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  • Rakarskiy
    replied
    This is no longer trivial, the conversion efficiency in the conducted experiment (approximately, as before) is greater than unity. In addition, the experiment was conducted roughly, without finding the optimal mode. Also, at first I took this to be a confirmation of the academic topology of the generators, but over time I looked and found a confirmation of the correct process topology.

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  • bistander
    replied
    Originally posted by Rakarskiy View Post
    You may be right that the resistive divider equation is wrong. The quote is taken from the electrician training website. Only I understood what is meant, the current that appears in the second part of the winding consists of the current of the primary winding at the output to the load circuit. Thus, there is no need for a secondary winding. This is implied.

    But your example is (input) 119V*0.87A = 103W, (output) 25V * 2.7A = 67W. Thus, the conversion efficiency is 0.65.

    If in the experiment we have 45 W at the input of the autotransformer, then at the output we will get no more than 0.6-0.7 of the input power:
    0.6*45W=27W (0.7 * 45 W = 31 W) - /this is at the input to the rotor winding/
    At the output we have 46 W. Conversion efficiency 46/27=1.7 (46/31 = 1.48).

    So, still more than one.
    Sincerely.​​
    Well I'm pleased you recognize your mistake. It is also a mistake to assume the single point efficiency shown in my test would apply elsewhere. The variac is sensitive (rating wise) to current and my test point was near the upper limit for the old variac. Other points which I ran yesterday at lower currents showed markedly higher efficiency. I've saved several meter snapshots. When I get a few minutes, I'll add some data here via edit.
    bi
    edit

    IMG_20230405_124750962_copy_800x600.jpg

    This was at the same load point but a bit sooner before the variac heated. I read 0.77A putting efficiency at 76%.

    And another, at lighter load.
    IMG_20230405_131032964_copy_800x600.jpg

    Looks like 89%.
    Last edited by bistander; 04-06-2023, 12:58 PM. Reason: Added 2nd pic and note

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  • Rakarskiy
    replied
    You may be right that the resistive divider equation is wrong. The quote is taken from the electrician training website. Only I understood what is meant, the current that appears in the second part of the winding consists of the current of the primary winding at the output to the load circuit. Thus, there is no need for a secondary winding. This is implied.

    But your example is (input) 119V*0.87A = 103W, (output) 25V * 2.7A = 67W. Thus, the conversion efficiency is 0.65.

    If in the experiment we have 45 W at the input of the autotransformer, then at the output we will get no more than 0.6-0.7 of the input power:
    0.6*45W=27W (0.7 * 45 W = 31 W) - /this is at the input to the rotor winding/
    At the output we have 46 W. Conversion efficiency 46/27=1.7 (46/31 = 1.48).

    So, still more than one.
    Sincerely.​​
    Last edited by Rakarskiy; 04-06-2023, 10:38 AM.

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  • bistander
    replied
    Originally posted by Rakarskiy View Post
    ​​​...
    [In other words, the output current is much less than the input current.]
    I think this phrase is enough not to think that more than I calculated was put on the generator rotor. The СOP ratio is maintained.
    ...
    Who wrote that quote (shown in bold)?
    bi
    {edit}
    Added test photo and data.
    IMG_20230405_124955912_copy_800x600.jpg

    Meters on left read input. 119.7 Vac & 0.87A. meters on right are output into resistor. 25.23Vac & 2.74A.
    120 step down to 25 Volts. 2.74A obviously higher than 0.87A.

    That quoted statement is wrong. And the variac or autotransformer cannot be modeled using a voltage divider (resistor network) as you claim.

    Test and photo taken 5April2023 by bi.
    ​​​​​​
    Last edited by bistander; 04-05-2023, 06:33 PM. Reason: Added photo and data

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  • Rakarskiy
    replied
    You didn't convince me. Firstly, this is the direction of current strength (I [A]) and voltage (U [V]) in a real circuit, they are opposite.
    What is hung in such educational books does not in any way reflect the real movement of current. In the experiment, there was just a divider effect (voltage reduction system)
    Take half the period of the voltage sine wave and see how it looks in all areas. Do not forget that this transformer has a zero terminal, it always has zero.

    2023-04-05_180601.jpg
    The ratio of EMF values is expressed by the formula: E1 / E2 = w1 / w2 = k, where E is the EMF, w is the number of turns, k is the transformation ratio.
    Considering that the voltage drop in the transformer windings is small, it can be ignored. In this case, the equalities: U1 = E1; U2 = E2 can be considered fair. Thus, the above formula becomes: U1/U2 = w1/w2 = k, that is, the ratio of voltages to the number of turns is the same as for a conventional transformer.
    Without going into details, we note that the ratio of the current strength of the upper coil to the load current, as for a conventional transformer, is expressed by the formula: I1/I2 = w2/w1 = 1/k. It follows that since in the step-down transformer w2 < w1, then I2 < I1. [In other words, the output current is much less than the input current.] Thus, less energy is spent on heating the wire, which allows the use of smaller wires.
    [In other words, the output current is much less than the input current.]
    I think this phrase is enough not to think that more than I calculated was put on the generator rotor. The СOP ratio is maintained.

    https://rakatskiy.blogspot.com/2022/...ransducer.html

    An autotransformer efficiency of 99% can only be achieved with a small difference between the input and output voltages.
    In the experiment 220/117 = 1.88 (almost double the difference), thus no more than 0.6 [45 W (in) * 0.6 = 27 W (out)]
    I think the rest of the discussion doesn't make sense anymore
    Last edited by Rakarskiy; 04-05-2023, 03:49 PM.

