Topic: IC Engines
A large diesel engine runs on a stroke cycle at 2000 rpm. The engine has a displacement of 25 litre and a brake mean effective pressure of 0.6 $MN/m²$. It consumes 0.018 kg/s of fuel (calorific value = 42000 kJ/kg). Determine the brake power.
- A. 250 KW
- B. 225 KW
- C. 275 KW
- D. none of the above
Answer: A. 250 KW
Explanation: The formula for brake power is $BP = P_{mep} \times V_s \times n$, where $n$ is the number of power cycles per second. For a 4-stroke cycle, $n = N / (2 \times 60) = 2000 / 120$. The brake mean effective pressure is $P_{mep} = 0.6 \times 10^6$ N/m$^2$ and stroke volume $V_s = 25 \times 10^{-3}$ m$^3$. Multiplying these gives $BP = 0.6 \times 10^6 \times 25 \times 10^{-3} \times \frac{2000}{120} = 250,000$ W, which equals 250 kW.