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issue in calculation of velocity u and v #5
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Hi, it happens that the codes included in both these chapters (as well as the worked out examples provided in the book for these chapters) adopt the incompressible LBM model. The reason is that complexity is decreased this way, as we can get rid of compressibility errors, which is possible since we are tackling steady-state problems. Now, if you want to re-write the codes in order to account for the standard (compressible) model then only including rho(x,y) in the momentum velocity moments is not sufficient. Besides this step, you also need to re-write the equilibrium f^(eq) accordingly (and on top of that, possibly it is also necessary to re-write the BCs and/or the force term, depending on the case you are considering). Please, take a more attentive look at the book and the codes where these elements are explained. Good luck. |
Thanks for replying so quickly. First I would like to point it out that I
am a student and my intervent's purpose was to ask to understand and not to
point it out an error. Then if i am not mistaken it is used not the
incompressible model, but the linear model. Question: the reason why there
is no rho in the denominator (both in just linear and incompressibIe model)
is beacuse the second moments presented in formulas 4.39 and 4.43 has
rho_0 instead of rho and in the codes rho_0 is considered to be 1? If not
how it is?
Also i would like to point it out i had already tried to modify the codes
using the incompressible model (pag 123) instead of linear one, modifying
f_eq, BCs and forces as you suggested in your reply and Ithink it worked
fine. Thanks for your time.
Best regards, Bianchi Domenico.
Il giorno gio 13 giu 2019 alle ore 20:47 GoncaloNunoSilva <
[email protected]> ha scritto:
… Hi, it happens that the codes included in both those chapters (as the
worked out examples provided in the book) adopt the incompressible LBM
model so that, because only steady-state solutions are considered, we can
get rid of compressibility errors.
If you want to re-write the codes to account for the standard
(compressible) model then only including rho(x,y) in the momentum velocity
moments is not sufficient. Besides this step, you also need to re-write the
equilibrium f^(eq) accordingly (and on top of that, also possibly the BCs
and/or the force term, depending on the case you are considering).
Please take a more attentive look at the book and the codes where these
elements are explained.
Good luck.
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Ah OK, I understand your question now. |
Mr Silva, i wrote you a message on ResearchGate. Please give it a look when you'll have some time to spare. |
Hi, in all codes from chapter 5 and 6 the velocity components u and v are calculated with c_i * f_i (x,t) (summation implied) while according to formula 3.1 the result should be divided by rho (x,t). Trying to add it the code doesn't give the correct result anymore.
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