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May 29, 2012, 08:57 
controlDict steady state

#1 
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Carles
Join Date: Jan 2012
Location: Karlsruhe
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Hi,
I would like to know what is the meaning of : endTime and deltaT for a steady state case. Itīs confusing. has it a relation with the max amount of iterations ? is it necessary to have a endTime and a deltaT that let the simulation go on until a convergence is reached ? Thanks a lot Carles 

May 29, 2012, 18:54 

#2 
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Hi CRT,
I've been trying to get a grip on the same thing, quoting from here "Residual/Convergence Control Solvers using the SIMPLE or PIMPLE algorithms now include convergence controls based on residuals of fields. The controls are specified through a residualControls subdictionary in the fvSolution file. The user specifies a tolerance for one or more solved fields and when the residual for every field falls below the corresponding residual, the simulation terminates. The following example sets tolerances for p, U and k and epsilon:" So on the face of it, it looks as if the time settings need to be long enough for the residuals to resolve and then the simulation will stop at that point. I've failed to make it work like that yet. 

May 30, 2012, 03:26 

#3 
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Location: Munich, Germany
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Hello,
Like you thought, endTime can be identified with the maximal amount of iterations. For problems, which donīt converge to a solution, the calculation would never stop. deltaT is necessary, because OpenFoam have to choose, how often the problem is solved. For example, if you choose startTime 0, endTime 5000 and deltaT 5, you have 1000 iterations. Then you could take a look at your solution, if the convergence criterion is full filled. In a mathematical approch, the time in a steady state problem can be seen as a "pseudo"time. Your solution is independent of it, but you need it to solve your problem. 

May 30, 2012, 09:18 

#4  
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Carles
Join Date: Jan 2012
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Thanks both for the replay.
Quote:
Code:
solvers { p { solver PCG; tolerance 1e05; relTol 0.001; maxIter 6000; preconditioner DIC; } "(UTkepsilonR)" { solver PBiCG; preconditioner DILU; tolerance 1e05; relTol 0.001; maxIter 6000; } } SIMPLE { nNonOrthogonalCorrectors 0; pRefCell 0; pRefValue 0; residualControl { p 1e7; U 1e7; T 1e7; "(kepsilonomega)" 1e5; } } Quote:
Code:
Time = 1 DILUPBiCG: Solving for Ux, Initial residual = 1, Final residual = 0.000936822, No Iterations 66 DILUPBiCG: Solving for Uy, Initial residual = 1, Final residual = 0.00094058, No Iterations 39 the deltaT is used to show how the residuals are going forward and also how you said, to guarantee that the program can do all the iterations that need to get a result, result under the SIMPLEīs condition Code:
residualControl { p 1e7; U 1e7; T 1e7; "(kepsilonomega)" 1e5; } 

October 15, 2014, 08:10 
Understanding Steady State

#5 
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Ali Kadar
Join Date: Oct 2014
Location: Delft
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For steady state cases the time step deltaT should be set to 1
since for steady state its role is effectively that of an iteration counter. source http://www.foamcfd.org/Nabla/guides/ProgrammersGuidese13.html My Conclusions For Steady State we use SIMPLE Algorithm which is essentially a guess and correct procedure for the calculation of pressure. So each time step involves a new guess followed by solving a linear system(to obtain the corrections). Code:
startTime 0; stopAt endTime; endTime 10000; deltaT 1; solvers { p tolerance 1e06; UkepsilonRnuTilda tolerance 1e05; } SIMPLE { p 1e2; U 1e3; kepsilonomega 1e3; } However the main stopping criteria is the tolerance 1e02 for p, 1e03 for U ..etc. The simple algorithm would run until the above tolerance is achieved for each of p, U etc or the limit of 10000 is reached(which would essentially mean no convergence). The tolerance for the solver is not really important and the values specified above(taken from pitzDaily simpleFoam tutorial) are good enough. There is no point considering smaller tolerance say 1e08 because we are trying to obtain corrections for pressure with guessed starting values. Some Observations(with a furnace geometry similar to pitzDaily but with a much refined mesh) p 1e3, U 1e2; kepsilonomega 1e2; SIMPLE solution converged in 1572 iterations p 1e2, U 1e1; kepsilonomega 1e1; SIMPLE solution converged in 281 iterations p 1e6, U 1e5; kepsilonomega 1e5; SIMPLE solution did not converge even after 10000 iterations. p 1e5, U 1e1; kepsilonomega 1e1; SIMPLE solution converged in 8910 iterations Code:
smoothSolver: Solving for Ux, Initial residual = 2.76007e06, Final residual = 2.76007e06, No Iterations 0 smoothSolver: Solving for Uy, Initial residual = 8.51021e06, Final residual = 8.51021e06, No Iterations 0 GAMG: Solving for p, Initial residual = 9.7978e06, Final residual = 8.63712e07, No Iterations 2 time step continuity errors : sum local = 3.83861e07, global = 6.94709e10, cumulative = 0.00612712 smoothSolver: Solving for epsilon, Initial residual = 9.99271e06, Final residual = 9.99271e06, No Iterations 0 smoothSolver: Solving for k, Initial residual = 9.86906e06, Final residual = 9.86906e06, No Iterations 0 ExecutionTime = 2808.42 s ClockTime = 2818 s The Final Residual shows the local improvement. For Ex Convergence after step 9000 would imply the sequence IR9001 ~ FR9001 ~ IR9002 ~ FR9002 .... Please email me at a.h.kadar@student.tudelft.nl if you doubt or find the above conclusions wrong.
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