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swept back blades
By asheets, Section Wind Posted on Wed Apr 04, 2007 at 05:53:39 PM MST
swept back blades

Perhaps this random thought belongs in the "Not Inventer Here" area...
Anyhow, I was reading up on how the Germans though of the swept wing concept to increase wing area and lifting power without increasing wing length. A really useful concept when trying to increase total aircraft speed and minimize parking space needs while at the same time trying to lift 400 fat Americans out of a high-altitude airport.
I wonder if this same approach has been tried in a windmill or generator before? I recently saw a decorative mill that appeared to have swept-back blades. Sticking straight out, I would bet the blades would have at least an 8 foot sweep -- however in swept back the way it was, it only took up about 3 foot of space. And it spun around like crazy.
Any comments from anyone?

swept back blades | 30 comments (30 topical)

Re: swept back blades (3.00 / 0) (#1)
by paradigmdesign on Wed Apr 04, 2007 at 01:26:34 PM MST

Yes, NREL and Knight and Carver are working on a design called the LWST (Low Wind Speed Turbine) / STAR (Swept Twist Adaptive Rotor). They are being tested, not to increase the output per foot, but to increase the overall wind capture without increasing the load envelope. They idea of this is to sweap the outboard sections so they will bend out of the wind and blead off some of the flapwise load, reducing the structeral cost.
"Don't worry about tomorrow, plan for it."

Re: swept back blades (3.00 / 0) (#2)
by Ungrounded Lightning Rod on Wed Apr 04, 2007 at 01:55:45 PM MST

The amount of power collected depends on the cross-section of the wind affected by the blades - "swept" in the other meaning of the word. Reducing the width seen by the oncoming wind can't be readily compensated for by making the depth along the wind direction greater.

There may be other advantages to sweeping the blades back. But reducing cross-section without reducing available power - and thus "beating the Betz limit" - isn't one of them.

Re: swept back blades (3.00 / 0) (#3)
by asheets on Wed Apr 04, 2007 at 02:54:02 PM MST

I'm sure you are correct about this, but thinking outside of the box a bit... what causes the torque in a wingen setup? Is it just the length of the blades, or is it the surface area of the blades (which happens to increase as either the blade length increases or as the sweep increases in a fixed blade length)?
_____________________________

Alan Sheets
[ Parent ]

Re: swept back blades (3.00 / 0) (#4)
by whatsnext on Wed Apr 04, 2007 at 03:10:54 PM MST

Once again someone is confusing torque with power. Please stop because there is no connection between the two and trying to increase torque will likely end up with a reduction in power. If we instead tried to "reduce" torque in order to get the blades spinning more freely we'd all be a lot happier.

[ Parent ]

Re: swept back blades (3.00 / 0) (#5)
by asheets on Wed Apr 04, 2007 at 03:28:53 PM MST

OK, please try to explain the confusion you accuse me of having having. Just saying I'm an idiot really doesn't help matters if you don't explain WHY I'm an idiot.

My understanding is that torque is the rotational force required to get an object started spinning about an axis. If I had a badly cogging generator, then more torque is what I'd need to get it spinning and generating, right?
_____________________________

Alan Sheets
[ Parent ]

Re: swept back blades (3.00 / 0) (#6)
by whatsnext on Wed Apr 04, 2007 at 03:38:17 PM MST

I never said you were an idiot. Not even close. If you have a heavily cogging alt that requires enormous torque to get started than you must be aware of the loss of efficiency once it does. Getting rid of the cogs so the alt could spin freely just makes more sense to me than building a poorly working prop to overcome an alt's flaws. If we all started designing blade sets that at least look like they should spin rapidly instead of trying to "create torque" we'd all have a lot more potential in our machines. Why worry about what the blades are doing when they are not making any power anyways? Swept area and rpm is where the real power is at unless the laws of physics have been repealed lately and I have not heard about it.

[ Parent ]

Re: swept back blades (3.00 / 0) (#22)
by Ungrounded Lightning Rod on Thu Apr 05, 2007 at 08:34:13 PM MST

My understanding is that torque is the rotational force required to get an object started spinning about an axis.

Torque is force in a twist context.

RPM is speed in a twist context.

