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Stability of an Erodible Bedin a Shear Flow
Kouame K. J. Kouakou and Pierre-Yves Lagree
[email protected]
Laboratoire de Modelisation en Mecanique,
UMR CNRS 7607, Boıte 162,
Universite Paris 6, 75252 Paris France
workshop on sand transport and dune formation, 9th - 11th June 2004 Carry Le Rouet – p.1
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outline
• Boundary layer solutions of the basic flows• Steady shear flow• Decelerated flow• Oscillating flow
• linear stability analysis of the bed• Steady shear flow• Decelerated flow case• Oscillating flow case
• Examples of long time evolution of the bed
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outline
• Perturbation of the basic flow• Equations near the wall• Linearised equations• Analytical law between the topography and
the skin friction
• linear stability analysis of the bed• Steady shear flow• Decelerated flow case• Oscillating flow case
• Examples of long time evolution of the bed
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outline
• linear stability analysis of the bed• Steady shear flow• Decelerated flow case• Oscillating flow case
• Examples of long time evolution of the bed
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outline
• linear stability analysis of the bed• Steady shear flow• Decelerated flow case• Oscillating flow case
• Examples of long time evolution of the bed
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every profile is linear near the wall
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every profile is linear near the wall
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Boundary layer solutions of the basic flows
• Steady basic flow
u∗ = U0y∗
δ+ · · ·
• Decelerated basic flow
u∗ = U0Erf(y∗
2√ν t∗
) =U0 y
∗
√π ν t∗
+ · · ·
t =π ν
δ2t∗ and δ =
√π ν T if O(t∗) = T
• Oscillating basic flow
u∗ = U0 [cos(ω t∗)− e−√
ω2 ν
y∗
cos(ω t∗ −√
ω
2 νy∗)]
u∗ = U0
√
ω
νcos(ω t∗ +
π
4) y∗ +O(y∗2).
t = ω t∗ +π
4and δ =
√
ν
ω
workshop on sand transport and dune formation, 9th - 11th June 2004 Carry Le Rouet – p.4
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Boundary layer solutions of the basic flows
• Steady basic flow
u∗ = U0y∗
δ+ · · ·
• Decelerated basic flow
u∗ = U0Erf(y∗
2√ν t∗
) =U0 y
∗
√π ν t∗
+ · · ·
t =π ν
δ2t∗ and δ =
√π ν T if O(t∗) = T
• Oscillating basic flow
u∗ = U0 [cos(ω t∗)− e−√
ω2 ν
y∗
cos(ω t∗ −√
ω
2 νy∗)]
u∗ = U0
√
ω
νcos(ω t∗ +
π
4) y∗ +O(y∗2).
t = ω t∗ +π
4and δ =
√
ν
ω
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Boundary layer solutions of the basic flows
• Steady basic flow
u∗ = U0y∗
δ+ · · ·
• Decelerated basic flow
u∗ = U0Erf(y∗
2√ν t∗
) =U0 y
∗
√π ν t∗
+ · · ·
t =π ν
δ2t∗ and δ =
√π ν T if O(t∗) = T
• Oscillating basic flow
u∗ = U0 [cos(ω t∗)− e−√
ω2 ν
y∗
cos(ω t∗ −√
ω
2 νy∗)]
u∗ = U0
√
ω
νcos(ω t∗ +
π
4) y∗ +O(y∗2).
t = ω t∗ +π
4and δ =
√
ν
ω
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Perturbation of the basic flow
with the scaling:
(x∗, y∗) = δ (x, y)
(u∗, v∗) = U0 (u, v) p∗ = ρ (U20 p− g y δ)
Navier Stokes equations:
∂u
∂x+∂v
∂y= 0
∂(u, v)
∂t+Reδ ([(u, v) · ∇](u, v) +∇p) = ∇2(u, v)
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Perturbation of the basic flow
• Rescaling
x = λx and y = εb y
Restricting the field of study to dimensions of a bump of thedisturbance of the bottom, one has:
u = U ′s(0) y +O(y2) = εbU
′s(0) y +O(y2)
whereU ′
s(0) = 1 for the steady case:U ′
s(0) = 1√t
for the decelerated flow and
U ′s(0) = cos(t) for the oscillating case
is a function of the alone variable t, hydrodynamic time.
