” in vector notation:

To proceed, we need information about the velocity v0

4

5-5

Dividing by x y z yields the continuity equation:

Overall (total) mass balance, considering all species

x xx x x

y yy y y

z zz z z

v y z v y z

x y z v x z v x zt

v x y v x y

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.

x y zv v v

t x y z

t

vor, in vector form (2)

5

5-6

Overall (total) mass balances for different coordinate systems:

6

5-7

Mass balance for species “1” in vector notation:

01 2

1 11( )c

D c rt

c

v (1)

Idea: Simplify the term in red with the help of the continuity equation (2)!

t

vContinuity equation in vector notation: (2)

Complication: Different reference velocities are often used!

An interesting case: When the density ρ is constant!

7

5-8

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

0

0 0 0

1

0 ( ) 0 ( ) 0

00 0

t

c

v

v v

v

v

We also note that 0 0 0

1 1 1c c c v v v →

Thus, eq. (1) becomes 1 0 2

1 1 1

cc D c r

t

v

This is the simpler form of the mass balance equation for species 1!!!

0 0

1 1c c v v

8

5-9

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

density systems :

0001 1 1 1

22 1 1 1

12 2 2 2 2

sin

1 1 1sin

sin sin

r

vvc c c cv

t r r r

c c cD r r

r r r r r

9

5-10

Examples

Fast diffusion through a stagnant film

1. Step: Select appropriate mass balance

1 1 1

1 1

1 1 z

r

c n nr n r

t r r r z

or for constant density:

01 1 1 10 0

2 2

1 1 1

12 2 2

1 1

r z

c c c cvv v

t r r z

c c cD r r

r r r r z

Remember few weeks back….y1l

10

5-11

2. Step: Simplify mass balance

1 1 1

1 1

1 1 z

r

c n nr n r

t r r r z

=0

steady state

=0

no reaction

=0

no flow in

r-direction

=0

symmetry

Thus: 10

zn

z

or:

20 1 1

2z

c cv D

z z

Then solve as before!!!

11

5-12

Fast diffusion into a semi-infinite slab

Remember again:1. Step: Select appropriate mass balance

Here: capillary 1 1 1

1 1

1 1 z

r

c n nr n r

t r r r z


• No flow in x-direction

• No flow in y-direction

• No chemical reaction

1 1,0

zc n

t z

or:

2

1 1 10

2z

c c cv D

t z z

Then solve as before!!!

12

5-13

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 balance0

1 1 1 10 0

2 2

1 1 1

12 2 2

1 1

r z

c c c cvv v

t r r z

c c cD r r

r r r r z

13

5-14

• Far from the rotating disk, the

fluid moves toward the disk

• In a thin layer immediately

adjacent to its surface, the fluid

acquires a rotating motion

• The angular velocity of the fluid

increases as the surface of the disk

is approached, until the angular

velocity of the rotating disk is

finally attained.

• Furthermore, the fluid also

acquires a radial velocity under the

influence of the centrifugal force.

ϴ

14

5-15

Velocity over a spinning disk

From: Levich, “Physiochemical Hydrodynamics”, Prentice-Hall, 1962.

ϴ

For sufficiently large disks where

edge effects are negligible, the

velocity profile suggests that there

is no concentration gradient in

radial direction!

This special flow behavior is taken

advantage of in the application of

homogeneous coatings, e.g. on

wafers.

15

5-16


2 201 1 1 1 1 1 10 0

12 2 2

1 1r z

c c c c c c cvv v D r r

t r r z r r r r z

=0

steady

state

=0

No reaction

=0 =0

angular

symmetry

angular

symmetry

See flow pattern

=0 =0

2

0 1 1

2z

c cv D

z z

See flow pattern

16

5-17

3. Step: Boundary conditions

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

z = c1 = 0

1 1 20 0

1exp ( )

z s

zc a v z dz ds aD

where a1 and a2 are integration constants calculated from the boundary

conditions. We find:

Solution:

0 01

1

0 0

1exp ( )

11( )

exp ( )

z s

z

s

z

v z dz dsc D

c satv z dz ds

D

ϴ

17

5-18

If we know the velocity vz we can get c1 (Fluid Mechanics Problem). From the

literature (Levich, 1962) we know that close to the disk:

320.51zv z

Inserting in the above equations gives

3

01

31

0

exp1

( ) exp

u duc

c sat u du

where

11/3 1/6

1/2

1.82 Dz

= kinematic viscosity of the liquid, = angular velocity of the disk.

Diffusion flux:

2/3 1/2

11 11/60

0

... 0.62 ( )z

z

c Dj D c sat

z

(Exercise …):

(Exercise …):

18

5-19

The flux near a spinning disk

• Such a constant flux is uncommon,

and it makes the interpretation of

experimental results unusually

straightforward.

• It is this feature that makes the

rotating disc a popular experimental

tool.

2/3 1/2

1 11/600.62 ( )

z

Dj c sat

• It does not depend on radius r. It has the same value near the disc’s center

and near its edge!!!

• It does not depend on the disc diameter d either!!!

19

5-20

Flux equation can be re-written as:

1/2 1/3

1 1

2

0.62 ( )D

jd

c satDd

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.

