simtk.org

when i run simulation,
"FATAL ERROR: Residual has exceeded tolerance" pop up even though i keep increasing the number of time steps.
how can i solve this problem?

also, when i create data files for simulation, sometimes details show like below
" ************************************************************
**************** w a r n i n g e r r o r ****************
************************************************************

routine zbrent"
what does it mean? and should i correct something?


===================================================
Mesh done. Statistics:

Number of Nodes: 60914
Number of Elems: 341852
Number of Edges: 34170
Number of Faces: 22780
=====================================================
The process ID for myrank (0) is (80329).


The number of processes is 1.

Solver Input Files listed as below:
------------------------------------
Local Config: solver.inp
Default Input File: Not Setup.

Parameter Values setup as below:
----------------------------------------------
Time Varying Boundary Conditions From File: True
BCT File Type: DAT
Number of BCT Files: 1
BCT Matching Type: Global Node ID
BCT Time Scale Factor: 1.0
Equation of State: Incompressible
Time Step Size: 0.002
Number of Timesteps: 3000
Viscous Control: Viscous
Number of Timesteps between Restarts: 10
Print Average Solution: True
Print Error Indicators: False
Number of Force Surfaces: 1
Surface ID's for Force Calculation: 1
Force Calculation Method: Velocity Based
Apply Wall Deformation: False
Number of Surfaces which Output Pressure and Flow: 0
Density: 1.06
Viscosity: 0.04
Solver Type: svLS
svLS Type: NS
Number of Krylov Vectors per GMRES Sweep: 100
Number of Solves per Left-hand-side Formation: 1
Tolerance on Momentum Equations: 0.05
Tolerance on Continuity Equations: 0.4
Tolerance on svLS NS Solver: 0.4
Maximum Number of Iterations for svLS NS Solver: 1
Maximum Number of Iterations for svLS Momentum Loop: 2
Maximum Number of Iterations for svLS Continuity Loop: 400
Residual Tolerance: 1000000.0
Time Integration Rule: Second Order
Time Integration Rho Infinity: 0.5
Flow Advection Form: Convective
Quadrature Rule on Interior: 2
Quadrature Rule on Boundary: 3
Number of Elements Per Block: 255
Number of Coupled Surfaces: 2
Pressure Coupling: Implicit
Number of Resistance Surfaces: 2
List of Resistance Surfaces: 3 4
Resistance Values: 1000 1000
Backflow Stabilization Coefficient: 0.2
Deformable Wall: False
Residual Control: True
Residual Criteria: 0.01
Minimum Required Iterations: 3
Step Construction: 0 1 0 1
Solver Task: Full Simulation


Time derivative of solution is set to zero (SAFE)
Element block size = 255
Domain size (x,y,z): 7.0278949738 3.5636596680 19.0669174194
Total number of nodes = 60914

Opening bct.dat
Process rank( 0) has 393 bct nodes.

