Author Topic: How many Turns do I need on a Coil to Charge the Battery?  (Read 25407 times)

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wooferhound

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How many Turns do I need on a Coil to Charge the Battery?
« on: July 28, 2010, 07:46:59 PM »
Build the machine without the coils, then put it out in the wind and let it spin. Determine the RPMs that it spins and take note. This will be "Unloaded" RPM. When you are making power the machine will be held back by drag while it is producing power. It will usually run at 60% of it's unloaded RPM when charging the batteries.

Letting a normal Horizontal Turbine run Unloaded is really dangerous. The speeds can reach self destruct, do not use that method for a hawt.

Take the machine down and make a single Test Coil and put it where the stator is going to be. Spin it up to 60% of your noted unloaded RPM and do measurements on your test coil. If needed make another test coil and repeat your measurements. When you get a coil that performs like you want it to, duplicate that coil 8 times so you end up with 9 coils to build a 3 phase stator. You can be very confident that your machine will perform as planned.

On a 3 phase machine you will have 3 coils per phase. After you rectify a phase you will get about 1.4 times the AC voltage of each phase. so you will want 3 to 4 volts per coil.

Also remember that Cut-In volts is the voltage where the battery will start to be charged. That will be 12 volts for a typical Lead Acid Battery. However that battery will not be fully charged till it reaches about 14.4 volts. Keep this in mind when determining your voltage and RPM requirements.

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« Last Edit: June 03, 2011, 01:47:06 PM by Bruce S »

wooferhound

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Re: How many Turns do I need on a Coil to Charge the Battery?
« Reply #1 on: July 28, 2010, 09:43:25 PM »
Somebody Help me here, This is a complicated answer, I would like to include formulas and Rule of Thumbs related to HAWTs.?

SparWeb

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Re: How many Turns do I need on a Coil to Charge the Battery?
« Reply #2 on: August 11, 2010, 02:16:05 PM »
I'll give it a try, Woof:

Unlike gravity, which is taught in high schools, electricity and magnetism are usually only subjects of university physics and engineering courses.  So very few of us get introduced to this stuff.  I count myself lucky to have had the class, and even so it took me a while to get from "theory" to "practice".

First, start with the "FLUX".  I've already written about that:  http://fieldlines.com/board/index.php/topic,143565.0.html    I hope it's a good start.

The flux is the amount of magnetic field that fits inside a closed loop.  You don't need to know a number or measurement of flux directly, but I'll add that it is measured in "Webers" (named after Max Weber who studied it).  All we need is the electro-motive force that can be generated.  We measure EMF with our multi-meters as a voltage, but ONLY when no current flows.

The formula you need:

EMF = N*F*f*2       Where:

   N = Number of turns in the coil
   F = the maximum Flux in the coil
   f  = frequency of change of polarity N-S

In the alternator, the poles pass N to S and N and so on.  One cycle N-S-N will flip the field direction, and hence the flux from N to S to N, and the EMF makes a sine wave.  Since the flux at the N pole was reduced to zero, then reversed to S, the total flux reversal is twice the amount Flux on one pole.  Each magnet is a pole, and most of these alternators have a dozen or so magnet poles.  Also, with increasing speed, the flux flips faster, and that makes much more EMF.

The other factor in the equation is N number of turns of wire.  Adding turns of wire increases the EMF (and open-circuit voltage).


Try an example:

I have wound a coil of 10 turns of wire and stuck it in the gap of the axial-flux alternator (with the magnets on).  Spinning at 60 RPM, I measure an AC voltage on the coil of about 1 volt.

Convert RPM to frequency:  (60 rev / minute) * ( 1 minute / 60 seconds) * (12 poles)  = 12 Hertz (cycle per second)

Convert measured RMS voltage on the meter to peak voltage (sine wave):   1.0 Volt * 1.41 = 1.41 volt peak
   (Note, multi-meters are not usually very accurate at low voltage or at low frequency, so be careful)

EMF = N*F*f*2       >  rearrange the equation...

F = EMF / ( N*f*2)   = 1.41 Volt / { (10turns)*(12Hz)*2 }   = 0.0059 Weber

If I want an open-circuit voltage of 12 Volts to cut-in at 180 RPM, then:

180 RPM => 36 Hz on a 12-pole alternator

N = EMF / (F*f*2)   = 12V / {0.0059Weber)*(36Hz)*2}   = 28 turns

Let's not forget that if the alternator is going to output 3 phase AC, then there must be 9 coils of wire, 3 in each phase.  Each of these coils must work together, and each is collecting the same amount of flux in phase with each other.  Divided among these 3 coils in the phase, we will distribute the needed 28 turns:

28 turns / 3 coil = 9.4 turns per coil   ... well actually we can either go with 9 or 10 turns.


If we wanted 48V, an also to cut-in at 180 RPM, then we can just scale it:

9.4 turns * (48/12) = 38 turns per coil   (if you calculate the whole thing again you get the same result)

If you used the same size of wire for the 48V coils as for the 12V coils, then obviously the coil would be much larger!  But actually to get the same amount of V*A power, you can use wire with a much smaller cross-section.  With 4x higher voltage, you need 4x less current, and the smaller wire can handle that.


There isn't really a simple (or cheap) way to measure flux directly, so I'd suggest the test-coil method, which is about as direct as you can get, and uses the alternator magnets exactly as you intend them to be used.  Also good for diagnosing problems, like one reversed magnet.
No one believes the theory except the one who developed it.  Everyone believes the experiment except the one who ran it.

System spec: 135w BP multicrystalline panels, regulated by Xantrex C40, DIY 8ft diameter wind turbine, regulated by Tri-Star TS60, 800AH x 24V AGM Battery, Xantrex SW4024

dlenox

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Re: How many Turns do I need on a Coil to Charge the Battery?
« Reply #3 on: August 11, 2010, 02:54:24 PM »
Steve,

Great example giving some good practical advice - but perhaps you can elaborate just a little more?

The last part of your example shows dividing the number of windings of a single coil by 3 to adjust for 3 coils per phase, but did not account for the coils being connected using star/wye vs delta vs IRP (independently rectified phases)

Since you simply divided by 3 my assumption is that the stator is delta?

Dan Lenox

SparWeb

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Re: How many Turns do I need on a Coil to Charge the Battery?
« Reply #4 on: August 12, 2010, 12:24:47 PM »
I stayed away from that just to keep it simple.  Perhaps too simple now that you bring it to my attention.

I should have added that this calculation work as-is only if the phases are connected in Delta, or if the phases are individually rectified.

If you are connecting the phases in Star (as they often are) then your target output voltage (say 12V) needs to be divided by 1.73, or the square root of 3, to get the EMF per phase. 

12V / 1.73 = 6.93 volts per phase of EMF

Then proceed with the calculation I've shown.  When going the other way, find the EMF per phase, and if you will wire it in Star, then multiply by 1.73.

Thanks for pointing that out Dan.
No one believes the theory except the one who developed it.  Everyone believes the experiment except the one who ran it.

System spec: 135w BP multicrystalline panels, regulated by Xantrex C40, DIY 8ft diameter wind turbine, regulated by Tri-Star TS60, 800AH x 24V AGM Battery, Xantrex SW4024

dlenox

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Re: How many Turns do I need on a Coil to Charge the Battery?
« Reply #5 on: August 12, 2010, 01:14:37 PM »
Steve,

That's why I pointed it out, if a newbie used the simple calculation they would end up with coils/stator that if connected in star/wye and would have the cut-in speed way too low for their intended application.

It seems that a large portion of people are connecting their coils in star/wye which makes it kinda the 'default' method as it takes less turns of copper to obtain cut-in speed.

Then the turbine would be running stalled due to too many windings in the coils.  In star/wye that multiplication by 1.73 throws it off enough to be very noticeable.

Thanks for the clarification,
Dan Lenox