Turning moment diagram and flywheel.

NOTE:

[1] Coefficient of fluctuation of energy.

$C_E = \frac{maximum \ fluctuation \ of \ energy}{Work \ done \ per \ cycle} = \frac{\triangle E}{W.D/cycle}$

Where $\triangle E$ = maximum energy – minimum energy.

i.e.

$\triangle E = E_1 – E_2$

$\rightarrow$ $W.D/Cycle = T_{mean} \times \theta$

Where,

$T_{mean} = mean \ torque$

$\theta =$ angle turned in radians in 1 revolution.

$= 2 \pi$ . . . .two stroke IC engine and steam engine.

$= 4 \pi$ . . . four stroke IC engine.

$\rightarrow$ $W.D/Cycle = \frac{P \times 60}{n}$

$n =$ number of working strokes per minute.

Where,

$n = N$ . . . . 2 stroke IC engine and steam engine.

$n = \frac{N}{2}$ . . . . 4 stroke IC engine.

{N = Speed in rpm}

[2] Coefficient of fluctuation of speed.

$C_s = \frac{maximum \ fluctuation \ speed}{mean \ speed} = \frac{N_1 – N_2}{N}$

Where, $N = \frac{N_1 + N_2}{2}$

[3] Energy stored in flywheel.

We know that, mean KE. Of flywheel is given by

$E = \frac{1}{2} I w^2$

Maximum fluctuation of energy,

$\triangle E = E_1 – E_2$

$= \frac{1}{2} Iw_1^2 = \frac{1}{2} Iw_2^2$

$\triangle E = \frac{1}{2} I (w_1^2 – w_2^2)$

$= \frac{1}{2} I (w_1 + w_2) (w_1 – w_2)$

$\triangle E = I. \omega (w_1 – w_2)$

Divide and multiply by $‘w’$

$\triangle E = I. w^2 (\frac{w_1- w_2}{w}) = I. \omega^2 C_s$

$\triangle E = I. (\frac{2 \pi N}{60}) [ \frac{2 \pi N_1}{60} - \frac{2 \pi N_2}{60}]$

$= I . \frac{4 \pi ^2}{3600} N (N_1 – N_2)$

$\triangle E = mk^2 (\frac{\pi ^2}{900}) N (N_1 – N_2)$

$= m k^2 (\frac{\pi ^2}{900}) N^2 C_s$