A methodology to quantify the nonlinearity of the Reynolds stress tensor

Abstract

Turbulent models provide closure equations that relate the Reynolds stress with kinematic tensors. In this study, we present a methodology to quantify the dependence of the Reynolds stress tensor on mean kinematic tensor basis. The methodology is based upon tensor decomposition theorems which allow to extract from the anisotropic Reynolds stress tensor the part that is linear or nonlinear in the strain rate tensor D, and the parts that are in-phase (sharing the same eigenvectors) and out-of-phase with the strain rate.

The study was conducted using direct numerical simulation (DNS) data for turbulent plane channel (from Re _τ=180 to Re _τ=1000) and square duct flows (Re _τ=160). The results have shown that the tensorial form of the linear Boussinesq hypothesis is not a good assumption even in the region where production and dissipation are in equilibrium. It is then shown that the set of tensor basis composed by D, D ² and the persistence-of-straining tensor D·(W−Ω^D )−(W−Ω^D )·D, where W is the vorticity tensor and Ω^D is the rate of rotation of the eigenvectors of D, is able to totally reproduce the anisotropic Reynolds stress.

With the proposed methodology, the scalar coefficients of nonlinear algebraic turbulent models can be determined, and the adequacy of the tensorial dependence of the Reynolds stress can be quantified with the aid of scaled correlation coefficients.

Keywords:

• Reynolds stress
• DNS
• channel flow
• square duct flow
• tensor decomposition
• Boussinesq hypothesis

Acknowledgements

We would like to acknowledge CNRS (Centre National de la Recherche Scientifique), CNPq (Conselho Nacional de Pesquisa e Desenvolvimento), FAPERJ (Fundação de Amparo à Pesquisa do Rio de Janeiro), and PETROBRAS for their financial support. The channel flow DNS database was produced through granted access to the HPC resources of (CCRT/CINES/IDRIS) under the allocation 2009-i2009022277 made by GENCI (Grand Equipement National de Calcul Intensif).

Notes

¹Two tensors are coaxial if they share the same eigenvectors.

²Two tensors A and B are orthogonal if (and only if) tr(A·B ^T )=0, where tr(·) is the trace operator and the superscript T denotes transposition.

³Two tensors are coaxial if (and only if) they share the same eigenvectors. This condition is satisfied if (and only if) they commute.

⁴Since is symmetric, , the so-called double dot product.

⁵This means special cases for are not considered, the most obvious one being if is an isotropic tensor, , then commutation will also occur

⁶Subsets are tensors maintaining the same null components.