partzmans board ATL

Author Topic: partzmans board ATL (Read 23884 times)

Peterae	partzmans board ATL « on: 2018-06-28, 21:02:08 »
Group: Administrator Hero Member Posts: 4635	Testing

partzman

Re: partzmans board ATL « Reply #1 on: 2019-08-04, 20:08:08 »

Group: Moderator
Hero Member

Posts: 2220

I scrapped the original topic for this thread and have decided to start a new subject rather than create a new private thread. The new topic will be Asymmetrical Transformers and a lot of initial info will come from Russian translated texts.

The first attachments will be from a Russian forum moderated by Alexander Abramovich.

~~I am asking that NONE of the data presented here be cross posted or shared on any other forum or with any other individual not listed as having access to this thread!~~

Regards,
Pm

Asymmetrical Transformers Pg1.pdf (760.94 kB - downloaded 78 times.)

Asymmetrical Transformers Pg2.pdf (681.33 kB - downloaded 54 times.)

« Last Edit: 2019-12-11, 19:53:05 by partzman »

partzman	Re: partzmans board ATL « Reply #2 on: 2019-08-04, 22:53:45 »
Group: Moderator Hero Member Posts: 2220	Next two pages- Asymmetrical Transformers Pg3.pdf (915.06 kB - downloaded 50 times.) Asymmetrical Transformer Pg4.pdf (586.08 kB - downloaded 42 times.)

partzman	Re: partzmans board ATL « Reply #3 on: 2019-08-04, 22:54:51 »
Group: Moderator Hero Member Posts: 2220	And the next- Asymmetrical Transformers Pg5.pdf (660.49 kB - downloaded 47 times.) Asymmetrical Transformers Pg6.pdf (639.87 kB - downloaded 43 times.)

partzman	Re: partzmans board ATL « Reply #4 on: 2019-08-04, 22:55:46 »
Group: Moderator Hero Member Posts: 2220	And the last two- Asymmetrical Transformers Pg7.pdf (664.46 kB - downloaded 48 times.) Asymmetrical Transformers Pg8.pdf (418.66 kB - downloaded 44 times.)

partzman	Re: partzmans board ATL « Reply #5 on: 2019-08-04, 22:58:56 »
Group: Moderator Hero Member Posts: 2220	This is a paper by Abramovich on the principles of the Karnaukhov generator which is similar to the Kapanadze device. It contains both part 1 and 2. Karnaukhova Generator Pt1.pdf (2102.3 kB - downloaded 57 times.)

partzman	Re: partzmans board ATL « Reply #6 on: 2019-08-05, 00:32:33 »
Group: Moderator Hero Member Posts: 2220	This is a paper of my own that demonstrates the reduction of the Lenz effect with constant current loads. Pm Reducing the Lenz Effect in a Transformer.pdf (1049.3 kB - downloaded 67 times.)

poynt99	Re: partzmans board ATL « Reply #7 on: 2019-08-05, 01:09:08 »
Group: Administrator Hero Member Posts: 3538 It's not as complicated as it may seem...	I think I understand your paper PM. Where might we go from there? --------------------------- "Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa

partzman	Re: partzmans board ATL « Reply #8 on: 2019-08-05, 01:58:55 »
Group: Moderator Hero Member Posts: 2220	Quote from: poynt99 on 2019-08-05, 01:09:08 I think I understand your paper PM. Where might we go from there? There are more details to be explained but basically this technique should be able to be used in a motor/generator design to reduce cogging or in a rotating magnetic field device. Whatever the means of induction, the energy produced would be stored in the constant current inductor. Pm

poynt99	Re: partzmans board ATL « Reply #9 on: 2019-08-05, 03:38:07 »
Group: Administrator Hero Member Posts: 3538 It's not as complicated as it may seem...	Looking forward to reading more. --------------------------- "Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa

Centraflow

Re: partzmans board ATL « Reply #10 on: 2019-08-05, 10:58:13 »

Group: Experimentalist
Hero Member

Posts: 3914

Thank you PM for starting this thread, still absorbing what you have written

Regards

Mike

---------------------------

"All truth passes through three stages. First, it is ridiculed, second it is violently opposed, and third, it is accepted as self-evident."
Arthur Schopenhauer, Philosopher, 1788-1860

As a general rule, the most successful person in life is the person that has the best information.

ion

Re: partzmans board ATL « Reply #11 on: 2019-08-05, 14:22:43 »

Group: Elite
Hero Member

Posts: 4127
It's turtles all the way down

Perhaps before getting into serious builds we should try to dissect the basic hypothesis as outlined in the Russian papers, specifically the opening of the Karnaukhova paper.

