Burke–Schumann flame

Example

Consider a fuel stream, in which the fuel mass fraction is $Y_{F,1}$, and an oxidiser stream, in which the oxidiser mass fraction is $Y_{O,2}$. The streams meet and react in one step in which $s$ units of mass of oxidiser burn one unit of mass of fuel. Assume that fuel and oxidiser have the same diffusivity $D$, and let $\mathbf{u}$ be the velocity field. The reaction removes oxidiser at $s$ times the rate at which it removes fuel, so the mixture fraction $$Z = \frac{s Y_F - Y_O + Y_{O,2}}{s Y_{F,1} + Y_{O,2}}$$ solves the reaction-free advection–diffusion equation $\partial_t Z + \mathbf{u} \cdot \nabla Z = D \Delta Z$. It equals $1$ in the fuel stream and $0$ in the oxidiser stream.

The Burke–Schumann flame is the limit of infinitely fast chemistry, in which fuel and oxidiser cannot coexist: $Y_F Y_O = 0$ everywhere. Then the flame is the surface $Z = Z_{st}$, with $$Z_{st} = \frac{Y_{O,2}}{s Y_{F,1} + Y_{O,2}}.$$ On the fuel side $Z > Z_{st}$ one has $Y_O = 0$ and $Y_F = Y_{F,1} (Z - Z_{st}) / (1 - Z_{st})$. On the oxidiser side $Z < Z_{st}$ one has $Y_F = 0$ and $Y_O = Y_{O,2} (1 - Z / Z_{st})$. Check: at $Z = Z_{st}$ the numerator $s Y_F - Y_O$ vanishes, and with $Y_F Y_O = 0$ this forces $Y_F = Y_O = 0$; at $Z = 1$ the formula gives $Y_F = Y_{F,1}$, and at $Z = 0$ it gives $Y_O = Y_{O,2}$.