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Chapter 4. Engineering Wave Properties

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(1)

Chapter 4. Engineering Wave Properties

4.1 Introduction

etc.

energy, wave

field, pressure ,

kinematics elevation Wave

Surface

potential Velocity

⎭→

⎬⎫

Wave transformation (Shoaling, Refraction, Diffraction) Wave breaking

4.2 Wave Kinematics for Progressive Waves

Particle velocity

) cosh sin(

) ( cosh

2 kx t

kh z h k g

H σ

φ =− σ + −

) sinh cos(

) ( cosh 2

) cosh cos(

) ( cosh 2

t kh kx

z h k H

t kh kx

z h k gk

H u x

σ σ

σ σ φ

+ −

=

+ −

∂ =

−∂

=

) sinh sin(

) ( sinh 2

) cosh sin(

) ( sinh 2

t kh kx

z h k H

t kh kx

z h k gk H w z

σ σ

σ σ φ

+ −

=

+ −

∂ =

−∂

=

where we used the dispersion relationship, . Note that and are out of phase.

kh gk tanh

2 =

σ u w

90°

Acceleration

) sinh sin(

) ( cosh 2

2 kx t

kh z h k H

t u dt

du = σ + −σ

≅∂

(2)

) sinh cos(

) ( sinh 2

2 kx t

kh z h k H

t w dt

dw =− σ + −σ

≅∂

At t=0

At x=0

(3)
(4)
(5)

Particle displacements

Consider a water particle moving around the mean position (x1,z1):

z dt u x

z u x u

dt z

x u t z x

z x z

x

⎥⎥

⎢⎢

⎡ +

∂ + ∂

∂ + ∂

=

+ +

=

L

) , ( )

, ( 1

1 1 1 1

1

1 1 1

1

) , (

) ,

( ) , , (

ξ ς

ξ ς ς

⎜ ⎞

T

H

⎜ ⎞

L

T H H/

) sinh sin(

) ( cosh 2

) , (

1 1

1 1

t kh kx

z h k H

dt z x u

σ + −

=

Similarly,

) sinh cos(

) ( sinh 2

) ,

( ) , , (

1 1

1 1 1

1

t kh kx

z h k H

dt z

x w t z x

σ ξ

ς ξ

+ −

=

+ +

=

Let

) cos(

) ( ) , , (

) sin(

) ( ) , , (

1 1

1 1

1 1

1 1

t kx z

B t z x

t kx z

A t z x

σ ξ

σ ς

=

=

(6)

in which

kh z h H k

z B

kh z h k z H

A

sinh ) ( sinh ) 2

(

sinh ) ( cosh ) 2

(

1 1

1 1

= +

= +

Then

1 ) (

cos ) (

sin2 1 2 1

2 2

=

− +

⎟ =

⎜ ⎞

⎝ +⎛

⎟⎠

⎜ ⎞

kx t kx t

B

A ξ σ σ

ς

which is the equation of an ellipse. Note that A is always greater than B because . Also note that

x x sinh

cosh > B=H/2 at z1 =0, that is, the water particle on the free surface moves vertically between crest and trough.

In shallow water,

) 1 (

2 f z1

kh

A= H ≠ but constant over the depth

⎟⎠

⎜ ⎞

⎝⎛ + + =

= h

z H kh

z h k

B H 1 1 1

2 ) (

2 varies from H/2 at surface to at bottom 0

In deep water,

1 1

2 2

kz kh

kh kz

H e e

e e

A= H =

(7)

A H e

e e e

B H kh kz

kh kz

=

=

= 1 1

2

2 : Circle with diameter decreasing exponentially with depth

(8)

4.3 Pressure Field under a Progressive Wave

Consider the Bernoulli equation:

( )

( )

2

1 2 2

t t C w

u p gz

∂ =

−∂ + +

+ φ

ρ

Neglecting small nonlinear terms,

0

0

∂ =

−∂

⎟⎟ ≅

⎜⎜ ⎞

−∂ +

⎟⎟ =

⎜⎜ ⎞

−∂ +

= z=

z

z g t

gz t p gz t

p φ η φ

ρ φ

ρ η

) cosh cos(

) ( cosh

2 kx t

kh z h g k

gz H gz t

p φ σ

ρ

+ +

∂ = + ∂

=

) (

) cosh cos(

) ( cosh 2

z K g gz

t kh kx

z h k g H

gz p

η p

ρ ρ

σ ρ

ρ +

=

+ − +

=

in which

kh z h z k

Kp

cosh ) ( ) cosh

( = +

is the pressure response factor, which varies from 1/coshkh at z=−h to 1 at . Note that

=0 z 1

) (z

Kp . In the equation for pressure, the first term represents the hydrostatic pressure, and the second term dynamic pressure due to waves. The hydrostatic pressure increases linearly with depth, but the dynamic pressure decreases exponentially with depth.

