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Stationary laminar flamelet model

Generally, near the stoichiometric surface, both terms in (5.268) are large in magnitude and opposite in sign (Peters 2000). A quasi-stationary state is thus quickly established wherein the accumulation term on the left-hand side is negligible. The stationary laminar flamelet (SLF) model is found by simply neglecting the accumulation term, and ignoring the time dependency of x(f, t)  [Pg.204]

given an appropriate form for /(f), the SLF model can be solved to find 7(f) with boundary conditions 7(0) = 7(1) = 0.116 Note that if, SV(0, f) = 0, the SLF model admits the solution 7(f) = 0, which corresponds to a non-reacting system (i.e., flame extinction ). Likewise, when /(f) Sy(Y, f), the solution will be dominated by the diffusion term so that 7(f) 0. In the opposite limit, the solution will be dominated by the [Pg.204]

114 This procedure is similar to the one discussed in Section 5.4 for equilibrium chemistry where the spatial information is contained only in the mixture fraction. [Pg.204]

115 In this equation, Y should be replaced by the conditional expected value of Y given f = f However, based on the structure of the flamelet, it can be assumed that the conditional PDF is a delta function centered at T(f, r). [Pg.204]

116 The SLF model generates a two-point boundary-value problem for which standard numerical techniques exist. [Pg.204]


See other pages where Stationary laminar flamelet model is mentioned: [Pg.11]    [Pg.223]    [Pg.204]    [Pg.11]    [Pg.223]    [Pg.204]    [Pg.437]   


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