# If RANS are time-averaged then what does timesteps mean in simulations?

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 July 21, 2021, 12:54 If RANS are time-averaged then what does timesteps mean in simulations? #1 New Member   Join Date: Feb 2019 Posts: 2 Rep Power: 0 Hi all! I am relatively new into the turbulence modeling. I started with ANSYS and tried to use the k-e model for the first time. The first thing I noticed was the existence of timesteps. This made me confused. If the RANS equations do not contain time derivatives then how do we get timesteps? My initial assumption from a RANS simulation was just to get a result somehow describing the average velocity and pressure in a domain but I got timesteps with snapshots. I googled it for a couple of hours and the best thing I found was URANS, however I haven't specified anything regarding URANS in my simplistic project. Thanks in advance! aerosayan and aero_head like this.

July 21, 2021, 14:10
#2
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Filippo Maria Denaro
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Quote:
 Originally Posted by orkhan78 Hi all! I am relatively new into the turbulence modeling. I started with ANSYS and tried to use the k-e model for the first time. The first thing I noticed was the existence of timesteps. This made me confused. If the RANS equations do not contain time derivatives then how do we get timesteps? My initial assumption from a RANS simulation was just to get a result somehow describing the average velocity and pressure in a domain but I got timesteps with snapshots. I googled it for a couple of hours and the best thing I found was URANS, however I haven't specified anything regarding URANS in my simplistic project. Thanks in advance!

Indeed, RANS formulation is by definition for a statistically steady state, what you have to read carefully is the URANS formulation. In this latter, the equations are acted on by a different averaging and they remain time-dependent. However, be aware that the meaning of this time dependence is somehow specific only for the case of external forcing.
In this forum you can find some posts with detailed answers

 July 21, 2021, 17:16 #3 Senior Member   Join Date: Oct 2011 Posts: 242 Rep Power: 16 Not an Ansys user but there are different ways to achieve steady-state solutions. If you discard temporal terms from the governing equations, after spatial discretisation, you end up in solving a (eventually big) system of non-linear equations. You can (try to) solve for it using Newton's method, I highly doubt ANSYS implemented that. Usually the discretized equations maintain a time-derivative, the objective beeing in the case of a steady-state computation to make it vanish as fast as possible as transient is of no interest. Then time becomes fictitious and is solely used to maintain a stable numerical integration through some stability criteria (CFL<1 for explicit schemes for examples). EyzehuChacham, aerosayan and aero_head like this.

 July 22, 2021, 01:56 #4 Senior Member   Joern Beilke Join Date: Mar 2009 Location: Dresden Posts: 501 Rep Power: 20 When you go from Navier-Stokes to (U)RANS you do some averaging regarding the turbulent fluctuations. But not every transient effect is related to turbulence :-) You can have an inlet velocity that changes with time. EyzehuChacham and aerosayan like this.

July 22, 2021, 03:44
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Paolo Lampitella
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Quote:
 Originally Posted by orkhan78 Hi all! I am relatively new into the turbulence modeling. I started with ANSYS and tried to use the k-e model for the first time. The first thing I noticed was the existence of timesteps. This made me confused. If the RANS equations do not contain time derivatives then how do we get timesteps? My initial assumption from a RANS simulation was just to get a result somehow describing the average velocity and pressure in a domain but I got timesteps with snapshots. I googled it for a couple of hours and the best thing I found was URANS, however I haven't specified anything regarding URANS in my simplistic project. Thanks in advance!
Formally, in the more general sense, the (U)RANS average can be regarded as an ensemble average, over an infinite number of experiments. In the steady case, ergodicity ensures that the result formally matches the time average. For the unsteady case, the formal understanding is that you should only see unsteady effects related to the boundary conditions/geometry, i.e., not stochastic at all.

Some texts relate RANS to the infinite time average and URANS to the average over a time step, requiring the latter to be somehow larger than any dynamic time scale related to the turbulent fluctuations, which can't be unless the unsteadiness really is slow, quasi-steady if you know what I mean.

From a practical perspective, URANS is RANS with the unsteady term. The real problem here is that the turbulence model is not really different with respect to the steady case, besides its own added unsteady term.

We discussed some aspects here (where, by the way, you can also see, in the comments, a link to a work by Prof. Layton that, indeed, tries to cure the URANS anomaly in the model formulation).

 Tags rans, urans