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Dual Rotor Toroid Core PMA build

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CraigM:
Ed,

Thank you for the reply. If I used a multimeter that reads true RMS then I'm okay with using 1.4 as the AC RMS to DC conversion? Then subtract 1.4 volts from DC volts for rectifier drop?



I repeated the test using a cheapo meter that doesn't read to as many decimal places and received a VAC reading of 1.0 to .9 on the 20 AWG coil, compared this with the VAC reading of 1.225 when using the true RMS meter. Believe this confirms the multimeter used in the first test does read true RMS.

I tried reading resistance (Ohms) with the good multimeter and with a Simpson analog meter I picked up cheap several years ago. I was getting readings that were not consistent and with not fully understanding what I was doing I didn't use this data.

I did research Ohms for magnet wire and found the following on several different websites;
15 AWG 3.18 ohms per 1000 feet. 3.18 / 1000 = .00318 ohms per foot
15 AWG .0104 ohms per meter
15 AWG using online calculator shows 6' of AWG 15 = .019 ohms. (each coil uses 6' of wire)

Circumference of coil is 6” (2” top, 2” bottom, .3125” radius half circle both ends is 1.96”.
15 AWG coil uses 12 turns x 6” per turn = 72 inches = 6 feet. .00318 x 6 = .019 ohms.
If using meters, 6 feet = 1.829 meters. 1.829 x .0104 = .019 ohms.
Online calculator showed .019 ohms per 6' of AWG 15.

So I feel pretty confident that each coil has a resistance of .019 ohms. Multiply this by 24 coils and I get .456 ohms. Checked this with online calculator for 144 feet (6' per coil x 24 coils) of 15 AWG and it gave a result of .458 ohms... close enough.

So now I know VAC, VDC and ohms per coil and for all 24 coils. A little extra wire will be needed for connections but I'm still in close enough mode.

Found a calculator online that will now give me amperage and watts. This looks really high to me... I was expecting watts to be much lower at cut in voltage.

Here's the calculator using 13.27 volts and .458 ohms... oops forgot, need to subtract 1.4 volts for rectifier loss... that gives me 11.87 volts. Only thing I'm not sure of is if I should enter VAC or VDC into the calculator.... thoughts on this?



My thoughts and/or questions on this testing.

* I'm seeing nearly 26 amps at 190 rpm, believe max amps for 15 AWG is around 28. Wire gauge seems to be too small.
* Using data I have can I calculate voltage if adding a second layer of winding? If 12 turns of 15 AWG = 11.87 VDC will 24 turns give me 23.74 VDC? This will also lower cut in speed for 12 volt system or will approach cut in speed for 24 volt system.
* For a HAWT how do I match blade size & tip speed ratio to this alternator?
* For a VAWT it appears I can bring cut in to a lower RPM and still use direct drive.
I'm open to anyone's thoughts. As I mentioned earlier I'm not in a good wind location and I'm not interested in spending the time and resources to put a HAWT 100 feet in the air. If I can feed a bit of wattage into my shop and run a smallish 24 volt system with a cheap inverter I'll enjoy the journey along the way and feel a bit of pride in the accomplishment.

Thank you much,
CM

MagnetJuice:
I'll attempt to answer some of your questions.

The voltages that you measured with the True RMS meter are probably correct. It was after midnight last night when I posted and I was just thinking of arriving at the voltages using an equation. So go with what your meter tells you.

A regular multimeter, even the good quality ones, cannot accurately measure very low resistances. For that you would need a milliohmeter. Average price for one is about $1000. I have an accurate milliohmeter that is good to measure from .1 to about 3 ohms. I built it myself using a small wallwart and a small constant voltage, constant current module. If you don't have a milliohmeter, then using those tables where they give the ohms per 1000 feet is good enough.

Now, concerning the wire size for the coils. With 3-phase alternators current flows through each phase for only 66% of the time. If total current is 30 amps, current through one phase is 30/1.73 = 17.3 Amps. That means that 15 AWG wire would be OK if the total Amps output of the alternator is about 50 Amps. You could see over 50 Amps if a hurricane hit Pensacola.  :)

If you don't have good winds in your area, it would be pointless to go with the big expense of a tower for a HAWT. And a VAWT would turn slow if it is a Savonius so you would have to use a lot of turns on your coils to produce good voltage at low RPM if it is going to be direct drive. A good thing is that if you increase your voltage, you can use a thinner wire, for example 17 AWG.

By seeing how meticulous you are in your construction and design of this system, I am confident that you will end up with a nice working windmill.

Ed

CraigM:
Hello Ed,

Thanks again for the help and direction you provide. It's greatly appreciated.

I spent the past few days researching HAWT blade design and tip speed ratio. Found that Hugh Piggott and if I remember correctly the “Dans” like a TSR of around 7. Also remembered seeing quite a bit of discussion about GOE 222, constant chord, no twist no taper blades. TSR for GOE 222 appears to be best in the 5-6 range. Chris Olson used these on most of his machines and strongly favored them... but then Chris had a strong opinion about many things... I miss Chris, Flux and many others, and the lively discussion from years ago, but nothing stays the same. Solar took over on price and the interest in wind power has waned.

