Author Topic: unsure about cut-in  (Read 3410 times)

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artv

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unsure about cut-in
« on: May 08, 2010, 12:42:13 PM »
here comes another one ...does the turbine start generating as soon as it begins to rotate (but very small voltages) but needs to reach a certian speed to achieve a high enough voltage to begin charging...is this cut -in speed.....thanks ...artv

wpowokal

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Re: unsure about cut-in
« Reply #1 on: May 08, 2010, 09:48:18 PM »
artv you are like a person hyperventilating you ask multiple question all variations on a theme and all in new threads, if you think of a new question or want a point clarified put in ONE post it will jump back to the top, just my opinion.

To answer your question, yes as soon as a magnet moves past a coil a voltage is generated, the voltage increases linear with speed, until this voltage exceeds battery volts plus diode  forward voltage drop, no current flows. Unfortunately it is AC although what you have against AC I can not work out.

allan
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artv

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Re: unsure about cut-in
« Reply #2 on: May 08, 2010, 11:42:39 PM »
Allan you say ..."as soon as a magnet moves past a coil a voltage is generated"....this is one set of magnets (in a dual rotor)...when the next set of mags move past the coil (the same coil) a voltage is generated ...but in the opposite direction.....this is AC ??.The way I see it ,it takes twice as much effort to make AC as it does DC ........I know theres' a reason I'm wrong but I can't see it................artv

wpowokal

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Flux

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Re: unsure about cut-in
« Reply #4 on: May 09, 2010, 02:53:17 AM »
It's just a basic of electromagnetic induction that things turn out to be ac. The flux has to be changing to induce a voltage. Oersted discovered the magnetic effect of a current in something like 1820 everyone soon realised that the reverse effect must be possible but no progress was made until Faraday's experiments. Like you the rest assumed that there would be some steady state situation where dc would be induced in a conductor.

Faraday proved conclusively that the effect only happened when there was a change of flux linkage. his experiments led directly to the alternator and the transformer. His first generator was a homopolar device with a disc and sliding contact, it is very difficult to extend that idea beyond a single conductor and to do so you need multiple sliding contacts. The concept never got anywhere and all real progress has been in the alternator direction.

It is no big issue having ac, all you have to do to charge a battery is to change the connections over every half cycle so the external current always flows in the same direction. No more effort is needed to generate ac, the work involved is not influenced by the direction of the current.

There are still chances for minor improvements but the concept will never change and I have come to the conclusion that you are chasing some sort of improvement that can only be obtained from ignoring basic principles. This is the over unity concept and it just doesn't work.

Have a go at defeating Betz that poses the biggest inefficiency to wind turbines. Again there may be room for manoeuvre , I doubt that the 0.593 figure is set in stone, it's based on a very simple concept, but even so it won't be far wrong.

Flux

artv

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Re: unsure about cut-in
« Reply #5 on: May 09, 2010, 07:24:42 PM »
Hi Flux.....Faradays' law states "The voltage induced in the conductor is directly proportional to the rate at which the conductor cuts the magnetic lines of force"....The four factors - strenght of field, speed of conductor with respect to field, angel which conductor cuts field (90 degrees optimum),and lenght of conductor. Are these statements true and are these not the basic principles? I haven't heard of Betz, but will  google it. As for over unity this will never happen ,we'll get close but it is immpossible to achieve can't get something thats not there. Also wasn't Edison setting up dc stations before ac came along and that Tesla actually worked for Edison Co. which inspired him to create ac??  not really sure though...........artv

Flux

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Re: unsure about cut-in
« Reply #6 on: May 10, 2010, 03:25:31 AM »
Hi Flux.....Faradays' law states "The voltage induced in the conductor is directly proportional to the rate at which the conductor cuts the magnetic lines of force"....The four factors - strenght of field, speed of conductor with respect to field, angel which conductor cuts field (90 degrees optimum),and lenght of conductor. Are these statements true and are these not the basic principles? .......artv

Yes sort of. That goes a bit beyond Faraday and includes others interpretation of the concept.

Faraday said it was rate of change of flux linkage. Obviously the strength of the field is a major factor and the other factor is the rate of change of that field. In a conventional alternator it is really the frequency that is the other factor.

