Question

Find the power transmitted by cross type belt drive connecting two pulley of 45.0cm and 20.0cm diameter, which are 1.95m apart. The maximum permissible tension in the belt is 1KN, coefficient of friction is 0.20 and speed of larger pulley is 100rpm.

Answer 1

 Given that 

D₁ = Diameter of the driver = 45cm = 0.45m 

R₁ = Radius of the driver = 0.225m 

D₂ = Diameter of the driven = 20cm = 0.2m 

R₂ = Radius of the driven = 0.1m 

X = Distance between the centers of two pulleys = 1.95m 

T₁ = Maximum permissible tension = 1000N 

µ = Coefficient of friction = 0.20 

N₁ = Speed of the driver(Larger pulley) in R.P.M. = 100RPM

Since we know that, 

Power Transmitted = (T₁–T₂).V/1000 Kw ...(i) 

Tension is in KN and V is in m/sec 

First ve find the velocity of the belt, 

V = Velocity of the belt in m/sec. 

Here we take diameter and RPM of larger pulley 

= πDN/60 m/sec, D is in meter and N is in RPM 

= (3.14 X0.45 X 100) /60 

= 2.36m/sec.......(ii)

Now Ratio of belt tension, T₁/T₂ = eµ.θ ...(iii) 

Here we don't know the value of θ, For θ, first find the value of α, by the formula, 

Angle of Lap for cross belt α = sin^–1(r₁ + r₂)/X 

= sin^–1(0.225 + 0.1)/1.95 

= 9.59° ...(iv) 

Now Angle of contact (θ) = Π + 2α ----- for cross belt 

θ = Π + 2 X 9.59° 

= 199.19° 

= 199.19°(Π/180°) = 3.47rad ...(v) 

Now putting all the value in equation 

(iii) We get 1000/T₂ =e(0.2)(3.47) T₂ = 498.9 N ...(vi) 

Using equation (i), we get 

P = [(1000 – 498.9) X 2.36 ]/1000 

P = 1.18KW.