A dissertation under topic titled on “AN ANALYTICAL STUDY OF BOUNDARY LAYER

THEORY”. In partial fulfilment of the requirements for the award of the degree of master
of science in mathematics on the elective paper Boundary Layer Theory- “DSE301B”.

SUBMITTED BY SUBMITTED TO
NAME: - SUVENDU SEKHAR GIRI UNIVERSITY DEPARTMENT OF
REGD.NO/YEAR: - KU2021030328/2021 MATHEMATICS, KOLHAN UNIVERSITY,
UNIVERSITY ROLL NO: -222606529940 CHAIBASA, 833201
COLLEGE ROLL NO: - 24
SUBJECT: - MATHEMATICS (M.Sc. 3 RD SEMESTER)
SESSION: - 2021-2023
 I hereby certify that the work which is being presented in the dissertation entitled "AN
ANALYTICAL STUDY OF BOUNDARY LAYER THEORY" in the partial fulfillment of the
requirement for the award of the degree of Master of Science, submitted in the department
of mathematics Kolhan University Chaibasa, bonafide work carried out under the
supervision of Mahendra Kumar Rana.

SUVENDU SEKHAR GIRI

ROLL NO. – 222606529940

REGD.NO. /YEAR –KU2021030328/2021

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my knowledge.

Signature of the external Signature of the Guide

 ACKNOWLEDGEMENT

I would like to express my special thanks gratitude to my mathematics

teacher Dr. B.K. Sinha Head of P.G. Department of Mathematics, Kolhan
University Chaibasa and Dhiraj Viswakarma who give me the excellent
opportunity to do this wonderful project on the topic of “AN ANALYTICAL
STUDY OF BOUNDARY LAYER THEORY”. which also helped me in doing a lot of
research and I came to know about so many new things.

I am really thankful to them. Secondly, I would also like to thank my

parents and friends who helped me a lot of finishing this project within the
limited times. I was able to create my project and make it good and enjoyable
experience.

THANKS AGAIN TO ALL WHO HELPED ME

NAME : SUVENDU SEKHAR GIRI

REGD. NO/ YEAR : KU2021030328/2021

UNIVERSITY ROLL NO. : 22266065229940

SUBJECT : MATHEMATICS (M.SC 3RD SEMESTER)
SESSION : 2021 - 2023
 DECLARATION

I here by declare that the project report entitled “AN ANALYTICAL STUDY OF
BOUNDARY LAYER THEORY " submitted by me to P.G. DEPARTMENT OF KOLHAN
UNIVERSITY, CHAIBASA. The requirement for the award of the degree of M.Sc. in
MATHEMATICS DEPARTMENT is a record of bonfide project work carried out by me under
the Supervision of Dhiraj Viswakarma

I further declare that the work reported in this Project has not been submitted
and will not be submitted, either in part or in full, for the award of any other degree or any
other institute or University.

Place: Signature of the Candidate

Date:
 CONTENT
Chapter ABSTRACT
Introduction 1
1.1 General Description 1
CHAPTER -01

1.2 Isotrophy 1
1.3 Some basic properties of the Fluid 1-5
1.4A Viscous and inviscid fluids 6
1.4B Viscosity 7-8
1.4C Newtonian and Non Newtonian fluids 8
1.4D Real and Ideal fluids 8
1.5 Some important types of flows 9 -10
Boundary Layer Theory 11 -33
2.1 Introduction 11
2.2 The main Limitation of Ideal ( non – viscous ) Fluid dynamics 11 - 12
CHAPTER- 02

2.3 Preandtl’s boundary layer theory 12 -14

2.4 Importance of preandtl’s boundary layer theory in fluid dynamics 14
2.5 Some Basic Defitinion 15
2.5A Boundary layer thickness 15 - 25
2.5B Displacement momentum and energy thickness for axially symmetric flow 25
2.6 The boundary layer equation in two dimension flow 26 - 32
2.7 Conclusion 33
Thermal Boundary Layer 34 - 49
3.1 Introduction 34
CHAPTER- 03

3.2 Forced convection definition 34

3.3 Thermal Boundary Layer Equation in Two Dimensional Flow 35 - 38
3.4 Temperature Distribution in The Spread of a Jet 38 - 39
3.5 Plane Free Jet 39 - 42
3.6 The Plane Wall Jet 43 - 48
3.7 Conclusion 49
Flow of compressible viscous fluids 50 - 59
4.1 Introduction 50
CHAPTER-04

4.2 One dimensional flow of a compressible viscous fluid 50 - 51

4.3 Laminar flow of a compressible viscous fluid through a circular pide 52 - 55
4.4 Laminar steady flow of a compressible viscous fluid through between two 55 - 58
concentric rotating cylinders
4.5 Conclusions
59
 Flow of Inviscid compressible Fluids, GasDynamics 60
5.1 Introduction 60
5.2 Some thermodynamics relations for a perfect gas 61 - 67
CHAPTER – 05

5.3 Basic equations of motion of gas 67 - 69

5.4 Basic equations for one-dimensional flow of a gas 69 - 71
5.5 The one dimensional wave equation 71 - 75
5.6 Wave equations in two and in three dimensions 75 - 77
5.7 Spherical waves 77 - 78
5.8 Conclusions 79
VISCOUS INCOMPRESSIBLE FLOW (FUNDAMENTAL ASPECTS) 80 - 87
CHAPTER – 06

6.1 Introduction 80
6.2 Laminar and turbulent flows
6.3 Fully developed flow 81 - 83
6.4 Laminar and turbulent shear
6.5 Turbulent velocity profile 83 - 87

BIBLIOGRAPHY
 CHAPTEER -01

INTRODUCTION
1.1 General Description

Field dynamic or Hydrodynamics is the branch of science which is concerned with the study of
the motion of fluids of that of bodies in contact with fluids. Fluids are classified as liquids and gases.
The former are not sensibly compressible expand to fill any closed spaces.

It is well knows that matter is made up of molecules or atoms which are always in a state of
random motion. In fluid dynamics the study of individuals is necessary nor appropriate from then
point of view of use of mathematical methods. Hence we consider the macroscopic (bulk) behaviour
of fluid by supposing the fluid to be continuously distributed in a given space. This assumption is
known as the continuum hypothesis. This continuum concept of matter allows us to sub-divide a fluid
element indefinitely. Furthermore, we defined a fluid particle as the fluid contained within the
physically infinitesimal volume.

1.2 Isotropy
A Fluid is said to be isotropic to some property (pressure, density, etc.) if that property is the
same in all direction at a point. A fluid is said to be an isotropic with respect a property if that property is
not the same in all directions.

1.3 Some basic properties of the Fluid

(1)Density, specific weight, specific volume.

The density of a fluid is defined as the mass per unit unit volume

Mathematically the density 𝜌 at a point P may be defined as

𝜹𝒎
𝝆 = 𝐥𝐢𝐦 ,
𝜹𝒗→𝟎 𝜹𝒗

1
Where𝜹𝒗is the volume element around P and 𝜹𝒎 is the mass of the fluid within 𝜹𝒗. The specific weight𝜸
of a fluid is defended as the weight per unit volume. Thus 𝜸 = 𝝆𝒈, where g is the acceleration due to
gravity.

The specific volume of a fluid is defined as the volume per unit mass and is clearly the
reciprocal of the density.

(ii) Pressure –
When a fluid is contained in a vessel, it exerts a force at each point of the inner side of the
vessel. Such a force per unit area is known as pressure.

Mathematically, the pressure 𝝆 at a point P may be defined as

𝜹𝑭
𝝆 =lim ,
𝜹𝑺

Where𝜹S is an elementary area around P and𝜹F is the normal force due to fluid on𝜹𝑺.

(iii) Temperature –

Suppose two bodies of different heat are brought into contact while isolated from all other
bodies. Then some thermal energy will move from one body into the other body. The body from where the
thermal energy moves is said to be at a higher temperature while the body into which the energy flow is
said to be at lower temperature. When two bodies are in thermal equilibrium then they are said to have a
common property, known as temperature T.

(iv) Thermal conductivity -

The well - known Fourier’s heat conduction law state that the conductive heat flow per unit
area (or heat flux)𝑞𝑛 , is proportional to the temperature decrease per unit distance in a direction normal to
the area through which the heat is flowing thus

Mathematically

𝝏𝑻
𝒒𝒏 ∝
𝝏𝒏
.

𝝏𝑻
So that, 𝒒𝒏 = -k
𝝏𝒏
2
 Where k is said to be the thermal conductivity.

(v) Specific heat –

The Specific heat C of a fluid is the amount of heat required to raise the of the temperature of a
unit mass of the fluid by one degree. Thus C =𝝏𝑸/𝝏𝑻, where𝝏𝑸is the amount of heat added to raise the
temperature by𝜹𝑻. The value of the specific heat depends on two well-known process – the constant
volume process and the constant pressure process. The specific heat of the above processes are denoted
and defined as

Specific heat at constantvolume =∁𝒗= (𝝏𝑸/𝝏𝑻 )𝑽,

Specific heat at constant pressure =∁𝑷=(𝝏𝑸/𝝏𝑻) 𝑷 ,

∁𝐏
Ratio of these two specific heats is denoted by .Thus,𝛄= .
∁𝐯

(vi) Incompressible and compressible fluids

Gases are compressible and their density changes readily with temperature and pressure
Liquids, on the other hand, are rather difficult to compress and for most problems we can trust them
incompressible. Only in such situations as sound propagation in liquids does one need to consider their
compressibility.

The density of a fluid is a thermodynamic property which depends on the state of the fluid. The density𝝆
can be expressed as a function of pressure and temperature. Such a relation is known as an equation of
state . For an ideal gas the equation of state may be expressed as

P = 𝝆RT,

Where R is the characteristic gas constant. The constant R has different values for various gases and its
units has the from

R = (energy) / (mass x temperature) =J /( kg * K)

We also, have R = ∁𝑷- ∁𝒗

Where ∁𝑷 is specific heat at constant pressure and∁𝒗 is specific heat at constant volume.

(vii) Compressibility and bulk modulus

The compressibility of a fluid is expressed by the quantity bulk modulus K,which is defined as the
ratio of volumetric stress to volumetric strain. If a small increase in pressure dp causes a change dv of the
specific volume v, then by definition

3
 𝒅𝒑 𝒅𝒑 𝒅𝒑 𝟏 𝒅𝒗 𝒅𝒑
K= = = 𝝆 , as v 𝜶 𝒔𝒊𝒏 ⇒ =
−( 𝒅𝒗/𝒗) (𝒅𝝆/𝝆) 𝒅𝝆 𝒑 𝒗 𝒑

The coefficient of compressibility𝜷 is defined as 𝜷 = 1/k we now proceed to

establish relationship between the bulk modulus and the local pressure for a perfect gas for two different
processes of compression.

Relationship for isothermal process.

We have p / 𝝆 = const = C₁ …………………………... (1)

From the equation of state for a perfect gas, P = 𝝆 RT ………………….. (2)

From (2), d𝝆 / dp = RT = p / 𝝆…………………… (3)

By definition, K = 𝝆 ( d𝝆 / dp ) =𝝆 × ( p / ) = p ,……………..(4) [ by ( 3 )]

Relationship for isentropic process.

We have p / 𝝆= const = 𝑪𝟐 = 𝜸𝑷𝜸 𝑪𝟐 = 𝜸p, [by ( 4]

Classification of fluids on the basis of density and viscosity:

In the following table, fluids have been classified on the basis of density and viscosity.

For complete understanding, read articles 1.4A,1.4B, 1.4C and 1.4D

Sl Types of Fluid Density Viscosity

No.
1. Ideal fluid Constant zero
2. Incompressible fluid Constant Zero or non- zero
3. In viscid fluid Constant or variable zero
4. Real fluid Variable Non- zero
5. Newtonian fluid Constant or variable Non- zero and 𝜏 =
𝜇 (𝑑𝑢/𝑑𝑦)
6. Non-Newtonian fluid Constant or variable 𝜏 ≠ 𝜇 (𝑑𝑢/𝑑𝑦)
7. Compressible fluid Variable Zero or non -zero

4
(viii) The Mach number. Subsonic and supersonic flows.
In compressible flow there is a great distinction between flow involving velocities less than that of
sound ( subsonic flow ) and flow involving velocities greater than that of sound ( supersonic flow ) . Main
difference between subsonic and. supersonic flowsbe discussed in chapter 20. The Mach number M is
defined as the ratio of the local speed of sound. Thus,

M=V/c
Where V is the fluid velocity and c is the local speed of sound.

Classification of compressible fluids on the basis of Mach number

The Mach number gives us important information about the type of compressible flow as shown in the
following table: For complete understanding. Read chapter 20.

Sl no. Types of fluid Mach number, M

1. Subsonic fluid M<1
2. Sonic fluid M =1
3. Supersonic fluid 1<M<6
4. Hypersonic fluid M>6
5. Transonic fluid M < 1 as well as M > 1

1.4 Remark -
For flows around objects, where M is less than about 0.3, the flow is regarded as approximately
incompressible.

5
1.4A. Viscous (orreal) and inviscid (non–viscous,frictionless, perfect or ideal) Fluids.

SURFACE FORCE

SHEARING STRESS NORMAL STRESS

P
SSSS
dS

An infinitesimal fluid element is acted upon by two types of forces, namely, body forces The
former is a type of force which is proportional to the mass (or possibly the volume) of the body or which it
acts while the letter is one which acts on a surface element and is proportional to the surface area.

Suppose that the fluid element be enclosed by the surface S. Let P be an arbitrary point of S and let ds be
the surface element around P. Then the surface force on ds is, in general, not in the direction of normal P
to ds. Hence the force may be resolved into components, one normal and the other tangential to the area
ds. The normal force per unit area is said to be the normal stress or pressure while the tangential force per
unit area is said to be the shearing stress.

A fluid is said to be viscous when normal as well as shearing stresses exist. On the other hand,
a fluid is said to be inviscid when it does not exert any shearing stress, whether at rest or in motion. Clearly
the pressure exerted by an inviscid fluid on any surface is always along the normal to the surface at that
point. Due to shearing stress a viscous fluid produces resistance to the body moving through as well as
6
between the particles of the fluid itself. Water and air are treated inviscid fluids whereas syrup and heavy
oil are treated as viscous fluids.

1.4B. Viscosity (or internal friction)

We know that the flow of water and air is much easier than syrup and heavy oil. This demonstrates the
existence of a property in the fluid, which controls its rate of flow. This property of a fluid is known as viscosity. Thus,
viscosity of a fluid is that property of fluid which exhibits a certain resistance to alteration of form. All existing fluids
possess the property of viscosity in varying degrees.

To explain the nature of viscosity considers a fluid which is initially at rest between two parallel plates separated by a
small distance h along the y - direction and extended infinitely in other direction. Suppose the upper plate is moving
with velocity U in x - direction whereas the lower plate kept at rest. By virtue of viscosity, the fluid will also be in
motion.

Figure showing fluid motion between a stationary plate and a moving plate. A small element shows the shear stress.

Fluid between the plates has a linear velocity profile (provided no pressure gradient exists along the plate
in the direction of motion). It is a fact supported by experimental observations that the relative velocity
between a solid surface and a fluid is zero for all fluids. Accordingly, the bottom layer of the fluid at y = 0
will be at rest and the upper layer at y = h will be moving with the plant having a velocity U

If we consider a small element of the fluid as shown in the figure, the shear stress 𝜏 on the top (which is
numerically the same as the bottom in this case) is given by -

𝝉 = 𝝁(du/dy)

Where𝜇 is a constant of proportionality which is called the coefficient of viscosity or the

coefficient of dynamic viscosity and is known as Newton's law viscosity.

7
The viscosity of a liquid decreases rapidly with increasing temperature whereas the viscosity of a gas
increases with temperature. The viscosity of fluids also depends on pressure, but this dependence is
usually of little importance to the temperature variation in problem of fluid dynamics.

1.4C. Newtonian and non - Newtonian fluids Newtonian fluid

The fluids that obeys Newton's law of viscosity (1) of Arts.1.4B is knows as Newtonian
fluid, for example, air and water. Viscous fluids such as tar and polymers do not obey Newton's law of
viscosity and the relation between stress shear and rate of shear strain ( refer chapter 13 fordeatils ) is non
- liner . Such fluids are known as non - Newton's fluids. The nature of relation between shear stress and
rate of shear strain for Newtonian and non - Newtonian fluids is shown in the following figure.

Figure showing classification of fluids

1.4D. Real and ideal fluids:

The concept of ideal fluid is based on theoretical consideration because all real fluids exhibit
viscous property. From relation (1) of Art.1.4B, it follows that 𝜋vanishes either for 𝝁 = 0or for du / dy =0.
Hence, we see that a real fluid with small viscosity and small velocity gradient can be regarded as
frictionless. For an ideal fluid, we shall take shear stress as zero and assume that, at the surface of contact
with solid, an ideal fluid can have relative velocity in the tangential direction although normal velocity must
be zero at the surface of contact.

8
1.5. Some important types of flows
(i) Laminar ( streamline ) and turbulent flows .

A flow , in which each fluid particle out a definite curve and the curves traced out by any two
different fluid particles do not intersect , is said to be laminar . On the other band, a flow, in which each
fluid particle does not trace out a definite curves traced out by fluid particles intersect, is said to be
turbulent. The following figure illustrates laminar and turbulent flows.

Figure lines indicate the paths of particles

(ii) Steady and unsteady flows:

A flow , in which properties and conditions ( P , say ) associated with the motion of the fluid
are independent of the time so that the flow pattern remains unchanged with the time , is said to be
steady . Mathematically, we may write 𝝏𝒑/𝝏𝒕= 0 Here P may be velocity,density,pressure, temperature
etc. On the other hand, a flow, in which properties and conditions associated with the motion of the fluid
depend on the time so that the flow pattern varies with time, is said to be unsteady.

(iii) Uniform and non - uniform flows:

A flow, in which the fluid particles possess equal velocities at each section of the channel
or pipe, is called uniform. On the other hand, a flow, in which the fluid particles different velocities at each
section of the channel or pipe, is called non uniform.These terms are usually used in connection with flow
in channels.

(iv) Rotation and irrationalflows:

A flow, in which the fluid particles go on rotating about their own axes, while flowing is said to be
rotational. On the other hand, a flow in which the fluid particles do not rotate about their own axes, while
flowing, is said to be irrational.

(v) BarotropicFlow:

The flow is said to be barotropic when the pressure is a function of density.

