Motor Generators

[bistander]

Any electromagnetic generator depends on the power of the load, or rather its magnetic system depends on it. Therefore, all generators are calculated based on the power of the core's ability to accommodate the field from the phase with the load and excitation. Magnetomotive force is an interesting parameter that design engineers use in their calculations. If you do not know how it is expressed in calculations, you can say that you are an empty talker.

Holcomb may have had difficulties with scaling, but it is quite possible to make a simple model for yourself with an output power of no more than 1 kW. I only recommend that everyone do this quietly and without publicity.

The "working rotating field" is established (produced) by the rotor's field from the the field coil using external excitation and it is present even without a load on the armature in the typical alternator. There is voltage on the armature terminals at no-load. It is generating this voltage from the rotating magnetic field. How can you say that depends on the load?

Sure, the design of the alternator depends on the intended load, but your statements are wrong. The rotating magnetic field in the typical alternator is present even without the load, open circuit.

Your statement "like a synchronous generator, depends largely on the connected load to create a working rotating field" is false.

it is quite possible to make a simple model for yourself with an output power of no more than 1 kW. I only recommend that everyone do this quietly and without publicity

Sure, but it won't do what you claim. You do it. Or hire it done. Prove it to us. Like Solarlab. Disappeared. Failure. Like Holcomb. Disappeared. Failure.
bi

[Rakarskiy]

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[bistander]

2022-05-26_21233111.jpg

Yes, I am correct. Certainly the load current in the armature will have an effect on the field, called armature reaction, but it doesn't "depend(s) largely on the connected load to create a working rotating field", as you said, born out by the fact that the rotating magnetic field is present in the total absence of load current.
bi

[Rakarskiy]

It is immediately obvious that you do not understand the subject at all or are deliberately misleading.

The excitation current forms a magnetic flux sufficient to induce an EMF, but the electromagnetic moment of the anchor reaction in idle mode is equal only to the mechanical costs of rotation. Only when the load is connected is a working magnetic field formed, which consists of the excitation magnetic field and the magnetic field of the phase currents under load; for the resulting magnetic system of the generator, this is the full flux. It is this flux that forms the electromagnetic moment of the generator; for the rotor to rotate, an appropriate mechanical force of rotation is required. We exchange the physical effort for the electromagnet control system and get the same effect without physical rotation.

Let's call the excitation magnetic flux differently - it is the control magnetic flux, and the magnetic flux of the phase currents is the working field. Ideally, these fields are equal, but in reality, especially for amateurs, the field from the phase currents is either less or more than the excitation field. If you do not know the specifics, you can not even start designing. So if we correctly organize the excitation work - control of the magnetic flux that controls the generation process, we get a stationary rotor with rotating magnetic poles.

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Two slides about generation, but such a different way of interpreting what is happening there.

967681910.jpg

[bistander]

It is immediately obvious that you do not understand the subject at all or are deliberately misleading.

The excitation current forms a magnetic flux sufficient to induce an EMF, but the electromagnetic moment of the anchor reaction in idle mode is equal only to the mechanical costs of rotation. Only when the load is connected is a working magnetic field formed, which consists of the excitation magnetic field and the magnetic field of the phase currents under load; for the resulting magnetic system of the generator, this is the full flux. It is this flux that forms the electromagnetic moment of the generator; for the rotor to rotate, an appropriate mechanical force of rotation is required. We exchange the physical effort for the electromagnet control system and get the same effect without physical rotation.

Let's call the excitation magnetic flux differently - it is the control magnetic flux, and the magnetic flux of the phase currents is the working field. Ideally, these fields are equal, but in reality, especially for amateurs, the field from the phase currents is either less or more than the excitation field. If you do not know the specifics, you can not even start designing. So if we correctly organize the excitation work - control of the magnetic flux that controls the generation process, we get a stationary rotor with rotating magnetic poles.

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Two slides about generation, but such a different way of interpreting what is happening there.

