limiting oxygen concentration - one way direction?

[kappie]

Hi,
the concept of LOC does it relate to adiabatic flame temperature?

LOC is the point in ppO2 where the gas mixture does not combust when you try to ignite it. Combustion defined as: no flames, no explosion. Slow (flame-less) oxidation is possible though below LOC.

When I have a combustible gas mixture with a vol% O2 slight above the LOC (say O2 = 12% where the LOC = 10%). When this mixture is ignited will the flames stop when the oxygen vol% has depleted to 10%? Or does the flame continue until all oxygen is depleted?
I would guess it continues because that would explain why for instance in a methane fed gas burner, the max adiabatic flame temperature is reached at stoichiometric air supply. So, apparently all methane is combusted in a burners despite the O2 falling below the LOC of methane (which is temperature dependent ofcourse) during burning.
In other words: Is the concept of LOC one-way directed?

thanks if you can shine some light here
Kappie

 

[zdas04]

Oxygen is not flammable. Oxygen enables fuel to burn when an ignition source is present, but without a fuel oxygen doesn't burn. I real life, there is always fuel present so a high oxygen concentration creates a significantly increased fire risk.

I usually see the discussion centered around the fuel concentrations in air instead of the oxygen combustion limit. For example, methane LEL (lower explosive limit) is 5% in air. The methane UEL is 15% in air. Since air is nominally 20% oxygen, I would expect the actual ratio for the onset of combustion is for methane to be between 1/4 and 3/4 of the oxygen percent. That means that in your 10% oxygen scenario, methane would be flammable between 2.5% and 7.5%.

David

 

[kappie]

Thanks David,

It is interesting that you mention the LEL and UEL here because they relate to flammability range when added with air.

the LOC (

http://en.wikipedia.org/wiki/Limiting_oxygen_concentration)

is the concentration of O2 where, no matter what amount of fuel present, no combustion will take place

kappie

 

[Latexman]

kappie,

Very interesting question. I don't know, but now that you have asked, I wish I did know. I knew a guy many years ago that may know. Laurence Britton. He knows a lot about fire, static ignition, dust exlosions and such. He's got a book and several articles on it too. Google him. He'd help you out if you needed to know. It may cost you though. He's a Consultant now.

Good luck,
Latexman

 

[unclesyd]

I can't find my process date on a process we had where we oxidized butane with air over a catalyst to make Maleic Anhydride. I can't give you the numbers but the operations had to stay below the stoichiometric ratio. If the mixed got to this ratio we were guaranteed a detonation.

The are several process where his happens. Below the stoichiometric ratio you get the desired product and at the ratio you get a bang. Anything way above this ratio you get soot.

Here is a plethora of information on gas + air from GEX-CON AS.

http://www.gexcon.com/handbook/GEXHBchap1.htm

 

[Latexman]

Have you researched the internet on Google or Bing, the periodicals on ScieceDirect, Wiley Interscience, or Infotrieve, or have you looked on Knovel? There's probably dozens of other services out there too.

Good luck,
Latexman

 

[Compositepro]

The answer to your question is in the Wikipedia article but not explicitly stated. In an intimately mixed gas mixture you cannot have a gradual depletion of oxygen. It will burn or it won't. If you have a burning candle, then you are gradually introducing fuel into the oxygen and burning will stop when O2 concentration drops below the LOC.

 

[kappie]

Latexman, Compositepro thanks for your replies.

I have checked sciencedirect, and the encyclopedia but could not find it that specific. But after reading the final reply it becomes clearer. So in a batch process the flame will stop after falling below below, but in a stocho fed continuously fed and intimiately fed gas burner it does not (Because on average the O2 concentration is probably > LOC). Thanks alot

kappie