Answer to Question #246936 in Mechanical Engineering for Hari

Question

A vertical double acting steam engine develops 100kW at 300 rpm. The maximum fluctuation of energy is 30
% of the work done per stroke. The maximum and minimum speeds are not to vary more than 1% on either
side of the mean speed. Find the mass of the flywheel required if the radius of gyration is 0.7 m

Answer 1

Given data:

• Power=75 kW
• N = 250 rpm
• ω1 - ω2 = 1%
• ω = 0.01ω
• k = 0.6
• Cs = ω1 - ω2/ ω = 0.01

Maximum fluctuation of energy "Δ E" = Work done per cycle × CE

WD/cycle = P x n/60

Where n = no of working strokes/min

P = Power developed in watts

WD/ cycle = (75 x 10^3 ) x 60/250 = 18,000 N-m^2

We know that Δ E = mk^2 ω^2Cs

Mass of the flywheel= 547 kg

Thus the mass of the flywheel is 547 kg.