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'Airy' (Linear) wave theory

This is a Linear wave theory first proposed by Laplace (1776) and Airy (1845) and is some times

known as Airy wave theory. The theory is applicable to waves of small amplitude in deep water.

The aim of a wave theory to obtain expressions for surface elevation, wave length, celerity, water

particle velocity, energy, power etc as a function of wave height, period and water depth.

First, we have to make some assumptions:

1. The fluid homogeneous and incompressible (density is constant)

2. Surface tension is neglected

3. Coriolis effect can be neglected

4. Pressure at the sea surface is uniform and constant

5. The fluid is ideal or inviscid (lacks viscosity)

6. The wave being considered does not interact with any other water motions

7. The sea bed is a horizontal, fixed, impermeable boundary which implies that the verticalvelocity at the sea bed is zero.

8. The wave form is invariant in time and space

9. The waves are two-dimensional

10. The wave height is small compared with both the wave length and water depth. Inparticular,

11. H / L << 1

12. HL2/h3 << 1 (Ursell parameter)

- 2 - 8/8/02

We define velocity potential φ as

u x z tx z tx

, , , ,( ) =

∂ ( )∂

φ

w x z tx z tz

, , , ,( ) =

∂ ( )∂

φ

Notation is shown on Figure 1.

We define, horizontal velocity u in the x- direction (ie direction of wave travel), vertical velocity

w (positive upwards). z is the vertical coordinate with the zero at mean sea level. Note that z

increases positive upwards which means all depth are negative. The sea bed is located at z = -h.

The surface profile excursion from the mean level at any given time is given by η(x,t) which has a

maximum at the crest level (ie ηmax = H/2).

Assumption (1) incompressible means that

∂∂

+∂∂

=ux

wz

0 continuity

or in terms of the velocity potential,

∂∂

+∂∂

=2

2

2

2 0φ φ

x z

Laplace equation

Standard solutions available.

The Bernoulli equation can be written

pu w gz

tρφ

+ +( )+ +∂∂

=12

02 2

We consider linear terms only and hence, neglect terms involving square of the velocity.

i.e.p

gztρφ

+ +∂∂

=0

η, the elevation at the surface where p = 0 (i.e. equal to atmospheric pressure) (assumption 4)

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Hence, we can write, for z = η,

ηφ

η=−

∂∂

=

1g t z

or

ηφ

=−∂∂

=

1

0g t z(A)

(small amplitude wave theory and η is small assumption 10).

This approximation leads to an error of the order of those already done in neglecting higher orderterms in the Bernoulli equation.

Since, wave slope is small, H/L << 1, we can assume that the vertical velocity of the surface hasthe same velocity as the water.

i.e.∂∂

= [ ] =ηt

w z 0

or∂∂

=∂∂

=

η φt z z

0

from (A) we can write

∂∂

= −∂∂

=

η φt g t

z

1 2

20

Hence, for z=0, we can write (from subtraction)

∂∂

+∂∂

=φ φz g t

.1

02

2

This is called the Cauchy-Poisson condition at the free surface.

In the solutions of the Cauchy-Poisson condition we assume a solution of the type

φ = f (z) sin (kx – σt)

then

∂∂

= −( )φσ

xf z k kx t ( ) cos

∂∂

= − −( )2

2φσ

xf z k kx t ( ) sin

- 4 - 8/8/02

Similarly

∂∂

=∂∂

−( )2

2

2

2φ

σz

f z

zkx t

( ) sin .

Laplace equation,

∂∂

+∂∂

=2

2

2

2 0φ φ

x z.

Substituting

∂∂

−( ) − −( ) =2

22 0

f

zkx t k f kx tsin sinσ σ

i.e.

∂∂

− =2

22 0

f

zk f

The general solution is of the form

f z Ae Bekz kz( ) = + −

where A and B are constants.

Thus we obtain

φ σ sin( ).= +( ) −−Ae Be kx tkz kz

We have boundary condition such that w = 0 at z = -h

wz z h

=∂∂

==−

φ0

which gives

Ae Bekh kh− − =0.

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The other boundary condition at z=0,

∂∂

+∂∂

=φ φz g t

10

2

2

which will produce, after some algebra

(σ2 – gk) A + (σ2 + gk) B = 0.

