Question

An open belt drive connects two pulleys 120cm and 50cm diameter on parallel shafts 4m apart. The maximum tension in the belt is 1855.3N. The coefficient of friction is 0.3. The driver pulley of diameter 120cm runs at 200rpm. Calculate (i) The power transmitted (ii) Torque on each of the two shafts.

Answer 1

D₁ = Diameter of the driver = 120cm = 1.2m 

R₁ = Radius of the driver = 0.6m 

N₁ = Speed of the driver in R.P.M. = 200RPM 

D₂ = Diameter of the driven or Follower = 50cm = 0.5m 

R₂ = Radius of the driven or Follower = 0.25m 

X = Distance between the centers of two pulleys = 4m 

µ = Coefficient of friction = 0.3 

T₁ = Tension in the tight side of the belt = 1855.3N 

Calculation for power transmitting: 

Let 

P = Maximum power transmitted by belt drive 

= (T₁–T₂).V/1000 KW ...(i)

Where, T₂ = Tension in the slack side of the belt 

V = Velocity of the belt in m/sec. 

= πDN/60 m/sec, D is in meter and N is in RPM ...(ii) 

For T₂, 

We use the relation Ratio of belt tension = T₁/T₂ = eµθ ...(iii) 

But angle of contact is not given, let 

θ = Angle of contact and, 

θ = Angle of lap for open belt, Angle of contact (θ) = Π – 2α ...(iv) 

Sinα = (r₁ – r₂)/X = (0.6 – 0.25)/4 α = 5.02° ...(v) 

Using the relation (iii), θ = Π – 2α = 180 – 2 X 5.02

= 169.96° = 169.96° X Π/180 = 2.97 rad ...(iv) 

Now using the relation (iii)1855.3/T₂ = e(0.3)(2.967)

T₂ = 761.8N ...(vii)

For finding the velocity, using the relation (ii) V 

= (3.14 X 1.2 X 200)/60 = 12.56 m/sec ...(viii) 

For finding the Power, using the relation (i) 

P = (1855.3 – 761.8) X 12.56 P = 13.73 KW

We know that, 1. Torque exerted on the driving pulley = (T₁ – T₂).R₁ 

= (1855.3 – 761.8) X 0.6 

= 656.1Nm

2. Torque exerted on the driven pulley 

= (T₁ – T₂).R₂ = (1855.3 – 761.8) X 0.25 

= 273.4.1Nm