Flu Mech Class Notes Slot g

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A Fundamental Course in Fluid Mechanics and

Heat Transfer

Srikanth Vedantam

November 27, 2011


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Contents

I Fluid Mechanics 5

1 Introduction 7

2 Fundamental concepts 112.1 Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.2 Motion and deformation of a fluid particle . . . . . . . . . . . . . 14

2.2.1 Linear translation of a fluid particle . . . . . . . . . . . . 142.2.2 Rotation of a fluid particle . . . . . . . . . . . . . . . . . 172.2.3 Linear strain rate of a fluid particle . . . . . . . . . . . . 182.2.4 Shear strain rate of a fluid particle . . . . . . . . . . . . . 19

2.3 System and Control Volume . . . . . . . . . . . . . . . . . . . . . 202.3.1 System . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202.3.2 Control volume . . . . . . . . . . . . . . . . . . . . . . . . 20

2.4 Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

3 Balance laws 25

3.1 Balance laws in integral form . . . . . . . . . . . . . . . . . . . . 263.1.1 Balance laws for a system . . . . . . . . . . . . . . . . . . 263.1.2 Balance laws for a control volume . . . . . . . . . . . . . 28

3.2 Balance laws in Local Form . . . . . . . . . . . . . . . . . . . . . 333.2.1 Mass balance . . . . . . . . . . . . . . . . . . . . . . . . . 333.2.2 Linear momentum balance . . . . . . . . . . . . . . . . . . 35

4 Navier-Stokes equations 37

4.1 Constitutive equations . . . . . . . . . . . . . . . . . . . . . . . . 374.2 Navier-Stokes equations . . . . . . . . . . . . . . . . . . . . . . . 384.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

5 Fluid Statics 41

5.1 Force balance for a static liquid . . . . . . . . . . . . . . . . . . . 415.2 Variation of pressure in a static liquid . . . . . . . . . . . . . . . 42

5.2.1 Manometers . . . . . . . . . . . . . . . . . . . . . . . . . . 435.3 Surface tension . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

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4 CONTENTS

5.4 Hydrostatic forces on submerged surfaces . . . . . . . . . . . . . 455.4.1 Planar surfaces . . . . . . . . . . . . . . . . . . . . . . . . 45

5.4.2 Curved surfaces . . . . . . . . . . . . . . . . . . . . . . . . 455.4.3 Example problems . . . . . . . . . . . . . . . . . . . . . . 45

5.5 Buoyancy and stability . . . . . . . . . . . . . . . . . . . . . . . . 455.6 Fluids under rigid body motion . . . . . . . . . . . . . . . . . . . 45

6 Dimensional Analysis 47

6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 476.2 Buckingham Pi theorem . . . . . . . . . . . . . . . . . . . . . . . 486.3 Non-dimensionalization . . . . . . . . . . . . . . . . . . . . . . . 496.4 Similarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 496.5 Non-dimensionalization of differential equations . . . . . . . . . . 50

7 Incompressible inviscid flow 53

8 Incompressible viscous flow 55

8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 558.2 Internal flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 558.3 External flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

9 Fluid machinery 57

10 Heat Transfer 59

10.1 Conduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5910.2 Convection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6010.3 R adiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6010.4 Steady state conduction in one dimension . . . . . . . . . . . . . 60

10.5 Fins . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6110.6 Unsteady conduction in one dimension . . . . . . . . . . . . . . . 6210.6.1 Lumped heat capacity system . . . . . . . . . . . . . . . . 63

10.7 Transient heat transfer from a semi-infinite solid . . . . . . . . . 6310.8 Finite difference method . . . . . . . . . . . . . . . . . . . . . . . 63


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Part I

Fluid Mechanics

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Chapter 1

Introduction

Suppose you have a glass bottle filled partially with water on a table. You mightrecognize that the glass bottle system consists of three states of matter: solidglass, liquid water and gaseous air. And further, you might classify the waterand air together as fluids.

What really makes the fluids fluid? At an atomic level, the glass will haveatoms arranged in an amorphous manner that is, not crystalline (in fact, fromthis point of view, glass is termed supercooled liquid). The amorphous distri-bution of atoms extends to the liquid water and the gaseous vapour materials aswell. The atoms in the gas are generally quite far apart whereas in water and inglass they tend to cluster quite close together. In fact, though we classify waterand the air as fluids, from an atomic viewpoint, glass and water look more alike.So why do we classify water and air together as fluids?

We define fluids as materials which flow under applied loads. That is, theytake the shape of surrounding material. Dont solids flow? In the presence of aforce (we need some force to be acting for example, even water will form aspherical drop in the absence of gravity), if we wait really long enough even forsmall forces, everything every material flows. It is the timescale which isdifferent for different materials. For convenience, we tend to think of materialswhich flow in a few seconds as fluids. Even a steel wire will flow given a longenough time (this phenomenon is called creep).

What exactly do we mean by flow? If you compressed a liquid equally fromall directions, the liquid may contract by a certain amount. For a constantcompressive force, the amount of compression will be constant. And this is trueof a solid as well. However, if you sheared a liquid with a constant appliedstress it will continue to deform without stopping, but a solid will only deform

to a certain extent. After some amount of shearing, the stress in the solidincreases and this resists the applied load. In a liquid this does not happen.This continuous deformation under shear stresses is called flow.

Mechanics is the study of forces and deformations of materials. Even amongfluids, the behaviour could be quite different among different fluids. Some fluidstend to resist greater forces when they are stationary than when they are moving.

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8 CHAPTER 1. INTRODUCTION

Other fluids do the opposite. Some fluid start to flow only after a certain level ofshear stress is applied. Characterising these sorts of behaviour is an important

aspect of Fluid Mechanics.In fact, it is useful to take a step back and look at the way we characterise

the mechanical behaviour of any material. Why would we want to do thischaracterisation in the first place? We would like to build models of materialbehaviour. In the ideal scenario, we would like to develop a model by usingdata from experiments in a simple setting (such as a one dimensional test) andthen predict what the material would do in a more complicated setting. Thiswill allow us to understand the physics better as well as develop more optimaldesigns.

How do we develop models of materials which we can then use to describethe material behaviour? Generally a few specific steps are followed which shallbe described below.

The first assumption we make is that the material is continuous. This seemsto be an error since we all know that all materials are composed of atoms withsignificant empty spaces between them and matter is thus NOT continuous!However, the assumption of continuity of matter makes life easy in terms ofdeveloping models and has proved to be quite accurate for most materials, evenfor nanomaterials which consist of only a few tens of thousands of atoms. Theassumption of continuity allows us to write differential equations to describematerials. Thus the study of mechanics of materials (including solids, liquidsand gases) under this assumption is called continuum mechanics.

The next step is to write down a description of the position and motion ofthe body. We can write this down in a very general manner. However, if weintend to study the material in very specific situations such as one dimensionalor two dimensional motions or small deformations, we can write down more

specific descriptions. This will simplify the model being developed greatly. Thisdescription of the motion of the body is called kinematics.The third step is to write down the physical laws governing the material.

These are some of the basic universal laws of nature such as the a. conservationof mass, b. conservation of linear momentum, c. conservation of angular mo-mentum, d. conservation of energy, e. conservation of charge. All the balancelaws may not be necessary for a particular problem. For example, unless theelectrical charge is expected to affect the forces or deformations of the body un-der consideration, it may not be necessary to list it. In addition to the balancelaws, we also write down the second law of thermodynamics which is an im-balance law. The balance laws and second law of thermodynamics are essentialcomponents for developing models of material behaviour.

The balance laws introduce various field quantities required to describe

the force, temperature, charge which are functions of position in the bodyand time. However, it will be seen that the number of balance laws introducedis always smaller than the number of variables to be solved for. This is becausethe balance laws are valid for any material in the universeand the informationdescribing a particular material has still not been introduced. Unless this specificinformation is given, the model does not know which material we are trying


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9

to describe. This information is introduced in a continuum model throughconstitutive equations. These give specific material relations and depend on

whether the material is a solid, liquid or gas. And even among solids, whetherthe material is an elastic solid or inelastic. There are many possible types ofconstitutive relations and choosing the appropriate one is the biggest challengein developing a material model. Appropriate constitutive equations can only bewritten by looking at the experimental behaviour of the material.

We provide enough constitutive equations to complete the system above(such that there are as many equations as there are unknowns). The next stepis to combine all the above equations and obtain the governing equations.

For a specific geometry we can then provide the boundary conditionsappro-priate to the problem and solve the governing equations to describe the material.However, the governing equations are usually complicated and we have to resortto numerical computational techniques to solve the system. The commonly usednumerical methods are finite difference or finite element methods.

