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Frequency spectrum2 Attachment(s)
Hello CFD-experts,
i have a question about transient turbomachinery simulations. I simulated a centrifugal compressor with vaneless diffusor and volute (360° case). The impeller has 6 main blades and 6 splitter blades. I used a massflow boundary condition at outlet. First i made a steady state simulation using the Frozen Rotor-Model. I initialized the transient simulation with the steady state results. First I dissolved 6°/timestep so that after 60 timesteps one revolution was completed. After 27 revolutions (1620 timesteps) the solution seemed engaged (massflow oscillations at impeller inlet). Then I refined the timestep dissolving 1°/timestep and again simulated 2 revolutions. Then I made a Fast Fourier Transformation (FFT) of the massflow at impeller inlet and static pressure at outlet over the last two revolutions to see if there are frequencies that interfere with each other, because oscillations still looked sinuslike but pretty unbalanced. The results are shown in the pics. I see a main frequency of 5000Hz and a little one of 10 000Hz. The impeller makes 1680rev/s, so with 6 main blade passages 10 080 main passages/sec are passing. Is this the way one could describe the little peak of 10 000Hz in the spectrum? Or is it to explain with the angular velocity (that is 10 600rad/s in this case)? And what about the big peak of 5000Hz? Sorry but I have no experience in using FFT for a transient simulation...Is anybody out there who got some experiences? Greetz from Susann |

Yes, the frequency you found at 10kHz looks like the blade passing frequency. Have a look at the resolution of your FFT - it probably only has a resolution of 1kHz so that is as accurate as it can go, so it cannot pick the difference between 10kHz and 10.08kHz. To pick that difference you will need a much higher FFT resolution.
The 5kHz peak looks like some harmonics on the main signal. It is also lost in the junk at the start of the FFT. Best to do another FFT aiming at lower frequencies and higher resolution so you can resolve those peaks more accurately, if that is what you want to do. |

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