About OC4 Free decay test #1414
Replies: 7 comments 20 replies
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Dear @icnosvbovwb, I'm not familiar with the paper you are referring to, but from a quick Google search, I believe you are referring to: https://www.sciencedirect.com/science/article/pii/S0029801822020285, which does not include NREL authors. I would suggest reaching out to the authors of that paper for questions related to it. That said, the three plots you circle ( Best regards, |
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Dear @jjonkman |
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Dear @jjonkman |
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Dear @jjonkman 3.Also, the TLP platform is invisible. I don't know where to set it |
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Dear @jjonkman |
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Dear @icnosvbovwb , I am a beginner of openfast, could you please tell me how to implement the initial conditions in the code when doing free-dacy? Best regards, |
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Dear @jjonkman , Sorry to bother, but I found another problem. My OpenFAST version is 3.5.1. In the case “5MW_OC4Semi_WSt_WavesWN” , I tried to reduce the water depth from 200m to 50m, and the length of the mooringline from 835.35 m to 735.35 m, all other conditions did not change, the strange thing is, the case was able to work with such a short length of chain(I was expecting an error), in fact, 735.35 m is less than the distance between the anchor and the floating platform, but as the result picture below, the mooringline was just stretched, and it seems to work fine. My question is: Why is the mooringline so short, but it still works? Is it because it can be stretched like a spring? ALL.windvel8.regwave102.35.waterdp200.line735.mp4windvel8.regwave102.35.waterdp200.line735.mp4Best regards, |
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Dear @jjonkman
"Numerical study on passive structural control of semi-submersible floating
wind turbine considering non-collinear wind and waves".
I see the free attenuation test of OC4 from this article and want to reproduce its process, but my results are a little different from them. The conditions and degrees of freedom I set are as follows:
CompInflow=0,CompAero=0, WaveMod=0, GenDOF=False, RotSpeed=0, BlPitch(1)=BlPitch(2)=BlPitch(3)=90.
I don't know why there are different results. Please indicate my mistakes. The calculation time step is 0.025, so the horizontal ordinate is 600/0.025=24000. I wonder if the tower is placed on one of the float bowl. If so, how to set the tower on the middle column
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