Question

The clutch system shown in Fig. P2–79 is used to transmit torque through a 2-mm-thick oil film with U = 0.38 N⋅s/m2 between two identical 30-cm-diameter disks. When the driving shaft rotates at a speed of 1200 rpm, the driven shaft is observed to rotate at 1125 rpm. Assuming a linear velocity profile for the oil film, determine the transmitted torque.

Answer 1

Given that :

absolute viscosity of SAE 30W oil, = 0.38 N.s/m²

thickness of oil film, h = 2 x 10^-3 m

angular speed of the driving shaft, ₁ = 1200 rpm = 125.6 rad/s

angular speed of the driven shaft, ₂ = 1125 rpm = 117.8 rad/s

diameter of a clutch, D = 0.3 m

Assume that one of the plate is stationary.

Then, linear velocity of other plate will be given as -

v = (₁ - ₂) r                                                                                              { eq.1 }

Consider a small area (dA) on the surface of a disc.

We know that, the shear force on a disc is given by -

dF = (dA) v / h

where, dA = (2r) dr

then, we get

dF = [ (2r) (₁ - ₂) r / h] dr                                                                       { eq.2 }

Torque on the small area (dA) will be given as -

d = (dF) r

d = [ (2) (₁ - ₂) r³ / h] dr                                                                     { eq.3 }

Therefore, the transmitted torque which will be given by -

integrating eq.3 on both sides with limit, we have

= [ (2) (₁ - ₂) r³ / h] dr

= [ (2) (₁ - ₂) / h] r³ dr

= [ (2) (₁ - ₂) / h] (r⁴ / 4) |^D/2₀

= (₁ - ₂) D⁴ / (32) h                                                                    

= {(3.14) (0.38 N.s/m²) [(125.6 rad/s) - (117.8 rad/s)] (0.3 m)⁴} / [(32) (2 x 10^-3 m)]

= [(0.00966492 Nm²) / (0.064 m)]

= 0.151 N.m