Experimenting with Windings: Maximum Voltage Output?

[Entrepreneur]

I attach an image of something I am experimenting with.

The picture itself (and added text) should be self-explanatory.

1) Which way should my windings (clockwise or counter-clockwise) go around P1; P2; P3 & P4 for maximum voltage (V) ?
[ 1 Ø ]
[ 1 x single enamelled copper conductor ]
[ 2.5mm² ]
[ 65 turns per P1 - P4 each ]
[rotational speed = 9 revs/sec]

2) Is there a method of calculating Vmax & I max outputs from the above ?

3) Feel free to email me at # # # # at # # # # dot com

 

[Baluncore]

I attach an image of something I am experimenting with.
The picture itself (and added text) should be self-explanatory.

Is it supposed to be a two phase alternator? Will you rectify the AC from the windings to produce DC?
If so, the gaps between the poles are much wider than sensible. Better utilisation of materials if closer together.
The coils should be on the stator, not the rotor. The coils will tend to be thrown off that rotor, so better to have a magnet that rotates with stationary windings and no slip rings.
At low RPM you will need many more poles. Fisher & Paykel smartdrive washing machines have 48 poles.

 

[Entrepreneur]

Is it supposed to be a two phase alternator? Will you rectify the AC from the windings to produce DC?
If so, the gaps between the poles are much wider than sensible. Better utilisation of materials if closer together.
The coils should be on the stator, not the rotor. The coils will tend to be thrown off that rotor, so better to have a magnet that rotates with stationary windings and no slip rings.
At low RPM you will need many more poles. Fisher & Paykel smartdrive washing machines have 48 poles.

1 Ø = single phase (given)
Yes, but AC rectification is not important at this stage (in Question 2 Vrms & Irms still need to be calculated)
the gaps between poles are not important at this stage
the coils ARE on the stator (rotating outer cylinder with magnets = rotor)
stator is where the windings are (P1 - P4) (stationary)

So back to my question :
Which way should my windings (clockwise or counter-clockwise) go around P1; P2; P3 & P4 for maximum voltage (V) ?

 

[Baluncore]

Vmax will be dependent on the number of turns per pole, the spacing and dimensions of the poles, the strength of the magnets and the specified RPM.
Your rotor and stator seem to be designed to produce two phases.

 

[Entrepreneur]

Vmax will be dependent on the number of turns per pole, the spacing and dimensions of the poles, the strength of the magnets and the specified RPM.
Your rotor and stator seem to be designed to produce two phases.

Yeah, it may look like it ?

So I guess you don't know about the winding direction ?

 

[jim hardy]

Six pole stator, four pole rotor... sounds more like a stepper motor than a generator.

So I guess you don't know about the winding direction ?

What does right hand rule tell you ?

 

[Baluncore]

Yeah, it may look like it ?

So I guess you don't know about the winding direction ?

I know it cannot be wound sensibly as a single phase alternator.

 

[Entrepreneur]

I know it cannot be wound sensibly as a single phase alternator.

I concede - it cannot be wound as a sensible alternator / generator, but that is all I have to work with.

A 6-pole rotor and a 4-pole stator.

 

[Entrepreneur]

Six pole stator, four pole rotor... sounds more like a stepper motor than a generator.

What does right hand rule tell you ?

Jim, the rotor is the outer part. [6-pole rotor & 4 pole stator.]

The 6 x neodymium magnets are fixed to the inside of a cylinder which rotates (9 rev/s or 540 rev/m) around the 4-pole stator which is fixed on a shaft.

OPTION A
The idea is to get the maximum voltage and current (within limitations) from this configuration (preferably single phase).
At this stage my stator is wound CCW-CW-CCW-CW and I get between 10Vrms & 15Vrms from the output - but the moment I put a load (10V globe) the output halves. I suspect that the windings at present is wrong and that a configuration of CW-CW-CCW-CCW would be more suited.

OPTION B
I suppose building a new alternator / generator would be the ideal (discarding the 4-pole stator & 6-pole magnet), but I would be limited by the rotation speed, diameter size of cylinder (usable stator plate size between shaft & inner cylinder). Rectification of the Vac is only done after I can establish a decent voltage and current output.

