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Maximum fluctuation of kinetic energy in an engine has been calculated to be 2600 J. Assuming that the engine runs at an average speed of 200 rpm, the polar mass moment of inertia (in kg.m2) of a flywheel to keep the speed fluctuation within ±0.5% of the average speed is _______

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Concept:

Flywheel:

  • A flywheel is used to control the variations in speed during each cycle of an engine.
  • A flywheel attached to the crankshaft makes the moment of inertia of the rotating parts quite large and thus, acts as a reservoir of energy.
  • The maximum fluctuation of the energy of flywheel is given by:

           ΔE = Iω2Cs

          where,

          I = Moment of inertia of the flywheel, ω = Mean speed

          Cs = Coefficient of fluctuation of speed = \(\frac{{{\omega _{max}}\; - \;{\omega _{min}}\;}}{\omega}\)

Calculation:

Given:

ΔE = 2600 J

N = 200 rpm

\(\therefore \omega = \frac{{2\pi N}}{{60}} = 20.944\frac{{rad}}{{sec}}\)

ω1 – ω2 = ± 0.5% ω

\(\begin{array}{l} ⇒ \frac{{{\omega _1} - {\omega _2}}}{\omega } = \pm 0.5{\rm{\% }} = 1{\rm{\% }} = 0.01\\ \therefore {C_s} = \frac{{{\omega _1} - {\omega _2}}}{\omega } = 0.01 \end{array}\)

ΔE = Iω2Cs

⇒ 2600 = I × (20.944)2 × 0.01

⇒ I = \(\frac{{2600}}{{4.38}} = 592.76\;{\rm{kg}}.{{\rm{m}}^2}\)

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