fluid dynamics

INTRODUCTORY LECTURES ON FLUIDDYNAMICS

Roger K. Smith

Version: June 13, 2008

Contents

1 Introduction 31.1 Description of uid ow . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Equations for streamlines . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Distinctive properties of uids . . . . . . . . . . . . . . . . . . . . . . 51.4 Incompressible ows . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.5 Conservation of mass: the continuity equation . . . . . . . . . . . . . 7

2 Equation of motion: some preliminaries 92.1 Rate-of-change moving with the uid . . . . . . . . . . . . . . . . . . 102.2 Internal forces in a uid . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.2.1 Fluid and solids: pressure . . . . . . . . . . . . . . . . . . . . 142.2.2 Isotropy of pressure . . . . . . . . . . . . . . . . . . . . . . . . 152.2.3 Pressure gradient forces in a uid in macroscopic equilibrium . 152.2.4 Equilibrium of a horizontal element . . . . . . . . . . . . . . . 162.2.5 Equilibrium of a vertical element . . . . . . . . . . . . . . . . 172.2.6 Liquids and gases . . . . . . . . . . . . . . . . . . . . . . . . . 172.2.7 Archimedes Principle . . . . . . . . . . . . . . . . . . . . . . . 18

3 Equations of motion for an inviscid uid 223.1 Equations of motion for an incompressible viscous uid . . . . . . . . 233.2 Equations of motion in cylindrical polars . . . . . . . . . . . . . . . . 263.3 Dynamic pressure (or perturbation pressure) . . . . . . . . . . . . . . 273.4 Boundary conditions for uid ow . . . . . . . . . . . . . . . . . . . . 28

3.4.1 An alternative boundary condition . . . . . . . . . . . . . . . 29

4 Bernoullis equation 314.1 Application of Bernoullis equation . . . . . . . . . . . . . . . . . . . 32

5 The vorticity eld 375.1 The Helmholtz equation for vorticity . . . . . . . . . . . . . . . . . . 41

5.1.1 Physical signicance of the term ( )u . . . . . . . . . . . 425.2 Kelvins Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

5.2.1 Results following from Kelvins Theorem . . . . . . . . . . . . 455.3 Rotational and irrotational ow . . . . . . . . . . . . . . . . . . . . . 47

1

CONTENTS 2

5.3.1 Vortex sheets . . . . . . . . . . . . . . . . . . . . . . . . . . . 475.3.2 Line vortices . . . . . . . . . . . . . . . . . . . . . . . . . . . . 485.3.3 Motion started from rest impulsively . . . . . . . . . . . . . . 49

6 Two dimensional ow of a homogeneous, incompressible, invisciduid 50

7 Boundary layers in nonrotating uids 567.1 Blasius solution (U = constant) . . . . . . . . . . . . . . . . . . . . . 587.2 Further reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

Chapter 1

Introduction

These notes are intended to provide a survey of basic concepts in uid dynamicsas a preliminary to the study of dynamical meteorology. They are based on a moreextensive course of lectures prepared by Professor B. R. Morton of Monash University,Australia.

1.1 Description of uid ow

The description of a uid ow requires a specication or determination of the velocityeld, i.e. a specication of the uid velocity at every point in the region. In general,this will dene a vector eld of position and time, u = u(x, t).

Steady ow occurs when u is independent of time (i.e., u/t 0). Otherwisethe ow is unsteady.

Streamlines are lines which at a given instant are everywhere in the directionof the velocity (analogous to electric or magnetic eld lines). In steady ow thestreamlines are independent of time, but the velocity can vary in magnitude along astreamline (as in ow through a constriction in a pipe) - see Fig. 1.1.

Figure 1.1: Schematic diagram of ow through a constriction in a pipe.

3

CHAPTER 1. INTRODUCTION 4

Particle paths are lines traced out by marked particles as time evolves. Insteady ow particle paths are identical to streamlines; in unsteady ow they aredierent, and sometimes very dierent. Particle paths are visualized in the laboratoryusing small oating particles of the same density as the uid. Sometimes they arereferred to as trajectories.

Filament lines or streaklines are traced out over time by all particles passingthrough a given point; they may be visualized, for example, using a hypodermicneedle and releasing a slow stream of dye. In steady ow these are streamlines; inunsteady ow they are neither streamlines nor particle paths.

It should be emphasized that streamlines represent the velocity eld at a specicinstant of time, whereas particle paths and streaklines provide a representation ofthe velocity eld over a nite period of time. In the laboratory we can obtaina record of streamlines photographically by seeding the uid with small neutrallybuoyant particles that move with the ow and taking a short exposure (e.g. 0.1 sec),long enough for each particle to trace out a short segment of line; the eye readilylinks these segments into continuous streamlines. Particle paths and streaklines areobtained from a time exposure long enough for the particle or dye trace to traversethe region of observation.

1.2 Equations for streamlines

The streamline through the point P , say (x, y, z), has the direction of u = (u, v, w).

Figure 1.2: Velocity vector and streamline

Let Q be the neighbouring point (x+ x, y+ y, z+ z) on the streamline. Thenx ut, y vt, z wt and as t 0, we obtain the dierential relationship

dx

u=

dy

v=

dz

w, (1.1)

between the displacement dx along a streamline and the velocity components. Equa-tion (1.1) gives two dierential equations (why?). Alternatively, we can represent thestreamline parameterically (with time as parameter) as


dx

u=

dt,

dy

v=

dt,

dz

w=

dt, (1.2)

Example 1

Find the streamlines for the velocity eld u = (y, x, 0), where is a constant.

Solution

Eq. (1.1) gives

dxy

=dy

x=

dz

0.

The rst pair of ratios give (x dx + y dy) = 0

orx2 + y2 = (z),

where is an arbitrary function of z. The second pair givedz = 0 or z = constant.

Hence the streamlines are circles x2 + y2 = c2 in planes z = constant (we havereplaced (z), a constant when z is constant, by c2).

Note that the velocity at P with position vector x can be expressed as u = kxand corresponds with solid body rotation about the k axis with angular velocity .

1.3 Distinctive properties of uids

Although uids are molecular in nature, they can be treated as continuous mediafor most practical purposes, the exception being rareed gases. Real uids generallyshow some compressibility dened as

=1

d

dp=

change in density per unit change in pressure

density,

but at normal atmospheric ow speed, the compressibility of air is a relative bysmall eect and for liquids it is generally negligible. Note that sound waves owe theirexistence to compressibility eects as do supersonic bangs produced by aircraftying faster than sound. For many purposes it is accurate to assume that uids are


incompressible, i.e. they suer no change in density with pressure. For the presentwe shall assume also that they are homogeneous, i.e., density = constant.

When one solid body slides over another, frictional forces act between them toreduce the relative motion. Friction acts also when layers of uid ow over oneanother. When two solid bodies are in contact (more precisely when there is anormal force acting between them) at rest, there is a threshold tangential force belowwhich relative motion will not occur. It is called the limiting friction. An exampleis a solid body resting on a at surface under the action of gravity (see Fig. 1.3).

Figure 1.3: Forces acting on a rigid body at rest.

As T is increased from zero, F = T until T = N , where is the so-calledcoecient of limiting friction which depends on the degree of roughness between thesurface. For T > N , the body will overcome the frictional force and accelerate.A distinguishing characteristic of most uids in their inability to support tangentialstresses between layers without motion occurring; i.e. there is no analogue of limitingfriction. Exceptions are certain types of so-called visco-elastic uids such as paint.

Fluid friction is characterized by viscosity which is a measure of the magnitudeof tangential frictional forces in ows with velocity gradients. Viscous forces areimportant in many ows, but least important in ow past streamlined bodies. Weshall be concerned mainly with inviscid ows where friction is not important, but itis essential to acquire some idea of the sort of ow in which friction may be neglectedwithout completely misrepresenting the behaviour. The total neglect of friction isrisky!

To begin with we shall be concerned mainly with homogeneous, incompressibleinviscid ows.

1.4 Incompressible ows

Consider an element of uid bounded by a tube of streamlines, known as a streamtube. In steady ow, no uid can cross the walls of the stream tube (as they areeverywhere in the direction of ow).

Hence for incompressible uids the mass ux ( = mass ow per unit time) acrosssection 1 (= v1S1) is equal to that across section 2 (= v2S2), as there can be no


accumulation of uid between these sections. Hence vS = constant and in the limit,for stream tubes of small cross-section, vS = constant along an elementary streamtube.

vS = constant along an elementary stream tube.

