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Find the power transmitted by a belt running over a pulley of 600mm diameter at 200r.p.m. The coefficient of friction between the pulleys is 0.25; angle of lap 160º and maximum tension in the belt is 2.5KN. 

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 Given data 

D1 = Diameter of the driver 

= 600mm = 0.6m N1 

= Speed of the driver in R.P.M. 

= 200RPM 

µ = Coefficient of friction 

= 0.25 θ = Angle of contact 

= 160º = 1600 X (π/180) = 2.79rad

(Angle of lap is always less than 10º, so it is angle of contact which is always greater than 150º, always in radian

T1 = Maximum Tension = 2.5KN Let 

T2 = Tension in the slack side of the belt 

V = Velocity of the belt in m/sec. = πDN/60 m/sec, 

D is in meter and N is in RPM 

P = Power transmitted by belt drive We know that 

Power Transmitted = (T1 – T2).V KW, T1 & T2 in KN 

Here T2 and V is unknown 

Calculation for V 

V = πDN/60 m/sec, 

D is in meter and N is in RPM Putting all the value, 

V = (3.14 X 0.6 X 200)/60 = 6.28m/sec ...(i) 

Calculation for T2 We also know that, Ratio of belt tension, T1/T2 = eµθ Putting all the value,

2.5/ T2 = e(0.25 × 2.79)

T2 = 1.24KN

Now, P = (2.5 – 1.24) × 6.28 ...(ii)

P = 7.92 KW.

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