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itsme_kit September 2, 2013 13:03

A silly question about turbulent flow
 
I wonder why we sometimes can apply steady state in the modelling of turbulent flow

Many thanks

Pauli September 4, 2013 09:42

Transient simulations cost more & steady state provides an adequate quality result.

itsme_kit September 4, 2013 11:54

Quote:

Originally Posted by Pauli (Post 449843)
Transient simulations cost more & steady state provides an adequate quality result.

Though the time expense in steady state is massively reduced, in what conditions we still need to use unsteady state

DaveyBaby October 14, 2013 10:31

Time dependent or not?
 
I will try to answer this generally!

Needing to do unsteady simulations or not depends on what information you need. I guess it's best to ask yourself "do I need to know what happens instantaneously, resolved to a very short time scale or not?" That will depend on the physical effects that are of interest to you.

ONE SUCH EXAMPLE
You would think, for example, that something seemingly as simple as pipe flow in a T-junction was well understood, and that a steady-state solution was all that was necessary. That turned out to not be true!
Pipe joins in industrial plants that contained hot fluid were becoming damaged and failing way quicker than expected, but nobody knew why, as they were engineered to deal with much more than the forces and temperatures experienced. These were well known from experimentally verified CFD.
It turned out to be something called "thermal fatigue". Pulsations in temperature due to turbulent fluctuations of hot and cold fluid were causing the pipe material to break down. You could call this "thermal vibration".

Time dependent CFD can predict not only the mean temperature, but the waveform of these thermal vibrations, as it predicts when these eddies of hot-cold fluid will arrive. So in this case, time dependent CFD was very useful.

afikr October 30, 2013 03:16

Quote:

Originally Posted by itsme_kit (Post 449403)
I wonder why we sometimes can apply steady state in the modelling of turbulent flow

Many thanks

If you take a course on turbulence, there is chapter covering RANS (Reynolds Averaged Navier-Stokes) equation. In this equation, velocity is composed of mean and fluctuating velocity.

After some substitutions, a term call reynold's stress will emerge. You can find this in most turbulence text books.

The RANS equation is not a closed equation since the reynold's stress term cannot be solve.

In order to solve the reynold's stress term turbulence modelling is required. the turbulence models are k-epsilon,k-omega and etc.

So, if you study the RANS equation, the reynolds term remains even if the time-dependent term is canceled out due to steady state. The reynold's term is associated with the turbulence phenomena.

that is why, we need to consider turbulence even for a steady state problem.

Please correct me if i'm wrong. I new to CFD too.

DaveyBaby October 30, 2013 08:38

As afikr has pointed out, turbulence is time-dependent in nature and the additional Reynolds stress terms are due to fluctuations around the mean velocities. The word "regime" is used to separate laminar from turbulent as they are so different in nature!

Laminar flow will reach a steady state where the flow-field at all points within the domain will remain unchanged with respect to time.

A steady-state with a turbulent flow is one where the mean properties are constant at every point within the domain. However, unlike the laminar case, the instantaneous properties are fluctuating rapidly!

Passing from laminar to turbulent has a large effect on these mean properties, so predictions can not be made based on assumptions applicable to laminar flow. A steady-state turbulent simulation will calculate the mean properties accurately, and not give you the time-dependent detail around it.

Hope this helps.


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