# Combustion calculations and the required amount of air

• moonunits
In summary: I don't know, it just seems like a lot of trouble for something that might not be necessary in the end. After reading more about excess oxygen combustion, I realize that this may not be the case for all gases. For instance, methane does not combust in excess oxygen, so I would have to calculate the air requirement for methane using the \lambda values of 1.1 and 1.2, which would be different than the air requirement for CO2 or N2. Do you have any suggestions on how to approach this?
moonunits
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-%):

Code:
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):

Code:
	       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).

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?

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).

## 1. What is combustion and why is it important?

Combustion is a chemical reaction that occurs when a fuel combines with oxygen to produce heat and light energy. It is important because it is the process used in many industries to generate heat and power, and is also the main source of energy for transportation and heating in homes.

## 2. What are the factors that affect the amount of air required for combustion?

The amount of air required for combustion depends on the type of fuel being burned, the composition of the fuel, and the desired temperature of the combustion reaction. Other factors such as humidity, altitude, and air pressure can also influence the amount of air needed.

## 3. How do you calculate the required amount of air for combustion?

The required amount of air for combustion can be calculated using the stoichiometric equation, which takes into account the chemical reaction between the fuel and oxygen. This equation can be solved using the molar mass and the amount of fuel being burned.

## 4. What is the ideal ratio of fuel to air for complete combustion?

The ideal ratio of fuel to air for complete combustion is known as the stoichiometric ratio. It is the exact amount of oxygen needed to completely burn a fuel without any leftover fuel or oxygen. This ratio varies depending on the type of fuel, but it is typically around 14.7 parts of air to 1 part of fuel by mass.

## 5. How does the presence of excess air affect combustion calculations?

The presence of excess air (more than the stoichiometric ratio) can increase the efficiency of combustion and reduce the emissions of pollutants. However, it can also result in wasted energy and increased costs. When calculating the required amount of air, it is important to consider the optimum ratio for the specific application to achieve the desired results.

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