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what p.relax() is actaully doing in simpleFOAM?

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Old   December 21, 2020, 13:46
Default what p.relax() is actaully doing in simpleFOAM?
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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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Old   December 21, 2020, 14:42
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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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