**Given data: **

Width of belt(b) = 100mm

Thickness of belt(t) = 8mm

Velocity of belt(V) = 160m/min = 2.66m/sec

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

Coefficient of friction(µ) = 0.3

Maximum permissible stress(f) = 2 X 10^{6} N/m^{2 }= 2N/mm^{2}

Mass of the belt material(m) = 0.9 Kg/m Power = ?

Initial tension (To) = ? We know that,

T_{max} = σ.b.t = 2 X 100 X 8 = 1600N ...(i)

Since m and velocity (V) is given, then

Using the formula, T_{C} = mV^{2}, For finding T_{c}

= 0.9(2.66)^{2}

= 6.4 N ........(ii)

Using the formula, Tmax = T_{1} + T_{c} , for finding T_{1}

1600 = T_{1} + 6.4

T_{1} = 1593.6N ......(iii)

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

1593.6/T2 = e^{(0.3)(2.88)}

T_{2 }= 671.69 N ........(iv)

Now Power Transmitted = (T_{1}–T_{2}).v/1000 Kw

P = (1593.6 – 671.69).2.66/1000 Kw

**P = 2.45KW**

Let To = initial tension in the belt

T_{o }= (T_{1 }+ T_{2})/2 + T_{C}

T_{o} = (1593.6 + 671.69)/2 + 6.4

**T**_{o} = 1139.045N.