DriveTrain Natural Frequency Calculation

Dear Jonkman & Buhl,

The simple hand calculation of Test14.fst gives yhe frequency of drivetrain as 66 Hz. The frequency we are getting from the FAST is ~18 Hz. Looks like we are doing some mistake in the hand calculation. Can you please advice on these calculations.

For your reference I have attached the calculation slide & FAST output file CampbellDiagram_24May_Test14.xls (Campbell Diagram & MBC3 output relation).

I would appreciate your help.

Regards,
Tejas
Frequency-Calculation.ppt (226 KB)

Dear Tejas,

The equation you’ve identified in your presentation represents the natural frequency of a drivetrain with a rigid rotor in a free-free condition. When comparing to the FAST results from http://forums.nrel.gov/t/campbell-diagram-mbc3-output-relation/687/1, I see a few problems:

*The FAST linearization from http://forums.nrel.gov/t/campbell-diagram-mbc3-output-relation/687/1 represents a fixed-free condition (the generator DOF is disabled, GenDOF = False).
*The FAST linearization from http://forums.nrel.gov/t/campbell-diagram-mbc3-output-relation/687/1 includes flexible blades (the rotor is not rigid, FlapDOF1 = FlapDOF2 = EdgeDOF = True).
*IA from your presentation should represent the entire rotor inertia; in the case of your hand calculation, you only used the hub inertia, HubIner.

Best regards,

Hello Jason,
I recalculated Drive train frequency using rotor inertia & got closer match with the FAST output. Thanks for your guidance.
One query I have now.

How this rotor inertia is calculated. I am using following formulation.
Rotor Inertia = Hub Inertia + (Number of blades * Inertia of each blade)

However, my hand calculation values of rotor inertia are much away from FAST (.fsm file) as under.
1.5 MW Turbine : FAST output = 2.96E6 kg/m2 & Hand Calculation = 2.44E6 kg/m2
2.5 MW Turbine : FAST output = 1.3E7 kg/m2 & Hand Calculation = 1.18E7 kg/m2
5 MW Turbine : FAST output = 3.88E7 kg/m2 & Hand Calculation = 1.54E7 kg/m2

Also, error goes on increasing as I go from 1.5MW to 5 MW turbine case. Am I doing some mistake?

I would appreciate your help.
With Regards,
Tejas

Dear Tejas,

I think I’ve answered your question in the forum topic found here: http://forums.nrel.gov/t/whats-the-exact-meaning-of-hubiner/595/1.

Best regards,

Dear Jason,

Thanks for your guidance. I got to know now how to calculate rotor inertia by hand calculation which is in agreement with results from FAST.

My interest is to calculate Drive Train Natural Frequency. As per your earlier comments in this post, I ran 4 different cases in FAST for 1.5 MW Turbine as follows.

Case1 : Free-Free Condition (GenDoF = True) & Rigid rotor (FlapDOF1 = FlapDOF2 = EdgeDOF = False)
Drivetrain natural frequency = 19.71 Hz (FAST o/p) ; 19.84 Hz (Hand calculation)

Case2 : Fixed-Free Condition (GenDoF = False) & Rigid rotor (FlapDOF1 = FlapDOF2 = EdgeDOF = False)
Drivetrain natural frequency = 6.91 Hz (FAST o/p) ; 6.92 Hz (Hand calculation)

Case3 : Free-Free Condition (GenDoF = True) & Flexible blades/rotor (FlapDOF1 = FlapDOF2 = EdgeDOF = True)
Drivetrain natural frequency = 25.41 Hz (FAST o/p) ; No Hand calculation

Case4 : Fixed-Free Condition (GenDoF = False) & Flexible blades/rotor (FlapDOF1 = FlapDOF2 = EdgeDOF = True)
Drivetrain natural frequency = 17.90 Hz (FAST o/p) ; No Hand calculation

The query I am having is as under:
As I move from Rigid rotor to Flexible rotor natural frequency of drive train increases (Case1 to Case3 OR Case2 to Case4). First of all, Am I doing some mistake while running FAST? If not, Is there any physical justification for this & Is there way to check natural frequencies of Case3 & Case4 by hand calculation?

Thanks in advance & I would appreciate your help.
With regards,
Tejas

Dear Tejas,

I don’t see a problem with your results.

When you make the rotor flexible, it is primarily the blade edgewise modes that will couple with the drivetrain rotational motion. If the edgewise motions move in-phase with the rotational motion, the drivetrain natural frequency is likely to decrease relative to the rigid-rotor case. But if the edgewise motions move out-of-phase with the rotational motion, the drivetrain natural frequency is likely to increase relative to the rigid-rotor case, as you are showing.

You could problably develop a hand calculation for a simplified representation (e.g., a rigid blade with “hinge” to characterize the edgewise mode instead of a generalized mode), but I would think it would be difficult to develop a hand calculation identical to the FAST formulation.

Best regards,

Dear Jason,

Thanks for your guidance. I really appreciate your help.

With Regards,
Tejas

Dear Jason,
I am calculating the equivalent torsional stiffness of the drive train for 5MW turbine. However the hand calculated value is quit away from the input given for 5MW turbine.

