Burke–Schumann flame

Consider a cylindrical duct with axis along ${\displaystyle z}$ direction with radius ${\displaystyle a}$ through which fuel is fed from the bottom and the tube mouth is located at ${\displaystyle z=0}$. Oxidizer is fed along the same axis, but in the concentric tube of radius Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): b outside the fuel tube. Let the mass fraction in the fuel tube be ${\displaystyle Y_{Fo}}$ and the mass fraction of the oxygen in the outside duct be ${\displaystyle Y_{Oo}}$. Fuel and oxygen mixing occurs in the region ${\displaystyle z>0}$. The following assumptions were made in the analysis:

• The average velocity is parallel to axis (${\displaystyle z}$ direction) of the ducts, ${\displaystyle \mathbf {v} =v\mathbf {e} _{z}}$
• The mass flux in the axial direction is constant, ${\displaystyle \rho v=\mathrm {constant} }$
• Axial diffusion is negligible compared to the transverse/radial diffusion
• The flame occurs infinitely fast (Burke–Schumann limit), therefore flame appears as a reaction sheet across which properties of flow changes
• Effects of gravity has been neglected

Consider a one-step irreversible Arrhenius law, ${\displaystyle \mathrm {Fuel} +s\mathrm {O} _{2}\rightarrow (1+s)\mathrm {Products} +q}$, where ${\displaystyle s}$ is the mass of oxygen required to burn unit mass of fuel and ${\displaystyle q}$ is the amount of heat released per unit mass of fuel burned. If ${\displaystyle \omega }$ is the mass of fuel burned per unit volume per unit time and introducing the non-dimensional fuel and mass fraction and the Stoichiometry parameter,

${\displaystyle y_{F}={\frac {Y_{F}}{Y_{Fo}}},\quad y_{O}={\frac {Y_{O}}{Y_{Oo}}},\quad S={\frac {sY_{Fo}}{Y_{Oo}}}}$

the governing equations for fuel and oxidizer mass fraction reduce to

${\displaystyle {\begin{aligned}{\frac {\rho D_{T}}{r}}{\frac {\partial }{\partial r}}\left(r{\frac {\partial y_{F}}{\partial r}}\right)-\rho v{\frac {\partial y_{F}}{\partial z}}={\frac {\omega }{Y_{Fo}}}\\{\frac {\rho D_{T}}{r}}{\frac {\partial }{\partial r}}\left(r{\frac {\partial y_{O}}{\partial r}}\right)-\rho v{\frac {\partial y_{O}}{\partial z}}=S{\frac {\omega }{Y_{Fo}}}\\\end{aligned}}}$

where Lewis number of both species is assumed to be unity and ${\displaystyle \rho D_{T}}$ is assumed to be constant, where ${\displaystyle D_{T}}$ is the thermal diffusivity. The boundary conditions for the problem are

${\displaystyle {\begin{aligned}{\text{at}}\,&z=0,\,0<r<a,\,y_{F}=1,\,y_{O}=0,\\{\text{at}}\,&z=0,\,a<r<b,\,y_{F}=0,\,y_{O}=1,\\{\text{at}}\,&r=b,\,0<z<\infty ,\,{\frac {\partial y_{F}}{\partial r}}=0,\,{\frac {\partial y_{O}}{\partial r}}=0.\end{aligned}}}$

The equation can be linearly combined to eliminate the non-linear reaction term ${\displaystyle \omega /Y_{Fo}}$ and solve for the new variable

${\displaystyle Z={\frac {Sy_{F}-y_{O}+1}{S+1}}}$,

where ${\displaystyle Z}$ is known as the mixture fraction. The mixture fraction takes the value of unity in the fuel stream and zero in the oxidizer stream and it is a scalar field which is not affected by the reaction. The equation satisfied by ${\displaystyle Z}$ is

${\displaystyle {\frac {1}{r}}{\frac {\partial }{\partial r}}\left(r{\frac {\partial Z}{\partial r}}\right)-{\frac {\rho v}{\rho D_{T}}}{\frac {\partial Z}{\partial z}}=0}$

(If the Lewis numbers of fuel and oxidizer are not equal to unity, then the equation satisfied by ${\displaystyle Z}$ is nonlinear as follows from Shvab–Zeldovich–Liñán formulation). Introducing the following coordinate transformation

${\displaystyle \xi ={\frac {r}{b}},\quad \eta ={\frac {\rho D_{T}}{\rho v}}{\frac {z}{b^{2}}},\quad c={\frac {a}{b}}}$

reduces the equation to

${\displaystyle {\frac {1}{\xi }}{\frac {\partial }{\partial \xi }}\left(\xi {\frac {\partial Z}{\partial \xi }}\right)-{\frac {\partial Z}{\partial \eta }}=0.}$

