UNIVERSITY OF EXTREMADURA

School of Industrial Engineering

Department of Mechanical, Energetic and Material Engineering

FLUID MECHANICS

Jose Mara Montanero

Area of Fluid Mechanics

Foreword

This subject aims at teaching the basic concepts of uid statics and dynamics, as well as their applications to industrialengineering. The subject contents are divided into two parts: (i) general fundamentals of uid mechanics, and (ii)applications of those fundamentals to the study of ows with constant density. In this last part, we shall considerproblems of hydrostatics, aerodynamics, and hydraulics.

General fundamentals.

Any course of uid mechanics must start by explaining the model adopted to study a uid. This involves distin-guishing between the microscopic and macroscopic descriptions, clarifying the relationships between them. These ideasconstitute the continuum hypothesis included in chapter 1. In this chapter, we also explain the dierence betweenliquids and gases.

In general, a uid-dynamic problem entails the existence of a ow as the response to the action of several typesof forces. Before analyzing the origin of the forces driving the uid movement, it is appropriate to acquire the basictools which allow one to describe the ow. This is the purpose of chapter 2. In this chapter, we introduce severalkinematic concepts to \visualize" the ow, and to extract relevant information contained in the velocity eld whichcharacterizes it.

We devote chapter 3 to the analysis of the forces acting on uid elements, and to the study of the energy transfertaking place in liquids and gases. This involves introducing and explaining the following concepts: surface force,external force, stress tensor, viscosity, heat ux, thermal conductivity, etc. These concepts allow one to derive themechanical equations that describe the evolution of a uid system. The understanding of this part of the subject is adicult task. It is necessary that students understand the origin of certain statements without spending a long timeon mathematical formalisms.

In chapter 3, we deduced the mechanical laws for a moving uid system. In most cases, one is not interested in theevolution of systems, but in the local eects that these systems cause when crossing a certain region (a machine, forinstance). The Reynolds transport theorem allows one to translate the equations for a uid system into the integralrelations which describe the evolution of the mechanical quantities in a control volume. Chapter 4 is devoted to getthose relations. The equations for a control volume can be applied to some specic cases to obtain interesting results;for example, the generalized Bernoulli equation for momentum, or the energy equation for a hydraulic machine. It isimportant to consider these applications for the student to notice the relevance and usefulness of the obtained results.

Because of the complexity of most technological uid-dynamic problems, it is unavoidable in many cases to rule outany theoretical approach. This is so not only in those situations in which the complexity is evident, but also in appar-ently simple problems. In this scenario, one must often consider information acquired from experiments exclusively.Experiments must be conducted obtaining the maximum information with the lowest number of experiments. Dimen-sional analysis (chapter 5) is the set of techniques and methods used to get information on a specic phenomenon,prior to the experiments, and just from the knowledge of the variables involved in that phenomenon.

Applications.

In chapter 6, we study the behavior of a uid at rest. We shall focus on the calculation of forces and torquesexerted on surfaces and volumes either partially or totally submerged in liquids.

In the rst section of chapter 7, the concept of turbulence is presented, and the consequences of this phenomenonon the dynamical behavior of uids is explained qualitatively. An important group of uid-dynamic applications isthat constituted by ows taking place next to solid walls for high enough Reynolds numbers (boundary layers). Inthis chapter, we provide a qualitative description of this kind of ow. The detachment of boundary layers requiresspecial attention due to its inuence on the aerodynamic behavior of blunt bodies and the performance of hydraulicdevices. We explain the origin of the boundary layer detachment and its eects on the form drag.

In chapter 8, we pay attention to the analysis of hydraulic applications of crucial importance in industrial engineer-ing. We shall start by showing Moody's diagram, which allows one to calculate the loss of reduced pressure in pipesworking in both laminar and turbulent modes. A practical approach suggests dealing with the problem qualitatively,avoiding mathematical demonstrations. In the second part of this topic, we analyze the loss of reduced pressurelocated in devices frequently connected to multiple-pipe systems. These devices are commonly used as ow meters

because the local loss of reduced pressure is an indirect measurement of the ow rate crossing them. In the third partof this chapter, we study the loss of reduced pressure taking place in multiple-pipe systems most commonly used inhydraulics. We classify multiple-pipe systems according to their structure, and provide the students with the basicrules to calculate ow rates and losses of reduced pressure. In the last section of this chapter, we study the role playedin multiple-pipe systems by pumps connected both in series and in parallel.

Chapter 9 is devoted to the analysis of some ows with free surfaces. Specically, we study the uniform turbulent

ow in channels. Because of the complexity of a rigurous theoretical approach to this problem, we restrict ourselvesto showing Manning's empirical formula to calculate the ow rate transported by a channel with arbitrary shape.Finally, we present a qualitative description of weirs and sluicegates, giving some useful empiric results.

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Table of Contents

I Essentials of Fluid Mechanics 1

1 INTRODUCTION 3

1 .- GENERAL COMMENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.1.- Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2.- Continuum hypothesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2 KINEMATICS 5

1 .- INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 .- LAGRAGIAN AND EULERIAN DESCRIPTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

3 .- STREAMLINE AND PATH . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

3.1.- Streamline . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

3.2.- Path . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

4 .- TYPES OF FLOWS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

4.1.- Steady ow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

4.2.- Uniform ows and streams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

4.3.- Two- and one-dimensional ows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

4.4.- Axisymmetric ows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

4.5.- Incompressible ows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

5 .- FLOW RATE AND MASS FLOW RATE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

5.1.- Flow rate and average velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

5.2.- Mass ow rate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

5.3.- Flux of a scalar or vector quantity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3 EQUATIONS FOR A FLUID SYSTEM 10

1 .- INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2 .- SURFACE FORCES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.1.- Surface forces versus external forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.2.- Stress tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.3.- Stress tensor symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.4.- Hydrostatic pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.5.- Viscous stress tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.6.- Navier-Poisson law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

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TABLE OF CONTENTS

2.7.- Viscosity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.8.- Surface forces exerted on a uid system . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3 .- HEAT CONDUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3.1.- Heat conduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3.2.- Heat ux vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3.3.- Fourier law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3.4.- Thermal conductivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3.5.- Heat ux in a uid system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

4 .- MECHANICAL EQUATIONS FOR A FLUID SYSTEM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

4.1.- Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

4.2.- Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

4.3.- Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

4 EQUATIONS FOR A CONTROL VOLUME 18

1 .- SYSTEMS AND CONTROL VOLUMES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2 .- REYNOLDS TRANSPORT THEOREM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

3 .- UNIFORM APPROXIMATION FOR THE FLUX TERM . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

4 .- CONTINUITY EQUATION FOR A CONTROL VOLUME . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

5 .- MOMENTUM EQUATION FOR A CONTROL VOLUME . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

6 .- BERNOULLI EQUATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

6.1.- Bernoulli and hydrostatics equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

7 .- ENERGY EQUATION FOR A CONTROL VOLUME . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

8 .- ENERGY EQUATION FOR A FLUID MACHINE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

8.1.- Application to the adiabatic case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

9 .- ENERGY EQUATION FOR A HYDRAULIC MACHINE . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

5 DIMENSIONAL ANALYSIS 27

1 .- INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

1.1.- Some denitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

1.2.- Dimensional homogeneity principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

2 .- BUCKINGHAM THEOREM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

2.1.- Formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

2.2.- Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

3 .- PHYSICAL SIMILARITY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

3.1.- Physical similarity. Dierent types of similarity . . . . . . . . . . . . . . . . . . . . . . . . 30

3.2.- Fluid-dynamic models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3.3.- Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3.4.- Partial similarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

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4 .- IMPORTANT DIMENSIONLESS GROUPS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

II Applications 35

6 HYDROSTATICS 37

1 .- INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

2 .- REDUCTION OF A SYSTEM OF FORCES IN HYDROSTATICS . . . . . . . . . . . . . . . . . . . . . . . . 37

2.1.- Equivalent system of parallel forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

2.2.- Application to hydrostatics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

3 .- FORCES AND TORQUES ON A FLAT SURFACE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

3.1.- Force exerted on a at surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

3.2.- Central axis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

4 .- FORCES AND TORQUES ON SUBMERGED BODIES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

4.1.- Forces exerted on a submerged body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

4.2.- Central axis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

7 ESSENTIALS OF FLUID DYNAMICS 42

1 .- INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

2 .- THE TURBULENCE PHENOMENON . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

3 .- BOUNDARY LAYER . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

4 .- BOUNDARY LAYER SEPARATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

4.1.- Form drag . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

4.2.- Boundary layer separation and form drag . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

4.3.- Origin of the BL separation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

8 HYDRAULICS 49

1 .- INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

2 .- FLOW IN PIPES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

2.1.- Loss of reduced pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

2.2.- Equivalent roughness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

2.3.- The Moody chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

3 .- LOCAL LOSSES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

3.1.- Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

3.2.- Formulation of the problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

3.3.- Dimensional analysis of the problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

3.4.- Drop of reduced pressure in pipes with connections . . . . . . . . . . . . . . . . . . . . . . 53

3.5.- Flow meters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

4 .- MULTIPLE- PIPE SYSTEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

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4.1.- Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

4.2.- Pipes in series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

4.3.- Pipes in parallel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

4.4.- Pipe networks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

4.5.- Pipeline with uniform draw-o . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

5 .- MULTIPLE- PIPE SYSTEMS WITH PUMPS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

5.1.- Head supplied by the pump . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

5.2.- Pumps in series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

5.3.- Pumps in parallel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

9 OPEN CHANNELS, WEIRS AND SLUICEGATES 60

1 .- INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

2 .- UNIFORM FLOW IN OPEN CHANNELS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

3 .- WEIRS AND SLUICEGATES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

3.1.- Weirs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

3.2.- Sluicegates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

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Part I

Essentials of Fluid Mechanics

Chapter 1

INTRODUCTION

1 .- GENERAL COMMENTS

Fluid mechanics studies the behavior of uids both at rest (hydrostatics) and in motion (hydrodynamics). In particular,one is interested in the uid-solid interaction occurring when the uid surrounds the solid body (aerodynamics), or itis inside the solid element (hydraulics).

