fluids notes

Fluid DynamicsDr. J.R. Lister

Michlmas 1996

These notes are maintained by Paul Metcalfe.Comments and corrections to [email protected].

Revision: 1.9Date: 2004/07/26 07:30:23

The following people have maintained these notes.

date Paul Metcalfe

Contents

Introduction v

1 Kinematics 11.1 Continuum Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Flow Visualization . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3 Material Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.4 Conservation of Mass . . . . . . . . . . . . . . . . . . . . . . . . . . 31.5 Kinematic Boundary Condition . . . . . . . . . . . . . . . . . . . . . 31.6 Incompressible Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . 31.7 Streamfunctions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2 Dynamics 52.1 Surface and volume forces . . . . . . . . . . . . . . . . . . . . . . . 52.2 Momentum Equation . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.2.1 Applications of integral form . . . . . . . . . . . . . . . . . . 62.3 Bernoullis Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.3.1 Application . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.4 Vorticity and Circulation . . . . . . . . . . . . . . . . . . . . . . . . 8

2.4.1 Vorticity Equation . . . . . . . . . . . . . . . . . . . . . . . 82.4.2 Interpretation of vorticity . . . . . . . . . . . . . . . . . . . . 92.4.3 Ballerina effect and vortex line stretching . . . . . . . . . . . 92.4.4 Kelvins Circulation Theorem . . . . . . . . . . . . . . . . . 92.4.5 Irrotational flow remains so . . . . . . . . . . . . . . . . . . 10

3 Irrotational Flows 113.1 Velocity potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113.2 Some simple solutions . . . . . . . . . . . . . . . . . . . . . . . . . 113.3 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

3.3.1 Uniform flow past a sphere . . . . . . . . . . . . . . . . . . . 123.3.2 Uniform flow past a cylinder . . . . . . . . . . . . . . . . . . 13

3.4 Pressure in irrotational flow . . . . . . . . . . . . . . . . . . . . . . . 143.5 Bubbles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3.5.1 General theory for spherically symmetric motion . . . . . . . 153.5.2 Small oscillations of a gas bubble . . . . . . . . . . . . . . . 153.5.3 Total collapse of a void . . . . . . . . . . . . . . . . . . . . . 16

3.6 Translating Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . 163.6.1 Steady motion . . . . . . . . . . . . . . . . . . . . . . . . . 163.6.2 Effects of friction . . . . . . . . . . . . . . . . . . . . . . . . 16

iii

iv CONTENTS

3.6.3 Accelerating spheres . . . . . . . . . . . . . . . . . . . . . . 173.6.4 Kinetic Energy . . . . . . . . . . . . . . . . . . . . . . . . . 18

3.7 Translating cylinders with circulation . . . . . . . . . . . . . . . . . 183.7.1 Generation of circulation . . . . . . . . . . . . . . . . . . . . 19

3.8 More solutions to Laplaces equation . . . . . . . . . . . . . . . . . . 203.8.1 Image vortices and 2D vortex dynamics . . . . . . . . . . . . 203.8.2 Flow in corners . . . . . . . . . . . . . . . . . . . . . . . . . 21

4 Free Surface Flows 234.1 Governing equations . . . . . . . . . . . . . . . . . . . . . . . . . . 23

4.1.1 Particle paths under a wave . . . . . . . . . . . . . . . . . . . 244.2 Standing Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

4.2.1 Rayleigh-Taylor instability . . . . . . . . . . . . . . . . . . . 244.3 River Flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

4.3.1 Steady flow over a bump . . . . . . . . . . . . . . . . . . . . 254.3.2 Flow out of a lake over a broad weir . . . . . . . . . . . . . . 264.3.3 Hydraulic jumps . . . . . . . . . . . . . . . . . . . . . . . . 27

Introduction

These notes are based on the course Fluid Dynamics given by Dr. J.R. Lister inCambridge in the Michlmas Term 1996. These typeset notes are totally unconnectedwith Dr. Lister.

Other sets of notes are available for different courses. At the time of typing thesecourses were:

Probability Discrete MathematicsAnalysis Further AnalysisMethods Quantum MechanicsFluid Dynamics 1 Quadratic MathematicsGeometry Dynamics of D.E.sFoundations of QM ElectrodynamicsMethods of Math. Phys Fluid Dynamics 2Waves (etc.) Statistical PhysicsGeneral Relativity Dynamical SystemsCombinatorics Bifurcations in Nonlinear Convection

They may be downloaded from

http://www.istari.ucam.org/maths/.

v

vi INTRODUCTION

Chapter 1

Kinematics

1.1 Continuum FieldsEveryday experience suggests that at a macroscopic scale, liquids and gases look likesmooth continua with density (x, t), velocity u(x, t) and pressure p(x, t) fields.

