[PierreLCV]

Dear CFD guys,
I have a general CFD question that i can not find the answer in the different books i have read. I will speak about the thermal wall function and the energy conservation equation but the question is roughly the same for the velocity wall function and the NS equation.
We know from the algebric version of the energy conservation that we need the temperature at the wall cells (for the convective term) and the temperature gradient at the wall cells (for the diffusive term). If we consider a cells from the flow (so a cell which is not in contact with the wall) these wall cells value and gradient will be assessed through different schemes (Linear, Upwind...).
My question is about the cells which are in contact with a wall.
Considering a coarse mesh with a Y+ value around 60 for instance, are we agree that the solver will apply the thermal wall function directly in the energy equation to assess as best as possible the temperature gradient at the boundary between the wall and the cell ?
This point is not very clear in my mind and i figure out that this link between the wall function and how they are linked with the conservation equation is not very clear...

Thank you for your help !
Pierre

[LuckyTran]

In my opinion, this is the wrong way to begin understanding wall functions. It is backwards.

Cells in the flow and near the wall are fundamentally no different. Each cell has a transport equation being solved, which eventually calls for face fluxes. These face fluxes are still discretized by generally the same upwind, MUSCL, whatever schemes. The difference is, cells adjacent to the wall have face fluxes determined by constraints on the boundary rather than neighbor cells. And these boundary conditions are provided by the user since they set up the specific problem. Note that this procedure is true no matter the transport equation being solved, even non-Navier-Stokes equations.

So wall functions. Why are they ever needed? Recognize that for a laminar case, there is no need for "wall functions." Wall functions begin with the turbulence model, how to apply BCs for (k and epsilon, k and omega) more specifically, the turbulent viscosity near the wall. The turbulent viscosity then propagates its way to other transport equations, because at y+>0, the turbulent viscosity at the cell center of the wall adjacent cell, may not be 0. If you ignore the fact that wall functions are tied to the turbulence then of course it is not clear how they are applied, but they're not needed if you look at "only" the momentum and energy equation.

[arjun]

Wall functions are used to calculate the heat flux through the wall in case of turbulent flow. This is how it is typically done:

Given

ustar   =   sqrt( wall_shear / density_fluid )

The heat flux per unit area would be

heat_flux =   (density_fluid * Cp_fluid * ustar ) / Tplus

wall_shear and Tplus off course come from wall functions.

[PierreLCV]

Dear LuckyTran & arjun,
Thank you very much for your answer !
Thank you particularly LuckyTran for your complete explanations. Indeed, you are right I should have started by remembered that wall function are needed ONLY in turbulent flows situations and are applied for U, T but also for k, epsilon or omega depending on the turbulence model chosen.
My question is I think simpler than that. I am just wondering if the solver will use the wall functions directly in the N-S (or NRJ) equation to estimate as best as possible the boundary gradient for the cells in contact with a wall ?
Indeed, i think i understand how the wall functions are needed, i am just wondering how they are used practically by the solver.

Thank you for your help !

Best regards,
Pierre

[arjun]

Originally Posted by PierreLCV

My question is I think simpler than that. I am just wondering if the solver will use the wall functions directly in the N-S (or NRJ) equation to estimate as best as possible the boundary gradient for the cells in contact with a wall ?

Solver does not use wall functions to estimate the gradients. Even if it did it is pointless since in turbulent flow wall functions is what is needed at wall not the gradients.

Gradients may be needed to calculate wall function quantities but this would be opposite of what you said.

Indeed, i think i understand how the wall functions are needed, i am just wondering how they are used practically by the solver.

As far as energy goes, they are used how i wrote in my first post.

For momentum wall shear rate is needed and is calculated by wall function. Again gradient may be needed here if yplus is smaller than 5.

[PierreLCV]

Dear Arjun,
Thank you very much for your quick reply !
Ok so gradient is calculated between the wall and the cell (in a linear way) when the cell border is such that Y+<5. I agree with that.
Considering a coarser mesh with a first cell such as Y+ > 30, in this case i can not calculate the heat flux as previously (linear temperature gradient) and i will apply the thermal wall function to have an accurate estimation of the wall heat flux. So if i take the NRJ conservation equation, for the border of the cell in contact with the wall, i will use the thermal wall function which will give an accurate estimation of the wall heat flux knowing the temperature of the wall and the temperature at the center of the cell. In other words, the wall function will be used directly in solving the energy conservation equation to evaluate wall fluxes?
Is that correct ?

Thank you for your patience

Pierre

[arjun]

Correct.


One can calculate the heat fluxes from gradients too but that would be wrong in case there is turbulence.

So solvers typically use wall functions to get correct heat fluxes.

The only case where gradients would make sense when one is doing dns and everything is resolved.

[PierreLCV]

Dear Arjun,
Thank you once again for your answer, it is very interesting !
In my first message, i spoke about temperature gradient because i thought that the solver worked only (for the NRJ equation) on the "temperature form" of the conservation energy:

Thus, i thought that the solver used the wall function to be able to access to the temperature gradient on the boundary cell which is contact with the wall...
Should i deduce that the solver is not working like this ?

Thank you VERY MUCH for your help !

Pierre

[arjun]

Yes.


The order in case of turbulent flow is:

1. Calculate yplus, uplus, ustar and shear stress with given flow field and turbulence quanties.

2. For momentum, shear rate is used to close the wall faces.

3. For turbulence, the wall turbulence production, omega or epsilon etc are calculated using ustar, uplus, yplus.


4. For energy, the heat flux through the wall is calculated using the Tplus, which is calculated from law of wall.


The gradients at the wall are not directly used (pretty much always). They are used indirectly in calculating wall turbulence production and hence are extremely important too.

[PierreLCV]

Dear Arjun,
Thank you very much for your complete answer ! It is much more clear now .
Last point, if i go throught with the NRJ equation:
If we consider the border of the cell in contact with a hot wall, knowing that the convective term is null at this border (velocity equal to 0), it will just remain the non permanent term and the conduction/diffusive term given by the wall function. Correct ?


Thank you very much !
Pierre

[arjun]

Originally Posted by PierreLCV

If we consider the border of the cell in contact with a hot wall, knowing that the convective term is null at this border (velocity equal to 0), it will just remain the non permanent term and the conduction/diffusive term given by the wall function. Correct ?

This is typical of CHT situations. That is one side is solid and another is fluid.

Here the situation is little bit more complicated. On the solid side the heat flux is calculated from temperature gradient -k(dT/dx) but on the other side the heat flux is function of turbulent quantities.

Here care shall be taken:

1. Heat flux across this interface is such that flux given (or taken) by solid is same is flux taken (or given) by fluid.

2. (1) allows you to find the temperature of the wall. That is used in Tplus calculation.

Then the heat flux is finally calculated. If there is a radiation loss present then good luck, it gets even more complicated.

[PierreLCV]

Dear Arjun,
Thank you very much for your answer ! It is more and more clear that... it is not so easy but it is at least clearer


I think I will close this thread. Thank you once again Arjun and speak you soon.


Pierre