Entropy in the Real World: Engines — Question 3

Question 3

The area of a reversible temperature-entropy cycle equals net work

A reversible cycle is a rectangle in a $T$–$S$ diagram. The upper and lower isotherms are at 600 K and 300 K; their entropy width is 4.00 J/K. The cycle moves right along the upper isotherm and left along the lower one.

1. Find absorbed heat, rejected heat, work and efficiency from the diagram.

2. Explain why the vertical legs transfer no heat, and whether the area rule $Q=\int T\,dS$ can be used unchanged for an irreversible path.

Question 3 - Solution

The area of a reversible temperature-entropy cycle equals net work

(a) Method and calculation

$$Q_h=600(4)=2400\,\mathrm J,\quad Q_c=300(4)=1200\,\mathrm J.$$$$W=Q_h-Q_c=(600-300)(4)=1200\,\mathrm J,\quad\eta=.500.$$

(b) Interpretation and further calculation

$$dS=0\text{ on a reversible vertical leg}\ \Rightarrow\delta Q_{\mathrm{rev}}=T\,dS=0.$$$$W_{\mathrm{cycle}}=\oint T\,dS\quad\text{for the reversible cycle}.$$For an irreversible process, $dS$ includes entropy generation, so actual heat is not generally $T\,dS$.

Physical check

The clockwise rectangle has positive $\oint T\,dS$ and positive work output.