To summarize, the basic steps of continuum mechanics are:

1. Describe the position and motion of the body: Kinematics.

2. Write down the basic conservation laws through: Balance laws andsecond law of thermodynamics.

3. Write down equations describing specific material behaviour through: Con-stitutive equations.

4. Combine all the above equations to obtain the Governing equations.

5. In order to describe a specific problem, specify the Boundary condi-tions.

6. Solve the governing equations and the boundary conditions, usually usinga Numerical method.

In this text we will restrict our discussion to describing the kinematics offluid systems, writing down the balance laws of linear and angular momentumand one simple constitutive equation. This simple equation describes whatis called a Newtonian fluid. We will combine this constitutive equation withthe balance laws of linear momentum to obtain the governing equations forfluids called the Navier-Stokes equations. In fully three dimensional form, theseequations are non-linear and complicated to solve. We will solve these equationsin some simple geometries with simplifying assumptions and try to understandthe physics of fluid flow in some very specific situations.


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10 CHAPTER 1. INTRODUCTION


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Chapter 2

Fundamental concepts

In Chapter 1 we noted that the first steps of continuum mechanics requiredthe description of the positions and velocities of the fluid body which we calledkinematics. In this chapter we will lay out the concepts connected with thekinematic description of the body. This will be connected to the deformationof the body and its motion which we will describe in detail. In addition we willalso introduce the concept of stress. It is important to note that all the conceptsin this chapter are quite general are not restricted to fluids alone. They applyto every form of matter.

2.1 Kinematics

Kinematics is the description of deformation and motion of matter. Thereare several possible ways of describing the deformation of a material. Thesedescriptions are very general and not restricted to any one form of matter.However, some are more convenient than the others for describing fluids.

First it is important to consider a particle of matter. A particle (not re-lated to atoms or molecules) is an infinitesimal (vanishingly small) element ofmatter. It is continuously connected to the neighbouring particles in accordwith the continuum viewpoint of matter. In order to give the position of aparticle we will first fix a coordinate frame. There are many possible coordinatesystems we can choose and the physical concepts do not depend on this choice.The form of the equations look different depending on the choice but are com-pletely interchangeable. For simplicity we will make the choice of a rectangularCartesian coordinate system. In the Appendix we will also list out the forms of

the equations in other coordinate systems.In Fig. 2.1, we show a particle located at a point with coordinates (x,y,z)

in a Cartesian coordinate frame with origin at O. The shape of the particle ischosen according to the coordinate frame with sides oriented along the axes ofthe coordinate frame. The extent of the particle is dx in the x direction anddy,dz along the y and z directions respectively. For convenience we choose

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12 CHAPTER 2. FUNDAMENTAL CONCEPTS

O x

y

z

dx

dy

dz

Figure 2.1: Differential fluid element

the center of the particle at (x,y,z). However, we can alternatively choose onecorner of the cube to be at (x,y,z) and this choice is completely equivalent sincethe eventual equations we derive will be for the limit of the particle size tendingto 0.

Before we consider a real material, let us first consider a special kind ofmaterial composed of students taking this course this semester. The studentsare particles of this material. Let us imagine a boundary which encloses all thestudents as shown by the dotted line in Fig. 2.2. As the students move about,the boundary of the body passes through the outermost students. The dashedline and the crosses show the students in their first class and the dashed dottedline and round particles shows the students as they are getting into their secondclassroom. Since Mechanics is the study of forces and motions of a body, theforce on this body may be the motivation of the students to get to the nextclass. The motion describes how fast they get there. Consider a case whereall the students are in one classroom for one class and have to all go to anotherclassroom for the next class. We are usually interested in asking two kinds ofquestions in describing this motion:

1. How fast was a particular student A able to move from one classroom tothe next? In other words what was the velocity of student A with time,vA(t)?

2. How fast are students able to move through the a particular region along


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2.1. KINEMATICS 13

A

A

Classroom 1 Classroom 2

Corridor

Figure 2.2: A student body in two different configurations.

the path, say, the doorway of the next classroom? What is the velocity ofthe particles at this doorway: v(doorway, t)?

In this example of the students, since we know the names of the studentsin the class, we are able to ask the first question meaningfully. In the case offluids we cannot label the particles except in special cases and so we generallyask the second question. The first question is usually posed in the cases of solidsundergoing small deformations and is actually very convenient in that setting.

As we mentioned earlier, in the context of the above example, the forceis the motivational level of the class and the acceleration is the eagerness ofthe students to step into the next classroom. The eagerness of a student at aparticular point such as the doorway to the classroom has two components: theacceleration of a particular student in the doorway to move into the classroom

(from the innate eagerness for the class) and the rush of her classmates behindwho are almost carrying the student in the doorway into the classroom bycrowding close behind her. We will see that the acceleration of a fluid elementhas two analogous components.

When we are tracking the velocity of a certain particle (as asked in the firstquestion above), we can write the student particles position at any given time

using a vector rA = xA(t)i + yA(t)j. Then her velocity and acceleration are

simply vA =dxA

dt i +dyA

dt j = vxA(t)i + vyA(t)j and aA =d2xA

dt2 i +d2yA

dt2 j. Thissort of description of the material is called a Lagrangian description.

On the other hand, describing the velocity of particles at every point in thedomain (such as in the second question above) is called the Eulerian descriptionof motion of the material. We will look at it in detail next in the context of

fluid flow.In the Eulerian description of fluid motion we consider a velocity field of

the fluid. That is we give the velocity of the fluid particles at every pointr = xi + yj + zk in the fluid as v(x,y,z,t) = vxi + vyj + vzk. The velocity fielddescribes the velocities of the particles flowing through the various points in thedomain.


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2.2 Motion and deformation of a fluid particle

A rigid particle experiences two kinds of motions: linear motion under externalforces and angular motion under external torques. In order to obtain the particletrajectory we solve the linear and angular momentum balance given by

F = ma, (2.1)T = I, (2.2)

where a is the linear and the angular acceleration. The total external forcesand torques on the particle are

F and

T respectively.

In the case of particles in a deformable body, in addition to the rigid particlemodes the particles also experience two modes of deformation: stretching andshearing. Thus a fluid particle can experience two types of motions and two

types of deformations in general as shown in Fig 2.3.We will derive the rates of linear and angular accelerations as well as the

rates of stretching and shearing next. It is important to note that just knowingthe velocity field, we can obtain all the rates of motion and deformation.

2.2.1 Linear translation of a fluid particle

In fluids the Lagrangian description can be employed to look at the position ofa drop of tracer dye in the fluid. In three dimensions, if the position of the dropof tracer dye is given by rp(t) = xp(t)i + yp(t)j + zp(t)k then the velocity and

acceleration of the tracer dye drop are given by vp(t) =dxp(t)

dti+

dyp(t)dt

j+dzp(t)

dtk

and ap(t) =d2xp(t)

dt2 i +d2yp(t)

dt2 j +d2zp(t)

dt2 k. These arise from the elementary

definitions of velocity and acceleration of a particle.When we want to calculate the acceleration from an Eulerian velocity field,

we cannot just differentiate the velocity with respect to time. We need tolook at the acceleration of the particles passing through the point of interestat the moment in time. At the time t let a particle p be at the point (x,y,z).Since we have the label of the particle at this point, we know that its velocity is(dxp/dt,dyp/dt,dzp/dt). But this should be the same as the velocity given at thepointp by the velocity field. Thus the velocity of the particle occupying the point(x,y,z) is v(x,y,z,t) = v(xp(t), yp(t), zp(t), t) = (vpx(t), vpy(t), vpz(t)). Thenthe acceleration of the particle occupying (x,y,z) at the instant t is obtained byapplying the chain rule of differentiation on the velocity of the particle flowingthrough the point

ap(x,y,z,t) = dvpdt

= vpt

+ vpxp

dxp(t)dt

+ vpyp

dxp(t)dt

+ vpzp

dxp(t)dt

. (2.3)

But at the instant t, the components of the particle velocity and the flow fieldvelocity match (this is the definition of the flow field velocity as described above).Thus we can replace the expressions dxp(t)/dt by vx and dyp(t)/dt and dzp(t)/dt


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2.2. MOTION AND DEFORMATION OF A FLUID PARTICLE 15

O x

y

O x

y

Ox

y

Ox

yTranslation Rotation

Shearing Stretching

Figure 2.3: The various modes of deformation of a fluid particle.


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by vy and vz respectively. Thus the acceleration in the Eulerian description isgiven by

a(x,y,z,t) = DvDt

= vt

+ vxp

vx +vyp

vy +vzp

vz. (2.4)

To emphasize the point that the derivative is of the velocities of the underlyingparticles in the flow, the special symbol D/Dt is used. It is called the materialderivativein continuum mechanics.

In component form the above equation can be written as

ax =vxt

+ vxvxx

+ vyvxy

+ vzvxz

(2.5)

ay =vyt

+ vxvyx

+ vyvyy

+ vzvyz

(2.6)

az =

vz

t + vxvz

x + vyvz

y + vzvz

z (2.7)

The reason for the presence of the last three terms is that the accelerationat a point is given by the local acceleration of the fluid element as well as themovement of surrounding fluid elements into the location. The movement of thesurrounding fluid elements into the location is called the convective componentof the acceleration.

