Help please with Windings

[gchap]

Hi

New to wind energy (fascinating subject). Been looking at Hugh's book and have a copy of his Axial plans. I intend to build the 4 foot (small) turbine. I have a question or two, as I like to try and understand what I am doing (this could be a long project!).

1. How do you calculate the number of windings necessary for a turbine. Your Wood-Ax unit uses 40 turns of AWG 18, but how did you calculate this? I have looked at the formula in Hugh's book, but am slightly confused. (Agap = area of magnets only or of steel disc? Crpm - what is the relationship between this variable and the speed of the wind.  Where does the actual gap distance between rotor and stator enter the equation (wood Ax = 0.25 inch).)
   
2. Hugh's plans for his 4 foot unit does not appear to have any laminations behind the coils, or am I reading the plans incorrectly.
   
   

Hope this is not too confusing.

Many thanks

Rgds

Graham

[Flux]

It's not that easy to calculate the number of turns for a winding.

If you have a standard magnetic circuit such as a dual rotor or a single rotor with a laminated core, you can with practice get quite close if you know the grade and area of the magnet and the air gap.

With an open magnetic circuit such as Hugh's 4 ft machine it is more difficult. you can not easily predict what flux density you have at a given point. It can be done by computer modeling and finite element analysis, but if you want to go that route it will take you even longer.

Most of us make a test coil, measure its voltage and resistance and predict the output from that.

That's the beauty of using a published design, someone else has done the hard work for you and you know it will work first time.

After you have built a few alternators you will be able to predict better using the results of previous experience.

You are right, that 4ft machine does not use laminations. The idea is for anyone to be able to build it. Suitable material for laminations is not always easy to come by.

I think the reasoning is that it uses the same size magnets as the big 8ft mill and they are a common magnet available reasonably cheaply.  It could have used smaller magnets with a dual rotor .

I am not sure that I have answered your question, if not come back.

Flux

[Flux]

It's not that easy to calculate the number of turns for a winding.

If you have a standard magnetic circuit such as a dual rotor or a single rotor with a laminated core, you can with practice get quite close if you know the grade and area of the magnet and the air gap.

With an open magnetic circuit such as Hugh's 4 ft machine it is more difficult. you can not easily predict what flux density you have at a given point. It can be done by computer modeling and finite element analysis, but if you want to go that route it will take you even longer.

Most of us make a test coil, measure its voltage and resistance and predict the output from that.

That's the beauty of using a published design, someone else has done the hard work for you and you know it will work first time.

After you have built a few alternators you will be able to predict better using the results of previous experience.

You are right, that 4ft machine does not use laminations. The idea is for anyone to be able to build it. Suitable material for laminations is not always easy to come by.

I think the reasoning is that it uses the same size magnets as the big 8ft mill and they are a common magnet available reasonably cheaply.  It could have used smaller magnets with a dual rotor .

I am not sure that I have answered your question, if not come back.

Flux

[Flux]

Sorry about the double post, it came up with "server error" first time, I didn't think it would work.

[LEXX]

I had the same question a little while ago, go here and see the comments http://www.fieldlines.com/story/2004/7/27/181631/793

LEXX

[gchap]

Hi Flux

Thanks for the comments. I have numerous questions 'spinning' around at the moment, so I shall re-read Hugh's generator section in his 'Windpower Workshop' prior to posting further questions.

What I tried to do was use the equations on pages 147 and 150 of Hugh's book to arrive at the number of windings on the Wood-Ax but could not, so it all started to baffle me.

Will let you know how I get on.

Thx

Rgds

Graham

[gchap]

Hi Lexx

Thanks for feedback. I will take a look.

Rgds

Graham