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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?

Respuesta :

Answer:

ΔP = 0.0360 Pa

Explanation:

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

Using Bernoulli's equation is given as

ΔP = (0.5) ρ (v²₂ - v²₁ )

ΔP = (0.5) (1.225) ((0.28)² - (0.14)² )

ΔP = 0.0360 Pa

Lanuel

By applying Bernoulli's equation, the pressure drop in the constriction is equal to 0.037 Pa.

Given the following data:

  • Speed of air in bronchus = 14 cm/s
  • Density of air = 1.225 kg/m
  • Speed of air in constriction = 2(14) = 28 cm/s

Conversion:

Speed of air in bronchus = 14 cm/s = [tex]\frac{14}{100} =0.14 \;m/s[/tex]

Speed of air in constriction = 2(14) = 28 cm/s =  [tex]\frac{28}{100} =0.28 \;m/s[/tex]

To determine the pressure drop in the constriction, we would apply Bernoulli's equation:

Mathematically, Bernoulli's equation is given by the formula:

[tex]\Delta P = \frac{1}{2} \rho (V_2^2 - V_1^2)[/tex]

Where:

  • [tex]\Delta P[/tex] is the change in pressure.
  • [tex]V_2[/tex] is the speed of air in constriction.
  • [tex]V_1[/tex] is the speed of air in bronchus.
  • [tex]\rho[/tex] is the density.

Substituting the given parameters into the formula, we have;

[tex]\Delta P = \frac{1}{2} \times 1.225 (0.28^2 - 0.14^2)\\\\\Delta P = 0.625 (0.0784 - 0.0196)\\\\\Delta P = 0.625 (0.0588)\\\\\Delta P = 0.037 \;Pa[/tex]

Pressure drop = 0.037 Pa

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