This depends on the number of poles and the rotational speed. If you start introducing the length of the length of conductor and the angle to the field you are getting back very close to that stupid e = Blv equation that causes so much confusion. The only way the conductor length really matters is in that it determines the loop size. The angle again is really confusing and really only implies that you link the loop in the shortest time.
 I would forget all about coil sides and angles. If you stick to absolutely conventional machines then it is sort of true but very confusing. When you stray into air gap machines then it will lead you down a silly path.

Flux

artv

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Re: unsure about cut-in
« Reply #7 on: July 24, 2010, 12:37:16 AM »
Another old thread .......so before cut-in the voltage thats produced is just wasted.?.......Why not just charge smaller load .then when mill increase's .....charge bigger load...........simple switching devise should sufice....artv

ghurd

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Re: unsure about cut-in
« Reply #8 on: July 24, 2010, 08:20:24 AM »
You are confusing open circuit voltage for power.
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artv

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Re: unsure about cut-in
« Reply #9 on: July 24, 2010, 10:05:05 AM »
ghurd I searched for open circut voltage but don't understand is it like a power supply with no load say 12v battery not hooked to anything....? If cut in-speed is 7mph 12.7volts....but the wind never gets above 6mph would the gen be putting out say 10volts but not enough to charge the 12v battery...... totally confused...artv

kcchow

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Re: unsure about cut-in
« Reply #10 on: July 26, 2010, 03:48:49 AM »
Hi!

Further to the above discussion, I would like to seek your advice on the cut-in issue. Does cut-in means the voltage/current is start to charge to battery or it is just sufficient to power up the controller itself?

wpowokal

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Re: unsure about cut-in
« Reply #11 on: July 26, 2010, 08:29:10 AM »
Open circuit voltage is the voltage the generator produces when there is no circuit (it's not connected to any load) when the generator is connected to batteries via a blocking diode there is no circuit until the generated voltage (same as terminal voltage)  exceeds battery voltage and that necessary to overcome forward voltage drop of the blocking diode (typically 0.6v for power diode).

Terminal voltage will rise with generator speed until a current flows (circuit is made) after which it is basically battery voltage, now if one was to disconnect the batteries/load the terminal voltage will be proportional to generator speed, a very good reason not to leave the generator connected to the inverter when the batteries are disconnected, should the generator be able to supply more power than the inverter needs then the voltage rises and magic smoke escapes.

Artv your analogy of a 12v battery is basically what happens but wind driven generators are seldom at a constant voltage. If you were to attempt to draw power from the generator before it reaches cut in it may well not reach cut in. Wind turbines will generally accelerate quite rapidly up to cut in if wind speed is enough to be useful, many factors of wind turbine design come into play here.

kcchow cut in is achieved when current begins to flow to the load (batteries) as described above, controllers would generally be powered direct from the batteries and therefore not dependent on generator voltage.

I believe I have adequately clouded the issue for now.

allan
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wpowokal

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Re: unsure about cut-in
« Reply #12 on: July 28, 2010, 08:57:01 PM »
It occurred to me that I have left part of the explanation out in my effort to keep it simple, that is what happens to generated voltage once current begins to flow, this is where purchasing one of the books written on these duel rotors is a great help to newbies.

Once current flows winding resistance comes into play and it gets a bit complicated but simplifying it by only considering resistance, the generated voltage needs to rise above terminal voltage sufficient to overcome winding resistance, this is where stator heating comes into play.

Lets say your turbine is outputting 10 amps at 25V(terminal volts because it saves discussing line and diode losses) into your 24V batteries, that is 250 watts (VxA=W), say the winding resistance(here allan leaves out how this is calculated) is 2.5 ohms so losses in the stator expressed as watts is 10A x 10A x 2.5 ohms =250 watts ( current squared x resistance). Here we can see that if current was to increase to 20 amps output to load would be 20 x 25 = 500 watts but stator losses would be 20 x 20 x 2.5=1000watts.

Using OHMS law  V (volts) =  I (amps) x R (resistance) we calculate the additional voltage over terminal volts that the generator will need to produce to make these 10 amps flow. Volts = 10 (A) x 2.5 (ohms) which is 25, so generated voltage when the terminal voltage is 25V and current is 10 amps is 50V.

