# What's the exact meaning of HubIner?

Hi,

I am trying to do a linearization of the one state wind turbine model: J*(d_omega/dt)=0.5CprhoAV^3/omega-Ntao_g. In this expression, J is the total inertia of all parts of rotation. It should be expressed like this:
J=J_L+J_H
N^2;
J_H is the inertia of the high speed side which is corresponding to the variable ‘GenIner’ in FAST Input file and N is the gearbox ratio. J_L is the inertia of low speed side. I think J_L should include the inertia of all blades, hub and low speed rotor. But from the explanation, the variable ‘HubIner’ seems only represent the inertia of Hub.

So my question is if J_L=HubIner is right? If not, how should I calculate the J_L?

Thanks,

Shu

Dear Shu,

In FAST, HubIner represents the inertia of the hub (without blades) about the low-speed shaft (LSS). In your equation, you want J_L to represent the inertia of the entire rotor about the LSS. This inertia would equal HubIner plus the inertia of each blade about the LSS. This rotor inertia is called RotIner in FAST and is calculated and reported in the FAST summary (*.fsm) file.

I hope that helps.

Best regards,

Thank you Jason! It does make sense now.

But when I check the relation among RotIner, inertia of each blade about the LSS and the HubIner, there is still a little confusion to me. Just taking the WindPACT 1.5MW Wind Turbine for example:
HubIner=34600(kg*m^2);RotIner=2962443.900;Inertia of each blade about the LSS=800799.916(it’s called second mass moment in .fsm file, I suppose it to be the blade internia), then:
34600+800799.916
3=2437000!=2962443.9

So where is the missing part in calculation?

Thanks,

Dear Shu,

In FAST’s .fsm file, the center of mass, 1st mass moment, and 2nd mass moment are all output relative to the blade root (not the rotor center). It is the transfer of the blade inertia from the blade root to the LSS that you are missing in your calculation.

Best regards,

Sorry to dig up this old thread, but I have similar questions regarding the rotational inertia of the 5MW offshore spar-buoy (OC3) rotor. I’m trying to determine the inertia about the LSS so I’m starting with the “Rotor Inertia” from my .fsm file (38759236 kg-m^2).

I see that this is ‘at the blade root’ and needs to be translated to the LSS, I assume via parallel-axis. However there is no orientation about this given Rotor Inertia at the root. Is it parallel to the LSS? And would my distance (in the parallel-axis theorem) simply be the radius of the hub?

Perhaps more straight to the point is: Is the rotor inertia (hub and blades as per http://forums.nrel.gov/t/rotor-inertia/850/1) reported in the .fsm file an inertia about the LSS? Thanks for any help!

Dear Sean,

The rotor inertia written to the FAST summary file of FAST v7 (.fsm) or the ElastoDyn summary file of FAST v8 (.ED.sum) is reported about the LSS, not about the blade root. Only the individual blade center of mass, 1st mass moment, and 2nd mass moment are output relative to the blade root.

Best regards,

Thank you Jason!

For the OC3-Hywind, starting with my rotor inertia of 38759236 kg-m^2 and adding in the generator inertia of 5025500 kg-m^2, I get a drivetrain inertia about the LSS of 43,784,736 kg-m^2. Can anyone verify that number? I’ve been attempting to utilize the baseline variable-speed controller (in separate, non-FAST code) and I suspect that my current inertia setting is wrong. Thanks for any help!

Dear Sean,

Yes, 43,784,700 kg-m^2 is the correct drivetrain (rotor + generator) inertia of the NREL 5-MW baseline turbine.

Best regards,

Dear Jason,

How is Rotor inertia calculated through .fsm file? I have calculated my

Rotor inertia = Hub inertia + 3*Blade inertia

through this formula i calculated my Rotor inertia = 5229.34 kgm2. But my rotor inertia calculated through FAST .fsm file is 5861.016. Even though Rotor blade mass is calculated correctly in .fsm file. What about Second Mass moment? is this rotor blade inertia? if it is then with formula i got rotor blade inertia = 1721.78kgm2, but with .fsm my Second Mass moment calculated as 1758.745 kgm2.

Second, i have attached code, My wind turbine is 50kw power but i am getting Power of around 80kw. Can you point me what i am doing wrong? i am stuck here. This will be very helpfull.

