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uidelines*****************************************************************************************************

nitial residual corresponds to the last result of the SIMPLE Algorithminal residual corresponds to the improvement Linear Solver

The solver stops if either of the following conditions are reached:1. The residual falls below the solver tolerance, tolerance(compulsory);

2. Tthe ratio of current to initial residuals falls below the solver relative tolerance, relTolcompulsory);

3. The number of iterations exceeds a maximum number of iterations, maxIter(optional);

In transient simulations, it is usual to set the solver relative tolerance to 0 to force theolution to

converge to the solver tolerance in each time step.

*****************************************************************************************************

onclusions*****************************************************************************************************

he tolerance for the Simple Algorithm will decide the no of iterations needed for convergence.

nly when the Initial residual for all the variables fall below there respective tolerancehe simulaton will stop(or when the endTime specified in the controlDict is reached).

n case the simulation runs till the specified endTime, that would imply the SIMPLE algorithm fails tochieve thepecified tolearnce for each variable in that many no of iterations( = (endTime-startTime) / timeStep)

bserve that the stopping criteria for the linear solver is always met with the relative tolerancepecified as 0.1o study the effect of solver tolerance put the rel tolerance = 0(to force the solver to achieve thebsolute tolerance).

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imulation Results#######################################################################################################

ptolerance 1e-06;relTol 0.1;

(U|k|epsilon|R|nuTilda)tolerance 1e-05;relTol 0.1;

residualControl{

p 1e-2;U 1e-1;"(k|epsilon|omega)" 1e-1;

}

ime = 280

moothSolver: Solving for Ux, Initial residual = 0.00135612, Final residual = 9.98628e-05, No Iteratio

moothSolver: Solving for Uy, Initial residual = 0.00375588, Final residual = 0.000287439, No Iteratio

AMG: Solving for p, Initial residual = 0.0100156, Final residual = 0.0006414, No Iterations 4ime step continuity errors : sum local = 0.000459646, global = 7.22828e-06, cumulative = 0.00325827moothSolver: Solving for epsilon, Initial residual = 0.00255638, Final residual = 0.000175623, No

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terations 2moothSolver: Solving for k, Initial residual = 0.00341547, Final residual = 0.000256861, No IterationxecutionTime = 19.69 s ClockTime = 20 s

ime = 281

moothSolver: Solving for Ux, Initial residual = 0.00135154, Final residual = 9.95587e-05, No Iteratio

moothSolver: Solving for Uy, Initial residual = 0.00374829, Final residual = 0.000286867, No Iteratio

AMG: Solving for p, Initial residual = 0.00997498, Final residual = 0.000590835, No Iterations 4ime step continuity errors : sum local = 0.00042311, global = 8.72907e-06, cumulative = 0.003267moothSolver: Solving for epsilon, Initial residual = 0.00254856, Final residual = 0.000175205, Noterations 2moothSolver: Solving for k, Initial residual = 0.003404, Final residual = 0.000256123, No Iterations xecutionTime = 19.75 s ClockTime = 20 s

IMPLE solution converged in 281 iterations-----------------------------------------------------------------------------------------------------

ptolerance 1e-06;relTol 0.1;

(U|k|epsilon|R|nuTilda)tolerance 1e-05;relTol 0.1;

residualControl{

p 1e-2;U 1e-3;"(k|epsilon|omega)" 1e-1;

}

moothSolver: Solving for Ux, Initial residual = 0.000261326, Final residual = 2.08759e-05, Noterations 2moothSolver: Solving for Uy, Initial residual = 0.000999202, Final residual = 8.32758e-05, No

terations 2AMG: Solving for p, Initial residual = 0.00202996, Final residual = 0.000190399, No Iterations 2ime step continuity errors : sum local = 0.000105138, global = -2.13266e-06, cumulative = 0.00352348moothSolver: Solving for epsilon, Initial residual = 0.00047137, Final residual = 3.80941e-05, Noterations 2moothSolver: Solving for k, Initial residual = 0.000722501, Final residual = 5.9995e-05, No IterationxecutionTime = 61.17 s ClockTime = 64 s

IMPLE solution converged in 1178 iterations-----------------------------------------------------------------------------------------------------

he Perfect Set of values for a Steady State Problem

ptolerance 1e-06;relTol 0.1;

(U|k|epsilon|R|nuTilda)tolerance 1e-05;relTol 0.1;

residualControl{

p 1e-6;U 1e-5;"(k|epsilon|omega)" 1e-5;

}

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IMPLE Algorithm does not converge even in 10000 iterations-----------------------------------------------------------------------------------------------------

ptolerance 1e-02;relTol 0;

(U|k|epsilon|R|nuTilda)tolerance 1e-02;

relTol 0;

residualControl{

p 1e-6;U 1e-5;"(k|epsilon|omega)" 1e-5;

}#######################################################################################################