1

Fluid Dynamics

C. NashMathematical Physics Department

National University of Ireland, Maynoothc© Charles Nash, 1998, 2006, all rights reserved.

§ 1. Introductory remarks: fluids and continua in general

Fluid dynamics is part of the wider subject of continuum mechanics whichincludes elasticity as one of its branches. There are also overlaps of thesubject matter of continuum mechanics with other branches of physics: forexample the subject of magnetohydrodynamics which combines the tech-niques of fluid dynamics with those of electromagnetic theory. The subjectmatter of the disciplines, soil mechanics, geology, polymer physics, the lowtemperature physics of superfluids, and the cosmological models of generalrelativity are still more examples where continuum mechanics plays a jointrole in conjunction with some other branch of condensed matter physics or,in the latter example, the physics of relativity.

However, in these lectures we are only concerned with fluids. We wantto emphasise at the beginning that the term fluid is a little misleading tosome since it may be taken to mean a liquid. In fact the term fluid, inthese lectures, and in the fluid dynamics literature in general, means eithera liquid or a gas.

Books on fluid dynamics

There are a huge number of books on such an old and large a subject as fluiddynamics. We have chosen to quote a few titles in the main body of thetext together with some comment on their content. In addition we providea supplementary list of some more titles without comment; the reader canreadily take matters from there by visiting the fluid mechanics section ofthe library.

Some selected books

(i) Rutherford D. E., Fluid dynamics, (Oliver and Boyd)This is a small concise book which more than covers the material of

these lectures; it is clear and well written.(ii) O’ Neill and Chorlton, F., Ideal and incompressible fluid dynam-

ics (Ellis Horwood)

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This is a more verbose and less concise version of Rutherford with alarger number of worked examples.(iii) McCormack, P. D. and Crane, L., Physical fluid dynamics (Aca-

demic Press)Finally this text is more advanced than the other two; however in

addition to the mathematics it contains good discussions of the physicsunderlying many fluid phenomena.

Further titles

(i) Lighthill, M. J., An informal introduction to theoretical fluid

mechanics (Oxford University Press)(ii) Fox, R, W. and McDonald, A. T., Introduction to fluid mechanics

(Wiley)(iii) Open University, Looking at fluids in motion. Understanding

fluid effects. Modelling fluid and thermodynamic systems

(Open University Press)(iv) Pedlosky, J., Geophysical fluid dynamics (Springer-Verlag)(v) Feynman R. P., Leighton R. B. and Sands M. L., The Feynman Lec-

tures on Physics vol. II (Addison–Wesley)

§ 2. Some basic terms and notions

The main mathematical task in fluid dynamics is to obtain and solve equa-tions for the velocity

v(x, y, z, t) (2.1)

where (x, y, z) is a point in the fluid and t is the time. Atomic structureis ignored and the fluid is regarded as a continuum. We shall denote thedensity of a fluid by

ρ = ρ(x, y, z, t) (2.2)

and the pressure byp = p(x, y, z, t) (2.3)

If a fluid has constant density—as is the case to a high degree of ap-proximation for many liquids—then it is called incompressible.

A fluid whose internal frictional forces are negligible is referred to asbeing non-viscous or inviscid. It is usual 1 to take the coefficient of viscosityof a viscous fluid to be a positive constant and denote it by

η (2.4)

1 Actually we shall see much later that, in some circumstances, there can be two

coefficients of viscosity for a viscous fluid; however if the fluid is also incompressible then

only one of these enters in the equation of motion.

3

Angular momentum and rotation in a fluid is often studied by focusingon the curl of the velocity vector i.e.

∇× v (2.5)

This quantity is known as the vorticity and is denoted by ω. In other wordsthe vorticity ω is defined by writing

ω = ∇× v (2.6)

If the flow given by some velocity distribution v(x, y, z, t) has

ω = 0, everywhere (2.7)

then the flow is called irrotational.A velocity distribution satisfying

∂v(x, y, z, t)

∂t= 0, everywhere (2.8)

is called a steady flow.

§ 3. The stream derivative D/Dt

In fluid dynamics it is very useful to introduce another rate of change withrespect to time: this rate of change is expressible as a derivative involvingthe time t and the fluid velocity v. It is called the stream derivative orsimply differentiation moving with the fluid. We shall denote it by

D

Dt(3.1)

but first we must see how the stream derivative arises naturally and thengive its definition.

To this end consider a completely arbitrary quantity Q associated witha fluid. Q can be scalar valued as it would be if it were temperature,pressure, density etc.; alternatively it could be vector valued, for examplethis would be the case if Q were the velocity itself. In any case the value ofQ will depend both on time t and position (x, y, z) within the fluid, i.e. wehave

Q ≡ Q(x, y, z, t) (3.2)

Now we imagine that we select an individual fluid particle located at

(x, y, z) (3.3)

4

at this timet (3.4)

Then, as time progresses and the fluid flows, this particle traces out a curvein the fluid which we shall denote by

(x(t), y(t), z(t)) (3.5)

The stream derivative of Q is simply defined to be the rate of change ofQ along this curve. In other words one first restricts Q(x, y, z, t) to be onthis curve and then differentiates this restricted Q with respect to t; theresult then is denoted by DQ/Dt. Let us carry out these two steps andthereby obtain a formula for DQ/Dt: Restricting Q(x, y, z, t) to the curve(x(t), y(t), z(t)) gives us the function

Q(x(t), y(t), z(t), t) (3.6)

and differentiating this function with respect to t gives us the equation

d

dtQ(x(t), y(t), z(t), t) =

∂Q

∂x

∂x

∂t+∂Q

∂y

∂y

∂t+∂Q

∂z

∂z

∂t+∂Q

∂t(3.7)

The RHS of 3.7 is thus the stream derivative of Q so that we can now write

DQ

Dt=∂Q

∂x

∂x

∂t+∂Q

∂y

∂y

∂t+∂Q

∂z

∂z

∂t+∂Q

∂t(3.8)

It is both useful and usual to abbreviate this equation somewhat by notingthat since one has the pair of equations

∇ = i∂

∂x+ j

∂

∂y+ j

∂

∂z

v =∂x

∂ti +

∂y

∂tj +

∂z

∂tk

(3.9)

then one sees at once that

v · ∇ =∂x

∂t

∂

∂x+∂y

∂t

∂

∂y+∂z

∂t

∂

∂z(3.10)

Hence we now have a much more compact form for the stream derivativenamely

DQ

Dt=∂Q

∂t+ (v · ∇)Q (3.11)

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or equivalently, and just as compactly, one can write

D

Dt=

∂

∂t+ (v · ∇) (3.12)

We shall always assume that matter is conserved when a fluid flows andit turns out that this conservation imposes some restrictions on the possiblevelocity distributions v(x, y, z, t) that can represent fluid flows. We explainthis in the next section.

The equation of continuity

In general the density ρ varies from point to point in a fluid. Howeverwhen ρ is a constant the fluid is incompressible—you cannot compress itor you would change the density. This brings us to our first importantequation for fluid flow which is known as the equation of continuity. Thisequation expresses the fact that, as a fluid flows, matter is neither creatednor destroyed: it states that

∇ · (ρv) +∂ρ

∂t= 0 (3.13)

We now embark on the proof of 3.13. Take a closed volume V of fluid withsurface S out of which fluid is flowing with velocity v, cf. Fig 1.

dS

dS

dSvv

v

Fig. 1: Fluid flowing out of the closed volume V

The mass of fluid flowing out per unit time through a surface patchlabelled by the vector dS is

ρv · dS (3.14)

so that the total mass of fluid flowing out of V per unit time is the integral∫

S

ρv · dS (3.15)

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Since matter is conserved this outflow is precisely equal to the correspondingdecrease of mass of the fluid in V . But this decrease is just

− ∂

∂t

∫

V

ρdV (3.16)

where the minus sign in the equation above compensates for the fact thatthe mass of the fluid is V is decreasing so that its time derivative is negative.In any case matter conservation has now given us the equation

∫

S

ρv · dS = − ∂

∂t

∫

V

ρdV

⇒∫

S

ρv · dS +∂

∂t

∫

V

ρdV = 0

⇒∫

V

∇ · (ρv)dV +∂

∂t

∫

V

ρdV = 0, using Gauss’s divergence theorem

⇒∫

V

∇ · (ρv) +∂ρ

∂t

dV = 0

(3.17)But since the volume V is arbitrary the integrand in the last line of 3.17must be zero and so we have obtained the result that

∇ · (ρv) +∂ρ

∂t= 0 (3.18)

which is the sought for equation of continuity.Now note that because 2

∇ · (ρv) = ρ∇ · v + v · ∇ρ (3.20)

the equation of continuity can be written as

ρ∇ · v + v · ∇ρ+∂ρ

∂t= 0 (3.21)

i.e. asDρ

Dt+ ρ∇ · v = 0 (3.22)

2 This fact is a vector identity: i.e. for any vector A and scalar function f one cancheck that

∇ · (fA) = f∇ · A + A · ∇f (3.19)

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This means that if we specialise to the case where the fluid density is con-stant that is the fluid is incompressible then the equation of continuitycollapses to just

∇ · v = 0 (3.23)

which is a very useful property to be borne in mind for an incompressiblefluid. It should not be forgotten, though, that

∇ · v = 0 6⇒ incompressibility (3.24)

rather it only implies thatDρ

Dt= 0 (3.25)

which definitely does not require the density to be constant 3.

§ 4. Euler’s equation of motion for non-viscous fluids

We are now ready to derive the equation of motion for an inviscid, or non-viscous, fluid. Since we are neglecting the frictional forces due to viscositythen there are only two types of force on the fluid particles and these are(i) Forces due to pressure differences(ii) External forces; e.g. gravity or perhaps a pair of magnetic field and

electric fields in the case of a charged fluid such as a plasma—a typicalplasma is the hot gas found on the Sun or other star.Let us now consider how these forces act on an infinitesimal cube within

the fluid such as that depicted in Fig. 2.

A

B

C

D

E

F

G

H

(x,y,z)

dxdy

dzP+dPP

(x+dx,y,z)

Fig. 2: An infinitesimal cube of fluid of volume dxdydz.

3 A word of caution on terminology: if a flow is such that ∇ · v = 0 then it is

sometimes referred to as an incompressible flow; by what we have just said this clearly

does not necessarily mean that the fluid itself is incompressible.

8

When talking about external forces on a fluid we shall always workwith the external force on a unit mass of the fluid; however we shall stilldenote this “force on a unit mass” by F even though it is not a force butan acceleration 4. Another piece of terminology is that the term body forceis also sometimes used to denote an external force.

In any case if F is the external, or body, force on the infinitesimal cubein Fig. 2 then , since the cube has volume dxdydz, the the force exerted byF on the cube in the x direction is just

F · i ρdxdydz (4.1)

where ρ is the density of the fluid.Next we come to the forces on the cube due to pressure differences. As

with the external force F we just consider the x direction.Now pressure differences will only produce a force in the x direction if

the pressure varies in the x direction. It should be clear from Fig. 2 that theforces in the x direction due to pressure differences are given by subtractingthe quantity pressure× area for the two faces EFGH and ABCD. So if,as is depicted in Fig 2, P denotes the point (x, y, z) on the face ABCD andP+dP denotes the point (x+dx, y, z) on the face EFGH, the force in thex direction due to pressure differences is

p(x, y, z)dydz − p(x+ dx, y, z)dydz = p(x, y, z)− p(x+ dx, y, z)dydz(4.2)

Now by Newton’s laws these two contributions 4.1 and 4.2 add up to givethe mass times the x component of the acceleration; i.e. we have

ρdxdydzDvx

Dt= F · i ρdxdydz + p(x, y, z)− p(x+ dx, y, z)dydz (4.3)

and on dividing by ρdxdydz we obtain

Dvx

Dt= F · i +

1

ρ

1

dxp(x, y, z)− p(x+ dx, y, z) (4.4)

Now Taylor’s theorem applied to p(x+ dx, dy, dz) gives

p(x+ dx, dy, dz) = p(x, y, z) +∂p

∂xdx+ negligible (4.5)

4 This confusing piece of notation has become a widespread convention in the fluid

mechanics literature and so we reluctantly follow it too.

