Re: Ethane combustion: model for CFD



Herman Family wrote:


"Allamarein" <matteo.diplomacy@xxxxxxxxx> wrote in message

news:ad854765-2059-41e0-ace1-e13a84c93445@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
I found a very accurate model to ethane combustion.
It runs very well in CHEMKIN.
However too many reactions and species don't allow me to run in a CFD
software as Fluent.

Anyone know an easy mechanism that predict pretty well flame
temperature and CO2, H2O, C2H6 and O2?

The answer is "It Depends". If you just want an overall flame
temperature,
then it isn't too hard. If you need the flame temperature at any
given point, then the answer is possibly "no".

Look at your basic reaction equation: fuel + oxygen = heat + CO2 +
H2O. I'll let you balance that out.

Next step. Recognize that oxygen is generally supplied with run of
the mill
air, which is about 79% nitrogen. There is also some water in the
air, but we'll forget that fact for now.

Add nitrogen to your equation in the proper quantities:

Fuel + O2 + N2 = heat + H2O + CO2 + N2 all hot)

Next, find the heat of combustion of your fuel. That is the heat of
formation of the fuel, minus the heat of formation of the CO2 and H2O.

Now, all that net heat is going to warm the products plus the nitrogen
up to their final temperature.
Using an equation or perhaps a set of equations for the enthalpy of
CO2, H2O, and N2, find the temperature at which the enthalpy increases
(from the starting temperate) equals the net heat of combustion, and
you basically have your flame temperature.

That's how to do it if you know that the reaction goes completely to H2O
and CO2, which may not be the case at high tempreatures. The next
better approximation will involve knowing the equilibrium composition
of the gas mixture as a function of T, and will require an iterative
calculation to find T. Finally, if your flame is operating under
conditions where equilibrium may not be reached for kinetic reasons,
you will have to involve the mechanism.


Note that if you add excess air, you have to heat that up too.

Now, if you want to do that for any point in the flame, then you
basically do the same thing, except recognize that the products of one
reaction are the reactants in the next one, and that the distance from
the nozzle at
constant velocity is basically a time dimension. At any given
distance (under a stable flame and continuous conditions) the same
reactions are taking place.

Michael

Michael

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