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-   -   Setting Initial Site Coverage (http://www.cfd-online.com/Forums/fluent/99500-setting-initial-site-coverage.html)

villager April 4, 2012 18:10

Setting Initial Site Coverage
 
Good day, dear all!
I study catalytic conversion of CO, NO & C3H6 on Pt/Rh, using the mechanism from detchem.com.
There's no convergence with C3H6-reactions on Pt after 100 iterations, N2 conc. falls to zero. Only reduced mech. seems to be converging.
But, there's another question. In mixture properties -> reaction mechanism I specify two surface sites: Pt and Rh with site density 2.04e-08 and 0.68e-08 respectively. I must specify Initial Site Coverage (Define.. button). When I set them to Pt/1 on Pt and Rh/1 on Rh, I have quick convergent solution - if no other species are mentioned in these dialog box. Solution is symmetric in a cylindric pipe. When I specify them, for example, with all surface species with equal (or unequal) fractions, I have asimmetric solution in a pipe, that converges slowly, and has questionable reliability.
How should I set Initial Site Coverage? Is there any guide rules?
And does it mean I turn off all surface species production, if no species except Pt and Rh were choosen in Initial Site Coverage dialog box?
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Thanks, Villager.

villager April 16, 2012 16:20

My thoughts
1) Initial Coverages are set by the known data - for ex., if I know, that catalyst is mostly coated with CO - then set CO fraction the largest. It's only initial guess. Real equilibrium (if it is) coverages are rapidly calculated by FLUENT.
2)
Quote:

And does it mean I turn off all surface species production, if no species except Pt and Rh were choosen in Initial Site Coverage dialog box?
Yes, it does. I've checked it.

villager October 16, 2013 11:01

Quote:

Originally Posted by villager (Post 354989)
2) Yes, it does. I've checked it.

No, certainly, it doesn't (the mechanism mentioned above works quite unstable in fluent, that was a case)
It's only initial guess for solution of the problem of finding a steady-state for surface coverage. If it's only one steady-state present, the problem will converge to that steady-state from different initial guesses.


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