# Derive the formula for maximum power transmitted by a belt when centrifugal tension in to account.

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Derive the formula for maximum power transmitted by a belt when centrifugal tension in to account.

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Let T1 = Tension on tight side T2 = Tension on slack side v = Linear velocity of belt Then the power transmitted is given by the equation

P = (T1–T2). V ...(i)

But we know that T1/T2 = eµθ

Or we can say that T2 = T1/ eµθ

Putting the value of T2 in equation (i)

P = (T1 - T1/ eµθ).v = T1(1 – 1/ eµθ). V...(ii)

Let (1–1/ eµθ) = K , K = any constant

Then the above equation is P = T1.K. V or KT1 V  ......(iii)

Let Tmax = Maximum tension in the belt

Tc = Centrifugal tension which is equal to m.v2

Then Tmax = T1 + Tc

T1 = Tmax – T

Putting this value in the equation (iii)

P = K(Tmax – Tc ).

V = K(Tmax – m.V2).

V = K(Tmax.v – m.V3

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(Tmax – 3.m.V2)=0

Tmax – 3mV2 =0

Tmax = 3mV2

V = (Tmax/3m)1/2

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

From equation (iv) Tmax = 3Tc ...(v)

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

We also know that Tmax = T1 + Tc

= T1 + Tmax/3 ........(vi)

T1 = Tmax - Tmax/3.......(vii)

Hence condition for the transmission of maximum power are:

Tc = 1/3 Tmax, and T1 = 2/3Tmax ...(viii)

NOTE: Net driving tension in the belt = (T1 – T2)

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 = mV2, For finding TC

5. Tmax = T1 + Tc , for finding T1

6. For T2, Using relation Ratio of belt tension = T1/T2 = eµθ

7. Power Transmitted = (T1–T2).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. TC =1/3 Tmax = mV2, For finding TC and velocity (If not given)

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

5. Tmax = T1 + Tc , for finding T

6. For T2, Using relation Ratio of belt tension = T1/T2 = eµθ

7. Maximum Power Transmitted = (T1 – T2).v/1000 Kw

Initial Tension in The Belt

Let To = initial tension in the belt

T1 = Tension in the tight side

T2 = Tension in the slack side

TC = Centrifugal Tension in the belt

To = (T1 + T2)/2 + TC