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Matching the load
By Flux, Section Diaries Posted on Fri Mar 17, 2006 at 06:56:46 PM MST
Variable speed fixed battery voltage.

There has been a load of discussion here lately more or less on the same thing but looked at from different angles. I thought it may be time to look at the basic problem then we may be in a position to look for solutions.
Man has been charging batteries with wind power for about a century so what is all the fuss about. What has changed to suddenly make this a big issue.
Almost certainly the thing that has raised the issue is the coming of neo magnets.
Let's start from scratch. We have a battery which is more or less a constant voltage sink, we put current into it and the volts stay the same. Not quite true but near enough for now.
We have a generator (or alternator and rectifier) consisting of a coil rotating in a magnetic field which produces a voltage. Somehow we make it dc to suit the battery with a commutator or rectifier( same thing).
If the magnetic field is constant and the winding has no resistance, current will flow into the battery when the generator voltage exceeds the battery voltage. The only thing trying to stop the current is the internal resistance of the battery. this is very low and let's ignore it.
If we increase the speed of the generator the voltage will rise and as there is no resistance anywhere the current will try to rise to infinity. The thing is destined to run at a fixed speed.
If we now add some winding resistance the current will cause a volt drop across this resistance and the current will be determined by how far the generator internal voltage( emf ) is above the battery voltage and by the resistance.
If we add lots of resistance we get a small current at cut in and the current will rise with speed. Part of the energy will go into the battery as something useful and the remainder will be lost as heat in the resistance.
There is no problem as long as we accept that.
The problem first came to light when people added dynamos to cars to charge batteries. Stationary lighting plant used constant speed engines and there was no problem. Now the car needs a variable speed engine and trouble started. If you made it charge at low speed the dynamo or the battery fried at high speed. If you made it work at high speed the battery went flat when travelling at low speed.
This was originally solved by some rather crude methods mostly based on slipping clutches or belts, a solution destined to failure.
The final solution was to change the magnetic field. As the speed rises, if we reduce the field strength the emf falls and the circuit resistance keeps current within limits.
What did we do with windmills? much the same really with one difference. The car has a big engine with unlimited power as far as the charging circuit is concerned and it runs at high speed.
The windmill runs at low speed and the big problem was to make a dynamo that had a low operating speed. The other method was to use a gearbox and regard it as a car with power dependent on speed.
Large low speed dynamos presented many challenges with only electromagnets to supply the field. To keep field consumption within limits the air gaps had to be short and flux was limited. This required large numbers of turns on the armature with a fair bit of resistance. The secondary effect was that the magnetic field of the armature reacted on and weakened the field with increase in output current.
The problem was now the reverse of our trouble, it required a massive and expensive generator to produce enough output to anyway near load the prop, we never met stall, the thing always ran away. Output was more limited by field weakening than circuit resistance and things tended not to burn out but speed ran away.
Magnet steels were available in place of wound fields but their energy density was low and they had the same problems as the wound field machine. Because of the large number of turns, small air gaps and the magnet characteristics their output was limited by field weakening ( armature reaction) and another characteristic caused by large numbers of turns. This is leakage reactance and behaves as if there is an inductor in series with the coil. The inductive reactance of a coil increases with frequency and the machine frequency increases with speed. These machines became reactance limited and the output finally became limited at constant current and stopped rising with speed. The problem was much worse than the wound field and props ran away in higher winds.
Ceramic magnets were significantly better but machines using them were generally reactance limited. This type of curve is totally wrong for wind power having a steep rise at low speed and bending over at high speed, but it gave us the extra watts wasted by the wound field and was such an improvement in light winds that few considered there characteristics a problem.
Then came Samarium cobalt, much better but too costly for normal use so it caused no problem to the wind power people.
Then modern science gave us the perfect answer to our problems with neo and solved the low speed issue. I think this is enough for now, time to take a break before we consider the present snag.
Flux

Matching the load | 63 comments (63 topical)

Re: Matching the load (3.00 / 0) (#1)
by Titantornado on Fri Mar 17, 2006 at 03:23:13 PM MST

OOh OOh!!!!! Cliff-hanger! ;-)

Rod

Re: Matching the load (3.00 / 0) (#2)
by electrondady1 on Fri Mar 17, 2006 at 03:58:41 PM MST

b.n. and a.n. the new age starts in chapter two !!

[ Parent ]

Re: Matching the load (3.00 / 0) (#3)
by Gary D on Fri Mar 17, 2006 at 07:22:54 PM MST

Yes Flux, if you look under the rants section, you'll see how much knowledge I don't have! I'd love to try understanding what you are willing to share. I'm a slow learner though, hope not to irritate you with stupid questions. Gary D.

Re: Matching the load (3.00 / 0) (#4)
by Flux on Sat Mar 18, 2006 at 02:18:47 AM MST

History lesson over but I thought it necessary to look at what has gone before.

We now have the biggest leap in magnet development since the introduction of Cobalt in about 1920.

We non have the capability of producing fields of strength equal to or greater than the electromagnets in much smaller spaces and with no consumption of power. Low speed without enormous size is a reality, winding turns are reduced and this combined with the effectively longer air gaps, lack of iron poles and the high field strength needed to demagnetise neo means that we can build alternators that do not have a curve that tails off with load.

Another feature of neo is that it is possible to still produce good flux density in substantial air gaps and this re-opened the possibility of reverting back to the air gap designs of Ferranti and Siemens, abandoned over 100 years ago.

For normal alternator design the things are designed to have maximum efficiency where they normally work ( near full load) and using iron cores has become the standard design. For us the iron core has certain features that make it less desirable. All iron has losses, both eddy current and hysteresis, the eddies can be reduced to low proportions but hysteresis is a loss at molecular level and increases with field strength and we cant avoid it.

To make the best use of iron cores the coils are wound in slots and this causes us more problems. This is the source of the cogging problem, there is also a loss from the sudden change of flux from tooth to tooth.

Now that neo can produce significant fields in practical gaps there is a real case for not using iron in wind generators. There can be no iron loss and cogging does not happen.

Now the elimination of the field supply, removal of iron loss and the absence of cogging and the ability of a prop to start in winds below 5 mph gives us a chance to think about extracting power from the low wind speeds.

With wound fields there was no real chance of supplying the field loss below 12 mph.

Now we have the perfect solution, why do we have a problem? Because we are greedy, the more we get, the more we want.

In the past we couldn't do much with low winds so we wound the alternator to work from about 12 mph upwards. We were on the part of the cube law curve that was not far removed from a straight line over the range we used ( say from 12 to 28 mph).

The perfect alternator has no slope, output rises vertically above cut in speed, but we can't make that and with about the best we can do the slope of a practical machine is a good fit to the wind power curve and man was happy.

Now the greedy one comes along and decides to harvest the small amount of power now available at say 6 mph, not a lot but useful, and by about 8 mph it is significant,

He winds his alternator to cut in at 6 mph and still tries for a high efficiency, now what happens.

At this point we really need curves but as Yet I haven't produced any but I think you can visualise a curve like a parabola but exaggerated. From 5 to 10 mph the curve has a very small slope, then between 10 to 15 mph the slope becomes about twice as steep and beyond about 15 mph it rises sharply. At low winds it is near horizontal but at the other end it is near vertical.

The efficient alternator has a steep slope and if we fit this at the 5 mph point it will be parallel to the curve at say 25 mph. Now we are in trouble, by possibly 8 mph our alternator will be producing the sort of power that the blades would like to make at say 15 mph, we do fairly well because the prop has some latitude in operating speed but we are pushing our luck at 8 mph and by the time we reach 12 mph the prop has given us all the slack it can offer and is now dragged down to a tsr where it is working well below peak power. We have hit stall and with wind speeds above this the prop falls so far off the cube law that we see no significant increase in power

To take an example, an 8ft prop could produce close to theoretical power from 6 to 10 mph and then level off to a maximum of say 150W for winds up to 25 mph.

