Levers, alternators and windmills

[rotornuts]

Devo's recent post regarding placing the magnets around the circumference of the rotor got me thinking about when I was thinking about doing that a year ago. At first I didn't remenber why I abandoned the idea but then it dawned on me. I remember now I realized a windmill is just a lever and the farther away from the fulcrum you place the load the more torque you require to do the same amount of work, granted at a higher rate.

I reasoned that placing the load on other side of the applied force would likely require require too much torque. It's funny how we can forget the reasons we come up with different ideas but this way of looking at rotors led me along a twisted path to the single blade thing.

Anyhow, I thought I'd kick this can into the circle and see what happens. It raises questions about alternator diameters and the importance of the root section in overall torque production given the root is so close to the load. Other questions arise as well.

See image below





I guess these days I wonder if maybe this line of reasoning is flawed as the total amount of work done may remain the same and only the velocity changes. I'm inclined to keep thinking it's relevant.

Give it a kick.

Mike

[rotornuts]

P.S. This is by no means a revolutionary concept I just wonder how often people look at it this way and what difference it may make?

Mike

[Devo]

This is an excellent subject that leads to the question where is the optimum point between the centre & the tip or is it just a trade off?

I am thinking farther out could use less wire because of the speed the mag passes the coil which should decrease coil resistance ,adding again to output. I guess experimening is the best way to answer.

What did you use to draw the diagram above? I often wish I could draw or create an image of my idea to show others what I am thinking but I don't know what to use to accomplish it?

Devo

[ghurd]

The 'average force' is all from the wind. There is more wind area in the outside half than the inside half.

So wouldn't the 'point of applied force' average be much closer to the outside?

G-

[electrondady1]

yes mike, you do come up with some nice graphics. but since you only showed me how to post a link yesterday i guess it will be awhile before i can respond with one of my own.but to the point , the longer the lever compared to the load the less wind you need to begin rotation but the closer to the end of the lever the load is the faster the velosity/greater output.my thing is vawt. lately i've been thinking to make the rotors/mag. disks the same dia. as the tip of the vanes to get max velosity.

[rotornuts]

Probably closer than shown but the above is just to illustrate the point.

Mike

[windstuffnow]

  The way I see it... its the same.  If you have a turbine producing 100 watts at 200 rpm in a given wind then its making approx .134 hp. Lets assume the lever is 4 ft.  If the magnet center radius is at 12 inches or 1 ft then the "lever" would see 3.52 ft lbs, If your drawing power from the outer edge the torque would be .88 ft lbs.  Basically you would have to change the alternator to react the same at a different torque level to match the power.  

  The only reason you need to increase the diameter of the alternator is to overcome certain limitations of a smaller diameter.  Such as magnetic area, more room for wire etc.  If you simply move the magnets from a small diameter to a larger diameter you will overall decrease its output.  For instance a 12 pole on an 8" disc, then moving the same 12 magnets to a 12 inch disc.  Granted the magnets will be moving faster but the timing is altered as well as making the coils larger and increasing the resistance.  If, on the other hand, you increase the diameter and add more magnets this reduces coil size, adds more magnetic area as well as other changes in variables that will increase your output, still having to match the output of the turbine at a given rpm.  

  Overall, If you can make the alternator smaller and still achieve the power required then this is the most efficient way.

 

[electrondady1]

i follow you ed, it's kind of pointless to increase dia. without adding more poles or increaseing the mag size. this is the route you have taken, baseing most of your projects on an 8" dia. disk with a high flux density, then concentrating on the most efficient stator to harvest maximum power.

[Norm]

  A lever yes....which is why the old pumper

windmills had many blades near the outside edge

...they didn't the shaft to turn fast, they wanted

something that had a lot of torque instead of

high speed even at that they needed gear reduction

to slow them down even more and increase the torque.

                  ( :>) Norm.

[rotornuts]

I guess I'm just thinking about the trade off here.

If you have a windmill that can produce a torque of 5ftlb at 6' and is doing so at 200 rpm then you have 2.28 hp available at .5 'radius for a 12" alt but only 1.95 hp available at .583' radius for a 14" alt. So then can I assume that the increase in velocity of the flux lines crossing the coils offsets the decrease in available power to yield an increase in overall work done or efficiency?(I do realize that there is a limit to the work potential dictated by the power produced by the rotor)

I must be missing something with regards to the velocity component.

Mike

[electrondady1]

it's all so clear to me now mike , you just turned my sexy new windmill into a bloody wheel barrow!!! ha ha

shawn

[rotornuts]

How about this new innovative Hawt design

Does that help with the visual?

[electrondady1]

looks like a book cover to me ! "how i got off the grid"

[windstuffnow]

  Looking at it that way it sounds like smaller is better.  If you had a 6" disc then the power would be doubled ( 4.?? hp).  

  In reality the 12" disc is 3.14 ft in circumference and the 14" disc is 3.66 ft.  At 200 rpm the 12" is traveling at a speed of 628 ft per minute and the 14" is moving at 732 ft per min.  

  So 5 lbs on a 6 ft lever = 60 lbs of force at .5 ft and only 51.46 lbs for the .583 so...

60. lbs x 628 ft = 37,680 ft lbs / 33000 = 1.14 hp
    
51. 6 lbs x 732 ft = 37,668 ft lbs / 33000 = 1.14 hp
    
    

  Ends up being the same...

