Here's the volts from the test coil

[elt]

I've read a lot while waiting for stuff to get here to put together; what I write is based on my (perhaps faultly) interpretation of what I've read so I'll definately appreciate any feedback very much!

I've been trying to follow the advice of "build the alternator, then size the blades to match." Still, I'm shooting for something like a 10 footer, 24 volt system. I think my test coil says that I should add a few turns but I'm not sure; when I "run the math" it doesn't look that bad but I don't have any feel for how much heat a rotor will withstand, whether blades stall or run away or what!

Here's what I have so far:

 2 rotors with 12 N42 2x1x.5" neos each

 7/8" air gap

test coil 60 turns, two in hand #15 wire, (2x 45 feet)

legs: .75 inch wide, .58 inch thick

60. rpm 1.04 vac rms
    
120. rpm 2.1 vac rms
     
     

Using the 60 rpm voltage to calculate cut-in rpm:

1.04(v rms) * 1.414 (for peak volts) * 3 (coils) * 1.7 (for star connection) = 7.5 v

rpm for 25 volts (24 volts - 1 for diode drop) 60rpm*25v/7.5v = 200 rpm

Flux estimated that 65 turns would have given a cut in at 180 rpm; scaling what I measure (with  60 turns) that looks right on the money...

With a 200 rpm cut in, I get 10 mph winds with a 10 foot TSR 7 blade. I read Flux say that blades behave like they have a higher TSR at cut in though I don't know what that effect is. (So ignoring that effect a 9 foot blade would cut in at about 9 mph... also, I could make the inside side of the coil a little smaller and get some more turns in.

Estimating power Flux'es technique in this thread - http://www.fieldlines.com/story/2006/1/9/205941/7206

(assuming 10 foot tsr 7 blade) if it makes 25 volts at 10 mph, then it will make 63 volts at 25 mph. I can't measure the resistance of the test coil but the wire table says .073 ohms, in star connection with 3 coils per phase and two phases active, that's 0.4374 ohms. Flux suggest of fudge factor of 1.3 so I use .57 ohms. Subtract the battery volts from alternator volts 63-25 = 38 volts; that divided by the resistance is the amps going into the battery = 38/.57 =  67 amps... times 24 volts = 1600 watts into the battery...

that's neearly 100% of Alton Moore's "power at blade" for a 10 footer, 68% efficiency according to the windpower spread sheet and about 80% of what's predicted on the windgenzen webpage so (once again!) I'm very confused by some of the stuff I've read...

My rectifier is rated at 800 volts at 110 amps at 100C, (150 amps at 60C) so if I can keep that cool, I don't see that burning up at 25 mph so I guess its a question of how much power the stator will shed without melting.

Is this right - the volts not going into the battery, 38 volts, times 67 amps = 2.5KW of heat? My gut says that's a lot but I've assumed that I would attempt to fold earlier than that to help insure that the mill was really, really furled by that time the winds got that high.

I appreciate any advice I get!

 - Ed.

[elt]

Okay, that was a rambling post; I fumble for understanding! The guess that questions are the same as usual:

1. how many turns,
   
2. what size, TSR blade
   
3. when to furl.
   
   

What I didn't write was that my winds are "pretty good" and I'm more interested in the mill being "robust" than reaping the lowest speed winds. (I realize that wind maps don't say what wind I have here but that's all I have to go on. Average speeds are listed as 15 mph in winter and spring; I calculate 12.5 mph average at blade height.)

 - Ed.

[SamoaPower]

What you're seeing with the numbers is the usual mismatch between the air rotor and the alternator. I wouldn't be too concerned because the rotor will be stalled long before you reach 25 mph.

A 10' diameter swept area has wind power of about 6.3kW at 25mph. With a rotor efficiency of 30% (typical), it will deliver about 1.9kW to the alternator. So, the 4.1kW (1.6+2.5) that your numbers predict won't happen.

If your alternator was 50% efficient at the stall speed, you would have about 950W to the load and in the stator.  I would not feel comfortable with that much in the stator for extended periods.

You may want to consider some means of matching in addition to furling.

