Author Topic: Volts/Coil  (Read 1623 times)

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Simen

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Volts/Coil
« on: October 14, 2008, 05:12:26 AM »
I just need some verification of my understanding of calculations for how many Volts each coil have to produce to start charging a battery... :)


What i got for 3-phase star:


Charge Volt / 1,41 / 1,7 / # of coils per phase = Single Coil Voltage.


Correct?


For 3-phase Delta, i would remove the 1,7 division. (?)

« Last Edit: October 14, 2008, 05:12:26 AM by (unknown) »
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Simen

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Re: Volts/Coil
« Reply #1 on: October 13, 2008, 11:37:44 PM »
Googled the board a bit... (Should have done that before posting, but i'm not quite awake yet... ;) )


I see most uses 1.73 for 3-phase star, not 1.7 as i wrote...


And i forgot to subtract the voltage drop at the rectifier... (ca. 1.4V.) :-o

« Last Edit: October 13, 2008, 11:37:44 PM by Simen »
I will accept the rules that you feel necessary to your freedom. I am free, no matter what rules surround me. If I find them tolerable, I tolerate them; if I find them too obnoxious, I break them. I am free because I know that I alone am morally responsible for everything I do. - (R. A. Heinlein)

Flux

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Re: Volts/Coil
« Reply #2 on: October 14, 2008, 01:28:57 AM »
You are near enough.

 The star/delta factor is root 3 but it doesn't matter if you use 1.7 or the full 1.731xxxxx to infinity. You will not have a perfect sine wave so it makes little difference if you approximate a bit.


Similarly the peak to mean of a sine wave is root 2 ( 1.413xxx to infinity) but again it only gives you the minute current at peaks of the 3 phase ripple. The more useful figure is 1.35 which is the mean dc from a 3 phase bridge and this is further complicated by the fact that the battery in some ways acts as a capacitor. Just use 1.4 as a compromise figure and it will be near enough.


What you use for rectifier drop is another grey area, allowing .6v per diode ( total drop 1.2) is fairly near for cut in. For higher currents diode drop can be up to or over 1v but except for cut in you might as well forget diode drop as the thing behaves as though it is current driven and the diode drop makes little difference to output.


Flux

« Last Edit: October 14, 2008, 01:28:57 AM by Flux »

Simen

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Re: Volts/Coil
« Reply #3 on: October 14, 2008, 03:10:30 AM »
Thank you, Flux. :)


While i'm at it: How does one know when a propeller reach cut-in speed (When the blades start to lift...) ?


I'm guessing tsr and windspeed are two of the parameters? :)

« Last Edit: October 14, 2008, 03:10:30 AM by Simen »
I will accept the rules that you feel necessary to your freedom. I am free, no matter what rules surround me. If I find them tolerable, I tolerate them; if I find them too obnoxious, I break them. I am free because I know that I alone am morally responsible for everything I do. - (R. A. Heinlein)

Flux

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Re: Volts/Coil
« Reply #4 on: October 14, 2008, 03:34:37 AM »
Not sure I understand the question.


During start up the blades will be severely stalled and as speed rises there will come a point when they come out of stall and speed will pick up. The maximum power will be developed at the design tsr of the blades but there will be enough lift at speeds lower than design tsr to produce useful power. For cut in with a normal direct battery connection you want the blades to be running higher than design tsr ( beyond the peak of the curve). As you cut in and load comes on you start coming down to design tsr very quickly and as even more load comes on in higher winds you come below the peak and eventually get close to stall.


If you don't cut in above design tsr you reach stall condition in lower winds and performance becomes bad.


Typically for a prop designed for running at tsr 6 you would want to base your cut in speed on the assumption that tsr was 7 or possibly 8.


Taking the case of blades with intended tsr 6, if you choose your cut in wind speed and base the blades on tsr 7.5, the speed you get will be a good figure for the alternator cut in. This is not very critical and a few rpm either way is not going to make a lot of difference, but choosing a cut in speed far too low for the blades will give only a very tiny improvement in the low wind region where power is very small. it will seriously muck up the performance in the more useful wind speed region. If the alternator is very powerful it will reach a limiting output and probably be safe but performing very badly. If it is not so powerful to hold the blades hard stalled you will have the thing running at very low efficiency in high winds with lots of stator heating and survival will depend on your ability to furl it in low enough winds not to fry it.


