Computing Alternator Back EMF as a function of load current

[Bcavender]

I am attempting to make a basic model of an alternator and determine terminal voltage levels at several different current levels.

The alternator design looks like this:

900 RPM fixed, 60hz, single phase
2000 gauss permanent magnet Flux Density, Magnet circumferential velocity 9.58 m/sec,
4 Coils w/220 turns each, 57.5 meters of 15awg/coil, 230m total
Total DC Resistance 2.4 ohms, Total Inductance 9.5 milliHenries

I compute open circuit voltage to be 121.1v RMS and 171.2 P-P.

As current begins to flow, I understand that back emf is generated as the load current is also moving in the magnetic field, but in opposition to the direct voltage generated.

I would like to calculate the back emf at 0.5, 1, 2, 3 and 5 amps load current.

I have searched the net for a method/formula that would allow me to calc the counter emf as the alternator load current increases, but I am not finding a good, reliable resource for this.

Any suggestions about resources or formulae to move this ahead would be greatly appreciated!Bruce

 

[CWatters]

I am attempting to make a basic model of an alternator and determine terminal voltage levels at several different current levels.

The alternator design looks like this:

900 RPM fixed, 60hz, single phase
2000 gauss permanent magnet Flux Density, Magnet circumferential velocity 9.58 m/sec,
4 Coils w/220 turns each, 57.5 meters of 15awg/coil, 230m total
Total DC Resistance 2.4 ohms, Total Inductance 9.5 milliHenries

I compute open circuit voltage to be 121.1v RMS and 171.2 P-P.

As current begins to flow, I understand that back emf is generated as the load current is also moving in the magnetic field, but in opposition to the direct voltage generated.

I would like to calculate the back emf at 0.5, 1, 2, 3 and 5 amps load current.

I have searched the net for a method/formula that would allow me to calc the counter emf as the alternator load current increases, but I am not finding a good, reliable resource for this.

Any suggestions about resources or formulae to move this ahead would be greatly appreciated!Bruce

Got a reference that discusses back EMF in a generator? Isn't that what actually produces the output voltage?

I think you just need to calculate the I*R loss.

 

[Bcavender]

Thanks for the thought. References that go into deep detail are seemingly rare ... or I just haven't hit the magic link yet :^)

You are right that I*R loss reduces output voltage as current loads up, but I believe back emf is on top of that.

From what I can determine in a generator no back emf is produced when the current is zero. You just have the Faraday induced voltage of the magnetism pressuring charges in the wire with no where to go ... just an open circuit voltage. I believe back emf comes into play when current begins to flow which creates a new, opposing magnetic field ... that bucks the direct magnetic field ... reducing the net magnetic field ... backing off the output voltage as load builds.

This back emf has to be the source of increased mechanical torque back to the prime mover as more electrical energy is delivered to the load ... to keep total energy in/out/lost balanced.

Still in search ...

 

[jim hardy]

From an old old thread at https://www.physicsforums.com/threads/armature-reaction-drop.826513/#post-5191404Let's really simplify our thinking down to a synchronous machine with a one turn armature and a permanent magnet rotor.
We'll allow the complication that the rotor is machined to produce sine shaped flux, though.
Recall that voltage and flux have a derivative/integral relationship meaning our sine shaped flux gives a cosine shaped voltage
because e = - n*dΦ/dt
and cosine is 90 degrees out of phase with sine. That's a quarter cycle.

so--- peak voltage appears at the flux zero crossing
and zero voltage appears at the flux peak
because of that sine-cosine relation.

Do we all accept that unity pf current is in phase with terminal voltage ?
and zero pf current is 90 degrees ahead or behind it ?
............

We'll make our simple one turn machine with only one pole pair so electrical and mechanical degrees are the same, 90 deg = ¼ turn.

Okay,
please -
pardon my crude sketch i just don't do well with computers.

The circles are my one-turn armature windings. I've freeze-framed my thinking as mentioned earlier, so these would be like high speed photographs.

