The Efficiencies of Real Engines — Question 8

Question 8

Regeneration recovers internal heat without adding cycle work

An idealized Stirling cycle uses one mole with $C_V=3R/2$ and $R=8.314$ J/(mol K), isotherms at 600 K and 300 K, and volume ratio 2. Its two constant-volume legs can exchange stored heat through a regenerator of effectiveness $\epsilon=.800$. The regenerator supplies fraction $\epsilon$ of the constant-volume heating requirement; unrecovered heat comes from the hot source. Assume no other losses.

1. Find net work and required external hot heat.

2. Calculate efficiency and compare with perfect regeneration. Explain why recovered internal heat must not be counted as a second external input.

Question 8 - Solution

Regeneration recovers internal heat without adding cycle work

(a) Method and calculation

$$W=nR(T_h-T_c)\ln2\approx1728.85\,\mathrm J.$$$$Q_{h,\mathrm{external}}=nRT_h\ln2+(1-\epsilon)nC_V(T_h-T_c)\approx4205.96\,\mathrm J.$$

(b) Interpretation and further calculation

$$\eta=W/Q_{h,\mathrm{external}}\approx.411048.$$$$\epsilon=1:\quad Q_h=nRT_h\ln2,\quad\eta=1-T_c/T_h=.500.$$The regenerator stores heat during cooling and returns it during heating. Its energy returns to the starting value each cycle, so the recovered share is internal to the full installation.

Physical check

Constant-volume legs do no boundary work; regeneration changes external heat demand rather than the ideal loop work.