MATH2620: Fluid Dynamics 1 School of Mathematics, University of Leeds

Lecturer: Dr Evy Kersal´e Office: Room 9.18, School of Mathematics Phone: 0113 343 5149 E-mail: [email protected] Website: http://www.maths.leeds.ac.uk/ kersale/2620 ∼ Discussion 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 Kersal´e([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 implications in 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 a formal mathematical description of fluid flows (e.g. in terms of ODEs) and the derivation of their governing equations (PDEs), using elementary techniques from calculus and vector calculus. This theoretical background is then applied to a series of simple flows (e.g. bath-plug vortex or stream past a sphere), giving students a feel for how fluids behave, and experience in modelling everyday phenomena. A wide range of courses, addressing more advanced concepts in fluid dynamics, with a variety of 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 physical phenomena can be understood using rigorous, yet relatively simple, mathematics. But, it also provides students with a general framework to devise models of real-world problems, using relevant theories. Students will learn how to use methods of applied mathematics to derive approximate 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... otherwise • you 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 minutes after lectures in order to answer questions.)

If you feel that the module is too difficult, or that you are spending too much time on • it, please come and talk to me.

Please, do not wait until the end of term to give a feedback if you are unhappy with • some aspects of the module.

Lecture notes:

Detailed lecture notes can be downloaded from the module’s website. You can print and • use them in the lecture if you wish; however, the notes provided should only be used as a 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 and • model 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 give • you practice on the types of questions that will be set in the examination. It is very important that you try them before the example classes.

There will be only two, yet quite demanding, pieces of coursework (mid and end of term • deadlines). 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 fluids3 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...... 79 A.4 Vector integral theorems...... 79 A.4.1 Alternative definitions of divergence and curl...... 79 A.4.2 Physical interpretation of divergence and curl...... 80 A.4.3 The Divergence and Stokes’ Theorems...... 80 A.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 we shall represent vectors and vector fields using bold fonts, e.g. A and u(x). Other commonly used notations for vectors include A~ or A (used in the lecture). A vector A, of components A , A and A in the basis ˆe1, ˆe2, ˆe3 , will interchangeably be written as a column or row 1 2 3 { } vector,   A1 A = A2 = (A1,A2,A3) = A1 ˆe1 + A2 ˆe2 + A3 ˆe3. A2

Three main systems of coordinates will be used throughout to represent points in a three dimensional space, Cartesian coordinates (x, y, z) with x, y and z in R; cylindrical polar coordinates (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 to ∈ ∞ ∈ ∈ plane polar coordinates (r, θ) in two dimensions. The vector position r x of a point in a ≡ three dimensional space will be written as

x = xˆex + yˆey + zˆex in Cartesian coordinates,

= rˆer + zˆez in cylindrical coordinates,

= rˆer in spherical coordinates, using the orthonormal basis ˆex, ˆey, ˆez , ˆer, ˆeθ, ˆez and ˆer, ˆeθ, ˆeϕ respectively. Notice p { } { } { } that x = x2 + y2 + z2 in Cartesian coordinates, x = √r2 + z2 in cylindrical coordinates k k k k and x = r in spherical coordinates. k k The variable t will represent time; for time derivatives (rates of change) we shall use the notation 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 ˆex + v ˆey + w ˆex in Cartesian coordinates,

= ur ˆer + uθ ˆeθ + uz ˆez in cylindrical coordinates,

= ur ˆer + uθ ˆeθ + uϕ ˆeϕ in spherical coordinates.

For the sake of simplicity, we shall write integrals of multivariable functions as single integrals, e.g. we shall use ZZZ Z ZZ Z f(x) dV f(x) dV or u(x) n dS u(x) n dS, V ≡ V S · ≡ S · for the volume integral of f in V and for the flux of u through the surface S with normal vector 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 the volume flux; g for the vector gravity and g for its magnitude; p for pressure and patm for atmospheric pressure; for the Bernoulli function. H 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 deformed to 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 the same physical laws (e.g. Newton’s law of motion), which can be mathematically formalised in terms of differential equations for scalar and vector fields.

1.1.1 Continuum hypothesis (Ref.: Paterson, 23 §III.1) One cubic centimetre of water contains of the order of 10 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 velocity ' of all molecules in a test volume δV (fluid particle) of length d (mesoscopic scale, i.e. between macroscopic and atomic/molecular scales).

3 4 1.1 Introduction

a d δV

If d l (molecular scale) then δV contains many molecules and the fluctuations due •  m to individual motions are averaged out. If d a (macroscopic scale) then δV is approximately a point in space. •  u¯

d lm a Hence, if l d a, the average velocityu ¯ is a smooth function of position, independent m   of d.

Continuum hypothesis. Molecular details can be smoothed out by assigning the velocity at a point P to be the average velocity in a fluid element δV centred in P . Thus, we can define the velocity field u(x, t) as a smooth function (shock waves break this assumption), differentiable and integrable. mass in δV Similarly, ρ(x, t) = is the local density of mass. δV

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 1.1 Shear flow: flow between two parallel plates when one is moved relative to the other, with a constant velocity U.

y = d U Uy  y  d    Uy u =   = ˆex.  0  d   x 0 y =0

The direction of the flow is indicated by an arrow and its magnitude by the arrow length. Chapter 1 – Mathematical modelling of fluids 5

Stagnation-point flow: flow with a point at which u = 0.

Consider the extensional flow  Ex  u =  Ey = Ex ˆex Ey ˆey, − − 0 where E is a constant. The point x = 0 where u = 0 is a stagnation point.

Vortex flow: flow in rotation about a central point.

 y  2 2  x + y   x  y x 2 2 2 u =   = ˆex ˆey, where r = x + y .  2 2  r2 − r2 −x + y  0

This flow is singular at x = 0: u 1/r as k k ∼ → ∞ r 0. →

1.2 Kinematics (Ref.: Paterson, §III.2) In the first four chapters of the course we shall concentrate solely on kinematic properties of fluid 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 particle be released at time t0 and at position (x0, y0, z0) into the velocity field

u(x, y, z, t) u(x, 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. Hence the particle position, x(t), satisfies the system of first order ODEs and Initial Conditions dx = u(x, t) with x = x at t = t , (1.1) dt 0 0 or, equivalently, in Cartesian components form 6 1.2 Kinematics

dx dy dz = u(x, y, z, t), = v(x, y, z, t), = w(x, y, z, t), (1.2) dt dt dt 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 variable time t.

u(x(t1), t1)

u(x0, t0) u(x(t2), t2)

Example 1.2 (Stagnation point flow) Consider the extensional flow u(x) = (Ex, Ey, 0) with constant E. From equation (1.2) one − gets dx dy dz = Ex, = Ey, = 0, dt dt − dt Et Et x(t) = x e , y(t) = y e− , z(t) = z if x = x at t = 0. ⇒ 0 0 0 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 x y y = 0 0 . x

1.2.2 Streamlines A streamline is a line everywhere tangent to the local fluid velocity, at some instant. At fixed time t, the equation of the streamline x(s, t) = (x(s, t), y(s, t), z(s, t)), parametrised by a parameter s (“distance” along the streamline, say), is dx dx dy dz = u(x, t) = u, = v, = w, (1.3) ds ⇔ ds ds ds in Cartesian components form or, equivalently, dx dy dz = = (= ds), (1.4) u v w since equations (1.3) are separable as u(x, t) is not explicitly function of s. (Notice that dx/ds is the vector tangent to the line x(s).)

Example 1.3 (Stagnation point flow)  x  Calculate the streamline passing through (x0, y0, 0) in the stagnation point flow u =  y. − 0

From equation (1.4), in 2D,

dx dy = ln x = ln y + C xy = A = x y , x y ⇒ | | − | | ⇒ 0 0 − where C is an integration constant and A = eC . So, | |x y streamlines are hyperbolae of equation y = 0 0 . x Chapter 1 – Mathematical modelling of fluids 7

  Example 1.4 U0 Consider the time dependent flow u(t) = kt, where U0 and k are constants. 0

Particle paths: • dx dy dz = U , = kt, = 0, dt 0 dt dt k x(t) = U t + x , y(t) = t2 + y , z(t) = z , if x = x at t = t . ⇒ 0 0 2 0 0 0 0 y

Eliminating t, the equation of the particle path through x0 becomes t > t0  2 x x0 2 (x(t),y(t)) − = (y y0) (parabola). U0 k − t0

(x0,y0) x Streamlines: the streamline through x is defined by • 0  x x dx dy  x = U s + x s = − 0 , kt = U , = kt 0 0 ⇒ U y = (x x ) + y . ds 0 ds ⇒ 0 ⇒ U − 0 0  y = kts + y0, 0

At t = 0, streamlines are horizontal lines y = y0; at later time t > 0, they are straight lines 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 path tracing), release a continuous stream of dye at that point. The dye will move around and define a curve given by x(t , t) , the position of fluid elements { 0 } that passed through x0 at some time t0, prior to the current time t.

A streakline is the line x(t0, t) for which ∂x = u(x, t) with x = x at t = t . (1.5) ∂t 0 0

At t fixed, streaklines are lines parametrised by t0. The point where the parameter t0 = t corresponds 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, particle paths, streamlines and streaklines are identical. This is not true for unsteady flows in general.

Example 1.5 Steady flow: •  x  ∂x ∂y ∂z u(x) =  y = x, = y, = 0, − ⇒ ∂t ∂t − ∂t 0 t t x(t , t) = A(t ) e , y(t , t) = B(t ) e− , z(t , t) = C(t ). ⇒ 0 0 0 0 0 0

At t = t0, x = x0 = (x0, y0, z0), so

t0 t0 x0 = Ae , y0 = Be− , z0 = C, t0 t0 A = x e− ,B = y e ,C = z . ⇒ 0 0 0 Hence, (t t0) (t t0) x(t0, t) = x0e − , y(t0, t) = y0e− − , z(t0, t) = z0.

Again, we can eliminate the parameter t0 to obtain the equation xy = x0y0 as for particle paths and streamlines.

Unsteady flow: •   U0 u(t) = kt with U0 and k constants, Particle path (particle 0 released at t=0) x(t , t) = x + U (t t ), ⇒ 0 0 0 − 0 k 2 2 and y(t0, t) = y0 + (t t ). (x ,y ) 2 − 0 0 0 Streamline at t=0 k At t = 0, x(t , 0) = x U t and y(t , 0) = y t2. 0 0− 0 0 0 0− 2 0 Eliminating t0, one obtains the equation of a parabola with negative curvature Streakline at t=0

k 2 y y0 = 2 (x x0) . − −2U0 −

1.2.4 Time derivatives Let f(x, t) be some quantity of interest (e.g. density, temperature, one component of the velocity, etc.).

Eulerian description of fluids. (At a fixed position in space.) ∂f  ∂ The partial derivative f(x, t) is the rate of change of f at fixed position x. ∂t x ≡ ∂t ∂ρ ∂u E.g., (scalar) is the rate of change of mass density; (vector) is not the acceleration of ∂t ∂t a fluid particle. Chapter 1 – Mathematical modelling of fluids 9

Lagrangian description of fluids. (Following fluid particles.) D The convective derivative (also Lagrangian derivative, or material derivative) f(x, t) is the Dt rate of change of f when x is the position of a fluid particle (i.e. x travels with the fluid along particle paths). Df So, = 0 implies that f remains constant along particle paths. Dt Example 1.6  x  For the flow u(x) =  y  a particle path is given by 2z − t t 2t x(t) = x0 e , y(t) = y0 e and z(t) = z0 e− , if x = x0 at t = 0.    t  x x0 e t So, along the particle path, upp =  y  =  y0 e  and 2z 2z e 2t − − 0 −  t    x0 e x Du d t ∂u upp =  y0 e  =  y  = 0 whereas = 0. Dt ≡ dt 2t 6 ∂t 4z0 e− 4z

Relation between D/Dt and ∂/∂t. There is no need to go through particle paths calculations to evaluate D/Dt. Consider a dx particle path x(t) = (x(t), y(t), z(t)) defined by = u. dt Using the chain rule, Df d ∂f ∂f dx ∂f dy ∂f dz f(x(t), t) = + + + , Dt ≡ dt ∂t ∂x dt ∂y dt ∂z dt Df ∂f ∂f ∂f ∂f ∂f = + u + v + w = + (u ) f. (1.6) ⇒ Dt ∂t ∂x ∂y ∂z ∂t · ∇ Dρ ∂ρ Hence, = + (u ) ρ is the Lagrangian derivative of the fluid density and the acceler- Dt ∂t · ∇ ation of a fluid particle is Du ∂u = + (u ) u. (1.7) Dt ∂t · ∇

Example 1.7  x  Again, consider the steady flow u =  y . 2z ∂u − Since = 0 (the velocity field u does not depend on time t for a steady flow), ∂t  ∂u ∂u ∂u  u + v + w  ∂x ∂y ∂z    Du ∂u ∂u ∂u  ∂v ∂v ∂v  = (u ) u = u + v + w =  u + v + w  , Dt · ∇ ∂x ∂y ∂z  ∂x ∂y ∂z   ∂w ∂w ∂w  u + v + w ∂x ∂y ∂z 1 0  0   x  ∂u ∂u ∂u = x + y 2z = x 0 + y 1 2z  0  =  y  , ∂x ∂y − ∂z − 0 0 2 4z − 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 f following a fluid particle becomes Df = (u ) f. Dt · ∇ Furthermore, since particle paths and streamlines are identical for time-independent flows, Df df = u (ˆes ) f = u Dt k k · ∇ k k ds where s denotes the distance along the streamlines x(s) and ˆes is the unit vector parallel to the streamlines, in the direction of the flow. So, in steady flows, (u ) f = 0 implies that f is constant along streamlines — the constant · ∇ can 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. n u The total mass in V is S ds Z MV = ρ dV, V V where ρ is the density of mass (mass per unit volume).

