UNSW Sydney · FACULTY OF ENGINEERING

MECH9761 Chap.3 Combustion Thermochemistry and Equilibrium

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Chapter 3 of 6 · MECH9761

Combustion Thermochemistry and Equilibrium

This chapter builds combustion calculations in a strict order: balance atoms, calculate the stoichiometric air-fuel ratio, classify the actual mixture, then apply energy or equilibrium relations. Formation enthalpy gives chemical-energy release, an adiabatic balance gives flame temperature, and Gibbs free energy plus the water-gas-shift relation closes incomplete-combustion product calculations.

Each later equation inherits the species coefficients from the atom balance, so a chemistry answer is only as reliable as its first carbon, hydrogen and oxygen ledger.

In this chapter

What this chapter covers

  • 01

    Complete hydrocarbon combustion with oxygen and inert nitrogen

  • 02

    Reaction enthalpy from formation enthalpies

  • 03

    Adiabatic flame temperature through an enthalpy balance

  • 04

    Stoichiometric air-fuel ratio for a general hydrocarbon

  • 05

    Lambda and phi classifications for lean and rich mixtures

  • 06

    Dissociation, Gibbs free energy and water-gas-shift equilibrium

Worked example · free

Classify a mixture from air-fuel ratio

Q [4 marks]. The marks shown in this worked check are for study planning, not a published university assessment scheme. A fuel has a stoichiometric air-fuel ratio of 15.0 and operates at an actual ratio of 18.0. Find lambda and phi and classify the mixture. (4 marks)
The plotted methane-air results peak near stoichiometric mixture, then fall on both the lean and rich sides in the course calculation.The plotted methane-air results peak near stoichiometric mixture, then fall on both the lean and rich sides in the course calculation.Equivalence ratioTemperature K0.91.01.12176 K2327 K2291 K
The plotted methane-air results peak near stoichiometric mixture, then fall on both the lean and rich sides in the course calculation.
  • +1Lambda = actual divided by stoichiometric = 18.0/15.0 = 1.20.
  • +1Phi = stoichiometric divided by actual = 15.0/18.0 = 0.833.
  • +1Lambda above one and phi below one both identify a lean mixture.
  • +1Excess air dilutes the released energy across more gas, tending to reduce adiabatic flame temperature from the stoichiometric peak.
Lambda is 1.20 and phi is 0.833, so the mixture is lean. The two reciprocal definitions give the same classification.
Sia tip — Calculate both ratios once. If their classifications disagree, one ratio has been inverted.
Glossary

Key terms

Formation Enthalpy
The enthalpy change assigned to forming a species from its elements in their standard states.
Reaction Enthalpy
Product formation enthalpy minus reactant formation enthalpy for a balanced chemical reaction.
Stoichiometric Mixture
A fuel-air mixture containing exactly the oxygen required for the stated complete-combustion products.
Chemical Equilibrium
A state with no net reaction tendency under the specified thermodynamic conditions.
Gibbs Free Energy
A thermodynamic potential whose minimum at fixed temperature and pressure identifies equilibrium.
Water-Gas Shift
The equilibrium reaction connecting carbon monoxide, water, hydrogen and carbon dioxide.
Air-Fuel Ratio
Air mass divided by fuel mass for a specified reacting mixture.
Air-Excess Ratio
Actual air-fuel ratio divided by its stoichiometric value, commonly denoted lambda.
Equivalence Ratio
Stoichiometric air-fuel ratio divided by actual air-fuel ratio, commonly denoted phi.
Adiabatic Flame Temperature
Product temperature obtained when reaction enthalpy raises sensible enthalpy without boundary heat loss.
Dissociation
High-temperature decomposition that changes equilibrium species and retains chemical energy in products.
FAQ

Combustion Thermochemistry and Equilibrium FAQ

Why must atoms be balanced before the energy equation?

Species coefficients are the mole numbers multiplying formation and sensible enthalpies. If the atom balance is wrong, the energy sum cannot conserve mass and its temperature or heat result is meaningless. Balance carbon, hydrogen and oxygen first, then bring those coefficients into the enthalpy equation.

Why does flame temperature peak near stoichiometric mixture?

Lean mixtures spread released energy across excess air, while rich mixtures lack enough oxygen for complete conversion and retain chemical energy in incomplete products. Under the course methane calculation, both sides lie below the stoichiometric value. The exact curve still depends on fuel, initial state and equilibrium assumptions.

When is the water-gas-shift equation needed?

A rich incomplete-combustion product set with carbon dioxide, carbon monoxide, water and hydrogen has four coefficients but only three elemental balances. The water-gas-shift equilibrium supplies the additional independent relationship needed to solve composition at the specified temperature and pressure.

How are lambda and phi related?

Lambda is actual air-fuel ratio divided by the stoichiometric ratio, while phi is the reciprocal. A lean mixture has lambda above one and phi below one; a rich mixture reverses those inequalities. Computing both from their definitions avoids relying on a memorised label.

Why does reaction enthalpy need a sign convention?

Products minus reactants gives a negative reaction enthalpy for an exothermic combustion reaction under the stated convention. Engineers often quote the positive magnitude of energy released. Writing whether a positive value means heat entering, heat rejected or chemical energy released prevents a correct magnitude from acquiring the wrong physical direction.

Does adiabatic mean the temperature stays fixed?

Adiabatic means no heat crosses the chosen boundary. It does not mean isothermal. In the flame calculation, the decrease in chemical enthalpy raises product sensible enthalpy, so temperature changes substantially. The balance must use consistent reference states for reactants and products.

Why can rich products require equilibrium?

Oxygen shortage makes a complete-product set of only carbon dioxide and water untenable. Adding carbon monoxide and hydrogen creates more unknown coefficients than elemental balances. An equilibrium relation such as water-gas shift provides the additional constraint at the specified temperature and pressure.

Study strategy

Exam move

Use the same worksheet order every time: write the fuel formula, calculate oxygen demand, carry nitrogen, balance products, convert molar amounts to mass and find the stoichiometric ratio. Only then calculate lambda or phi. For flame temperature, write the adiabatic enthalpy equality and keep formation and sensible terms separate.

For rich products, count unknown coefficients and independent balances before adding the equilibrium constant. Finish by interpreting oxygen availability and likely composition. Rebalance the same fuel under lean, stoichiometric and rich labels and explain which product assumptions change. Check every species coefficient by recounting atoms on both sides.

For enthalpy work, keep a two-column ledger of formation and sensible terms. After solving, use the sign and mixture classification to predict the qualitative temperature and composition direction before accepting the number.

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