Question

The maximum allowable tension, in a V-belt of groove angle of 30º, is 2500N. The angle of lap is 140º and the coefficient of friction between the belt and the material of the pulley is 0.15. If the belt is running at 2m/sec, Determine: (i) Net driving tension (ii) Power transmitted by the pulley, Neglect effect of centrifugal tension.

Answer 1

Given data: 

Angle of groove(2α) = 30º, α = 15º 

Max. Tension(T_max) = 2500N 

Angle of lap(contact) (θ) = 140º = 140º X (Π/180º) = 2.44 rad 

Coefficient of friction (µ) = 0.15 Speed of belt(V) = 2m/sec We know that, 

T_max = T₁ =2500N 

(TC is neglected, since belt speed is less than 10m/sec) 

Ratio of Tension in V-Belt:: T₁/T₂ = e^{µ.θ.cosec α}

2500/T₂ = e^{(0.15).(2.44).cosec15}

T₂ = 2500/4.11 

T₂ = 607.85N ...(i) 

(i) Net driving tension = (T₁–T₂) 

 = 2500 – 607.85 = 1892.2 N 

 (ii) Power transmitted = (T₁ – T₂)X V W 

 = (2500 – 607.85) X 2 = 3784.3 Watt