Motor Generators

[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]

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.

466753451.jpg
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]

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.

466753451.jpg
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]

Yeah! There's no cure.

What does normal alternator speed mean?
The garage foreman has correct measurements on the alternator terminals at idle (no load) and with load at different RPMs.
Second, what is the excitation? If the rotor has permanent magnets, and this is the maximum magnetic induction of excitation (constant).
Third, the calculation I made for one measurement position, you can do it yourself for all the others.

I do not recommend anyone to deal with you in terms of analysing electrical circuits and electromechanisms.

Regards!

[bistander]

Yeah! There's no cure.

What does normal alternator speed mean?
The garage foreman has correct measurements on the alternator terminals at idle (no load) and with load at different RPMs.
Second, what is the excitation? If the rotor has permanent magnets, and this is the maximum magnetic induction of excitation (constant).
Third, the calculation I made for one measurement position, you can do it yourself for all the others.

I do not recommend anyone to deal with you in terms of analysing electrical circuits and electromechanisms.

Regards!

"What does normal alternator speed mean?"

Rated RPM at rated load.
Look at nameplate or manufacturer's specification.

"what is the excitation"

On typical generator, it is field current. On permanent magnet generator it is the magnet(s).

What was the garage foreman's measured no-load terminal voltage measured at open circuit?

Like I say many times and always mean, don't take my word for it, look it up, using reputable sources. You'll find that I post truth and fact.
bi

[Turion]

https://www.youtube.com/watch?v=nnKYlchUm5w




Just for reference.
This is a simple setup with a generator coil and a motor coil run on a bit over 12 volts. These are the principles I have used in my big generator. Like I said, it's a simple 7th grade science experiment to see the principles in action. If you bother to build it. The machine in the video has all north magnets on the rotor, so I bias the coils with the correct polarity of magnet on the BACK of the coil to get the best effect as a motor coil and a generator coil. I also use magnets in OPPOSITION to both the generator coil and the rotor coil to show the effect there. YOU be the judge of whether it has an effect or not. I know what I see on the bench despite what individuals here have said. By the way, the amp draw goes DOWN when the opposition magnets are put in place.

The PROPER size magnet at the PROPER distance will show a better result than what I demonstrate in the video, but it is the principles that are important.

My two friends who were having work done on the two versions of the big generator ran up a $5,000 bill at the machinist, so the gens have been in the shop for months, and won't be out until late October. One of them, who is a doctor, had a stroke, and he was providing the capital for what they were doing. Eventually I will have an actual machine to show. My buddy in Florida, who built my black machine, has his up and running, and is working on getting the amperage out of the coils that I was able to get with my coils. He was recently successful, so I may eventually have some video of HIS machine running also. Only time will tell.

The systems I am devoting my time to have no moving parts and are QUIET. After years of working with the generator and having some near misses with exploding rotors, it is nice to have a quiet setup that outputs power. There is always a danger when working with electricity, but I feel SO much safer!

[Rakarskiy]

A very clear example of magnetic Anapol!

Магнитный хранитель - это Невероятно! Самый загадочный магнитный эффект #Shorts Игорь Белецкий (youtube.com)
https://www.youtube.com/shorts/1ZAMosm2M54

Invention of the electromagnetic generator | Patreon

------------------------------------
Addendum:
At one of the physics sites, the topic of anapole and anapole moment (the correct modern name is toroidal moment) was discussed. Where the physicist pointed out that all this is not new and is used in electrical engineering with closed magnetic circuits. Naturally, she cited that everything is in wikipedia:

Toroidal moment - Wikipedia

She also noted that there are differences between the existing curricula of schools and technical universities. Specific knowledge on the subject is needed for a narrow circle of specialists. In one of the universities in the USA all this is taught in a detailed form, but exactly for those who will be engaged in it.


967681910.jpg
In modern physics, a variant of electromagnetic induction (Figure B) is described as toroidal toroidal moment.

Magnetic toroidal toroidal toroidal moment and its relation to the magnetoelectric effect.
The presence of a magnetic toroidal dipole moment T in condensed matter is due to the magnetoelectric effect: the application of a magnetic field H in the plane of the toroidal solenoid leads, through the Lorentz force, to the accumulation of current loops and thus to an electric polarisation perpendicular to both T and H. The resulting polarisation is Pi = εijkTjHk (where ε is the Levy-Civitas symbol). Thus, the resulting magnetoelectric tensor describing the mutually correlated response is antisymmetric.

Functioning of the device described in my publication, is based on the principle of EMF in closed magnetic circuits like all synchronous generators. Dr Holcomb's generator works on the same principle:

Electromagnetic generator, without rotation of the magnetic rotor in self-propelled mode. | Patreon