Energy amounts to force times distance. Power is energy per unit time, which amounts to force times speed.

So power is proportional to torque times RPM.

You can have a lot of power by having a lot of torque at a low RPM or a little torque at a high RPM.

The higher the RPM, the more power a generator with a given amount of magnets and wire can convert from mechanical to electrical form. So higher RPM is good on the genny side.

With horizontal-axis lift-type turbine blades the higher the tip speed ratio (TSR) the less energy they leave in the air as downstream rotation. So fast is good for blades, too, even though it means lower torque at a given power level. (TSRs of 6ish are good.) To get higher torque you have to lower the speed, which lowers the TSR, which lowers the efficiency.

(This is not surprising, since the torque comes from "twisting the wind". By Newton's action-reaction law, the more torque the wind puts on the blade, the more torque the blade puts on the wind.)

If I had a badly cogging generator, then more torque is what I'd need to get it spinning and generating, right?

Starting is a separate issue. Lift-type turbine blades don't have any torque from lift when they're stopped. In that case they act as drag-type turbine blades - "airscrews". A high-TSR blade is nearly flat to the wind (except near the hub if it has proper twist - but the hub section doesn't intercept much wind). So an efficient blade set may not provide much starting torque at all.

Cogging is like a ball sitting in a piece of corrugated sheet metal. It starts out sitting at a low spot. To get it moving you have to push it hard enough to get it up and over the first high spot. Once you do that, it picks up enough energy going down the next down slope to get it up the next up slope, and you only have to apply enough force to keep it going against friction and load. So cogging is an issue ONLY for startup. After you're started all it does is give you a bit of vibration.

Of course if you have a lot of cogging you need a lot of torque to get started. So you either sacrifice efficiency in your blade to get the torque, or you fix the darned alternator to reduce the cogging until the standing torque of the blade at a windspeed that would just achieve cutin is enough to break the cogging.

One other thing: Near the hub a lift-type blade needs to be twisted to a much lower speed ratio and to be much fatter and wider to achieve efficiency. But it sweeps little area and so contributes little power. Thus the airfoil shape is typically attenuated or just left off, to save material, weight, and cost. But because of its low speed ratio it does provide significant starting torque. So if your genny has a cogging problem it may make sense to continue the "proper" twist and blade widening down to near the hub, to get enough starting torque.

[ Parent ]

Re: swept back blades (3.00 / 0) (#28)
by Brian H on Sun Apr 08, 2007 at 03:35:55 PM MST

Excellent explanation ULR!

Cheers!

Brian H
"Remember, I'm pullin' for ya'! We're all in this together!" - Red Green
[ Parent ]

Re: swept back blades (3.00 / 0) (#8)
by thefinis on Wed Apr 04, 2007 at 05:47:39 PM MST

Torque and power do have a connection. This is a lot like the kw kwh confusion. KW and Torque both refer to measurements of power available at this moment. Both need a qualifier kw/time and torque/movement to produce power. To put this very simply with out going into the formulas torque times rpms equals work produced. You can have the same shaft power coming from setups one with high torque/low rpms and one with low torque/high rpms. The confusing part comes when you multiply by 0 rpms. You can have any amount of torque and no power will be made at 0 rpms. It is considered a stalled condition or a wind load like on a tower. Now if it blows the tower over it will have done some work.

The efficiency losses tend to be in the conversion(gearbox) to desired rpms. It seems to me that once your tsr increases above normal ranges(3-5) that the drag/lift ratio can get delicate. Often power made is decreased/sacrificed for higher rpms to match generator needs. No matter the tsr as diameter increases the rpms drop so all large commerical type turbines will be high torque low rpm beasts.

Finis
Texas born and bred
[ Parent ]

Re: swept back blades (3.00 / 0) (#20)
by whatsnext on Thu Apr 05, 2007 at 10:54:15 AM MST

This is nothing like the Kw and Kwh confusion unless you too are confused. Torque is a measure of force and nothing more. Kw is a measure of power. Kwh is a measure of total power over a given amount of time. The only torque that a mill builder should care about is one that occurs at a relatively high RPM. This is not new science so there is no need whatsoever to look "outside the box" and doing so confuses people and provides zero benefit.

