Belt Drive Design

Problem Statement: Design a belt drive to transmit 2000 watts of power over a distance of 3 meters. The driver pulley is connected to a motor running at 1500 RPM, and the belt material is made of rubber. The coefficient of friction between the belt and pulley is assumed to be 0.3. Calculate the appropriate dimensions for the belt and pulleys.

Solution:

Step 1: Power to be transmitted:

The power transmitted by the belt drive is given as:

\( P = 2000 \, \text{W} \)

Step 2: Driver Pulley Speed and Pulley Diameters:

Assuming the driver pulley speed \(N_1 = 1500 \, \text{RPM}\), we can choose the driver and driven pulley diameters as:

Driver pulley diameter \( D_1 = 200 \, \text{mm} \)
Driven pulley diameter \( D_2 = 400 \, \text{mm} \)

The velocity ratio \( VR \) is given by:

\( VR = \frac{D_2}{D_1} = \frac{400}{200} = 2 \)

Step 3: Belt Speed:

The belt speed \( v \) can be calculated using the formula:

\( v = \frac{\pi D_1 N_1}{60} \)

Substituting the known values:

\( v = \frac{\pi \times 0.2 \times 1500}{60} = 15.7 \, \text{m/s} \)

Step 4: Power Transmission Capacity of the Belt:

The power transmitted by the belt can be expressed as:

\( P = (T_1 - T_2) v \)

Where \( T_1 \) is the tension on the tight side and \( T_2 \) is the tension on the slack side. Using the equation for the ratio of tensions:

\( \frac{T_1}{T_2} = e^{\mu \theta} \)

Assuming \( \mu = 0.3 \) (coefficient of friction) and \( \theta = \pi \, \text{radians} \) (angle of contact between belt and pulley):

\( \frac{T_1}{T_2} = e^{0.3 \times \pi} \approx 2.57 \)

Step 5: Tensions in the Belt:

The power equation becomes:

\( 2000 = (T_1 - T_2) \times 15.7 \)

Substitute \( T_1 = 2.57 T_2 \) into the equation:

\( 2000 = (2.57T_2 - T_2) \times 15.7 \)

Solving for \( T_2 \):

\( T_2 = \frac{2000}{1.57 \times 15.7} \approx 81.27 \, \text{N} \)

Now, calculate \( T_1 \):

\( T_1 = 2.57 \times 81.27 \approx 208.87 \, \text{N} \)

Step 6: Belt Dimensions:

The cross-sectional area \( A \) of the belt is related to the tension by the formula:

\( T_1 = \sigma \cdot A \)

For rubber, the allowable tensile stress \( \sigma \) is around 3 MPa. Therefore:

\( A = \frac{T_1}{\sigma} = \frac{208.87}{3 \times 10^6} = 69.62 \, \text{mm}^2 \)

Assuming a rectangular cross-section with width-to-thickness ratio of 2:1:

\( 2x \cdot x = 69.62 \quad \Rightarrow \quad x^2 = 34.81 \quad \Rightarrow \quad x \approx 5.9 \, \text{mm} \)

Thus, the belt dimensions are:

Width \( b = 2 \times 5.9 = 11.8 \, \text{mm} \)
Thickness \( t = 5.9 \, \text{mm} \)

Summary of Belt Drive Design:

Belt type: V-belt made of rubber.
Driver pulley diameter: \( D_1 = 200 \, \text{mm} \)
Driven pulley diameter: \( D_2 = 400 \, \text{mm} \)
Belt speed: \( v = 15.7 \, \text{m/s} \)
Belt cross-sectional dimensions: Width \( b = 11.8 \, \text{mm} \), Thickness \( t = 5.9 \, \text{mm} \)