Wind Turbine 'Battery-charge' algorithm

[Usman]

I am investigating the 'load-effects' of battery over a PMA. The question pop-up when I was ensuring that the rotor is free enough (not over-geared to spin at any wind speed) to start-up, build momentum rpm (inertia) until it reaches the battery-voltage or 'cut-in' voltage when it the batteries pose a load to the turbine to charge up.

I am not sure if there is a standard formulating or technique to confirm if a certain designed turbine is free enough to spin in light winds and wouldn't stall when batteries exert a working load over it's turning force in light winds. My quests:

1-Am I correct in my conception about the batteries loading nature i.e. they exert load or a resistive force to rotor motion only when the preexisting battery voltage is surpassed- and the batteries start receiving charge? If so, how strong is that resistive force on the turbine i.e. is it proportional to the charge status of battery, or size of the battery-pack, or varies with the type of battery etc....I am only worried that my gearing might be too high to pose a stalling at any point, especially in light winds as the spinning rotor reaches the battery cut-in voltage!

Even if the rotor doesn't stall, it may experience a sudden jerk and the rotor rpm would immediately cut down from 30 rpm (for example) to 10 rpm and may never exceed that value until the wind surpasses certain speed. So, no useful charge may be received in low-medium wind speeds, even if the rotor was turning or struggling to pass that limiting force.

Thanks.

[Ungrounded Lightning Rod]

Excluding some minor losses (such as eddy current drag):

Essentially all the torque resisting the shaft in a PMA is the direct result of (and directly proportional to) the current delivered to the load.  This current produces a magnetization in the stator which attracts the permanent magnets toward their previous positions, resisting their motion as they attempt to move forward.

It's easy to compute:

 - Generated voltage (NOT voltage at the terminals!) is directly proportional to RPM.

 - That voltage drives current through a circuit consisting of:

    - The resistance of the coils, wiring, and battery internals.

    - The voltage drop of the rectifiers.

    - The voltage of the battery's chemistry.

 - Coil and wiring resistances are significant and essentially independent of current (so current is directly proportional to voltage across the resistance).  Battery internal resistances are very small.

 - Rectifier voltage drop is a function of current, but it's got a very steep slope in the operating region, so you can treat it as a constant voltage drop of about 1.2V for a full-wave bridge and the diode's behavior as an otherwise ideal rectifier (one-way valve).

 - Battery chemistry voltage drop varies essentially only with state of charge.  (There's a current-dependent component but you can treat that as a very small resistance and lump it with the wiring.)

So treat your system as:

 - A generator with output voltage proportional to RPM feeding a series combination of

 - A resistor

 - A fixed voltage drop, and

 - A one-way valve

You'll see that you have no torque fighting the rotor until you reach cutin RPM (which varies slightly with state of charge), then the torque rises in essentially direct proportion to the RPM EXCESS above the cutin speed.

If you could wire it with superconductors and ideal rectifiers and batteries the mill would free-wheel up to cutin speed and then never go ANY faster - resisting harder (and charging more) as the wind pushed harder.  With no resistance the current would rise to arbitrarily high levels (and thus produce arbitrarily strong resisting torque).  But because there is nontrivial resistance in the circuit, you must generate excess voltage to drive the current through that resistance.  The current (and thus the torque fighting the rotor) is proportional to the excess voltage (and thus excess speed) of the prop above the cutin speed.

Got it?

[DamonHD]

Lovely simple explanation...

Rgds

Damon

[TomW]

usman;

I have to wonder where you come up with these questions that seem to hop from one system to another on every post and no linearity I can see?

Are you designing a commercial product or something?

None of it seems geared to DIY.

Just curious myself.

Cheers.

TomW

[Usman]

Thanks for the very precise feedback; the last paragraph seemed more understandable to me than the above physics. I am aware of most of that but was questioning towards the behaviors of the wind turbine in a battery-charging applications.

A few more bullet points on my project, that also answers TomW quest, I am just designing a turbine for low wind regime and that's also a part of my academic interest. I need to be able to answer the most miniature questions, even not-thought of technical points, so that I could avail my masters. Anything that comes across my mind or from my fellow colleagues, juniors or seniors, I attempt to put that in a sensible format for comments from the board members. Does that answer your question, TomW.

