Subtopic: Calculation of power, torque, fuel consumption, heat load and power losses
Question Statement:
The brake power of a six-cylinder, four-stroke diesel engine running at 3000 rpm is 125 kW. Its brake specific fuel consumption is 200 g kW⁻¹ h⁻¹. Assuming specific gravity of fuel as 0.85, the volume of fuel to be injected per cycle per cylinder in ml is _____.
Official Answer Key:
0.05 to 0.06
Step-by-Step Solution & Explanation
Total fuel mass consumption rate = BP × bsfc = 125 × 200 = 25000 g/h = 25 kg/h.
Volumetric fuel rate = mass rate/density = 25000 g/h ÷ 0.85 g/cm³ = 29412 cm³/h (= ml/h).
For a 4-stroke engine, one complete cycle occurs every 2 crankshaft revolutions, so cycles per hour = (3000/2)×60 = 90000 cycles/h.
Fuel volume injected per cycle (whole engine) = 29412/90000 = 0.327 ml. Divide by 6 cylinders: 0.327/6 = 0.0545 ml per cylinder per cycle. This lies within the accepted range 0.05–0.06.
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