Two routes can be followed to analyze a uid-mechanical problem: the theoretical and experimental ones. Thetheoretical analysis is necessarily restricted to very simple problems, or to those for which global results (resultantforces, average pressures, ow rates, etc.) are to be obtained. In more complex phenomena, it is necessary to resortto experimentation in order to get useful information of the system considered. Experiments, guided by dimensionalanalysis, provide data for designing uid machinery.

Recently, numerical methods have become a third route to analyze uid-dynamical problems based on the hugecapability of current computers. They can provide reliable results in situations of technological interest. The simulationof uid-dynamical problems essentially reduces to the numerical integration of the motion equations with certain initialand boundary conditions. This integration allows one to obtain detailed information on the uid behavior. The useof this kind of technique is beyond the scope of the present introductory subject.

1.1.- Fluids

Fluid denition

Normal and shear stresses are the forces per unit area exerted on a surface in the directions normal and tangential tothat surface, respectively (see gure 1.1). Fluid is the state of matter which deforms continuously under the actionof shear stresses. In other words, any shear stress causes a dynamical deformation of the uid, which continues untilthat stress disappears. Therefore, shear stresses must vanish in hydrostatics. The above denition distinguishes uidsfrom solids. Solids show a static deformation when shear stresses are applied to them.

Gases y liquids

Fluids are divided into gases and liquids. Gases easily compress/expand due to variations of pressure, while liquidsare almost incompressible. By way of illustration, consider the following example: one must double the air pressureto double the density (for a constant temperature), while the water pressure must be raised 100 times to increase thedensity by just 0.5%.

Density variations associated with temperature changes may be noticeable in some cases. For instance, the waterdensity decreases by 0.5-1% (for a constant pressure) if the temperature increases about 30 C. In most of the problemsconsidered in this subject, we shall not observe signicant temperature variations. Therefore, the liquid density willbe assumed to be constant.

3

INTRODUCTION

fluid element

normal stresses

shear stresses

fluid element

fluid element

Figure 1.1: Normal and shear stresses.

1.2.- Continuum hypothesis

Fluids consist of molecules moving in a vacuum. If the number of molecules is high enough at any postion ~r andinstant t, then the uid can be described as a continuum medium whose behavior is characterized from its density ,velocity ~v, pressure p, and temperature T , which depend on the position ~r and time t. In this case, we shall considerthat the uid is made of elements or particles, which are very small (innitesimal) portions of this medium.

This description is possible in almost all applications. For instance, there are of the order of 107 molecules in109 mm3 of air under pressure and temperature standard conditions. Only in exceptional situations, the continuumhypothesis does not verify (for example, in nanouidics of gases or around spacecrafts entering the atmosphere). Inthis case, uid mechanics is replaced with other disciplines (kinetic theory, for instance) that take into account the

uid microscopic structure.

4

Chapter 2

KINEMATICS

1 .- INTRODUCTION

The full resolution of a uid-dynamical problem involves the calculation of the uid density, velocity, pressure, andtemperature at any times and for each position. In other words, one has to obtain the scalar and vector elds (~r; t),~v(~r; t), p(~r; t), and T (~r; t). In general, the velocity eld ~v(~r; t) is a three-dimensional vector function that dependson both the spatial coordinates and time. It provides the velocity of a uid element (particle) located at a certainposition ~r at an instant t. In Cartesian coordinates, the velocity eld reads:

~v = u(x; y; z; t)~i+ v(x; y; z; t)~j + w(x; y; z; t)~k : (2.1)

In this chapter, the function ~v(~r; t) is assumed to be known, and some tools will be developed to analyze it.

2 .- LAGRAGIAN AND EULERIAN DESCRIPTIONS

In the Lagragian description, attention is paid to the evolution of uid particles. One is interested in the variationof the hydrodynamic quantities (density, presure, . . . ) of a given uid element as a function of time. In the Euleriandescription, one focuses on the temporal evolution of the hydrodynamic quantities at a xed point of the regionoccupied by the uid. In this case, one does not monitorize the evolution uid particles which crossed that point.

In order to clarify these denitions, we shall resort to a simile from our everyday experience. Suppose that trac ona highway can be compared with a uid motion. Cars are the uid particles in this simile. The Lagragian descriptionwould entail to pay attention to a certain car, observing how its velocity changes. On the contrary, the Euleriandescription involves to focus on a given segment of the highway, watching the variation of the number of cars, averagevelocity, etc. As will be seen throughout this course, the Eulerian description is the technique most frequently adoptedin uid mechanics.

3 .- STREAMLINE AND PATH

3.1.- Streamline

Streamlines are those tangential to the velocity eld at each point of the uid domain for a given instant. In otherwords, the lines that verify the dierential equations

dx1v1(~r)

=dx2v2(~r)

=dx3v3(~r)

; (2.2)

where v1, v2, and v3 are the components of the velocity vector ~v along the orthogonal axes x1, x2, and x3, respectively1.

The integration of equations (2.2) provides two relationships which dene the streamline passing through a given point.

1Hereafter, the axes x1, x2, and x3 will refer to any (Cartesian, cylindrical, or spherical) orthogonal system of coordinates. The unitvectors ~e1, ~e2, and ~e3 indicate the corresponding directions.

5

KINEMATICS

In Cartesian coordinates, equation (2.2) reads

dx

u(~r)=

dy

v(~r)=

dz

w(~r); (2.3)

where u, v, and w are the components of the velocity vector ~v along the axes x, y, and z, respectively. In cylindricalcoordinates, the equations dening the streamline are:

dr

vr= r

d

v=

dz

vz; (2.4)

where vr, v, and vz are the components of the velocity vector ~v along the axes r, , and z, respectively. Finally, theyare written in spherical coordinates as

dr

vr= r

d

v= r sen

d

v; (2.5)

where vr, v, and v are the components of the velocity vector ~v along the axes r, , and , respectively.

Two streamlines can pass across each other only at singular points where the uid velocity is zero (stagnationpoint) or innite (sink or source). Otherwise, the velocity vector would be tangential to two lines simultaneously.

A streamtube is a surface in the shape of a tube which is made up by streamlines (see gure 2.1). A commonexample is any imaginary cylinder coaxially located inside a pipe. No matter which ow is, the uid contained insidea streamtube at a given instant moves without getting out of it at that instant. Otherwise, two streamlines wouldcross at the point where the uid gets out of the streamtube.

Figure 2.1: Streamtube.

3.2.- Path

A path is the trajectory followed by a uid particle over time. To get its parameter (as a function of the parametert) equations, one has just to integrate the equations of motion, i.e.,

dx1dt

= v1(x1; x2; x3; t) ;dx2dt

= v2(x1; x2; x3; t) ;dx3dt

= v3(x1; x2; x3; t) : (2.6)

In gure 2.2, we sketch two streamlines at an instant t, and the path traced by a uid particle over time.

4 .- TYPES OF FLOWS

4.1.- Steady ow

In a steady ow, the velocity eld does not depend on time at any point, i.e., ~v = ~v(~r). It can be readily seen thatstreamlines and paths coincide in steady ows. In unsteady ows, they can coincide or not. In addition, in a steady

ow the uid contained in a streamtube moves without getting out of it over time.

6

4 .- TYPES OF FLOWS

Paths

t=t

2

t=t

3

t=t

1

Streamlines

t=t

0

Figure 2.2: Stream line and path.

It must be noted that uid particles can undergo accelerations in steady ows. In fact, if a uid particle movesaccording to an inhomogeneous velocity eld, its velocity will change (acceleration) despite the fact that the velocityvector remains constant at each point (steady ow). For instance, in the steady ow taking place in a converging nozzle(see gure 2.3), the uid particle speeds up as approaches the nozzle outlet because the magnitude of the velocity eldis larger in that region.

t=t2t=t1

x

Figure 2.3: Acceleration of a uid particle in a nozzle.

4.2.- Uniform ows and streams

In a uniform ow, the velocity eld does not depend on the spatial coordinates, i.e.,, ~v = ~v(t). If the ow is steadytoo, it is called uniform stream.

