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Wind turbine help Hi, I’m new to the site and forums and wind turbines in general, and was hoping to get some much needed help from some veterans of wind movement! A quick bit about me, I’m a qualified painter, welder and carpenter, I guess what I’m saying is that I’m hands on and can build almost anything and make it pretty to boot, but when it comes to mathematical equations my eye starts to twitch and I keep thinking ,did I do that right, that can’t be right can it! And in saying that was after clarification on the calculations I’ve done thus far, so where I’m at is as follows. I have determined the type of wind turbine I believe to be the most appropriate for my site, by taking into account average wind speeds, type of wind (turbulent) ,and council restriction a long with other fundamentals and have decided on a modified 6 air foil Lenz type, I have done a cad drawing from the bearing shaft up so far and before going further won’t to confirm power, torque and RPM calculations, as there is formulas for hawt and not much in the way of vawt I did it in the only way that made sense to me and that was to take the m2 of the surface area that faced the wind and apply the mechanical force of the radius Wind Calculations Wundowie elevation= 235 mamsl Air density at 235m= 1.17 Average wind speed from Australian bureau of meteorology statistics Morning average 4.3m/s Afternoon average 5.4m/s Daly average wind speed of 4.8m/s Wind turbine dimensions 3m radius air foils x6 3mx1.5m Power potential calculations Power potential of 1m3 air mass at different wind villosities ½ air density=.585 x drag/lift area 13.5m2 x wind speed m/s3 x Betz's law of coefficients 20% 4.5m/s3 = 91.125 (x7.8975) =719.659 x 20% coefficient =143.931 watt/newtons 5.5m/s=166.375 (x7.8975) =1313.946 x 20% coefficient =262.789 watt/newtons 6.5m/s=274.625 (x7.8975) =2168.850 x 20% coefficient =433.77 watt/newtons 13.9 m/s=2685.619 (x7.8975) =21209.676 x 20% coefficient =4241.935 watt/newtons Mechanical advantage Mechanical advantage off 2900mm @25mm fulcrum point 143.931watt/newtons + applied Mechanical advantage=16695.996 watt/newtons 262.789watt/newtons + applied Mechanical advantage=30483.524 watt/newtons 433.77 watt/newtons + applied Mechanical advantage=50317.32 watt/newtons 4241.935 watt/newtons + applied Mechanical advantage=492064.46 watt/newtons Mechanical advantage watt/newtons to kg 16695.996 watt/newtons=1702.517Kg 30483.524 watt/newtons=3108.454Kg 50317.32 watt/newtons=5130.938Kg 492064.46 watt/newtons=50176.610Kg Revolutions per minute Rpm calculation,{ (v=2PiR x m/s)( Wrad=v÷r)(÷2Pi)x time} meaning wind speed m/s x 2PiR =circumference 18.85m ÷radius 3m ÷2Pie =360 x time 60sec 4.5m/s =4.712RPM 5.5m/s=5.759RPM 6.5m/s=6.806RPM 13.9m/s=14.556RPM
govertical Guru Joined: 11/12/2008 Location: United StatesPosts: 383
Posted: 07:38am 06 May 2017
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http://www.windstuff.org/calc/calc.php
Hi, here is a online power calculator.
I built and tested a VAWT for two years and found out I live in a low wind area:( It was designed to produce 5 amp @ 12 volts at a wind speed of 10 MPH.
I found out you really need a Buck converter to reduce (I squared R) at the stator or most of the power produced is lost as heat at the stator.
https://youtu.be/uTxD7x0WJNM
Here is a configuration that may help.
https://youtu.be/QESgfGq1CeA
https://youtu.be/8ydsU1CAsZg
Best of luck with your project. Cheers and welcome :)Edited by govertical 2017-05-07just because your a GURU or forum administer does not mean your always correct :)
Thank you for your input Govertical I have found every time I input my data to the calculators I seem to get different answers!!! it's a shame your wind speeds are low, I'm lucky I guess my average is over 10 mph
DaveP68 Senior Member Joined: 25/11/2014 Location: New ZealandPosts: 292
Posted: 03:27pm 06 May 2017
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Hi Hatman
Welcome to the forum.
