In order to be able to efficiently burn fuel in an internal combustion engine, you need to know at which exact air-fuel ratio all the fuel will be burnt and all air be used up. Learn how to calculate this ratio for different fuels.
What is the stiochiometric air-fuel ratio?
Internal combustion engines burn fuel to create kinetic energy. The burning of fuel is basically the reaction of the fuel with the oxygen in the air. The amount of oxygen present in the cylinder is the limiting factor for the amount of fuel can be burnt. If there’s too much fuel present, not all fuel will be burnt and un-burnt fuel will be pushed out through the exhaust valve.
When building an engine, it’s very important to know the air-fuel ratio at which exactly all the available oxygen is used to burn the fuel and all the fuel is burnt completely. This ratio is called the stoichiometric air-fuel ratio.
Calculation
As already stated, the stoichiometric air-fuel ratio is the ratio at which all oxygen is used up and all fuel is completely burnt. This ratio is a basic property of a fuel and is the result of its chemical composition. Let’s for example look at natural gas (methane). When burning any carbon-based fuel, carbon dioxide and hydrogen are formed. Going back to the octane example, the following reaction equation describes the oxidation of the fuel:
CH4 + 2O2 -> CO2 + 2H20
Following from the equation, we need 2 molecules of oxygen for every octane molecule.
If we look up the atomic weights of the atoms that make up octane and oxygen, we get the following numbers:
Carbon (C): 12,01
Oxygen (O): 16
Hydrogen (H): 1,008
- So 1 molecule of methane has a molecular weight of: 1 * 12,01 + 4 * 1,008 = 16,042
- One oxygen molecule weighs: 2 * 16 = 32.
- The oxygen-fuel mass ratio is then: 2 * 32 / 1 * 16,042 = 64 / 16,042 = .
- So we need 3,99 kg of oxygen for every 1 kg of fuel.
- Since 23,2 mass-percent of air is actually oxygen, we need : 3,99 * 100/23,2 = 17,2 kg air for every 1 kg of methane.
So the stoichiometric air-fuel ratio of methane is 17.2.
Common fuels
When the composition of a fuel is known, this method can be used to derive the stoichiometric air-fuel ratio. For the most common fuels, this, however, is not necessary because the ratios are known:
Gasoline: 14.7
Natural gas: 17.2
Propane: 15.5
Ethanol: 9
Methanol: 6.4
Hydrogen: 34
Diesel: 14,6
You may find it interesting that methanol and ethanol both have a very low air-fuel ratio, while the carbon chain length is comparable to methane and ethane. The reason for that is alcohols like methanol and ethanol already carry oxygen themselves, which reduces the need for oxygen from the air.
Conclusion
In order to be able to judge if an air-fuel mixture has the correct ratio of air to fuel, the stoichiometric air fuel ratio has to be known. If the composition of a fuel is known, this ratio can be calculated rather easily.