Motor Generators

[bistander]

Are you serious, where is the calculation? Solving the problem involves operating with real results.
An equation is not everything, there must be a confirmation. Dear opponent, your statement is a profanation that has nothing to do with solving the problem. You are not a professional, I would not trust you with any analysis.

SOLUTION BASED ON AVAILABLE DATA:

We take the data from the notes of the master (who does not care about my concept or your equations), he does this based on the fact of the measurements taken, and very correctly.

rpm - 400; E = 262.1 V; U = 85 V; I = 9 A; Rload = 9.0 Ohm.

(Load resistance - rheostat 9 Ohm, with all connections and connecting wires approximately 9.4-9.5 Ohm).

The resulting voltage drop is: ∆U = E - U = 262.1 V - 85 V = 177.1 V. Knowing the current, we can determine the resistance at which this voltage drop occurs. R = ∆U/I = 177.1 V / 9 A = 19.6 Ohm.
Let's check the load resistance: Rload = U/I = 85 V / 9 A = 9.44 Ohm.
Let's subtract the load resistance from the total voltage drop resistance to get the source resistance: r = R - Rload = 19.6 Ohm - 9.44 Ohm = 10.2 Ohm.

Let's check the solution with my method

∆U / (Rload + r) = I = U/Rload
177.1 V / (9.44 Ohm + 10.2 Ohm) = 85 V / 9.44 Ohm = 9 A.

------
The author does not measure the resistance of the windings of the three-phase stator of the asynchronous motor with a power of 1.1 kW. (he did not rework the windings, but only made a rotor with permanent magnets, turning the AC motor into a synchronous generator). We can clarify the resistance of the windings, according to known data. For example, there is a lot of data on connecting three-phase motors to one phase, where the resistances of the windings are indicated.

A screenshot of such data for a single-phase motor indicating the resistances for a 1 kW motor.


Resistances are taken from this resource: https://imperia-comforta.ru/soprotiv...gatelya-1-kvt/

One winding from the star for a 1 kW motor is 6 Ohms. For a 1.1 kW motor, this resistance will be within 5 - 5.3 Ohms.
With a series connection, like a star, like the author of the video, the resistance will be 10 - 10.6 Ohms (in our case, 10.2 Ohms).

So everything is as if by notes, a real example and my solution.

You can shove the theory of electrons in electricity up your ... well, you get the idea. Which of us is lying? You are a liar.

___________________________________

For those interested in what the debate is about, here is my post about the essence of current strength in a closed circuit

EMF, CURRENT, VOLTAGE, RESISTANCE. | Patreon

Hello Rakarskiy,

Your numerical example sounds like the same that you used a few days ago. Still the numbers don't make sense. I pointed out some serious issues which I see have not been addressed. You make it far to complex.

Simply measure the generator terminal voltage at rated speed and excitation. Then connect your known load resistance, still at rated speed and excitation. Measure terminal voltage and current.

Armature resistance = (no load voltage - loaded voltage) / current

What's your problem?

Don't call me a liar for posting fact and truth.
bi

[Rakarskiy]

Rakarskiy,

Refer to the attached diagrams and equations, found in so many textbooks.

The explanation you seek concerns the difference in the terminal voltage between loaded and no-load conditions. It is obvious that difference is the two terms containing the armature current, jI_AX_S and I_AR_A. For the no-load or open circuit terminal voltage, V_t, equals E_A. Under load, I_A is present and there is a voltage drop in the armature winding and reactance so V_t is lower than E_A.
bi

Are you serious, where is the calculation? Solving the problem involves operating with real results.
An equation is not everything, there must be a confirmation. Dear opponent, your statement is a profanation that has nothing to do with solving the problem. You are not a professional, I would not trust you with any analysis.

SOLUTION BASED ON AVAILABLE DATA:

We take the data from the notes of the master (who does not care about my concept or your equations), he does this based on the fact of the measurements taken, and very correctly.

rpm - 400; E = 262.1 V; U = 85 V; I = 9 A; Rload = 9.0 Ohm.

(Load resistance - rheostat 9 Ohm, with all connections and connecting wires approximately 9.4-9.5 Ohm).

The resulting voltage drop is: ∆U = E - U = 262.1 V - 85 V = 177.1 V. Knowing the current, we can determine the resistance at which this voltage drop occurs. R = ∆U/I = 177.1 V / 9 A = 19.6 Ohm.
Let's check the load resistance: Rload = U/I = 85 V / 9 A = 9.44 Ohm.
Let's subtract the load resistance from the total voltage drop resistance to get the source resistance: r = R - Rload = 19.6 Ohm - 9.44 Ohm = 10.2 Ohm.

Let's check the solution with my method

∆U / (Rload + r) = I = U/Rload
177.1 V / (9.44 Ohm + 10.2 Ohm) = 85 V / 9.44 Ohm = 9 A.

