magnet and coil advice.

[theTinker]

My magnets should arrive within the next few days and im nearly finished building my stator backing with laminates so i would like advice on the coil windings, wire size, and magnets.

Can people suggest the best arrangement of magnets and coil windings please? i can experiment a little but a good starting point would be very helpful.

--------------Details of Turbine:-------------------

12" Single rotor, //Break disc

Laminates, //Steel strapping from pallet wrappings

24. Magnets available, //Size 2" * 1" * 0.5"
    
3. Phase config(star most likely)
   
5. ' 10" diameter blades, //they are pretty newbie built so dont count on them too much
   
   

Low Winds most of the time.

expect about 1/4" gap between mags and coils(slight wobble)

Goal: charge a 12 volt battery pretty constantly

If i use 12 i have 1" space between magnets on their inner circumference.

If i use 16 i have 0.57" space between magnets on their inner circumference.

(not sure if this is too close or not)

for coils

if i use 12 mags, ill have legs 0.9" wide,

if i use 16 mags, ill have legs 0.55" wide,

user "troy" did sum really great experiments on the old board which convinced me that the coil former should be the same size as the mags or just slightly smaller.

The coils will be resting on top of the laminates as in dans wood AX machine.

The blades are being built atm as TSR n stuff are way above my skills at the moment.

The came out fair i think for a first timer and they have that twist, air foil is not as pronounce as i would like and its a little further back than a 1/3 as wanted. They obviously won't be extremely efficient. I'll gladly sacrifice amps for volts as a slow charing battery is better than a no charge

Any other info needed i can dig up quickly,

Thanks for the help.

[willib]

you have 12" dia rotors , have you placed the mags to the perimeter?

[WXYZCIENCE]

Don't play around with two of them. I found the neo multiplier principal works well. 1+1=6 magnets. Joe.

[hiker]

you won"t save on weight with a single rotor type--the lams will weigh about the same as a single rotor--so just go with a dual rotor-9&12 combo--3 phase...

with those mags you will have plenty of volts and amps if wired up right..

try #14 wire maybe 65-75 turns should work out okay..

[hvirtane]

Whatever your arrangement will be,

two magnet rotors or one magnet

rotor, I think that it is best

to make a test coil.

There is a good explanation how

to use a test coil:

http://www.fieldlines.com/story/2006/2/2/22716/60628

- Hannu

[theTinker]

no i Havent placed any mags yet, they wont come for a few days. but in my calculations i have placed the mags at the edge yes

[theTinker]

I cant do a dual yet. not enough bearings, nor can i find another 12 inch brake disc,

the gargages said that its from a large trailer or sumthing.

ive only got 2 of regular car brake discs.about 8-9 diameter i think. and even then no 2 have the same bolt pattern

ill be doing another mill with dual rotor and proper blades in january.

for now i just wanted to learn as from as many mistakes as i can and test my woodworkings skills.

[theTinker]

Thats a very post. numbers are a bit tricky but quite helpful.

If i build my mill without magnets and wire and put it in the air, could i use this RPM(As i dont know what it will spin) as the rpm in them calculations?

[hvirtane]

It isn't a really good idea to try measuring

RPM with the freely spinning wind rotor.

When spinning freely the wind rotor

will take very high RPM. It is better

to try designing the wind rotor so that

TSR will be... let's say 6, and then

try fixing the desired cut-in speed.

You can calculate RPM according to the selected TSR.

(1) RPM = (60 x TSR x V)/(3,14 x D)

, where

RPM is turns in minute,

TSR is tip speed ratio

V is the wind speed (meters per second)

D is the diameter of the wind rotor (meters)

You can also calculate the approximate

available power from the wind rotor

according to the formula:

(2) P (watts) = 0,15 x (DxD) x (VxVxV)

, where

D is the diameter of the wind rotor (in meters)

V is the wind speed (in meters per second).

Then you can try matching your generator

with the power available from the wind rotor

at a desired wind speed...

- Hannu

[theTinker]

Flux Quote:

"For 12v dc you need 12/1.4 volts ac line volts. ( 8.6v ac )

This is  8.6/1.73 volts per phase  =5v ac.

One phase will have 3 coils, so you would expect 5/3 v for one coil.( 1.67v ac )"

This has to be the most concise reasoning ever. it totally clears things up.

I had the 1.4 reversed in my head too.

I can make and reason out a test coil then for my 12v dc with this info.

Ill start out with a #16 wire 70 turns and work from there.

