A single plate clutch, with both sides effective, has outer and inner diameters 300 mm and 200 mm respectively. The maximum intensity of pressure at any point in the contact surface is not to exceed 0.1 N/mm^2. If the coefficient of friction is 0.3, determine the power transmitted by a clutch at a speed 2500 r.p.m.

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A single plate clutch, with both sides effective, has outer and inner diameters 300 mm and 200 mm respectively. The maximum intensity of pressure at any point in the contact surface is not to exceed 0.1 N/mm^2. If the coefficient of friction is 0.3, determine the power transmitted by a clutch at a speed 2500 r.p.m.

(a) 61.693 kW

(b) 71.693 kW

(c) 81.693 kW

(d) 91.693 kW

This question was addressed to me during an interview.

This key question is from Laws of Fluid Friction in chapter Friction of Machine Kinematics

by (106k points)
Right option is (a) 61.693 kW

For explanation I would say: Given : d1 = 300 mm or r1 = 150 mm ; d2 = 200 mm or r2 = 100 mm ; p = 0.1 N/mm^2 ; μ = 0.3 ; N = 2500 r.p.m. or ω = 2π × 2500/60 = 261.8 rad/s

Since the intensity of pressure ( p) is maximum at the inner radius (r2), therefore for uniform wear,

p.r2 = C or C = 0.1 × 100 = 10 N/mm

We know that the axial thrust,

W = 2 π C (r1 – r2) = 2 π × 10 (150 – 100) = 3142 N

and mean radius of the friction surfaces for uniform wear,

R = r1 + r2/2 = 150 + 100/2 = 125 mm = 0.125m

We know that torque transmitted,

T = n.μ.W.R = 2 × 0.3 × 3142 × 0.125 = 235.65 N-m           …( n = 2,for both sides of plate effective)

∴ Power transmitted by a clutch,

P = T.ω = 235.65 × 261.8 = 61 693 W = 61.693 kW.