Author Topic: help with axial flux [CLOSED] (Read 12405 times)

xclr82xtc · « **Reply #33 on:** December 07, 2009, 02:34:09 AM »

see i dont understand how you getting the power rating...i did faradays law...and this is what i got...

1 coil, 1 rpm, 1 magnet = .104v ...no thats gonna happen 30 times per coil (30 magnets) this equals 3.12v.... now i get sorta lost again, but not...so 7 coils per phase, we say 3.12v*7= 21.84v...so this is the output voltage at 1rpm, for 1 phase...now i need the amperage..

so right now we got 21.84v per phase...give me a sec and let me do somemore math

xclr82xtc · « **Reply #34 on:** December 07, 2009, 02:51:57 AM »

ok got an average of .75 inch diamter coil..80 turns...this equal 2.35 inch circum. x 80 turns...give me 15.6 feet or wire per coil...

times 7 coils...= 109.2 feet per phase..

ohm per lb = .0314

31.48 ft/lb

109.2/31.48 = 3.46lbs

3.46lbs*.0314ohms = .109 ohms per phase...

oh so per phase we have .109 ohms...at 1 rpm we have 21.84volts...now for the magic lol

ohms law

amps=volts/ohms

so we get

A=21.84/.109

that give me 200 amps for a single phase... at 1....ONE rpm....

check my math...please

xclr82xtc · « **Reply #35 on:** December 07, 2009, 03:09:20 AM »

redid with 3phase..it looks to be 115 amp out at 21v...

however i think 980 turns is unrealistic, im goign to redo all the math after i wrap a coil to see how many turns i can get with 12 gauge

xclr82xtc · « **Reply #36 on:** December 07, 2009, 03:43:45 AM »

ok i did mess up..since the magnets are pairs..it would be 15 magnets not 30..easy fix...it would be 10.92 v per phase...

this reduces the amps to 52 amps for 1rpm. or 1094 watts for 1rpm...

however as said..i think my coils arent feasible so i need to redo all the math anyway

Perry1 · « **Reply #37 on:** December 07, 2009, 03:58:35 AM »

I don't know what to say at this point. I think I will sign off this thread for now as you are not listening to anything anyone is saying. I will leave you with this, cut and pasted from my previous post.

*****

At 15 mph your rotor will be exposed to 131 watts of wind power. The rotor will convert it to 65.5 watts of mechanical power and the alternator will turn that into 33 watts of electrical power.

******

Until you understand this very basic engineering fundamental you will never get it. Until then you have a bunch of wrong equations trying to prove over-unity perpetual energy machines. Something frowned upon on any credible forum.

Perry

Gary D · « **Reply #38 on:** December 07, 2009, 06:17:26 AM »

Perhaps a link to the available power in the wind is in order? What the posters are trying to tell you in laymans terms is that your unit will stall if you try to extract more power than the "free wind" can supply. Simply put the blades will not spin, just the same as if you had a waterwheel trying to extract more power than the amount of water given. The water (or air in this case) would just fall around the wheel (or vwat in this case). The folks are looking at the total system for you, you are looking at the individual parts. This isn't a long article from our hosts here, but maybe will help you to understand where the posters are coming from....http://www.otherpower.com/bottom_line.shtml

Gary D · « **Reply #39 on:** December 07, 2009, 06:21:30 AM »

Perhaps this link and post will work this time. So that everyone is on the same page. Stall is an issue due to trying to take more power from the wind http://www.otherpower.com/bottom_line.shtmlthan is there.

ghurd · « **Reply #40 on:** December 07, 2009, 06:25:52 AM »

You messed up a lot of stuff.

Not sure what all. The order of magnitude is 'big'.

Maybe interchanging square meters vs square inches.

Or RPM vs frequency.

Maybe start by looking closely at something that works (and exists).

Adjust the turns per phase by the ratio of magnet surface area between theirs and yours (you lose).

Adjust the turns per phase by the ratio of magnets per disk (you win).