Figure 1.4:

It follows that, where streamlines contract the velocity increases, where they ex-pand it decreases. Clearly, the streamline pattern contains a great deal of informationabout the velocity distribution.

All vector elds with the property that

(vector magnitude) (area of tube)remains constant along a tube are called solenoidal. The velocity eld for an incom-pressible uid is solenoidal.

1.5 Conservation of mass: the continuity equation

Apply the divergence theoremV

u dV =

S

u n ds

to an arbitrarily chosen volume V with closed surface S (Fig. 1.5).Let n be a unitoutward normal to an element of the surface ds u.c. If the uid is incompressibleand there are no mass sources or sinks within S, then there can be neither continuingaccumulation of uid within V nor continuing loss. It follows that the net ux ofuid across the surface S must be zero, i.e.,

S

u n dS = 0,

whereupon

V u dV = 0. This holds for an arbitrary volume V , and therefore

u = 0 throughout an incompressible ow without mass sources or sinks. This isthe continuity equation for a homogeneous, incompressible uid. It corresponds withmass conservation.


Figure 1.5:

Chapter 2

Equation of motion: somepreliminaries

The equation of motion is an expression of Newtons second law of motion:

mass acceleration = force.

To apply this law we must focus our attention on a particular element of uid,say the small rectangular element which at time t has vertex at P [= (x, y, z)] andedges of length x, y, z. The mass of this element is x y z, where is theuid density (or mass per unit volume), which we shall assume to be constant.

Figure 2.1: Conguration of a small rectangular element of uid.

The velocity in the uid, u = u(x, y, z, t) is a function both of position (x, y, z)and time t, and from this we must derive a formula for the acceleration of the elementof uid which is changing its position with time. Consider, for example, steady owthrough a constriction in a pipe (see Fig. 1.1). Elements of uid must accelerateinto the constriction as the streamlines close in and decelerate beyond as they openout again. Thus, in general, the acceleration of an element (i.e., the rate-of-changeof u with time for that element) includes a rate-of-change at a xed position u/t

9

CHAPTER 2. EQUATION OF MOTION: SOME PRELIMINARIES 10

Figure 2.2:

and in addition a change associated with its change of position with time. We derivean expression for the latter in section 2.1.

The forces acting on the uid element consist of:

(i) body forces, which are forces per unit mass acting throughout the uid becauseof external causes, such as the gravitational weight, and

(ii) contact forces acting across the surface of the element from adjacent elements.

These are discussed further in section 2.2.

2.1 Rate-of-change moving with the uid

We consider rst the rate-of-change of a scalar property, for example the temperatureof a uid, following a uid element. The temperature of a uid, T = T (x, y, z, t),comprises a scalar eld in which T will vary, in general, both with the position andwith time (as in the water in a kettle which is on the boil). Suppose that an elementof uid moves from the point P [= (x, y, z)] at time t to the neighbouring point Q attime t + t. Note that if we stay at a particular point (x0, y0, z0), then T (x0, y0, z0)is eectively a function of t only, but that if we move with the uid, T is a functionboth of position (x, y, z) and time t. It follows that the total change in T betweenP and Q in time t is

T = TQ TP = T (x + x, y + y, z + z, t + t) T (x, y, z, t),and hence the total rate-of-change of T moving with the uid is

limt0

T

t= lim

t0T (x + x, y + y, z + z, t + t) T (x, y, z, t)

t.

For small increments x,y,z,t, we may use a Taylor expansion

T (x + x, y + y, z + +z, t + t) T (x, y, z, t) +[T

t

]P

t +


[T

x

]P

x +

[T

y

]P

y +[T

z

]P

z + higher order terms in x, y, z, t .

Hence the rate-of-change moving with the uid element

= limt0

[T

tt +

T

xx +

T

yy +

T

zz

]/t

=T

t+ u

T

x+ v

T

y+ w

T

z,

since higher order terms 0 and u = dx/dt, v = dy/dt, w = dz/dt, where r = r(t) =(x(t), y(t), z(t)) is the coordinate vector of the moving uid element. To emphasizethat we mean the total rate-of-change moving with the uid we write

DT

Dt=

T

t+ u

T

x+ v

T

y+ w

T

z(2.1)

Here, T/t is the local rate-of-change with time at a xed position (x, y, z), while

uT

x+ v

T

y+ w

T

z= u T

is the advective rate-of-change associated with the movement of the uid element.Imagine that one is ying in an aeroplane that is moving with velocity c(t) =(dx/dt, dy/dt, dz/dt) and that one is measuring the air temperature with a ther-mometer mounted on the aeroplane. According to (2.1), if the air temperaturechanges both with space and time, the rate-of-change of temperature that we wouldmeasure from the aeroplane would be

dT

dt=

T

t+ c T. (2.2)

The rst term on the right-hand-side of (2.2) is just the rate at which the tem-perature is varying locally ; i.e., at a xed point in space. The second term is therate-of-change that we observe on account of our motion through a spatially-varyingtemperature eld. Suppose that we move through the air at a speed exactly equal tothe local ow speed u, i.e., we move with an air parcel. Then the rate-of-change ofany quantity related to the air parcel, for example its temperature or its x-componentof velocity, is given by

t+ u (2.3)

operating on the quantity in question. We call this the total derivative and oftenuse the notation D/Dt for the dierential operator (2.3). Thus the x-component ofacceleration of the uid parcel is


Du

Dt=

u

t+ u u, (2.4)

while the rate at which its potential temperature changes is expressed by

D

Dt=

t+ u . (2.5)

Consider, for example, the case of potential temperature. In many situations,this is conserved following a uid parcel, i.e.,

D

Dt= 0. (2.6)

In this case it follows from (2.5) and (2.6) that

t= u . (2.7)

This equation tells us that the rate-of-change of potential temperature at a point isdue entirely to advection, i.e., it occurs solely because uid parcels arriving at thepoint come from a place where the potential temperature is dierent.

For example, suppose that there is a uniform temperature gradient of 1 C/100km between Munich and Frankfurt, i.e., the air temperature in Frankfurt is cooler. Ifthe wind is blowing directly from Frankfurt to Munich, the air temperature in Munichwill fall steadily at a rate proportional to the wind speed and to the temperaturegradient. If the air temperature in Frankfurt is higher than in Munich, then thetemperature in Munich will rise. The former case is one of cold air advection (coldair moving towards a point); the latter is one of warm air advection.

Example 2

Show that

DF

Dt=

F

T+ (u ) F

represents the total rate-of-change of any vector eld F moving with the uid velocity(velocity eld u), and in particular that the acceleration (or total change in u movingwith the uid) is

Du

Dt=

u

t+ (u ) u.

Solution

The previous result for the rate-of-change of a scalar eld can be applied to each ofthe component of F, or to each of the velocity components (u, v, w) and these resultsfollow at once.


Example 3

Show thatDr

Dt= u.

Solution

Dr

Dt=

r

t+ (u ) r = 0 +

(u

x+ v

y+ w

z

)(x, y, z) = (u, v, w)

as x, y, z, t are independent variables.

2.2 Internal forces in a uid

An element of uid experiences contact or internal forces across its surface due tothe action of adjacent elements. These are in many respects similar to the normalreaction and tangential friction forces exerted between two rigid bodies, except, asnoted earlier, friction in uids is found to act only when the uid is in non-uniformmotion.

Figure 2.3: Forces on small surface element S in a uid.

Consider a region of uid divided into two parts by the (imaginary) surface S,and let S be a small element of S containing the point P and with region 1 belowand region 2 above S. Let (X, Y, Z) denote the force exerted on uid in region 1by uid region 2 across S.

This elementary force is the resultant (vector sum) of a set of contact forces actingacross S, in general it will not act through P ; alternatively, resolution of the forces


will yield a force (X, Y, Z) acting through P together with an elementary couple

with moment of magnitude on the order of (S)1/2 (X2 + Y 2 + Z2)1/2

.The main force per unit area exerted by uid 2 on uid 1 across S,[

X

S,Y

S,Z

S

]is called the mean stress. The limit as S 0 in such a way that it always containsP , if it exists, is the stress at P across S. Stress is a force per unit area. The stressF is generally inclined to the normal n to S at P , and varies both in magnitude anddirection as the orientation n of S is varied about the xed point P .

