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A pulley used to transmit power by means of ropes, has a diameter of 3.6m and has 15 groove of 45º angle. The angle of contact is 170º and the coefficient of friction between the ropes and the groove side is 0.28. The maximum possible tension in the ropes is 960N and the mass of the rope is 1.5Kg per m length. What is the speed of the pulley in rpm and the power transmitted if the condition of maximum power prevails?

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Given data:

Dia. Of pulley(D) = 3.6m 

Number of groove(or ropes) = 15 

Angle of groove(2a) = 45°, α = 22.50º 

Angle of contact(θ) = 170° = 1700 X (Π/180°) = 2.97 rad 

Coefficient of friction(µ) = 0.28 

Max. Tension(Tmax) = 960N 

Mass of rope(m) = 1.5Kg per m length 

For maximum power: Tc = 1/3Tm 

= 1/3 X 960 = 320N ...(i)

Tm = T1 + TC

960 = T1 + 320 

T1 = 640N ...(ii)

Now Tc = (1/3)Tm = mV2

V = (Tm/3m)1/2

= [960/(3 X 1.5)]1/2

= 14.6m/sec ...(iiii)

Since V = πDN/60 = 14.6, 

N = 77.45R.P.M.

Now, Ratio of Tension in V-Belt:: T1/T2 = eµ.θ.cosec α 

640/T2 = e(0.28).(2.97).cosec22.5 

T2 = 73.08N ...(iv) 

Maximum power transmitted(P) = (T1–T2).v/1000 Kw 

P = [(640 – 73.08) X 14.6]/1000 KW 

P = 8.277KW 

Total maximum power transmitted = Power of one rope X No. of rope 

P = 8.277 X 15 = 124.16KW

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