Chapter 9. Waves over Real Seabeds We have to consider the followings in real sea: -

Chapter 9. Waves over Real Seabeds

We have to consider the followings in real sea:

- viscosity (fluid)

- Roughness, rigidity, permeability (seabed) Effect of viscosity

Inside the bottom boundary layer, - Viscosity is important.

- No-slip condition at bottom:

=0

∂

−∂

=

−

= h

x z

u φ

Linear laminar Navier-Stokes equation for viscous fluid:

⎟⎟⎠

⎜⎜ ⎞

⎝

⎛

∂ + ∂

∂ + ∂

∂

− ∂

∂ =

∂

2 2 2

1 2

z u x

u x

p t

u ν

ρ

z g w x

w z

p t

w ⎟⎟⎠−

⎜⎜ ⎞

⎝

⎛

∂ + ∂

∂ + ∂

∂

− ∂

∂ =

∂

2 2 2

1 ν 2

ρ

To evaluate relative sizes of various terms, introduce nondimensional variables:

x'=kx z'=kz t'=σt

ga p p

= ρ '

( )

1 ( , )

' ,

' u w

w a

u ⁼ σ

By chain rule,

' '

' k x

x x x

x ∂

= ∂

∂

∂

∂

= ∂

∂

∂

2 2 2 2

2

' '

' '

' k x

k x k x x k x

k x x

x ∂

= ∂

⎟⎠

⎜ ⎞

⎝

⎛

∂

∂

∂

= ∂

⎟⎠

⎜ ⎞

⎝

⎛

∂

∂

∂

= ∂

⎟⎠

⎜ ⎞

⎝

⎛

∂

∂

∂

= ∂

∂

∂

' '

' t t

t t

t ∂

= ∂

∂

∂

∂

= ∂

∂

∂ σ

The x -momentum equation becomes

⎟⎟⎠

⎜⎜ ⎞

⎝

⎛

∂ + ∂

∂ + ∂

∂

− ∂

∂ =

∂

2 2 2 2 2

2

' ' '

' '

' '

'

z u x

a u x k

ga p k t

a u ρ ν σ

σ ρ

⎟⎟⎠

⎜⎜ ⎞

⎝

⎛

∂ + ∂

∂ + ∂

∂

− ∂

∂ =

∂

2 2 2 2 2

2 '

' '

' '

' '

'

z u x

u k

x p gk t

u

σ ν σ

where

2 1 2

1 2

2

2 1 (1);

−

− = = =

=

=

gk O C

C gk k C

gk

gk F

σ F

) 10 ( and

);

10

~ 10

1 ( _{7 8} ¹ ₅

2

2 − −

−

− = =

=

=

=

= Ck O

C O k Ck

k

k ν

ν ν

ν σ

ν R

R

which implies that frictional stresses are negligible so that there is a slip boundary condition at the bottom (u≠0 at z=−h). However, physically, there is no flow at the bottom in the viscous flow. Hence, our argument above must be modified.

Near the bottom, u varies rapidly with z . Therefore, the vertical length scale must be different from the horizontal length scale. Let us take the vertical length scale as the boundary layer thickness, δ , so that

δ z'= z

and

' 1 '

' z z

z z

z ∂

= ∂

∂

∂

∂

= ∂

∂

∂

δ ;

2 2 2 2 2

' 1

z

z ∂

= ∂

∂

∂ δ

Using this scale but the same scales as before for other parameters,

2 2 2 2

2 2 2

' ' 1

' ' '

' '

'

z a u x

a u x k

ga p k t a u

∂ + ∂

∂ + ∂

∂

− ∂

∂ =

∂ σ

ν δ σ

ν ρ ρ

σ

2 2 2 2 2

2 '

' '

' 1 ' ' 1 '

'

z u x

u x

p t

u

∂ + ∂

∂ + ∂

∂

− ∂

∂ =

∂

σ δ

ν F R

The second term on the RHS is again very small compared with other terms, but the last term is of O(1) with δ ~ ν/σ . Now, the x -momentum equation can be approximated to

2

1 2

z u x

p t

u

∂ + ∂

∂

− ∂

∂ =

∂ ν

ρ

Assuming u=u_p +u_r, where and , respectively, represent the potential and rotational part of the flow, we can separate the equation into

up u_r

x p t

u_p

∂

− ∂

∂ =

∂

ρ 1

2 2

z u t

u_{r r}

∂

= ∂

∂

∂ ν

each of which has a form of Euler equation and diffusion equation, respectively.

