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April 23, 2013, 06:23 |
Transient flow in fluent
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#1 |
Member
Linyx
Join Date: Mar 2011
Posts: 31
Rep Power: 15 |
I am modoling a multi element airfoil at large angle of attack, basiclly it is a transient problem and I use the steady solver to iteration for some time and change to transient. The solver is implicit and use simple method.
but I have a problem when determine the time step size and max iteration step. In fluent guide it says 'A good way to judge the choice of time step size is to observe the number of iterations FLUENT needs to converge at each time step. The ideal number of iterations per time step is 10-20. If FLUENT needs substantially more, the time step is too large. If FLUENT needs only a few iterations per time step, may be increased' but what is the convergence criteria in the transient problem? In steady flow when the parameter we care is not changing, we call it has converged. Like lift and drag for the airfoil. In transient solver, I can no longer observe the lift and drag but rasidual in one time step: the lift and drag would print only after one time step iteration is finished. As read from fluent guide, I can set the residual as 10-5, and time step as (chord length/100*free stream velocity), the iteration seem to converge quickly, if I set the residual as 10-7, the solver iterates about 100 times and do not meet this critiera, the residual line goes straight with iteration. So I should lower the time step size? gosh, it would take too much of time. and set the residual manually would change the result? for example, the result of 10-5 is 10-8 is not the same. Ps:In steady flows, when the problem is converged, the residual line always go straigh in the plot, is it applicable in the transient solver? Since after several time the residual seems go straight and not lower again...I wanna know whether it is applausible in fluent transient solver? |
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April 23, 2013, 09:06 |
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#2 | |
Senior Member
Rick
Join Date: Oct 2010
Posts: 1,016
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Quote:
I'm not expert in airfoil but maybe this can be usefull to you: - in transient solver, as in the steady solver, you can monitor one or more parameter as you want; in steady solver, when these parameters don't change anymore you have the converged solution; in unsteady solver, your parameters are changing with time, but in every time step they have to reach convergence, i.e. these parameters don't change anymore within this time step. So even in unsteady solver you can judge convergence looking at these parameters. - in unsteady solver, when solution is converged, within each time step, residual curve goes straigh as in steady solver; when the time changes you have a peak, but as you see the residual curve goes straight you have your converged solution. Judge your convergence looking at monitor parameters and residual curve. Once you have observed when your solution converges at each time step, you can change it and see if it converges in 20-30 iterations. Daniele |
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April 23, 2013, 09:16 |
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#3 |
Member
James Willie
Join Date: Mar 2009
Posts: 81
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Hi,
There is no convergence criteria for a transient problem. What you can do is define a plane for a three D problem or a line for a 2D problem at a position of interest in your domain. Monitor a parameter (say pressure, vel or temp) that is of interest. Usuallly, the monitoring is done at the outlet or close to the outlet if you have an internal flow. When you see that statistically the variable is not changing then you can use that to say you have statistical convergence. The variable will still be changing but the peak to peak variation would remain unchanged. The same can be said of the residuals.I hope this helps. Jimmy |
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April 23, 2013, 09:37 |
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#4 | |
Senior Member
François Grégoire
Join Date: Jan 2010
Location: Canada
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Quote:
Thanks ghost and jwillie for the useful comments. |
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April 23, 2013, 10:55 |
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#5 | |
Senior Member
Rick
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Quote:
I meant setting the timestep once, at the start of the simulation and keep it constant, trying to optimize it at the start of simulation, or after few iterations. With this time step you have to reach convergence, then if number of iterations in one or more time steps is 70 instead of 20, I think it's ok! |
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April 23, 2013, 12:19 |
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#6 |
Senior Member
François Grégoire
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Location: Canada
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Hi Daniele,
What I meant is: 1. Set a time step size that won't change during the simulation, or maybe just in the first few time steps. 2. Set a high enough number of iterations at each time step in order to obtain flat residuals. 3. Set very low residuals limits that will never be achieved, this way the solver will go to next time step when it reaches the max number of iterations/time step, and at that time the residuals should be flat. Critical calibration takes place at step 2 because - if the number of iterations is too low, it's not converged. - if the number of iterations is too high, you waste a loooot of calculation time. Maybe there's a way to implement a UDF that analyzes the residuals mean, variance, etc. in order to go to next time step? I suppose the answer is yes, I'll have to check the UDF forum for that. |
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April 23, 2013, 23:22 |
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#7 |
Member
Linyx
Join Date: Mar 2011
Posts: 31
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Hi macfly, currently my way of dealling with transient convergence is like the way you did, set the residual criteria low enough, and set the maxium iteration large enough to obtain flat residuals at first few time steps, then lower the maximum iteration so that the residual plot still can get flat at the end of the iteration. And I could even increase the time step.
