A single plate dry clutch transmits 15 kW power at 1200 rpm. The clutch sustains a maximum axial load of 2.65 kN. The ratio of outer to inner diameter of the frictional surface is 1.25:1. Considering uniform wear with a coefficient of friction of 0.3 on both the frictional surfaces, the outer diameter of the clutch plate is ________ mm.
Official Answer Key:
166.00 to 168.00
Step-by-Step Solution & Explanation
Torque: T = 60P/(2πN) = 60 × 15000/(2π × 1200) = 119.37 N m.
Uniform wear, single plate with two friction surfaces: T = 2 × μW × (R + r)/2 = μW(R + r).
119.37 = 0.3 × 2650 × (R + r) ⇒ R + r = 119.37/795 = 0.15015 m.
With R/r = 1.25 ⇒ r(1.25 + 1) = 0.15015 ⇒ r = 0.06673 m, R = 0.08342 m.
Outer diameter D = 2R = 0.16683 m = 166.8 mm — within the official range 166.00–168.00.
Common error: taking one friction surface only, which doubles (R + r) and gives D ≈ 334 mm.
Section 4 — Soil and Water Conservation Engineering
6 questions · 10 marks
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