Hyperloop: Shock waves

As the speed of the pod is constant, its length $l$ is equal to its speed $v$ times time $t$

Therefore, in the tube, the volume of air displaced by the pod per unit time is given by

$\frac{V_{displaced}}{t}=\frac{\pi r_p^{2}l}{t}=\frac{\pi r_p^{2}vt}{t}=\pi r_p^{2}v$ ,

Denote the mass of air sucked out per unit time as S,

Volume of air sucked out per unit time is thus $\frac{S}{\rho_{air}}$

The volume of air flows around the pod per unit time is

$\frac{dV}{dt}=\pi r_p^{2}v - \frac{S}{\rho_{air}}$

The open area between the tube and the pod can be expressed as

$A=\pi(r_t^{2}-r_p^{2})$

the velocity of the air flowing past is hence

$v_{air}=\frac{dV}{A*dt}=\frac{\pi r_p^{2}v-\frac{S}{\rho_{air}}}{\pi (r_t^{2}-r_p^{2})}$

In order to keep the air flow relative to the pod below speed of sound, the following equation needs to hold,

$v+v_{air} \leq v_{sound}$

Rearranging the equations we have obtained,

$v_{air} \pi r_t^{2} = \pi r_p^{2}v -\frac{S}{\rho_{air}}$

$\frac{S}{\rho_{air}}+v_{air}\pi r_t^{2}= \pi r_p^{2} (v+v_{air}) = \pi v_{sound} r_p^{2}$

$r_p=\sqrt{\frac{\frac{S}{\rho_{air}}+v_{air}\pi r_t^{2}}{v_{sound}\pi}}$

Using The Ideal Gas Law $PV=nRT$

$\rho_{air}=\frac{m}{V}=\frac{n\mu}{V}=\frac{PV\mu}{RTV}=\frac{P\mu}{RT}$

plugging in the expression for $\rho_{air}$ into the equation, we get

$r_p=\sqrt{\frac{\frac{SRT}{P\mu}+(v_{sound}-v)\pi r_t^{2}}{v_{sound}\pi}}$

Subbing in the numerical values, assuming

$R=8.31J/molK, v_{sound}=340m/s$

we find the radius of the pod to be $\boxed{0.729}$

In an empty section of the tube, prior to the pod passing through, air fills the volume at a molar density given by the ideal gas law $\displaystyle n/V = P/RT$ . The density of air is then given by $\displaystyle\rho_{air}=\mu P/RT$ .

The volume displaced by the pod is given by $\displaystyle \pi r_{pod}^2l_{pod}$ . As the pod travels at constant speed, we can write $\displaystyle l_{pod} = v_{pod}t$ . Then, the mass of air displaced by the pod per unit time is given by $\displaystyle\rho_{air}\pi r_{pod}^2v_{pod}$ . At the same time, the pod can clear air at the rate $v_{suck}$ .

The volume of air that has to travel down the sides of the tube per unit time is then

$\displaystyle \frac{\Delta V}{\Delta t} = \frac{\rho_{air}\pi r_{pod}^2v_{pod}-v_{suck}}{\rho_{air}}$

The area of the gap between the sides of the tunnel and the pod is given by

$\displaystyle \Delta A = \pi\left(r_{tunnel}^2-r_{pod}^2\right)$

The velocity of the volume of air moving down the cracks is then $\displaystyle \frac{\Delta V}{ \Delta t}\frac{1}{\Delta A} = v_{air} = \frac{\rho_{air}\pi r_{pod}^2v_{pod}-v_{suck}}{\rho_{air}\pi\left(r_{tunnel}^2-r_{pod}^2\right)}$

This gives the velocity of air relative to the walls of the tunnel as a function of the radius of the pod. We'd like to know when the velocity of air relative to the pod, which is given by $v_{pod}+v_{air}$ , is equal to the speed of sounds, $v_{sound}$ = 340.29 m/s .

Setting $v_{sound} = v_{air} + v_{pod}$ and solving for $\displaystyle r_{pod}$ we find

$\displaystyle \boxed{r_{pod} = \sqrt{\displaystyle\frac{RT}{P\pi\mu}\frac{v_{suck}}{v_{sound}} + r_{tunnel}^2\left(1-\frac{v_{pod}}{v_{sound}}\right)}}$