An Otto cycle engine develops 50 kW at 150 r.p.m. with 75 explosions per minute. The change of speed from the commencement to the end of power stroke must not exceed 0.5% of mean on either side. Design a suitable rim section having width four times the depth so that the hoop stress does not exceed 4 MPa.

Assume that the flywheel stores 16/15 times the energy stored by the rim and that the work done during power stroke is 1.4 times the work done during the cycle.

Given Data

$\mathrm{P}=50 \mathrm{kW}=50 \times 10^{3} \mathrm{W}, \mathrm{Speed}-\mathrm{N}=150 \mathrm{rpm} \therefore \omega=15.70 \mathrm{rad} / \mathrm{sec}, \mathrm{n}=75$ $k_{s}=1 \%, \mathrm{Wp}=1.4 \mathrm{W}, \sigma_{\mathrm{t}}=4 \mathrm{MPa}$

Assumptions- It is four stroke engine as n=N/2, flywheel material- CI with density $7200 kg/m^3$ , width-b=D/5

Solution