Smudge would be the guy to delve deep into this I suppose.

Or I guess it could also be a parallel activity with bench tests, nevertheless a brief outline, discussion, and critique of the basic hypothesis would be useful.

Regards.

---------------------------

"Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy

Smudge

Re: partzmans board ATL « Reply #12 on: 2019-08-05, 18:00:27 »

Group: Professor
Hero Member

Posts: 2306

I've started to read the papers but have difficulty following the arguments as the (English) language is somewhat peculiar. My initial reaction is based on my understanding on how transformers work as analysed in the magnetic domain where the coil currents create mmf (magnetic "voltage"), the core reluctance acts like magnetic resistance and flux acts like electric current. The loaded secondary creates mmf at 90 degrees to the flux hence acts like magnetic "inductance".

The trap most people seem to fall into is to consider the secondary as producing Lenz flux, and then to dream up ways of negating that Lenz flux. In fact the secondary does not produce any flux, it only produces mmf. The alternating flux in the transformer changes very little between no-load and full-load conditions (in the perfect transformer it doesn't change at all). The primary magnetizing current creates that flux, and that remains at a constant AC value. Somehow the transformer acts to draw load current from the primary whose mmf exactly opposes the mmf from the secondary, so those load currents do not create flux. In my view trying to stop the secondary from creating Lenz flux is the wrong approach.

If there is flux leakage because primary and secondary are separated from each other, then that opposition of mmfs is no longer exact, and the difference between primary and secondary mmf is what drives that leakage flux through the reluctance of that leakage path. That leakage flux does not represent a loss since it is generally through air, so it can be used to store energy that can be recouped. It strikes me that the asymmetrical transformer, if it works at all, must use that leakage flux to create a primary load current that is not 90 degrees shifted from the magnetizing current. Then it is possible to argue that the atomic dipoles responsible for the high mu of the core can actually supply anomalous power. So I already have a biased view that is at variance to anything written in those Russian papers.
Smudge

partzman	Re: partzmans board ATL « Reply #13 on: 2019-08-05, 21:23:48 »
Group: Moderator Hero Member Posts: 2220	Here is a paper that goes into a little more detail and explains how the concept can be applied to a PM motor/generator. Regards, Pm Applications for the RLE.pdf (897.78 kB - downloaded 63 times.)

partzman

Re: partzmans board ATL « Reply #14 on: 2019-08-08, 20:47:10 »

Group: Moderator
Hero Member

Posts: 2220

The simulation attached below shows the gain that could be achieved from typical external induction to a coil biased with constant current.

In this example, L1 could be a pickup air coil that is biased by the current from L3, the constant current source. L2 is the source of induction in this case and could be a passing PM.

Prior to the start of the sim, L3 and L1 are biased with 300ma as a starting current. The buildup of current in L2 is the induction source in this case and is seen to reach an energy level of 34.32uJ.

The ending current in L3 and L1 after 50us is 303.44ma which results in an energy increase of ((303.44e-3)^2- (300.00e-3)^2)* .10243/2 = 106.3uJ. Based on our input induction energy above, the apparent overall gain is 106.3/34.32 = 3.1 .

The equivalent wattage for this time period is U/dt =106.3e-6/50e-6 = 2.126 watts.

Again, the input energy source 'sees' the equivalent input inductance of L2 which using L = E*dt/di = 20*50e-6/68.48e-3 = 14.602mh as compared to the actual L1 inductance of 14.77mh. IOW, the outside induction source 'sees' a much higher impedance than would be experienced with a conventionally loaded pickup coil.

Regards,
Pm

partzmans board ATL

RLE Induction Sim.png (38.5 kB, 1090x778 - viewed 496 times.)

Smudge

Re: partzmans board ATL « Reply #15 on: 2019-08-09, 16:43:47 »

Group: Professor
Hero Member

Posts: 2306

Quote from: partzman on 2019-08-08, 20:47:10

The simulation attached below shows the gain that could be achieved from typical external induction to a coil biased with constant current.

In this example, L1 could be a pickup air coil that is biased by the current from L3, the constant current source. L2 is the source of induction in this case and could be a passing PM.

