Stoichiometric Air-Fuel Ratio
Calculate the stoichiometric air-fuel ratio (AFR) by mass of a hydrocarbon C_xH_y, AFR = (x + y/4) × 137.93 ÷ M, from the number of carbon atoms x, hydrogen atoms y and the fuel molar mass M (g/mol). The result (kg air per kg fuel) is the exact amount of air needed for complete combustion, with no leftover air or fuel. Methane (CH₄) has AFR ≈ 17.2; gasoline ≈ 14.7. The 137.93 constant comes from the air mass per mole of O₂ (32 ÷ 0.232). It is the base parameter of combustion control and mixture in engines and burners. Enter x, y and the fuel molar mass.
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Relação ar-combustível estequiométrica
Para queimar completamente um combustível, é preciso a quantidade exata de ar — nem mais, nem menos. Essa proporção é a relação ar-combustível (AFR) estequiométrica. Para um hidrocarboneto genérico C_xH_y, a reação completa é C_xH_y + (x + y/4)·O₂ → x·CO₂ + (y/2)·H₂O, ou seja, cada mol de combustível precisa de (x + y/4) mols de oxigênio. Como o ar tem só 21% de O₂ (em volume) e 23,2% em massa, calcula-se a massa de ar necessária e divide-se pela massa do combustível: AFR = (x + y/4) × 137,93 ÷ M, onde M é a massa molar do combustível e 137,93 = 32 ÷ 0,232 é a massa de ar por mol de O₂. O resultado é kg de ar por kg de combustível. Valores conhecidos: metano (CH₄): AFR ≈ 17,2; gasolina (~C₈H₁₈): ≈ 14,7; etanol: ≈ 9,0 (menos ar, por já conter oxigênio); hidrogênio: ≈ 34,3. A AFR estequiométrica é fundamental no controle de combustão: motores a gasolina modernos usam sensores lambda para manter a mistura próxima de 14,7:1 (ideal para catalisadores); queimadores industriais trabalham com algum excesso de ar (AFR maior) para garantir queima completa. Misturas com menos ar que o estequiométrico (ricas) geram CO e fuligem; com mais ar (pobres), desperdiçam energia. Informe x, y e a massa molar do combustível.
Related Tools
Actual Air-Fuel Ratio
Calculate the actual air-fuel ratio, AFR_actual = AFR_stoichiometric × (1 + excess air ÷ 100), from the stoichiometric air-fuel ratio and the excess air (%). The result (kg air per kg fuel) is the amount of air actually supplied in practice, always greater than stoichiometric, because real combustion needs excess air to ensure complete burning (the mixture is never perfect). This value sizes the air supply (fans), the exhaust gas flow and influences the flame temperature and efficiency. Enter the stoichiometric air-fuel ratio and the excess air.
Excess Air (from Flue Gas)
Calculate the excess air of a combustion from the oxygen in dry flue gas, EA = O₂ ÷ (20.9 − O₂) × 100%, from the measured O₂ percentage in the stack. The result, in %, shows how much air was supplied beyond stoichiometric — measured by the leftover oxygen in the exhaust gases. Some excess air (10-30%) is needed to ensure complete combustion (avoid CO and soot), but too much wastes energy heating useless air that leaves hot through the stack. Gas analyzers measure O₂ and compute the excess air to optimize combustion efficiency. Enter the O₂ percentage in the gases.
Theoretical Combustion Air Volume
Calculate the theoretical air volume needed for complete combustion of a hydrocarbon C_xH_y, V_air = (x + y/4) × 22.4 ÷ (0.21 × M), from the carbon x, hydrogen y atoms and the molar mass M. The result, in Nm³ of air per kg of fuel (at normal conditions), is the stoichiometric air — the basis for sizing fans, burners and combustion air systems. The factor 22.4 is the ideal gas molar volume (L/mol at STP) and 0.21 the volume fraction of O₂ in air. Methane needs ~13.3 Nm³/kg. Multiplied by the excess air, it gives the actual air supplied. Enter x, y and the molar mass.
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