Need help with blade sizing

[Tony Savage]

I am starting to think about blade fabricating and I'm not sure how big I need to go with them. I know that I saw a formula that calculated output power, in watts, with blade size, wind speed and efficiency used as variables but I can't find it. I would like to aim for around 1000 watts at 10 mph. Another question is one of the width of the blades. I have never seen any mention of this before. Is there a width to length ratio that is preferred?

Because I work with fibreglass my thoughts are to shape something from wood and then cast a mold from that. Then lay up the blades in that mold and use either balsa or foam for the core material. This way the blades will all be identical in shape. They should also be quite stiff and strong. Does the weight of the blades have much of an effect on performance?

One more question that relates to feathering. I'm not sure if I am going to go that route yet but am confused as to which way the blades would feather. Do they rotate so that long axis of the cross section of the blade is more parallel to the wind direction or more perpendicular? In other words does the wind just push on the narrow edge when feathered or on the flat area of the blade?

[Rob Beckers]

Hi Tony,

Others will no doubt jump in, but I'll give you my Q&D guide to sizing: The average small turbine on the market has an overall efficiency of around 20 - 25% from kinetic energy in the wind to electrical power out of the inverter. If you're going to do direct battery charging it'll be worse.

So, to make 1000 Watt at rated wind speed you'll need to have 4000 Watt "in the wind". Your rated speed is 10 mph, or 4.5 m/s. The equation for swept area vs. energy in the wind is:

A = P / (0.6125 * V^3)

This is for a standard atmosphere, at sea level. With all the other hand-waving already going on there's no point really to try and be more specific at this point.

So, for your case, A = 4000 / (0.6125 * 4.5^3) = 71 square meters
That makes a diameter of 9.5 meters = 30 feet, or 15 feet per blade, in other words, it's going to be big! Then again, you're asking for quite a bit of output power at a very low wind speed (I'm scratching my head at this point, wondering if I made a mistake, because these are huge blades, but you really are asking for large power at a very low wind speed)...

Stew will likely jump in to talk about 'solidity' which relates to width (chord) of the blades vs. length, as well as the number of blades.

As to the direction of blade pitch change, take a look at my pictures, at http://www.solacity.com/pictures.htm, it'll help visualize how they move. The turbine there turns clockwise, and the two fly-weights will move outwards. This will pitch those blades 'flatter' when viewed from the front. The blades in those pictures move towards stall, you asked for feather, so you want your blades to move in the opposite direction. That means they'll need to move towards more parallel with the wind direction, showing a smaller profile when viewed from the front.

Hope this helps (and I didn't screw up in my calculations :o ).

-Rob-

[Stewart Corman]

I have reposted my xls calculator which answers all your questions including the blade cord dimension

just plug in a number anywhere you see a "?"

note the example on sheet 3 of the Sandia design with twist and taper

bottom of sheet 1 has a metric converter from mph -> m/s

Stew Corman from sunny Endicott

[Tony Savage]

Hi Rob,

Thanks for the formula and the calculations. It does end up with a fairly large blade but I was expecting at least 12 feet anyway. That's a great looking unit that you're working with. I wish I could afford something like that. How long are the blades for it? You can really get a good idea as to how the blades rotate to stall. I'm still a little fuzzy on the terminology so excuse me if I screw up. As I understand what you are saying then this is what happens. The blades start off at quite a angle (feathered?) when viewed from the front. As speed of rotation picks up they rotate to becomes flatter and then if rotation becomes too fast they turn more to become quite flat (stalled?) Is this the preferred way to control blade speed or is the opposite action a better way to go? I'm sure that there are pros and cons to both methods.

Hi Stewart,

I've seen that XLS file in your other posts but am unable to view it on this computer. I'll try to look at it at work and print it off.

[Rob Beckers]

Hi Tony,

This may be redundant, pardon me if that is the case, first a bit about stall vs. feather:

Envision an airplane wing (it's a lot easier than envisioning a rotating turbine with wind blowing at an odd angle, it's aerodynamically identical though). Normally a wing is set to have a little bit of an angle to the wind, another term for this is "angle of attack". Now slowly rotate that wing so the angle of attack gets larger and larger. This will increase lift of the wing (and drag), up to a point where the air can no longer negotiate the sharp bend over the (now) angled wing. That's stall. The lift drops off sharply at stall, and drag very suddenly increases a great deal. If you've ever flown an airplane, this really is how it feels, you drop out of the sky very suddenly, and it is very lethal if you're close to the ground. In short, stall is when the angle of the wind vs. the chord of the wing is too large to sustain airflow over it. The wind looses its "bite" and in case of a wind turbine it stops accelerating the blades (the increased drag helps slow them too).

