Co-Current and Counter-Current
Exchange in Dialysis
Steven A. Jones
BIEN 501
Monday, May 5, 2008
Louisiana Tech University
Ruston, LA 71272
Slide 1
Co-Current Exchange
M b  z  M b z  dz 
Qb
dM z 
Qd
M d z  M d  z  dz 
Conservation of Mass
dM  M d z  dz   M d z   Qd Cd z  dz   Cd z   Qd dCd
Membrane Diffusion:
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dM  k0 CB  CD dAm
Slide 2
Different Notation
I personally do not find the notation dM to be
completely intuitive. For me, a better way is to talk
about the mass flux per unit area along the z-direction.
The conservation of mass then becomes:
 b z  dz 
m b z  m
Qb
M  z 
Qd
mb dzW   Cb  z   Cb  z  dz   Qb
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Slide 3
Definitions
Variable
Definition
Qb
Flow rate of blood

m
Mass flux per unit area of membrane from
blood to dialysate
Concentration of the solute in the blood
Cb
Cd
Concentration of the solute in the dialysate
W
Width of the membrane
h
Height of the membrane
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Slide 4
The Differential Equation
With:
mdzW   Cb  z   Cb  z  dz   Qb
Divide through by dz
1  Cb  z   Cb  z  dz  
m
Qb
W
dz
Take the limit as dz goes to 0.
1 dCb  z 
m
Qb
W dz
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Slide 5
Interpretation
Consider:
mb  
dCb  z 
dz
1
Qb
W
This equation has the simple interpretation that concentration
decreases along the length of the membrane as a result of
mass loss through the membrane. Because it uses welldefined derivatives, its meaning is much more clear than the
equation in the book:
dM  Qb dCb
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Slide 6
Dialysate Side, Mass Conservation
On the dialysate side, the amount of mass coming in is
equal to the amount of mass leaving the blood. Thus,
1 dCb  z 
1 dCd  z 
m
Qb , m 
Qd
W dz
W dz
And these two equations can be combined to obtain:
1 dCb  z 
1 dCd  z 
m
Qb 
Qd
W dz
W dz
dCd  z 
dCb  z 
Qd  
Qb
dz
dz
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Slide 7
The Two Basic Equations
Conservation of Mass is then:
Qb dCb
m
W dz
And Membrane Diffusion is:
m  k0 Cb  Cd   k0C  z 
Where:
C  z   Cb  z   Cd  z 
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Slide 8
Compare to the Book’s Notation
A comparison of our membrane equation:
To the equation in the book:
  k0 Cb  Cd 
m
dM  k0 Cb  Cd dAm
Clarifies what the book means by dM and dAm , for if we
divide by dAm:

dM
 k0 Cb  Cd 
dAm
 dA  m is the rate of change of total mass
Thus, dM
m
flux with membrane area, and it is equivalent to the mass
flux per unit area as we have defined it.
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Slide 9
Combine the Two Equations
dCb
m  Qb h
dz
  k0 Cb  Cd   k0 C
m
dCb
dCd
Qb
 k0W  Cb  Cd  , Qd
 k0W  Cb  Cd 
dz
dz
Blood Side
Dialysate Side
Divide the blood side by Qbh and divide the dialysate side by Qdh
dCb k0W  Cb  Cd  dCd k0W  Cb  Cd 

,

dz
Qb
dz
Qd
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Slide 10
Subtract
With:
dCb k0W  Cb  Cd  dCd k0W  Cb  Cd 

,

dz
Qb
dz
Qd
dCb dCd  k0W  Cb  Cd  k0W  Cb  Cd 



dz
dz
Qb
Qd
 1
d C
1 

  k0W C 


dz
 Qb Qd 
So:
 1
d C
1 
 k0W  
 C  0
dz
 Qb Qd 
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Slide 11
Combine the Two Equations
 k0W
C  C0 exp  
 Q
From

z

We need to determine the transport across the
membrane. We will integrate the equation below with
respect to z:
L
M   mWdz  
0
M 
L
0
L
0
dAm
k0 C
dz
dz
dAm
dAm
k0 C
dz  M  k0
dz
dz
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
L
0
 k0W
C0 exp  
 Q