967681910.jpg

Okay Rakarskiy,
You've thrown in a term with which I was unfamiliar and I assumed it was just an artifact of language differences. "a working magnetic field" as you call it can infer a magnetic field interacting with a moving charge, in which case there is, obviously, armature current in the dynamo.

But by the same token,

appropriate mechanical force of rotation is required

you admit that mechanical power is necessary for conversion to electrical power via the

electromagnetic moment

. That is the truth which I insist negates

exchange the physical effort for the electromagnet control system and get the same effect without physical rotation.

The "same effect" will not be realized. The apparatus becomes a transformer, not a generator. Either way, OU or free energy is not possible.
bi

[Rakarskiy]

The mechanical force required to rotate the rotor by breaking the magnetic connection between the rotor and the stator is a condition, not an energy for the transformation. These two excitation fields and the phase winding have sources, one in the rotor and the other in the stator. As long as there is no source in the stator, there is no braking, only idle braking.

If we move the magnetic field in the rotor by switching the excitation electromagnets, like the anchor of a DC motor, we get the same effect as a mechanical rotor. We get generation. If we perform switching by a pulse method, breaking the system of work of permanent magnets, you get a transformation - mutual induction.

You are in a stupor because you do not know how a transformer works. How interturn mutual induction works. This is the key point for the problem of implementing a magnetic rotor with a rotating field without rotating the anchor core.

Hippolytus Pixia's first generator was an alternator. | Patreon

[bistander]

The mechanical force required to rotate the rotor by breaking the magnetic connection between the rotor and the stator is a condition, not an energy for the transformation. These two excitation fields and the phase winding have sources, one in the rotor and the other in the stator. As long as there is no source in the stator, there is no braking, only idle braking.

If we move the magnetic field in the rotor by switching the excitation electromagnets, like the anchor of a DC motor, we get the same effect as a mechanical rotor. We get generation. If we perform switching by a pulse method, breaking the system of work of permanent magnets, you get a transformation - mutual induction.

You are in a stupor because you do not know how a transformer works. How interturn mutual induction works. This is the key point for the problem of implementing a magnetic rotor with a rotating field without rotating the anchor core.

Hippolytus Pixia's first generator was an alternator. | Patreon

No, it is you who are misinformed and promoting falsehoods because you fail to realize there is the electromagnetic moment at the air gap of both cases, diagram A and diagram B, shown below.

From your previous post, it indicates that you realize the involvement of the electromagnetic moment or mechanical force and displacement or mechanical power for the conversation process of mechanical power into electrical power, or what is known as generation, as depicted in diagram A. But somehow, you cannot recognize it in the case of diagram B. Are you blind to the obvious?

Also, in the example of diagram B, it is obvious that a continuous line of force from the rotor N pole to S pole that remains unbroken, will indeed cross the conductors of the coil passing through the magnetic core as that rotor rotates. Your argument about lines not cutting the conductors in your closed magnetic circuit is bunk because the magnetic circuit is not closed at the air gaps which allows for rotation and the magnetic moment facilitating mechanical to electrical energy conversion. Your devices without mechanical motion do not allow this and therefore are transformers, not generators.

I've got millions and millions of proofs, you have zero. Who is right?
bi

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Two slides about generation, but such a different way of interpreting what is happening there.

967681910.jpg

{edit}

If we perform switching by a pulse method, breaking the system of work of permanent magnets, you get a transformation - mutual induction.

"you get a transformation"
Yes, a transformer, electric to electric, not generation, or conversion of energy.

[Rakarskiy]

A difficult case! Well, okay - Transformer EMF, which is used to calculate the EMF of an alternator at idle.

emf-equation-of-a-transformer-l.jpg

An example of its justification in a transformer: EMF Equation of Transformer | Turns Voltage Transformation Ratio of Transformer | Electrical4U

Let's take the core in its classical sense and make the corresponding calculations. The dimensions of the core are made of M-19 steel (a = 15 mm, h = 50 mm; window 40 * 40 mm); windings with a hundred turns, primary and secondary. The maximum excitation current of the primary winding is 1 Ampere ( U=12V, r=0.89ohm), with a switching frequency of 60 Hz.

