Session/Time	Poster/Thu, 4:30 - 6:30 PM
Title	Increase in Airway Resistance Due to Endotracheal Intubation and the Required Compensatory Pressure Support to Offset It.
Author	M Rom, K Bock, P Silver, M Sagy
Affiliation	Schneider Children's Hospital, New Hyde Park, New York
Introduction	Any increase in airway resistance entails a larger pressure gradient (AP) between the airway opening and the alveoli in order to maintain the same imtiratory flow. (dP--VxR; V-flow, R- resitance). Since flow in the trachea is tubuien the resistance and consequently the pressure are inversely related to the radius of the airway to the fifth power. Thus, a reduction in the tracheal radius by a fraction of "X" will cause an increase by (1/X)5 in the pressure gradient required to maintain a given flow across the trachea. To test the hypothesis that sizes of endotracheal tubes (ETT) derived from the formula (age + 16)14 signifiantly increases the transairway pressure gradient
Method	Fourteen consecutive pediatric patients, mean age 7.4y + 7.8y, who had been ndotracheally intubated for respiratory failure and'who underwent a CT scan of their lungs, were included in the studv. MeasurementsandMai.,tResults:TheETTdiameters CF,TTD) and the tracheal diameters (Td) were measured at various CT cuts and the average values were used to calculate the reduction fraction (X) in the tracheal radius (ETTd/Td=X). This value ranged from 0.33 to 0.65 (0.54+-0.08). Sine e "XI' Nvas similar in ird nts as in young adults (table), we used it to calculate variables of adult pulmonary physiology. Thus, the value (1/X)5 was calculated for each patient and then multiplied by die kno, vn normal tracheal resistance for adults (0.07cm H20/I/scc) to obtain the new tracheal resistance for incubated adults. The pressure gradient to overcorie this resistance for an inspiratory flo ,v rate of I I/sec (60 I/min) would be (1/x)5x 0.07-,I. Since in nonintubated patients the overall normal transairway pressure gradient is close to l cm H2O, the new estimated total transairway pressure gradient (8P) secondary to the ETT for adults will be dP' = [(1/x)5 x 0.07 x 1 ] + 1
Result

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