How to solve excess air problems in Chemical Engineering?

In learning Chemical process calculations you might have encountered with the excess air or excess oxygen problems. This is more common in combustion type of problems, where oxygen is used as a aid to combustion. One can easily solve this kind of problems by simply noting down the four points. they are,

1. How many compounds are there which can be combustible ?
2. How much of oxygen is already there ?
3. How much actual oxygen is to be supplied ?
4. What is the percentage excess given ?

Consider a example where methane ( CH₄ ) undergoes combustion with air, according to the reaction,

CH₄     +         2O₂      =         CO₂     +          2H₂O

CH₄     +          Air (O₂ + N₂)                        =         CO₂  +          2H₂O  +          N₂

One mole of methane burns in the presence of two moles of oxygen giving one mole carbon dioxide and two moles water. If the above combustion occurs at 100% conversion rate the output stream contains only Carbon dioxide and water, if the conversion rate is less than 100%  then the output contains unreacted methane ( CH₄ ), unreacted oxygen, carbon dioxide and water. ( If we are considering air as the source for oxygen then output contains nitrogen also, because air contains 21% Oxygen and 79% Nitrogen and this 21% oxygen should be sufficient to achieve the given case ).

Let us consider the case where 2 Lb mole methane is burnt with 20% excess air in order to ensure complete combustion of methane. then what is the output gas composition ?

 Now apply the four steps.

1. Methane is the only compound which undergoes combustion.
2. Along with methane there is no oxygen , so oxygen already present is zero.
3. For 2 Lb mole methane 4 Lb mole oxygen is to be supplies as per stoichiometry.
4. But 20% excess air is supplied so indirectly it means that 20% excess oxygen is considered, so total oxygen moles = 4 Lb moles + 4*20/100  Lb moles = 4 *1.2 Lb moles.

Output gas composition :

Methane ( CH₄ ) is completely burnt so for 2 Lb moles methane 2 Lb moles carbon dioxide will form along with 4 Lb moles water as per stoichiometry. But the unreacted oxygen = 4*1.2 - 4 Lb moles = 0.8 Lb moles. Nitrogen in output is calculated by

Total air in = total oxygen/0.21 = 4*1.2/.21 = 22.857 Lb moles

Total nitrogen in inlet air  =  total air * 0.79 =22.857*.79 = 18.057 Lb Moles.

( If there is oxygen along with methane then net oxygen required for combustion = oxygen required for combustion as per stoichiometry - oxygen already present . Now apply percent excess to this quantity and proceed further )

Now start your own calculation. All the best.