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 Z dM ρu n dS = V . − S · dt

So, Z d Z Z ∂ρ ρu n dS = ρ dV = dV, since V is fixed. − S · dt V V ∂t Applying the divergence theorem, Z ∂ρ Z Z ∂ρ  dV = (ρu) dV + (ρu) dV = 0. V ∂t − V ∇· ⇔ V ∂t ∇·

11 12 2.2 Incompressible fluids

Since V is arbitrary, this equation must hold for all volume V . Thus, the continuity equation ∂ρ + (ρu) = 0 (2.1) ∂t ∇· holds at all points in the fluid. Expand the divergence as (ρu) = ρ u + u ρ to derive ∇· ∇· ·∇ the Lagrangian form of the continuity equation Dρ + ρ u = 0. (2.2) Dt ∇· 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) of ∇· ∇· the flow. u < 0 u > 0 ∇· ∇·

2.2 Incompressible fluids

In an incompressible fluid, the density of each fluid particle (i.e. fluid element following the motion 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 flow ∇ · ∇ × defined 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 incompressible and therefore u = 0.) This is a good approximation in many circumstances, e.g. water ∇ · and very subsonic air flows.

2.3 Two-dimensional flows

Simplifications arise in the mathematical modelling of fluid flows when one considers systems which possess symmetries. We shall first consider flows confined to a plane, expressed in Cartesian coordinates as

u(x, y, t) u = v(x, y, t) = u(x, y, t) ˆex + v(x, y, t) ˆey. 0 Chapter 2 – Mass conservation & streamfunction 13

2.3.1 Streamfunctions

For two-dimensional incompressible flows, we define

S = ψ(x, y, t) ˆez, (2.5) where ψ is a (scalar) streamfunction, such that, from equation (2.4),

∂ψ ∂ψ u = (ψ ˆez) u = and v = . (2.6) ∇× ⇒ ∂y − ∂x

Clearly the incompressibility condition is then automatically satisfied:

∂u ∂v ∂2ψ ∂2ψ u = + = = 0. ∇ · ∂x ∂y ∂x∂y − ∂y∂x

Streamfunctions and streamlines. A key property of streamfunctions comes from con- sidering ∂ψ ∂ψ ∂ψ ∂ψ ∂ψ ∂ψ u ψ = u + v = = 0, · ∇ ∂x ∂y ∂y ∂x − ∂x ∂y 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 constant along streamlines. Conversely, consider a parametric curve x(s) = (x(s), y(s)) of constant ψ, so that

dψ ∂ψ dx ∂ψ dy dx dy dx = 0 + = v + u = u = 0, ds ⇔ ∂x ds ∂y ds − ds ds × ds where dx/ds is a vector tangent to the curve x(s). So, u is tangent to the curve where ψ remains constant which is therefore a streamline.

Example 2.1 U  Consider the 2-D flow u = 0 , with U and k constants. Since u = 0, one can define kt 0 ∇· the streamfunction ψ(x, y, t) = U y ktx + C, 0 − where C is an arbitrary constant of integration. The value assigned to C does not affect the flow, so we shall usually take C = 0 without loss of generality. The streamfunction is constant along the lines ψ = U y ktx = C, where C is a constant 0 − identifying streamlines. So, the streamlines are lines of equation

kt C dy kt y = x + with gradient = . U0 U0 dx U0

 y  Example 2.2 x2 + y2 The flow defined as u =   is incompressible since  x  − x2 + y2

∂u ∂v 2xy ( 2y)( x) u = + = − + − − = 0. ∇· ∂x ∂y (x2 + y2)2 (x2 + y2)2 14 2.3 Two-dimensional flows

∂ψ y 1 2y 1 So, = u = = ψ(x, y) = ln x2 + y2
+ α(x). ∂y x2 + y2 2 x2 + y2 ⇒ 2 ∂ψ x dα dα Then v = = + = 0. So, α is constant and − ∂x −x2 + y2 dx ⇒ dx 1 ψ(x, y) = ln x2 + y2
(choosing α = 0). 2

2.3.2 Polar coordinates We now change from Cartesian coordinates, (x, y), in the (X,Y )-plane to polar coordinates, (r, θ), defined by Y

 x(r, θ) = r cos θ, ˆe y(r, θ) = r sin θ, θ y ˆer and consider again a two-dimensional flow, u, indepen- M r dent of z, with uz = 0, such that θ X x u(r, θ) = ur(r, θ) ˆer + uθ(r, θ) ˆeθ.

So, in plane polar coordinates, substituting S = ψ(r, θ, t) ˆez in u = S for 2-D incompress- ∇ × ible flows gives 1 ∂ψ ∂ψ u = (ψ ˆez) u = , u = . (2.7) ∇× ⇒ r r ∂θ θ − ∂r Example 2.3 p For the streamfunction ψ = ln x2 + y2 = ln r, u = 0 and u = 1/r. The streamlines are r θ − circles about the origin with u decreasing as r increases. This is a reasonable model for a | θ| bath-plug vortex. ˆe θ u θ r

2.3.3 Physical significance of the streamfunction We noted earlier that the streamfunction is constant on streamlines. So, we consider the two streamlines defined by ψ(x, y) = ψP and ψ(x, y) = ψT .

T ψ = ψT n

C

P ψ = ψP The flow rate or the volume flux through an arbitrary curve C :(x(s), y(s)), parametrised by s [0, 1] and connecting P and T , is ∈ Z 1 Q = u n ds. (2.8) 0 · Chapter 2 – Mass conservation & streamfunction 15

dx dy  Let dl = dxˆex + dyˆey = ds ˆex + ˆey be an infinitesimal ds ds displacement along the curve C. The infinitesimal vector normal dl (x(s),y(s)) dy to dl is therefore   C dx dy dx n ds = dyˆex dxˆey = ˆex ˆey ds. nds − ds − ds So, Z 1 ∂ψ dy ∂ψ dx Z 1 dψ Z ψT Q = + ds = ds = dψ = ψT ψP . 0 ∂y ds ∂x ds 0 ds ψP − Hence, the flux of fluid flowing between two streamlines is equal to the difference in the streamfunction. 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 the k k k∇ k gradient 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) ˆez), invariant along the z-axis, with no z-component. ∇×

Similarly, axisymmetric flows, such as

u(r, z) = ur(r, z) ˆer + uz(r, z) ˆez (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.

ˆer

ˆez

2.4.1 Stokes streamfunctions

For axisymmetric incompressible flows, we define 1 S = Ψ(r, z, t) ˆeθ, (2.10) r where Ψ is a (scalar) Stokes streamfunction, such that, from equation (2.4),

1  1 ∂Ψ 1 ∂Ψ u = Ψ ˆeθ u = and u = . (2.11) ∇× r ⇒ z r ∂r r −r ∂z (We use Ψ to distinguish Stokes streamfunctions from planar streamfunctions denoted ψ.) Clearly the incompressibility condition is then again automatically satisfied: 1 ∂ ∂u 1 ∂2Ψ 1 ∂2Ψ u = (ru ) + z = + = 0. ∇ · r ∂r r ∂z −r ∂r∂z r ∂z∂r 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. ∂Ψ ∂Ψ 1  ∂Ψ ∂Ψ (u ) Ψ = u + u = ru + ru , · ∇ r ∂r z ∂z r r ∂r z ∂z 1  ∂Ψ ∂Ψ ∂Ψ ∂Ψ + = 0. r − ∂z ∂r ∂r ∂z Thus, Ψ is constant in the direction of the flow.

For axisymmetric flows it is useful to think of streamtubes: surface of revolution spanned by all the streamlines through a circle about the axis of symmetry.

Streamtubes are surfaces of constant Ψ.

ii. Relation between volume flux and streamtubes. The volume flux, or fluid flow, between two streamtubes with Ψ = Ψi and Ψ = Ψo is Z Q = u n dS = 2π(Ψ0 Ψi). (2.12) S · −

Ψo Co S

Ψ i Ci n

Proof. Z Z 1  Q = u n dS = Ψ ˆeθ n dS, (definition of Ψ) S · S ∇ × r · I 1 I 1 = Ψ ˆeθ dl + Ψ ˆeθ dl, (Stokes’ theorem) Co r · Ci r · I 1 I 1 = Ψo ˆeθ dl + Ψi ˆeθ dl, (Ψ Ψ o,i onto C o,i ) Co r · Ci r · ≡ { } { } Z 2π Z 0 = Ψo dθ + Ψi dθ = 2π(Ψo Ψi) (recall, dl = dr ˆer + rdθ ˆeθ). 0 2π −  Example 2.4 For a uniform flow parallel to the axis, ur = 0 and uz = U, ∂Ψ ∂Ψ 1 = rU and = 0 Ψ(r) = Ur2. ∂r ∂z ⇒ 2 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 Ψ(0) = 0 z

S u 1 Ψ(a)= Ua2 2 The volume flux Z Z Z Z 2 Q = u n dS = u ˆez dS = uz dS = U dS = πUa . S · S · S S Also, 1  2π(Ψ Ψ ) = 2π(Ψ(a) Ψ(0)) = 2π Ua2 0 = πUa2 as required. o − i − 2 −

Example 2.5 Consider a flow in a long pipe a radius a:  U 2 2
uz = 0 on r = a, ur = 0, uz = 2 a r with a − uz = U at r = 0.

n a z

S

∂Ψ = ru = 0 Ψ Ψ(r), ∂z − r ⇒ ≡ dΨ U U a2r2 r4  and = ru = (a2r r3) Ψ(r) = + C. dr z a2 − ⇒ a2 2 − 4

Hence,

Ur2 Ψ(r) = (2a2 r2) (choose C such that Ψ(0) = 0). 4a2 − So, Ψ(0) = 0 and Ψ(a) = Ua2/4, and the volume flux Z π Q = u n dS = 2π (Ψ(a) Ψ(0)) = Ua2. S · − 2 Indeed, Z Z 2π Z a Z a 2πU 2 3 u n dS = uzr drdθ = 2 (a r r ) dr, (since dS = rdθdr) S · 0 0 a 0 − 2πU a2r2 r4 a πUa2 = 2 = , as required. a 2 − 4 0 2 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 of ∇ · the 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 importance in fluid dynamics.

3.2 Physical meaning

 u(x, y, t)  In the simplest case of a 2-D flow, u(x, y, t) =  v(x, y, t) , the vorticity ω = u = ∇ × 0   0  0  ∂v ∂u   is perpendicular to the plane of motion; its magnitude is ω = .  ∂v ∂u  ∂x − ∂y ∂x − ∂y

Consider the two short fluid line-elements AB and AC. The vertical differential velocity is

19 20 3.2 Physical meaning

B δu ∂v δy δv = vC vA = v(x+δx, y) v(x, y) δx (Taylor theorem). δv − − ' ∂x ∂v So, is the angular velocity of the fluid line-element AC. ∂x A δx C

Similarly, the horizontal differential velocity

∂u δu = u u = u(x, y + δy) u(x, y) δy . B − A − ' ∂y

∂u So, is the angular velocity of the fluid line element AB. − ∂y

ω 1  ∂v ∂u Thus, = represents the average angular 2 2 ∂x − ∂y velocity of the two fluid line-elements AB and AC. This could be measured using a crossed pair of small vanes that float with the fluid.

Example 3.1 (Solid body rotation)  x  For u = Ω r, with constant angular velocity Ω and r =  y , × z

ω = (Ω r) = Ω( r) (Ω )r, ∇ × × ∇ · − · ∇ Ω = 3Ω Ω = 2Ω. − So, the vorticity is twice the local rotation rate — which is also global here.

