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VAWT swept area clarification.


By cr8zy1van, Section Wind
Posted on Thu Oct 20th, 2005 at 07:05:40 AM MST
Question for vawt community.

I am building my first VAWT, based on a Darrius H rotor design. I am trying to calculate the total swept area to see if I am heading in the right direction.. For HAWT's you simply calculate swept area as (Area = Pi x r2) but VAWT's have me a little stumped!

The cord of my blades measure 12 inches. Both are 34 inches tall and spaced 21 inches from the center of rotation. (If someone wants to give me the formula I would love to add it to my little book!)

I have been watching this: http://www.fieldlines.com/story/2005/10/7/63930/5558 post closely but it would appear like it has dried up. Anthony Chessick calculated 10 Watts at 15 Mph, about half way down that post. Can anyone tell me how that number was calculated?

I look forward to a day of testing this up-coming Saturday, so your prayers for wind will be greatly appreciated!! If a crow farted we would have more air movement than we do at this moment. I will start a diary with video and pictures if this proves a worth while endeavor.
VAWT swept area clarification. | 14 comments (14 topical, 0 editorial)

Re: VAWT swept area clarification. (3.00 / 0) (#1)
by windstuffnow (elenz(at)windstuffnow(dot)com) on Thu Oct 20th, 2005 at 06:55:32 AM MST
(User Info) http://www.windstuffnow.com/main

  The swept area of an "H" type  VAWT is simply width x height.  In the case of an "eggbeater" darrieus its width x 1/2 height.
.
Have Fun! Windstuff Ed


Re: VAWT swept area clarification. (3.00 / 0) (#2)
by MountainMan (jp@jeffSPAMPREVENTIONpritchard.com) on Thu Oct 20th, 2005 at 09:57:16 AM MST
(User Info) http://www.real-home-based-business-opportunity.com

To further clarify Ed's reply, for your H that would be a width of 21 inches and a height of 34 inches.

jp
MountainMan, Julian California
http://www.real-home-based-business-opportunity.com
My ultimate hobby project is to build a ten foot tall robot.



Re: VAWT swept area clarification. (3.00 / 0) (#3)
by MountainMan (jp@jeffSPAMPREVENTIONpritchard.com) on Thu Oct 20th, 2005 at 09:58:05 AM MST
(User Info) http://www.real-home-based-business-opportunity.com

oops, width equals 21 inches times 2 (total diameter rather than radius).

jp
MountainMan, Julian California
http://www.real-home-based-business-opportunity.com
My ultimate hobby project is to build a ten foot tall robot.



Re: VAWT swept area clarification. (3.00 / 0) (#4)
by cr8zy1van (salsa117@hotmail.com) on Thu Oct 20th, 2005 at 10:49:44 AM MST
(User Info)

OK just to make sure that I'm understanding. Here is a diagram of my setup:





From Ed's posting I take width(12) * height(34) = 408 sq in.


From Mountainman's first posting I take radius(21) * height(34) = 714 sq in.


And lastly Mountainman's second: (radius*2)(21) or diameter(42) * height(34) = 1428 sq in.


Am I over complicating things? Are we talking width of each blade? Or radius, or total diameter of blade path?


[ Parent ]


Re: VAWT swept area clarification. (3.00 / 0) (#6)
by windstuffnow (elenz(at)windstuffnow(dot)com) on Thu Oct 20th, 2005 at 12:06:55 PM MST
(User Info) http://www.windstuffnow.com/main

  I would call it 1428 sq inches as Mountain Man.   I should have said diameter x height ( by width I meant diameter )   This is the area of wind the machine sees.  Basically the same as if you put a flat board in the wind that was 34" tall and 42" wide this is the area that the machine sees.  

  The actual power collection area at any given point in rotation would vary as well does the efficiency of collection.  There are points where the wings are at full power and other points where it is completely stalled and only along for the ride.  Also, there are certain positions where the wings are actually taking power from the machine, the later two are the ones I like to work with... thats were the fun is !  