9
1.6 Some useful results of vector analysis.
In the study of fluid dynamic the use of vector and tensor notations is become more frequent their
use not only simplifies and condenses the proposed results but also makes physical concepts easier to
understand. It is the purpose of this article to present a summary of the properties of vectors. Throughout
this book, bold face type is used to denote vector quantities.

Let a , b , c , denote vector quantities such that a =i 𝑎1 + 𝑗𝑎2 + k𝑎3 = (𝑎1 , 𝑎2 , 𝑎3 ) and |𝑎 | = a =
√𝑎1 2 + 𝑎2 2 + 𝑎3 2 etc. Then, we have

(i) Vector addition and subtractions etc.

• a + b = i (𝑎1 + 𝑏1) + j (𝑎2 + 𝑏2 ) + k (𝑎3 + 𝑏3)

• a - b = i (𝑎1 − 𝑏1 ) + j (𝑎2 − 𝑏2 ) + k (𝑎3 - 𝑏3 )
• a+b=b+a (Commutative Law)
• a + ( b + c ) = ( a + b ) + c ( Associative Law )

(ii) Scalar (dot) product of two vectors

a.b = abcos𝜃 = 𝑎1 𝑏1 + 𝑎2 𝑏2 + 𝑎3 𝑏3 ,

Where 𝜃 is the angle between the vectors a andb.

a. (b + c) =a.b + a.c(Distributive law)

i.j = j.k = k.i = 0 ,i.i = j.j = k.k = 1

(iii) Vector (cross) product of two vectors.

a× b =ab sin𝜃 n

Where𝜃 is the angle measured from a to b and n is unit vector normal to the plane consisting of a and b
such that a,b and n form a right handed system.

10
 CHAPTER – 02

BOUNDARY LAYER THEORY

2.1 Introduction

The Navies - Stokes equations were obtained in Art. 14.1. In Art. 16. 1. it was observed that a
complete solution of these equations has not been accomplished to date. This is particularly true when
friction and inertia forces are of the same order of magnitude in the entire flow system, so that neither can
be neglected in chapter 16, we discussed some very special CASES OF FLOW PROBLEMS FOR WHICH EXACT
OLUTIONSOF THE Navies – Stokes equations are possible, in those cases, the equations were made linear
by taking a simple geometry of flow and assuming the fluid to be incompressible.

In chapter 17 we dealt with a case of the approximate solutions of the Navies Stokes equations
for very small Reynolds’s number. In that chapter the friction forces far overshadowed the inertia forces,
and the equations became linear by omitting the convective acceleration. The present chapter discusses
the opposite, i.e., flow characterized by very large Reynolds’s numbers.

2.2 The main limitations of ideal (non - viscous) fluid dynamics,

(i) The theory is unable to predict flow separation. In other words, by neglecting viscosity of the fluid, we
are not in positions to explain why fluid which flows close to the cylinder's surface around the upstream
side would tend to move away from the surface on the downstream side.

(ii) The theory is unable to explain the existence of a wake.A Non - viscous theory applied to flow over a
symmetrical blunt - nosed body like a cylinder predicts a symmetrical flow pattern. In other words, the
flow downstream of the cylinder is similar to the flow upstream which is contrary to the observed facts.

(iii) The theory predicts that the pressure distribution will produce no total force. Since the flow pattern is
symmetrical, the pressure forces on the upstream side exactly balance those on the downstream side and
so the net force on the cylinder due to fluid pressure must vanish. This is contrary to actual observations.

(iv) The theory assumes that viscous forces are absent. Since ideal fluid theory assumes to possess no
viscosity, ideal fluid exerts no force on the cylinder, which is again contrary to experience.
(v) The theory makes the analysis of heat transfer unsatisfactory. We consider the predictions of heat
transfer from the following two points of view

(a) An ideal fluid provides no mechanism for heat transfer. Hence the theory predicts zero rates of
heat transfer.

(b) Since there is nothing between the fluid and the solid surface which could give rise to a
resistance to heat transfer, we would expect infinite transfer tats. But these both facts are contrary to our
experience.

(vi) The solutions based on the theory are unable to fulfil the desired boundary conditions.

11
 Since the ideal fluid has zero coefficient of viscosity, the order of the momentum equation of a viscous
fluid is decreased by omitting the viscous terms. Now the order of a differential equation determines the
number of conditions to be satisfied by its solution.

Hence, we conclude that the desired solution of an ideal fluid can satisfy fewer boundary
conditions than that of a viscous fluid.

We now propose to modify the theory of ideal fluids so that it may explain the most
casual observations. An important contribution to fluid dynamics was made by L. Prandtl in 1904 by
introducing the concept of boundary layer. He clarified the essential influence of viscosity in flows at high
Reynold's numbers by showing how the Navies - Stokes equations could be simplified to yield proximate
solutions for overcoming the limitation just listed.

Remarks: Note that the solution for an ideal fluid is not the limiting form of the solution for a viscous
fluid when μ→ 0(re → ∞) because the order of the differential equation if higher than that of ideal fluid
and so requires more boundary conditions to be satisfied for its solution. Moreover, there exists an
essential difference in the boundary conditions in the two cases, namely an ideal fluid slips freely over a
solid boundary whereas a viscous fluid adheres to the boundary and has no velocity relative to it.

2.3 Prandtl’s boundary layer theory –

For convenience, consider laminar two - dimensional flow of fluid of small viscosity (rage
Reynolds's number) over a fixed semi - infinite plate. It is observed that, unlike an ideal (non - viscous) fluid
flow, the fluid does not slide over the plate, but " sticks " to it. Since the plate is at rest, the fluid in contact
with it will also be at rest. As we move outwards along the normal, the velocity of the fluid will gradually
increase and at a distance far from the plate the full stream velocity U is attained. Strictly speaking this is
approached asymptotically. However, it will be assumed that the transition from zero velocity at the plate
to the full magnitude U takes place within a thin layer of fluid in contact with the plate. This is known as
the boundary
layer.
There is no definite line between the potential flow region where friction is negligible and
the boundary layer. Therefore, in practice, we define the boundary layer as that region where the fluid
velocity, parallel to the surface, is less than 99 % of the free stream velocity which is described by potential
flow theory. The thickness of the boundary layer, 𝛿grows along a surface (over which fluid is flowing) from
the leading edge.

12
 Figure-(i)

The shape of the velocity profile and the rate of increase of the boundary layer thickness,
depend on the pressure gradient, 𝜕𝑝 ⁄ 𝜕𝑥𝑇ℎ𝑢𝑠if the pressure increases in the direction of flow, the
boundary layer thickness increases rapidly and the velocity profiles will take the form as shown in Fig (ii)
When adverse pressure gradient is large, then separation will occur

BOUNDARY LAYER THEORY

Followed by a region of reversed flow. The separation point S is defined as the point where

(𝜕𝑢/𝜕𝑦) 𝑦=𝑜 = 0 (separation)

Where u is the velocity parallel to the wall in the x direction and y is the coordinate normal to
the wall. Due to the reversal of flow, there is a considerable thickening of the boundary layer, and
associated with it, there is a flow of boundary layer fluid into the outside region. The exact location of the
point of separation can be determined only with the help of integration of the boundary layer equations.
13
 The method of dividing the fluid in two regions was first proposed by Prandtl in 1904. He
suggested that the entire field of flow can be divided, for the sake of mathematical analysis, into the
following regions.

(i) A very thin layer (boundary layer) in the vicinity of the plate in which the velocity gradient normal to the
wall (i.e.,⁄𝜕𝑦) is very large. Accordingly, the viscous stress μ (𝜕𝑢⁄𝜕𝑦) becomes important even when μ is
small. Thus, the viscous and inertial forces are of the same order within the boundary layer.

(ii) In the remaining region (i.e., outside the boundary layer) (𝜕𝑢⁄𝜕𝑦) is very small and so, the viscous
forces may be ignored completely. Outside the boundary layer, the flow can be regarded non - viscous and
hence the theory of non - viscous fluids offers a very god approximation there.

Remark: 1 the above discussion equally holds even if a blunt body (i.e., a body with large radius of
curvature such as air foil etc.) is considered in place of a flat plate.

Remark:2 The following three conditions must be satisfied by any velocity distribution in boundary
layer.
(i) At y = 0, u = 0 and 𝒅𝒖⁄𝒅𝒚has some finite value
(ii) At y = 𝜹, u = U,
(iii) y = 𝜹, 𝒅𝒖⁄𝒅𝒚= 0

2.4. Importance of Prandtl's boundary layer theory in fluid dynamics.

Although the boundary layer is thin, it plays a vital role in fluid dynamics. It has become a very powerful
method of analysing the complex behaviour of real fluids. The concept of a boundary layer can be utilized
to simplify the Navies - Stokes equations to such an extent that it becomes possible to tackle many
practical problems of great importance. The drag on ships and missiles, the efficiency of compressors and
turbines in jet engines, the effectiveness of air intakes for ram and turbojets and so on depend on the
concept of the boundary layer and its effects on the main flow.

While solving any equation and depending on the arguments in physical terms, the boundary
layer theory is capable of explaining the difficulties encountered by ideal fluid dynamics [refer Art.18.2].
The boundary layer theory is able to predict flow separation. It can explain the existence of a wake. The
pressure distribution produces a net force in the direction in which the stream flows. There exists a viscous
stress on the boundary region and it acts in the direction of flow. The analysis of heat transfer is
satisfactory. The solution of a particular problem is able to satisfy the required boundary conditions. We
shall impose the following conditions on the distribution of velocity in the boundary layer:

(i) The no - slip condition,

(ii) That no mass shall flow through the wall, and

(iii) That the velocity at the outer edge of the boundary layer shall approach that predicted by an
appropriate non - viscous theory.

14
2.5. Some basic definitions.

(a) Boundary layer thickness.

In a qualitative manner, the boundary layer thickness is defined as the elevation above
the boundary which covers a region of flow where there is a large velocity gradient and consequently non-
negligible viscous effects. Since transition from velocity in the boundary to that outside it takes place
asymptotically, there is no obvious demarcation for permitting the measurement of a boundary layer
thickness in a simple quantitative manner. For distance from the solid boundary thickness of the boundary
layer is generally defined as that distance from the solid boundary where the velocity differs by 1 per cent
from the external velocity U (i.e., free - flow velocity). It is easily seen that the above definition of the
boundary layer is to a certain extent arbitrary. Because of this arbitrary and somewhat ambiguous
definition of 𝛿, we employ following three other types of thicknesses which are based on physically
meaningful measurements.

Displacement thickness.

Because of viscosity the velocity on the vicinity of the plate is smaller than in the free - flow
region. The reduction in total flow rate cause by this action is
∞
∫𝟎 (𝐔 − 𝐮)dy

If this integral is equated to a quantity U𝛿􀬵, 𝛿􀬵can be considered as the amount by which the potential flow
has been displaced from the plate. Thus, for displacement thickness 𝛿1 , we have the definition.

∞
𝑼𝜹𝟏= ∫𝒐 (𝐔 − 𝐮)𝐝𝐲 ………... (1)

∞
or𝜹𝟏= ∫𝒐 𝟏 − (𝐮 − 𝐔)𝐝𝐲 ………... (2)

(ii) Momentum thickness.

It is defined by comparing the loss of momentum due to wall - friction in the boundary
to the momentum in the free flow region. Thus, for the momentum thickness 𝛿2 ,we have the definition

∞
𝝆𝑈 2𝜹𝟐= 𝝆∫𝒐 𝐮(𝐔 − 𝐮)𝐝𝐲 ………………. (3)

∞𝒖
𝜹𝟐 = ∫𝒐 𝑼
𝟏 − (𝐮 − 𝐔)𝐝𝐲 ………………… (4)

15
(iii) Energy thickness or dissipation energy thickness or kinetic energy thickness.

There is always a loss in energy because of the viscosity of the fluid. Now the loss of
Kinetic energy in the boundary layer at a distance y from the plate is (p/2) x (𝑼𝟐 - 𝒖𝟐 ) and consequently
the total rate at which the kinetic energy is being lost is

𝝆∫ (𝑼𝟐 − 𝒖𝟐 )𝐮𝐝𝐲
𝟏 ∞
𝟐 𝒐

If this integral is equated to a quantity (p/2) × (𝑼𝟑 𝜹𝟑) 𝜹𝟑can beconsidered as thickness of a layer which
has the same kinetic energy flux as the rate at which has the kinetic energy is being lost in the boundary
region. Thus, for the energy thickness 𝛿3 , we have the definition

𝟏 𝟏 ∞
𝝆𝑼𝟑 𝜹𝟑 = 𝝆 ∫𝟎 (𝑼𝟐 -𝒖𝟐 )udy …………… (5)
𝟐 𝟐

∞𝒖 𝒖𝟐
𝜹𝟑 = ∫𝟎 (1- )dy ……………... (6)
𝑼 𝑼𝟐

(b) Drag and lift:

Let a solid body be present in the flow region of a fluid flow. Then the body will experience
two types of forces. The force component exerted on the body from a moving fluid in the direction of free
- stream of the fluid far from the body is defined as drag and the force component of the body normal to
the free - stream of the fluid far from the body is defined as lift.

Since the measurement of drag D and lift L depend on the transition in boundary
layer, separation of the boundary layer and so on, it is a very difficult task to measure them. We,
therefore, employ experimental data and define them as follows.

D = Drag = (𝑪𝑫 𝑨𝝆 𝑼𝟐 )/2 …………. (7)

L = Lift = (𝑪𝑳 𝑨𝝆 𝑼𝟐)/2 …………. (8)

16
And

where 𝑪𝑫 is the coefficient of drag, 𝑪𝑳 is the coefficient of lift, A is the projected area in the direction of
flow and U is the free - flow velocity, notice that the coefficients 𝑪𝑫 and 𝑪𝑳 are both dimensionless
quantities.

(c) Local skin coefficient

The shearing stress on the plane boundary layer is given by,

𝒅𝒖
𝝅𝝎 = 𝝁 ( )
𝒅𝒚 𝒚=𝟎

Then, the dimensionless shear stress, which is known as local skin coefficient is denoted a defined as

𝝆𝑼𝟐
𝑪 𝒇 = 𝝉𝝎 / ( )
𝟐

2.5 A. Derivation of different types of thicknesses

Boundary layer thickness (𝜹).

The velocity u within the boundary layer increases from zero at the boundary surface to the
velocity U of the main stream stream asymptotically. So, the boundary layer thickness is arbitrarily defined
as that distance from the boundary in which the velocity reaches 99 per cent of the velocity of the free
stream (u = 0.99U). It is denoted by 𝛿.

The above definition gives an approximate value of the boundary layer thickness. For
better accuracy, the boundary layer thickness if defined in terms of certain mathematical expressions
which are the measure of the boundary layer of the flow. We now give three commonly used definitions
of the boundary layer thickness.

(i) Displacement thickness Definition.

It is the distance measured perpendicular to the boundary of the solid body, by which the
main/free stream is displaced on account of boundary layer.

17
 Another definition.
It is an additional "wall thickness" that would have to be added to compensate for the reduction
in flow are on account of boundary layer formation.

The displacement thickness is denoted by 𝜹𝟏 or 𝜹∗

Derivation of expression for 𝜹𝟏

Let an incompressible fluid of density 𝜌 flow past a stationary plate with velocity U as shown in the
above figure. At a distance x from the leading edge, consider a section 1-1. The velocity of fluid at B is zero
and at C, which lies on the boundary layer, is U. Thus, here BC = boundary thickness = , At the section 1 1,
consider an elementary strip. Let y - distance of the elementary strip from the plate, dy = thickness of the
elementary strip, u = velocity of fluid at the elementary strip and b = width of plate.

Mass flow per second through the elementary strip

= 𝝆 × velocity × area of elementary strip = 𝝆 × 𝝁 × (bdy) ……………. (1)

If there had been no plate then the fluid would have been flowing with a constant velocity
equal to free stream velocity U at the section 1-1

Hence, mass flow per second through the elementary strip if the plate were not there

= 𝝆 × velocity × area = 𝝆 × 𝑼 × (𝒃𝒅𝒚) … … … …. (2)

18
 Since U >𝝁, it follows that due to the presence of the plate and consequently due to the
formation of the boundary layer, there will be a reduction in mass flowing per second through the
elementary strip.

The reduction of mass flow rate through the elementary strip

= mass / sec given by equation (2) mass / sec given by equation (1)

= 𝝆Ub dy – 𝝆ub dy - 𝝆b (U - u) dy.

So, total reduction of mass flow rate through BC due to introduction of plate

𝜹 𝜹
= ∫𝟎 𝝆𝒃 (𝑼 − 𝒖)𝒅𝒚 = 𝝆𝒃 ∫𝟎 (𝑼 − 𝒖)𝒅𝒚

[ 𝒇𝒍𝒖𝒊𝒅 𝒊𝒔 𝒊𝒏𝒄𝒐𝒎𝒑𝒓𝒔𝒔𝒊𝒃𝒍𝒆 ⤇ 𝝆 𝒊𝒔 𝒄𝒐𝒏𝒔𝒕. ]

Let the plate be displaced by a distanced 𝛿1 , and velocity of flow for the distance 𝛿 be the main / free
stream velocity (i.e., U). Then,

Loss of mass of the fluid / sec flowing through the distance 𝜹₁

= 𝝆 × 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚 × 𝒂𝒓𝒆𝒂 = 𝝆 × 𝑼 × (𝜹𝟏b) ……………… (4)

Equating the expressions obtained in (3) and (4), we have.

𝜹
𝝆𝑼𝜹𝟏b = 𝝆b∫𝟎 (𝑼 − 𝒖)dy

𝜹 𝜹 (𝑼−𝒖)𝐝𝐲
𝜹𝟏= 1/U∫𝟎 (𝑼 − 𝒖)dy = ∫𝟎 ,
𝑼

19
 as u is constant.

𝜹 𝒖
𝜹𝟏 = ∫𝟎 (𝟏 − )dy
𝑼

Another modified expression for 𝜹𝟏

From boundary layer theory,

u = UI for y >𝜹

∞ 𝒖 𝒖 𝑼
∴ ∫𝟎 (𝟏 − )dy = 0 as 1 – = 1- =0 for y ≥ 𝛿
𝑼 𝑼 𝑼

Thus,

𝜹 𝒖 ∞ 𝒖 ∞ 𝒖
𝜹𝟏 = ∫𝟎 (𝟏 − )dy + ∫𝜹 (𝟏 − )dy = ∫𝟎 (𝟏 − )dy
𝑼 𝑼 𝑼

(ii) Momentum thickness.

Definition.

