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MATH2620: Fluid Dynamics 1
School of Mathematics, University of Leeds
Lecturer: Dr Evy Kersale
Office: Room 9.18, School of Mathematics
Phone: 0113 343 5149
E-mail: [email protected]
Website: http://www.maths.leeds.ac.uk/∼kersale/2620Discussion group on the VLE if needed
Lectures: Tuesday 13:00–14:00, Roger Stevens LT 24
Friday 12:00–13:00, Roger Stevens LT 24
Workshops: Wednesday [Group 1] 11:00–12:00, Roger Stevens LT 02
Jenny Wong ([email protected])
Wednesday [Group 1] 13:00–14:00, Roger Stevens LT 06
Evy Kersale ([email protected])
Thursday [Group 2] 13:00–14:00, Roger Stevens LT 07
Mouloud Kessar ([email protected])
Office Hours: Open door policy
Assessment: 85% final examination and 15% coursework (10 credits).
Textbooks & Picture books:
• A.I. Ruban & J.S.B. Gajjar: Fluid dynamics. Part 1, Classical fluid dynamics, OUP,2014. (Recommended)
• P.S. Bernard, Fluid dynamics, CUP, 2015. (Recommended)
• A.R. Paterson: A first course in fluid dynamics, CUP, 1983. (Recommended)
• P.K. Kundu & I.M Cohen: Fluid mechanics, AP, 2002. (Electronic resource)
• D.J. Acheson: Elementary fluids dynamics, OUP, 1990.
• G.K. Batchelor: An introduction to fluid dynamics, CUP, 2000. (Advanced)
• M. van Dyke: An album of fluid motion, Parabolic Press, 1982.
• M. Samimy et al.: A gallery of fluid motion, CUP, 2003.
ii
Module summary:Fluid dynamics is the science of the motion of materials that flow, e.g. liquid or gas. Under-standing fluid dynamics is a real mathematical challenge which has important implicationsin an enormous range of fields in science and engineering, from physiology, aerodynamics,climate, etc., to astrophysics.This course gives an introduction to fundamental concepts of fluid dynamics. It includes aformal mathematical description of fluid flows (e.g. in terms of ODEs) and the derivationof their governing equations (PDEs), using elementary techniques from calculus and vectorcalculus. This theoretical background is then applied to a series of simple flows (e.g. bath-plugvortex or stream past a sphere), giving students a feel for how fluids behave, and experiencein modelling everyday phenomena.A wide range of courses, addressing more advanced concepts in fluid dynamics, with a varietyof applications (polymers, astrophysical and geophysical fluids, stability and turbulence),follows on naturally from this introductory course.
Objectives:This course demonstrates the importance of fluid dynamics and how interesting physicalphenomena can be understood using rigorous, yet relatively simple, mathematics. But, it alsoprovides students with a general framework to devise models of real-world problems, usingrelevant theories. Students will learn how to use methods of applied mathematics to deriveapproximate solutions to given problems and to have a critical view on these results.
Pre-requisites: Calculus, vector calculus, ODEs.
Course Outline:
• Mathematical modelling of fluids.
• Mass conservation and streamfunctions.
• Vorticity.
• Potential flow.
• Euler’s equation.
• Bernoulli’s equation.
• Flow in an open channel.
• Lift forces.
iii
Lectures:
• You should read through and understand your notes before the next lecture... otherwiseyou will get hopelessly lost.
• Please, do not hesitate to interrupt me whenever you have questions or if I am inaudible,illegible, unclear or just plain wrong. (I shall also stay at the front for a few minutesafter lectures in order to answer questions.)
• If you feel that the module is too difficult, or that you are spending too much time onit, please come and talk to me.
• Please, do not wait until the end of term to give a feedback if you are unhappy withsome aspects of the module.
Lecture notes:
• Detailed lecture notes can be downloaded from the module’s website. You can print anduse them in the lecture if you wish; however, the notes provided should only be used asa supplement, not as an alternative to your personal notes.
• These printed notes are an adjunct to lectures and are not meant to be used indepen-dently.
• Please email me ([email protected]) corrections to the notes, examples sheets andmodel solutions.
Example sheets & homework:
• Five example sheets in total to be handed out every fortnight.
• Examples will help you to understand the material taught in the lectures and will giveyou practice on the types of questions that will be set in the examination. It is veryimportant that you try them before the example classes.
• There will be only two, yet quite demanding, pieces of coursework (mid and end of termdeadlines). Your work will be marked and returned to you with a grade from 1-100.
• Model solutions will be distributed once the homework is handed in
iv
Contents
Notations 1
1 Mathematical modelling of fluids 3
1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.1.1 Continuum hypothesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.1.2 Velocity field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.2 Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.2.1 Particle paths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.2.2 Streamlines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.2.3 Streaklines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.2.4 Time derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
2 Mass conservation & streamfunction 11
2.1 Conservation of mass: the continuity equation . . . . . . . . . . . . . . . . . . . 11
2.2 Incompressible fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.3 Two-dimensional flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.3.1 Streamfunctions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.3.2 Polar coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
2.3.3 Physical significance of the streamfunction . . . . . . . . . . . . . . . . . 14
2.4 Axisymmetric flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2.4.1 Stokes streamfunctions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2.4.2 Properties of Stokes streamfunctions . . . . . . . . . . . . . . . . . . . . 16
3 Vorticity 19
3.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
3.2 Physical meaning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
3.3 Streamfunction and vorticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
3.4 The Rankine vortex . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
3.5 Circulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
3.6 Examples of vortex lines (vortices) . . . . . . . . . . . . . . . . . . . . . . . . . 23
4 Potential flows 27
4.1 Velocity potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
4.2 Kinematic boundary conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
4.3 Elementary potential flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
4.3.1 Source and sink of fluid . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
4.3.2 Line vortex . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
4.3.3 Uniform stream . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
4.3.4 Dipole (doublet flow) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
4.4 Properties of Laplace’s equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
v
vi CONTENTS
4.4.1 Identity from vector calculus . . . . . . . . . . . . . . . . . . . . . . . . 31
4.4.2 Uniqueness of solutions of Laplace’s equation . . . . . . . . . . . . . . . 31
4.4.3 Uniqueness for an infinite domain . . . . . . . . . . . . . . . . . . . . . . 31
4.4.4 Kelvin’s minimum energy theorem . . . . . . . . . . . . . . . . . . . . . 32
4.5 Flow past an obstacle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
4.5.1 Flow around a sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
4.5.2 Rankine half-body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
4.6 Method of images . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
4.7 Method of separation of variables . . . . . . . . . . . . . . . . . . . . . . . . . . 35
5 Euler’s equation 39
5.1 Fluid momentum equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
5.1.1 Forces acting on a fluid . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
5.1.2 Newton’s law of motion . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
5.2 Hydrostatics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
5.3 Archimedes’ theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
5.4 The vorticity equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
5.5 Kelvin’s circulation theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
5.6 Shape of the free surface of a rotating fluid . . . . . . . . . . . . . . . . . . . . 44
6 Bernoulli’s equation 47
6.1 Bernoulli’s theorem for steady flows . . . . . . . . . . . . . . . . . . . . . . . . 47
6.2 Bernoulli’s theorem for potential flows . . . . . . . . . . . . . . . . . . . . . . . 50
6.3 Drag force on a sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
6.4 Separation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
6.5 Unsteady flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
6.5.1 Flows in pipes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
6.5.2 Bubble oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
6.6 Acceleration of a sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
7 Steady flows in open channels 61
7.1 Conservation of mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
7.2 Bernoulli’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
7.3 Flow over a hump . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
7.4 Critical flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
7.5 Flow through a constriction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
7.6 Transition caused by a sluice gate . . . . . . . . . . . . . . . . . . . . . . . . . . 69
8 Lift forces 71
8.1 Two-dimensional thin aerofoils . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
8.2 Kutta-Joukowski theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
8.3 Lift produced by a spinning cylinder . . . . . . . . . . . . . . . . . . . . . . . . 72
8.4 Origin of circulation around a wing . . . . . . . . . . . . . . . . . . . . . . . . . 73
8.5 Three-dimensional aerofoils . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
A Vector calculus 75
A.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
A.2 Suffix notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
A.3 Vector differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
A.3.1 Differential operators in Cartesian coordinates . . . . . . . . . . . . . . 76
A.3.2 Differential operators polar coordinates . . . . . . . . . . . . . . . . . . 78
CONTENTS vii
A.3.3 Vector differential identities . . . . . . . . . . . . . . . . . . . . . . . . . 79A.4 Vector integral theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
A.4.1 Alternative definitions of divergence and curl . . . . . . . . . . . . . . . 79A.4.2 Physical interpretation of divergence and curl . . . . . . . . . . . . . . . 80A.4.3 The Divergence and Stokes’ Theorems . . . . . . . . . . . . . . . . . . . 80A.4.4 Conservative vector fields, line integrals and exact differentials . . . . . 80
viii CONTENTS
Notations
In fluid dynamics, it is crucial to distinguish vectors from scalars. In these lecture notes weshall represent vectors and vector fields using bold fonts, e.g. A and u(x). Other commonlyused notations for vectors include ~A or A (used in the lecture). A vector A, of componentsA1, A2 and A3 in the basis e1, e2, e3, will interchangeably be written as a column or rowvector,
A =
A1
A2
A2
= (A1, A2, A3) = A1 e1 +A2 e2 +A3 e3.
Three main systems of coordinates will be used throughout to represent points in a threedimensional space, Cartesian coordinates (x, y, z) with x, y and z in R; cylindrical polarcoordinates (r, θ, z) with r ∈ [0,∞), θ ∈ [0, 2π) and z ∈ R; and spherical polar coordinates(r, θ, ϕ) with r ∈ [0,∞), θ ∈ [0, π] and ϕ ∈ [0, 2π). Cylindrical polar coordinates reduce toplane polar coordinates (r, θ) in two dimensions. The vector position r ≡ x of a point in athree dimensional space will be written as
x = xex + yey + zex in Cartesian coordinates,
= rer + zez in cylindrical coordinates,
= rer in spherical coordinates,
using the orthonormal basis ex, ey, ez, er, eθ, ez and er, eθ, eϕ respectively. Notice
that ‖x‖ =√x2 + y2 + z2 in Cartesian coordinates, ‖x‖ =
√r2 + z2 in cylindrical coordinates
and ‖x‖ = r in spherical coordinates.
The variable t will represent time; for time derivatives (rates of change) we shall use thenotation df/dt ≡ f , where f is a function of t.
The velocity of a fluid element, defined by u = dx/dt, will be written as
u = u ex + v ey + w ex in Cartesian coordinates,
= ur er + uθ eθ + uz ez in cylindrical coordinates,
= ur er + uθ eθ + uϕ eϕ in spherical coordinates.
For the sake of simplicity, we shall write integrals of multivariable functions as single integrals,e.g. we shall use∫∫∫
Vf(x) dV ≡
∫Vf(x) dV or
∫∫S
u(x) · n dS ≡∫S
u(x) · n dS,
for the volume integral of f in V and for the flux of u through the surface S with normalvector n, respectively.
1
2 CONTENTS
Unless otherwise stated, we shall use the following naming conventions: ψ for planar stream-functions and Ψ for Stokes’ streamfunctions; φ for the velocity potential; Ω for angular ve-locity; ω for the vector vorticity and ω for its magnitude; Γ for fluid circulation; Q for thevolume flux; g for the vector gravity and g for its magnitude; p for pressure and patm foratmospheric pressure; H for the Bernoulli function.
Chapter 1
Mathematical modelling of fluids
Contents
1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.2 Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.1 Introduction
Fluid dynamics (or fluid mechanics) is the study of the motion of liquids, gases and plasmas(e.g. water, air, interstellar plasma) which have no large scale structure and can be deformedto an unlimited extent (in contrast with solids).
It is an old subject (Newton, Euler, Lagrange) but still a very active research area, e.g.
• astrophysical fluids: galaxies, stars, interstellar medium;
• geophysical fluids: earth’s core, atmosphere and ocean, weather;
• environmental fluids: pollution, water and wind power;
• biological fluids: blood and air flows, swimming organisms;
• Aerodynamics and hydrodynamics: aeroplanes, ships.
These fluids differ widely in their physical properties (e.g. density, temperature, viscosity)and in the time- and length-scales of their motion. However, they are all governed by thesame physical laws (e.g. Newton’s law of motion), which can be mathematically formalisedin terms of differential equations for scalar and vector fields.
1.1.1 Continuum hypothesis(Ref.: Paterson,§III.1)One cubic centimetre of water contains of the order of 1023 molecules of typical size lm '
10−10 m, in continuous motion — even in still water (thermal agitation).
It is impossible to calculate the motion — velocity and position — of individual particles.
Instead, one tries to concentrate on ”bulk properties” of fluids, i.e. to look at the motion,mass, etc., of a “blob” of fluid called a fluid particle.
For instance, in the section of a pipe of radius a ' 1 cm, one calculates the average velocityof all molecules in a test volume δV (fluid particle) of length d (mesoscopic scale, i.e. betweenmacroscopic and atomic/molecular scales).
3
4 1.1 Introduction
a
δVd
• If d lm (molecular scale) then δV contains many molecules and the fluctuations dueto individual motions are averaged out.
• If d a (macroscopic scale) then δV is approximately a point in space.
dlm
u
a
Hence, if lm d a, the average velocity u is a smooth function of position, independentof d.
Continuum hypothesis. Molecular details can be smoothed out by assigning the velocityat a point P to be the average velocity in a fluid element δV centred in P . Thus, we candefine the velocity field u(x, t) as a smooth function (shock waves break this assumption),differentiable and integrable.
Similarly, ρ(x, t) =mass in δV
δVis the local density of mass.
1.1.2 Velocity field
The fluid velocity is defined, within the continuum hypothesis, as the vector field u(x, t),function of space and time.
Example 1.1Shear flow: flow between two parallel plates when one is moved relative to the other, witha constant velocity U .
u =
Uy
d
0
0
=Uy
dex.
y
x
y = 0
y = d U
The direction of the flow is indicated byan arrow and its magnitude by the arrowlength.
Chapter 1 – Mathematical modelling of fluids 5
Stagnation-point flow: flow with a point at which u = 0.
Consider the extensional flow
u =
Ex−Ey
0
= Ex ex − Ey ey,
where E is a constant. The point x = 0 where u = 0 isa stagnation point.
Vortex flow: flow in rotation about a central point.
u =
y
x2 + y2
− x
x2 + y2
0
=y
r2ex−
x
r2ey, where r2 = x2 + y2.
This flow is singular at x = 0: ‖u‖ ∼ 1/r → ∞ asr → 0.
1.2 Kinematics(Ref.: Paterson,§III.2)
In the first four chapters of the course we shall concentrate solely on kinematic properties offluid flows, i.e. on properties of fluid velocity fields, ignoring the causes of motion. Dynamics(effects of forces) will be covered in subsequent chapters.
We shall first answer the question “How can we visualise fluid motion?”
1.2.1 Particle paths
This method consists in following the motion of a “tracer” particle in the flow. Let a particlebe released at time t0 and at position (x0, y0, z0) into the velocity field
u(x, t) =
u(x, y, z, t)v(x, y, z, t)w(x, y, z, t)
.
Since the particle is suspended in the fluid, its velocity will be equal to that of the fluid. Hencethe particle position, x(t), satisfies the system of first order ODEs and Initial Conditions
dx
dt= u(x, t) with x = x0 at t = t0, (1.1)
or, equivalently, in Cartesian components form
6 1.2 Kinematics
dx
dt= u(x, y, z, t),
dy
dt= v(x, y, z, t),
dz
dt= w(x, y, z, t), (1.2)
with the initial conditions x = x0, y = y0 and z = z0 at t = t0.
A particle path is a line of equation x(t) = (x(t), y(t), z(t)), parametrised by the variabletime t.
u(x(t1), t1)
u(x(t2), t2)u(x0, t0)
Example 1.2 (Stagnation point flow)Consider the extensional flow u(x) = (Ex,−Ey, 0) with constant E. From equation (1.2) onegets
dx
dt= Ex,
dy
dt= −Ey, dz
dt= 0,
⇒ x(t) = x0eEt, y(t) = y0e
−Et, z(t) = z0 if x = x0 at t = 0.
Note that particles at the stagnation point x0 = y0 = z0 = 0 do not move since u = 0.
The time variable, t, can be eliminated to show that particle paths are hyperbolae of equation
y =x0y0x
.
1.2.2 Streamlines
A streamline is a line everywhere tangent to the local fluid velocity, at some instant. Atfixed time t, the equation of the streamline x(s, t) = (x(s, t), y(s, t), z(s, t)), parametrised bya parameter s (“distance” along the streamline, say), is
dx
ds= u(x, t)⇔ dx
ds= u,
dy
ds= v,
dz
ds= w, (1.3)
in Cartesian components form or, equivalently,
dx
u=
dy
v=
dz
w(= ds), (1.4)
since equations (1.3) are separable as u(x, t) is not explicitly function of s. (Notice that dx/dsis the vector tangent to the line x(s).)
