Braking a VAWT ?

[topspeed]

I have used short circuiting as a brake.

Is that valid in a bigger system ?

[makenzie71]

All PMAs work the same.  With a bigger system there's more danger because of the additional power you're basically putting right back into the coils.

If you're building a vertical turbine that's actually capable of making real power there's ZERO reason to not install a physical brake.  The sacrifices in performance should come with the benefit of easier operation.

[topspeed]

Roger that McKenzie !

[Adriaan Kragten]

I have used short circuiting as a brake.

Is that valid in a bigger system ?

It depends on the type of VAWT. If you use a Darrieus rotor, it has region in the Cq-lambda curve for which the Cq is negative. Only if lambda is almost zero it becomes positive because then the rotor works as a drag machine. So if the generator is short-circuited, it needs only a low torque to keep the rotor turning at a very low rpm. However, if you use a Savonius rotor, the torque at low rpm will be very high and then there will be a wind speed for which the braking torque of the generator will be lower than the rotor torque en then the rotor will speed up.

[topspeed]

I have used short circuiting as a brake.

Is that valid in a bigger system ?

It depends on the type of VAWT. If you use a Darrieus rotor, it has region in the Cq-lambda curve for which the Cq is negative. Only if lambda is almost zero it becomes positive because then the rotor works as a drag machine. So if the generator is short-circuited, it needs only a low torque to keep the rotor turning at a very low rpm. However, if you use a Savonius rotor, the torque at low rpm will be very high and then there will be a wind speed for which the braking torque of the generator will be lower than the rotor torque en then the rotor will speed up.

I have a darrieus that has a high torque at low rpm ....so does it need a mechanical brake or not ?

[Adriaan Kragten]

I have a darrieus that has a high torque at low rpm
[/quote]


How is that possible? The blades of a normal Darrieus rotor stall at the front and at the back side if it runs at a low tip speed ratio. A stalling symmetrical airfoil has a very high drag coefficient and this causes a negative torque coefficient at low tip speed ratios. Normally this effect is compensated by using a small Savonius rotor at the top and the bottum of the Darrieus rotor. I have described another methode in my report KD 601.

You can get a small positive torque coefficient at lambda = 0 if you give the blades a small positive blade angle. However, a positive blade angle when the blade is at the front side means a negative blade angle when the blade is at the back side for a fixed blade with a symmetrical airfoil. As the wind speed at the front side will be a little higher than at the back side, this can result in a small positive starting torque coefficient. However, if the blade is running at its optimum tip speed ratio, this positive blade angle at the front side results in decrease of the angle of attack and so in decrease of the lift coefficient. Decrease of the lift coefficient results in decrease of the generated torque if the blade is at the front side. A negative blade angle at the back side side results in increase of the angle of attack. This increase of the angle of attack at the back side may result in stalling and this results in increase of the drag coefficient. So both effects together will result in reduction of the maximum Cp. A normal Darrieus rotor therefore uses a symmetrical airfoil and a blade angle of 0°. So if you have a high torque at low rpm, you don't have a normal Darrieus rotor.

But whatever you may have done to get a positive starting torque coefficient at low tip speed ratios, if you really have it, you can't trust on the braking torque of a PM-generator. So in this case you need a mechanical brake but this requires an automatic braking procedure if the rotational speed becomes too high. Manual activation of the brake is too dangerous.

[makenzie71]

darrieus turbines are typically high rpm/low torque.  I'd like to see pictures of your setup and know what PMA and blade profile you're using to get a low rpm/high torque setup.

Like Kragten says, you have to have a mechanical brake...there's no other way to go about it.  You can put a motorcycle brake rotor on the turbine shaft at the bottom and mount a caliper very easy.  You can use a generic electric switch and a coil to actuate a master cylinder automatically.

[topspeed]

darrieus turbines are typically high rpm/low torque.  I'd like to see pictures of your setup and know what PMA and blade profile you're using to get a low rpm/high torque setup.

Like Kragten says, you have to have a mechanical brake...there's no other way to go about it.  You can put a motorcycle brake rotor on the turbine shaft at the bottom and mount a caliper very easy.  You can use a generic electric switch and a coil to actuate a master cylinder automatically.

