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I have a question about the 2nd Carnot Corollary.

According to Moran, Fundamentals of Engineering Thermodynamics, the 2nd Carnot Corollary stated that "all reversible power cycles operating between the same two thermal reservoirs have the same thermal efficiency."

However, later in some examples on Vapor Power System and Gas Power System, the efficiency of ideal Rankine cycle and ideal Brayton cycle are lower than that of Carnot cycle.

I am confused because this fact is contradicting with the 2nd Carnot Corollary. Can someone please explain me where did I understand wrong?

Thank you :)

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  • $\begingroup$ Do you mean that the efficiency of these engines is less than $1-\frac{T_{cold}^{}}{T_{hot}^{}}$? Are they working in the reversible limit? $\endgroup$
    – Sunyam
    Commented Oct 13, 2017 at 12:31
  • $\begingroup$ Thus for sure the ideal Rankine and Brayton cycles operating between two sources are not reversible. $\endgroup$
    – Diracology
    Commented Oct 13, 2017 at 12:36
  • $\begingroup$ Does the Rankine cycle operate between just two reservoirs, each at fixed temperature? $\endgroup$
    – Philip Wood
    Commented Oct 13, 2017 at 12:39
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    $\begingroup$ Ahh! I see! So you mean that the corollary is true for two reservoirs with fixed temperature, am I right? If it's so, then this makes sense bcs there is a temperature change in T-s diagram for ideal Rankine and Brayton cycle $\endgroup$
    – Raihan
    Commented Oct 13, 2017 at 12:58

2 Answers 2

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Because the Rankine cycle does not operates between two reservoirs (that is , one of the steps involves heat absorption during a change in temperature). Same with the Brayton cycle.

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  • $\begingroup$ I don't think it can be as efficient as the Carnot cycle. The Bryton cycle exchanges heat through two isobaric processes and those heat exchanges through finite temperature difference imply that the engine is irreversible. Hence, as correctly mentioned by the OP, Bryton cycle is less efficient than Carnot cycle and that is in agreement with Carnot theorem. $\endgroup$
    – Diracology
    Commented Oct 13, 2017 at 12:52
  • $\begingroup$ my god, I missed that the temperature also varies. But you can always imagine that a cycle that can be drawn in a PV diagram is quasistatic and reversible, at least ideally. $\endgroup$
    – user167013
    Commented Oct 13, 2017 at 13:13
  • $\begingroup$ Yes, you can and that is the whole point! The cycle constructed in that way may be reversible. However, an engine operating between only two sources cannot follow that infinitesimal sequence of quasistatic and reversible processes. $\endgroup$
    – Diracology
    Commented Oct 13, 2017 at 13:16
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A Carnot cycle is always defined as one that has 4 reversible stages (1) isothermal heat absorption from a thermostat at temperature (heat source) $T_{high}$, (2) adiabatic work absoprtion (compression) from a work source, (3) isothermal heat release into a thermostat (heat sink) at $T_{low}$, (4) adiabatic work release (exapansion) in to a work sink. The reversible Carnot cycle's efficiency is always $1 - \frac {T_{low}}{T_{high}}$. A reversible adiabatic process is isentropic.

In the Rankine cycle, the work stages (2) and (4) are nearly but not exactly isentropic, hence slightly irreversible. Stage (3) is essentially isothermal as it involves a phase transition from steam to liquid water. Stage (1) that starts with liquid water and ends with steam is not isothermal, it is in fact isobaric, and it is only during the water's phase transition is isothermal, not in the beginning nor towards the end. The Rankine cycle is not a Carnot cycle unlike the Brayton cycle that is in fact a Carnot cycle. See the diagrams on https://en.wikipedia.org/wiki/Brayton_cycle and https://en.wikipedia.org/wiki/Rankine_cycle

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