That should get you a ball park guesstimate number of turns per phase that will make the same open voltage at the same RPM as the data from the existing PMA.

I am not saying it should be built that way, though the numbers should be pretty close.

G-

Beaufort · « **Reply #41 on:** December 07, 2009, 06:27:22 AM »

Here, here! The dude's just not getting it. And if I read one more meaningless lol, I'll puke after lmfho. WTF?

I don't want to sound like an old coot, but this makes me want to volunteer at the local college to calm these kids down and teach them some problem solving methods. I was fortunate to have several people like this throughout my career "set me straight" and it wasn't a candy coated lesson wrapped up in a Disney production. I had to lurk on this board and others for about 2-3 years before I finally accumulated all the pieces to build a wind turbine from scratch. It's not easy, and you have to listen and read, then read again. But the first step is to listen.

As I said before, buy the latest book from Mr. Piggott and turn to page 54; it's all there.

wdyasq · « **Reply #42 on:** December 07, 2009, 08:24:19 AM »

I have a saying used in cases like this:

"Ignorance can be cured. Stupid is forever."

The nicer version is:

"Stupid is a condition. Ignorance is a choice."

Take your choice.

Ron

electrondady1 · « **Reply #43 on:** December 07, 2009, 08:50:06 AM »

well xclr82xtc,

you didn't crack under questioning and that's good.

you have a fat enough budget but not much time for such an ambitious project.

perhaps you could break the project into stages.

during that j leno garage video it comes out that the unit they are demonstrating does in fact, use conventional bearings.

so for this semester it would be quite a feat for you to just design and build your maglev style machine with out getting involved in the generating aspect.

why not talk to your teacher about it.

you can learn alot about windmill alternators from some of the people who make up this forum.

but they are mostly interested in hawt machines.

a vawt is a different animal.

you need a large swept area to harvest energy on the scale you envision

to fully understand the forces involved,

hold a 2'x8' piece of styrofoam insulation up into your 8 mph north Carolina wind .

that's about it!

your mill will produce a fraction of that force you feel.

take it one step at a time

it's not easy to make electricity ,

if it was every one would do it

good luck.

electrondady1 · « **Reply #44 on:** December 07, 2009, 09:18:35 AM »

as i said,

you can learn a great deal from SOME of the members of the forum .

while others are here to simply be sarcastic , negative ,and contribute nothing to the overall good.

feeding their own impotent egos by attempting to poison those that have a genuine enthusiasm.

morglum · « **Reply #45 on:** December 07, 2009, 11:33:31 AM »

xclr82xtc

You can never generate more electricity than what is present in the wind. You're studying in engineering, you probably have heard of the laws of thermidynamics:

(http://en.wikipedia.org/wiki/Laws_of_thermodynamics)

The first law of thermodynamics, which mandates conservation of energy, and states in particular that heat is a form of energy.

The second law of thermodynamics, which states that the entropy of an isolated macroscopic system never decreases, or (equivalently) that perpetual motion machines are impossible.

In this case:

Wind (kinetic energy) --> moves magnets (kinetic energy) --> generate current in coil (electrical energy) and heat through friction.

The first law says that if you started with 100 J of energy in the wind, then at the end we will have 100 J of energy spread between slower wind, heat and electricity.

The second law says that no matter how efficient your "machine" is, you will never get 100% of the wind's energy converted to electrical energy. At least some of it will be converted to heat.

Now, physics rule say that in an area of 5' by 36", a 15 mph wind will generate X watts. You cannot hope to get more than X watts of electricity.

*That's like saying you will feed 2 drops of gas to your massive car engine and expecting it to produce 200hp for an hour. The gas' (or the wind's) energy just isnt enough. *

ghurd · « **Reply #46 on:** December 07, 2009, 11:45:07 AM »

Maybe THIS is what you are looking for:

For a 3-phase PMA, wired in Star/Wye,

Turns per phase = 17000 x (System voltage V +1.4) / ( Desired cut in speed RPM) / (Strength of flux in Tesla) / (total area of magnet face in square inches on one disk)

I did try to go through some things.

