Mass Transport
Steven A. Jones
BIEN 501
Friday, April 13, 2007
Louisiana Tech University
Ruston, LA 71272
Slide 1
Mass Transport
Major Learning Objectives:
1. Obtain the differential equations for mass
transfer.
2. Compare and contrast these equations
with heat transfer equations.
3. Apply the equations to nitric oxide
transport in the body.
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Ruston, LA 71272
Slide 2
Mass Transport
Minor Learning Objectives:
1. Review Continuum.
2. Describe various definitions of concentration.
3. Write down the equation for conservation of mass for a
single species and a multi-component system.
4. Obtain the solution for 1-dimensional diffusion of a
substance with a linear reaction.
5. Linearize a nonlinear partial differential equation.
6. Apply the Laplace transform to a partial differential
equation.
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Slide 3
Continuum Concept
Properties are averaged over small regions of space.
The number of blue circles moving left is larger
than the number of blue circles moving right.
We think of concentration as being a point value,
but it is averaged over space.
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Slide 4
Time Derivative
The Eulerian time derivative still applies.
dB B

 v  B 
dt
t
1. For mass transport, B is usually a concentration.
2. There are various ways to refer to concentration.
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Slide 5
Expressions for Concentration
Mass per Unit Volume.   A  g/cm3
Moles per Unit Volume. c A  moles/cm3
Example: NO has molecular weight of 30 (one nitrogen,
molecular weight 14 g/mole, one oxygen, molecular
weight 16 g/mole). A 1 mg/liter solution of NO has a
molar concentration of 1 mg/liter / 30 mg/mmole, or 33
nmole/liter.
Mass fraction.   A  is the mass of species A divided by the
total mass of all species.
Molar fraction x(A) is the number of moles of species A
divided by the total number of moles of all species.
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Slide 6
Expressions for Concentration
To convert mass concentration to molar concentration, it is
necessary to divide by the molar mass of the species.
Two species may have the same mass concentration, but
vastly different molar concentrations (e.g. NO vs. a large
protein. If they had the same mass concentration, the
molar concentration of the protein would be much smaller
than the molar concentration of the NO).
If we have a pure solution of something, then the mass
fraction is 1.
Similarly, molar fraction may be vastly different from the
mass fraction.
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Slide 7
Properties of a Multicomponent
Mixture
N
Density:      A
A1
1
N
Mass-Averaged Velocity: v      A v  A
A1
N
  A

v  A    A v  A
A1 
A1
N
N
Molar-Averaged Velocity:
v   x A  v  A 
o
A1
Rate of reaction (rate of production): r A  g/(cm3-s)
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Slide 8
Density of the System
N
To convince yourself that
     A
A1
Consider the following thought experiment. Add 1 liter of
water to 1 liter of another liquid with density 1.2 g/cm3.
Although the density of water is 1 g/cm3, the density of
water in the mixture is (1 liter x 1 g/cm3)/(2 liters). I.e., in
the mixture, there is 0.5 g/cm3 of water. Similarly, there is
0.6 g/cm3 of the other liquid. The density of the mixture is
then 0.5 g/cm3 + 0.6 g/cm3, or 1.1 g/cm3, as expected.
It is important to remember that (A) is the density in the
mixture (0.5 g/cm3 for water), not the density of the
substance itself (1 g/cm3).
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Slide 9
Density
+
=
1 Liter
1 Liter
2 Liter
Mass 1 Kg
Mass 1.2 Kg
Mass 2.2 Kg
=1 Kg/L
=1.2 Kg/L
=1.1 Kg/L
a =1 Kg/(2 L) = 0.5 Kg/L
b =1.2 Kg/(2 L) = 0.6 Kg/L
 = a + b
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Slide 10
Conservation of Mass
In terms of mass-averaged velocity, for a single species:
  A
t
     A v  A   r A
if n  A     A  v  A 
c A 
t
   c A  v  A   
  A
t
r A 
M  A
   n  A  r A
For a the entire system:
N

    v    r A   0
t
A1
Consequence of the individual
mass balances, not an
independent equation.
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The last equality is a result
of conservation of mass.
It would not hold for molar
concentration.
Slide 11
Species Velocity
In
c A 
t
   c A  v  A   
r A 
M  A
The difference between species velocity and mass
averaged velocity adds an increased level of complexity.
Consider two species, where one is dominant.
N
v    A v  A  1 v 1   2 v  2
A1


 1 v 1  1  1 v  2   v 1 because  2   0
 v 2 
v 1  1 v 1
1    

1
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v (1) 1  1 
1  1 
 v 1
I.e. velocity of both species
will be about the same.
Slide 12
Mass Flux
With respect to a fixed coordinate system.
With respect to a mass-averaged velocity.
j A     A  v  A   v 
Which leads to
  A
t
     A v     j A  r A
Compare to slide 11:
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Tells us how rapidly
substance A is moving
with respect to the
other substances.
  A
t
     A v  A   r A
Slide 13
Subtle Point
If I have the following:
z=0
Will there ever be any NO
upstream of z=0?
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Slide 14
Fick’s Law of Diffusion
In
  A
t
j A   D
  A
t
 A
o
 AB 
Use Fick’s Law
To get
     A v     j A  r A
     A v     DoAB   A   r A
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Slide 15
Fick’s Law of Diffusion
If density and diffusion coefficient are constant:
  A
Becomes
     A v     D
t
  A
t
o
 AB 
  A   r A
     A v   DoAB  2  A  r A
Which is like the equation for heat transfer. In many
cases, the mass-averaged velocity will be uncoupled
from the mass transport, so the equations can be
solved independently of one another.
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Slide 16
Sources of Nonlinearity
1. Reaction rates may be nonlinear. I.e., reaction
may depend on higher powers of concentration.
r A  k
2
 A
2. When concentrations are high, changes in mass
fractions may affect the overall density. E.g. for a
2-component system:
  

