Engine displacement and mean piston speed

Given: $D = 0.1$ m, $L = 0.12$ m, $z = 4$, $r = 9$, $N = 2400$ rpm.

(a) Swept volume and displacement

$$\begin{aligned} V_s &= \frac{\pi}{4} D^2 L = \frac{\pi}{4}(0.1)^2(0.12) = 9.425 \times 10^{-4}\ \text{m}^3 = 942.5\ \text{cm}^3 \\ V_{engine} &= z V_s = 4 \times 942.5 = 3770\ \text{cm}^3 = 3.77\ \text{L} \end{aligned}$$

(b) Clearance volume

$$r = \frac{V_c + V_s}{V_c} \Rightarrow V_c = \frac{V_s}{r - 1} = \frac{942.5}{8} = 117.8\ \text{cm}^3$$

(c) Mean piston speed

The piston travels $2L$ in each revolution:

$$\bar{S}_p = 2LN = 2 \times 0.12 \times \frac{2400}{60} = 9.6\ \text{m/s}$$

(d) Power strokes per second

In a four-stroke engine each cylinder fires once in two revolutions:

$$\text{Firing rate} = z \times \frac{N}{2 \times 60} = 4 \times \frac{2400}{120} = 80\ \text{per second}$$

Answer: (a) $V_s = 942.5$ cm$^3$, displacement 3.77 L; (b) $V_c = 117.8$ cm$^3$; (c) 9.6 m/s; (d) 80 power strokes per second for the engine (20 per cylinder).