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Elementary Fluid Dynamics D.J.Acheson, Clarendon Press (90) Notes by CKWong

Elementary Fluid Dynamics by Acheson

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Elementary Fluid Dynamics

D.J.Acheson, Clarendon Press (90)

Notes by CKWong

1 IntroductionSources:

D.J.Acheson, “Elementary Fluid Dynamics”, Clarendon Press (90)

R.E.Meyer, ”Introduction to Mathematical Fluid Dynamics”, Wiley (71).

Experiment:

Aerofoil & Starting Vortex

Applications:

Waves (§3.1)

Flow Instability (§9.1)

Hydraulic Jumps (§3.10)

Smoke Ring Interactions (§5.4)

Atmospheric Jet Stream (§9.8)

Quantum Vortices (§5.8)

Sperm Movement (§7.5)

Spindown of Stirred Tea (§8.5)

1.1 Preliminary

Flow velocity: , , ,t u v w u u x (1.1)

ut

0 (1.2)

2-D Flow: , , , , , ,0u x y t v x y t u (1.3)

2-D Steady Flow: , , , ,0u x y v x y u (1.4)

Streamline: Curve with tangent equal to u everywhere.

Let curve be: ( )x x s , ( )y y s , ( )z z s

Thendx ds

u

dy ds

v

dz ds

w

/ / / (1.5)

Streak photographs: illuminated polystyrene beads of neutral buoyancy.

Material derivative, or, rate of change “following the fluid”:Df

Dt

d

dtf x t y t z t t [ ( ), ( ), ( ), ]

rc

fvv

f f dx f dy f dz

t x dt y dt z dt

f f f fu v w

t x y z

( )f

ft

u (1.6)

Acceleration:

D

Dt t

uu u

u

( ) (1.7)

Uniform rotation: ( , ,0)y x u

u

So that

D

Dt

uu u

, ,0y x y xx y

2 , ,0x y

Df ff f

Dt s

u u s u

where ss s is the tangent vector to a streamline.

0f u means f is constant along a streamline but it can vary between different

streamlines.

0Df

Dt means f is constant for a fluid particle but it can differ between different

particles.

A material (dyed) volume is a collection of designated fluid particles.

It may deform during the course of motion.

1.2 Ideal Fluids

An ideal fluid is

Incompressible: ρ = constant, or 0D

Dt

, which implies 0 u .

Inviscid (no viscosity): p n .

Mass of a volume V fixed in space & bounded by surface S is:

Vm dV

Rate of change is

V V

dm ddV dV

dt dt t

Net rate of mass leaving the volume is

S V

dS dV u n uConservation of mass means

V V

dV dVt

u (A)

Actually, one can show that eq.(A) is true even if V denotes a volume moving along

with the fluid. [ see R.E.Meyer, Chap 1 ]. More generally, the convection theorem

( Meyer, p.10) states that

V V

D DffdV f dV

Dt Dt u

Going back to eq.(A), since it is true for all V, we have

0t

u

which is called the equation of continuity.

Using

u u u

and

D

Dt t

u

We can writing the equation of continuity as

0D

Dt

u

For an incompressible fluid, ρ = constant, or 0D

Dt

, so that

0 u (incompressible condition)

Force on dyed blob in inviscid fluid:

S Vp dS pdV n

which means the force density is p .

1.2.1 Euler’s Equation

Newton’s law applied to a fluid volume:

V V

DdV p dV

Dt u g

Using the convection theorem, the left side is

V V

D D DdV dV

Dt Dt Dt

u uu u u u u

V

DdV

Dt

u

where the last equality was obtained with the help of the eq. of continuity.

We thus obtain Euler’s eqs.:

Dp

Dt

ug (1.12)

where g is the gravitational acceleration.

Writing

g (1.13)

we have

Dp

Dt t

uu u

or

p

t

u u

Using

21

2 u u u u u

we have

21

2

p

t

uu u u (1.14)

1.2.2 Bernoulli Streamline Theorem

H u u

where

21

2

pH

u (1.15)

Using

0 u u u

we have

0H u (1.16)

ie, H = const along streamline of an ideal fluid in steady flow.

Irrotational flow means

0 u (1.17)

Bernoulli Streamline Theorem for irrotational flow:

H = const everywhere in a steady irrotational flow of an ideal fluid.

1.3 Irrotational Flow

Vorticity: ω u (1.18)

2-D flow:

[ ( , , ), ( , , ),0]u x y t v x y tu

0,0, 0,0,

0

v u

x y z x y

u v

i j k

ω

v u

x y

(1.19)

Consider 2 mutually perpendicular line segments AB & AC (fig.1.3):

( , , ) ( , , )B A

vv v v x x y t v x y t x

x

= CCW angular velocity of AB

( , , ) ( , , )C A

uu u u x y y t u x y t y

y

= CW angular velocity AC

Hence

1 1

2 2

v u

x y

= average angular velocity of AB & AC.

Note: ω donotes local ‘spin’ & has nothing to do with global rotations. (Fig.1.6)

Eg. Shear flow (fig.1.4)

,0,0yu (1.20)

has no global rotation but

0v u

x y

Eg. Line vortex flow

ˆk

ru

θ(1.21)

has

1 10

0 0r z

r r

r r z r r zu ru u k

rθ z r θ z

ω u

but is obviously rotating as a whole.

Eg. Flow

ˆr uθ

spin has same rate as global rotation.

Eg. Rankine Vortex (fig.1.7)

2

r

u a

r

forr a

r a

(1.23)

0r zu u

1.4 Vorticity Equation

Euler’s eq. (1.14):

Ht

u

u u

can be written as

Ht

u ω u

both sides gives

0t

ω ω u (1.24)

Using

a b b a a b a b b a

we have

ω u u ω ω u ω u u ω

uω ω u

where

0 u for incompressible fluid

0 ω u for any vector u.

Thus (1.24) becomes

0t

ω

u

ω ω u

or

D

Dt

ω ω u (1.25)

which is called the vorticity equation.

For 2-D flow

, , , , , ,0u x y t v x y t u 0,0,ω

we have

0z

ω u u

so that

0D

Dt

ω(1.27)

Thus, for a 2-D flow of ideal fluid under conservative forces, ω is conserved for each

fluid element.

0 uω

so that ω is conserved along each streamline.

Thus, steady 2-D flow pass an aerofoil is irrotational since 0 for x .

1.5 Steady Flow Past Fixed Wing

Lift of wing in steady flow is due to (fig.1.9; also demonstrated by Wright program)

upperp p lowerp p

Flow is irrotational.

Bernoulli theorem means 21

2p u is constant everywhere.

Thus lift means upper loweru u .

1.5.1 Circulation

Circulation Г around closed curve C is

Cd u x (1.31)

Stokes theorem:

C S

d dS u x u n (1.32)

SdS ω n

where C is the boundary of S.

For the irrotational flow past an aerofoil,

0 for any C not enclosing the aerofoil.

0 for any C that encloses the aerofoil

(surface of foil is another boundary of S).

Furthermore, for fig.1.8, lift means 0 .

1.5.2 Kutta-Joukowski Hypothesis

For wing with sharp trailing edge, u can be finite everywhere only for one value ofcirculation K . Otherwise, u at the sharp edge.

That K describes the actual flow is the Kutta-Joukowski hypothesis.

For thin, symmetric wing with small α, (see Chap 4)

sinK UL (1.34)

where

U u L = length of wing

1.5.3 Lift

For ideal flow,

Drag = 0

Lift U L (1.35)

For K , we have

2 sinU L L (1.36)

For large angles, 1.36 overestimates badly (fig.1.11).

1.6 Conclusion

Near wing tip, flow is not 2-D & can’t be irrotational.

This leads to trailing vortices & hence drag. (fig.1.12)

Generation of Γ is a response to starting vortices (Chap 5), which in turn requires

viscosity.

2 Viscous FlowSource: D.J.Acheson, “Elementary Fluid Dynamics”, Chapter 2, Clarendon Press (90)

2.1 Introduction

Theoretically, inviscid, and in particular, irrotational (potential) flow is easiest to deal

with. Naively, we may expect it to describe flows with low viscosity.

In practice, we do find the theory useful in describing phenomena such as water

waves, sound waves, and slow flow past streamlined objects.

For flow past a general object, the down-stream side becomes turbulent.

The root cause of this may be traced to the viscosity related no-slip boundary

condition which a potential flow usually cannot satisfy. Thus, flows from the inviscid

theory are valid only if they can be made to reconcile with the no-slip condition

within a sufficiently thin transitional unseparated boundary layer.

If the inviscid theory predicts an increase of pressure along the flow direction on the

boundary, separation of boundary layer is expected and turbulence results. (see

chapters 4 & 8).

On the other hand, theory of very viscous fluid is very successful (see chapter 7).

Newtonian fluid:

du

dy (2.1)

where τ is the shear stress, and μ is the coefficient of viscosity.

Navier-Stokes Eqs (for incompressible Newtonian fluids):

21p

t

u

u u u g (2.3)

0 u

See Landau for a derivation.

The kinematic viscosity is defined as

(2.2)

No-slip condition:

0u at stationary boundary.

Reynolds Number:

ULR

(2.4)

where U and L are the characteristic flow velocity and dimension of the system,

respectively. And ν is the kinematic viscosity.

Using estimates

O Uu UO L u

on the Navier-Stokes eqs. (2.3), we have

Inertia term: 2UO L u u

Viscous term: 22

UOL

u (2.5)

so that

2

2

inertia term

viscous term

ULO O R

UL

(2.6)

High R flows:

Corresponds to flows with small viscosity.

Under favorable circumstances, eg. flow past streamlined bodies, it corresponds

to the flow of ideal fluid except for a thin boundary layer (required for the no-slip

condition) and a narrow trailing wake.

Typical thickness of such a boundary layer is

12O RL

(2.7)

In general, high R flows are unstable & easily become turbulent.

Low R flows:

0.01R .

Flow is well-ordered & reversible.

2.2 Navier-Stokes Eqs.

Consider the Euler eq

D Dp

Dt Dt

u uu u

Using

D

Dt t

u u

u u u u u u

t

u

uu

we can re-write it as

0pt

u

uu

or

0

t

u

(A)

where the momentum flux density tensor П is

ij i j iju u p

Eq(A) is just the “eq of continuity for momentum” which expresses the conservation

of momentum.

Since we expect the principle of the conservation of momentum to be valid even in

viscous fluids, eq(A) should apply there with a suitable modification of П. We thus set

ij i j iju u

where the stress tensor ij is related to the viscous stress tensor 'ij by

'ij ij ijp

[Note: Acheson used ijT instead of ij (see chap 6) ]

What this means is that the generalization of the Euler’s eq to viscous fluids can be

done by replacing p with 'p so that, for example,

Dp

Dt

u

becomes

'D

pDt

u

or

'pt

u

u u (B)

Next, we need to determine the form of ' .

Now, viscous effects are due to “friction” between adjacent fluid elements moving

with different velocities. Hence ' should depend on i

j

u

x

but not on u itself.

Furthermore, there’s no friction if the fluid rotates as a whole with uniform angular

velocity Ω so that uΩ r

or i ijk j ku x .

Now:

i mijk j km ijm j mji j

m i

u u

x x

and

0iiji j

i

u

x

Hence, the choice

' 'ji kij ij ji

j i k

uu ua b

x x x

where a, b are constants, will satisfy the above requirements.

A more common form is

2'

3ji k k

ij ij ijj i k k

uu u u

x x x x

(C)

where the positive constants

a 2 2

3 3b a b

are called coefficients of viscosity. (Landau used η instead of μ)

This also defines the Newtonian fluid. [cf eq(2.1)].

Using (C), (B) becomes

' jii ij

j i j

u u pu

t x x x

2

3ji k k

ij iji j j i k k

up u u u

x x x x x x

2

3jk i

i k j j i

uu up

x x x x x

1

3k i

i k j j

u up

x x x x

or in vector form:

'pt

u

u u

21

3p

u u

21

3p

u u (D)

For incompressible fluids, we have the Navier-Stokes eqs.:

2pt

u

u u u (2.3)

0 u

Note also that for incompressible fluids,

' jiij

j i

uu

x x

[cf (2.1)]

2.3 Simple Flows

2.3.1 Plane Parallel Shear Flow

Plane parallel shear flow is defined as

, ,0,0u y t u (2.8)

Thus

0u

x

u (incompressible)

Navier-Stokes eqs becomes2

2

1u p u

t x y

0p p

y z

(2.9)

The last eq means p and hencep dp

x dx

is a function of ,x t only.

Hence, the right and left hand side of2

2

1 p u u

x t y

is a function of ,x t and ,y t , respectively. Thus, either can only be function of t

only.

2.3.2 Impulsively Moved Boundary

Let fluid lies at rest in region 0 y at 0t .

At 0t , boundary at 0y is set in motion in x-direction with constant speed U.

No-slip condition means ,0,0Uu at boundary for 0t .

In general, we expect , ,0,0u y t u so that it is a plane parallel shear flow.

According to the last section, pf t

x

so that p f t x C .

Let there be no externally applied pressure gradient, ie., p at x are equal.

This means we must have 0f t and p C .

Eq(2.9) thus becomes2

2

u u

t y

(2.12)

which is simply the 1-D diffusion eq.

This must be solved with the initial condition

,0 0u y for 0y

and boundary conditions

0,u t U for 0t

, 0u t for 0t

Now, (2.12) is invariant under the scale transformation

y y 2t twhere α is a constant.

This means we can attempt a similarity solution:

u f where yt

(2.13)

so that

32

y

t t

1

y t

3' ' '

22

u yf f f

t t tt

1' '

uf f

y y t

2

2

1 1'' ''

uf f

y y tt

where 'df

fd

. Hence (2.12) becomes

1' ''2

f ft t

or

'' 1

' 2

f

f

so that

21ln '

4f const

or

2

4'f Be

whence

2

4

0

sf A B e ds

where A, B are constants.

Using

for 0t or y 0 for 0y

the boundary & initial conditions become

0f and 0f U

Hence2

4

0

10

sA B e ds A B

U A

whence

2

4

0

11

sf U e ds

(2.14)

or

2

4

0

1, 1

y stu y t U e ds

[see fig 2.8]

The vorticity is22

44y

tu U Ue e

y y t

(2.15)

which indicates a diffusion with standard deviation 2 t . In other words,

viscous diffusion time 2LO (2.16)

where L is a distance.

2.3.3 Flow Down Inclined Plane

With reference to fig. 2.9, the Navier-Stokes eqs

21p

t

u

u u u g

becomes

2 2

2 2

1sin

u pu v u u g

t x y x x y

2 2

2 2

1cos

v pu v v v g

t x y y x y

where , ,0u vu .

The no-slip condition means 0u for 0y .

Hence u must vary with y if it’s not identically 0.

The simplest solution possible is therefore a steady flow of the form

, ,0u y v y u

The incompressibility condition

0u v

x y

becomes

0v

y

so that v const .

However, 0 0v so that 0v for all y.

The steady state Navier-Stokes eqs. thus become2

2

10 sin

p d ug

x dy

10 cos

pg

y

(2.17)

The 2nd eq gives

cosp gy f x (A)

Since all streamlines are in the x direction, the free surface must be y h where h is

a constant. For a flat interface, p must be equal on both sides and the tangential stress

must vanish. Let the atmospheric pressure be 0p , we have

0p p and 0du

dy at y h (2.18)

Putting these into (A) gives

0 cosp gh f x

so that (A) becomes

0cos cosp gy p gh

or

0 cosp p g h y

whence

0p

x

and the remaining Navier-Stokes eq reduces to2

20 sin

d ug

dy

which gives

sindu gy A

dy

2sin

2

gu y Ay B

Putting in the boundary conditions

0u at 0y

and

0du

dy at y h

we have

0 B

sin0

gh A

so that

2sin

2

g yu hy

(2.19)

The volume flux per unit length in the z-direction is

2 3 23

0 0

sin sin sin

2 6 2 3

h hg y g h h gQ udy hy dy h h

2.3.4 Another Example

Consider the impulsively moved plane boundary problem with an added stationary

plane boundary a distance h above it. (see fig 2.10)

The problem is to solve , ,0,0u y t u from

2

2

u u

t y

(2.20)

with the initial condition

,0 0u y for 0h y

and boundary conditions

0,u t U and , 0u h t for 0t

The form of (2.20) clearly suggests the use of the method of separation of variables.

Let

u Y y T t

(2.20) becomes2

2

dT d YY T

dt dy

or2

2

dT d Y

Tdt Ydy const

Thus, for 0 ,tT Ae y yY Be Ce

where A, B, C are constants and

For 0 , we have

T A ' 'Y B y C

where ', 'B C are constants.

Hence,

0 00

, t y yu y t a y b e a e b e

(A)

is a general solution of (2.20).

The next task is to adjust the constants to satisfy the initial & boundary conditions:

0 00

,0 0y yu y a y b a e b e

for 0h y

00

0, tu t b e a b U

for 0t

0 00

, 0t h hu h t a h b e a e b e

for 0t

The last 2 eqs can be satisfied only if the time dependent sums vanish identically.

Hence, we have

0b U

0 0 0a h b so that 0

Ua

h

and

0

0te a b

0

0t h he a e b e

for 0t

Since the te terms are independent, each term in these sums must vanish

individually so that

0a b

0h ha e b e

which gives

b a

2sinh 0h he e h so that

h n i

and2 2

22

0n

h

Eq(A) becomes

2 2

2

1

, sinn

th

nn

U nu y t y U e A y

h h

where 2nA ia 0 are to be determined by the initial condition, namely,

1

,0 sin 0nn

U nu y y U A y

h h

for 0h y

Using

0

sin sin2 nmnx mxdx

we have

0sin 0

2

h

n

U ndy y U y A

h h h

or

0

21 sin

h

n

y nA U dy y

h h h

0

21 sin

n zU dz z

n n

where

nz y

h

0

2 1cos cos sin

nU

z z z zn n

2U

n

so that, finally

2 2

2

1

2 1, 1 sin

nt

h

n

y nu y t U e y

h n h

(2.21)

2.4 Flow with Circular Streamlines

In cylindrical coordinates , ,r z , (see Appendix A.5)

, ,r z r r z zu u u u u u u e e e

1 2cos sinr e e e 1 2sin cos e e e 3z e e

r

e

e r

e

e (all other partials vanishes) (2.23)

r zf fr r z

e e e

r zf u u u fr r z

u

zr

V VrV

r r r z

V

1r z

r z

r

r r zV rV V

e e e

V

2 22

2 2 2f r f

r r r r z

To write the Navier-Stokes eqs

21p

t

u

u u u

in cylindrical coordinates, we 1st consider the term u u :

r r r r r ru u u u e u e e u

r r z r r ru u u u ur r z

e e u

r r r ru u ur

e e u

rr r

u uu

r

e e u

u u u u e u e e u

r zu u u u ur r z

e e u

u u ur

e e u

2

r

uu

r

e e u

z z z z z zu u u u e u e e u

z zu e u

so that

2

rr r r z z

u u uu u u

r r

u u e e u e e u e u

2

rr r z z

u u uu u u

r r

e u e u e u

Next, we turn to 2 u :

2 2

22 2 2r r r ru r u

r r r r z

e e

2 2

2 2 2r r r rr u ur r r z r

e e

Now

2

2r r

r r r r

uu u

ee e

2 2

2 22r r r r

r r

u uu

e ee

2

22 r r

r r r

u uu

e e e

so that

2 2

22 2 2

12 r r

r r r r r r r

u uu r u u

r r r z r

e e e e e

22

12 r

r r r r

uu u

r

e e e

22 2

2r rr r

u uu

r r

e e

Similarly

2

2

uu u

ee e

2 2

2 22

u uu

e ee

2

22 r

u uu

e e e

so that

2 2

22 2 2

u r u ur r r z r

e e e

2 2

2 2 2

12 r

u ur u u

r r r z r

e e e e

22

12 r

uu u

r

e e e

Finally, with

2 2z z z zu u e e

we have

2 2 2 22 2 2

12 2r r

r r r z z

u u uu u u u

r r r

u e e e e e e

2 2 22 2 2 2

2 12r r

r r z z

u u uu u u u

r r r r

e e e

Hence

2

22 2

1 2r rr r

u u p u uu u

t r r r r

u

22 2

1 12r ru u u p u

u u ut r r r r

u (2.22)

21zz z

u pu u

t z

u

where

r zf u u u fr r z

u

2 22

2 2 2f r f

r r r r z

The incompressibility condition is simply

0zr

u uru

r r r z

u

2.4.1 Differential Eq.

Consider a 2-D circular flow

,u r t u e (2.27)

The incompressibility condition

0zr

u uru

r r r z

u

is automatically satisfied.

The Navier-Stokes eqs (2.22) becomes

ru :2 1u p

r r

u :2

1 1u p ur u

t r r r r r

zu :1

0p

z

Now, u is a function of ,r t only.

The u eq can therefore be written as

,p

f r t

Integrating, we have, since p is independent of z:

, , , ,p r t f r t C r t

where C is another arbitrary function of ,r t .

Now, p must be single valued. This means

, , , 2 ,p r t p r n t

This is possible only if 0f , ie, p is a function of ,r t only.

The Navier-Stokes eqs can be further simplify to

,u r t u e ,p p r t

ru :2 1u p

r r

u :2

1u ur u

t r r r r

(2.30)

0d du

r r udr dr

Setting lny r , it becomes

0d du

udy dy

so that

y y Bu Ae Be Ar

r (2.31)

If the fluid occupies region 1 2r r r between 2 cylinders rotating with angular

velocities 1 and 2 , respectively, the no-slip condition gives

1 1 11

Br Ar

r 2 2 2

2

Br Ar

r

Solving for A and B gives2 2

1 1 2 22 2

1 2

r rA

r r

2 2

1 2 1 21 22 2

2 12 2

1 2

1 1r r

Br r

r r

(2.32)

This formula will be of use in the study of Taylor vortices. (see Fig 9.8)

2.4.3 Spin-Down

Let fluid occupies the interior of an infinitely long cylinder of radius a.

Initially, the whole system is rotating with uniform angular velocity Ω so that

u r for r a 0t

The cylinder is suddenly stopped at 0t .

We thus need to solve

2

1u ur u

t r r r r

with the above initial condition as well as the (no-slip) boundary condition

0u at r a for all 0t .

Using the separation of variables method:

,u r t R r T t

we have

2

1dT d dRr R

Tdt R rdr r r

so that for 0 ,

tT t e

1R r J r

where nJ is the Bessel function defined by

2

22

0n

d d nr k J kr

rdr dr r

[see M.Abramowitz, I.A.Stegun, “Handbook of Mathematical Functions”]

Hence

1, tu r t c e J r

To satisfy the no-slip condition at r a , we need

na

where n are roots of 1J , ie., 1 0nJ .

Therefore

2

n

a

and

2

11

,n t

an n

n

ru r t c e J

a

The initial condition

u r for r a 0t

means

11

n nn

rr c J

a

Using the orthogonality of the Bessel functions:

2

2

1

0 2

a

n m nm n

r r aJ J rdr J

a a

where n is the nth root of J , we have

2

221 2

0 2

a

n n n

r aJ r dr c J

a

Using

1

11

0

1J ar r dr J a

a

( see I.S.Gradshteyn, I.M.Ryzhik, “Table of Integrals, Series, and Products”, formula

6.561.5)

we have

2

232 2

1

2n n nn

aa J c J

so that

2

2n

n n

ac

J

whereupon

2

11 2

2,

n ta

nn n n

a ru r t e J

J a

Finally, using

1 1

2J x J x J x

x

we have

0 2 1

20n n n

n

J J J

( n are roots of 1J )

so that

2

11 0

2,

n ta

nn n n

a ru r t e J

J a

(2.33)

This equation overestimate the spin down time of a cup of tea by an order of

magnitude since there’s no account for the effect of the bottom (see chapter 5).

2.4.4 Viscous Decay of Line Vortex

A line vortex

0

2 r

u e (2.34)

The vorticity singularity at 0r is expected to diffuse in a viscous fluid.

Consider the circulation of a circular path C of radius r,

2

0, , 2 ,

Cr t d u r t rd ru r t

u r (2.35)

Eq(2.30)

2

1u ur u

t r r r r

can be rewritten in terms of as follows:

1

2u

r

1

2

u

t r t

2

1 1 1

2

u

r r r r

2 2

2 3 2 2

1 2 2 1

2

u

r r r r r r

2

2

1u u ur

r r r r r r

2

3 2 2 2

1 2 2 1 1 1 1

2 2r r r r r r r r r

2

3 2 2

1 1 1 1

2 r r r r r

so that

2

3 2 2 3

1 1 1 1 1 1

2 2 2r t r r r r r r

or

2

2

1

t r r r

(2.36)

For the line vortex decay, the initial condition is

0,0r

The requirement of u be finite at 0r afterwards means

0, 0t for 0t

Eq(2.36) is invariant under the same scale transformation as (2.12), namely,

r r 2t twhere α is a constant. We therefore seek a similarity solution:

,r t f wherer

t

As in the case of (2.13), we have

32

r

t t

1

r t

3' ' '

22

rf f f

t t tt

1' 'f f

r r t

2

2

1 1'' ''f f

r r tt

so that (2.36) becomes

1 1 1' ' ''2

f f ft r tt

or

1' ' ''2

f f f

' 1

' 2

dfd

f

2

ln ' ln4

f C

or2

4'fCe

2 2

4 4f C e d ae b

Hence

2

4r

tae b

Putting in the initial & boundary conditions, we have

0 b 0 a b so that 0a b

Hence2

40 1

rte

(2.37)

The circulation thus diffuses with a standard deviation of 2 t . (see fig.2.12)

For very small r, ie., 2r t , we have2

0 4

r

t

so that

0 8

ru

t (2.38)

which corresponds to uniform rotation with angular velocity 08 t

.

2.5 Convection & Diffusion of Vorticity

Applying the same technique that begets vorticity eq (1.25) to the Navier-Stokes eq

2pt

u

u u u

we obtain the viscous vorticity eq.

2D

Dt

ω ω u ω (2.39)

For 2-D flow

, , , , , ,0u x y t v x y t u

0,0,ω

we have (cf 1.27),

2D

Dt

ω ω

or

2 2

2 2t x y

u (2.40)

Obviously, the term uω

describes the vorticity convection while 2 ω its

diffusion.

2.5.1 2-D Flow Near Stagnation Point

Consider the irrotational flow pattern near a stagnation point as shown in fig.2.13 with

u x v y (2.41)

where 0 is a constant.

Obviously, the no-slip condition is not satisfied at the boundary 0y .

However, the mainstream flow speed | | | |u x increases in the flow direction

along the boundary. According to the Bernoulli’s theorem, the mainstream pressure

decreases in the flow direction along the boundary. Therefore, we expect a thin,

unseparated boundary layer with thickness 1OL R

.

Using2UL L

R

, we have O .

The no-slip condition creates a vortex sheet at the boundary which diffuses into the

fluid. The existence of an unseparated boundary layer implies the convective effects

are sufficient to balance this diffusion and confine it within the layer. (Do Ex.2.14)

2.5.2 High R Flow Past Flat Plate

3. WavesRef: D.J.Acheson, “Elementary Fluid Dynamics”, Clarendon Press (90)

3.1 Introduction

Consider a simple harmonic surface wave:

cosA kx t (3.1)

Wave speed is dispersive:

gc

k k

(3.2)

so that waves with longer wavelengths 2k

travel faster.

The group velocity is defined as

g

dc

dk

(3.3)

For a dispersion given by (3.2), we have

1 1

2 2g

d gc gk c

dk k (3.5)

Hence, individual wave crests travel twice as fast as the group so that they continually

appear at the back of the group & disappear at the front.

Dispersion is also responsible for wake pattern behind moving boat.

Stationary wave patterns of flow past object contains both up & down stream

disturbances because of surface tension effects.

Waves at interface between 2 fluids caused by buoyancy effects have speed

1 2

1 2

gc

k

(3.6)

where 1 and 2 are the densities of the fluids.

A variant of this is internal gravity waves that travel in a fluid with density that

varies with height (eg. the atmosphere).

Sound waves are due to compressibility and have speed

00

0

pa

(3.7)

which is non-dispersive.

3.2 Surface Waves on Deep Water

2-D water waves:

u [ ( , , ), ( , , ), ]u x y t v x y t 0

Irrotational motion:

u

i j k

x y zu v 0

0

v

x

u

y0

Thus, there exists a velocity potential φ :

ux

vy

Incompressibility:

u 0

2

2

2

2 0 x y

Let wave on free surface be described by

y t ( , )x

3.2.1 Condition at Free Surface

Surface condition: Fluid particles on surface remain there.

Let

, , ,F x y t y x t

the surface condition means that

, , , 0F x x t t

ie, 0F on surface. Thus

0DF F

FDt t

u on ,y x t (3.17)

Using

F

t t

F

x x

1

F

y

(3.17) becomes

0u vt x

on ,y x t (3.18)

3.2.2 Bernoulli’s Eq. for Unsteady Irrotational Flow

Consider the Euler eq. for incompressible, irrotational flow:

21

2

p

t

uu where gy

Writing u and integrate, we have

21

2

pG t

t

u (3.19)

which is called the Bernoulli’s eq. for unsteady irrotational flow.

Note that the arbitrary function G is doesn’t affect the value of u.

The incompressibility condition in Eq(3.19) can be removed as follows. (see Landau)

Consider the specific (per unit mass) enthalpy h,

1dh Tds vdp Tds dp

we have, for adiabatic processes ( 0ds ),

1p h

Therefore, (3.19) becomes

21

2h G t

t

u

3.2.3 Pressure Condition at Free Surface

For inviscid fluid, there is no surface tension.

Hence, the condition at a free surface is simply

0p p on ,y x t

where 0p is the (usually constant) atmospheric pressure.

The unsteady Bernoulli eq thus becomes, with a proper choice of G,

2 2 0

0

10

2

pu v g G t

t

on ,y x t (3.20)

3.2.4 Linearization of Surface Conditions

To 1st order in u, v, and η, (3.20) becomes

0gt

on ,y x t

Now

2

2

0 0

, , , , , ,

y y y

x y t x y t x y t

t t t

Sincet

is already of order u, the true 1st order approximation of (3.20) is

0gt

on 0y (3.22)

Similarly, for (3.18):

0vt

on 0y

Using vy

, we have

0t y

on 0y (3.21)

3.2.5 Dispersion Relation

Eliminating η from (3.21-22) gives2

2

10

g t y

on 0y

which is clearly separable. Setting

, , ,x y t f y x t

we have

0

' 0 ,y

f x ty

2 2

2 2

0

0y

ft t

so that

2

2

10 ' 0 0f f

g t

or

22

20

t

where0

'

y

fg

f

On the other hand, incompressibility means satisfies the Laplace eq. inside the

fluid for all t:2 2

2 20

x y

ie

2

2'' 0f f

x

Since f is a function of y only, we must have2

22

''const

fk

x f

which, combining with the t eq., gives

2 2

2 2 2 2

1 1

t k x

Thus, both the time and spatial components of Φ are harmonic oscillators while

together, they obey the 1-D wave eq. with propagation speed ck

.

A general solution is of the form

, sin sinx t C kx t C kx t

where C, δ are arbitrary constants.

If we consider a simple wave propagating in the x direction, we have

, sinx t C kx t

Now, f itself satisfies2'' 0f k f

so that

exp expf y B ky D ky (3.A)

Since the wave should die out towards the bottom, we must have

0f y (3.B)

so that 0D and

expf y B ky

whereupon

0

'

y

fg gk

f

(3.26)

giving a wave speed

gc

k k

The potential itself is

, , exp sinx y t C ky kx t (3.25)

From (3.22), we have

0

1, cos cos

y

x t C kx t A kx tg t g

(3.23)

where

kA C C

g

3.2.6 Meaning of Small Amplitude

From (3.23,5), we have

exp cos exp cosu Ck ky kx t A ky kx tx

exp sin exp sinv Ck ky kx t A ky kx ty

sin sinCk kx t Ak kx tx g

2

sin sin exp sinC kx t Ck kx t A ky kx tt g

Thus, the linearization condition that leads to (3.21)

u vx

means

1x

ie.

