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Net sediment transport and formation of bedforms in the coastal zone results from

- asymmetric velocity profile of waves- joint action of waves/tides and steady currents

Crescentic bottom forms ( near Algeria)from Bowen et al. (1962)

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Questions:

1. Why do velocity profiles become asymmetric?

2. What mechanisms cause generation ofsteady currents?

Ad 2: • wind• pressure gradient• … but also nonlinear processes

In case dominant flow is time-periodic =>(partial) conversion of periodic→steady flow: rectification

Ad 1: nonlinear terms in hydrodynamic equations

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Aim of this session:

1. Simple model to demonstrate generation of

• wave asymmetry• rectified flow (streaming)

in the bottom boundary layer

2. Consequences for

• bottom stress• sediment transport• formation of bottom patterns

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Situation sketch (notation Mei et al.)

z=0

z=H+ζ

z=H

u

w

x

z

• flat, horizontal bottom• density ρ = constant• no variation in y-direction• shallow water waves• weakly nonlinear waves (amplitude a

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Equations of motion and boundary conditions:

2

2

0

1

10

u wx zu u u P uu wt x z x z

P gz

νρ

ρ

∂ ∂+ =

∂ ∂∂ ∂ ∂ ∂ ∂

+ + = − +∂ ∂ ∂ ∂ ∂

∂= − −

∂

Here, ν~10-6 m2s-1 molecular viscosity coefficient

Boundary conditions:

at : , , 0

at 0 : 0 (noslip)

atmd uz H w P Pdt z

z u w

ζζ ν ∂= + = = =∂

= = =

hydrostaticbalance

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In this case: consider laminar flow(no turbulence)

and 0function ( / )cRe A z= critical Reynolds number

z0 : roughness length

Observations + theory: in case of oscillatory flow (frequency ω, near-bed orbital amplitude A) this assumption is valid if

cRe Re<

where 2ARe ω

ν= Reynolds number

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For smooth bed: z0 = d50/12 (d50 median grain size)For rippled bed: z0 ~ (ripple height)2 / ripple length

In practise bottom roughness is measured by

030sk z= Nikuradse roughness

104

105

106Rec

A/ks101 102 103 104

Typical dependence of Rec on A/ks

rough

smooth

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Example:

; smooth bed : d50~ 4⋅10-4 m

Near-bed orbital amplitude :

Wave amplitude/frequency : a=0.3 m, ω ~ 0.6 s-1

=> A/ks = 103, Rec ~ 106

while Re=5 ⋅105 ; so laminar flow

104

105106

Rec

A/ks101 102 103 104

rough

smooth

( )1/ 2 ~ 1mgHaAH ω

=

Depth H=2.5m

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Analysis of the model

From hydrostatic balance:

( , )P g z p x tρ= − +

no z-dependencesign error p42

Furthermore:2

2viscous ~inertia

uzut

ν ∂∂∂∂

⎡ ⎤⎢ ⎥⎣ ⎦

⎡ ⎤⎣ ⎦

So1/ 22νδ

ω⎛ ⎞=⎜ ⎟⎝ ⎠

scale on which viscous termsare important

2 2

2 2

[ ] / ( / )[ ] 2

U HU H H

ν ν ω δω

= = ≡

velocity scale

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For ν =10-6 m2s-1 and ω =0.5 s-1, δ ~ 2 mm, so δ

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In the inviscid core region : u = UI(x,t)

where1I I

IU U pUt x xρ

∂ ∂ ∂+ = −

∂ ∂ ∂

gxζ∂

= −∂

(shallow water waves)

independent of z

Here, assume UI to be prescribed.

{ }0 0Re ( ) | ( ) |cos[ ( )]i tIU U x e U x t xω ω ϕ−= = −

with |U0(x)| : velocity amplitudeϕ(x)=arg(U0): phase (time of maximum UI)

Note: velocity profile is symmetrical (!)

