Quote:
|
I wonder if you can calculate at what M/N so that the mixture is 100% self-supporting for complete combustion?
|
I don't think you can do that.
There will always be a need for addtional oxygen to burn both Nitromethane and methanol. They just need much less oxygen than gasoline for complete combustion.
Net result is you can burn much more fuel for a given amount of combustion air.
Nitromethane Gram molar weight = 61.0406
Oxygen content by mass = 52.4221584%
Assuming the following reaction is correct;
1 ( CH3-NO2 ) + 2(O2) = HNO3 + CO2 + H2O
Then you need 2x 31.9988 grams oxygen = about 63.9976 grams oxygen to burn every 61.0406 grams on nitro methane, instead of the approx 3.381 grams of oxygen needed to completely burn 1 gram of gasoline.
The stoich mixture for Nitro then is about 1.04844 :1 oxygen to nitro.
Since Air is 23.16% oxygen by weight, then the fuel air ratio for complete combustion of Nitro would be about 4.52696:1. This means Nitro only needs .3079 x the oxygen as gasoline to burn completely.
Larry
In rough terms for each gram of gasoline fuel you replace with Nitro, you free up about 2.3 grams of oxygen that is still available to burn something else.