Air–fuel ratio

Mixture is the predominant word that appears in training texts, operation manuals, and maintenance manuals in the aviation world.

Air–fuel ratio is the ratio between the mass of air and the mass of fuel in the fuel–air mix at any given moment. The mass is the mass of all constituents that compose the fuel and air, whether combustible or not. For example, a calculation of the mass of natural gas—which often contains carbon dioxide (CO
₂), nitrogen (N
₂), and various alkanes—includes the mass of the carbon dioxide, nitrogen and all alkanes in determining the value of m_fuel.[2]

For pure octane the stoichiometric mixture is approximately 15.1:1, or λ of 1.00 exactly.

In naturally aspirated engines powered by octane, maximum power is frequently reached at AFRs ranging from 12.5 to 13.3:1 or λ of 0.850 to 0.901.

The air-fuel ratio of 12:1 is considered as the maximum output ratio, whereas the air-fuel ratio of 16:1 is considered as the maximum fuel economy ratio.

Fuel–air ratio is commonly used in the gas turbine industry as well as in government studies of internal combustion engine, and refers to the ratio of fuel to the air.

${\displaystyle \mathrm {FAR} ={\frac {1}{\mathrm {AFR} }}}$

Air–fuel equivalence ratio, λ (lambda), is the ratio of actual AFR to stoichiometry for a given mixture. λ = 1.0 is at stoichiometry, rich mixtures λ < 1.0, and lean mixtures λ > 1.0.

There is a direct relationship between λ and AFR. To calculate AFR from a given λ, multiply the measured λ by the stoichiometric AFR for that fuel. Alternatively, to recover λ from an AFR, divide AFR by the stoichiometric AFR for that fuel. This last equation is often used as the definition of λ:

${\displaystyle \lambda ={\frac {\mathrm {AFR} }{\mathrm {AFR} _{\text{stoich}}}}}$

Because the composition of common fuels varies seasonally, and because many modern vehicles can handle different fuels when tuning, it makes more sense to talk about λ values rather than AFR.

Most practical AFR devices actually measure the amount of residual oxygen (for lean mixes) or unburnt hydrocarbons (for rich mixtures) in the exhaust gas.

The fuel–air equivalence ratio, Φ (phi), of a system is defined as the ratio of the fuel-to-oxidizer ratio to the stoichiometric fuel-to-oxidizer ratio. Mathematically,

${\displaystyle \phi ={\frac {\mbox{fuel-to-oxidizer ratio}}{({\mbox{fuel-to-oxidizer ratio}})_{\text{st}}}}={\frac {m_{\text{fuel}}/m_{\text{ox}}}{\left(m_{\text{fuel}}/m_{\text{ox}}\right)_{\text{st}}}}={\frac {n_{\text{fuel}}/n_{\text{ox}}}{\left(n_{\text{fuel}}/n_{\text{ox}}\right)_{\text{st}}}}}$

where m represents the mass, n represents a number of moles, subscript st stands for stoichiometric conditions.

The advantage of using equivalence ratio over fuel–oxidizer ratio is that it takes into account (and is therefore independent of) both mass and molar values for the fuel and the oxidizer. Consider, for example, a mixture of one mole of ethane (C
₂H
₆) and one mole of oxygen (O
₂). The fuel–oxidizer ratio of this mixture based on the mass of fuel and air is

${\displaystyle {\frac {m_{{\ce {C2H6}}}}{m_{{\ce {O2}}}}}={\frac {1\times (2\times 12+6\times 1)}{1\times (2\times 16)}}={\frac {30}{32}}=0.9375}$

and the fuel-oxidizer ratio of this mixture based on the number of moles of fuel and air is

${\displaystyle {\frac {n_{{\ce {C2H6}}}}{n_{{\ce {O2}}}}}={\frac {1}{1}}=1}$

Clearly the two values are not equal. To compare it with the equivalence ratio, we need to determine the fuel–oxidizer ratio of ethane and oxygen mixture. For this we need to consider the stoichiometric reaction of ethane and oxygen,

C₂H₆ + 7⁄2 O₂ → 2 CO₂ + 3 H₂O

This gives

${\displaystyle ({\text{fuel-to-oxidizer ratio based on mass}})_{\text{st}}=\left({\frac {m_{{\ce {C2H6}}}}{m_{{\ce {O2}}}}}\right)_{\text{st}}={\frac {1\times (2\times 12+6\times 1)}{3.5\times (2\times 16)}}={\frac {30}{112}}=0.268}$