Mei et al. choose

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Use assumption of weakly nonlinear waves=> x-momentum equation is

Solution procedure

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2I I

IUu uu w U

x zUu

zxu

t tν∂∂ ∂+ = + +

∂ ∂∂∂

+∂ ∂ ∂

∂∂

Near bottom: black terms are of order ω [U] = ω2 A

The red terms are of order k [U]2 (k : wave number),i.e., factor kA

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21 1

2Iu U u

t t zν∂ ∂ ∂= +

∂ ∂ ∂

First order system is linear (retain only black terms):

for given UI .Boundary +matching condition:

1 10 at 0 , for 1Izu z u UH

ξ= = → ≡ >>

Recall,

{ }0 0Re ( ) | ( ) |cos[ ( )]i tIU U x e U x t xω ω ϕ−= = −So, also solution for u1 will be periodic in timewith frequency ω:

{ }1 0 1Re ( ) ( ) i tu U x F e ωξ −=

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Substitution yields2

11 0 0 02 2

i t i t i td Fi F U e i U e U ed

ω ω ωνω ωδ ξ

− − −− = − +

12 ωor

21

12

12

d F i F idξ

+ =

Solution that obeys the boundary + matching condition:

[ ]1 1 exp (1 )F i ξ= − − −

{ }1 0 1Re ( ) ( ) i tu U x F e ωξ −=Thus

{ }( ) ( )0 1Re ( ) ( )i x i i tU x e F e eϕ ψ ξ ωξ −= ⋅ ⋅0 1( ) ( ) cos[ ( ( ) ( ))]U x F t xξ ω ϕ ψ ξ= − +

amplitude phase

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Results:

dimensionless amplitude

phase Ψ(in units of wave period)

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Time behaviour of u1/U0 profile

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Now the first order vertical velocity component

From continuity equation:

So

1 1 0u wx z

∂ ∂+ =

∂ ∂

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0

z uw dzx

∂= −

∂∫

01 1

0

Re ( ') ' i tdUw F d edx

ξωδ ξ ξ −

⎧ ⎫⎪ ⎪= −⎨ ⎬⎪ ⎪⎩ ⎭

∫

Substitute known solution for u1 :

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Bottom stress due to linear waves:

01

0

Re (1 ) i twz

Uu i ez

ωρντ ρ νδ

−

=

∂ ⎧ ⎫≡ = −⎨ ⎬∂ ⎩ ⎭

So magnitude:1/ 2

0[ ] ( )w Uτ ρ ν ω=

Identification of (*) and (**) yields:1/ 2

0

2 ( )wf U

ν ω=

1/ 2

1/ 2

2 2A Re

νω

= = (laminar flow !)

(*)

In practise, [τw] is often related to wave friction factor fw :

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1[ ]2w w

f Uτ ρ= (**)

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The second order system

2

2I I

IUu uu w U

x zUu

zxu

t tν∂∂ ∂+ = + +

∂ ∂∂∂

+∂ ∂ ∂

∂∂

Approximate solutions:

2 21 1.... , ....u u wu w w= + + = + +

Recall, the x-momentum equation:

Black terms ~ ω2 A; Red terms factor kA smaller than black terms

Collect terms that are proportional to A2=>

1 11

22 2

21I

Iu u Uu w U

zx xu

zu

tν∂ ∂ ∂+

∂ ∂∂

+∂

∂+ =

∂ ∂

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1 11

22 2

21I

Iu u Uu w U

zx xu

zu

tν∂ ∂ ∂+

∂ ∂∂

+∂

∂+ =

∂ ∂

Recall,

Red terms are known (solutions of first order problem)Unknown here is u2

Question: what is the structure of u2?

Answer: that is determined by the red terms

( )1 1 1 1( )u u u wx z∂ ∂

+∂ ∂

from continuity

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1 1 1

22 2

21( ) ( )I

IUu u u w U

x zu u

zxtν∂∂ ∂+

∂∂ ∂

+ = +∂ ∂∂ ∂

Recall,

and the first order solutions

{ } { }0 1 0 1Re ( ) , Re ( ) ( )i t i tIU U x e u U x F eω ωξ− −= =

01 1

0

Re ( ') ' i tdUw F d edx

ξωδ ξ ξ −

⎧ ⎫⎪ ⎪= −⎨ ⎬⎪ ⎪⎩ ⎭

∫

constant in time periodic with frequency 2ω

Other ‘red terms’ : same behaviour, but z-dependent !