For the potential flow part, we have

) (

cosh ) (

cosh _{i kx t}

p e

kh z h k

u gak ^σ

σ ⁻

= +

For the rotational part, assume

)

) (

(z e^{ikx t} Af

u_r = ^−σ

Substituting into the diffusion equation,

0 )

' '

(−iσAf −νAf e^{i(kx− tσ)} =

or

0 '

'+i f =

f ν

σ

Solving this equation and keeping only the term that decays away from bed (or

as ),

0 ) (z →

f (h+ )z →∞

) ( 2 / ) 1 ( )

(

) /

(z Be ^{i h z} Be ^{i h z} f = ^{−σ ν +} = ^{− − σ ν +}

Time-out

De Moivre theorem: z^1/n =ⁿ re^iθ/n

) 1 ( 2 1 2 1 2 1 4

sin3 4

cos3

1 ⁴

2 3 2 / 3

2 e e i i i

i = ⁱ = ⁱ = + =− + =− −

− ^{π π} π π

ν σ ν

ν σ σ ν

σ (1 ) (1 ) 2

2 / 1

/ i i i

i = − =− − =− −

−

Now u_r becomes

)) ( 2 / (

) ( 2 /

) ( ) ( 2 / ) 1 (

z h t

kx i z h

t kx i z h i r

e Ae

e Ae

u

+ +

− +

−

− +

−

−

=

=

ν σ σ ν

σ

σ ν

σ

Using u=u_p +u_r =0 at z=−h,

cosh 0

1 eⁱ(^{kx− t}) +Aeⁱ(^{kx− t}) = kh

gak _{σ σ}

σ

or

kh A gak

cosh 1

− σ

=

Therefore,

)) ( 2 / (

) ( 2 /

cosh

1 h z i kx t h z

r e e

kh

u =−gak ^{− σ ν + −σ+ σ ν +} σ

and

[

^{cosh ( )cos( ) cos( /2 ( ))}

]

cosh

) ( 2

/ kx t h z

e t kx z

h kh k

u= gak + −σ − ^{− h+z} −σ + σ ν +

σ ^{σ ν}

See Figure 9.1 for the horizontal velocity profile near the bed.

The vertical velocity in the bottom boundary layer can be obtained from the continuity equation:

x dz w u

z w x u

∂

−∂

=

∂

→

∂ = + ∂

∂

∂ 0

∫

∫

∫

∫

− − − − ⁺ =

∂

− ∂

∂ =

− ∂

=

∂

⎟=

⎠⎞

⎜⎝

⎛ ∂ = =

= ^{z h z}

h z

h z

h z

h ds

x dz u

x w u

z w w w z

w( ) ( ) 0 Eq.(9.12)

Bed shear stress

) 4 / (

) 4 / (

) (

cosh

2 2 cosh

2 2

cosh )

(

π σ

π σ

σ

σ ν ρ

ν σ σ

μ

ν σ ν

σ σ

μ

μ μ

τ

−

−

−

−

−

=

=

⎟⎟⎠

⎞

⎜⎜⎝

⎛ −

=

⎟⎟⎠

⎜⎜ ⎞

⎝

⎛

∂ + ∂

∂

≅ ∂

⎟⎠

⎜ ⎞

⎝

⎛

∂ +∂

∂

= ∂

−

=

t kx i

t kx i

t kx i p r zx

kh e gak

kh e gak

e kh i

gak

z u z u x

w z h u

z

where the relationship, A(1−i)= 2Ae^−iπ/4, was used by De Moivre theorem. In terms of flow velocity, we have