[ Parent ]

Re: swept back blades (3.00 / 0) (#10)
by vawtman on Wed Apr 04, 2007 at 06:53:28 PM MST

So are you saying the blades should just be foam and spin like crazy.Sounds pretty weak to me.

[ Parent ]

Re: swept back blades (3.00 / 0) (#19)
by whatsnext on Thu Apr 05, 2007 at 10:37:26 AM MST

"So are you saying the blades should just be foam and spin like crazy.Sounds pretty weak to me.
I keep searching but not sure what im looking for.Have fun Vawtman"

It's weak because it springs from your own imagination. I never said any such thing.

[ Parent ]

Re: swept back blades (3.00 / 0) (#12)
by Nando on Wed Apr 04, 2007 at 07:46:13 PM MST

PEACE, PEACE ON EARTH!!

Read to calm ourselves

Nando

torque1(tôrk)
n.
The moment of a force; the measure of a force's tendency to produce torsion and rotation about an axis, equal to the vector product of the radius vector from the axis of rotation to the point of application of the force and the force vector.
A turning or twisting force.
tr.v.,torqued,torqu·ing,torques.
To impart torque to.

[From Latin torqure, to twist.]

torquer torqu'er n.
torquey torque'y adj.
torque2 (tôrk)
n.
A collar, a necklace, or an armband made of a strip of twisted metal, worn by the ancient Gauls, Germans, and Britons.

[French, from Old French, from Latin torqus, from torqure, to twist.]
Torque
Concept

Torque is the application of force where there is rotational motion. The most obvious example of torque in action is the operation of a crescent wrench loosening a lug nut, and a close second is a playground seesaw. But torque is also crucial to the operation of gyroscopes for navigation, and of various motors, both internal-combustion and electrical.

How It Works

Force, which may be defined as anything that causes an object to move or stop moving, is the linchpin of the three laws of motion formulated by Sir Isaac Newton (1642-1727.) The first law states that an object at rest will remain at rest, and an object in motion will remain in motion, unless or until outside forces act upon it. The second law defines force as the product of mass multiplied by acceleration. According to the third law, when one object exerts a force on another, the second object exerts on the first a force equal in magnitude but opposite in direction.

One way to envision the third law is in terms of an active event--for instance, two balls striking one another. As a result of the impact, each flies backward. Given the fact that the force on each is equal, and that force is the product of mass and acceleration (this is usually rendered with the formula F = ma), it is possible to make some predictions regarding the properties of mass and acceleration in this interchange. For instance, if the mass of one ball is relatively small compared to that of the other, its acceleration will be correspondingly greater, and it will thus be thrown backward faster.

On the other hand, the third law can be demonstrated when there is no apparent movement, as for instance, when a person is sitting on a chair, and the chair exerts an equal and opposite force upward. In such a situation, when all the forces acting on an object are in balance, that object is said to be in a state of equilibrium.

Physicists often discuss torque within the context of equilibrium, even though an object experiencing net torque is definitely not in equilibrium. In fact, torque provides a convenient means for testing and measuring the degree of rotational or circular acceleration experienced by an object, just as other means can be used to calculate the amount of linear acceleration. In equilibrium, the net sum of all forces acting on an object should be zero; thus in order to meet the standards of equilibrium, the sum of all torques on the object should also be zero.

Real-Life Applications

Seesaws and Wrenches

As for what torque is and how it works, it is best discuss it in relationship to actual objects in the physical world. Two in particular are favorites among physicists discussing torque: a seesaw and a wrench turning a lug nut. Both provide an easy means of illustrating the two ingredients of torque, force and moment arm.

In any object experiencing torque, there is a pivot point, which on the seesaw is the balance-point, and which in the wrench-and-lug nut combination is the lug nut itself. This is the area around which all the forces are directed. In each case, there is also a place where force is being applied. On the seesaw, it is the seats, each holding a child of differing weight. In the realm of physics, weight is actually a variety of force.

Whereas force is equal to mass multiplied by acceleration, weight is equal to mass multiplied by the acceleration due to gravity. The latter is equal to 32 ft (9.8 m)/ sec2. This means that for every second that an object experiencing gravitational force continues to fall, its velocity increases at the rate of 32 ft or 9.8 m per second. Thus, the formula for weight is essentially the same as that for force, with a more specific variety of acceleration substituted for the generalized term in the equation for force.