Since fluids are made of molecules this is of course only an approximate descrip-tion. On large lengthscales we can define these fields by averaging over a volume Vsmaller than the scale of interest but large enough to contain many molecules. Theeffect of this averaging is to exchange an enormous number of ODEs that describe themotion of each molecule for a few PDEs that describe the averaged fields.

This continuum approximation is not always appropriate. For instance the veloc-ity structure about a spacecraft during re-entry has a lengthscale comparable with themolecular mean free path. Similarly, blood flow in capillaries must take the red bloodcells into account.

1.2 Flow VisualizationThere are many experimental techniques for obtaining a description of the velocity fieldu(x, t). Three simple visualisation techniques give rise to the ideas of streamlines,pathlines and streaklines. We will illustrate these ideas by application to the simpletwo-dimensional example u(x, t) = (t, y).

Streamlines are curves that are everywhere parallel to the instantaneous flow. Theyare visualised experimentally by the short-time exposure of many brightly-lit particles the streamlines are obtained by joining the resulting short segments in a manneranalogous to obtaining magnetic fields from iron filings.

Mathematically, a streamline is a curve x(s;x0, t) at a given fixed time t with svarying along the curve and passing through a given point x0 that satisfies

xs

= u(x, t) x(0;x0, t) = x0.

For our example we have

x(s;x0, t) = x0 + ts y(s;x0, t) = y0es.

1

2 CHAPTER 1. KINEMATICS

This gives the curvex x0t

= logy

y0.

Pathlines are particle paths: paths traversed by particles moving with the flow. Theyare visualised experimentally by the long-time exposure of a few brightly-lit particles.

Mathematically, a pathline is a curve x(t;x0, t0) corresponding to a particle re-leased from x = x0 at t = t0. The differential equation is

xt

= u(x, t) x(t0;x0, t0) = x0.

Our example gives

x(t;x0, t0) = x0 +t2 t20

2y(t;x0, t0) = y0ett0 .

For a particle released at t0 = 0 this gives a curve

y = y0e2(xx0).

Streaklines give the position at some fixed time of dye released over a range ofprevious times from a fixed source (e.g. an oil spill).

Mathematically, a streakline is a curve x(t0;x, t) with t0 varying along the curveand a fixed observation time t. To obtain it, we still solve

xt

= u(x, t) x(t0;x0, t0) = x0,

but then fix t instead of t0. Suppose we observe our flow at t = 0 we can useour previous solution to get a streakline

y = y0et0 = y0e2(x0x).

Note that for this unsteady flow we get different results from each method of visu-alisation. The different methods give the same result if the flow is steady.

1.3 Material DerivativeThis is a rate of change moving with the fluid. For any quantity F , the rate of changein that quantity seen by an observer moving with the fluid is the material or Lagrangianderivative DFDt .

F =F (x+ x, t+ t) F (x, t)

t

= (x )F + tFt

+ smaller terms.

HenceDFDt

= (u )F + Ft.

1.4. CONSERVATION OF MASS 3

1.4 Conservation of MassConsider an arbitrary (at least smooth this is applied maths) volume V , fixed in spacewith bounding surface A and outward normal n. The mass inside V is

M =V

dV,

and the mass changes due to the flow over the boundary, so

M

t=

A

u ndA.

Application of the divergence theorem givesV

tdV +

V

(u) dV = 0.Since V is arbitrarily small,

t+ (u) = 0,

or rewritten using the material derivative

DDt

+ u = 0.

1.5 Kinematic Boundary ConditionThis is an expression of mass conservation at a boundary. If the velocity of the bound-ary is uA, the condition for no mass flux is

(u(x, t) uA(x, t)) nAt = 0,

which gives that u n = uA n. For a fixed surface, uA = 0, so the surface is astreamline.

1.6 Incompressible FluidsFor this course, we restrict ourselves to fluids with = const. Mass conservationreduces to u = 0. Such a velocity field u is said to be solenoidal.

1.7 StreamfunctionsThis gives a representation of the flow satisfyingu = 0 automatically. For example,in 2D Cartesians, any velocity field u = (u, v, 0) is solenoidal if there exists (x, y, t)such that u = y and v = x .

In 2D polars, we want such that ur = 1r and u = r .

In axisymmetric cylindrical polars, we want such that uz = 1rr and ur =

1r z . is called a Stokes streamfunction.In axisymmetric spherical polars, we want such that ur = 1r2 sin

and u =1

r sin r .