1

2A

k

the wave amplitude of the surface wave is much small than its wavelength λ.

For (3.22), we need2 2u v g

which means2 2A gA

ie

2

1

2

gA

k

3.2.7 Particle Path

In the Eulerian picture, the equations

exp cosu A ky kx t

exp sinv A ky kx t

give the velocity at each point ,x y at time t.

In the Lagrangian picture, the corresponding velocity of a fluid particle whose

trajectory is ,x t y t is described by

exp cosdx t

A ky t kx t tdt

exp sindy t

A ky t kx t tdt

which can be integrated to give the particle path.

It is straightforward to prove that the ansatz

exp sinx t x A ky kx t

exp cosy t y A ky kx t

is a solution to 1st order in A.

It describes a circular motion of radius expA ky about the average position

,x y .

3.3 Dispersion

Given a dispersion relation

k (3.28)

the group velocity is defined as

g

dc

dk

(3.29)

Properties:

1. Waves of a single wavelength travels with phase velocity ck

.

2. Whenever superpositions of nearby wavelengths are substantial, the packet

travels with the group velocity g

dc

dk

. This is true whether the packet is

isolated or just a Fourier component of a complicated excitation.

3. Energy is transferred at group velocity. (see Ex.3.12)

3.3.1 Motion of a Wave Packet

A general disturbance can be written as

, expx t a k i kx t dk

(3.30)

where a k is the Fourier amplitude and it is understood that only real quantities are

physically meaningful.

When a k is narrowly centered at 0k , the disturbance is called a wave packet.

Writing

0 0' gk k c (3.31)

where

0 'k k k , 0 0k ,0

0gk k

dc

dk

(3.32)

we have

0 0 0' ' gkx t k k x k c t

0 0 0' gk x t k x c t

so that (3.30) becomes

0 0 0 0, exp ' exp ' 'gx t i k x t a k k ik x c t dk

(3.33)

3.3.2 Gaussian Packet

Consider a Gaussian wave packet with standard deviation σ:

2

01exp

22

k ka k

The integral in (3.33) becomes

2

0

1 'exp ' '

22g

kI ik x c t dk

22

0 0

1 1exp exp ' '

2 22g gx c t k i x c t dk

2

0exp2 gx c t

so that

2

0 0 0, exp2 gx t i k x t x c t

which shows a wave packet with an envelope moving with velocity 0gc .

3.3.3 Large Time Response to Localized

Disturbance

After a sufficiently long time, different Fourier components of a local disturbance will

be greatly dispersed. What’s left is a sinusoidal wavetrain with both k and ω slowly

varying functions of ,x t .

For surface waves on deep water,g

ck

so that waves with longer wavelengths

travel faster, one expects the wavelength λ to increase with x. (see fig 3.8)

Let the slowly varying wave train be

, , exp ,x t A x t i x t (3.34)

where ,x t is a phase function.

The local wavenumber k and frequency ω are defined by

kx

t

(3.35)

Now2 2k

t t x x t x

or

0k

t x

(3.36)

Hence

0k d k

t dk x

or

0gc kt x

(3.37)

ie., k is constant for an observer moving with velocity gc .

This can also be seen from the fact that

gk f x c t (3.38)

satisfies (3.37) and is obviously constant for constgx c t .

3.3.4 Surface Waves on Deep Water

gk (3.39)

1

2g

gc

k

After a long time, any local disturbance becomes locally sinusoidal with wavenumber

k in the neighborhood of distance

gx c t (3.40)

from the initial disturbance region so that

2

24 4g

g g tk

c x

(3.41)

2

gt

x

and hence2

24

gt

x x

2

gt

t x

2

4

gt

x

where ε is a constant. Eq(3.34) thus becomes

2

, , exp4

gtx t A x t i

x

(3.42)

3.4a Laplace’s Formula

Source: Landau

Consider 1st the interface between 2 inviscid fluids.

Let the principal radii of curvature at a point on the interface be 1R and 2R . ( 0R

if it is drawn into fluid 1)

The corresponding line elements there would be 1 1 1dl R d and 2 2 2dl R d where

dθ denotes the angle subtended by the line element. A surface element there is

1 2dS dl dl .

Let the interface be moved by an infinitesimal amount r .

Both 1R and 2R would be changed by an amount n̂ r , where n̂ the interface

normal pointing from fluid 1 to fluid 2.

Line element 1dl would become

11 1 1 1

1 1

11

dlR n d R n dl n

R R

r r r

and similarly for 2dl

The surface element dS then becomes

1 2

1 2

1 11 1dl n dl n

R R

r r

The change induced in dS is therefore

1 2

1 1ndS

R R

r

This will change the surface energy by

1 2

1 1ndS

R R

r

where α is the surface tension coefficient.

On the other hand, the work required to be done against the pressure difference is

2 1 ˆS

p p ndS r

In equilibrium, the total change of energy must be zero so that we have

2 1

1 2

1 10p p ndS

R R

r

ie.

1 21 2

1 1p p

R R

For viscous fluids, this is easily generalized to

2 1

1 2

1 1n n

R R

where the stress tensor is defined by

'pI

3.4b Curves

Source: T.M.Apostol, “Calculus”, vol I.

Consider a curve C t with parameter t in an n-dimensional space.

If C is the path of a particle, it is usually denoted in vector notation as tr .

The tangent (velocity) of C is defined by

ˆdvv

dt

rv

where v v is the speed (magnitude of v) and v̂ is the unit tangent.

Assuming Euclidean metric for all vectors, the path length can be written as

0

ts t v t dt

so that

dsv

dt

The principal normal is defined by

ˆdv

dtn

or

2

ˆ 1 1 ˆdv d d dv dvv nn

dt dt v v dt v dt v dt

v v vn a

where n̂ is the unit normal and a the acceleration.

A more familiar way to write this is

ˆ ˆ ˆd vvd dv dv

v vdt dt dt dt

v

a

Consider the rate of v̂ with respect to s:

1 ˆd v dt d v nn n

ds ds dt v v n

where

ˆ1n dv

v v dt

is called the curvature.

1R

3.4b.1 Circle

Consider curve cos , sint a t a t r which describes a circular path of radius a

in the x-y plane.

The tangent (velocity) is sin , cost a t a t v .

The unit tangent is ˆ sin ,cosv t t t .

The speed is v a .

The path length is 0

ts t a dt a t .

The normal is ˆcos , sin

dvt t t

dt n .

The unit normal is ˆ cos , sinn t t t

1 1 1 ˆcos ,sind v

t t nds v a a

n

The curvature is1

a .

a .

3.4b.2 Graph

Consider a graph of f t vs t.

The foregoing analysis can be applied here using

,t t f t r

Hence

The tangent (velocity) is 1, 't fv where 'df

fdt

.

The unit tangent is 2

1ˆ 1, '1 '

v t ff

.

The speed is 21 'v f .

The path length is 21 's t f dt .

The normal is 2

ˆ 11, '

1 '

dv dt f

dt dt f

n .

2

3 3 22 22 2

' '' ' '' '',

1 '1 ' 1 '

f f f f f

ff f

3

2 2

''',1

1 '

ff

f

The unit normal is 2

1ˆ ',11 '

n t ff

2 32 2 2

1 '' '' ˆ',11 ' 1 '

d v f ff n

ds v f f

n

The curvature is

32 2

1 ''

1 '

f

R f

.

32 21 '1

''

fR

f

.

3.4 Surface Tension & Capillary Waves

For our 2-D wave problem for an inviscid fluid, the boundary condition on the free

surface when surface tension effects are included is (see 3.4a and b)2

0 2

1p p

R x

(

2

x

neglected)

where α is the coefficient of surface tension, 0p the (constant) atmospheric pressure

and R the radius of curvature of the surface ( 0R if it is drawn inside the fluid).

Eq(3.22) thus generalized to2

20g

t x

on 0y (3.44)

while the other eqs remain unchanged, ie.

0t y

on 0y (3.21)

2 2

2 20

x y

Eliminating η gives

2 3

2 20g

t y x y

on 0y

Setting

, , ,x y t f y x t

gives

2 2

2 20 ' 0 0f f g

t x

(A)

while the Laplace eq. still gives2

22

''const

fk

x f

with solution expf y B ky as before.

Hence (A) becomes2 2 2

22 2 2 2

0g

k g k kt t k x

and the wave speed is

gc k

k

(3.46)

so that

3kc k gk

(3.45)

2

3

3

2g

k gd

cdk

k gk

(3.47)

For large k (short wavelengths), we have

kc

3 3

2 2g

kc c

Such waves are called capillary waves or ripples. In contrast with the gravity waves,

they travel faster if their wavelengths are shorter. Furthermore, the group velocity is

greater than the corresponding phase velocity. Thus, wave crests in a wave packet

travel backwards.

In general, we have capillary-gravity waves. (see fig 3.11 for dispersion)

Salient points are:

1. there is a minimum phase velocity

14

min

4gc

atg

k

.

2. there is both a capillary & a gravity wave for each phase velocity.

3.5 Finite Depth Effects

The effect of a finite depth h is to change the boundary condition satisfied by

exp expf y B ky D ky (3.A)

to

0y h

dfv h

dy

(3.C)

ie

exp exp 0B kh D kh

exp 2D

khB

exp exp 2f y B ky kh ky

exp expkhBe k y h k y h

2 coshkhBe k y h

so that (3.25) is replaced by

, , cosh sinx y t C k y h kx t (3.D)

The dispersion relation (3.26) becomes

0

sinh'tanh

coshy

k khfg g gk kh

f kh

(3.51)

so that the phase speed is (see fig 3.12)

tanhg

c khk k

(3.52)

which, for h large, becomes

gc

k (3.53)

as it should.

For small h, we have

gc kh gh

k (3.54)

which is non-dispersive.

3.6a Thermodynamics of Ideal Gas

An ideal gas is defined by 2 eqs.

A BPv RT N k T

0vu c T

where v is the molar volume, u the molar energy. The molar heat capacity at constant

volume 0vc is a function of T only. (Note: specific heat is heat capacity per unit

mass)

Joules8.3144 universal gas constant

mole KelvinR

236.023 10 Avogardro's numberAN

23 Joules1.381 10 Boltzmann's constant

KelvinBk

From 1st law:

du Tds Pdv or

du Pds dv

T T

which, for an ideal gas, becomes

0vc Rds dT dv

T v

and integrated to

0

00

0

lnT

v

T

c vs s dT R

T v

The T integral cannot be done until the functional form of 0vc is known.

Assuming 0vc to be a constant, we have

0

00 0

ln

vcRT v

s s RT v

Next, consider the molar enthalpy

h u Pv which, for an ideal gas can be written as

0ph c T

where the molar heat capcity at constant pressure 0pc is a function of T.

Thus

0 0p vc T c T RT

so that

0 0p vc c R

or

0

1v

R

c where 0

0

p

v

c

c

Hence, for 0vc independent of T:1

11

00 0 0 0

ln ln1

T v R T vs s R

T v T v

For an adiabatic process, 0s so that1Tv const

Other forms of this are

Pv P const

3.6 Sound Wave

For an inviscid, compressible fluid, we have

Euler’s eq.: pt

u

u u (3.55)

Eq. of continuity: 0t

u (3.56)

Note that an implicit assumption for inviscid fluid is that all processes are adiabatic.

(see Landau, §2).

Let the fluid be an ideal gas. Adiabatic processes satisfy

p const 0

0

p

v

c

c

or

0D

pDt

(3.57)

3.6.1 Small Amplitude Waves

Let the unperturbed state be a homogeneous one at rest.

For small perturbations, we can write

1u u 0 1p p p 0 1 (3.58)

where subscript 0 denotes unperturbed quantities and 1 the (small) perturbations.

Since the unperturbed state is homogeneous, 0 0p is independent of position.

For adiabatic perturbations, p remains 0 0p for each fluid element and is

therefore also independent of position. Hence

0 1 0 1 0 0p p p

ie

1 1

0 0

1 1 1p

p

which, to 1st order of perturbation, gives

1 1 1 1

0 0 0 0

1 1 1 1p p

p p

ie

1 1

0 0

0p

p

which, in terms of the sound velocity (see later),

00

0

pa

(3.59)

can be written as

21 0 1p a (3.60)

To 1st order of perturbation, the Euler eq (3.55) becomes

10 1p

t

u

(3.61)

and the eq of continuity (3.56),

10 1 0

t

u (3.62)

Eliminating 1u from (3.61-62), we have

2

21 0 1 12

pt t

u

which, with the help of (3.60), becomes2

21 12 2

0

10p p

a t

(3.63)

or2

21 12 2

0

10

a t

Thus, both the pressure and density variation are governed by the well–known wave

equation.

22

2 2

10

c t

whose general solution is of the form

, t t x x vξ

so that

ii

i i

vt t

2 2

2 i j i ji j i j

v v v vt

jij

i i j j ix x

2 2

i i i ix x

1 1p p t x v with 0av

so that the perturbation, called sound, travels with speed 0a .

3.6.2 Spherical Waves

For spherically symmetric waves, ,r t . The wave eq. becomes

22

2 2 2

1 10r

r r r c t

(3.65)

Setting

1, ,r t h r t

r

we have

2

1 1h

r r r r

2 2

2 2 2 2 3

1 1 1 2 2h h

r r r r r r r r r r

2 22

2 2 2

1 2 1r h

r r r r r r r r

so that (3.65) simplifies to a 1-D wave eq. for h:

2 2

2 2 2

10

h h

r c t

with solution

,h r t h r ct h r ct

so that

1,r t h r ct h r ct

r (3.66)

The h and h obviously denote outgoing and incoming waves, respectively.

The radiation condition simply sets 0h .

3.7 Supersonic Flow Past a Thin Aerofoil

Consider a 2-D inviscid, compressible, adiabatic flow past an aerofoil of the form

1 1, ,0U u v u 0 1p p p 0 1 (3.67)

As in §3.6.1, we consider only relations up to 1st order in perturbed quantities.

The adiabatic relation is the same as in §3.6.1:

21 0 1p a (3.68)

The Euler eqs.

0 1 1 1 1 0 1U u v U u p pt x y x

0 1 1 1 1 0 1U u v v p pt x y y

simplify to

10 1

pU u

t x x

10 1

pU v

t x y

1 10

u pU

x x

1 10

v pU

x y

Combining these 2 eqs gives2 2 2

1 1 10 0 2

u p vU U

x y x y x

ie.

1 10 0

u vU

x y x

which says the vorticity 1 1u v

y x

is independent of x.

On the other hand, 0 for the uniform flow far away from the aerofoil. Hence,

0 everywhere. The flow is irrotational (potential) so that we can write

1ux

1v

y

(3.69)

The Euler eqs thus become

0 1 0U px x

0 1 0U py x

which means

1 0p U Cx

(3.70)

where C is a constant. Since 1 0p when 1 0ux

, we must have 0C .

Similarly, the eq of continuity

0t

u

ie

0 10 1 1 0 1 1 0U u v

t x y

becomes, to 1st order perturbation,

1 1 1 10 0

u vU

t x x y

which, for steady state, simplifies to

1 1 10 0

u vU

x x y

which, in terms of , becomes

2 21

0 2 20U

x x y

Putting in the adiabatic condition (3.68), we have

2 21

02 2 20

0U p

a x x y

which, by (3.70), becomes

2 2 2 20

02 2 2 20

0U

a x x y

ie

2 2

22 2

1 0Mx y

(3.71)

where

0

UM

a

is called the Mach number for the flow.

3.8 Internal Gravity Waves

Incompressible, inviscid, fluids with stratified density variation can support internal

gravity waves which are driven by buoyancy forces.

One example is salty water with vertical salt concentration variation.

Consider the 2-D case where

, , , , , ,0u x y t v x y t u

In the static equilibrium state, we assume

0 y with 0 ' 0

so that the density increases as one goes down toward the bottom.

The corresponding pressure distribution is given by the steady state Euler eqs.,

00p

x

000

pg

y

so that by a suitable choice of the origin, we have

0 0p y gy

Since the fluid is incompressible & inviscid, the relevant eqs are

pt

u u g

0 u (3.80)

0t t

u u

which, in our choice of coordinates, becomes

0 10 1 1 1 1

p pu v u

t x y x

0 10 1 1 1 1 0 1

p pu v v g

t x y y

1 1 0u v

x y

1 1 0 1 0u vt x y

For 1st order perturbations, these simplify to

1 10

u p

t x

1 10 1

v pg

t y

1 1 0u v

x y

(3.81)

1 01 0v

t y

Consider a Fourier mode of the form

, , expa x y t A y i kx t (3.82)

where a can be 1 1 1, ,u v p , or 1 . Eq(3.81) becomes

0 1 1i U ikP

10 1 1

dPi V R g

dy

11 0

dVikU

dy (3.83)

01 1 0

di R V

dy

Using ' to denoted

dy, the 3rd, 1st & last eqs give

1 1 'i

U Vk

0 01 1 12

'P U i Vk k

11 0 '

VR

i

so that the 2nd become

10 1 0 1 02

' ' 'V

i V i V igk

which simplifies to

01 0 1 0 1 12 2

0 0

1 '' ' ''V V V g V

k

or

20 01 1 12

0 0

' ''' ' 1 0V V k g V

2 22

1 1 12'' ' 1 0

N NV V k V

g

(3.84)

where the buoyancy number N is defined by

20

0

' 0g

N

( 0 ' 0 )

The simplest case is when 0 is exponential, ie

0 expy

y AH

so that

2 gN

H

is a constant. So are the coefficients Eq(3.84), the solution of which is simply

1 expV B y

where ξ satisfies

2 22 2

21 0

N Nk

g

or

2 22

11 0

gk

H H

ie.

22

1 11 1 4

2 2

gHk H il

H H

where

22

14 1

2

gl Hk H

H

(3.84A)

The solution (3.82) thus become

1

1, , exp

2v x y t B il y i kx t

H

exp exp2

yB i kx ly t

H

(3.86)

which is a plane wave with wavevector , ,0k lk and amplitude that decays

exponentially as it rises.

Using 1 11 0 ' exp

V V yR i

i H H

, we have

1 , , exp exp2

B yx y t i i kx ly t

H H

(3.88)

Note that 1 is usually more accessible to experimental observation.

Solving for ω from (3.84A), we get the dispersion relation:2

2 2 2

1

4

g l HH

Hk k

2

2 22

22 2 2 2

2 22 2

1 114 44

gkg N kH

l H k l k lHH HHk k

(3.87)

which is clearly anisotropic.

If the wavelength

2 2

2 2

k l

k

is small compared with the scale height H, (3.87) simplifies to2 2

22 2

N k

k l

(3.89)

2 2

Nk

k l

which gives a group velocity

, ,g k k l m

c , ,k l mk (3.90)

2

2 2 3 32 2 2 2

1, ,0

k klN

k l k l k l

32 2, ,0

Nll k

k l

2 2, ,0

ll k

k l k

(3.91)

Now, the phase velocity is

2 2

ˆ , ,0p k k lk k k l

c

so that

0g p c c

ie., g pc c . (see fig.3.13)

3.9a Partial Differential Eqs.

Source: F.B.Hildebrand, “Advanced Calculus for Applications”, 2nd ed, Chapter 8,

Prentice Hall (1976)

nth order ODE:

n independent arbitrary constants.

linear ODE: General solution = linear combination of n independent functions.

Non- linear ODE: may exist singular solution with no arbitrary constants.

nth order PDE:

General solution = arbitrary functions of specific functions (arguments).

In case of specific boundary conditions, it’s usually easier to determine a set of

particular solutions & combine them to satisfy the BCs.

3.9a1 Quasi-Linear Eq. of 1st Order

Most general quasi-linear eq. of 1st order:

, , , , , ,z z

P x y z Q x y z R x y zx y

(15)

Linear case:

1 2, , , ,z z

P x y Q x y R x y z R x yx y

(16)

Let

, ,u x y z c (17)

defines a solution (integral surface) of (15). Treating z as ,z x y , the chain rule

gives:

0u u z

x z x

0

u u z

y z y

(18)

or

uz x

uxz

uz y

uyz

(19)

which, when substituted into (15), gives

0u u u

P Q Rx y z

(20)

Introducing a vector , ,P Q RV in the , ,x y z space, (20) becomes

0u V (21)

Since u is normal to the integral surface (17), V is tangent to a curve on the latter.

Conversely, a curve with tangent parallel to V everywhere must be on the integral

surface. Such curves are called characteristic curves of the differential eq.

Any curve can be denoted by the position vector sr r where s is the arc length.

The unit tangent to the curve is

d dx dy dz

ds ds ds ds

ri j k

If it is a characteristic curve, we must have

dxP

ds dy

Qds

dzR

ds (22)

where μ can be any function of , ,x y z .

Eq(22) can be written as

dx dy dz

P Q R (23)

Let the independent solutions to (23) be

1 1, ,u x y z c 2 2, ,u x y z c (24)

where 1 2,c c are constants. Any surface specified as

1 2, 0F u u (25a)

or

2 1u f u (25b)

will be an integral surface of (15).

3.9a2 Initial Conditions

For the linear case

1 2, , , ,z z

P x y Q x y R x y z R x yx y

(35)

One of the eqs in (23), namely,

dx dy

P Q

is an ordinary differential eq.

Let its solution be

1 1,u x y c (35a)

which is called a characteristic cylinder. (The intersection of this cylinder with the

xy plane is called a characteristic base curve.)Eq(35a) can be used to express y in terms of x and 1c so that another eq of (23), say,

1 2

dx dz

P R z R

ie.

1 2dz R Rz

dx P P

is just an ODE involving z x .

The solution to this can be written as

2 1 1 2 1 2, ,u z x c x c c

or

2 1 2 2, ,u z x y x y c

The general solution to (35) is therefore

21

1 1

,1,

, ,

x yz f u x y

x y x y

or

1 2 3, , ,z s x y f s x y s x y (36)

For a 1st order ODE, the arbitrary integration constant can be determined by requiring

the integral curve to go through a specific point in the xy plane.

For a 1st order PDE, the arbitrary integration function in (25) can be determined by

requiring the integral surface to include a specific curve in the xyz space.

Let this curve be given by the intersection of 2 surfaces

1 , , 0x y z 2 , , 0x y z (42)

and the independent solutions to (23) are

1 1, ,u x y z c 2 2, ,u x y z c (43)

The elimination of , ,x y z then results in an eq of the form 1 2, 0F c c .

The required integral surface is then simply 1 2, 0F u u .

3.9a3 Characteristics of Linear 1st Order Eqs.

Any curve curve in the xyz space can be specified as

C: x x , y y , z z (90)

The projection of this onto the xy plane is

0C : x x , y y , 0z (91)

The prescribed curve (42) can alternatively be given as a specification of z on 0C in

the form of (90).

The question is whether the PDE has an integral surface that actually contains C.

Now, on 0C , everything, including z, is a known function of λ. Hence

dz z dx z dy

d x d y d

(94)

Also, the PDE itself becomes an eq of independent variable λ.

1 2

z zP Q R z R

x y

(93)

These 2 eqs can be considered as simultaneous eqs forz

x

andz

y

on 0C .

The condition for the existence of a unique solution is

0P Q

dy dxP Qdx dy

d dd d

(95)

ie.

dx dy

P Q on 0C (96)

In other words, a unique solution that satisfies the boundary condition prescribed on a

curve 0C exists only if 0C is no where tangent to a characteristic base curve.

Now, in case 0C is itself a characteristic base curve, the determinant in (95) vanishes

identically:

0P Q

dx dy

d d (96a)

There’ll either be no or infinitely many solutions to (93) & (94). For the latter case,

we need, according to Cramer’s rule,

1 2

0R z R Q

dz dy

d d

or

1 2

0P R z R

dx dz

d d

(96b)

This means a curve C that satisfies eqs (96a,b) must be a characteristic curve.

Indeed, this provides a way to find characteristic curves for more complicated PDEs.

3.9 Finite-Amplitude Waves in Shallow Water

Let the bottom of the fluid be at 0y ; its free surface, ,y h x t .

Let 0h be some typical value of ,h x t , and L be a typical horizontal length scale.

The shallow water approximation means

0h L (3.92)

The full 2-D eqs for ideal fluid are

1u u u pu v

t x y x

(3.93)

1v v v pu v g

t x y y

(3.94)

0u v

x y

(3.95)

The shallow water approximation means (see §3.9.1 for justification)

Dvg

Dt

so that (3.94) simplifies to

10

pg

y

which can be integrated immediately to give

,p gy f x t

The integration constant ,f x t can be determined by the boundary condition that

we have atmospheric pressure 0p at the free surface ,y h x t . Hence

0 ,p p g y h x t

With this, (3.93) becomes

Du hg

Dt x

which means the change of u for each fluid element is independent of y.

Hence, if u is independent of y at some time, it will be so for all time.

In which case, (3.93) simplifies to

u u hu g

t x x

(3.96)

and (3.95) can be integrated to give

,u

v y f x tx

The integration constant ,f x t can be determined by the boundary condition at the

bottom, ie.,

0v at 0y so that 0f and

uv y

x

(3.96a)

As shown in §3.2.1, the kinematic condition at the free surface is

h hu v

t x

on ,y h x t (3.18)

Combining with (3.96a), we have

0h h u

u ht x x

on ,y h x t (3.97)

Eqs(3.96-7) are known as the shallow water equations.

3.9.1 Shallow Water Equations

The shallow water eqs are (see Acheson §3.9)

0t x xu uu gh (3.96)

0x t xhu h uh (3.97)

where subscripts denote partial derivatives.

Let u, h be specified on curve 0C .

We have

' ' 't xu u t u x ' ' 't xh h t h x

where ' denotes derivative with respect to λ.

Rewriting these as a set of simultaneous eqs for , , ,t x t xu u h h , we have

1 0 0

0 1 0

' ' 0 0 '

0 0 ' ' '

t

x

t

x

uu g

uh u

ht x u

ht x h

The determinant of coefficients is

1 0

' 0 0 ' 1

0 ' ' 0 ' '

h u u g

D x t h u

t x t x

' ' ' ' ' ' 'x x ut t u x ut ght

2 2 2' 2 ' ' 'x x t u t gh u

Setting 0D gives

2 2 2 2' ' ' 'x ut u t t gh u

' 'x t u gh

dxu gh

dt (a)

which give us the characteristic base curves 0C .

According to Cramer’s rule, the condition for infinitely many solutions for xu is

1 0 0

0 0 10

' ' 0 0

0 ' ' '

g

u

t u

h t x

ie

0 1 0 0

' 0 0 ' 0 1 ' ' ' ' ' ' 0

' ' ' ' ' '

u g

u t u u x uu t t gh

h t x h t x

'' '

'

hx t u g

u

dx dhu g

dt du (b)

Combining (a) and (b) gives

dhgh g

du

ie

2dh

du g d ghh

2u gh const (c)

which are satisfied on the characteristic base curves.

Following Acheson, we define

c gh (3.98)

so that the characteristic base curves (a) become

dxu c

dt (3.103)

on which 2u c const .

Consider the functions

2u c and 2u c

so that

1

2u 1

4c

221 1

16h c

g g

and

1

8t t thg

The shallow water eq (3.96) thus become

1 1 10

2 4 8t t x x x x

which simplifies to

13 3 0

4t t x x x x

4 3 3 0t t x x (d)

Similarly, (3.97) becomes

21 1 10

32 8 16x x t t x xg g g

4 2 0x x t t x x

4 3 3 0t t x x x x

4 3 3 0t t x x (e)

(d)+(e): 4 3 0t x

13 0

4t x

2 0u c u ct x

(3.99)

(d)-(e): 4 3 0t x

13 0

4t x

2 0u c u ct x

(3.100)

3.9.2 Flow Caused By Dam Break

Let static 2-D water be contained between 00 y h for 0x by a dam at 0x .

(see fig.3.15a)

The problem is to find the water flow after the dam breaks at 0t .

Consider 1st the initial conditions.

For 0x , 0t , we have 0u , 0h h , so that 0 0c c gh .

The characteristic base curves

dxu c

dt (3.103)

become

0

dxc

dt

ie.,

0 0x c t x (3.104a)

which are straight lines that intercept the x axis at 0x . (see fig.3.15c)

After the dam breaks, these characteristic base curves will extend continuously,

wherever possible, into the 0t region. On such extensions, 02 2u c c .

Now, some points may be reachable by extensions of both types of base curves. (Point

P in fig.3.15c) Both eqs 02 2u c c must then be satisfied, which means

0u and 0c c

ie., the water there are not yet disturbed and the characteristic base curves are still

straightlines given by (3.104a). As is obvious from fig.3.15d, the upper boundary of

this undisturbed region is the line 0x c t .

Beyond the line 0x c t , a point (eg., Q in fig.3.15.d) can be reached only by the

extension of characteristic base curves of the ‘positive’ typedx

u cdt

on which

02 2u c c (3.104)

The other, ‘negative’ type, characteristic base curve that pass through Q , ie.

dxu c

dt (3.104b)

requires

2u c k (3.104c)

where 02k c is a constant along the ‘negative’ base curve (3.104b).

The solution of (3.104) & (3.104c) is

0 2

ku c and 0

1

2 2

kc c

(3.105a)

which are both constants on the ‘negative’ base curve (3.104b) which itself becomes a

straight line

0

1 3

2 2

dxc k

dt

ie.

0 0

1 3

2 2x c k t x

where 0x is the value of x at 0t .

Now, different ‘negative’ base curves are characterized by different k so that they have

different slopes. To avoid crossing of such curves, we need 0 0x so that the

‘negative’ base curves all eminate from the origin:

0

1 3

2 2x c k t

(3.105b)

The ‘positive’ base curves are given by

0

1 13

2 2

dxc k

dt

(3.105c)

which are not straightlines since k is not constant on such curves.

From (3.105b), we have

0

22

3

xk c

t

so that the ‘positive’ base curve (3.105c) becomes

0 0 0

1 1 13 2 4

2 3 3

dx x xc c c

dt t t

while (3.105a) becomes

0

2

3

xu c

t

and 0

12

3

xc c

t

(3.105,6)

Since 0c gh , these relations are valid only for

02x c t

Using (3.106) we have22

0

12

9

c xh gh

g g t

for 0 02c t x c t

At 02x c t , we have 0h , which should remain so for 02x c t on physical

grounds. (see fig.3.15b).

3.9.3 Non-Linear Wave Distortion

The dam break flow may be taken as a smoothing out of the initial discontinuity in

,h x t through non-linear mechanism.

2 0u c u ct x

(3.99)

2 0u c u ct x

(3.100)

Consider the region 0 02c t x c t where 02 2u c c .

Eq (3.99) is automatically satisfied.

Eliminating u from (3.100) gives

02 3 0c c ct x

which can be rewritten as

0z z

zt x

(3.109)

with

03 2z c c (3.108)

The characteristics eqs of (3.109) are (see §3.9a1)

1 0

dt dx dz

z

so that

1z c

and

dxz

dt ie. 2x zt c

The general solution is therefore

z F x zt (3.110)

where F is an arbitrary differentiable function.

Note that (3.110) is an implicit solution.

To show that (3.110) is indeed a solution of (3.109), we need to evaluate the various

partials of (3.110) with x and t taken as the independent variables. Setting

x zt

the partials on (3.110) are

'z dF z

z t Ft t d t

1 'z dF z

t Fx x d x

where 'dF

Fd

.

Solving for the partials of z, we have

'

1 '

z zF

t tF

(3.111)

'

1 '

z F

x tF

(3.112)

which obviously satisfies (3.109) identically for arbitrary differentiable F.