 B    x B    1   A   
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 B 
1   A
Slide 17
Fick’s Law of Diffusion
Then:
 1 

 A
o

     A v   
D AB  A   r A
 1 

t
 A


 A

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


Slide 18
Example – Transport of NO
Consider the diffusion of Nitric Oxide from a monolayer
of platelets:
cNO 0, z   0

NO  O2  ONOO 
2nd order reaction depends on O2
concentration.
NO
J NO t,0  J 0 ut  (where J0 is constant)
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Slide 19
Example – Transport of NO
The reaction between NO and O2- is nonlinear in that it is
the product of the NO concentration and the O2concentration, and as NO is consumed, so is O2-.
 
rA  k1 O2 NO
However, if we assume that the O2- concentration is
constant, then we can “linearize” the equation such that:
 
k  k1 O2
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 rA  k NO
Slide 20
Transport of NO
Modeling Assumptions
1. Platelets adhere in a monolayer along a wall placed at z=0 and
simultaneously begin to produce NO with constant flux in the positive
z – direction.
2. Diffusion obeys Fick’s law.
3. All densities and diffusion coefficients are constant.
4. The wall is infinite in width and height so that diffusion occurs in the
z – direction only.
5. No convection.
6. Consumption of NO by O2- follows a first order reaction (not really).
7. Constant flux of NO from the monolayer.
8. Initial NO concentration is zero.
9. Concentration of NO at infinity is zero.
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Slide 21
Fick’s Law of Diffusion
General Law:
J  Dc
One-dimensional Diffusion:
c
J  D
z
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Slide 22
Fick’s Law of Diffusion
Conservation of Mass:
C A (t , z )
 2 C A (t , z )
 D AB
 kCA (t , z )
2
t
z
Rate of increase
r A 
Diffusive transport
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Slide 23
Initial and Boundary Conditions
1. Initial concentration of NO throughout the medium is zero:
C (0, z )  0
2. The concentration must go to zero for large values of z.
C (t , )  0
3. The flux of NO through the surface at z=0 is constant and
equal to J0 for all t > 0, i.e.:
 D AB
C (t ,0)
 J 0 u t  at z  0.
z
Where u(t) is the unit step function.
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Slide 24
Laplace Transform Property
Recall that:
 df t 






L 

s
L
f
t

f
0

dt


We will use L(s,z) to represent the Laplace transform
of CA(t,z).
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Slide 25
Solution to the Governing Equation
Use the Laplace transform to transform the time
variable in the governing equations and boundary
conditions. The governing PDE transforms from:
C A (t , z )
 2 C A (t , z )
 D AB
 kCA (t , z )
2
t
z
To: sL( s, z )  C(0, z )  DAB L( s, z )  kL( s, z )
sk
 L( s, z )  0
Or: L( s, z )  
 DAB 
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d 2 Ls, z 
L( s, z ) 
dz 2
Slide 26
Transformed BC’s
1. Initial condition was already used when we transformed
the equation.
2. The concentration must go to zero for large values of z.
C(t, )  0  Ls,   0
3. The flux of NO through the surface at z=0 is constant and
equal to J0 for all t > 0, i.e.:
 D AB
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J0
C (t ,0)
Ls,0
 J 0 u t   D AB

z
z
s
J0

or L s,0  
D AB s
Slide 27
Match Boundary Conditions
With:

sk
L( s, z )  A exp   z
D AB


 sk
  B exp  z


D AB






The 2nd term will go to infinity for infinite z, so B=0.
Thus:

sk

L( s, z )  A exp   z
D AB





To apply constant flux, we must differentiate:
sk
L( s; z )   A
e
D AB
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z
s k
DAB
z 0
J0
J0

 A
D AB s
s D AB s  k
Slide 28
Invert the Laplace Transform
The solution for the Laplace transform is:
L( s, z ) 
J0
s DAB s  k
z
e
sk
D AB
So the solution for the concentration is the inverse
Laplace transform:
C A (t , z ) 
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J0
D AB
  z s k 
D AB


e
1
L 

s s  k 


Slide 29
Invert the Laplace Transform
In the Laplace transform, division by powers of s is the
same as integration in the time domain from 0 to t. The
result of this application is:
C A t , z  
J0
D AB
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  z s k 
DAB


e
1
L 

s s  k 


K
D AB
  z s k
t
DAB

e
1
L
0  s  k




 du


Slide 30
Invert the Laplace Transform
Now we can use the time shifting property of Laplace
transforms:
L
1
F ( s  k )  e ktL 1 F ( s)
C A t , z  
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J0
D AB
 z s
t
D AB

e
 ku
1
e
L

0
s




 du


Slide 31
Invert the Laplace Transform
Compare the form of the inverse Laplace transform in:
C A t , z  
J0
D AB
 z s
t
D AB

e
 ku
1
e
L

0
s




 du


To the following form from a table of Laplace transforms:
a2

4t
a s 

e
e
1
L 

t
 s 
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with a 
z
D AB
Slide 32
Invert the Laplace Transform
Then:
 z s
DAB

e
L1 
s


z2


 e 4 DABt

t


So:
J0
C A (t , z ) 
D AB
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t

0
e

z2 

 ku 

4
D
u
AB 

u
du
Slide 33
Example Concentration Profiles
0.006
5s
0.005
Increasing time
uM
0.004
0.003
0.002
0.001
0
0.1
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1
10
um
100
1 10
3
Slide 34