Species Concentration Initialization for a Methane reduced chemical kinetic mechanism

abdelrazik · June 20, 2020, 17:30

Hello all,
I'm writing a quasi 1d MATLAB code to simulate pulse jet engine and i'm stuck at the combustion phase. I have to use the reduced 4 reactions-8 species mechanism that was mentioned in "REDUCED REACTION SCHEMES FOR METHANE, METHANOL AND PROPANE FLAMES" by PACZKO. In that mechanism the main species are: CH4,O2,H2,H,H2O,CO,CO2 and M, and to calculate the rates of the reactions i need the concentrations of intermediate species that are OH, O, CHO, HO2, CH3, C2H5, C2H2 and CH. To calculate these intermediates, PACZKO managed to get some algebric equations using the main species for each of them based on some approximations, like steady-state for example (as indicated in the attached 2 images).
My question is how to initialize the domain by stoichiometric ratio of fuel/oxidizer and the other species are zeros while they are needed in the denominator of the intermediate species ?

Eifoehn4 · June 20, 2020, 20:44

Perhaps you can assume that all species exist up to a small number, e.g. $\epsilon=1\times 10^{-8}$ ?

I have a little trouble to understand your problem. Perhaps you can simplify your question. Or you can use the latex functionality instead of posting some parts of a book.

Good questions often correlate with good answers.

Regards

abdelrazik · June 21, 2020, 06:22

Ok, let me simplify this question.
Initially I have the following concentrations for the main species [Kmol/m^3]:
C_CH4 = 0.0039
C_O2 = 0.0077
C_M = 0.029 % Third Body treated as Nitrogen
C_H2 = C_H = C_H2O = C_CO = C_CO2 = 0

To calculate the rates of reactions for the first time step, I need the concentrations of some intermediate species, let's take C_OH as an example:
C_OH = (C_H * C_H2O) / (C_H * Kc3), where Kc3 is a partial-equilibrium constant.

So, here C_H, C_H2O and C_H are zeros, then C_OH is undefined number. and some other species tends to infinity because of the existing of ZERO in the denominator. How can I deal with this suitation >

Eifoehn4 · June 21, 2020, 08:20

Thank you, now it is easier to follow your problem! However, I think you have a typo in your formulas.

At a first step I would ignore mathematical formulas and consider what physically happens. If only $x_{H},x_{H2O},x_{H2}$ are part to create $x_{OH}$ , the answer is zero.

However if you want a mathematical proof, the rule of L'Hôpital may help you for problems like $\frac{0}{0}$ at

. Here

is your concentration. If you assume that all concentration approach with the same speed to zero it holds:

$x \equiv x_{H},x_{H2O},x_{H2}$ .

$\lim_{x \to 0} \frac{x x}{x k} = \lim_{x \to 0} \frac{\frac{d}{dx}(x x)}{\frac{d}{dx}(x k)} = \lim_{x \to 0} \frac{(2 x)}{(k)} =0.$

However i do not know if this is the answer to your question. Perhaps you have to solve the problem iteratively if other intermediate chemical reactions create $x_{H},x_{H2O},x_{H2}$ in the meanwhile.

Regards

abdelrazik · June 21, 2020, 09:36

Thank you so much for your help. Actually i thought of using L'hopital too, but the other species have nested algebric equations to calculate so this method is complicating.

Another point is, if we think physically, we find it acceptable that these intermediate species have zero concentrations in the beginning, but that assumption will give us zero m_dot fir the main species, i.e, there will be NO change in the initial concentrations and the problem will be the same at the next time step.

Eifoehn4 · June 21, 2020, 09:43

I'll quote myself if you don't mind.

Quote:

Originally Posted by Eifoehn4

Perhaps you have to solve the problem iteratively if other intermediate chemical reactions create $x_{H},x_{H2O},x_{H2}$ in the meanwhile.

Are there any other intermediate chemical reactions, which create other intermediate or main concentrations, e.g. $x_{H},x_{H2O},x_{H2}$ . If this is the case, i think (but i am not sure) you have to solve a system iteratively to get the intermediate solution.

Regards

abdelrazik · June 21, 2020, 09:56

Yes, there are $x_{CO},x_{CO2}$ as major species needed to calculate other intermediate species like $x_{O}, x_{CHO}, x_{HO2}, x_{CH3}, x_{C2H5}, x_{C2H2} and x_{CH}$

abdelrazik · June 21, 2020, 10:00

May i ask you another question please that may help us in thinking differently ?
PACZKO mentiones in his paper that for calculating intermediate species, he used partial equilibrium and steady state approximations. But, we know that these approximations relates to fast buil-up then slow consumtion for these species. So, how can we use the final reduced mechanism with these approximations at the beginning for the combustion phase ?

Eifoehn4 · June 21, 2020, 10:08

I am not deep in the combustion topic. There may be some other guys here in the forum which are more familiar with chemical reactions.

