/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 A journal bearing \(120 \mathrm{... [FREE SOLUTION] | 91Ó°ÊÓ

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A journal bearing \(120 \mathrm{~mm}\) in diameter and \(60 \mathrm{~mm}\) long has a diametral clearance of \(0.20 \mathrm{~mm}\). The journal rotates \(3000 \mathrm{rpm}\) and is lubricated with SAE 20 oil at an average temperature of \(70^{\circ} \mathrm{C}\). Using Petroff's equation, estimate the power loss and the friction torque.

Short Answer

Expert verified
The estimated power loss is \(2.04 \, \text{kW}\) and the friction torque is \(6.497 \, \text{N.m}\).

Step by step solution

01

Finding Out the Values of Different Variables

The first step involves calculating the relevant parameters. The journal bearing diameter is \(d = 120 \, \text{mm}\), the length \(l = 60 \, \text{mm}\), the diametral clearance \(c = 0.20 \, \text{mm}\), the rotation speed \(N = 3000 \, \text{rpm}\), and the oil viscosity is estimated to be approximately \(0.029 \, \text{Pas}\) for SAE 20 oil at \(70^{\circ}C\). Now, it is needed to convert the diameter and clearance to meters for future calculations, so \(d = 0.12 \, \text{m}\), \(l = 0.06 \, \text{m}\), \(c = 0.00020 \, \text{m}\). The radius \(r = d/2 = 0.06 \, \text{m}\), and the angular speed \(\omega = 2 \pi N / 60 = 314.16 \, \text{rad/s}\).
02

Calculating Friction Torque Using Petroff's Equation

Now, use the Petroff's equation which is given by \(T = \eta \omega r^2 l / c\). Substituting the values into the Petroff's equation, you get \(T = 0.029 \times 314.16 \times (0.06)^2 \times 0.06 / 0.00020 = 6.497 \, \text{N.m}\). So the friction torque is \(6.497 \, \text{N.m}\).
03

Calculating Power Loss

The power loss in a bearing can be calculated using the equation \(P = T \omega\). Using the calculated friction torque and the angular speed, the power loss can be calculated as \(P = 6.497 \times 314.16 = 2040.9 \, \text{W}\) or \(2.04 \, \text{kW}\).

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