# A multiple disc clutch has five plates having four pairs of active friction surfaces. If the intensity of pressure is not to exceed 0.127 N/mm^2, find the power transmitted at 500 r.p.m. The outer and inner radii of friction surfaces are 125 mm and 75 mm respectively. Assume uniform wear and take coefficient of friction = 0.3.

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A multiple disc clutch has five plates having four pairs of active friction surfaces. If the intensity of pressure is not to exceed 0.127 N/mm^2, find the power transmitted at 500 r.p.m. The outer and inner radii of friction surfaces are 125 mm and 75 mm respectively. Assume uniform wear and take coefficient of friction = 0.3.

(a) 17.8 kW

(b) 18.8 kW

(c) 19.8 kW

(d) 20.8 kW

I have been asked this question by my college professor while I was bunking the class.

Asked question is from Laws of Fluid Friction topic in division Friction of Machine Kinematics

by (106k points)
The correct option is (b) 18.8 kW

For explanation: n1 + n2 = 5 ; n = 4 ; p = 0.127 N/mm^2 ; N = 500 r.p.m. or ω = 2π × 500/60 = 52.4 rad/s ; r1 = 125 mm ; r2 = 75 mm ; μ = 0.3

Since the intensity of pressure is maximum at the inner radius r2, therefore

p.r2 = C or C = 0.127 × 75 = 9.525 N/mm

We know that axial force required to engage the clutch,

W = 2 π C (r1 – r2) = 2 π × 9.525 (125 – 75) = 2990 N

and mean radius of the friction surfaces,

R = r1 + r2/2 = 125 + 75/2 = 100 mm = 0.1 m

We know that torque transmitted,

T = n.μ.W.R = 4 × 0.3 × 2990 × 0.1 = 358.8 N-m

∴ Power transmitted,

P = T.ω = 358.8 × 52.4 = 18 800 W = 18.8 kW.

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