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Plagiarism claim

SEVERE PLAIGERISM!

This entire article, even including formatting, was ripped from . I'm not sure exactly how to do this, but someone should at least cite it...

-Leif

Actually , it's the other way around :) Look at the small print at the bottom of the page you referenced. Duk 16:40, 1 Mar 2005 (UTC)

Questionable claim

I question this claim under the references:

Van Ness, H.C. Understanding Thermodynamics, New York: Dover, 1969, starting on page 53. A discussion of the vortex tube in terms of conventional thermodynamics, which is able to explain it completely.

For this to be true in 1969 is very unlikely as there are many papers in the 1990s that cannot even claim this. I haven't read the reference in question, however, so I cannot disbute it directly. X2-PB 10:29, 19 August 2005 (UTC)

The following discussion was written by the author of the GAMBLE theory: 24.62.112.163 (talk) 05:57, 15 September 2010 (UTC)John G. Gamble jggcmg @Comcast.net

The physics is absolutely sound and will be proved by demonstration shortly. 24.62.112.163 (talk) 05:58, 17 September 2010 (UTC)John G.Gamble

I don't think it's unreasonable to remove the text, at least on the simple grounds that it does not substantially aid the reader in understanding the topic. -- Taral 19:59, 12 September 2005 (UTC)

Please give some references that claim it is not understood. The point is that it does not contradict the laws of thermodynamics. GangofOne 00:41, 13 September 2005 (UTC)

It is true that it doesn't contradict the laws of thermodynamics, but those laws do not explain the mechanism behind the temperature separation within the tube. This is the point I was making. The latest authors atribute the effect to a number of different causes and it is likely that it is the result of a summation of different mechanisms all occuring at once. X2-PB 12:09, 9 October 2005 (UTC)

Link to Slashdot discussion?

Is the link to a Slashdot discussion really relevant? I don't think it's "encyclopedical"... or else we would go around adding Slashdot links to Misplaced Pages articles in whatever subject they discuss there. I suggest removing it. -- LodeRunner 22:23, 12 September 2005 (UTC)

I agree. Let's archive the link to the Slashdot discussion here. After all, it's just another discussion  Quicksilver 05:06, 9 October 2005 (UTC)

References

The Scientific American article reference is apparently from the 1960 book by that title (out of print since 1972), compiled by C.L. Stong and published by Simon and Schuster. Stong served as editor of the Amateur Scientist column from 1954 until his death in 1977. If anyone has access to a copy of the book, a complete, correct citation for the article would be appreciated. A new anthology of Amateur Scientist articles from the 1920's to 1999 is now available on CD ROM, published by Tinkers Guild (2002) ISBN: 0970347626. Available from Amazon.com  Quicksilver 05:06, 9 October 2005 (UTC)

Done. No one should be without a copy of this excellent book. -- Derek Ross | Talk 20:27, 9 October 2005 (UTC)

There's a date discrepancy between the article and the Reference about Rudolf Hilsch's article. The article states 1945, the Reference 1947. Which is correct?  Quicksilver 05:15, 9 October 2005 (UTC)

The referenced article is a translation of the original. The translation was indeed published in 1947, but I don't know about the original. Itsallgroovy 15:21, 15 January 2007 (UTC)

how it works

So, how does it work? --DavidCary 07:20, 6 January 2006 (UTC)

the outer vortex spins toward the "hot" end of the tube, picking up energy from the inner vortex which is returning from the hot end and going toward the cold end, there are plenty of theories why the inner vortex loses energy while the outer vortex gains it when the law of conservation of angular momentum means the inner one should gain energy, the easiest to understand is an ultrasonic "hum" forms from the vortices moving at arguably a million RPM, and it is this "hum" that shuffles higher energy molecules out to the outer vortex and lower energy molecules to the inner vortex due to the difference in sound moving through warmer and/or colder materials. But like I said, Theory, in reality.... no one REALLY knows how these damned things work, lot of good ideas, but in the end it's a mystery, especially since the phenomenon doesn't occur with water, although that might be more of an issue with thermodynamics, or viscosity, or a thousand other laws of reality. --Munky--68.124.106.93 00:16, 4 February 2006 (UTC)

