[vidyadhar]

Hello All,

I would like to know if there is any relation between simulation time (time taken by a machine to execute the code) and Reynolds number for a given mesh.

It is taking 50 hrs of time to complete my simulation at Re 400 using 20 cores.

I would like to know how much it would take for higher and lower Re.

Thanks!
Vidyadhar

[piu58]

It depends on your model.

If you use a model which covers turbulency to some extend the solvers need more steps to converge if there is more turbulency in the flow. For LES simulations the time for calculation increases considerably with turbulency and therewith with the Reynold number.

If you use a steady state solver the dependency between Re and the calculation time is not as steep. But you may get problems with convergence at very high Re numbers.

[LuckyTran]

For a fixed grid, there's no direct correlation. The time needed to do 1 iteration is dependent only on number of cells and the number of equations you solve, etc. The solver doesn't care what the flow Reynolds number is, it just cranks the algorithm the same.

For a steady case and fixed grid, the number of iterations you need to converge to the solution varies somewhat with Reynolds number due to non-linearities and it is hard to say whether you need more or less.

For transient simulations, you usually decrease your time-step size as velocity increases in order to maintain the same temporal bandwidth. I.e. you double the velocity, you half the time-step size and do twice as many time-steps. In this case it's linearly proportional to Reynolds.

[piu58]

Dear LuckyTran

> The time needed to do 1 iteration is dependent only on number of cells a

If you use a direct solver for the system of equation this is true. But if you use an iterative solver like conjugated gradients the number of iterations depend on the condition of the system. This condition is more worse with much changing in the flow.

Some more advanced solvers for the fluid dynamics may be controlle by the Co number and select the time step by theirself. With more turbulence in the flow they choose lower time steps and need more computational time. This is equivalent to your statement

> you usually decrease your time-step size as velocity increases in order to maintain the same temporal bandwidth

[LuckyTran]

The automatic Co is just the software picking time-step for the user because the user is too lazy to do it for themself. There are a lot of acceleration schemes, there are a lot of tricks that can likely speed up and speed down the process. My claim is that these are indirect behaviors, and that the only meaningful difference in solution time is the number of time-steps you do in a transient simulation.

Originally Posted by piu58

If you use a direct solver for the system of equation this is true. But if you use an iterative solver like conjugated gradients the number of iterations depend on the condition of the system. This condition is more worse with much changing in the flow.

Even if you use a direct solver, the discretized advection operator in the N-S introduces the problem where you have to do many iterations. Your solver may look direct to the discretized N-S over a single-cell, but it is still effectively an iterative/transient-solver, you must do many iterations/time-steps/sweeps to inject non-linearities into your linear system (you must update the advection operator).

As for the pre-conditioning matrix, it is also nearly equivalent to saying the solution time now depends on good/bad is your initial guess. So if you change the Reynolds number, you should also change your initial guess so that the iterative solvers do the same number of iterations.

Now where in the algorithm (direct/indirect) is there a switch that says, ah-hah for a different Reynolds number I must do something different? My claim, although it is not 100% true, is that there's not really such a switch because the linear system you are trying to solve is the same (the discretized N-S with all the time and space discretizations). You have these little quirks that make a little bit of difference here and there, but most of the time you are just solving a linear system Ax=b