Answer to Question #180225 in Mechanical Engineering for nnaemeka

Question #180225

A vertical diesel engine running at 420 rpm develops 650kw and has 4 impulses per revolution. If the fluctuation of energy is 28% of the work done during each impulse, estimate the cross sectional area of the rim


1
Expert's answer
2021-04-12T02:32:54-0400

w = (2\pi*420)/60 = 14\pi rad/s

Work done per cycle = (650 000 x 60)/180 = 216 667 J

e = 0.28 x 216 667 J = 60 667 J

60 667 = 1/2 * I * (w12 - w22) = I * w2 * ((w1 - w2)/w) = I * (12\pi)2 ∗ 0.01

I = 4273 kg m2

Centrifugal stress, σ = ρ * V2 = ρ * ω2 * R2

where R is the mean rim radius, 

5.5* 106 = 7200 * 144π2 * R2

R = 0.732 m

I = mk2 = ρ * A * 2πR * R2

4273 = 7200 * A * 2\pi *(0.732 m)2

A = 0.176 m2

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