Hyperloop: Air can be unpleasant

The drag force will be proportional to the density of the medium (air). This density is proportional to the number of air molecules in it, since volume is constant. The number of molecules in the pod, $n$ , is then proportional to the pressure since temperature and volume are constant. This can be seen by the formula $PV=nRT$ . Therefore the drag force is proportional to the pressure, so we can write $F_1/P_1 = F_2/P_2$ . Then $320/99 = F_2/101325$ , so $F_2 = ma =327515.15 N$ . Therefore $a = F_2/m = 21.834 = 2.228g$ $m/s^2$ , or $2.228$ $g$ 's.

First we must use the pressure formula to determine the amount of surface area the force is acting on under standard conditions in the Hyperloop.

$P = \frac{F}{A}$

$A = \frac{F}{P} = \frac{320N}{99Pa} \approx 3.23m^{2}$

After the hole in the tunnel is created the only information that has been directly given to us is the pressure. We know that the surface area of the pod does not change after the hole is made in the tunnel and therefore it is constant. We can use the value obtained after solving the last equation to determine the force now acting on the Hyperloop pod.

$P = \frac{F}{A}$

$F = PA = (101325Pa)(3.23m^{2}) \approx 327515N$

Now we can use Newton's second law of motion to determine the acceleration the passengers in the Hyperloop are experiencing. Remember the mass of the pod is provided in the details and assumptions.

$F = ma$

$a = \frac{F}{m} = \frac{327515N}{15000kg} \approx 21.8ms^{-2}$

Now we must convert this acceleration from $ms^{-2}$ to $g$ and state the answer to three significant figures.

$a = \frac{21.8ms^{-2}}{9.8ms^{-2}} \approx 2.23g$

Consider the drag equation: $F_D\, =\, \tfrac12\, \rho\, v^2\, C_D\, A$ , note that a change in pressure would only affect $\rho$ . To see what happens to $\rho$ , we rearrange $Pv=nRT$ to give $n=\frac{PV}{RT}$ , $V, T$ and $R$ are all constant, so $n\propto P$ ; noting also that $n\propto\rho$ , we deduce that $\rho\propto P$ .

The original drag force was $320~N$ , so the new drag force is $320\times \frac{101325}{99}$ , subtracting the Hyperloop propulsion force we have $320\times \frac{101325}{99}-320=327195~N$ . $a=\frac{F}{m}=\frac{327195}{15000}=21.8~ms^{-2}$ . Dividing $21.8$ by $9.8$ gives us our answer of $2.23~g$ .

We know that the pressure of a gas is directly proportional to its density and the density of a gas is directly proportional to the drag force exerted on a body moving in the gas at a constant velocity. This means that the pressure of a gas is directly proportional to the drag force exerted on a body moving in the gas.

Therefore the Drag Force Before is given by the pressure before multiplied by a constant value $k$ : $$ 320=k \cdot 99 $$ $$ k=\frac{320}{99} $$

The Dag Force After is given by $k$ multiplied by the pressure after: $$ D_a=\frac{320}{99}101,325 $$

Knowing that Hyperloop is moving at a constant velocity (0 acceleration) gives us that the force that pushes it is equal in magnitude to the Drag Force Before.

After the change in pressure the Drag Force gets stronger and therefore the net force on the train is given by the difference between the force that pushes it and the Drag Force After: $$ F=320-\frac{320}{99}101,325 $$

The acceleration is given by force divided by mass: $$ a=\frac{320-\frac{320}{99}101,325}{15000} $$

The acceleration in g's is given by the acceleration in $ms^{-2}$ divided by $g$ :

$a_g=\frac{320-\frac{320}{99}101,325}{15000 \cdot 9.8}=-2.22581 \qquad \mbox{g's}$

q.e.d.

We can first use the equation Area = Force/Pressure to calculate the area of the front of the train. The drag of 320N is being applied to the front of the train with a pressure of 99Pa.( 99Pa = 99N/m^2). Thus: 320/99 = 3.23m^2. Now we can find the force applied to the front of the train with a new pressure of 101,325N. We use the same formula but switched around to get: F = 3.23*101,325 = 327,280N. Now we can find the deceleration using the formula a = Force/mass (F=ma). Thus: a = 327,280/15000 = 2.228m/sec^2

We know that P=F/A 99=320/A So, A=320/99=3.2323... So now, 101325=F/3.2323... F=327515.151514 mg'=327515.151514 g'=327515.151514/15000 =21.834343 ag=21.834343 {g'=ag ; a is a constant} a=21.834343/9.8 =2.228 approx

* F = $Pressure\times Area$ *

Initially, F = 320N, therefore, * Area = F/Pressure = (320/99) = 3.23 $m^{2}$ *

Pressure is changed by 101325-99 = 101226 pascals.

*Therefore, *

F = $Change in Pressure\times Area$ Acceleration = F/mass

Change in pressure = 101325-99 = 101226

Acceleration = ( $101226\times Area$ = $101226\times 3.23$ )/150000 = 2.228 m/ $m^{2}$

Since the area is the same: $F_i\times P_i$ = k

Then: $F_2 = \frac{F_1\times P_2}{P_1}$

You can say that $F=M\times g_{ap}$

Then: $g_{ap} = \frac{F_1\times P_2}{m\times P_1}$

Finally: $\frac{g_{ap}}{g} = \frac{F_1\times P_2}{mg\times P_1} = 2.228$

Drag force is directly proportional to the pressure. Let the new drag force be F. Therefore, $\frac{320}{F}$ = $\frac{99}{101,325}$ which yields F=327,515.15N. Now, acceleration= $\frac{F}{m}$ = $\frac{327,515.15}{15000}$ =21.83m/ $s^{2}$ . Thus, acceleration in terms of g= $\frac{21.83}{9.8}$ =2.228 times of g.