LPG to Oxygen ratio for combustion

[saldanha]

Hi all
what would be the Oxygen to LPG ratio for complet combustion of LPG. Our Oxygen is 95% pure.
what will happen if combution at high temperature (i.e.800 deg.C) is not complet. would cracking take over?

Thanks

 

[josephn16]

I will try to give a partial answer, because I am not sure if LPG means low pressure gas, or something different. If you are using natural gas, the ratio should be around 2:1, O2 to gas, but I am only basing that on the fact that the ratio is 10:1 will standard air, and there is around 5x the O2 in pure O2. That will not be exactly correct, because the N in air absorbs heat of combustion and lowers the flame temp. Someone else may know more about this than I do, you might want to try the Combustion Engineering forum if you haven't already.

But, I do know that the result of incomplete combustion depends on why the combustion is incomplete (i.e. xs air or xs gas.) Cracking could occur with xs gas (or poor mixing, but let's not get into that) but it would have to be quite a bit of xs gas.

 

[Montemayor]

saldanha:

As one ChemE to another, please employ units to identify specifically what "ratio" you are referring to. There are mole/mole, lb/lb, ft3/mole, etc., etc. ad infinitum ratios. Otherwise, you'll get a multitude of different responses - and each may be right(but without units!)

Go to "Liquefied Petroleum Gases"; Williams & Lom; Ellis Horwood Ltd.; Sussex, England. On Page 64 you will find:

m3 air req'd per m3 gas for combustion = 24 for commercial propane & 30 for commercial butane;

Ignition temp.(oC) in air = 450 for commercial propane & 420 for commercial butane;

Flame temperature (oC) = 1970 for commercial propane & 1975 for commercial butane.

LPG (Liquefied Petroleum Gas) is approximately 50% propane + 50% butane mixture. If you do an elementary stoichiometry combustion calculation you will easily get the theoretical amount of pure Oxygen required for complete combustion and you should compare that with the above figure (corrected for Oxygen).

Why you would not complete the combustion is anybody's guess. But the results would include CO and yes, some "cracking" down to CH4 and possibly C2H2 & others.


Art Montemayor
Spring, TX

 

[mbeychok]

I assume that you are asking about Liquified Petroleum Gas (LPG). If so, I would point out that LPG could be pure propane, pure butane, or mixtures of propane and butane which may or may not contain isobutane or even some butylenes.

You should first determine the actual component analyses of your LPG (preferable in mol percentages). Then, as Art Montemayor has pointed out, perform a stoichiomtric combustion calculation to determine the theoretical pure oxygen required for complete combustion.

Milton Beychok
mbeychok@xxx.net (replace xxx with cox)
(Visit me at www.air-dispersion.com)