Krogh Cylinder

Louisiana Tech UniversityRuston, LA 71272

Slide 1

Krogh Cylinder

Steven A. Jones

BIEN 501

Wednesday, May 7, 2008


Slide 2

Announcements

1. All homeworks have been assigned.2. Final exam will be taken from parts of the

homework.3. No homework on Krogh cylinder.4. Krogh cylinder will not be on the exam.5. Friday – finish Krogh and do Comparmental

Models6. Monday – Review and Course Evaluations7. Wednesday - Exam8. Friday – No class, will be available for

questions.9. Today – Office Hours will start at 10:30.


Slide 3

Energy Balance

Major Learning Objectives:

1. Learn a simple model of capillary transport.


Slide 4

The Krogh Cylinder

Capillary

Tissue


Slide 5

Assumptions

• The geometry follows the Krogh cylinder configuration– Radial symmetry– Transport from capillary

– Capillary influences a region of radius Rk.

• Reactions are continuously distributed

• There is a radial location at which there is no flux


Slide 6

Capillary Transport

Consider the following simple model for capillary transport:

Reactive Tissue

Capillary Interior

Matrix


Slide 7

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)

Not metabolic (no consumption)

continuous2Oc


Slide 8

Diffusion Equation

x

x

rr

cr

rrD

t

c

rcDt

c

1

2

For steady state:

rcrrr

rcr

rrD x ,

1


Slide 9

Constant Rate of Reaction

Mr

cr

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

r

RcDRJ

cRc

kk

c

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


Slide 10

Constant Reaction: Roadmap

• Integrate the differential equation once to obtain a solution for flux as a function of r.

• Use the zero flux boundary condition at Rk to determine the first constant of integration.

• Substitute the constant back into the previously-integrated differential equation.

• Integrate again to obtain the form for concentration.• Use the constant concentration boundary condition to

determine the second constant of integration.

The roadmap for solving the steady state problem is as follows:


Slide 11

First Integration

Because there is only one independent variable, :

aD

Mr

dr

dcr

D

Mr

dr

dcr

dr

d

2

2

Mdr

dcr

dr

d

rDM

r

cr

rrD

11

Integrate once:

2

dc Mr aDJ D

dr r Write in terms of flux:

d

We will use this form to satisfy the no-flux boundary condition at Rk.


Slide 12

Flux Boundary Condition at Rk

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

2

02 2

k

k k

r R k

MR MRdc aDD a

dr R D

22

222

2

222

rRD

M

dr

dcr

D

MR

D

Mr

dr

dcra

D

Mr

dr

dcr

k

k

Substitute back into the (once-integrated) differential equation (boxed equation in slide 10):


Slide 13

Second Integration for Concentration

br

rRD

Mc

rr

R

D

M

dr

dc

rRD

M

dr

dcr

k

k

k

2ln

2

2

2

22

2

22

With the previous differential equation:

Divide by r:

Integrate:


Slide 14

Boundary Condition at Capillary Wall

From the problem statement (Slide 9) c(Rc) = c0. Evaluate the previous solution for c(r) at r=Rc.

0

22

2ln

2cb

RRR

D

MRc c

ckc

2ln

2

22

0c

ck

RRR

D

Mcb

2ln

22ln

2

22

0

22 c

ckk

RRR

D

Mc

rrR

D

Mrc

b

Solve for b.


Slide 15

Simplify

Combine like terms, recalling that :

2ln

2

222

0c

ck

Rr

R

rR

D

Mcrc

Or, in terms of partial pressures:

2ln

2

222 c

ckc

Rr

R

rR

D

MPrP

cc RrRr lnlnln


Slide 16

Krogh Cylinder Solution

-30-20-10

0102030405060

0 0.01 0.02 0.03r (cm)

Co

nc

en

tra

tio

n

(nm

ole

s/L

)

Critical Supply

Starvation

Ample Supply

Efffect of Different M Values

cR

kR


Slide 17

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)

Co

nce

ntr

atio

n

(nm

ole

s/L

)


Slide 18

Finding Rk

The steady state equation is a function of Rk, but an important question is, “What is Rk, given a certain metabolic rate?”

