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Forum Index : Windmills : about windmills furling system
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| Aris Newbie  Joined: 14/11/2010 Location: IndonesiaPosts: 7 |
in web have explained. about furling system, how “The turbine thrust or force can be worked out with - Turbine Thrust = Diameter2 * WindSpeed2 / 24”. i’m question is what mean 24 ??? please help me…with detail.thanks…. |
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MacGyver Guru  Joined: 12/05/2009 Location: United StatesPosts: 1329 |
Aris Welcome to the 4m. Click this link and read the explanation. The math is in the lower portion. . . . . . Mac Nothing difficult is ever easy! Perhaps better stated in the words of Morgan Freeman, "Where there is no struggle, there is no progress!" Copeville, Texas |
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scoraigwind Newbie Joined: 23/09/2009 Location: United KingdomPosts: 21 |
I have to admit that the formula is mine, and the number 24 is intended to give you the answer in kg of thrust. I published several such formulae in the back of my book 'Windpower Workshop' so as to help people make rough design decisions. Although based on theoretical principles, this is not a precise equation but more like a 'rule of thumb' that gives a useful answer. Various people have taken my design equations and popularised them for example Ed Lenz at windstuffnow.com which is how this formula arrived at the helpful page on this domain http://www.thebackshed.com/Windmill/Docs/Furling.asp which helps you find the right design of tail for a particular situation. I use the same equations for my own design work but i have also learned to tune them based on experience. I would not use them to two places of decimals and expect a precise result. Often there are many factors at work that make reality different from the theory. Hugh Hugh Piggott |
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| Aris Newbie Joined: 14/11/2010 Location: IndonesiaPosts: 7 |
@MacGyver i already read that link.. but i still confuse where is the "24" come from in the "turbine thrust" formula?? and any one can help me with free body diagram of this furling system..?? thanks before... |
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| scoraigwind Newbie Joined: 23/09/2009 Location: United KingdomPosts: 21 |
I got it from the Betz analysis of optimum action on the wind. Speed in the wake should ideally be slowed down to 1/3 of the upstream speed. Speed through the rotor disk can then be shown to be 2/3 of upstream V. Thrust force = rate of change of momentum = mass flow rate * change of speed Mass flow rate is (density)*(Area)*(speed through disk) = 1.2 * pi/4 * D^2 * 2/3 * V change of speed is also 2/3 V So finally the thrust in Newtons is 1.2 * pi/4 * D^2 * 2/3 * V * 2/3 * V = D^2 * V^2 /2.4 N = D^2 * V^2 /24 kg Voila. Rough and ready. 23 might be more accurate to be honest, but this is a rule of thumb, and 24 is easier for mental calcs. Real world data shows that a factor of 30 works better for a lot of machines. But this will depend on tip speed ratio etc. And the centre of thrust is not always at the centre of a skewed rotor anyway..... I have published plenty of plans for tail furling systems that work well, but other than this equation I have not published a design procedure. You can use the procedure in the above link but take care and allow for some adjustment of the weight of the tail. Also I would recommend angling the tail beyond 90 degrees so that it actively opposes the thrust of the rotor blades and you will hold the rotor closer to facing the wind properly. In other words give the tail vane an angle of attack to the wind when the rotor is squarely facing it. Hugh Piggott |
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Gizmo Admin Group  Joined: 05/06/2004 Location: AustraliaPosts: 5078 |
The "24" is a constant, recycled maths. It would have been calculated using a combination of air pressure at sea level, imperial to metric units conversion, etc, values that wont change much, or at all. Basically, to save the work of recalculating values that never change, we use recycle old calcualtions, and in this case, use 24 in our maths to represent all the other stuff, other than wind speed and diameter. Hope that made sense. There's another example on this post http://www.thebackshed.com/windmill/forum1/forum_posts.asp?T ID=252. It uses Force on the turbine = .00492 x area x windspeed^2, same thing, the .00492 is our constant. Hugh has made an important point to, you can do the maths, but in real life it will perform differently. Furling can be a bit tempermental to set up, and you will usually need to make a few changes in the field to get it right. Glenn The best time to plant a tree was twenty years ago, the second best time is right now. JAQ |
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| petanque don Senior Member  Joined: 02/08/2006 Location: AustraliaPosts: 212 |
Is it a coincidence that “24” is “42” transposed???? Spooky isn’t it.  |
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| Aris Newbie Joined: 14/11/2010 Location: IndonesiaPosts: 7 |
how about with free body diagram of this furling system..??  |
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| MacGyver Guru Joined: 12/05/2009 Location: United StatesPosts: 1329 |
[Quote=Aris]Is it a coincidence . . . I think it interesting to note what Albert Einstein had to say about this; "Coincidence is God in disguise". Could be true, could be not; a curious thought either way. . . . . . Mac Nothing difficult is ever easy! Perhaps better stated in the words of Morgan Freeman, "Where there is no struggle, there is no progress!" Copeville, Texas |
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powerednut Senior Member Joined: 09/12/2009 Location: AustraliaPosts: 221 |
Thanks for that Hugh - very informative. |
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| Aris Newbie Joined: 14/11/2010 Location: IndonesiaPosts: 7 |
oke thanks all.... 
i have questions again....about pivot angle i have windmills my diameter turbine = 1.129 m so my Turbine Thrust = 21.244 kg The turbine moment ( torque )= 2.124 kgm tail weight = 2.893 kg tail Length = 69.5 cm So what about the pivot angle results calculation ?? please help me...step by step...thanks.... |
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| Gizmo Admin Group Joined: 05/06/2004 Location: AustraliaPosts: 5078 |
You left out the important bit, what speed do you want the tail to start furling? Remember, the calculations will give you a approximate result, other factors can also affect furling, like alternator loading, changes in air density, etc. Once you have an angle, you might still need to experiment to get it spot on. Glenn The best time to plant a tree was twenty years ago, the second best time is right now. JAQ |
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| Aris Newbie Joined: 14/11/2010 Location: IndonesiaPosts: 7 |
@Gizmo.... i want the tail to start furling with speed 20 m/s (72 kmh)too. help me... |
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GWatPE Senior Member Joined: 01/09/2006 Location: AustraliaPosts: 2127 |
Start furling at 20m/s? This is quite small, almost a windmill ornament size. To start furling at 20m/s, implies fully furled at approx 25m/s. Most windmills fitted with furling, would start furling at under 10m/s. Check, most commercial windmill of this small size don't bother with furling. The relatively high rpm needed to extract power from these windmills dictates skinny blades. These are usually low torque, and any furling system will be difficult to setup. For all practical purposes, any calculated parameters will be a guess at best, so extensive in the wind testing will be required. Gordon. become more energy aware |
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| Aris Newbie Joined: 14/11/2010 Location: IndonesiaPosts: 7 |
yes, 20 m/s please give me solution and results calculation about my pivot angle... |
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| Aris Newbie Joined: 14/11/2010 Location: IndonesiaPosts: 7 |
i have questioan again about formula furling system in furmula have : Furl resistance = Tail Weight * Sin ( Pivot angle in degrees) * Sin 45 and Tail pivot angle = Sin^-1 ( Turbine Moment / Tail Length / Sin 45 / Tail weight ) What mean is Sin 45 ?? thank before..... |
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