Note: Comparing starting torques of constant chord and tapered blades, available

[Adriaan Kragten]

The note: "Comparing starting torques of constant chord and tapered blades" can be copied for free from my website www.kdwindturbines.nl at the bottom of the menu KD-reports. This note shows that the starting torque coefficient of a rotor with blades with a constant chord and blade angle is somewhat lower than the starting torque coefficient of a rotor with tapered and twisted blades if both rotors have the same design tip speed ratio. But the reduction of the starting torque coefficient is much less than the increase of the lift for a tapered blade at the blade root because of the larger chord and the larger blade angle. Constant chord blades normally also have a somewhat lower maximum Cp but they use much less material and are much easier to manufacture than tapered and twisted blades.

[SparWeb]

Hello Adriaan,
Thank you for writing that.  When the blades are not rotating you can simplify the math.  It helps to set aside so many variables and just focus on a static blade. 

-then I come along and ask for complications!

Would you be interested in adding consideration of blade twist in this comparison?  In my experience, I see many constant-chord blades that are not twisted, but almost all tapered blades have some twist to their angle of incidence.  The effect of twist at the root is substantial, as you know.  It might be best to leave the first section as you've written it because it gets to the point quickly, then add a section for the consideration of twist and its effects.

In your comparison, would it be fair to show two blades with equal surface area?  I believe the tapered blade in your example has 35% more surface area than the straight one.

[Adriaan Kragten]

The reason why most constant chord blades are not twisted is that they need a low lift coefficient at the blade tip and a high lift coefficient at the blade root. A low lift coefficient means a small angle of attack and a high lift coefficient means a large angle of attack. It appears for design tip speed ratios in between about 5 and 8, that the increase of the angle of attack is about the same as the increase of the angle phi at decreasing radius r and this results in a constant angle beta if a normal airfoil is used. However, for cambered airfoils, mostly some blade twist is needed but a constant chord cambered steel blade can be twisted easily linear.

I have used the two examples given in KD 35 because these blades have the same design tip speed ratio. This resulted in a total blade area which is somewhat larger for the tapered blade than for the constant chord blade for the chosen lift coefficients. If you would compare two blades with the same blade area, I think that the constant chord blade will have the largest starting torque coefficient. This is because of the great influence of the outer blade section on the total starting torque coefficient. In my note, you can see that the influence of the outer blade section is five time stronger than for the inner blade section for a constant chord blade.

The derivation of the formula for the starting torque coefficient of a constant chord blade with no blade twist is rather simple and only this derivation is given in KD 35. But it must be possible to find a rather simple formula for a blade which is tapered and twisted linear. But you will need a differential equation for this derivation and I would have no differential equations in KD 35. However, if you divide a blade into enough blade sections, you can use the formula for a constant chord and blade angle if you the take the average values for each section. This procedure is also valid if the blade isn't twisted linear.  As the blade twist of the blade used in example 1 of KD 35 isn't linear, I have used this method in my note.

In my report KD 97 I explain more about how the torque coefficient changes if the rotor is rotating slowly. Formula 11 and 12 of this report give the contribution of a blade section to the starting torque coefficient for a tapered and twisted blade.

I don't understand your question: "Would you be interested in adding consideration of blade twist in this comparison"?
This is just what I have done. I have compared a constant chord blade with a constant blade angle, so with no blade twist, to a tapered blade with a blade angle which is increasing at decreasing radius. So the tapered blade is twisted.

Considering a constant chord blade in combination with a certain blade twist isn't logic if the blade is provided with a normal airfoil. This is because in this case the rotor is no longer designed according to the aerodynamic theory if the design tip speed ratio lies in between about 5 and 8. A constant chord simply results in a constant blade angle if you want to realize the correct product of the lift coefficient times the chord. The only point is that the inner part of the blade is not very effective as there the blade is stalling.

In public report KD 484, I give the rotor calculations of the VIRYA-3B3 rotor. The result of the calculations is given in table 1 at page 4. In this table you can see that the theoretical beta is 6.5° for station A and 7.6° for station E. So the increase in beta is only 1.1°. Finally I have chosen an average linearised value for beta of 7° for the whole blade length. In table 1 you can also see that the airfoil is stalling for stations F, G and H as alpha is larger than 10°. But the Cp is corrected for this non effective inner part of the blade as it is assumed that only the outer blade length k' = 1.05 m is effective. Even with this non effective inner blade part, a maximum Cp of 0.4 must be possible.

[SparWeb]

But you will need a differential equation for this derivation and I would have no differential equations in KD 35.

