Maximum Cp of traditional multi bladed rotors

[topspeed]

double

[Adriaan Kragten]

How many percent does the ANEW company lie about their produced electricity coeffiecient when they are at 70% ?

(Attachment Link)

Unrealistic high measured Cp values are mostly the result of incorrect measuring of the wind speed. As I have emphasized in my previous post, one has to measure the undisturbed wind speed V at a height of the rotor shaft. Theoretically this wind speed must be measured at an infinitive distance before the rotor but in practice one can take three times the rotor diameter. But this gives a problem, especially for big rotors. Assume that the rotor diameter is 10 m. So this means that the wind speed meter must be placed at a distance 30 m in front of the rotor. But this also means that it takes a rather long time for the air molucules for which the speed was measured to reach the rotor plane. In this time, the wind speed can change a lot. So one can only measure accurately if the wind speed is constant for a long time. And still, if one selects only measuring points for which the wind speed was constant for a long time and for which the wind direction was in line with the wind meter and the rotor shaft, one will find a cloud of measuring points if one measures a wind turbine in real wind. This might be a stimulation to measure the wind speed at a smaller distance in front of the rotor but this gives a strong increase of the measured Cp.

I will demonstrate this with an example. Assume that the undisturbed wind speed is 6 m/s. Betz has found that for maximum power, the wind speed in the rotor plan must be reduced to 2/3 V, and so it must be 4 m/s. This results in an expanding wake. Assume that the wind meter was placed at a distance of one rotor diameter in front of the rotor. At this place, the wake has already expanded substantially and so the measured wind speed will be about 5 m/s. The Cp is now determined for a wind speed of 5 m/s but the real undisturbed wind speed was 6 m/s. The power increases proportional to the cube of the wind speed and so the power at 6 m/s is a factor (6/5)^3 = 1.73 higher than at 5 m/s. This means that the measured Cp is a factor 1.73 too high! This demonstrates that a small mistake in the measured wind speed results in a large mistake in the measured Cp.

Another reason for a too high Cp can be that the wind speed meter was't properly calibrated. All cup anemometers have a certain small bearing friction and the V-n curve is therefore not going through the origin of the V-n graph. Sometimes a wind speed of more than 1 m/s is needed to make the cups starting to rotate. So the V-n curve has to be corrected for bearing friction. If this isn't done properly, one measures a lower wind speed than the real wind speed.

[topspeed]

How many percent does the ANEW company lie about their produced electricity coeffiecient when they are at 70% ?

(Attachment Link)

Unrealistic high measured Cp values are mostly the result of incorrect measuring of the wind speed. As I have emphasized in my previous post, one has to measure the undisturbed wind speed V at a height of the rotor shaft. Theoretically this wind speed must be measured at an infinitive distance before the rotor but in practice one can take three times the rotor diameter. But this gives a problem, especially for big rotors. Assume that the rotor diameter is 10 m. So this means that the wind speed meter must be placed at a distance 30 m in front of the rotor. But this also means that it takes a rather long time for the air molucules for which the speed was measured to reach the rotor plane. In this time, the wind speed can change a lot. So one can only measure accurately if the wind speed is constant for a long time. And still, if one selects only measuring points for which the wind speed was constant for a long time and for which the wind direction was in line with the wind meter and the rotor shaft, one will find a cloud of measuring points if one measures a wind turbine in real wind. This might be a stimulation to measure the wind speed at a smaller distance in front of the rotor but this gives a strong increase of the measured Cp.

I will demonstrate this with an example. Assume that the undisturbed wind speed is 6 m/s. Betz has found that for maximum power, the wind speed in the rotor plan must be reduced to 2/3 V, and so it must be 4 m/s. This results in an expanding wake. Assume that the wind meter was placed at a distance of one rotor diameter in front of the rotor. At this place, the wake has already expanded substantially and so the measured wind speed will be about 5 m/s. The Cp is now determined for a wind speed of 5 m/s but the real undisturbed wind speed was 6 m/s. The power increases proportional to the cube of the wind speed and so the power at 6 m/s is a factor (6/5)^3 = 1.73 higher than at 5 m/s. This means that the measured Cp is a factor 1.73 too high! This demonstrates that a small mistake in the measured wind speed results in a large mistake in the measured Cp.