4.3.- Two- and one-dimensional ows

A two-dimensional ow is referred to as that whose velocity eld has only two non-zero Cartesian components, andthose components depend on their corresponding coordinates (and time) exclusively; for instance,

~v = u(x; y; t)~i+ v(x; y; t)~j : (2.7)

In a one-dimensional ow, these conditions are met for one of the velocity components.

4.4.- Axisymmetric ows

An axisymmetric ow is a three-dimensional ow which possesses a symmetry axis, so that one angular coordinateand the corresponding velocity component do not take part in the problem.

In cylindrical coordinates, the velocity eld of an axisymmetric ow can be written as

~v = vr(r; z; t)~ur + vz(r; z; t)~uz ; (2.8)

where vr and vz are the radial and axial velocity components, respectively (v = 0 is the angular component), and zis the symmetry axis. In spherical coordinates, the velocity eld is given by the expression

~v = vr(r; ; t)~ur + v(r; ; t)~u ; (2.9)

7

KINEMATICS

where vr and v are the radial and polar velocity components, respectively (v = 0 is the azimuthal component). Inthis case, = 0 is the symmetry axis of the problem.

The classical examples of axisymmetric ows are the uniform streams that strike solid bodies with an axis ofrevolution which coincides with that of the uniform stream. To describe this class of ows, either cylindrical orspherical systems of coordinates are used depending on the solid body shape. For instance, a cylindrical system ofcoordinates is used to describe the ow around a cylinder immersed in a uniform stream (see gure 2.4 (left)), whilethe ow around a sphere is analyzed in terms of a spherical system of coordinates (see gure 2.4 (right)2).

symmetry axis

z

u

symmetry axis

r

Figure 2.4: Axisymmetric ow around a cylinder (left) and a sphere (right).

4.5.- Incompressible ows

In an incompressible ow, the density of any uid particle remains constant over time. Any liquid motion, as well asthose gas ows driven by relatively low-pressure gradients, can be regarded as incompressible.

5 .- FLOW RATE AND MASS FLOW RATE

5.1.- Flow rate and average velocity

The ow rate Q is the volume of uid crossing a surface S per unit time. To calculate the ow rate from the velocityeld, we shall consider a dierential surface element of area dS, and the volume dV built from it as shown in gure2.5. The volume of uid that passes through dS within a period of time dt equals dV . This last quantity can becalculated as

dV = dS cos v dt ; (2.10)

where is the angle formed by the unit vector ~n normal to the surface element and the velocity vector at that point.Also, v is the magnitude of that velocity vector. Then, one can obtain the ow rate Q crossing S from the integrationof (2.10):

Q =

ZS

dV

dt=

ZS

~v ~n dS =ZS

~v d~S : (2.11)

dVdS

v dt

vn

Figure 2.5: Flow rate that crosses a surface element.

In addition, the average velocity vm at which the uid crosses the surface S is dened as:

vm =Q

S: (2.12)

2In the gures, vectors are represented in bold letters.

8

5 .- FLOW RATE AND MASS FLOW RATE

5.2.- Mass ow rate

The mass ow rate m0 is the mass of uid that passes through a surface S per unit time. The mass ow rate iscalculated analogously to the ow rate. Let us consider, for instance, the uid element in gure 2.5. The mass dm ofthis element is obtained from the expression

dm = dV = dS cos v dt ; (2.13)

where the meaning of these variables is that established in the previous section. The volume dV of this uid elementequals the volume of uid that crosses dS within the time interval dt. Therefore, the mass transported across dS perunit time is

dm

dt= ~v ~n dS : (2.14)

The mass ow rate m0 that crosses the entire surface S is calculated from the integral

m0 =ZS

~v ~n dS : (2.15)

If the uid is incompressible (liquid), the density is constant and can be brought out of the integral. In this case,the following relation also holds:

m0 = Q : (2.16)

5.3.- Flux of a scalar or vector quantity

Let B be any scalar or vector quantity used to describe the ow. Let dB(~r) be its value for a uid element locatedat the position ~r. Suppose that dB(~r) is proportional to the mass dm of the uid element. The quantity (~r) is theamount of B per unit mass, i.e., = dB=dm.

The quantity dB is transported by the uid element owing to its motion. Consider, for instance, the uid elementshown in gure 2.5. The value dB corresponding to this element can be obtained from the equation

dB = dm = dV = dS cos v dt ; (2.17)

where the meaning of these variables is that established in previous sections. The volume dV of this uid element isthe same as the volume of uid that crosses dS within the time interval dt. Therefore, the amount of B transportedacross dS per unit time is

dB

dt= (~v ~n) dS : (2.18)

The ux of the quantity B across the surface S is calculated from the integral

=

ZS

(~v ~n) dS : (2.19)

The ux represents the amount of B transported by the uid across S per unit time. The expressions for the

ow rate Q and the mass ow rate m0 can be regarded as two particular cases of equation (2.19). Thus, = Q whenB = V (uid volume) or = 1= (specic volume), and = m0 when B =M (uid mass) or = 1.

If the surface S encloses a volume V and the vector ~n is dened \outwards", then > 0 implies that the amountof B contained in V decreases over time; while < 0 implies the opposite. We will come back to this point when theReynolds transport theorem is presented.

To close this chapter, we will introduce Einstein's notation for repeated subindexes, which we will resort to lateron. Two repeated subindexes indicate a sum over those subindexes. For example, ajbj = a1b1 + a2b2 + a3b3, or

@ai@xj

@aj@xi

=@a1@x1

@a1@x1

+@a1@x2

@a2@x1

+@a1@x3

@a3@x1

+@a2@x1

@a1@x2

+@a2@x2

@a2@x2

+@a2@x3

@a3@x2

+@a3@x1

@a1@x3

+@a3@x2

@a2@x3

+@a3@x3

@a3@x3

: (2.20)

As can be seen, Einstein's notation allows one to simplify enormously equations involving vectors and/or matrixes.

9

Chapter 3

EQUATIONS FOR A FLUID SYSTEM

1 .- INTRODUCTION

A uid system is a given mass (portion) of uid. A uid system is delimited by a uid surface, which separates itfrom the surroundings. The uid surface consists of elements of uid surface that move with the uid velocity at thepoint they are.

The momentum and energy of a uid system can vary over time. This variation has two origins. On one side,the exchange of momentum and energy with the surroundings across the uid surface. On the other side, the exter-nal sources (external forces and sources of energy). In the course of this chapter, we shall calculate the variationscorresponding to the rst case.

System

Surroundings

Figure 3.1: Fluid-surroundings interaction.

2 .- SURFACE FORCES

2.1.- Surface forces versus external forces

As mentioned above, the forces acting on a uid system have two origins. First, the external forces exerted on the

uid system. Gravity and forces exerted by solid elements in contact with the system are the most frequent cases.Second, the surroundings exerts through the uid surface forces on the system. These forces are generically calledsurface forces. The surface forces are calculated from the stress tensor, which we shall focus on immediately.

2.2.- Stress tensor

Consider a uid surface element located at a point 0, and that separates two uid regions [(1) and (2)] (see gure 3.2).

This element is characterized by the vector d~S0, normal to the surface, with a magnitude equal to the element area,and pointing from (2) towards (1). It is possible to demonstrate (we will not do it here) that the component i of the

force d~F0 exerted by (1) on (2) through the surface element is given by the expression

dF0i = ij(~r0)dS0j ; (3.1)

10

2 .- SURFACE FORCES

where ij is the stress tensor, which depends only on the position ~r0 where the surface element is located. Einstein'snotation1 was introduced in equation (3.1). This equation can also be written as0@ dF01dF02

dF03

1A =0@ 11 12 1321 22 23

31 32 33

1A0@ dS01dS02dS03

1A : (3.2)

(1)

(2)

n

0

0

(2)

(1)

dF

0

dS

0

r

0

Figure 3.2: Surface forces.

2.3.- Stress tensor symmetry

As shown above, surface forces are calculated in terms of the stress tensor. In this section, we shall demonstrate thatthis tensor is symmetric. For this purpose, let us consider the uid element shown in gure 3.3. The torque dM0 withrespect to the axis 0 exerted by the surface forces is

dM0 = 2(21x2x3)x12 2(12x1x3)x2

2: (3.3)

By denition, dM0 is positive if produces a rotation as indicated in the gure, and negative otherwise. The rstaddend on the right side of equation (3.3) corresponds to the contribution to dM0 of the front and back faces of the

uid element, while the second reects the contribution of the two lateral faces. The external forces must be appliedin the uid element center, and thus they do not contribute to the torque dM0. Therefore,

Id!

dt= dM0 ; (3.4)

where I is the moment of inertia of the uid element.

The uid element volume scales as 3, being a characteristic length of the uid element (for instance, x1). Thedistance of an element portion to the axis 0 scales as , and thus the moment of inertia I scales as 4. However, thetorque dM0 scales as

3 [see equation (3.3)]. Therefore, equation (3.4) can not be satised for an innitesimal uidelement ( ! 0) unless

dM0 = 0 : (3.5)

From equations (3.3) and (3.5), one gets

12 = 21 : (3.6)

It is easy to verify that if one had chosen the axis 0 as that parallel to ~e1 or ~e2, then one would have concluded that

23 = 32 or 13 = 31 ; (3.7)

respectively. In this way, it is demonstrated that the stress tensor is symmetric.

