Hyperloop vacuum
A recent proposal for rapid inter-city transit is the Hyperloop, popularized by Elon Musk in 2012. The idea is to transport people inside of a levitating pod, by way of a partially evacuated tube running from city to city. Among the advertised benefits are its high speed (close to the speed of sound) and the potential to power the Hyperloop with solar energy.
Inside the tube, the pod enjoys a reduced atmosphere with a pressure of 0 . 1 k Pa , about 1 0 0 0 1 th the pressure at sea level. Thanks to the evacuated tunnel and the magnetic levitation used to lift and propel the train, the energy needed to maintain the train's cruise velocity against the opposing forces of drag is very low.
However, this boon requires an up-front energy expenditure in the form of evacuating a 6 0 0 k m -long tunnel large enough to fit the train.
Approximately how many one-way trips of a comparable open-air (i.e. not in a tube), high-speed train would it take to recoup the initial energy expenditure to draw the vacuum?
Assume
- the pods and the Hyperloop tube have a radius of r = 2 . 5 m , the density of air is ρ air = 1 . 2 2 k g / m 3 , the distance from Los Angeles to San Francisco is d LA to SF ≈ 6 0 0 × 1 0 3 m , the drag coefficient is C D ≈ 0 . 1 5 , and when run as a high-speed train, the pods travel at 1 5 0 m / s ;
- once drawn, the Hyperloop tube maintains the partial vacuum state.
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2 solutions
The problem raises so many strange questions. Could you just make the first pod the size of the tunnel and just push all the air out? Would that take 100000J? Does the pod need to be aerodynamic as pictured since it's in a near-vacuum? Why the kangaroo crossing sign in SoCal?
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...trolls put that sign there, Jeremy
As for why the pods need to be so aerodynamic, the suprising answer is that they still need to be, even in a near vacuum! I don't think Musk had expected that when he first started pushing the hyperloop concept.
Here's an engraving of an 19th century pneumatic subway that comes close to what you're talking about....it worked for a while but the idea never took off.
Well, the question tells us what the distance between SF and LA is, but does not tell us that the high-speed train is travelling between these two cities, so it is possible that it is in fact running in Australia. Either that, or the sign and/or the kangaroos are a hallucination.
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The red dirt has to be imported from Australia, since there's nothing like that in the I-5 corridor in California. As anyone who has seen the Mad Max movies, Australia has a lot of that red stuff.
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@Michael Mendrin – Based on this soil colour, the train could also be in Jamaica (plenty of bauxite, but perhaps a bit small to find a flat 150 km stretch) and southern Africa. Of course, neither of these parts of the world are famous for their kangaroos.
I think we are back to the hallucination option...
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@Mark Hennings – Yet another practical problem with the hyperloop concept involving airliner speeds is curves, which will have to be much larger. For most of the I-5 corridor it should work out, but what about the Tehachapi Mountains? The pods/trains will have to slow down, because as it is already, the I-5 through those mountains have plenty of curves that forces the traffic to be a bit careful...at only 70 mph. Passengers on the hyperloop will feel like they're in a rolller coaster if they're going at 300 m/s or almost 10 times faster.
It definitely is a very creative artist impression. I'm curious when they will develop this transparent material that is sturdy, has no issues with changing temperature (contraction and expansion of the pipe due to daily temperature oscillations in the desert will be enormous) and can safely hold an atmosphere of pressure without leaking. So far, the best place to put the hyperloop in my opinion is underground, at a depth where temperature is more or less constant, so contraction and expansion are eliminated. And probably multiple walls around the tube... It will be a very expensive piece of infrastructure, but once it is installed hopefully the long-term energy savings compensate for it.
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@Roland van Vliembergen – It is a very creative artist's impression, kind of like the illustrations done for sci-fi novels. In fact, more recent proposals for hyperloops call for non-transparent steel pipes, which can handle such stresses routinely, the engineering of which is already very well known. Then the pods are to have LED "window" displays that allows the passengers to see the outside world go by in real time.
The vacuum behind the pod would pull the pod back (more accurately the air in front would push it back without the air behind it pushing it forwards)
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The idea of a pneumatic tube transport is to force air behind the cars as the air is being pushed out...see the engraving of a 19th century pneumatic subway I've posted in my comments. It does work, it isn't improbable, but it couldn't compete with non-pneumatic subway systems on a practical basis.
Can you explain why the work needed to evacuate a tunnel is AD[(100-0.1)1000 + 0.1x1000ln(10^-3)] ?
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You would need to evacuate a volume A D ( 1 − 1 0 0 0 . 1 of gas (work required for that is A D ( 1 − 1 0 0 0 . 1 ) × 1 0 0 × 1 0 0 0 = A D ( 1 0 0 − 0 . 1 ) 1 0 0 0 J. But then you would allow the remaining 1 0 0 0 . 1 A D of gas expand isothermally to fill the original volume of A D , regaining A D × 0 . 1 × 1 0 0 0 × ln ( 1 0 3 ) J of energy.
Can we not use a differential volume approach to derive a mathematical equation for the work done to create a low pressure zone?
I'm aware that this is very non-rigorous approach, but it turns out to work quite well in this case.
I assumed that the pod is travelling at constant speed all the time, neglecting starting acceleration and final deceleration. In order to keep the pod moving constantly at speed v , there must be a force equal to drag force, with same magnitude but opposite direction. Hence:
F = 2 1 C d ρ a i r v 2 A
The energy required to apply force F all the way from LA to SF is equal to work done by that force, namely E = W = F ⋅ l . On the other hand, I assumed that energy required to draw a partial vacuum state with pressure p v is equal to the energy difference between systems with pressures p i and p v and volume V , respectively:
E = p i V − p v V = ( p i − p v ) ⋅ V
We equate these two energies and plug in our values:
2 1 C d ρ a i r v 2 A ⋅ l 2 1 C d ρ a i r v 2 A ⋅ n d LA to SF n = 2 C d ρ a i r v 2 0 . 9 9 9 p a t m = ( p i − p v ) ⋅ V = ( p a t m − 0 . 0 0 1 p a t m ) ⋅ A d LA to SF = 4 8 . 5 2 4 6 ≈ 5 0 trips
This is very close to the value @Mark Hennings got in his solution.
The work needed to evacuate a tunnel of cross-sectional area A and length D to 0 . 1 kPa is A D [ ( 1 0 0 − 0 . 1 ) 1 0 0 0 + 0 . 1 × 1 0 0 0 ln ( 1 0 − 3 ) ] = 9 . 9 2 × 1 0 4 A D Joules.
A comparable high-speed train is travelling a distance D at speed v = 1 5 0 m/s, and has cross-sectional area A as well, so it experiences an air resistance force 2 1 ρ A C v 2 , where C is the drag coefficient. Thus the work done against air resistance each trip is 2 1 ρ A C D v 2 . Thus the number of trips that can be done with the energy used to (almost) evacuate the tunnel is ρ C v 2 2 × 9 . 9 2 × 1 0 4 = 4 8 . 1 8 Thus roughly 5 0 trips can be made by the comparable high-speed train.