Another reason for a too high Cp can be that the wind speed meter was't properly calibrated. All cup anemometers have a certain small bearing friction and the V-n curve is therefore not going through the origin of the V-n graph. Sometimes a wind speed of more than 1 m/s is needed to make the cups starting to rotate. So the V-n curve has to be corrected for bearing friction. If this isn't done properly, one measures a lower wind speed than the real wind speed.

Yes...seems that their cup system is 8 m below the center of the wing...and possibly in turbulent air because of the legs.

[mbouwer]

LIDAR provides many options for wind measurements.
It is possible at a great distance in front of the windmill, but also on top of the nacelle.

[Adriaan Kragten]

How many percent does the ANEW company lie about their produced electricity coeffiecient when they are at 70% ?

(Attachment Link)

Unrealistic high measured Cp values are mostly the result of incorrect measuring of the wind speed. As I have emphasized in my previous post, one has to measure the undisturbed wind speed V at a height of the rotor shaft. Theoretically this wind speed must be measured at an infinitive distance before the rotor but in practice one can take three times the rotor diameter. But this gives a problem, especially for big rotors. Assume that the rotor diameter is 10 m. So this means that the wind speed meter must be placed at a distance 30 m in front of the rotor. But this also means that it takes a rather long time for the air molucules for which the speed was measured to reach the rotor plane. In this time, the wind speed can change a lot. So one can only measure accurately if the wind speed is constant for a long time. And still, if one selects only measuring points for which the wind speed was constant for a long time and for which the wind direction was in line with the wind meter and the rotor shaft, one will find a cloud of measuring points if one measures a wind turbine in real wind. This might be a stimulation to measure the wind speed at a smaller distance in front of the rotor but this gives a strong increase of the measured Cp.

I will demonstrate this with an example. Assume that the undisturbed wind speed is 6 m/s. Betz has found that for maximum power, the wind speed in the rotor plan must be reduced to 2/3 V, and so it must be 4 m/s. This results in an expanding wake. Assume that the wind meter was placed at a distance of one rotor diameter in front of the rotor. At this place, the wake has already expanded substantially and so the measured wind speed will be about 5 m/s. The Cp is now determined for a wind speed of 5 m/s but the real undisturbed wind speed was 6 m/s. The power increases proportional to the cube of the wind speed and so the power at 6 m/s is a factor (6/5)^3 = 1.73 higher than at 5 m/s. This means that the measured Cp is a factor 1.73 too high! This demonstrates that a small mistake in the measured wind speed results in a large mistake in the measured Cp.

Another reason for a too high Cp can be that the wind speed meter was't properly calibrated. All cup anemometers have a certain small bearing friction and the V-n curve is therefore not going through the origin of the V-n graph. Sometimes a wind speed of more than 1 m/s is needed to make the cups starting to rotate. So the V-n curve has to be corrected for bearing friction. If this isn't done properly, one measures a lower wind speed than the real wind speed.

Yes...seems that their cup system is 8 m below the center of the wing...and possibly in turbulent air because of the legs.

To prevent the time difference in between the measured signal of the wind speed meter and the rotor plane, one sometimes measures the wind speed in the rotor plane above or below the rotor. However, this has as disadvantage that one doesn't measure the wind speed at the height of the rotor shaft. The wind speed increases at increasing height because of the wind shear. How much depends on the turbulence caused by the roughness of the earth surface and by the sun. So if you measure the wind speed below the rotor, you will get a substantially lower wind speed than the undisturbed wind speed far before the rotor and this results in a maximum Cp which is much too high.

One must also measure far enough from the rotor to stay outside the tip wake. This is at least half the rotor radius from the tip. So the best place seems to position one wind meter at the right and one wind meter at the left side of the rotor and to average the signals of both wind meters. But if the positions of both wind meters are fixed, one should take only measuring points for which the wind direction is about perpendicular to the line through both wind meters. If you want to measure the whole Cp-lambda curve, it must also be possible to strongly vary the load of the wind turbine.

