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

T₁/T₂  = 1.2

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