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-   -   High order scheme vs Specified Blend factor 1 (http://www.cfd-online.com/Forums/cfx/65588-high-order-scheme-vs-specified-blend-factor-1-a.html)

realanony87 June 19, 2009 06:58

High order scheme vs Specified Blend factor 1
 
I have read that the High order scheme in Ansys CFX v11 is more robust, but less accurate than using a specified blend factor of 1.
However in my simulations of flow around a 3D wing it seems that I have it the other way round
For example my largest max residual with the high order scheme is stuck at around 5e-4 for the momentum in the "lift" direction,
while with a specified blend factor of 1 all momentum residuals go below 1e-4 and with a faster convergence rate towards the end of the simulation ( 0.8 vs 0.9)

Could it be that the default high order scheme's adjustment of the blend factor be causing problems with convergence ?

Other details
My mesh is structured hex, double precision is turned on.
Minimum Orthogonality Angle 30.9
Maximum Aspect Ratio 219.5
Maximum Mesh Expansion Factor 15.3

heron1 November 10, 2015 11:36

High order scheme vs Specified Blend factor 1
 
I quote oj.bulmer


Quote:

Originally Posted by oj.bulmer (Post 418168)
Yes you do need to have a second (or higher order) solution in order to be able to use the blending factor. In very simplistic terms, blending factor B can be defined as:

\phi  =  \phi_{first-order} +  B*(\phi_{higher-order}-\phi_{first-order})

Here \phi is the quantity like velocity, pressure etc.

Now, if you have a look at the equation, if B=0, we have results of first order and if B=1, we have results of higher order. But if you don't have a higher order solution, then \phi_{higher-order} becomes \phi_{first-order}, and for any value of B, you get the solution the same as \phi_{first-order}.

Hence it is necessary to have a second or higher order solution in order to use blending factor.

It is healthy to go gradually from first order to different blending factors (0.4, 0.7) etc depending on your unstability and then switch to second order, if you are having problems with direct second order solutions.

OJ

It could depend by this part:

"But if you don't have a higher order solution, then \phi_{higher-order} becomes \phi_{first-order}, and for any value of B, you get the solution the same as \phi_{first-order}."


Regards


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