Question

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

Answer 1

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.