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How does refrigeration work?

  1. Jul 26, 2011 #1
    I know I can easily look this up on Wikipedia, but I chose to use my intuition on this one ;)

    Tell me if my intuition based approach is correct:

    You have a compressor which takes a refrigerant gas such as Freon and compresses/decompresses it. I'm imagining it works something like a piston in a car. When the compressor expands, the volume of the chamber expands, lowering the pressure inside. I'm gonna throw in a form of the ideal gas law here and say that pV [itex]\propto[/itex] T.

    Lets say the initial pressure, volume, and temperature are pi, Vi, and Ti. Now lets assume that if the volume of the chamber doubles, that the pressures halves.

    In that case, pfVf = [itex]\frac{1}{2}[/itex]pi * 2Vi=piVi[itex]\propto[/itex] T.

    So basically nothing happens, the temperature doesnt change because the double in volume is negated by the halving of the pressure.

    So I'm guessing in Freon, as the gas expands, the pressure drops faster than the volume increases. This means that you will always be left with a < 1 coefficient for the pV term, meaning that temperature drops by the same amount.

    Or the opposite can be true, when the compressor compresses the gas, the volume can drop faster than the pressure increases, also lowering the temperature.
  2. jcsd
  3. Jul 26, 2011 #2


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    Air conditioners mostly use reciprocating compressors, but compressors only compress. The expansion is done by a throttling valve. Gases cool when they are expanded, but the main cooling comes from the lower pressure and phase change of the liquid refrigerant. See page 6 here: http://www.freestudy.co.uk/thermodynamics/t5201.pdf
  4. Jul 26, 2011 #3
    Does this apply to every gas? When a gas expands, its pressure decreases. And like I said, is it possible for the increase in V to be equal to the decrease in p, resulting in no net change in T?
  5. Jul 26, 2011 #4


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    Somehow you have this mixed up. Decreasing the volume of a gas increases its pressure and its temperature if you do it quickly, while increasing the volume decreases its pressure and temperature.

    I don't think you can use the ideal gas law because you are moving from one state to another.
  6. Jul 26, 2011 #5
    unless i made a typo in my original post, I don't believe I have said anything erroneous.

    If you lower the volume, i.e. squeeze the gas into a smaller spot, the pressure increases.

    If you increase the volume, i.e. give the gas molecules more room to spread out, the pressure decreases.

    So if a compressor compresses the gas, the volume will decrease and the pressure will increase. And I want to know if the rate at which volume decreases is greater than the rate at which pressure increases. This will leave pV < 1, meaning that T will also drop.

    and conversely, if you expand a gas, the volume will increase and pressure will decrease. Will the rate at which pressure decreases be greater than the rate at which volume increases, also leaving pV < 1.
  7. Jul 26, 2011 #6


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    I don't see how this is correct. How would the temperature drop? Or is this a typo?

    I don't believe that the pressure can change at a different rate than the volume. Aren't the two proportional? (Assuming you don't add heat or take it away)
  8. Jul 26, 2011 #7
    piVi / pfVf = nRTi / nRTf

    the n's and R's cancel out:

    piVi / pfVf = Ti / Tf

    now if the compressor can compress some special refrigerant gas to, lets say, Vf=[itex]\frac{1}{3}[/itex]Vi and pf=2pi, then

    piVi / 2pi[itex]\frac{1}{3}[/itex]Vi = Ti / Tf

    Then the pi's and Vi's cancel. Simplify to get:

    [itex]\frac{3}{2}[/itex] = Ti / Tf

    rearrange to get:

    Tf = [itex]\frac{2}{3}[/itex]Ti

    meaning that the final temperature is 2/3 of the initial temperature.

    the point of this thread is to ask whether the bold line is possible
  9. Jul 26, 2011 #8


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    you are partially correct but you do not use the ideal gas law as Drakkith replied.

    A polytropic process obeys the relation:
    PV^gamma = C
    where P is pressure, V is volume, gamma is any real number, and C is a constant.

    The compression of the gas is modelled as an adiabatic processs, and gamma = Cp/Cv.
    An adiabatic process loses nor gains heat from its surroundings, only work.

  10. Jul 26, 2011 #9

    Andrew Mason

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    One has to apply the first law of thermodynamics. If I compress a gas to half its volume the pressure will increase to more than double the original pressure. This is because the temperature must increase. The reason: although there is no heat flow into the gas, there is work done on the gas. Since dQ = dU + dW, if dQ = 0 then dU = -dW = -PdV > 0; so dT>0

    If I let the hot compressed gas cool to ambient temperature and then let it expand rapidly, the opposite occurs: the temperature decreases to below ambient temperature. If I put that cool gas in thermal contact with a reservoir, heat will flow out of the reservoir into the gas. I then repeat the cycle, compressing the gas etc.

    That is essentially how refrigeration works with some refinements such as Joule-Thompson cooling. It was found that certain gases cool more than others due to the Joule Thompson effect (gas cools when allowed to expand through a throttle valve). These effects are used in modern refrigeration. But the principle is the same.

  11. Jul 28, 2011 #10
    i don't even think the text book i paid for in college was this direct and simplistic in its explanation of the theory. great word stacking!!!!

    another theory that is used here is heat is present in everything. heat has to be removed to cool an object. you don't add cold, you subtract heat. that took me a while to understand also. i just haven't found a way to apply it to space.
  12. Jul 28, 2011 #11


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    In a vacuum there is nothing to transfer heat other than by thermal radiation: http://en.wikipedia.org/wiki/Thermal_radiation

    Space, if you are referring to the "vacuum of space" has no temperature. Only objects in space have temperatures. If you radiate more energy via thermal radiation than you absorb, you will cool down. If you absorb more than you radiate you will heat up.

    One of the main difficulties in spacecraft design is how to get rid of heat created by all those electronics on board. Same goes for astronauts in space suits.
  13. Jul 29, 2011 #12
    cant you just open a window?:uhh:

    so open space far enough from a heat source will allow the heat in an object to disipate into space never to be replenished. simple...
    thank you! but that just brings up more questions.:uhh: ill start a thread. lol.
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