So e.g.,*

20 00 0

1 1Re Re2 2

iI tI

UUU UU Ux

ex x

ω−⎧ ⎫∂ ∂⎧ ⎫= +⎨ ⎬ ⎨ ⎬∂ ∂⎩ ⎭⎩ ⎭

∂∂

complex conjugate

22M0-M2:net current

u3

M2-M4: wave asymmetry

u3

Thus, solution of second order problem of the form

{ }22 2 2ˆ( , , ) ( , ) Re ( , ) i tu x z t u x z u x z e ω−= +

Time series of velocity u2 are asymmetric !Implications for net sediment transport:

steady (‘M0’) higher harmonic (‘M4’)

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From continuity equation: vertical velocity at second order:

{ }22 2 2( , , ) ( , ) Re ( , ) i tw x z t w x z w x z e ω−= +steady higher harmonic

Procedure:

• substitute solutions u2, w2 in equations at second order

• separate problems for steady parts and time-harmonic parts

• solve equations by straightforward methods

Here focus on the steady parts of the solutions,which describe (viscous) streaming

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*0

2 2 01( , ) Re ( ) Uu x F U

xξ ξ

ω⎧ ⎫∂

= − ⎨ ⎬∂⎩ ⎭

The solution for the horizontal steady velocity:

and F2(ξ ) in Eq. (2.15a) of Mei et al.

Note:

So streaming extends beyond the bottom boundary layer (!)

23( 1) (1 )4

F iξ >> → −

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Choice in Mei et al.:

[ ]( , ) sin( ) sin( )IU x t A kx t R kx tω ω ω= + + −

wave to left wave to right

withω A = [U] : velocity amplitudek : wave numberR : reflection coefficient

symmetrical (!)

Analysis of streaming for a specific UI

In particular R=0 : travelling wave (to left)R=1 : standing wave

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Eulerian streaming for R=0 (travelling wave)

independent of x;and in direction ofwave propagation

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3( 1)4

u k Aξ ω>> → −

Note

This expression is often used as input insediment transport formulations

(minus sign: wave to left)

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Eulerian streaming for R=1 (standing wave)

Streaming depends on x and z and changes sign

Contour plot u(x,z)=constantA: antinode of free surface, N: node

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Assume: sediment is moved by streaming near the bed=> sediment converges underneath nodes of ζ=> waves force generation of longshore bars

Eulerian streaming for R=1 (standing wave)

Below: plot of streamlines Ψ=constant, where

2 2,u wz x∂Ψ ∂Ψ

= − =∂ ∂

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Example of template bottom pattern: submarine longshore bars

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1. Eulerian laminar streaming for 0 < R < 1Analysis in Exercise 2.1 (Matlab script)

Remarks and outlook:

2. Free forced generation of bedformsVelocity field 0 sin( )IU U tω=

U0 independent of xPerturbations in bed level: ( ) cos( )bz x Kxα=

wavenumber K is arbitrary

bar trough

streaming:

bar

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3. Flow in the turbulent boundary layer

Options:

Remarks and outlook:

• Direct simulation of Navier Stokes equations(lectures G. Vittori)

• Replace molecular viscosity ν byturbulent eddy viscosity νe (P. Blondeaux/G. Vittori)

Zeroth order approach: νe=constant,where νe >> ν (→ Exercise 2.2)

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Topic of next week: Tidal sand banks(P. Blondeaux)

Literature (download from website):Huthnance, J. 1982. On one mechanism forming linear sand banks. Est. Coastal Shelf Sci. 14, 77-97.

Important aspects: * tidal rectification → net sediment transport* morphodynamic self-organisation

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