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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