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Proving that a heat engine cannot exceed the carnot efficiency

  1. Aug 13, 2013 #1
    In many textbooks, a proof is provided where the work output of a super-efficient heat engine is provided to a carnot refrigerator, with the net result that a spontaneous heat transfer occurs from the cold reservoir to a hot reservoir.

    Let's use some numbers, TH = 600 K and TL = 300 K, so that means the carnot efficiency is 50% and the carnot COP is 1.
    Between these temperatures, By connecting a heat engine of efficiency 60% to the carnot fridge of COP = 1, then one can show that the impossible occurs.

    But, between these temperatures (TH = 600 K and TL = 300K) what if we connect a super-efficient heat engine of efficiency 60% to a fridge of COP = 0.5?

    Though a super-efficient heat engine can't exist, coupling the two together gives a net heat transfer from the hot reservoir to the cold reservoir, which could occur. What is wrong with this line of argument?
  2. jcsd
  3. Aug 13, 2013 #2

    Andrew Mason

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    You end up with more mechanical work output from the heat engine than is needed to simply return the original Qh. If you use that excess mechanical work to move more heat from the cold reservoir you end up moving more than Qh heat back to the hot reservoir.

    For the heat engine: W/Qh = (Qh-Qc)/Qh = .6; ∴W=.6Qh

    For the reverse cycle: Qc/W = Qc/(Qh-Qc) = Tc/(Th-Tc) = 300/(600-300) = 1; ∴W = Qc'; Qh' = W + Qc' = 2W

    So let the heat engine produce W = .6Qh of work. This results in .4Qh flowing to the cold reservoir.

    Then use that work to move heat back to the hot reservoir: Qh' = 2W = 1.2Qh. This consists of .6Qh coming from the cold reservoir and .6Qh from the work produced from the heat engine. So there is .2Qh more heatflow (.6Qh - .4Qh) out of the cold reservoir than the heat flow into the reservoir.

    So there is a net heat flow from cold to hot with no work being added to the system (all work is generated from the heat flow within the system).

  4. Aug 14, 2013 #3


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    I think you mean COP of 2.

    Plug the numbers into the efficiency equations. You'll see that a 60% efficient heat engine and COP 2.0 refrigerator don't operate at the same temperatures.
  5. Aug 14, 2013 #4

    Andrew Mason

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    What you are trying to do is see if a heat engine of efficiency greater than (Th-Tc)/Th violates the second law.

    You could simply not run the reverse cycle at all and have a net flow of heat to the cold reservoir. But that would prove nothing.

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