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Perturbation of the basic flow
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Perturbation of the basic flow
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Perturbation of the basic flow
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Perturbation of the basic flow
εb
δ� 1 and
λ
δ� 1
y
PSfrag replacements
·· · ·
λ
δ
a
x
εb
U0
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Equations near the wall
• As εb = O(λ)
∂u
∂x+∂v
∂y= 0
ε2b Reδ {[(u, v) · ∇](u, v) +∇p} = ∆(u, v)
• As εb � λ∂u
∂x+∂v
∂y= 0
u∂u
∂x+ v
∂u
∂y= −∂p
∂x+
λ
ε3b Reδ
∂2u
∂y2.
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Equations near the wall
• As εb = O(λ)
∂u
∂x+∂v
∂y= 0
ε2b Reδ {[(u, v) · ∇](u, v) +∇p} = ∆(u, v)
• As εb � λ∂u
∂x+∂v
∂y= 0
u∂u
∂x+ v
∂u
∂y= −∂p
∂x+
λ
ε3b Reδ
∂2u
∂y2.
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Equations near the wall
• Finally
x∗ = δ ε3b Reδ x, y∗ = δ εb y and εb � 1
∂u
∂x+∂v
∂y= 0
u∂u
∂x+ v
∂u
∂y= −∂p
∂x+∂2u
∂y2
0 =∂p
∂y.
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Equations near the wall
The fact of having
λ ∼ ε3b Reδ =2Aε3bδ
gives r =2A
λ' δ
ε3b
r is the aspect ratio between the characteristic scales uses in formerstudies. However
δ
εb� 1 and
1
ε2b� 1
one thus has well
r =2A
λ' δ
εb(
1
ε2b)� 1
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Equations near the wall
Rousseaux et al. (2003) workshop on sand transport and dune formation, 9th - 11th June 2004 Carry Le Rouet – p.10
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Linearised equations
f = af1
u = U ′s(0) y +O(y2)
that gives us the variables of the problem in the form
u = U ′s(0) [y + au1(x, y, t) + · · ·]
v = U ′s(0) av1(x, y, t) + · · ·
p = U ′s(0) ap1(x, y, t) + · · ·
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Linearised equations
we keep the equations with the 1st order in a
∂u1
∂x+∂v1∂y
= 0
U ′s(0) y
∂u1
∂x+ U ′
s(0) v1 = −∂p1
∂x+∂2u1
∂y2
0 =∂p1
∂y
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Linearised equations
Decomposing in modes of Fourier, taking into account the continuityequation
f1 = fk e−i k x+σ tL
u1 = φ′(y) e−i k x+σ tL
v1 = (i k)φ(y) e−i k x+σ tL
p1 = ψ(y) e−i k x+σ tL
, ψ,y = 0
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Perturbated solutions in Fourier space
• Pression
p1 = 3 aAi′(0) (U ′s(0))
5/3 (−i k)−1/3 f1
• friction
τ1 =∂u1
∂y= 3 aAi(0)U ′
s(0) (−i k U ′s(0))
1/3 f1
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validation of linear friction
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validation of linear friction
here, taking simply U ′s(0) = 1 (steady shear), the friction (τ − 1)
calculated by CASTEM 2000 (Navier-Stokes)
Handing the scale of the theory
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validation of linear friction
here, taking simply U ′s(0) = 1 (steady shear), the friction (τ − 1)
calculated by CASTEM 2000 (Navier-Stokes) and rescaled is
compared to the linearised solution
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Decelerated basic flow
U ′s(0) =
1√t,
the bottom friction is
τTotal =1√t
+ TF−1{3Ai(0) (−i k)1/3 [t]−2/3 e−i k x+σ tL}(x, t)
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Oscillating basic flow
For one period of oscillation
0.2 0.4 0.6 0.8 1t
-0.4
-0.2
0
0.2
0.4
k
k
PSfrag replacements
τ(±
)Total
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Oscillating basic flow
For one period of oscillation
0.2 0.4 0.6 0.8 1t
-0.4
-0.2
0
0.2
0.4
PSfrag replacements
τ(±
)Total
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Oscillating basic flow
during one cycle the topography does not changewe take the mean value of all the quantities
Multiscale analysis...
