Quote:
meth is 49% oxygen by weight
ethanol is 34% oxygen by weight.
gasoline is zero% oxygen
|
I am not an expert in combustion chemistry (though I am a chemist by education), I do not seem to remember a mechanism that accounts for non-molecular oxygen playing a direct role in combustion, and certainly not one that adds to the overall energy of the system.
you have to remember: a flame is an extremely complicated chemical system...
gas phase chemistry is never as simple and straight forward as you would expect from a cursory knowledge of the chemicals involved...
combustion involves reaction of O- radical anions (atomic oxygen, NOT molecular oxygen), R radicals (aliphatic/carbon radicals) H radicals, O2 radicals, R-OO radicals (peroxide radicals, including hydrogen peroxide)...
at the end of days, that OH in methanol and other alcohols DOESNT behave like oxygen (molecular)...
its function in the overall combustion mechanism is DIFFERENT, and thus its inclusion in the initial fuel has no impact on the combustion process itself.
in other words: the fact that methanol molecules include an atom of oxygen has absolutely no relationship with the amount of oxygen fuel (O2 from the air) that is used in the combustion process, nor does it affect the ACTUAL stoichiometry of a REAL flame (which is NOT the same thing as a general chemistry balanced equation of nCH4OH + mO2 -> xCO2 + yH2O).
the real stoichiometry of a real flame takes into account the mechanism of the chemical reaction, which the simple grade school thermodynamic representation DOES NOT.