Example 1.3 (Stagnation point flow)Calculate the streamline passing through (x0, y0, 0) in the stagnation point flow u =
x−y0
.
From equation (1.4), in 2D,
dx
x=
dy
−y ⇒ ln |x| = − ln |y|+ C ⇒ xy = A = x0y0,
where C is an integration constant and |A| = eC . So,
streamlines are hyperbolae of equation y =x0y0x
.
Chapter 1 – Mathematical modelling of fluids 7
Example 1.4Consider the time dependent flow u(t) =
U0
kt0
, where U0 and k are constants.
• Particle paths:
dx
dt= U0,
dy
dt= kt,
dz
dt= 0,
⇒ x(t) = U0t+ x0, y(t) =k
2t2 + y0, z(t) = z0, if x = x0 at t = t0.
Eliminating t, the equation of the particle paththrough x0 becomes(
x− x0U0
)2
=2
k(y − y0) (parabola).
y
x
(x(t), y(t))
t > t0
t0
(x0, y0)
• Streamlines: the streamline through x0 is defined by
dx
ds= U0,
dy
ds= kt⇒
x = U0s+ x0 ⇒ s =x− x0U0
,
y = kts+ y0,⇒ y =
kt
U0(x− x0) + y0.
At t = 0, streamlines are horizontal lines y = y0; at later time t > 0, they are straightlines of gradient kt/U0.
t = 0 t1 > 0 t2 > t1
1.2.3 Streaklines
Instead of releasing a single particle at x0 = (x0, y0, z0) at t = t0, say (case of particle pathtracing), release a continuous stream of dye at that point.
The dye will move around and define a curve given by x(t0, t), the position of fluid elementsthat passed through x0 at some time t0, prior to the current time t.
A streakline is the line x(t0, t) for which
∂x
∂t= u(x, t) with x = x0 at t = t0. (1.5)
At t fixed, streaklines are lines parametrised by t0. The point where the parameter t0 = tcorresponds to x0, the locus of the source of dye.
8 1.2 Kinematics
Steady flows: For time-independent flows, which therefore satisfy ∂u/∂t = 0, particlepaths, streamlines and streaklines are identical. This is not true for unsteady flows in general.
Example 1.5• Steady flow:
u(x) =
x−y0
⇒ ∂x
∂t= x,
∂y
∂t= −y, ∂z
∂t= 0,
⇒ x(t0, t) = A(t0) et, y(t0, t) = B(t0) e−t, z(t0, t) = C(t0).
At t = t0, x = x0 = (x0, y0, z0), so
x0 = Aet0 , y0 = Be−t0 , z0 = C,
⇒A = x0e−t0 , B = y0e
t0 , C = z0.
Hence,
x(t0, t) = x0e(t−t0), y(t0, t) = y0e
−(t−t0), z(t0, t) = z0.
Again, we can eliminate the parameter t0 to obtain the equation xy = x0y0 as forparticle paths and streamlines.
• Unsteady flow:
u(t) =
U0
kt0
with U0 and k constants,
⇒ x(t0, t) = x0 + U0(t− t0),
and y(t0, t) = y0 +k
2(t2 − t20).
At t = 0, x(t0, 0) = x0−U0t0 and y(t0, 0) = y0−k
2t20.
Eliminating t0, one obtains the equation of aparabola with negative curvature
y − y0 = − k
2U20
(x− x0)2.
(x0, y0)
Streakline at t=0
Streamline at t=0
Particle path (particlereleased at t=0)
1.2.4 Time derivatives
Let f(x, t) be some quantity of interest (e.g. density, temperature, one component of thevelocity, etc.).
Eulerian description of fluids. (At a fixed position in space.)
The partial derivative
(∂f
∂t
)x
≡ ∂
∂tf(x, t) is the rate of change of f at fixed position x.
E.g.,∂ρ
∂t(scalar) is the rate of change of mass density;
∂u
∂t(vector) is not the acceleration of
a fluid particle.
Chapter 1 – Mathematical modelling of fluids 9
Lagrangian description of fluids. (Following fluid particles.)
The convective derivative (also Lagrangian derivative, or material derivative)D
Dtf(x, t) is the
rate of change of f when x is the position of a fluid particle (i.e. x travels with the fluid alongparticle paths).
So,Df
Dt= 0 implies that f remains constant along particle paths.
Example 1.6For the flow u(x) =
xy−2z
a particle path is given by
x(t) = x0 et, y(t) = y0 et and z(t) = z0 e−2t, if x = x0 at t = 0.
So, along the particle path, upp =
xy−2z
=
x0 et
y0 et
−2z0 e−2t
and
Du
Dt≡ d
dtupp =
x0 et
y0 et
4z0 e−2t
=
xy4z
6= 0 whereas∂u
∂t= 0.
Relation between D/Dt and ∂/∂t.There is no need to go through particle paths calculations to evaluate D/Dt. Consider a
particle path x(t) = (x(t), y(t), z(t)) defined bydx
dt= u.
Using the chain rule,
Df
Dt≡ d
dtf(x(t), t) =
∂f
∂t+∂f
∂x
dx
dt+∂f
∂y
dy
dt+∂f
∂z
dz
dt,
⇒ Df
Dt=∂f
∂t+ u
∂f
∂x+ v
∂f
∂y+ w
∂f
∂z=∂f
∂t+ (u · ∇) f. (1.6)
Hence,Dρ
Dt=∂ρ
∂t+ (u · ∇) ρ is the Lagrangian derivative of the fluid density and the acceler-
ation of a fluid particle isDu
Dt=∂u
∂t+ (u · ∇) u. (1.7)
Example 1.7Again, consider the steady flow u =
xy−2z
.
Since∂u
∂t= 0 (the velocity field u does not depend on time t for a steady flow),
Du
Dt= (u · ∇) u = u
∂u
∂x+ v
∂u
∂y+ w
∂u
∂z=
u∂u
∂x+ v
∂u
∂y+ w
∂u
∂z
u∂v
∂x+ v
∂v
∂y+ w
∂v
∂z
u∂w
∂x+ v
∂w
∂y+ w
∂w
∂z
,
= x∂u
∂x+ y
∂u
∂y− 2z
∂u
∂z= x
100
+ y
010
− 2z
00−2
=
xy4z
,
as before (see example 1.6).
10 1.2 Kinematics
Steady flows. In steady flows (i.e. flows such that ∂u/∂t = 0), the rate of change of ffollowing a fluid particle becomes
Df
Dt= (u · ∇) f.
Furthermore, since particle paths and streamlines are identical for time-independent flows,
Df
Dt= ‖u‖ (es · ∇) f = ‖u‖df
ds
where s denotes the distance along the streamlines x(s) and es is the unit vector parallel tothe streamlines, in the direction of the flow.
So, in steady flows, (u · ∇) f = 0 implies that f is constant along streamlines — the constantcan take different values for different streamlines however.
Chapter 2
Mass conservation &streamfunction
Contents
2.1 Conservation of mass: the continuity equation . . . . . . . . . . . . 11
2.2 Incompressible fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.3 Two-dimensional flows . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.4 Axisymmetric flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2.1 Conservation of mass: the continuity equation
In any situation the mass of a fluid must be conserved. For continuous media, such as a fluids,this fundamental principle is expressed mathematically in the form of the continuity equation.
Consider a volume V , fixed in space, with surface S and outward normal n.
ds
nu
S
V
The total mass in V is
MV =
∫VρdV,
where ρ is the density of mass (mass per unitvolume).
MV can only change if mass is carried inside or outside the volume by the fluid.
The mass flowing through the surface per unit time (i.e. the mass flux) is
−∫Sρu·n dS =
dMV
dt.
So,
−∫Sρu·n dS =
d
dt
∫Vρ dV =
∫V
∂ρ
∂tdV, since V is fixed.
Applying the divergence theorem,∫V
∂ρ
∂tdV = −
∫V∇·(ρu) dV ⇔
∫V
[∂ρ
∂t+∇·(ρu)
]dV = 0.
11
12 2.2 Incompressible fluids
Since V is arbitrary, this equation must hold for all volume V . Thus, the continuity equation
∂ρ
∂t+∇·(ρu) = 0 (2.1)
holds at all points in the fluid. Expand the divergence as ∇·(ρu) = ρ∇·u + u·∇ρ to derivethe Lagrangian form of the continuity equation
Dρ
Dt+ ρ∇·u = 0. (2.2)
The density of a fluid particle, which moves with the fluid, only changes if there is an expansion(i.e. divergence such that ∇·u > 0) or a contraction (i.e. convergence such that ∇·u < 0) ofthe flow.
∇ · u < 0 ∇ · u > 0
2.2 Incompressible fluids
In an incompressible fluid, the density of each fluid particle (i.e. fluid element following themotion of the fluid) remains constant, so that Dρ/Dt = 0. Thus, the continuity equation (2.2)reduces to
ρ∇ · u = 0.
So, incompressible flows must satisfy the constraint
∇ · u = 0, (2.3)
which means that the fluid velocity can be expressed in the form
u = ∇× S, (2.4)
for some vector field S(x, t). Indeed, since ∇ · (∇ × F) = 0 for any vector field F, the flowdefined by equation (2.4) satisfies the continuity equation (2.3).
For the rest of this course we shall assume that ρ is constant. (So, the fluid is incompressibleand therefore ∇ · u = 0.) This is a good approximation in many circumstances, e.g. waterand very subsonic air flows.
2.3 Two-dimensional flows
Simplifications arise in the mathematical modelling of fluid flows when one considers systemswhich possess symmetries. We shall first consider flows confined to a plane, expressed inCartesian coordinates as
u =
u(x, y, t)v(x, y, t)
0
= u(x, y, t) ex + v(x, y, t) ey.
Chapter 2 – Mass conservation & streamfunction 13
2.3.1 Streamfunctions
For two-dimensional incompressible flows, we define
S = ψ(x, y, t) ez, (2.5)
where ψ is a (scalar) streamfunction, such that, from equation (2.4),
u = ∇× (ψ ez) ⇒ u =∂ψ
∂yand v = −∂ψ
∂x. (2.6)
Clearly the incompressibility condition is then automatically satisfied:
∇ · u =∂u
∂x+∂v
∂y=
∂2ψ
∂x∂y− ∂2ψ
∂y∂x= 0.
Streamfunctions and streamlines. A key property of streamfunctions comes from con-sidering
u · ∇ψ = u∂ψ
∂x+ v
∂ψ
∂y=∂ψ
∂y
∂ψ
∂x− ∂ψ
∂x
∂ψ
∂y= 0,
which shows that the gradient of the streamfunction is orthogonal to the velocity field, im-plying that ψ remains constant in the direction of flow. So, the streamfunction ψ is constantalong streamlines.
Conversely, consider a parametric curve x(s) = (x(s), y(s)) of constant ψ, so that
dψ
ds= 0⇔ ∂ψ
∂x
dx
ds+∂ψ
∂y
dy
ds= −vdx
ds+ u
dy
ds= u× dx
ds= 0,
where dx/ds is a vector tangent to the curve x(s). So, u is tangent to the curve where ψremains constant which is therefore a streamline.
Example 2.1Consider the 2-D flow u =
(U0
kt
), with U0 and k constants. Since ∇·u = 0, one can define
the streamfunction
ψ(x, y, t) = U0y − ktx+ C,
where C is an arbitrary constant of integration. The value assigned to C does not affect theflow, so we shall usually take C = 0 without loss of generality.
The streamfunction is constant along the lines ψ = U0y − ktx = C, where C is a constantidentifying streamlines. So, the streamlines are lines of equation
y =kt
U0x+
C
U0with gradient
dy
dx=kt
U0.
Example 2.2The flow defined as u =
y
x2 + y2
−xx2 + y2
is incompressible since
∇·u =∂u
∂x+∂v
∂y=
−2xy
(x2 + y2)2+
(−2y)(−x)
(x2 + y2)2= 0.
14 2.3 Two-dimensional flows
So,∂ψ
∂y= u =
y
x2 + y2=
1
2
2y
x2 + y2⇒ ψ(x, y) =
1
2ln(x2 + y2
)+ α(x).
Then v = −∂ψ∂x
= − x
x2 + y2+
dα
dx⇒ dα
dx= 0. So, α is constant and
ψ(x, y) =1
2ln(x2 + y2
)(choosing α = 0).
2.3.2 Polar coordinates
We now change from Cartesian coordinates, (x, y), inthe (X,Y )-plane to polar coordinates, (r, θ), defined by
x(r, θ) = r cos θ,y(r, θ) = r sin θ,
and consider again a two-dimensional flow, u, indepen-dent of z, with uz = 0, such that
u(r, θ) = ur(r, θ) er + uθ(r, θ) eθ.
y
Y
Xx
eθ
er
θ
r
M
So, in plane polar coordinates, substituting S = ψ(r, θ, t) ez in u = ∇×S for 2-D incompress-ible flows gives
u = ∇× (ψ ez)⇒ ur =1
r
∂ψ
∂θ, uθ = −∂ψ
∂r. (2.7)
Example 2.3For the streamfunction ψ = ln
√x2 + y2 = ln r, ur = 0 and uθ = −1/r. The streamlines are
circles about the origin with |uθ| decreasing as r increases. This is a reasonable model for abath-plug vortex.
eθ uθ r
2.3.3 Physical significance of the streamfunction
We noted earlier that the streamfunction is constant on streamlines. So, we consider the twostreamlines defined by ψ(x, y) = ψP and ψ(x, y) = ψT .
C
n
P
T ψ = ψT
ψ = ψP
The flow rate or the volume flux through an arbitrary curve C : (x(s), y(s)), parametrised bys ∈ [0, 1] and connecting P and T , is
Q =
∫ 1
0u · n ds. (2.8)
Chapter 2 – Mass conservation & streamfunction 15
dl dy
dxC
nds
(x(s), y(s))
Let dl = dxex + dyey = ds
(dx
dsex +
dy
dsey
)be an infinitesimal
displacement along the curve C. The infinitesimal vector normalto dl is therefore
n ds = dyex − dxey =
(dy
dsex −
dx
dsey
)ds.
So,
Q =
∫ 1
0
(∂ψ
∂y
dy
ds+∂ψ
∂x
dx
ds
)ds =
∫ 1
0
dψ
dsds =
∫ ψT
ψP
dψ = ψT − ψP .
Hence, the flux of fluid flowing between two streamlines is equal to the difference in thestreamfunction. Consequently, if streamlines are close to one another the flow must be fast.
Equivalently, the equation ‖u‖ = ‖∇ψ‖ shows that the speed of the flow increases with thegradient of the streamfunction, that is when the distance between two streamlines decreases.
2.4 Axisymmetric flows
So far, we have considered flows confined to a plane, with velocity fields of the form u(x, y, t) =∇× (ψ(x, y, t) ez), invariant along the z-axis, with no z-component.
Similarly, axisymmetric flows, such as
u(r, z) = ur(r, z) er + uz(r, z) ez (2.9)
in cylindrical polar coordinates, have only two non-zero components and two effective coor-dinates.
A flow in a circular pipe or a flow past a sphere are examples of flows with axial symmetry.
er
ez
2.4.1 Stokes streamfunctions
For axisymmetric incompressible flows, we define
S =1
rΨ(r, z, t) eθ, (2.10)
where Ψ is a (scalar) Stokes streamfunction, such that, from equation (2.4),
u = ∇×(
1
rΨ eθ
)⇒ uz =
1
r
∂Ψ
∂rand ur = −1
r
∂Ψ
∂z. (2.11)
(We use Ψ to distinguish Stokes streamfunctions from planar streamfunctions denoted ψ.)
Clearly the incompressibility condition is then again automatically satisfied:
∇ · u =1
r
∂
∂r(rur) +
∂uz∂z
= −1
r
∂2Ψ
∂r∂z+
1
r
∂2Ψ
∂z∂r= 0.
16 2.4 Axisymmetric flows
2.4.2 Properties of Stokes streamfunctions
The Stokes streamfunctions have properties analogous to planar streamfunctions.
i. Ψ is constant on streamlines.
(u · ∇) Ψ = ur∂Ψ
∂r+ uz
∂Ψ
∂z=
1
r
(rur
∂Ψ
∂r+ ruz
∂Ψ
∂z
),
1
r
(−∂Ψ
∂z
∂Ψ
∂r+∂Ψ
∂r
∂Ψ
∂z
)= 0.
Thus, Ψ is constant in the direction of the flow.
For axisymmetric flows it is useful to think of streamtubes: surface of revolution spannedby all the streamlines through a circle about the axis of symmetry.
Streamtubes are surfaces ofconstant Ψ.
ii. Relation between volume flux and streamtubes.The volume flux, or fluid flow, between two streamtubes with Ψ = Ψi and Ψ = Ψo is
Q =
∫S
u · n dS = 2π(Ψ0 −Ψi). (2.12)
Ψo
Ci
Co
n
S
Ψi
Proof.