U ′s(0) = cos(t),
τ(+)Total = cos(t) + TF−1{3Ai(0) (−i k)1/3 [cos(t)]4/3 e−i k x+σ tL}(x, t)
τ(−)Total = −cos(t)− TF−1{3Ai(0) (−i k)1/3 [cos(t)]4/3 e−i k x+σ tL}(x, t)
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Oscillating basic flow
< τ >Total=1
T[
∫ tp
0
τ(+)Total dt+
∫ T
tp
τ(−)Total dt ]
< τ >Total=9Ai[0] [(−i k)1/3 − (i k)1/3] Γ( 7
6 )
4√π Γ( 2
3 )
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Fluid
Up to now, we have for any initial profile, the skinfriction,
need for a law of matter flux.
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Fluid
Up to now, we have for any initial profile, the skinfriction,need for a law of matter flux.
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Laws of matter flux
In the majority of their work, B. Sumer (1984), P. Blondeaux (1990), G.
Parker (1995), K. Richards (1999), F. Charru (2002), K. Kroy, Hermann
Sauermann (2002) , established that
q ∝ τ 3
2 .
Asu = U ′
s(0) [y + u1(x, y, t) + · · ·],τ = U ′
s(0) [1 + τ1(x, y, t) + · · ·] with |τ1| � 1
so
q ∝ (1 + τ1)3
2 ≈ 1 +3
2τ1
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Laws of matter flux
if τ > τth q = τ − τthelse q = 0
PSfrag replacements
qs
ττth
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Laws of matter flux
• Linear form (Yang (1995), Fredsøe and Deigaard (1992))
q = τ − τth − Λ∂f
∂x
• An another form (Andreotti and al. (2002) simplified Kroy and al
(2002) Sauermann and al (2001))
lK∂q
∂x+ q = τ − τth
with lK proportional to 1U ′
s(0) .
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Fluid/ bed coupling
Up to now, we have for any initial profile, the skinfriction, and then the flux of matter
q ← τ ↔ f
∂f
∂t= −∂q
∂x
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Fluid/ bed coupling
Up to now, we have for any initial profile, the skinfriction, and then the flux of matter
q ← τ ↔ f
∂f
∂t= −∂q
∂x
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Linear stability analysis
• steady shear case
FT [τ ] = FT [f ](3Ai(0))(−(ik))1/3 ∂f
∂t= − ∂q
∂x
So, for a mode k, looking to f = eσt+iωte−ikx,
σ + iω =3
1
3
Γ( 23)
(1/2 + i√
3/2)(k)4/3 − Λ k2
With Λ = 0 all waves are always instable
slope effect Λ 6= 0 give an amplification for long waves; short waves
always instable.
Or length of saturation effect give an amplification for long waves which
are always stable; short waves always instable.
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Linear stability analysis
• steady shear case
0.1 0.2 0.3 0.4 0.5
0.01
0.02
0.03
0.04
0.05
PSfrag replacements
k
σ
Constant shear, U ′s(0) = 1, amplification factor σ as function of number
k (here q = τ − τth − Λ∂f∂x with Λ = 1), decreasing Λ increases the cut
off value of k.workshop on sand transport and dune formation, 9th - 11th June 2004 Carry Le Rouet – p.22
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Linear stability analysis
• steady shear case
0.1 0.2 0.3 0.4 0.5
0.01
0.02
0.03
0.04
0.05
PSfrag replacements
k
σ
Constant shear, U ′s(0) = 1, amplification factor σ as function of number
k (here lK∂q∂x + q = τ − τth with lK = 1), decreasing lK increases the
cut off value of k.workshop on sand transport and dune formation, 9th - 11th June 2004 Carry Le Rouet – p.22
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Examples of long time evolution : steady shear flow
coarsening
lK∂q∂x + q = τ − τth with lK = 1
U ′
s(0) = 1
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Examples of long time evolution : steady shear flow
Number of dunes and maximal height versus time,
lK∂q∂x + q = τ − τth with lK = 1
U ′
s(0)
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Linear stability analysis
• Decelerated shear case
∂f
∂tL= − ∂q
∂x
whiletL ' O(t)
f = fk(t) e−i k x, u1 = uk(t) e−i k x · · ·
∂fk(t)
∂t= − 3Ai(0) (−i k) (−i k)1/3 t−2/3 − Λ k2 fk(t).