( )

max ) 4 / ( max

max

max 8 2 cosh

)

( ⎟⎟

⎠

⎞

⎜⎜⎝

=⎛

⎟⎠

⎜ ⎞

⎝

=⎛

⎟⎠

⎜ ⎞

⎝

=⎛

−

= ^−σ−π

σ ν ρ ρ

τ_zx ρ _{b b} ^w _{b b} e^{ikx t}

kh u gak

f u u

f u h

z

where = potential flow velocity outside boundary layer. Solving for the friction coefficient, ,

ub

f

ν σν ς

σ ν σ

σ σ

ν

/ 8 /

8 cosh

8 cosh

8

max max max

max

2 2 2

b b b

b

b kh u u

gak u

kh gak

f =u = = =

or

2 / 1

8

b

f =R ; / ² / 10³ ~10⁴ for smooth bottom

max = <

= _b ς_b ν ς_bσ ν

b u

R

See Figure 9.2 of textbook for linear relation between and for laminar flow.

f

log logR_b

Energy dissipation

Mean rate of energy dissipation is given by (Fluid Dynamics, Batchelor)

kh E k

z h z d u

z h z d

w z

u x w x

u

h r

h D

2 sinh

2 /

) (

) ( 2

2

0

2 0

2 2

2

ν σ ν

ρν ρν ε

=

⎟ +

⎠

⎜ ⎞

⎝

⎛

∂

≅ ∂

⎪⎭ +

⎪⎬

⎫

⎪⎩

⎪⎨

⎧ ⎟

⎠

⎜ ⎞

⎝

⎛

∂ + ∂

⎟⎠

⎜ ⎞

⎝

⎛

∂ + ∂

∂ + ∂

⎟⎠

⎜ ⎞

⎝

⎛

∂

= ∂

∫

∫

x AE EC_g

D =

∂

−∂

= ( )

ε where A=νk σ /2ν /sinh2kh.

Consider waves on a flat bottom. Since C_g ≠ f(x), we have

C E A x

E

g

−

∂ =

∂

x C

A _g

e E

E= ₀ ^{−( / )}

x C

A _g

e a

a² = ₀^{2 −( / )}

x kh C

k x

C

A _{g g}

e a e

a

a ^⎟

⎟

⎠

⎞

⎜⎜

⎝

−⎛

− =

= ^{2 sinh2}

2 /

0 ) 2 / ( 0

ν σ ν

which states that the wave amplitude decays exponentially with x .

Turbulent boundary layer

Large waves + Rough bottom → Turbulent boundary layer (most cases in nature)

In turbulent boundary layer,

) / (k_{e b} f

f = ς

as shown in Figure 9.2 and where

2d90

k_e =

considering only skin friction without form drag due to ripples.

The bed shear stress is given by

t t f u

u f u h

z _{b b m}

zx ρ ρ σ σ

τ cos cos

8 ) 8

( =− = = ²

where

t u

u_b = _mcosσ with

kh a kh u_m gak

sinh cosh

σ

σ ⁼

=

Note that the time average of shear stress is zero. The mean rate of energy dissipation is not zero and is given by

3 3 3

4 / 0

3 3

4 / 0

3 3

2

sinh 6

6 3 2 2

4 8

) cos 4 / (

1 8

4 cos / 1 8

) 8 (

⎟⎠

⎜ ⎞

⎝

= ⎛

=

=

=

=

=

−

=

∫

∫

kh a f

f u f u

T tdt f u

T tdt f u u

f u u h

m m

T m

T m

b b b

zx D

σ π

ρ

π ρ π ρ

σ σ σ

ρ

ρ σ τ ρ

ε

Energy loss with distance

kh a f dx

EC d

D g

3 3 3

sinh 6 )

