[NREL 5 MW RWT] Tower mass, fore-aft inertia and parallel axis theorem

Wind & Water Structural Analysis
1

Dear wind turbine enthusiasts,

Maybe the topic i am gonna talk has already been discussed but frankly, i cannot find the good answer.

In fact, i am looking to compute only the tower (without RNA) mass and inertia of the NREL 5 MW Reference wind trbine.

To do so, i used the information in the technical report and i get the following:
mass=Tower_height*trapz(fraction,mass_per_unit_length) = 347460 kg

inertia_in_fore_aft=Tower_height*trapz(fraction,inertia_per_unit_length)=1.12e+6 kg.m^2

CM_location_along_z_direction=(Tower_height^2*trapz(fraction,mass_per_unit_length.*fraction))/(Tower_height*trapz(fraction,mass_per_unit_length)) = 38.0658 m

However, when i apply theorem of parallel axis, i obtain negative value of inertia:
inertia_around_CM=inertia_in_fore_aft - mass*CM_location_along_z_direction^2=1.12e+6-347460*(38.0658)^2 = -5.02e+8 kg.m^2

Could you please tell me what i have did wrong ?

Thanks in advance,

Best Regards,

Riad

2

Dear @Riad.Elhamoud,

I agree with your calculation of mass and CM_location_along_z_direction. However, your calculation of inertia_in_fore_aft misses the contribution of the mass_per_unit_length offset from the tower base.

Best regards,

1 Like
3

Dear @Jason.Jonkman,

Thank you for your reply.

Therefore:

inertia_in_fore_aft_around_tower_base=87.6*trapz(fraction,inertia)+87.6^3*trapz(fraction,mass*fraction^2) = 7.2222e+08 kg.m^2

inertia_around_CM=inertia_in_fore_aft - mass*CM_location_along_z_direction^2=7.2222e+08-347460*(38.0658)^2 = + 2.1875e+08 kg.m^2

Do you agree ?

Thanks in advance.

Best Regards,

Riad

4

Dear @Riad.Elhamoud,

Yes, I agree.

Best regards,

1 Like