9

and substituting this into our equation for Dvx/Dt gives the result

Dvx

Dt= F · i +

1

ρ

1

dx

p(x, y, z)− p(x, y, z)− ∂p

∂xdx

⇒ Dvx

Dt= F · i− 1

ρ

∂p

∂x

(4.6)

and this latter equation is the equation of motion for the x direction. Ex-actly similarly the equations of motion in the y and z directions are, respec-tively, the pair

Dvy

Dt= F · j − 1

ρ

∂p

∂y

Dvz

Dt= F · k − 1

ρ

∂p

∂z

(4.7)

We now combine these three scalar equations of motion into a single vectorequation of motion. To do this we simply note that we have if we decomposethe vectors v, F and ∇p into their components we have

v = vxi + vyj + vzk

F = Fxi + Fyj + Fzk

∇p =∂p

∂xi +

∂p

∂yj +

∂p

∂zk

(4.8)

This allows us to rewrite the x equation of motion as

D(v · i)Dt

= F · i − 1

ρ(∇p · i)

i.e.

(

Dv

Dt

)

· i =

(

F − 1

ρ∇p)

· i(4.9)

It is now clear that the vector form of the equation that we are after istherefore simply

Dv

Dt= F − 1

ρ∇p (4.10)

and this equation is called Euler’s equation of motion for an inviscid fluid(subject to a body force F).

§ 5. Bernoulli’s equation

We now come to an important special case of Euler’s equation known asBernoulli’s equation.

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To obtain Bernoulli’s equation we simply start with Euler’s equation(so we still have a non-viscous fluid) and assume(i) The flow is irrotational(ii) Any body force F present is conservative; this, by definition, means

that F = −∇K for some function K.Using (i) we have

∇× v = 0 ⇒ v = −∇φ, for some function φ (5.1)

This function φ is then called the velocity potential. Now Euler’s equationsays that v obeys

Dv

Dt= F − 1

ρ∇p (5.2)

or, using 3.12 for D/Dt and F = −∇K from (ii), v obeys

∂v

∂t+ (v · ∇)v = −∇K − 1

ρ∇p (5.3)

Next we have the fact (which we quote without proof) that (v ·∇)v satisfiesthe important vector identity

(v · ∇)v = ∇(

v2

2

)

+ (∇× v) × v (5.4)

Using this identity Euler’s equation now takes the form

∂v

∂t+ ∇

(

v2

2

)

+ (∇× v) × v = −∇K − 1

ρ∇p (5.5)

But since, by assumption (i), ∇× v = 0 we immediately have

∂v

∂t+ ∇

(

v2

2

)

= −∇K − 1

ρ∇p (5.6)

Also if we use the fact that v = −∇φ in the term ∂v/∂t we find that

− ∂

∂t(∇φ) + ∇

(

v2

2

)

= −∇K − 1

ρ∇p (5.7)

Interchanging ∇ with ∂/∂t and rearranging terms gives us

∇(

−∂φ∂t

)

+ ∇(

v2

2

)

+ ∇K +1

ρ∇p = 0 (5.8)

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In a moment we shall integrate this equation along a line but first weneed to recall simple piece of calculus which is this: If f(x, y, z) is anyfunction then we know that

df =∂f

∂xdx+

∂f

∂ydy +

∂f

∂zdz (5.9)

Now if we choose an infinitesimal element of length dl given by

dl = dxi + dyj + dzk (5.10)

then taking the dot product of dl with the vector ∇f gives precisely df , i.e.we have

∇f · dl = (∂f

∂xi +

∂f

∂yj +

∂f

∂zk) · (dxi + dyj + dzk)

=∂f

∂xdx+

∂f

∂ydy +

∂f

∂zdz

so ∇f · dl = df, as claimed

(5.11)

Returning to 5.8 we take the dot product of both sides with dl and use5.11 on each of the four terms so that we obtain

∇(

−∂φ∂t

)

· dl + ∇(

v2

2

)

· dl + ∇K · dl +1

ρ∇p · dl = 0

⇒ d

(

−∂φ∂t

)

+ d

(

v2

2

)

+ dK +dp

ρ= 0

(5.12)

Finally we integrate both sides of this equation along the line of which dl

is the line element, and use that fact that∫

df = f , so that we produce theequation

−∂φ∂t

+v2

2+K +

∫

dp

ρ= C(t) (5.13)

where C(t) is a constant of integration; also we note that C(t) is allowedto depend on t since t was not an integration variable. This last equation,that is to say 5.13, is Bernoulli’s equation.

Bernoulli’s equation for incompressible fluids

If the fluid is incompressible so that ρ is constant then we have

∫

dp

ρ=

1

ρ

∫

dp =p

ρ(5.14)

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This means that Bernoulli’s equation simplifies to

−∂φ∂t

+v2

2+K +

p

ρ= C(t) (5.15)

If we further assume that there is no body force and that the flow is steadythen K = 0 and ∂φ/∂t = 0; it will also be the case that the constant C(t)will be t-independent so we have an extremely simple form of Bernoulli’sequation namely

v2

2+p

ρ= C

(incompressible fluid, steady flow)

(5.16)

This equation 5.16 is simple but very instructive: To see why first notethat that since all quantities on the LHS are positive then the LHS itself ispositive. This, in turn, means that the constant C is positive. Now we seethat neither v2/2 nor p/ρ can be arbitrarily large since their sum has to beC: in fact neither term can exceed the value C.

The most useful way of looking at this is to note that if |v| increasesthen p must decrease and vice-versa 5. Reasoning further we can say that,in a fluid, a region of high pressure is a region of low velocity and vice-versa.Those readers who pay attention to television weather maps may well havenoticed this as being property of high and low pressure regions in the Earth’satmosphere

This relationship between p and v is also the explanation of the curvingof the trajectory of spinning golf, tennis or ping-pong balls—the so calledMagnus effect—as well as the mechanism of the lift under an airplane’swing; or indeed the method of propulsion of the famous Atlantic crossingboats of the 1920’s designed by the German engineer Flettner, cf. Fig. 3.

5 Remember ρ does not change since, by assumption it, is constant

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Fig. 3: The boats designed by Anton Flettner

Example Water draining out of a circular tank

We can use Bernoulli’s equation to derive the shape of the trumpet shapedsurface created by the draining of water out of the bottom a circular tank—cf. Fig. 4.

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Fig. 4: Fluid flowing out of a circular tank

We can assume that the flow is steady and so Bernoulli’s equation takesthe form

K +v2

2+p

ρ= C (5.17)

but since the body force is just the downward pull of gravity then

F = −∇K, with K = gz (5.18)

so our equation is now

gz +v2

2+p

ρ= C (5.19)

The key idea is now to evaluate this equation on the trumpet shaped surfaceof the water since, on this surface, since it is the interface between the waterand the air the pressure of the water and the air are both equal and hence

p = pA (5.20)

where pA = atmospheric pressure. So now we have on the trumpet shapedsurface

gz +v2

2= D, where D = C − pA

ρ= a constant (5.21)

next if r =√

x2 + y2 is the distance from a point on the surface to thez-axis, and we simply quote the fact

|v| =E

r, E a constant (5.22)

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then we find that

gz +E2

2r2= D

⇒ z =1

g

(

D − E2

2r2

) (5.23)

or equivalently

r =E√

2√D − gz

(5.24)

which the reader can easily verify by plotting is the graph of a trumpetshaped surface such as that depicted in Fig. 4.

For example if we choose

E =√

2, D = 50, g = 9 (5.25)

then those readers who use the mathematics computer utility Maple cantype in the following lines which create the postscript file trumpet.eps whichis shown below in Fig. 5—it may help to remember that the r and z of theformulae are the r and z of the r, θ, z cylindrical coordinates; the fact thatθ is absent means that the equation of the surface is independent of θ i.e. itcan be obtained by rotating a curve about an axis (the z axis in this case).

trumpet := r− > 1/sqrt((50 − 9 ∗ z));

plotsetup(ps, plotoutput = ‘trumpet.eps‘, plotoptions = ‘portrait, noborder‘);

plot3d(trumpet(r), theta = 0..2 ∗ Pi, z = 0..5, coords = cylindrical);

Fig. 5: The trumpet surface for the draining tank

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§ 6. Adiabatic flow and the Mach number

In gases one sometimes has to consider what is called adiabatic flow: Math-ematically speaking 6 this is simply a flow where pressure and density arerelated by the equation

p = kργ, with k and γ constant, γ > 1 (6.5)

We want to study the effect of this adiabatic relationship between p and ρusing Bernoulli’s equation. Hence we start with

−∂φ∂t

+v2

2+K +

∫

dp

ρ= C(t) (6.6)

but we shall study steady flow so that ∂φ/∂t = 0 and the constant C(t) istime independent and equal to C, say; we shall also have no body force sothat K = 0. This leaves us with

v2

2+

∫

dp

ρ= C (6.7)

But sincep = kργ (6.8)

6 In physics an adiabatic change of a system is one which takes place without exchangeof energy between the system and its surroundings. The example here comes from thekinetic theory of gases where one uses perfect gas law

PV

T= C, C a constant (6.1)

to obtain, at constant temperature, the special case

PV = const. (6.2)

But if the sample of gas has mass M and density ρ then we find that

P = Dρ D a cosntant (6.3)

All this is for changes which keep the gas in thermodynamic equilibrium and necessitateexchange of energy with its surroundings. For changes that are adiabatic one finds thatP and ρ are related by the equation

P = kργ (6.4)

which is what we have here.

17

thendp = γkργ−1dρ

⇒∫

dp

ρ= γk

∫

ργ−2dρ

=γ

(γ − 1)kργ−1

(6.9)

Now sound is a longitudinal pressure wave travelling through a fluidand, if the speed of sound in a fluid is denoted by c then c is given by

c2 =dp

dρ

= γkργ−1

= γkργ

ρ

⇒ c2 =γp

ρ

(6.10)

So now Bernoulli’s equation gives us

v2

2+

γ

(γ − 1)kργ−1 = C (6.11)

which, if we use the speed of sound c, we can write as

v2

2+

c2

(γ − 1)= C (6.12)

Now define for convenience the constant c0 by writing

C =c20

γ − 1(6.13)

then Bernoulli’s equation, written as an equation for v, becomes

v2 =2

(γ − 1)

(

c20 − c2)

(6.14)

This equation allows us to find an expression for the Mach number M ofthe flow. M is defined as the ratio of the speed of the flow to the speed ofsound, i.e.

M =|v|c

(6.15)

18

Dividing 6.14 by c2 we get

M2 =2

(γ − 1)

(

c20c2

− 1

)

⇒M =

√

2

(γ − 1)

√

(

c20c2

− 1

)

(6.16)

so that we now have an equation for the Mach number; incidentally, for air,the constant γ has the value 1.4 approximately.

Finally we have the well known division of the flow into three types: If

|v| < c the flow is called subsonic|v| = c the flow is called sonic|v| > c the flow is called supersonic

(6.17)

In addition, as long as |v| > c, there is a shock wave produced in the fluid;this is what is usually referred to as a sonic boom.

§ 7. Circulation and Kelvin’s theorem

For an inviscid fluid there are no ‘frictional forces’ present to slow down arotating mass of fluid. This means that rotation, once present will persistforever. Hence, in this case, angular momentum is conserved.

This fact is usually established indirectly by proving what is calledKelvin’s theorem; this theorem states, roughly, that the angular momentumper unit mass is conserved. More precisely Kelvin’s theorem states thatwhat is called the circulation Γ round any curve C is constant for an inviscid,barotropic fluid subject to a conservative body force F.

We must first define the circulation and its definition goes as follows:Select any closed curve C and integrate the velocity v around it. Thisquantity is the circulation round C and is denoted by Γ. In other words wehave

Γ =

∫

C

v · dl (7.1)

If S denotes the surface which is surrounded by the curve C, and we useStokes’ theorem, then we have

∫

C

v · dl =

∫

S

∇× v · dS

=

∫

S

ω · dS, where ω is the vorticity

Hence Γ =

∫

S

ω · dS

(7.2)

19

and the relation of Γ to angular momentum is now obvious. The readercan quickly check that Γ has the dimensions of angular momentum per unitmass.

Now we shall formally state and prove Kelvin’s theorem.

Theorem (Kelvin’s theorem) The circulation

Γ =

∫

C

v · dl (7.3)

is a constant of the flow, i.e.