(9)

Above the mean water level, i.e., in the range of 0<z1<η,

( ) ( )

) (

) 0 ( ) (

1 1

1 0 0 0 1 1

z g

gz g

K g z gz

z K

g gz

z z p p z p

p z p z

z

=

=

+ +

∂ − + ∂ +

=

∂ + + ∂

=

=

=

=

η ρ

ρ η ρ

η ρ ρ η

ρ

ρ L

L

That is, the pressure is hydrostatic for 0< z<η.

(10)

Wave measurement using a pressure transducer

) ) (

( gK z

z p K g p

p d p

d =ρ η ⇒ η= ρ

If the pressure transducer is located at the bottom (z=−h),

) gK ( h

p

p d

= − η ρ where

⎩⎨

≅⎧

=

− 0 for deep water (short-period wave) wave) period - (long water shallow for

1 cosh

) 1

( h kh

Kp

Dynamic pressure due to short-period wave is hardly recorded by a pressure transducer located at the bottom. It is difficult to distinguish real signal and noise of the gauge.

Therefore, pressure should be measured near the water surface when water depth is large.

4.4 Wave Kinematics for Standing Waves

(11)

t kh kx

z h k H g

t kh kx

z h k H g

t kx t

kx t

kx t

kh kx z h k H g

t kh kx

z h k H g

t kh kx

z h k H g

s p p

p p

σ σ σ σ

σ σ

σ σ σ

σ σ σ σ

φ

sin cosh cos

) ( cosh 2

sin cosh cos

) ( cosh

) sin cos cos

sin sin

cos cos

sin cosh (

) ( cosh 2

) cosh sin(

) ( cosh ) 2

cosh sin(

) ( cosh 2

= +

= +

+ +

+ + −

=

+ + +

+ −

=

t kh kx

z h H k

t kh kx

z h k gk

H u x

s s

σ σ

σ σ φ

sin sinh sin

) ( cosh 2

sin cosh sin

) ( cosh 2

= +

= +

−∂

=

t kh kx

z h H k

w φz sσ σ

sin sinh cos

) ( sinh 2

− +

∂ =

−∂

=

t Hs kx σ

η cos cos

= 2

At node, coskx=0, sinkx=1, w=0, π

⎜ ⎞

⎝⎛ +

= 2

n 1 kx

At antinode, coskx=1, sinkx=0, u=0, kx=nπ

At vertical wall,

=0

−∂

= x

u φ

⇒ Antinode at the wall

(12)

Local accelerations:

t kh kx

z h H k

t

u sσ sin cosσ

sinh ) ( cosh 2

2 +

∂ =

∂ ;

t u

∂ is max at node and zero at antinode.

t kh kx

z h H k

t

w s σ cos cosσ

sinh ) ( sinh 2

2 +

∂ =

∂ ;

t w

∂ is max at antinode and zero at node.

Particle displacements:

t z

x A

t kh kx

z h k H

dt z x u dt z

x u z x

s

σ

σ ξ

ς ς

cos ) , (

cos sinh sin

) ( cosh 2

) , ( )

, ( ) , (

1 1

1 1

1 1 1

1 1

1

=

− +

=

≅ + +

=

∫ ∫

t z

x B

t kh kx

z h H k

z

x s

σ

σ ξ

cos ) , (

cos sinh cos

) ( sinh ) 2

, (

1 1

1 1

1 1

=

= +

ς ς ξ

ξ ⎟

⎜ ⎞

−⎛

=

= A

B A

B : Straight line with slope ⎟

⎜ ⎞

−⎛ A B

(13)

4.5 Pressure Field under a Standing Wave

Consider the Bernoulli equation:

( )

( )

2

1 2 2

t t C w

u p gz

∂ =

−∂ + +

+ φ

ρ

By the same way as for progressive waves,

) (

cos cosh cos

) ( cosh 2

z K g gz

t kh kx

z h H k

g gz gz t p

p s

η ρ ρ

σ ρ

ρ ρ φ ρ

+

=

+ +

=

∂ + ∂

=

Again

η η

ρ − < <

= 1 1

1) ( ) for 0

(z g z z

p

Wave force acting on a vertical wall

∫ ∫

⎥⎦ +

⎢⎣ ⎤

⎡ +

+

=

=

=

w w

η

-h w w

h w

z)dz ρg(η

kh dz z) ρgη k(h

gz -

dz z p F

0 0

cosh cosh

wall at the )

( ρ

η

η η

The first integral is a first-order term, and the second integral is a second-order term.