4 CHAPTER 1. KINEMATICS

Chapter 2

Dynamics

2.1 Surface and volume forcesTwo types of force are considered to act on a fluid: those proportional to volume (e.g.gravity) and those proportional to area (e.g. pressure). This is a simplification appro-priate to the continuum level description e.g. surface forces in a gas are the averageresult of many molecules transferring momentum by collision with other moleculesover the very short distance of the mean free path.

Volume forcesWe denote the force on a small volume element V by FV (x, t)V . The volume forceis often conservative, with a potential energy per unit volume , so that FV = (or potential energy per unit mass , so that FV = .

The most common case is

FV (x, t)V = gV.

Surface forcesWe denote the force on a small surface element nA by FA(x, t,n)A, which dependson the orientation n of the surface element. A full description of surface forces includesthe effects of friction of layers of water sliding over each other or over rigid boundaries(viscosity).

Viscous effects are important when the Reynolds number

UL

1,

where U is a typical velocity, L a typical length and is the dynamic viscosity, whichis a property of the fluid.

In many cases fluids act as nearly frictionless and in this course we neglect frictionalforces completely. For a treatment of viscous fluids see the Fluid Dynamics 2 coursein Part IIB.

For inviscid (frictionless) fluids the surface force is simply perpendicular to thesurface with a magnitude independent of orientation:

5

6 CHAPTER 2. DYNAMICS

FA(x, t)A = p(x, t)nA,where p is the pressure. The minus sign is so that pressure is positive.

2.2 Momentum EquationConsider an arbitrary (at least smooth this is applied maths) volume V , fixed in spacewith bounding surface A and outward normal n. The momentum inside V is

V

udV,

and the momentum changes due to the flow over the boundary, surface forces andvolume forces, so

ddt

V

udV = A

u(u n) dA+A

pndA+V

FV dV, (2.1)

which is the momentum integral equation. Written in component form

ddt

V

ui dV = A

uiujnj dA+A

pni dA+V

FVi dV

uiuj is called the momentum flux tensor.V is fixed, so LHS is

Vuit dV , and using the divergence theorem on the RHS,

then letting V be arbitrarily small, we achieve the Euler momentum equation

(ut

+ (u )u)

= p+ FV . (2.2)

The associated dynamic boundary condition is that given forces are applied at theboundary (i.e. p is given).

2.2.1 Applications of integral formUncoiling of hosepipes

Assume a steady uniform flow U through a pipe of constant cross-section A. Neglectgravity. Now (2.1) becomes

walls+

ends(u(u n) + pn) dA = 0.

2.3. BERNOULLIS THEOREM 7

The integral over the walls iswalls

pndA = force on pipe,

since u n = 0 on the walls. The integral over the ends is

(U2 + p)A(2 1)

and so the force on the pipe is F = (U2 + p)A(1 2).

Pressure change at abrupt junction

Apply a momentum balance to the sketched shape. Neglect gravity, and also neglectthe time derivative, which is zero on average.

The momentum integral equation (2.1) becomes(u(u n) + pn) dA = 0.

The horizontal component gives

u21A1 + p1A1 = u22A2 + p2A2,

and mass conservation gives u1A1 = u2A2. Then we see that

p2 p1 = u21A1A2

(1 A1

A2

)> 0.

2.3 Bernoullis TheoremFor a steady flow (ut = 0) with potential forces (FV = ),

(u )u = (p+ ),


which can be written

(12u2 u ( u)

)= (p+ ). (2.3)

We define the vorticity = u and then let H = 12u2 + p + . ThenH = u. Now u H = 0, so H is constant on streamlines. This is Bernoullistheorem.

Note also that H = 0 and that H is constant on vortex lines.The constancy of H means that p is low at high speeds.

2.3.1 ApplicationConsider a water jet hitting an inclined plane.

Neglect gravity, so that on the surface streamline p = pa the speed is constant. Letthis speed be U .

Now apply the momentum integral equation (2.1) to getA

uu n+ (p pa) ndA = 0.

Now u n = 0 except at the ends. Mass conservation gives

aU = a1U + a2U.

Now balance the momentum parallel to the wall to get

aU2 cos = a2U2 a1U2.

Thusa1 =

1 + cos2

a and a2 =1 cos

2a.

Balancing the momentum perpendicular to the wall we get F = aU2 sin.

2.4 Vorticity and Circulation2.4.1 Vorticity EquationStart with the Euler momentum equation (2.2) with potential forces

(ut

+ (u )u)

= (p+ )

2.4. VORTICITY AND CIRCULATION 9

and take its curl to obtain

t= ( )u (u ) + u u.

