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running unsteady k-omega sst
Posted: Sat Oct 23, 2010 1:10 pm
by Mubashir Ali
hi,
Could anyone tell me what does this warning means and what could the impact on the simulation if this warning is not headed (I am using restart files obtained from steady state k-omega simulation)
WARNING: WHEN READING THE AUXILIARY RESTART FILE
@ *******
@ THE RUN RESTARTED WITH IDTVAR = 1
@ FROM RUN CONDUCTED WITH IDTVAR = -1
@
@ The variable time step method has been modified.
@ The (uniform) value of the time step is
@ DTREF = 0.1000E+00 given in usini1.
@
@ It is advised however in this case to
@ verify the value of IDTVAR in usini1.
@
@ Verify that the auxiliary restart file being used
@ corresponds to the present case.
@
@ The run will continue...
Thnx in advance
Kind Regards
Mubashir
Re: running unsteady k-omega sst
Posted: Sat Oct 23, 2010 3:40 pm
by Alexandre Douce
This warning means that you have changed you setup between a previous computation and a restart. In your case you have switched the velocity-pressure coupling algorithm from a steady SIMPLE to an unsteady SIMPLEC.
It could be seen as a computational strategy: searching a steady state solution and then use it as an initialization step for a transient algorithm.
In a general manner, Code_Saturne always checks the compatibility of the setup when the user restarts a computation. It makes a warning when a difference is detected. It continues the computation when the change is compatible, it stops otherwise.
Re: running unsteady k-omega sst
Posted: Mon Oct 25, 2010 9:12 am
by Mubashir Ali
Thnx for reply. Yes, i am using results of steady state to start transient soultions.
I am using UPWIND scheme for u,v,w and CENTERED (flux reconstruction is off) for Turbenergy and omega (k-omega SST model)
Time step is variable in time and uniform in space with
maximal Courant no. = 10
maximal Fourier no. = 10
minimal time step = 0.01
maximal time step variation = 0.01
Gradient calculation = IMRGRA = 0 (as i have hexahederal mesh )
The problem is after sometime step i am getting very high vaules of pressure , velocity and courant no, and other parameters which is beyond imagination even though i dont get warning for divergence or convergence problem.
Please find my listing file with msg.
Could you please suggest me bttr setting to run unsteady simulations
Re: running unsteady k-omega sst
Posted: Mon Oct 25, 2010 12:00 pm
by Alexandre Douce
At the initialization you have already huge values for the velocity (results from the steady computation), but perhaps it is OK for your case?
** VARIABLES INITIALIZATION
------------------------
---------------------------------
Variable Min. value Max. value
---------------------------------
Pressure -0.1882E+04 0.7589E+03
VelocitU -0.1565E+02 0.5698E+02
VelocitV -0.4572E+02 0.4572E+02
VelocitW -0.4598E+02 0.4598E+02
TurbEner 0.2143E+01 0.7642E+02
omega 0.4253E+02 0.6473E+06
---------------------------------
but you did not adapt the time step (i.e. DTREF) to this velocity and the mesh.
At the end of the first iteration of your restart, you have a too huge Courant number:
Courant max 0.60263E+03 0.22369E-01 0.13858E+00 -0.45029E-01
Could you check if your initialization step is OK, and then try a DTREF = 0.0001?
Re: running unsteady k-omega sst
Posted: Mon Oct 25, 2010 2:01 pm
by Mubashir Ali
I think the initialization is okay as I am simulating a bluff body somewhat like hemisphere but not exactly, with its flat face facing free stream velocity 36 m/sec and Reynolds no. is of the order of 10e6.
Now i am trying with DTREF = 0.0001 . But still at the very first step i got
Courant min 0.81645E+00 0.24516E+00 -0.20914E-01 -0.98391E-01 Courant max 0.60263E+03 0.22369E-01 0.13858E+00 -0.45029E-01
which is exactly the same as the last one (copy pasted in your reply)
Isnt it at the first few iteration the values fluctuate a lot but with time they reduce as the flow attains steadiness ???
Re: running unsteady k-omega sst
Posted: Mon Oct 25, 2010 10:48 pm
by Alexandre Douce
o you have adapted DTMIN and DTMAX
to the new DTREF? For example:
DTREF = 0.0001
DTMIN = 0.000001
DTMAX = 0.01
Re: running unsteady k-omega sst
Posted: Tue Oct 26, 2010 8:17 am
by Mubashir Ali
Ok. I got it now .... in the user guide there is a relation given b/w DTREF and DTMIN and DTMAX . Now with
DTREF = 0.0001
DTMIN = 0.000001
DTMAx= 0.01
i get courant no. as below for the very first iteration.
Courant min 0.95512E-03 0.24516E+00 -0.20914E-01 -0.98391E-01 Courant max 0.70499E+00 0.22369E-01 0.13858E+00 -0.45029E-01
Thnx a lot for your help :D