[Flux]

Samoa has answered most of your issues.

You will be well stalled at 25mph, that is why the blade calculator figures don't tie up.

If you furl at 25 mph you should be ok. You could most likely do better by adding resistance in the lines to let the alternator volts rise instead of being clamped to battery volts.

Despite the loss you are introducing you will run the prop above stall and actually gain power in the lower winds. Stator heating will decide on the final power output and I would furl at a speed that keeps the peak power into your battery at about 1kW.

It is better to have losses in series resistance ( especially if you can use the heat) rather than have them in the stator. You can have the same battery current for the same heating in lower winds.

Without going to a power matching scheme you have probably got the best compromise.

If you want more power with less stator heating it would be possible to reduce turns and use a boost converter for low winds.

If you have enough power in normal winds you may want to furl earlier rather than risk frying your stator when the batteries are already charged.

If you make it furl at 1kW peak it will most likely reduce output when it goes hard into furl. The occasional peaks will not cause you a problem, but I wouldn't want to see a continuous power out of 1kW for minutes at a time in high wind, even though it is well within what others are doing.

Flux

[elt]

Thanks Samoa, Flux,

I can run the numbers for amps, winding, volts, etc. but when it gets to stalling a furling, I'm totally in the dark.

Flux wrote:

I wouldn't want to see a continuous power out of 1kW for minutes at a time in high wind,

What do you recommend that the max continuous power be limited to?

All I've found in searching is DanB's experiment, (http://www.fieldlines.com/story/2006/3/17/19253/1723). He's smoking single coils with 400 watts. I assume that would be equivalent to 2400 watts in a 9 coil (six at a time) star configuration. The story has some nice info on resins and fillers but I couldn't find info on that max current/power that doesn't smoke coils ...

Once I have a notion of what power limit to set, then I run the numbers and use the furling calculator to get a first pass at the tail geometry.

Also, I don't understand what value to use for the resistors in series. Hugh's booklet suggests using 300 watt, .5 ohm heating resistors in series if the (24 volt) mill stalls. I can do something like that on faith but I wish I understood the reason choosing .5 ohm as opposed to some other value... perhaps .6 ohm (or .4 ohm) would be better but I have no understanding of it to help me know To tell the truth, I'm not even sure what a "heating resistor" is. Is that nichrome wire or a power resistor or something else. Here, heat is good. I was even thinking of using a ceramic heater on an inverter (with a remote-able switch) as the dump load.

If you want more power with less stator heating it would be possible to reduce turns and use a boost converter for low winds.

I've given some thought to that. I could see a mill making lower volts and then boosting the volts for the battery. (I even use a boost converter in one of my telescope controllers but I designed that from app notes, not understanding.) One reason I was targeting a 24 volt system, though, was to reduce the amps into the rectifier. More power and lower volts would mean more rectifier amps, wouldn't it? Maybe I'm being over conservative, but a blown rectifier was blamed for at least one mill blowing apart and I thought that that was a risk that could be minimized.

...

All in all, adding series resistance seems the most straight-forward thing to do but that still leaves me with a few x's without values: mostly what resistance to use and what furl speed (or max power) to shoot for.

Thanks again,

- Ed.

[Flux]

"Flux wrote:

I wouldn't want to see a continuous power out of 1kW for minutes at a time in high wind,

What do you recommend that the max continuous power be limited to?"

I doubt that anyone has done tests that would answer this. Dan's tests were not in the wind with extra cooling.

Also the time at maximum power is rather uncertain. I have no idea how long a machine in clean air free of turbulence stays at full power. With low towers and turbulence here, even in a significant gale the power is below maximum furling for significant periods and this helps cooling.

In some wind areas things could be much worse. Anyone using pitch control would likely need to de-rate the alternator.

My suspicion is that if your furling prevents you going over 1kW you are likely to be running cooler than most. The failure rate depends on time and temperature, the hotter you run the shorter the life. If you can keep the centre of the coil below 160C it will likely last for many years.

The line resistance you need is not easy to determine, You will probably need to choose a suitable value when you have it running. It depends on alternator resistance, cable resistance, prop tsr, efficiency stall, characteristics and how fast and noisy you want to run.

