Synchronous generator excitation

[cnh1995]

I recently read about synchronous generator and some things really confuse me..
How does the excitation affect reactive power? I don't understand their relation.. For constant load and variable excitation, how does the power factor change? Icosø remains the same but both I and cosø change with excitation. I understood this behavior for synchronous motor but not for generator. Please connect the dots for me..Also, why is synchronous impedance obtained from OCC and SCC taken as constant for any pf at full load? Doesn't synchronous reactance vary with pf?

 

[anorlunda]

In approximate terms, the AC real power from one node to the next is proporitonal to the phase angle between the voltages. The reactive power between the nodes is proportional to the RMS voltage magnitude difference. This has nothing to do with generators; it applies to any two AC nodes separated by a single brance with a fixed impedance.

Raising the exitation raises terminal voltage magnitude, which raises reactive power flow in the out direction, which raises voltages on nearby nodes on the net.

 

[jim hardy]

see if this old thread is any help...?

https://www.physicsforums.com/threa...ance-vary-with-frequency.567856/#post-3713324

 

[cnh1995]

see if this old thread is any help...?

https://www.physicsforums.com/threa...ance-vary-with-frequency.567856/#post-3713324

In emf method (synchronous impedance method), we calculate Zs at short circuit condition,which is assumed to be almost purely inductive due to large Xs and small armature resistance. But how come the same value of Zs is taken for same current (full load current) but different pf? In purely resistive case, I read that the armature reaction is only distorting. How does that demagnetize the field due to the poles and produce "armature reaction drop"? If there's a capacitive load, armature amp turns are added to the field amp turns,raising the terminal voltage. Why is that treated as an impedance?
Also, if the load is kept constant, how does the pf change with excitation? I saw a phasor diagram describing the effect of variable excitation. There too, Zs is taken to be constant and Ia, Φ change such that I_acosΦ remains constant and generated emf E and power angle δ change such that Esinδ remains constant. I find it very non-intuitive..

 

[anorlunda]

If your desire is to start with the basics, then a synchronous generator connected to an impedance load is not it.

A synchronous generator connected to an ideal AC voltage source (constant magnitude, constant frequency) thorough a series inductance is the simplest case. It also closely resembles real life where the ideal source is the power grid.

Single machine to infinite bus is the way we teach power engineers.

The reason that is is simpler is that the machine is best described by direct and quadrature axes. But diect with respect to what? That is where the infinite bus comes in. Tomorrow, I'll post a block diagram that may help.

 

[cnh1995]

If your desire is to start with the basics, then a synchronous generator connected to an impedance load is not it.

A synchronous generator connected to an ideal AC voltage source (constant magnitude, constant frequency) thorough a series inductance is the simplest case. It also closely resembles real life where the ideal source is the power grid.

Single machine to infinite bus is the way we teach power engineers.

So how does that way work? I am new to this stuff and I have worked very little on an actual machine, that too on a single separate one in our lab. So whatever I asked, does it apply only to a grid connected alternator and not to an isolated one (like the one in our lab)?

 

[anorlunda]

Yes. I have only my phone today, so I can't write much, Tomorrow.

 

[cnh1995]

Yes. I have only my phone today, so I can't write much, Tomorrow.

All right..

 

[jim hardy]

In purely resistive case, I read that the armature reaction is only distorting. How does that demagnetize the field due to the poles and produce "armature reaction drop"? If there's a capacitive load, armature amp turns are added to the field amp turns,raising the terminal voltage.

the secret lies in phasor diagrams, perhaps.
This sketch from link referenced earlier showed physically why mmf from reactive armature amps add directly to(or subtract from) field 's mmf

here's another old thread that's related.

https://www.physicsforums.com/threads/ac-generator-on-grid-governor-and-amp.759387/#post-4785338
Unfortunately most of the images have disappeared, though. Gone with the e-wind?
This one survived...

armature mmf phasor is just the green one painted black and slid up there, preserving length and direction.
In your mind rotate green armature amps to any pf you like
near unity field mmf and armature mmf are perpendicular so effect of armature reaction doesn't change magnitude of total mmf very much, only shifts phase. Hence term "Distorting" used by some authors

near zero pf the mmf's add directly, try it by rotating green arrow straight up or down... remember armature mmf follows

Why is that treated as an impedance?

Because viewed from outside the machine it gives the same effect as would an impedance.
Establish 100% terminal volts, V= 1.0per unit
short circuit the machine, measure current as fraction of rated
divide 1.0 volts by measured current, result is named X_a

The actual phenomenon that's happening is opposing mmf's cancelling one another
which is indistinguishable from opposing voltages cancelling one another,,,,,
I_aZ_a is equal and opposite internal EMF at zero pf
and at any other pf it's still a phasor addition
so most authors present armature reaction as if it were an internal impedance.

