# Step size for sliding mesh...

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 April 4, 2000, 09:41 Step size for sliding mesh... #1 Jack Keays Guest   Posts: n/a Hi, I am running a 2d simulation of a centrifugal pump. The flow is unsteady and I am using a sliding mesh technique. Would be a good idea to use the MRF in Fluent to get an initial guess and then continue from there with the sliding mesh?? ALso, what would be considered an intelligent step size? Does anyone know of any article/paper dealing with this? I am currently using 15 degrees. Probably too large. I hope someone can help me, Jack.

 April 4, 2000, 10:16 Re: Step size for sliding mesh... #2 Glenn Price Guest   Posts: n/a Based on my experience, it can take a long time to reach time-period state if you run from start up (~5 - 50 complete revolutions - depending on the impeller Re). This number often can be reduced by a factor of 2 - 5 using MRF to get a reasonable initial guess. An intelligent time step size depends on the details on the internal geometry for your problem. An additional computational criterion is that the number of sub-iterations per time step shouldn't exceed 30-40 iterations. If you have the time, you may want to: 1. Run MRF to get an initial guess and then run sliding mesh with 15 degree step, plotting several variables (e.g. impeller torque, pressure at a point, etc) to monitor the time history and determine when the solution reaches a periodic state. 2. Repeat step 1 with a smaller time step, perhaps 7.5 degrees, to evaluate the effect of time step. Make sure you are using the second-order temporal integration scheme. Good luck. hadeside likes this.

 April 4, 2000, 10:25 Re: Step size for sliding mesh... #3 Jack Keays Guest   Posts: n/a Do you mean that my number of iterations per time step should not exceed 40-50?? Should I obey this for the sliding mesh case also? Is that not very low? I am currently using time step of up to 600. If I were to run my solution at 40-50 iterations per time step would I eventually begin to reach convergence after perhaps 30-40 revolutions using the MRF? Could I then use that at an initial guess for the sliding mesh scheme?. ( 0.001 is my scaled residual convergence criterion.).

 April 4, 2000, 11:33 Re: Step size for sliding mesh... #4 Jonas Larsson Guest   Posts: n/a I haven't done any simulations of centrifugal pumps, but I have a lot of experience from axial turbines and compressors - in Fluent you should in that case use around 50 to 100 time-steps per blade-passing period. If you do this each time-step usually converges in something like 10 to 20 iterations. If you need 600 iterations you are surely running with a much too long time-step. 15 degs sounds huge to me!

 April 4, 2000, 14:56 Re: Step size for sliding mesh... #5 Glenn Price Guest   Posts: n/a 1. I mean a maximum of 40-50 iterations per time step for a sliding mesh calculation (depending on the convergence criterion). 2. Why aren't your running the MRF simulation as steady state calculation? The rotor's frozen in an MRF calc. right, so you don't need to worry about time step here. 3. As Jonas mentions, generally only 10-20 iterations should be necessary to convergence the solution at time step. I like to convergence the solution down to a tighter tolerance than 0.001 - generally an order of magnitude lower (1.0e-4). However, a convergence of 0.001 or even higher can be employed in the steady MRF calculation since it is only an initial guess.

 April 5, 2000, 00:49 Re: Step size for sliding mesh... #6 chris Guest   Posts: n/a Hy Jonas, some questions: 1. blade-passing-period: is it 360°/number of blades ? 2. Do you take scaled residuals and 1e-3 as convergence ? 3. Which time-discretization do you use ? 4. how much cells do you have at the interface ? is one timestep much more or less than the size of a cell there ? Thanks a lot chris

 April 5, 2000, 04:08 Re: Step size for sliding mesh... #7 Jack Keays Guest   Posts: n/a Thanks for the help guys! So you reckon that I should have between 50 and 100 time steps per passing period and that each time step should have 20-50 iterations? I take it, that it will take at least 20-30 passes to converge the solution?? Can some answer this please...I am talking about the sliding mesh case.

 April 5, 2000, 04:11 Re: Step size for sliding mesh... #8 Jack Keays Guest   Posts: n/a Sorry, in the MRF step I have the blade frozen. The problem is that I need to remesh the model because the sliding interface wil not work for MRF...Isn't that correct??? Could you also tell me of any books/papers/articles which deal with time step size, the number of iterations etc...thanks!

 April 5, 2000, 05:05 Re: Step size for sliding mesh... #9 Jack Keays Guest   Posts: n/a Another question! How can I insure that my sliding mesh model completes say, 25 revolutions before convergence?? Do I need to simply reduce the convergence criterion to something very small like 1e-06, so that the solution has reached a good level of convergence but has not yet fulfilled the criterion. Is there another way to do this?? Thanks again.

 April 5, 2000, 11:07 Re: Step size for sliding mesh... #10 Joern Beilke Guest   Posts: n/a Forget about the convergence level. Just do a postprocessing on the fly and watch the results (mass flux and pressure at the inlet and outlet and the impeller-torque) and wait until they become periodic. Normally this requires at least 2 revolutions. If you want to be able to compare these results with some measurements you might have to do a complete 3d calculation.