What has our greed for low end power cost us? Say that was a 12v mill and we now connect it to a 24v battery. Cut in will now be raised to roughly 12 mph and we are up on the steep part of the power curve and the alternator line is now much more on the power curve ( with the same slope but along the power curve not just parallel to it).

We may pick up a bit of power at about 10 mph and by 12 we are coming on song. At 28 mph with the same prop and alternator we should be in the 1kW region.

That is the issue, be greedy at the bottom, pay the price at the top. That is why I keep trying to stop people going for desperately low cut in. For normal wind areas you suffer a serious price if you aim to cut in below about 8 mph. You will not often see the 1kW winds but if you kill your performance in the 15 mph range you will come off badly.

Can we be greedy and have it at both ends? yes but it doesn't come easy. That is what this is all about.

Time for a break and perhaps a few curves.
Flux

[ Parent ]

Re: Matching the load (3.00 / 0) (#5)
by Flux on Sat Mar 18, 2006 at 05:36:25 AM MST

Here are the curves for a 6 ft machine. Machine input output and prop output at best tsr.

With axial machines the output curve is linear and the input includes the losses and curves upwards.

Points to the left of the prop power curve are towards stall ones to the right are towards over speed.

You can see that the right hand track is good and even at 10 mph the prop will speed up to compensate to a large extent.

With the other case we go into stall from the beginning. and you don't get the alternator output suggested by its output curve because there is no power from the prop to supply the input

[ Parent ]

Re: Matching the load (3.00 / 0) (#6)
by Flux on Sat Mar 18, 2006 at 08:00:25 AM MST

If we try to make the alternator input fit along the low wind part of the power curve you can see that the output will be very low in high winds because the alternator is not stiff enough to load the prop and to do this we have made the efficiency very low.

What is normally done is to make a compromise, choose the cut in for the best prop tsr at say 10 mph, the prop will run fast enough to allow cut in at about 7 mph. We make the alternator less efficient so that it has a lower slope and we ask the prop to run on the edge of stall in mid winds. I high winds it will be less stalled or even near the peak. Alternator efficiency is about 50% at say 20 mph and falls lower beyond this. We have good low wind performance with reasonable efficiency in the most common winds and we accept less than the ideal in high winds when the batteries are likely to be dumping.

We get results below 10 mph as the prop will run fast enough at good power. We don't quite make the output curve in the higher winds because the prop is running slow but not far off peak as to cause serious loss.

So to sum up, the first step is to choose a suitable cut in speed and if too low increase air gap to raise it. Then make the alternator just stiff enough to avoid stall and if it does stall, increase line resistance to lower the slope

Not a perfect solution but a simple one that gives the cheapest alternator and has good energy capture in the common winds. We have far better energy capture than the old machines with cut in about 12 mph with field and iron losses but we do loose the high peaks in the occasional high winds. For a while most have been happy to go this way.

If we don't like it we have to do clever things. I will only consider one way now and that is to include a variable ratio mechanical transmission to let the prop speed get ahead of the alternator in high winds. Possible but for our size of machine not really practical and it leaves the alternator efficiency determined by the resistance at cut in.
Flux

[ Parent ]

Re: Matching the load (3.00 / 0) (#7)
by Flux on Sat Mar 18, 2006 at 08:42:50 AM MST

Before looking at other options this seems a good place to look at some actual results from a 6 ft machine with different options.

This graph shows what happens with the cut in set for low speed and keeping alternator efficiency high. Curve marked stall.

The next case has resistance added to the line to let the prop speed up and shows clearly that lowering the alternator efficiency is more than compensated for by bringing the prop out of stall.

The final curve is the holy grail where the load is matched to keep the prop happy and done without adding losses. the curve is called buck.

The wind site was very bad and results are below what is possible but it is a reasonable comparison. At higher wind speeds the difference is much more dramatic, stall stays where it is, resistive match tails off like stall but at a nmch higher value and buck keeps rising until the thing furls.

[ Parent ]

Re: Matching the load (3.00 / 0) (#8)
by kitno455 on Sat Mar 18, 2006 at 10:58:30 AM MST

thank you Mr Flux!

so now i am on the edge of my seat- what is used to make the 'buck' curve :)

allan

[ Parent ]

Re: Matching the load (3.00 / 0) (#9)
by Flux on Sat Mar 18, 2006 at 02:10:37 PM MST

The buck curve is using a dc-dc buck(step down ) converter. As far as I can tell the MX60 is a buck converter.

Perhaps I shouldn't have thrown in that graph at this point, I did intend to look at many options and look more closely at a few that show most promise.

At the moment I will continue on that route but if there is a general feeling that the less promising ones are a waste of time I will skip them, but you bet someone will propose them next week.

What about weakening the field ad has been done on car alternators for years. This is not impossible but is a tricky one with neo. It may be a possibility for those with mechanical skill and don't want to mess with electrics. It is not easily possible to alter the strength of field with a winding and current as the neo is so difficult to demagnetise, it would take enormous currents into a winding on the magnet but it could be done in other parts of the iron circuit. I seem to remember someone( jimovonz?) has done this with a car alternator with magnets added to the claw rotor.

The other possibility is to alter the air gap mechanically, not easy but not more difficult than a pitch control and there may be a chance to combine both.

The best reason not to do it is that you have to make an efficient alternator at the lowest speed and it makes it big and costly. The winding resistance does not change and has to be low at the start even when little power is produced.

What we need is a winding that starts with lots of turns of thin wire and changes to fewer turns of thick wire for the higher winds. Simple concept but not so easy to do. The nearest I can think of is to use a variac (variable ratio transformer) between the alternator and the rectifier. This would be driven by a servo motor to maintain the ideal voltage for each speed. It is not a practical idea, it would work well enough but the cost would be crazy and brushes would not last many days, so that one is not for me.

This brings us to tap changers and re-connection ideas such as series/parallel and star delta, constantly being proposed but not often used. On the face of things they seem simple and logical so why not. Let's consider star/delta ( series parallel is very similar).

If we design the winding to match the high speed end from 15 mph upwards we get a good match. Change to star and volts increase by 170% so cut in now comes nicely down to about 8 mph. The first snag is that the slope of the machine in star is far too steep but over a limited range the prop will manage fairly well but is dragged to stall. Change to delta and the machine now wants to run fast on the curve towards overspeed. The prop now picks up speed and soon has the load back on with good performance. When wind drops and you change back into star there is a sudden deceleration as the thing goes back into stall. Using speed as a sensor is stable so it is a working scheme. If we had high wind and low wind days it would be quite a good way to go, but wind is not that convenient. On most days the thing spends its life changing gear up and down slowing down and accelerating. Reasonable on a small machine but far from pleasant on a big one, I suppose some could live with it.

The next issue is the actual switching, it is an absolute mess to do electronically at low voltages as the only common ac switch is the triac and the drive circuits are messy and volt drops make it poor for less than about 48V. Relays, contactors etc have a hard life and it may not be a long one. I have never run a scheme for long enough to find out how long it is.

One thing that did come out of the discussion on Jerry connection is that star Jerry could be done with a relay and the contacts would only need to carry the low wind current and never change over under load. This may well make the thing more of a possibility for smaller machines. Speed switching from frequency with a 2917 tacho chip is easy to do.

Tap changing by switching out turns from the coils as speed rises is just about as messy and if you cut out part of the winding when it is working hardest you make it less easy to keep the efficiency up. Ultimately you end up going down the route of switching on the dc side with mosfets and when you go this far you might as well go the full electronic route and have a smooth and more efficient result.

