8.2 · Calculate theoretical air and excess-air fractions

University of Alberta MEC E 340: Applied Thermodynamics

Chemical Reactions and Combustion

A stoichiometric method for introductory combustion calculations

What you will learn

• Set a fuel basis and balance a complete-combustion reaction.
• Calculate the theoretical oxygen and dry air required for complete combustion.
• Determine actual air supplied and express excess air as a fraction or percentage.
• Check air–fuel calculations using atom balances and clear units.

1. Set the basis and balance complete combustion

• Choose and state a fuel basis before calculating.
• Balance carbon and hydrogen into $\mathrm{CO_2}$ and $\mathrm{H_2O}$, then balance oxygen.
• Use the stated dry-air composition; do not count air nitrogen as oxygen.

2. Convert oxygen demand to theoretical air

• Theoretical air is based on the oxygen needed for complete combustion.
• The air amount depends on the assumed oxygen fraction in air.
• A molar air–fuel ratio and a mass air–fuel ratio are different quantities.

3. Measure actual air and excess air

• Compare actual and theoretical air on identical bases.
• Excess-air fraction uses theoretical air in the denominator.
• State whether a result is a fraction, a percentage, or percent theoretical air.

4. A reliable calculation and its checks

• Balance atoms before comparing actual with theoretical air.
• Use one consistent fuel basis and one consistent air basis.
• Do not infer product temperature or combustion performance from excess air alone.

Worked example

Theoretical air for methane

1. Choose the basis
   Use 1 kmol of fuel, so all calculated amounts are per kmol of methane. Assume complete combustion and the stated dry-air composition.
   $$n_{\mathrm{CH_4}}=1\ \mathrm{kmol}$$
2. Balance the reaction
   One carbon atom forms one carbon dioxide molecule, and four hydrogen atoms form two water molecules. These products need four oxygen atoms, or two oxygen molecules.
   $$\mathrm{CH_4+2O_2\rightarrow CO_2+2H_2O}$$
3. Find theoretical air
   For each kmol of oxygen, the dry air contains $79/21$ kmol of nitrogen. Divide the oxygen requirement by the oxygen mole fraction to obtain total theoretical air.
   $$n_{\mathrm{air,th}}=\frac{2}{0.21}=9.52\ \mathrm{kmol\ air/kmol\ CH_4}$$

Worked example

Excess air for propane

1. Balance complete combustion
   On a 1 kmol propane basis, three carbon atoms form three carbon dioxide molecules and eight hydrogen atoms form four water molecules. The products require ten oxygen atoms, or five oxygen molecules.
   $$\mathrm{C_3H_8+5O_2\rightarrow 3CO_2+4H_2O}$$
2. Calculate theoretical air
   The given dry-air oxygen fraction converts the 5 kmol oxygen requirement to the theoretical air supply.
   $$n_{\mathrm{air,th}}=\frac{5}{0.21}=23.81\ \mathrm{kmol\ air/kmol\ C_3H_8}$$
3. Compare actual and theoretical air
   Both air amounts are on the same basis. Subtract the theoretical amount from the actual amount and divide by theoretical air; multiply by 100 to report a percentage.
   $$f_{\mathrm{excess}}=\frac{30.0-23.81}{23.81}=0.260=26.0\%$$

Worked example

Convert percent theoretical air to excess-air fraction

1. Interpret the percentage
   115% theoretical means actual air is 1.15 times the theoretical amount. The excess is the amount above 100% theoretical, not 115% of theoretical.
   $$\frac{n_{\mathrm{air,act}}}{n_{\mathrm{air,th}}}=1.15$$
2. Find actual air
   Multiply the given theoretical air by the actual-to-theoretical ratio, keeping the same fuel basis.
   $$n_{\mathrm{air,act}}=1.15(8.00)=9.20\ \mathrm{kmol\ air/kmol\ fuel}$$
3. Calculate excess air
   The actual amount is 1.20 kmol above theoretical. Dividing that difference by theoretical air gives the dimensionless excess-air fraction.
   $$f_{\mathrm{excess}}=\frac{9.20-8.00}{8.00}=0.150=15.0\%$$

Common mistakes and how to avoid them

Lesson summary

• Balance complete combustion to find theoretical oxygen demand.
• Convert oxygen demand to air using the stated oxygen fraction in dry air.
• Calculate excess-air fraction as actual air minus theoretical air, divided by theoretical air.
• Percent theoretical air equals 100% plus excess-air percentage when actual air exceeds theoretical.

Check your understanding

Key terms

Theoretical air

The amount of air that supplies the stoichiometric oxygen required for complete combustion under stated assumptions.

Excess-air fraction

The difference between actual and theoretical air divided by theoretical air.

Percent theoretical air

Actual air expressed as a percentage of theoretical air.

Complete combustion

The idealized model in which fuel carbon forms carbon dioxide and fuel hydrogen forms water.

Continue through MEC E 340

View the complete MEC E 340 University of Alberta MEC E 340: Applied Thermodynamics curriculum and lessons

• 8.1 · Balance complete combustion with air
• 8.3 · Use formation enthalpies in reaction energy balances
• 8.4 · Distinguish higher and lower heating values
• 8.5 · Estimate adiabatic flame temperature under stated assumptions
• 1.1 · Define the system, working fluid, state, and process
• 1.2 · Read property tables and identify phase regions

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