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Equivalence ratio in partially premixed flame system

  1. Jun 4, 2015 #1
    Hi, I'm learning the basics of plasma assisted combustion so I've been reading up on a few articles related to it.
    My question is regarding this article http://enu.kz/repository/2011/AIAA-2011-971.pdf
    1-s2.0-S0010218011002203-gr1.jpg
    Using a setup consisting of a pair of counterflow burners, the composition of the oxidizer stream was fixed at O2/Ar/He/CH4(0.26:0.32:0.4:0.02) while the fuel stream was CH4 diluted by Ar (fuel mole fraction varied from 0.2 to 0.4). In this experiment, both oxidation and fuel reforming occurs. How do I find the equivalence ratio for both cases?

    Oxidation: CH4 + 2O2→ CO2 + 2H2O
    Fuel reforming: 2CH4 + O2→ 4H2 + 2CO

    The equivalence ratio is defined as (fuel-to-oxidizer ratio)/ (fuel-to-oxidizer ratio)st and usually methane will be the fuel and oxygen as oxidizer, but in this case the oxidizer itself is premixed with methane so how do I calculate the equivalence ratio? Please help. I get so confused.
     
  2. jcsd
  3. Jun 5, 2015 #2
    The equivalence ratio is different everywhere in your flame. In partially premixed cases, it is more useful to use the Takeno flame index as a measure for the combustion regime, or use the mixture fraction together with the equivalence ratio to characterize the flame.
     
  4. Jun 12, 2015 #3
    I see! I never thought about different parts having different equivalence ratios before. No wonder I couldn't get a constant value. It makes more sense now. Thank you so much!
     
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