Equilibrium states

Some authors use simplifying assumptions to derive more compact forms of the expressions for the solution in (76). In the following, a few examples, which are special cases of the general solution discussed here, are reviewed.

In deriving their version of the general solution (76), Canuto et al. (2001) e.g. assumed $ P_b=\epsilon_b$ and constant $ r$. Under these conditions, because of (74), the dependence on $ \overline{T}$ dissapears, and the counter-gradient term $ \Gamma_B$ in (76) drops. It was further assumed that $ P+G=\epsilon$ in (67) only, leading to $ {\cal N} = (c_1 + c_1^*)/2$ and $ {\cal N}_b=c_{b1}$. These particularly simple expressions linearize the system, and a fully explicit solution can be obtained, provided $ k$ and $ \epsilon$ are known. Burchard and Bolding (2001) adopted the solution of Canuto et al. (2001) and complemented it by $ k$ and $ \epsilon$ computed from dynamical equations (`$ k$-$ \epsilon$ model').

In contrast, Canuto et al. (2001) and Cheng et al. (2002) decided for a further simplification. They solved (76) with $ k$ and $ \epsilon$ from algebraic expressions. In their case, $ k$ followed from the approximation $ P+G=\epsilon$ of (152) (see section 4.17), and $ \epsilon$ from a prescribed length-scale.

Using (76), (79), and (80), it is easy to show that the assumption $ (P+G)/\epsilon$ leads to

$\displaystyle N_n \overline{S}^2 - N_b \overline{N}^2 - D = 0 \; , \quad$ (81)

which is polynomial equation in $ \overline{S}$ and $ \overline{N}$. This expression can be used to replace one of the latter two variables by the other. An interesting consequence is the fact that all non-dimensional turbulent quantities can be expressed in terms of the Richardson number $ Ri = \overline{N}^2 / \overline{S}^2$ only. Replacing $ \overline{N}^2$ by $ \overline{S}^2 Ri$ in (81), a quadratic equation for $ \alpha_M = \overline{S}^2$ in terms for $ Ri$ can be established (see e.g. Cheng et al. (2002). Using the definitions given in section 4.26, this equation can be written as

$\displaystyle \alpha_M^2 \left( -d_5 + n_2 - \left( d_3 - n_1 + n_{b2}\right) R...
..._M \left( -d_2 + n_0 - \left( d_1 + n_{b0} \right) Ri \right) - d_0 = 0 \quad .$ (82)

The solution for $ \alpha_M$ can, via (81), be used to expressed also $ \overline{N}^2$ in terms of $ Ri$. This implies that also the stability functions and hence the complete solution of the problem only depends on $ Ri$.

Investigating the solution of the quadratic equation (82), it can be seen that $ \alpha_M$ becomes infinite if the factor in front of $ \alpha_M^2$ vanishes. This is the case for a certain value of the Richardson number, $ Ri=Ri_c$, following from

$\displaystyle -d_5 + n_2 - \left( d_3 - n_1 + n_{b2}\right) Ri_c - \left( d_4 + n_{b1} \right) Ri_c^2 = 0 \quad .$ (83)

Solutions of this equation for some popular models are given in table 1. For $ Ri=Ri_c$, equilibrium models predict complete extinction of turbulence. For non-equilibrium models solving dynamical equations like (152), however, $ Ri_c$ has no direct signifcance, because turbulence may be sustainned by turbulent transport and/or the rate term.

Table 1: Critical Richardson number for some models
$ 0.47$ $ 0.24$ $ 0.85$ $ 1.02$ $ 0.96$

Karsten Bolding 2012-12-28