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Derive the energy required for an ideal refrigerator

  1. Sep 24, 2016 #1
    1. The problem statement, all variables and given/known data
    Show that the energy, E, that needs to be supplied to an ideal refrigerator to cool a mass M with specific heat C by [itex]\Delta[/itex]T from an initial temperature [itex]T_i[/itex] is:
    [tex]E\sim \frac{MC\Delta T^2}{2T_i}[/tex]

    2. Relevant equations
    Carnot efficiency where T_i is the starting higher temperature: [itex]e=\frac{\Delta T}{T_i}[/itex]

    3. The attempt at a solution
    I do not know how to marry the Carnot efficiency with Q=MCdeltaT, or why the approx. equal to appears with the 0.5 coefficient!
     
  2. jcsd
  3. Sep 24, 2016 #2

    Andrew Mason

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    What can you say about the total change in entropy (ie. ΔSM + ΔSsurr) from the initial to final states?

    AM
     
  4. Sep 25, 2016 #3
    Assuming an *ideal* refrigerator involves an irreversible cooling process of mass M, then [itex]\Delta S_M+\Delta S_{surr}>0[/itex]
    ...but how does this help?
     
  5. Sep 26, 2016 #4

    Andrew Mason

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    I think the concept behind the "ideal refrigerator" is that it runs on a Carnot cycle - the reverse of the Carnot engine cycle.

    Write out the equations:

    ##\Delta S_{Mass} = \int_{T_i}^{T} \frac{dQ_{Mass}}{T} = \int_{T_i}^{T} MC\frac{dT}{T} ##

    ##\Delta S_{Surr} = -\frac{(\Delta Q_{Mass} + W)}{T_i}## where W is the work required to run the reverse Carnot cycle (or the energy E that must be supplied to the refrigerator to do the work W),

    Does that help?

    AM
     
    Last edited: Sep 26, 2016
  6. Sep 27, 2016 #5

    rude man

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    Oddball problem.
    Because the answer clearly depends on the refrigerator's (constant) environment temperature T1.

    Call the instantaneous temperature of the mass T2.
    So, make the following assumption: initially, T2 = Ti = T1. In other words, the mass was at T1 for a long time so its initial temperature is T1.
    Then, over a very small but complete carnot cycle, the average T2 = T1 - dT2/2. (It starts at T2 = Ti and ends up at T2 = Ti - dT2).
    You know what we mean by "very small", right? So you need to work with the 1st and 2nd laws in their differential form.
     
  7. Sep 28, 2016 #6
    Not really I'm afraid. I'm conscious that I don't appear lazy so wish to stress that I am keen to have a good go myself, but thermodynamics has never been my strongest suit, and I cannot see an intuitive route to get to their final result.
    I computed the integral as: [itex]\Delta S_{mass}=MC\ln \frac{T_i+\Delta T}{T_i}[/itex] but there are no logarithms in it....?
    So would you mind guiding me further please.
     
  8. Sep 28, 2016 #7

    rude man

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    W is not a part of the change in entropy of the surroundings.
     
  9. Sep 28, 2016 #8
    Yes I did think that. Can you offer some guidance please?
     
  10. Sep 28, 2016 #9

    rude man

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    I did. See my post #5.
     
  11. Sep 29, 2016 #10

    Andrew Mason

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    ?? W contributes to Qh, the heat flow to the hot reservoir. The difference between Qh and Qc is W. In the forward direction, W = Qh-Qc. where W is the work produced by the system, Qh is the heat flow from the hot reservoir and Qc is the heatflow to the cold. In the reverse direction, W + Qc = Qh, where Qh is the heat flow to the hot reservoir (ie the surroundings). So W contributes to Qh and Qh/Th is the change in entropy of the surroundings.

    I assume in this case that Th, the temperature of the hot reservoir, is Ti. But I am not entirely clear on what the set-up is. I was assuming that a Carnot cycle operated between the mass and a hot reservoir at a constant temperature, Ti . Otherwise it would have to state the temperatures that the refrigerator is operating between. But perhaps we do not have the entire question.

    AM
     
  12. Sep 29, 2016 #11

    rude man

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    I think you have the right interpretation, which is what I wrote in my post. And OK the way you wrote the change in entropy of the hot side is correct. I was misled by the appearance of W in the entropy expression (the idea of W/T being entropy threw me).

    The mass and the ambient are initially at the same temperature, then the refrigeration cycle is started and in very short order an infinitesimal Carnot cycle is completed and the temperature of the mass is reduced by an amount dT. The Carnot cycle is infinitesimally small in all four of its segments so the equations have to be in differential form. So integrations are inapposite.

    This problem is interesting in that it involves an infinitesimanl carnot cycle in all four segments. If you remember, back on Aug. 23, 2014 in the intro. physics forum there was a similar problem involving another infinitesimal carnot cycle but there the initial temperature difference between hot & cold reservoirs was initially finite and ended up → 0, so the isotherms were initially widely separated but with the adiabatics → 0. Here all four segments are infinitesimally small and stay that way.
     
  13. Sep 29, 2016 #12
    Thanks both. Yes I posted the entire question.
     
  14. Oct 2, 2016 #13

    Andrew Mason

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    I am still having problems with the question.

    If this was a Carnot refrigerator operating between constant temperatures T and Ti, the COP would be: COP = Qc/W = T/(Ti-T) so W = Qc/COP = MCΔT/(T/ΔT) = MCΔT2/T. Since 2Ti >> T, we know that this is not the setup.

    If, then, this is a Carnot cycle operating between the object initially at Ti and a hot reservoir at Ti (because there is no other temperature mentioned), and a cold reservoir being the object itself (initially at Ti), the energy required (work that must be supplied) in order to effect a change in temperature dT would be: W = Qc/COP = MCdT/(Ti/dT) = MCdT2/Ti. For the answer to be correct, the work supplied would be one half of this amount, which means that the refrigerator would require less work than a Carnot refrigerator, which is not possible. So I think the OP is correct. There is something wrong with the question.

    AM
     
  15. Oct 2, 2016 #14

    rude man

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    The infinitesimal cycle starts with Tc = Ti and ends up with Tc = Ti - dTc. So during the cycle the AVERAGE temperature difference is not dTc but dTc/2.
    The problem statement and answer look right to me.
     
  16. Oct 2, 2016 #15

    Andrew Mason

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    Ok. That may be what they are thinking. But this only works for an infinitesimal change in temperature and is approximately true only when T and Ti are very close.

    AM
     
  17. Oct 2, 2016 #16

    rude man

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    OK, I recomputed this with a finite ΔT and verified the given answer again. Only requirement is that ΔT << Ti.
    Since the statement does not say ΔT → 0 I do agree that a differentially small carnot cycle doesn't have to be assumed and finite integrations are perfectly OK (as I did here).
     
  18. Oct 2, 2016 #17

    rude man

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    OK, I recomputed this with a finite ΔT and verified the given answer again. Only requirement is that ΔT << Ti.
    Since the statement does not say ΔT → 0 I do agree that a differentially small carnot cycle doesn't have to be assumed and finite integrations are then needed (as I did this time). I also agree that the problem should have stated ΔT << Ti. I think some people associate a delta with a small change but that is not of course always the case.
     
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