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16 ft. blades...
By jondecker76, Section Wind Posted on Tue Jun 13, 2006 at 01:10:30 AM MST
Any help on carving them?

Hello. I'm ready to start carving my blades for a 16 ft. dual rotor machine. I have the tools I need (draw knife, hand planer. etc...). As for the wood - i do have some 2"x10"x8' dimensional lumber (which actually measures 1.5" x 9.5" x 8') and I was wondering if this would be good enough to carve my blades from? At 16 foot, i've seen some blade calculators suggesting some rather large numbers that would make this lumber unusable, but looking at other blades in posts on here, they look like they were carved smaller than these blade calculators suggest also. I mean warlocks blade calculator recomments a 32.5" chord at the first station for 16' blades. For some reason that does not seem right to me...
So anyways - any advice on building a 16' set of blades at 7 to 8 TSR (i'm leaning more towards 8 tsr) using standard 2x10x8 lumber? Any thoughts or advice would be greatly appreciated!
thanks,
Jon

16 ft. blades... | 9 comments (9 topical)

Re: 16 ft. blades... (3.00 / 0) (#1)
by scottsAI on Mon Jun 12, 2006 at 08:05:45 PM MST

Hello Jondecker76,
Do not worry about the root of the blade.
Sweep area is what is important. Missing the first foot or two of the blade will only require extending the blade a few inches to make up for the missing area. Big difference in requirements at the root of the blade.
Check out the math of a circle, then remove center area that makes it difficult to fit the wood to, extend blade (circle area) to compensate.
Have fun,
Scott.

Re: 16 ft. blades... (3.00 / 0) (#2)
by hvirtane on Tue Jun 13, 2006 at 03:39:13 AM MST

I would like to second the previous comment.
You can also glue with epoxy or with other good
glue more wood on the root area of your wood lumbers.

- Hannu

Re: 16 ft. blades... (3.00 / 0) (#3)
by finnsawyer on Tue Jun 13, 2006 at 08:59:57 AM MST

If you sit down and draw some pictures showing the effective wind at each station for a TSR of 8 it should become quickly apparent that a 1.5 times 9.5 inch piece of lumber can not allow you to get enough twist in the blade to keep the blade acting as a proper air foil for any radius much below 50 to 75 percent of the maximum. Once the section of blade has an attack angle greater than stall (about 12 degrees), the drag increases greatly. While the effective wind speed drops considerably toward the root compared to the tip, the wide chord toward the root would give you only drag and no lift. So, why make it so wide? Stick with the 9.5 inches. You're off the curve for the air foil there anyway. It's also been pointed out recently that at the tip the blade also loses effectiveness, so you're stuck with a blade that produces power only from 60%? to 95%? of the radius. Not only that, you're basically dragging chunks of material through the air near the root. In the past people realized this, and designed the blades with more twist, but that now seems to have been given over to cheapness and convenience.

This brings up the question of why don't these blade design programs flag those values of radius when the blade is no longer acting like an air foil? The basic aim of blade design is to match the power available in the wind at all radii of interest keeping the drag as low as possible. In point of fact one could design a blade having constant attack angle that has lift at all useful radii if one puts in enough twist. Or one could allow the attack angle to vary as well the twist angle to design a blade of constant width that matches the power available for a wide range of radii. Restricting the blade twist to what you can get from a 9 by 10 is definitely going to limit performance.
GeoM

Re: 16 ft. blades... (3.00 / 0) (#6)
by Ungrounded Lightning Rod on Tue Jun 13, 2006 at 09:26:05 PM MST

Or you could accept the twist and let the tail of the blade "disappear" into the no-wood area near the root.

Though I'd be inclined to laminate on additional wood for the rootward portion and make it a proper airfoil most of the way to the hub: The extra depth also provides strength against the bending load from slowing the air, which goes up faster than the inverse of radius. IMHO this is far more important than just improving efficiency - or reducing drag - in the small swept area near the hub.

