Difference between 'RtAeroMxh' and 'RotTorq' in an onshore turbine

Dear Jason,
Checking the outputs for an onshore test (Test 24 with all DOFs = False) I noticed a difference between the ‘RtAeroMxh’ and ‘RotTorq’ trends (Figure 1). Since the ‘RtAero’ loads are in a frame rotating around x, I would expect the torque around this axis to be the same in the transition from rotating frame to fixed, instead I notice that ‘RotTorq’ is much less noisy by ‘RtAeroMxh’.
Could you tell me the reason?
Since to get ‘TwBsFx’ you have to use ‘RotTorq’ plus the transport moments, could you tell me the relationship between ‘RotTorq’ and ‘RtAeroMxh’?
Thanks for your help,
Lorenzo.

Dear Lorenzo,

AeroDyn v15 output RtAeroMxh is the aerodynamic applied torque. ElastoDyn output RotTorq is the reaction torque within the low-speed shaft, and so, includes both the aerodynamic applied torque, as well as the rotor inertia. I would expect these outputs to be the same of the rotor is rigid and spinning at a constant speed. When you say all DOFs = False, do you mean only the platform DOFs are disabled? If so, the difference between RtAeroMxh and RotTorq is the result of the rotor inertial contribution.

Best regards,

Dear Jason,
Thanks for your reply.
I have implemented my MATLAB model for an onshore turbine, obtaining trends of forces and moments due to aerodynamics consistent with FAST (Figure 1, my rotor thrust in blu, RotThr of FAST in red). However, zooming in you can see that my outputs are noisier than FAST’s (Figure 2). I was wondering if FAST used filters and, if not, if you knew how to explain this phenomenon.
Thanks for your help,
Best regards,
Lorenzo.


Dear Lorenzo,

The outputs of FAST are not filtered before they are written out to the file, if that is what you are asking.

But it is impossible to guess what causes the differences between FAST and your solution without understanding more about your model or the simulation set up.

Best regards,

Dear Jason,
One more question: performing an offshore simulation in still water, with aerodynamics off and initial rotor speed 0, I get a RotorTorque that is small but different from 0.
What causes this torque, given that the rotor speed remains 0?
Best regards,
Lorenzo.

Dear Lorenzo,

Presumably you are referring to a nonzero output of “RotTorq” given that you’ve disabled aerodynamics? Any acceleration of the blades–not just from rotor speed variations, but also from platform-induced motion–can result in inertial loading. Unless all initial conditions are defined at the equilibrium state, there will be a small transient at the beginning of each simulation due to the gravitational loading in the system (e.g., the tower will tilt in the direction of the overhanging weight of the rotor / nacelle assembly).

Best regards,

Dear Jason,
To understand why my outputs are much noisier than FAST I would like to investigate the pT and pN values on the blade nodes. Please confirm that these values correspond to ‘BαNβFt’ and ‘BαNβFn’ in Aerodyn outputs?
Thanks for the reply,
Best regards,
Lorenzo.

Dear Lorenzo,

I don’t know what you mean by “pT and pN values”, but the AeroDyn v15 outputs are well documented on OpenFAST readthedocs: openfast.readthedocs.io/en/mast … t-channels.

Best regards,

Dear Jason,
I mean tangential and normal loads per unit of length (N/m) for every node along a blade (t and n in your doc).
Best regards,
Lorenzo.

Dear Lorenzo,

Yes, AeroDyn v15 outputs BαNβFt and BαNβFn are the tangential and normal aerodynamic applied forces per unit length.

Best regards,

Dear Jason,
I was wondering the exact formula to derive ‘RotSpeed’, now I’m using the following formula:

‘RtAeroMxh’ - ‘GenTq’ * 97 = Irot * d (‘RotSpeed’)/dt

from which we obtain d(‘RotSpeed’)/dt that, integrated, gives ‘RotSpeed’.
I use the gearbox ratio N=97 and the rotor inertia Irot=43858156.919 kg m^2.
However I get small differences compared to the FAST output ‘RotSpeed’. (my output blu, FAST output yellow)


I suspect my formula is incomplete, can you tell me what’s missing or wrong?
Thanks for your help,
Best regards,
Lorenzo.

Dear Lorenzo,

Is this an otherwise completely rigid model with only the generator DOF enabled? If so, then I agree with your equation except that Irot should be replaced with the combined inertia of the rotor plus generator relative to the low-speed shaft (GenIner*GBRatio^2).

Best regards,

Dear Jason,
Yes, it’s a completely rigid model with only the generator DOF enabled. In my previous post I was referring to the moment of inertia of the drivetrain when talking about Irot. To use a more precise notation,
Idt = Irot + Ngear^2*Igen;
Said Ngear=97 and Igen=534.116kg•m^2, I don’t know the exact formula to obtain Irot.
Where can I find it?
Best regards,
Lorenzo

Dear Lorenzo,

The rotor inertia computed by ElastoDyn is written to the ElastoDyn summary file. Combined with the generator inertia, I calculate about 43700000 kgm^2, which is slightly different then the number you posted.

Best regards,

Dear Jason,
I calculated Irot as follows:

Ihub = 115926; % [kgm²] hub moment of inertia around rotation axis
Rhub = 1.5; % [m] hub radius
Mblade = 17740; % [kg] single blade mass
CMblade = 20.475; % [m] blade center of mass distance from root
Iblade = 11776047; % [kgm²] single blade moment of inertia at the root
IbladeG = Iblade - MbladeCMblade^2; % [kgm²] single blade moment of inertia at its COM
IbladeR = IbladeG + Mblade * (Rhub + CMblade)^2; % [kgm²] single blade moment of inertia at rotor axis
Irot = Ihub + 3
IbladeR; % [kgm²] rotor moment of inertia

Do you see an error in a formula or in turbine data?
Thanks a lot,
Lorenzo.

Dear Lorenzo,

I agree with your general approach except that there is a small blade precone, which I don’t see that you accounted for. Your values are slightly different than what is found in the ElastoDyn summary, which is likely the main reason for the small difference between our numbers.

Best regards,

Dear Jason,
In my simulation I setted Precone=0.
Do I still have to take this into account when calculating Irot?
Best regards,
Lorenzo.

Dear Lorenzo,

When PreCone = 0, then your equations are correct.

Best regards,

Dear Jason,
Thanks for your reply.
I still have a little doubt: in a steady test, in order to have constant rotor speed, according to the formula reported above, after the initial transient it should be RtAeroMxh = GenTqNgear.


Instead, as you can see from the graph, there is a small negative offset at steady state which then results in a negative rotor acceleration, hence a slightly decreasing rotor speed.

What causes this difference between RtAeroMxh and GenTq
Ngear? I found a value of Ngear=96.9379 instead of 97 to overlap the two graphs satisfactorily. Could be this?
Best regards,
Lorenzo.

Dear Lorenzo,

I’m not sure I really understand your question or how you are setting up your simulation. What generator torque model are you using?

If GenTq*GBRatio does not exactly equal RtAeroMxh, then the rotor will accelerate (or decelerate). If you are setting GenTq directly from an output of RtAeroMxh, numerical round off could play a role if you RtAeroMxh is truncated in the output file (true for ASCI output, but full precision is maintained in binary output).

Best regards,