Example 3.2 (Shear flow)  0  For u = ky, v = 0, the vorticity ω =  0  with ω = k. − k −

Vorticity rollers The vorticity is not a measure of global rotation: a shear flow has no global rotation but a non-zero vorticity.

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 a bath-plug vortex — see examples 2.2 and 2.3) Chapter 3 – Vorticity 21

∂ψ k Streamfunction: = ψ = k ln r (const. = 0). So, ∂r − r ⇒ − the streamlines are circles.

However, from 1 ∂ 1 ∂  1 ∂k ω = (ru ) u ˆez = ˆez = 0, r ∂r θ − r ∂θ r r ∂r one finds that the vorticity is zero everywhere in the flow except at r = 0 where the functions u 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

 0  ∂v ∂u In a two-dimensional flow, the vorticity ω =  0  where ω = in Cartesian ∂x − ∂y ω ∂ψ ∂ψ coordinates. If, in addition, the fluid is incompressible, u = and v = , so that ∂y − ∂x ∂2ψ ∂2ψ ω = = 2ψ. (3.2) − ∂x2 − ∂y2 −∇ More generally (Cartesian or plane polar coordinates),

u = (ψˆez) ω = ( (ψˆez)), ∇ × ⇒ ∇ × ∇ × 2 = ( (ψˆez)) (ψˆez), ∇ ∇ ·  − ∇ ∂ 2 = ψ ψ ˆez. ∇ ∂z − ∇

Since ∂ψ/∂z = 0 for a 2-D flow, one finds again

2 ω = ψ ˆez (3.3) −∇

3.4 The Rankine vortex

Example 3.3, with uθ = k/r, is a crude model for the bath-plug vortex: infinite vorticity concentrated 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 if r a (Ω constant), ω = ≤ 0 if r > a 22 3.5 Circulation

( 1 d  dψ  Ω if r a, Hence, ψ ψ(r) and 2ψ = r = − ≤ ≡ ∇ r dr dr 0 if r > a.

d  dψ  dψ Ω B r a: r = r Ω = r + = u . • ≤ dr dr − ⇒ dr − 2 r − θ We require u to be bounded at r = 0. So, B = 0 and, for r a, θ ≤  Ω  u = r,  θ 2 Ω  ψ = r2. − 4

d  dψ  dψ A r > a: r = 0 u = = . • dr dr ⇒ θ − dr r Ω The continuity of u at r = a implies that A = a2. θ 2 dψ Ωa2 Ωa2 So, = ψ = ln r + D, and applying the continuity of ψ at r = a now dr − 2r ⇒ − 2 Ωa2  r  gives ψ = 1 + 2 ln . − 4 a So,  Ω  r2 if r a, − 4 ≤ ψ = 2  Ωa  r   1 + 2 ln if r > a. − 4 a and Ω  r if r a : solid body rotation,  2 ≤ uθ = 2 Ωa  if r > a : bath-plug flow. 2r

uθ ω Ω r 1/r ∝ ∝

a r a r There is still a discontinuity in vorticity but the flow is quite adequate for predicting the shape of the water surface.

3.5 Circulation

Consider a closed curve in the flow. The circulation around is the line integral of the C C tangential velocity around : C I Γ = u dl. (3.4) · C Chapter 3 – Vorticity 23

n

C















S







dl u By Stokes theorem, for any surface spanning the curve , S C I Z Z Γ = u dl = ( u) n dS = ω n dS. (3.5) · ∇ × · · C S S So, the circulation around is equal to the vorticity flux through the surface : it is the C S strength of the vortex tube.

Example 3.4 (The Rankine vortex) The circulation around a circle of radius r is given by

( Z 2π πΩr2 if r a, Γ = uθ(r) r dθ = 2π r uθ(r) = 2 ≤ 0 πΩa if r > a.

When the vorticity is concentrated in thin filaments (tubes) as it is in the Rankine vortex, it is 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 decay very slowly and are a danger for small aircraft flying behind large ones.

iii. Horseshoe vortex & downwash behind chimneys.

Vortex lines in shear flows above ground level travel with the air and can be stretched and bent by tall buildings and chimneys. This results in a downwards flow behind chimneys, dragging pollutant down to ground level.

iv. Vortex rings.

Smoke rings and underwater bubble rings are examples of vortex rings. Chapter 3 – Vorticity 25

v. von K´arm´anvortex street.

In certain conditions, a flow past an obstacle (e.g. a cylinder, an island) produces a series 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, such that ω = u = 0. (4.1) ∇ ×

If a velocity field u is irrotational, that is if u = 0, then there exists a velocity potential ∇ × φ(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; boundary ∇ − ∇ conditions on ψ and φ are different however.)

27 28 4.2 Kinematic boundary conditions

4.2 Kinematic boundary conditions

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, U then, locally, the surface advances (i.e. moves in the direction 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, ∇ ∂φ n φ = = U n; (4.4) · ∇ ∂n · 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, 1 d  dφ dφ m 2φ = r = 0 = , ∇ r dr dr ⇔ dr 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 m u = φ = ˆer ∇ r corresponding to a source (m > 0) or sink (m < 0) of fluid of strength m. Notice that the constant C in φ is arbitrary and does not affect u. By convention the constant m = Q/2π where 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 spherical ≡ polar coordinates, 1 d  dφ dφ m 2φ = r2 = 0 = , ∇ r2 dr dr ⇔ dr r2 one finds m φ(r) = + C, (4.6) − r where m and C are integration constants. This potential produces the three-dimensional radial velocity m u = φ = ˆer. ∇ r2 corresponding to a source (m > 0) or sink (m < 0) of fluid of strength m. By convention the constant 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 equation in plane polar coordinates, one has ∂φ 1 ∂φ k u = = 0 and u = = , r ∂r θ r ∂θ r where the strength of the flow k is a constant. By convention k = 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 zero vorticity 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. A dipole is the superposition of a sink and an source of equal but opposite strength next to each other. 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 the − position (0, 0, δ).

Sink Source O z

δ The velocity potential of the flow is formed by adding the potentials of the source and sink, m m φ = p p , x2 + y2 + z2 − x2 + y2 + (z δ)2 m m − = , r − √r2 2zδ + δ2 − where m > 0 is constant and r = (x2 + y2 + z2)1/2. Expanding the potential to first order in δ, m m  z  φ = 1 + δ + O(δ2) , r − r r2 and taking the limit δ 0, leads to the potential of a dipole → z φ = mδ , − r3 where m as δ 0 so that the strength of the dipole µ = mδ remains finite. Thus, → ∞ → µ r 1 φ = · = µ , − r3 · ∇ r where µ = µˆez and r x is the vector position. The three components of the fluid velocity, ≡ u = φ, are for a dipole of strength µ, ∇

xz u = 3µ , x r5 yz u = 3µ , y r5 µ  z2  u = 1 3 . z −r3 − r2

Planar flow Similarly, combining a line sink at the origin with a line source of equal but opposite strength at (δ, 0) gives m m (x δ)2 + y2  φ = ln(x2 + y2) ln (x δ)2 + y2
 = ln − , m > 0. − 2 − − 2 x2 + y2 As in the three-dimensional case, we consider the limit δ 0, with µ = mδ fixed. The → expression of the potential for a two-dimensional dipole of strength µ then becomes µ r φ = · = µ ln r, − r2 − · ∇ where µ = µˆex 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. From vector calculus,

(f f) = f 2f + f 2 S Z ∇ · ∇ Z∇ |∇ | Z n (f f) dV = f 2f dV + f 2 dV. ⇒ V ∇ · ∇ V ∇ V |∇ | V So, using the divergence theorem Z Z Z f( f) n dS = f 2f dV + f 2 dV. (4.7) S ∇ · V ∇ V |∇ |

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. the · boundary 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 = φ ∇ 1 and u2 = φ . Let f = φ φ , then ∇ 2 1 − 2

2f = 2φ 2φ = 0 ∇ ∇ 1 − ∇ 2 in the domain V and

( f) n = ( φ ) n ( φ ) n = u1 n u2 n = 0 ∇ · ∇ 1 · − ∇ 2 · · − · Z 2 on the boundary S. Hence, from identity (4.7), φ1 φ2 dV = 0. However, since V |∇ − ∇ | 2 φ φ 0 one must have φ = φ everywhere. Therefore u1 = u2 and the |∇ 1 − ∇ 2| ≥ ∇ 1 ∇ 2 solution to the boundary value problem is unique. 

4.4.3 Uniqueness for an infinite domain

S′ The proof above holds for flows in a finite domain. What about flows in an infinite V n

S domain — e.g. flow around an obstacle?



The above argument holds by considering the

volume V as shown and letting S . (See, 0 → ∞ e.g. Patterson p. 211.) 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, ∇ · 1 Z K = ρ u 2 dV. 2 V | |

Proof. Let u be another incompressible but non vorticity-free flow such that u n = u n 0 · 0 · on S and u = 0 in V but with u = 0. ∇ · 0 ∇ × 0 6 The fluid flow u is potential, so let u = φ such that ∇ Z Z Z ρ u 2 dV = ρ φ 2 dV = ρ φ 2 dV, V | | V |∇ | V |∇ | Z = ρ φ u n dS (by identity (4.7) with f = φ), S · Z = ρ φ u0 n dS (boundary condition), S · Z = ρ (φ u0) dV (divergence theorem), V ∇ · Z = ρ u0 φ dV ( u0 = 0), V · ∇ ∇ · Z = ρ u0 u dV. (4.8) V · So, Z Z 2 2 2
ρ (u u0) dV = ρ u 2ρu u0 + ρ u0 dV, V − V | | − · | | Z 2 2
= ρ u0 ρ u dV (from (4.8)). V | | − | |

2 Therefore, since (u u0) 0, − ≥ Z Z Z Z 2 2 2 2 ρ u0 dV = ρ u dV + ρ (u u0) dV ρ u dV. V | | V | | V − ≥ V | |



4.5 Flow past an obstacle

Since the solution to Laplace’s equation for given boundary conditions is unique, if we find a solution, we have found the solution. (This is only true if the domain is simply-connected; if the domain is multiply connected, multiple solutions become possible.) One technique to calculate non elementary potential flows involves adding together simple known 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 = urˆer + uzˆez in cylindrical polar coordi- nates (r, θ, z). Chapter 4 – Potential flows 33

n At large distances from the sphere of ra- r U dius a the flow is asymptotic to a uni- form stream, u = 0, u = U, and at the a r r z sphere’s surface, r = a, the fluid velocity O Z must satisfy u n = 0 since the solid body z · forms a non-penetrable boundary.

The unit vector normal to surface of the sphere is r z n = n ˆer + n ˆez with n = and n = . r z r a z a So, the boundary condition u n = 0 implies that · r z u + u = 0 ru + zu = 0 r a z a ⇔ r z 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. ' k k  Now, add to the uniform stream a dipole velocity field of strength µ = µˆez at the origin, µz φ(r, z) = Uz , − (r2 + z2)3/2 so that ∂φ 3µrz ∂φ µ  3z2  ur = = and uz = = U + 1 . ∂r (r2 + z2)5/2 ∂z (r2 + z2)3/2 r2 + z2 − Thus, at the sphere’s surface, r z z  3µ(r2 + z2) µ  u n = ur + uz = U + , · a a a (r2 + z2)5/2 − (r2 + z2)3/2 z  2µ  z  2µ = U + = U + , a (r2 + z2)3/2 a a3 since r2 + z2 = a2. Hence the boundary condition u n = 0 at the sphere’s surface determines · the strength of the dipole, Ua3 µ = . − 2 The velocity potential for a uniform flow past a stationary sphere is therefore given by  a3  φ(r, z) = Uz 1 + . (4.9) 2(r2 + z2)3/2

The corresponding Stokes streamfunction is given by

Ur2  a3  Ψ(r, z) = 1 . (4.10) 2 − (r2 + z2)3/2

Outside the sphere Ψ > 0, but we also obtain a solution inside the sphere with Ψ < 0. This flow is not real; it is a “virtual flow” that allows for fluid velocity to be consistent with the boundary 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 a point source (m > 0), so that ! m mr mz φ(r, z) = Uz u = φ = ,U + . − (r2 + z2)1/2 ⇒ ∇ (r2 + z2)3/2 (r2 + z2)3/2 p This flow has a single stagnation point u = u = 0 at r = 0 and z = m/U. r z − To find the streamlines of the flow we calculate the Stokes streamfunction using 1 ∂Ψ 1 ∂Ψ u = and u = . r −r ∂z z r ∂r Thus, ∂Ψ mrz Ur2 mz = Ur + Ψ = + α(z), ∂r (r2 + z2)3/2 ⇒ 2 − (r2 + z2)1/2 1 ∂Ψ m mz2 α (z) = + + 0 , ⇒ r ∂z −r (r2 + z2)1/2 r (r2 + z2)3/2 r
m r2 + z2 z2 α (z) mr α (z) = − + 0 = + 0 , − r (r2 + z2)3/2 r −(r2 + z2)3/2 r mr = ur = . − −(r2 + z2)3/2

