Is there a heat engine (except Carnot ones), which gets the heat at the temperature $T=T_H$ and exhausts its waste heat at $T=T_C$, having an efficiency of $\mu=1-\frac{T_c}{T_H}$?
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$\begingroup$ can you specify what is Tc andTh $\endgroup$– SourabhSep 29, 2018 at 10:32
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$\begingroup$ $T_c$ is the temperature which the heat engine gives heat, and $T_H$ is the temperature when the heat engine gets heat. Think of a Carnot Engine.. $\endgroup$– lminslSep 29, 2018 at 10:34
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$\begingroup$ so Tc is temperature of sink and Tʰ is temperature of Source $\endgroup$– SourabhSep 29, 2018 at 10:38
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1$\begingroup$ @Sourabh This is standard notation for heat engines... $\endgroup$– BioPhysicistSep 29, 2018 at 10:57
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$\begingroup$ You are asking about Carnot's Theorem. You should read up on that. $\endgroup$– Paul T.Sep 29, 2018 at 11:47
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In general, we have the theorem of Clausius stating that in any thermodynamic cycle a system may go through during which arbitrary amounts of $\delta q_k$ heat is absorbed from sources that are at fixed temperatures $T_k$ this inequality holds: $\sum_k \frac{\delta q_k}{T_k} \le 0$.
At the same time it does not matter what the system does, and we implicitly assume that any other stages of the process are adiabatic work transfers. If the inequality is strict then the process is called irreversible, otherwise when strict equality holds the process is called reversible.
By definition, a Carnot cycle is a reversible one that has only two heat exchangers and then $\frac{\delta q_1}{T_1} + \frac{\delta q_2}{T_2} = 0$. In between the two heat transfers we have adiabatic work stages, and since $\delta q_1+\delta q_2 + \delta w=0$ its efficiency is $\frac{\delta w}{\delta q_1}=1-\frac{T_2}{T_1}$. So any reversible cycle with two heat sources is performed by a "Carnot engine".
answered Sep 29, 2018 at 12:23
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$\begingroup$ Regarding your last statement, what about an ideal Rankine cycle? It is reversible and operates between two heat sources, yet it is not a Carnot engine. $\endgroup$– Bob DSep 29, 2018 at 12:35
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$\begingroup$ @Bob_D the ideal Rankine cycle is not a Carnot cycle because it also absorbs heat at varying temperatures during which the pressure is constant, see stage 2-3 in en.wikipedia.org/wiki/Rankine_cycle. In 2-3 the temperature is constant only during the liquid to vapor phase transition, the horizontal line. $\endgroup$ Sep 29, 2018 at 13:00
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1$\begingroup$ I understand that, but shouldn't you say "with two fixed temperatures" instead of two heat sources? The boiler is a single "heat source", albeit the mean temperature at which heat is delivered varies is less than the maximum temperature, thus the lower thermal efficiency. $\endgroup$– Bob DSep 29, 2018 at 13:51
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$\begingroup$ @Bob_D I think I said something like that, see above "... heat is absorbed from sources that are at fixed temperatures $T_k$", and when there are only two such temperatures we have the Carnot cycle. In this context "fixed" means that the temperature of the heat source does not change during heat transfer, i.e., it has a heat capacity that is ideally infinite, in practice, much larger than anything else. $\endgroup$ Sep 29, 2018 at 19:12
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$\begingroup$ @Bob_D It can be shown that any cycle, even a reversible cycle, that has more than two temperatures has lower efficiency than the Carnot efficiency over two temperatures, see physics.stackexchange.com/questions/300347/…. $\endgroup$ Sep 29, 2018 at 19:12