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Contents

1 Basic Fluid Concepts 51.1 Introduction to the Course . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.1.1 What Chemical Engineers Do ? . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.1.2 How do Chemical Engineers Carry out the Process ? . . . . . . . . . . . . . . . 51.1.3 Product and Service oriented disciplines . . . . . . . . . . . . . . . . . . . . . . 61.1.4 Unit Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.2 Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.3 Dimensions and Units, and Measurements . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.3.1 Measurement errors and propagation of errors . . . . . . . . . . . . . . . . . . . 91.4 Index Notation of Vectors and Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.4.1 Algebra with Index notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.4.2 Physical Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171.4.3 Comparisons with Bold face notation . . . . . . . . . . . . . . . . . . . . . . . . 18

1.5 Velocity and Stress Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201.5.1 Origin of surface forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201.5.2 Stress Tensor Convention . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211.5.3 Tutorial on Obtaining Local Stress Direction . . . . . . . . . . . . . . . . . . . . 22

1.6 Dimensional Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251.6.1 Steps for obtaining dimensionless groups . . . . . . . . . . . . . . . . . . . . . . 261.6.2 Example: Drag on a sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

1.7 Fluid Statics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281.7.1 Horizontal and Vertical forces, and Rotation . . . . . . . . . . . . . . . . . . . . 28

2 Differential Analysis of Fluid Flow 392.1 Types of ows and Methods of Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . 39

2.1.1 System, Conservation Laws, and Control Volume . . . . . . . . . . . . . . . . . 392.2 Useful Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

2.2.1 Reynolds Transport Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412.2.2 Divergence Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

2.3 Fundamental Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 442.3.1 Continuity Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 452.3.2 Derivation of Navier-Stokes Equation . . . . . . . . . . . . . . . . . . . . . . . . 452.3.3 Energy balance or Heat equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 472.3.4 Species Balance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

2.4 Boundary Layer Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 512.4.1 Integral Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 522.4.2 Laminar and Turbulent Boundary Layers . . . . . . . . . . . . . . . . . . . . . . 572.4.3 Differential Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

2.5 Simple Viscous Flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

2.5.1 Poiseuille and Couette Flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

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4 CONTENTS

2.5.2 Torsional ow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 802.5.3 Free surface ows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 812.5.4 Wire Coating . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 822.5.5 Tube ow of Power law uid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

2.6 Creeping Flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 852.6.1 Flow past a sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

2.7 Inviscid ows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 862.7.1 Stagnation point ow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 932.7.2 Flow past a cylinder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 932.7.3 Porous Media Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

2.8 Turbulent Flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 962.8.1 What They Said . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 962.8.2 Origin of Turbulence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 972.8.3 Boussinesq Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

2.8.4 Prandtl Mixing Length Hypothesis . . . . . . . . . . . . . . . . . . . . . . . . . . 1002.8.5 Velocity Proles in Turbulent Flow . . . . . . . . . . . . . . . . . . . . . . . . . . 101

3 Integral Analysis of Fluid Flow 1013.1 Mass Balances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

3.1.1 Multiport Device . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1013.1.2 Tank Draining: Unsteady Flows . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

3.2 Momentum Balances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1033.2.1 Jet Impingement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

3.3 Energy Conservation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1053.3.1 Energy Balance for a Flow through Equipment . . . . . . . . . . . . . . . . . . . 1053.3.2 Pumping Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

4 Flows in Equipments 1134.1 Pipe Flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

4.1.1 Major loss . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1134.1.2 Minor losses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1154.1.3 Piping Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1164.1.4 Non-Circular Ducts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

4.2 Flow Past Objects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1184.2.1 Terminal (Settling) Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118

4.3 Fixed Bed of Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1214.3.1 Pressure Drop Calculations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122

4.4 Fluidised Beds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

Bibliography 127

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Chapter 2

Differential Analysis of Fluid Flow

2.1 Types of ows and Methods of Analysis

2.1.1 System, Conservation Laws, and Control VolumeLecture 2.1 Control Volume Analysis in uid ow is like Free Body diagrams in Mechanics

Concepts You Must Know

1. What is a system, how is it different from a control volume ?

2. Is a control mass equivalent to a system or a control volume?

3. What are the basic conservation laws that apply to a system?

The system is dened as an identiable mass of a body. In thermodynamics we have closed andopen systems. Here we use system to identify with the closed system, in which there can be heat orwork transported in and out of the system, but there mass remains the same. It is also sometimescalled as a Control Mass (CM). For most engineering purposes, where there is no conversion of massto energy, it is easily possible to make a identication of a system.

The basic conservation laws applies to an identiable mass, or a system.

Mass ConservationdM system

dt = 0 (2.1)

Newtons second Law:d( M V)system

dt

=

F (2.2)

Energy Conservation

dE systemdt

= 0 (Isolated) (2.3)

dE systemdt

= Q W (Closed) (2.4)

The conservation laws in the form presented above are applicable only for a isolated or closedsystems. But for most engineering interest involving uid ow, it is not useful to consider an iso-lated system, especially when we are interested in calculating the effects of ow on machinery, oronly a particular part of it. Therefore, we introduce models of ow, which consider the transport of

conserved quantities across a restricted volume. We call this volume as control volume(CV).

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40 Chapter 2. Differential Analysis of Fluid Flow

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1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

M

M V

E

Isolated System

Conserved Quantities

Figure 2.1: Conserved quantities in an isolated system

Consider a case of a water tap connected to an overhead tank through a pipe. The control massor the system would be the entire identiable mass of water, i.e., that in the tank, pipe, and thebucket. This is of little use, if we were to nd out the local effect of the force applied by water owin the pipe. We can instead consider a small cylindrical volume of uid in the pipe, this would bethe control volume. But we have a problem here: The conservation equations, the only governingequations that we know of the material, applies to the whole system. The goal then is to derive asuitable form of the equations that applies to the control volume.

Control volume is a concept introduced in continuum mechanics. We are dealing with a limitedvolume of a continuum of a uid (of identiable mass). The boundary of the volume is called the con-trol surface. Some of these surfaces could be real, identiable interfaces: such as the interface betweenthe uid and the solid wall. Some surfaces are purely imaginary: such as any arbitrary cross-section

of the pipe. Identication of a control volume is an important step in the solution of uid mechanicsproblems. Many a times the choice can result in simpler algebra. Depending on how the volume istaken we can have four different models of ow (see Figure 2.2). Note that we use the notation V orv to denote velocities, and V (with a cross) to denote volume.

In the following lecture we will derive relationships that will relate the derivatives of a systemto equivalent derivative terms that apply to a control volume of a continuum. These relationships,in combination with the conservation laws are fundamental to uid mechanics. These equationscan be applied to arbitrary control volumes. When the CV is a nite volume, such as that of anequipment, the resulting equations are called as integral balances because it applies to the whole of the equipment, and when the volume is differential elemental volume, we recover the dif ferential bal-ances of uid ow, such as the Navier-Stokes equations. Integral balances are used to obtain grosseffects of the ow on the object it interacts with. Differential balances are used to obtain the detailed

continuum variables such as the velocity and stress prole.



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Lecture 2.2. Reynolds Transport Theorem 41

V V d

V d

Finite Volume Differential Volume

Fixed in space

Moving with flow

v

V

Figure 2.2: Models of ow: Finite volume or differential volume.

2.2 Useful TheoremsLecture 2.2 Reynolds Transport Theorem and Divergence theorem are the two building blocks

for deriving equations governing uid ow.


