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Flammability of fuels 1

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Joyer1

Chemical
Aug 4, 2014
4
Hi readers,
Please can someone help me out with interpreting LFL and UFL values
I have UFL values of 1.00E+06 ppm which is about 100% of it, what does this mean?
Can I still consider the mixtures to be flammable?
Some have LFL values and then UFL of this value but some of them do not have any LFL values just the UFL of 1.00E+06 ppm

Thanks
 
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An upper limit of 100% would mean the material has it's own oxidizer or doesn't need one. Acetylene for example will auto ignite with no O2. It decomposes into H2 and C

A LFL of 0 would be a similar situation.

I'd like to know what materials you are dealing with. H2 has one of the widest ranges I'm familiar with.
 
Thanks for the reply,
The mixture I am dealing with comprises of hydrocarbons (C1 to C10) and water. I am working on flammability of HC mixtures with water, to investigate the point at which it can be considered non-flammable
 
What exactly you mean by a mixture of C1 to C10 hydrocarbons with water? The only way you'll get any of these materials to "mix" with water is to add a surfactant to stabilize an emulsion.

C1 through about C4 are gases at room temperature. They are sparingly soluble in water and come out of solution as soon as the pressure is let down, concentrating in the air space above the water. Surfactants aren't going to stop that from happening.

C5-C10 are liquids at room temperature, are all sparingly soluble in water (perhaps a few hundred ppm at most), and are all less dense than water, and so tend to form a film or sheen to a separate layer on top of water when allowed to contact with air- unless they are thoroughly emulsified.

In any case- dissolved gas, emulsified liquid or free product, you're dealing with pure component vapour pressures to determine what matters, i.e. the air/fuel mixture concentration in the headspace. It's as if the fuel and the water were in separate containers inside a common air space. Even in an emulsion there is no true solvation- you just have stabilized droplets of free product liquid hydrocarbon with its surface made "friendly" to the water by the surfactant. You get zero benefit from the presence of water, i.e. no vapour pressure suppression to reduce the concentration of flammables, other than by the presence of water vapour as an inert.
 
Thanks

Its a crude oil sample composition ,includes small proportions of N2, H2S and CO2 also and I am modelling with the consequence model tool which is giving results for pool fires.

Trying to understand the chemistry of this interaction of the components and its effect on flammability because at some point the mixtures are non-flammable with about 95% water by mass
 
Somebody has used a simplified model of the mixtures in your software then. You can definitely get a 5% layer of oil on water to burn as long as the oil film is thick enough and there are sufficient lights in the oil to give a flammable vapour. Can you get an emulsion of 5% oil in water to burn? If the oil isn't crude oil but rather is something like octane or ethylbenzene (both C8), it would surprise me greatly if the resulting emulsion didn't have enough of those components in the vapour space to form a flammable concentration.
 
This doesn't sound like a good idea.

you must get smarter than the software you're using.
 
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