Answer to Question #111881 in Classical Mechanics for Ram

Question

If the radius of the tunnel is 1.1 m, the speed of the pod is 300 m/s and the air in the tunnel is at a pressure of 99 Pa and temperature of 293 K, then what is the maximum radius of the pod in m that will keep the air flow relative to the pod below the speed of sound?

Treat air as incompressible. This isn't true at high mach number, but it will make the solution easier.
The molar mass of air is \mu= 29μ=29 g/mol.
Neglect the effects of gravity and viscosity.
Assume the air flow is perfectly laminar.

Answer 1

The density of air:

$$\rho=\frac{\mu p}{RT}.$$

The pod will displace the volume of air:

$$V=\pi r^2l,$$

where r and l - radius and length of the pod. Since the pod moves at constant speed, $l=vt.$ It gives the mass of air per unit time

$$m=\rho\pi r^2v,$$

and also the pod may suck air at a speed $u$. Therefore, the volume of air per unit time that is displaced in the tube is

$$\frac{V}{t}=\frac{\rho\pi r^2v-u}{\rho}$$

The area of the gap between the tunnel and the pod:

$$A=\pi(R^2-r^2).$$

The velocity of displaced air becomes

$$\frac{V}{tA}=v_\text{air}=\frac{\pi r^2(v-u)}{\pi(R^2-r^2)}.$$

This is the velocity of air as a function of the radius of the pod.

When we set

$$v_\text{sound}=v_\text{air}+v_\text{pod},$$

and express the radius of the pod, we get

$$r=\sqrt{\frac{RT}{p\pi\mu}\frac{u}{v_\text{sound}}+R^2\bigg(1-\frac{v}{v_\text{sound}}\bigg)}=0.734\text{ m}.$$