Synchronous Generator: Questions & Answers

In summary: I think you are talking about an alternator, and the brushes are in contact with a slip ring. So the rotor has a dc current, and an ac voltage is induced in the stator. In summary, the current in the rotor will lag behind the induced emf in the stator. This is because the stator has coils of its own, which are inductors, and the current in the rotor will lag behind the induced emf.
  • #1
Bassalisk
947
2
Few questions about Synchronous generator.

A rotor is rotating with angular frequency ω. In this rotor we have windings, which produce magnetic field, through a DC supply.

This supply is connected to a rotor with brush contacts.

Now this field induces a emf in windings, which are in stator.

Few questions:

If i connect a load to a stator(generator), current will start to run. This current creates its own changing magnetic field. How come this field doesn't affect the field produced by rotor? Shouldn't we at one point have 0 induced emf? Since 2 fields cancel out. I am probably getting this wrong.

And it says here in my book, that the current will lag behind induced emf in stator. Why?

Thank you.
 
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  • #2
This is an interesting question. I think you are talking about an alternator, and the brushes are in contact with a slip ring. So the rotor has a dc current, and an ac voltage is induced in the stator.

If the stator is connected to a resistive load, the ac current is in phase with the stator voltage. This current will in turn induce an emf in the rotor, which may result in an ac current component in the dc rotor current, which in turn can induce a phase shifted emf in the stator.

If the dc current in the rotor comes from a current source, the induced ac emf in the rotor will not produce any ac current in the rotor. End of story. If the source of current in the rotor does not come from a current source, then the induced ac emf in the rotor (which is phase shifted 90 degrees from the current in the stator by Faraday's Law) will produce an ac current in the rotor, also phase shifted by 90 degrees. Now, this current will induce an ac emf in the stator, phase shifted by another 90 degrees. So the net effect is to produce a back emf in the stator, that is 180 degrees out of phase with the main (primary) induced emf. Does this sound correct?

The basis for this lies in Faraday's Law. In Faraday's Law, ∫E·dl = V = -(d/dt)N∫B·dA, where V is the induced emf, N is the number of turns, A is the area of the coil, and B is the magnetic field. The (d/dt) produces a 90 degree phase shift. The magnetic field B is produced by a current, and the induced quantity is a voltage (emf).
 
  • #3
Bob S said:
This is an interesting question. I think you are talking about an alternator, and the brushes are in contact with a slip ring. So the rotor has a dc current, and an ac voltage is induced in the stator.

If the stator is connected to a resistive load, the ac current is in phase with the stator voltage. This current will in turn induce an emf in the rotor, which may result in an ac current component in the dc rotor current, which in turn can induce a phase shifted emf in the stator.

If the dc current in the rotor comes from a current source, the induced ac emf in the rotor will not produce any ac current in the rotor. End of story. If the source of current in the rotor does not come from a current source, then the induced ac emf in the rotor (which is phase shifted 90 degrees from the current in the stator by Faraday's Law) will produce an ac current in the rotor, also phase shifted by 90 degrees. Now, this current will induce an ac emf in the stator, phase shifted by another 90 degrees. So the net effect is to produce a back emf in the stator, that is 180 degrees out of phase with the main (primary) induced emf. Does this sound correct?

The basis for this lies in Faraday's Law. In Faraday's Law, ∫E·dl = V = -(d/dt)N∫B·dA, where V is the induced emf, N is the number of turns, A is the area of the coil, and B is the magnetic field. The (d/dt) produces a 90 degree phase shift. The magnetic field B is produced by a current, and the induced quantity is a voltage (emf).


It make sense to me, almost fully. Just few more things.

I feel like it doesn't matter if the load is purely resistive or not, i feel like the current will always lag behind emf in stator, because stator has coils of its own. Which are inductors, making the current lag. Am I correct?

And this induced emf in rotor which induces emf in stator, how does this affect the alternator?
 
  • #4
indeed an interesting question.

one needs to go to OLD books to get the physical explanation for many things. I love my old 1901 Thompson dynamo book for that reason

Sigh i wish i were computer savvy enough to make drawings.

