As a person inhales, air moves down the windpipe (bronchus), through a constriction where the air...
As a person inhales, air moves down the windpipe (bronchus), through a constriction where the air speed doubles. If the air is travelling 13 cm/s before the constriction and we treat air as an incompressible fluid,
determine the pressure drop in the constriction. Use the density
of air as 1.29 kg/m3.
Pa
Homework Answers
From the question the initial speed of the air is v1 = 13 cm/s =
0.13 m/s
final speed of the air flow is v2 = 2*v1 = 2*0.13 = 0.26 m/s
density of air is ρ = 1.29 kg/m ^3
Using Bernouli's theoremP1+(1/2)ρ(v1)2 =P2+(1/2)ρ(v2)2
So pressure drop P1-P2 =(1/2)ρ(v2)2 -(1/2)ρ(v1)2
-----1
putting the values in equation 1 and solve for P1 -P2 .
P1 -P2 = (1/2)*1.29*[(0.26)^2 -( 0.13 )^2]
=
0.033 Pa
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