The corresponding boundary conditions become

${\displaystyle {\begin{aligned}{\text{at}}\,&\eta =0,\,0<\xi <c,\,Z=1,\\{\text{at}}\,&\eta =0,\,c<\xi <1,\,Z=0,\\{\text{at}}\,&\xi =1,\,0<\eta <\infty ,\,{\frac {\partial Z}{\partial \xi }}=0.\end{aligned}}}$

The equation can be solved by separation of variables

${\displaystyle Z(\xi ,\eta )=c^{2}+2c\sum _{n=1}^{\infty }{\frac {1}{\lambda _{n}}}{\frac {J_{1}(c\lambda _{n})}{J_{0}^{2}(\lambda _{n})}}J_{0}(\lambda _{n}\xi )e^{-\lambda _{n}^{2}\eta }}$

where ${\displaystyle J_{0}}$ and ${\displaystyle J_{1}}$ are the Bessel function of the first kind and ${\displaystyle \lambda _{n}}$ is the nth root of ${\displaystyle J_{1}(\lambda )=0.}$ Solution can also be obtained for the planar ducts instead of the axisymmetric ducts discussed here.

[3][4]

In the Burke-Schumann limit, the flame is considered as a thin reaction sheet outside which both fuel and oxygen cannot exist together, i.e., ${\displaystyle y_{F}y_{O}=0}$. The reaction sheet itself is located by the stoichiometric surface where ${\displaystyle Sy_{F}=y_{O}}$, in other words, where

${\displaystyle Z=Z_{s}\equiv {\frac {1}{S+1}}}$

where ${\displaystyle Z_{s}}$ is the stoichiometric mixture fraction. The reaction sheet separates fuel and oxidizer region. The inner structure of the reaction sheet is described by Liñán's equation. On the fuel side of the reaction sheet (${\displaystyle Z>Z_{s}}$)

${\displaystyle y_{F}={\frac {Z-Z_{s}}{1-Z_{s}}},\,y_{O}=0}$

and on the oxidizer side (${\displaystyle Z<Z_{s}}$)

${\displaystyle y_{F}=0,\,y_{O}=1-{\frac {Z}{Z_{s}}}.}$

For given values of ${\displaystyle Z_{s}}$ (or, ${\displaystyle S}$) and ${\displaystyle c}$, the flame shape is given by the condition ${\displaystyle Z(\xi ,\eta )=Z_{s}}$, i.e.,

${\displaystyle Z_{s}=c^{2}+2c\sum _{n=1}^{\infty }{\frac {1}{\lambda _{n}}}{\frac {J_{1}(c\lambda _{n})}{J_{0}^{2}(\lambda _{n})}}J_{0}(\lambda _{n}\xi )e^{-\lambda _{n}^{2}\eta }.}$

When ${\displaystyle Z_{s}\rightarrow 0}$ (${\displaystyle S\rightarrow \infty }$), the flame extends from the mouth of the inner tube and attaches itself to the outer tube at a certain height (under-ventilated case) and when ${\displaystyle Z_{s}\rightarrow 1}$ (${\displaystyle S\rightarrow 0}$), the flame starts from the mouth of the inner tube and joins at the axis at some height away from the mouth (over-ventilated case). In general, the flame height is obtained by solving for ${\displaystyle \eta }$ in the above equation after setting ${\displaystyle \xi =1}$ for the under-ventilated case and ${\displaystyle \xi =0}$ for the over-ventilated case.

Since flame heights are generally large for the exponential terms in the series to be negligible, as a first approximation flame height can be estimated by keeping only the first term of the series. This approximation predicts flame heights for both cases as follows

${\displaystyle {\begin{aligned}\eta &={\frac {1}{\lambda _{1}^{2}}}\ln \left[{\frac {2cJ_{1}(c\lambda _{1})}{(Z_{s}-c^{2})\lambda _{1}J_{0}(\lambda _{1})}}\right],\quad {\text{under-ventilated}}\\\eta &={\frac {1}{\lambda _{1}^{2}}}\ln \left[{\frac {2cJ_{1}(c\lambda _{1})}{(Z_{s}-c^{2})\lambda _{1}J_{0}^{2}(\lambda _{1})}}\right],\quad {\text{over-ventilated}},\end{aligned}}}$

where ${\displaystyle \lambda _{1}=3.8317.}$