The ideal (theoretical) air-fuel ratio, for a complete combustion, is called stoichiometric air-fuel ratio. For a gasoline (petrol) engine, the stoichiometric air-fuel ratio is around 14.7:1. This means that, in order to burn completely 1 kg of fuel, we need 14.7 kg of air. The combustion is possible even is the AFR is different than stoichiometric. For the combustion process to take place in a gasoline engine, the minimum AFR is around 6:1 and the maximum can go up to 20:1.
When the air-fuel ratio is higher than the stoichiometric ratio, the air-fuel mixture is called lean. When the air-fuel ratio is lower than the stoichiometric ratio, the air-fuel mixture is called rich. For example, for a gasoline engine, an AFR of 16.5:1 is lean and 13.7:1 is rich.
In the table below we can see the stoichiometric air-fuel ratio for several fossil fuels.
For example, in order to burn completely 1 kg of ethanol, we need 9 kg of air and to burn 1 kg of diesel fuel, we need 14.5 kg of air.
Spark ignition (SI) engines usually run on gasoline (petrol) fuel. The AFR of the SI engines varies within the range 12:1 (rich) to 20:1 (lean), depending on the operating condition of the engine (temperature, speed, load, etc.). Modern internal combustion engines operate as much as possible around the stoichiometric AFR (mainly for gas after-treatment reasons). In the table below you can see an example of a SI engine AFR, function of engine speed and torque.
Compression ignition (CI) engines usually run on diesel fuel. Due to the nature of the combustion process, CI engines always run on lean mixtures, with AFR between 18:1 and 70:1. The main difference, compared with SI engines, is that CI engines run on stratified (non homogeneous) air-fuel mixtures, while SI run on homogeneous mixtures (in case of port-injection engines).
The table above is entered in a Scilab script and a contour plot is generated.
In order to understand how the stoichiometric air-fuel ratio is calculated, we need to look at the combustion process of the fuel. Combustion is basically a chemical reaction (called oxidation) in which a fuel is mixed with oxigen and produces carbon dioxide (CO2), water (H2O) and energy (heat). Take into account that, in order for the oxidation reaction to occur we need an activation energy (spark or high temperature). Also, the net reaction is highly exothermic (with heat release).
For a better understanding, let's look at the oxidation reaction of methane. This is a pretty common chemical reaction, since methane is the primary component of natural gas (in proportion of around 94 %).
Step 4. Calculate the mass of fuel, which is 1 mol of methane, made up from 1 atom of carbon and 4 atoms of hydrogen.
Step 5. Calculate the mass of oxygen, which consists of 2 moles, each mol made up from 2 atoms of oxygen.
Step 6. Calculate the necessary mass of air which contains the calculated mass of oxygen, taking into account that air contains around 21 % oxygen.
The calculated AFR for methane is not exactly as specified in the literature. The difference might come from the fact that, in our example, we made several assumptions (air contains only 21 % oxygen, the products of the combustion are only carbon dioxide and water).
The same method can be applied for the combustion of gasoline. Considering that gasoline is made up from isooctane (C8H18), calculate the stoichiometric air-fuel ratio for gasoline.
Step 4. Calculate the mass of fuel, which is 1 mol of isooctane, made up from 8 atoms of carbon and 18 atoms of hydrogen.
Step 5. Calculate the mass of oxygen, which consists of 12.5 moles, each mol made up from 2 atoms of oxygen.
Again, the calculated stoichimetric air-fuel ratio for gasoline is slightly different that the one provided in literature. Thus, the result is acceptable since we made a lot of assumptions (gasoline contains only isooctane, air contains only oxygen in proportion of 21 %, the only products of combustion are carbon dioxide and water, the combustion is ideal).
We have seen what is and how to calculate the stoichiometric (ideal) air-fuel ratio. In reality, internal combustion engines do not work exactly with ideal AFR, but with values close to it. Therefore we'll have an ideal and a actual air-fuel AFR. The ratio between the actual air-fuel ratio (AFRactual) and the ideal/stoichiometric air-fuel ratio (AFRideal) is called equivalence air-fuel ratio or lambda (λ).
Depending on the value of lambda, the engine is told to work with lean, stoichiometric or rich air-fuel mixture.
Depending on the type of fuel (gasoline or diesel) and the type of injection (direct or indirect), an internal combustion engine can function with lean, stoichiometric or rich air-fuel mixtures.
For example, the Ford Ecoboost 3-cylinder engine runs with stoichiometric air-fuel ratio for idle to medium engine speed and complete load range, and with rich air-fuel mixture at high speed and load. The reason for which it runs with rich mixture at high engine speed and load is engine cooling. The additional fuel (which will remain unburnt) is injected to absorb heat (through evaporation), reducing this way the temperature in the combustion chamber.
