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: Find the swept volume of one cylinder.
Bore \(D = 0.1\) m and stroke \(L = 0.12\) m, so the swept volume is
\[ V_s = \frac{\pi}{4} D^2 L = \frac{3.14}{4} (0.1)^2 (0.12) = 0.000942\ m^3 \]

Step 2: Find the indicated work per crankshaft revolution.
Work done in one power stroke of one cylinder is \(W = p_{mi} V_s = 0.9 \times 10^6 \times 0.000942 = 847.8\) J.
In a four-cylinder four-stroke engine the cylinders fire \(180^{\circ}\) apart, so two power strokes occur in every single crankshaft revolution.
Indicated work per revolution \(= 2 \times 847.8 = 1695.6\) J.

Step 3: Find the energy fluctuation.
The flywheel constant relates fluctuation to work per revolution: \(\Delta E = C_k \times 1695.6 = 0.30 \times 1695.6 = 508.68\) J.

Step 4: Find the mean angular speed and the speed fluctuation coefficient.
\(\omega = \dfrac{2\pi N}{60} = \dfrac{2 \times 3.14 \times 1800}{60} = 188.4\) rad/s.
A fluctuation of \(\pm 1\%\) about the mean gives a total range of \(2\%\), so \(C_s = 0.02\).

Step 5: Solve for the flywheel inertia.
\[ \Delta E = I \omega^2 C_s \implies I = \frac{\Delta E}{\omega^2 C_s} = \frac{508.68}{(188.4)^2 \times 0.02} = \frac{508.68}{709.89} \]

Final Answer:
The required flywheel mass moment of inertia is \[ \boxed{0.72\ kg.m^2} \]