For feather the wing moves the other way. Instead of increasing the angle of attack, it is decreased. Lift (and drag) decrease, right up until the air pressure on top and on the bottom of the wing are equal, and there's no lift left. Depending on the airfoil, it's now almost "in line" with the wind (the chord line that is), and the wind blows over it with very little resistance. Once again, the wind has no "grip" on the wing, and in case of a turbine it won't be able to accelerate the blades. Your windmill comes to a (grinding) halt.

Now if you look at just about any wind turbine you'll notice that at rest the blades are in fact just about perpendicular to the wind. Their flat side faces the wind, the curved (convex) side of the blade is away from the wind. This makes for a very high angle of attack, and at stand-still wind turbines are normally completely stalled (yeah, even those with a variable angle blade, where the root angles towards the wind). So how do they get started you ask? I'll admit this isn't too clear to me yet. Probably a combination of the always present turbulence that makes just a little lift, stalled blades still having a little bit of lift, and because the blades work as a drag device, causing them to start turning. Once they start turning the angle the wind makes over the blade is no longer the same as 'just' the angle of the wind as you perceive it on the ground. The blade's movement gets added to the wind direction. If you bent your mind around the effective angle of the wind over the blades it'll come at a much shallower angle of attack, and the faster they spin, the more the wind seems to be coming from the direction of the blade's leading edge (and the plane of rotation).

So, we went from stall at standstill, to a reasonable angle of attack when the rotor is spinning. Now, if we keep increasing the rate of rotation in proportion to the wind speed, the angle of attack will stay the same. Ideally you want to keep it at the blade's optimum angle of attack (highest lift vs. drag ratio), by loading up the blade just enough for each rotational speed. That's what Maximum Power Point Tracking does, but that's another story. For stall regulated turbines this is pretty much what happens, until the rotational speed gets large enough to generate enough force through fly-weights to start depressing a spring. Pushing that spring simultaneously moves the blade angle of attack, and in case of a stall regulated machine this will increase the angle of attack so the blades are at the edge of stalling. The harder the wind blows, the more they get twisted into stall (it's a very non-linear process, a small RPM range makes for a large change in angle of attack). What helps too, is that the harder the wind blows, the more the effective angle changes towards stall, even if one didn't move the blades, just as the blades are stalled at startup.

For a feather regulated machine, when the fly-weights start depressing the spring they move to a shallower angle of attack. Lift decreases, and if the angle of attack keeps decreasing lift will go all the way to zero. Once again, the dynamics of fly-weights, spring, and blade angle will keep the rotor RPM at an almost constant value. There is a little problem with feather regulating through; Just as the effective angle of attack increased when the wind blows harder for stall regulated machines, it will do the same for feather regulated machines. In stall an increased angle of attack is less lift. Not in feather. Increased angle of attack in feather means more lift! More force on the blades to spin harder!

That brings us to the disadvantage of feather regulated machines. Because in really high winds the apparent wind direction over the blades is almost from the front, we need to twist the blade angle such that they are almost edge-on to the wind. It is a much larger angle change for feather than what is needed for stall, making it harder to design and build.

The advantage of feather regulation is that in high winds the blades generate very little drag, so the horizontal thrust of the turbine is small, and you can get away with a much less beefy tower (cheaper!) than for stall regulation. In stall the blades generate a great deal of drag, so that is high horizontal thrust. For the Scirocco (stall regulated) it's 5000 Newton (1123 lbs) in the horizontal direction at 215 km/h winds (133 mph). Blade size for the Scirocco is 5.6 meters diameter, or 18.3 feet. About 9 feet per blade.

This turned into a story that's quite a bit longer than I intended. It needs pictures to make it easier, but I'm no artist. At the time I started in wind turbines it took me a long time to piece all this together, so hopefully it'll help others understand a bit better how wind turbines work!

-Rob-

[Mark Parsons]

Rob,

A great post. Very informative.

Could you provide reference information and data (or math functions) on the non-linear aspects of both stall and feather regulated schemes?

Thanks.
Mark

[Tony Savage]

Hi Rob,

Thank you for that excellent post. It makes much more sense to me now. I'm glad that you pointed out the comparative horizontal thrusts for stalled and feathered blades. It is one of the things that have pondering to great length. With blades as large as I anticipate using, they will generate a large force on the blades themselves with considerable flexing when in a stalled position. It seems that feathering will be the way to go if I can overcome the mechanical design of the mechanism.

