How I calculate the molar concentration of [O2], [N2] and [NO]. As formula is concentration=(% increase in volume% ppm * molar wt)/(10^6 * Volume@T,P). My question is that is the increase in volume after the combustion would be applied equally to O2 and N2 and what is increase in Volume? and also is the standard volume would be applied i.e 22.41 litre?
The ideal gas law indicates the overall molar volume as in
n/V =P/(RT) (moles/unit volume = absolute pressure/(gas constant times absolute temperature)
The molar concentration (moles of species/unit volume) of each gas species is the mole fraction times this number.
@Larry L. Baxter As mole fraction of oxygen is nearly 0.19 and nitrogen is 0.71 in Air-fuel ratio concentration. My question is that during NOx mechanism nitrogen is constant while oxygen mole fraction decreases. So, in mole fraction what concentration of oxygen should be used. Is take the avg. of mole fraction before the combustion and after the combustion or take the mole fraction which is before the combustion? and also I want to know that as Temperature and Pressure in the engine varies. Do I upload the Temperature and pressure data in the above formula which is varies with crank angle?
You should use the mole fractions that exist at each stage of the reaction. That is, the O2 mole fraction should start at 0.19 and decrease as you step through the reaction.
If you are trying to use these results in some sort of NOx prediction scheme, you should be aware that the NOx concentration is very sensitive to details of the mixing and turbulent chemical kinetics and very difficult to predict accurately.
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