Find out the safe power transmitted by pulley, Mechanical Engineering

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

Posted Date: 10/18/2012 5:27:59 AM | Location : United States

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