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Matching the alternator to the blades


By DanB, Section Wind
Posted on Tue Feb 13th, 2007 at 02:00:52 PM MST
Some discussion about the alternators we build for 10' machines

This is sort of an interesting chart we made up yesterday - based upon the tests we did last week.  Interesting to think about matching the alternators to the blades.

Alternator 1 uses the 1" x 2" x 1/2" blocks (N40 grade) and the coils are wound with  140 turns of #17 gage wire.  The cutin speed as tested was 140 rpm.  The resistance of this stator is around 3Ohms, so you can figure how much power is consumed overall - the curve only reflects what we get out - the power into the shaft is always more, especially at higher outputs.  This one is about 50% efficient at 700 Watts.

Alternator 2 uses teh 2" diameter x 1/2" discs (N40 grade) and the coils are wound with 110 turns of #15 gage wire.  The cutin speed of this one was around 130 rpm.  Resistance in this one is about half (1.5 ohms or so) of alternator 1 - so at any given output there's about half the heat in the stator.

'Betz' (the blue line) is based on a 10' diameter blade running at TSR 6.  So that is the power curve of a 'perfect' 10' diameter blade.  To get wind speed you can divide rpm by 16.8.  

Of course our blades are not nearly as good as they could be, so the 'betz' curve there doesn't reflect reality.  The TSR of the blades is also bound to change at different wind speeds because it's impossible to perfectly match the power curve of the blades with the alternator (maybe you could with an electronic MPPT controller.  In order to maintain the right TSR all the time, the power produced by the alternator (Both useful power and wasted power) would have to exactly match the power curve of the blades.  

The power curve of the alternator is linear.  It slumps off at the higher end on the graph, that's because of the wasted power (heat in the stator) is not shown there. The power curve of the blades is related to the cube of rpm and windspeed.  So a good match is tricky and we shoot to line things up best we can between about 6 and 25 mph or so.  (an intelligent MPPT controller should be able to match things up nearly perfectly I suppose)

Any time the power produced by the alternator (again, we have to consider power wasted in the stator and the line) is less than the power produced by the blades, then the blades will run at a higher TSR, and if the power produced by the alternator is less than that of the blades, then the blades will run at a lower TSR  - if it's too low, the blades will stall.

When we consider the power wasted in the stator, then Alternator 1 is a pretty good match to the blades - assuming our blades are somewhat less efficient than the betz limit, and so long as we furl at 600 - 700 Watts output or so.  This graph seems to agree with the sort of performance we've seen from those machines.

Alternator 2 would stall the blades and as it is - this would happen in very low winds because the cutin speed is too low.  It should be a pretty good match though if we open the airgap and add about 1 ohm to the line I figure.  Some don't like the idea of a wide airgap, but I sort of like the wide mechanical clearance - it's a bit wasteful of magnets but adds a safety factor I think.  The other safety factor in this alternator is greatly reduced heat in the stator.  Instead, we'll have that heat  in a resistor, between the rectifiers and the battery.

This is a 1 Ohm 1KW resistor which should be a good start in matching alternator #2 to a 48V battery.  Of course, if it were a 24 V machine we'd use a .25 ohm resistor and at 12 Volts it'd be 1/4 of that (about .06 ohm).  For a lower voltage machine one might want to build the heavier alternator just to make up for line loss - use the line as the resistor.

This is all sort of simple stuff but I thought I'd post it because I think the heavier alternator is a bit safer for 10' machines.  It might also be useful to some people building new machines to understand why you cant stick a big blade on a tiny alternator.  Getting the alternator matched to the blades is the trick with all this stuff.  

Matching the alternator to the blades | 21 comments (21 topical, 0 editorial)

Re: Matching the alternator to the blades (3.00 / 0) (#1)
by Titantornado on Tue Feb 13th, 2007 at 08:10:49 AM MST
(User Info)

Thanks Dan. Matching is the one thing I still don't understand.  At this point, I am strictly guessing at my prop/alternator match, as there just doesn't seem to be definitive "this goes with that" numbers and a lot of it comes right down to trial and error.  Hopefully, I'll be fine.

