I am working on a 2DQ9 fortran code, where I am trying to apply de MEM to get Fx and Fy of the square block.
I am setting a Poiseuille flow channel on inlet with equilibrium conditions in inlet and outlet.
the bounceback is placed after the stream routine, so this means it is half way bounce back isn’t it?
anyway, my BB routine by now does invert all the fs and nothing more.
The hard part goes when I try to understand the MEM equation and to apply it in the code. I am not fully understanding any of the papers I have reviewed…
one example of the clearest paper is from wen et all (PHYSICAL REVIEW E 85, 016704 (2012)
does the equation looks like this?:
loop all the square boundary nodes
do all “i” directions 1 to 8 (0 is center); ONLY the “i” directions that go from solid to liquid and viceversa
F_x = F_x + [(fi(y,x)*e(i,x)) + fî(y+e(i,y),x+e(i,x))*e(î,x)] -> drag
F_y = F_y + [(fi(y,x)*e(i,y)) + fî(y+e(i,y),x+e(i,y))*e(î,y)] -> lift
Is it correct? What it pretends to be is to sum up the momentum of all the outgoing particles of the solid plus the particle momentum of all the ingoing particles from the fluid to solid.
I look forward to hearing an answer, if you need any information just say it. I have been too much time with this
The other question that I have is, where does this method takes place exactly? before/between collision and streaming, after collision or after streaming? Before or after computing Fs and Feqs…?? Should it be the last routine of my main loop??
I have actually solved the problem of force evaluation.
I am not sure 100%. But I think that the source of the problem was that I was using a full way bounce back. I changed the routine to a half way bounceback and now the forces are well evaluated.
I was wondering how can a Momentum Exchange Method be implemented whether the bounce-back is full way.
Analyzing the state. A MEM is really simple, as you just sum up all the fluid to solid fs, and you sum up also all their opposite solid to fluid fs. Considering that the wall is exactly placed at the middle of the two nodes. It ends up to be two times the sum of all fs that go from fluid to solid.
How is it in fullway bounce back? Can I apply the same method or I’ll have the change it a bit? my intuition tells me that it is not as simple…
is there any MEM that uses the full way bounce back?