If you inject cold and hot air in outlets of a vortex tube, do you obtain a greater pressure at inlet?

  • $\begingroup$ This is a good question, but not really about engineering, so you might get better answers on the physics stackexchange. $\endgroup$ – fibonatic Jun 1 '15 at 14:04
  • $\begingroup$ @fibonatic Do not advise users to cross-post to other sites. If the question is off-topic, flag it as off-topic -> belongs on another site. If the question is on-topic, but might be better answered somewhere else, you can advise the user to flag their post for migration. $\endgroup$ – Air Jun 1 '15 at 15:26
  • $\begingroup$ @Air flagging for belonging to another stackexchange only allows for choosing meta-engineering, at least on the android app. $\endgroup$ – fibonatic Jun 1 '15 at 16:37
  • $\begingroup$ @fibonatic We can look into adding more options, but if the site you want doesn't show up in the list, just use a custom flag. We have plenty of capacity to handle those given the present levels of traffic. $\endgroup$ – Air Jun 1 '15 at 16:39
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    $\begingroup$ @fibonatic I disagree; while it may be on-topic at physics, this is definitely an engineering question. $\endgroup$ – Chris Mueller Jun 1 '15 at 20:23

No; air will not flow from low pressure to high pressure, at least not in a vortex tube. The momentum equation for an inviscid fluid flowing at steady-state in one dimension is $$ \frac{\partial \left( \rho u^2 \right)}{\partial x} = -\nabla p $$ where $\rho$ is the fluid density, $u$ is velocity, and $\nabla p$ is the pressure gradient. The negative sign in front of the pressure gradient signifies that, in the absence of any other forces, the flow direction will always be opposite that of the pressure gradient.

In your hypothetical reverse vortex tube, you want the outlet pressure to be higher than the cold/hot inlet pressures, but you also want the flow direction to be from the inlet to the outlet. These two conditions contradict each other via the momentum equation. In other words, you want the flow direction to be in the same direction of the pressure gradient, which is not allowed (without some internal machinery like a compressor, etc.)

  • $\begingroup$ @Carlton Ram compressors and ejectors are counter examples to your affirmation. If you have a shock wave or a suitable method to transfer momentum to a fluid (a vortex tube is a momentum transfer device), you can obtain a flow with a positive pressure gradient. $\endgroup$ – user3368561 Jun 1 '15 at 20:55
  • $\begingroup$ @user3368561 Good point. There are cases where you can have flow opposing the pressure gradient. My answer assumes that the vortex tube is operating under more-or-less normal conditions, just in reverse. I was specifically imagining two reservoirs (hot air and cold air) connected to the two inlets of the reverse vortex tube. $\endgroup$ – Carlton Jun 2 '15 at 11:38

No. What you will do is simply mix your gasses, arriving at the mean temperature. You loose all the potential useful work in the temperature difference without producing any work. In a way it's hard to explain, in another way terribly easy: there's simply no mechanism, no process there that might transform this heat difference into a higher pressure at the outlet.


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