Furling fin mass calculation for a horizontal axis wind turbine

[abdurahman]

I would be thankful to if anyone help me in the calculation of Furl mass.

I am working on the designing of a furling mechanism for a horizontal axis wind turbine.
Turbine diameter is 5.6 m. I did following calculation:

Turbine thrust is 1.628 kN (using WT_perf) at 12 m/s wind speed (speed at which furling should start)

Offset between generator and tower axes is 0.28                           (5% of Turbine diameter)

Turbine Moment is 0.4558 (kN-m)                                                 (Turbine Thrust x offset)

Boom length is 2.8 m                                                                     (Turbine radius)

Fin Area is 1.23 m^2                                                                      (Turbine swept area x offset = 2 x Boom Length x Fin Area)

Angle of Fin bracket from vertical is 20 degree   
Angle of Fin bracket from the plane of the blades 45

Fin mass is calculated from following relations:
                         Turbine Moment = Fin Moment
                          Fin Moment = Furl Resistance x Boom Length
                          Furl Resistance = Furl Mass (kg) x 9.8 x sin20 x sin45
From above three equations we can write that:

Furl Mass = Turbine Moment / Boom Length / sin20 / sin45 / 9.8

Furl Mass = 0.4558 x 1000 / 2.8 / sin20 / sin45 / 9.8
So,                           Furl Mass = 68.7 kg
Its look too much.

I go through from data available for different HAWT furling system, and found that they use around 30 kg furl mass for 5 to 6 m diameter HAWT.

Is this Furl mass OK (68.7 kg for 5.6 m diameter HAWT)? Or can you identify the mistake.

Waiting for your reply.
Thanks

[Flux]

I am not good enough at maths to answer this but basically you are equating the moment turning the prop out of the wind against the restoring force of the tail trying to keep it in the wind.

The prop thrust x alternator offset seems correct but I don't follow the rest.

The force trying to keep it in the wind is the horizontal torque moment derived from the tail boom length x mass multiplied by the vectorconverting it to rotation in the horizontal plane. This comes about as a result of the 20 deg tail hinge angle. I suspect the component is sine 20 ( someone here will no doubt correct me).

The 45deg angle is not involved in this, it is just to set the tail restoring moment at the correct point. If the tail was at the back there could be no restoring moment and at 90 deg it would be maximum. 45deg puts the thing in the working range so it is a reasonable compromise between start of furl and restoring when fully furled.

I can't see if you have included the fin area in your calculation, again it only matters in as much as the fin is big enough to let the tail vane act as an anchor down wind. If the vane is too small the tail will run at a big angle to the wind at all times, it doesn't affect the furling point.

Sadly when you have done all the maths you will still have an unknown factor you can't deal with and that is the tendency for the prop to seek the wind. This effectively reduces the moment from the thrust of the prop. The tail weight you calculate will be too high and only experience of similar machines with similar prop and loading will get you the right figure.

Apart from including the 45deg angle your figures may be correct and practical weights will also include a correction for the seeking force.

Flux

[Flux]

Having had a bit more time to look at this, I think you are ok to include the 45 deg angle. As I said the restoring force with hinge at the back is zero and maximum at 90 deg. It is therefore reasonable to expect it to be sin 45 at the start of furling with the angle at 45deg ( assuming you have not offset the tail away from furl to make it steer nearer to the wind in normal operation.)

I think the difference between your calculated value and the figures others have used is purely due to the prop seeking factor. Reducing the thrust to about a half is in line with some of the measurements I have made, but it is very dependent on the prop profile, tsr and method of loading, it really makes detailed calculations pointless but I would suggest you start at half your calculated value and work up if necessary to be safe.

Flux

[ChrisOlson]

Hi abdurahman and welcome to the wonderful world of turbine building!

I have never been able to get the published calculations to work for figuring furling moments.  Over many turbine builds, taking many yaw torque measurements and making observations I have, over time, arrived at a different set of calculations that I use to determine the rotor axial thrust component.  My method takes into account turbine efficiency (how much power it actually extracts from the wind) and coefficient of drag of the rotor blades.