It is the distance, measured perpendicular to the boundary of the solid body, by which
the boundary should be displaced to compensated for reduction in momentum of the flowing fluid on
account of boundary layer formation.

Another definition.–

It is the distance through which the total loss of momentum per second be equal to it if
it were passing a stationary plate the momentum thickness is denoted by 𝜹𝟐 𝜽

20
Derivation of expression for 𝜹𝟐 .

Refer figure given on page 18.5. Let incompressible fluid of density p flow past a
stationary plate with velocity U as shown in the figure. At a distance x from the leading edge, consider a
section 1-1. The velocity of fluid at B is zero and at C, which lies on the boundary layer, is U. Thus, here BC
= boundary layer thickness = 𝜹,at the section 1-1 consider an elementary strip. Let y distance of the
elementary strip from the plate, dy = thickness of the elementary strip. u = velocity of fluid at the
elementary strip, and b = width of plate.

Mass flow per second through the elementary strip

= p × velocity × area of elementary strip = p ×u( bdy ).

Momentum/sec of this fluid inside the boundary layer

= (𝝆ubdy) × 𝒖 = 𝒑𝒖𝟐 bdy.

∴ Loss of momentum/sec through elementary strip

= p𝒖𝟐 bdy - puUbdy = pbu(U-u) dy

So total loss of momentum/sec through BC due to introduction of plate

𝜹 𝜹
= ∫𝟎 𝝆𝒃𝒖 ( 𝑼 − 𝒖)dy = 𝝆𝒃 ∫𝟎 𝒖(𝑼 − 𝒖)dy. ………. (1)

[∴ 𝒇𝒍𝒖𝒊𝒅 𝒊𝒔 𝒊𝒏𝒄𝒐𝒎𝒑𝒓𝒆𝒔𝒔𝒊𝒃𝒍𝒆 ⤇ 𝝆 𝒊𝒔 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕]

Let 𝜹𝟐 = distance by which plate is displaced when the fluid is flowing with constant velocity U of main/free
stream.
21
∴ Loss of momentum/sec of fluid through distance 𝛿2 with velocity U

= (P × 𝒂𝒓𝒆𝒂 × 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚) × velocity = [𝒑 × (𝜹𝟐 ) × 𝑼 ] ×U = p𝜹𝟐 b𝑼𝟐

Equating the expressions obtained in (1) and (2), we have

𝜹
𝝆𝜹𝟐b𝑼𝟐 = 𝝆𝒃 ∫𝟎 𝒖(𝑼 − 𝒖)dy

𝟏 𝜹 𝜹 𝒖(𝑼−𝒖)
or 𝜹𝟐 = ∫𝟎 𝒖(𝑼 − 𝒖)dy = ∫𝟎 dy, as U is constant.
𝑼𝟐 𝑼𝟐

Or

𝜹𝒖 𝒖
𝜹𝟐 = ∫𝟎 (𝟏 − )dy.
𝑼 𝑼

Another modified expression for 𝜹𝟐

Force boundary layer theory, u = U for y > 𝛿.

Hence

∞ 𝒖 𝒖 𝒖 𝑼
∫𝜹 (𝟏 − )dy = 0 as 1 - = 1 - = 0, for y≥ 𝜹.
𝑼 𝑼 𝑼 𝑼

Thus,

𝜹𝒖 𝒖 ∞ 𝒖 𝒖 ∞ 𝒖 𝒖
𝜹𝟐= ∫𝟎 (𝟏 − )dy + ∫𝜹 (𝟏 − )dy = ∫𝟎 (𝟏 − )dy …………. (4)
𝑼 𝑼 𝑼 𝑼 𝑼 𝑼

22
(iii) Energy thickness (or dissipation energy thickness or kinetic energy thickness):

It is defined as the distance, measured perpendicular to the boundary of the

solid. body, by which the boundary should be displaced to compensate for the reduction in kinetic energy
of the flowing fluid on account of boundary layer formation. It is denoted by 𝜹𝟑 , or 𝜹∗∗ or𝜹𝒆 .

Derivation of expression for 𝜹𝟑 ,

Refer figure of Art 18.5A given on page 18.5. Let an incompressible fluid of density p flow
past a stationary plate with velocity U as shown in the figure. At a distance x from the leading edge,
consider a section 1-1. The velocity of fluid at B is zero and at C, which lines on the boundary layer, is U.
Thus, here BC= boundary layer thickness =𝜹. At the section 1-1, consider an elementary strip. Let y =
distance of the elementary strip form the plate, dy thickness of the elementary strip, velocity of fluid at the
elementary strip and b = width of plate.

Mass flow per second through the elementary strip

= px velocity area of elementary strip

= p × 𝒖 × ( b dy).

Then, kinetic energy of this fluid inside the boundary layer

= (1/2) × (mass) x (velocity) = (1/2) (pub dy) x u²

Again, kinetic energy of the same fluid in the absence of boundary layer.

= (1/2) × (mass) x (velocity) ² = (1/2) × (pub dy) X U²

23
 ∴ Loss of kinetic energy through elementary strip

= (1/2) × (pup dy) 𝑼𝟐 - (1/2) × (pub dy) 𝒖𝟐 - (1/2) × pub ( 𝑼𝟐 - 𝒖𝟐 ) dy.

∴ Total loss of kinetic energy through BC due to introduction of plate

𝜹𝟏 𝟏 𝜹
= ∫𝟎 𝝆𝒖𝒃(𝑼𝟐 − 𝒖𝟐 )𝒅𝒚 = 𝝆𝒃 ∫𝟎 𝒖(U²- u²) dy ……………... (1)
𝟐 𝟐

(fluid is incompressible > P is constant)

Let 𝜹𝟑 , = distance by which the plate is displaced to compensate for the reduction in kinetic energy.
Hence, loss of kinetic energy through 𝜹𝟑 , of flowing with velocity U

= (1/2) × mass x (velocity) ² = (1/2) x (p x area x velocity) x (velocity) ²

= (1/2) × [p x (b x 𝜹𝟑 ) U² = (1/2)× pb𝜹𝟑 U²

Equating the expressions obtained in (1) and (2), we have

𝜹
pb𝜹𝟑 𝑼𝟑 = 𝝆bu ∫𝟎 𝒖(𝑼𝟐 − 𝒖𝟐 )dy

𝟏 𝜹 𝜹 𝒖(𝑼𝟐 −𝒖𝟐 )
or 𝜹𝟑= ∫ 𝒖(𝑼𝟐 − 𝒖𝟐 )dy = ∫𝟎
𝑼𝟑 𝟎 𝑼𝟑
dy as U is constant

𝜹 𝑼𝟐
or 𝜹𝟑 = ∫𝟎 (𝟏 − 𝟐 )dy ……… (3)
𝒖

Another modified expression for 𝜹𝟑

24
 From boundary layer theory, we have

∞𝒖 𝒖𝟐 𝒖𝟐 𝑼𝟐
∴ ∫𝜹 (𝟏 − 𝟐
) 𝒅𝒚 = 𝟎,as 1- =𝟏− =𝟎, for y ≥ 𝛿
𝑼 𝑼 𝑼𝟐 𝑼𝟐

u = U for y ≥ 𝜹.

𝜹 𝑼𝟐 ∞𝒖 𝒖𝟐 ∞𝒖 𝒖𝟐
Thus, 𝜹𝟑 = ∫𝟎 (𝟏 − 𝟐 )dy + ∫𝜹 (𝟏 − 𝟐
) dy = ∫𝟎 (𝟏 − )dy ……… (4)
𝒖 𝑼 𝑼 𝑼 𝑼𝟐

2.5 B. Displacement, momentum and energy thicknesses for axially symmetric flow:

In art 18.5A we have presented definitions of displacement, momentum and

energy thicknesses for two-dimensional boundary layer flows. In the present article, we wish to define
displacement, momentum and energy thicknesses, for axially symmetric boundary layer flows. In order to
achieve these thicknesses, it can be shown that we have to replace by 2πr dn in expressions for 𝜹𝟏 , 𝜹𝟐 and
𝜹𝟑 as obtained in Art. 18.5A. here is the axial distance and dn is the line element along the normal to the
surface of the body. From dimensional consideration we drive the resultant expressions by 2πa, where a is
a reference radius, which may be a function of the axial distance. Accordingly, for axially symmetric
boundary layer flows, we have

∞ 𝐮 𝐫
(i) Displacement thickness = 𝛅𝟏 = ∫𝟎 (𝟏 − ) dn
𝐔 𝐚

∞ 𝒖 𝒓𝒖
(ii) Momentum thickness = 𝜹𝟐 = ∫𝟎 (𝟏 − ) dn
𝑼 𝒂𝑼

∞ 𝒖𝟐 𝒓𝒖
(iii)Energy thickness = 𝜹𝟑 = ∫𝟎 (𝟏 − ) d
𝑼𝟐 𝒂𝑼

In above result, n is the normal distance from the surface of the body.

25
 2.6 The boundary layer equations in two - dimensional flow:

In what follows, we shall obtain the boundary layer equations in two dimensional
flows by using the following two methods;

(i) order of magnitude approach

(ii) asymptotic approach.

Method L. Order of magnitude approach

For simplicity we shall first derive the boundary layer equations for the flow over a
semi-infinite flat plate. We take rectangular Cartesian coordinates (x, y) with x measured in the plate in
the direction of the two - dimensional laminar incompressible flow, and y measured normal to the
plate, and (u, u) are the velocity components. Let viscosity of the fluid be small and let 𝛿 be small
thickness of the boundary layer. Let U be the velocity in the main stream just outside the boundary
layer. Then the Navies Stokes equations, without body forces for two-dimensional flow are:

𝝏𝒖 𝝏𝒖 𝝏𝒖 𝟏 𝝏𝒑 𝝏𝟐 𝒖 𝝏𝟐 𝒖
X-direction: +u +v =- + v( 𝟐
+ ) ……… (1)
𝝏𝒕 𝝏𝒙 𝝏𝒚 𝝆 𝝏𝒙 𝝏𝒙 𝝏𝒚𝟐

𝝏𝒗 𝝏𝒗 𝝏𝒗 𝟏 𝝏𝒑 𝝏𝟐 𝒗 𝝏𝟐 𝒗
Y-direction: +u +v =- + v( 𝟐
+ ) ………. (2)
𝝏𝒕 𝝏𝒙 𝝏𝒚 𝝆 𝝏𝒙 𝝏𝒙 𝝏𝒚𝟐

The continuity equation is

𝛛𝐮⁄𝛛𝐱 + 𝛛𝐯⁄𝛛𝐲 = 𝟎. …………. (3)

The boundary conditions are: u = 0, v = 0 when y = 0, u = U(𝒙, 𝒕) 𝒘𝒉𝒆𝒏 𝒚 = ∞ …… (4)

We now determine the order of magnitude of each term in (1), (2) and (3) to enable us to drop
small terms and thus to arrive at the simplified boundary layer equations. We shall designate the order of
any quantity (say) by O (g).The orders.of magnitude are shown in (1) to (3) under the individual terms.
26
Let O (u) = 1 and O (𝝏𝒖⁄𝝏𝒙 ) = 1 within the boundary layer. Then 𝛛𝐮⁄𝛛𝐲 is large as O (𝛛𝐮⁄𝛛𝐲 )=
1/𝛅,udecreasing from a finite value U at the outer boundary of the layer to zero at the flat plate. Again
O(𝝏𝟐𝒖⁄𝝏𝒚𝟐)= 1/𝜹𝟐 in the boundary layer. Further O (𝝏𝒖⁄𝝏𝒕) =O (𝝏𝟐 𝒖⁄𝝏𝒙𝟐 ) = 1. Since (𝛛𝐮⁄𝛛𝐱) = 1, (3)
shows that O (𝛛𝐯 ⁄𝛛𝐲 ) = 1.

Since v= 0 when y = 0, O (u) =𝜹. Again, O(𝝏𝒖/𝝏𝒕) = O(𝝏𝒖/𝝏𝒙) =0 (𝝏𝟐𝒖/𝝏𝒙𝟐 ) = 𝜹 whereas

since the viscous force is taken as of the same order as the inertia forces within the boundary layer, (1)
implies that we must have

O (v / 𝜹𝟐 ) = 1 so that 0 ( 𝜹 ) = √𝒗= √𝝁/𝝆 .

Showing that smaller the viscosity of the fluid, the thinner the boundary layer. By neglecting the
terms of the order & and smaller form (1) to (3), we obtain

𝝏𝒖 𝝏𝒖 𝝏𝒖 𝟏 𝝏𝒑 𝝏𝟐 𝒖
+u +v = - +v
𝝏𝒕 𝝏𝒙 𝝏𝒚 𝝆 𝝏𝒙 𝝏𝒙𝟐

𝝏𝒖/𝝏𝒙 + 𝝏𝒗/𝝏𝒚 = 0

The above equations (6) and (8) are known as Prandtl's boundary layer equations with
boundary conditions (4).

Now (7) shows that the pressure distribution is a function of x only, i.e. for a given x.p is constant
throughout the boundary layer. Thus, the pressure throughout the boundary layer possesses the same
value of p as in the potential flow It follows that we can now regard u and u as unknown variables in (6)
and (8) instead of three unknown variables u, v and p in (1), (2) and (3). This simplifies the process of
finding solution to a great extent because we now have a system of two simultaneous equations for the
two unknownu and v.

At the outer edge of the boundary layer, we have u = U (x, t). Since large velocity gradients are
absent there, the viscous terms in (1) vanish for small values of v (v = 𝜇/𝜌is small because 𝜇 is small by
assumption). Hence for the flow outside the boundary, we have
27
 𝝏𝑼 𝝏𝑼 𝟏 𝒅𝒑
+ U = - ……………... (9)
𝝏𝒕 𝝏𝒙 𝝆 𝒅𝒙

For the case of steady flow, (9) reduces to

𝒅𝑼 𝟏 𝒅𝒑 𝒅𝒑 𝒅𝑼
U = , so that = 𝝆U ……………... (10)
𝒅𝒙 𝝆 𝒅𝒙 𝒅𝒙 𝒅𝒙

Further, for the steady flow boundary layer equations (6) and (8) reduce to still more simplified form:

𝝏𝒖 𝝏𝒖 𝟏 𝒅𝒑 𝝏𝟐 𝒖
u +v = +v
𝝏𝒙 𝝏𝒚 𝝆 𝒅𝒙 𝝏𝒚𝟐

𝝏𝒖/𝝏𝒙 + 𝝏𝒗/𝝏𝒚 = 0

Where 𝒅𝒑/𝒅𝒙 is given by(10)

Remark1. Since the same argument can be carried out at all places along the boundary layer (except at
the front edge), the equations deduced so far are general and hold for any two - dimensional boundary
layer flow. It is also to be pointed out that, for a slightly curved boundary, these equations are valid if we
use curvilinear coordinates fitting the boundary.

Mothed II.Asymptotic approach.

In what follows, we shall prove that Prandtl boundary layer equations may be treated
as the asymptotic from of the Navies - Stokes equations at large Reynolds number.

Let 𝒕′ 𝒙′ 𝒚′ 𝒖′ 𝒗′ and𝒑′ dimensionless quantities such that

𝒕 𝒙 𝒚 𝒖 𝒗 𝒑
𝒕′ = , 𝒙′ = , 𝒚′ = , 𝒖′ = , 𝒗′ = and 𝒑′ = ……… (1)
𝑻 𝑿 𝒀 𝑼 𝑽 𝑷

28
Where T. X, Y, U, V and P are the units of measurement of the corresponding quantities.And The Navier -
Stokes equations of motion for a viscous incompressible fluid in the absence of body forces, in a two -
dimensional flow are given by

𝝏𝒖 𝝏𝒖 𝝏𝒖 𝟏 𝝏𝒑 𝝏𝟐 𝒖 𝝏𝟐 𝒖
And +u +v = - +v( 𝟐
+ ) …………………. (2)
𝝏𝒕 𝝏𝒙 𝝏𝒚 𝝆 𝝏𝒙 𝝏𝒙 𝝏𝒚𝟐

𝝏𝒗 𝝏𝒗 𝝏𝒗 𝟏 𝝏𝒑 𝝏𝟐 𝒗 𝝏𝟐 𝒗
+u +v = - +v( 𝟐
+ ) …………………. (3)
𝝏𝒕 𝝏𝒙 𝝏𝒚 𝝆 𝝏𝒚 𝝏𝒙 𝝏𝒚𝟐

The continue equation is

𝝏𝒖/𝝏𝒙 + 𝝏𝒗/𝝏𝒚 = 0 ……………. (4)

Here the xy - plane is the plane of motion, the x - axis is along the wall of the plate and y - axis
perpendicular to it. Substituting (1) in equations (2), (3) and (4), we get

𝐗 𝛛𝐮′ 𝛛𝐮′ 𝐗𝐕 𝛛𝐮′ 𝐏 𝛛𝐩′ 𝐯 𝛛𝟐 𝐮′ 𝐯𝐗 𝛛𝟐 𝐯 ′

+ 𝐮′ + 𝐯′ = - + + , ……………… (5)
𝐔𝐓 𝛛𝐱 ′ 𝛛𝐱 ′ 𝐘𝐔 𝛛𝐲 ′ 𝛒𝐔𝟐 𝛛𝐱 ′ 𝐗𝐔 𝛛𝐱 ′𝟐 𝐘𝟐 𝐔 𝛛𝐲 ′𝟐

𝐗 𝛛𝐯 ′ 𝛛𝐯 ′ 𝐗𝐕 𝛛𝐯 ′ 𝐏 𝛛𝐩′ 𝐯 𝛛𝟐 𝐯 ′ 𝐯𝐗 𝛛𝟐 𝐯 ′
+ 𝐮′ + 𝐯′ = + + ………………. (6)
𝐔𝐓 𝛛𝐱 ′ 𝛛𝐱 ′ 𝐘𝐔 𝛛𝐲 ′ 𝛒𝐘𝐔𝐕 𝛛𝐲 ′ 𝐗𝐔 𝛛𝐱 ′𝟐 𝐘𝟐 𝐔 𝛛𝐲 ′𝟐

𝛛𝐮′ 𝐗𝐕 𝛛𝐮′
+ =0 ……………………. (7)
𝐱′ 𝐘𝐔 𝛛𝐲 ′

Treating X and U as the fundamental units, the Reynolds number for the given flow is taken as

Re = (XU)/v …………………. (8)

29
 Then the units of time and pressure can be expressed in terms of X and U follows

T = X/U and P = 𝝆𝑼𝟐 …………… (9)

We now derive the units of measurement of Y and V by applying the condition that the system of
equations (5), (6) and (7) must possess only a single flow parameter Re given by (8) Accordingly, we must
have

𝐗𝐕 𝐯𝐗 𝑿 𝑼
= 1 and = 1 so that y= and V= ………. (10)
𝐘𝐔 𝐘𝟐 𝐔 √𝑹𝒆 √𝑹𝒆

Hence equation (5), (6) and (7) may be re - written as

𝛛𝐮′ 𝛛𝐮′ 𝐗𝐕 𝛛𝐮′ 𝛛𝐩′ 𝟏 𝛛𝟐 𝐮′ 𝛛𝟐 𝐯 ′

+ 𝐮′ + 𝐯′ = - + + …………………. (11)
𝛛𝐭 ′ 𝛛𝐱 ′ 𝐘𝐔 𝛛𝐲 ′ 𝛛𝐱 ′ 𝑹𝒆 𝛛𝐱 ′𝟐 𝛛𝐲 ′
𝟐

𝟏 𝛛𝐯 ′ 𝛛𝐮′ 𝐗𝐕 𝛛𝐯 ′ 𝛛𝐩′ 𝟏 𝛛𝟐𝐮′ 𝛛𝟐𝐯 ′

( ′
+ 𝐮′ + 𝐯′ )= - + (𝑹𝒆)𝟐 𝟐 + 𝟐 ……………. (12)
𝐑𝐞 𝛛𝐱 𝛛𝐱′ 𝐘𝐔 𝛛𝐲 ′ 𝛛𝐲 ′ 𝛛𝐱 ′ 𝛛𝐲 ′

𝝏𝒖′ /𝝏𝒙′ + 𝝏𝒗′ /𝝏𝒚′ = 0 …………………………………... (13)

If Re is large then 1 / √𝑹𝒆 can be treated as small parameter. We now obtain the solution of (11), (12) and
(13) with help of a power series containing 1 / Re as follows:

𝟐
𝒖′ = 𝒖𝟎 + (𝟏 / √𝑹𝒆)𝒖𝟏 + (𝟏 / √𝑹𝒆) 𝒖𝟐 + …..
𝟐
And 𝒗′ = 𝒖𝟎 + (𝟏 / √𝑹𝒆)𝒗𝟏 + (𝟏 / √𝑹𝒆) 𝒗𝟐 + ………………………. (14)
𝟐
𝒑′ = 𝒑𝟎 + (𝟏 / √𝑹𝒆)𝒑𝟏 + (𝟏 / √𝑹𝒆) 𝒑𝟐 + …..

