am new here of course,..I have only one question,....I have been kicking this idea around for quite a while and I just want to know if you think it may work as I invisioned here.
I have a fan (electric motor) in a ac ground unit.
It operates at 1075 rpms @ 2.6 amps.
I was thinking about setting a PMA above it set on brackets, with the shafts lined up and coupled.
The pma will generate 12.7 amps@950 rpms and is nearly coggless.
The weight of the pma(including shaft)would hover a 1/2 inch above the electric motor shaft on a bracket, thereby, the weight of the PMA is not adding weight to the electric motor.
My thoughts are that the electric motor that operates @ 1075 rpms,...when coupled with the PMA,..the rpm would round off to about 1000 rpm.
My question is,..can this work without burning up the electric motor?
If this will not burn up the electric motor,...then even though it takes 2.6 amps to operate the electric motor from the grid,..I am gaining basicly 10.1 amps excess that can be attached to a grid-tie inverter.
Your thoughts would be appreciated..thankyou
@returnedintime:
Isn't the 2.6 amp draw on the fan motor an "unloaded" draw measurement? I would expect that once you have the PMA coupled to the motor that it will pull ALOT more current.
truesearch
Thank You for the response.
Once the PMA reaches 950 RPMS,..it is nearly coggless(Very little resisance),...also, the PMA is brushless.
I would think that if the Electric fan motor were to draw more amps for operation,..it would not be much............sorta like a household box fan that has dust that has built up on the fan blades over time,...it does not blow as well,..but it still works for months before it burns the motor.
I would think that the fan would draw away heat from the electric motor when in operation,..and with the fan blowing on the PMA,...helps keep it cooled down.
If this would work,...one could turn this on at night and cause the meter to turn backwards
I would give it a try. Rotating magnets have been known to do strange things.
So it might just work.
.
Yes,..I have every intention of trying this,..just as soon as I can afford the money to put this together.
Hi,
Although the PMA can become cogless, unfortunately it does not mean it will represent a dragless load to your fan motor when you start loading the PMA with real (I mean heavier) load, to really take out 8-10 Ampers. Your PMA is an off the shelf device so that Lenz law surely manifests in it, unfortunately.
Gyula
Thank you for replying:
Could you please expand on your statement of..."Heavier Load"?
The electric motor operates @ 1075 rpm,....the PMA is designed for windmill generator,..it produces electricity at low speeds,..but it has been tested @950 rpms and will put out 12.7 amps
It is brushless,...the only restraint is the weight of the shaft and not the outter core,...but at 950 RPM,..it is nearly coggless.
I have no doubt that there will be a slight cog present,..but I do not believe it is enough to hinder the output.
The electric motor(Fan) Is @ 1075 rpm,......considering I figured there would be a slight drag,...that is why I anticipate the rpms on the Electric motor fan to round out @ 1000 rpms.
Electric motors have shown to continue running without burning up when drag is present.
Quote from: returnedintime on January 28, 2013, 06:39:45 PM
am new here of course,..I have only one question,....I have been kicking this idea around for quite a while and I just want to know if you think it may work as I invisioned here.
I have a fan (electric motor) in a ac ground unit.
It operates at 1075 rpms @ 2.6 amps.
I was thinking about setting a PMA above it set on brackets, with the shafts lined up and coupled.
The pma will generate 12.7 amps@950 rpms and is nearly coggless.
The weight of the pma(including shaft)would hover a 1/2 inch above the electric motor shaft on a bracket, thereby, the weight of the PMA is not adding weight to the electric motor.
My thoughts are that the electric motor that operates @ 1075 rpms,...when coupled with the PMA,..the rpm would round off to about 1000 rpm.
My question is,..can this work without burning up the electric motor?
If this will not burn up the electric motor,...then even though it takes 2.6 amps to operate the electric motor from the grid,..I am gaining basicly 10.1 amps excess that can be attached to a grid-tie inverter.
