Acceleration of the tips of wind turbine blades

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Homework Help Overview

The discussion revolves around a wind turbine's rotor, specifically focusing on the calculations related to the acceleration of the tips of the blades, torque, rotational kinetic energy, and angular acceleration. The problem involves concepts from rotational dynamics and power output in the context of wind energy conversion.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between torque and power output, questioning how to account for friction and counter-torque in the context of the problem. There is discussion about the assumptions regarding constant rotation and the forces acting on the rotor.

Discussion Status

Participants are actively engaging with the problem, raising questions about the assumptions made in the original post. Some have suggested that the counter-torque from the motor is relevant to the calculations, while others emphasize the need to focus on the torque caused by the wind. There is a recognition of the complexity involved in interpreting the problem's requirements.

Contextual Notes

There is a noted lack of information regarding friction and counter-torque, which some participants argue is necessary for a complete understanding of the system. The problem also specifies to ignore air resistance, which adds to the complexity of the discussion.

asaspades
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Homework Statement


A wind turbine has a rotor (its rotating section) which has a moment of inertia I = 1.26 × 10^7 \text{kgm}^2. At peak output, with the rotor completing 0.25 \text{ revolutions per second}, the turbine produces a power of P = 3\text{MW}. The tips of the rotor blades sweep out a circle of diameter 90 \text{m}.
Calculate the magnitudes of
(i) the acceleration of the tips of the blades,
(ii) the torque acting on the rotor shaft due to the wind,
(iii) the amount of rotational kinetic energy stored in the rotor,
(iv) the rotor’s angular acceleration should the wind suddenly cease to blow. [Ignore any torque on the blades due to air resistance.]
(v) Draw a sketch to show the direction of the acceleration of the blade tips.

Homework Equations


a=\omega^2r


The Attempt at a Solution


The equation given above in obviously the centripetal acceleration, but this is not helpful for part ii. In fact, since I can only assume the rotor is traveling at constant \omega, surely the torque in part ii is 0. I'm just lost on this entire question.
 
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Yes, the total torque on the system is zero. But there is friction that the wind must overcome in order to keep the turbine moving. And the wind does work as it pushes the blades around in a circle.
 
paisiello2 said:
Yes, the total torque on the system is zero. But there is friction that the wind must overcome in order to keep the turbine moving. And the wind does work as it pushes the blades around in a circle.
I don't see any information which allows us to take friction into account, so it's just the counter torque from the motor that we need to consider. Asaspades, use the given power output to calculate that.
 
haruspex said:
I don't see any information which allows us to take friction into account, so it's just the counter torque from the motor that we need to consider.
And what causes this "counter torque"? I don't see any information which allows us to take the counter torque from the motor into account.
 
paisiello2 said:
And what causes this "counter torque"? I don't see any information which allows us to take the counter torque from the motor into account.
The power output and the rotation rate.
 
I put it to you that that counter-torque of the motor is actually friction.
 
paisiello2 said:
I put it to you that that counter-torque of the motor is actually friction.
No. Friction would be loss of work. Counter torque from the motor relates to the useful work done in turning the motor. You mentioned this in your own post. My point is that we have enough information to quantify that, but no information on friction.
 
Friction might be technically loss of work but it's also the force that keeps the wind force from accelerating the turbine. The problem also doesn't specify the counter-torque of the motor either (whatever that might be).

Regardless, I think we would both agree that the question is asking specifically about the torque caused by the wind and therefore we don't need to know what causes the counter-torque.
 
Last edited:
paisiello2 said:
Friction might be technically loss of work but it's also the force that keeps the wind force from accelerating the turbine. The problem also doesn't specify the counter-torque of the motor either (whatever that might be).

Friction has nothing to do with the question.

The basic design of a wind turbine is a rotating shaft with the blades at one end, and an electrical generator at the other end. If the generator is producing 3MW of electrical power, as the question says, if takes 3 MW of power to keep the shaft turning at constant speed. That power comes from wind creating aerodynamic forces on the blades.

The extra power to overcome friction in the bearings of the shaft etc would probably be less than 1% of the 3MW, so it can be ignored.

In part (v) of the question, the generator doesn't "know" that the wind has stopped blowing, so it continues to convert 3MW of power into electrical energy. With no wind, the only place that power can come from is by using up the kinetic energy in the rotating blades, as the rotation speed decreases.

From other threads on PF, there seems to be a fairly common misunderstanding that because you can't "see" any work being done when a generator is producing electricity, you don't have to do any work to turn the shaft. The easy way to disprove that is to get a generator that can produce a reasonable amount of power, for example a car alternator, and set it up so you can turn it with a hand crank. With nothing connected to the electrical output, you can turn it easily, which shows the friction forces are small. But if you connect it to some car headlights that take say 100W of power, you will have to work a lot harder to turn it!
 
  • #10
Ok, I agree with what you said. The counter-torque is caused by a Lorentz force from the current in a magnetic field. I was under the mistaken impression that this would only occur when the rotor was accelerating. I see now I was wrong about that.

Regardless, it doesn't affect the answer to the OP's question.
 
  • #11
paisiello2 said:
Ok, I agree with what you said. The counter-torque is caused by a Lorentz force from the current in a magnetic field. I was under the mistaken impression that this would only occur when the rotor was accelerating. I see now I was wrong about that.

Regardless, it doesn't affect the answer to the OP's question.
It most certainly is relevant to the answer. The provided information (rotation rate and power output) allows us to calculate the counter torque, and that's what we need for part (ii).
 
  • #12
Question asked for the torque caused by wind. Didn't ask for the counter-torque.
 
  • #13
paisiello2 said:
Question asked for the torque caused by wind. Didn't ask for the counter-torque.
Since it is not accelerating, the two are the same. How else would you propose we calculate the torque with the information provided?
 
  • #14
Thank you for the help. I was confused since I'd never seen a relation between torque and power before, but it wasn't hard to derive. I guess I should have tried before posting here! Thanks again
 

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