Answer to Question #146644 in Mechanical Engineering for siphosethu

Question #146644
A vertical double acting steam engine develops 80 kW at 240 rpm. The maximum
fluctuation of energy is 25% of the work done per stroke. The maximum and minimum
speeds are not to vary more than ±1% of the mean speed. Find the mass of the flywheel,
if the radius of gyration is 0.65 m
1
Expert's answer
2020-12-07T03:47:11-0500

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.


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