The stress F may be resolved into a normal reaction N , or tension, acting normalto S and shearing stress T , tangential to S, each per unit area.

Figure 2.4: The stress on a surface element S can be resolved into normal andtangential components.

Note that in the limit S 0 there is no resultant bending moment as

limS0

M

SlimS0

(S)1/2[(

X

S

)2+

(Y

S

)2+

(Z

S

)2]1/2= 0

provided that the stress is bounded.The stress and its reaction (exerted by uid in region 1 on uid in region 2) are

equal and opposite. This follows by considering the equilibrium of an innitesimalslice at P ; see Fig. 2.5.

2.2.1 Fluid and solids: pressure

If the stress in a material at rest is always normal to the measuring surface forall points P and surfaces S, the material is termed a uid ; otherwise it is a solid.Solids at rest sustain tangential stresses because of their elasticity, but simple uidsdo not possess this property. By assuming the material to be at rest we eliminatethe shearing stress due to internal friction. Many real uids conform closely to this


Figure 2.5: The stress and its reaction are equal and opposite.

denition including air and water, although there are more complex uids possessingboth viscosity and elasticity. A uid can be dened also as a material oering noinitial resistance to shear stress, although it is important to realize that frictionalshearing stresses appear as soon as motion begins, and even the smallest force willinitiate motion in a uid in time. The property of internal friction in a uid is knownas viscosity.

Although the term tension is usual in the theory of elasticity, in uid dynamicsthe term pressure is used to denote the hydrostatic stress, reversed in sign. In a uidat rest the stress acts normally outwards from a surface, whereas the pressure actsnormally inwards from the uid towards the surface.

2.2.2 Isotropy of pressure

The pressure at a point P in a continuous uid is isotropic; i.e., it is the same forall directions n. This is proved by considering the equilibrium of a small tetrahedralelement of uid with three faces normal to the coordinate axes and one slant face.The proof may be found in any text on uid mechanics.

2.2.3 Pressure gradient forces in a uid in macroscopic equi-

librium

Pressure is independent of direction at a point, but may vary from point to point ina uid. Consider the equilibrium of a thin cylindrical element of uid PQ of lengths and cross-section A, and with its ends normal to PQ. Resolve the forces in thedirection P for the uid at rest. Then pressure acts normally inwards on the curved


cylindrical surface and has no component in the direction of PQ (2.6). Thus theonly contributions are from the plane ends.

Figure 2.6: Pressure forces on a cylindrical element of uid.

The net force in the direction PQ due to the pressure thrusts on the surface ofthe element is

pA (p + p)A = ps

A s = ps

V,

where dV is the volume of the cylinder. In the limit s 0, A 0, the net pressurethrust (p/s) dV, or p/s = s p per unit volume of uid (s being aunit vector in the direction PQ). It follows that p is the pressure gradient forceper unit volume of uid, and n p is the component of pressure gradient forceper unit volume in the direction n.

Figure 2.7: A horizontal cylindrical element of uid in equilibrium.

2.2.4 Equilibrium of a horizontal element

The cylindrical element shown in Fig. 2.7 is in equilibrium under the action ofthe pressure over its surface and its weight. Resolving in the direction PQ, thex-direction, the only force arises from pressure acting on the ends


pA (p + s) A = Apx

x = px

V

and hence in equilibrium in the limit V 0,

px

= 0.

Alternatively, the horizontal component of pressure gradient force per unit volumeis i p = p/x = 0, from the assumption of equilibrium.

Thus p is independent of horizontal distance x, and is similarly independent ofhorizontal distance y. It follows that

p = p(z)

and surfaces of equal pressure (isobaric surfaces) are horizontal in a uid at rest.

2.2.5 Equilibrium of a vertical element

For a vertical cylindrical element at rest in equilibrium under the action of pressurethrusts and the weight of uid

k p V + g V = 0, where k = (0, 0, 1).

Thus 1 dpdz

= g, per unit volume, since p = p(z) only (otherwise we would write

p/z!). Hence = 1g

dpdz

is a function of z at most, i.e., = (z).

2.2.6 Liquids and gases

Liquids undergo little change in volume with pressure over a very large range ofpressures and it is frequently a good assumption to assume that = constant. Inthat case, the foregoing equation integrates to give

p = p0 + gz,

where p = p0 at the level z = 0.Ideal gases are such that pressure, density and temperature are related through

the ideal gas equation, p = RT , where T is the absolute temperature and R isthe specic gas constant. If a certain volume of gas is isothermal (i.e., has constanttemperature), then pressure and density vary exponentially with depth with a so-called e-folding scale H = RT/g (see Ex. 4).

1Here z measures downwards so that sgn(z) = sgn(p). Normally we take z upwards whereupondp/dz = g.


Figure 2.8: Equilibrium forces on a vertical cylindrical element of uid at rest.

2.2.7 Archimedes Principle

In a uid at rest the net pressure gradient force per unit volume acts verticallyupwards and is equal to dp/dz (when z points upwards) and the gravitational forceper unit volume is g. Hence, for equilibrium, dp/dz = g. Consider the vertically-oriented cylindrical element P1P2 of an immersed body which intersects the surfaceof the body to form surface elements S1 and S2. These surface elements havenormals n1, n2 inclined at angles 1, 2 to the vertical.

The net upward thrust on these small surfaces

= p2 cos 2 S2 p1 cos 1 S1 = (p2 p1) S,where S1cos1 = S2cos2 = S is the horizontal cross-sectional area of the cylinder.Since

p1 p2 = z1

z2

g dz ,

the net upward thrust

=

( z1z2

g dz

)S

= The weight of liquid displaced by the cylindrical element.

If this integration is now continued over the whole body we have Archimedes Principlewhich states that the resultant thrust on an immersed body has a magnitude equal


Figure 2.9: Pressure forces on an immersed body or uid volume.

to the weight of uid displaced and acts upward through the centre of mass of thedisplaced uid (provided that the gravitational eld is uniform).

Exercises

1. If you suck a drink up through a straw it is clear that you must accelerateuid particles and therefore must be creating forces on the uid particles nearthe bottom of the straw by the action of sucking. Give a concise, but carefuldiscussion of the forces acting on an element of uid just below the open endof the straw.

2. Show that the pressure at a point in a uid at rest is the same in all directions.

3. Show that the force per unit volume in the interior of homogeneous uid isp, and explain how to obtain from this the force in any specic direction.

4. Show that, in hydrostatic equilibrium, the pressure and density in an isothermalatmosphere vary with height according to the formulae

p(z) = p(0) exp(z/HS) ,(z) = (0) exp(z/HS) ,

where Hs = RT/g and z points vertically upwards. Show that for realisticvalues of T in the troposphere, the e-folding height scale is on the order of 8km.


5. A factory releases smoke continuously from a chimney and we suppose that thesmoke plume can be detected far down wind. On a particular day the windis initially from the south at 0900 h and then veers (turns clockwise) steadilyuntil it is from the west at 1100 h. Draw initial and nal streamlines at 0900and 1100 h, a particle path from 0900 h to 1100 h, and lament line from 0900to 1100 h.

6. Show that the streamline through the origin in the ow with uniform velocity(U, V,W ) is a straight line and nd its direction cosines.

7. Find streamlines for the velocity eld u = (x,y, 0), where is constant,and sketch them for the case > 0.

8. Show that the equation for a particle path in steady ow is determined by thedierential relationship

dx

u=

dy

v=

dz

w,

where u = (u, v, w) is the velocity at the point (x, y, z). What does thisrelationship represent in unsteady ow?

9. A stream is broad and shallow with width 8 m, mean depth 0.5 m, and meanspeed 1ms1. What is its volume ux (rate of ow per second) in m3 s1? Itenters a pool of mean depth 3 m and width 6 m: what then is its mean speed?It continues over a waterfall in a single column with mean speed 10ms1 at itsbase: what is the mean diameter of this column at the base of the waterfall?Will the diameter of the water column at the top of the waterfall be greater,equal to, or less at its base? Why?

10. Under what condition is the advective rate-of-change equal to the total rate-of-change?

11. Express u andu in Cartesian form and show that they are quite dierent,one being a scalar function and one a scalar dierential operator.

12. Some books use the expression df/dt. Would you identify this with Df/Dt orf/t in a eld f(x, y, z, t)?

13. The vector dierential operator del (or nabla) is dened as

[

x,

y,

z

]

in rectangular Cartesian coordinates. Express in full Cartesian form the quan-tities: u, u, u , and identify each.