The ideal value will put you at the top of the tsr curve at furling, too much will take you over the top and will reduce heat but also reduce output and add excessive speed and noise.

The snag with line resistors is that they will be very low resistance, fine if you want air heating but awkward if you want to pre-heat water.

Any resistors will do as long as they don't burn out or cause fires. Wire wound glass covered industrial resistors are excellent, but pieces of iron wire will do if you limit surface temperature.

If you don't actually want the heat then the cheapest method is to use thinner lines to the rectifier and save cost. For water heating you may have to pay a high price to a specialist manufacturer or make your own.

You have missed the point with the boost converter.

At the higher power you will not be operating at lower volts, the alternator will run faster to track the prop power but you will still generate 24v, the only increased loading on the rectifier is from extra power you can gain from better matching and higher efficiency.

Below about 12 mph your alternator will now be too fast and that is the time you use the boost converter. At cut in you will be boosting from perhaps 10V and the converter should phase back to let the prop track wind speed so that at about 12 mph you will be up to the speed where the main rectifier can take over. The total power in the boost region will not be much over 200W for your size of machine.

Flux

[elt]

The line resistance you need is not easy to determine, You will probably need to choose a suitable value when you have it running. It depends on [... just about everything else.]

Okay. I priced 300 watt variable resistors in the .5 ohm range at about $16 USD new; not bad but I've got time to hope some come up on ebay for less. (Or, as I understand, even though my mill-to-rectifier distance is short, I can also have a pile of feed wire hanging on the wall.)

You have missed the point with the boost converter.

I'm not surprised at all! I had two guesses and your reply wasn't either one of them

I am very interested in this.

What I thought was my best guess was that the mill would put out battery volts or a little more at furling speed and otherwise use the boost converter at lower speeds. (The 80 to 90% efficiency of the converter beat the heat losses that would have been in the stator.) The snag with that (which I read from googling the site) was something about inductors only being good for a 10 to 1 power range. Dreaming of high power at the high end, it didn't seem at first that the inductor would offer much at the lower end. But, as I understanding it, doubling the wind speed is a 9x increase in power so if I was furling at 20 mph, then half the speed and 1/9 of the power at 10 mph sounds like about what my original cut-in speed was anyway and would be in the 10::1 power range of an inductor...

If I understand your reply correctly (and I wouldn't be surprised if I don't), you're only recommending that the boost circuit operate in much lower winds. If that's the case, do you switch it in and out with relays, use something like barrier diodes, or would the higher voltage and current pass through?

I'm certainly not capable of designing anything like that, though I did incorporate a 90 watt boost circuit (based on a max641) in my telescope controller by mostly following app notes so if I knew what to build I wouldn't be afraid to try.

- Ed.

[Flux]

I am not sure where this 9:1 power ratio about inductors came from, certainly not from me.

If you can find the article I did about matching the load I showed a method of using the boost converter in low winds without it having to take the high current in higher winds, by using auxiliary diodes on the main rectifier.

It was rather basic and intended as a starting point but was a workable scheme.

I left it rather basic as I think only serious electronic experimenters would be prepared to attempt it.

To provide something more complete would require detailed circuit layouts as they are quite critical. I had hoped that it would provoke someone with detailed knowledge of inverter layout to develop it a stage further.

At the time I suspected that air gap alternators did not have enough inductance to boost directly by chopping the ac current. I have since found that they have more than enough and it opens up another approach that may achieve even higher convreter efficiency. I did give the basic circuit for motor conversion people to try.

I did a fair bit of experimenting a few months ago but then it got forgotten to concentrate on other things. Perhaps it is time to have another look at it.

The main snag as I saw it then was that it needed a mixed diode and mosfet bridge to carry the whole output current, something that from data sheets presents no problem but I just don't have any faith in things in a TO220 package to take the total output of a 1kW 24v mill.

Again it would be possible to use a main bridge and let that take over in higher winds but it would need introducing extra volt drops in the converter bridge to let the main bridge take over. I am suspecting that a little bit of resistance would do the same thing with less low wind loss.

