How to set a Windmaster for 600 black?

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dedeleco
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by dedeleco » 14/10/12, 17:00

I looked and thought about the little information given on this wind turbine, to have coherent information between them:

first we must pay attention to the different types Black 600 12v or 24V or 48V because the voltages produced vary according to the same report.

In your posts it is necessary to remember this tension to be sure that you have the same model, and to avoid possible confusion.

The values ​​are probably the optimal operating values ​​at mppt with maximum power given.

This mppt is that given by the wind to the wind turbine, which includes the power lost in the winding of the wind turbine is RxI ^ 2 which is not negligible, to see closely with the internal resistance, about R = 0,5 to 0,6 Ohms for that at V = 12V, 4 times more for model 24V and 8 times for that at 48V, as far as I guess the value if the total volume of copper wound remains constant.
With V multiplied by 2, I is divided by 2 and RI ^ 2 unchanged by R multiplied by 4 because it believes that the square
The 24V model is indicated as giving 12V to 1,8 at 2m / s of wind, without any vorticity, and therefore not at the top of an ordinary flag roof.

Another parameter is the speed of rotation which has the optimum must be proportional to that of the wind, more precisely, with the speed of the tip of the tapered blades, about 5 times that of the wind.
this is imposed by the blades to have the relative wind with the right incidence, without stalling (blade too slow) or negative impact which slows down, (blade too fast).
The vortices destroy the good incidence and the speed decreases to about that of the wind at least and the yield decreases of this factor 5 at least.

So this type of wind turbine with houses around, is often a scam as explained on the report:
https://www.econologie.com/forums/post242163.html#242163
http://www.youtube.com/watch?v=qLTCZMlgtJI


So the frequency of the generated current gives the speed of rotation which must be wedged to that of the wind for this optimum, with the obtained tension also proportional to this wind speed, about 30 to 35V for 10 to 11m / s of wind for the model 12V.
2tour / s = 120 per minute gives 10Hz as indicated in German, and for a diameter of 1,6m is 5m of circumference, we get 11m / s for max power 600W on paper sellers, 11m / sx5 / 5m = 11 turns / s is 55 Hertz which corresponds to a voltage of 29V on the 12V model, thanks to an indication of 24Hz for 12,6V in the forum German.which also allows to guess that the current flow is about 20 to 25 A (with the battery charge current of 12V function of this frequency).

The model 12V gives its max power to 30V for 600W = 30x20 and a little above, with at least I = 20 amps which gives the power lost in the coils of 0,6Ohmsx20x20 = 240W, or even more that quickly believes like the square of I .
And the alternator is going heat very strongly, even in this wind !!
So at 600W the wind actually delivers to the blades a mechanical energy of 600 + 250 = 850W

This power is obtained only at a minimum frequency of 55Hz, and a rotation speed of the blades of 11 revolutions / second.

If the wind turbine is working well under optimum mppt, it must run at a speed in turn / seconds equal to the wind speed Ve in m / s coarsely (with the ratio of 5 approximately, speed at the end of the blade of 5 times the wind speed ) .
It is impossible as soon as there are eddies that change the direction of the wind from 5 to 10 °, as on the video of the scams, because stalls (excessive incidence) or braking (negative impact).

To set, the optimum point mmpt is the one where the current I is believed to be the square of the optimum voltage U, which gives the obtained power P = UxI = U ^ 3 increasing as the cube of that Ve of the wind (Betz constant%) with I = ctxVe ^ 2 U = CtvVe

This U must be decreased by RxI of the voltage lost in the coils, which cuts high voltage, to properly evaluate.

So the 24V ends up at 11m / s of wind stable at U 60 at 70V, Imax 10 at 12A, and the 4 points approaching the curve in I = ct U ^ 2 roughly with straight segments, the first 2 zero to be able to start unloaded blades unhooked at low speed by compared to that of the wind.

So at half voltage 35V I can be Imax / '4 is 2,5A can be less
and at Imax / rac (2) Imax / 2 is 50V about 5 to 6A

This is valid for a place without whirlwinds at all, more than 10 times the height of the different disturbances, and if it is impossible, as for the ransacked, the value will be much lower, the blades remaining constantly unhooked at slow speed and yield divided by 5 at least.
Then the blades are totally unsuitable, too often unhooked.

