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100
7.7. The short contact time solutions for concentration, Equation (7.6.26) and flux, Equation
(7.5.28), are valid as long as zDij/vmaxR2 <0.01. For oxygen Dij = 1.10 x 10-5 cm2 s-1. From data in
Table 2.4 for the canine cardiovascular system, the radii of capillaries and venules are 0.0003 and
7.8. By substituting various values for y/R, the approximation of the velocity profile by vmaxy/R is
within 10% of the true value for y/R less than 0.19.
y/R
vz
vmax
=y
R2−y
R
⎛
⎝
⎜ ⎞
⎠
⎟
vz
vmax
=2y
R
Error
0.05
0.0975
0.1
2.56%
0.10
0.19
0.20
5.26%
7.9. Short contact time release in a rectangular channel. The geometry is shown in the schematic
101
where
vz=3
2
v 1−4y2
H2
⎛
⎝
⎜
⎞
⎠
⎟
7.10. The local mass transfer coefficient for laminar flow along a flat plate is:
kloc z
Dij
=0.323 Reloc
1 / 2 Sc1 / 3
(S7.10.1)
z=0
102
kf =
0.646Dij
L
vL
ν
⎛
⎝
⎜
⎞
⎠
⎟
1/ 2
Sc1 / 3 =
0.646Dij
L
Re1/ 2 Sc1/ 3
(S7.10.6)
7.11. From the data in Example 7.6, the diffusion coefficients of urea and albumin in aqueous
solution are 1.81 x 10-5 cm2 s-1 and 6.34 x 10-7 cm2 s-1, respectively.
Shear rate, s-1
Urea
Albumin
1
0.960 x 10-5 cm2 s-1
0.0341x 10-5 cm2 s-1
7.12. The flux across the membrane equals the flux through the fluid to the membrane surfaces
Nix = kf1(Cb1-C1) = Pi(C1-C2) = kf2(C2 - Cb2) where Pi = ΦDij/L.
Equating the first two expressions and solving for C1:
C1=kf 1Cb1 +PiC2
kf 1 +Pi
kf 1 +Pi
⎝
⎠
kf 1 +Pi
kf 2
Pi
kf 2
103
1
7.13. For countercurrent exchange, the dialysate enters at z = L and exits at z = 0. The clearance and
the extraction ratio are:
The reciprocal of the concentration ratio on the right hand side of Equation (S7.13.2) is expanded as
follows:
CiB(L) −CiD(L)
CiB (0) −CiD(L)
=CiB(L) −CiB (0) +CiB (0) −CiD(L)
CiB (0) −CiD(L)
=1−E
Thus, the term in the logarithm of Equation (S7.13.1) becomes:
CiB(0) −CiD(0)
CiB(L) −CiD(L) =1+E(1−Z) CiB (0) −CiD(L)
CiB(L) −CiD(L)
⎛
⎝
⎜
⎜
⎞
⎠
⎟
⎟ =1+E(1−Z)
1−E=exp E 1−Z)
( )
[ ]
Rearranging this expression one obtains:
104
E(1 −Z) =1−E
( )
exp E 1−Z)
( )
[ ]
−1
( )
Collecting terms multiplied by E and rearranging yields equation (7.8.18b):
E=exp E 1 −Z)
( )
[ ]
−1
exp E 1−Z)
( )
[ ]
−Z
7.14. (a) The extraction fraction E = K/QB. From the data given:
Urea E = 200/208 = 0.96
E = 136/208 = 0.65 without protein
105
0.65 0.28 −exp[FR(1 −Z)]
( )
=1−exp[FR(1 −Z)]
0.182 =1−0.35exp[FR(1 −Z)]
exp[FR0.72] =2.337
FR = k0Am/QB =1.18
k0Am = 245 ml/min
Now for the case of protein binding, Z = 0.28*8.304 = 2.33
FR = k0Am/QB(1+KA) = 245/208*8.304= 0.142
E=1−exp[FR(1 −Z)]
Z−exp[FR(1 −Z)]
=1−exp[−0.189]
2.33 −exp[−0.189]
=0.115
Note: An alternative way to obtain KA is to compare E obtained with and without proteins. For this
case:
K
QB1+KA
( )
=Kwith protein
QB
Rearranging:
KA=K
Kwith protein
−1=4.667
7.15. (a) First determine the average velocity and Reynolds number per channel as well as the
diffusion coefficient in blood and the Schmidt number.
