/*! 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 47 A shaft \(6.00 \mathrm{cm}\) in ... [FREE SOLUTION] | 91Ó°ÊÓ

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A shaft \(6.00 \mathrm{cm}\) in diameter is being pushed axially through a bearing sleeve \(6.02 \mathrm{cm}\) in diameter and \(40 \mathrm{cm}\) long. The clearance, assumed uniform, is filled with oil whose properties are \(\nu=0.003 \mathrm{m}^{2} / \mathrm{s}\) and \(\mathrm{SG}=0.88 .\) Es timate the force required to pull the shaft at a steady velocity of \(0.4 \mathrm{m} / \mathrm{s}.\)

Short Answer

Expert verified
The force required is approximately 398 N.

Step by step solution

01

Calculate the Clearance

First, determine the clearance between the shaft and the bearing sleeve. Subtract the diameter of the shaft from the diameter of the sleeve: \(6.02 \text{ cm} - 6.00 \text{ cm} = 0.02 \text{ cm}\). Since the clearance is needed in meters, convert it: \(0.02 \text{ cm} = 0.0002 \text{ m}\).
02

Determine the Dynamic Viscosity

The dynamic viscosity \(\mu\) of the oil is calculated using the kinematic viscosity \(u\) and the specific gravity (SG). Use the formula \(\mu = u \cdot \rho\), where \(\rho = SG \cdot 1000 \text{ kg/m}^3\). Calculate the density: \(\rho = 0.88 \cdot 1000 = 880 \text{ kg/m}^3\). Thus, \(\mu = 0.003 \text{ m}^2/\text{s} \cdot 880 \text{ kg/m}^3 = 2.64 \text{ kg/m/s} = 2.64 \text{ Pa}\cdot\text{s}\).
03

Calculate the Shear Stress

The shear stress \(\tau\) is determined by the formula \(\tau = \mu \frac{du}{dy}\), where \(du/dy\) is the velocity gradient \(\frac{0.4 \text{ m/s}}{0.0002 \text{ m}} = 2000 \text{ s}^{-1}\). Thus, \(\tau = 2.64 \text{ Pa}\cdot\text{s} \times 2000 \text{ s}^{-1} = 5280 \text{ Pa}\).
04

Determine the Area for Force Application

The area \(A\) through which the shear force acts is the lateral surface area of the shaft, which is \(\pi\) times the diameter of the shaft times the length of the sleeve, \(A = \pi \times 0.06 \text{ m} \times 0.4 \text{ m} = 0.0754 \text{ m}^2\).
05

Calculate the Force Required

The force \(F\) required to pull the shaft is the shear stress times the area, \(F = \tau \cdot A = 5280 \text{ Pa} \times 0.0754 \text{ m}^2 = 397.632 \text{ N}\). Rounding this gives approximately \(398 \text{ N}\).