But sometimes it is quite not robust since as the iteration goes, the residual plot was not flat any more, and I need to change the time step and max iteration again and again, its quite annoying. I wonder, is this method the usual way cfders deal with transient problems?? I see a lot of cfder just set the residual limits and when the criteria is met, let it jump the next iteration. |
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April 23, 2013, 23:31 |
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#8 |
Member
Linyx
Join Date: Mar 2011
Posts: 31
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hi ghost, about the parameter I care, like lift and drag, it only print out when one time step is over, I do not know whether this is a converged solution, the only one parameter I can obersve during one time step is residuals
So I think only residuals make sense to determine I have a converged result or not during one time step |
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April 24, 2013, 05:11 |
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#9 | |||
Senior Member
OJ
Join Date: Apr 2012
Location: United Kindom
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2) The residuals should be flat towards the end of the timestep. But the flatness should not be proloonged, which indicates that timestep can be further increased to accelerate the solution. 3) The global volumetric mass imbalance should reach small values (say 1e-6 etc) at the end of every timestep. OJ |
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April 24, 2013, 05:35 |
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#10 | |
Senior Member
Rick
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Quote:
monitor some other parameters, create surfaces in your domain and monitor velocities, pressure, turbulent kinetic energy, etc. About OJ comment, it's true that variable time step can be better, however in 90% of my cases I need to make animation of the transient run, so I need a fixed time step. Daniele |
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April 24, 2013, 05:38 |
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#11 | |
Senior Member
OJ
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Quote:
OJ |
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April 24, 2013, 05:48 |
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#12 | |
Senior Member
Rick
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Quote:
So If I have fixed time step of 1 s for example, I have dat files at 0s, 1s, 2s, etc. but if I have variable time step my animation will be not good, as for example I will have dat files at 0s, 1s, 1.2s, 2.5s. This is a little off topic, but can you elaborate a bit more what do you mean by "but at a fixed frequency of time"?I ever used save after x iterations. Thank you Daniele |
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April 24, 2013, 06:42 |
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#13 | |
Member
Linyx
Join Date: Mar 2011
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Quote:
why in time averaged values problem don't need to converge each timestep to flatten?? Since residual flatten is the character of convergence, don't we need a converged solution at each time at time averaged problem? |
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April 24, 2013, 09:54 |
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#14 | ||
Senior Member
OJ
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OJ |
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April 24, 2013, 10:40 |
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#15 |
Senior Member
François Grégoire
Join Date: Jan 2010
Location: Canada
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My project is a baking process that has a duration of about 2 weeks. I have turbulent combustion and radiation in a somewhat complex geometry. I fixed my time step at 5 minutes and it takes about 200 iterations at each time step to reach the default residual values.
If I reduce the time step < 1 s, I can get maybe 50 iterations per time step but the number of time steps will be ridiculously high for a 2 weeks simulation. I think it's not profitable in my case. |
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April 24, 2013, 10:54 |
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#16 |
Senior Member
OJ
Join Date: Apr 2012
Location: United Kindom
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200 inner iterations for 5 minutes timestep and 50 for 1 sec! Are you sure of these numbers?
OJ |
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