one-step reaction mechanism of hydrogen/methane

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 July 30, 2004, 01:14 one-step reaction mechanism of hydrogen/methane #1 Sheng Chen Guest   Posts: n/a Sponsored Links Hi,everyone, I am now simulating premixed laminar flame. The fuel is hydrogen or methane and the oxidizer is air. But I can not find their one-step reaction mechanism. Could you tell me? Thanks a lot!

 July 30, 2004, 02:28 Re: one-step reaction mechanism of hydrogen/methan #2 matej Guest   Posts: n/a the best reference would be: Westbrook Ch., Dryer F.L., Chemical Kinetic modeling of Hydrocarbon Combustion, Prog.Energy Sci.,1984,vol.10, pp 1-57 for the immediate kick-start, here is what he says: for CH4 + 2O2 -> CO2 + 2H2O and rate= A*T^n *exp(-Ea/RT)* [CH4]^a * [O2]^b the values are: A=1.3e9 [s^-1]; Ea= 48.4 [kcal/mol];a=-0.3;b=1.3 for eq. ratio from 0.5 to 1.6. you may find the values for A warying at some references and sources. do not forget the limitation of this single step chemistry! matej

 July 30, 2004, 23:38 Re: one-step reaction mechanism of hydrogen/methan #3 Sheng Chen Guest   Posts: n/a Hi, matej, Firstly thanks for your message. There is a question about the single step reaction that what does n mean in the formula.

 August 2, 2004, 16:47 Re: one-step reaction mechanism of hydrogen/methan #4 Dean Guest   Posts: n/a As I recall, n=0 for the Westbrook-Dryer rates, but check the paper to be sure. Also, be aware that these rates can produce flame simulations that are not very accurate. Just look at the flame speed vs equivalence ratio plots in the paper. Also, post-flame temperatures come out high if you don't account for the fact that the products contain dissociation products of water and carbon dioxide, such as CO and OH. There is also a problem with the negative exponent on the fuel concentration under lean conditions, the rate goes to infinity as the fuel is depleted. This is not physical. User beware!

 August 3, 2004, 03:26 Re: one-step reaction mechanism of hydrogen/methan #5 matej Guest   Posts: n/a Dean is right. n=0 for wesbrook. What you can test with is pre-exponential factor A. YOu can fix with this the laminar flame speed. You should definetelly test it with your code to compare the flame speed with the literature, preferebly cited westbrook article. The solution will be sensitive to the discretisation scheme you use. matej

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