Question

A power of 4.5 kW must be transmitted from a pulley of an effective diameter of 0.25 m. The pulley is running at 1200 RPM. The angle of contact is 160° and the groove angle is 40°. The coefficient of friction is 0.2. If the allowable tension in each belt is 100 N, calculate the number of V-belts that will be required to transmit the power.

Answer 1

Solution;

Given;

Power transmitted,P=4.5kW

Effective diameter,$d_{eff}=0.25m$

Speed of pulley,N=$1200$ r.p.m

Angle of contact,$\theta$ =160°

Maximum tension,$T_1$ =100N

Co-efficient of friction,$\mu$ =0.2

Groove angle,2$\beta$ =40°; $\beta$ =20°

Contact angle,$\theta$ =$\frac{π×160}{180}=2.792rad$

By using the relation;

$2.3log(\frac{T_1}{T_2})=\mu×\theta×cosec\beta$

$log(\frac{T_1}{T_2})=\frac{0.2×2.792×cosec(20°)}{2.3}$

$log(\frac{T_1}{T_2})=0.7098$

$\frac{T_1}{T_2}=e^{0.709}$ $=2.0336$

$T_2=\frac{T_1}{2.0336}=\frac{100}{2.0336}$

$T_2=49.172N$

Torque transmitted;

$T=(T_1-T_2)×\frac{d_{eff}}{2}$

$T=(100-49.172)×0.125$

$T=6.3535Nm$

Power transmitted by a single belt;

$P_s=\frac{2πNT}{60}$

$P_s=\frac{2π×1200×6.35}{60}$ =$798.40W$

Number of belts;

$n=\frac{P}{P_s}=\frac{4500}{798.4}=5.6$

$n=5.636\approx6$

Use 6 belts.

Answer;

6 belts.