.
Have Fun! Windstuff Ed
[ Parent ]



Re: VAWT swept area clarification. (3.00 / 0) (#5)
by IntegEner (mail@int.......com) on Thu Oct 20th, 2005 at 11:14:16 AM MST
(User Info) http://www.integener.com

Someone with a handle like "cr8zy1van" shouldn't be asking questions with a mathematical content but then someone with a handle like mine, "Knucks the Knucklehead" (aka Anthony Chessick), shouldn't be answering such questions either. Verticals are fun mostly because there is still a lot not known about them. It is clear that these machines have two rotor surfaces seen by the wind, the front and the back, the front having a convex shape and the rear a concave one. These complicate the calculation of the blade swept area since, especially for the back one, it can be as large as the entire half circumference of the rotor circle or one half of pi times the rotor diameter. But this requires several assumptions including that the rear surface is taking the entire load and this is not the case. Having constructed several small verticals rotors of about 3 feet in diameter I see that it is often factors such as the air drag on the upwind side that has some controlling influence (at least on the efficiency) also.

If the assumption is made that both the front and the back take the load equally there is still a problem in that when the blades are near the outer fringe of their circuits (like a sailboat sailing nearly directly upwind or downwind) they are not as able to convert energy as when they are at the points where they are crossing at right angles to the wind (helped by the fact that the entire blade is at work at the rotor arm length rather than a declining amount over its length as with the horizontals blades).

To do the math for this, some calculus is necessary but it works out surprisingly well. Here is my cut at it:



I know it is a great disappointment for the fans of the verticals to hear that a factor of 2/3 should be applied to the rotor front aspect area, i.e. the rotor diameter times the blade length, but a serious look at what is involved leads to this conclusion. Just think that the horizontals have not done much better, few to none of them being able to produce even half of the energy that their swept areas indicate is possible under Betz at rated conditions and able sometimes to meet only 75% anytime other.

The 10 watts at 15 mph was found this way. The rotor diameter of the project in the message is 12" and the blade length is 18". This aspect area, 12" x 18", is 1.5 square feet. In square meters this is = .13935 m^2. This divided by a standard factor of .114 yielded the energy in kilowatts at the Betz limit for 30 mph winds = .1222 kws = 122.2 watts. Divided by 8 for 15 mph winds this came to 15.3 watts. But then the factor of 2/3 derived above was applied and this then resulted in 10.2 watts. For a device as small as this, that is quite good energy. For your device, 42" diameter x 34" blade length, this same calculation for a 15 mph windspeed yields 67.3 watts.

My verticals device shown on the IntegEner-W website has a blade length of 22" and rotor diameter of 36" resulting in a potential at 15 mph windspeed of some 37.4 watts.

Watch your blade thickness-to-chord ratio. I prefer thin blades that, even so, have difficulty starting up. With two blades, some "hemming and hawing" will take place when stopped but they will eventually start. Mine with the four doubled blade configurations shown starts up in winds almost as light as the crow's aforementioned gaseous emission and certainly hardly even sensed by observers, some 3 mph.

This has been fun writing this and I hope your project turns out well.

Knucks, Tehachapi, www.integener.com



Re: VAWT swept area clarification. (3.00 / 0) (#7)
by IntegEner (mail@int.......com) on Thu Oct 20th, 2005 at 03:06:20 PM MST
(User Info) http://www.integener.com

I see a misplaced decimal point in my last. The standard factor for converting the square meters to kilowatts at 30 mph wind speed is 1.14 and not .114 . I have done this so often it is all committed to memory.

The verticals can be said to have a "physical blade swept area" as others have mentioned and also an "effective blade swept area" which is less and is equal to the former times the factor of 2/3 that I have been suggesting.

I think most everyone would agree in general terms with this.

Knucks, www.integener.com

[ Parent ]



Re: VAWT swept area clarification. (3.00 / 0) (#8)
by IntegEner (mail@int.......com) on Fri Oct 21st, 2005 at 09:08:59 AM MST
(User Info) http://www.integener.com

Just so everyone knows where my sympathies lie, I believe the horizontals with Jerry blades and those "gutter" blades mentioned (both of plastic) and the aluminum tlgwindpower.com blades have it all over everything else, including all the verticals so far mentioned in these postings. They have their faults but they do what is necessary, shovel the wind with the thinnest profile possible from one vector direction to another. That's all blades are supposed to do and they accomplish it wonderfully well as others have attested. The rest of all that I see on this discussion list does not hit the mark and reveals a love for fastening shiny pieces of metal or plastic together in complex patterns that add nothing very new or substantial to wind energy.

The verticals need something thin like those mentioned above for their own purposes. The thickness-to-chord ratios must be less than 10% for them to avoid the terror of high air drag to which all verticals are subjected when traveling on the part of their paths heading straight into the teeth of the oncoming wind. This factor is often not given its due as if it doesn't exist and everyone finds out, sooner or later, sometimes after spending a lot of money and time on their projects, that it does. Look at what has happened to Mass Megawatts (www.massmegawatts.com/index.htm) on the East Coast with their drag-based verticals. They even built their own wind tunnel to test out their designs, all without a satisfactory understanding of basic physics and vector algebra. It is a shame some of these efforts are not better directed.