DΓ

Dt= 0 (7.4)

provided the fluid is inviscid, barotropic and the external force is conserva-tive

Proof: First we explain the term barotropic: a barotropic fluid is one wherethe pressure is a function of (only) the density, i.e.

p = f(ρ), for some function f (7.5)

The relevance of barotropicity to the proof will emerge shortly.We now make a start by writing

DΓ

Dt=

D

Dt

∫

C

v · dl

=

∫

C

D(v · dl)

Dt

=

∫

C

Dv

Dt· dl + v·D(dl)

Dt

(7.6)

Note that the reason that the term

D(dl)

Dt(7.7)

is not zero and has to be included is that dl is an element of length of thecurve C which is a curve moving with the fluid and hence dl depends ontime; all this matters of course because D/Dt is the stream derivative andnot the ordinary partial derivative ∂/∂t. But

D(dl)

Dt= d

Dl

Dt

(7.8)

20

andDl

Dt= v (7.9)

since a piece l of the curve moves with the fluid velocity v. Substitutingfrom 7.8 and 7.9 in 7.6 gives

DΓ

Dt=

∫

C

Dv

Dt· dl + v · dv

(7.10)

Now for an inviscid fluid Euler’s equation 4.10 says that

Dv

Dt= F − 1

ρ∇p (7.11)

hence we have

DΓ

Dt=

∫

C

F · dl− 1

ρ∇p · dl + v · dv

(7.12)

But the external force F is conservative so

F = −∇K (7.13)

and

d

(

v2

2

)

= v · dv (7.14)

and we obtainDΓ

Dt=

∫

C

−dK − 1

ρdp+ d

(

v2

2

)

(7.15)

where we have used the fact that ∇f ·dl = df (cf. 5.11) on K and p.Now it is time to use the barotropicity of the fluid. To this end we

suppose that, for some f ,p = f(ρ) (7.16)

Then if we define F (ρ) by

F (ρ) =

∫

dp

ρ(7.17)

one can see immediately that

dF =dp

ρ≡∫

1

ρ

dp

dρdρ (7.18)

21

allowing us to express DΓ/Dt as

DΓ

Dt=

∫

C

−dK − dF + d

(

v2

2

)

(7.19)

Finally we note that for any function g, say, single valuedness forces dg tohave zero integral round any closed curve 7; hence

∫

C

dg = 0, if C is a closed curve (7.21)

So we have∫

C

dK =

∫

C

dF =

∫

C

d

(

v2

2

)

= 0 (7.22)

and therefore we do haveDΓ

Dt= 0 (7.23)

and the theorem is proved.

§ 8. Two dimensional flow and complex variable methods

In two dimensional flows the velocity v only has two components

v = vx i + vy j (8.1)

and these are functions of the two variables x and y. It turns out that, forsome flows, much insight and calculational improvements are obtained bychanging variables from x and y to the complex variable

z = x+ iy (8.2)

We now give a short summary of what are called the Cauchy–Riemannequations; these are central to complex variable theory and we need them

7 It may help to recall that

∫ b

a

df =

∫ b

a

df(x)

dxdx = f(b) − f(a) (7.20)

and if the line segment [a, b] is bent round to form a curved path with endpoints a and

b this result still holds. Hence if the endpoints of the path are joined rendering the path

closed then a = b and so the integral is zero.

22

for our fluid mechanical discussion. The reader who is familiar with themcan skip on to the next section.

The Cauchy–Riemann equations

The point of the Cauchy–Riemann equations for a function f(x, y) is that iff is expressed in terms of z and z instead of x and y giving f = f(z, z), thenthe Cauchy–Riemann equations express the fact that f(z, z) is independentof z so that we can write just f = f(z).

We can easily derive the Cauchy–Riemann equations for f as follows.We write

f(x, y) ≡ f(z, z) = u(x, y) + iv(x, y) (8.3)

where u and v are real-valued functions. The condition we want to imposeis 8

∂f

∂z= 0 (8.4)

Expanding this condition more fully gives

∂f

∂z= 0

⇒ ∂u

∂z+ i

∂v

∂z= 0

⇒ ∂u

∂x

∂x

∂z+∂u

∂y

∂y

∂z+ i

(

∂v

∂x

∂x

∂z+∂v

∂y

∂y

∂z

)

= 0

(8.5)

But

z = x+ iy, z = x− iy,⇒ x =1

2(z + z), y =

1

2i(z − z)

⇒ ∂x

∂z=

1

2,

∂y

∂z= − 1

2i=i

2

(8.6)

This means that the equation ∂f/∂z = 0 now becomes

1

2

∂u

∂x− ∂v

∂y

+i

2

∂u

∂y+∂v

∂x

= 0 (8.7)

8 This is only a formal calculation which serves to obtain the Cauchy–Riemann equa-

tions quickly and to expose the ideas that underly them. In fact z cannot be regarded

as independent of z because knowing z one can also write down z. A rigorous way of

obtaining the Cauchy–Riemann equations is to define df/dz appropriately and exploit

the fact that one can compute the derivative by taking the limit in any direction in the

complex z plane: then one chooses different directions and equates the two derivatives,

the resulting equations are the Cauchy–Riemann equations; the two chosen directions

being parallel to the x and y-axes respectively.

23

and, equating real and imaginary parts to zero on both sides, we obtain

∂u

∂x=∂v

∂y,

∂u

∂y= −∂v

∂x(8.8)

which are indeed the Cauchy–Riemann equations for u and v.

§ 9. The complex potential W for a two dimensional flow

Everything centres round a single function W (z) known as the complexpotential and we now embark on its study.

Let an incompressible fluid (flowing in two dimensions) have a velocitypotential φ(x, y) so that

v = −∇φ = −∂φ∂x

i − ∂φ

∂yj = vx i + vy j (9.1)

Then the complex potential W (z) of this flow is the analytic function of zgiven by

W (z) = φ(x, y) + iψ(x, y) (9.2)

i.e.Re(W (z)) = φ(x, y), Im(W (z)) = ψ(x, y) (9.3)

The function ψ is called the stream function and it is obtained mathemat-ically from φ by solving the celebrated Cauchy–Riemann equations for Wwhich are of course

∂φ

∂x=∂ψ

∂y,

∂φ

∂y= −∂ψ

∂x(9.4)

One can also calculate the z-derivative of W (z) by differentiating W alongany direction: for example the x-direction. This means that we can write

dW

dz=∂φ

∂x+ i

∂ψ

∂x(9.5)

which can be useful. This last equation means that |dW/dz| has a simplephysical meaning: it turns out to be the speed |v| of the flow. Checking thiswe have

∣

∣

∣

∣

dW

dz

∣

∣

∣

∣

=

√

∂φ

∂x

2

+

∂ψ

∂x

2

=√

(−vx)2 + (vy)2

=√

(vx)2 + (vy)2

= |v|

(9.6)

24

Summarising the properties of the complex potential W , so as to be able tosee them all at once, we have

W = φ+ iψ

∂φ

∂x=∂ψ

∂y,

∂φ

∂y= −∂ψ

∂x

vx = −∂φ∂x

= −∂ψ∂y

vy = −∂φ∂y

=∂ψ

∂x

dW

dz=∂φ

∂x+ i

∂ψ

∂x= −vx + ivy

∣

∣

∣

∣

dW

dz

∣

∣

∣

∣

= |v|

(9.7)

One of the most useful properties of W is that the function ψ, whichwe recall is called the stream function, gives the actual lines of flow of thefluid—the so called streamlines of the flow. All we have to do is to set ψequal to a constant; i.e. we simply choose a constant C and write

ψ = C (9.8)

The point is that ψ is a function of x and y and so equating ψ to a con-stant fixes x in terms of y and hence defines some curve in the plane. Westate without proof the fact that all curves of flow of the fluid—i.e. allstreamlines—arise in this way by choosing a suitable value for C; converselyall curves of the form ψ = C correspond to streamlines.

All this becomes much easier to understand if we work with some ex-amples and so now proceed to an example.

Example Uniform flow at an angle to the x-axis

Choose

W = −U exp(−iα)z, U > 0 and α real constants (9.9)

Now we know that z = x+ iy and that W = φ+ iψ so we can say that

W = −U exp(−iα)z = φ+ iψ

⇒ −U(cos(α) − i sin(α))(x+ iy) = φ+ iψ

⇒ −U cos(α)x− U sin(α)y = φ, U sin(α)x− U cos(α)y = ψ(9.10)

25

So we already have isolated the velocity potential φ and the stream functionψ. Let us now find the streamlines. Setting ψ equal to a constant we get

U sin(α)x− U cos(α)y = C

⇒ y = tan(α)x− C

U cos(α)

(9.11)

But this is instantly recognisable as the equation of a straight line withslope tan(α) and intercept −C/U cos(α). Also by varying the constant Call we do is to change the intercept but not the slope. Hence the entire flowis a set of parallel lines which make an angle α with x-axis—cf. Fig. 6.

Finally we can calculate the speed of the flow by computing |dW/dz| =|v|. To this end we have

W = −U exp(−iα)z

⇒ dW

dz= −U exp(−iα)

⇒∣

∣

∣

∣

dW

dz

∣

∣

∣

∣

= U = a constant

(9.12)

Hence the speed |v| is just given by the constant U . This is why the flowis described as being uniform—the phrase uniform flow being just anotherterm for a flow of constant speed and direction.

α

Fig. 6: Streamlines: Uniform parallel flow at an angle α to the x-axis.

Example A source or sink

Let W be the potential

W = −m

2πln z, m a real constant (9.13)

26

The key trick here is to use the polar representation z = r exp(iθ) insteadof the Cartesian representation z = x+ iy. So we start off and obtain

W = −m

2πln z = φ+ iψ

⇒ −m

2πln(r exp(iθ)) = φ+ iψ, using z = r exp(iθ)

⇒ −m

2π(ln(r) + iθ) = φ+ iψ

⇒ φ = −m

2πln(r), ψ = −m

2πθ

(9.14)

and so we have very easily found φ and ψ. The streamlines ψ = C aretherefore given by

−m

2πθ = C

⇒ θ = −2πC

m= a constant

(9.15)

In other words the streamlines are just lines of constant θ i.e. they are radiallines cf. Fig. 7; such a flow is, for obvious reasons, called a source (if theflow is outwards from the origin) or a sink (if the flow is inwards towardsthe origin).

x

y

Fig. 7: Streamlines: A source

27

Example A vortex

Consider the potential

W = iκ

2πln z (9.16)

We notice that ln z has appeared again and, as we did when working withthe previous example, we should write z as z = r exp(iθ). Steaming aheadwe have

W = iκ

2πln z = φ+ iψ

⇒ iκ

2πln(r exp(iθ)) = φ+ iψ, (z = r exp(iθ))

⇒ iκ

2π(ln(r) + iθ) = φ+ iψ

⇒ φ = − κ

2πθ, ψ =

κ

2πln(r)

(9.17)

Having found φ and ψ the streamlines are clearly given by

κ

2πln(r) = C

⇒ ln(r) =2πC

κ= a constant

⇒ r = a constant

(9.18)

Thus the streamlines have constant r so they are circles centred at the origini.e.. we have a vortex, cf. Fig. 8.

Fig. 8: Streamlines: A vortex

28

If κ is positive the rotation is anti-clockwise and if it is negative therotation is clockwise.

Example The circulation of the vortex

Finally we shall calculate the circulation or, angular momentum per unitmass, of the vortex round a closed curve C.

We just choose C to be the circle of radius a centred at the origin sothat it is given by

r = a (9.19)

Then we want to calculate

Γ =

∫

C

v · dl (9.20)

Now dl is tangential to the circle C whose radius is a and so, if eθ is a unittangent to the circle, then

dl = r dθ eθ (9.21)

Hence, if vθ = v · eθ, we have

v · dl = rvθ dθ (9.22)

But in polar coordinates we have

vθ = −1

r

∂φ

∂θ

⇒ vθ = −1

r

(

− κ

2π

)

, since φ = − κ

2πθ

=κ

2πr

(9.23)

Putting this collection of modest facts together we find that

Γ =

∫ 2π

0

rvθ dθ

=

∫ 2π

0

κ

2π=[ κ

2πθ]2π

0

⇒ Γ = κ dθ

(9.24)

and so the circulation Γ round r = a of the vortex W = i κ2π

ln z is verysimple it is just given by the constant κ. Having spent some time discussingangular momentum of fluids we now move on to discuss another centralquantity which is the energy of the flow.

29

§ 10. Kinetic energy

Let an incompressible fluid undergo potential flow so that its velocity v isgiven by

v = −∇φ

then we shall derive a simple formula for its kinetic energy.Suppose that the fluid occupies a (finite or infinite) volume V then its

kinetic energy E is given by

E =1

2

∫

V

ρv2dV

=1

2

∫

V

ρ∇φ2dV

(10.1)

Now consider the vectorA = φ∇φ (10.2)

Evidently∇ · A = ∇ · (φ∇φ) = (∇φ) · (∇φ) + φ∇2φ

⇒ (∇φ)2 = −φ∇2φ+ ∇ · (φ∇φ)

⇒ E =ρ

2

∫

V

−φ∇2φ+ ∇ · (φ∇φ)

dV

(10.3)

But the fluid is incompressible so

∇ · v = 0 ⇒ ∇2φ = 0 (10.4)

Hence

E =ρ

2

∫

V

∇ · (φ∇φ) dV

=ρ

2

∫

S

φ∇φ · dS, using Gauss’ theorem

(10.5)

where the volume V has surface S. The last formula is the one we want sowe quote it again by itself giving us

E =ρ

2

∫

S

φ∇φ · dS (10.6)

We shall now use this in a practical way in a rather novel example which ac-tually looks, at first sight, to be completely outside the range of applicabilityof this formula.

