Question

A belt 100mm wide and 8.0mm thick are transmitting power at a belt speed of 160m/minute. The angle of lap for smaller pulley is 165º and coefficient of friction is 0.3. The maximum permissible stress in belt is 2MN/m2 and mass of the belt is 0.9Kg/m. find the power transmitted and the initial tension in the belt.

Answer 1

Given data: 

Width of belt(b) = 100mm 

Thickness of belt(t) = 8mm 

Velocity of belt(V) = 160m/min = 2.66m/sec 

Angle of contact(?) = 165° = 165° X Π/180 = 2.88rad 

Coefficient of friction(µ) = 0.3 

Maximum permissible stress(f) = 2 X 10⁶ N/m² = 2N/mm² 

Mass of the belt material(m) = 0.9 Kg/m Power = ? 

Initial tension (To) = ? We know that, 

T_max = σ.b.t = 2 X 100 X 8 = 1600N ...(i)

Since m and velocity (V) is given, then

Using the formula, T_C = mV², For finding T_c

= 0.9(2.66)²

= 6.4 N ........(ii)

Using the formula, Tmax = T₁ + T_c , for finding T₁

1600 = T₁ + 6.4

T₁ = 1593.6N ......(iii)

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

1593.6/T2 = e^(0.3)(2.88)

T₂ = 671.69 N ........(iv)

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

P = (1593.6 – 671.69).2.66/1000 Kw

P = 2.45KW

Let To = initial tension in the belt

T_o = (T₁ + T₂)/2 + T_C

T_o = (1593.6 + 671.69)/2 + 6.4

T_o = 1139.045N.