So measuring of the rotor characteristic in real wind always gives problems. It is much more accurate to build a scale model and test it in the wind tunnel at the correct Reynolds numbers. But this must be in front of an open wind tunnel to prevent tunnel blockage. Measuring in a closed wind tunnel only results in a realistic maximum Cp if the model is very small with respect to the dimensions of the wind tunnel.

[topspeed]

I wonder if ANEW expressed their "efficiency" as  the power harnessed from the wind ( typical 40%) as a relation to the Betz limit as follows...: 40/59.3 = 67.45 ??

[Adriaan Kragten]

I wonder if ANEW expressed their "efficiency" as  the power harnessed from the wind ( typical 40%) as a relation to the Betz limit as follows...: 40/59.3 = 67.45 ??

The aerodynamic efficiency is a factor 1.5 higher than the Cp. So a Cp of 0.4 means an aerodynamic efficiency of 0.6. But mostly people don't know the difference in between the Cp and the aerodynamic efficiency and when they talk about the efficiency, they mean the Cp. So I think that this company really believes that they have measured a Cp of 0.7. But if they have measured the wind speed some metres below the rotor, they have measured a wind speed which is much lower than the undisturbed wind speed and this explaines the unrealistic high Cp.

But even if they really mean an aerodynamic efficiency of 0.7, this means a Cp of 0.467 and this is also much too high for a Darrieus rotor

[joestue]

mostly people don't know the difference in between the Cp and the aerodynamic efficiency and when they talk about the efficiency, they mean the Cp. So I think that this company really believes that they have measured a Cp of 0.7. But if they have measured the wind speed some metres below the rotor, they have measured a wind speed which is much lower than the undisturbed wind speed and this explaines the unrealistic high Cp.

But even if they really mean an aerodynamic efficiency of 0.7[...]

i think you give them way too much credit, i doubt they know what aerodynamic efficiency means.

[topspeed]

I wonder if ANEW expressed their "efficiency" as  the power harnessed from the wind ( typical 40%) as a relation to the Betz limit as follows...: 40/59.3 = 67.45 ??

The aerodynamic efficiency is a factor 1.5 higher than the Cp. So a Cp of 0.4 means an aerodynamic efficiency of 0.6. But mostly people don't know the difference in between the Cp and the aerodynamic efficiency and when they talk about the efficiency, they mean the Cp. So I think that this company really believes that they have measured a Cp of 0.7. But if they have measured the wind speed some metres below the rotor, they have measured a wind speed which is much lower than the undisturbed wind speed and this explaines the unrealistic high Cp.

But even if they really mean an aerodynamic efficiency of 0.7, this means a Cp of 0.467 and this is also much too high for a Darrieus rotor

Is Darrieus somehow inferior to HAWT ?   

If so then why ?

[Adriaan Kragten]

I wonder if ANEW expressed their "efficiency" as  the power harnessed from the wind ( typical 40%) as a relation to the Betz limit as follows...: 40/59.3 = 67.45 ??

The aerodynamic efficiency is a factor 1.5 higher than the Cp. So a Cp of 0.4 means an aerodynamic efficiency of 0.6. But mostly people don't know the difference in between the Cp and the aerodynamic efficiency and when they talk about the efficiency, they mean the Cp. So I think that this company really believes that they have measured a Cp of 0.7. But if they have measured the wind speed some metres below the rotor, they have measured a wind speed which is much lower than the undisturbed wind speed and this explaines the unrealistic high Cp.

But even if they really mean an aerodynamic efficiency of 0.7, this means a Cp of 0.467 and this is also much too high for a Darrieus rotor

Is Darrieus somehow inferior to HAWT ?   

If so then why ?

Yes, a Darrieus rotor is inferior to a well designed HAWT and the discussion why, is conducted already many times on this forum. I give all the disadvantages of Darrieus rotors in my public report KD 215. Altough this question about Darrieus rotors is completely out of topic, I will give the main points again and a hope that this ends the ever lasting discussion with people who favour Darrieus rotors.

