H-Darreius wind turbines ( VAWT ) !

[topspeed]

New blades for 6th version are on the way.

They are:

1. 45% thinner foil than in 4th version.
2. 15% more capture area by widening the moment arm
3. 10% more capture area by lenghtening the blades
4. fully laminar flow wing
5. blade's aspect ratio is 20% greater than in the 5th blade
6. the connection to moment arm is aerodynamically 90% more efficient than previous contraption

Also I will make solid foundation to the tower...tower is now really solid.

Wish me luck !

 

[topspeed]

I made experiments with white light styrofoam.

I you spread laminating epoxy on it and let it cure...it becomes sturdy as a rock...weighs a little more than styrofoam...but way more harder that I tought it would be.

Anyone tried it ?

I only need to glue plywood sheet on it and it is extremely strong.

Will it become weaker when in use for week, months, years...decades ?

[mbouwer]

If you have covered the styrofoam bladeform with epoxy and fibreglass,
Why would you need to put another covering around it?

[topspeed]

If you have covered the styrofoam bladeform with epoxy and fibreglass,
Why would you need to put another covering around it?

(Attachment Link)

I have a mixed structure on it...lighter styrofoam foils are in the tip.

I am aiming on a laminar flow on it.

[MattM]

Sounds a lot like model plane building.  As long as the spar and leading edges hold up it should be interesting.  As tough as fiberglass is to our skin it is weak against leading edge abuse.  May have to touch it up once in awhile.  The only real concern I would have it sealing everything up from vapor penetration.  It needs to repel water while allowing it to breathe vapors out.

[topspeed]

Sounds a lot like model plane building.  As long as the spar and leading edges hold up it should be interesting.  As tough as fiberglass is to our skin it is weak against leading edge abuse.  May have to touch it up once in awhile.  The only real concern I would have it sealing everything up from vapor penetration.  It needs to repel water while allowing it to breathe vapors out.

Yes it is a bit like model plane wing building. I also dream about building a real aeroplane. No one ever ( well maybe ) ever built a mostly wood wing in extreme laminar flow as an aim.

[topspeed]

New blade has both sides covered of the main part....tips are still without the covering. 

[topspeed]

Wouldn't TSR of 10 be better for higher efficiency ?

[Adriaan Kragten]

Wouldn't TSR of 10 be better for higher efficiency ?

(Attachment Link)

The optimum tip speed ratio of a HAWT can be chosen in between about 1 and 10. There is an aerodynamic theory (see report KD 35) which gives the chord and the blade angle as a function of the rotor diameter, the radius, the number of blades and the chosen lift coefficient. The higher the tip speed ratio, the smaller the chord and the blade angle and the lower the Cd/Cl ratio must be to get an acceptable Cp. A very high tip speed ratio like 10, requires a very low Cd/Cl ratio of about 0.02 for a 3-bladed rotor to get a theoretical maximum Cp of 0.45 (see KD 35 figure 4.7).

For a normal H-Darrieus rotor, the blade angle is fixed at 0° and you have to use a symmetrical airfoil because the angle of attack varies in between a large positive angle if the blade is just at the front side and a large negative angle if the blade is just at the back side. The fact that the blade angle is fixed at 0°, results in only one optimum tip speed ratio. In my public report KD 601, I have calculated that this optimum tip speed ratio is about 4.2. If the tip speed ratio is smaller, the airfoil will stall, especially at low Reynolds numbers. If the tip speed ratio is bigger, the angles of attack will become too small to generate sufficient lift. The variation of the angle of attack is given in figure 2 of KD 601 for twelve positions of the blade if the tip speed ratio is 4.2. So it isn't true that you can chose the optimum tip speed ratio for a H-Darrieus rotor and certainly you can't chose for a very high optimum tip speed ratio of 10.

The fact that the angle of attack varies in between a large positive value and a large negative value, means that only for a small part of a revolution, the airfoil is working for the angle of attack for which the Cd/Cl ratio is minimal. The average Cd/Cl ratio will therefore be higher than for a HAWT which can be designed such that the Cd/Cl ratio is minimal for every part of the blade. The maximum Cp must therefore be lower than for a well designed HAWT.

So I don't believe that your presented Cp-lambda curves are correct. If these curves are the result of measurements, the wind speed must have been measured incorrectly. The Cp is defined with respect to the undisturbed wind speed V and for that wind speed you must measure at least three rotor diameters in front of the rotor. If you measure in the field close to the rotor plane or in a closed wind tunnel, the Cp will be much too high.

[topspeed]

Wouldn't TSR of 10 be better for higher efficiency ?

(Attachment Link)

The optimum tip speed ratio of a HAWT can be chosen in between about 1 and 10. There is an aerodynamic theory (see report KD 35) which gives the chord and the blade angle as a function of the rotor diameter, the radius, the number of blades and the chosen lift coefficient. The higher the tip speed ratio, the smaller the chord and the blade angle and the lower the Cd/Cl ratio must be to get an acceptable Cp. A very high tip speed ratio like 10, requires a very low Cd/Cl ratio of about 0.02 for a 3-bladed rotor to get a theoretical maximum Cp of 0.45 (see KD 35 figure 4.7).

