CFD Online Discussion Forums

CFD Online Discussion Forums (https://www.cfd-online.com/Forums/)
-   Main CFD Forum (https://www.cfd-online.com/Forums/main/)
-   -   unsteady flow, when do you use steady when do you switch to transient (https://www.cfd-online.com/Forums/main/108848-unsteady-flow-when-do-you-use-steady-when-do-you-switch-transient.html)

jpc November 2, 2012 11:46

unsteady flow, when do you use steady when do you switch to transient
 
this "philisophical" debate seems to be coming up more and more often.

flow is almost always turbulent and in essence, all problems are transient.

but generally for most problems you're interested in trends and are ok with a steady state solution.

when do you make the cross-over to transient?

some ideas:
1. at a certain reynolds number?
2. when you know a specific phenomena may occur? (vortex shedding for example...are there others? how do you know )
3. for any particular case of interest from the trend studies?
4. for final analyses? (for all or just the ones of interest?)

side question, when working with someone that isn't tuned into the world of CFD or fluid dynamics. how do you best explain turbulent flow vs unsteady flow? I found a really good example using a river, but am interested in any others people might have.

vicarious November 2, 2012 13:14

Hello jpc,
As you mentioned, some flow behaviors should be considered unsteady all the time, for example vortex shedding, combustion, chemical reaction, multi-phase flow, turbulent studies etc. In my opinion, When sth specific is changing in your domain you need to consider unsteady simulation, But sometimes you want to determine a performance of a system, for example drag/drift coefficients of an airfoil, then you can run a steady simulation. Although some cases could be treated unsteady at the first place and then become steady. briefly it depends to what goal you are after in a numerical simulation.
Regarding to your last question, turbulent flow is always a unsteady phenomena. In some turbulent cases, you cannot run a steady simulation because of high turbulent intensity and the solver cannot resolve the high gradients of parameters even with relaxations. Other cases of turbulent flows result in a steady pattern of flow when you need to determine a specific performance, pressure drop in a straight pipe at igh Re, but inside is a high turbulent flow, high pressure gradients and etc.
Hope it could help.

jpc November 2, 2012 13:22

hi vicarious, thank you for your input.

regarding the situations that need to be run as unsteady (transient), how do you determine which need to be run that way?

do you just have to be an expert in the phenomena?

are there specific parameters to look at..etc.

vicarious November 2, 2012 13:58

As I said, It mostly depends on to what exact purpose you are interested in. Of course expertise is important. Parameters do not determine the unsteadiness of flow. In a low Re fluid flow, just because the flow is laminar, does not mean it is steady. You need to realize that the phenomena you are tracking has what kind of effect in your flow field. Another simple example, Filling a tank with a fluid cannot be simulated steady because the pressure is always changing, or all the problems involve a secondary flow, turbomachinery studies & etc.

FMDenaro November 2, 2012 16:38

Steady simulations can be performed also for turbulent flow that are in energetic equilibrium. This is however a statistical meaning of the steady state, and is usually associated with RANS formulation.
URANS is a transient simulation in statistical sense.
LES and DNS are transient simulations with different meaning in the resolved fields.
A deterministic steady state is valid only for laminar flow (low Re number)

cfd seeker November 3, 2012 01:27

Some flows are inherently transient e.g Wing/Aerofoil at High angle of attacks, shock-shock interaction, shock-boundary layer interaction. I always explain the steady and transient phenomenon to some like this....The flow around a wing or aerofoil at low angle of attacks(0,1,2,3 degree) can be considered and solved with steady solver because there is always negligible amount of flow separation(of course flow separation is highly unsteady) at these angle of attacks however with increase in angle of attacks the flow separation becomes prominent especially near stall so you will switch over to transient solver to properly capture the physics of the problem. So in short you should master physics of the problem in order to correctly approach it.
Regards

jpc November 9, 2012 16:45

thank you everyone for your input

from a practical standpoint, how do you imploy steady state solutions in your every day CFD usage?

for example you want to evaluate 20 different airfoils, at 20angles of attack at a high enough velocity that would be considered beyond transition (turbulent flow guaranteed), it really isn't practical to run 400 CFD solutions as transient. would you approach the problem from a steady state first and then move to transient for the cases that are of the most interest?