Left sketch:
Flux from armature zero pf(real) current is perpendicular to flux from field
so there is little interaction beyond a phase shift, and that's why there is a power angle.
At this instant, zero pf current is zero so it does nothing.

Right sketch:
A quarter turn later (or earlier) when unity pf (real?) current is zero
and reactive(zero pf) current is maximum,
look where the rotor is !
That's why zero pf current in the armature adds or subtracts directly to field flux. It peaks when rotor pole is right underneath the armature turns.

You should stick with the notation in your textbook

but i find this mental image useful to understand why there's such a thing as synchronous reactance. It's not same as leakage reactance.
It does affect terminal voltage by reducing flux in the machine(,or raising it )
but the external characteristics are same as if it were another circuit element
and that's how most authors present it.

Is that any help ?

old jim

 

[dlgoff]

@Bcavender
I don't know if there's anything that may be useful for your modeling but take a look at this page from ANIMATION OF ELECTRIC MACHINES.

 

[Bcavender]

Hi Jim,

I appreciate your response.

Help me catch up to where you are. I am confused by the reference to power factor. If the alternator had a purely resistive load, wouldn't the apparent power be equal to the real power?

Maybe I am missing something simple here ... what current assumptions are you making for the winding?

Thank you!

Bruce

-----------------------------------------------------------------------------------

From an old old thread at https://www.physicsforums.com/threads/armature-reaction-drop.826513/#post-5191404Let's really simplify our thinking down to a synchronous machine with a one turn armature and a permanent magnet rotor.
We'll allow the complication that the rotor is machined to produce sine shaped flux, though.
Recall that voltage and flux have a derivative/integral relationship meaning our sine shaped flux gives a cosine shaped voltage
because e = - n*dΦ/dt
and cosine is 90 degrees out of phase with sine. That's a quarter cycle.

so--- peak voltage appears at the flux zero crossing
and zero voltage appears at the flux peak
because of that sine-cosine relation.

Do we all accept that unity pf current is in phase with terminal voltage ?
and zero pf current is 90 degrees ahead or behind it ?
............

We'll make our simple one turn machine with only one pole pair so electrical and mechanical degrees are the same, 90 deg = ¼ turn.

Okay,
please -
pardon my crude sketch i just don't do well with computers.

The circles are my one-turn armature windings. I've freeze-framed my thinking as mentioned earlier, so these would be like high speed photographs.
View attachment 223462
Left sketch:
Flux from armature zero pf(real) current is perpendicular to flux from field
so there is little interaction beyond a phase shift, and that's why there is a power angle.
At this instant, zero pf current is zero so it does nothing.

Right sketch:
A quarter turn later (or earlier) when unity pf (real?) current is zero
and reactive(zero pf) current is maximum,
look where the rotor is !
That's why zero pf current in the armature adds or subtracts directly to field flux. It peaks when rotor pole is right underneath the armature turns.

You should stick with the notation in your textbook

but i find this mental image useful to understand why there's such a thing as synchronous reactance. It's not same as leakage reactance.
It does affect terminal voltage by reducing flux in the machine(,or raising it )
but the external characteristics are same as if it were another circuit element
and that's how most authors present it.

Is that any help ?

old jim

 

[jim hardy]

Help me catch up to where you are. I am confused by the reference to power factor. If the alternator had a purely resistive load, wouldn't the apparent power be equal to the real power?

Yes. That's unity power factor.

As current begins to flow, I understand that back emf is generated as the load current is also moving in the magnetic field, but in opposition to the direct voltage generated.

i am confused by your term "back emf" .
Resistance of the windings means there'll be an ohmic drop proportional to the current and in phase with it.
Inductance of the windings means there'll be a reactive drop proportional to the current and out of phase with it .

That's why you see phasor diagrams in all electric machinery textbooks .

Generators are described as an internal voltage in series with an inductance and resistance.

This back emf has to be the source of increased mechanical torque back to the prime mover as more electrical energy is delivered to the load ... to keep total energy in/out/lost balanced.

