GATE AG 2025 • Q52
NAT • 2 Marks
Section 3: Farm Power
Farm Power
Subtopic: I.C. engine systems – fuel, cooling, lubrication, ignition, electrical, intake and exhaust
Question Statement:
A 4-stroke diesel engine: displacement 6.0 L, 2300 rpm, 75% mechanical efficiency, indicated mean effective pressure 800 kPa, brake-specific fuel consumption 320 g/h/kW, calorific value 44.6 MJ/kg. Find the fuel equivalent power (kW). (Rounded to 2 decimal places)
Official Answer Key:
273.00 to 274.00
Step-by-Step Solution & Explanation
Indicated power IP = IMEP×Vd×(N/2)/60 = 800000×0.006×(2300/2)/60 = 92.0 kW (N/2 since 4-stroke gives 1 power stroke per 2 revolutions). Brake power BP = ηmech×IP = 0.75×92.0 = 69.0 kW. Fuel consumption rate = bsfc×BP = (320×10⁻³/3600 kg/s per kW)×69.0 kW = 0.006133 kg/s. Fuel equivalent power = fuel-mass-rate × calorific value = 0.006133×44.6×10³ kW = 273.55 kW.
Practice This in Real CBT Exam Interface
Test yourself with the official TCS-iON palette, virtual calculator, and immediate performance analytics.
Launch CBT Mock Test