( σ

π ε =−ρ

−

=

On flat bottom,

kh a f dx

gC_g da

3 3 3 2

6 sinh 2

1 σ

π ρ =− ρ

kh a f dx agC_g da

3 3 3

6 sinh σ

− π

=

khdx gC

f a

da

g 3 3

2 6π sinh

− σ

=

Defining A= fσ³/(6πgC_g sinh³kh) and integration give

1

1 Ax C

a=− +

−

Using the boundary condition a=a₀ at x=0, we get C₁=−1 a/ ₀. Therefore

0

1 1

Ax a a=− −

−

Finally we have

x Aa a a

0 0

1+

=

which is not exponential decay with x . For small Aa₀, we have

x

e Aa

a x Aa a

a≅ ₀(1− ₀ )≅ ₀ ^{− 0} ← exponential decay

Waves over viscous mud bottom

At the bottom, we need the kinematic and dynamic matching conditions:

φ χ

χ φ _{=− +}

∂

−∂

∂ =

−∂

∂ =

∂ z h

z z

t ^{2 1} on

χ +

−

=

= p z h

p_{1 2} on

In Region 1, we can assume the solution as

(

Acoshk(h z) Bsinhk(h z)

)

e^{i(kx σt)}

φ₁= + + + ⁻

where the second term is included because the vertical velocity is no longer zero at the bottom. Using LDFSBC,

) sinh cosh

0 (A kh B kh

g

a =−iσ +

Using LCFSBC,

0 ) sinh cosh

( )

cosh sinh

(

2 + =

−

+ A kh B kh

kh g B

kh A

k σ

Multiplying g cosh/ kh on both sides,

0 ) tanh (

) tanh

(gk kh− ² +B gk − ² kh =

A σ σ

From LDFSBC,

kh kh B

iga kh

kh A ⁱ B

ga

cosh tanh cosh

sinh ₀

0 − = −

= −

σ σ Plug in LCFSBC,

) tanh

( cosh tanh

) tanh

( ^{2 0} σ²

σ σ ⎟ −

⎠

⎜ ⎞

⎝

⎛ − +

=

− B kh gk kh

kh B iga

kh gk

) tanh cosh (

) tanh tanh

tanh

( ^{2 2 2 0 2} gk kh

kh kh iga

kh gk

kh gk

B − − + = σ −

σ σ σ

) tanh cosh (

cosh

1 _{0 2}

2 gk kh

kh iga

Bgk kh = σ −

σ

) tanh cosh (

) tanh cosh (

cosh² _{0 2 0 2}

kh k gk

kh kh ia

kh gk iga gk

B= kh − = σ −

σ σ σ

Now,

) tanh cosh (

tanh cosh tanh

cosh

cosh )sinh tanh cosh (

cosh

0 2

2 2

2 0

0 2 0

kh k gk

kh ia

kh gk

kh kh gk k

kh ia

kh kh kh

k gk kh ia

kh A iga

σ σ σ σ

σ σ σ

−

=

⎟⎠

⎜ ⎞

⎝

⎛ − +

=

−

−

=

Note: On rigid bottom, B=0 or σ² =gk tanhkh.

In Region 2, assuming infinite depth,

) ( ) (h z ikx t

k e

de ^σ

φ₂ = ^{+ −}

which indicates exponential decay as h+ z→−∞. The horizontal velocity is given by

r r

p u

u x u

u_{2 2 2} ² + ₂

∂

−∂

= +

= φ

where is the boundary layer correction given by (cf. Eq. 9.9) u₂r

) ) ( ( 2 / ) 1 ( 2

2 h z ikx t

i e

fe u r

ν σ

σ + −

= −

Assuming is small and using the linearized kinematic matching condition at the boundary:

wr

h z z

z

t =−

∂

−∂

∂ =

−∂

∂ =

∂χ φ² φ¹ on

Bk dk m

i =− =−

− σ ₀

B d =

∴ ,

σ σ

σ

idk iBk

i

m₀ = Bk =− =−

Applying dynamic matching condition:

χ +

−

=

= p z h

p_{1 2} on

χ ρ

φ ρ ρ φ ρ

ρ + + − − =− +

∂

= ∂

∂ −

∂ gh g z h z h

gz t

t¹ _{1 2} ² _{1 2} ( ) on

1

where the last two terms represent hydrostatic pressures due to water and mud, respectively.