[ Parent ]

Re: 16 ft. blades... (3.00 / 0) (#4)
by SparWeb on Tue Jun 13, 2006 at 12:33:26 PM MST

If you were to take a look at my blade carving "lesson":

http://www3.telus.net/faheydumas/Turbine_Conventional.html#Blade%20Carving

you would see a 6-foot prop carved out of a 2x6 piece of lumber. Like yours, it's actually 1.5" thick.

This prop is a horrible performer, because the airfoil twist at the root is only about 15 degrees. The twist corresponds to a ridiculous TSR of about 15, when the prop actually turns at about 7, max. As a lesson in carving, it was very valuable. As a device for extracting wind energy, it's disappointing, because it has trouble starting itself.

Compare this to your lumber, and the longer chord reduces your twist even more, further preventing start-up. If you can bond the lumber together, to get about 4 layers thick (6" deep x 9.5" chord) you might do better. The calculator you used may have used a number of blades different from what you intend. A 9.5" chord might work on a 4-blade prop, but mind you, it won't turn as fast as a 3-bladed prop with the same diameter.

Maybe before you start carving, you could try other "calculators" (search for Alton Moore and Hugh Piggott), and try to predict the performance of what you actually will make. Make sure that it matches the alternator you intend to bolt it onto.

Steven Fahey

Re: 16 ft. blades... (3.00 / 0) (#5)
by jondecker76 on Tue Jun 13, 2006 at 09:02:58 PM MST

thanks guys -

looks like the dimensional lumber isn't my best option. I think i'll avoid using it and talk to my amish neighbors down the road about getting some more appropriately sized rough-cut lumber from one of their logging operations. Sounds like it will save me some headaches! Thanks for all of the insight!

Jon

Re: 16 ft. blades... (3.00 / 0) (#7)
by Ungrounded Lightning Rod on Tue Jun 13, 2006 at 09:29:12 PM MST

See if they have some stuff that's been seasoned for a few years. That will hold its shape and weight far better than green lumber.

[ Parent ]

Re: 16 ft. blades... (3.00 / 0) (#8)
by Flux on Wed Jun 14, 2006 at 12:45:40 AM MST

Like ULR I would like to see something thicker than 1.5" at the root but that is purely from the strength point of view.

From practical experience I have never found it necessary ( or even desirable) to use the dimensions predicted by the calculators at the root. The root section does help a bit with start up if the alternator has drag, but if this is a dual rotor then it won't have any.

The few blades that I have taken the root to these dimensions on have not performed better. They seem less capable of operating over as wide a speed range.

The calculators give a good result for the outer part of the blade, but as long as your piece of wood is strong enough then take what comes for the centre bit. Until someone can produce measured evidence I shall ignore the theory about extracting power from the centre bit, I like to use theory that fits the observed results.

The calculators are based on simple theory which may not even apply, I know that some useful work has been done on aerofoils for small wind turbines and shown benefit, but those who have spent money on this research still don't use blades the size of a house at the root. If you make it run at a phenominal speed like the Air machines then the speed at the root may be high enough to do something.

Flux

[ Parent ]

Re: 16 ft. blades... (3.00 / 0) (#9)
by Dave B on Thu Jun 15, 2006 at 12:16:57 AM MST

Just like you to hear the other side of the story. I am not charging batteries but rather running hot water heating elements with direct AC from my mill. I have a variable load controller also and a 12' 3 blade set approx. TSR 6 with 4" drop at the root, twist and tapered to a couple degrees at the tips. I see it written here time and time again that only the outer portion of the blade is what really matters and that except for strength the root of the blade profile is of little consequence. Please realize that first of all most on this discussion board who use wind power are likely charging batteries (a totally different game from utilizing heating elements through out a wide RPM range.) Also, a flatter blade is easier to carve and 2x stock is reasonably priced and readily available. Couple this with no load until the necessary cut-in speed above stall to effectively charge batteries and there you have it. An easy to carve, reasonably
inexpensive,inefficient but functional blade set for charging batteries. If your application demands more power at the lower end RPM and yet you desire the same TSR then carve as much drop and twist as your stock (and budget) will allow from looking at the blade calculators. The numbers work, take a look at the blades on the mega watt units. Dave B.

[ Parent ]

16 ft. blades... | 9 comments (9 topical)

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