However, note that this F is neither first-order nor second-order. For a complete second order solution, the nonlinear terms (i.e., terms) should be included from the beginning.

u2

(14)

To the first order,

2

0 2

0

2 tanh 1

cosh ) ( sinh 2

1

cosh cosh

gh k kh

g

kh z h k k

gz g

kh dz z) ρgη k(h

gz - F

w

h w

-h w

ρ ρ η

ρ η ρ

ρ

+

=

⎥⎦⎤

⎢⎣⎡ +

+

=

⎥⎦⎤

⎢⎣⎡ +

+

=

The first term is the wave force due to dynamic pressure, while the second term is the hydrostatic pressure force.

Since

Hs t

w σ

η cos

= 2 we have

2 2

2 max

2 1 2

2 tanh 1 2

gh C

H

gh kh

k H F g

s s

ρ ρ ρ ρ

+

=

+

=

(15)

4.6 Partial Standing Waves

Partial reflection from beaches or breakwaters: where = incident wave height, and = reflected wave height. The wave potential is

r

i H

H > Hi

Hr

) cosh sin(

) ( cosh ) 2

cosh sin(

) ( cosh

2 σ ε

σ σ

φ σ + + +

+ + −

= kx t

kh z h k H g

t kh kx

z h k

Hi g r

where ε = phase lag.

) 2 cos(

) 2 cos(

1

0

ε σ φ σ

η = − + + +

= ∂

=

t H kx

t H kx

t g

r i

z

Since , there is no true node. Instead, we have wave envelopes within which the surface elevation varies (see Fig. 4.7 of textbook). In other words, the upper and lower envelopes are the upper and lower boundaries of surface elevation, i.e.

r

i H

H

ηmax and η . Therefore, min ∂η/∂t=0 on envelopes.

Separation of incident and reflected waves

t x F t x I

t H kx

H kx t H kx

H kx

t kx

t H kx

t kx t

H kx

r i

r i

r i

σ σ

σ ε

σ ε

σ ε σ

ε σ

σ η

sin ) ( cos

) (

sin ) 2 sin(

2 sin cos

) 2 cos(

2 cos

) sin ) sin(

cos ) (cos(

) 2 sin sin cos

2 (cos

+

=

⎭⎬

⎩⎨

⎧ − +

⎭ +

⎬⎫

⎩⎨

⎧ + +

=

+

− +

+ +

=

(16)

To find the envelopes,

t x F t x

t σI σ σ σ

η =0=− ( )sin + ( )cos

t x F t x

I( )sinσ = ( )cosσ

) (

) ) (

tan( I x

x t m = F σ

2 2

2 2

2 2

2 2

2 max 2

) ( ) (

) ( ) (

) ( ) (

) ( ) (

) ) (

( ) ( ) (

) ) (

(

x F x I

x F x I

x F x I

x F x I

x x F

x F F x I

x x I

I

+

=

+

= +

+ +

= + η

2 2

2 2

2 min 2

) ( ) (

) ( ) (

) ) (

( ) ( ) (

) ) (

(

x F x I

x F x I

x x F

x F F x I

x x I

I

+

=

+ + −

+

= − η

) 2 2 cos(

2 2

) 2 sin(

2 sin )

2 cos(

2 cos

2 2

2 2

min max

ε

ε ε

η

+

⎟ +

⎜ ⎞

⎝ +⎛

⎟⎠

⎜ ⎞

± ⎛

=

⎭⎬

⎩⎨

⎧ − +

⎭ +

⎬⎫

⎩⎨

⎧ + +

±

=

H kx H H

H

H kx H kx

H kx H kx

r i r

i

i r i r

(17)
(18)

1 ) 2

cos( kx+ε =− cos(2kx+ε)=1 π

ε (2 1)

2kx+ = n− 2kx+ε =2nπ

(

i r

)

r i r

i

r i r

i

H H

H H H

H

H H H

H

=

− +

=

⎟ −

⎜ ⎞

⎝ +⎛

⎟⎠

⎜ ⎞

= ⎛

2 1 2 2 1

2 2

2

2 2

2 2

ηmax

(

i r

)

r i r

i

r i r

i

H H

H H H

H

H H H

H

+

=

+ +

=

⎟ +

⎜ ⎞

⎝ +⎛

⎟⎠

⎜ ⎞

= ⎛

2 1 2 2 1

2 2

2

2 2

2 2

ηmax

Distance between x1 and x2 = x2x1

k n k

n

2 ) 1 2 ( 2

2 π −ε − − π −ε

=

k 2

= π

4

= L same as perfect reflection

( )

=

(

HiHr

)

2 1

max node

η

( )

=

(

Hi +Hr

)

2 1

antinode

ηmax

( ) ( )

(

ηmaxmax

)

antinodeantinode

(

ηmaxmax

)

nodenode

η η

=

+

=

r i

H H

To measure the wave envelopes, we have to measure waves as slowly moving a wave gauge along the wave tank (see Fig. 4.9 of textbook). This technique for separation of incident and reflected waves is applicable only to regular waves. For irregular waves,

(19)

use multiple fixed gauge method (Suh et al. 2001, Coastal Engineering, 43, 149-159).

