CFD Online Discussion Forums

CFD Online Discussion Forums (http://www.cfd-online.com/Forums/)
-   Main CFD Forum (http://www.cfd-online.com/Forums/main/)
-   -   Negative mass Fraction on cold boundary (http://www.cfd-online.com/Forums/main/112555-negative-mass-fraction-cold-boundary.html)

Joshua_X January 30, 2013 18:33

Negative mass Fraction on cold boundary
 
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.

michujo January 31, 2013 07:55

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.

Joshua_X January 31, 2013 18:21

Quote:

Originally Posted by michujo (Post 405268)
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....

michujo February 1, 2013 18:41

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.

Joshua_X February 19, 2013 14:18

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.

Joshua_X February 20, 2013 01:20

Any big guy could tackle this problem further?

Joshua_X February 23, 2013 00:51

up up. hope there are people could help


All times are GMT -4. The time now is 03:05.