Linearizing,

h z t g

t g − =−

∂

= ∂

∂ −

∂ ¹ _{1 2} ² ₂ on

1 φ ρ χ ρ φ ρ χ

ρ

0 2 2 0

1

1A gm i d gm

iσρ −ρ =−σρ −ρ

−

⎟⎠

⎜ ⎞

⎝⎛−

− +

= σρ ρ ρ σ

σρ A i B g ^iBk

i _{1 2} ( _{2 1})

⎥⎦

⎢ ⎤

⎣

⎡ ⎟+

⎠

⎜ ⎞

⎝⎛ −

− =

−

= _{2 2}

1 2 2

1 1 2 1

2 ( ) 1

σ σ

ρ ρ σ

ρ ρ ρ ρ

ρ gk gk

gkB B B

A

Now,

kh gk

kh gk

gk gk

B A

tanh

1 ₂ tanh

2 2

2 2

−

= −

⎟+

⎠

⎜ ⎞

⎝⎛ −

= σ

σ σ

σ ρ

ρ

1

where Eqs. (9.51) and (9.52) were used. Rearranging gives the dispersion relationship:

kh gk

kh gk gk

kh gk gk

tanh )

tanh (

) tanh (

1 ₂ ² ₂ ^{2 2}

1

2 σ σ

σ σ σ ρ

ρ ⎟ − + − = −

⎠

⎜ ⎞

⎝⎛ −

(

^{2 gk}

)

^{( 2 gktanhkh) gk( 2 gktanhkh) gk 2 4 tanhkh}

1

2 σ σ σ σ σ

ρ

ρ − − + − = −

(

1 tanh

)

(1 tanh ) 1 ( )²tanh 0

1 2 2 1

2 4 1

2 ⎟⎟ =

⎠

⎜⎜ ⎞

⎝

⎛ −

+ +

−

+ kh gk kh gk kh

ρ σ ρ

ρ σ ρ ρ

ρ

0 tanh

1 tanh

) (

1 2 1

2 2

2 ⎥=

⎦

⎢ ⎤

⎣

⎡ ⎟⎟

⎠

⎜⎜ ⎞

⎝

⎛ −

⎟⎟−

⎠

⎜⎜ ⎞

⎝

⎛ +

−gk kh gk kh

ρ ρ ρ

σ ρ σ

We have two possible cases:

=gk

σ2 ← surface wave case

or

kh kh gk

tanh tanh 1

1 2 1 2 2

+

⎟⎟⎠

⎜⎜ ⎞

⎝

⎛ −

= ρ ρ ρ ρ

σ ← interfacial wave case

Using Eqs. (9.52) and (9.59),

) tanh (

cosh

) tanh (

cosh

2 2 0 2

0

kh gk

kh

kh iBk gk

kh i

kB m

a

= −

⎟⎠

⎜ ⎞

⎝⎛−

−

=

σ σ σ σ

σ

For surface wave case,

sinh 1 cosh

1 )

tanh 1 (

0 cosh

0 = >

= −

= − e^kh

kh kh

kh kh

gk

gk m

a

The free surface and interface are in phase, and the free surface amplitude is greater than the interface amplitude.

For interfacial wave case,

e kh

m

a ₋

⎟⎟⎠

⎜⎜ ⎞

⎝

⎛ −

−

= 1

1 2 0

0

ρ ρ

0 0 / m

a depends of ρ₂/ρ₁ and kh . a₀ /m₀ <1 if ρ₂ /ρ₁<2 for kh=0 .

