BP is found from net tension and circumference as

392.4 N x (3.0 m x 1800/60) = 35.3 kW

Which pair is correct?

eta_bth = BP/Q = 19.26%; eta_ith = IP/Q = 22.66%

Asked 5/22
2069 Chaitra · 8 marks

Engine test: four-cylinder, 85% mechanical efficiency

Short answer

Brake thermal efficiency = 19.26%, indicated thermal efficiency = 22.66%, imep = 4.08 bar, bsfc = 0.425 kg/kWh.

In a test for four-cylinders, four-stroke engine has a diameter of 120 mm, stroke = 150 mm, speed of engine = 1800 rpm, fuel consumption of 0.25 kg/min, calorific value of fuel is 44000 kJ/kg. Difference in tension on either side of brake pulley = 40 kg, Brake circumference is 300 cm. If the mechanical efficiency is 85 %. Determine (a) Brake-thermal efficiency, (b) Indicated thermal efficiency, (c) Indicated mean effective pressure and (d) Brake specific fuel consumption.

Data: 4 cylinders, four-stroke, $D = 0.12$ m, $L = 0.15$ m, $N = 1800$ rpm, $\dot m_f = 0.25$ kg/min, CV = 44 000 kJ/kg, net tension = 40 kgf, brake circumference = 3.0 m (taken as the effective circumference, no rope size given), $\eta_m = 0.85$ .

Brake and indicated power

\begin{aligned} F &= 40 \times 9.81 = 392.4\ \text{N},\quad R = \frac{3.0}{2\pi} = 0.4775\ \text{m} \\ BP &= \frac{2\pi N F R}{60} = 392.4 \times \frac{3.0 \times 1800}{60} = 35\,316\ \text{W} = 35.316\ \text{kW} \\ IP &= \frac{BP}{\eta_m} = \frac{35.316}{0.85} = 41.548\ \text{kW} \end{aligned}

Heat supplied:

\dot Q = \frac{0.25}{60}\times 44\,000 = 183.33\ \text{kW}

(a) Brake thermal efficiency

\eta_{bth} = \frac{35.316}{183.33} = 0.1926 = 19.26\%

(b) Indicated thermal efficiency

\eta_{ith} = \frac{41.548}{183.33} = 0.2266 = 22.66\%

(c) Indicated mean effective pressure

\begin{aligned} A &= \frac{\pi}{4}(0.12)^2 = 0.011310\ \text{m}^2,\quad n = \frac{N}{2} = 900,\ k = 4 \\ p_{mi} &= \frac{60\,IP}{L A n k} = \frac{60 \times 41\,548}{0.15 \times 0.011310 \times 900 \times 4} \\ &= 4.082\times10^5\ \text{N/m}^2 = 4.08\ \text{bar} \end{aligned}

(d) Brake specific fuel consumption

bsfc = \frac{0.25 \times 60}{35.316} = \frac{15}{35.316} = 0.4247\ \text{kg/kWh}

Answer: $\eta_{bth}$ = 19.26 %, $\eta_{ith}$ = 22.66 %, imep = 4.08 bar, bsfc = 0.425 kg/kWh.

✓ Answer checked by Claude Opus 5.5 (high effort)

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