Hi all i am simulating the combustion of CO and dry air using the finite-rate-model. In FLUENT the reaction rate expression is used with water steam as follows:

k=2.239E+12exp(-1.7E+8/T)(CO)(O2)^0.25(H2O)^0.5

but í´d like to get the expression WITHOUT WATER STEAM. please give me a hand.

thanks in advance.

use the DEFINE_VR_RATE MACRO to program your reaction rate into each cell in the domain.

thanks sivakumar,i´d like to know the reaction rate expression without water stream, not how to implemente the expression in the code.

This is the standard global form for this equation. The dependence on steam is due to reaction of radicals such as OH if a multi-step scheme is used (See Westbrook & Dryer, Comb Sci Tech., vol 27, 1981).

If your system doesn't have steam you might try leaving off the steam dependence. I haven't seen a form like this in the literature though. Perhaps others can comment on this.

Physically though CO and O2 react almost instantaneously at high temps (1500-3000K). Depending upon your setup, you might find Fluent has difficulty with these reactions using only the finite rate model. I've had to use some numerical limiting using a udf to get solutions and then test the sensitivity of these limiters with respect to the solution, for laminar diffusion flames, since Fluent can't handle the stiff kinetics. I believe other software (CFD-ACE?) can handle this better than Fluent.

Have you managed to get any converged solution using the rates above. What solver and settings have you tried?

Greg

According to Westbrook & Dryer

Hi Greg Perkins, thank you for your comments. Yes, only using the finite reate model in Fluent is very difficulty to get a reasonalbe solution. I hve used the eddy-dissipation/finite-rate model with segrated solver. The converged solution i can get. The simulated Temperature, CO2 and O2 content are very good agreement with the experimental results. But the calculated CO content is 5 to 10 times higher or lower than the experiment data.

Is your problem turbulent or laminar?

Greg

Hi,the gas flow in my system is turbulent.