krogh cylinder

Louisiana Tech UniversityRuston, LA 71272

Slide 1

Krogh Cylinder

Steven A. JonesBIEN 501

Friday, April 20, 2007


Slide 2

Energy Balance

Major Learning Objectives:1. Learn a simple model of capillary

transport.


Slide 3

The Krogh Cylinder


Slide 4

Assumptions

• The geometry follows the Krogh cylinder configuration

• Reactions are continuously distributed• There is a radial location at which there is

no flux


Slide 5

Capillary Transport

Consider the following simple model for capillary transport:

Reactive Tissue

Capillary Interior

Matrix


Slide 6

Capillary Transport

0

0

2

2

O

O

J

c or

00 22JJCc OO or

What are appropriate reaction rates and boundary conditions?

Constant rate of consumption (determined by tissue metabolism, not O2 concentration

No reaction

continuous2Oc


Slide 7

Diffusion Equation

x

x

rrcr

rrD

tc

rcDtc

1

2

For steady state:

xrrcr

rrD

1


Slide 8

Constant Rate of Reaction

Mrcr

rrD

1

Assume the rate of reaction, rx, is constant:

And that the concentration is constant at the capillary wall and zero at the edge of the Krogh cylinder

0

0

k

c

RccRc

(M will be numerially negative since the substance is being consumed).


Slide 9

Constant Rate of ReactionBecause there is only one independent variable:

braD

Mrcra

DMr

drdc

aD

Mrdrdcr

DMr

drdcr

drd

ln42

22

2

Mdrdcr

drd

rDM

rcr

rrD

11

(Note the change from partial to total derivative)

Integrate once:

Divide by r and integrate again:


Slide 10

Constant Rate of ReactionBecause there is only one independent variable, :

aD

Mrdrdcr

DMr

drdcr

drd

2

2

Mdrdcr

drd

rDM

rcr

rrD

11

Integrate once:

raMr

drdcDJ

2Write in terms of flux:

d


Slide 11

Flux Boundary Condition at Rk

Since flux is 0 at the edge of the cylinder (Rk),

DMRa

RaMR

drdcD k

k

k

Rr k2

02

2

22

222

2

222

rRD

Mdrdcr

DMR

DMr

drdcra

DMr

drdcr

k

k

Substitute back into the differential equation:


Slide 12

Solution for Concentration

brrRD

Mc

rr

RD

Mdrdc

rRD

Mdrdcr

k

k

k

2ln

2

2

2

22

2

22

Substitute back into the differential equation:

Divide by r:

Integrate:


Slide 13

Boundary Condition at Capillary Wall

From the problem statement (Slide 9) cc(Rc) = c0

0

22

2ln

2cbRRR

DMRc c

ckc

2ln

2

22

0c

ckRRR

DMcb

2ln

22ln

2

22

0

22 c

ckkRRR

DMcrrR

DMrc

b


Slide 14

Simplify

Combine like terms and recalling that :

2ln

2

222

0c

ck

RrRrR

DMcrc

Or, in terms of partial pressures:

2ln

2

222 c

ckc

RrRrR

DMPrP

cc RrRr lnlnln


Slide 15

Plot of the Solution

Krogh Cylinder Solution

0

10

20

30

40

50

60

0 0.005 0.01 0.015 0.02 0.025

r (cm)

Con

cent

ratio

n (n

mol

es/L

)


Slide 16

Plot of the SolutionNote that the solution is not valid beyond rk.

