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what p.relax() is actaully doing in simpleFOAM? |
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December 21, 2020, 13:46 |
what p.relax() is actaully doing in simpleFOAM?
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#1 |
Senior Member
Join Date: Dec 2017
Posts: 153
Rep Power: 8 |
Hello guys,
I am a bit puzzled about p.relax() you find in simpleFOAM. At the first glace I was thinking it was relaxing the pressure as: p = pold +urf*(p-pold) But it does not look so. This is the relaxationdictionary in my fvSolution: relaxationFactors { fields { p 0.3; } equations { U 1.0; } } I have built a silly 3x3 lid driven cavity case, (nu=1, dx=dy=1) and I can see that "p" is the same before and after p.relax is called. Here the code output: DILUPBiCG: Solving for Ux, Initial residual = 1, Final residual = 1.5878e-08, No Iterations 5 DILUPBiCG: Solving for Uy, Initial residual = 0, Final residual = 0, No Iterations 0 DICPCG: Solving for p, Initial residual = 1, Final residual = 3.24343e-18, No Iterations 7 p before relax= 1.54988e-16, aC= -0.733333, bC= 0.05 p before relax= 0.748609, aC= -0.591667, bC= -3.98836e-10 p before relax= 1.49722, aC= -0.366667, bC= -0.05 p before relax= -0.475881, aC= -0.591667, bC= 0.2 p before relax= 0.748609, aC= -0.9, bC= -1.59535e-10 p before relax= 1.9731, aC= -0.591667, bC= -0.2 p before relax= -1.36364, aC= -0.366667, bC= 0.55 p before relax= 0.748609, aC= -0.591667, bC= -3.98836e-10 p before relax= 2.86085, aC= -0.366667, bC= -0.55 time step continuity errors : sum local = 9.09766e-17, global = 1.54198e-18, cumulative = 1.54198e-18 pressure= 1.54988e-16 pressure= 0.748609 pressure= 1.49722 pressure= -0.475881 pressure= 0.748609 pressure= 1.9731 pressure= -1.36364 pressure= 0.748609 pressure= 2.86085 ExecutionTime = 0.01 s ClockTime = 0 s Can somebody please shed some light on this? Many thanks! |
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December 21, 2020, 14:42 |
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#2 |
Senior Member
Join Date: Dec 2017
Posts: 153
Rep Power: 8 |
Ok I am done, I was putting the relaxation factors in the wrong place (oh boy!).
Actaully it is exactly as expected. Cheers |
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