Non-starvation: Halfway between capillaries.

Starvation: Is the solution still valid?


Slide 19

Non Steady State

Mr

cr

rrD

t

c

McDt

c

1

2Diffusion equation:

rtc allforat 00 Initial Condition:

0,

),( 0

r

trcD

tuctrc

k

cBoundary Conditions:


Slide 20

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,

dr

trf

trf

and/or

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

0),(, 0 tuctrctrf c


Slide 21

Homogeneous Boundary Conditions

However, if we define the following new variable:

0

0

,c C

r tC

The boundary condition at rc becomes:

, 0 for 0cr t t

And the boundary condition at rk is still homogeneous:

0

,

r

trk


Slide 22

Non-Dimensionalization

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

Let:2

,cc r

Dt

r

r

The boundary conditions become:

0,1 0

,

rk

allforat 00 The initial condition becomes:

Why are these forms obvious?


Slide 23

Non-Dimensionalization

Mr

cr

rrD

t

c

1

The diffusion equation can now be non-dimensionalized:

Use: 0 0, ,c r t C r t C

So that:

Mr

rrr

DCt

C

Mr

CCr

rrD

t

CC

1

1

00

0000


Slide 24

Non-Dimensionalization (Continued)

Use:

To determine that:

2cc r

tD

r

r

2

1

c

c

r

D

tt

rrr


Slide 25

Non-Dimensionalization (Continued)

Now apply:

To:

D

rtrr c

c

2

,

Mr

rrr

DCr

DC

cc

ccc

111

020To get:

2,

1

cc r

D

trr

Mr

rrr

DCt

C

1

00


Slide 26

Non-Dimensionalization (Cont)

Mr

rrr

DCr

DC

cc

ccc

111

020Simplify

Mr

DCr

DC

cc

2020 1

Multiply by :0

2

DC

rc

0

21

DC

Mrc


Slide 27

Non-Dimensionalization (Cont)

Examine

0

21

DC

Mrc

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 28

The Mathematical Problem

The problem reduces mathematically to:

0

21

DC

Mrc

Differential Equation

Boundary Conditions

Initial Condition

0,1 t 0

,

r

tk

00,


Slide 29

Change to Homogeneous

Diffusion equation:

,, gf Let:

0

21,1,

DC

Mrf

r

gg 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

21

DC

Mrc


Slide 30

Divide the Equation

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

0

21

0,1,

DC

Mrf

r

gg

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 31

Solution to the Spatial Part

M

r

rfr

rD

1

We already know that the solution to:

Is:

ck

cc r

rr

rr

D

Mcrc ln

222

22

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 32

Non-Dimensionalize the Spatial Part

In terms of the non-dimensional variables:

ln

222

22

0

2

kcc

DC

Mrf

And this form also becomes zero at the two boundaries.


Slide 33

Transient Part

We 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

DC

Mrg


Slide 34

Separation of Variables

TRthatAssume ,g

0

1

R

TT

R

0

1

R

RT

T'

Homogeneous diffusion equation:

constant

R

RT

T'

21

Function of onlyFunction of only


Slide 35

The Two ODEs

Solutions:

0

0

1

222

22

2

2

2 2

R

RR

R

R

T

T'

d

dR

d

Rd

d

d

dr

d

d

d

d

d

AeT

What does this equation remind you of?


Slide 36

Radial Dependence

Could it perhaps be a zero-order Bessel Function?