The topic is :"power potential"
I like how you've gone into a lot of detail in your illustration and the numbers from my limited experience look good from what is covered of so far. Please don't be put off with what I'm about to explain.
The one thing I've learn't from some of "veterans of wind movement" as you've stated above on this website is W/hrs of accumulated power.
Take what you have calculated above as the amount of power that can be extracted from the 4 different wind speeds. This will be the maximum amount of power at each wind speed that can be taken from the wind.
This Mechanical advantage only applies to putting energy into a form of gearing system to increase your RPM to a usable range to drive some form of PMA (permanent magnet alternator).
The RPM you have calculated is too low to get any meaningful power from most PMA's. The power that can extracted from the "Mechanical advantage" needs to drive a gearing system to get up to a minimum of 80 RPM at start up.
It comes down to the Faraday’s Motional emf Expression The magnitude of the electromagnetic induction is directly proportional to the flux density, β the number of loops giving a total length of the conductor, l in meters and the rate or velocity, ν at which the magnetic field changes within the conductor in meters/second or m/s, giving by the motional emf expression:
E=-β.l.ν
It's all about rate of change when driving a PMA which a higher working RPM range is very important. That's where your "Mechanical advantage" calculations become very useful to work out how best to set up a gearing system.
Cheers
DavidEdited by DaveP68 2017-05-08There are realities if you do not accept, will lead to frustration because you will be spending time on wrong assumptions and the results cannot follow! The Dunning Kruger Effect :)
Hi, David thanks, yes the plan is to gear up to meet RPM of PMA just trying to work out how many horses I got to play with or more accurately lose in gearing. I have already built the MPA with 8 fisher/p rotor & stators 60 series in 7x2 poles star configuration, they take less than 15Nm (my meter starts at 15) to over come cogging and 34Nm at 65RPM to produce 300w loaded. So was trying to work both how much power I have to use and also stop with brakes (manual and dynamic systems).
DaveP68 Senior Member Joined: 25/11/2014 Location: New ZealandPosts: 292
Posted: 07:05pm 06 May 2017
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Hi Hatman
Sounds like an impressive set up with 8x F&P 60 series rotors & stators! That would have taken a while to rewire them 7x 2 pole star?
Can I ask what the Volts/Amps readings are at 65 RPM to get 300 W of power?
Have you come across the 36 pole factory decogged 60 series copper stators used with the newer black 48 magnet rotor caps? Your set up would only require 6x 36 pole stators to output 325 W at 65 RPM.
They would also have much less cogging thus starting force would be much lower than 15 Nm. As I can't measure Nm stating force, I just measure the power required to turn one from start up on a cordless drill. It's less than 6 W per unit compared to 70 W on the same drill for a non decogged 42 pole stator.
Reason I'm asking is they give 40 % more power output for the exact same RPM. For this reason I no longer use the 42 pole stators (got lots of them still gathering dust on the shelf).
There's anything wrong with the 42 pole stators. It's just that they require decogging and put out 40 & less power per unit than the newer 36 pole copper version (stay away from the latest aluminium 36 pole version).
Not trying to say you need to rebuild what you have put together. Twisting the poles on the 42 pole stators removes a lot of their coggimng. Hope this information is of use.
Cheers
DavidThere are realities if you do not accept, will lead to frustration because you will be spending time on wrong assumptions and the results cannot follow! The Dunning Kruger Effect :)
David, In terms of voltes/amp not completely certain my volt meter met an ugly demise so I strung a bunch of 12v lights together, 300w is what I just happened to have (car headlights and such) so at least 12v which is what 24amp
DaveP68 Senior Member Joined: 25/11/2014 Location: New ZealandPosts: 292
Posted: 09:24pm 06 May 2017
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I did the same thing to a 60 VDC 100 Amp power meter just yesterday testing the dynamic braking function of an F&P stator. From what I can work out volts went just above 60 V and display went blank! No more power meter, so now resort to separate volts & amps meters. Just have to do the math to get power output...There are realities if you do not accept, will lead to frustration because you will be spending time on wrong assumptions and the results cannot follow! The Dunning Kruger Effect :)
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