Q =
∫S
u · n dS =
∫S∇×
(1
rΨ eθ
)· n dS, (definition of Ψ)
=
∮Co
1
rΨ eθ · dl +
∮Ci
1
rΨ eθ · dl, (Stokes’ theorem)
= Ψo
∮Co
1
reθ · dl + Ψi
∮Ci
1
reθ · dl, (Ψ ≡ Ψo,i onto Co,i)
= Ψo
∫ 2π
0dθ + Ψi
∫ 0
2πdθ = 2π(Ψo −Ψi) (recall, dl = dr er + rdθ eθ).
Example 2.4For a uniform flow parallel to the axis, ur = 0 and uz = U ,
∂Ψ
∂r= rU and
∂Ψ
∂z= 0⇒ Ψ(r) =
1
2Ur2.
Chapter 2 – Mass conservation & streamfunction 17
(We choose the integration constant such that Ψ = 0 on the axis, at r = 0).
Now consider a streamtube of radius a.
a
uS
zΨ(0) = 0
Ψ(a) =1
2Ua2
The volume flux
Q =
∫S
u · n dS =
∫S
u · ez dS =
∫Suz dS = U
∫S
dS = πUa2.
Also,
2π(Ψo −Ψi) = 2π(Ψ(a)−Ψ(0)) = 2π
(1
2Ua2 − 0
)= πUa2 as required.
Example 2.5Consider a flow in a long pipe a radius a:
ur = 0, uz =U
a2(a2 − r2
)with
uz = 0 on r = a,uz = U at r = 0.
z
S
a
n
∂Ψ
∂z= −rur = 0⇒ Ψ ≡ Ψ(r),
anddΨ
dr= ruz =
U
a2(a2r − r3)⇒ Ψ(r) =
U
a2
(a2r2
2− r4
4
)+ C.
Hence,
Ψ(r) =Ur2
4a2(2a2 − r2) (choose C such that Ψ(0) = 0).
So, Ψ(0) = 0 and Ψ(a) = Ua2/4, and the volume flux
Q =
∫S
u · n dS = 2π (Ψ(a)−Ψ(0)) =π
2Ua2.
Indeed,∫S
u · n dS =
∫ 2π
0
∫ a
0uzr drdθ =
2πU
a2
∫ a
0(a2r − r3) dr, (since dS = rdθdr)
=2πU
a2
[a2r2
2− r4
4
]a0
=πUa2
2, as required.
18 2.4 Axisymmetric flows
Chapter 3
Vorticity
Contents
3.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
3.2 Physical meaning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
3.3 Streamfunction and vorticity . . . . . . . . . . . . . . . . . . . . . . 21
3.4 The Rankine vortex . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
3.5 Circulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
3.6 Examples of vortex lines (vortices) . . . . . . . . . . . . . . . . . . . 23
3.1 Definition
We have already considered ∇ ·u, which is a measure of the local expansion or contraction ofthe fluid. (∇ · u = 0 for incompressible fluids.)
The vorticity,ω = ∇× u, (3.1)
is a measure of the local rotation — or spin — in a flow. It is a concept of central importancein fluid dynamics.
3.2 Physical meaning
In the simplest case of a 2-D flow, u(x, y, t) =
u(x, y, t)v(x, y, t)
0
, the vorticity ω = ∇ × u =00
∂v
∂x− ∂u
∂y
is perpendicular to the plane of motion; its magnitude is ω =∂v
∂x− ∂u
∂y.
Consider the two short fluid line-elements AB and AC. The vertical differential velocity is
19
20 3.2 Physical meaning
A
B
δy
δx C
δv
δu
δv = vC−vA = v(x+δx, y)−v(x, y) ' δx∂v∂x
(Taylor theorem).
So,∂v
∂xis the angular velocity of the fluid line-element AC.
Similarly, the horizontal differential velocity
δu = uB − uA = u(x, y + δy)− u(x, y) ' δy∂u∂y.
So, −∂u∂y
is the angular velocity of the fluid line element AB.
Thus,ω
2=
1
2
(∂v
∂x− ∂u
∂y
)represents the average angular
velocity of the two fluid line-elements AB and AC.
This could be measured using a crossed pair of small vanesthat float with the fluid.
Example 3.1 (Solid body rotation)For u = Ω× r, with constant angular velocity Ω and r =
xyz
,
ω = ∇× (Ω× r) = Ω(∇ · r)− (Ω · ∇)r,
= 3Ω−Ω = 2Ω.
So, the vorticity is twice the local rotation rate — which isalso global here.
Ω
Example 3.2 (Shear flow)For u = ky, v = 0, the vorticity ω =
00−k
with ω = −k.
Vorticity rollers The vorticity is not a measure of global rotation:a shear flow has no global rotation but a non-zerovorticity.
Example 3.3 (Line vortex flow)Let ur = 0, uθ = k/r and uz = 0 where k is a positive constant. (This is a crude model for abath-plug vortex — see examples 2.2 and 2.3)
Chapter 3 – Vorticity 21
Streamfunction:∂ψ
∂r= −k
r⇒ ψ = −k ln r (const. = 0). So,
the streamlines are circles.
However, from
ω =
[1
r
∂
∂r(ruθ)−
1
r
∂
∂θur
]ez =
1
r
∂k
∂rez = 0,
one finds that the vorticity is zero everywhere in the flow except at r = 0 where the functionsu and ω are not defined. (In fact ω can be defined as a Dirac delta distribution.)
Although the flow is rotating globally, there is no local rota-tion. Crossed vanes placed in the flow would move in a circle,but without spinning.
3.3 Streamfunction and vorticity
In a two-dimensional flow, the vorticity ω =
00ω
where ω =∂v
∂x− ∂u
∂yin Cartesian
coordinates. If, in addition, the fluid is incompressible, u =∂ψ
∂yand v = −∂ψ
∂x, so that
ω = −∂2ψ
∂x2− ∂2ψ
∂y2= −∇2ψ. (3.2)
More generally (Cartesian or plane polar coordinates),
u = ∇× (ψez)⇒ ω = ∇× (∇× (ψez)),
= ∇(∇ · (ψez))−∇2(ψez),
= ∇(∂
∂zψ
)−∇2ψ ez.
Since ∂ψ/∂z = 0 for a 2-D flow, one finds again
ω = −∇2ψ ez (3.3)
3.4 The Rankine vortex
Example 3.3, with uθ = k/r, is a crude model for the bath-plug vortex: infinite vorticityconcentrated in r = 0 (singularity). In real vortices the vorticity is spread over a small area.
Consider an azimuthal flow, ur = uz = 0 and uθ = f(r), in cylindrical coordinates, such that
ω =
Ω if r ≤ a (Ω constant),
0 if r > a
22 3.5 Circulation
Hence, ψ ≡ ψ(r) and ∇2ψ =1
r
d
dr
(r
dψ
dr
)=
−Ω if r ≤ a,0 if r > a.
• r ≤ a:d
dr
(r
dψ
dr
)= −rΩ ⇒ dψ
dr= −Ω
2r +
B
r= −uθ.
We require uθ to be bounded at r = 0. So, B = 0 and, for r ≤ a,uθ =
Ω
2r,
ψ = −Ω
4r2.
• r > a:d
dr
(r
dψ
dr
)= 0 ⇒ uθ = −dψ
dr=A
r.
The continuity of uθ at r = a implies that A =Ω
2a2.
So,dψ
dr= −Ωa2
2r⇒ ψ = −Ωa2
2ln r +D, and applying the continuity of ψ at r = a now
gives ψ = −Ωa2
4
(1 + 2 ln
r
a
).
So,
ψ =
−Ω
4r2 if r ≤ a,
−Ωa2
4
(1 + 2 ln
r
a
)if r > a.
and
uθ =
Ω
2r if r ≤ a : solid body rotation,
Ωa2
2rif r > a : bath-plug flow.
r
∝ r ∝ 1/r
a
uθ
ra
ω
Ω
There is still a discontinuity in vorticity but the flow is quite adequate for predicting the shapeof the water surface.
3.5 Circulation
Consider a closed curve C in the flow. The circulation around C is the line integral of thetangential velocity around C:
Γ =
∮C
u · dl. (3.4)
Chapter 3 – Vorticity 23
C
dlu
n
S
By Stokes theorem, for any surface S spanning the curve C,
Γ =
∮C
u · dl =
∫S
(∇× u) · n dS =
∫Sω · n dS. (3.5)
So, the circulation around C is equal to the vorticity flux through the surface S: it is thestrength of the vortex tube.
Example 3.4 (The Rankine vortex)The circulation around a circle of radius r is given by
Γ =
∫ 2π
0uθ(r) r dθ = 2π r uθ(r) =
πΩr2 if r ≤ a,πΩa2 if r > a.
When the vorticity is concentrated in thin filaments (tubes) as it is in the Rankine vortex, itis useful to think in term of vortex.
3.6 Examples of vortex lines (vortices)
i. Bath-plug vortex.
The shape of the free surface of water can be modelled using the Rankin vortex.Note that the sense of rotation is not determined by the rotation of the Earth!
ii. Vortices behind aeroplanes.
24 3.6 Examples of vortex lines (vortices)
The characteristic vapour trails left by aircraft are vortex lines shed from the wing tips.(The vortices have low pressure, so vapour water condenses there.) These vortices decayvery slowly and are a danger for small aircraft flying behind large ones.
iii. Horseshoe vortex & downwash behind chimneys.
Vortex lines in shear flows above groundlevel travel with the air and can bestretched and bent by tall buildings andchimneys. This results in a downwardsflow behind chimneys, dragging pollutantdown to ground level.
iv. Vortex rings.
Smoke rings and underwater bubble rings are examples of vortex rings.
Chapter 3 – Vorticity 25
v. von Karman vortex street.
In certain conditions, a flow past an obstacle (e.g. a cylinder, an island) produces aseries of line vortices.
26 3.6 Examples of vortex lines (vortices)
Chapter 4
Potential flows
Contents
4.1 Velocity potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
4.2 Kinematic boundary conditions . . . . . . . . . . . . . . . . . . . . . 28
4.3 Elementary potential flows . . . . . . . . . . . . . . . . . . . . . . . . 28
4.4 Properties of Laplace’s equation . . . . . . . . . . . . . . . . . . . . 31
4.5 Flow past an obstacle . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
4.6 Method of images . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
4.7 Method of separation of variables . . . . . . . . . . . . . . . . . . . . 35
4.1 Velocity potential
We shall now consider the special case of irrotational flows, i.e. flows with no vorticity, suchthat
ω = ∇× u = 0. (4.1)
If a velocity field u is irrotational, that is if ∇× u = 0, then there exists a velocity potentialφ(x, t) defined by
u = ∇φ. (4.2)
This is a result from vector calculus; the converse is trivially true since ∀φ, ∇×∇φ ≡ 0.
If in addition the flow is incompressible, the velocity potential φ satisfies Laplace’s equation
∇2φ = 0. (4.3)
Indeed, for incompressible irrotational flows one has ∇ · u = ∇ · ∇φ = ∇2φ = 0.
Hence, incompressible irrotational flows can be computed by solving Laplace’s equation (4.3)and imposing appropriate boundary conditions (or conditions at infinity) on the solution.(Notice that for 2-D incompressible irrotational flows, both velocity potential, φ, and stream-function, ψ, are solutions to Laplace’s equation, ∇2ψ = −ω = 0 and ∇2φ = 0; boundaryconditions on ψ and φ are different however.)
27
28 4.2 Kinematic boundary conditions
4.2 Kinematic boundary conditions
U
n
Consider a flow past a solid body moving at velocity U.If n is the unit vector normal to the surface of the solid,then, locally, the surface advances (i.e. moves in thedirection of n) at the velocity (U·n) n.
Since the fluid cannot penetrate into the solid body, its velocity normal the surface, (u·n) n,must locally equal that of the solid,
u · n = U · n.So, since u = ∇φ,
n · ∇φ =∂φ
∂n= U · n; (4.4)
the velocity potential satisfies Neumann boundary conditions at the solid body surface.
4.3 Elementary potential flows
4.3.1 Source and sink of fluid
Line source/sink
Consider an axisymmetric potential φ ≡ φ(r). From Laplace’s equation in plane polar coor-dinates,
∇2φ =1
r
d
dr
(r
dφ
dr
)= 0⇔ dφ
dr=m
r,
one finds
φ(r) = m ln r + C, (4.5)
where m and C are integration constants. This potential produces the planar radial velocity
u = ∇φ =m
rer
corresponding to a source (m > 0) or sink (m < 0) of fluid of strength m. Notice that theconstant C in φ is arbitrary and does not affect u. By convention the constant m = Q/2πwhere Q is the flow rate. This flow could be produced approximately using a perforated hose.
Source (m > 0) Sink (m < 0)
Chapter 4 – Potential flows 29
Point source/sink
Consider a spherically symmetric potential φ ≡ φ(r). From Laplace’s equation in sphericalpolar coordinates,
∇2φ =1
r2d
dr
(r2
dφ
dr
)= 0⇔ dφ
dr=m
r2,
one finds
φ(r) = −mr
+ C, (4.6)
where m and C are integration constants. This potential produces the three-dimensionalradial velocity
u = ∇φ =m
r2er.
corresponding to a source (m > 0) or sink (m < 0) of fluid of strength m. By convention theconstant m = Q/4π where Q is the flow rate or volume flux.
Source (m > 0) Sink (m < 0)
4.3.2 Line vortex
For the potential φ(θ) = kθ, solution to Laplace’s equationin plane polar coordinates, one has
ur =∂φ
∂r= 0 and uθ =
1
r
∂φ
∂θ=k
r,
where the strength of the flow k is a constant. By conventionk = Γ/2π if Γ is the circulation of the flow.This represents a rotating fluid (bath-plug vortex) around a line vortex at r = 0; it has zerovorticity but is singular at the origin.
4.3.3 Uniform stream
For a uniform flow along the z-axis, u = (0, 0, U), the velocity potential
φ(z) = Uz.
(The integration constant is set to zero.)
4.3.4 Dipole (doublet flow)
Since Laplace’s equation is linear we can add two solutions together to form a new one. Adipole is the superposition of a sink and an source of equal but opposite strength next to eachother.
30 4.3 Elementary potential flows
Three-dimensional flow
Consider a point sink of strength −m at the origin and a point source of strength m at theposition (0, 0, δ).
z
Sink Source
δ
O
The velocity potential of the flow is formed by adding the potentials of the source and sink,
φ =m√
x2 + y2 + z2− m√
x2 + y2 + (z − δ)2,
=m
r− m√
r2 − 2zδ + δ2,
where m > 0 is constant and r = (x2 + y2 + z2)1/2. Expanding the potential to first orderin δ,
φ =m
r− m
r
(1 +
z
r2δ +O(δ2)
),
and taking the limit δ → 0, leads to the potential of a dipole
φ = −mδ zr3,
where m→∞ as δ → 0 so that the strength of the dipole µ = mδ remains finite. Thus,
φ = −µ · rr3
= µ · ∇(
1
r
),
where µ = µez and r ≡ x is the vector position. The three components of the fluid velocity,u = ∇φ, are for a dipole of strength µ,
ux = 3µxz
r5,
uy = 3µyz
r5,
uz = − µr3
(1− 3
z2
r2
).
Planar flow
Similarly, combining a line sink at the origin with a line source of equal but opposite strengthat (δ, 0) gives
φ = −m2
[ln(x2 + y2)− ln
((x− δ)2 + y2
)]=m
2ln
[(x− δ)2 + y2
x2 + y2
], m > 0.
As in the three-dimensional case, we consider the limit δ → 0, with µ = mδ fixed. Theexpression of the potential for a two-dimensional dipole of strength µ then becomes
φ = −µ · rr2
= −µ · ∇ ln r,
where µ = µex and r ≡ x is the vector position.
Chapter 4 – Potential flows 31
4.4 Properties of Laplace’s equation
4.4.1 Identity from vector calculus
Let f(x) be a function defined in a simply connected domain V with boundary S. Fromvector calculus,
V
n
S∇ · (f∇f) = f∇2f + |∇f |2
⇒∫V∇ · (f∇f) dV =
∫Vf∇2f dV +
∫V|∇f |2 dV.
So, using the divergence theorem∫Sf(∇f) · n dS =
∫Vf∇2f dV +
∫V|∇f |2 dV. (4.7)
4.4.2 Uniqueness of solutions of Laplace’s equation
Given the value of the normal component of the fluid velocity, u ·n, on the surface S (i.e. theboundary condition), there exists a unique flow satisfying both ∇ · u = 0 and ∇× u = 0 (i.e.incompressible and irrotational).
Proof. Suppose there exists two distinct solutions to the boundary value problem, u1 = ∇φ1and u2 = ∇φ2. Let f = φ1 − φ2, then
∇2f = ∇2φ1 −∇2φ2 = 0
in the domain V and
(∇f) · n = (∇φ1) · n− (∇φ2) · n = u1 · n− u2 · n = 0
on the boundary S. Hence, from identity (4.7),
∫V|∇φ1 − ∇φ2|2 dV = 0. However, since
|∇φ1 − ∇φ2|2 ≥ 0 one must have ∇φ1 = ∇φ2 everywhere. Therefore u1 = u2 and thesolution to the boundary value problem is unique.