0222

22

ypzz

yz

z

yz

0222

22

R

RR

d

d

d

d

zLet


Slide 37

Radial Dependence

2

22

2

2

dz

d

dz

d

dz

d

d

d

d

d

d

d

dz

d

dz

d

d

dz

d

d

zlet

022

22

2

2

zz

dz

zdz

dz

zdzR

RR

Differential equation for radial dependence becomes:

So the solution is: 00 BYAJ R


Slide 38

Radial Dependence

022

22

2

2

zz

dz

zdz

dz

zdzR

RR

This is Bessel’s equation, so the solution is:

zzBYzAJz withR 00

022

22 zz

dz

zdz

dz

zdz R

RR

00 BYAJ Ror


Slide 39

Bessel Functions

-1

0

1

0 2 4 6 8

z

Be

ss

el F

un

cti

on

J0Y0J1Y1


Slide 40

Derivatives of Bessel Functions

The derivatives of Bessel functions can be obtained from the general relations:

xYxYx

n

dx

xdY

xJxJx

n

dx

xdJ

nnn

nnn

1

1

Specifically:

xYdx

xdYxJ

dx

xdJ 1

01

0 ,


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 k. I.e.

2

00 eYBJA

02

11 eYBJA kk

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

011 kk YBJA

For every value of .


Slide 42

Relationship between A and B

In other words:

k

k

J

YBA

1

1

011 kk YBJA

And for the boundary condition at the capillary wall:

0

011

)1(0,

001

1

00

YJJ

YB

BYAJ

c

k

k

cc recall


Slide 43

Characteristic Values

000

1

1

YJJ

YB

k

kFrom

We conclude that the allowable values of are those which satisfy:

00101 YJJY kk

This is not as simple as previous cases, where the Y0 term became zero, but it is possible to find these values from MatLab or Excel, given a value for k.


Slide 44

The Characteristic Function (k=10)

-0.4

-0.2

0

0.2

0.4

0.6

0.8

0 1 2 3 4 5

f( )

0101 YJJYf kk


Slide 45

To Calculate (and Plot) in Excel

0.1 =BESSELY(A1*etak,1)*BESSELJ(A1,0)

- BESSELJ(A1*etak,1)*BESSELY(A1,0)








Slide 46

Finding the Roots

Use “Tools | Goal Seek” to find the roots of the equation. For example, the plot indicates that one root is near =1.2. With the value 1.2 in cell A1 and the formula in cell A2:

Set cell: A2To value: 0By changing cell: A1

Goal Seek

OK Cancel

“OK” will change the value of cell A1 to the 4th root, 0.110266.


Slide 47

First Six Roots

0.110266

0.497895

0.855445

1.208701

1.560322

1.8

This process gives the following value for the first 6 roots:

n1

2

3

4

5

6

n


Slide 48

Complete Solution

The complete solution has the form:

100

1

12

,n

nnkn

knn

neYJJ

YBc

Where the values of n are obtained as described above.


Slide 49

A Prettier Form

We obtain the somewhat more aesthetically pleasing form:

1

0101

2

,n

nknnknnneYJJYc

If we define:

kn

nn J

B

1


Slide 50

Initial Condition

Now apply the initial condition:

ln

220, 2

22

0

2

kcc

DC

Mrg

1

01010,n

nknnknn YJJYc

ln22

222

0

2

10101

kcc

nnknnknn

DC

Mr

YJJY

To:

So:


Slide 51

Orthogonality

For simplicity, define:

0 01, , 0 if

k

n m d m n

C C

0 1 0 1 0, n n k n n k nY J J Y C

These are called “Cylindrical Functions.” They satisfy the radial Laplacian operator, are zero for =1, and have zero derivative for =. According to the Sturm-Liouville Theorem, the following orthogonality relationship holds for these functions:


Slide 52

Sturm-Liouville

2

2

, ,, 0,

d x d xp x p x q x w x x

dx dxa x b

2

2 2 22

R 0d R dR

d d

Compare:

To:

Note that is not the first derivative of 2, but if we divide by :

2

22

R 0d R dR

d d

and 1 is the

1st derivative of


Slide 53

Sturm-Liouville

2

2

, ,, 0,

d dp p q w

dx dxa b

Compare:

To:

2

22

R 0d R dR

d d

, 1, 0,p p q w

The w() is particularly important because it tells us the weighting factor.


