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,
D_{1} = Diameter of the driver = 200mm = 0.2m
T_{1}/T_{2} = 1.2
Since between T_{1} and T_{2}, T_{1} is greater than T_{2},
Hence T_{1} = 240KN
N_{1} = Speed of the driver in R.P.M. = 60PRM
P = ?
We know that
T_{2} = T_{1}/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 = (T_{1} - T_{2}) X V
P = (240 - 200) X 0.628
P = 25.13KW .......ANS