# High order scheme vs Specified Blend factor 1

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 June 19, 2009, 06:58 High order scheme vs Specified Blend factor 1 #1 New Member   Join Date: May 2009 Posts: 21 Rep Power: 17 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

November 10, 2015, 10:36
High order scheme vs Specified Blend factor 1
#2
New Member

R.Maleci
Join Date: Oct 2009
Location: Italy
Posts: 3
Rep Power: 16
I quote oj.bulmer

Quote:
 Originally Posted by oj.bulmer 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: Here 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 becomes , and for any value of B, you get the solution the same as . 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 becomes , and for any value of B, you get the solution the same as ."

Regards