If we introduce a finite temperature difference between source/drain and that of the working fluid in thermal contact, we would make the process irreversible since the direction of heat flow cannot reverse as it is only dependant on the temperature difference. This means that the direction of $Q_H/Q_L$ cannot change.

If we also ensure (for the forward cycle) that the working fluid traces the same path as a carnot cycle, the work done as well as the heat taken and lost matches that of the carnot engine. The process would merely be faster than a carnot cycle. Thus we would have an irreversible engine that has same efficiency of a reversible engine (which shouldn't be possible according to second law of thermodynamics), hence is also one that violates the following variation of Clausius' theorem:

$S_{irreversible} > \int \frac{\delta Q}T$

Where am I wrong in this?

  • $\begingroup$ I think you want to read about the so-called endo-reversible cycle; start here en.wikipedia.org/wiki/Endoreversible_thermodynamics then read the references $\endgroup$
    – hyportnex
    Commented May 15, 2021 at 23:31
  • 1
    $\begingroup$ Clausius inequality uses the reservoir temperature, not the system temperature. This is not the same. $\endgroup$ Commented May 16, 2021 at 0:20
  • $\begingroup$ @ChetMiller Ah, I should've remembered that properly from the derivations! But two follow-up questions: should the above engine that does less work for more temperature difference be considered less efficient and wouldn't that need change in definition of efficiency somehow? By that I mean the basic n = (Qh-Ql)/Qh equation and I'm hoping such an irreversible engine is indeed less efficient. Thanks! $\endgroup$
    – 372191
    Commented May 16, 2021 at 0:48

1 Answer 1


The maximum possible Carnot efficiency is determined by the reservoir temperatures. If you had operated a reversible cycle using the same reservoir temperatures as you had to use for your irreversible cycle, you could have obtained the same amount of work with higher efficiency.

  • $\begingroup$ How can the work be the same? Doesn't the entropy generated in the gas during the heat addition have to be transferred to the cold reservoir, leaving less heat to do work? What am I missing here? $\endgroup$
    – Bob D
    Commented May 16, 2021 at 15:55
  • $\begingroup$ No need to do the math. I know $W=Q_{H}-Q_{L}$. But if you do that to make the work the same how then are the efficiencies different? For both reversible and irreversible cycles the efficiency is $W/Q_H$. $\endgroup$
    – Bob D
    Commented May 16, 2021 at 19:54
  • $\begingroup$ This solves it, I have compared in my post an irreversible engine operating on higher temperature difference and a reversible engine on lower temperature difference. I understand that the second law compares efficiency of engines of different kinds on the same environmental conditions. Thank you! $\endgroup$
    – 372191
    Commented May 16, 2021 at 20:06
  • $\begingroup$ @BobD N1N74's comment answers your question. $\endgroup$ Commented May 16, 2021 at 22:21
  • $\begingroup$ @ChetMiller Not really, but that's OK. $\endgroup$
    – Bob D
    Commented May 17, 2021 at 12:19

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