I'm watching some brushless outrunners with low kv as a generator for a simple wind turbine that I'm going to build soon. It might not be a scale or fully useful turbine, because I want to build it for testing and prototype purposes for now; I will test it with a fan(s) probably, however I want it to be as useful as possible. I have found the motors that are linked below:

1) Turnigy CA80-80 Brushless Outrunner

2) Turnigy Aerodrive SK3 - 6374

3) Turnigy RotoMax50

4) Turnigy RotoMax80

5) Turnigy RotoMax100

6) Turnigy RotoMax150

As you see there are lots of details rather than just kv and I`m not such educated to understand these all. Could you clarify:

(a) which of these is the best as a wind generator? why?

(b) which of these is the best for price? why?

(c) and do you recommend something else than of these?

If you ask me how much output I need, I don't know, I would be additionally thankful if you could clarify how much these things can produce, what is the optimal load, and other information you think is useful.

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    \$\begingroup\$ Some good information at scoraigwind.com including this set of plans, which also includes formulae for power prediction for different rpm, windspeed, blade size. scoraigwind.com/pirate%20oldies/… \$\endgroup\$ – Brian Drummond Jun 26 '15 at 21:13
  • \$\begingroup\$ Are you planning on using some type of gearbox? Otherwise I can't imagine the wind will spin these motors fast enough for you to pull anything significant from them unless you live on an advanced research base in Antarctica. \$\endgroup\$ – Matt Jun 26 '15 at 21:38
  • \$\begingroup\$ Hi Matt, I havent planned any gearbox so far, I just want to know if these types of generators will be good to start with. Because the instruction prepared by windgenkit recommends to use pitsco motor 500 which is even worse than the ones I listed above. As much as I understand, you see the listed motors as not efficient. Which motor do you recommend then? I plan the diameter of the blade area 0.7-1 meter. \$\endgroup\$ – Hasan A. Jun 27 '15 at 8:29
  • \$\begingroup\$ @MattAnderson's answer did not mean they were inefficient when used as intended but that the RPM they are designed to work at is >> RPM at usual wind speeds. eg a 1m dia blade has a ~~ 3m circumference. At wind speed of 5 m/s and a TSR of 5:1 that gives 25m/s tip speed so RPS = 25/3 ~+ 8 RPS ~= 500 RPM. Thos nice but VERY dear considering motors start at 150 kV = 150 RPM/Volt so at 500 RPM you'd get about 500/150 ~= 3.3V. That's for motors designed for 40-60V range operation so overall not very suitable at all. Alas. | Treadmill motors can be good to start playing with. \$\endgroup\$ – Russell McMahon Jul 1 '15 at 3:25
  • \$\begingroup\$ This is a good question. The people who closed it do not realise that it is a design question. I suggest you consider changing the wording to show that you are looking at what motor characteristics are suitable and why with some examples for people to comment or wrt relevance. \$\endgroup\$ – Russell McMahon Jul 1 '15 at 3:28

Quoting from Hugh Piggott

Blade power = 0.15 x Diameter^2 x windspeed^3
= 0.15 x (2.4 metres)^2 x (10 metres/second)^3
= 0.15 x 6 x 1000 = 900 watts approx. (2.4m diameter rotor at 10 metres/sec or 22 mph)

Plugging in your numbers :

(1) 0.7m diameter
= 0.15 x (0.7 metres)^2 x (10 metres/second)^3
= 0.15 x 0.5 x 1000 = 75 watts approx.

(2) m diameter
= 0.15 x (1 metres)^2 x (10 metres/second)^3
= 0.15 x 1 x 1000 = 150 watts approx.

This is the power available from the wind. If you work the same out for 11mph (5m/s) you should find 1/8 of the power is available, or just shy of 10 and 20W. You haven't told us what your local wind speed profile is, so you'll have to revisit this calculation yourself.

Now before we can work out how to extract some of that power, we need to know how fast the blades spin.

Rpm = windspeed x tsr x 60/circumference
=3 x 7 x 60 /(2.4 x 3.14)= 167 rpm

Assuming you follow his recommendation of a tip speed ratio of 7, and you aren't interested in speeds below 5 m/s (11mph) where there's less than 10W available:

(1) 0.7m diameter = 5 * 7 * 60/(0.7*Pi) = 954 rpm

(2) 1m diameter = 5 * 7 * 60/(1.0*Pi) = 668 rpm and these speeds potentially double at 11m/s.

