91Ó°ÊÓ

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

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

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

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

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

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

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

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$0.96v_s.$