**Given date: **

Angle of contact(θ) =165° = 165° X Π/180 = 2.88rad

Velocity of belt(V) = 1700m/min = 28.33m/sec

Width of belt(b) = 20cm

Thickness of belt(t) = 8mm 0.8cm

density of belt = 1gm/cm^{3}

Maximum permissible stress(f) = 250 N/cm^{2}

Coefficient of friction(µ) = 0.25 Maximum Power = ?

We know that, Tmax = σ.b.t

= 250 X 20 X 0.8 = 4000N ......(i)

Since Unit mass (m) = ρ.b.t.L

= 1/1000 X 20 X 0.8 X 100 = 1.6Kg .... (ii)

Since velocity(V) is given,

So we don't find the velocity using formula T_{C} =1/3 Tmax = mV^{2}, then Using the formula, TC = mV^{2}, For finding T_{C.}

= 1.6(28.33)2 = 1284 N ...(iii)

Using the formula, T_{max}

= T_{1} + T_{c} , for finding T_{1}

4000 = T_{1} + 1284

T1 = 2716N ...(iv)

Now, For T_{2}, Using relation Ratio of belt tension = T_{1}/T_{2} = e^{µθ }

2716/T_{2 }= e(0.25)(2.88) T_{2} = 1321 N ...(v)

Now Maximum Power Transmitted = (T_{1}–T_{2}).V/1000 KW

P = (2716 – 1321) X 28.33/1000 KW

**P = 39.52KW.**