Find out power transmitted by belt:
A belt is running over pulley of 1.5m diameters at 250RPM. The angle of contact is 120º and coefficient of friction is 0.30. If maximum tension in belt is 400N, find out power transmitted by belt.
Sol: Given data
Diameter of pulley(D) = 1.5m
Speed of driver(N) = 250RPM
Angle of contact(?) = 1200 = 1200 X (Π/180º) = 2.09 rad
Coefficient of friction(µ) = 0.3
Maximum tension(T_{max}) = 400N = T_{1}
Power (P) = ?
As P = (T_{1} - T_{2}) X V Watt ...(i)
T1 is given, and for finding value of T_{2}, using formula
Ratio of belt tension = T_{1}/T_{2} = e^{µθ}
400/T_{2} = e(0.3)(2.09)
T_{2 }= 213.4N ...(ii)
We know that V = ΠDN/60 m/sec
V = [3.14 X 1.5 X 250]/60 = 19.64m/sec ...(iii)
By putting all the value in equation (i)
P = (400 - 213.4) X 19.64 watt
P = 3663.88Watt or 3.66KW ......ANS