Building the sickest ® VAWT ever. Brilliant minds unite please!!

brandnewb

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--- TL;DR update
we are getting somewhere now. managed to reduce diameter while increasing flux to play with. I am not yet so far as to give output data. Only predictions can be found in this long thread.

ALSO!! (unfunded) snag remarks are NOT welcome here. constructive criticisms is both welcome and needed though.

I think I finally cracked the code. I am going this route

--- end edit

Ok boys and girls,

After having seen many commercial turbines fail, underperform and otherwise not life up to expectation I have decided it is time to put an end to this sad state of affairs.
For an analysis of my latest disappointment please see

I see no other option than to take it upon my self to DIY a VAWT and alternator that actually works i.e. that outputs an insane ® amount of power at low (safe) rpm. But as you might have guessed by the title I probably can't do it alone. The title has changed but that is not to say I do not need/respect help

I will be referring to the components in the thread above with a prefix of crap. So when I say e.g. crap coil I mean the coil of that cheap Chinese alternator.

The crap stator is a 3 phase 36 coil (12 coils per phase) radial flux variant.
The crap coils consist roughly of 68 - 70 winds (+-136 - 140 turns) at +- 16cm each totaling of about 11m per coil x 12 coils = +-132m of copper wire per phase.
The crap wire it self has a diameter of 0.4mm.

There are 12 poles in the form of sector magnets with an, as of yet, unknown tesla strength. I will be figuring out a way to measure it sooner or later. I changed my mind, I will not try that as it will mean extra equipment that I am not interested in for now. NO the magnets we are using here are far!!!! more powerful

I will be using an rpm for testing of 60 so that I can easily spin the alternator by hand with a reasonable consistency. At that rpm I hope to get a usable output already if at all possible but in any case not faster than 120 rpm, which should output at insane (TM) levels, else risk injury or worse if the, yet to be constructed, blades fling off.

when spining the crap alternator at 60 rpm there will be a 6 Hz oscillation in the crap coils. With the crap rotor having a diameter of 8 cm.

Now here comes the fun part. I am contemplating building the following 3 phase axial flux alternator
1638089194608.png

Having 48 x 60x10x5mm magnetic poles at a 35 - 40 cm radius.
I have 1mm enameled copper wire that is about 71m long.
1 winding (2 turns) of each coils is about 30cm totaling in about 3.6 7.2 meters of 1mm coil wire per phase per 1 winding.
The alternator will produce an oscillation of 24 Hz while running at the reference speed of 60 rpm.

Now should I try and match or supersede the total crap phase wire length or should +- 71m be adequate for the new phase coil wire length?
 
Last edited:

wholybee

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You cannot get more energy out of the alternator than you put into it. So, while you might be able to build an alternator that produces more power at 60RPM, the amount of wind energy you put into the alternator to get a given wattage will be about the same. You will just need to reach that energy with larger blades or higher windspeed, to overcome a larger force required to spin the alternator. To be clear, if you have 2 alternators, one that produces 500W at 60 RPM, and another 500W at 1000 RPM, it will take just as much energy to spin the 60 RPM version as the 1000 RPM version.

It may be possible to build a more efficient alternator with less energy loss, but it will be difficult to get very much improvement over a commercial wind turbine, and any increase in performance will be limited to the law set forth in my first statement. You cannot get insane output without insane input.

I do not have any expertise in building this. But I expect to get a higher output at lower RPM, you need, more turns (greater length of wire), and a greater diameter of the alternator/coils. To improve efficiency, use stronger magnets, closer to the coils, better bearings, better cooling of the coil, better layup/winding of the wire, and better design of the blades/airfoil. Weight of the rotating mass, composition of the copper wire, magnets, and any ferrite core material will all affect it as well. I might be wrong in how these parameters affect performance, but certainly, they are all things you need to pay careful attention to if you are to match or beat the performance of a commercial unit.
 

chrisski

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I’m going to recommend two things:

1) Move this project to fieldline.com and you will get a better answers. That is a forum dedicated to alternate energy where most of the posters are wind guys. This is a solar forum which is very active with solar, but not so much with wind.

