Dear Sir,

I need to know how can I get the amount of O2, CO and CO2 at the exit point of a Coal Fired Furnace.Does anybody know how to calculate them using CFX4 Post Processor.

Any response will be highly appreciated.

Monir

I can just talk about O2: You can see results for mass fraction 2 which is oxidant mass fraction. In the case of a coal fired boiler, oxidant is air, therefore: % O2 = 0.23*Mass fraction 2*100 (0.23 is the mass fraction of oxygen in air). I hope this can help, if you find something about CO and CO" please put in on this forum, I am also interested in it.

Thanks IIN for your prompt response. I will definitely post it if I get something about CO and CO2 calculations.

Monir

Dear IIN,

This morning I got an email from Dr. David Fletcher who is the CFX Technical Support Manager for Australasia and SE Asia. He has mentioned that from output file we can see the Product Composition which includes CO2, Nitrogen and water. From Post Processor View we can calculate the Mass fraction39Product) at any particular position.Then just multiplying Massfraction3 by CO2 composition,we can get the CO2 content of the Product for that specific position.He has also mentioned that from the current CFX Code we can not calculate CO without having a separate CO model enabled and this is not part of the standard code.

Hope, this will help you.

Monir

Dear Monir, thank you for your help. From your answer I wonder a new qustion: what about mass fraction 4 (char products mass fraction)? I think that it also contributes to the CO2 production, therefore if you want to know the total CO2 content you should sum both terms: (MF3 + MF4)*CO2 mass fraction in gases (from the output file). What do you think about this? Best IIN

Dear IIN,

You are right. I have checked your opinion with Dr. David Fletcher. I also did some calculation on the basis of MF3+MF4. And the data matches well with the measured data. But then one question came to my mind, that is the Molecular Weight of Char Product which I got from Output File. The Molecular weight is 30.4691. If the Char product is only CO2 then MW should be 44. It means there is some CO also. In that case what should we do?

What do you think about it?

Monir

Dear Monir,

At the moment, I am also trying to understand how CFX calculates combustion values. The CFX support team has answered me regarding your question and the answer was:

"Molecular Weight of Char Product (CHP) is the mean molecular weight of the char products which are a mixture of CO2 and NO2. The molar fraction of N2 is the same as the molar fraction of N2 in the combustion Air"

When I calculate this CHP molecular weight I get 31.37 considering 79% N2 + 21% CO2 (in molar basis) while CFX gives in the output file: 30.75.

I don´t understand why the molar fraction of N2 in Char Products is 0.79 and in volatile products is 0.64 (given by CFX) when no model of NOx is used. I thought N2 mass fraction should remain constant

Moreover, I am burning coal with a 25% excess of air, and I don't understand why Oxygen is not considered as a product while N2 is.

What do you think about this?

I am very interesting in such this questions. Do you mind if I send you an e-mail ?

Best IIN

Monir, I said in my previous message that CHP are a mixture of CO2 and NO2. It should be:

CO2 + N2

Sorry! IIN

Here is how it works! The mole fraction of N2 is the same for oxidant and CHAR products because one mole of O2 (gas) reacts with one mole of C (solid carbon) to give one mole of CO2. If it were some other reaction that gave two moles of something else, the mole fractions would change, but as the reaction is one to one the fractions will not change.

Turning to volatile reactions, the more complex volatile molecule CxHyOz reacts with O2 to form multiple molecules, changing the mole fractions in the products c.f. the reactants. The actual amount of N2 remains the same, but now it must share the volume with more molecules so its mole fraction is reduced.

Also, dont get confused between MASS fraction and MOLE fraction, they are two completely different things. I dont know what the advantage is for specifying mass fraction, but the disadvantage is it leads to confusion like this. Better that you convert mass fraction to mole fraction, then think about it.

As for the 25% excess air question. You cannot think of these species as individual molecules reacing with each other. We are specifying global reactions of OXIDANT + FUEL gives PRODUCTS. Oxidant, not O2 is converted to product, Whatever O2 goes into making product must be accompanied by an amount of N2 (based on the ratios in the oxidant). Any excess oxidant (air) stays as oxident and will never show up in the products. The N2 in the products shows you how much O2 was required to completely react the fuel. It is a difficult concept because you tend to start thinking of individual molecular reactions.

Dear IIN,

James Hart has well explained everything you wanted to know.

If you like to contact with me via e-mail, you are most welcome.

My email is mhossain@swin.edu.au

With regards,

Monir

Dear Mr Hart, thank you very much for your help, but I still have some points to clarify, and perhaps you can help me again. As you say, for CHAR PRODUCTS "...The mole fraction of N2 is the same for oxidant and CHAR products because one mole of O2 (gas) reacts with one mole of C (solid carbon) to give one mole of CO2...". In this case, if I consider CO2 + N2 as char products, is it correct to say: Mole fraction of N2=0.79 (as in air), Mole Fraction of CO2=1-0.79=0.21?? In this case, the molecular weight of Char Products should be: 0.79*28+0.21*44=31.36, while the output file is giving me: 30.75. Where am I making a mistake?

Regarding volatile gases (fuel) composition, my output file is giving me the following:

EMPIRICAL FORMULA OF GAS EVOLVED CARBON =*** HYDROGEN =*** OXYGEN =*** NUMBER OF MOLECULES =841

I think it is due to a Fortran programming error regarding formats for reals, but now I need the empirical formula of gas evolved. Do you now how can I calculate it?

Thank you in advance. Best IIN

Yes, that is a tricky one. By the way the default composition is 78%N2 not 79%, but that doesn't account for the discrepency. All I can say is the answer is more than likely in the manual. Just play around with some numbers and eventually it may become clear.

As far as I know there is no way to get around this formatting problem. But I never pursued it very far. If AEA are reading this then it might be about time it was fixed. Sorry I cant be of more help!

James