So, since α0(z) = 0, α is a constant (set to zero). The Stokes streamfunction is therefore Ur2 mz Ψ(r, z) = . 2 − (r2 + z2)1/2 p 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 ! Ur2 z Ψ(r, z) = m = m 1 + . ⇔ 2 (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 rm 2m r 2 . 2 ' ⇒ ' U Thus, the velocity potential and the Stokes streamfunction ! ! a2 U a2z φ(r, z) = U z and Ψ(r, z) = r2 − 4 (r2 + z2)1/2 2 − 2 (r2 + z2)1/2 p 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 the domain 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 a solid wall at x = 0. In free space (no wall), the potential of the y source is m φ = p , ∞ − (x d)2 + y2 + z2 − O x ∂φ m(x d) u = ∞ = − . ⇒ ∞ ∂x [(x d)2 + y2 + z2]3/2 − d So that, at x = 0, md u = = 0, ∞ −(d2 + y2 + z2)3/2 6 which is inconsistent with the boundary condition u n = u ˆex = u = 0 at the wall. · · To rectify this problem, (i.e. for the flow to satisfy the boundary condition at the wall), we add a source of equal strength m outside the domain, at ( d, 0, 0). By symmetry, this source − will produce an equal but opposite velocity field at x = 0, so that the boundary condition for the combined flow can be satisfied. The velocity potential for both sources becomes m m φ = p p , − (x d)2 + y2 + z2 − (x + d)2 + y2 + z2 − and the velocity field along the x-axis, ∂φ m(x d) m(x + d) u = = − + . ∂x [(x d)2 + y2 + z2]3/2 [(x + d)2 + y2 + z2]3/2 − Clearly, at x = 0, now u = 0 as required. The fluid can slip along the wall however as, for x = 0, S 2my v = , (d2 + y2 + z2)3/2 2mz w = . (d2 + y2 + z2)3/2 d 0 d x − 4.7 Method of separation of variables

This is a standard method for solving linear partial differential equations with compatible boundary conditions. We shall seek separable solutions to Laplace’s equations, of the form φ(x, y) = f(x)g(y) in Cartesian 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) in Laplace’s equation expressed in plane polar coordinates,

1 ∂  ∂φ 1 ∂2φ 2φ = r + = 0, ∇ r ∂r ∂r r2 ∂θ2 g d  df  f d2g r + = 0, ⇒ r dr dr r2 dθ2 r d  df  1 d2g r + = 0, (division by f(r)g(θ)/r2) ⇒ f dr dr g dθ2 r d  df  1 d2g r = . ⇒ f dr dr −g dθ2

Since the terms on the left and right sides of the equation are functions of independent variables, r and θ respectively, they must take a constant value, k2 say. Thus we have transformed a partial differential equation for φ into two ordinary differential equations for f and g,

r d  df  d  df  r = k2 r r k2f = 0, f dr dr ⇒ dr dr − 1 d2g d2g = k2 + k2g = 0. g dθ2 − ⇒ dθ2 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 d2f df r2 + r n2f = 0. dr2 dr − 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’s ± equation in plane polar coordinates is therefore

n n
n n
φ(r, θ) = Ar + Br− cos(nθ) + Cr + Dr− 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 a similar manner, but involves Legendre polynomials Pl(cos(θ)).

Example 4.2 (Cylinder in an extensional flow) Consider the velocity potential

2 2
φ(r, θ) = Ar + Br− 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

∂φ  B  u = = 2r A cos(2θ). r ∂r − r4 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

 a4   a4  u = 2Ar 1 cos(2θ) and u = 2Ar 1 + sin(2θ) r − r4 θ − r4 produced by the potential  a4  φ(r, θ) = Ar2 1 + cos(2θ) r4 represents a fluid flow past a solid cylinder placed in an extensional flow.

Notice that at large distances, i.e. if r a, the  fluid velocity is that of an extensional flow

u 2Ar cos(2θ) and u 2Ar sin(2θ), r ' θ ' − 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 incompressible and irrotational fluid flows.

We shall now study the dynamics of fluid flows and consider changes S in motion due to forces acting on a fluid.

We derive an evolution equation for the fluid momentum by consider- V ing forces acting on a small blob of fluid, of volume V and surface S, containing many fluid particles.

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 ,

Z Fv = ρ g dV. V

Surface forces are caused by interactions at the surface S. For the rest of this course we shall 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 the S normal n. The force exerted on the fluid into V by the fluid on the other side of S is, by convention, written as n Z Fs = p n dS, − S V where p(x) > 0 is the fluid pressure.

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 V is equal to the rate of change of its momentum. Since Du/Dt is the acceleration of the fluid particles, or fluid elements, within V , one has

Z Du Z Z ρ dV = p n dS + ρ g dV. V Dt S − V

We now apply the divergence theorem,

Z Du Z ρ dV = ( p + ρ g) dV, V Dt V −∇ and notice that both integrands must be identical, since V is arbitrary. So, the evolution of fluid momentum is governed by Euler’s equation

Du ∂u  ρ = ρ + (u ) u = p + ρ g. (5.1) Dt ∂t · ∇ −∇ 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

Du ρ = p + ρ g + µ 2u. (5.2) Dt −∇ ∇

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 reduced to the equation of hydrostatic balance,

p = ρg p(x) = ρg x + C, (5.3) ∇ ⇔ · where C is a constant.

Example 5.1 The density of mass in the ocean can be considered as constant, ρ , and the gravity g = gˆez. 0 − (The coordinate z is the upward distance from sea-level.) Chapter 5 – Euler’s equation 41

From Euler’s equation one has z O dp = ρ0g p(z) = p0 ρgz. ρ g dz − ⇒ − g 0 p Hence the pressure increases linearly with −∇ depth (z < 0). Taking typical values for the physical constant, g 10 m s 2, ρ 103 kg m 3 and a pressure ' − 0 ' − of one atmosphere at sea-level, p p = 105 Pa = 105 N m 2 gives p(z) 105(1 0.1z); 0 ' atm − ' − 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 = 0 and a given depth z = d, is equal to (p p )S = ρ gdS, which is the weight of a column of − − 0 0 water 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 S totally submerged in a fluid of density ρ . The total 0 F = ρ V g force on the body caused by the fluid surrounding − 0 g S it is Z F = p n dS, V n − S where p is the fluid pressure. (Notice that the pressure distribution on the surface S is the same whether the fluid contains a solid or not.) So, using successively the divergence theorem and the equation of hydrostatic balance, P = ρg, we find ∇ Z Z F = p dV = ρ0g dV = ρ0V g. − V ∇ − V −

The buoyancy force is equal the weight of the mass of fluid displaced, M = ρ0V , and points in the direction opposite to gravity.

If the fluid is only partially submerged, then we need to split it into parts above and below the water surface, and apply Archimedes’ theorem to the lower section only.

Consider a solid body of volume V and density ρs ρ V g − 0 2 partially submerged in a fluid of density ρ0 > ρs. V1 Let V1 be the volume of solid above the fluid surface ρ0 ρs and V2 the volume underneath. Since, the solid is in V2 ρsV g equilibrium, its weight is balanced by the buoyancy force, so that ρsV g = ρ0V2g. Hence, the fractions of the volume of solid immersed in the fluid and not immersed are V ρ V V ρ ρ 2 = s and 1 = 1 2 = 0 − s , V ρ0 V − V ρ0 respectively. For an iceberg of density ρ 0.915 kg m 3 floating in salted water of density s ' − ρ 1.025 kg m 3, V /V 89.3% and V /V 10.7%. s ' − 2 ' 1 ' 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 ∂u = + (u ) u, Dt ∂t · ∇ the nonlinear term can be rewritten using the vector identity

 u 2   u 2  u ( u) = k k (u ) u (u ) u = k k u ω, × ∇ × ∇ 2 − · ∇ ⇒ · ∇ ∇ 2 − × where u 2 = u u. So, we can rewrite Euler’s equation (5.1) as k k · ∂u 1 + (u ) u = p + g, ∂t · ∇ −ρ∇ ∂u  u 2  p + k k u ω = + g (ρ constant), ∂t ∇ 2 − × −∇ ρ and take its curl

∂ω (u ω) = 0, ∂t − ∇ × × ∂ω + (u ) ω (ω ) u + ( u) ω ( ω) u = 0. ∂t · ∇ − · ∇ ∇ · − ∇ ·

For incompressible flows u = 0 and, in addition, ω = 0 as ω = u. Hence, ∇ · ∇ · ∇ × Dω ∂ω = + (u ) ω = (ω ) u. (5.4) Dt ∂t · ∇ · ∇ This is the vorticity equation. It shows that the vorticity of a fluid particle changes because of 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 irrotational remain so.

ii. In a two-dimensional planar flow, u = (u(x, y), v(x, y), 0), the vector vorticity has only one non-zero component, ω = (∂v/∂x ∂u/∂y)ˆez, so that − d (ω ) u = ω u(x, y) = 0. · ∇ dz

Hence, the vorticity equation, reduced to

Dω ∂ω = + (u ) ω = 0, (5.5) Dt ∂t · ∇

shows that the vorticity of a fluid particle remains constant. If, in addition the flow is steady, ∂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 increase uϕ of its vorticity. r Consider, for example, an incompressible steady flow in a converging cone, function of ur the radial distance only in spherical polar co- ordinates,

u(r) = ur(r)ˆer + uϕ(r)ˆeϕ. O

The component ur represents a radial inflow and uϕ the swirling of the fluid. Since (u (r)ˆer) = 0, only the swirling motion contributes to a non-zero vorticity ∇ × r

ω(r) = u = (u (r)ˆeϕ) = ω ˆer + ω ˆeθ. ∇ × ∇ × ϕ r θ From mass conservation in an incompressible spherically symmetric inflow, we find

1 d k u = r2u
= 0 u = , ∇ · r2 dr r ⇔ r −r2

where k > 0 is constant. Since the flow is steady, ∂ω/∂t = 0, the evolution equation for the radial component of the vorticity, ωr, becomes

Dωr dωr dur d ωr ωr = ur = ωr ln = 0 = α, constant. Dt dr dr ⇔ dr ur ⇔ ur

Thus, αk w = αu = , r r − r2 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 amount of background vorticity is amplified by the flow converging into a small hole. (This mechanism can be interpreted as the conservation 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 of uniform density, subject to conservative forces. Hence, dΓ d I = u dl = 0, (5.6) dt dt C(t) · 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 parametric representation 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 d Z 1 ∂ Z 1 d  ∂x = u(x(s, t), t) x(s, t) ds = u(x, t) ds, dt dt 0 · ∂s 0 dt · ∂s Z 1 ∂u ∂x   ∂x ∂2x  = + u + u ds, 0 ∂t ∂t · ∇ · ∂s · ∂s∂t Z 1 ∂u  ∂x ∂u = + (u ) u + u ds, 0 ∂t · ∇ · ∂s · ∂s since ∂x/∂t = u is the velocity of the fluid particle at x(s, t). So, using Euler’s equation in the form ∂u  p  + (u ) u = + g x , ∂t · ∇ ∇ −ρ · we find dΓ Z 1   p  ∂x ∂  u 2  Z 1 ∂  u 2 p  = + g x + k k ds = k k + g x ds. dt 0 ∇ −ρ · · ∂s ∂s 2 0 ∂s 2 − ρ · Thus, since the curve C(t) is closed,

dΓ I  u 2 p  = d k k + g x = 0, dt C(t) 2 − ρ · as required.  Recall that the circulation around a closed curve C is equal to the flux of vorticity through an arbitrary surface S that spans C. So, from Kelvin’s circulation theorem, dΓ d I d Z = u dl = ω n dS = 0; dt dt C(t) · dt S(t) · this demonstrates that the flux of vorticity through a surface that spans a material curve is constant. Thus, many properties of the vorticity equation could equivalently be derived from the 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 of rotation. This phenomenon, which can be observed when stirring coffee in a mug, results from a radial pressure-gradient balancing the centrifugal force acting within the fluid.