Now, (3.110) is a wave of z that travels with velocity –z. Assuming positive amplitude

( 0z ), the wave travels to the left with speed that is proportional to its amplitude,

thus providing the mechanism for smoothing out the discontinuity in the dam break

problem.

Consider now the eq

0z z

zt x

(3.113)

Applying the foregoing analysis, we see that its solution is

z F x zt (3.114)

which describes a wave of z that travels with velocity z. Assuming positive amplitude

( 0z ), the wave travels to the right with speed that is proportional to its amplitude,

thus steeping the wave profile as shown in fig.3.16.

From the analog of (3.112):

'

1 '

F Xz

x tF X

X x zt (3.115)

we see thatz

x

goes to infinity at time ct defined by

1 ' 0c ct F x zt (3.116)

In case of multiple solutions, only the minimum value will be physically meaningful

since the wave will collapse after that. (see fig.3.16c)

3.9.4 The Formation of a Bore

Let fluid be contained by a vertical plate in the region 0x , 00 y h for 0t .

At 0t , the plate at 0x begins to move in the x direction with speed U t ,

where α is a constant. (see fig.3.17)

Once again, we need to solve the shallow water eqs (3.96-7) or (3.99-100).

The treatment is analogous to that of the dam break problem so we’ll simply highlight

the salient points.

For 0t , the fluid occupies 0x with 0u and 0h h , 0 0c c gh .

The corresponding characteristic base curves

0 0x c t x (3.104a)

are straight lines in the 4th quadrant that intercept the x axis at 0x . (see fig.3.18)

The undisturbed region ( 0u and 0c c ) is found by extending the above straight

lines into the 1st quadrant. The upper boundary of this is obviously the line 0x c t .

(see fig.3.18)

As is clear from fig.3.18, only ‘negative’ type of base curvesdx

u cdt

can be

extended across the boundary 0x c t . On such base curves,

02 2u c c (3.117)

so that (3.100) is automatically satisfied. Eliminate c from (3.99) gives

0

30

2u c u

t x

(3.118)

whose characteristics are given by

0

31 02

dt dx du

u c

so that

1u c

and

0

3

2

dxu c

dt ie. 1 0 2

3

2x c c t c

The general solution of (3.118) is therefore

0

3

2u F x u c t

(3.119)

Now, we expect u to have a maximum at the plate and decrease to zero at 0x c t .

Eq(3.118) is then of the profile steepening type (3.113) discussed in §3.9.3.

The actual form of F is of course determined by the boundary conditions.

At time t, the vertical plate is at 21

2x t moving with speed u t .

Eq(3.119) becomes

2 20 0

1 3

2 2t F t t c t F t c t

0t (3.120)

with

0 0F

Writing

20t c t

we have

20 0

14

2t c c

so that (3.120) becomes

20 0

14

2F c c 0 (3.120)

where the + root is chosen so that 0 0F .

Eq(3.119) is evaluated by putting 0

3

2x u c t

in (3.120) so that

20 0 0

1 34

2 2u c c x u c t

Solving fo u, we have

2 20 0 0

32 4

2u c c x u c t

20 04 4 6 4 0u c t u x c t

20 0

30

2u c t u x c t

so that

2

0 0 0

1 3 34

2 2 2u c t c t x c t

(3.121)

where the + root is chosen to make 0u at the undisturbed boundary 0x c t .

Eq(3.117) can be used to find c:

2

0 0 0 0 0

1 3 34

2 4 2 2

uc c c c t c t x c t

(3.122)

Hence

1

2 2

0 0

34

2 2

cc t x c t

x

At the boundary 0x c t ,

0

32

2

c

x c t

which explodes at time

02

3c

ct

(3.123)

At this moment, the vertical plate is at position

20

1 1

2 3c cx t c t

whereupon (3.122) becomes

0 0 0 0 0

1 1 8 44

4 4 3 3c cc c x c t c c t c

which means the water level at the gate at time ct is

20

1 16

9h c h

g

3.10 Hydraulic Jumps & Shock Waves

3.10.1 Hydraulic Jumps

A bore (§3.9.4) that is stationary to the observer is called a hydraulic jump.

Both are shallow water phenomena which exhibit an abrupt jump of water level.

One way to produce it is to turn on the kitchen tap. Water in the sink splays outward

radially in a thin layer which exhibits a sudden increase in height at a certain radius

that depends on the flow rate.

With reference to fig.3.19 for notations, we see that

1. Conservation of mass means

1 1 2 2U h U h (3.124)

2. Conservation of momentum means

2 2 2 21 1 1 2 2 2

1 1

2 2gh hU gh h U (3.125)

where terms proportional to g come from pressure, and those to U from

convective contributions to the stress tensor.

3. Conservation of energy means turbulence at the jump will dissipate energy in the

form of heat. The energy loss at the jump is (Ex.3.20)

312 1

24

gUh h

h

(3.128)

where

2 1h h (3.129)

Introduce now the Froude number

p

U UF

cgh (3.126)

where pc gh is the phase speed of water waves in shallow water.

It plays a similar role as the Mach number in compressible flow.

Flows with 1F ( 1F ) are called sub- (super-) critical.

Now, eliminating 2U using (3.124), (3.125) becomes2 2

2 2 2 1 11 1 1 2

2

1 1

2 2

h Ugh hU gh

h

so that

2 22 2 12 2 2 1

11 2 1 12 2

gh h h gh h hU

h h h h

Interchanging indices 1 2 gives

2 1 2 12

22

gh h hU

h

Hence, the corresponding Froude numbers are

2 1 211 2

11

h h hUF

hgh

1 1 222 2

22

h h hUF

hgh

(3.127)

Since 2 1h h , we have

1 1F and 2 1F (3.130)

ie., the jump changes a supercritical flow to a subcritical one.

Consider the sound wave problem (§3.6) without the small amplitude approximation.

The relevant eqs are

Euler eq.: pt

u

u u (3.55)

Eq. of continuity: 0t

u (3.56)

pDt

(3.57)

Using the Leibniz rule, (3.57) becomes

1 0Dp D

Dt Dt

or

2Dp p D Da

Dt Dt Dt

(3.133a)

where

2 pa

(3.133)

Taking the material derivative of (3.133) gives

2

12

Da Dp p Da

Dt Dt Dt

2 2a a D

Dt

[from (3.133a)]

2

1a D

Dt

(3.133b)

Rewriting (3.56) as

D

Dt

u

(3.133b) becomes

2 1Da

u

or

20

1

aa a

t

u u (3.133c)

Now, if the gas is initially at rest, (3.57) becomes

p const so that, for example,

1 0p p ie

2pp a

Also, analogous to (3.133b), we have

2

2 1a

a a

so that

2

1

ap a

The Euler eq. then becomes

20

1

aa

t

u

u u (3.133d)

For 1-D gas flow with , ,0,0u x t u , (3.133c,d) simplify to [cf. Ex.3.22]

20

1

a a uu a

t x x

20

1

u u a au

t x x

(3.133e)

These are similar to, but not the same as, the shallow water eqs(3.96-7).

The characteristics are obtained by considering

1 0 102

2 00 11 '

' ' 0 0 '0 0 ' '

t

x

t

x

au

a

a auu a

t x u ut x

This set of eqs is indeterminate if 2 conditions hold:

I.

1 0 12

20 1 01

' ' 0 0

0 0 ' '

au

au

t x

t x

ie.,

2 0 11 212

' 0 0 ' 11

0 ' '0 ' '

aa uu

ax t u

t xt x

2' ' ' ' ' ' 'x x ut t u x ut a t

2 2 2' ' ' 0x ut a t

which requires

' 'x u a t

dxu a

dt (3.133f)

II.

1 0 0 12

0 0 1 0' ' 0 0

0 ' ' '

a

u

t a

u t x

ie.

0 0 10 1 2' 0 0 ' 0 1

' ' ' ' ' '

au

a t u

u t x u t x

' ' ' ' ' 12

aa x ut t u

' ' ' ' ' 1 02

aa x ut t u

so that

2

1

du dxu

da a dt

Combining with (3.133f), we have

2

1

du

da

ie.

2

1

au c

To summarize. The characteristic base curves are

dxu a

dt (3.133f)

on which

2

1

au c

(3.133g)

where c are constants.

These can be used to simplify (3.133e) as follows. [cf. §3.9.1]

Consider the functions

2

1

ab

u b and u b

so that

1

2u 1

2b

1 1

2 4a b

The 1st eq in (3.133e) thus become

1 1 10

2 2 8t t x x x x

which simplifies to

4 1 3 3 1 0t t x x (d)

Similarly, the 2nd one becomes

1 1 10

2 2 8t t x x x x

4 1 3 3 1 0t t x x (e)

(d)+(e): 4 1 3 0t x

Now:

2 21 3 1 3

1 1

a au u

4 u a

so that we have

0t xu a

20

1

au a u

t x

(3.131)

Similarly, (d)-(e): 4 3 1 0t x

Now:

2 23 1 3 1

1 1

a au u

4 u a

so that we have

0t xu a

20

1

au a u

t x

(3.132)

Consider the case of a piston moving with speed U t into a long tube containing

gas at rest with sound speed 00

0

pa

.

The situation is analogous to the formation of a bore (§3.9.4).

For 0t , the gas occupies 0x with

0u and 0a a

The characteristic base curves are straightlines in the 4th quadrant:

0

dxa

dt ie. 0

0

1t t x

a

where 0t is the intercept with the t axis.

The + type lines must have 0 0t . The line

0

1t x

a ie., 0x a t

thus form an upper boundary for them.

For the – type lines, only those with 0 0t can be extended across this boundary.

Thus, inside the bore region, we have, along the – type characteristic base curves

02 2

1 1

a au

(3.134)

and (3.132) is automatically satisfied.

For the + type characteristic base curves, eliminating a using (3.134) turns (3.131)

into

0

11 0

2u a u

t x

(3.135)

The characteristics of this are given by

0

11 012

dt dx du

u a

The general solution of (3.135) is therefore

0

11

2u F x u a t

(3.136)

where F is an arbitrary differentiable function.

The function F is determined by the boundary condition at the piston:

u t at 21

2x t

ie.,

0

1

2t F t a t

0t

Setting

0

1

2t a t

we have

20 0

12t a a

so that

20 0

12a a F

where the + root was chosen to make 0 0F .

Thus

20 0 0

1 12 1

2u a a x u a t

Solving for u, we have

2 2 2 20 0 0 0

12 2 1

2u a u a a x u a t

20 02 1 2 0u a t u x a t

2

0 0 0

12 1 2 1 8

2u a t a t x a t

where the + root is chosen to make 0u at 0x a t .

Hence

2

0 0

2

2 1 8

u

x a t x a t

Now, (3.136) is of the wave steepening type,u

x

will eventually become singular,

which happens when

2

0 02 1 8 0a t x a t

At 0x a t , this occurs at time

02

1c

at

whence (3.135) breaks down and a shock propagates down the tube.

3.10.3 Normal Shock Waves

A shock normal to the flow is similar to a hydraulic jump (§3.10.1).

With reference to fig.3.20 [cf. fig.3.19], we have the Rankine-Hugoniot equations:Conservation of mass: 1 1 2 2U U

Conservation of momentum: 2 21 1 1 2 2 2p U p U (3.137)

Conservation of energy:2 2

2 21 21 2

1 1

2 1 2 1

a aU U

[The last is different from the hydraulic jump case which substains an energy loss

across the jump.]

Furthermore, the 2nd law of thermodynamics

2 1 0S S

means that

2 2 1 1log logp p

ie.,

2 2 1 1p p or 2 2

1 1

p

p

(3.138)

We leave as an exercise to show that

1. 2 0 1U a U , ie., 2 11M M 2. 2 1p p , 2 1

3.11 Viscous Shocks & Solitary Waves

3.11.1 Weak Viscous Shocks

For a weak viscous shock propagating in x direction into a gas at rest, the velocity

,u x t satisfies Burger’s equation

2

0 2

1 21

2 3

uu a u

t x x

(3.140)

where0

if both μ and are small.

For 0 , (3.140) reverts to (3.135).

Consider the ansatz

u f x Vt (3.141)

such that

1f U 0f

Now:

'u

Vft

'u

fx

2

2''

uf

x

where

'df

fd

x Vt

Eq(3.140) becomes

0

1 21 ' ''

2 3V f a f f

which, using

' ' ''' '

df df df dff f

d d df df

becomes

0

2 1' 1

3 2df V f a df

ie

20

2 1' 1

3 4f f V a f Const

or

2 048

'3 1 1

a Vf f f C

where C is a constant.

Hence

2 0

3 148

1

dfa V

f f C

In order to the formula

2

1ln

dx x p

ax bx c a p q x q

where p, q are the roots of 2ax bx c , we set

1a 04

1

a Vb

c C

21, 4

2p q b b C

so that

2

3 1 1ln

8 4

f p

f qb C

Now, the points require ,f p q .

The condition 0f thus require either

2 4 0b b C or 2 4 0b b C

Only the 1st eq has a solution with 0C . Therefore, we have

3 1 1ln

8

f b

b f

The condition 1f U then means

1b U

so that

1

1

3 1 1ln

8

f U

U f

which can be solved for u f to give

113 1

exp8

UU u

u

( 1u U )

ie.,

1 1

13 1 1 exp1 exp8

U Uu

x VtU

(3.142)

where the shock thickness is given by

1

8

3 1 U

(3.144)

Thus, u decays from 1U to 0 within a distance of order .

Finally, the condition on b means

01

4

1

a VU

Solving for the shock speed V gives

0 1

11

4V a U (3.143)

3.11.2 Solitary Waves in Shallow Water

We now look for a nonlinear shallow water wave of permanent form whose

steepening effects are caused by weak dispersion.

Let

0 be typical magnitude of , the vertical displacement of water surface.

0h be that for h, the depth of water.

L be that for horizontal length scale.

If

0

0h

and2

02

h

L (3.145)

are small & of the same order of magnitude, then satisfies the Korteweg- de Vries

equation:

320

0 0 0 30

3 10

2 6

cc c h

t h x x

(3.146)

where the last term describes dispersive effects.

We seek a traveling solution of the form

f x Vt f

with boundary conditions

( ) 0nf 0,1,2,n (3.145a)

Thus, (3.146) becomes

200 0 0

0

3 1' ' ''' 0

2 6

cVf c f f c h f

h

or

2 200 0 0

0

3 1' ' ''' 0

4 6

cc V f f c h f

h

which, on integrating, becomes

2 200 0 0

0

3 1''

4 6

cc V f f c h f C

h (3.146a)

where C is a constant.

Putting in the boundary conditions (3.145a) gives 0C .

Writing

21'' '

2

df f

df

(3.146a) becomes

2 2 200 0 0

0

1 1 3'

6 2 4

cc h d f c V f f df

h

2 2 2 300 0 0

0

1 1'

12 2 4

cc h f c V f f C

h (3.146b)

where C is another constant.

Putting in the boundary conditions (3.145a) gives 0C .

Rearranging terms in (3.146b) gives

3 2 2 20 0

0

1' 2 1

3

Vh f h f f a f f

c

(3.146c)

where

00

2 1V

a hc

(3.146d)

Taking the square root of (3.146c):

30

3'f f a f

h

so that

30

3 dfC

h f a f

which, upon the use of the formula

11 2ln tanh

dx a bx a a bx

ax a bx a a bx a a

gives

13

0

3 2tanh

a f

h aa

where 0C according to the boundary conditions (3.145a).

Solving for f gives

30

3tanh

4

a f a

a h

2 23 3

0 0

3 31 tanh sech

4 4

a af a a

h h

ie.,

23

0

3sech

4

aa x Vt

h

(3.147)

Eq(3.146d) can also be rewritten as

00

12

aV c

h

(3.148)

Since sech ranges between 0 and 1, a is the maximum of the wave height 0 . [see

fig.3.23]. Owing to the assumptions (3.145), these eqs are valid only for 0a h .

According to (3.148), waves with larger amplitude a travels with greater velocity V.

Thus, a higher wave can overtake a lower one. What seems astonishing is that after

such a collision, both waves emerge unscathed. [see fig.3.24] Such waves are called

solitons.

4. Classical Aerofoil TheorySource: D.J.Acheson, “Elementary Fluid Dynamics”, Chapter 4, Clarendon Press (90)

4.1 Introduction

4.2 Velocity Potential & Stream Function

4.2.1 Velocity Potential

For irrotational flow,

0 u

so that one can define a velocity potential at point P by

( )P

OP d u x (4.2)

u (4.3)

where O is some reference point.

In a simply connected region, is path independent & hence single- valued.

In a multiply connected region, can be path dependent & hence multi- valued.

The circulation around a closed contour C is

CC Cd d u x x (4.4)

where C is the change of in going around C.

Examples:

1. Uniform flow ,0,0Uu has Ux const .

2. Irrotational flow , ,0x y u has stagnation point at the origin.

To solve

xx

y

y

0

z

we see that the 3rd eq means is a function of x and y only.

Next, we integrate the 1st eq to get

21( )

2x f y

Takingy

of this & equating the result to the 2nd eq gives

dfy

dy

which gives

21

2f y C C const

so that

2 21

2x y C

which is single valued so that 0 for any circuit.

3. Line vortex flow

k

r u e

is irrotational except at origin.

Since u is not defined at origin, any domain that covers the origin is multiply

connected.

In cylindrical coordinates,

ˆˆ ˆr r z

r

θ z

so that we need to solve

0r z

and

k

r r

This gives

k C

which is multi-valued as expected.

Circulation for a circuit that goes round the origin n times is

2 nk

4.2.2 Stream Function

For 2-D incompressible irrotational flow, we can write

, ,u vy x

u (4.5)

so that

0u v

x y

u (4.6)

is automatically satisfied.

Now,

0u vx y y x x y

u (4.7)

so that ψ is constant along a streamline.

ψ is called a stream function since streamlines are simply curves with const .

Another way to write (4.5) is

,

0 0

x y z y x

i j k

u k (4.8)

In cylindrical coordinates, we have

ˆ ˆˆ1 ˆˆ

0 0

r

r r z r r

rθ k

u k r

θ

or

rur

u

r

(4.9)

which obviously satisfies the incompressibility condition trivially

0rru ur r r

(4.10)

4.3 Complex Potential

Consider a 2-D incompressible flow.

0u v

x y

we have

uy

vx

(4.11)

If it is also irrotational,

0u v

y x

we have

ux

vy

(4.11a)

Thus, and satisfy the Cauchy- Riemann conditions

x y

y x

(4.11b)

so that they are the real & imaginary parts of an analytic function, ie.,

w i (4.12)

where w z , with z x iy , is known as the complex potential.

By definition

0

limz

f z z f zdf

dz z

We say f is analytic at z ifdf

dzis independent of the path z approaches 0.

Thus, if f a ib is analytic, we have

df a b a bi i i

dz x x y y

for the 2 cases z x and z i y .

Another useful property for f to be analytic at z is that it has a Taylor expansion there.

Since w is analytic, we have

dwi u iv

dz x x

(4.15)

The flow speed is therefore

2 2 dwq u v

dz (4.16)

Either of the real & imaginary parts of an analytic function satisfies the Laplace eq.

individually. Hence2 2

2 20

x y

(4.13)

2 2

2 20

x y

(4.14)

4.3.1 Uniform Flow At Angle to x-Axis

Here

cosu U sinv U so

cos sin idwu iv U i Ue

dz

and, integrating givesiw Uze (4.17)

4.3.2 Line Vortex

2 r

u e (4.18)

From §2.4.4, we have

2

(4.19)

From (4.9), we have

0rur

2u

r r

so that

ln2

f r r

Hence

ln ln ln2 2 2

w i i r i r i i z

(4.20)

whereiz re

This can immediately generalized to a line vortex at 0z z :

0ln2

w i z z

(4.21)

4.3.3 Flow Near Stagnation Point

The Taylor expansion of w near 0z is

2

0 0 0 0 0

1' ''

2w z w z z z w z z z w z

The constant term 0w z is inconsequential.

If 0z is a stagnation point, 0' 0w z .

Hence, the flow near a stagnation point is determined, to the lowest order, by the

Writing

0'' iw z e and 0 1z z z

we have

21

1

2iw const z e

22

1

2const z

where

22 1

i

z z e

Dropping the constant term & subscripts, the complex potential near a stagnation

point is therefore

2 2 21 12

2 2w z x y ixy (4.22)

where α is real.

The velocity potential is

2 21Re

2w x y

so the equipotential lines are rectangular hyperbolae.

The stream function is

Im w xy (4.24)

so the streamlines (constant ψ) are rectangular hyperbolae perpendicular to the

equipotentials [see fig.4.1]

The corresponding flow is

Re Redw

u z xdz

Im Imdw

v z ydz

(4.23)

4.4 Method of Images

Let there be a rigid plane wall at 0x .

Consider a line vortex of strength at ,0dr [see fig.4.2].

The boundary condition

0, 0u y or0

0xx

can be satisfied by replacing the wall with a mirror image of the vortex.

As in electrostatics, the image should be at ,0d r with strength .

According to (4.21), the complex potential is therefore

ln ln2 2

w i z d i z d

(4.25)

ie.,

ln2

z di i

z d

(4.26)

ln2

z di i

z d

where

iz d z de

z d z d

Hence

ln2

z d

z d

so that the streamlines are

ln2

z dconst

z d

or simply

z dR

z d

(4.27)

where R is a constant.

Now, (4.27) can be written as

2 22 2 2x d y R x d y

or, on collecting terms:

2 2 2 2 21 2 1 0R x d y R dx

22 2 2

2

12 0

1

Rx dx d y

R

2 2 22 22 2

2 2 2

1 1 21

1 1 1

R R Rx d y d d

R R R

which is a circle centered at2

2

1,0

1

Rz d

R

2

1

Ra d

R

.

Such circles are known as coaxial circles.

The distances between the center of a coaxial circle and those of the vortices at

x d are

22

2 2

1 21

1 1 1

RR dc d

R R

so that

22 2

2

2

1

dc c R a

R

(4.27a)

Consider now the flow inside a circular cylinder z a due to a line vortex of

strength at ,0z c a . The boundary condition of no flow in the normal

direction is satisfied if z a is a streamline. This is easily accomplished by placing

an image vortex of strength at a point which according to (4.27a) is

2

,0a

z ac

. The corresponding complex potential is [cf. (4.25)]:

2

ln ln2 2

aw i z c i z

c

(4.28)

4.4.1 Milne-Thomson’s Circle Theorem

Consider ( )w f z where all singularities of f lie in z a . Then

2a

w f z fz

(4.29)

is the complex potential of a flow with

1. same singularities as f z in z a .

2. z a as a streamline.

Proof:

1. Since all singularities of f z are in z a , those of2a

fz

and hence

2af

z

are in2a

az ; ie., a z . Therefore,

2af

z

is regular z a and

w has the same singularities as f z there.

2. On z a , we have

iz ae ( real)2

iaae z

z

so that

2 Rew z f z f z w z

Hence,

0 on z a

which means z a is a streamline.

4.5 Irrotational Flow Past A Circular Cylinder

IR flow in x direction, with speed U at infinity, past fixed cylinder z a .

Complex potential for unperturbed flow is: f z Uz (singular only at infinity).

Applying the circle theorem (§4.4.1), we have

2 2a af U Uz

z z

2 2a af U

z z

so

2a

w z U zz

(4.31)

is the complex potential with z a as a streamline.

Any superposition with a line vortex of strength Г is also a solution with z a as a

streamline so that a more general form of (4.31) is

2

ln2

aw z U z i z

z

(4.32)

Consider first (4.31). Writing iz re we have

2

i iaw z U re e i

r

so that

2

cosa

U rr

(4.33)

2

sina

U rr

(4.34)

Hence

2

21 cosr

au U

r r

2

21 sin

au U

r r

(4.35)

The flow is symmetric fore & aft (see fig4.4a)

At r a , 2 sin 0u U for 0, , ie, there is slip everywhere except for 2

points.

Define the slip velocity as 0su u at r a .

The arc length along the top of the cylinder from the forward stagnation point is

s a .

Eq(4.35) then gives (with r a ),

2 sin 2 sins

s su U U

a a

(4.37)

2cossdu U s

ds a a

The slip velocity therefore rises from 0 at front stagnation point to a maximum at

2 , then falls to 0 at rear stagnation point.

For finite Г, we deal with (4.32), which becomes

2

ln2

i iaw z U re e i r i

r

so that

2

cos2

aU r

r

2

sin ln2

aU r r

r

and

2

21 cosr

au U

r r

2

21 sin

2

au U

r r r

(4.38)

For aerofoil problems, one usually deal with 0 ; in which case, let

02

BUa

(4.39)

Now we search for the stagnation points at which 0u .

On r a , (4.38) gives

0ru

2 sin 2sin2su u U U B

a

Thus, for 2B , there are 2 stagnation points at [see fig.4.4.b],

1sin2

B and 1sin2

B (4.39a)

For 2B , these 2 stagnation points coalesce into 1 at 1 3sin 1 or

2 2

.

For 2B , there is no real solution to (4.39a), ie., the stagnation points, if exist, are

off the cylinder. This means 0ru must now be satisfied by setting cos 0 in

the 1st eq in (4.38). We put3

2

so that our result may converge properly to that

of the 2B case. The 2nd eq in (4.38) thus becomes

2 2

2 21 1 0

2

a a Bau U U

r r r r

ie.,2 2 0r Bar a

with solutions

2 42

ar B B

However, the – root gives 0r at 2B so that it must be discarded.

Therefore, there is only 1 stagnation point at

2 42

ar B B ,

3

2

(4.40)

Since we have irrotational steady flow, the Bernoulli theorem applies and we have

21

2

pconst

u everywhere.

Note: if the flow is not irrotational, the above still applies to the surface of the

cylinder since it is a streamline.

On the cylinder, r a so that2

2 2 2 2 2 sin2ru u u U

a

u

2 24 sin 2 sinU U consta

The Bernoulli eq. thus becomes

2 22 sin sinp U

const Ua

on r a (4.40a)

The net force F on the cylinder is

2

0ˆˆpndS a p d

F r

whereˆ cos sin r i j

Now, (4.40a) is symmetric fore & aft, ie., invariant under transformation .

Since cos changes sign under the same transform, the x-component of F vanishes.

For the y-component, we have

2

0sinyF a p d

2 2 2

02 sin sin sin

Ua const U d

a

U (4.41)

where only the 3rd term in the integrand contributes.

Thus, we have no drag but an upward lift L U for 0 , as mentioned in

§1.6.3.

If the oncoming stream makes an angle α with the x-axis, the unperturbed complex

potential is given by (4.17) as iw Uze . Eqs(4.31,2) are accordingly modified to

2

i iaw z U ze e

z

2

ln2

i iaw z U ze e i z

z

(4.42)

Fig.4.4 should likewise be turned anticlockwise through angle α to describe the

present situation.

4.6 Conformal Mapping

Let w z be the complex potential of some 2-D irrotational flow in the z-plane with

w i .

Consider an analytic function

Z f z (4.43)

with inverse

z F Z (4.44)

which is also analytic. Then

W Z w F Z w z (4.45)

is analytic.

Let

Z X iY (4.46)

and write

, ,W Z X Y i X Y (4.47)

where X, Y , Φ, Ψ are real.

Since W Z is analytic, the Cauchy-Riemann conditions give

* ,u X YX Y

* ,v X YY X

(4.48)

where * * andu v are the flow velocity components in the Z-plane.

Thus,

* *

dWi u iv

dZ X X

Similarly,

dwu iv

dz

Writing

dwdW dz

dZdZdz

(4.49)

we have

* * '

u ivu iv

f z

(4.50)

where, from (4.43), we’ve used ' 'Z f .

Thus, if we want the mapping to preserve the uniform flow at infinity, we must have

' 1f z . as z

Consider a point 0z with 0 0Z f z .

Let ( )0

nf z be the 1st non-vanishing derivative at 0z .

The Taylor expansion of Z around 0z gives

1

0 0 0!

nnnz

Z z z Z f z O zn

so that

1

0 0 0!

nnnz

Z Z z z Z f z O zn

(4.50a)

Any complex function f can be written as

exp argf f i f

For the product of 2 complex functions,

exp arg argfg fg i f g

so that

arg arg argfg f g

Applying this to (4.50a) gives

0arg arg arg nZ n z f z

where we’ve dropped the higher order terms and made use of the fact that the arg of a

real function vanishes.

Consider now 2 such expansions around 0z , namely,

1 1 0arg arg arg nZ n z f z

2 2 0arg arg arg nZ n z f z

2 1 2 1arg arg arg argZ Z n z z (4.51)

Hence, for 1n , the angle between small line segments is preserved under the

mapping. Such mappings are called conformal.

The Joukowski transformation is given by

2c

Z f z zz

(4.52)

Now,2

2' 1

cf

z

2

3

2''

cf

z

so that at z c , we have 2n .

The inverse of (4.52), ie., 1f Z z , is obtained by solving for z from

2 2 0z Zz c which gives

2 214

2z Z Z c

so that 1f has 2 branches with branch points at 2Z c . As is the usual practice,

we take the line joining these banch points to be the branch cut.

Furthermore, to ensure

z Z as Z

we choose the + branch so that

2 214

2z Z Z c (4.53)

4.7 Irrotational Flow Past An Elliptical Cylinder

We now apply the Joukowski transformation2c

Z zz

(4.52)

to the circleiz ae

with 0 c a .

Thus2

i icZ X iY ae e

a

2 2

cos sinc c

a i aa a

ie.,

2

cosc

X aa

2

sinc

Y aa

Eliminating gives

2 2

2 22 21

X Y

c ca a

a a

(4.54)

which is an ellipse with semi-axis2c

aa

. [see fig.4.5.b].

The complex potential of a flow passing an elliptical cylinder is therefore obtained by

substituting the inverse

2 214

2z Z Z c (4.53)

of (4.52) into the complex potential

2

ln2

i iaw z U ze e i z

z

(4.42)

of a flow past a circular cylinder. Here is the angle that the uniform flow makes

with the x-axis.

Using

2 2

22 2

1 2 4

24

Z Z c

z cZ Z c

we have

2 2

2 2 22

1 44

2 2i iZ Z c

W Z U Z Z c e a ec

2 21ln 4

2 2i Z Z c

(4.55)

4.8 Irrotational Flow Past A Finite Flat Plate

If we choose c a so that (4.52) becomes2a

Z zz

(4.56)

The ellipse

2

cosc

X aa

2

sinc

Y aa

collapses to

2 cosX a 0Y which describes a flat plate of length 4a on the X-axis.

From

2

ln2

i iaw z U ze e i z

z

(4.42)

we have

2

2 2i idw a

U e e idz z z

From (4.56), we have2

21

dZ a

dz z

Therefore

* *

dwdW dzu iv dZdZ

dz

12 2

2 21

2i ia a

U e e iz z z

(4.57)

Using

2 214

2z Z Z c (4.53)

eq(4.57) can be rewritten as a function of Z.

At the ends of the plate where 2Z a or z a , the flow speed given in (4.57)

usually becomes infinite. [see, eg., fig.4.6a]

One exception is when the numerator of (4.57) also vanishes, ie.,

02

i iU e e ia

at z a

or

2 4 sini iaiU e e aU

Thus, to remove the singularity at the trailing edge ( z a ), the circulation must be

4 sinaU (4.58)

in which case, (4.57) is evaluated by the L’Hospital rule to be

12 2

* * 3 2 3

2 2lim

2

i

z a

Ua e au iv i

z z z

12 2 sin 2iUe U

ia a a

siniUe iU cosU

so that

* cosu U * 0v

which means the flow leaves the trailing edge horizontally.