I believe it does work with water, or atleast there is evidence that it does, but because of the higher viscosity of water it means that a water vortex tube has to be pretty huge.WolfKeeper 23:32, 24 October 2006 (UTC)

It works with water to a very small extent. I believe the maximum temperature seperation is only a few degrees whereas, with air, a variation of over 60 degrees can be achieved. It is thought that the heating effect reported with water is actually due to a different mechanism altogether. The main reason that the mechanism (for air) is not known for certain is that is is the result of a combination of a number of different mechanisms, some of these are laminar, some are turblent. It is possible, however, to accurately model the effect using CFD codes and so, with enough research, it may be possible to exactly discover the presice mechanism(s) causing the effect. X2-PB 19:25, 1 June 2007 (UTC)

I didn't see specs on performance (if they exist). What diameter & length would be required to get the 45°C Δ_t quoted? Trekphiler 16:55, 26 July 2007 (UTC)

Question: Douglas Cotton (talk) 03:18, 3 April 2014 (UTC) Do you have any citation regarding your "theory" here? It would be better to say there are a number of "hypotheses" but none that you know of can be explained with the laws of physics - or words to that effect, wouldn't you agree. It takes a long time and much evidence before a hypothesis becomes accepted as a theory. Douglas Cotton (talk) 03:18, 3 April 2014 (UTC)

It works because of the gravito-thermal effect  Douglas Cotton (talk) 22:16, 2 April 2014 (UTC)

The Second Law of Thermodynamics explains how a state of thermodynamic equilibrium evolves and has maximum entropy within the constraints of an isolated system. In a gravitational field (such as in Earth's troposphere) thermodynamic equilibrium is also hydrostatic equilibrium because of the fact that each is the state of maximum entropy. When molecules are in free path motion between collisions, kinetic energy (KE) is interchanged with gravitational potential energy (PE). Temperature is based on the mean KE per molecule, as explained in Kinetic Theory. This means that gravity sets up both a density gradient and a temperature gradient. (The pressure is then a corollary, being proportional to the product of density and temperature, and it also has a Pressure-gradient force at hydrostatic equilibrium which is the same state of maximum entropy that is thus also thermodynamic equilibrium.)

Now, by equating KE gain with PE loss, we deduce that the thermal gradient is the quotient of the acceleration due to the gravitational force and the weighted mean specific heat of the gases, as derived under lapse rate. For the vortex tube, the effective gravitational force is between about 10^6 and 10^7g, so let's say 5 * 10^6. The approximate distance (internal radius) is about 5mm. The above quotient gives 9.8 * 5 * 10^6 degrees per kilometer, and that reduces to about 250 degrees in 5mm, as is observed according to the article. If a particular tube only generates 10^6g we would expect 50 degree temperature difference. So the hypothesis appears to be well supported by the data in this article. Douglas Cotton (talk) 22:16, 2 April 2014 (UTC)

VQuakr (talk) 23:51, 2 April 2014 (UTC)

Additional links added above Douglas Cotton (talk) 03:31, 3 April 2014 (UTC)

Citation: Misplaced Pages - standard physics as linked, and this Vortex tube article for data relating the 250 degrees temperature difference. Douglas Cotton (talk) 03:11, 3 April 2014 (UTC)

Heat Pump

The definition of a heat pump is a device which uses work to move heat from one resevoir to another. The vortex tube, however, uses neither work nor heat transfer in its system. Given this, and the recent modifications to this article, I question whether or not the vortex tube is a heat pump.

In order for it to run, it needs a compressor to blow the air into it. That takes work. If you run a tube from the cold pipe leading from the vortex tube, then you can use that tube to cool things. Meanwhile the hot pipe can be used to heat things. Hence it is a heat pump.WolfKeeper 23:29, 24 October 2006 (UTC)

You beat me to it, Wolfkeeper. I was about to say the same thing. I'll just add that without a pump to compress the input gas, a vortex tube is just a pipe with a hole in the side. -- Derek Ross | Talk 23:49, 24 October 2006 (UTC)

These arguments show two glaring misconceptions about thermodynamics. One, you claim that the compressed air requires work. This is true, it does require work to compress air. This means that there is a compressor linked up to this tube compressing air, and that the compressor is doing work on the air before it enters the vortex tube. The tube itself exerts no work on the air.