I have to make some assumptions with your turbine because I do not know the design details.  But assuming typical efficiencies of homebrew blades and generator, I calculate a more likely rotor axial thrust component to be 88 kg @ ~4 kW output (32% rotor efficiency, 50% generator efficiency) with Cd 1.28 @ 12 m/s.

Try plugging that number into the rest of your calculations and see if you arrive at a more reasonable value for tail tip weight.
--
Chris

[abdurahman]

Dear ChrisOlson

How you calculated 88 kg?   
(I have to make some assumptions with your turbine because I do not know the design details.  But assuming typical efficiencies of homebrew blades and generator, I calculate a more likely rotor axial thrust component to be 88 kg @ ~4 kW output (32% rotor efficiency, 50% generator efficiency) with Cd 1.28 @ 12 m/s.)

[abdurahman]

I did some refine calculation ang got following results:

Turbine diameter is 5.6 m.
Turbine thrust is 1.127 kN (using WT_perf) at 10 m/s wind speed (speed at which furling should start)
Offset between generator and tower axes is 0.28             (5% of Turbine diameter)
Turbine Moment is 0.31556 (kN-m)                                (Turbine Thrust x offset)
Tail Arm Length is 2.8 m                                                 (Turbine radius)
Tail Arm Mass is 10 kg                                                    (A guess)
Fin Area is 1.2315 m^2                                                   
(Turbine swept area x offset = 2 x Boom Length x Fin Area)

Angle of Fin bracket from vertical is 20 degree 
Angle of Fin bracket from the plane of the blades 55 degree

Tail Arm Moment 0.0384 kN-m     
(Tail Arm Moment = Tail Arm Mass x 9.8 x Tail Arm Length/2 x sin20 x sin55)   
       
 Tail Van mass is calculated from following relations:
                         Turbine Moment / 1.5 = Tail Vane Moment + Tail Arm Moment
                         Tail Vane Moment = Turbine Moment/1.5 – Tail Arm Moment = 0.172 kN-m
                         Tail Van Moment = Tail Van Mass x 9.8 x Tail Arm Length x sin20 x sin55
From above three equations we can write that:
Tail Van Mass = Tail Van Moment / Tail Arm Length / sin20 / sin55 / 9.8
Tail Van Mass = 0.172 x 1000 / 2.8 / sin20 / sin55 / 9.8
So,                           Tail Van Mass = 22.4 kg
Its still look too much.

I go through from data available for different HAWT furling system and found that others use smaller Tail Van Mass for 5 to 6 m diameter HAWT.
Is this Tail Van Mass OK (22.4 kg for 5.6 m diameter HAWT)? Or can you identify the mistake.
 

Thanks

[abdurahman]

What is the physical significance of these angles:

Angle of Fin bracket from vertical (20 degree)   
Angle of Fin bracket from the plane of the blades (45 degree)

[ChrisOlson]

What is the physical significance of these angles:

Angle of Fin bracket from vertical (20 degree)  
Angle of Fin bracket from the plane of the blades (45 degree)

This first one is very important.  The second one is immaterial.

There are only two things you must look at here
The torque moment at the yaw shaft that is caused by rotor thrust working on the offset
The torque moment of the tail acting in the opposite direction on the yaw shaft.

I think you are making it too complicated when the basic concept is simple.  When the torque moment from rotor thrust exceeds the countering torque from the tail, the machine furls.  So break it down into those two components and totally forget about any sort of relationship between the two when figuring each one separately.

What is important is the weight at the tip of the tail when calculated the tail moment.  Have an assistant balance it on a block of wood or something on the hinge end and weigh the tail way out at the tip.  I calculate you will need 26 kg of tail tip weight given the other dimensions you have supplied.

Yes, I am sure many others you will see have used less tip weight on the tail.  The turbine's generator will burn up otherwise.  These furling calculations assume a generator that is able to handle the continuous output at those wind speeds.  Most of the generator designs I have seen cannot.  Therefore you have to lighten the tail to prevent the machine from putting out too much power and destroying itself.