1 Let as remember that two repeated subindexes indicate a sum over those subindexes. For instance, ajbj = a1b1 + a2b2 + a3b3

11

EQUATIONS FOR A FLUID SYSTEM

x2

x3x1

0e3

e2e1

Figure 3.3: Torque exerted by the surface forces with respect to an axis.

2.4.- Hydrostatic pressure

We have just shown that the component i of the force d~F0 exerted by a uid region on another across the surfaceelement dS0 that separates them (see gure 3.2a) is

dF0i = ij(~r0)dS0j ; (3.8)

where ij(~r0) is the stress tensor that depends only on the position ~r0 where the surface element is located.

As explained in chapter 1, the dening feature of uids is the fact that they deform continuously under the actionof shear stresses. Therefore, shear stresses must vanish in uids at rest. In other words, surface forces must beperpendicular to the surfaces in hydrostatics. In this case,

d~F0 = p(~r0) d~S0 ; (3.9)where the scalar quantity p(~r0) is called the hydrostatic pressure

2. This equation is equivalent to

ij(~r0) = p(~r0) ij ; (3.10)ij being the identity tensor.

The hydrostatics equation (3.9) is referred to as Pascal's principle. Pascal's principle establishes that forces inhydrostatics are perpendicular to the surface elements, and their magnitudes are independent of the orientation ofthose surface elements. We will show in chapter 6 how to calculate the hydrostatic pressure eld p(~r0).

2.5.- Viscous stress tensor

Equation (3.10) provides the value of the stress tensor for a uid at rest. It is natural to assume that its value in a

ow is given by (3.10) plus a contribution ij associated with the uid motion:

ij(~r) = p(~r)ij + ij(~r) : (3.11)This contribution is due to viscous friction caused by the displacement of some uid layers with respect to others. Forthis reason, ij is called the viscous stress tensor. For uids at rest or inviscid uids, ij = 0, and ij reduces to thestress tensor (3.10).

2.6.- Navier-Poisson law

In this section, we shall introduce an empirical law which allows one to calculate the viscous stress tensor ij fromthe velocity eld. To simplify the analysis, we will focus on the particular (and most common) case of incompressible

ow. This condition holds for liquids always, and for gases as long as pressure (density) variations are small enough.

2Taking into account how the vector d~S0 is dened (see gure 3.2a), it is obvious that the negative sign in equation (3.9) ensures thatp always takes a positive value.

12

2 .- SURFACE FORCES

As mentioned in the previous section, viscous forces are caused by the displacement of some uid layers withrespect to others. The strain rate tensor

ij =1

2

@vi@xj

+@vj@xi

(3.12)

is a measure of the displacement magnitude. In this regard, it is natural to hypothesize that ij is proportional to ij .Symmetry considerations (not included here) allow one to assert that, for incompressible ows,

ij = 2 ij ; (3.13)

where is the uid viscosity . Its value depends on the uid as well as its pressure and temperature at the consideredpoint. Equation (3.13) is referred to as the Navier-Poisson law. In a Cartesian system of coordinates, the elements ofthe viscous stress tensor (3.13) are:

xx = 2@u

@x; xy =

@u

@y+@v

@x

; xz = zx ;

yx = xy ; yy = 2@v

@y; yz =

@v

@z+@w

@y

;

zx =

@u

@z+@w

@x

; zy = yz ; zz = 2

@w

@z:

(3.14)

In a cylindrical system of coordinates, they are:

rr = 2@vr@r

; r =

1

r

@vr@

+@v@r

vr

; rz = zr ;

r = r ; = 2

1

r

@v@

+vrr

; z =

@v@z

+1

r

@vz@

;

zr =

@vz@r

+@vr@z

; z = z ; zz = 2

@vz@z

:

(3.15)

Liquids commonly used in industrial applications (water, oils, . . . ) obey the Navier-Poisson law. However, theresome substances for which the linear relationship between the viscous stress and strain rate tensors is no longer valid.Among those substances, one can mention plastic, dilatational, or pseudoplastic liquids. Rheology is the disciplinedevoted to the analysis of such liquids. Newtonian uids are those which verify the Navier-Poisson (Newton) law,while non-Newtonian liquids are those whose behavior can not be described by that law. In this course, we shallconsider Newtonian uids exclusively.

2.7.- Viscosity

Viscosity can play an essential role in the dynamics of uids. Its inuence can be measured through the ratio =,where is the uid density. This ratio indicates the relation between the viscous force (proportional to the viscosity) and inertia or the uid resistance to acceleration (proportional to the density ). Therefore, is an indicator of theeects of the viscous force on the uid dynamics. Because its units are kinematics (m2/s), it is called the kinematicviscosity. In order to better distinguish the two viscosities, the quantity is often referred to as dynamic viscosity.

As mentioned above, both the dynamic and kinematic viscosities depend on the uid considered. There is a largerange of viscosity values. For instance, for standard pressure and temperature conditions,

glicerine 1400water 80000air and glicerine 80air 1400water : (3.16)The dynamic viscosity could also depend on the uid pressure and temperature. However, variations of pressure havenegligible inuence, while temperature changes may produce considerable eects. The dependence of the viscosityupon temperature is very dierent in gases and liquids. In fact, viscosity increases with temperature in gases, whilethe opposite occurs in liquids.

13

EQUATIONS FOR A FLUID SYSTEM

2.8.- Surface forces exerted on a uid system

We have introduced the stress tensor to calculate the force exerted by a uid region on another through the uidsurface element that separates them from each other. Consider a uid system S enclosed by the system (uid) surface

SS. The component i of the resultant surface force ~Fs exerted by the surroundings on the uid system S through SS is

Fsi =

ZSS

ijdSj =

ZSS

ijnjdS ; (3.17)

where (as always) the vectors d~S and ~n are dened \outwards". If the stress tensor is split into the pressure andviscous contributions, one gets

Fsi = Fpi + Fvi ; (3.18)

where

Fpi = ZSS

p ij nj dS = ZSS

p ni dS (3.19)

and

Fvi =

ZSS

ij nj dS (3.20)

are the components i of the pressure and viscosity forces exerted on the system, respectively.

3 .- HEAT CONDUCTION

3.1.- Heat conduction

The energy transfer due to heat can take place by means of three mechanisms: convection, radiation, and conduction.Convection is the transfer of energy between two regions due to the uid motion from one to the other. It is not anexchange of energy between the uid system and its surroundings, but the spatial transport of energy associated withthe uid displacement. Therefore, it must not be considered in this chapter. In radiation, energy coming from externalsources is transmitted by electromagnetic waves (photons), crossing the surroundings and reaching the uid system,where it is partially absorbed. Therefore, it must not be considered here either. Heat conduction is the transfer ofenergy between the two sides of a uid surface (the uid system and its surroundings) because they are at dierenttemperatures. This phenomenon does correspond to the scope of this chapter, and will be analyzed in the next section.

3.2.- Heat ux vector

Consider a uid surface element located at a point 0, and that separates two uid regions [(1) and (2)] (see gure

3.4). This element is characterized by the vector d~S0. This vector is normal to the surface, has a magnitude equal tothe element area, and points from (2) towards (1). It is possible to demonstrate (we will not do it here) that the heattransfer of energy per unit time dq00 between (1) and (2) through the surface element is given by the expression

dq00 = ~q(~r0) d~S0 ; (3.21)where ~q is heat ux vector, which depends only on the position ~r0 where the surface element is located. By denition,the quantity dq00 is positive when the region (1) transfers energy to the region (2), and negative otherwise. Takinginto account the direction of d~S0 (see gure), it is obvious that ~q indicates the direction in which energy ows.

3.3.- Fourier law

Heat conduction is caused by the temperature gradient in the uid, so that energy ows from the hotter to the colderregion. It is natural to assume that ~q is a vector proportional to and with the opposite direction of the temperaturegradient ~rT ; i.e.,

~q = ~rT ; (3.22)

14

4 .- MECHANICAL EQUATIONS FOR A FLUID SYSTEM

dq'

(1)

(2)

n

0

0

(2)

(1)

dS

0

r

0

Figure 3.4: Heat ux.

where is the thermal conductivity, which depends on the uid considered as well as its pressure and temperature.Equation (3.22) is the Fourier law.

3.4.- Thermal conductivity

As happens with the viscosity value, pressure variations have a slight inuence on the thermal conductivity, whiletemperature changes may produce a signicant conductivity variation. The dependence of the conductivity withrespect to the temperature value is similar to that of the viscosity: it increases with temperature in gases, whilst theopposite occurs in liquids.

