A multi-disc clutch has 4 steel discs and 3 bronze discs. The outside and inside diameters of contact surfaces are 250 mm and 180 mm, respectively. The coefficient of friction is 0.3 and axial force is 400 N. Assuming uniform wear, the power in kW that the clutch can transmit at 1000 rpm is _____.
Official Answer Key:
8.0 to 8.2
Step-by-Step Solution & Explanation
Number of active friction (contact) surfaces, n = (total discs) − 1 = (4+3) − 1 = 6.
Mean radius (uniform-wear theory): R_m = (R_o + R_i)/2 = (0.125 + 0.090)/2 = 0.1075 m.
Friction torque: T = n·μ·F·R_m = 6 × 0.3 × 400 × 0.1075 = 77.4 N·m.
ω = 2πN/60 = 2π(1000)/60 = 104.72 rad/s.
Power = T·ω = 77.4 × 104.72 = 8105 W ≈ 8.11 kW, within the accepted range.
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