Now u = 0 and = 0, so we obtain

t= ( )u (u )

orDDt

= ( )u. (2.4)

2.4.2 Interpretation of vorticityConsider a material line element (ie a line element moving with the fluid). Then in atime t, l l + (l )ut, which gives that lt = (l )u. Hence the tensoruixj

determines the local rate of deformation of line elements.

uixj

=12

(uixj

+xjui

)+

12

(uixj

xjui

)= eij +

12jikk.

The local motion due to eij is called the strain. The motion due to the second term12jikklj =

12 ( l) is rotation with angular velocity 12.

2.4.3 Ballerina effect and vortex line stretchingThe vorticity equation (2.4) can be interpreted as saying that vorticity changes just likethe rotation and stretching of material line elements. This is just the conservation ofangular momentum.

Consider a rotating fluid cylinder, initially with angular velocity 1, radius a1 andlength l1. Conservation of mass gives a21l1 = a22l2 and conservation of angular mo-mentum gives a41l11 = a42l22. These combine to give

12

=l1l2,

which says that vorticity increases as the fluid is stretched. This explains the bathtubvortex.

2.4.4 Kelvins Circulation TheoremAssume constant and FV = . Define the circulation C(t) around a closedmaterial curve (t) by

C(t) =(t)

u(x, t) dl.

ThenC(t)t

=(t)

DuDt

dl+ u ddtdl


since moves with the fluid. But from the momentum equation DuDt = p+ andddtdl = (dl )u. Hence

C(t)t

=(t)

((12u2 p+

)) dl

= 0 since is closed

So, for an inviscid fluid of constant density with potential forces, the circulationaround a closed material curve is constant.

2.4.5 Irrotational flow remains soA flow with = 0 is said to be irrotational. If = 0 everywhere at t = 0, then thevorticity equation (2.4) becomes DDt = 0, implying that = 0 for all times t 0.

This isnt quite true, vorticity can leave a stagnation point, especially at sharp trail-ing edges.

Chapter 3

Irrotational Flows

You will want to find a table of vector differential operators in various co-ordinatesystems. There is one in the back of Acheson.

3.1 Velocity potentialThe vorticity equation tells us that an initially irrotational flow remains so for all time.Sinceu = 0 for all time there exists a velocity potential (x, t) such that u = .Note both the + sign and that any function f(t) can be added to . Given u, we canfind from

= xx0

u(x, t) dl.

The result is independent of path since u = 0. Note that can be multivaluedin 2D if there are holes with

holeu dl 6= 0.

Mass conservation for an incompressible fluid reduces to = 0, and so if u = we have to solve 2 = 0.1

The kinematic boundary condition uA n = u n is

uA n = n n

.

Thus solving the Euler momentum equation (2.2) reduces to solving the more fa-miliar Laplaces equation with Neumann boundary conditions. The flow is only non-zero because of the applied boundary conditions on n (on bodies, surfaces and atinfinity).

3.2 Some simple solutionsFor simple geometries it is often possible to write down solutions of 2 = 0 as asum of separable solutions in suitable co-ordinate systems. See the Methods course fordetails.

1Hence irrotational flows are sometimes called potential or harmonic flows.

11

12 CHAPTER 3. IRROTATIONAL FLOWS

Cartesians = U x gives u = U. This is uniform flow with velocity U (e.g. flow in a

straight pipe). = (ekz or cosh kz or sinh kz) (ekx or cos kx or sin kx) gives a flow

which is periodic in x (e.g. waves).

Spherical polarsThe general axisymmetric solution of 2 = 0 in spherical polars is

=n0

(Anr

n +Bnrn1)Pn(cos ),

where the Pn are the Legendre polynomials. We will only use the first few modes.

= m4pir gives u = m4pi rr2 . This is a radial flow. The total outflow over a sphereof radius R is 4piR2ur = m, which is independent of R by mass conservation.This is a point source of strength m. (If m < 0 it is a point sink.)

= Ur cos Uz gives uniform flow again. r2 cos gives dipole flow.

Plane polarsThe general solution of 2 = 0 in plane polars is

= K +A0 log r +B0 +n1

(Anr

n +Bnrn)en.

We again use only the first few modes.

= m2pi log r gives u = m2pi rr . This is a radial flow, a line source of strength m.

= 2pi gives u = 2pi r . The circulation about a circle of radius R is , which isindependent of R by u = 0. This is a line vortex of circulation .

r1 cos is a 2D dipole. r2 cos x2 y2 is a 2D straining flow.