From my quick point of view you have to solve the whole system iteratively to get the steady state solution for intermediate as also final concentration solutions.

abdelrazik · June 21, 2020, 16:50

Thank you so much for your effort.

June 20, 2020, 20:44		#2
Eifoehn4 Senior Member - Join Date: Jul 2012 Location: Germany Posts: 184 Rep Power: 14	Perhaps you can assume that all species exist up to a small number, e.g. $\epsilon=1\times 10^{-8}$ ? I have a little trouble to understand your problem. Perhaps you can simplify your question. Or you can use the latex functionality instead of posting some parts of a book. Good questions often correlate with good answers. Regards __________________ Check out my side project: A multiphysics discontinuous Galerkin framework: Youtube, Gitlab.

June 21, 2020, 08:20		#4
Eifoehn4 Senior Member - Join Date: Jul 2012 Location: Germany Posts: 184 Rep Power: 14	Thank you, now it is easier to follow your problem! However, I think you have a typo in your formulas. At a first step I would ignore mathematical formulas and consider what physically happens. If only $x_{H},x_{H2O},x_{H2}$ are part to create $x_{OH}$ , the answer is zero. However if you want a mathematical proof, the rule of L'Hôpital may help you for problems like $\frac{0}{0}$ at . Here is your concentration. If you assume that all concentration approach with the same speed to zero it holds: $x \equiv x_{H},x_{H2O},x_{H2}$ . $\lim_{x \to 0} \frac{x x}{x k} = \lim_{x \to 0} \frac{\frac{d}{dx}(x x)}{\frac{d}{dx}(x k)} = \lim_{x \to 0} \frac{(2 x)}{(k)} =0.$ However i do not know if this is the answer to your question. Perhaps you have to solve the problem iteratively if other intermediate chemical reactions create $x_{H},x_{H2O},x_{H2}$ in the meanwhile. Regards __________________ Check out my side project: A multiphysics discontinuous Galerkin framework: Youtube, Gitlab.

June 21, 2020, 10:08		#9
Eifoehn4 Senior Member - Join Date: Jul 2012 Location: Germany Posts: 184 Rep Power: 14	I am not deep in the combustion topic. There may be some other guys here in the forum which are more familiar with chemical reactions. From my quick point of view you have to solve the whole system iteratively to get the steady state solution for intermediate as also final concentration solutions. __________________ Check out my side project: A multiphysics discontinuous Galerkin framework: Youtube, Gitlab.

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June 21, 2020, 06:22		#3
abdelrazik New Member Abdelrazik Essam Join Date: Nov 2018 Posts: 8 Rep Power: 8	Ok, let me simplify this question. Initially I have the following concentrations for the main species [Kmol/m^3]: C_CH4 = 0.0039 C_O2 = 0.0077 C_M = 0.029 % Third Body treated as Nitrogen C_H2 = C_H = C_H2O = C_CO = C_CO2 = 0 To calculate the rates of reactions for the first time step, I need the concentrations of some intermediate species, let's take C_OH as an example: C_OH = (C_H * C_H2O) / (C_H * Kc3), where Kc3 is a partial-equilibrium constant. So, here C_H, C_H2O and C_H are zeros, then C_OH is undefined number. and some other species tends to infinity because of the existing of ZERO in the denominator. How can I deal with this suitation >

June 21, 2020, 09:36		#5
abdelrazik New Member Abdelrazik Essam Join Date: Nov 2018 Posts: 8 Rep Power: 8	Thank you so much for your help. Actually i thought of using L'hopital too, but the other species have nested algebric equations to calculate so this method is complicating. Another point is, if we think physically, we find it acceptable that these intermediate species have zero concentrations in the beginning, but that assumption will give us zero m_dot fir the main species, i.e, there will be NO change in the initial concentrations and the problem will be the same at the next time step.

June 21, 2020, 09:56		#7
abdelrazik New Member Abdelrazik Essam Join Date: Nov 2018 Posts: 8 Rep Power: 8	Yes, there are $x_{CO},x_{CO2}$ as major species needed to calculate other intermediate species like $x_{O}, x_{CHO}, x_{HO2}, x_{CH3}, x_{C2H5}, x_{C2H2} and x_{CH}$

June 21, 2020, 10:00		#8
abdelrazik New Member Abdelrazik Essam Join Date: Nov 2018 Posts: 8 Rep Power: 8	May i ask you another question please that may help us in thinking differently ? PACZKO mentiones in his paper that for calculating intermediate species, he used partial equilibrium and steady state approximations. But, we know that these approximations relates to fast buil-up then slow consumtion for these species. So, how can we use the final reduced mechanism with these approximations at the beginning for the combustion phase ?

June 21, 2020, 16:50		#10
abdelrazik New Member Abdelrazik Essam Join Date: Nov 2018 Posts: 8 Rep Power: 8	Thank you so much for your effort.