1. What derivatives does Reynolds Transport Theorem (RTT) relate?

2. Is RTT valid for arbitrary control volume ?3. What is Divergence theorem in one dimension?

We rst dene some notation. Let P be a scalar conserved extensive quantity (one of mass,momentum, or energy). We also dene a corresponding scalar conserved intensive quantity . Ina given control volume (CV) V , could in general vary with space and time = (x, y, z, t ). Theextensive property for the entire CV is related to by

P CV = CV

dV (2.5)

where is the local mass density. Note that the quantities appearing in the numerator in the LHS of Equations ( 2.1) to (2.4) are each an extensive conserved quantity. In the case of momentum, we havethree conserved quantities u, v,w, in each of the three coordinate directions x, y, z. The complete list isgiven in Table 2.1. In order to interpret Equations ( 2.1) to (2.4) for a CV, we have to evaluate the LHSof these equations for each of the CVs. Therefore we need an expression for

dP systemdt

= ? ( in a CV) (2.6)

Here is where the Reynolds Transport Theorem (RTT) comes in.

2.2.1 Reynolds Transport TheoremConsider a identiable mass of a uid (control mass, or system) that is displaced in space and time,

as shown in Figure 2.3. For simplicity we consider a two dimensional gure, but the terms area



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Table 2.1: Conserved quantities in extensive and intensive form, as dened in Equation (2.5)

Equation Physical quantity P (2.1) Mass M 1(2.2) Momentum M V u = u, v,w(2.4) Energy E e

+ t t

V

n

dACV

System

I

III

dA

Vn

II

CS

CSIII

I

l

t

dV

Figure 2.3: The system (identiable mass) is displaced in time. The dashed line is the system bound-ary and the shaded region with solid line boundary is the control volume. The System and ControlVolume coincide as t 0.

and volume will be used to identify the edge and area. Consider a xed control volume shownby the shaded region. If we track all the particles in the volume, then after a period t they can beidentied to be present in a region in the neighbourhood (shown by the dashed curve in the gure).

When t 0, the system coincides with the control volume. The following identities apply

V CV = V I + V II (2.7)V sys = V II + V III (2.8)

The total change in the quantity P in the system is

dP sysdt

= lim t 0

P sys (t + t ) P sys (t ) t

(2.9)

In terms of the individual region quantities

P sys (t + t ) = (P II + P III )(t + t ) = (P CV P I + P III )(t + t ) (2.10)P sys (t ) = (P I + P II )(t ) = P CV (t ) (2.11)Then Equation (2.9) can be written as:

dP sysdt

= lim t 0

P CV (t + t ) P CV (t ) t

x

+P III (t + t )

t

y

P I(t + t )

t

z

(2.12)

The rst term on the RHS x can be written as

Term x =P CV

t =

t

CV

dV (2.13)



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Lecture 2.2. Reynolds Transport Theorem 43

For the second term y , recognise that the term P III is the total amount of P present in the volume III .To nd this we consider the dependence of a differential amount dP of the quantity P in the volumedV dP III = dV (2.14)The volume dV is swept by the system boundary from the boundary of the CV in the time interval t .This volume depends on the direction of the local velocity of the boundary V as well as its directionn . From the volume of the parallelogram we have

dV = dA l cos where l = V t is the length swept by the volume. In general we can write

dV = (dA V) t = (dA n V) t

Term y can now be written from Equation ( 2.14), by integrating over the complete surface CS III

Term y = CS III

dP III / t = CS III

dA V = CS III

dA (n V) (2.15)

A similar expression can be derived for z in CS I . The only difference being, in I the direction of Vand that of n are opposite, i.e., the expression for the differential volume will have a negative n tokeep the volume positive:

dV = (dA (n ) V) t Therefore we have,

Term z = CS I dPI / t = CS I dA V

= CS I dA (n V) (2.16)Combining all the three terms in Equations ( 2.13), (2.15),and( 2.16) we get twoequivalent expressionsfor the Reynolds Transport Theorem

dP sysdt

=t

CV

dV + CS

dA V (2.17)

=t

CV

dV + CS

dA (n V) (2.18)

Here we have used the complete integral over the control surface that bounds the control volume tobe

CS dA = CS I dA + CS III dAReynolds transport theorem basically is an expression for the rate of change of a conserved quan-

tity in a given system (or controlled mass) in terms of variables dened in a control volume (inside aswell as its surface). It simply says that the total derivative (ie, the system derivative) of a conservedquantity is equal to the sum of that due to the partial time derivative in the CV and the net inuxin to the CV through the CS. Note that this form of the derivative is applicable to control volumesthat is xed or moving with constant velocity with respect to a inertial frame of reference. We havederived the model equation for the nite volume case. When the CV is xed, the form of equationis called as conservative form. In order to derive the differential form of the conservative equations,

we need to invoke the divergence theorem.



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2.2.2 Divergence Theorem

For any vector eld F dened in the volume V , divergence theorem relates the volume integral di-vergence of the eld to the surface integral of the normal component of the eld.

CV

dV ( F ) = CS

dA (n F ) (2.19)

This is a generalisation of the one dimensional integral of a function.

b

a

dx f (x) = F (b) F (a )

where dF dx = f (x), i.e.,b

a

dxdF dx

= F (b) F (a)

Here the one dimensional integral over a line is reduced to a sum over the boundary points, ac-counted for the direction. In two dimensions the divergence theorem will relate an integral over anarea to an integral over the peripheral edge (boundary). Along with Equation ( 2.19), the ReynoldsTransport Theorem (RTT) Equation ( 2.18) can be written as

dP sysdt

=t

CV

dV + CV

dV ( u ) (2.20)

Note that the divergence includes the scalar terms . We have also used u for the velocity in place of V, for convenience of familiarity of the equations to be derived. These both are equivalent, and maybe used interchangeably. V is typically used when the velocity is evaluated on a surface (integral),and u is used when it is evaluated in a volume (integral).

In order to derive the differential form of the governing equations, we consider an innitesimallysmall control volume, V , where the eld variables ,, u are uniform in space. Applying RTT to thisvolume we get

dP sysdt

=t

+ ( u ) V (2.21)

2.3 Fundamental EquationsLecture 2.3 The fundamental equations for the conserved quantities can be derived from the

conservation laws for the system and RTT


1. What are the integral and differential forms of the mass conservation law as applied to a con-tinuum? What is the differential form called as?

We now have the general relationship between derivatives of a conserved quantity on a systemto that in a control volume, we can express the LHS of the conservation laws Equations ( 2.1) to (2.4)on a control volume. This is used to derive the familiar equations of uid mechanics, the derivation

of which you may be familiar with using a shell balance.



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Lecture 2.4. Navier Stokes Equation 45

2.3.1 Continuity Equation

The continuity equation can be derived by using the mass conservation law and the RTT Equa-tion (2.21). From Table 2.1, we see for P = M , = 1. Therefore we get Integral Form:

dM sysdt

=t

CV

dV + CS

dA (n V) = 0 (2.22)

The differential form written for an elemental volume V can be simplied tot

+ (u ) = 0

t + j(u j ) = 0(2.23)

Note the interchangeable use of u and V, again.

2.3.2 Derivation of Navier-Stokes Equation

Lecture 2.4 Navier Stokes Equation is a force balance


1. What are the integral and differential forms of the Newtons second law of motion as appliedto a continuum? What is the differential form called as?

