Combustion Calculator

Stoichiometric Air-Fuel Ratio (mass basis)

AFR_stoich = (a / 0.21) · 28.97 / M_fuel | with | a = x + y/4

Variables: AFR_stoich = stoichiometric air-fuel ratio (kg air per kg fuel), a = stoichiometric moles of O₂ required per mole of fuel (dimensionless; 2 for CH₄, 5 for C₃H₈, 12.5 for C₈H₁₈, 0.5 for H₂), x = carbon atoms per fuel molecule (dimensionless; 1, 3, 8, 0), y = hydrogen atoms per fuel molecule (dimensionless; 4, 8, 18, 2), 0.21 = mole fraction of O₂ in dry air (dimensionless), 28.97 = molar mass of dry air (kg/kmol), M_fuel = molar mass of the fuel (kg/kmol; 16.04, 44.10, 114.23, 2.016 respectively). Gives 17.20, 15.64, 15.10 and 34.21 kg air per kg fuel for methane, propane, octane and hydrogen. Dry air is modelled as 21% O₂ / 79% N₂ by mole with argon lumped into the N₂ and no humidity. Source: Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 15 (Chemical Reactions) — Fuels and Combustion, Theoretical and Actual Combustion Processes

Actual Air-Fuel Ratio and Percent Excess Air from Equivalence Ratio

AFR_actual = AFR_stoich / φ = ((a / φ) / 0.21) · 28.97 / M_fuel | excess air (%) = (1/φ − 1) · 100

Variables: φ = equivalence ratio (dimensionless; φ = 1 stoichiometric, φ < 1 lean with genuine excess air, φ > 1 rich with genuinely deficient air), AFR_actual = actual air-fuel ratio (kg air per kg fuel), AFR_stoich = stoichiometric air-fuel ratio (kg air per kg fuel), a/φ = moles of O₂ actually supplied per mole of fuel (dimensionless), a = stoichiometric moles of O₂ per mole of fuel (dimensionless), 0.21 = mole fraction of O₂ in dry air (dimensionless), 28.97 = molar mass of dry air (kg/kmol), M_fuel = fuel molar mass (kg/kmol). Percent excess air is POSITIVE for lean mixtures (φ < 1, e.g. +25% at φ = 0.8) and NEGATIVE for rich mixtures (φ > 1, e.g. −20% at φ = 1.25); the calculator shows a rich mixture's negative excess air as a positive deficiency, e.g. 20.0 % deficient at φ = 1.25. Source: Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 15 (Chemical Reactions) — Theoretical and Actual Combustion Processes; Turns, An Introduction to Combustion, 3rd ed., Ch. 2 (Combustion and Thermochemistry)

Product Balance for Lean and Stoichiometric Mixtures (φ ≤ 1), per mole of fuel

C_xH_y + (a/φ)·(O₂ + 3.76·N₂) → x·CO₂ + (y/2)·H₂O + a·(1/φ − 1)·O₂ + 3.76·(a/φ)·N₂ | [valid for φ ≤ 1]

Variables: x = carbon atoms per fuel molecule (dimensionless; 1, 3, 8, 0 for CH₄, C₃H₈, C₈H₁₈, H₂), y = hydrogen atoms per fuel molecule (dimensionless; 4, 8, 18, 2, so the H₂O yield y/2 is 2, 4, 9, 1), a = x + y/4 = stoichiometric moles of O₂ per mole of fuel (dimensionless), φ = equivalence ratio (dimensionless), 3.76 = 0.79/0.21 = moles of N₂ carried with each mole of O₂ in dry air (dimensionless; the tool applies the exact ratio 3.762). All product quantities are moles per mole of fuel (dimensionless molar ratios), which is the basis the calculator reports. This balance holds for φ ≤ 1; for φ > 1 see the rich-mixture balance below. Source: Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 15 (Chemical Reactions) — combustion with excess air and incomplete combustion; Turns, An Introduction to Combustion, 3rd ed., Ch. 2 (Combustion and Thermochemistry)