 April 5, 2000, 12:52 Re: Step size for sliding mesh... #11 Jonas Larsson Guest   Posts: n/a Quick answers: 1. yes 2. Something like that, 1E-3 is minimum. I prefer to get 1E-4 - usually what I do is to run with a larger number in the beginning in order to obtain a periodic flow. Then when the flow starts to look periodic you can sharpen your convergence criteria to improve the accuracy. 3. 2:nd order 4. Varies a lot from case to case and what you are after. If you want to have wakes passing the interface with some accuracy I'd use 50 cells or something like this - this is really very much dependent on the case. With 100 time-steps per blade-passing this gives 0.5 cells per time-step. If you have this your iterations most often converge within 10 to 20 iterations. I'm talking about 3D simulations here. If I'd run 2D I'd use much more - simulation time isn't a problem then so why save grid-points then.

 April 5, 2000, 13:09 Re: Step size for sliding mesh... #12 Jonas Larsson Guest   Posts: n/a I'd prefer to get convergence in 10 to 20 iterations in each time-step, sometimes more is necessary though. I don't know how many blade-passes you need in your application. In axial turbines you need at least 10, sometimes 20. You can easily monitor this by plotting the time-history of variables like velocities in a few strategic points. Also remember to check that mass-flows etc. are consistant.

 April 5, 2000, 13:11 Re: Step size for sliding mesh... #13 Jonas Larsson Guest   Posts: n/a Eh, I'm not sure I understand your question - convergence is just in each time-step. How many revolutions you get is not dependent on this - you control this by how many time-steps you run and how long they are. The convergence-criteria just determines how accurate each time-step is.

 April 6, 2000, 05:44 Re: Step size for sliding mesh... #14 Joakim Majander Guest   Posts: n/a Is your result time-step converged even at this time-step? I haven't done any pump/compressor simulations, but a lot of stirred tank simulations. I once did a test with different grids, time-steps and MRF vs Sliding Mesh. I found out that I was not able to get time-step convergence. I stopped at 1/3 cell/time-step (I had about 40 cells/90 degrees, which was the symmetry for 4 bladed pitched blade turbine). I still saw quite a large jump at some integrated values (k, epsilon, kinetic energy etc) when I reduced the time-step. Infact small enough time-step seemed to be very far away. In this case I had to run about 100 revolutions (400 blade passages) to get convergence. Starting from a coarser time-step/MRF solution didn't help much. I found out I could get much better results with MRF, since denser grid helped more than time accuracy, but impeller-blade interactions are much weaker in stirred tank vessel than rotor-stator ones in compressors/pumps. MRF is atleast 10 times faster with the same grid. If you go to 1/2 cell or less/time step, make that 100. Joakim Joakim

 April 6, 2000, 06:38 Re: Step size for sliding mesh... #15 Jonas Larsson Guest   Posts: n/a Further reducing the time-step is a good check - if you use say 200 time-steps per blade-passage you should get the same flow as with 100. This is also the case most often, ie 100 is enough for one blade-passing period, at least that is my experience from axial turbines - I haven't seen a lot of problems with discontinuities at the sliding interfaces. I am always very carful with the grid on these interfaces though - I only use hexahedrals and I make sure that I have matching grid-lines in the radial direction and try to have no overly coarse regions in the tangential direction.

 April 6, 2000, 07:02 Re: Step size for sliding mesh... #16 Joakim Majander Guest   Posts: n/a I was not talking about "discontinuities at the sliding interfaces". I meant that the results still changed after reducing the time-step. By integrated variables I meant the volume (or mass) integral of k, epsilon etc. over the whole domain. I use hexas as well and in this case a totally even distribution in tangential direction, which seems to be the best approach for stirred tanks. I'm using wall functions. It's nice to hear there really is a small enough time-step for sliding mesh. Joakim

 April 6, 2000, 07:20 Re: Step size for sliding mesh... #17 Jack Keays Guest   Posts: n/a Do you reckon that if the solution is not periodic after 3-4 revolutions that I require more revolutions or should I refint the grid? The people at fluent said that I might need 20 revolutions......is this possible?? I feel that the number of iterations per time step need to be quiet large...Is it best to keep it small...( 40-60)?? My solution does not seem to be converging for these small iteration numbers. The solution levels out and oscillates between e-01 and e-02. What should I do? Simply continue to run for more revolutions, increase the number of iterations per time step, reduc the time step or refine the grid??

 April 6, 2000, 07:25 Re: Step size for sliding mesh... #18 Jack Keays Guest   Posts: n/a I know convergence is in each time step. What I mean is that I want to ensure that each residual goes below ,say, e-03 at each time step and for the solution to save the result and go on to the next time step. At the moment if I set the model running for say 36 time steps and I set it up to save the results at each step it will do this provided the solution does not reach convergence at any given time step. If it does, then the model stops and forgets about continuing with the further time steps. Maybe I am mistaken, but I don't think so. This seems to a problem with the sliding mesh situation.

 April 6, 2000, 08:45 Re: Step size for sliding mesh... #19 Jonas Larsson Guest   Posts: n/a I've never had this problem. When the iterations for one time-step converge (the residual goes below the limit that you've set) then the solver just goes on to the next time-step.

 April 6, 2000, 08:53 Re: Step size for sliding mesh... #20 Jonas Larsson Guest   Posts: n/a Which solver are you using? Which pressure-scheme? How fine is your grid? Have you tried to adjust solver parameters etc.? Can you locate the problematic region where you have high residuals? Getting these things to run well often takes a lot of tweaking. It is difficult to give any general advice.