Consider, for example, a cylindrical container z partially filled with fluid and mounted on a horizontal turntable. When the flow reaches Ω a steady state, ∂u/∂t = 0, the fluid rotates uniformly, with a constant angular velocity Ω about the vertical z-axis. (The fluid rotates h(r) with the container as a solid body.) h0 In order to calculate the height of the free sur- O face of fluid, z = h(r), we shall solve Euler’s r equation, ∂u   1  ρ u ω = ρu ω = p + ρ u 2 + ρg, (5.7) ∂t − × − × −∇ 2 k k Chapter 5 – Euler’s equation 45 in cylindrical polar coordinates. The velocity and vorticity fields of the fluid in uniform rotation are u = uθˆeθ and ω = ωzˆez, where uθ = rΩ and ωz = 2Ω. Hence, the nonlinear term in Euler’s equation 2 u ω = u ω ˆer = 2rΩ ˆer. × θ z The radial component of the vector equation (5.7) (i.e. its scalar product with ˆer) is therefore

∂p ρ du2 ∂p 2ρrΩ2 = θ = ρrΩ2, − −∂r − 2 dr −∂r − ∂p = ρrΩ2. (5.8) ⇔ ∂r This shows that the pressure must vary with the radius inside the fluid, in order to balance the 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 ˆez) we find ∂p 0 = ρg p(r, z) = ρgz + A(r), −∂z − ⇒ − where A(r) is a function to be determined using appropriate boundary conditions. At the free surface the fluid pressure should match the atmospheric pressure. Hence, p = patm at z = h(r), so that p = ρgh(r) + A(r) A(r) = p + ρgh(r). atm − ⇔ atm So, substituting the expression of the pressure,

p(r, z) = p + ρg (h(r) z) , atm − in equation (5.8) leads to

∂p dh r2Ω2 = ρg = ρrΩ2 h(r) = h + , ∂r dr ⇒ 0 2g where h0 is the height of the free surface at r = 0, on the rotation axis. This result shows that the free surface of a uniformly rotating fluid in a cylindrical container is a paraboloid.

Such rotating cylindrical containers filled with highly-reflecting liquids (e.g. mercury or ionic liquid coated with silver) have been used to build mirrors with a large (up to 6 m) smooth reflecting paraboloid surface, which could be used 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 with constant density) in the form ∂u 1  p u ω + u 2 = + g. ∂t − × ∇ 2k k −∇ ρ So, since the constant gravity g = (g x), one has ∇ · ∂u u ω + = 0, (6.1) ∂t − × ∇H where p 1 (x, t) = + u 2 g x (6.2) H ρ 2k k − · 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 ) = 0, (6.3) · ∇ H since u (u ω) 0. · × ≡ Hence, for an ideal fluid in steady flow, p 1 (x) = + u 2 g x (6.4) H ρ 2k k − · 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, is a function of ψ ( (x) (ψ)). H H ≡ H 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.

(a) (c) h1 h3 V (b) 1 S1 S3 h2 A A1 1 S2 A2

(Three narrow vertical tubes, (a), (b) and (c), are used to measure the pressure at different points.) The fluid velocity is assumed inform on cross sections, S. Upstream the fluid velocity is V . Z 1 Mass conservation implies ρu n dS = constant for any cross-section S, so S · Z Z Z Z ρu n dS = ρu n dS u n dS = u n dS (ρ = constant), S1 · S2 · ⇒ S1 · S2 · A1 A1V1 = A2V2 V2 = V1 > V1 (since A1 > A2). ⇒ ⇒ A2 Neglecting gravity, we apply Bernoulli’s equation to any streamline, p 1 p 1 ρ 1 + V 2 = 2 + V 2 p = p V 2 V 2
, ρ 2 1 ρ 2 2 ⇒ 2 1 − 2 2 − 1 2 ρV1 2 2
p2 = p1 2 A1 A2 < p1. ⇒ − 2A2 − Thus, in the constriction the speed of the flow increases (conservation of mass) and its pressure decreases (Bernoulli’s equation). This can be measured by the thin tubes where there is fluid but no flow (i.e. fluid in hydrostatic equilibrium). If h1 is the height of fluid in the tube (a) then

p = p + ρgh (p p ). 1 0 1 0 ≡ atm

If h2 is the height of fluid in the tube (b) then

2 p2 p0 p1 p0 V1 2 2
p2 = p0 + ρgh2 h2 = − = − 2 A1 A2 , ⇒ ρg ρg − 2gA2 − 2 V1 2 2
h2 = h1 2 A1 A2 < h1. ⇒ − 2gA2 −

In tube (c), V3 = V1 since A3 = A1 (mass conservation). So, Bernoulli’s equation gives 1 1 p + ρV 2 = p + ρV 2 p = p , 3 2 3 1 2 1 ⇒ 3 1 and so h3 = h1. (In practice, h3 will be slightly less than h1 due to viscosity but the effect is small.) Chapter 6 – Bernoulli’s equation 49

Example 6.2 (Flow down a barrel.) How fast does fluid flow out of a barrel? z

A(h) g

h

0 a U

Let h be the height of fluid level in the barrel above the outlet, which has cross-sectional area a. If a A(h), then the flow can be treated as approximately steady.  dh dh Mass conservation: A = aU (with U > 0). So, if a A then U . − dt  dt  | | Bernoulli’s theorem: consider a streamline from the surface of the fluid to the outlet, 1 p + ρ u 2 + ρgz = const. 2 k k dh At z = 0: p = p and u = U; at z = h: p = p and u = . So, atm atm dt 1 1 dh2 p + ρU 2 = p + ρ + ρgh, atm 2 atm 2 dt  2 dh p dh U 2 = + 2gh U 2gh since U . ⇒ dt ⇒ ' dt  | | We could have guessed this result from conservation of energy ) with KE 0 and PE = ρgh at z = h 1 '1 ρU 2 ρgh. and KE = ρU 2 and PE = 0 at z = 0 ⇒ 2 ' 2 Example 6.3 (Siphon.) A technique for removing fluid from one vessel to another without pouring is to use a siphon tube. z B

L A 0 g

H

C 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 will continue 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 from the surface to the exit of the pipe.

At point A: p = patm, z = 0. We shall assume that the container’s cross-sectional area is much larger than that of the pipe. So, U 0 (from mass conservation; see example 6.2 A ' A dh/dt = aU). − At point C: p = p , z = H, u = U U. atm − c ≡ Bernoulli’s equation:

patm 1 patm 1 p + U 2 = + U 2 gH U 2gH. ρ 2 A ρ 2 − ⇒ ' 0 '

If B is the highest point:(U = U U from mass conservation) B C ≡ p 1 p 1 B + U 2 + gL = atm + U 2 gH p = p ρg(L + H) < p . ρ 2 ρ 2 − ⇒ B atm − atm

p 105 For p > 0, we need H + L < atm = 10m. B ρg ≈ 103 10 ×

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 p 1 u ω = where (x, t) = + u 2 g x. ∂t − × −∇H H ρ 2k k − ·

If the fluid flow is irrotational, i.e. if ω = u = 0, then u ω = 0 and u = φ; so, the ∇ × × ∇ equation above becomes ∂φ  + = 0, ∇ ∂t H ∂u ∂ φ ∂φ since = ∇ = . ∂t ∂t ∇ ∂t Thus, for irrotational flows, ∂φ ∂φ p 1 + = + + φ 2 g x f(t) (6.5) ∂t H ∂t ρ 2k∇ k − · ≡ is a function of time, independent of the position, x.

If, in addition, the flow is steady, p 1 = + φ 2 g x, (6.6) H ρ 2k∇ k − · is constant; has the same value on all streamlines. H 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 planar z fluid flow, (ur(r), uθ(r)) in cylindrical polar coor- Ω dinates, we wish to calculate the height of the free surface of the fluid near to the rod, h(r). We also assume that the solid rod is an impenetrable sur- face on which the fluid does not slip, so that the boundary conditions for the velocity field are g h(r)

ur = 0 and uθ = aΩ at r = a. r a From mass conservation, one has 1 d C u = (ru ) = 0 u (r) = , ∇ · r dr r ⇔ r r where C is a constant of integration. However, the boundary condition ur = C/a = 0 at r = a implies that C = 0. So, ur = 0 and the fluid motion is purely azimuthal. As we assume an irrotational flow, 1 d k u = (ru ) ˆez = 0 u (r) = , ∇ × r dr θ ⇔ θ r where k is an integration constant to be determined using the second boundary condition. At 2 r = a, uθ = k/a = aΩ which implies that k = a Ω. So, the fluid velocity near to the rod is

a2Ω u = 0 and u = . r θ r Notice that the velocity potential, function of θ, can be determined using

1 dφ a2Ω u = φ = φ(θ) = a2Ωθ. ∇ ⇒ r dθ r ⇒ By applying Bernoulli’s theorem for steady potential flows to the free surface (which is not a streamline, as streamlines are circles about the rod axis) we obtain,

patm 1 2 patm = + uθ(r) + gh(r) = + gh , H ρ 2 ρ ∞ | {z } | {z } near rod at large r where the constant pressure p = patm is the atmo- spheric pressure and lim h(r) = h . (Notice also r ∞ that u 1/r 0 as r→∞ .) 1/r2 θ ∝ → → ∞ ∝ Thus, the height of the free surface is h 4 2 h(r) ∞ 1 2 a Ω h(r) = h uθ(r) = h , (6.7) ∞ − 2g ∞ − 2gr2 which shows that the free surface dips as 1/r2 near to the rotating rod.

Alternatively, Euler’s equation could be solved directly (i.e. without involving Bernoulli’s theorem) as in 5.6 with an azimuthal flow, now potential, of the form u = a2Ω/r. We § θ 52 6.3 Drag force on a sphere can then explain the result (6.7) in terms of centripetal acceleration; since the fluid particles move in circles, there must be an inwards central force producing the necessary centripetal acceleration (i.e. balancing the centrifugal force). Indeed, from the radial component of the momentum equation, one has

u2 ∂p ∂p a4Ω2 ρ θ = = ρ . − r −∂r ⇒ ∂r r3 However, since the fluid is in vertical hydrostatic equilibrium, the pressure satisfies

∂p = ρg p(r, z) = p ρg(z h(r)). ∂z − ⇒ atm − − Hence, we have ∂p dh a4Ω2 a4Ω2 = ρg = ρ h(r) = h , ∂r dr r3 ⇒ ∞ − 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.

In 4.5.1 we obtained the velocity potential of a § uniform stream, Uˆez, past a stationary sphere of n r radius a, U a r  a3  φ(r, z) = Uz 1 + , θ ϕ 2(r2 + z2)3/2 z in cylindrical polar coordinates (r, θ, z). In spherical polar coordinates, (r, θ, ϕ), this velocity potential becomes

 a3  φ(r, θ) = U cos θ r + . (6.8) 2r2

The non-zero components of the fluid velocity, u = φ, are then ∇ ∂φ  a3  1 ∂φ  a3  u = = U cos θ 1 and u = = U sin θ 1 + . (6.9) r ∂r − r3 θ r ∂θ − 2r3

Hence, at r = a, on the solid sphere’s surface, ur = 0 as required by the kinematic boundary conditions and 3 u (θ) = U sin θ. θ |r=a −2 To express the pressure force on the sphere in terms of the fluid velocity, we use Bernoulli’s theorem for steady potential flows, = p/ρ + u 2/2 = constant, ignoring gravity. At r = a H k k the fluid pressure, p(θ), therefore satisfies

p(θ) 1 2 p 1 2 + u = ∞ + U , ρ 2 θ r=a ρ 2 where p is the pressure as r . ∞ → ∞ Chapter 6 – Bernoulli’s equation 53

Thus, the pressure distribution on the sphere is 1  9  p(θ) = p + ρU 2 1 sin2 θ , (6.10) ∞ 2 − 4 and the total pressure force is the surface integral of p(θ) on the sphere r = a, Z Z π Z 2π 2 F = p n dS = p(θ) ˆer a sin θ dϕdθ, (6.11) − S − 0 0 where ˆer = sin θ cos ϕ ˆex + sin θ sin ϕ ˆey + cos θ ˆez. As the flow is axisymmetric, the only non-zero component of the force should be in the axial direction, z. Indeed, Z 2π Z π 2 2 Fx = F ˆex = a cos ϕ dϕ p(θ) sin θ dθ = 0, · − 0 0 and Z 2π Z π 2 2 Fy = F ˆey = a sin ϕ dϕ p(θ) sin θ dθ = 0. · − 0 0 However, after substituting for p(θ) in Z π 2 Fz = F ˆez = 2πa p(θ) sin θ cos θ dθ, · − 0 we find that   Z π Z π  2 1 2 9 2 3 Fz = 2πa p + ρU sin θ cos θ dθ ρU sin θ cos θ dθ = 0, − ∞ 2 0 − 8 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 body moving 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 low p

high p S1 S2 high p

low p

The flow is clearly fore-aft symmetric (symmetry about z = 0); the front (S1) and the back 1 2 (S2) of the sphere are stagnation points at equal pressure, PS1 = PS2 = p + 2 ρU . At the 2 ∞ side, ur = 0 and uθ > 0, so from Bernoulli’s theorem, the pressure there is lower than at the stagnation points but it must have the same symmetry as the flow. Notice that, from Bernoulli’s theorem, the pressure does not depend on the direction of the flow, but on its speed u only. However,k k the real flow past a sphere is not symmetric and, as a consequence, the fluid exerts a 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,