We model the magnetic induction in the core in the FEMM program. We get a value equal to 1.42 Tesla.

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Now, using the formula for the transformer EMF [E = 4.44NФf], we calculate the EMF of the secondary winding, which has the same number of turns equal to 100 turns.

Е = 4.44 * 100 turns * 1,42 Т * (0.015 m * 0.05 m) * 60 Hz = 28.37 V

Untitled-1.png

As you can see, the statement that this formula is equivalent for the primary and secondary windings in terms of voltage in turns is not true.

Further, everyone knows that the directions of current in the turns of the primary and secondary windings are opposite, we will model this situation.
As we can see, the magnetic flux in the core is actually equal to zero and does not have a closed state in a closed core. This is already the second statement of the formation that is false.

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I highly recommend studying the operation of a magnetic amplifier, when and what triggers it. I've been through all of that.

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Good luck in serving the *******s of Silicon Valley.

[Rakarskiy]

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The invention of the electromagnetic generator | Patreon

HOW A TRANSFORMER WORKS (MUTUAL INDUCTION) | Patreon

Hippolytus Pixia's first generator was an alternator. | Patreon

Motionless ALTERNATOR / Нерухомий АЛЬТЕРНАТОР / ALTERNADOR estacionario | Patreon

[bistander]

You say:

The mechanical force required to rotate the rotor by breaking the magnetic connection between the rotor and the stator is a condition, not an energy for the transformation.

and this:

But with an electromagnetic generator this principle is not completely fulfilled. First, mechanical power ( P=Fv ) can in no way be transformed into electrical power (W=IU).

But then elsewhere you say this:

When the load is connected, a current is formed in the secondary circuit, which forms its own source of magnetic field. Breaking two fields between the rotor and stator costs mechanical costs equal to the generated electrical power.

from here:
https://overunitymachines.com/index....ic,13.165.html
post #166

bi

[Rakarskiy]

Re: Electromagnetic Generator Theory #170 Electromagnetic Generator Theory - Page 35

The very first time the system (the coal lobby with the nascent oil lobby) opposed free energy technology that could change the world back in the late 19th century was - Keele's ethereal steam generator. Generator (Keely) – AnthroWiki

At the expense of evidence? Everyone has to make their own choice. I'm not new to electricity, but I've had to rethink electricity many times to understand how things really work.

First, you need to understand what electricity is in your outlet and how it works. Where does the component called - CURRENT - come from?

EMF, CURRENT, VOLTAGE, RESISTANCE. | Patreon

Learn to design a typical electromagnetic generator with stator and rotor cores (closed magnetic circuits) with a ratio of 0.89-0.92 between the output power and the mechanical power required to rotate the rotor.

Next, you need to learn how to control the operation of this generator with an electrical circuit with different loads that are constantly turned on or off according to their own operating algorithm.

When you master everything and know exactly what you're talking about, you'll look at all the offers for great gadgets differently.

But I still do not advise anyone, if you build yourself a self-propelled source, to show it to everyone. Firstly, you are putting yourself and your family at risk, and secondly, get ready to receive an endless stream of garbage at your address. This stream will be fed by the "system itself", for which your device is undesirable and increases the risk of losing dominance in the electricity sales market.

In my posts on Patreon, I try to explain everything and even give advice on how to do it. But everyone has their own choice, their own level of education and opportunities, their own skills - technical, design, etc.

[bistander]

Re: Electromagnetic Generator Theory #170 Electromagnetic Generator Theory - Page 35

The very first time the system (the coal lobby with the nascent oil lobby) opposed free energy technology that could change the world back in the late 19th century was - Keele's ethereal steam generator. Generator (Keely) – AnthroWiki
At the expense of evidence? Everyone has to make their own choice. I'm not new to electricity, but I've had to rethink electricity many times to understand how things really work.

First, you need to understand what electricity is in your outlet and how it works. Where does the component called - CURRENT - come from?

