open TELEMAC-MASCARET The mathematically superior suite of solvers

Hello,

Generally we like to refine the mesh near the bottom, but in your case the problem is probably the tidal flats with elements totally flat on dry zones and highly distorted on the shoreline. You have an example in the test-case 009_wesel. You could try with 2 or 3 levels only and see what happens.

With best regards,

Jean-Michel Hervouet

Hello,

I have a rather general question about the vertical mesh transformation. Let me explain my question by the following example.

I use a SIGMA transformation with given proportions. In the cas file I specify to use 10 horizontal layers. In the CONDIM subroutine I specify the following z* levels:
ZSTAR%R(1)=0.00D0
ZSTAR%R(2)=0.05D0
ZSTAR%R(3)=0.10D0
ZSTAR%R(4)=0.15D0
ZSTAR%R(5)=0.35D0
ZSTAR%R(6)=0.50D0
ZSTAR%R(7)=0.65D0
ZSTAR%R(8)=0.80D0
ZSTAR%R(9)=0.90D0
ZSTAR%R(10)=1.0D0

I only specify 10 z* levels, so I think the equations are only solved in 9 vertical prisms throughout the domain. As such, I would only expect results in 9 vertical layers, the first one being at a level of 0.025D0 (middle of the first prism), the second one at a level of 0.075D0 (middle of the second prism), .... However, in my result file I can find 10 vertical layers! As such I don't know to which local water depth the calculated layers correspond. Does layer 1 correspond with the values at 0.0D0, being the bed level?

Can someone clarify this vertical mesh transformation please? Thank you very much in advance!

Best regards,
Stefaan

Hello,

In fact, our unknowns are defined on the mesh nodes, so we do have in your case 10 planes (i.e. 10 2D meshes piled up on the vertical) where we have results. Only in rare cases the variables are defined per layer, and in this case we would have 9 results. This is for example the case of the vertical velocity in the transformed mesh (W*), which is defined at mid-layer, or the diffusion given by the mixing-length model. Even the dynamic pressure is computed on mesh nodes, unlike what is done for staggered grids.

With best regards,

Jean-Michel Hervouet

Dear Jean-Michel,

Thanks for your quick answer which clarifies a lot. One small remaining question: how to interpret the physical meaning of the lowest mesh with z*=0.0D0? In fact this mesh corresponds with the bottom, and as such the flow velocities should be 0. However, this is not the case.
Thanks!

Best regards,
Stefaan

Hello,

This small question is a big question. Plane 1 is indeed the bottom, but the velocity is not zero because it is supposed to be given at a small distance from the bottom,outside the viscous layer (think that it is subjected to turbulent viscosity), and also above the grain size. In fact the friction velocity is determined with the velocity of the second plane (see e.g. tfond.f), so this value of velocity on the bottom and the exact altitude over the bottom is not very important. My own theory is that it should be taken at delta(Z)/e**2, height of the first layer divided by the square of 2.71828.. In this case the discharge in the first layer computed with a linear interpolation of velocity gives the same discharge than with the integral of the logarithmic profile over the same layer... but this is a very mathematical point of view. We do not set this velocity to 0 because in the first layer and with linear interpolation it would give a wrong (too small) discharge.

With best regards,

Jean-Michel Hervouet