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