   Chapter 9, Problem 38P

Chapter
Section
Textbook Problem

When a person inhales, air moves down the bronchus (windpipe) at 15 cm/s. The average How speed of the air doubles through a constriction in the bronchus. Assuming incompressible flow, determine the pressure drop in the constriction.

To determine
The pressure drop in the constriction.

Explanation
Use Bernoulli’s equation for the pressure drop in constriction (ignoring the very small change in vertical position) is P1P2=(1/2)ρ(v22v12)=(3/2)ρv12 since the flow speed in the constriction is 2v1 .

Given info: The density of air is 1.29kg/m3 and the flow speed of the air in the bronchus is 15×102m/s .

The formula for the pressure drop in the constriction is,

ΔP=32ρv12

• ρ is density of air.
• v1 is flow speed of air.

Substitute 1

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