The foundations are now in place in wind energy. One of the problems that need to be solved is the problem of blade parasitic drag, which exists everywhere, even on the multi-megawatt turbines of the big wind energy companies. It is hard to imagine but true that 50% of the power in the wind seen by these big machines, thousands of kilowatts, is being wasted in blade parasitic drag as revealed by a check of their power curves. It also can be seen in the efforts to go out on a limb with blades bending back as far as they do and with high stresses when under load. This is serious business.

I would also have to say that the small horizontals, the Bergeys and the like, tend to spin too fast as if they must give a good accounting of themselves visually. I would imagine that they only are creating more noise than necessary and it comes as no surprise that it is due to the power wasted in the parasitic drag of the fast moving blades. The energy produced has never been highly satisfactory, either. Four kwhs a day at the 7 cents we used to pay for electrical power in Minnesota comes to just what it used to cost to buy the daily paper before the price went up. Something better is possible.

New ideas are welcomed but it is easy to avoid responding to those who clearly do not have a good conception of what all this is about.

Knucks, Tehachapi, www.integener.com/



Re: VAWT swept area clarification. (3.00 / 0) (#9)
by windstuffnow (elenz(at)windstuffnow(dot)com) on Fri Oct 21st, 2005 at 06:27:42 PM MST
(User Info) http://www.windstuffnow.com/main

  That has to be the most unique way that I've ever heard to tell people that what their doing here is a useless waste of time and money.

  Very tunnel visioned indeed.

.
Have Fun! Windstuff Ed
[ Parent ]



Re: VAWT swept area clarification. (3.00 / 0) (#10)
by rotornuts on Fri Oct 21st, 2005 at 09:42:59 PM MST
(User Info)

Anthony, many of us have visited the subject of parasitic drag extensively. I think your making a critical error in attributing so much to parasitic drag, besides, parasitic drag is an exteamely vague term used to cover the issues of parasitic drag(loosely defined as friction drag plus induced drag) and induced drag(drag "induced" or a result of the creation of lift). Parasitic drag increases at the square of velocity so a low rpm machine such as a Vawt should actually be viewed as an inprovement for someone fixated on drag. I could scream every time I se someone blinded by drag. DRAG IS RELATIVE and that's it, I don't know what else to say.

I don't want to have a system that produces dick all for lift and dick all for drag. I'd far rather a system that produces significant lift VS drag. As I said it's relative and I could care less what anyone else says. Thin airfoils are not the way to go for the system your trying to design.

You need to bring your aerodynamic knowledge into the verrrrry subsonic world where it belongs.

Mike

[ Parent ]



Re: VAWT swept area clarification. (3.00 / 0) (#13)
by elvin1949 (elvin1949@yahoo.com) on Sat Oct 22nd, 2005 at 06:14:26 AM MST
(User Info)

Morning Mike
 AMEN
later
elvin

[ Parent ]


Re: VAWT swept area clarification. (3.00 / 0) (#11)
by rotornuts on Fri Oct 21st, 2005 at 09:46:24 PM MST
(User Info)

I guess before we get hung up on terminology we can substitute friction drag for "form" drag.

Just wanted to preclude an opportunity to point out a trivial error in terminology.

Mike



Re: VAWT swept area clarification. (3.00 / 0) (#12)
by CG on Sat Oct 22nd, 2005 at 03:34:23 AM MST
(User Info)

This site may be of a little interest, eurowind-uk.net. They have been going a long while and getting nowhere fast, it is probably more an example of how not to run a windpower company . The Uk put all its eggs in the H turbine basket, and didn't far. This is not to say I am agains this type of turbine, it's just an historical fact.



Re: VAWT swept area clarification. (3.00 / 0) (#14)
by schnyder (IsaacQuah@yahoo.com) on Wed Mar 29th, 2006 at 04:04:58 PM MST
(User Info)

Guys, I'm currently designing a H-type darrieus turbine. Well, I've already calculated the swept area of the turbine which is 6 meters square. So I was wondering what would be a good length for the rotor diameter and blade? and what would be the chord length and thickness for the blade?



VAWT swept area clarification. | 14 comments (14 topical, 0 editorial)
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· http://www.fieldlines.com/stor y/2005/10/7/63930/5558
· Also by cr8zy1van

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