2. Interpret each of the terms in the Navier-Stokes equation with respect to the term obtainedfrom the Newtons Law.

3. How is a gradient of a stress tensor related to a net force ?

In Equation ( 2.2), for P = MV , = {u, v,w}. We now have to identify the forces. The forces inEquation (2.2) acted on the system. What are the forces acting on the control volume? Recall that theimplication of the differential equation is an instantaneous balance at time t , that is the LHS and RHSof conservation equations are evaluated at time t . But at time t the system coincides with the controlvolume. Therefore the forces acting on the system (that appear on the RHS) can also be taken to bethe forces on the control volume

F sys

=

F CV

The forces acting on the uid volume can be of two types:

F = F Surface + F Body

The total surface force can be obtained from Equation ( 1.55) by integrating over the control surface(or the control volume, using divergence theorem):

F s = CS

dA (n ) = CV

dV ( ) (2.24)

The local body force is to be integrated over the volume

F B =

CV

dV g

http://-/?-http://-/?-


13/72


For simplicity we consider only one component of the momentum, viz., in the x-direction. For this,from Table 2.1 we have P = Mu ; = u. The integral form for the x-momentum conservation reads

dM sys udt

=t

CV

dV u + CS

dA (n V) u = CS

dA (n )x + CV

dV g x (2.25)

Note the usage of V , which is the surface velocity and the x-component of the velocity u. The x-momentum integral depends not only on u, but also on the other components coming through thedot product n V. To derive the differential form of the momentum balance, we consider an elementalcontrol volume V inside which the eld variables can be assumed to be uniform. We also use thedivergence form for the surface ux and the surface force to convert them to volume integrals. Thiswill result in all the terms in Equation ( 2.25) being multiplied by V , resulting in the differential form

u

t +

(V u) = (

)x + g x (2.26)

or a more general form in index notation

t ( ui) + j(u j ui) = j ji + g i (2.27)

Notice the ease of conversion from a scalar form of u to ui: both denote a scalar component of avector, but the equations should be read as a collection of three equations as i = 1, 2, 3. The stresstensor can be expanded in terms of the thermodynamic and viscous stresses i j = pi j + i j , to givethe Navier-Stokes equations (NSE).

t ( ui) + j(u j ui) = i p + j ji + g i (2.28)This LHS of this equation can be further simplied along with Equation ( 2.23).

t ( ui) + j(u j ui) = t ui + uit + u j jui + ui j(u j)

= t ui + u j jui + $ $ $ $ $ $

$ X 0t + j(u j) ui

This gives another familiar form of the Navier-Stokes equations

t ui + u j jui D u iD t

= i p + j ji + g i (2.29)We will spend some time recalling how this equation was derived. The LHS came from the masstimes acceleration of Newtons second law (converted to control volume variables by RTT). The LHSis therefore a term related to the acceleration of the uid packet V , and is called as the inertial term.The RHS is from the sum of forces acting on a elemental volume. i p, or for simplicity x p is thenormal pressure difference (in x-direction) across a uid packet. If the pressure gradient is non-zero,then there will be a net normal force acting on the uid packet. Similarly if there is an appropriategradient in the viscous stress, it leads to a net force. A simple illustration would be to considery yx 0, (stress acting on the horizontal surface) which will lead to a net force in the x-direction.The last term is simply due to the gravitational force.

In the case of Newtonian uids, the viscous stress is related to the local velocity gradients by

i j = j ui + i u j + 23

k uk i j (2.30)

where is the Newtonian (shear) viscosity and is called as dilatational viscosity, bulk viscosity, vol-ume viscosity, or second viscosity. The bulk viscosity is identically zero for dilute mono-atomic

gases.



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Lecture 2.5. Forms of energy 47

2.3.3 Energy balance or Heat equation

Lecture 2.5 The forms of energy can be simply stated to be potential and kinetic, each applied toa microscopic and macroscopic systems.

In order to derive the continuum version of the energy equation, we need to identify the variousforms of energy associated with matter. Because energy is transferable among these various forms,and only the sum total is conserved. There are various physical forms of energy: kinetic, potential,thermal, gravitational, light, elastic, electromagnetic. However, all these can be simply categorisedbroadly under two heads: Kinetic (energy due to motion) and Potential (energy due to interactionwith a force eld). Remember the known force elds are just four: Gravitational, Electromagnetic,Strong, and Weak nuclear forces. All the example energy forms cited above can be classied underone of these heads.

The energy of the system E sys in Equation ( 2.4) therefore consists of

Internal Energy, U (microscopic Kinetic and Potential) Macroscopic Kinetic Energy, M V 2 / 2

Macroscopic Potential Energy, M g xwhere the microscopic forms of energy have been clubbed in to one head called as the internal en-ergy, U . In a macroscopic or a continuum system, the microscopic forms cannot be distinguished asseparate. The total energy is therefore:

E sys = U +MV 2

2+ M g x (2.31)

In terms of the specic local energy density e

E sys = CV dV e (2.32)we associate specic (intensive) quantities

e = U +u2

2+ g x (2.33)

Here U represents the internal energy per unit mass. This completes the identication of forms of energy on the LHS of Equation (2.4). We now have to identify the constituent terms on the RHS.

The heat input to the system Q can be due to two forms

Surface transport (Molecular conduction),

Q s Bulk (Radiation)

The surface term can be written as an integral over the system boundary (or the control volumeboundary, since they both coincide at time t ).

Qs = CS

dA n q = CS

dA n i qi (2.34)

qi is the local heat ux vector. Note that the -ve sign is so that the integral is positive when heat isadded to the system, ie, n (outward normal) and q are in opposite directions.

The work done by the system is due to the forms of work done by the system on its surroundings.

Some typical work are



15/72


Shaft work (due to impellers or other drives), W shaft

Due to Surface forces (normal and tangential) (Molecular origin), W s Other forces: Electrical, Electromagnetic, W other ,

We will ignore the other work for simple uids (that do not carry any charge). The shaft work cannotbe quantied using the system variables, but only by the properties of the surroundings on which itacts, so we retain the term as W shaft . The surface work is very important. Remember that the systemis closed, but could be acted upon by surface forces. The system can change its shape against theseforces, therefore it does a work against these forces. The local rate of work done by an elemental areais dF V = dA i ui. Expanding the stress vector in terms of the local stress tensor and integrating overthe control surface, we get

W s =

CS

dA n V =

CS

dA n j jiui

= CS

dA (n j ji ui ni ui p)(2.35)

The RTT for energy conservation can be written from Table 2.1 and Equation (2.17) as

dE sysdt

=t

CV

dV e + CS

dA V e (2.36)

Inserting this and expressions for the work done from Equation ( 2.35) and heat input from Equa-tion (2.34) into Equation ( 2.4) we get

t CV dV e + CS dA V e = CS dA ni qi + ni i j u j p n i ui W shaft W other (2.37)

which in complete bold face (vector) notation is

t

CV

dV e + CS

dA n V e = CS

dA n (q + u p u ) W shaft W other (2.38)

Simplifying using divergence theorem, Equation ( 2.19)

t CVd

V e +

CVd

V

V e =

CVd

V

(

q +

u

p u )

W shaft

W other (2.39)

Here we have ignored bulk heat input due to radiation. For further simpler forms of the equationwe will ignore the shaft and other forms of work. It is often convenient to take the pressure term tothe LHS, and writing the density as reciprocal of specic volume V and simplifying we get:

t

CV

dV e + CV

dV

xiui e + p V =

CV

dV

xi qi + i j u j (2.40)

