Mass Transport
description
Transcript of Mass Transport
Louisiana Tech UniversityRuston, LA 71272
Slide 1
Mass Transport
Steven A. Jones
BIEN 501
Friday, April 13, 2007
Louisiana Tech UniversityRuston, LA 71272
Slide 2
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.
Louisiana Tech UniversityRuston, LA 71272
Slide 3
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.
Louisiana Tech UniversityRuston, LA 71272
Slide 4
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.
Louisiana Tech UniversityRuston, LA 71272
Slide 5
Time Derivative
The Eulerian time derivative still applies.
Bt
B
dt
dB
v
1. For mass transport, B is usually a concentration.
2. There are various ways to refer to concentration.
Louisiana Tech UniversityRuston, LA 71272
Slide 6
Expressions for Concentration
Ac
AMass per Unit Volume. g/cm3
Moles per Unit Volume. 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 g/liter solution of NO has a molar concentration of 1 g/liter / 30 g/mole, or 33 nmole/liter.
Mass fraction. 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.
A
Louisiana Tech UniversityRuston, LA 71272
Slide 7
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.
Louisiana Tech UniversityRuston, LA 71272
Slide 8
Properties of a Multicomponent Mixture
Density:
N
AA
1
Mass-Averaged Velocity:
N
AAA
N
AA
AN
AAA
111
1vvvv
Rate of reaction (rate of production): Ar
Molar-Averaged Velocity:
N
AAAx
1
vvo
g/(cm3-s)
Louisiana Tech UniversityRuston, LA 71272
Slide 9
Density of the System
To convince yourself that
N
AA
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).
Louisiana Tech UniversityRuston, LA 71272
Slide 10
Density
+
1 Liter
Mass 1 Kg
=1 Kg/L
1 Liter
Mass 1.2 Kg
=1.2 Kg/L
=
2 Liter
Mass 2.2 Kg
=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
Louisiana Tech UniversityRuston, LA 71272
Slide 11
Conservation of Mass
In terms of mass-averaged velocity, for a single species:
01
N
AArt
v
For a the entire system:
AAA
A rt
v
The last equality is a result of conservation of mass.
It would not hold for molar concentration.
AAA vn if
AAA rt
n
Consequence of the individual mass balances, not an independent equation.
A
AAA
A
M
rc
t
c
v
Louisiana Tech UniversityRuston, LA 71272
Slide 12
Species Velocity
A
AAA
A
M
rc
t
c
v
The difference between species velocity and mass averaged velocity adds an increased level of complexity. Consider two species, where one is dominant.
In
1 1 2 21
1 1 1 2 1 2
1 1 1 (1) 12 1
11
1 because 0
1
11
N
A AA
v v v v
v v v
v v vv v
I.e. velocity of both species will be about the same.
Louisiana Tech UniversityRuston, LA 71272
Slide 13
Mass Flux
With respect to a fixed coordinate system.
With respect to a mass-averaged velocity.
vvj AAA
Which leads to
AAA
A rt
jv
Tells us how rapidly substance A is moving with respect to the other substances.
AAA
A rt
v
Compare to slide 11:
Louisiana Tech UniversityRuston, LA 71272
Slide 14
Subtle Point
If I have the following:
z=0Will there ever be any NO upstream of z=0?
Louisiana Tech UniversityRuston, LA 71272
Slide 15
Fick’s Law of Diffusion
AAA
A rt
jv
AABA Dj o
In
Use Fick’s Law
AAABA
A rDt
ovTo get
Louisiana Tech UniversityRuston, LA 71272
Slide 16
Fick’s Law of Diffusion
If density and diffusion coefficient are constant:
Becomes
AAABA
A rDt
ov
AAABA
A rDt
2ov
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.
Louisiana Tech UniversityRuston, LA 71272
Slide 17
Sources of Nonlinearity
2AA kr
1. Reaction rates may be nonlinear. I.e., reaction may depend on higher powers of concentration.
2. When concentrations are high, changes in mass fractions may affect the overall density. E.g. for a 2-component system:
1
1B
B B AA
x
Louisiana Tech UniversityRuston, LA 71272
Slide 18
Fick’s Law of Diffusion
Then:
o
1
1
A A
A AB A A
A
D rt
v
Louisiana Tech UniversityRuston, LA 71272
Slide 19
Example – Transport of NO
Consider the diffusion of Nitric Oxide from a monolayer of platelets:
ONOOONO 2
NO
2nd order reaction depends on concentration.
