## Appendix: Maximal power of heat machines

#### Abstract

When a motor generates work by exchanging heat with two sources at tempera-tures T 1 > T 2 , Carnot's maximum efficiency 1 − T 2 /T 1 can be approached only for reversible processes; but these are too slow to yield significant power. At the other extreme, rapid processes are strongly irreversible and cannot produce much work. A compromise is needed to maximise the power output. Jacques Yvon tackled this problem in 1955, focusing on a major cause of irreversibility, the delay required for heat transport. A simple approach is presented here. During the first step of a cycle, between the times t 0 and t 1 = t 0 + τ 1 , the motor M evolves at a temperature Θ 1 (t), receiving the heat Q 1 from the hot source S 1 at temperature T 1 . We assume that heat transfers obey Fourier's law, so that the heat flux Φ 1 (t) from S 1 to M has at times t (t 0 ≤ t ≤ t 1) the form Φ 1 (t) = C 1 [T 1 − Θ 1 (t)] where C 1 is supposed to be given. During the second step, between t 1 and t 2 = t 1 + τ A , M evolves adiabatically, its temperature going down from Θ 1 (t 1) to Θ 2 (t 2). During the third step, between t 2 and t 3 = t 2 + τ 2 , M evolves at a temperature Θ 2 (t), giving the heat Q 1 to the cold source S 2 at temperature T 2 , through a flux Φ 2 (t) = C 2 [Θ 2 (t) − T 2 ](t 2 ≤ t ≤ t 3). Finally, between t 3 and t 4 = t 3 + τ A , M returns adiabatically to its initial state, with a temperature rising from Θ 2 (t 3) to Θ 1 (t 4) = Θ 1 (t 0). The cycle has the total duration τ = t 4 − t 0 = τ 1 + τ A + τ 2 + τ A . The work produced by M during a cycle is the difference between the heat Q 1 = t1 t0 dt Φ 1 (t) that it received from S 1 and the heat Q 2 = t3 t2 dt Φ 2 (t) that it yielded to S 2 , so that the average power output is P = 1 τ C 1 t1 t0 dt [T 1 − Θ 1 (t)] − C 2 t3 t2 dt [Θ 2 (t) − T 2 ] . (1) The system M evolves in a closed cycle, so that its change ∆S of entropy between the times t 0 and t 4 vanishes: ∆S = C 1 t1 t0 dt [T 1 − Θ 1 (t)] Θ 1 (t) − C 2 t3 t2 dt [Θ 2 (t) − T 2 ] Θ 2 (t) = 0 . (2) We wish to maximise P as function of the temperatures Θ 1 (t), Θ 2 (t) and of the durations τ 1 , τ A , τ 2 , τ A , for given values of T 1 , T 2 , C 1 , C 2 , τ , under the constraint ∆S = 0. Introducing a Lagrange multiplier λ/τ , and writing that P − λ∆S/τ is stationary with respect to Θ 1 (t) and Θ 2 (t), we obtain Θ 1 (t) = λT 1 , Θ 2 (t) = λT 2 . (3)

#### Domains

Physics [physics]
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