tidal sand banks - universiteit utrecht · 2011-05-18 · 2 norfolk banks flemish and hinder banks...

1

TIDAL SAND BANKS

BEDFORMS IN SHALLOW TIDAL SEAS

2

Norfolk banks

Flemish and Hinder banks

• Crests slightly counter-clockwise or clockwise rotated with respect to the principal axis of the tidal ellipse

• Almost static bed forms (however they have a dynamics)

• Typical heights of the order of tens of metres

Schole Bank (Length 50 km ) – Normandy – Bisquay & Ledu 1999

• Crest to crest distances of the order of kilometres

• Lengths of the order of several tens of kilometres

BEDFORM CHARACTERISTICS

3

SAND BANKS (Norfolk banks (UK))

Counter-clockwise crests (with respect to the direction of the tidalcurrent). The wavelength is about 8-10 Km and the ridges are tens of kilometres long.

Tidal current

SAND BANKS (Zeeland banks (NL))

Clockwise crests (with respect to the direction of the tidal current).The wavelength is about f 4-5 Km and the ridges are tens of kilometres.

Tidal current

4

Shadowed regions=clockwise rotating tidal ellipseBlank regions= counter-clockwise rotating tidal ellipseA=clockwise oriented sand

banksB=counter-clockwise

oriented sand banks

B BA

AB

SAND BANKS AND TIDE ROTATION IN THE NORTH SEA

Adapted from Dyer & Huntley (1999) Adapted from Soulsby (1983)

B

Simultaneous presence of sand banks and sand waves

5

WHAT ARE THE MOTIVATIONS TO STUDY SAND BANKS ?

Dunwich, important port in the medioeval England - All Saints church: a) 1907

a)

Example of the problems induced by coastal erosion: what can we make ?

6

All Saints church: b) 1914

All Saints church: actual situation

7

EXAMPLE OF HUMAN INTERVENTION

Benidorm beach near Alicante.a) Before and b) after a beach nourishment made with sand extracted by submarine pits near Sierra Helada

A FURTHER EXAMPLE OF HUMAN INTERVENTION

Jesolo beach after a nourishment (1 million mc along 10 Km of coast) and the construction of transverse piers.

8

Jesolo beach during the nourishment works

Volumes of beach nuorishments along the Lagoon of Venice (total volume equal to about 7 million of m3)

9

KWINTEBANK (Belgium)

Sand extraction has been suspended and the bottom evolution is monitored in

order to understand the bed forms dynamics.

Left: Actual sea bottom topography.

The depressed areas have been evidenced by the

yellow-red colours.Right: Initial bottom

configuration (data given by the UMARSAND Project).

Which are the mechanisms controlling sand banks formation, development and their interaction with human activities?

10

Which are the mechanisms controlling sand banks formation, development and

their interaction with human interventions?

THEORETICAL ANALYSIS OF SAND BANK APPEARANCE

Objectives: 1) To demonstrate that sand banks can formspontaneously as free instabilities of the coupled water-bottomsystem, 2) Predict their geometrical characteristics Methodology: Stability analysis1 step: Model formulation2 step: Evaluation of the basic state (tidal current over a flat bottom,no bedforms)3 step: Introduction of a small bottom perturbation4 step: Evaluation of the flow responce5 step: Evaluation of the sediment transport rate induced by the bottom waviness6 step: Evaluation of the convergence/divergence of the sedimentflux over the crests

11

Northern Emisphere

Anticyclonic Residual Circulation

TIDAL

Sand

bankcrest

TIDAL

San

dba

nk

cres

t

CoriolisCoriolis

Hulscher etal. 1993

Sand banks counter-clockwise rotated Sand banks clockwise rotated

Velocity increases upstream of the crest Velocity decreases downstream of the crest

Velocity decreases upstream Velocity increases downstream

CREST AGGRADATION CREST DAMPING

Coriolis Coriolis

Sand

bankcrest

Hulscher etal. 1993

TIDALCoriolis

TIDAL

San

dba

nk

cres

t

Coriolis

clockwise rotated sand

banks EXIST!!

