Question

i) Find the power transmitted by a belt running over a pulley of 600 mm diameter

at 200 r.p.m. The coefficient of friction between the belt and the pulley is 0.25,

angle of lap 160° and maximum tension in the belt is 2500 N.​

Answer 1

Given that   D1 = Diameter of the driver  = 600mm = 0.6mN1  

= 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  

T1 = Maximum Tension = 2.5KN  

Explanation:    

• We know that  Power Transmitted = (T1 – T2)*V    
• 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 R.P.M 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 = e^µθ 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.