Example Problem 2.2.1. Let the velocity field in a tube be given byv(x,y,z,t) = Ux/Li.

1. Find the acceleration of a particle moving along thex

axis.

2. Find the positionxp(t) of a particle which was initially at x = 0 at t = 0.Calculate its velocity vp(t) and acceleration ap(t).

Solution. 1. Substituting the given velocity field vx = Ux/L, vy = 0, vz = 0into eqs. (2.5)(2.7), we obtain ax = U

2x/L2, ay = 0, az = 0.

2. Recall that the definition of the velocity field at each point in the domainis nothing but the velocities of particles at those points. Therefore, whenthe particle p is at a point xp its velocity is dxp/dt = vx. Thus

dxpdt

=U xp

L

Integrating the above equation and using the fact that the particle wasat x = 0 at t = 0, we get xp(t) = L(eUt/L 1). This is the Lagrangiandescription of the fluid flow. (Note that the acceleration of the particleis d2xp/dt

2 = U2x/L2 which matches with the acceleration field from theEulerian description.)


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A

B

C

DA

B

C

vy

vx

vx +vxyy

y

x

vxdt

vydt

vy +

vyxx

dt

vx

vxyydt

d

d

At t At t+ dt

Figure 2.4: Angular velocity of a fluid particle.

2.2.2 Rotation of a fluid particle

The rotation of the fluid particle and its corresponding angular velocity of theparticle = xi + yj + zk are important in characterizing the flow. Theangular velocity is defined as the average of the angular velocities of two per-pendicular lines on the fluid particles. From Fig 2.4 showing a fluid particle attimes t and t + dt, we can see that the average of the angular velocities of thesides of the fluid element oriented along the x and y directions is

z = 12

ddt + ddt

= 12dt 1y vxy ydt+ 1x vyx xdt

=1

2

vyx

vxy

. (2.8)

The negative sign in the term vx (vx/x)x arises since the line AB is tothe left of AB in order to account for the positive (counter-clockwise) angularvelocity.

You can see for yourself that the other two components of the angular ve-locity are given by

x =1

2

vzy

vyz

, (2.9)

y =1

2v

xz

vz

x

. (2.10)

The vorticity is defined to be twice the angular velocity. Thus the vorticityof a fluid particle is the curl of the velocity vector:

= v. (2.11)


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vxdt

A BA B

vx +

vxx

x

dt

x

vx +vxx

xvx

At t At t + dt

Figure 2.5: Linear strain rate.

Example Problem 2.2.2. Let the velocity field in a two dimensional domainbe given by v(x, y) = x

x2+y2i + y

x2+y2j. Find the angular velocity of fluid

particles in this flow.

Solution. Using Eq. (2.8), we see that the z component of the angular ve-locity, z =

12

12 y(x2+y2)3/2 2x

12 x(x2+y2)3/2 2y

= 0. The x and y

components can easily be verified to be 0 as well. Thus the flow has zero angu-lar velocity. How is that possible? It can be seen that the flow field is radial:that is every particle is moving outward from the centre. In such a flow theangular velocity is zero.

2.2.3 Linear strain rate of a fluid particle

The linear deformation (strain) rate of a fluid element is described by the ve-locity of stretching or contraction of perpendicular lines on the fluid element.As can be seen in Fig. 2.5, AB = AB + BB AA. Thus the rate of changeof length per unit length of a fluid element along the x axis is

1

x

d

dt(x) =

1

dt

AB ABAB

=1

dt

1

x

dx +

vxx

xdt x

=vxx

. (2.12)

Similar arguments in the y and z directions give the linear strain rates inthose directions as

vyy and

vzz .

The volumetric strain rateis the rate of change of volume of the fluid element

per unit volume of the particle is given by

vxx

+vyy

+vzz

. (2.13)

To see this, note that the volume of the fluid particle is V= xyz and


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A

B

C

DA

B

C

vy

vx

vx +vxyy

y

x

vxdt

vydt

vy +

vyxx

dt

vx +

vxyydt

d

d

At t At t+ dt

Figure 2.6: Derivation of the shear strain rate.

the volumetric strain rate is

1

Vd

dt(V) = 1

xyz

d

dt(xyz)

=1

x

d

dt(x) +

1

y

d

dt(y) +

1

z

d

dt(z), (2.14)

which, from the linear strain rates derived above is the expression given in (2.13).

2.2.4 Shear strain rate of a fluid particle

The change in shape of a fluid particle is given by the shear strain rate. Thisis defined as the rate of decrease of the angle between two initially mutuallyperpendicular lines. As seen in the derivation of the angular velocity of thefluid particle, the change in angle in a time dt is clearly equal to (see Fig 2.6)

d

dt

2

=

d

dt+

d

dt=

vyx

+vxy

(2.15)

It is important to note that the direction of d is opposite to that in the deriva-

tion of the angular velocity. This is because, for the angular velocity we wantthe average counter-clockwise rotation of the particle whereas here we want thedecrease in the angle from the initially perpendicular lines.


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Example Problem 2.2.3. Let the velocity field in a two dimensional domainbe given by v(x, y) = x

x2

+y2

i + y

x2

+y2j as in Ex. (2.2.2). Find the shear

strain rate for this flow.

Solution. Using Eq. (2.15), we see that the component of the shear strain rate,

xy =1

2y

(x2+y2)3/22x +

1

2x

(x2+y2)3/22y

= 2xy(x2+y2)3/2

. The x and ycomponents can easily be verified to be 0. The flow has zero angular velocitybut a finite shear rate.

2.3 System and Control Volume

In the last section we described the velocity of a fluid at a point. In many cases,it is very useful to look at an entire volume of fluid. In this case we have two

choices, we can follow a fixed mass of fluid or we can consider a fixed region ofspace through which fluid flows. These will be described in this section.

It is worth emphasizing the difference between this section and the last onceagain. In the last section we looked at infinitesimal fluid particles and describedthe position, velocity and acceleration of such particles. Depending on whetherwe followed a particular particle or looked at the overall velocity field in thefluid domain, the description was called Lagrangian or Eulerian respectively.Here we are looking at finite volumes of fluid domains. Depending on whetherthe boundary allows exchange of fluid mass or not, we call the domain a systemor control volume respectively. In both cases, the domain itself may have anoverall motion or deformation into a different shape.

2.3.1 System

A system is defined a fixed, identifiable mass of material. That is, we cannotexchange a part of the mass of the fluid with another of equal mass. Such asituation arises naturally in some situations such as in a mass of helium in aparty balloon. The rubber balloon serves as the system boundary in this case.As the balloon floats around the system moves but the mass of helium is fixedand identifiable. The balloon may be squeezed into different shapes and thesystem boundaries may deform.

2.3.2 Control volume

Situations in which a well defined system is identifiable are few. In most casesof interest, the mass of material is constantly flowing in and out of regions ofspace. A region of space into which material is flowing in and out is called acontrol volume. The boundaries of a control volume called control surfaces may be mobile as well. These control surfaces may be real surfaces such asthe surface of a solid over which liquid is flowing or imaginary.


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2.4. STRESS 21

O x

y

z

fx

fy

fz

Figure 2.7: Two differential elements showing the resolved forces on the inter-face.

2.4 Stress

In mechanics we are interested in forces in a body and how they affect thebody. If we want to consider how forces are transmitted within a body, wewill naturally think about a particle such as that in Fig. 2.7 and its adjacentparticle. As shown in the figure let us consider the adjacent particle along thexaxis first. In general, the particle on the right applies a force on the particleon the left on the interface between the two particles. The components of thevector force can be resolved in the x, y and z directions as seen in thefigure. The x component of the forces affects the adjacent particles differentlythan the y and z component of the forces. The fx component is trying topull apart (or push together depending on the sign) the surfaces whereas fy and

fz are trying to slide the surfaces apart (in different directions). These forcesare acting on the surface area Ax = dydz whose normal is in the x direction.Similarly, on the surface of the particle whose normal is in the y direction withsurface area Ay = dxdz, fy is the force component which pulling the surfaceapart whereas fx and fz are the sliding components. Finally, fz is the pullingapart force on the area Az = dxdy with the normal in the z direction whereasthe sliding components are fx and fy.

When we are looking at the force on an interface and its effect on the particle,it is not the actual magnitude of the force acting on the surface, but rather the

force per unit area which is important. After all we can have a relatively smallforce acting on a very small area which may cause a greater effect than a largerforce on a much larger area. Thus we consider the force per unit area which weterm stressas the important quantity in the mechanics of deformable bodies. As

we saw in the previous paragraph we can identify several important componentsof forces for each surface: those that act to pull apart surfaces and those thatact to slide apart surfaces. Thus the stresses acting on the surface with normalin the x direction has stresses xx = fx/Ax and xy = fy/Ax, xz = fz/Ax ofwhich xx is the normal stress and xy, xz are the shear stresses. Similarly, onthe surface with normal in the y direction, yy = fy/Ay is the normal stress


22/64


and yx = fx/Ay,yz = fz/Ay are the shear stresses. To complete the full setof stresses, the surface with normal in the z

direction has zz = fz/Az as

the normal stress and zx = fx/Az,zy = fy/Az the shear stresses. These ninestresses are written in matrix form as

=

xx xy xzyx yy yz

zx zy zz

(2.16)

Depending on the coordinate frame chosen, the values of the stresses may bedifferent but the matrices of stresses in the various frames are related througha simple transformation rule. Any such matrix object which transforms underthese geometric rules is called a tensor. The nine components of the stress tensorare shown on the differential element in Fig. 2.8.