The direction of the velocity at

the bottom differs from the direction of the depth-averaged

velocity

THE MECHANISM OF SAND BANK FORMATION(Huthnance, 1982, Estuarine, Coastal and Shelf Science, 14)

A SIMPLE APPROACH : 2DH MODEL(Huthnance 1982, Estuarine, Coastal and Shelf Science, 14)

AIMInvestigation of the growth of sand banks due to the forcing of tidal currents

MAIN ASSUMPTIONS 1) A 2DH model (x-y plane), which follows by averaging the 3D shallow water equations over the vertical2) Coriolis effects are taken into account3) No suspended load

12

The tidal current far from the bottom waviness is supposed to be unidirectional, to form an angle α with the x*-axis and to be characterized by a strength U*0 I(t*) (I(t) is a periodicfunction, moreover a start denotes dimensional quatities).

U*

V*

*0

~D

*~D

Now, let us consider a wavy sea bed like that shown in the figure. Let us chose the (x,y,z)-axes such that the water depthD* depends only on y* but not on x*.

THE HYDRODYNAMICS

The problem is solved using dimensionless variables. The angular frequency ω* of the tide is used to scale time and U*0 is used to scale the velocities. Moreover, U*0/ω* and D*0 are used as horizontal and vertical length scales.

The amplitude of the tide is assumed to be much smaller thanthe local water depth. Hence the oscillations of the water surface play a minor role into the problem and are neglected. Then mass conservation provides

αsin~ IVD =In the previous relationship, dimensionless variables have beenused which are indicated without the star

13

Dimensionless momentum equation in the x-direction reads

used. is drag bottom quadratic a Moreover, value.- large itsby replaced becan and oft independen isgradient surface free theIndeed

~wheresin)(cos||cos

~

*0

*

*0

22

yy

DUrtfIIIrC

dtdI

fVD

UVUrCyUV

tU

D

D

ωααα =−+=

=−+

+∂∂

+∂∂

( ) ( )

( )( ) parameter Coriolis22

)11755 (e.g. roughness bottom

on the dependes which less),(dimensionparameter Chezy where

1 such that

10

2

22

20

=Ω=Ω=

=

=

=+

=

φωϕ

ρττ

sin/sinf

z/D~log.C

CC

CD~

V,UVUCU,

**

*r

*

DD**

*y

*x

( )

given is ~ once

ics,hydrodynam thedescribes completelyequation This

~sin~

cos||

1~1sin~

sincos

followsit

,~sinequation continuity from obtainingBy

2

222

D

UD

IUDrCIIrC

DfI

yU

DI

dtdI

tU

DIVV

DDα

α

ααα

α

+−=

=⎟⎠⎞

⎜⎝⎛ −−

∂∂

+−∂∂

⎟⎠⎞

⎜⎝⎛ =

14

Huthnance determined the sediment transport by means of the empirical formula

( )

1Or

length typical where2

such thattransport

sediment ofcomponent slope-down for the 1factor

t enhancemenan 2) rate,ransport sediment t thescales

such that factor scale essdimensionl a 1) esincorporatwhich

0

0

0

30

230

ϕϕωλ

πωλλ

λ

λ

tgrtgUh

LUL

yh

gUS

SyhVV,UV

gUSQ,Q

*

**

**

**

Huthnance

*

*

*

**y

*x

==

==

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

−

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

−=

THE MORPHODYNAMICS

( )

( )

( )

)s(CS

dg)s(CVdgs

gVS

dg)s(udgs

UV

gUS

dg)s(/dgsV

gUSQ

S

/

**

***

*

*

/

*****

*

*

*

*

/

**

****

*

**

c

18

1181

1181

followsit 1

81

)0( formulaMuller -Peter-Meyerwith predictor ransport sediment tthecomapringobtained of Estimate

3

23

3

33

3

2333

3

0

30

2333

30

−≈

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−≈

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−≈⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−≈≈