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  • bistander
    replied
    Originally posted by Rakarskiy View Post

    I am attaching an autotransformer circuit (LATR - laboratory autotransformer).
    If you look at the circuit and label the winding parts as resistances, you get exactly what I'm talking about.
    There is practically no reactance under load. Only the active resistance of the wires remains. Thus, it is absolutely acceptable to use the system as a voltage divider.
    Sincerely.
    2023-04-04_231249.jpg

    The fact that the directions of the currents are slightly different in reality than in the diagram (which I took from Wikipedia) from a resistive divider is not reflected in any way in the voltage division in the sections.
    Hi Rakarskiy,
    You are mistaken. You say:
    "Thus, it is absolutely acceptable to use the system as a voltage divider."
    This is untrue. Take a few minutes and learn about it. The device is a transformer and must be treated as such.
    bi

    https://www.electronics-tutorials.ws...ansformer.html

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  • Rakarskiy
    replied
    Originally posted by bistander View Post

    Hi Rakarskiy,
    I noticed you say "LATR is essentially a voltage divider". I am not familiar with term LATR, but from some limited Google info and appearance of component in your diagram, it would seem to be what is commonly referred to as a variac, which is an adjustable autotransformer. Therefore it can not be treated as a voltage divider.
    Rregards,
    bi
    I am attaching an autotransformer circuit (LATR - laboratory autotransformer).
    If you look at the circuit and label the winding parts as resistances, you get exactly what I'm talking about.
    There is practically no reactance under load. Only the active resistance of the wires remains. Thus, it is absolutely acceptable to use the system as a voltage divider.
    Sincerely.
    2023-04-04_231249.jpg

    The fact that the directions of the currents are slightly different in reality than in the diagram (which I took from Wikipedia) from a resistive divider is not reflected in any way in the voltage division in the sections.
    Last edited by Rakarskiy; 04-04-2023, 08:30 PM.

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  • bistander
    replied
    Originally posted by Rakarskiy View Post
    Hi Rakarskiy,
    I noticed you say "LATR is essentially a voltage divider". I am not familiar with term LATR, but from some limited Google info and appearance of component in your diagram, it would seem to be what is commonly referred to as a variac, which is an adjustable autotransformer. Therefore it can not be treated as a voltage divider.
    Rregards,
    bi

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  • Rakarskiy
    replied
    Article about Clement Figuera

    https://rakatskiy-blogspot-com.trans..._x_tr_pto=wapp

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  • Rakarskiy
    replied
    I think it makes sense to start an interesting topic. generators that engineering thought developed and even produce. They are not suitable for the production of industrial frequency electricity, but they have great advantages and opportunities.
    I know for sure that these devices include the generators and motors of Joe Flynn (USA) and Kornilov (Russia, who died in December 2021, while his laboratory was looted)
    Even Bedina's impulse twister can somehow be attributed to this section. These are impulse energy devices.
    My material may puzzle many people, but it contains the most cherished formula, OverUnity.

    You can find my material at the links. Unfortunately, in the near future I will not be able to do this closely because of the cannibalistic Russian aggression against my country. I don't want such an interesting moment to be lost.

    https://rakatskiy-blogspot-com.trans..._x_tr_pto=wapp


    http://rakarskiy.narod.ru/_ld/0/43_RIZU.pdf

    Sincerely, Serge Rakarsky

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  • Matthew Jones
    replied
    OK Now I know what your doing. Sometime tomorrow I will start a new thread and we go through that Simple Motor. That's almost 15 years old now. You gotta nice rig going there.We'll go over that. I do not have much more than I had but I do have good ideas to hone that one in for better results, With both LA batts and lithium.
    Be patient.

    Cheers
    Matt

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  • i_ron
    replied
    Hi Matt

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  • i_ron
    replied
    Hi Matt
    MJSCwriteup.pdf

    Ron

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  • Matthew Jones
    replied
    To post a picture you look in the tool bar above the text area. About 3/4 of the way down there is an icon for "Image". In the image popup you can hit the tab for upload, then link to it or something. Or you post a link to the MJSC write up which I do not have a copy of. And would like. I lost a lot of stuff at one time.
    I am trying to catch up with a few older things I have done and start trying to update them. David brought up the TS motor so I started with that one. I am working on the design now that can be printed or CNC'd and boards can be ordered with a part reference. Anyway if you can post that picture or email it to me, I think you have my email...I haven't made any progress lately on older stuff but maybe we can. I'll start a new thread soon to cover some of it.

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  • i_ron
    replied
    Hi Matt

    OK, it was from your "MJSC Write up" and All the numbers are there, so let me rephrase my question, have you progressed this any further?

    Somebody would have to explain to me how to post a picture on here.

    Thanks, Ron

    Leave a comment:

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