How does that work? Water will dissociate into hydrogen and oxygen at very high temperatures, but the mixture of hydrogen and oxygen so produced is already stoichiometrically balanced. There's no surplus oxygen left to be reacted with additional fuel.
The confusion was on my part, I used release of oxygen but it is probably better explained as intake of oxygen, the cooling effect of the water leads to a higher air density in the mix of atomized air. The water is not releasing more oxygen rather there is more air in the cooler denser mixture of air, water and methanol, the water vaporizes thus creating more volume and then the methanol flashes and consumes the additional oxygen that is brought in via a cooler intake, it is my understanding that this oxygen is only available after the flashing of atomized water into gas, thus I used the term release (which is a poor choice of words for the process). It works similar to an inter-cooler, but unlike an intercooler the mix seems to preserve some oxygen for the methanol combustion cycle so it is more targeted then just forcing more oxygen into the intake via a denser volume of air (e.g intercooling). Sorry for the confusion on what is actually happening, you are correct it is not separating hydrogen atoms from oxygen atoms. If that where the case then as you said, there would be no need for the methanol as it would be generating the additional fuel via hydrogen and oxygen in the generated browns gas to burn it.