=

τ

ρτ

ϑ

15

⎥⎦

⎤⎢⎣

⎡∂

∂+

∂∂

−=∂∂

−

⎥⎥⎦

⎤

⎢⎢⎣

⎡

∂

∂+

∂∂

−=∂∂

−

yQ

xQ

D~gUS

th)n(

gUS

n

yQ

xQS

th)n(

yx**

*

*

*

*

*y

*

*x

*

*

0

20

30

1

toleadsequation continuitysediment ,introducedalready quantities ldimensiona theuse and

with rateransport sediment t thescaleyou If

porosity.sediment is where

1

readsequation continuitySediment

introduced is ~)1( scale timemicmorphodyna thewhere

||||

||||~)1(

followsit Then

||||~)1(

requires sand ofon conservati the,on dependnot does rateransport sediment t at theaccount th into Taking

*0

*

2*0

2

2*0

*

2*0

2*0

*

2*0

tDgn

UST

yhVVV

yTh

yhVVV

yDgnUS

th

yhVVV

yDgUS

thn

x

−=

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

−∂∂

−=∂∂

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

−∂∂

−−=

∂∂

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

−∂∂

−=∂∂

−

λ

λ

λ

16

THE STABILITY ANALYSISNow, let us consider a perturbation of small amplitude ε

( ) ( )

( )

( )

direction. in theequation momentumby provided is which determine toand

)! estimate toneed (we expand toalso need wecourse Of

where

thatfollowsit

Since

1Costant

−

++=

=

++=

=

+−=++=

x,u

.c.cue)T(Acos)t(IU

VU

sin)t(Iv

.c.cve)T(Asin)t(IV

,D~sin)t(IV

.c.ce)T(AD~;.c.ce)T(Ah

iky

iky

ikyiky

εα

α

εα

α

εε

ref. level

bottom

water levels.w.l.

h

DD

( )

( ) ( )( )

( )

α

ε

εεε

εα

sinHence

..)(1

)(..)(11..)(1

1~1

and ..)(1~with ~sin

2

Iv

ccETA

OccETAccETAD

eEccETADD

IV iky

=

++=

=++−⋅−=+−

=

=+−==

17

( )

[ ]

D**

*D

*

**

D

D

iky

rCD~UCF

Ukk

)sin(cos)t(I)t(IrCsin)t(fI

u)cos()t(IrCsin)t(ikIdtdu

.c.cue)T(Acos)t(IUx

==

=

+−=

=+++

++=

0

0

0

2

2

parameter thedefined Huthnance

ison perturbati bottom theofr wavenumbeessdimensionl theHere

1

1

rise givesequation momentum the

as velocity theofcomponent - theWriting

ω

ω

ααα

αα

εα

( )

( )

( )[ ] ( )[ ]( )[ ] ( )

( )[ ]

[ ][ ] ..)sin1(cos)cos1(cos

..)cos(coscos2cos

..)cos(cos..cos2cos

sin..coscos

..21..cos

sin~~

evaluate usLet

~sin~

cos||

1~1sin~

sincos

22

2

22

2222

2

222

ccIuIAEII

ccIuIIIIuAEII

ccEIuAIIIccIAEAEuI

IccEAIAuEI

ccAEccAuEI

IUDDU

UD

IUDrCIIrC

DfI

yU

DI

dtdI

tU

DD

+++++=

+−+++=

=⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+−++++=

=++−+

++++=

=+

+−=

=⎟⎠⎞

⎜⎝⎛ −−

∂∂

+−∂∂

αααεα

αααεα

ααεαεα

ααεα

εεα

α

αα

ααα

18

( )

[ ] ( )ααααα

αα

ααα

ε

22

2

222

11

are

11oforder of terms theTherefore

sincosIIrCsinfIu)cos(IrCsinikIdtdu......

