The Carnot cycle and maximum efficiency

The Carnot cycle is made up of two isotherms and two reversible adiabats. It is the most efficient cycle possible between two given temperatures, and this is no coincidence: it is proven by imposing that all transformations be reversible, i.e. that the total entropy change be zero in each stage ($\Delta S_\text{tot} = 0$).

The two isotherms. Along an isotherm of an ideal gas the internal energy does not change, so all the exchanged heat equals the work:

• $A \to B$ (isothermal expansion at $T_H$): $Q_H = nRT_H\ln(V_B/V_A)$
• $C \to D$ (isothermal compression at $T_C$): $Q_C = nRT_C\ln(V_C/V_D)$

The two adiabats. Along a reversible adiabat $TV^{\gamma-1} = \text{const}$ holds, and the entropy of the gas stays constant:

$\begin{aligned} B \to C:\quad & T_H V_B^{\gamma-1} = T_C V_C^{\gamma-1} \\ D \to A:\quad & T_C V_D^{\gamma-1} = T_H V_A^{\gamma-1} \end{aligned}$

The key step. Dividing the two adiabatic relations side by side, the temperature factors cancel and we get $V_B/V_A = V_C/V_D$. The two logarithms in the expressions for $Q_H$ and $Q_C$ are therefore equal, and only the temperatures remain:

$\frac{|Q_C|}{Q_H} = \frac{T_C}{T_H}$

Substituting into the general efficiency $\eta = 1 - |Q_C|/Q_H$ leads to Carnot’s law.

Law — Carnot efficiency

$\ev{\eta_\text{Carnot} = 1 - \frac{T_C}{T_H}}$ No real engine, however refined, can exceed this value between the same two temperatures.

The remarkable fact is that the working fluid does not appear in the final result: the maximum efficiency depends only on the two temperatures, exactly as Carnot had sensed in 1824.

Warning — kelvin and the third law

The temperatures in the formula must be in kelvin, not degrees Celsius. A $100\%$ efficiency would require $T_C = 0\;\text{K}$, an absolute zero physically unreachable due to the third law of thermodynamics. Carnot’s wall therefore always stays below $100\%$.

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Topics: Macchine termiche Concepts: Ciclo di Carnot · Secondo principio della termodinamica Objects: Gas ideale

Related exercises: Problema — Una macchina troppo efficiente · Problema — Vero o falso su macchine e frigoriferi · Problema — Carnot rispetto agli altri cicli