30
 Substituting (14) in equations (11), (12) and (13) and then comparing the coefficient of the
zeroth power of Re, we get

𝛛𝒖𝟎 𝛛𝒖𝟎 𝐗𝐕 𝛛𝒖𝟎 𝛛𝒑𝟎 𝛛𝟐 𝒖𝟎

+ 𝒖𝟎 + 𝒗𝟎 = - + 𝟐 ………………………. (15)
𝛛𝐭 ′ 𝛛𝐱 ′ 𝐘𝐔 𝛛𝐲 ′ 𝛛𝐱 ′ 𝝏𝒚′

𝛛𝒑𝟎
0=- ………………………. (16)
𝛛𝐲 ′

𝛛𝒖𝟎 𝛛𝒗𝟎
And + ………………... (17)
𝛛𝐱 ′ 𝛛𝐲 ′

Equations (15), (16) and (17) are the same resulting equations as if we would have allowed directly in
equations (11), (12) and (13) the Reynolds number tending to infinity, reverting to dimensional form and
dropping the index zero in equations (15), (16) and (17), we obtain

𝝏𝒖 𝝏𝒖 𝝏𝒖 𝝏𝒑 𝝏𝟐 𝒖 𝝏𝟐 𝒖
+u +v = - + 𝟐
+ ……………. (18)
𝝏𝒕 𝝏𝒙 𝝏𝒚 𝝏𝒙 𝝏𝒙 𝝏𝒚𝟐

0 = - 𝝏𝒑/𝝏𝒚 ………………………... (19)

And𝝏𝒖/𝝏𝒙 + 𝝏𝒗/𝝏𝒚 = 0 ……………… (20

Now equation (19) shows that P is a function of x only. Hence the pressure is constant in a direction
normal to the boundary layer and may be equal to that at the outer edge of the boundary layer where it is
determined by the in viscid flow (potential flow). Thus, we have

𝟏 𝒅𝒑 𝝏𝒖 𝝏𝒖
- = +U ………………… (21)
𝝆 𝒅𝒙 𝝏𝒕 𝝏𝒙

Where U is the potential flow velocity

31
Hence the Prandtl boundary layer equations for a two-dimensional unsteady incompressible flow are

𝝏𝒖 𝝏𝒖 𝝏𝒖 𝝏𝑼 𝝏𝟐 𝒖
+u +v =U + ……………. (22)
𝝏𝒕 𝝏𝒙 𝝏𝒚 𝝏𝒙 𝝏𝒚𝟐

and 𝝏𝒖/𝝏𝒙 + 𝝏𝒗/𝝏𝒚 = 0 ………………. (23)

Now due to no - slip condition and the wall being solid (non - porous), both u and v must vanish at y = 0.
Hence equations (22) and (23) are solved under the following boundary conditions.

U=v=0 when u = o and u=U and y → ∞ ……………… (24)

32
 CONCLUSION – 2

2.7 The boundary layer is an important concept in aerodynamics; its discovery advanced the
science of aerodynamics to a useful tool in predicting flows. Today it is an important part of engineering.

Parameter such as boundary layer thickness displacement thickness and momentum thickness
characterize the boundary layer's size and shape.

The boundary layer equation provides a significant simplification to the Navies in viscid
solution to be used outside the boundary layer and by changing the equations from elliptic to parabolic
inside the boundary layer. The equations are valid for high Reynolds number flows, but not when close to,
the leading edge of the wall.

33
 CHAPTER- 03

Thermal Boundary Layer

3.1 Introduction.

When a fluid flows past heated or cooled bodies the heat is transferred by conduction,
convection and radiation Heat transfer by radiation is negligible unless the temperature is very high.
Accordingly, we shall confine our present discussion to heat transfer by conduction and convection only.
The conductivity of ordinary fluids is small For such fluids the heat transport due to conduction is
comparable to that due to convection only across a thin layer near the surface of the body . It follows that
the temperature field which spreads from the body expends only over a narrow zone in the immediate
vicinity of its surface, whereas the fluid at a large distance from the surface is not materially affected by
the heated body. This thin layer (narrow region) near the surface of the body is called thermal boundary
layer. This concept is analogous to the concept of velocity boundary layer described in Art. 18.3 Of chapter
18.

There are two types of problems of thermal boundary layers , namely.

(I) Forced convection and

(II) Free ( or natural ) convection

3.2 Forced convection Definition:

The forced convection is the flow in which the velocities, arising from variable
density (i.e., due to the force of buoyancy) are negligible in comparison with the velocity of the main of
forced flow.

Free or natural convection:

Definition:- The free convection is the flow in which the motion is essentially caused by the effect of
gravity on the heated fluid of variable density.

34
3.3 The thermal boundary layer equations in two - dimensional flow.

We propose to derive the thermal - boundary layer equations for the flow over semi -
infinite flat plate . We take rectangular Cartesian co - ordinates (x, y ) with x measured in the plea in the
direction of the two - dimensional laminar incompressible flow, and y measured normal to the plate.

Let (u,v) e the velocity components in x and y - directions Let viscosity of the fluid be
small and let 𝛿 be small thickness of the boundary layer.

Let 𝛿𝑡 be small thermal boundary layer. Let 𝑈∞ be the velocity in the main
stream just outside the boundary layer. Then the equation of energy for the steady flow of a viscous
incompressible fluid in two - dimensional is given by

𝝏𝑻 𝝏𝑻 𝝏𝟐 𝑻 𝝏𝟐 𝑻
𝝆𝑪𝒑 (𝒖 +𝒗 )=K( 𝟐
+ ) +∅ ………. (1)
𝝏𝒙 𝝏𝒚 𝝏𝒙 𝝏𝒚𝟐

1 1 1 𝜹 𝟏⁄𝜹 𝜹𝟐𝒕 1 𝟏⁄
𝒕 𝜹𝟐𝒕

Where the dissipation function is ∅ given by

∅ = 𝟐𝛍{(𝛛𝐮⁄𝛛𝐱)𝟐 + (𝛛𝐯 ⁄𝛛𝐲 )𝟐} + 𝛍{(𝛛𝐯 ⁄𝛛𝐱)𝟐 + (𝝏𝒖⁄𝝏𝒚)𝟐 + 𝟐(𝛛𝐯 ⁄𝛛𝐱) × (𝛛𝐮⁄𝛛𝐲 )} ..... (2)

𝜹𝟐 1 𝜹𝟐⁄ 𝜹𝟐 𝜹𝟐 𝟏⁄ 𝜹 × (𝟏⁄𝜹 )
𝜹𝟐𝒕 𝜹𝟐𝒕 𝒕

We shall follow a method similar to that used in Art . 18.6 In order to simplify (1). We
now determine the order of magnitude of each term in (1) to enable us to drop small terms and thus to
arrive at the simplified thermal boundary layer equations. We shall designate the order of any quantity
(q,say) by O (q). The orders of magnitude are shown in (1) under the individual terms. Let 𝜹and 𝜹𝒕 , be
velocity boundary layer thickness and thermal boundary layer thickness respectively .

35
 Let O ( x ) = 0 ( u ) = O ( T ) =1 and O (y ) = 𝛿1 . As proved in Art .18.6 , O ( v ) = 𝛿 and O ( v ) =𝛿 2 .

From (1), it follows that the term 𝜕 2 𝑇/𝜕𝑥 2 may be neglected in comparison with
𝜕 2 𝑇/𝜕𝑦 2 . Again, the conduction terms become of the same order of magnitude as the convectional
term, only if O (k)= 𝛿𝑡2 and 𝛿and 𝛿𝑡 , are of the same order of magnitude.

𝑲/ 𝝆𝑪𝒑 𝜹𝟐𝒕
∴ O( )= that is,
𝒗 𝜹𝟐

Where𝑃𝑟 = Prandtl number = ( 𝜇𝐶𝑝 )/k and v = /𝜌 .

Since O ( 𝑃𝑟 ) = 1 for gases, so O ( 𝛿𝑡 ) = 0 ( 𝛿) approximately for gases .On the other

hand for liquids, O ( 𝛿𝑡 ) < O (𝛿 ).

Suppose that 𝛿𝑡 and 𝛿 be of the same order of magnitude. Then observing (1) and (2), we
note that the only term which can be retained in ∅ is 𝜇(𝜕𝑥 ⁄𝜕𝑦 )2 . Hence the equation of the thermal
boundary layer for an incompressible constant properties fluid in two - dimensional steady flow is given by

𝝏𝑻 𝝏𝑻 𝝏𝟐 𝑻 𝝏𝒖 𝟐
𝝆𝑪𝒑 {𝒖 ( ) + 𝒗 ( )} = k ( 𝟐
) + 𝝁( )
𝝏𝒚 𝝏𝒚 𝝏𝒚 𝝏𝒚

2 (𝑈
The first term on the R.H.S of (3) is of order 𝑇𝑤 - 𝑇∞ , where 𝑈∞ ∞ and
𝑇∞ being velocity and temperature of the tree steam and 𝑇𝑤 the temperature of the plate). Hence both
2
are of the same order when 𝑈∞ / ( 𝑇𝑤 − 𝑇∞) of order unity. To make it dimensionless we divide by 𝐶𝑃 ,
which is of order unity. Hence the second term on R.H.S of (3) representing heat due to friction if of the
same order as the other terms when

𝟐 / 𝛒 ( 𝐓 − 𝐓 ) = Ec = Eckert number
𝐔∞ ……………. (4)
𝐰 ∞

Is of order unity.

36
For example, temperature difference of 10 ° F, O (Ec) = 1 when 𝑈∞= 500 ft. / sec.

Usually (3) is solved under the following boundary conditions

When y = 0, T = 𝑇𝑤 , or 𝜕𝑡/𝜕𝑦 = 0; when y = ∞, T = 𝑇∞ ……………………… (5)

The temperature of the plate may be constant or be a function of the distance along
the wall. The boundary condition 𝜕𝑡/𝜕𝑦= 0 when y = 0, which is condition of an adiabatic plate. Is valid
only when the frictional heat is taken into consideration.

In Art 18.6 , we have already obtained boundary layer equations ( Refer ( 10 ) , ( 11 ) and ( 12
) , in the following form :

𝝏𝒖/𝝏𝒙 + 𝝏𝒗/𝝏𝒚 = 𝟎 …………………… (6)

𝝏𝒖 𝝏𝒖 𝒅𝑼 𝝏𝟐 𝒖
u( )+𝒗( ) = U( )+v( ) …….. (7)
𝝏𝒙 𝝏𝒚 𝒅𝒙 𝝏𝒚𝟐

Subject the boundary conditions:

u = v = 0 when y = 0 and u = U (x) when y = ∞

Thus the basic equations which govern the velocity and temperature distribution in a boundary layer past a
solid body, in forced convection, are given by:

𝝏𝒖 𝝏𝒖 𝒅𝑼 𝝏𝟐 𝒖
U( )+𝒗( ) = U( )+v( ) …….. (ii)
𝝏𝒙 𝝏𝒚 𝒅𝒙 𝝏𝒚𝟐

37
 𝝏𝑻 𝝏𝑻 𝝏𝟐 𝑻 𝝏𝒖 𝟐
𝝆𝑪𝒑 {𝒖 ( ) + 𝒗 ( )} = k ( 𝟐
) + 𝝁 ( ) ………….(iii)
𝝏𝒚 𝝏𝒚 𝝏𝒚 𝝏𝒚

With the boundary conditions

u = v = 0, T = 𝑻𝒘 (x) or 𝒅𝑻/𝒅𝒚 = 0 when y =0 ……….(iv)

u = U(x) and T =𝑻∞ when y = ∞ ………….(v)

Note 1. In an incompressible flow the velocity is usually small and the temperature difference is
moderate. Hence the second term of the R.H.S of (iii) which is the heat due to friction, can be neglected
and then (iii) reduces to

u (𝝏𝑻/𝝏𝒙) + v (𝝏𝑻/𝝏𝒚) = (𝒌/𝝆𝑪𝑷 ) × (𝝏𝟐 𝑻/𝝏𝟐 𝒚)

Note 2. The velocity boundary equation

(i) is non - linear in character and is independent of the temperature field and hence
can be solved independently to the thermal boundary layer equation.

(ii) On the other hand, the thermal boundary layer equation.

(Iii) is liner in character and it depends on the velocity field and hence it can be
solved only when the velocity field is known. In what fellows we shall use solutions of (i) and (ii) as already
obtained and then solve (iii)

Note 3. Since (iii) is linear in character hence its solution is essay form. Mathematics point of view because
the law or superposition of known solutions holds.

3.4 Temperature distribution in the spread of a jet:

In chapter 18, we have already studied the velocity distribution in the spread of a jet, for three
cases (i) plane free jet in Art 18.12 (ii) lane wall jet in Art. 18.13 And (iii) circular jet in Art 18.14. We now
propose to study the corresponding temperature distribution, neglecting dash heat due to fiction. Note
38
that these are those examples, where compact solutions of the thermal boundary layer equation are
possible.

3.5 Plane free jet (Two - dimensional jet):

Refer figure of Art. 18.12 in chapter 18 consider the flow of an incompressible fluid passing
through a slit O in the wall AB and the mixing with the surrounding fluid which is at rest and has a
temperature 𝑇∞. With the origin in the slit O, we choose x - axis and y - axis along and normal to the plane
wall respectively neglecting the heat due to friction , the velocity and thermal boundary layer equations
are given by

𝛛𝐮/𝛛𝐱 + 𝛛𝐯/𝛛𝐲 = 0 …………… (1)

u ( 𝛛𝐮/𝛛𝐱) + v ( 𝛛𝐯/𝛛𝐲 ) = v(𝛛𝟐 𝐮/𝛛𝐲 𝟐 ) ………… (2)

And u ( 𝛛𝛉/𝛛𝐱) + v ( 𝛛𝛉/𝛛𝐲 ) = (𝐯/𝐏𝐫 ) × (𝛛𝟐 𝛉/𝛛𝐲 𝟐 ) ………(3)

Where 𝛉 = (𝐓 − 𝐓∞ ) / 𝐓∞

The boundary conditions for the given problem are

y =0:𝜕𝑢/𝜕𝑦 = 0, v =0, 𝜕𝜃/𝜕𝑦 = 0, (due to symmetry) ……… (4)

y=±∞ ; u = 0, 𝜃 =0 ………………………… (5)

In addition to these boundary conditions, for a non - trivial solution, the following integral conditions
should also be satisfied

∞
∫−∞ 𝝆𝒖𝟐 dy = const = 𝑱𝟎 (say) …………….(6)

39
 ∞ 𝑯
And ∫−∞ 𝜽udy = const = 𝑻 𝟎 (say) …………(7)
∞

Where 𝑱𝟎 and 𝑯𝟎 are to be prescribed.

For proof of (5), refer Art. 18.12. We now prove (6)

From (1), we have

𝛉(𝛛𝐮/𝛛𝐱) + 𝛉(𝛛𝐯/𝛛𝐲)= 0 ……………(8)

𝝏𝒖 𝝏𝜽 𝝏𝒗 𝝏𝜽 𝒗 𝝏𝟐 𝜽
Adding (3) and (7) (𝜽 + 𝝁 ) + (𝜽 + 𝒗 ) = ×
𝝏𝒙 𝝏𝒙 𝝏𝒚 𝝏𝒚 𝒑𝒓 𝝏𝒚𝟐

𝝏(𝜽𝒖) 𝝏( 𝜽𝒗) 𝒗 𝝏 𝝏𝜽
Or + = × ( )
𝝏𝒙 𝝏𝒚 𝒑𝒓 𝝏𝒚 𝝏𝒚

Integrating it with respect to y between the limits -∞ 𝑡𝑜 ∞, we get

𝒅 ∞ 𝒗 𝝏𝜽 ∞
∫ 𝜽𝒖𝒅𝒚 + [𝜽 𝒗]∞
−∞ = × [ ] ……………… (8)
𝒅𝒙 −∞ 𝒑𝒓 𝝏𝒚 −∞

Since at the edge of the boundary 𝝏𝜽/𝝏𝒚 = 0 and𝜽 = 𝟎 when y = ±∞ …….. (8)

𝒅 ∞ ∞ 𝑯𝟎
∫ 𝜽𝒖𝒅𝒚 = 0
𝒅𝒙 −∞
so that ∫−∞ 𝜽𝒖𝒅𝒚= const =
𝑻∞
(say).

