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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
2. Step: Simplify mass balance
• 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. Step: Simplify mass balance
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
2. Step: Simplify mass balance
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.
1. Step: Select appropriate mass balance
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. Step: Simplify mass balance
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
1. Step: Select appropriate mass balance
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