30

Example The kinetic energy of flow outside a solid sphere

Let a solid sphere of radius r0 be embedded in an inviscid incompressiblefluid. The fluid is at rest at infinity and has a velocity potential φ given, inspherical polar coordinates with origin at the centre of the sphere, by

φ =1

2v0r30r2

cos(θ), v0 constant (10.7)

Hence find the total kinetic energy E of the fluid.At first it seems that the expression 10.1 we derived above for the

kinetic energy is not applicable; however this is not so. Note that our fluidis in three dimensional space which we can approximate (temporarily) by alarge hollow ball BR, say, of radius R.

When we do the fluid occupies the volume inside BR minus the vol-ume Br0

of the solid sphere; hence the region occupied by the fluid has asboundary the two separate surfaces SR and Sr0

of the two spheres.Now we can use our expression 10.1 for the energy E giving us

E =ρ

2

∫

SR

φ∇φ · dS− ρ

2

∫

Sr0

φ∇φ · dS (10.8)

where the second integral has a minus sign in front of it because dS alwayspoints outwards from the volume to which it belongs and this means thatdS is inwards on the surface Sr0

. However note that the integrand of eachintegral contains the term φ∇φ and it is easy to see that this is proportionalto r−3 on the surface of the sphere. For the large sphere this means thatas R → ∞ the entire integral over SR tends to zero (this also follows fromthe fact that the fluid is at rest at infinity). The upshot is that, in the limitwhere R→ ∞, which we must take in order to return to the whole of threedimensional space, the integral over SR tends to zero. Hence E is actuallygiven by the formula

E = −ρ2

∫

Sr0

φ∇φ · dS (10.9)

We now begin the calculation. If er denotes a unit vector in the radialdirection, then on Sr0

we have

dS = |dS| er = r20 dΩ er (10.10)

where we used the definition of solid angle. But for the component of ∇φin the radial direction we have

∇φ · er =∂φ

∂r(10.11)

31

So on Sr0we find that

∇φ · dS =∂φ

∂r

∣

∣

∣

∣

r=r0

r20dΩ (10.12)

Now

φ =1

2v0r30r2

cos(θ)

⇒ ∂φ

∂r= −v0

r30r3

cos(θ)

⇒ ∇φ · dS = −v0r30r3

∣

∣

∣

∣

r=r0

cos(θ)r20dΩ

(10.13)

But it is standard that dΩ is given by

dΩ = sin(θ)dθdφ (10.14)

So we find that the entire integrand of the energy expression is just

ρ

2φ|r=r0

∇φ · dS = −ρ2

(

1

2v0r30r20

)

cos(θ)2 sin(θ)v0r20dθdφ (10.15)

Hence E is given by

E =ρ

4

∫

v20r

30 cos2(θ) sin(θ)dθdφ

=2πρv2

0r30

4

∫ π

0

cos2(θ) sin(θ)dθ

= −2πρv20r

30

4

[

cos3(θ)

3

]π

0

= −πρv20r

30

2

−1

3− 1

3

⇒ E =πρv2

0r30

3

(10.16)

and so the energy is found.The next section marks a return to the subject of vorticity and angular

momentum.

§ 11. A vorticity equation and its shortcomings

Consider Euler’s equation for an inviscid, incompressible, fluid in the pres-ence of a body force F. Hence we have the equation

Dv

Dt= F − 1

ρ∇p (11.1)

32

Let F be conservative so that F = −∇K giving us

Dv

Dt= −∇K − 1

ρ∇p (11.2)

Using the full form for the stream derivative Dv/Dt we get

∂v

∂t+ (v · ∇)v = −∇K − 1

ρ∇p (11.3)

We also recall that

(v · ∇)v = ∇(

v2

2

)

+ (∇× v) × v (11.4)

and using this we obtain

∂v

∂t+ ∇

(

v2

2

)

+ (∇× v) × v = −∇K − 1

ρ∇p (11.5)

Now, in an effort to obtain an equation for the vorticity ω = ∇×v, wetake the curl of both sides yielding

∇× ∂v

∂t+ ∇×∇

(

v2

2

)

+ ∇× (ω × v) = −∇×∇K − 1

ρ∇×∇p (11.6)

But since ∇×∇f = 0 for any f this reduces at once to the simpler equation

∇× ∂v

∂t+ ∇× (ω × v) = 0 (11.7)

and, if we use the fact that ∇ × ∂v/∂t = ∂(∇ × v)/∂t = ∂ω/∂t, we thenget our unsatisfactory equation for the vorticity ω, which is

∂ω

∂t+ ∇× (ω × v) = 0 (11.8)

The reason that this equation is unsatisfactory is the following: In generalthere is a vector identity for ∇× (ω × v) giving

∇× (ω × v) = (v · ∇)ω − (ω · ∇)v (11.9)

But in two dimensions (we state this fact without proof but it is easy tocheck) it is the case that

(ω · ∇)v = 0 (11.10)

33

Hence, in two dimensions, our vorticity equation becomes

∂ω

∂t+ (v · ∇)ω = 0

i.e.Dω

Dt= 0

⇒ ω is conserved??

(11.11)

Now ω cannot be conserved because this would mean that vorticity, oncecreated, would never disappear from a fluid, or, once absent could neverbe created. This state of affairs is in flat contradiction with experiment sothere is something amiss.

It turns out that if we include the viscosity η of the fluid then thecontradiction is avoided and we return to agreement with experiment. Thisis what we turn to in the next section.

§ 12. Viscous flow, the Navier-Stokes equation and the satisfactory

vorticity equation

We shall now simply quote 9 that the viscous forces in an incompressiblefluid are given by adding the term

η

ρ∇2v (12.1)

to the Euler equation yielding the

Dv

Dt= F− 1

ρ∇p+

η

ρ∇2v (12.2)

This equation 12.2 is the celebrated Navier–Stokes equation for viscous flow.It is now very simple indeed to check that if we go through the same

steps as we did in the previous section to derive an equation for the vorticityω the only new feature is the viscous term (η/ρ)∇2v and the effect ofthis term on the derivation is to add to 11.8 its curl—i.e. the quantity(η/ρ)∇×∇2v so that our vorticity derivation now gives the equation

∂ω

∂t+ ∇× (ω × v) =

η

ρ∇×∇2v =

η

ρ∇2(∇× v)

⇒ ∂ω

∂t+ ∇× (ω × v) =

η

ρ∇2

ω

(12.3)

9 In the longer version of this course we also provide a derivation of this fact.

34

So our new vorticity equation is

∂ω

∂t+ ∇× (ω × v) =

η

ρ∇2

ω (12.4)

and this has no drawbacks.We want to point out that, for small velocities, the second term on the

LHS of 12.4 is smaller than the first one since it is quadratic in v. If weapproximate by neglecting this term then the vorticity equation for smallvelocities is

∂ω

∂t=η

ρ∇2

ω (12.5)

and this is exactly like the heat flow equation

∂T

∂t= κ∇2T (12.6)

where T is the temperature inside a solid slab, of thermal conductivity κ,through which heat is flowing. Hence we obtain the useful conclusion that,for small velocities, vorticity diffuses through a fluid in the same way asheat does through a solid.

§ 13. The Reynolds number

In this section we apply the technique of dimensional analysis to viscous flowand obtain a formula for a very interesting number R known as the Reynoldsnumber of the flow. The Reynolds number allows one to understand manyfeatures of viscous flow and provides an understanding of why wind tunnelexperiments work.

We begin with viscous incompressible flow past an upright cylinder, cf.Fig.9.

a

Fig. 9: Viscous flow past an upright cylinder of radius a

35

Suppose that the cylinder has radius a and that far away from thecylinder the speed of the flow tends to the constant value V . Then the fourparameters

η, ρ , a and V (13.1)

characterise the flow and combine to form the number R known as theReynolds number whose definition is

R =ρV a

η(13.2)

It is possible to let the Reynolds number emerge naturally from a discus-sion of the flow which involves a change of variables: First remember thatone is interested in factors which particularly depend on shape rather thansize—think of the wind tunnel where one builds models of the same shapebut usually not of the same size as the real life object—one changes to adimensionless set of variables and, when one does this, it transpires out thatR turns up naturally.

To this end let us change the variables x, y, z, t and v to the dimen-sionless set of variables

x′, y′, z′, t′ and v′ (13.3)

where

x′ =x

a, y′ =

y

a, z′ =

z

a, t′ =

V

at and v′ =

v

V(13.4)

The next step is to make this change of variables in our vorticity equation

∂ω

∂t+ ∇× (ω × v) =

η

ρ∇2

ω (13.5)

If we note that, for any function f ,

∂f

∂x=

∂f

∂x′∂x′

∂x=

1

a

∂f

∂x′(13.6)

then it is clear that

∇ =1

a∇′ (13.7)

where ∇′ is the same as ∇ except for the substitution of derivatives in x′, y′

and z′ for the derivatives in x, y and z. Now we see that, for the the vorticity

36

ω , we have

ω = ∇× v

=1

a∇′ × v

=V

a∇′ × v′

⇒ ω =V

aω

′

(13.8)

where, as should be expected, ω′ denotes ∇′ × v′. In a similar fashion we

can calculate that∂f

∂t=V

a

∂f

∂t′(13.9)

Thus∂ω

∂t=V

a

∂ω

∂t′

=V 2

a2

∂ω′

∂t′

(13.10)

Also

ω × v =

(

V

aω

′

)

× (V v′) =V 2

aω

′ × v′ (13.11)

from which we readily deduce that

∇× (ω × v) = ∇×(

V 2

aω

′ × v′

)

=1

a∇′ ×

(

V 2

aω

′ × v′

)

=V 2

a2∇′ × (ω′ × v′)

(13.12)

This means that we have dealt with all the terms on the LHS of our vorticityequation; it is completely straightforward to verify that

∇2ω =

V

a3(∇′)

2ω

′ (13.13)

Hence, in the new dimensionless variables, we now have the vorticity equa-tion

V 2

a2

∂ω′

∂t′+V 2

a2∇′ × (ω′ × v′) =

η

ρ

V

a3(∇′)

2ω

′ (13.14)

37

which we proceed to rewrite in the form

∂ω′

∂t′+ ∇′ × (ω′ × v′) =

η

ρV a(∇′)

2ω

′ (13.15)

that is

∂ω′

∂t′+ ∇′ × (ω′ × v′) =

1

R(∇′)

2ω

′, with R =ρV a

η(13.16)

and we see that, as promised above, the Reynolds number R has emergednaturally.

First of all one should think of R loosely as being given by

R =ρ

η× a typical length × a typical velocity (13.17)

We now want to explain the significance of the Reynolds number R. Let usconsider the example of a wind tunnel: One takes the real full size aircraftwith parameters η1, ρ1, V1 and a1 giving rise to a Reynolds number R1,say; then one takes the model aircraft in the wind tunnel with parametersη2, ρ2, V2 and a2 giving rise to a Reynolds number R2. The vital fact aboutthe wind tunnel, though, is that one arranges that its Reynolds number R2

is the same as the original Reynolds number R1. Hence we always have

R1 = R2

orρ1V1a1

η1=ρ2V2a2

η2

(13.18)

So the individual values of η1, ρ1, V1, a1 and η2, ρ2, V2, a2 can, and do,differ but only in a manner which keeps R1 = R2.

Now we can also see why we can use the phrase a typical length and atypical velocity in our expression 13.17 above for R. The point is that if wereplace the length a by 2a then R increases by a factor of 2; but if we nowlook at the wind tunnel discussion this would have the effect of replacingR1 and R2 by 2R1 and 2R2 respectively, and we see that it doesn’t matterwhether we require 2R1 to equal 2R2 or just R1 to equal R2 we get thesame relation among the two sets of parameters. A similar remark appliesto the parameter V .

This method of matching Reynolds numbers is also used in marinetechnology when designing boats. The model boats are floated in tankswhich sometimes contain the liquid mercury—a liquid metal—so we see

38

that very different fluids can be used in the two situations as long as theReynolds numbers match.

The size of the Reynolds number also controls the production of vorticesand the consequent onset of turbulence as we shall illustrate diagrammati-cally in the next section.

§ 14. Turbulence and the Reynolds number

Consider the set of five diagrams shown below in Fig. 10; they show viscousflow past our cylinder for steadily increasing values of the Reynolds numberR.

