The turning moment diagram of a flywheel fitted to a fictitious engine is shown in the figure.

The mean turning moment is 2000 Nm. The average engine speed is 1000 rpm. For fluctuation in the speed to be within ±2% of the average speed, the mass moment of inertia of the flywheel is _________ kgm².

The correct answer is 3.58.

Here is the step-by-step derivation and explanation of why this answer is correct:

1. Identification of Given Values from the Diagram and Text:
From the problem description and the turning moment diagram:

• Mean turning moment ($T_{\mathrm{mean}}$) = 2000 Nm
• Average engine speed ($N$) = 1000 rpm
• Maximum speed fluctuation is within ±2% of the average speed. Therefore, the coefficient of fluctuation of speed ($C_s$) is:
  $C_s=2\times 2\%=0.04$
• Looking at the diagram, the turning moment in each crank angle interval is:
  • From $0$ to $\frac{\pi }{2}$: $T=1500\text{Nm}$
  • From $\frac{\pi }{2}$ to $\pi$: $T=3000\text{Nm}$
  • From $\pi$ to $\frac{3\pi }{2}$: $T=1000\text{Nm}$
  • From $\frac{3\pi }{2}$ to $2\pi$: $T=2500\text{Nm}$

2. Determining the Maximum Fluctuation of Energy ($\Delta E_{\max }$):
The fluctuation of energy is the area of the turning moment diagram loop relative to the mean line ($T_{\mathrm{mean}}=2000\text{Nm}$). Let the energy at crank angle $\theta =0$ be $E_0$.

At $\theta =\frac{\pi }{2}$:
$E_1=E_0+(1500-2000)\times \frac{\pi }{2}=E_0-250\pi$

At $\theta =\pi$:
$E_2=E_1+(3000-2000)\times \frac{\pi }{2}=E_0-250\pi +500\pi =E_0+\mathrm{250<mi>\pi </mi>}$

At $\theta =\frac{3\pi }{2}$:

At $\theta =2\pi$:
$E_4=E_3+(2500-2000)\times \frac{\pi }{2}=E_0-250\pi +250\pi =E_0$

Comparing these energy values:

• Maximum Energy ($E_{\max }$) = $E_0+250\pi$ (at $\theta =\pi$)
• Minimum Energy ($E_{\min }$) = $E_0-250\pi$ (at $\theta =\frac{\pi }{2}$ and $\theta =\frac{3\pi }{2}$)

3. Computing Mean Angular Velocity (${\omega }_{\mathrm{mean}}$):
Using the average engine speed:
${\omega }_{\mathrm{mean}}=\frac{2\pi N}{60}=\frac{2\times \pi \times 1000}{60}\approx 104.72\text{rad/s}$

4. Calculating the Flywheel Mass Moment of Inertia ($I$):
The relation between maximum energy fluctuation, moment of inertia, angular velocity, and coefficient of speed fluctuation is given by:
$\Delta E_{\max }=I\times {{\omega }_{\mathrm{mean}}}^2\times C_s$

Substituting the calculated parameters into the equation:
$500\pi =I\times {\left(104.72\right)}^2\times 0.04$

$1570.80=I\times 10966.28\times 0.04$

$1570.80=I\times 438.65$

$I=\frac{1570.80}{438.65}\approx 3.58{\text{kgm}}^2$