Note that the singularity at the leading edge ( z a ) is still present. [see fig.4.6.b]

4.9 Flow Past A Symmetric Aerofoil

A mapping of a circular cylinder to an aerofoil is obtained by applying2a

Z zz

(4.56)

to a circle centered at ,0z , of radius r a , where 0 . Note that the

circle goes through the point ,0z a but encloses ,0z a . [see fig.4.7a]

Points on the circle are given by

iz a e (4.59a)

where is the phase angle [see fig.4.7a].

Putting it into (4.56) gives

2

ii

aZ a e

a e

(4.59)

The positions of the trailing / leading edges are giving by 1ie , so that

2a

Z aa

2

2

22

a

aa

a

Turning to the potentials, we must 1st adapt

2

ln2

i iaw z U ze e i z

z

(4.42)

to the present situation. This is easily done by the substitution

z z a a whereupon (4.42) becomes

2

ln2

i iaw z U z e e i z

z

Doing the same to

2

2 2i idw a

U e e idz z z

gives

2

2i idw a

U e e idz z z

Hence,

* *

dwdW dzu iv dZdZ

dz

12 2

21

2i ia a

U e e iz z z

(4.60)

The most important difference from the flat plate case is that the singularity at

z a is inconsequential since it now resides inside the aerofoil. The othe

singularity at the trailing edge, z a , is eliminated as before by setting the numerator

to zero, ie.,

0

2i iU e e i

a

which gives

4 sina U (4.61)

When a , the aerofoil given by (4.59) is thin & symmetric.

Keeping only 1st order terms in , (4.59) becomes

1 1i i iZ a e ae ea

2i i ia e a e e

2 cos 1 cos2 2sin sin 2a i

The length of the aerofoil is

0Re Re 2 2 2 2 4L Z Z a a a

(4.61a)

The thickness of the foil is

Im Im 2 2sin sin 2T Z Z

The extremum point satisfies

2 2cos 2cos2 0dT

d

ie.,2cos 2cos 1 cos

which gives

11

cos 1 1 8 14

2

or

0

2

3

0

32 3 3 3

2

T

Hence, the maximum thickness is 3 3 .

Using (4.61a) and dropping the term, (4.61) becomes

sinLU

4.10 Blasius’s Theorem

Consider a steady irrotational 2-D flow about a fixed body with boundary C.

Let w be the complex potential of the flow, ,x yF F be components of net force on

body, then2

1

2x y C

dwF iF i dz

dz

(4.62)

This is known as the Blasius’s theorem.

Proof:

Reminder: a complex number z x iy is equivalent to a 2-D vector ,x y .

Consider an infinitesimal segment z on the boundary C.

We can write

i iz z e e s

where s is the length, and the angle the segment made with the x-axis.

As 0s , the segment, which points in the direction cos ,sin , becomes

tangent to C.

The (outward) normal to C is therefore in the direction sin , cos .

This means the pressure force on the segment is sin , cosp s . [see fig.4.8]

In other words,

sinxF p s cosyF p s

so that

sin cos ix yF i F p i s pie s (4.62a)

For inviscid fluid, C is a streamline so that

, cos ,sinu v q u on C

where 2 2q u v u is the speed.

Hence,

cos sin idwu iv q i qe

dz on C. (4.62b)

Since the flow is steady & irrotational, the Bernoulli theorem applies so that

21

2p q k where k const

Eq(4.62a) thus becomes

21

2i

x yF i F q k ie s

which, with the help of (4.62b) to eliminate q, becomes

221

2i i

x y

dwF i F e k ie s

dz

21

2i idw

i e s ike sdz

Now

ie s z x i y

so that

2

1

2i

x y

dwF i F i e s ik x i y

dz

2

1

2

dwi z ik z

dz

Integrating over C then gives the theorem (4.62) since the integration of the 2nd term

depends only on the end points and hence vanishes over any closed contour.

We leave it as an exercise (Ex.4.5) to show that the moment about the origin is

21

Re2 C

dwz dz

dz

(4.63)

4.10.1 Uniform Flow Past A Circular Cylinder

This problem was already treated in §4.5 so it serves as a check on the Blasius’stheorem.

The complex potential is

2

ln2

aw z U z i z

z

(4.32)

so that

2

21

2

dw aU i

dz z z

Applying the theorem gives2

1

2x y C

dwF iF i dz

dz

2

Resdw

dz

Ui i U

Hence

0xF yF U (4.64)

as before.

4.10.2 Uniform Flow Past An Elliptical Cylinder

The complex potential W Z is assumed to be obtained by conformal mapping from

the circular cylinder result w z as described in §4.7.

We shall calculate the torque using

21

Re2 ellipse

dWZ dZ

dZ

(4.63)

In terms of the circular cylinder quantities:2c

Z zz

(4.52)

dwdW dz

dZdZdz

2

21

dZ cdZ dz dz

dz z

we have2 2 1

dW dw dZdZ dz

dZ dz dz

Eq(4.63) becomes

21

Re2 z a

dw dzZ dz

dz dZ

For 0 ,

2

i iaw z U ze e

z

(4.42)

so that

2

2i idw a

U e edz z

21Re

2U I (4.65a)

where

2 12 2 2

2 21i i

z a

c a cI z e e dz

z z z

22 2 2 2

2 2 3

i i

z a

z c z e a edz

z c z

so the poles are at 0,z c . Since the pole at 0z is of 3rd order, direct

evaluation of the residue there will involve 2nd derivatives of the other parts of the

integrand. This is rather tedious. We thus choose an alternative way & evaluate the

residue by Laurent’s expansion. Using

12 2 4

2 2 2 2 2 2 4

1 1 11 1

c c c

z c c z c z z

( z c )

the integrand of I becomes

2 4

2 2 4 2 2 2 4 22 3 2 4

1 12 1i i c c

z c z e a z a ec z z z

The coefficient of 1z is

2

2 2 2 2 2 22 2

12 2i i ia

c e a c e ec c

Therefore

2 22

2 24 4 cos2 4 sin 2ia a

I i e ic c

so that (4.65a) becomes2 22 sin 2U c (4.65)

which is clockwise & tends to turn the ellipse broadside to the stream.

4.11 The Kutta-Joukowski Lift Theorem

Consider a steady irrotational 2-D flow past a body with contour boundary C.

Let the flow at infinity be uniform with ,0Uu , and the circulation around the

body be . Then

0, U F (4.66)

This is the Kutta-Joukowski Lift Theorem.

Proof:

1st, choose the coordinates so that the body lies entirely inside a circle z R .

Assuming the absence of any singularities, the flowdw

dzis analytic for z R ,

inclusive of the infinity where limz

dwU

dz

. Hence,dw

dzhas a Taylor series in

R

z

so that

1 22

dw a aU

dz z z (4.67)

Consider now the Blasius’s theorem2

1

2x y C

dwF iF i dz

dz

(4.62)

Sincedw

dzis analytic for z R , the contour integral is path independent there.

Putting in (4.67) gives2

1 22'

1

2x y C

a aF iF i U dz

z z

1

12 2

2i i Ua

12 Ua (4.68)

where C’ satisfies z R .

Now, (4.67) implies

1'2

C

dwdz ia

dz

Thus

1 ' '2

C Ca w i (4.68a)

where CF denotes the difference of the value of F going once around C.

By means of analytic continuation, (4.68a) can be extended to C so that

12C C

a w i

Now, since C is a streamline, const there; hence,

12C C

a d u l (4.69)

Eq(4.68) thus becomes

x yF iF i U

which is just the theorem.

4.12 Lift: The Deflection of Airstream

We’ve explained the lift of an aerofoil in terms of the pressure difference due to the

negative circulation around the aerofoil (K-J theorem).

Another way to look at it is to say that the aerofoil imparts a downward momentum to

the oncoming stream & thereby, according to Newton’s law of action-reaction, attains

an upward lift.

For a single aerofoil, the deflection of the airstream is not apparent since it becomes

undetectible at infinity.

One thus needs to study the flow past an infinite stack of airfoils for such effects.

5. Vortex Motion

5.1 Kelvin’s Circulation Theorem

5.2 Persistence Of Irrotational Flow

5.3 Helmholtz Vortex Theorem

5.4 Vortex Rings

5.5 Axisymmetric Flow

5.6 Motion Of A Vortex Pair

5.7 Vortices In Flow Past A Circular Cylinder

5.8 Instability Of Vortex Patterns

5.9 Steady Viscous Vortex Maintained By A

Secondary Flow

5.10 Viscous Vortices: Prandtl-Batchelor

Theorem

5.1 Kelvin’s Circulation Theorem

Theorem

Circulation is conserved along particle path for an ideal fluid subject to

conservative forces.

Proof

Let the circulation be

C td u x (5.1)

To avoid confusion, let us denote derivatives of coordinates by δ and reserve the

symbol d for time derivatives. Hence [see Ex.5.2 for a more formal approach],

C t u x

C t C t

d d d

dt dt dt

u xx u (5.1a)

Now

21

2

d d

dt dt

x x

u u u u u

so that

2 21 1

02 2 CC t C t

d

dt

x

u u u since u is single valued.

Hence, (5.1a) becomes

C t

d d

dt dt

u

x

For an ideal fluid under conservative forces f , the Euler eq. becomes

d p

dt t

u uu u

where

p

Hence

0

CC t

d

dt

xsince ξ is single valued.

Notes on the Theorem

1. The theorem only applies to circuits that ‘move’ with the fluid; not to stationary

ones.

2. The theorem applies whenever we can write

d

dt

u(a)

Hence, the listed assumptions on the fluid can be relaxed as long as (a) is valid.

3. Similarly, the theorem applies wherever (a) holds on a circuit C; whether it holds

for other regions of the fluid is immaterial.

4. The theorem does not require the fluid domain to be simply connected since we

did not invoke the Stokes theorem.

Generation of Lift on Aerofoil

As mentioned in §1.6 and Chapter 4, the uplift of an aerofoil requires a negative

circulation Γ around the aerofoil.

If the aircraft is initially at rest, 0 . If the air is an ideal fluid, Γ will always be

zero according to Kelvin’s circulation theorem. Hence, no flight is possible.

If air is viscous, a wake of separated boundary layers and vortices will always be

present downstream of the aircraft. In particular, if the starting vortex is positive, its

shedding will give the aerofoil a negative Γ. [see Fig.5.2]

Encarsia Formosa

This tiny wasp employs the Weis-Fogh mechanism of lift generation that works even

for inviscid fluid. [see Fig.5.3 & caption for explanation]

5.2 Persistence Of Irrotational Flow

Cauchy-Lagrange Theorem

Consider an ideal fluid under conservative forces.

Any portion of fluid that is irrotational at some time will remain so afterwards.

Proof

Suppose ω u is not identically 0 for that portion V of fluid at some later time.

Then, by Stoke’s theorem, there exists a circuit C bounding the surface S of V such

that

0C S S

d dS dS u x u nω n

However, this violates Kelvin’s theorem since 0 at some time before. QED.

As discussed in §1.5, the vorticity eq.

D

Dt

ω ω u (1.25)

simplifies to

0D

Dt

ω(1.27)

for 2-D flows, so that the above result is rather obvious.

The usefulness of the Cauchy-Lagrange theorem is that it applies also to 3-D flows.

For an irrotational flow, we can write

u (5.3)

where is single-valued if the flow region is simply connected.

If the flow is incompressible,

0 u ie., 2 0 (5.4)

[See Ex.5.23-29 for applications in this respect]

5.3 Helmholtz Vortex Theorem

A vortex line is, at one instance, a curve

, ,s x s y s z s x

whose tangent at each point is parallel to the vorticity vector

, ,x y z ω u (5.5)

ie.,

x y z

dydx dzds ds ds

or

x y z

dx dy dz

(5.5a)

A vortex surface is a surface to which ω is tangent at each point. [fig.5.4b]

The set of all vortex lines that pass through a closed curve in space forms the

boundary of a vortex tube. [See fig.5.4a]

Theorems

Consider an ideal fluid under conservative forces so that Kelvin’s circulation theorem

applies. The Helmholtz vortex theorems state that

1. Vortex lines, & hence vortex tubes, move with the fluid.

2. The strength of a vortex tube, as defined by,

SdS ω n (5.6)

is time independent and the same for all cross section of the tube.

Proof of (1)

Consider a circuit C that lies, at time 0t , completely on a vortex surface S.

Let C be bounded by a simply connected subsurface *S of S.

The circulation around C is, with the help of Stoke’s theorem,

* *C S S

d dS dS u x u nω n

which is simply the strength of a vortex tube that goes through C.

Since *S is a vortex surface, 0 ω n there. Hence

0 on C

Now, let C t be the circuit composed of, at time t, the same fluid particles as C at

0t . [In short, C moves with the fluid flow to C t at time t.]

Let t the circulation of C t . According to Kelvin’s theorem, 0t for all

t.

Since C is arbitrary, this means

0 ω n on S t

where S t is the surface composed of, at time t, the same fluid particles as S at

0t . [In short, S moves with the fluid flow to S t at time t.]

Thus, S t is also a vortex surface.

Another way to say it is that a vortex surface moves with the flow.

A vortex line can be considered as the intersection of 2 vortex surfaces.

Since both surfaces moves with the flow, so is their intersection, the vortex line.

Proof of (2)

Consider the volume V inside a vortex tube bounded by 2 of its cross sections [see

fig.5.4a].

Denote the section of the vortex tube surface involved by TS ; the cross sections by

1S and 2S ; their bounding circuits, 1C and 2C , respectively.

The Gauss theorem gives

1 2TS S S V

dS dV

ω n ω

Now, since TS is a vortex surface, 0 ω n everywhere on it so

0TS

dS ω n

Next,

0 ω u

so

0V

dV ω

Hence,

1 2

0S S

dS

ω n

Noting that n here are outward normals of the surface bounding V, this becomes

1 2 where i is the strength of the tube at iS , as defined by (5.6). QED.

Thin Vortex Tube

Let the cross section S of a thin vortex tube be small enough so that is

practically constant over it. Thus, S .

Since the surface of the tube is a vortex surface, it moves with the flow.

Thus, the volume of fluid inside the tube and bounded by 2 cross sections will remain

thus bounded during the flow. Since the flow is incompressible, this volume is

conserved. Therefore, stretching the tube will decrease S . Since is a constant,

must increase to compensate.

To summarize, stretching a vortex tube will intensify local vorticity.

[see fig.5.5 for application to a tornado; fig.5.6 for that to spin down of a cup of tea].

5.3.1 Vorticity Equation

The Helmholtz theorems were originally proved starting with the vorticity equation

D

Dt

ω ω u (5.7)

Some of the conclusions drawn from the Helmholtz theorems are also obtainable

directly through (5.7)

One such example is the local intensification of vorticity due to the stretching of a

vortex tube.

Let the vortex lines be mostly in the z-direction so that ω k and

D

Dt z

ω u (5.8)

The z-component of which is

D w

Dt z

Now, stretching the fluid in the z direction makes 0w

z

so that 0

D

Dt

as

promised.

Another simple example is that of 2-D flow.

Vortex tubes are all straight & parallel to the z-axis and 0w .

There is no stretching of tubes and

0D

Dt

(5.9)

Still another example is the axisymmetric flow, which, in terms of cylindrical

coordinates , ,R z , is of the form

, , , ,R R z zu R z t u R z t u e e (5.10)

Since u is independent of , the streamlines all lie in planes const .

The vorticity is

1

0

R z

R z

R z

R

u u

R R z z R

u u

e e e

ω e (5.11)

The vortex lines are rings centered around the z axis.

Vortex tubes are surfaces of doughnuts similarly centered.

According to the 1st Helmholtz theorem, they move with the fluid, which means they

can only expand or contract about the z axis.

Incompressibility of the fluid means the volume enclosed by each tube is conserved.

Let the radius of a thin tube be R, its cross section S , so that its volume is 2 R S .

As R varies in time with the flow, S also changes to maintain a constant volume.

However, according to the 2nd Helmholtz theorem, S const so that

const R

In more precise language, we write

0D

Dt R

(5.12)

which can be deduced more formally from (5.7) using (5.10).

When, in axisymmetric flow, an isolated vortex tube surrounded by irrotational flow

is called a vortex ring. [see fig.5.7].

5.3.2 Proof of eq(5.12)

Using

, ,R R z zu u R z t u e e u

ω e where , ,R zu uR z t

z R

we have

R R z z Ru u uR R

ω u e e e

where we’ve used

R

e e

Thus, the vorticity eq (5.7) becomes

R

Du

Dt R

ω

e

Writing

D D DR

Dt Dt Dt R

ωe e

D DR R

Dt R R Dt

e e

Now

DR R

Dt e u e

R zu u RR z

e

Ru e

so that

R R

D DR u u

Dt Dt R R R

ωe e e

ie.,

0D

Dt R

5.4 Vortex Rings

Collision

Leapfrog

5.5 Axisymmetric Flow

5.5.1 The Stokes Stream Function

5.5.2 Irrotational Flow Past A Sphere

5.5.3 Hill’s Spherical Vortex

5.5.1 The Stokes Stream Function

For an incompressible 2-D flow,

z u e (4.8)

ensures that the incompressibility condition 0 u is automatically satisfied while

the stream function , ,x y t is constant along streamlines, ie.,

0 u (4.7)

The question is whether an analogous form ' u e , with ' ' , ,R z t ,

exists for axisymmetric incompressible flows.

Now

' 'R zu u uR R z

u

'R zu uR z

Using

1 ' ' '

0 ' 0

R z

R z

R

R R z z R R

R

e e e

u e e

we have

' ' ' ' ' ' ''

z R R R z R z

u

so that ' is not generally constant along streamlines.

Now, the offending term originates from the curl of ' in the expression of u. It is

easily removed by setting 'R

so that

R

u e (5.13)

1

0 0

R z

R z

R

R R z z R

e e e

e e

whereupon

0z R R z

u

ie., the Stokes stream function is constant along streamlines.

In spherical coordinates , ,r ,

2

sin

1

sin

sin

r

r

r r

r r

A rA r A

e e e

A

The analog (5.13) of is obviously

sinr

u e (5.14)

so that

2

sin

1 1

sin sin

0 0

r

r

r r

r r r r r

e e e

u e e (5.15)

while , ,r t is constant along streamlines:

sinr

uuu

r r r

u

r

uu

r r

1 10

sinr r r r r

5.5.2 Irrotational Flow Past A Sphere

In a steady flow, the vorticity eq. for axisymmetric flow,

0D

Dt R

(5.12)

reduces to

0sinr

u (5.16)

where sinR r .

Thussinr

is constant along a streamline.

Consider a uniform inviscid flow past a rigid sphere with surface r a .

If there are no closed streamlines, all streamlines will start and end at the infinities,

where 0 . Hence, according to (5.16), 0 everywhere and the flow is

irrotaional.

Now, for an axisymmetric flow in spherical coordinates:

, , , ,r ru r t u r t u e e

2

sin

1

sin

0

r

r

r r

r r

u ru

e e e

ω u

1 ruru

r r

e e

where

1 1r ru u u uru

r r r r r

(5.17)

In terms of the Stokes stream function Ψ defined by

1

sin sin rr r r r

u e e e (5.14-5)

we have

2

2 2

1 1 1 1

sin sinr r r

(5.18)

Thus, for our irrotational flow past a sphere, we need to solve2

2 2

sin 10

sinr r

r a (5.19)

with boundary conditions

0ru on r a [impenetrable wall] (5.20a)

and

cosru U , sinu U as r

which, by (5.15), means

2

1cos

sinU

r

,1

sinsin

Ur r

as r

or

2 sin cosr U

, 2sinrUr

as r (5.20b)

which, upon integration, becomes

2 21

1sin

2r U C r

2 22

1sin

2r U C as r

In order for these 2 expressions to be compatible, the arbitrary functions 1 2,C C must

be constants that can be conveniently set to zero. Hence

2 21sin

2r U as r (5.20)

Eq (5.19), as well as the boundary conditions (5.20,a,b), suggests that we can look for

a separable solution, ie.,

,r R r

whereupon, (5.19) becomes 2 ODEs2

2 20

d RR

dr r

(5.21a)

1sin

sin

d d

d d

(5.21b)

where α is a constant.

In order to satisfy (5.20), we have

21

2R r r U

2sin (5.21c)

Substituting into (5.21b), we have

2. . . 2sin cos 2sind

L H Sd

2. . . sinR H S so that (5.21b) is satisfied only if 2 ; whereupon, (5.21a) becomes

2

2 2

20

d RR

dr r (5.21d)

Since this is homogeneous, we look for a power solutionnR r

which turns (5.21d) into

1 2 0n n

with solutions

2n , or 1so that

2 BR Ar

r with A, B constants.

From (5.21c), we have

1

2A U

so that

2 21, sin

2

Br Ur

r

Finally, to satisfy (5.20a), we have

22 2 3

1 1 1 22sin cos cos

sin sin 2r

B Bu Ur U

r r r r

so that at r a

3

2cos 0

BU

a

which gives

31

2B Ua

so that

3

2 21, sin

2

ar U r

r

for r a (5.21)

3

31 cosr

au U

r

3 3

2 3

1 sin2 1 sin

sin 2 2

U a au r U

r r r r r

The streamlines const are shown in Fig.5.12a.

The slip velocity on the sphere surface is

0ru r a

3sin

2u r a U (5.22)

Thus, the fore & aft points 0, are stagnation points.

Maxima are at2

.

According to the Bernoulli theorem, this implies a severe adverse pressure gradient

over the downstream side of the sphere. ( 02

). A large wake with separated

boundary layer is expected.

5.5.3 Hill’s Spherical Vortex

In §5.5.2, we solved the steady axisymmetric flow outside a solid sphere r a .

Now, let the surface r a to be an imaginary one in the fluid. We shall show that

there is a flow pattern, called the Hill’s spherical vortex, inside the sphere that can

match up at r a with the one given in §5.5.2.

In other words, one can find vortices embedded in a steady, uniform flow. [see

Fig.5.12]

The problem is to solve

0sinr

u for r a (5.16)

with the boundary conditions

0ru r a

3sin

2u r a U (5.22)

Now, (5.16) means thatsinr

is a constant along each streamline, a property

shared by the Stokes stream function Ψ. Hence, (5.16) implies

sin

cr

where c is an arbitrary function.

Using the spherical coordinate relation (5.18) between ω and Ψ, we have

2

2 2

1 1 1 1sin

sin sinc r

r r r

or

2

2 22 2

sin 1sin

sinc r

r r

(5.23)

Now, in terms of , (5.22) become

2

10

sin r aa

1 3sin

sin 2r a

Ua r

(5.24)

Now, the easiest way to match (5.24) is to try the same type of ansatz as in the r aregion, namely,

2, sinr g r for r a

Plugging into (5.23) gives

2

2 2 2 22 2

sin 2 sin sind g g

c rdr r

or

2

2 2 4

12

d g gc

r dr r (5.24a)

Since the R.H.S. is independent of θ, we must have c const .

The homogeneous eq to (5.24a) the same as the radial eq. for r a . Thus, the

homogeneous solution is

2 Bg Ar

r

To find the particular solution, we again try a power solutionng r

so that (5.24a) becomes

41 2 nn n r c

which gives

4n and10

c

Hence, the general solution for (5.24a) is

2 4

10

B cg Ar r

r

so that

2 4 2, sin10

B cr Ar r

r

Now, to keep Ψ regular at 0r , we must set 0B so that

2 4 2, sin10

cr Ar r

and

2 4 22

2sin cos 2 cos

sin 10 10r

c cu Ar r A r

r

3 2 21 4 12 sin 2 sin

sin 10 5

cu Ar r A cr

r

Matching the conditions (5.24) at r a requires

22 cos 010

cA a

and

21 32 sin sin

5 2A ca U

ie.,

2 010

cA a

21 3

5 4A ca U

with solution

3

4A U

2

15

2

Uc

a

so that

2

2 22

3, 1 sin

4

rr Ur

a

for r a (5.25)

The corresponding streamlines, which are all closed, can be found in Fig.5.12b.

The circulation around each streamline differs for different streamlines.

The maximum is the streamline that goes around a full hemispherical cross section.

By Stokes theorem,

max 0 0

a

hemi

dS d drr

Using

2

15sin sin

2

Ucr r

a

we have

3

2max 2 20 0

15 15sin 2 5

2 2 3

a a

Thus, a Hill spherical vortex will travel through stationary fluid with a speed

max

5U

a

5.6 Motion Of A Vortex Pair

A line vortex is defined as

2 r

u e

Consider a vortex pair of strength and a distance 2d apart as shown in

Fig.5.13a. Each will induce a downward flow / 4 d at the instantaneous position

of the other. Thus, the pair, as well as the accompanying streamline pattern, moves

downward at that speed.

In a reference frame moving with the vortices, an additional uniform upward flow

/ 4 d is experienced. The full complex potential is therefore

ln ln4 2 2

i z i iw z d z d

d

(5.26)

When calculating the velocity of one of the vortices, the contribution of that vortex to

w must be subtracted. Thus, the velocity ,U V of the vortex at z d is given by

ln4 2 z d

d i z iU iV z d

dz d

0

4 2z d

i i

d z d

(5.27)

ie., it’s stationary as stated before.

Separating the real and imaginary parts in (5.26), we have

ln4 2 4 2

y x z dw i

d d z d

where

iz d z de

z d z d

Thus, the stream function is

2 ln4

x z d

d z d

Now,

z d x iy d

z d x iy d

22 2

2 2

x d iy x d iy x d yz d

z d x d iy x d iy x d y

so that

2 2

2 2ln

4

x d yx

d x d y

The corresponding streamlines are sketched in Fig.5.13b.

5.7 Vortices In Flow Past A Circular Cylinder

Consider a circular cylinder of radius a initially at rest in a fluid of kinematic

viscosity ν. Let it be suddenly set in uniform motion with speed U perpendicular to its

axis so that the Reynolds number

2aUR

(5.28)

5.7.1 The Initial Phase: Almost Irrotational Flow

5.7.2 Flow At Later Stage: The von Karman Vortex

Street

5.7.3 A Simple Model

5.7.1 The Initial Phase: Almost Irrotational Flow

If the fluid were inviscid, our 2-D flow would be governed by

0D

Dt

(5.9)

Since the fluid is initially at rest, 0 .

According to (5.9), it will remain so for all subsequent time so the flow will always be

irrotational.

Consider now our viscous fluid.

During a very short initial phase before the cylinder has moved a distance comparable

to its radius, the flow is predominantly irrotational (see fig.4.4a).

There is intense vorticity in the rapidly thickening boundary layer.

However, there is not yet time for separation to occur so the mainstream is still free of

vorticity.

At this stage, results of the (inviscid) irrotational flow give the velocity at the edge of

the boundary layer.

Positions at which reversed flow 1st occur are those where the velocity decreases most

rapidly with distance along the boundary in the flow direction.

In case of a circular cylinder, this happens at the rear stagnation point. (see Fig.5.14a)

5.7.2 Flow At Later Stage: The von Karman Vortex

Street

Once reversed flow is initiated, the flow begins to differ substantially from the

irrotational results.

1st, the 2 attached eddies behind the cylinder grow in size (Fig.5.14b).

Then the flow becomes asymmetric about the centerline (Fig.5.14c).

Finally, it settles into an unsteady but highly structured flow in which vortices are

shed alternatively from the 2 sides of the cylinder, giving rise to the von Karman

vortex street (Fig.5.14d,e).

Note that the wake is entirely turbulent for 2000R while for 30R , the 2 eddies

remain symmetrically attached.

5.7.3 Von Karman Vortex Street: A Simple Model

We now model a fully formed vortex street (Fig.5.14d,e) by 2 sets of line vortices as

shown in fig.5.15.

Members of the lower set of vortices, all of strength Г, are at z na , with

0, 1, 2,n .

Those of the upper set, of strength , are at1

2z n a ib

.

As in §5.6, we assume each line vortex to move with the local velocity due to the

combined flow of everything other than itself.

Consider any vortex.

Only the x-component U of velocity due to vortices in the other row is finite.

All else vanishes since contributions from other vortices cancel in pairs.

By symmetry, U is the same (and points to the left for 0 ) for all vortices.

Since the complex potential w z due to a line vortex at 0z is

0ln2

iw z z z

(4.21)

the total complex potential due to the lower line of vortices is

ln2 n

iw z z na

1

1

ln ln ln2 n n

iz z na z na

1

1

ln ln ln2 n n

iz z na z na

2 2 2

1

ln ln2 n

iz z n a

2

2 22 2

1 1

ln ln 1 ln2 n n

i zz n a

n a

2

2 21

ln 12 n

i zz C

n a

where

2 2

1

ln2 n

iC n a

is a constant and can be dropped.

Using the identity (see Carrier et al)

2

21

sin 1n

zz z

n

we have

ln sin2

i zw const

a

(5.29)

Hence

cot2

dw i z

dz a a

For an upper row vortex at1

2z n a ib

, we have

1cot tan tanh

2 2 2 2

dw i b i b bn i i

dz a a a a a a

Thus, the whole street moves with velocity

tanh ,02

b

a a

u

5.8 Instability Of Vortex Patterns

Consider 2 line vortices of equal strength Г and distance 2a apart.

Each vortex induces, at the center of the other, a flow of speed4 a

perdendicular

to the line joining them. Hence, both rotates about the mid-point of that line with

angular velocity 2/ 4 a .

Next, consider n line vortices of strength Г placed symmetrically on a circle of radius

a. (see Fig.5.16) The angular separation between adjacent vortices is2

n

. The

distance between the mth neighbors is therefore 2 sin2m

md a

, where2

m

m

n

with2

nm , and the floor a of a is the largest integer that is equal or less than a.

Note that all neighbors occur in pairs except for the2

nth one, which is unique, when

n is even.

Consider a vortex and the pair of its mth neighbors. The line joining a vortex to its

mth neighbor makes an angle2

m with the radius that passes through the vortex.

The flow at the vortex, induced by that neighbor, then makes an angle2m m

n

with the radius. By symmetry, the radial components of the flows induced by a pair

of the neighbors cancels each other. The tangential component is equal to

2 sin2 2 2

m

md a

For n even, there is only one2

nth neighbor situated at the other end of the diameter

that goes through the vortex. The induced flow is again tangential with magnitude

4 a

.

For n odd, there are 1 / 2n pairs of neighbors so that the total flow is

11

2 2 4

nn

a a

.

For n even, there are / 2n pairs and a unique2

nth neighbor. The total flow is

12 2 4 4

nn

a a a

In both case, the vortex rotates with angular velocity

21

4n

a

(5.32)

Stability of a dynamical system can be investigated by the introduction of a small

perturbation. If the state of the system remains close to the unperturbed one, it is

stable. Otherwise, it is unstable.

There are cases that do not quite fit in this scheme of classification. For example,

some systems are stable under infinitesimal perturbation but unstable under finite

perturbation. Others can be stable under any instantaneous perturbation but unstable

under a prolonged one. These cases are sometimes called metastable.

An introduction to the subject can be found in chapter 9.

Here, we list, without proof, some results of interest concerning the stability of vortex

patterns.

1. The double row vortex pattern in fig.5.15, §5.7.3, is unstable except for the case

cosh 2b

a

ie., 0.281

b

a (5.31)

2. The equally spaced vortices on a circle pattern is stable for 7n , unstable for

7n and metastable for 7n . This is observed in liquid helium.

Finite Patches of Concentrated Vorticity

An elliptical path of uniform vorticity ω rotates with angular velocity

2

a b

(5.33)

where a and b are the semi-axes of the ellipse.

This motion is stable for1

33

b

a .

A circular array of n patches of vorticity is unstable even for 7n , if the patches are

big enough. The critical patch size decreases with increasing n.

For the finite patch version of the vortex street, there is again only 1 spacing ratio for

which the pattern is stable. This ration is close to the von Karman value and only

weakly dependent on the patch size.

A patch of 2-D vorticity distribution

v u

x y

is still governed by the vorticity eq.

