Equilibrium model
A simple verion of the presumed PDF model involves mapping all chemical state variables, generically represented by $\eta$, to the mixture fraction $Z$ via adiabatic equilibrium:
$$\eta = \eta_{eq}(Z).$$
In RANS or LES, the Favre-mean $\eta$ is evaluated by integrating over the mixture fraction PDF:
$$\widetilde{\eta} = \int_Z\eta_{eq}(Z)\widetilde{P}(Z)dZ.$$
The mixture fraction PDF is modeled as a $\beta$ function, parameterized by the Favre-mean mixture fraction $\widetilde{Z}$ and its variance $\widetilde{\sigma^2}$:
$$\widetilde{\eta} = \int_Z\eta_{eq}(Z)P_\beta(Z;\widetilde{Z}, \widetilde{\sigma^2})dZ. \qquad(1)$$
A two-dimensional table is created to give
$$\widetilde{\eta} = \widetilde{\eta}(\widetilde{Z}, \widetilde{\sigma^2}).$$
This table can be interpolated for any pairs of $\widetilde{Z}$ and $\widetilde{\sigma^2}$. A RANS or LES transports $\widetilde{Z}$ and provides $\widetilde{\sigma^2}$ via a transport equation or an algebraic relation.
Variance
The mixture fraction variance ranges from $0$, indicating a fully mixed cell with a single composition, to $\widetilde{\sigma^2}_{max}$, indicating a cell with pure fuel and air and no mixing between them. Note that $\widetilde{\sigma^2}_{max}$ depends on the mean mixture fraction. Because of this, it is convenient to tabulate
$$\widetilde{\eta} = \widetilde{\eta}(\widetilde{Z}, f_{\sigma^2}),$$
Instead of $\widetilde{\eta} = \widetilde{\eta}(\widetilde{Z}, \widetilde{\sigma^2}),$ where $f_{\sigma^2}\in[0,1]$ and
$$f_{\sigma^2}=\widetilde{\sigma^2}/\widetilde{\sigma^2}_{max}.$$
This facilitates a Cartesian grid for the table, not necessarily uniform, with $n_Z$ and $n_{\sigma^2}$ points in the mean and variance directions.
Algorithm
- Build table
- Choose $n_Z$ and $n_{\sigma^2}$ that gives good resolution.
- Loop over table indices $i_Z$ and $j_v$ with corresponding values of $\widetilde{Z}$ and $f_{\sigma^2}$
- For each value of $\widetilde{Z}$ and $\widetilde{\sigma^2}=f_{\sigma^2}\widetilde{\sigma^2_{max}}$, evaluate Equation 1 for $\widetilde{\eta}$ using $\eta_{eq}(Z)$ and $P_\beta(Z;\widetilde{Z}, \widetilde{\sigma^2})$.
- Fill the table at the given point.
- Table lookup
- CFD transports or provides $\widetilde{Z}$, $\widetilde{\sigma^2}$.
- Evaluate $\widetilde{\sigma^2}_{max}$ from $\widetilde{Z}$.
- Evaluate $f_{\sigma^2}=\widetilde{\sigma^2}/\widetilde{\sigma^2}_{max}.$
- Interpolate the table to find $\widetilde{\eta}$.
Heat loss
For non-adiabatic flames, enthalpy is not a function of mixture fraction alone, due to radiative heat losses, and the equilibrium state is set by the mixture fraction (defining elemental composition), and enthalpy:
$\eta = \eta_{eq}(Z,h)$.
The Favre-mean could be evaluated by
$$\widetilde{\eta} = \iint\eta_{eq}(Z,h)\widetilde{P}(Z,h)dZdh,$$
but then the joint PDF of $Z$ and $h$ is needed. A common assumption is to assume independence of $Z$ and $h$ and use $\widetilde{P}(Z,h)\approx \widetilde{P}(Z)\widetilde{P}(h)$. But this is not a good approximation since clearly the enthalpy variations around the hot stoichiometric flame will be higher than near the fuel and oxidizer regions. Similarly, if we assume independence, then a given value of $h$ would apply to all mixture fractions when evaluating $\widetilde{P}(Z,h)\approx \widetilde{P}(Z)\widetilde{P}(h)$ in the integration. But applying an enthalpy deficit at the flame (due to, e.g., radiation), would depress the enthalpy at the fuel and oxidizer to below ambient values, which is not physical. For example, consider a hydrogen-air flame. The adiabatic enthalpy is zero for all mixture fractions. Applying a uniform negative enthalpy at all mixture fractions results in unphysically small temperatures on the lean and rich sides of the flame where the sensible enthalpy approach zero at $Z=0$ and $Z=1$. Loosly, the $T(Z)$ curve is uniformly decreased (assuming constant $c_p$) resulting in temperatures lower than ambient in the fuel and air.
