Question

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

Answer 1

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 = e^µθ

Or we can say that T₂ = T₁/ e^µθ

 Putting the value of T₂ in equation (i)

P = (T₁ - T₁/ e^µθ).v = T₁(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 KT₁ V  ......(iii)

Let Tmax = Maximum tension in the belt

T_c = Centrifugal tension which is equal to m.v²

Then T_max = T₁ + T_c

T₁ = T_max – T

Putting this value in the equation (iii)

P = K(T_max – T_c ).

V = K(T_max – m.V₂).

V = K(T_max.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(T_max – 3.m.V²)=0

T_max – 3mV² =0

T_max = 3mV²

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 = T₁ + T_c

= T₁ + T_max/3 ........(vi)

T₁ = T_max - T_max/3.......(vii)

Hence condition for the transmission of maximum power are: 

T_c = 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 + T_c , 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. T_C =1/3 T_max = mV₂, 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 = T₁ + T_c , for finding T₁ 

6. For T₂, Using relation Ratio of belt tension = T₁/T₂ = e^µθ 

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 

T_C = Centrifugal Tension in the belt 

T_o = (T₁ + T₂)/2 + T_C