# Combustion calculations and the required amount of air

by moonunits
Tags: air, combustion, natural gas, stoichiometric
 P: 2 I'm working as an intern at a factory that produces refractory bricks, mainly doing measurements on a tunnel kiln they use for firing the bricks. The bricks are heated with several natural gas burners in a firing zone. To determine the required air for both stoichiometric and excess-air burning ($\lambda$=1,1 and 1,2), I've done some basic combustion calculations we used to do back at the University. However, I'd like to check whether I've made correct assumptions and/or if I've simplified the problem too much. The composition of the used natural gas is roughly the following (with mole-%): Methane CH4 89,51 Ethane C2H6 5,8 Propane C3H8 2,25 Butane i-C4H10 & n-C4H10 0,9 Pentane i-C5H12 & n-C5H12 0,21 Hexane C6H14 0,06 Carbon dioxide CO2 0,85 Nitrogen N2 0,42 and hence $M_{NG} = \Sigma x_{i} M_{i}$ = 17,822 g/mole I've assumed that the reactants combust completely and that both CO2 and N2 do not react. I've also assumed the following combustion reactions (is this oversimplifying things?): 1) CH4 + 2 O2 -> CO2 + 2 H2O 2) 2 C2H6 + 7 O2 -> 4 CO2 + 6 H2O 3) C3H8 + 5 O2 -> 3 CO2 + 4 H2O 4) 2 C4H10 + 13 O2 -> 10 H2O + 8 CO2 5) C5H12 + 8 O2 -> 5 CO2 + 6 H2O 6) 2 C6H14 + 19 O2 -> 14 H2O + 12 CO2 While considering, for instance, combusting 1 mole NG I've then calculated (with the reactions above) the required amount of O2 (for methane $n_{O2}=2 * n_{CH4}$, ethane $n_{O2}=7/2 * n_{C2H6}$ etc):  moles O2 required (moles) CH4 0,8951 1,7902 C2H6 0,058 0,203 C3H8 0,0225 0,1125 C4H10 0,009 0,0585 C5H12 0,0021 0,0168 C6H14 0,0006 0,0057 CO2 0,0085 0 N2 0,0042 0 =============== 1 2,1867 N2 in the air = 3,77 * 2,1867 = 8,243859 moles => tot. required air for stoichiometric combustion = = (2,1867 + 8,243859) mole = 10,430559 mole(air)/mole(natural gas) So then the molar stoichiometric AF ratio would be 10,43 (with $\lambda$=1,1 it would be 11,47 and with $\lambda$=1,2 12,52). And then the stoich AF ratio using masses would be $\frac{ m_{air} }{ m_{NG} } = 10,43 * \frac{ M_{air} }{ M_{NG} }$ Now the question is: are these calculations correct or should I approach this in some completely different way? Am I simplifying things too much / am I not taking something essential into account (that might render these calculations useless)? If I measure the volumetric flow of natural gas to a burner (and convert it to massflow), can I then use $\dot{m}_{air} = 11,47 * \dot{m}_{NG} * \frac{ M_{air} }{ M_{NG} }$ to set the desired inlet airflow (for $\lambda$=1,2)? How well do the theoretical calculations work in practice? (and by the way, if someone has any tips regarding adjusting/checking the adjustments of gas burners, I'd love to discuss the subject more). Thanks in advance! :)
 Sci Advisor P: 1,395 Those calculations seem like a reasonable first order estimate, but I am not that familiar with the practical aspects of combustion. In particular, your balanced equations are really for combustion processes in excess oxygen, which is typically taken to mean a large excess (as in combustion analysis). I am not sure if your $\lambda$ values of 1.1 and 1.2 are enough to ensure that you are in that limit. The reason I bring that up is that the kinetics of combustion for each of those hydrocarbons are likely to be different, and so you may get complete combustion of some of them, but not others. How well you approach the limit of complete combustion for a given $\lambda$ in your system likely depends on the specific design of your burner, which will determine how well the fuel and oxygen are mixed, and how long they remain in the hot portion of the flame before exiting to the exhaust. I am not an engineer, so I don't really know how these issues are handled in practice. You might try asking over on one of the engineering forums if you don't get the answers you are looking for here. Anyway, I would probably shoot for trial and error, assuming that is possible/reasonable .. how are you planning to measure the extent of combustion? Do you know the losses in your system well enough to extrapolate the total heat released in the combustion process?
 P: 2 Thanks for your quick reply, SpectraCat. Yeah, that was exactly what I was wondering too. I recall our professor telling us, that by doing calculations assuming complete combustion and then adding the extra air with a $\lambda$ of 1,1-1,2 would (in most cases?) be sufficient for complete combustion "in real life". But calculating the theoretical minimum amount of air for complete combustion assuming there is a large excess of air, just to then afterwards add excess air seems a little dodgy.. (whoa, that could probably have been written a lot clearer, hope you understand what I'm going after). Any thoughts on this? So far I have only estimated air and natural gas flow rates by measuring the differential pressure over orifice assemblies prior to the burners. Apparently we've had some reducing atmosphere on one side of the kiln, and I'm trying to figure out if this could be caused by poor adjustment of a/some burner/burners (you manually set the (constant) air and gas flow rates). I suppose this would be easier with a flue gas analyzer, but unfortunately we don't have one (I think we're contemplating on getting one at some point though).