Given data

D_{1} = Diameter of the driver

= 600mm = 0.6m N_{1}

= Speed of the driver in R.P.M.

= 200RPM

µ = Coefficient of friction

= 0.25 θ = Angle of contact

= 160º = 1600 X (π/180) = 2.79rad

(Angle of lap is always less than 10º, so it is angle of contact which is always greater than 150º, always in radian

T_{1} = Maximum Tension = 2.5KN Let

T_{2} = Tension in the slack side of the belt

V = Velocity of the belt in m/sec. = πDN/60 m/sec,

D is in meter and N is in RPM

P = Power transmitted by belt drive We know that

Power Transmitted = (T_{1} – T_{2}).V KW, T_{1} & T_{2} in KN

Here T_{2} and V is unknown

**Calculation for V **

V = πDN/60 m/sec,

D is in meter and N is in RPM Putting all the value,

V = (3.14 X 0.6 X 200)/60 = 6.28m/sec ...(i)

Calculation for T_{2} We also know that, Ratio of belt tension, T_{1}/T_{2} = e^{µθ} Putting all the value,

2.5/ T_{2} = e(0.25 × 2.79)

T_{2} = 1.24KN

Now, P = (2.5 – 1.24) × 6.28 ...(ii)

**P = 7.92 KW.**