Given that my blades are "newbie built"(they look like other people blades but TSR, angles, n such are way above me atm), 5' 10 inches, in low winds(most are blocked out by buildings) what kind of rpm would be good to put into the calculations, 30? 80?

thanks.

[hvirtane]

Given that my blades are "newbie built"(they look like other people blades but TSR, angles, n such are way above me atm), 5' 10 inches, in low winds(most are blocked out by buildings) what kind of rpm would be good to put into the calculations, 30? 80?

--

Why not as low as even 60?

(How big is the blade tip angle?)

If the wind speed is 2 m/s, the available power

would be about:

0,15 x (3,5 x 3,5) x (2 x 2 x 2) = 14,7 (W)

(I'm assuming that your wind rotor

diameter is 3,5 m)

If the wind rotor TSR is 6, at 2 m/s

RPM = (60 x 6 x 2)/(3.14 x 3,5) = 65,5

- Hannu

[theTinker]

Is picking up a common 12 volt light bulb from a DIY store fine for a load on a test coil and for testing the whole alternator connected?

shortened from

"I've been collating all the data i have on testing coils etc and i think i have everything about wrapped up about coil shape, size, and how to calculate what i need from each coil, phase and everything else, except one thing. When im winding my test coil I see i have to load it with sumthing or else the voltage i get is "open circuit voltage" which is far different than what i will get with a battery connected to it.

now since im desiring voltage of about 1.7 volts per coil to get my volts for charging a battery, im unsure about what to load it with.

the Dans loaded theirs with a light bulb, but others say this is a really bad load. then people suggested 1 ohms and 10 ohms which gave very different results."

[Flux]

To determine cut in speed you don't need a load. If you are doing it with a rectifier on the full winding sometimes some meters like a bit of load but that is very small and a few milliamps will do, analog meters are happy without it and so are many digitals.

I have never got any useful answers by loading coils into a resistor or a short circuit.

I can do far better determining cut in speed from open circuit volts and deducing the results into a battery from the winding resistance.

If you were intending to use it for direct heating then a resistive load may have some meaning but with the whole winding, not a test coil.

Flux

[theTinker]

So if im looking for a 1.7volt from a coil I can just wind it and use my meter directly on it? Can I take a 1.7volt unloaded as being accurate enough or do i need to run it through some function?

-----------

As a side note, just to clear up some thinking for me on blade stuff.

The TSR will decide how many RPMs my mill does, so it must be high enough for my mill to get the RPM needed for cut in at a desired wind speed, but it also must not allow my alternator to take more than the betz limit or else my mill will stall. If it does take more power than the betz limit, it will stall. To solve this i must increase the mill diameter so more power is available or reduce the tsr so it takes less power from the wind, or i can reduce the rpm needed of my alternator?

is this correct so far?

as a second thought, why do mills claim 35% efficency from a given diameter, why dont they just use a higher TSR? im assuming that 35%  efficency means that they take 35% from blade span power available.

[Flux]

Yes you seem to be on the right track but I am not sure how experienced you are at doing things.

Measuring 1.7 v ac is not without its problems at low frequency if you have a cheap meter. If you can manage to drive it at a higher known speed it would help. you could then scale the figures to your cut in speed. If you have to hand crank it and time the revs, then try at 120 and 180 rpm. If the 180 figure is not 50% more than the 120 figure smell a rat.

Again with your other questions you are on the right lines but life may not be as simple as you imagine.

Aim for a realistic tsr at modest wind speed (12 to 15 mph). This will depend on your abilities to make a prop. For a wood prop tsr 5 to 7 is sensible. Lower wastes too much wood, higher makes too much noise.

At cut in you could expect the prop to run fast. tsr6 should run about 8 for cut in speed decisions.

You need to be able to predict the alternator load at various speeds and check that it matches the power available from the prop at various speeds. You will never extract any near Betz limit without stall. Taking prop and alternator efficiency into account, a cP of 0.2 is more realistic than Betz figure 0.59.

If the alternator is too powerful for the prop you need a bigger prop or ways to reduce alternator output.

Take care if the prop is too powerful for the alternator, you will have to be very careful not to over speed.

These things are fairly easily sorted but do need an in depth knowledge or plenty of experience. When you say you don't have enough bearings to make a dual rotor I think you may be well advised to look carefully at what others have done and get some real experience with something that will work first time.

Flux

[willib]

check this out..

it looks like a wood frame alt.

http://www.fieldlines.com/story/2004/7/20/142837/061

http://www.otherpower.com/images/scimages/1911/PICT0081.jpg

thanks Sasa!!