Transcript MODELISATION DES ECOULEMENTS EN LITS COMPOSES

‫الــمدرســــة الــوطــــنــيــة للــمــهــنــدســيــن بــتــونــس‬
école nationale d'ingénieurs de Tunis
Laboratoire de Modélisation en Hydraulique et Environnement
SECOND ORDER
MODELLING OF COMPOUND
OPEN CHANNEL-FLOWS
Prepared by :
Olfa DABOUSSI
Presened by
Zouhaïer HAFSIA
Plan
Introduction.
Secondary currents in compound open channel flow.
Experimental results.
Turbulence model.
Numerical results.
Conclusions.
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INTRODUCTION
After strong rain.
The compound channel is composed from many stages.
Mean channel
In laboratory, compound channel are represented by the main channel and one
floodplain with rectangular sections.
It is interesting to study the compound channel flow to understand main
channel – floodplain interaction.
The turbulence model : second order model Rij.
We use the CFD code PHOENICS for numerical simulations.
Numerical results are compared to experimental data of Tominaga and al.
(1989).
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THE RECTANGULAR OPEN COMPOUND
CHANNEL FLOW
I – Rectangular compound channel
Symmetric
Asymmetric
Free surface
B

b
 Hh
H
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II – Tested cases
Three cases of β values are simulated (Mesures of Tominaga and al., 1989).
β = 0.5
β = 0.242
β = 0.343
λ = 2.07
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Tominaga and al. (1989) data
Cas
B (m)
b (m)
λ = B/b
h (m)
H (m)
β = (H-h)/H
Wmoy (m
s-1)
1
0.195
0.094
2.07
0.0501
0.1001
0.500
0.315
2
0.195
0.094
2.07
0.0501
0.0661
0.242
0.32
3
0.195
0.094
2.07
0.0501
0.0763
0.343
0.273
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NUMERICAL SIMULATIONS
I – Governing Equations
In incompressible Newtonian fluid and parabolic flow through the z direction, the
Reynolds stress turbulence model of Launder and al. (1975) is written as :
- Continuity :
U
V

 0 (1)
x
y
- Momentum :
U
U
² U
² U  u '²
U
V



x
y
 x²
 y²
x
(2)
V
V
1 P
² V
² V  v '²
U
V




 g (3)
x
y
 y
 x²
 y²
y
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W
W
² W
² W  u ' w '  v ' w '
U
V
 gI  



x
y
 x²
 y²
x
y
(4)
- Kinetic equation :
U
k
 k  t  k
  k
 U V 2  W 2  W 2
V

(
) ( t
)   t ((

) (
) (
) )   (5)
x
 y  x k  x  y  k  y
y x
x
y
- ε equation :

   t  
 t  
  U V 2  W 2  W 2
2
U
V

(
) (
)  C1 ((

) (
) (
) )  C 2
x
 y  x   x  y   y
k  y x
x
y
k
- Reynolds stress :
u i ' u j '
u i ' u j '
u i ' u j '

k u i ' u j '
2
 Ul

(C s u ' k u ' k

)  P ij  ij
t
xl
 xl
 xl
xl
3

2
2
i, j, k = 1, 2, 3
 C1 (u i ' u j '  ijk)  C 2(P ij  ijP k )
(7)
k
3
3
8
(6)
 Ul


'
'
ul uk 
Pk
xk
U j
U


(
'
'
ui u l   u j'u l'  i )
Pij
xl
xl
C1, C2 and Cs are constants.
- Boundary conditions :
- Smooth wall logarithmic law :
1

U  ln y  A



U 
U

y 
U*
U* y

  0.41
- Near walls, k , ε and Rij are :
2
kw
U*

C
3
U*
w 
yw
- The free surface is considered as a symmetric plane :
 (U, W, k, )
0
y
u ' v'  0
v'w '  0
- On the vertical symmetric plane :  (V, W, k, )  0
x
V 0
u ' v'  0 u 'w '  0 U  0
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I – In PHOENICS
There are four derivations of the Rij model : IPM, IPY, QIM and SSG.
Where :
• IPM is the Isotropisation of production model.
• IPY is the IPM model of Younis (1984).
• QIM is the quasi-isotropic model
• SSG is the model of Speziale, Sarkar and Gatski
PHOENICS use the finite volume numerical method.
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PHOENICS take the z axis as the main flow direction for the parabolic ones. The cell
along z is a slab.
The general form of the transport equations is :
 



( U j   
)  S (8)
t
 xj
xj
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RESULTS OF SIMULATIONS
k-ε model :
Using the CFD code PHOENICS, we have tested two cases for β = 0.5.
Results with coarse grid :
The k- can not
reproduce the
isovelocity bulging
shown
experimentally
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Results with thin grid :
k-
do
not
reproduce
the
isovelocity
bulging
The k-ε model is isotropic.
Experience shows a strong anisotropy
The second order model Rij take account of the turbulence anisotropy.
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Results with the Rij model:
The four derivations of Rij give the same results.
• Secondary currents :
Free surface vortex
Main channel vortex
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Tranversal velocity profils
on X = 0.055 m :  = 0.5
on X = 0.087 m :  = 0.5
on X = 0.101 m :  = 0.5
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Longitudinal velocity variations :
Wall law at X = 0.02 m : β = 0.5
Vertical averaging of the velocity : β = 0.5
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Comparaison of numerical isovelocity with Tominaga et al. (1989) data : β = 0.5
The Rij model reproduce
well the isovelocity.
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Wall shear stress variation :
The shear stress on the floodplain raises
near the main channel- floodplain
junction.
Momentum transfer from the main
channel to the floodplain.
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THE ADIMENSIONAL DISPERSION
COEFFICIENT
The adimensional dispersion coefficient appears after integration of the momentum equation
through the transversal section.
Needs a closure law
 (SW)  (SWW)
1 (S P)
²(SW) (S  w"w" ) (S  w ' w ' )





 Mz
2
t
z
 z
z
z
z
Two approches :
Gradient closure :
 w "w "    Dzz
W
z
Correlation based on the momentum distribution
 WW

WW
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Calculations show that α is different from 1. It depends of  : from 1.04 to 1.35
Whith λ = 2.07, α varies as second degrees polynomial in function of  as follow :
  1.05  0.3   0.66 2
α variation in function of β :
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CONCLUSIONS
Secondary currents modify longitudinal iso-velocity.
The first order k-ε model do not reproduce the isovelocity bludging
The second order turbulence model can reproduce the interaction between the main
channel and flood-plain (momentum transfer)
Numerical computation show that the dispersion coefficient α is expressed as a
polynomial function of β
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