Not really.  You could if you want, but a more practical approach is just an additional column in a table.  Make a table of the stations and repeat the same calculations for each segment of the span.  The twist would just be an additional list of values in such a table.  Easy.
For an example, you can look at the extract from Hugh Piggott's table, which I posted in my other thread.

I think your comparison paper would stand on its own quite well if you wanted to just carry the reference information from KD 35 and KD 97 into it.  It would add a page but would make it easier to read.  That would also give you an opportunity to make a fair comparison between two types of blade with equal surface area.  At the moment the tapered blade has more area than the straight blade, just clouding the comparison.

[Adriaan Kragten]

But you will need a differential equation for this derivation and I would have no differential equations in KD 35.

  At the moment the tapered blade has more area than the straight blade, just clouding the comparison.

I don't agree with this statement. I have compared two rotors with the same design tip speed ratio. If you design a rotor with an optimum lift coefficient, you will get a blade shape according to the curve "theoretical chord" as given in figure 5.4 of KD 35. If you linearise the chord, you get the blade as given in figure 5.3 of KD 35. Linearisation of the chord gives only a small change of the lift coefficient and therefore only a very small difference in Cd/Cl ratio (compare last two columns of table 5.2). So a blade with a linearised chord has a maximum Cp which is lying very close to that of a blade which is designed with the optimum lift coefficient. The blade of figure 5.3 has simply a larger blade area than the constant chord blade as given in figure 5.6. If you design a tapered blade which would have the same blade area as for a constant chord blade, this blade must have a higher design tip speed ratio and this would be comparing apples with pears.

A problem with tapered blades is that the angle of attack is almost constant and about 3°. But the angle phi is increasing strongly at decreasing radius. The angle phi is given by formula 5.3 of KD 35 which is shown in the lowest curve of figure 5.2 of KD 35. So this means that the change of the angle beta is about the same as the change in angle phi but that beta is only about 3° smaller than phi. So the smaller the radius, the stronger the blade twist. This gives an extra problem for the manufacture of tapered blades from massive wood.

If you linearise the twist for tapered blades (dotted line figure 5.5 of KD 35), you will deviate from the design theory near the blade root. So in this case you will get too large angles of attack near the blade root and this means that the airfoil will stall. This problem of stalling is less than for a constant chord blade but it will give some reduction of the maximum Cp.

In figure 2 of my note, I give the blade angles beta for cross section a, b, c, d and F. So in this figure you can clearly see that the blade is twisted and that the twist increases at decreasing radius.

When I was working at the University of Technology Eindhoven, a student has derived the formula for the starting torque coefficient for a linear tapered blade with a linear twist. I can't find the report in which this was done but I am sure that he used differential calculus because two factors are changing being the chord and the blade angle. The final formula was rather simple but I see it as useless to put effort in it to find it again as most tapered blades have no linear twist. In this case, following my procedure by dividing the blade in five sections and using the formula for a constant chord blade, gives a result which is accurate enough.

[SparWeb]

I'm very confused by your answer.
There's nothing "linear" in the blade shape I'm talking about.  I didn't mean to imply it, I didn't write it, and I don't even carve my own blades with linear twist, so I don't know where that idea came from.  But if somehow that came across, then I apologize for confusing you.  I read what you wrote anyway.  It's all good.  Just not relevant to what I asked.
The graphic I was referring to is this one, below.  The twist is not linear.



Perhaps you would appreciate looking at a copy of Hugh's blade geometry spreadsheet, to get a better understanding of how this can be calculated.  Without differential equations.

https://scoraigwind.com/sheets/bladedesign.xls

But this also distracts from my previous question/request:
Can you offer a comparison that is fair, by comparing two blades of the same area?

If you design a tapered blade which would have the same blade area as for a constant chord blade, this blade must have a higher design tip speed ratio and this would be comparing apples with pears.

That's a problematic statement.  There's no need to compromise the TSR just because a tapered area is selected.

It may help if I explain a little better why I am pressing the subject.  When estimating a performance parameter like startup torque, you tend to be looking at the rotor as a whole, not just the detail in the analysis.  If you are a designer choosing the blade taper, you probably already have a TSR selected, but you probably also have weight and airfoil and manufacturing to consider.  In that situation I would be likely to make as many factors equal as possible (TSR, span, area, airfoil, L/D) before making the starting torque comparison with varying taper ratios.  Do 3 or 4 choices of taper ratio and see what's optimised.  That would serve as a design comparison of start-up behaviour, useful for guidance when turning around to attempt to optimize a different parameter.

