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Engine test: 4-cylinder, B.P. 31.4 kW

Short answer

BSFC = (0.2×60)/31.4 = 0.382 kg/kWh. IP = 31.4/0.8 = 39.25 kW, and for four-stroke n=13.333, giving IMEP = 7.81 bar. ηbth = 31.4/146.67 = 21.41%.

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 = 0.2 kg/min, B.P. = 31.4 kW, Mechanical efficiency = 80%, Calorific value of fuel = 44000 kJ/kg. Determine: i) BSFC (Brake Specific Fuel Consumption) ii) IMEP (Indicated Mean Effective Pressure), and iii) Brake thermal efficiency.

Data: 4 cylinders (k=4k=4), four-stroke, D=0.1D = 0.1 m, L=0.12L = 0.12 m, N=1600N = 1600 rpm, m˙f=0.2\dot m_f = 0.2 kg/min, BP=31.4BP = 31.4 kW, ηm=0.8\eta_m = 0.8, CV=44,000CV = 44{,}000 kJ/kg.

i) Brake specific fuel consumption

BSFC=m˙f (kg/h)BP=0.2×6031.4=1231.4=0.382 kg/kWhBSFC = \frac{\dot m_f\ (\text{kg/h})}{BP} = \frac{0.2\times60}{31.4} = \frac{12}{31.4} = 0.382\ \text{kg/kWh}

ii) Indicated mean effective pressure

IP=BPηm=31.40.8=39.25 kWn=N2×60=1600120=13.333 power strokes/s per cylinderA=π4(0.1)2=7.854×10−3 m2pmi=IPL A n k=39.25×1030.12×7.854×10−3×13.333×4=7.809×105 N/m2=7.81 bar\begin{aligned} IP &= \frac{BP}{\eta_m} = \frac{31.4}{0.8} = 39.25\ \text{kW} \\ n &= \frac{N}{2\times60} = \frac{1600}{120} = 13.333\ \text{power strokes/s per cylinder} \\ A &= \frac{\pi}{4}(0.1)^2 = 7.854\times10^{-3}\ \text{m}^2 \\ p_{mi} &= \frac{IP}{L\,A\,n\,k} = \frac{39.25\times10^3}{0.12\times7.854\times10^{-3}\times13.333\times4} \\ &= 7.809\times10^5\ \text{N/m}^2 = 7.81\ \text{bar} \end{aligned}

iii) Brake thermal efficiency

Q=0.260×44,000=146.67 kWηbth=BPQ=31.4146.67=0.2141=21.41 %\begin{aligned} Q &= \frac{0.2}{60}\times44{,}000 = 146.67\ \text{kW} \\ \eta_{bth} &= \frac{BP}{Q} = \frac{31.4}{146.67} = 0.2141 = 21.41\ \% \end{aligned}

Answer: BSFC = 0.382 kg/kWh, IMEP = 7.81 bar, ηbth\eta_{bth} = 21.41 %.

✓ Answer checked by Claude Opus 5.5 (high effort)

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