As for moment arm, this is the distance from the pivot point to the vector on which force is being applied. Moment arm is always perpendicular to the direction of force. Consider a wrench operating on a lug nut. The nut, as noted earlier, is the pivot point, and the moment arm is the distance from the lug nut to the place where the person operating the wrench has applied force. The torque that the lug nut experiences is the product of moment arm multiplied by force.

In English units, torque is measured in pound-feet, whereas the metric unit is Newtonmeters, or Nm. (One newton is the amount of force that, when applied to 1 kg of mass, will give it an acceleration of 1 m/ sec2). Hence if a person were to a grip a wrench 9 in (23 cm) from the pivot point, the moment arm would be 0.75 ft (0.23 m.) If the person then applied 50 lb (11.24 N) of force, the lug nut would be experiencing 37.5 pound-feet (2.59 Nm) of torque.

[ Parent ]

Re: swept back blades (3.00 / 0) (#24)
by thefinis on Fri Apr 06, 2007 at 10:15:50 AM MST

Thanks for trying to keep me on the right path Nando

I guess this is where I am confused about torque. I know that in most cases torque is only measured when there is movement but with a torque wrench you read the torque after it stops moving. Well with an old style anyway where the bending of the wrench shaft caused a needle to show the torque being applied. So what is the rotational force called on a set of stalled blades that have not overcome clogging?

The turbine setups that I am trying now are all low rpm high torque vawts. LOL If I had been building hawts I probably would be making electricity now but I am saving hawts to fall back on if I give up on the vawts.

Dang it wish I had been able to go to college for more than one year. Aw well in 35+ years much has changed and much of what I learned has been buried too deep to find. They were not pleased with me anyway as all I wanted to take and pay for was math, computers and science courses. Computers were mainly card readers and key punch in those days.

Finis
Texas born and bred
[ Parent ]

Re: swept back blades (3.00 / 0) (#7)
by Photic on Wed Apr 04, 2007 at 04:49:10 PM MST

Maybe I'm having a hard time visualizing this, do you mean a prop like that of a pinwheel?

Re: swept back blades (3.00 / 0) (#9)
by southpaw on Wed Apr 04, 2007 at 05:49:21 PM MST

I believe the swept wing concept allowed similar wing lengths with a smaller wing span by sweeping them back. Sweeping the "wings" on a turbine back would be similar to raising the tips of an airplane wing which wouldn't increase lift only stability.

Re: swept back blades (3.00 / 0) (#11)
by Countryboy on Wed Apr 04, 2007 at 07:08:44 PM MST

Anyhow, I was reading up on how the Germans though of the swept wing concept to increase wing area and lifting power without increasing wing length.

The Germans DID increase wing length though. What they didn't increase was wingspan.

Sweeping back a wing only decreases the amount of force any given portion of the wing sees. By sweeping the wing, you spread out the load over a larger area.

It might be an idea if we weren't using furling blades, and routinely had blades snapped because of wind sheer. Making the blade longer, but without increasing the blade diameter just helps the structural integrity of the blade.

Re: swept back blades (3.00 / 0) (#13)
by southpaw on Wed Apr 04, 2007 at 10:29:26 PM MST

It's hard to compare an aircraft wing with a turbine blade because of the fact that air craft wings are travelling the same speed at all parts of the wing while turbine blades travel faster as you go out from the root, hence the twist in the blade. Perhaps a better comparison would be to helicopter rotors which tend to be nearly level and a great deal of research has gone into thier efficiency to obtain maximum lift with minimum drag.

Re: swept back blades (3.00 / 0) (#16)
by finnsawyer on Thu Apr 05, 2007 at 10:27:54 AM MST

There is a problem comparing an airplane propeller or helicopter blades with the action of wind mill blades. In the case of the two former most of the lift is in the desired direction, forward or up. In the wind mill's case most of the lift is toward the tower, at least near the tip. Only a small fraction is in the direction of rotation. On the other hand, most of the drag is in the direction of rotation. So, we need a different analysis to maximize the output force vesus the drag.

