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

Expert's answer

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 * (w_{1}^{2} - w_{2}^{2}) = I * w^{2} * ((w1 - w_{2})/w) = I * (12\pi)^{2} ∗ 0.01

I = 4273 kg m^{2}

Centrifugal stress, σ = ρ * V^{2} = ρ * ω^{2} * R^{2}

where R is the mean rim radius,

5.5* 10^{6} = 7200 * 144π^{2} * R^{2}

R = 0.732 m

I = mk^{2} = ρ * A * 2πR * R^{2}

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

A = 0.176 m^{2}

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