Question 3 of 4: Square V6 Engine Geometry

Problem 3: Square V6 Engine Geometry (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Total displacement $V_{\text{total}}=3\,\text{L}$, $n_{\text{cyl}}=6$ (V6), compression ratio $r_c=9.5$, engine speed $N=3600$ rpm, square engine ($B=S$).

Find. (1) Bore $B$ and stroke $S$. (2) Mean piston speed $\bar{S}_p$. (3) Clearance volume $V_c$ of one cylinder.

Approach. Split the total displacement evenly over six cylinders, then use the square-engine constraint $B=S$ to solve the single-unknown cylinder-volume equation for bore/stroke; the mean piston speed follows from stroke and rev/s, and the clearance volume from the compression-ratio definition.

1. Displacement per cylinder. $$V_d=\frac{V_{\text{total}}}{n_{\text{cyl}}}=\frac{3000\ \text{cm}^3}{6}=500\ \text{cm}^3$$
2. (1) Bore and stroke. For a square cylinder, $V_d=\dfrac{\pi}{4}B^2S=\dfrac{\pi}{4}B^3$: $$B=\left(\frac{4V_d}{\pi}\right)^{1/3}=\left(\frac{4(500)}{\pi}\right)^{1/3}=\boxed{8.60\ \text{cm}}=S$$
3. (2) Mean piston speed. $\bar{S}_p=2SN$ (two stroke-lengths per revolution), with $N$ in rev/s: $$\bar{S}_p=2(0.0860\ \text{m})\left(\frac{3600}{60}\right)=\boxed{10.3\ \text{m/s}}$$
4. (3) Clearance volume. $r_c=\dfrac{V_d+V_c}{V_c}\Rightarrow V_c=\dfrac{V_d}{r_c-1}$: $$V_c=\frac{500}{9.5-1}=\boxed{58.8\ \text{cm}^3}$$