If you are prepared to risk using one bridge the circuit becomes quite simple and is significantly less critical on layout.

If you are in a good wind area and are worried about stator heating then it is seriously worth a try. If all your efforts with the converter failed you would have excellent results in higher wind.

A compromise winding with a bit lower cut in speed than I would choose may give a reasonable compromise that would give quite good results without the boost and still give you less stator heating and if you could manage the converter then you would gain the better low wind performance.

I am reluctant to encourage you down this route without a fair bit of electronic experience, but it's about time someone tried it. At the very worst it may cost you a stator if you went for the best compromise and couldn't build the converter.

Flux

[elt]

Okay, let me test for understanding...

Samoa wrote:

If your alternator was 50% efficient at the stall speed, you would have about 950W to the load and in the stator.

This suggests to me that if the cut in speed were 10 mph (that I calculated from the coil test) then the stall speed would be around 20 mph.

There's concern about the stator disapating 1KW so flux suggest one option is to raise the cut in speed to, perhaps, 12 mph and use a boost converter at lower speeds.

If I extrapolate from my current numbers:

A coil leg with a (fudge-factored) .57 ohms gave me 25 volts at 10 mph so 10/12 of that would be .47 ohms of coils for a cut in at 12 mph. The excess voltage at 20 mph would be 20/12*25 - 25 = 16.7 volts divided by .47 volts would be 35 amps. Times 24 volts into the battery gives 850 watts into the battery. 16.7 excess volts times 35 amps = 580 watts in stator heat... about 100 watts less into the battery but 400 watts less stator heat, that seems like a very good trade off and the stall speed will be raised (that's a question!) so there's more power to be reaped if I want to furl at a slightly higher windspeed.

I think I have a superficial understanding of the boost circuit in the "matching the load" thread and I'm not scared to give it a try with help... First question I have is that I'm not sure of how many watts the the boost circuit has to produce. Assuming that the power at 12 mph is about 12^3/20^3 = 21% of the power at 20 mph, is it 21% of 850 watts or 21% of 850+550 = 1400 watts? ... 180 watts or 295 watts?

Thanks!

- Ed.

[Flux]

As you are thinking about a 10ft machine I can give you some figures.

Designing as most do for low wind speed cut in, you would choose to cut in at about 170 rpm. With an alternator matched to just clear bad stall it would be down to 50% at 20 mph and half the power would be heating the stator.

This machine of mine has to work with line resistance that I had no control over so I choose a cut in of 220 rpm. The alternator with no line resistance produces 1kW at 330 rpm. This is low and if there was no line resistance I would have raised cut in to about 250 rpm.

As it is, it cuts in below 10 mph but is not really well up on the prop curve until 10 mph. If I had used the higher cut in it would have started about 10mph and been reasonable by 12 mph.

With the converter it starts to produce at about 6 mph and the power in low winds is steady. Without the converter it is lumpy and not consistent and the total enrgy capture will be considerably less. The converter phases out completely at about 10A (24v nominal). It doesn't make much difference if the converter comes off a bit earlier or later, the prop accommodates a bit of speed change so assume the converter supplies a max of 250W

If I had no line resistance and chose the higher cut in speed (250) the converter would possibly need to hold to 300W for best results but if it came off at 250 it wouldn't matter much.

By going from 170 to 250 rpm I reduce turns in the ratio of 170/250. Resistance is reduced much more than this as you gain from less turns and from having more room for thicker wire. I probably get 1/3 of the resistance with an overall efficiency of about 75%.

You are a bit short of magnet for a seriously higher output from a 10ft machine and to benefit you will need very low line resistance to be able to raise your cut in to 250rpm but you may be able to do it. Certainly you can come up to about 220.

You should get the 1kW without having to think about stator heating and if you are willing to push things as hard as others do then think 2kW.

My machine dissipates heat better than you can hope for, it is not an air gap axial, but a slotless radial drum so I would suggest you not be greedy and limit to 1kW with peaks up to perhaps 1.5 and stay safe.

You will reach your 1kW at lower wind speed than if you matched by stall and resistance.

