Efficiency of cycles bounded by two isotherms

[Philip Koeck]

I want to consider all possible reversible cycles that consist of an isothermal expansion at T_H and an isothermal compression at T_C.
The other two processes can be isochoric, isobaric, adiabatic or anything else, but they should never leave the temperature range between the two isotherms.
I also want to explicitly exclude heat recycling using a regenerator.
Pressure and volume of the system should always remain finite and the temperature is always finite and larger than 0 K.

Is there a general proof that the Carnot cycle has a higher efficiency than all other cycles considered above?

 

[Chestermiller]

Let ##dQ_H## be the differential increments of heat received by the engine during the heating part of the cycle, ##dQ_C## be the increments of heat rejected by the engine during the cooling part of the cycle, ##T_{max}## be the maximum temperature during the heating part of the cycle, and ##T_{min}## be the minimum temperature during the cooling part of the cycle. Then, $$\int{\frac{dQ_H}{T}}=\frac{Q_H}{T_{max}}+\delta_H$$where ##\delta_H## is positive (since T < ##T_{max}##). Similarly, $$\int{\frac{dQ_C}{T}}=\frac{Q_C}{T_{min}}-\delta_C$$ where ##\delta_C is positive (## since T > ##T_{min}##). So, $$\Delta S=\frac{Q_H}{T_{max}}+\delta_H-\frac{Q_C}{T_{min}}+\delta_C=0$$Therefore, $$Q_C=\frac{T_{min}}{T_{max}}Q_H+T_{min}(\delta_H+\delta_C)$$So the efficiency is $$e=\frac{Q_H-Q_C}{Q_H}=\frac{T_{max}-T_{min}}{T_{max}}-\frac{T_{min}(\delta_H+\delta_C)}{Q_H}$$

 

[Philip Koeck]

Let ##dQ_H## be the differential increments of heat received by the engine during the heating part of the cycle, ##dQ_C## be the increments of heat rejected by the engine during the cooling part of the cycle, ##T_{max}## be the maximum temperature during the heating part of the cycle, and ##T_{min}## be the minimum temperature during the cooling part of the cycle. Then, $$\int{\frac{dQ_H}{T}}=\frac{Q_H}{T_{max}}+\delta_H$$where ##\delta_H## is positive (since T < ##T_{max}##). Similarly, $$\int{\frac{dQ_C}{T}}=\frac{Q_C}{T_{min}}-\delta_C$$ where ##\delta_C is positive (## since T > ##T_{min}##). So, $$\Delta S=\frac{Q_H}{T_{max}}+\delta_H-\frac{Q_C}{T_{min}}+\delta_C=0$$Therefore, $$Q_C=\frac{T_{min}}{T_{max}}Q_H+T_{min}(\delta_H+\delta_C)$$So the efficiency is $$e=\frac{Q_H-Q_C}{Q_H}=\frac{T_{max}-T_{min}}{T_{max}}-\frac{T_{min}(\delta_H+\delta_C)}{Q_H}$$

Thanks! That's a great proof.