The answer to your question is - basically yes. the concept neesds to be grasped, but in the today's culture most textbooks will just give you equations. if you have the mental picture you can figure out the formula instead of remembering it..

the flux from current flowing in the armature indeed is real so the total flux in the machine is a vector sum of the flux from field and flux from stator current.
most books I've seen describe mmf's instead of flux, and that's fine they add too..

The KEY point to this is that voltage is derivative of flux
e = n d(phi)dt
and they go to a lot of trouble to make sure the rotor produces a sinusoidal flux so that generator voltage will be a sinewave - it's the only function i know of whose derivative looks just like itself.
So - since flux is a sinewave so is terminal voltage, but terminal voltage is NOT at sinewave peak when north pole of field is directly under the coil and flux is peak.
recall derivative of sinewave shifts 90 degrees
which means terminal voltage is at peak when flux is crossing zero, ie when field is mechanically perpendicular to coil

but for a resistive load, armature current is in phase with terminal volts
so the mmf(flux) from armature current is perpendicular to that from field
and since the addition is a vector one
the total mmf(flux) isn't affected very much by contribution from load current.
Pythagors rules -add 1 +j( .1) and you get not 1.1 but 1.005 and that's half percent change not ten.

But the effect is there.
It is called "Armature Reaction" in the books and that seems a good name.
a machine with 10% armature reaction as described above would regulate pretty well.

If the load current were instead reactive, say maybe the machine is feeding a capacitor,
load current would be 90 degrees ahead of terminal voltage
which means its mmf would mechanically align directly with the rotating field
and armature reaction's mmf would add directly to field's mmf.
for that reason real and reactive components of load current are called "Direct" and "Quadrature" current, once again not bad names for them.
Reactive current does affect flux and terminal voltage way more than real current.
In the power plant we control the watts with amount of steam to turbine, and the vars(volt-amps-reactive) with voltage regulator.

you can see this yourself if you take a pencil and paper and draw yourself a simplified alternator with only two coils of one turn each and permanent magnet field.
Surely there's a hobby site with such a drawing, i'd be surprised if Bill Beatty doesn't have one.

That is why you can throw a short circuit directly on an alternator-- Load current becomes reactive and cancels out the field, limiting short circuit current.
That's also why alternators work so well in automobiles - old Chrysler voltage regulators didnt even measure current.

draw that sketch
and may you go well in your machinery classes
they were my favorites

old jim
 
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  • #5
Bassalisk;3548929I said:
feel like it doesn't matter if the load is purely resistive or not, i feel like the current will always lag behind emf in stator, because stator has coils of its own. Which are inductors, making the current lag. Am I correct?

And this induced emf in rotor which induces emf in stator, how does this affect the alternator?

If the load is resistive, then the current in the load is always in phase with the applied emf across the load.

Emfs never induce emfs. Currents and magnetic fields induce emfs.
 
  • #6
jim hardy said:
indeed an interesting question.

one needs to go to OLD books to get the physical explanation for many things. I love my old 1901 Thompson dynamo book for that reason

Sigh i wish i were computer savvy enough to make drawings.

The answer to your question is - basically yes. the concept neesds to be grasped, but in the today's culture most textbooks will just give you equations. if you have the mental picture you can figure out the formula instead of remembering it..

the flux from current flowing in the armature indeed is real so the total flux in the machine is a vector sum of the flux from field and flux from stator current.
most books I've seen describe mmf's instead of flux, and that's fine they add too..

The KEY point to this is that voltage is derivative of flux
e = n d(phi)dt
and they go to a lot of trouble to make sure the rotor produces a sinusoidal flux so that generator voltage will be a sinewave - it's the only function i know of whose derivative looks just like itself.
So - since flux is a sinewave so is terminal voltage, but terminal voltage is NOT at sinewave peak when north pole of field is directly under the coil and flux is peak.
recall derivative of sinewave shifts 90 degrees
which means terminal voltage is at peak when flux is crossing zero, ie when field is mechanically perpendicular to coil

but for a resistive load, armature current is in phase with terminal volts
so the mmf(flux) from armature current is perpendicular to that from field
and since the addition is a vector one
the total mmf(flux) isn't affected very much by contribution from load current.
Pythagors rules -add 1 +j( .1) and you get not 1.1 but 1.005 and that's half percent change not ten.