3.3 Applications3.3.1 Uniform flow past a sphereConsider a hard sphere of radius a in a fluid having a uniform velocity U at infinity.

We have to solve the equations

2 = 0 in r > a,

r= 0 at r = a,

and Ur cos as r .

3.3. APPLICATIONS 13

We can satisfy the Laplace equation and boundary condition at infinity with

= U cos (r +

B

r2

).

The boundary condition at r = a yields B = a3

2 . Thus

u =(U cos

(1 a

3

r3

),U sin

(1 +

a3

2r3

), 0)

in spherical polars (r, , ).

3.3.2 Uniform flow past a cylinderConsider a hard cylinder of radius a in a fluid with a uniform velocity U at infinity andwith a circulation .

We have to solve

2 = 0 in r > a,

r= 0 at r = a,

and Ur cos as r .


We further need r=a

u dl = = []r=a

to obtain a unique solution. These conditions give

= U cos (r +

a2

r

)+

2pi.

In plane polars (r, ), we therefore have

u =(U cos

(1 a

2

r2

),U sin

(1 +

a2

r2

)+

2pir

).

3.4 The pressure in irrotational potential flow with po-tential forces

The momentum and vorticity equations (2.2, 2.4) become

(t

+ (u )u)

= (p+ )

and(u )u =

(12u2)

respectively, and these combine to give

(

t+

12u2 + p+

)= 0,

which integrates to give

t+

12u2 + p+ = f(t) independent of x. (3.1)

Application

We can apply this theory to the free oscillations of a manometer. We need to calculate

(x, t) = xx0

u dl.

3.5. BUBBLES 15

Let s be the arc length from the bottom, with the equilibrium points at s = l1, l2.From mass conservation the flow is uniform, so that u = h everywhere. Therefore = hs, and so

t

x

= hs.

At the interfaces the pressure is constant. Using the equation for potential flow(3.1) we get(l1 + h)h+ 12h2 + pa gh sin = (l2 + h) + 12h2 + pa + gh sin.This simplifies to give

h =g(sin+ sin)

l1 + l2h,

and so we have SHM (even for large oscillations).

3.5 Bubbles3.5.1 General theory for spherically symmetric motionThe pressure is p(r, t), and the far-field pressure is p(, t). We have radial motion,u 1r2 . If the radius of the bubble is a(t), then a = ur at r = a, which gives thatu = rr2 aa

2 = . So = aa2r and tr= aa2+2a2ar . Putting all this together

gives

aa2 + 2a2ar

+

2a2a4

r4+ p(r, t) = p(, t).

At r = aaa 3

2a2 = p(, t) p(a, t),

orddt

(12a3a2

)= a2a (p(a, t) p(, t)) .

This can be interpreted as rate of change of kinetic energy equals rate of working bypressure forces.

Another rewrite gives

p(r, t) p(, t) = (p(a, t) p(, t)) ar+

12a2(a

r a

4

r4

)

3.5.2 Small oscillations of a gas bubbleAssume a(t) = a0 + a(t) and that a(t) is small, p is constant in time and that thegas in the bubble has pressure such that pgas = p 3aa0 (**which can be obtainedfrom PV constant for ideal adiabatic gas**). Neglect surface tension. Linearising

aa 32a2 = p(, t) p(a, t)

about a = a0 and a = 0 we obtain

a0a = 3paa0

,

which is SHM with =(3pa2o

) 12

.


3.5.3 Total collapse of a voidBernoulli implies that the pressure decreases as the speed increases. If this makes thepressure negative, then the liquid will break to form a void filled only with vapour. Thishas important consequences for valves and propellors (cavitation).

Consider a spherical void (ie p(a, t) = 0) at rest with a = a0 and a = 0 at t = 0with a constant background pressure p. Now

ddt

(12a3a2

)= a2a (p(a, t) p(, t)) ,

so12a3a2 =

13p

(a30 a3

).

Integrate this again (numerically!), to get

tcollapse = 0.92(a20p

) 12

.

3.6 Translating sphere & inertial reaction to accelera-tion

3.6.1 Steady motionUse the inertial frame moving steadily with the sphere. It was shown in 3.3.1 thatuniform flow past a fixed sphere has

= U cos (r +

a3

2r2

)and

u =(U cos

(1 a

3

r3

),U sin

(1 +

a3

2r3

), 0,

).

Hence at r = a, |u| = 32U sin . Potential flow means we can apply

t+

12u2 + p+ independent of x,

with t = 0 (since steady motion) to compare r = a and r =. This gives

p(a, ) = p +12U2

(1 9

4sin2

).

