Would a vortex tube still work if instead of a cone plugged into the 'hot' end you had a smaller hole on the 'cold' end? As I understand it, the point of the cone on the hot end is to only allow the fluid to escape from the outer-most part of the tube where it is hottest, so if you were to instead make the hole on the cold end smaller, as to only allow fluid to escape from the inner-most part of the tube where it is coldest, would it still work?


I've been thinking of building one of these for fun, but I was really hoping someone might be able to answer this question beforehand so I could avoid unnecessary work. As I understand it, these things have to be made very, very well to get a measurable effect.

Oh well, I guess I'm just charging ahead on my own. Wish me luck. I'll post my results back here when I'm done.

  • $\begingroup$ I should think that there are only two sound paths an answer to your question, given the uncertainty in the theoretical underpinnings of this device: (1) are there any numerical / analytical models of this (nearly) axisymmetric system? If so, do they confirm the "undertainty" described on the Wikipedia page or can they yield insight? (I'd suggest adding this to your question and tagging with (in descending priority) "navier stokes", "simulation", "numeric", "mathematical physics" ) (2) Build what you suggest and see what happens. My gut feeling is that your suggestion will not work for ..... $\endgroup$ Nov 14, 2013 at 1:26
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    $\begingroup$ ...two reasons: (1) It seems that this device has been designed pretty much wholly experimentally, so I'd be surprised if simple changes to the configuration along the lines you suggest wouldn't have been tried; (2) there seems to be some need for the flow to be nearly laminar and translationally invariant (along the length of the stovepipe): conical outlets would help keep it so. What you may well find experimentally is that your suggestion would work at much lower efficiency: a sudden rather than tapered block to flow along the lines you suggest would render the flow nonlaminar and .... $\endgroup$ Nov 14, 2013 at 1:29
  • $\begingroup$ ... with complicated end effects, so that a significant length of the tubes would not be working in the laminar, Tx-invariant way the handwaving explanations assume it is (if indeed they are right). $\endgroup$ Nov 14, 2013 at 1:31
  • $\begingroup$ I'm sure posting your findings would be a great self answer. You may have to make your tube really long. $\endgroup$ Nov 15, 2013 at 1:39

1 Answer 1


It seems that the exact exit conditions of a Vortex doesn't influence too much on the Vortex functionality. For example a water Vortex; it doesnt matter if you take the water out from the top, or from the bottom. The basical funcionality of the Vortex remains very much the same.

Please look this film Secondary flows. in 1:20-2:55 and 11:00-12:00 as a reference.

According to this paper the radius of the exit defines the functionality linarily. The coolest air is coming from the middle and the hottest from the biggest radius. Still, the pipe/vortex radialsize doesn't actually make any difference. As long as the peripheral velocity is equal. This is limited by the speed of sound, and is $\Delta T = 115.2 K$ in sea level with the speed of sound $340 m/s$

It can be noted, that this tube works also with incompressible fluids, but then it just separates the heat, without any cooling effect. As the cooling effect needs the fluid to be compressed on the outer radius, and then decompressed in the middle. This causes the cooling simply because of ideal gas law.

So to Conclude the answer; I think it doesn't matter too much how you exactly organize the inflow /outflow holes as long as the basical principe works; Hot comes out from the outer radius, and cold from the middle.

Another interesting aspect was that according to this theory the vortex inside the tube must be a rigid body vortex. This must be also the reason why there is the typical whistling sound; if there is no sound, also the rigid body vortex hasn't formed yet. And when it has formed, the fluid rotates like a solid inside the tube and is "grinding" the tube walls which produces the sound. (this is my own idea)


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