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Chemical reaction UDF in Fluent

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Old   April 5, 2015, 07:29
Default Chemical reaction UDF in Fluent
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Londo
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Hello all,

Let's say that I have a reaction that goes as following:

C2H4 + 2O2 -> aCO + bH2O + cH + dOH

where a, b, c, d are not constants that you would usually have in a chemical mechanism, but are functions of concentrations and reaction rates.

This obviously has to go through UDF. The problem is that I can't seem to find the udf macro that allows to define non-constant stochiometric values. Can you help me with that?
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Old   April 5, 2015, 08:14
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Cees Haringa
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Hi Londo,

I don't know of such a UDF either, but you can choose to skip the whole reaction formalism and go for source-UDFs. There, you can simply set the local source values (kg/m3-s) based on whatever equation you want, both positive and negative, hence it should not be difficult to combine with varying stoichiometry.

Best,
Cees
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Old   April 5, 2015, 08:52
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Cees,

Thank you for you reply. That's an interesting idea. The question is whether Fluent will calculate temperatures in this case, or I'll have to write that as well?
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Old   April 5, 2015, 09:27
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You can include a sourceterm for energy in your UDF as well. It won't happen automatically indeed.
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Old   April 5, 2015, 09:38
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The question is whether Fluent will be able to calculate the temperatures without it. Naturally, my aspiration is to keep the simulation as simple as possible, hence I wish to make as little changes as possible. Is it possible?
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Old   April 5, 2015, 09:53
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I'm not sure what you mean. If you use UDF sources instead of the built-in reaction model, there will be no calculation of reation enthalpy unless you explicitly add it to your source UDFs as an energy source. So unless there are other energy sources, your simulation will be isothermal. If temperature matters, you will need to add an enthalpy UDF as well. But as far as I know, that's the only way to do it. I know of UDFs to edit reaction rates, but not stoichiometry.
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Old   April 5, 2015, 10:10
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Actually, this completely answers my question. Thank you very much!
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Old   April 5, 2015, 10:14
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Great! Glad to be of help
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