Hope that helps you make a choice.

Flux

[elt]

to benefit you will need very low line resistance to be able to raise your cut in to 250rpm but you may be able to do it. Certainly you can come up to about 220.

I guess one of the benefits of a short tower is a short lead in; I estimate 40 feet from mill to batteries.

I do have a spool of #15 that I'm going to have use. Right now I'm getting 200 rpm cut in with 60 turns two in hand. I could reduce turns to get up to 220 rpm cut in but I couldn't fit in more wire with a 220 rpm cut in. That would let me make a little bit thinner coil and perhaps close my air gap some. (Currently .6" thick coil and 7/8" air gap)

I figure that I can get a 250 rpm cut in with 48 turns three in hand and that I can fit that in with two more winding layers on the coil; I'd have to pinch in the bottom of the coil to about .8" (1"x2" wide magnets) and make the coil one layer thicker.

I'm not sure which way to go ... perhaps because I'm not looking to get more (high rpm) power out of the alternator, just less heat, then just using few turns of #15 two in hand would be adequate. (But then, if that lets me close up the air gap a little bit, wouldn't that give a little more power anyway?)

...

What value for the inductor did you use in the boost circuit? I think I'm on the low end of "adequate" designing digital circuits but have no real experience with analog... I did look at a lot of app notes last night and made a gut-feeling approximation of 5 to 6 uh for 12 to 27 volts at 30KHz ... did I come close? I got bags of schottky diodes, power mosfets and 5 amp 22uh ferite core coils for my current 80 watt boost converter, was thinking that I could parallel them to get get the inductance down and current handling up for a prototype.

 - Ed.

[Flux]

If you only have #15 wire, I think you will really struggle with 3 in hand.

My thought would be to stick with 2 in hand, go for 48 turns and reduce the thickness a bit to keep the coils about touching. Thinner will dissipate heat better.

Reducing the air gap a little will make it about equal to 50 turns now and give a cut in of near 230 rpm. you should have close on 60% efficiency and about 650W in the stator for 1kW out.

It should just run clear of stall.

I didn't measure the inductor of the converter at the time, it was well more than the minimum and I suspect it was about 500uH. I can make something similar and measure it.

I will try to give you the core dimensions in the hope that you can find something similar. It was an old core design, very similar to the early valve line output transformers used in TVs in this country.

The wire was three in hand .071mm, slightly twisted together and about 20 turns.

The core had a whopping big air gap in the centre limb (from an industrial inverter).

The inductance was very adequate as I seem to remember trying it with one half core off.

Take a look at the new stuff in my diary. I will post the control circuit using a hall sensor, it's easier to use than the shunt version I gave before.

Flux

[elt]

stick with 2 in hand, go for 48 turns [...] Reducing the air gap a little will make it about equal to 50 turns now and give a cut in of near 230 rpm

Will do, thanks!

Re. the 200/300 watt booster inductor:

The wire was three in hand .071mm, slightly twisted together and about 20 turns.

Did you mean ".71mm" ? .071mm sounds very thin to me, will 3x .071mm carry 8 or 10 amps?

Take a look at the new stuff in my diary. I will post the control circuit using a hall sensor, it's easier to use than the shunt version I gave before.

I've been looking at what's there so far and can see all kinds of advantages including the ability for it to evolve into a full-fledged charge controller.

...I have zero experience at windpower system design but one thing I liked about making a full power circuit was the ability to make a buck converter. I liked the 24 volt alternator for the reduced line loss and amps in the rectifier but didn't/don't like that 24 volt inverters are (mostly) much more expensive than what you can get for 12 volt systems...

Personally, I'm more comfortable with the low speed (only) booster. As you point out in your "load matching" diary with the full power version, reasonably priced components may have marginal power dissipation at 1kw and if I combine that idea with the idea that "stuff happens", then I think that any system that I might be able to afford to build would be more likely to crash and burn...

Thanks again,

- Ed.

[Flux]

Yes 0.71, sorry about that, brain works faster than fingers, and I am sure some of it is the software. I invariably get i instead of I unless I do it dead slowly.

Flux