Your thoughts would be appreciated..thankyou
Hi returnedintime,
it would be nice if it could work. However, you have not mentioned how many
Volts your PMA puts out when it reaches 12.7 Amps at 950 rpm. It could be in the 12 volts range to keep a battery charged?... that would be about 150 Watts.
Now your fan motor probably works on 120 Volts AC and at 2.6 Amps it would come to a little over 300 Watts.
If you're low on cash I would not invest money in this, as many have tried this before. I wish it would be so simple :-\
Nothing wrong in thinking of alternative energy solutions but using off the shelf items chances it would work are next to none.
I wish you all the best in your search for a solution
Luc
Quote from: returnedintime on January 29, 2013, 06:59:54 PM
Thank you for replying:
Could you please expand on your statement of..."Heavier Load"?
The electric motor operates @ 1075 rpm,....the PMA is designed for windmill generator,..it produces electricity at low speeds,..but it has been tested @950 rpms and will put out 12.7 amps
It is brushless,...the only restraint is the weight of the shaft and not the outter core,...but at 950 RPM,..it is nearly coggless.
I have no doubt that there will be a slight cog present,..but I do not believe it is enough to hinder the output.
The electric motor(Fan) Is @ 1075 rpm,......considering I figured there would be a slight drag,...that is why I anticipate the rpms on the Electric motor fan to round out @ 1000 rpms.
Electric motors have shown to continue running without burning up when drag is present.
Hi,
On 'heavier load' I mean as follows: you use say a 12 Ohm power resistor as a load across the output of your alternator, the current draw will be 1 Amper if the alternator output voltage is around 12V. Now suppose you replace the 12 Ohm resistor with a 2 Ohm power resistor, this is what I call a heavier load because the current draw goes up to at least 6 Amper provided the alternator keeps its output voltage at 12V as earlier.
Some more notes: it would be good for you to feel the torque by your gloved hand on the shaft this PMA must receive from the wind or from any other prime mover when it rotates at 950 RPM and you take the 12.7 Amper out of it. Will the fan motor be able maintain that much torque on its shaft when the PMA is tied to it? The only correct answer you can get is to test this and see for yourself. I understand you do not belive us, this is natural.
rgds, Gyula
Here is some thing you all may find quite interesting.
In the video below,a guy has a standard generator and a standard motor.
He show's the motor and coupled generator winding up,then disconecting the power-with the generator output disconected.
Then he show's the motor and generator winding up ,then disconecting the batteries and conecting the generator output to the motor.
Now one would think that if the motor and generator both have losses,then the thing should wind down to a stop quicker?
But watch the video.
One has to wonder just how hard it would actualy be to get this to work-because this standard motor-generator setup is very close already
http://www.youtube.com/watch?v=hl57y-c_bWc
Quote from: tinman on January 30, 2013, 06:18:10 AM
...
Now one would think that if the motor and generator both have losses,then the thing should wind down to a stop quicker?
...
Hi tinman,
My take on why they do not wind down much quicker is that they are nearly "matched".
I mean the motor have
T torque at a load and at an
I input power and the generator is able to supply around
I output power when rotated by the
T torque on its shaft. These are basic requirements for a longer wind-down to occur but eventually the setup winds down due to the copper and friction losses.
There might be one hope I can see here for such setup to maintain operation: once we have such well matched motor and generator then we could try to achieve electrical resonance by tuning capacitors in parallel with them so that the reactive currents in the coils could be enhanced Q times as much, like Hector Perres's rotoverter principle but in this case the useful torque output (if any) would be tied to a given RPM and whenever the load would increase, the chances of falling the system apart and eventually stop would be high. I do not say this will surely maintain operation because I have not tried such setup but I do not reject it off hand if resonant operation would be involved. Here I mean that say the motor-generator combination has an L resultant inductance when they are hooked up (and this L would fluctuate in a range in the function of RPM, unfortunately) and for this L value you find a C (run cap type) capacitor which would give resonance with L at the frequency of the RPM.
rgds, Gyula
PS In my answer with respect to the fan motor - PMA case of the first post above I considered the huge power differences involved between the two. There are several patents from the past 50 or more years which show exactly this: connecting a motor shaft directly or via gear to a generator and hooking them parallel electrically too and claiming a continuous operation. However, I have been aware of failures only. OF course we should try and do much more testings.