It is good to measure the frequency of the alternator current (or that of the blades) with oscilloscope or multimeter, as well as the voltages and current, which gives internal R exact, to refine these rough estimates from partial measurements on the German site, charging on variable resistance of power or battery (a simple heater of contactor can do the case of resistance).
The fluctuation of the frequency gives a clear indication on the wind vortices.

Hermann's values ​​seem to me inconsistent with those of the Black 600 48V model ???????????????
by comparing to other info on this forum !!
It will never happen to 600W.
It must be the reduction of U by RI intenren in the coils.

Edit: Hermann am 4.10.09 Wie sich jetzt herausstellt ist mein Black 600 nicht wie jeder andere 600er, deshalb ist es fraglich ob diese Einstellung
also für andere 600er gut ist.Bitte weiterlesen.
Vielleicht kann Generator un mal sagen bei welcher Umdrehung der 600er48V die 28,5 Volt erzeugt.
Das sind auf jeden Fall nur wenige Umdrehungen.
Die Einstellung 27 V für den unteren Punkt würde ich persönlich nicht wählen, Weil dies die eigentliche Abschaltspannung ist, dh bei deiner Einstellung fallen Ein- und Ausschaltpunkt zusammen, lässt ihm keinen Puffer.
Die max. Leistung wirst of the nicht erreichen, weil dein Black bei der von gewählten Spannung noch nicht die erforderliche Leistung hat.
Vielleicht kann Generator a bad sagen bei welcher Umdrehung der 600er48V die 28,5 Volt erzeugt.


Logically the 48V rises more than twice as the 12V, to manage to charge the battery of 48V (or 12V respectively), but the internal resistance R (to measure !!) decreases in appearance U of RI, high voltage, this which is to be evaluated, probably more than 300W of lost in the coils for a high internal resistance.
On the forum we find a lot of inconsistencies!
Understanding by measuring characteristics carefully avoids mistakes.
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Ruthenian
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by Ruthenian » 15/10/12, 07:19

Hi Dedeleco and thank you very much for these explanations.

Can you tell me exactly what to measure (in situation)?

My version is 48 V.
Yesterday on a burst of 39.2 Km / h is almost 11 m / s so the nominal value (normally to 600 W), the windmaster has produced instantly 187 W peak.

I changed the setting this morning to let the blades get faster.
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dedeleco
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by dedeleco » 15/10/12, 14:46

I had written long and I thought I was clear?
But physics is not so simple.

It is necessary to measure the frequency of the voltage of the alternator (or the speed of rotation of the blades and roughly check the factor 5 between the two, like 10Hz => 2tours / s to be sure that what the German site indicates is correct )

Measure the internal resistance of the windings, also the global one, either variation of voltage of alternator output rectified, on variation of current with a change of resistance of load at fixed rotation speed.

This change fixed wind on the contrary leads to a decrease in the speed of rotation which indicates the coupling with the wind, and which makes it possible to determine the optimum point mppt, ie the frequency (or t / s) for each wind, in principle proportional to the wind, at the speed of the wind. tip of the blades about 5 times that of the wind !!

So Frequency (or t / s), voltage, and current, alternator output, on a variable resistive load, depending on the variable wind speed will give full info.

Having the same thing with the alternator at different rotational speeds is also a plus, but it can be removed from previous measurements.


At 11m / s we should have wind 600W without vortices (stable wind vane orientation better than 5 ° to 10 °) !!.

The measurement of the characteristics on variable resistive load should make it possible to know if it is possible or impossible.

Variations DV / DI allow to measure the resistances directly (on any heating apparatus).
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coccigro
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by coccigro » 16/10/12, 18:16

Hello,

my windmaster comes from airechologic, like br3. when the inverter does not run, it displays a consumption of 1 mA; as soon as it starts, it displays about 0,250 A and it increases when the wind turbine accelerates. for me it is therefore about production (well, I hope).

for my coubre, I prefer to return amp from the beginning because if I do not exploit the weak winds (for which the black 600 is intended) I lose much of my potential wind! if these values ​​do not exceed the actual values ​​of the wind turbine, I should not slow it down?

I specify that I also have an 48V version.

I still have trouble understanding where are the 600W maxi this wind. is it the direct current or alternating current coming out of the inverter? and why do we say 48V? it is obvious that we do not have 600W at 48V, I missed something.

what i would like to understand now is the relationship between what enters the windmaster and what comes out of it. with x V and XA in incoming DC, how many A's should 230V output in alternative?