On a side note, my project is currently stalled.  Too many necessary expenses are consuming any normally disposable income right now, along with an unstable job after a recent company buyout.  Hopefully, I can get the steel needed soon, so I can weld this thing up and finally see some electric come out of it.

Rod



Re: Matching the alternator to the blades (3.00 / 0) (#2)
by outback (dogs3-at-sitestar-dot- net) on Tue Feb 13th, 2007 at 08:31:25 AM MST
(User Info)

I was just wondering were those resitors came from or out of?
You can lead a horse to water,but you cant make him eat the trough.


Re: Matching the alternator to the blades (3.00 / 0) (#3)
by DanB (danb@*no spam*otherpower.com) on Tue Feb 13th, 2007 at 09:03:55 AM MST
(User Info) http://www.otherpower.com/

I dont really know - they're new (never used).  I got a bunch off ebay.

[ Parent ]


Re: Matching the alternator to the blades (3.00 / 0) (#4)
by willib (willibur at comcast dot net) on Tue Feb 13th, 2007 at 09:10:22 AM MST
(User Info) http://www.njwind.com/webcam.html#

i had a similar question
but i was wondering how big they were.
what size bolts and nuts are those ,that may help me give it a size relationship.
Carpe Ventum (seize the wind)
[ Parent ]


Re: Matching the alternator to the blades (3.00 / 0) (#5)
by DanB (danb@*no spam*otherpower.com) on Tue Feb 13th, 2007 at 09:13:34 AM MST
(User Info) http://www.otherpower.com/

They're about 10" long, 2" diameter - made by 'ohmite'.  They mount nicely on 1/2" hardware.  

Ebay is a really good place to find bargains on big power resistors.

[ Parent ]



Re: Matching the alternator to the blades (3.00 / 0) (#13)
by Drives (drivesdeanw at yahoo.com) on Tue Feb 13th, 2007 at 03:45:41 PM MST
(User Info)

Dan:

In the Drives industry this is a common resistor.  You find this value resistor typically used on DC drive applications to allow fast stopping of the load.  For example a crane's bridge or hoist motor drive, a printing press, ski lift, etc.  

We know that DC motors can act as generators.  As a generator, we can send power from the armature into a large resistor to actually turn the DC motor into a brake.

This feature is used to stop a load by taking the load's motion (inertia) and turning it into generated power creating braking torque.

This function to connect the armature to the resistor is performed by use of a "dynamic" brake contactor.  Once stopped, the load is held with a mechanical disc or drum brake.

AC drive dynamic brakes are the same wattage per HP size, but are higher resistance values, and probably not of much use for windmill applications.

Dean
---------Dean-----------
[ Parent ]



Re: Matching the alternator to the blades (3.00 / 0) (#6)
by Nando (nando37-at-tx-dot-rr-dot-com Correct theanti-spam) on Tue Feb 13th, 2007 at 10:27:47 AM MST
(User Info)

DAN:

Your idea of doing that chart is very good.

Check Betz curve that defines the power harvesting of a 10' (feet) machine and compare the root initial power and the discrepancy of the harvested power by the two mills.

The error maybe due to different setups without correlation to the 10 feet mill and the Betz of the two mills that may have a different power curve.

To have a GOOD correlation, I think, that instead of watts a initial cut in voltage and current is needed, then with the stator resistance a total power can be calculated to assign power lost due to the stator and power harvested by the load, though the stator-rotor gap voltage is not available. ( not measured I presume) a close total efficiency can be defined.

Determining the cut in voltage and current plus the mill size, one can calculate the Betz for such wind mill, assuming constant TSR, then data ( current and voltage) taken of the generated power with the stator resistance one can get two curves,
. 1) curve for harvested power into the load and
. 2) the power lost due to the stator resistance
which adding both, will show a total power that is below the Betz Curve, taking the difference from the Betz, One will find the stator-gap power loss ( also absorbed by the generator)

This way of taking data is too general and error prone due to lack of accurate Betz power definitions, which really requires a torque measuring device plus an accurate RPM reader attached to the mill shaft, but for this experimental basic conditions, I presume it is good enough to give a basic generalized power curves.

