# Perfectly Reversible Heat Engine

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 $$\mu=1-\frac{T_c}{T_H}$$?

• can you specify what is Tc andTh Sep 29, 2018 at 10:32
• $T_c$ 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.. Sep 29, 2018 at 10:34
• so Tc is temperature of sink and Tʰ is temperature of Source Sep 29, 2018 at 10:38
• @Sourabh This is standard notation for heat engines... Sep 29, 2018 at 10:57
• You are asking about Carnot's Theorem. You should read up on that. Sep 29, 2018 at 11:47

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: $$\sum_k \frac{\delta q_k}{T_k} \le 0$$.
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".
• @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. Sep 29, 2018 at 19:12