Question

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 increases by 1.8 times. if the air is traveling 36 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. 0.15 pa 0.19 pa 19 pa 1500 pa 0.084 pa

Answer 1

The pressure drop in the constriction is 0.410 Pa

Use the Bernoulli's equation to find the pressure difference.

$\begin{aligned}P_1+\frac{1}{2} \rho v_1^2+\rho g h & =P_2+\frac{1}{2} \rho v_2^2+\rho g h \\P_1+\frac{1}{2} \rho v_1^2 & =P_2+\frac{1}{2} \rho v_2^2 \\P_1-P_2 & =\frac{1}{2} \rho v_2^2-\frac{1}{2} \rho v_1^2\end{aligned}$

Here, P_1 and P_2 are the absolute pressures, v_1 and v_2 are the speeds of the air in two different locations, $\rho$ is the density of air and $h$ is the height.

If $v_2=2 v_1$, then the above equation becomes as,

$\begin{aligned}P_1-P_2 & =\frac{1}{2} \rho\left(\left(2 v_1\right)^2-v_1^2\right) \\& =\frac{1}{2} \rho\left(3 v_1^2\right) \\& =\frac{3}{2} \rho v_1^2\end{aligned}$

Therefore, the pressure drop is,

$\begin{aligned}P_1-P_2 & =\frac{3}{2} \rho v_1^2 \\& =\frac{3}{2}\left(1.29 \mathrm{~kg} / \mathrm{m}^3\right)\left(46 \times 10^{-2} \mathrm{~m} / \mathrm{s}\right)^2 \\& =0.409446 \mathrm{~Pa} \\& \approx 0.410 \mathrm{~Pa}\end{aligned}$

The definition of absolute pressure is the pressure of having no matter inside a space, or a perfect vacuum. Measurements taken in absolute pressure use this absolute zero as their reference point.

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