Bray–Moss–Libby model

Let us define a non-dimensional scalar variable or progress variable ${\displaystyle c}$ such that ${\displaystyle c=0}$ at the unburnt mixture and ${\displaystyle c=1}$ at the burnt gas side. For example, if ${\displaystyle T_{u}}$ is the unburnt gas temperature and ${\displaystyle T_{b}}$ is the burnt gas temperature, then the non-dimensional temperature can be defined as

${\displaystyle c={\frac {T-T_{u}}{T_{b}-T_{u}}}.}$

The progress variable could be any scalar, i.e., we could have chosen the concentration of a reactant as a progress variable. Since the reaction sheet is infinitely thin, at any point in the flow field, we can find the value of ${\displaystyle c}$ to be either unity or zero. The transition from zero to unity occurs instantaneously at the reaction sheet. Therefore, the probability density function for the progress variable is given by

${\displaystyle P(c,\mathbf {x} ,t)=\alpha (\mathbf {x} ,t)\delta (c)+\beta (\mathbf {x} ,t)\delta (1-c)}$

where ${\displaystyle \alpha (\mathbf {x} ,t)}$ and ${\displaystyle \beta (\mathbf {x} ,t)}$ are the probability of finding unburnt and burnt mixture, respectively and ${\displaystyle \delta }$ is the Dirac delta function. By definition, the normalization condition leads to

${\displaystyle \alpha (\mathbf {x} ,t)+\beta (\mathbf {x} ,t)=1.}$

It can be seen that the mean progress variable,

${\displaystyle {\bar {c}}(\mathbf {x} ,t)=\int _{0}^{1}cP(c,\mathbf {x} ,t)\,dc=\beta (\mathbf {x} ,t)}$

is nothing but the probability of finding burnt gas at location ${\displaystyle \mathbf {x} }$ and at the time ${\displaystyle t}$. The density function is completely described by the mean progress variable, as we can write (supressing the variables ${\displaystyle \mathbf {x} ,t}$)

${\displaystyle P(c)=(1-{\bar {c}})\delta (c)+{\bar {c}}\delta (1-c).}$

Assuming constant pressure and constant molecular weight, ideal gas law can be shown to reduce to

${\displaystyle {\frac {\rho }{\rho _{u}}}={\frac {T_{u}}{T}}={\frac {1}{1+\tau c}}}$

where ${\displaystyle \tau }$ is the heat release parameter. Using the above relation, the mean density can be calculated as follows

${\displaystyle {\frac {\bar {\rho }}{\rho _{u}}}=1-\beta +{\frac {\beta }{1+\tau }}.}$

The Favre averaging of the progress variable is given by

${\displaystyle {\tilde {c}}\equiv {\frac {\overline {\rho c}}{\bar {\rho }}}={\frac {\rho _{u}}{\bar {\rho }}}{\frac {\beta }{1+\tau }}.}$

Combining the two expressions, we find

${\displaystyle {\bar {c}}=\beta ={\frac {(1+\tau ){\tilde {c}}}{1+\tau {\tilde {c}}}}}$

and hence

${\displaystyle \alpha ={\frac {1-{\tilde {c}}}{1+\tau {\tilde {c}}}}.}$

The density average is

${\displaystyle {\bar {\rho }}={\frac {\rho _{u}}{1+\tau {\tilde {c}}}}.}$

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If reaction sheet is not assumed to be thin, then there is a chance that one can find a value for ${\displaystyle c}$ in between zero and unity, although in reality, the reaction sheet is mostly thin compared to turbulent scales. Nevertheless, the general form the density function can be written as

${\displaystyle P(c,\mathbf {x} ,t)=\alpha (\mathbf {x} ,t)\delta (c)+\beta (\mathbf {x} ,t)\delta (1-c)+\gamma (\mathbf {x} ,t)f(c,\mathbf {x} ,t)}$

where ${\displaystyle \gamma (\mathbf {x} ,t)}$ is the probability of finding the progress variable which is undergoing reaction (where transition from zero to unity is effected). Here, we have

${\displaystyle \alpha (\mathbf {x} ,t)+\beta (\mathbf {x} ,t)+\gamma (\mathbf {x} ,t)=1}$

where ${\displaystyle \gamma }$ is negligible in most regions.