FAST_v8.15 OutList parameters

Hi,

I haven’t been able to clearly understand the meanings of certain OutList parameters. I’ve copy-pasted the description given in the OutList file that is included in FAST.

  1. RtAeroPwr- Rotor aerodynamic power
    Is this equal to the aerodynamic power captured by the rotor or is it equal to the aerodynamic power incident on the rotor?

  2. Rotor aerodynamic torque isn’t one of the outlisted parameters. Can it be obtained by dividing RtAeroPwr by the RtSpeed (converted to rad/s)?

  3. Is there any relationship between RotPwr and GenPwr? I’m working with gearbox efficiency of 100%; however, RotPwr is not equal to GenPwr. Do they always have the same sign?

  4. Am I right in this interpretation: RotPwr and GenPwr always have the same sign.

If both are negative, it means the RotTorq is trying to increase the RtSpeed (or RotSpeed, both are almost exactly equal) by using input power from the outside. The value of GenTq can be controlled by us to control the RotTorq and hence control the RtSpeed as well.

If both are positive, it means the rotor is actually generating power by driving the LSS and hence the generator.

What are the two torques acting on the rotor ultimately? Is one of them the aerodynamic torque (calculated as mentioned in point 2 above) and is the other GenTq?

My apologies for some rather trivial questions. I just wanted to make sure I am interpreting the outputs correctly.

Many thanks,

Sarat

Dear Sarat,

Here are my answers to your questions:

  1. The RtAeroPwr output of AeroDyn v15 is the aerodynamic power captured by the rotor.

  2. Actually, the rotor aerodynamic torque is available in the RtAeroMxh output of AeroDyn v15. But yes, you could derive RtAeroMxh from RtAeroPwr and RtSpeed, as you stated.

  3. The RotPwr output of ElastoDyn is the mechanical power in the low-speed shaft i.e. the reaction torque in the low-speed shaft times its speed. The GenPwr output of ServoDyn is the electrical power output from the generator. Under steady state conditions with 100% gearbox efficiency while producing power, GenPwr would be smaller than RotPwr by the generator electrical-conversion efficiency. But under transient conditions (where the shaft is accelerating or decelerating), the relationship is more complicated.

  4. Under steady state conditions, yes, RotPwr would have the same sign as GenPwr–both are positive when generating power and both are negative when motoring. But under transient conditions (where the shaft is accelerating or decelerating) this need not be true.

For clarification of how the drivetrain model in FAST works, please see the following forum post: http://forums.nrel.gov/t/resistant-moment-of-the-rotor-and-of-the-electric-generator/408/1.

I hope that helps.

Best regards,

Hello,

I am working with FAST and with the DTU 10MW reference wind turbine and I am trying to get the aerodynamic torque of the rotor in order to achieve a better understood of my system. So, I wanted to compare by a plot the outputs RtAeroMxh and GenTq. However, when drawing this plot, I found out that those signals where completely different, the RtAeroMxh signal was senseless, so I tried to get the aerodynamic torque by doing the following operations using the rotor aerodynamic power and the rotor speed given by FAST:

aerodynamic rotor torque = RtAeroPwr/(RtSpeed2pi/60)

From that checking, I have doubts about how FAST is calculating those signals since RtAeroPwr and RtAeroMxh are exactly equal and the RtSpeed signal also has the same shape as them but with a smaller magnitude order. Do you have any idea of what the reason for this could be?

Thank you in advance,

Leire Gandiaga

Dear Leire,

I’m not sure I understand your results, but I would expect the following relationship between AeroDyn v15 outputs aerodynamic power (RtAeroPwr), aerodynamic torque (RtAeroMxh), and rotor speed (RtSpeed):

RtAeroPwr = RtAeroMxhRtSpeed(2*pi/60)

Is this not what you are seeing?

Best regards,