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Question about cooling due to expansion of compressed air

  1. Dec 18, 2015 #1
    Hi! I'm a teacher in Mechanical Engineering, and I organize a (yearly) student's project where they design and create a cooling machine. The machine works by compressing air, cooling the compressed air, and then rapidly expanding it. This can lead to temperatures of minus 7 degrees Celsius.

    I want to understand the physics behind this completely. When you compress the air, it becomes denser and hotter. If you cool the compressed bottle for a short time, temperature drops back to near room level. If you release the air, temperature drops. Sounds logical.

    The problem: some of my (physicist) collegues say that the temperature only drops when the expanding air is doing work. What is considerd work in this regard, and why would an expanding gas only drop in temperature when doing work?

    Some say they have tested making a small hole at the end of a high-pressure pnuematic pipe (10 bar) and letting air escape. This supposedly doesn't lead to a temperature drop. Others say that they heard that pneumatic tools can become very cold when used.

    It's also unclear to me if you can create a continues process of cooling by compressing and expanding air.

    It seems that the physics behind cooling with expanding air are not so simple, and my collegues don't seem to agree on this either. So what I'm looking for is someone with a lot of knowledge in this field, who can explain the physics behind it.

    I would be greatful for any replies, or suggestions where to post if this isn't the right place!

    Thanks, Daan Haeyen
     
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  3. Dec 18, 2015 #2

    SteamKing

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    Don't any of your ME colleagues know how a refrigeration system works?

    What you seem to be describing is what is known as the Joule-Thomson effect, which is used quite often in the cryogenic field to liquefy various gases:

    https://en.wikipedia.org/wiki/Joule–Thomson_effect

    In common refrigeration applications, the vapor-compression cycle is often used, but the refrigerant experiences two phases, saturated liquid and superheated vapor, during the operation of this cycle:

    https://en.wikipedia.org/wiki/Vapor-compression_refrigeration
     
  4. Dec 18, 2015 #3
    Hi, thanks for your reply. Of course we know how a refrigerator works. In my situation, there is no phase change.

    But your link to the Joules-Thomson effect is very helpful. Although I still don't understand why it's important for an expanding gass to do work or not. Is simply expanding of air to atmospheric pressure considered work or not? And what is the best way of expanding in our case?

    Thanks! Daan
     
  5. Dec 18, 2015 #4

    SteamKing

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    It takes a certain amount of work to expand the gas through the outlet from the high pressure reservoir. That's explained in the article on the Joule-Thomson Effect.

    Different means of throttling the gas can be used, but one essential feature is that the throttling mechanism should be well insulated to prevent the transfer of heat to/from the gas while it's expanding. A simple valve can be used.

    This process is used to produce industrial gases like oxygen and nitrogen from the air. You should look for additional topics on air separation and cryogenic engineering.
     
  6. Dec 18, 2015 #5
    It is correct to state that the gas only cools if it does work when expanding. If the gas expands into a vacuum no work is done and there is no change in temperature.
    Temperature is a measure of the KE (velocity) of molecules, expansion into a vacuum has no effect on the KE of the molecules.
    Expansion into, say the atmosphere, means that the gas does work...pushing back the atmosphere, therefore the molecules lose KE, ie are slower, ie..are colder
     
  7. Dec 18, 2015 #6
    For the case of a gas contained under pressure in an insulated cylinder in which gas is allowed to escape through a hole or a valve, picture an imaginary membrane surrounding the gas that remains within the cylinder. This gas expands and does work on the gas ahead of it as it forces that gas out of the cylinder. From the first law of thermodynamics, if the gas within the membrane does work, its internal energy must decrease. For an ideal gas, the internal energy is a monotonic function of temperature. So, when the gas inside the membrane expands and does work, its temperature must decrease.
     
  8. Dec 18, 2015 #7
    Thanks for all the replies. So an expanding pressure cilinder does work by pushing the lower-pressure gas away and therefore decreases in temperature. The most temperature drop would occur inside the cilinder, I assume?

    Does it make a difference if the expanding gas does more work, say in the form of driving a wheel or something?

    Also, why doesn't the temperature decrease when simply openening a valve of a high-pressure air pipe? Or should it?

    Thanks, Daan
     
  9. Dec 18, 2015 #8
    Yes.
    Sure. The more work it does, the cooler it gets.
    Until the gas is expanded, no work is done.
     
  10. Dec 18, 2015 #9

    SteamKing

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    It's not clear what you mean here. Are you talking about a piston in a cylinder?
    Yes, this is quite clearly explained in the article. The expansion of the gas must be free and not expanded thru a turbine, for example.
    If the throttling conditions are as described above in the article, the J-T effect will take place, and the escaping air will cool. If the valve is not insulated (and most probably are not), then heat can be exchanged during the throttling process, and the temperature drop may not occur.
     
  11. Dec 19, 2015 #10
    Sorry, I mean a simple pressure cilinder, without a piston (actually, students use a Coca Cola bottle).