O x

y

z

yy

xx

zz

yxyz

xz

xyzy

zx

Figure 2.8: Differential element showing the nine stress components.

2.5 Problems

1. Describe an example of a Langrangian description of a material and anexample of an Eulerian description of a material.

2. For a velocity field given by v = xiyj at what time does a particle whichstarted at x = 1 at t = 0 reach x = 1.5? At what time does a particlewhich started at y = 2 at t = 0 reach y = 1.5?

3. For a velocity field given by v = yi xj what is the location of a particlewhich started at (1, 1) at t = 0 at time t = 1.5?

4. Consider four particles initially located at the corners of a unit square(1, 1), (1, 2), (2, 1), (2, 2) in a flow field given by

(a) v = xi + yj

(b) v = xi yj


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2.5. PROBLEMS 23

(c) v = yi + xj

(d) v = yi x

j

What are the locations of these particles at t = 1? What is the shapeobtained by joining the particles with straight lines? What is the areaof this shape? What are the angles between the sides of this shape? Dothese answers match with the calculation of the volumetric strain rate andshearing rate from the given velocity field?

5. For the following two dimensional flows, find the acceleration, angularvelocity components, volumetric strain rate and shearing rates.1

(a) v = (3x + y)i + (2x 3y)j(b) v = t2 i + t3j

(c) v = t(x + y)i + t(x y)j6. For the following three dimensional flows, find the acceleration, angular

velocity components, volumetric strain rate and shearing rates.

(a) v = (at + bx)i + (ct + dy)j + (et + f z)k

(b) v = axt2 i + byt3j + czt4k

(c) v = t(x + y 3z)i + t(x y + z)j + (x2 + y2 + z2)k7. A flow field in the xy plane has a velocity field given by v(x, y) =

y

x2+y2i x

x2+y2j

. What is the angular velocity of particles in this

flow? What is the shear rate?

8. If the y component of the velocity in a flow field is given by vy = (xy)give all the possible steady (i.e. time-independent) x components suchthat the flow is irrotational (i.e. angular velocity is zero).

9. What is an incompressible flow? Can an apparently easily compressiblefluid (such as a gas) experience incompressible flow?

10. Which of the following flow fields represent two-dimensional incompress-ible flows?

(a) v = (4x + y)i + (x 4y)j(b) v = (xt2 + y)i + (xt2 + x)j

(c) v = (x + 2y)ti + (x 2y)tj11. Which of the following flow fields represent three-dimensional incompress-

ible flows?

1In this and following problems, you must assume that there are unit coefficients which

allow each term of the velocity components to have the units of velocity.


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(a) v = (4x + y)i + (x 4y)j + xyk

(b) v = (x + y + zt)i + (xt + y + z)

j + (x + yt + z)

k

(c) v = (x + 2y)ti + (x 2y)tj12. The three components of velocity in a flow field are given by vx = Ax +

By + Cz , vy = Dx + Ey + F z and vz = Gx + Hy + Jz. Determine arelationship between the constant coefficients A , B , . . . , J such that theflow field would represent an incompressible flow.

13. If the x component of the velocity in a flow field is vx = Ax(By), find apossible steady y component of the velocity such that the flow is incom-pressible. How many y components are possible to allow incompressibleflow? Are there unsteady y components for incompressible flow?

14. Would the case of the student body example discussed in Section 2.1,

be a system or control volume?


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Chapter 3

Balance laws

The basic balance laws or conservation laws of nature are merely statementsof bookkeeping. They do the job of an accountant for a company who tries totrack income and expenditure. Money is not really conserved due to the ficklenotion of value, but the principle that is followed is that whatever money isearned minus the money spent must be a part of the companys assets. Thecomplexity comes from the fact that money comes in different forms such ascash, credit and assets. Similarly, the conservation laws are at their basis verysimple, but careful bookkeeping is required.

As we mentioned in the introduction, the basic quantities which are nei-ther created nor destroyed in the universe are mass, linear momentum, angularmomentum, energy and charge. In this text we will write down the balancelaws for the first four quantities. Actually, you may be aware that the generaltheory of relativity allows for interchange between mass and energy. So reallywe must write a balance law for mass-energy. However, in most applicationsthat we deal with, these relativistic effects do not play a significant role andthe mass-energy interchange does not occur. Hence we can write separate equa-tions for mass and energy. Charge also does not affect the flow for the types offluids we will be studying in this text. It may be necessary to write the conser-vation of charge equations in situations where it affects the fluid flow such asmagnetohydrodynamics.

In some instances we may recognize that other higher order quantities maybe conserved for the time duration of interest. In such a case it may be moreuseful to write down the balance laws for these higher order quantities. Forexample, the ushers for a museum recognize that if a certain number of peoplewent into the museum since it opened, the same number should exit by closing

time. The museum guards do not weigh each person and write down the law ofconservation of mass for the museum.This bookkeeping can be performed for an entire system or at a particle

level. When done at the system level or control volume level, the balance lawsarise in the form of integrals over the volume and surface areas of the systemor control volume. In this chapter we will derive the equations for the basic

25


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26 CHAPTER 3. BALANCE LAWS

conservation laws of nature in integral form.

3.1 Balance laws in integral form

When local variations in fluid velocities and accelerations are not important, theglobal or integral forms of the balance laws become important. However, effectssuch as the influence of a wall on the neighbouring fluid cannot be accountedfor by this approach. There are many important problems that can be solvedusing the integral form as we will see in this section.

3.1.1 Balance laws for a system

Recall that a system is a fixed, identifiable mass of material. It is useful tounderstand the balance laws as described in words first and only then look at

the equations describing these statements.

1. Balance of mass: The rate of change of mass of a system is zero.

dM

dt

system

= 0 (3.1)

where M is the mass of the system

Msystem =

M

dm =

V

dV. (3.2)

where M is the system mass and V is the system volume. We used thestandard definition of density to relate dm = dV. Thus,

d

dt

M

dm =d

dt

V

dV = 0 (3.3)

2. Balance of momentum: The rate of change of momentum of a system isequal to the sum of all the external forces acting on the system.

dP

dt

system

= F (3.4)

where the linear momentum of a system is defined as

Psystem

= M

vdm = V

vdV. (3.5)

Putting these two together, the balance of linear momentum is given by

F =d

dt

M

vdm =d

dt

V

vdV (3.6)


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3.1. BALANCE LAWS IN INTEGRAL FORM 27

3. Balance of angular momentum: The rate of change of angular momentumof a system is equal to the sum of all the external torques acting on the

system.dH

dt

system

= T (3.7)

where the angular momentum of a system H is defined as

Hsystem =

M

r vdm =V

r vdV. (3.8)

Thus the balance of angular momentum is

T =d

dt

M

r vdm = ddt

V

r vdV. (3.9)

4. Balance of energy: The rate of change of energy of a system is equal tothe total heat power supplied to the system minus the work power done bythe system.

dE

dt

system

= Q W (3.10)

where the total energy of the system E is the sum of the internal energy,and kinetic and potential energies

Esystem =

M

u +

|v|22

+ gz

dm =

V

u +

|v|22

+ gz

dV. (3.11)

Thus the balance of energy is given by

Q W = ddt

M

u +

|v|22

+ gz

dm =

d

dt

V

u +

|v|22

+ gz

dV.

(3.12)

5. Imbalance of entropy: The rate of increase of entropy of a system is greaterthan or equal to the rate of heat supplied divided by absolute temperature.

dS

dt

system

QT

(3.13)

where the total entropy of the system is given by

Ssystem =

M

sdm =

V

sdV. (3.14)

Thus the second law of thermodynamics can be written as

d

dt M sdm =d

dt V sdV Q

T(3.15)

In most cases in Fluid Mechanics, a system of fixed identifiable mass cannotbe identified easily. Instead, the control volume is a preferred approach. How-ever, since mass enters and leaves a control volume, the bookkeeping shouldinclude the flux of the quantity carried in and out by the fluid mass. We willsee how this is done next.


