About the viscosity in PDE

[mathscope29]

Hello,

I have a question about the viscosity in the considered PDE described in the paper of AirfRans (https://arxiv.org/pdf/2212.07564.pdf): $(u\nabla)u + \dfrac{1}{\rho}p = (\nu + \nu_t)\nabla^2u $

Is the viscosity in the output of the benchmark data $\nu$ or $\nu_t$, or $\nu + \nu_t$? I see that in the Notebook 1, it is the turbulent kinematic viscosity, i.e (as I understand) $\nu_t$. So how can we define $\nu$ is this case?

[daviddanan]

Hello, sorry for the delay, i have only noticed you issue now.

I asked the authors directly, here is their answer: "It is the turbulent kinematic viscosity that is predicted while nu is only a caracteristic known constant at a pressure of 1.025hPa and a temperature of 298.15K (approximately 1.55e-6 m²/s)".

Closing the issue.