open TELEMAC-MASCARET The mathematically superior suite of solvers

Hi

First of all, the law differ from weir treatment because in this case we deal with small structure (smaller than the mesh size) where the weir treatment requires at least 2 nodes on each side of the weir. So we should take in account the width of the structure.

For the other point, the main aim of this subroutine is to allow different type of flows (free surface or in charge). If you're sure that the flow is allways in charge, you could use siphon.f instead.

In our mind, we decided to separate the case where the charge is only upstream or only downstream. Both is also a possibility which is treated as the first case and you could modify this if you're disagree with our solution.
For us, we're not following a real orifice law because a tube is a linear infrastructure and this create a sort of nozzle which limite the output head-loss...

Regards

Hi!

Thank you for your reply.
I was actually pointing out that in the validation case 039_bridge, the results show that the flow is in charge in the structure, whereas the computed flow was obtained using the formula defined for the case when the structure is in charge only at the upstream end.

By giving a closer look at the subroutine it appears that the IF condition used seems to be actually pointing at the wrong formula (see ***):
I think it should rather be ***IF(S2.LT(RD2+HAUT))***. Otherwise the last formula (used in siphons) is never used.
I am not familiar with programming, but if a condition IF((A<C).AND.(B<C)) is not fulfilled, do the action defined by ELSE corresponds to the case when (A>C).AND.(B>C)?

I made a test by adding reformulating the conditionning in the orifice law part the following way:
The result gives S1 = 10,43, S2 = 7,04, and the computed discharge is now coherent if I check it by hand using the siphon formula.

Regards
PL

Hi again,

Just to say that a made a copy/paste mistake in my previous post, I indeed modified buse.f using the following section (S2 and not S1 in **):
Regards
PL

Hello,

question regarding the bridge modelling:
does the tube treatment in v6p2 distinguish between pressurized / free flow below bridge deck and also bridge deck overflow?

Thanks in advance!
Clemens

Hi,

You can either include your deck features in the geometry file if you want it to be computed or add a second tube on nearby nodes in which you set a very large height (Hau) to make sure it functions as a weir.

Regards,
PL

Hi Clemens

As already said, this solution is mainly for small structure (size lawer than the mesh size).
In our mind this is a functionnality to compute the discharge below a bridge when we mesh the top of the bridge itself.
But you could also describe the overflowing part as a tube which is allways a free flow as PL wrote

Regards

Hi

NO !!!
This is not a programming rule but a logical rule. The contrary of (A<C AND B<C) is ((A>=C) OR (B>=C))

The formula is correct assuming the consideration describe in my previous post.
Your version could be a potential evolution but in your case, you forgot one situation : (S1.LT.(RD1+HAUT).AND.S1.GE.(RD2+HAUT)).
In such case, you doesn't compute any discharge!

Regards

Hi Christophe and thanks for your reply.

You are right, I forgot to add this condition.
I read again the code, and it is still a bit unclear for which S1 conditions are the 2 formulas applied...

Anyway, there is still something I find strange. According to the tube geometry given in the bridge file (039_bridge), both tubes are submerged completely (invert elevation = 1, height = 1,5 so the top of the section lies at 2,5 whereas the computed water levels are above 7 m...).

In my mind, it is the last formula that should be used, the one used in siphon.f: Q = SECT * SQRT( 2.D0*GRAV*(S1-S2)/(L+CS2+CE1) )
But the discharge is computed using Q = SECT * SQRT( 2.D0*GRAV*(S1-S2)/(CE1+L) ) instead.

I made a test by replacing the tubes by siphons, and this time the computed discharge and water levels are coherent, with the same results as those detailed in my previous post, with my own IF/THEN conditions.

Finally, could you please give more details about the formula Q = SECT * SQRT( 2.D0*GRAV*(S1-S2)/(CE1+L) )? If I understand right the logical conditions, it should be used when S1 > RD1+HAUT and S1 < RD2+HAUT ? It seems anyway that the tube geometry is controlling the flow, ie that Q does not depend on S2, since CS2 does not appear. Lastly, I am also surprised by the head loss coefficient term, because I would have expected a "1" to be included, corresponding to the kinetic energy at the outlet. If I am right, this "1" is the minimum head loss factor according to Bernouilli if neglecting contraction and friction in a tube emptying a reservoir flowing free.
The term (CE1+L)^-0,5, with classical head loss coefficients (ie 0,5 and 0,2 for a rather short tube such as a bridge) is > 1. Whereas when neglecting head losses in Bernouilli, we have at maximum Q = S(2gH)^0,5.

I am maybe wrong, but I would like to understand exactly what is telemac doing before implementing it or using it with my own modifications.

Thanks in advance,
PL