14. Are the two x-components in rectangular Cartesian coordinates,

(u u)x and(1

2u2)x

the same or dierent? Note that (u )u = (12u)2u, where = u.

Chapter 3

Equations of motion for an invisciduid

The equation of motion for a uid follows from Newtons second law, i.e.,

mass acceleration = force.

If we apply the equation to a unit volume of uid:

(i) the mass of the element is kg m3;

(ii) the acceleration must be that following the uid element to take account bothof the change in velocity with time at a xed point and of the change in positionwithin the velocity eld at a xed time,

Du

Dt=

u

t+ (u ) u = u

t+ u

u

x+ v

u

y+ w

u

z;

(iii) the total force acting on the element (neglecting viscosity or uid friction)comprises the contact force acting across the surface of the element p perunit volume, which is a pressure gradient force arising from the dierence inpressure across the element, and any body forces F, acting throughout the uidincluding especially the gravitational weight per unit volume, gk.

The resulting equation of motion or momentum equation for inviscid uid ow,

Du

Dt= p + F , per unit volume,

oru

t+ (u ) u = 1

p + F , per unit mass,

22

CHAPTER 3. EQUATIONS OF MOTION FOR AN INVISCID FLUID 23

is known as Eulers equation. In rectangular Cartesian coordinates (x, y, z) withvelocity components (u, v, w) the component equations are

u

t+ u

u

x+ v

u

y+ w

u

z= 1

p

x+ X ,

v

t+ u

v

x+ v

v

y+ w

v

z= 1

p

y+ Y ,

w

t+ u

w

x+ v

w

y+ w

w

z= 1

p

z+ Z ,

where F = (X, Y, Z) is the external force per unit mass (or body force). These arethree partial dierential equations in the four dependent variables u, v, w, p and fourindependent variables x, y, z, t. For a complete system we require four equations inthe four variables, and the extra equation is the conservation of mass or continuityequation which for an incompressible uid has the form

u = 0, or u x

+ v

y+

w

z= 0.

3.1 Equations of motion for an incompressible vis-

cous uid

It can be shown that the viscous (frictional) forces in a uid may be expressed as2u = 2u where the coecient of viscosity and = / the kinematicviscosity provide a measure of the magnitude of the frictional forces in particularuid, i.e., and are properties of the uid and are relatively small in air or waterand large in glycerine or heavy oil. In a viscous uid the equation of motion for unitmass,

u

t+ (u ) u = 1

p + F + 2u

local accel-eration

advectiveaccelera-tion

pressuregradientforce

body force viscousforce

is known as the Navier-Stokes equation. We require also the continuity equation,

u = 0,to close the system of four dierential equations in four dependent variables. Thereis no equivalent to the continuity equation in either particle or rigid body mechanics,because in general mass is permanently associated with bodies. In uids, however,we must ensure that holes do not appear or that uid does not double up, and wedo this by requiring that u = 0, which implies that in the absence of sources


or sinks there can be no net ow either into or out of any closed surface. We mayregard this as a geometric condition on the ow of an incompressible uid. It is not,of course, satised by a compressible uid (c.f. a bicycle pump). We say that anyincompressible ow satisfying the continuity equation u = 0 is a kinematicallypossible motion.

The Navier-Stokes equation plus continuity equation are extremely important,but extremely dicult to solve. With possible further force terms on the right, theyrepresent the behaviour of gaseous stars, the ow of oceans and atmosphere, themotion of the earths mantle, blood ow, air ow in the lungs, many processes ofchemistry and chemical engineering, the ow of water in rivers and in the permeableearth, aerodynamics of aeroplanes, and so forth....

The diculty of solution, and there are probably no more than a dozen or sosolutions known for very simple geometries, arises from:

(i) the non-linear term (u )u as a result of which, if u1 and u2 are two solutionsof the equation, c1u1 + c2u2 (where c1 and c2 are constants) is in general not asolution, so that we lose one of our main methods of solution;

(ii) the viscous term, which is small relative to other terms except close to bound-aries, yet it contains the highest order derivatives(

2u/ x2 , 2u/ y2 , 2u/ z2),

and hence determines the number of spatial boundary conditions that must beimposed to determine a solution.

The Navier-Stokes equation is too dicult for us to handle at present and weshall concentrate on Eulers equation from which we can learn much about uidow. Eulers equation is still non-linear, but there are clever methods to bypass thisdiculty.

Example 4

Find the velocity eld u = (y,x, 0) for constant as a possible ow of anincompressible liquid in a uniform gravitational eld F g = (0, 0, g).

Solution

(i) This is a kinematically-possible steady incompressible ow, as u satises thecontinuity equation

u = u x

+ v

y+

w

z= 0 + 0 + 0 = 0.


(ii) We nd the corresponding pressure eld from Eulers equation.

u u = 1p + g.

If the given velocity eld is substituted in the Eulers equation and it is rear-ranged in component form,

p

x= 2x,

p

y= 2y,

p

z= g.

We may now solve these three equations as follows.

p

x[ p

x

]y,z constant

= 2x p = 122x2 + constant

where constant can include arbitrary functions of both y and z (Check: p/ x = 2x + 0). We continue in like manner with the other componentequations:

p

y= 2y p = 1

22y2 + g(z, x),

p

z= g p = gz + h(x, y),

where f(y, z), g(z, x) and h(x, y) are arbitrary functions. By comparison of thethree solutions we see that f(y, z) must incorporate 1

22y2 and gz and so

forth. Hence the full solution is

p =1

22(x2 + y2) gz + constant,

and we nd that this solution does in fact satisfy each of the component Eulerequations. On a free surface containing the origin O(x = 0, y = 0, z = 0), p =po the constant = po, where po is atmospheric pressure, and r2 = x2 + y2,

p = po +1

22r2 gz.

(iii) The equation for the free surface is now given by p = p0 over the whole liquidsurface, which therefore has equation

z =2

2 gr2 =

2

2gr2.


(iv) Streamlines in the ow are given by

dx

u=

dy

v=

dz

w, or

dx

y =dy

x=

dz

0,

yielding two relations

x dx +

y dy = 0 x2 + y2 = constant

dz = 0 z = constant

and streamlines are circles about the z-axis in planes z = constant. The velocityeld represents rigid body rotation of uid with angular velocity about theaxis Oz (imagine a tin of water on turntable!).

3.2 Equations of motion in cylindrical polars

Take the cylindrical polars (r, , z) and velocity (vr, v, vz). These are more com-plicated than rectangular Cartesians as vr, v change in direction with P (in factOP rotates about Oz with angular velocity v/r). Suppose that r, n, z are the unitvectors at P in the radial, azimuthal and axial directions, as sketched in Fig. 3.1.Then zis xed in direction (and, of course, magnitude) but r and n rotate in theplane z = 0 as P moves, and it follows that dz/dt = 0, but that

dr

dt= n,

dn

dt= (r) = r

Hence, as = v/r,

v = (vrr+ vn + vzz) ,

v = vr r+ vrdr

dt+ vn + v

dn

dt+ vzz = (vr vr/r) r+ (v + vrv/r) n+ vzz.

Recalling also that d/dt must be interpreted here as D/Dt, the acceleration is[DvrDt

v2

r,DvDt

+vrvr

,DvzDt

].

If we now write (u, v, w) in place of (vr, v, vz), Eulers equations in cylindricalpolar coordinates take the form


Figure 3.1: Velocity vectors and coordinate axes in cylindrical polar coordinates.

u

r+

1

r

v

+

z= 0

u

t+ u

u

r+

v

r

u

+ w

u

z v

2

r= 1

p

r+ Fr ,

v

t+ u

v

r+

v

r

v

+ w

v

z+

uv

r= 1

r

p

+ F ,

w

t+ u

w

r+

v

r

w

+ w

w

z= 1

p

z+ Fz .

3.3 Dynamic pressure (or perturbation pressure)

If in Eulers equation for an incompressible uid,

Du

Dt= 1

p + g, (3.1)

we put u = 0 to represent the equilibrium or rest state,


0 = 1p0 + g (3.2)

This is merely the hydrostatic equation

p0 = g or p0 x

= 0, p0 y

= 0, p0 z

= g,

where p0 is the hydrostatic pressure. Subtracting (3.1)-(3.2) we obtain

Du

Dt= 1

(p p0) = 1

pd

where pd = pp0 = (total pressure) - (hydrostatic pressure) is known as the dynamicpressure (or sometimes, especially in dynamical meteorology, the perturbation pres-sure). The dynamic pressure is the excess of total pressure over hydrostatic pressure,and is the only part of the pressure eld associated with motion.

