NREL 1.5 MW

Dear Jonkman,
Thanks for your reply , but i want to do the analysis in which blades are not rotating but wind is coming. so wind feels that the blades are obstruction.

According to your reply when i make the Flag Compaero = FALSE , FAST will not calculate the aerodynamic force but at the same time it will not also taking the wind.

Can there is any way to do the analysis which i am asking i.e blades are not rotating but wind is coming.

Thanking you.

Dear Anuj,

Disable the structural DOFs, but leave CompAero = True enabled.

Best regards,

Dear Jonkman,
sir i tried to calculate the moment arm using the relations :

Arm 1 = TwrBsMxt/ TwrBsFyt and Arm 2 = TwrBsMyt/ TwrBsFxt .

As per my logic both the arms must be same and should be equal to the hub height, but i am not getting it.

Then i tried
Arm 3 = TwrBsMxt/ YawBrFyn and Arm 4 = TwrBsMyt/ YawBrFxn .

Again Arm 3 should be equal to Arm 4 and should be approximately equal to hub height. That also i didn’t get.
Can you please clarify which moment arm have been considered to calculate the moments at the tower base.

Thanks in advance.

Dear Anuj,

A similar question has been answered in the forum topic found here: RotThrust, LSSTipMya, TwrBsMyt (particularly my Mar 13, 2013 post). Of course, this forum topic applies to a turbine with a rigid tower and fixed platform. For a flexible tower or moveable platform, inertia forces (due to tower deflection or platform motion) and additional moment arms due to tower deflection or platform motion would also contribute to the tower-base moments.

Best regards,

Dear Jonkman,
Thanks for your reply.
Now, i want to know that for a 3- bladed wind turbine as we can gave three different sectional property files or same for all blades as mention in .fst file.Baseline_blade.dat.
Can we do the same for aerodynamic files for three different blades???

Actually i want to eliminate the aerodynamic load from the tip of one blade.(i.e assume if the blade length is 31 m, so i have to eliminate the aerodynamic force at the tip having length of 3 m so for this blade only 28 m blade length will calculate the aerodynamic forces )

And i want this only for one blade, actually i tried but it is taking for all the blades.

Dear Anuj,

In the current version of FAST, the only way to model an aerodynamic imbalance is to set the blade pitch independently for each blade as discussed in the forum topic found here: Varied Pitch angles:Incorporating blade prod'n tolerance. Without modifications to the source code, it is not possible to eliminate the aerodynamic load from the tip of one blade.

Best regards,

Dear Jonkman,
Thanks for your reply . Now , i need a clarification. I am doing an analysis on 1.5 MW wind turbine .
I analyzed for two wind speeds 14.6 m/s and 20.6 m/s .
But the results (tower base forces and moments ) are coming less in case of wind speed 20.6 compare to 14.6 .
Why ?
Please have a look and clarify the doubt.

Thank you in advance.

Dear Anuj,

In a conventional variable-speed, active blade pitch-to-feather controlled wind turbine, the mean aerodynamic rotor thrust peaks at rated wind speed, then drops with increasing wind speed. So the mean rotor thrust would be higher at 14.6 m/s than at 20.6 m/s.

Best regards,

Dear Jonkman,
Thanks for your reply but can you tell me what is the rated speed for 1.5 MW baseline wind turbine. means after which wind speed it starts decreasing (rotor thrust). And how can we know that what is the optimum wind speed or rated speed for any wind turbine.

Thank you in advance.

Dear Anuj,

The rated wind speed for the WindPACT 1.5-MW baseline wind turbine is about 11.5 m/s.

The “rated” wind speed is not typically an “optimal” wind speed (i.e., a turbine does not typically operate at peak aerodynamic efficiency at the rated wind speed). Instead the “rated” wind speed is the wind speed at which the turbine first produces rated power for its generator. The rated wind speed of a turbine is a design decision and is influenced by the physical components of the turbine and controller settings.

Best regards,

Dear jonkman,
Thanks for your reply one more question i want to ask . In the fast manual co-ordinate system is there in which “a” represents what i mean which component axis.??
And the rotation of wind turbine gives us what ? Forces or Forces and moment both .?

Thanks in advance.

Dear Anuj,

I’m sorry, but I don’t understand your question regarding the FAST coordinate systems.

FAST has force and moment outputs available at points throughout the wind turbine.

Best regards,

Dear Jonkman,
I am talking about the FAST manual Page no.7 Subtitle heading ("Coordinate system). There it is written additional coordinate system i,p,a,s and c. so i am asking what a represents on the wind turbine. Because the output at Hub “LSShftFxa” is difined at “a” coordinate.

Thank you.

Dear Anuj,

The “a” or “azimuth” coordinate system is described on page 9 of the FAST User’s Guide.

Best regards,

Dear Jonkman,
I want to know can we lock the Blades . Means they are not rotating !! If yes then how ??

Dear Anuj,

Please see the “Simulating Special Events” Section of the “Controls” Chapter of the FAST User’s Guide for information on how to simulate parked or idling rotor conditions.

Best regards,

Dear Jonkman,
Thanks for the reply i now read the control chapter and then check .!! I have one more doubt know .
I have disabled the CompAero to False , Made the wind speed to 0.01 m/sec.
But in output RotThrust is coming ?? How this can be possible and if it is correct then what this RotTrust significe ??

Dear Anuj,

A similar question has been answered in the forum topic found here: The effect of Tilt Angle. Basically, FAST output RotThrust not only includes the applied aerodynamic thrust, but also the gravity and inertial loads of the rotor. See the link for more information.

Best regards,

Dear Jonkman,
Thanks for your reply i understood that . But now i am running a Parking condition as you mention “special events” in control chapter. Everything is coming correct but why Rotor Torque is coming . We make the blade angle to 0. and the blade are locked then why we are getting the Rotor Torque.
Waiting for your reply.

Dear Anuj,

I would expect a rotor torque if the blades are fixed at 0 degrees. To reduce torque when parked or idling, blades are normally fixed at around 90 degrees (leading edge into the wind).

Best regards,