8.1 · Balance complete combustion with air

University of Alberta MEC E 340: Applied Thermodynamics

Chemical Reactions and Combustion

MEC E 340 Applied Thermodynamics — Study topic 8.1

What you will learn

• Represent dry air using an oxygen–nitrogen mole ratio.
• Balance complete combustion by conserving carbon, hydrogen, and oxygen atoms.
• Determine the theoretical oxygen and air required for a fuel.
• Include excess air and identify the remaining oxygen in the products.

1. Set the basis and represent the air

• Choose a basis, commonly one kilomole of fuel.
• For the dry-air model here, nitrogen accompanying oxygen is $3.76$ times the oxygen amount.
• Balance atoms before using any energy calculation.

2. Balance complete combustion in a reliable order

• Set carbon dioxide from the carbon balance.
• Set water from the hydrogen balance.
• Use the oxygen balance to find oxygen supplied by air, accounting for oxygen already in the fuel.
• Carry nitrogen from the air into the products unchanged.

3. Add excess air and check the result

• Actual oxygen with excess air equals theoretical oxygen times $1+e$.
• Unconsumed oxygen appears among the products.
• Verify every element balance and keep the basis consistent.

4. Scope and notation

• The reaction balance is a conservation calculation, not a prediction of flame behaviour.
• State the excess-air assumption and the basis for any reported product composition.

Worked example

Methane with theoretical air

1. Set the basis
   Use one kilomole of fuel. The control-volume reaction is assumed complete, and air is represented by oxygen with 3.76 kilomoles of nitrogen per kilomole of oxygen.
   $$1\ \mathrm{kmol\ CH_4}$$
2. Balance the atoms
   One carbon atom per fuel molecule gives one carbon dioxide. Four hydrogen atoms give two water molecules. The products contain four oxygen atoms, so two oxygen molecules are required. The accompanying nitrogen is 3.76 times the oxygen amount.
   $$\mathrm{CH_4}+2(\mathrm{O_2}+3.76\,\mathrm{N_2})\rightarrow\mathrm{CO_2}+2\,\mathrm{H_2O}+7.52\,\mathrm{N_2}$$
3. Report air supplied
   The air model counts oxygen and nitrogen together. Thus, two kilomoles of oxygen correspond to 7.52 kilomoles of nitrogen and 9.52 kilomoles of air per kilomole of methane.
   $$n_{\mathrm{air}}=2+7.52=9.52\ \mathrm{kmol}$$

Worked example

Propane with 20% excess air

1. Find theoretical oxygen
   Balance carbon and hydrogen first: one kilomole of propane forms three kilomoles of carbon dioxide and four kilomoles of water. These products require ten oxygen atoms, or five kilomoles of oxygen gas.
   $$n_{\mathrm{O_2,stoich}}=3+\frac{8}{4}=5\ \mathrm{kmol}$$
2. Apply excess air
   Twenty percent excess means supplying 1.20 times the theoretical oxygen. The air nitrogen is based on this actual oxygen supply, not the theoretical amount.
   $$n_{\mathrm{O_2,actual}}=1.20(5)=6\ \mathrm{kmol},\qquad n_{\mathrm{N_2}}=3.76(6)=22.56\ \mathrm{kmol}$$
3. Balance product oxygen
   Carbon dioxide and water use five kilomoles of oxygen gas in total. Of the six supplied, one kilomole remains as oxygen in the products.
   $$\mathrm{C_3H_8}+6\,\mathrm{O_2}+22.56\,\mathrm{N_2}\rightarrow3\,\mathrm{CO_2}+4\,\mathrm{H_2O}+1\,\mathrm{O_2}+22.56\,\mathrm{N_2}$$

Worked example

Ethanol with 30% excess air

1. Find the theoretical requirement
   Two carbon atoms give two carbon dioxide molecules; six hydrogen atoms give three water molecules. These products require seven oxygen atoms. The fuel supplies one oxygen atom, so air must supply six oxygen atoms, or three kilomoles of oxygen gas.
   $$n_{\mathrm{O_2,stoich}}=2+\frac{6}{4}-\frac{1}{2}=3\ \mathrm{kmol}$$
2. Calculate actual air and products
   Thirty percent excess gives 3.90 kilomoles of oxygen. The accompanying nitrogen is 3.76 times that amount. The complete-combustion products still contain two carbon dioxide and three water molecules; the unused oxygen is 0.90 kilomole.
   $$\mathrm{C_2H_6O}+3.90\,\mathrm{O_2}+14.664\,\mathrm{N_2}\rightarrow2\,\mathrm{CO_2}+3\,\mathrm{H_2O}+0.90\,\mathrm{O_2}+14.664\,\mathrm{N_2}$$
3. Find dry product composition
   Exclude water from the dry total. The dry products total 17.564 kilomoles. Divide each dry species amount by this total to obtain its dry-basis mole fraction.
   $$y_{\mathrm{CO_2}}=0.1139,\qquad y_{\mathrm{O_2}}=0.0512,\qquad y_{\mathrm{N_2}}=0.8349$$

Common mistakes and how to avoid them

Lesson summary

• Use one consistent molar basis and state the dry-air model.
• For complete combustion, balance carbon to CO₂, hydrogen to H₂O, then oxygen; carry air nitrogen through unchanged.
• For oxygenated fuel CₓHᵧO_z, theoretical oxygen is x + y/4 − z/2 kmol O₂ per kmol fuel.
• With excess air, increase supplied oxygen and nitrogen accordingly; show unused oxygen among the products.
• Check every element and state whether product composition is wet or dry.

Check your understanding

Key terms

Theoretical air

The amount of air that supplies exactly the oxygen required for the specified complete-combustion products.

Excess air

Air supplied beyond the theoretical amount, commonly stated as a percentage of theoretical air or oxygen.

Complete combustion

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

Dry product basis

A product-composition basis that excludes water from the total amount used to calculate mole fractions.

Continue through MEC E 340

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

• 8.2 · Calculate theoretical air and excess-air fractions
• 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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