MECH6019 Chap.2 Combustion Stoichiometry and Fuels
Combustion Stoichiometry and Fuels
Stoichiometric combustion provides the theoretical oxygen demand used to define excess air, equivalence ratio and exhaust composition. Start by choosing a fuel basis. One kilomole is convenient for a pure hydrocarbon, while one hundred kilograms turns an ultimate analysis by mass into direct kilogram amounts. Convert each element to kilomoles of atoms before assigning products.
Carbon forms carbon dioxide, hydrogen forms water and sulphur forms sulphur dioxide under the complete-combustion assumption.
Oxygen demand is built element by element. Each kilomole of carbon atoms needs one kilomole of oxygen molecules, hydrogen atoms need one quarter per kilomole of atoms, and sulphur atoms need one. Oxygen already bound in the fuel reduces external demand by half its atomic amount.
Nitrogen and ash do not demand oxygen in the simplified balance. Moisture contributes water to wet products but is not combustible hydrogen. Keeping these roles separate prevents mass percentages from being inserted directly into a molar reaction.
Air supplies oxygen together with nitrogen. The course basis uses the atmospheric oxygen-to-nitrogen relationship when carrying nitrogen into products.
Nitrogen dominates the dry exhaust denominator even when it is treated as inert, so omitting it makes every dry percentage wrong. After theoretical air is known, actual air is obtained using the stated excess-air factor. Unused oxygen then appears in lean complete-combustion products.
Equivalence ratio points in the opposite direction from the air ratio under the convention used here.
Actual air divided by theoretical air above one describes a lean mixture; equivalence ratio is its reciprocal and falls below one. State the convention whenever using the symbol because other contexts may invert it. Do not change the theoretical oxygen demand when excess air is introduced: the fuel chemistry establishes demand, while the operating choice establishes supply.
A product calculation closes three checks.
Carbon atoms in products must equal carbon in fuel. Hydrogen atoms must match after water is counted. Oxygen atoms entering through fuel and air must equal those in carbon dioxide, water, sulphur dioxide and unused oxygen. On the selected wet or dry basis, product mole fractions must sum to one.
These audits are quicker and more reliable than rereading a long arithmetic chain.
When converting to an air-fuel mass ratio, first obtain oxygen moles, then total air moles, then air mass using component molecular weights. Divide by fuel mass only at the end. For a coal analysis, retain ash in the fuel mass denominator even though it is not burned.
The resulting ratio, exhaust composition and equivalence ratio should all describe the same basis and actual-air condition.
What this chapter covers
- 01
Fuel basis
- 02
Elemental atom balance
- 03
Theoretical oxygen
- 04
Air nitrogen
- 05
Excess air
- 06
Equivalence ratio
Worked example · free
Oxygen demand for methane
- 1Balance carbon and hydrogen.
- 1Obtain 2 kmol oxygen.
Key terms
- Stoichiometric air
- The exact air required for complete ideal combustion.
- Excess air
- Air supplied above theoretical demand.
- Equivalence ratio
- Actual fuel-air ratio divided by the stoichiometric ratio.
- Ultimate analysis
- Fuel elemental composition reported by mass.
Combustion Stoichiometry and Fuels FAQ
Why choose a fuel basis first?
A fixed basis converts composition data into definite moles or masses and makes atom conservation checkable. Without it, ratios can be mixed accidentally. One kilomole suits a pure fuel, while one hundred kilograms makes an ultimate analysis easy to translate before oxygen demand is calculated.
Why carry nitrogen through the balance?
Nitrogen dominates the exhaust denominator even when treated as inert. Omitting it makes dry gas percentages and ppm calculations incorrect. Carry the nitrogen accompanying actual air, not only theoretical air, because excess air changes both unused oxygen and the total dry products.
What does a lean mixture imply?
Under the course convention, actual air above theoretical produces an equivalence ratio below one and normally leaves unused oxygen in complete-combustion products. State the convention explicitly because some sources define the reciprocal ratio, and check that the product balance reflects the same air supply.
Exam move
Balance pure fuels first, then progress to ultimate analyses. Check every answer by counting atoms on both sides before calculating percentages. For a pure hydrocarbon, write carbon dioxide and water coefficients directly from carbon and hydrogen, calculate oxygen last, and carry atmospheric nitrogen only after the oxygen demand is secure. Repeat with methane, propane and one unfamiliar formula until the pattern is automatic.
Next use a one-hundred-kilogram ultimate-analysis basis. Convert element masses to atomic kilomoles, list oxygen demand by element, subtract fuel-bound oxygen and keep ash and moisture in their correct roles. Create a separate line for theoretical air, actual air, excess-air percentage and equivalence ratio so their directions cannot be confused.
After obtaining products, audit carbon, hydrogen, sulphur and oxygen atoms and require wet or dry fractions to sum to one. Redo one problem by mass and by kilomoles to verify both routes agree. When practising exhaust composition, keep nitrogen in the denominator and unused oxygen in lean products.
Annotate each step with the physical meaning rather than only algebra: demand comes from fuel atoms, supply comes from actual air, and residual oxygen reflects the difference. In timed revision, use a five-line template: basis, elemental moles, oxygen demand, air supply, products. Do not move to ppm or emission rate until those lines close.
A wrong final percentage is then traceable to one section rather than buried in a long calculation.