1) The angle of attack varies in between a large positive angle if the blade is at the front side and a large negative angle if the blade is at the back side. Therefore one has to use a symmetrical airfoil which has a lower Cl value for the minimal Cd/Cl ratio than an asymmetrical airfoil with the same thickness. The average Cd/Cl ratio during a whole revolution is a lot higher than for a well designed HAWT which can be designed such that the airfoil is used at the lift coefficient for which the Cd/Cl ratio is minimal. The maximum Cp decreases at increasing Cd/Cl ratio and the maximum Cp of a Darrieus rotor is therefore a lot lower. A maximum Cp of 0.35 is already very high for a big Darrieus rotor.
2) A symmetrical airfoil is very sensible for the Reynolds number (see KD 601 figure 1). So a low Cd/Cl ratio is only possible for a rather large chord and for a rather large wind speed. That is why small Darrieus rotors perform very badly.
3) The blades of a H-Darrieus rotor are connected to the hub by spokes and these spokes have aerodynamic resistance. The power needed for this resistance reduces the Cp.
4) The lift coefficient varies such that the thrust points inwards the rotor if the blade is at the front side and outwards the rotor if the blade is at the backside and this shakes the blade terribly.
5) If the rotor is running at a low tip speed ratio, the angle of attack is that large that airfoil stalls if it is at the front and at the back side. Then the airfoil has a very high resistance resulting in a negative Cp. So the rotor must be started by an electric motor to pass the lambda area for which the Cp is negative.
6) The power at high wind speeds can't be reduced by turning the rotor out of the wind. So a Darrieus rotor needs a big mechanical brake to reduce the maximum rotational speed and if this brake isn't working automatically, the rotor can be destroyed easily in a big storm
7) The optimum tip speed ratio of a Darrieus rotor is about 4.2 (see derivation given in KD 601). Big HAWT's have an optimum, tip speed ratio of about 8. The required chord decreases about with the square of the local tip speed ratio and the needed solidity for a big HAWT is therefore much lower. This results in a much lighter and so a much cheaper rotor for the same power. So Darrieus rotors will always loose the battle with big HAWT's if the economics are taken into account. That is the main reason why you only see big HAWT's and no big Darrieus rotors.

[topspeed]

I wonder if ANEW expressed their "efficiency" as  the power harnessed from the wind ( typical 40%) as a relation to the Betz limit as follows...: 40/59.3 = 67.45 ??

The aerodynamic efficiency is a factor 1.5 higher than the Cp. So a Cp of 0.4 means an aerodynamic efficiency of 0.6. But mostly people don't know the difference in between the Cp and the aerodynamic efficiency and when they talk about the efficiency, they mean the Cp. So I think that this company really believes that they have measured a Cp of 0.7. But if they have measured the wind speed some metres below the rotor, they have measured a wind speed which is much lower than the undisturbed wind speed and this explaines the unrealistic high Cp.

But even if they really mean an aerodynamic efficiency of 0.7, this means a Cp of 0.467 and this is also much too high for a Darrieus rotor

Is Darrieus somehow inferior to HAWT ?   

If so then why ?

Yes, a Darrieus rotor is inferior to a well designed HAWT and the discussion why, is conducted already many times on this forum. I give all the disadvantages of Darrieus rotors in my public report KD 215. Altough this question about Darrieus rotors is completely out of topic, I will give the main points again and a hope that this ends the ever lasting discussion with people who favour Darrieus rotors.

