power transmitted by a belt

POWER TRANSMITTED BY A BELT

Consider a belt drive, as shown in Fig.8.20.

Let

P = Power transmitted by a belt in watts,

T1 = Tension on the tight side of the belt in newtons,

T2 = Tension on the slack side of the belt in newtons,

v = Velocity of the belt in m/s,

d1 and d2 = Diameter of driver and driven pulleys respectively, and

N1 and N2 = Speed of driver and driven pulleys respectively.

The effective tension or force acting at the circumference of the driven pulley is the difference between these two tensions (i.e., T1 - T2).

⸫ Effective driving force = T1 - T2

Work done per second = Force × Velocity

Example 8.4 

A pulley of diameter 700 mm is driven by a flat belt running at a speed of 300 rpm. The coefficient of friction between the pulley and the belt is 0.3 and the angle of lap is 160°. If the maximum tension in the belt is 2.453 kN, find the power transmitted by a belt.

Given data: 

d = 700 mm = 0.7 m; N = 300 rpm; μ = 0.3; θ = 160° = 160° × (π/180°) = 2.792 rad; T1 = 2.453 kN = 2453 N

Solution: 

Example 8.5 

An open belt drive connects two pulleys of 1.2 m and 0.5 m diameters on parallel shafts 4 m apart. The maximum tension in the belt is 1800 N. The coefficient of friction is 0.3. The driven pulley of diameter 1.2 m runs at 250 rpm. Calculate: (i) the length of the belt required, (ii) the power transmitted, and (iii) the torque on each of the two shafts.

Given data: 

Open belt drive; d1 = 1.2 m or r1 = 0.6 m; d2 = 0.5 m or r2 = 0.25 m; x = 4m; T1 = 1800N; μ = 0.3; N2 = 250 rpm.

Solution:

(i) Length of the belt required (L):

We know that length of the open belt,

(ii) Power transmitted (P): We know that to find power transmitted, first we need to determine the values of T1, T2 and v.

T1 = 1800'N (given)

The value of T2 can be found using tension ratio equation as below. In order to use tension ratio equation, calculate the value of angle of contact (θ) first.

To find θ: We know that angle of contact for the open belt drive,

To find T2: We know that the tension ratio for flat belt drive,

To find P: We know that the power transmitted by the belt,

(iii) Torque on each of the shafts:

Torque on the shaft of driver pulley:

We know that torque on the shaft of driver pulley,

Torque on the shaft of driven pulley:

We know that torque on the shaft of driven pulley,

Example 8.6 

For the above problem (Example 8.5), determine (i) the power lost in friction, and (ii) the efficiency of the belt drive.

Given data: Refer Example 8.5.

Solution: In this case, ddriver = 0.5 m; ddriven = 1.2 m; Ndriver = 250 rpm    …(Given)

(i) Power lost in friction:

We know that input power,

(ii) Efficiency of the belt drive: 

We know that efficiency of the belt driye,

Example 8.7 

Two pulleys, one 450 mm diameter and the other 200 mm diameter are on parallel shafts 2.1 m apart and are connected by a crossed belt. The larger pulley rotates at 225 rpm. The maximum permissible tension in the belt is 1 kN and the coefficient of friction between the belt and the pulley is 0.25. Find: (i) the length of the belt required, and (ii) the power that can be transmitted. 

[A.U., Nov/Dec 2013]

Given data: 

Cross belt drive; d1 = 450 mm = 0.45 m or r1 = 0.225 m; d2 = 200 mm = 0.2 m or r2 = 0.1 m; x = 2.1 m; N1 = 225 rpm; T1 = 1 kN 1000 N; μ = 0.25.

Solution:

(i) Length of the belt required (L):

We know that length of the crossed belt,

(ii) Power transmitted (P): We know that to find power transmitted, first we need to determine the values of T1, T2 and v.

T1 = 1000 N (given)

The value of T2 can be found using tension ratio equation as below. In order to use tension ratio equation, calculate the value of angle of contact (θ) first.

To find θ: We know that angle of contact for the crossed belt driven,

To find T2: We know that the tension ratio for flat belt drive,

To find P: We know that the power transmitted by the belt,