Question

A four-cylinder four-stroke engine (100 mm bore and 120 mm stroke) runs at 1800 rpm mean speed with 0.9 MPa indicated mean effective pressure (IMEP).
The flywheel constant (ratio of the energy fluctuation to the indicated work per crankshaft revolution) is 0.30. All other losses are neglected.
For an allowable speed fluctuation of \(\pm 1\%\) about the mean, the required flywheel mass moment of inertia (in \(kg.m^2\)) is ________. (Rounded off to two decimal places)
(Take \(\pi = 3.14\))

Hint:

Find the indicated work done per crankshaft revolution, then use \(\Delta E = I \omega^2 C_s\) to get the flywheel inertia.

Answer 1

Correct Answer: 0.72

Step 1: Convert speed to revolutions per second.
$N = 1800$ rpm means $n = 1800/60 = 30$ rev/s, so $\omega = 2\pi n = 2 \times 3.14 \times 30 = 188.4$ rad/s.

Step 2: Get the cylinder swept volume in convenient units.
Bore $= 10$ cm, stroke $= 12$ cm, so $V_s = \frac{\pi}{4}(10)^2(12) = 0.785 \times 100 \times 12 = 942\ cm^3 = 0.000942\ m^3$.

Step 3: Get the work done by one cylinder in one power stroke.
$W_1 = p_{mi} V_s = (0.9 \times 10^6\ Pa)(0.000942\ m^3) = 847.8$ J.
This engine has 4 cylinders firing every half revolution apart, so each single crankshaft turn carries 2 such strokes worth of work: $W_{rev} = 2(847.8) = 1695.6$ J.

Step 4: Scale by the flywheel constant to get the fluctuation.
$\Delta E = 0.30 \times 1695.6 = 508.68$ J.

Step 5: Use the coefficient of speed fluctuation to size the flywheel.
A $\pm 1\%$ swing about the mean speed is a total swing of $2\%$, so $C_s = 0.02$.
From $\Delta E = I\omega^2 C_s$:
\[ I = \frac{508.68}{(188.4)^2(0.02)} = \frac{508.68}{709.89} = 0.7166 \]

Final Answer:
Rounded off, the flywheel needs a mass moment of inertia of \[ \boxed{0.72\ kg.m^2} \]