Nope. I'm still having trouble. I asked about this before, but it's still not gelling for me.
There's a wind turbine on the "front lawn" of the city of Toronto that I pass every day. Its blades are never more than ~20 degrees from flat, i.e the blades are almost perpendicular to the flow of air. If my model of lift generated by the blades is correct*, that represents a ~70 degree angle of attack. Surely that's beyond stall angle.
What am I missing? Perhaps there's something about 'relative' angle of attack based on slow wind speed and fast blade movement?
* my understanding of wind turbine blades is that you can liken it to a plane, with the turbine shaft as the fuselage and each blade as a wing. The lift generated by each blade is perpendicular to the shaft and is parallel to the plane of the rotor i.e. the lift acts to increase the rotation of the rotor.
FredGarvin
Jun26-08, 12:14 PM
There is a very good chance that it is not really a turbine in the classical sense. It could be similar to a Pelton wheel which is a reaction turbine. With such a steep AoA, this makes a bit more sense that they are not relying on aerodynamics of lift but rather just trying to get as much frontal area perpendicular to the wind.
DaveC426913
Jun26-08, 01:36 PM
Nope. It's definitely a wind turbine (http://upload.wikimedia.org/wikipedia/commons/thumb/f/f4/Windshare-toronto_hydro-wind_turbine-20030223.jpg/450px-Windshare-toronto_hydro-wind_turbine-20030223.jpg).
DaveC426913
Jun26-08, 01:41 PM
I found this:
"The rotation of the blades delays the static stall, because the centrifugal pumping results in Coriolis acceleration terms that induce favourable pressure gradients. The result is that boundary layer separation is delayed. "
http://aerodyn.org/Rotors/wturbine.html
...which is sort of what I was alluding to in the OP - even though I didn't know what I was tlaking about.
FredGarvin
Jun26-08, 04:33 PM
Ahh. I see. OK. Next theory.
russ_watters
Jun26-08, 04:54 PM
Nope. I'm still having trouble. I asked about this before, but it's still not gelling for me.
There's a wind turbine on the "front lawn" of the city of Toronto that I pass every day. Its blades are never more than ~20 degrees from flat, i.e the blades are almost perpendicular to the flow of air. If my model of lift generated by the blades is correct*, that represents a ~70 degree angle of attack. Surely that's beyond stall angle.
If the blades are spinning, their motion contributes to the wind speed and the resulting relative wind is the actual angle of attack, not the angle of the blades wrt the wind.
http://www.physclips.unsw.edu.au/jw/sailing.html
Both a sailboat and a wind turbine, when starting from a dead stop will have a normal looking angle of attack that then increases as it accelerates. Ie, the blades will point toward the wind and as it accelerates, they will rotate to flatten-out. This is actually computer controlled and constantly being adjusted.
If you count the rpms and estimate the diameter and pitch angle at the tip and wind speed, you can calulate the tip speed, relative wind, and actual angle of attack.
russ_watters
Jun26-08, 05:11 PM
* my understanding of wind turbine blades is that you can liken it to a plane, with the turbine shaft as the fuselage and each blade as a wing. The lift generated by each blade is perpendicular to the shaft and is parallel to the plane of the rotor i.e. the lift acts to increase the rotation of the rotor. Yes, the resultant vector is always perpendicular to the chord of the airfoil. But the resultant vector isn't the lift vector, it is a combination of the lift and drag vectors. The lift vector is perpendicular to the relative wind direction.
But, while in an airplane you can change your aoa to make sure your lift vector points in the direction you want it to, in a sailboat and in a wind turbine, you are limited by your dependence on the wind (in an airplane, you make the wind go in the direction you want it to). So sometimes you have to accept a lift vector that isn't parallel to your direction of motion. Again, in the sailing website from above, you can see that when the boat is moving upwind, the force vector pushing the boat forward is tiny compared to the force vector pushing it sideways. But the centerboard of the boat (and the shaft of the windmill) constrains the motion to the direction you want.
DaveC426913
Jun26-08, 10:29 PM
Yes, the resultant vector is always perpendicular to the chord of the airfoil. But the resultant vector isn't the lift vector, it is a combination of the lift and drag vectors. The lift vector is perpendicular to the relative wind direction.
But, while in an airplane you can change your aoa to make sure your lift vector points in the direction you want it to, in a sailboat and in a wind turbine, you are limited by your dependence on the wind (in an airplane, you make the wind go in the direction you want it to). So sometimes you have to accept a lift vector that isn't parallel to your direction of motion. Again, in the sailing website from above, you can see that when the boat is moving upwind, the force vector pushing the boat forward is tiny compared to the force vector pushing it sideways. But the centerboard of the boat (and the shaft of the windmill) constrains the motion to the direction you want.
Yeah, I'm a sailor. So I get the apparent wind thing.
If you count the rpms and estimate the diameter and pitch angle at the tip and wind speed, you can calulate the tip speed, relative wind, and actual angle of attack. Right, so, if some section of the blade is slicing the air at, say 50ft/s, and its angle were 20 degrees, then I can conclude that the wind at that point is moving past at 50*sin(20)?