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Grade 9Modern Physics

Design a rubber belt to drive a dynamo generating 20KW at 2200RPM. It is fitted with a pulley of diameter 200mm. Allowable stress for belt= 2.4MPa.Density of rubber=1000kg/m^3.Angle of contact for dynamo pulley= 165 degreeCoefficient of friction between belt and pulley= 0.3Assume dynamo efficiency to be 80%

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12 Years agoGrade 9
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ApprovedApproved Tutor Answer1 Year ago

Designing a rubber belt to drive a dynamo generating 20 kW at 2200 RPM involves several steps, including calculating the required belt dimensions, tension, and other parameters. Let’s break this down step by step to ensure we cover all necessary aspects.

Understanding the Requirements

First, we need to clarify the specifications:

  • Power output of the dynamo: 20 kW
  • Speed of the dynamo: 2200 RPM
  • Pulley diameter: 200 mm (0.2 m)
  • Allowable stress for the belt: 2.4 MPa
  • Density of rubber: 1000 kg/m³
  • Angle of contact: 165 degrees
  • Coefficient of friction: 0.3
  • Dynamo efficiency: 80%

Calculating the Required Torque

To find the torque required to drive the dynamo, we first need to determine the effective power input considering the efficiency:

Effective Power Input (P) = Output Power / Efficiency = 20 kW / 0.8 = 25 kW

Next, we convert this power into torque using the formula:

Torque (T) = (Power × 60) / (2π × RPM)

Substituting the values:

T = (25000 W × 60) / (2π × 2200) ≈ 216.5 Nm

Determining the Tension in the Belt

To find the tension in the belt, we can use the relationship between torque and tension:

Torque (T) = (Tension (T1) - Tension (T2)) × (Radius of Pulley)

Here, the radius of the pulley (r) is half of the diameter:

r = 0.2 m / 2 = 0.1 m

Rearranging the formula gives us:

T1 - T2 = T / r = 216.5 Nm / 0.1 m = 2165 N

Calculating the Maximum Tension

The maximum tension in the belt can be calculated using the allowable stress:

Maximum Tension (Tmax) = Allowable Stress × Cross-sectional Area

To find the cross-sectional area, we need the width and thickness of the belt. Assuming a belt width (b) of 0.05 m and a thickness (h) of 0.01 m:

Cross-sectional Area (A) = b × h = 0.05 m × 0.01 m = 0.0005 m²

Now, calculating Tmax:

Tmax = 2.4 MPa × 0.0005 m² = 1200 N

Finding the Required Tension Ratio

Using the relationship between T1 and T2:

T1 = T2 + 2165 N

We also know that T1 must not exceed Tmax:

T1 ≤ 1200 N

Thus, we can set up the equation:

T2 + 2165 N ≤ 1200 N

This indicates that the design needs to be adjusted, as the maximum tension exceeds the allowable limit. We may need to increase the belt width or thickness to accommodate the required tension.

Friction and Angle of Contact

The frictional force can be calculated using the coefficient of friction and the normal force:

Frictional Force (F) = Coefficient of Friction × T2

Using the angle of contact (θ) in radians (165 degrees = 2.88 radians), we can calculate the relationship between T1 and T2:

T1 = T2 × e^(μθ)

Substituting the values:

T1 = T2 × e^(0.3 × 2.88)

Solving these equations will help us find the appropriate dimensions for the belt.

Final Considerations

In summary, designing a rubber belt for this application involves careful calculations of torque, tension, and material properties. Adjustments to the belt dimensions may be necessary to ensure that the design meets the required performance without exceeding the allowable stress. Always consider safety factors and real-world conditions when finalizing your design.