Subcritical case 1.

Subcritical case 2.

Transcritical case 1.

Transcritical case 2.

Transcritical case 3.



Debit

amont(m3/s)

Debit

aval (m3/s)

Hauteur

amont (m)

Hauteur

avale (m)

Conditions

Initiales

Hauteur critque

pour canal horizontal

Type d'Ecoulement

attendu

    8.859

********* ********

    2

cote constante

    0.966

Fluvial

we calculate the critical heigh under the asumption of a horizontal channel. this is true for the left and the right sides of the channel. the information get is very important for imposing the regim we want.

the simulation take about half an hour, and the solution converge after 60s in terme of real time.

the froud number is under one. so it's a subcritical regim every where in the channel. this show that we have used the good boundary conditions.

the hump in the midele of the channel increase the velocity of the flow, but not enought so as to chane the regime. the flow regime remains subcritical every where in teh channel.

Return.


Subcritical case 2



Debit

amont(m3/s)

Debit

aval (m3/s)

Hauteur

amont (m)

Hauteur

avale (m)

Conditions

Initiales

Hauteur critque

pour canal horizontal

Type d'Ecoulement

attendu

    8.859

********* ********

    2

solution analytique

    0.966

Fluvial

we start the simulation with an initial condition that have been computed by the classical Newton method in the CONDIN soubroutine of the FORTRAN file.

So we start with th analytical solution. the simulation converge very fastely. this shows the importance of the initial condition for the time of convergence. we have got the following graphic.

Return.


Transcritical case1


Debit

amont(m3/s)

Debit

aval (m3/s)

Hauteur

amont (m)

Hauteur

avale (m)

Conditions

Initiales

Hauteur critque

pour canal horizontal

Type d'Ecoulement

attendu

    2

********* ********

    0.6

cote = 0.6 m

    0.47

Flu. / Tor. /Flu.

In This case we fixed one boundary condition in each side of the channel

At the inflow boundary we impose 2.(m3/s) for the flow, and at the outflow a 0.6 m for the height.

The flow is taken immobile at the start time. An initial condition of a fixed cote have been fixed to 0.6m.

Then we track the solution trought the time. We have increased the steps of iterations until we reach the steady regim.

This graph shows that there are three regions in the channel.

from the left to the right we have succecively, Fr<1, Fr>1and Fr<1. So it's a Transcritical regim.

As the velocity increase near the hump the water heigh decrease. and we see that we have a jump just after the hump. this is due to the passage from the supercritique regim to a subcritical regim.(cf. second year cours, Suzane)

Return.


Transcritical case 2


Debit

amont(m3/s)

Debit

aval (m3/s)

Hauteur

amont (m)

Hauteur

avale (m)

Conditions

Initiales

Hauteur critque

pour canal horizontal

Type d'Ecoulement

attendu

    2

********* ********

    0.4

cote = 0.4 m

    0.47

Fluvial / Torentiel

In This case we fixed one boundary condition in each side of the channel. the height value imposed at the right side (0.4) of the channel is lower then the critical height value (0.47). so we exspect to get a supercritical regim in the right side of the channel. At the inflow boundary we impose 2.(m3/s) for the flow.

The flow is taken immobile at the start time. An initial condition of a fixed cote have been fixed to 0.4m.

Then we track the solution trought the time. We have increased the steps of iterations until we reach the steady regim.

the solution converge after almost half an hour.

we see from the result that we get , as it was expected thwo regions refering to subcritical and supercritical regims. .

Return.


Transcritical case 3


Debit

amont(m3/s)

Debit

aval (m3/s)

Hauteur

amont (m)

Hauteur

avale (m)

Conditions

Initiales

Hauteur critque

pour canal horizontal

Type d'Ecoulement

attendu

    2

********* 0.4

    0.6

cote = 0.4

    0.47

Torentiel / Fluvial

In this case we fixe three boundaru conditions : two for the inflow and one for the out flow. this has been done in order to obtain a supercritical regim at the left side of the channel and a subcritical one at te right side.

contrary of what we were expecting we get a subcritical regime at the left side. this shows that we can control the flow regime only by changing the boundary conditions.

above this the convergent solution present oscillation at the left side. this can be explain by the fact that we should not impose more then one condition at te inflow if it's a subcritical regime. this result is is phase with theory.