[Tony Savage]

Hi Stewart,

I finally am able to view your XLS file. There is some great info there. Are the maximum power available numbers calculated on 100% efficiency? One final question, how does one figure out Tip Speed Ratio?

[Rob Beckers]

I'm glad my lengthy masterpiece helped a bit!:)
Tip speed ratio is real easy to calculate, because it's exactly what it means: The ratio between the blade tip speed and the wind speed.

For example, that Scirocco I described earlier rotates at 245 RPM at rated power and at rated wind speed of 11.5 m/s. How fast are the blade tips going at 245 RPM? Well, the distance of one revolution, 2 * pi * R = 2 * 3.14 * (5.6/2) = 17.6 meters. In one minute they travel a distance of 17.6 * 245 = 4310 meters, or per second that is 4310 / 60 = 72 meters/second. So, the tip speed ratio is 72 / 11.5 = 6.2 (since it's a ratio it's dimensionless).

It works the same way if you calculate everything in miles/hour or some other measure of speed, as long as you keep the units consistent.

Just so you get some idea of how fast blade tips are actually moving, that 72 m/s is the same as 260 km/h, or 161 mph!!! And the Scirocco is a "slow" wind turbine, many of the small ones out there have blade tips that move a whole lot faster. That should give you a feeling why tip speed has such an effect on the amount of noise a turbine produces...

-Rob-

[Tony Savage]

I guess I should re-phrase my question. How will I know what the rpm will be? Is it just something that is inherent of the blade design and you'll find out when the bird flies. Wouldn't a variable pitch unit have different TSR's?

[Rob Beckers]

I guess I should re-phrase my question. How will I know what the rpm will be? Is it just something that is inherent of the blade design and you'll find out when the bird flies. Wouldn't a variable pitch unit have different TSR's?

Those are tough questions! :eek:
I'll start with the easy one, about variable pitch turbines. All of the small variable pitch wind turbines that I'm aware off only vary their pitch at the very end of their RPM range, as a mechanism to protect themselves and to produce power beyond rated wind speed. So, from 0 to almost rated RPM they are in essence fixed pitch. There is a little bit of an exception to this for some that have multi-stage pitch control. For example the Scirocco uses a "start" pitch at standstill, to make it easier for the turbine to get started, then pitches the blades to run angle at 60 RPM. It just uses two springs of different strength, and in effect it lowers cut-in speed by 1 m/s.

For large turbines that use an induction generator things are different (this is an area I know little about, so please correct me if I'm wrong). Since they have to keep their turbines running at a fixed RPM (in sync with the grid) I believe they do use active pitch changes throughout the whole wind speed regime to keep the angle-of-attack optimal.

RPM and TSR are (ideally) design parameters. You design the turbine for a specific RPM range, in practice usually based on the alternator you have available, and power you want to get out of it. For an optimal design the TSR is based on getting the most efficiency out of the number of blades, and parasitic drag of the blades (related to tip speed and thus blade size); Spinning blades faster is more efficient, but it also increases drag, so there's an optimum somewhere. There are empirical tables that show the optimum based on number of blades. I believe Stew has lots of info on this.

RPM and TSR relate to each other through wind speed and blade size, if you have one, you can calculate the other (for any given wind speed and blade size).

So how does TSR tie in with my previous posting about angle-of-attack? Remember that a given blade profile is most efficient (highest lift-to-drag ratio) at a specific angle-of-attack. The design goal is to run the turbine as much as possible at this angle-of-attack. Since we can fix the blades at any angle we want with respect to the blade hub, we can choose at what RPM (and thus TSR) the apparent wind meets the blade at our desired (optimal) angle-of-attack. Here's the crux; If we can keep the blade moving at just the right RPM to keep that angle-of-attack optimal for the wind speed, then the TSR is going to be constant throughout our wind speed regime. Why is that TRS constant you ask? Because to keep that angle of the wind over the blade the same the blade (tip) speed has to change linearly, at the same ratio, as the wind speed. It's simple vector arithmetic. Twice the wind speed, twice the blade RPM, same TSR, and same angle-of-attack.

The next question would then be, how do you keep the TSR (and thus angle-of-attack) constant throughout the working wind speed range of your turbine? That's done by loading it up just right, by putting the right load on the alternator, or in other words, extracting just the right amount of power for each wind speed. MPPT does exactly that.

In practice it's all one big trade off. Direct battery charging will not present the right load, rather it'll correctly load it at only one wind speed. Alternators rarely work at the very low RPMs that optimal blade speeds and TSR would suggest, the list of compromises is endless.

