Flywheel sizing comparison at equal rated power: Which engine type will require the heaviest flywheel for smooth operation, given the same 30 kW rating?

Introduction / Context:
A flywheel stores rotational energy to smooth cyclic speed variation caused by non-uniform torque over an engine cycle. The required mass moment of inertia depends on the number of power strokes per revolution, torque pulsation severity, operating speed, and acceptable coefficient of fluctuation of speed.

Given Data / Assumptions:

• All engines rated at 30 kW.
• Lower rpm increases required inertia for the same energy smoothing.
• Four-stroke has one power stroke per two revolutions per cylinder; two-stroke has one per revolution.
• Diesels generally exhibit higher compression and larger torque pulsations than comparable petrol engines.

Concept / Approach:

Energy fluctuation per cycle must be buffered by the flywheel: I * (omega_max^2 - omega_min^2) ≈ energy variation per cycle. For the same power, lower speed implies higher work per cycle and thus larger energy fluctuations. Fewer power strokes (four-stroke vs two-stroke) also increase fluctuation. Diesel torque waves are typically harsher than petrol, further increasing required smoothing capacity.

Step-by-Step Solution:

Verification / Alternative check:

Textbook flywheel sizing problems consistently show I ∝ energy fluctuation / omega^2. Halving speed approximately quadruples the inertia needed for similar speed regulation, strengthening the conclusion.

Why Other Options Are Wrong:

Higher rpm options need less inertia. Two-stroke engines have more frequent power strokes, reducing fluctuation. Petrol engines have smoother pressure rise, reducing inertia demand compared to diesels.

Common Pitfalls:

Comparing only fuel type without considering rpm and stroke; ignoring that power equality does not imply equal per-cycle work.

Final Answer:

30 kW four-stroke diesel engine at 750 rpm