1 2/ρ

ρ can be larger for larger kh to satisfy a₀ /m₀ <1. Also the free surface and

interface are 180° out of phase because a₀ /m₀ <0 always.

Because the free surface and interface are out of phase, between them a quasi- bottom exists where there is no vertical flow. If we denote its vertical distance from the SWL as

180°

z0 so that

0

1 0 on z z

z = =−

∂

∂φ

the dispersion relationship becomes

0 2 =gktanhkz σ

Waves over rigid, porous bottom

At the bottom, we need the kinematic and dynamic matching conditions:

h z

z =w =−

∂

−∂φ¹ ₂ on

h z p

p₁ = ₂ on =−

In the porous medium, we use the equation for unsteady Darcy’s flow:

2 2

2 1

1 u p

K t

u

n =− − ∇

∂

∂

ρ ν _r r

where n = porosity and K = permeability constant. Using ν =μ/ρ, the above equation can be written as

² ₂ K p₂

t u u n

K =− − ∇

∂

∂

μ

ν ^r

r

Since ur ∝e^i(kx−σt), the above equation becomes

2

2 2

2 K p

u K u

n

i =− − ∇

− ν μ

σ _{r r}

12 9 ~10 10

~ ^{− −}

K m² and ν ~10⁻⁶ m²/s. Therefore, . Thus, the preceding equation can be approximated to

6 3 ~10 10

~ ) /

(σK ν ^{− −}

2

2 K p

u =− ∇ r μ

Using the continuity equation, ∇ ur⋅ ₂ =0, we get

2 =0

∇ p₂

1

2

which is the governing equation in the porous medium.

As with the waves over a mud bottom, assume the solutions as

(

Acoshk(h z) Bsinhk(h z)

)

e^{i(kx σt)}

φ = + + + ⁻

and

) ( ) (h z ikx t

k e

De

p = ^{+ −σ}

which give non-zero vertical velocity at the bottom and exponential decay of pore pressure as h+ z→−∞.

As done for viscous mud bottom, LKFSBC and LDFSBC give

) tanh

cosh ( ₂

0 gk kh

k kh

A ia σ

σ ⁻

=

) tanh cosh ( ₂

0 gk kh

k kh

B= ia σ −

σ

Dynamic matching condition at the bottom:

h z

t = p =−

∂

∂φ¹ ₂ on ρ

or

D A

i =

− σρ ρσ i A D

=−

Kinematic matching condition at the bottom:

h z z

K p

z =−

∂

− ∂

∂ =

−∂ ^{1 2} on

μ φ

or

K kD Bk⁼⁻ μ

−

K D B⁼ μ

Now

kh gk

kh gk

K i i K

B A

tanh tanh 1

2 2

−

= −

=

⎟⎟⎠

⎜⎜ ⎞

⎝

− ⎛

= σ

σ ν

σ ρσ μ

which gives

) tanh (

) tanh

( ² K gk ² kh

kh gk

i σ

ν

σ − = σ −

The LHS and RHS are imaginary and real, respectively. Therefore we get

kh gk tanh

2 =

σ and σ² =gk cothkh

which are in conflict. To satisfy the dispersion relationship, the wave number should be complex:

i

r ik

k k = +

The real part is related to the wave length, while the imaginary part determines the spatial damping:

) ( 0 ) ( 0

t x k i x k t

kx

i a e i e r

e

a ^{σ σ}

η= ⁻ = ^{− −}

In intermediate depth, where we can assume that Kσ /ν<<1 and (or no significant damping), Reid and Kajiura (1957) obtained

<<1 h k_i

h k gk tanh

2 ≅

σ _{r r}

h k h

k K k k

r r

r

i 2 sinh2

) / ( 2

≅ + σ ν

which is shown in Figure 9.6 of textbook.

In shallow water where kh <π /10, and are given by Eqs. (9.95) and (9.96), respectively.

ki k_r