106
4Deff L
vH2=
4 4.5x10−6
( )
25
0.926 0.01
( )
2=4.89 >> 0.01
For the dialysate,
vdialysate =Q
NW 2H=750 / 60
90 20
( )
0.01 =1.39 cm/s
Redialysate = 1.39(0.01998)/0.009 = 3.09 and Sc =
0.009/3 x 10-6 = 3000.
4Deff L
vH2=
4 3x10−6
( )
25
1.39 0.01
( )
2=2.16 >> 0.01
For the dialysate and blood, the following result should be used for Sh
k2H
Dij
=3.770
For blood, kB = 3.77(4.53 x 10-6 cm2/s)/(0.01 cm) = 1.71 x 10-3 cm/s
For dialysate, kD = 3.77(3.0 x 10-6 cm2/s)/(0.01 cm) = 1.13 x 10-3 cm/s
The overall mass transfer coefficient k0 is: 6.0 x 10-4 cm/s
FR = k0Am/QB = (6.0 x 10-4 cm/s)(25*10*2*90)/(500/60)= 3.234
Where Am = LW*2*90, since each channel has an upper and lower surface for exchange.
Z = QB/QD = 2/3.
E=1−exp[3.234(0.333)]
0.667 −exp[3.234(0.333)]
=0.853
K = EQB = 0.853(500) = 426.5 cm3 min-1
To determine the concentration in blood, use the single compartment model (equation (7.9.24)):
CiB(t)=CiB0exp −Kt
V
⎛
⎝
⎜⎞
⎠
⎟.
Solving for time
t=−V
Kln CiB(t) / CiB0
⎡
⎣⎤
⎦=−10000
426.5 ln 10−12 / 3 x 10−9
( )
=187.72 min
.
(b.1) If the dialysate were recycled, then the concentrations at the end of dialysis can be determined
from a mass balance.
Initial mass = mass in blood + mass in dialysate
The mass is the product of the concentration times the volume times the MW.
C0VB = CBVB+CDVD
At equilibrium, CB= CD = C. So, C = C0VB/(VB+VD) = (3 x 10-9 mole/cm3)10L/(10L + 50L) = 0.5 x
10-9 mole/cm3. Thus, even if the process were fast enough recycling would not permit sufficient
107
volume to reduce the concentration in the blood to the desired level. In fact, the recycle volume
needed to reduce the concentration to 1 x 10-12 mole/cm3 is 29,990 L!
(b.2) Setting t = 360 minutes, then one can solve
CiB(t)=CiB0exp −Kt
V
⎛
⎝
⎜⎞
⎠
⎟
for K which equals
Examining the table below for E as a function of Z, we see that E = 0.4446 for Z =1.875. This
corresponds to QD = 266.7 ml/min. This corresponds to 80 L in 3 hours. This is the smallest
volume that can be exchanged in the dialysate for the conditions provided.
Z
E
0.1
1
0.3
1
0.5
1.000
7.16. (a) At steady state the solute conservation relation near the membrane surface is:
This is a simplification in that we assume that the balance of convection and diffusion can be
characterized in terms of a mass transfer coefficient, kf = Dij/d. This result indicates that the flux is
constant. Thus the flux through the fluid to the membrane surface equals the flux leaving the
108
membrane, -vfCf. The negative sign arises because the velocity is in the negative y direction. The
flux through the fluid is.
The effect of concentration polarization is to raise the concentration at the wall and thereby increase
osmotic pressure. Consequently, net pressure difference will be reduced resulting in a decrease in
the filtration velocity.
7.17. (a)
109
Accumulation of solute in = rate of transport into - rate of transport from
110
Ci=G
Vexp −KR+K
( )
V
t
⎛
⎝
⎜⎞
⎠
⎟exp
KR+K
( )
V
t
⎛
⎝
⎜⎞
⎠
⎟dt
∫+Aexp −KR+K
( )
V
t
⎛
⎝
⎜⎞
⎠
⎟
