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August 30, 2004, 14:17 
Equilibrium approach.

#1 
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I wondering if someone could explain what people mean when they say they used an equilibrium approach in combustion.
Thanking you all. 

August 30, 2004, 16:44 
Re: Equilibrium approach.

#2 
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"Equlibrium approach" refers to a model of the chemical reactions that assumes zero reaction time, or conversely an infinite rate of reaction for all relevant chemistry. All species are present in amounts described by the equilibrium and stoichiometric equations.


August 31, 2004, 10:44 
Re: Equilibrium approach.

#3 
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"All species are present in amounts described by the equilibrium and stoichiometric equations."
What are these equations? Can you have equilibrium approach in premixed and nonpremixed combustion? How does one solve for combustion in such a problem? Thank you. 

September 1, 2004, 11:15 
Re: Equilibrium approach.

#4 
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You have to write the continuity, momentum, energy and species conservation. Find the mass rate production of each reactant in your chemical mechanism, by means of Arrhenius law.


September 2, 2004, 03:06 
Re: Equilibrium approach.

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The equilibrium approach is the chemical composition obtained for a given initial composition at a given thermodynamic state after infinite time, i.e. when the chemical kinetics of all steps reached the asymtotic state.
Basically, this can be solved in two ways. The more straightforward is to find the minimum free energy for the initial composition and state. The initial composition may be given as species moles or mass or merely as the atomic population. The thermodynamic state may be prescribed by the themperature and pressure (but other pairs of state variables, e.g. volume and entropy are also possible). Many packages were developed for this approach, and I think at least some of them are free and have decent documentation. A less direct approach is through the kinetics, solving for a long enough time to allow all variables to reach their asymptotic values. This method requires more input and is less efficient. In many combustion applications the chemical kinetics is faster than the mixing ("mixingcontrolled combustion"), and therefore equilibrium approach for the chemistry is sometimes used. Other possibilities, however, are also used in such cases e.g., the Eddy Break Up model. 

September 3, 2004, 13:32 
Re: Equilibrium approach.

#6 
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does solving the mass production of each species using Arrhenius equations, form part of the first or second approach you speak of???
thanks 

September 5, 2004, 03:04 
Re: Equilibrium approach.

#7 
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ingigo,
When you use Arrhenius equations, you are definitely solving chemical kinetics, i.e., the second approach mentioned in my former posting. 

September 7, 2004, 07:03 
Re: Equilibrium approach.

#8 
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thankyou rami,
could you also explain what are asymptotic studies and where the assumption of a very large activation energy fits in and why it is useful. how do these 2 terms fit in with what you previuosly said? thanks 

September 7, 2004, 08:49 
Re: Equilibrium approach.

#9 
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I am afraid you misunderstood me. I did not use the term "asymptotic studies". Rather, I suggested (in the second approach) that you can reach equilibrium if you allow the kinetics develop for a sufficiently long time (instead of infinite) to asymptotically reach the equilibrium state.
I also don't think I used "assumption of a very large activation energy". Hopefully this one is clearer. 

September 7, 2004, 11:32 
Re: Equilibrium approach.

#10 
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hi rami, the terms asymptotic studies and the use of a large activeation energy are both ones i have seen from other sources (actually from peters and williams work).
i think the idea is something like if you use a large actuvation energy in the arrenhuis equation you get an assumption of a thin flame and can get from it the equilibrium solutions?!? is this correct? and the previous question was really does this idea fit in at all with your 2 descriptions of reaching the equilibrium solution. in your description of asymptotically reaching the equilibrium solution what values in the arrenhius equation are used? and what exactly is changing as we got through each step towards the asymptotic solution? thanks again. hope this is clearer. 

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