In DC machines, back EMF X armature current does equal power being converted

AC machinery texts use "power angle" .
Consider the rotor just a rotating magnet, as i drew up above. Its field rotates with it.
The stator creates a rotating magnetic field.
If the rotor is behind by the stator field and being dragged forward by it then the machine is acting as a motor.
If instead the rotor is ahead of the stator field and being held back by it then the machine is acting as a generator.
So it's that power angle between rotor and stator fields that engineers use as a measure of power.
That's a physical angle and you can watch it by synchronizing a stroboscope to terminal voltage and shining it on the shaft keyway.
Armature current will flow to make the sum of the two fields align with terminal volts.

"Back Emf" in an alternator is a new concept for me. Maybe it's just a change in terminology , my machinery classes were after all fifty years ago...
Can you show from where you got the term ? Maybe it'll jostle my thinking into synch with the times.old jim

 

[Tom.G]

@jim hardy
Perhaps the OP is considering the magnetic field being distorted by the load current. As in the same reason brushes in a DC motor are rotationally offset from their 'neutral' position.

 

[jim hardy]

Perhaps the OP is considering the magnetic field being distorted by the load current. As in the same reason brushes in a DC motor are rotationally offset from their 'neutral' position.

Possibly. just can't make the words 'Back EMF" work out to any mental picture of a synchronous machine that's in my repertoire.

It is a fact that in a DC machine the product of counter emf and armature current is the electric power being converted into mechanical or vice versa.
In an AC machine armature current distorts the field similarly, adding to it in the fashion of vectors and phasors.

If by "Back EMF" he means the voltage drop across what's called in traditional machinery courses "Synchronous Impedance"
he'll benefit from this online lecture or a similar one
https://www.slideshare.net/mkazree/chapter-4-synchronous-machine

Synchronous Impedance is sum of R_S and X_S .
Internal voltage E_G is proportional to field strength which for a permanent magnet rotor is constant ,
he gave open circuit voltage of 121.1 volts so that's E_G

Resistance of his stator winding he gave as 2.4 ohms so that's R_S
He calculated stator inductance of 9.5 millihenries which at 60 hz is j3.58 ohms so that's a first estimate of X_S
sum of those is 2.4 +j3.58 = 4.31∠56.2° ohms

so he could try solving the above circuit with those values.

I'd want a cross check though. A couple things about the original post confuse my alleged brain..
1.

900 RPM fixed, 60hz, single phase
...
4 Coils w/220 turns each,..

900 RPM 60 hz requires EIGHT coils to make 4 pole pairs
so do we have 220 turns per coil or 220 turns per pair of coils ? or is this an 1800 RPM machine with four coils ?

2.

2000 gauss permanent magnet Flux Density,

Inductance L = NΦ/I
flux per amp Φ/I = L/N
so flux due to armature inductance is L/N = 0.0095 /220 = 4.32 X 10^^-5 Webers per amp
and that is how much the field flux should be affected by armature flux.

But the Gauss he gave us is flux density not flux. His 2000 Gauss is 0.2 Tesla, or 0.2 Weber per square meter. Without area we don't know flux.

so i don't know how to cross check whether the synchronous reactance estimated from his armature inductance is reasonable.
Need areas of poles, air gap., magnetic path length and permeability to do it right. Last one i remember doing was winter of '65 so excuse my rustiness.

Perhaps @Bcavender will show us his calculations and a sketch.

old jim

 

[Bcavender]

Hi Jim,

Most likely I am misusing the "back emf" term as a total electrical machinery noob/experimenter and it may not be applicable at all in the alternator lexicon. I assumed that it was a similar interaction affect as when dc rotor/stator fields work on each other. I think I am beginning to get a grasp on the induction/excitation voltage part, but my knowledge of how the magnetic fields interact is deficient (among many other things :^)

Your lead/lag between the rotor and stator for motor/gen situations makes it clearer. Good analogue. That helps.

~~~~~~~~~~~~~~~~

I had conceived a 4" radius rotor, 2" thick with eight permanent magnet poles on 45 degree centers around the circumference with four, single phase series connected coils such that the working length of each coil side would be magnetically 180 degrees out of phase from the other side of the coil (one magnet pushing charge/one pulling charge on opposite sides of a coil). My thinking was 90 degrees rotation of the 8 magnet rotor would create a full cycle, one full rotation would create 4 cycles and 15 revolutions per sec (900rpm) would create 60 cycles/sec. This is my first try at a concept. Am I missing something in my thinking?