${\displaystyle ({\text{fuel-to-oxidizer ratio based on number of moles}})_{\text{st}}=\left({\frac {n_{{\ce {C2H6}}}}{n_{{\ce {O2}}}}}\right)_{\text{st}}={\frac {1}{3.5}}=0.286}$

Thus we can determine the equivalence ratio of the given mixture as

${\displaystyle \phi ={\frac {m_{{\ce {C2H6}}}/m_{{\ce {O2}}}}{\left(m_{{\ce {C2H6}}}/m_{{\ce {O2}}}\right)_{\text{st}}}}={\frac {0.938}{0.268}}=3.5}$

or, equivalently, as

${\displaystyle \phi ={\frac {n_{{\ce {C2H6}}}/n_{{\ce {O2}}}}{\left(n_{{\ce {C2H6}}}/n_{{\ce {O2}}}\right)_{\text{st}}}}={\frac {1}{0.286}}=3.5}$

Another advantage of using the equivalence ratio is that ratios greater than one always mean there is more fuel in the fuel–oxidizer mixture than required for complete combustion (stoichiometric reaction), irrespective of the fuel and oxidizer being used—while ratios less than one represent a deficiency of fuel or equivalently excess oxidizer in the mixture. This is not the case if one uses fuel–oxidizer ratio, which takes different values for different mixtures.

The fuel–air equivalence ratio is related to the air–fuel equivalence ratio (defined previously) as follows:

${\displaystyle \phi ={\frac {1}{\lambda }}}$

The relative amounts of oxygen enrichment and fuel dilution can be quantified by the mixture fraction, Z, defined as

${\displaystyle Z=\left[{\frac {sY_{\mathrm {F} }-Y_{\mathrm {O} }+Y_{\mathrm {O,0} }}{sY_{\mathrm {F,0} }+Y_{\mathrm {O,0} }}}\right]}$,

where

${\displaystyle s=\mathrm {AFR} _{\mathrm {stoich} }={\frac {W_{\mathrm {O} }\times v_{\mathrm {O} }}{W_{\mathrm {F} }\times v_{\mathrm {F} }}}}$,

Y_F,0 and Y_O,0 represent the fuel and oxidizer mass fractions at the inlet, W_F and W_O are the species molecular weights, and v_F and v_O are the fuel and oxygen stoichiometric coefficients, respectively. The stoichiometric mixture fraction is

${\displaystyle Z_{\mathrm {st} }=\left[{\frac {1}{1+{\frac {Y_{\mathrm {F,0} }\times W_{\mathrm {O} }\times v_{\mathrm {O} }}{Y_{\mathrm {O,0} }\times W_{\mathrm {F} }\times v_{\mathrm {F} }}}}}\right]}$[3]

The stoichiometric mixture fraction is related to λ (lambda) and Φ (phi) by the equations

${\displaystyle Z_{\text{st}}={\frac {\lambda }{1+\lambda }}={\frac {1}{1+\phi }}}$,

assuming

${\displaystyle \mathrm {AFR} ={\frac {Y_{\mathrm {O,0} }}{Y_{\mathrm {F,0} }}}}$[4]

In industrial fired heaters, power plant steam generators, and large gas-fired turbines, the more common terms are percent excess combustion air and percent stoichiometric air.[5][6] For example, excess combustion air of 15 percent means that 15 percent more than the required stoichiometric air (or 115 percent of stoichiometric air) is being used.

A combustion control point can be defined by specifying the percent excess air (or oxygen) in the oxidant, or by specifying the percent oxygen in the combustion product.[7] An air–fuel ratio meter may be used to measure the percent oxygen in the combustion gas, from which the percent excess oxygen can be calculated from stoichiometry and a mass balance for fuel combustion. For example, for propane (C
₃H
₈) combustion between stoichiometric and 30 percent excess air (AFR_mass between 15.58 and 20.3), the relationship between percent excess air and percent oxygen is:

${\displaystyle {\begin{aligned}\mathrm {Mass\%\ O_{2}\ in\ propane\ combustion\ gas} &\approx -0.1433(\mathrm {\%\ excess\ O_{2}} )^{2}+0.214(\mathrm {\%\ excess\ O_{2}} )\\\mathrm {Volume\%\ O_{2}\ in\ propane\ combustion\ gas} &\approx -0.1208(\mathrm {\%\ excess\ O_{2}} )^{2}+0.186(\mathrm {\%\ excess\ O_{2}} )\end{aligned}}}$