Biomedical Engineering Reference
In-Depth Information
When delivering anesthesia, a similar input is used as in Example Problem 7.10—that
is, a bolus plus constant infusion. The reason for this type of input is to quickly raise
the anesthesia to a desired level (bolus) and then to maintain the level for the operation
(step).
7.6.2 Sink Compartment
A sink compartment is one that has only inputs and no output. Similar to the source
compartment, a sink acts like an integrator and has a zero root. Moreover, the solution of
the nonsink compartment is independent of the sink compartment in the two-compartment
case. Once solved, the quantity in the nonsink compartment is used to solve the quantity in
the sink compartment. Using the model shown in Figure 7.15, a sink compartment exists if
either
K
12
and
K
10
,or
K
21
and
K
20
are zero.
EXAMPLE PROBLEM 7.11
Consider the two-compartment model shown in Figure 7.15 with
K
12
¼
K
10
¼
0,
K
21
¼
0.2,
K
20
¼
1,
f
1
(
t
)
¼
0, and
f
2
(
t
)
¼
10d(
t
). Assume that the initial conditions are zero. Solve for the quan-
tity in each compartment.
Solution
Since
0, this compartment model has a sink for compartment 1. As
before, rather than solving the problem with a bolus input, the initial condition is changed
in compartment 2 to
K
¼
0
:
2and
K
¼
21
12
q
2
ð
0
Þ¼
10, with zero input. The conservation of mass for each compart-
ment is
q
1
¼
K
21
q
2
¼
0
:
2
q
2
ð
7
:
71
Þ
q
2
¼
K
21
þ
K
20
ð
Þ
q
2
¼
1
:
2
ð
7
:
72
Þ
q
2
e
1:2
t
u
ð
t
Þ:
Since Eq. (7.72) involves only
q
2
, we solve directly to get
q
2
¼
10
Next, substitute
q
2
into (Eq. 7.71), yielding
e
1:2
t
q
1
¼
0
:
2
q
2
¼
2
ð
7
:
73
Þ
q
1
f
¼
B
2
e
1:2
t
, which
Eq. (7.73) gives a single root at
s
¼
0, and
q
1
n
¼
B
1
. The forced response is
when substituted into Eq. (7.71) gives
B
2
¼
1
:
67
:
The complete response is
e
1:2
t
q
1
¼
q
1
n
þ
q
1
f
¼
B
1
1
:
667
ð
7
:
74
Þ
With
q
1
ð
0
Þ¼
0, we have
B
1
¼
1
:
667 from Eq. (7.74), and the complete response is
u
ð
t
Þ
e
1:2
t
q
1
¼
1
:
667 1
ð
7
:
75
Þ
This result indicates that more than 80 percent of the solute has moved from the system into
the environment. If
u
ð
t
Þ:
e
0:2
t
u
ð
t
Þ
e
0:2
t
K
20
¼
0, then
q
2
¼
10
and
q
1
¼
10 1
Here, all the solute
exponentially moves from compartment 2 to 1 as expected.