This pressure distribution is symmetric fore and aft and around the equator, sono force is exerted on a steadily moving sphere (**or indeed any 3D body**). Thissurprising result is called DAlemberts paradox.

3.6.2 ** Effects of friction **DAlemberts paradox can be understood only by analogy with Newtonian dynamics:in the absence of friction forces are needed only for acceleration and not for uniformmotion.

3.6. TRANSLATING SPHERE 17

This potential flow result is a good approximation for bubbles in steady motion,because they have slippery surfaces. It is a bad approximation for rigid spheres insteady motion. What is observed in experiments is separation and a wake.

The small amount of friction produces a thin layer of fluid next to the rigid surfacethat is slowed down from the potential flow. This thin boundary layer is sensitive to theslowing down of the surrounding potential flow from the equator to the rear stagnationpoint, and detaches from the sphere into the body of the fluid to produce a shear layerof concentrated vorticity (this is called separation). Separation also occurs behindother bluff bodies in steady translation but can be suppressed to a certain extent behindstreamlined/tapered bodies. Boundary layers are covered in more detail in Part IIB andPart III courses.

We can estimate the drag force as proportional to the projected area (A) times thepressure difference: applying Bernoulli gives

drag force = 12CDU2A,

where CD is a dimensionless coefficient that must be measured experimentally (0.4 fora sphere, 1.1 for a disc, 1.0 for a cylinder).

3.6.3 Accelerating spheresPotential flow is useful for slippery bubbles, rapid accelerations of small (rigid) parti-cles and small oscillations.

For the accelerating sphere, it is best to have the fluid at at rest. Consider asphere of radius a and centre x0(t) and velocity u(t) = x0(t). The velocity potentialproblem is then

2 = 0 in r a, 0 as r ,

r

t

= u(t) n on r = a.

with solution

= U cos a3

2r2

= u (x x0(t)) a3

2 |x x0(t)|3

To calculate the force, we need

t

r

=u ra3

2r3+ u (terms linear in x0) .


Now = terms linear in u. These linear terms must be the same as for steadymotion. Hence the pressure on the sphere is p = p + ur2 +

12U

2(1 94 sin2

).

The force on the sphere is given byr=a

pndA = 2

r=a

(u r) radA

Now by the isotropy of the sphere,r=a

rirj dA = 4pia4

3 ij , so

F = 12u

4pia4

31a= mu

where m = 12fluid4pia3

3 , the added (or effective or virtual) mass.

3.6.4 Kinetic EnergyThe kinetic energy of fluid motion in a volume V is

T =V

12u2 dV

=

2

V

()2 dV

=

2

V

() 2dV

=

2

S

() dS

=S

2 (n )dS

For the translating sphere we get T = 12mU2.

3.7 Translating cylinders with circulationThe potential for flow past a uniform cylinder with circulation is

= U cos (r +

a2

r

)+

2pi,

withu =

(U cos

(1 a

2

r2

),U sin

(1 +

a2

r2

)+

2pir

).

On r = a, |u| = 2 sin U + 2pia , thus

p(a, ) = p +12

(U2

( 2pia

2U sin )2)

and so (by doing the integral), F = (0,U) (in Cartesians). This is a lift forceperpendicular to the velocity. The same lift force applies to arbitrary aerofoils (at leastto the first approximation).

3.7. TRANSLATING CYLINDERS WITH CIRCULATION 19

3.7.1 ** Generation of circulation **

In order to calculate the lift on an aerofoil we thus need to know the circulation .Potential flow without circulation would have very large velocities around the sharptrailing edge.

This is strongly opposed by friction, and the flow avoids these high velocities byshedding an eddy as the aerofoil starts to move. This eddy is of such a size as tostreamline the flow on the aerofoil.

The circulation around C1 + C2 is initially zero and so remains zero at all times(by Kelvins theorem). Hence the circulation around the aerofoil (the bound vortex) isequal in magnitude to the circulation in the shed starting vortex (but opposite in sign).The strength is chosen to avoid a singularity at the sharp trailing edge, a conditionwhich gives pilU sin, provided < 14. For > 14 the flow separates andthe aerofoil stalls.2

Since = 0 vortex lines cannot end in the fluid, and in fact the bound vortexon the wing is connected to the starting vortex by two vortices shed from the wingtips.These are responsible for the observed vapour trails.

2See Fluids IIB or Acheson for more details.


3.8 More solutions to Laplaces equation

3.8.1 Image vortices and 2D vortex dynamicsRecall that a single line vortex has

=

2pi, with u =

2pir(y, x).

Since the potential flow equations are linear we can superpose solutions.

