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Old   September 19, 2000, 02:02
Default Time step for rotating frame
Ales Skotak
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In my turbine runner computation I used multiple rotating frames of reference. I used only one rotating blade segment and stationary bladeless parts upstream and downstream. Unsteady solution converged well. When I solved the task as an unsteady one with relatively large time step (1*number_of_blades*rotating_speed) and task again converged well. When I used the shorter time step (0.1*number_of_blades*rotating_speed) the solution could not converged. Is there any restriction for the critical shortest time step in the unsteady computation?
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Old   October 28, 2000, 07:59
Default Re: Time step for rotating frame
Peter Staykov
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Hi Ales, I have solved similar tasks with Multiple References Frame (MRF) and Sliding Mesh (SM) computational techniques for different type of turbines - Rushton, Pitch Blade and NS turbine - and I have solutions for these types of impellers with one and two turbine stirred tanks. In your question I do not understand what dimention do you use for rotating turbine, more precisly,

in your formula : 1*number_of_blades*rotating_speed

what is the dimention of the rotating speed - rpm, rps or Hz ?

Best regards, Peter Staykov
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Old   November 1, 2000, 03:11
Default Re: Time step for rotating frame
Ales Skotak
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Hi Peter, Thanks for your comment. The formula for the critical time step is dt=1/(rotational_speed * number_of_blades). The rotational speed is in round per second. It does mean, that the critical time step value is the same as the blade passing frequency. When I use the lower time step then the blade passing frequency, I could not obtain converged solution.

Best Regards.

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