Since the velocity field given by (1) and (2) is independent of the temperature field, therefore we
find first the velocity field and use this to find the temperature

40
 u = 6v𝜶𝟐 𝒙−𝟏/𝟑sec𝒉𝟐 𝝃 and v = 2𝜶v𝒙−𝟐/𝟑 (2 𝝃 sec𝒉𝟐 𝝃 - tanh 𝝃) …………… (9)

Where 𝝃 = 𝜶 y 𝒙−𝟏/𝟑 and 𝜶 = ( 𝑱𝟎 / 𝟒𝟖 𝝆 𝒗𝟐 )𝟏/𝟐 ……………………….. (10)

Field from (3). In Art 18.12 we have found an exact solution of the velocity distribution. The necessary
results which we shall require are given by a solution of the thermal boundary layer equation( 3 ) can be
found by first reducing it into an ordinary differential equation with help of the transformation

𝜽 =( 𝑯𝟎 / 6v𝜶𝑻∞ ) × 𝒙−𝟏/𝟑 g ( 𝝃) …………………………… (11)

Substituting u, vand𝜽 from (9) and (10) in (3), we get

𝒅𝒈′ 𝒅
𝒈′′ + 2Pr{𝒈′ 𝒕𝒂𝒏𝒉𝝃 + 𝒈 𝒔𝒆𝒄𝒉𝟐 𝝃 } = 0 or + 2Pr (g tanh 𝝃 ) = 0 …....... (12)
𝒅𝝃 𝒅𝝃

With the boundary conditions 𝝃 = 0: 𝒈′ = 0 and 𝝃 = ±∞; g=0 ………. (13)

∞
the integral condition (6) reduces to ∫−∞ 𝒈 𝒔𝒆𝒄𝒉𝟐 𝝃 𝒅 𝝃 = 𝟏 ………………. (14)

Where the prime denotes differentiation with respect to𝝃.

Integrating (11),

𝒈′ + 𝟐𝑷𝒓𝒈 𝐭𝐚𝐧𝐡 𝝃 = 𝑪 ,C being an arbitrary constant ……. (15)

Putting 𝝃 = 0 in (14) and using (12), we get C = 0. Then, (14) becomes

𝒅𝒈 𝒅𝒈
= - 2Pr g tanh 𝝃 or = - 2 Pr tanh 𝝃 d𝝃
𝒅𝝃 𝒈

Integrating, log g – log C = - 2Pr log cosh 𝝃 = 2 Pr log( 𝒄𝒐𝒔𝒉 𝝃 ) −𝟏 = 2 Pr log sech𝝃
41
 Or log (g/C) = log (𝒔𝒆𝒄𝒉 𝝃)𝟐𝑷𝒓 or g = C (𝒔𝒆𝒄𝒉 𝝃)𝟐𝑷𝒓 ……………. (16)

Relation (15) automatically satisfies the second boundary condition (namely, g = 0 when (𝝃 = ± ∞ ). Hence
C can obtain by suing (13). This leads to

∞ −𝟏
C = { 𝟐 ∫𝟎 (𝒔𝒆𝒄𝒉𝝃) 𝟐+𝟐 𝑷𝒓 𝒅 𝝃 } ………………….. (17)

As particular case, when Pr = 1, (17) reduces to

∞ −𝟏
C = { 𝟐 ∫𝟎 𝒔𝒆𝒄𝒉𝟒 𝝃 𝒅𝝃} ……………………… (18)

∞ ∞
But {∫𝟎 𝒔𝒆𝒄𝒉𝟒 𝝃 𝒅𝝃} = ∫𝟎 𝒔𝒆𝒄𝒉𝟐 𝞷 (1 - tan𝒉𝟐 𝞷 ) d𝞷,

As sec𝒉𝟐 𝞷 = 1 – tan 𝒉𝟐 𝞷
𝟏
=∫𝟎 ( 𝟏 − 𝒕𝟐 ) dt,* putting tanh𝝃 = t and sec𝒉𝟐𝞷 d𝞷
𝟏
𝒕𝟑
= [𝒕 − ( )] = 1 – (1/3) = 2/3
𝟑 𝟎

𝟐 −𝟏
Then, (17) gives C = [𝟐 × ( )] = 3/4 for Pr = 1. ……………… (19)
𝟑

𝑯𝟎
𝜽 = (C ) 𝒙−𝟏/𝟑 (𝒔𝒆𝒄𝒉𝞷)𝟐𝑷𝒓 ……………… (20)
(𝟔𝒗𝜶𝑻∞)

Thus the temperature distribution in a plane free jet flow in the absence of frictional heat is given with
help of (11) and (16) in the following form

Where C is given by (17)

42
3.6 The plane wall jet
Refer figure of Art. 18.13 In chapter 18. The flow in a viscous jet bounded on one side by a wall
and on the other by fluid at rest in known as a plane wall jet. Let an incompressible viscous fluid be
discharged through a narrow slit at O in half space along a plane and mix with the surrounding fluid which
is at rest and has a temperature𝑻∞. The wall is also maintained at the same constant temperature 𝑻∞ with
the origin in the slit O, we choose x – axis andy - axis along and normal to the plane wall respectively.

Neglecting the heat due to friction, the velocity and thermal boundary layer equations are given by

𝛛𝐮/𝛛𝐱 + 𝛛𝐯/𝛛𝐲 = 0 …………… (1)

u ( 𝛛𝐮/𝛛𝐱) + v ( 𝛛𝐯/𝛛𝐲 ) = v(𝛛𝟐 𝐮/𝛛𝐲 𝟐 ) ………… (2)

And u ( 𝛛𝛉/𝛛𝐱) + v ( 𝛛𝛉/𝛛𝐲 ) = (𝐯/𝐏𝐫 ) × (𝛛𝟐 𝛉/𝛛𝐲𝟐 ) ……… (3)

Where 𝛉 = (𝐓 − 𝐓∞ ) / 𝐓∞ ………………………. (4)

The corresponding boundary conditions are

y =0: 𝒖 = v =0, 𝜽 =0 and y= ∞ ; u = 0, 𝜽 = 0 ……. (5)

In addition to the boundary condition, for a non-trivial solution, the following integral should also
satisfied

∞ 𝒚
∫𝟎 𝒖𝟐 (∫𝟎 𝒖𝒅𝒚)dy = const. = E (say) ……………………. (6)

∞ 𝒚
And ∫𝟎 𝜽𝒖 (∫𝟎 𝒖𝒅𝒚)dy = const. = I (say) ……………………. (7)

For proof of (6), refer Art. 18.13. Following the similar method, we can easily prove (7) since the
velocity field given by (1) and (2) is independent of the temperature field; therefore we find first the

43
velocity distribution and use this to find the temperature field from (3). In Art 18.13, we have found an
exact solution of the velocity distribution. The necessary results which we shall require are given by

U = (𝑬/𝒗𝒙)𝟏/𝟐 𝒇′ (𝜼) , v = (𝟏/𝟒) × (𝒗𝑬/𝒙𝟑 )𝟏/𝟒 × {𝟑𝜼𝒇′ (𝜼) − 𝒇 (𝜼)} ,

Where 𝝶 = (𝑬/𝒗𝟑 )𝟑/𝟒 × y𝒙−𝟑/𝟒 and f(𝜼) = 𝒇∞ F(𝜼), 𝒇∞= (𝟒𝟎)𝟏/𝟒…………. (8)

And F (𝝶) is given by the equation

𝟐 𝟏 +𝑭+√𝑭 √𝟑𝑭
𝝶 = {𝒍𝒐𝒈𝒂 𝟐 + 𝟐√𝟑 𝒕𝒂𝒏−𝟏 }…………………. (9)
𝒇∞ (𝟏−√𝑭) 𝟐+√𝑭

Before solving (3) for arbitrary values of Pr we find that Pr = 1,

𝜽 = (𝟏/𝑬)u …………………… (10)

Is a solution of (3) . It is known as croc’s integral.

In order to solve (3) for arbitrary values Pr, keeping the croc’s integral (10) in mind, we take

𝟏
𝜽 = (𝟏/𝑬) × (𝑬/𝒗𝒙)𝟐 × h(𝜼) ………………………… (11)

Substituting u and v from (8) and 𝜃 from (1)in(3), we see that (3) reduces to an ordinary differential
equation

4𝒉′ + Pr(𝒇𝒉′ + 𝟐𝒇′ 𝒉) = 0 …………………. (12)

Within the boundary condition

𝜼 = 𝟎, h=0 and 𝜼 = ∞,h = 0 ……………. (13)

44
Where prime denotes the differentiation with respect to 𝝶

∞
∫𝟎 𝐡𝐟𝐟 ′d𝝶 = 1, ……………………….. (14)

In order solve (12), the following transformations of the independent and dependent variable will be sued;

𝟐 𝟐
S = {𝑭(𝒏)}𝟑/𝟐 and H(s) = × ∫∞ 𝒉(𝒏) ………. (15)
𝟑

Then (12) reduces to

𝒅𝟐 𝑯 𝟐 𝟐 𝒅𝑯 𝟒
S (1-s) 𝟐
+{ − ( − 𝑷𝒓 + 𝟏) 𝒔} + 𝑷𝒓𝑯 = 𝟎, ……….. (16)
𝒅𝒔 𝟑 𝟑 𝒅𝒔 𝟑

With the boundary conditions:

S = 0, H = 0 and s→ 𝟏, H = 0 …………. (17)

And the integral condition

𝟏
∫𝟎 𝑯 𝒔𝟏⁄𝟑 𝒅𝒔 = 𝟏. …………. (18)

Solution of the hyper geometric equation (16) is given by

H(s)=𝑨𝟐 𝑭𝟏 (𝒂, 𝒃; 𝒄; 𝒔) + 𝑩 𝒔𝟏⁄𝟐 𝟐𝑭𝟏 (𝒂 − 𝒄 + 𝟏, 𝒃 − 𝒄 + 𝟏; 𝟐 − 𝒄; 𝒔). ……….(19)

Where a + b = (𝟐⁄𝟑) - Pr, ab = -(𝟒⁄𝟑) × Pr and c = (𝟐⁄𝟑) ………. (20)

45
Since Pr> 0 for a fluid, it follows that the conditions of absolute convergence of the series are satisfied
here. Putting s = 0 is (19) and using (17), we find A = 0.

Now, re-writing the solution of (19), we have

𝟏⁄ (𝟏−𝒔)𝒇𝟏
𝒔 𝟑 2F1 ( a –c + 1, b-c +1; 2-c) = 𝒔𝟏/𝟑(𝟏 − 𝒔)𝒇𝟏 2F1 (𝟏 − 𝒂, 𝟏 − 𝒃, 𝟐 − 𝒄; 𝒔) …. (21)

From (21), we see that the second boundary condition (namely, s→ 0; H = 0 is identically satisfied by (19),
Thus, B still an unknown constant Using (18), B is given by

𝟏 −𝟏
B = {∫𝟎 𝒔𝟐/𝟑 𝟐𝑭𝟏 (𝒂 − 𝒄 + 𝟏, 𝒃 − 𝒄 + 𝟏; 𝟐 − 𝒄, 𝒔)𝒅𝒔}

𝟏 −𝟏
= (𝟓/𝟑) × {∫𝟎 𝒔𝟐/𝟑 𝟑𝑭𝟏 (𝒂 − 𝒄 + 𝟏, 𝒃 − 𝒄 + 𝟏; 𝟐 − 𝒄, 𝒔)𝒅𝒔} …… (22)

H (S) = B𝑺𝟏/𝟐 2𝑭𝟏 (𝒂 − 𝒄 + 𝟏, 𝒃 − 𝒄 + 𝟏; 𝟐 − 𝒄; 𝒔) ……………. (23)

Where B is given by (22)

The dimensionless temperature gradient at the wall can be obtained by using (11), (15), and (23)
and it is given by

𝝏𝜽 𝟏
( ) = (𝒗𝑬𝒙)𝟏/𝟐
𝒂𝟏(𝑷𝒓) , ………………. (24)
𝝏𝜼 𝜼=𝟎

𝒂𝟏 (𝑷𝒓) = B/𝜹𝒇∞. …………….. (25)

46
Where

The heat flux through a across section of the boundary layer is given by

∞ 𝑩𝑰
𝑾(𝒙) = ∫𝟎 𝜽𝒖𝒅𝒚 = (𝒗𝑬𝒙)𝟏/𝟒 3𝑭𝟐 (𝒂 − 𝒄 + 𝟏; 𝒃 − 𝒄 + 𝟏; 𝟐 − 𝒄, 𝟐; 𝟏) …….. (26)
𝒇 ∞

Again from Art, 18.13, the volume flux through a cross- section of the jet is given by

Q = 𝒇∞ (𝒗𝑬𝒙)𝟏/𝟒 …………………… (27)

From (26) and (27) , we have

QW = (𝟓/𝒍𝟑)𝒃𝟏(𝑷𝒓) …………………….. (28)

𝟑𝑭𝟐 (𝒂−𝒄+𝟏,𝒃−𝒄+𝟏;𝟐−𝒄 ,𝟐;𝟏)

𝒃𝟏 (𝑷𝒓) = 𝟓 …………… (29)
𝟑𝑭𝟐 (𝒂−𝒄+𝟏,𝒃−𝒄+𝟏,𝟑,;𝟐−𝒄 ,𝟐,𝜹/𝟑;𝟏)

Where

When Pr = 1, the series on the R.H.S. of (29) terminate and we easily get = (20/9) I, which is in conformity
with equation = (20/9) E of Art 18.13, on noting the relation (10)

The relation (28) I show that the product of the volume and heat flux through # cross -
section of the boundary layer , is constant for a given Prandtl number and this may be taken as the physical
meaning of the integral condition (7).

In the following figure we have potted H, which is proportional to the

dimensionless temperature distribution, against the similarity variable 𝝶 for different values of Pr.

47
 Fig.Temperature distribution in a plan wall jet.

From the above figure, we find that the curve corresponding to Pr = 1 is the same as
for the velocity distribution for Pr> 1, the thermal boundary layer is thinner than the velocity boundary
layer whereas for Pr< 1, the thermal boundary layer is thicker than the velocity boundary layer.

48
 CONCLUSIONS–3
3.7 The transient thermal boundary - layer response due to a convicted no uniform temperature fluid was
numerically investigated for an incompressible laminar flow under the assumption of a steady hydraulic
boundary layer on a semi - infinite plate. Far-field temperature changes that result in changes in the
magnitude and/or direction of the fluid-plate heat transfer were considered, including the case where no
thermal boundary layer exists initially. Unlike the problems of uniform change of fluid or plate
temperature, the boundary layer experiences no changes until the convicted far-field variation or contact
surface arrives at a given plate location. A time-accurate boundary-layer flow and temperature
computation methodology was developed, verified against some analytical solutions and prior work, and
then applied to predict the boundary-layer temperature profiles in response to a convicted step change in
fluid temperature. The analysis of the local instantaneous Nussle number shows that the use of steady
state heat transfer correlations in transient thermal states can cause both magnitude and direction of heat
transfer rate to be incorrectly predicted during the transition. The response time and thermal boundary
layer thickness are correlated with the fluid Prandtl number. It is anticipated that the numerical
methodology developed here will be extended to compressible and turbulent flow. The errors related to
quasisteady assumptions may be less significant for turbulent flow in gases but must be quantified for
either justification or relaxation of this assumption. Practical no steady devices such as the wave rotor
subject to complex temperature variations of a flowing gas may then modelled with a realistic transient
heat transfer prediction. This will enable important technologies to be improved for energy conversion
efficiency and other benefits.

49
 CHAPTER - 04
FLOW OF COMPRESSIBLE VISCOUS FLUIDS

4.1 Introduction.

While discussing flow of an incompressible viscous fluid, we determine only the velocity and
pressure at each point of the fluid. On the other hand, while discussing flow of a compressible viscous fluid,
we have to determine density and temperature in addition to the velocity and pressure at each point of
the fluid. For ready reference, the complete list of basic equations governing the flow of compressible
viscous fluid is given below both for Cartesian and cylindrical coordinates:

4.2 One - dimensioned flow of a compressible viscous fluid .

Consider the laminar steady one-dimensional flow, without body forces, of a

compressible viscous fluid moving parallel to x-axis. Assume that all the variables are functions of x alone.
Under the present flow conditions, we have

u = u(𝑿), v = 0, w = 0, p = p(𝑿), 𝝏/𝝏𝒚 = 0, 𝝏/𝝏𝒛 =0

In view of ( 1 ) , the basic equations for flow of one - dimensional steady compressible
viscous fluid are given by ( refer equation (1) , (2a), (2b), (2c), (3) and (4) of Art. 21.1)

d(𝝆𝒖)/dx = 0 ………………………. (1)

𝒅𝒖 𝒅𝒑 𝟒 𝒅 𝒅𝒖
𝝆𝒖 = - + (𝝁 ) ……………….. (2)
𝒅𝒙 𝒅𝒙 𝟑 𝒅𝒙 𝒅𝒙

50
 𝒅𝑻 𝒅𝒑 𝟒 𝒅𝒖 𝟐 𝒅 𝒅𝑻
𝝆𝒖𝑪𝒑 -u - 𝝁( ) - (𝒌 ) =0 ……………… (3)
𝒅𝒙 𝒅𝒙 𝟑 𝒅𝒙 𝒅𝒙 𝒅𝒙

P = 𝝆RT ……………………… (4)

Integrating (1) 𝝆𝒖 = 𝝆𝟎 𝒖𝒐 ……………..… (5)

Where 𝝆𝟎 , is the density at the reference velocity 𝒖𝟎, Using (5), (2) becomes

𝒅𝒖 𝒅𝒑 𝟒 𝒅 𝒅𝒖
𝝆𝟎 𝒖𝟎 + - (𝝁 )=0
𝒅𝒙 𝒅𝒙 𝟑 𝒅𝒙 𝒅𝒙

Integrating it, 𝝆𝟎 𝒖𝟎 𝒖 + 𝝆 − (𝟒𝝁/𝟑) × (𝒅𝒖/𝒅𝒙) = 𝑪, 𝑪 𝑏𝑒𝑖𝑛𝑔 𝑎𝑛 𝑎𝑟𝑏𝑖𝑡𝑟𝑎𝑟𝑦 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

Or 𝝆𝝁𝟐 +𝝆-(𝟒𝝁/𝟑) × (𝒅𝒖/𝒅𝒙) = C, using (5)

Hence, 𝛒 = ( 4𝛍/𝟑) × (𝐝𝐮/𝐝𝐱) - 𝛒𝛍𝟐 + C …………….. (6)

𝝆𝑹𝑻 = ( 4𝝁/𝟑) × (𝒅𝒖/𝒅𝒙) - 𝝆𝝁𝟐 + C, using (4)

𝟒𝝁 𝒅𝒖 𝑪𝒖
Or RT = ( 4𝝁/𝟑𝝆) × (𝒅𝒖/𝒅𝒙) - 𝝁𝟐 + C/𝝆 or RT = u - 𝒖𝟐 + ,
𝟑𝝆𝟎 𝒖𝟎 𝒅𝒙 𝝆𝟎 𝒖𝟎

using (5) …. (7) Equations (4) and (7) can be solved numerically by using relationship between µ and T
given by

𝝁 𝑻 𝒎
=( ) , where 0.5 < m < 1 ……….. (8)
𝝁𝟎 𝑻𝟎

51
4.3 Laminar flow of a compressible viscous fluid through a circular pipe.

