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Temperature changes of gas under centrifugal acceleration

  1. Sep 12, 2011 #1
    What happens to gas temperature in different parts of closed chamber (let's say for arguments sake, a closed cylinder made out of material that transduces heat minimally) that is rotated centrifugally so, that one end of the chamber experiences much greater centrifugal acceleration than the other end?

    Since the gas pressure obviously raises in the outer end (end spinning farther from the rotation axis) of the chamber, is it so that the gas temperature at the outer end raises and stays hotter than the temperature at the inner end?

    Can such a device be categorized a heat pump (transmits thermal energy from colder inner end of the chamber to hotter outer end of the chamber) and if so, how does it draw the energy from rotating motion of the chamber?
     
  2. jcsd
  3. Sep 13, 2011 #2

    Jano L.

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    Hello Jovain,
    in the first instant the adiabatic compression of the air will cause increase in the temperature. However, if the angular velocity remains the same, after a while the heat conduction and possibly convection will cause temperature to equilibrate, until there is the same temperature everywhere. The pressure, of course, will be highest at the surface of the cylinder.
    Jano
     
  4. Sep 13, 2011 #3
    Thanks Jano L, but:

    that's what I thought at first too. But looking at the ideal gas equations, your answer looks impossible. The gas occupying uniform space, having different pressures at different physical coordinates, must have also different temperatures. This is, as I understand, exactly the result of the heat transfer within gas- when gas particles from the lower temperature end of cylinder eventually circulate towards the higher pressure end, the temperature raises, necessarily. To my understanding, if it was not so then the acoustic wave refrigerators would not work - and they do work as known.

    Also it is claimed in many sources that part of the atmosphere cooling when going to higher altitudes is result of dropping air pressure. I do not see the difference between gravitational pull and centrifugal push in this case - these should have same effect to the gas experiencing them.

    So in any case I would need more precise answer and the relation of the said situation to gas equations. Also, I doubt if your answer is correct, because of the reasons stated above.
     
  5. Sep 14, 2011 #4

    Jano L.

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    Hello Jovain,
    if your cylinder is isolated and you wait long enough, the temperature will be the same everywhere. This is because the particles are interacting by collisions and they are in external field; for such cases statistical physics predicts constant temperature and Boltzmann 's distribution of energies.

    "The gas occupying uniform space, having different pressures at different physical coordinates, must have also different temperatures."

    Not necessarily. Gas equation

    pV = NkT

    with constant T implies constant p/n, where n = N/V is density of molecules. That is, at the wall the pressure is higher because the density of particles is higher (they are repelled by the centrifugal force).

    Earth's atmosphere is not isolated (Sun) and therefore not in equilibrium, so you can have temperature gradients.

    I do not know much about acoustic refrigeration, but it is a nonequilibrium process, quite different from the example with cylinder.

    Jano
     
  6. Sep 14, 2011 #5
    Thank you again Jano L!

    After digging deeper into the subject myself, I found out that basically the same thing has been discussed here before, with very many posts:

    https://www.physicsforums.com/showthread.php?t=423654

    It seems even the senior commentors have quite mixed understanding of this issue of thermal gradient in ideal gas under gravitational field/centrifugal acceleration...

    The only actual hard material trying to prove with kinectic theory that the temperature will be same on all parts of this gas will be the same:

    http://iopscience.iop.org/0143-0807/17/1/008

    This paper supposedly only shows the math for the 1-dimensional version, which I can see will support the claim that temperature will be the same on all parts of the gas. However with more degress of freedom (3 in case of ideal gas) I don't see that 3D-version showing the same average kinetic energy distribution would be trivial.

    Rather than quoting the laws that may or may not apply to this situation, I would like to see actual (at least) 3-dimensional proof with kinetic theory and ideal gas. Actually, I am not even sure the said centrifugal system is in equilibrium because of the effects happening between the walls of the rotating cylinder and the gas particles inside the sylinder.
     
  7. Sep 14, 2011 #6
    Hi.

    In the sense of gas under acceleration field, your case and atmosphere on Earth is similar. Higher the altitude, lower the temperature as you are aware. In texts of Earth science you may find the solution.

    Regards.
     
  8. Sep 14, 2011 #7
    Thanks for answer sweet springs, but:

    According to this published paper:

    http://iopscience.iop.org/0143-0807/17/1/008

    and this lenghty discussion:

    https://www.physicsforums.com/showthread.php?t=423654

    and comments from Jano L, your statement (thermal gradient forms in gas because of acceleration field) is supposedly wrong. However I feel inclined to the same intuitive viewpoint than you!

    Anyway atmosphere of the earth is not similar situation, because it is not a closed system.

    Like said and like visible in the discussion mentioned, it seems even the senior commentors have quite mixed understanding of this issue of thermal gradient in ideal gas under gravitational field/centrifugal acceleration...
     
  9. Sep 14, 2011 #8
    Thanks for answer sweet springs, but:

    According to this published paper:

    http://iopscience.iop.org/0143-0807/17/1/008

    and this lenghty discussion:

    https://www.physicsforums.com/showthread.php?t=423654

    and comments from Jano L, your statement (thermal gradient forms in gas because of acceleration field) is supposedly wrong. However I feel inclined to the same intuitive viewpoint than you!

    Anyway atmosphere of the earth is not similar situation, because it is not a closed system.

    Like said and like visible in the discussion mentioned, it seems even the senior commentors have quite mixed understanding of this issue of thermal gradient in ideal gas under gravitational field/centrifugal acceleration...
     
  10. Sep 14, 2011 #9

    Jano L.

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    The standard reason the temperature should be the same everywhere is as follows:

    The gravitational force is given by external field, i.e. the gas itself is not significant contributor to the gravitational force.

    Then potential energy of each molecule is a function of particle's position and therefore linearly proportional to the density of molecules. Energy is then additive, which is supposed to justify Boltzmann's statistics with one temperature everywhere (see Landau Lifgarbagez, Statistical Physics I, sections 4, 34, 37, 38).

    But I agree with you that these arguments are not entirely convincing. It would be nice to have some direct calculation or perhaps numerical simulation to show that temperature will equilibrate.

    Jano
     
  11. Sep 14, 2011 #10
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