4.4.3 Uniqueness for an infinite domain
The proof above holds for flows in a finitedomain. What about flows in an infinitedomain — e.g. flow around an obstacle?
The above argument holds by considering thevolume V as shown and letting S′ →∞. (See,e.g. Patterson p. 211.)
S ′
SV n
32 4.5 Flow past an obstacle
4.4.4 Kelvin’s minimum energy theorem
Of all possible fluid motions satisfying the boundary condition for u · n on the surface S and∇ · u = 0 in domain V , the potential flow is the flow with the smallest kinetic energy,
K =1
2
∫Vρ |u|2 dV.
Proof. Let u′ be another incompressible but non vorticity-free flow such that u · n = u′ · non S and ∇ · u′ = 0 in V but with ∇× u′ 6= 0.The fluid flow u is potential, so let u = ∇φ such that∫
Vρ |u|2 dV =
∫Vρ|∇φ|2 dV = ρ
∫V|∇φ|2 dV,
= ρ
∫Sφu · n dS (by identity (4.7) with f = φ),
= ρ
∫Sφu′ · n dS (boundary condition),
= ρ
∫V∇ · (φu′) dV (divergence theorem),
= ρ
∫V
u′ · ∇φ dV (∇ · u′ = 0),
= ρ
∫V
u′ · u dV. (4.8)
So, ∫Vρ (u− u′)2 dV =
∫V
(ρ |u|2 − 2ρu · u′ + ρ |u′|2
)dV,
=
∫V
(ρ |u′|2 − ρ |u|2
)dV (from (4.8)).
Therefore, since (u− u′)2 ≥ 0,∫Vρ |u′|2 dV =
∫Vρ |u|2 dV +
∫Vρ (u− u′)2 dV ≥
∫Vρ |u|2 dV.
4.5 Flow past an obstacle
Since the solution to Laplace’s equation for given boundary conditions is unique, if we find asolution, we have found the solution. (This is only true if the domain is simply-connected; ifthe domain is multiply connected, multiple solutions become possible.)
One technique to calculate non elementary potential flows involves adding together simpleknown solutions to Laplace’s equation to get the solution that satisfies the boundary condi-tions.
4.5.1 Flow around a sphere
We seek an axisymmetric flow of the form u = urer + uzez in cylindrical polar coordi-nates (r, θ, z).
Chapter 4 – Potential flows 33
a
O
U
Z
r
r
n
z
At large distances from the sphere of ra-dius a the flow is asymptotic to a uni-form stream, ur = 0, uz = U , and at thesphere’s surface, r = a, the fluid velocitymust satisfy u ·n = 0 since the solid bodyforms a non-penetrable boundary.
The unit vector normal to surface of the sphere is
n = nrer + nzez with nr =r
aand nz =
z
a.
So, the boundary condition u · n = 0 implies that
urr
a+ uz
z
a= 0⇔ rur + zuz = 0
at the spherical surface of equation r2 + z2 = a2.
At large distances, the flow is essentially uniform along the z-axis,
φ ' Uz, for ‖r‖ a.
Now, add to the uniform stream a dipole velocity field of strength µ = µez at the origin,
φ(r, z) = Uz − µz
(r2 + z2)3/2,
so that
ur =∂φ
∂r=
3µrz
(r2 + z2)5/2and uz =
∂φ
∂z= U +
µ
(r2 + z2)3/2
(3z2
r2 + z2− 1
).
Thus, at the sphere’s surface,
u · n = urr
a+ uz
z
a=z
a
(U +
3µ(r2 + z2)
(r2 + z2)5/2− µ
(r2 + z2)3/2
),
=z
a
(U +
2µ
(r2 + z2)3/2
)=z
a
(U +
2µ
a3
),
since r2 + z2 = a2. Hence the boundary condition u ·n = 0 at the sphere’s surface determinesthe strength of the dipole,
µ = −Ua3
2.
The velocity potential for a uniform flow past a stationary sphere is therefore given by
φ(r, z) = Uz
(1 +
a3
2(r2 + z2)3/2
). (4.9)
The corresponding Stokes streamfunction isgiven by
Ψ(r, z) =Ur2
2
(1− a3
(r2 + z2)3/2
). (4.10)
Outside the sphere Ψ > 0, but we also obtain a solution inside the sphere with Ψ < 0. Thisflow is not real; it is a “virtual flow” that allows for fluid velocity to be consistent with theboundary condition on a solid sphere.
34 4.5 Flow past an obstacle
4.5.2 Rankine half-body
Suppose that, in the velocity potential of a flow past a sphere, we replace the dipole with apoint source (m > 0), so that
φ(r, z) = Uz − m
(r2 + z2)1/2⇒ u = ∇φ =
(mr
(r2 + z2)3/2, U +
mz
(r2 + z2)3/2
).
This flow has a single stagnation point ur = uz = 0 at r = 0 and z = −√m/U .
To find the streamlines of the flow we calculate the Stokes streamfunction using
ur = −1
r
∂Ψ
∂zand uz =
1
r
∂Ψ
∂r.
Thus,
∂Ψ
∂r= Ur +
mrz
(r2 + z2)3/2⇒ Ψ =
Ur2
2− mz
(r2 + z2)1/2+ α(z),
⇒ 1
r
∂Ψ
∂z= − m
r (r2 + z2)1/2+
mz2
r (r2 + z2)3/2+α′(z)r
,
= −mr
(r2 + z2 − z2
)(r2 + z2)3/2
+α′(z)r
= − mr
(r2 + z2)3/2+α′(z)r
,
= −ur = − mr
(r2 + z2)3/2.
So, since α′(z) = 0, α is a constant (set to zero). The Stokes streamfunction is therefore
Ψ(r, z) =Ur2
2− mz
(r2 + z2)1/2.
At the stagnation point (r = 0, z = −√m/U), Ψ = m. Hence, the equation of the streamline,
or streamtube, passing through this stagnation point is
Ψ(r, z) = m⇔ Ur2
2= m
(1 +
z
(r2 + z2)1/2
).
Notice that the straight line r = 0 with z < 0 satisfies the equation of the streamline Ψ = m.For large positive z, the equation of the streamtube Ψ = m becomes
Ur2
2' 2m⇒ r ' 2
√m
U.
Thus, the velocity potential and the Stokes streamfunction
φ(r, z) = U
(z − a2
4 (r2 + z2)1/2
)and Ψ(r, z) =
U
2
(r2 − a2z
2 (r2 + z2)1/2
)provide a model for a long slender body of radius a = 2
√m/U .
Chapter 4 – Potential flows 35
4.6 Method of images
In previous examples we introduced flow singularities (e.g. sources and dipoles) outside of thedomain of fluid flow in order to satisfy boundary conditions at a solid surface.
This technique can also be used to calculate the flow produced by a singularity near a bound-ary; it is then called method of images.
Example 4.1 (Point source near a wall)Consider a point-source of fluid placed at the position (d, 0, 0) (Cartesian coordinates) near asolid wall at x = 0.
In free space (no wall), the potential of thesource is
φ∞ = − m√(x− d)2 + y2 + z2
,
⇒ u∞ =∂φ∞∂x
=m(x− d)
[(x− d)2 + y2 + z2]3/2.
y
x
d
O
So that, at x = 0,
u∞ = − md
(d2 + y2 + z2)3/26= 0,
which is inconsistent with the boundary condition u · n = u · ex = u = 0 at the wall.
To rectify this problem, (i.e. for the flow to satisfy the boundary condition at the wall), weadd a source of equal strength m outside the domain, at (−d, 0, 0). By symmetry, this sourcewill produce an equal but opposite velocity field at x = 0, so that the boundary condition forthe combined flow can be satisfied. The velocity potential for both sources becomes
φ = − m√(x− d)2 + y2 + z2
− m√(x+ d)2 + y2 + z2
,
and the velocity field along the x-axis,
u =∂φ
∂x=
m(x− d)
[(x− d)2 + y2 + z2]3/2+
m(x+ d)
[(x+ d)2 + y2 + z2]3/2.
Clearly, at x = 0, now u = 0 as required.The fluid can slip along the wall howeveras, for x = 0,
v =2my
(d2 + y2 + z2)3/2,
w =2mz
(d2 + y2 + z2)3/2.
x−d 0 d
S
4.7 Method of separation of variables
This is a standard method for solving linear partial differential equations with compatibleboundary conditions.
We shall seek separable solutions to Laplace’s equations, of the form φ(x, y) = f(x)g(y) inCartesian coordinates or φ(r, θ) = f(r)g(θ) in polar coordinates.
36 4.7 Method of separation of variables
Plane polar coordinates. We substitute a potential of the form φ(r, θ) = f(r)g(θ) inLaplace’s equation expressed in plane polar coordinates,
∇2φ =1
r
∂
∂r
(r∂φ
∂r
)+
1
r2∂2φ
∂θ2= 0,
⇒ g
r
d
dr
(r
df
dr
)+f
r2d2g
dθ2= 0,
⇒ r
f
d
dr
(r
df
dr
)+
1
g
d2g
dθ2= 0, (division by f(r)g(θ)/r2)
⇒ r
f
d
dr
(r
df
dr
)= −1
g
d2g
dθ2.
Since the terms on the left and right sides of the equation are functions of independentvariables, r and θ respectively, they must take a constant value, k2 say. Thus we havetransformed a partial differential equation for φ into two ordinary differential equations for fand g,
r
f
d
dr
(r
df
dr
)= k2 ⇒ r
d
dr
(r
df
dr
)− k2f = 0,
1
g
d2g
dθ2= −k2 ⇒ d2g
dθ2+ k2g = 0.
Thus, g(θ) = A cos(kθ) +B sin(kθ). For a 2π-periodic function g, such that g(θ) = g(θ+ 2π),k must be integer. So
g(θ) = A cos(nθ) +B sin(nθ), n ∈ Z,
and f is solution to
r2d2f
dr2+ r
df
dr− n2f = 0.
Substituting nontrivial functions of the form f = arα gives,[α(α− 1) + α− n2
]arα = 0⇔ α2 = n2.
The two independent solutions have α = ±n; the general separable solution to Laplace’sequation in plane polar coordinates is therefore
φ(r, θ) =(Arn +Br−n
)cos(nθ) +
(Crn +Dr−n
)sin(nθ), n ∈ Z, (4.11)
where A, B, C and D are constants to be determined by the boundary conditions.
Separable solutions to Laplace’s equation in spherical polar coordinates can be obtained in asimilar manner, but involves Legendre polynomials Pl(cos(θ)).
Example 4.2 (Cylinder in an extensional flow)Consider the velocity potential
φ(r, θ) =(Ar2 +Br−2
)cos(2θ)
corresponding to a particular solution to Laplace’s equation of the form (4.11), with n = 2.The radial velocity of this flow is
ur =∂φ
∂r= 2r
(A− B
r4
)cos(2θ).
Chapter 4 – Potential flows 37
It vanishes at the surface of a solid cylinder of radius a placed at the origin if B = a4A.Therefore the velocity field
ur = 2Ar
(1− a4
r4
)cos(2θ) and uθ = −2Ar
(1 +
a4
r4
)sin(2θ)
produced by the potential
φ(r, θ) = Ar2(
1 +a4
r4
)cos(2θ)
represents a fluid flow past a solid cylinder placed in an extensional flow.
Notice that at large distances, i.e. if r a, thefluid velocity is that of an extensional flow
ur ' 2Ar cos(2θ) and uθ ' −2Ar sin(2θ),
in polar coordinates, or equivalently
u ' 2Ax and v ' −2Ay,
in Cartesian coordinates.
38 4.7 Method of separation of variables
Chapter 5
Euler’s equation
Contents
5.1 Fluid momentum equation . . . . . . . . . . . . . . . . . . . . . . . . 39
5.2 Hydrostatics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
5.3 Archimedes’ theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
5.4 The vorticity equation . . . . . . . . . . . . . . . . . . . . . . . . . . 42
5.5 Kelvin’s circulation theorem . . . . . . . . . . . . . . . . . . . . . . . 43
5.6 Shape of the free surface of a rotating fluid . . . . . . . . . . . . . . 44
5.1 Fluid momentum equation
So far, we have discussed some kinematic properties of the velocity fields for incompressibleand irrotational fluid flows.
We shall now study the dynamics of fluid flows and consider changesin motion due to forces acting on a fluid.
We derive an evolution equation for the fluid momentum by consider-ing forces acting on a small blob of fluid, of volume V and surface S,containing many fluid particles.
V
S
5.1.1 Forces acting on a fluid
The forces acting on the fluid can be divided into two types.
Body forces, such as gravity, act on all the particles throughout V ,
Fv =
∫Vρg dV.
Surface forces are caused by interactions at the surface S. For the rest of this course weshall only consider the effect of fluid pressure.
39
40 5.2 Hydrostatics
Collisions between fluid molecules on either sides of the surface S pro-duce a flux of momentum across the boundary, in the direction of thenormal n.The force exerted on the fluid into V by the fluid on the other side ofS is, by convention, written as
Fs =
∫S−pn dS,
where p(x) > 0 is the fluid pressure.
V
S
n
5.1.2 Newton’s law of motion
Newton’s second law of motion tells that the sum of the forces acting on the volume of fluid Vis equal to the rate of change of its momentum. Since Du/Dt is the acceleration of the fluidparticles, or fluid elements, within V , one has∫
Vρ
Du
DtdV =
∫S−pn dS +
∫Vρg dV.
We now apply the divergence theorem,∫Vρ
Du
DtdV =
∫V
(−∇p+ ρg) dV,
and notice that both integrands must be identical, since V is arbitrary.
So, the evolution of fluid momentum is governed by Euler’s equation
ρDu
Dt= ρ
[∂u
∂t+ (u · ∇) u
]= −∇p+ ρg. (5.1)
This equation neglects viscous effects (tangential surface forces due to velocity gradients)which would otherwise introduce an extra term, µ∇2u, where µ is the viscosity of the fluid,as in the Navier-Stokes equation
ρDu
Dt= −∇p+ ρg + µ∇2u. (5.2)
For the rest of the course we shall only consider perfect fluids which are idealised fluids,inviscid and incompressible with constant mass density.
5.2 Hydrostatics
We first consider the case of a fluid at rest, such that u = 0. Euler’s equation is then reducedto the equation of hydrostatic balance,
∇p = ρg⇔ p(x) = ρg · x + C, (5.3)
where C is a constant.
Example 5.1The density of mass in the ocean can be considered as constant, ρ0, and the gravity g = −gez.(The coordinate z is the upward distance from sea-level.)
Chapter 5 – Euler’s equation 41
From Euler’s equation one has
dp
dz= −ρ0g ⇒ p(z) = p0 − ρgz.
Hence the pressure increases linearly withdepth (z < 0).
z
O
gρ0g
−∇p
Taking typical values for the physical constant, g ' 10 m s−2, ρ0 ' 103 kg m−3 and a pressureof one atmosphere at sea-level, p0 ' patm = 105 Pa = 105 N m−2 gives p(z) ' 105(1 − 0.1z);the pressure increases by one atmosphere every 10 m.(Notice that the change in pressure force on a surface S, between the ocean surface z = 0and a given depth z = −d, is equal to (p− p0)S = ρ0gdS, which is the weight of a column ofwater of height d and section S.)
5.3 Archimedes’ theorem
The force on a body in a fluid is an upthrust equal to the weight of fluid displaced.
Consider a solid body of volume V and surface Stotally submerged in a fluid of density ρ0. The totalforce on the body caused by the fluid surroundingit is
F = −∫Spn dS,
where p is the fluid pressure.
g
V n
S
F = −ρ0V g
(Notice that the pressure distribution on the surface S is the same whether the fluid containsa solid or not.) So, using successively the divergence theorem and the equation of hydrostaticbalance, ∇P = ρg, we find
F = −∫V∇pdV = −
∫Vρ0g dV = −ρ0V g.
The buoyancy force is equal the weight of the mass of fluid displaced, M = ρ0V , and pointsin the direction opposite to gravity.
If the fluid is only partially submerged, then we need to split it into parts above and belowthe water surface, and apply Archimedes’ theorem to the lower section only.
Consider a solid body of volume V and density ρspartially submerged in a fluid of density ρ0 > ρs.Let V1 be the volume of solid above the fluid surfaceand V2 the volume underneath. Since, the solid is inequilibrium, its weight is balanced by the buoyancyforce, so that ρsV g = ρ0V2g.
ρ0
V1
ρsV g
−ρ0V2g
V2
ρs
Hence, the fractions of the volume of solid immersed in the fluid and not immersed are
V2V
=ρsρ0
andV1V
= 1− V2V
=ρ0 − ρsρ0
,
respectively. For an iceberg of density ρs ' 0.915 kg m−3 floating in salted water of densityρs ' 1.025 kg m−3, V2/V ' 89.3% and V1/V ' 10.7%.
Question: A glass of water with an ice cube in it is filled to rim. What happens as ice melts?