This configuration would give

=i

ii2pi

.

This gives us a new method of finding solutions in nice geometries. For instance,consider a vortex near a plane wall as shown.

The velocity field is the same as if there was an image vortex of opposite strengthso that normal velocities cancel. This image produces a velocity 4pid at the real vortex,and so the vortex trundles along parallel to the wall as shown.

3.8. MORE SOLUTIONS TO LAPLACES EQUATION 21

Application to dispersal of wingtip vortices

We have vortices at (x0(t), y0(t)) and so we need to add images at (x0(t),y0(t)).Now

x0 =

4pix20

y0(x20 + y20)

and y0 = 4piy20

x0(x20 + y20).

This gives dy0dx0 = y30x30

, or

1x20

+1y20

= C.

3.8.2 Flow in cornersWe use = r sin, which satisifies

= 0 on = pi

2.

Now |u| r1, and so there are infinite velocities as r 0 if < 1. The effectof friction here is to introduce a circulation.

Chapter 4

Free Surface Flows

4.1 Governing equationsThe flow is assumed to start from rest and is thus irrotational (and remains so). Let thefree surface be at (x, y, t), || sufficiently small for nice things to happen. So

2 = 0 z h,

t+

12 ||2 + p+ gz = f(t),

p = pair at z = ,

t+

x

x

z=

+

y

y

z=

=

z

z=

,

z

z=h

= 0.

We will restrict to the 1-D case but even so, to have any hope of solving this wehave to linearise it.

1. Throw out the nonlinear terms.

2. Evaluate at z = using information at z = 0.

3. Throw out new nonlinear terms.

2 = 0 0 z h,

t

z=0

+ g = f(t),

p = pair at z = ,

t=

z

z=0

,

z

z=h

= 0.

23

24 CHAPTER 4. FREE SURFACE FLOWS

We seek separable solutions and obtain (x, z, t) = A cosh k (z + h) ei(kxt).Using the dynamic boundary condition t

z=0

+ g = f(t), we obtain the disper-sion relation

2 = gk tanh kh.

In deep water, h 1k , so tanh kh 1, giving 2 = kh and c =

gk

1. In shallow

water, tanh kh kh giving = kgh and c = gh.

4.1.1 Particle paths under a waveBy linearising, we get that

x(t) = x0 ia cosh k (z0 + h)sinh kh ei(kx0t)

z(t) = z0 +a sinh k (z0 + h)

sinh khei(kx0t),

which is elliptic motion. x0 and z0 are the mean positions of the particles in the(x, z) plane.

In the deep water limit we get circular motion and in the shallow water limit we getmainly horizontal motion.

4.2 Standing WavesConsider waves in a deep rectangular box, 0 x a, 0 y b and z 0. Look forlinearised waves with displacement (x, y, t). Try separable solutions to obtain

(x, y, z, t) = A cosmpix

scos

npiy

bekzeit.

The Laplace equation determines k from m and n by

k2 = pi2(m2

a2+n2

b2

),

and the dynamic boundary conditions determine 2 = gk. This is to be expected, sincea standing wave is the sum of progressive waves.

4.2.1 Rayleigh-Taylor instabilityTurn the box in the previous section upside down (or equivalently replace g with g).This makes imaginary, leading to a e

gkt term, which quickly magnifies any devia-

tions from = 0. The largest growth rate is for large k, but this is cancelled by surfacetension in the real world.

4.3 River FlowsThese are nonlinear problems, but can be solved since the shallowness of the river incomparision to its length means that the flow is nearly unidirectional.

1The wave speed...

4.3. RIVER FLOWS 25

4.3.1 Steady flow over a bump

What happens to the free surface at the bump?We assume that

the river has vertical sides and constant width; it varies only slowly in the x direction, so that the flow is (to a good approxima-

tion) horizontal and uniform across any vertical section (since uz = y = 0); the flow is steady; far from the bump (x ), 0, h h and U U.

Then mass conservation gives

U( + h) = Uh, (4.1)

whilst applying Bernoullis equation (2.3) to the surface streamline gives12U2 + patm + g =

12U2 + patm. (4.2)

Now eliminate between (4.1) and (4.2) to obtain12U2 +

gUhU

=12U2 + gh. (4.3)

We can now extract information from (4.3) graphically.

There are thus two roots and three possibilities.

If the bump is too big then the assumption of steady/slowly varying flow mustfail. This gives a hydraulic jump see 4.3.3.


If the bump is just right then the flow can pass smoothly from one root the other(e.g. flow over a weir in 4.3.2).

If the bump is not too large then the flow stays on the same root. This has twosubcases.