Now you know as much as I do about wind turbine aerodynamics... :cool:

-Rob-

[Rob Beckers]

There's still a little bit missing from my story, so for the sake of completeness here it is:

From the previous parts you may have the idea that running the blades at optimal lift-to-drag ratio (and its resulting angle-of-attack) is all that matters, but that is not so. You can run those blades at the right angle-of-attack for any value of TSR, depending on how you mount them on the hub and how fast you allow the turbine to spin (by extracting power from it). So what makes one TSR value better than another?

There are two types of drag involved in airfoils, induced drag, and parasitic drag. Induced drag comes from the angle-of-attack. For an ideal airfoil at zero angle, there would be no induced drag. However, the larger the angle-of-attack, the more induced drag (the more lift too, lift is good, drag is bad). The best lift-to-drag ratio and corresponding angle-of-attack for an airfoil optimizes for induced drag. Parasitic drag is always present when we yank something solid through the air, it simply increases with speed. The faster the blade (especially the tip) goes through the air, the more parasitic drag. It increases with the square of the airspeed. It gets even worse when the blades move so fast that the next blade tip of a rotor hits the wake turbulence of the previous one before the turbulent air has time to move downstream (which happens with less than the wind speed, since we're slowing down the air as it passes through the rotor).

At this point you're probably wondering, so why not move that blade through the air really slowly? We could, but we wouldn't be extracting much energy from the wind. We extract energy from the air by slowing it down and converting part of its kinetic energy into torque and rotational energy, and the more air we can slow down the more energy is being extracted. Moving a blade through the air very slowly will move most of the air through the turbine downstream before it has a chance of being affected by the blades. It just "blows through". So, once again, there's an optimum between going fast enough to extract the most energy from the wind, and going slow enough so parasitic drag doesn't eat up most of that extracted energy.

You can now also see why a one-bladed prop (yes, they exist) has to spin faster than a two-bladed one. And why a two-bladed prop has to move faster than a three-bladed one for optimal energy extraction. You can also see now why a prop with more blades will eat up more of the extracted energy in parasitic drag, versus a prop with less blades (a bit simplified, since more blades also means it can spin more slowly, but by and large I believe this is true).

There probably are equations for the optimal TSR vs. number of blades, but I don't know them (If anyone has references for these, please let me know!). There are graphs that show this, and one thing that jumps out from them is that there's not much difference for quite a range of TSRs. It's not all that critical. There are other TSR considerations (the understatement of the day), a large TSR and thus high RPM makes for a cheaper alternator, a large TSR also makes for more noise. In practise, the small turbines I know about almost all run at far higher TSRs than optimal.

-Rob-

P.S. In case anyone gets the idea that I claim to be the authority on wind turbines, Absolutely not! What I've written is my understanding of how these things work, and how to design them. I've been doing some studying lately because I have an interest in designing one. Anyone that knows more, or has corrections, please feel free to comment and add to this. I'd love to learn more!

[Stewart Corman]

Tony,

Are the maximum power available numbers calculated on 100% efficiency? yes, that is all you can get out of the wind w/o building a perpetual motion machine :rolleyes:

One final question, how does one figure out Tip Speed Ratio?Rob has attempted a long dissertation on explaining this, but I will provide a very short answer.

Firstly, you don't "figure out" a TSR in a computer or using graphs ..it happens ...you choose a "good" profile from the experience of others and the NACA plane wing profiles developed in the late 30's aren't very much different in performance then the best simulations such as the Sandia design (twisted, tapered SG6050). You then choose a length of blade, cord width (tapered?) and pitch angle ...TSR "happens" when you fly it! ...if you screw up, then you won't get as much power out as you expected at any particular WS. You would not believe how good a turbine you can make with thin flat planks! I agree with one of Rob's conclusions that many of the factors are not really that critical in the final design.

From the limited benchtop experiment so far, I am drawing this conclusion ( and it isn't much diff from common knowledge from H Piggot) ... shape of the blade and cord width determines the max power you will get (Lift), while pitch as well as blade number, determines TSR . Optimum TSR is around 3.5 ...fastest is NOT bestest!! That said, TSR gets limited by how fat the tip is ie where rotor is spinning the fastest (Drag).
Make the tip too fat, and drag will kill you as it slows everything down.

