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Chapter 7: Problem 6

A V-belt of \(6.0 \mathrm{~cm}^{2}\) cross-section has a groove angle of \(40^{\circ}\) and an angle of lap of \(165^{\circ} .\) The mass per unit length of belt is \(1.2 \mathrm{~kg} /\) s. The coefficient of friction between belt and pulley is \(0.1\). The maximum allowable stress in the belt is \(850 \mathrm{~N} / \mathrm{cm}^{2}\). Determine the power that can be transmitted at a belt speed of \(30 \mathrm{~m} / \mathrm{s}\)

Short Answer

Expert verified
The power transmitted is 85.89 kW.

Step by step solution

01

Convert Angles to Radians

The angle of lap is given as \(165^{\circ}\). First, we need to convert this angle from degrees to radians using the formula: \( \text{radians} = \text{degrees} \times \frac{\pi}{180} \). Thus, \( 165^{\circ} = 165 \times \frac{\pi}{180} = 2.88 \, \text{radians} \).
02

Calculate the Tension Ratio

The tension ratio in a V-belt is given by \( \frac{T_1}{T_2} = e^{(\mu \theta / \sin \beta)} \), where \( \mu = 0.1 \) is the coefficient of friction, \( \theta = 2.88 \) radians is the angle of lap, and \( \beta = \frac{\alpha}{2} = \frac{40^{\circ}}{2} = 20^{\circ} \). Converting \( \beta \) to radians gives \( 20 \times \frac{\pi}{180} = 0.349 \). Hence, \( \frac{T_1}{T_2} = e^{(0.1 \times 2.88 / \sin(0.349))} = e^{0.826} \approx 2.28 \).
03

Calculate Maximum Tension

The maximum allowable stress in the belt is given as \( \sigma = 850 \, \text{N/cm}^2 \). If the cross-section area \( A = 6 \, \text{cm}^2 \), the maximum tension is \( T_{1_{\text{max}}} = \sigma \times A = 850 \times 6 = 5100 \, \text{N} \).
04

Calculate Minimum Tension

Using the tension ratio \( \frac{T_1}{T_2} = 2.28 \) and \( T_{1_{\text{max}}} = 5100 \, \text{N} \), the minimum tension \( T_2 \) can be calculated as \( T_2 = \frac{T_1}{2.28} = \frac{5100}{2.28} \approx 2237 \, \text{N} \).
05

Calculate Power Transmitted

The power transmitted by the belt is given by \( P = (T_1 - T_2) \times v \), where \( v = 30 \, \text{m/s} \) is the belt speed. Therefore, \( P = (5100 - 2237) \times 30 = 2863 \times 30 = 85890 \, \text{W} = 85.89 \, \text{kW} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tension Ratio
The tension ratio is a crucial factor in understanding how V-belts transmit power. It is determined by the relationship between the tight side tension ( T_1) and the slack side tension ( T_2) of the belt. The formula for the tension ratio is given as \( \frac{T_1}{T_2} = e^{( \mu \theta / \sin \beta )}\) . Here, \( \mu \) represents the coefficient of friction, \(\theta\) is the angle of lap in radians, and \(\beta\) is half the groove angle in radians. The ratio shows how much easier it is for the belt to grip on the pulley due to friction and the design of the belt and pulley. In simple terms, it helps understand how much force the drive pulley can apply before the belt starts slipping. A higher tension ratio indicates a better grip on the pulley, allowing for more efficient power transmission. Reviewing these calculations helps ensure the system operates efficiently without slippage.
Coefficient of Friction
The coefficient of friction between the belt and the pulley is a vital parameter in determining the efficiency of power transmission in a V-belt system. It is symbolized as \(\mu\) and is a measure of how adequately the belt grips the pulley surfaces without slipping. In the given exercise, the coefficient of friction is 0.1 , which might seem low. However, even small values can significantly impact the tension ratio and the belt's ability to transmit power efficiently.
  • A higher coefficient of friction means better gripping and less risk of slippage.
  • It directly affects the tension ratio, and thus, the performance of the belt.
Choosing materials with the right friction properties for the belt and pulley surfaces can improve system performance and ensure that the power is transmitted as desired without any loss.
Maximum Allowable Stress
Understanding the maximum allowable stress is crucial for any V-belt system. In this context, stress refers to the force exerted on the belt per unit area. The maximum allowable stress dictates how much tension (or force) the belt can handle before failing. In our exercise, this is mentioned as \(850 \text{ N/cm}^2\). Given the cross-section area of the belt is \(6.0 \text{ cm}^2\), the maximum tension the belt can withstand is calculated using the formula \( T_{1_{\text{max}}} = \sigma \times A \), resulting in 5100 \text{ N}.
  • This value is crucial in determining the belt's efficiency and longevity.
  • Overloading the belt beyond this stress limit can lead to catastrophic failure, comprising system safety.
It's essential to design the system to ensure that the operational tensions remain within these limits to avoid early wear or damage.
Belt Speed
The belt speed is an integral component of calculating the power transmitted by the V-belt. It refers to the velocity of the belt as it wraps around the pulleys in the system. This value is represented as \(v\), and in the given exercise, it's 30 \text{ m/s}. Belt speed contributes to the overall power as determined by the formula \( P = (T_1 - T_2) \times v\), where \(P\) is the power transmitted. The belt speed directly impacts the rate at which power is transferred. Faster-moving belts can transmit more power.
  • Higher speeds mean greater power transfer but could lead to increased wear and stress on the belt.
  • Optimal speed is crucial for maintaining efficiency while prolonging the belt's life.
Understanding how belt speed interacts with tension forces can help optimize system design for both performance and longevity.

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Most popular questions from this chapter

A flat belt is used to transmit a torque from pulley A to pulley B. The radius of each pulley is \(75 \mathrm{~mm}\) and the coefficient of friction is \(0.30\). Determine the largest torque that can be transmitted if the allowable belt tension is \(4 \mathrm{kN}\).

A flat belt connects a \(1.20 \mathrm{~m}\) diameter pulley on a shaft running at \(25 \mathrm{rad} / \mathrm{s}\) with another pulley running at \(50 \mathrm{rad} / \mathrm{s}\), the angle of lap on the latter pulley being \(150^{\circ} .\) The maximum permissible load on the belt is \(1200 \mathrm{~N}\) and the coefficient of friction between the belt and the pulley is \(0.25\). If the initial tension in the belt may have any value between \(800 \mathrm{~N}\) and \(960 \mathrm{~N}\), what is the maximum power that the belt can transmit?

The driving pulley of an open belt drive is of \(720 \mathrm{~mm}\) diameter and rotates at \(300 \mathrm{rpm}\) while transmitting power to a driven pulley having \(225 \mathrm{~mm}\) diameter. The modulus of elasticity of the belt material is \(110 \mathrm{~N} / \mathrm{mm}^{2}\). Determine the speed lost by the driven pulley due to creep if the stresses in the tight and slack sides of the belt are found to be \(0.8 \mathrm{~N} / \mathrm{mm}^{2}\) and \(0.32 \mathrm{~N} / \mathrm{mm}^{2}\), respectively.
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