second order modelling of compound open channel-flows laboratoire de modélisation en hydraulique et...
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SECOND ORDER MODELLING OF COMPOUND
OPEN CHANNEL-FLOWS
école nationale d'ingénieurs de Tunis
ــس ـون ــت ـب ــن ـي ـدـس ــن ــه ــم لـل ـة ــي ــن ــ ـوـط اـل ـة ــ ـمدرـس اـل
Laboratoire de Modélisation en Hydraulique et Environnement
Prepared by :
Olfa DABOUSSI
Presened by
Zouhaïer HAFSIA
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Plan
Introduction.
Secondary currents in compound open channel flow.
Turbulence model.
Numerical results.
Conclusions.
Experimental results.
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INTRODUCTION
In laboratory, compound channel are represented by the main channel and one floodplain with rectangular sections.
After strong rain.
The compound channel is composed from many stages. Mean channel
The turbulence model : second order model Rij.
We use the CFD code PHOENICS for numerical simulations.
It is interesting to study the compound channel flow to understand main channel – floodplain interaction.
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
B
b
HhH
Free surface
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Three cases of β values are simulated (Mesures of Tominaga and al., 1989).
β = 0.5
β = 0.343
β = 0.242
λ = 2.07
II – Tested cases
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Cas B (m) b (m) λ = B/b h (m) H (m) β = (H-h)/HWmoy (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
Tominaga and al. (1989) data
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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 :
U V0 (1)
x y
- Continuity :
- Momentum :
U U ² U ² U u ' ²U V (2)
x y x² y² x
V V 1 P ² V ² V v ' ²U 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 (4)
x y x² y² x y
2 2 2t tt
k k
k k k k U V W WU V ( ) ( ) (( ) ( ) ( ) ) (5)
x y x x y y y x x y
- Kinetic equation :
22 2 2t t
1 2
U V W WU V ( ) ( ) C (( ) ( ) ( ) ) C (6)
x y x x y y k y x x y k
- ε equation :
i j i j i j i jijs k k ijl
l l ll
ij k1 i j ij 2 ij
' ' ' ' ' ' ' 'k 2u u u u u u u u( )C u ' u ' PU
t x 3x x x
2 2( ' ' k) ( ) (7)C u u C P P
k 3 3
- Reynolds stress :
i, j, k = 1, 2, 3
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)xU'u'u
xU'u'u(P
l
ilj
l
jliij
xU'u'uP
k
lklk
C1, C2 and Cs are constants.
- Boundary conditions :
1ln AyU
- Smooth wall logarithmic law :
yU
y * 0.41
- Near walls, k , ε and Rij are :
3*
w
w
Uy
*
UU
U
2*
wU
kC
- The free surface is considered as a symmetric plane :
(U, W,k, )0
y
u 'v ' 0 v 'w ' 0 V 0
- On the vertical symmetric plane : (V, W,k, )0
x
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.
• 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
Where :
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.
j
j j
( U ) S (8)t x x
The general form of the transport equations is :
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k-ε model :
Using the CFD code PHOENICS, we have tested two cases for β = 0.5.
The k- can not
reproduce the
isovelocity bulging
shown experimentally
RESULTS OF SIMULATIONS
Results with coarse grid :
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The k-ε model is isotropic.
The second order model Rij take account of the turbulence anisotropy.
k- do not
reproduce the
isovelocity
bulging
Results with thin grid :
Experience shows a strong anisotropy
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Results with the Rij model:
The four derivations of Rij give the same results.
• Secondary currents :
Main channel vortex
Free surface vortex
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Tranversal velocity profils
on X = 0.087 m : = 0.5
on X = 0.101 m : = 0.5
on X = 0.055 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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z2
(SW) (SWW) 1 (S P) ² (SW) (S w"w" ) (S w 'w ' )M
t z z z z z
The adimensional dispersion coefficient appears after integration of the momentum equation through the transversal section.
Needs a closure law
Two approches :
Gradient closure :zz
Ww"w" D
z
Correlation based on the momentum distribution
W W
W W
THE ADIMENSIONAL DISPERSION COEFFICIENT
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Whith λ = 2.07, α varies as second degrees polynomial in function of as follow :
α variation in function of β :
21.05 0.3 0.66
Calculations show that α is different from 1. It depends of : from 1.04 to 1.35
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CONCLUSIONS
Numerical computation show that the dispersion coefficient α is expressed as a polynomial function of β
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)