The two boundary conditions could be written in matrix form as

σ σ2 2

0− +

=− +

gk gk

e e

A

Bkh kh

A and B are non-zero. Hence,

σ σ2 2

0− +

−=

− +

gk gk

e ekh kh

e gk e gkkh khσ σ2 2 0−( )− +( ) =−

σ2 0e e gk e ekh kh kh kh+( ) − −( ) =− −

or

σ2 =−+

−

−gke e

e e

kh kh

kh kh

e e khkh kh− =− sinh

e e khkh kh+ =− cosh

σ2 =gk khtanh . DISPERSION RELATIONSHIP

- 6 - 8/8/02

Now,

σπ π

= =2 2T

and kL

Substitute into the dispersion relationship

4 2

2

2

2

2

2

π π

π

π

Tg

Lkh

LgT

kh

CgT

kh as CLT

=

=

= =

tanh( )

tanh( )

tanh( )

Now consider values of tanh kh or tanh r

where

r = 2π

Lh = 2π h

L.

Hence the value of tanh r depends on the ratio

Lh

–water depthwave length .

As r becomes large tanh r → 1

i.e.

LgT

or L T0

2

02

21 56= =

π .

Refers to deep water

as hL

must be large.

and

CgT

or C T0 021 56= =

π .

Lh

> 14

gives 5% error

Lh

> 12

gives 0.37% error .

Consider the other extreme,i.e.

hL

is small

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then, tanh r → r

or

L = gT2 . L h

L2 = ghT2

L T ghS =

C = gT . L h

L = CT

C2 = gh

and

C gh =

for waves in shallow water. In shallow water, celerity of waves only depend on the water depth

when

hL

< 120

the error is 5%.

Hence, the limits of the deep, intermediate and shallow water waves from the ratio hL

,

we have,

L =

gT2

2π tanh kh

L 0 =

gT2

2π

and also

C =gT2π tanh kh

C0 =gT2π

We can write,

LL

CC

kh0 0

tanh .= =

Hence, knowing offshore conditions can calculate inshore conditions and vice versa.

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Explicit solution of the wave dispersion relationship

σ2 = gk tanh kh

Non-dimensional form

σ2hg

kh kh= tanh .

ασ π π

ππ

= =( )

= =2 2

2 20

2 22

2hg T

hg gT

hh

L

= k0h.

Eckart (1952) has shown that

kh ~ tanh

%/α

α[ ]±1 2 5

Velocity Potential

Dispersion relationship was obtained by assuming a solution of the form

φ = ƒ(z) sin (kx — σt)

for the Cauchy-Poisson condition at the free surface,

∂∂

+∂

∂=

φ φz g t

10

2

2

ƒ was described by,

∂ ƒ∂

− ƒ =2

22 0

zk .

Solution for this is given by,

φ σ cosh

sinh sin

'' ( )

=+( )[ ]

( ) −( )

ƒ

HC

k z h

khkx t

z

21 2444 3444

we had

ηφ

=−∂∂

=1

0g t

at z

∂∂

= −+( )[ ]

( ) −( ) −( )φσ σ

t gH

Ck z h

khkx t

cosh

sinh cos

12

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i.e. z = 0

η = H2

Cg

. σ 1tanh kh

cos (kx – σt)

= H2

. 1 tanh khtanh kh

cos (kx – σt)

η = H2

cos (kx – σt)

Horizontal and Vertical Velocities

φ σ=+( )[ ] −( )HC k z h

khkx t

2 cosh

sinh( ) sin

Ux

HCk

k z h

khkx t=

∂∂

=+( )[ ] −( )φ

σ2

.cosh

sinh( ) cos

Consider,

HCk H LT L

HT2 2

2= ⋅ ⋅ =

π π

UHT

k z h

khkx t=

+( )[ ] −( )πσ

cosh

sinh( ) cos

Similarly

∂∂

= =+( )[ ] −

φ πσ

zw

H k z h

khkx t

T

sinh

sinh( )sin( )

- 10 - 8/8/02

Particle Orbits

Particle orbits can be written

x u dtx

dt

t t

=⌠

⌡

∂∂

⌠

⌡

. .