On the left hand root: U

gh flow faster than shallow water

waves we have

bump up h RHS u .

i.e. fast shallow flow converts KE to PE to maintain the mass flux.

Normal rivers are in the slow deep state.

4.3.2 Flow out of a lake over a broad weirThis is the same as 4.3.1 but with a smooth change of branch. We need the bump justright.

How fast is the outflow as a function of the minimum of h(x)?The lake is large and deep, so we take the limit h , U 0 with Uh =

Q fixed. Mass conservation gives

U( + h) = Q

and Bernoulli on the surface streamline gives

12U2 + g = 0.

Eliminating as before gives

12U2 +

gQ

U= gh(x), (4.4)

which is a cubic for unknown U(x) given h(x).

4.3. RIVER FLOWS 27

The flow starts high on the LH branch in the lake, but comes out of the weir highon the RH branch. The minimum of h(x) (the crest of the weir) must be at the join ofthe branches, and so

Q =(

827gh3min

) 12

,

giving the flow rate as a function of the height of the weir.At the crest of the weir, U =

(23gh

3min) 12 and mass conservation gives

crest = 13hmin

and the fluid depth is 23hmin. Also, U2crest = g(fluid depth)crest, so that the flow is

travelling at the speed of shallow water waves at the crest. There can therefore be nocommunication from after the weir back to the lake.

4.3.3 Hydraulic jumps

Given U1 and h1 from river data and h2 from tidal theory can we predict the speed Vof the jump (and the flow U2)?

The energy loss in the turbulent jump makes Bernoulli inapplicable (friction is im-portant in the small-scale unsteady motions). However the flat-bottomed case providesan application of the momentum integral equation.

Change to a frame moving with the jump. Away from the jump there is no verticalacceleration and so the pressure is hydrostatic there.

Taking the horizontal component of


ddt

V

udV = A

(u(u n) + pn) dA+V

FV dV,

and noting that the flow is steady on average we obtain (from the area integral)

(V + U1)2h1 + (pah1 +12gh21) + pa(h2 h1)

= (V U2)2h2 + (pah2 + 12gh22),

which boils down to

(V + U1)2h1 (V U2)2h2 = 12g(h22 h21).

Mass conservation gives (V + U1)h1 = (V U2)h2, and eliminating V U2between these equations gives (eventually)

(V + U1)2 =g(h1 + h2)h2

2h1.

We can solve this for V and then solve

(V U2)2 = g(h1 + h2)h12h2for U2.

Note that if h2 > h1 then V U2

gh1 the jump

travels faster than shallow water waves on the river side and overtakes all informationof its future arrival.

References

D.J. Acheson, Elementary Fluid Dynamics, OUP, 1990.This is an excellent book, easy to read and with everything in. It is also good for FluidsIIB. Highly recommended.

G.K. Batchelor, An Introduction to Fluid Dynamics, CUP, 1967.The lecturer recommended this, and it is a reasonably good book for Fluids IIB. Per-sonally I think youd be wrong in your head to buy it for this course, but YMMV.

M. van Dyke, An Album of Fluid Motion, The Parabolic Press, 1982.Lots of pictures of flows. An excellent book. Go out and buy it. Now.

Related coursesIn Part IIA there are courses on Transport Processes and Theoretical Geophysics. TheIIB fluids courses are Fluid Dynamics 2 and Waves in Fluid and Solid Media, both ofwhich use the material in this course to some extent.

29

IntroductionKinematicsContinuum FieldsFlow VisualizationMaterial DerivativeConservation of MassKinematic Boundary ConditionIncompressible FluidsStreamfunctions

DynamicsSurface and volume forcesMomentum EquationApplications of integral form

Bernoulli's TheoremApplication

Vorticity and CirculationVorticity EquationInterpretation of vorticityBallerina effect and vortex line stretchingKelvin's Circulation TheoremIrrotational flow remains so

Irrotational FlowsVelocity potentialSome simple solutionsApplicationsUniform flow past a sphereUniform flow past a cylinder

Pressure in irrotational flowBubblesGeneral theory for spherically symmetric motionSmall oscillations of a gas bubbleTotal collapse of a void

Translating SphereSteady motionEffects of frictionAccelerating spheresKinetic Energy

Translating cylinders with circulationGeneration of circulation

More solutions to Laplace's equationImage vortices and 2D vortex dynamicsFlow in corners

Free Surface FlowsGoverning equationsParticle paths under a wave

Standing WavesRayleigh-Taylor instability

River FlowsSteady flow over a bumpFlow out of a lake over a broad weirHydraulic jumps

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fluids notes