1  9  p(θ) = p + ρU 2 1 sin2 θ , ∞ 2 − 4

2 reaches its minimum, pmin = p 5/8 ρU , at θ = π/2. So, the pressure gradient in the ∞ − ± direction of the flow, (u )p, is a positive from θ = 0 to θ = π/2 and negative beyond. ·∇ ± pmin (u )p< 0 (u )p> 0 ·∇ ·∇ U θ U pmax pmax

(u )p< 0 (u )p> 0 ·∇ pmin ·∇ An adverse pressure gradient,(u )p > 0 (i.e. pressure ·∇ increasing in the direction of the flow along the surface), is “bad news” and causes the flow to separate, leaving a turbulent wake behind the sphere. Very roughly one can estimate the pressure difference upstream and downstream as 1/2 ρU 2, so that the drag force F 1/2 ρU 2 A, where A is the cross-sectional ∝ × area. The ratio F C = (6.12) D 1 2 2 ρU A is called drag coefficient and depends, e.g., on the shape of 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 of • solid bodies. These can be reduced by using more “streamlined” shapes, that avoid di- verging streamlines (e.g., aerodynamic bike helmets (time trial cyclist), ships, aeroplanes and 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 flow out of a narrowing tube, opened to the atmosphere at both ends, where the exit is not much smaller than the cross-section (i.e. the fluid flow cannot be assumed steady). z

A(h)

h(t) g

a

Let A(z) be the smoothly varying cross-sectional area of the pipe at height z, such that A A as z and A(0) = a. → ∞ → ∞ We assume that the flow is potential and purely in the z-direction, u = ∂φ/∂z w. z ≡ By conservation of mass the volume flux, Q(t) = w(z, t)A(z), must be independent of height. − Hence, ∂φ Q(t) Z z dµ = w(z, t) = φ(z, t) = φ(0, t) Q(t) . ∂z −A(z) ⇒ − 0 A(µ) (Note that we could set φ(0, t) = 0 without loss of generality.) Applying Bernoulli’s theorem for potential flows, p/ρ + u 2/2 + ∂φ/∂t g x = F (t), at the free surface and the exit gives, k k − · p 1 Q2(t) d at z = 0, atm + + φ(0, t) = F (t), ρ 2 a2 dt p 1 dh2 d dQ Z h dz and at z = h, atm + + φ(0, t) + gh = F (t). ρ 2 dt dt − dt 0 A(z) Equating both expressions gives " # 1 dh2 Q2(t) dQ Z h dz 2 + gh = 0, 2 dt − a − dt 0 A(z) 1  A2(h) dh2 d2h Z h dz dh 1 2 + A(h) 2 + gh = 0 since Q(t) = A(h) . ⇔ 2 − a dt dt 0 A(z) − dt The fluid height, h(t), is then solution to the nonlinear second order ordinary differential equation  Z h dz  d2h 1  A2(h) dh2 A(h) 2 + 1 2 + gh = 0. (6.13) 0 A(z) dt 2 − a dt

Far from the exit this equation becomes approximately

 2  1 A 2 hh¨ + 1 ∞ h˙ + gh = 0, 2 − a2 since, as h , → ∞ Z h dz Z h dz h A(h) A and = . ∼ ∞ 0 A(z) ∼ 0 A A ∞ ∞ 56 6.5 Unsteady flows

Using the chain rule, h¨ = dh/˙ dt = dh˙ /dh dh/dt = h˙ dh˙ /dh, one finds

˙  2  ˙ 2  2  ˙ 2 dh 1 A 2 1 dh 1 A h hh˙ + 1 ∞ h˙ + gh = 0 + 1 ∞ + g = 0 dh 2 − a2 ⇔ 2 dh 2 − a2 h which can be written as a linear differential equation for Z = h˙ 2/2,

dZ  A2  Z + 1 ∞ + g = 0. dh − a2 h

6.5.2 Bubble oscillations The sound of a “babbling brook” is due to the oscillation (compression/expansion) of air bubbles entrained into the stream. The pitch of the sound depends on the size of the bubbles. da Consider a bubble of radius a(t); the velocity of the fluid at the bubble surface, u = a.˙ r dt ≡

a(t)

gas

liquid

We can model the oscillations of the bubble of air using a potential flow due to a point source/sink of fluid at the centre of the bubble,

k(t) ∂φ k φ(r, t) = u = = . − r ⇒ r ∂r r2 k The boundary condition at the bubble’s surface, r = a, is u = =a ˙. So, r a2 aa˙ 2 aa˙ 2 ∂φ aa¨ 2 aa˙ 2 k =aa ˙ 2 u = and φ = = 2 ⇒ r r2 − r ⇒ ∂t − r − r Applying Bernoulli’s theorem (ignoring gravity) as r , → ∞

p 1 2 ∂φ p + φ + = F (t) = ∞ (as r , φ 0 and u 0: the fluid is stationary). ρ 2k∇ k ∂t ρ → ∞ → k k → At the bubble’s surface,

p(a) 1 aa¨ 2 aa˙ 2 p(a) 3 + a˙ 2 2 = aa¨ a˙ 2 = F (t). ρ 2 − a − a ρ − − 2 Combining the two expressions above, one gets

p(a) p 3 2 − ∞ =aa ¨ + a˙ , (6.14) ρ 2 where p(a) is the fluid pressure at the bubble’s surface. Now, if the gas inside the bubble of mass m is subject to adiabatic changes, its equation of state is 3m p = Kργ where ρ = , g g g 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 of pressure pg = p(a) must be satisfied at the surface r = a(t).

Now, for a bubble in equilibrium, such that a = a0 anda ˙ =a ¨ = 0, equation (6.14) gives p = p and, imposing pressure continuity pg = p at r = a0, one gets ∞  γ  3 γ γ 3m 4πa0 pg = Kρg = K 3 = p K = p . 4πa0 ∞ ⇒ ∞ 3m

So, pressure continuity at the bubble’s surface r = a(t) implies

 3 γ  γ 3γ γ 4πa0 3m a0  p(a) = pg = Kρg = p = p . ∞ 3m 4πa3 ∞ a

Then, equation (6.14) becomes

3γ ! p a0 3 2 ∞ 1 =aa ¨ + a˙ . ρ a3γ − 2

For small amplitude oscillations about the equilibrium a(t) = a +(t) where  a , so that 0 | |  0 a˙ = ˙,a ¨ = ¨ anda ˙ 2 = ˙2 0; the nonlinear terms are negligible at first approximation. Thus, '   3γ p a0 p  ∞   ∞ a0¨ = 3γ 1 3γ , ρ  3γ   −  ' − ρ a a 1 +  0 0 a0 3γp ∞ ¨+ 2  = 0. ⇒ ρa0

3γp 1/2 ∞ The bubble undergo periodic small amplitude oscillations with frequency ω = 2 . ρa0 Note that the frequency scales with the inverse of the (mean) radius of the bubbles. E.g. for 5 3 3 γ = 3/2, p = 10 Pa and ρ = 10 kg m− , ∞ ω 1 r3γp f = = ∞ f a0 3 kHz mm. 2π 2πa0 ρ ⇒ × ' For bubbles of size a = 0.2 mm, f 15 kHz (G9). 0 ' 6.6 Acceleration of a sphere

We have already shown that a sphere moving with a steady velocity under a potential flow has 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

Ua3 φ = cos θ. − 2r2 (This flow satisfies the following boundary conditions: u = φ 0 as r together with ∇ → → ∞ ur = U cos θ ˆer at r = a.) 58 6.6 Acceleration of a sphere

Rather than calculating the pressure via Bernoulli’s theorem, we calculate the work done by the 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

1 Z 1 T = mU 2 + ρ ( φ)2 dV, 2 V 2 ∇   1 1 Z = mU 2 + ρ  (φ φ) φ 2φ dV, (using (fA) = A f + f A) 2 2 V ∇· ∇ − ∇|{z} ∇· · ∇ ∇· 0 1 1 Z = mU 2 + ρ φ φ n dS, by divergence theorem. 2 2 S ∇ ·

2 Here S is the surface of the sphere of radius a. So n = ˆer and dS = a sin θ dθ dϕ, such that − Z π 1 2 1 ∂φ 2 T = mU ρ φ r=a 2πa sin θ dθ, 2 − 2 0 | ∂r r=a 1 πa3 Z π = mU 2 + ρU 2 cos2 θ sin θ dθ, 2 2 0 1 πa3 Z π 1 Z π d cos3 θ 2 = mU 2 + ρU 2 since cos2 θ sin θ dθ = dθ = . 2 3 0 −3 0 dθ 3

1 2 So T = (m + M)U 2, where M = πa3ρ is called the added mass and represents the mass of 2 3 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,

dT dU FU = = (m + M)U . dt dt

Hence, the force required to accelerate the sphere is given by

dU F = (m + M) . dt

Thus, the acceleration of a bubble (mass m and radius a) rising under gravity (see 5.3 on § Archimedes theorem) satisfies

4 3 dU F = πa ρg mg = (2M m)g = (m + M) , z 3 |−{z } − dt | {z } weight buoyancy force buoyancy dU 2M m 4πa3ρ 3m = − g = − g. ⇒ dt M + m 2πa3ρ + 3m As mass density is much less for a gas than for a liquid, we can assume m M, so that mg  dU 2g. dt '

dz Alternatively: Consider a bubble of mass m rising under gravity with speed U = . dt Chapter 6 – Bernoulli’s equation 59

z

U

t At height z the potential energy is 4 V = mgz πa3ρgz . |{z} − 3 weight | {z } buoyancy

In absence of dissipative processes the total energy remains constant; hence, 1 4 T + V = (m + M)U 2 + mgz πa3ρgz = const. 2 − 3 Differentiating this expression with respect to time gives

dU  4  (m + M)U + m πa3ρ gU = 0, dt − 3 dU 2M m 4πa3ρ 3m = − g = − g. ⇒ dt M + m 2πa3ρ + 3m dU Again, for a bubble of gas in a liquid M m, so 2g; the bubble accelerates at twice  dt ' 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 simplified models of smooth steady flows in open channels, such as rivers and channels with weirs.

7.1 Conservation of mass

z ξ(x)

g h(x) w u Z(x) z = 0 x x + δx x Consider a channel of large width which can be approximated as a two dimensional flow in the (x, z)-plane and whose base lies on z = Z(x) (topography), with water of depth h(x). We define 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 ! Z ξ(x+δx) Z ξ(x) d Z ξ(x) ρ u dz ρ u dz = 0 ρ u dz = 0. Z(x+δx) − Z(x) ⇒ dx Z(x)

So the volume flux Z ξ(x) Q = u dz Z(x) 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. Recall that, for steady flows, p 1 (x, t) = + u 2 g x H ρ 2k k − · is constant along a streamline. p 1 As the free surface is a streamline, = atm + (u2 + w2) + gξ is independent of x. H ρ 2

Free surface (streamline) u

dl dξ dx ξ(x)

x  1  Let dl = dx be an infinitesimal line element along the free surface. Since u is parallel dξ/dx to streamlines u  1  dξ u dl = dx = 0 w = u . × w × dξ/dx ⇒ 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  dξ/dx to H(x, z) = z ξ(x) = 0. Hence, H = − must be perpendicular the isosurface − ∇ 1 H = 0, i.e. to the free surface. So u H = 0, leading again to · ∇ dξ dξ u + w = 0 w = u , − dx ⇒ 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 1 u2 + gξ = u2 + g(h + Z) = const. (7.2) 2 2 Equivalently, F 2 Z  gh + 1 + = const., (7.3) 2 h u where the Froude number F = has no dimension and determines how flows react to √gh 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 hump of height Z(x). We assume dZ/dx 1, so that the flow is approximately unidirectional (the | |  argument for the free surface holds for the base of the channel) and dξ/dx 1. | | 

H U h(x) Zmax z =0 Z(x) x Let H be the depth of water and U its velocity far upstream. Hence, the upstream Froude U number F = is given. √gH Conservation of mass gives H Q = u(x) h(x) = UH u = U , ⇒ h and Bernoulli’s equation

1 1 u2(x) + g(h(x) + Z(x)) = U 2 + gH. 2 2 H2 Substituting for u2 = U 2 leads to h2 1 H2 1 U 2 + g(h(x) + Z(x)) = U 2 + gH, 2 h2 2 which can be rearranged as an equation for Z,

H U 2  H2  Z = 1 + H h, 2 gH − h2 − Z F 2  H2  h = 1 + 1 , ⇒ H 2 − h2 − H U where F = is the upstream Froude number. This gives the relationship between the √gH height of the base of the channel, Z, and the depth of water, h, when F is fixed.

