[becklei]

Hi! I recently met a headache problem about a laminar model. I assembled the interFoam with a energy equation and defined phase1 with cross-WLF viscosity model. unfortunately, as shown in figure, I got asymmetric results with Symmetry conditions. please help me!

[sepidecent]

Hi there,

I have exactly the same problem in CFX. I was wondering if you have found any answer for this problem.

Regards,
Sep

[becklei]

I try to find the reason from the discrete methods and the Algebraic process, and I have no answers now.

[RodriguezFatz]

Hi,

Can you post, where in your model you have what kind of symmetry b.c.?

[becklei]

Originally Posted by RodriguezFatz

Hi,

Can you post, where in your model you have what kind of symmetry b.c.?

thanks for your attention! To simulated the polymer injection flow field, I choose the no slip boundary condition(u=0) for the left/right boundary.

[RodriguezFatz]

What is on top / bottom, what is plotted in your picture?

[becklei]

Originally Posted by RodriguezFatz

What is on top / bottom, what is plotted in your picture?

The bottom boundary is inlet(alpha=1, u=(0,0,0.08)), and the top as same as left/right. The chart on the right side is plotted according to the data on the white line.

[RodriguezFatz]

Does that mean you have u=0 at top, left, right and u=0.08 into the domain at the bottom?

[becklei]

Originally Posted by RodriguezFatz

Does that mean you have u=0 at top, left, right and u=0.08 into the domain at the bottom?

yes, It is common for injection molding simulation

[RodriguezFatz]

So this is time-dependent?

[becklei]

yes,this is transient

[RodriguezFatz]

Can you show the log file of the simulation?

[becklei]

Originally Posted by RodriguezFatz

Can you show the log file of the simulation?

Create time

Create mesh for time = 0

// using new solver syntax:
T
{
solver PBiCG;
preconditioner DILU;
tolerance 1e-07;
relTol 0;
}

// using new solver syntax:
T+T
{
solver smoothSolver;
smoother ILU;
nSweeps 3;
minIter 0;
maxIter 1000;
tolerance 1e-06;
relTol 0;
}


Reading g
Reading transportProperties

Reading TK＝273.15

Reading field T

Reading field pd

Reading field alpha1

Reading field U

Reading/calculating face flux field phi

Reading transportProperties

Selecting incompressible transport model CrossWLF
Selecting incompressible transport model Newtonian
Calculating field g.h

Selecting turbulence model type laminar
time step continuity errors : sum local = 0.0016, global = -0.0016, cumulative = -0.0016
DICPCG: Solving for pcorr, Initial residual = 1, Final residual = 7.465223e-11, No Iterations 27
time step continuity errors : sum local = 1.194432e-13, global = -1.582459e-14, cumulative = -0.0016
Courant Number mean: 0.001270323 max: 0.4439866 velocity magnitude: 0.1109966

Starting time loop

Courant Number mean: 0.00141147 max: 0.4933184 velocity magnitude: 0.1109966
deltaT = 0.001111111
Time = 0.001111111

MULES: Solving for alpha1
Liquid phase volume fraction = 0.0008888889 Min(alpha1) = 0 Max(alpha1) = 1
MULES: Solving for alpha1
Liquid phase volume fraction = 0.001739616 Min(alpha1) = 0 Max(alpha1) = 1
DICPCG: Solving for pd, Initial residual = 1, Final residual = 0.02363721, No Iterations 6
DICPCG: Solving for pd, Initial residual = 1.341937e-07, Final residual = 4.663726e-08, No Iterations 1
DICPCG: Solving for pd, Initial residual = 1.669412e-07, Final residual = 5.153588e-08, No Iterations 1
time step continuity errors : sum local = 0.001140456, global = 0.0008470537, cumulative = -0.0007529463
DILUPBiCG: Solving for T, Initial residual = 1, Final residual = 5.301002e-08, No Iterations 17
ExecutionTime = 0.54 s ClockTime = 1 s

