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DPM Iteration, incomplete

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Old   June 9, 2003, 14:08
Default DPM Iteration, incomplete
  #1
rookie
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Hello Folks,

Do you know what "incomplete" means in the following message? Thanks a lot.

DPM Iteration .... number tracked = 2100, escaped = 0, aborted = 0, trapped = 32, evaporated = 0, incomplete = 2068

Rookie

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Old   June 9, 2003, 20:36
Default Re: DPM Iteration, incomplete
  #2
xiangrb
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Incomplete means that your trajectory caculations are not completed. In this case, you should increase the max.number of steps in the Discrete Phase Model Panel if you are sure that the particle is not recirculating somewhere in your domain. Good Luck!
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Old   June 10, 2003, 12:49
Default Re: DPM Iteration, incomplete
  #3
Alex Munoz
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Hi

Xian gave you in my opinion unacurate answers.

Imcoplete trajectory means that a particle or particles were traped by the flow. Therefore the number of time steps that you specify were not enough to allow the particle(s) to reach the outlet.

As a result, you can increase the number of time step to allow Fluent follow those particle in a longer path that the actual one. However, This approach is not going to solve the problem because always some particles will remain trapped.

Therefore, You have to determine a statistical criteria such as 95 or 98% of the particles should leave the domain to consider a satisfactory simmulation of the discrete phase.

Best regards

Alex Munoz
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Old   June 10, 2003, 20:35
Default Re: DPM Iteration, incomplete
  #4
xiang
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Hello, I think "trapped" and "incomplete" are different. If you set particle boudary type for the walls as "trapped", the particle will be trapped whenever this particle collide with the wall and the trajectory calculation for this particle will be terminated. Therefore, those particles which touch the wall will not reach the outlet. However, the trajectory calculation for these particles are completed.

"incomplete" means the particle doesn't collide with any boundary, the trajectory calculation terminates somewhere inside the domain upon reaching the max. number of time steps. If you increase the number of steps, the particle will move further.

In my opion,if no particle touches the other boundary type and no particle is recirculating in the domain, all particles should reach the outlet and leave the domain with sufficient number of time steps.

Thanks.
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Old   June 10, 2003, 23:53
Default Re: DPM Iteration, incomplete
  #5
Alex Munoz
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Hi rockie

Perhaps I wasn't clear enough. When I wrote particles trapped by the flow means particles that get inside a vortex or eddy for a time period longer that you allow to the particles to exit the domain. In any instance I wrote particles traped by the boundaries of the domain, However, it seems that some people understand the word "trap" in the terms that fluent use it.

BTW, Some people like me expect that you reply a thank you note as a simbol of message read, and also as form of curtesy. I am aware of that some culture are not use to reply a thank you note, but keep in mind that I and other CFD user take a few minute of their time to read your question. Therefore, they deserve a note of acknolegde

Best regards

Alex Munoz

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Old   June 12, 2003, 20:31
Default Re: DPM Iteration, incomplete
  #6
rookie
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Hello Alex, Xiang and All,

Thank you very much for your kindness replies. If the particle can not reach the outlet (incomplete), does the particle keep moving in the next trajectory periods?

Thanks again, Alex and Xiang.

Rookie.
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Old   June 12, 2003, 21:44
Default Re: DPM Iteration, incomplete
  #7
winnie
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Hi, rookie

During each dpm iteration, the particles' trajectories are calculated based on the current continuous field from their injections to either trapped by the boundary or complete at the outlet or incomplete in the field, so whether they complete or incomplete during the last iteration, they will be calculated from injection to get their another path again.

But why don't you give a larger Max. Number of Steps to complete the trajectory. In my opinion, if the particles don't reach the outlet, the exchange(mass, momentum...) between discrete phase and continuous phase are not completely calculated which effect the correctness of the ultimate result.

Regards

winnie
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