4.7 Wave Energy

Total energy (E) = potential energy (Ep) due to displacement of free surface + kinetic energy (Ek) due to water particle movement

Potential energy

Potential energy per unit surface area:

(20)

( )

2 2

2 2

2 2

2

16 1 2

1

2 1 2

1 2 2 1

) 2 (

1

gH gh

dx dx

h L L h

g

dx h

L h g

dx g h

E L

L x x L

x x L

x x L x p x

ρ ρ

η ρ η

η ρ η

ρ η

+

=

⎭⎬

⎩⎨

⎧ + +

=

+ +

=

+

=

+ +

+ +

The first term is the potential energy in still water, while the second term is the potential energy due to wave. Therefore, the potential energy of only wave is

2

16 1 gH Ep = ρ

Another view:

L g H H

g HL 2

2 16 16 energy 2

Potential π ρ

ρ π × × =

=

∴ For unit area, 2

16 1 gH Ep = ρ

Kinetic energy

(21)

( )

{

[ ]

2

2 0

2 2

2 2

2 2 2

16 1

) (

2 cos ) ( 2 2 cosh 1 cosh

1 2

2

)}

( sin ) ( sinh

) (

cos ) ( cosh cosh

1 2

2

2 1

gH

dzdx t kx z

h kh k

gk H L

dzdx t kx z

h k

t kx z

h kh k

gk H L

dzdx w L u

E

L x

x h

L x

x h

L x

x h

k

ρ σ σ ρ

σ σ σ

ρ

ρ

η η

=

− +

⎟ +

⎜ ⎞

≅ ⎛

− +

+

⎟ +

⎜ ⎞

= ⎛

+

=

∫ ∫

∫ ∫

∫ ∫

+

+

+

Now, the total energy per unit surface area is

⎟⎠

⎜ ⎞

= ⎛ ⋅

= +

= 2 2 2 2

m Joule m or

m N 2

1 8

1 gH ga

E E

E p k ρ ρ

Energy flux

Energy flux, or rate of work done by wave, is

= η

h pdudz F

Average energy flux over one wave period is

(22)

g T t

t h

T t

t h d

EC ECn

kh kh gH k

dzdt t kh kx

z h k gk

H kh

z h g k

T

udzdt T p

F

=

=

⎥⎦

⎢ ⎤

⎡ ⎟

⎜ ⎞

⎝⎛ +

=

⎭⎬

⎩⎨

⎧ + −

⎭⎬

⎩⎨

⎧ +

=

=

∫ ∫

∫ ∫

+

+

2 sinh 1 2 2 1 8

1

) cosh cos(

) ( cosh 2

cosh ) ( cosh 1

1

2 σ ρ

σ σ η

η ρ

η

where Cg = group velocity at which wave energy is transmitted.

(23)

dk nC d Cg = = σ

where σ2 =gk tanhkh is used for the last step.

In shallow water, n→ ,1 ∴ Cg =C In deep water,

, 2 2

1 0

0

C C

n→ ∴ g =

So far, we considered the dynamic view of the group velocity.

Kinematic view of group velocity

Consider two waves with slightly different frequencies and wave numbers, which gives a wave group.

) 2 cos(

)

2 cos( 1 1 2 2

2 1

t x H k

t x

H k σ σ

η η η

− +

= +

=

where

, 2 2

, 2 2

2 2

1 1

k k k

k k k

+ Δ Δ =

+

=

−Δ Δ =

= σ σ σ

σ σ σ

such that

1 ,

1 Δ <<

<<

Δ

k k σ

σ

Now, using the relationship

(24)

cos 2 cos 2

2 cos

cos x y x y

y

x+ = + −

we have

( ) ( ) ( ) ( )

⎥⎦

⎢ ⎤

⎡ ⎟

⎜ ⎞

⎛ Δ

−Δ Δ

=

⎟⎠

⎜ ⎞

⎛− Δ + Δ

=

⎥⎦⎤

⎢⎣⎡ − − −

⎥⎦⎤

⎢⎣⎡ + − +

=

k t x k t

kx H

t kx

t kx H

t x

k k t

x k k H

σ σ

σ σ

σ σ σ

σ η

2 cos 1 ) cos(

2 1 2

cos 1 ) cos(

2 1 2

cos 1 2

1 2

cos 1 1 2 1 2 1 2 1 2

In the last term of the above expression, the first part represents the individual wave propagating at the speed of C=σ /k, while the second part represents the modulation or envelope propagating at the speed of Cg =Δσ /Δk.

The speed of propagation of envelope or wave group is

Cg k Δ

≡Δσ

This is why we call Cg “group velocity”.