Courant Number mean: 0.003166069 max: 0.61741 velocity magnitude: 0.1389173
deltaT = 0.0008888889
Time = 0.002

MULES: Solving for alpha1
Liquid phase volume fraction = 0.002372943 Min(alpha1) = 0 Max(alpha1) = 1
MULES: Solving for alpha1
Liquid phase volume fraction = 0.002980165 Min(alpha1) = 0 Max(alpha1) = 1
DICPCG: Solving for pd, Initial residual = 0.04042854, Final residual = 0.0007290781, No Iterations 2
DICPCG: Solving for pd, Initial residual = 0.2796866, Final residual = 0.006973493, No Iterations 5
DICPCG: Solving for pd, Initial residual = 0.2282629, Final residual = 5.127153e-08, No Iterations 18
time step continuity errors : sum local = 1.623944e-06, global = -3.650373e-07, cumulative = -0.0007533113
DILUPBiCG: Solving for T, Initial residual = 0.1983379, Final residual = 4.778879e-08, No Iterations 17
ExecutionTime = 0.8 s ClockTime = 1 s

Courant Number mean: 0.9925878 max: 141.3327 velocity magnitude: 39.74981
deltaT = 3.144654e-06
Time = 0.002003145

MULES: Solving for alpha1
Liquid phase volume fraction = 0.00298302 Min(alpha1) = 0 Max(alpha1) = 1
MULES: Solving for alpha1
Liquid phase volume fraction = 0.002985876 Min(alpha1) = 0 Max(alpha1) = 1
DICPCG: Solving for pd, Initial residual = 0.1967935, Final residual = 0.007452839, No Iterations 4
DICPCG: Solving for pd, Initial residual = 0.1448237, Final residual = 0.004567321, No Iterations 4
DICPCG: Solving for pd, Initial residual = 0.1081815, Final residual = 9.174754e-08, No Iterations 16
time step continuity errors : sum local = 5.25175e-10, global = -1.027555e-10, cumulative = -0.0007533114
DILUPBiCG: Solving for T, Initial residual = 0.0001458371, Final residual = 2.35284e-08, No Iterations 3
ExecutionTime = 1.03 s ClockTime = 1 s

Courant Number mean: 0.001923476 max: 0.1389821 velocity magnitude: 13.58828
deltaT = 3.772102e-06
Time = 0.002006917

[RodriguezFatz]

I am not pretty familiar with the interFoam syntax, but it looks like there are no inner iterations each time step. So you can not expect each time step to be converged !? This could be a reason. Also, in general, I would not expect to get symmetric results just because I have symmetric b.c.

[becklei]

Originally Posted by RodriguezFatz

I am not pretty familiar with the interFoam syntax, but it looks like there are no inner iterations each time step. So you can not expect each time step to be converged !? This could be a reason. Also, in general, I would not expect to get symmetric results just because I have symmetric b.c.

Thanks for your possible reasons, and I'll think about it. I would like to promptly post any progress!

[rcastilla]

Hello,

i am getting a similar problem. I get nonzero Uz in a x-y plane defined as symmetryPlane in U. I don't understand the sentence: "in general, I would not expect to get symmetric results just because I have symmetric b.c.". What is the symmetry b.c. for then? Does it mean that the simulation has to be fully converged in each timestep to get symmtric results?

Thanks

Robert

[RodriguezFatz]

Hi Robert,

He doesn't use the "symmetry boundary condition", which means zero gradient by definition, but he uses symmetric boundary conditions on the left an right hand side of his domain. This is a huge difference. In a bluff body flow the bluff body can be symmetric, but that does not mean, that the flow is also symmetric all the time!

[rcastilla]

Hi, Philipp,

Yes, you are right! My apologizes for the confusion.

Nevertheless, is it normal that I get Uz in a symmetryPlane? Well, in fact maybe it is better if I open a new thread with this question...

Thank you for your answer

Robert

[RodriguezFatz]

Uz should be zero, because this is the actual b.c.