Also, if the load is kept constant, how does the pf change with excitation? I saw a phasor diagram describing the effect of variable excitation. There too, Zs is taken to be constant and Ia, Φ change such that IacosΦ remains constant and generated emf E and power angle δ change such that Esinδ remains constant. I find it very non-intuitive..

In that phasor diagram observe that total mmf must be perpendicular to terminal volts, and must have same radius..
So, for any terminal volts , total mmf point lies on top of same circle
now in your mind (or on paper) lengthen or shorten field mmf phasor
what must happen to armature mmf to bring you back to point of total mmf phasor?
Now re-align green armature amps to same length and angle as armature mmf.

That made it intuitive for me.

As i recall it wasn't too excruciating to make those phasors with Paint...



old jim

 

[cnh1995]

the secret lies in phasor diagrams, perhaps.
This sketch from link referenced earlier showed physically why mmf from reactive armature amps add directly to(or subtract from) field 's mmf

here's another old thread that's related.

https://www.physicsforums.com/threads/ac-generator-on-grid-governor-and-amp.759387/#post-4785338
Unfortunately most of the images have disappeared, though. Gone with the e-wind?
This one survived...

armature mmf phasor is just the green one painted black and slid up there, preserving length and direction.
In your mind rotate green armature amps to any pf you like
near unity field mmf and armature mmf are perpendicular so effect of armature reaction doesn't change magnitude of total mmf very much, only shifts phase. Hence term "Distorting" used by some authors

near zero pf the mmf's add directly, try it by rotating green arrow straight up or down... remember armature mmf follows
Because viewed from outside the machine it gives the same effect as would an impedance.
Establish 100% terminal volts, V= 1.0per unit
short circuit the machine, measure current as fraction of rated
divide 1.0 volts by measured current, result is named X_a

The actual phenomenon that's happening is opposing mmf's cancelling one another
which is indistinguishable from opposing voltages cancelling one another,,,,,
I_aZ_a is equal and opposite internal EMF at zero pf
and at any other pf it's still a phasor addition
so most authors present armature reaction as if it were an internal impedance.
In that phasor diagram observe that total mmf must be perpendicular to terminal volts, and must have same radius..
So, for any terminal volts , total mmf point lies on top of same circle
now in your mind (or on paper) lengthen or shorten field mmf phasor
what must happen to armature mmf to bring you back to point of total mmf phasor?
Now re-align green armature amps to same length and angle as armature mmf.

That made it intuitive for me.

As i recall it wasn't too excruciating to make those phasors with Paint...
old jim

If I rotated the green arrow so as to get unity pf, total mmf vector should be the hypotenuse, with other two sides as field mmf vector and armature mmf vector. But if the field is constant (radius of the circle), how could I get total mmf vector as the hypotenuse? Its the field mmf that becomes the hypotenuse in that case and not the total mmf.. I don't know what I am missing...

 

[anorlunda]

OK, answering your post #6.

1. I suggest that your students can learn easier, and more directly useful knowledge if you connect your laboratory machine to the grid instead of a fixed resistance. For safety's sake, just make sure that it is connected to a compatible voltage, and connected through an inductance with a big enough impedance that no matter how the machine is handled (or mishandled), the currents will be small enough that nothing will be damaged, and nobody gets injured. Students can learn just as much with small currents and powers as they can with big ones.
   
   Alternatively, you can connect it to a second synchronous generator/motor. Call the larger of the two machines "the grid".
2. Before I post a lot of stuff, let me ask. Are you familiar with the D-Q (direct-quadrature axes) representation of synchronous machines? That is the great simplification that makes things easy. The D-Q quanties are related to the terminal quantities via Parks Transformation (google "Parks Equations"). There is a good description, plus equivalent circuits at https://www.eal.ei.tum.de/fileadmin...AUSIGA/lecture/2011-2012-W/wuppertal-park.pdf
3. Once you work with the D-Q axes, things like the effect of field voltage become more obvious, and words like armature or armature reaction become unnecessary.

If this is not clear, post again.

 

[jim hardy]

If I rotated the green arrow so as to get unity pf, total mmf vector should be the hypotenuse, <<<<<see 2 below jh with other two sides as field mmf vector and armature mmf vector. But if the field is constant (radius of the circle), how could I get total mmf vector as the hypotenuse? Its the field mmf that becomes the hypotenuse in that case and not the total mmf.. I don't know what I am missing...

hmm i found those missing images

unity power factor, zero load

same excitation but load applied by increasing torque supplied to shaft

^^note this is with machine connected to grid so terminal volts are constant . As anorlunda says, that's how it's usually taught ,so it's become intuitive to us ..^^



this is fixed terminal volts, same power as 2, but excitation tweaked to restore pf to unity.