We shall usually omit the sux d since it is fairly clear that if g is included weare using total pressure, and if no g appears we are using the dynamic pressure,

Du

Dt=

u

t+ (u ) u = 1

p.

3.4 Boundary conditions for uid ow

(i) Solid boundaries : there can be no normal component of velocity through theboundary. If friction is neglected there may be free slip along the boundary,but friction has the eect of slowing down uid near the boundary and itis observed experimentally that there is no relative motion at the boundary,either normal or tangential to the boundary. In uids with low viscosity, thistangential slowing down occurs in a thin boundary layer, and in a number ofimportant applications this boundary layer is so thin that it can be neglectedand we can say approximately that the uid slips at the surface; in many othercases the entire boundary layer separates from the boundary and the inviscidmodel is a very poor approximation. Thus, in an inviscid ow (also called theow of an ideal uid) the uid velocity must be tangential at a rigid body, and:

for a surface at rest n u = 0;for a surface with velocity us n (u us) = 0.

(ii) Free boundaries: at an interface between two uids (of which one might bewater and one air) the pressure must be continuous, or else there would bea nite force on an innitesimally small element of uid causing unboundedacceleration; and the component of velocity normal to the interface must becontinuous. If viscosity is neglected the two uids may slip over each other. Ifthere is liquid under air, we may take p = p0 = atmospheric pressure at theinterface, where p0 is taken as constant. If surface tension is important theremay be a pressure dierence across the curved interface.


3.4.1 An alternative boundary condition

As the velocity at a boundary of an inviscid uid must be wholly tangential, itfollows that a uid particle once at the surface must always remain at the surface.Hence for a surface or boundary with equation

F (x, y, z, t) = 0,

if the coordinates of a uid particle satisfy this equation at one instant, they mustsatisfy it always. Hence, moving with the uid at the boundary,

DF

Dt= 0

or F

t+ u F = 0,

as F must remain zero for all time for each particle at the surface.

Exercises

3.1 Describe briey the physical signicance of each term in the Euler equation forthe motion of an incompressible, inviscid uid,

u

t+ (u )u = p + g,

explaining clearly why the two terms on the left are needed to express the massacceleration fully. To what amount of uid does this equation apply?

3.2 The velocity components in an incompressible uid are

u = 2xyz(x2 + y2)2

, v =(x2 y2) z(x2 + y2)2

, w =y

x2 + y2.

Show that this velocity represents a kinematically possible ow (that is, thatthe equation of continuity is satised).

3.3 Find the pressure eld in the inviscid, incompressible ow with velocity eld

u = (nx,ny, 0).

3.4 If r, n are the unit radial and azimuthal vectors in cylindrical polars (r, , z)show that

dn

dt=

r


3.5 State the boundary conditions for velocity in an inviscid uid at (a) a stationaryrigid boundary bisecting the 0x, 0y axes; (b) a rigid boundary moving withvelocity V j in the direction of the y axis.

3.6 Write down Eulers equation for the motion of an inviscid uid in a gravitationaluniform eld: (i) in terms of the total pressure p, and (ii) in terms of thedynamic pressure pd. Relate p and pd.

3.7 Explain briey why DF/Dt = 0 provides an alternative form of the bound-ary condition for ow in a region of inviscid uid bounded by the surfaceF (x, y, z, t) = 0. Find the boundary condition on velocity at a xed planey + mx = 0 and show that the equation y = m(x + y Ut) represents a cer-tain inclined plane moving with the speed U in a certain direction. Find thisdirection and obtain the boundary condition at this plane.

Chapter 4

Bernoullis equation

For steady inviscid ow under external forces which have a potential such thatF = Eulers equation reduces to

u u = 1p,

and for an incompressible uid

u u+ 1(p + ) = 0.

We may regard p + p as a more general dynamic pressure; but for the particularcase of gravitation potential, = gzand F = = (0, 0, g) = gk.

We note that

u (u )u = u(u ) u + v(u ) v + w(u )w= u 1

2

(u2 + v2 + w2

)= (u ) 1

2u2,

using the fact that u is a scalar dierential operator. Hence,

u [u u+ (p/ + )] = u [12u2 + p/ +

]= 0,

and it follows that(

12u2 + p/ +

)is constant along each streamline (as u is

proportional to the rate-of-change in the direction u of streamlines). Thus for steady,incompressible, inviscid ow

(12u2 + p/ +

)is a constant on a streamline, although

the constant will generally be dierent on each dierent streamline.

31

CHAPTER 4. BERNOULLIS EQUATION 32

4.1 Application of Bernoullis equation

(i) Draining a reservoir through a small hole

If the draining opening is of much smaller cross-section than the reservoir (Fig.4.1), the water surface in the tank will fall very slowly and the ow may beregarded as approximately steady. We may take the outow speed uA as ap-proximately uniform across the jet and the pressure pA uniform across the jetand equal to the atmospheric pressure p0 outside the jet (for, if this were notso, there would be a dierence in pressure across the surface of the jet, andthis would accelerate the jet surface radially, which is not observed, althoughthe jet is accelerated downwards by its weight). Hence, on the streamline AB,

12u2A + p0/ =

12u2B + p0/ + gh,

and as uB


Figure 4.2: Flow round a blu body in this case a cylinder.

This provides the basis for the Pitot tube in which a pressure measurement isused to obtain the free stream velocity U0. The pressure p = p0 +

12U20 is

the total or Pitot pressure (also known as the total head) of the free stream,and diers from the static pressure p0 by the dynamic pressure

12U20 . The

Figure 4.3: Principal of a Pitot tube.

Pitot tube consists of a tube directed into the stream with a small centralhole connected to a manometer for measuring pressure dierence p p0 (Fig.4.3). At equilibrium there is no ow through the tube, and hence the left handpressure on the manometer is the total pressure p0+

12U20 . The static pressure

p0 can be obtained from a static tube which is normal to the ow.

The Pitot-static tube combines a Pitot tube and a static tube in a single head(Fig. (4.4). The dierence between Pitot pressure (p0 +

12U20 ) and static pres-

sure (p0) is the dynamic pressure12U20 , and the manometer reading therefore

provides a measure of the free stream velocity U0. The Pitot-static tube canalso be own in an aeroplane and used to determine the speed of the aeroplanethrough the air.

(iii) Venturi tube

This is a device for measuring uid velocity and discharge (Fig. 4.5). Supposethat there is a restriction of cross-sections in a pipe of cross-section S, with


Figure 4.4: A Pitot-static tube.

Figure 4.5: A Venturi tube.

velocities v, V and pressures p, P in the two sections, respectively, the pipebeing horizontal. Then

p

+

1

2v2 =

P

+

1

2V 2

or

v2 V 2 = 2(P p) = 2

m gh = 2gh

m

.


The dischargeQ = vs = V S,

and substitution gives [Q

s

]2[Q

S

]2= 2gh

m

,

i.e.,

Q =sS

S2 s2

2ghm

V =Q

S=

sS2 s2

2gh

m

.

Exercises

1. Hold two sheets of paper at A and B with a nger between the two at top andbottom, and blow between the sheets as illustrated in Fig. 4.6. The trailingedges of the sheets will not move apart as you might have anticipated, buttogether. Explain this in terms of Bernoullis equation, assuming the ow tobe steady.

Figure 4.6:

2. Explain why there is an increase in pressure on the side of a building facingthe wind.

3. A uniform straight open rectangular channel carries a water ow of mean speedU and depth h. The channel has a constriction which reduces its width by halfand it is observed that the depth of water in the constriction is only 1

2h. By

applying Bernoullis theorem to a surface streamline nd U in terms of g andh.

4. Using Bernoullis equation (often referred to as Bernoullis theorem):

(i) show that air from a balloon at excess pressure p1 above atmospheric willemerge with approximate speed

2p1/p;


(ii) nd the depth of water in the steady state in which a vessel, with a wastepipe of length 0.01 m and cross-sectional area 2 105 m2 protrudingvertically below its base, is lled at the constant rate 3 105 m3 s1.