Krogh Cylinder Solution

0

10

20

30

40

50

60

0 0.005 0.01 0.015 0.02 0.025 0.03

r (cm)

Con

cent

ratio

n (n

mol

es/L

)


Slide 17

Non Steady State

Mrcr

rrD

tc

McDtc

1

2Diffusion equation:

rtc allforat 00 Initial Condition:

0,),( 0

rtrcD

tuctrc

k

cBoundary Conditions:


Slide 18

Homogeneous Boundary Conditions

The problem will be easier to solve if we can make the boundary conditions homogeneous, i.e. of the form:

0,

0,

drtrf

trf

and/or

Our boundary condition at r = rc is not homogeneous because it is in the form:

0),(, 0 tuctrctrf c


Slide 19

Homogeneous Boundary Conditions

However, if we define the following new variable:

0

0,C

CCtr

The boundary condition at rc becomes:

0, trc

And the boundary condition at rk is still homogeneous: 0,

rtrk


Slide 20

Non-DimensionalizationThe new concentration variable also has the advantage of being dimensionless. We can non-dimensionalize the rest of the problem as follows:

Let:2,

cc rDt

rr

The boundary conditions become:

0,1 0,

rk

allforat 00 The initial condition becomes:


Slide 21

Non-Dimensionalization

Mrcr

rrD

tc

1

The diffusion equation can now be non-dimensionalized:

Use: 00 ,, CtrCtrC

So that:

Mr

rrr

DCt

C

Mr

CCrrr

Dt

CC

1

1

00

0000


Slide 22

Non-Dimensionalization (Continued)

Use:

To determine that:

2cc r

tDrr

2

1

c

c

rD

tt

rrr


Slide 23

Non-Dimensionalization (Continued)

Now apply:

To:

Drtrr c

c

2

,

Mr

rrr

DCr

DC

cc

ccc

111

020To get:

2,1

cc rD

trr

Mr

rrr

DCt

C

1

00


Slide 24

Non-Dimensionalization (Cont)

Mr

rrr

DCr

DC

cc

ccc

111

020Simplify

Mr

DCr

DC

cc

2020 1

Multiply by :0

2

DCrc

0

21DCMrc


Slide 25

Non-Dimensionalization (Cont)

Examine

0

21DCMrc

The term on the right hand side must be non-dimensional because the left hand side of the equation is non-dimensional. Thus, we have found the correct non-dimensionalization for the reaction rate.


Slide 26

The Mathematical Problem

The problem reduces mathematically to:

0

21DCMrc

Differential Equation

Boundary Conditions

Initial Condition

0,1 t 0,

rtk

00,


Slide 27

Change to Homogeneous

Diffusion equation:

,, gf Let:

0

21,1,DCMrf

rgg c

Then:

Follow the approach of section 3.4.1 (rectangular channel) to change the non-homogeneous equation to a homogeneous equation and a simpler non-homogeneous equation.

0

21DCMrc


Slide 28

Divide the EquationThe equation is solved if we solve both of the following equations:

0

21

0,1,

DCMrf

rgg

c

In other words, we look for a time-dependent part and a time independent (steady state) solution. Note that since the reaction term does not depend on time, it can be satisfied completely by the time-independent term.


Slide 29

Solution to the Spatial Part

Mrrfr

rD

1

We already know that the solution to:

Is:

ck

cc r

rrrrD

Mcrc ln22

222

And that this form satisfies the boundary conditions at rc and rk. It will do so for all time (because it does not depend on time).


Slide 30

Non-Dimensionalize the Spatial Part

In terms of the non-dimensional variables:

ln

222

22

0

2

kcc

DCMrf

And this form also becomes zero at the two boundaries.


Slide 31

Transient PartWe therefore require that:

0,1,

r

gg

With Boundary Conditions

And the Initial Condition

0, tg c 0,

r

tg k

00,0, gf

ln

220, 2

22

0

2

kcc

DCMrg


Slide 32

Boundary conditions

Recall what we did:

trfrrrrr

DMctrc

ck

cc ,ln

22, 2

2

And that:

trgrftrc ,,

rtc allforat 00 Initial Condition:

0,),( 0

rtrcD

tuctrc

k

cBoundary Conditions:


Slide 33

Boundary conditions for g

rrfrg

rrgrfallforallfor

0,00,Initial Condition:

0,0,

1,),( 00

rtrg

rtrg

rrf

tuctrgtuctrgrf

kkk

ccc

Boundary Conditions:

0

c0


Slide 34

It follows that we must solve

0,

1, 0

rtrg

tuctrg

k

c

r

trgrr

Dt

trg ,1,

The equation is solved if we solve both of the following equations:

With the following initial/boundary conditions:

crrrfrg allfor0,


Slide 35

Separation of Variables

trtrg TRthatAssume ,

01

rrr

rrtD

ttr RTTR

01

rrr

rrrD

tt R

RTT'

Homogeneous diffusion equation:

constantRRT

T'

srrr

rrrD

tt 1


Slide 36

The Two ODEs

Solutions:

0

0

1

22

22

rrDs

drrdRr

drrRdr

rrDs

drrdr

drd

sdr

rdrdrd

rrD

AetTstt st

R

RR

RR

TT'

What does this equation remind you of?


Slide 37

Radial Dependence

Could it perhaps be a zero-order Bessel Function?

0222

22

ypz

zyz

zyz

022

22 rr

Dq

drrdr

drrdr RRR

sDrzlet


Slide 38

Radial Dependence

2

2

2

2

dzdsD

dzdsD

dzdsD

drd

drd

drd

dzdsD

dzd

drdz

drd

sDrz let

022

22

zz

dzzdsD

sDz

dzzdsD

sDz RRR


Slide 39

Radial Dependence

022

22

zz

dzzdsD

sDz

dzzdsD

sDz RRR

0022

22 zRz

dzzdRz

dzzdz R


Slide 40

Radial Dependence

002

22 zRz

dzzdRz

dzzdz RSo since:

zBYzAJz 00 R

sDrBYsDrAJr 00 R

Note: When we used Bessel’s equation in Womersley flow, we did not use the Y0 term because it goes to

At z=0. However, in this case, we do not need to go to r=0, so we will keep it in the solution.


Slide 41

Flux Boundary Condition

In the solution we will have terms like:

We will be requiring the gradient of these terms to go to zero at rk. I.e.

stesDrYsBsDrJsA 00

011 stkk esDrYsDBsDrJsDA

The only way these terms can go to zero for all t is if:

011 sDrYsDBsDrJsDA kk

For every value of s.


Slide 42

Relationship between A and B

In other words:

sDrJ

sDrYsB

sDrJsDsDrYsD

sBsA

sDrYsDsBsDrJsDsA

k

k

k

k

kk

1

1

1

1

11 0

In contrast to problems we have seen before, in which the separation variable could take on only discrete values, here we can satisfy the boundary condition for any value of s.


Slide 43

Bessel Functions

-2

-1

0

1

2

0 2 4 6 8

x

Bes

sel F

unct

ion

J0Y0J1Y1


Slide 44

s = 0

We must also consider the case for s = 0.

rrrdr

rddr

rdrdr

rdrdrd

drrdr

drd

rrD

tTtt

ln

0

01

0

RR

RR

RR

TT'

But this expression is already a part of the steady state solution.


Slide 45

Complete Solution

m

tsmmmm

ck

cc

mesDrYBsDrJA

rrrrr

DMctrc

00

22

ln22

,

If s could take on only discrete values, the complete solution would be:


Slide 46

Complete Solution

dsesDrYsBsDrJsA

rrrrr

DMctrc

st

ck

cc

0 00

22

ln22

,

However, since s can take on any value, the complete solution must be:


Slide 47

Complete Solution

dsesDrYsDrJ

sDrJsDrY

sB

rrrrr

DMctrc

st

k

k

ck

cc

0 001

1

22

ln22

,

And since: sDrJ

sDrYsBsA

k

k

1

1


Slide 48

Initial Condition

0

ln22

0,

0 001

1

22

dtsDrYsDrJsDrJsDrY

sB

rrrrr

DMcrc

k

k

ck

ccThe initial condition is:

ck

cc

k

k

rrrrr

DMc

dssDrYsDrJsDrJsDrY

sB

ln22

22

0 001

1


Slide 49

Boundary Condition at r = rc

tucdsesDrYsDrJ

sDrJsDrY

sB

rrrrr

DMctrc

cst

cck

k

c

ck

cccc

0 001

1

22

ln22

,

tucdsesDrYsDrJ

sDrJsDrY

sBc cst

cck

kc

0 001

1


Slide 50

Boundary Condition at r = rc

1

0 001

1

tucdtesDrYsDrJsDrJsDrY

sB cst

cck

k

11

001

1

tucsDrYsDrJ

sDrJsDrY

sB ccck

ktL


Slide 51

What about the similarity solution

As it turns out, we did not need to abandon the similarity solution. We could have done the same thing we did with the separation of variables solution. I.e. we could have said that the complete solution is the sum of the particular solution and a sum of similarity solutions.


Slide 52

Counter-Current Exchange

zM d dzzM d

zM b dzzM b

zMd

ddddddd dCQzCdzzCQzMdzzMMd

mDB dACCkMd 0

Conservation of Mass

Membrane Diffusion:


Slide 53

Counter-Current Exchange

dzCdQ

dzMd

d

dzdACk

dzMd m 0

Conservation of Mass might better be written as:

And Membrane Diffusion as:

DB CCC Where:


Slide 54

Combine the Two Equations

dzdACk

dzCdQ m

d

0

z

QWkCCd

00 exp

Solution:

Where: WdzdA

constant


Slide 55

Integrate w.r.t. z

dzCdQ

dzMd

d

dzdACk

dzMd m 0


Slide 56


Slide 57


Slide 58

Differential Form

xx

capacityheatpC

xz

y

x

zvDescribes how much a volume of material will increase in temperature with a given amount of heat input.

I.e., if I add x number of Joules to a volume 1 cm3, it will increase in temperature by T degrees.

TJoules in


Slide 59

Thermal Conductivity

k – defines the rate at which heat “flows” through a material.

Fourier’s law (A hot cup of coffee will become cold). Fourier’s law is strictly analogous to Fick’s law for diffusion.

Tkq

q is flux, i.e. the amount of heat passing through a surface per unit area.

Gradient drives the flux.


Slide 60

Heat Production, Example

VIP

RVP

2

LR2

Consider the case of a resistor:

V1 V2

The resistor dissipates power according to ,

(where I is the current) or, equivalently

The volume of the resistor is . Therefore, at any spatial location within the resistor, it is generating:

LRaV2

2

Joules/s/cm3


Slide 61

Boundary Conditions

• Constant temperature (T=T0)• Constant flux

• Heat transfer:

• More general:– Surface Condition or

on the closed surface.– Initial condition within the

volume.

0JTk n

TThq w

),,(0,,, 0 zyxTzyxT

tzyxTT ,,, tzyxJJ ,,,


Slide 62

Boundary Conditions

• Constant temperature (T=T0)• Constant flux

• Heat transfer:

• More general:– Surface Condition or

on the closed surface.– Initial condition within the

volume.

0JTk n

TThq w

),,(0,,, 0 zyxTzyxT

tzyxTT ,,, tzyxJJ ,,,


Slide 63

Semi-Infinite Slab (of marble)

2z

02 z

Flow of Heat

T=T1

T=T0 uniform

How does temperature change with time?

Initial Temperature Profile


Slide 64


tzatTTztatTT

0

00

21

20

TktTC p

Differential Equation (no source term):



Slide 65


tzTTztTT

0

00

21

20

atat

22

2

zTk

tTC p

For 1-dimensional geometry and constant k:

This problem is mathematically identical to the fluid flow near an infinitely long plate that is suddenly set in motion.


