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Old   March 15, 2012, 04:31
Default Grid Independency Variation
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akj
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anil
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Friends, i have few doubts regarding the grid independency. In my case i am taken 3 meshes one is 6 lakh, 8 lakh & 9 lakh.
For 6 lakh m getting value of 1.59 and for 8 lakh m getting a value of 1.45 whee as for 9 lakh m getting a value og 1.56. So i would like to know as the variation is not unidirectional so whether is it compulsory to have a unidirectional change or not. Other thing which i wanna know is how much % of variation is generally allowed.
And what are the possible reasons for this kind of peculiar behaviour.?

Any kind of suggestion will be of big help.

Regards,
Anil
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Old   March 15, 2012, 07:59
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Filippo Maria Denaro
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Quote:
Originally Posted by akj View Post
Friends, i have few doubts regarding the grid independency. In my case i am taken 3 meshes one is 6 lakh, 8 lakh & 9 lakh.
For 6 lakh m getting value of 1.59 and for 8 lakh m getting a value of 1.45 whee as for 9 lakh m getting a value og 1.56. So i would like to know as the variation is not unidirectional so whether is it compulsory to have a unidirectional change or not. Other thing which i wanna know is how much % of variation is generally allowed.
And what are the possible reasons for this kind of peculiar behaviour.?

Any kind of suggestion will be of big help.

Regards,
Anil
Monotone convergece is ensured monotonically... try using finer meshes, otherwise your results could indicate some errors in the code
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Old   March 15, 2012, 17:21
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Paolo Lampitella
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The numerical error should reduce monotonically as long as the code is correctly written, the solution derivatives actually exist and, as stated by Filippo, when the grid is sufficiently fine for the Taylor series approximation being valid. Also, its absolute change from the coarse-medium to the medium-fine grid couple is supposed to reduce.

Besides some error in the code, the lack of monotone convergence could also indicate the presence of an additional error source not considered in the previous approach. A typical issues is the convergence of the overall system of equations; consider that a certain residual does not imply at all that your error in the equation solver is at the same level. Actually, it usually is some orders of magnitude higher. As a consequence, the level of convergence in the solver has to be carefully set when the convergence of the solution is looked for (that is, Fluent's default 1e-3 is out of question)
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Old   March 15, 2012, 21:51
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IMO, it depends what one is looking at. IMO, there is nothing that says how an integrated quantity, such as lift or drag, should behave as the grid converges. Yes, a flow value in space, lets say pressure, will probably converge monotonically with finer grids. Though that is not guaranteed since the value in space is a result of a trajectory. Unfortunately, the pressure at one point may converge in one direction and the pressure at another location may converge from another direction. Therefore, the lift and drag convergence, which is based on pressure differences, could be anything.

So, a better bet is to look at pressures or, even better, the flow quantities explicitly being solved for (such as u, v, w, etc.). Assuming, of course, that pressure isn't being explicitly solved for. For example, for a compressible flow solver, pressure is a by product and the values being explicitly solved for are rho, rhou, rhov, rhow, and rhoe0.

Grid refinement must be done after all the physics, which can be captured, have been captured. If refining the grid captures new physics, such as a separation point, then the answer will be very much different.

I also agree with an earlier post in the sense that it is my opinion that the residuals should be driven to machine zero.

In general, it takes a very fine grid to believe grid convergence values.

http://www.hegedusaero.com/examples/.../Vassberg.html
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