And I learned you overlook some very basic stuff.

For one thing it took me half an hour to decipher that whole 'ohms per pound' thing.

Life would be easier if you just said "#10", if that is what it was.

The 80 turns of #10 is 15.6'? Meaning the coil is going to be 9" thick?

And how thick (long) are the magnets to achieve 0.64 Tesla in a 9" long coil?

Remember there is a opposite polarity magnet very near it. On both sides.

Path of Least Resistance kind of idea. The flux path will go to the magnets that are an inch away on both sides instead of fly 9" (tight and focused) to the magnet on the other disk.

Then after these 9" thick/long coils are in there, the plan is to add more in the unused area for increased efficiency.

A couple other things that cause me some concern,

"the steel plate the magnets will mount to will be and intregral part of the base"

Brushes? That's antique thinking.

"i want the magnetic field to be captured inside of the steel casing"

I have no idea why you want the flux shorted.

You seem to understand the redirection of flux in this case, but not if the magnets are mounted on a steel disk.

I strongly suggest several holes are drilled in the casing with nuts welded over then so the magnets can be removed from the casing so they can be resold on ebay later.

From the same sentence, "i hope this will have the effect of compressing alot of magnetic field in the inner most ring of coils."

It will not. It will be counter productive.

I wonder if Jay Leno said "What? $1000 extra for the maglev version? I do not want to spend that much more!"?

Or if they simply couldn't make it work as advertised.

Jay was already in for more $$$ than would ever be recovered in his lifetime, so he must not have wanted to spend the extra on the maglev stuff just to increase efficiency?

The more I type and read, the more I want to say, but this is the last of it.

G-

xclr82xtc · « **Reply #47 on:** December 07, 2009, 12:11:15 PM »

i rechecked my math. i dont know if im doing faradays law right, but i keep getting the same answer...i think the magnets may be to powerful and causethe stall...im going to get some magnet wire right now, and wrap a coil so i can get the exact turns and ohms.

xclr82xtc · « **Reply #48 on:** December 07, 2009, 12:13:49 PM »

i aggree with your statement...so what i need to figure out now is how much wind power is going to be needed to break the stall.

you can always use weaker magnets...the magents i was thinking about using are .6400tesla that seems pretty strong to me. cutting that back to 3000 would give a more realistic output.

xclr82xtc · « **Reply #49 on:** December 07, 2009, 12:21:54 PM »

as i said i dont know exactly how to do the math, im shooting blind here...i di the calculations with 12 gague wire...with an average diameter of .75 inchs...4 layers, 20 turns per layer,...after doing the math that gave me a total of 15feet of wire. (as i said i dont know if this is accurate im going to wrap a coil later and see whati can get into a 1x2inch cube with a 1/2 inch air core

"the steel plate the magnets will mount to will be and intregral part of the base"

Brushes? That's antique thinking.....

- i meant the steel plate will carry the magnets on one side and the blades on the other, while the center will be machined for the large magnet...thats what i meant by integral

I strongly suggest several holes are drilled in the casing with nuts welded over then so the magnets can be removed from the casing so they can be resold on ebay later.

- actually the design consist of magnet "holders" that attach to the steel plate. because i wanted to use both poles in the coil n s coil s n configuration, i didnt want them attached directly to the plates....

also the magents im looking at ar e n42 3/4inch by 2 inch cylinder at 6403gauss..for the calculations i just used .6403 tesla, becuase yes even though you lose gauss when you get distance...im also using two magnets...but i dont know if this is the proper way.

xclr82xtc · « **Reply #50 on:** December 07, 2009, 12:22:58 PM »

its ok...im the first here to admit i dont know what im talking about. all these equations are news to me. lol im tryin to get a grasp on it as quickly as possible.

ghurd · « **Reply #51 on:** December 07, 2009, 01:23:01 PM »

Suddenly it occurs to me this may be a high school class,

instead of a college class?

