Engine test: four-cylinder at 1800 rpm

Data: 4 cylinders ($k = 4$), four-stroke, $D = 0.1$ m, $L = 0.12$ m, $N = 1800$ rpm, $\dot m_f = 0.2$ kg/min, $CV = 44000$ kJ/kg, net brake load $(W-S) = 40$ kg, brake circumference $\pi D_b = 3$ m, $\eta_m = 0.9$. Take $g = 9.81\ \text{m/s}^2$.

Brake power

$$\begin{aligned} (W-S) &= 40\times9.81 = 392.4\ \text{N} \\ BP &= \frac{(W-S)\,\pi D_b\,N}{60} = \frac{392.4\times3\times1800}{60} = 35.32\ \text{kW} \end{aligned}$$

Heat supplied by fuel

$$\dot m_f = \frac{0.2}{60} = 0.003333\ \text{kg/s},\qquad Q = \dot m_f\times CV = 0.003333\times44000 = 146.67\ \text{kW}$$

i) Brake thermal efficiency

$$\eta_{bth} = \frac{BP}{Q} = \frac{35.32}{146.67} = 24.08\ \%$$

ii) Indicated thermal efficiency

$$\begin{aligned} IP &= \frac{BP}{\eta_m} = \frac{35.32}{0.9} = 39.24\ \text{kW} \\ \eta_{ith} &= \frac{IP}{Q} = \frac{39.24}{146.67} = 26.75\ \% \end{aligned}$$

(Check: $\eta_{ith} = \eta_{bth}/\eta_m$.)

iii) Indicated mean effective pressure

For a four-stroke engine, power strokes per second per cylinder $n = N/(2\times60) = 15$.

$$\begin{aligned} A &= \frac{\pi}{4}(0.1)^2 = 7.854\times10^{-3}\ \text{m}^2 \\ IP &= p_{mi}\,L\,A\,n\,k \\ p_{mi} &= \frac{39.24\times10^3}{0.12\times7.854\times10^{-3}\times15\times4} = 6.939\times10^5\ \text{N/m}^2 = 6.94\ \text{bar} \end{aligned}$$

iv) Brake specific fuel consumption

$$BSFC = \frac{\dot m_f\ (\text{kg/h})}{BP} = \frac{0.2\times60}{35.32} = \frac{12}{35.32} = 0.340\ \text{kg/kWh}$$

Answer: $\eta_{bth}$ = 24.08 %, $\eta_{ith}$ = 26.75 %, IMEP = 6.94 bar, BSFC = 0.340 kg/kWh (BP = 35.32 kW).