SST turbulence model question in ANSYS CFX

drsidd10 · January 9, 2015, 08:59

Hello everyone

I have simulated a tidal turbine blade geometry using SST turbulence model for steady state analysis.

I was wondering what is the difference in results between SST steady state analysis and SST transient for the same geometry?

Kind regards

Siddharth

JuPa · January 9, 2015, 10:35

My understanding is:

There isn't really a difference in turbulence models between steady state and transient. There is a difference in the conservation equations solved due to the addition of the transient terms in the continuity, momentum and energy equations.

In transient runs steady RANS equations become unsteady URANS equations. The SST model is applied just the same to RANS as is URANS.

Edit: I've just re-read the SST turbulence equations. It turns out I'm slightly wrong. The SST turbulence model does contain unsteady terms. Namely
d(rho k) / dt and d(rho omega) / dt
Where rho is density, k is turbulence kinetic energy, omega is turbulence dissipation rate and t is time.

In steady state these partial derivatives go to zero. The rest of what I wrote sounds correct.

ghorrocks · January 18, 2015, 06:38

As Mr CFD says the only real modification to the equations is the addition of the normal transient term.

But an important distinction is implicitly changed - RANS assumes that there is a turbulent flow component which averages out over time to zero, and a mean flow component. In steady state this separation of turbulent and mean flow components is simple. But in transient flows the distinction is not straight forward. You have to assume that the simulated, resolved flow represents the mean flow and that the turbulent component is on a much faster time scale so it can still average out over time to zero. This results in some issues for transient flows:
1) If the time step size in the simulation is in the turbulent time scales then you cannot make a clear distinction between mean flow and turbulent time scales and the RANS assumption becomes less appropriate; and
2) Some flows do not have clear distinctions between flow and turbulent time scales. Vortex shedding is an example - is the vortex street turbulent or mean flow? There are many other examples. When there is no clear distinction then you should consider LES and related models.

January 9, 2015, 08:59	SST turbulence model question in ANSYS CFX	#1
drsidd10 New Member Siddharth Kulkarni Join Date: Oct 2010 Location: Birmingham, UK Posts: 6 Rep Power: 16	Hello everyone I have simulated a tidal turbine blade geometry using SST turbulence model for steady state analysis. I was wondering what is the difference in results between SST steady state analysis and SST transient for the same geometry? Kind regards Siddharth

Similar Threads
Thread	Thread Starter	Forum	Replies	Last Post
Wrong calculation of nut in the kOmegaSST turbulence model	FelixL	OpenFOAM Bugs	27	March 27, 2012 10:02
Low Reynolds k-epsilon model	YJZ	ANSYS	1	August 20, 2010 14:57
Fluent 12 k-w SST turbulence model	DarrenC	FLUENT	0	December 13, 2009 09:33
Turbulence model for CFX moving mesh	songxguan	CFX	7	June 28, 2009 22:05
Urgent: How to model a stationary sphere in a pressure driven flow using Ansys CFX?	farhan	OpenFOAM Running, Solving & CFD	1	April 14, 2009 15:34

January 9, 2015, 10:35		#2
JuPa Senior Member Mr CFD Join Date: Jun 2012 Location: Britain Posts: 361 Rep Power: 15	My understanding is: There isn't really a difference in turbulence models between steady state and transient. There is a difference in the conservation equations solved due to the addition of the transient terms in the continuity, momentum and energy equations. In transient runs steady RANS equations become unsteady URANS equations. The SST model is applied just the same to RANS as is URANS. Edit: I've just re-read the SST turbulence equations. It turns out I'm slightly wrong. The SST turbulence model does contain unsteady terms. Namely d(rho k) / dt and d(rho omega) / dt Where rho is density, k is turbulence kinetic energy, omega is turbulence dissipation rate and t is time. In steady state these partial derivatives go to zero. The rest of what I wrote sounds correct.

January 18, 2015, 06:38		#3
ghorrocks Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,871 Rep Power: 144	As Mr CFD says the only real modification to the equations is the addition of the normal transient term. But an important distinction is implicitly changed - RANS assumes that there is a turbulent flow component which averages out over time to zero, and a mean flow component. In steady state this separation of turbulent and mean flow components is simple. But in transient flows the distinction is not straight forward. You have to assume that the simulated, resolved flow represents the mean flow and that the turbulent component is on a much faster time scale so it can still average out over time to zero. This results in some issues for transient flows: 1) If the time step size in the simulation is in the turbulent time scales then you cannot make a clear distinction between mean flow and turbulent time scales and the RANS assumption becomes less appropriate; and 2) Some flows do not have clear distinctions between flow and turbulent time scales. Vortex shedding is an example - is the vortex street turbulent or mean flow? There are many other examples. When there is no clear distinction then you should consider LES and related models.