I do usually go for apples, though 

[Adriaan Kragten]

There must have been some misunderstanding from both sides but I now think that I have understood your original question. I have cancelled the note and replaced it by report KD 697. In stead of comparing the starting torques of the two rotors for a certain wind speed, I have now calculated the starting torque coefficients for both rotors. This was possible because I introduced a new formula for the contribution of a blades segment of a tapered blade to the total starting torque coefficient. The new method resulted in a much larger difference in between the starting torques than found in the note. I am sure that the method given in KD 697 is correct. I have found the mistake in the note. In the note it is written that L decreases with the factor of the ratio in between the angles of attack alpha. This is wrong. It must be that L increases with the factor of the ratio in between the blade angles beta. This mistake resulted in wrong values for L and so for Q.

I still find that you should compare only rotors with the same design tip speed ratio as only these rotors will run at the same rotational speed for the same wind speed. But we can discuss what happens if both blade areas are taken the same.

The rotor of example no. 2 has a constant chord of 0.2 m. The rotor of example no. 1 has a chord at the middle of the blade, so at station c, of 0.27 m. So the blade area of the tapered rotor is a factor 0.27 / 0.2 = 1.35 larger. If you would design a rotor with tapered blades which has the same total blade area as for the rotor with constant chord blades, it means that all chords must be a factor 1 / 1.35 = 0.74 smaller. If the design tip speed ratio is maintained at five, it means that all lift coefficients Cl must be a factor 1.35 larger because the product of Cl times c must have a certain value for a certain design tip speed ratio.

The original rotor with the theoretical chords was designed for the optimum Cl value which is 0.8 for the Gottingen 623 airfoil. So this means that a new rotor with theoretical chords must be designed for a lift coefficient of 0.8 * 1.35 = 1.08. This is much too high to call this an optimum design. If you want an optimum design with a chord which is a factor 0.74 smaller than for the rotor of example no. 1, I expect that the design tip speed ratio must be about 5.8 in stead of 5. This is because the tip speed ratio increases about with the square root of the ratio in between the chords if the Cl value is maintained at 0.8.

This new rotor has smaller chords but it will also have smaller blade angles than the rotor of example no. 1. The starting torque coefficient will therefore be reduced much stronger than only according to the ratio for which the chords are reduced. The result of smaller chords and smaller blade angles together may be that the starting torque coefficient of this new rotor with lambda design is 5.8 may be about the same as for the constant chord rotor with lambda design = 5. But if you want to be sure, you must design the new rotor and calculate the starting torque coefficient with the method as given in KD 697.

I don't say that you need differential equations if you want to design a rotor with an optimum lift coefficient. I only say that you need differential equations if you want to make one formula for the starting torque coefficient if the blade chord and twist changes linear. If the blade chord and blade twist doesn't change linear, like in your green pictures, it will even be impossible to find one formula for the starting torque coefficient. In this case, the blade has to be divided into several blade sections and you find the starting torque coefficient by the method as given in KD 697.

I really don't understand why so many people and companies make blades with a geometry which follows from the design procedure to use one lift coefficient for the whole blade. If the chord is linearised, this gives almost no reduction of the maximum Cp but manufacture of the blade is much simpler. If the chord and blade angle are taken constant, this gives a reduction of the maximum Cp from about 0.45 to 0.4 but constant chord blades are even more simple to manufacture and need much less material.

[SparWeb]

Oh, I think I see where our misunderstanding may have come from.
Has your goal been to simplify blade construction with linear blade twist? 

I use the "hyperbolic" curve for twist because the math just spits it out, and it's also an aesthetic choice.  You are probably right when you say it uses more material, as far as wooden blades are concerned.  If speaking of fiberglass blades, either twist form would use about the same amount of material.  In the case of a metallic blade with a flat-plate profile, I can't see this changing the material quantity, but it definitely would be more difficult to form a hyperbolic twist profile.

There is also a report from the NREL where they compared a blade with linear (simple) twist and a blade with hyperbolic (curved) twist and found very little difference between the two rotors Cp or thrust coefficients.  It's a very detailed report - I should dig it up for you...

[Adriaan Kragten]

Linearisation of the blade chord is done to simplify production. I can't say that I find a curved blade more beautiful than a linearised blade but beauty is in the eye of the beholder (if this is the right expression in English). I linearised the chord but I didn't linearise the blade twist in my example no. 1 of KD 35. I show linearised twist in figure 5.5 of KD 35 as the dotted line. In this figure you can see that the difference in blade angles in between a blade for which only the chord is linearised and a blade with the theoretical chord is very small but that the difference becomes very large at small values of r, if the twist is linearised too.