When written for an elemental control volume V , as for the Navier-Stokes and continuity equations,we get the differential form of the energy conservation equation for the continuum:

t ( e) + i ui e + p

V = i

qi + i j u j (2.41)



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Lecture 2.6. Species balance 49

Special forms of Energy Equation The form of equation given above is not suitable for most of theheat transfer applications involving uid ow. They can however be simplied. One of the mostcommon form is that expressed in terms of the temperature T . To obtain this, the mechanical energyequation is subtracted from the above equation. The mechanical energy is not a conserved quantity,but can be obtained from the momentum balance by multiplying each term by ui, and simplifying itwith the help of the continuity equation. The resulting equation, after subtraction, is the equation of change for the internal energy: Internal energy equation (Total Mechanical):

t U + i uiU = D U D t

= piui iqi + i j iu j (2.42)Here we have expressed the LHS in terms of the substantial derivative, or the material derivative DD t .For any scalar we can show using continuity equation that

t ()+

j(u j)=

t +

u j j+

$ $ $ $ $

$ $ X 0

t +

j(u j) D

D t (2.43)We can make further simplications to Equation ( 2.42), in terms of the thermodynamic functionenthalpy (per unit mass) H . In terms of Enthalpy,

D U D t

= D H D t

D pD t

(2.44)

dH = cpdT + V T V t p

dp (2.45)

The most useful form of the energy equation is written in tems of the temperature, which is ob-tained by substituting Equation ( 2.45) in Equation ( 2.44) and further in Equation ( 2.42). In terms of Temperature:

c p D T D t = iqi + i j iu j ln ln T p

D pD t

(2.46)

Simplied forms of Temperature Equation The temperature equation Equation ( 2.46) can be sim-plied further for various special cases.

1. Ideal gas ( p/ = R T / M w, giving ln ln T p= 1; qi = k iT )

cpD T D t

= k 2i T + iu j iu j +D pD t

(2.47)

2. Constant Pressure System, D pD t = 0 or Constant density systemT

= 0 and Newtonian viscosity

c pD T D t

= k 2i T + iu j iu j (2.48)

2.3.4 Species BalanceLecture 2.6 Species balance is similar to continuity equation but written only for one particular

species

In the case of multicomponent systems, the chemical species is conserved as well. Chemical reactionscan take a species from one compound to another. In this case we can write a mass balance for aspecies alone.

dM dt sys

= r where, M =

CV

dV (2.49)



17/72


The left hand side can be expanded by the Reynolds transport theorem, Equation ( 2.20), just as it wasdone for the continuity equation, except that the local velocity of the control surface will be u of thatof the species , and not the mass averaged velocity u = u / .

dM dt

=t

CV

dV + CS

dA (n u ) (2.50)

Adding and subtracting a term proportional to the mass average velocity u , we get for the speciesbalance an integral form

dM dt

=t

CV

dV + CS

dA [n u ] + CS

dA [n (u u )] = r (2.51)

Differential Form of Species Balance The differential form is obtained, as before, by consideringan elemental volume V where the properties are uniform. t + iui = i (u i ui)

j

+ r (2.52)

In terms of mass fraction = / , and using Equation ( 2.43)

t ( ) + i(ui ) = D D t

= i( D i ) + r (2.53)

where, Ficks law for diffusion ux has been used

j , i = D i (2.54)

Lecture 2.7 Problems in Process Fluid Mechanics can be taken to be two broad types: Macro-scopic and Microscopic.

Problems in process uid mechanics can be broadly classied under two categories: Macroscopic orsimply Macro-, and Microscopic or Micro-. Macro-problems deal with gross properties of the ow,such as total drag, work done, energy requirement, equipment sizes, etc, without requiring to ndout the variation in the velocity and stress elds. Macro-problems are usually simple to solve math-ematically, but may involve several unknowns, which need to be determined either by experimentsor by solving the microscopic problem. Micro-problems involve nding the detailed velocity andstress elds and using them to calculate other properties, which might be used in macro-problems.Micro-problems are often difcult to solve mathematically to obtain a closed form analytical solution.In such cases, a numerical (or computational) solution of the microscopic equations are obtained.Navier-Stokes equations accurately describes models the behaviour at the normal densities of New-tonian uids (only in very low densities, such as that of a plasma, or sudden variation density, as ina shock wave, does it break down). In such cases, with the availability of computational power, thecomputational route of solving the equation is fast emerging as tool to obtain the behaviour of theuids. This is called as computational uid dynamics (CFD), where experimental conditions (geometryand boundary conditions) are simulated on a computer, and the microscopic balance equations aresolved numerically.

Some macroscopic problems include

Determination of a pump specication (RPM and size), and energy required to transport a uidacross a height at a given ow rate.

Determination of the pressure drop across a bed packed with solids through which a uid has

to pass at a given ow rate.



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Lecture 2.8. Introduction Boundary Layer 51

Determination of the energy required to stir (agitate) a tank containing a liquid which is to bemixed with other components, which may be a bubbling gas, another liquid, or solids (slurry).

Determination of the rate at which a lter used to separate solids from a liquid can produce theltrate.

Microscopic problems are not microscopic in the strict sense of the word. We are still dealingwith continuum here. However, we need to nd out how the velocity and stress elds vary ordevelop inside (or outside) a solid object (equipment).

Determination of the velocity proles around a rough sphere (a cricket ball?) as it movessteadily in a uid.

Determination of local heat transfer coefcient, which is dependent on the local velocity varia-tions. A convective current towards the wall increases the heat transfer as well.

Determination of net rate of reaction on a catalyst surface which is suspended (uidised) in airby its upward ow.

The equations derived in the last section can be applied to macro- as well as micro-problems. Theintegral form (over the control volume) is used to set up a macro-problem and the differential formis used to solve the velocity, stress, temperature and concentration elds. In the next lecture we willconsider an important concept in uid mechanics, namely the boundary layer, in which both themacroscopic and the microscopic methods will be applied.

2.4 Boundary Layer TheoryLecture 2.8 Boundary layers occur in High Reynolds number ows, close to a solid boundary


1. Boundary layers is a relatively small zone close to a solid surface, in which the velocities varyrapidly in comparison with a slow variation just outside.

2. Boundary layers occur in high Reynolds number ows, i.e. when the viscous terms are negli-gible in comparison with the inertial terms.

3. A laminar boundary layer is one in which the velocities are in streamlines.

4. A Turbulent boundary layer has velocities in all directions without any uniformity.

5. The requirement of a Boundary layer is the condition of no slip imposed on the solid surface.

6. The thickness of a Boundary layer (from the solid wall) can vary along the length of the ow,often very slowly, and therefore a thickness is usually adopted by some convention.

7. The boundary layer inuences the transport from the solid to the bulk and therefore is crucialin determination of drag, heat transfer, and mass transfer coefcients.