But the effect is there.
It is called "Armature Reaction" in the books and that seems a good name.
a machine with 10% armature reaction as described above would regulate pretty well.

If the load current were instead reactive, say maybe the machine is feeding a capacitor,
load current would be 90 degrees ahead of terminal voltage
which means its mmf would mechanically align directly with the rotating field
and armature reaction's mmf would add directly to field's mmf.
for that reason real and reactive components of load current are called "Direct" and "Quadrature" current, once again not bad names for them.
Reactive current does affect flux and terminal voltage way more than real current.
In the power plant we control the watts with amount of steam to turbine, and the vars(volt-amps-reactive) with voltage regulator.

you can see this yourself if you take a pencil and paper and draw yourself a simplified alternator with only two coils of one turn each and permanent magnet field.
Surely there's a hobby site with such a drawing, i'd be surprised if Bill Beatty doesn't have one.

That is why you can throw a short circuit directly on an alternator-- Load current becomes reactive and cancels out the field, limiting short circuit current.
That's also why alternators work so well in automobiles - old Chrysler voltage regulators didnt even measure current.

draw that sketch
and may you go well in your machinery classes
they were my favorites

old jim

Wow. only that word I have for this reply. Makes a lot more sense now. I personally don't like "big currents". I mean I love EE in general, just transistors, telecommunications, fields etc, those are my main interests.

I have this class, which these questions arise from, Fundamentals of Electro-energetic systems. This is a class by choice. I chose this class, because I wanted to have a better knowledge how I get power and energy from my power plant to home. I do realize, that I probably will never use this knowledge in my EE practice as Telecommunicationalist ( :D ), but its one of those things, that are nice to know.

As everything in my life, I want best results, and although I probably won't need this much, I want perfection. Thats why I bust my head with questions like these. Everything has to be in place.

Thank you mr. Jim you have been more than helpful. I will surely explore this concept more, as you and the other mr. Bob gave me a lot to cover.
 
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  • #7
""I personally don't like "big currents". I mean I love EE in general, just transistors, telecommunications, fields etc, those are my main interests. ""

i felt same way until junior year.

since you are curious about how you get power i suggest that you find your local power plant, knock on the door, tell them you're an engineering student soon to be on the job market. you'll probably get a great tour.

a power plant is a mix of most every type of machine known to man -
boilers, steam turbines, water treatment, chemisty for the boiler water, big pumps to move it, electic motors from huge to tiny, electronics to control them, computers to report on them, communications, meteorology, you name it. A tinkerer's paradise.


old jim
 
  • #8
During the tour, ask about how oil pressure is maintained on the generator bearings during a sudden shutdown. The million kilowatt "Big Allis" generator in New York City burned up about $30 million in journal (main) bearings because the hydraulic pumps keeping the bearings lubricated, which ran on house electric power, shut down during the big blackout of 1965.

http://www.nytimes.com/2003/08/15/u...nts-city-s-history.html?pagewanted=all&src=pm

http://www.practicalmachinist.com/v.../machinery-builders-plates-178619/index2.html
 
  • #9
ours had a main oil pump on the front end of turbine shaft. Big bronze impeller almost three feet across.
There was an AC driven pump for normal shutdown
and a backup AC pump
and a DC emercency pump for lubrication in a blackout
that DC pump was considered so important its fuses were short lengths of copper pipe, we'd gladly sacrifice it to protect those big journal bearings
if they wear enoough that the blades rub the case you're into multimillion dollar repair.
sounds like that's what got "Big Allis" - she was an industry legend.

old jim
 
  • #10
Not a bad idea about that tour. I will definitely try to do just that.
 
  • #11
Reviving this thread:

Can somebody explain to me what does first sentence mean?

[PLAIN]http://pokit.org/get/7c7977d375e6d2d33cf262f9b31fefcb.jpg [Broken]

I mean I really think I have good English, but this I cannot really understand.