42 5.4 The vorticity equation
5.4 The vorticity equation
In the expression of the acceleration of a fluid particle,
Du
Dt=∂u
∂t+ (u · ∇) u,
the nonlinear term can be rewritten using the vector identity
u× (∇× u) = ∇(‖u‖2
2
)− (u · ∇) u⇒ (u · ∇) u = ∇
(‖u‖22
)− u× ω,
where ‖u‖2 = u · u. So, we can rewrite Euler’s equation (5.1) as
∂u
∂t+ (u · ∇) u = −1
ρ∇p+ g,
∂u
∂t+∇
(‖u‖22
)− u× ω = −∇
(p
ρ
)+ g (ρ constant),
and take its curl
∂ω
∂t−∇× (u× ω) = 0,
∂ω
∂t+ (u · ∇)ω − (ω · ∇) u + (∇ · u)ω − (∇ · ω) u = 0.
For incompressible flows ∇ · u = 0 and, in addition, ∇ · ω = 0 as ω = ∇× u. Hence,
Dω
Dt=∂ω
∂t+ (u · ∇)ω = (ω · ∇) u. (5.4)
This is the vorticity equation. It shows that the vorticity of a fluid particle changes becauseof gradients of u in the direction of ω
Properties of the vorticity equation.
i. If ω = 0 everywhere initially, then ω remains zero. Thus, flows that start off irrotationalremain so.
ii. In a two-dimensional planar flow, u = (u(x, y), v(x, y), 0), the vector vorticity has onlyone non-zero component, ω = (∂v/∂x− ∂u/∂y)ez, so that
(ω · ∇) u = ωd
dzu(x, y) = 0.
Hence, the vorticity equation, reduced to
Dω
Dt=∂ω
∂t+ (u · ∇)ω = 0, (5.5)
shows that the vorticity of a fluid particle remains constant. If, in addition the flow issteady, ∂ω/∂t = 0 then the vorticity is constant along streamlines.
Chapter 5 – Euler’s equation 43
iii. Vortex stretching.
The stretching of a vortex leads to the increaseof its vorticity.
Consider, for example, an incompressiblesteady flow in a converging cone, function ofthe radial distance only in spherical polar co-ordinates,
u(r) = ur(r)er + uϕ(r)eϕ.O
r
uϕ
ur
The component ur represents a radial inflow and uϕ the swirling of the fluid. Since∇× (ur(r)er) = 0, only the swirling motion contributes to a non-zero vorticity
ω(r) = ∇× u = ∇× (uϕ(r)eϕ) = ωrer + ωθeθ.
From mass conservation in an incompressible spherically symmetric inflow, we find
∇ · u =1
r2d
dr
(r2ur
)= 0⇔ ur = − k
r2,
where k > 0 is constant. Since the flow is steady, ∂ω/∂t = 0, the evolution equation forthe radial component of the vorticity, ωr, becomes
DωrDt
= urdωrdr
= ωrdurdr⇔ d
drln
∣∣∣∣ωrur∣∣∣∣ = 0⇔ ωr
ur= α, constant.
Thus,
wr = αur = −αkr2,
which demonstrates that the vorticity ωr increases as ur in-creases; the initial vortex is stretched by the inflow.This is the reason for the bath-plug vortex. A small amountof background vorticity is amplified by the flow converginginto a small hole. (This mechanism can be interpreted as theconservation of angular momentum of fluid particles.)
5.5 Kelvin’s circulation theorem
The circulation around a closed material curve remains constant — in an inviscid fluid ofuniform density, subject to conservative forces. Hence,
dΓ
dt=
d
dt
∮C(t)
u · dl = 0, (5.6)
if C(t) is a closed curve formed of fluid particles following the flow.
Proof. Let C(t) be a closed material curve, hence formed of fluid particles, of parametricrepresentation x(s, t) with s ∈ [0, 1], say. Using this parametric representation, the rate of
44 5.6 Shape of the free surface of a rotating fluid
change of the circulation around C(t) can be written as
dΓ
dt=
d
dt
∫ 1
0u(x(s, t), t) · ∂
∂sx(s, t) ds =
∫ 1
0
d
dt
[u(x, t) · ∂x
∂s
]ds,
=
∫ 1
0
[∂u
∂t+
(∂x
∂t· ∇)
u
]· ∂x
∂s+ u · ∂
2x
∂s∂t
ds,
=
∫ 1
0
[∂u
∂t+ (u · ∇) u
]· ∂x
∂s+ u · ∂u
∂s
ds,
since ∂x/∂t = u is the velocity of the fluid particle at x(s, t). So, using Euler’s equation inthe form
∂u
∂t+ (u · ∇) u = ∇
(−pρ
+ g · x),
we find
dΓ
dt=
∫ 1
0
[∇(−pρ
+ g · x)· ∂x
∂s+
∂
∂s
(‖u‖22
)]ds =
∫ 1
0
∂
∂s
(‖u‖22− p
ρ+ g · x
)ds.
Thus, since the curve C(t) is closed,
dΓ
dt=
∮C(t)
d
(‖u‖22− p
ρ+ g · x
)= 0,
as required.
Recall that the circulation around a closed curve C is equal to the flux of vorticity throughan arbitrary surface S that spans C. So, from Kelvin’s circulation theorem,
dΓ
dt=
d
dt
∮C(t)
u · dl =d
dt
∫S(t)
ω · n dS = 0;
this demonstrates that the flux of vorticity through a surface that spans a material curve isconstant. Thus, many properties of the vorticity equation could equivalently be derived fromthe circulation theorem (e.g. vortex stretching, persistence of irrotationality.)
5.6 Shape of the free surface of a rotating fluid
The surface of a rotating liquid placed in a container is not flat but dips near the axis ofrotation. This phenomenon, which can be observed when stirring coffee in a mug, resultsfrom a radial pressure-gradient balancing the centrifugal force acting within the fluid.
Consider, for example, a cylindrical containerpartially filled with fluid and mounted on ahorizontal turntable. When the flow reachesa steady state, ∂u/∂t = 0, the fluid rotatesuniformly, with a constant angular velocity Ωabout the vertical z-axis. (The fluid rotateswith the container as a solid body.)In order to calculate the height of the free sur-face of fluid, z = h(r), we shall solve Euler’sequation,
z
r
h0h(r)
O
Ω
ρ
(∂u
∂t− u× ω
)= −ρu× ω = −∇
(p+
1
2ρ‖u‖2
)+ ρg, (5.7)
Chapter 5 – Euler’s equation 45
in cylindrical polar coordinates. The velocity and vorticity fields of the fluid in uniformrotation are u = uθeθ and ω = ωzez, where uθ = rΩ and ωz = 2Ω. Hence, the nonlinear termin Euler’s equation
u× ω = uθωzer = 2rΩ2er.
The radial component of the vector equation (5.7) (i.e. its scalar product with er) is therefore
−2ρrΩ2 = −∂p∂r− ρ
2
du2θdr
= −∂p∂r− ρrΩ2,
⇔ ∂p
∂r= ρrΩ2. (5.8)
This shows that the pressure must vary with the radius inside the fluid, in order to balancethe centrifugal force. Moreover, the pressure must also satisfy the vertical hydrostatic balance(i.e. balance between pressure and gravity).
From the vertical component of the vector equation (5.7) (i.e. its scalar product with ez) wefind
0 = −∂p∂z− ρg ⇒ p(r, z) = −ρgz +A(r),
where A(r) is a function to be determined using appropriate boundary conditions. At thefree surface the fluid pressure should match the atmospheric pressure. Hence, p = patm atz = h(r), so that
patm = −ρgh(r) +A(r)⇔ A(r) = patm + ρgh(r).
So, substituting the expression of the pressure,
p(r, z) = patm + ρg (h(r)− z) ,
in equation (5.8) leads to
∂p
∂r= ρg
dh
dr= ρrΩ2 ⇒ h(r) = h0 +
r2Ω2
2g,
where h0 is the height of the free surface atr = 0, on the rotation axis. This result showsthat the free surface of a uniformly rotatingfluid in a cylindrical container is a paraboloid.
Such rotating cylindrical containers filled withhighly-reflecting liquids (e.g. mercury or ionicliquid coated with silver) have been used tobuild mirrors with a large (up to 6 m) smoothreflecting paraboloid surface, which could beused as primary mirrors for telescopes.
46 5.6 Shape of the free surface of a rotating fluid
Chapter 6
Bernoulli’s equation
Contents
6.1 Bernoulli’s theorem for steady flows . . . . . . . . . . . . . . . . . . 47
6.2 Bernoulli’s theorem for potential flows . . . . . . . . . . . . . . . . 50
6.3 Drag force on a sphere . . . . . . . . . . . . . . . . . . . . . . . . . . 52
6.4 Separation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
6.5 Unsteady flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
6.6 Acceleration of a sphere . . . . . . . . . . . . . . . . . . . . . . . . . 57
In section (5.4) we obtained the momentum equation for ideal fluids (i.e. inviscid and withconstant density) in the form
∂u
∂t− u× ω +∇
(1
2‖u‖2
)= −∇
(p
ρ
)+ g.
So, since the constant gravity g = ∇(g · x), one has
∂u
∂t− u× ω +∇H = 0, (6.1)
where
H(x, t) =p
ρ+
1
2‖u‖2 − g · x (6.2)
is called the Bernoulli function.1
6.1 Bernoulli’s theorem for steady flows
In the case of steady flows, i.e. when ∂u/∂t = 0, taking the scalar product of equation (6.1)with the fluid velocity, u, gives the Bernoulli equation
(u · ∇)H = 0, (6.3)
since u · (u× ω) ≡ 0.
Hence, for an ideal fluid in steady flow,
H(x) =p
ρ+
1
2‖u‖2 − g · x (6.4)
is constant along a streamline.
1Not to be mistaken for Bernoulli’s polynomials.
47
48 6.1 Bernoulli’s theorem for steady flows
So, if a streamfunction ψ(x) can be defined, H is a function of ψ (H(x) ≡ H(ψ)).
Example 6.1 (The Venturi effect.)Consider a flow through a narrow constriction of cross-section area A2; upstream and down-stream the cross-sectional area is A1.
h1
h2
h3
A1
S1
S2
A2
A1
S3V1
(a)
(b)
(c)
(Three narrow vertical tubes, (a), (b) and (c), are used to measure the pressure at differentpoints.)
The fluid velocity is assumed inform on cross sections, S. Upstream the fluid velocity is V1.
Mass conservation implies
∫Sρu · n dS = constant for any cross-section S, so
∫S1
ρu · n dS =
∫S2
ρu · n dS ⇒∫S1
u · n dS =
∫S2
u · n dS (ρ = constant),
⇒ A1V1 = A2V2 ⇒ V2 =A1
A2V1 > V1 (since A1 > A2).
Neglecting gravity, we apply Bernoulli’s equation to any streamline,
p1ρ
+1
2V 21 =
p2ρ
+1
2V 22 ⇒ p2 = p1 −
ρ
2
(V 22 − V 2
1
),
⇒ p2 = p1 −ρV 2
1
2A22
(A2
1 −A22
)< p1.
Thus, in the constriction the speed of the flow increases (conservation of mass) and its pressuredecreases (Bernoulli’s equation).
This can be measured by the thin tubes where there is fluid but no flow (i.e. fluid in hydrostaticequilibrium). If h1 is the height of fluid in the tube (a) then
p1 = p0 + ρgh1 (p0 ≡ patm).
If h2 is the height of fluid in the tube (b) then
p2 = p0 + ρgh2 ⇒ h2 =p2 − p0ρg
=p1 − p0ρg
− V 21
2gA22
(A2
1 −A22
),
⇒ h2 = h1 −V 21
2gA22
(A2
1 −A22
)< h1.
In tube (c), V3 = V1 since A3 = A1 (mass conservation). So, Bernoulli’s equation gives
p3 +1
2ρV 2
3 = p1 +1
2ρV 2
1 ⇒ p3 = p1,
and so h3 = h1. (In practice, h3 will be slightly less than h1 due to viscosity but the effect issmall.)
Chapter 6 – Bernoulli’s equation 49
Example 6.2 (Flow down a barrel.)How fast does fluid flow out of a barrel?
A(h) g
U
h
a
z
0
Let h be the height of fluid level in the barrel above the outlet, which has cross-sectional areaa. If a A(h), then the flow can be treated as approximately steady.
Mass conservation: −Adh
dt= aU (with U > 0). So, if a A then
∣∣∣∣dhdt∣∣∣∣ |U |.
Bernoulli’s theorem: consider a streamline from the surface of the fluid to the outlet,
p+1
2ρ‖u‖2 + ρgz = const.
At z = 0: p = patm and u = U ; at z = h: p = patm and u =dh
dt. So,
patm +1
2ρU2 = patm +
1
2ρ
(dh
dt
)2
+ ρgh,
⇒ U2 =
(dh
dt
)2
+ 2gh⇒ U '√
2gh since
∣∣∣∣dhdt∣∣∣∣ |U |.
We could have guessed this result from conservation of energy
with KE ' 0 and PE = ρgh at z = h
and KE =1
2ρU2 and PE = 0 at z = 0
⇒ 1
2ρU2 ' ρgh.
Example 6.3 (Siphon.)A technique for removing fluid from one vessel to another without pouring is to use a siphontube.
z
0
C
B
L
H
A
g
50 6.2 Bernoulli’s theorem for potential flows
To start the siphon we need to fill the tube with fluid, but once it is going, the fluid willcontinue to flow from the upper to the lower container.
In order to calculate the flow rate, we can use Bernoulli’s equation along a streamline fromthe surface to the exit of the pipe.
At point A: p = patm, z = 0. We shall assume that the container’s cross-sectional area ismuch larger than that of the pipe. So, UA ' 0 (from mass conservation; see example 6.2−Adh/dt = aU).
At point C: p = patm, z = −H, u = Uc ≡ U .
Bernoulli’s equation:
patmρ
+1
2U2A
' 0
=patmρ
+1
2U2 − gH ⇒ U '
√2gH.
If B is the highest point: (UB = UC ≡ U from mass conservation)
pBρ
+1
2U2 + gL =
patmρ
+1
2U2 − gH ⇒ pB = patm − ρg(L+H) < patm.
For pB > 0, we need H + L <patmρg≈ 105
103 × 10= 10m.
6.2 Bernoulli’s theorem for potential flows
In this section we shall extend Bernoulli’s theorem to the case of irrotational flows.
Recall that Euler’s equation can written in the form
∂u
∂t− u× ω = −∇H where H(x, t) =
p
ρ+
1
2‖u‖2 − g · x.
If the fluid flow is irrotational, i.e. if ω = ∇ × u = 0, then u × ω = 0 and u = ∇φ; so, theequation above becomes
∇(∂φ
∂t+H
)= 0,
since∂u
∂t=∂∇φ∂t
= ∇(∂φ
∂t
).
Thus, for irrotational flows,
∂φ
∂t+H =
∂φ
∂t+p
ρ+
1
2‖∇φ‖2 − g · x ≡ f(t) (6.5)
is a function of time, independent of the position, x.
If, in addition, the flow is steady,
H =p
ρ+
1
2‖∇φ‖2 − g · x, (6.6)
is constant; H has the same value on all streamlines.
Chapter 6 – Bernoulli’s equation 51
Example 6.4 (Shape of the free surface of a fluid near a rotating rod)We consider a rod of radius a, rotating at constant angular velocity Ω, placed in a fluid.
Assuming a potential, axisymmetric and planarfluid flow, (ur(r), uθ(r)) in cylindrical polar coor-dinates, we wish to calculate the height of the freesurface of the fluid near to the rod, h(r). We alsoassume that the solid rod is an impenetrable sur-face on which the fluid does not slip, so that theboundary conditions for the velocity field are
ur = 0 and uθ = aΩ at r = a.
h(r)
Ω
r
g
z
a
From mass conservation, one has
∇ · u =1
r
d
dr(rur) = 0⇔ ur(r) =
C
r,
where C is a constant of integration. However, the boundary condition ur = C/a = 0 at r = aimplies that C = 0. So, ur = 0 and the fluid motion is purely azimuthal.As we assume an irrotational flow,
∇× u =1
r
d
dr(ruθ) ez = 0⇔ uθ(r) =
k
r,
where k is an integration constant to be determined using the second boundary condition. Atr = a, uθ = k/a = aΩ which implies that k = a2Ω. So, the fluid velocity near to the rod is
ur = 0 and uθ =a2Ω
r.
Notice that the velocity potential, function of θ, can be determined using
u = ∇φ⇒ 1
r
dφ
dθ=a2Ω
r⇒ φ(θ) = a2Ωθ.
By applying Bernoulli’s theorem for steady potential flows to the free surface (which is not astreamline, as streamlines are circles about the rod axis) we obtain,
H =patmρ
+1
2u2θ(r) + gh(r)︸ ︷︷ ︸
near rod
=patmρ
+ gh∞︸ ︷︷ ︸at large r
,
where the constant pressure p = patm is the atmo-spheric pressure and lim
r→∞h(r) = h∞. (Notice also
that uθ ∝ 1/r → 0 as r →∞.)
