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V. G. Ferreira December 15, 2000 15:10

About Reynolds Averaged Navier-Stokes Equations
 
Hi Dear friends. In many and many papers, I have seen that the Reynolds averaged Navier-Stokes equations contains a temporal derivative for mean velocity. But, in the Reynolds decomposition, the mean velocity is time-independent. Please, can someone explain this confusion?

Gabriel Usera December 15, 2000 16:02

Re: About Reynolds Averaged Navier-Stokes Equation
 
Hi:

Reynolds treatment of turbulence involves a time averaging procedure. If your mean (in time) flow is stationary, everything is clear, and the mean velocity is time independent.

But, more often than not, your mean flow is unsteady. So, how do we define an average in time for such flows ?. Somehow one must define some time length at which the averaging is done. The code will simulate the variations in time that are slower than this reference time scale. Variations that are faster than this scale would be averaged out, and its influence over the mean flow must be modelled. That is where RANS turbuelence models come into play.

So the time derivative for the mean flow in the RANS equations accounts for variations at time scales larger than those of turbulence.

Hope that helps.

Gabo.

Ghanshyam Singh December 16, 2000 00:56

Re: About Reynolds Averaged Navier-Stokes Equation
 
You are right Gabo that the time length must be smaller then the "time scale" one is looking for to capture. This is very important when we are having pseudo time marching scheme. In that if we want to capture a certain "time scale", in my view one must FIRST get a steady state solution and again run the code for the time length lesser than the "time scale" we are looking for. Of course with a suitable small "deata-t". Am I correct?

GS

John C. Chien December 16, 2000 03:33

Re: About Reynolds Averaged Navier-Stokes Equation
 
(1). The answer to your question is well illustrated in Figure 2.3 and Figure 2.4 , Section 2.1 Reynolds Averaging, of the book "Turbulence Modeling for CFd" by David Wilcox. (2).For steady-state mean motion, it is written as: ui(x,t)=Ui(x)+ui'(x,t). In this case, the mean flow is not a function of time, that is Ui(x). (3). For transient mean flow, it is written as: ui(x,t)=Ui(x,t)+ui'(x,t). In this case the mean flow Ui(x,t) is transient. (4). In the transient mean flow case, the time period used to obtain the Reynolds average must be relatively short when compared with the transient mean motion. (the other way to say is that the mean motion must be slow varying motion when compared with the turbulence fluctuations.) In this case, ui'(x,t) is measured from the Ui(x,t), which is a slow varying function of time.(relative to the turbulence fluctuation)

Lars Ola Liavåg December 18, 2000 02:33

Re: About Reynolds Averaged Navier-Stokes Equation
 
In the case of transient mean flow, you might look at the averaging procedure as an ensemble averaging, i.e. an average over multiple repeatitions of a case. Then the time-dependent mean does not present a problem, and the averaging procedure is formally the same.

Best regards,

Lars Ola

sylvain December 18, 2000 06:52

Re: About Reynolds Averaged Navier-Stokes Equation
 
That's perfectly true.

Moreover, time averaging procedure was introduce in turbulence modelling just because it tends to the same result than ensemble averaging procedure for steady flows.

For example if you are looking to the flow throught a channel of half-wide 0.1m, at speed U equal to 10m/s and with turbulence intensity (sqrt(u'u')/U) equal to 20% then :

- you will have to wait 8seconds before the time averaging speed tends to the ensemble averaging speed with an error less than 1%.

- you will have to wait 200seconds for double correlation, and you will have to wait 530s for fourth order correlation.

Best regards,

Sylvain


Wilhelm April 26, 2012 11:36

Reynolds averaging makes only sense for stationary flows regardless whether the averaging operator is a time average or an ensemble average.
The time derivative N-S term vanishes for the time average operator because U is not a function of time (by definition of the operator).

But a stationary flow is ergodic what means that the ensemble mean converges to the time average (ergodic theorem).
From that follows that the time derivative of an ensemble mean vanishes too.
After all that's what one expects of a stationary flow - vanishing of all time derivatives.

However if the flow is not stationary then it may be non ergodic and the ensemble mean no longer converges to the time average. It is not known whether N-S is ergodic or not and it probably isn't for all flows.
Formally the time derivative of the time average vanishes again but the ensemble average does not necessarily do so.
That's why if one wanted to use RANS, then it would be necessary to use the ensemble average operator but RANS isn't famous to deal well with non stationary flows anyway.
LES or hybrids is better.


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