1) The angle of attack varies in between a large positive angle if the blade is at the front side and a large negative angle if the blade is at the back side. Therefore one has to use a symmetrical airfoil which has a lower Cl value for the minimal Cd/Cl ratio than an asymmetrical airfoil with the same thickness. The average Cd/Cl ratio during a whole revolution is a lot higher than for a well designed HAWT which can be designed such that the airfoil is used at the lift coefficient for which the Cd/Cl ratio is minimal. The maximum Cp decreases at increasing Cd/Cl ratio and the maximum Cp of a Darrieus rotor is therefore a lot lower. A maximum Cp of 0.35 is already very high for a big Darrieus rotor.
2) A symmetrical airfoil is very sensible for the Reynolds number (see KD 601 figure 1). So a low Cd/Cl ratio is only possible for a rather large chord and for a rather large wind speed. That is why small Darrieus rotors perform very badly.
3) The blades of a H-Darrieus rotor are connected to the hub by spokes and these spokes have aerodynamic resistance. The power needed for this resistance reduces the Cp.
4) The lift coefficient varies such that the thrust points inwards the rotor if the blade is at the front side and outwards the rotor if the blade is at the backside and this shakes the blade terribly.
5) If the rotor is running at a low tip speed ratio, the angle of attack is that large that airfoil stalls if it is at the front and at the back side. Then the airfoil has a very high resistance resulting in a negative Cp. So the rotor must be started by an electric motor to pass the lambda area for which the Cp is negative.
6) The power at high wind speeds can't be reduced by turning the rotor out of the wind. So a Darrieus rotor needs a big mechanical brake to reduce the maximum rotational speed and if this brake isn't working automatically, the rotor can be destroyed easily in a big storm
7) The optimum tip speed ratio of a Darrieus rotor is about 4.2 (see derivation given in KD 601). Big HAWT's have an optimum, tip speed ratio of about 8. The required chord decreases about with the square of the local tip speed ratio and the needed solidity for a big HAWT is therefore much lower. This results in a much lighter and so a much cheaper rotor for the same power. So Darrieus rotors will always loose the battle with big HAWT's if the economics are taken into account. That is the main reason why you only see big HAWT's and no big Darrieus rotors.

Well I have to chew on that for a while.

At least my h-Darrieus and the ex-Nokia folks Darrieus with single blade don't have symmetrical air foil.

Furthermore my HD has pitch control that enables it to create gyroscopicforces that stabilize it...more or less.

Yes TSR 4 is about right.....but it affects the whole wing...in HAWT it affects the tip.

Things to consider...at this stage...as I am continuing my tests soon.

 

ANEW had at least one 1,5 MW HD built. Not so small.

[Adriaan Kragten]

In figure 2 of my report KD 601, I give the speed diagram for a H-Darrieus rotor for twelve positions of the blade if it runs at a tip speed ratio of 4.2 and if the wind speed at the rotor is 2/3 V. In reality, the wind speed at the front blade will be a little higher than 2/3 V and at the back blade it will be a little lower than 2/3 V because of the expanding wake but taking this into account would make the whole description too complicated. The angles of attack alpha have been measured from figure 2 and are given in table 1. In this table you can see that the angle at the front position 1 is 9° and that the angle at the back position 7 is -9°. A symmetrical airfoil like the NACA 0015 can have these angles without stalling if the Reynolds value is large enough (see figure 1). But if an asymmetrical airfoil is used with the flat side to the outside of the rotor, it can have a larger angle of attack when the blade is at the front side without stalling but a much lower negative angle of attack when the blade is at the back side. So an asymetrical airfoil will stall when the blade is at the back side and this will cause a lot of drag.

In table 1 you can see how alpha changes for all twelve positions of the blade. It is 0° for positions 4 and 10. This means that the lift coefficient is zero for these positions. So around positions 1 and 4, the lift coefficient is very small and this means that at these positions of the blade, the lift will have no component in the direction of rotation and there is only a drag component opposite to the direction of rotation. So around positions 4 and 10 the blade will work as an aerodynamic brake and this is a reason for the rather low maximum Cp.

But the main reason is that the average Cd/Cl ratio for a symmetrical airfoil is much higher than that of an asymmetrical airfoil for the same ratio in between the thickness and the cord and for the same Reynolds value. In the last colum of table 1, I give the Cd/Cl ratio for all twelve positions. Taking into account that the Cd/Cl ratio is infinitive for positions 4 and 10, the average value will be about 0.04. For an airfoil like the Gö 711, a minimum Cd/Cl ratio of 0.015 is realized for an angle of attack of 4° and for a lift coefficient Cl = 1 (see figure 3 and 4 of report KD 285). The Gö 711 has a thickness of 14.85 % of the chord and was measured for Re = 4 * 10^5 and so these values are about the same as used for the NACA 0015.