EDIT: In the video Robert33, he uses a series resistor (he calls it a low resistance, attenuator) this further reduces the wind down time...
Quote from: gotoluc on January 29, 2013, 11:38:40 PM
Hi returnedintime,
it would be nice if it could work. However, you have not mentioned how many Volts your PMA puts out when it reaches 12.7 Amps at 950 rpm. It could be in the 12 volts range to keep a battery charged?... that would be about 150 Watts.
Now your fan motor probably works on 120 Volts AC and at 2.6 Amps it would come to a little over 300 Watts.
If you're low on cash I would not invest money in this, as many have tried this before. I wish it would be so simple :-\
Nothing wrong in thinking of alternative energy solutions but using off the shelf items chances it would work are next to none.
I wish you all the best in your search for a solution
Luc
I appreciate your feedback,....which gives me reason for pause,....my thinking is that I am on the right track,..but using the wrong PMA,...with that thought in mind,..what are your thoughts of using a pma that produces 1000 Watts at 999 rpm, using the same electric fan motor as mentioned in earlier posts?
The PMA uses a slant core system that makes the PMA vertually coggless.
Quote from: gyulasun on January 30, 2013, 05:30:08 AM
Hi,
On 'heavier load' I mean as follows: you use say a 12 Ohm power resistor as a load across the output of your alternator, the current draw will be 1 Amper if the alternator output voltage is around 12V. Now suppose you replace the 12 Ohm resistor with a 2 Ohm power resistor, this is what I call a heavier load because the current draw goes up to at least 6 Amper provided the alternator keeps its output voltage at 12V as earlier.
Some more notes: it would be good for you to feel the torque by your gloved hand on the shaft this PMA must receive from the wind or from any other prime mover when it rotates at 950 RPM and you take the 12.7 Amper out of it. Will the fan motor be able maintain that much torque on its shaft when the PMA is tied to it? The only correct answer you can get is to test this and see for yourself. I understand you do not belive us, this is natural.
rgds, Gyula
Thanks for your input,.....I'm already convinced that I was targeting the wrong PMA.....I'm thinking of a much more powerful PMA that uses slant core technology(Makes it coggless) and by using a 12 Volt, 5 amp. (60 Watts) halogen light bulb. Just hand cranking the PMA brings this light bulb to maximum power.
returnedintime:
Cogging has nothing to do with Lenz drag, you seem to worry about cogging but it can quasi completely be elliminated mainly by using odd - even ratio in the rotor - stator poles or possibly with the slant core technology you refer to but Lenz drag remains. And in general the heavier load you apply to a generator, the higher the Lenz drag becomes.
Good luck!
Quote from: gyulasun on January 30, 2013, 01:47:48 PM
returnedintime:
Cogging has nothing to do with Lenz drag, you seem to worry about cogging but it can quasi completely be elliminated mainly by using odd - even ratio in the rotor - stator poles or possibly with the slant core technology you refer to but Lenz drag remains. And in general the heavier load you apply to a generator, the higher the Lenz drag becomes.
Good luck!
So if I understand you correctly,..you are stating that regardless how cogless a alternator is,...it is still going to have resistance present?
Quote from: returnedintime on January 30, 2013, 03:37:58 PM
So if I understand you correctly,..you are stating that regardless how cogless a alternator is,...it is still going to have resistance present?
Yes, resistance to rotor rotation
whenever you connect a load across the generator output coil.
You are expected to do your homework first in conventional electrical engineering and when you know what obstacles you are to face with in energy conversion you may start doing something to circumvent them.