0 0

φ

z w dtz

dt

t t

=⌠

⌡

∂∂

⌠

⌡

. .

0 0

φ

by considering the position of a particle at φ a point (x + α, y + β) where α and β is small, the

following expression is obtained.

α2

A2+

β2

B2= 1

where

A = H2

cosh k (z + h)sinh kh

B = H2

sinh k (z + h)sinh kh

The ratio BA

= tanh k (z + h)

as z → –h (i.e. at the bottom)

tanh k (z + h) → 0

i.e.BA

→ 0

As A is finite B → 0

i.e. the particles oscillate in a degenerate ellipse which is a horizontal line at the sea bed.

When the water is deep h → ∞ can show by taking the limit of the equation for the ellipse that

α β2 2 2 2

2+ = =a e a

Hkz

Equation of a circle with radius a ekz at z=0, radius is a2 and decreases exponentially with depth.

For Airy wave theory no mass transport particles move in closed orbits

- 11 - 8/8/02

Pressure

Bernoulli equation neglecting higher order terms is given by :

pgz

tρφ

+ +∂∂

=0

pgz

tρφ

+ − −∂∂

=0

i.e.

φ σ=+( )[ ] −( )H

Ck z h

khkx t

2

cosh

sinhsin

substitute

pgz

HC

k z h

khkx t

ρσ σ=− +

+( )[ ] −2

cosh

sinhcos( )

=− ++( )[ ] −gz

HC

k z h

khkx t

2

cosh

coshcos( )σ

Hence,

p gzgH k z h

khkx t Pa=− +

+( )[ ] − +ρ σ2

cosh

coshcos( )

= hydrostatic less dynamic pressure, + atmos.(neglected)

η σ cos= −( )Hkx t

2

hence, p becomes

p gz gk z h

kh

cosh

cosh= − +

+( )[ ]ρ ρ η

The ratio

Kk z h

khz cosh

cosh=

+( )[ ]

is called the pressure response factor.

Now, the equation for p could be written as

p g K zz= −( )ρ (A)

when,

z h ,= − i.e. at the sea bed

kkh

cosh

=1

- 12 - 8/8/02

Tabulated in wave tables. Hence, if we record pressure at the sea bed then, by using A cancalculate η.

Energy

Need to know the energy of waves as all design criteria depend on it. Also important incalculating longshore sediment transport.

The energy contained in the waves is compose of two parts :

(1) Potential energy : due to the deviation of its profile from the mean level.

(2) Kinetic energy : due to the oscillatory motion of the particles.

The potential energy of a wave per unit area is given by

EL

gzdzdxp

L

=⌠

⌡

⌠

⌡

1

0 0

η

ρ

= 116 ρg H2

The kinetic energy is given by (per unit area).

EL

u w dzdzk

L h

=⌠

⌡

⌠

⌡

+( )−

1 12

0 0

2 2ρ

=1

162ρgH

Hence,

potential energy = kinetic energy (for Airy waves)

The total energy is given by

E E E gHp k = + =18

2ρ Joule m-2

The energy is averaged over a wave length hence, is energy per unit area. Sometimes termedspecific energy or energy density. Sometimes, energy is expressed as per unit wave crest lengthwhich is equivalent to EL.

Energy flux or power the rate at which energy is transmitted in the direction of the wave is givenby,

- 13 - 8/8/02

PT

p u dx dt

T h

=

⌠

⌡

⌠

⌡

−

1

0 0

integration leads to

P g H Ckh

kh= +

18

12

12

22

sinhρ

writing

nkh

kh= +

12

12

2sinh

we get,

P gH Cn=18

2ρ

P ECn =

Energy flux has units of power, Joule s-1 m-1 or Watts m-1

In deep water v = 12 and increases as the depth decreases and in shallow water n = 1

Values of n are tabulated

P gives the power transmitted across a plane perpendicular to wave advance. If a plane otherthan that is used then the component must be taken.

Also associated with wave advance is a flux or transmission of momentum.

Longuet-Higgins and Stewart (1964) define "radiation stress" as excess flow of momentum due tothe presence of waves.

