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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Volume de ar teórico de combustão
Para dimensionar o suprimento de ar de um queimador, forno ou caldeira, é preciso saber quanto ar (em volume) cada quilo de combustível precisa. O volume de ar teórico (estequiométrico) de um hidrocarboneto C_xH_y é V_ar = (x + y/4) × 22,4 ÷ (0,21 × M), onde (x + y/4) é o número de mols de O₂ por mol de combustível, 22,4 é o volume molar de um gás ideal nas condições normais (L/mol, ou Nm³/kmol), 0,21 é a fração volumétrica de O₂ no ar, e M é a massa molar do combustível. O resultado vem em Nm³ de ar por kg de combustível (Nm³ = metro cúbico normal, nas CNTP). Valores típicos: o metano precisa de ~13,3 Nm³/kg; o óleo combustível, ~10,5 Nm³/kg; o gás natural, similar ao metano. Esse volume é o mínimo teórico; na prática, multiplica-se pelo fator de excesso de ar (1 + EA/100) para obter o ar real a fornecer. Conhecer o volume de ar é essencial para: dimensionar os ventiladores (insufladores de ar de combustão e exaustores de gases), calcular a vazão de gases de exaustão (ar + combustível, que define o tamanho de chaminés, dutos e sistemas de limpeza de gases), avaliar a capacidade de um queimador (quanto combustível pode queimar com o ar disponível), e fazer o balanço de massa da combustão. Um suprimento de ar insuficiente leva a combustão incompleta (CO, fuligem, perigo); ar demais desperdiça energia de ventilação e resfria a chama. Informe x, y e a massa molar.
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.
CO₂ Volume Produced
Calculate the CO₂ volume produced in complete combustion of a hydrocarbon C_xH_y, V_CO₂ = x × 22.4 ÷ M, from the number of carbon atoms x and the fuel molar mass M (g/mol). The result, in Nm³ of CO₂ per kg of fuel, is the carbon dioxide generated by complete burning — information for emission inventories, exhaust system sizing and gas analysis. Each carbon atom in the fuel becomes one CO₂ molecule. Methane produces ~1.4 Nm³/kg. Fuels with more carbon per unit mass (coal, heavy oils) produce more CO₂. Enter x and the fuel molar mass.
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.
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