Consider the Laminar steady flow, without body forces of a compressible fluid through an infinite
circular pipe of radius a with axial symmetry as shown in the following figure

For the present problem, we consider all basic equations for compressible fluid flow in
cylindrical co - ordinates. Let Z be the direction of the flow along the axis of the pipe and r denote the
radial direction measured outwards form the z -axis. For the present axis - symmetrical flow (𝜕/𝜕𝜃 = 0),
the velocity components 𝒒𝒓 and𝒒𝜽 in radial and tangential directions, respectively, are zero.

Under these flow conditions the equation of continuity (refer equation (1) of Art. 21.1) reduces to
𝝏(𝝆𝒒𝒛 )/𝝏𝒛 =0 so that 𝐪𝟐 = 𝐪𝟐(𝐫)

Thus, for the present configuration, we have

𝒒𝒓 = o, 𝒒𝟎 = 0, 𝒒𝟐 = 𝒒𝟐 (r), 𝝏/𝝏𝝑 = 0 ……… (1)

In view of (1), the Navies - Stokes equations (refer equation (2a), (2b) and (2c)

For the

𝜽= - (𝝏𝒑/𝝏𝒓) …………… (2)

𝜽= -(1/r)(𝝏𝒑/𝝏𝜽) …………. (3)

52
 And

𝝏𝒑 𝒅 𝒅𝒒𝒛 𝝁 𝒅𝒒𝒛
𝜽=- + (𝝁 )+ …………. (4)
𝝏𝒛 𝒅𝒓 𝒅𝒓 𝒓 𝒅𝒓

From (2) and (3) 𝝏𝒑/𝝏𝒓 = 0 and 𝝏𝒑/𝝏𝜽 = 0

So that the pressure depends on z only. Hence, (4) can be re-written as

𝒅𝒑 𝟏 𝒅 𝒅𝒒𝒛
= (𝝁 ) ……………… (5)
𝒅𝒛 𝒓 𝒅𝒓 𝒅𝒓

Differentiating both sides of (5) w. r. t. z, we get

𝒅 𝒅𝝆
𝒅𝟐 p/d𝒛𝟐 = 0 so that ( ) = 0
𝒅𝒛 𝒅𝒛

Integrating (12) ,dp/dz = P, where P is a constant ………….. (6)

From (5) and (6), we have

𝒅
(r 𝒅𝒒/𝒅𝒛) = Pr ………….. (7)
𝒅𝒓

Integrating (7),

𝜇r (d𝒒𝒓 / dr) =Pr² /2 + A, A being an arbitrary constant … (8)

Let 𝜏𝑤 be the skin friction at the surface of the cylinder so that when r = a,

53
we have

[𝛍((𝛛𝐪𝐫)/𝐝𝐫)]𝐫=𝐚 = 𝛕𝐰 ……………… (9)

Putting r = a in (8) and using (9), we get at ,

a𝝉𝒘 = P𝒂𝟐 /2 + A so that A = a𝝉𝒘 – P𝒂𝟐 /2 …………. (10)

From (8) and (10),

𝜇r (d𝒒𝒓 / dr) = a 𝝉𝒘 + P (𝒓𝟐 − 𝒂𝟐 )/ 2 ………………… (11)

Finally, for the present flow satisfying (1), energy equation (refer equation ( 3 ) Art 21.1 ) reduces

to

𝟏 𝐝 𝐝𝐓 𝐝𝐪𝛕 𝟐 𝒅 𝒅𝑻 𝐝𝐪𝛕 𝐝𝐪𝛕

𝛉= (𝐫𝐤 ) + 𝛍( ) Or (𝒓𝒌 ) = (𝝁𝒓 ( )) × ( )
𝐫 𝐝𝐫 𝐝𝐫 𝐝𝐫 𝒅𝒓 𝒅𝒓 𝐝𝐫 𝐝𝐫

𝒅 𝒅𝑻 𝟏 𝒅𝒒𝒛
Or (𝒓𝒌 ) = {𝒂 𝝉𝒘 + 𝑷( 𝒓𝟐 − 𝒂𝟐 )} × , using (11) ………….. (12)
𝒅𝒓 𝒅𝒓 𝟐 𝒅𝒓

Integrating (12), we have

𝒅𝑻 𝒓 𝒅𝒒𝒛 𝑷 𝒓 𝒅𝒒𝒛
𝒓𝒌 = ∫𝟎 𝒂𝝉𝒘 𝒅𝒓 + ∫( 𝒓𝟐 − 𝒂𝟐 ) × 𝒅𝒓 + B, B being an arbitrary constant ... (13)
𝒅𝒓 𝒅𝒓 𝟐 𝟎 𝒅𝒓

𝒅𝑻 𝑷 𝒓 𝒅𝒒𝒛
Or 𝒓𝒌 + a 𝝉𝒘 𝒒𝒛 = B - ∫𝟎 (𝒂𝟐 − 𝒓𝟐 ) × dr ……………. (14)
𝒅𝒓 𝟐 𝒒𝒓

While solving the above equation, we assume that prandtl number 𝑷𝒓 and 𝑪𝑷 , are constants. Clearly, (14)
can be integrated provided we know the variation of the coefficient of viscosity 𝝁 and the coefficient of

54
thermal conductivity k. From either the simple kinetic theory of gases or empirical data 𝝁 is usually
expressed with sufficient accuracy as a power of the absolute temperature is the form:

𝝁 𝑻 𝒎
=( ) , 0.5 < m < 1 ………………. (15)
𝝁𝟎 𝑻∞

Where 𝝁∞ is the value of 𝝁 at the moving plate? For air ordinary temperature, we usually take m= 0.76 .
As the temperature increases, m decreases towards 05 .

(𝝁𝑪𝒑 ) (𝝁𝑪𝒑 )
Now, 𝑷𝒓 = ⇒ k= ……………… (16)
𝒌 𝑷𝒓

Using (15) and (16), equations (11) and (14) can be integrated numerically to computer 𝒒𝒛 and T .
Equation of state is given by P = k 𝝆T ……………………… (17)

From which pressure can be obtained.

4.4 Laminar steady flow of a compressible viscous fluid between two concentric rotating
cylinders.

Consider two infinitely long, concentric circular cylinders of radii 𝑟1 and𝑟2 , ( 𝑟1 < 𝑟2 ) rotating with
constant angular velocities 𝝎𝟐 , respectively. Let there be a compressible viscous fluid in the annular space.
Then the cylinders induce a steady,

55
Axi-symmetrical, tangential motion in the fluid so that we have only the tangential component of velocity
𝑞0 . Thus here 𝒒𝒓 = 𝒒𝒛 = 0. Let z-axis be taken along the common axis of the cylinders.

Let the flow depend on only r so that for the present flow, we have

𝒒𝒓 = 0𝒒𝒓 = 𝒒𝟎 (z), 𝒒𝒛 = 0, P = P (r), T = T(r), 𝝏/𝝏𝜽 = 0, 𝝏/𝝏𝒛 = 0 ………… (1)

In view of ( 1 ) , the equation of continuity ( see equation ( 1 ) of Art 21.1 ) is satisfied identically and the
Navies Stokes equations reduce to ( refer equation ( 2a ) and ( 2b ) of Art 21.1 )

𝒒𝟐 𝜽 𝒅𝒑 𝒅𝒑 𝝆𝒒𝟐 𝜽
−𝝆 =− so that = ………………….. (2)
𝒓 𝒅𝒓 𝒅𝒓 𝒓

𝒅 𝒅𝒒𝜽 𝒒𝜽 𝟐𝝁 𝒅𝒒𝜽 𝒒𝜽
And 0= {𝝁 ( − )} + ( − )
𝒅𝒓 𝒅𝒓 𝒓 𝒓 𝒅𝒓 𝒓

or

𝒅 𝒅𝒒𝜽 𝒒𝜽 𝒅𝒒𝜽 𝒒𝜽
𝒓𝟐 {𝝁 ( − )} + 𝟐𝒓𝝁 ( − ) = 0 ………………….. (3)
𝒅𝒓 𝒅𝒓 𝒓 𝒅𝒓 𝒓

The boundary conditions are

At r = 𝒓𝟏 𝒒𝜽 = 𝒓𝟏𝝎𝟏 T = 𝑻𝟏; at r = 𝒓𝟐𝒒𝜽 = 𝒓𝟐 𝝎𝟐 , T = 𝑻𝟐; …..(4)

Let 𝜔 be the angular velocity at a point ( (𝑟 , 𝜃, 𝑧 ) in the fluid. Then we have 𝑞𝜃 = 𝑟𝜔 and hence (3)
reduces to

𝐝 𝐝(𝐫𝛚) 𝐝 𝐝(𝐫𝛚)
{𝐫 𝟐 𝛍 ( − 𝛚)} = 0 0r ( 𝐫𝟑𝛍 ( )) = 0
𝐝𝐫 𝐝𝐫 𝐝𝐫 𝐝𝐫

56
Integrating,

𝐫 𝟑 𝛍((𝐝𝛚)/𝐝𝐫) = A, A being an arbitrary constant given by (refer equation (3) of Art.21.1)

𝟏 𝒅 𝝁𝑪𝒑 𝒅𝑻 𝒅𝒒𝜽 𝒒𝜽 𝟐
0= ( 𝒓 ) + 𝝁( − )
𝒓 𝒅𝒓 𝑷𝒓 𝒅𝒓 𝒅𝒓 𝒓

𝟐
𝟏 𝒅 𝝁𝑪𝒑 𝒅𝑻 𝒅𝒒𝜽 𝒒𝟐𝟎 𝝁𝑪𝒑 𝝁𝑪𝒑
Or 0= ( 𝒓 )+ 𝝁 ( − ) , as 𝑷𝒓 ⇒k=
𝒓 𝒅𝒓 𝑷𝒓 𝒅𝒓 𝒅𝒓 𝒓 𝒌 𝑷𝒓

𝑪𝒑 𝟏 𝒅 𝒅𝑻 𝒅𝒒𝜽 𝒒𝜽 𝟐
(𝒓𝝁 ) + 𝝁( − ) =0
𝑷𝒓 𝒓 𝒅𝒓 𝒅𝒓 𝒅𝒓 𝒓

𝑪𝒑 𝟏 𝒅 𝒅𝑻 𝒅𝝎 𝟐
Or (𝒓𝝁 ) + r𝝁 (𝒓 ) =0
𝑷𝒓 𝒓 𝒅𝒓 𝒅𝒓 𝒅𝒓

𝑪𝒑 𝟏 𝒅 𝒅𝑻 𝒅𝝎
Or (𝒓𝝁 ) + A = 0, using (5)
𝑷𝒓 𝒓 𝒅𝒓 𝒅𝒓 𝒅𝒓

𝑪𝒑 𝒓𝝁
Integrating, ( ) × (𝑑𝑇/𝑑𝑟 ) + A𝜔 = B, B being an arbitrary constant ……. (6)
𝑷𝒓

And k from either the simple kinetic theory of gases or empirical data 𝜇 is usually expressed with sufficient
accuracy as a power of the absolute temperature in the form.

𝝁/𝝁𝟎 = (𝑻/𝑻∞ )𝒎, where 0.5 < m < 1 ……………. (7)

Where is the value of u at the moving plate? For air at ordinary temperature, we usually take m = 0.76. As
the temperature increase, m decreases towards 0.5.

Since𝑃𝑟 , is very nearly constant for all common gases and𝐶𝑝, is also nearly constant for a
fairly wide range of temperatures around ordinary temperature,

𝑷𝒓 = (𝑪𝒑 𝝁/)k ⇒ k is proportional to 𝝁

57
 Using the above facts equation (5) and (6) can be solved numerically to compute to and T. finally,
pressure can be obtained by using the equation of state, namely,

P = 𝒌𝝆𝑻

58
 CONCLUSION -4

4.5 The charactenstics of turbulent flows of air in the presence of solid particles in pipes and under condition of
flow past belies have been treated. The basic result given in the book includes.

( i ) The suggested classification of turbulent heterogeneous flows by the volume concentration and inertia of
particles in averaged large scale and small scale fluctuation motion .

(ii) The solution of large complex of metrological problems associated with the diagnostics of the structure of
particle - leaden turbulent flows of gas.

(iii)Detailed analysis of the characteristics of motion of particles and of their inverse effect on the parameters of
carries gas under conditions of flow in vertical pipes and of flow past bellies.

59
 CHAPTER - 05
FLOW OF INVISCID COMPRESSIVE FLUID. GAS DYNAMIC:

5.1 Introduction.
Flow of compressible fluids involves appreciable variation in density throughout a flow field
compressibility becomes significant at high flow speeds or for large temperature changes. Large changes
in velocity produces large pressure changes; for gas flows these pressure changes are produced along
with appreciable variations in both density and temperature Gases are treated as compressible fluids and
study of this type of flow is often referred to as 'gas dynamics’. Some important problems where
compressibility effect has to be considered are

(i) flights of aero plane at high altitude with high velocities

(ii) flow of gases through orifices and nozzles

(iii) Flow of gases in compressors.

Another important property of compressible fluids their ability to support wave motion. We shall
show that a small disturbance in a compressible fluid is propagated throughout the fluid as a wave
motion and at speed known as the speed of sound. On the other hand, a small disturbance created at
any point of an incompressible fluid is transmitted everywhere in the fluid.

We know that compressions and expansions of the gas involve work done on and by the gas,
respectively. This in turn produces changes in temperature of the gas. Hence the concepts of
thermodynamics will play an important role to determine the physical quantities involved in a given
problem.

In this chapter, large number of problems based on compressible fluid motion will be studied
with the assumption of steady one dimensional flow. Hence the variation of fluid properties will occur
only in one direction (namely, the direction of motion). It should be noted carefully, that this direction of
motion need not always be a straight - line motion.

It can be the motion along the curved axis of a stream filament. However, in order that fluid
property at each cross - section of the filament to be constant, the curvature of the filament should not
be too large. Although many real flows of interest are more compels, the above mentioned restrictions
allow us to concentrate on the effects of basic flow process.

60
5.2 Some thermodynamic relations for a perfect (or ideal) gas

(i) Perfect or ideal gas. In some cases, the molecules of a gas have but negligible volume and there are
almost no mutual attractions between the individual molecules. Such a gas is known as a perfect (or ideal) as.

(ii) Specific volume. The specific volume of as a gas is denoted by v and is defined as the volume
occupied by a unit mass of the gas. Thus, the specific volume v is the reciprocal of the density, i.e., v = 1 /ρ.

𝒅𝒑 𝒅𝒑
K= or K =𝝆 ,
−(𝒅𝒗/𝒗) 𝒅𝝆

Where we have used the fact that v = 1 /ρ so that dv = −(𝟏/𝝆𝟐 )𝒅𝝆.

(iii) Equation of state of a gas. Equation of state is defined as the equation which gives the relationship
between the pressures P, density ρ and temperature T of a gas. For a perfect gas the equation of state is
given by

P = ρRT…. (1)

Since p = 1/ v, (1) may be re-written as pv = RT … (2)

The constant R has different values for various gases. Its unit has the form

R = (energy) / (mass × tem) = J/ kg – K

Note: Here T is measure in absolute (or Kelvin) scale.Thus,if temperature of a body a body is t°C, then
its temperature on the absolute scale is T = (t + 273) K.

(iv) Specific heats of a gas. The specific heat C of a gas is defined as the amount of heat required to raise
the temperature of a unit mass of a gas by one degree. Thus C =𝝏⁄𝝏𝑻 , where 𝛿 is the amount of heat added
to raise the temperature by 𝜹T. The value of specific heat depends on two well-known processes, namely,
the constant volume process and constant pressure process. The specific heats of the above processes are
denoted and defined as follows:

Specific heat of a constant volume = CV= (𝝏𝑸⁄𝝏𝑻)v

Specific heat of a constant pressure = Cp= (𝝏𝑸⁄𝝏𝑻)p

61
(v) Internal energy. It is denoted by E and is defined as heat stored in a gas. When a certain amount of heat
is supplied to a gas, then temperature of gas may increase or volume of gas may increase thereby doing
some extremely work or both temperature and volume may increase. It has been observed that whenever
temperature of gas increases during its heating, its internal energy also increases. For a perfect gas, E is
function of temperature T only; i.e., E = E (T).

(vii) Enthalpy. The sum of internal energy E and pressure specific volume product pave is denoted by h and
is known as enthalpy per unit mass of the system. It is also called the total heat content. This, we have
h = E + pv.

(viii) Relations between C, and C, for a perfect gas. Let be the total thermal energy per unit mass, E the
internal energy per unit mass; p the pressure and v the specific volume of a perfect gas. Then, by the first
law of thermodynamics, we have.

dQ = dE + p dv … (1)

Note that for a perfect gas, E is function of T only. Hence, from (1),
We have

Cv = (𝝏𝑸⁄𝝏𝑻)v = dE/dT ... (2)

And Cp = (𝝏𝑸⁄𝝏𝑻)p = dE/dT + p(𝝏𝒗⁄𝝏𝑻)p,

Or Cp = Cv, + p (𝝏𝒗⁄𝝏𝑻 ) p using (2) … (3)

From (2) and (3) Cp = Cv + R … (4)

Or Cp - Cv = R … (5)

Pv = RT so that v = RT/p ... (6)

62
 Form (4), ( 𝝏𝒗⁄𝝏𝑻 ) p = R / p ... (7)

From the kinetic energy theory of a perfect gas we know that the internal energy is directly
proportional to the temperature T. Hence (2) shows that Cv, must be a constant. Again R is a constant for a
particular perfect gas. Hence (5) shown that Cp , is also constant.

