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 ±1% about the mean, the required flywheel mass moment of inertia (in kg·m²) is ________. (Rounded off to two decimal places) (Take π = 3.14)
Official Answer Key:
0.70 to 0.74
Step-by-Step Solution & Explanation
Step 1 — swept volume per cylinder:
Vₛ = (π/4)D²L = (3.14/4)(0.10)²(0.12) = 9.42 × 10⁻⁴ m³
Step 2 — indicated work per cylinder per cycle:
W = IMEP × Vₛ = 0.9 × 10⁶ × 9.42 × 10⁻⁴ = 847.8 J
Step 3 — indicated work per crankshaft revolution. For a four-stroke engine one cycle occupies two crankshaft revolutions; there are 4 cylinders:
W_rev = (4 × 847.8)/2 = 1695.6 J per revolution
Step 4 — energy fluctuation:
ΔE = flywheel constant × W_rev = 0.30 × 1695.6 = 508.7 J
Step 5 — coefficient of speed fluctuation. “±1% about the mean” means the total swing is from 0.99ω to 1.01ω:
C_s = (ω_max − ω_min)/ω_mean = 0.02
Step 6 — mean angular speed:
ω = 2πN/60 = 2 × 3.14 × 1800/60 = 188.4 rad·s⁻¹
Step 7 — flywheel inertia from ΔE = I ω² C_s :
I = 508.7/(188.4² × 0.02) = 508.7/(35,494.6 × 0.02) = 508.7/709.9 = 0.72 kg·m²
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