** Given that **

D_{1 }= Diameter of the driver = 45cm = 0.45m

R_{1} = Radius of the driver = 0.225m

D_{2} = Diameter of the driven = 20cm = 0.2m

R_{2} = Radius of the driven = 0.1m

X = Distance between the centers of two pulleys = 1.95m

T_{1} = Maximum permissible tension = 1000N

µ = Coefficient of friction = 0.20

N_{1} = Speed of the driver(Larger pulley) in R.P.M. = 100RPM

Since we know that,

Power Transmitted = (T_{1}–T_{2}).V/1000 Kw ...(i)

Tension is in KN and V is in m/sec

First ve find the velocity of the belt,

V = Velocity of the belt in m/sec.

Here we take diameter and RPM of larger pulley

= πDN/60 m/sec, D is in meter and N is in RPM

= (3.14 X0.45 X 100) /60

= 2.36m/sec.......(ii)

Now Ratio of belt tension, T_{1}/T_{2} = eµ.θ ...(iii)

Here we don't know the value of θ, For θ, first find the value of α, by the formula,

Angle of Lap for cross belt α = sin^{–1}(r_{1} + r_{2})/X

= sin^{–1}(0.225 + 0.1)/1.95

= 9.59° ...(iv)

Now Angle of contact (θ) = Π + 2α ----- for cross belt

θ = Π + 2 X 9.59°

= 199.19°

= 199.19°(Π/180°) = 3.47rad ...(v)

Now putting all the value in equation

(iii) We get 1000/T_{2} =e(0.2)(3.47) T_{2} = 498.9 N ...(vi)

Using equation (i), we get

P = [(1000 – 498.9) X 2.36 ]/1000

**P = 1.18KW.**