Urans and rans

bohluly · April 4, 2009, 04:57

HI dears
are there any different between terms of equations of RANS and URANS ? only in apparent not in application!

sbaffini · April 4, 2009, 05:50

The URANS have a time derivative term in each equation, the RANS don't have it.

So, the first one are unsteady, the second one not. It's much more than a visual difference

bohluly · April 4, 2009, 10:02

tks
but in RANS eq. We have to use this time derivations !!

sbaffini · April 4, 2009, 12:58

No, RANS stands for Reynolds Averaged Navier-Stokes while URANS stands for Unsteady-RANS. So, in the RANS equations no time derivative exists, not at all.

In RANS equations the velocity field is decomposed in a time averaged part and a turbulent fluctuation and both of them, when time averaged, gives two costant terms (in time) whose time dirivative is zero.

In URANS equations the time average is defined in a different way such that the time averaged part of the velocity field is not costant in time so its time derivative is not zero.

Even if questionable, this is the only difference in the two set of equations.

bohluly · April 5, 2009, 03:40

that is true!
but in text books (Ferziger (pp 293) or ...) in RANS Eqs. there is
@(ro u) /@t in momentum Eq.

sbaffini · April 5, 2009, 14:44

It's probably due to the fact that i'm not a native english speaker but i'm not a huge fan of the Feriziger's book...it's always quite compressed and obscure, explaining things in a very nonlinear fashion.

However, in this point, it is relatively clear and the real problem is just a matter of definition. In fact:

9.4 (pp 292): "In Reynolds-averaged approaches to turbulence, all of the unsteadiness is averaged out i.e. all unsteadiness is regarded as part of the turbulence"

But, few lines down:

9.4.1: "In a statistically steady flow, every variable can be written as the sum of a time-averaged value and a fluctuation about that value... Here t is the time and T is the averaging interval. This interval must be large compared to the typical time scale of the fluctuations; thus we are interested in the limit of T -> inf, see Fig. 9.10..."

"If the flow is unsteady, time averaging cannot be used and it must be replaced by ensemble averaging...This type of averaging can be applied to any flow"

So, what he pratically says is that under the name of RANS he has included the cases where the flow is statistically steady and a simple time average can be used and the cases where it isn't and an ensemble averaged is required. He explains this with the help of Fig 9.10 and saying that performing the ensemble average covers both of the two approaches and it is necessary to include the time derivative term

What effectively changes between RANS and URANS is the presence of the time derivative term and is up to you to choose if your case has to be run in steady or unsteady mode. You can check this wherever is present the definition of both the terms.

bohluly · April 6, 2009, 04:09

thanks.
It is correct! "In a steady condition @/@t is zero"
but I have another question.
In simulation of a steady condition can we didn't use @/@t ? if "yes" are you think that this type of Eq. converged to steady condition?
I didn't test that myself, but I think for stability (or for more stability), even in steady condition, we have to use terms of @/@t.
If you have good reference for this discussion please send it for me "a_bohluly@yahoo.com"

sbaffini · April 6, 2009, 09:08

Actually, steady computations make use of the unsteady equations integrated in time with a simple backward euler scheme (so implicit) and very big time steps.

The equations are integrated in time until the steady state is reached.

It's easy to understand that if the flow is such that a steady condition isn't possible than there will not be any convergence

shyamdsundar · April 6, 2009, 09:31

As mentioned in the previous message, if a flow has unsteady phenomena, it might not converge in RANS computation. In my case, I did a axisymmetric jet case, and for a particular pressure ration, I was not able to get a convergence with the RANS model, and the computations showed something like vortex shedding (though not time accurate). To my surprise, the corresponding experimental results reported a screetching tone!!

April 4, 2009, 04:57	Urans and rans	#1
bohluly New Member Asghar Join Date: Apr 2009 Posts: 16 Rep Power: 17	HI dears are there any different between terms of equations of RANS and URANS ? only in apparent not in application!

April 4, 2009, 05:50		#2
sbaffini Senior Member Paolo Lampitella Join Date: Mar 2009 Location: Italy Posts: 2,192 Blog Entries: 29 Rep Power: 39	The URANS have a time derivative term in each equation, the RANS don't have it. So, the first one are unsteady, the second one not. It's much more than a visual difference aero_head likes this.