Since Cp, and Cv, are both constants, we have

Cp / Cv = γ or Cp = γ Cv, … (8)

Where y is a constant known as the adiabatic constant (or adiabatic index). From (6), we have Cp>Cv, and
hence (8) shows that y > I. Solving (7) and (8), we have

Cv = R/ (𝜸 – 1) and C = 𝜸R/ (𝜸 - 1) … (9)

Now, from (2), E = CvT + C … (10)

Where C is constant of integration. If we choose the temperature scale in such a manner that E = O when T =
O. Then (10) given C = O and hence (10) reduces to

E = CvT. … (11)
Now, by definition of enthalpy (vii), we have

h = E + pv or h = E + RT, by equation of state

 dh/dT = dE/dT + R or dh/dT = C v + R, by (2)

dh/dT = Cp, using (6) … (12)

63
 (ix) Functions of state, entropy and isentropic flow, using the equation of state we can express any
thermodynamic function  in terms of any two of the measurable quantities p, ρ0 and T.

Since v = 1/ρ, we write  (p,v) as function of p and v. If, in changing from an initial state p to another state
p, the value of  (p, v) depends only on the conditions at p , and p and not at all on the various paths joining
them, then ( p,v ) is known as a function of state.

Let (p, v)=M (p, v) dp + N (p, v) dv, (1)

Where M (p, v) and N (p, v) are functions of p and v only. If ( - p) is independent of the path joining p , and
p, then d must be an exact differential. The necessary and sufficient condition for d to be exact differential
is

So when (2) holds, (p, v) must be a function of state.

From the first law of thermodynamics, we have dQ = dE + p dv … (3)

For a perfect gas, we know that dE = CvdT… (4)

And pv = RT so that dT = (1/R)  d (p … (5)

From (4) and (5), dE = (cv/R)  d (pv) … (6)

 From (3) and (6), dQ = (cv/R)  (pdv + vdp) + pdv

Or dQ = (CV / R)vdp + (1+Cv/R)pdv

Or dQ = (Cv /R) vdp + (R + Cv) (p/R) dv

64
 Or dQ = (Cv /R) vdp + (Cp /R)pdv, as Cp = R + CV

Or dQ =T (Cv/ p)dp + T (Cp / v)  dv, as R = (pv)/T

Or dS = (Cv/p)dp + (Cp/v)  dv … (7)

Where dS = (1 / T)dQ … (8)

Comparing (7) with (1), here M (p, v) = Cv / p and N (p, v) = Cp/ v

Hence (∂M / ∂v)p = O = (∂N / ∂p)v.

Showing that ds is an exact differentia land so S is an function of state.

Integrating (7),

S = Cvlog p + Cp log v + S0 ……….… (9)

Where S0 is constant of integration. Since Cp / Cv = 𝜸so that Cp = 𝜸Cv,(9) can be re-written as,

Cv log p + 𝜸Cv log v = S - S0 or log (p𝒗𝜸 ) = (S - S0 ) / Cv

Or p 𝒗𝜸 = exp [ (S – S0) /Cv] … (10)

65
 The quantity S is known as entropy per unit mass. Flows for which S is constant are called
isentropic. Since S0 and C0 are constants and S is constant for an isentropic flow, it follows that R.H.S. of
(10) must also be a constant. Thus, isentropic flows satisfy the relation

p𝒗𝜸 = constant … (11)

Since v =1/ρ, (10) gives S – S0 = Cv log (p / 𝝆𝜸 ) … (12)

(x) Isothermal, isentropic, adiabatic and home tropic processes:

(a)Isothermal process:

This is the process in which a gas is compressed or expanded while the temperature T is kept
constant. In the case of a perfect gas, we have VP=RT. Hence an isothermal process is characterized by
p v =constant, which is known as Boyle's law.

(b)Isentropic process:

If a change from a state P , to another state P takes places in such a manner that the entropy of
very single particle of gas stays constant, then such a change is known as an isentropic process. Hence an
isentropic process is characterized by pv = constant.

(c) Adiabatic process:

If the compression and expansion of a gas takes place in such a way that the gas neither gives heat nor
takes heat from its surroundings, then the process is said to be adiabatic. For a reversible adiabatic
process, we have pv² = constant.

(d) Home tropic process:

If the entropy of every single quantity of a gas of fixed mass is the same and stays constant in any change,
the, such a change is known as homentropicprocess.

Remark: Consider relation pv = constant

The constant on R.H.S. of (1) is the same for each considered small quantity of a gas in isentropic
flow but a different constant must be taken corresponding to each such quantity. For hemitropic flow,
however, the constant on R.H.S. of (1) is always the same throughout the entire volume of gas.

66
Example. Find the values of gas constant and adiabatic index for methane. Whose values of specific heat
at constant pressure and constant volume are 2.17 and 1.65respectively.

Sol. Let R be the required gas constant. Here Cp= 2.17 and Cv= 1.65. Hence, we have

R = Cp- C =2.17 – 1.65 = 0.52 J/kg – K

Also, adiabaticindex=𝛄 = Cp/ Cv = 2.17 / 1.65 = 1.32 .

5.3 Basic equations of motion of a gas.

(i) The equation of continuity:

In the case of steady motion, the equation continuity (see Art. 2.8, Chapter 2) is given by

 . (p, q) = 0,

Where q and ρ are velocity vector and density at any point p of a gas.

(ii) The equation of motion:

In the case of steady motion under no body forces, the vector equation of motion ( see Art 3.1,
Chapter 3 ) is given by

(q . )q = -(1 / ρ) p … (2)

Where p is the pressure at any point p of a gas.

67
(iii) Bernoulli's equation (Energy equation):

In the case of steady motion under no body forces, the Bernoulli's equation (See
Art 4.1, Chapter4) is

𝟏 𝐝𝐩
𝐪𝟐 + ∫ = constant, q = |q |= magnitude of velocity … (3)
𝟐 𝛒

We now re-write (3) for an isotropic flow-for which the entropy of each particle
remains constant along any streamline.

For such a flow, we have

𝟏⁄
P / 𝛒𝛄 constant= K, say  ρ = (𝐩/𝐤) 𝛄 … (4)

Substituting the value of p from (4), we obtain

𝐝𝐩 𝐝𝐩 𝐩 𝟏− 𝟏⁄𝛄 𝐩 𝟏⁄ 𝛄 𝛄 𝛄 𝐩
∫ 𝛒 =∫ (𝐩⁄𝐤)𝟏⁄𝛄 = 𝐤 𝟏⁄𝛄 ∫ 𝐩 −𝟏⁄𝛄 dp= 𝐤 𝟏⁄𝛄 𝟏− 𝟏⁄𝛄=(𝛒𝛄 ) 𝛄− 𝟏
𝐩𝟏− 𝟏⁄𝛄 =
𝛄−𝟏
×
𝛒
…. (5)

From (3) and (5),

𝟏 𝛄 𝐩
𝐪𝟐 + × =constant …………… (6)
𝟐 𝛄−𝟏 𝛒

Where the constant is the same along any streamline, but unless the flow is homentropic it will
vary from one streamline to another.

(iv) Equation of state:

For a perfect (ideal) gas, the equation of state is given by

P = ρRT … (7)
68
Where R is a constant for the particular gas under consideration.

Remark. Equations ( 1 ) , ( 2 ) , ( 6 ) and ( 7 ) can be solved to find p ,ρ and T. Solutions of these equations
can be obtained only by numerical methods. Closed form (analytical) solutions, however, exist for a
limited class of problems. Therefore, in what follows, we shall make some simplifying approximations. We
shall consider steady one-dimensional inviscid compressible flows.

5.4 Basic equations for one - dimensional flow of a gas.

Fig. Flow through stream tube

69
 In this chapter, large number of problems based on compressible motion will be studied with the
assumption of one dimensional flow. Accordingly, the variation of fluid properties will occur only in one
direction, namely, the direction of motion. It should be noted carefully that this direction of motion need
not always be a straight-line motion. It can be the motion along the curved axis of a stream filament.
However, in order those fluid properties at each cross-section of the filament to be constant, the
curvature of the filament should not be too large.

Although the restriction on one-dimensional flow appears to be too much this physical model is found to
be good approximation to many actual flows.

Let a gas flow parallel to OL through a pipe whose cross-section at any point p on OL be A. Let pressure
p, density ρ, temperature T and velocity V of the flowing gas be constant throughout the cross–section A.

(i) The equation of continuity:

Mass of gas moving through the cross - section at p along OL is ρAV. Hence by the law of
conservation of mass, the required equation of continuity is

ρAV = constant = k, say … (1)

Differentiating (1), we get AV dρ + ρVdA + ρAdV = 0

Dividing by ρAV, we get (1 / ρ) dρ + (1 / A) dA + (1/V) dV = 0 … (2)

Which is known as equation of continuity in the differential form.

(ii) The equation of motion :

It is obtained by equating the rate of change of momentum of the flowing gas in the direction of OL
to the net force in the direction of OL.

70
 (iii) Bernoulli’s equation (Energy equation):

Refer Art.20.3 Take q = V. Thus, we have

(iv) Equation of state:

𝛄 𝐩 𝐕𝟐
× + = 𝐩 = 𝐑𝛒𝐓
𝛄−𝟏 𝛒 𝟐

5.5 The one-dimensional wave equation:

Let x denote a distance associated with a time t, and c a constant. Then consider a function  (x, t)
defined by

In order that (1) may have any physical significance, the dimensions of c must be those of a speed.
Again, from (1), we have

(𝐱 + 𝐜𝐓 , 𝐭 + 𝐓) = 𝐟{(𝐱 + 𝐜𝐓) − 𝐜(𝐭 + 𝐓)} = 𝐟(𝐱 − 𝐜𝐭) … (2)

From (1) and (2), we get (𝐱 + 𝐜𝐓, 𝐭 + 𝐓) =  (𝐱, 𝐭) …(3)

Showing that the values of  at distance x at time t is the same as that at distance (x + cT) at time ( t + T )

71
 Fig. (i) Profile at t Fig. (ii) Profile at (t + T)

When  is plotted against x at time t , then as arbitrary shape so graph is drown as shown in figure ( i ).
Similarly the plot of  against x at time (t + T) has been shown in figure (ii). We find that the second figure
gives a profile congruent to first but shifted along the positive x-axis a distance cT.As the displacement
occurs is time T, we conclude that the profile must be moving with velocity c, in the direction of positive x-
axis. It follows that a possible physical interpretation of (1) is that (x, t) represents a disturbance moving in
the positive x-direction with velocity c, its profile remaining undistorted. Thus (1) provides us with a
mathematical model of one-dimensional wave propagation without distortion.

Combining (1) and (4), it follows that

Represents a disturbance moving without distortion in the negative x-direction with speed c. combining
(1) and (4), it follows that

 (𝐱, 𝐭 ) = 𝐟 (𝐱 − 𝐜𝐭) + 𝐠 (𝐱 + 𝐜𝐭) … (5)

Represents the superposition of a forward and a backward travelling wave, each moving with speed c.

Differentiating (5) partially w.r.t x twice, we have

𝛛⁄𝛛𝐱 = 𝐟 ′ (𝐱 − 𝐜𝐭) + 𝐠 ′ (𝐱 + 𝐜𝐭)

72
 And 𝛛𝟐 ⁄𝛛 𝐱 𝟐 = 𝐟 ′′ (𝐱 − 𝐜𝐭) + 𝐠 ′′ (𝐱 + 𝐜𝐭)…(6)

Again

Differentiating (5) partially w.r.t.‘t twice, we get

∂𝛛𝐭 = 𝐟 ′ (𝐱 − 𝐜𝐭). (−𝐜) + 𝐠 ′ (𝐱 + 𝐜𝐭). 𝐜

𝛛𝟐 ⁄𝛛𝐭 𝟐 =𝐟 ′′ (𝐱 − 𝐜𝐭). (−𝐜)𝟐 + 𝐠 ′′ (𝐱 + 𝐜𝐭). 𝐜 𝟐…(7)

From (6) and (7), we get

𝛛𝟐 ⁄𝛛𝐭 𝟐 = 𝐜 𝟐 (𝛛𝟐 /𝛛𝐱 𝟐 ).…(8)

Which is known as one-dimensional wave equation. The form of  given by (5) is its general solution.

Example.
Find the profile  (x, t) of a dimensional wave propagation if at

t=0,  = 𝐅 (𝐱) 𝐚𝐧𝐝 𝛛/𝛛𝐭 = 𝐆(𝐱).

Sol. One dimensional wave equation is given by

𝛛𝟐 /𝛛𝐭 𝟐 = 𝐜 𝟐 (𝛛𝟐 /𝛛𝐱 𝟐 ). … (1)

73
Given that

(𝐱, 𝟎) = 𝐅(𝐱) and (𝛛/𝛛𝐭)t=0 = G(𝐱) …….(2)

We know that the general solution of (1) is given by

(𝐱, 𝐭) = 𝐟 (𝐱 − 𝐜𝐭) + 𝐠 (𝐱 + 𝐜𝐭), …………. (3)

Where f and g are arbitrary functions.

Putting t = 0 in (3) and using (2), we have

f (x) + g (x) = F (x) …(4)

Differentiating (3) partially w.r.t.‘t’, we get

𝛛⁄𝛛𝐭 = 𝐟 ′ (𝐱 − 𝐜𝐭). (−𝐜) + 𝐠 ′ (𝐱 + 𝐜𝐭). 𝐜 ….. (5)

Putting t = 0 in (5) and using (2), we get

𝐆(𝐱) = −𝐜 𝐟 ′ (𝐱) + 𝐜𝐠 ′ (𝐱) 𝐨𝐫 − 𝐟 ′ (𝐱) + 𝐠 ′ (𝐱) = (𝟏/𝐜) 𝐆(x) … (6)

Integrating (6),

𝐱
−𝐟(𝐱) + 𝐠(𝐱) = 𝟐𝐀 + (𝟏/𝐜) ∫𝟎 𝐆(𝐱)𝐝𝐱, ………. (7)

Where 2A is an arbitrary constant.

Solving (4) and (7) for f (x) and g (x), we get

74
 𝟏 𝟏 𝐱 𝟏 𝟏 𝐱
𝐟(𝐱) = 𝐅(𝐱) − 𝐀 − ∫ 𝐆(𝐱)𝐝𝐱,and𝐠(𝐱) = 𝟐 𝐅(𝐱) + 𝟐𝐜 ∫𝟎 𝐆(𝐱)𝐝𝐱
𝟐 𝟐𝐜 𝟎

From these values of f(x) and g(x), we have

𝟏 𝟏 𝐱−𝐜𝐭
𝐟(𝐱 − 𝐜𝐭) = 𝐅(𝐱 − 𝐜𝐭) − 𝐀 − ∫ 𝐆(𝐱)𝐝𝐱,
𝟐 𝟐𝐜 𝟎

𝟏 𝟏 𝐱+𝐜𝐭
𝐠(𝐱 + 𝐜𝐭) = 𝐅(𝐱 + 𝐜𝐭)𝟔 + 𝐀 + ∫ 𝐆(𝐱)𝐝𝐱,
𝟐 𝟐 𝟎

Substituting the above values of f(x − ct) and g(x + ct) in (3), we get

𝟏 𝟏 𝟎 𝟏 𝐱+𝐜𝐭
(𝐱, 𝐭) = {𝐅(𝐱 − 𝐜𝐭) + 𝐅(𝐱 + 𝐜𝐭)} + ∫ 𝐆(𝐱)𝐝𝐱 + ∫ 𝐆(𝐱)𝐝𝐱
𝟐 𝟐𝐜 𝐱−𝐜𝐭 𝟐𝐜 𝟎

𝟏 𝟏 𝐱+𝐜𝐭
Or (𝐱, 𝐭) = {𝐅(𝐱 − 𝐜𝐭) + 𝐅(𝐱 + 𝐜𝐭)} + ∫ 𝐆(𝐱)𝐝𝐱, … (8)
𝟐 𝟐𝐜 𝐱−𝐜𝐭

5.6 Wave equations in two and in three dimensions:

Suppose a disturbance occurs in three dimensions in such a manner that the disturbance stays
constant over any plane perpendicular to the direction of propagation. Then the wave is known as a plane
wave and any such plane is known as wave front. Let n be unit vector and l , m , n be its direction cosines
so that

𝐧= 𝐥𝐢+𝐦𝐣+𝐧𝐤 Where f + m² + n² = 1 …(1)

75
 Suppose a wave front travels with speed c in direction n. Then the function f(lx + my + nz −
ct)will satisfy the necessary requirements because the wave fronts have equations lx + my + nz =
constant, at any considered time, Likewise g(lx + my + nz + ct) will represent a disturbance travelling in
the direction-n with the same speed. It follows that the function

(𝐱, 𝐲, 𝐳, 𝐭) = 𝐟(𝐥𝐱 + 𝐦𝐲 + 𝐧𝐳 − 𝐜𝐭) + 𝐠(𝐥𝐱 + 𝐦𝐲 + 𝐧𝐳 + 𝐜𝐭) … (2)

Will represent the superposition of plane waves travelling with speeds e in the directions± n .
Differentiating (2) partially w.r.t. x,y,z, and t , twice by turn , we finally obtain

𝛛𝟐 ⁄𝛛𝐱 𝟐 = 𝐥𝟐 𝐟′′ (𝐥𝐱 + 𝐦𝐲 + 𝐧𝐳 − 𝐜𝐭) + 𝐥𝟐 𝐠 ′′(𝐥𝐱 + 𝐦𝐲 + 𝐧𝐳 + 𝐜𝐭) … (3)

𝛛𝟐 ⁄𝛛𝐲 𝟐 = 𝐦𝟐 𝐟 ′′(𝐥𝐱 + 𝐦𝐲 + 𝐧𝐳 − 𝐜𝐭) + 𝐦𝟐 𝐠 ′′(𝐥𝐱 + 𝐦𝐲 + 𝐧𝐳 + 𝐜𝐭) … (4)

𝛛𝟐 ⁄𝛛𝐳𝟐 = 𝐧𝟐 𝐟 ′′ (𝐥𝐱 + 𝐦𝐲 + 𝐧𝐳 − 𝐜𝐭) + 𝐧𝟐 𝐠 ′′ (𝐥𝐱 + 𝐦𝐲 + 𝐧𝐳 + 𝐜𝐭)… (5)

𝛛𝟐 ⁄𝛛𝐭 𝟐 = 𝐜 𝟐 {𝐟 ′′(𝐥𝐱 + 𝐦𝐲 + 𝐧𝐳 − 𝐜𝐭) + 𝐠 ′′ (𝐥𝐱 + 𝐦𝐲 + 𝐧𝐳 + 𝐜𝐭)}

Or (𝟏/𝐜 𝟐 ) × (𝛛𝟐 ⁄𝛛𝐱 𝟐 ) = 𝐟 ′′ (𝐥𝐱 + 𝐦𝐲 + 𝐧𝐳 − 𝐜𝐭) + 𝐠 ′′ (𝐥𝐱 + 𝐦𝐲 + 𝐧𝐳 + 𝐜𝐭) … (6)

Adding (3), (4) and (5) and using (1), we get

𝛛𝟐
⁄𝛛𝐱 𝟐 + 𝛛𝟐 ⁄𝛛𝐲 𝟐 + 𝛛𝟐 ⁄𝛛𝐳 𝟐 = 𝐟 ′′ (𝐥𝐱 + 𝐦𝐲 + 𝐧𝐳 − 𝐜𝐭) + 𝐠 ′′ (𝐥𝐱 + 𝐦𝐲 + 𝐧𝐳 + 𝐜𝐭)

Thus 𝛛𝟐 ⁄𝛛𝐱 𝟐 + 𝛛𝟐 ⁄𝛛𝐲 𝟐 + 𝛛𝟐 ⁄𝛛𝐳 𝟐 = (𝟏/𝐜 𝟐 ) × 𝛛𝟐 ⁄𝛛𝐭 𝟐, using (7) …(7)

Which is the equation of wave equation is three dimensions.