UD~

sinIUD~rCcosI|I|rC

D~sinfI

yU

D~sinI

dtdIcos

tU

DD

DD

+−=+++

+−=

=⎟⎠⎞

⎜⎝⎛ −−

∂∂

+−∂∂

[ ]

))()(

)()(2

timesfixedat determined is )(unknown the, at assuming and

)()(

Writingapproach. Kutta-Runge a (e.g.

means. numerical use tonecessary isit current), sinusoidal (e.g. currents tidaldifferent For

1982) Huthnance, (see 222for 1

220for 1)( if

means analyticalby integrated becan equation previous The

212

21

2111

2121

00

21

⎥⎦

⎤⎢⎣

⎡+−+=

+−+=

===

=+

⎩⎨⎧

+<≤+++<≤+−

=

++++

+

nnnnn

nnnnn

n

tfutfdtuu

tfutfdtuu

ndtttuttuu

tfutfdtdu

ntnntn

tIπππππππ

19

Real and imaginary parts of u versus t for k=10 and α=20and a sinusoidal current

Real part of u

Imaginary part of u

[ ]

e)(

issolution The

)(sin)()()2sin()sin21(sin)()(

)()()(out that It turns

||||

equation continuitysediment gconsiderinby on perturbatibottom theoft developmen time thedetermine to

possibleisit known,isfieldflow perturbed theOnce

0)(

0

22

2

∫=

−++−

=Γ=

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

−∂∂

−=∂∂

ΓT

dttATA

TAtIiktutItikI

TAtdT

TdA

yhVVV

yTh

αλααα

λ

20

( )[ ] ( )[ ][ ]( )[ ]( )[ ]

...

... toleadsequation continuitysediment Then,

..sincos2..sincos2

..sincos2

..sin..cos

22

2

2

222

ccEIuIAIccEvuIAI

ccAvEAuEII

ccAvEIccAuEIV

+++=

+++=

+++=

=+++++=

ααε

ααε

ααε

εαεα

0.005) 0.,f 100),(r 1(rC results theof Example

21

rategrowth theofpart real theof valueby the controlled isonperturbati theof pinggrowth/dam the,e Because

20

00

====

∫ Γ=Γ

∫=Γ

λπ

πr

dt)t(

dt)t(

A)T(AT

α

Growth rate versus k and α

k

21

3**

*0

*

2*0

*2*

0

3*0

**

1-

)1(8

~)1( that Note

years 100010028

)1(~)1(10

order of is timefolding-e the10order of be out to turnsrategrowth

theand is scale timemicmorphodyna theSince

CsSt

DgnUST

TU

CsDgnt

T

tidee

−=

−=

−≈−−

≈

ω

π

A REFINED APPROACH

1) Introduction of a tidal ellipse2) Introduction of an angle between the bed shear stress and

the depth averaged velocity3) Analytical solution for a sinusoidal current4 ) Sediment transport rate with a threshold value of θ5) Finite amplitude sand banks

22

Non linear relationship

HydrodynamicsContinuity equation

Momentum equations

****

**

*

**

*

*

D~Dy

)VD(x

)UD(tD ζ+==

∂∂

+∂

∂+

∂∂ 0

****

*

*

**

*

**

*

**

*

*

****

*

*

**

*

**

*

**

*

*

UfDy

gyVV

xVU

tV

VfDx

gyUV

xUU

tU

y

x

−−∂∂

−=∂∂

+∂∂

+∂∂

+−∂∂

−=∂∂

+∂∂

+∂∂

ρτζ

ρτζ

where the Coriolis parameter is defined by ϕsin2 ** Ω=f

Non linear relationship

By introducing dimensionless variables using the tidal excursion (U0

*/ω*) as length scale, the inverse of the tideangular frequency (ω*-1) as time scale, the maximum value(U0

*) of the depth averaged velocity as velocity scale, theamplitude a* of the tide to scale the free surface elevation andD0* to scale the vertical coordinate, the problem reads:

ζaD~Dy

)DV(x

)DU(tDa +==

∂∂

+∂

∂+

∂∂ where0

1

1

fUDyay

VVxVU

tV

fVD

rxay

UVxUU

tU

y

x

−−∂∂

−=∂∂

+∂∂

+∂∂

+−∂∂

−=∂∂

+∂∂

+∂∂

τζ

τζ

THE DIMENSIONLESS PROBLEM

23

*0

*

*0

*0

*

~~Dg

UDaa ==

*0

*

*0~D

Urω

=

The hydrodynamic problem is characterised by the followingdimensionless parameters

where a* is defined as function of U0*

fVD

rXt

UaxX

DgL

U

fVD

rxat

U

x

x

+−∂∂

−=∂∂

=

=

+−∂∂

−=∂∂

τζ

ω

ω

τζ

and introduce should weHence ./~but

/not is scalelength horizontal eappropriat theflow basic for the because balanced is problem The

1by described is flow basic ion theapproximat oforder

leading at the tide,amplitude small a Assuming

**0

**

**0

24

THE BED SHEAR STRESS

2

22)cossin,sincos(),(C

VUVUVUyx

++−=

φφφφττ

which accounts for Coriolis effects which deviate the bed shear stress with respect to the depth averaged velocity (de Swart & Hulscher, 1995). The values of Ф can be obtained from 3D models. The resistance coefficient C, which depends on thebottom roughness zr

*, can be evaluated by meansof heuristic formulae, e.g. C=5.75log(11D0*/z*

r)

A relationship between the bottom shear stress and the depthaveraged velocity should be introduced to close the problem. Different formulae can be adopted, e.g a linear relationship or a more complex constitutive relationship like

( ) 01

1*

*

*

*

*

*

=⎥⎥⎦

⎤

⎢⎢⎣

⎡

∂

∂+

∂∂

−+

∂

∂

yQ

xQ

nth yx

Morphodynamics

Sediment continuity equation

By using dimensionless variables

**

*

)1( dgsQQ

yQ

xQ

Th yx

−=

∂

∂+

∂∂

=∂∂

where the morphodynamic time scale is introduced

~ , )1(

)~(with )1( *

0

*

**

2*0

Dd

ddgs

DrpdtT d

d

=−

=−

=ωψ

ψ

25

( ) ( ) ( )

( ) ( )

( )( )1 value,critical its is parameter, Shield theis

7.030)(

)(,

,

formula) Deigaard & (Fredsoe bed horizontal aover load bed The

,,, :slope bed theby induced load bed theand bed horizontalan over

load bed the:onscontributi woidentify t topossibleisit ,negligible is load suspended that theAssuming

RATE TRANSPORT SEDIMENT THE

)()(

)()()()(

≈

−−=Φ

Φ=

+=

μϑϑ

ϑϑϑϑπμ

ϑ

ϑϑϑϑ

c

cc

yxby

bx

py

px

by

bxyx

QQ

QQQQQQ

Sedi

men

ttra

nspo

rtra

te

Shields parameter

26

( )

( )

( )

( )⎥⎥⎦

⎤−

∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛+

∂∂

−∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛+⎢⎣

⎡∂∂

Φ=

==−=Φ

Φ−=

⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

Φ=

nnssyxx

nny

ss

nnssyxy

nnx

ss

py

px

nssnnnc

ss

py

px

GGxhGG

yh

GGyhGG

xhr

QQ

GGGddG

xh

xh

rGQQ

22

2

2

2

22

2

2

2

)()(

)()(

;

,

0,55.0,7.1

,,

israteransport sediment t slope-down The

ϑϑϑ

ϑϑ

ϑϑ

ϑϑϑ

ϑϑ

ϑϑ

ϑϑϑ

( )

⎥⎥⎦

⎤⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

−∂∂

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

−∂∂

=∂∂

−⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂∂

=∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

=∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

=•∇

=∂∂

⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

=∇

20

000

20

000

20

0000

0000

0

,,

,

1

,

ϑ

ϑϑϑ

ϑϑϑϑ

ϑ

ϑϑϑϑ

ϑϑϑϑ

ϑ

yyx

xyx

yxyx

yx

yh

xh

yh

yh

xh

xh

sh

yh

xh

nh

yh

xh

sh

yh

xhh

sh

nh

shh

27

[ ]