~~~~~~~~~~~~~~~~

On the flux, I was using the B field value being swept by the length of wire at a circumferential velocity which may be a different route to getting at total flux:

emf = Tesla * Wire Length * Velocity So one wire on one side of a coil would induce ~ 2000 gauss /10,000 * 0.0508 m * 9.576 m/sec = 0.0975 v

Or 0.195 v per loop X 220 loops/coil = 42.9 v Times four coils = 171.6 v (peak) 121 v (rms)

~~~~~~~~~~~~~~~~

I did run across an interesting tutorial in my continuing search for generator output voltage vs load current model. This passage was interesting:

V = E-IZ where E is the Excitation voltage, I the current and Z the impedance of the Generator winding. Here V is the terminal voltage across the load.
Thus you see how Generated voltage and load terminal voltage is related.​

Open circuit at zero current delivers the direct emf to the output. The second term indicates a lower voltage production as the fields interact and load current drawn. I am not sure how accurate/practical this formula would be in real life, but it seems to make some sense.

~~~~~~~~~~~~~~~~

At least that is my first shot at a conceptual approach to my experiment based on what I have picked up here and there from different websites.

I greatly appreciate the great assistance and suggestion/guidance here. Self teaching is a bit of a slow process sometimes.

All suggestions/ideas welcome!

THANK YOU!

Bruce
--------------------------------------------------------------------------------------------------------------------------

Yes. That's unity power factor.

i am confused by your term "back emf" .
Resistance of the windings means there'll be an ohmic drop proportional to the current and in phase with it.
Inductance of the windings means there'll be a reactive drop proportional to the current and out of phase with it .

That's why you see phasor diagrams in all electric machinery textbooks .

Generators are described as an internal voltage in series with an inductance and resistance. In DC machines, back EMF X armature current does equal power being converted

AC machinery texts use "power angle" .
Consider the rotor just a rotating magnet, as i drew up above. Its field rotates with it.
The stator creates a rotating magnetic field.
If the rotor is behind by the stator field and being dragged forward by it then the machine is acting as a motor.
If instead the rotor is ahead of the stator field and being held back by it then the machine is acting as a generator.
So it's that power angle between rotor and stator fields that engineers use as a measure of power.
That's a physical angle and you can watch it by synchronizing a stroboscope to terminal voltage and shining it on the shaft keyway.
Armature current will flow to make the sum of the two fields align with terminal volts.

View attachment 223606 "Back Emf" in an alternator is a new concept for me. Maybe it's just a change in terminology , my machinery classes were after all fifty years ago...
Can you show from where you got the term ? Maybe it'll jostle my thinking into synch with the times.old jim

 

[jim hardy]

Build one and try it out ?

I'm accustomed to poles being adjacent one another not diametrically opposite.
http://www.marineengineering.org.uk/page55.html

Your coil area appears to be 4 square inches , 2X2 if i read your post correctly. That's 4/(39.37)^{^2} = 2.5806^e-3m²

If flux density in your as-built configuration is 0.2 Tesla (which i question)
then each coil links 0.2 Wb/m² X 2.5806e^-3m² = 5.161 e-⁴ Webers

if we assume it's sine shaped , ie flux Φ = 5.161e-⁴ sin(120pi X t)
volts per turn would be dΦ/dt = 5.161e-⁴ X 120pi cos(120pi X t) = a sinwave of 0.1946 volts peak, about 0.138 volts RMS per turn.
That's same as you got by blv... What Great News !

