SimScale CAE Forum

CFD - Symmetry Condition (Drag)


#1

Hi All,

when using a symmetry boundary condition - for example around X axis - and we seek the drag fore of a Body (the flow stream in X direction ), Shall we multiply the resultant pressure force in X axis direction by 2 in order to get the magnitude forces acting on both (the actual body & mirrored body )?


#2

Hi @Hesham_Reda!

Yes! You can test it as well :slight_smile: And we would be happy if you share your results with us in form of plots or if possible as a paper/thesis or simply document it by using the comment function of the forum.

Best,

Jousef


#3

In the same vein, I made the assumption that for, Force and Moment Coefficients, when using a 1/2 model (the left or right side of an aircraft), that the reference area should be 1/2 the total wing area (or only the wing area of the 1/2 model), is that correct?

And I assumed the reference length in ‘Force and Moment Coefficients’ was the same for the 1/2 model as the full because that axis length is the same in both.

And, while I am remembering this, I am guessing that the Freestream velocity magnitude in ‘Force and Moment Coefficients’ cannot be obtained from the Boundary Conditions for some reason but it would be nice that if when you change a boundary condition velocity to analysis a different airspeed and you also have a ‘Force and Moment Coefficient’ Result control, then you would be reminded that you should also change your Freestream velocity in the results control, or is it ok to have different velocities in the boundary condition and the the ‘Force and Moment Coefficient’ parameter?


#4

Hi @DaleKramer!

That’s correct. Did you get anything different? And good impetus from your side, if you have any ideas or points that could be optimized feel free to post them inside the Vote For Features section. @Get_Barried & @vgon_alves : Anything else you want to add?

Best,

Jousef


#5

Hi everyone,

Just to reinforce what Jousef said, that is correct.

The theorical subject is, in crude words, that if you are seeking for getting drag force it is needed to be multiplied by 2. But in drag coefficient its not needed because increase of the frontal area compensate this (remember the drag coefficient is the dimensionless value of the force total drag, which results from the integration of the local voltage across the entire).

About ‘Force and Moment Coeffients’, when you achieve results of symmetric solution, the forces are only half that. But if you work with dimensionless numbers (such as drag coefficient, as I said before), in that case the results are independent of the area, so in that case they will be equal to half of your solution domain or all that. So remember that it depends on your unit of interest case-by-case.

Cheers,

VinĂ­cius


#6

’ in that case the results are independent of the area,’

But in order to have a coefficient be dimensionless, does it not do so by using a reference area or length depending on the coefficient? If that is the case, then the coefficients are dependent on area (or length).

Also, I am still unsure of the ‘Force and Moment Coefficient’ Result control Freestream Velocity usage. For example if in Boundary Control, I input the x,y,z velocities as 1000,100,100 (in/sec), should I put the Freestream velocity in ‘Force and Moment Coefficient’ Result control as the magnitude of the vector sum of those 3 velocities?


#7

Hi @DaleKramer,

By definition, the drag coefficient is a function of several parameters like shape of the body. Being the symmetry condition a ‘mirror’ of the shape of the body, it will not affect de Cd.
In other words, that is what I said before: let’s imagine 3 different bodies with the same frontal area:

  1. a box
  2. a drop of water
  3. a sphere

Now, looking to the equation:

Cd = (2 Fd) / (ρ v2 A)

We can see that essentially is the ‘dynamic pressure multiplying the frontal area’ scaled by the ‘drag coefficient’ that is specific to the shape of the body, in our case a box, a drop of water and a sphere has the same value. A box and a sphere and a drop of water all with the same frontal area will have different Cd in the same flow.
You can visualize it better in THIS table, when some objects with the same frontal area do not have the same Cd.
To conclude our example, a drop of water clearly would cut through the flow easier than a box or sphere. So yes, the body along a plane of symmetry would reduce its frontal area by half, but the efficiency of the shape isn’t necessarily doubled (Cd cut in half) by cutting the body in half.

About the Freestream velocity in Result Control,: Yes, you should provide this magnitude as an input of the velocity.

Cheers,

VinĂ­cius


#8

Hi Vinicius,

I was just trying to clarify that coefficient formulas are dependent on areas or lengths, which I think we agree on.

I am familiar with the coefficient formulas but I want to make sure that my coefficients that I get using your formulas are based on the same areas and lengths that I am used to. For example XFLR5 gives these options for reference dimensions:
coefficient_areas

So (it would really help to get a Yes or No for these with explanation if needed):

  1. XFLR5 is not using a 1/2 model so it is using the complete planform wing area but for a 1/2 model in simscale I use 1/2 the complete planform wing area for ‘Reference Area Value’, is this correct for simscale?

  2. XFLR5 uses the main wing MAC chord length for its Ref. Chord length. Is this the value I should use in simscale for ‘Reference Length Value’?

  3. I am not sure there is a simscale parameter that matches the XFLR5 Ref. Span Length, is there?

  4. For Cd measuements is simscale using the 1/2 model projected frontal area perpendicular to the freestream vector for its Cd calculation.