Figure 4.7 out of report KD 35 shows that the maximum Cp which can be realize for a 3-bladed rotor depends very much on the Cd/Cl ratio if the design tip speed ratio is higher than 4. For a Cd/Cl ratio of 0.04 and a lambda of 4.2 it can be read that Cpth is about 0.43. However, a H-Darrieus rotor has tip losses at the upper and the lower side of the blade and it has spokes with which the blades are connected to the hub which have a lot of resistance and this gives a reduction of the maximum Cp to about 0.35. And this is the value for a rotating rotor. The rotor has to be started to bridge the part of the Cq-lambda curve for which the Cq is negative. This starting consumes energy and this energy reduces the generated yearly energy. Tests on the Dutch test field in Schoondijke on the Dutch 3-bladed Turby H-Darrieus rotor with screwed blades have shown that this starting energy has a considerable negative influence on the yearly output. 

[topspeed]

I had no Idea you could put the flat side of the foil outside of the rotor.

[Adriaan Kragten]

I had no Idea you could put the flat side of the foil outside of the rotor.

If you put the flat side of the airfoil at the inside the rotor, the blade will stall if it is at the front side of the rotor. This is even worse than stalling at the back side because at the front side, the airflow is undisturbed and so a blade will extract more energy out of the wind when it is at the front side than when it is at the back side. This is because at the back side, the blade feels the turbulence caused by the front blade and by the rotor shaft.

Realize that an airfoil can stall if the angle of attack becomes too large but it can also stall if the angle of attack becomes too small. If you would use the Gö 711 airfoil with the flat side at the inside of the rotor and with a blade angle of 0°, you get a negative angle of attack of -9° when the blade is at the front side of the rotor and when the rotor runs at a tip speed ratio of 4.2. The Cd/Cl ratio for this large negative angle of attack is very large.

Another aspect of stalling is that there is hysteresis in the Cl-alpha curve, especially at low Reynolds values. This means that starting and ending of stalling occur not at the same angle of attack. So if you have a symmetrical airfoil like the NACA 0015, stalling may start at an angle of for instance 14° if you go from small to large angles but stalling may end at an angle of 10° if you go from large to small angles. This means that if stalling has started somewhere during the revolution of a blade, it can be maintained for a rather large angle of rotation.

[MattM]

Someone will get creative and turn those vertical blades into sharrow hoops.  They'll explain away every shortcoming of every other design.  And the real world performance will be meh.

[kitestrings]

I enjoyed reading through this explanation Adriaan.  Thank you.

I've always liked how they look, and the aesthetics may be in part, part of the appeal.  They are often shown in sleek, urban backgrounds with complimentary architecture around them.

In very simplistic terms, I think of them as having several real-world obstacles:  Not being self-starting is probably the biggest in my mind.  I think also that above a certain size (small to mid-scale), they just become too difficult to construct and get up in the air where the wind is -  They are rarely seen on terrain with forestation of any sort.  And, you end up with very high, concentrated stresses at the base.  There are examples of smaller units with guying to the top, but that would seem impractical at any large scale.  Then there is the overspeed control issue.

Adriaan and others have explained things in much more detail, but these are the initial reactions that I always have when the next "break-thru" model surfaces.

[topspeed]

I had no Idea you could put the flat side of the foil outside of the rotor.

If you put the flat side of the airfoil at the inside the rotor, the blade will stall if it is at the front side of the rotor. This is even worse than stalling at the back side because at the front side, the airflow is undisturbed and so a blade will extract more energy out of the wind when it is at the front side than when it is at the back side. This is because at the back side, the blade feels the turbulence caused by the front blade and by the rotor shaft.

Realize that an airfoil can stall if the angle of attack becomes too large but it can also stall if the angle of attack becomes too small. If you would use the Gö 711 airfoil with the flat side at the inside of the rotor and with a blade angle of 0°, you get a negative angle of attack of -9° when the blade is at the front side of the rotor and when the rotor runs at a tip speed ratio of 4.2. The Cd/Cl ratio for this large negative angle of attack is very large.