0D

u vDt t x y

(5.9)

For incompressible flow, we also have

0u v

x y

(5.35)

Now, (5.9) implies ω is conserved for a fluid element while (5.35) implies the

cross-section of that element in the x-y plane is conserved. Thus

dS const (5.36)

where the integration is over the whole flow plane.

There are other relationships of this kind:

x dS const y dS const (5.37)

Vortex Merging

Let 2 circular patches of uniform and equal vorticity, each of radius R, have centers a

distance d apart at 0t .

For 3.5d

R , the patches end up deformed and rotating about a common center.

For 3.5d

R , the patches quickly merge. To satisfy the conservation laws, they do

this by wrapping around each other with a layer of irrotational flow between them.

When the vorticity are of opposite sign, a vortex pair may be formed.

5.9 Steady Viscous Vortex Maintained By A

Secondary Flow

In the presence of viscosity, the vorticity eq. becomes

2

t

ω u

ω ω u ω(5.38)

The processes controlled by the various terms are roughly:

: convection of vortex lines.

ω u : intensification of vorticity (when vortex lines are stretched).

2 ω : diffusion of vorticity.

Burgers Vortex

One known situation that involves all 3 processes and still admits an exact solution to

the Navier Stokes eqs is known as the Burgers vortex.

Basically, it is a line vortex whose viscous decay (§2.4.4) is countered by a secondary

flow which

1. sweeps the vorticity back towards the axis.

2. intensifies the vorticity by stretching fluid element along the axis direction.

The result is a steady vortex of the form

1

2ru r 2 / 412

ru er

zu z (5.39)

where 0 and Г are constants.

The velocity profile is shown in fig.5.18.

The vorticity

2 / 4

1

11

2 2

r z

r

r

r r z

r e z

e e e

ω

2 / 4

4r

ze

e (5.40)

is concentrated within a core of radius of order

.

In (5.39), the secondary flow, as parametrized by α, and the rotary flow, as

parametrized by Г are not coupled since both parameters can vary independently.

For a real flow, the secondary and rotary flows are usually coupled due to the

presence of rigid boundaries.

5.10 Viscous Vortices: Prandtl-Batchelor

Theorem

For a 2-D steady inviscid flow, the vorticity eq. reduces to

0 u where ω k

ie., ω is constant along a streamline so that we can write

(5.41)

since the stream function Ψ is, by definition, constant along a streamline.

If, as in fig.1.8, all flow lines can be traced to the infinities where the flow is uniform,

we can deduce immediately that 0 everywhere.

If closed streamlines are present, no simple conclusion can in general be drawn for

inviscid flows.

For steady viscous flow, the Navier-Stokes eq become

2p

u u u

Using

21

2 u u u u u

2 u u u

we have

21

2

p

u u u u u

or

21

2

p

u u u u u

Integrating over a closed contour C, we have

C C

d d u x u u x If C is a streamline, dx is parallel to u so that the right hand side vanishes. Thus

0C

d u x if 0

ie.,

0C

d ω x (5.42)

For our 2-D flow,

, ,0

0 0

x y z y x

i j k

ω

, ,0d

d y x

, ,0d d

u vd d

u

so that (5.42) becomes

0C

dd

d

u x (5.43)

or

0C

d

d

whered

d

can be taken outside the integral because ψ is constant on the streamline

C. In case of a finite circulation C , we have

0d

d

which is the Prandtl-Batchelor theorem: in a steady 2-D viscous flow, the vorticity

is constant throughout any region of closed streamline in the limit 0 .

6. The Navier-Stokes Equations

6.1 Introduction

6.2 The Stress Tensor

6.a Navier-Stokes Eqs.

6.B Convection Theorem

6.2 The Stress Tensor

Let x be the position vector of some point fixed in the fluid, and S a small

geometrical surface element, with unit normal n, that passes through x.

Let the force exerted on this surface by the fluid on the side pointed by n be

St (6.1)

where t is called the stress vector.

For an inviscid fluid, we have

,p t t x n

so that the stress aligns with the unit normal.

For viscous fluid, we expect t to have non-vanishing components tangent to the

surface. This is described by the stress tensor T whose Cartesian component ijT is

defined as the i-component of stress on a surface with unit normal jn e .

Thus, for a surface with unit normal j jnn e , the stress vector is

i ij jt T n or T t n (6.3)

For a more fastidious proof, see Acheson.

6.3 Cauchy’s Equation Of Motion

Convection Theorem

Ref: R.E.Meyer, “Introduction to Mathematical Fluid Dynamics”, Wiley (71)

Theorem 3.1

If 1, tf t C x ,

D

DtfdV

Df

Dtf dV

t t z z FHG

IKJv (3.5)

Corollary 3Let 1 1( )t t ,

D

DtfdV

tfdV f dS

t z z z

1 1

v n (3.6)

Proof

0t tH

So that

fdV g t J t dVt z z ( , ) ( , )a a 0

0

where

, , ,g t f t ta x a

( ,0)a x a

det i

j

xJ

a

Thus

D

DtfdV

tgJdV

tgJ dV

t z z z

0 0

0 0

b g

FHG

IKJz J

Df

Dtf

J

tdV0

0

FHG

IKJz Df

Dtf

J

J tdV

t

FHG

IKJz Df

Dtf dV

t

v

if

J

tJ v

Now:

i nik kn nk

n i

x xJ A A

a a

where knA are the cofactors (Caution: our ijA is Meyer’s jiA ),

( )k nkn knA

where kn is the determinant of the minor matrix obtained by striking out the kth

row & nth column of the matrix i

j

x

a

.

Writing the ith row of J as a vector ir with components i jr , we have

1 2 3 1 2 3 1 2 3 1 2 3det , , det , , det , , det , , r r r r r r r r r r r r

1 1 2 2 3 3k k kk k kA A A r r r

where ∂ is the differential operator, ie.,

jjk

k

xJ A

a

[see Determinant.doc for a better derivation]

Hence:

k k k i kkn kn kn ik

n n i n i

J x v v x vA A A J J

t t a a x a x

v QED

For the corollary:

Df

Dtf

f

tf

v vb g

so that

z zf dV f dSv v nb g b g1 1

6.4 A Newtonian Viscous Fluid: The

Navier-Stokes Equations

6.a Navier-Stokes Eqs.

Consider the Euler eq

D Dp

Dt Dt

u uu u

Using

D

Dt t

u u

u u u u u u

t

u

uu

we can re-write it as

0pt

u

uu

or

0

t

u

(A)

where the momentum flux density tensor П is

ij i j iju u p

Eq(A) is just the “eq of continuity for momentum” which expresses the conservation

of momentum.

Since we expect the principle of the conservation of momentum to be valid even in

viscous fluids, eq(A) should apply there with a suitable modification of П. We thus set

ij i j iju u

where the stress tensor ij is related to the viscous stress tensor 'ij by

'ij ij ijp

[Note: Acheson used ijT instead of ij (see chap 6) ]

What this means is that the generalization of the Euler’s eq to viscous fluids can be

done by replacing p with 'p so that, for example,

Dp

Dt

u

becomes

'D

pDt

u

or

'pt

u

u u (B)

Next, we need to determine the form of ' .

Now, viscous effects are due to “friction” between adjacent fluid elements moving

with different velocities. Hence ' should depend on i

j

u

x

but not on u itself.

Furthermore, there’s no friction if the fluid rotates as a whole with uniform angular

velocity Ω so that uΩ r

or i ijk j ku x .

Now:

i mijk j km ijm j mji j

m i

u u

x x

and

0iiji j

i

u

x

Hence, the choice

' 'ji kij ij ji

j i k

uu ua b

x x x

where a, b are constants, will satisfy the above requirements.

A more common form is

2'

3ji k k

ij ij ijj i k k

uu u u

x x x x

(C)

where the positive constants

a 2 2

3 3b a b

are called coefficients of viscosity. (Landau used η instead of μ)

This also defines the Newtonian fluid. [cf eq(2.1)].

Using (C), (B) becomes

' jii ij

j i j

u u pu

t x x x

2

3ji k k

ij iji j j i k k

up u u u

x x x x x x

2

3jk i

i k j j i

uu up

x x x x x

1

3k i

i k j j

u up

x x x x

or in vector form:

'pt

u

u u

21

3p

u u

21

3p

u u (D)

For incompressible fluids, we have the Navier-Stokes eqs.:

2pt

u

u u u (2.3)

0 u

Note also that for incompressible fluids,

' jiij

j i

uu

x x

[cf (2.1)]

6.B Convection Theorem

Ref: R.E.Meyer, “Introduction to Mathematical Fluid Dynamics”, Wiley (71)

Theorem 3.1

If 1, tf t C x ,

D

DtfdV

Df

Dtf dV

t t z z FHG

IKJv (3.5)

Corollary 3Let 1 1( )t t ,

D

DtfdV

tfdV f dS

t z z z

1 1

v n (3.6)

Proof

0t tH

So that

fdV g t J t dVt z z ( , ) ( , )a a 0

0

where

, , ,g t f t ta x a

( ,0)a x a

det i

j

xJ

a

Thus

D

DtfdV

tgJdV

tgJ dV

t z z z

0 0

0 0

b g

FHG

IKJz J

Df

Dtf

J

tdV0

0

FHG

IKJz Df

Dtf

J

J tdV

t

FHG

IKJz Df

Dtf dV

t

v

if

J

tJ v

Now:

i nik kn nk

n i

x xJ A A

a a

where knA are the cofactors (Caution: our ijA is Meyer’s jiA ),

( )k nkn knA

where kn is the determinant of the minor matrix obtained by striking out the kth

row & nth column of the matrix i

j

x

a

.

Writing the ith row of J as a vector ir with components i jr , we have

1 2 3 1 2 3 1 2 3 1 2 3det , , det , , det , , det , , r r r r r r r r r r r r

1 1 2 2 3 3k k kk k kA A A r r r

where ∂ is the differential operator, ie.,

jjk

k

xJ A

a

[see Determinant.doc for a better derivation]

Hence:

k k k i kkn kn kn ik

n n i n i

J x v v x vA A A J J

t t a a x a x

v QED

For the corollary:

Df

Dtf

f

tf

v vb g

so that

z zf dV f dSv v nb g b g1 1

7. Very Viscous Flow

7.1 Introduction

7.2 Low R Flow Past A Sphere

7.3 Corner Eddies

7.4 Uniqueness & Reversibility of Slow Flows

7.5 Swimming At Low R

7.6 Flow in a Thin Film

7.7 Flow in a Hele-Shaw Cell

7.9 Thin Film Flow Down a Slope

7.10 Lubrication Theory

7.1 Introduction

For a steady viscous flow, the Navier-Stokes eqs become

21p

u u u g (7.1)

The term ‘very viscous flow’ is used to denote situations in which the convective

(inertia) term u u is negligible, whereupon (7.1) is linearized.

There are 2 rather different ways to achieve this, as described in the following.

Slow Flow Equations

This happens when

1UL

R

(7.2)

Since

2U

OL

u u 2

2

UO

L

u

we have

21

ULO O R

u u

u

Furthermore, if gravity is negligible, (7.1) simplifies to20 p u (7.3)

which, together with the incompressible condition

0 u (7.3a)

are known as the slow flow eqs.

Thin Film Equations

This applies to motion of thin films of fluid.

If the thickness of the film is sufficiently small, the velocity gradient across it may be

enhanced to a level where the convective term is overcome by the viscous term, even

though R is quite large in the conventional sense.

The governing eqs, called the thin-film eqs, willl be derived in §7.6.

7.2 Low R Flow Past A Sphere

We now seek a solution to the slow flow eqs (7.3) for the uniform flow past a sphere

7.2.1 Equation for the Stokes Stream Function

7.2.2 Boundary Conditions

7.2.3 Solutions

7.2.4 Drag

7.2.5 Further Considerations

7.2.6 Navier-Stokes Eqs For Ψ

7.2.7 Matched Asymptotic Expansions

7.2.1 Equation for the Stokes Stream Function

Let the uniform flow be in the z direction.

The flow is axisymmetric about the z-axis, so that in terms of spherical coordinates

, ,r , it takes the form

, , , ,0ru r u r u

The incompressibility condition 0 u is

sinr

u e (5.14)

so that

2

sin

1

sin

0 0

r r r

r r

e e e

u

1

sin rr r r

e e (5.15)(7.6)

and

2

sin

1

sin

1 10

sin sin

r r r

r r

r r r

e e e

u

2

2 2

1 1 1 1

sin sinr r r

e

21

sinE

r e

where2

22 2

sin 1

sinE

r r

2 2

2 2

Q

r r

and

2 1sin

sinQ

Hence

2

2

sin

1

sin

0 0

r r r

r r

E

e e e

u

2 22

1

sin r E r Er r

e e

Using

2 2 u u u u

eq(7.3) can be written as

p u

2 22 sin r E r E

r r

e e

ie.,

22 sin

pE

r r

21

sin

pE

r r r

Eliminating p gives

2

2 22 2

1 1 10

sin sinE E

r r

or

22

2 2

sin 10

sinE

r r

2 2 0E E

22

2 2

sin 10

sinr r

(7.7)

7.2.2 Boundary Conditions

The no-slip boundary condition means

2

1, 0

sinrr a

u aa

1, 0

sin r a

u aa r

ie.,

0r a r ar

(BC1)

As the flow becomes uniform at infinity, ie.,

zUu e for r

we have

, cosru U , sinu U

so that

2

1lim cos

sinrU

r

1

lim sinsinr

Ur r

(7.7a)

which is possible only if Ψ is of the form

2r f

whereupon (7.7a) become

1' cos

sinf U

and

2sin

sinf U

ie.,

21sin

2f U

2 21sin

2r U for r (BC1a)

Finally, 0u on the surface of the sphere so that every curve on it is a streamline

with zero flow velocity. Thus, Ψ is a constant on r a and, for convenience, we can

set

, 0a (BC1b)

7.2.3 Solutions

In view of the form of (7.7) & (BC1), a simple ansatz could be

2sinf r

To see if it works, we calculate:

2 1sin 2 sin cos 2

sinQ f

2 2 22 2 2

2 2 2 2 2

2 2''sin sin

Q dE f f

r r r dr r

22 2

2 22 2

QE E

r r

2 2 22

2 2 2 2

2sin

Q df

r r dr r

222

2 2

2sin

df

dr r

so that (7.7) is satisfied if

22

2 2

20

df

dr r

(7.7b)

with the boundary conditions

' 0f a f a 21lim

2rf r U

(BC2)

(BC2) suggests the ansatz

f rso that

2

22 2

21 2

df r

dr r

22

42 2

21 2 2 3 2

df r

dr r

Thus, (7.7b) is satisfied if

1 2 2 3 2 0

which means

1 2 0 or 2 3 2 0

so that

211 1 8

12

or

415 25 16

12

A general solution to (7.7b) is therefore

2 4Af Br Cr Dr

r

where A, B, C, D are constants.

Putting in (BC2), we have

2 40A

Ba Ca Daa

32

0 2 4A

B Ca Daa

0D 1

2C U

Substituting the last 2 into the first 2 gives

21

2

ABa Ua

a

2

AB Ua

a

which gives

3 31 1 11

2 2 4A Ua Ua

21 1 31

2 2 4B Ua Ua

a

Hence

323 2

4

U af ar r

r

(7.8a)

32 23 2 sin

4

U aar r

r

(7.8)

See Fig.7.2 for the corresponding streamlines.

7.2.4 Drag

Now, from (7.8a),

2 3

2 2 3

2 3 32 2 0 2 2 2 2

4 2

d U a a Uaf

dr r r r r

so that

2 23sin

2

UaE

r

2 3sin cos

UaE

r

2 22

3sin

2

UaE

r r

and

3

3cos

p Ua

r r

(7.8a)

3

1 3sin

2

p Ua

r r

(7.8b)

Integrating (7.8a) gives

2

3cos

2

Uap c

r (7.8c)

Taking the partial with respect to θ:

2

3sin '

2

p Uac

r

which, on comparison with (7.8b), gives

' 0c ie., c const .

Thus, (7.8c) becomes

2

3cos

2

Uap p

r

wherer

p p .

Since the unit normal of the spherical surface is

rn e

the stress vector

T t n

can be written in spherical coordinates as

, , , ,r rr r rt t t T T T t

where, according to eq(A.44)

2 rrr

uT p

r

1r rr

u u u u uT r

r r r r r r

0rT

Now, from

32 23 2 sin

4

U aar r

r

(7.8)

we have

32

2 2

13 2 cos

sin 2r

U au ar r

r r r

3

2

13 4 sin

sin 4

U au a r

r r r r

so that, on r a , 0ru u as expected.

Further differentiation gives

3 2

4 2 2 2

33 3 cos 1 cos

2 2ru U a a Ua a

r r r r r

32

23 2 sin

2ru U a

ar rr r

3 2

4 2 2 2

33 3 sin 1 sin

4 4

u U a a Ua a

r r r r r

which, on r a , become

0r ru u

r

3sin

2

u U

r a

Hence, on the surface of the sphere, the stress vector is

3cos

2r

Ut p p

a

3sin

2

Ut

a

0t

The net force on the sphere should, by symmetry, in the direction of the uniform

stream (z-axis). The corresponding component of t beingcos sinz rt t t

2 23 3cos cos sin

2 2

U Up

a a

3cos

2

Up

a

The drag on the sphere is therefore

2 12

0 1

coszD t a d d

12

1

2 cosza t d

12

1

32 cos cos

2

Ua p d

a

2 32 6

Ua aU

a

(7.9)

This formula was confirmed experimentally by measuring the terminal velocity TU

of a steel ball dropped into a pot of glycerine. Thus, the viscous drag exactly balances

the buoyancy reduced pull of the gravity so that

346

3T shere fluidU a a g

7.2.5 Further Considerations

The above theory was due to Stokes.

It was found to be flawed since, for example, its application to the 2-D case fails in

the sense that the solutions it provides cannot satisfy all the required boundary

conditions (see Ex.7.4).

The trouble lies in the fact that the non-linear term u u dropped to obtain the

slow flow eqs (7.3) is not negligible in the transitional region where the deflection due

to the sphere begins to die out and the flow is more or less, but not quite, uniform.

One way to handle the problem is by means of the mathched asymptotic expansions

due to I. Proudman and J.R.A. Pearson. [J.Fluid.Mech. 2, 237-62 (1957)].

7.2.6 Navier-Stokes Eqs for Ψ

7.2.6 Navier-Stokes Eqs for Ψ

We now re-write the full steady state Navier-Stokes eqs

21p

u u u

in terms of the Stokes stream function Ψ.

Using

21

2 u u u u u

2 2 u u u u

we have

21 1

2p

u u u u

Using

21

sinE

r u e

we have

21

0 0sin

r

ru u u

Er

e e e

u u

2 2

sin sinr

r

u uE E

r r

e e

2 22 2 3 2

1 1

sin sinr E Er r r

e e

22 2

1 1

sinr Er r r

e e

From §7.2.1, we have

2 22

1

sin r E r Er r

u e e

Hence

21 1

2p

u

22 2

1 1sin sin

sin r r Er r r r

e e

ie.,

2 22 2

1 1 1sin

2 sinp E

r r r

u

2 22 2

1 1 1 1sin

2 sinp r E

r r r r

u

Eliminating 21

2p u by cross differentiation gives

2 22 2 2

1 1 1sin sin

sin sinE r E

r r r r r r

Terms proportional to ν were already worked out in §7.2.1. Thus

2 2 2 22 2 2 2

1 1

sin sin sinE E E E

r r r r

2 22 2 2 2

1 1

sin sinE E

r r r r

22 3 2

1 2cos 1

sin sinE

r r

22 3 2

1 2 1

sinE

r r r

22 2

1 22cot

sinE

r r r r

ie.,

2 2 22

1 22cot

sinE E E

r r r r

(7.11a)

Shifting to dimensionless variables, we set

'r

ra

'U

u

u2

'Ua

so that

2 2 2 22

' ' 'U

E E E Ea

22 2

2 2

1 1 '' '

' '

UE E

r r r r a

Substituting these into (7.11a) and then leave out the primes for clarity, we have

2 2 22

22cot

sin

RE E E

r r r r

(7.11)

where the Reynolds number is defined by

UaR

(7.12)

7.2.7 Matched Asymptotic Expansions

The method is rather involved.

Here, we are contended with only a summary of the salient results.

Interested readers should consult the original paper:

I. Proudman and J.R.A. Pearson. [J.Fluid.Mech. 2, 237-62 (1957)],

where they obtained solution to the Navier-Stokes eqs (7.11) in 2 parts.

Near the sphere

They found

2 211 sin

4r

2

3 1 3 1 1 1 2 2 cos

8 8R R

r r r

(7.13)

which is just the sum of the 1O and O R terms in an asymptotic expansion for

Ψ that is exact in the limit 0R for fixed r.

If we take only the 1O terms, we have

2 21 11 2 sin

4r

r

2 21 12 3 sin

4r r

r

which is just Stokes solution (7.8) in dimensionless quantities.

The O R term improves the accuracy of the solution so that, for example, the drag

(7.9) becomes

36 1

8D Ua R

Far from the sphere

With

*r Rr (7.15)

they found that

2

2**2

1 3 1 1sin 1 cos 1 exp 1 cos

2 2 2

rr

R R

(7.14)

which is just the sum of the 2O R and 1O R terms in an asymptotic expansion

for Ψ that is exact in the limit 0R for fixed *r .

Thus, by ‘far away’ from the sphere, we mean a distance r of order 1R or greater as

0R .

Matching

Solutions (7.13) & (7.14) are matched in the following sense.

Rewriting (7.13) in terms of *r , we have2

2*11 sin

4

r

R

2

2* * *

3 3 1 2 2 cos

8 8

R R RR R

r r r

Keeping only terms of 2O R and 1O R ,

2

2*

*

1 3 11 2 1 cos sin

4 4

rR

R r

2 2

2* *2

*

1 3 32 1 cos sin

4 4

r r

R r R

2

2*2

*

1 3 32 1 cos sin

4 4

rR

R r

(7.15a)

Rewriting (7.14) in terms of r, we have

2 21 3 1 1sin 1 cos 1 exp 1 cos

2 2 2r rR

R

Keeping only the 1O and O R terms,

22 2 2 21 3 1 1 1sin 1 cos 1 cos 1 cos

2 2 2 8r rR r R

R

2 2 2 2 21 3 3sin sin 1 cos sin

2 4 16r r r R

2 21 3 3sin 2 1 cos

4 4r R

r

(7.15b)

In view of (7.15), we see that (7.15a) & (7.15b) are identical.

This is what we mean by saying that (7.13) & (7.14) are matched.

7.3 Corner Eddies

In a 2-D slow flow past a symmetric obstacle, eddies may occur symmetrically for

and aft of the body (see Fig.7.1a).

If the shape of the obstacle is a circular arc, eddie-free flow is still possible if the

corner angle is larger than 146.3C .

Below C , an infinite sequence of eddies of alternating senses occurs (see Fig.7.4).

Here, we follow an elementary approach to the problem due to Moffatt.

7.3.1 Governing Equations

7.3.2 Solutions

7.3.3 Boundary Conditions

7.3.4 Discussions

7.3.1 Governing Equations

We begin with the stream function representation

uy

vx

ie.,

u k or 3i ijj

ux

so that

u k

or

2

3ijk klij lx x

u

2

3 3il j i jlj lx x

2 2

33

il j jx x x x

Since ψ is a function of x and y only, we have

2

23 3i ii

j jx x

u

or

23 u e

Consider now the slow flow equation2p u (7.3)

Taking the curl gives

2 2 2 230 u u e

ie.,

2 2 0 (7.16)

In terms of cylindrical coordinates,

1ru

r

ur

(7.17)

2 22

2 2 2

1 1

r r r r

so that (7.16) becomes

22 2

2 2 2

1 10

r r r r

(7.18)

7.3.2 Solutions

Seeking for a separable solution of (7.18), we write

g r f

It is easy to see that separation can be achieved if2

22

d fconst f m f

d (7.18a)

where the particular form of the constant is chosen for later convenience.

Substituting into (7.18) gives

22 2

2 2

10

d d mg

dr r dr r

(7.18b)

which is homogeneous in r. A natural ansatz is therefore

g r

Now

2 2

2 22 2

11

d d mr m r

dr r dr r

22 2

2 2 22 2

11 2 3 2

d d mr m m r

dr r dr r

22 2 2 22m m r

so that (7.18b) is satisfied if

m or 2 m

Now, a general solution to (7.18a) is

1 2cos sinf c m c m

For a given λ, the general solution to (7.18) is therefore

cos sin cos 2 sin 2r A B C D (7.18c)

where A, B, C, D are constants.

7.3.3 Boundary Conditions

The system of interest is shown in Fig.7.4.

We shall put the origin at the vertex with the x-axis bisecting the corner angle.

The no-slip condition is therefore

0ru u on

where 2α is the angle of the corner.

Applying them on the solution (7.18c) gives

u : cos sin cos 2 sin 2 0A B C D

cos sin cos 2 sin 2 0A B C D

ru :

sin cos 2 sin 2 2 cos 2 0A B C D

sin cos 2 sin 2 2 cos 2 0A B C D

They are easily decoupled to give

cos cos 2 0A C (7.19a)

sin 2 sin 2 0A C (7.19b)

sin sin 2 0B D (7.19c)

cos 2 cos 2 0B D (7.19d)

Eqs(7.19a,b) are consistent only if

cos cos 20

sin 2 sin 2

ie.,

2 cos sin 2 cos 2 sin 0

which can be simplified as

tan 2 tan 2

sin 2sin2

cos cos 2

sin cos 2 2 sin 2 cos

sin cos 2 sin 2 cos

2sin 2 cos (7.19e)

sin 2 sin 2 1 sin 2

1 sin 2 sin 2 1

where we’ve used the trigonometric identities

sin cos cos sin sinA B A B A B

2sin cos sin sinA B A B A B

Setting

2 1x

we have

sin 2 sin2

xx

sin 2 sin

2

x

x

(7.19)

Now, eqs(19.c,d) can be obtained from eqs(19.a,b) by interchanging sine and cosine.

Hence, the counterpart of (7.19e) is

cos sin 2 cos 2 sin

2cos 2 sin

which can be simplified to

sin 2 sin 2 1 sin 2

1 sin 2 sin 2 1

sin 2 sin

2

x

x

(7.19’)

Obviously, (7.19) & (7.19’) cannot be satisfied simultaneously.

Thus, we have 2 independent solutions, namely,

1. cos cos 2r A C

withsin 2 sin

2

x

x

(7.19)

2. sin sin 2r B D

withsin 2 sin

2

x

x

(7.19’)

7.3.4 Discussions

For a given corner subtending an angle 2α, we can solve eq(7.19) or eq(7.19’) to get x

and hence

12

x

Both ru and u , as calculated from (7.17), vary with r as1 / 2xr r .

Thus, for x real and positive (λ real and greater than 1),

0u at 0r Furthermore, for a fixed θ, neither ru nor u changes sign as r varies.

The flow is therefore of the simple form with no eddies, as shown in Fig.7.3a.

For λ complex, ie., of the form

p iq with p, q real

the measured value of any physical quantity is defined to be the real part of the

corresponding complex function.

For a fixed θ, both ru and u are of the form

1 1 1 lnRe Re Rep iq p iq rcr cr cr e

where c is some complex constant (different ones for ru and u , of course), and

we’ve used

exp ln exp lniq iqr r iq r

Writing ic c e , where ε is real, we have

ln1 1Re cos lni q rp pc r e c r q r

which changes sign twice with r whenever

lnq r changes by 2π.

Note that each sign change indicates the presence of a vortex (eddie).

Since0

limlnr

r

, the sign changes occur with increasing rapidity as one

approaches the vertex. Thus, an infinite sequence of eddies is indicated.

Now, eq(7.19) is usually solved by graphical method as shown in fig.7.5.

Thus, the soloution 0x is simply the ordinate of the intersection of the curve

sin xy

x and the straight line

sin 2

2y

.

As is clear from fig.7.5, no intersect is possible if

sin 20.217

2

ie.,

sin 20.217

2

Solving

sin 20.217

2C

C

gives

2 146.3C

Eddies occur if 02 2 .

7.4 Uniqueness & Reversibility of Slow Flows

Theorem

Consider viscous fluid in region V with closed surface S.

There is at most one solution to the slow flow eqs (7.3) which satisfies a given

boundary condition

Bu u x on S.

Proof

Let there be another solution *u that satisfies the same boundary condition.

Define the ‘difference flow’ by

* v u u

and the ‘difference pressure’ by

*P p p

The linearity of the slow flow eqs2p u 0 u

implies2P v 0 v

with the boundary conditions

0v on S.

In component form, these eqs become

2i

i j j

P v

x x x

0i

i

v

x

Multiplying the 1st with iv , we have

2i

i ii j j

P vv v

x x x

Using

i ii i

i i i i

v P P v Pv P v

x x x x

we have

2i i

ii j j

Pv vv

x x x

Integrating over V and applying the divergence theorem gives

2i

i ij jS V

vPv dS v dV

x x

The left hand side vanishes since 0v on S. Together with

2i i i i

i ij j j j j j

v v v vv v

x x x x x x

we get

0 i i ii

j j j jV

v v vv dV

x x x x

Applying the divergence theorem to the 1st term shows that it vanishes on account of

the boundary condition on S. Thus

2

,

0 i i i

i jj j jV V

v v vdV dV

x x x

Since the integrands are all non-negative, we must have

0i

j

v

x

for all i, j at all Vx

which means

constv for all Vx .

Since 0v on S, we have 0v everywhere so that * u u . QED.

Reversibility

Let 1u x , together with the attendant pressure field 1p x , be the solution to the

slow flow eqs2p u 0 u

with boundary conditions Bu x u x on S.

Next, we change the boundary condition to B u x u x on S.

By inspection, we see that 1u x , together with a pressure field 1p c x , c

being a constant, is a solution that satisfies the new boundary condition.

Furthermore, according to the uniqueness theorem, it is the only such solution.

Thus, we see that for flows governed by the slow flow eqs,

Reversed boundary conditions leads to reversed flow.

One example of this is the concentric cylinder experiment shown in Fig.2.6.

Note that for real fluids, reversal of flow is only partial since

1. The slow flow eqs themselves are only approximations.

2. Due to thermal and mechanical fluctuations, no boundary condition can be

exactly reversed in practice.

7.5 Swimming At Low R

Consider a battery powered mechanical fish consisting of a cylindrical body and a

plane tail which flaps as shown in Fig.7.6a.

It swims easily in water but makes no progress in syrup.

The problem it encounters in the latter case is caused by the reversibility of flow for

low Reynolds numbers as discussed in §7.4.

Swimming in low R situations thus requires motions that is not time-reversible.

This can be achieved by replacing the plane tail with a rotating helical coil.

Spermatozoa also used this mechanism, sending helical waves down their tails.

7.5.1 Swimming of a Thin Flexible Sheet

7.5.2 Basic Eqs

7.5.3 Dimensionless Form

7.5.4 Iterative Equations

7.5.5 1st order Solution

7.5.6 2nd order Solution

7.5.1 Swimming of a Thin Flexible Sheet

A simple model for ciliary propulsion is a thin extensible sheet which flexes

according to

sx x sinsy a k x t (7.21)

where ,s sx y denote the coordinates of a particle on the sheet (see Fig.7.7).

Thus, a wave travels along the x direction with speed

ck

while the particles on the sheet move only in the y direction with velocity

cossdya k x t

dt

This motion is not time-reversible since ‘running the film backward’ entails the wave

moving in the opposite direction.

We shall show that, for /a small, 2 / k being the wavelength, the flow

induced by this oscillatory flexing contains a steady component2

22a

U c

(7.22)

in the x-direction. Thus, the sheet swims to the left in a fluid that is otherwise at rest.