Then we come to the 2 machine idea, a small one with low cut in speed to deal with light winds and a big one to deal with the high winds. Simple idea but not so nice in the end. There all the derivitives based on un-balanced windings that seem to offer promise but are not so attractive in the end.

The snag is similar to star delta in some respects, the little machine needs to be inefficient to have a low slope for the low winds, not a serious problem as the efficiency at low currents is reasonable, but what do you do in high winds. Best switch it out otherwise it becomes so inefficient that it robs significant power for the big machine and most likely it will be so inefficient that it will burn out.

Once again you need a speed switch and it is a step change and not smooth. In this form I don't like it, but if you use a little buck converter to control it you can keep it efficient and avoid stall without added resistance. It will phase out smoothly as the big winding takes over or in some cases you may choose to leave it contributing part of the load, depends which matches better.

Having gone so far you can make it one machine with 2 windings one wound for cut in and the other which will occupy most of the space wound for a nice track from about 15 mph.

I call this a hybrid so if I come back to it remember what it is about. It works well, is pretty efficient and only uses electronics for a small converter to handle the low wind power. If the converter fails you still have a machine that performs perfectly well in the higher winds and the main machine will stop the prop over speeding and can be used to stop it with a brake switch.

Time to stop, it's bed time.
Flux

[ Parent ]

Re: Matching the load (3.00 / 0) (#10)
by kitno455 on Sat Mar 18, 2006 at 03:37:08 PM MST

wow. thanks again flux. i think i have my head around the need for high turn count at low speed to give usable voltage (cut-in), and low turn count at high speed to reduce stator heating (inefficiency).

but i dont understand how any kind of post-rectifier converter is going to effect that change? is the problem that E cannot rise when it is clamped by a battery, so instead I rises, and hence I^2R rises? but the buck converter allows E to rise instead?

allan

[ Parent ]

Re: Matching the load (3.00 / 0) (#11)
by oztules on Sat Mar 18, 2006 at 07:40:35 PM MST

Kitno455

Luckily, flux has already described this for us above:

"What we need is a winding that starts with lots of turns of thin wire and changes to fewer turns of thick wire for the higher winds. Simple concept but not so easy to do. The nearest I can think of is to use a variac (variable ratio transformer) between the alternator and the rectifier. This would be driven by a servo motor to maintain the ideal voltage for each speed"

Think of it as ...In australia we need a transformer to transform 240v down to 24v for batt charging.....In the USA, they will require a different transformer with different ratios to transform from 110v to 24v

In both instances, we use a different transformer to achieve this impedance matching. If we plug the batts directly (via rectifier) to the mains voltage..... well it's not pretty. Either the power station has to drop down to 24v, the batteries have to rise to meet the mains voltage, or the rectifier explodes as these scenarios cannot be accomadated.

None of the above are acceptable. So we use a transformer to match the two competing interests 240-24v or 110-24v (one for each countries voltage regime) or use a solid state "black box" to achieve the same thing. In this latter case, if we design it well, we may even be able to use the same "blackbox" for both countries power grids.

In the case of "post rectification", transformers will not work with the ac side, and so now we need another method...dc:dc converters, that will "transform" the high voltage-high impedance input to the low voltage-low impedance of the load. Just as in the two countries two voltage problem, we now have a dynamic generator(variable, not just two voltages) and a static load...the batteries, essentially only a single voltage.

Flux's black box must do the work of that variable transformer-servo motor driven device he described above...or some reasonable approximation thereof.

Well it's Flux's show, and he will want to explain the "black-box/s"

........oztules

[ Parent ]

Re: Matching the load (3.00 / 0) (#12)
by kitno455 on Sat Mar 18, 2006 at 07:47:05 PM MST

which is a very long winded way of saying 'yes' to my question?

allan

[ Parent ]

Re: Matching the load (3.00 / 0) (#14)
by oztules on Sat Mar 18, 2006 at 08:30:31 PM MST

Yes

I'm sorry about that, but i figured that others may have the same question, but not your understanding....so i went very much overboard.

the intention was good....

.....oztules

[ Parent ]

Re: Matching the load (3.00 / 0) (#20)
by kitno455 on Sun Mar 19, 2006 at 08:43:57 AM MST

oh, yes. i dont want to shut anyone else out or suggest that your post was not helpful- its just that we are on the edge of my understanding, and too much information makes it hard for me to figure out the next step to take.

thanks-
allan

[ Parent ]

Re: Matching the load (3.00 / 0) (#63)
by Boondocker on Sun Oct 15, 2006 at 05:58:49 PM MST

Thanks for this post.

You explain "increase line resistance to lower the slope". How is line reistance increased? Coil design?

Thanks

Duane

[ Parent ]

Re: Matching the load (3.00 / 0) (#13)
by commanda on Sat Mar 18, 2006 at 08:17:05 PM MST

Great going Flux; I love it.

Would it clarify it for some to say that the stright-line power vs rpm curve of the generator must be below the cube-law parabolic curve of the prop at all rpm's, to avoid stall?

Hopefully you're headed where I think you're headed with this, so won't say any more.

Amanda

Re: Matching the load (3.00 / 0) (#15)
by phil b on Sat Mar 18, 2006 at 08:38:18 PM MST

Thanks for all the info Flux!!! This post went into my hotlist immediately!
You have clarified several points and raised more new questions. I'll save them for later. I'm looking forward to your next post!
Phil

[ Parent ]

Re: Matching the load (3.00 / 0) (#16)
by Flux on Sun Mar 19, 2006 at 02:00:23 AM MST

Amanda
Strictly speaking it is the input power line( which curves upwards) not the straight line output one that we must consider, but for our discussion the input line is nearly straight compared with the cube .

I did say somewhere that points to the left of the cube are towards stall and left it at that because the prop power curve has a significantly flat top and going either side of the cube in moderation is not going to hurt . Stall( separation of air on the blade) comes when we go too far and run off the flat top.

If I gave the impression that we can't operate to the left of the line I couldn't justify the way we design the present machines for best operation. In that explanation I said that in mid winds winds we were pushing the prop towards stall as far as we can reasonably go. It is the fact that the prop curve has this flat top that lets us get a reasonable performance from most machines.

That said, you are right that folks must grasp the idea that if we go too far to the left of the line we stall and if we go too far to the right we run away and come off the peak
Perhaps it would help to think of the power curve as a very thick line and we work without falling off the edge.

[ Parent ]

Re: Matching the load (3.00 / 0) (#17)
by Flux on Sun Mar 19, 2006 at 03:04:54 AM MST

I think most of you have now got to grips with the problem. We need a voltage that rises directly with speed to keep the prop on the top of its curve and we have a battery that stays at constant voltage. Our best hope lies with an electronic converter to match the voltages on the dc side. It is the equivalent of the variac on the ac side but with no moving parts to wear out.

The voltage we decide to start from will decide the type of converter we use. From previous methods we have started with an alternator voltage suitable to start charging at cut in wind speed. If we do this we need a step down converter( buck).

We have also seen that from 15 mph upwards a conventional alternator of good efficiency tracks the power curve very well so why not consider this as the starting point and use dc boost converter to produce power in low winds. If for some odd reason we chose an alternator that was too fast in low winds and too slow in high winds we could still use it with a buck-boost converter.

These ideas so far assume we keep a standard diode rectifier but again we are not forced to do so. A controlled rectifier can be used to match a high voltage alternator to a lower dc load. This is good old technology proven since the 60's and very robust so it would be foolish to ignore it. Modern methods have advantages but are not as tough and in this application that may have a large effect on the final choice.

In the last few years there has been a lot of work on rectifiers that draw a sine wave current and cause less distortion of the line. These can also buck and boost and the effect of loading the alternator with a sine wave is known to result in higher efficiency.

Not a black box! just a whole supermarket shelf stacked with black boxes.

