The areas of the turning moment diagram for one revolution of a multi-cylinder engine with reference to the mean turning moment, below and above the line, are – 32, + 408, – 267, + 333, – 310, + 226, – 374, + 260 and – 244 $mm^2$.

The scale for abscissa and ordinate are: $1 mm = 4.2°$ and $1 mm = 850 N-m$ respectively.

A rimmed flywheel is required to keep down fluctuations in speed from 200 to 220 rpm. The mean diameter of the flywheel should not exceed 2m. It can be assumed that the rim contributes 90% of the required moment of inertia. Determine the dimensions of the rim.

Given data

$N_{\max }=220 \mathrm{rpm}, N_{\min }=200 \mathrm{rpm}, D_{\max } \leq 2 m, \Delta E=30000 \mathrm{Nm}$

Assumptions - The cross-section of the rim is a rectangular, flywheel material - FG200 with density $7100kg/m^3$ , width-b=D/5

Solution:

Step 1 - Energy analysis of prime mover and flywheel

Average speed is 210 $\mathrm{rpm}$ and $\omega=21.99 \mathrm{rps}$

Coefficient of fluctuation of speed- $k_{s}=\frac{N_{\max }-N_{\min }}{N}=\frac{220-200}{210}=0.095$

since the scale of crank angle is $1 \mathrm{mm}=4.2^{\circ}=4.2 \times \pi / 180=0.0745 \mathrm{rad},$ and the scale of the turning moment is $1 \mathrm{mm}=650 \mathrm{N}-\mathrm{m},$ therefore 1 $\mathrm{mm}^{2}$ on the turning moment diagram $=850 \times 0.0745=63.29 \mathrm{N}-\mathrm{m}$

Let the total energy at $\mathrm{A}=\mathrm{E}$

Therefore, from Fig, Energy at $\mathrm{B}=\mathrm{E}-32$

Energy at $\mathrm{C}=\mathrm{E}-32+408=\mathrm{E}+376$

Energy at $\mathrm{D}=\mathrm{E}+376-267=\mathrm{E}+109$

Energy at $\mathrm{E}=\mathrm{E}+109+333=\mathrm{E}+442$

Energy at $\mathrm{F}=\mathrm{E}+442-310=\mathrm{E}+132$

Energy at $\mathrm{G}=\mathrm{E}+132+226=\mathrm{E}+358$

Energy at $\mathrm{H}=\mathrm{E}+358-374=\mathrm{E}-16$

Energy at $\mathrm{I}=\mathrm{E}-16+260=\mathrm{E}+244$

Energy at $\mathrm{J}=\mathrm{E}+244-244=\mathrm{E}=$ Energy at $\mathrm{A}$

From above, is seen that the energy is maximum at $\mathrm{E}$ and minimum at $\mathrm{B}$.

$\therefore$ Maximum energy $=\mathrm{E}+442$ and minimum energy $=\mathrm{E}-32$

The maximum fluctuation of energy,

$\Delta \mathrm{E}=$ Maximum energy - Minimum energy

$=(\mathrm{E}+442)-(\mathrm{E}-32)=474 \mathrm{mm}^{2},=474 \times 63.29=30000 \mathrm{N}-\mathrm{m}$

Step 2 - Flywheel Dimensions

Maximum fluctuation of energy is $\Delta E=I k_{s} \omega^{2}$

Radius of gyration $-\mathrm{k}$ and $k^{2}=\frac{D^{2}}{4},$ Moment of inertia- $\mathrm{I}=\mathrm{m} k^{2}$

As rim contribute 90$\%$ of the required moment of inertia,

$\quad \therefore E_{r i m}=0.9 \times 30000=27000=m \frac{D^{2}}{4} \times 0.095 \times 21.99^{2}$

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

And width-b $=\mathrm{D} / 5$

$\quad \therefore 27000=\frac{\pi}{4} D^{2} \frac{D}{5} \times 7100 \times \frac{D^{2}}{4} \times 0.095 \times 21.99^{2}$

$\mathrm{D}=1.16 \mathrm{m}\lt2 \mathrm{m}, \mathrm{mass}-\mathrm{m}=1740.81 \mathrm{kg}, \mathrm{b}=0.232 \mathrm{m}$

Mass of $\mathrm{rim}=\mathrm{m}=\pi \mathrm{D} \mathrm{bh} \rho $

$\therefore 1740.81=\pi \times 1.16 \times 0.232 \times h \times 7100$

Rim thickness- $\mathrm{h}=0.289 \mathrm{m}$