So I wanted to relate TSR to wind speed and to alternator cut in speed. On Hugh's first wind turbine recipe book there is this formula for calculating TSR;
~ Rpm = windspeed (in m/s) x tsr x 60 / circumference.
My best guess at blade diameter to match my alternator is 2.4 meters or 8 feet. Using Hugh's formula to determine cut in speed at first useful wind speed of 7 mph (or 3 m/s) I get the following.
~ 3 (m/s) x 7 (tsr) x 60 / 7.539 (2.4 x 3.14 = 7.539) = 167 rpm
To get to 190 rpm... this is where I reach 24 volt cut in with two layers, 24 turns of 15 AWG I need to bump the m/s wind speed to 3.5 or 8 mph. So I may lose ever so slightly in the lowest wind by needing 190 rpm for cut in. And just for comparison in a 10 m/s (22 mph) wind the formula gives me 557 rpm.

I ran the same formula for a GOE 222 blade and used 6 for the TSR. They may be able to run at 5 without stalling but I have no hands on experience to validate this.
~ 3 (m/s) x 6 (tsr) x 60 / 7.539 (2.4 x 3.14 = 7.539) = 143 rpm
~ 10 (m/s) x 6 (tsr) x 60 / 7.539 (2.4 x 3.14 = 7.539) = 477 rpm
My understanding of GOE 222 is that they deliver greater torque and work well with a stiff alternator. With no experience I can't say if my alternator will be stiff or not. Will there be magnetic drag on the core? I don't know. The ferrite magnets don't feel strong enough to produce much drag. Anyone have an opinion on this? Chris?  :)

So now I'm at a spot of what gauge wire to use, how many winds and even if I should be moving forward with a 24 volt or 12 volt alternator.

When using 15 AWG wire, 24 turns (two full layers) I get 25.10 volts including 1.4 volt rectifier drop at 190 rpm. Using the same wire size and turns I get 12.6 volts at 90 rpm. So I'm a little high on rpm at 24 volts and will not cut in until 8 mph, 3.5 m/s winds and quite low rpm for 12 volts.

When using 17 AWG wire, 32 turns (two full layers) I reach 26 volts at 142 rpm which is just right for 8 foot GOE 222 blades and 12.8 volts at 70 rpm. Almost feels like I could use the alternator to run a HAWT at 24 volts or a VAWT at 12 volts.

So with regards to wire gauge, number of turns and tsr I just need to make a decision... $hit or get off the pot is what I hear my father saying.


Now for something that is frustrating me and can't seem to understand. How do I determine amperage from my voltage testing? Do I need to test with a load? Sparweb mentioned doing this. Sparweb? Using a formula from the chart posted earlier I can determine a number but what does it represent? Amperage = volts / ohms, looks simple enough. I have a voltage reading at a certain rpm and I know resistance, ohms for the total length of wire that will be used.
~ Voltage is 25.1 at 190 rpm, resistance is .917 ohms at 288 feet of wire = 27.37 amps which is 687 watts.

Where are my wheels falling off the bus? Using another handy formula in Hugh Piggotts recipe book he says blade power in the wind equals .15 x rotor diameter (squared 2) x windspeed (squared 3). This gives me .15 x 6 (2.4 meter squared 2) x 42.9 (3.5 m/s squared 3) = 39 watts.

If I only have 39 watts available with 2.4 meter blades at 3.5 m/s wind speed but I can get 190 rpm at 3.5 m/s at tsr 7 how is it the calculation above shows 687 watts?

I know a battery clamps alternator voltage down to battery voltage so is 687 watts the practical maximum available at .917 ohms resistance?

Voltage x amps = watts so at cut in with 25.1 volts I should see amps of around 1.5 to equal the 39 watts available to the blades.

Can I measure / plot volts and amps along a rpm range using only volts and resistance I now have?

Many thanks to all. It's now 5:11 am and I've been up for awhile... couldn't sleep from thinking of this.... way too much at times.
CM

SparWeb:
Chris,
Very busy at work trying to win a bid for the coolest project I've ever done - and may ever do in my career.
I have made you wait for a response. 
Maybe in the meantime you've figured it out - or just done your business and got off the pot, as they say.

The balance between blades and alternator comes from mechanical power.  The only connection they make is the steel shaft between them, which transmits only torque at some speed or other.  As you have gathered, this pinch point in the variables makes it possible to solve the problem of matching blades to alternators, but it forces you to solve BOTH systems before you can see where the curve for one crosses the curve for the other.

So whether the blades are GOE222 or NACA0012, the power in the blade disk is generally about the same.  The airfoil choice alters subtle things, like the angle as-carved to get the needed angle of attack, and the torque to get started when they aren't turning.  There are some big things affected by airfoil, such as the L/D ratio, but when talking about un-twisted blades that doesn't matter because there is only a small portion of the blade running at max L/D.  The rest of the blade span is way off the ideal angle and the root is nearly stalled.

So to get rough matching, just use (power in) = (power out)
Power In = aerodynamic power that the blades can sustain a given RPM
Power Out = Mechanical power required to drive the alternator at a given RPM

You've got what you need to work out the alternator side.  The blades power can be worked out approximately enough now, too.

Number crunching will take some time for me to get into.  I might prefer to just open up the calculations I did years ago and run them with your test numbers.
I've got errands to do this morning but I'll try to get back soon. 

CraigM:
Sparweb,

Thank you for your input, as always, GREATLY appreciated!

Good luck with your bid, sounds like a very exciting time for you. Focus yourself on the task at hand.

I'll review your original calculations and see what I can come up with.

Thanks again, don't over concern yourself with getting back right away. No hurry on this end.

Take care and win that bid!
CM

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