All of the above is simplified in the interest of the explanation but is adequate to give an understanding of the principles, others can give more depth if they wish. And of course correct me if I have strayed from the truth.

allan
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Ungrounded Lightning Rod

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Re: unsure about cut-in
« Reply #13 on: July 28, 2010, 09:30:42 PM »
.....Faradays' law states "The voltage induced in the conductor is directly proportional to the rate at which the conductor cuts the magnetic lines of force"....The four factors - strenght of field, speed of conductor with respect to field, angel which conductor cuts field (90 degrees optimum),and lenght of conductor. Are these statements true and are these not the basic principles?

True but:  If you work them all comes down to how much flux goes from one side to the other of the conductor (i.e. between inside and outside of a coil times number of turns) in a given amount of time that determines induced voltage.

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I haven't heard of Betz, but will  google it.

Basic idea is similar to the Laffer Curve with taxes vs. government revenue.
 - You get power from slowing down the wind.
 - The more you slow down a given mass of air, the more power you get from it.
 - But the more you slow down the air going through your mill, the lower the mass of air you can pull power from.
 - If you don't slow it at all you get no power because you left it all in the air.  If you stop it you get no power because no air goes through the mill - you have to take less than all there is to keep the mill running.  In between there's power, and at some point there's a peak in the curve.

Betz is the guy who figured out where the peak is (air leaves at 1/3 the speed it entered) and what percentage of power it represents (16/27, about 59.3%).  This is the "Betz Limit".

His simplifying assumptions are very close to the real world case and actual mills can essentially only deviate from them by being not as good (finite number of blades, nonzero air friction, air leaves with some spin that represents more lost energy, etc.).  About the only way they can do better than the assumptions is by affecting a little air just outside the "swept area", acting like they were slightly wider than they actually are.

So a "perfectly efficient" mill would turn 59.3% of the wind power into HP on the shaft and real mills fall somewhere below that.  "Beating Betz" in a windmill is like beating the Carnot cycle efficiency limit in a heat engine.  Don't bother trying.  There's more energy there but it can't be captured.

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Also wasn't Edison setting up dc stations before ac came along

Yep.  The generators used commutators (rotating switches on the shaft) to rectify the genny's AC to DC.  Downside:  Voltage is limited because they'll arc over if you go too high.  So you have to generate near the load (using the technology of that time) to avoid excessive resistive losses in the transmission system.  AC had the advantage that it could be stepped 'way up (and back down) with simple transformers, so it could be shipped long distances.  (Tesla teamed with Westinghouse to build what engineers had been dreaming about for decades:  A system to power New York City with waterpower from Niagra Falls.)

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... and that Tesla actually worked for Edison Co. which inspired him to create ac??

Tesla worked for Edison at first.  But Edison was heavily invested in DC and wasn't interested in developing and marketing Tesla's ideas.  (Foolish:  He could have used rotary converters to turn it into lower-voltage DC for his existing distribution system after it arrived.)

Ungrounded Lightning Rod

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Re: unsure about cut-in
« Reply #14 on: July 28, 2010, 09:33:51 PM »
Hi!

Further to the above discussion, I would like to seek your advice on the cut-in issue. Does cut-in means the voltage/current is start to charge to battery or it is just sufficient to power up the controller itself?

Cutin is where you just start to charge.

Ungrounded Lightning Rod

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Re: unsure about cut-in
« Reply #15 on: July 28, 2010, 09:47:24 PM »
Allan you say ..."as soon as a magnet moves past a coil a voltage is generated"....this is one set of magnets (in a dual rotor)...when the next set of mags move past the coil (the same coil) a voltage is generated ...but in the opposite direction.....this is AC ??.The way I see it ,it takes twice as much effort to make AC as it does DC ........I know theres' a reason I'm wrong but I can't see it................artv

Power is voltage times current.

If your genny is putting out some voltage but you don't pull any current it's like the outlet in the wall with nothing plugged in.  There's voltage there but your meter doesn't spin.  Similarly with a windmill:  If the generator doesn't put out a high enough voltage to push current through the diodes into the batteries, it's like your battery is not hooked up to it.

Also:  With no current flowing there's no "electrical drag" on the shaft.  So no mechanical power is pulled from the rotor.  (Except for some {typically very small} losses:  Bearing friction, heating from "eddy currents" {little parasitic circular currents in conductors near the magnets}, hysteresis losses if you have an iron core through your coils, etc.)