Carrying this on a little and stating that I don't care much for using torque, since the torque from each value of the radius will vary and we must integrate it anyway, I propose the following approach:

Start with the expression for energy: E = 1/2MV^2

Differentiate with time: P = dE/dt = MVdV/dt = FV, where F is now the component of force in the direction of V and, of course, P is the power. If one starts with the velocity squared as the dot product of the vector velocity with itself and then differentiates with time, one gets the result of the vector force dot the vector velocity, which is just the component of force in the direction of the velocity. It is assumed that the force and velocity do not change during at least one revolution of the mill. We need the desired component of force on a small section of the blade at radius R (force per unit length times dr)and the speed of the blade at radius R. The force, of course, will be the component of the lift force (Flxdr) in the direction of rotation. The resulting FV will then be equated to the power available from the wind for the annular ring of width dr at radius R, which is:

Pw = 0.59x0.5xpx(2xPixRxdr)xV^3 = FlxdrxV

From which we can calculate the total width of the blades (divide by three for a three bladed design). Note that the drag force does not enter into this equation. Some of the lift force is needed to counteract the drag force. Once we know the width we can find the component of the drag force in the direction of rotation and subtract it from the component of lift to get the net force in the direction of rotation at that radius.

I didn't say anything about how you got to an expression for the lift force, as that depends on the blade angle, the AOA, and so on. Perhaps the best approach would be to find the blade particulars for the best lift to drag ratio. I would be interested in what people think of this approach.
GeoM
[ Parent ]

Re: swept back blades (3.00 / 0) (#17)
by finnsawyer on Thu Apr 05, 2007 at 10:29:36 AM MST

Start with the expression for energy: E = 1/2MV^2

Pw = 0.59x0.5xpx(2xPixRxdr)xV^3 = FlxdrxV

Re: swept back blades (3.00 / 0) (#21)
by southpaw on Thu Apr 05, 2007 at 06:38:59 PM MST

I agree with your comment. But (aren't those buts annoying) in the case of a glider wing the lift is enough to keep it in the air and the main purpose is forward propulsion similar to a turbine blades function of rotation. I'm by no means an expert on any of these topics but enjoy debating ideas and concepts even though there are those who would have no debate, many including myself do not have a complete grasp of electronics, electrical generation or aerodynamics but learn more from these debates than having someone cram theory down our throats. "I'm right and were not going to discuss it anymore" doesn't appeal to many. It's a fine balance between genius and lunacy but the moderators are doing a great job of keeping the board balanced.

[ Parent ]

Re: swept back blades (3.00 / 0) (#23)
by finnsawyer on Fri Apr 06, 2007 at 08:07:57 AM MST

Gliders stay in the air because the air mass through which they are moving is continually rising. That is, the glider is actually falling through the upward moving air mass. The process does produce some power, which must balance the power lost to the various drag forces. If one wanted to charge a battery by some means, the glider would have to fall faster. I guess there are some similarities to a wind mill blade, but I really don't see much coming of it.
GeoM
[ Parent ]

Re: swept back blades (3.00 / 0) (#14)
by kamikaze on Thu Apr 05, 2007 at 04:07:26 AM MST

Hi,

The reason that high speed aircraft have swept back wings is only to improve their efficiency in the cruise, at high percentages of the speed of sound. Because modern jet airliners spend most of their time in the cruise, that is when the wing should perform optimally. For take off and landing a swept back wing is less efficient than a straight wing. As such a swept back wing needs to be larger in area than a straight wing to produce the same lift.

Having said that, I personally like a swept back blade for small wind turbines, because of the reduced noise, and because they look cool.

I also prefer the low tsr/ high torque setup, because, assuming the gen is suitable for low rpm, the wider blades will be better at overcoming any cogging, will produce more power in light winds, produce less noise and general wear and tear.

At higher wind speeds the high torque/low rpm turbine should produce similar power to the high tsr machine.

Inter Ventos Laboramus ( in the winds we work )

Re: swept back blades (3.00 / 0) (#15)
by paradigmdesign on Thu Apr 05, 2007 at 06:11:41 AM MST

I should have elaborated in my first post. The outboard section of the blade is swept back, flaps out of the wind, reduces blade loading, thus a LARGER BLADE can be made with the same amount of loading on the rest of the turbine, tower etc. There are actually aerodnamic losses (though small until bending) for using sweapt blades.
"Don't worry about tomorrow, plan for it."