5. A vertical round post stands in a river, and it is observed that the water level atthe upstream face of the post is slightly higher than the level at some distanceto either side. Explain why this is so, and nd the increase in the height interms of the surface stream speed U and acceleration of gravity g. Estimatethe increase in height for a stream with undisturbed surface speed 1 ms1.

Chapter 5

The vorticity eld

The vector = u curl u is called the vorticity (from Latin for a whirlpool).The vorticity vector (x, t) denes a vector eld, just like the velocity eld u(x, t).In the case of the velocity, we can dene streamlines that are everywhere in thedirection of the velocity vector at a given time. Similarly we can dene vortex linesthat are everywhere in the direction of the vorticity vector at a given time. We willshow that the vorticity is twice the local angular velocity in the ow.

Figure 5.1:

(i) Bundles of vortex lines make up vortex tubes.

(ii) Thin vortex tubes, such that their constituent vortex lines are approximatelyparallel to the tube axis, are called vortex laments (see below).

(iii) The vorticity eld is solenoidal, i.e. = 0. This very important resultresult is proved as follows:

= ( u)

37

CHAPTER 5. THE VORTICITY FIELD 38

=

x

[ w

y v

z

]+

y

[ u

z w

x

]+

z

[ v

x u

y

]= 0.

Figure 5.2:

From the divergence theorem, for any volume V with boundary surface SS

n ds =

v

dV = 0,

and there is zero net ux of vorticity (or vortex tubes) out of any volume:hence there can be no sources of vorticity in the interior of a uid (cf. sourcesof mass can exist in a velocity eld!).

(iv) Consider a length P1P2 of vortex tube. From the divergence theoremS

n ds = dV = 0.

We can divide the surface of the length P1P2 into cross-sections and the tubewall,

S = S1 + S2 + Swall,

or S

n ds =

S1

n ds +

S2

n ds +

Swall

n ds = 0. (5.1)


Figure 5.3:

However, the contribution from the wall (where n) is zero, and henceS2

n ds =

S1

(n) ds

where the positive sense for normals is that of increasing distance along thetube from the origin. Hence

Ssec tion

n ds

measured over a cross-section of the vortex tube with n taken in the same senseis constant, and taken as the strength of the vortex tube.

In a thin vortex tube, we have approximately:S

n dS n

S

dS = S

and area = constant along tube (a property of all solenoidal elds). Here, = ||.

(v) CirculationCu dr

From Stokes theorem S

( u) n dS =

C

u dr

Hence the line integral of the velocity eld in any circuit C that passes onceround a vortex tube is equal to the total vorticity cutting any cap S on C, andis therefore equal to the strength of the vortex tube. We measure the strengthof a vortex tube by calculating

Cu dr around any circuit C enclosing the

tube once only. The quantityCu dr is termed the circulation.

Vorticity may be regarded as circulation per unit area, and the component inany direction of is

limS0

1

S

c

u drwhere C is a loop of area S perpendicular to the direction specied.


Figure 5.4:

Example 5

Show that 12u2 + p/ + = constant along a vortex line for steady, incompressible,

inviscid ow under conservative external forces.

Solution

As beforeu u + (p/ + ) = 0,

where

u u = (1

2u2) u ( u) =

(1

2u2) u .

Henceu = [1

2u2 + p/ +

].

u. u (u ) 0 = u [12u2 + p/ +

](5.2)

(u ) 0 = [12u2 + p/ +

](5.3)

From Eq.(5.2) 12u2 + p/ + = constant along a streamline, and from Eq.(5.3)

12u2 + p/ + = constant along a vortex line. Thus we have a Bernoulli equation

for vortex lines as well as for streamlines.


Exercises

1. Dene the circulation round a closed circuit C and show that it is equal to thenet vorticity cutting any cap on that circuit.

2. Show that vorticity may be interpreted as circulation per unit area of section.

3. Does uid with velocity

u =

[z 2x

r, 2y 3z 2y

r, x 3y 2z

r

]

possess vorticity (where u = (u, v, w) is the velocity in the Cartesian framer = (x, y, z) and r2 = x2 + y2 + z2)? What is the circulation in the circlex2 + y2 = 9, z = 0? Is this ow incompressible?

4. Find the vorticity passing through the circuit x2+y2 = a2, z = 0 in the velocityeld u = U(z, x, y)/a.

5.1 The Helmholtz equation for vorticity

From Eulers equation for an incompressible uid in a conservative force eld.

u

t+ u u = 1

p

oru

t+

(1

2u2) u =

(p

+

);

taking the curl,

ut (u ) +

[

(1

2u2 +

p

+

)]= 0.

Using (i) () for all , and (ii)

(u ) = u ( ) ( u) + ( )u (u )= ( )u (u )

as is always solenoidal and u is solenoidal in an incompressible uid; we obtain

D

Dt=

t+ (u ) = ( )u,

which is the Helmholtz vorticity equation.


Figure 5.5:

5.1.1 Physical signicance of the term ( )uWe can understand the signicance of the term ( )u in the Helmholtz equationby recalling that is a directional derivative and ( )u is proportional to thederivative in the direction of along the vortex line (see example 7).

D

Dt= ( )u = || u = u

s,

where s is the length of an element of vortex tube. We now resolve u into compo-nents u parallel to and u at right angles to and hence to s. Then

s

D

Dt=

s(u + u) s

=us

s +us

s

[u (r + s) u(r)] rate of stretching of element

+ [u (r + s) u(r)] rate of turning of element

(a) (b)

Figure 5.6:


stretching along the length of the lament causes relative amplication of thevorticity eld;

turning away from the line of the lament causes a reduction of the vorticityin that direction, but an increase in the new direction.

Example 6

Discuss properties of the directional derivative.

Solution

Suppose that P is a point on the level surface of a scalar function, and that N andP are points on the neighbouring surface + in the direction of the normal atP (n) and a specied curve (s). Then

Figure 5.7:

s= lim

n0

s= lim

n0

n

n

s=

ncos .

= n /n is the largest of the directional derivatives at P (as n is the minimumseparation distance between the surfaces, , + ) and has the direction n of theoutward normal at P . Then

s = s nn

=

ncos =

s,

and u = || u = us

where s is distance along the vortex line.


5.2 Kelvins Theorem

The ideas of vorticity and circulation are important because of the permanence ofcirculation under deformation of the ow due to pressure forces. We next look at therate-of-change of circulation round a circuit moving with an incompressible, invisciduid:

D

Dt

u dr =

D

Dt(u dr)

=

Du

Dt dr +

u D

Dtdr .

The rst integral on the right may be written (1

p

) dr, and the

second one

u DDt

dr =

u du (see Example 7). HenceDu

Dt dr = D

Dt

u dr =

[1

p

] dr+

u du

=

[1

dp d + d

(1

2u

2

)]

=

d

(p

+ 1

2u2)

= 0

as p/ + 12u2 returns to its initial value after one circuit since it is a single

valued function.

Example 7

Show thatu D

Dtdr =

u du.

Solution

Suppose that the elementary vector P Q = r at t is advected with the ow toPQ = r (t + t) at t + t. Then

r (t + t) u (r) t + r(t) + u (r+ r) t,or

r (t + t) r(t) u (r+ r) t u (r) t,or

limt0

r (t + t) r(t)t

= lims0

u (r+ r) u (r)s

s

D

Dt(r) u

ss u


Figure 5.8:

in a xed reference frame 0xyz, where |r| = s and s is arc length along the pathP. In the limit as r dr , u du,

D

Dt(dr) = du.

5.2.1 Results following from Kelvins Theorem

(i) Helmholtz theorem: vortex lines move with the uid

Consider a tube of particles T which at the instant t forms a vortex tube ofstrength k. At that time the circulation round any circuit C lying in the tubewall, but not linking (i.e. embracing) the tube is zero, while that in an circuitC linking the tube once is k. These circulations suer no change moving withthe uid: hence the circulation in C remains zero and that in C remains k,i.e. the uid comprising the vortex tube at T continues to comprise a vortextube (as the vorticity component normal to the tube wall - measured in C - isalways zero), and the strength of the vortex remains constant. A vortex lineis the limiting case of a small vortex tube: hence vortex lines move with (arefrozen into) inviscid uids.