Slide 54

Initial Condition

With the initial condition:

2 222

01 0

, ln2 2

ccn n k

n

Mr

DC

C

Multiply both sides by and integrate:

0 011

2 222

010

, ,

ln ,2 2

k

k

n n mn

cck m

d

Mrd

DC

C C

C

0 , m C


Slide 55

Initial Condition

Now use the orthogonality condition:

201

2 222

010

,

ln ,2 2

k

k

n n

cck n

d

Mrd

DC

C

C

These integrals can be found in tables and worked out to determine the values for the n’s, from which the final answer is obtained.


Slide 56

Fourier Bessel Series

For Example:

2 2

2 2 2 22 2 2 2 2 2

2 2 2 2

b

n na

n n n n n n

d

b a b ab a b a

C

C C C C


Slide 57

Integral of Cylindrical Functions

To obtain the previous result, take Bessel’s Equation, substituting in the cylindrical function, and multiply by and then integrate. /nd d C

2

2 2 22

0n nn n

d d

d d

C CC

22

2

2 2 0

b bn n n n

a a

b nn na

d d d dd d

d d d d

dd

d

C C C C

CC


Slide 58

Norm of Cylindrical Functions

22 22

2 2

b bn n n

a a

d d ddd d

d d d d

C C C

For the first term:

Integrate by parts with:

2

2 2

2

n n

u du d

d dddv d v

d d d

C C


Slide 59

Integration by Parts, First Term

22

1

2 22

2

2

b n

a

b

bn n

a

a

ddI d

d d

d dd

d d

C

C C

22 2

2 2

2 2

b nn n a

db ab a d

d

C

C C


Slide 60

Integration by Parts, Second Term

2

2

b n

a

dI d

d

C

22

22 21 2

b nn n a

dI b a d

d

CC C

The second integral of Slide 54 is:

But:

So the second integral of Slide 54 is cancels with the last term of I1, leaving:

2 2

2 21 2 2 2n n

b aI I b a C C


Slide 61

Last Term

222 2 2

3 2

b bn nn n na a

d dI d d

d d

C C

C

For the last term:

22

2, ; ,2

nn

du du d dv d v

d

C

C

22 2

2 2 2 2 2

2 2

bb bn

n n n n na aa

dd d

d

C

C C


Slide 62

Last Term

2 2

2 2 2 2 2 23 2 2

b

n n n n n na

b aI b a d C C C


Slide 63

Combine the Three Integrals

2 22 2

1 2 3

2 22 2 2 2 2 2

2 2

02 2

n n

b

n n n n n na

b aI I I b a

b ab a d

C C

C C C

2 2

2 2 2 22 2 2 2 2 2

2 2 2 2

b

n na

n n n n n n

d

b a b ab a b a

C

C C C C

The expression will simplify further, given that we will be working with problems for which either the function or its derivative is zero at each boundary.


Slide 64

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 65


Slide 66

Similarity Solution?

Dtrr

Dt

r

tt

DtrDt

r

tDt

r

4

142

1

4

1,

42

1

4

3

3We could attempt a similarity solution:

Mc

Dtr

DtrD

c

Dt

r

Mr

cr

rrD

t

c

4

1

4

11

42

1

1

3

Which transforms the equation to:


Slide 67

Transform Variables

D

rM

cc 23

Mc

Dtr

DtrD

c

Dt

r

4

1

4

11

42

13

From Previous:

Multiply by r2/D

D

rM

cc

Dt

r

Mc

rrD

c

t22

2

1

2

1


Slide 68

The Problem

D

rM

cc 23

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.

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Krogh Cylinder