Carnot engine

Question posed by Sadi Carnot, a French engineer, in 1824: with a hot reservoir at T₁ and a cold one at T₂, what is the largest possible efficiency of a heat engine, and what cycle gives it? He found the right answer before the basic ideas of heat were firmly settled.

The ideal engine must be reversible. Heat exchange with a finite temperature difference is not quasi-static, so heat must be taken in isothermally at T₁ and given out isothermally at T₂.

To move the working substance between T₁ and T₂ without other reservoirs, only reversible adiabatic steps will do; any other process, e.g. isochoric, would need a whole series of reservoirs between T₂ and T₁.

A reversible engine working between two temperatures is a Carnot engine. The Carnot cycle, with an ideal gas: 1→2 isothermal expansion at T₁, absorbing Q₁ = W₁₂ = μRT₁ ln(V₂/V₁); 2→3 adiabatic expansion from T₁ to T₂, work by the gas μR(T₁ − T₂)/(γ − 1).

3→4 isothermal compression at T₂, releasing Q₂ = W₃₄ = μRT₂ ln(V₃/V₄) (work done on the gas); 4→1 adiabatic compression from T₂ back to T₁, work on the gas μR(T₁ − T₂)/(γ − 1).

The two adiabatic works cancel. Net work W = μRT₁ ln(V₂/V₁) − μRT₂ ln(V₃/V₄), and the adiabatic steps (TV^(γ−1) = constant) give V₃/V₄ = V₂/V₁.

Efficiency: η = W/Q₁ = 1 − Q₂/Q₁ = 1 − T₂/T₁, with temperatures in kelvin. For example, between 500 K and 300 K, η = 1 − 300/500 = 0.40 = 40%.

Every step can be reversed. Run backwards, the cycle takes Q₂ from the cold reservoir, has work W done on it, and delivers Q₁ to the hot reservoir: a reversible refrigerator.

Carnot's theorem: (a) no engine working between two given temperatures is more efficient than a Carnot engine; (b) the Carnot efficiency does not depend on the working substance.

Proof of (a): couple an irreversible engine I, used as an engine, to a Carnot engine R, run as a refrigerator that returns the same Q₁ to the source. If I were more efficient (W′ > W), the pair would take W′ − W of heat from the cold reservoir and turn it wholly into work with no other change, which the Kelvin–Planck statement forbids.

Since Q₂/Q₁ = T₂/T₁ holds for every Carnot engine, whatever the substance, it can define a universal thermodynamic temperature scale; with an ideal gas as the working substance this scale is the same as the ideal-gas temperature.