Problem — Four cycles in the T-S plane

Problem

Four reversible cycles are drawn in the $T$-$S$ plane between the same $T_c$ and $T_f$: (A) rectangle (Carnot), (B) trapezium, (C) tilted parallelogram, (D) ellipse. Which has the highest efficiency? Which the lowest? Why?

Solution

The $T$-$S$ plane as a tool. In a temperature-entropy diagram the heat exchanged is the area under the curve: $Q = \int T\,\dd S$. The heat absorbed is the area under the section where $S$ increases; the heat rejected is under the section where $S$ decreases. The net work is the enclosed area of the cycle.

Why the rectangle (Carnot) wins. The efficiency is $\eta = L/Q_c = \text{enclosed area}/\text{area under the absorption branch}$. Only the rectangle absorbs all its heat at the maximum temperature $T_c$ (top horizontal branch) and rejects all of it at the minimum $T_f$ (bottom branch). Every other cycle absorbs part of the heat at temperatures below $T_c$ and rejects part at temperatures above $T_f$: both of these worsen the efficiency.

$\ev{\eta_A = \eta_\text{Carnot} = 1 - \frac{T_f}{T_c}\ \text{(maximum)}}$

The minimum: the ellipse. The ellipse (D) is the opposite case to the rectangle: it exchanges heat over a continuous range of temperatures, with very little actual exchange at the extreme temperatures. Its enclosed area is small relative to the heat absorbed, so

$\ev{\eta_D\ \text{(ellipse)} = \text{minimum}}$

Ordering. Trapezium (B) and parallelogram (C) sit in between, closer to Carnot the more “angular” and pressed towards the two extreme temperatures they are:

$\eta_A > \eta_B,\ \eta_C > \eta_D$

Moral: among all reversible cycles within the same range $[T_f, T_c]$, Carnot is unbeatable precisely because it’s the only one that concentrates every heat exchange at the temperature extremes.

Collegamenti

Argomenti: Macchine termiche · Entropia e secondo principio Concetti: Ciclo di Carnot · Rendimento · Entropia Competenze: Ragionamento di ranking · Lettura dei grafici