TSR is never constant (it peaks), since the optimal L/D ratio only occurs at one station along the blade at one AOA, but that L/D curve varies with WS ( RE#). The following two charts show that nothing stays the same.
Remember that AOA changes dramatically and non-linearly as you move down the stations towards the root. The USNPS profile nontwist, non tapered, is what Jacobs has been using for years (not the bestest, but works!)

Hope this helps,
Stew Corman from sunny Endicott

note: higher RE# for each color are on top

http://i145.photobucket.com/albums/r203/scorman1/Experiment/data%20charts/LDratio.jpg
http://i145.photobucket.com/albums/r203/scorman1/Experiment/data%20charts/ClvsAOA.jpg

[Rob Beckers]

Hi Stew,

A couple of notes/questions:


... TSR "happens" when you fly it! ...if you screw up, then you won't get as much power out as you expected at any particular WS ...

While I agree that this is traditionally the way small wind turbines have been designed, especially with direct battery charging, I really don't believe that it is the best way to go about it. If you have no control over the load (or very little), then it's a matter of getting the combination of rotor, alternator, and load "just right" so you get decent performance. It'll likely loose a lot in the process though (I only have to refer to the stories from people I know that get better performance out of their genny by adding a resister to the battery cable so it doesn't load the turbine quite so much, meanwhile burning power in that resister). With MPPT you actually control how the turbine is loaded, and therefore you control TSR.


... TSR is never constant (it peaks), since the optimal L/D ratio only occurs at one station along the blade at one AOA, but that L/D curve varies with WS ( RE#) ...

Agreed that TSR varies because the best angle-of-attack varies with Reynolds number and therefore blade speed (If I remember right from publications TSR doesn't change all that much though. Numbers anyone?). Why would optimal lift-to-drag ratio only have to occur at one station along the blade though? That would depend on blade design, and proper twist of the blade, right? Moreover, you'd want optimal TSR such that the outer 1/3 or 1/2 of the blade runs at optimal lift-to-drag, no? That's where most of the power comes from, because it sweeps so much more area than the rest of the blade. Ideally, wouldn't you want to control TSR through MPPT so it stays at its optimum?

It would seem to me one would. Of course, in real life and without an MPPT controller you take what you get, that's what the compromise part was about... ;)

-Rob-

[Rob Beckers]

For the more practical side to my rather qualitative explanation, see Hugh Piggott's publication, at http://www.scoraigwind.com/wpNotes/bladeDesign.pdf. It has equations for optimal blade width, RPM, TSR, blade angle etc. (With thanks to Stew for pointing this out).

-Rob-

[Stewart Corman]

Rob's q:
Why would optimal lift-to-drag ratio only have to occur at one station along the blade though? That would depend on blade design, and proper twist of the blade, right?Geometry kills you here ... IF, if you could presume the TSR you will actually get, then you can design the non-linear twist to get an exactly uniform AOA along the entire blade length.. OK so far

Now look at the tip velocity vs a station 1/4 down ...only 75% the m/s ..also means 3/4 RE# ...just moved down the chart of L/D numbers.

BUT, but lift is proportional to cord width AND Cl (from chart) and apparent WS (not just rotational rpm) ...geometry bites you in butt again

So you have to make blade wider or else that station isn't pushing as hard ( lacking in torque). Think of the blade as rigid and think about the parallel of two people manning a long oar and one is pulling harder than the other....one doesn't do any real work.
Trick here is not to have exactly uniform AOA, but to get the tip design at the shallow AOA whereby Cl it is still increasing as you go down the blade.
Now add the fact that the tip creates vortexes (reality bites you here) which rob the total power, which is why they spec a blade's model at the 75% station. But all the above presumes good sophisticated models which they aren't !

Then you get Bergey who sells the XL-1 w/thin blades, w/o twist and w/o taper ...and it works! Jacobs have been working for 30+ years!

Furthermore, IF, if you used the current momentum equations ( Betz' stuff) on Allison's triangular non-twist, non-tapered blades ...just like the proverbial bee ..it can't fly!

look at the following ...it is TERRIBLE! yuk!

this is a turbine that GENERATES drag ....no lift, pitiful L/D, stalls at <5 degrees where at 7.5 degrees DRAG>LIFT



α Cl Cd L/D
[°] [-] [-] [-]
1 0.498 0.02703 18.433
3 0.603 0.03161 19.061
5 0.253 0.0424 5.964
7 0.072 0.06604 1.093
9 0.027 0.11557 0.234
11 0.013 0.09299 0.136
13 0.007 0.12786 0.054
15 0.004 0.1761 0.024
17 0.003 0.23683 0.011
19 0.002 0.3129 0.006
21 0.001 0.39844 0.003
but stick 4 rotors of these in a helix and you get a Cp (efficiency) = 56%
http://s145.photobucket.com/albums/r203/scorman1/Wind/Allison%20article/

