As a person breathes, during the inhale part of the cycle air moves down the windpipe (bronchus) and through a constriction where the air speed doubles. If the air is traveling 46 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. in Pa

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Answer 1

If the air is traveling 46 cm/s before the constriction and we treat air as an incompressible fluid, the pressure drop in the constriction is 106.8 Pa.

If we treat air as an incompressible fluid, the pressure drop in the constriction can be determined using Bernoulli’s equation. Bernoulli’s equation states that the sum of the pressure, kinetic energy, and potential energy of a fluid stream is constant.

The equation can be expressed as:

P1 + (1/2)ρv12 + ρgh1 = P2 + (1/2)ρv22 + ρgh2

Where P1 and P2 are the pressures before and after the constriction, ρ is the density of the fluid, v1 and v2 are the velocities before and after the constriction, g is the acceleration due to gravity, and h1 and h2 are the heights before and after the constriction.

Given that the air velocity before the constriction is 46 cm/s and the density of air is 1.29 kg/m3, we can calculate the pressure drop in the constriction as follows:

P1 + (1/2)*(1.29 kg/m3)*(46 cm/s)2 + (1.29 kg/m3)*(9.8 m/s2)*h1 = P2 + (1/2)*(1.29 kg/m3)*(92 cm/s)2 + (1.29 kg/m3)*(9.8 m/s2)*h2

Since h1 and h2 are equal and the acceleration due to gravity is constant, we can simplify the equation to:

P1 + (1/2)*(1.29 kg/m3)*(46 cm/s)2 = P2 + (1/2)*(1.29 kg/m3)*(92 cm/s)2

Substituting the given numerical values, we get

P1 + (1/2)*(1.29 kg/m3)*(46 cm/s)2 = P2 + (1/2)*(1.29 kg/m3)*(92 cm/s)2

P1 + 35.6 Pa = P2 + 142.4 Pa

Therefore, the pressure drop in the constriction is 106.8 Pa.

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