Laminar flamelet equations¶

Using shorthand notation, e.g.: $$\left(\frac{\partial y_i}{\partial t}\right)_x\equiv(\partial_ty_i)_x.$$ The $x$ subscript is sometimes left off when it is understood what is being held constant.

Species transport equation (assuming all $D_i$ are equal: $$\rho\partial_ty_i + \rho v\partial_xy_i - \partial_x(\rho D\partial_xy_i) = \dot{m}_i^{\prime\prime\prime}.$$ Mixture fraction transport equation: $$\rho\partial_t\xi + \rho v\partial_x\xi - \partial_x(\rho D\partial_x\xi) = 0.$$

Flamelet transformations¶

$$y_i(t,x) \rightarrow y_i(t,\xi(x,t)).$$ Total derivative: $$dy_i = (\partial_ty_i)_xdt + (\partial_xy_i)_tdx \rightarrow (\partial_ty_i)_\xi dt + (\partial_\xi y_i)_td\xi.$$ The final $d\xi$ is $d\xi = (\partial_t\xi)_xdt + (\partial_x\xi)_tdx$, hence,

$$dy_i = (\partial_ty_i)_\xi dt + (\partial_\xi y_i)_t \left[(\partial_t\xi)_xdt + (\partial_x\xi)_tdx \right].$$

Now, divide through by $dx$ and take constant $t$: $$(\partial_xy_i)_t = (\partial_\xi y_i)_t(\partial_x\xi)_t.$$ This gives the flamelet transformation of the spatial coordinate $x$, where we are swapping the order of the last two terms:

$$(\partial_x)_t = (\partial_x\xi)_t(\partial_\xi)_t.$$

Now, divide the blue equation by $dt$ and take constant $x$: $$(\partial_ty_i)_x = (\partial_ty_i)_\xi + (\partial_\xi y_i)_t(\partial_t \xi)_x.$$ This gives the flamelet transformation of the time coordinate $t$, where we are swapping the order of the last two terms:

$$(\partial_t)_x = (\partial_t)_\xi + (\partial_t\xi)_x(\partial_\xi)_t.$$

Note that it is important to keep track of what is being held constant in the partial derivatives for this equation to make sense. In the literature, we often see the following definition (which is never explicitly stated, $\tau$ just appears and is used): $$(\partial_t)_\xi\equiv\partial_\tau.$$

Apply flamelet transformation to the species equation¶

$$\rho\partial_\tau y_i + \rho(\partial_t\xi)\partial_\xi y_i + \rho v(\partial_x\xi)\partial_\xi y_i - (\partial_x\xi)\partial_\xi(\rho D(\partial_x\xi)\partial_\xi y_i) = \dot{m}_i^{\prime\prime\prime}.$$

Apply the product rule to the last term on the left hand side:

$$\rho\partial_\tau y_i + \rho(\partial_t\xi)\partial_\xi y_i + \rho v(\partial_x\xi)\partial_\xi y_i - (\partial_x\xi)(\partial_\xi y_i)\partial_\xi(\rho D(\partial_x\xi)) - \rho D(\partial_y\xi)^2 \partial_{\xi\xi}^2y_i = \dot{m}_i^{\prime\prime\prime}.$$

Factor out $\partial_\xi y_i$:

$$\rho\partial_\tau y_i + \partial_\xi y_i\left[ \rho\partial_t\xi + \rho v\partial_x\xi - \partial_x(\rho D\partial_x\xi)\right] - \rho D(\partial_x\xi)^2\partial_{\xi\xi}^2y_i = \dot{m}_i^{\prime\prime\prime}.$$

The term in brackets is the mixture fraction transport equation, which is zero. This gives

$$\rho\partial_\tau y_i - \rho D(\partial_x\xi)^2\partial_{\xi\xi}^2y_i = \dot{m}_i^{\prime\prime\prime}.$$

Now, the factor $\rho D(\partial_x\xi)^2$ is defined as $\rho\chi/2$, where $\chi$ is given by $\chi = 2D\partial_{xx}^2\xi.$ Finally, divide through by $\rho$ to get

$$\frac{\partial y_i}{\partial\tau} = \frac{\chi}{2}\frac{\partial^2 y_i}{\partial\xi^2} + \dot{m}_i^{\prime\prime\prime},$$

$$\chi = 2D\left(\frac{\partial\xi}{\partial x}\right)^2.$$

$\chi$ is called the scalar dissipation rate.

• The units on $\chi$ are 1/s, so that $\chi$ is a mixing rate.
• In a turbulent flow, the scalar dissipation rate is the rate at which fluctuations in the mixture fraction are dissipated.
• $\chi$ is what couples the physical coordinate $x$ to the mixture fraction coordinate $\xi$.
• The shape of $\chi(\xi)$ is like a bell curve, with a value of zero at $\xi=0$ and $\xi=1$, and peaking at $\xi=0.5$.
• $\chi$ is often modeled as $$\chi = \chi_0\exp[-2(\text{erf}^{-1}(2\xi-1))^2],$$ or $$\chi = \chi_0(1-(2\xi-1)^2)^2.$$
• Normally, one assumes $\alpha = D$, or that $Le=\alpha/D=1$.