Efficiency of cycles bounded by two isotherms

  • #1
Philip Koeck
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I want to consider all possible reversible cycles that consist of an isothermal expansion at TH and an isothermal compression at TC.
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?
 
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  • #2
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 $$e=\frac{Q_H-Q_C}{Q_H}=\frac{T_{max}-T_{min}}{T_{max}}-\frac{T_{min}(\delta_H+\delta_C)}{Q_H}$$
 
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  • #3
Chestermiller said:
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.
 
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