Let the stoichiometric volume of oxygen consumed by burning the unit volume of combustible (*), at some conveniently defined temperature and pressure, be denoted by VO2. The corresponding stoichiometric volume of (dry) air is: Vair = VO2 / 0.21, and that of N2 would be: VN2 = VO2·(0.79 / 0.21). For a typical combustible, the stoichiometric volume of combustion exhaust gas (stack gas) should be: VS ≈ VN2 + VCO2 + VH2O, while neglecting other possible contributions; such as soot particles, volatile inerts, nitrogen oxides, or sulphur oxides. Let the excess air (%) be defined as: X = [(Vair* - Vair)/Vair]·100%, were Vair* stands for the actual volume of air admitted to combustion. The corresponding volume of stack gas, would be: VS* = X·Vair + VS = X· VO2 / 0.21 + VO2·(0.79 / 0.21) + VCO2 + VH2O = (X + 0.79)·(VO2 / 0.21) + VCO2 + VH2O, while accepting complete combustion (hence neglecting CO emission). Here; VO2, VCO2, and VH2O, are combustible-dependent variables, defined for the stoichiometric ratio between air and combustible. Furthermore, VO2 was defined for admitted air, while VCO2 and VH2O are for exhaust gas. 

While considering the inverse problem, we may want to rewrite last equation to yield the excess air in terms of the volume of stack gas: X = [21·(VS* - VCO2 - VH2O - (79 / 21)·VO2) / VO2]%.

(*) or per unit mass of combustible, when the mass basis is found more convenient, as is often the case with liquid or solid combustibles.  

Thank you for your anwser.

There is a quick and dirty approach to estimate the actual air excess ratio, based on the measured oxygen content in the exhaust gas, which works quite well:

Air Ratio = actual air volume (or mass) flow / min. required air flow

                = 21 % / (21 %-O2meas),

where O2meas is the measured dry (!) oxygen concentration in the exhaust gas.

It is important to keep in mind in these discussions that measured concentrations are usually given as dry values, i.e. with the water removed from the flue gas. This is done because many measurement technologies have problems with moisture, therefore the flue gas sample is cooled down below the boiling point of water before it is fed into the flue gas analyzer.

Thus, if you calculate a stoichiometric methane combustion with air, the actual (wet) concentrations of CO2 and H2O in the flue gas will be 9.5 vol.-% CO2 and 19 vol.-% H2O, while the dry value for the CO2 concentration will be 11.73 vol.-%. The latter value is the one you need to compare with measured data. The same goes for excess oxygen in the flue gas or any other species you might want to measure.

Hope this helps.

Thank you for your answer.

Actually, since I now have two approaches to estimate the actual air excess ratio (one with O2 and CO2 contents in flue gas on dry and the other only with O2 content on dry), I am wondering which one of those is accurate and more reliable ! 

Just to remind the two relationships that I have for the moment (with your help too) : 

• Air Ratio = (79*O2)/(21*(1-CO2)-O2)
• Air Ratio (2) = 21 / (21 -O2)

where : O2 and CO2 are the measurement at the flue gas (exhaust gas : at the outlet of the furnace or boiler )

Thank you in advance ! 

The easiest way is to simply test it:

if you assume to have pure methane as fuel and an air ratio of 1.2, then an 1-step reaction equation gives that the dry CO2 concentration in the flue gas is 9.59 vol.-%, and the dry oxygen concentration 3.84 vol.-%. if you use these numbers for the 2 equations, you can calculate:

1) 1+0.79*0.0384/(0.21*(1-0.0959)-0.0384)=1+0.030336/0.151461=1.2002

2) 0.21/(0.21-0.0384)=1.22

If you do this calculation for other fuels or air ratios, you will get similar results.

The first equation is more accurate, but you only need one measured value for the 2nd equation and are less likely to make a mistake... as I said before... a quick and dirty approach. In this case, it's off from the true value by about 1 %. As both equations are derived directly from the 1-step equation, the reliability is the same.

In many larger applications, the excess oxygen is routinely monitored for control and safety purposes. CO2 measurements are not that common, at least in industrial boilers.

Personally, I think that the second equation is more than sufficient for most real-life applications, and much simpler.

Ok, I will take this into consideration, take you so much for your anwser !

For a related case study; see:

https://www.researchgate.net/post/200_mol_of_ethane_are_burned_in_a_furnace_with_50_excess_air_A_conversion_of_95_is_achieved_Calculate_the_composition_of_the_stack_gases