Correct Answer - Option 3 : Ratio of maximum fluctuation of energy to the work done per cycle
Explanation:
Coefficient of fluctuation of energy:
It is the ratio of the maximum fluctuations of energy to the work done per cycle.
CE = \(\frac{Δ E}{W_{cycle}}\)
ΔE = CE × Wcycle ....... eq (1)
where CE = permissible limits of the coefficient of fluctuation of energy, Wcycle = Indicated work per revolution = W
The kinetic energy of the rotational body is:
K = \(\frac{1}{2}\)Iω2
Iω2 = 2K ....... eq (2)
Change in kinetic energy of the flywheel is:
ΔE = \(\frac{1}{2}I(\omega_1^2-\omega_2^2)=I\left(\frac{\omega_1+\omega_2}{2}\right)(\omega_1-\omega_2)\)
\(\Delta E=I\omega(\omega_1-\omega_2)=I\omega^2\left(\frac{\omega_1-\omega_2}{\omega}\right)=I\omega^2C_S\)
ΔE = ICsω2 ....... eq (3)
where, I = moment of inertia of the flywheel, ω = mean rotational speed, Cs = coefficient of fluctuation of speed.
substitute equation 1 and 2 in equation 3, we get
CE × W = 2K × Cs
K = \(\frac{WC_E}{2C_s}\)
