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.

 

[thulsidass]

The temperatures of the isothermal processes determine the efficiency of cycles bounded by two isotherms, like a Carnot cycle. As the temperature difference between the hot and cold reservoirs grows, efficiency rises.

 

[Philip Koeck]

The temperatures of the isothermal processes determine the efficiency of cycles bounded by two isotherms, like a Carnot cycle. As the temperature difference between the hot and cold reservoirs grows, efficiency rises.

I would say only the Carnot engine and certain other engines with regenerators have an efficiency that's only determined by the temperatures of the isotherms.

 

[Philip Koeck]

The temperatures of the isothermal processes determine the efficiency of cycles bounded by two isotherms, like a Carnot cycle. As the temperature difference between the hot and cold reservoirs grows, efficiency rises.

Can you explain more precisely? What do you mean by "cycles bounded by two isotherms, like a Carnot cycle"?