The Efficiencies of Real Engines — Question 9

Question 9

Overall efficiency must weight operating modes by energy input

A plant operates for one hour at net power 80 kW and efficiency .400, then for three hours at net power 20 kW and efficiency .200. Each operating interval is steady; transition energy is negligible.

1. Find total useful energy, total input energy and overall efficiency.

2. Compare with the time-weighted arithmetic average of efficiencies. Derive the correct weighted-average expression and identify its weights.

Question 9 - Solution

Overall efficiency must weight operating modes by energy input

(a) Method and calculation

$$W_1=80\,\mathrm{kWh},\quad Q_1=80/.400=200\,\mathrm{kWh}.$$$$W_2=20(3)=60\,\mathrm{kWh},\quad Q_2=60/.200=300\,\mathrm{kWh}.$$$$\eta_{\mathrm{overall}}=(80+60)/(200+300)=.280.$$

(b) Interpretation and further calculation

$$\eta_{\mathrm{time\ average}}=(.400+3(.200))/4=.250.$$$$\eta_{\mathrm{overall}}=\frac{\sum_i\eta_iQ_i}{\sum_iQ_i}=.400(.400)+.200(.600)=.280.$$The correct weights are fractions of input energy, not elapsed time alone or output energy alone.

Physical check

A mode with low efficiency can consume a large share of fuel even at modest output power.