The purpose of this post is to study the universal properties of fully developed turbulence \(\text{Re}\gg\text{Re}^*\sim 10^3\). Thanks to direct numerical simulation (DNS), there is strong evidence to suggest that the nonlinear advective term \(\left(\textbf v\cdot\frac{\partial}{\partial\textbf x}\right)\textbf v\) in the Navier-Stokes equations correctly captures turbulent flow in fluids. However, rather than trying to find analytical solutions \(\textbf v(\textbf x,t)\) that exhibit turbulence (which is clearly pretty hopeless), it makes sense to decompose \(\textbf v=\bar{\textbf v}+\delta\textbf v\) into a mean velocity field \(\bar{\textbf v}(\textbf x,t)\) superimposed by some fluctuations \(\delta\textbf v(\textbf x,t)\). The precise meaning of the word “mean” in the phrase “mean velocity field” for \(\bar{\textbf v}\) is time-averaged over some “suitable” period \(T\), also known as Reynolds averaging:
\[\bar{\textbf v}(\textbf x,t):=\frac{1}{T}\int_{t}^{t+T}\textbf v(\textbf x,t’)dt’\]
By construction, this implies that the Reynolds time average of the fluctuations vanishes \(\overline{\delta\textbf v}=\overline{\textbf v-\bar{\textbf v}}=\bar{\textbf v}-\bar{\textbf v}=\textbf 0\).
One can also check that \(\textbf v\) is incompressible if and only if both the Reynolds averaged flow \(\bar{\textbf v}\) and the fluctuations \(\delta\textbf v\) are also incompressible. One similarly works with the Reynolds averaged pressure \(p=\bar p+\delta p\) so that by design \(\overline{\delta p}=0\).
Substituting \(\textbf v=\bar{\textbf v}+\delta\textbf v\) and \(p=\bar p+\delta p\) into the Navier-Stokes equations and Reynolds averaging both sides of the equation yields the well-named Reynolds-averaged Navier-Stokes (RANS) equations:
\[\rho\left(\frac{\partial\bar{\textbf v}}{\partial t}+\left(\bar{\textbf v}\cdot\frac{\partial}{\partial\textbf x}\right)\bar{\textbf v}\right)=-\frac{\partial\bar p}{\partial\textbf x}+\eta\left|\frac{\partial}{\partial\textbf x}\right|^2\bar{\textbf v}-\rho\overline{\left(\delta\textbf v\cdot\frac{\partial}{\partial\textbf x}\right)\delta\textbf v}+\bar{\textbf f}\]
Or in a Cartesian basis:
\[\rho(\dot{\bar{v}}_i+\bar v_j\partial_j\bar v_i)=\partial_j\bar{\sigma}_{ij}+\bar{f}_i\]
where the Reynolds averaged stress tensor \(\bar{\sigma}\) now includes an additional turbulent contribution \(\bar{\sigma}_{\text{Reynolds}}=-\rho\overline{\delta\textbf v\otimes\delta\textbf v}\) known as the Reynolds stress:
\[\bar{\sigma}_{ij}=-\bar p\delta_{ij}+\eta(\partial_j\bar v_i+\partial_i\bar v_j)-\rho\overline{\delta v_i\delta v_j}\]
(this can be quickly checked using the incompressibility conditions \(\partial_j\bar v_j=\partial_j\delta v_j=0\)).
At this point, assuming that the external body forces have no fluctuations \(\delta\textbf f=\textbf f-\bar{\textbf f}=\textbf 0\), one can subtract the RANS equations from the original Navier-Stokes equations to obtain:
\[\rho\left(\frac{\partial\delta\textbf v}{\partial t}+\left(\bar{\textbf v}\cdot\frac{\partial}{\partial\textbf x}\right)\delta\textbf v+\left(\delta\textbf v\cdot\frac{\partial}{\partial\textbf x}\right)\bar{\textbf v}+\delta\left(\left(\delta\textbf v\cdot\frac{\partial}{\partial\textbf x}\right)\delta\textbf v\right)\right)=-\frac{\partial\delta p}{\partial\textbf x}+\eta\left|\frac{\partial}{\partial\textbf x}\right|^2\delta\textbf v\]
Taking the outer product of both sides with \(\delta\textbf v\) and then Reynolds averaging yields (to be added: closure problem, Boussinesq approximation as a closure model).
Dimensional Analysis