Find optimum angle for turbine blades

  • Thread starter Thread starter iScience
  • Start date Start date
  • Tags Tags
    Angle Turbine
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 3K views
iScience
Messages
466
Reaction score
5
This isn't a homework problem, but i embarrassingly keep getting stuck on this simple problem..

Suppose my turbine's blades are flat surfaced.

FeGBvxw.png


the green lines represent the wind towards the black turbine blade at the right. the blade is drawn as an edge view, such that you'd be looking towards the center motor.
i've tilted my blade at some angle theta in reference to the plane perpendicular to the direction of the wind (as depicted by the blue angle). finally, the purple arrow is just to represent the direction that the turbine's blade will end up moving towards.

consider the next image

ISsk174.png


(Assumptions being made:

* Elastic collision
* Smooth surface (such that no tangential component of force is transferred to the surface) )

This is a force diagram as well as a path diagram just put into one.

Green & orange arrows: represent the incident and reflected path of a air particle (since no tangential force component transferred, angle of incidence (green angle) should equal angle of reflection (orange angle))

Red arrow:Transferred normal force component by collided particle.

$$F(norm)=F(air)cos\theta$$

now breaking the F(norm) into its components WRT the blade's motion...

PkYdKLH.png


The purple vector F(horizontal), being the force comp. responsible for propelling the turbine blade.

to explicitly show what [itex]\phi[/itex] is:

HIYaZ1u.png


the blade's surface is superimposed on the previous image with all the angles drawn

so then [itex]\phi[/itex]= 90-[itex]\theta[/itex]

$$F_{h}=F_{norm}cos(90-\theta)=F_{norm}sin\theta=F(air)cos\theta sin\theta$$

but the flux must also be considered, since greater angle means less orthogonal surface area on blade surface.

Flux:

$$\Phi=\int F_{h}\cdot dA$$

since area is flat and surface dot product is constant..

$$\Phi=F_{h}Acos\theta=F(air)cos^{2}(\theta)sin(\theta)A$$

to find the maximum i just graphed the [itex]\theta[/itex] terms on my calculator, and here's what bothers me:

it gives me multiple peaks at intervals of 35.26° in both + & - directions. Moreover all the peaks are @ the same height.

peaks @:...
...
...
-144.74°
-35.26°
35.26°
144.74°
...
...
...What am i doing wrong?thanks guys
 
Last edited:
Physics news on Phys.org
There are several problems with the way you're modelling it.
The blades usually have an aerofoil cross-section, and the motive force arises in the same way as wing lift.
Ignoring that, you have to consider the rate at which the blades are moving across the airstream. This changes the relative velocity of the air to the blade. This is why blades are more edge-on to the wind close to the rotor shaft. It also means that even after fixing the blade shape you can optimise the power output by adjusting the load (resistive torque).
Thirdly, if the air were bouncing elastically off the blade (which it doesn't - it flows over the blade) that would increase the normal force.
 
Apart from the unrealistic model, when I graph [itex]cos^2 \phi sin \phi[/itex] I get a maximum at +35.26 degrees and a minimum at -35.26 degrees. This is because if you angle the blades the other way, the force will be in the opposite direction.
The other maxima/minima are ithe same because the turbine doesn't change if you rotate the blades 180 degrees.