Slide 66


01

0*

TTTTT

0TT

Dimensionless Temperature:

Dimensionless temperature describes the difference between temperature and a reference temperature with respect to some fixed temperature difference, in this case T1 – T0. 01 TT


Slide 67


pCk

zT

tT

where,2

2

*2*

0TT

0,01

00

2*

*

tzT

tT

for

for

Equations in Terms of Dimensionless Temperature:


01 TT

T* is valuable because it nondimensionalizes the equations and simplifies the boundary conditions.


Slide 68


tz

2

Assume a similarity solution, and define:

. We make use of the following relationships:

23212

23212

22 tz

tz

tt

ttzz11

22

2

2

22222

2 1111 ttttzzzz


Slide 69

Semi-Infinite Slab (of marble)With these substitutions, the differential equation becomes:

012 2

*2*

23212

Tt

Tt

z

This can be divided by 2321

2 2 tz

to yield:02

2

*2

2

*

T

ztT

The combination tz 2 can now be replaced with

to give: 022

*2*

TT

.


Slide 70

Semi-Infinite Slab (of marble)Instead of trying to solve directly for T*, try to solve for the first derivative:

*T

.

02

dd

,exp2

41

12

41

*

CeC

ddT

The equation is rewritten as:

which is separated as: dd

2

.

Thus: 12

41ln C

.

So:


Slide 71

Semi-Infinite Slab (of marble)Integrate:

.

,exp2

41

12

41

*

CeCddT

.

To obtain:

0041 2

CdCec

.

This integral cannot be evaluated in closed form by standard methods. However, it is tabulated in handbooks and it can be evaluated under standard software. The integral is called the Error function (because of it’s origins in probability theory, where Gaussian functions are important).


Slide 72

Newton’s Law of Cooling

Often we must evaluate the heat transfer in a body that is in contact with a fluid (e.g. heat dissipation from a jet engine, cooling of an engine by a radiator system, heat loss from a cannonball that is shot through the air). The boundary between the solid and fluid conforms neither to a constant temperature, nor to a constant flux. We make the assumption that the rate of heat loss per unit area is governed by:

TTh snq


Slide 73

Newton’s Law of Cooling

The heat transfer coefficent, h, is a function of the velocity of the cannonball. I.e. the higher the velocity, the more rapidly heat is extracted from the cannonball.

One generally assumes that the heat transfer coefficient does not depend on temperature. However, in free convection problems, it can be a strong function of temperature because the velocity of the fluid depends on the fluid viscosity, which depends on temperature.

TTh snq


Slide 74

Free Convection

In free convection, the fluid moves as a result of heating of the fluid near the body in question.

Fluid becomes less dense near the body. Bouyency causes it to move up, enhancing transfer of heat.


Slide 75

Forced Convection

• In forced convection (vs. free convection), the velocity is better controlled because it does not depend strongly on the heat flow itself. A fan, for example, controls the velocity of the fluid.


Slide 76

Convection vs. Conduction

• Convection enhances flow of heat by increasing the temperature difference across the boundary.

Because “hot” fluid is removed from near the body, fluid near the body is colder, therefore the temperature gradient is higher and heat transfer is higher, by Fourier’s law. In other words, Fourier’s law still holds at the boundary.

Small gradient, small heat transfer.

Large gradient, large heat transfer.


Slide 77


Slide 78

Similarity Solution?

Dtrr

Dtr

tt

DtrDt

rtDt

r

4142

141,

421

4

3

3We could attempt a similarity solution:

McDt

rDtr

Dc

Dt

r

Mrcr

rrD

tc

41

411

421

1

3

Which transforms the equation to:


Slide 79

Transform Variables

DrMcc 2

3

McDt

rDtr

DcDtr

41

411

421

3From Previous:

Multiply by r2/D

DrMcc

Dtr

Mcrr

Dct

22

2

121


Slide 80

The Problem

DrMcc 2

3

From Previous:

If this had been a “no reaction” problem, the method would work. Unfortunately, the reaction term prevents the similarity solution from working, so we need to take another approach.

krogh cylinder

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