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Hey,

I tried to use siphons in Telemac 2d, but there is no flow in my siphon. So I read the subroutine siphon.f to understand the functioning of the culvert in Telemac and I don't understand why I have no discharge in my siphon. At each time step, the conditions S1.GE.S2, S1.GT.ALTSCE(I1) and S1.GT.ALTSCE(I1) are fulfilled and the elevation of pipes are equal to the elevation of the bottom. So when I calculate by myself the flow with the data obtaining in my simulation I obtain a discharge greater than zéro...
The only value I don't know in the subroutine is V2DPAR%R which is necessary to calculate Qmax. What is this variable?

Thank for your help

Regards

Marine

Hello,

V2DPAR is the integral of test functions (also an area around every point), in the case of V2DPAR it is assembled in parallel.

You can check your point numbers, and follow in subroutine siphon why the discharge ends up in 0. Note that there is a threshold depth required.

There was also a bug corrected some time ago, so make sure that the version of telemac2d.f is the latest issued. If I remember well, you must have DSCE in the call to SIPHON, whereas a former version had DSCE2.

With best regards,

Jean-Michel Hervouet

Thank you
I think the problem comes to V2DPAR because it is very small (10^-269).
I am not really understand your sentence "Note that there is a threshold depth required"
I have another problem at the beggining of the simulation, Telemac write :

POINT SOURCE 1 HORS DOMAINE
POINT SOURCE 2 HORS DOMAINE
POINT SOURCE 3 HORS DOMAINE
POINT SOURCE 4 HORS DOMAINE
POINT SOURCE 5 HORS DOMAINE
POINT SOURCE 6 HORS DOMAINE
POINT SOURCE 7 ASSIMILE AU POINT 38422
SITUE A 0.000000000000000E+000 METRES
POINT SOURCE 8 ASSIMILE AU POINT 55581
SITUE A 0.000000000000000E+000 METRES

I made my siphon with Fudaa so the sources are on an existing mesh and in my model so I don't understand why it writes "HORS DOMAINE"
Maybe my "FICHIER DE DONNEES FORMATE" is not right

please find my project in attachments maybe this can help to understand my problem

and with the files...

Hello,

You are in parallel mode, so for a given sub-domain, a point source may be out of the domain. You should try in scalar mode first to see what really happens. V2DPAR cannot be 0, it is the integral of a test function, so if 0 it would mean that one element is reduced to a point or a line. In subroutine siphon, if IR1 or IR2 is strictly positive, it means that the point is inside the sub-domain and then you should have V2DPAR%R(IR1 or IR2) about the area of elements. If not there is something wrong in the mesh.

JMH

Thank you

Which parameters are used to calculate V2DPAR? What is test function?
I have not flat mesh but I have small meshes with edge length equal to 1 m. It is too small?
Another question, I made small meshes because my culvert have a diameter equal to 1 m so this seems to be logical to have edge length equal or smaller to the diameter of my culvert.

In the subroutine siphon.f you need to call DSCE2 instead of DSCE?

Regards

Marine

Hello,

V2DPAR is purely geometric, it cannot be 0 if you have a mesh size of about 1 m.

In siphon.f, what you must check is the formula giving Q and why it gives 0 (lines 162 to 174). Then you have the relaxation that mixes Q and DSCE(I2) and there it may be set to 0 again, but look first if the first computation of Q, so after line 174, gives 0.

After relaxation you have the limitation with the QMAX and there comes the V2DPAR. If there is one V2DPAR%R(IR1) equal to 0 then you have an error in your mesh at point IR1, which is strange, it would mean that all elements around point IR1 have no area...

Regards,

JMH

Hello,

Thank for your help.
I have not really find why there is no flow in my culvert...
If I understand well the role of Qmax, it impose to have a edge length near to the breadth of the culvert to have a real maximum flow
But I don't really understand the functioning of the relaxation and how this can set the flow equal to zero

Regards

Marine

Hi
Qmax exist to limit the discharge in the culvert to the available volume of water on the upstream side. Imagine you have 10 m3 available and the computation of the discharge require a volume of 12 m3 on the time step. You will have a problem ...

The relaxation exist to limit the variation of discharge in time. At each time you consider 90% of the previous value (at the previous time step) and 10% of the computed value. This is not mandatory but increase the numerical stability because sometimes when the water level upstream and downstream are really close you could observe a sort of ping-pong in the direction of flow which change at every time step.

As you mention the computation you made manually should give a positive value, maybe you could try to print the value used during the computation and see where the problem comes from.
It's difficult for us to find a reason without anymore information

Regards