/*! 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 17 The wall shear stress, \(\tau_{w... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 17

The wall shear stress, \(\tau_{w},\) in a boundary layer depends on distance from the leading edge of the body, \(x\), the density, \(\rho\) and viscosity, \(\mu,\) of the fluid, and the freestream speed of the flow, \(U\). Obtain the dimensionless groups and express the functional relationship among them.

Short Answer

Expert verified
The two dimensionless groups are \(\pi_{1} = \frac{\tau_{w}}{\rho U^{2}}\) (also known as the drag coefficient) and \(\pi_{2} = \frac{\mu}{\rho U x}\) (also known as the Reynolds number). These can be related by a functional relationship \(\pi_{1} = F(\pi_{2})\).

Step by step solution

01

Identify the parameters and their dimensions

Firstly, we have to identify the given parameters and their dimensions in the problem. There are five parameters: \(\tau_{w}\) (shear stress), \(x\) (distance), \(\rho\) (density), \(\mu\) (viscosity), and \(U\) (speed). Their dimensional formulae are: \(\tau_{w}\) - \(ML^{-1}T^{-2}\), \(x\) - \(L\), \(\rho\) - \(ML^{-3}\), \(\mu\) - \(ML^{-1}T^{-1}\), and \(U\) - \(LT^{-1}\). M = Mass, L = Length and T = Time are the basic dimensions.
02

Apply Buckingham pi theorem

According to the Buckingham Pi theorem, the number of dimensionless groups is equal to the total number of variables minus the number of fundamental dimensions. Here, the total number of variables is 5, and there are 3 fundamental dimensions (M, L, T). Therefore, there will be \(5-3 = 2\) dimensionless groups.
03

Form the dimensionless groups

To form the dimensionless groups, choose \(x\), \(\rho\), and \(U\) as the repeating variables, because they contain all the fundamental dimensions M, L and T.\nThe first dimensionless group can be formed by combining \(\tau_{w}\) (shear stress) and the repeating variables. The dimensionless group is \(\pi_{1} = \frac{\tau_{w}}{\rho U^{2}}\).\nThe second dimensionless group can be formed by combining \(\mu\) (viscosity) and the repeating variables. The dimensionless group is \(\pi_{2} = \frac{\mu}{\rho U x}\).
04

Express the functional relationship

According to the Buckingham pi theorem, all of the variables can be expressed as a functional relationship between these dimensionless groups: \(\pi_{1} = F(\pi_{2})\), where F is some function, which is not specified by the method of dimensional analysis.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In a fluid mechanics laboratory experiment a tank of water, with diameter \(D,\) is drained from initial level \(h_{0}\). The smoothly rounded drain hole has diameter \(d\). Assume the mass flow rate from the tank is a function of \(h, D, d, g, \rho,\) and \(\mu,\) where \(g\) is the acceleration of gravity and \(\rho\) and \(\mu\) are fluid properties. Measured data are to be correlated in dimensionless form. Determine the number of dimensionless parameters that will result. Specify the number of repeating parameters that must be selected to determine the dimensionless parameters. Obtain the \(\Pi\) parameter that contains the viscosity.

The equation describing motion of fluid in a pipe due to an applied pressure gradient, when the flow starts from rest, is $$\frac{\partial u}{\partial t}=-\frac{1}{\rho} \frac{\partial p}{\partial x}+\nu\left(\frac{\partial^{2} u}{\partial r^{2}}+\frac{1}{r} \frac{\partial u}{\partial r}\right)$$ Use the average velocity \(\bar{V}\), pressure drop \(\Delta p\), pipe length \(L\) and diameter \(D\) to nondimensionalize this equation. Obtain the dimensionless groups that characterize this flow.

A 1: 50 -scale model of a submarine is to be tested in a towing tank under two conditions: motion at the free surface and motion far below the surface. The tests are performed in freshwater. On the surface, the submarine cruises at 24 knots. At what speed should the model be towed to ensure dynamic similarity? Far below the surface, the sub cruises at 0.35 knot. At what speed should the model be towed to ensure dynamic similarity? What must the drag of the model be multiplied by under each condition to give the drag of the full-scale submarine?

An ocean-going vessel is to be powered by a rotating circular cylinder. Model tests are planned to estimate the power required to rotate the prototype cylinder. A dimensional analysis is needed to scale the power requirements from model test results to the prototype. List the parameters that should be included in the dimensional analysis. Perform a dimensional analysis to identify the important dimensionless groups.

The energy released during an explosion, \(E,\) is a function of the time after detonation \(t,\) the blast radius \(R\) at time \(t,\) and the ambient air pressure \(p,\) and density \(\rho .\) Determine, by dimensional analysis, the general form of the expression for \(E\) in terms of the other variables.
See all solutions

Recommended explanations on Physics Textbooks

Nuclear Physics

Read Explanation

Turning Points in Physics

Read Explanation

Force

Read Explanation

Modern Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Medical Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.