Fig. 10: Viscous flow past a cylinder as R increases

We note that in Fig. 10 (a) where R = 10−2, there is no vorticity butin Fig. 10 (b) where R has been increased to 20 there is now a pair ofvortices behind the cylinder but staying close to the solid boundary. Then,as R increases, the vortices break away from the cylinder and move off down

39

stream cf. Fig. 10 (c) where R = 100. The final stages are attained whenR goes through the values 104–106 which we see in the last two figures. InFig. 10 (d) the vortices are huge in number and move together in clumpsthat look a bit like egg shells; then when R = 106 there is a continuouswake of turbulent vortices starting at the solid boundary (i.e. the cylinder)and stretching indefinitely far downstream. This flow is no longer steadyand indeed has been unsteady since R increased above the value 40 or so.

We are now ready to look at some solutions to the Navier-Stokes equa-tions.

§ 15. Some exact solutions to the Navier–Stokes equations

The first problem in viscous flow that we shall solve is the flow down acircular pipe.

Example Viscous flow in a pipe or Poiseuille flow

We shall take a horizontal circular pipe of radius a and length L down whichincompressible viscous fluid is flowing. The assumptions that we make tosimplify the calculation somewhat are(i) The flow is steady.(ii) The flow is parallel to the axis of the pipe.(iii) There is no body force.(iv) There is a constant pressure difference across the ends of the pipe.

A vital extra experimental fact about viscous flow past a solid object isthat the fluid velocity dies away to zero on the surface of the object. Thisis called the no slip boundary condition and must not be forgotten.

We begin with the full Navier–Stokes equation

∂v

∂t+ (v · ∇)v = F− 1

ρ∇p+

η

ρ∇2v (15.1)

Now since there is no body force and the flow is also steady we have at oncethat

F = 0,∂v

∂t= 0 (15.2)

so we are left with the equation

(v · ∇)v = −1

ρ∇p+

η

ρ∇2v (15.3)

Next we choose to make the x-axis coincide with the axis of the pipe and,having done this, the fact that the flow is parallel means that v only has anx-component i.e.

v = vx(x, y, z)i (15.4)

40

Steaming on we note that incompressibility means that

∇ · v = 0

⇒ ∂vx

∂x= 0, since vy = vz = 0

⇒ vx = vx(y, z)

(15.5)

so vx is independent of x. This has the very useful consequence that (v ·∇)vwhich in full is given by

(v · ∇)v = (vx

∂

∂x+ vy

∂

∂y+ vz

∂

∂z)(vxi + vyj + vzk) (15.6)

reduces to

(v · ∇)v = vx

∂vx

∂xi

= 0(15.7)

Hence the Navier–Stokes equation has now become just

0 = −1

ρ∇p+

η

ρ∇2v

⇒ ∇2v =1

η∇p

(15.8)

We also know that there is a constant pressure difference across the ends ofthe pipe and we shall further assume 10 that p only depends linearly on xand so is given by

p = Ax+B, A and B constants (15.9)

Now if we let the beginning of the pipe be at x = 0 and label the pressurethere by P0 then the end of the pipe must be at x = L and we denote thepressure there by P1. Thus we have

p(x = 0) = P0, p(x = L) = P1 (15.10)

10 We don’t have to do this but it simplifies things. In fact the Navier Stokes equation

for the components vy and vz , which we know are zero, show that ∂p/∂y = ∂p/∂z = 0 so

that p only depends on x. To show that this dependence is linear we take the divergence of

the Navier–Stokes equation for vx which incompressibility reduces for us to the statement

∇2p = 0; this then immediately gives p = Ax + B since ∇2 reduces to d2/dx2 in this

case.

41

from which we deduce that

A =(P1 − P0)

L, B = P0 (15.11)

giving

p =(P1 − P0)

Lx+ P0

⇒ ∇p =(P1 − P0)

Li

(15.12)

Now we turn to the Navier–Stokes equation for vx which is

∇2vx(y, z) =(P1 − P0)

ηL

⇒(

∂2

∂y2+

∂2

∂z2

)

vx(y, z) =(P1 − P0)

ηL

(15.13)

This equation is simple enough to guess its solution. Simply note that, if Cand D are constants, then a solution is

vx = C −D(y2 + z2) (15.14)

if we choose D appropriately. Substituting in to the Navier–Stokes equationgives us the requirement that

4D = −(P1 − P0)

ηL(15.15)

so D is found. This must be the unique solution if we can make it satisfythe no slip boundary condition

v = 0, on the surface of the pipe (15.16)

But the pipe has radius a and so its surface has equation

y2 + z2 = a2 (15.17)

so setting y2 + z2 = a2 in the equation vx = C −D(y2 + z2) and requiringvx to vanish gives

0 = vx = C −Da2 (15.18)

42

so that C = Da2 and our solution is complete. Summarising we have foundthat

v = vxi

with vx =(P1 − P0)

4ηL(y2 + z2 − a2)

(15.19)

Poiseuille’s law The quantity of fluid delivered by a pipe of radius a.

It is very interesting to calculate the total mass Q of fluid delivered by theflow down the pipe per second.

For convenience we set

r2 = y2 + z2 (15.20)

and now we consider a thin annulus of radius r (r < a) and thickness drinside the pipe. This annulus flows a distance precisely vx(r) in one secondand thereby traces out a volume 2πrdrvx(r). The mass of fluid in thisvolume is just

ρ 2πrdrvx(r) (15.21)

and so the total mass Q flowing through the pipe per second is given byintegrating over r. We have

Q =

∫ a

0

ρ 2πrvx(r)dr

Using

vx(r) =(P1 − P0)

4ηL(r2 − a2) (15.22)

we find that

Q = 2πρ(P1 − P0)

4ηL

∫ a

0

(r2 − a2)rdr

= 2πρ(P1 − P0)

4ηL

[

r4

4− r2a2

2

]a

0

= 2πρ(P1 − P0)

4ηL

(

−a4

4

)

⇒ Q = πρ(P0 − P1)

8ηLa4

(15.23)

This formula

Q = πρ(P0 − P1)

8ηLa4 (15.24)

43

for Q is known as Poiseuille’s law.We note that Q has a very strong dependence on the radius a of the

pipe: we see thatQ ∝ a4 (15.25)

This law applies to the flow of gases and liquids in all kinds of differentcircumstances. The a4 dependence has some dramatic consequences—weprovide one illustration.

Let us consider blood flow in an artery of radius a. If cholesterol depo-sition has narrowed the artery by 10%—this is a conservative supposition,in individuals with severe heart disease an artery can be blocked—then a isreduced to 0.9a. This 10% narrowing means that Q is reduced by a factor

(0.9a)4 = 0.6561a4 (15.26)

which amounts to a reduction in blood flow of approximately 35%. So wehave found that a loss of more than a third of the blood throughput comesfrom a reduction in the artery radius of 10%. This is somewhat idealisedas, actually, arteries have a certain amount of elasticity. The reader canreadily construct other examples of this formula in action.

Example Viscous flow in an infinitely deep straight channel

Our next example models flow in a deep straight canal. We shall takethe flow of an incompressible viscous fluid down a straight channel. Ourassumptions this time are(i) The flow is steady.(ii) The channel is infinitely long, infinitely deep and has parallel flat walls.(iii) The flow is parallel to the walls.

We also recall that, since the flow is viscous, we automatically have theno slip boundary condition which asserts that

v = 0, on the walls (15.27)

We take the width of the channel to be 2b and show its orientation relativeto the x− y plane in Fig. 11 below.

b

2b

parabolic velocity profile

i

j

Fig. 11: Viscous flow through a straight channel

44

Now, since the flow is parallel, we have

v = vxi (15.28)

Now just as we had in the previous example incompressibility gives

∇ · v = 0

⇒ ∂vx

∂x= 0, since vy = vz = 0

⇒ vx = vx(y, z)

(15.29)

This also gives again(v · ∇)v = 0 (15.30)

When we take account of the fact that the flow is steady, and that there isno body force, the Navier–Stokes equation becomes

0 = −∇pρ

+η

ρ∇2v (15.31)

So far this is very close to the previous example; a further simplificationhere is that, although we have,

vx = vx(y, z) (15.32)

in fact vx cannot depend on z since the channel has infinite depth. Henceall we have for vx is

vx = vx(y) (15.33)

The resulting Navier–Stokes equation is now very simple being

∇2vx =1

η

dp

dx

⇒ d2vx

dy2=

1

η

dp

dx= a constant

(15.34)

where we note that dp/dx is constant as it was in the Poiseuille example.All this means that we can integrate the Navier–Stokes equation twice andwe have its solution which is

vx =1

2η

dp

dxy2 + Ay +B, A and B constants (15.35)

45

We find A and B by imposing the no slip boundary condition on the walls,i.e.

vx = 0, when y = ∓b (15.36)

So when y = b we have vx = 0 yielding

1

2η

dp

dxb2 + Ab+B = 0 (15.37)

and when y = −b we also have vx = 0 so that

1

2η

dp

dxb2 − Ab+B = 0 (15.38)

We easily solve these two equations for A and B obtaining the result

A = 0, B = − 1

2η

dp

dxb2 (15.39)

and so our final solution for the velocity v down the channel is

v = vxi

where vx =1

2η

dp

dx(y2 − b2)

(15.40)

We note that v has a parabolic velocity profile as indicated in Fig. 11.

Example Viscous flow through a channel with one moving wall

Our last example is a refinement of the previous one where we allow oneof the walls to move while the other remains stationary—cf. Fig. 12. Thissituation models the flow between the side of a super tanker as it comes into dock.

i

j

Vx =

Vx =

h

0

U

Fig. 12: Viscous flow past a moving wall

46

We see from Fig. 12 that the upper wall moves parallel to the lowerone with speed U . We note, too, that for convenience we have moved theposition of the x− y axes and also renamed the width of the channel to beh.

Mathematically the only new feature, as compared with the previousexample, is that the boundary conditions change. They are now

vx = U, at y = h

vx = 0, at y = 0(15.41)

The boundary condition at the upper wall ensures that the fluid does notslip when it touches this wall. Again we have a solution of the form

vx =1

2η

dp

dxy2 + Ay +B (15.42)

and the boundary conditions give the simultaneous equations

1

2η

dp

dxh2 + Ah+B = U, at y = h

B = 0, at y = 0

(15.43)

from which we deduce that

A =U

h− 1

2η

dp

dxh, B = 0 (15.44)

so that our final solution is

v = vxi

where vx =1

2η

dp

dx(y2 − hy) +

U

hy

(15.45)

The shape of the velocity profile, in this super tanker case, is determinedby the quantity S where

S = −h2

2η

dp

dx(15.46)

If we use S we can write vx as

vx = Sy

h

(

1 − y

h

)

+U

hy (15.47)

Then we can distinguish three distinct cases, namely,

47

(i) S > 0(ii) S = 0(iii) S < 0

If S > 0 then dp/dx < 0 and p decreases as we move to the right alongthe channel giving forward flow.

If S = 0 than dp/dx = 0 and we have what is called simple Couetteflow giving a straight velocity profile.

Finally, if S < 0, then dp/dx > 0 and p increases as we move to theright along the channel giving the possibility of backward flow cf. Fig. 13.

