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How to Calculate Percentage of Excess Air in Combustion Process

In this post, I want to share how to calculate percentage of excess air flow in combustion process. The known data are:

  1. Fuel composition
  2. Flue gas composition

Before we jump into the calculation, we need to understand about excess air and why we need it.

To ensure complete combustion of the fuel used combustion chambers are supplied with excess air. Excess air increases the amount of oxygen to the combustion and the combustion of fuel. When fuel and oxygen from the air are in perfect balance – the combustion is said to be stoichiometric.

The combustion efficiency increases with increased excess air – until the heat loss in the excess air is larger than the heat provided by more efficient combustion.

Let’s see example below.

In a test on a furnace fired with natural gas (composition 96% methane, 4% nitrogen), the following flue gas analysis was obtained:

  • Carbon dioxide 9%
  • Carbon monoxide 0.3%
  • Oxygen 4.6%
  • Nitrogen 86.1%

All percentages by volume.

Calculate the percentage excess air flow (percentage above stoichiometric).

Excess air calculation in combustion process
Excess air calculation in combustion process


Reaction of combustion:

CH4 + 2O2 → CO2 + 2H2O

The flue gas analysis is reported on dry basis. Any water formed having been condensed out.

Use basis 100 mol of dry flue gas. As the analysis of the flue gas is known, the mols of each element in the flue gas (flow out) can be easily calculated and related to the flow into the system.

Let the quantity of fuel per 100 mol dry flue gas be X.

Quantity of air per 100 mol dry flue gas be Y.

Balance of carbon: mols in fuel = mols in flue gas

0.96 X = 9% (100) + 0.3% (100)

X = 9.3 / 0.96 = 9.69 mol

Balance of nitrogen = mols in fuel + mols in air = mols in flue gas

4% X + 79% Y = 86.1% (100)

0.04 (9.69) + 0.79 Y = 86.1

Y = 108.5 mol

  • Flow of fuel per 100 mol dry flue gas is 9.69 mol.
  • Flow of air per 100 mol dry flue gas is 108.5 mol.

Therefore, flow of CH4 in fuel = 96% (9.69) = 9.30 mol.

Based on stoichiometry of combustion ratio of CH4 and O2 is 1:2. Therefore, stoichiometry mols of oxygen is 2/1 x 9.30 = 18.6 mol.

Air required (stoichiometry) = 100/21 x 18.6 = 88.6 mol.

Therefore, excess air = (air supplied – stoichiometric air)/stoichiometric air = (108.5 – 88.6)/88.6 = 22.50%

Feel confused?

Don’t worry. In this spreadsheet, I attached the calculation method above. I hope you find this simple post useful.