Analyse expérimentale et modélisation du transfert de matière et du ...

Annexe Chapitre 22.521.512.521.51DNS(Rogers et al.[1])model (a)model (b)model (c)this worku'θ'0.5v'θ'0.500-0.5DNS(Rogers et al.[1])model (a)model (b)-1model (c)this work-1.50 5 10 15St-0.5-1-1.50 5 10 15StFigure 2: Development of the turbulent passive scalar flux. Comparison between the numerical results and DNS dataof Rogers et al. [1].3. First order modelling of the scalar fluxequationsRodi [13] suggested that the transport ofthe Reynolds stress tensor is proportionalto the transport of the turbulent energy.This assumption is written as:DDt' 'u u − Diffij' 'u u' ' i j ⎛ Dk ⎞( u u ) = ⎜ − Diff (i jk)⎟ ⎠k⎝ Dt' 'u=iu j( P − ε )k(16)Using the formulations (8), (9) and (16),the components of Reynolds stress can bewritten as:⎛ P 2 P11 ⎞( 1−γ ) ⎜ −'2⎟u 2 ⎝ ε 3 ε= +⎠(17)k 3PC1−1+ε⎛ 2 P ⎞( 1 γ )'2 − ⎜ − ⎟v 2 ⎝ 3 ε= +⎠(18)k 3PC1−1+ε⎛ P12⎞( 1−γ ) ⎜ ⎟' 'u v ε=⎝ ⎠(19)kPC −1+1εThe formulation of the transversal normalcomponent of the Reynolds stress tensor'2v (Eq. (18)), together with equation(19) allow us to obtain an explicitexpression of the turbulent shear stress inthe form:' ' 2u v =3( 1−γ )PC −1+γ1ε2⎛ P ⎞⎜C−1+1 ⎟⎝ ε ⎠2k ∂Uε ∂y(20)Boussinesq hypothesis allows expressingthe Reynolds stress using the concept ofturbulent viscosity νtas:⎛⎜ ∂U⎝∂U⎞⎟⎠' 'ij− uiuj= νt+ − kδij(21)⎜ ∂x∂x⎟ 3jiOne can thus deduce, from equation (20),the following expression of the turbulenteddy viscosity, [10]:Pγν2 C −1+12= ( 1−γ ) ε k(22)t23 ⎛ P ⎞ ε⎜C−1+1 ⎟⎝ ε ⎠2170