3.5.- Heat ux in a uid system

We have introduced the heat ux vector to calculate the heat exchange between two uid regions across the surfaceelement that separates them. Consider a uid system S delimited by a system (uid) surface SS. The energy per unittime exchanged by S with the surroundings across SS due to heat conduction is

Q0=

ZSS

~q d~S = ZSS

qjnjdS ; (3.23)

where the vectors d~S are ~n point \outwards" and, therefore, Q0> 0 if the system gains energy, and Q

0< 0 otherwise.

4 .- MECHANICAL EQUATIONS FOR A FLUID SYSTEM

The momentum and energy of a uid system change over time owing to the action of both external factors and thesurroundings. In this chapter, we have analyzed the role played by the surroundings. We present in this section theequations for the evolution of the mechanical quantities characterizing a uid system from the results obtained above.

4.1.- Mass

The mass m of a uid system can be calculated as

m =

ZVS

(~r) dV ; (3.24)

where VS is the volume occupied by the uid system at a given instant. As mentioned previously, this quantityremains constant over time. Therefore, the conservation equation

d

dt

ZVS

(~r) dV = 0 (3.25)

15

EQUATIONS FOR A FLUID SYSTEM

always holds.

4.2.- Momentum

The momentum ~P of a uid system,

~P =

ZVS

(~r) ~v(~r) dV ; (3.26)

can vary over time because of the action of external ~Fext and surface ~Fs forces. Taking into account the expressions(3.18){(3.20), the momentum equation for a uid system reads

d

dt

ZVS

(~r) ~v(~r) dV = ~Fext ZSS

p(~r) ~n dS +

ZSS

(~r) ~n dS ; (3.27)

where SS is the uid surface that encloses the system. This equation corresponds to Newton's second law for a uidsystem.

4.3.- Energy

In uid mechanics, the total energy per unit mass e is dened as the sum of the thermal energy and the mechanical one;specically, the sum of the internal energy associated with temperature, the kinetic energy due to the uid motion,and the potential energy owing to the gravitational eld:

e = u^+ 12v2 + gz ; (3.28)

where g is the gravitational acceleration, and z the upward axis.

The total energy E accumulated in a uid system,

E =

ZVS

(~r) e(~r) dV ; (3.29)

evolves because of the following two factors: (a) the work done by external forces, (b) heat transferred by externalsources, (c) work done by surface forces, and (d) heat conduction. The rst two factors correspond to external sources,while the last two ones origin from the interaction with the surroundings. The symbols W 0ext and Q

0ext stand for the

energy transfer per unit time due to the rst and second factors, respectively.

The uid layer in contact with an impermeable solid wall moves with the same velocity as that of the wall. Thisis the so-called non-slip boundary condition, and it is the result of the mechanical equilibrium established almostinstantaneously between the layer and the wall. Any mechanical system exchanges energy due to work only if moveswhile exerting or withstanding a force. Therefore, the uid system can only exchange energy due to work with movingsolid elements (blades, propellers, pistons, . . . ) in contact with it. Solid walls at rest (pipes, valves, . . . ) exert andwithstand forces but can not exchange energy with the uid system due to work. On the other side, the term W 0extmust not include the action of gravity. The potential energy associated with the gravitational eld is included in thedenition of total energy e. Therefore, gravity does not alter the total energy value, it only transforms a kind of energyinto another.

The internal energy quanties both the kinetic energy associated with the thermal motion of the uid molecules,and the potential energy accumulated in the intermolecular interactions. It must not be included here the chemicalenergy associated with the intramolecular bonds. In an exothermic chemical reaction (combustion), part of the energycontained in the intramolecular bonds is released and absorbed by a uid system as internal or kinetic energy. Inthis regard, combustion modies the total uid energy e, and must be considered as an external source (Q0ext) in theenergy equation, although it is physically originated in the uid itself.

Taking into the above comments, the evolution of the uid system energy is given by the equation

d

dt

ZVS

(~r) e(~r) dV = Q0ext ZSS

~q(~r) d~S ZSS

p(~r) ~v(~r) ~n dS

W 0ext +ZSS

~v(~r) [(~r) ~n] dS : (3.30)

16

4 .- MECHANICAL EQUATIONS FOR A FLUID SYSTEM

Equation (3.30) can be regarded as the generalization of the rst law of thermodynamics for a moving uid system. Itincludes both the mechanical terms and viscous energy dissipation. The sign convention used in this equation coincideswith that of the rst law of thermodynamics for the corresponding terms: the heat exchange is positive if the systemgains energy, and negative otherwise; the work done by the surroundings is positive when the uid system compresses(~v ~n < 0), and negative otherwise. Because of the minus sign in front of W 0ext, W 0ext < 0 if energy is transmitted tothe system and W 0ext > 0 otherwise.

17

Chapter 4

EQUATIONS FOR A CONTROL VOLUME

1 .- SYSTEMS AND CONTROL VOLUMES

In the previous chapter, we obtained the laws of mechanics for a uid system. In most cases, one is not interestedin the evolution of uid systems, but in the local eects that those systems cause when crossing a certain region (forinstance, a machine).

To understand this twofold perspective, we must distinguish uid systems from control volumes. Let us remainthat a uid system is a certain part (or portion) of the uid under consideration. This part is separated from thesurroundings by a uid surface. A uid system can move, rotate, and deform over time. Each element of the uidsurface moves with the same velocity as that of the uid at the point where the element is.

A control volume (CV) is a xed spatial region which a uid system crosses, while the control surface (CS) is theone delimiting that volume (see gure 4.1). CVs are used to enclose hydraulic machines, mobile solid bodies, hydraulicdevices, etc. Some considerations to appropriately choose CVs are:

Use must be made of a frame of reference solidly moving with the CV, so that this volume remains at rest withrespect to that frame of reference.

Fluid systems must completely ll CVs. Therefore, the CV and CS coincide with the volume and surface,respectively, of the uid system contained in that CV.

In some applications, CVs contain uid systems made up by two phases (for instance, water and air) (see gures4.4 or 4.6).

System 1

System 2

Control

volume

Figure 4.1: Systems and CVs.

Equations (3.25), (3.27), and (3.30) describe the evolution of mechanical quantities referred to a uid system. Aswill be shown in the next section, the Reynolds transport theorem allows one to get the equations for CVs from theircounterparts for the uid systems contained in those volumes. Here, this theorem will be formulated in mathematicalterms, and applied to uid mechanics. Its demonstration is beyond the scope of this introductory subject.

2 .- REYNOLDS TRANSPORT THEOREM

Let V (t) and V 0(t) be two arbitrary volumes, and S(t) and S0(t) the corresponding limiting surfaces (see gure 4.2).Assume that these two volumes merge at a given instant t = t0, i.e., V (t0) = V

0(t0) and S(t0) = S0(t0). Let (~r; t) be

18

3 .- UNIFORM APPROXIMATION FOR THE FLUX TERM

a scalar or vector function dened over the volume V (t) (or V 0(t)). The Reynolds transport theorem establishes that,for t = t0,

d

dt

ZV 0(t)

(~r; t) dV =d

dt

ZV (t)

(~r; t) dV +

ZS(t0)

(~r; t0) [~vr(~r; t0) ~n] dS ; (4.1)

where ~vr(~r; t0) is the relative velocity of the S0(t0) element with respect to that of the S(t0) element, both located at

the point ~r. In addition, ~n is the \outwards" unit vector normal to the surface S(t0) (or S0(t0)).

V'

v

r

S'

S

t=t

0

n

r

0

V

Figure 4.2: Reynolds transport theorem.

Equation (4.1) has a clear interpretation as applied to uid mechanics. Let B be a scalar or vector mechanical

quantity whose evolution in a CV we want to calculate (for instance, momentum ~P ). Let (~r; t) be the amount of Bper unit mass at the point ~r and instant t (for example, ~v(~r; t)). Suppose that V 0 corresponds to the system volume(SV) while V is the CV which contains it at the instant t = t0. In this case, S

0 and S are the system surface (SS)and the CS, respectively. Because CV is a xed volume, the relative velocity ~vr is simply the uid velocity ~v at thepoint under consideration. In addition, one can identify the variable (~r; t) with the product (~r; t) (~r; t). Takinginto account the above considerations, the Reynolds transport theorem for the quantity B leads to

d

dt

ZSV

dV =d

dt

ZCV

dV +

ZCS

(~v ~n) dS : (4.2)

The left side of this equation represents the variation per unit time of B referred to the uid system. The rst addendof the right side of the equation corresponds to the variation per unit time of the amount of B contained in theCV, while the second is the ux (equation (2.19)) of this quantity through CS. Therefore, the Reynolds transporttheorem establishes that the variation of B in the CV equals the sum of the variation associated with the contained

uid system plus the dierence between the amount of B entering and getting out of the CV (the ux).

The equations for the conservation of mass, momentum, and energy in a CV will be obtained in this chapter.To this end, we shall consider equation (4.2) as applied to the appropiate values of . Besides, we shall account forequations (3.25), (3.27), and (3.30) for a uid system.