Flux

« Last Edit: October 14, 2008, 03:34:37 AM by Flux »

Simen

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Re: Volts/Coil
« Reply #5 on: October 14, 2008, 06:43:01 AM »
Excellent answer! Thank you! :)


It cleared up the term Stall and when it occurs, which have always been a bit vague for me... ;) And yes, you did understood the question better than i managed  to phrase it. ;)


I've just bought a bicycle computer which actually have an rpm display, and it works rather good. I guess all that i need now is a anemometer, and i'll be all set to find the actual tsr of my 7' 3-bladed rotor (made after instructions on Otherpower out of 2"x6" boards), and calculate the rest that i'll need. :)

« Last Edit: October 14, 2008, 06:43:01 AM by Simen »
I will accept the rules that you feel necessary to your freedom. I am free, no matter what rules surround me. If I find them tolerable, I tolerate them; if I find them too obnoxious, I break them. I am free because I know that I alone am morally responsible for everything I do. - (R. A. Heinlein)

Ungrounded Lightning Rod

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Re: Volts/Coil
« Reply #6 on: October 16, 2008, 05:26:46 PM »
"Stall" is "aerodynamic stall".


It occurs when the angle of attack is such that the air flowing over the heavily-curved (downwind) side of the blades doesn't stay "attached" all the way to the trailing edge.  Instead it lifts away, allowing air to come in from the trailing edge and raise the pressure there, reducing the "lift" force on the blades.  (There's a positive feedback effect as the replacement air stream comes in faster than the main air stream can suck it out and the "sucking out" slows the main airstream, leading the detachment to quickly move forward over a significant part of the blade to a region of different angle.)


In an airplane this occurs on the top of the wings when it tries to climb too steeply and at too low an air speed.  The loss of lift causes the nose to drop and the airplane to dive, until the falling and angle change gets the airspeed up and the airstream reattached to the top of the wings.  Then the plane can level out (if it hasn't started spinning...)  This is what makes toy gliders do a series of scallops if the wings aren't adjusted correctly for stable flight.


In a wind turbine the angle of attack is the angle of the APPARENT wind at the blade.  (Apparent wind is the vector sum of the actual wind and the blade's motion.)  Putting load on the shaft slows the blade, which raises the angle of attack.  Do this sufficiently and the blade starts to stall.  This happens at two times:


 - At startup, when the blade is stopped.  You have to have enough excess of torque over shaft load when operating as a DRAG rather than a LIFT turbine to get you spinning, get the wind attached, and get the resulting wind-produced torque up.  (Fortunately that's easy with a magneto turbine and a battery charging load, which doesn't significantly load the shaft until the voltage is up to the battery voltage.  It's tougher with resistive loads, which load the shaft as soon as it turns.  But set your cutin too low, even with a windcharger, and you may find your blades never get out of stall and up to operating power.)


 - At high winds (IF the alternator is strong enough to load the blades sufficiently).  The battery's load clamps the output voltage, which puts an increasing load on the prop as the wind continues to speed up and the charging current increases.  This causes the mill's RPM to increase more slowly than the wind, raising the angle-of-attack on the blades.  IF (BIG IF) the alternator is strong enough this eventually cause the blades to stall, reducing the torque and slowing the blades further in a positive-feedback loop, until the blades are turning slowly as a drag turbine.


Some commercial mills are designed to furl by stalling (rather than by turning away from the wind using a pivoting tail, air brakes, centrifugal brakes, pitch adjustment, or some other mechanism.)  But getting that tuned right is a tough design job - with thrown blades if you get it wrong on the high side.  And your mill runs away if you lose your load in a high wind.  But the big problem, IMHO, is that when a blade stalls it sheds vortices downwind, turning much of the lost power into sound.  And a TINY amount of power is a LOT of sound!  A stall-furling wind turbine sounds like a helicopter landing on your head.  Not nice for family and neighbor relations.  (The furling tail design is much more peaceful and easy to tune.  B-) )

« Last Edit: October 16, 2008, 05:26:46 PM by Ungrounded Lightning Rod »

Ungrounded Lightning Rod

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Re: Volts/Coil
« Reply #7 on: October 16, 2008, 05:30:28 PM »
... this eventually cause the blades to stall, reducing the torque and slowing the blades further in a positive-feedback loop, until the blades are turning slowly as a drag turbine.


Or, if the detachment doesn't go all the way to the front of the blade, until the blades are turning significantly slower as a much lower-powered lift turbine.

« Last Edit: October 16, 2008, 05:30:28 PM by Ungrounded Lightning Rod »