(25)

4.8 Wave Transformation Due to Depth Change

In this section, we deal with wave shoaling, refraction, and breaking.

Conservation of waves equation

{

( , ,)

}

Re AeiS xyt η=

where

(

k x σt

) (

k x k y t

)

S≡scalar phasefunction = ⋅ − = x + y −σ

t

∂ / and ∇ are independent operators so that the change of order does not make any difference. Thus

0 )

( ⎟=

⎜ ⎞

∇ ∂

∂ ∇

t S S

t

(26)

or

=0

∂ +

∂ σ

t k

This is the equation for conservation of waves, which states that the temporal variation of wave number k should be balanced by the spatial variation of frequency σ .

Consider 1-D case:

Number of waves coming in at section 1 per unit time =

π σ 1 = 2

T .

xdx

⎜ ⎞

∂ + ∂

⎟⎠

⎜ ⎞

=⎛

⎟⎠

⎜ ⎞

π σ π

σ π

σ

2 2

2 2 1

2

1 2

2

2 ⎟

⎜ ⎞

−⎛

⎟⎠

⎜ ⎞

=⎛

⎟⎠

⎜ ⎞

− ∂

π σ π

σ π

σ dx x

Number of waves in dx = k dx L

dx π

= 2 . The rate at which the number of waves changes in dx = ⎟

⎜ ⎞

k dx t 2π .

(27)

x dx k dx

t

⎜ ⎞

− ∂

⎟=

⎜ ⎞

∴ ∂

π σ

π 2

2

=0

∂ + ∂

x t

k σ

Conservation of waves equation in 1-D

For a steady state,

0

0 =

⇒ ∂

∂ =

x t

k σ

that is, σ (or T) is constant along x in a steady state.

Refraction

0 ) (∇ =

×

=

×

k S by vector identity (curl of gradient = 0)

ˆ 0 0

ˆ ˆ ˆ

⎟⎟ =

⎜⎜ ⎞

−∂

= ∂

= ∂

×

k

y k x k k

k

z y x

k j i

k y x

y x

) 0 cos ( ) sin

( =

−∂

y k x

k θ θ

(28)

Simple example: straight and parallel contours

constant sin

0 ⇒ =

∂ =

k θ

y

Dividing by σ ,

constant sin =

C θ

or

0

sin 0

sin

C C

θ θ

= Snell’s law

Since C<C0, θ < θ0

(29)

Variable bathymetry:

) 0 cos ( ) sin

( =

−∂

y k x

k θ θ

Since k =k(x,y) and θ =θ(x,y),

0 cos

sin sin

cos =

− ∂

∂ + ∂

∂ + ∂

y k k y

x k

k θ θx θ θ θ θ

This is a 1st-order nonlinear partial differential equation for θ, because is given by for given

k kh

gk tanh

2 =

σ σ and . This equation was solved numerically by Noda et al. (1974).

h

A simpler way by Perlin and Dean (1983), Proc. Coastal Structures ’83, ASCE 988-999:

Defining

θ θ cos )

, (

sin )

, (

k y x B

k y x A

=

=

we have

x A y B

=∂

∂ ← solved numerically

(30)

2 2

2 B k

A + = ⇒ θ is calculated.

Wave height H is calculated by the similar way using the conservation of energy equation. Also see Dalrymple (1988). JWPCOE, ASCE, 114(4): 423-435, which includes wave-current interaction and nonlinearity.

(31)

Conservation of energy

Assume a steady state and no energy input or loss.

Assume wave energy is not transmitted across wave rays. Then

( )

=0

bECg

s

In Cartesian coordinates,

(

cos

) (

sin

)

=0

∂ + ∂

g θ ECg θ

EC y x

or

( )

=0

ECg Conservation of energy equation

(32)

For straight and parallel contours, ∂ y/∂ =0. Therefore,

(

cos

)

=0

ECg θ x

( )

ECg 1b1=

( )

ECg 2b2 2 2 2 2 1 1

2

1 8

1 8

gH Cg b = ρgH Cg b

2 1

2 1 1 2 1

2 1 1

2 cos

cos θ θ

g g

g g

C H C b b C H C

H = =

If position 1 is in deep water,

r s g

g

K K C H

H C b b C H C

H 0

2 0

2 0 0 2 0

2 0 0

2 cos

cos 2

2 = =

= θ

θ

where Ks = shoaling coefficient, and Kr = refraction coefficient.

(33)

Ray tracing method

1) Graphical method: Use Snell’s law over two adjacent bottom contours.

L sin ,

, sin sin

sin

3 3

3 2

2 2

2 2 1

1 θ θ θ θ θ

θ = → = →

C C

C C

Draw → refraction diagram

⎭⎬

⎩⎨

crests s orthogonal

Measure b ⇒ Calculate

b

Kr = b0H =H0KrKs

(34)

2) Numerical method Munk and Arthur (1952)

Griswold (1963), Journal of Geophysical Research, 68(6): 1715-1723.