[rotornuts]

Seems like I was missing something, namely the right equation for calculating horsepower for this application. I used rpm which remains constant for both scenarios. The equation I found says HP = torque x rpm/5252. Our torque calculations were the same but our horsepower fiqures differ. Why are there different equations and why do these two differ?  

Now I really am confused!

Mike

[windstuffnow]

  The lever formula ( L x W / l = ft lbs) is what throws it off I think.  If you place 5 lbs on the end of a 6 ft lever with a .5 overhang from the center point then your balancing the weight with 60 lbs.  It doesn't actually include the amount of work it's actually doing.  It seems logical to use the number in "ft/lbs" as if you were cranking it 200 times per minute but the actual work done is different.  You would end up putting the 60 lb weight on a rope that was 628 ft long.  So your raising the 60 lb weight 628 ft in one minute.  The same with the 5 lb weight at the 6 ft radius.  The circumference being 37.7 ft per revolution with 5 lbs or 37.7 x 200 x 5 = 37700 / 33000 = 1.14 hp.  So by hanging the weight on the center .5 radius and running it backwards so to speak is the same as well.  

  If you have a spinning machine with a prony brake or dyno showing the force in ft/lbs and measuring the rpm then the HP=torque x rpm is correct.  The difference, one is the measure of force and the other represents the amount of work done.

  Fun stuff.. I actually confused myself while calculating and came up with some real odd numbers...

[rotornuts]

Thanks Ed, This is exactly why I posted this. I was trying to clear it up in my head one way or the other. The right side of my brain kept telling me that the torque was different at different points on the lever so it couldn't end up the same but the left side of my brain kept telling me that the amount of work done was likely the same becuase of the difference in velocity. I couldn't find the info I needed on the net because I was unable to associate the info available to this application.

I'm still a little hazy on the subject, I understand the numbers and the differences between what I did and what you did but I usually take a few days to reconcile new understanding.

Mike

[electrondady1]

sorry my math is rusty, i don't understand were the 60lb or 51.6lb figures  come from

[electrondady1]

ok, i got it  !

[benjamindees]

If you simply move the magnets from a small diameter to a larger diameter you will overall decrease its output.  For instance a 12 pole on an 8" disc, then moving the same 12 magnets to a 12 inch disc.  Granted the magnets will be moving faster but the timing is altered as well as making the coils larger and increasing the resistance.

I don't see how this follows.  Why wouldn't you use larger gauge wire when making the coils larger?

larger gauge wire = less resistance (not more?!)

better cooling = less resistance

increased magnet spacing = less flux leakage

Granted, without altering the coil wiring, there would be gaps in the output.  With the right arrangement and/or more copper, you should get at least the same amount of power from a larger diameter, and probably more.  By reducing flux leakage, you get more output for the same amount of magnets.  And if, by providing more space for coils, you can make the coils thinner, then you can decrease the magnetic gap as well, also increasing output.

Isn't this right?  Or, is flux leakage a non-issue?  Am I missing something?

[electrondady1]

i'm just a novice builder benjamin put i think ed was making the point that increaseing the linear velosity of the mags alone will do little to increase power.by making the changes you have listed it might make up for the wider coils/mag spacing.  

[windstuffnow]

  You right, if you increase wire size then you can break even or come out ahead but at a higher cost.  Less efficient use of materials so to speak.  

  For instance using an example of a 12/9 arrangement on an 8" disc with 40 turns.  The coils on the 8" would be 7 inches per turn.  Assume all other aspects didn't change such as magnets, air gap etc.  Placing the same magnets on a 12" disc would increase the coil size to 10" per turn.  The top and bottoms of the coils are wasted and there is still only 4" of wire in flux per coil.  The 8" wastes 3" of wire where the 12" wastes 6" of wire.

  It would be far more efficient to use 16/12 on the 12" disc.  This would allow you to increase wire size, reduce the turns needed, ( both would increase output considerably ), as well as increasing magnetic area and wasting less wire.  

  As far as flux leakage, I believe it works both ways... If their to close you loose and if their to far apart you loose.  So there is an optimum area but you can fudge a little either way and still come out good.  

  One last point... If the magnet has to travel a farther distance before encountering a coil leg then that is work wasted in my opinion.  If the magnet leaves one coil leg and encounters the next it's more of a continuous load.

  These are simply my theories on the subject based on my experience of designing and building literally 100's of them.  I don't consider myself an expert on the subject but I have a reasonable understanding of what works and what doesn't, I'm still learning....

[benjamindees]

Ah, I see what I'm missing.  You use a ferrous rotor to basically turn two magnets next to each other into one horseshoe magnet?  In that case, then I can see how large spacing would be offset by other losses.

Since copper seems to be cheaper than neodymium, I'm playing with the idea of using more wire and less magnets.  It seemed like the best way to do that was to increase spacing.  But if the steel rotor provides more benefit than wider spacing, then I guess I'll scrap that idea.

If the magnet has to travel a farther distance before encountering a coil leg then that is work wasted in my opinion.  If the magnet leaves one coil leg and encounters the next it's more of a continuous load.

Yes, I've seen you do a great job here explaining that the most optimal design will have copper in front of all magnets at all times.  That's something that is not obvious from many of the project write-ups.

Thanks for the explanation.