View
2
Download
0
Category
Preview:
Citation preview
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
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
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
UVx
UUt
Uy
DVx
DUyV
xU
VU
yy
xx
−+−∂∂
−=∂∂
+∂∂
+∂∂
++−∂∂
−=∂∂
+∂∂
+∂∂
=∂∂
+∂∂
+⎥⎦
⎤⎢⎣
⎡∂∂
+∂∂
ττζ
ττζ
( ) ( )
( ) ( ) ( )
50
12
02002
00
0
0101001000011111
000000
11
111111
5ˆ,,1where
ˆˆˆ,ˆ,ˆ,
,,out kedeasily wor becan expansions
stressshear bed theon time,only depend ... ,ˆ,ˆsuch that
,...ˆ,ˆ write weIf
CTVUR
CT
ER
VVUUTRTVURTVU
VURT
VU
EVVEUU
yx
yx
=+==
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛ +++Π=
=
Π=Π=
ττ
ττ
29
( ) ( ) [
( ) ( )
n
n
tinn
xy
yxxyyxyx
yx
e
tGtFdt
d
TRViUiRTVVUUTR
iirVUifVUidt
d
UiVi
,1
,11
1
11
100000
0101010
00111001
111
ˆ unknowns for the systemlinear
algebraican solving and ˆˆ timeof series
Fourier a as ˆ expandingby obtained becan solution The
)(ˆ)(ˆ
form in the written becan which
ˆˆˆˆ
ˆˆˆˆ
ˆˆˆ vorticityfor the
equationobtain theweequationsmomentum twotheCombining
η
ηη
η
ηη
ηαα
ταταααηααη
ααη
⎟⎠
⎞⎜⎝
⎛=
=+
⎥⎦
⎤++−⎟⎟
⎠
⎞⎜⎜⎝
⎛+++
+−−+−=++
−=
∑∞
−∞=
( )
( )12222
001
1222200
1
111
0011
ˆˆˆ
ˆ
ˆˆˆ
ˆ
determined be tocomponents velocity two theallow
ˆˆˆ
definitionicity with vortalong
ˆˆˆˆequation Continuity
ηαα
αααααα
ηαα
αααααα
ααη
αααα
yx
x
yx
yxy
yx
y
yx
yxx
yx
yxyx
iVUV
iVUU
UiVi
VUVU
+−
+
+=
++
+
+=
−=
+=+
30
( ) ( ) ( )
( )( )
( )( ) obtained) be tosimplebut ...,
long very are functions (these ...,
24.1330
7.030;)(
,,, write topossible also isit
paramenter small theof presence theof Because
)(1
)(1
)(0
)(0
0
0011
000
10
)(1
)(1
)(0
)(0
)()(
=
=
−−=Φ
−−=Φ
Φ+Φ=Φ
+=
by
bx
by
bx
cc
cc
by
bx
by
bx
by
bx
QQQQQQ
ϑϑϑϑϑ
ϑπμ
ϑϑϑϑπμ
εϑ
ε
ε
( )( )
( )
( )⎥⎥⎦
⎤−
∂∂
+⎟⎟⎠
⎞⎜⎜⎝
⎛+
∂∂
−∂∂
+⎟⎟⎠
⎞⎜⎜⎝
⎛+⎢⎣
⎡∂∂
Φ=
=
0020
0012
0
20
020
20
01
0020
0012
0
20
020
20
01
0)(1
)(1
)(0
)(0
;
,
)0,0(,
nnssyxx
nny
ss
nnssyxy
nnx
ss
py
px
py
px
GGxhGG
yh
GGyhGG
xhr
ϑϑϑ
ϑϑ
ϑϑ
ϑϑϑ
ϑϑ
ϑϑ
31
[ ]
( ) ( )
[ ] ΠΓ=Π+−=∂Π∂
Π=
+Π+= +
)(ˆˆ
toleads order at equation continuitysediment theemployed, is
ˆ,ˆ,notation theand considered are
rateransport sediment t theofexpansion with thealong..)(coston perturbati bottom theIf
11
1111
)(
tQQiT
EQQQQ
cceTh
yyxx
yxyx
yxi yx
αα
ε
ε αα
( ) ( )
∫
∫Γ=Γ
ΓΠ=Ππ2
0
0
)( rategrowth averaged The
)(exp
dtt
dttT
Growth rateRΓReal part
IΓMigration speed of the perturbation
Imaginary part
followsit
)(
From
ΠΓ=Π t
dTd
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
Growth rate versus the wavenumbers of the bottomperturbation (the x-axis is aligned with the major axis of the tidal ellipse)
αy
αx
f=0.8, r = 120, zr=0.001, a/b=1, Ψd=0.0058
Growth rate versus the wavenumbers of the bottomperturbation (the x-axis is aligned with the major axis of the tidal ellipse)
αx
αy
f=0.8, r = 120, zr=0.001, a/b=0.2, Ψd=0.0058
34
Growth rate versus the wavenumbers of the bottomperturbation (the x-axis is aligned with the major axis of the tidal ellipse)
αx
αy
f=0.8, r = 120, zr=0.001, a/b=0.2, Ψd=0.0058
Growth rate versus the wavenumbers of the bottomperturbation (the x-axis is aligned with the major axis of the tidal ellipse)
αy
αx
f=0.8, r = 90, zr=0.001, a/b=0.2, Ψd=0.0058
35
Growth rate versus the wavenumbers of the bottomperturbation (the x-axis is aligned with the major axis of the tidal ellipse)
αy
αx
f=0.8, r = 80, zr=0.001, a/b=0.2, Ψd=0.0058
Growth rate versus the wavenumbers of the bottomperturbation (the x-axis is aligned with the major axis of the tidal ellipse)
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
Recommended