# Is this a viable mini wind turbine setup?

I'm going to build my first wind turbine. I have already constructed the prop, 5 blades 50cm long from PVC pipes. I've searched for a decent sized motor and charge controller. The turbine will be located in the city, so winds will hardly ever go above 10-12 m/s. The battery i'm planning on having permanently connected is a 12v lead-acid 6-7Ah

How is this for a setup?

I also have a boost/buck converter lying around (No current regulation) , would it be a good idea to connect that first, in order to boost the voltage and get less power loss through the wire and then connecting to the buck converter (with CC) as listed above.

Note: The buck converter has a really low minimum input voltage of 2.8v (max 32v input) so using this as a first step would make sure that the battery is charging at 13-14v even though the voltage from the motor is between 2.8v and 14v.

• How about a circuit diagram with details of the calculations you have done to design this (where appropriate). Jan 8, 2016 at 14:20
• The main problem with small homebuilt wind generators is that there is very little power available (due to little swept area) until wind speed becomes quite high (and with all wind generators, the power available goes as a cube function of the windspeed.) That, and mechanical realities (about the time the wind is fast enough to make some real power, the thing starts to fly apart.) otherpower.com is a good read (particularly the failures they learned from...) Jan 8, 2016 at 15:12

Check your available power and rotation rate using the formula on page 9 of Hugh Piggott's handy guide...

I recommend doing this in a spreadsheet for several different windspeeds.

(1) The blade design you chose determines your tip speed ratio. That gives you the rotation rate at any given windspeed.

(1a) Blade diameter and windspeed give the available power.

(2) The Kv (speed constant) of your chosen motor will then tell you the unloaded voltage at that speed.

(3) The available power divided by that voltage gives you the maximum current you can expect.

(4) Now multiply the motor resistance by that current : this voltage is lost as heat inside the motor. (NOTE: Kv and winding resistance will be specified for any motor worth buying.)

(5) Subtract the voltage loss (4) from the unloaded voltage (2) to get the expected output voltage.

(6) Divide output voltage (5) by output current (3) to give the ideal load resistance. Note that this resistance is different at different windspeeds. Too high a load resistance extracts less power than is available. Too low tries to extract too much, which will stall the blades.

In a battery charging application, the charge controller could use something like a MPPT (Maximum Power Point Tracking) algorithm to regulate the charging current, to find the best load impedance to place on the generator. Or simply adjust charging power to match the windspeed. I have no idea if your chosen charge controller does this, perhaps its datasheet can answer that.

(7) Multiply output voltage (5) by output current (3) to give the expected power output.

If power output (7) is bigger than the motor's rated power you need a bigger motor. If power output (7) is much smaller than the motor's rated power, beware : too big a DC motor may have too much friction to turn at all at low windspeeds, and will waste power at any windspeed.

You can eliminate brush friction (but not cogging or bearing friction) by using a BLDC model aircraft motor. This generates 3-phase AC so you need a rectifier to get DC. Having an oversized BLDC motor is not a problem and usually helps efficiency by reducing the winding resistance (step 4) - the remaining downsize is cost. If that's an issue, details on building your own here...

(You can eliminate cogging losses too, by building a custom ironless generator, which is what Hugh's book above is about).

Many more resources on Hugh Piggott's website http://www.scoraigwind.com/

Now to the actual question : is it viable?

Knowing the overall project cost, the likely generated output, and the price of getting that power elsewhere, you can work out the payback time :-)

But the real value (unless your city derives all its electric power from AA cells!) is learning and experience.

• Thank you for that very detailed answer. I feel like my project is however now leaning towards a divition waaay to high for my abilities :D I don't need the perfect setup, just one that works decently. Does adding a 36w generator to a 5-blade 1m diameter prop sound stupid? Would it be over, or underpowered? Using the formula for blade power (Blade power = 0.15 x Diameter2 x windspeed3) 0,15 x 1^2 x 10^3=150W (sounds like alot of power for that sized prop??) Jan 8, 2016 at 15:51
• forgot to tag you... @brian Jan 8, 2016 at 16:03
• I can't answer that. I don't know your blade design or TSR but something low (3 or 4) would be about right for 5 blades. So crunch the numbers and see if rotation speed works for your generator. And learn from that book. If 150W sounds a lot, learn NOW that you REALLY don't want to stop a 1m dia blade with your hand! At what rotation speed (ditto windspeed) is 36W (or 40 or 50W) available? If your goal is 36W would you be better off with shorter blades (and higher speeds) or using the current blade and getting 36W at lower speeds? Just design the furling system (see book) for that speed.
– user16324
Jan 8, 2016 at 17:19
• And in a city I'd guess 10m/s is a rarity. Got a cheap weather station?
– user16324
Jan 8, 2016 at 17:20