Momentum Exchange Algorithm

Hi all!

I am actually working with a BGK 2DQ9 model of LBM, a channel with a square obstacle.

I have calculated lift and drag and Strouhal by means of the stress tensor integration, and the results seem to be somehow accurate. However, the peak-to-peak lift coefficient doesn’t resemble the bibliography.

I am trying to solve the problem, but I find it impossible. So I want to try the called Momentum Exchange Method to contrast the data.

I would appreciate If someone can facilitate me a FORTRAN routine or a complete code of a similar case where the forces on the surface of the obstacle are evaluated.

In advance, Thanks a lot!!

Send me your code and i have done similiar what you have done , will go through it or you may explain what you have done if you don’t want to send your code.

respected Sir,
I am also facing same problem in evaluating FX and Fy by momentum exchange method. I am using C++ for coding …
my code for F is…what Improvenment should i do for getting correct values of Fx and Fy for flow past a square obstacle… pls reply sooon…

Thanks and Regards

// calculation of force:
for (j=ys; j<=yn; j++)
for (i=xw; i<=xe; i++)



Are you using half-way BB or full way BB?

Are you considering the evaliation of all square forces or only the boundary wall forces?

I actually do not know to work with C++ yet.

Albert P

I am using full way bb and and summation of forces on all over square surface…
sir, u can tell me in fortran…

Thanks and regards,

Actually krunal,

I was using full way bounceback and the drag and strouhal was ok. But lift had more than 10% error. Always 10% lower than references of Breuer et al. 2000.

Since I have changed the algorythm of BB to half-way I have solved that problem.

you should look at THIS PAGE

there are good resources and they solve analitically a MEM with drag and lift, but with half way BB

I am not sure why the problem occurs with full way, I probably lack of some concepts of error propagation or extrapolation.

Anyway, I wish you luck.

[size=medium]Thank you Pujgar sir …
Sir one more thing i wann ask you…
In calculation of Cd from D=15 to D=19, and grid size "50D8D" i am getting appropriate value of Cd…but as value of D is increased from 19 to 20, value of Cd is equal to 0.003452 for Re=10.0… I am using reference value of U=0.2, D=20.0 and alpha=(UD/Re)…
what i am doing wrong…so that i am not getting grid independence…

Thanks and regards,[/size]

Try working with a grid test that fits perfectly with your model. If you increase you D by one (from 19 to 20 e.g.) it would probably dont the same exact dimensions. maybe it hassomething to do with odds and evens in your D that affect to your total flow development. but this phenomena occurs particularly with the lift because of the x-axismetric modification if you work with an horizontal channel.

Try to do your grid test according to S. Izquierdo 2009 references, they just multiply their diameter with an integer number of obstacle each time.

I hope this might help you.