Answer to Question #180357 in Mechanical Engineering for nnaemeka

Question #180357

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.


1
Expert's answer
2021-04-21T07:05:47-0400

"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, "\u03c3 = \u03c1 \\times V^2 = \u03c1 \\times \u03c9^2 \\times R^2"

where R is the mean rim radius, 

"5.5 \\times106 = 7200 \\times144\u03c0^2 * R^2"

R = 0.732 m

"I = mk^2 = \u03c1 \\times A \\times2\u03c0R \\times R^2"

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

A = 0.176 m2

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