7.5.2 Basic Eqs

As in §7.3, we work in the stream function representation with

uy

vx

(7.23)

so that

23 u e

and the slow flow eq becomes

2 2 0

which, in Cartesian coordinates, is simply

22 2

2 20

x y

(7.24)

The boundary condition on the sheet is simply su u .

In terms of ψ, we have

0s

s

y y

uy

coss

sy y

v a k x tx

(7.25)

on sinsy y a k x t .

7.5.3 Dimensionless Form

Without loss of generality, we shall solve eq(7.24) only at 0t .

Solutions at other times can be obtained by replacing kx with k x t .

To further facilitate manipulations, we introduce the dimensionless quantities:

'x kx 'y ky 'k

a

(7.26)

On substituting into eqs(7.24-25) and then dropping the primes for clarity, we have,

(7.24):22 2

2 20

x y

(7.27)

(7.25): 0s

s

y y

uy

coss

sy y

v xx

(7.28)

on sinsy y x .

7.5.4 Iterative Equations

For ε small, the boundary conditions (7.28) can be approximated by a Taylor series

expansion:

2

20 0

sin 0sy y y y

xy y y

2

0 0

sin cossy y y y

x xx x y x

(7.30)

Similarly, we write ψ as a power series in ε,

21 2 3 (7.31)

where the n ’s are independent of ε.

On substituting into

(7.27)

and using the fact that the coefficient of each power of ε must vanish, we have

22 2

2 20nx y

for all n.

By successive differentiations on (7.31), we have

21 2 3

0 0 0 0

n n n n

n n n n

y y y yy y y y

21 2 31 1 1 1

0 0 0 0

n n n n

n n n n

y y y yy x y x y x y x

On substituting into the 1st boundary conditions in (7.30), we have

21 2 3

0 0 0y y yy y y

2 2 221 2 3

2 2 2

0 0 0

siny y y

xy y y

21 2 3

0 0 0

1sin

!

m m mm

m m m

y y y

xm y y y

0

Setting the coefficients of each power of ε to zero, we have

1

0

0y

y

22 1

20 0

sin 0y y

xy y

1 1

11

0 0 0

sin sin0

! 1 !

n n m mm m n

n my y y

x x

y n y m y

Doing the same to the 2nd boundary conditions in (7.30), we have

1

0

cosy

xx

22 1

0 0

sin 0y y

xx y x

1 1

11

0 0 0

sin sin0

! 1 !

n n m mm m n

n my y y

x x

x n y x m y x

To summarize, to get m , we must solve

22 2

2 20mx y

with the boundary conditions

1 1

11

0 0 0

sin sin0

! 1 !

n n m mm m n

n my y y

x x

y n y m y

1 1

11

0 0 0

sin sin0

! 1 !

n n m mm m n

n my y y

x x

x n y x m y x

which requires previous knowledge of 1 1, , m .

7.5.4 Solution

7.5.5 1st order Solution

The 1st order solution 1 satisfies

22 2

12 20

x y

(7.32)

with boundary conditions

1

0

0y

y

1

0

cosy

xx

A separable solution

1 X x Y y

together with the boundary conditions suggest X to be sinusoidal so that we set2''X X

so that

1 2cos sinX c x c x

In order to satisfy the 2nd boundary condition, we need

sinX x 0 1Y

so that

''X X and

22

21 0

dY

dy

SettingyY e

gives

221 0

which has 2 pairs of double roots

1

The general solution is therefore

y yY A By e C Dy e

If u is to be finite as y , we must have

0C D

With 0 1Y , we have

1 yY By e

Finally, 1

0

0y

y

means

1 sin 0yBy B e x at 0y

ie., 1B so that

1 1 sinyy e x (7.34)

1 1 1 sin siny yy e x ye xy

(7.34a)

7.5.6 2nd order Solution

The 2nd order solution 2 satisfies

22 2

22 20

x y

(7.33)

with boundary conditions

22 1

20 0

sin 0y y

xy y

22 1

0 0

sin 0y y

xx y x

Using the 1st order result

1 1 sinyy e x (7.34)

we have

1 1 1 sin siny yy e x ye xy

2

12

1 sinyy e xy

21

2

0

siny

xy

21 cosyye x

x y

21

0

0y

x y

so that the boundary conditions become

22

0

siny

xy

2

0

0yx

(7.35)

A separable solution

2 X x Y y

with

1 2cos sinX c x c x

yY A By e (7.35a)

obviously cannot satisfy (7.35)

To proceed, we rewrite the 1st eq. in (7.35) as

2

0

11 cos2

2y

xy

so that the cosine term is of the manageable type (7.35a).

On the other hand, the constant term requires 2 to be of the form

2 X x Y y f y (7.35b)

so that the boundary conditions become

1' 0 0 1 cos2

2X x Y f x

0 sin 2 0Y x

which requires

1cos2

2X x x ' 0 1Y 1

' 02

f

0 0Y (7.35c)

Plugging into (7.35a) gives

0 0Y A

' 0 1Y A B B

so thatyY ye

Furthermore, for (7.35b) to remain a solution, f itself must be a solution of (7.33), ie.,(4) 0f

so that f is of the form3 2f Ay By Cy D

and2' 3 2f Ay By C

Now, u is proportional to f’ so that to keep the former finite at y , we need

0A B Plugging in the boundary condition gives

1'

2f C

so that, finally, we have

1

2f y

where D, which is of no physical significance, was dropped for convenience.

Putting everything together, we have

22

1 1cos2

2 2yye x y (7.36)

so that

22 1 12 1 cos2

2 2yy e x

y

Together with

1 1 1 sin siny yy e x ye xy

(7.34a)

we have

1 2

y y y

21 1sin cos2

2 2y yye x y e x

(7.37)

Reverting to unprimed quantities, this becomes

21 1 1sin cos2

2 2ky kykye kx ky e kx

a y

or

uy

21 1sin cos2

2 2ky kyakye kx a ky e kx

2 21 1sin cos2

2 2ky kyye kx c ky e kx

Generalizing to arbitrary time, we have

2 21 1sin cos2

2 2ky kyu ye kx t c ky e kx t

(7.38)

21

2U c (7.22)

7.6 Flow in a Thin Film

7.6.1 Basic Conditions

7.6.2 Basic Equations

7.6.1 Basic Conditions

Consider a viscous fluid in steady flow between 2 rigid boundaries

0z and ,z h x y

Let U and L be typical flow speed and length scale in the horizontal ( ) direction.

The condition for a thin film flow is

h L (7.39)

Now, the no-slip condition must be satisfied at 0z and z h .

As z increases from 0 to h, u 1st increases to order U, then drops back to 0, thus

UO

z h

u

Meanwhile,z

u also experiences a sign change somewhere mid-stream so that

2

2 2

UO

z h

u

By comparison, the horizontal gradients u and 2 u , which are ofU

OL

and

2

UO

L

, respectively, are seen to be much weaker owing to (7.39).

Although the magnitude of zw u e is not specified, the no-slip condition still

applies, so that its horizontal gradients are also expected to be much weaker than the

vertical one.

Thus, the viscous term can be approximated as2

22z

uu

From the incompressibility condition,

0u v w w

x y z z

u

we see that

w UO

z L

and hence

Uhw O O U

L

u

Order of magnitude estimates also give

U Uh Uw

z L Lh L

u u

, , 1,1,Uh h

U U UL L

u

so that

2

1,1,U h

L L

u u

2

2 21,1,

U h

z h L

u

Thus, u u may be neglected when compared to2

2z

uif

2

2

U U

L h ie.,

2

1Uh

L

or2

1UL h

L

(7.40)

which is the 2nd condition for a thin film flow. [the 1st is (7.39)]

7.6.2 Basic Equations

Under the conditions eqs(7.39,40) and in the absence of external forces, the steady

state Navier-Stokes eqs simplify to2

2

10 p

z

u

0 u

i.e.,2

2

p u

x z

2

2

p v

y z

2

2

p w

z z

0u v w

x y z

(7.41)

As shown in the last section, w and2

2

w

z

are both smaller by a factor /h L than

their horizontal counterparts. According to (7.41), so isp

z

.

Setting p to be independent of z, the 1st 2 eqs in (7.41) can be integrated to give

1u pz A

z x

1v pz C

z y

21

2

pu z Az B

x

21

2

pv z Cz D

y

(7.42)

where A, B, C, D are functions of x, y only.

From

2

2 2

UO

z h

u

eq(7.41) implies

2

p p UO

x y h

so that

2

ULp O

h

(7.44)

Consider now the stress tensor

jiij ij

j i

uup

x x

(7.43)

The largest term in the parenthesis is

2

U ULO O p

z h h

u

Hence

ij ijp (7.45)

which means the thin-film flow behaves like an inviscid flow in this aspect.

7.7 Flow in a Hele-Shaw Cell

A Hele-Shaw cell flow is a thin film flow between 2 parallel plane boundaries.

Usually, fluid flow is driven by horizontal pressure gradient to past by cylindrical

objects in the gap as shown in Fig.7.8.

Applying the no-slip condition on 0z and z h to (7.42) gives

0 B 210

2

ph Ah

x

0 D 210

2

ph Ch

y

so that

1

2

pA h

x

1

2

pC h

y

and

1

2

pu h z z

x

1

2

pv h z z

y

(7.46)

Note that

pv y

pux

and hence the direction of flow and the streamline patterns, are all independent of z.

Eliminating p in (7.46) by cross differentiation, we have

0u v

y x

(7.47)

so that at a given z, the flow is similar to a 2-D irrotational flow.

However, if we calculate the circulation around any closed curve C, we find

C

udx vdy

1

2 C

p pz h z dx dy

x y

1

2 C

z h z p d

x

1

2 Cz h z p

0 (7.48)

where we’ve used the fact that p is a physical quantity and hence must be

single-valued.

Now, (7.48) is valid irrregardless of the global topology of the fluid domain.

This behavior is markedly different from that of true 2-D flows for which can take

on finite values.

Thus, if we place a flat plate at an angle of attack to the oncoming stream, the

streamline patterns will be as shown in Fig.4.6a but never as in Fig.4.6b.

Another example is the flow into a rectangular opening as shown in Fig.7.9b, which is

to be contrasted with that of high R flow shown in Fig.7.9a.

It is well known that a rather large force is required to pull a disc away from a rigid

plane if the two are separated by a thin film of viscous fluid.

Here we study the problem in terms of the steady state thin film flow eqs and attribute

all the time dependence of the flow to the changing boundary conditions. Justification

of this assumption will be left as an exercise.

Let the radius of the disk be a and the height of the fluid be h t . [see Fig.7.10]

By symmetry of the setup, we expect the (unsteady) flow to be of the axisymmetric:

, , , ,r r z zu r z t u r z t u e e

In terms of cylindrical coordinates, the relevant thin film eqs are (cf.7.41)2

2rp u

r z

2

20 zu

z

10z

r

uru

r r z

Using the fact that p is independent of z, we can integrate the 1st eq. and get

1ru pz A

z r

21

2r

pu z Az B

r

where A, B are functions of r & t only.

Sticking in the no-slip boundary condition at 0z and z h t , we have

0 B

210

2

ph Ah

r

so that

1

2r

pu h z z

r

Thus

2

2

1

2ru p

h z zr r

and the incompressibility condition becomes

2

2

1 10

2 2zp p u

h z z h z zr r r z

i.e.,

2

2

1 1

2zu p p

h z zz r r r

which integrates to

22 3

2

1 1 1 1

2 2 3z

p pu hz z C

r r r

No-slip conditions at 0z then gives

22 3

2

1 1 1 1

2 2 3z

p pu hz z

r r r

The boundary condition at the moving disk is

z

dhu

dt at z h t

so that

23

2

1 1

12

dh p ph

dt r r r

which can be rearranged to give2

2 3

1 1 12p p p dhr

r r r r r r h dt

(7.49)

Integrating, we get

23

6p dhr r D

r h dt

23

3ln

dhp r D r E

h dt

where D, E are functions of t only.

To avoid a singularity at 0r , we set 0D .

With the boundary condition 0p p at r a , 0p being the atmospheric pressure,

we have

2 203

3 dhp r a p

h dt

The force exerted on the disk by the fluid is therefore

2

00 0

aF p p rdrd

2 23 0

h dt

4 43

6 1 1

4 2

dha a

h dt

43

3

2

dha

h dt

(7.50)

Thus, if one tries to pull up the disk, i.e., 0dh

dt , then 0F , so that the pull by the

fluid is downward. Furthermore, the 3h dependence makes F rather large for small

h.

7.9 Thin Film Flow Down a Slope

Consider the 2-D problem of a layer of viscous fluid spreading down a slop under the

action of gravity.

[See Fig.7.11 for the definitions of the coordinate system and various quantities.]

As in §7.8, we neglectt

u

and attribute all time dependencies to that of the

boundaries. In the presence of gravity, the thin film flow eqs become2

2

10 sin

p ug

x z

10 cos

pg

z

(7.52)

The 2nd eq can be integrated to give

cos ,p gz f x t (7.52a)

Consider the free surface

,z h x t

Its unit tangent and normal vectors in the x-z plane are (see §3.4b.2)

2

1ˆ 1,

1

hv

xh

x

2

1ˆ , 1

1

hn

xh

x

Under the thin film approximation

h L (L is horizontal length scale)

we have

1h h

x L

so that

ˆ 1, 0v x̂ ˆ ˆ0, 1n z

Consider now the stress tensor

jiij ij

j i

vvp

x x

The stress vector at the surface is

ˆ ˆn z t

1 13

u w u

z x z

t

3 33 2w

p pz

t [see (7.45)]

Hence. the continuity of the stress across an interface gives

0u

z

(7.53)

0p p (7.53a)

on ,z h x t .

Putting (7.53a) into (7.52a) gives

0 cosp gh f

so that

0cosp g h z p

cosp h

gx x

The 1st eq. in (7.52) thus becomes2

2cos sin

u hg g

z x

(7.54)

Now,

1h h

Ox L

so that unless α is very small, eq(7.54) can be approximated by2

2sin

ug

z

Integrating, we get

sinu g

z Az

21sin

2u z g Az B

(7.54a)

where A, B are functions of ,x t .

Applying (7.53) at z h , we have

0 sing

h A

Applying the no-slip condition at 0z , we have

0 BHence, (7.54a) becomes

1sin

2

zu h z g

(7.55)

sinu z h

gx x

so that the incompressibility condition gives

sinw z h

gz x

which can be integrated to give2

sin2

z hw g A

x

Putting in the no-slip condition at 0z , we have

0A so that

2

sin2

z hw g

x

Putting this into the surface wave condition,

h hw u

t x

at z h [see eq(3.18)]

we have

2

sin 02

h h hu g

t x

at z h (7.55a)

Now, (7.55) gives

2

sin2

hu g

at z h

so that (7.55a) becomes2

sin 0h h h

gt x

(7.56)

The characteristics of this obey (see §3.9a3)

21 sin 0

dt dx dh

h g

so that

1h c

21

sindx gc

dt

21 2

singx c t c

Hence, the general solution of (7.56) is

2singh f x h t

(7.56a)

where f is an arbitrary function.

Now, (7.56a) is of the wave-steepening type discussed in §3.9.3.

Thus, taking its partial, we have

2 sin1 '

h g hht f

x x

'2 sin

1 '

fg

tff

which blows up at Ct t , where

2 sin1 ' 0

Ct t

gtff

In practice, surface tension effects set in to prevent instability of the type depicted in

Fig.3.16c. The wave profile at the front therefore remains in the form of a ‘nose’shown in Fig.3.16b and Fig.7.11.

According to Huppert, beyond Ct , h is described by the similarity solution

sin

xh

g t

(7.57)

[see H.E.Huppert, J.Fluid Mech 173, 557-94 (1986)]

The position Nx of the front edge is obtained from the constant volume condition

0

Nx

hdx A (7.58)

where A is the area of the cross section of the fluid in the x-z plane.

Putting in (7.57), we have

3/ 2

0

2

sin sin 3

Nx

NA xdx xtg tg

or

1/ 329 sin

4N

A gx t

(7.59)

7.10 Lubrication Theory

When a solid body slides over another one, the frictional resistance it experiences is

usually comparable in magnitude to the normal force it exerts on the other.

However, if there is a thin film of fluid between them, the frictional resistance can be

reduced drastically. This is called lubrication.

The main reason for this is that, in thin film flows, the ratio between the magnitudes

of the tangential and normal stresses is of the order [see §7.6]

2

1U

hhUL L

h

7.10.1 Slider Bearing

7.10.2 Flow Between Eccentric Rotating Cylinders

7.10.1 Slider Bearing

Consider the 2-D system shown in Fig.7.12.

The rigid lower boundary at 0z moves with velocity U past a stationary block of

length L. The space between them is filled with viscous fluid, and the pressures at

both ends of the bearing are equal to the atmospheric value 0p .

Working with the 2-D version of the thin film flow eqs (7.41), we can begin at the

solution

21

2

dpu z Az B

dx (7.42)

where p is a function of x only.

The boundary conditions are

u U at 0z 0u at z h

which, when plugged into (7.42), gives

U B

210

2

dph Ah U

dx

so that

11

2

dp zu h z z U

dx h

1

2

dp Uz h z

dx h

(7.60)

The integral form of the incompressibility condition is that the mass flux Q is the

same across all cross-sections of the film. In terms of (7.60), we have

0

h

Q udz

0

1

2

h dp Uz h z dz

dx h

3 21 1 1 11

2 2 3 2

dp Uh h

dx h

31

12 2

dp Uh h

dx (7.61)

Rearranging terms gives

3

12

2

dp Uh Q

dx h

which integrates to

0 30

12

2

x Up p h Q dx

h

2 30 0

1 16 2

x x

U dx Q dxh h

(7.61a)

where 0p p at 0x .

Now, we also have 0p p at x L , so that

2 30 0

1 10 6 2

L L

U dx Q dxh h

ie.,

2 20 0

3 30 0

6 1

212 1

L L

L L

Udx dxh hU

Q

dx dxh h

(7.62)

In the special case of a plane slider bearing, h vary linearly from 1h at 0x to 2h

at x L , ie.,

2 1 1

xh x h h h

L

Using the formulae

2

1dx

b a bxa bx

3 2

1

2

dx

a bx b a bx

we have

22 1 2 1

1

1dxh h h hh h x

L L

2

2 1 1 2 1

L

h h h L h h x

2 1

L

h h h

23

2 1 2 11

1

2

dx

h h h h hh x

L L

3

2

2 1 1 2 12

L

h h h L h h x

22 12

L

h h h

so that

22 1 2 1 2 10

1 1L dx L L

h h h h h h h

2 13 2 2 2 22 1 2 1 2 10

1 1L dx L Lh h

h h h h h h h

whereupon eq(7.62) gives

2 1

2 12

U h hQ

h h

Also, Eq(7.61a) becomes

0 2 12 2

2 1 1 2 1 2 1 1

1 1 1 1

6

p p UL h h LU

h h h h h h h h h h

2 12 2

2 1 1 2 1 1

1 1 1 1UL h h

h h h h h h h h

2 12 2 2 22 1 1 2 1 1

1 1 1 1 1 1ULh h

h h h h h h h h

2 12 2 22 1 2 1 1 2

1 1 1 1 1ULh h

h h h h h h h h

22 1 1 22 2 2

2 1

ULh h h h h h

h h h

1 22 2 22 1

ULh h h h

h h h

(7.63)

For 2 1h h h , the last expression is positive so that 0p p and we have an

upward lift. Hence, lubrication occurs if h decreases in the direction of flow as

shown in Fig.7.12.

7.10.2 Flow Between Eccentric Rotating Cylinders

Consider the setup shown in Fig.7.13.

Viscous fluid fills the narrow gap between a fixed outer cylinder of radius

1r a , and an inner, offset, cylinder of radius a which rotates with peripheral

velocity U.

This serves as a simple model for an axle rotating in its housing.

Geometry

Let the centers of the outer and inner cylinders be at 0,0x and ,0x ,

respectively.

The quantity h then provides a convenient measure of the ‘height’ of the fluid.

Simple trigonmetric maneuver gives

1h a

where2 2 2 2 cosa a

sin sin

a

Now, ε and hence η are small. Keeping only 1st order terms in them, we have

1 sin sinsin

a a

cos cos

cos cos sin sin

cos cos sin sin

2cos sina

1 cos 1 cosa aa a

and

cos 1 cosh a aa

(7.64)

where

offset of centers

a

The minimum of h is at 0 :

0 1 0h a

Hence, 0 1 .

For 0 , the cylinders are coaxial.

For 1 , they are in contact at 0 .

Solution

For small gaps, curvature effects may be neglected so that the results for the plane

slider bearing are applicable here with straightforward adaptations such as x a ,

u u , etc.

Thus, eq(7.60) becomes

1

2

dp Uu z h z

a d h

(7.65)

where z is the distance across the gap, measured along the radial direction.

Similarly, eq(7.61) becomes

31

12 2

dp UQ h h

a d (7.66)

Using

1 cosh a (7.64)

the counterparts to the integrals in (7.62) are

2

2 22 20 0

1

1 cos

d dI a

h a

2 2

3 33 2 30 0

1

1 cos

d dI a

h a

These can be evaluated by contour integral as follows.

1st, letiz e

so that

1 1 1cos

2 2i ie e z

z

idz ie d izd

whereupon

2 2 22 2 22

1 4

21 112

C C

dz zdzI

a iaz ziz z

z

2

3 3 32 3 2 3 32

1 8

21 112

C C

dz z dzI

a iaz ziz z

z

where C is the unit circle centered at the origin.

Now, the poles are the roots of

2 21 0z z

i.e.,

22

1 1 11 1 1z

Since 0 1 , they are both real.

Furthermore, 1z z so that only one of them is inside C.

Obviously, z z , so that only the pole at z that contributes to the integrals.

With the help of

2z z

221z z

we have

2 2 22 2

4

C

zdzI

ia z z z z

22 2

42

z z

d zi

ia dz z z

2 32 2

8 1 2z

a z z z z

32 2

8 z z

a z z

3/ 22 2

2

1a

2

3 3 32 3 3

8

C

z dzI

ia z z z z

2 2

32 3 3 2

8 12

2z z

d zi

ia dz z z

2

3 42 3 3

8 2 3

z z

d z z

a dz z z z z

42 3 3

28

z z

z z zd

a dz z z

4 52 3 3

4 28 2 2 z z zz z

a z z z z

2 2

52 3 3

162 2z z z z z

a z z

2 2

52 3 3

164z z z z

a z z

2

52 3 3

162z z z z

a z z

2

5/ 2 22 3 2

42

2 1a

2

5/ 22 3 2

2

1a

Thus, (7.62) becomes

2

32

UIQ

I

5/ 22 3 2

3/ 2 22 2

12

2 21

aU

a

2

2

1

2Ua

(7.67)

From (7.66), we see thatdp

dis an even function of θ since h is.

Hence, p, and hence cosp , are odd functions of θ.

The horizontal force on the inner cylinder

cosx

S

F p dS therefore vanishes.

The calculation of the vertical force is too involved to be included here.

We merely list the result:

2 2 2

12

1 2

UF

(7.68)

Of particular interest is the factor 1/ 221

which makes

F as 1 .

In other words, by moving horizontally towards the outer cylinder, the inner one can

Let us consider the flow at .

From (7.64), we see that

1h a at

From (7.67), we have

2

1

2

QU

h

at

We can therefore write (7.66) as

2 312

dp U Qa

d h h

2 2

112 1

2

Ua

h

2

2 2

112

2

Ua

h

0 at which implies an adverse pressure and hence reversed flow in that neighborhood.

Differentiating (7.65), we have

1 1

2 2

u dp dp Uh z z

z a d a d h

12

2

dp Uh z

a d h

3

62

2

Uh UQ h z

h h

2

3 2

6 12

2 2

U h Ua h z

h h

Hence

2

2 2

6 1

2 2z h

u U h Ua

z h h

2

2 2

6 1

3 2

U ha

h

2

2 2

6 1 11 cos

3 2

Ua a

h

2

2 2

2 3 31 cos

2

Ua

h

2

2 2

2 1 4cos

2

Ua

h

2

2 2

2 1 4cos

2

Ua

h

(7.69)

For this to be negative, we need

2

2

1 4cos 0

2

(7.69a)

Now, reversed flow, if occurs, is expected around where cos 1 or

cos 1 . Rewriting (7.69a) to reflect this, we have

2

2

1 41

2

0z h

u

z

so that no reversed flow occurs.

Rewrite the criterion as

3 2 20 4 2 1 1 3 1 (7.69b)

The critical values of λ are the roots of the polynomial on the right side, i.e.,

1 13 13

2

However, since 0 1 , the only valid value is

13 13 0.30

2

Writing

2 3 1

we see that near so that , eq(7.69b) becomes

0 13 13

so that to satisfy (7.69b), we must have 0 , i.e.,

Conversely, if

(7.70)

there will exist a range of θ such that 0z h

u

z

so that reversed flow is expected.

This feature, though of theoretical interest, is usually overwhelmed by other

complications in practice.

8. Boundary Layers

8.1 Prandtl’s Paper

8.2 Steady 2-D Boundary Layer Eq.

8.3 Boundary Layer On Flat Plate

8.4 High R Flow In Converging Channel

8.5 Rotating Flows Controlled By Boundary

Layers

8.6 Boundary Layers Separation

8.1 Prandtl’s Paper

8.2 Steady 2-D Boundary Layer Eq.

We now derive the eq. for a steady, incompressible, 2-D boundary layer adjacent to a

rigid wall at 0y .

The no-slip condition for a stationary boundary means

0u v at 0y (8.2)

which differs from the inviscid flow boundary condition only by the additional

demand of 0u . Our main concern is therefore to see how u merges from 0 to the

inviscid flow value.

For a thin boundary layer, we expect

u u

y x

(8.4a)

To get an order of magnitude estimates, let 0U be some typical value of u.

Let the distance for u to change by an amount 0U be of order L in the x-direction,

and δ in the y-direction. Hence, (8.4a) means

0 0U U

L

ie.,

L (8.4)

The Navier-Stokes eqs. are

2 2

2 2

1u u p u uu v

x y x x y

2 2

2 2

1v v p v vu v

x y y x y

(8.5)

0u v

x y

The 3rd eq. in (8.5) means

0v u U

y x L

so that v is of order 0U

L

. Hence v u .

The other 2 eqs. in (8.5) then imply

p p

x y

which means we may write p p x in the boundary layer.

Similarly, from

20

2 2

u U

x L

20

2 2

u U

y

we deduce

2 2

2 2

u u

y x

These estimates can be used to simplify (8.5) into the boundary layer eqs.,2

2

1u u dp uu v

x y dx y

(8.1)

0u v

x y

(8.2)

Eq(8.1) provides the means to estimate the magnitude of the layer thickness δ.

Now,2 2

0 0 0 0. . .U U U U

L H SL L L

Assuming viscosity effect to be significant within the layer, we have2

0 02

U U

L

ie.,

0

1

L U L R

where R is the Reynolds number.

Hence, thin boundary layer needs large R.

By interpreting x as the direction tangent, y as that normal, to the boundary, (8.1-2)

are applicable to curved boundaries. What’s worth mentioning is that although

p p

x y

, we still have p p x .

Let U x be the slip velocity at 0y for the corresponding inviscid flow.

For a thin boundary layer, we expect u U x at the “edge” y , where,

according to the Bernoulli theorem, 21

2p U const along a streamline. Hence,

dp dUU

dx dx (8.8)

so that a rise in U is accompanied by a drop in p and vice versa.

To ensure the boundary layer merges smoothly into the mainstream inviscid flow, we

impose the boundary condition on (8.1-2)

u U x as ory y

(8.9)

8.2.1 Example

Consider the problem

'' ' 1u u ; 0 0u , 1 2u (8.10)

where 0 is a small constant.

Integrating, we have

1

'ln 1 '

1 '

duy u C

u

or,

/21 ' yu C e

so that

/ /2 3 4

y yu y C e dy y C e C

Putting in the boundary conditions, we have

3 40 C C

1/3 42 1 C e C

which is easily solved to give

3 1/

1

1C

e

and 4 1/

1

1C

e

so that/

1/

1

1

yeu y

e

(8.11)

Now, 1/ 1e .

Also / 1ye unless y .

Hence, u as given by (8.11) may be splitted into 2 parts:

1. Mainstream part: 1Mu y for y .

2. Boundary layer part: /1 yBLu e for 0 y .

Note that

0limMu u

with y fixed.

0limBLu u

with /y fixed.

and

/ 0lim limBL M

y yu u

which is equivalent to (8.9).

Alternative Approach

Another approach which is more akin to that taken in fluid dynamic problems is to

take advantage of the smallness of ε at the earliest stage. Thus, (8.10) simplifies to

' 1ou with solution

ou y c

There is now only 1 arbitrary constant so that only 1 of the 2 boundary conditions can

be satisfied. On making 1 2ou , we obtain an ‘outer’ solution

1ou y

What we did so far is similar to obtaining the mainstream inviscid flow:

The problem is simplified by invoking the smallness of a parameter to drop the

highest derivatives.

However, this lowering the order of the system reduces the number of arbitrary

constants in the solution so that not all boundary conditions can be satisfied.

Thus, an ‘inner’ solution, or boundary layer, is needed to satisfy the other boundary

conditions.

To accentuate the relatively rapid change of u inside the boundary layer, we shift to

the variable

yY

so that (8.10) becomes2

2

d u du

dY dY

On invoking 1 , we have the ‘inner’ (boundary layer) eq.,2

20i id u du

dY dY

which, on integrating, gives

1ln iduY C

dY

ie.,

2Yidu

C edY

so that

3 4Y

iu C e C

On making this satisfy the ‘no-slip’ boundary condition 0 0u at 0y , we

get

3 40 C C

so that

3 1Yiu C e

Matching the inner & outer solutions with

0lim limi oY y

u u

we get

3 1C

so that the full solution is

/

1

1 y

yu

e

as 0 for fixed/

y

y

as found before.

8.3 Boundary Layer On Flat Plate

If the fluid is inviscid, a uniform stream approaching a flat plate at zero incidence

angle suffers no deflection. Hence U x const . According to (8.8), 0dp

dx so that

the boundary layer eqs (8.1-2) reduce to2

2

u u uu v

x y y

(8.12)

0u v

x y

(8.13)

For a semi-infinite plate that extends from 0x to x , it seems natural to seek a

similarity solution:

uh

U with

y

g x (8.14)

and g is to be determined.

1st of all, (8.13) can be satisfied if we introduce the stream function ψ so that

uy

vx

(8.15)

With the help of (8.14), we have

, ,x y u x y dy k x

dyUh d k x

d

Ug x h d k x If the plate is a streamline, ψ is constant on it. For convenience, we can set

( , 0) , 0 0x y x

and get rid of k x by writing

0

,x Ug x h d Ug x f

(8.16)

where

0

f h d

so that 0 0f

Thus,

1 df dfu Ug x U

y g x d d

(8.17)

Using

2

y dg dg

x g dx g dx

we have

dg dg dfv U f g

x x dx g dx d

df dgU f

d dx

(8.18)

Similarly,2 2

2 2

u d f dg d fU U

x x d g dx d

2 2

2 2

u d f U d fU

y y d g d

(8.18a)

2 3 3

2 3 2 3

u U d f U d f

y g y d g d

so that (8.12) becomes

2 2 3

2 2 2 3

df dg d f df dg U d f U d fU U U f

d g dx d d dx g d g d

or

2 3

2 3

dg d f d fUg f

dx d d

(8.18b)

which is self-consistent only if

1

dgg c

dx

3 2

13 20

d f d fc Uf

d d

where 1c is a constant.

Integrating the 1st eq. gives

21 2

1

2g c x c (8.18c)

where 2c is another constant.

Now, eqs(8.18a) show that the partials of u are all proportional to 1g , which blows

up as 0g . The only place such extremes can be tolerated is at the leading edge of

the plate where 0x . Hence, we set 2 0c in (8.18c).