Really? So you are seriously claiming that energy due to the pressure of the hot and cold sides of the vortex tube is exactly equal to the energy of the supply? I find that difficult to believe. I think that the hot side of the vortex tube is hotter than the cold side is cold. That's because of waste heat generated due to viscosity in the tube. So work has been done there at the very least. And from the reversibility of the Carnot cycle we know that it cannot exceed the efficiency of the Carnot cycle (otherwise you could build a perpetual motion machine). In practice, the vortex tube is *very* inefficient, so it doesn't get anywhere near that point.WolfKeeper 05:41, 28 October 2006 (UTC)

Two, you claim that because "the hot pipe can be used to heat things" and the cold pipe can be used to cool things, there is heat transfer. This is not heat transfer into or out of the system. You seem to be mistaking heat with temperature. Yes the temperature at the inlet and the two outlets is different, but the overall heat of all the air in the control volume of the vortex tube remains constant. This is the thermodynamic definition of heat transfer.

Also, if the hot air were to heat another system, that would be after it was out of the system of the vortex tube. You can't consider effects of what the mass does once its out of the system (or before it's in the system, from your compressor argument) to be heat transfer or work done while in the system.

But we don't have to let the air exhaust to atmosphere and thus leave the system. We just have to pipe it from the ends of the vortex tube (after it's heated/cooled the external reservoirs via conduction through the walls of the vortex tube) and feed it back into the pump inlet. That makes the system much easier to analyse and removes your objection about mass leaving the system. Now only heat and work can enter and leave the system: the former entering/leaving via the cold/hot end walls of the vortex tube; the latter via the pump. -- Derek Ross | Talk 05:33, 28 October 2006 (UTC)

69.174.226.202 if you want to be really anal about it, the vortex tube isn't in itself, on its own, a heat pump, but you can add the cold sink heat exchanger, the hot source heat exchanger and plumbing to mix the hot and cold airstreams back together again before going through the air pump and the vortex tube, and then it is. But everybody knows all that, "so quit buggin' me man".WolfKeeper 05:46, 28 October 2006 (UTC)

I've revised the into to remove pro/anti heat pump claims. Let's find good references and cite them properly before re-adding (I don't have access the current references or I would look myself). --Duk 20:21, 9 November 2006 (UTC)

Heat transfer issue: Of course there is a transfer of kinetic (thermal) energy across the internal (imaginary) boundary between the inner and outer regions. This is in accord with what the second law of thermodynamics says will happen as the state of thermodynamic equilibrium with maximum entropy is approached. This is basic thermodynamics, but not well understood here it seems. The transfer stops only when thermodynamic equilibrium is attained, this being the same as hydrostatic equilibrium in this example. Douglas Cotton (talk) 05:17, 3 April 2014 (UTC)

Real World Observations

Understanding the operation of this device has been an interest to me for quite a while. It is the core device in solving both the energy needs and global warming problems we face. The fluid dynamics of it are quite awesome and deserve much more attention. Some of the observations I have discovered is that the temperatures reached on the high side is in relation to the friction generated against the inner walls of the tube. This was verified by using a floating inner wall. The inner wall was supported by magnets such that it spun with the vortex. The high side outlet temperature tracked the inlet temperature. One other feature is that if the low side is used to pre-cool the inlet gas, along with the magnetic floating inner wall, that the temperature of the cold side drops even more radically. If the inlet temperature id 70F then the outlet tends to be below the condensation point of CO2 and you produce nice chunks of CO2 ice. I am not going to get into my best application for this device, but one that I will throw out there, is that buy using the vortex tube as a second stage device on existing HVAC applications, that you can double the cooling capacity of standard HVAC equipment. This is because HVAC equipment takes a liquid to a gas. As the gas exits from the primary cooling coil on return to the compressor, you pass it through a vortex tube stage. This stage is placed in front of the primary cooling coil much like an econimizer on boilers. ( Thank You USS Constellation for my education on 1200PSI Steam ) There is a balance of the high side and low side that must be set via valves. The high side is run through a secondary cooling coil, and the low side on my next experiment is still cold enough to cool the high side outlet of the compressor. But this may be abandoned if I get trenching done to my AG well to use it as an energy well to draw off the unwanted heat. The end result is 10 tons of cooling on a 5 ton compressor, and the compressor runs at a cooler temp and lower load even on hot days. — Preceding unsigned comment added by 134.187.128.41 (talk) 17:39, 29 June 2011 (UTC)

