III.1 Exact solution
The real solution is given by the following
The meshing of the domain was realized with GAMBIT mesher as pictures below.
Domain Extend :
x-coordinate (m) : min = 0 , max = 20
y-coordinate (m) : min = -0.2 , max = 2
define by the following equation y = -0.05*(x-10)^2 for x in [8,12] (E1)
Grid size :
Number of nodes : 2202
Number of faces : 4522
Number of cells : 5040
of the meshing
Such a problem is resolved with FLUENT 5.0 by using V.O.F
The first phase is water_liquid.
The second phase is air.
Repartition of the two phases at T = 0 :
water for x in [0,20] and y in [-0.2,1.8]
air for x in [0,20] and y in [1.8,2]
Boundary conditions :
Superieurs and inferior sides are walls
right side is an outflow
left side is divided in two parts :
for y in [-0.2,1.8] velocity inlet of water v1 = 4.429 m/s
for y in [1.8,2] velocity inlet of air v2 = 0.01 m/s
Initial repertition of pressure:
the gauge pressure in the air is P = 0
the pressure in the water is given by the equation P(y) = 10000*(1.8-y)
Time Step :
In all the cases, the highest time step necessary to have convergence of the method is Dt = 0.001
It will be the time step choosen.
III.3 Test-case 1
In this case the initial speed of the flow is taken contant
at V = 4.429 m/s
The V.O.F scheme choosen is the geo_reconstruc scheme
As we will realize it is two big mistakes.
Let's see the results
simulation at t=1s
simulation at t=2s
simulation at t=3s
As we can see the solution is far from the real solution and furthermore it is not steady.
- the first mistake made is the choise
of the V.O.F sheme :
with V.O.F method if the result is steady you have to choice the IMPLICIT scheme and not the default_scheme which is
with the IMPLICIT scheme you have two possibilities:
make the simulation in steady if the result does not depend on the initial flow
make the simulation in unsteady if the result depends on the initial flow
- the second mistake is the choise of the initial water velocity :
in fact the velocity is constant but the flow is not constant because of the obstacle.
III.4 Test-case 2
In this case the V.O.F scheme is the IMPLICIT scheme.
the initial condition is not on the velocity of the wather but on the flow which is taken constant :
As to leave of the equation (E1) is not difficult to find the relation that has to respect the initial velocity to keep constant the flow :
V(x) = Q/(1.8+0.059x-10)^2 for x in [8,12] Q = 8.859 is the flow
Let's see the results
simulation at t=0.3
simulation at t= 0.1
The result is better than in the first case but it is
not the real solution.
I think do perform better there is two parameters upon which we will have to play:
impose the outlet height of the two phase.
impout a more realistic initial condition of the water.