So the question to me becomes ' how was your magnet specified ?' If it makes 2000 Gauss in free air, what will it make when surrounded by your magnetic structure ? In other words how much flux can it push through your air gap ? from https://www.ibsmagnet.com/pdf/en/ibs_e.pdf

Plodding along one step at a time,

old jim

 

[jim hardy]

I did run across an interesting tutorial in my continuing search for generator output voltage vs load current model. This passage was interesting:

V = E-IZ where E is the Excitation voltage, I the current and Z the impedance of the Generator winding. Here V is the terminal voltage across the load.
Thus you see how Generated voltage and load terminal voltage is related.

The concept is called "Armature Reaction" and "Synchronous Impedance". It's why alternators are inherently current limiting . That sure simplified automobile voltage regulators in the early 60's, they no longer needed to incorporate an armature current limiter.

Keep on with your studies. I hope you build a machine.

 

[Bcavender]

Jim,

Thank you for your efforts! It IS great news that our computations arrived independently at the same value. I am not used to getting an answer correct without at least several iterations :^) That's dang near encouraging!

I will dig into the literature you referenced. Good looking stuff.

On the two magnet examples ... the first makes intuitive sense that with double the magnets, you will double the B field. The second with the iron hitch for concentrating the field into a magnetic circuit is less clear from the text there ... but it seems to say that the iron magnetic path will "again" double the field over the previous example ... or achieve 4X the field density of a single magnet alone? Am I following that correctly? (I was able to get my hands on a DC Gaussmeter to experiment with. A quick test gave me just over 2000g about 3mm above the surface of a test Neo. That is where the 2000 came from.)

I need to experiment with the different current levels to estimate the I * Z components to predict the terminal voltage drop as it loads up. Going to have to brush up on my real/imaginaries. I remember enough to recognize that notation, but it has been several decades since I touched that particular math ... LOL!

The next big leap after that is coming to grips with the method of how to create the requisite number of sinusoidal magnetic fields with magnets around a rotor ... and possibly harder is how arrive at the physical dimensions of some number of coils will interact with those varying fields' to come close to the predicted total emf. At the extremes, I can visualize that at too small a rotor diameter everything gets too tight to build and too large a diameter could turn the device into a pulse generator ... far from anything like a sine wave. With little experience it is difficult to gage just how critical this really is. I would guess that it is likely more critical than I perceive it to be right now.

I need to put together some type test rig and just experiment I guess. I have seen several software simulation packages that seem to do something like this ... MotorSolve/magNET/etc, but these are targeted at high end companies bound for large markets ... and priced accordingly well out of an independent student's affordable range. I was hoping some enterprising academic with some time on their hands might have developed some type entry level software (they could use to supplement their retirement years) with very rough modeling power, but I have been unable to find such an animal as yet.

Plodding here too ... I appreciate your comments and help!Bruce

 

[jim hardy]

but it seems to say that the iron magnetic path will "again" double the field over the previous example ... or achieve 4X the field density of a single magnet alone? Am I following that correctly?

Yes. We study circuits where what flows, current, is constrained to the wires. And it clogs our thinking.
Magnetic fields know no such constraint. What flows, flux, is free to roam. What pushes flux is MMF , measured in Oersteds or Amp-Turns.
It takes considerable MMF to push flux through air. But not much at all to push it through iron.
So if your magnetic path includes some iron , that iron is almost a short circuit for flux.
That means your MMF source has leftover 'oomph' to push more flux through the remaining air.
That's why the iron stirrup gives you more flux .

In circuits we don't usually deal with current density, just current ..
And we don't often use electric field strength just volts.

Physics majors and antenna engineers work more in fields, current density J and volts per meter. Physics guys actually apply them to circuits !

Magnetics is taught as fields because that's how magnetism behaves, it's a field not constrained to wires. So the terminology will be flux density and MMF . Magnetics still uses a mix of cgs units like Gauss and Oersted, as well as SI units like Tesla and Amp per meter.

In your case you are pushing flux with a permanent magnet , not an electromagnet. That you have a Gauss meter is awesome. Experiment away until you get "that intuitive feel" . .

I remember enough to recognize that notation, but it has been several decades since I touched that particular math ... LOL!

It'll come back quickly. Draw pictures and work neatly one step at a time.