z dz

Slide 12
Combine the Two Equations
dAm

M  k0
dz

C0

M 
Qd
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L
0
 k0W
C0 exp  
 Qd

z dz


 k0W 
L 
1  exp  
 Qd 

Slide 13
Eq. 7.8.10
In our notation, the book’s equation is:
C 0  C L

M  k 0 Am
 C 0 

ln 
 C L 
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Slide 14
Compare Co- and Counter-Current
Co-Current
Counter-Current
 C0 
 1
1 
  k0 Am 

ln 

 CL 
 QD QB 
 C0 
 1
1 
  k0 Am 

ln 

 CL 
 QB QD 
The concentration differences (inlet vs. outlet) are closer
to one another for counter-current exchange. What
about the net mass transfer?
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Slide 15
Total Mass Transfer
From Equations 7.8.1 and 7.8.2:
L dC
dCd
dM
d
 Qd
 M  Qd 
dz  Qd Cd L   Cd 0
0 dz
dz
dz
And similarly:
M  QB CB L   CB 0
M
QB  
CB L   CB 0
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Note sign
change.
M
QD 
CD L   CD 0
Slide 16
Total Mass Transfer
Combine to eliminate flow rates (co-exchange):
 C0 
 1
1 
  k0 Am 

ln 

 CL 
 QD QB 




 C0 
1
1




ln 
 k0 Am 





M
M

C
L 



 C D L   C D 0 C B L   C B 0 
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Slide 17
Total Mass Transfer
 C0 
 CD L   CD 0 CB L   CB 0 
  k0 Am 
ln 




M
M


 CL 




 C 0  C L  

M  k0 Am 

 ln  C0  
 C  

L 



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Slide 18
Total Mass Transfer
Combine to eliminate flow rates (counter-exchange):
 C0 
 1
1 
  k0 Am 

ln 

 CL 
 QB QD 




 C0 
1
1




ln 
 k0 Am 





M
M

C
L 



 C B L   C B 0 C D L   C D 0 
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Slide 19
Total Mass Transfer
 C0 
 CB L   CB 0 CD L   CD 0 
  k0 Am 
ln 




M
M


 CL 




 C 0...L   C 0...L  

M  k0 Am 

 C0 




ln
 C 


Sort of a cross L


difference w.r.t 0
and L.
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Slide 20
Look at the Plots
Look at CB 0  CD L and CB L  CD 0
z=0
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z=L
z=0
z=L
Slide 21
Clearance (K)
M
K
C B ,in  C D ,in
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Slide 22
Efficiency
K
E
QB
Co-current
Counter-current
Where
QB
Z
,
QD
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1
1  exp  FR1  Z 
E
1 Z
1  exp FR1  Z 
E
Z  exp FR1  Z 
k0 Am
FR 
QB
Are the same for coand counter-current.
Slide 23
Questions
1
1  exp  FR1  Z 
E
1 Z
Co-current
1  exp FR1  Z 
E
Z  exp FR1  Z 
Counter-current
Where
QB
Z
,
QD
k0 Am
FR 
QB
What do these functions look like?
If QB = QD, do you get infinite efficiency?
For what value of Z is E a maximum?
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Slide 24
Efficiency as a function of Z
1
Counter-flow
1
0.8
Counter-flow
Co-flow
Co-flow
0.8
0.6
FR = 3
E
0.6
E
0.4
FR = 3
0.2
0.4
0
0
2
4
0.2
6
8
10
Z
0
0
2
4
6
8
10
Z
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Slide 25
Question
Does it help to have a large QD?
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Slide 26
Counter-Current Exchange
M b  z  M b z  dz 
dM z 
M d z  M d  z  dz 
Conservation of Mass
dM  M d z  dz   M d z   Qd Cd z  dz   Cd z   Qd dCd
L dC
dCd
dM
d
 Qd
 M  Qd 
dz  Qd Cd L   Cd 0
0 dz
dz
dz
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Slide 27
Compare Co- and Counter-Current
Co-Current
Counter-Current
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C 0  C L

M  k 0 Am
 C 0 

ln 
 C L 
C 0  C L

M  k 0 Am
 C 0 

ln 
 C L 
Slide 28
Integrate w.r.t. z
dM
dC
 Qd
dz
dz
dAm
dM
 k0 C
dz
dz
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Slide 29