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Louisiana Tech UniversityRuston, LA 71272
Slide 1
Krogh Cylinder
Steven A. Jones
BIEN 501
Wednesday, May 7, 2008
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
Slide 3
Energy Balance
Major Learning Objectives:
1. Learn a simple model of capillary transport.
Louisiana Tech UniversityRuston, LA 71272
Slide 4
The Krogh Cylinder
Capillary
Tissue
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
Slide 6
Capillary Transport
Consider the following simple model for capillary transport:
Reactive Tissue
Capillary Interior
Matrix
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
Slide 8
Diffusion Equation
x
x
rr
cr
rrD
t
c
rcDt
c
1
2
For steady state:
rcrrr
rcr
rrD x ,
1
Louisiana Tech UniversityRuston, LA 71272
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).
Louisiana Tech UniversityRuston, LA 71272
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:
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
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):
Louisiana Tech UniversityRuston, LA 71272
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:
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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
)
Louisiana Tech UniversityRuston, LA 71272
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?
Louisiana Tech UniversityRuston, LA 71272
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:
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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?
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
Slide 24
Non-Dimensionalization (Continued)
Use:
To determine that:
2cc r
tD
r
r
2
1
c
c
r
D
tt
rrr
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
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,
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
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).
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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?
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
Slide 39
Bessel Functions
-1
0
1
0 2 4 6 8
z
Be
ss
el F
un
cti
on
J0Y0J1Y1
Louisiana Tech UniversityRuston, LA 71272
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 ,
Louisiana Tech UniversityRuston, LA 71272
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 .
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
Slide 45
To Calculate (and Plot) in Excel
0.1 =BESSELY(A1*etak,1)*BESSELJ(A1,0)
- BESSELJ(A1*etak,1)*BESSELY(A1,0)
0.15 =BESSELY(A2*etak,1)*BESSELJ(A2,0)
- BESSELJ(A2*etak,1)*BESSELY(A2,0)
0.2 =BESSELY(A3*etak,1)*BESSELJ(A3,0)
- BESSELJ(A3*etak,1)*BESSELY(A3,0)
0.25 =BESSELY(A4*etak,1)*BESSELJ(A4,0)
- BESSELJ(A4*etak,1)*BESSELY(A4,0)
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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:
Louisiana Tech UniversityRuston, LA 71272
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:
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
Slide 65
Louisiana Tech UniversityRuston, LA 71272
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:
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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.