I have a question. Any turbulent simulation which is performed in fluent is it RANS or URANS. Please also tell me the difference between RANS and URANS. Thanks alot for your reply. prasat

The difference resides in the concept of the time step, and in fact some people would and do argue the validity of unsteady RANS. In essence if you solve your equations with one global time step, which is used in every cell, and if value of the time step is small enough then you will be able to capture fluctuations, or unsteady behaviour in the MEAN quantities. In other words your solution is time accurate. Steady RANS, or RANS, marches the solution with a local optimised time step for each cell, and hence is not time accurate, you will get a faster solution, but it will be steady state.

Hope this helps

Andy

I have a question here.all the turbulence unsteady simulation performed by Fluent or any other cfd software is it RANS or not. If yes how do they select the time step for averaging. Prasat

Reynolds averaging consists in averaging over realizations. So the mean value (Reynolds way) of the physical variable G(x,t) is [G(x,t)], where x is the spatial position and t the time since the beginning of the experiment. Thus, Reynolds averaging does not mean time averaging. This point is used for ICE computation : the time dependance is given by the cranck angle. The fact is that under certain conditions, time averaging will converge - when the period of averaging is large enough - to the Reynols averaging value. So people often forget the time dependence of the RANS value.

The RANS solution of the wake behind a cylinder must be symetrical : nobody knows where the first Karman's vortex will appear, then both sides of the cylinder are equiprobables. That doesn't mean that the wake must be steady : the size of the recirculating pocket may grow with time before reaching its asymptotical value.

From this point of view, URANS can not be just unsteady RANS computation : reducing the time step can not be enougth to get the correct solution since the consistancy of the turbulence model is not conserved.

Thus, Reynolds averaging does not mean time averaging Is this correct comment which you have made sylvain

I think a few people are missing the point, there is a large difference between time averaging and ensemble averaging, and thus there is a large difference between the definition, both physically and mathematically, of the mean value in a RANS solution. The previous post talked about averaging with respect to IC engines this is then the average over many realisations of the flow, and is an ensemble average, not to be confused with a time average. With respect to your question Reynolds averaging does mean time averaging if it is time averaged, you are correct, you should however be aware of the fact that there is also an ensemble average which is different.

Yes, Reynolds averaging does not mean time averaging.

In the case of the wake behind a cylinder, time averaging converges to Reynols averaging, and time dependent RANS solutions have no physical meaning.

For ICE, time averaging does not have sens, and time dependent RANS solutions have physical meaning.

[tushar]

Reynolds averaging is a time averaging (http://www.cfd-online.com/Wiki/Reynolds_averaging). People have developed URANS model by separating time scale of mean motion and time scales of turbulent motion. They have assumed that "turbulent time scale" << "mean flow time scale". The basic assumption in URANS is that averaging time (DT) is higher than turbulent time scale but much lower than mean flow time scale. Many physical cases does not obey this rule for example jet flows have approximately same order of time scale for both, at jet boundaries. URANS really didnt work here. we do separate turbulent scales in LES but there we resolve the large scale and model the small scale motions. URANS is not resolving mean flow variation.. its complected to believes on URANS

[Sandeep Nain]

I guess people here are getting confused with the Reynolds Averaging procedure. Reynolds averaging is actually time average only but further it can be subdivided into categories like running time averaging, grid-cell averaging and ensemble averaging. Each one of these are time averaged, but as explained by David C Wilkox in "Turbulence Modeling for CFD" Reynolds averaging can also be done over spatial coordinates for the Homogeneous turbulence. And mostly used ensemble averaging is actually nothing but the time averaging of many identical processes at certain time. This can be understood by referring to 2nd chapter of above mentioned text book as well as following paper
http://kiwi.atmos.colostate.edu/grou..._Averaging.pdf

Though its quite an old thread but I hope it will be helpful for new readers to understand this..

[randolph]

Originally Posted by Andy
;25884

The difference resides in the concept of the time step, and in fact some people would and do argue the validity of unsteady RANS. In essence if you solve your equations with one global time step, which is used in every cell, and if value of the time step is small enough then you will be able to capture fluctuations, or unsteady behaviour in the MEAN quantities. In other words your solution is time accurate. Steady RANS, or RANS, marches the solution with a local optimised time step for each cell, and hence is not time accurate, you will get a faster solution, but it will be steady state.

Hope this helps

Andy

It looks like I may have some misunderstanding. Could you explain in more details that why RANS is not steady state solution? Or whats your definition of steady-state?

[randolph]

Originally Posted by tushar

Reynolds averaging is a time averaging (http://www.cfd-online.com/Wiki/Reynolds_averaging). People have developed URANS model by separating time scale of mean motion and time scales of turbulent motion. They have assumed that "turbulent time scale" << "mean flow time scale". The basic assumption in URANS is that averaging time (DT) is higher than turbulent time scale but much lower than mean flow time scale. Many physical cases does not obey this rule for example jet flows have approximately same order of time scale for both, at jet boundaries. URANS really didnt work here. we do separate turbulent scales in LES but there we resolve the large scale and model the small scale motions. URANS is not resolving mean flow variation.. its complected to believes on URANS

Could you explain more why URANS is not resolve the mean flow variation?

Thanks in advance

[LuckyTran]

Originally Posted by randolph

Could you explain more why URANS is not resolve the mean flow variation?

Thanks in advance

This explanation is fairly subjective because it depends on how you define "mean flow" and "turbulence". It arises because people assume turbulence is fast and mean-flow slow based on physical grounds. If you consider the mean flow as the coherent part and turbulence as random part of the flow (via statistical arguments), then the story is completely different.