It is convenient to start with what is known as a at-plate boundary layer in which a uid movingat a uniform velocity (therefore laminar) approaches a at plate whose normal is perpendicular tothe ow, as shown in Figure 2.4. At the up-stream end (leading-edge) of the plate the velocities areuniform, and in the down stream, very far away from the plate the velocities is again uniform. Onlyclose to the plate there is a layer where the inuence of the plate is felt. The plate exerts a drag on theuid (no-slip at the walls), which consequently inuences (drags) other layers away from the plate.The laminar boundary layer develops along the ow, and its thickness increases. The ow remains

viscous and laminar. Beyond a critical thickness, turbulence sets in: due to some disturbances in the



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y

x

Flat Plate

Turbulent Boundary Layer

Buffer Layer

Laminar sublayerLaminar Boundary Layer

Figure 2.4: Development of boundary layer over a at plate

y

x

a

d Flat Plate

b

c

Figure 2.5: Compute net ow across control surface ab .

ow or roughness in the solid surface. The boundary layer, called as the turbulent boundary layer isthen dominated by turbulent patterns (3D ows, uctuations, non-uniformity, eddies).

Even in the turbulent BL, a still smaller layer persists where the ow is still laminar. This layer iscalled as the laminar sub-layer. This is important to heat and mass transfer because this is the regionof dominant resistance: There is no cross ow, and all the transport must occur through moleculartransport: diffusion and conduction. In between the viscous layer and the turbulent layer is a layerknown as the buffer layer, where the viscous and turbulent transport are equally important.

2.4.1 Integral Analysis

Lecture 2.9 Computing the net mass ow in a boundary layer problem.

Tutorial Example A at plate of width 3 mm is approached by air ( = 1kg/m 3) having an uniformvelocity of 2 m/s, as shown in Figure 2.5. Calculate the net mass ow rate across the surfaceab . Assume that the velocity inside the boundary layer is given by u/ U = 2(y/ ) (y/ )2 , andthe boundary layer thickness at c is = 5 mm.

Given Data

= 1 kg / m3

w = 3 mm = 3E3 m = 5 mm = 5E3 m

U = 2 m/ s

Governing Equations :



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Lecture 2.9. Integral Analysis of BL 53

Mass balance Equation ( 2.22)

& &

& &

& b xt

CV

dV + CS

dA (n V) = 0

Assumptions :

x Steady owy Laminar owz FM convention for normal (outward){ Incompressible ow (constant ).

Detailed calculations : The integral over the control surface can be evaluated in four parts over thefour surfaces

CS = ab + bc + cd + ad Before evaluating each of the integrals, let us do a qualitative analysis of the problem. Whyshould there be a ux across ab ? If at all there is one, in which direction will the net mass owbe? The total mass ow across ad (uniform velocity) is more than that in bc (reduced from theuniform). There is no ux of mass across cd (no-slip), therefore there must be a net out uxacross ab . This is only a guess, we could be wrong.

To evaluate the integral we need to know the variation of the local velocity vector V and thenormal n . We rst choose the origin and direction of the coordinate system, as shown in Fig-

ure 2.5.Across ab we do not know the velocity distribution (not given, so do not assume any). All wecan say is that ux = U across ab . We do not know about uy. Across bc , we only know ux, not uy.Across cd , V = 0 because of no-slip, and nally across ad , V = {U , 0}. With this information andwith the direction of the local normal we can evaluate the four integrals

1. Across ab , n = {0, 1}, points in the y direction, so the only component that will matter is uy,but since this is not known, we denote the integral as the total mass ow rate

mab ab

dA uy

If it is positive (or negative) then the specic mass ux V is in the same (or opposite)direction to the outward normal n , respectively.

2. Across bc , n = {1, 0}, and only ux matters. uy, even though unknown, does not contributeto the integral. Substituting the given expression for the variation in the velocity andintegrating we get

bc

dA ux = w

0

dy U 2y

y

2=

2 U w 3

Note that since the outward normal and the velocity are in the same direction, this integralevaluates to a positive quantity.

3. The integral over cd evaluates to zero because velocity is zero due to the no-slip boundary.



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y

x 0

Flat Platec

b

a

d

Figure 2.6: Net mass ux across control surface ab .

4. The integral over ad also has contribution only from ux = U , however n = {1, 0}in thenegative x direction. The part integral is given by

ad

dA (n V) = ad

dA U = w

0

dy U = U w

Putting all the integrals together

mab + bc + cd + ad = 0we get,

mab = U w 23

U w

=13

U w

Since the mass ux across ab evaluates to be positive, it ( V) is in the direction of the normal n ,and there is a net out ux of mass from the surface ab , in agreement with our initial guess.

Symbolic Expression for the Unknown :

mab =13

U w

Numerical : Substituting the values the net mass ux out of the surface bc is

mbc = 10E6 m3 / s = 10cc / s

Lecture 2.10 There is a net inux of mass into the boundary layer

In the same situation as considered in the previous problem, consider a slightly different problem of the mass ux across the boundary layer, as shown in Figure 2.6. In this problem we wish to nd outthe net mass ux across the boundary layer. This is different in that when the control surface ab isalong the boundary layer, we get the net ux across the curved BL, and not a horizontal surface asbefore.

To solve this problem, we consider two thickness 0 at x = d and at x = c. The imaginary control

volume is enclosed by is abcd . Most of the derivation is same as that for the previous case. We will



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Lecture 2.11. Integral Momentum for Boundary Layer 55

+d

d

w

p(x)

p(x+dx)

dx

x x + dx

b

a

d c

n

V

Figure 2.7: Combination of differential and integral analysis in Boundary Layer: Differential lengthalong x and nite length along y

only point out the difference here. We are still wish to obtain an expression for the mass ux acrossab , denoted by mab . The normal to this surface changes with x, but the following expression is stillcorrect

mab ab

dA n j u j

The difference is that we cannot write it in terms of known n j as before. The integral across bc andcd are the same, but across ad , the integral is similar to that of bc but over a different boundary layerthickness 0 .

ad

dA u x = 2 U w 0

3

Combining all the uxes we get

mab =23

U w (0 ) (2.55)Taking > 0 as an observable fact, we nd that mab < 0. Recall that mass ow rate is positive alongthe direction of the outward normal (mass ux per unit area V is in the general direction of thenormal). Therefore the above equation implies mab < 0, and there is a net inux into the boundarylayer. This would seem immediately surprising, since we had just derived a relationship where therewas a ux outside surface ab . But on hindsight we can see that in the earlier case we had integrateda at velocity prole (upstream) against a decient velocity prole (downstream). Here however, weare comparing a more decient (upstream) against a lesser decient (downstream).

Lecture 2.11 The mass and momentum balances is done on a differential element along the di-rection of ow and a nite length along the boundary layer thickness.

We will now carry out an analysis which is differential in one direction and integral in another. This isparticularly useful in interpreting some qualitative (but very important) features of boundary layers.Consider control volume shown in Figure 2.7, which is a section of the boundary layer between

lengths x and x + dx, with the thickness going from to + d. We list the directions and magnitudes



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of known vector quantities across the four surfaces. n ab = {sin , cos }, nbc = {1, 0}, n cd = {0, 1},n ad =

{1, 0

}and Vab =

{U , ?

}, Vbc =

{u, ?

}, Vcd =

{0, 0

}, Vad =

{u, ?

}.