Here is a demo on Lenz law.
http://www.youtube.com/watch?v=WHCwgc_xs3s (http://www.youtube.com/watch?v=WHCwgc_xs3s)
As an analogy, think gravity as the prime mover for this virtual generator and the Lenz drag in this case works against gravity. In case of real generator Lenz drag works agains the prime mover of the generator. In the video the load is the self-impedance of the output coil because it is shorted. If you load this coil with say 2 Ohm, or 5 Ohm or 100 Ohm resistors instead of the short (i.e. instead of the self-impedance of the output coil) the falling speed of the coil would gradually increase, meaning the Lenz drag reduces as the loading current gets smaller and smaller, this is just like in a real generator.
Quote from: gyulasun on January 30, 2013, 05:01:04 PM
Yes, resistance to rotor rotation whenever you connect a load across the generator output coil.
You are expected to do your homework first in conventional electrical engineering and when you know what obstacles you are to face with in energy conversion you may start doing something to circumvent them.
Here is a demo on Lenz law.
http://www.youtube.com/watch?v=WHCwgc_xs3s (http://www.youtube.com/watch?v=WHCwgc_xs3s)
As an analogy, think gravity as the prime mover for this virtual generator and the Lenz drag in this case works against gravity. In case of real generator Lenz drag works agains the prime mover of the generator. In the video the load is the self-impedance of the output coil because it is shorted. If you load this coil with say 2 Ohm, or 5 Ohm or 100 Ohm resistors instead of the short (i.e. instead of the self-impedance of the output coil) the falling speed of the coil would gradually increase, meaning the Lenz drag reduces as the loading current gets smaller and smaller, this is just like in a real generator.
If you look at a generator in depth, it is not an energy conversion device. It is a poorly designed power generator that fights itself. The more load on the generator, the more counter magnetic field that causes magnetic drag on the rotor, which takes proportionally more power to turn the generator. It is not due to power conversion at all.
Quote from: Liberty on January 30, 2013, 11:51:05 PM
If you look at a generator in depth, it is not an energy conversion device. It is a poorly designed power generator that fights itself. The more load on the generator, the more counter magnetic field that causes magnetic drag on the rotor, which takes proportionally more power to turn the generator. It is not due to power conversion at all.
Hi Liberty
Ok here is a question for you.
If a generator's lenz force drag is greater than the power output that the generator is giving,then in the below video(that i will attach again)the motor should roll to a stop much quicker than it would with the generator unhooked-as a generator is a loss.
Now we know that this is the case(a generator is a loss)or we would have self running devices all over the place.
But here you have a motor coupled to a generator,and with the generator disconected there would be now lenz drag created within the generator-and when the motor is disconected from the batteries,it rolls to a stop in a given amount of time.
Now when a loss(the generator) is added to that system,one would think that the motor would roll to a stop in a shorter amount of time-as we are adding another loss to the system.
We also know that to get that motor to run on for a longer amount of time when the batteries are disconected,we would need some sort of energy input.
So how is it that by adding a loss to the system,we get a gain in run time?
If the lenz force drag is greater than the power coming out of the generator being routed back into the motor-the motor should run down to a stop quicker.
But here we see the run time is increased dramaticly when the generator is conected to the motor,and batteries disconected.
http://www.youtube.com/watch?v=hl57y-c_bWc
Quote from: Liberty on January 30, 2013, 11:51:05 PM
If you look at a generator in depth, it is not an energy conversion device. It is a poorly designed power generator that fights itself. The more load on the generator, the more counter magnetic field that causes magnetic drag on the rotor, which takes proportionally more power to turn the generator. It is not due to power conversion at all.
Well I may have used loose term to include 'energy conversion', I was focusing on Lenz drag effects in normal generators. On energy conversion I thought of using mechanical energy input to get electrical energy output. that is all.
Thanks, Gyula