Question

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

Answer 1

Initial speed v = 15 cm / s = 0.15 m / s

speed after v ' = 30 cm / s = 0.30 m / s

Pressure drop = ( 1/ 2) ρ[ v ' ^ 2- v ^ 2]

                      = 0.5 * 1.225 * [ 0.0675 ]

                     = 0.043 Pa

(or)

By Bernoullis principle:

     Δp   = (1/2)density (v₂² - v₁² )   

             = (1/2)density (2v₁² - v₁² )  

              = (3/2)density ( v₁² )

so:   Δp   = (3/2) * 1.29kg/m³ * (0.17 ² )

              = 0.0559 Pa

                      = 33.75 Pa