Running Numbers for a VAWT

[silentblue1987]

Just running some numbers, feel free to pitch in.

Target Peak Wattage = 700

Target Voltage = 49.5
TSR of Lenz2 = .8
Cut in RPM = 80
Coil Count = 9
Magnet Count = 12 (pairs)
Turns Per Coil = 230

Now to figure out how big to make the blades.... Any ideas?

[Warrior]

Hello Silentblue,

I'm no expert in wind energy, but I'd say you'll need a very large area to get 700 watts. That is a very ambitious output and you didn't mention at what wind speed you want to make that power.

HAWT have the advantage that the area varies to the square of the diameter. Where as a VAWT is simply height x width.

I'd say you'll need something 10 feet high by 8 feet wide. That's about equal to a 10 foot HAWT.

Take into account the VAWT has lower efficiency and running it close to the ground you won't get much power with a small area. It will also spin at low rpm so you will either need a huge alternator or a gearbox.

Good luck,

Warrior

[silentblue1987]

That's the thing about the way I'm building mine. I understand 700watt will almost never happen, i just keep that as a reference to hugh piggot's core design. I'm basing it from his 8' alternator but using 230 turns for low rpm. Cut-in rpm should be around 80 with 7mph on a Lenz2, max power might be achieved at 200 rpm or so (guessing 18mph). As for turbulent winds, I plan on mounting it to a 30' tower with no trees around and an average wind speed year round of 7.

I'm wondering though if there's  a way to find out what size blades a vawt would need to match his 8' blades albeit much lower rpm and much higher torque.

I was thinking use the same "swept area" fundamentals, but still unsure since the swept area of an 8' hawt is 50ft^2.
[Area = pi x r^2]

Assume we have 8' tall, 3' diameter vawt. (Swept area of a vawt is width x height)

8' x 3' is only 24ft^2 but that doesn't seem accurate at all since vawt blades don't travel in 2 dimensions like a hawt's blades.

8' x 3' x 3' gives 72ft^3 which sounds more accurate since it incorporates all 3 dimensions, but also seems much higher than realistically possible.

8' x (3' x pi) gives 75ft^2 which is the most accurate travel path of a vawt (8ft x perimiter of 3ft circle), yet still sounds much too high.

My head hurts  

[GoVertical]

photos of 1k vawt

http://www.globalwindgroup.com/projects?album=1&gallery=1

What size wire and magnets are you using?

[Warrior]

Assume we have 8' tall, 3' diameter vawt. (Swept area of a vawt is width x height)

8' x 3' is only 24ft^2 but that doesn't seem accurate at all since vawt blades don't travel in 2 dimensions like a hawt's blades.

Think of the swept area of a VAWT as a solid cylinder facing the wind. If you view the cylinder's profile then its just a rectangle facing the wind. I doubt a VAWT will take wind in three dimensions all at the same time. Even if the wind changes it's still the same area facing the wind.

Hope that helps...

Warrior

[electrondady1]

lots of formulas on the net.
just google power in the wind
to get an idea of what sort of kinetic energy is contained in the moving air

the lens2 harvests something like 30% -35%
the alternator , perhaps 50% of that.
maybe a bit better if your not holding the voltage down with a battery.

[silentblue1987]

photos of 1k vawt

http://www.globalwindgroup.com/projects?album=1&gallery=1

What size wire and magnets are you using?

Thanks for the pic, gives me an idea of how small they made theirs and still achieved 1kw. Mine was going to be taller with a smaller diameter to increase rpm slightly without compromising too much torque.

19 awg heavy build.
2" x 1" x .5" N42, Quad coat (70lbs pullforce, tested)

Think of the swept area of a VAWT as a solid cylinder facing the wind. If you view the cylinder's profile then its just a rectangle facing the wind. I doubt a VAWT will take wind in three dimensions all at the same time. Even if the wind changes it's still the same area facing the wind.

.

Yes its very unlikely that vawt's capture power from every direction at once. Which sounds like we should change the formula to encompass the perimeter but only the portion that's actually producing power.