1/2 1/3

1 1

( )min

0.62 ( )Re

k mass transfer coefficientMore in the upco g chapters

Dj c sat

dSc

or

20

5-21

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

(in a spherical shell outside the pill)

0001 1 1 1

22 1 1 1

12 2 2 2 2

sin

1 1 1sin

sin sin

r

vvc c c cv

t r r r

c c cD r r

r r r r r

21

5-22


0001 1 1 1

22 1 1 1

12 2 2 2 2

sin

1 1 1sin

sin sin

r

vc c v c cv

t r r r

c c cD r r

r r r r r

=0

no reaction

=0

symmetry

=0

symmetry

=0=0

symmetry

=0

“Sparingly” soluble → stagnant surroundings

1 2 1

2

c cDr

t r r r

22

5-23

3. Step: Boundary conditions

t = 0: r c1 = 0

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

r = c1 = 0 (drug is consumed at the gut wall)

To solve it, we introduce the combined variable:

Then, equation reduces to that describing unsteady diffusion through a semi-

infinite slab, thus

0 01

1

1( ) 4

R r Rcerf

c sat r Dt

1c r

(Exercise …):

23

5-24

4. Flux for such a sparingly soluble pill:0

01 11 1

0

( )1

r R

unsteady state

Rc Dc satn j D

r R Dt

Now, we can calculate how long

you need to reach steady-state by

calculating when the term

0 0.1 1R

Dt

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

80 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…

24

5-25

Example: Diffusion through a Polymer Membrane

A diaphragm-cell separates olefins (ethylene) from aliphatic hydrocarbons.

Upper compartment: vacuum, lower: ethylene.

We can measure the ethylene concentration in the

upper compartment, as a function of time.

Experimentally: Initially, the pressure in the upper

compartment varies in a complex way, but it

eventually approach that in the lower

compartment. At the moderate times of most of our

experiment, the pressure in the upper compartment

is proportional to time, with a known slope and a

definite intercept.

Analysis: from these, compute D!

25

5-26

Measure the ethylene concentration inside the

upper cell as a function of time. Find the

diffusivity.


01 1 1 10 0

2 2

1 1 1

12 2 2

1 1

r z

c c c cvv v

t r r z

c c cD r r

r r r r z

Example: Diffusion through a Polymer Membrane

26

5-27

Under which other assumptions this eq. becomes this?


2

1 1

2

c cD

t z

3. Step: Boundary conditions for the polymer film:

t = 0: z c1=0

t > 0: z=0 c1= H p0

z=L c1= H pL0

L = film (membrane) thickness, H = partition coefficient relating ethylene

conc’n (partial pressure) in the gas to ethylene conc’n in the membrane.

4. Solution (J. Crank, The mathematics of diffusion, 2nd ed., 1975, p.50, eq. 4.22)!

27

5-28

• This is the concentration profile c1(z, t) in the membrane. 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:

1dN V dp

dt R T dt

(1)

moles entering

upper compartment at z=L1

1 z L

z L

cA j A

z

(2)

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

compartment, we find:

2 2 2

0

2 2 21

cos21 exp

n

nARTp HL Dn tp HDt

VL n L

4. Solution:2 2

1

210

sin2

1 expn

n z

c z Dn tL

H p L n L

(Exercise …):

28

5-29

At long times, the exponential term 0, thus:

2

0

6known

ARTp HLp HDt

VL

• intercept related to Henry’s constant H

• slope related to D.H, the permeability of ethylene through the membrane.

From another point of view: We can obtain H from the intercept, and then we

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.

2

0

2 21

cos2

n

nARTp HLp HDt

VL n

2

2 21 1

cos 1...

12

n

n n

n

n n

(Exercise …):

29

5-30

(a) What physical system is implied?

(b) What is the differential volume on which this mass balance is written?

(c) What is the meaning of each of the four terms?

Example: From the diffusion equation back to the Physical Problem

2

1 1 1 1

2z

c c c cDv r D

t z r r r z

(a) Mass transfer in cylindrical

coordinates

(b) Unsteady state

(c) Convection in the z direction

(d) Diffusion along r and z

30

5-31

Example: Effective diffusion coefficients in a porous catalyst pellet

(spherical shape)

Imagine that we have a porous catalyst pellet

containing a dilute gaseous solution.

We want to measure the effective diffusion of solute by dropping this

pellet into a small, well-stirred bath of a solvent gas and measuring how

fast the solute appears in this bath.

How can we plot these measurements to find the effective diffusion

coefficient?

31

5-32

Example: Effective diffusion coefficients in a porous catalyst pellet

(spherical shell)

0001 1 1 1

22 1 1 1

12 2 2 2 2

sin

1 1 1sin

sin sin

r

eff

vc c v c cv

t r r r

c c cD r r

r r r r r

=0

no reaction

=0

symmetry

=0

symmetry

=0=0

symmetry

=0

Slow dissolution

through diffusion,

no convection


32

5-33

2. Step: Simplify mass balance2

1 1

2eff

c cD

t z

3. Step: Boundary conditions:

• R is the pellet

radius

• c1(t) is the bath

concentration, a

function of time.

1 1,0

1

1 1

0, ,

0 0, 0

, ( )

t r c c

ct r

r

r R c c t

4. Mass balance of the

solute in the bath of

volume VB.

11

21 14

B r R

B eff

r R

dcV A n

dt

dc cV R D

dt r

10, 0t c

• Coupling of sphere

and bath

concentrations.

• This makes this

problem

interesting!!!

33

5-34

5. Solution [1. Carslaw and Jaeger (1986), 2. Crank (1975)]

The most useful result given is that for the concentration in the bath:

2

1,0

1 1,0 2 2 21

2 2

3

61 9 1

3tan

3

4

3

eff nD a t

n n

nn

n

B

c ec t Bc

B B R a B

R aR a

B R a

VB

R

ε = void fraction of the pellet

34

5-35

6. Numerical Methodology:

- We first calculate:

- We then read:

- And thus we obtain Deff!

1

1,0

and 1c

B Bc

2

effD t

R

35

lecture 5: generalized mass balances

Documents