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TOPIC: Time Step

Time Step 14 years 3 months ago #391

  • konsonaut
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Hello,

I'm simulating the River confluence case from the validation-document (2000). In the validation case the time step was 0.1 sec.

For comparing purposes I reduced the time step to a value of 0.01 sec.
In both the simulations I used the k-e turbulence model, rough walls and roughness coefficent of boundaries same as friction of bottom. The other parameters are the same as in the validation document.

Comparing the two simulations in some nodes I get quite different velocity magnitudes, see attached image. And why does the velocity oscillate when using a smaller time step?

I would be glad for an answer.
Clemens
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Re:Time Step 14 years 3 months ago #392

  • c.coulet
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Hi clemens
There is no attached image (maybe too large?)
A too small time step is not really efficient (time consuming). It's also possible to find the solution without any iteration because the accuracy is automaticaly satisfied...
See the number of iterations at each time step!
Good luck
Christophe
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Re:Time Step 14 years 3 months ago #394

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Sorry, now the image should be attached.

I'm wondering why in some nodes (like in the node in the image) there appear such velocity oscillations when using a time step of 0.01 sec? And why the velocity magnitudes from the two simulations are so different.

The other question is: checking the fluxes trough the liquid boundaries gives me the following values after convergence:

VOLUME IN THE DOMAIN : 2.004673 M3
FLUX BOUNDARY 1: -0.1032049 M3/S
FLUX BOUNDARY 2: 0.5000000E-01 M3/S
FLUX BOUNDARY 3: 0.4999999E-01 M3/S
RELATIVE ERROR IN VOLUME AT T = 1000. S : -0.1604878E-03

Why there is a flux defect of 3 l/s between inflows(2+3) and outflow(1)?

Clemens
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Re:Time Step 14 years 3 months ago #402

  • jmhervouet
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On mass conservation : yes, CONTINUITY correction is necessary to see a perfect mass conservation. It is due to the fact that on exits with prescribed depth or elevation, the continuity equation is not solved (it is replaced by the equation depth = prescribed value). When continuity correction is activated, the flux at the exit is deduced from the continuity equation, regardless of the computed velocities. With that option we can check that at steady state the flux at the exit is equal to the flux at the entrance.

The mass conservation may however depend on the solver accuracy. Try SOLVER = 8 (direct solver) if you want to see a mass-conservation at the accuracy of the machine. Another trick is to use the key-word :

TREATMENT OF NEGATIVE DEPTHS : 2

because with this option the continuity equation is recomputed at the end, considering that the velocities are correct, and changing the depths to ensure the continuity. This algorithm is used to have positive depths in presence of tidal flats, but has this interesting side effect.

Regards,

Jean-Michel
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Re:Time Step 14 years 3 months ago #406

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Thanks for the illuminating answers!

Maybe someone is interested in some details:
when I set the type of advection for U,V and h to SUPG (2;2) and continuity correction, so I get perfect mass conservation and plausible velocity distribuitons with recirculation zones in the domain. Using characteristic method (1;1) and continuity correction, I have quite a big error in mass conservation. It is also to say that with the latter option I don't get a recirculation zone downstream of the confluence, which should appear when using the k-e model.

In another forum topic there was the question if there is the possibility to account for wall friction. It seems that when using the k-e model and rough wall you can account for wall friction via the keyword ROUGHNESS COEFFICIENT OF BOUNDARIES. Please correct me if I'm not correct.

Regards,
Clemens
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Re:Time Step 14 years 3 months ago #408

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Yes, the k-epsilon takes into account the rougness coefficient of boundaries.

Regarding your previous message, I have to add that the TYPE OF ADVECTION is a set of integers, the first one for U and V and the second one for depth. The second one should now always be equal to 5. Other values are obsolete and can give non conservative options, this is why you find big mass error when trying 1;1. Just try 1;5 and you will see that it is also mass-conservative.

On the choice of advection scheme for velocities, I like the method of characteristics that is the less diffusive with large Courant numbers. I don't like very much SUPG because it implies tuning the upwinding with the extra key-word SUPG OPTION.
Distributive schemes 4 or 5 are OK but diffusive. The edge by edge form of scheme 4 (number 14) is the only viable alternative to characteristics when there are dry zones, and it gives nice solutions in the dam break Ritter test-case.

Regards,

Jean-Michel
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Re:Time Step 14 years 3 months ago #395

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Sorry, now the uploading should work
RiverConfluence_TimeStepInfluence.JPG
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Re:Time Step 14 years 3 months ago #396

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Clemens
About the fluxes, the difference should come from the computation of the fluxes. On the boundary where the prescribed elevation (downstream), the coninuity equation is not solved.
Try with the following keywords activate:
    CONTINUITY CORRECTION
    COMPATIBLE COMPUTATION OF FLUXES
Christophe
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Re:Time Step 14 years 3 months ago #401

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Hello,

In the theory of finite elements, it is said that you tend to the solution if both mesh size and time step tend to zero, so reducing the time step without reducing the mesh size is not always good. The method of characteristics for example is better with large time steps. I would say that the important parameter is the Courant number based on velocity. A value around one is OK with finite elements and can go much further in the case of Telemac, maybe at the cost of accuracy. A very small Courant number may accumulate truncation errors. It is even said that the method of characteristics may become unstable if the time-step becomes too small.

To be absolutely sure of a result, we should try to test the convergence by reducing both the time-step and the mesh-size, e.g. by splitting every triangle into 4 with STBTEL, and dividing the time-step by 2. However the computer time increases very rapidly when we do so.

Regards,

JM
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