Problem — Carnot cycle in the T-S plane

Problem

On the temperature-entropy diagram, a Carnot cycle is a rectangle. Why are the isotherms horizontal and the reversible adiabatics vertical? What does the enclosed area correspond to?

Solution

Isotherms = horizontal segments. Along an isotherm the temperature is by definition constant. In the plane with $T$ on the ordinate and $S$ on the abscissa, $T = \text{const}$ is a horizontal line. During the isothermal exchange the entropy changes (heat enters or leaves), so the segment is horizontal and shifts in $S$.

Reversible adiabatics = vertical segments. For a reversible transformation the entropy change is:

$\dd S = \frac{\dd Q_{\text{rev}}}{T}$

In an adiabatic there is no heat exchange ($\dd Q = 0$), so $\dd S = 0$: the entropy stays constant (isentropic transformation). $S = \text{const}$ is a vertical line, along which only $T$ varies.

The four phases. The Carnot cycle combines two isotherms (horizontal, at $T_c$ and $T_f$) and two reversible adiabatics (vertical): the result is exactly a rectangle with base $\Delta S$ and height $T_c - T_f$.

Enclosed area. The heat exchanged is $Q = \int T \, \dd S$, i.e. the area under the curve in the $T$-$S$ plane. Over the closed cycle:

$\oint T \, \dd S = Q_c - Q_f = W_{\text{net}}$

The area of the rectangle is therefore the net work produced:

$\ev{W_{\text{net}} = (T_c - T_f)\,\Delta S = \text{area of the rectangle}}$

The $T$-$S$ diagram makes the Carnot cycle geometrically trivial: a rectangle whose area is the work, the most direct way to “see” $\eta = 1 - T_f/T_c$.

Collegamenti

Argomenti: Entropy and the second law · Heat engines Concetti: Carnot cycle · Entropy · Thermodynamic transformations Competenze: Reading graphs Oggetti: Heat engine