Applying the continuity equation to this system, and without assuming any specic velocityprole (unlike the previous examples) the general expression for the mass ux across ab can be shownto be

mab ab

dA (n V) = ad

dA (n V) bc

dA (n V) = w

0

dy u w+ d

0

dy u (2.56)

Here u denotes the x-component of the uid velocity. The second term can be expanded in Taylorsseries about x

mab = w

0

dy u w

0

dy u +

xw

0

dy u dx = w dx

x

0

dy u (2.57)

Note the x-direction is a differential balance and the y direction is an integral equation.For the integral momentum balance at steady state, we write down each of the terms from Equa-

tion (2.25). The surface integral on the LHS

CS

dA (n V) u = ab + bc + cd + ad (2.58)The integral across the two surfaces ad and bc can be written, similar to the continuity equation, as

ad

+

bc

= w

0

dy u u + w

0

dy u u +

xw

0

dy u u dx = w dx

x

0

dy u u (2.59)

This is because, across ad and bc the only component in V that is effective is u, along the direction of the normal. The integral over cd is zero as the velocities are zero. Across ab , we note the following.The direction of the normal (the angle ) is not known a priori, and the velocity V over the surface hasboth the components u and v non-zero in general. Therefore no simplication can be done on n V.However the other velocity u, appearing in the integral on its own, is equal to the local free-streamvelocity U on the boundary layer surface (by denition), therefore

ab

dA (n V) u = U ab

dA (n V) = U mab (2.60)

where the second equation has been obtained from Equation ( 2.57). Note that in general U = U (x)

can be a function of x outside the boundary layer. Only for a at plate is it a constant. Even then,for a differential element we can assume it to be a constant across ab . This completes the LHS of Equation (2.25). Next we move on to the forces.

We neglect the body forces in the x-direction. For the surface force in the x-direction

CS

dA n j jx = CS

dA (pn x + n j jx ) (2.61)

This implies p will contribute only in surfaces where nx is non-zero, namely on ab , bc , ad . On abn = {sin , cos }, and assuming the pressure at the midpoint acts uniformly on the whole of ab , weget

ab

dA (pn x) = wdx

cos p(x) +

12

dx px

(sin ) (2.62)



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Lecture 2.11. Integral Momentum for Boundary Layer 57

This can be further simplied with the local slope tan

ab dA (pn x) = w dx pddx

+

B 0

O px

ddx

(2.63)

Across the other two surfaces the pressure term can be simplied to

ad

dA (pn x) + bc

dA (pn x) = w p w + dx x p + dx px

= w dx p x w dx p

x

+

B 0

Odx2(2.64)

Combining the three integrals we have for the pressure force term

CS dA (pn x) = w dx p x

(2.65)

where, higher order derivatives have been neglected as the leading term in the nal expression is

O(dx x p).The viscous stress term can be expanded to give nx xx + ny yx. For an incompressible uid xx = 0.So the only component we have to consider is the shear stress yx, which has no effect on ad and bcwhere ny = 0. On ab , ny is not zero, but outside the boundary layer yu = 0, as u saturates to U .Therefore yx = yu = 0 on the surface ab . So the shear stress has contribution only in the bottomsurface cd , where ny = 1.

CS

dA n j jx =

cd

dA ny yx w dx w (2.66)

where w is the local average wall shear stress over the length dx, dened by the above equation.Combining the LHS (from Equations ( 2.58) to (2.60)), the expression for mab from Equation ( 2.57),

the pressure term Equation ( 2.65), and the viscous term Equation ( 2.66) we get

w dx

x

0

dy u u U w dx

x

0

dy u = w dx px w dx w

which can be simplied, for incompressible ow, to

x

0

dy u2 U (x)

x

0

dy u =

px

w

(2.67)

This is the general momentum balance equation of the boundary layer, where there can be a variationin the free stream velocity U (x), and the pressure p(x) outside the boundary layer.

Find Out Yourselves

1. When can xx have nonzero values?

2. List all the assumptions made when simplifying the momentum balance to the boundary layercase in arriving at Equation (2.67)

2.4.2 Laminar and Turbulent Boundary Layers



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Lecture 2.12 Integral Momentum analysis across a boundary layer gives the scaling behaviourof the boundary layer thickness and the wall drag with length.

Equation (2.67) can be simplied further for a at plate boundary. In this case, the velocity distribu-tion outside the boundary layer is uniform U (x) = U = constant. In the absence of velocity gradients,the NSE Equation (2.29) shows that pressure distribution is uniform as well ( i p = 0). From Equa-tion (2.67) we then have

x

0

dy u2 U

x

0

dy u = w

(2.68)

To solve this integro-differential equation, we need to know the function u = u(x, y), which requiresthe solution of the NSE Equation ( 2.29). However, we will adopt a simplied approach, but will getsome important qualitative predictions. We assume that the function u(x, y) can be written of theform

u(x, y) = U f (x, y) = U f y

(x)

uU

= f y

(x) f ()(2.69)

where the dependence on x is through , and = y/ ( is pronounced as /eeta/, and is a variantof the normal /fai/) . This form is dimensionally consistent. On the LHS we have a dimensionlessnumber, it has to be a function only of dimensionless parameters, as the resultant has to be a number.A function of this form is called as a similarity function, because the form of the function remains thesimilar for all x. Though it is difcult to justify this assumption now, we take it as a plausible guess.The momentum balance Equation (2.68) can be simplied as

ddx

1

0

d (1 ) =

U dd = 0

(2.70)

where we have used the shear stress at the wall to be

w = uy y= 0

(2.71)

We now assume a velocity distribution function that satises the boundary conditions:

y = 0, u/ U = 0

y = , u/ U = 1

y = , yu = 0

A simple choice is a quadratic function

=uU

= 2y

y

2= 2 2

Substituting in Equation ( 2.70) we getddx

=15 U

Solving for we get

2 =30

U x



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A value of n = 7 is commonly used. In order to nd the scaling behaviour of the boundary layerthickness and the skin friction coefcient, we can use this expression.

= 1/ 7

This will not be a problem in evaluating the integral terms appearing on the LHS of Equation ( 2.70),however, the wall stress term will result in

dd

=1

7 6/ 7(2.73)

which diverges for = 0 (at the wall). Therefore we need to look for alternate expressions for thewall shear stress. One way to obtain this is to use experimental correlations observed for a smoothcircular cross section pipe and use it for a at plate. For a smooth pipe the wall stress is given bythe Fanning friction factors correlations, that relates the dimensionless wall stress to the Reynoldsnumber. This is empirically obtained from the Blasius equation

f F w

12 U

2= 0.079 Re 1/ 4 = 0.079

D U 1/ 4

(2.74)

where U is the mean velocity in the pipe. In order to use this expression for a at plate, someapproximations have to be made. Firstly recognise that there is no mean velocity U for a at plate.The skin friction coefcient is given in terms of the maximum velocity outside the boundary layer U .These two are related by

U =1

R2 dA u 1 R2R

0

dr 2 r 1 r R

1/ n

For a power law velocity prole, assuming the prole to extend from wall to centerline (the viscous

boundary layer is small) the relationship can be shown to beU U

=2 n2

(n + 1)(2 n + 1)

For n = 7 this evaluates to U = 0.817 U . The expression for w can be written down in terms of theBlasius equation Equation (2.116) as

w12 U

2= 0.0466

R U

1/ 4

Assuming that this law applies to the at plate boundary layer with R = , we get

w1

2 U 2

= 0.0466

U

1/ 4

(2.75)

This expression is directly substituted in Equation ( 2.68), and the integrals on the LHS are evaluatedusing the power law prole Equation (2.73) to give:

772

ddx

= 0.0233

U

1/ 4

which can be solved for to givex

= 0.381 Re 1/ 5xThe skin friction can be obtained from Equation (2.75)

C f = 0.0593 Re 1/ 5x

The boundary layer thickness for laminar and turbulent ows is compared in Figure 2.8.



