Engine displacement, compression ratio and power

Given: $D = 80$ mm, $L = 90$ mm, $n = 4$ cylinders, $V_c = 55$ cc, $N = 4000$ rpm, $T = 120$ N m.

(i) Swept volume and displacement

$$\begin{aligned} V_s &= \frac{\pi}{4} D^2 L = \frac{\pi}{4}(8)^2(9)\ \text{cm}^3 = 452.4\ \text{cm}^3 \\ V_d &= n V_s = 4 \times 452.4 = 1809.6\ \text{cm}^3 \approx 1.81\ \text{L} \end{aligned}$$

(ii) Compression ratio

$$r = \frac{V_s + V_c}{V_c} = \frac{452.4 + 55}{55} = 9.23$$

(iii) Brake power

$$BP = \frac{2\pi N T}{60} = \frac{2\pi \times 4000 \times 120}{60} = 50\,265\ \text{W} \approx 50.27\ \text{kW}$$

(iv) Brake mean effective pressure

For a four-stroke engine, $BP = \dfrac{p_{mb} V_d N}{2 \times 60}$ (one power stroke per two revolutions).

$$p_{mb} = \frac{BP \times 120}{V_d N} = \frac{50\,265 \times 120}{1.8096\times10^{-3} \times 4000} = 8.33\times10^{5}\ \text{Pa} = 8.33\ \text{bar}$$

(v) Mean piston speed

$$\bar{S}_p = 2LN/60 = \frac{2 \times 0.09 \times 4000}{60} = 12\ \text{m/s}$$

Answer: $V_s = 452.4$ cc, $V_d = 1809.6$ cc, $r = 9.23$, $BP = 50.27$ kW, bmep $= 8.33$ bar, $\bar S_p = 12$ m/s.