Your thoughts (and Baluncore's) would be appreciated.

 

[Asymptotic]

I get between 10Vrms & 15Vrms from the output - but the moment I put a load (10V globe) the output halves.

Out of curiosity, what is a 10V globe? What was the loaded voltage, and how much current was flowing?

 

[Entrepreneur]

Out of curiosity, what is a 10V globe? What was the loaded voltage, and how much current was flowing?

Yes, I was as surprised as you are - my wife happened to find a small incandescent 10Vac globe. The loaded (output V) was 8.5Vac - made the globe glow - when I measured the V while glowing it was 4.5V ?

 

[Entrepreneur]

Yes, I was as surprised as you are - my wife happened to find a small incandescent 10Vac globe. The loaded (output V) was 8.5Vac - made the globe glow - when I measured the V while glowing it was 4.5V ?

i didn't measure AC current because of the limitations of my multimeter.

 

[Asymptotic]

Incandescent lamp resistance is very non-linear. When the filament is cold, resistance is only a few ohms, and it rises to a much higher value at full temperature. Voltage collapse suggests more current is flowing than the alternator as presently configured can provide.

 

[Entrepreneur]

Incandescent lamp resistance is very non-linear. When the filament is cold, resistance is only a few ohms, and it rises to a much higher value at full temperature. Voltage collapse suggests more current is flowing than the alternator as presently configured can provide.

Totally off topic, but interesting nonetheless. Thank you.

 

[Asymptotic]

Totally off topic, but interesting nonetheless. Thank you.

Your alternator load is an incandescent lamp. In what way is it off topic to consider how an incandescent lamp operates?

For example, filament resistance of a 100W, 120V tungsten bulb is approximately 144 ohms (I=E/R, 0.83 amps. R=E/I, 144 ohms) at normal operating temperature. Measured filament resistance at room temperature is 9.5 ohms. When 120V is connected to it, for a very brief time (in the low millisecond range) current surge approaches 120V/9.5 ohms (12.6 amps), then as the tungsten heats up, resistance increases, and current drops off.

You didn't provide the test lamp's power rating. For example, if it is 10 watts, then current will be 1 amp (I=P/E, 10W/10V = 1A), with a hot resistance of R=E/I = 10/1 = 10 ohms. Resistance decreases, and current demand increases the closer the filament gets to room temperature (where it is probably less than an ohm).

The loaded (output V) was 8.5Vac - made the globe glow - when I measured the V while glowing it was 4.5V ?

What was the difference between when output was 8.5V, and when it was 4.5V?

If your multimeter doesn't have an AC current range then wire a small value power resistor in series with the load, measure voltage drop across it, and calculate load current using Ohm's law. For instance, a 0.1 ohm, 10 watt resistor will drop 500 mV across it at 5 amps of current flow (E=IR, 5A*0.1Ω = 0.5V).

 

[Baluncore]

I concede - it cannot be wound as a sensible alternator / generator, but that is all I have to work with.
A 6-pole rotor and a 4-pole stator.

Then wind it as a two phase alternator, with (P1+P3) and (P2+P4).
Rectify each phase independently with a bridge rectifier, sum the DC outputs by connecting the DC outputs in parallel.
The DC output voltage will not fall to zero like a rectified single phase alternator.

 

[CWatters]

At this stage my stator is wound CCW-CW-CCW-CW

How are you viewing the coils and how are they connected together? Can you mark up the drawing?

 

[jim hardy]

Here's the thought line i was playing with

There will be leakage flux through tips of P2 & P4 , between adjacent magnets, that i didnt draw because it won't couple your windings.

so it seems there's likelihood it'd work as pair of two pole rotors.
Connect opposing poles in series , wound same direction. Note a single winding is CW looking into one end, CCW looking into the other.

Energy conversion will flip-flop between even and odd poles.

I think that's what you have .I'd sure like to see the waveform that thing makes. Have you an oscilloscope?

old jim

 

[Entrepreneur]

Your alternator load is an incandescent lamp. In what way is it off topic to consider how an incandescent lamp operates?