April 4, 2009, 12:58		#4
sbaffini Senior Member Paolo Lampitella Join Date: Mar 2009 Location: Italy Posts: 2,192 Blog Entries: 29 Rep Power: 39	No, RANS stands for Reynolds Averaged Navier-Stokes while URANS stands for Unsteady-RANS. So, in the RANS equations no time derivative exists, not at all. In RANS equations the velocity field is decomposed in a time averaged part and a turbulent fluctuation and both of them, when time averaged, gives two costant terms (in time) whose time dirivative is zero. In URANS equations the time average is defined in a different way such that the time averaged part of the velocity field is not costant in time so its time derivative is not zero. Even if questionable, this is the only difference in the two set of equations. aero_head likes this.

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April 4, 2009, 10:02		#3
bohluly New Member Asghar Join Date: Apr 2009 Posts: 16 Rep Power: 17	tks but in RANS eq. We have to use this time derivations !!

April 5, 2009, 03:40		#5
bohluly New Member Asghar Join Date: Apr 2009 Posts: 16 Rep Power: 17	that is true! but in text books (Ferziger (pp 293) or ...) in RANS Eqs. there is @(ro u) /@t in momentum Eq.

April 5, 2009, 14:44		#6
sbaffini Senior Member Paolo Lampitella Join Date: Mar 2009 Location: Italy Posts: 2,192 Blog Entries: 29 Rep Power: 39	It's probably due to the fact that i'm not a native english speaker but i'm not a huge fan of the Feriziger's book...it's always quite compressed and obscure, explaining things in a very nonlinear fashion. However, in this point, it is relatively clear and the real problem is just a matter of definition. In fact: 9.4 (pp 292): "In Reynolds-averaged approaches to turbulence, all of the unsteadiness is averaged out i.e. all unsteadiness is regarded as part of the turbulence" But, few lines down: 9.4.1: "In a statistically steady flow, every variable can be written as the sum of a time-averaged value and a fluctuation about that value... Here t is the time and T is the averaging interval. This interval must be large compared to the typical time scale of the fluctuations; thus we are interested in the limit of T -> inf, see Fig. 9.10..." "If the flow is unsteady, time averaging cannot be used and it must be replaced by ensemble averaging...This type of averaging can be applied to any flow" So, what he pratically says is that under the name of RANS he has included the cases where the flow is statistically steady and a simple time average can be used and the cases where it isn't and an ensemble averaged is required. He explains this with the help of Fig 9.10 and saying that performing the ensemble average covers both of the two approaches and it is necessary to include the time derivative term What effectively changes between RANS and URANS is the presence of the time derivative term and is up to you to choose if your case has to be run in steady or unsteady mode. You can check this wherever is present the definition of both the terms.

April 6, 2009, 04:09		#7
bohluly New Member Asghar Join Date: Apr 2009 Posts: 16 Rep Power: 17	thanks. It is correct! "In a steady condition @/@t is zero" but I have another question. In simulation of a steady condition can we didn't use @/@t ? if "yes" are you think that this type of Eq. converged to steady condition? I didn't test that myself, but I think for stability (or for more stability), even in steady condition, we have to use terms of @/@t. If you have good reference for this discussion please send it for me "a_bohluly@yahoo.com"

April 6, 2009, 09:08		#8
sbaffini Senior Member Paolo Lampitella Join Date: Mar 2009 Location: Italy Posts: 2,192 Blog Entries: 29 Rep Power: 39	Actually, steady computations make use of the unsteady equations integrated in time with a simple backward euler scheme (so implicit) and very big time steps. The equations are integrated in time until the steady state is reached. It's easy to understand that if the flow is such that a steady condition isn't possible than there will not be any convergence

April 6, 2009, 09:31		#9
shyamdsundar New Member Shyam Join Date: Apr 2009 Posts: 29 Rep Power: 17	As mentioned in the previous message, if a flow has unsteady phenomena, it might not converge in RANS computation. In my case, I did a axisymmetric jet case, and for a particular pressure ration, I was not able to get a convergence with the RANS model, and the computations showed something like vortex shedding (though not time accurate). To my surprise, the corresponding experimental results reported a screetching tone!!