0
$\begingroup$

I am trying to estimate the amount of torque and power required to rotate a 3 bladed Horizontal Axis Wind Turbine (HAWT) about a vertical axis("yaw" motion). I would like to estimate the yawing torque and power required to properly size the motors required for yawing. Aerodynamic factors have been neglected for this dynamics calculation.

I am modelling the wind turbine's motion as follows: The 3 (rotor) blades, the rotor hub and the main shaft all move at the same rotational speed $\Omega$ about the main shaft's axis. I am ignoring the contribution of the bearings, brake, gearbox and the generator in this model. Lets say the moment of inertia of this whole assembly about the main shaft's axis is known to be equal to $I_r$.

Now on top of this rotor motion, the turbine would be required to yaw about the vertical axis (to face the oncoming wind for maximum power capture) with some known yaw rotational speed equal to $\omega _{yaw}$. Lets say the moment of inertia of the rotor and nacelle assembly about the tower's yaw axis be known to be equal to $I_{yaw}$.

What I am trying to do is to estimate the amount of torque and power required to yaw the blade-hub-shaft assembly (which is rotating wrt the shaft axis at a constant $\Omega$) at a constant $\omega _{yaw}$. Please note that I am not considering any angular accelerations here, although I would love to hear from your thoughts on that.

Obviously some application of 3D rigid body dynamics will be required. I was trying to use $\overrightarrow\tau = \frac{d\overrightarrow L}{dt}$. I can write the angular momentum as $\overrightarrow L = I \overrightarrow \omega$. The vector $\overrightarrow \omega$ can be written as $(\Omega,0,\omega _{yaw})$ with the $x$-axis along the main shaft axis, $z$-axis as the vertical axis and the y-axis fixed accordingly as being perpendicular to $x$ and $z$ axes according to the right hand thumb rule. The moment of inertia tensor can be assumed to be fully known (all 9 components known). After getting the $\overrightarrow L$ vector I can find its time derivative. Now since all the components of the inertia tensor as well as the $\overrightarrow \omega$ as constant, the only contribution to the time derivative is due to the rotation of the $x$ and $y$ axes themselves. The time derivative of the unit vectors along $x$ and $y$ axes can be found by $\frac {d\overrightarrow i}{dt} = \overrightarrow\omega_{yaw} \times \overrightarrow i$. (Similarly for y axis also.)

I simply do not understand why there is a non-zero $\tau_2$ appearing in the calculation. The HAWT assembly does not change its inclination with respect to the vertical axis - so I do not undrstand why a $\tau_2$ appears. Also in reality only a $\tau_3$ about the vertical axis is provided by the motors for the yawing.

The above approach is a general 3D dynamics approach, but I feel that this problem is similar to a spinning top precessing about a vertical axis passing through a fixed point with its spin axis inclined with $z$ axis at an angle of $\frac{\pi}{2}$. In that case I think the torque would come out to be $I_r \Omega \omega_{yaw}$ - is this the same torque that is supposed to be provided by the motor (gear ratio factors can be assumed to known).

I could not get any reference for yaw motor sizing of wind turbines anywhere on the internet and hence I am asking this about this simplified model on this website. This is my thought process regarding the torque calculation. I am sorry for my messy and disorganized presentation of my ideas. Can anyone guide me regarding this and tell me whether I am correct or not?

Will this gyroscopic stuff even contribute to the yaw motor's sizing calculations? There is a non-zero torque required for precession but will the yaw motor provide that additional torque?

Can anyone provide any resources where I can read more about yaw motor sizing? Thank You in advance.

$\endgroup$
3
  • $\begingroup$ Have you looked at a real machine to see the size of motor used? $\endgroup$
    – Solar Mike
    Mar 25 at 7:19
  • $\begingroup$ @SolarMike I have but for this case I am finding it difficult to size the yaw motor - I am not even sure if the gyroscopic torque (assuming it's not very small) will contribute to the yaw torque requirement. $\endgroup$
    – Quadro
    Mar 26 at 8:06
  • $\begingroup$ Curious, if you have then that gives a good starting point: size and mass of nacelle,gearbox, blades etc bearing support, diameter of gear needed and the ratio which leads to speed of rotation. $\endgroup$
    – Solar Mike
    Mar 26 at 8:20

1 Answer 1

0
$\begingroup$

Most machines would keep the blades parked until the blades are close to facing the wind direction, then the brakes would be removed to allow rotation.

This is done to prevent a varying load on the blades especially as they rotate into the wind.

$\endgroup$
1
  • $\begingroup$ Thanks a lot for mentioning that. Can you also help me with this torque estimation? Even a basic guideline would be helpful. $\endgroup$
    – Quadro
    Mar 22 at 11:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.