( ) ( ) ( )1100

10

1010

)(11010

)(11

,,,

Moreover, , ,1 with

~ , 1 since

..costcost

bottomflat theofon perturbatibottom aconsider usLet

yxyxyx

yxi

yxi

aVVVUUU

eEEDDDDD

DDa

ccehh

yx

yx

ττεττττζεζζ

εε

ε

εε

αα

αα

+=+=

+=+=

=Π−==+=

=<<

+Π+=+=

+

+

THE TIME DEVELOPMENT OF A SMALL BOTTOM PERTURBATION

..

..

0

0

cceabV

cceU

it

it

+=

+=

;

THE BASIC STATE (flat bottom) We simulate the M2 constituent with a given tidal ellipse

The tidal ellipse is forced by appropriate surface slopes

ayY,axX

fUY

rt

VY

fVY

rt

UX

y

x

==

−−∂∂

−=∂∂

−−∂∂

−=∂∂

where

00

000

00

000

τζ

τζ

28

THE PERTURBATION PROBLEM

[ ]

[ ] 11011

01

01

11011

01

01

10

10

11 0

have weequations, momentum andcontinuity into ... , , of expansions thepluggingBy

fUryy

VVxVU

tV

fVrxy

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32

STUDY CASESSTUDY CASES

UNIDIRECTIONAL TIDES (e=0):

LOW ELLIPTICITY TIDES:

“Sand banks are likely to occur where the tidal currents are rotary or have low ellipticity”Dyer & Huntley, Est. CoastalShelf Science, 1999

e=-1

clock-wise rotating velocity vector

e=0.8

counterclock-wise rotating velocity vector

αx

αy

Growth rate versus the wavenumbers of the bottomperturbation (the x-axis is aligned with the major axis of the tidal ellipse)

f=0.8, r = 120, zr=0.001, a/b=0, Ψd=0.0058

33


αy

αx

f=0.8, r = 120, zr=0.001, a/b=1, Ψd=0.0058


αx

αy

f=0.8, r = 120, zr=0.001, a/b=0.2, Ψd=0.0058

34


αx

αy

f=0.8, r = 120, zr=0.001, a/b=0.2, Ψd=0.0058


αy

αx

f=0.8, r = 90, zr=0.001, a/b=0.2, Ψd=0.0058

35


αy

αx

f=0.8, r = 80, zr=0.001, a/b=0.2, Ψd=0.0058


αy

αx

f=0.8, r = 70, zr=0.001, a/b=0.2, Ψd=0.0058

36

Growth rate as a function of the wavenumbers

Largest growth rate: counter clockwise rotated

sandbanks with wavelength equal to 7 km

Semidiurnal tide U*0=0.55 m/s D*0= 30 m e= 0.9 (clockwise rotating) d*=0.2 mm z*r=3 cm φ=7.5°

Contour lines ΓR>0

As r tends to a critical value, ranging around 93, the growth rate

decreases to 0.

Although the most unstable wave number decreases as r tends to its critical value, its limit

value is still finite!!

It is possible to perform a weakly non linear stability

analysis !!!

SAND BANKS OF FINITE AMPLITUDE

Semidiurnal tide U*0=0.55 m/s D*0= 30 m e= 0.9 (clockwise rotating) d*=0.2 mm z*r=3 cm φ=7.5°

37

As r tends to a critical value, ranging around 93, the growth rate

decreases to 0.

Although the most unstable wave number decreases as r tends to its critical value, its limit

value is still finite!!

It is possible to perform a weakly non linear stability

analysis !!!

SAND BANKS OF FINITE AMPLITUDE

Semidiurnal tide U*0=0.40 m/s h*0= 30 m e= 0.9 (clockwise rotating) d*s=0.2 mm z*r=3 cm φ=7.5°

,4

tidal sand banks - universiteit utrecht · 2011-05-18 · 2 norfolk banks flemish and hinder banks...

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