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How to Calculate Excess Air in a Boiler

Combustion requires both oxygen and fuel. If the amount of oxygen present is greater than what the fuel needs to burn, the surplus is called excess air. For a boiler, a certain amount of excess air is desirable to maximize the boiler's efficiency. Too much excess air, however, can actually decrease efficiency because the surplus air carries away heat in the exhaust. The ideal amount of excess air is determined empirically, although you can calculate the amount of excess air you are using based on the type of fuel.

It's important to use the right amount of excess air.

Step 1

Write out a chemical equation for the combustion reaction you expect will take place and balance it. If you need some review on balancing chemical equations, refer to the first link under the Resources section; alternatively, you can use the second link under the Resources section to balance your reaction equation for you automatically. For methane, for example, the reaction is as follows:

CH4 + 2(O2 + 3.76 N2) ----> CO2 + 2 H2O + 7.52 N2

Notice that you include nitrogen in this equation even though it's not a direct participant in the chemical reaction. The main reason you include nitrogen here is because air is largely nitrogen by volume. For every molecule of oxygen you feed into the combustion chamber, you get 3.76 molecules of nitrogen along for the ride. Also notice that with this as with any other combustion reaction equation, you have two things on the left -- fuel and air. On the right, you will always have carbon dioxide, water and nitrogen.

Step 2

Multiply the number in front of the parentheses on the left of the equation by 4.76, then multiply it by 29 kg per kmol. In the example, for instance, you have 2 kmol of air, so you multiply 4.76 by 29 to get 276.08 kg of air.

Step 3

Multiply the number in front of the fuel on the left side of the equation by its molecular weight in kg per kmol. To find the molecular weight, enter the molecular formula of your fuel into the calculator at the third link under the Resources section below. The molecular weight of methane, for example, is 16.04 g per mole or 16.04 kg per kmol, which rounds to 16.

Step 4

Divide the answer from step 2 by the answer from step 3. This gives you the stoichiometric air-fuel ratio. In the example, 276 / 16 = 17.3 kg of air per kg of fuel.

Step 5

Multiply the stoichiometric air-fuel ratio by the number of kilograms of fuel you burn. If you burned 4 kg of methane, for example, you would multiply 17.3 by 4 to get 69.2 kg of air required.

Step 6

Subtract this result from the amount of air you actually supplied. If you supplied 75 kg of air, for example, you would subtract 69.2 to get 5.8 excess.

Step 7

Divide this answer by the amount of air required and multiply by 100 to convert to a percent. In the example, you would divide 5.8 by 69.2 to get 8.4 percent excess air.

About the Author

Based in San Diego, John Brennan has been writing about science and the environment since 2006. His articles have appeared in "Plenty," "San Diego Reader," "Santa Barbara Independent" and "East Bay Monthly." Brennan holds a Bachelor of Science in biology from the University of California, San Diego.

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