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Forum Index : Windmills : 10 kw Wind turbine: probs and Sols
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| Abhinav India Newbie  Joined: 26/07/2006 Location: IndiaPosts: 8 |
Hi, I am Abhinav, an Electrical Engg. based in Mumbai, India. Encouraged by recent reports on small wind turbine and perenial shortage of power in India, I decided to build a wind turbine. I am making a 10KW water pumping and battery charging model. I am facing some problem with Autofurl. As the size is big(Rotor dia is 11meter, as India as very low winds)I am unable to calculate the dimension and eight of the Tail in order to acheive the AutoFurl. Can you suggest me how I should do it. Thanks Abhinav Wind is Free, wind energy is Freedom |
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makourain Senior Member  Joined: 19/04/2006 Location: Posts: 111 |
being near the equator i would have thought you would get plenty of wind. what do u mean by autofurl? what do u mean by "dimension and eight" |
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| Abhinav India Newbie Joined: 26/07/2006 Location: IndiaPosts: 8 |
Hi makourain, India is a divided in two by tropic of cancer. Most villages are far from coast. Autofurl is a system that stabilizez the turbine in High winds (see the video in pic section of this website) Pardon my spelling error. I am looking for a means to find the Dimension and Weight of the Tail and Tail Fin. Abhinav Wind is Free, wind energy is Freedom |
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Gizmo Admin Group  Joined: 05/06/2004 Location: AustraliaPosts: 5078 |
Hi Abhinav, Thats an impressive windmill your building. The furl point of a windmill depends on the blade diameter, tail hinge angles, tail length and weight. I dont have a formular to work the exact details out, but I think I may have seen some calculations on the Fieldlines web site www.fieldlines.com. My tails were always guess work, they looked right and worked right. My tail booms are usually half the blade diameter, and the tail fin is about 1/8 the swept area of the blades. Maybe someone else here has some hard maths to work it out. Glenn The best time to plant a tree was twenty years ago, the second best time is right now. JAQ |
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| Abhinav India Newbie Joined: 26/07/2006 Location: IndiaPosts: 8 |
Hi Glenn, Thanks for your guess estimates. I will try to find more. Couldn't get much on otherpower.com. Another interesting thing about my turbine is we are using Junk Automobile wheel axle for making the Hub and Yaw. another sticky point I see coming is the twisting of cables. But for now, I am grappling with the Tail. regards Abhinav Wind is Free, wind energy is Freedom |
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windstuffnow Newbie Joined: 30/06/2006 Location: United StatesPosts: 31 |
Hi Abhinav, The tail weight, angle of the tail pivot and offset of the turbine head all act together on the gravity system. I normally use a tail pivot of around 20 degrees and it seems to work out fairly well to date. As I said all the variables interact. So suppoose your rotor shaft is offset by 125mm. The thrust for furling is 40kg. Then the moment is 40 x .125 = 5Kg/m Find the moment of weight of the tail by weighing the tip. If the tip weighs 10Kg and the tail is 2 meters long then you have a 20kg moment of weight. The angle determines how much of that weight will be opposing the furling. Furling moment is moment of weight times sine(angle). If moment of weight of the tail is 20kg/m and yu want 5kg/m then the ange should be 14.5 degrees (sine(5/20)= 14.5 You can make up a spreadsheet using the following to find the numbers you need... Input: rotor offset (meters) Diameter of Rotor (meters) Wind speed (m/s) Tail tip weight (kg) Length of tail (meters) Calculations: rotor thrust =Diameter of rotor^2 x Windspeed^2/24 Tail moment = Tail tip weight x length of tail Rotor moment = Rotor thrust x Rotor offset Tail mount angle =sin-1(Rotor moment/Tail moment) To find tail weight at a given tail mount angle; Tail moment = Rotor moment/ (Sin(angle in degrees) Tail weight = Tailmoment/Length of tail Hope that helps Have Fun! windstuff Ed |
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Megawatt Man Senior Member Joined: 03/05/2006 Location: AustraliaPosts: 119 |
Hello Abhinav, I don't know what voltage you intend to generate, but at 415V 3 phase, 10kW is only 13 amps per phase. Use slip rings and forget the worry of tangling your cables. Motor car starter motor brushes last for years carrying 200 amps DC so even if you generate at 41.5 volts you would have spare brush capacity. Megawatt Man |
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| Megawatt Man Senior Member Joined: 03/05/2006 Location: AustraliaPosts: 119 |
Hello windstuffnow, any chance of a vector diagram? I am visualising the rotor moment acting horizontally and the tail moment acting vertically, hence no furling correction, so I am not seeing something correctly. Maybe words will help... Megawatt Man |
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| windstuffnow Newbie Joined: 30/06/2006 Location: United StatesPosts: 31 |
It works the same as the caster angle on your car or even as simple as the angle of the forks holding the front wheel on a bike. Because of the angle the weight of the car or bike tends to move the wheel straight or in balance. When the balance is changed, turning the wheel right or left, it tries to lift the the car/bike when you let it go gravity and the angle try to find a balance once again. If you visualize one person holding the tail in place but allowing it to move up or down and another pushing on the center of the rotor, both people representing the wind. The weight of the tail is resisting because of gravity holding it in place, once the person pushing on the rotor overcomes the force of the tail it simply folds lifting the tail up. When the pressure is released from the rotor gravity will pull the tail back down forcing the rotor back into the wind. A quick simple test would be to turn the bike over and mount a board on the forks (reprsenting the tail), move the board off center and it will return to center when released as long as the angle of the fork is off center, the larger the angle the more reactive it is and the less weight you need to keep it in place. Hope that helps create a better picture... WindstuffEd |
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| Abhinav India Newbie Joined: 26/07/2006 Location: IndiaPosts: 8 |
Hi friends, Been working on this 10KW water pumping rural electricity proj on and off for past few months. I have got the blades ready. No I am on the generator. Making a Permanent Magnet generator , using the car wheel as a rotor and mounting magnets on it. I am buying Neo magnets of Grade N33 I want some help in testing the magnets. I need to know how much power they have. There is a link on otherpower, but the images are not showing up. http://www.fieldlines.com/story/2006/5/2/17039/94666 Thanks in advance Abhinav Wind is Free, wind energy is Freedom |
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| Megawatt Man Senior Member Joined: 03/05/2006 Location: AustraliaPosts: 119 |
Hello Abhinav, I guess you don't know the magnetising force that cahracterises the Neos? If you knew that, you would just use the fundamental theory of the magnetic circuit with air gap to establish the flux and then use the voltage/time rate change of flux/number of turns formula to find induced voltage. You would know these things, so it must be that you don't know about the magnets. The manufacturer must be able to tell you. All the best mate! Megawatt Man |
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| Abhinav India Newbie Joined: 26/07/2006 Location: IndiaPosts: 8 |
Hi Megawatt Man, I have done the calculations on paper and the numbers look good. Now I am also planning to crunh the numbers and the design into a FE s/w. But when I have the magnet peices in hand, how do I know that the manufacturer has delivered what he had prmoised? Thats the problem, I can't figure out. Abhinav Wind is Free, wind energy is Freedom |
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