How to calculate the forces due the pressure in navier-stoke

Post processing utility for Elmer
rodrigodga
Posts: 16
Joined: 04 May 2010, 18:20

Re: How to calculate the forces due the pressure in navier-stoke

Post by rodrigodga » 31 May 2010, 17:40

Hi Peter,

My doubt is about the results I found. My fin is very similar to a wing, where I have a flat surface on the botton and a curved surface on top. I would like to know the drag and the lift forces generated by the fin when subjected to a water flow (interely submerged in the fluid). The pressure profile over the fin seems consistent to me, but I not knowing how to see the vector forces (in newton) resulting from the pressure around the fin.
Hamed taught me a procedure (described bellow) that worked but I think the values are too high to be the forces exerted on a fin of 110 mm with speed of 1 m\s.

The procedure a used was:

Solver 2
Procedure = "SaveData" "SaveScalars"
Filename = dragforce.dat
Operator 1 = boundary sum
Variable 1 = Flow Solution Loads 1
Variable 2 = Flow Solution Loads 2
End

and the results I found was:

Variables in columns of matrix: dragforce.dat
1: boundary sum: flow solution loads 1 over bc 1
2: boundary sum: flow solution loads 1 over bc 2
3: boundary sum: flow solution loads 1 over bc 3
4: boundary sum: flow solution loads 1 over bc 4

.dat file
-2.153568617542E+003 -2.644249973207E+005 -2.980330116155E+005 0.000000000000E+000

So I Thought that this results was the sum of the pressures boundary and not the forces.

Thank you,

Best regards,
Rodrigo

rodrigodga
Posts: 16
Joined: 04 May 2010, 18:20

Re: How to calculate the forces due the pressure in navier-stoke

Post by rodrigodga » 02 Jun 2010, 02:47

Hi Hamed, Peter,

I was commiting a mistake. I was forgetting to put a command on the sif file. The two procedures described above are exatly what I was looking for.

Thank you very much.

Best regards,
Rodrigo

kanuk
Posts: 14
Joined: 20 Jul 2010, 02:51

Re: How to calculate the forces due the pressure in navier-stoke

Post by kanuk » 22 Oct 2010, 03:54

Hello,

I was wondering how you managed to resolve this issue. I am trying to validate with a NACA 0012 airfoil. For a zero angle of attack, I get the following in my "dragforce.dat" file (see my attached .sif to see what this means)

-5.029592858741E+000 5.218212250599E-001 3.275385306949E-009

This leads me to believe that the third parameter has something to do with the lift (which should be zero in theory). But, when I change the AoA to say 10 degrees, I get:

-9.434822856886E+000 -1.246319643055E+001 -2.997330903905E-009

This is what confuses me. What, exactly do the parameters mean? And is it possible to get the lift and drag coefficients without digging into the results file?

PS - I have one more question: when a boundary condition is not specified explicitly in the .sif file for Navier-Stokes flow, what becomes the default BC?

Best,
Dave


Code: Select all

Header
  Mesh DB "airfoil"
End

Simulation
   Coordinate System = Cartesian 2D
   Simulation Type = Steady
   Steady State Max Iterations = 20
   Output Intervals = 1
   Post File = "rigid.ep"
End

Body 1
  Material = 1
  Equation = 1
End

Material 1
  Viscosity = 0.00002
  Density = 1.2
End

Solver 1
  Exec Solver = "Always"
  Equation = "Navier-Stokes"
  Variable = "Flow Solution"
  Variable Dofs = 3
  Linear System Solver = "Iterative"
  Linear System Iterative Method = "TFQMR"
  Linear System Max Iterations = 5000
  Linear System Convergence Tolerance = 1.0e-8
  Linear System Abort Not Converged = True
  Linear System Preconditioning = "ILU0"
  Linear System Residual Output = 1
  Steady State Convergence Tolerance = 1.0e-02
  Stabilize = True
  Nonlinear System Convergence Tolerance = 1.0e-05
  Nonlinear System Max Iterations = 8
  Nonlinear System Newton After Iterations = 3
  Nonlinear System Newton After Tolerance = 1.0e-02
  Nonlinear System Relaxation Factor = 1
  Linear System Precondition Recompute = 1
  Calculate Loads = logical true
End

Solver 2
  Procedure = "SaveData" "SaveScalars"
  Filename = dragforce.dat
  Operator 1 = boundary sum
  Variable 1 = flow solution loads 1
  operator 2 = boundary sum
  variable 2 = flow solution loads 2
  operator 3 = boundary sum
  variable 3 = flow solution loads 3  
End