3 .- UNIFORM APPROXIMATION FOR THE FLUX TERM

In many applications, the uid under consideration crosses the CS only through at inlet and outlet sections. In thiscase, only those sections contribute to the ux . Very often, the ow can be regarded as uniform over those sections(see gure 4.3). In this case, the ux can be calculated as:Z

CS

(~v ~n) dS =Xk

k k (~vk ~nk) Ak =Xk

k m0k ; (4.3)

where k, k, and ~vk stand for the values of the corresponding quantities for the section k. Also, ~nk and Ak are theunit vector and the area of section k, respectively. In the above equation, m0k is the magnitude of the mass ow rategoing through the surface k. The positive sign applies if the uid gets out of the CV across that surface, and thenegative one otherwise.

If the CS is chosen so that the inlet and outlet sections are perpendicular to the ow (see gure 4.3), then equation(4.3) can be written as Z

CS

(~v ~n) dS =Xk

k k vk Ak ; (4.4)

19

EQUATIONS FOR A CONTROL VOLUME

CS

1

1

v

1

A

1

3

3

v

3

A

3

2

2

v

2

A

2

Figure 4.3: Uniform approximation for the ux term.

where the positive and negative signs apply to the outlet and inlet sections, respectively.

Figure 4.3 represents, for instance, the ow inside a hydraulic machine. The CV is chosen so that CS coincideswith the machine walls. As can be seen, the uid crosses CS through three at sections, where the ow is uniformand perpendicular to CS. In this case, the ux can be calculated asZ

CS

(~v ~n) dS = 1 1 v1 A1 + 2 2 v2 A2 + 3 3 v3 A3 : (4.5)

4 .- CONTINUITY EQUATION FOR A CONTROL VOLUME

In this section, we will obtain the equations which describe the temporal evolution of the uid mass contained in aCV. Consider equation (4.2) with = 1. Accounting for equation (3.25) for the mass of a uid system, one gets theexpression

d

dt

ZCV

dV = ZCS

(~v ~n) dS ; (4.6)

which is referred to as the continuity equation for a CV.

If the mass contained in a CV remains constant over time (steady regime), thenZCS

(~v ~n) dS = 0 : (4.7)

By using the approximation of uniform ow at the inlet and outlet sections [equation (4.3)], the above expression canbe written as X

k

k (~vk ~nk) Ak =Xk

m0k = 0 ; (4.8)

where m0k is the (magnitude of the) mass ow rate that crosses section k. The positive sign applies if the uid getsout of CV through section k, and the negative sign otherwise. If the ow is incompressible and there are only two(the inlet and outlet) sections, then

Qinl = Qout ; (4.9)

where Qinl and Qout are the (magnitude of the) inlet and outlet ow rates, respectively.

5 .- MOMENTUM EQUATION FOR A CONTROL VOLUME

In this section, we will obtain the equations which describe the temporal evolution of the momentum contained in aCV. Consider equation (4.2) with = ~v. Assuming the second Newton's law (3.27) for a uid system, one gets thevector equation

~Fext ZCS

p ~n dS +

ZCS

~n dS = ddt

ZCV

~v dV +

ZCS

~v (~v ~n) dS ; (4.10)

20

5 .- MOMENTUM EQUATION FOR A CONTROL VOLUME

where the fact that SS and CS coincide has been taken into account. The term associated with viscous stresses canbe neglected in most practical applications because those stresses are much smaller than the hydrostatic pressure overthe inlet and outlet sections of CS. It must be noted that viscous friction between solid walls and the uid system donot correspond to that term because those walls are not inlet or outlet sections. On the contrary, that viscous frictionmust be included in the term ~Fext.

If the momentum contained in CV remains constant over time (steady regime), then

~Fext + ~Fp =

ZCS

~v (~v ~n) dS ; (4.11)

where ~Fp denotes the force exerted by the surroundings on the uid system due to pressure.

By using the approximation of uniform ow at the inlet and outlet sections [equation (4.3)], the above expressioncan be written as

~Fext + ~Fp =Xk

m0k~vk ; (4.12)

where the positive sign applies if the uid gets out of CV through section k, and the negative sign otherwise. In thisequation, m0k denotes the (magnitude of the) mass ow rate crossing section k. If the ow is incompressible,

~Fext + ~Fp = Xk

Qk ~vk ; (4.13)

Qk being the (magnitude of the) ow rate crossing section k.

In spite of the important approximations leading to equation (4.13), this expression is used in many practicalapplications. If there are only two CS sections crossed by the uid (the inlet and outlet), then

~Fext + ~Fp = Q(~vout ~vinl) ; (4.14)where Q is the (magnitude of the) ow rate crossing CV, and ~vout and ~vinl are the uid velocities at the outlet andinlet sections, respectively.

External forces are typically gravity and those exerted by solid elements in contact with the uid system containedin CV. The latter are the unknowns in most problems. In addition, if the pressure is constant over CS, then ~Fp = ~0.The demonstration of this assertion is easy:

~Fp = ZCS

p ~n dS = pZCS

~n dS = ~0 : (4.15)

In the problem sketched in Fig. 4.4, a water jet strikes the screen located in a moving cart. One wants to calculatethe force F driving the cart motion. The CV used to analyze this problem is that represented in the gure, and the

uid system is made up by both the water and air contained in that volume. In this way, CV and the system volumecoincide. The frame of reference must move solidly with the cart, so that CV remains still for that frame of reference.The equation for the horizontal component of momentum allows one to calculate the force exerted by the screen on the

uid system (water and air) in the CV. That force, with the opposite sign, is the reaction exerted by the uid system(water and air) on the screen, which drives the cart motion. In this example, both the water jet and the surrounding

air are at the (constant) atmospheric pressure, and thus ~Fp = ~0.

F

p=p

at

CS

Figure 4.4: Force exerted by a water jet on a screen.

21

EQUATIONS FOR A CONTROL VOLUME

6 .- BERNOULLI EQUATION

Consider a streamtube element of innitesimal length ds in a steady ow. As shown in chapter 2, the uid moveswithin the streamtube without getting out of it (see gure 4.5). Suppose that the ow is parallel to the direction sperpendicular to the inlet and outlet sections, and uniform over those sections. We will use the symbol A to denotethe inlet section area, and , v, p, and m0 to refer to the density, velocity, pressure, and mass ow rate at that section,respectively. These quantities take the values A + dA, + d, v + dv, p + dp, and m0 + dm0 at the outlet section,respectively. The quantities dA, d, dv, dp, and dm0 are innitesimals because the distance between the inlet andoutlet section is innitesimal as well.

p+dp/2

A+dA +d

p+dp v+dv m'+dm'

z

A

p v m'

ds

s

Figure 4.5: Innitesimal streamtube element.

To analyze this ow, one chooses a CV which matches the streamtube element, and neglects second-order innites-imals. In this case, the continuity equation is

dm0 = 0 ; (4.16)

and the projection along the axis s of the momentum equation becomes

dFext + dFp + dFv = m0dv + vdm0 : (4.17)

In this equation, dFext and dFp are the components along the s axis of the external and pressure forces, respectively.Also, dFv is the s-component of the force associated with the viscous stresses applied onto CS.

In absence of solid elements, dFext is the result of the gravitational eld:

dFext = Adsgs = Adzgz = Adzg : (4.18)

To calculate the pressure force, one has to consider its action on the inlet and outlet sections, and on the lateralsurface. The pressure on the lateral surface can be calculated as the average of those of the inlet and outlet sections.

The projection along the axis s of the pressure force on the lateral surface is (p + 12dp)dA. Therefore, the resultant

pressure force is

dFp = pA+ (p+12dp)dA (p+ dp)(A+ dA) = dpA : (4.19)

Taking into account (4.18){(4.19), the momentum equation (4.17) yields

dp

+ v dv + g dz dFv

dmds = 0 ; (4.20)

where dm = A ds is the mass of the streamtube element.

Any streamtube is made up by an innite number of elements as that sketched in gure 4.5. One can relate theproperties of the streamtube beginning (section 1) to those of the end (section 2) by integrating the two sides ofequation (4.20) along the axis s. The result isZ 2

1

dp

+ 12 (v

22 v21) + g(z2 z1) + wv = 0 ; (4.21)

22

7 .- ENERGY EQUATION FOR A CONTROL VOLUME

where wv is the work per unit mass done by the viscous force dFv along the tube, i.e.,

wv = Z 21

dFvdm

ds : (4.22)

Because the viscosity force opposes the uid motion, dFv < 0 and wv > 0 always. The energy wv dissipated by viscousstresses increases with the distance between 1 and 2.

In deducing equation (4.21), use has been made of the uniform ow approximation at the spanwise sections. Thisapproximation always holds for a streamtube innitely thin (streamline). Equation (4.21) is usually referred to as thegeneralized Bernoulli equation, and is valid for any streamline of a steady ow. However, its usefulness is very limiteddue to the diculties associated with the calculation of the viscous term (4.22).

In a steady ow, the generalized Bernoulli equation for a streamline (path) describes the evolution of the mechanicalenergy associated with a uid particle over its motion. This mechanical energy is the sum of the pressure, potential,and kinetic energies. The viscous dissipation term wv represents the loss of mechanical energy between 1 and 2 dueto the work done by the friction force exerted by the surroundings on the uid particle.