The logarithm of each mode of Fourier of f
log(fk(t)) = − 9Ai(0) (−i k ) (−i k )1/3 t−1/3 − Λ k2 t
workshop on sand transport and dune formation, 9th - 11th June 2004 Carry Le Rouet – p.25
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Linear stability analysis
• Decelerated shear case
With Λ = 0 all waves are always instable
slope effect Λ 6= 0 give an amplification for long waves; short waves
always instable.
Or length of saturation effect give an amplification for long waves which
are always stable; short waves always instable.
workshop on sand transport and dune formation, 9th - 11th June 2004 Carry Le Rouet – p.25
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Linear stability analysis
• decelerated shear case, law of q with saturation effect
0.5 1 1.5 2 2.5 3k
0.25
0.5
0.75
1
1.25
1.5
1.75
2
t
PSfrag replacements
log(fk)
No saturation effect (q = τ − Λ ∂f∂x ) with Λ = 0.4
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Linear stability analysis
• decelerated shear case, law of q with saturation effect
0.5 1 1.5 2 2.5 3k
0.25
0.5
0.75
1
1.25
1.5
1.75
2
t
PSfrag replacements
log(fk)
Saturation effect lK∂q∂x + q = τ − τth with lK = 1
U ′
s(0)
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Linear stability analysis
• decelerated shear case, law of q with saturation effect
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8k
0
0.05
0.1
0.15
0.2
t
PSfrag replacements
log(fk)
kc
Saturation effect with lK∂q∂x + q = τ − τth with lK = 1
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Examples of long time evolution : decelerated shear case, with saturation effect
Saturation effect with lK∂q∂x + q = τ − τth with lK = 1
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Linear stability analysis: Oscillating Flow
• Multiscale analysis for the flux relation
∂f
∂t= −θ ∂q
∂x
with θ << 1, t0 = t, and t1 = θt the long time.∂
∂t=
∂
∂t0+ θ
∂
∂t1
Let f = f0(t0, t1) + θf1(t0, t1)
and q = q0(t0, t1) + θq1(t0, t1)∂f0∂t0
= 0
ie the topology is quasisteady
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Linear stability analysis: Oscillating Flow
• Multiscale analysis for the flux relation
∂f0∂t1
+∂f1∂t0
= − ∂q∂x
so f0(t0, t1) = F0(t1), decomposition: q is Q+ q′ where Q =< q > and< q′ >= 0 so
∂f1∂t0
= (−∂q′
∂x) + (−∂Q
∂x− ∂F0
∂t1)
secular term: (−∂Q∂x −
∂F0
∂t1) must be 0, q′ must be borned.
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Linear stability analysis: Oscillating Flow
0.1 0.2 0.3 0.4 0.5 0.6k
0
0.005
0.01
0.015
0.02
PSfrag replacements
σ
no saturation effect q = τ − Λ ∂f∂x with Λ = 0.01
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Linear stability analysis: Oscillating Flow
0.1 0.2 0.3 0.4 0.5 0.6k
0
0.005
0.01
0.015
0.02
PSfrag replacements
σ
with saturation effect and lK∂q∂x + q = τ − τth with lK = 1
U ′
s(0)
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Examples of long time evolution : Oscillating Flow
lK∂q∂x + q = τ − τth with lK = 1/U ′
s(0)
animationworkshop on sand transport and dune formation, 9th - 11th June 2004 Carry Le Rouet – p.30
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Examples of long time evolution : Oscillating Flow
lK = 1/U ′s(0) number of ripples and maximum height
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Examples of long time evolution : Oscillating Flow
lK = 1/U ′s(0) and slope limitation (very simple avalanche)
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conclusion
• Analytical solution of skin friction in an asymptotical framework
• Stability analysis of different flows with various linear matter flux
• Long time numerical evolution leading to coarsening
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perspectives
• An full avalanche model upstream and downstream from each
bumpy
PSfrag replacements
·
x
U0Avalanche
• comparison with experiments (G. Rousseaux and H. Caps)
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