--- title: "Diesel engine 75 kW: fuel consumption and efficiencies" subject: "Power Plant Equipment" chapter: 5 key: "semester_7/power_plant_equipment:diesel-75kw-fuel-consumption" source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/diesel-75kw-fuel-consumption site: BCT Engineering --- # Diesel engine 75 kW: fuel consumption and efficiencies Power Plant Equipment · Chapter 5: Diesel Power Plant - Asked in: 2079 Bhadra, 4+4 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_plant_equipment/diesel-75kw-fuel-consumption - Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-5 ## Question A diesel engine consumes fuel at the rate of 5.5 gm/sec and develops a power of 75 kW. The mechanical efficiency is 85%. The lower heating value of the fuel is 44 MJ/kg. Determine: a) Brake specific fuel consumption b) Indicated specific fuel consumption c) Brake thermal efficiency and d) Indicated thermal efficiency ## Short answer IP=BP/eta_m=75/0.85=88.24 kW and fuel consumption is 19.8 kg/h. Thus bsfc=0.264 kg/kWh, isfc=0.2244 kg/kWh. With heat input 242 kW, brake thermal efficiency is 30.99% and indicated thermal efficiency is 36.46%. ## Answer **Data:** $\dot m_f = 5.5$ g/s $= 0.0055$ kg/s, $BP = 75$ kW, $\eta_m = 0.85$, LHV $= 44$ MJ/kg $= 44\,000$ kJ/kg. Indicated power: $$ IP = \frac{BP}{\eta_m} = \frac{75}{0.85} = 88.24\ \text{kW} $$ Fuel used per hour: $0.0055 \times 3600 = 19.8$ kg/h. #### a) Brake specific fuel consumption $$ bsfc = \frac{\dot m_f\ (\text{kg/h})}{BP} = \frac{19.8}{75} = 0.264\ \text{kg/kWh} $$ #### b) Indicated specific fuel consumption $$ isfc = \frac{19.8}{88.24} = 0.2244\ \text{kg/kWh} $$ #### c) Brake thermal efficiency Heat supplied by fuel: $\dot m_f \times LHV = 0.0055 \times 44\,000 = 242$ kW. $$ \eta_{bth} = \frac{BP}{\dot m_f\,LHV} = \frac{75}{242} = 0.3099 = 30.99\% $$ #### d) Indicated thermal efficiency $$ \eta_{ith} = \frac{IP}{\dot m_f\,LHV} = \frac{88.24}{242} = 0.3646 = 36.46\% $$ Check: $\eta_{bth} = \eta_m \times \eta_{ith} = 0.85 \times 36.46 = 30.99\%$. **Answer:** bsfc = 0.264 kg/kWh, isfc = 0.2244 kg/kWh, $\eta_{bth}$ = 30.99 %, $\eta_{ith}$ = 36.46 %. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is IP found from BP = 75 kW and mechanical efficiency 0.85? A. IP = 75 x 0.85 = 63.75 kW B. IP = 75 + 0.85 C. IP = 75 / 0.85 = 88.24 kW D. IP = 75 - 0.85 Answer: C. Mechanical efficiency = BP/IP, so IP = BP/eta_m = 88.24 kW. (Key point: Indicated power) 2. What must be done to the fuel rate of 5.5 g/s before computing bsfc in kg/kWh? A. Convert to 19.8 kg/h and divide by BP B. Divide 5.5 by 3600 C. Multiply by the calorific value only D. Divide by IP in watts Answer: A. 0.0055 kg/s x 3600 = 19.8 kg/h, and bsfc = 19.8/75 = 0.264 kg/kWh. (Key point: bsfc calculation) 3. Why is isfc smaller than bsfc for the same engine? A. IP is larger than BP while fuel flow is the same B. Fuel flow is lower for isfc C. isfc uses the higher heating value D. Calorific value changes Answer: A. isfc = 19.8/88.24 = 0.2244 versus bsfc = 0.264 because IP > BP. (Key point: isfc vs bsfc) 4. Which relation links brake and indicated thermal efficiency? A. eta_bth = eta_ith / eta_m B. eta_bth = eta_m x eta_ith C. eta_bth = eta_m + eta_ith D. eta_bth = eta_ith - eta_m Answer: B. 0.85 x 36.46% = 30.99%, matching BP/(m_f LHV) = 75/242. 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