QED

I have a document from Clarkson's experiment where they make flat plate blades go at TSR from 1 -> 4 ?
(too big to attach)


Stew Corman from sunny Endicott

[Rob Beckers]

Geometry kills you here ... IF, if you could presume the TSR you will actually get, then you can design the non-linear twist to get an exactly uniform AOA along the entire blade length.. OK so far

Now look at the tip velocity vs a station 1/4 down ...only 75% the m/s ..also means 3/4 RE# ...just moved down the chart of L/D numbers.


Eh, not sure about that...
Reynolds number has Air Velocity * Characteristic Length (ie. blade chord) in it. So as long as the 'ideal' blade compensates by getting a wider chord (and good blades do just that) the Reynolds number doesn't change. I was talking about an optimal design. As always, real-life will be more of a trade-off but we have to start somewhere.

Let's assume Reynolds number does change. That's not unreasonable in reality, since we can only compensate in blade chord so much. All it takes to keep angle-of-attack optimal for that station is to adjust its twist to make it so. No need to change TSR to make that station run at its optimum, TSR can stay the same throughout the entire wind speed regime IMO.

BUT, but lift is proportional to cord width AND Cl (from chart) and apparent WS (not just rotational rpm) ...geometry bites you in butt again


Exactly!
Chord width needs to adapt to station. That's what makes a good blade. Call me thick, but I'm still missing why geometry is biting me here???

So you have to make blade wider or else that station isn't pushing as hard ( lacking in torque). Think of the blade as rigid and think about the parallel of two people manning a long oar and one is pulling harder than the other....one doesn't do any real work.

But good blades are wider the closer you get to the root. The parts closer to the hub will provide a lesser part of the total torque in any event, because the airspeed over that station is less (hence less lift, despite the wider station, since that goes with the square of the airspeed), and because the arm of the force is smaller than those stations further out on the blade, no? Torque is Arm * Force.

Trick here is not to have exactly uniform AOA, but to get the tip design at the shallow AOA whereby Cl it is still increasing as you go down the blade.
Now add the fact that the tip creates vortexes (reality bites you here) which rob the total power, which is why they spec a blade's model at the 75% station. But all the above presumes good sophisticated models which they aren't !

Agreed! Though I still don't understand how this would necessitate or result in a variable TSR, or how geometry is biting me...

By the way, a friend of mine at Duke does CFD simulations of turbine blades (for jet engine turbines). Those actually are very accurate in predicting performance and behavior. His simulations actually couple aerodynamics with a structural model of the blades so flexing and resonance show up. The same could also be done for wind turbines, just that very few manufacturers do this, and even fewer hobbyists.

Then you get Bergey who sells the XL-1 w/thin blades, w/o twist and w/o taper ...and it works! Jacobs have been working for 30+ years!


Yeah, but the Bergey turbines are really very inefficient for their size! It works, just not very well. Jacobs is another story, those work far better than Bergey's blades.


Furthermore, IF, if you used the current momentum equations ( Betz' stuff) on Allison's triangular non-twist, non-tapered blades ...just like the proverbial bee ..it can't fly!

look at the following ...it is TERRIBLE! yuk!

this is a turbine that GENERATES drag ....no lift, pitiful L/D, stalls at <5 degrees where at 7.5 degrees DRAG>LIFT



α Cl Cd L/D
[°] [-] [-] [-]
1 0.498 0.02703 18.433
3 0.603 0.03161 19.061
5 0.253 0.0424 5.964
7 0.072 0.06604 1.093
9 0.027 0.11557 0.234
11 0.013 0.09299 0.136
13 0.007 0.12786 0.054
15 0.004 0.1761 0.024
17 0.003 0.23683 0.011
19 0.002 0.3129 0.006
21 0.001 0.39844 0.003
but stick 4 rotors of these in a helix and you get a Cp (efficiency) = 56%
http://s145.photobucket.com/albums/r203/scorman1/Wind/Allison%20article/

QED

The table shows a profile that stalls at >5 degrees, that's where drag will be (much) larger than lift. That's a pretty good efficiency! So the question is why it's that high, and if there is room for improvement by using a different blade profile. Possibly one benefit of a better profile that doesn't stall so quickly would be to be less sensitive to TSR and loading. Speed up a bit with this profile and angle-of-attack goes to low, ie. no output, slow down just a bit and it stalls, no output again...