I understand the first half of it, but after magnetizes I get lost.

Really ashamed of my English, I thought I knew it well.

http://pokit.org/get/442589c9e021c6a4dc31189c8e11d9d7.jpg [Broken]
[picture for this text]

http://media.wiley.com/product_data/excerpt/75/04716144/0471614475.pdf
[source, page 24]
 
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  • #12
try this simple thought and see if it helps...

go back to our paragraph on synchronous machine and armature reaction.

this time your author has set us up with NO flux from the field, by defining there IS no field current.

Remember the total flux in machine is sum of fluxes from (currents in) field and armature

when there's no field current,
only flux that can exist is that from armature current

and armature current will be WHATEVER IS REQUIRED to create dphi/dt enough to oppose applied voltage. Just like any other inductor.

remember - leading reactive current adds to field's mmf. It'll make as much flux as is necessary to support counter emf, all of it if it has to.

As i heard a grizzled old electricial lecturing an apprentice::
"if you don't give a synchronous machine a field it'll make its own".

that has some ramifications but basically it's how an induction motor works. Except in the induction motor the field is shorted so that current can flow in it and that'll make torque.

when you're going to start a synchronous motor across the line you short its field.
Else the dphi/dt from armature reaction would make horrendous voltage in field windings and overcome their insulation, hence the old electrician's statement. often on medium size generators you'll see a big transient absorber across the field to protect against such accidents.
Think about a squirrel cage induction motor - there's a shorted rotor for you.


any help?

fwiw i found his wording awkward.
As Lavoisier says 'science is but language well arranged.'
we have to polish our thought processses like a fine work of poetry or prose.


trouble is I'm real awkward too.

old jim
 
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  • #13
jim hardy said:
and armature current will be WHATEVER IS REQUIRED to create dphi/dt enough to oppose applied voltage. Just like any other inductor.

So if rotor is in "off" mode, and we apply some voltage to generator, nothing happens, we only get the field from stator(armature). Ok I think I got it.

So when we excite just a little bit rotor, with that DC current, it will try to create its own magnetic field, which is opposite from armature field.

Will this create a little torque on the rotor? Just a little? and will it start spinning(rotating)?And back to my original post, can you tell me from the material that I provided, what is E1, V1 and counter-EMF in those pictures/text? Where does that happen?

Thank you very much for your help, I am aiming high for this course.
 
  • #14
i think you have it

and the mental picture will solidify as you contemplate various thought experiments.

IF you removed the rotor from a three phase machine
AND excited the stator (at reduced voltage because there's now a larger airgap and you want only reasonable magnetizing current to flow)
there would be a rotating magnetic field caused by current in the stator.
IF you tossed a steel ball in there
would it move around following that field?
A pocket compass needle would certainly follow it

so would a rotor with small DC excitation
.

usually we use a generator the other way round -we make the rotor produce the magnetric field,
but as the grizzled old electrician said to the apprentice...the stator can make one.

i appreciate the plight of teachers , searching for the words that'll paint a physically accurate mental picture in somebody else's mind.

and i applaud your dogged persistence at forming such a mental picture. i too struggle.

Einstein said he thought in pictures. I suppose there are people who think in math formulas, maybe even words(computer programmers?)... i envy them.
 
  • #15
jim hardy said:
i think you have it

and the mental picture will solidify as you contemplate various thought experiments.

IF you removed the rotor from a three phase machine
AND excited the stator (at reduced voltage because there's now a larger airgap and you want only reasonable magnetizing current to flow)
there would be a rotating magnetic field caused by current in the stator.
IF you tossed a steel ball in there
would it move around following that field?
A pocket compass needle would certainly follow it

so would a rotor with small DC excitation
.

usually we use a generator the other way round -we make the rotor produce the magnetric field,
but as the grizzled old electrician said to the apprentice...the stator can make one.

i appreciate the plight of teachers , searching for the words that'll paint a physically accurate mental picture in somebody else's mind.

and i applaud your dogged persistence at forming such a mental picture. i too struggle.

Einstein said he thought in pictures. I suppose there are people who think in math formulas, maybe even words(computer programmers?)... i envy them.

