A pulley is driven by a flat belt running at a speed of 600 m/min. the coefficient of friction between the pulley and the belt is 0.3 and the angle of lap is 160°. if the maximum tension in the belt is 700 n; find the power transmitted by a belt.
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the power is transmitted by the belt is approximately 750/min.
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Answer:
Flat belt speed = V = 600 m/min = 600/60 m/sec = 10 m/sec
Coefficient of friction =µ = 0.3
Angle of lap = θ =160
Belt tension ratio = T1/ T2 = eµθ = e 0.3(160x π/180) = 2.31; T1/ T2 = 2.31;
T1= T2 x 2.311
T2 = 303.03
P = ( T1 - T2) x V
P = 4KW
Step-by-step explanation:
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