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A Vortex Tube takes a pressurized input stream, most typically of a gas, and creates two output streams with a temperature differential. Apparently, it has been described as a Maxwell's Demon.

Both linked sources are scarce with information about how and why this works. Now, I have two questions:

  • Why does it work, specifically why should the situation in the vortex lead to a transfer of thermal energy from the inner stream to the outer one?

  • How efficient can it be?

  • How do you define the efficiency of a device that may be closer to Maxwell's Demon than to a heat pump? My feeling is that any analysis should not only take into account the sum of input energy (thermal and mechanical energy instream) and sum output (thermal energies and pressures of both gas streams), but also the Temperature differential that is created - since that contains an ability to create work.*

  • If course it's pointless to create heat from high grade 8mech.) energy to transform it back to mech. energy - but it gives an idea on the worth of the output.

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  • $\begingroup$ This is a very interesting question - I hadn't heard of these things before. $\endgroup$
    – N. Virgo
    May 7, 2013 at 9:07

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I have performed few tests with Vortex tube to find its efficiency comparing with a refrigeration system.objective was to determine whether we can use compressed air or CO2 to replace refrigeration system.vortex tube works only with pressurised fluid.when sudden drop on pressure reduces temperature and the spiral form separator inside the vortex tube which circulates the fluid around it causing it to separate hot fluid and cold and directs it to 2 different direction.using cold air which got separated. I could cool water up to .5 degree in c scale.but it require high flow rate and pressure to drops down any further. enter image description here

Video on air temperature delta in a vortex tube- Testing Video

Video on cooling water using cold air from outlet of vortex tube.Testing Video

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  • $\begingroup$ GReat to find someone with actual experience! do you have actual figures? $\endgroup$
    – mart
    Sep 24, 2013 at 9:50
  • $\begingroup$ Yes. I have temp diff between inlet and outlet.mass flow rate diff between inlet and outlet.and various temp drop at different pressure and flow rate.i could get finally at 80psig inlet pressure and mass flow rate of 260lpm.a temp drop of .2 degree C and cold outlet.i have data on doe.the noise level is higher at hot air outlet. $\endgroup$
    – Vaishakh Rajan K
    Oct 1, 2013 at 15:03
  • $\begingroup$ And how is the efficiency compared to peltier elements? $\endgroup$
    – fibonatic
    Jan 2, 2014 at 0:28
  • $\begingroup$ I dint do any tests with peltier element.but I can test it.let me try and get back to you.may be in few months. $\endgroup$
    – Vaishakh Rajan K
    Jan 12, 2014 at 15:34
  • $\begingroup$ Ping! Did you get around to do your tests? I'd greatly appreciate if you can expand your answer $\endgroup$
    – mart
    May 26, 2014 at 10:48
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Why does it work? One needs to understand static gas temperature, total gas temperature and propulsion if a proper physical picture of the effect is to be constructed. This is an article by me and co-authors that explains the fundamental law of rotational cooling (also known as Euler's turbine equation), it assumes a sophomore-level math and physics:

Polihronov, J. et al, A.Thermodynamics of angular propulsion in fluids, Phys Rev Lett 109 054504 2012

Also, see this web site, I am putting together an easy-to-read explanation of the vortex tube effect
https://sites.google.com/site/vortextubeeffect

In more detail -

Consider the concept of vortex flow "discretization": let's simplify the vortex flow by introducing a simple flow system, which still exhibits the physics of temperature separation. The simple flow system comprises a rotating adiabatic duct and a tank of compressed gas attached to the inlet of the duct. The outlet of the duct is at $r=0$, while the inlet is at $r=R$, the point $0$ is the rotation center.

Set the system into uniform rotation. Let the linear speed at the inlet is $c= \omega R$. Then, in the stationary frame of reference, the total temperature of the gas at the inlet (at periphery) is $T=T_0 + c^2/2c_p$, where $T_0$ is the static temperature of the gas in the tank. From rothalpy conservation we get the total temperature at the outlet (at center) to be $T=T_0 -c^2/2c_p$, $c_p$ is the isobaric heat capacity of the gas. Thus, the total temperature separation is $\Delta T=c^2/c_p$. What happens with the static temperature $T_s$? At inlet (at periphery), $T_s=T_0$; at outlet, it is $T_0-c^2/2c_p$. Thus, the static temperature separation is $\Delta T_s=c^2/2c_p$.

Temperature separation is observed in a rectilinearly moving system as well. Consider an elemental system, comprising an adiabatic duct and a tank of compressed gas attached to the leading end of the duct. Set the system with uniform linear velocity $c$. Let gas leave the system with velocity $0$ in the stationary frame of reference, the gas comes out the trailing end of the duct.

The temperature separations $\Delta T$ and $\Delta T_s$ are exactly the same as in the rotation case; only now conservation of enthalpy needs to be applied to solve for the temperatures.

To sum up, the vortex tube phenomenon is an example of Euler's turbine equation at work.

When it comes to the vortex tube effect, this analysis is a very good place to start.

I hope this helped!