A compression ignition (diesel) engine runs all the time with lean air-fuel mixture, the value of the equivalence factor (λ) depending on the engine's operating point (speed and torque). The reason for this is the working principle of a diesel engine: controlling load not through air mass (which is always in excess) but through fuel mass (injection time).
Remember that a stoichiometric equivalence factor (λ = 1.00) means an air-fuel ratio of 14.7:1 for gasoline engines and 14.5:1 for diesel engines.
The engine performance in terms of power and fuel consumption is highly dependent on the air-fuel ratio. For a gasoline engine, the lowest fuel consumption is obtained at lean AFR. The main reason is that there is enough oxygen available to burn completely all the fuel which translates in mechanical work. On the other hand, the maximum power is obtained with rich air-fuel mixtures. As explained before, putting more fuel in the cylinder at high engine load and speed, cools down the combustion chamber (through fuel evaporation and heat absorption) which allows the engine to produce maximum engine torque thus maximum power.
In the figure above we can see that we can not get the maximum power of the engine and the lowest fuel consumption with the same air-fuel ratio. The lowest fuel consumption (best fuel economy) is obtained with lean air-fuel mixtures, with an AFR of 15.4:1 and an equivalence factor (λ) of 1.05. The maximum engine power is produced with rich air-fuel mixtures, with an AFR of 12.6:1 and an equivalence factor (λ) of 0.86. With a stoichiometric air-fuel mixture (λ = 1), there is a compromise between maximum engine power and minimum fuel consumption.
Compression ignition (diesel) engines always run on lean air-fuel mixtures (λ > 1.00). Most of the modern diesel engines run with λ between 1.65 and 1.10. The maximum efficiency (lowest fuel consumption) is obtained around λ = 1.65. Increasing the fuel amount above this value (going towards 1.10) will produce more soot (unburned fuel particles).
There is an interesting study performed by R. Douglas on 2-stroke cycle engines. In his doctoral thesis "Closed Cycle Studies of a Two-Stroke Cycle Engine", R. Douglas comes with a mathematical expression of the combustion efficiency (ηλ) function of equivalence factor (λ).
For diesel engines, if the equivalence factor goes above 2.00, the combustion efficiency is maximum (1.00 or 100 %).
We can use a Scilab script to plot the variation of the combustion efficiency function of the equivalence factor.
Running the Scilab instructions above outputs the following graphical window.
As you can see, the compression ignition (diesel) engine, at stoichiometric air-fuel ratio has a very low combustion efficiency. The best combustion efficiency is obtained at λ = 2.00 for diesel and λ = 1.12 for spark ignition (gasoline) engines.
Internal combustion engine exhaust gas emissions depend heavily on the air-fuel ratio (equivalence factor). The main exhaust gas emissions in ICE are summarized in the table below.
For a gasoline engine, CO, HC and NOx exhaust gas emissions are heavily influenced by air-fuel ratio. CO and HC are mainly produced with rich air-fuel mixture, while NOx with lean mixtures. So, there in no fixed air-fuel mixture for which we can obtain the minimum for all exhaust emissions.
A three way catalyst (TWC), used for gasoline engines, has the highest efficiency when the engine operates in a narrow band around stoichiometric air-fuel ratio. The TWC converts between 50 … 90 % of hidrocarbons and 90 … 99 % of carbon monoxide and nitrogen oxides, when the engine runs with λ = 1.00.
In order to meet the exhaust gas emissions regulations, it is critical for internal combustion engines (especially gasoline) to have an accurate control of the air-fuel ratio. Therefore, all of the modern internal combustion engines have closed-loop control for air-fuel ratio (lambda).
The critical component for the system to work is the lambda (oxygen) sensor. This sensor measures the level of oxygen molecules in the exhaust gas and sends the information to the engine electronic control unit (ECU). Based on the value of the oxygen sensor reading, the gasoline engine ECU will adjusts the level of fuel mass in order to keep the air-fuel ratio around the stoichimetric level (λ = 1.00).
For example (gasoline engines), if the level of oxygen molecules is above the threshold for stoichimetric level (therefore we have a lean mixture), at the next injection cycle, the injected fuel amount will be increased in order to make use of the excess air. Bear in mind that the engine will always transition from lean mixture to rich mixture between injection cycles, which will give an "average" of stoichiometric air-fuel mixtures/ratio.
For diesel engines, since it always runs on lean air-fuel ratio, lambda control is performed in a different manner. The end goal being still the same, control of the exhaust gas emissions.
How about plotting a graph of the stoichiometric air fuel ratio for petrol mixed with ethanol from 10-30% showing how the varying percentage of ethanol changes the stoichiometric A/F/R compared to the other percentages & ethanol free petrol?
Excellent U done have good job.
You've made an error in the calculation of AFR for methane in example 1. You've used 63.996 as the mass for fuel instead of 16.043.
Thanks for noticing, I've corrected it.
Excellent graphs descriptions etc …however in last image the word 'motor' should be replaced with 'engine'.