Our problem is to pick the best compromise between cost, reliability, performance and the prospects of folks building it.

We are in an awkward situation, normally we buy electronic things, in theory we could make a TV but given the circuit diagram could we get the bits. If we did and we didn't have the board layout would it work in the end? That is why we buy these things.

At the moment we can't buy this black box for our windmill. We know exactly what to do and for a few expert electronics designers it should be easy but for the rest of us it is not. Starting from scratch to build a converter to work reliably under the extreme conditions of a windmill handling a few kW is a far leap from some crap power supply thing producing 10W at 5v. Lots of information on what you do but none on how to do it.

Soon someone is going to ask the question, why worry about the converter that's the easy bit, what about the circuit to track the peak prop power and control the inverter?

I shall no doubt upset some by saying that we need to walk before we can run. Make the converter work and we can fiddle its characteristic to keep it close to the power curve. We don't have to be that accurate. What we have now is miles out and it works. if we get a lot nearer the prop will forgive the little errors and we shall have something that works. See my buck curve. Mppt is difficult with wind and at best I doubt you will do any better with it. The most you can hope to gain is the fact that it will find the correct curve most of the time. I can't visualise this going on with the wind doing frantic things. Maybe the best idea would to be to store the best long time track in some sort of look up table and use that. If you changed something such as blade size it would slowly find its new curve.

Time for thought, time to catch up some other urgent things. More soon

[ Parent ]

Re: Matching the load (3.00 / 0) (#18)
by SamoaPower on Sun Mar 19, 2006 at 07:33:34 AM MST

Well done Flux.

I applaud your efforts to enlighten us all on the vagaries of wind power PMA's. I'm sure you had a whole bunch of people sitting on the edge of their seats awaiting the next installment of the saga. I know you have passed along tidbits via your many lucid comments (with many repeats) to various posts, but I strongly suspect this anthology will go down as a major input to the "bible". You should really consider more original posts (I believe you've done two) as a significant part of your repertoire. Not only do you obviously have a well-grounded understanding of the basics and more, you have actually done the work to show your contentions. Few can say the same (myself included).

"We have also seen that from 15 mph upwards a conventional alternator of good efficiency tracks the power curve very well so why not consider this as the starting point and use dc boost converter to produce power in low winds."

Why not, indeed? I've thought all along that the best efficiency point should coinside with the energy peak wind speed for a given site (16 mph in my case). By selecting coil parameters to provide cut-in at this speed, we should have the best starting point. Coil DC resistance will be lower to help the high end problem.

I believe a common misconception that is assumed by many is that when MPPT and/or converter electronics are mentioned that they are considered to be one in the same, which is far from the truth. MPPT is simply one algorithm that can be applied to an electronic controller. There are others such as constant voltage, constant current, etc.

"Starting from scratch to build a converter to work reliably under the extreme conditions of a windmill handling a few kW is a far leap from some crap power supply thing producing 10W at 5v."

You're right, of course. People who want to do this need to educate themselves to what is required. It doesn't come easy. One thing I've noticed is that current data sheets are giving more information on circuit layout which is certainly helpful.

One thing I've noticed Flux, is that you tend to steer people away from variable pitch, citing mechanical difficulties. I assume that you have had problems in implementing variable pitch. It offers another degree of freedom in controlling our machines and although it certainly increases mechanical complexity, it has a lot to offer. I'm not referring to flyball regulated machines but to active pitch controlled machines. Is this another case deemed to be too complicated for the uninitiated? I would hope that those with knowledge and/or experience would present their findings and let us make up our own minds.

Having said that, I still feel a debt of graditude to you Flux, for having helped so many.

[ Parent ]

Re: Matching the load (3.00 / 0) (#19)
by Flux on Sun Mar 19, 2006 at 08:18:57 AM MST

I don't want to digress to much at this stage, but I am not against pitch control.
It is a serious engineering challenge but for those capable it is great. I wouldn't depart from the flyball type scheme to a servo. The flyball scheme works perfectly and if well built and maintained is perfectly reliable. Adding a servo is like the motorised variac, yes it will work but equally certain is the fact that it will fail.

The complexity is in the design of a bearing system that will hold up under load without weighing a ton and also the problem of fixing shafts to blades reliably. It is fairly easy to make a reliable hub to hold wooden blades to flat discs but to attach a shaft to a wooden blade that is strong and reliable needs mathematical analysis , experience or good luck. Some of the pictures I have seen here look fine as long as they are 1/2 a mile from me.

This issue is no easier if you use a servo.

If you can produce a perfectly reliable pitch control it would be great but I would use it to get rid of the tail and all the gyroscopic forces. If you want a servo, use it to steer the machine into the wind, if that fails and the pitch control works you are ok.
Flux

[ Parent ]

Re: Matching the load (3.00 / 0) (#21)
by Flux on Sun Mar 19, 2006 at 08:46:46 AM MST

Now back to converters.

There are so many choices and many have their good and bad points. If I was going to invest money to develop a commercial scheme I would not choose the same way as if I wanted a simple reliable and effective scheme for a one off.

There is little doubt that using some form of buck scheme will give the highest efficiency and if you have to contend with high line losses between the mill and the battery this may decide the choice. In this case you could start with a high voltage generator and forget the line loss problem but it adds other issues.

For 12v again a buck scheme is likely to give considerably better results but most larger schemes are not 12v and this is not always a major issue. The scheme works as well for small machines as big ones but we must never loose site of the fact that a 6ft mill with optimum tracking will not perform better than a reasonable one with without it and with an 8 ft prop. We can't exceed the Betz limit, if we are stuck with a fixed size of blade then this is great but never forget that if we have a choice, swept area defines a windmill better than any other thing.

If we have a boat or a caravan and are stuck with blade size then this is great. Otherwise it is likely to be bigger machines that will benefit most from the reduction in losses in the stator and the heating problem.

The issue that is certain is that a buck scheme will handle the total machine power. this is generally true for boost to some extent as often proposed but there are ways to keep the converter power down to that produced in the light winds and you can bypass it in high winds.

For this reason for an introduction to the process the boost converter seems a better starting point although in some ways it has limitations.

I have had more experience with this method and also the Bergey XL1 does it this way and has been a good machine. Perhaps Air X does it the other way, I really don't know what they do, but it is mostly to stop noise.

Before I go on to look at boost circuits I will throw in one thing in favour of the buck converter. You are starting with a machine designed to stall and making it not do so. When the battery is charged you have the option to drive it back into stall as the first stage of charge control and if that is not enough you only need a small dump regulator to bleed off the machines stalled output and it sits there turning slowly. I like that bit.

[ Parent ]

Re: Matching the load (3.00 / 0) (#22)
by Flux on Sun Mar 19, 2006 at 09:43:19 AM MST

Now let's look at what we could do with a boost converter. I said previously that if we took a machine intended to cut in at say 7 mph and ran it on a system of double the voltage it would most likely track the high winds with an efficiency of over 70% depending on the alternator design and the line losses.

For anyone who wants to experiment this could be a starting point that would cost nothing except for an extra battery. It could be a good way to upgrade a 12v machine to 24v or a 24v system to 48V. If anyone is serious about trying it, try your alternator on a system of twice voltage and see if it is suitable. You will loose all output below 10 mph at this stage but it should perform exceedingly well above 12 to 15 mph and you should probably see the same amps into the new system at twice the volts.

How can we boost the volts at low speed to make use of the low winds?. There are many ways, some good and others not so good. Voltage doublers crop up here from time to time using capacitors charged at the low voltage and added to the incoming ac volts to double it. They do work but beyond a few watts they are not nice things and they do their own thing, we need to decide what they do.