[/code]

Thank you
best regards
Muddassir Nawaz

Dear Muddassir,

I’m not sure why your hand calculation of the rotor blade inertia does not match the FAST v7 summary file, but your equation for the rotor inertia (RotIner) is not correct because you miss (1) the influence of the hub radius (HubRad) and (2) the influence of the rotor precone (PreCone(k)). The former is important because the second mass moment of inertia of the blade (SecondMom(k)) reported in the FAST v7 summary file is calculated about the blade root, not about the centerline of the low-speed shaft. With these adjustments, the correct equation is:

RotIner = HubIner + SUM( BldMass(k)*( ( HubRad + BldCg(k) )^2 - BldCg(k)^2 )*COS( PreCone(k) )^2, k = 1,2,3 )

(That said, I see that you’ve set PreCone(k) = 0 in your FAST v7 primary input file, so, COS( PreCone(k) )^2 = 1.)

I see that you are using the Simple Induction Generator (SIG) model (VSContrl=0, GenModel=1). My guess is you are getting more than 50 kW because your generator speed is spinning faster than 1500*(1.02222) rpm (and thus your generator torque is higher than 318.31 Nm).

Best regards,

Hello jason,

What is "BldCg(k)" is it Blade Center of Mass?

best regards
Muddassir Nawaz

Dear Muddassir,

Yes, BldCg(k) in the equation I stated in my posted above is the distance along the pitch axis from the root to the blade center of mass.

Best regards,

hello jason,

Thank you so much for your help. i solved my rotor inertia. I still have  problem with my Generator speed. I am using (VSContrl = 0 and GenModel = 1).

---------------------- DRIVETRAIN ---------------------------------------------- 98.0 GBoxEff - Gearbox efficiency (%) 92.8 GenEff - Generator efficiency [ignored by the Thevenin and user-defined generator models] (%) 23.23 GBRatio - Gearbox ratio (-) False GBRevers - Gearbox reversal {T: if rotor and generator rotate in opposite directions} (flag) 9999.9 HSSBrTqF - Fully deployed HSS-brake torque (N-m) 9999.9 HSSBrDt - Time for HSS-brake to reach full deployment once initiated (sec) [used only when HSSBrMode=1] "unused" DynBrkFi - File containing a mech-gen-torque vs HSS-speed curve for a dynamic brake [CURRENTLY IGNORED] (quoted string) 1.76E5 DTTorSpr - Drivetrain torsional spring (N-m/rad) 1.0E5 DTTorDmp - Drivetrain torsional damper (N-m/(rad/s)) ---------------------- SIMPLE INDUCTION GENERATOR ------------------------------ Crude approximation of torque/speed curve. 1e-2 SIG_SlPc - Rated generator slip percentage (%) [used only when VSContrl=0 and GenModel=1] 1500.0 SIG_SySp - Synchronous (zero-torque) generator speed (rpm) [used only when VSContrl=0 and GenModel=1] 318.31 SIG_RtTq - Rated torque (N-m) [used only when VSContrl=0 and GenModel=1] 1.0 SIG_PORt - Pull-out ratio (Tpullout/Trated) (-) [used only when VSContrl=0 and GenModel=1]

My rotor speed is 1500/23.23 = 64.5 rpm. i put inital condition " 64.5 RotSpeed - Initial or fixed rotor speed (rpm)". Now Generator speed should be (1500*1.00001 ) But when simulate my model with FAST. i got rotor speed (LSSGaV) around 100-120 rpm and GenSpeed(HSShftV) around 2500 rpm!!!

Why is there so much difference? How are LSSGaV & HSShftV are calculated with FAST?

Thank you
Best regards
Muddassir Nawaz

Dear Muddassir,

You’ve set your rated torque (SIG_RtTq) to 318.31 Nm and the pull-out ratio (SIG_PORt) to 1.0, so, the generator torque can’t exceed 318.31 Nm. if the aerodynamic torque (relative to the high-speed shaft) ever exceeds 318.13 Nm, then the generator / rotor speeds will accelerate well above the rated speed.

Best regards,

Dear Dr Jason
I need some clarification about the red part in the equation below:
RotIner = HubIner + SUM(( SecondMom(K) + BldMass (K)( HubRad + BldCg(k) )^2 - BldCg(k)^2)( CosPreC(K)**2 )), k = 1,2,3 )
According to the Parallel Axis Theorem, the red part of equation should represent the square distance between the blade root and the centerline of the low-speed shaft. I think this distance may be equal to HubRad^2.
Why you add and subtract BldCg(k)? OR
Where is the location of blade root? (Is it at point along pitch axis with a distance of HubRad from Apex of cone rotation?)

Dear Mohamed,

Yes, the blade root is located a distance of HubRad from the apex along the pitch axis, so ( HubRad + BldCg(K) ) is the distance from the apex to the blade center of mass. The -BldMass(K)*BldCg(K)^2 term is included so as not to double count the inertia effect already accounted for in SecondMom(K).

I hope that helps.

Best regards,

Dear Dr Jason

Thank you for clarification

Best regards