I have a document from Clarkson's experiment where they make flat plate blades go at TSR from 1 -> 4 ?
(too big to attach)


Hmm.... This falls under the heading "I can make a brick fly. Just bolt on a big enough engine.". Sure you can make any rotor spin, the question is if you can extract meaningful amounts of power from it in relation to the available power in the wind. The problem I have with these examples is that they are anecdotal, they don't make a pattern nor help understand why this is happening. The understanding why turbines behave the way they do is what I'm after.

Stew, I'm still missing your point about the need for variable TSR and the biting of geometry. Please bear with me, I can be a bit slow. :D

-Rob-

[Tony Savage]

Gulp!!! and thought I was confused before. LOL. No, really though, thanks for all of the great input, believe it or not I think I'm starting to get a handle on all of this. Let's find out. As I understand what you guys have said, a blade that has a twist in it, is optimized along it's length to provide the correct AOA at each station to compensate for the variances of the apparent wind speed as we move from the root of the blade out to the tip. That sentence ending up sounding a lot more technical than I had intended. Having said all of this, it seems like that there is not much of an advantage going to a twisted blade especially when you consider how much easier it is to make a "straight" blade. (Like the ones you made Stewart). Stewart, would you be able to give me an idea what the chord should be at the root and tip of a 15 ft blade.

[Stewart Corman]

Tony,
I'll go over the design choice procedure to hopefully make it easier.

If you look at lower part of sheet 2 on xls calculator, is the chart form of Hugh's formula which is same as the one that Claus Nybroe has on his site as well.

You can plug in blade number, lift coefficient Cl ( at 75% station?) and look on chart for respective diameter and TSR to get tip cord in inches. Root is simply as wide as reasonable, but a 1:4 ratio at 25% station gives you constant velocity/cord relationship (similar RE#).
Play whith the xls as an exercise by randomly plugging in numbers to see how quickly cord changes.

Sounds simple except you have to make decisions, like which profile to give you what Cl ...I'd pick the Jacobs with a flat bottom and small cord to thickness ratio and small radius on leading edge ( easy to cut with router bit). For TSR = 3.5 , you can choose a pitch that gives you Cl=1.0

Then you have to decide which physical construction method you want to build.
If you choose the standard flat plate bolt-on mounting, then the pitch has to be carved into the wood.
If you choose the tubing spar method I used, the blade shaping can be automated more easily and an electric hand plane is the key tool ...all lines from tip to root are straight lines.
Tubing spar with exhaust clamps allows a key feature ....infinite adjustment of pitch angle. You can vary the TSR setpoint to accomodate the typical WS to get the max power out....if TSR is too low, then pitch the blades shallower.

If you choose more blades than three ie 5 or 6 blades, then the tip width cord is reduced and you need to choose a TSR around 3.5 -> 4 rather than 6, so figure around 6 inch tip cord for 15 ft diameter.
A smaller cord tip is thinner which is goodness.
Blade root has to be thick enough to accomodate a 1 3/8" spar tubing, or about 2 1/2 -> 3 inch max thickness, but that makes it 20 inches wide.
Need to compromise here and easiest choice is to thicken the root profile from 12 % to 24% so that a 10-> 11 inch root is obtained.

If I choose this design for my 15+ footer, I am planning to grind an 11 inch profile blade for my moulder/shaper so these can be cut in a single setup (multiple passes). Only other carving is router bit for leading edge undercut and all blades are identical...BTW, I'd use poplar as wood choice since it is straight grained, free of knots, and lightweight, and I'd probably fiberglass/polyester resin/gelcoat the surface for permanence



OT - spar mounting has an additional design capability to extend blade out from hub. I am contemplating testing a long spar ie forget the center 1/2 area and make the diameter an extra foot larger. example, for your 15 ft diameter a conventional would have blades carved 7 1/2 feet long (176sqft), but to make a 16 foot out of 4 ft "paddles" (200sqft- 50sqft center) , would have almost identical real swept area, but you eliminate the size/weight of the blades and extra wide cord of the root which stalls/drags at high AOA ..interesting ...would like some comments here

Stew Corman from sunny Endicott

[Mark Parsons]

Hi Stew,

How would you dress the center of a 16' diameter swept area using only 4' paddles leaving a 8 foot diameter center hole? Big nose cone?

2' diameter hub with 2 U-bolt exhaust brackets on each blade spar still leaves a 8' minus 2' diameter annulus. Leave it open? Not worth filling?

