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: T_{c} = 1/3Tm

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

T_{m} = T_{1} + T_{C}

960 = T_{1} + 320

T_{1 }= 640N ...(ii)

Now T_{c} = (1/3)T_{m} = mV^{2}

V = (T_{m}/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:: T_{1}/T_{2} = e^{µ.θ.cosec} α

640/T_{2} = e(0.28).(2.97).cosec22.5

T_{2} = 73.08N ...(iv)

Maximum power transmitted(P) = (T_{1}–T_{2}).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**