Prior to the start of the sim, L3 and L1 are biased with 300ma as a starting current. The buildup of current in L2 is the induction source in this case and is seen to reach an energy level of 34.32uJ.

The ending current in L3 and L1 after 50us is 303.44ma which results in an energy increase of ((303.44e-3)^2- (300.00e-3)^2)* .10243/2 = 106.3uJ. Based on our input induction energy above, the apparent overall gain is 106.3/34.32 = 3.1 .

Hi PM,
That L value you used in the L3 plus L1 energy calculation is 0.10243. How have you arrived at that value?

Smudge

Smudge	Re: partzmans board ATL « Reply #16 on: 2019-08-09, 16:50:15 »
Group: Professor Hero Member Posts: 2306	Here are some thoughts on asymmetrical transformers. As I use PowerPoint to create my drawings I wrote this paper in PowerPoint, whereas I usually use Word and copy the images into Word. PowerPoint doesn't automatically do spell check so please forgive any typos. Smudge Asymmetrical transformer.pdf (48.15 kB - downloaded 58 times.)

partzman	Re: partzmans board ATL « Reply #17 on: 2019-08-09, 17:53:44 »
Group: Moderator Hero Member Posts: 2220	Quote from: Smudge on 2019-08-09, 16:43:47 Hi PM, That L value you used in the L3 plus L1 energy calculation is 0.10243. How have you arrived at that value? Smudge Hi Smudge, Since L3 and L1 are not coupled, this value is simply the sum of their inductances. Therefore both L1 and L3 share the starting and ending currents. Regards, Pm

partzman

Re: partzmans board ATL « Reply #18 on: 2019-08-09, 18:00:33 »

Group: Moderator
Hero Member

Posts: 2220

Quote from: Smudge on 2019-08-09, 16:50:15

Here are some thoughts on asymmetrical transformers. As I use PowerPoint to create my drawings I wrote this paper in PowerPoint, whereas I usually use Word and copy the images into Word. PowerPoint doesn't automatically do spell check so please forgive any typos.

Smudge

Thanks for posting this paper! I will study it carefully. I have already run many bench experiments and sims with a core arrangement similar to your E core arrangement but still come up conservative up to now. However I will continue to pursue this avenue.

Regards,
Pm

PhysicsProf	Re: partzmans board ATL « Reply #19 on: 2019-08-09, 19:14:38 »
Group: Professor Hero Member Posts: 3215	Interesting, smudge - thanks for posting your ideas. I've wondered along similar lines, glad to see experiments proceeding...

partzman

Re: partzmans board ATL « Reply #20 on: 2019-08-10, 18:07:43 »

Group: Moderator
Hero Member

Posts: 2220

Here is some food for thought which is somewhat redundant to my previous papers, but a simple two winding transformer can be considered asymmetrical if looked at from the right perspective.

I will use the sim below for an example.

In the schematic we see a transformer T1 with a primary L1 and secondary L2 with a coupling of 0.8 . Notice the measured aid and buck inductances of L1 and L2 but in particular the aid which is 10.46mh. This is larger than the sum of L1 and L2 due to the coupling factor of the coils which is primarily determined by their shape and position on the core relative to one another. With equal windings and a perfect coupling of K=1, the aiding inductance would be double the sum of their inductances (neglecting coil resistances). So, in this case if we had K=1, we would measure 11.6mh in the aid mode.

Now let's consider what we would have if we could somehow charge each winding independently.

IOW with perfect coupling, let's charge L1 to a point of 100ma to obtain an energy level of .1^2*2.9e-3/2 = 14.5uJ and save this energy temporarily. Now charge L2 in the same manner for a total energy stored or consumed of 29uJ. If we now connect L1 and L2 in the aid mode with a combined inductance of 11.6mh with 100ma stored in each, we can now realize a combined energy stored of .1^2*11.6e-3/2 = 58uJ or twice the original. The key is to somehow accomplish the independent charging in order to reach a theoretical maximum COP=2.

I've attached the sim below which charges L1 over a precise 13.591us period which allows the ramping current in L1 to equal the relatively constant current in L2 and L3. This is an attempt to use the constant current LRE to achieve independent charging but when all energies are considered, the COP<1.

