# Error message disappears

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 August 17, 2005, 02:03 I am using simpleFoam with kEp #1 ghanshyam Guest   Posts: n/a I am using simpleFoam with kEpsilon model. After few iterations it looks like it stops solving equations of k and epsilon as I don't see error on my terminal. Does this mean that these equations diverged? How to make it visible all the time. I have experimented with "tolerences" as well as "relaxation factors" no improvement. Regards GS

 August 17, 2005, 02:07 Maybe the initial residual is #2 Senior Member   Xiaofeng Liu Join Date: Mar 2009 Location: State College, PA, USA Posts: 118 Rep Power: 9 Maybe the initial residual is less than tolerences, so these equations don't need to be solved. Steady state? __________________ Xiaofeng Liu, Ph.D., P.E., Assistant Professor Department of Civil and Environmental Engineering Penn State University 223B Sackett Building University Park, PA 16802 Web: http://water.engr.psu.edu/liu/

 August 17, 2005, 02:10 Yes, steady state, I was guess #3 ghanshyam Guest   Posts: n/a Yes, steady state, I was guessing the same, but I still want to solve. How do I do that. Regards

 August 17, 2005, 04:30 Do you see the initial residua #4 Senior Member   Join Date: Mar 2009 Posts: 854 Rep Power: 14 Do you see the initial residual fall? Does it drop below your convergence tolerance? Have you tried making the convergence tolerance very tight?

 August 17, 2005, 04:40 >Do you see the initial residu #5 ghanshyam Guest   Posts: n/a >Do you see the initial residual fall? Yes, initially there is reduction in residual. >Does it drop below your convergence tolerance? Yes >Have you tried making the convergence tolerance very tight? Yes Currently I am compiling the code by making following change in "tests.C" //converged_ = true; converged_ = false; Will this help? Regards GS

 August 17, 2005, 04:53 I would be able to help you mo #6 Senior Member   Join Date: Mar 2009 Posts: 854 Rep Power: 14 I would be able to help you more if I had the answers to my questions or better still a section of the log file from just before k-epsilon stop solving to just after.

 August 17, 2005, 04:55 Sorry, I missed the answers bu #7 Senior Member   Join Date: Mar 2009 Posts: 854 Rep Power: 14 Sorry, I missed the answers buried there. Anyway, I would still like to see the section of log file. Have you looked the the field valeus for k-epsilon? Are they VERY large?

 August 17, 2005, 07:18 BICCG: Solving for Ux, Initia #8 ghanshyam Guest   Posts: n/a BICCG: Solving for Ux, Initial residual = 0.0114222, Final residual = 0.000931753, No Iterations 1 BICCG: Solving for Uy, Initial residual = 0.0227955, Final residual = 0.00190204, No Iterations 1 BICCG: Solving for Uz, Initial residual = 0.00691902, Final residual = 0.000593082, No Iterations 1 ICCG: Solving for p, Initial residual = 0.0216008, Final residual = 0.000214944, No Iterations 116 time step continuity errors : sum local = 0.00871451, global = 0.000437168, cumulative = 0.0171293 BICCG: Solving for epsilon, Initial residual = 0.021943, Final residual = 2.2619e-08, No Iterations 1 BICCG: Solving for k, Initial residual = 1.78372e-05, Final residual = 1.1457e-06, No Iterations 1 ExecutionTime = 201 s Time = 257 BICCG: Solving for Ux, Initial residual = 0.0114393, Final residual = 0.000938528, No Iterations 1 BICCG: Solving for Uy, Initial residual = 0.0228175, Final residual = 0.0019204, No Iterations 1 BICCG: Solving for Uz, Initial residual = 0.00691351, Final residual = 0.00059707, No Iterations 1 ICCG: Solving for p, Initial residual = 0.0229688, Final residual = 0.000222948, No Iterations 115 time step continuity errors : sum local = 0.00902571, global = 0.000449829, cumulative = 0.0175791 BICCG: Solving for epsilon, Initial residual = 0.0225105, Final residual = 2.18334e-08, No Iterations 1 bounding epsilon, min: -3.5104e-12 max: 822122 average: 4651.92 ExecutionTime = 205.23 s Time = 258 BICCG: Solving for Ux, Initial residual = 0.0114469, Final residual = 0.00095594, No Iterations 1 BICCG: Solving for Uy, Initial residual = 0.0226974, Final residual = 0.00194632, No Iterations 1 BICCG: Solving for Uz, Initial residual = 0.00688176, Final residual = 0.000596608, No Iterations 1 ICCG: Solving for p, Initial residual = 0.0220571, Final residual = 0.00021926, No Iterations 118 time step continuity errors : sum local = 0.00883914, global = 0.000396233, cumulative = 0.0179754 Will this help to clarify myself? You can see after Time(iteration) # 257, k and epsilon disappears forever. Regards GS

 August 17, 2005, 07:27 What convergence tolerance hav #9 Senior Member   Join Date: Mar 2009 Posts: 854 Rep Power: 14 What convergence tolerance have you set for k and epsilon?

 August 17, 2005, 07:31 For both 1e-5(Tolerance), 0.1( #10 ghanshyam Guest   Posts: n/a For both 1e-5(Tolerance), 0.1(RelativeTolerance)

 August 17, 2005, 07:35 For cases will wall-functions #11 Senior Member   Join Date: Mar 2009 Posts: 854 Rep Power: 14 For cases will wall-functions I use 1e-8.

 August 17, 2005, 07:38 Now I am running with 1e-15 an #12 ghanshyam Guest   Posts: n/a Now I am running with 1e-15 and 0.1 for both. It appears again. This way what is the lower tolerence limit? Thanks and regards GS

 August 17, 2005, 07:43 Check the k and epsilon fields #13 Senior Member   Join Date: Mar 2009 Posts: 854 Rep Power: 14 Check the k and epsilon fields, it sounds like they are becoming VERY large which affecting the residual normalisation.

 August 17, 2005, 08:03 Max epsilon is 676404.2 m^2/s^ #14 ghanshyam Guest   Posts: n/a Max epsilon is 676404.2 m^2/s^3 and k is 72 m^/s^2, which is not too large value. It has disappeared again even after giving tolerance 1e-15 for both k and epsilon. Regards GS

 August 17, 2005, 08:08 It appears k and epsilon have #15 Senior Member   Join Date: Mar 2009 Posts: 854 Rep Power: 14 It appears k and epsilon have converged for your case.

 August 17, 2005, 08:12 But I do not want it to stop s #16 ghanshyam Guest   Posts: n/a But I do not want it to stop solving k and epsilon, as flow field is still developing. How do I do that? Regards GS

 August 17, 2005, 08:19 For some reason, probably due #17 Senior Member   Join Date: Mar 2009 Posts: 854 Rep Power: 14 For some reason, probably due to something in your setup/BCs, the k-epsilon system has converged and nothing you do to the tolerances will make it converge further. You need to find out what is wrong with your setup. We would be happy to check and correct your case as part of your support contract with OpenCFD.

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