Assume a two-stream problem having one portion of the boundary the fuel stream with fuel mass fraction Y_F=Y_{F,F} and another portion of the boundary the oxidizer stream with oxidizer mass fraction Y_{O}=Y_{O,O}. For example, if the oxidizer stream is air and the fuel stream contains only the fuel, then Y_{O,O}=0.232 and Y_{F,F}=1. In addition, assume there is no oxygen in the fuel stream and there is no fuel in the oxidizer stream. Let s be the mass of oxygen required to burn unit mass of fuel (for
hydrogen gas, s=8 and for \mathrm{C}_m\mathrm{H}_n
alkanes, s=32(m+n/4)/(12m+n)). Introduce the scaled mass fractions as y_F=Y_F/Y_{F,F} and y_O = Y_O/Y_{O,O}. Then the mixture fraction is defined as :Z = \frac{Sy_F-y_{O}+1}{S+1} where :S = \frac{sY_{F,F}}{Y_{O,O}} is the
stoichiometry parameter, also known as the overall
equivalence ratio. On the fuel-stream boundary, y_F=1 and y_O=0 since there is no oxygen in the fuel stream, and hence Z=1. Similarly, on the oxidizer-stream boundary, y_F=0 and y_O=1 so that Z=0. Anywhere else in the mixing domain, 0. The mixture fraction is a function of both the spatial coordinates \mathbf{x} and the time t, i.e., Z=Z(\mathbf{x},t). Within the mixing domain, there are level surfaces where fuel and oxygen are found to be mixed in stoichiometric proportion. This surface is special in combustion because this is where a diffusion flame resides. Constant level of this surface is identified from the equation Z(\mathbf{x},t)=Z_s, where Z_s is called as the stoichiometric mixture fraction which is obtained by setting Y_F=Y_{O}=0 (since if they were react to consume fuel and oxygen, only on the stoichiometric locations both fuel and oxygen will be consumed completely) in the definition of Z to obtain :Z_s = \frac{1}{S+1}. ==Relation between local equivalence ratio and mixture fraction==