Let = travel time along a ray of a wave moving with speed . Let the wave is located at at time . Then

t C(x,y)

[

x(t),y(t)

]

t θ

Ccos

dx =dt (1) θ

Csin

dy =dt (2)

? ) , (x y = θ

θ

θ sin

cos dn

ds

dx= =− ds

n C C dn

d dCdt

d

− ∂

=

=

≅ 1

tan θ θ

θ

θ cos

sin dn ds

dy = =

Cdt ds=

(35)

⎟⎟⎠

⎜⎜ ⎞

− ∂

= ∂

⎟⎟⎠

⎜⎜ ⎞

∂ + ∂

− ∂

⎟⎟=

⎜⎜ ⎞

∂ +∂

− ∂

∂ =

− ∂

∂ =

y C x

C C

y C x

C C

n y y C n x x C C n C C s

θ θ

θ θ θ

cos 1 sin

cos 1 sin

1 1

y C x

C C s

t s s

t

− ∂

= ∂

= ∂

=∂

∂θ θ θ θ θ

cos

sin (3)

Solve (1), (2), and (3) to find a ray using a numerical method (ray by ray).

Let β =b/ b0 = ray separation factor such that

2 / 0 = 1

= β

b Kr b

β can be found analytically (Munk and Arthur, 1952)

θ θ sin

cos n

s

x= −

θ

θ cos

sin n s

y= +

By chain rule,

y x

s y y s x x

s

+ ∂

= ∂

∂ + ∂

= ∂

∂ cosθ sinθ (4.112b)

(36)

y x

n y y n x x

n

+ ∂

− ∂

∂ =

∂ + ∂

= ∂

∂ sinθ cosθ (4.112c)

n C C n k k x k k

y k k

y x

s

− ∂

∂ =

= ∂

− ∂

= ∂

∂ + ∂

= ∂

∂ 1 1

1sin 1cos

sin

cosθ θ θ θ θ θ

θ (4.113)

θ θ θ θ

sin sin ,

cos cos

, C

dt ds dt C dy

dt ds dt C dx dt

ds = = = = =

At the point A,

ds d db nb

d =

=∂θ θ

θ ,

n s b

b

=∂

∂ θ

1

n s b

b

=∂

∂β θ

β

0 0

1

n s

=∂

∂β θ

β

1 (4.120b)

(37)

Now

2

2 2

2

2 2

2 2

2 2 2

2 2

2

2 2 2

2 2 2

2 2 2

2 2

2

2 2 2

2 2 2

1 1

cos sin sin

sin cos

sin

cos cos

sin cos

sin cos

cos sin cos

cos sin

cos

sin cos

sin cos

sin sin

cos sin

sin cos

sin cos

cos sin

cos sin

sin cos

⎟⎠

⎜ ⎞

∂ + ∂

⎟⎠

⎜ ⎞

= ∂

⎟⎠

⎜ ⎞

∂ + ∂

⎟⎠

⎜ ⎞

= ∂

⎟⎟⎠

⎜⎜ ⎞

∂ + ∂

⎟⎠

⎜ ⎞

= ∂

− ∂

⎟⎟⎠

⎜⎜ ⎞

∂ + ∂

∂ + ∂

− ∂

− ∂

∂ + ∂

∂ + ∂

⎟⎠

⎜ ⎞

∂ + ∂

∂ + ∂

⎟⎟⎠

⎜⎜ ⎞

∂ + ∂

∂ + ∂

− ∂

− ∂

− ∂

− ∂

⎟⎠

⎜ ⎞

= ∂

⎟⎟⎠

⎜⎜ ⎞

∂ + ∂

− ∂

⎟⎟⎠

⎜⎜ ⎞

∂ + ∂

− ∂

⎟⎟⎠

⎜⎜ ⎞

∂ + ∂

⎟⎟ ∂

⎜⎜ ⎞

∂ + ∂

− ∂

=

⎟⎟⎠

⎜⎜ ⎞

∂ + ∂

− ∂

− ∂

⎟⎟⎠

⎜⎜ ⎞

∂ + ∂

= ∂

− ∂

s n

C C

n s

y x

y y

y x x

y

y x y

x x

x

y y

y x x

y

y x y

x x

x

y x

y x

y x

y x

y x

s y x

n n s s n

β β θ θ

θ θ

θ θ θ θ

θ θ θ θ

θ θ θ

θ θ θ

θ θ θ θ

θ θ θ

θ

θ θ θ θ

θ θ θ θ

θ θ θ

θ θ θ

θ θ θ θ

θ θ θ

θ

θ θ θ θ

θ θ

θ θ θ θ

θ θ

θ θ θ θ

θ θ θ θ

θ θ

On the other hand, directly from (4.113) and (4.120b),

2 2 2

2 2 2 2

2

1 1

1 1

1 1

s s

n C C n

C C

s s

n C C n n s s n

− ∂

⎟⎠

⎜ ⎞

∂ + ∂

− ∂

⎟⎠

⎜ ⎞

= ∂

⎟⎟⎠

⎜⎜ ⎞

− ∂

⎟⎠

⎜ ⎞

− ∂

= ∂

− ∂

β β β

β β β θ

θ

(38)

