Theoretical Maximum CO₂
Calculate the theoretical maximum CO₂ percentage in dry flue gas from complete combustion of a hydrocarbon C_xH_y, CO₂max = x ÷ (x + (x + y/4) × 3.762) × 100%, from the carbon x and hydrogen y atoms. The result, in %, is the CO₂ obtained with perfect stoichiometric combustion (no excess air) — the reference value of gas analyzers. Methane has CO₂max ≈ 11.7%; coal, ~18-20%. Comparing the measured CO₂ with the theoretical maximum indicates the excess air: the lower the measured CO₂ relative to the maximum, the more excess air diluting the gases. Enter x and y.
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Theoretical maximum CO₂
When a hydrocarbon C_xH_y burns completely with the exact amount of air (stoichiometric, no excess), the dry flue gas holds the highest possible percentage of CO₂. This theoretical maximum CO₂ (CO₂max) is a characteristic property of each fuel: CO₂max = x ÷ (x + (x + y/4) × 3.762) × 100%, where x is the number of carbon atoms, (x + y/4) is the stoichiometric O₂, and 3.762 (= 79/21) turns that O₂ into the nitrogen that comes with it in air. The denominator counts every molecule in the dry gas: the CO₂ (x molecules) plus the N₂ carried in by the air. Typical values: methane (CH₄): CO₂max ≈ 11.7%; propane: ~13.8%; fuel oils: ~15-16%; coal: ~18-20%; pure carbon: ~21%. Fuels with more carbon and less hydrogen have a higher CO₂max (the hydrogen turns into water, which leaves the dry gas). CO₂max is the reference used by combustion analysers. Since in practice there is always excess air, the measured CO₂ always comes out lower than the theoretical maximum — the extra air dilutes the gas. That very difference lets the excess air be worked out: by comparing the measured CO₂ with CO₂max (excess air ≈ CO₂max/CO₂measured − 1, approximately), or by measuring the residual O₂. Tuning combustion so the measured CO₂ approaches the theoretical maximum (without going so far as to make CO from lack of air) is therefore the goal of optimisation: it maximises CO₂ and minimises both the excess air and the stack loss. Enter x and y for the fuel.
Related Tools
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.
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.
Adiabatic Flame Temperature
Estimate the adiabatic flame temperature, T_ad = T_initial + LHV ÷ (m × cp), from the lower heating value LHV (kJ/kg fuel), the mass of combustion products per kg fuel m, the average specific heat of the gases cp (kJ/kg·K) and the initial temperature. The result, in °C, is the maximum theoretical temperature the gases would reach if all the combustion energy heated the products, with no heat loss. It is an upper bound: real flames are cooler (radiation losses, dissociation, excess air). It sets the thermal severity on materials and NOx formation. Enter the LHV, the gas mass, the cp and the initial temperature.
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.
Stack Heat Loss (Siegert)
Calculate the heat loss through the exhaust gases by the Siegert formula, loss = K × (T_gas − T_air) ÷ CO₂, from the fuel factor K (~0.5 for natural gas, ~0.6 for oil), the gas and combustion air temperatures (°C) and the CO₂ percentage in the gases. The result, in %, is the largest energy loss of a boiler or furnace — the heat escaping hot through the stack. Lowering the gas temperature (with economizers and preheaters) and adjusting the excess air (which dilutes CO₂) minimizes this loss. The combustion efficiency is approximately 100% minus this loss. Enter the K factor, the temperatures and the CO₂.
Equivalence Ratio (φ)
Calculate the combustion equivalence ratio, φ = AFR_stoichiometric ÷ AFR_actual, dividing the stoichiometric air-fuel ratio by the actual one. The dimensionless result classifies the mixture: φ = 1 is stoichiometric (exact air); φ < 1 is lean (excess air, typical of industrial burners and diesel engines); φ > 1 is rich (lack of air, produces CO and soot, but more power in gasoline engines). The equivalence ratio is the preferred dimensionless parameter in combustion science, linked to excess air (φ = 1/(1+EA)). Enter the stoichiometric and actual air-fuel ratios.
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