Most of the practical converters involve storage of energy in inductors. All the common boost circuits worth looking at use this inductive approach. I don't feel inclined to spend time drawing and explaining these devices at the moment, do a Google and they are everywhere. If this link works it seem a as good a starting place as any,
http://www.powerdesigners.com/InfoWeb/design_center/articles/DC-DC/converter.shtm

It has all the common forms of converter and the explanation is fairly simple, you should get the general idea at least. If you get carried away with more complicated things I must warn you that there is an error in his drawing of the CUK converter but otherwise it it is fine.

If you look at the boost converter you will see that the input is to an inductor. This raises an interesting point for those playing with motor conversions. Your alternator has 3 inductors in it ( part of its synchronus reactance for the technical) and it is possible to use this to do the boosting but I will come back to that later with some more ideas. For those with an air gap alternator you will need to rectify the dc and smooth it a bit with a capacitor and treat that as the input to your boost converter. If you turn the transistor switch on and off at about 50% duty cycle the thing will double the volts. This is the starting point at cut in and you will find that the output rises much too fast with wind speed. You need to reduce the pulse length to reduce the boost ratio of the converter and when you reach 12 to 15mph you no longer need the converter and if you remove the transistor pulse the machine carries on without the converter.

In this simple form the converter inductor and series diode both have to carry the full machine current at full load but the transistor is not involved.

There are ways to avoid this and if you take that approach you have a low wind converter that is redundant in high winds and if it failed you still have good performance above 12 mph. For experimental things this may be of more importance than the slight gain in efficiency if we boost in another way.

[ Parent ]

Re: Matching the load (3.00 / 0) (#23)
by willib on Sun Mar 19, 2006 at 11:12:43 AM MST

flux i dont like the buck converter on that page , it seems to me the diode is backwards.
http://www.otherpower.com/images/scimages/2965/bad_buck_boost.gif 2Kb

i like this one much better..
a buck and a buck boost .
http://www.otherpower.com/images/scimages/2965/buck_buck_boost.jpg 28Kb
please continue..

Carpe Ventum (seize the wind)
[ Parent ]

Re: Matching the load (3.00 / 0) (#24)
by Flux on Sun Mar 19, 2006 at 11:29:10 AM MST

His buck converter is ok, so is the boost. He has made a cockup with the buck -boost and the CUK. both of these invert the polarity, it's not the diode that is wrong it's the output voltage marked with the wrong polarity. I spotted the Cuk but not the buck boost. It may not have been a wise choice for a link. Yours is fine but doesn't have the waveforms.

Well spotted.

[ Parent ]

Re: Matching the load (3.00 / 0) (#25)
by Flux on Sun Mar 19, 2006 at 03:11:28 PM MST

Here is a curve of a small alternator with boost at bottom end ( green is input red is output) and also a circuit of how to feed a converter so that it only carries the boost current.

For reasonable size and efficiency the inductor needs to have a ferrite core and the switching frequency can be about 30kHz. If you use an iron core for the inductor the frequency will have to be kept to a few kHz but it will need more turns and have more resistance and even at these frequencies it will have significant iron loss, For the high frequency chopper the output diode of the converter must be as fast as possible so it must be a schottky or an ultra fast diode. The 3 input diodes can be ordinary slow ones but schottky would help with volt drop.

Ideally the wire for the inductor should be multi strand ( Litz) but for this power level you may get away without it.

The circuit shows a bipolar transistor( easy to draw) but the thing to use is a mosfet. Although they have high gate impedance you must drive them from a low impedance source to prevent capacitively coupled spikes from the drain circuit from breaking the gate insulation. There are plenty of proper drivers around and the practice of using logic gates or 555's is not good enough even for a converter of this size.

Although 30kHz is not that fast, the edges of the switching waveforms are in low radio frequencies and there are no bits of wire at this frequency, they are normally inductors, can be capacitors or transmission lines so layout is critical. the input part with the inductor is no problem but the loop round the mosfet, fast diode and output caoacitor must be as short as possible and it is a good idea to mount the whole thing close to a ground plate to act as a ground plane. The capacitor ought to be low esr type and a small low inductance polypropylene or similar is worth adding in parallel.

I will suggest some things for the pwm generator and the control signal later.

[ Parent ]

Re: Matching the load (3.00 / 0) (#26)
by jimovonz on Sun Mar 19, 2006 at 03:25:03 PM MST

Flux, I appreciate the effort you are going to presenting this information. I've been playing with air-cored inductors in the belief that I'd run into problems with saturation using (pulsed) DC. At what point (if any?) does this become an issue?

[ Parent ]

Re: Matching the load (3.00 / 0) (#27)
by Flux on Sun Mar 19, 2006 at 03:58:18 PM MST

If you include a substantial air gap in the ferrite core ( 1 - 2mm) it can't hard saturate. You must have enough inductance to make sure the current doesn't rise to saturation during the switch on time. If you get this right you are ok. there is a flux reset during the energy transfer to the load, the core does not progressively saturate as if it was on dc.

Keep well above minimum inductance it seems to do no harm. small power supplies seem to keep inductors to the absolute minimum but cost is likely to be the issue, on a 1 off we can be generous.

I think you will have a hard time without a core and your L/R ratios will not be good.
Flux

[ Parent ]

Re: Matching the load (3.00 / 0) (#28)
by jimovonz on Sun Mar 19, 2006 at 04:07:33 PM MST

Much appreciated Flux, I will give ferrite a go - it would be nice to save a bit of size/weight. My inductors have been looking something similar to the coils in our turbines...

[ Parent ]

Re: Matching the load (3.00 / 0) (#29)
by SamoaPower on Sun Mar 19, 2006 at 04:32:59 PM MST

What would be your controlling feedback parameter to the PWM drive?

[ Parent ]

Re: Matching the load (3.00 / 0) (#30)
by Flux on Mon Mar 20, 2006 at 12:58:14 AM MST

The original units used speed (frequency). One is still in use with a big inductor and switching at audio frequency.

I had a lot of trouble with the 2917 tacho chip picking up the chopped waveform from the boost circuit. I think I ended up with some sort of opto isolator in the end.

Later units have used battery input current and this has been far easier and the curve shape is better.

I have managed with a shunt but I far prefer A Hall effect current sensor, where you can pass turns through it to change the ratio.

You need just enough gain to make the converter run in the low speed region. Too little gain and you have a steep curve that keep you stalled and you stay on the converter when you should be on the main machine. Too much and you come off before the main machine has enough volts. High gain may also cause instability with the various time constants hanging around.

By using battery current the main machine current as well as the converter output current pass through the current transformer and there is no risk of the converter operating in the high winds. With the speed scheme it did overlap for a while

[ Parent ]

Re: Matching the load (3.00 / 0) (#31)
by Flux on Mon Mar 20, 2006 at 01:48:35 AM MST

Most of my circuits never get drawn properly and the bits of paper get lost.

This is one that I did draw, but it used a shunt as the control. You could feed the Hall CT into the input and change the gain to suit. I will try to find a more normal circuit but may take a while.

The IR 4426/7 series of mosfet drivers are convenient.

Various switch mode power supply chips will do the pwm. Nearly all have things built in for their intended use and some are a nuisance for our job. The MC33060A seems a convenient one although you can find some with mosfet drivers (3525 has but is not single ended).

Not sure how legible this will be as a gif.

[ Parent ]

Re: Matching the load (3.00 / 0) (#32)
by willib on Mon Mar 20, 2006 at 01:42:15 PM MST

i found the 33060 for those wanting to view it.

http://www.onsemi.com/pub/Collateral/MC34060A-D.PDF

i couldnt find the 4467 and this one seemed to be the only Int. Rectifier , gate driver they sell.although i could be wrong, its the only one i could find..
http://www.irf.com/product-info/datasheets/data/ir1167aspbf.pdf
Carpe Ventum (seize the wind)
[ Parent ]

Re: Matching the load (3.00 / 0) (#33)
by SamoaPower on Mon Mar 20, 2006 at 10:09:06 PM MST

Quite clever Flux. I see how this could do the job pretty well. Since I started the design phase of a larger machine, I've wanted to find a way to reduce alternator losses. Choosing a higher cut-in speed and being able to consequently reduce the coil losses seems attractive. It looks like the boost converter can fill in the low end nicely.