------
The author does not measure the resistance of the windings of the three-phase stator of the asynchronous motor with a power of 1.1 kW. (he did not rework the windings, but only made a rotor with permanent magnets, turning the AC motor into a synchronous generator). We can clarify the resistance of the windings, according to known data. For example, there is a lot of data on connecting three-phase motors to one phase, where the resistances of the windings are indicated.

A screenshot of such data for a single-phase motor indicating the resistances for a 1 kW motor.


Resistances are taken from this resource: https://imperia-comforta.ru/soprotiv...gatelya-1-kvt/

One winding from the star for a 1 kW motor is 6 Ohms. For a 1.1 kW motor, this resistance will be within 5 - 5.3 Ohms.
With a series connection, like a star, like the author of the video, the resistance will be 10 - 10.6 Ohms (in our case, 10.2 Ohms).

So everything is as if by notes, a real example and my solution.

You can shove the theory of electrons in electricity up your ... well, you get the idea. Which of us is lying? You are a liar.

___________________________________

For those interested in what the debate is about, here is my post about the essence of current strength in a closed circuit

EMF, CURRENT, VOLTAGE, RESISTANCE. | Patreon

[bistander]

Well, electronic nonsense of modern science. Forum members are trying to sell you this as the truth.
A very fresh video from a master who makes generators and wind turbines in his garage. In the video, the master converted a regular asynchronous motor (1.1 kW) into a synchronous generator with a rotor on permanent magnets. He carried out control measurements of the no-load voltage and under load of 9 - 9.5 Ohms.
Dear defender of electrons, explain where the part of the EMF that was measured at no-load went, and under load it is no longer there.
...

Rakarskiy,

Refer to the attached diagrams and equations, found in so many textbooks.

The explanation you seek concerns the difference in the terminal voltage between loaded and no-load conditions. It is obvious that difference is the two terms containing the armature current, jI_AX_S and I_AR_A. For the no-load or open circuit terminal voltage, V_t, equals E_A. Under load, I_A is present and there is a voltage drop in the armature winding and reactance so V_t is lower than E_A.
bi

[Rakarskiy]

Що стосується Холкомба, то у нього дуже цікавий дизайн. Я ще не розібрався, як це працює, але це дуже цікаво.

http://www.energeticforum.com/forum/energetic-forum-discussion/renewable-energy/509697-holcomb-energy-systems-the-holy-grail-may-have-arrived?p=515313#post515313

[Rakarskiy]

Well, electronic nonsense of modern science. Forum members are trying to sell you this as the truth.
A very fresh video from a master who makes generators and wind turbines in his garage. In the video, the master converted a regular asynchronous motor (1.1 kW) into a synchronous generator with a rotor on permanent magnets. He carried out control measurements of the no-load voltage and under load of 9 - 9.5 Ohms.
Dear defender of electrons, explain where the part of the EMF that was measured at no-load went, and under load it is no longer there.

If you cannot do this in the form of a calculation (at the same time we will check how you can calculate the resistance of an asynchronous motor with a power of 1.1 kW), then I will call you a liar. It was fair.
When I assert something, I cite facts. What is written in textbooks is not a fact, these are different interpretations, as well as the interpretation of the extraneous forces for Ohm's law of a complete circuit.

9B: Electric Current, EMF, and Ohm's Law - Physics LibreTexts

If everything is ok for you, then it is not ok for me, there is no analysis of the electromagnetic generator in the circuit with the load.

I found the most logical interpretation, but still not straightforward. Physicists are afraid of "losing face" in front of other physicists and the shadow government. By the way, it doesn't matter in America, Ukraine or Europe. But I assert my conception on practical examples and logic.

It will be more interesting further on, in fact, this is a prelude.

Генератор с асинхронника 1,1 кВт, почти под ключ. - YouTube



Once again, a link to the material, in case someone is interested in the essence of the confrontation.

EMF, CURRENT, VOLTAGE, RESISTANCE. | Patreon

[bistander]

Here's another opinion:

ChatGPT

You said:
"During the passage of current, there is a continuous loss of charges, more precisely, the neutralization of positive and negative electricity." Do you agree with this statement?
ChatGPT said:
ChatGPT
The statement is somewhat unclear and can be misleading if taken at face value. Let me break it down:

Continuous loss of charges: This part seems incorrect if referring to electric current in a conductor. In a typical circuit, there isn't a continuous loss of charges (i.e., electrons). Electrons move through the conductor, but they are not "used up" or lost in the process. Current is essentially the flow of electrons (or charge carriers) that move in a closed loop, driven by a potential difference (voltage).

