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Obtain Excess Air From Flue Gas Analysis?


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#1 fatma

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Posted 19 April 2012 - 02:04 PM

hi,, how can i obtain excess air %
c=87.6% wt%
H2=10.27%
S=0.7%
H2O=0.05%

and want to draw figures between %O2 in flue gas and % excee air ?

anyone can help me?

Edited by fatma, 19 April 2012 - 03:12 PM.


#2 Art Montemayor

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Posted 19 April 2012 - 02:47 PM

Fatma:

If you are a chemical engineering student in Egypt, you should have already taken the first engineering course in your curriculum that deals with stoichiometry, material and heat balances, etc. Part of that course was to show you how to do combustion equations and what you need to generate a process balance.

If you took that course, you should know that it is not possible to have an orsat analysis for fuel oil. Are you serious, or have you made a gross mistake in writing that?

If you have been told to find the excess air supplied to a combustion of fuel oil, then you need the complete fuel oil analysis AND a flue gas orsat analysis.

With these two inputs you can then do a routine stoichiometry combustion calculation and find the amount of excess air that caused the flue gas analysis to come out as it did.

#3 fatma

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Posted 19 April 2012 - 03:13 PM

thank you so much for your reply
iam still student and just start this course so i don't have all principles
AND sorry for this mistake .
thank you again

#4 Shivshankar

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Posted 20 April 2012 - 01:05 PM

Fatma,

Quantity of excess air can be determined by measuring the % Air in the flue gas. With the measured volume of % Air, the excess air can be calculated by the following formula

% excess air = ((Theoretical Air - Actual Air)/(Actual Air)) X 100

Regards
Shivshankar

Edited by Shivshankar, 20 April 2012 - 01:05 PM.


#5 fatma

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Posted 20 April 2012 - 02:16 PM

Shivshankar:

thank you so much

#6 ankur2061

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Posted 21 April 2012 - 09:27 AM

Shivshankar,

You do not measure air in flue gas. You measure the oxygen in the flue gas and based on that you calculate the excess air. For example if the oxygen content in flue gas is 1.5% mole percent then the excess air calculated would be:

1.5*100 / 21 = 7% excess air

based on the fact that air contains 79% by volume or moles of Nitrogen and 21% by volume or moles of oxygen.

Regards,
Ankur.

#7 Shivshankar

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Posted 21 April 2012 - 09:37 AM

Thanks Ankur for correcting it.

Regards
Shivshankar

#8 fatma

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Posted 21 April 2012 - 06:22 PM

thank you so much

#9 kkala

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Posted 24 April 2012 - 11:27 AM

Post No 6 presents an example of quick rough estimate of excess air. Logic could be that air had initially 21% O2 (mol/mol), versus 1.5% in flue gases. Assuming no change in moles between air and flue gases, excess air is 1.5/(21-1.5)=0.077 or 7.7%. Probably it is generally more precise to neglect remaining air (1.5%) in denominator, so that excess air = 1.5/21=0.071 or 7.1%. In other words, % excess air multiplied by 0.21 would give a rough estimate of % O2 in flue gases (mol / mol). Let us check applicability of this rough rule on a specific gas fuel, composed of CH4=85%, C2H6=11%, C2H8=2.5%, CO2=1.5% (mol /mol). Calculations are presented in attached NGr0.xls.
1. For 7% excess air, resulting O2 in flue gases is 1.3% (wet basis) - 1.5% (dry basis); so rule is verified.
2. The higher the excess air, the higher the deviation from the rule, as the diagram on the NGr0.xls indicates (for the specific gas). For e.g. 100% excess air, rule indicates O2 in stack gases ~ 20%, versus ~10% calculated.
3. However for excess air up to ~20% (usual in gaseous fuels), rule gives reasonable rough estimates, at least for the selected fuel.
4. In the general case, one had better consider fuel composition and excess air, making stoichiometric calculations. Liquid fuel composition is usually given as % C, H, S, O, N, H2O, w / w. Taking 100 kg of fuel as basis, one can estimate flue gas composition and volume per kg (or 100 kg) of fuel. And one can make diagrams between excess air - % O2 if flue gases by varying excess air (query of post No 1).
5. Formula of post No 4 can be probably expressed as: excess air =
[(actual O2 in - stoichiometric O2)/stoichiometric O2] (stoichiometric instead of theoretical).

Attached Files

  • Attached File  NGr0.xls   36.5KB   526 downloads


#10 Shivshankar

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Posted 25 April 2012 - 03:27 AM

Fatma,

FInd attached PDF.

http://eyrie.shef.ac...30/comfun1.html
http://www.gutenberg...rs/fluegas.html

Regards
Shivshankar

Attached Files


Edited by Shivshankar, 04 June 2012 - 08:45 AM.


#11 fatma

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Posted 12 May 2012 - 06:14 AM

thank you so much

#12 jmukund

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Posted 31 May 2012 - 04:54 AM

Dear All,

Please find attched workbook for quick estimation of excess air based on flue gas analysis

Attached Files






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