Comparing the two results,

1 0 1

2 2 2

2 =

∂ + ∂

s n

C C

β β 1 0

2 2 2

2 =

∂ + ∂

∂ β β

n C C s

where

2 2 2 2

2 2 2

2 2 2 2

2 2 2

2 2 2 2

2 2 2

2 2 2 2

2 2 2

2 2 2 2

2

2 2

2 2 2

2 2

2 2

cos cos

sin 2 1 sin

sin cos

2

cos cos

sin 2 sin

2

cos cos

sin 2 sin

sin cos

cos cos

sin 2 cos

sin sin

sin cos

sin cos

cos cos

sin cos

sin cos

cos sin sin

sin cos

sin

cos sin

cos

cos sin

sin sin

cos

cos sin

sin cos

cos sin

y C y

x C x

C s

y C x

C

y C y

x C x

C n

s C

y C y

x C x

C y

C x

C n

n s C

y C y

x C y

x y

C

x C y

x x

C n

s C

y C y

C y y

x C x

C y

y x

C y

C x x

C x

C x n

s C

y C x

C y

y C x

C x

n y C x

C

n y

C y

C n n

x C x

C n

y C x

C n

n C

∂ + ∂

− ∂

∂ + ∂

⎟⎟ ∂

⎜⎜ ⎞

∂ + ∂

− ∂

=

∂ + ∂

− ∂

∂ + ∂

− ∂

=

∂ + ∂

− ∂

∂ + ∂

⎟⎟⎠

⎜⎜ ⎞

∂ + ∂

− ∂

−∂

=

∂ + ∂

− ∂

⎟⎟⎠

⎜⎜ ⎞

− ∂

∂ + ∂

∂ + ∂

⎟⎟⎠

⎜⎜ ⎞

− ∂

∂ + ∂

−∂

=

∂ + ∂

− ∂

− ∂

− ∂

− ∂

∂ + ∂

∂ + ∂

∂ + ∂

−∂

=

⎟⎟⎠

⎜⎜ ⎞

∂ + ∂

− ∂

∂ + ∂

⎟⎟⎠

⎜⎜ ⎞

∂ + ∂

− ∂

− ∂

⎟⎟∂

⎜⎜ ⎞

∂ + ∂

− ∂

=

− ∂

− ∂

− ∂

− ∂

=

⎟⎟⎠

⎜⎜ ⎞

∂ + ∂

− ∂

= ∂

θ θ

θ β θ

θ β θ

θ θ

θ θ θ

θ θ

θ θ

θ θ θ

θ

θ θ

θ θ θ θ

θ θ

θ θ θ θ

θ θ θ

θ θ θ θ θ

θ θ θ

θ θ θ

θ θ θ

θ θ θ

θ θ

θ

θ θ

θ θ θ

θ

θ θ θ θ θ

θ

θ θ

(39)

Now, we have an ODE for β(s):

2 0

2β + β + β =

ds q pd ds d

where

⎟⎟⎠

⎜⎜ ⎞

∂ + ∂

− ∂

= y

C x

C s C

p 2 cosθ sinθ )

(

⎟⎟⎠

⎜⎜ ⎞

∂ + ∂

− ∂

= 1 sin222 2sin cos 2 cos2 22 )

( y

C y

x C x

C s C

q θ θ θ θ

Initial condition:

β =1, =0 dt dβ

at t=0 in deep water

which means that no refraction occurs, i.e. Kr =1, in deep water

A ray tracing method gives θ and H along rays not at grid points. So, it is difficult to use the results from ray tracing method as input data for numerical models (e.g.

sediment transport model) using finite difference method. We need interpolation.

Furthermore, infinite wave heights are calculated at the ray-crossing points. To resolve these problems, finite-difference refraction models have been developed, e.g. Noda et al.

(1974), Perlin and Dean (1983), Dalrymple (1988).

(40)

Wave breaking

(1) Finite, constant depth Miche (1944):

⎟⎠

⎜ ⎞

= ⎛

⎟⎠

⎜ ⎞

L h L

H

b

π tanh 2 142 . 0

As h/L→0, H /h=0.89 at breaking.