Neutralization of positive and negative electricity: If this is referring to the interaction between positive and negative charges, such as recombination of ions in certain contexts (like in electrolytes or semiconductors), then "neutralization" can happen in that sense. However, in conductors where electrons flow freely, we don’t usually talk about "neutralization" in this manner. The charges in a conductor don't cancel out but instead move under the influence of an electric field.

In typical electrical circuits, charges (electrons) do not disappear or neutralize; they flow in response to the electric field and return to the source, completing the circuit.

So, I would not agree with the statement as written. It might be trying to refer to specific cases like recombination of ions or semiconductors, but not to a general description of electric current in conductors.

[bistander]

This is a quote from a physics textbook for universities published in the Soviet Union in 1977. I have studied this point in different editions. So this is exactly the most academic physics. Who can I get a review from? I'm an independent researcher, which means I don't ask anyone for permission on what to research or how to publish my research. Can you counter anything about the results of my research? I haven't seen any of your scientific or engineering publications. So. "Whose mouth is mooing and whose mouth is silent."

Rakarskiy,
I offer up helpful criticism, take it or leave it. I usually state "don't believe me; look it up". Which I suggest you do. Re. The Law of Conservation of Charge.
bi

[Rakarskiy]

This is a quote from a physics textbook for universities published in the Soviet Union in 1977. I have studied this point in different editions. So this is exactly the most academic physics. Who can I get a review from? I'm an independent researcher, which means I don't ask anyone for permission on what to research or how to publish my research. Can you counter anything about the results of my research? I haven't seen any of your scientific or engineering publications. So. "Whose mouth is mooing and whose mouth is silent."

[bistander]

Rakarskiy,
You write "During the passage of current, there is a continuous loss of charges, more precisely, the neutralization of positive and negative electricity. "
You do believe some strange things. I think you have not learned fundamentals and have serious misconceptions. Then you criticize our science which has proven to be correct and then develop your erroneous theories. And believe in frauds like Holcomb.
I've tried to help you but you continue to ignore truth, logic and fact. I come to an end with you. Please have peer review prior to publishing.
Respectfully,
bi

[Rakarskiy]

My explanation of what current strength in a closed circuit is, because the orthodox still have electrons and particles running along wires. They also teach this "stuff", defend their doctrines, etc.

Wise Eye OverUnity: CURRENT IN A CLOSED CIRCUIT (rakatskiy-blogspot-com.translate.goog)

[Rakarskiy]

Every motor and generator (they're the same) behaves per orthodox physics and none produce Overunity or Free Energy. That is proof.
bi

motore di flynn 12 poli (youtube.com)

[bistander]

Greetings!

If you can prove to me with an example on a physical generator, then I will accept your argument. But anyway, leave your wishes to yourself.
And then I was already told "that they agree with me, but they cannot declare it, because they will lose their teaching accreditation." It is good that designers and garage masters accept everything that helps them.

Bi.

Every motor and generator (they're the same) behaves per orthodox physics and none produce Overunity or Free Energy. That is proof.
bi

[Rakarskiy]

Greetings!

If you can prove to me with an example on a physical generator, then I will accept your argument. But anyway, leave your wishes to yourself.
And then I was already told "that they agree with me, but they cannot declare it, because they will lose their teaching accreditation." It is good that designers and garage masters accept everything that helps them.

Bi.

[bistander]

Hi!

You are not the first to try to convince me that I am wrong!
That's why I have supplemented the post with very vivid examples of practical measurements. There are times when a postulate from a physics textbook is perfect, but in real generators it does not work. Such is the paradox of your orthodox physics.
One ‘convincing me’ is confused about what he knows and what he sees, because, all his life he trusted postulates.

Respectfully.

Wise Eye OverUnity | Serge Rakarskiy Independent researcher on overunity systems | Patreon

Dear Rakarskiy,

You should put your ego aside. Replace your Wise Eye with a Smart Ear. What I tell you results from life and career experience after "orthodox physics" textbook education. Study my criticisms carefully and you'll find reality, no paradox.
bi

[Rakarskiy]

Hi Rakarskiy,

"How is the force of current in a closed circuit formed?"
Obviously the potential difference.


I took time and read your article thoroughly.

In your Wise Eye article EMF, Current, Voltage, Resistance, you say "Ohm's law doesn't say anything about voltage drop, only about voltage or applied voltage." This statement of yours is false. Ohm's Law always applies, including the potential difference across a circuit element, such as "voltage drop."
There are also some other issues with your article on which I'll elaborate.

*****************

Hi!

You are not the first to try to convince me that I am wrong!
That's why I have supplemented the post with very vivid examples of practical measurements. There are times when a postulate from a physics textbook is perfect, but in real generators it does not work. Such is the paradox of your orthodox physics.
One ‘convincing me’ is confused about what he knows and what he sees, because, all his life he trusted postulates.

Respectfully.

Wise Eye OverUnity | Serge Rakarskiy Independent researcher on overunity systems | Patreon