You mention that by doing this that the higher end matches up pretty well but I wonder if there's still room for improvement there (20-25 mph). I'm sure you've considered cascaded buck and boost converters and I wonder what your conclusions were. An interesting recent chip release by Linear Technologies, the LTC3780 controller seems to offer a rather simple way to do this. It's meant for medium power applications but does lend itself to power up-grade with larger drivers and FETs. It uses a single inductor and claims seamless transition between modes. It also claims efficiencies around 95% and uses synchronous rectification. It might be worth a look.

I look forward to more installments of the saga.

[ Parent ]

Re: Matching the load (3.00 / 0) (#34)
by Flux on Tue Mar 21, 2006 at 02:15:51 AM MST

The biggest nuisance with the scheme I have described is that for good high wind efficiency you need to keep line losses to an absolute minimum. Ok for a short run but for a long wire run it does become an issue.

Alternator efficiency problems are the same whatever voltage you choose but the line loss is worse with low volts and buck schemes will come out better if you cant keep the line loss negligible.

For conventional machines we need line loss( it's better there than in the alternator) but once we go for high efficiency schemes loss is not needed to match the prop and the limiting factor will be how much expense you are prepared to throw at it to gain a few %.

With a 24v system with the same size of alternator as normally resistive matched, If you can avoid significant line loss and match the alternator winding at 25 mph, you should see over 70% and that is with no increase in cost. If you want much better then alternator cost goes up in big leaps for every few % you gain.

With a well designed system with boost converter at 24v you can achieve 70% from cut in to furling with the same alternator cost as the resistive match.

At 12v things are not so good, with the scheme as I have shown it, you will be boosting from about 6v and the diode losses will have a significantly greater effect. If you include the rectifiers in the boost circuit you may do a little better but then you no longer have a scheme that can work without the converter if it should fail.

Once we decide this is the way to go and stop worrying about failure then going for a buck scheme will give better efficiency on low voltages and also reduce the effect of line loss.

Starting from scratch I would not choose a buck-boost scheme, but if you have an alternator already and it was too slow in low winds and too fast in high winds then as long as it was efficient enough then a buck-boost scheme would be perfectly satisfactory.

The actual converter circuitry like everything else will be a compromise between simplicity, reliability, the bits you can get and and your ability to manage it.

Refinements such as synchronus rectification are logical moves but for our voltages they are not a big issue. A 3.3v computer supply is a different matter.

It is not realistic price wise to attempt to build alternators with upwards of 90% efficiency and if you keep conventional rectifiers I don't even think you can do it.

It's back to the greedy thing, if it's possible you want it.

Moving from an efficiency of 40% to one of 70% will make a dramatic reduction in stator loss and it seems to me to be about the most cost effective region to aim for.

If you go for motor conversions you may get a higher top end efficiency at a cost effective price but you will have to trade a bit of low wind performance as you can't recover the iron loss.

The snag with all these things is that there is no one case that will meet every individual need and even if I gave specific circuits people in various parts of the world may not be able to get the specified bits. It's not quite like magnets and wire.

All I can really hope to do is give general ideas of the various schemes . I may produce a reasonably documented boost circuit based on the ideas I have given, but even then anyone not involved in electronics may not get it to work. Ok if it works first time, but fault finding without experience and test equipment makes it a doubtful starter.

Probably general advice to steer those capable in the right direction is much more useful. .
Flux

[ Parent ]

Re: Matching the load (3.00 / 0) (#35)
by Flux on Tue Mar 21, 2006 at 03:33:46 AM MST

For anyone wanting to boost a motor conversion or a smartdrive you may like to try this way.

The inductors are the winding leakage recatance of the machine so you don't need to add anything. You will have to switch at a fairly low frequency to prevent serious iron losses and to cope with the slow bridge rectifiers (less than 5kHz at a guess).

If you want to avoid the drop of the series diode you can leave it out and replace the 3 bottom diodes of the bridge with mosfets and gate all three together. Your mosfets now have to be rated to carry full load bridge current.

I don't think you have enough inductance to do it with an air gap machine. You will have to add inductors. To do it without introducing lots of resistance the inductors will need to he high frequency ferrite ones and you will need fast diodes in the bridge and chop at a high frequency. You may need capacitors across the machine output but before the inductors to stop the high frequency producing losses in the magnet faces. I haven't tried it, it means a far less robust bridge but should be possible and you have the prospects of a lower harmonic rectifier. With clever sinewave pwn to each mosfet you could make a true low harmonic force commutated rectifier, but I think driving all fets together would be more practical.

More ideas to think about.

[ Parent ]

Re: Matching the load (3.00 / 0) (#36)
by oztules on Tue Mar 21, 2006 at 03:46:01 AM MST

I'm missing something here, If we took out the series diode and replaced the lower three diodes with fets, wouldn't the "main" fet short out the battery, and as such turn rather sickly.........

What did I miss here?.............oztules

[ Parent ]

Re: Matching the load (3.00 / 0) (#37)
by Flux on Tue Mar 21, 2006 at 05:28:22 AM MST

Sorry. I was trying to save having to do another drawing. I should have made it clear that the 3 fets in the bridge replace the single one.
Flux

[ Parent ]

Re: Matching the load (3.00 / 0) (#40)
by commanda on Tue Mar 21, 2006 at 02:21:01 PM MST

This might be an appropriate point for me to outline what I'm working on, and why. After Flux's excellent explanation, my approach might make sense to some people.

F&P smartdrive (80S) unmodified. Transmission is high voltage, less line losses.
First converter is a 3 phase boost PFC (power factor correction). Output 340 volts DC. PFC's essentially draw sine-wave current in phase with the voltage, so no nasty harmonics to waste power and heat the stator.
Second converter is a buck converter down to 24 volts to charge the batteries. The control signal for this converter is mppt derived, and also uses a tacho input signal in its algorithm to help with gusty winds.
I chose 340 volts dc because that's the rectified and filtered voltage of the Australian 240 volt mains. This means I could use modified computer power supplies, or an off-the-shelf switchmode battery charger as the second converter if I wished to.

Amanda

[ Parent ]

Re: Matching the load (3.00 / 0) (#46)
by Flux on Wed Mar 22, 2006 at 02:07:12 AM MST

Amanda
From the look of the F & P it must have a lot of leakage reactance. I think that keeping volts up and using the pfc rectifier should let you extract a lot more power from it than loading it at constant voltage.

Keep up the good work.
Flux

[ Parent ]

Re: Matching the load (3.00 / 0) (#41)
by SamoaPower on Tue Mar 21, 2006 at 03:34:09 PM MST

Many thanks Flux.

". . . you should see over 70% and that is with no increase in cost. If you want much better then alternator cost goes up in big leaps for every few % you gain."

Maybe we should take a little closer look at the economics since for many, the wind generator is an investment. The energy we harvest has value whether we play at the game or depend on it. Using my site (10.3 mph mean) and proposed machine as an example, the energy available to the alternator from the air rotor (16'-42%) calculates to 7900 kWh per year (the integral of the energy distribution curve). If we assume a machine lifetime of 10 years, the total collectable energy could be 79,000 kWh. If we assign a value equal to the local grid energy rate (US$0.25/kWh) we have a potential harvest value of US$19,750. At an alternator efficiency 0f 70% and ignoring other losses, our relizable value is US$13,825. If we improved the alternator efficiency to 75% the energy value would be US$14,813 for a difference of US$988. What this says to me is that I could spend up to about 1000 bucks for a 5% improvement in alternator efficiency and not lose in the long term. I think it does pay to spend more (within reason) to improve efficiency even for a few percent. Of course, there are variables not accounted for in this simple analysis but I think one thing we can say with near certainy, is that energy rates will go up.

