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Help a Newbie: Twist and Angle of Attack

Hi folks,

I am a student trying to get to grips with QBlade for a project. I have been researching airfoils and their aerodynamics but I  have one key question relating to QBlade’s twist function. I apologise if it seems to be a stupid error on my part.

On the “Polar Extrapolation to 360” module I used the Glide Ratio – Angle of attack graph feature to find that the optimal angle of attack was roughly 12.5 degrees. I then built four identical turbines // HAWT Blade designs only changing the twist to be 5 degrees, 10 degrees, 12.5 degrees and 15 degrees. I then built turbines and tested them, but the Cp value for 5 degrees was higher for 5 degrees than the other.

My question is thus: using the polar extrapolation graph, for the optimal angle of attack A in degrees, how can I calculate the best twist angle T in degrees? I originally presumed that A = T. I know there is an optimization feature, but how does this select the best twist angle using the optimal angle of attack?

 

Many thanks,

 

Toby

Hello Toby,

the optimum twist angle is found by locking at the “velocity triangle” of every blade station.

velocity triangle

The velocity that the blade “sees” is made up from the contributions of rotational (relative) blade velocity, inflow (wind) velocity and the axial (a) and tangential (a’) induction in the rotor plane. The blade twist angle Phi (Φ) is then chosen so that at a given inflow angle phi (ϕ) (resulting from Vx and Vy in this image) the angle of attack alpha (α) is at the optimum glide ratio.

For optimization purposes the axial induction is often assumed to be 1/3, according to Betz. For simplification you can set a’ to zero.

For a more thorough explanation of this you will find much more information on the topic of blade design and optimization in the textbook of your choice on wind energy basics.

BR,

David