0,0 zcNO
tuJtJ NO 00, (where J0 is constant)
2O
Louisiana Tech UniversityRuston, LA 71272
Slide 20
Example – Transport of NO
The reaction between NO and O2- is nonlinear in that it is
the product of the NO concentration and the O2-
concentration, and as NO is consumed, so is O2-.
NOOkrA 21
However, if we assume that the O2- concentration is
constant, then we can “linearize” the equation such that:
NOkrOkk A 21
Louisiana Tech UniversityRuston, LA 71272
Slide 21
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.
Louisiana Tech UniversityRuston, LA 71272
Slide 22
Fick’s Law of Diffusion
z
cDJ
cDJ
One-dimensional Diffusion:
General Law:
Louisiana Tech UniversityRuston, LA 71272
Slide 23
Fick’s Law of Diffusion
ArRate of increase
Conservation of Mass:
),(),(),(
2
2
ztkCz
ztCD
t
ztCA
AAB
A
Diffusive transport
Louisiana Tech UniversityRuston, LA 71272
Slide 24
Initial and Boundary Conditions1. Initial concentration of NO throughout the medium is zero:
0),0( zC
2. The concentration must go to zero for large values of z.
0),( tC3. The flux of NO through the surface at z=0 is constant and
equal to J0 for all t > 0, i.e.:
.0)0,(
0
ztuJ
z
tCDAB at
Where u(t) is the unit step function.
Louisiana Tech UniversityRuston, LA 71272
Slide 25
Laplace Transform Property
Recall that:
0ftfsdt
tdf
LL
We will use L(s,z) to represent the Laplace transform of CA(t,z).
Louisiana Tech UniversityRuston, LA 71272
Slide 26
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:
),(),(),0(),( zskLzsLDzCzssL AB
),(),(),(
2
2
ztkCz
ztCD
t
ztCA
AAB
A
To:
Or: 0),(),(
zsLD
kszsL
AB
2
2 ,),(
dz
zsLdzsL
Louisiana Tech UniversityRuston, LA 71272
Slide 27
Transformed BC’s1. Initial condition was already used when we transformed the equation.
2. The concentration must go to zero for large values of z.
0,0),( sLtC
3. The flux of NO through the surface at z=0 is constant and equal to J0 for all t > 0, i.e.:
sD
JsL
s
J
z
sLDtuJ
z
tCD
AB
ABAB
0
00
0,
0,)0,(
or
Louisiana Tech UniversityRuston, LA 71272
Slide 28
Match Boundary Conditions
With:
ABAB D
kszB
D
kszAzsL expexp),(
The 2nd term will go to infinity for infinite z, so B=0.
To apply constant flux, we must differentiate:
Thus:
ABD
kszAzsL exp),(
ksDs
JA
sD
Je
D
ksAzsL
ABABz
D
ksz
AB
AB
00
0
);(
Louisiana Tech UniversityRuston, LA 71272
Slide 29
Invert the Laplace Transform
ABD
ksz
AB
eksDs
JzsL
0),(
The solution for the Laplace transform is:
So the solution for the concentration is the inverse Laplace transform:
kss
e
D
JztC
ABD
ksz
AB
A10),( L
Louisiana Tech UniversityRuston, LA 71272
Slide 30
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:
duks
e
D
K
kss
e
D
JztC
t D
ksz
AB
D
ksz
AB
A
ABAB
0
110, LL
Louisiana Tech UniversityRuston, LA 71272
Slide 31
Invert the Laplace Transform
Now we can use the time shifting property of Laplace transforms:
)()( 11 sFeksF kt LL
t D
sz
ku
AB
A dus
ee
D
JztC
AB
0
10, L
Louisiana Tech UniversityRuston, LA 71272
Slide 32
Invert the Laplace Transform
Compare the form of the inverse Laplace transform in:
t D
sz
ku
AB
A dus
ee
D
JztC
AB
0
10, L
To the following form from a table of Laplace transforms:
t
e
s
e t
asa
41
2
L
ABD
za with
Louisiana Tech UniversityRuston, LA 71272
Slide 33
Invert the Laplace Transform
Then:2
41
AB AB
s zzD D te e
Ls t
So:
t uD
zku
AB
A duu
e
D
JztC
AB
0
40
2
),(
Louisiana Tech UniversityRuston, LA 71272
Slide 34
Example Concentration Profiles
0.1 1 10 100 1 1030
0.001
0.002
0.003
0.004
0.005
0.006
um
uM
Increasing time
5 s