Thus, the height of the free surface is
h(r) = h∞ −1
2gu2θ(r) = h∞ −
a4Ω2
2gr2, (6.7)
which shows that the free surface dips as 1/r2 nearto the rotating rod.
h∞h(r)
∝ 1/r2
Alternatively, Euler’s equation could be solved directly (i.e. without involving Bernoulli’stheorem) as in § 5.6 with an azimuthal flow, now potential, of the form uθ = a2Ω/r. We
52 6.3 Drag force on a sphere
can then explain the result (6.7) in terms of centripetal acceleration; since the fluid particlesmove in circles, there must be an inwards central force producing the necessary centripetalacceleration (i.e. balancing the centrifugal force). Indeed, from the radial component of themomentum equation, one has
−ρu2θ
r= −∂p
∂r⇒ ∂p
∂r= ρ
a4Ω2
r3.
However, since the fluid is in vertical hydrostatic equilibrium, the pressure satisfies
∂p
∂z= −ρg ⇒ p(r, z) = patm − ρg(z − h(r)).
Hence, we have∂p
∂r= ρg
dh
dr= ρ
a4Ω2
r3⇒ h(r) = h∞ −
a4Ω2
2gr2,
as in equation (6.7).
6.3 Drag force on a sphere
We wish to calculate the pressure force exerted by a steady fluid flow on a solid sphere.
ϕ
U
z
n r
θ
ra
In § 4.5.1 we obtained the velocity potential of auniform stream, U ez, past a stationary sphere ofradius a,
φ(r, z) = Uz
(1 +
a3
2(r2 + z2)3/2
),
in cylindrical polar coordinates (r, θ, z). In sphericalpolar coordinates, (r, θ, ϕ), this velocity potentialbecomes
φ(r, θ) = U cos θ
(r +
a3
2r2
). (6.8)
The non-zero components of the fluid velocity, u = ∇φ, are then
ur =∂φ
∂r= U cos θ
(1− a3
r3
)and uθ =
1
r
∂φ
∂θ= −U sin θ
(1 +
a3
2r3
). (6.9)
Hence, at r = a, on the solid sphere’s surface, ur = 0 as required by the kinematic boundaryconditions and
uθ(θ)|r=a = −3
2U sin θ.
To express the pressure force on the sphere in terms of the fluid velocity, we use Bernoulli’stheorem for steady potential flows, H = p/ρ+ ‖u‖2/2 = constant, ignoring gravity. At r = athe fluid pressure, p(θ), therefore satisfies
p(θ)
ρ+
1
2u2θ∣∣r=a
=p∞ρ
+1
2U2,
where p∞ is the pressure as r →∞.
Chapter 6 – Bernoulli’s equation 53
Thus, the pressure distribution on the sphere is
p(θ) = p∞ +1
2ρU2
(1− 9
4sin2 θ
), (6.10)
and the total pressure force is the surface integral of p(θ) on the sphere r = a,
F = −∫Spn dS = −
∫ π
0
∫ 2π
0p(θ) er a
2 sin θ dϕdθ, (6.11)
where er = sin θ cosϕ ex + sin θ sinϕ ey + cos θ ez.As the flow is axisymmetric, the only non-zero component of the force should be in the axialdirection, z. Indeed,
Fx = F · ex = −a2∫ 2π
0cosϕdϕ
∫ π
0p(θ) sin2 θ dθ = 0,
and
Fy = F · ey = −a2∫ 2π
0sinϕdϕ
∫ π
0p(θ) sin2 θ dθ = 0.
However, after substituting for p(θ) in
Fz = F · ez = −2πa2∫ π
0p(θ) sin θ cos θ dθ,
we find that
Fz = −2πa2[(p∞ +
1
2ρU2
)∫ π
0sin θ cos θ dθ − 9
8ρU2
∫ π
0sin3 θ cos θ dθ
]= 0,
so that the total drag force on the sphere, due to the fluid flow around it, is zero!
D’Alembert’s paradox: it can be demonstrated that the drag force on any 3-D solid bodymoving at uniform speed in a potential flow is zero (see, e.g., Paterson, § XI.9, p. 240).This is not true in reality of course, as flows past 3-D solid bodies are not potential.
We can see why a potential flow past a sphere gives zero drag by looking at the streamlines.
U
S1 S2
low p
high phigh p
low p
The flow is clearly fore-aft symmetric (symmetry about z = 0); the front (S1) and the back(S2) of the sphere are stagnation points at equal pressure, PS1 = PS2 = p∞ + 1
2ρU2. At the
side, ur = 0 and u2θ > 0, so from Bernoulli’s theorem, the pressure there is lower than atthe stagnation points but it must have the same symmetry as the flow. Notice that, fromBernoulli’s theorem, the pressure does not depend on the direction of the flow, but on itsspeed ‖u‖ only.However, the real flow past a sphere is not symmetric and, as a consequence, the fluid exertsa net drag force on the sphere.
54 6.4 Separation
6.4 Separation
The pressure distribution on the surface a solid sphere placed is a uniform stream,
p(θ) = p∞ +1
2ρU2
(1− 9
4sin2 θ
),
reaches its minimum, pmin = p∞ − 5/8 ρU2, at θ = ±π/2. So, the pressure gradient in thedirection of the flow, (u·∇)p, is a positive from θ = 0 to θ = ±π/2 and negative beyond.
(u·∇)p < 0
(u·∇)p < 0
U
(u·∇)p > 0
(u·∇)p > 0
pmin
pmaxU θ
pmin
pmax
An adverse pressure gradient, (u·∇)p > 0 (i.e. pressureincreasing in the direction of the flow along the surface),is “bad news” and causes the flow to separate, leaving aturbulent wake behind the sphere.Very roughly one can estimate the pressure differenceupstream and downstream as 1/2 ρU2, so that the dragforce F ∝ 1/2 ρU2 × A, where A is the cross-sectionalarea.The ratio
CD =F
12ρU
2A(6.12)
is called drag coefficient and depends, e.g., on the shapeof the body (see Acheson §4.13, p. 150).
The way to reduce drag (i.e. resistance) is to reduce separation:
• Streamlining: separation occurs because of adverse pressure gradients on the surface ofsolid bodies. These can be reduced by using more “streamlined” shapes, that avoid di-verging streamlines (e.g., aerodynamic bike helmets (time trial cyclist), ships, aeroplanesand cars).
• Surface roughness: paradoxically, a rough surface can reduce drag by reducing separa-tion (e.g. dimple pattern of golf balls and shining of cricket ball on one side).
Chapter 6 – Bernoulli’s equation 55
6.5 Unsteady flows
6.5.1 Flows in pipes
In example 6.2 we consider a flow out of a barrel through a small hole. Now, consider a flowout of a narrowing tube, opened to the atmosphere at both ends, where the exit is not muchsmaller than the cross-section (i.e. the fluid flow cannot be assumed steady).
gh(t)
z
a
A(h)
Let A(z) be the smoothly varying cross-sectional area of the pipe at height z, such thatA→ A∞ as z →∞ and A(0) = a.
We assume that the flow is potential and purely in the z-direction, uz = ∂φ/∂z ≡ w.
By conservation of mass the volume flux, Q(t) = −w(z, t)A(z), must be independent of height.Hence,
∂φ
∂z= w(z, t) = −Q(t)
A(z)⇒ φ(z, t) = φ(0, t)−Q(t)
∫ z
0
dµ
A(µ).
(Note that we could set φ(0, t) = 0 without loss of generality.) Applying Bernoulli’s theoremfor potential flows, p/ρ+ ‖u‖2/2 +∂φ/∂t−g ·x = F (t), at the free surface and the exit gives,
at z = 0,patmρ
+1
2
Q2(t)
a2+
d
dtφ(0, t) = F (t),
and at z = h,patmρ
+1
2
(dh
dt
)2
+d
dtφ(0, t)− dQ
dt
∫ h
0
dz
A(z)+ gh = F (t).
Equating both expressions gives
1
2
[(dh
dt
)2
− Q2(t)
a2
]− dQ
dt
∫ h
0
dz
A(z)+ gh = 0,
⇔ 1
2
[1− A2(h)
a2
](dh
dt
)2
+A(h)d2h
dt2
∫ h
0
dz
A(z)+ gh = 0 since Q(t) = −A(h)
dh
dt.
The fluid height, h(t), is then solution to the nonlinear second order ordinary differentialequation (
A(h)
∫ h
0
dz
A(z)
)d2h
dt2+
1
2
[1− A2(h)
a2
](dh
dt
)2
+ gh = 0. (6.13)
Far from the exit this equation becomes approximately
hh+1
2
(1− A2
∞a2
)h2 + gh = 0,
since, as h→∞,
A(h) ∼ A∞ and
∫ h
0
dz
A(z)∼∫ h
0
dz
A∞=
h
A∞.
56 6.5 Unsteady flows
Using the chain rule, h = dh/dt = dh/dh dh/dt = h dh/dh, one finds
hhdh
dh+
1
2
(1− A2
∞a2
)h2 + gh = 0 ⇔ 1
2
dh2
dh+
1
2
(1− A2
∞a2
)h2
h+ g = 0
which can be written as a linear differential equation for Z = h2/2,
dZ
dh+
(1− A2
∞a2
)Z
h+ g = 0.
6.5.2 Bubble oscillations
The sound of a “babbling brook” is due to the oscillation (compression/expansion) of airbubbles entrained into the stream. The pitch of the sound depends on the size of the bubbles.
Consider a bubble of radius a(t); the velocity of the fluid at the bubble surface, ur =da
dt≡ a.
a(t)
gas
liquid
We can model the oscillations of the bubble of air using a potential flow due to a pointsource/sink of fluid at the centre of the bubble,
φ(r, t) = −k(t)
r⇒ ur =
∂φ
∂r=
k
r2.
The boundary condition at the bubble’s surface, r = a, is ur =k
a2= a. So,
k = aa2 ⇒ ur =aa2
r2and φ = − aa
2
r⇒ ∂φ
∂t= − aa
2
r− 2
aa2
r
Applying Bernoulli’s theorem (ignoring gravity) as r →∞,
p
ρ+
1
2‖∇φ‖2 +
∂φ
∂t= F (t) =
p∞ρ
(as r →∞, φ→ 0 and ‖u‖ → 0: the fluid is stationary).
At the bubble’s surface,
p(a)
ρ+
1
2a2 − aa2
a− 2
aa2
a=p(a)
ρ− aa− 3
2a2 = F (t).
Combining the two expressions above, one gets
p(a)− p∞ρ
= aa+3
2a2, (6.14)
where p(a) is the fluid pressure at the bubble’s surface. Now, if the gas inside the bubble ofmass m is subject to adiabatic changes, its equation of state is
pg = Kργg where ρg =3m
4πa3,
Chapter 6 – Bernoulli’s equation 57
and K is a constant to determine — the adiabatic index γ depends on the gas considered.
Moreover, since the bubble of gas is in balance with the surrounding fluid, continuity ofpressure pg = p(a) must be satisfied at the surface r = a(t).
Now, for a bubble in equilibrium, such that a = a0 and a = a = 0, equation (6.14) givesp = p∞ and, imposing pressure continuity pg = p at r = a0, one gets
pg = Kργg = K
(3m
4πa30
)γ= p∞ ⇒ K = p∞
(4πa303m
)γ.
So, pressure continuity at the bubble’s surface r = a(t) implies
p(a) = pg = Kργg = p∞
(4πa303m
)γ (3m
4πa3
)γ= p∞
(a0a
)3γ.
Then, equation (6.14) becomes
p∞ρ
(a3γ0a3γ− 1
)= aa+
3
2a2.
For small amplitude oscillations about the equilibrium a(t) = a0 + ε(t) where |ε| a0, so thata = ε, a = ε and a2 = ε2 ' 0; the nonlinear terms are negligible at first approximation. Thus,
a0ε =p∞ρ
a3γ0
a3γ0
(1 + ε
a0
)3γ − 1
' −3γp∞ρ
ε
a0,
⇒ ε+3γp∞ρa20
ε = 0.
The bubble undergo periodic small amplitude oscillations with frequency ω =
(3γp∞ρa20
)1/2
.
Note that the frequency scales with the inverse of the (mean) radius of the bubbles. E.g. forγ = 3/2, p∞ = 105 Pa and ρ = 103 kg m−3,
f =ω
2π=
1
2πa0
√3γp∞ρ⇒ f × a0 ' 3 kHz mm.
For bubbles of size a0 = 0.2 mm, f ' 15 kHz (G9).
6.6 Acceleration of a sphere
We have already shown that a sphere moving with a steady velocity under a potential flowhas no drag force. What about an accelerating sphere?
The velocity potential for a sphere of radius a moving with velocity U in still water is
φ = −Ua3
2r2cos θ.
(This flow satisfies the following boundary conditions: u = ∇φ→ 0 as r →∞ together withur = U cos θ er at r = a.)
58 6.6 Acceleration of a sphere
Rather than calculating the pressure via Bernoulli’s theorem, we calculate the work done bythe forces acting on the sphere as it moves at speed U , function of time, through the fluid.
The total kinetic energy of the system sphere of mass m plus fluid is
T =1
2mU2 +
∫V
1
2ρ (∇φ)2 dV,
=1
2mU2 +
1
2ρ
∫V
∇·(φ∇φ)− φ∇2φ︸︷︷︸0
dV, (using ∇·(fA) = A · ∇f + f∇·A)
=1
2mU2 +
1
2ρ
∫Sφ∇φ · n dS, by divergence theorem.
Here S is the surface of the sphere of radius a. So n = −er and dS = a2 sin θ dθ dϕ, such that
T =1
2mU2 − 1
2ρ
∫ π
0φ|r=a
∂φ
∂r
∣∣∣∣r=a
2πa2 sin θ dθ,
=1
2mU2 +
πa3
2ρU2
∫ π
0cos2 θ sin θ dθ,
=1
2mU2 +
πa3
3ρU2 since
∫ π
0cos2 θ sin θ dθ = −1
3
∫ π
0
d cos3 θ
dθdθ =
2
3.
So T =1
2(m+M)U2, where M =
2
3πa3ρ is called the added mass and represents the mass of
fluid that must be accelerated along with the sphere.
The rate of working of the forces F acting on the sphere equals the change of kinetic energy,
FU =dT
dt= (m+M)U
dU
dt.
Hence, the force required to accelerate the sphere is given by
F = (m+M)dU
dt.
Thus, the acceleration of a bubble (mass m and radius a) rising under gravity (see §5.3 onArchimedes theorem) satisfies
F =4
3πa3ρg︸ ︷︷ ︸
buoyancy force
−mg︸ ︷︷ ︸weight
= (2M −m)g = (m+M)dU
dt,
⇒ dU
dt=
2M −mM +m
g =4πa3ρ− 3m
2πa3ρ+ 3mg.
As mass density is much less for a gas than for a liquid, we can assumemM , so that
dU
dt' 2g.
buoyancy
mg
z
Alternatively: Consider a bubble of mass m rising under gravity with speed U =dz
dt.
Chapter 6 – Bernoulli’s equation 59
z
U
t
At height z the potential energy is
V = mgz︸︷︷︸weight
− 4
3πa3ρgz︸ ︷︷ ︸
buoyancy
.
In absence of dissipative processes the total energy remains constant; hence,
T + V =1
2(m+M)U2 +mgz − 4
3πa3ρgz = const.
Differentiating this expression with respect to time gives
(m+M)UdU
dt+
(m− 4
3πa3ρ
)gU = 0,
⇒ dU
dt=
2M −mM +m
g =4πa3ρ− 3m
2πa3ρ+ 3mg.
Again, for a bubble of gas in a liquid M m, sodU
dt' 2g; the bubble accelerates at twice
the gravitational acceleration.
60 6.6 Acceleration of a sphere
Chapter 7
Steady flows in open channels
Contents
7.1 Conservation of mass . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
7.2 Bernoulli’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
7.3 Flow over a hump . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
7.4 Critical flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
7.5 Flow through a constriction . . . . . . . . . . . . . . . . . . . . . . . 67
7.6 Transition caused by a sluice gate . . . . . . . . . . . . . . . . . . . 69
In this chapter we shall use conservation of mass and Bernoulli’s equation to study simplifiedmodels of smooth steady flows in open channels, such as rivers and channels with weirs.
7.1 Conservation of mass
x
u
xx+ δx
h(x)
zξ(x)
Z(x)
z = 0
w
g
Consider a channel of large width which can be approximated as a two dimensional flow inthe (x, z)-plane and whose base lies on z = Z(x) (topography), with water of depth h(x). Wedefine ξ(x) = Z(x) + h(x), the height of the free surface.
For a steady flow (∂h/∂t = 0), the net mass flux through a volume δV between x and x+ δx(i.e. the mass flows in and out of δV ) must be zero. This leads to
ρ
∫ ξ(x+δx)
Z(x+δx)udz − ρ
∫ ξ(x)
Z(x)u dz = 0⇒ ρ
d
dx
(∫ ξ(x)
Z(x)u dz
)= 0.
So the volume flux
Q =
∫ ξ(x)
Z(x)udz
is constant in space (and time).
61
62 7.2 Bernoulli’s theorem
Let us further assume that u is independent of z, so that
Q = u(x) [ξ(x)− Z(x)] = u(x)h(x) (7.1)
is independent of x.