Unfortunately it is classified.

It is not that easy as the generator is in the moving part of the of the assembly....only stator is stationary.

It is manageable though.

[topspeed]

How is that possible? The blades of a normal Darrieus rotor stall at the front and at the back side if it runs at a low tip speed ratio. A stalling symmetrical airfoil has a very high drag coefficient and this causes a negative torque coefficient at low tip speed ratios. Normally this effect is compensated by using a small Savonius rotor at the top and the bottum of the Darrieus rotor. I have described another methode in my report KD 601.

You can get a small positive torque coefficient at lambda = 0 if you give the blades a small positive blade angle. However, a positive blade angle when the blade is at the front side means a negative blade angle when the blade is at the back side for a fixed blade with a symmetrical airfoil. As the wind speed at the front side will be a little higher than at the back side, this can result in a small positive starting torque coefficient. However, if the blade is running at its optimum tip speed ratio, this positive blade angle at the front side results in decrease of the angle of attack and so in decrease of the lift coefficient. Decrease of the lift coefficient results in decrease of the generated torque if the blade is at the front side. A negative blade angle at the back side side results in increase of the angle of attack. This increase of the angle of attack at the back side may result in stalling and this results in increase of the drag coefficient. So both effects together will result in reduction of the maximum Cp. A normal Darrieus rotor therefore uses a symmetrical airfoil and a blade angle of 0°. So if you have a high torque at low rpm, you don't have a normal Darrieus rotor.

But whatever you may have done to get a positive starting torque coefficient at low tip speed ratios, if you really have it, you can't trust on the braking torque of a PM-generator. So in this case you need a mechanical brake but this requires an automatic braking procedure if the rotational speed becomes too high. Manual activation of the brake is too dangerous.

I have made an innovation.

It is kinda like "a shape shifter".

[Adriaan Kragten]

Determination of the torque coefficient at a certain lambda is much more difficult for a Darrieus rotor than for a HAWT. That is because the angle of attack alpha changes continiously. To find alpha for a certain position of the blade, one has to determine the speed diagram for that position. This means that one has to determine the direction of the absolute wind speed, the direction of the blade speed and the direction of the relative speed W which is felt by the blade. The direction of W with respect to the zero line of the airfoil detemines alpha. If alpha is known, one can determine the lift coefficient Cl and the drag coefficient Cd. Both coefficients have to be resolved in a tangential component in the direction of rotation and a component perpendicular to the direction of rotation. The tangential component of the drag is always pointing opposed to the direction of rotation. The differece in between the tangential component of the lift in the direction of rotation and the tangential component of the drag opposed to the direction of rotation determines the torque coefficient for that position of the blade.

This procedure is described in chapter 4 of my report KD 601 for a symmetrical NACA 0015 airfoil if the blade moves with a lambda of 4.2 and if the wind speed in the rotor plane is 2/3 V. The speed diagrams are given in figure 2 for twelve positions of the blade. The angles of attack alpha, the Cl, the Cd and the Cd/Cl ratio found this way are given in tabel 1. If this procedure is followed for a low lambda of for instance 1, you will find very large angles alpha for the front and the back stations. This means that the airfoil stalls and this results in a very high drag coefficient especially at low Reynolds values (see figure 1). The estimated Cq-lambda curve for a normal Darrieus rotor is given as the lowest curve in figure 4. The highest curve is the curve for a rotor for which the blades can oscillate. So getting a high torque coefficient at a low lambda is simply impossible for a Darrieus rotor with fixed blades. Any modification of a fixed blade to increase the starting torque coefficient at a low lambda will go at the cost of a strong reduction of the maximum Cp at a high lambda. If the starting torque coefficient is really high at a low lambda, I am afraid that it concerns a drag machine in stead of a Darrieus rotor. But a drag machine has an optimum lambda of about 0.15 but this is about 4.2 for a well designed Darrieus rotor (see figure 3).