2) Get the book “A WInd TUrbine Recipe book.” This goes into great deal about building a wind turbine, especially te part you are describing. There’s a lot of math to answer your energy question, but in one chart refers to a stator diameter of 1200 as 200 watts. Also goes into a little bit of how when you build it and test it, if you get overcurrent, you adjust the wires closer to the magnet. The answer you’re looking for will probably come on fieldlines.com
 

upnorthandpersonal

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Wind doesn't scale to small turbines. You're dealing with two fundamental issues:

- The Betz Limit
- The Power in Wind equation

The Betz Limit is basically a theoretical number of the maximum efficiency you can possibly get. At most, only 59.3% of the kinetic wind energy can be used to spin the turbine and generate electricity. Remember this is a theoretical limit; in practice, you're going to be closer to 40%.

The Power in Wind equation is given as:

P = 1/2 x ρ x A x V³

Where:
P = power in Watts
ρ = air density (kg/m³, at about 1.2 at sea level)
A = Swept area of the blades (m²)
V = Velocity of the wind

So, no matter how good your turbine is, you will get in practice at most 40% of the wind energy converted to electricity. To capture the wind energy in the first place, you have two variables to increase (one in your control, the other not): swept area and wind velocity. The smaller you make the turbine, the faster you need to spin to make any meaningful energy. The only variable you control is the swept area, which means making the blades as big as possible. Also notice that the velocity is cubed in that equation, so you'll generate much, much less power at low wind speeds; don't assume this is linear or even exponential!

In other words, you're trying to improve something that doesn't really matter if you don't have access to the other variables, and even then, to get any meaningful energy out of it, you need to increase the size of the blades - even in areas with high winds all of the time, you'll just rip the thing apart at those velocities.
 

brandnewb

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I’m going to recommend two things:

1) Move this project to fieldline.com and you will get a better answers. That is a forum dedicated to alternate energy where most of the posters are wind guys. This is a solar forum which is very active with solar, but not so much with wind.

2) Get the book “A WInd TUrbine Recipe book.” This goes into great deal about building a wind turbine, especially te part you are describing. There’s a lot of math to answer your energy question, but in one chart refers to a stator diameter of 1200 as 200 watts. Also goes into a little bit of how when you build it and test it, if you get overcurrent, you adjust the wires closer to the magnet. The answer you’re looking for will probably come on fieldlines.com
1) still waiting to be accepted there
2) Can you tell me what specific page of that book as I bought it and did not find anything relevant first time I scanned through it.
 

brandnewb

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You cannot get more energy out of the alternator than you put into it. So, while you might be able to build an alternator that produces more power at 60RPM, the amount of wind energy you put into the alternator to get a given wattage will be about the same. You will just need to reach that energy with larger blades or higher windspeed, to overcome a larger force required to spin the alternator. To be clear, if you have 2 alternators, one that produces 500W at 60 RPM, and another 500W at 1000 RPM, it will take just as much energy to spin the 60 RPM version as the 1000 RPM version.

It may be possible to build a more efficient alternator with less energy loss, but it will be difficult to get very much improvement over a commercial wind turbine, and any increase in performance will be limited to the law set forth in my first statement. You cannot get insane output without insane input.

I do not have any expertise in building this. But I expect to get a higher output at lower RPM, you need, more turns (greater length of wire), and a greater diameter of the alternator/coils. To improve efficiency, use stronger magnets, closer to the coils, better bearings, better cooling of the coil, better layup/winding of the wire, and better design of the blades/airfoil. Weight of the rotating mass, composition of the copper wire, magnets, and any ferrite core material will all affect it as well. I might be wrong in how these parameters affect performance, but certainly, they are all things you need to pay careful attention to if you are to match or beat the performance of a commercial unit.
As can be read in the original thread. The wind here can be brutal. Anyway the blades of the turbine will be a lift type vertical helix. I will make them out of carbon fiber in a stackable manner. So that if one unit does not do enough we can keep stacking more of them until we can get enough wind.