Actual applications

Units - is BTU being used here to talk about power (BTU/h but with /h left off to keep things short?) if so then this needs to be converted to kW rather than kJ. I don't want to make the change my self as I'm british, and we havn't used BTU's for 30+ years and so i'm out of my depth. — Preceding unsigned comment added by 81.106.111.84 (talk) 23:22, 23 July 2012 (UTC)

There's a section on Proposed applications, but I don't see any mention of its actual use in the Helikon vortex separation process despite the fact that this article links to that page in the See also section.

And hasn't South Africa also used this technology to provide relatively cool air for workers in deep mines to breathe? I was led to believe that this was where they got the idea for the enrichment plant. Andrewa (talk) 08:27, 29 January 2012 (UTC)

See also Valindaba and South African Nuclear Energy Corporation. Andrewa (talk) 08:32, 29 January 2012 (UTC)

Changed '...a temperature drop of about 26.6 °C (80 °F).' to '...a temperature drop of about 26.6 °C' (48 °F)'. 26.6 °C = 80 °F on a thermometer but in absolute terms 1°C = 1.8°F. Pawprintoz (talk) 02:04, 25 January 2013 (UTC)

p.s. Unless of course the 80 °F is correct and the 26.6 °C was wrong. This may be true in light of the lengthy quote in italics under 'Questionable claim' above. Please correct it if I'm wrong. Pawprintoz (talk) 09:22, 28 January 2013 (UTC)

These are incorrect statements in the article

(a) It is also usually agreed upon that there is a slight effect of hot air tending to "rise" toward the center, but this effect is negligible — especially if turbulence is kept to a minimum.

(b) One simple explanation is that the outer air is under higher pressure than the inner air (because of centrifugal force). Therefore the temperature of the outer air is higher than that of the inner air.

Reasons:

(a) Whilst in the Earth's troposphere convection appears slow, the rate at which thermodynamic equilibrium with maximum entropy is approached in this vortex tube is almost instantaneous because we have a radial gravitational force field (created by centrifugal force) which is several orders of magnitude greater than Earth's gravitational force. Molecules that happen to move towards the outside gain very significant kinetic energy, whilst those that move towards the center lose much of their kinetic energy as they gain equivalent (and huge) gravitational potential energy. So, rather than being "insignificant" this does in fact explain the whole observed temperature difference, and no other valid physics can.

(b) High pressure is not what causes or maintains high temperatures, because this is not a conclusion that can be drawn from the ideal gas law or any other physics. The onus is upon the author of this statement to establish this claim, which is impossible to do using valid physics.

Recommendation: In that there probably is not as yet any suitably reviewed and published explanation, I suggest this article should either ..

(a) just state the observations of different temperatures and explain the mechanism, or

(b) do as in (a) and then discuss what are clearly called "hypotheses" that the editor(s) can also explain using valid physics, not "guesses" as in these two paragraphs.

Douglas Cotton (talk) 03:54, 3 April 2014 (UTC)

For (a), I agree that this seems dubious. It is also unsourced in the article. I went ahead and removed the sentence. Adiabatic processes do indeed change the temperature of a gas, but sentence is unreferenced and at a minimum seems an oversimplification. Note that the onus is not on the author to "establish" this claim; we need sources, not logic or original synthesis. I tagged it as needing sources, which of course the entire section on the tube's operation needs. VQuakr (talk) 04:09, 3 April 2014 (UTC)

Just a comment: Basically I agree, but as I said there will be no "sources" until the world of science comes to grips with the reality of the gravito-thermal gradient (which the vortex tube demonstrates so well with observations matching the expected "dry adiabatic lapse rate" of -g/Cp) and the consensus accepts that Josef Loschmidt was right after all. I'll be spending the rest of my life submitting papers on this until some editor recognizes the validity of the physics, which of course rebuts the greenhouse conjecture that is based on isothermal assumptions with unbalanced energy potentials, contrary to what the Second Law says would be the state of thermodynamic equilibrium. Yes, 99.9% don't understand thermodynamics: I happen to be in the 0.1% who do. Douglas Cotton (talk) 05:28, 3 April 2014 (UTC)

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