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Vortex tube question

  1. Apr 6, 2017 #1
    I hope most here are familiar with vortex tube i.e. Ranque-Hilsch tube. I just want to guess if humid air used. We can get some idea about functionality of vortex tube from this page. Now take the example for the lowest pressure level mentioned in the page. As per this page, when the cold: hot ratio is 4:!, then the temperature of the hot flow is 107°C above the inlet temperature. That means if the inlet temperature is at around 30°C, then the temperature at the hotter end is 137°C and the colder end will be at 2.5°C. Suppose 100% humid air compressed isothermally to 20 psiG (the pressure level mentioned) and then used in the vortex tube.
    Now, at 30°C, the moisture content of 100% humid air is 30.32 g/m³ while at 2.5°C, it's 5.74 g/m³. That simple means that the moisture content of the hot flow is 24.58 g/m³. All the information given about moisture content of humid air has been collected from this page.
    What made above is just an assumption because I am not sure whether the cold air can be 100% humid or not. It's highly possible that all the moisture will be concentrated at the hotter end as water vapour molecules are lighter than both Nitrogen and Oxygen. I am requesting those who have done some work on vortex tube can show me light in this regards,
     
  2. jcsd
  3. Apr 8, 2017 #2

    RonL

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    I hope to see some input from others, as the tube function has always interested me. Long ago I benefited from the use of a unit to provide cold air inside of a sandblast hood, when building a metal push boat, years later I found the explanation of their operation in an air conditioning service manual, the biggest surprise was the mention of a 5,000,000 rpm spin speed of the air inside????
    We had very good dryers on our Sulair Compressor, so moisture did not come into play, I suspect water inside a tube would present a few difficulties and because of the motion of air in the tube, the calculator might not be usable.

    My impression from the drawings in both links, is that the tube is little understood even by those manufacturing them......maybe we are close to a time when some computer software program can give more cross section numbers?

    Again I hope to see some input here....:smile:
     
  4. Apr 8, 2017 #3
    does it simply mean that?
    Why did the moisture content drop, and where did the difference go?
     
  5. Apr 9, 2017 #4

    Baluncore

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    Compressed air is dried before use. If wet compressed air is used, injection nozzles may ice-up due to adiabatic cooling.

    If the cold fraction reaches 100% RH, denser liquid condensate will move out to dissolve in the hot fraction, while water vapour, being lighter than air will continue to move to the centre. If sufficient water was present in the compressed air, then I would expect the cold fraction to reach 100% RH before exit from the tube.
     
  6. Apr 9, 2017 #5
    The air already contains the moisture inside even before being compressed.
    I want to mean that the difference will be concentrated at the hotter end. As it's hotter, therefore it can contain far more amount of vapour than the amount of mentioned. And at that point I am not sure whether all the vapour will be concentrated at the hotter end or there will be some left in the cooler flow.
    Not possible! This will happen when this compressed air will either pass through a Joule-Thompson nozzle or rotate a turbine.
    My knowledge of physics and common sense tells the opposite. As far as I know (I am sure many here will agree with me) that lighter particles will go to the periphery and heavier particles concentrate at the centre if the rotation is happening in uniform angular momentum.
     
  7. Apr 12, 2017 #6
    Actually, I am thinking about one point. Vortex tubes has been considered to be less efficient than compressors based coolers. But that can be altered. Nobody so far has given attention to a fact that both the hot and cold flow coming out of a vortex tube has velocities. If the cold flow has been used to rotate a turbo expander, than more of its internal energy will be converted into power and the flow will become colder.
     
  8. Apr 13, 2017 #7

    RonL

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    It's been too many years ago, but I remember a pleasant stream of cool air into my sandblast hood and a muffled, shielded hot air exhaust that had little velocity, not anything that would make any useful power. About 30% of the air's energy is lost to internal friction, while the air volume is split into roughly 75% cold and 25% hot discharges. (that split is based on a 170 degree hot and a 40 degree cold discharge)
    I am curious about the comments of where particles would go :rolleyes: would a sand particle and water do the same thing? I have my doubts :smile:.
    Extending the vortex tube upward in scale (size and volume flow) I believe would lead you to a Tesla Turbine or the real principles of jet engine technology in order to convert airflow and heat exchange into mechanical working power.
     
  9. Apr 17, 2017 #8
    What's the velocity of the cold air flow with the mentioned i.e. 75:25 ratio?
     
  10. Apr 17, 2017 #9
    Do you mean that with this level of separation, the temperature of the hot exhaust is at 170°C above the inlet temperature and the colder is -40°C below the inlet?
     
  11. Apr 17, 2017 #10

    RonL

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    Sorry, I should have made it clear the numbers are Fahrenheit temperatures. (40 F cold and 170 F hot) I don't know how to get a degree symbol to appear in my typing.:sorry:

    As for velocities at discharge points, I never have had thoughts or reason to quantify them.

    From the wiki link in your first post, the vortex tube is said to have been discovered in 1933....I was using one in about 1975 and I think their application as cooling devices were just coming into practice.
    You might find it interesting that Tesla patented his turbine in 1916 and some of his talking points in his application are about the thermodynamic conversion to energy that takes place as air flows through his design.
     
  12. Apr 17, 2017 #11
    Well! In that case kindly convert those into °C and tell me the hot and cold outlet temperature. By the way, what's the pressure of the inlet?
     
  13. Apr 24, 2017 #12
    I am currently using a exair vortex system and am having problems with the condensation on the generator. If anyone knows how to get the right fraction so I can get colder air that would be great. As far as what you are saying I have been working on keeping the pressure down and decreasing diameter and length to get more laminar flow... I doubt that helps I haven't seen a lot of vortex discussions and am new to using them.
     
  14. Apr 25, 2017 #13
    At which part the condensation occurs? What's the input pressure and how much is the exit velocity of both hot and cold?
     
  15. Apr 27, 2017 #14
    Condensation occurs around 80 PSI. We use an input pressure of 40 psi because we cannot exceed a certain decibel level. That gives us an cold air exit of 28 ft/s and a hot air exit of 33 ft/s.There is three junctions and about a foot of .25" diameter tubing with the cold air exit. I did a rough calculation and we are having a heat loss of about 500 joules due to this tubing and junctions.
     
  16. May 2, 2017 #15
    In that case, I want to know what's the cold and hot exit temperature in °C.
     
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