Find out the safe power transmitted by pulley:
A belt drives pulley of 200mm diameter such that ratio of tensions in tight side and slack side is 1.2. If maximum tension in belt is not to exceed 240KN. Find out the safe power transmitted by pulley at the speed of 60rpm.
Sol: Given that,
D1 = Diameter of the driver = 200mm = 0.2m
T1/T2 = 1.2
Since between T1 and T2, T1 is greater than T2,
Hence T1 = 240KN
N1 = Speed of the driver in R.P.M. = 60PRM
P = ?
We know that
T2 = T1/1.2 =240/1.2 = 200KN ...(i)
V = Velocity of belt in m/sec.
= pDN/60 m/sec, D is in meter and N is in potation per minute
= (3.14 X 0.2 X 60)/60 = 0.628 m/sec (ii)
P = (T1 - T2) X V
P = (240 - 200) X 0.628
P = 25.13KW .......ANS