Equation (2) is the general solution of (7).

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 Likewise two-dimensional wave equation is given by

𝛛𝟐 ⁄𝛛𝐱 𝟐 + 𝛛𝟐 ⁄𝐲 𝟐 = (𝟏⁄𝐜 𝟐 ) × (𝛛𝟐 ⁄𝛛𝐳 𝟐 ),

Whose general solution is given by

(𝐱, 𝐲, 𝐳) = 𝐟(𝐥𝐱 + 𝐦𝐲 − 𝐜𝐭) + 𝐠(𝐥𝐱 + 𝐦𝐲 + 𝐜𝐭), Where 𝐥𝟐 + 𝐦𝟐 = 𝟏

5.7 Spherical waves:

Wave equation 𝛁𝟐 𝐮 = (𝟏/𝐜 𝟐 ) × (𝛛𝟐 𝐮⁄𝛛𝐭 𝟐)in three dimensions in terms of spherical polar co -
ordinates (r,,) is given by

𝛛𝟐 𝐮 𝟐 𝛛𝐮 𝟏 𝛛 𝛛𝐮 𝟏 𝛛𝟐 𝐮 𝟏 𝛛𝟐 𝐮
+ + (𝐬𝐢𝐧 𝛉 ) + 𝟐 =
𝛛𝐫 𝟐 𝐫 𝛛𝐫 𝐫 𝟐 𝐬𝐢𝐧 𝛉 𝛛𝛉 𝛛𝛉 𝐫 𝐬𝐢𝐧 𝛉 𝛛𝟐 𝐜 𝐩 𝛛𝐭 𝟐

When there is spherical symmetry, we have 𝐮 = 𝐮(𝐫, 𝐭) and the above equation reduces to

𝛛𝟐 𝐮 𝟐 𝛛𝐮 1 𝛛𝟐 𝐮 𝛛𝟐 (𝐫𝐮) 𝟏 𝛛𝟐 (𝐫𝐮)
+ = 𝐨𝐫 = 𝟐 … (𝟏)
𝛛𝐫 𝟐 𝐫 𝛛𝐫 c 2 𝛛𝐭 𝟐 𝛛𝐫 𝟐 𝐜 𝛛𝐫 𝟐

Which is of the form of a one dimensional wave equation (compare equation (1) with equation
(8) of Art 20.5, on replacing x by r and  by (Ru). Hence the general solution of (1) is given by

77
 𝐫𝐮 = 𝐟(𝐫 − 𝐜𝐭) + 𝐠(𝐫 + 𝐜𝐭) 𝐨𝐫 𝐮(𝐫, 𝐭) = (𝟏/𝐫 ) × {𝐟(𝐫 − 𝐜𝐭) + 𝐠(𝐫 + 𝐜𝐭)}

Which represents concentric spherical wave fronts, with centre at origin O and having radii
which increase or decrease at speed c. Due to presence of the factor I/r in the solution, the wave profiles
will change with time.

78
 CONCLUSION - 5

5.8. Compressible fluids or gas dynamics is the branch of fluid mechanics that deals with glows having
significant charge in fluid density. While all flows are compressible flows are usually treated as being.

Incompressible when the Mach number, the study of compressible flow is relevant to high speed entry
into a planetary atmosptiere, gas pipelines commercial applications such as abrasive blasting and many
other fields.

79
CHAPTER – 06
VISCOUS INCOMPRESSIBLE FLOW
(FUNDAMENTAL ASPECTS)

Introduction
Being highly non-linear due to the convective acceleration terms, the Nervier-Stokes equations are
difficult to handle in a physical situation. Moreover, there are no general analytical schemes for solving the
nonlinear partial differential equations. However, there are few applications where the convective
acceleration vanishes due to the nature of the geometry of the flow system. So, the exact solutions are
often possible. Since, the Nervier-Stokes equations are applicable to laminar and turbulent flows, the
complication again arises due to fluctuations in velocity components for turbulent flow. So, these exact
solutions are referred to laminar flows for which the velocity is independent of time (steady flow) or
dependent on time (unsteady flow) in a well-defined manner. The solutions to these categories of the flow
field can be applied to the internal and external flows. The flows that are bounded by walls are called as
internal flows while the external flows are unconfined and free to expand. The classical example of
internal flow is the pipe/duct flow while the flow over a flat plate is considered as external flow. Few
classical cases of flow fields will be discussed in this module pertaining to internal and external flows.

Laminar and Turbulent Flows-

The fluid flow in a duct may have three characteristics denoted as laminar, turbulent
and transitional. The curves shown in Fig. 5.1.1, represents the x-component of the velocity as a function
of time at a point 'A' in the flow. For laminar flow, there is one component of velocity 𝐯⃗ = 𝐮𝑖̂and random
component of velocity normal to the axis becomes predominant for turbulent flows i.e. 𝐯⃗ = u𝐢̂ + v𝐣̂ + 𝐰𝐤 ̂
when the flow is laminar, there are occasional disturbances that damps out quickly. The flow Reynolds
number plays a vital role in deciding this characteristic. Initially, the flow may start with laminar at
moderate Reynolds number. With subsequent increase in Reynolds. Number, the orderly flow pattern is
lost and fluctuations become more predominant. When the Reynolds number crosses some limiting value,
the flow is characterized as turbulent. The changeover phase is called as transition to turbulence. Further,
if the Reynolds number is decreased from turbulent region, then flow may come back to the laminar state.
This phenomenon is known as relaminarization

80
 Fig 5.1.1: Time dependent fluid velocity at a point

The primary parameter affecting the transition is the Reynolds number defined as,
𝝆𝑼𝑳
RE = where, U is the average stream velocity and L is the characteristics Q PUL
𝝁
μ length / width .The flow regimes may be characterized by the following approximate ranges

0 Re < 1: Highly viscous laminar motion

1 < Re < 100: Laminar and Reynolds number dependence

𝟏𝟎𝟐< Re <𝟏𝟎𝟑 : Laminar boundary layer

𝟏𝟎𝟑< Re <𝟏𝟎𝟒: Transition to turbulence

𝟏𝟎𝟒 < Re <𝟏𝟎𝟔:Turbulent boundary layer

Fully Developed Flow

The fully developed steady flow in a pipe may be driven by gravity and/or pressure forces. If the pipe is
held horizontal, gravity has no effect except for variation in hydrostatic pressure. The pressure difference
between the two sections of the pipe essentially drives the flow while the viscous effects provide the
restraining force that exactly balances the pressure forces. This leads to the fluid moving with constant
velocity (no acceleration) through the pipe. If the viscous forces are absent, then pressure will remain
constant throughout except for hydrostatic variation. In an internal flow through a long duct is shown in
Fig. 5.1.2. There is an entrance region where the inviscid upstream flow converges and enters the tube.

81
The viscous boundary layer grows downstream, retards the axial flow [u(r x)] at the wall and accelerates
the core flow in the centre by maintaining the same flow rate.

Q = ∫ 𝒖𝒅𝑨 = Constant

Fig. 5.1.2 Velocity profile and pressure changes in a duct flow

82
 At a finite distance from entrance, the boundary layers form top and bottom wall merge as
shown in Fig. 5.1.2 and the in viscid core disappears, thereby making the flow entirely viscous. The axial
velocity adjusts slightly till the entrance length is reached ( x = 𝐿𝑒 ) and the velocity profile no longer
changes in x and u ≈ 𝑢( r ) only . At this stage, the flow is said to be fully-developed for which the velocity
profile and wall shear remains constant. Irrespective of laminar or turbulent flow, the pressure drops
linearly with x. The typical velocity and temperature profile for fully developed laminar flow in a pipe is
shown in Fig. 5.1.2. The most accepted correlations for entrance length in a flow through pipe of diameter
(d), are given below;

Laminar and turbulent shear

In the absence of thermal interaction, one needs to solve continuity and momentum.
Equation to obtain pressure and velocity fields. If the density and viscosity of the fluids is assumed to be
constant, then the equations take the following form;

This equation is satisfied for laminar as well as turbulent flows and needs to be subjected to no-
slip condition at the wall with known inlet/exit conditions. In the case of laminar flows, there are no
𝜕𝑢 𝜕𝑢 𝜕𝑢
random fluctuations and the shear stress term such as 𝜇 ,𝜇 and 𝜇 In x- direction.
𝜕𝑥 𝜕𝑦 𝜕𝑧

83
 For turbulent flows, velocity and pressure varies rapidly randomly as a function of space and time as
shown in Fig. 5.1.3.

Fig; 5.1.3: Mean and fluctuating turbulent velocity and pressure

One way to approach such problems is to define the mean/time averaged turvariables. The time mean of a
turbulent velocity (u) is defined by ,

𝟏 𝑻
𝒖
̅ = ∫𝟎 𝒖𝒅𝒕 …………. (5.1.4)
𝑻

Where, T is the averaging period taken as sufficiently longer than the per fluctuations. If the
fluctuation u ' (=u - ̅
𝒖) is taken as the deviation from it’s a value, then it leads to zero mean value.
However, the mean squared of fluctuation ̅̅̅̅̅
(𝑢 ′ 2 )is not zero and thus is the measure of turbulent intensity.

In order to calculate the shear stresses in turbulent flow, it is necessary to know the fluctua ting
components of velocity. So, the Reynolds time - aver concept is introduced where the velocity components
and pressure are split into and fluctuating components;
84
 u = 𝒖 + 𝒖′ ; v = 𝒗 + 𝒗′ ; w = 𝒘 + 𝒘′ ; p = 𝒑 + 𝒑′ ; ………….. 5.1.6

Substituting Eq. (5.1.6) in continuity equation (5.1.3) and take time mean of each equation;

𝟏 𝑻 𝝏𝒖 𝝏𝒖′ 𝟏 𝑻 𝝏𝒗 𝝏𝒗′ 𝟏 𝑻 𝝏𝒘 𝝏𝒘′

∫ ( +
𝑻 𝟎 𝝏𝒙 𝝏𝒙
)dt + ∫ ( +
𝑻 𝟎 𝝏𝒚 𝝏𝒚
)dt + ∫ ( +
𝑻 𝟎 𝝏𝒛 𝝏𝒛
)dt = 0 … (5.1.7)

Let us consider the first term of eq.(4.1.7)

𝟏 𝑻 𝝏𝒖 𝝏𝒖′ 𝟏 𝝏𝒖 𝑻 𝟏 𝝏 𝑻 𝝏𝒖 𝝏𝒖
∫ ( +
𝑻 𝟎 𝝏𝒙 𝝏𝒙
)dt =
𝑻
(
𝝏𝒙
) ∫𝟎 𝒅𝒕 + ∫ 𝒖′ 𝒅𝒕 = 𝝏𝒙 + 0 = 𝝏𝒙 …….. (5.1.8)
𝑻 𝝏𝒙 𝟎

consider the similar analogy for other terms, Eq. (5.1.7) is written as,

𝝏𝒖 𝝏𝒗 𝝏𝒘
+ + = 0 …………… (5.1.9)
𝝏𝒙 𝝏𝒚 𝝏𝒛

This equation is very much similar with the continuity equation for laminar flow except the fact that
the velocity components are replaced with the mean values of velocity components of turbulent flow. The
momentum equation in x-direction takes the following form;

𝝏𝒖 𝝏𝒑 𝝏 𝝏𝒖 2 𝝏 𝝏𝒖 2
𝜌
𝝏𝒙
= + 𝜌𝑔𝑥 + (𝝁 − 𝜌𝑢/ ) + (𝝁 − 𝜌𝑢/ 𝑤′ ) …… (5.1.10)
𝝏𝒙 𝝏𝒙 𝝏𝒙 𝝏𝒛 𝝏𝒛

𝟐
The terms - 𝝆𝒖/ , −𝝆𝒖′ 𝒗′ , and −𝝆𝒖′ 𝒘′ RHS of Eq.(5.1.3) have same dimensions as that of
stress and called as turbulent stresses.For viscous flow in ducts and boundary layer flows, it has been
observed that the stress terms associated with the y direction (i.e. normal to the wall) is more
dominant.So, necessary approximation can be made by neglecting other components of turbulent stresses
and simplified expression may be obtained for Eq.(5.1.10)

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 𝝏𝒖 𝝏𝒑 𝝏𝝉 𝝏𝒖
𝜌
𝝏𝒕
≈ − + 𝜌𝑔𝑥 + ; 𝝉=𝝁 - −𝝆𝒖′ 𝒗′ = 𝝉𝒍𝒂𝒎 + 𝝉𝒕𝒖𝒓 ……… (5.1.11)
𝝏𝒙 𝝏𝒚 𝝏𝒚

It may be noted that n'and v' are zero for laminar flows while the stress terms - 𝜌𝑢 ′ 𝑣 ′ , is positive
for turbulent stresses.Hence the shear stresses in turbulent flow are always higher than laminar flow.The
terms in the form of

- 𝝆𝒖′ 𝒗′ , −𝝆𝒖′ 𝒘′ , - 𝝆𝒖′ 𝒗′

Are also called as Reynolds stresses

Turbulent velocity profile

A typical comparison of laminar and turbulent velocity profiles for wall turbulent flows, are shown
in Fig. 5.1.4 (A-b). The nature of the profile is parabolic in the case of laminar flow and the same trend is
seen in the case of turbulent flow at the wall. The typical measurements across a turbulent flow near the
wall have three distinct zones as shown in Fig. 5.1.4(c). The outer layer (r) is of two or three order
magnitudes greater than the wall layer (m) and vice versa. Hence, the different sub - layers of
Eq . (5.1.11) may be defined as follows;

Wall layer (laminar shear dominates)

Outer layer (turbulent shear dominates)

- Overlap layer (both types of shear are important).

{ Fig. 5.1.4; velocity and shear layer distribution: (a) velocity profile in laminar flow ; (b) velocity profile in turbulent f low; (c) shear
layer in a turbulent flow given below ;-}

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 In a typical turbulent flow, let the wall shear stress, thickness of outer layer and velocity at the
edge of the outer layer be 𝜏𝑤 , 𝛿 and U, respectively. Then the velocity profiles (u) for different zones may
be obtained from the empirical relations using dimensional analysis.

Wall layer: In this region, it is approximated that is independent of shear layer thickness so that the
following empirical relation holds good.

𝒖 𝝆𝒚𝒖∗ 𝝉 𝟏/𝟐
u = f (𝝁, 𝝉𝒘 , 𝝆 , 𝒚 ); 𝒖 += ∗
=F( ) and 𝒖∗ = ( 𝒘 ) ……………. (5.1.12)
𝒖 𝒖 𝝆

Eq. (5.1.12) is known as the law of wall and the quantity is called as friction velocity. It should
not be confused with flow velocity.

Outer layer: The velocity profile in the outer layer is approximated as the deviation from the free
stream velocity and represented by an equation called as velocity - defect law

𝑼−𝒖 𝒚
(𝑼 − 𝒖).𝒐𝒖𝒕𝒆𝒓 = g (𝜹, 𝝉𝒘 , 𝝆 , 𝒚 ); = G( ) ……. (5.1.13)
𝒖∗ 𝜹

Overlap layer - Most of the experimental data show the very good validation of wall law and velocity
defect law in the respective regions. An intermediate layer may be obtained when the velocity profiles
described by Eq. ( 5.1.12 & 5.1.13 ) overlap smoothly . It is shown that empirically that the overlap layer
varies logarithmically with y (Eq. (5.1.14). This particular layer is known as overlap layer.

𝒖 𝟏 𝒚𝝆𝒖∗
= ln ( )+5 …… (5.1.14)
𝒖∗ 𝟎.𝟒𝟏 𝝁

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BIBLIOGRAPHY :-

[1]. Raisinghania M. D., M.SC., Ph.D., Formerly Reader & Head Department of Mathematics S.D. College,
Muzaffarnagar, U.P

[2]. Chung. T. J (2002) Computational fluid dynamics. New York USA: Cambridge University Press.

[3]. Clarke. C.J. & Cars well. B (2007) Principles of astrophysical fluid dynamics. Cambridge: Cambridge
University Press.

[4]. Criminale, W. O (2003) Theory and computation of hydrodynamic stability. Cambridge, England:
Cambridge University Press.

[5]. Darby , R. (2001) Chemical engineering fluid mechanics ( 2 ed . , rev ) , new York : Marcel Dekker.

[6]. Das, M. M. (2010) Fluid mechanics and turbo machines New Delhi, India: PHIL earning.

[7]. Davidson, P.A (2004) Turbulence: an introduction for scientists and engineers. Oxford: Oxford
University Press.

[8]. Davidson, P.A (Ed) (2011) Voyage through turbulence. Cambridge, Cambridge University Press.

[9]. Denn, M.M. (1980) Process fluid mechanics. Upper saddle River, USA: Prentice - hall.

[10]. Denny, M. W. (1993) Air and water: the biology and physics of life's media .Princeton, New Jersey:
Princeton University press

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