The flywheel does not maintain a constant speed, it simply reduces the fluctuation of speed. In other words, a flywheel controls the speed variations caused by the fluctuation of the engine turning moment during each cycle of operation.

Following terms are useful in the design on the flywheel.

1. Coefficient of Fluctuation of Speed

   The difference between the maximum and minimum speeds during a cycle is called the maximum fluctuation of speed. The ratio of the maximum fluctuation of speed to the mean speed is called the coefficient of fluctuation of speed.

   $k_{s}=\frac{N_{\max }-N_{\min }}{N}=\frac{\omega_{\max }-\omega_{\min }}{\omega}=\frac{v_{\max }-v_{\min }}{v}$

   The coefficient of fluctuation of speed is a limiting factor in the design of flywheel. The fluctuation of energy may be determined by the turning moment diagram for one complete cycle of operation.

2. Fluctuation of Energy

   The variations of energy above and below the mean resisting torque line are called fluctuation of energy. The turning moment diagram for a multi-cylinder engine is shown in the figure below.

   

   The difference between the maximum and the minimum energies is known as a maximum fluctuation of energy. If the maximum and minimum energy occurs at points E and B, Then, The maximum fluctuation of energy is defined as the difference between the maximum kinetic energy and minimum kinetic energy in the cycle. It is denoted by $\Delta E$.

   $\Delta E=\mathrm{E}_{\mathrm{E}}-\mathrm{E}_{\mathrm{B}}=(\mathrm{E}-\mathrm{a} 1+\mathrm{a} 2-\mathrm{a} 3+\mathrm{a} 4)-(\mathrm{E}-\mathrm{a} 1) = a2 - a3 + a4$

3. The coefficient of fluctuation of energy is defined as the ratio of the maximum fluctuations of energy to the work done per cycle. It is denoted by $k_e$.

   $k_{e}=\frac{\Delta E}{\text {Work done per cycle}}$

   The work done per cycle may be obtained by ($T_{mean} \times \theta $)

   Where $\theta$ -Angle turned in radians per revolution ($2\pi$, for two-stroke engines and $4\pi$, in case of four-stroke engines)

   $T_{mean}$ - Mean torque

   Also $\text{work done / cycle} = P \times 60 / n$, where n=N for two-stroke engines and N/2 for four stroke engines

4. Energy stored in Flywheel

   A flywheel is shown in fig. Let,

   • m = Mass of the flywheel in kg
   • k = radius of gyration of the flywheel in meters
   • I = Mass moment of inertia of the flywheel about the axis of rotation in $kg-m^2 = m.k^2$,
   • $N_1$ and $N_2$ = Maximum and minimum speeds during the cycle in r.p.m.
   • $\omega_1$ and $omega_2$ = Maximum and minimum angular speeds during the cycle in rad/s

   The mean kinetic energy of the flywheel,

   $E=\frac{1}{2} I \omega^{2}=\frac{1}{2} m k^{2} \omega^{2}$

   As the speed of the flywheel changes from $\omega_1$ and $omega_2$, the maximum fluctuation of energy,

   $\begin{aligned} \Delta E &= \text{Maximum KE} - \text{Minimum KE} \\ &=\frac{1}{2} I \omega_{1}^{2}-\frac{1}{2} I \omega_{2}^{2} \\ &=\frac{1}{2} I\left(\omega_{1}-\omega_{2}\right)\left(\omega_{1}+\omega_{2}\right) \\ &=I \frac{\left(\omega_{1}-\omega_{2}\right)\left(\omega_{1}+\omega_{2}\right)}{2} \times \frac{\omega}{\omega} \\ &=I \omega^{2} k_{s}\\ &=m k^{2} \omega^{2} k_{s} \end{aligned}$

   The radius of gyration (k) may be taken equal to the mean radius of the rim (R) because the thickness of rim is very small as compared to the diameter of rim. Therefore, substituting $k = R$

   $\therefore \Delta E=m R^{2} \omega^{2} k_{s}=m v^{2} k_{s}$

5. Design of rim flywheel

   The flywheel may be of the disc type or rim type as shown in fig.

   

   Let D - Diameter of the flywheel, R-Radius of the flywheel, m-Mass of the flywheel, b-Width of the flywheel, h-Thickness of rim

   The diameter of flywheel may be obtained by using following equation

   $\Delta E=m k^{2} \omega^{2} k_{s}$

   Where, m - mass of flywheel = Volume $\times$ Density = $\frac{\pi}{4} D^{2} b \rho$

   $k^{2}=\frac{D^{2}}{4}$ when his small, $k^{2}=\frac{D^{2}}{8}$ for disc type flywheel

   Also $\Delta E_{r i m}=\frac{1}{2} m_{r i m} v^{2}$

   $\Delta E_{\text {rim}}$ may be assumed as $0.95 \Delta E,$ as rim contributes to 95$\%$ of the total energy stored by the flywheel.

   The rim thickness is determined by $m_{rim} = \pi D b h \rho$ ------ PSG7.120

   Width of rim may be assumed as $b=D/5$