So for now focusing on the alternator, and getting that to be awesome at max 120 rpm, is a good strategy ;)

So basically what I did is increase the diameter, the magnet size and strength and the coil wire thickness in relation to the crap alternator but the coil wire length per phase is more than half of the crap coil wire length. I am asking should I also increase the coil wire length or can I already expect massive gains compared to the crap alternator?
 
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chrisski

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For the page, I have a E-Book so its a little different. My reader shows me 76% complete and its on the same page as the picture of the lug and terminal block.

I do wish that there was a little more math in that book that would give you specifics. The chart I’m referring to seems to call a windmill that has a Diameter of 1200 a 200 watt windmill. Looking at it again, that seems to be a generic term of what you use to call a windmill of that size.

THe book talks of a cut in speed, or certain RPMs the windmill needs to reach to produce electricity. I can’t find the formula for that, but at around 90% it gives some discussion for RPMs and how it needs to go fast enough to get to the DC voltage to charge the battery and from there, amps are added.
 

Hedges

I See Electromagnetic Fields!
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The alternator will have to be massive to produce much power at low speed like 60 RPM.
Our synchronous motors are typically 3600 RPM at 60 Hz, and pretty massive like 30 lbs for 2 HP.
For aircraft, 400 Hz is commonly used to reduce weight of motors and other magnetic components.

Force from an electromagnet is related to mass of core, so increasing velocity is a way to increase power for a given weight.

Some wind turbines use gearing or belts to increase alternator RPM, but that is more drag and more things to fail.
 

boondox

Chief Engineer, RedNeckTech Industries
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Messages
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Have you heard nothing from Fieldlines? Strange, the mod got back to me quickly and referred it to an admin. Anyway...

You really can't just start with the alt. The alt is part of the whole system and you need to know RPM and torque. So a windmill that is actually functional and productive for you area needs to be designed as a whole.

As noted above, the swept area and the Betz limit dictate the maximum power out. If you really want to build this go search all of the alternator build threads on Fieldlines. A ton of the information you are asking for here is over there.
 

brandnewb

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Wind doesn't scale to small turbines. You're dealing with two fundamental issues:

- The Betz Limit
- The Power in Wind equation

The Betz Limit is basically a theoretical number of the maximum efficiency you can possibly get. At most, only 59.3% of the kinetic wind energy can be used to spin the turbine and generate electricity. Remember this is a theoretical limit; in practice, you're going to be closer to 40%.

The Power in Wind equation is given as:

P = 1/2 x ρ x A x V³

Where:
P = power in Watts
ρ = air density (kg/m³, at about 1.2 at sea level)
A = Swept area of the blades (m²)
V = Velocity of the wind

So, no matter how good your turbine is, you will get in practice at most 40% of the wind energy converted to electricity. To capture the wind energy in the first place, you have two variables to increase (one in your control, the other not): swept area and wind velocity. The smaller you make the turbine, the faster you need to spin to make any meaningful energy. The only variable you control is the swept area, which means making the blades as big as possible. Also notice that the velocity is cubed in that equation, so you'll generate much, much less power at low wind speeds; don't assume this is linear or even exponential!

In other words, you're trying to improve something that doesn't really matter if you don't have access to the other variables, and even then, to get any meaningful energy out of it, you need to increase the size of the blades - even in areas with high winds all of the time, you'll just rip the thing apart at those velocities.
Ok, I'll, parallel to the alt, will also make a start of the turbine components so that I can get some actual theoretical values
 

boondox

Chief Engineer, RedNeckTech Industries
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The order is more like this. Determine how many watts out you need. Look at your local windspeeds. Note that wind conditions change widely even if you move a small distance. Once you have an average windspeed and how many watts you can use swept area and the Betz limit to determine how big your mill needs to be.