So lets assume the original 8' x 3' again but instead of all 3 directions only the surface area facing into the wind/producing power (50% for example).
[8' x ((3' x pi) x .5) = 37.6'^2]

the lens2 harvests something like 30% -35%

If what you're saying is accurate about 30-35% for a Lenz2 then.
[8' x ((3' x pi) x .3) = 22.4^2]
[8' x ((3' x pi) x .35) = 25.6^2]

This makes me wonder now how much is legitimately captured from a 8' hawt considering betz limit is 59%
[(pi x 4^2) x .59 = 29.5'^2]

Does anyone have some tested and proven numbers for a hawt they'd like to throw in?

[electrondady1]

the swept area of vertical mill is width x height
just as a horizontal mill is the area the blades sweep p.x r.x r.

 verticals don't get no pie ,but we've known that from the beginning.

[silentblue1987]

verticals don't get no pie ,but we've known that from the beginning.

Even if verticals don't get no pie I'm still looking for something that makes sense, which seems to play nicely if you use 30-35% perimeter multiplied by height. It's the closest so far to the plain jane width x height = 24^2.

Is there anyone that can dish out hawt blade efficiency?
Vawt: [8' x ((3' x pi) x .35) = 25.6^2]
Hawt: [(pi x 4^2) x .59 = 29.5'^2]
It's too similar (even theoretically) to be overlooked in my opinion, and surprisingly not a lot of posts about hawt blade efficiency.

Its like the all time dispute between Intel and AMD.
[Intel favored raw hertz speed (similar to high TSR) vs AMD favoring more flops per hertz (similar to high torque)]

[Rob Beckers]

Hi Silent,

Blade efficiency of the average small HAWT runs around 30% - 40%. Overall efficiency for a good HAWT hits about 30% for MPPT loaded turbines (including alternator and inverter losses), the average small HAWT probably does a little less than that. For a pure drag device VAWT, such as a Savonius, you're lucky to get 10% blade efficiency (one of my customers just tested one that they designed/made at 8%). I know next to nothing about the Lenz2 turbines, so feel free to educate me. It looks like a turbine that's between drag and lift, using a bit of both, though the below-unity TSR makes me wonder if it will do much better than a drag device. A somewhat optimistic guestimate would be 15% blade efficiency(?). Multiply that by about 0.8 for grid-tie (MPPT) with a good alternator, less for direct battery charging (voltage clamped).

As mentioned, for a VAWT the swept area is simply its frontal area (width x height for the 'rectangular' types).
So, for an 8' (2.44m) HAWT the power from the blades at 11 m/s (24.6 mph) works out to about 1,200 Watt, and from that you'll get about 900 Watt to the grid.
For a Lenz2 measuring 8'x3' = 24 ft^2 (2.22 m^2), that equates to the swept area of a 5.5 ft HAWT, and assuming 15% efficiency you'd get around 240 Watt out of the rotor at 11 m/s wind speed.

Turning that around: To get 700 Watt out to the grid at 11 m/s you'd need 875 Watt out of the rotor (and a pretty decent alternator). That requires a swept area of about 8.05 m^2, or about 87 ft^2. That's almost 10' by 9'.

Of course, if my 15% blade efficiency guess is off and you have a better number just scale accordingly...
Hope this helps!

-RoB-

[silentblue1987]

Excellent work Rob, you answered the call!  

All of the numbers you've found are quite accurate except with the Lenz2, not surprising...

Lenz2 is a hybrid design combining Drag and Lift at the same time which is why the creator said its most efficient at .8 TSR. The torque of the drag style Cup shape on the back-sweep gets power from the lift style Wing shape on the forward-sweep.

The cup shape is similar to the PVC pipe savonius prototypes you see on youtube, some of the really well made ones approached 30% efficiency for the complete system with a load. They had to put extra turns in the coils to reach charging voltages in 7mph winds and have the torque available to overcome a stator at 80rpm. Most people will treat them like trash due to extra cost for more wire, stronger bearings, and much more material to make the blades, housing, wind vanes, etc. However, with a whopping 1.6 TSR at best, a thrown blade in 20mh winds is going 26 mph... Don't gotta worry about decapitating your neighbour.  
http://www.youtube.com/watch?v=zzRpDSKQmdY

The wing shape is what the hawts use and its not a pushover either. I've seen just how beasty they can be and yes they are more cost effective from a material perspective, yes they are more complex on perfecting the blade shape, and furling, the braking system, air gap, but you also get what you pay for in the short and long run. The only thing I don't necessarily like is at TSR 7 in 20 mph wind, the blades are going 140mph. If they aren't made well, or something malfunctions, that blade becomes a lethal projectile when it snaps.  
http://www.youtube.com/watch?v=oAWMpxX60KM

This is not a bash on hawts and is not intended to be one in any shape, fashion, or form. Hawts are a tested and proven technology and demands the respect it has earned.
So after that said you can see why the Lenz2 is unique in it's own respect. It takes less construction material than a savonius, and still reaches similar power to a lift style hawt albeit much lower rpm.