    So say Yes, but then you say that the expansion must be free and not through a turbine. I've read this in the article, and I don't get it. Expanding through a turbine surely makes the air do more work?

    Thanks!
     
  12. Dec 19, 2015 #11

    SteamKing

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    The expansion of the gas must be isenthalpic in order to obtain the decrease in temperature. Expanding the gas thru a device like a turbine means the gas is no longer isenthalpic coming out the exhaust of the turbine; the enthalpy of the gas must decrease in order to furnish the work driving the turbine.

    The decrease in temperature of the gas occurs because the kinetic energy of the gas molecules is reduced by being converted into potential energy. Running the gas thru a turbine would not allow this conversion of kinetic to potential energy in the gas to take place.
     
  13. Dec 20, 2015 #12
    SteamKing and I have had some discussions in pm's about this thread, but were not able to come to a consensus regarding the proper answer, or even about what the question was. SteamKing felt that most of the focus of your question was on what is happening to the temperature of the gas as it passes through the hole in the wall of the cylinder (Joule-Thompson effect), and I felt that most of the focus was on the change in temperature of the gas expanding within the cylinder, prior to exiting the hole. Actually, in real life both these things are occurring, so I'm going to discuss both of them.

    Inside the cylinder, the gas is experiencing close to an adiabatic reversible expansion, where the gas remaining within the cylinder expands and does work to force the air ahead of it out the hole. In an adiabatic reversible expansion, a gas undergoes a decrease in internal energy equal to the amount of work that it does. This results in a decrease in the temperature of the gas. So the gas exiting the cylinder is already partially cooled when it arrives at the hole.

    Passing through the hole in the cylinder, the gas experiences an irreversible adiabatic expansion called the Joule-Thompson effect. This expansion takes place at constant enthalpy (per unit mass) of the gas. In Joule-Thompson, the cooling that would have taken place if the expansion were ideally reversible is nearly offset by the irreversible viscous frictional heating that occurs in passing through the hole. So a smaller temperature drop would occur than if the expansion were reversible. If the gas entering the hole were at relatively low pressures were ideal gas behavior prevails (nominally less than 10 bars), practically all the expansion cooling would be cancelled by the frictional heating and there would be virtually no temperature change whatsoever. However, if the pressure of the gas entering the hole were at higher pressures, beyond the ideal gas region, there would be some additional cooling, resulting from the effect of pressure on enthalpy for a real gas (non-ideal gas effect).

    So, in summary, for the gas that has exited the hole, the net effect of both these steps is a decrease in temperature, compared to its original temperature that prevailed inside the cylinder before the hole existed. Part of this decrease is due to the expansion within the cylinder and the remainder is due to the Joule-Thompson effect in the hole.

    Chet
     
  14. Dec 21, 2015 #13
    Thanks for your reply Chester and Steamking.

    The reason behind my question is this: for a few years we have this project where students have to design and build a cooling machine, just using air and water as media and powered just by hand. This for cooling a small chamber for storing medicines in third-world countries.

    Until now, they also had to research what method to use. After a few weeks, this leads to a choice for Peltier, vortex tubes, reversed Stirling engines and compression-expansion machines. After the first year, we discarded Peltier as a choice because of the limited mechanical challenge.

    The only method that has ever worked (when build by the students) is the compression, cooling and expansion of air.

    Now, these being mechanical engineering students, I would like to skip the research into different cooling methods and get straight to business by providing them with a method or even a concept.

    There are enough engineering difficulties when doing this. For example: one of the specs is that the machine is operated by one continues motion only, so valves are opened and closed automatically. So far, there has been no group that really solved this, mainly because of too little time.

    So, that's why I want to know exactly what happens with the compression-expansion method. As I understand it, the most cooling happens inside the bottle. (By the way, pressures are limited to 3 bars above atmospheric.)

    So it would make sense to think about a concept where the air doesn't expand through a hole, but by pushing away a piston. If you could use part of that work to compress another cilinder, you have an interesting system, that starts to look like a reverse Stirling, only a bit simpler (hopefully ;)

    Thanks, Daan
     
    Last edited: Dec 21, 2015
  15. Dec 22, 2015 #14

    cjl

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    So it does, and expanding gas through a turbine causes an even larger temperature drop, hence the extremely low temperature at the exit of a turboexpander (https://en.wikipedia.org/wiki/Turboexpander).
     
  16. Dec 26, 2015 #15
    330px-Refrigerator-cycle.svg.png
    basically the bottle represent area 1.
    the coolest part is right after expansion valve where the most pressure drop occurs. in your case it is right after the hole.
     
  17. Dec 26, 2015 #16
    Another common example is CO2 fire extinguisher. You can easily find air moisture turns into frost around the nozzle.
     
  18. Dec 28, 2015 #17
    Temperature is a measure of the KE of the molecules in the rest frame of the center of mass and expansion into a vacuum may change the velocity of the center of mass. However, this effect can't be used in a cooling machine.
     
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