FMDenaro November 10, 2012 04:20

Quote:

Originally Posted by jpc (Post 391331)
thank you everyone for your input

from a practical standpoint, how do you imploy steady state solutions in your every day CFD usage?

for example you want to evaluate 20 different airfoils, at 20angles of attack at a high enough velocity that would be considered beyond transition (turbulent flow guaranteed), it really isn't practical to run 400 CFD solutions as transient. would you approach the problem from a steady state first and then move to transient for the cases that are of the most interest?

From a pratical (and rigorous) point of view, you can run only the RANS formulation in a steady state solution as you look for a statistic solution. No other formulations (URANS/LES/DNS) can be run by fixing to zero the time-derivative of the variables.

Far November 10, 2012 11:07

Quote:

Originally Posted by FMDenaro (Post 389975)
URANS is a transient simulation in statistical sense.

What is the physical significance of running the URANS?

In laminar case, for example cylinder at Re= 200 is laminar flow, but you consider it unsteady due to vertex shedding. And solving through URANS will give the trend of vortex shedding behind cylinder. Now few questions arises:

1. URANS in laminar flow regime (statistical as suggested) will be equal to LES as there is only single mode of unsteadiness? Am I right?

2. How do we compare the URANS and LES results? Which parameters we must consider e.g. mean.

3. Why transition models are successful for turbomachinery? They are able to predict the all modes of transition including natural transition.

4. Why there is very less difference in RANS, URANS and LES for mean quantities for the practical flows?

FMDenaro November 10, 2012 17:48

Quote:

Originally Posted by Far (Post 391401)
What is the physical significance of running the URANS?

In laminar case, for example cylinder at Re= 200 is laminar flow, but you consider it unsteady due to vertex shedding. And solving through URANS will give the trend of vortex shedding behind cylinder. Now few questions arises:

1. URANS in laminar flow regime (statistical as suggested) will be equal to LES as there is only single mode of unsteadiness? Am I right?

2. How do we compare the URANS and LES results? Which parameters we must consider e.g. mean.

3. Why transition models are successful for turbomachinery? They are able to predict the all modes of transition including natural transition.

4. Why there is very less difference in RANS, URANS and LES for mean quantities for the practical flows?


1. No, my idea is that URANS have a model which acts also on resolved part of the unsteadiness. Hence, on laminar flows LES is able to reduce to a DNS, URANS no.

2. Yes, zero order statistics should be comparable. High order statistics can be computed only in LES.

3. I have no experience in turbomachinery

4. No, I don't think that the differences in the mean between RANS/URANS and LES solution is always disregardable.... It depends on many factors..

Far November 10, 2012 17:55

Quote:

3. I have no experience in turbomachinery
I wanted to ask that prviously transition prediction was only possible with LES, now RANS models can also predict the transition from laminar to turbulent with good success.

Quote:

No, I don't think that the differences in the mean between RANS/URANS and LES solution is always disregardable.... It depends on many factors..
At design condition. For example wing at 4 deg AOA. Obviously at the stall flow there would great difference among them.

iman November 11, 2012 18:40

thank you guys for your valuable discussion. i got alot of hints.
i want to simulate sewer pipelines for a certain study and analysis of sewage flow properties and i'm standing in between transiet, steady, turbulent characteristics. and it is really difficult to make the right decision and be confident about it!!
i want to start from simple and change the complexity of my design step by step to ease the study!! but i'm a little worried about the effect of over simplification!!

cfd seeker November 12, 2012 11:33

Quote:

Why transition models are successful for turbomachinery? They are able to predict the all modes of transition including natural transition
Do you mean that transition models are best only for turbomachinery applications? Do you have any supporting material for this claim?

