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-   -   reactingFoam constant/reactions file (http://www.cfd-online.com/Forums/openfoam/90954-reactingfoam-constant-reactions-file.html)

Cyberholmes July 26, 2011 11:45

reactingFoam constant/reactions file
 
Hello,

I wanted to check to make sure I was using the correct units for my Arrhenius values in the reactions file for reactingFoam. Apparently, back when CHEMKIN was used, the prefactor A was given in units of (cm^3/mol)^(r-1) s^-1, and activation energy (given as Ta??) was expressed in units of cal/mol. These seem like odd units to use. Are these still the units used in the reactions file? Thank you.

l_r_mcglashan July 26, 2011 11:59

If there are no units specified after REACTIONS then the Act. energy is in cal/mole. A is in CGS units.

The whole chemkin format is a pain. Make sure when you read in a mechanism to check it has been read in properly. You can use chemkinToFoam to check it.

Cyberholmes July 26, 2011 12:02

I'm not actually using chemkin, I'm using the new format that is not chemkin (I'm using OpenFOAM 2.0.0). I just wanted to check if the units chemkin used, which I mentioned in the above post, are still the same ones used in the new format, in the reactions file in the constant directory.

Thanks.

Cyberholmes July 27, 2011 11:24

Does anyone know the units of the reactions file?

l_r_mcglashan July 27, 2011 11:49

You can either ask them directly or look through the code, which is will be somewhere in src/thermophysicalModels/, probably in reactionThermo or specie.

I would *assume* it's chemkin standard (looks like it), so cal/mole for act energy, but most of my assumptions are usually incorrect.

As I said earlier, an easy way of checking would be to use chemkinToFoam on your chemkin files and see if the numbers that are in the foam files match.

megacrout July 28, 2011 10:00

Quote:

Originally Posted by l_r_mcglashan (Post 317745)
I would *assume* it's chemkin standard (looks like it), so cal/mole for act energy, but most of my assumptions are usually incorrect.

Aha, I really like this one!

There is an answer here: http://www.openfoamworkshop.org/6th_...ist_slides.pdf
on slide 27. It is a bit confusing though as he made a typo while writing the last unit (twice K instead of K, kJ or K, J) and he gave 2 different values of A even though they should be equal except for the order of magnitude.

So this is no absolute answer and
Quote:

Originally Posted by l_r_mcglashan (Post 317745)
ask them directly or look through the code, which will be somewhere in src/thermophysicalModels/, probably in reactionThermo or specie.

is probably a better option. IŽll give it a shot right now.
Thanks for the advice Laurence.

Tibo

Cyberholmes July 28, 2011 13:20

Ah, I figured out something at least! The K is not a typo in the slides you posted, because the third value in the native format is the activation energy divided by the gas constant, so the units are (cal/mol)/(cal/(mol * K)) = K. I checked the units using chemkinToFoam and they work out.

What I can't figure out is what the deal is with A, the first value. It says the units in chemkin are (cm^3/mol)^(r-1) s^-1, and the units in native are (m^3/kmol)^(r-1) s^-1, and so the numbers in that field should differ between the two formats only by a factor of a power of ten. However, they don't. In the slides, they differ by a factor of ~31.6, and in the reactinFoam tutorial on the wiki, they differ by a factor of 177.8.

What does this mean?

megacrout October 10, 2011 05:06

They differ by those amounts because the guys who did those computations did them wrong. It is pretty easy to check with chemkinToFoam (thanks to Laurence for the advice, 2nd post of this thread).

A factor 31.6 results from converting units of a reaction of the type A + 2B -> products while considering it as a reaction of the type 0.5A + B -> products. Instead of multiplying the power of ten by 3 (1+2 from 1A + 2B) while converting the units from mol to kmol or conversely, they used a power of 1.5 (0.5 + 1 from 0.5A + 1B). The difference is a factor of 10^1.5 = 31.62.

From there on, I tried to understand where the factor 177.8 you found could come from. You solve 10^x = 177.8 and get x = 2.250. It might thus come from a reaction of the type 2A + 2.5B -> products being considered as A + 1.25B -> products. Difference is (2+2.5)-(1+1.25) = 2.25. I donŽt know what the reaction you looked at was, though.

Hope that helps.

Tibo


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