A two cylinder four stroke diesel engine develops 15 kW power at 2400 rpm. Its brake specific fuel consumption is 0.268 kg kW⁻¹ h⁻¹. If the specific gravity of fuel is 0.85, quantity of fuel injected per cylinder per cycle in m³ is
Options:
(A) 0.028 × 10⁻⁶
(B) 0.033 × 10⁻⁶
(C) 0.056 × 10⁻⁶
(D) 0.065 × 10⁻⁶
Official Answer Key:
B
Step-by-Step Solution & Explanation
Total fuel mass flow = BSFC × Power = 0.268 × 15 = 4.02 kg/h. Fuel volume flow = mass/density = 4.02/(0.85×1000) = 4.729×10⁻³ m³/h. For a 4-stroke engine, each cylinder fires once every 2 crankshaft revolutions, so power (firing) cycles per cylinder per minute = N/2 = 1200, i.e. per hour = 1200×60 = 72,000 cycles/h per cylinder. Volume injected per cylinder per cycle = (4.729×10⁻³ m³/h)/(2 cylinders × 72,000 cycles/h) = 3.28×10⁻⁸ m³ ≈ 0.033×10⁻⁶ m³, matching option (B).
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