Equation 1
Active Solvers(2) = 1 2
End

!inlet
Boundary Condition 1
  Target Boundaries(1) = 2
  Velocity 1 = 20.0000
  Velocity 2 = 0.00000
End

!outlet
Boundary Condition 2
  Target Boundaries(1) = 4
  Velocity 2 = 0.00000
End

!airfoil
Boundary Condition 3
   Target Boundaries(1) = 5
   Velocity 1 = 0.00000
   Velocity 2 = 0.00000
   Save Scalars = Logical True
end

raback
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Re: How to calculate the forces due the pressure in navier-stoke

Post by raback » 22 Oct 2010, 11:13

Hi

1) In 2D the third component (in 3D the 4th) of "Flow Solution Loads" is the residual coming from the continuity equation. So it does not have much relevance. The two first components are Fx and Fy.

2) Natural BC i.e. zero traction.

-Peter

Captain Slow
Posts: 72
Joined: 10 Nov 2009, 19:17

Re: How to calculate the forces due the pressure in navier-stoke

Post by Captain Slow » 09 Nov 2010, 19:27

May I resurrect this thread.... (it seemed more apt to add my query here rather than open a new topic)

I am using the fluidic force, with the boundary sum over surfaces. Now the output force is in x y and z directions, as opposed to normal (or parallel) to the surface (which is fine, one can calculate the normal).

My question now relates to the normal used by elmer in the calculation of the loads, that in equation 33.1 of the ElmerModelsManual. My results seem to indicate that this ''normal'' is not normal to the surface, rather it is a unit vector (or unit normal) along the axes. Is this the correct conclusion?

Thanks,
John

raback
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Re: How to calculate the forces due the pressure in navier-stoke

Post by raback » 09 Nov 2010, 20:36

Hi

If you're referring to the "calculate loads" method, there is no explicit forming of the normal vector. Rather what you get as the residual is known to correspond to the term \sigma \cdot \vec{n}. It's actually difficult to assign this force to different the boundary elements when computing the force in the corners. Maybe somebody with more insight in the FE method can comment how the normal that is only present implicitely in the weak form should optimally be computed?

In the fluidic force (currently 36.1.?) the normal is formed for each boundary element for which the integration is performed. The product within the integral is a tensor product so the outcome is still a vector with three components. Or perhaps I don't understand the question correctly...

-Peter

Captain Slow
Posts: 72
Joined: 10 Nov 2009, 19:17

Re: How to calculate the forces due the pressure in navier-stoke

Post by Captain Slow » 10 Nov 2010, 15:17

Apologies-I am using an old (about 1 year old) version of the models manual-hence the 33.1 rather than 36.1 in the most recent version.

So, as I understand, the fluid force is computed normal to each boundary element, then when it is displayed in the user specified output file for the boundary sum one can get 3 components. It is my understanding that the components were the fluid force in the x, y, and z directions (IE not normal to the surface).

Maybe my question is best served by an example. Below is a screengrab from elmer, the compass shows the direction of the axes (red cone is x, green y) I have a flow of fluid coming from the boundary to the rightmost of the screen (therefore flow along negative x).

setup.png
(240.84 KiB) Not downloaded yet
fluidforceh100_2.dat
(269 Bytes) Downloaded 151 times
I wish to calculate the fluid force on the cantilever beam. One part of the cantilever is selected. I take into account the top, bottom, and edges of the beam and use the boundary sum of the fluid force for the total fluid force on each area. Attached is the output file using the fluid force/boundary sum method for the ''flow solution loads 1'' 2 and 3 using the following code in the sif file:

Code: Select all

Solver 2
 Procedure = "SaveData" "SaveScalars"
 Filename = fluidforceh100.dat
 Operator 1 = boundary sum
 Variable 1 = Flow Solution Loads 1
End
fluidforceh100_2.dat
(269 Bytes) Downloaded 151 times
Now, if in the output file we take the second line corresponding to ''flow solution loads 2'', which I believe to be that along the y axis, and we wish tocalculate the total fluidic force in that direction. If the results displayed are along the y axis the we must add all the values (IE side1+side2+bottom+top= -2.231074766653E-008 + (-2.336941529793E-008) + (-2.397792564864E-008) + (-9.446020778595E-010) = -7.06E-08). But, If the output is somehow normal to the surface one has to take account of this, then one presumably has to do total=side1+side2+bottom-top= -6.87E-08. If this latter is the case then I would have expected that side1 and 2 would have the same magnitude but different signs-their values should be the same as the setup is symmetrical with fluid flowing along x.

I hope I have managed to explain this, though I suspect not in the clearest manner. The nub of the matter is how to treat those 4 numbers in the output file.

Thankyou, John

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