Someone dropped a number in an earlier post saying the lenz2 has 30-35% efficiency.
This is decent when compared to a hawt with 30-40% efficiency which you so kindly provided.
Meaning I'd have to use a little bit more size than 8' to gain an equivalent blade for a vawt.

Let's run the numbers on a 8' x 3' Vawt at 35%, and compare results against an 8' Hawt at 40%.
Vawt: [8' x ((3' x pi) x .35) = 25.6'^2]
Hawt: [(pi x 4^2) x .40 = 20'^2]

These are probably horribly wrong since this is based solely around swept area.. But if they are even slightly correct this would destroy the community...  

Proceed to picking me apart gents.  


 

[Perry1]

No need to try to invent new math to describe something that's been well understood for almost a hundred years. When calculating your power just use the width x height (the hole in the air so to speak) that the wind control volume will see.

Concerning efficiency, that is not a constant. While you quote a number (which I think is way high for the Lenz2) at what windspeed are you talking? The Lenz design is very aerodynamically dirty. Above 20 mph, the efficiency drops off dramatically and above 25 mph that figure will be down around 10-15%. Take your alternator and other losses and your turbine is down around 10% efficient.

I don't think you can say that a Lenz make similar lift and power as a traditional HAWT. Especially across all the windspeeds you will encounter.

Lenz style VAWT's are interesting but they are pretty much a fun hobby project style turbine. I have yet to see anybody that has been running them reliably for a while and making any appreciable energy. Just because I'm not aware of it doesn't mean they don't exist. If anyone out there has had one up over a year and made any real energy with it, let us know.

Don't discount the HAWT's. Remember, VAWT's existed long, long, long before HAWT's. Their shortcomings are what lead to HAWT development.

Don't take my word on all these facts. I'm just parroting what Ed says about his design here. He'll walk you through the math as well.

http://www.windstuffnow.com/main/lenz2_turbine.htm

Perry

[electrondady1]

ed built some Darius types but found they would take off and blow up
so he built one mimicking STOL aircraft.
a fat wing that won't over speed.
it's a clever design.
lots have been built around the world .
ed and any one else who talked vertical mills used to get slammed pretty hard around here.
i'm glad things have changed
carry on .

[silentblue1987]

The entire goal for making this thread was to find an equivalent size Lenz2 for an 8' Hawt
I think after spending a week poking this thread that I now have enough info to match blade sizes with relative accuracy.

I'm in the final stages of completion so I might post the entire project here and see what the community thinks.
It's not easy to find a cnc laser cutter near town..

[XXLRay]

In the end everything is about the efficiency (or call it power coefficient). Assume the HAWT has an efficiency of 35% and the Lenz2 20% you get only 20 / 35 = 0.57 times the power from the Lenz2 compared to the HAWT. Thus you have to make it 35 / 20 = 1.75 times bigger to get the same power out of it.

Here is an example (in metrics - sorry for that) at wind speed 5m/s for a HAWT diameter of 2,4m (which is a bit less then 8').

HAWT power: 0,5 * 1.2 kg/m³ * (5m/s)³ * 3.14 * (1.2m)² * 0.35 = 0.6 * 125 * 4.5216 * 0.35 * W = 119W

From that the area of the Lenz2 has to be 4.5216m² * 1.75 = 7.9128

Lenz2 power: 0.6 * 125 * 7.9128 * 0.2 * W = 119W

You see that the power is the same although the efficiency has changed. This is compensated by teh bigger area. Of course you are right that the Lenz2 and the HAWT efficiency have to be determined first but I think I made an educated guess.

[lohearth]

"Above 20 mph, the efficiency drops off dramatically "
Isn't this where you would start to consider furling anyway.