open TELEMAC-MASCARET The mathematically superior suite of solvers

Dear all
It seems to me that there is a problem with the tube routine, if I take the example of telemac2d bridge in the "fo1_bridge.txt" file, I change the dimensions of the first tube width = 0.5m and height 0.5m knowing its hydraulic capacity not exceeding 0.5m3 / s. After simulation (bridge exemple) with a step of 1 second I see that the flow through the tube is 2.34 m3 / s. This rate is higher than the capacity of the tube 0.5 x 0.5m !
Do you have any explanation for this anomaly?

Hi
What are the water level upstream and downstream in your simulation?

You said the hydraulic capacity couldn't exceed 0.5 m3/s which means the velocity couldn't exceed 2 m/s.
In the subroutine tube, we don't have any consideration on the velocity inside the tube, only a calculation of the discharge...

Look in the subroutine buse.f to understand the computation method of the discharge

hope this helps

PS: Please think to update your profile

Mr. Coulet
Thank you for your response.
In my case I only tested the example given when downloading télémac2d. in the validation file, there is a "Bridge" exemple.
In the "cas" file, the initial side is 7m, and prescribed elevations 7m, 7m.
When Simulating the tube switch from free flow area to close-driven flow
and level deph in upstream becomes 9m.
For buse.f file, I confess I do not know FORTRAN, but I wrote a routine in vb dot net that allows me to calculate the hydraulic capacity of a tube.
I can only see that the flow in the tube is overestimate by buse.f routine!

Hello Mohamed

I can not understand how the same law which is implemented in buse.f can give different flowrate when it is programmed with VB. If you have reproduced the same law with the same hydraulic conditions, you have to retrieve exactly the same result in both cases.
did you verify that you reprocuce exatly the same conditions? and the same law?
Kind regards and happy Aid :)

Riadh

Dear Riadh
Thank you for your reply, I either I do't understand how a tube 0.5 x 0.5m can have a capacity of 2.3 M3/s.

good aid for you too.

Hi

The value is high, but it's just the result of the application of the formula entered in buse.f
Maybe in your case, the coefficient CE, L12 and CS are well adapted and they should be modified for this tube...

Could you share with us the law you programmed in your vb code? and also the data and the result you have?

Regards

Thank you Mr. Coulet
I am in the process of understanding your program Fartran I compare formulas and results and get back to you with the VB program.

best regards