COMBUSTION ENGINEERING / ENGINEERING GUIDE

How to Calculate Combustion Air for Solid Fuels

Reviewed by WattCostLab Editorial Team · Updated September 2, 2026

Combustion-air demand can be calculated from the fuel ultimate analysis by first determining stoichiometric oxygen and then converting that oxygen requirement to theoretical dry or wet air.

Start with the ultimate analysis

The WattCostLab solid-fuel model uses elemental and balance inputs expressed as mass percentages: carbon C, hydrogen H, nitrogen N, oxygen O, sulfur S, fuel moisture W and M. The page checks that these seven percentages sum to 100% before the result is treated as a coherent ultimate-analysis input set.

Stoichiometric oxygen is calculated from the carbon, hydrogen, sulfur and oxygen terms. In the original model, the percent values are converted by division by 100 in the numerical implementation.

The result is expressed as normal cubic metres of oxygen per kilogram of fuel after the percentage conversion used by the calculator.

Convert oxygen requirement to dry combustion air

The model uses an oxygen fraction of 0.21 for dry air:

This is the theoretical dry-air volume required per kilogram of fuel under the stated stoichiometric model.

Account for humid combustion air

The calculator also includes an air-moisture input x and preserves the humidity factor 1.61 from the original equations:

This gives the theoretical wet-air volume. Do not confuse the air-moisture parameter x with the fuel-moisture term W. They affect different parts of the model.

Worked example from the calculator

For the default ultimate analysis C = 75.2%, H = 4.6%, N = 0.5%, O = 8.8%, S = 0.9%, W = 8.0% and M = 2.0%, the balance is exactly 100%. With x = 10%, the calculator gives:

• Stoichiometric oxygen: 1.6067503125 m³N O₂/kg fuel.
• Stoichiometric dry air: 7.6511919643 m³N/kg fuel.
• Stoichiometric wet air: 8.8830338705 m³N/kg fuel.

If the fuel flow rate is 1 kg/h, the wet-air flow in this example has the same numerical value in m³N/h. At other fuel flow rates, the air-flow result scales linearly with the entered fuel flow.

Where excess air enters

The theoretical air calculations above define the stoichiometric baseline. The excess-air coefficient λ enters the real-flue-gas equation. At λ = 1, the real and stoichiometric gas-volume calculations coincide for the same input set. At λ > 1, additional air increases the air-derived nitrogen, humid-air water contribution and residual oxygen terms in the real gas.

Why the input balance matters

Changing only one percentage without adjusting the rest can make the ultimate analysis exceed or fall below 100%. That may be useful for a sensitivity thought experiment, but it is no longer a normalized composition. WattCostLab therefore shows the sum and warns about a mismatch.

What the calculation does not predict

The model is deliberately narrower than a combustion-equilibrium program. It does not predict flame temperature, dissociation, NOx formation, carbon monoxide from incomplete combustion, unburned hydrocarbons or ash chemistry. Its purpose is transparent calculation of combustion-air demand and the major flue-gas volume terms present in the cited source model.

Scientific basis

The equations are based on Paraschiv, Serban and Paraschiv, “Calculation of combustion air required for burning solid fuels (coal / biomass / solid waste) and analysis of fluegas composition,” Energy Reports 6(Suppl. 3), 36–45 (2020). DOI: 10.1016/j.egyr.2019.10.016.

What each ultimate-analysis term does in the air calculation

Carbon and hydrogen contribute strongly to theoretical oxygen demand. Sulfur contributes through its term, while oxygen already present in the fuel reduces the external oxygen requirement in the implemented equation. Nitrogen, W and M are part of the normalized input balance but do not all enter the stoichiometric oxygen equation directly. Their roles appear elsewhere in the gas calculation or in the balance check.

From air per kilogram to air flow rate

Air requirement per kilogram is useful for comparing fuels independently of plant size. To estimate a flow rate, multiply the wet-air volume per kilogram by fuel mass flow B. If V°_au = 8.883 m³N/kg and B = 5 kg/h, the model gives about 44.415 m³N/h of wet combustion air at the stoichiometric baseline. The calculation is linear in B.

Keep the composition basis consistent

Ultimate analyses can be reported on different bases in technical work. WattCostLab does not automatically convert among as-received, dry or other reporting bases. Enter a composition that is internally consistent and sums to 100% according to the basis you intend to model. When comparing two fuels, use the same conceptual basis if the goal is a like-for-like comparison.

Useful sensitivity checks

At fixed composition, changing fuel flow should scale air flow but not the per-kilogram oxygen or air volumes. At fixed fuel composition and x, changing λ should not change the stoichiometric dry-air result because λ enters the real-flue-gas calculation, not V°_a. Changing air humidity x should change wet-air volume while leaving dry-air volume unchanged.

These relationships make it easier to audit a spreadsheet or another implementation. The WattCostLab calculator exposes the intermediate values rather than showing only one final air-flow number.

Engineering use versus burner or furnace design

The calculated air volume is a stoichiometric/gas-balance result. Actual burner, grate, furnace or boiler design also depends on mixing, pressure drop, fan characteristics, leakage, temperature, staging and control strategy. Use the model as a transparent calculation layer, not as a complete combustion-system sizing package.

Documenting a reproducible calculation

For a reproducible engineering note, record the complete ultimate analysis, the analysis basis, x, λ, fuel flow B and the model constants. Report both the per-kilogram air volume and the flow-rate result. This makes it clear whether a difference between two cases comes from fuel composition or simply from a different firing rate.

It is also useful to retain intermediate V°O₂ and V°a values. If another implementation produces a different final wet-air flow, those intermediate numbers quickly reveal whether the discrepancy comes from the oxygen equation, the 0.21 air conversion, the humidity factor or the final multiplication by B.