For a normal H-Darrieus rotor, the blade angle is fixed at 0° and you have to use a symmetrical airfoil because the angle of attack varies in between a large positive angle if the blade is just at the front side and a large negative angle if the blade is just at the back side. The fact that the blade angle is fixed at 0°, results in only one optimum tip speed ratio. In my public report KD 601, I have calculated that this optimum tip speed ratio is about 4.2. If the tip speed ratio is smaller, the airfoil will stall, especially at low Reynolds numbers. If the tip speed ratio is bigger, the angles of attack will become too small to generate sufficient lift. The variation of the angle of attack is given in figure 2 of KD 601 for twelve positions of the blade if the tip speed ratio is 4.2. So it isn't true that you can chose the optimum tip speed ratio for a H-Darrieus rotor and certainly you can't chose for a very high optimum tip speed ratio of 10.

The fact that the angle of attack varies in between a large positive value and a large negative value, means that only for a small part of a revolution, the airfoil is working for the angle of attack for which the Cd/Cl ratio is minimal. The average Cd/Cl ratio will therefore be higher than for a HAWT which can be designed such that the Cd/Cl ratio is minimal for every part of the blade. The maximum Cp must therefore be lower than for a well designed HAWT.

So I don't believe that your presented Cp-lambda curves are correct. If these curves are the result of measurements, the wind speed must have been measured incorrectly. The Cp is defined with respect to the undisturbed wind speed V and for that wind speed you must measure at least three rotor diameters in front of the rotor. If you measure in the field close to the rotor plane or in a closed wind tunnel, the Cp will be much too high.

Chart is from Research Gate publication.

This was the first time I read about Cd/Cl ratio...as it is usually otherway round. I have to check my numbers on this.

I have to chew on this for a little while.

Hopefully my next attempt will shed new light on the matter with my original pitch control system.

This blade is almost symmetrical....and produces less lift than symmetrical NACA 0015...but has way less drag. So it actually when flying right might be the answer what you and I have been looking for.

Yes Adriaan....and thank you again for you thorough answer.

[topspeed]

Yes...I checked my number.

I got 0.014 for general Cd/Cl...could be even less.

My biggest problem is to make the blades ( 2 ) equal in form and weight.

[Adriaan Kragten]

It is true that instead of using a fully symmetrical airfoil you can also use an airfoil which is slightly asymmetrical. If a symmetrical airfoil is bent such that the radius of the zero line coincides with the radius of the rotor, aerodynamically it will work if it was fully symmetrical. However, the radius of the rotor is very large with respect to the chord of the airfoil and so such a slightly bent airfoil will have almost the same Cl-alfa curve and Cl/Cd curve as a fully symmetrical airfoil.

The only other thing you can change is the thickness of the airfoil. The thicker the airfoil, the higher the maximum lift coefficient but thick airfoils are more sensible to stalling at low Reynolds numbers than thin airfoils. Thick airfoils also have a higher minimal Cd/Cl ratio than thin airfoils. An advantage of thick airfoils is that the bending moment of resistance and the bending moment of inertia is larger and so such an airfoil is stronger and stiffer which is very important for a H-Darrieus rotor as the centrifgal force gives a large bending moment in the blade at the points where the blade is connected to the spokes.

To find the minimum Cd/Cl ratio of a certain arfoil, you have to take the Cl/Cd curve for a certain Reynolds value and draw a line through the origin which touches the curve. The touching point gives the optimum lift coefficient. The ratio in between Cd and Cl for the touching point gives the minimum Cd/Cl ratio. In the Cl-alfa curve you can see which optimum angle of attack alfa belongs to the optimum lift coefficient. The straight line through the origin which is touching the curve is intersecting with the horizontal line for Cl = 1. If you read the Cd value for Cl = 1, you directly get the minimum Cd/Cl ratio.

[topspeed]

It is true that instead of using a fully symmetrical airfoil you can also use an airfoil which is slightly asymmetrical. If a symmetrical airfoil is bent such that the radius of the zero line coincides with the radius of the rotor, aerodynamically it will work if it was fully symmetrical. However, the radius of the rotor is very large with respect to the chord of the airfoil and so such a slightly bent airfoil will have almost the same Cl-alfa curve and Cl/Cd curve as a fully symmetrical airfoil.

The only other thing you can change is the thickness of the airfoil. The thicker the airfoil, the higher the maximum lift coefficient but thick airfoils are more sensible to stalling at low Reynolds numbers than thin airfoils. Thick airfoils also have a higher minimal Cd/Cl ratio than thin airfoils. An advantage of thick airfoils is that the bending moment of resistance and the bending moment of inertia is larger and so such an airfoil is stronger and stiffer which is very important for a H-Darrieus rotor as the centrifgal force gives a large bending moment in the blade at the points where the blade is connected to the spokes.

To find the minimum Cd/Cl ratio of a certain arfoil, you have to take the Cl/Cd curve for a certain Reynolds value and draw a line through the origin which touches the curve. The touching point gives the optimum lift coefficient. The ratio in between Cd and Cl for the touching point gives the minimum Cd/Cl ratio. In the Cl-alfa curve you can see which optimum angle of attack alfa belongs to the optimum lift coefficient.

Ok...as H-Darrieus in general...then mine is not a general type as it is not affected by the specific vibration...as the general type..

[Adriaan Kragten]

Ok...as H-Darrieus in general...then mine is not a general type as it is not affected by the specific vibration...as the general type..

How would that be possible. Any Darrieus rotor of whatever type will have an angle of attack which varies strongly as a function of the position of the blade. So this results in a variation of the lift coefficient and so in variation of the total lift on each blade. This results in vibrations.