(ii) A ow which is initially irrotational remains irrotational Circulation isadvected with the uid in inviscid ows, and vorticity is circulation per unitarea. If initially for all closed circuits in some region of ow, it must remainso for all subsequent times. Motion started from rest is initially irrotational(free from vorticity) and will therefore remain irrotational provided that it isinviscid.

(iii) The direction of the vorticity turns as the vortex line turns, and itsmagnitude increases as the vortex line is stretched.


The circulation round a thin vortex tube remains the same; as it stretches thearea of section decreases and

circulationarea

= vorticity

increases in proportion to the stretch.

Figure 5.9:

Exercises

1. Explain the physical signicance of each term in the Helmholtz equation forvorticity in inviscid incompressible ow.

2. Show that in two-dimensional ow, with u = (u(x, y), v(x, y), 0) vorticity isnecessarily normal to the xy-plane, = (0, 0, ). Hence show that in two-dimensional inviscid incompressible ow the Helmholtz vorticity equation re-duces to the form

D

Dt= 0,

so that if the distribution of vorticity is initially uniform it must remain so,and if the motion is initially irrotational (free from vorticity) it must remainso.

3. Explain the statement that in inviscid ows vorticity is frozen into the uid.

4. Show that the circulation in any circuit embracing a vortex tube (i.e. passingonce round it) in otherwise irrotational uid is equal to the strength of thevortex tube

s

n dS


taken over any section of tube. Hence, or otherwise, show that a vortex tubecannot terminate in the interior of a uid region.

5.3 Rotational and irrotational ow

Flow in which the vorticity is everywhere zero ( u = 0) is called irrotational.Other terms in use are vortex free; ideal ; perfect. Much of uid dynamics used tobe concerned with analysing irrotational ows and deciding where these give a goodrepresentation of real ows, and where they are quite wrong.

We have neglected compressibility and viscosity. It can be shown that the neglectof compressibility is not very serious even at moderately high speeds, but the eectof neglecting viscosity can be disastrous. Viscosity diuses the vorticity (much asconductivity diuses heat) and progressively blurs the results derived above, theerrors increasing with time.

There is no term in the Helmholtz equation

D

Dt= ( )u

corresponding to the generation of vorticity: the term u represents processingby stretching and turning of vorticity already present. It follows, therefore, that inhomogeneous uids all vorticity must be generated at boundaries. In real (viscous) u-ids, this vorticity is carried away from the boundary by diusion and is then advectedinto the body of the ow. But in inviscid ow vorticity cannot leave the surface bydiusion, nor can it leave by advection with the uid as no uid particles can leavethe surface. It is this inability of inviscid ows to model the diusion/advection ofvorticity generated at boundaries out into the body of the ow that causes most ofthe failures of the model.

In inviscid ows we are left with a free slip velocity at the boundaries which wemay interpret as a thin vortex sheet wrapped around the boundary.

5.3.1 Vortex sheets

Consider a thin layer of thickness in which the vorticity is large and is directedalong the layer (parallel to 0y), as sketched. The vorticity is

=u

z w

x,

where u/z is large (but not w/x, which would lead to very large w). We cansuppose that within the vortex layer

u = u0 + z

changing from u0 to u0 + between z = 0 and , with mean vorticity

=(u0 + ) u0

= .


Figure 5.10:

This vortex layer provides a sort of roller action, though it is not of course rigid, andit also suers high rate-of-strain.

If we idealize this vortex layer by taking the limit 0, , such that remains nite, we obtain a vortex sheet, which is manifest only through the free slipvelocity. Such vortex sheets follow the contours of the boundary and clearly maybe curved. They are innitely thin sheets of vorticity with innite magnitude acrosswhich there is nite dierence in tangential velocity.

5.3.2 Line vortices

We can represent approximately also strong thin vortex tubes (e.g. tornadoes, wa-terspouts, draining vortices) by vortex lines without thickness. The circulation in acircuit round the tube tends to a denite non-zero limit as the circuit area (S) zero. If the ow outside the vortex is irrotational then all circuits round the vortexhave the same circulation, the strength of the vortex:

C

u dr as C 0.

As a consequence, the velocity as the line vortex is approached, like (distance)1.

The eect of viscosity is to thicken vortex sheets and line vortices by diusion;however, the eect of diusion is often slow relative to that of advection by the ow,and as a result large regions of ow will often remain free from vorticity. Vortexsheets at surfaces diuse to form boundary layers in contact with the surfaces; orif free they often break up into line vortices. Boundary layers on blu bodies oftenseparate or break away from the body, forming a wake of rotational, retarded owbehind the body, and it is these wakes that are associated with the drag on the body.


Figure 5.11:

5.3.3 Motion started from rest impulsively

Viscosity (which is really just distributed internal uid friction) is responsible forretarding or damping forces which cannot begin to act until the motion has started;i.e. take time to act. Hence any ow will be initially irrotational everywhere except atactual boundaries. Within increasing time, vorticity will be diused form boundariesand advected and diused out into the ow.

Motion started from rest by an instantaneous impulse must be irrotational. For,if we integrate the Euler equation over the time interval (t, t + t) t+t

t

Du

Dtdt =

t+tt

F dt t+t

t

1

p dt

or

[u]

b+tt

=

t+tt

F dt 1 t+t

t

p dt .

In the limit t 0 for start-up by an instantaneous impulse, the impulse of the bodyforce 0 (as the body force is unaected by the impulsive nature of the start) and

u u0 = 1P,

where the uid responds instantaneously with the impulsive pressure eld P = tp dt, and the impulse on a uid element is P per unit volume, producing

a velocity from rest of

u0 = 1P.

This is irrotational as

v = 1 (P ) 0.

Chapter 6

Two dimensional ow of ahomogeneous, incompressible,inviscid uid

In two (x, z) dimensions, the Euler equations of motion are

u

t+ u

u

x+ w

u

z= 1

p

x, (6.1)

w

t+ u

w

x+ w

w

z= 1

p

z, (6.2)

and the continuity equation is

u

x+

w

z= 0. (6.3)

The vorticity has only one non-zero component, the y-component, i.e., =(0, , 0), where

=u

z w

x. (6.4)

Taking (/z) (6.1) (/x) (6.2) and using the continuity equation we can showthat

D

Dt= 0. (6.5)

This equation states that uid particles conserve their vorticity as they movearound. This is a powerful and useful constraint. In some problems, = 0 for allparticles. Such ows are called irrotational.

Consider, for example, the problem of a steady, uniform ow U past a cylinderof radius a. All uid particles originate from far upstream (x ) where u = 0,

50

CHAPTER 6. TWO DIMENSIONAL FLOWOFA HOMOGENEOUS, INCOMPRESSIBLE, INVIS

Figure 6.1:

w = 0, and therefore = 0. It follows that uid particles have zero vorticity for alltime.

The inviscid ow problem can be solved as follows. Note that the continuityequation (6.3) suggests that we introduce a streamfunction , dened by the equa-tions

u =

z, w =

x. (6.6)

Then Eq. (6.3) is automatically satised and it follows from (6.4) that

=2

x2+

2

z2(6.7)

In the case of irrotational ow, = 0 and satises Laplaces equation:

2

x2+

2

z2= 0. (6.8)

Appropriate boundary conditions are found using (6.6). For example, on a solidboundary, the normal velocity must be zero, i.e., u n = 0 on the boundary. Ifn = (n1, 0, n3), it follows using (6.6) that n1

z n3 x = 0, or n = 0 on the

boundary. We deduce that is in the direction of n, whereupon is a constanton the boundary itself.

Figure 6.2:

Let us return to the example of uniform ow past a cylinder of radius a: seediagram below.

The problem is to solve Eq. (6.8) in the region outside the cylinder (i.e. r > a)subject to the boundary condition that


Figure 6.3:

u =

(

z, 0,

x

) (U, 0, 0) as r , (6.9)

and

u n = 0 on r = a, (6.10)where r = (x2 + y2)

1/2. For this problem it turns out to be easier to work in cylin-

drical polar coordinates centred on the cylinder.It is easy to check that the solution of (6.8) satisfying (6.9) and (6.10) is

= U

(r a

2

r

)sin . (6.11)

Note that for large r, Ur sin = Uz, whereupon u = /z U as required.