That is why I have gone to using a different method to figure this.  The amount of thrust that the turbine rotor develops is not a function of size or swept area - it is a function of how much power is being extracted from the wind.  I have a 3.2 meter turbine that develops 30% more thrust at 12 m/s than a 3.8 meter turbine that is on another tower only a short ways from it.  The reason is that that 3.2 meter makes more power than the 3.8 meter, and that's why it makes more thrust.

So let me be clear here - my calculations assume your generator at 50% efficiency, which is going to be pretty typical for most homebrew generators @ 4 kW output.  That means the stator is dissipating 4 kW in heat.  In reality it would take a very elaborate cooling system on that stator for it dissipate 4 kW in heat without damaging it.  If I would re-do my calculations to consider what your generator can actually (probably) handle, then the tail is going to be a lot lighter than 26 kg.
--
Chris

[ChrisOlson]

How you calculated 88 kg? 

I can only answer by saying I will respond in a PM if you really want to know how I arrive at that.  You are using PROP or WT_Perf here, and there is a big difference between theory and what most of these homebrew machines with hand carved blades are capable of.

I do not wish to cause confusion with the "published" methods in the homebrew books used to calculate this.  This furling issue has been a pet peeve of mine for a long time.  I spent a lot of time building turbines and taking measurements to arrive at what the real relationships are.  But I am somewhat of a renegade that does not follow the book, or even refer to it.  So please PM me if you wish to review my method of calculating it.
--
Chris

[TimS]

Awesome discussion, with great details.  Thanks!

[abdurahman]

A:  In books and papers both angles (Angle from vertical and angle to the rotor plane) are specified and used in tail moment calculation. If angle to rotor plane is not use in tail moment calculation then what its importance?

B:    How to calculate tail fin area:

1    Tail fin area should be 7% of the rotor swept area (as recommended by different authors) or
2   Rotor Volume = 2 x Tail Volume
Rotor swept area x offset = 2 x Tail area x tail arm length (This give fin area which is 5% of swept area)

I can make tail fin of same mass (22.4 kg as calculated) with different area (larger than 7% or smaller than 7% of swept area).

1. What tail fin area you recommend?
2. May you please explain effect of smaller and larger area tail fin?
3. What do you mean by PM? (I can only answer by saying I will respond in a PM if you really want to know how I arrive at that)

[ChrisOlson]

A PM is Private Messages - you should see it in your profile thing at the top.

All mounting the tail hinge at an angle to the rotor plane does is allow you to add some preload to the tail boom so it's not swinging around in back of the turbine like a limp noodle.  In other words it will have a positive return to the full flying position.  Some folks have incorporated dampers to cushion the return of the tail boom, such as this one on my 3.2 meter machine:



Otherwise, furling boils down to two torque calculations on the yaw shaft - the rotor applying torque in one direction, the tail applying torque in the other.
--
Chris

[abdurahman]

A.  Do you mean that angle of Fin bracket from the plane of the blades (55 degree) is just to set the tail at correct position by providing restoring moment and do not have any effect on the tail fin mass calculation.??

B.  If we have offset the tail away from furl to make it steer nearer to the wind in normal operation then we do not require this angle.
Or we can incorporated dampers to cushion the return of the tail boom, such as in your 3.2 meter machine???
 

C. Please explain how to calculate tail fin area:

i)    Tail fin area should be 7% of the rotor swept area (as recommended by different authors) or
ii)    Rotor Volume = 2 x Tail Volume
Rotor swept area x offset = 2 x Tail area x tail arm length (This give fin area which is 5% of swept area)

I can make tail fin of same mass (22.4 kg as calculated) with different area (larger than 7% or smaller than 7% of swept area).

1. What tail fin area you recommend?
2. May you please explain effect of smaller and larger area tail fin?

[Flux]

I thought I covered the two angles, but I assumed you understood why they are present.

The 20 deg from the vertical is the angle that causes the weight of the tail  to provide a moment to oppose the thrust moment of the prop on the alternator offset. If the hinge was vertical the tail would produce no force in the horizontal plane. If the hinge was at the back and you inclined the top backwards the tail would then drop to the lowest point and that would be straight back and again there would be no horizontal moment.

