# Zero iteration step, why??

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 March 18, 2013, 18:32 Zero iteration step, why?? #1 Senior Member   Join Date: Nov 2012 Posts: 168 Rep Power: 6 Hi All, In one of my simulations, I used k-epsilon model, and got the following output: ............... pEqn.H_1: rho max/min : 1.1966 1.19659 diagonal: Solving for rho, Initial residual = 0, Final residual = 0, No Iterations 0 time step continuity errors (Low Ma): sum local = 8.61908e-16, global = -2.16302e-17, cumulative = -2.87666e-15 pEqn.H_2: rho max/min : 1.19661 1.1966 GAMG: Solving for phd, Initial residual = 0.00121095, Final residual = 8.07744e-06, No Iterations 8 GAMG: Solving for phd, Initial residual = 0.00275727, Final residual = 2.14566e-05, No Iterations 5 GAMG: Solving for phd, Initial residual = 0.00166235, Final residual = 9.90116e-06, No Iterations 4 pEqn.H_1: rho max/min : 1.19661 1.1966 diagonal: Solving for rho, Initial residual = 0, Final residual = 0, No Iterations 0 time step continuity errors (Low Ma): sum local = 7.21639e-16, global = -2.3332e-17, cumulative = -2.89999e-15 pEqn.H_2: rho max/min : 1.1966 1.19659 DILUPBiCG: Solving for epsilon, Initial residual = 5.56717e-07, Final residual = 5.56717e-07, No Iterations 0 DILUPBiCG: Solving for k, Initial residual = 4.73169e-07, Final residual = 4.73169e-07, No Iterations 0 Pimple Loop End: rho max/min : 1.1966 1.1966 Pimple Loop End: nu max/min : 1.528e-05 1.528e-05 ExecutionTime = 1907.55 s ClockTime = 1915 s .................... From above, we can see the number of the iterations is zero. What does it mean? Does it mean k and epsilon are solved using explicit method? (I have this think because in the compressible solvers, the density is solved using explicit method with zero iteration step). Why are they (k and epsilon) are solved with zero iteration steps? In the beginning of the simulations, they are not zero, but 3-5 steps. Does anybody give me some hints about this? best H

 March 18, 2013, 19:05 #2 Senior Member   Julien de Charentenay Join Date: Jun 2009 Location: Australia Posts: 230 Rep Power: 10 Hi H, The number of iteration is 0 since the initial residual is less than the convergence criteria you set. So for all intent and purpose, the solver is assuming that the epsilon and k fields are "converged" and do not need iterations. If you do want to have more iteration, decrease the convergence criteria for k and epsilon. Julien __________________ --- Julien de Charentenay

March 18, 2013, 19:57
#3
Senior Member

Join Date: Nov 2012
Posts: 168
Rep Power: 6
Hi Julien,

When I change the tolerance in system/fvSolution as follows:

k
{

solver smoothSolver;
smoother GaussSeidel;
tolerance 1e-9;
relTol 0;
nSweeps 1;

/* solver PBiCG;
preconditioner DILU;
tolerance 1e-010;
relTol 0.1;
*/
};

epsilon
{

solver smoothSolver;
smoother GaussSeidel;
tolerance 1e-9;
relTol 0;
nSweeps 1;

/* solver PBiCG;
preconditioner DILU;
tolerance 1e-010;
relTol 0.1;
*/
};

The output from the solver is like the followings:

diagonal: Solving for rho, Initial residual = 0, Final residual = 0, No Iterations 0
time step continuity errors (Low Ma): sum local = 7.207e-16, global = -8.28698e-18, cumulative = -7.88457e-15
pEqn.H_2: rho max/min : 1.1966 1.19659
DILUPBiCG: Solving for epsilon, Initial residual = 4.13716e-07, Final residual = 4.13716e-07, No Iterations 0
DILUPBiCG: Solving for k, Initial residual = 5.42019e-07, Final residual = 5.42019e-07, No Iterations 0

It is a little strang that changing the tolerance does not change the iteration number. Besides, I found that linear algebraic solver DILUPBiCG is not the one I specified in the fvSolution file (smoothSolver). When I change the linear algebraic solver for the velocity U, it indeed changes. So I am a little confused about the controling parameter for k and epsilon. Why can they not be changed in system/fvSolution although they should be?

Thank you so much!
hz283

Quote:
 Originally Posted by julien.decharentenay Hi H, The number of iteration is 0 since the initial residual is less than the convergence criteria you set. So for all intent and purpose, the solver is assuming that the epsilon and k fields are "converged" and do not need iterations. If you do want to have more iteration, decrease the convergence criteria for k and epsilon. Julien

March 18, 2013, 20:08
#4
Senior Member

Join Date: Nov 2012
Posts: 168
Rep Power: 6
Hi Julien,

If during the simulation the iteration number of k and epsilon is always zero, does it mean that actually the values of k and epsilon are not updated by the solver? Thanks.

Quote:
 Originally Posted by julien.decharentenay Hi H, The number of iteration is 0 since the initial residual is less than the convergence criteria you set. So for all intent and purpose, the solver is assuming that the epsilon and k fields are "converged" and do not need iterations. If you do want to have more iteration, decrease the convergence criteria for k and epsilon. Julien

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