Lecture 14 on Mass Transport

Louisiana Tech UniversityRuston, LA 71272

Slide 1

Mass Transport

Steven A. Jones

BIEN 501

Friday, April 13, 2007


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.


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.


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.


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.


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


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.


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)


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).


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


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


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.


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:


Slide 14

Subtle Point

If I have the following:

z=0Will there ever be any NO upstream of z=0?


Slide 15

Fick’s Law of Diffusion

AAA

A rt

jv

AABA Dj o

In

Use Fick’s Law

AAABA

A rDt

ovTo get


Slide 16


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.


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


Slide 18


Then:

o

1

1

A A

A AB A A

A

D rt

v


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


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


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.


Slide 22


z

cDJ

cDJ

One-dimensional Diffusion:

General Law:


Slide 23


ArRate of increase

Conservation of Mass:

),(),(),(

2

2

ztkCz

ztCD

t

ztCA

AAB

A

Diffusive transport


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.


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).


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


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


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

);(


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


Slide 30


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


Slide 31


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


Slide 32


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


Slide 33


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

),(


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

Lecture 14 on Mass Transport

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