Dear @Nitin.Sivakumar,
If I understand correctly, you are comparing the rotor inertia computed by the structural model of FAST v7 against the results from ElastoDyn in OpenFAST. I’m not exactly sure I can pinpoint why this change exists, but regardless, the difference is about 0.2%, and so, is not too concerning either.
Best regards,
Dear Jason sir,
I want to calculate the drive train damping coefficient for 5 MW wind turbine through openfast linearization. I got from the forum that I need to activate the GenDOF and DrTrDOF and linearize . But I don’t understand how to identify and get the value exactly from the linearized matrices. Please excuse me if the doubt is trivial and asked before.
Thanks and regards
Dear @Nitin.Sivakumar,
The damping coefficient of the drivetrain torsion model is actually an input to ElastoDyn (DTTorDmp). If you are referring to the damping ratio of the free-free mode of the drivetrain with rigid rotor, you can calculate this analytically based on the knowledge of the generator inertia expressed relative to the low-speed shaft (GenIner*GBRatio^2), the rotor inertia (RotIner), and the drivetrain torsional stiffness (DTTorDmp). That is:
natural frequency of drivetrain free-free mode = SQRT( DTTorSpr/RotIner + DTTorSpr/( GenIner * GBRatio^2 ) )/( 2 * pi )
damping ratio of drivetrain free-free mode = DTTorDmp*SQRT( 1/( DTTorSpr * RotIner ) + 1/( DTTorSpr * GenIner * GBRatio^2 ) )/2
Of course, DTTorSpr, DTTorDmp, RotIner, and GenIner * GBRatio^2 will also appear in the linearized state matrix “A” that you can obtain through an OpenFAST linearization analysis, and you can obtain the same natural frequency and damping ratio by computing the eigenvalues of this matrix.
Best regards,
Thanks for your quick reply.
Best Regards
Hello,
I have another questions about the HubIner parameter.
The parameter is described as “Hub inertia about rotor axis [3 blades] or teeter axis [2 blades] (kg m^2)”.
For a 2-bladed turbine, how do we take into account the hub inertia about the rotor axis?
Thanks.
Abhineet
Dear @Abhineet.Gupta,
For a 2-bladed rotor, ElastoDyn will assume that the hub inertia about the rotor axis equals the hub inertia about the teeter pin, except that the inertia is shifted (by the parallel axis theorem) from the teeter pin to the hub mass center.
Best regards,
Thanks for the reply @Jason.Jonkman.
I am a little confused here.
The rotor axis lies in the plane but the teeter axis is perpendicular to the plane of the following figure. How can we use parallel axis theorem in this case?
Or, more likely, does this mean that the description should say “Hub inertia about rotor axis [3 blades] or teeter pin [2 blades] (kg m^2)” (instead of axis).
Regards,
Abhineet
Dear @Abhineet.Gupta,
Similar to the figure you share, the hub for 2-bladed rotor is assumed to be a uniform rod directed along the axis normal to the teeter pin, passing through the hub center of mass. After tranferring the hub inertia from the teeter pin to the hub center using the parallel axis theorem, the hub inertia is assumed to be the same about the two transverse axes of this uniform rod–one parallel to the teeter pin and one about the shaft axis (when the teeter angle is zero).
Best regards
Thanks @Jason.Jonkman,
The formulation makes sense with the assumption of hub being a uniform rod.
For my current application, this assumption might not be reasonable and if so, I’ll raise a feature request in OpenFAST to add a separate ‘HubIner_Teeter’ parameter.
Regards
Abhineet