S> 0S= 0S< 0

Fig. 13: The velocity profile and the parameter S

§ 16. ‘Derivation’ of the Navier–Stokes equation

We start with the Euler equation with an unknown viscous term F added

ρDv

Dt= −∇p+ F (16.1)

and the idea is that we want to derive the form of this term F.From now on we use the summation convention for indices. To make

any further progress we have to assume that the fluid is what is calledNewtonian (almost all fluids are Newtonian) and this means that the viscousterm F is given in terms of a derivative as follows

F = Fiei

where Fi = −∂jQij , for some Qij

(16.2)

Qij is called the shear stress. We keep this form for F in storage for a fewlines and turn to do a little work on the other terms in the equation ofmotion. Using the definition of Dv/Dt this equation can be written as

ρ∂v

∂t= −ρ(v · ∇)v −∇p+ F (16.3)

48

But∂

∂t(ρv) =

∂ρ

∂tv + ρ

∂v

∂t

⇒ ρ∂v

∂t=

∂

∂t(ρv) − ∂ρ

∂tv

(16.4)

the continuity equation says that

∂ρ

∂t+ ∇ · (ρv) = 0 (16.5)

and substituting for ∂ρ/∂t in the equation above we find that

ρ∂v

∂t=

∂

∂t(ρv) + ∇ · (ρv)v (16.6)

Now we substitute for ρ∂v/∂t in 16.3 obtaining thereby the equation

∂

∂t(ρv) + ∇ · (ρv)v = −ρ(v · ∇)v −∇p+ F

⇒ ∂

∂t(ρv) = −∇ · (ρv)v − ρ(v · ∇)v −∇p+ F

⇒ ∂

∂t(ρv) = −∇ · (ρv) + ρ(v · ∇)v −∇p+ F

(16.7)

But ∇ · (ρv) expands according to the identity

∇ · (ρv) = ρ∇ · v + v · ∇ρ (16.8)

and using this we obtain the equation

∂

∂t(ρv) = −ρ∇ · v + v · ∇ρ+ ρ(v · ∇)v −∇p+ F (16.9)

However if we adopt the summation convention we can simply check bydifferentiation that the following is true

ei∂j(ρvivj) = ρ∇ · v + v · ∇ρ+ ρ(v · ∇)v (16.10)

and thus we obtain

∂

∂t(ρv) = −∇p− ei∂j(ρvivj) + F (16.11)

49

But recall that we had, for F = Fiei, the Newtonian fluid property Fi =−∂jQij ; if we add to this the fact that

v = vi ei, and ∇p = ∂ip ei (16.12)

then we can write the equation of motion in component form as

∂

∂t(ρvi) = −∂j(p δij + ρvivj +Qij) (16.13)

where we used the easily verified property that ∂ip = ∂jp δijWe point out in passing (we shall not actually make any use of this fact

here) that we can combine together the first and third terms on the RHSof this equation by introducing what is called the stress tensor Sij whosedefinition is simply that 11

Sij = −pδij −Qij (16.16)

For a Newtonian fluid we make the assumption that Qij is symmetric ini and j and constructed entirely from derivatives of the velocity components.To this end we write 12

Qij =a

2(∂ivj + ∂jvi) + b(∂kvk)δij , a and b constants (16.19)

11 The key property of Sij being, as one can check in one differentiation, that

∂jSij = −∂ip − ∂jQij (16.14)

so the Navier–Stokes equation 16.13 takes on the compact form

∂

∂t(ρvi) = −∂j(ρvivj) + ∂jSij (16.15)

12 This somewhat arbitrary looking assumption can be elaborated on though we donot have the space to give much more detail here. Nevertheless we add that if one thinksof the evolution of the velocity vector from v at the point (x, y, z) to v + dv at the point(x + dx, y + dy, z + dz), then the chain rule for partial differentiation says that

dvi =∂vi

∂xjdxj

=1

2

(

∂vi

∂xj+

∂vj

∂xi

)

+

(

∂vi

∂xj−

∂vj

∂xi

)

dxj

= MSijdxj + MA

ijdxj

where MSij =

1

2

(

∂vi

∂xj+

∂vj

∂xi

)

, MAij =

1

2

(

∂vi

∂xj−

∂vj

∂xi

)

(16.17)

50

Now the constants a and b, as defined, are not in general positive, so, to getrid of this inconvenience we replace them by the two constants

η, and ζ (16.20)

which are always positive. The relation between the two sets of constantsis just that

a = −2η and b =2

3η − ζ (16.21)

The constants η and ζ are known as coefficients of viscosity—of course wehave already met η but ζ is a new one. Using η and ζ we find that

Qij = −η (∂ivj + ∂jvi) +

(

2

3η − ζ

)

(∂kvk)δij (16.22)

Now we compute ∂jQij and find that

∂jQij = −η∂j (∂ivj + ∂jvi) +

(

2

3η − ζ

)

∂j(∂kvk)δij

= −η∂j∂jvi − η∂i(∂jvj) +

(

2

3η − ζ

)

∂i(∂kvk)

= −η∂j∂jvi −η

3∂i(∂kvk) − ζ∂i(∂kvk), since ∂jvj = ∂kvk

= −η∂j∂jvi −(

ζ +η

3

)

∂i(∂kvk)

(16.23)

Now if we insert this expression for Qij back where it came from, i.e. into16.13, it yields the equation

∂

∂t(ρvi) = −∂ip− ∂j(ρvivj) + η∂j∂jvi +

(

ζ +η

3

)

∂i(∂kvk) (16.24)

It turns out that MAijdxj only rotates an infinitesimal cube of fluid since it is clear that

eiMAijdxj = −

1

2ω × dr (16.18)

but the term MSij

dxj strains the cube of fluid—i.e. changes its volume or shape and it is

this action which is resisted by viscous forces. This makes it plausible that MSij should

be part of the tensor Qij and this is one of the main parts of the assumption that a fluid

is Newtonian. This footnote is not an explanation but just a pointer to where one can

go for further material.

51

The last step in this calculation is to put 16.24 back into the usual vectorform. To accomplish this multiply both sides of 16.24 by ei and note that,from our earlier manipulations, we know that

ej∂j = ∇, ∂j∂j = ∇2

ei

∂

∂t(ρvi) + ei∂j(ρvivj) = ρ

∂v

∂t+ ρ(v · ∇)v = ρ

Dv

Dt

(16.25)

The result of this is to give us the equation

ρDv

Dt= −∇p+ η∇2v +

(

ζ +η

3

)

∇(∇ · v) (16.26)

This equation is the full Navier–Stokes equation for a viscous Newtonianfluid and we see that it involves the two coefficients of viscosity η and ζ.However we shall only do detailed calculations for a viscous incompressiblefluid so that ∇ · v = 0; and we see at once that this makes the termcontaining ζ in 16.26 disappear leaving us with the simpler version of theNavier–Stokes equation that we have met before namely

ρDv

Dt= −∇p+ η∇2v (16.27)

and so we are back to having only one coefficient of viscosity which is, ofcourse, η.

As an illustration of the viscosity in action we now look at the dragcreated as a viscous fluid flows past a solid boundary.

Example The viscous drag on a flat plate

We return to the problem of the flow between two flat plates one of whichis moving cf. Fig. 12.

We already know that the solution of this problem is

v = vx i, vx = Sy

h

(

1 − y

h

)

+U

hy (16.28)

where

S = −h2

2η

dp

dx= a constant (16.29)

The shear stress Qij has only one non-zero component namely Q12 and ifwe use 16.22 we have

Q12 = −η(

∂v1∂x2

+∂v2∂x1

)

(16.30)

52

Now clearly we have the correspondences

(vx, vy, vz) = (v1, v2, v3), (x, y, z) = (x1, x2, x3) (16.31)

So we know at once that

v1 = Sx2

h

(

1 − x2

h

)

+U

hx2, v2 = v3 = 0 (16.32)

and so we easily compute that

Q12 = −η(

∂v1∂x2

+ 0

)

= −η

S1

h

(

1 − x2

h

)

+ Sx2

h

(

− 1

h

)

+U

h

= −η

−h2

2η

dp

dx1

1

h

(

1 − x2

h

)

− h2

2η

dp

dx1

x2

h

(

− 1

h

)

+U

h

=h2

2

dp

dx1

1

h

(

1 − x2

h

)

+h2

2

dp

dx1

x2

h

(

− 1

h

)

− ηU

h

(16.33)

Now the viscous drag is given by the contribution to Q12 which is left if weset the pressure gradient dp/dx1 to zero: doing this gives just the term

ηU

h= the viscous drag on the upper plate (16.34)

which we also note is the only part of Q12 which depends on the viscosityη.

§ 17. Dissipation of energy in a viscous incompressible fluid

We are already familiar with the expression for the kinetic energy T of anincompressible fluid since we met it in 10.1; in any case we know that

T =ρ

2

∫

V

v2 d3r (17.1)

The rate of dissipation of energy on the fluid is just T so our task is tocalculate

T =∂

∂t

ρ

2

∫

V

v2 d3r

=ρ

2

∫

V

∂v2

∂td3r, since ρ is constant

(17.2)

53

We see that the main task is to calculate ∂v2/∂t and we set about doingthis right away. We have

ρ

2

∂v2

∂t= ρv · ∂v

∂t

= −ρv · (v · ∇)v − v · ∇p+ ηv · ∇2v, using 16.27

(17.3)

But

(v · ∇)v = ∇(

v2

2

)

+ (∇× v) × v

⇒ v · (v · ∇)v = v · ∇

v2

2

, since v · (∇× v) × v = 0

(17.4)

Thus we can now write

ρ

2

∂v2

∂t= −ρv · ∇

v2

2+p

ρ

+ ηv · ∇2v (17.5)

The next step requires us to do some manipulations on the viscous termηv · ∇2v. First note that

∇2vi = ∂j∂jvi (17.6)

and also that since we have assumed that the fluid is incompressible we have∇ · v = ∂jvj = 0 and so we can write

∇2vi = ∂j∂jvi = ∂j(∂jvi + ∂ivj)

= ∂jτij, where τij = ∂jvi + ∂ivj

(17.7)

Hence the viscous term ηv · ∇2v is given by

ηvi∂jτij (17.8)

With this information 17.5 becomes

ρ

2

∂v2

∂t= −ρv · ∇

v2

2+p

ρ

+ ηvi∂jτij (17.9)

Now note that

η∂j(viτij) = η∂jviτij + ηvi∂jτij

⇒ ηvi∂jτij = η∂j(viτij) − η∂jviτij

= η∇ · (θiei) − η∂jviτij , where θi = viτij

(17.10)

54

Now the latest version of our equation for (ρ/2)∂(v2)/∂t is

ρ

2

∂v2

∂t= −ρv · ∇

v2

2+p

ρ

+ η∇ · (θiei) − η∂jviτij

= −v · ∇

ρ

(

v2

2+p

ρ

)

+ η∇ · (θiei) − η∂jviτij , since ρ is constant

(17.11)The last step in our workings is to use the vector identity

∇ · (fA) = A · ∇f + f∇ · A (17.12)

with

A = v and f = ρ

v2

2+p

ρ

(17.13)

With this choice of f and A we find that

v · ∇

ρ

(

v2

2+p

ρ

)

= ∇ ·

ρ

(

v2

2+p

ρ

)

v

− rho

(

v2

2+p

ρ

)

∇ · v

= ∇ ·

ρ

(

v2

2+p

ρ

)

v

, since ∇ · v = 0

(17.14)Substituting into our last equation for (ρ/2)∂(v2)/∂t and combining all the‘div’ terms into one we obtain the result that we were after, namely

ρ

2

∂v2

∂t= −∇ ·

ρ

(

v2

2+p

ρ

)

v + ηθiei

− η∂jviτij (17.15)

Finally substituting 17.15 into 17.2 yields the equation

T = −∫

V

∇ ·

ρ

(

v2

2+p

ρ

)

v + ηθiei

dV − η

∫

V

∂jviτijdV

= −∫

S

ρ

(

v2

2+p

ρ

)

v + ηθiei

· dS − η

∫

V

∂jviτijdV

(17.16)

where S is the surface of the volume V and we have used Gauss’s divergencetheorem. However we now impose the no slip boundary condition on thesurface S which means that the entire integrand of the surface integralvanishes when v = 0 so we are left with the dissipation formula

T = −η∫

V

∂jviτijdV (17.17)

55

and this formula was our goal.We finish by remarking that we would expect, on physical grounds,

that T should decrease with time as a viscous liquid flows. This means thatT should be negative. In fact we can easily see that this is indeed correctfor note that we can write ∂jviτij as

∂jviτij = (∂jvi)(∂jvi + ∂ivj)

=1

2(∂jvi + ∂ivj)

2(17.18)

where multiplying out the last expression explicitly and dividing by the 1/2will soon convince the reader that the last equality is correct. This meansthat

T = −η∫

V

1

2(∂jvi + ∂ivj)

2dV (17.19)

and since η is positive, and the integrand is now explicitly positive, one seesimmediately that T < 0 as conjectured.

§ 18. Assorted problems on inviscid and viscous fluids

Problem 1. An infinite mass of inviscid fluid of constant density ρ isinitially at rest and has a spherical cavity of radius

a(t) = a+ cos(bt)

embedded in it where a and b are constants. The fluid moves radially out-wards, the pressure at infinity is zero, and the velocity potential φ at a pointa distance r from the centre of the cavity is given by

φ(t) =f(t)

r

Calculate the pressure p on the surface of the cavity and find when p attainsits maximum.

Solution sketch: The fluid is inviscid and there is a velocity potential andthis suggests that we use Bernoulli’s equation to calculate the pressure.