Excess momentum is because in the derivation, dynamic pressure is used, with hydrostaticpressure subtracted from the absolute pressure.

i.e.

p g k zz

dynamic hydrostatic

= −ρ η( )123 123

- 14 - 8/8/02

y

wave crests

xwave advance

The radiation stress (momentum flux) across the plane x = constant (parallel to shore say) is givenby

Sxx = E 2 khsinh (2kh)

+ 12

n = 12

1 + 2 khsinh kh

2 khsinh kh

= 2n – 1

substitute

Sxx = E 2n – 12

The flux across plane y - constant

In deep water, n = 12 ,

then

Sxx = E2

and Syy = 0

In shallow water, n = 1

Sxx = 3E2

and Syy = E2

Radiation stress is important as many applications use it

longshore velocity

surf beat

interaction between waves and currents etc.

Group Velocity

Consider the sum of two wave forms give by,

- 15 - 8/8/02

η σ σ

η η

cos cos ' '= −( ) + −( )Hkx t

Hk x t

2 2

1 2

1 244 344 1 244 344

where, k' and σ' are small variations from k and σ from trigonometry

cos x + cox y = 2 cos 12

( x + y) cos 12

(x – y)

12

(x + y) = 12

(kx – σt + k′x – σ′t) = 12

(k + k′) x – 12

(σ + σ') t

12

(x – y) = 12

(k – k′) x – 12

(σ – σ') t

Hence,

η = H cos 12

k + k′ x – 12

σ + σ′ t cos 12

k + k′ x – 12

σ – σ' t

Hence, the summation of two waves have produced the multiplication of two waves with doublethe wave height.

Consider the 1st term

cos ' '12

12

k k x t−( ) − +( )

σ σ

now as k ≈ k' and σ ≈ σ' this is the ......... has same waveley and period as original η1.

Consider the 2nd wave form

cos ' '12

12

k k x t−( ) − −( )

σ σ

Thus has wavelength

4πk k− '

and period

4πσ σ− '

and celerity

= LT

= σ – σ′k – k′ = δσ

δk.

Hence, in the limit,

Cg = dkdσ = d

dk(kC) = C + k dc

dk

Cg is the propagation speed of the envelope define as group velocity

- 16 - 8/8/02

Cg = C + K ddk

gk

tanh kh 1/2

differentiating and after some algebra we get

Ch L

h LCg

n

= +

12

14

4

/sin /

π

π1 2444 3444

Cg = Cn

In deep water

n = 12

Cg = 12

C

Hence, the wave group is half the wave celerity

In shallow water n = 1

Cg = C

i.e. celerity is same

- 17 - 8/8/02

Standing Waves

So far we looked at progressive waves and obtained equation for celerity and wave length etc.from linear theory i.e. first order solution.

We now look at standing waves - 1st order solutions.

Standing waves may be thought of as the sum of two progressive waves of equal wavelength andperiod travelling in opposite directions.

We have surface elevation given by

η1 = H2

cos (kx – σt)

for opposite direction, x = –x

η2 = H2

cos – (kx – σt) = H2

cos (kx + σt)

η η σ σ1 2 2+ = −( ) + +( )[ ] cos cos

Hkx t kx t

= H2

2 cos kx cos –σt

cos A + cos B = 2 cos A + B2

cos A – B2

hence

η σ= ′Hkx t

2cos cos

Hence, surface wave form has wave height H' which is twice the wave height of the progressivewaves as H'= 2H.

Equations for C and L are the same for standing waves (obtained from assumptions and Laplaceequation)

φ= HCz

cosh [k (z + h)]sin kh

coskx sinσt

velocity components

∂∂

= =+( )[ ]

∂∂

= =+( )[ ]

φ πσ

φ πσ

xu

HT

k z h

khkx t

zw

HT

k z h

khkx t

cosh

sinh sin sin

cosh

sinh cos sin

wave energy = E = Ep + Ek = 1/8 ρgH2.

- 18 - 8/8/02

The main difference is the orbits.

We have a system of nodes and antinodes set up

Nodes - No motion - same level as S.W.L.

Anti Nodes- Maximum excursion 2 OHP's

Rectilinear paths as opposed to circles and ellipses

Can calculate all other parameters such as pressure, etc.