When you cough, you expel air at high speed through the trachea and upper bronchi so that the air will remove excess mucus lining the pathway. You produce the high speed by this procedure: You breathe in a large amount of air, trap it by closing the glottis (the narrow opening in the larynx), increase the air pressure by contracting the lungs, partially collapse the trachea and upper bronchi to narrowthe pathway, and then expel the air through the pathway by suddenly reopening the glottis. Assume that during the expulsion the volume flow rate is$\mathbf{7}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\frac{\mathbf{}{\mathbf{m}}^{\mathbf{3}}}{\mathbf{s}}$.What multiple of$\mathbf{343}\mathbf{}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$(the speed of sound vs) is the airspeed through the trachea if the trachea diameter (a) remains its normal value of$\mathbf{14}\mathbf{}\mathbf{}\mathit{m}\mathit{m}$and (b) contracts to$\mathbf{5}\mathbf{.}\mathbf{2}\mathbf{}\mathbf{}\mathit{m}\mathit{m}$?

The volume flow rate is$R=7\times {10}^{-3}{\text{m}}^{\text{3}}\text{/s}$.

An ideal fluid is incompressible and lacks viscosity, and its flow is steady and irrotational. A streamline is a path followed by an individual fluid particle. A tube of flow is a bundle of streamlines. The flow within any tube of flow obeys the equation of continuity:

${\mathit{R}}_{\mathbf{v}}\mathbf{=}\mathit{A}\mathit{v}\mathbf{=}\mathbf{\text{Constant}}$

Where ${\mathit{R}}_{\mathbf{v}}$is the volume flow rate,A is the cross-sectional area of the tube of flow at any point, and is the speed of the fluid at that point.

We can find the airspeed through the trachea by using the formula for volume flow rate.

The volume flow rate is

$\begin{array}{rcl}R& =& vA\\ & =& v\pi r^2\\ & =& \frac{v\pi d^2}{4}\\ & & \end{array}$

Thus, the airspeed through the trachea, if the trachea’s diameter is 14 mm is

$\begin{array}{rcl}v& =& \frac{4R}{\pi d^2}\\ & =& \frac{4\left(7\times {10}^{-3}{\text{m}}^{\text{3}}\text{/s}\right)}{3.142{\left(14\times {10}^{-3}{\text{m}}^2\right)}^2}\\ & =& 45.4\text{m/s}\\ & \approx & 45\text{m/s}\\ & & \end{array}$

But, the speed of the sound is $v_s=343\text{m/s}$. Thus,

$\begin{array}{rcl}\frac{v}{v_s}& =& \frac{45.4\text{m/s}}{343\text{m/s}}\\ & =& 0.13\\ v& =& 0.13v_s\\ & & \end{array}$

Therefore, the airspeed through the trachea if the trachea’s diameter is 14 mm is$0.13v_s.$

The airspeed through the trachea if the trachea’s diameter is 5.2 mm is

$\begin{array}{rcl}v& =& \frac{4R}{\pi d^2}\\ & =& \frac{4\left(7\times {10}^{-3}{\text{m}}^{\text{3}}\text{/s}\right)}{3.142{\left(5.2\times {10}^{-3}{\text{m}}^2\right)}^2}\\ & =& 329.6\text{m/s}\\ & \approx & 330\text{m/s}\\ & & \end{array}$

But the speed of the sound is $v_s=343\text{m/s}$. Thus,

$$\begin{gathered} \frac{v}{v_s}=\frac{330}{343} \\ v=0.96v_s \end{gathered}$$

Therefore, the airspeed through the trachea if the trachea’s diameter is 5.2 mm is$0.96v_s.$