Calculation of the Excess Air of Combustion Process

The amount of air needed to completely burn a certain amount of fuel can be calculated theoretically using the basic principles of stoichiometry (see following article). In other words, if every fuel molecule precisely contact with oxygen in the air and reacted, then the entire fuel must be burned, and there will not be a certain amount of excess air at the exhaust. But in reality, it is impossible. Fuel molecules can not 100% met directly with the oxygen. It takes some excess air to make sure all the fuel molecules can burn completely. This is what we know as excess air (read the following article).

What is the right amount of excess air for a burning process?

Determining the amount of excess air in the boiler burning process depends on several main factors such as fuel type, boiler design, burner design, and boiler load. Generally, coal fired boilers use excess air as much as 15% to 30%. For boilers with gas or petroleum as its fuel, requires less excess amount of water. Gas-fired boilers require excess air of 5% to 10%, while petroleum-fired boilers require excess air of 3% to 15%. This condition indicates that the gas and liquid phase fuels more easily mix and react with oxygen, compared to solid phase fuel.

How much the boiler load has a great impact on excess air. The design of the diagonal cross-section of the boiler’s combustion chamber should be able to bear the flow of gas flow when the boiler is in full load. The opposite condition occurs when the boiler load is lower, where the flow of gas decreases so that mixing the fuel with air becomes more difficult. Therefore, when the boiler is below full load, the amount of excess air required increases to ensure a complete combustion. In coal-fired boilers for example, at 50% load, takes two times excess air than when the load is 100%.


Although the excess air is important to ensure complete combustion, excess air has a negative impact on boiler efficiency. The higher amount of excess air will make the heat energy wasted following the exhaust gas. Therefore, in terms of efficiency, the amount of excess air should be kept as low as possible.

To keep the excess air stay at the optimum value, modern boilers equipped with sensors of amount of oxygen and carbon monoxide on the boiler exhaust side. Both of these parameters can be use to keep the excess air amount stay at the optimum level throughout the boiler operating time.

Excess Air Calculations

Let’s use the example of coal data at the previous article as follows:

Which the stoichiometric combustion reaction equation is as follows:

CH0,74O0,061N0,018S0,026 + 1,211(O2 + 3,762N2) → CO2 + 0,37H2O + 0,026SO2 + 4,565N2

Furthermore, if determined boiler using excess air of 15%, then the chemical reaction of combustion becomes as follows:

CH0,74O0,061N0,018S0,026 + 1,393(O2 + 3,762N2) → CO2 + 0,37H2O + 0,026SO2 + 5,24N2 + 0,212O2

From the above reaction we can calculate the percentage of excess oxygen in boiler flue gas:

O2 excess = \dfrac {0,212}{1+0,37+0,26+5,24+0,212}\times 100\%

O2 excess = 3,096%

While the air-fuel ratio corrected to:

AFR = \dfrac {1,393\left( 32+3,762\times 28\right)}{12+1\times 0,74+16\times 0,061+14\times 0,018+32\times 0,026}

AFR = 12,926

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Stoichiometric Calculation of Coal Burning Process

Generally, coal is composed of several important chemical elements. They are carbon (C), hydrogen (H), sulfur (S), oxygen (O), and some other elements. These elements interconnect chemically to form new hydrocarbon compounds. The hydrocarbon chemical bonds stored energy. When the bond is cut off through the combustion process, the stored energy will be released into the environment. If we write into a chemical reaction, then coal combustion will look like the below:

Coal + O2 → Product + Heat Energy

The carbon burning process, produced the most heat energy for the entire process. Carbon burning also allows the formation of carbon monoxide if incomplete combustion occur. The content of hydrogen and sulfur in coal also contributes a small part of the heat energy when the combustion process takes place.

To facilitate our understanding, let us consider the following example calculations.

For example, a coal content analysis mention that the coal has the following composition:

  • Carbon: 73%
  • Hydrogen: 4.5%
  • Oxygen: 5.9%
  • Nitrogen: 1.5%
  • Sulfur: 5%
  • Water: 2.1%
  • Ash: 8%

From the above data, we can determine the mol/100 gram value of each coal component, as well as we specify the mol/mol of carbon to be able to determine the chemical structure of the coal molecule.

So the chemical formula of coal molecule is:


With a slight rounding up the air composition, which is 79% nitrogen and 21% oxygen, so for one mole of oxygen, there are 3,762 nitrogen. With that data let us make a perfect burning stoichiometry reaction from the related coal.

CH0,74O0,061N0,018S0,026 + 1,211 (O2 + 3,762N2) → CO2 + 0,37H2O + 0,026SO2 + 4,565N2

From the equation of the perfect combustion above, then we can determine the air / fuel ratio, so we know how much air is needed to burn 1 kg of coal.

AFR = \dfrac {1,211\left( 32+3,762\times 28\right)}{12+1\times 0,74+16\times 0,061+14\times 0,018+32\times 0,026}

AFR = 11,237

Air Molecular Weight Calculation

The molecular weight (molar mass) of a substance is the mass of one mole of the substance, which can be calculated based on the molar mass of the constituent atoms. Dry air is composed of two important chemical elements: 78% nitrogen and 21% oxygen, and about 1% are mixtures of carbon dioxide, neon, helium, methane, krypton, hydrogen, and xenon.

Before further calculating the dry air molar mass, we must understand the notion of the mole. Mol is a unit of measurement for the number of substances, which denotes the number of representative particles equal to the number of atoms in 12 grams of carbon-12 atoms (12 C). The number of particles is expressed in Avogadro Numbers which is equivalent to 6.022140857 × 10 23 particles / mol. One thing to note is that one mole of nitrogen, has a size of the number of constituent atoms equal to one mole of oxygen, one mole of carbon dioxide, and one mole of other substances present in the world.

Now let’s calculate how much dry air molar mass goes through the table below:

The molar masses of each of the above air constituent components are calculated according to the standard data from the periodic table of chemical elements. Nitrogen for example, with the chemical formula N2, has an atomic molar mass of 14.007 g/mol. Then the molar mass of 2 nitrogen atoms (which make up N2) is 28.014 g/mol. The same calculation is also performed for other dry air constituent elements.

Furthermore, the molar mass of each constituent element is multiplied by the percentage value of the element content in the dry air. From this we get the elemental element molar value. After the elemental molar mass of the portion is summed, we obtain a molar mass for air of 28.9647 gram/mol.