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I learned that the theoretical Carnot cycle has the highest possible efficiency:

$e=1-\frac{T_{cold}}{T_{hot}}$

What about the ideal Stirling cycle? Does it also create the Carnot efficiency?

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Any heat engine, as along as it is reversible, has the same efficiency as a Carnot engine. A proof relies on the fact that if a heat energy is reversible, we can run it backwards with the same efficiency; hence by connecting two heat engines between reservoirs and employing energy conservation, we can show that the efficiency cannot be higher than a Carnot engine. Therefore since an ideal Stirling engine is also reversible, it has the same efficiency as a Carnot engine.

EDIT: in response to the comment below, let us assume we have aStirling engine using a monatomic gas as a heat exchanger, operating between temperatures $T_{hot}$ and $T_{cold}$, undergoing isochoric heating and cooling at volumes $V_1$ and $V_2$ respectively. It can be shown that in this case, the efficiency $\eta$ is given by

$\frac{1}{\eta} = \frac{3}{2ln(\frac{V_2}{V_1})}+\frac{T_{hot}}{T_{hot}-T_{cold}}$.

Now in the idealized limit as $\frac{V_2}{V_1} \rightarrow \infty$, then we can see the efficiency tends towards a Carnot efficiency. Of course this is impossible to achieve in practice, but the maximum possible efficiency of a Stirling engine is the Carnot efficiency because the entropy change around the circuit tends towards zero.

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  • $\begingroup$ This is totally wrong. Only the Carnot-cycle, isothermal-adiabatic-isothermal-adiabatic has the efficiency of the $1-\frac{T_{low}}{T_{high}}$, for proof see physics.stackexchange.com/questions/300347/… $\endgroup$
    – hyportnex
    Commented Mar 24, 2023 at 2:31

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