6.1.- Bernoulli and hydrostatics equations

If the ow is incompressible, equation (4.21) reduces to

p1g

+v212g

+ z1 =p2g

+v222g

+ z2 + hp ; (4.23)

where all the addends were divided by g for them to have length dimension. The sum p=g + v2=2g + z is called thehead. The term hp = wv=g represents the head loss between 1 and 2 due to viscous friction.

If the uid is not very viscous, hp ' 0, equation (4.23) becomesp1g

+v212g

+ z1 =p2g

+v222g

+ z2 ; (4.24)

known as the Bernoulli equation.

In hydrostatics, the kinetic energy terms can be neglected, which leads to the fundamental equation of hydrostatics

p1 + gz1 = p2 + gz2 ; (4.25)

applicable to whichever the points be.

7 .- ENERGY EQUATION FOR A CONTROL VOLUME

Consider equation (4.2) with = e. Taking into account the energy conservation equation (3.30) for a uid system,one gets the expression

Q0ext ZCS

~q ~n dS ZCS

p ~v ~n dS W 0ext +ZCS

~v ( ~n) dS = ddt

ZCV

e dV +

ZCS

e (~v ~n) dS ; (4.26)

where we have accounted for that CV and SV coincide.

If the energy contained in CV is constant, then

Q0 ZCS

p ~v ~n dS W 0sol W 0v =ZCS

e (~v ~n) dS : (4.27)

In this equation, Q0 stands for the total energy exchanged due to heat (conduction plus external sources) per unittime; W 0sol is the power transferred through the moving solid elements in contact with the uid system; and W

0v is the

23

EQUATIONS FOR A CONTROL VOLUME

power associated with viscous stresses exerted on CS. One must remember that Q > 0 means an increase of energyin CV, while W > 0 means the opposite. The work done by gravity must not be included in W 0ext because it does notalter the total energy value (it only transforms potential energy into other types of energy or vice versa).

It is important to point out that only moving solid elements contribute to W 0sol. The uid layer in contact witha solid surface remains attached to that surface (non-slip boundary condition), and thus the solid and uid velocitiescoincide over that surface. Therefore, if the solid wall remains at rest, the uid layer will be still as well, and exchangeof energy due to work will not take place. The viscous term W 0v can be neglected in many applications. The reasonlies in the fact that viscous stresses exerted on CS are much smaller than pressure.

To simplify equation (4.27), the term due to the pressure force is moved to the right-side of that equation. Takinginto account that h = u^+ p=, equation (4.27) becomes

Q0 W 0sol =ZCS

h+ 12v

2 + gz(~v ~n) dS : (4.28)

This equation can be interpreted in the following way. The energy exchanged with the external sources (plus conduc-tion) is the dierence between the hydraulic energy transported by the uid at the inlet and outlet sections of the CV.This hydraulic energy is the sum of the total energy e and that associated with the hydrostatic pressure p=. In thisway, one eliminates from the analysis the work done by the surroundings.

By considering the uniform ow approximation at the inlet and outlet sections [equation (4.3)], and neglecting thedierence between the heights of those two sections, the above equation can be written as

Q0 W 0sol =Xk

m0khk +

12v

2k + gzk

; (4.29)

where the positive sign applies if the uid gets out of CV through section k, and the negative sign otherwise. In thisequation, m0k, hk, vk, and zk are the mass ow rate, enthalpy, velocity, and height of section k, respectively.

If the ow is incompressible,

Q0 W 0sol = Xk

Qkhk +

12v

2k + gzk

; (4.30)

Qk being the (magnitude of) ow rate crossing section k.

8 .- ENERGY EQUATION FOR A FLUID MACHINE

In this section, we will examine the energy balance taking place in a machine which processes a uid. One candistinguish two working regimes. At the initial stage, there is always a transient regime through which the machineadapts to the working conditions. This rst stage is usually very short, and gives rise to a steady regime in which theamount of mass, momentum, and energy contained in the machine remains constant on time. In this last stage, theterm d=dt

RCV

in the CV equations vanishes. In this section, this condition will be assumed.

Let us give an example to illustrate the above comment. When a pump is turned on, the propeller starts rotating,and the water inside the machine begings moving. The momentum and energy contained inside the pump increasesharply. After a short transient regime, the propeller rotation speed reaches a constant value, and thus the momentumand energy of the water inside the pump no longer vary.

A CV enclosing the machine is used to study the energy balance in that device. Suppose that uid gets in andgets out of CV across two surfaces normal to the ow (see gure 4.6), i.e., the sections of the inlet and outlet pipes.The ow is assumed to be uniform over these sections. The surroundings exerts viscous stresses on the uid systemonly through the inlet and outlet sections. As mentioned previously, and apart from exceptional situations, this eectis negligible, and thus W 0v can be neglected in the energy equation.

Under the above conditions, the continuity equation establishes that the mass ow rate m0 at the inlet and outletsections coincide. The energy equation reads

Q0 W 0sol = m0ho hi + 12v2o 12v2i + gzo gzi

; (4.31)

24

9 .- ENERGY EQUATION FOR A HYDRAULICMACHINE

machine

CV

Figure 4.6: Sketch of a uid machine.

where the subindexes i and o refer to the inlet and outlet values, respectively.

If all the terms of equation (4.31) are divided by m0, one gets the expression

hi +12v

2i + gzi = ho +

12v

2o + gzo + wsol q ; (4.32)

where wsol = W0sol=m

0 is the work done by the moving solid elements, and q = Q0=m0 is the total heat exchanged bythe uid system, both per unit mass.

The sum h+ 12v2+gz is the stagnation (total) enthalpy h0 per unit mass, and thus equation (4.32) can be written as

h0i = h0o + wsol q : (4.33)

It must be pointed out that equations (4.32) and (4.33) are valid only if the machine is working in the steadymode, the ow is uniform over the inlet and outlet sections, and the power associated with viscous stresses on thosesections is negligible. Despite these restrictions, equations (4.32) and (4.33) are valid in the majority of applications,regardless the shape, size, or working principle of the considered machine. This degree of generality confers a greatrelevance on this equation.

8.1.- Application to the adiabatic case

Most of the ows taking place inside uid machines are so fast that heat exchange mechanisms can be neglected(adiabatic ow, q = 0), and thus equation (4.33) reads

h0 = wsol ; (4.34)where h0 is the variation of stagnation enthalpy taking place between the inlet and outlet sections. In pumps andfans, wsol < 0 and h0 > 0, while the opposite occurs in turbines. The usefulness of the above equation is enormousin industrial engineering.

9 .- ENERGY EQUATION FOR A HYDRAULIC MACHINE

The energy equation (4.32) can be re-written as

pii

+ 12v2i + gzi =

poo

+ 12v2o + gzo + wsol + u^o u^i q ; (4.35)

where use has been made of the relationship between internal energy and enthalpy h = u^ + p=. By denition, the

ow in a hydraulic machine is incompressible (i = o = ), and equation (4.35) reduces to

pig

+v2i2g

+ zi +HB =pog

+v2o2g

+ zo ; (4.36)

where

HB = 1g(wsol + u^o u^i q) : (4.37)

25

EQUATIONS FOR A CONTROL VOLUME

It must be mentioned again that wsol and q are the energies per unit mass that the system exchanges with externalfactors in the shape of work and heat, respectively. The quantity HB is the head added to or subtracted from the uidby the hydraulic machine. In pumps and fans, wsol < 0 and HB > 0, while the opposite occurs in turbines.

Suppose now that the ow inside a hydraulic machine is almost inviscid. In this case, energy dissipation can beneglected, and thus there is no internal heat production mechanism. Therefore, all the energy absorbed (or released)by the uid system in the shape of heat is included in the quantity q (energy exchange with outside). The rst law ofthermodynamics establishes for an incompressible ow that the increase (or decrease) of the internal energy of a uidparticle is the same as the heat absorbed (or released) by that uid particle. Therefore, u^o u^i q = 0 in the inviscidand incompressible case, and hence

HB = wsolg

= W0sol

m0g= W

0sol

gQ: (4.38)

In other words, the energy wsol transmitted (or absorbed) by a machine in almost inviscid and incompressible owstranslates into an increase (or decrease) of head. On the contrary, if viscous eects were not negligible, there wouldbe an energy dissipation, u^s u^e q, which would cause a decrease of the head supplied to the uid. This decreaseis quantied by the hydraulic performance h = 1 (u^s u^e q)=wsol.

26

Chapter 5

DIMENSIONAL ANALYSIS

1 .- INTRODUCTION

Due to the complexity of most uid-dynamic problems of technological interest, one has to rule out in many occasionsany theoretical approach. This is so not only in applications whose complexity is apparent, but also in those lookingsimple. Therefore, experimentation constitutes most of the time the only route to get useful information. Experi-mentation is generally very expensive, and thus it must be conducted in a rationale manner, obtaining the maximuminformation possible from the minimum number of experiments.

In a broad sense, dimensional analysis is the set of techniques, rules, and procedures used to get information ona certain phenomenon just from the knowledge of the quantities (variables) involved in that phenomenon. The stepstypically followed in a dimensional analysis can be arranged as:

1. Following physical arguments, one proposes the dimensional quantities involved in the problem. Contrarily towhat one may think, this rst step may become really dicult, because requires having a certain background inthe discipline the problem belongs to.