Another aspect of stalling is that there is hysteresis in the Cl-alpha curve, especially at low Reynolds values. This means that starting and ending of stalling occur not at the same angle of attack. So if you have a symmetrical airfoil like the NACA 0015, stalling may start at an angle of for instance 14° if you go from small to large angles but stalling may end at an angle of 10° if you go from large to small angles. This means that if stalling has started somewhere during the revolution of a blade, it can be maintained for a rather large angle of rotation.

How come mine does not stall ?

Campered foils loose a bit of the Cp...compared to symmertical...in stiff models where the pitch control is missing.

[Adriaan Kragten]

I had no Idea you could put the flat side of the foil outside of the rotor.

If you put the flat side of the airfoil at the inside the rotor, the blade will stall if it is at the front side of the rotor. This is even worse than stalling at the back side because at the front side, the airflow is undisturbed and so a blade will extract more energy out of the wind when it is at the front side than when it is at the back side. This is because at the back side, the blade feels the turbulence caused by the front blade and by the rotor shaft.

Realize that an airfoil can stall if the angle of attack becomes too large but it can also stall if the angle of attack becomes too small. If you would use the Gö 711 airfoil with the flat side at the inside of the rotor and with a blade angle of 0°, you get a negative angle of attack of -9° when the blade is at the front side of the rotor and when the rotor runs at a tip speed ratio of 4.2. The Cd/Cl ratio for this large negative angle of attack is very large.

Another aspect of stalling is that there is hysteresis in the Cl-alpha curve, especially at low Reynolds values. This means that starting and ending of stalling occur not at the same angle of attack. So if you have a symmetrical airfoil like the NACA 0015, stalling may start at an angle of for instance 14° if you go from small to large angles but stalling may end at an angle of 10° if you go from large to small angles. This means that if stalling has started somewhere during the revolution of a blade, it can be maintained for a rather large angle of rotation.

How come mine does not stall ?

Campered foils loose a bit of the Cp...compared to symmertical...in stiff models where the pitch control is missing.

If you deviate from a symmetrical airfoil with a blade angle of 0°, you should do the same as I did in figure 2 of my report KD 601. So you should determine the angle of attack and the lift coefficient for twelve positions of the blade. For a symmetrical airfoil with a blade angle of 0°, the angle of attack varies is between +9° at the front side and -9° at the back side if the tip speed ratio is 4.2. So the difference is 18°. If you use an asymmetrical airfoil, you will also get a difference of 18° for a tip speed ratio of 4.2. Assume that you chose the airfoil and the blade angle such that the angle of attack is for instance 5° at the front side of the rotor. This means that the blade angle must be 9° - 5° = 4°. The angle of attack will be 5° - 18° = -13° at the back side and an asymetrical airfoil will stall at this large negative angle of attack. Realize that a positive angle of attack at the front side means that the flat side of the airfoil must be at the outside of the rotor. Only then you get a direction of the lift such that the tangential component of the lift points in the direction of rotation.

Only a rather thick symmetrical airfoil accepts the very large difference in angle of attack of 18° without stalling and having an acceptable low Cd/Cl value for the whole alpha range. If you use a symmetrical airfoil with a blade angle of for instance 5° at the front side, the angle of attack will be 9° - 5° = 4°. The angle of attack at the back side of the rotor will be 4° - 18° = -14°. So even a symmetrical airfoil will stall at the back side if the blade angle isn't choosen close to 0°.

In the aerodynamics of rotor design there are three different angles. These angles are specified in figure 3.2 of my report KD 35. Phi is the angle in between the relative wind W and the rotor plane. Beta is the blade angle and alpha is the angle of attack. The angle of attack is defined with respect to the zero-line of the airfoil. The zero line of symmetrical airfoils lays at the heart of the airfoil. The zero line of Gö airfoils with a flat lower side lays at the lower side. The zero line of NACA airfoils with a flat lower side lays at the line through the airfoil nose and the tailing edge. It is valid that phi = beta + alfa. This figure 3.2 is valid for VAWT's and for Darrieus rotors but for a H-Darrieus rotor, the rotor plane is the cylindrical plane in which a blade moves at the position which is taken into account. What I have determined in figure 2 of KD 601 is the angle phi. However, if beta = 0°, the angle phi becomes equal to the angle alpha. But if beta isn't 0°, one finds alfa by using the given formula.

