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Engine test: fuel measured in litres

Short answer

bsfc = 0.382 kg/kWh, imep = 7.81 bar, brake thermal efficiency = 21.43%.

A 4-cylinder, 4-stroke cycle engine having cylinder diameter 100 mm and stroke 120 mm was tested at 1600 rpm and the following readings were obtained. Fuel consumption = 0.27 liter/minute, Specific gravity of fuel = 0.74, B.P = 31.4 kW, Mechanical efficiency = 80%, Calorific value of fuel = 44000 kJ/kg. Determine: i) bsfc ii) imep and iii) Brake thermal efficiency

Data: 4 cylinders, four-stroke, D=0.1D = 0.1 m, L=0.12L = 0.12 m, N=1600N = 1600 rpm, fuel 0.27 L/min, specific gravity 0.74, BP=31.4BP = 31.4 kW, ηm=0.80\eta_m = 0.80, CV = 44 000 kJ/kg.

Mass of fuel:

m˙f=0.27 L/min×0.74 kg/L×60=11.988 kg/h (0.00333 kg/s)\dot m_f = 0.27\ \text{L/min} \times 0.74\ \text{kg/L} \times 60 = 11.988\ \text{kg/h}\ (0.00333\ \text{kg/s})

i) Brake specific fuel consumption

bsfc=m˙fBP=11.98831.4=0.3818 kg/kWhbsfc = \frac{\dot m_f}{BP} = \frac{11.988}{31.4} = 0.3818\ \text{kg/kWh}

ii) Indicated mean effective pressure

IP=BPηm=31.40.80=39.25 kWA=π4(0.1)2=7.854×10−3 m2,n=N2=800, k=4pmi=60 IPLAnk=60×39 2500.12×7.854×10−3×800×4=7.809×105 N/m2=7.81 bar\begin{aligned} IP &= \frac{BP}{\eta_m} = \frac{31.4}{0.80} = 39.25\ \text{kW} \\ A &= \frac{\pi}{4}(0.1)^2 = 7.854\times10^{-3}\ \text{m}^2,\quad n = \frac{N}{2} = 800,\ k = 4 \\ p_{mi} &= \frac{60\,IP}{L A n k} = \frac{60 \times 39\,250}{0.12 \times 7.854\times10^{-3} \times 800 \times 4} \\ &= 7.809\times10^5\ \text{N/m}^2 = 7.81\ \text{bar} \end{aligned}

iii) Brake thermal efficiency

Q˙=11.9883600×44 000=146.52 kWηbth=31.4146.52=0.2143=21.43%\begin{aligned} \dot Q &= \frac{11.988}{3600}\times 44\,000 = 146.52\ \text{kW} \\ \eta_{bth} &= \frac{31.4}{146.52} = 0.2143 = 21.43\% \end{aligned}

Answer: bsfc = 0.382 kg/kWh, imep = 7.81 bar, ηbth\eta_{bth} = 21.43 %.

✓ Answer checked by Claude Opus 5.5 (high effort)

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