It is actually more complex that that though. Startup speed, tip speed ratio and other factors come into play to decide on a profile for your blades. Root and tip size, all sorts of things determine how much power at what windspeed you will produce. Before you start buying or designing I would recommend a deep dive into reading on the subject.
 

brandnewb

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Determine how many watts out you need
50 Terrawatts ;)
But more realistically I am looking to gather 10000 KWh per year as to be able to go off grid. I can realistically put 3 turbines on my property without obstructing/shading my solar array. If it turns out it makes sense to do that anyway then I can put an additional 2 turbines up.
I am having a bit difficulties finding the wind speed average for my location as different sources give different values.
Low end is an annual avg of 4m/s while the high avg is 7.5m/s so I will need more time to find a good source for good numbers.

My local code allows for a turbine diameter of 2m and for the ones on the roof I can make them 3m high. The one at the far end of the property can be stacked as high as 11m.
With stacked I mean placing multiple units of helix VAWT on top of each other as to be able to collect more wind.

I am looking for a good airfoil shape to design the VAWT unit with at the moment.
The Power in Wind equation is given as:

P = 1/2 x ρ x A x V³
I could be mistaken but my intuition tells me that this formula better applies to HAWT's than a helix shaped VAWT like I'd like to build. The reason being is that with such a turbine not all of the blade area is facing the wind at the same time. For rough estimates I will half any results I get.

I am going to change the title of this thread to indicate we are no longer only talking about the alternator.
 

brandnewb

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ugg ;( this is disheartening.
When using the power in wind formula with the lower avg wind speed I found I get 39,36W when aiming an 1m2 swept area.
0,5×1×1,23×4×4×4
now apply betz limit at 40% = +- 15.7W divide in half for VAWT and we have +- 8W left.
8 x 24 x 365 = +- 70Kwh annually.

So it all boils down to increasing the sweapt area of the turbine units and of course work with more realistic numbers in the calculation.

You know what they say. It aint over till the fat lady sings.
Pss. I made sure she can't sing for the time being ;)
 

brandnewb

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NOt sure I need to divide by2 to account for VAWT.

When looking at the orientation of the blades of a randomly found model I think it's more like 80% left.
1638163699498.png

Any thoughts on this?
 

upnorthandpersonal

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I could be mistaken but my intuition tells me that this formula better applies to HAWT's than a helix shaped VAWT like I'd like to build.

The only difference will be the swept area calculation. For a horizontal axis, this is A = π * L², for a vertical axis, it's A = D * H.
  • L is the blade length
  • D is the diameter
  • H is the turbine height
 

brandnewb

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The only difference will be the swept area calculation. For a horizontal axis, this is A = π * L², for a vertical axis, it's A = D * H.
  • L is the blade length
  • D is the diameter
  • H is the turbine height

Well that makes me very happy ;) With the below HAWT at 2m diam. and 3m height.
0,5×6×1,23×4×4×4 = +-236W * .4(betz) = +- 94 W * 24 * 365 = 823440 = +- 823 KWh anually
2 of those on the roof makes +- 1650KWh annually
1 big one with 2 times stacked at also 1650 KWh anually
Makes for +- 3300 KWh year

Now if I learn how to reduce my energy usage. i.e. get a military education in discipline. I could one day go off grid hopefully.

Today the magnets will arrive btw ;) and then it's game on!

1638176445118.png
 

brandnewb

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even more sweet is the 4m/s avg wind speed seems very pessimistic.
Based on
1638178091772.png
it looks like most of the year it is above 11m/s (40km/s) so I am going to use that as an optimistic value knowing full well that going the optimistic route leads to disappointments but I need something to keep the motivation high.

0,5×6×1,23×11×11×11 = +- 4911W * .4 betz = +- 1964W * 24 * 365 = +- 17209 KWh per year.

Now does not this make one's heart beat rise?
 

Hedges

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2 of those on the roof makes +- 1650KWh annually

On the roof? It will experience high force in high wind (also noise). 2 m^2 is a large sail. (no larger than a 500W PV panel, but like one facing the wind.
Can the blades furl? The helical blades you show should provide more uniform torque, but can't be adjustable pitch without stretching.
If vertical or diagonal, they could rotate, just break away from normal position when they overcome a spring-loaded detent.
 
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