Dear Swapnil,

Usually mixture fraction is used for non-premixed flame, and stchiometric mixture fraction is the location of maximum flame temperature. The composition mentioned by you in on mass/molar basis not clear, but in both cases the it is a rich premixed(partially premixed) flame. Can you send the link so that I can have a closer look

Thank you for replying.

It is 25% CH4, 75% dry air by volume.

This is the link:

https://www.sandia.gov/TNF/DataArch/FlameD.html

Hello there,

stoichiometry is something which is case independent, just a simple way to say that at this given mixture, the fuel and oxidizer are fully consumed.

For methane this would lead to:

CH4 + 2O2 -> CO2 + 2H2O (with pure oxygen as oxidizer).

Using dry air:

CH4 + 2/0.23 (0.23O2 + 0.77N2) -> CO2 + 2H2O + 2*0.77/0.23 N2

From that you can see that at stoichiometry the volumetric ratio of CH4 to dry air is around 8.7 (1 mol CH4, 8.7 mol air). With that knowledge, you can calculate the stoichiometric mixture fraction value following eq. 1.46 in https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=2ahUKEwjAj7a-l7XnAhURDuwKHepDBAgQFjAAegQIBhAC&url=https%3A%2F%2Fwww.itv.rwth-aachen.de%2Ffileadmin%2FDownloads%2FSummerschools%2FPeters_Summerschool_reference.pdf&usg=AOvVaw1vxdyeZVdn-eKb2M2iVZrj .

From the equivalence ratio definition, the value 0.77 given for the pilot is telling you that it's a lean mixture, since values smaller than one correspond to mixtures below stoichiometry. Besides, phi = 0.77 corresponds to a Bilger mixture fraction of around 0.43 (see eq. 1.55 in the linked document).

The main jet is composed of 25 vol% CH4 and 75 vol% of dry air. With that information, and the respective molecular weights, you can calculate the mixture fraction and or convert it to an equivalence ratio if needed (again, using eq. 1.55, I did obtain a mixture fraction of ~ 0.156 or phi ~ 3.175).

Good luck!