An alternative to parameterizing the flame by enthalpy is to use a fraction of sensible enthalpy termed a heat loss parameter $\gamma$, defined by
$$\gamma = \frac{h_a(Z) - h}{h_s(Z)},$$ giving $$h = h_a(Z) - \gamma h_s(Z).$$
Here, $h_a(Z)=h_0\cdot(1-Z) + h_1\cdot Z$ is the adiabatic enthalpy, and $h_s(Z)$ is the sensible enthalpy, defined as the difference in enthalpy between the adiabatic equilibrium products at a given $Z$ and the same products at the cold mixing temperature between the fuel and air streams. The heat loss $\gamma$ is the fraction of sensible enthalpy lost at a given $Z$. $h_s$ and $h_a$ both depend only on $Z$. Note that $h_s(Z=0)=h_s(Z=1)=0$ so that $h=h_a$ for $Z=0$ and $Z=1$ for all $\gamma$.
Temperature and enthalpy for a methane-air flame are plotted below for $\gamma=0.3$
We now write
$$\widetilde{\eta} = \iint\eta_{eq}(Z,h(Z,\gamma))\tilde{P}(Z,\gamma)dZd\gamma.$$
We assume independence of $Z$ and $\gamma$ and use a $\beta$-PDF for $\tilde{P}(Z)$, as before. $P(\gamma)$ is approximated as $\delta(\gamma-\hat{\gamma})$, where $\hat{\gamma}$ is an appropriate mean value, defined below as Equation 3. Upon integrating over $\gamma$, we obtain
$$\widetilde{\eta}(\widetilde{Z},\widetilde{\sigma^2},\hat{\gamma}) = \int\eta_{eq}(Z,h(Z,\hat{\gamma}))P_\beta(Z;\widetilde{Z},\widetilde{\sigma^2}))dZ. \qquad(2)$$
For $\widetilde{\eta} = \widetilde{h}$, this equation gives
$$\widetilde{h} = \int \left(h_a(Z) - \hat{\gamma}h_s(Z)\right)P_\beta(Z;\widetilde{Z},\widetilde{\sigma^2}))dZ,$$
or,
$$\widetilde{h} = h_a(\widetilde{Z}) - \hat{\gamma}\widetilde{h_s},$$
and
$$\hat{\gamma} = \frac{h_a(\widetilde{Z}) - \widetilde{h}}{\widetilde{h_s}}, \qquad(3)$$
where $\widetilde{h_s}$ is a function of $\widetilde{Z}$, $\widetilde{\sigma^2}$.
Algorithm
- Build table
- Select $n_Z$, $n_{\sigma^2}$, $n_\gamma$ that give appropriate resolution.
- Set 1D grids for $\widetilde{Z}$, $f_{\sigma^2}$, and $\hat{\gamma}$.
- For each $\widetilde{Z}i$, $f{v,j}$, and $\hat{\gamma}_k$:
- evaluate $\widetilde{\sigma^2}_{max},$
- evaluate $\widetilde{\sigma^2} = f_{v,j}\widetilde{\sigma^2}_{max}$,
- evaluate $\widetilde{\eta}_{i,j,k}$ using Equation 2.
- Build a 2D table $\widetilde{h_s}(\widetilde{Z},f_{\sigma^2})$ from $\widetilde{h_s} = \int h_s(Z)P_\beta(Z;\widetilde{Z},\widetilde{\sigma^2})dZ$.
- (This is used to evaluate $\hat{\gamma}$ from $\widetilde{h}$ in the lookup step.)
- Table lookup
- CFD transports or provides $\widetilde{Z}$, $\widetilde{\sigma^2}$, $\widetilde{h}$.
- Evaluate $\widetilde{\sigma^2}_{max}$ from $\widetilde{Z}$.
- Evaluate $f_{\sigma^2} = \widetilde{\sigma^2}/\widetilde{\sigma^2}_{max}$.
- Lookup $\widetilde{h_s}(\widetilde{Z},\widetilde{\sigma^2}).$
- Evaluate $\hat{\gamma}$ from Equation 3.
- Lookup $\widetilde{\eta}(\widetilde{Z},f_{\sigma^2},\hat{\gamma})$.
Progress variable
Relaxing the equilibrium assumption requires accounting for flame chemistry. This can be done using strained one-dimensional flames. For nonpremixed flames, solution is computed either directly in a mixture fraction coordinate, or in a spatial coordinate that is monotonic in mixture fraction for which a direct mapping applies. The strain variable can be a domain length $L$, scalar dissipation $\chi_0$ at some given $Z$, or a strain rate.