AFR Calculator (Air-Fuel Ratio)

• $m_{\mathrm{air}} = \mathrm{AFR} \times m_{\mathrm{fuel}}$ — air needed for a given mass of fuel
• $m_{\mathrm{fuel}} = \frac{m_{\mathrm{air}}}{\mathrm{AFR}}$ — fuel that a given mass of air can burn
• $\mathrm{AFR} = \frac{m_{\mathrm{air}}}{m_{\mathrm{fuel}}}$ — ratio from known masses (in “Other” mode)

Where:

• $\mathrm{AFR}$ — air-fuel ratio on a mass basis
• $m_{\mathrm{air}}$ — mass of air
• $m_{\mathrm{fuel}}$ — mass of fuel

Sizing air for a small engine

A small generator burns about 0.5 kg of octane per hour. How much air does it need?

The engine needs about 7.5 kg of air per hour for complete combustion. Real engines run slightly rich or lean of this theoretical value, but it's a useful starting point for estimating intake requirements.

Air supply for a propane heater

A workshop heater burns 10 kg of propane. How much air is required for complete combustion?

Complete combustion of 10 kg of propane needs about 156 kg of air. This is the kind of figure used when checking that a room or duct can supply enough air for a burner.

Does the calculator account for excess air?

No. It gives the theoretical (stoichiometric) AFR — the minimum air for complete combustion. Real systems often add excess air to make sure all the fuel burns, so the actual air supplied is usually higher than the value shown here.

Is the AFR the same as lambda?

Not quite. Lambda ($\lambda$) is the ratio of the actual AFR to the stoichiometric AFR. $\lambda = 1$ means stoichiometric, $\lambda > 1$ is lean (excess air), and $\lambda < 1$ is rich (insufficient air).

Do the mass units affect the AFR result?

No. The AFR is a ratio of two masses, so it is the same whether you enter grams, kilograms, or pounds — as long as both masses use the same unit.