Page 369 - Clinical Application of Mechanical Ventilation

Ventilator Waveform Analysis 335

1.0

V (L) 0.5 25%

75%
1 2 3 4 5 6 7 8
V (L/min) 60

75%
25%
1 2 3 4 5 6 7 8

260
P (cm H 2 O) 80

40 25%
75%
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1 2 3 4 5 6 7 8
Time (sec)
Figure 11-17 Effects of inspiratory time using descending ramp flow on the delivered tidal
volume (V T ) and peak inspiratory pressure (PIP). An increase of inspiratory time from 1 (left) to
2 sec (right) raises the V T and PIP.

fast and becomes proportionately reduced over time as flow descends. In the first
example, peak flow is set at 60 L/min and T is 1 second, and the volume delivered
I
is 0.5 liter [1/2 (peak flow 3 I time)]. The volume delivered in the first half of inspi-
ration is 0.375 L (75% of 0.5 L). The volume delivered in the second half of inspira-
tion is 0.125 L (25% of the remaining V ). See Table 11-7 for calculation of volume
T
delivered in the first half of inspiration during true DRFW.
The pressure-time pattern in the first example in Figure 11-17 shows the de-
velopment of P ALV in correspondence with delivery of the V . Determining the
T
P ALV curve requires knowing the volume delivered per time and the total C .
LT
Because there is no flow at end-inspiration, there is no P at end-inspiration,
TA
and the PIP represents the peak P ALV , which is 40 cm H O. Total C is 0.0125
LT
2
L/cm H O or 12.5 mL/cm H O (C 5 volume/P ALV 5 0.5 L/40 cm H O).
LT
2
2
2
If there is no change in C during V delivery, P ALV halfway through inspira-
T
LT
tion will depend on the volume delivered by that time. The example shows, as
discussed earlier, that 75% of the V is delivered in the first half of inspiration;
T
consequently, 75% of the peak P ALV will be attained, which is 30 cm H O (75% 3
2
40 cm H O), or P ALV is equal to volume calculated for that time divided by total
2
C (P ALV 5 0.375 L/0.0125 L/cm H O 5 30 cm H O). The remaining 25%
LT
2
2
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