-J. Polihronov

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    $\begingroup$ Unfortunately the link provided has been "moved", and the new location is a dead end. I would very much like to see it being updated. As of the time of this comment, another paper by this author can be found on ResearchGate $\endgroup$
    – Floris
    Dec 26, 2017 at 17:42
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The input stream does not only have a thermal energy - it also has a mechanical one. Mechanical energy can be used for work, and the gas temperature is easily changed by work - in adiabatic processes it rises when gas is pressed, and falls when gas is able to expand. This gives a general idea why this tube could work and at the same time not be a Maxwell's Demon. Though detailed picture can be very complex and ask for advanced understanding of gas dynamics.

If we calculate total thermal and mechanical energies of both output streams, we would find that some mechanical energy is lost, and total entropy has risen. Mixing those streams back in one would give us a slower and somewhat hotter gas than it was initially. This is the same effect as simple slowing down the stream on some obstacles.

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  • $\begingroup$ 1) can you expand on, um, why part of the gas is expanding in the tube? 2) for any heat engine or heat pump $\Delta T$ plays a large role ... why not here? $\endgroup$
    – mart
    May 7, 2013 at 9:44
  • $\begingroup$ 1) Please think of this not only as of heat engine but also as of a mechanical system. Gas mechanics here is highly relevant. Gas has inertia, and it can hit walls, thus making regions of higher and lower pressure. Also there are other streams of gas besides walls, and possibly some acoustic phenomena. 2) $\Delta T$ plays large role here too, as in the heat pump: it poses constraints on what can be done and what cannot be done with such a device. But $\Delta T$ is not the source of the work here - at least, basically. $\endgroup$
    – firtree
    May 7, 2013 at 10:11
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    $\begingroup$ we are getting closer to the truth... this is fun... look at the tube this way: the compressed air enters the tube through a set of angled nozzles which impart spin to the air. think of the air spinning around the inner circumference of the tube as the rim wall in a rapidly spinning mechanical centrifuge, which is urging the air interior to the centrifuge wall to rotate with it via shear forces. at some radius interior to the rotating rim wall, the remainder of air (the "core") rotates as a single mass, which compresses the air on the periphery of that core and rarifies that in the center... $\endgroup$
    – niels nielsen
    Dec 29, 2017 at 21:20
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    $\begingroup$ via centrifugal forces. this heats the air on the outside of the core and chills the air on the inside of the core, nearest the tube's central axis. the tube peels off the hot peripheral air surrounding the core and dumps it out one end of the tube, leaving the chilled core air behind. $\endgroup$
    – niels nielsen
    Dec 29, 2017 at 21:25
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Gas temperatures rise with compression and fall with decompression. The outer layers of the vortex the is compressing the gas (heat) due to centrifugal force. The center of the vortex has a low pressure comparitavely. (Cool).

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Here is a paper in which the authors compare two open-cycle refrigeration systems: the bell-coleman cycle and the vortex tube, and calculate coefficients of performance for both. along with Nemu Rozario's work, I think this might answer the OP's questions.

https://www.google.com/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&cad=rja&uact=8&ved=0ahUKEwi9svfCg7HYAhUIslQKHRTWBfsQjhwIBQ&url=https%3A%2F%2Fwww.slideshare.net%2Faparnamalyala%2Fair-refrigeration-system-by-bell-coleman-cycle-and-vortex-tube&psig=AOvVaw1LDm7iitvP7LGIoXbr07e5&ust=1514698972087380

see this in SlideShare on-line:

Air refrigeration system by Bell Coleman cycle and Vortex tube 1. By: Project Guide: E.Nikhil Kumar(12000T0338) Anil Kumar M.Aparna (12000T0304) P.Jagan (12000T0314) k.Sirisha (12000T0353) M.Ram Kumar(12000T0321) 2. Aim  The aim of our project is to produce the refrigeration effect using both Bell-Coleman cycle and Vortex tube.  In this project we mainly concentrated on the cold end temperatures of Vortex tube through which refrigeration effect is produced. We fabricated four different vortex tubes with different dimensions, number of nozzles, orifice & Venturi and compared their COP’s and cooling rates.  Our project gives the scope of replacing conventional refrigeration systems with air refrigeration system.

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  • $\begingroup$ regarding the separation of flow into a hot and a cold stream: the "centrifuge" rotational mode apparently exists in the inner core of the air mass inside the device. as noted above, the periphery of the rotating mass of air will be hotter than that in the center; the structure of the tube is such as to peel off that hotter portion of the air and conduct it out one port and send the cooler core air out a different port. In any case, it is most definitely NOT a "maxwell's demon" device. $\endgroup$
    – niels nielsen
    Dec 29, 2017 at 20:58
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This research paper on "The Maximum COP of Vortex Tubes" provides a very clear and thorough explanation.

The introduction explains the vortex tube as an Angular Propulsion Engine and further goes on to derive the maximum theoretical efficiency based on this.

The Maximum COP of Vortex Tubes, Jeliazko G. Polihronov and Anthony G. Straatman

Section 4 on page 4 lists the theoretical maximum efficiency for a vortex tube using air to be 42%. It then sites reference #3 for empirical efficiency valves that consider friction and other inefficiencies (M. O. Hamdan, A. Alawar, E. Elnajjar and W. Siddique. “Feasibility of Vortex Tube Air-Conditioning System”, Proc. ASME, AJTEC2011, (2011).)

The published empirical efficiency for a vortex tube is between 3% to 5%.

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