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Key Concepts

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

Dynamic Viscosity Calculation
Calculating the dynamic viscosity of a fluid is a crucial step in understanding its behavior when subjected to external forces and velocities. Dynamic viscosity, denoted as \( \mu \), measures a fluid's internal resistance to flow. In the context of fluid mechanics, it is calculated using the kinematic viscosity \( u \) and the fluid's density \( \rho \). The relationship is expressed as:\[ \mu = u \cdot \rho \]To find the density of the fluid, we multiply the specific gravity (SG) by the density of water, which is approximately \( 1000 \, \text{kg/m}^3 \). For instance, if the oil's kinematic viscosity is \( 0.003 \, \text{m}^2/\text{s} \) and its specific gravity is \( 0.88 \), the density \( \rho \) would be \( 880 \, \text{kg/m}^3 \). Thus, the dynamic viscosity would be:\[ \mu = 0.003 \, \text{m}^2/\text{s} \times 880 \, \text{kg/m}^3 = 2.64 \, \text{Pa} \cdot \text{s} \]This calculation is fundamental for predicting how easily a fluid can be moved under force.
Shear Stress
Shear stress in fluid mechanics is the force per unit area exerted by a fluid against a surface, like a shaft moving through a bearing. It represents how much force is needed to "slide" one layer of fluid over another. To calculate the shear stress \( \tau \), we use the dynamic viscosity \( \mu \) and the velocity gradient \( \frac{du}{dy} \). The formula is:\[ \tau = \mu \frac{du}{dy} \]In this exercise, the velocity gradient is the change in velocity over the clearance between the shaft and the bearing:\[ \frac{du}{dy} = \frac{0.4 \, \text{m/s}}{0.0002 \, \text{m}} = 2000 \, \text{s}^{-1} \]Substituting the values, the shear stress becomes:\[ \tau = 2.64 \, \text{Pa} \cdot \text{s} \times 2000 \, \text{s}^{-1} = 5280 \, \text{Pa} \]Understanding shear stress is vital for ensuring mechanical components can handle the forces encountered in operation without failing.
Kinematic Viscosity
Kinematic viscosity \( u \) is a measure of a fluid's intrinsic flow characteristic. It's calculated as the dynamic viscosity \( \mu \) divided by density \( \rho \). It gives insight into how easily a fluid flows under its own weight. The formula is:\[ u = \frac{\mu}{\rho} \]In applications involving fluids in mechanical environments, knowing the kinematic viscosity helps predict behavior under different flow conditions. For example, in an oil-lubed bearing:- A lower kinematic viscosity indicates easier flow but might result in less friction reduction.- Conversely, a higher kinematic viscosity points to thicker, more resistant flow which can reduce slippage and provide more cushioning.This balance is essential when selecting appropriate lubricating oils for mechanical systems.
Force Estimation
Estimating the force required to move an object through a viscous fluid involves several parameters, notably shear stress and the effective area over which the force is applied. To determine this force \( F \), we use the formula:\[ F = \tau \cdot A \]Where:- \( \tau \) is the shear stress calculated from the viscosity and velocity gradient.- \( A \) is the surface area in contact with the fluid.In the shaft example, the force is calculated using the shaft’s surface area:- The cylindrical surface area \( A \) is derived from the shaft’s circumference (\( \pi \times \text{diameter} \)) and its length:\[ A = \pi \times 0.06 \, \text{m} \times 0.4 \, \text{m} = 0.0754 \, \text{m}^2 \]In this context, understanding force estimation is crucial for designing systems that must move objects efficiently through lubricating fluids.
Bearing Clearance
Bearing clearance is the space between the shaft and the bearing sleeve. It's crucial in determining how easily a shaft can rotate or slide within a bearing, based on the fluid's lubricating ability. The clearance directly affects lubrication performance since it dictates the thickness of the lubricating film.In practice, bearing clearance impacts:- The ease of movement: smaller clearances entrain less oil, which might lead to increased wear and tear.- Load distribution: uniform clearances ensure even pressure distribution across the shaft surface.For accurate force calculations in mechanical systems, it's necessary to compute and convert the clearance accurately.In the example:- A small clearance of \( 0.02 \, \text{cm} \) translates to \( 0.0002 \, \text{m} \), significantly influencing the lubrication layer and the applied forces.Overall, understanding bearing clearance helps in optimizing mechanical design for longevity and efficiency.

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Most popular questions from this chapter

The laminar-pipe-flow example of Prob. 1.12 can be used to design a capillary viscometer [27]. If \(Q\) is the volume flow rate, \(L\) is the pipe length, and \(\Delta p\) is the pressure drop from entrance to exit, the theory of Chap. 6 yields a formula for viscosity: $$\mu=\frac{\pi r_{0}^{4} \Delta p}{8 L Q}$$ Pipe end effects are neglected [27]. Suppose our capillary has \(r_{0}=2 \mathrm{mm}\) and \(L=25 \mathrm{cm} .\) The following flow rate and pressure drop data are obtained for a certain fluid: $$\begin{array}{l|l|l|l|l|l} Q, \mathrm{m}^{3} / \mathrm{h} & 0.36 & 0.72 & 1.08 & 1.44 & 1.80 \\ \hline \Delta p, \mathrm{kPa} & 159 & 318 & 477 & 1274 & 1851\end{array}$$ What is the viscosity of the fluid? Note: Only the first three points give the proper viscosity. What is peculiar about the last two points, which were measured accurately?

A two-dimensional steady velocity field is given by \(u=\) \(x^{2}-y^{2}, v=-2 x y .\) Derive the streamline pattern and sketch a few streamlines in the upper half plane. Hint: The differential equation is exact.

Algebraic equations such as Bernoulli's relation, Eq. (1) of Ex. \(1.3,\) are dimensionally consistent, but what about differential equations? Consider, for example, the boundary-layer \(x\) -momentum equation, first derived by Ludwig Prandtl in 1904 : $$\rho u \frac{\partial u}{\partial x}+\rho v \frac{\partial u}{\partial y}=-\frac{\partial p}{\partial x}+\rho g_{x}+\frac{\partial \tau}{\partial y}$$ where \(\tau\) is the boundary-layer shear stress and \(g_{x}\) is the component of gravity in the \(x\) direction. Is this equation dimensionally consistent? Can you draw a general conclusion?

A soap bubble of diameter \(D_{1}\) coalesces with another bubble of diameter \(D_{2}\) to form a single bubble \(D_{3}\) with the same amount of air. Assuming an isothermal process, derive an expression for finding \(D_{3}\) as a function of \(D_{1}, D_{2}\) \(p_{\text {atm }},\) and \(Y.\)

The kinematic viscosity of a fluid is the ratio of viscosity to density, \(\nu=\mu / \rho .\) What is the only possible dimensionless group combining \(\nu\) with velocity \(V\) and length \(L ?\) What is the name of this grouping? (More information on this will be given in Chap. 5.)

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