I want to design a small wind turbine that can be easily carried out of the classroom into the sports field in relatively low-wind conditions (5-15km/h) that can power a small ultra-bright 5V LED - just enough to show that it works.

How do I calculate the power I can get from different blade diameters in these wind conditions, and the power needed to drive a small DC motor/generator enough to light the LED?

  • $\begingroup$ Power ~~~= Z x 600 x A x Cd x (V/10)^3. watts | Cd = drag coeff - set to 1 to start. 0<Z<0.6 (Betz limit) is efficiency. Good DIY = 0.2. Likely ~= 0.1 lower too easy :-). V in m/s A = swept area in m^2. | At 2 m/s you'll typically have problems with "stiction" when starting. Also cogging from many alternators. It's easy to make a large disk with light blades that will spin up in low wind. $\endgroup$ – Russell McMahon May 28 '16 at 14:36

As it happens, I just recently went through that calculation myself for a different site.

Given the following facts from a quick web search, it isn't difficult to work out the numbers.

  • The maximum efficiency of a (large) windmill is about 40%.
  • The density of air is 1.225 kg/m3
  • You need about 50 mW (10 mA at 5V) to light up an LED

First, we'll need about 50 mW / 0.40 = 125 mW of air power flowing through the windmill to create the electricity we need (ignoring other factors such as the actual efficiency of a small windmill and the efficiency of the generator).

The power of the air flowing through the windmill is 0.5mv2, where m is the mass rate of the air flowing through the "disk" defined by the diameter of the blades. For example, suppose we have a disk of 0.03m2 (about 20cm in diameter). The mass rate of the air is the area of the disk multiplied by the air velocity, multiplied by the density of the air:

$$\text{Mass rate} = 0.03 \text {m}^2 \cdot v \cdot 1.225 \text{ kg/m}^3 = v \cdot 0.03675 \text{ kg/m}$$

The power of that air is therefore:

$$P = 0.5 \cdot \text{Mass rate} \cdot v^2$$

Substituting and solving for $v$:

$$v = \sqrt[3]{\frac{0.125 \text{ W}}{0.5 \cdot 0.03675 \text{ kg/m}}} = 1.9 \text{ m/s}$$

... or about 7 km/h.

Taking into account the efficiencies we ignored earlier, plus the losses in a gear train that might be needed to get the generator RPMs up to a usable level, I would probably shoot for about 4× the area, or about 2× the diameter (40-50 cm), in order to get reasonable results.

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  • $\begingroup$ Great answer, and I'm intrigued.Does this assume 3 or 4 blades? What sort of material should be used for a turbine that the poster intends to carry out to a football field? $\endgroup$ – KTM Jan 22 '15 at 22:16
  • $\begingroup$ @KTM: It doesn't assume anything. Such details are hidden in the wind turbine efficiency number I found online. $\endgroup$ – Dave Tweed Jan 22 '15 at 22:18
  • $\begingroup$ Woops, it looks like I missed the last sentence in the OP (and can see why you drew the line of scope where you did). $\endgroup$ – KTM Jan 22 '15 at 22:25
  • $\begingroup$ This is perfect, I was concerned that if I built an experiment and and it doesn't work then I wouldn't know if it's too small or something else was wrong. Now I have enough information to get started. $\endgroup$ – jhabbott Jan 22 '15 at 23:39

Try to cut a few from balsa wood to see what you get. Experimentation is probably the best method on such a small scale, and the math won't get you very far without an extensive knowledge of the materials (i.e. the cheaper the material, the less likely the blade will look like the big ones).

I worked summer activities at my university using materials from KidWind, and it's actually a ton of fun to build your own and experiment with it. We had always set up a wind tunnel and turbine stands/rotor that the kids would then design and attach blades to. Tons of fun, and you'll be amazed at what kinds of shapes actually work worth a darn. The activity even took less than an hour to complete.



A page with a photo of a tunnel (we modeled ours off of this): http://challenge.kidwind.org/?/national/tunnel

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