Hence we start with

−∂φ∂t

+v2

2+K +

∫

dp

ρ= C(t) (18.1)

56

Since φ(t) = f(t)/r and the pressure at ∞ is zero then taking r → ∞ inBernoulli’s equation shows that C(t) = 0. Also here, K = 0, and the fluidis incompressible so we obtain

−∂φ∂t

+v2

2+p

ρ= 0

⇒ p = ρ

∂φ

∂t− v2

2

(18.2)

The key idea of the solution is to realise that, on the surface of the cavityone can equate da(t)/dt and the radial component of velocity vr = −∂φ/∂r.More explicitly we have

da(t)

dt= − ∂φ

∂r

∣

∣

∣

∣

r=a(t)

, with a(t) = a+ cos(bt)

⇒ −b sin(bt) =f(t)

a(t)2, using φ(t) =

f(t)

r

⇒ f(t) = −b sin(bt)(a+ cos(bt))2

(18.3)

Now all the ingredients are in place to calculate the pressure p since weknow f(t) and so can calculate ∂φ/∂t, in addition v2 = a2(t); substitutingall this into our expression above for p we find that

p = ρ

−b2 sin2(bt)(a+ cos(bt))4

2((a+ cos(bt))4− b2 cos(bt)

(a+ cos(bt))(a+ cos(bt))2

+2b2 sin2(bt)(a+ cos(bt))

(a+ cos(bt))

(18.4)

which simplifies to

p = ρ

3b2

2sin2(bt) − b2 cos(bt)(a+ cos(bt))

(18.5)

and so we have now calculated the pressure on the surface of the cavity. Weare also asked to find when p attains its maximum: to do this we simplycompute dp/dt and equate it to zero. The reader will easily find that thisgives the equation

cos(bt) = −a5

(18.6)

so that p attains its maximum when

t =1

bcos−1

(

−a5

)

(18.7)

57

Problem 2. Incompressible viscous fluid flows steadily parallel to the axisin the space between two fixed coaxial cylinders of radii a and b, a < b,under a constant pressure gradient P. Show that the velocity v is given by

v =P

4η

[

r2 − (a2 − b2)

ln(a/b)ln r − a2 +

(a2 − b2)

ln(a/b)ln(a)

]

e

where e is a unit vector in the direction of the axis.

Solution sketch: This time we have a viscous fluid so we must use theNavier–Stokes equation which, since there is no body force and the fluid isalso incompressible, takes the form

∂v

∂t+ (v · ∇)v = −1

ρ∇p+

η

ρ∇2v (18.8)

We shall take the common axis of the cylinders to coincide with the z-axisso that e = k. But the flow steady and parallel to the axis so

∂v

∂t= 0, and v = vz k (18.9)

Incompressibility plus parallel flow means that

(v · ∇)v = 0 (18.10)

for exactly the same reasons as it did in some earlier works, cf. eq. 15.7 forexample. So we are left with the task of solving

(

∂2

∂x2+

∂2

∂y2

)

vz(x, y) =P

η(18.11)

To do this we use cylindrical coordinates (r, θ, z) in which we have

(

∂2

∂x2+

∂2

∂y2

)

=

(

∂2

∂r2+

1

r

∂

∂r+

1

r2∂2

∂θ2

)

(18.12)

So we have(

∂2

∂r2+

1

r

∂

∂r+

1

r2∂2

∂θ2

)

vz(r, θ) =P

η(18.13)

Circular symmetry means that vz(r, θ) is independent of θ so that

∂vz

∂θ= 0 (18.14)

58

and we are left with

(

∂2

∂r2+

1

r

∂

∂r

)

vz =P

η

⇒(

r∂2

∂r2+

∂

∂r

)

vz =Pr

η

(18.15)

The obvious thing to do is to set u = ∂vz/∂r thereby obtaining the equation

r∂u

∂r+ u =

Pr

η

⇒ ∂(ru)

∂r=Pr

η

⇒ ru =Pr2

2η+A

⇒ u =Pr

2η+A

r

⇒ vz =Pr2

4η+A ln r +B

(18.16)

where A and B are constants. To find A and B we just implement the noslip boundary conditions on the surfaces of the inner and outer cylinders.In other words we demand that vz vanish at r = a and at r = b; this easilydetermines A and B and one finds that

v =P

4η

[

r2 − (a2 − b2)

ln(a/b)ln r − a2 +

(a2 − b2)

ln(a/b)ln(a)

]

e

as required.

Problem 3. A compressible inviscid fluid with p = kργ flows steadily rounda solid core of radius b with |v| inversely proportional to the distance fromthe centre of the core. Obtain an equation for v of the form

2c2

(γ − 1)+ v2 = B

where B is a constant and c2 = dp/dρ. Show also that there is a minimumvalue b0 that can be specified for the core radius.

59

Solution sketch: This is a two dimensional calculation where we treat theflow of a compressible fluid round a solid core whose radius we take to be

b (18.17)

We want to show that the velocity satisfies the equation

2c2

(γ − 1)+ v2 =

A2

b20(18.18)

where b0 is the minimum value that can be taken for the core radius b, andγ and c have their usual meanings for the case of adiabatic compressibleflow.

The appropriate equation to use is Bernoulli’s equation so that we have

−∂φ∂t

+K +v2

2+

∫

dp

ρ= C(t) (18.19)

Now we know that the flow is steady and that it is circulating roundthe core with velocity components given by

vθ =A

r, (A a constant), vr = 0

and v = vrer + vθeθ

(18.20)

where r is the distance from the centre of the core. The fact that the flowis steady means that

∂φ

∂t= 0 (18.21)

and that C(t) = C a constant independent of t. We also take K = 0 sincethere is no body force. Using these facts Bernoulli’s equations gives us

1

2

A2

r2+

∫

dp

ρ= C (18.22)

at a distance r from the centre of the core.Now the flow is compressible and

p = kργ (18.23)

so that we now obtain

A2

2r2+

γk

(γ − 1)ργ−1 = C (18.24)

60

Recalling that the speed of sound c (which of course is not constant) is givenby

c2 = γkργ−1 (18.25)

Hence we can now write

2c2

(γ − 1)+A2

r2= 2C (18.26)

If r shrinks to b, the radius of the core this equation becomes

2c2b(γ − 1)

+A2

b2= 2C (18.27)

where we write c as cb to emphasize that this the value of c at r = b. Nowthe maximum value that the term A2/b2 can have is attained when cb = 0i.e. when

2C =A2

b2(18.28)

but this is also the minimum value of the core radius b. If we denote thisvalue of b by b0, then we have

2C =A2

b20(18.29)

and, at a general distance r from the core centre, we have the equation

2c2

(γ − 1)+ v2 =

A2

b20(18.30)

which is what we wished to derive.

Problem 4. A stationary solid sphere of radius a is placed in an incom-pressible fluid. The velocity potential Φ is given, in spherical polar coordi-nates, with origin at the centre of the sphere by

Φ = U cos θ

(

r +1

2

a3

r2

)

, r ≥ a.

By using Bernoulli’s equation find the pressure p on the sphere’s surface.Show that on the equator r = a, θ = π/2 p is a minimum, show furtherthat this minimum pressure is zero if

U =

√

8p∞5ρ

61

Where p∞ is the pressure at infinity.

Solution sketch: For this problem, as indicated by their appearance inthe wording of the problem itself, we must use spherical polar coordinates(r, θ, φ). Doing this we have three velocity components vr, vθ and vφ andthese are given by

vr = −∂Φ

∂r= −U cos θ

(

1 − a3

r3

)

vθ = −1

r

∂Φ

∂θ= U sin θ

(

1 +1

2

a3

r3

)

vφ = 0

(18.31)

We use Bernoulli’s equation to find p. Bernoulli’s equation gives us

p

ρ+

v2

2= C (18.32)

where C is a time independent constant. But v = vrer + vθeθ so

v2 = v2r + v2

θ = U2 cos2 θ

(

1 − a3

r3

)2

+ U2 sin2 θ

(

1 +1

2

a3

r3

)2

(18.33)

Hence, on the sphere’s surface where r = a, we have

v2 =

(

3

2U sin θ

)2

(18.34)

while at ∞, where r → ∞, we have v2 = U2 and p = p∞.So using Bernoulli’s equation to equate things at r = a and r = ∞ we

get

p

ρ+

1

2

(

3

2U sin θ

)2

=p∞ρ

+1

2U2

⇒ p = p∞ +1

2ρU2

(

1 − 9

4sin2 θ

)(18.35)

so the pressure p has been found.To find the minimum pressure we solve

dp

dθ= 0 (18.36)

62

which yields the result that

θ =π

2(18.37)

and we readily calculate that, at this minimum, p takes the value pmin

where

pmin = p∞ − 5

8ρU2 (18.38)

and it is clear immediately that pmin = 0 if

U =

√

8p∞5ρ

(18.39)

as required.

Problem 5. Viscous fluid flows steadily under no forces along a pipe oflength l in the direction of the z-axis. The cross-section of the pipe is theellipse x2/4 + y2 = 1 and a pressure difference P is maintained across theends. If the velocity v is such that

v = wk

Find a soln in the form

w =(

x2 + 4y2 − 4)

f(x, y)

and show that the total mass of fluid Q delivered through the pipe/second isgiven by

Q =2πPρ

5ηl

Solution sketch: This is the same as the example we dealt with when consid-ering Poiseuille flow the only difference is that the pipe now has an ellipticalcross section. This means that we can skip ahead to the equation for w whichis

(

∂2

∂x2+

∂2

∂y2

)

w(x, y) =1

η

∂p

∂z= C, a constant (18.40)

But1

η

∂p

∂z= C

⇒ p = ηCz +B

(18.41)

63

Now, using w = (x2 + 4y2 − 4)f , we find that f obeys the equation

(x2 + 4y2 − 4)(fxx + fyy) + 4xfx + 16yfy + 10f = C (18.42)

But this is clearly a solution if f is just a constant, say

f = D (18.43)

where we must choose

D =C

10(18.44)

So we now have our solution in the form

v = (x2 + 4y2 − 4)C

10k (18.45)

we also know that the pressure difference across the pipe is P and

P = p(z = 0) − p(z = l) = −ηCl (18.46)

so we have found C: it is given by

C = −Pηl

(18.47)

and the final expression for the velocity v is

v = −(x2 + 4y2 − 4)P

10ηlk (18.48)

The tricky part is to calculate Q because the elliptical cross sectionmakes things difficult. However the basic form for Q is still an integral overshells, it is just that they are now elliptical shells instead of circular ones.hence we can write, in analogy with what we wrote in 15.22

Q =

∫

elliptical shellsρ|v|dxdy (18.49)

It is easier to consider a general ellipse

x2

a2+y2

b2= 1 (18.50)

64

and then specialise to the one that we have which has a = 2 and b = 1so we shall do things that way. The idea is to change variables so thatthe ellipse becomes a circle: we accomplish this by changing from (x, y) to(x′, y′) where

x = ax′, y = by′ (18.51)

The equation of the ellipse then becomes

(ax′)2

a2+

(by′)2

b2= 1

⇒ (x′)2 + (y′)2 = 1

(18.52)

so the ellipse has become a circle of radius 1. Hence all we have to do, isto implement the change of variable in the integral and then integrate overcircular shells whose radius varies in size from 0 to 1. Implementing thechange of variable in the integral is easy: one knows, in this simple case,that

dxdy = abdx′dy′ (18.53)

and dx′dy′ for a circular shell of radius λ is given by

dx′dy′ = 2πλdλ (18.54)

So the integral for Q is given by

Q = ρ

∫ 1

0

|v|ab2πλdλ (18.55)

But, now we set a = 2 and b = 1, note that

|v| = −(x2 + 4y2 − 4)P

10ηl(18.56)

and in terms of x′, y′ which satisfy

(x′)2 + (y′)2 = λ2 (18.57)

we get

Q = −∫ 1

0

ρ(4λ2 − 4)P

10ηl4πλdλ (18.58)

which one easily checks gives

Q =2πPρ

5lη(18.59)

65

as claimed.

Problem 6. A complex potential W is given by

W = Uz +Ua2

z+ ik ln z, U, a > 0.

and represents the 2-dim flow of an incompressible fluid of density ρ. Showthat(a) The flow is due to a cylinder of radius a, centre the origin placed in a

uniform stream.(b) The circulation about the cylinder is 2πk.(c) The cylinder experiences a lifting force/unit length of 2πρkU .

Solution sketch: For (a) we just need to show that the circle r = a is astreamline. Put

z = reiθ (18.60)

and we readily compute that

W = Ureiθ +Ua2

re−iθ + ik(ln r + iθ)

= φ+ iψ

(18.61)

So we see that

ψ = Ur sin θ − Ua2

rsin θ + k ln r (18.62)

Now if we set r = a in the expression for ψ we find that

ψ = Ua sin θ − Ua sin θ + k ln a

= k ln a = a constant(18.63)

so r = a is a streamline as required.For (b) one simply has to calculate

∫

C

v · dl (18.64)

where C is a circle of radius b with b > a. Now, on C

dl = b dθ eθ

v = vr er + vθ eθ = −∇φ

= −∂φ∂r

er −1

r

∂φ

∂θeθ

(18.65)

66

This gives

v · dl = −1

b

∂φ

∂θ

∣

∣

∣

∣

r=b

b dθ = −∂φ∂θdθ

⇒ Γ = −∫ 2π

0

∂φ

∂θdθ

= −φ(2π) + φ(0)

(18.66)

But it is simple to compute φ from W obtaining the result that

φ = Ur cos θ +Ua2

rcos θ − kθ (18.67)

from which we find that

Γ = 2πk (18.68)

which is what we were supposed to get.