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Lecture 2.13. Turbulent Boundary Layers 61

0

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

0 0.2 0.4 0.6 0.8 1

LaminarTurbulent

x

Figure 2.8: Developing boundary layers as computed from the laminar and turbulent ow cases, forRe x = 1E5 at x = 1.



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Lecture 2.14. Displacement Thickness 63

Fluid Flow in a Channel

Cross SectionVelocityProfiles

Figure 2.10: Illustration of displacement thickness in a channel ow.

where U is dened as U = 0 for 0 y , and U = U for y . Equating the two deciencies(Equations ( 2.76) and (2.77)) we get an expression for in terms of the local velocity prole

=

0

dy 1 uU

is a measure of the boundary layer thickness, which depends on the local velocity prole u and nota single point value (such as u/ U = 0.99). For example for a quadratic velocity prole we get = / 3.

The illustration of the displacement thickness is better illustrated in a uid ow in a channel.Consider ow in a channel as shown in Figure 2.10; for simplicity we take the channel to have onlytop and bottom wallsand is innitely wide down the depth. The actual velocity prole, with twoboundary layers on the top and bottom of the channel, is shown in blue colour. The red colouredcurve has the same amount of mass ow rate deciency at each of the walls as does the blue curve.Note that the red and blue curves have the same free stream velocity. The displacement thickness is dened such a way that the loss horizontal of mass ow rate in the boundary layer is equal to thesame loss that happens in an ideal uid with the same free stream velocity but with the boundarydisplaced by .



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2.4.3 Differential Analysis

Lecture 2.15 Ordering analysis is an method to obtain approximate solutions to algebraic anddifferential equations.

So far we were only concerned with the general behaviour of the boundary layer thickness. Weassumed some general prole for the velocity in the boundary layer. In this section we seek theexact solution of the velocity prole. For this we have to turn to the differential balance of mass andmomentum.

The governing differential equationsof uid mechanics, the Navier-Stokes equation, Equation ( 2.29)is a non-linear equation in the velocity, because of the terms u j jui. In some situations, this term isidentically zero, as in the case of unidirectional laminar ow and can be neglected in low Reynoldsnumber ows. But in most other situations this term is the origin of the difculties in obtaining a so-lution to the problem. Particular examples are boundary layer, diverging, converging, and turbulentows. In these situations there are two ways to obtain the solution: one is a numerical method and

the other is to use approximate methods. Here, we will introduce one such approximation technique,which was initially applied to the viscous boundary layer problem. Similar techniques used in otherphysical phenomena are also called as boundary layer solutions.

2.4.3.1 Asymptotics and Scaling Analysis

The basic idea is to obtain a simplied set of equations, for which exact solutions can be computed.First the governing equations are simplied by neglecting terms. This renders the equations tractable(meaning a solution can be obtained). The underlying assumption is that an exact solution to theapproximate equations is also an approximate solution to the exact equations. This is mostly true, andis valid only in the parameter space where the approximations are valid, but there are interestingcounter examples where it fails (See Denn , 1980, Chapter 11, for example).

Example 1: Algebraic Equation We will illustrate the concept by a simple algebraic equation. Weconsider determining the roots of a cubic equation

x3 4.001 x + 0.002 = 0 (2.78)This solution to equation is equation is formidable, unless you can recall the formula for the cubicroots, but you will see that an approximate solution can be obtained quickly and at the back of anenvelope! Though we know the solution to the roots of the equation ( x = 2, 2.0005 , 0.0005 ), wewill adopt an approximate scheme to simplify the equation and then obtain an exact solution to it(approximated equation).

Firstly, we observe that there are some decimal numbers with some kind of insignicance. Forexample, 4.001

4, 0.002

0 are some initial guesses. But such approximations need to be carried out

carefully. One way to do it is to try and parametrise the equation. In this case only one parameter isrequired, let us call it as (pronounced /epsilon/). is a small parameter, we rewrite Equation ( 2.78)as

x3 (4 + ) x + 2 = 0 (2.79)and in this case = 0.001 . Some points about the expected solution are in order

The solution to x is a function of the parameter : x = f ( ).

is a small parameter in comparison to 1, denoted as 1.

Since the complete problem is difcult to solve, we seek an approximate solution in the limit 0.

In the asymptotic approximation scheme, we



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Lecture 2.15. Scaling and Ordering Analysis 65

1. Make a guess on the behaviour of x or equivalently of f ( ) as 0.2. Carry out simplications of the equation3. Seek an exact solution

4. Check if the solution is consistent with the initial guess.

We will use the notation O(.) (read as order of) to denote the asymptotic behaviour of a function,for example:1. x = f ( ) O(1) as 0, meaning x approaches a constant of order unity.2. x = f ( ) O( ) as 0, meaning x approaches zero in a similar rate as does . If we Taylorexpand the solution about zero, then the rst term would be proportional to . Eg: sin O( )but cos O(1).3. x = f ( ) O1 as 0, meaning x diverges to in a similar rate as does 1/ .4. In general we can writex = f ( ) O( n), where means asymptotically equal to, if

lim 0

f ( ) n

= Constant

If f ( ) is written as a Taylor series in , then it would be

f ( ) = n f n + n+ 1 f n+ 1 + . . .

where f n and f n+ 1 are all constants. f n is called as the leading term or the dominant term in theseries. A pictorial illustration of the asymptotic behaviour of a function is shown in Figure 2.11.

5. If x O( ), then x2 O 2 : it is the leading term in a product of two Taylor expansions.6. So far we used O() for the asymptotic behaviour of dimensionless numbers x = f ( ) in termsof some other small dimensionless number . For a dimensional physical quantity, O(.) is alsoused to indicate the physical order of magnitude. When we say the velocity u is u O(1) m/s,we imply that u is within one order of magnitude from 1 m/s, i.e., 0.3 m/s u 3 m/s. In a log

scale one order of magnitude represents one linear unit, and 3 ( 100.5) is approximately half way between 1 and 10 orders. In linear scale just as we round off any 0.5 x < 1.5 as x 1, inlog scale or in physical order of magnitude analysis we say for any 3 u 30 as u O(10).Let us carry out an asymptotic analysis of the Equation ( 2.79). In asymptotic analysis we have to

make a series of guesses if required (if we have not found all solutions, or if our guess is inconsistent)

1. Let x = f ( ) = O(1) as 0, then the terms in Equation ( 2.79) have orders as under.Terms: x3 4x 4 x + 2 O(.) : 1 1

Since 1, we can neglect the third and forth term in comparison with the rst two and weget

x3 4x = 0

x (x2 4) = 0which can be easily solved. x = 0 and x = 2. The rst solution is not acceptable accordingto our assumption. as x approaches only a constant O(1). But the other two solutions areconsistent.



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0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

Function

Asymptotic

f()

Figure 2.11: Asymptotic behaviour of a function. Here the function is f ( ) = 1e 2

and its asymptoticbehaviour is 2 . Close to zero, the asymptotic closely matches with the function.

2. Let x O( ), then Terms: x3 4x 4 x + 2 O(.) : 3 2

Neglecting the rst and third terms in comparison with the second and forth since ( 1 3 2) we get

4x + 2 = 0x = 2

This is consistent with our initial assumption. We have therefore found out all the roots of theequation.

