Biomedical Engineering Reference
In-Depth Information
Expiration
dV
(n)
e
dt
+
V
(n)
+ P
(n)
ex
= P
peep
;
(n1)
+ t
(n
i
t
(n)
R
e
e
C
tot
tot
(5)
V
(n)
(
(n)
tot
) = 0
e
for n = 1; 2; 3; ::: .
In our study, we restrict our analysis to settings where there is no leakage
by the mask i.e., R
m
!1. In this case, equation (3) becomes
!
ex
V
(n)
1
R
i
(t)
Q
(n)
;
(n1)
tot
t
(n1)
+ t
(n)
i
P
set
P
(n1)
i
vent
(t) =
:
tot
C
(6)
The inspiratory time, t
(n
i
, is determined by the ventilator cut-o. If we
assume that the ventilator cuts o at some predetermined fraction of the
initial ow, Q
(n)
vent
(
(n1)
), which we denote by , 0 < < 1, then, since
tot
we assume that V
(n)
i
(
(n1)
tot
) = 0, the inspiratory time during the nth cycle
can be computed as
!
P
set
P
(n1)
t
(n)
i
ex
K
(n)
R
i
= CR
i
log
n = 1; 2; 3; :::
(7)
to which there is a possible solution of Q
(n)
vent
(t
(n)
) = Q
(n)
vent
(
(n1)
)
i
tot
K
(n)
.
4. Model Simulations
The inspiratory initial value problem (4) can be solved for V
(n)
i
(t), yielding
1e
k
i
(t
(n
1)
V
(n)
i
)
P
set
P
(n1)
(t) = C
(8)
tot
ex
1
where k
i
=
CR
i
. We may also solve the expiratory initial value problem (5)
for V
(n)
(t), obtaining
e
1e
k
e
(
(n)
V
(n)
e
tot
t)
P
peep
P
(n)
(t) = C
(9)
ex
CR
e
. Finally, we determine P
(n)
by setting V
(n)
i
(
(n)
tot
+ t
(n)
i
1
where k
e
=
) =
ex
V
(n)
(
(n)
tot
+ t
(n)
i
) and obtain:
e
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