Engine dimensions for 5 kW at 2000 rpm

Assumptions: single-cylinder engine; the 5 kW is the power developed with the given mean effective pressure (brake mep basis), so $P = \dfrac{p_m L A n}{60}$ with $n = N/2$ for a four-stroke engine.

Data: $P = 5$ kW $= 5000$ W, $N = 2000$ rpm, $n = 1000$ per min, $p_m = 7.5$ bar $= 7.5\times10^5$ N/m², $L = 1.25D$.

$$\begin{aligned} P &= \frac{p_m \, L \, \frac{\pi}{4}D^2 \, n}{60} = \frac{p_m (1.25D)\frac{\pi}{4}D^2 n}{60} \\ 5000 &= \frac{7.5\times10^5 \times 1.25 \times 0.7854 \times D^3 \times 1000}{60} \\ D^3 &= \frac{5000 \times 60}{7.5\times10^5 \times 1.25 \times 0.7854 \times 1000} = 4.074\times10^{-4}\ \text{m}^3 \\ D &= 0.07413\ \text{m} = 74.1\ \text{mm} \\ L &= 1.25 \times 74.13 = 92.7\ \text{mm} \end{aligned}$$

Check: $\dfrac{7.5\times10^5 \times 0.09267 \times \frac{\pi}{4}(0.07413)^2 \times 1000}{60} = 5000$ W.

Answer: Bore $D \approx 74.1$ mm, stroke $L \approx 92.7$ mm.