Recent content by agm

For this example, we have ##\frac{F}{O} = \frac{1}{4}##, assume we're using ##\ce{CH4}##. If we need to find the amount of air that corresponds to ##4## units of oxygen, we perform the following based on gas composition of air $$ (4)_{\ce{O2}} \times \frac{(100)_{\text{Air}}}{(21)_{\ce{O2}}}...

So I believe I've managed to find a solution to my problem. There's some nuance we dig through before agreeing on what "equivalence ratio" means. It seems like air is a super-common oxidizer in the combustion-industry, so it makes sense for some people/researchers to use an equivalence ratio...

Okay so I think I see what the crux of my understanding. There seem to be different ways to represent equivalence ratio, it seems like one way is a definition of Fuel to Oxidizer (the oxidizer part in air is oxygen, so out of the total volumetric flow of air, I have to extract the oxygen part...

So apparently you can also find the equivalence ratio on a volumetric flow rate basis - paper This should work out well, as I'd like to know what the equivalence ratio is for the flowing gasses in a methane + air flame. If I were to add an inert gas into the flow like Nitrogen, how can I factor...

So I was mostly referencing this Section 3.2 book summary on Science Direct So it would not be correct to say the Fuel to AIR ratio is 1/2. Instead the Fuel to Oxidizer (oxygen) is 1/2?

I'd like to do some experiments with flames at different Equivalence Ratios - but I'm confused as to how I can find the Equivalence Ratio for different conditions. Wiki-article So the stoichiometric balance for Methane + Air is CH4 + 2(O2 + 3.76N2) -> CO2 + 2H2O + 7.52N2 Referencing equations...