Perhaps your professor knows what you think will happen, and he knows what will happen, because it ain't gonna happen.

"the magents im looking at ar e n42 3/4inch by 2 inch cylinder"

What?

Is that a 1/5" or 3.141592654" square inches of surface area?

"average diameter of coil is .75 inch"

Why would anyone make a coil 0.44^2" when the magnet has (maybe) 3.14159^2"???

Why would anyone make a coil 0.44^2" when the magnet has (maybe) 0.2^2"???

"...thats what i meant by integral"

That is called shorting the flux.

.

.

.

Here is exactly, Exactly, EXACTLY how to do the math for a 3 phase PMA connected in Star/Wye:

"Turns per phase = 17000 x (System voltage V +1.4) / ( Desired cut in speed RPM) / (Strength of flux in Tesla) / (total area of magnet face in square inches on one disk)"

oztules · « **Reply #52 on:** December 07, 2009, 01:27:38 PM »

Your right Ron..... he's pressing hard... My #1 simpleton title is starting to look shaky now..

I didn't realize how much competition there would be for the title.

Your education system has some answering to do.... and I suspect ours is no better..... sigh...

.............oztules

xclr82xtc · « **Reply #53 on:** December 07, 2009, 01:55:22 PM »

i did you one better..i went a bought a couple physics books, that are very much intuned to electromagnetism etc...ill read through these books and try to get a good grasp of whats going on...

unfortunatly the local radio shack didnt have any magent wire bigger than 20, so im stuck with that for now..i have some regular 2 gauge wire, so i will wrap it instead..should give me a decent base line

xclr82xtc · « **Reply #54 on:** December 07, 2009, 02:04:07 PM »

i see your point..but i would just like to completely forget about the wind turbine at this point...the whole point of this thread was to get axial flux help...all i need to know is how much power i can get out of one coil at 1 rpm with a 34 inch diameter centerline through the coil, usin 2 6400gauss magnets, then i need to multiply that into 7 coils and 3phases to get total power output at 1 rpm....thats it..thats all i need...if the wind turbine wont turn it, then i will use smaller magnets or create a bigger mill.

TomW · « **Reply #55 on:** December 07, 2009, 02:12:06 PM »

If you wind "2 in hand" of #20 you get an effective #17 when you connect them in parallel to make one bigger wire. This is very common, in fact. You could go 3 or 4 or more "in hand" and get an effectively larger "wire" for the coil.

Tom

xclr82xtc · « **Reply #56 on:** December 07, 2009, 02:13:14 PM »

thanks for the link...

you guys keep giving me the power output of the windturbine...you guys said what 32watts or something like that?

what YOU guys are saying is that the alternator will produce x amount of electrical power from 32 watts..that just doesnt seem right to me...what IM saying is we have 32 watts to TURN the rotor...thats how much torque we have..the actual power output comes from the stregth of the magnet field and coil output.. how much torque does it take to pass the field thru the coil? unless your using stupid large magnets, i cant see it using all 32 watts or mechanical power.

there is no "free" energy here...you are converting magnetism to electricity.. how is my logic wrong?

i know you guys know more than me...maybe im just missing something small and once i get it the bell will ring...but as of right now, im not comprehending.

xclr82xtc · « **Reply #57 on:** December 07, 2009, 02:19:55 PM »

neo 42 magents that are cylinder shaped 3/4inch diameter 2 inches long.

you allso have to remember the coil has 4 layers, .75 is just the average diameter. you have to do more math to get the actual surface area.

i designed it like that becuase as i said i was planning on using one pole of the magnet on one coil and the other end of the magnet on the other coil.

i think we should start a new thread...this is to much of a cluster %%%% now..

ghurd · « **Reply #58 on:** December 07, 2009, 02:21:33 PM »

You have some "regular 2 gauge" which should give you a decent base line?

You going to wrap a test coil with 25 turns of #2?

