Answer to Question #180357 in Mechanical Engineering for nnaemeka

Question

A vertical diesel engine running at 420 rev/min develops 650 kW and has 4 impulses per revolution. If the fluctuation of ener is 28 % of the work done during each impulse, estimate the cross-sectional are 1. f the rim of the flywheel required to keep the speed within 2.5 rev/min of the mean speed when the mean peripheral speed of the rim is 1440 m/min. Cast iron has a density of 7.2 Mg/m3.

Answer 1

$w = \frac{(2\pi \times420)}{60} = 14\pi rad/s$

Work done per cycle = $\frac{(650 000 x 60)}{180} = 216 667 J$

e = 0.28 x 216 667 J = 60 667 J

$60 667 = \frac{1}{2} \times I \times (w_{12} - w_{22}) = I \times w_2 \times ((w_1 - \frac{w_2}{w})) = I \times (12\pi)2 \times 0.01$

I = 4273 kg m²

Centrifugal stress, $σ = ρ \times V^2 = ρ \times ω^2 \times R^2$

where R is the mean rim radius, 

$5.5 \times106 = 7200 \times144π^2 * R^2$

R = 0.732 m

$I = mk^2 = ρ \times A \times2πR \times R^2$

$4273 = 7200 \times A \times 2\pi \times(0.732 m)^2$

A = 0.176 m²