Determination of the closure coefficients
In the theory section, it was mentionned that the closure coefficients were obtained empirically. In this example, the calibration process will be explained in more details for three canonical flows.
Shear flow in local-equilibrium
From dimensional analysis, remember that the turbulent viscosity νT was given by
(1) νT = Cμ(k2)/(ϵ)
As a first example, the closure coefficient Cμ will be extracted from experimental data. The production of kinetic energy in the flow is
from the Boussinesq assumption. Introducing Eq.(1↑) into Eq.(2↑), the following equation is obtained.
For shear layers in local-equilibrium (production P = dissipation ϵ), experimental results [2] suggest that uv ⁄ k≃0.3. Hence, it directly follows from Eq.(3↑) that Cμ = 0.09.
Decaying isotropic turbulence (grid turbulence)
The modeled equation for the TKE and dissipation of TKE are given in the theory section by
(4) (Dk)/(Dt) = (∂)/(∂xj)⎡⎣⎛⎝ν + (νT)/(σk)⎞⎠(∂k)/(∂xj)⎤⎦ + ⎡⎣νT⎛⎝(∂Ui)/(∂xj) + (∂Uj)/(∂xi)⎞⎠ − (2)/(3)kδij⎤⎦(∂Ui)/(∂xj) − CD(k3 ⁄ 2)/(l)
and
(5) (∂ϵ)/(∂t) + Uj(∂ϵ)/(∂xj) = (∂)/(∂xj)⎡⎣⎛⎝ν + (νT)/(σϵ)⎞⎠(∂ϵ)/(∂xj)⎤⎦ − Cϵ1(k)/(ϵ)ui’uj’(∂Ui)/(∂xj) − Cϵ2(ϵ2)/(k)
In the case of decaying isotropic turbulence, there is no mean velocity gradient in the flow and no transport phenomenon. Eq.(4↑) and Eq.(5↑) are respectively reduced to
It appears that only the closure coefficient Cϵ2 plays a role in this flow. According to experimental data [1], the turbulent kinetic energy decays as a power of time.
On the other hand, the modeled dissipation equation (Eq.(6↑)) becomes
(8) (dϵ)/(dt) ∝ − Cϵ2n2t − (n + 2)
In order to match these two expressions, it is required that n(n + 1) = n2Cϵ2, or Cϵ2 = 1 + 1 ⁄ n. Most experimental data [1] suggest that n≃1.3, so that Cϵ2≃1.77. However, this value somehow depends on the initial and boundary conditions of the experiment. In the standard k − ϵ model, Cϵ2 = 1.92 was chosen, which represents a value of roughly 1.1 for n.
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