7.2 Bernoulli’s theorem
We can apply Bernoulli’s theorem to the free surface where the pressure p = patm. Recallthat, for steady flows,
H(x, t) =p
ρ+
1
2‖u‖2 − g · x
is constant along a streamline.
As the free surface is a streamline, H =patmρ
+1
2(u2 + w2) + gξ is independent of x.
Free surface
(streamline)
dξ
x
ξ(x)
dx
dl
u
Let dl =
(1
dξ/dx
)dx be an infinitesimal line element along the free surface. Since u is parallel
to streamlines
u× dl =
(uw
)×(
1dξ/dx
)dx = 0⇒ w = u
dξ
dx.
This is similarly derived from the equation of streamlines dx/ds = u⇔ dξ/w = dx/u.(Alternative derivation: the free surface is defined by the equation z = ξ(x) which is equivalent
to H(x, z) = z − ξ(x) = 0. Hence, ∇H =
(−dξ/dx
1
)must be perpendicular the isosurface
H = 0, i.e. to the free surface. So u · ∇H = 0, leading again to
−udξ
dx+ w = 0⇒ w = u
dξ
dx,
on the free surface.)
Thus, if the free surface is smooth, i.e. if |dξ/dx| 1, then w u can be neglected and,from Bernoulli’s equation, we obtain
1
2u2 + gξ =
1
2u2 + g(h+ Z) = const. (7.2)
Equivalently,
gh
(F 2
2+ 1 +
Z
h
)= const., (7.3)
where the Froude number F =u√gh
has no dimension and determines how flows react to
disturbances. (Recall that Z = 0 when the topography is flat.)
Chapter 7 – Steady flows in open channels 63
7.3 Flow over a hump
Suppose that the fluid flowing along a channel with base at z = 0 encounters a smooth humpof height Z(x). We assume |dZ/dx| 1, so that the flow is approximately unidirectional (theargument for the free surface holds for the base of the channel) and |dξ/dx| 1.
H U h(x)
z = 0 x
Zmax
Z(x)
Let H be the depth of water and U its velocity far upstream. Hence, the upstream Froude
number F =U√gH
is given.
Conservation of mass gives
Q = u(x)h(x) = UH ⇒ u = UH
h,
and Bernoulli’s equation
1
2u2(x) + g(h(x) + Z(x)) =
1
2U2 + gH.
Substituting for u2 = U2H2
h2leads to
1
2U2H
2
h2+ g(h(x) + Z(x)) =
1
2U2 + gH,
which can be rearranged as an equation for Z,
Z =H
2
U2
gH
(1− H2
h2
)+H − h,
⇒ Z
H=F 2
2
(1− H2
h2
)+ 1− h
H,
where F =U√gH
is the upstream Froude number. This gives the relationship between the
height of the base of the channel, Z, and the depth of water, h, when F is fixed.
For simplicity, let Z =Z
Hand h =
h
H. (Z and h are nondimensional measures of the bump
height and depth of water respectively.) Hence,
Z = f(h) where f(h) =F 2
2
(1− 1
h2
)+ 1− h. (7.4)
The function f(h)→ −∞, both as h→ 0 and as h→ +∞. Its derivative is
df
dh=F 2
h3− 1,
⇒ df
dh= 0 at hc = F 2/3 with
d2f
dh2= −3
F 2
h4< 0.
64 7.3 Flow over a hump
So, f has a unique maximum at hc = F 2/3 — or equivalently at hc = HF 2/3; its values isgiven by
Zc = max (f) = f(hc) =F 2
2
(1− F−4/3
)+ 1− F 2/3 =
F 2
2+ 1− 3
2F 2/3,
=1
2
(F 2/3 − 1
)2 (F 2/3 + 2
)> 0. (7.5)
The function f reaches its maximum Zc at hc = F 2/3. (Note that Zc and hc are determinedby the upstream Froude number, i.e. by the upstream properties of the fluid flow.)
h
Zc
hc h2
Z
h1
The equation f(h) = 0 has two solutions, h1 and h2, one of which is h = 1 (i.e. h = H);indeed, far upstream Z/H = f(1) = 0.
Note also that, using equation (7.4), we find that the height of the free surface ξ = Hξ = Z+his given by
ξ = Z + h = 1 +F 2
2
(1− 1
h2
).
Thus ξ > 1 if h > 1 and ξ < 1 if h < 1; the free surface goes up when the depth of water hincreases and goes down when the depth of water decreases.
Let the size of the hump (i.e. its maximum height) be less than the critical height, Zmax < Zc.
• If F > 1, then h1 = 1 (since hc = F 2/3 > 1).
Zc
Z
Zmax
1 hch
Supercritialbranch
Starting from h = 1, the depth of water first increases then decreases with Z and returnsto h = 1 (i.e. h = H). This is called a supercritical flow
Chapter 7 – Steady flows in open channels 65
UH H
U
• If F < 1, then h2 = 1 (since hc = F 2/3 < 1).
Z
Zmax
Zc
1 hhc
branchSubcritical
Here, as Z increases the water surface goes down and when Z returns to zero, the depthof water returns to h = 1 (i.e. h = H). This is called a subcritical flow.
UH H
U
Why these two different behaviours? The equationu2
2+ g(h + Z) = const. expresses
the conservation of energy (kinetic energy, KE, and gravitational potential energy, PE). Inorder to flow over the hump, the fluid has two choices. (i) To increase its KE by reducing itsPE, i.e. decreasing h; (ii) to increase its PE by reducing its KE, i.e. increasing h. The Froudenumber, F = U/
√gH is the ratio of kinetic to potential energy. If F > 1 then KE > PE
upstream, so that it is easier to reduce KE as the fluid flows over the bump; on the contrary,if F < 1 then PE > KE upstream and conversion of PE into KE is preferred.
7.4 Critical flows
In both previous cases (supercritical and subcritical flows), the fluid flow returns to its originalheight and speed after passing over the bump.
However, if the maximal height of the hump Zmax = Zc the flow can move across from onebranch of the solution to the other.
Stable transitions occur only in one direction, from subcritical to supercritical.
UH hc
Zmax = Zc
66 7.4 Critical flows
In this case, a subcritical flow upstream (F < 1) is transformed smoothly into a supercriticalflow downstream (F > 1). The condition for this to happen, Zmax = Zc, leads to h = hc =HF 2/3 at the top of the hump.
From conservation of mass uh = UH ⇒ u = UH/h. So, the local Froude number f satisfies
f2 =u2
gh=U2
gH
(H
h
)3
= F 2
(H
h
)3
.
At the top of the bump, hc/H = F 2/3, so that f = 1. The local Froude number must beequal to 1 at the top of the bump. (Note that the local f is a continuous function of x, lessthan 1 upstream, greater than 1 downstream and equal to 1 at the top of the hump.)
Example 7.1A flow along a uniform channel encounters a bump of height Z(x). Downstream the height offluid h0 is a half of the height upstream. Find the upstream and downstream fluid velocitiesand the height of the bump.
Let U1 and U2 be the upstream and downstream fluid velocities respectively.
2h0U2h0
h(x)U1
Zmax
Z(x)
From conservation of mass,
Q = u(x)h(x) = 2h0U1 = h0U2 ⇒ U2 = 2U1;
and from Bernoulli’s equation,
1
2u2 + g(h+ Z) =
1
2U21 + 2gh0 =
1
2U22 + gh0.
Combining the two equations above leads to
1
2U21 + 2gh0 = 2U2
1 + gh0 ⇒3
2U21 = gh0 ⇒ U1 =
√2
3gh0.
The upstream Froude number F 21 =
U21
2gh0=
1
3< 1 (subcritical flow); so, U2 = 2
√2
3gh0 and
the downstream Froude number F 22 =
U22
gh0=
8
3> 1 (supercritical flow). The flow is critical
as it is smoothly transformed from subcritical upstream to supercritical downstream.
To find the height of the bump, Zmax(= Zc), for a critical flow, use f2 = 1 (local Froudenumber) at the top of the bump.
We get Zmax from Bernoulli’s equation, using
f2 =u2cghc⇒ u2c = ghc.
Chapter 7 – Steady flows in open channels 67
Indeed
1
2U21 + 2gh0 =
1
2u2c + g(hc + Zmax) =
ghc2
+ g(hc + Zmax),
⇒ gZmax =1
2U21 + 2gh0 −
3
2ghc but U2
1 =2
3gh0,
⇒ Zmax =7
3h0 −
3
2hc.
Next, hc can be eliminated using mass conservation, uchc = 2U1h0, together with u2c = ghc
and U21 =
2
3gh0:
u2ch2c = gh3c = 4U2
1h20 =
8
3gh30 ⇒ hc =
(8
3
)1/3
h0 =2
31/3h0 = 2h0F
2/31 .
So, the size of the hump (i.e. its maximal height) is
Zmax = Zc =
(7
3− 32/3
)h0 ' 0.253h0.
7.5 Flow through a constriction
An alternative to varying the height of the base of the channel is to vary its breadth b(x).
UB b(x)
UHh(x)
Side view
Top view
What does happen to the height of water as b varies?
Conservation of mass gives
Q = u(x)b(x)h(x) = UBH = const.,
and Bernoulli’s theorem,
1
2u2(x) + gh(x) =
1
2U2 + gH = const.
So,
u2 = U2 + 2g(H − h), (7.6)
⇒[U2 + 2g(H − h)
]b2h2 = U2B2H2.
68 7.5 Flow through a constriction
Dividing by gH3B2,[U2
gH+ 2
(1− h
H
)]b2
B2
h2
H2=U2
gH⇒ h2
H2
[F 2 + 2
(1− h
H
)]= F 2B
2
b2,
where F =U√gH
is the upstream Froude number.
Again, let us make use of the nondimensional variables h =h
Hand b =
b
B:
h2
F 2
[F 2 + 2
(1− h
)]=
1
b2.
Now, let1
b2− 1 ≡ K(h) where K(h) =
h2
F 2
[F 2 + 2
(1− h
)]− 1. (7.7)
The function K ∼ − 2
F 2h3 → −∞ as h→ +∞ and K(0) = −1. Its derivative
dK
dh= 2
h
F 2
(F 2 + 2− 3h
)= 0
if either h = 0 or h ≡ hc =F 2 + 2
3. Furthermore,
d2K
dh2=
2
F 2
(F 2 + 2− 6h
);
so, K has a local minimum at h = 0 sinced2K
dh2=
2F 2 + 4
F 2> 0 at h = 0; and K has a local
maximum at h = hc sinced2K
dh2= −2F 2 + 4
F 2< 0 at h = hc.
hch1 h2
−1
h
K
b−2c − 1
supercriticalbranch
subcriticalbranch
The equation K(h) = 0 has two solutions, h1 and h2, one of which is h =h
H= 1. Note also
that hc =F 2 + 2
3> 1 if F > 1 and hc < 1 if F < 1.
• If F > 1 (supercritical), then h1 = 1 and h = h/H increases as b = b/B decreases (i.e.K increases). The height of the free surface rises through the constriction.
• If F < 1 (subcritical), then h2 = 1 and h = h/H decreases as b = b/B decreases.
Chapter 7 – Steady flows in open channels 69
Once again, a smooth transition from a subcritical flow to a supercritical flow can occur ifthe narrowest point in the constriction reaches a critical breadth, bc, defined such that
1
b2c− 1 = max
(1
b2− 1
)= K(hc) =
h2cF 2
[F 2 + 2
(1− hc
)]− 1,
where hc =F 2 + 2
3. In this case the local Froude number f2 =
u2
ghcsatisfies
f2 =U2 + 2g(H − hc)
ghc=U2 + 2g(H − hc)
gH
H
hc,
using equation (7.6). So,
f2 =
(U2
gH+ 2− 2
hcH
)H
hc=
(F 2 + 2− 2
hcH
)H
hc,
=
(F 2 + 2− 2
F 2 + 2
3
)1
F 2/3 + 2/3=F 2/3 + 2/3
F 2/3 + 2/3= 1.
Thus, the local Froude number f = 1 at the narrowest point in the constriction for a criticalflow.
7.6 Transition caused by a sluice gate
Another way to generate a transition between a subcritical and a supercritical flow is via asluice gate.
U1
U2H2
H1
Using conservation of mass and Bernoulli’s equation for the free surface we obtain
U1H1 = U2H2 and1
2U21 + gH1 =
1
2U22 + gH2,
dividing by gH1 ⇒1
2
(F 21 + 2
)=
1
2
(F 22 + 2
) H2
H1.
Then, sinceH2
2
H21
=U21
U22
=F 21H1
F 22H2
⇒ H32
H31
=F 21
F 22
,
one has (F 21 + 2
)3=(F 22 + 2
)3 F 21
F 22
⇔(F 21 + 2
)3F 21
=
(F 22 + 2
)3F 22
.
Thus, G(F1) = G(F2) where G(f) =
(f2 + 2
)3f2
. The function G(f) → +∞ as f → 0 and as
f →∞. Its derivative
dG
df=
6f2(f2 + 2
)2 − 2(f2 + 2
)3f3
= 0⇒(f2 + 2
)2 (6f2 − 2f2 − 4
)= 0⇒ f2 = 1.
70 7.6 Transition caused by a sluice gate
G
C
ff < 1 1 f > 1
The equation G(f) =
(f2 + 2
)3f2
= C (where C > 27) has two solutions, one corresponding to
a subcritical flow and one corresponding to a supercritical flow.
Chapter 8
Lift forces
Contents
8.1 Two-dimensional thin aerofoils . . . . . . . . . . . . . . . . . . . . . 71
8.2 Kutta-Joukowski theorem . . . . . . . . . . . . . . . . . . . . . . . . 72
8.3 Lift produced by a spinning cylinder . . . . . . . . . . . . . . . . . . 72
8.4 Origin of circulation around a wing . . . . . . . . . . . . . . . . . . 73
8.5 Three-dimensional aerofoils . . . . . . . . . . . . . . . . . . . . . . . 73
How do aeroplanes fly? An Airbus A380 weights 560 tonnes (5.6× 105 kg) at take-off and sorequires a lift force in excess of 5.6 × 106 N. The lift force is provided by the wings (span≈ 80 m, area ≈ 845 m2) and is generated by aerodynamic forces.
In this section we shall give a brief discussion of the lift forces on aerofoils.
8.1 Two-dimensional thin aerofoils
Consider a 2-D flow past a thin aerofoil and assume that the flow does not separate and canbe modelled as a potential flow (i.e. aerofoil is smooth).
We can use Bernoulli’s theorem to calculate the pressure — recall that there is no drag forceacting on a solid body placed in a potential flow.
x
y
U pT
pB
For a flat aerofoil, the force in the upward vertical direction will be the difference betweenpressure forces on the bottom and on the top of the aerofoil. This force per unit length is
F =
∫(pB − pT ) dx.
Using Bernoulli’s theorem,pBρ
+1
2u2B =
pTρ
+1
2u2T ,
F =ρ
2
∫(u2T − u2B) dx =
ρ
2
∫(uT + uB)(uT − uB) dx.
71
72 8.2 Kutta-Joukowski theorem
For a thin aerofoil, both uT and uB will be close to U (the free stream velocity), so that
uT + uB ' 2U ⇒ F ' ρU∫
(uT − uB) dx = −ρU∮C
u · dl,
where C is the curve around the aerofoil.
CThus, the force acting on the aerofoil,
F = −ρUΓ where Γ =
∮C
u · dl,
is proportional to the circulation around the wing.
8.2 Kutta-Joukowski theorem
The above result is an example of a general exact general result of inviscid irrotational flowtheory.
Theorem 8.1 (Kutta-Joukowski)Any 2-D body in relative motion to the ambient fluid with velocity U has a lift force, per-pendicular to U, of magnitude
F = −ρUΓ where Γ =
∮C
u · dl. (8.1)
L
α
For a flow around a flat plate, Γ = −πUL sinα(using conformal mapping).
Nose up (α) leads to more lift and nose down(α) to less lift.
8.3 Lift produced by a spinning cylinder
φ = U
(r +
a2
r
)cos θ +
Γ
2πθ,
ur = U
(1− a2
r2
)cos θ,
uθ =Γ
2πr− U
(1 +
a2
r2
)sin θ.
FL
Low velocityHigh pressure
Low pressureHigh velocity
Flow direction Top spin
On r = a, ur = 0 and uθ =Γ
2πa− 2U sin θ; the circulation is Γ =
∫ 2π
0a uθ|r=a dθ.
So,
u2θ∣∣r=a
=
(Γ
2πa− 2U sin θ
)2
=
(Γ
2πa
)2
− 2ΓU
πasin θ + 4U2 sin2 θ.
The pressure at the cylinder surface can now be calculated using Bernoulli’s theorem,
p = p∞ +1
2ρU2 − ρΓ2
8π2a2+ρΓU
πasin θ − 2ρU2 sin2 θ.
Chapter 8 – Lift forces 73
The pressure force per unit length
F =
∮−p n dl =
∫ 2π
0−p n a dθ, where n = er = cos θ ex + sin θ ey.