[MattM]

How you measure angles on a H-Darrieus rotor is not around a circle, which is already a bit misleading  All the space in a VAWT is curved.  Employing a completely flat blade is choosing a chord that is optimal at perpindicular angles with the wind.  The lever arm would be from the center of rotation to center of connection to the chord, rather than a radius relative to the outer most point, instead making the circumference of rotation relative to that connection.  Tilting a blade on H-Darrieus rotor means increasing the size of the chord, moving the rotor arm off the center of chord, and using a sub-section of the true chord length.  A simple chord is a cup shape relative to curved space, and relative camber is between total circumference of your volume and chord.  Tilting your blade increases relative camber to your airflow and depends on the relative sub-section's width within the actual chord.

Lenz actually makes use of the radius in complete contrast to an H-Darrieus rotor, making better use of a given volume.  And your Lenz blade uses distance off the rotational circumference at the end of each arm to figure camber.  So your Lenz curved blade may look thick compared to an H-Darrieus but in reality enjoy a much thinner relative camber at all points as it rotates.  Your total volume will be slightly larger than your circumference of rotation.  While a Lenz might rotate fewer RPMs in a given space, the length of lever arm and relative tangential velocity should be better for similar relative mass put into its construction.

[Adriaan Kragten]

How you measure angles on a H-Darrieus rotor is not around a circle, which is already a bit misleading 

Lets take the rotor as given in KD 601 as an example. This rotor has a diameter D = 2 m and so R = 1 m = 1000 mm. The rotor has three blades with a chord c = 200 mm and uses a NACA 0015 airfoil. It is assumed that the blade angle beta = 0°. Lets draw a circle segment with R = 1000 mm through the airfoil nose and the airfoil tailing edge. If we compare this circle segment with the symmetry line of the airfoil, there is only a small difference. The camber C of the circle segment is only about 2.5 %. The fact that the blade is rotating and not moving in a flat plane has therefore a small influence on the generated lift. This influence can be neutralised if the symmetry line of the airfoil is cambered with a radius of 1000 mm but this gives a strange deformation of the airfoil.

I think that this effect can be neglected if the chord c is relatively small with respect to R. The NACA 0015 airfoil has the largest thickness at 0.3 * c. What I would do is to draw a line through this 0.3 * c point and the centre of the rotor and take the symmetry line of the airfoil perpendicular to this radial line. This is also done in figure 2 of KD 601. The angle phi is now taken with respect to the symmetry line of the non modified airfoil. Most asymmetrical airfoils also have the thickest point at 0.3 * c and then you can follow the same procedure.

[MattM]

But your chord is only flat with respect to incoming and receding wind at right angles to wind direction.

It is not flat relative to the circular rotation path.  From the perspective of the curve, a thin chord must form along the curve.  A flat blade is curved relative to that circular path, so your flat chord, aerofoil shape or not, is never a relatively thin camber.  At best you get a pulsing as your blade ebbs and flows, relative to two positions.  And this pulsing is only relative to those two positions so you have very little to control rotational velocity.  Honestly, a trainwreck waiting to happen.

[Adriaan Kragten]

This out off topic discussion about Darrieus rotors has a tendency to go on forever. Every answer I give results in new questions. I have never made a secret about my opinion about Darrieus rotors. These windturbines have that many disadvantages that it is a waste of time to try to develope them. Any attempt to solve one of the disadvantages, like the negative Cq-lambda curve for low values of lambda, creates new problems. Many companies which tried to develop a Darrieus rotor went bankrupt. The yearly output of Darrieus rotors is always disappointing.

The fact that the blade moves along a cylinder in stead of a flat plane is only a minor problem and not worth solving by cambering of the symmetry line of a symmetrical airfoil. For each of the twelve positions as given in figure 2 of report KD 601, the relative wind speed W is mainly determined by the speed of the rotor and only a little by the local wind speed 2/3 V. So cambering of the symmetry line will be not more or less useful for positions 1 and 7 than for positions 4 and 10. This is my last comment to this discussion. If someone wants to discuss Darrieus rotors, please do that in a new post.