When a person inhales, air moves down the bronchus (windpipe) at 14 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. (enter the magnitude.) pa?

v₁ = speed of air in bronchus = 14 cm/s = 0.14 m/s

v₂ = speed of air in constriction = 2 v₁ = 2 x 14 = 28 cm/s = 0.28 m/s

ρ = density of air = 1.225 kg/m³

ΔP = pressure drop in the constriction = ?

pressure drop in the constriction using Bernoulli's equation is given as

ΔP = (0.5) ρ (v²₂ - v²₁)

inserting the values

ΔP = (0.5) (1000) ((0.28) ² - (0.14) ²)

ΔP = 29.4 Pa