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3.1.2 Balance laws for a control volume

We first note that in each of the balance laws for the system we had an extensiveproperty which was the integral of an intensive property over the system. Thuswe had

Nsystem =

M

dm =

V

dV, (3.16)

where, by comparing with Eqs. (3.2), (3.29), (3.8), (3.11) and (3.14) we see thatif,

N = M, then = 1 (3.17)

N = P, then = v (3.18)

N = H, then = r v (3.19)N = E, then = u +

|v|22

+ gz (3.20)N = S, then = s (3.21)

The strategy to derive the balance laws for a control volume is to comparethe system and control volume at an instant in which both coincide. The rateof change of any quantity in a system is related to the change in the quantityin a corresponding control volume by

dN

dt

system

=

t

CV

dV +

CS

v da (3.22)

This equation is called the Reynolds Transport Theorem. It is very importantto note that here the velocity v is measured relative to the control surface. Wewill now look at the specific forms for each of the balance laws.

Mass balance

Using Eqs. (3.3) and (3.22) and setting = 1 from (3.17) we can obtain thebalance of mass for a control volume

t

CV

dV +

CS

v da = 0 (3.23)

It should be noted that m =

CSv da is the mass flow rate through the

control surface and q =

CS v da is the volume flow rate through the controlsurface.

When the flow is steadythere are no time varying terms in the problem andthe mass balance becomes

CS

v

da = 0. (3.24)

When the fluid is incompressible, = constant and the mass balance becomes

Vt

+

CS

v da = 0 (3.25)

where V= CV dV is the volume of the control volume.


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A1

A2

A3

A4

45

60

Figure 3.1: Problem 3.1.1

Example Problem 3.1.1. Consider the steady flow of water through the mul-tiport device shown in Fig. 3.1. The areas are A1 = 20 cm

2, A2 = 15 cm2,

A3 = 12 cm2, and A4 = 5 cm

2. The mass flow rate into the device throughA1is 2 kg/s and the volume flow rate out of the device through A3=0.001 m

3/s.

The velocity of water throughA2 is v2 = 4i m/s. If the velocity of water may beassumed to be in the direction of the outlet ports, calculate the velocity of waterthrough A4.

Solution. The control volume is shown using a dotted line in the Fig. 3.1. Sincethe flow is steady, we can use Eq. (3.24). Since the fluid is water, we will alsoassume that its density is constant. If, as usual, we assume that the velocityprofile across each inlet/outlet port is constant, the integrals reduce to

v1 A1 + v2 A2 + v3 A3 + v4 A4 = 0

The quantity m1 = v1 A1 is the mass flow rate through area A1 which isgiven as 2 kg/s.

Since the velocity through A2 is given as v2 = 4i m/s and the area A2 = 15icm2, the mass flow rate is m2 = v2 A2 = 6 kg/s. (The density of water is1000 kg/m3)The volume flow rate out of A3 is q3=0.001 m

3/s (given). Thus the mass flowrate out of A3 is q3 = 1 kg/s.Substituting all the mass flow rates through Areas 1,2 and 3, we obtain the massflow rate through 4 to be m4 = 5 kg/s into the device. Since the area vector

A4 = 5(cos 30i +sin 30j) cm2, the velocity vector is v4 = 1(cos 30i +sin 30j)m/s.


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Linear momentum balance

Using Eqs. (3.6) and (3.22) and setting = v from (3.18) we can obtain thebalance of linear momentum for a control volume

t

CV

vdV +

CS

vv da = F (3.26)

where F is the total external force acting on the system. The external forcemay be composed of a body force FB and a surface force FS where the bodyforce can be written as

FB =

bdm =

CV

bdV (3.27)

When the force of gravity is the only body force then b = g.It is very important to note that the velocity v is measured relative to the

control volume and that Eq. (3.26) holds only for inertial or nonacceleratingcontrol volumes. In the case of rectilinear acceleration of the control volume,the derivation must be done with more care. We begin again Newtons secondlaw for a system

dP

dt

system

= F (3.28)

where the linear momentum is

Psystem =

M

vdm =

V

vdV. (3.29)

This equation is valid only for velocities measured relative to an inertial referenceframe. If we denote an inertial reference frame by XY Z, then Newtons secondlaws is

F =dPXY Z

dt

system

=d

dt

M

vXY Zdm =

M

dvXY Zdt

dm (3.30)

The velocities of fluid particles measure relative to the inertial reference framevXY Z and the accelerating CV reference frame vxyz can be related by

vXY Z = vxyz + vCV (3.31)

where vCV is the velocity of the control volume relative to the XY Z frame.Differentiating the above equation with time we get

dv

XY Zdt = a

XY Z = dv

xyzdt + d

vCV

dt = axyz + aCV (3.32)

Substituting Eq. (3.32) into Eq. (3.30) gives

F =

M

dvCVdt

dm +

M

dvxyzdt

dm (3.33)


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or

F MdvCV

dt dm =

dPxyz

dtsystem (3.34)

Now we can use the Reynolds Transport Theorem for the term on the RHS ofthe above equation to obtain

F

CV

aCV dV =

t

CV

vxyz dV +

CS

vxyz vxyz da (3.35)

where we have defined aCV =dvCV

dt and used the fact that the system and CVare coinciding at the time of consideration.

Angular momentum balance

Using Eqs. (3.9) and (3.22) and setting = r v from (3.19) we obtain thebalance of angular momentum for a control volume

T =

t

CV

r vdV +

CS

r vv da (3.36)

where the total external torque on the control volume may be written as

T = rFs +

CV

r bdV + Tshaft (3.37)

the first two terms on the RHS being the torques due to surface and body forcesrespectively. Combining the above equations we get the angular momentumbalance for an inertial control volume

rFs +

CV

rbdV + Tshaft = t

CV

rvdV +

CS

rvv da (3.38)

Again it must be emphasized that this equation is valid only for an inertialcontrol volume.

We will now present the angular momentum balance for a rotating controlvolume. Consider a fixed inertial fram XY Z and a rotating and translatingreference frame xyz as shown in Fig. ??. Let the angular velocity of xyz be and r be the position vector of a fluid particle in the rotating control volume.

The balance of angular momentum is in the rotating control volume xyz is

rFs + CV r bdV + Tshaft

CV

r [2 vxyz + ( r + r]dV

=

t

CV

r vxyz dV +

CS

r vxyz vxyz da (3.39)


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Energy balance

The first law of thermodynamics for the system states that

Q W = dEdt

system

(3.40)

where the total energy of the system is given by

Esystem =

M

edm =

V

edV (3.41)

with

e = u +|v|2

2+ gz (3.42)

As usual we use the Reynolds Transport Theorem with N = E and = e toobtain

dE

dt

system=

t

CVedV +

CS

ev da (3.43)Thus the first law of thermodynamics may be stated as

Q W = t

CV

edV +

CS

ev da (3.44)

For convenience of problem solving we write W = Wshaft + Wnormal + Wshear +

Wother. Here Wshaft is the work transferred in and out of the control volume

through a shaft. Wnormal is the work done by normal stresses acting on theboundary of the control volume. This can be rewritten as

Wnormal =

CS

nnv a (3.45)

Viscous stresses can make nn different from the hydrostatic pressure p as wewill see in the constitutive equations. However for most cases of engineeringinterest nn p. Thus the normal stresses term becomes

Wnormal =

CS

nnv a =

CS

p

v a (3.46)

in which we just multiplied and divided by .Note that if we choose the control surfaces perpendicular to the flow then

the shear work Wshear = 0. This is because the shear stresses act tangential tothe control surfaces whereas the fluid flow is normal. Thus there will no poweradded by the shear stresses. For a general choice of a control surface, there willbe a shear power input term.

Putting everything together we obtain the energy balance for a control vol-ume

QWshaftWshearWother = t

CV

edV+

CS

u +

|v|22

+ gz +p

vda(3.47)


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3.2. BALANCE LAWS IN LOCAL FORM 33

Entropy imbalance

The second law for a system is

dS

dt

CV

1T

Q (3.48)

where the total entropy of the system is

Ssystem =

M

sdm =

V

sdV (3.49)

The Reynolds Transport Theorem to the system formulation of the secondlaw to obtain

dSdt

CV

=t

CV

sdV +

CS

sv da (3.50)

Since the system and CV coincide at the time instant under consideration

1

TQ

system

=1

TQ

CV

=

CS

1

T

Q

A

dA (3.51)

Combining all of the above we get the second law of thermodynamics for acontrol volume

t

CV sdV +

CS sv da CS1

T Q

A

dA (3.52)

3.2 Balance laws in Local Form

The integral balance laws are useful when we are looking at the overall behaviourof the interaction of a fluid with a control volume. However, we cannot getdetailed information about the fluid from point to point. Besides the interactionwith surfaces cannot be treated in a proper fashion. For this we need to lookat the differential form of the balance laws. In this section we will derive thebalance laws for mass and linear momentum.