MPPT indeed will show a more accurate harvesting power curve with reduced stator losses, though the stator-rotor gap losses still be the same with equal currents levels.

Nando




Re: Matching the alternator to the blades (3.00 / 0) (#7)
by DanB (danb@*no spam*otherpower.com) on Tue Feb 13th, 2007 at 10:40:23 AM MST
(User Info) http://www.otherpower.com/

Hi Nando -
Im not clear on what you mean by 'stator-rotor gap losses'.

I also think you may misunderstand the chart...
I put the betz limit on there just to show the cubic power curve of the wind at the sort of power the blades might make if they're perfect.

The alternators were not tested as wind turbines - those numbers come from testing alternators on the ground with the tractor driving them.  So there is not very much error there.  

[ Parent ]



Re: Matching the alternator to the blades (3.00 / 0) (#9)
by Flux on Tue Feb 13th, 2007 at 11:37:40 AM MST
(User Info)

Dan, that has me lost as well . I have a feeling that it is something that gets lost in translation, many technical terms translate very strangely.

I think Nando has had more experience with iron cored alternators and I think this gap term has something to do with synchronous impedance. Iron cored alternators have drops due to leakage reactance and armature reaction in addition to the resistive losses.

The whole effect is generally lumped together as synchronous impedance and it is quite an elaborate procedure if you want to separate armature reaction from leakage reactance.

With our air gap construction these factors are small enough to ignore and resistance alone is good enough to determine the fall in voltage with load.

flux

[ Parent ]



Re: Matching the alternator to the blades (3.00 / 0) (#10)
by Nando (nando37-at-tx-dot-rr-dot-com Correct theanti-spam) on Tue Feb 13th, 2007 at 12:19:04 PM MST
(User Info)

Dan:
Your testing and the Betz got me confused !!!.

The Stator-rotor gap is an existing fact that represents a voltage drop (in essence it is a reduction of the magnetic field) that can be represented as a resistance ( which causes power losses).

Nando

[ Parent ]



Re: Matching the alternator to the blades (3.00 / 0) (#11)
by DanB (danb@*no spam*otherpower.com) on Tue Feb 13th, 2007 at 12:28:44 PM MST
(User Info) http://www.otherpower.com/

Yes, sorry to be confusing.
the 'betz' curve is based on a perfect 10' blade running at a constant TSR6 - so it's sort of the 'ideal' power curve.  The alternators are just data based on tests we did with the tractor into a 50V battery.

Im still confused about the stator rotor gap...

[ Parent ]



Re: Matching the alternator to the blades (3.00 / 0) (#12)
by Flux on Tue Feb 13th, 2007 at 12:44:15 PM MST
(User Info)

From Nando's comment , he does mean armature reaction.

With a conventional alternator with iron circuits and particularly with iron poles and wound fields, the magnetic field from current in the stator reacts on the field on the rotor and modifies the field.

With a purely resistive load the armature reaction just distorts the main field, but if any part of the iron is saturated it effectively results in field weakening.

Inductive loads result in field weakening, that is one of the reasons that small alternators are not capable of starting reasonable sized motors.

Capacitive loads cause field strengthening and a rise of volts.

Neo more or less acts as a large air gap ( saturated material) so pma's even with iron cores are much less affected by this effect ( magnet steels could be demagnetised)

The air gap alternators with neo show negligible change in gap flux with load so we can ignore it.

flux

[ Parent ]



Re: Matching the alternator to the blades (3.00 / 0) (#8)
by Nando (nando37-at-tx-dot-rr-dot-com Correct theanti-spam) on Tue Feb 13th, 2007 at 10:59:50 AM MST
(User Info)

DAN:

I would like to add about the RESISTORS.

I have a power resistor that is 0.6 ohms good for at least 4 KW, weight at least 12 pounds of so, the unit has two elements, spot welded, in parallel and if needed one can be cut and remove from the ceramic body.

The units looks like yours, but it is oval instead of rounded.

Size : (Inches) 5 * 2.5 * 20 and including the ceramic ends 25 inches

I believe that there are more than 100 available and if there is interest, I will photograph the resistor plus weight it to determine the shipping costs.

