Subtopic: Selection, operation, maintenance and repair of I.C. engines
Question Statement:
A 3-cylinder 4-stroke CI engine coupled with a turbocharger has bore 120 mm and stroke 130 mm. Engine runs at 1600 rpm with volumetric efficiency 150%. Air-fuel ratio (weight basis) = 14.9:1, density of air entering the cylinder = 1.2 kg/m³. Fuel consumption in kg/h is
Options:
(A) 17.05
(B) 25.57
(C) 33.33
(D) 51.14
Official Answer Key:
B
Step-by-Step Solution & Explanation
Swept volume per cylinder, Vs = (π/4)D²L = (π/4)(0.12)²(0.13) = 1.4703×10⁻³ m³.
For a 4-stroke engine, power (induction) strokes occur once every 2 revolutions, so induction events per minute = N/2 = 1600/2 = 800.
Volumetric air drawn (at volumetric efficiency ηv = 150% = 1.5, meaning the turbocharger boosts the air charge to 1.5× the swept volume) per cylinder per minute = Vs × ηv × (N/2) = 1.4703×10⁻³ × 1.5 × 800 = 1.7644 m³/min, for 3 cylinders: × 3 = 5.293 m³/min.
Mass of air = 5.293 × 1.2 kg/m³ = 6.352 kg/min = 381.1 kg/h.
Fuel consumption = mass of air / (A/F ratio) = 381.1/14.9 = 25.58 ≈ 25.57 kg/h. Answer: (B) 25.57.
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