Entropy in the Real World: Engines — Question 8

Question 8

A triangular reversible cycle has heat input at a range of temperatures

A reversible cycle in $T$–$S$ space follows A$(1\,\mathrm{J/K},300\,\mathrm K)$ to B$(5\,\mathrm{J/K},500\,\mathrm K)$ on a straight line, then vertically to C$(5\,\mathrm{J/K},300\,\mathrm K)$, then horizontally back to A. Use $\delta Q_{\mathrm{rev}}=T\,dS$.

1. Find heat on every leg and net work.

2. Calculate efficiency and explain why it is below $1-300/500$ even though every leg is reversible.

Question 8 - Solution

A triangular reversible cycle has heat input at a range of temperatures

(a) Method and calculation

$$Q_{AB}=\frac{300+500}{2}(5-1)=1600\,\mathrm J,\quad Q_{BC}=0.$$$$Q_{CA}=300(1-5)=-1200\,\mathrm J,\quad W_{\mathrm{net}}=400\,\mathrm J.$$

(b) Interpretation and further calculation

$$\eta=400/1600=.250<1-300/500=.400.$$$$T(S)=300+50(S-1)\quad\text{along AB, with the stated SI coordinates}.$$Heat enters over a range of temperatures, not solely from a 500 K reservoir. Reversibility alone does not make this cycle a two-reservoir Carnot cycle.

Physical check

The triangle area is $\tfrac12(4\,\mathrm{J/K})(200\,\mathrm K)=400$ J.