AFR Calculator
Calculate the stoichiometric air-fuel ratio for common fuels including methane, propane, octane, hydrogen, and more with customizable fuel composition.
What is Air-Fuel Ratio (AFR)?
The air-fuel ratio (AFR) is the mass ratio of air to fuel present during combustion. It is a critical parameter in internal combustion engines, gas turbines, heating systems, and industrial burners. The stoichiometric AFR represents the exact amount of air needed for complete combustion of a fuel, where all fuel is burned and no oxygen remains.
The AFR is calculated as:
$$AFR = \frac{mass_{air}}{mass_{fuel}}$$
For hydrocarbon fuels, the stoichiometric combustion reaction with air (21% O2, 79% N2) is:
$$C_{\alpha}H_{\beta} + a(O_2 + 3.76N_2) \rightarrow \alpha CO_2 + \frac{\beta}{2}H_2O + 3.76a N_2$$
where $a = \alpha + \beta/4$ is the moles of oxygen required per mole of fuel.
How to Calculate Stoichiometric AFR
Follow these steps to calculate the stoichiometric air-fuel ratio:
- Determine the fuel's molecular formula ($C_{\alpha}H_{\beta}O_{\gamma}$).
- Calculate the moles of oxygen needed: $a = \alpha + \beta/4 - \gamma/2$.
- Calculate the molar AFR: $AFR_{molar} = a \times 4.76$ (since air is 21% O2).
- Calculate the mass AFR: $AFR_{mass} = AFR_{molar} \times M_{air} / M_{fuel}$.
- The result gives the mass of air required per unit mass of fuel.
Common Fuel AFR Values
Different fuels have different stoichiometric AFR values. For example, methane requires 17.19 kg of air per kg of fuel, while hydrogen requires 34.21 kg of air per kg of fuel. Gasoline (octane) has a stoichiometric AFR of approximately 14.7:1, which is why many engine control systems target this ratio.
Frequently Asked Questions
What does a higher AFR mean?
A higher AFR means more air is required per unit of fuel for complete combustion. Hydrogen has a high AFR (34.21:1) because it is very light and energy-dense. Rich mixtures have lower AFR (less air), while lean mixtures have higher AFR (more air).
What is the difference between stoichiometric and actual AFR?
The stoichiometric AFR is the theoretical ideal ratio for complete combustion. The actual AFR may differ due to engine operating conditions. A ratio lower than stoichiometric is called "rich" (excess fuel), while higher is "lean" (excess air).
Why is nitrogen included in the air calculation?
Air is composed of approximately 21% oxygen and 79% nitrogen by volume. While nitrogen does not participate in combustion (it is inert), it makes up most of the mass of air and must be accounted for in AFR calculations. The 3.76 factor represents the molar ratio of N2 to O2 in air.
Can I calculate AFR for oxygenated fuels like ethanol?
Yes. For fuels containing oxygen (like alcohols), the formula adjusts to $a = \alpha + \beta/4 - \gamma/2$, where $\gamma$ is the number of oxygen atoms. The oxygen in the fuel reduces the external oxygen required for combustion.
What is lambda in relation to AFR?
Lambda ($\lambda$) is the ratio of actual AFR to stoichiometric AFR. $\lambda = 1$ means stoichiometric, $\lambda < 1$ means rich, and $\lambda > 1$ means lean. For example, if stoichiometric AFR is 14.7:1 and actual AFR is 13.2:1, then $\lambda = 0.9$.
For related combustion and conversion tools, see the Chemical Equation Balancer and the Temperature Converter.
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