The asymptotic roots of equation are: x = 2, 2,/ 2, with = 0.01 . Compare this with the exact roots:x = 2, 2.0005 , 0.0005 . The rst root and the third roots are exact; the second root is also correct withinthe order of the approximation used. When we assumed a root to be O(1), we had ignored terms of O( ), and 0.0005 = /2 is O( ).Lecture 2.16 Singular perturbations are asymptotic problems, where by the very nature of the

assumption of smallness, one of the solutions lost

Example 2: Differential Equation Now we will take up the case of a differential equation. We arenot considering a simple equation here but one that is called as a singular perturbation problem. Inthis the small parameter multiplies the highest derivative, therefore the as 0, if that term has tobe neglected then all the boundary conditions cannot be satised.

Consider the differential equation for u(x)

u + u = 1 subject to u(0) = 0, u(1) = 2 (2.80)

where u denotes dudx . We are seeking solutions of the form u = u(x; ) as 0. For functions, u O(1)implies at constant x, u = u0(x) as 0, without any leading dependence on . This means thefunction u(x; ) can be written asu(x; ) = u0 (x) + u1 (x) + 2 u2(x) + . . . (2.81)

Similarly, u O( ) implies we can expand the function u(x; ) for all x asu(x; ) = u1(x) +

2u2(x) +

3u3(x) + . . .



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Lecture 2.16. Singular Perturbation 67

andlim 0

u(x; )

= u

1(x)

With this brief introduction to the asymptotic behaviour of functions, we begin the guess work forthe solution of the differential equation.

1. Guess u O(1). This implies u, u O(1). This can be easily seen by taking the derivatives of Equation (2.81). The orders of the terms at constant x in Equation (2.80) read:Terms: u + u = 1O(.) : 1 1

Neglecting terms of O( ) in comparison with 1, we see u = 1, or u = x + c. Notice immediatelythat the second order differential equation has been reduced to a rst order, and there is onlyone boundary condition to be satised. The two possible solutions are

uouter = x (2.82)uouter = x + 1 (2.83)

(2.84)

For the second boundary condition to be satised, the second order term must be retained inthe equation, and this cannot happen if u O(1). Let us try other possibilities.

2. Guess u O( ), this gives u, u O( ). The order of each of the terms give.Terms: u + u = 1

O(.) : 2 1

Neglecting 1, this leads to a inconsistent equation 0= 1. Similar inconsistencies occur for

u O( n ), for n > 1.3. For u O(1/ ), we get u = 0, which again leaves out the highest order derivative, and leads toa solution u O(1), inconsistent with the initial assumption.

So far we have been seeking solutions in a parameter separated form: and u(x) were occurring asproducts in the solution. Another possible guess that is usually made is to assume a form in which xand occur as products, and the solution is a function of this product. We seek solution of the form

u(x; ) = u(X ; )

where X = x n is a new independent variable.

1. We rst guess X = x/ . The new variable is X O(1), this implies the old variable is in therange {0, }. The new variable is the stretched independent coordinate, as though we arezooming in a small region close to zero (Compare Figure 2.12 and Figure 2.13). Next, weguess u(X ; ) O(1) implying

u(X ; ) = U 0(X ) + U 1(X ) + 2 U 2(X ) + . . .

= U 0x

+ U 1x

+ . . .(2.85)

so that

lim 0u(x; )

=

U 0x

at constantx



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Note that x/ is kept constant, whereas earlier it was only x that was kept constant as 0.With this we have for the higher order derivatives of u from Equation (2.85) asu(x) = 1

U 0

x

+ U 1x

+ H.O.T O1

u(x) = 1 2

U 0x

+1

U 1x

+ H.O.T O1 2

Carrying out a balance of terms of equal magnitude in Equation (2.80)

Terms: u + u = 1O(.) : 1 1 1

we getu + u = 0

This solution to this equation is sought in the stretched independent coordinate (because theequation is valid only under that assumption)

X x

which gives a solutionuinner = A eX + C

This solution is consistent with our initial assumption u(X ; ) O(1) . This solution is calledthe inner solution because the new the variable x = X goes from 0 to . Satisfying the innerboundary condition u(0) = 0 we get

uinner = A (1 eX ) = A (1 ex/ ) (2.86)Since the boundary condition at zero is satised by this solution, we take the other bound-ary condition to be satised by the solution found earlier in Equation ( 2.83) (discarding Equa-tion (2.82)). Note that the constant A for in inner solution cannot be determined by the outerboundary condition u(1) = 2, as the solution is itself valid only in a small region close to theboundary, x O( ).

We have thus got two solutions: outer in Equation (2.83) and inner in Equation (2.86), with onearbitrary constant A. Note that

the outer solution is valid for 0 with xed x, and the inner solution is valid for 0 with xed x/ .

One way to determine this constant is to do a matching at a boundary. Here, we simply set thefar eld value of the inner solution to the zero eld value of the outer solution. The outer solutionapproaches 1 as x 0. For the inner solution the far-eld X gives uinner (X ) = A. Matchingthese two we get A = 1 and the matched solutions are

uinner = 1 ex/ uouter = x + 1

This matching of two asymptotic solutions are called as, you guessed right, Matched Asymptotics.Figure 2.12 and Figure 2.13 show the extent to which the matched solutions agree with the exactsolution. The exact solution to this case is analytically solvable and is given by

uexact = x +1 ex/ 1

e1/



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Lecture 2.16. Singular Perturbation 69

0

0.5

1

1.5

2

0 0.2 0.4 0.6 0.8 1

Outerinner A=0.5inner A=1.0Exact

x

u

Figure 2.12: The inner, outer, and the exact solution to Example 2 differential equation. Here = 0.01 .

0

0.2

0.4

0.6

0.8

1

1.2

0 0.02 0.04 0.06 0.08 0.1

Outerinner A=0.5inner A=1.0Exact

x

u

Figure 2.13: The inner, outer, and the exact solution to Example 2 shown zoomed in a region close tothe boundary x = 0. Here = 0.01



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L

D

x

yU

Figure 2.14: A general boundary layer ow past a solid object.

Lecture 2.17 Differential Analysis of Boundary Layer Problem results in a Singular Perturbationproblem.


1. Differential Analysis of Boundary layer gives the scaling of the boundary layer thickness, with-out actually fully solving the differential equation:

x / U .2. Inside the boundary layer, the pressure variation along the boundary layer (along x) is much

larger than that across the boundary layer (along y).

3. px inside the boundary layer equals px outside.

4. The pressure outside the boundary layer is obtained by the solution of the inviscid (Euler)equations.

We now apply the governing equations of uid ow to the boundary layer problem. We willconsider a general two-dimensional boundary layer ow, in which the external ow eld need notbe uniform (This can happen for example for a ow past a sphere, converging/diverging duct, airfoiletc). In the Figure 2.14 L and D represent the two length scales in the x and y directions respectively.L = D corresponds to a sphere, L > D corresponds to an air-foil, and L

D corresponds to a atplate,

The governing equations at steady state can be written down

xu + yv = 0 (2.87)

u xu + v yu = 1

x p + 2xu + 2yu (2.88)

u xv + v yv = 1

y p + 2xv + 2yv (2.89)

There are two boundary conditions to be satised for each velocity eld

u (surface ) = 0 inner boundary condition (2.90)u (

far eld) =

{U , 0

} outer boundary condition (2.91)



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Lecture 2.17. Boundary Layer Equations 71

In general the velocity is two-dimensional, and varies in both x and y.

http moodle.iitb.ac.in moodle 2009 file.php file= 4522 viscous fm lectures

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