[topspeed]

Determination of the torque coefficient at a certain lambda is much more difficult for a Darrieus rotor than for a HAWT. That is because the angle of attack alpha changes continiously. To find alpha for a certain position of the blade, one has to determine the speed diagram for that position. This means that one has to determine the direction of the absolute wind speed, the direction of the blade speed and the direction of the relative speed W which is felt by the blade. The direction of W with respect to the zero line of the airfoil detemines alpha. If alpha is known, one can determine the lift coefficient Cl and the drag coefficient Cd. Both coefficients have to be resolved in a tangential component in the direction of rotation and a component perpendicular to the direction of rotation. The tangential component of the drag is always pointing opposed to the direction of rotation. The differece in between the tangential component of the lift in the direction of rotation and the tangential component of the drag opposed to the direction of rotation determines the torque coefficient for that position of the blade.

This procedure is described in chapter 4 of my report KD 601 for a symmetrical NACA 0015 airfoil if the blade moves with a lambda of 4.2 and if the wind speed in the rotor plane is 2/3 V. The speed diagrams are given in figure 2 for twelve positions of the blade. The angles of attack alpha, the Cl, the Cd and the Cd/Cl ratio found this way are given in tabel 1. If this procedure is followed for a low lambda of for instance 1, you will find very large angles alpha for the front and the back stations. This means that the airfoil stalls and this results in a very high drag coefficient especially at low Reynolds values (see figure 1). The estimated Cq-lambda curve for a normal Darrieus rotor is given as the lowest curve in figure 4. The highest curve is the curve for a rotor for which the blades can oscillate. So getting a high torque coefficient at a low lambda is simply impossible for a Darrieus rotor with fixed blades. Any modification of a fixed blade to increase the starting torque coefficient at a low lambda will go at the cost of a strong reduction of the maximum Cp at a high lambda. If the starting torque coefficient is really high at a low lambda, I am afraid that it concerns a drag machine in stead of a Darrieus rotor. But a drag machine has an optimum lambda of about 0.15 but this is about 4.2 for a well designed Darrieus rotor (see figure 3).

Adrian...you are very knowledgeable indeed.

Since you don't know how my system works and I cannot reveal it until it has been patented we can only speculate...but one thing is for sure in the tests...even without bearing in the wing moment arm and wing conjunction it was able to rotate the very stiff reduction geared generator at 2 m/s wind ( and kept rotating at 1.5 m/s )...which must tell you something about the torque being generated.

[Adriaan Kragten]

Since you don't know how my system works and I cannot reveal it until it has been patented we can only speculate...but one thing is for sure in the tests...even without bearing in the wing moment arm and wing conjunction it was able to rotate the very stiff reduction geared generator at 2 m/s wind ( and kept rotating at 1.5 m/s )...which must tell you something about the torque being generated.

Yes, what it tells me is that you have designed a drag machine (see KD 416) or a rotating blade machine (see KD 417). I don't believe in fairy tails about Darrieus rotors. People have spent a lot of money in patenting drag machines which finally appear to generate only very little power for the required investment costs. You should not bring a design at this public forum if you don't want to explain how it works.

I have several patents on my name and I know the procedure. It is only checked if the idea is new and not if the idea really fulfills the claims. But checking if the idea is new requires reading thousands of reports and books especially if it concerns wind turbines. So often it appears later that the idea wasn't new even if the patent is gained. So then you can get problems with original inventer. Taking a patent is only a good option for big companies which have their own patent officers and which have enough money to fight on court if their patent is questioned.

[topspeed]

Yes, what it tells me is that you have designed a drag machine (see KD 416) or a rotating blade machine (see KD 417). I don't believe in fairy tails about Darrieus rotors. People have spent a lot of money in patenting drag machines which finally appear to generate only very little power for the required investment costs. You should not bring a design at this public forum if you don't want to explain how it works.

I have several patents on my name and I know the procedure. It is only checked if the idea is new and not if the idea really fulfills the claims. But checking if the idea is new requires reading thousands of reports and books especially if it concerns wind turbines. So often it appears later that the idea wasn't new even if the patent is gained. So then you can get problems with original inventer. Taking a patent is only a good option for big companies which have their own patent officers and which have enough money to fight on court if their patent is questioned.

Adrian,

Suspicion is always good.

I keep testing as the weather here permits to prove my point.

Modest world records hitherto certainly prove nothing.