Data: 5 MW Turbine (Reference : nrel.gov/docs/fy09osti/38060.pdf)
G= Modulus of Rigidity =8.08E10 N/m2
GBRatio = 97
HSSRad = 0.2 m
HSSLength = 1.195 m
LSSRad = 0.4 m
LSSLength= 4.78 m

Calculation:
J_HSS = 3.14*(HSSRad)^4/32 = 0.0025 m^4
K_HSS = GJ_HSS/HSSLength = 1.7E8 Nm/rad
K_HSS_to_LSS = K_HSS
GBRatio^2 = 1.6E12 Nm/rad
J_LSS = 3.14*(LSSRad)^4/32 = 0.04 m^4
K_LSS = GJ_LSS/LSSLength = 6.8E8 Nm/rad
Equivalent torsional Stiffness =(K_LSS
K_HSS_to_LSS)/(K_LSS + K_HSS_to_LSS) = 6.79E8 Nm/rad
As per the reference the, the equivalent torsional stiffness of the LSS is 8.67E8 Nm/rad
Hand calculated Equivalent torsional stiffness doesn’t match with FAST reference input (5 MW)

Can you please let me know what is going wrong in the hand calculation?

I would appreciate your help.
With Regards,
Tejas

Dear Tejas,

As far as I can tell, you’ve assumed some values for G, HSSRad, HSSLength, LSSRad, and LSSLength (these are not specified in the NREL 5-MW specifications report). Given the number of assumptions you’ve made, I’m actually surprised how well your hand-calculated value of the drivetrain torsional stiffness for the NREL 5-MW matches the value specified. Why do you expect them to be closer?

Best regards,

Dear Jason,

Very good topic! I understand that free-free drivetrain condition gives quite similar results to fixed-free condition considering as “fixed” the hub side. On the other hand, free-fixed condition (considering as “fixed” the generator side) gives you a very low frequency due to the high rotor inertia.

My question is about power production boundary conditions. I understand that people assumes that when wind turbine is in power production the boundary condition at generator side is considered as free. This would mean, from a physical point of view, that you have a resistive torque due to generator electromagnetic interaction but that interaction can’t be assimilated as one spring. I have tried to measure experimentally this drivetrain natural frequency in field during operation. But when I perform an spectral analysis over Mx drivetrain signal I’m not able to see anything around the expected natural frequency. I understand that wind turbine control damp out that frequency and that’s the reason why I’m not able to see it. During a grid loss or equivalent event where control doesn’t act over the generator you can clearly see the drivetrain natural frequency since it’s a purely free-free condition.

Moreover, when you perform one “classic” modal analysis in simulation, external forces over the system are neglected. Only pre-stressed conditions could be taken into account. Then you get one 1st Side-to-Side tower natural frequency higher (faster) than the 1st Fore-Aft tower mode since drivetrain doesn’t follow Side-to-Side motion. That rotational DOF is free. So, it’s clear that computational approaches consider this generator boundary condition as free, but in reality it can be considered as free? There would be any difference between DFIG or permanent magnet generators?

Thank you very much for your time.

Regards,

Roger

Dear Roger,

I’m not sure I fully understand your questions. I would just like to clarify two things:

*In our aero-elastic simulations, we typically represent the electromagnetics of the generator as a rotational damper - i.e., as a rotational speed-dependent torque. The generator is not modeled as a spring -i.e., there is no azimuth dependence to the torque.
*At NREL we’ve collected measurements on real wind turbines that confirm the “free-free” natural frequencies of the drivetrain as predicted by FAST.

Best regards,

Dear Jason,

Thanks for your quick response. Nice to know that you have seen this behavior in field. I was not able to determine it in real wind turbines. I understand that it was due to the control algorithm used in that wind turbines.

According to you I understand that DFIG and Permanent Magnet Generators (PMG) can be considered as free boundary condition. And only cogging torque in case of PMG due to pol-magnet interaction should be considered as excitation source of the system.

Do you agree?

Thanks once again for your support.

Regards,

Roger

Dear Roger,

I’ve never explicitly modeled the cogging torque, so, I can’t answer your question. My previous answer explains how we typically model a generator.

Best regards,

Hi All,

       I am also calculating the equivalent torsional stiffness of the drive train for my wind turbine. I followed above 

. And i got

Data: Krogmann15-50 Wind turbine
G= Modulus of Rigidity =8.08E10 N/m2
GBRatio = 23.213
HSSLength = 0.37 m
LSSLength = 0.596 m

Calculation:
K_HSS = 1.8028E6 Nm/rad
K_HSS_to_LSS = 9.7143E8 Nm/rad
K_LSS = 8.0777E7 Nm/rad
Equivalent torsional Stiffness =(K_LSS* K_HSS_to_LSS)/(K_LSS + K_HSS_to_LSS) = 7.456E7 Nm/rad , But what about DTTorSpr value? As i have all the properties including HSSRad, LSSRad available for Drivetrain, what would be Drivetrain torsional spring (N-m/rad) value for my wind turbine? i cannot figure this out.

Best Regards
Muddassir Nawaz

Dear Muddassir,

The equivalent torsional stiffness you are calculating with Tejas’ equation is DTTorSpr.

Best regards,