Now

z=

r

r

z+

z

and z = r sin r/z = 1/ sin and 1 = r cos /z, whereupon

z=

1

sin

r+

1

r cos

.(6.12)

Similarly,

x=

r

r

x+

x

and x = r cos r/x = 1/cos and 1 = r sin /x, whereupon

x=

1

cos

r 1

r sin

. (6.13)

The boundary condition on the cylinder expressed by (6.10) requires that

zcos

xsin = 0


at r = a and for all and, using (6.12) and (6.13), this reduces to

= 0 at r = a. (6.14)

This equation implies that is a constant on the cylinder; i.e., the surface of thecylinder must be a streamline. Substitution of (6.11) into (6.14) conrms that 0on the cylinder.

It remains to show that satises (6.8). To do this one can use (6.12) and (6.13)to transform (6.8) to cylindrical polar coordinates; i.e.,

r

(r

r

)+

1

r

2

z2= 0. (6.15)

It is now easy to verify that (6.11) satises (6.15) and is therefore the solution forsteady irrotational ow past a cylinder. Note that the solution for is unique onlyto within a constant value; if we add any constant to it, it will satisfy equation (6.8)or (6.15), but the velocity eld would be unchanged.

It is important to note that we have obtained a solution without reference tothe pressure eld, but the pressure distribution determines the force eld that drivesthe ow! We seem, therefore, to have by-passed Newtons second law, and haveobviously avoided dealing with the nonlinear nature of the momentum equations(6.1) and (6.2). Looking back we will see that the trick was to use the vorticityequation, a derivative of the momentum equations. For a homogeneous uid, thevorticity equation does not involve the pressure since p 0. We infer from thevorticity constraint [Eq. (6.7)] that the ow must be irrotational everywhere and usethis, together with the continuity constraint (which is automatically satised whenwe introduce the streamfunction) to infer the ow eld. If desired, the pressure eldcan be determined, for example, by integrating Eqs. (6.1) and (6.2), or by usingBernoullis equation along streamlines.

Figure 6.4:

Now the solution itself. The streamline corresponding with (11) are sketched inthe gure overleaf. Note that they are symmetrical around the cylinder. ApplyingBernoullis equation to the streamline around the cylinder we nd that the pressuredistribution is symmetrical also so that the total pressure force on the upstream side


of the cylinder is exactly equal to the pressure on the downwind side. In other words,the net pressure force on the cylinder is zero! This result, which in fact is a generalone for irrotational inviscid ow past a body of any shape, is known as dAlembertsParadox. It is not in accord with our experience as you know full well when you tryto cycle against a strong wind. What then is wrong with the theory? Indeed, whatdoes the ow round a cylinder look like in reality? The reasons for the breakdownof the theory help us to understand the limitations of inviscid ow theory in generaland help us to see the circumstances under which it may be applied with condence.First, let us return to the viscous theory.

Figure 6.5:

The Navier-Stokes equation is the statement of Newtons second law of motionfor a viscous uid. It reads

Du

Dt= 1

p + 2u. (6.16)

The quantity of is called the kinematic viscosity. For air, = 1.5105 m2 s1;for water = 1.0 106 m2 s1. The relative importance of viscous eect ischaracterized by the Reynolds number Re, a nondimensional number dened by

Re =UL

,

where U and L are typical velocity and length scales, respectively. The Reynoldsnumber is a measure of the ratio of the acceleration term to the viscous term in(6.16). For many ows of interest, Re >> 1 and viscous eects are relatively unim-portant. However, these eects are always important near boundaries, even if only ina thin boundary-layer adjacent to the boundary. Moreover, the dynamics of thisboundary layer may be crucial to the ow in the main body of uid under certaincircumstances. For example, in ow past a circular cylinder it has important con-sequences for the ow downstream. The observed streamline pattern in this case atlarge Reynolds numbers is sketched in the gure overleaf. Upstream of the cylinderthe ow is similar to that predicted by the inviscid theory, except in a thin viscous


boundary-layer adjacent to the cylinder. At points on the downstream side of thecylinder the ow separates and there is an unsteady turbulent wake behind it. Theexistence of the wake destroys the symmetry in the pressure eld predicted by theinviscid theory and there is net pressure force or form drag acting on the cylinder.Viscous stresses at the boundary itself cause additional drag on the body.

Chapter 7

Boundary layers in nonrotatinguids

We consider the boundary layer on a at plate at normal incidence to a uniformstream U as shown.

Figure 7.1:

The Navier Stokes equations for steady two-dimensional ow with typical scaleswritten below each component are:

uu

x+ w

u

z= 1

p

x+

[2u

x2+

2u

z2

], (7.1)

U2

L

UW

H

P

L

U

L2U

H2] (7.2)

uw

x+ w

w

z= 1

p

z+

[2w

x2+

2w

z2

], (7.3)

UW

L

W 2

H

P

H

W

L2W

H2(7.4)

and the continuity equation is

56

CHAPTER 7. BOUNDARY LAYERS IN NONROTATING FLUIDS 57

u

x+

w

z= 0,

U2

L

W

H(7.5)

From the continuity equation we infer that since |u/x| = |w/z|, W UH/Land hence the two advection terms on the left hand sides of (7.1) and (7.3) are thesame order of magnitude: U2/L in (7.2) and (U2/L)(H/L) in (7.4). Now, for a thinboundary layer, H/L > 1

P

H/W

H2 U

2

H/U

HL UL

= Re >> 1.

But if both the inertia terms and friction terms in (7.3) are much less than thepressure gradient term, the equation must be accurately approximated by

p

z= 0.

This implies that the perturbation pressure is constant across the boundary layer.It follows that the horizontal pressure gradient in the boundary layer is equal to thatin free stream.

Collecting these results together we nd that an approximate form of the Navier-Stokes equations for the boundary layer to be

uu

x+ w

u

z= U

dU

dx+

2u

z2, (7.6)

1Note that if this were not true, steady flow would not be possible as the large pressure gradientwould accelerate the flow further.


with

u

x+

w

z= 0, (7.7)

and U = U(x) being the (possible variable) free stream velocity above the boundarylayer, Equations (7.6) and (7.7) are called the boundary layer equations.

7.1 Blasius solution (U = constant)

Equation (7.6) reduces to

uu

x+ w

u

z=

2u

z2, (7.8)

and we look for a solution satisfying the boundary conditions u = 0, w = 0 at z = 0,u U as z and u = U at x = 0. Equation (7.7) suggests that we introduce astreamfunction such that

Figure 7.2:

whereupon must satisfy the conditions = constant, /z = 0 at z = 0, Uz as z and = Uz at x = 0. It is easy to verify that a solutionsatisfying these conditions is

u =

z, w =

x, (7.9)

where

= (U/2x)1/2 z, (7.10)


f() satises the ordinary dierential equation

f + ff = 0, (7.11)

subject to the boundary conditions

f(0) = f (0) = 0; f () = 1. (7.12)Here, a prime denotes dierentiation with respect to . It is easy to solve Eq.

(7.11) subject to (7.12) numerically (see e.g. Rosenhead, 1966, Laminar BoundaryLayers, p. 222-224). The prole of f which characterizes the variation of u acrossthe boundary layer thickness is proportional to and we might take = 4 ascorresponding with the edge of the boundary layer. Then (7.10) shows that thedimensional boundary thickness (x) = 4(2x/U)1/2; i.e., increases like the squareroot of the distance from the leading edge of the plate. We can understand thethickening of the boundary layers as due to the progressive retardation of more andmore uid as the ctional force acts over a progressively longer distance downstream.

Often the boundary layer is relatively thin. Consider for example the boundarylayer in an aeroplane wing. Assuming the wing to have a span of 3 m and that theaeroplane ies at 200 ms1, the boundary layer at the trailing edge of the wing (as-suming the wing to be a at plate) would have thickness of 4 (2 1.5 105 3/200)1/2 =2.7103 m, using the value = 1.5105m2s1 for the viscosity of air. The calcu-lation assumes that the boundary layer remains laminar; if it becomes turbulent, therandom eddies in the turbulence have a much larger eect on the lateral momentumtransfer than do random molecular motions, thereby increasing the eective valueof , possibly by an order of magnitude or more, and hence the boundary layerthickness.

Note that the boundary layer is rotational since = (0, , 0), where = uz w

x,

or approximately just u/z.

7.2 Further reading

Acheson, D. J., 1990, Elementary Fluid Dynamics, Oxford University Press, pp406.

Morton, B. R., 1984: The generation and decay of vorticity. Geophys. Astrophys.Fluid Dynamics, 28, 277-308.

fluid dynamics

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