# What is the turbulent Navier-Stokes equation for cylindrical coordinates?

I am looking for turbulent Navier Stokes equation for cylindrical coordinates.

I know that RANS (Reynolds Averaged Navier Stokes) eq. is the solution, I understood the point of it but only for Cartesian coordinates.

Without killer mathematical expressions, can I ask the formula ? You will see what I mean by saying "killer mathematical expressions" by looking at the picture I post here.

Perhaps you are looking for a coordinate-free RANS equation, which you can then write in any coordinate system of your choice. The actual incompressible NS equation, with usual choice for symbols, is: $$\nabla\cdot\tilde{\mathbf{u}}=0\\ \frac{\partial\tilde{\mathbf{u}}}{\partial t}+\tilde{\mathbf{u}}\cdot\nabla\tilde{\mathbf{u}}=-\frac{\nabla \tilde{p}}{\rho}+\nu\nabla^2\tilde{\mathbf{u}}$$ in which the quantities with the tilde are total quantities. We split the total quantity into the sum of mean and fluctuating parts: $$\left<\tilde{\mathbf{u}}\right>=\mathbf{U},\left<\tilde{p}\right>=P\\ \tilde{\mathbf{u}}=\mathbf{U}+\mathbf{u},\quad\tilde{p}=P+p\\ \left<\mathbf{u}\right>=\left<p\right>=0$$ in which $\left<\right>$ is the averaging operator, and being an ensemble average, commutes with the time and spatial derivative operators. Averaging the continuity equation gives: $$\left<\nabla\cdot\tilde{\mathbf{u}}\right>=0\\ \nabla\cdot\left<\tilde{\mathbf{u}}\right>=0\\ \nabla\cdot\mathbf{U}=0$$ Therefore $$\nabla\cdot\tilde{\mathbf{u}}=\nabla\cdot(\mathbf{U}+\mathbf{u})=0\\ \nabla\cdot\mathbf{u}=0$$

Similarly averaging the NS equation gives: $$\frac{\partial\mathbf{U}}{\partial t}+\left<\tilde{\mathbf{u}}\cdot\nabla\tilde{\mathbf{u}}\right>=-\frac{\nabla P}{\rho}+\nu\nabla^2\mathbf{U}$$ Now $$\left<\tilde{\mathbf{u}}\cdot\nabla\tilde{\mathbf{u}}\right>=\left<(\mathbf{U}+\mathbf{u})\cdot\nabla (\mathbf{U}+\mathbf{u})\right>\\ =\mathbf{U}\cdot\nabla\mathbf{U}+\left<\mathbf{u}\cdot\nabla\mathbf{u}\right>$$ Since $\nabla\cdot(\mathbf{u}\mathbf{u})=\mathbf{u}\cdot\nabla\mathbf{u}+\mathbf{u}\nabla\cdot\mathbf{u}=\mathbf{u}\cdot\nabla\mathbf{u}$, the RANS equation in coordinate-free notation is: $$\nabla\cdot\mathbf{U}=0\\ \rho\left( \frac{\partial\mathbf{U}}{\partial t}+\mathbf{U}\cdot\nabla\mathbf{U}\right)=-\nabla P+\mu\nabla^2\mathbf{U}+\nabla\cdot (-\rho\left<\mathbf{u}\mathbf{u}\right>)$$ in which $-\rho\left<\mathbf{u}\mathbf{u}\right>\equiv\sigma_R$ is the Reynolds stress tensor, which is obviously symmetric.

In the picture that you have posted $\sigma_R$ is written in Cartesian coordinates, whose orthonormal coordinate vectors are $\{\mathbf{e}_x,\mathbf{e}_y,\mathbf{e}_z\}$. The box you have drawn on the picture actually shows the matrix representation of $\sigma_R$ in that coordinate system. To write $\sigma_R$ in cylindrical coordinates you must adopt, at a given point in the flow field, orthonormal coordinate vectors $\{\mathbf{e}_r,\mathbf{e}_\theta,\mathbf{e}_z\}$ but the procedure of multiplying out to get the components of the tensor $\mathbf{u}\mathbf{u}$ is the same. While taking divergence however beware the fact that $\{\mathbf{e}_r,\mathbf{e}_\theta,\mathbf{e}_z\}$ are not constant vectors.