Entropy in the Real World: Engines — Question 1

Question 1

Heat-flow rates determine power efficiency and cooling demand

For a cyclic heat engine, $W=Q_h-Q_c$ and thermal efficiency is $\eta=W/Q_h$, with heat and work symbols denoting positive magnitudes. A steady cyclic engine receives heat at 250 kW and delivers net mechanical power of 90.0 kW. It runs for one hour with no net energy storage.

1. Find thermal efficiency, rejected heat rate and energies transferred over the hour.

2. If output rises to 100 kW at the same heat-input rate, find the new efficiency and reduction in cooling demand. Distinguish percentage points from relative efficiency improvement.

Question 1 - Solution

Heat-flow rates determine power efficiency and cooling demand

(a) Method and calculation

$$\eta=90/250=.360,\quad\dot Q_c=250-90=160\,\mathrm{kW}.$$$$Q_h=250(3600)=900000\,\mathrm{kJ},\quad W=324000\,\mathrm{kJ},\quad Q_c=576000\,\mathrm{kJ}.$$

(b) Interpretation and further calculation

$$\eta_{\mathrm{new}}=100/250=.400,\quad\dot Q_{c,\mathrm{new}}=150\,\mathrm{kW}.$$$$\Delta\eta=.0400=4.00\text{ percentage points},\quad\Delta\eta/\eta=.0400/.360=11.111\%.$$Cooling demand falls by 10.0 kW; net power already accounts for any specified internal auxiliaries.

Physical check

Energy-rate accounting must use consistent time units before integrating to energy.