2. Following a certain procedure, the relevant dimensionless numbers are constructed.

3. Taking into account the relevant dimensionless numbers, one draws up an experiment with a prototype or amodel.

In this chapter, we shall explain in detail these ideas.

1.1.- Some denitions

Before starting, we must make sure that some denitions are correctly understood.

Fundamental dimensions: they are the basic dimensions from which the quantities describing a phenomenon aredened. The number of fundamental dimensions is two (length [L] and time [T ]) if the problem is kinematic,three (length, time, and mass [M ]) if it is dynamical, and four (length, time, mass, and temperature [T ]) if itis thermo-dynamical. Systems of units (for instance, the International Unit System) are used to measure thefundamental dimensions and their combinations.

Dimensional variables: they are variables measured in terms of fundamental dimensions (density, pressure,viscosity, . . . ). Therefore, constant or dimensionless parameters such as angles, for instance, are not includedin this category. Naturally, the value taken by a dimensional variable depends on the system of units used toquantify the fundamental dimensions; for example, water = 10

3 kg/m3 and 1 gr/cm3.

Dimensionless group: they are dimensionless monomials formed by dimensional variables. A monomial is theproduct of several variables raised to certain powers. The Reynolds number Re= vL= is a very importantexample. Because dimensionless groups have no dimension, their values do not depend on the system of unitsused to to quantify the fundamental dimensions.

27

DIMENSIONAL ANALYSIS

1.2.- Dimensional homogeneity principle

The dimensional homogeneity principle establishes that any equation relating dimensional variables must be dimen-sionally homogeneous; i.e., all the addends involved in such relation must have the same fundamental dimensions. Ifan equation is dimensionally homogenous, then it is invariant under a change of the system of units, which constitutesa necessary condition for that equation to have physical meaning. By way of illustration, let us consider the Bernoulliequation (4.25):

p1g

+v212g

+ z1 =p2g

+v222g

+ z2 : (5.1)

It can be readily veried that all the addends of this equation have dimensions of length.

Based on the dimensional homogeneity principle, one can perform an elemental dimensional analysis of simpleproblems. Consider the problem aiming at measuring the pendulum period for small-amplitude oscillations. Firstly,assume that the dimensional variables intervening in this phenomenon are the period , mass m, and length ` of thependulum, as well as the gravitational acceleration g. We have allowed ourselves to make an error here, because does not depend on m. Secondly, one can expect that these variables are related through an equation of the form

= C ma `b gc ; (5.2)

where the factor C and the exponents a, b, and c are unknown dimensionless constants. The dimensional homogeneityprinciple forces the two sides of the equation to have the same dimensions. It can be easily veried that this conditionis equivalent to the following system of equations for the exponents:

a = 0

b+ c = 0

2c = 1 ; (5.3)whose solution is a = 0, b = 1=2, and c = 1=2. The rst, second, and third equalities must hold for the dimensionsof mass, length, and time, respectively, to be the same on both sides of equation (5.2). Taking into account (5.2) andthe solution of (5.3), one concludes that obeys the expression

= C

s`

g: (5.4)

As can be seen, this elemental analysis, simply based on the dimensional homogeneity principle, shows that the massm must be ruled out of the pendulum formula, thus correcting our intentional error1. Finally, it is necessary to performexperiments to conrm hypothesis (5.2), and to determine the value of C. In gure 5.1, the values of the dimensionless

group =(`=g)12 are plotted for several experiments. The experimental results conrm hypothesis (5.2) and allows one

to calculate the C value.

4

6

8

/(l/g)

1/2

tests ( ,m,l,g)

C=2

Figure 5.1: Oscillation period of a pendulum for dierent experiments.

The kind of analysis presented here is applicable only to simple problems, for which a law of type (5.2) is valid. Forthe rest of cases, it would be desirable to have a systematic and general method at one's disposal. As will be shownimmediately, the Buckingham theorem provides the appropriate procedure.

1In fact, a dimensional quantity can intervene in a phenomenon only if its fundamental dimensions appear in at least another quantityof the problem.

28

2 .- BUCKINGHAM THEOREM

2 .- BUCKINGHAM THEOREM

2.1.- Formulation

Suppose that a certain phenomenon can be described in terms of the n dimensional variables a1; a2; : : : ; an:

f(a1; a2; : : : ; an) = 0 : (5.5)

Assume that these variables involve the m fundamental dimensions d1; d2; : : : ; dm, and that the dimensions [a1],[a2]; : : :,[an] of the n dimensional variables are given by

2

[a1] = d111 d

122 : : : d

1mm

[a2] = d211 d

222 : : : d

2mm

...

[an] = dn11 d

n22 : : : d

nmm : (5.6)

The Buckingham theorem establishes that it is possible to form n j, and not more than n j, independentdimensionless groups3 from the n dimensional variables, j being the rank of the matrix0BB@

11 12 : : : 1m21 22 : : : 2m

n1 n2 : : : nm

1CCA : (5.7)In addition, the theorem establishes that equation (5.5) is equivalent to

g(1; 2; : : : ; nj) = 0 ; (5.8)

where 1, 2, . . . , nj are the dimensionless groups mentioned above. The natural number j is the reduction of theproblem and, obviously, is equal to or smaller than the number m of fundamental dimensions (columns of (5.7)).

2.2.- Application

The application of the Buckingham theorem to any problem consists of the following steps:

1. To propose the n dimensional variables involved in a phenomenon

2. To form the matrix (5.7) from the fundamental dimensions of those variables

3. To calculate the rank j of that matrix

4. To obtain the n j independent dimensionless numbers. There are always alternatives to obtain these groups.It is convenient to choose those which can be easily recognized by other experimenters. We shall mention someof them in section 4.

The equation which describes the considered problem involves these dimensionless groups.

For the sake of illustration, let us consider again the measurement of the pendulum oscillation period. Supposeagain that the dimensional quantities involved in this problem are the period , mass m, and length ` of the pendulum,

2For instance, if a1 = , [a1] = d1 d32 , d1 and d2 being the mass [M ] and length [L] fundamental dimensions, respectively. In this case,

11 = 1, 12 = 3, and 13 = 14 = ::: = 1m = 0.3Independent dimensionless groups are those which can not be written as a function of the others.

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DIMENSIONAL ANALYSIS

as well as the gravitational acceleration g. The involved fundamental dimensions are [M ], [L], and [T ]. The matrix oftype (5.7) for this problem is

[L] [M ] [T ]

m`g

0BB@0 1 01 0 01 0 20 0 1

1CCA : (5.9)The determinant of the submatrix formed by the rst three rows and columns of (5.9) is dierent from zero. Therefore,the matrix rank is j = 3. The only dimensionless group of the problem is

1 p`=g

: (5.10)

Consequently, the theorem establishes that the pendulum period veries

p`=g

= C ; (5.11)

C being a dimensionless constant. Again, gure 5.1 can represent the values of the dimensionless group =(`=g)12 for

dierent experiments. In this case, the experimental results allow one to determine the C value: C = 2.

Contrarily to what occurred in the previous analysis, here we did not formulate any (5.2)-type hypothesis; we onlyhad to propose the dimensional variables involved in the phenomenon. This simple example also shows how the theorem allows one to discard variables which do not intervene in a certain physical process. Hovever, the validityof the hypothesis made in step 1 determines the validity of the nal result most of the time. The experimenter willpropose correctly the relevant dimensional variables only if he/she possesses sucient knowledge on the topic.

3 .- PHYSICAL SIMILARITY

As mentioned in the introduction of this chapter, the last step in the dimensional analysis of a phenomenon is todesign an appropriate experiment with a prototype (real conditions) or by using a physically similar model. In thissection, concepts such as similarity and model will be dened, and their usefulness in experimentation will be claried.

3.1.- Physical similarity. Dierent types of similarity

Consider two phenomena [(1) and (2)] which involve the same dimensional quantities. The theorem allows one toobtain the dimensionless groups in both cases. Because the dimensional variables are the same in both phenomena,so are the dimensionless groups. Let as denote by

(1)1 ;

(1)2 ; : : : ;

(1)k and

(2)1 ;

(2)2 ; : : : ;

(2)k (5.12)

the values taken by the dimensionless numbers in the phenomena (1) and (2), respectively. Phenomena (1) and (2)are physically similar if

(1)i =

(2)i for i = 1; 2; : : : ; k ; (5.13)

i.e., the dimensionless groups take the same values in both phenomena.

The geometrical conguration necessarily intervenes in the phenomenon considered. Therefore, all the relevantdistances of the problem must be included in the set of governing dimensional variables. Thus, the sets of dimensionlessnumbers (5.12) involve all the independent ratios which can be obtained from those relevant distances. If all thosedimensionless groups take the same values in (1) and (2), those two phenomena are geometrically similar. In otherwords, two phenomena are geometrically similar if their geometries dier just in a global scale. Geometrical similarityis a necessary but not sucient condition for physical similarity.

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