"Moving from an efficiency of 40% to one of 70% will make a dramatic reduction in stator loss and it seems to me to be about the most cost effective region to aim for."

Amen!

[ Parent ]

Re: Matching the load (3.00 / 0) (#42)
by oztules on Tue Mar 21, 2006 at 04:02:57 PM MST

"If we improved the alternator efficiency to 75% the energy value would be US$14,813 for a difference of US$988. What this says to me is that I could spend up to about 1000 bucks for a 5% improvement in alternator efficiency and not lose in the long term"

Whilst this does hold water, another important consideration from the "world according to Flux" is that nothing quite defines the power available to us more than the swept area of the mill .........or words to that effect. A thousand bucks could buy you a rotor size increase which would send the 5% increase you mention into the land of insignificance.

So it depends if bang for buck or perfection of energy capture is the measuring stick.
Flux has defined a procedure to improve from 40% to 70 % with a scheme involving very little outlay to an economicaly justafiable best point. Beyond that point,it would seem that power electronics and other complexities start to play against perceived gains.

From that point on, it is probably better to spend your extra power dollars on increasing the size, and still applying Flux's modifications.

However, if it more a personal vendetta against system losses, then thats a different parameter to measure success by.

one is for fun/pleasure/curiosity, the other is pragmatic bang for buck....I think i'm somewhere in between, so i'll shoot for 60-70%,......... or a bigger rotor if I want more power over that.

On a different note, have you found anything useful yet from the asymetric windings on your air403's yet.

.........oztules

[ Parent ]

Re: Matching the load (3.00 / 0) (#43)
by SamoaPower on Tue Mar 21, 2006 at 06:08:12 PM MST

Thanks for the comments oztules.

"Whilst this does hold water, another important consideration from the "world according to Flux" is that nothing quite defines the power available to us more than the swept area of the mill .........or words to that effect. A thousand bucks could buy you a rotor size increase which would send the 5% increase you mention into the land of insignificance."

While I certainly don't disagree with you or Flux on this point, I think you reach a point with machine size (?) where the additional associated difficulties and problems tend to outweigh the benefits. Increasing the rotor diameter by 25% of a 8' machine to 10' is trival compared to the same increase for a 16' rotor to 20'. For a small degree improvement, you're right. But, take the guy who has a modest size convential machine who wishes to improve his energy harvest. Does he haul it down, make new blades and perhaps a new stator and put it back up only to find that the match is worse? Or, would it be better to employ electronics to do the job? And, if he spent somewhat more on the electronics to get another 5%, would it be worthwhile? I think so.

I admit to having a few biases. I have some components of my proposed machine which I want to utilize. I also have a 'thing' about efficiency. A more 'elegant' solution to a problem is something I find much more satisfying than applying a 'brute force' approach (and possibly burning up stators). I think you had me pegged pretty well, oztules. I also try to avoid strenuous physical activity where possible. Running machines up and down the tower isn't my idea of fun. If I can play with the electronics in the comfort of my air conditioned shop (this is the tropics - I don't have to contend with that funny, cold, white stuff), I'm happy.

Re: Air machines. The tests so far have generated more questions than answers. Tests into a resistive load showed about a 30% improvement for a star connection over the original configuration. However, the same tests into a battery showed just the opposite. Until I can explain this, it's on hold but still on my long 'to do' list.

[ Parent ]

Re: Matching the load (3.00 / 0) (#51)
by SamoaPower on Sun Mar 26, 2006 at 02:06:03 PM MST

Flux,

I've taken some of your ideas and plugged them into my design spreadsheet for my proposed 16' machine and run through a number of iterations in an attempt to optimize parameters. I must say that your contentions are confirmed quite nicely except for one point.

You indicated that to achieve efficiences of better than 70% or so is going to require a lot more in alternator size and cost. Although I agree that a larger one is necessary, the degree of increase is tolerable. In my example, it required a magnet rotor diameter change of 1.5", from 18 to 19.5" (but same magnets) to give an indicated efficiency of 82% at 125 RPM (16 mph) and 73% at 160 RPM (20 mph) with cut-in set at 95 RPM (12 mph). Not a major size/cost factor. This change was indicated to accomodate the necessary stator coil changes. These changes also resulted in an ouput improvement of about 20% to 1.2kW to the batteries (line loss accounted for) at 16 mph, my target speed.

The real surprise was the sensitivity to very small changes in coil DC resistance. One milliOhm change per coil in this 16/12 12 volt machine changed the curves dramatically. This set me on a course to look closely at coil design. This effort resulted in coils of 19 turns of 7 in hand #14, a fair departure from the usual seen here. It's going to take an interesting winding jig to do these well. The DC resistance of these are a calculated 6.9 milliOhm each.

Per your suggestion, I've decided to curb the greed factor and limit the machine at 20 mph. This would be easily done by the pitch servo. The pitch control curve would be constant TSR below 20 mph, constant speed between 20 and 25 mph (2.1 kW out) and feathering shutdown at 25.

I want to thank you again Flux for getting me off the traditional mind-set and on an improved course. I'm much happier with the axial flux machine now.

[ Parent ]

Re: Matching the load (3.00 / 0) (#52)
by willib on Mon Mar 27, 2006 at 02:32:31 PM MST

i've wound a coil with less than 0.005833 ohms , single strand #10 wire..
with 23 turns :)
Carpe Ventum (seize the wind)
[ Parent ]

Re: Matching the load (3.00 / 0) (#53)
by SamoaPower on Mon Mar 27, 2006 at 03:35:43 PM MST

Must be a mighty small coil since that indicates an average turn length of about three inches or a total coil wire length of 5.8 feet. #10 solid is about one milliOhm per foot.

[ Parent ]

Re: Matching the load (3.00 / 0) (#54)
by willib on Mon Mar 27, 2006 at 04:10:36 PM MST

yep 5.8 feet is real close
its 1.75 dia x .510 with a 7/8" hole :-)
but as thick as the wire is (.106 thick enamel)it has to be easier to wind than seven in hand #14 wire. no?
Carpe Ventum (seize the wind)
[ Parent ]

Re: Matching the load (3.00 / 0) (#55)
by SamoaPower on Mon Mar 27, 2006 at 05:41:43 PM MST

Aah, I see. Yes indeed, small coils.

My latest coil design is about 6.5" x 3.6" outside and has an average turn length of 13" (slighty tapered) - about 20.6' total length for 19 turns.

I think winding #10 to efficiently use the available space would be much more difficult. I'd need more like #5.5 to do it with a single conductor. Also, there are eddy current losses to consider with a single larger wire.

Seven in hand #14 allows this coil to be wound with a single turn per layer resulting in a coil that's still only .5" thick and space efficient.

[ Parent ]

Re: Matching the load (3.00 / 0) (#56)
by oztules on Mon Mar 27, 2006 at 05:48:09 PM MST

You are right about the departure from the norm...

how many coils/phase for what voltage?

maybe time to split some starter motors and get the ribbon wire they use in the field coils.