I have added a new chapter 7 to report KD 532 in which calculations of the VIRYA-3.1 rotor are given. This rotor has two blades made out of Roofmate, glass fibre and epoxy. As Roofmate isn't strong, the blade must be tapered. The starting torque was originally calculated by using the formula for a constant chord blade and by taking the average chord and blade angle halfway the blade. In chapter 7, I have calculated the starting torque coefficient again but now using formula 3 out of KD 697 for a tapered blade.  This calculation shows that it is allowed to use the formula for a constant chord blade if the calculated value is reduced somewhat.

[MattM]

fillm at thebackshed extruded blades out of aluminum which he called the 'GOE222 profile'.  Extruded blades are nice and uniform.

http://www.ozwindengineering.com/uploads/5/500750W.jpg

[Adriaan Kragten]

If I click on the link, I only get a photo and no film. But may be this is right. I have looked in report R-443-D: "Catalogue of Aerodynamic Characteristics - - -" if I could find this airfoil but it isn't there. But I found the Gö 227 measured by Riegels which looks similar. Information about this airfoil is given on page 3-88, 3-90 and 3-92 of R-443-D. The Cl, Cd and Cm values are measured for a Reynolds value of 0.96 * 10^5 which is rather low and this airfoil can therefore be used very well for a small wind turbine. The maximum Cl value is rather high (about 1.68) and this must be caused by the hollow lower side of the airfoil. The optimum lift coefficient is about 1.3 (at alpha = 2°) which is also rather high. The minimum Cd/Cl ratio is about 0.03. These are very good values for such a low Reynolds value.

If you can buy this airfoil as endless aluminium bar and if it is provided with a circular hole for mounting of a pipe, making rotors is really easy. I am curious what chord it has. The chord determines what maximum rotor diameter can be used for a certain design tip speed ratio. You will get a rotor with a constant chord and blade angle but such a rotor can have a maximum Cp of about 0.4 and this is certainly possible for aluminium blades as they have a very accurate geometry and a low surface roughness. Is there any information about the supplier?

[MattM]

fillm had the dimensions posted at thebackshed.com when he was proposing the idea.  I give him credit, he paid for a Chinese manufacturer to produce the equipment necessary to build the blades, and IIRC he found them to be considerably cheaper to actually do the labor rather than do it in Australia.  When the manufacturer went into financial difficulty they sold him the equipment to do the manufacturing.  I am not sure if he now possesses the extraction equipment or has a third party do the manufacturing.  But the guy really showed a full commitment to this project.  And he was selling them for $35AU per linear meter last I heard.

https://up.picr.de/17877144gv.jpg
https://up.picr.de/17897684jf.jpg
https://up.picr.de/17877145qo.jpg
HAWT https://www.youtube.com/embed/OD4Kky91VjM?rel=0
VAWT https://www.youtube.com/embed/5rb7G7H38aE?rel=0

[Adriaan Kragten]

Nice information. Today I have written report KD 698 about the Gö 227 airfoil (see separate post). This airfoil looks about the same as the the airfoil used for the aluminium blades, only the tail is different. Such airfoils with partly convex and partly concave lower sides have high optimum lift coefficients and low Cd/Cl ratios, even for low Reynolds values.

The last video shows a VAWT using this airfoil. To my opinion this is a very good airfoil to use it for a HAWT but a very bad airfoil to use it for a VAWT. This is because for a VAWT you should use a symmetrical airfoil (see report KD 601 figure 2). The angle of attack is positive when the blade is at the front side and negative when the blade is at the backside. When you use this airfoil you can take the hollow side at the outside or at the inside. Assume that you take it at the outside and you make the blade angle zero degrees.

In table 1 of KD 601 you can see that at the front side (position 1) the angle of attack is 9° and that at the backside (position 7) the angle of attack = -9° for a rotor with a design tip speed ratio of 4.2. If you look at the Cl-alpha curve of the Gö 227 airfoil given in report KD 698, you can see that this airfoil is almost stalling for alpha = 9° and that Cl = 1.68. You can also see that Cl = 0° for alpha is -9°. So only if a blade is at the front side of the rotor, it is contributing positively to the torque. In figure 4 of KD 698 it can be seen that Cd = 0.095 for alpha = 9° and that Cd = 0.1 for alpha = -9°, So for both positions, the airfoil will have a lot of drag which is much too high for a design tip speed ratio of 4.2. So using such an airfoil for a VAWT is a very bad choice.