# Better understanding aerodynamic results

Hi, this is my first attempt to use CFD to characterize a lifting surface aerodynamic coefficient.

I’ve setup a simple simulation where I evaluated a wing at cruise speed & 0deg AOA Force et Moment and it’s aerodynamic coefficients.
https://www.simscale.com/projects/borowczyk_alex/test_wing/
However, I’m a bit confused, I’m getting a lift coefficient of 0.3834.
Using the ref Surface of 0.408 m^2, ref free stream velocity of 13.9 m/s and air density of 1.1965 kg/m^3, get a lift force of 18.08 N. However, looking at the force résultat, the Z force is computed at 15.11

Additionally, the lift coefficient much different of the one provided by openvsp which use a vortex lattice method. The VLM evaluates the lift at 0.2422

I am misunderstanding something?

ps. I’m also concerned of my trailing edge meshing, what could I do to fix it?

Thanks

I see so far your mesh refinements are alright. Have you considered doing a steady-state analysis first to iron out and possible mesh issues before moving on to compressible? 13.9m/s is less than 0.3 Mach and your flow can be deemed incompressible which will simply your simulation, allowing you to identify any possible issues.

As for the mesh itself, you need to consider your Y+ value. Assuming you’ve elected to keep using the same turbulence model k-omega SST, you will typically need a Y+ of less than 1 to capture the wall effects. Do consider this before re-meshing but I do recommend switching to incompressible flow first.

Sorry what do you mean? Is the mesh showing weird distortions at the trailing edge? I can’t see it now as you’ve started to re-mesh. A good way to mitigate this issue would be to increase surface refinement and probably a feature refinement. That would help the mesh conform to your geometry.

Do let me know how it goes.

Cheers!

Regards,
Barry

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Oh I am embarrassed, I don’t know why, but I believed compressible flow assumption started at 0.03.

For the turbulence model, I choose the k-omega SST due to the fact it’s been use in all the exemple I saw. Is it a appropriate chose?

When you mention the Y+, you mean the mesh box?

Thanks for the help

Don’t worry I’ve made that mistake as well and after looking at all the setting was utterly confused as to what to do

K-omega SST is widely used because it is relatively accurate and computationally not very expensive compared to other models like k-epsilon (less accurate to some degree but computationally cheaper) or LES (very accurate but very computationally expensive). K-omega is a good combination of effects away and near the wall hence its popularity. For your case, its totally fine to use it.

Y+ is the dimensionless distance from the wall for you first layer of the boundary layer inflation (under mesh refinements). You can calculate your desired y+ here with your parameters to deduce the first layer distance (if I recall correctly). That value (estimated wall distance m) is what the maximum size of that first layer of your boundary layer inflation should be in order to maintain your y+ value (of <1).

Oh do note that your “boundary layer length” is your chord length of the foil, so since yours is tapered, maybe consider the the largest chord length? I’m not too sure about this so maybe someone can correct me but for now just assume your distance as the 1/4 chord length.

Hope this helps!

Cheers.

Regards,
Barry

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Thanks for all the information.

For the Y+ using the calculator I’ve gotten the estimated wall distance to 0.000022 m (for a Y+ of 1, should it be lower?)
If I understand correctly this is the desired height of the first layer right?

So my the final layer thickness would be:

first layer thickness * (expansion rate ^ number of layer) = final layer Thickness

With a expansion rate of 1.3 and 3 layer, this gives me a final thickness of ~0.03m. Does that makes sense?
however I’m confused by the minimum overall layer thickness. Isn’t redundant.

Finally, I’ve seen the default Use relative size for layers option is set to “true” how does it impact my choose of values>

Thanks again

With an estimated wall distance if 0.000022 m shouldn’t your final layer be 0.000048334 m?

Minimum overall layer thickness will be the lowest possible thickness of the inflation layer. If this number is higher than any layer, that inflation layer will not be inflated.

Lastly, the relative size take reference from the nearest grid size. If the option is set to “true” your inflation layer will be scale with the most adjacent grid. Example 0.03 final thickness layer will be 0.03*size of nearest grid.

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You are right I miss calculated the wall distance.
So know what would be the strategy to chose all those values?

What value are you talking about?

The number of layer and Y+ depends on the requirement of you turbulence model.
Final layer thickness will depend on your Y+ value and your number of layers.
For the minimum overall layer thickness, choosing a value that is smaller than your first layer should be fine.

Cheers

I guess I’m confused on how to determine these parameter if relative size is activate.
And also if it’s better on or off?

If it’s on, you estimate your cell size. Every refinement will decrease the size by 2. Example the biggest grid is 0,1 if your refinement level is 4. it will be 0.1/(2^4). You will take you inflation thickness divide by that number.

For simplicity sake, I would recommend you to off the relative size if you want accuracy.

3 Likes

What about Simscales LES Spalart-Allmaras. I found this quote from this study

In the current work the SST and Spalart-Allmaras
turbulence models generated results that were nearly identical to
each other and most closely matched the experimental data.
There is a considerable difference in the runtime of the two
models when allocated the same resources. The SpalartAllmaras
model was able to generate almost identical results as
the SST model in just under one-ninth the wall clock time

Sounds pretty good to me but I don’t want to waste core hours to confirm if the same ratio happens here…

Dale

@DaleKramer, I wouldn’t spend them just to test this, LES is immensely expensive, and this ratio will be very dependent upon courant number, mesh size etc. It takes a lot of efficient thinking to do it well.

Best,
Darren

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BUT it says that their LES simulations took 9 times LESS than the SST, are you saying that their REDUCTION over SST can only be achieved with LES if you know what you are doing

Ah sorry, I misread, I think They ran a steady state SA model which I believe is possible, however, our platform only has a transient SA model similar to LES but with better wall treatment. Because it is transient it will take significantly longer than K-w SST (Steady)

Apologies for the confusion.
Darren

Ah… at least I am tried…
Thanks a bunch