MECH3610 · Advanced Thermofluids
Gas Mixtures & Combustion
The course closes with the thermodynamics of non-reacting gas mixtures and then combustion (CLO3). This chapter covers mass and mole fractions, Dalton's and Amagat's laws and mixture properties, then balanced stoichiometric reactions, the air-fuel ratio and equivalence ratio, enthalpy of formation and heating value, and the adiabatic flame temperature from a reactor energy balance. This block is examined in part of the 30% open-book final exam; standard combustion canon is stated plainly.
What this chapter covers
- 01Mass fractions mf_i = m_i/m and mole fractions y_i = N_i/N for a mixture
- 02Dalton's law of additive pressures and Amagat's law of additive volumes
- 03Apparent molar mass M_m = sum(y_i*M_i) and mixture gas constant R_m = R_u/M_m
- 04Mixture properties as mass- or mole-weighted sums (u, h, c_p, c_v, s)
- 05Balancing combustion with air (3.76 mol N2 per mol O2) and the products of complete combustion
- 06Air-fuel ratio AF = m_air/m_fuel, stoichiometric (theoretical) air, and equivalence ratio phi
- 07Excess air / percent theoretical air; lean (phi < 1) vs rich (phi > 1) mixtures
- 08Enthalpy of formation, heating value, and the adiabatic flame temperature from an energy balance
Stoichiometric air-fuel ratio for methane, then equivalence ratio of a lean mixture
- +1Balanced with theoretical air: CH4 + 2(O2 + 3.76 N2) -> CO2 + 2 H2O + 7.52 N2. (One carbon needs one CO2, four hydrogens need two H2O, which fixes 2 mol O2, carrying 2*3.76 = 7.52 mol N2.)
- +1Air per mole of fuel = 2*(1 + 3.76) = 2*4.76 = 9.52 mol air per mol CH4.
- +1Mass basis: AF_stoich = (9.52*28.97)/(1*16.04) = 275.8/16.04 = 17.2 (kg air per kg fuel).
- +1The equivalence ratio compares stoichiometric to actual AF: phi = AF_stoich/AF_actual = 17.2/22 = 0.78, so phi < 1 means a lean (excess-air) mixture. Percent excess air = (22 - 17.2)/17.2 * 100 = 28%.
Key terms
- Mole fraction (y_i)
- y_i = N_i/N, the fraction of total moles that is species i. Mole fractions weight mole-based mixture properties and, by Dalton's law, equal the partial-pressure fraction p_i/p.
- Apparent molar mass (M_m)
- M_m = sum(y_i*M_i), the mole-weighted average molar mass of a mixture; it gives the mixture gas constant R_m = R_u/M_m used in the ideal-gas law for the mixture.
- Air-fuel ratio (AF)
- AF = m_air/m_fuel, the mass of air supplied per unit mass of fuel. The stoichiometric value uses exactly the theoretical air for complete combustion (about 17.2 for methane).
- Equivalence ratio (phi)
- phi = AF_stoich/AF_actual (equivalently the fuel-air ratio over its stoichiometric value). phi < 1 is lean (excess air), phi = 1 is stoichiometric, phi > 1 is rich (excess fuel).
- Enthalpy of formation
- The enthalpy change to form one mole of a compound from its elements in their reference states; it lets a reacting-flow energy balance account for the chemical energy released or absorbed.
- Adiabatic flame temperature
- The product temperature when all the reaction's energy heats the products with no heat loss or work (Q = 0); found by iterating the product enthalpy in a steady-flow energy balance. Excess air lowers it.
Gas Mixtures & Combustion FAQ
Why do I carry 3.76 mol of nitrogen per mol of oxygen?
Dry air is about 21% O2 and 79% N2 by mole, and 79/21 = 3.76. So every mole of O2 drawn from air brings 3.76 mol of N2 along. Nitrogen is largely inert in the reaction but must appear on both sides of the balance because it absorbs energy and lowers the adiabatic flame temperature.
What is the difference between air-fuel ratio and equivalence ratio?
The air-fuel ratio AF = m_air/m_fuel is the actual mass ratio supplied. The equivalence ratio phi = AF_stoich/AF_actual normalises it against the stoichiometric requirement, so phi tells you at a glance whether the mixture is lean (phi < 1, excess air), stoichiometric (phi = 1) or rich (phi > 1, excess fuel), independent of the fuel.
How is the adiabatic flame temperature found?
Set Q = 0 and no work in the steady-flow energy balance, so the total enthalpy of the products (including their enthalpies of formation and sensible enthalpies) equals that of the reactants. Because the product sensible enthalpies depend on the unknown temperature, you iterate: guess a temperature, evaluate product enthalpies, and adjust until the balance closes. More excess air lowers the result.
How is combustion examined in MECH3610?
It forms part of the 30% open-book final exam (CLO3), typically a stoichiometric balance with an air-fuel-ratio or equivalence-ratio calculation and a first-law energy release or adiabatic-flame-temperature step. The equation sheet and property tables are permitted, so marks favour setting up the balance correctly. Confirm the exam coverage on Moodle.
Exam move
Practise balancing combustion reactions with theoretical air until the 3.76-N2 bookkeeping is automatic, then convert between mole and mass bases to get the air-fuel ratio. Fix the direction of the equivalence ratio in your head — phi < 1 lean, phi > 1 rich — and be able to convert between phi, percent theoretical air and percent excess air. For energy release, keep the reactants-minus-products enthalpy balance (with enthalpies of formation) and the adiabatic-flame-temperature iteration clear and separate. Since this is the last block before the open-book final, rehearse selecting the right relation from the sheet under time pressure. Confirm the exam format on Moodle.
Working through Gas Mixtures & Combustion in MECH3610? Sia is AskSia’s AI Engineering tutor — ask any MECH3610 Gas Mixtures & Combustion question and get a clear, step-by-step explanation grounded in how MECH3610 is taught and assessed. Read this chapter free, then take your hardest questions to Sia.
