/*! 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 62 A shaft with outside diameter of... [FREE SOLUTION] | 91Ó°ÊÓ

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A shaft with outside diameter of \(18 \mathrm{mm}\) turns at 20 revolutions per second inside a stationary journal bearing \(60 \mathrm{mm}\) long. A thin film of oil \(0.2 \mathrm{mm}\) thick fills the concentric annulus between the shaft and journal. The torque needed to turn the shaft is \(0.0036 \mathrm{N} \cdot \mathrm{m}\). Estimate the viscosity of the oil that fills the gap.

Short Answer

Expert verified
The viscosity of the oil is calculated as \(0.0845 Ns/m^2\) or \(0.0845 Pas\).

Step by step solution

01

List Given Information

The outside diameter of the shaft, the length of the journal bearing, the thickness of the oil, and the torque needed to turn the shaft are given, respectively as: 18mm, 60mm, 0.2mm and 0.0036 N·m. Also the shaft turns at 20 revolutions per second which is converted to angular velocity as \(2\pi n = 2\pi x 20 = 40\pi rad/sec\).
02

Convert Units

First, convert the units of the given quantities to SI units. The length L, the radius R of shaft and thickness of oil in meters are: \(L = 60mm = 60/1000 m = 0.06m\), \(R = \frac{18mm}{2} = 9mm = 9/1000 m = 0.009m\), \(t = 0.2mm = 0.2/1000 m = 0.0002m\).
03

Apply the Formula for Torque

Now apply the formula for torque on a rotating cylinder \(T=2\pi L\eta R^{2}n\). Solving this equation for viscosity \(\eta\), we get \(\eta=\frac{T}{2\pi L R^{2}n}\).
04

Substitute the Given Values

Substitute the given values into the formula: \(\eta=\frac{0.0036}{2\pi*0.06*0.009^{2}*40\pi}\) .
05

Calculate the Viscosity

Calculate the expression above to find the numerical value of the viscosity \(\eta\) of the oil.

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