Pivoting Blades Vertical Axis Wind Turbine

Vertical axis wind turbines have some advantages, for example they don't need to be oriented towards the wind. They have also disadvantages, the blades have to go upstream for a certain part of their trajectory. This lower the yield.

There are two types of vertical axis wind turbine, the slow ones like Savonius, who have a tip speed ratio (blade_speed / wind_speed) smaller than one. And the fast ones, who have a tip speed ratio greater than one like the Darrieus. The Savonius has a bad power coefficient because of the big blade surface moving upstream. The Darrieus has a better power coefficient but can't start alone because no torque is created at low speed.

This prototype is an attempt to combine the best of both worlds.

I used a laser cutter to cut the parts but making a hand made version should not be a problem. Download the attached part files and cut them.

You need:

• Plywood 3mm
• Paper tape (50mm wide)
• Small stepper motor (mitsumi m42sp-4np) (used as a generator)
• 5 Screws M3x16
• 10 Nuts M3
• 10 Braces bearing (I cut a tube of 3mm inside diameter into 3mm pieces)

• Cover the five blades with the paper tape on both sides (image 1, 2).
• Cut a window on one side around the slots (image 3).
• Insert the brass bearings in the hinges
• Plug the hinges in the blade (image 4).
• Attach the blades to the rotor with the screws and nuts. The parts order is shown in image 5. The two nuts are tighten together, they pinch the rotor.

Wind turbine power

The stepper motor is unipolar with five pins:

1. Com
2. Coil 1 left side
3. Coil 1 right side
4. Coil 2 left side
5. Coil 2 right side

I measured the tension between the pins 3 and 5 or 2 and 4 with no load (just take two pins with a sin-like output and the highest tension).

The first images shows a peak tension of 1.5 V at 40.9 Hz and the second image shows at peak tension of 2.5 V at 67.7 Hz. Which gives a linear relation with a slope of 0.0368 V/Hz. As this stepper has 48 steps for one turn, I guess we can divide by 48 / 2 = 24 (because I measure between 3 and 5) to obtain the turbine speed: 67.7 / 24 = 2.82 rps.

The resistance between the pins 3 and 5 (or 2 and 4) is 80 Ω so if I add a load of 80 Ω and rotate at 2.82 rps:

• U = Û / sqrt(2) = 2.5 / 1.414 = 1.77 V
• I = U / (Rint + Rload) = 1.77 / 160 = 11 mA
• Pload = Rload * I^2 = 10 mW

As I have two coils, I can extract the double: 20 mW

With this load (80 Ω) the yield is 50% because the half power is dissipated inside the generator. To increase the generator yield, use a bigger load. But a bigger load generates less braking torque so the extracted power might be smaller or bigger.

I don't know which wind speed is required to make the turbine and this load generating this power but I am working on it (making my anemometer). If the blade tip speed is around 1 then 2.82 rps correspond to a wind speed of:

rotation_speed * rotor_rayon = 2π * 2.82 * 0.1 = 1.77 m/s

Wind Power

Pwind = rho * S * v^3 * 1/2

• rho: air density = 1.2 kg/m^3
• S: surface reachable by the turbine = 0.15 * 0.2 = 0.03
• v: wind speed

The maximum theorettically extractible power is only 59% of Pwind because of the Betz limit. The best wind turbines reach 50%.

So for example if the wind is 1.77 m/s, Pwind = 100 mW and Pmax_turbine = 59mW

Conclusion

The wind required to generate 20 mW is probably faster than 1.77 m/s. So this stepper motor is probably not powerful enough to extract all the power made by this wind turbine. I will try with my Nema 17 because it has a smaller internal resistance.

I wrote a Matlab script (sorry for the french in commentaries) to visualize the apparent wind vector in function of the true wind speed and the rotation speed. Indeed when the turbine spins, the wind appears from the blade's point of view with a different angle from the true wind. This will allow me to determine if the blades are in a position who creates lift or drag. Actually I don't have a way to measure the wind speed but I started to make an anemometer...

When the rotation speed is low, the optimal blade position can be found on the first image: Select the boat with the same moving direction relatively to the wind direction and you find the blade angle. The second image show the apparent speed for a wind speed of 1 m/s and a rotation speed of 1 rps. The third image shows the vectors for a wind speed of 1 m/s and a rotation speed of 3 rps (situation with tip speed ratio around one).

When the rotating speed is high, it is more complicated to select the good boat on the first image because we don't know the wind angle. All I can say is that the blade should stay between the "close reach" and the " close hauled" position. The fourth image illustrate this situation with the vectors for a wind speed of 1 m/s and a rotation speed of 5 rps.

In both case, there is a part of the trajectory where the blade is in "no sailing possible" which means that in this part, the blade creates only drag because they move upstream.

Matlab script

vvent=1; %vitesse du vent m/s
omega=[0 0 3*2*pi]; %vitesse angulaire de l'helice rad/s

r=0.05; % rayon de l'helice m

nbrPts=10;%nombes de points sur un tour

vecVent=[vvent*ones(1,nbrPts);zeros(1,nbrPts)];

angles=0 : 2*pi/nbrPts : 2*pi-pi/nbrPts;

d=r*[sin(angles) ; cos(angles) ; zeros(1,nbrPts)]; %vecteur allant du centre ? la pale

vapparent=cross(repmat(-omega,nbrPts,1)',d);%vecteur du vent apparent

scale=0.03;

quiver(d(1,:),d(2,:),(vapparent(1,:)+vecVent(1,:))*scale,(vapparent(2,:)+vecVent(2,:))*scale,0)%plot la resultante

hold on

axis equal

quiver(d(1,:),d(2,:),vapparent(1,:)*scale,vapparent(2,:)*scale,0)%plot les vecteurs vent apparent, le dernier parametre est l'echelle

%quiver(d(1,:),d(2,:),vecVent(1,:)*scale,vecVent(2,:)*scale,0)%plt les vecteurs vent reel

%quiver(zeros(1,nbrPts),zeros(1,nbrPts),d(1,:),d(2,:),0)

How the blades can create lift if they can freely pivot? The same way this women can windsurf without using her hands (cf. image 1): If the pivoting axis is at the lifting center, no torque is needed to keep the good angle.

Also there is a condition for good operation of the blades: They have to be balanced. When hold vertically, the blades should stay horizontally (cf. image 2). This means that when the turbine is spinning, the blades tend to stay tangent to their trajectory. I wanted the blades pivoting according to the wind angle and the centrifugal force bringing the blades back to their neutral (no drag) position.

Where is the lifting center? As for planes, the lift center is around one third of the profile chorde (25% says wikipedia).

We can observe on this slow-motion video that the blades behave as expected: In this video, the wind is stronger and no load is connected. The result is that the turbine spin faster and the blades pivot less (smaller angle). The centrifugal force is stronger so it brings back the blades to the neutral position with a stronger force. Finally we can tune this force (torque) by modifying the hinge length and hopefully obtain best attack angles all around the trajectory.

The ultimate dream would be to obtain a wind turbine that behaves like a slow type wind turbine (tip speed ratio < 1) when the wind speed is low and a fast type wind turbine when the wind is stronger (tip speed ratio > 1).