Biomedical Engineering Reference
In-Depth Information
Q
KA
Bin
,
. Parameters
E
and
N
T
provide the performance parameters for the dialyzer. Value of
E
varies from 0
to 1 and depends on the blood flow rate, and dialysate flow rate on a day-to day-
operation. It is used to monitor the patient along with clearance rate or adjust the
flow rate of the dialysate. The extraction ratio will decrease if the blood flow rate is
increased. At any given blood flow rate, a dialysate flow rate increase will increase
the extraction ratio, and thereby also increase the clearance.
where
N
=
U
is the number of transfer units and
z
=
T
v
Q
Bin
,
Din
,
EXAMPLE 10.5
A clinical engineer has two dialyzer units. One is a small, with
K
U
A
value of 400 mL/min
and the other with a
K
0
A
value of 1,000 mL/min. Two patients (
v
b
= 200 mL/min) are con-
nected with these dialyzers in countercurrent mode with a dialysis rate of 500 mL/min.
(a) Which one has higher efficiency of extraction?
(b) If the blood urea level of a patient is 100 mg/dL and dialysate inlet urea content is
zero, what is the urea concentration of blood exiting the dialyzer in a single
pass?
(c) The blood volume of that patient is 4L. If the second dialyzer is used, how much
time will the patient save sitting in a dialysis center for 90% clearance of urea.
Solution:
200[mL/min]
400[mL/min]
1,000[mL/min]
(a)
z
0.4;
N
2;
N
5
=
=
=
=
=
=
T
1
T
2
500[mL/min]
200[mL/min]
200[mL/min]
For countercurrent flow, extraction ratio for
(
)
exp 2 1
0.4
1
⎡
−
⎤
−
⎣
⎦
The first dialyzer,
E
0.794
=
=
1
(
)
exp 2 1
0.4
0.4
⎡
−
⎤
−
⎣
⎦
(
)
exp 5 1
0.4
1
⎡
−
⎤
−
⎣
⎦
The second dialyzer,
E
0.970
=
=
2
(
)
exp 5 1
0.4
0.4
⎡
−
⎤
−
⎣
⎦
Since
E
2
is significantly higher than
E
1
, it has a better extraction efficiency.
(b) From (10.39),
C
C
(1
E
)
C
E
=
−
+
BU out
,
BU in
,
DU in
,
C
D,in
=
0 and
C
b,in
=
100 mg/dL.
Hence in the first dialyzer,
C
C
,
(1
0.794)
20.6 mgdL
=
−
=
BU out
,
BU in
In the second dialyzer,
C
100(1
0.970)
3 mgdL
=
−
=
Bout
,
(c)
V
B
=
4L.
C
BU
,
out
depends on
C
BU,in
, which will change upon reentering the blood.
Hence
C
BU
,
out
should be expressed as a function of
C
BU
, in which is assumed to
be
C
BU
.
C
C
,
(1
0.794)
0.206
C
rom above, for the first dialyzer,
=
−
=
BU out
,
BU in
BU
rom (10.32) with the assumption that
Q
B
,
in
=
Q
B
,
out
,
dC
200[mL/min]
dC
(
)
1
BU
0.206
CC
BU
0.0397[min
−
]
C
=
−
→
= −
BU
BU
BU
dt
4,000[mL]
dt
Integrating and using the limits of
C
BU
(
t
=
0)
=
100 mg/dL,
C
BU
(
t
=
t
) 10 mg/dL.
t
=
58 minutes for the first dialyzer.
C
0.03
C
and
t
47.476 minutes
For the second dialyzer,
=
=
BU out
,
BU
Time saved
=
58
−
47.476
=
11.524 minutes.