Z h For simplicity, let Z˜ = and h˜ = .(Z˜ and h˜ are nondimensional measures of the bump H H height and depth of water respectively.) Hence,

F 2  1  Z˜ = f(h˜) where f(h˜) = 1 + 1 h.˜ (7.4) 2 − h˜2 −

The function f(h˜) , both as h˜ 0 and as h˜ + . Its derivative is → −∞ → → ∞ df F 2 = 1, dh˜ h˜3 − 2 2 df 2/3 d f F = 0 at h˜c = F with = 3 < 0. ⇒ dh˜ dh˜2 − h˜4 64 7.3 Flow over a hump

2/3 2/3 So, f has a unique maximum at h˜c = F — or equivalently at hc = HF ; its values is given by

2 2 F  4/3 2/3 F 3 2/3 Z˜ = max (f) = f(h˜ ) = 1 F − + 1 F = + 1 F , c c 2 − − 2 − 2 1  2   = F 2/3 1 F 2/3 + 2 > 0. (7.5) 2 −

2/3 The function f reaches its maximum Z˜c at h˜c = F . (Note that Z˜c and h˜c are determined by the upstream Froude number, i.e. by the upstream properties of the fluid flow.)

Z˜

Z˜c

˜ ˜ ˜ h1 hc h2 h˜

The equation f(h˜) = 0 has two solutions, h˜1 and h˜2, 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 +h is given by F 2  1  ξ˜ = Z˜ + h˜ = 1 + 1 . 2 − h˜2

Thus ξ˜ > 1 if h˜ > 1 and ξ˜ < 1 if h˜ < 1; the free surface goes up when the depth of water h˜ increases 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, Z˜max < Z˜c.

If F > 1, then h˜ = 1 (since h˜ = F 2/3 > 1). • 1 c Z˜

Z˜c

Z˜max Supercritial branch

˜ 1 ˜ h hc

Starting from h˜ = 1, the depth of water first increases then decreases with Z˜ and returns to h˜ = 1 (i.e. h = H). This is called a supercritical flow Chapter 7 – Steady flows in open channels 65

U U H H

If F < 1, then h˜ = 1 (since h˜ = F 2/3 < 1). • 2 c Z˜

Z˜c ˜ Zmax Subcritical branch

˜ hc 1 h˜

Here, as Z˜ increases the water surface goes down and when Z˜ returns to zero, the depth of water returns to h˜ = 1 (i.e. h = H). This is called a subcritical flow.

U U H H

u2 Why these two different behaviours? The equation + g(h + Z) = const. expresses 2 the conservation of energy (kinetic energy, KE, and gravitational potential energy, PE). In order to flow over the hump, the fluid has two choices. (i) To increase its KE by reducing its PE, i.e. decreasing h; (ii) to increase its PE by reducing its KE, i.e. increasing h. The Froude number, 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 original height and speed after passing over the bump.

However, if the maximal height of the hump Zmax = Zc the flow can move across from one branch of the solution to the other. Stable transitions occur only in one direction, from subcritical to supercritical.

H U hc

Zmax = Zc 66 7.4 Critical flows

In this case, a subcritical flow upstream (F < 1) is transformed smoothly into a supercritical flow 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 ⇒ u2 U 2 H 3 H 3 f 2 = = = F 2 . gh gH h h

2/3 At the top of the bump, hc/H = F , so that f = 1. The local Froude number must be equal to 1 at the top of the bump. (Note that the local f is a continuous function of x, less than 1 upstream, greater than 1 downstream and equal to 1 at the top of the hump.)

Example 7.1 A flow along a uniform channel encounters a bump of height Z(x). Downstream the height of fluid h0 is a half of the height upstream. Find the upstream and downstream fluid velocities and the height of the bump.

Let U1 and U2 be the upstream and downstream fluid velocities respectively.

U1 2h0 h(x) Zmax U2 h0 Z(x) From conservation of mass,

Q = u(x)h(x) = 2h U = h U U = 2U ; 0 1 0 2 ⇒ 2 1 and from Bernoulli’s equation,

1 1 1 u2 + g(h + Z) = U 2 + 2gh = U 2 + gh . 2 2 1 0 2 2 0

Combining the two equations above leads to r 1 3 2 U 2 + 2gh = 2U 2 + gh U 2 = gh U = gh . 2 1 0 1 0 ⇒ 2 1 0 ⇒ 1 3 0

2 r 2 U1 1 2 The upstream Froude number F1 = = < 1 (subcritical flow); so, U2 = 2 gh0 and 2gh0 3 3 2 2 U2 8 the downstream Froude number F2 = = > 1 (supercritical flow). The flow is critical gh0 3 as it is smoothly transformed from subcritical upstream to supercritical downstream.

2 To find the height of the bump, Zmax(= Zc), for a critical flow, use f = 1 (local Froude number) at the top of the bump. We get Zmax from Bernoulli’s equation, using

2 2 uc 2 f = uc = ghc. ghc ⇒ Chapter 7 – Steady flows in open channels 67

Indeed 1 1 gh U 2 + 2gh = u2 + g(h + Z ) = c + g(h + Z ), 2 1 0 2 c c max 2 c max 1 3 2 gZ = U 2 + 2gh gh but U 2 = gh , ⇒ max 2 1 0 − 2 c 1 3 0 7 3 Z = h h . ⇒ max 3 0 − 2 c

2 Next, hc can be eliminated using mass conservation, uchc = 2U1h0, together with uc = ghc 2 and U 2 = gh : 1 3 0  1/3 2 2 3 2 2 8 3 8 2 2/3 u h = gh = 4U h = gh hc = h0 = h0 = 2h0F . c c c 1 0 3 0 ⇒ 3 31/3 1

So, the size of the hump (i.e. its maximal height) is

7  Z = Z = 32/3 h 0.253h . max c 3 − 0 ' 0

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).

Side view

H U h(x)

Top view

B U b(x)

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 1 u2(x) + gh(x) = U 2 + gH = const. 2 2 So,

u2 = U 2 + 2g(H h), (7.6) − U 2 + 2g(H h) b2h2 = U 2B2H2. ⇒ − 68 7.5 Flow through a constriction

Dividing by gH3B2,

 U 2  h  b2 h2 U 2 h2   h  B2 + 2 1 = F 2 + 2 1 = F 2 , gH − H B2 H2 gH ⇒ H2 − H b2

U where F = is the upstream Froude number. √gH h b Again, let us make use of the nondimensional variables h˜ = and ˜b = : H B

h˜2 h  i 1 F 2 + 2 1 h˜ = . F 2 − ˜b2 Now, let 1 h˜2 h  i 1 K(h˜) where K(h˜) = F 2 + 2 1 h˜ 1. (7.7) ˜b2 − ≡ F 2 − − 2 The function K h˜3 as h˜ + and K(0) = 1. Its derivative ∼ −F 2 → −∞ → ∞ −

dK h˜   = 2 F 2 + 2 3h˜ = 0 dh˜ F 2 − F 2 + 2 if either h˜ = 0 or h˜ h˜ = . Furthermore, ≡ c 3 d2K 2   = F 2 + 2 6h˜ ; dh˜2 F 2 − d2K 2F 2 + 4 so, K has a local minimum at h˜ = 0 since = > 0 at h˜ = 0; and K has a local dh˜2 F 2 d2K 2F 2 + 4 maximum at h˜ = h˜c since = < 0 at h˜ = h˜c. dh˜2 − F 2 K 2 ˜b− 1 c − supercritical subcritical branch branch

˜ ˜ ˜ ˜ h1 hc h2 h

1 − h The equation K(h˜) = 0 has two solutions, h˜ and h˜ , one of which is h˜ = = 1. Note also 1 2 H F 2 + 2 that h˜ = > 1 if F > 1 and h˜ < 1 if F < 1. c 3 c

If F > 1 (supercritical), then h˜ = 1 and h˜ = h/H increases as ˜b = b/B decreases (i.e. • 1 K increases). The height of the free surface rises through the constriction.

If F < 1 (subcritical), then h˜ = 1 and h˜ = h/H decreases as ˜b = b/B decreases. • 2 Chapter 7 – Steady flows in open channels 69

Once again, a smooth transition from a subcritical flow to a supercritical flow can occur if the narrowest point in the constriction reaches a critical breadth, ˜bc, defined such that   ˜2 1 1 hc h 2  i 1 = max 1 = K(h˜c) = F + 2 1 h˜c 1, ˜2 ˜2 2 bc − b − F − − 2 2 F + 2 2 u where h˜c = . In this case the local Froude number f = satisfies 3 ghc U 2 + 2g(H h ) U 2 + 2g(H h ) H f 2 = − c = − c , ghc gH hc using equation (7.6). So,

 U 2 h  H  h  H f 2 = + 2 2 c = F 2 + 2 2 c , gH − H hc − H hc  F 2 + 2 1 F 2/3 + 2/3 = F 2 + 2 2 = = 1. − 3 F 2/3 + 2/3 F 2/3 + 2/3 Thus, the local Froude number f = 1 at the narrowest point in the constriction for a critical flow.

7.6 Transition caused by a sluice gate

Another way to generate a transition between a subcritical and a supercritical flow is via a sluice gate.

U1 H1

U2 H2

Using conservation of mass and Bernoulli’s equation for the free surface we obtain 1 1 U H = U H and U 2 + gH = U 2 + gH , 1 1 2 2 2 1 1 2 2 2 1 2
1 2
H2 dividing by gH1 F1 + 2 = F2 + 2 . ⇒ 2 2 H1 Then, since 2 2 2 3 2 H2 U1 F1 H1 H2 F1 2 = 2 = 2 3 = 2 , H1 U2 F2 H2 ⇒ H1 F2 one has 2 2
3 2
3 2
3 2
3 F1 F1 + 2 F2 + 2 F1 + 2 = F2 + 2 2 2 = 2 . F2 ⇔ F1 F2 f 2 + 2
3 Thus, G(F ) = G(F ) where G(f) = . The function G(f) + as f 0 and as 1 2 f 2 → ∞ → f . Its derivative → ∞ 2 2
2 2
3 dG 6f f + 2 2 f + 2 2 = − = 0 f 2 + 2
6f 2 2f 2 4
= 0 f 2 = 1. df f 3 ⇒ − − ⇒ 70 7.6 Transition caused by a sluice gate

G

C

f < 1 1 f > 1 f f 2 + 2
3 The equation G(f) = = C (where C > 27) has two solutions, one corresponding to f 2 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 so × requires 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 can be 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 force acting on a solid body placed in a potential flow. y

U pT x

pB For a flat aerofoil, the force in the upward vertical direction will be the difference between pressure forces on the bottom and on the top of the aerofoil. This force per unit length is Z F = (p p ) dx. B − T

p 1 p 1 Using Bernoulli’s theorem, B + u2 = T + u2 , ρ 2 B ρ 2 T

ρ Z ρ Z F = (u2 u2 ) dx = (u + u )(u u ) dx. 2 T − B 2 T B T − B

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 Z I uT + uB 2U F ρU (uT uB) dx = ρU u dl, ' ⇒ ' − − C · where C is the curve around the aerofoil.

C Thus, the force acting on the aerofoil, I F = ρUΓ where Γ = u dl, − C · 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 flow theory. 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 I F = ρUΓ where Γ = u dl. (8.1) − C · For a flow around a flat plate, Γ = πUL sin α − L (using conformal mapping). α Nose up (α ) leads to more lift and nose down % (α ) to less lift. & 8.3 Lift produced by a spinning cylinder

 a2  Γ φ = U r + cos θ + θ, Low velocity r 2π High pressure Flow direction Top spin  a2  ur = U 1 2 cos θ, − r High velocity F Γ  a2  L Low pressure u = U 1 + sin θ. θ 2πr − r2

Γ Z 2π On r = a, ur = 0 and uθ = 2U sin θ; the circulation is Γ = a uθ r=a dθ. 2πa − 0 | So,  2  2 Γ Γ ΓU u2 = 2U sin θ = 2 sin θ + 4U 2 sin2 θ. θ r=a 2πa − 2πa − πa The pressure at the cylinder surface can now be calculated using Bernoulli’s theorem, 1 ρΓ2 ρΓU p = p + ρU 2 + sin θ 2ρU 2 sin2 θ. ∞ 2 − 8π2a2 πa − Chapter 8 – Lift forces 73

The pressure force per unit length

I Z 2π F = p n dl = p n a dθ, where n = ˆer = cos θ ˆex + sin θ ˆey. − 0 −

The force can be decomposed into its components parallel and perpendicular to the free stream velocity (in the x direction): F = F ˆex +F ˆey, with F = 0 (no drag force) and the lift force F = ρΓU (Kutta-Joukowski theorem).k ⊥ This is calledk 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 the § absence of viscosity. Thus the flow should remain vortex free. (Recall that the circulation is equal 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 that downstream separation occurs at the trailing edge.