Product Balance for Rich Mixtures (φ > 1), per mole of fuel

C_xH_y + (a/φ)·(O₂ + 3.76·N₂) → n_CO₂·CO₂ + n_CO·CO + n_H₂O·H₂O + n_H₂·H₂ + 3.76·(a/φ)·N₂ | n_CO₂ + n_CO = x, n_H₂O + n_H₂ = y/2, 2·n_CO₂ + n_CO + n_H₂O = 2a/φ, n_CO·n_H₂O = K·n_CO₂·n_H₂ with K = 3.5

Variables: n_CO₂, n_CO, n_H₂O, n_H₂ = moles of each product per mole of fuel (dimensionless), x, y = carbon and hydrogen atoms per fuel molecule (dimensionless), a = x + y/4 = stoichiometric moles of O₂ per mole of fuel (dimensionless), φ = equivalence ratio (dimensionless, above 1), 3.76 = 0.79/0.21 = moles of N₂ per mole of O₂ in dry air (the tool applies 3.762), K = equilibrium constant of the water–gas shift CO₂ + H₂ ⇌ CO + H₂O, taken as 3.5, the value Heywood uses for fuel-rich engine exhaust (the NIST-JANAF tables give K = 3.5 at about 1720 K). The three element balances (C, H, O) and the water–gas shift fix the four amounts: with B = 2a/φ − x, n_CO₂ is the non-negative root of (K − 1)·n² + [K·(y/2 − B) + x + B]·n − x·B = 0, and at φ = 1 this gives complete combustion. No O₂ is left, and no soot or unburned fuel is assumed; when 2a/φ < x the air cannot turn all the carbon into CO and the calculator shows no products. Source: Heywood, Internal Combustion Engine Fundamentals (McGraw-Hill, 1988), for K = 3.5 in fuel-rich exhaust; NIST-JANAF Thermochemical Tables, 4th ed. (Chase, 1998), tables C-093 (CO), C-095 (CO₂), H-050 (H₂) and H-064 (H₂O), whose Gibbs energies of formation give K = 3.5 at 1723 K; Turns, An Introduction to Combustion, 3rd ed., Ch. 2 (water–gas shift equilibrium)

Adiabatic Flame Temperature — Energy Balance (lean and stoichiometric mixtures, up to 3000 K)

Σ n_i · [H_i(T_ad) − H_i(298.15 K)] = LHV · M_fuel | H_i(T) − H_i(298.15 K) from the NIST-JANAF tables | n_i from the complete-combustion balance (φ ≤ 1)

Variables: T_ad = adiabatic flame temperature (K; also displayed in °C as T_ad − 273.15), n_i = moles of CO₂, H₂O, O₂ and N₂ per mole of fuel from the lean product balance above (dimensionless), H_i(T) − H_i(298.15 K) = each product's enthalpy above its value at 298.15 K (J/mol) from the NIST-JANAF Thermochemical Tables, tables C-095 (CO₂), H-064 (H₂O, gas), N-023 (N₂) and O-029 (O₂), between tabulated temperatures by the cubic polynomial that matches the tabulated enthalpy and heat capacity at both ends, LHV = the fuel's lower heating value as listed (MJ/kg), M_fuel = fuel molar mass (kg/kmol); the reactants enter at 298.15 K. The balance assumes complete combustion with no dissociation, so real flames near stoichiometric, where some CO₂ and H₂O dissociate and take up heat, are cooler than it gives. The tool solves for T_ad by bisection up to 3000 K, where the excerpt of the NIST-JANAF tables it uses ends (the tables themselves run to 6000 K; none of its fuels comes near 3000 K). Rich mixtures (φ > 1) get no estimate, since their temperature depends on how the products split and dissociate at flame temperature. Methane gives about 1481 K at φ = 0.5 and about 2328 K at φ = 1. Source: NIST-JANAF Thermochemical Tables, 4th ed. (M. W. Chase, J. Phys. Chem. Ref. Data Monograph 9, 1998; janaf.nist.gov); Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 15 (Adiabatic Flame Temperature); Turns, An Introduction to Combustion, 3rd ed., Ch. 2