Biomedical Engineering Reference
In-Depth Information
Setting Eq. (7.59) equal to zero and
t
¼
t
max
gives
ð
0
Þ
K
q
2
3
K
1
M
þ
K
1
U
K
21
K
21
e
K
21
t
max
Þ
e
K
1
M
þ
K
1
U
ð
Þ
t
max
þ
K
1
M
þ
K
1
U
ð
¼
0
or
K
21
e
K
21
t
max
Þ
e
K
1
M
þ
K
1
U
ð
Þ
t
max
¼
K
1
M
þ
K
1
U
ð
e
K
1
M
þ
K
1
U
ð
Þ
t
max
and dividing by
Multiplying both sides of the previous equation by
K
21
gives
¼
K
1
M
þ
K
1
U
K
e
K
1
M
þ
K
1
U
ð
Þ
t
e
K
21
t
max
¼
e
K
1
M
þ
K
1
U
K
21
ð
Þ
t
21
Taking the logarithm of both sides gives
K
M
þ
K
1
1
U
ð
K
M
þ
K
U
K
Þ
t
¼
ln
1
1
21
max
K
21
Solving for
t
max
yields
M
þ
K
K
1
1
U
ln
K
21
¼
ð
7
:
60
Þ
t
max
ð
K
1
M
þ
K
1
U
K
21
Þ
It should be clear from Eq. (7.59) that the smaller the term
K
1
M
þ
K
1
U
compared to
K
21
, the
more time it takes to reach the maximum concentration or quantity in the plasma.
EXAMPLE PROBLEM 7.9
Suppose 50 g of solute is ingested. Find the maximum amount of solute in the plasma if the
compartmental model in Figure 7.17 is used with
0.005 min
1
and
0.02 min
1
.
K
1
M
þ
K
1
U
¼
K
21
¼
Solution
Using Eq. (7.60) gives
K
1
M
þ
K
1
U
K
21
0
:
005
ln
ln
0
:
02
ln 0
ð
:
25
Þ
t
max
¼
Þ
¼
02
¼
015
¼
92
:
42 min
ð
K
1
M
þ
K
1
U
K
21
0
:
005
0
:
0
:
The maximum amount of solute in compartment 1 at
t
max
is therefore
t
¼92:42
Þ¼
q
ð
0
Þ
K
21
K
1
M
þ
K
1
U
K
21
2
e
K
21
t
e
K
1
M
þ
K
1
U
ð
Þ
t
q
1
ð
t
max
¼
50
0
:
02
e
0:0292:42
e
0:00592:42
¼
31
:
5g
0
:
005
0
:
02
The next example introduces an encapsulated pill input that releases a portion immedi-
ately and the remainder continuously until the pill completely dissolves. Mathematically
this input is approximated as zd(
t
)
þ
(1
z)(
u
(
t
)
u
(
t
t
0
)). To estimate the gastric transfer
