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Will the results of steady state solver and transient solver be same?

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Old   September 26, 2019, 04:01
Question Will the results of steady state solver and transient solver be same?
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Hi, foamers,

I have this question quite a long time ago.
I think the steady state solver will give the same results as the transient solver does.
However, what will it be when a system won't have a steady state but a steady state solver is applied?
will the results not converge? or it will converge but give a non-reasonable result?

I simulated a flow using the steady state chtmultiregion solver, and all the initial residuals go below 1e-6 at last. I also simulated the same case use the transient chtmultiregion solver. However I found the flow pattern is quite different.

here is the simulation domain, on pipe for inlet, one pipe for outlet.


near the bottom surface I made a slice and vector glyph
this is for steady state solver near where the outlet pipe is:
the flow swirls and flows outside.


however, for the result of transient solver,
the flow won't swirl and flows out directly:


the fluid will flow through the domain in about 10s, and I calculated for 25s using the transient solver. The k-e RAS model is applied.

So, does anyone know why the results are different? Does this mean that this flow do not have a steady state?

Thanks!
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Old   December 3, 2019, 19:49
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Anyone has any ideas?
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Old   December 5, 2019, 02:37
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I suggest k-e model to be used only for steady state simulation without swirling motion. The model dissipates a large fraction of turbulent kinetic energy into turbulent viscosity, hence you will always observe non-physical behavior of swirling motion regardless whether you run it in steady (without dt term) or unsteady way (with dt term).



It is correct that you could always observe convergence, the model you choose is easy to converge, but results are often non-physical.


Kai.
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Old   December 5, 2019, 03:10
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Hi dear Kai,

Thank you for the reply.
And could you kindly give me more details about your reply?

You think the k-e model should not be used while in the flow system the swirling motion exists?
which means you think the transient results without the swirling motion in my simulation is correct?
Am I understood you right?

In my simulations, I don't have experiment results so I don't know the flow swirling flows out or not.
The k-e model is applied in both the steady state and transient simulations, but steady state gives swirling flowing out and transient gives direct flowing out.

Thank you.
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Old   December 6, 2019, 07:03
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Hi.


This is interesting.

I've found that sometimes (using simple and pimpleFoam. I don't know much about multiregionFoam), when the flow is physically unsteady, the steady-state solver shows the same results as the transient one (ie. varying solution with iterations and high residuals) and sometimes, the solution is "trapped into an attractor" and an unphysical steady solution is found. The solvers are highly sensitive to initial conditions and the results can change dramatically.

I suggest you to try running a transient simulation starting from the steady state solution. It might just continue swirling. And you may wanna do the opposite, try to start a steady state calculation from the transient solution.

You can also run your transient simulation for longer to see if there's any change.
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Old   December 8, 2019, 16:47
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Hi,

This problem might be slightly complicated. Let's consider a classic problem of flow over a cylinder.

For Reynolds number less than 40 (I do not remember exactly) where the flow is symmetric and stable (no oscillation), the steady and transient solution with laminar solver should converge to the same status.

For Reynolds number is larger than 40 but less than 150 approximately, the flow starts to oscillate, though the flow is still laminar. the transient solution with a laminar solver should reproduce the same oscillation results. One MAY be able to converge your steady-state solver to a "steady-state solution". But physically, I do not know what that "steady-state solution" represents.

As you increase the Reynolds number to approximately 150-3e5, you start getting into the transitional regime. In my opinion, the only tool that is appropriate to study the flow characteristic in this region is the high-order DNS code. And even with the high-order DNS code, one will still have an issue of how to numerically trigger the instability (transient growth in eigenvalue make things complicate). Let's just ignore computing the flow in this region for this discussion.

If one increase the Reynolds number more, one will start to get into a turbulent regime. For the simplicity of this discussion, let's just assume we arrive to the "fully-developed turbulent regime", which is also the interest for the majority engineering applications.

If one uses the transient laminar solver of OpenFOAM to compute this flow with sufficient spatial and temporal accuracy (a lot of grid), one will start to resolve turbulence and observe the fluctuation in your U file (velocity). In some sense, you are doing "DNS" now. However, due to the fundamental construction of OpenFOAM, it requires a lot of grid for a second-order code (OpenFOAM/Fluent etc) to compute the right turbulence statistics in comparison to the high-order method. For example, a turbulence channel flow at bulk Reynolds number of 2800 (Retau 180, KMM), OpenFOAM probably need approximately 10 million cells to get the right Reynolds stress.

If one uses the steady-state laminar solver to compute the flow at this Reynolds number, I will be very surprised if you can get the solver to converge.

As mentioned, in order to get some useful flow solution at the turbulent regime, one needs a lot of computation with the laminar solver. Meanwhile, for many engineering applications, people do not care about the instantaneous flow velocity. Some statistical quantities are rather sufficient and useful. This is the motivation of turbulence modeling. People try to compute the time-averaged flow quantity (mean flow) directly instead of "do the DNS and then compute the time-average".

By doing the Reynolds average, people find that they can predict the mean flow (time-averaged flow) directly as long as knowing the Reynolds stress. However, one can only know the true Reynolds stress by doing DNS (dead loop)... But people are smart, they come up with some formulas to approximate/model the Reynolds stress. This gives the birth of famous K-epsilon and K-omega SST, etc...blabalaabala

Come back to the discussion, if you compute the flow at the turbulent regime with RANS solver (steady-state), you are computing the time-averaged flow. Although the instantaneous flow velocity of this problem at the turbulent regime is chaotic and oscillatory. The time-averaged flow (mean flow) is stationary (not dependent on time).

At high Reynolds number, the chaotic and oscillatory fluid motion is the natural solution of the Navier-Stokes equation. If one has tons of computational power, directly solve the NS equation with sufficient spatial and temporal resolution will give you the "exact" solution to the problem (aka DNS).