Now if you tie a rope to the tail fin and pull it around to 45 deg then the tail is still trying to go to the lowest point, but as it can't therte is a horizontal component of force at the hinge which is tail vertical moment x (sin20) trying to move the pivot in the horizontal plane.

You want this condition to apply when the tail is straight back and resting on its stop so that is why you move the tail pivot round to 45deg in the horizontal plane. Now if you were to remove the tail stop the tail would go to the lowest point and hang at 45 deg to the prop axis ( away from furl).

The important angle in developing the tail restoring moment is the 20 deg vertical inclination. The 45 deg angle is a secondary one and effectively defines how the restoring force varies during the furl. According to the ideal theory the restoring force ought to be constant and you would get constant thrust ( constant power out). The simple hinge won't give this constant moment ( it doesn't matter as the thing fails to obey the ideal theory anyway)

If you calculate the thrust on a solid disc then it will depend on the area of the disc. If you replace the disc with a prop then the thrust will be less. If stationary it will be the drag on the blades only. When rotating at ideal tsr and extracting maximum power it will be less than the disc loading. above or below ideal tsr the thrust will fall, so you can fairly easily see where Chris is coming from in saying that it depends on the power extracted. After all it is the change of momentum of the wind through the prop that is producing the power so thrust must depend on the amount of energy you extract from the wind.

In addition there is a wind seeking force trying to hold the prop into the wind, so you have a moment from the thrust turning it out and a seeking force holding it up wind. The result is that the thrust moment is reduced below the value you calculate and you will need a lighter tail.

The final weight you use must be such that it will furl within the alternator limits and that becomes the limiting wind speed you use for the calculation.

Now vane area, this has no direct effect on furling. The purpose of the vane is to anchor the tail boom directly down wind, you can remove the vane and tie a rope on the boom and tie it to a tree down wind, for sorting out your furling calculations. If the vane is infinitely large you will have the ideal case of the rope, but as you reduce the vane area the reaction of the tail restoring moment balancing the prop moment will push the vane away from the down wind position. If the vane is very small then you will have a balance with the thing running at a large angle to the wind below furl, This always happens but the vane is chosen to be big enough to keep this angle reasonable and you can set the tail away from furling by 20 or 30 degrees to get the thing to track fairly well .

The only direct effect the vane area will have on the furling is that it affects the tail weight and if the vane is too big you won't get the weight down to a safe figure. Everything is a compromise and not directly capable of mathematical solutions.

I hope this helps explain where Chris is coming from. His figures have been derived for his own machines but it should work for you if you have a similar prop.

My advice is to use the lowest weight you can manage with your vane and work upwards if it furls too early, don't use the theoretical figures as it will be too heavy.

Flux

[abdurahman]

After all these discussions Finally I did following calculations:

Turbine diameter is 5.6 m (5 kW output power).
Turbine thrust is 1.626 kN (using BEM Method) at 12 m/s wind speed    (speed at which furling should start)
Offset between generator and tower axes is 0.28                          (5% of Turbine diameter)
Turbine Moment is 0.455 (kN-m)                                                  (Turbine Thrust x offset)
Tail Arm Length is 2.8 m                                                             (Turbine radius)
Tail Arm Mass is 10 kg                                                                 (A guess)
Fin Area is 1.48 m^2                                                                  (A guess)
(6 %Turbine swept)

Angle of Fin bracket from vertical is 20 degree 
Angle of Fin bracket from the plane of the blades 55 degree

Tail Arm Moment 0.0384 kN-m     
(Tail Arm Moment = Tail Arm Mass x 9.8 x Tail Arm Length/2 x sin20 x sin55)   
       
                         Turbine Moment / 2.0 = Tail Vane Moment + Tail Arm Moment
                         Tail Vane Moment = Turbine Moment/2 – Tail Arm Moment = 0.189 kN-m
                         Tail Van Moment = Tail Van Mass x 9.8 x Tail Arm Length x sin20 x sin55

So:

Tail Van Mass = Tail Van Moment / Tail Arm Length / sin20 / sin55 / 9.8
Tail Van Mass = 0.189 x 1000 / 2.8 / sin20 / sin55 / 9.8
So,                           Tail Van Mass = 24.6 kg

So Finally I got:
Tail Arm Length = 2.8 m 
Tail Arm Mass    = 10 kg
Tail Van Mass     =  24.6 kg
Tail Van Size        = 1.48 m^2

For a 5 kW HAWT (rotor dia 5.6 m) these values are OK or required some modifications?

