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Negative mass Fraction on cold boundary

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Old   January 30, 2013, 18:33
Default Negative mass Fraction on cold boundary
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I am trying to simulate the chemically reacting flow> And I found that the formula of reactive NS always gives the negative mass fraction (for product) or over unity mass fraction (for reactant), if the mass diffusion is taken into account. For instance, in 1-d steady case.

Be grateful to any body could give me some hint.
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Old   January 31, 2013, 07:55
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Hi, this is usually due to the numerical dispersion associated to your numerical scheme. This kind of schemes lead to undershoots and overshoots. If you are mixing, say, pure fuel with air then the code will encounter a pretty stiff gradient at the interface, which will result in some wiggles as the fuel diffuses into the oxidiser (Yfuel>1 and Yox<0).

Are you using a centered scheme? Try then using a higher order scheme with lower numerical dispersion.

I hope it helps.

Cheers,
Michujo.
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Old   January 31, 2013, 18:21
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Quote:
Originally Posted by michujo View Post
Hi, this is usually due to the numerical dispersion associated to your numerical scheme. This kind of schemes lead to undershoots and overshoots. If you are mixing, say, pure fuel with air then the code will encounter a pretty stiff gradient at the interface, which will result in some wiggles as the fuel diffuses into the oxidiser (Yfuel>1 and Yox<0).

Are you using a centered scheme? Try then using a higher order scheme with lower numerical dispersion.

I hope it helps.

Cheers,
Michujo.
-------------------------------------------------
Thank you for your answer. I believe what you said is quite often based on your experience, and I appreciate that. But what I am facing is the kind of like the cold boundary difficulty (e.g. described in 'Combustion Theory' [F. A. William]).

Let me try to clarify this problem. We have mass fraction Y and let's just talk about reactant. Then from species eqn. (1-d, steady state), we have,:

dY/dx-(D/V)d_sup{2}Y/dx_sup{2}=w,

where D, V and w are diffusion coefficient, velocity and reaction rate respectively. For the reactant, RHS w must be <0 and then based on the eqn above, the calculated Y must be larger than original value. This is my confusion. In

MAdding....
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Old   February 1, 2013, 18:41
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Hi, I misunderstood your question.

As for the cold boundary difficulty I quote C.L. Law's book "Combustion Physics", page 249.:

"The cold boundary difficulty can be removed by simply freezing the chemical reaction at a distance sufficienty far upstream of the flame. Such an approach can be interpreted on the basis of an ingnition temperature, T_{ig}. That is, we simply set \omega=0 for T<T_{ig}."

This way you're supposed to remove the ill-posedness of the species transport equation at the cold boundary (where both dY/dx and d2Y/d2x are zero, so that w should be zero as well, but it is not unless you explicitly freeze it).

You might have a go at it and see if it works, although I wonder if you'll just be transporting the ill-posedness somewhat downstream ..., no idea.

Cheers,
Michujo.
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Old   February 19, 2013, 14:18
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Quote:
You might have a go at it and see if it works, although

""" I wonder if you'll just be transporting the ill-posedness somewhat downstream ..., no idea. """



Michujo.
Yes, you are correct. In my calculation, if I freeze the reactant, I am just transporting the ill-posedness downstream. The situation can't be saved this way, especially when I am dealing with detailed chemistry where maybe tens of transport equations are present.
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Old   February 20, 2013, 01:20
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Any big guy could tackle this problem further?
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Old   February 23, 2013, 00:51
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up up. hope there are people could help
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