If you use the RANS turbulent model, you are "solving" for statistical quantity (time-averaged flow) of the turbulent flow. To the numerical solver, there is nothing turbulent in the RANS model. In fact, to the numerical solver, the RANS model is some sort of non-newtonian laminar flow. Your eddy viscosity model will dissipate all the fluctuation that otherwise will blow your steady-state solution up. This is why one can compute the "steady-state solution" of turbulent flow with RANS. You are computing a strategically constructed non-Newtonian laminar flow to MIMIC the mean-flow equation of the turbulent flow. This is also why for RANS simulation, since you are solving some sort of laminar equation, you do not have small-eddy/scales/fluctuation in your solution and hence high-order non-dissipative methods are not needed. In my limited experience, most time first-order for RANS is most robust. Although, the second-order scheme theoretically should be more accurate, sometimes (in fact, a lot of time) introduce strong sensitivity on mesh quality and become destructive to your solution.

For URANS ("transient solver for RANS"), the solution should agree to RANS solution. UNLESS, the numerical Reynolds number that seen by the solver [UL/(nu+nut)] reaches finite value where the instability kicks in (analog to the laminar flow oscillation), the URANS solution will become unsteady and start to oscillate. Some researchers argue the oscillation from the URANS is the "large scale" of the turbulence. And in fact, you can find URANS will give you the right the oscillation frequency for the problem of flow over a cylinder. But personally, I do not have enough experience to make a judgment or justify why URANs works for this case. It might associate with the trigger condition for the Kelvin–Helmholtz instability.


I really wish I am helping in some sense.
Rdf

Last edited by randolph; December 9, 2019 at 16:09.
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Old   December 9, 2019, 21:45
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dear Rotidpor,

thank you for sharing your experience!
In fact I have tried to start the transient solver from the swirling results of the steady-state-solver, and the results was swirling and parameters such as velocities did not oscillate very much.
I will try another thing that you said , start the steady-state-solver from the no-swirling results of the transient solver to see what will happen.

best regards.

Quote:
Originally Posted by Rotidpor View Post
Hi.


This is interesting.

I've found that sometimes (using simple and pimpleFoam. I don't know much about multiregionFoam), when the flow is physically unsteady, the steady-state solver shows the same results as the transient one (ie. varying solution with iterations and high residuals) and sometimes, the solution is "trapped into an attractor" and an unphysical steady solution is found. The solvers are highly sensitive to initial conditions and the results can change dramatically.

I suggest you to try running a transient simulation starting from the steady state solution. It might just continue swirling. And you may wanna do the opposite, try to start a steady state calculation from the transient solution.

You can also run your transient simulation for longer to see if there's any change.
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Old   December 9, 2019, 22:04
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Hi, Randolph

thank you for your reply!
it definitely helps me a lot!

can you help me to see if I understanding is right:

a) if we just use the laminar model to calculate a real turbulent flow, then actually we are doing a "very coarse DNS" and the results mean nothing because the grid is not fine enough to capture the real flow structure.

b) because the RANS equations are time-averaged equations, so if we use a steady-state-solver, apply the RANS model and if we got a converged result, then this result represents the time-averaged turbulent flow field.

c) however, in b), if the result did not converge, then what does this result mean? or is it just meaningless?

d) if the RANS model is applied to a transient solver in OF, what will it mean? because the RANS are time-averaged equations, in the transient solver what will the "time" in "time-averaged" mean? does it mean the flow field is time-averaged from the 0 to now? if not, does the variation of the results represents the flow developing during time?


I really learned a lot from your reply.
In fact I'm always doing DNS in very small scales in the past and never touched the RANS and LES, so it is a little difficult for me to understand the applications of these models.
it is also the first time I heard URANS, I will search the internet to learn it.

Thank you for taking your time

best regards.



Quote:
Originally Posted by randolph View Post
Hi,

This problem might be slightly complicated. Let's consider a classic problem of flow over a cylinder.

......

I really wish I am helping in some sense.
Rdf
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Old   December 19, 2019, 11:03
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Hi,

a) I think so. There is also something associated with the energy-preserving property of the solver/scheme, someone on this forum told me that native OF solver is not energy-preserving.


b) Correct. RANS results are the representation of time-averaging/ensemble averaging for a statistically stationary turbulent flow.

c) if the result did not converge, I think the results are meaningless.

d) the transient solver for RANS is referred as URANS (unsteady RANS). I do not read enough material to justify the physical meaning behind URANS. In fact, in my own experiment, I often struggle with the interpolation of the URANS. one of my professors told me that you can think URANS as "filtering in time/frequency". But generally, I only use URANS for 2d simulation. if I need 3d simulation, I always go for RANS and LES.
__________________

I went the other way around. I learn the RANS and then DNS/LES. Personally, I found it is much easier for me to understand the RANS model if I had experience with DNS first.

Thanks,
Rdf
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Old   December 28, 2019, 05:21
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Hey,
Sorry I was too busy to reply to this thread after my 1st post.
I'd like to answer your last post now based on my experience (haven't read the rest):
a) I think yes, a laminar model gives you DNS result if your grid is fine enough. (considering high order numerical scheme)

b) Your steady-state RANS (say k-e) solution presents you mean flow field + clipped fluctuation. Often this fluctuation is incorrect if k-e is used for swirling flow. (try backward step simulation with k-e and other RANS model, backflow occurs at different positions)

c) My feelings is unless your mesh is terrible, k-e will always give you a non-physical converged result.
d) Time averaging in URANS is equivalent to steady-state RANS result if you have done averaging for long sufficient time. (at least 20 several swirling time or flow through time). Any difference you observe only indicate insufficient averaging time.


Kai.
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