[ChrisOlson]

My advice is to use the lowest weight you can manage with your vane and work upwards if it furls too early, don't use the theoretical figures as it will be too heavy.

Flux covered this well.  I refuse to tell you if your calculations are "correct" because my experience has told me I need to know a lot more about your machine to be able to even get close.

The important thing to realize is that there is only, probably, one chance in a thousand you will get it perfect first time on the tower.  And if you do it will be pure luck more than attempting to design it with calculations.  All the calculations provide is an understanding of how it works, and a basic starting point.

Every new design, that does not follow proven designs that have been adjusted to work, is finalized thru trial and error when it comes to furling.  There are too many variables to attempt to arrive at the correct solutions with calculations - some variables you cannot even measure until you get the machine on the tower to see how it acts.

As an example, the machine I showed you the photo of above will furl fine @ 12.5 m/s with the rotor running at 400 rpm and reaches full furl around 15 m/s @ 450 rpm.  If I speed the rotor up so it is running at 500+ rpm, the machine refuses to furl at all, even at 20 m/s, and it puts out way excessive levels of power.  This is an example of why it is impossible to calculate furling to the last digit.  It won't work.  You need to arrive at a starting point, and fly it to see what it does, assuming you are working with a new design.

Tail fin area is also not a precise thing.  I prefer to use long tail booms on my machines with a small, sleek sexy looking tail.  Some folks like to use a shorter boom with a big club of a tail.  If you make the tail boom long enough, you don't need a fin at all because aerodynamic drag on the boom structure will steer it.  If you want to use 7% of the swept area, I would say that will work fine in probably all cases of boom length.

Turbine diameter is 5.6 m (5 kW output power).
Turbine thrust is 1.626 kN (using BEM Method) at 12 m/s wind speed    (speed at which furling should start)

Now, looking at your calcs, @ 5 kW output your generator is running at about 54% efficiency, which I would say is reasonable.  I would calculate rotor thrust to be .981 kN using the CO (Chris Olson) method, assuming 36% rotor efficiency @ 5 kW output.  I do not know what the BEM method is, but it is wrong in my experience and will yield a tail that is way too heavy.

The second thing is, will your generator actually handle sustained output at 5 kW while dissipating 4,200 watts in the stator winding?  I highly doubt it.  That means you need to furl it earlier at the power level your stator can handle for heat dissipation.  So while I refuse to tell you that your calculations are correct, I will tell you they are NOT correct.  I know beyond a shadow of a doubt that if the above machine was built and flown it would burn up the generator in sustained 13 m/s wind.  Even if I drop the rotor efficiency to 30% in my calcs, which raises gen efficiency to 65% @ 5 kW, it doesn't figure out very well because you would still be dissipating 2.7 kW in the stator.  That could be done with a very elaborate cooling system on the stator in sustained 13 m/s wind.  But it would be very questionable.

My point in all this is that the machine cannot be furled at outputs beyond what your generator can handle or you will burn it up.  So what are the design details of your generator?  What voltage does it operate at?  How much surface area does it have to dissipate heat?  What is the resistance of the windings in it?  All those factors determine how much power it can actually produce, sustained, and therefore affect furling calculations.
--
Chris

[abdurahman]

My rotor efficiency is 0.48, rotor radius is 2.8.
If I take drag coefficient is 1.36 (1.28+0.8=1.36) then,
thrust by using CO method is:

Thrust = ½ x 1.24  x 12 x 12 x pi x 2.8 x 2.8 x 0.48 x 1.36 = 1.435 kN

Which is near to BEM method 1.652 kN.

Is it OK?