Help to calculate rotation speed of rotor

In summary: Wells turbine. The blades in a Wells turbine are symmetrical and operate initially at a high angle of attack. It is assumed that the rotor keeps spinning between the bidirectional gusts, which lowers the angle of attack and makes it possible to get a little energy.1-Sin(a) is really only applicable to lift at low angles of attack, near optimum, before the stall. It is an approximation, probably NOT applicable to the complexity of a Wells turbine.
  • #1
HauTo
5
3
Hello everyone, i have a question, how to find wind turbine rotor rotation speed based on freewheel rotation speed of rotor (RPM)(torque = 0)? Thanks for your attention.
 
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  • #2
Welcome to PF.

The unloaded speed will be when the angle of attack of the blade airfoil is close to zero. As power is extracted from the rotor, the angle of attack will increase to between about 5° and 15°, and the rotor will slow down. If you try to extract more energy from the rotor, the airfoil will begin to stall at higher angles of attack. You need to operate the rotor at an optimum angle of attack.

Note that there is a twist along the turbine blades. That allows for the blade velocity due to rotation increasing with radius, and the angle of attack remaining close to optimum along the blade. The angle of attack relates the ratio of the blade velocity to the wind velocity.

Without more blade airfoil details your question can only be answered approximately. I will have to check my maths, but I believe the relationship between angle of attack, a, loaded RPM, Va, and unloaded RPM, Vo, goes something like;
Va = Vo * ( 1 - Sin( a ) )

As an example, let us assume that the optimum angle of attack is 12°,
For a = 12°, 1 - Sin(12°) = 0.792 = 79.2 % of the unloaded speed.
 
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  • #3
Baluncore said:
Welcome to PF.

The unloaded speed will be when the angle of attack of the blade airfoil is close to zero. As power is extracted from the rotor, the angle of attack will increase to between about 5° and 15°, and the rotor will slow down. If you try to extract more energy from the rotor, the airfoil will begin to stall at higher angles of attack. You need to operate the rotor at an optimum angle of attack.

Note that there is a twist along the turbine blades. That allows for the blade velocity due to rotation increasing with radius, and the angle of attack remaining close to optimum along the blade. The angle of attack relates the ratio of the blade velocity to the wind velocity.

Without more blade airfoil details your question can only be answered approximately. I will have to check my maths, but I believe the relationship between angle of attack, a, loaded RPM, Va, and unloaded RPM, Vo, goes something like;
Va = Vo * ( 1 - Sin( a ) )

As an example, let us assume that the optimum angle of attack is 12°,
For a = 12°, 1 - Sin(12°) = 0.792 = 79.2 % of the unloaded speed.
Thank you very much for your answer. It is very helpful for me.
 
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  • #4
Baluncore said:
Welcome to PF.

The unloaded speed will be when the angle of attack of the blade airfoil is close to zero. As power is extracted from the rotor, the angle of attack will increase to between about 5° and 15°, and the rotor will slow down. If you try to extract more energy from the rotor, the airfoil will begin to stall at higher angles of attack. You need to operate the rotor at an optimum angle of attack.

Note that there is a twist along the turbine blades. That allows for the blade velocity due to rotation increasing with radius, and the angle of attack remaining close to optimum along the blade. The angle of attack relates the ratio of the blade velocity to the wind velocity.

Without more blade airfoil details your question can only be answered approximately. I will have to check my maths, but I believe the relationship between angle of attack, a, loaded RPM, Va, and unloaded RPM, Vo, goes something like;
Va = Vo * ( 1 - Sin( a ) )

As an example, let us assume that the optimum angle of attack is 12°,
For a = 12°, 1 - Sin(12°) = 0.792 = 79.2 % of the unloaded speed.
Just another question, sir. In Well Turbines situation, Is the above method still valid? This is my blade airfoil below. Thanks for your attention.
airfoil.jpg
 
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  • #5
HauTo said:
In Well Turbines situation, Is the above method still valid? This is my blade airfoil below.
"Well Turbines", maybe it is confused in translation ?
https://en.wikipedia.org/wiki/Wells_turbine

Your asymmetrical airfoil is not suited to a Wells turbine. The blades in a Wells turbine are symmetrical and operate initially at a high angle of attack. It is assumed that the rotor keeps spinning between the bidirectional gusts, which lowers the angle of attack and makes it possible to get a little energy.

1-Sin(a) is really only applicable to lift at low angles of attack, near optimum, before the stall.

It is an approximation, probably NOT applicable to the complexity of a Wells turbine.
 
  • #6
Thanks for your opinion, sir. My airfoil are symmetrical about the horizontal axis, the picture just shows only half. Sorry, my English is not very good. If the above formula does not apply in this case. Can I use other factors to calculate rotational speed with the existing unload rotational speed?
 
  • #7
HauTo said:
Can I use other factors to calculate rotational speed with the existing unload rotational speed?
You might use 1-Sin(a) as a crude or approximate design guide for an initial computer model, but you will have to do some real experiments, to verify it gives a sensible prediction. I see no other way to predict the maximum power point, MPP.

You have a very complex situation with a Wells' turbine. They are not particularly easy or efficient to operate. The airflow is varying in direction sinusoidally, so you cannot optimise power extraction from the flywheel rotor in a steady state, as can be done with a normal wind turbine in a steady wind.

There must be a flywheel, and then you must control energy extraction by changing the excitation of an alternator. There would be an advantage in having a mass-air-flow, MAF, sensor in the tunnel, along with a manometer to measure the pressure drop, and so the flow direction across the turbine.

The changing unloaded RPM cannot be found while you are actually generating with a spinning flywheel. The unloaded speed and the optimum power RPM will be continuously changing, so you will need to sample flywheel RPM over several full water-wave cycles, to identify if you are taking more or less power than can be sustained. Ignore the thoughts that follow if it is too much, or ask more questions.
To optimise the system while it is running, one could very slowly vary the power extraction sinusoidally about the mean. Then record the changing RPM of the flywheel. Analyse that with an FFT, looking for the second harmonic of the power extraction frequency. The FFT must reject or filter out the higher water wave frequencies. Adjust the mean power extraction to make the varying RPM waveform symmetrical, with an equally flattened top and bottom, which should happen near the MPP. When the recorded RPM wave is symmetrical, analysis will show little 2'nd harmonic, but with some 3'rd harmonic that lets you know your extraction variation is being detected in the RPM record. The presence of a significant 2'nd harmonic component tells you that it is non-optimum. The phase of the 2'nd harmonic tells you which way to change the mean rate of energy extraction to approach the MPP.
 
  • #8
I am much obliged to you for your answer, sir.
 
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