Re: swept back blades (3.00 / 0) (#25)
by Dave B on Fri Apr 06, 2007 at 11:11:55 AM MST

Admin,
Could you please push this and all future similar posts of "rants and opinions" into the rants and opinions section ? Thank you for considering it. Dave B.

[ Parent ]

Re: swept back blades (3.00 / 0) (#26)
by TomW on Fri Apr 06, 2007 at 11:53:38 AM MST

Dave;

Admin,
Could you please push this and all future similar posts of "rants and opinions" into the rants and opinions section ? Thank you for considering it. Dave B.

Well, it didn't start out as one that I see and other than strict "on topic" comment enforcement, I don't see how to do it fairly?

Always open to ideas, however. The whole "topicalness" of comments thing should be observed by comenters but thread jacking [redirecting a thread] seems to be very common and acceptable to most, however unfair it is to the thread initiator. Either a hard nosed on topic policy or courtesy are the only options and lately here anything but total acceptance of any behavior is not O.K. Lots of good discussion actually happens when a thread deviates from the original topic so it is not an easy thing to decide on a course of action.

Cheers.

TomW

The Truth is the Truth, even if no one believes it; and a lie is a lie even if everyone believes it

[ Parent ]

Re: swept back blades (3.00 / 0) (#27)
by Dave B on Fri Apr 06, 2007 at 12:34:39 PM MST

Thanks Tom for the comment. I can appreciate how tuff it can be to edit and I was hoping that others even more so than Admin. would maybe read this and think about what's happening here.
This in itself should be in the rants and opinions but to be seen and a point made I'll reply here. I will post a related to the rants and opinions to try to pull it away from here.
Think about it, this is a wind section. Does anyone else care to keep it that so we can see and talk about projects ? I hate to suggest another topic, maybe just changing the name ? (wind projects) I don't have the answers but respect seems to be going away and with it the fun of sharing. Dave B.

[ Parent ]

Re: swept back blades (3.00 / 0) (#29)
by paradigmdesign on Sun Apr 08, 2007 at 04:53:15 PM MST

Are you asking this from the admin about my particular comment or the whole topic in general? I feel my response was quite "on-topic".
"Don't worry about tomorrow, plan for it."
[ Parent ]

Re: swept back blades (3.00 / 0) (#30)
by Dave B on Mon Apr 09, 2007 at 11:38:06 AM MST

No one in particular, the whole topic is interesting but I just wish there was a way to keep the rants and harsh opinions at times out of project posts. I was excited to post my first project. Suggestions to possibly help are one thing but the bashing by those with egos so large they forgot we all had to walk before running is hurtful. Especially hard to take are these comments from those who have never scuffed one knuckle building something of their own. I have posted over in the rants and opinons trying to bring these discussions over there and away from the new project posts that seem to be attacked and hijacked lately. Dave B.

[ Parent ]

Re: swept back blades (3.00 / 0) (#31)
by ghurd on Mon Apr 09, 2007 at 03:14:47 PM MST

I do not like attacks to anyone. Pointing out an error or easier way is one thing, attacks

I like my posts to be hijacked with anything relative... to anything remotely relative.
Even if it is only relative to the board, and not the project.
IE: resizing diagrams hijack taught me a lot from an unrelated post, and I know for certain it helped others (and me). It must have saved bandwidth, my time and many others time. A jpg (or 10,000 future jpgs) under 4K has it's benefits to everyone.

Alan asked about swept back blades in `wind'. Seems reasonable to me.
Should all nichrome, TSR, furling, blade carving or regional power outages questions and posts be in `homebrew electricity'?
Deleting all of them would cause good people not to come back?
I don't intend to bash, argue or criticize. Just saying some areas are grey.

ADMIN does a fine job, IMHO. Leniency has its virtues.

(sorry Alan, it didn't fit, out of context, in the other post)
G-

Ghurd.info
[ Parent ]

swept back blades | 30 comments (30 topical)

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