Now we can look at one of your motors: the CA-80-80, first on your list

This has Kv=160rpm/V and a motor resistance of 0.011 ohms.

Kv=160 means that, driven directly, it should generate 954/160 = 6V (AC) from the 0.7m rotor.

As Kv is defined in terms of the driving DC voltage, this may turn out to be the peak AC voltage generated, rather than the RMS voltage. Which you get isn't clear from the motor specs. If so, you'll get just shy of 5V after the rectifier, but if that 6V is the RMS voltage, the AC peak is 8.3V and you'll see about 7V after rectification.

From the 1m rotor you'll get 668/160 = 4.1V (AC) at 5m/s (same considerations apply).

Now 10W at 6V means you can extract 1.66A, by setting the load resistance to 3.6 ohms. Taking more current than that will simply stall the blades. Alternatively, 20W at 4V means you can extract 5A with an 0.8 ohm load.

You'll lose some power in the motor's own resistance : at 5A, you'll lose I^2*R = 25*0.011 = 0.275W (out of 20W : negligible).

For interest, let's see how the 1m rotor performs at 10m/s : speed will be 1336rpm, voltage 8.35V. Power available is 150W, so current = P/V = 18A. So to extract full power you need to tune the load resistance to 8.35/18 = 0.46 ohms and you'll lose 3.5W in the motor resistance.

Tuning the load resistance to best extract the power available is outside the scope of this answer : it would usually be done by a switching converter such as an intelligent battery controller, like the "MPPT" chargers in solar power systems. But to demonstrate power generation you can simply switch power resistors in and out of circuit and measure voltage and current.

Clearly this motor will work, quite efficiently, at extracting the power available from any reasonable windspeed with the rotor sizes you suggest. Equally clearly, at £99.53 it's an outrageously expensive way of generating 10-150W. As a motor it's rated easily in excess of 30V and 100A so this is barely 5% of its rating.

But now you can repeat this exercise with the key parameters for the other motors and see if another fits your definition of "best" or best value.

(One point about "alternators" from Hugh Piggot's writings : he reminds us that because they aren't permanent magnet based, they require power to generate the magnetic field. About 40W in the case of car alternators, which makes them less attractive for smaller wind turbine applications)

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  • \$\begingroup\$ It was very helpful. It appears these motors are expensive and their ratings are too low for such activity. I might be thinking about a mini axial flux for this purpose. Thanks. \$\endgroup\$ – Hasan A. Jun 27 '15 at 14:50
  • \$\begingroup\$ Probably I should have said "pull anything significant compared to the motor capability". My area of knowledge is the motors/generators themselves, I was amazed that you could even get speeds that fast from wind. Good answer \$\endgroup\$ – Matt Jul 1 '15 at 13:36

The "best" is the one that can produce the most watts continuously. Watts = volts * amps, and none of the specifications list this under loaded conditions, so it is impossible to determine. Perhaps you could try a smaller one as an initial test, say from eBay. Take some measurements, make some calculations, and then you'll have an idea how a bigger generator will perform. But the bottom line is watts - if you want to use this generator to power something, it takes a certain number of watts averaged over the course of so many hours (watt-hours.) So the generator needs to replenish more than that many watt-hours over the course of the day, to keep the batteries charged.

It is worth noting that not all motors will work as generators. Permanent-magnet motors will work, but "induction" and "brushless" may not. A modern alternator (such as is used in automobiles) are designed specifically for this purpose, but they are more difficult to use. Instead of having a permanent magnet, they have an electromagnet. So they need some power initially applied to this "field winding" to start generating electricity. And this is nowadays usually done from the vehicle's ECU computer, which introduces way too much complexity. But is it possible? Sure. A typical passenger car alternator can deliver 50-70A at 14v, or 700-1000 watts.

One of the advantages of alternators is the lack of a permanent magnet. Permanent magnets will lose all of their magnetism (permanently) if they get too hot. Alternators have a temperature limit also, but it may be higher. Some of those aircraft motors do have higher wattage ratings, so might work better, but have permanent magnets, so their temperature will have to be monitored.

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  • \$\begingroup\$ Thanks for answer rdtsc. So which generator on ebay do you recommend to start with? \$\endgroup\$ – Hasan A. Jun 27 '15 at 8:31

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