I will try to decipher dedeleco, it will take me some time.

good evening to all.
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dedeleco
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by dedeleco » 16/10/12, 23:21

coccigro wrote:Hello,

my windmaster comes from airechologic, like br3. when the inverter does not run, it displays a consumption of 1 mA; as soon as it starts, it displays about 0,250 A and it increases when the wind turbine accelerates. for me it is therefore about production (well, I hope).

for my curve, I prefer to return amp from the beginning because if I do not exploit the weak winds (for which the black 600 is intended) I lose much of my potential wind! if these values ​​do not exceed the actual values ​​of the wind turbine, I should not slow it down?


As soon as you ask for energy from the alternator, energy coming from the wind, the wind turbine slows down, even to the optimal value for which the wind turbine was designed, with a power proportional to the cube of the wind speed ( Betz), up to a max, in principle P = 600W at V = 11m / s = 39,6km / h, with approximately the speed at the end of the blades TSR = 5 times that of the wind, info not specified and evaluated, crucial for the performance, and blade design. (see the basic courses in the links I put on econology, which I summarize)
In a vacuum the wind turbine turns very fast, enormous, just limited by the incidence of wind on the end blades which becomes negative, and brakes instead of pushing, and the blades can break, even at V = 11m / s, ( rather weak wind, the bike, without charging at all as in the Tour de France, 45 55km / h).
This is the reason for the anti-runaway brake.
So at the optimum of production the wind turbine is always slowed by the power taken.
The Amperes of I in function of the volts of U must in principle be those at the optimum mmpt of the wind turbine, ie I as the square of U, these volts U as the speed of rotation of the wind turbine, and it such that the end of the blades darken like 5 times the wind (5 is approximate, because not indicated, and fixed by the shape of the blades).
I specify that I also have an 48V version.

I still have trouble understanding where are the 600W maxi this wind. is it the direct current or alternating current coming out of the inverter?

The three-phase rectifier has a good efficiency and therefore the Power P supplied to 3 three-phase resistors is identical to that supplied to one of a single continuous resistance, after rectification, apart from the approximately half volt of the diodes, which becomes negligible at U> > 0,5 to 0,7V from the diodes.
read the courses:
http://fr.wikiversity.org/wiki/Redresse ... phas%C3%A9
http://fr.wikipedia.org/wiki/Courant_triphas%C3%A9
http://en.wikipedia.org/wiki/Three-phase_electric_power
in infinite number on google:
https://www.google.fr/search?num=100&hl ... fubn8v1kus

So at 5 10% is the same, the difference is what is heating the rectifier!
Attention 3 resistors mounted in three-phase, according to the prices, before the rectifier and only one continues after.

and why do we say 48V? it is obvious that we do not have 600W at 48V, I missed something.

To read the forum German, the 600-48V starts to give a little current to a voltage half of 48V, at U = 24V with wind at V = 1,8m / s and 48V significantly recharges a battery above 48V, at a few watts, for V of the wind, twice 3,6m / s = 2x1,8 approximately, but not to 600W, with a frequency of 24Hz = 4,8t / s by reading the scattered infos of forum.

To guess by cross-checking, on this forum, the 600W should be produced, for V = 11m / s (or 10) is, by rule of three, passing from V = 3,6m / s which gives Ui = 48V to V = 11m / s which then gives Ui = 48 / 3,6x11 = 146V induced in the alternator, to which, it is necessary to subtract the voltage lost in the resistance R of the winding of the alternator is Up = RxI which gives the U measured at the output of the alternator U = Ui-RxI

AV = 11m / s, the frequency must be 24x11 / 3,6 = 73Hz = 14tours / s very roughly, because the speed at the end of the blade also gives a lower frequency on diameter 1,6m, 11 turns / s is 55Hz, probably more exact?
But then the voltage Ui evaluated by the frequency ratio (instead of the wind) gives Ui lower with 24Hz ---> 48V then Ui = 48x55 / 24 = 110Volts much lower.

So you have to know R and I at V = 11m / s on the optimal load which gives the speed of the end of the blades to 5 times that of the wind.
This ratio TSR = lamda = r = 5 is taken because typical often, to have a valid% of Betz.

There is a problem on the 600W, is that which comes out of the alternator P = UxI = UixI-RxIxI usable, or is this UxI which is given by the wind to the alternator including the RxIxI lost by the law of Ohm to heat the windings of the alternator ????????