You saw the #20 at rat shak?

You have #2?

Maybe you work for that UK 'bang' show?

Always amazes me how the smartest guys never bother to respond to this crap,

and how the editors always see it coming.

Here is where the post is officially past needing removed from public view.

This is the poster child of why things need removed from public view.

ghurd · « **Reply #59 on:** December 07, 2009, 03:14:57 PM »

How old are you?

What year were you born?

Are you in high school, Jr Hi, or college?

I was cutting you slack because I thought you were in Jr Hi.

"you have to do more math to get the actual surface area."

Really?

Enlighten us about this magic math we are ignorant of,

which you know so much about,

but you do not understand.

I have to do more math?

You think we are idiots because we don't have all the information?

I came up with the surface area when you were still throwing random numbers around.

Perry1 · « **Reply #60 on:** December 07, 2009, 03:17:17 PM »

I would be an advocate of NOTstarting another thread. So many people have tried to help you and you refuse to listen to any of it. I don't believe you even tried to comprehend what others are telling you. Maybe if you spend a couple weeks reading up on very basic engineering precepts before reposting.

Perry

xclr82xtc · « **Reply #61 on:** December 07, 2009, 03:23:03 PM »

i wasnt saying YOU have to do more math..i was just sayin to get the actal numbers, more math would have to be done.... not tryin to insult you guys or anything, i just dont think what you guys are seeing corallates to what im trying to say.

ghurd · « **Reply #62 on:** December 07, 2009, 03:57:53 PM »

The math has been show.

"i was just sayin to get the actal numbers, more math would have to be done."

Why?

I did the math, backwards, and more than that I did the math with "Secret Crap".

It is exactly the same as frontwards.

The crap you are saying, in the context you are saying it, is either OU or BS.

xclr82xtc · « **Reply #63 on:** December 07, 2009, 04:09:36 PM »

no one is forcing you to help me...you can either try to break it down for me or you dont have to..i would appreciate the help but throwing a few equaions up and saying "this is what you will get" doesnt help me very much.

not only that...but the math you guys are doing, doesnt seem to coincide at all with the formulas i am finding for coil output..so once again, im missing someone, or someone is wrong.

Perry1 · « **Reply #64 on:** December 07, 2009, 04:39:10 PM »

You hit the nail on the head...you are wrong

I don't know why I keep responding but here goes.

You are trying to build a 2-3 kW alternator. Fine, fair enough, noble cause and you are trying to perform those calcs. Personally, I would listen to most everything Flux, Oz, Ghurd, etc say about their design as they will save you countless hours and dollars.

The problem none of us can get around is that you are trying to power it with a 63 watt wind 'engine'

You are trying to put 63 watts into something and have it produce 3000 watts out the back end.

I think you still believe it is the magnets making the energy, not the wind.

I am trying to think of different ways to explain it because you are not responding to the math.

Imagine if you had a 3kw Honda genset. Big internal combustion engine in it, right. You are advocating taking that big engine out and replacing it with a small model airplane engine and expecting it to still put out 3 kW.

Does this help?

Perry

xclr82xtc · « **Reply #65 on:** December 07, 2009, 04:50:31 PM »

yes it does...thank you..can you look at the other thread "axial flux 2"

I have always been under the impression that the magnetic field induced into the coil determines the power output.

.ie... if you use Big magnets you get big power...but they are harder to turn...if you use smaller magenets you get smaller power, but it uses less mechanical power to turn.

so one thing we DO agree on is that the wind generator will put out 63watts of mechanical power. what im trying to get across is, "how big of a magnet can i push around, through 21 coils, to get the most out of those 63 watts of power...

an example PMA video i saw on a windturbine site made lije 20 watts of power just being turned by hand.. thats is basically why im not undestanding how there is only 63 watts of electrical power as well...you live in the US? you want my number..this might be easier on the phone.

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Author Topic: help with axial flux [CLOSED] (Read 12405 times)

morglum