# Numerical - Power transmitting capacity, given allowable tension, velocity & mass

Q.  A V-belt pulley has belt velocity 20 m/s and mass 0.7 kg per meter. If allowable tension in the belt is 600 N then what will be the power transmitting capacity of belt?
(Assume μ = 0.5 & θ = 2.5 rad)

- Published on 19 Oct 15

a. 3.75 kW
b. 3.2 kW
c. 4.5 kW
d. 5.23 kW

#### Discussion

• Sravanthi   -Posted on 05 Oct 15
Given: V = 20 m/s, m = 0.7 kg, Ft1 = 600 N, μ = 0.5 & θ = 2.5 rad

Formula: 1) P = V (F1 – F2) / 1000
2) Ratio of effective tension in tight side & slack side of belt F1 / F2 = eμθ
3) Centrifugal tension (Fc) = mV2
4) F1 = Allowable tension (Ft1) - centrifugal tension (Fc)

1) F1 / F2 = eμθ
Substituting the given values we get,
F1 / F2 = 3.490

F1 = 320 N & F2 = 91.69 N

2) P = V(F1 – F2) / 1000
Substituting the given values we get,

Power transmitting capacity = 4.5 kW

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