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Reynolds-Averaged Navier-Stokes formulation in the rotating frame.

Acceleration term expressed for a rotating frame around the z axis ($\vec \Omega$).

Notation: I: inertial, R: rotating

General vector:

$\left [ \frac{d \vec A}{dt} \right ]_I = \left [ \frac{d \vec A}{dt} \right ]_R + \vec \Omega \times \vec A$

Position vector:

$\left [ \frac{d \vec r}{dt} \right ]_I = \left [ \frac{d \vec r}{dt} \right ]_R + \vec \Omega \times \vec r$

$\vec u_I = \vec u_R + \vec \Omega \times \vec r$

For the accelration, the velocity vector is:

$\left [ \frac{d \vec u_I}{dt} \right ]_I = \left [ \frac{d \vec u_I}{dt} \right ]_R + \vec \Omega \times \vec u_I$

$\left [ \frac{d \vec u_I}{dt} \right ]_I = \left [ \frac{d \left [ \vec u_R + \vec\Omega \times \vec r \right ] }{dt} \right ]_R + \vec \Omega \times \left [ \vec u_R + \vec \Omega \times \vec r \right ]$

$\left [ \frac{d \vec u_I}{dt} \right ]_I = \left [ \frac{d \vec u_R}{dt} \right ]_R + \frac{d \vec \Omega}{dt} \times \vec r + \vec \Omega \times \underbrace{ \left [ \frac{\vec r}{dt} \right ]_R }_{\vec u_R} + \vec \Omega \times \vec u_R + \vec \Omega \times \vec \Omega \times \vec r$

$\left [ \frac{d \vec u_I}{dt} \right ]_I = \left [ \frac{d \vec u_R}{dt} \right ]_R + \frac{d \vec \Omega}{dt} \times \vec r + 2 \vec \Omega \times \vec u_R + \vec \Omega \times \vec \Omega \times \vec r$ Eqn [1]

Navier-Stokes equations in the inertial frame

$\begin{cases} \frac {D \vec u_I}{D t} = - \nabla (p/\rho) + \nu \nabla \cdot \nabla (\vec u_I) \\ \nabla \cdot \vec u_I = 0 \end{cases}$ Eqn [2]

$\begin{cases} \frac {\partial \vec u_I}{\partial t} + \vec u_I \cdot \nabla \vec u_I = - \nabla (p/\rho) + \nu \nabla \cdot \nabla (\vec u_I) \\ \nabla \cdot \vec u_I = 0 \end{cases}$

$\begin{cases} \frac {\partial \vec u_I}{\partial t} + \nabla \cdot (\vec u_I \otimes \vec u_I) - \underbrace{\nabla \cdot \vec u_I}_{0} \vec u_I= - \nabla (p/\rho) + \nu \nabla \cdot \nabla (\vec u_I) \\ \nabla \cdot \vec u_I = 0 \end{cases}$

$\begin{cases} \frac {\partial \vec u_I}{\partial t} + \nabla \cdot (\vec u_I \otimes \vec u_I) = - \nabla (p/\rho) + \nu \nabla \cdot \nabla (\vec u_I) \\ \nabla \cdot \vec u_I = 0 \end{cases}$ Eqn [3]

Let's look at the left-hand side of the momentum equation of Eqn [2], by taking into account Eqn [1] for the acceleration term:

$\frac {D \vec u_I}{D t} = \frac{D \vec u_R}{Dt} + \frac{d \vec \Omega}{dt} \times \vec r + 2 \vec \Omega \times \vec u_R + \vec \Omega \times \vec \Omega \times \vec r$

$\frac {D \vec u_I}{D t} = \frac {\partial \vec u_R}{\partial t} + \vec u_R \cdot \nabla \vec u_R + \frac{d \vec \Omega}{dt} \times \vec r + 2 \vec \Omega \times \vec u_R + \vec \Omega \times \vec \Omega \times \vec r$

$\frac {D \vec u_I}{D t} = \frac {\partial \vec u_R}{\partial t} + \nabla \cdot (\vec u_R \otimes \vec u_R) + \frac{d \vec \Omega}{dt} \times \vec r + 2 \vec \Omega \times \vec u_R + \vec \Omega \times \vec \Omega \times \vec r$ since $\nabla \cdot \vec u_R = \nabla \cdot \vec u_I = 0$

$\nabla \cdot \vec u_I = \nabla \cdot \left [ \vec u_R + \vec \Omega \times \vec r \right ]$

$\nabla \cdot \vec u_I = \nabla \vec u_R + \underbrace{\nabla \left [ \vec \Omega \times \vec r \right ]}_{0}$

$\nabla \cdot \vec u_I = \nabla \vec u_R$