[Raymond_1981]

Hello!

I would like to receive some help with understanding which integration time step is used in the unsteady dpm model in Fluent.



I have the following case. I’m modeling Stairmands High-Efficiency cyclone. I start with a steady state k-epsilon RNG simulation after which I change the turbulence model to RSM. One converged I switch to a transient simulation and let the solution converge for a period of time. I use a fluid time step of 1e-05s and a CFL of approximately 0.5.

The next phase is to inject particles in the transient simulation. These particles are tracked unsteady and I use one-way coupling. To account for turbulent dispersion, I use the random walk model with a random Eddy lifetime. The number of steps are set to 1000 and the step length is set to 5. The accuracy settings are default and I use the trapezoidal rule for integration. The simulation runs fine and the results are as expected.

Now I have some trouble understanding the time step used in the integration of the dpm equations. I wasn't able to find an answer in the theory guide unfortunately. Is the time step automatically set to the random Eddy lifetime? And is this integration (with a random Eddy lifetime) repeated for a number of steps until the continuous phase time step of 1e-05s is reached, after which it solves for the continuous phase again and repeats the process? Or is the time step based on the step length? But how is then the random Eddy lifetime incorporated?

Who can help me understanding the procedure used in Fluent? Any help is appreciated. Thanks!

Best regards, Raymond

[vinerm]

Time-step for integration is always based on the either Step Length Factor or Length Scale, depending upon what is specified by the user.

Turbulent dispersion is an optional model. Even if it is disabled, particles are to be tracked. DRW uses multiple tracks based on turbulent kinetic energy and a Gaussian distribution around it.

[Raymond_1981]

Hi Vinerm,


Thank you for your reply! I do have some follow up questions:

1. In the user manual the integration time step is defined as the (transit time)/(step length factor). Is this also the case with an unsteady simulation or is then the fluid time step used in stead of the estimated transit time? How is the fluid time step considered in this integration, since a particle can only move a certain distance in the given fluid time step?
2. The theory guide states that the instantaneous velocity field is used in the particle force balance is stead of the mean fluid phase velocity in case of stochastic tracking. If the integration time differs from the eddy lifetime (or crossover time), how is the updated instantaneous velocity field (after reaching the eddy lifetime or crossover time) used in the particle force balance?

Thanks!

[vinerm]

Whether particles are tracked in a steady (which is actually a misnomer) or unsteady manner, time-step is always based on same approach. The time given in Unsteady Tracking is used for injection of particles.

Integration time is used for integration of the Lagrangian equation while the eddy cross-over or its lifetime marks the update of random velocity field used in equation.

[Raymond_1981]

Hi Vinerm,


Thank you for your answer! This makes sense.



The only thing I haven't got clear yet is how the new position of the particles are determined in relation to the fluid time step. You can see that with every fluid time step the particles advance in the domain. How is this done?



Kind regards,


Raymond

[vinerm]

The primary effect of fluid time-step is via update of flow field; the flow-field most likely is changing every instant in a transient simulation. The other effect is if the particles are injected at the fluid time-step, however, this is user specified. Otherwise, fluid time-step and particle tracking are independent. You can simulate fluid flow for 10 s, however, it can have some particles that move out of the domain in less than a second and some particles that will spend more than an hour in there.

[Raymond_1981]

Hi Vinerm,


Everything makes sense now. Thanks for your help, it is much appreciated!



Kind regards, Raymond