Why does the peak adiabatic temperature usually happen for an equivalence ratio a bit higher than 1? I think it had something to do with the probability of each fuel molecule finding a corresponding oxidizer, and the fact that the molecular weight of the fuel is generally higher than that of the oxidizer.

  • $\begingroup$ Just a guess, but is it because nitrogen in the air is inert, and you have to "waste" a bit of the heat from combustion to heat it up? It's a small effect, relative to the heat of combustion, so my guess is the the max temperature would be ever-so-slightly to the carbon-rich side of the equivalence ratio. $\endgroup$
    – Curt F.
    Jul 10, 2015 at 6:05
  • $\begingroup$ > [...] $\phi = 1.1$ was chosen because the flame temperature for many fuel peaks not at the stoichiometric value but between $\phi = 1.0$ and $\phi = 1.1$ owing to lower mean specific heats of the richer products. The maximum temperature for acetylene-air peaks, for example, at a value of $\phi = 1.3$.$^{[1]}$ ## References 1. Irwin Glassman, Richard A. Yetter, Combustion, 4th ed; Academic Press: New York, 2008 $\endgroup$
    – Jo Elsakka
    Apr 13, 2021 at 13:07

1 Answer 1


I actually found the answer in this 2011 paper, here is a quote from the abstract:

The characteristic rich shifting of the maximum adiabatic flame temperature from the stoichiometric value for hydrocarbon/air mixtures is demonstrated to be caused by product dissociation and hence reduced amount of heat release. Since the extent of dissociation is greater on the lean side as a result of the stoichiometry of dissociated products, the peaking occurs on the rich side.


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