To summarize, we now have

12g c x 1

2

dg c

dx x

12

y

c x

12U c x f (8.19)

dfu U

d 1

2

df cv U f

d x

(8.17,18)

3 21

3 20

d f c U d ff

d d (8.20)

Note that the value of η at the origin 0x y depends on the path one chooses to

take the limit. Thus, if we approach along the y-axis but 0 if we

approach along any other straight line y mx on which12

xm

c .

Since (8.20) is a 3rd order ODE, there’ll be 3 arbitrary integration constants (besides

1c ) in a general solution. Their determination require 3 boundary conditions.

0 0f .

Next, as the flow approaches the leading edge of the plate along the x-axis ( 0 ),

we expect 0u at 0 so that (8.17) gives

0

' 0 0df

fd

Finally, far away from the plate ( y or ) the flow should be ,0Uu .

Hence, (8.17) gives

' 1f

while (8.18) gives

lim ' 0f f

Finally, inspection of the eqs shows that 1c serves only as a scale for x so that its role

is more or less redundant for a similarity solution. For convenience, we can set it to

1cU

so that (8.20) becomes3 2

3 20

d f d ff

d d (8.20’)

which is free of any parameters of the problem.

Solution to this problem must be obtained numerically.

The results are shown in fig.8.8.

Of interest is that 0.97u

U at 3 .

The boundary layer thickness δ can be estimated from2

Uy

x

to give

xO

U

(8.22)

as shown in Fig.2.14.

The horizontal stress on the plate is

0 0

xy

y y

u v u

y x y

where0

0y

v

x

since ,0 0v x for all 0x .

Uisng (8.18a) and , we have

2

2

0

'' 0xy

U d f Uf

g d g

Since

1

22

xg c x

U

we have

2

2

0

'' 02 2xy

U d f UU U f

x d x

(8.23)

which decreases as x increases.

Plate of Length L

Assuming (8.23) to hold for a plate of finite length L, the drag on it is

0 0

2 2 '' 0 2 2 '' 02

L L

xy

U dxD dx U f U UL f

x

where the factor 2 accounts for the equal contributions from both the upper & lower

side of the plate. In terms of the Reynolds numberUL

R

, we have

2 1/ 22 2 '' 0D U Lf R (8.24)

Note that D is proportional to L . Hence

0lim 0D

Numerically, '' 0 0.4696f .

Both (8.24) & the velocity profile agrees with experiment unless R is very high, at

which case, the boundary layer becomes unstable and turbulence ensues. The critical

value of R for this onset is about 5 610 ~ 10 .

8.4 High R Flow In Converging Channel

We consider the high R, 2-D flow between 2 plane walls

0y and tany x

A narrow slit at the origin provides the means for the necessary intake or discharge to

For the mainstream flow, the simplest ansatz is purely radial:

,r ru r u e

The incompressibility condition

0r

uru

r r r

u

then becomes

0rrur

Therefore

r

Qu

r (8.25a)

where Q is a constant and the – sign is inserted so that 0Q represents inflow.

Next, the 2-D Euler eqs in polar coordinates are [cf. eq(2.22), §2.4]

2 1r

r

u u pu

t r r

u

1ru u u pu

t r r

u

which, for our steady flow ansatz, become,

1rr

u pu

r r

p p r

Thus, ru is also a function of r only and, upon integration,

21

1 1

2 ru p c

Using (8.25a), we have2

21 12

1

2 r

Qp u c c

r

Boundary Layer on 0y

The mainstream flow, according to (8.25a), is

QU x

x (8.25)

The boundary eqs

dp dUU

dx dx (8.8)

2

2

1u u dp uu v

x y dx y

(8.1)

0u v

x y

(8.2)

thus become2

2 3

dp Q QU

dx x x

2 2

3 2

u u Q uu v

x y x y

(8.26)

Following the approach of §8.3, we try a similarity solution

Qu U x h h

x with

y

g x

Introducing the streamline function ψ to satisfy (8.2), we have

uy

vx

and

, ,x y u x y dy k x

Q dyh d k x

x d

Qg x h d k x

x

Setting the wall 0y to be a streamline, ψ is constant on it. For convenience, we

can set

( , 0) , 0 0x y x

and get rid of k x by writing

0

,Q Q

x g x h d g x fx x

where

0

f h d

so that 0 0f

Thus

' 'Q Q

u g f fy x y x

and, with

2

''

y gg

x g g

we have

2

' ''

g g g gv Q f f f

x x x x g

2

' '1 '

Qg g gx f x f

x g g

2

''

Qg gf x f f

x g

Similarly,

2 2

1 1 ' '' '' ' ''

u g Q gQ f f f x f

x x x g x g

''u Q

fy xg

2

2 2'''

u Qf

y xg

so that (8.26) becomes

2 2

' '' ' '' ' ''

Q Q g Qg g Qf f x f f x f f f

x x g x g xg

2

3 2'''

Q Qf

x xg

which simplifies to

22

2

'' 1 '' 1 '''

g xf x ff f

g g Q

(8.26a)

This can be consistent by setting

1g c x

so that (8.26a) becomes

22

1

' 1 '''f fc Q

To make this parameter free, we set 21c

Q

, where Q is used to ensure the reality

of 1c . Hence2' 1 '''f f (8.27a)

where the +/- sign corresponds to Q positive/negative, ie., to in/out flow.

One of the boundary conditions associated with (8.27a) was already found to be

0 0f

Another one should be provided by the no-slip condition

0u v for all 0y

Since0

0y

v is automatically satisfied by (8.27), we get only one new condition

' 0 0f

The 3rd boundary condition is obtained by merging the solution with the mainstream

flow:

y

Qu

x

ie.,

' 1f

Setting

'F f

eq.(8.27a) can be transformed into a 2nd order eq.,2 1 ''F F (8.28)

with boundary conditions

0 0F 1F (8.29)

Summary

Taking stock of progress so far, we have

g xQ

Q y

x

Q f (8.27)

'Q

u fx

2

' 'QQg

v f fx x

2' ''

u Qf f

x x

2

''Qu Q

fy x

'F f

where the +/- sign corresponds to in/out flow.

The ODE to solve is

2'' 1 0F F

with boundary conditions

0 0F ; 1F

Solutions

Using

21'' '

2

dF F

dF

the ODE becomes

2 21' 1 0

2

dF F

dF

which is integrated to

2 32

1 1'

2 3F F F c

(8.29a)

From the mainstream flow (8.25a), we have

2rdu Q

dr r

so that

20

r

y

du Q

dr x

The boundary layer quantity that should merge to this is

2 2'

u Q QF F

x x x

Since 1F , this means

' 0F

Substituting back to (8.29a) gives

2

2

3c

so that (8.29a) becomes

2 31 1 2' 0

2 3 3F F F

which can be rearranged to give

22 33' 2 3 2 1

2F F F F F

For the outflow case, we have

3' 2 1

2F F F

Since 0 0F , the right hand side is imaginary at 0 . Hence, there is no

solution that can satisfy the required boundary conditions.

For the inflow case, we need to solve

3' 2 1

2F F F (8.29b)

To get rid of the square root, we set2 2G F

so that

2 ' 'GG Fand (8.29b) becomes

232 ' 3

2GG G G

or

21' 3

6G G

Integrating gives

132

1tanh

36 3 3

dG Gc

G

or

43 tanh2

G c

243tanh 2

2F c

(8.30)

where the ic ’s are constants.

Now, (8.30) satisfies 1F automatically. To satisfy 0 0F , we need

14

2tanh 1.14

3c

That 4c can have 2 possible values is an example of the non-uniqueness of flow at

high R. [see §9.7]

The flow pattern for 4 1.14c is shown in fig.8.9 and is typically observed in

experiment.

The flow pattern for 4 1.14c involves reversed flow close to the wall.

Using the fact that 1 1F , the layer thickness δ can be estimated from

1Q

x

ie.,

xQ

(8.31)

which decreases as one approaches the corner.

8.5 Rotating Flows Controlled By Boundary

Layers

8.5.1 Almost Uniform Rotation: Basic Eqs.

8.5.3 Ekman Boundary Layers

8.5.4 The ‘Interior’ Flow

8.5.1 Almost Uniform Rotation: Basic Eqs.

The time rate of change of a vector Q as seen from a fixed and a rotating frame are

related by

F R

d d

dt dt

Q Q Ω Q

where is the angular velocity vector of the rotation.

Setting Q x , we have

F R u uΩ x

Taking the time derivative of the whole eq. gives

F R

F F F

d d d

dt dt dt

u u xΩ

RR

R R

d d

dt dt

u xΩ u Ω Ω Ω x

2RR

R

d

dt

u Ω u Ω Ω x

For fluid motion,d

dtis replaced with

D

Dt. Thus, we have

2F RF F R R R

F Rt t

u u

u u u u

Ω u Ω Ω x

where the 3rd and 4th terms on the right are the Coriolis and centrifugal acceleration,

respectively.

[For a more rigorous derivation, see R.E.Meyer, “Introduction to Mathematical Fluid

Mechanics”, Chap 5, Wiley (71)]

At a given instance of time, quantities in the 2 frames are related by simple coordinate

transformations. Hence, provided the 2 frames are kept in the same orientation, we

have

F R

Q Q

Dropping the subscript R for clarity, the Navier-Stokes eqs in the rotating frame are

simply

212 p

t

u

u u

Ω u Ω Ω x u(8.34)

0 u

Using

a b c d a c b d a d b c

we have

22 2 Ω x Ω x Ω x Ω x

Using

a b a b b a a b b a

we have

21

2 a a a a a [cf. 21

2f f f ]

Ω x Ω x Ω x Ω x Ω

22 2 Ω x Ω x Ω x Ω x Ω

2 2 2 2 2 x x x x x x x x

where we have used

0kijk ijk kj ijji

j

x

x

x

ij j ij ii

j

x

x

Ω x

ij j ij ii

j

xx x x

x

x x

Thus

2 22 2 Ω x Ω x Ω x

22 2 2 Ω x Ω x Ω Ω x Ω

or

21

2 Ω Ω x Ω x (8.35)

Thus, for const , (8.34) becomes

2*

12 p

t

u

u u

Ω u u(8.34a)

where

2

*

1

2p p Ω x (8.36)

which contains the centrifugal forces, is called the reduced pressure.

With the understanding that only reduced pressures will be used in the following, we

can drop the subscript * for the sake of clarity.

Let

U be typical value of u ,

L be typical length scale of flow.

We are interested in cases where the flow u is small compared to the rotation of the

system, ie.,

1U

L

(8.36a)

Now,

2U

OL

u u

O U Ω u

Eq.(8.36a) therefore implies

1

UO

L

u u

Ω u

Eq(8.34a) thus simplifies to

212 p

t

u Ω u u (8.37)

0 u (8.38)

The relevant Reynolds number is2L

R

[notUL

]

and the flow will be considered inviscid if 1R .

Let the coordinate system attached to the rotating frame be , ,x y z with the z-axis

parallel to the axis of rotation, so that 0,0, Ω .

With , ,u v wu , we have

0 0 , ,0v u v u

u v w

i j k

Ω u i j

For a steady, inviscid flow, Eq(8.37) becomes

12

pv

x

(8.39)

12

pu

y

(8.40)

10

p

z

(8.41)

while (8.38) becomes

0u v w

x y z

(8.42)

From (8.41), we see that p is independent of z.

So are u and v by way of (8.39-40).

Eq(8.42) then put w also independent of z.

Thus, u is independent of z, which is known as the Taylor-Proudman theorem.

8.5.3 Ekman Boundary Layers

Consider the steady flow between 2 boundaries that rotate with slightly different

angular velocities.

In the rotating frame attached to 1 of the boundaries, the fluid flow is obviously small

so that the almost uniform rotation eqs. (8.37-38) apply.

We assume the flow consists of 2 components.

One is the inviscid interior (mainstream, high R) flow that obeys eqs(8.39-42).

The other is the so called Ekman boundary layer flow for which the viscous term

2 u can be approximated by 2 u .

Let the boundaries be planes perpendicular to the z-axis.

The angular velocities of the boundaries at 0z and z L are Ω and 1 ,

respectively. As stated earlier, 1 .

Obviously,2

22z

u

u (8.43a)

Consider 1st the boundary layer on 0z .

Borrowing the results from §8.5.2, and the help of (8.43a), eqs. (8.37-38) become2

2

12

p uv

x z

(8.43)

2

2

12

p vu

y z

(8.44)

2

2

10

p w

z z

(8.45)

0u v w

x y z

(8.46)

The arguments presented in §8.2 can be applied here by the transformations

,x x y and y z

Thus, (8.4a) becomes

,u u u

z x y

,

v v v

z x y

(8.4a’)

To get an order of magnitude estimates, let 0U be some typical value of u and v.

Let the distance for u or v to change by an amount 0U be of order L in the x-y plane,

and δ in the z-direction. Hence, (8.4a’) means

0 0U U

L

ie.,

L (8.4)

Eq. in (8.46) means

0w u v U

z x y L

so that w is of order 0U

L

. Hence ,w u v .

Eq(8.43-5) then imply

orp p p

z x y

so that we may write ,p p x y in the boundary layer.

Hence,p

x

andp

y

remain close to their inviscid ‘interior’ values, which obey

12 I

I

pv

x

(8.39)

12 I

I

pu

y

(8.40)

The boundary layer eqs(8.43-4) can therefore be written as

2

22 I

uv v

z

(8.47)

2

22 I

vu u

z

(8.48)

8.47 8.48i gives

2

22 I I

u ivv iu v iu

z

2 I Ii iv u iv u

Setting

I If u iv u iv

we have2

22

fi f

z

(8.48a)

Consider the ansatza zf e

where a is a constant. Eq(8.48a) is satisfied if2 2a i

or

/ 42 2 ia i e

[ / 2ii e ]

2cos sin 1

4 4i i

Setting

*z z

the general solution to (8.48a) is

* *1 1i z i zf Ae Be

where A and B are arbitrary functions of x and y.

Hence

I Iu iv u iv f

* *1 1i z i zI Iu iv Ae Be

To match the interior flow, we need

0f as z

which requires 0B , so that

*1 i zI Iu iv u iv Ae

The no-slip condition 0u v at the wall 0z gives

0 I Iu iv A

so that

*11 i zI Iu iv u iv e

** *1 cos sinz

I Iu iv e z i z

Equating real and imaginary parts, we have

* ** *1 cos sinz z

I Iu u e z v e z

* ** *1 cos sinz z

I Iv v e z u e z

or

** *cos sinz

I I Iu u e u z v z (8.49)

** *cos sinz

I I Iv v e v z u z (8.50)

Taking the partials

** *cos sinzI I Iu u u v

e z zx x x x

** *cos sinzI I Iv v v u

e z zy y y y

and putting them into the incompressibility condition (8.46), we have

w u v

z x y

* ** *1 cos sinz zI I I Iu v u v

e z e zx y y x

From (8.39-40), we have21

2I Iv p

y x y

21

2I Iu p

x x y

so that

0I Iu v

x y

and

**sinzI Iw u v

e zz y x

or

**

*

sinzI Iw u ve z

z y x

With* 0

0z

w , we have

*

** *

0

sinz

zI Iu vw e z dz

y x

At the edge of the Ekman layer where *z , we have

** *

0

sinzI IE

u vw e z dz

y x

Now,

** *

0

sinzI e z dz

1

00

1 1 1Im Im Im

1 1 2i zz ize dz e

i i

so that

1

2I I

E

u vw

y x

(8.51)

1

2 I

where I is the z-component of the vorticity of the interior flow.

If the boundary is rotating with angular velocity B relative to the rotating frame,

(8.51) is generalized to

1

2E I Bw

(8.52)

Similarly, if the upper boundary at z L rotates with angular velocity T relative

to the rotating frame, we have

1

2E T Iw

(8.53)

8.5.4 The ‘Interior’ Flow

According to §8.5.2, Iu , Iv , and Iw are all independent of z.

This means Ew at the top and bottom of the interior flow must match.

Eqs(8.52-3) thus imply

1 1

2 2I B T I (8.54)

or

I T B

In the case of Fig.8.10,

0B , T

so that

I II

v u

x y

or, in terms of cylindrical coordinates , ,r z ,

1

0 0

r z

I

I

r

r r zru

e e e

ω

1z I z

dru

r dr e e

where we’ve used the fact 0Iu

z

.

Integrating, we have

21

1

2Iru r c

or

11

2I

cu r

r

If Iu is to be regular at 0r , we must set the constant 1 0c , so that

1

2Iu r

Thus, the angular velocity of the fluid is the average of those of the boundaries.

In other words, the motion of the fluid is entirely controlled by the boundary layers.

We now obtain the rest of the solution. Using (8.51), we have

1 1

2 2zI Iu

The incompressibility condition in cylindrical coordinates

1 10I zI

rI

u uru

r r r z

then reduces to

10rIru

r r

which gives

2rIru cHowever, if rIu is to be regular at 0r , we must set the constant 2 0c , so that

0rIu

The secondary flow is therefore purely in the z direction. (see fig.8.10).

Consider the situation depicted in fig.8.11.

Initially, the fluid, together with its boundaries at z L , rotate about the z-axis with

angular velocity 1 .

At 0t , the angular velocities of both boundaries are reduced to Ω.

Obviously, the fluid will eventually ‘spin-down’ to the same angular velocity.

What interests us is the relevant time-scale.

Physically, we expect the spin down to begin with the formation of Ekman layers on

both boundaries. At this point, the interior inviscid flow still rotates essentially with

angular velocity 1 . As is the case in §8.5.4, Iu and Iv are independent of z

but not Iw . In fact, according to eqs(8.52-3), 0Iw ( 0Iw ) near the lower

(upper) Ekman layer.

Thus, as time goes by, the Ekman layers extend towards the interior flow, which is the

essence of the spin-down process.

The almost uniform rotation eqs(8.37-8) for the inviscid interior flow are

12I I

I

u pv

t x

(8.56)

12I I

I

v pu

t y

(8.57)

1I Iw p

t z

(8.57a)

0I I Iu v w

x y z

(8.46)

Eliminating p from (8.56-7) gives

2 0I I I Iu v v u

t y x y x

which, with the help of (8.46), becomes

2 0I I Iu v w

t y x z

(8.58)

As discussed earlier, 0Iw

z

so that the vorticity in the interior decreases with time:

0I I Iu v

t t y x

in agreement of the Helmholtz vortex theorem (§5.3).

Now, Iu and Iv are independent of z so that (8.58) can be integrated over the

vertical span of the interior fluid to give

2 0IIL w

t

(8.58a)

where L is the height of the interior fluid and Iw is the difference of Iw at its top

& bottom.

By eq(8.51),

1

2I Iw

ontop

bottom

of thelower

upper

Ekman layer (8.59)

Hence, (8.58a) becomes

2 0IIL

t

(8.60)

with solution

/exp 2 t TI A t Ae

L

where A is an arbitrary function of ,x y and

2

LT

(8.61)

is called the spin down time.

Applying the initial condition

2I at 0t

we have

2 A and

/2 t TI e

In terms of cylindrical coordinates , ,r z and assuming axisymmetry, we have

1

0 0

r z

I

I

r

r r zru

e e e

ω

/12 t T

z I z

dru e

r dr e e

which is easily integrated to give

2 /1

t TIru r e c

Regularity at 0r then sets 1 0c so that

/t TIu r e (8.62)

which decribes the main ‘dissipation’ process of the excess rotation.

The other components of Iu are obtained from (8.58) and (8.55) as

/t LrI

ru e

L

/2 t LzI

zu e

L (8.63)

[cf. the last part of §8.5.4]

The streamlines of this secondary flow are described by

2zI

rI

dzdz u zdt

drdr u rdt

which integrates to

ln 2 lnz r const or

2zr const(see Fig.8.11).

8.6 Boundary Layers Separation

9. Instability

9.1 The Reynolds Experiment

9.2a Interface Waves

9.2 Kelvin-Helmholtz Instability

9.3 Thermal Convection

9.4 Centrifugal Instability

9.5 Instability of Parallel Shear Flow

9.6 General Theorem on Stability of Viscous

Flow

9.7 Uniqueness & Non-Uniqueness Of Steady

Viscous Flow

9.8 Instability, Chaos, And Turbulence

9.9 Instability At Very Low Reynolds Number

9.1 The Reynolds Experiment

9.2 Kelvin-Helmholtz Instability

Consider a deep layer of inviscid fluid of density 2 moving with uniform speed U

over another stationary deep layer of density 1 2 . [see Fig.9.4]

Let the interface be ,y x t .

A small traveling wave disturbance there can be described by

, i k x tx t Ae (9.2)

with the understanding that only the real part of any complex quantity has physical

meaning.

The dispersion relation is (detailed derivation is given in §9.2a),

2 2 21 2 2 1 2 1 2 1 2kU k U k g k (9.3)

where α is the coefficient of surface tension.

This can be written as

R Ii

where

2

1 2R

kU

2 2 21 2 1 2 1 2

1 2

1I k U k g k

Thus, ω is complex if I is real, ie.,

2 2 21 2 1 2 1 2 0k U k g k

or

2

1 21 2

1 2

U gk

k

(9.4)

in which case, the dispersion indicates an exponential growth with time of any

disturbance. This is known as the Kelvin-Helmholtz instability.

The minimum flow CU that can induce such an instability is given by

2

1 21 2

1 2

minC

k

U gk

k

Setting

1 2

gf k k

k with 0k

we have

1 22

df g

dk k

The extremum of f are therefore at

1 2C

gk

with

1 22Cf k g

which is obviously a minimum.

Hence

2

1 21 2

1 2

2CUg

(9.6)

which says that both gravity & surface tension serves to stabilize the system by

raising the value of CU .

Kelvin-Helmholtz instability can also occur in a continuously stratified fluid with

0 0d

dy

The buoyancy frequency [see §3.8]

2 0

0

g dN

dy

(9.7)

then serves as a measure of the stabilizing effects of the density distribution.

We leave it as an exercise (Ex.9.2) to show that the system becomes unstable only if

the Richardson number

2

2/

NJ

dU dy (9.8)

is less than1

4somewhere in the flow.

Such instabilities are observed in the atmosphere, sometimes in the form of ‘clear air

turbulence’ and sometimes marked by distinctive cloud patterns.

9.2a Interface Waves

We now generalize the surface waves results in §§3.2-4 to waves at the interface of 2

fluids.

All quantities related to the upper and lower fluids are labeled 2 and 1, respectively.

Mechanical equilibrium in the unperturbed state then requires 1 2 .

The interface is at ,y x t .

For ease of reference, we list below the basic eqs governing surface waves.

0ut x y

on ,y x t (3.18)

21

2

pgy G t

t

u (3.19)

2 2

2 20

x y

(3.24)

Eqs(3.18,19) are linearized as:

t y

on 0y (3.21)

pgy G t

t

(3.19a)

with the boundary condition

0p p on y and setting

0pG t

Eq(3.19a) becomes

0gt

on 0y (3.22)

For the interface waves, eqs(3.21,19a,24) become

1

t y

2

t y

on 0y (3.21a)

1 11

1

pgy G t

t

(3.19b)

2 22

2

pgy G t

t

(3.19c)

2 21 1

2 20

x y

(3.24a)

2 22 2

2 20

x y

(3.24b)

The boundary conditions are

1 0y

y

2 0

yy

(A1)

1 2p p p , on y (A2)

and from (3.21a),

1 2

y y

on 0y (A3)

Consider the ansatz

i k x tAe (3.23)

1 1

i k x tf y e 2 2

i k x tf y e (B)

To satisfy (3.24a,b) and (A1), we need

21 1'' 0f k f 2

2 2'' 0f k f

with the boundary conditions

1 0f 2 0f

Thus

1 1k yf c e 2 2

k yf c e

where 1 2,c c are constants.

The interface condition (A3) then demands

2 2c c C

so that (B) becomes

1

k y i k x tCe 2

k y i k x tCe

Eq(3.21) gives

kC i A (3.21b)

1

1

i k x t pi C gA e G

22

i k x t pi C gA e G

(C)

Since 1 2,G G are functions of t only, they must vanish identically.

Consistency then demands

i k x tp Pe

where P is a constant, so that (C) become

1

Pi C gA

2

Pi C gA

Eliminating P gives

1 2 1 2 0i C gA (3.21c)

Eq(3.21b,c) are compatible only if

1 2 1 2

0k i

i g

ie.,

21 2 1 2 0kg

or

2 1 2

1 2

kg

1 2

1 2

kg

The phase velocity is therefore

1 2

1 2p

gv

k k

and the group velocity:

1 2

1 2

1 1

2 2g p

d gv v

dk k

9.2a.1 Surface Tension

The effect of the surface tension at the interface is to modify the interface condition

(A2) to2

2 1 2p p

x

at ,y x t

where α is the coefficient of surface tension. Thus, eq(C) is modified into

11

1

i k x t pi C gA e G

and

22 1

2

1i k x t i k x ti C gA e G p k Ae

As before, 1 2 0G G and

1

i k x tp Pe

so that

1

Pi C gA

and

2

2 2

k Pi C g A

Eliminating P gives

21 2 1 2 0i C A g k

which is compatible with (3.21b) only if

21 2 1 2

0k i

i g k

ie.,

3 21 2 1 2 0k kg

or

32 1 2

1 2 1 2

kkg

9.2a.2 Uniform Flow

We now consider the case where the flow of the upper fluid tends to a uniform speed

U sufficiently far away from the interface.

To emphasize this fact, we write

1 11 1 1, ,u v

x y

u 2 22 2 2, ,U u v U

x y

u

where

1 1 2 2 0u y v y u y v y

The linearized version of eqs(3.18) is

1

t y

2Ut x y

on 0y (3.21’)

which combines to give an interface condition

1 2 Uy y x

on 0y

to supplement2

2 1 2p p

x

at ,y x t

The linearized version of eqs(19,24) are

1 11

1

pgy G t

t

(3.19’)

22 2 22

2

1

2

pU U gy G t

t x

(3.19’’)

2 21 1

2 20

x y

(3.24a)

2 22 2

2 20

x y

(3.24b)

Consider the ansatz

i k x tAe (3.23)

1 1

i k x tf y e 2 2

i k x tf y e (B)

The solutions to (3.24a,b) are still

1 1

k y i k x tC e 2 2

k y i k x tC e

Eq(3.21’) gives

1 0i A kC

2 0i kU A kC (3.21b’)

Eq(3.19’,’’) become

11 1

1

i k x t pi C gA e G

2 22 2 2 1

2

1 1

2i k x t i k x ti C ikUC gA e G U p k Ae

(C’)

Since 1 2,G G are functions of t only, we must have

1 0G and 22

1

2G U

Consistency also demands

1

i k x tp Pe

where P is a constant, so that (C’) become

11

Pi C gA

2

22 2

k Pi kU C g A

Eliminating P gives

21 1 2 2 1 2 0i C i kU C g k A (3.21c’)

which becomes, with the help of the 1st eq in (3.21b’),

2

2 12 2 1 2 0i kU C g k A

k

Eq(3.21b’,c’) are compatible only if

2

2 12 1 2

0k i kU

i kU g kk

ie.,

23 21 2 1 2 0kg k kU

or

2 2 2 31 2 2 2 1 22 0kU k U kg k [cf. Ex.3.6]

so that

2 2 2 2 2 32 2 1 2 2 1 2

1 2

1kU k U k U kg k

2 2 22 1 2 1 2 1 2

1 2

1kU k U k g k

or

2 2 21 2 2 1 2 1 2 1 2kU k U k g k (9.3)

9.3 Thermal Convection

Consider a viscous fluid resting between 2 horizontal rigid boundaries at 0z and

z d . A temperature difference T is maintained between the boundaries, with

the lower one being hotter. If the density of the fluid decreases with rising

temperature, the fluid becomes top-heavy. Nonetheless, if T is increased from 0

slowly by small steps, the fluid can remain stable up to a critical value whereupon an

organized cellular motion sets in. (see fig.9.6)

9.3.1 Linear Stability Theory

9.3.2 Stability to Finite-Amplitude Disturbances

9.3.3 Experimental Results

9.3.1 Linear Stability Theory

For small temperature changes, the fluid density ρ at temperature T can be

approximated by

1 T T (9.9)

where α is the volume coefficient of thermal expansion, and the density at

temperature T .

Since the change in ρ is usually slight, the fluid is still approximately incompressible:

0D

Dt

and 0 u (9.10)

Assuming the viscosity to be temperature independent, the conservation of

momentum still takes the form of the Navier-Stokes eqs.

2Dp

Dt

uu g (9.11)

Thermal conduction in a fluid is a rather complicated process. For an introductory

discussion, see Landau & Lifshitz, “Fluid Mechanics”, chapter V.

Here, we simply generalize the elementary heat diffusion eq.

2TT

t

to include convective effects, ie.,

2DTT

Dt (9.12)

Here, κ is the thermal diffusivity of the fluid. It is related to the thermal conductivity χby

pc

where pc the specific heat at constant pressure.

What (9.12) describes is a situation where the change of the amount of heat inside a

fluid element is due solely to the exchange of heat by conduction with the surrounding

fluid. Neglected are all other energy sources (eg., work done on fluid element by

stresses exerted by the surrounding fluid) and sinks (eg., energy dissipation due to

viscosity).

In the unperturbed state of rest, the temperature 0T z must satisfy the steady state,

no flow version of (9.12), namely2

02

0d T

dz (9.12a)

of which solution is

0 1 2T z c z c

Using the boundary conditions (see Fig.9.6)

0 0 lT T and 0 lT d T T

we have

2lT c and 1 2lT T c d c

which is easily solved to give

0 l

zT z T T

d (9.13)

The corresponding density distribution eq(9.9) thus becomes

0 01z T z T (9.14)

with the accompanying hydrostatic pressure is given by the static, stationary version

of (9.11):

000

dpz g

dz (9.15)

Consider a slight perturbation to the system such that

0 1T T z T 0 1z

0 1p p z p 1 1 1 1, ,u v w u u (9.16)

where the perturbations, distinguished by the subscript 1, are assumed small and

functions of , , ,x y z t .

Eqs(9.9) thus become

0 1 0 11 T T T

which, with the help of (9.14) becomes

1 1T (9.17)

Eqs(9.10) is now

11 1 0

t

u

1 0 u (9.18)

Eq(9.11) is

210 1 0 1 0 1 1

Dp p

Dt

uu g

which, with the help of (9.15) becomes

210 1 1 0 1 1 1

Dp

Dt

uu g

which in term can be linearized to

210 1 0 1 1p

t

u

u g (9.19)

Finally, eq(9.12) becomes

20 1 0 1

DT T T T

Dt

which, with the help of eqs(9.12a,13), becomes

20 11 1

dT DTw T

dz Dt

which in term can be linearized to

211 0 1'

Tw T T

t

(9.20)

where, according to (9.13)

00 '

dT TT const

dz d

For small density changes, 0 z in eq(9.19) may be approximated by the constant

. Eqs(9.17-20) then become a set of PDE’s with constant coefficients.

We now aim to obtain an equation involing 1w alone.

To begin, we eliminate 1p in (9.19) by taking the curl of the whole equation:

21 1 1t

u g g

Using (9.17) to replace 1 , we have

21 1T

t

u g

Taking the curl again and using the vector identities

2 a a a

a b b a a b a b b a

we have

2 2 21 1 1T T

t

u g g (9.20a)

where we’ve also used (9.18).

Since 0,0, g g , the z-component of (9.20a) is

2

2 2 211 12

Tw g g T

t z

2 21 1

2 2

T Tg

x y

(9.21)

Rewriting (9.20) as

21 1 0 'T w T

t

1T can be eliminated from eq(9.21) by operating on the whole equation with

2

t

, so that

2 22 2 2 2

1 12 2w g T

t t x y t

2 2

0 12 2'gT w

x y

(9.22)

Since derivatives of x and y are always in the combination2 2

22 2x y

, we see

that 1w is isotropic in the x-y plane. The independent variables thus fall into 3

groups: t, ,x y , and z. A separable ansatz thus takes the form

1 ,w W z f x y g t (9.23)

Now, a successful separation means that we can write

F

Fc

F

D(9.23a)

where F is any one of the separated functions, Fc a constant, and D an ordinary

differential operator that involves only functions and derivatives of the independent

variables of F.

The complicated form of (9.22) means that the most general form of separation is not

easily achieved. However, the situation will be greatly simplified if the operators

D for 2 of the separated functions take the simplest possible forms. Since the

primary changes are in the z direction, we should simplify f and g so that

2 21f a

f (9.24)

1 dgs

g dt (9.24a)

where the peculiar form of the proportionality constant for f is chosen to conform with

Acheson’s notation, as well as for future convenience.

Note that (9.24a) can be solved immediately to give

1s tg t c e

One useful property of eq(9.23a) is

11 1F

ww F c w

F D D (9.23b)

Letd

Ddz

..

With the help of eqs(9.24, 24a, 23b), we have,

2 2 21 1w D a w

so that eq(9.22) becomes

2 2 21w

t t

2 2 2 2 2 21s D a s D a D a w

20 1'gT a w

Cancelling fg from both sides gives

2 2 2 2 2 2 20 's D a s D a D a W gT a W (9.25)

Eq(9.25) is a 6th order ODE. Its general solution thus contains 6 arbitrary constants.

However, since the equation is homogeneous, only 5 of them are independent. (The

6th, taken as the overall constant multiplication factor, is immaterial).

Since the boundary conditions always appear in pairs (1 each for the upper & lower

boundary), the integration constants and boundary conditions can never be matched

entirely by themselves. As will be shown later, there are 6 boundary conditions. We

therefore have an eigenvalue problem whereby a solution satisfying all the boundary

conditions exists only for some special values of the parameters in (9.25).

The no-slip condition gives us

1 1 1 0u v w (BC1)

1 1 1 0u v w

t t t

(BC2)

1 1 1 0n n n

n n n

u v w

x x x

(BC3)

1 1 1 0n n n

n n n

u v w

y y y

(BC4)

for all n, at 0,z d .

Since the boundaries are kept at fixed temperatures, we have

1 1 11 0

n n

n n

T T TT

t x y

(BC5)

for all n, at 0,z d .

We must now translate these conditions into ones on W so that they can be applied to

(9.25).

(BC1) together with (9.23) gives us

0W at 0,z d . (BCa)

(BC3,4), with 1n , and the incompressibility condition (9.18) gives us

0DW at 0,z d . (BCb)

(BC5), with 2n , gives

2 2

12 20T

x y

on 0,z d

so that (9.21) becomes

2 21 0w

t

on 0,z d

ie.,

2 2 2 2 0s D a D a W at 0,z d

On expansion, we get

4 2 2 22 0D a s D a W at 0,z d

which, on applying (BCa), becomes

4 2 22 0D a s D W at 0,z d (BCc)

Eqs(BCa-c) thus provide 6 boundary conditions to (9.25).

Now, (9.25) is an ODE with constant coefficients.

Its solution is therefore of the formze

which, on substituting into (9.25), gives

2 2 2 2 2 2 20 's a s a a gT a (9.25a)

For given values of a and s, this is a 3rd degree polynomial of 2 .

Let the roots be 2i , 1,2,3i .

The general solution to (9.25) is

3

1

i iz zi i

i

W Ae B e

As mentioned before, only 5 of the 6 constants ,i iA B are independent.

The determination of , ,i iA B s so that the boundary conditions as well as (9.25a)

are satisfied is a straightforward but tedious task which is not particularly

illuminating.

For illustrative purposes, it may be more fruitful to attack another related problem

where the boundary conditions are simplified to2 4 0W D W D W at 0,z d (BC)

which is easily seen to be satisfied by the ansatz

sinN z

Wd

1,2,3,N

The secular equation for the eigenvalue s is simply (9.25a) with2

2 N

d

.

Setting2

2 2 2 2*

Na a a

d

(9.25a) becomes

2 2 2 2* * * 0 's a s a a gT a

or

2

2 2 4* * 0 2

*

' 0a

s a s a gTa

with solution

2

22 4 4* * * 0 2

*

14 '

2

as a a a gT

a

Using

0 'T

Td

we have

2

22 4* * 2

*

14

2

T as a a g

d a

Hence, for 0T , the factor inside the square root is always positive so that s is

always real.

Furthermore, if

2

2 24 4* *2

*

4T a

(9.26a)

we have 0s , so that the perturbation grows exponentially with time and the

system is unstable.

Eq(9.26a) can be simplified as2

4*2

*

T ag a

d a

36 2 22*

2 2 2

1g T a Na

d a a d

(9.27b)

Now, a is not a parameter related to the system. It is simply some unknown constant

related to the horizontal length scale via eq(9.24). Hence, (9.27b) is satisfied as soon

as the left side is greater than the minimum of the right side with respect to both a and

N. The minimum with respect to N can be found by inspection to be at 1N . That for

a satisfies

3 22 2

2 23 2 2 2

2 32 0a a a

a d a d

or

22 2

23 0a a

d

22

22a

d

so that the instability criterion is

32 2 4

2 2 4

2 3 27

2 4

g T d

d d d

which, in terms of the Rayleigh number3g Td

R (9.27)

is simply427

4

R

For the boundary conditions (BCa-c), the instability criterion is

1708Rwhich corresponds to 3.1/a d .

9.3.2 Stability to Finite-Amplitude Disturbances

We have just shown that a critical value of T exists, above which disturbances

grow exponentially with time so that even infinitesimal ones may become finite as

t .

A related question is whether a critical value of T also exists, below which any

disturbances die out eventually.

In our idealized system of thermal convection, the answer is yes and the 2 critical

values are equal. Thus, for 1708R , all disturbances subside eventually and the

9.3.3 Experimental Results

Criteria for the onset of instability as calculated from the linear theory are usually in

good agreement with experiment. However, after the system become unstable, the

predicted exponential growth of the disturbances cannot be substained indefinitely

since non-linear effects will eventually become substantial enough to halt the growth.

The system then reaches a steady state which cannot be described by the linear theory.

With respect to the thermal convection discussed earlier, this means only the critical

value cR of the Rayleigh number agrees with observation.

Other interesting observations that are beyond the reach of the linear theory include:

1. Convection patterns such as the 2-D rolls and hexagonal cells in the x-y plane

(see fig.9.7).

2. Shifts of the unperturbed steady state convection to other, often time- dependent,

patterns as one continues to increase R to values well beyond cR .

3. Effects due to variation of surface tension in case of a free upper surface.

(Benard instabilities).

9.4 Centrifugal Instability

Consider 2 concentric, rotating, circular cylinders with the gap between them filled

with a viscous fluid.

Let the inner and outer cylinders, together with their associated quantities, be labeled

1 and 2, respectively.

As discussed in §2.4.2, the steady flow

Bu Ar

r (2.31) (9.28)

with2 2

1 1 2 22 2

1 2

r rA

r r

2 2

1 21 22 2

2 1

r rB

r r

(2.32) (9.29)

is an exact solution to the Navier-Stokes eqs that satisfies the no-slip boundary

condition.

If the cylinders rotates in the same sense, instability occurs if 1 c , where c

is some critical value that depends on 2 1 2, ,r r , and ν. [see eq(9.41)]. The ensuing

flow consists of counter-rotating Taylor vortices superimposed on the main rotary

flow. [see Fig.9.8]

9.4.1 Linear Stability Theory

9.4.2 Inviscid Theory: the Rayleigh Criterion

9.4.3 Experiments

9.4.1 Linear Stability Theory

Since the flow is axisymmetric, all quantities related to the unperturbed, steady

floware functions of r only.

Small quantities describing the perturbation will be distinguished by a prime. They

are assumed to be functions of r, z and t.

For example, we write

', ', 'r zu U u u u (9.30)

0 'p p p

The Navier-Stokes eqs. in cylindrical coordinates were derived in §2.4:

2

22 2

1 2r rr r

u u p u uu u

t r r r r

u

22 2

1 12r ru u u p u

u u ut r r r r

u (2.22)

21zz z

u pu u

t z

u

where

r zf u u u fr r z

u

2 22

2 2 2f r f

r r r r z

The incompressibility condition is:

0zr

u uru

r r r z

u

0 U r u e 0 0p p r

201U dp

r dr

2

10

d dUr U

rdr dr r

(2.30)

Note that

0 Ur

u

so that

0 0 0 0r zu u u u u u

Using (2.30) to cancel out the unperturbed parts, eq(2.22) become

2 22

' 1 1 ' '' ' 2 ' ' 'r r

r r

u p uu U u u u

t r r r

u

22

' '' 1' ' ' ' 'r

r

u U uu dUu u u u

t dr r r

u

2' 1 '' ' 'z

z z

u pu u

t z

u

and

'' 0z

r

uru

r r z

u

where2 2

22 2r z

The linearized version is

22

' 2 ' 1 ' ''r r

r

u U u p uu

t r r r

22

' 1' ' 'r

u dU Uu u u

t dr r r

2' 1 ''z

z

u pu

t z

(9.31)

and

'' 0z

r

uru

r r z

u

To simplify the mathematics, we assume the gap between the cylinder to be small, ie.,

2 1 1d r r r

Now

22

'' r

r

uu O

d 2 2

1

' 'r ru uO

r r

so that

22

'' r

r

uu

r

and similarly for 'u .

Eq(9.31) thus simplify to

2 2 ' 1 ''r

U u pu

t r r

(9.33a)

2 ' ' 0r

dU Uu u

t dr r

(9.33b)

2 1 ''z

pu

t z

(9.33c)

Using

1

' ' ' 'r r r ru u u uO O

r d r r

the incompressibility condition becomes

' '0r zu u

r z

(9.33d)

Finally, from (9.28), we see that

2 22

dU U B BA A A

dr r r r

so that (9.33b) is simplified to

2 ' 2 ' 0ru Aut (9.33b’)

Eliminating 'p between (9.33a,c) gives

2 ' ' 2 '0r zu u U u

t z r r z

Further eliminating 'zu using (9.33d), we have

2 2 22

2 2 2

' ' 2 '0r ru u U u

t z r r z

or2

2 22

2 '' 0r

U uu

t r z

(9.34)

Still further eliminating 'u using (9.33b’), we have

2 22 2

2

4 '' 0r

r

AU uu

t r z

(9.38a)

which is a 6th order linear PDE for 'ru alone.

Now, the coefficients of (9.38a) are all constants except for the termU

r .

Thus, for 1 2 , we can write

1 2

1

2

U

r

so that all the coefficients in (9.38a) are constants. In which case, we have

2 22 2

2

'' 4 0r

r

uu A

t z (9.38b)

We must now establish and then satisfy the relevant boundary conditions.

No-slip condition obviously requires

' ' ' 0r zu u u (BC1)

' '0

n nr rn n

u u

z

(BC2)

' '0

n n

n n

u u

z

' '0

n nz zn n

u u

z

for all n on the surfaces 1 2,r r r .

That is, only partials with respect to r can be finite.

The incompressibility condition then implies

'0ru

r

on 1 2,r r r (BC3)

From (9.34), we have

2 2 ' 0rut

on 1 2,r r r (BC4)

Since2 2

22 2r z

we have

2 2 2 2 4 4 42 2

2 2 2 2 4 2 2 42

r z r z r z r z

so that (BC4) becomes

2 4 4

2 4 2 22 ' 0ru

t r r r z

on 1 2,r r r (BC4a)

Our task is to find a solution to (9.38b) that satisfy (BC1,2,3,4a).

Eq(9.38b) is obviously separable.

Since the partials with respect to z are all of even order, a periodic solution can be

supported. Thus, we consider the ansatz

' coss tr ru u r e nz (9.35)

so that

''r

r

usu

t

22

2

''r

r

un u

z

2

22

'coss tr

r

ue nz D u

r

whered

Ddr

.

2 2 2' coss tr ru e nz D n u

Thus, (9.38b) becomes

22 2 2 2 24 0r rs D n D n u A n u

(9.38)

while the boundary conditions at 1 2,r r r simplify to

(BC1,2): 0ru (BC5)

(BC3): 0rDu (BC6)

(BC4a): 2 4 2 22 0rsD D n D u

ie., 4 2 22 0r

sD n D u

(BC7)

Since (9.38) is homogeneous and of even order, there is an odd number (=5) of

independent integration constants. The number of boundary conditions are even (=6).

Therefore, it is an eigenvalue problem for which a parameter, eg. s, can take on only

special values.

In general, s can be complex.

Let Rs s i , where both Rs and ω are real.

The time dependence of 'ru is then Rs t i te e .

Instability occurs if 0Rs .

The threshold to instability is therefore at 0Rs .

For non-oscillatory evolution, 0 , so that the threshold is at 0s .

The equations decribing this marginal state are

(9.38): 32 2 2 24 0r rD n u A n u (9.38a)

subject to boundary conditions at 1 2,r r r :

(BC5): 0ru

(BC6): 0rDu

(BC7): 4 2 22 0rD n D u (BC7a)

Shifting to a more natural coordinate defined by

1 1

2 1

r r r rx

r r d

we have

1dx d dD

dr dx d dx

Eq(9.38a) thus becomes

322 2 2

2 2

14 0r r

dn u A n u

d dx

which can be rearranged to give

32 2 6

2

2 2

40r r

d A n dnd u u

dx

On setting

a nd4

2

4A dT

it becomes

322 2

20r r

da u Ta u

dx

(9.38b)

subject to boundary conditions at 0,1x :

(BC5): 0ru

(BC6): 0rdu

dx

(BC7a):4 2

24 2

2 0r

d da u

dx dx

(9.40)

Using,2 2

1 1 2 22 2

1 2

r rA

r r

(2.32)

the Taylor number becomes3 2 2 3 2 2

1 1 2 2 1 1 2 22 2

1 2 1

4 2d r r d r rT

r r r

(9.41)

where the last equality made use of the narrow gap appoximation 1 2r r .

For a given a, eq(9.38b) and the boundary conditions (9.40) constitute an eigen

problem with eigenvalues T a . Let

minlT a T a for given a.

The threshold of instability corresponds to the minimum of lT ranging over all a.

Remarkably, eq(9.38b) & (9.40) take the same form as the thermal instability eqs

(9.25-6) with 0s . Hence, the threshold for centrifugal instability is also

1708T (9.42)

Likewise, the critical value of n is approximately3.1

d.

The streamlines of the secondary flow are shown in Fig.9.8.

Since the period of the flow in the z direction is2

n

, the height of each cell is

3.1

dH

n

Although the final steps of our derivation invoke the assumption 1 2 , eqs(9.41-2)

turns out to quite adequate whenever 1 2, are of the same sign.

Note that for 1 2, 0 , eq(9.41-2) implies the necessary condition for instability is

that 2r decreases sufficiently quickly with r.

9.4.2 Inviscid Theory: the Rayleigh Criterion

9.4.3 Experiments

9.5 Instability of Parallel Shear Flow

9.5.1 The Inviscid Theory

9.5.2 The Viscous Theory

9.5.3 Experimental Results

9.5.1 The Inviscid Theory

Consider the 2-D, inviscid, incompressible, flow between 2 flat plates at y L .

(see Fig.9.10).

The basic eqs are the Euler eqs

1u u u pu v

t x y x

(E1)

1v v v pu v

t x y y

(E2)

and the incompressibility condition

0u v

x y

(C1)

Consider the parallel shear flow

0 ,0U y u (9.46)

where U is an arbitrary function of y.

Substituting (9.46) into the basic eqs, we have

(E1):1

0p

x

(E2):1

0p

y

(C1): 0 0

Hence, 0u is a solution to the basic eqs with 0p p const .

Consider a general and presumably small 2-D disturbance.

All quantities related to the disturbance will be distinguished by the subscript 1.

They are in general functions of , ,x y t .

For example,

0 1 1 1' ,U y u v u u u

0 1p p p

The basic eqs becomes

1 1 1 11 1

1u u dU u pU u v

t x dy y x

(E3)

1 1 1 11 1

1v v v pU u v

t x y y

(E4)

1 1 0u v

x y

(C2)

of which the 1st two can be linearized to

1 1 11

1'

u u pU U v

t x x

(E5)

1 1 11v v pU

t x y

(E6)

where a prime denotes derivative with respect to y.

Since all coefficients in these eqs are independent of t and x, the natural ansatz is

1 , , i k x tf x y t f y e

(9.47)

where f can be u, v, or p. Thus

(E5): 1'i kU u U v ikp

(E7)

(E6): 1'i kU v p

(E8)

(C2): ' 0iku v (C3)

Eliminating p

from (E7,8) gives

' ' '' ' 'ikU u i kU u U v U v k kU v

Further eliminating of u

through (C3) gives

1' ' '' '' ' 'U v kU v U v U v k kU v

k

which can be simplified to

1'' '' 0kU v U k kU v

k

or

2'''' 0

kUv k v

kU

(9.48)

Boundary conditions to (9.48) are the impenetrable wall condition

0v at y L (9.49)

Eq(9.48) is therefore an eigenvalue problem for .

The following trick is due to Lord Rayleigh (1880).

Let the complex conjugate of v

be denoted by *v

.

* 9.48L

L

dyv

gives

2* 2'''' 0

L L

L L

kUv v dy v k dy

kU

(9.50)

Now, integration by part gives, with the help of (9.49),

2* * *'' ' ' ' 'L L L

L

LL L L

v v dy v v v v dy v dy

so that (9.50) becomes

2 2 2''' 0

L L

L L

kUv dy v k dy

kU

(9.50a)

Writing

R Ii

the imaginary part of (9.50a) is

2

2

''0

L

I

L

Uk v dy

kU

(9.51)

For 0I , this means

2

2

''0

L

L

Uv dy

kU

which is possible only if ''U changes sign somewhere in the interval.

Since instability results from exponential growth with time, which in turn requires

0I , this gives us the Rayleigh’s inflection point theorem, which says:

A necessary condition for the linear instability of an inviscid shear flow U y is

that ''U y changes sign somewhere in the flow.

9.5.2 The Viscous Theory

Consider now the case where the fluid is viscous.

The discussion in §9.5.1 is modified as follows.

The basic eqs are now the Navier-Stokes eqs instead of the Euler eqs.

2 2

2 2

1u u u p u uu v

t x y x x y

(NS1)

2 2

2 2

1v v v p v vu v

t x y y x y

(NS2)

and the incompressibility condition

0u v

x y

(C1)

With the same ansatz as (9.47), the counterpart of eqs(E7,8) are

21' ''i kU u U v ikp k u u

(NS7)

21' ''i kU v p k v v

(NS8)

while (C3) remains unchanged:

' 0iku v (C3)

7 'NS gives

21' ' '' ' ' ' ' '''ikU u i kU u U v U v ikp k u u

(NS7a)

8ik NS gives

21' ''k kU v ikp ik k v v

(NS8a)

7 8NS a NS a gives

' ' '' ' 'ikU u i kU u U v U v k kU v

2 2' ''' ''k u u ik k v v

Eliminating u

through (C3) gives

1' ' '' '' ' 'U v kU v U v U v k kU v

k

2'' '''' ''i

ikv v ik k v vk

which can be simplified to

1'' ''kU v U k kU v

k

2 4'''' 2 ''i

v k v k vk

(NS9)

Introduce the stream function

i k x ty e

so that

1uy

1v

x

and

'u v ik

Eq(NS9) becomes

'' ''i kU ik U k kU

2 4'''' 2 ''k k

which can be rearranged to give

2 4'''' 2 '' '' ''i k k kU k U k kU

2'' ''kU k kU (9.53)

In terms of ψ, the no-slip boundary conditions are

' 0 at y L (9.54)

We leave it as an exercise to show that for a plane Poiseuille flow defined as

2

max 21

yU y U

L

(9.55) (see Ex.2.3)

solutions of (9.53-4) leads to a curve of marginal stability shown in fig.9.11.

Of particular interest is the fact that instability occurs for a band of wavenumbers k

(slashed region in fig.9.11) if

max 5772U L

R

(9.56)

In contrast, the flow (9.55) is stable for all wavenumbers k in an inviscid flow.

Thus, viscosity plays a dual role.

According to (9.56), it is stabilizing since it raises the value of threshold value of

maxU to reach instability.

On the other hand, it is de-stabilizing because the flow would have been stable if

viscosity is absent altogether.

9.5.3 Experimental Results

Criterion (9.56) for the Poiseuille flow has been confirmed experimentally.

However, it applies only to systems with extremely low level of background

turbulence. Furthermore, non-linear effects are significant so that the criterion applies

only when the amplitudes of the disturbances are sufficiently low.

9.6 General Theorem on Stability of Viscous

Flow

Theorem:

Consider an incompressible viscous fluid occupying region V t that is enclosed

within a sphere of diameter L.

Let , tu x be a bounded solution to the Navier-Stokes eqs in V t satisfying the

boundary condition ,B tu u x on the boundary S t of V t . Thus, there exists

Mu such that

,Mu t u x for all V tx at all t.

Let * , tu x be another solution satisfying the same boundary condition but different

initial conditions at 0t .

The “difference flow”

* v u u (9.59)

thus satisfies

0v on S t (9.60)

The kinetic energy E of v defined as

21

2 V t

E dV v (9.61)

satisfies

2 22

0 2

3exp M

tE E u

L

(9.57)

where 0E is its initial value.

Thus if

3Mu LR

(9.58)

then

0E as t and the flow is unstable.

Before the actual proof of the theorem, we 1st establish the following relation:

2

1 i ij i

j jV t

dE v vv u dV

dt x x

(9.62)

9.6.1 Proof of Eq(9.62)

9.6.2 Proof of the Theorem

9.6.1 Proof of Eq(9.62)

Both u and *u are solutions of the Navier-Stokes eqs., therefore

21p

t

u

u u u

2** * * *

1p

t

u

u u u

Subtracting, we have

2* * *

1p p

t

v

u u u u v

Now,

* * * u u u v u v u u v u u v

so that by setting

*

1P p p

we have

2* P

t

v

v u u v v

or, in index notations,

2

*i i i i

j jj j i j j

v u v P vv u

t x x x x x

Multiplying by iv and sum over i gives

22 2

*

1 1

2 2i i

i j j i ij j i j j

u P vv v u v v

t x x x x x

v v (A)

We now try to put as many as possible the spatial derivatives in divergence form.

Thus

j jii j i i j i j i i

j j j j

u vvv v u v v u v u v

x x x x

j ii j i j

j j

u vv v u v

x x

where 0j

j

v

x

v

so that

j ii j i j i i j

j j j

u vv v v v u u v

x x x

(B)

Similarly

2 2

*2 2* * *

jj j j

j j j j

uu u u

x x x x

v v

v v (C)

where ** 0j

j

u

x

v

ii i i

i i i i

P v Pv P v P v

x x x x

(D)

i i i ii i

j j j j j j

v v v vv v

x x x x x x

so that

i i i ii i

j j j j j j

v v v vv v

x x x x x x

(E)

Puuting (C-E) into (A), we have

2 2*

1 1

2 2i i i i

i j i j i i i jj j j j j

v v v vv v u u v P v u v

t x x x x x

v v

Upon integrating with V t

dV the divergence term vanishes since

2*

1

2i

i j i j i ij jV t

vv v u u v P v dV

x x

v

2*

1

2i

i j i j i i ijS t

vv v u u v P v n dS

x

v

*

1

2i

j i i j i jjS t

vv u v u P v n dS

x

0 since 0v on S t

Hence

21

2i i i

i jj j jV t V t

v v vdV u v dV

t x x x

v

The convective (Reynold’s) theorem gives

2 2 21 1 1

2 2 2V t V t S t

ddV dV dS

dt t

v v v u n

21

2V t

dVt

v since 0v on S t

Thus

21

2i i i

i jj j jV t V t

d v v vdV u v dV

dt x x x

v

which is (9.62).

9.6.2 Proof of the Theorem

From

2

0iij

j

vA

x

for any ijA

we have

2

22 0i iij ij

j j

v vA A

x x

(9.63)

which holds even when, as will be assumed hereafter, summation over repeated

indices is implied.

Choosing i jij

u vA

, we have

2 2

2 0i j i ji i

j j

u v u vv v

x x

or

2

21

2 2i i

i j i jj j

v vu v u v

x x

On substituting into (9.62), we have

22 2

2i

i jjV t

dE vu v dV

dt x

Using

2 2 2 2

2 2i j M j M

V t V t

u v dV u v dV u E

we have

2

2 21

2i

MjV t

dE vu E dV

dt x

(9.64)

On the other hand, setting ij j iA h v in (9.63) gives

2

22i i

j i j ij j

v vh v h v

x x

which, upon using

2 22 jij i j i i

j j j

hvh v h v v

x x x

becomes

2

22 2 jij i i j i

j j j

hvh v v h v

x x x

On integrating over V t , the divergence term vanishes since 0v on its boundary

S t . Hence

222 ji

i j ij jV t V t

hvdV v h v dV

x x

2 2

V t

dV h h v (9.64a)

Now, (9.64a) holds for arbitrary h.

If we manage to find an h such that2 0C h h (9.65)

for all x in V t for all t, where C is some real positive constant, (9.64a) becomes

2

2 2i

jV t V t

v CdV C dV E

x

v

which, when substituted into (9.64), gives

2 21M

dEu C E

dt

(9.66)

Integrating, we have

2 20 exp M

tE E u C

which is our theorem eq(9.57) with

2

2

3C

L

(9.66a)

1st, we must show that such an h do exist and find the maximum C it affords.

2nd, we need to find the optimal h that can maximize C among all h’s.

At this point, the dimension of the system enters into consideration.

As stated in the theorem, we shall assume V t always lies inside a sphere, center at

the origin, of diameter L. Hence, all points of the system satisfy Lx .

The easiest construction is perhaps of the type

rh rh e

so that

2 2 22

1 dr h h

r dr h h

Setting h r ar , where a is a constant, we have

2 2 23f a a r h h where2

Lr .

Since

22 0f

a rr

for all r,

f is a monotonically decreasing function of r.

Its minimum is therefore at2

Lr where

2 2

34

a Lf a

The value of a that maximize this is given by

213 0

2aL

ie.,2

6a

L

so that

2

6h r r

L

2 22 2

6 63f r

L L

h h

and

2 22

6 6 93

4L

rC f

L L

which is only slightly less than the value (9.66a) given in the theorem.

Eq(9.66a) is achieved with

tanr

h rL L

so that2

2sec 0dh r

dr L L

for2

Lr

Thus, h is monotonically increasing with minimum at 0r .

Furthermore

22dh h

f hdr r

2 22 22

sec tan tanr r r

L L r L L L L

22

tanr

L r L L

22

1 tanL r

L r L

22

1 tan xL x

wherer

xL

The minimum of f is given by

22

2 2tan sec 0x x

x x

which implies

sin cos 0x x x ie.,

2 sin 2x xwhose solution is

0x or 0r

At which point, (9.66a) is achieved and the theorem is proved.

9.7 Uniqueness & Non-Uniqueness Of Steady

Viscous Flow

Theorem

Consider a fixed region V of fluid that is enclosed in a sphere of diameter L.

Let u and *u be 2 steady solutions of the Navier-Stokes eqs in V with the same value

Bu x on the boundary of V.

Let Mu be an upper bound to u in V.

If

3Mu LR

(9.68)

the 2 flows are identical, ie.,

*u u

In other words, bounded steady viscous flow with 5R is unique.

Proof

The proposition of this theorem is just the steady state version of the theorem in §9.6.

Thus, the t limit of (9.57) is satisfied, ie.,

2 22

0 2

3lim exp 0Mt

tE E u

L

where (9.68) ensured the coefficient of t is negative. QED.

9.7.1 An Example of Non-Uniqueness of Steady

Flow

9.7.2 Hysteresis

9.7.1 An Example of Non-Uniqueness of Steady

Flow

We now consider some variants to the Taylor experiment of §9.4 involving viscous

fluid occupies the gap 1 2r r r between 2 coaxial cylinders.

The differences we introduce are:

1. 2 0 .

2. Fluids are also bounded by stationary plane walls at 0,z L with L adjustable.

(see fig.9.13).

The system is characterized by 3 dimensionless numbers:

2

r

r.

2. Reynolds number 1 1rdR

where 2 1d r r .

3. Aspect ratioL

d .

Benjamin & Mullin (1982) had performed measurements on such an apparatus with

1

2

0.6r

r 12.61 359R (9.71)

The salient points were:

1. Some 20 different stable steady flows were observed.

2. On theoretical grounds, they inferred the existence of 19 other steady flows that

were unstable and hence unobservable.

3. All observed flows were of an axisymmetric cellular nature as shown in fig.9.8.

4. Different flows are distinguished by different number of cells and/or different

sense of rotation within each cell.

5. Appearance of a particular flow pattern depends on the way the boundary

conditions (9.71) were achieved from an initial state of rest.

6. If 359R were achieved by small steps from 0, the same flow consisting 12

cells was always observed.

For more details, see

T.B.Benjamin, T.Mullin, J.Fluid Mech 121, 219-30 (1982).

9.7.2 Hysteresis

Measurements on very short cylinders, with 4 , indicated that transitions between

2-cell & 4-cell modes were decribed by a state diagram shown in fig.9.14.

Such diagrams can be studied using catastrophic theory, a good reference of which is

J.M.T.Thompson, “Instabilities & Catastrophes in Science & Engineering”,

Wiley (1982).

Folds in the surface implies non-uniqueness of solutions in parts of the R plane;

hence hysteresis.

The middle sheet of the fold corresponds to unstable solutions which are not

observable.

9.8 Instability, Chaos, And Turbulence

Consider again the Taylor vortex apparatus with 1 2 and a temperature

difference 2 1 0T T T maintained between the cylinders.

This may serve as a crude model for the atmosphere as a rotating fluid under

differential heating.

Ref: R.Hide, Q.J.R.Met.Soc. 103,1-28 (1977).

If Ω is sufficiently small, a weak differential rotation is observed (see Fig.9.16a).

As Ω is increased in small steps past a critical value C that depends on T ,

baroclinic instabilities set in which amplify non-axisymmetric waves.

Further increasing Ω increases the amplitude of these waves leading to a meandering

jet structure reminiscent of atmospheric jet streams (see Fig.9.16b).

Amplitude, shape, and wavenumber of this jet can be either steady or oscillatory.

At still higher values of Ω, complicated, aperiodic fluctuations set in and the system

become chaotic (see Fig.16c).

9.9 Instability At Very Low Reynolds Number

The theorem of §9.6 guarantees stability for flows of sufficiently low R.

However, it applies only if u is prescribed on some, possibly varying, boundary.

For flows with free boundaries, instabilities are possible for arbitrarily small R.

Viscous Fingering in Hele-Shaw Cell.

A Hele-Shaw cell (see §7.7) is formed by pressing 2 flat sheets of transparent plastics

together with syrup filling the small gap (~2mm) in between.

A hole is drilled on the top sheet for the insertion of a syringe.

Air is then injected by the syringe.

In principle, one might expect the air to displace the syrup in a symmetrical manner,

so that the air-syrup interface is circular.

However, such an interface is found to be unstable. Ripples soon form and develop

into fingers as shown in Fig.9.20.

Such behavior is observed whenever a more viscous fluid is displaced by a less

viscous one. It is called the Saffman- Taylor instability.

Buckling of Viscous Jets

A falling jet of viscous fluid is approximately symmetrical about a vertical axis if the

height H is below a critical value CH . (see Fig.9.21a).

If CH H , the jet becomes unstable and buckles (see Fig.9.21b).

What makes it interesting is that, in contrast with the other instabilities discussed so

far, it occurs only if R is less than some critical value.