The power resistor is good for dump controllers, and of course your idea of series resistor for the mill.

Nando



Re: Matching the alternator to the blades (3.00 / 0) (#14)
by commanda (alwynne at unwired dot com dot au) on Tue Feb 13th, 2007 at 04:33:54 PM MST
(User Info)

I think another way of interpreting that graph, is to say that alternator 2 requires slightly bigger blades, to get the Betz curve above the alternator output at any given rpm.

Amanda



Re: Matching the alternator to the blades (3.00 / 0) (#15)
by tljones (tljones@venturecomm.net) on Tue Feb 13th, 2007 at 06:50:06 PM MST
(User Info)

Hate to be the dope in the midst of a really good thread. But if I understand correctly, the blades will stall because the alternator is harvesting more power than there is available from the blades harvesting the wind right????

So the resistor reduces that, shifting the cut in and the curve to the right and dropping it below the Betz curve.
Or as amanda says add larger blades to lift the curve vertically and getting it above the betz curve.

But as is any RPM between cutin and where it falls below the betz curve will stall the blades.

OK so far I think.....

What happens when it stalls???? I realize this is the dope question but indulge me will you??? If the aerfoil stalls it will lose power, hence power and rpms, until it drops below cut in right??? Then it will accelerate above cut in and fall back. If enough wind was applied would it ever accelerate through the magic RPM line and take off into full efficiency???

Thanks all, hope I added something to the thread for the other non EE's out there.

Tom
Lord, help me be half the man my kids and dogs think I am, or a quarter the man my wife wishes I was.....
[ Parent ]



Re: Matching the alternator to the blades (3.00 / 0) (#18)
by Flux on Wed Feb 14th, 2007 at 02:20:42 AM MST
(User Info)

tljones

Life is far from simple so don't feel bad about being confused. Blades produce their designed output at one specific tip speed ratio. run too slow or too fast and the power falls off. There is a band of tsr where the thing works fairly well and if you keep within that you will have good results. What happens when you go too fast depends a great deal on the aerofoil but we needn't worry about that condition.

As you load the prop the power starts to fall off the peak of the curve, quite slowly at first and as you load things more the fall becomes steeper. When you reach true stall (angle of attack of wind above about 12 deg) then power falls off very rapidly indeed. That is the region that you need to avoid.

If the cut in speed is too low, you will start to fall below optimum tsr almost immediately and by 10 mph you may approach true stall and it then becomes so drastic that you may increase wind speed from 10 to 25 mph with negligible increase in power.

The trick is to start with the prop above design tsr at cut in. ( A prop with nominal tsr of 7 will still give near 80% output at over tsr 8 and possibly up to 8.5)

You will find that tsr drops very rapidly and you may be down to the intended 7 by 8 mph. At 10 mph you may be down to tsr6. With the cubic power of the wind you may still be able to hold the tsr above tsr5 up to 25 mph if the alternator is matched just right.

If the alternator is too powerful you will fall below tsr5 and hit hard stall.

Playing with the air gap alters the cut in speed so you would select that to give a cut in with tsr over 8 at 7 mph. If you stall higher up the curve then that is where the series resistor comes into play. It doesn't affect cut in speed.

It seems odd that wasting power in a resistor can give you an increased output, but you have to remember that if you can keep the prop out of stall you have a cube law power increase with wind speed. Loss in a resistor is proportional to I^2, so when you consider the small percentage of resistor loss compared with the power into the battery you can afford to loose significant electrical power if it gets you more input power from the prop.

Because of this flexibility of the prop to run at a band of tsr reasonably efficiently it becomes extremely difficult to compare input and output curves.

Dan's curves give a good idea of what happens with a prop that would only work at design tsr. Real life makes things less critical that it seems from those graphs.

Some days ago someone suggested switching out the resistor to keep the low wind performance, if you are reading this you will find that it makes negligible difference and may even not be an advantage, it is certainly not worthwhile.

The prop tsr falls so rapidly in the first few mph wind speed increase that you are even loosing prop power. Also the power lost in the resistor at low currents is very small. The two factors combine to make it a waste of effort to switch the resistor out.