3.2.1 Mass balance

Consider the differential control volume shown in Fig. 3.2. Fluid is entering andleaving all the surfaces. Using Taylors series expansion about the center of thecubical element, the density and velocities of the fluid on the various surfaces


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O x

y

z

dx

dy

dz

Figure 3.2: Differential control volume

are

(x dx/2) = x

dx

2(3.53)

(x + dx/2) = +

x

dx

2 (3.54)

vx(x dx/2) = vx vxx

dx

2(3.55)

vx(x + dx/2) = vx +vxx

dx

2(3.56)

(3.57)

We state the law of conservation of mass as the net rate of mass flux out ofthe control volume plus the rate of increase of mass of the control volume equalszero. The net rate of mass flux out of the control volume can be seen to be

vx

x+

vyy

+vz

z dxdydz (3.58)

The rate of change of mass in the control volume is given by

tdxdydz (3.59)

Combining these two terms into the word state of the law of conservation of


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3.2. BALANCE LAWS IN LOCAL FORM 35

mass we obtain

t+

vx

x+

vy

y+

vz

z= 0 (3.60)

which can be written in vector form as v = 0. For an incompressible fluid =constant in space and time and we get

vxx

+vyy

+vzz

= 0 (3.61)

which is v = 0 in vector form.

3.2.2 Linear momentum balance

The balance of linear momentum of the differential control volume gives

vxt + vx vxx + v

y vxy + vz vxz

= gx + xxx +

xyy +

xzz (3.62)

vyt

+ vxvyx

+ vyvyy

+ vzvyz

= gy +

yxx

+yy

y+

yzz

(3.63)

vzt

+ vxvzx

+ vyvzy

+ vzvzz

= gz +

zxx

+zyy

+zz

z(3.64)

where gx, gyandgz are body force components in the x,y, and z directions re-spectively.

Angular momentum balance tell us that the stress matrix is symmetric:xy = yx, xz = zx, yz = zy


36/64



37/64

Chapter 4

Navier-Stokes equations

We derived the mass balance and linear momentum balance equations for adifferential control volume in the previous chapter. We have six unknowns ofthe stress components and three unknowns of the velocity components and thepressure. We have just four equations mass balance and three components ofthe linear momentum balance. In order to be able to solve the problems usingthis approach we need to apply constitutive equations.

4.1 Constitutive equations

For a Newtonian fluid the stresses may be expressed in terms of the velocity

gradients thus:

xy = yx =

vyx

+vxy

(4.1)

xz = zx =

vzx

+vxz

(4.2)

yz = zy =

vyz

+vzy

(4.3)

xx = p 23

v + 2 vxx

(4.4)

yy = p 23

v + 2 vyy

(4.5)

zz = p 23

v + 2 vzz

(4.6)

(4.7)

where is the viscosity and p is the thermodynamic pressure.

37


38/64

38 CHAPTER 4. NAVIER-STOKES EQUATIONS

4.2 Navier-Stokes equations

Substituting the constitutive equations in the linear momentum balance statedin the previous chapter, we obtain

vxt

+ vxvxx

+ vyvxy

+ vzvxz

= gx p

x+

x

2

vxx

23 v

+

y

vyx

+vxy

+

z

vzx

+vxz

(4.8)

vyt

+ vxvyx

+ vyvyy

+ vzvyz

= gy p

y+

x

vyx

+vxy

+

y

2

vyy

23 v

+

z

vyz

+vzy

(4.9)

vz

t + vxvzx + vy

vzy + vz

vzz

= gz p

z +

x

vz

x +

vxz

+

y

vyz

+vzy

+

z

2

vzz

23 v

(4.10)

In this course we will focus only on the incompressible Navier Stokes equa-tions:

vxt

+ vxvxx

+ vyvxy

+ vzvxz

= gx p

x+

2vxx2

+2vxy2

+2vxz2

(4.11)

vyt

+ vxvyx

+ vyvyy

+ vzvyz

= gy p

y+

2vyx2

+2vyy2

+2vyz2

(4.12)

vz

t + vx

vz

x + vy

vz

y + vz

vz

z

= gz p

z + 2vz

x2 +

2vz

y2 +

2vz

z2(4.13)

These equations in addition to the incompressible mass balance equation

vxx

+vyy

+vzz

= 0 (4.14)

constitute the governing equations for incompressible fluid flow. In conjunctionwith appropriate boundary conditions the velocity and pressure field in a fluidmay be solved.

4.3 Problems

1. Solve for the fully developed, steady, two dimensional velocity field fora fluid flow between two plane horizontal surfaces. The bottom plane isat rest whereas the top surface is moving at a constant speed U in thex-direction. The separation between the plates is h.

(a) Calculate the shear stress at the top and bottom plates.


39/64

4.3. PROBLEMS 39

(b) What is the force required to drag the top plate at the constantvelocity U if the total area of the plate is A?

2. Solve for the fully developed, steady, two dimensional velocity field for afluid flow down an incline. The incline is at rest and the top surface ofthe fluid of layer is exposed to vacuum. If the mass flow rate is m what isthe height of the fluid layer?

3. Solve for the fully developed, steady, two dimensional velocity field for afluid flow between two plane horizontal surfaces. Both the top and bottomplates are at rest but there is constant pressure gradient p/x drivingthe flow. If the outlet is to the atmosphere, what is the power required todrive the fluid at a mass flow rate of m?


40/64

40 CHAPTER 4. NAVIER-STOKES EQUATIONS


41/64

Chapter 5

Fluid Statics

There are many instances in which structures that we design and build willneed to take into account fluid pressures. An example would be any under seastructure on which water exerts tremendous pressure. There are instances wheresurface tension driven phenomena need to be accounted for. In this chapter wewill discuss these aspects of fluid mechanics.

A note on the notation: vectors will be given in boldface. That is, a vectora will be written in the text as a.

5.1 Force balance for a static liquid

Consider a fluid which is not deforming but which may be in rigid body motion.In such a motion, if you put a marker dye in any location of the fluid, the

dye will not move within the fluid. Let us derive the equation governing suchmotion.

For a differential fluid element such as in Fig 2.1, the body force is given by

dfB = gdm = gdV = gdxdydz (5.1)

The pressure in the y-direction on the left face of the differential element is

py = p + py

dy2

= p p

y

dy

2(5.2)

Similarly the pressure acting on the right face of the differential element is

py+ = p +

p

ydy

2

= p +

p

y

dy

2 (5.3)

The force due to the pressure on the left face is

dfpy =

p py

dy

2

dxdz ey (5.4)

41


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42 CHAPTER 5. FLUID STATICS

and the force due to the pressure on the right face is

dfpy+ =

p py

dy2

dxdz (ey) (5.5)

Thus the net force due to pressure in the y-direction is given by

dfpy = py

dxdydz ey (5.6)

Similarly the other components due to pressure force can be derived and thetotal pressure force in vector form is given by

dfp =

p

xex +

p

yey +

p

zez

dxdydz (5.7)

Recall from Physics that grad p is written as p and given by

p =

p

xex +

p

yey +

p

zez

(5.8)

Thus the balance of forces on the fluid element per unit volumecan be writtenas

p + g = a (5.9)where a is the acceleration experienced by the fluid element and is a rigid bodyacceleration. Since we are not considering relative motion between fluid elementsa is a constant.

5.2 Variation of pressure in a static liquid

If we neglect all other body forces other than gravity and consider a static liquid,the variation of pressure in a liquid is given by

dp

dz= g (5.10)

where g is the acceleration due to gravity.This principle is used in measuring the pressure difference between two fluids

in a manometer.Integrating eq. (5.10) using a presssure p0 at some reference level z0, we can

obtain

pp0 = g(z z0). (5.11)It is convenient to measure the distance downward from the free surface of

the liquid and thuspp0 = gh (5.12)

where h is measured positive down from the free surface into the liquid.


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5.2. VARIATION OF PRESSURE IN A STATIC LIQUID 43

5.2.1 Manometers

Consider a simple U-tube manometer shown in fig. 5.1. A is connected toa tank containing liquid with density A and B is connected to another tankcontaining liquid with density B . The manometeric liquid is of density M.The difference in pressure between A and P is given by pP pA = Aga. Thepressure difference between P and Q is given by pP pQ = Mgb. Finally thepressure difference between B and Q is given by pQ pB = Bgc. Putting allthese together we get

pA pB = g(cB + bM aA) (5.13)

The problem with these simple manometers is that the sensitivity is not very

good. That is, small difference in pressures translates to small changes in heightand this makes it hard to read. In order to increase the sensitivity, inclined tubereservoir manometers are usually used. How do these work? Consider theinclined tube manometer shown in fig. ??.

Figure 5.1: U-tube manometer


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5.3 Surface tension

When a static liquid comes in contact with a solid surface, it makes an angleto the surface as shown in Fig. ??. This contact angle is determined by therelative energies of the liquid-vapour interfaces, solid-liquid interface and thesolid-vapour interface. The drop contact angle is such that it minimizes thetotal free energy of the system.

Consider a hemispherical drop on a solid surface with contact angle . Thefootprint of the drop is a circle of radius, say, R. The volume of the drop isgiven by

V =1

3R3

(2 3cos + cos3 )sin3

. (5.14)

The liquid vapour surface area of the drop is given by ALV = 2R2(1

cos )/ sin2 and the solid-liquid surface area is given by ASL

= R2.

