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From the answers to the following question (1), I gather that it is theoretically possible to operate an alternator / dynamo / generator:

  1. Without permanent magnets.

  2. With an external energy source for the electromagnet which substitute themselves to the permanent magnets.

  3. Even without an external energy source, provided that the current needed for the electromagnets can be tapped from the energy provided by te device itself (via either a temporary storage (battery or capacitor) or via the use of the residual magnetism of the electromagnets's cores).

My question is:

Is there a way to derive theoretically the proportion of the current (or of the power) produced by the alternator which is lost (has to be re-injected) in order to supply the electromagnets of the device with power?

In other words, at what costs in terms of percentage of the output does the choice of operating an alternator without electromagnet come?

(1) Source of initial magnetic field in an alternator

Serge Hulne
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    Did you mean "without permanent magnet" in the last sentence? | The excitation field MUST come from somewhere. Choices: Permanent magnets / fed back from alternator output / from undependent source eg battery. | I cannot think of any proper alternatives. Some few machines have an internal alternator to provide field energy for the main alternator. IF you can "bootstrap" the machine on externakl magnetism it will then run and you then do not need brushes or slip rings. | Cost of self excitation seems to be about 20%-30% in car alternators BUT they are optimised for cost, not efficiency. – Russell McMahon Jun 20 '21 at 08:03

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Is there a way to derive theoretically the proportion of the current (or of the power) produced by the alternator which is lost (has to be re-injected) in order to supply the electromagnets of the device with power?

The proportion is not a fixed relationship. There are a lot of details of the machine design that determine how much power is required to furnish the excitation current or power. Somewhere in the design process, you determine the required magnetic flux. You then design the excitation and regulation system. During that process, you determine the power required. You may find some proportion estimates based on previous designs. Estimates may give typical values for various sizes of machines and types of excitation and regulation systems.

  • So, according to you, there is no rule of the thumb, it’s totally on a case-to-case basis? – Serge Hulne Jun 18 '21 at 19:29
  • @SergeHulne I can't speak for Charles, but I suspect he's just saying that empirical results, themselves born from real cost and technology limits in manufacturing, together with theoretical considerations are highly inter-related and that a truly useful rendition would go far beyond what's reasonable here. I know something about flash tubes. If you exclude radiation transport and atomic interactions -- wrong, but simplifying -- then apprehending gas discharge requires at least 2 spatial and one time dimension of PDEs coupled to at least 6-dim ODEs. There's no way I'd write that here. – jonk Jun 18 '21 at 20:28
  • @SergeHulne But if you wanted to design a flash tube, you'd need that (and much more.) It's not that there isn't a methodology. It's just that writing it up here would exceed all reasonable bounds. And failing to write that much, would mean that there's no real understanding about the methods used in designing a flash tube that meets design specifications. So what's the point in trying, at this site? I think that is more where Charles was pointing. But that's just my opinion. You don't need to always assume the worst about the writer, or the situation itself, when reading an answer. – jonk Jun 18 '21 at 20:30
  • @jonk My question is about estimating the efficiency of a kind of dynamo, basically, not about discharges in gases… – Serge Hulne Jun 18 '21 at 21:21
  • @ Serge Hulne: There may be "rules of thumb," but I doubt that there is just one that covers the many orders of magnitude that separate the smallest alternator / dynamo / generator and the largest. In addition, electromagnets have the advantage of electronic adjustability. There are also several methods for transferring the power from the stationary world to the rotor of the machine. Yes, you can derive theoretically the power required for excitation and state that as a percentage of the machine output, but that is no easy task and it can not be stated simply; easier to look at historical data –  Jun 18 '21 at 22:01
  • I guess one way to minimise the loss of energy by Joule effect is to minimise the resistance of the windings of the electromagnets by using : different alloys, different diameters for the winding’s conductors and some kind of heat management scheme (three parameter which have an impact on the energy dissipation in the windings owing to Joule’s effect). – Serge Hulne Jun 18 '21 at 22:21
  • @SergeHulne You misunderstand what I wrote. Or, at least, why I wrote it. I had no intention of discussing gas discharge, except by way of making a point. Which, apparently, you missed. Oh, well. – jonk Jun 19 '21 at 02:50
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Did you mean "without permanent magnet" in the last sentence?

The excitation field MUST come from somewhere.

Choices: Permanent magnets / Fed back from alternator output / From independent source eg battery.

I cannot think of any other proper alternatives (but they may exist :-).

Some few machines have an internal alternator to provide field energy for the main alternator.
IF you can "bootstrap" the machine on external magnetism it will then run and you then do not need brushes or slip rings.

Cost of self excitation seems to be about 20%-30% in car alternators BUT they are optimised for cost, not efficiency.

Russell McMahon
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