This vortex, called the starting vortex, remains behind on the runway. Its circulation is equal and 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 sin α). This is true up to a point: if − the angle is too steep, the flow separates so the drag force on the aeroplane increases signifi- cantly and it partially looses its lift force. This is 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

pB

Since pB > pT , there is a pressure gradient driving a flow around the edge of the wing. This leads to a vortex from the edge of the wing. Trailing vortex

Starting vortex

Trailing vortex These trailing vortices are parts of a single vortex tube formed by the wings, the trailing vortices and the starting vortex. (Vortex tubes must be closed as they cannot start or end in an 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 vector when it is determined by its magnitude and direction. It is crucial to distinguish vectors from scalars; standard notations for vectors include ~u u u. A unit vector, commonly denoted ≡ ≡ by ˆı ˆı, has magnitude one. The coordinates of a vector a are the scalars a , a and a such ≡ 1 2 3 that   a1 a = a1e1 + a2e2 + a3e3 (a1, a2, a3) a2 , ≡ ≡ a3 in the basis e1, e2, e3 . If ˆe1, ˆe2, ˆe3 are orthonormal (i.e. ˆei ˆej = δij) the magnitude of { p } { } · the vector is a = a2 + a2 + a2. | | 1 2 3 Scalar or dot product: a b = a b cos θ = a b + a b + a b · | || | 1 1 2 2 3 3 is a scalar.

Vector or cross product:

a b = (a b a b )e1 + (a b a b )e2 + (a b a b )e3. × 2 3 − 3 2 3 1 − 1 3 1 2 − 2 1 is a vector with magnitude a b sin θ and a direction perpendicular to both vectors a and b | || | in 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 may appear once or twice in any term in an equation; a suffix that appears just once is called a free 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 and 3. Hence, free suffices must be the same on both sides of an equation whereas the names of dummy suffices are not important (e.g. aibick = ajbjck).

Tensors used in suffix notation. Kronecker Delta: ( 1 0 0 1 if i = j,
δij = δij = 0 1 0 . 0 if i = j, ⇔ 6 0 0 1 The Kronecker Delta is symmetric, δij = δji, and δijaj = ai. Alternating Tensor:  0 if any of i, j or k are equal;  ijk = 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,  =  , and it is invariant under cyclic per- ijk − jik mutations of the indices, ijk = jki = kij. The tensors δ and  are related to each other by   = δ δ δ δ . ij ijk ijk klm il jm − im jl Examples. Standard algebraic operations on vectors can be written in a compact form using the suffix notation. A scalar product can be written as a b = a b + a b + a b = a b and the ith component · 1 1 2 2 3 3 j j of a vector product as (a b) =  a b . × i ijk j k 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, where x = (x , x , x ) (x, y, z) are Cartesian coordinates in the orthonormal basis ˆe1, ˆe2, ˆe3 1 2 3 ≡ { } ≡ ˆı, ˆ, kˆ . We also define the vector differential operator { }  ∂ ∂ ∂  , , . ∇ ≡ ∂x1 ∂x2 ∂x3 ( is pronounced grad, nabla or del.) ∇ The gradient of a scalar field f(x , x , x ) is given by the vector field • 1 2 3  ∂f ∂f ∂f  ∂f ∂f ∂f grad f f = , , = ˆe1 + ˆe2 + ˆe3. ≡ ∇ ∂x1 ∂x2 ∂x3 ∂x1 ∂x2 ∂x3 f is the vector field with a direction perpendicular to the isosurfaces of f with a ∇ magnitude 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 • ∂F ∂F ∂F div F = F = 1 + 2 + 3 . ∇ · ∂x1 ∂x2 ∂x3 A vector field F is solenoidal if F = 0 everywhere. ∇ · The curl of a vector field F is given by the vector field • ∂F ∂F ∂F ∂F ∂F ∂F  curl F = F = 3 2 , 1 3 , 2 1 , ∇ × ∂x2 − ∂x3 ∂x3 − ∂x1 ∂x1 − ∂x2       ∂F3 ∂F2 ∂F1 ∂F3 ∂F2 ∂F1 = ˆe1 + ˆe2 + ˆe3, ∂x2 − ∂x3 ∂x3 − ∂x1 ∂x1 − ∂x2

ˆe ˆe ˆe 1 2 3 ∂ ∂ ∂ = . ∂x ∂x ∂x 1 2 3 F1 F2 F3

A vector field F is irrotational if F = 0 everywhere. ∇ × The directional derivative (F ) is a differential operator which calculates the derivative • ·∇ of 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 ) f = F f = F1 + F2 + F3 . · ∇ · ∇ ∂x1 ∂x2 ∂x3

If ˆn is a unit vector, (ˆn )f gives the rate of change of f in the direction of ˆn. · ∇ The directional derivative of a vector field G is given by the vector field

(F )G = ((F ) G1, (F )G2, (F )G3) · ∇ · ∇ · ∇ · ∇ ∂G1 ∂G1 ∂G1 = F1 + F2 + F3 , ∂x1 ∂x2 ∂x3  ∂G2 ∂G2 ∂G2 ∂G3 ∂G3 ∂G3 F1 + F2 + F3 ,F1 + F2 + F3 . ∂x1 ∂x2 ∂x3 ∂x1 ∂x2 ∂x3

2 2 2 2 ∂ ∂ ∂ The Laplacian = 2 + 2 + 2 is a differential operator which can act on scalar • ∇ ∂x1 ∂x2 ∂x3 or vector fields: 2 2 2 2 ∂ f ∂ f ∂ f f = 2 + 2 + 2 ∇ ∂x1 ∂x2 ∂x3 is a scalar field and 2F = ( 2F , 2F , 2F ) ∇ ∇ 1 ∇ 2 ∇ 3 is a vector field.

The Lagrangian derivative of scalar and vector fields are • D d ∂ f(x, t) = f(x(t), t) = f(x, t) + (u )f(x, t) Dt dt ∂t · ∇ 78 A.3 Vector differentiation

and D d ∂ F(x, t) = F(x(t), t) = F(x, t) + (u )F(x, t), Dt dt ∂t · ∇

respectively, with u(x, t) = dx/dt and where acts upon the variables (x , x , x ) only. ∇ 1 2 3

dF dF dF dF  Notice that if F = F(t) then = 1 , 2 , 3 . dt dt dt dt

Differential operators in Cartesian coordinates using suffix notation

∂f (gradf)i = ( f)i = , ∇ ∂xi ∂F div F = F = j , ∇ · ∂xj ∂Fk (curl F)i = ( F)i = ijk , ∇ × ∂xj ∂f (F )f = Fj . · ∇ ∂xj

Notice that the operator ∂/∂xj cannot be moved around as it acts on everything that follows it.

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 = urˆer +uθˆeθ +uzˆez be scalar and vector fields respectively, functions of (r, θ, z).

∂p 1 ∂p ∂p p = ˆer + ˆeθ + ˆez, ∇ ∂r r ∂θ ∂z 1 ∂ 1 ∂u ∂u u = (ru ) + θ + z , ∇· r ∂r r r ∂θ ∂z       1 ∂uz ∂uθ ∂ur ∂uz 1 ∂ ∂ur u = ˆer + ˆeθ + (ru ) ˆez, ∇ × r ∂θ − ∂z ∂z − ∂r r ∂r θ − ∂θ ∂p u ∂p ∂p (u )p = u + θ + u , · ∇ r ∂r r ∂θ z ∂z 1 ∂  ∂p 1 ∂2p ∂2p 2p = r + + . ∇ r ∂r ∂r r2 ∂θ2 ∂z2 Chapter A – Vector calculus 79

Spherical polar coordinates

Let p and u = urˆer +uθˆeθ +uϕˆeϕ be scalar and vector fields respectively, functions of (r, θ, ϕ). ∂p 1 ∂p 1 ∂p p = ˆer + ˆeθ + ˆeϕ, ∇ ∂r r ∂θ r sin θ ∂ϕ 1 ∂ 1 ∂ 1 ∂u u = r2u
+ (u sin θ) + ϕ , ∇· r2 ∂r r r sin θ ∂θ θ r sin θ ∂ϕ     1 ∂ ∂uθ 1 1 ∂ur ∂ u = (u sin θ) ˆer + (ru ) ˆeθ ∇ × r sin θ ∂θ ϕ − ∂ϕ r sin θ ∂ϕ − ∂r ϕ   1 ∂ ∂ur + (ru ) ˆeϕ, r ∂r θ − ∂θ ∂p u ∂p u ∂p (u )p = u + θ + ϕ , · ∇ r ∂r r ∂θ r sin θ ∂ϕ 1 ∂  ∂p 1 ∂  ∂p 1 ∂2p 2p = r2 + sin θ + . ∇ r2 ∂r ∂r r2 sin θ ∂θ ∂θ r2 sin θ ∂ϕ2

A.3.3 Vector differential identities The simple vector identities that follow can all be proved with suffix notation. Note that these 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) , ∇ ∇ ∇· − ∇× ∇× 1 F ( F) = (F F) (F ) F. × ∇× 2∇ · − ·∇ A.4 Vector integral theorems

A.4.1 Alternative definitions of divergence and curl An alternative definition of divergence is given by • 1 ZZ F = lim F n dS, ∇ · δV 0 δV · → δS where δV is a small volume bounded by a surface δS which has a normal n, pointing outwards. 80 A.4 Vector integral theorems

An alternative definition of curl is given by • 1 I n F = lim F dx, · ∇ × δS 0 δS · → δC 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 contraction • there is in the field.

The curl of a vector field gives a measure of how much rotation or twist there is in the • field.

A.4.3 The Divergence and Stokes’ Theorems The divergence theorem states that • ZZZ ZZ F = F n dS, ∇ · · V S where S is the closed surface enclosing the volume V and n is the outward-pointing normal from the surface.

Stokes’ theorem states that • ZZ I ( F) n dS = F dx, ∇ × · · S C where C is the closed curve enclosing the open surface S and n is the normal from the surface.

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. Z ·Q iv. F dx is independent of the path of integration from P to Q. P · I v. F dr = 0 around every closed curve in the region. · C 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 = f(y)g(x), dx we can separate variables to find

Z 1 Z dy = g(x)dx, f(y) for which each side can now be integrated independently .

Linear equations

An equation of the form dy a(x) + b(x)y = f(x) dx can be integrated using an integrating factor. First put the equation into standard form by dividing through by a(x), dy b(x) f(x) + y = . dx a(x) a(x)

The integrating factor is then Z b(x) p(x) = exp dx a(x) which can be calculated. Multiply through by the integrating factor and if everything has been done correctly the equation can now be written

d p(x)g(x) (p(x) y) = dx 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 d2y dy a + b + cy = f(x), dx2 dx where a, b and c are constants can be solved using the complimentary function and particular integral method. First, consider the homogeneous equation by setting the RHS to zero:

d2y dy a + b + cy = 0. dx2 dx

λx Seeking solutions of the form y1 = e leads to the quadratic auxillary equation

aλ2 + bλ + c = 0.

λ1x λ2x So, y1 = Ae +Be where λ1 and λ2 are the roots of the quadratic and A and B constants. λx The solution becomes y1 = (A + Bx)e if λ is a double root. (Notice that solutions can be written 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 trying functions y2 that look like the desired RHS. The solution sought is the sum of the general solution to the homogeneous equation with the particular integral, y = y1 + y2.

Cauchy equations Equations of the general form

d2y dy ax2 + bx + cy = 0. dx2 dx can be solved by seeking solutions of the form y = xλ, giving, as above, an algebraic auxillary equation 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 the form,

dx = f(x, y, z, t), dt dy = g(x, y, z, t), dt dz = h(x, y, z, t). dt The method of solution depends upon how the equations are coupled. Chapter B – Ordinary differential equations 83

Example 1: • dx dy = x + y, = y. dt dt − Here the equation for dx/dt depends upon knowing y(t), however since dy/dt does not t contain x(t) we can be compute y(t) first. This has the general solution, y(t) = y0e− . We can now substitute this expression for y(t) to give

dx t = x + y e− . dt 0

t This is a linear equation so it can be solved using an integrating factor p(t) = e− to give d t
2t xe− = y e− , dt 0 so that y0 t t x(t) = e− + Ce . − 2 Example 2 • dx dy = 2x + y, = x. dt dt − Here, dx/dt depends on y(t) and dy/dt depends on x(t) therefore, we can’t solve either equation 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). Differentiating dx/dt = 2x + y with respect to t on both sides shows that

d2x dx dy = 2 + . dt2 dt dt Then, substituting dy/dt by its expression results in

d2x dx = 2 x. dt2 dt − This is a constant coefficient second-order differential equation and can be solved via an auxillary 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 then be solved to find dx y(t) = 2x = (A + At + B)et 2(At + B)et = (A B At)et. dt − − − −