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.
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
The clutch system shown in Fig. P2–79 is used to transmit torquethrough a 2-mm-thick oil...