Quote:

Why there is very less difference in RANS, URANS and LES for mean quantities for the practical flows?
Again I want toa sk do you have any supporting material for this because I think there should be an appreciable difference between the URANS and LES results

vicarious November 12, 2012 14:58

Guys,
What do you think about the difference between turbulence phenomenon and turbulence modeling?
I believe that the turbulence phenomenon lies in our classic laws naturally. These laws are a series of nonlinear partial differential equations. This turbulence is not completely known to humanity and the only strong solution for these equations is numerical methods. This is the difference between turbulence phenomenon and turbulence modeling. Numerical correlations consists of simple math procedure based on the computational domain we create. When the regime is turbulent, these simple math procedure (summing, multiplying, etc.) fail to handle the high extreme fluctuations in parameters. Thus, turbulence modeling is created because we need them in numerical simulation.
I suppose that when intensity is high, such flows in turbomachinery including large separations, large passage vortexes, high amplitude oscillations; only unsteady solvers will give the correct answer. DNS and LES methods are definitely more accurate specially in turbomachinery.

jpc November 16, 2012 00:59

for those that responded, how many of you use RANS tools and how many use LES/DNS tools?

those that use RANS tools, how often do you move from steady state to transient?

for transient solutions, how do you determine the length of time to run the solution? (in cases that aren't transient specific...ie where something is turned on and you want to see it develop over time)

for those that are responding, what types of problems are you solving? (transient and/or steady)

overall it seems like the answer to my original question is that the key is understanding the physics of the problem before accepting a steady state solution. most cases are transient in nature and should be run as transient and that one must be careful about using a steady state solution.

i'm not 100% sure that i am a proponent of that because of the wide variety of applications i'm exposed to, but i have a feeling that those that responded likely perform simulations in the same application group type and/or based on the tools that you use. see questions above.

FMDenaro November 16, 2012 04:09

I am not sure to understand the real sense of this topic...
In nature, steady laminar flows are quite rare ... in the most part of cases, the natural state of the flow motion is unsteady and turbulent

jpc November 16, 2012 12:05

FM, that is exactly the point. Hence, how do you explain a steady state solution in that case and how do you apply them. The extension of the question is more about the application of transient because in general, having used and supported solidworks flow simulation for over 8 years, I have seen only a handful of people use transient.

FMDenaro November 16, 2012 13:10

Quote:

Originally Posted by jpc (Post 392577)
FM, that is exactly the point. Hence, how do you explain a steady state solution in that case and how do you apply them. The extension of the question is more about the application of transient because in general, having used and supported solidworks flow simulation for over 8 years, I have seen only a handful of people use transient.


in laminar flows is a physical condition in deterministic sense, unlike in turbulence where is a statistically steady state. In the latter case, you need of an energy equilibrium as well, otherwise you have no steady state.

Far November 16, 2012 13:25

Just take an example of Turbomachinery. We all know that the flow in the turbo-machinery is highly 3d, transient, turbulent and complex, but still we do the steady state simulations. Why? And question arises when to use the steady state and when to use unsteady?

In my point of view it depends on the requirment. For example if we are interested in the mean values, like pressure ratio, mass flow rate and efficiency then we use steady state. But if we are interested that what be the effect of interaction of components on the pressure ratio, mass flow rate and efficiency during different relative position of upstream and downstream components then we use the transient simulation. Moreover in the case of stall, where flow is not taking a single value of these variables then we are forced to use the transient simulation, as we dont have any single answer and it keeps on changing with each iteration and we wont get the acceptable convergence.

m zahid November 17, 2014 11:17

hi
 
hi, can anyone list the steps to perform unsteady analysis by standard k-epsilon model. (fluent 14.0)

thanks


All times are GMT -4. The time now is 06:54.