Original question was winding direction.

For example, filament resistance of a 100W, 120V tungsten bulb is approximately 144 ohms (I=E/R, 0.83 amps. R=E/I, 144 ohms) at normal operating temperature. Measured filament resistance at room temperature is 9.5 ohms. When 120V is connected to it, for a very brief time (in the low millisecond range) current surge approaches 120V/9.5 ohms (12.6 amps), then as the tungsten heats up, resistance increases, and current drops off.

You didn't provide the test lamp's power rating. For example, if it is 10 watts, then current will be 1 amp (I=P/E, 10W/10V = 1A), with a hot resistance of R=E/I = 10/1 = 10 ohms. Resistance decreases, and current demand increases the closer the filament gets to room temperature (where it is probably less than an ohm).

it was a Fong Kong make - no instructions / brochure. ...and it is now obsolete (removed by the garbage collectors - broken)

What was the difference between when output was 8.5V, and when it was 4.5V?

I only measured V prior to load and V with the load.

If your multimeter doesn't have an AC current range then wire a small value power resistor in series with the load, measure voltage drop across it, and calculate load current using Ohm's law. For instance, a 0.1 ohm, 10 watt resistor will drop 500 mV across it at 5 amps of current flow (E=IR, 5A*0.1Ω = 0.5V).

Didn't consider that - good tip!

 

[Entrepreneur]

Then wind it as a two phase alternator, with (P1+P3) and (P2+P4).
Rectify each phase independently with a bridge rectifier, sum the DC outputs by connecting the DC outputs in parallel.
The DC output voltage will not fall to zero like a rectified single phase alternator.

Sound like a do-able idea - thanks!

 

[Entrepreneur]

Here's the thought line i was playing with

View attachment 204913

There will be leakage flux through tips of P2 & P4 , between adjacent magnets, that i didnt draw because it won't couple your windings.View attachment 204914

so it seems there's likelihood it'd work as pair of two pole rotors.
Connect opposing poles in series , wound same direction. Note a single winding is CW looking into one end, CCW looking into the other.

Energy conversion will flip-flop between even and odd poles.

I think that's what you have .I'd sure like to see the waveform that thing makes. Have you an oscilloscope?

old jim

Jim

This was a good proposal and indeed this is what I will do. After measurement etc. I will post the results.

I will take a picture of my oscilloscope waveform (I am just as curious to see the wave form) and submit.

I am NOT a physics / electrical boffin, and many have told me the "experiment" will not work (taking a conveyor idler/roller and retrofitting it so that it can deliver power) - yet I persisted - and so far I have disproved them (albeit a miniscule output) - my "bulldog" approach is driven by my persistence to start manufacturing these "rollers" for the mining industry supplying them with an alternate source of electricity along conveyors.

I am NOT giving up - yet. Ha ha ha

The parameters (limitations) of the rollers have thus far proven to be a GREAT challenge and it seems I may have to RE-Design from scratch - a 12-pole rotor?

The mathematics is incredible and I had to brush up on ALL my studies (30+ years ago) and design a more efficient electrical machine.

Anyway Jim, I'LL keep this site posted with my re-designed experiment.

Thanks for your (and others') much valued input.

 

[NTL2009]

... I am NOT a physics / electrical boffin, and many have told me the "experiment" will not work (taking a conveyor idler/roller and retrofitting it so that it can deliver power) - yet I persisted - and so far I have disproved them (albeit a miniscule output) - my "bulldog" approach is driven by my persistence to start manufacturing these "rollers" for the mining industry supplying them with an alternate source of electricity along conveyors.

I am NOT giving up - yet. Ha ha ha ...

I can't imagine why they would say that it cannot work. But you have not addressed the most important parts.

How much power do you need to produce for this to be effective? A few watts for night lights, or hundreds of watts for equipment?

How much power can these idler rollers provide before they lose traction with the conveyor or load the driving motor/engine?

A review of those numbers will help inform you if this is realizable and how to size the equipment. And why build your own generator? You should be able to find one and gear it or direct drive it to the roller.