III.1 Exact solution

    The real solution is given by the following figure :

                                                                                                                                  figure 1


III.2 Methodology

    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

detail of the meshing

Such a problem is resolved with FLUENT 5.0 by using V.O.F methods :
    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.