EMF, CURRENT, VOLTAGE, RESISTANCE. | Patreon

Learn to design a typical electromagnetic generator with stator and rotor cores (closed magnetic circuits) with a ratio of 0.89-0.92 between the output power and the mechanical power required to rotate the rotor.

Next, you need to learn how to control the operation of this generator with an electrical circuit with different loads that are constantly turned on or off according to their own operating algorithm.

When you master everything and know exactly what you're talking about, you'll look at all the offers for great gadgets differently.

But I still do not advise anyone, if you build yourself a self-propelled source, to show it to everyone. Firstly, you are putting yourself and your family at risk, and secondly, get ready to receive an endless stream of garbage at your address. This stream will be fed by the "system itself", for which your device is undesirable and increases the risk of losing dominance in the electricity sales market.

In my posts on Patreon, I try to explain everything and even give advice on how to do it. But everyone has their own choice, their own level of education and opportunities, their own skills - technical, design, etc.

Hello Mr. Rakarskiy,
Here you refer to your essay:
https://www.patreon.com/posts/emf-current-112414335

Near the beginning you write:

but first try to solve the problem :
Task : It is necessary to calculate the EMF [ E =? ] of a single-phase generator, with a connected load of power: Pz = 1 kW (heating element, active load); frequency in the circuit: f = 50Hz ; voltage at the generator terminals; U= 220V , active resistance of the generator phase: r = 5 Ohm . The resistance of the connecting wires does not need to be taken into account.

You have all the data for the decision, I'm just sure that many will make a blunder.

It is apparent that you have made several significant blunders.

91466693 (2).jpg
After the diagram you write:

we simulate a circuit with an alternating generator, the current of which is rectified through a diode bridge, we also plan a capacitor C1 in the circuit, which acts as a storage device for smoothing pulses. As a result, we get a constant voltage of 220V on the load.

When you previously stated the generator output voltage of 220V at 50Hz, then you do not see a constant voltage of 220V on the load.

brcap.gif
Vac = Vdc x 0.71
________
Ref:
https://www.sowter.co.uk/rectifier-t...alculation.php
&
https://www.electronics-tutorials.ws/diode/diode_6.html

The correct voltage on the load is 220V / 0.71 = 310V.

Later in the essay, you insert a battery into the load circuit, saying:

Let's assume that the EMF of the generator phase is 242V, the battery voltage is 220V,

then:

It remains only to calculate the EMF: E = U + ∆U = 220V + 242.703V = 462.703V.

and then:

We solved the problem that the EMF of the generator phase is equal to 462.7 Volts,

Which is quite wrong.
I've mentioned before that you really need a good technical editor/proof reader. I see similar issues with everything that I read from you. And then you have the audacity to proclaim that scientific and technical education is a lie, when it's obvious that you don't understand or know what you're talking about. Read a few good textbooks and take an engineering course or two.
bi

[Rakarskiy]

You're the one hallucinating, I gave an example in my piece EMF, CURRENT, VOLTAGE, RESISTANCE. | Patreon with the measurements of a garage handyman.

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.
145188291.jpg
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.
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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.
http://www.energeticforum.com/fileda...0&d=1727417932
2024-09-27_084708.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.

[bistander]

You're the one hallucinating, I gave an example in my piece EMF, CURRENT, VOLTAGE, RESISTANCE. | Patreon with the measurements of a garage handyman.

Rakarskiy,
Your excuse has nothing to do with the two blunders on which I called you, namely voltage out of the bridge rectifier and inserting a battery into the load circuit. Do you not see the errors of those calculations?
bi

[Rakarskiy]

You have proven nothing because you have never made a controlled bridge rectifier. If you had, you wouldn't be writing this nonsense.

But I have not only done it, but also taught others. For one mini-hydropower plant.
The voltage at idle (no load) was 60-65V, and the battery voltage at which the charge occurred was 24V. So, the thyristor was given a switch-off signal when the voltage reached 27V, which corresponds to a full charge of the 24V battery. Before that, where did the rest of the voltage, or rather the EMF, go?