Yes I always think in pictures. I cannot understand something, and do math problem about it(say calculate the generators inductance) without having a clear thought of what I am calculating.

Thank you very much kind sir.

Little bit more of work and equation backup and I am ready to go!
 
  • #16
""And back to my original post, can you tell me from the material that I provided, what is E1, V1 and counter-EMF in those pictures/text? Where does that happen?""

Author tells us in paragraph 17.1 that V1 represents applied voltage (from an external source i presume)

and further down, just below fig 1.24 he tells us E1 is counter-emf due to stator currents (magnetizing current)

he never states why he reversed the sign of applied voltage V1 in the third column of fig 1.24
and it seems counter-intuitive to me to do so.

go back to the physical - reactive current in the stator either aids or opposes rotor current in making flux. it happens in the magnetic circuit. Draw your simple one turn generator again or look at his fig 1.19- it is helpful to imagine a permanent magnet machine like that one.
 
  • #17
jim hardy said:
""And back to my original post, can you tell me from the material that I provided, what is E1, V1 and counter-EMF in those pictures/text? Where does that happen?""

Author tells us in paragraph 17.1 that V1 represents applied voltage (from an external source i presume)

and further down, just below fig 1.24 he tells us E1 is counter-emf due to stator currents (magnetizing current)

he never states why he reversed the sign of applied voltage V1 in the third column of fig 1.24
and it seems counter-intuitive to me to do so.

go back to the physical - reactive current in the stator either aids or opposes rotor current in making flux. it happens in the magnetic circuit. Draw your simple one turn generator again or look at his fig 1.19- it is helpful to imagine a permanent magnet machine like that one.

I almost fully got it. I will investigate the effects of counter-emf. And I improved my English in process too !
 
  • #18
i'm going to suggest you take apart an old car alternator. They are three phase multi-pole synchronous machines. Follow the magnetic circuit. Then realize that once the diodes begin to conduct, armature current is limited by amount of field current available.
In a 12 volt automobile , by giving field enough resistance that full 12 volts on field can only allow so much armature current, you have achieved an inherently current limited generator. That's why automobile voltage regulators got so much simpler around 1963...no current measuring coil anymore.

but i sort of enjoyed the old three-coil contraptions...

keep it simple -

old jim
 
  • #19
jim hardy said:
i'm going to suggest you take apart an old car alternator. They are three phase multi-pole synchronous machines. Follow the magnetic circuit. Then realize that once the diodes begin to conduct, armature current is limited by amount of field current available.
In a 12 volt automobile , by giving field enough resistance that full 12 volts on field can only allow so much armature current, you have achieved an inherently current limited generator. That's why automobile voltage regulators got so much simpler around 1963...no current measuring coil anymore.

but i sort of enjoyed the old three-coil contraptions...

keep it simple -

old jim
I might as well do that. Science needs experiments after all.
 

1. What is a synchronous generator?

A synchronous generator is a device that converts mechanical energy into electrical energy. It is also known as an alternator and is commonly used in power plants to generate electricity.

2. How does a synchronous generator work?

A synchronous generator works by using the principle of electromagnetic induction. When a conductor is rotated within a magnetic field, an electric current is induced in the conductor, producing electricity.

3. What are the main components of a synchronous generator?

The main components of a synchronous generator include a rotor (rotating part), stator (stationary part), exciter, and prime mover. The rotor is the part that rotates within the stator and is connected to the prime mover, which can be a steam turbine, gas turbine, or diesel engine. The exciter supplies the DC current to the rotor to create the magnetic field.

4. What are the advantages of using a synchronous generator?

One of the main advantages of using a synchronous generator is its ability to maintain a constant output voltage and frequency. It also has a high efficiency and can operate at various speeds. Additionally, synchronous generators are relatively easy to maintain and have a long lifespan.

5. What are the applications of synchronous generators?

Synchronous generators are commonly used in power plants to generate electricity. They are also used in various industries for backup power, as well as in renewable energy systems such as wind turbines and hydroelectric power plants. They can also be found in smaller applications such as backup generators for buildings and homes.

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