Two-stroke motoring test: 5 kW motoring

Data: two-stroke engine; motoring power (friction power) $FP = 5$ kW; $T = 300$ N·m; $N = 1200$ rpm; fuel used = 6.5 kg in 1 h; $CV = 40$ MJ/kg.

In a motoring test the engine is driven by a motor with fuel cut off; the power drawn equals the friction power, so $IP = BP + FP$.

Brake power

$$BP = \frac{2\pi NT}{60} = \frac{2\pi\times1200\times300}{60} = 37{,}699\ \text{W} = 37.70\ \text{kW}$$

Indicated power

$$IP = BP + FP = 37.70 + 5 = 42.70\ \text{kW}$$

Heat supplied

$$Q = \dot m_f\times CV = \frac{6.5}{3600}\times40{,}000 = 72.22\ \text{kW}$$

i) Mechanical efficiency

$$\eta_m = \frac{BP}{IP} = \frac{37.70}{42.70} = 0.8829 = 88.29\ \%$$

ii) Indicated thermal efficiency

$$\eta_{ith} = \frac{IP}{Q} = \frac{42.70}{72.22} = 0.5912 = 59.12\ \%$$

iii) Brake thermal efficiency

$$\eta_{bth} = \frac{BP}{Q} = \frac{37.70}{72.22} = 0.5220 = 52.20\ \%$$

(Check: $\eta_{bth} = \eta_m\times\eta_{ith} = 0.8829\times0.5912 = 0.522$.)

Answer: $\eta_m$ = 88.29 %, $\eta_{ith}$ = 59.12 %, $\eta_{bth}$ = 52.20 %. (These efficiencies are higher than in a real diesel engine, about 30–40 %; they follow from the given data.)