Simulation Type: RIGID WALL

Number of global nonzeros 889224
maxfill= 25
nnz ok nnz= 35 actually needed 15
0 is my rank and my nnz_tot is: 889224
1 2.000E+00 6.607E-01 ( 0) 1.699E+01 2.292E+29 < 53051- 1| 17> [ 0 - 0]
1 4.000E+00 5.831E-01 ( 0) 3.916E-01 4.426E-01 < 53051- 1| 16> [ 0 - 0]
2 7.000E+00 4.990E+00 ( 8) 3.937E+00 3.342E+00 < 32829- 1| 13> [ 0 - 0]
2 8.000E+00 2.741E+00 ( 6) 4.768E-01 1.827E-01 < 32818- 1| 17> [ 0 - 0]
3 9.000E+00 2.865E+01 ( 16) 6.768E-01 1.816E-01 < 34271- 1| 8> [ 0 - 0]
3 1.100E+01 3.921E+00 ( 7) 3.154E-02 2.218E-04 < 36078- 1| 16> [ 0 - 0]
4 1.200E+01 1.972E+01 ( 14) 2.961E-01 1.264E-01 < 32829- 1| 11> [ 0 - 0]
4 1.300E+01 2.617E+00 ( 5) 3.497E-02 1.748E-03 < 32829- 1| 15> [ 0 - 0]
5 1.500E+01 1.804E+01 ( 14) 1.624E-01 1.335E-01 < 36240- 1| 9> [ 0 - 0]
5 1.700E+01 2.833E+00 ( 6) 2.263E-01 1.041E+00 < 32794- 1| 15> [ 0 - 0]
6 1.800E+01 1.650E+01 ( 13) 2.619E-01 4.190E-01 < 36844- 1| 9> [ 0 - 0]
6 1.900E+01 2.153E+00 ( 5) 5.026E-02 9.573E-03 < 32829- 1| 16> [ 0 - 0]
7 2.000E+01 1.255E+01 ( 12) 2.642E-01 1.042E-01 < 32926- 1| 12> [ 0 - 0]
7 2.100E+01 1.888E+00 ( 4) 5.534E-02 1.613E-02 < 33023- 1| 15> [ 0 - 0]
8 2.300E+01 1.198E+01 ( 12) 3.218E-01 8.979E-02 < 48286- 1| 10> [ 0 - 0]
8 2.400E+01 1.763E+00 ( 4) 3.311E-01 8.129E-01 < 32944- 1| 14> [ 0 - 0]
9 2.500E+01 9.385E+00 ( 11) 2.457E-01 1.549E-01 < 38450- 1| 10> [ 0 - 0]
9 2.600E+01 1.508E+00 ( 3) 3.404E-02 1.163E-02 < 33023- 1| 15> [ 0 - 0]
10 2.800E+01 7.625E+00 ( 10) 2.147E-01 2.095E+00 < 32973- 1| 11> [ 0 - 0]
10 2.900E+01 1.138E+00 ( 2) 2.884E-02 5.326E-03 < 32829- 1| 15> [ 0 - 0]
~
~
~
~
~
155 4.310E+02 4.544E+04 ( 48) 1.420E+00 1.262E+00 < 42340- 1| 16> [ 0 - 0]
155 4.320E+02 1.031E+05 ( 51) 8.639E-01 6.145E-01 < 51387- 1| 14> [ 0 - 0]
156 4.340E+02 1.676E+05 ( 54) 1.508E+00 1.552E+00 < 52213- 1| 17> [ 0 - 0]
156 4.350E+02 3.999E+05 ( 57) 9.870E-01 1.521E+00 < 43284- 1| 16> [ 0 - 0]
157 4.370E+02 6.607E+05 ( 60) 2.050E+00 1.904E+00 < 34771- 1| 16> [ 0 - 0]
FATAL ERROR: Residual has exceeded tolerance 1000000.0000000000

Hello,

The log file shows that the non-linear residual (3rd column) is 4.544E+04 which indicates that your simulation is not setup correctly, residual should be around 1e-3.

Most likely your time step is too large and violates the CFL condition.

Cheers,
Dave

Hello All

When I run simulation I get also the same error message (cf infra)
I've tried to modify time step size and number but it still doesn't work.
Does anyone have any ideas on how to resolve this issue ?
Thank you in advance for your help.

Antoine

Is there a link with meshing quality of the geometry ? I can provide my test file if needed.

Hi Antoine,

The error I see is FATAL ERROR: Residual has exceeded tolerance 1000000.0000000000. The residual tolerance is sometimes exceeded at the start of a simulation but improves later. You can increase the tolerance by using the solver.inp Residual Tolerance: parameter.

Cheers,
Dave

In fact, I saw in a previous topic how to modify the solver inp file. By increasing the tolerance from 10e6 to 10e7, it works. However, I wonder to what extent this affects the results of the simulation and how to estimate the increase in the margin of error by increasing this parameter ?
Thanks again for your help,

Best regards,

Antoine

Hi Antoine,

The initial large residue just means there there is a large difference in the initial state of the system. The residue should decrease substantially in the following iterations. You do need to make sure that the residue is small (~1e-03) for the results to be trusted.

Cheers,
Dave