Flux

[ Parent ]



Re: Matching the alternator to the blades (3.00 / 0) (#19)
by s4w2099 (movlw0x13h@yahoo.com) on Wed Feb 14th, 2007 at 08:50:17 AM MST
(User Info) http://www.s4wsbox.com/

I think that a good analogy would be a turbocharged car. Using a turbo will take some of the engine's power by increasing the resistance of the exhaust but when driven at the right RPM the gains in power are far more than what you loose adding resistance to the exhaust.

Hope it helps

[ Parent ]



Re: Matching the alternator to the blades (3.00 / 0) (#20)
by jmk on Thu Feb 15th, 2007 at 08:31:23 AM MST
(User Info)

 Thank you Flux, That was very understandable. You are a great asset to this board. We could never say enough. You put in so much!
jmk
[ Parent ]


Re: Matching the alternator to the blades (3.00 / 0) (#16)
by DanB (danb@*no spam*otherpower.com) on Tue Feb 13th, 2007 at 09:21:48 PM MST
(User Info) http://www.otherpower.com/

Hi Amanda - yes, with the low cutin speed and higher output it could take larger blades (but then we might be back on risky ground with regard to burning out).  I prefer - with this one, to open the airgap and add resistance back so it's a good match for 10' blades.  But yes - as it is I expect 11' or slightly larger blades should work well if it furls early enough.

[ Parent ]


Re: Matching the alternator to the blades (3.00 / 0) (#17)
by SparWeb (sparweb at ANTISPAM_hotmail_com) on Tue Feb 13th, 2007 at 09:25:40 PM MST
(User Info)

Dan,

I love graphs.  Thank you for posting that.  I have been wondering for a long time why you are using so many turns of wire in your alternators for so long (Alt #1 in your example) and here, finally, I can see why.

I pulled out my calculator, and I can see the ~50% efficiency of your alternator on a 48V battery.  I was very surprized to see how much worse the efficiency becomes when the battery voltage is set lower, even 24V: at 700W the current doubles, the lost power quadruples, and thus the required power input jumps by 2000 Watts!  (If I've done my math right).  That has a very large impact on the required prop size.

I see...

In the coming week or two, I'm planning to put my alternator through some lathe run-up tests, and I'll make a chart just like that.

I regularly find 300W resistors with low values (1.0 ohm, etc.) at surplus stores like Princess Auto (a Canadian chain, but there ought to be equivalents in other countries).  I've connected one, with a resistance of 0.16 Ohm, as a shunt to measure current when I do my lathe tests.

Note: for those having difficulty interpreting the graph, bear in mind that Dan derived the "Betz" curve by simple calculation - it is not a test result from an actual set of blades.  The formula is: P= (rho/2)x(V^3)x(pi/4)x(D^2)xCp

Dan would probably also admit that the power coefficient of his blades is below 40% (What would you say Dan?) so you can't assume that an alternator that just fits below a Betz limit curve won't stall the blades.

TLJones: the Alternator 2 requires too much torque for a set of 10' blades to turn.  From 150 to 300 RPM the prop can't overcome the load.  You can see this from the chart because at, say, 250 RPM, the alternator produces 1200W, but the blades can only put in 800W.  Just like hitting the deep wet grass with your lawnmower - stall.

If you have a set of blades that you know has a power coefficient of, say 30%, then you could re-create the curve Dan made and see how close your "Power In" curve gets to your alternator load curve.

Again, thanks for posting this, Dan.
Steven Fahey



Re: Matching the alternator to the blades (3.00 / 0) (#21)
by Devo on Tue Feb 20th, 2007 at 03:51:53 PM MST
(User Info)

If you were to use less "wraps" in the coil to not only increase the cuttin but also reduce the resistance would this make the stall go away as it cuts in at a higher rpm or worse because the resistance is less? This is the part I still have trouble with when varying from proven methods. Just Curious.

I can't wait to see the actual test results from the mill on the tower.....

Keep up the good work

Devo.  



Matching the alternator to the blades | 21 comments (21 topical, 0 editorial)
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