Autoignition v. Flammibility v. Flash Point

[M.D. Holloway]

I know the text book definitions:

Flashpoint is the temperature at which the vapors produced from a fluid will ignite (flash off) with the presence of an ignition source.

Firepoint is the temperature at which the fluid will sustain a fire if ignited by an outside ignition source.

Autoignition is the minimum temperature at which a substance must be heated to cause ignition without application of flame or spark.

I was interested in writing an algorithm for the relationship of the three - of at least two of the three - say Autoignition and Flashpoint but there seems to be more variables to tackle than I want.

Does anyone know if there is a simple, see Jane Run rule of thumb for the delta relationship between Autoignition and Flashpoint?

The following is a chart of some simple gases from CH3 to C12H26 I put together to get a visual. The algorithm is easy at STP with simple gases but more complex molecules - it gets whack.

Any ideas?

[Hugh R]

Try starting off with Kelvin temperature. Celsius is just an arbitrary temperature set at the freezing point of distilled water at STP. You might also try a natural log or log 10 on the temperature and/or molecular weight. Also, it's the H-C bonds that provide the oxidation, not the C-C bonds. 4 H-C bonds for methane and 26 for tetradecane. Good luck because up to pentane, I believe, you have a gas and beyond that they're liquids at STP, I think. You may want to try dividing into two algorithms one for gases and one for liquids.

I'll give it some more thought tonight.

[RWebb]

take the square root of the Gibbs Free Energy and then divide by the activation energy

after that.... you're on your own

[M.D. Holloway]

Originally Posted by Hugh R

Try starting off with Kelvin temperature. Celsius is just an arbitrary temperature set at the freezing point of distilled water at STP. You might also try a natural log or log 10 on the temperature and/or molecular weight. Also, it's the H-C bonds that provide the oxidation, not the C-C bonds. 4 H-C bonds for methane and 26 for tetradecane. Good luck because up to pentane, I believe, you have a gas and beyond that they're liquids at STP, I think.

I'll give it some more thought tonight.

yup - the slope will be relative right? doing a log will make it linear which is easier on the eyes I guess. I am actually trying to figuer out something for oils. The gases were just easy to use to try to get a simple expression down.

Thanks for your help.

[M.D. Holloway]

Originally Posted by RWebb

take the square root of the Gibbs Free Energy and then divide by the activation energy

after that.... you're on your own

Throw in a Hamiltonian operator and before you know it all sorts of thermodynamics should be jumping out at me!


I wanna beer...

[MotoSook]

You can probably develop a general equation (your rule of thumb) for the oils you are considering. The equation would be based on your plots and only work for those oils.

E.g. in your chart above, curve fit, and they solve for delta-y (x)