Mark

[Laurie Forbes]

Tony, I have run your requirements (1,000 W at 10 mph wind) through a Hugh Piggot based spreadsheet (that I have modified to make more useful) and the result ain't pretty. The spreadsheet indicates a mill diameter of about 35 feet (which ties in pretty well to what Rob calculated). It would also require a 21 to 1 gear up assuming a gen speed of 1000 rpm for 1,000 W output (giving a prop speed of about 48 RPM).

A 35 foot turbine is no trivial job for a home builder (I've never heard of any anywhere close to that large) so you might want to relax your specs a bit. For example, 1,000 W at 15 mph would reduce the size to about 19 ft.

I have attached a copy of the spreadsheet if you wish to try it. As you will note, it also calculates the blade dimensions.

[Stewart Corman]

Tony,

The spreadsheet indicates a mill diameter of about 35 feet (which ties in pretty well to what Rob calculated).


I don't think so! if you are figuring power AT 10mph running speed at 40% eff..then it is only 25 feet diameter anyway

BUT, as I said to Mark in another thread ...the Weibull distribution is almost 2x, SO, an AVG 10mph WS is equivalent to running at 12.5mph = 18 footer, which is quite reasonable ...probably what I am ultimately going to build with 6 foot blade "paddles" on 3 ft tubing spars

Stew Corman from sunny Endicott

[Laurie Forbes]

Tony,
I don't think so! if you are figuring power AT 10mph running speed at 40% eff..then it is only 25 feet diameter anyway

I get 25 ft also if 40% effic is assumed however I was using 20% which was recommended IIRC by Hugh P (and seems more realistic to me(?)), esp with a non-direct drive gen, capacitor excited gen. FWIW, my 12 footer was calculated to produce about 1 KW which is about what it did.

BUT, as I said to Mark in another thread ...the Weibull distribution is almost 2x, SO, an AVG 10mph WS is equivalent to running at 12.5mph = 18 footer, which is quite reasonable ...probably what I am ultimately going to build with 6 foot blade "paddles" on 3 ft tubing spars

Stew Corman from sunny Endicott

[Stewart Corman]

Mark,
In response to the question:

How would you dress the center of a 16' diameter swept area using only 4' paddles leaving a 8 foot diameter center hole? Big nose cone?

2' diameter hub with 2 U-bolt exhaust brackets on each blade spar still leaves a 8' minus 2' diameter annulus. Leave it open? Not worth filling?
IMHO think that deflecting the center wind could be an enhancement and the bullet shape is used in plane turbines would be my choice, however there were AWEA threads claiming that it was a detriment. I will add that to my "to do list" in the models after I improve the wind generator.

Here is a + recent thread on Otherpower that directly relates to your question:
"the best results, according to theory, occur when the nose piece diameter is half that of the entire mill. In fact, at that radius you get a 22% boost in power over what there is in the incident wind."

http://www.fieldlines.com/story/2006/4/18/41516/1677

Back to original premise:
After my post on the subject of "paddles", I built a three blade model and it didn't perform that well, so I looked at my xls calculator and noted that since TSR determines AOA, so the trick for non-twist blades design would be finding that station one would assume is at stall and "cut it off there" would be appropriate. At TSR = 6 you'd be cutting off a functioning part of the blade at 50%, whereby at TSR= 3.5 it should have worked, so my 66% original design is probably close. Again pitch angle becomes an important design criteria as it directly affects TSR. There are some who claim that stall models at very low RE# are incorrect, since those models are for airplanes that don't fly at 10mph! and that root sections at 20 degrees AOA still produce + force/power.

BTW, for Tony, here is a recent shot from Otherpower of a 18 footer (wooden blades), which IMHO has too thin cord ..I would put 6 of those blades on that hub and make it look like a Windflower which has been shown to have Cp=47% at TSR = 3.6 . When you just add extra blades, you don't have to adjust the pitch to steeper angle, but there is some optimum setpoint to get max power from lowest WS. ( it also becomes self stall regulating and easier to start at pitch = 12 degrees rather than 7 degrees)

Stew Corman from sunny Endicott

http://www.otherpower.com/images/scimages/57/18_foot_assembled.jpg

[Ralph Day]

Hello Stewart

The unit you're standing in front of...is it for show and the one attached to the anchor to your left is for go?

ralph

[Stewart Corman]

Ralph,
that one ain't mine ..just referencing it so people could get an idea of size
http://www.fieldlines.com/story/2007/5/1/6102/21635

he is sticking that toy on top of an 85 ft tower:
http://www.fieldlines.com/comments/2007/4/21/7231/85739/4#4