Finally for (c) and the calculation of the force per unit length we useBernoulli’s equation

p

ρ+

v2

2= C (18.69)

67

and on a streamline 13

v2dz =

(

dW

dz

)2

dz (18.71)

so we obtain(

dW

dz

)2

dz =

(

2C − 2p

ρ

)

dz (18.72)

Then we integrate this expression around the cylinder, i.e. around the circleC above. We get

∫

C

(

dW

dz

)2

dz =

∫

C

(

2C − 2p

ρ

)

dz

= 0 − 2

ρ

∫

C

p dz

= −2

ρ

∫

p dx+2i

ρ

∫

p dy

(18.73)

13 Here we use the fact that the velocity is tangential on a streamline C so that(vx + ivy)/|v| is a unit tangent vector to C. Thus, if dl is an element of length along C,we have

dz = dl(vx + ivy)

|v|

so that also

dz = dl(vx − ivy)

|v|

from which we deduce that

dz =dz

dzdz

=(vx − ivy)

(vx + ivy)dz

However, recall that |v|2 = v2x + v2

y = (vx − ivy)(vx − ivy), and

dW

dz= −vx + ivy

⇒ |v|2dz = (vx + ivy)(vx − ivy)(vx − ivy)

(vx + ivy)dz

⇒ |v|2dz = (vx − ivy)2dz

=

(

dW

dz

)2

dz

(18.70)

as claimed.

68

Let us define Fx and Fy by

Fx = −∫

p dy, and Fy =

∫

p dx (18.74)

then these are known as the lifts or thrusts per unit length. So now we have

∫

C

(

dW

dz

)2

dz = −2

ρFy − 2i

ρFx (18.75)

The last step is to calculate the integral on the LHS using Cauchy’s residuetheorem. We have

W = Uz +Ua2

z+ ik ln z

⇒ dW

dz= U − Ua2

z2+ik

z

⇒(

dW

dz

)2

= U2 − 2U2a2

z2+U2a4

z4+

2ikU

z− 2ikUa2

z3− k2

z2

(18.76)

Cauchy’s theorem says that the only term in the integrand which con-tributes is the 1/z pole term

2ikU

z(18.77)

and its contribution is (2πi)(2ikU). Hence we have done the integral aroundC and deduced that

∫

C

(

dW

dz

)2

dz = (2πi)(2ikU) = −4πkU

⇒ −4πkU = −2

ρFy − 2i

ρFx

⇒ Fx = 0, Fy = 2πkρU

(18.78)

and this is the required lifting force per unit length exerted on the cylinderby the flow.

Problem 7. Incompressible viscous fluid rotates in the space between twoinfinite coxial rotating cylinders of radii a and b and angular velocities Ω1

and Ω2 respectively (a < b). Find the resulting fluid velocity v. Also, ifb = 2a, Ω1 = −2Ω and Ω2 = Ω, show that the fluid is at rest at a distance√

2a from the axis of the two cylinders.

69

Solution sketch: This is a viscous fluid so we must use the Navier–Stokesequation again. The geometry of the situation is crying out for us to usecylindrical coordinates and so we shall do so.

Let (r, θ, z) be the cylindrical coordinates and (er, eθ, ez) be their as-sociated orthonormal basis vectors. Note first that, in these cylindricalcoordinates, we have

(v · ∇)v =(

vr∂r +vθ

r∂θ + vz∂z

)

(vrer + vθeθ + vzez)

=vθ

r∂θ(vθeθ), since fluid is rotating so vr = vz=0

(18.79)

Further, by cylindrical symmetry, and the fact that the cylinders are in-finitely long, we can infer that vθ is independent of θ and z respectively.Hence we write

vθ = vθ(r) (18.80)

So now we have 14

(v · ∇)v =vθ

r∂θ(vθeθ)

=v2

θ(r)

r∂θeθ

⇒ (v · ∇)v = −v2θ(r)

rer, since ∂θeθ = −er

(18.82)

Consider next the pressure p which also only depends on r so that p = p(r).Hence ∇p reduces to just

∇p =∂p

∂rer (18.83)

Thus the Navier–Stokes equation

∂v

∂t+ (v · ∇)v = −1

ρ∇p+

η

ρ∇2v (18.84)

14 To see why ∂θeθ = −er we reason as follows: (er, eθ, ez) are an orthonormal triadwith ez (which is identical to k) constant then, denoting ∂θ by a dot, we have

ez × er = eθ

⇒ ez × er = eθ

but er = eθ

⇒ eθ = ez × eθ

= −er

(18.81)

70

becomes1

ρ

∂p

∂r=v2

θ

r(18.85)

in the er direction and

(

∂2

∂r2+

1

r

∂

∂r+

1

r2∂2

∂θ2

)

(vθ(r)eθ) = 0 (18.86)

in the eθ direction. We now note that since eθ = −er then

eθ = −eθ (18.87)

so the equation of motion in the eθ direction becomes

(

∂2vθ(r)

∂r2+

1

r

∂vθ(r)

∂r− 1

r2vθ(r)

)

eθ = 0

⇒ r2∂2vθ(r)

∂r2+ r

∂vθ(r)

∂r− vθ(r) = 0

(18.88)

and this is recognisable as an equation of Euler type with solutions of theform rn for some n. The reader can easily check that the allowed values ofn are n = ∓1. Thus the general solution is

vθ = Ar +B

r, A and B constants (18.89)

The no slip boundary conditions that fix A and B are that

vθ = aΩ1, on inner cylinder, i.e. when r = a

vθ = bΩ2, on outer cylinder, i.e. when r = b(18.90)

It is routine to verify that this gives

A =(b2Ω2 − a2Ω1)

(b2 − a2), B =

(Ω1 − Ω2)

(b2 − a2)a2b2 (18.91)

So v is now completely determined.To see when v can vanish we simply demand that

vθ = 0

⇒ Ar = −Br

⇒ r2 = −BA

(18.92)

71

and now if we set b = 2a and Ω2 = Ω1, Ω1 = −2Ω we find at once that

r =√

2a (18.93)

as it should.

Problem 8. Calculate the pressure difference between the inner and outercylinders in the preceding example

Solution sketch: To calculate the pressure difference in the previous problemwe go to the equation of motion in the er direction which we recall is

1

ρ

∂p

∂r=v2

θ

r(18.94)

But now we know vθ so we have

1

ρ

∂p

∂r=

1

r

(

Ar +B

r

)2

⇒ 1

ρ

∂p

∂r= A2r +

2AB

r+B2

r3

⇒ p(r) = ρ

(

A2r2

2+ 2AB ln r − B2

2r2+ C

)

, C a constant

(18.95)

So the desired pressure difference is simply the difference

p(b) − p(a) (18.96)

which is a piece of arithmetic we leave to the reader; note that the value ofC is immaterial since it will cancel in this difference.

Problem 9. Examine again the problem of the first example and showthat the pressure can be written in the form

p = p0 +ρ

2(3R2 + 2RR)

where R(t) is the radius of the bubble.

Solution sketch: Recall from the first problem that, if R(t) is the radius ofthe bubble, we had

R(t) = − ∂φ

∂r

∣

∣

∣

∣

r=R(t)

(18.97)

72

But∂φ

∂r= −f(t)

r2

⇒ R(t) =f(t)

R2(t)

⇒ f = R2(t)R(t)

(18.98)

Thus Bernoulli’s equation gives

− 1

R

d

dt(R2R) +

v2

2+p

ρ=po

ρ(18.99)

where note we have written the constant as p0/ρ, p0 being the pressureat infinity. Remember that on the surface of the bubble |v| equals R soBernoulli gives

− 1

R

d

dt(R2R) +

R2

2+p

ρ=po

ρ

⇒ p = p0 + ρ

(

1

R

d

dt(R2R) − R2

2

)

⇒ p = p0 +ρ

2

(

3R2 + 2RR)

(18.100)

and we are finished.

73

§ 19. A simple example of waves in a liquid

The complex potential method we studied earlier can be used to study wavemotion in a liquid. In particular it can be employed to describe surfacewaves in a tank of liquid. The bulk of the treatment is in the example thatnow follows.

Example A wave travelling from left to right

We take an infinitely long rectangular tank filled with liquid to a depth h.Note carefully that the y-axis coincides with the vertical direction (we cannotuse z as is usual since we are using z as the complex variable z = x+ iy).

The complex potential W that describes the waves is given by

W = A cos

2π

λ(z + ih− ct)

, (z = x+ iy) (19.1)

and we now decompose W into real and imaginary parts in the usual mannergiving

W = φ+ iψ (19.2)

Soφ = Re W

= A cos

2π

λ(x− ct)

cosh

2π

λ(y + h)

(19.3)

where we used the fact that

cos(a+ ib) =1

2(ei(a+ib) + e−i(a+ib))

=1

2(eiae−b + e−iaeb)

=1

2

(cos a+ i sin a)e−b + (cos a− i sin a)eb

= cos aeb + e−b

2− i sin a

eb − e−b

2

= cos a cosh b− i sin a sinh b

(19.4)

and we took

a+ ib =2π

λ(z + ih− ct) =

2π

λ(x+ iy + ih− ct) (19.5)

and chosea = x− ct and b = y + h (19.6)

74

Similarly the stream function ψ is given by

ψ = Im W

= −A sin

2π

λ(x− ct)

sinh

2π

λ(y + h)

(19.7)

Now we resort to Bernoulli’s equation. We also note that, because thecomplex potential W depends on t, the flow will not be steady. In any caseBernoulli’s equation gives us

−∂φ∂t

+ gy +v2

2+p

ρ= C (C a constant) (19.8)

where g is the acceleration due to gravity. Now we assume that we havewaves on the surface of an almost static liquid so that we can neglect v2

giving

−∂φ∂t

+ gy + +p

ρ= C (19.9)

The wave-shaped surface of the liquid is a surface of constant (atmospheric)pressure pA so that on this wave surface y satisfies

− ∂φ

∂t+ gy + +

pA

ρ= C

⇒− ∂φ

∂t+ gy = C − pA

ρ= D, say

(19.10)

Using our expression for φ above this immediately gives us a formula for y,namely

y =2πcA

gλsin

2π

λ(x− ct)

cosh

2π

λ(y + h)

+D

g(19.11)

Next we assume that the wave is small and that y varies by a smallamount so that we can neglect the variation of the quantity

cosh

2π

λ(y + h)

(19.12)

and approximate it by its value at y = 0, i.e. by

cosh2πh

λ(19.13)

75

Now if we take the undisturbed surface of the liquid to correspond to y = 0,then setting D to zero makes y coincide with the displacement of the freesurface of the liquid from y = 0. So, with our approximation, the equation

y =2πcA

gλcosh

2πh

λsin

2π

λ(x− ct) (19.14)

represents a wave of wavelength λ, velocity c, and amplitude

2πcA

gλ(19.15)

It is clear, too, that the wave travels from left to right along the x-axis.

Example A standing wave

If we modify W slightly, but otherwise retain the facts and assumptionsabove, we can describe a standing wave.

To this end choose

W = A cos

2π

λ(z + ih)

cos2πct

λ, (z = x+ iy) (19.16)

Now if we use

cos a cos b =1

2cos(a+ b) + cos(a− b) (19.17)

we can write W as

W =A

2cos

2π

λ(z + ih+ ct)

+A

2cos

2π

λ(z + ih− ct)

(19.18)

and this leads, by exactly the same sequence of steps in the previous exam-ple, to an expression for y which is

y =πcA

gλcosh

2πh

λsin

2π

λ(x+ ct) +

πcA

gλcosh

2πh

λsin

2π

λ(x− ct) (19.19)

This is a superposition of a left-right moving and a right-left moving waveof identical amplitudes, wavelengths and velocities. Hence it is a standingwave.

§ 20. The wind chill factor

The formula, sometimes attributed to Hill, for heat loss from human skinat temperature into dry air is

H =(

10.45 + 10√v − v

)

(33 − T ) (20.1)

where v is the wind speed, T is the temperature of the air, and H denotesthe heat loss per unit area per unit time. The units used for these quantitiesare as follows: H is measured in kilo−cal/m2/hour, v is measured in m/secand T is in 0C; the significance of the number 33 is that 330C is the normaltemperature for bare skin (370C is body temperature).