The total gibbs free energy of the system is given by

G = ASLSL + ALVLV + ASV SV . (5.15)

To minimize the Gibbs free energy, we require

dG = dASL SL + dALVLV + dASV SV = 0 (5.16)

Differentiating the above expressions for ALV and ASL we get

dALV = 4R(1 cos )

sin2 dR + 2R2

1

sin sin

1 + cos d (5.17)and

dASL = 2RdR. (5.18)

Note that

dASL = dASV , (5.19)

that is, a change in solid-liquid surface area can only come at the expense of thechange in solid-vapour surface area. Finally by differentiating the expression forthe volume of the drop (and setting it to 0 since there is no change in volume)we get,

Rd = 2(sin + sin 2/4)dR. (5.20)

Substituting the above expression into (5.17) we see that

dALV = dASL cos . (5.21)

Using


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5.4. HYDROSTATIC FORCES ON SUBMERGED SURFACES 45

5.4 Hydrostatic forces on submerged surfaces

5.4.1 Planar surfaces

5.4.2 Curved surfaces

5.4.3 Example problems

5.5 Buoyancy and stability

5.6 Fluids under rigid body motion


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Chapter 6

Dimensional Analysis

6.1 Introduction

Consider a simple spring mass system shown in Figure ??. From high schoolphysics we know that the equation governing the system is given by

md2x

dt2+ kx = 0, x(0) = x0 (6.1)

were m is the mass, k is the spring constant and x is the elongation of thesping from its rest length at any time t and x0 is its initial extension. Then theextension of the spring at any time t is given by

x(t) = x0 sink

m t (6.2)

If we want to plot this function we have to choose a mass, a spring constant,initial extension and the units to express these quantities in. For various choicesof the parameters we will get various curves as seen in Figure ??(a). However,if we define x = x/x0 and t = t

k/m as non-dimensional position and time, we

see that whatever the choice of m,k,x0 we get the curve given by x(t) = sin t(Figure ??(b)). From this we see the underlying form of the solution muchbetter.

In fact, we did not need the solution of this equation to be able to obtain thenon-dimensional forms for x and t. We know that the solution of the equation(6.1) can only depend on m,k,t,x0 and therefore

x = f(m,k,t,x0). (6.3)

Now let us look at the units of the quantities in the above equation.We know that we need to have the same dimensions for every additive term

of an equation. That is, it makes no sense to say 5 m = 4 s +1 kg or anythinglike that. The dimensions ofx and x0 are those of length which we will indicate

47


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48 CHAPTER 6. DIMENSIONAL ANALYSIS

as [L]. Similarly the dimensions of m is [M], k has dimensions of [M]/[t]2 andt has dimension [t].

Since the left side of (6.3) has dimension [L] so must the overall dimensionsof right side. All other dimensions must somehow cancel out. Since only x0 has[L] in it, f must be a linear function of x0. Thus f(m,k,t,x0) = x0g(m,k,t)where g is itself a dimensionless quantity. We can see that the only way g canbe dimensionless is if g(m,k,t) = h(t

k/m). If any other combination comes

up we will be have some dimensional quantities left over. Thus we see that

x/x0 = h(t

k/m). (6.4)

We, of course, wont know what functional form h is from this analysis but nowwe can do experiments to find out.

This analysis gives us a way of reducing our equation to a simpler form whosenature we can see more clearly, but more importantly it allows us to reduce thenumber of experiments we need to perform to get our function.

If we had started out with the original problem and tried to find out f, wewould need to vary m,k,t,x0 independently and get the value of x. If we did10 experiments for each parameter, we will need to do 104 experiments. Andwe would still not be able to see the underlying form easily. Using (6.4) we needto do only about 10 experiments to start seeing the form.

Lastly, and perhaps the most important reason for non-dimensionalization,is the fact that we can perform tests with masses and stiffnesses which are ex-perimentally convenient. For example, if we are asked to measure the frequencyof oscillation of a micron sized spring mass oscillator, we can do a scaled up testwith convenient masses of and spring stiffnesses. If we want to design a newcar with good aerodynamic characteristics, we can build and work with scalemodels.

6.2 Buckingham Pi theorem

Suppose for some physical problem we have a dependent variable x1 which is afunction ofn1 independent variables x2, x3, . . . , xn, we can write the functionalform as

x1 = f(x2, x3, . . . , xn). (6.5)

Alternatively we can write this as

g(x1, x2, . . . , xn) = 0. (6.6)

The Buckingham Pi theorem states that we can group these dimensional formsinto n m independent dimensionless ratios, or parameters which can bewritten in a functional form

G(1, 2, . . . , nm) = 0. (6.7)

The number m is the number of independent dimensions required to specify theproblem. The dimensions mentioned here need not be only the primary dimen-sions [M], [L], [t], temperature [T] or charge [Q]. They can be combinations of


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6.3. NON-DIMENSIONALIZATION 49

the primary dimensions if only those combinations occur. The next section willexplain a systematic process of obtaining the parameters in detail.

6.3 Non-dimensionalization

There are several methods of obtaining the dimensionless parameters. I preferthe simple and straightforward method described below.

1. List all the dimensional parameters involved.

2. Select a fundamental list of dimensions. Make sure that these dimensionsappear uniquely.

3. List the dimensions of the parameters in terms of the dimensions involved.

4. Select a set of parameters which contain the dimensions involved uniquely.

5. Divide the remaining parameters with the chosen parameters such thatnon-dimensional groups are obtained.

The above procedure is demonstrated using two examples. In the first theprimary dimensions are independent, whereas in the second they are not.

Example Problem 6.3.1. Consider the drag force on sphere moving throughair. The drag force on the sphere can be written as a function of the ball diameterD, velocity V, density of air and viscosity . What are the dimensionlessgroups for this problem?

6.4 SimilarityWe mentioned in the introduction that an important reason for nondimensional-ization is the ability to work with scale models. In order to enforce a correlationbetween scale model tests and life sized models, we need to ensure similarity.The similarity between the scale models and life size prototype models is ofthree kinds:

1. Geometric similarity

2. Kinematic similarity

3. Dynamic similarity

Geometric similarity requires that the model and the prototype have similarshape and dimensions ratios. Kinematic similarity requires that velocities incorresponding points should be scaled by the same scale factor. Finally, dynamicsimilarity requires that the forces at corresponding points should be scaled bythe same ratio. Dynamic similarity ensures kinematic similarity but not theother way around.


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In actual tests, it may not be possible to achieve complete dynamicallysimilar cases. However, it may still be useful to perform the scale model tests

since they show qualitative trends about how increasing one parameter affectsother parameters of interest. Sometimes, a correction factor is calculated andapplied to get quantitatively useful results.

6.5 Non-dimensionalization of differential equa-

tions

So far weve looked at physical problems and tried to guess the important de-pendent and independent parameters. In describing most common fluids, thegoverning equations are known: the Navier-Stokes equations. It is possible and

necessary to non-dimensionalize the governing equations even before solvingthem. It is then possible to consistently simplify the non-dimensional equa-tions.

We will non-dimensionalize the Navier Stokes equations now. The conserva-tion of mass is

vxx

+vyy

+vzz

= 0 (6.8)

and the incompressible Navier-Stokes equations are

vxt

+ vxvxx

+ vyvxy

+ vzvxz

= gx px

+

2vxx2

+2vxy2

+2vxz2

(6.9)

vyt

+ vxvyx

+ vyvyy

+ vzvyz

= gy p

y+

2vyx2

+2vyy2

+2vyz2

(6.10)

vzt

+ vxvzx

+ vyvzy

+ vzvzz

= gz p

z+

2vzx2

+2vzy2

+2vzz2

(6.11)

To non-dimensionalize these equations we will take a reference length Land a reference velocity V. The reference length and velocity could be somecharacteristic of the system such as the diameter and inlet velocity into a pipe.Then the basic quantities can be non-dimensionalized as

x =x

L, y =

y

L, z =

z

L, vx =

vxV

, vy =vy

V, vz =

vzV

, t =tL

V, and, p =

p

V2(6.12)

The non-dimensional counterparts to the above equations can clearly be seenas

vxx

+vyy

+vzz

= 0 (6.13)


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6.5. NON-DIMENSIONALIZATION OF DIFFERENTIAL EQUATIONS 51

and the incompressible Navier-Stokes equations are

vxt

+ vx vx

x+ vy v

x

y+ vz v

x

z= gxL

V2 p

x+ 1

Re

2vxx2

+ 2vxy2

+ 2vxz2

(6.14)

vyt

+ vxvyx

+ vyvyy

+ vzvyz

=gyL

V2 p

y+

1

Re

2vyx2

+2vyy2

+2vyz2

(6.15)

vzt

+ vxvzx

+ vyvzy

+ vzvzz

=gzL

V2 p

z+

1

Re

2vzx2

+2vzy2

+2vzz2

(6.16)

where we have set Re = VL . Re is an important parameter called theReynolds number. We can see that for a model system and a prototype tobe dynamically similar, we only require the non-dimensional parameters, theReynolds number and the Froude numbers, F r = giL/V2, for i = x,y,z to bethe same. In situations where the body forces are not important (such situations

are characterised by small values ofF r), only the Reynolds number needs to bematched.

