SU2 is an open-source CFD suite that is built around a RANS-solver. The main PDE that is solved, is the following:

$$ \frac{\partial}{\partial t} \mathbf{U} + \nabla \cdot \mathbf{F^c} - \nabla \cdot \mathbf{F^v} = \mathbf{Q} \qquad \mathrm{in}\ \Omega, t > 0 $$

In this equation, $\mathbf{U}$ is the state vector and $\mathbf{F^c}$ and $\mathbf{F^v}$ represent the convective and viscous fluxes, respectively.

Palacios et al. (2013) state the following (p. 13):

Naturally, both the laminar Navier-Stokes and Euler equations are also available in the code as subsets of the RANS equations by disabling turbulence modeling and by completely removing viscosity, respectively.

This, however, is not so 'natural' to me. My thoughts/ideas so far:

Turbulent RANS to laminar NS

In the RANS-case, part of $\mathbf{F^v}$ is the term $v_j \tau_{ij} + \mu^*_{tot} C_p \frac{\partial}{\partial i} T$ (part of the energy equation). As $\mu_{tot}$ (as well as $\mu^*_{tot}$) is a combination of $\mu_{dyn}$ (dynamic viscosity) and $\mu_{turb}$ (turbulent viscosity). Given the quote on 'disabling turbulence modelling', my guess would be that that would mean setting $\mu_{turb} = 0$ (such that $\mu_{tot} = \mu_{dyn}$). Is that indeed correct?

Turbulent RANS to inviscid Euler

I get that setting $\mathbf{F^v} = 0$, the inviscid Euler equations are obtained from the main PDE above. However, taking a more theoretical approach, I would think that the process of averaging flow variables (into mean and time-fluctuating parts) that one uses to obtain RANS from NS, somehow interferes with that. Put differently: if we can go from RANS to Euler, that would imply that all viscous effects are contained in these time-fluctuating parts. On one hand that makes sense, on the other hand, I'm not so sure.

Follow-up question: does this hold for RANS formulations in general, or only the specific implementation used in SU2?


1 Answer 1


The only difference between RANS and Navier Stokes equations is the Reynolds' stress tensor: $$\tau_R = -\overline{v'\otimes v'}$$ This term is modeled using a turbulent viscosity and therefore: $$\tau_{R}\propto \mu_t$$ In the limit when $\mu_t\to 0$ the RANS equations converge to NS.

Viscous effects are not within any part: nor fluctuating neither averaged. Viscous effects are owed by the nature of the fluid itself. They are relevant when the Reynolds number is high and negligible when low. In turbulence these are always relevant (dissipative effects).

From this, it is also natural to set $\mu\to 0$ to obtain the Euler equations from the NS. Note that the process would be RANS$\to$ NS $\to$ Euler. But the natural process would have been: NS $\to$ RANS or NS$\to$ Euler.

Please ask whatever is not clear.


The regime of the fluid (Laminar-Transition-Turbulent) depends on the Reynolds number. I mean the term of study is: $$\frac{1}{Re}\triangle \vec{v}$$ This term is always dissipative (it takes kinetik energy from the fluid and transforms it into heat) makes the fluid to turn, creating boundary layers and recirculation bubbles.

The higher the reynolds the lower the scale of energy dissipation (more or less the size of the recirculation bubble or eddy. This length scale has a minimum threshold in which the all kinetic energy is dissipated into heat, called Kolmogorov length scale (check it).

The fact in which $\mu_t\to 0$ yields to nothing. We agreed with that in this case we recovered the full Navier Stokes equations in which the effect of turbulence is implicitly defined and it depends on the value of the Reynolds number.

  • $\begingroup$ If I understand you correctly, my thoughts on going from RANS $\rightarrow$ laminar NS are indeed correct. Just to check: $\mu_t \rightarrow 0$ yields the laminar NS, right? For the Euler part (I also feel NS $\rightarrow$ Euler to be much more logical), setting $\mu \rightarrow 0$ still leaves the $v_j \tau_{ij}$-term, which is a viscous term, correct? $\endgroup$
    – Bram
    Commented Aug 29, 2018 at 9:48
  • $\begingroup$ Okay, let us dive deeper. I will explain it in the edit. $\endgroup$
    – HBR
    Commented Aug 29, 2018 at 9:50
  • $\begingroup$ If you put $\mu\to 0$ then $\tau\to 0$, even in the energy equation. This is for an Euler fluid, there are no heat losses because they are isentropic $s=constant$. $\endgroup$
    – HBR
    Commented Aug 29, 2018 at 10:03
  • $\begingroup$ Note that for some authors (and packages) RANS is only valid when time averaging on the longest scales, so that the transient $\frac{\partial}{\partial t}$ term vanishes. $\endgroup$
    – origimbo
    Commented Aug 29, 2018 at 14:36

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