My design a 48V~5 KW turbine with a 3-bladed rotor diameter of 6.3m (20-feet), that I availed from a local fiberglass firm -professionally built as per the airfoil patterns, which has a unique drive train design of turning two PMAs, one 3KW(rated 400 rpm -Asian origin) and the second 2KW (rated 300rpm-also Asian product), which are connected to the rotor shaft with a 1 to 3.5 rpm gearing . I am supposing that the turbine would achieve a rating of 5KW at only 100 rotor rpm (at around 7.5 m/s) AND up to 8KW at 9m/s when the rotor is turning at 150 rpm.

As you say that the rotor speed stays constant after the cutin, then why do we see a battery charger turbine's speed increase and decrease so frequently? It doesn't stay the same and keeps on increasing and decreasing smoothly.

My understanding is that beyond the cutin rpm ( I would name it as an overhead rpm value -  any ideas on calculating that) , the rpm is directly proportional to the wind speed.

Regarding the information on my turbine, do you see any sensibility in the ratings? If these don't tele much, then how would you rate or imagine my turbine in terms of rotor rpm, cut-in wind speed, general idea of gearing (if its right), battery cutin rotor rpm etc...I know there is a lot more information needed on the PMAs and the blade profile etc...but generally any reference or presumptions would help.

Generally a 3-bladed 20-footer commercial battery charger turbine would be rated at 7 to 9-KW in 12 m/s winds, I am driving my 5KW @ 7.5 to 8m/s -calculations or (more appropriately) presumptions by comparing is to commercially available turbines.

Any ideas, suggestions and recommendation would help.

Thanks.

[Ungrounded Lightning Rod]

As you say that the rotor speed stays constant after the cutin, then why do we see a battery charger turbine's speed increase and decrease so frequently? It doesn't stay the same and keeps on increasing and decreasing smoothly.

Please re-read the end of my post.

I said it would stay constant after cutin IF IT WERE MADE OF SUPERCONDUCTORS.  Because it's made of copper, which has SIGNIFICANT RESISTANCE, you need some extra speed to make extra voltage (beyond the battery voltage and rectifier voltage drop) to push current through the resistance.

So the graph of rotor speed to wind speed has a knee in it.  Up to cutin it rises with wind speed.  After cutin it rises more slowly than wind speed.

(Also:  Because the drag from the shaft changes the direction of apparent wind at the rotor, aerodynamic effects cause the behavior of the rotor to be non-linear with load, making the above-cutin graph something other than a straight line.  In particular: At high enough windspeed the airstream detaches from the rotor blades and you lose most of the lift.  So the curve actually drops again.  This is how "stall furling" works.)

[Flux]

I am going out now and don't have time to reply.

ULR has covered the basic questions that you asked very clearly. If you want to understand a bit more then I did a diary entry about matching the load and although it was a long time ago I think I covered all this in the first part of it. Sorry I can't find a reference but others have posted links to it so google search may pick it up.

Flux

[Usman]

Thanks ULR,

Is there a way to determine the cutin voltage of a generator, which I sourced from Chinese supplier? No ratings on it except it says rated rpm 300, rated power 3KW and rated volts 48V?

Is 300 rpm the cut-in speed as well, if we apply the same formulation of your description, but it also mentions on the nameplate; max.power-4.5KW. I contacted the supplier but no information is available. Any idea on how to calculate or practically determine cutin volts?

Thanks.

[Usman]

Thanks FLUX for your input.

Google search brought up no rsults, any clues?

Manufacturers rate their turbines at a certain rpm but mention that the battery charging starts at (for example) 500 rpm and rated rpm at rated wind speed of 12m/s is 1800 rpm. I presume that 500 rpm is the generator cutin with the battery-voltage- right? So why doesn't it clamps upto that cutin battery voltage and rpm rather than exceeding two or three folds?

Please also lookin into my new positing (related to this subject);"How to rate 'RPM' for any given Wind Turbine"

Thanks.