Courant Number

NickolasPl · August 30, 2011, 03:38

Hello everyone,

I m working on a steady state problem in simpleFoam folder with turbulence off, as I study steady-state, incompressible and laminar flows (polymers to be exact)
with power - law or Bird - Carreau viscosity models. As simpleFoam runs in "steady state" does the Courant number comes into the solution? I mean, the timesteps performed in simpleFoam are actually the time of iterations until the solution is obtained?
And based on the above, should I choose a small timestep (e.g. 0.0001) and perform simulations from 0 to 1 or should I broaden it from 0 to say 100??

Thank you in advance!

Nickolas

eysteinn · August 30, 2011, 04:33

Hi Nickolas.

The CFL number is used for unsteady flow.
And yes you are correct the timesteps are actually the iterations.
( For steady flow your deltaT should be 1 and then end time is your
max number of iterations. )

/Eysteinn

NickolasPl · August 30, 2011, 04:48

Ah okay....I understand now, thanks a lot!
Did you read this on OF manual or somewhere else?

Nickolas

eysteinn · August 30, 2011, 04:59

I can't remember where I got this from.

Did a very quick search in the documentation and the best
I could find now was this:
http://openfoam.com/docs/user/plateHole.php
see
2.2.1.4 Control

The time step is not important since this is a steady state case; in this situation it is best to set the time step deltaT to 1 so it simply acts as an iteration counter for the steady-state case. The endTime, set to 100, then acts as a limit on the number of iterations.

/Eysteinn

August 30, 2011, 03:38	Courant Number	#1
NickolasPl Member Nickolas P Join Date: Oct 2010 Location: Greece Posts: 30 Rep Power: 15	Hello everyone, I m working on a steady state problem in simpleFoam folder with turbulence off, as I study steady-state, incompressible and laminar flows (polymers to be exact) with power - law or Bird - Carreau viscosity models. As simpleFoam runs in "steady state" does the Courant number comes into the solution? I mean, the timesteps performed in simpleFoam are actually the time of iterations until the solution is obtained? And based on the above, should I choose a small timestep (e.g. 0.0001) and perform simulations from 0 to 1 or should I broaden it from 0 to say 100?? Thank you in advance! Nickolas

Thread Tools	Search this Thread
Show Printable Version Email this Page	Search this Thread: Advanced Search
Display Modes
Linear Mode Switch to Hybrid Mode Switch to Threaded Mode

Similar Threads
Thread	Thread Starter	Forum	Replies	Last Post
[mesh manipulation] Mesh Refinement	Luiz Eduardo Bittencourt Sampaio (Sampaio)	OpenFOAM Meshing & Mesh Conversion	42	January 8, 2017 12:55
Upgraded from Karmic Koala 9.10 to Lucid Lynx10.04.3	bookie56	OpenFOAM Installation	8	August 13, 2011 04:03
DecomposePar unequal number of shared faces	maka	OpenFOAM Pre-Processing	6	August 12, 2010 09:01
Unaligned accesses on IA64	andre	OpenFOAM	5	June 23, 2008 10:37
Could anybody help me see this error and give help	liugx212	OpenFOAM Running, Solving & CFD	3	January 4, 2006 18:07

August 30, 2011, 04:33		#2
eysteinn Member Eysteinn Helgason Join Date: Sep 2009 Location: Gothenburg, Sweden Posts: 53 Rep Power: 16	Hi Nickolas. The CFL number is used for unsteady flow. And yes you are correct the timesteps are actually the iterations. ( For steady flow your deltaT should be 1 and then end time is your max number of iterations. ) /Eysteinn

August 30, 2011, 04:48		#3
NickolasPl Member Nickolas P Join Date: Oct 2010 Location: Greece Posts: 30 Rep Power: 15	Ah okay....I understand now, thanks a lot! Did you read this on OF manual or somewhere else? Nickolas

August 30, 2011, 04:59		#4
eysteinn Member Eysteinn Helgason Join Date: Sep 2009 Location: Gothenburg, Sweden Posts: 53 Rep Power: 16	I can't remember where I got this from. Did a very quick search in the documentation and the best I could find now was this: http://openfoam.com/docs/user/plateHole.php see 2.2.1.4 Control The time step is not important since this is a steady state case; in this situation it is best to set the time step deltaT to 1 so it simply acts as an iteration counter for the steady-state case. The endTime, set to 100, then acts as a limit on the number of iterations. /Eysteinn