But, Nelson (1983, 6th Australian Conf. on Coastal and Ocean Eng.) found based on experimental results that

55 .

≤0

⎟⎠

⎜ ⎞

h b

H on a constant depth

(41)

(2) Shoaling waves in shallow water

where hb = breaking depth, and Hb = breaker height.

⎟⎟⎠

⎜⎜ ⎞

= ⎛ , 2 gT m H h

H b

b

b κ as in Fig. 12.7 of textbook,

where m = beach slope.

Surf similarity parameter based on (Battjes, 1974, Proc. 14th Coastal Eng. Conf., ASCE, 466-480):

Hb

/ L0

H m

b b = ξ

Breaker types depending on ξb:

Spilling ξb ≤0.4 smaller beach slope and steeper wave Plunging 0.4≤ξb ≤2

(42)

Collapsing 2≤ξb ≤3.5

Surging 3.5≤ξb larger beach slope and milder wave

For other breaker height formulas, see Rattanapitikon and Shibayama (2000) Coastal Eng. Journal, 42(4): 389-406.

Analyzing many laboratory results, Weggel (1972) proposed

) 2

( )

( gT

m H a m

bb

κ =

where

(

e m

)

m

a( )=43.81.0− 19

(

1.0 19.5

)

1

56 . 1 )

(m = +e m b

On a beach with straight and parallel contours,

r s g

K K C H

H C

H 0 0 0 0

cos cos

2 =

= θ

θ

For breaking wave in shallow water, Cgghb and θb ≅0. Therefore

2 / 1

0 0

0 cos

2 ⎟⎟

⎜⎜

≅ ⎛

=κ θ

b b

b gh

H C h H

0 2 0

0 2

2 cos

2 θ

κ

b

b gh

H C h =

(43)

2 1 02 0 cos 0

2 2 / 1 2 /

5 θ

κ

C H hb = g

5 / 2 0 0 2 0 5 / 4 5 /

1 2

cos

1 ⎟⎟⎠

⎜⎜ ⎞

= ⎛ θ

κ

C H hb g

5 / 2 0 0 2 0 5 / 1

2

cos ⎟⎟⎠

⎜⎜ ⎞

⎟⎟ ⎛

⎜⎜ ⎞

=⎛

=κ κ H C θ

h g Hb b

For a plane beach with h=mx, where x = distance from shoreline,

5 / 2 0 0 2 0 5 / 4 5 /

1 2

cos

1 ⎟⎟⎠

⎜⎜ ⎞

= ⎛ θ

κ

C H mxb g

5 / 2 0 0 2 0 5 / 4 5 /

1 2

cos

1 ⎟⎟⎠

⎜⎜ ⎞

= ⎛ θ

κ

C H xb mg

where xb = distance from shoreline to breaker point.

(44)

4.9 Wave Diffraction

High energy region →(Energy transfer)→ Low energy region

Examples:

(45)

Diffraction behind a breakwater: Penney and Price (1952) Philosophical Transaction of Royal Society of London, A244: 236-253.

BBC, DFSBC, and KFSBC are same as before. Far offshore from breakwater,

) cosh sin(

) ( cosh

2 ky t

kh z h k g

H σ

φ =− σ + − as y→−∞

Assume

{

F x y Z z e iσt

}

φ =Im ( , ) ( )

where

) ( cosh )

(z k h z

Z = +

Substituting into the Laplace equation,

2 0

2 2

2 2 2

2 ⎟⎟=

⎜⎜ ⎞

∂ + ∂

∂ + ∂

= ∂

z F Z y

ZF x

e iσt ZF φ

2 0

2 2 2

2 + =

∂ + ∂

Fk Z

y Z F x Z F

2 0

2 2 2

2 + =

∂ + ∂

∴ ∂ k F

y F x

F Helmholtz equation for F(x,y)

(46)

The solution is given by Eq. (4.135) of textbook.

4.10 Combined Refraction-Diffraction

Assume locally

t

e i

kh z h y k

x

F σ

φ = +

cosh ) ( )cosh , (

Berkoff (1972), Proc. 13th International Conference on Coastal Eng., 471-490:

(

)

+ 2 =0

F

C F C

CCg h g

h σ mild-slope equation

The mild-slope equation is an elliptic equation, which gives a boundary value problem in horizontal space and can be solved when all the boundary conditions are specified.

Radder (1979, Journal of Fluid Mechanics, 95: 159-176) proposed the parabolic approximation, which gives an initial value problem so that the parabolic equation can be solved by marching toward shore, with given offshore boundary condition.

Also, Copeland (1985, Coastal Engineering, 9: 125-149) proposed a hyperbolic equation, and Massel (1993, Coastal Engineering, 19: 97-126) proposed an extended refraction-diffraction equation, which includes higher-order bottom effect terms compared with the mild-slope equation.

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