It should be emphasised that the boundary conditions also need to be similarin order for the model and prototype flows to be similar. In this process, wemay get some further non-dimensional numbers which need to be matched. Forexample, if the boundary condition is given by

vbc = V sin t (6.17)

then the non-dimensional form is

vbc =vbcV

= sin

L

Vt

(6.18)

Thus for the model and prototype to be similar, the Strouhal number St =L/V should also match.


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Chapter 7

Incompressible inviscid flow

53


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54 CHAPTER 7. INCOMPRESSIBLE INVISCID FLOW


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Chapter 8

Incompressible viscous flow

8.1 Introduction

8.2 Internal flows

8.3 External flows

55


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56 CHAPTER 8. INCOMPRESSIBLE VISCOUS FLOW


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Chapter 9

Fluid machinery

57


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58 CHAPTER 9. FLUID MACHINERY


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Chapter 10

Heat Transfer

10.1 Conduction

In a body in which there is a temperature gradient, heat flows from regions ofhigh temperature to the low temperature regions. This rate of heat transferper unit area is empirically determined to be proportional to the temperaturegradient. Thus,

Q = kA Tx

(10.1)

where k is the thermal conductivity of the body (W/m.K) which always pos-itive as a consequence of the second law of thermodynamics. Thus when thetemperature gradient is negative, there is heat transfer in the positive direction.

The problem of conduction heat transfer is to determine the temperature

field in a body given either fixed temperatures or heat fluxes at the boundariesof the body. Energy balance in this case gives the heat conduction equation.Consider a one-dimensional infinitesimal element as shown in Fig. ??. Thebalance of energy can be stated as:

Energy conducted in left face + heat generated within element

= change in internal energy + energy conducted out right face

These quantities may be represented mathematically as:

Energy conducted in left face = Qx = kA Tx

(10.2)

Heat generated within element = qAdx (10.3)

Change in internal energy = cA

T

t dx (10.4)

Energy out right face = Qx+dx = A

kT

x+

x

k

T

x

dx

(10.5)

where q is the heat generated per unit volume (W/m3), c is the specific heat ofthe material, (J/kg.K) and is the density (kg/m3).

59


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60 CHAPTER 10. HEAT TRANSFER

Combining the above terms and simplifying we get

x

k Tx

+ q = c Tt

(10.6)

If we do a similar calculation for a three-dimensional element we get thethree-dimensional heat conduction equation

x

k

T

x

+

y

k

T

y

+

z

k

T

z

+ q = c

T

t(10.7)

If the thermal conductivity is constant we can rewrite this equation as

2T

x2+

2T

y2+

2T

z2+

q

k=

1

T

t, (10.8)

where = k/c is called the thermal diffusivity (m2/s). The larger the value

of the faster the heat will diffuse through the material. This is because high can come through high k in which case, the heat will conduct away fasteror through low heat capacity c in which case less energy is used up to storeheat and raise temperature and more energy can be conducted away. Thus amaterial with higher feels cooler to touch than a material with a lower evenif the two materials are at the same temperature.

10.2 Convection

The convection heat transfer from a solid surface is given by

Q = hA(Tw T) (10.9)

where h is the convection coefficient (W/m2.K), Tw is the temperature of thesolid wall and T is the temperature of the bulk liquid far from the surface.

10.3 Radiation

The net heat exchange between two solid grey surfaces at temperatures T1 andT2 is given by

Q = (T41 T42 ) (10.10)Thus when T1 > T2 heat is transferred from high temperature surface to thelow temperature surface.

10.4 Steady state conduction in one dimensionConsider a plane wall of constant thermal conductivity k, area A, width L andsurfaces temperatures T1 and T2. The heat transfer in the wall is

Q = kAL

(T2 T1) (10.11)


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10.5. FINS 61

If there are n such walls each of width Li and thermal conductivity ki as inthe figure, the heat transfer is

Q = k1AL1

(Ti2 T1) = k2AL2

(Ti3 Ti2) = . . . = knALn

(T2 Tin) (10.12)

The intermediate temperatures Ti2, Ti3, . . . , T in are unknown and have to besolved from the above equations. If the intermediate temperatures are notrequired but only the overall heat transfer rate, then we can use

Q =T1 T2

LikiA

. (10.13)

Li

kiAis the thermal resistanceof the system. If the end walls are not at fixed

temperature but experience convection with convection coefficients h1 and h2

with bulk fluid temperatures T1 and T2 respectively, then the heat transferis given by

Q =T1 T2

1h1A

+ Li

kiA+ 1h2A

. (10.14)

10.5 Fins

A fin is a device which enhances heat transfer by increasing the surface areaavailable for convection. Fins are only effective when the conduction heat trans-fer rate is far greater then the convection heat transfer rate. If the opposite istrue (i.e. convection heat transfer is greater then conduction), then fins decreasethe overall heat transfer rate.

Consider a fin shown in Fig. ??. The energy balance gives

Energy in left face = energy out right face + energy lost by convection

Noting that the energy lost by convection is hPdx(T T), where P is theperimeter of the cross-section and the conduction in left face is kA(dT/dx)and out right face is kA(dT/dx + dxd2T/dx2), the fin equation can be writtenas

d2T

dx2 hP

kA(T T) = 0 (10.15)

Letting = T T and m2 = hP/kA, the equation becomes

d2dx2

m2 = 0 (10.16)

The solution of the fin equation can be written as

= C1 exp(mx) + C2 exp(mx) (10.17)


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At the base of the fin, let the temperature be To so that = o = To T.The boundary conditions at the fin tip can be of several types: (1) The fin may

be considered long in which case the boundary conditions are

= o at x = 0 (10.18)

is bounded at x = (10.19)

The solution to the fin equation is

o= exp(mx) (10.20)

(2) The fin tip is insulated and so dT/dx = 0 at x = L so that the boundaryconditions are

= o at x = 0 (10.21)d

dx= 0 at x = L (10.22)

The solution is

o=

exp(mx)1 + exp(2mL) +

exp(mx)

1 + exp(2mL)(10.23)

which can be written more conveniently using the hyperbolic functions, cosh x =(ex + ex)/2,

o=

cosh m(L x)cosh mL

(10.24)

(3) Fin loses heat from the tip through convection so that the boundary condi-tions are

= o at x = 0 (10.25)

k ddx

x=L

= hL at x = L (10.26)

The solution in this case is

o=

cosh m(L x) + (h/mk)sinh m(L x)cosh mL + (h/mk)sinh mL

(10.27)

10.6 Unsteady conduction in one dimensionThe partial differential equation governing unsteady conduction in one dimen-sion is given by

2T

x2=

1

T

t(10.28)


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10.7. TRANSIENT HEAT TRANSFER FROM A SEMI-INFINITE SOLID63

10.6.1 Lumped heat capacity system

If a solid object is immersed in a fluid and if the heat conduction rate is muchhigher than the convection from the surface we can assume that the entire bodyis at the same temperature T. Then the heat transfer from the body is givenby

Q = hA(T T) = cV dTdt

(10.29)

for which the solution is

T TTo T = e

[hA/cV]t (10.30)

It is common to write this solution as

T TTo

T= eBiFo (10.31)

where Bi = hV/kA is the Biot number and F o = kA2t/cV2 is the Fouriernumber.

The lumped capacity of the system is valid when the Biot number is small:

Bi =h(V /A)

k< 0.1

10.7 Transient heat transfer from a semi-infinite

solid

When the surface of a semi-infinite solid at initial temperature Ti is suddenlychanged to To its temperature is given by

T(x, t) ToTi To = erf

x

2

t(10.32)

where erf(y) is the error function defined by

erf(y) =2

y0

e2

d (10.33)

10.8 Finite difference method

The finite difference method is a useful numerical technique to solve partialdifferential equations whose solutions cannot be obtained in closed form. In

order to establish a difference solution, we divide the domain into (usually equal)discrete pieces. We then look to evaluate the solution of the equation only atthe discrete locations. The differential terms are replaced by difference termsusing Taylors series expansions such as

dT

dt=

Tk+1 Tkt

(10.34)


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where the subscript k represents the time t = kt. Similarly, the second ordertime derivative can be represented by

d2T

dt2=

Tk+1 2Tk + Tk1t2

(10.35)

which can be obtained by using Eq.( 10.36) at two time steps t apart.Similarly the spatial derivatives

dT

dx=

Ti+1 Tix

(10.36)

where the subscript i represents the point x = ix. The second order derivativecan be represented by

d2T

dx2=

Ti+1 2Ti + Ti1x2

(10.37)

Convince yourself that the finite difference form of the equation

2T

x2=

1

T

t(10.38)

can be written as

Ti+1,k 2Ti,k + Ti1,kx2

=1

Ti,k+1 Ti,kt

(10.39)

If the v

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