Single plate clutch transmits 25 Kw at 900rpm. The maximum pressure intensity between the plates is $85KN/m^2$ .The outer diameter of the plate is 360mm. Both sides of the plate are effective and the coefficient of friction is 0.25. - Determine the - (i) Inner diameter of the plate - (ii) Axial force to engage the clutch.
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Mumbai University > MECH > Sem 5 > Theory Of Machines 2

Marks: 8 M

Year: May 2015

Given:

pmax = 85 kpa

R0 = 0.36m

P=25 kW

N= 900 rpm

μ= 0.25

n = 2

Solution:

Torque $= T = P/w = P \times 60/ 2πN = 25000 \times 60 / 2 π \times 900 \\ = 265.4 Nm$

T $=n \times μ \times F \times R_m \\ = 2 \times 0.25 \times (p \times A) \times R_m \\ = 0.5 \times (85 \times 10^3) \times π ( R^2_0-R^2_i ) \times R_m$

$265.4 = 0.5 \times (85 \times 10^3) \times π ( R^2_0-R^2_i) \times 2/3 \times (R^3_0-R_i^3) / ( R^2_0-R^2_i ) \\ 265.4 = 0.5 \times (85 x 10^3) \times π \times 2/3 \times (R^3_0-R^3_i) \\ 265.4 = 0.5 \times (85 x 10^3) \times π \times 2/3 x (0.36^3-R^3_i) \\ R_i = 0.26 m = 260mm$

F $= p A = 85 \times 10^3 \times π ( R^2_0 – R^2_0 ) \\ = 85 \times 10^3 \times π ( 0.36^2 – 0.24^2) \\ = 19.22 kN$