........oztules

[ Parent ]

Re: Matching the load (3.00 / 0) (#57)
by willib on Mon Mar 27, 2006 at 06:38:45 PM MST

litz wire comes in some very thick versions.
Carpe Ventum (seize the wind)
[ Parent ]

Re: Matching the load (3.00 / 0) (#58)
by oztules on Mon Mar 27, 2006 at 07:26:40 PM MST

I was thinking more for coil winding and packing. The ribbon wire would be rectangular wire, 12mm by 1.6mm for 19 turns would use 100% of the winding space less insulating thickness.Samoapower is an efficiency wizard, maybe that would help in this pursuit. The thin cross section would be no worse for eddy currents, but the 12mm width would aid in current carrying, and reduce waste packing /air space, and be the same size coils physically (hopefully with lower resistance than 7 in hand #14)

.........oztules

[ Parent ]

Re: Matching the load (3.00 / 0) (#60)
by DanB on Wed Mar 29, 2006 at 10:19:39 AM MST

if its an axial flux airgap machine then the ribbon wire is probably not a good idea based upon our experiences building scotties little diesel last year. Not all flux is in a perfectly straight line from magnet to magnet and in my experience, using fairly thin ribbon wire (But 1/2" thick) - eddy currents are a killer. Our rpm were high - but the destruction from eddy currents was amazing and it happened quickly. I think that leaves us with our best bet being round wire in the axial flux machines with big airgaps. I could be misunderstanding what your all talking about though...

[ Parent ]

Re: Matching the load (3.00 / 0) (#61)
by oztules on Wed Mar 29, 2006 at 03:46:45 PM MST

No Danb,

I'm thinking you are understanding well enough to keep us out of trouble, and since I have availed myself of your story regarding the unhappy demise of Scotties gennie, I I'm understanding that my suggestion should have been /has been relegated to the dustbin.

Samoapower was on it straight away fortunately, as I was going to pillage some starter motors and do some testing myself until I read your tale of woe....no more dreaming of thickwire here.... (lets see a hundred in hand of spiderweb should do it.)

I was very surprised with the results, and they were contrary to what I would have expected at these frequencies.

But nothing proves/disproves theory like real world experiment, and luckily for the rest of us, you have done this.

Thanks for your experimentation, you have my and others gratitude for this (this includes a dozen startermotors I had recovered from the tip for this purpose, they can live on in one peace now)

Well done as usual................oztules

[ Parent ]

Re: Matching the load (3.00 / 0) (#59)
by SamoaPower on Mon Mar 27, 2006 at 09:54:17 PM MST

These are for a 16/12 12 volt machine so four coils/phase.

Re: strap conductors
DanB did some tests using wide conductors - didn't work out well.

http://www.fieldlines.com/story/2005/5/10/151616/052

Think I'll stick with round.

[ Parent ]

Re: Matching the load (3.00 / 0) (#62)
by seanchan00 on Sun Apr 30, 2006 at 09:41:41 AM MST

Hi Flux,

Due to poor wind, my windmill is currently generating less than 11V worth of power 70% of the time, and i can't use this to charge my 12V batteries. I'm considering using your Boost circuit above to boost my output voltage to above 13V. However, i do not know what Capacitor and Inductor values to use, or what type of Mosfet to use. Can you please advise?

also, would it be necessary to put in a cut-off feature in my boost circuit when the wind picks up?

Regards,
SeanChan

[ Parent ]

Re: Matching the load (3.00 / 0) (#38)
by willib on Tue Mar 21, 2006 at 08:00:57 AM MST

flux nice work , riviting reading !..
a few posts back you mentioned litz wire in the inductor,i have an ulterior motive for asking this question , i think you know what it is :)
What is the benefit of using litz wire in the inductor?
Carpe Ventum (seize the wind)

Re: Matching the load (3.00 / 0) (#39)
by Flux on Tue Mar 21, 2006 at 10:07:12 AM MST

At these frequencies thick wire suffers from eddy currents in much the same way as when in the field of rotating magnets. At 30 kHz, wire thicker than about 1/2 mm is better avoided, several thin wires in hand works a lot better,

As well as eddy currents, the skin effect is also present on thick wire, current is carried on the surface but not the centre part. More thin wires have more surface and less centre.

No it is not worth the trouble for alternator coils unless the wire is over 1/8" thick.

Flux

[ Parent ]

Re: Matching the load (3.00 / 0) (#44)
by Gary D on Tue Mar 21, 2006 at 09:15:40 PM MST

Very informative Flux. I don't have the expertise to duplicate your circuits, but do have friends that can.
A few thoughts though. If you do upgrade from 12 to 24 volts, and get the same amps at a higher voltage,at a slightly higher cutin... you need to double your battery bank size to allow it to accept the same/more current as it matches the power curve better. This wasn't mentioned, however it's an added cost(no free lunch). And/or you must have an automated higher wattage dump ready to get rid of the excess, which is a great thing for water/space heating in winter as winds get higher(especially on larger machines). Also, on the larger machines I do wonder how they would handle the increased rpm's(or if that's a non issue). Thought possibly some might ponder some of this(or maybe/probably I'm just a little "off").
Nice presentation, hope to see more! Gary D.

Re: Matching the load (3.00 / 0) (#45)
by Flux on Wed Mar 22, 2006 at 01:21:57 AM MST

Gary
You raise a significant issue. The present machines are quite slow even when reasonably matched to avoid bad stall. Large ones such as Dan's 17 ft are basically stall limited from about 10 mph upwards. They are as a consequence rather docile and quiet.

To operate on the peak of the power curve they will have to run faster. Most commercial machines are fast and noisy to keep costs low. Higher speeds will result in increased noise but if you start with reasonable tsr and keep it at design tsr at top speed I haven't found the noise to be an issue, but that is only with machines up to 10 ft. Some of the forces on the machine increase rather a lot with speed and it may be that a 17ft machine operating at optimum speed may need to be more strongly constructed, but I am not sure this is beyond our means but it needs thinking about.

As I said previously, with the right converter you have the option to bring the machine back into stall and for a bigger machine I would be very tempted to let it track the wind speed to perhaps 20 mph and then stall it out in the high winds so that it ran at low aerodynamic efficiency but high electrical efficiency so that it would still produce modest power at a low speed with no danger of burn out and much less stress on things generally. The greedy will have to live with the frantic high speed and stresses to capture enormous power in the 30 mph range.
Flux

[ Parent ]

Re: Matching the load (3.00 / 0) (#47)
by willib on Wed Mar 22, 2006 at 10:57:41 AM MST

Flux are there any dissadvanteges to using a buck boost , if your volts need boosting , but you have plenty of available current in your alt.?
for example say a person wants to go from 12 to 24 volts or from 12 to 48V ?
Carpe Ventum (seize the wind)

Re: Matching the load (3.00 / 0) (#48)
by Flux on Wed Mar 22, 2006 at 01:56:37 PM MST

Sorry I don't follow the question. Any voltage conversion involves losses. If you gain more from the other factors than you loose in loss it must be an advantage.

If you have a 12v system and you decide to uprate it to 48v then if you do it right you will gain. If you charge a 12v battery and boost this to charge a 48v battery you will have worse results than using power at 12v because you have done nothing to improve the 12v system and you have added another loss.
Flux

[ Parent ]

Re: Matching the load (3.00 / 0) (#49)
by willib on Wed Mar 22, 2006 at 03:16:23 PM MST

i'm talking about a gen that has low volts output but lots of current..at say 155 rpm..
Carpe Ventum (seize the wind)
[ Parent ]

Re: Matching the load (3.00 / 0) (#50)
by Flux on Thu Mar 23, 2006 at 12:39:41 AM MST

yes that is what this is about. For the scheme as I proposed it the machine would be slow enough to work with the system voltage in high winds but need boosting for the low winds.

If your machine is too slow even in high winds you can boost through the complete speed range but the converter now needs to handle all the power.

Not a point I would start from, but if you were in that situation it would still work but at lower efficiency than choosing a machine with more volts.

You still have the advantage of tracking wind speed.
Flux

[ Parent ]

Matching the load | 63 comments (63 topical)

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