The force can be decomposed into its components parallel and perpendicular to the free streamvelocity (in the x direction): F = F‖ ex+F⊥ ey, with F‖ = 0 (no drag force) and the lift forceF⊥ = −ρΓU (Kutta-Joukowski theorem). This is called the “Magnus effect” (e.g. football,tennis, table tennis).
8.4 Origin of circulation around a wing
When the plane is stationary on the runway, there is no circulation around the wings. In§ 5.4, we showed that vorticity cannot be created in an initially vorticity free fluid, in theabsence of viscosity. Thus the flow should remain vortex free. (Recall that the circulation isequal to the flux of vorticity.)
A potential flow past an inclined wing is of the form:
However, small viscous effects allow the aerofoil to shed a vortex off the trailing edge, so thatdownstream separation occurs at the trailing edge.
This vortex, called the starting vortex, remains behind on the runway. Its circulation is equaland opposite to the circulation around the wing.
Note: the greater the angle of inclination(or attack angle), the greater the circula-tion and hence the lift (e.g. flat plate Γ =−2πUL sinα). This is true up to a point: ifthe angle is too steep, the flow separates so thedrag force on the aeroplane increases signifi-cantly and it partially looses its lift force. Thisis called a stall.
8.5 Three-dimensional aerofoils
No wings is infinitely long (i.e. 2-D). Special care needs to be taken with wings tips.
74 8.5 Three-dimensional aerofoils
pT
pBSince pB > pT , there is a pressure gradient driving a flow around the edge of the wing. Thisleads to a vortex from the edge of the wing.
Trailing vortex
Trailing vortex
Starting vortex
These trailing vortices are parts of a single vortex tube formed by the wings, the trailingvortices and the starting vortex. (Vortex tubes must be closed as they cannot start or end inan inviscid fluid.)
Appendix A
Vector calculus
We shall only consider the case of three-dimensional spaces.
A.1 Definitions
A physical quantity is a scalar when it is only determined by its magnitude and a vectorwhen it is determined by its magnitude and direction. It is crucial to distinguish vectors fromscalars; standard notations for vectors include ~u ≡ u ≡ u. A unit vector, commonly denotedby ı ≡ ı, has magnitude one. The coordinates of a vector a are the scalars a1, a2 and a3 suchthat
a = a1e1 + a2e2 + a3e3 ≡ (a1, a2, a3) ≡
a1a2a3
,
in the basis e1, e2, e3. If e1, e2, e3 are orthonormal (i.e. ei · ej = δij) the magnitude of
the vector is |a| =√a21 + a22 + a23.
Scalar or dot product:
a · b = |a||b| cos θ = a1b1 + a2b2 + a3b3
is a scalar.
Vector or cross product:
a× b = (a2b3 − a3b2)e1 + (a3b1 − a1b3)e2 + (a1b2 − a2b1)e3.
is a vector with magnitude |a||b| sin θ and a direction perpendicular to both vectors a and bin a right-handed sens.
Triple scalar product:
[a,b, c] = a · b× c = a× b · c = b · c× a
is a scalar.
Triple vector product:
a× (b× c) = (a · c)b− (a · b)c
is a vector.
75
76 A.2 Suffix notation
A.2 Suffix notation
It is often very convenient to write vector equations using the suffix notation. Any suffix mayappear once or twice in any term in an equation; a suffix that appears just once is called afree suffix and a suffix that appears twice is called a dummy suffix.
Summation convention.Dummy suffices are summed over from 1 to 3 whilst free suffices take the values 1, 2 and3. Hence, free suffices must be the same on both sides of an equation whereas the names ofdummy suffices are not important (e.g. aibick = ajbjck).
Tensors used in suffix notation.Kronecker Delta:
δij =
1 if i = j,
0 if i 6= j,⇔(δij)
=
1 0 00 1 00 0 1
.
The Kronecker Delta is symmetric, δij = δji, and δijaj = ai.
Alternating Tensor:
εijk =
0 if any of i, j or k are equal;
1 if (i, j, k) = (1, 2, 3), (2, 3, 1) or (3, 1, 2);
−1 if (i, j, k) = (1, 3, 2), (3, 2, 1) or (2, 1, 3).
The Alternating Tensor is antisymmetric, εijk = −εjik, and it is invariant under cyclic per-mutations of the indices, εijk = εjki = εkij .
The tensors δij and εijk are related to each other by εijk εklm = δilδjm − δimδjl.Examples.Standard algebraic operations on vectors can be written in a compact form using the suffixnotation.A scalar product can be written as a · b = a1b1 + a2b2 + a3b3 = ajbj and the ith componentof a vector product as (a× b)i = εijk ajbk.
A.3 Vector differentiation
A.3.1 Differential operators in Cartesian coordinates
We consider scalar and vector fields, f(x) and F(x) = (F1(x), F2(x), F3(x)) respectively, wherex = (x1, x2, x3) ≡ (x, y, z) are Cartesian coordinates in the orthonormal basis e1, e2, e3 ≡ı, , k. We also define the vector differential operator
∇ ≡(
∂
∂x1,∂
∂x2,∂
∂x3
).
(∇ is pronounced grad, nabla or del.)
• The gradient of a scalar field f(x1, x2, x3) is given by the vector field
grad f ≡ ∇f =
(∂f
∂x1,∂f
∂x2,∂f
∂x3
)=
∂f
∂x1e1 +
∂f
∂x2e2 +
∂f
∂x3e3.
∇f is the vector field with a direction perpendicular to the isosurfaces of f with amagnitude equal to the rate of change of f in that direction.
Chapter A – Vector calculus 77
• The divergence of a vector field F is given by the scalar field
div F = ∇ · F =∂F1
∂x1+∂F2
∂x2+∂F3
∂x3.
A vector field F is solenoidal if ∇ · F = 0 everywhere.
• The curl of a vector field F is given by the vector field
curl F = ∇× F =
(∂F3
∂x2− ∂F2
∂x3,∂F1
∂x3− ∂F3
∂x1,∂F2
∂x1− ∂F1
∂x2
),
=
(∂F3
∂x2− ∂F2
∂x3
)e1 +
(∂F1
∂x3− ∂F3
∂x1
)e2 +
(∂F2
∂x1− ∂F1
∂x2
)e3,
=
∣∣∣∣∣∣∣∣e1 e2 e3∂
∂x1
∂
∂x2
∂
∂x3F1 F2 F3
∣∣∣∣∣∣∣∣ .A vector field F is irrotational if ∇× F = 0 everywhere.
• The directional derivative (F·∇) is a differential operator which calculates the derivativeof scalar or vector fields in the direction of F. (It is not to be confused with the scalar∇ · F.)
The directional derivative of a scalar field is given by the scalar field
(F · ∇) f = F · ∇ f = F1∂f
∂x1+ F2
∂f
∂x2+ F3
∂f
∂x3.
If n is a unit vector, (n · ∇)f gives the rate of change of f in the direction of n.
The directional derivative of a vector field G is given by the vector field
(F · ∇)G = ((F · ∇)G1, (F · ∇)G2, (F · ∇)G3)
=
(F1∂G1
∂x1+ F2
∂G1
∂x2+ F3
∂G1
∂x3,
F1∂G2
∂x1+ F2
∂G2
∂x2+ F3
∂G2
∂x3, F1
∂G3
∂x1+ F2
∂G3
∂x2+ F3
∂G3
∂x3
).
• The Laplacian ∇2 =∂2
∂x21+
∂2
∂x22+
∂2
∂x23is a differential operator which can act on scalar
or vector fields:
∇2f =∂2f
∂x21+∂2f
∂x22+∂2f
∂x23
is a scalar field and
∇2F = (∇2F1,∇2F2,∇2F3)
is a vector field.
• The Lagrangian derivative of scalar and vector fields are
D
Dtf(x, t) =
d
dtf(x(t), t) =
∂
∂tf(x, t) + (u · ∇)f(x, t)
78 A.3 Vector differentiation
and
D
DtF(x, t) =
d
dtF(x(t), t) =
∂
∂tF(x, t) + (u · ∇)F(x, t),
respectively, with u(x, t) = dx/dt and where ∇ acts upon the variables (x1, x2, x3) only.
Notice that if F = F(t) thendF
dt=
(dF1
dt,dF2
dt,dF3
dt
).
Differential operators in Cartesian coordinates using suffix notation
(gradf)i = (∇f)i =∂f
∂xi,
div F = ∇ · F =∂Fj∂xj
,
(curl F)i = (∇× F)i = εijk∂Fk∂xj
,
(F · ∇)f = Fj∂f
∂xj.
Notice that the operator ∂/∂xj cannot be moved around as it acts on everything that followsit.
A.3.2 Differential operators polar coordinates
The differential operators defined above in Cartesian coordinates take different forms for differ-ent systems of coordinates, e.g. in cylindrical polar coordinates or spherical polar coordinates.
Cylindrical polar coordinates
Let p and u = urer+uθeθ+uzez be scalar and vector fields respectively, functions of (r, θ, z).
∇p =∂p
∂rer +
1
r
∂p
∂θeθ +
∂p
∂zez,
∇·u =1
r
∂
∂r(rur) +
1
r
∂uθ∂θ
+∂uz∂z
,
∇× u =
(1
r
∂uz∂θ− ∂uθ
∂z
)er +
(∂ur∂z− ∂uz
∂r
)eθ +
1
r
(∂
∂r(ruθ)−
∂ur∂θ
)ez,
(u · ∇)p = ur∂p
∂r+uθr
∂p
∂θ+ uz
∂p
∂z,
∇2p =1
r
∂
∂r
(r∂p
∂r
)+
1
r2∂2p
∂θ2+∂2p
∂z2.
Chapter A – Vector calculus 79
Spherical polar coordinates
Let p and u = urer+uθeθ+uϕeϕ be scalar and vector fields respectively, functions of (r, θ, ϕ).
∇p =∂p
∂rer +
1
r
∂p
∂θeθ +
1
r sin θ
∂p
∂ϕeϕ,
∇·u =1
r2∂
∂r
(r2ur
)+
1
r sin θ
∂
∂θ(uθ sin θ) +
1
r sin θ
∂uϕ∂ϕ
,
∇× u =1
r sin θ
(∂
∂θ(uϕ sin θ)− ∂uθ
∂ϕ
)er +
1
r
(1
sin θ
∂ur∂ϕ− ∂
∂r(ruϕ)
)eθ
+1
r
(∂
∂r(ruθ)−
∂ur∂θ
)eϕ,
(u · ∇)p = ur∂p
∂r+uθr
∂p
∂θ+
uϕr sin θ
∂p
∂ϕ,
∇2p =1
r2∂
∂r
(r2∂p
∂r
)+
1
r2 sin θ
∂
∂θ
(sin θ
∂p
∂θ
)+
1
r2 sin θ
∂2p
∂ϕ2.
A.3.3 Vector differential identities
The simple vector identities that follow can all be proved with suffix notation. Note thatthese vector identities are true for all systems of coordinates.
Let F and G be vector fields and ϕ and ψ be scalar fields.
∇ · (∇ϕ) = ∇2ϕ,
∇ · (∇× F) = 0,
∇× (∇ϕ) = 0,
∇(ϕψ) = ϕ∇ψ + ψ∇ϕ,∇ · (ϕF) = ϕ∇ · F + F · ∇ϕ,∇× (ϕF) = ϕ∇× F +∇ϕ× F,
∇× (∇× F) = ∇(∇ · F)−∇2F,
∇(F ·G) = F× (∇×G) + G× (∇× F) + (F · ∇)G + (G · ∇)F,
∇ · (F×G) = G · (∇× F)− F · (∇×G),
∇× (F×G) = F(∇ ·G)−G(∇ · F) + (G · ∇)F− (F · ∇)G,
∇2F = ∇ (∇·F)−∇× (∇×F) ,
F× (∇×F) =1
2∇ (F · F)− (F·∇) F.
A.4 Vector integral theorems
A.4.1 Alternative definitions of divergence and curl
• An alternative definition of divergence is given by
∇ · F = limδV→0
1
δV©∫∫δS
F · n dS,
where δV is a small volume bounded by a surface δS which has a normal n, pointingoutwards.
80 A.4 Vector integral theorems
• An alternative definition of curl is given by
n · ∇ × F = limδS→0
1
δS
∮δC
F · dx,
where δS is a small open surface bounded by a curve δC which is oriented in a right-handed sense.
A.4.2 Physical interpretation of divergence and curl
• The divergence of a vector field gives a measure of how much expansion and contractionthere is in the field.
• The curl of a vector field gives a measure of how much rotation or twist there is in thefield.
A.4.3 The Divergence and Stokes’ Theorems
• The divergence theorem states that∫∫∫V
∇ · F =©∫∫S
F · n dS,
where S is the closed surface enclosing the volume V and n is the outward-pointingnormal from the surface.
• Stokes’ theorem states that ∫∫S
(∇× F) · n dS =
∮C
F · dx,
where C is the closed curve enclosing the open surface S and n is the normal from thesurface.
A.4.4 Conservative vector fields, line integrals and exact differentials
• The following five statements are equivalent in a simply-connected domain:
i. ∇× F = 0 at each point in the domain.
ii. F = ∇φ for some scalar φ which is single-valued in the region.
iii. F · dx is an exact differential.
iv.
∫ Q
PF · dx is independent of the path of integration from P to Q.
v.
∮C
F · dr = 0 around every closed curve in the region.
• If ∇ · F = 0 then F = ∇×A for some A. (This vector potential A is not unique.)
Appendix B
Ordinary differential equations
B.1 First order equations
Separable equations
If an equation is of the form
dy
dx= f(y)g(x),
we can separate variables to find ∫1
f(y)dy =
∫g(x)dx,
for which each side can now be integrated independently .
Linear equations
An equation of the form
a(x)dy
dx+ b(x)y = f(x)
can be integrated using an integrating factor. First put the equation into standard form bydividing through by a(x),
dy
dx+b(x)
a(x)y =
f(x)
a(x).
The integrating factor is then
p(x) = exp
∫b(x)
a(x)dx
which can be calculated. Multiply through by the integrating factor and if everything hasbeen done correctly the equation can now be written
d
dx(p(x) y) =
p(x)g(x)
a(x)
which can be integrated up with respect to x.
81
82 B.2 Second order equations
B.2 Second order equations
Equations with constant coefficients
Equations of the form
ad2y
dx2+ b
dy
dx+ cy = f(x),
where a, b and c are constants can be solved using the complimentary function and particularintegral method.
First, consider the homogeneous equation by setting the RHS to zero:
ad2y
dx2+ b
dy
dx+ cy = 0.
Seeking solutions of the form y1 = eλx leads to the quadratic auxillary equation
aλ2 + bλ+ c = 0.
So, y1 = Aeλ1x+Beλ2x where λ1 and λ2 are the roots of the quadratic and A and B constants.The solution becomes y1 = (A + Bx)eλx if λ is a double root. (Notice that solutions can bewritten in terms of sine and cosine functions when roots have complex values.)
Next, find a particular integral — a special case y2 that gives the correct RHS — by tryingfunctions y2 that look like the desired RHS.
The solution sought is the sum of the general solution to the homogeneous equation with theparticular integral, y = y1 + y2.
Cauchy equations
Equations of the general form
ax2d2y
dx2+ bx
dy
dx+ cy = 0.
can be solved by seeking solutions of the form y = xλ, giving, as above, an algebraic auxillaryequation
aλ2 + (b− a)λ+ c = 0.
B.3 System of coupled equations
In calculating the path of a fluid particle we have to solve a set of differential equations of theform,
dx
dt= f(x, y, z, t),
dy
dt= g(x, y, z, t),
dz
dt= h(x, y, z, t).
The method of solution depends upon how the equations are coupled.
Chapter B – Ordinary differential equations 83
• Example 1:dx
dt= x+ y,
dy
dt= −y.
Here the equation for dx/dt depends upon knowing y(t), however since dy/dt does notcontain x(t) we can be compute y(t) first. This has the general solution, y(t) = y0e
−t.We can now substitute this expression for y(t) to give
dx
dt= x+ y0e
−t.
This is a linear equation so it can be solved using an integrating factor p(t) = e−t togive
d
dt
(xe−t
)= y0e
−2t,
so thatx(t) = −y0
2e−t + Cet.
• Example 2dx
dt= 2x+ y,
dy
dt= −x.
Here, dx/dt depends on y(t) and dy/dt depends on x(t) therefore, we can’t solve eitherequation directly as it depends on the solution to other equation, which we don’t know.
Instead, we can eliminate y(t) to form a second-order equation for x(t). Differentiatingdx/dt = 2x+ y with respect to t on both sides shows that
d2x
dt2= 2
dx
dt+
dy
dt.
Then, substituting dy/dt by its expression results in
d2x
dt2= 2
dx
dt− x.
This is a constant coefficient second-order differential equation and can be solved via anauxillary equation, to give
x(t) = (At+B)et.
Once x(t) has been found, this can be plugged into the equation for y(t) which can thenbe solved to find
y(t) =dx
dt− 2x = (A+At+B)et − 2(At+B)et = (A−B −At)et.
