Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook