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PM windgen questions | 14 comments (14 topical, editorial)
Re: PM windgen questions (3.00 / 0) (#5)
by Cinosh07 (sylviec@m.csvdc.qc.ca) on Wed Jun 8th, 2005 at 09:53:12 AM MST
(User Info) none

Hello,
There is a copy of an e-mail i've receive from ed from windstuff website. Couldbe usefull for you.

Cinosh07

The following formula will get you in the ball park for how many turns of wire you'll need...

Voltage= NARBP*2

where:
N= number of turns
A= area of 1 magnet in square meters (  the triangle neo's are .001525 )
R= revolutions per second ( rpm/60 )
B= flux passing the coils in Tesla's ( gauss / 10000 )
P= total number of poles

To find number of turns:

Voltage / ( P * R * B * A)/2 = Number of turns

B is confusing because not every one has a Magnetometer ( gaussmeter)  so I'll give you what I've come up with while working with different versions:

On a dual rotor system, not exceeding an air gap of .75" between the magnets you'll see 0.5Tesla through the coils
A simple coil over laminant will get around 0.4 Tesla throught the coils
And a slotted silicon laminant with an air gap of .030 between the magnet and stator will see slightly over 1 Tesla through the coils with these magnets.

As an example my 8 inch dual rotor shown on my site, using 0.5T in the formula to find the amount of turns per coil would go like this...j

Number of turns = 14volts / ( 12poles * 3.33rps * 0.5 Tesla * .001525 area) / 2 = 229 turns

There are 36 coils in that arrangement total and each phase has 12 coils....   229/12 = 19 turns per coil

Since its wired in star you can divide the 19 by 1.732 which equals 11 turns per coil

If you were using a 9 coil arrangement ( 3 coils in series per phase)  then 229/3/1.732 = 44 turns per coil

That should get you in the ball park for designing an alternator that will work with the blade you want.   You want to use the largest size wire you can fit into the space allowed for maximum power.   Its best to keep the resistance in the wire as low as possible to achieve maximum output.

As far as the relationship of coils per magnets... On a single phase machine you would have 1 pole per 1 coil ( so if you have 12 magnets you would have 12 coils for single phase ).   The normal 3 phase is 4 magnets per 3 coils.   So you can have 8 magnets and 6 coils or 12 magnets and 9 coils... the magnets are always added in multiples of 4 and coils are added in multiples of 3.   Another way is 2 poles per 3 coils where 8 poles would have 12 coils or 12 poles would have 18 coils.   And finally you can overlap the coils using a single phase arrangement with 3 phases where each pole would have 3 coils so with 8 poles you would have 24 coils and 12 poles you would have 36 coils.   The 4pole/3 coil arrangement is the most popular because of its simplicity and the coils are laid out in a single layer.

Hope this helps
Have Fun
Ed
www.windstuffnow.com/main

Cinosh07 The French Canadian R/E guy



Re: PM windgen questions (3.00 / 0) (#6)
by Flux on Wed Jun 8th, 2005 at 10:23:23 AM MST
(User Info)

Yes that's quite a convenient little formula. Remember that the flux densities quoted are for Ed's magnets so you will have to alter the Tesla to suit your magnets and air gap length.

The 2 x1 x 1/2" magnets will be nearer 0.7 T  with about 3/4" gap for dual rotor.

Flux

[ Parent ]



Re: PM windgen questions (3.00 / 0) (#7)
by scoraigwind (magnet@scoraigwind.co.uk) on Wed Jun 8th, 2005 at 05:13:40 PM MST
(User Info) http://www.scoraigwind.co.uk

I am not sure exactly what is meant by voltage in ed's equations.

Here are some useful equations for coil design

Average open circuit voltage (emf) for an alternator will be

Eave = 2*n*A B(rpm/60) volts

where Eave is average emf per phase
n is turns per coil (multiplied by number of coils connected in series)
A*B is flux in Webers
A is total area of poles in square metres
B is average flux density at poles in Tesla
(rpm/60) is revolutions per second

The peak voltage will be about 50% higher Epeak=1.56* Eave

If a three phase arrangement of coils is used then they are usually
connected Star (Wye) giving root(3)1.73 higher output volts.

If this is fed through a rectifier, then the output is approximately
the peak with the forward biassing voltage of the diodes subtracted

Edc = Epeak*root(3)-1.4 volts for silicon diodes.

In this way you can establish the cut in speed of an alternator.
Where Edc = Vbat=battery voltage, the rpm can be found.

rpm = (Vbat+1.4)60/(1.73*1.56*2*n*A *B)

You can also use E to predict the output current.  There is no
current until Edc>Vbat.  After that, current I will be

Idc(Edc-Vbat)/R amps neglecting reactive effects (self induction)

Where the coils are wound on laminations, the inductance is strong
and it is hard to predict output current.  The current will be
limited by the inductive reactance at higher speeds.  With air-core
coils in a stator which does not contain iron, we can look at the
resistance R in ohms as the main impedance.

R =L/Aw*0.022(1+0.004*(temp-70)) ohms

where L is the length of copper wire in the coils of one phase in metres
Aw is the wire cross sectional area in mm (pi()*square(diameter)/4)
temp is wire operating temperature in degrees C

Copper Loss = square(Idc)*R watts
Hugh Piggott http://www.scoraigwind.co.uk
[ Parent ]



Re: PM windgen questions (3.00 / 0) (#8)
by kenputer on Wed Jun 8th, 2005 at 07:33:28 PM MST
(User Info)

Couldn't have said it better myself,ain't it great that we know so much and our exsplaination are so easy to understand! What I mean is I'm lost with that one:-(

[ Parent ]


Re: PM windgen questions (3.00 / 0) (#9)
by Chris05 on Wed Jun 8th, 2005 at 10:35:03 PM MST
(User Info)

Hey, i know those magnets are pretty small :(
Its hard to find stuff within Australia at a reasonable price, i usually resort to overseas buying for stuff like this which i might end up doing.

Those formula's are great, accept i dont have a magnetometer... My dad might have one, ill ask him when he gets home.

Thanks for your help guys.

[ Parent ]



Re: PM windgen questions (3.00 / 0) (#10)
by Flux on Thu Jun 9th, 2005 at 01:00:41 AM MST
(User Info)

Good point, Ed's formula should give ac average volts per phase, for rms the factor should be 2.2.

I see that in his example he is treating it as though it predicts the dc or peak ac volts.

Perhaps he has found it gives the right answers with what he thinks is the right flux value.

Flux

[ Parent ]



PM windgen questions | 14 comments (14 topical, 0 editorial)

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