rUA inside/outside PISO loop
 

Hi,

I have a question regarding the position of the computation of rUA in les codes. In do not understand if it is important to have it take place inside or outside the PISO loop, and to what extent this affect the solver.

In channeloodles:

In oodles:

My hypothesis is that, since UEqn uses phi from last timestep to create its coefficients, rUA is never modified inside the PISO loop (the new phi computed here is never used to modify the UEqn), and can be calculated outside as well.

1. Is this correct ?
2. If yes, why isn't rUA calculated once and for all outside the PISO loop, in oodles too ?

Did you now an answer up to now? I am also wondering why rUa is one time included in the PISO loop and one time not. I stored UEqn.A() in a tmp variable outside the PISO loop and compared UEqn.A() with my tmp variable every PISO loop step. There was no change. The difference was zero. So I think its equal to write it inside or outside of PISO loop :-D

kind regards,
Fabian

Hi,

You might find this note useful: http://www.tfd.chalmers.se/~hani/kur...7/rhiechow.pdf

That site is well worth looking at in general -- there's a lot of useful slides and student papers on Op-nFOAM.

/Ola

Hello,

yes I know this "paper" but in my opinion it is more usefull to read the book of ferziger and peric "Computational Methods for Fluid Dynamics". They describe the PISO algorithm much more in detail and its fundamentals. But the main problem is to find the different equations in the source code. The paper is nice but I missed a little bit the link between theory and source code :-D

Fabian

fs82: I just moved on other problems :), still have no answers on that one...

As I mentioned above, there is no change in the UEqn.A() Operator during the PISO loop. I think is correct, because you only evaluate the correction terms in a PISO approach. So in my opinion its equal to write it inside or outside of the loop.