Excess Air Ratio in combustion
Hello everyone. This is K in Hong&Kang Environment
For these days, I saw the news about dating abuse, teen age murderer killed his grandmother and dad raped and murdered his daughter. (Korean news)
Sincerely sorry to hear that. I hope that kinds of these incidents are broadcasted by press without hiding anything.
The life of parents has a great impact on the life of their children than anything else
- Carl Jung, Swiss psychiatry (defined complex, introvert and extrovert) -
For this time, I am going to explain about Air Ratio
Air ratio means the ratio of theoretical and actual air requirement when fuel is combusted.
When fuel is combusted, the combustion is not executed in ideal condition because of the change of combustion temperature, unequable air injecton etc.
So we are able to calculate Air Ratio.
Then, why do we have to know Air Ratio?
<Low Air Ratio>
1) Exceeding of legal emission standard (THC, CO..)
2) Increase of smoke and soot
3) Decrease of combustion efficiency
<High Air Ratio>
1) Increase of thermal losses by combustion temperature's decrease
2) Exceeding of legal emission standard (SOX, NOX) or the cause of corossion(erosion)
3) The capacity of environmental prevention facility should be increased(installment/operation expense should be increased)
4) Decrease of combustion product concentration by dillution effect
Air ratio calculation
1. When theoretical(Ao) and actual(A) air requirements are given
2. Emission gas analysis by Orsat method
*Orsat method: the method to analyze gas concentration of sample by passing oxygen/carbon dioxide in sample through absorbent and then by checking volume reduction
(Atmospheric process test standards, ES01314.2a - Korean law..)
*Formula derivation
In my opinion, this formula follows the assumption that the composition of air are 79% N2 and 21% O2 (other materials < 1%)
1) Actual air requirement(A): since 79% N2 exists in the air, it needs to N2 → Actual air requirement
So, actual air requirement = Air(100) / N2(79) * N2 volume(Sm3)
2) Theoretical air requirement(Ao): since all oxygen is reacted lideally, has to subtract oxygen volume of emission gas on actual air requirement (oxygen is not included in emission gas)
Check the point that carbon dioxide(CO2) is only produced in ideal reaction not carbon monoxide(CO)
However, CO is produced in actual air requirement.
Therefore, it has to subtract oxygen volume (which needs to convert CO(actual combustion)→CO2(theoretical combustion)) on oxygen volume in emission gas.
* More oxygen needs to produce CO2 from CO because reaction formula is CO + 1/2O2 = CO2, O2=0.5CO(based on volume)
3. When O2% exists in emission gas
*Formula derivation
1) O2 = 0.21(m-1)Ao: Subtract used oxygen amount(0.21Ao) in combustion on excess oxygen amount(0.21mAo), because oxygen concentration in emission gas
2) At the end of the formula, When denominator divided by N2 and numerator divided by N2, "3." is completed
* Let you think how O2% in 3. formula is same with 79O2/N2
4. The method using CO2max(%) and CO2(%)
- CO2: CO2 volume percentage in emission gas when fuel is combusted without excess oxygen
- CO2max: CO2 maximum volume percentage in emission gas when only CO2 is emitted in process using excess oxygen without CO production process by incomplete combustion of fuel
Thank you so much