Download - Krogh Cylinder
Louisiana Tech UniversityRuston, LA 71272
Slide 1
Krogh Cylinder
Steven A. JonesBIEN 501
Friday, April 20, 2007
Louisiana Tech UniversityRuston, LA 71272
Slide 2
Energy Balance
Major Learning Objectives:1. Learn a simple model of capillary
transport.
Louisiana Tech UniversityRuston, LA 71272
Slide 3
The Krogh Cylinder
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
Slide 5
Capillary Transport
Consider the following simple model for capillary transport:
Reactive Tissue
Capillary Interior
Matrix
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
Slide 7
Diffusion Equation
x
x
rrcr
rrD
tc
rcDtc
1
2
For steady state:
xrrcr
rrD
1
Louisiana Tech UniversityRuston, LA 71272
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).
Louisiana Tech UniversityRuston, LA 71272
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:
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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:
Louisiana Tech UniversityRuston, LA 71272
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:
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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
)
Louisiana Tech UniversityRuston, LA 71272
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
)
Louisiana Tech UniversityRuston, LA 71272
Slide 17
Non Steady State
Mrcr
rrD
tc
McDtc
1
2Diffusion equation:
rtc allforat 00 Initial Condition:
0,),( 0
rtrcD
tuctrc
k
cBoundary Conditions:
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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:
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
Slide 22
Non-Dimensionalization (Continued)
Use:
To determine that:
2cc r
tDrr
2
1
c
c
rD
tt
rrr
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
Slide 26
The Mathematical Problem
The problem reduces mathematically to:
0
21DCMrc
Differential Equation
Boundary Conditions
Initial Condition
0,1 t 0,
rtk
00,
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
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).
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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:
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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,
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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?
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
Slide 39
Radial Dependence
022
22
zz
dzzdsD
sDz
dzzdsD
sDz RRR
0022
22 zRz
dzzdRz
dzzdz R
Louisiana Tech UniversityRuston, LA 71272
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.
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 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.
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
Slide 43
Bessel Functions
-2
-1
0
1
2
0 2 4 6 8
x
Bes
sel F
unct
ion
J0Y0J1Y1
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
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:
Louisiana Tech UniversityRuston, LA 71272
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:
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
Slide 50
Boundary Condition at r = rc
1
0 001
1
tucdtesDrYsDrJsDrJsDrY
sB cst
cck
k
11
001
1
tucsDrYsDrJ
sDrJsDrY
sB ccck
ktL
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
Slide 52
Counter-Current Exchange
zM d dzzM d
zM b dzzM b
zMd
ddddddd dCQzCdzzCQzMdzzMMd
mDB dACCkMd 0
Conservation of Mass
Membrane Diffusion:
Louisiana Tech UniversityRuston, LA 71272
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:
Louisiana Tech UniversityRuston, LA 71272
Slide 54
Combine the Two Equations
dzdACk
dzCdQ m
d
0
z
QWkCCd
00 exp
Solution:
Where: WdzdA
constant
Louisiana Tech UniversityRuston, LA 71272
Slide 55
Integrate w.r.t. z
dzCdQ
dzMd
d
dzdACk
dzMd m 0
Louisiana Tech UniversityRuston, LA 71272
Slide 56
Louisiana Tech UniversityRuston, LA 71272
Slide 57
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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 ,,,
Louisiana Tech UniversityRuston, LA 71272
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 ,,,
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
Slide 64
Semi-Infinite Slab (of marble)
tzatTTztatTT
0
00
21
20
TktTC p
Differential Equation (no source term):
Boundary Conditions:
Louisiana Tech UniversityRuston, LA 71272
Slide 65
Semi-Infinite Slab (of marble)
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.
Louisiana Tech UniversityRuston, LA 71272
Slide 66
Semi-Infinite Slab (of marble)
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
Louisiana Tech UniversityRuston, LA 71272
Slide 67
Semi-Infinite Slab (of marble)
pCk
zT
tT
where,2
2
*2*
0TT
0,01
00
2*
*
tzT
tT
for
for
Equations in Terms of Dimensionless Temperature:
Boundary Conditions:
01 TT
T* is valuable because it nondimensionalizes the equations and simplifies the boundary conditions.
Louisiana Tech UniversityRuston, LA 71272
Slide 68
Semi-Infinite Slab (of marble)
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
Louisiana Tech UniversityRuston, LA 71272
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
.
Louisiana Tech UniversityRuston, LA 71272
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:
Louisiana Tech UniversityRuston, LA 71272
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).
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
Slide 77
Louisiana Tech UniversityRuston, LA 71272
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:
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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.