I'm creating a world with different oxygen density, about 15%. When given same fuel and air pressure, how will the fire burn up in this 15% oxygen density atmosphere compared to our world? And the world atmosphere is composed with 0.2% of Carbon Dioxide, and 1% of Methane. (Nitrogen? Same.)
1$\begingroup$ Look here: en.wikipedia.org/wiki/Limiting_oxygen_concentration#/media/… $\endgroup$– Alexander von WernherrFeb 17, 2017 at 12:12
$\begingroup$ @AlexandervonWernherr Well, that was one nice chart, although it was bit hard to understand. $\endgroup$– Seori LeeFeb 17, 2017 at 12:14
$\begingroup$ Any open flame will burn the methane too. $\endgroup$– JDługoszFeb 17, 2017 at 12:27
1$\begingroup$ It should be noted that without a methane source, that 1% of methane would degrade to CO2 and H2O rather quickly. $\endgroup$– rangerike1363Feb 17, 2017 at 14:12
Considering that oxygen is still present in the atmosphere, the fuel will definitely ignite. However, due to 35% lesser oxygen, the combustion reaction of fuel and oxygen will be 35% slower.
That is to say, that the flame will be 35% smaller in that environment, than the flame produced on Earth now. However, you also have 1% methane in the environment and the flame will burn atmospheric methane too. It is difficult to say exactly how much positive effect it will have on the flame, so let us do some math here.
CH$_4$ (16 grams) + 2O$_2$ (64 grams) ---------> CO$_2$ (44 grams) + 2H$_2$O (36 grams)
The reaction posted above (with given quantity of methane and oxygen) produces 810 KJ of energy. Also note that every one molecule of methane requires 2 molecules of oxygen to burn. However, since methane is present as only 1% of the atmosphere, it will use up only 2% of the oxygen which would be otherwise available for burning the fuel.
It depends on the type of fuel to decide if the fuel combustion process will take priority over methane combustion or methane combustion will take priority. For most types of common fuels in use today (gasoline, diesel, kerosine etc), methane combustion takes priority.
So, all in all we could say that the flame produced will be 65% (fuel burning part) + 2% (methane burning part) = 67% of the flame we have in our atmosphere.
$\begingroup$ density is mass ÷ volume. density of 15% is meaningless. The oxygen concentration of air is 21%. 78% N2 + 15% O2 + 1% CH4 = 94%. It must add up to 100%. Reducing oxygen % will slightly reduce heat given off by burning at the same time lengthen the time of burning and also increase the amount of residue, especially materials which could burn off in a more intense fire. In most fires, oxygen is not the limiting ingredient. All combustible materials will be harder to ignite, this would be especially noticeable with wet and humid materials (wood, paper, cloth). $\endgroup$– Li ZhiFeb 17, 2017 at 21:25