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May 12, 2009, 18:36 
A question for unsteady state

#1 
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Kostas
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With other words, if i have a model witch converged at 1000 iteration on steady state, must be put 1000 iteration for each time step on unsteady state? Or with more less iteration, i will be ok? And what diference there is? 

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May 12, 2009, 22:09 

#2 
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Martin Gariepy
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I will suppose that you're running an explicit scheme. Usually, you should set you're time step to be able to converge in less than 15 iterations on each time step. So first of all:
1) choose you're convergence level (residuals) 2) choose a time step 3) iterate. You must be able to convergence inside 15 iterations. If it's not the case, lower your time step. 4)iterate. About a certain time, you will converge a lot faster, maybe 5 iterations. Then it's time to increase youre time step 5)Continue until you reach steady state or unsteady cyclic state. Hope it's help 

May 13, 2009, 05:03 

#3 
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Kostas
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ok. i do this. I decrease the time step and the solution converge in each time step. But the time step in witch take converged solution in 1520 iteration is t=1e07.
This time is much small... Untill arrive approximately 1 sec. What i do now???? If i put more big time step and the solution do not converged in 20 iteration, there is problem??? Thus, ever, must be converge the sollution? 

May 13, 2009, 09:36 

#4 
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Martin Gariepy
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If you increase you're time step, crucial information will propagate faster than a cell length leading to a decrease in accuracy. If you're patient, you will soon see that it taking less iteration to converge, than you can try to increase your time step.
Or a second option is increase your time step and your number of iterations to let it converge but that's not recommanded du to the lack of accuracy. 

May 15, 2009, 15:51 

#5 
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Akour
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Hi Kosta,
How many iterations each timestep will take to reach convergence will also depend on the residual values you use...these define how accurate each timestep will be, make them a bit smaller then the fluent defaults (i.e. make mass and continuity 1e4...energy at 1e6 is probably ok) and how many timesteps you use will define how accurate your entire problem is. It is not a good idea to just randomly choose a timestep, there must be a rough calculation you can do to determine what your timestep should be, i.e. a kolmogorov timescale to use as a maximum timestep value (see wikipedia) would be a good start if you were solving a turbulent problem (with the kepsilon model turned OFF). If you dont know enough about the physics, then I recommend starting at a timestep of about =total time you want to simulate/N where N is the number of timesteps, start at 100...then go up to 500...1000...until average values in your solution domain stop changing a lot (5% change is acceptable in most industrial settings)...this means that your solution is independent of finer timesteps...which will save you time...there is no point having a timestep of .0000001 seconds if you only need .01 seconds to resolve accurate information. Hope that helps. Akour 

May 15, 2009, 16:05 

#6 
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Kostas
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ok thanks. i understand...
I write this issue because i have a problem with cavitation which i want simulate. And i beliaved that there is problem with choice of time step.. But it isnt there my problem.. My problem is that when i run the mixture model with no vapor equation in "solution" i take a converged solution and when i put vapor equation i have disconvergence. Anyway... 

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