URANS is a Reynolds-Averaged equation. When you do this reynolds-averaging you assume that you average long enough so that you resolve the mean. That is, you assume you CAN actually find the average. If the flow time-scales and turbulent time-scales are similar and you don't average long enough the turbulent contribution will not be zero and you therefore get a mean-flow that is contaminated by turbulence statistics (which is not the true mean flow). This to me in a user-error and not a problem with URANS. If you don't sample long enough to get the correct statistics then you simply get the wrong statistics.

[randolph]

Originally Posted by LuckyTran

URANS is intended to resolve the mean-flow variation. However, under some flow situations it doesn't quite accomplish this feat.

URANS is a Reynolds-Averaged equation. When you do this reynolds-averaging you assume that you average long enough so that you resolve the mean. That is, you assume you CAN actually find the average. If the flow time-scales and turbulent time-scales are similar, this assumption fails because when you take the average, the turbulent contribution will not be zero and you therefore get a mean-flow that is contaminated by turbulence statistics (which is not the true mean flow).

This explanation is fairly subjective because it depends on how you define "mean flow" and "turbulence". It arises because people assume turbulence is fast and mean-flow slow. If you consider the mean flow as the coherent part and turbulence as random part of the flow, then the story is completely different.

Sorry I do not understand.

1.Seemly you are talking about RANS. For a stationary ergodic dynamical system, your infinite time-average is equivalent to ensemble averaging. So I do not understand "When you do this reynolds-averaging you assume that you average long enough so that you resolve the mean".

2. I do not understand "If the flow time-scales and turbulent time-scales are similar". What do you mean by flow time-scale and turbulent time-scales?

3. "If the flow time-scales and turbulent time-scales are similar and you don't average long enough the turbulent contribution will not be zero and you therefore get a mean-flow that is contaminated by turbulence statistics (which is not the true mean flow)." I think you are talking about that, if, in URANS, one use time scale that is larger than the integral length scale, one will force turbulence model to model problem-depended flow structure and lead to unreliable results. Instead, the time step in URANS should be chosen small enough (smaller than integral time scale) so that it will resolve problem-depended flow structure and model the problem-independed fluctuation with turbulence model.

[FMDenaro]

I think that the question about RANS vs. URANS was already answered in other posts.
However, it is quite simple to address the difference by the definitions

RANS: <f>(x) = lim T->+Inf (1/T) Int [t0,T] f(x,t) dt
URANS <f>(x,t) = lim N->+Inf (1/N) Sum [k=1,..N] fk(x,t)

[randolph]

I totally agree. However, the physical interpolations that associated are not always obvious (At least to me).

[einstein_zee]

Originally Posted by randolph

Sorry I do not understand.

1.Seemly you are talking about RANS. For a stationary ergodic dynamical system, your infinite time-average is equivalent to ensemble averaging. So I do not understand "When you do this reynolds-averaging you assume that you average long enough so that you resolve the mean".

2. I do not understand "If the flow time-scales and turbulent time-scales are similar". What do you mean by flow time-scale and turbulent time-scales?

3. "If the flow time-scales and turbulent time-scales are similar and you don't average long enough the turbulent contribution will not be zero and you therefore get a mean-flow that is contaminated by turbulence statistics (which is not the true mean flow)." I think you are talking about that, if, in URANS, one use time scale that is larger than the integral length scale, one will force turbulence model to model problem-depended flow structure and lead to unreliable results. Instead, the time step in URANS should be chosen small enough (smaller than integral time scale) so that it will resolve problem-depended flow structure and model the problem-independed fluctuation with turbulence model.

I think what he meant is that when you want to do the time averaging you need to pick it (T) much bigger than the frequency (or better to say period) of the fluctuations, but it has to be smaller than the system time scale so that you don't lose information. And for sure steadiness of RANS doesn't imply that turbulence is steady, we do know that it is completely unsteady. But it means that the behaviour of the mean flow is steady since we solve the equations for the mean quantities.

[Rachaelgraves]

RANS and URANS are two of the most popular numerical methods used to solve problems in fluid dynamics. They are both based on the Navier-Stokes equations, which govern the motion of fluids. RANS is an acronym for Reynolds Averaged Navier-Stokes, while URANS is an acronym for Unsteady Reynolds Averaged Navier-Stokes. Both methods are used to approximate the solutions to the Navier-Stokes equations.

RANS is a statistical approach that averages the fluid flow over time. This method is suitable for problems with a steady state flow, such as incompressible laminar flow. URANS is a more sophisticated approach that takes into account the unsteady nature of turbulent flow. It is capable of handling both compressible and incompressible flows.

[FMDenaro]

Originally Posted by Rachaelgraves

RANS and URANS are two of the most popular numerical methods used to solve problems in fluid dynamics. They are both based on the Navier-Stokes equations, which govern the motion of fluids. RANS is an acronym for Reynolds Averaged Navier-Stokes, while URANS is an acronym for Unsteady Reynolds Averaged Navier-Stokes. Both methods are used to approximate the solutions to the Navier-Stokes equations.

RANS is a statistical approach that averages the fluid flow over time. This method is suitable for problems with a steady state flow, such as incompressible laminar flow. URANS is a more sophisticated approach that takes into account the unsteady nature of turbulent flow. It is capable of handling both compressible and incompressible flows.

That has nothing to do with steady laminar flows!