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MRF steady state mixing time

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Old   August 11, 2016, 05:14
Default MRF steady state mixing time
Thomas Sprich
Join Date: Mar 2015
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Dear Fluent Forum,

I have done some CFD work before, but never in Fluent so please be gentle with me. A third party is simulating my problem and I am questioning the results.

The problem is as follows: I wish to determine the mixing time for a liquid in a tank that contains a mixer. The rotation of the mixer is handled using MRF methods. The third party has provided what they believe to time taken to mix the tank.

My understanding of MRF methods is that it is used with a steady state solver (i don't know which); whereas the problem of mixing from an unmixed tank state to a homogeneously mixed tank state is a transient problem. Therefore, determining the mixing time should be impossible unless a transient solver with sliding meshes is used (which requires greater computational effort).

I'm not sure I understand the method of how the third party determined when they had reached a state of complete mixing so I cannot provide more details with regards to this. I suspect that there is a comparison done between iteration steps and when these are unchanging, the tank is said to be mixed. To my understanding, this means the solution has converged (and possibly mixed) which doesn't necessarily mean that this is the mixing time.

In summary, this is how I understand my situation:
'time step' is concept that only exists in a transient solver. In a steady state solver, there are only iteration steps. Sorry, this question might be trivial to someone who has used Fluent. Unfortunately I have not. There is therefore no way to relate iteration steps to time to calculate the mixing time. Is this correct?

I have tried looking up the different solvers Fluent has but it is not yet clear to me.

Thanks for your help.

Swift is offline   Reply With Quote


iteations step, mrf, steady state, time step, transient

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