Let T_{1} = Tension on tight side T_{2} = Tension on slack side v = Linear velocity of belt Then the power transmitted is given by the equation

P = (T_{1}–T_{2}). V ...(i)

But we know that T_{1}/T_{2} = e^{µθ}

Or we can say that T_{2} = T_{1}/ e^{µθ}

Putting the value of T_{2} in equation (i)

P = (T_{1 }- T_{1}/ e^{µθ}).v = T_{1}(1 – 1/ e^{µθ}). V...(ii)

Let (1–1/ e^{µθ}) = K , K = any constant

Then the above equation is P = T_{1.}K. V or KT_{1} V ......(iii)

Let Tmax = Maximum tension in the belt

T_{c } = Centrifugal tension which is equal to m.v^{2}

Then T_{max }= T_{1} + T_{c}

T_{1} = T_{max} – T

Putting this value in the equation (iii)

P = K(T_{max} – T_{c} ).

V = K(T_{max} – m.V_{2}).

V = K(T_{max}.v – m.V_{3})

The power transmitted will be maximum if d(P)/dv = 0

Hence differentiating equation w.r.t. V and equating to zero for maximum power, we get

d(P)/dv = K(T_{max} – 3.m.V^{2})=0

T_{max} – 3mV^{2 }=0

T_{max} = 3mV^{2}

V = (T_{max}/3m)^{1/2}

Equation (iv) gives the velocity of the belt at which maximum power is transmitted.

From equation (iv) T_{max} = 3T_{c} ...(v)

Hence when the power transmitted is maximum, centrifugal tension would be 1/3rd of the maximum tension.

We also know that T_{max} = T_{1} + T_{c}

= T_{1} + T_{max}/3 ........(vi)

T_{1 }= T_{max} - T_{max}/3.......(vii)

Hence condition for the transmission of maximum power are:

T_{c} = 1/3 Tmax, and T_{1} = 2/3Tmax ...(viii)

**NOTE:** Net driving tension in the belt = (T_{1 }– T_{2})

**STEPS FOR SOLVING THE PROBLEM FOR FINDING THE POWER**

EPS FOR SOLVING THE PROBLEM FOR FINDING THE POWER

1. Use the formula stress (σ) = force (Maximum Tension)/Area Where;

Area = b.t i.e., Tmax = σ.b.t

2. Unit mass (m) = ρ.b.t.L Where; ρ = Density of a material b = Width of Belt t = Belt thickness L = Unit length Take L = 1m, if b and t are in meter Take L = 100cm, if b and t are in cm Take L = 1000mm, if b and t are in mm

3. Calculate V using V = πDN/60 m/sec (if not given)

4. TC = mV_{2}, For finding TC

5. Tmax = T_{1} + T_{c} , for finding T_{1}

6. For T_{2}, Using relation Ratio of belt tension = T_{1}/T_{2} = eµθ

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

**Steps for Solving the Problem for Finding the Maximum Power**

1. Use the formula stress (σ) = force (Maximum Tension)/Area Where; Area = b.t i.e. Tmax = σ.b.t

2. Unit mass (m) = ρ.b.t.L

Where ρ = Density of a material

b = Width of Belt t = Belt thickness

L = Unit length Take L = 1m, if b and t are in meter

Take L = 100cm, if b and t are in cm

Take L = 1000mm, if b and t are in mm

3. T_{C} =1/3 T_{max} = mV_{2}, For finding T_{C} and velocity (If not given)

We don't Calculate Velocity using V = πDN/60 m/sec (if not given)

5. Tmax = T_{1} + T_{c} , for finding T_{1 }

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

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

**Initial Tension in The Belt **

Let To = initial tension in the belt

T_{1} = Tension in the tight side

T_{2 }= Tension in the slack side

T_{C} = Centrifugal Tension in the belt

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