If it's the one given by the wind, 600 = UixI with Ui = 146Volts to 11m / s
then gives I = 600 / 145 = 4,14A and a dissipation in an R to verify, but estimated at 0,5x4 ^ 2 = 8ohms (variation of DU / DI after rectifier on variable resistive load, to measure it and the 0,5 are the R = 0,5Ohms of the Black600-12V given on the forum )
which gives a power dissipated in the windings of RxI ^ 2 = 8x4,14x4,14 = 137W

With Ui = 110V this gives I = 600 / 110 = 5,45A and a much higher winding dissipation of RxI ^ 2 = 8x5,45x5,45 = 238W much stronger, in my opinion probable that warms up the coils.

If the 600W given at the output of the alternator, the wind is forced to give more 600W + RxI ^ 2 = 600 + 230 = 630W roughly, significantly stronger, which gives I = 830 / 110 = 7,55A and then a corrected dissipation in the windings of RI ^ 2 = 8x7,55 ^ 2 = 455W in the windings, and then the output drops with much less than 600W at the output (830-455 = 374W)

This is a detective work, to specify with measurements in real situation, R windings, UI frequency f (or t / s) with different resistive loads after recovery, depending on the wind speed , which makes it possible to check the factor 5 = lamnda = estimated TSR, namely !!
read with grans care the basic course:
http://eolienne.f4jr.org/eolienne_etude_theorique

The higher this TSR is, the better, but less, the slightest whirlwind is tolerable, and the TSR plummets to 1 in the vortex regime and the resulting power.
So much to take anything for the blades, even sails!

A machine without this minimum of information is almost a scam.

More serious theses:
http://pastel.archives-ouvertes.fr/docs ... ourieh.pdf
http://pastel.archives-ouvertes.fr/docs ... dobrev.pdf
etc ...

what i would like to understand now is the relationship between what enters the windmaster and what comes out of it. with x V and XA in incoming DC, how many A's should 230V output in alternative?

I will try to decipher dedeleco, it will take me some time.

good evening to all.


The power is kept between input and output with a probable return from 70 to 80%:
IxV continues to enter 0,8 = Iacc / 2 x Vacc / 2 x x1 / 2

Iacc and Vacc AC, the peak-to-peak DC values ​​on oscilloscope = 2 times the amplitude Ia or Va and rac (2) of the mean of the sinusoid squared of I ^ 2 in time.
600W to 230V effective gives 600 / 230 = 2,6A effective.

see the database:
http://fr.wikipedia.org/wiki/Courant_alternatif

Do not have an allergy to wikipedia, like somebecause wiki and internet avoid buying a lot of textbooks, expensive to 50 € to 100 € each, or more, or to go to class years!


Finally having listened to A5 and Séralini, if you are with GMOs in your body, you must take in excess, you will die less, at least men like male rats!
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by Ruthenian » 19/10/12, 22:28

Coccigro, do you have results?
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coccigro
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by coccigro » 23/10/12, 21:43

Good evening,

no result with as little wind.

I am now comparing the amps that go to those coming out of the UPS. I feel that it will be interesting.

The power is kept between input and output with a probable return from 70 to 80%:
IxV continues to enter 0,8 = Iacc / 2 x Vacc / 2 x x1 / 2


IxV x 0,8, I understand but the rest, I have trouble. I continue to search.

for now, I have a big difference at startup. Could it be that the 0,252A displayed by my ammeter at the start of the fix are not a usable current? what would explain the difference between my ammeterre and the meter?

good night.
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Ruthenian
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by Ruthenian » 23/10/12, 22:01

I noticed something really weird with the EDF approved Sagem meter. At startup and until 80-100VA is generated, the counter displays these values.
When the solar panels inject current via another inverter but counted on this same meter, the shift between them does not take place.

Explanation if the panels then produce 150 VA, the counter displays 150 VA and not by 240 VA extrapolation.
Subsequently, if there is wind production, absolute cumulation is consistent.

Do not understand this shot!

FYI, when the PV inverter starts (it's not a mastervolt) the Sagem does not show me 80-100VA:

Statements for explanation:

Image
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dedeleco
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by dedeleco » 24/10/12, 11:43

Mystery ??????

Unless an inverter gives a 220V not at all sinusoidal displeased EDF, which can no longer count quietly and goes on strike ?????????????

To be checked with an oscilloscope by measuring everything, V and I.
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Ruthenian
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by Ruthenian » 24/10/12, 21:25

Would it not be by chance reactive power?
How can we shake it?
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