Regards,
Pm

partzmans board ATL

Trans Sym1.png (20.82 kB, 773x460 - viewed 447 times.)

partzmans board ATL

Trans Sym1 Schematic.png (15.2 kB, 771x391 - viewed 421 times.)

partzman

Re: partzmans board ATL « Reply #21 on: 2019-08-12, 16:37:35 »

Group: Moderator
Hero Member

Posts: 2220

In consideration of an actual PM mo/gen build, the sim below represents a coupled 2 coil topology with the Pm passing in between a coil pair for maximum efficiency. The PM is represented again by L2 as the external induction source and L4 is added to make the coil pair with a coupling of .7 which should be reasonably close.

We see at the end of a 50us period, the input energy from the external induction source is 36.228uJ. The output current in the L1, L4 coil pair in series with the current source L3 has reached 306.96ma from a starting current of 300ma in the same time period.

The series inductance of L1 and L2 is 8.33mh so the energy gain in the L1, L4, and the current source L3 network is ((.30696e^2)-(0.300^2))*.10833/2 = 229uJ . The apparent gain is 229/36.23 = 6.29 .

Again for comparison, L2 reaches a peak current in this period of 72.29ma which equates to an effective input inductance of L=E*dt/di = 20*50e-6/.07229 = 13.83mh.

The wattage equivalent for this energy/time period is W = dU/dt = 229e-6/50e-6 = 4.58 watts.

Regards,
Pm

partzmans board ATL

RLE PM Induction x2.png (40.46 kB, 956x918 - viewed 395 times.)

partzman	Re: partzmans board ATL « Reply #22 on: 2019-08-12, 21:04:36 »
Group: Moderator Hero Member Posts: 2220	Smudge, Osamu Ide says in his paper "Increased Voltage Phenomenon in a Resonance Circuit of Unconventional Magnetic Configuration" there exists some controversy over the fact some authors have stated that motional emf and induced emf are independent phenomena. Do you have any thots on this? Several references are given but I can't seem to find them. Regards, Pm

partzman	Re: partzmans board ATL « Reply #23 on: 2019-08-13, 15:16:57 »
Group: Moderator Hero Member Posts: 2220	FWIW, Here are the Kikuchi and Moon references- Pm kikuchi1984.pdf (395.71 kB - downloaded 33 times.) moon1955.pdf (1058.24 kB - downloaded 31 times.)

Smudge

Re: partzmans board ATL « Reply #24 on: 2019-08-13, 16:53:01 »

Group: Professor
Hero Member

Posts: 2306

High PM,
I will come back to your question about Osama Ide later. Meanwhile I have solved the conundrum in your sims showing apparent energy gain. I take your sim posted here on 8th August as the example. You start with the series combination of L1 and L3 being precharged to 300mA from some external source. That puts 4.5mJ into the 100mH L3 and 109.35uJ into the 2.43mH L1. L1 is the secondary of the input transformer while L3 is the load. Now you apply a fixed voltage to the primary L2 for 50uS which produces a linear current amp from zero to 68.48mA. That is an input energy of 34.32uJ. That increases the 300mA around the secondary circuit to 303.443mA. You then work out the increase in energy of that series combination of L1 and L3, which would be true if that external source provided the increased current. But it didn't come that way, it came from the secondary L1, and as such the original 109.35uJ stored there, stored as magnetic energy in the core, does not increase. In fact it deceases because the flux there decreases. Establishing how much it decreases is a little involved so I hope you can follow this.

You need to know the reluctance of the core, and that is easily obtained from the inductance value and the number of turns via Rel=N²/L. You give L values for the primary and the secondary. For the primary we have N=325 and L=14.77mH yielding a reluctance of 7.1513E6. For the secondary we have N=130 and L=2.43mH yielding a reluctance of 6.9574E6. The reason these are slightly difference is because you modeled the transformer with a coupling factor of 0.9. So let's take 7E6 as a value to use. We want to know the energy stored in that reluctance when the primary is carrying 68.48mA which with 325 turns is a forward mmf of 22.256AT. and the secondary is carrying 303.443mA which with 130 turns as a back mmf of 39.447AT. Thus the core is effectively driven by the difference between forward and back mmf which is 17.191AT. Energy stored in the core is given by U²/Rel where U is the mmf, and that gives us 42.219uJ. That is a loss of 67.131uJ from the original energy of 109.35uJ stored there. That amount of energy when added to the input to the primary accounts for the increase in energy of the load L3. There is no OU here, sorry.

Smudge