Boundary Conditions

Establishing the boundary conditions for the elemental flame requires separate consideration of each of the flame configurations previously described.

Single Opposed Jet

For the single opposed jet flame, the boundary conditions for the species and energy equations consist of defining the temperature and mass fraction of the two incoming streams. For the momentum equation, spatial gradients in $U$ must vanish on both sides of the flame, and we solve the simplified form of the momentum equation:

$$\rho \frac{\partial U}{\partial t}+\rho U^2-{\rho }_{\mathrm{\infty }}(\frac{da}{dt}+a^2)=0$$

for each boundary point. For steady strain rates, this equation requires that $U=a$ on the reactants side and $U=a\sqrt{{\rho }_u/{\rho }_b}$ on the products side.

For a premixed flame with reactants supplied from $-\mathrm{\infty }$, the boundary conditions for the species and energy equations are:

$$\begin{array}{r}\begin{array}{ccc}r=-\mathrm{\infty }:& Y_k=Y_{k,u}& T=T_u\\ r=+\mathrm{\infty }:& Y_k=Y_{k,b}& T=T_b\end{array}\end{array}$$

The burned gas temperature and composition correspond to the unburned mixture brought to equilibrum at constant enthalpy and pressure. The boundary condition for the continuity equationis taken at the left side of the computational domain, where $V$ is held fixed.

Central Control Volume

For curved flames at finite stagnation radius, or flames specified in terms of the unified formulation given in Unified Formulation, the $r=\mathrm{\infty }$ boundary condition is the same as for the curved flame at zero stagnation radius. The boundary conditions at $r=0$, however, require special attention. First, the mass flux at the center must be expressed in terms of $rV$ because $V\to \mathrm{\infty }$ as $r\to 0$ in the potential flow solution. If we specify the non-reacting stagnation point radius $R$, then the boundary condition for the continuity equation is:

$$r=0:rV=\frac{1}{\alpha +1}\rho a\left|R\right|R$$

If the boundary mass flux ${\left(rV\right)}_0$ is positive, indicating the presence of source at the boundary, special care must be taken in specifying the boundary conditions for the energy, species and momentum equations. While the zero-gradient condition must still hold because of the symmetry at that boundary, it is important to retain the effect of the mixture being introduced, which may not be at the same state as the mixture in the vicinity of $r=0$. To this end, we consider an integral, control volume approach to the $r=0$ boundary condition. Beginning with the species conservation equation, we multiply through by $r^{\alpha }$ and integrate from 0 to some small radius $R_i$:

$${\int }_0^{R_i}{r}^{\alpha }\rho \frac{\partial Y_k}{\partial t}+r^{\alpha }V\frac{\partial Y_k}{\partial r}+\frac{\partial }{\partial r}\left[r^{\alpha }{j}_k\right]-r^{\alpha }{\dot{\omega }}_k{W}_{k}dr=0$$

Because this volume is small, we assume that variations of $Y_k$, $\rho$ and ${\dot{\omega }}_k$ are negligible, so the unsteady term and the production term may be taken out of the integral:

$$\frac{R_i^{\alpha +1}}{\alpha +1}(\rho \frac{\partial Y_k}{\partial t}-{\dot{\omega }}_k{W}_k)+{\int }_0^{R_i}{r}^{\alpha }V\frac{\partial Y_k}{\partial r}+\frac{\partial }{\partial r}\left[r^{\alpha }{j}_k\right]dr=0$$

The convection term may be integrated by noting that variations in $r^{\alpha }V$ are negligible across this small distance. Furthermore, we recognize that $Y_k{|}_{r=0}=Y_{k,left}$ is the mass fraction corresponding to the inlet mixture (either reactants or products). The diffusion term may also be integrated, noting that $j_k{|}_{r=0}=0$ by the symmetry condition. We now have an ODE for the mass fraction of species $k$ in the vicinity of the symmetry boundary:

$$\frac{R_i^{\alpha +1}}{\alpha +1}(\rho \frac{\partial Y_k}{\partial t}-{\dot{\omega }}_k{W}_k)+{\left(r^{\alpha }V\right)}_0(Y_k-Y_{k,left})+R_i^{\alpha }{j}_k=0$$

Finally, we divide by the leading coefficient so that this equation scales similarly to the species equation in the rest of the domain:

$$\rho \frac{\partial Y_k}{\partial t}-{\dot{\omega }}_k{W}_k+\frac{\alpha +1}{R_i^{\alpha +1}}{\left(r^{\alpha }V\right)}_0(Y_k-Y_{k,left})+\frac{\alpha +1}{R_i}{j}_k=0$$

A similar analysis for the energy equation yields:

$$\rho \frac{\partial T}{\partial t}+\frac{1}{c_p}\sum _{k=1}^K{\hat{h}}_k{\dot{\omega }}_k+\frac{\alpha +1}{R_i^{\alpha +1}}{\left(r^{\alpha }V\right)}_0(T-T_{left})-\frac{\alpha +1}{R_i{c}_p}\left(\lambda \frac{\partial T}{\partial r}\right)=0$$

In the case of the momentum equation, we neglect the term associated with the boundary mass flux, and obtain:

$$\rho \frac{\partial U}{\partial t}+\rho U^2-{\rho }_{\mathrm{\infty }}(\frac{\partial a}{\partial t}+a^2)-\frac{\alpha +1}{R_i}\mu \frac{\partial U}{\partial r}=0$$

If the boundary mass flux ${\left(rV\right)}_0$ is negative, the term including it in each of these boundary conditions is eliminated.