Your stipulated condition, constant load, means power factor is fixed which means the angles don't change just the lengths of the sides
so changing excitation changes only voltage (the diameter of your circle) and armature current

caveat: assuming that by "constant load" you mean "some fixed impedance" and not "some fixed KVA"
because if you lock both KVA and excitation then you've fixed two sides and an angle of the triangle so nothing is free to change.

Fire up Paint and draw yourself some circles ?

 

[cnh1995]

OK, answering your post #6.

1. I suggest that your students can learn easier, and more directly useful knowledge if you connect your laboratory machine to the grid instead of a fixed resistance. For safety's sake, just make sure that it is connected to a compatible voltage, and connected through an inductance with a big enough impedance that no matter how the machine is handled (or mishandled), the currents will be small enough that nothing will be damaged, and nobody gets injured. Students can learn just as much with small currents and powers as they can with big ones.
   
   Alternatively, you can connect it to a second synchronous generator/motor. Call the larger of the two machines "the grid".
2. Before I post a lot of stuff, let me ask. Are you familiar with the D-Q (direct-quadrature axes) representation of synchronous machines? That is the great simplification that makes things easy. The D-Q quanties are related to the terminal quantities via Parks Transformation (google "Parks Equations"). There is a good description, plus equivalent circuits at https://www.eal.ei.tum.de/fileadmin...AUSIGA/lecture/2011-2012-W/wuppertal-park.pdf
3. Once you work with the D-Q axes, things like the effect of field voltage become more obvious, and words like armature or armature reaction become unnecessary.

If this is not clear, post again.

I've studied D-Q axes representation for salient pole machine but we don't have Parks equations in our syllabus. I guess I'd have to start with that?

 

[cnh1995]

hmm i found those missing images

unity power factor, zero load

same excitation but load applied by increasing torque supplied to shaft

^^note this is with machine connected to grid so terminal volts are constant . As anorlunda says, that's how it's usually taught ,so it's become intuitive to us ..^^
this is fixed terminal volts, same power as 2, but excitation tweaked to restore pf to unity.

Your stipulated condition, constant load, means power factor is fixed which means the angles don't change just the lengths of the sides
so changing excitation changes only voltage (the diameter of your circle) and armature current

caveat: assuming that by "constant load" you mean "some fixed impedance" and not "some fixed KVA"
because if you lock both KVA and excitation then you've fixed two sides and an angle of the triangle so nothing is free to change.

Fire up Paint and draw yourself some circles ?

So can I say that the grid keeps the terminal volts constant and not the machine? So whatever I asked doesn't apply to the alternators in our lab?

 

[jim hardy]

So can I say that the grid keeps the terminal volts constant and not the machine? So whatever I asked doesn't apply to the alternators in our lab?

hmmm
usually a machine is equipped with a voltage regulator
in the lab the voltage regulator is probably you , adjusting the field rheostat or whatever you use to control field current

Is this "whatever you asked" ?

I recently read about synchronous generator and some things really confuse me..
How does the excitation affect reactive power? I don't understand their relation..

In order for reactive current to flow, your machine must be connected to something that will accept reactive current.
A pure resistive load won't allow any reactive current, pf will be 1.0 and terminal volts will follow excitation.
A fixed Z load will allow both real and reactive current to flow in accordance with ohm's law, and volts will still follow excitation.

Has teacher yet introduced to you the concept of "Infinite Bus" ?
it's just an electrical node with so much capacity that no single machine connected to it can affect its voltage.
an "Infinite Bus" load will accept real and reactive power an any amount your machine can deliver,
The Grid is almost an infinite bus, and for learning theory we consider how would a machine behave if connected to a genuine infinite bus.

For constant load and variable excitation, how does the power factor change?

When you say " constant load and variable excitation" it leaves me wondering...
because in utility parlance "Load" means Megawatts being passed into the grid,
reactive power is called "Megavars"
If you ask operator "What load do you have right now? " he'll tell you what his megawatt meter reads
to find out reactive power you'll have to ask him "How much reactive?" or "How many megavars"
He adjusts megavars with excitation
and megawatts with the steam valve so to answer your question it is necessary to define "constant load"
are you asking about a machine connected to infinite bus, or connected to some fixed impedance ?

 

[cnh1995]

hmmm
usually a machine is equipped with a voltage regulator
in the lab the voltage regulator is probably you , adjusting the field rheostat or whatever you use to control field current

Is this "whatever you asked" ?In order for reactive current to flow, your machine must be connected to something that will accept reactive current.
A pure resistive load won't allow any reactive current, pf will be 1.0 and terminal volts will follow excitation.
A fixed Z load will allow both real and reactive current to flow in accordance with ohm's law, and volts will still follow excitation.

Has teacher yet introduced to you the concept of "Infinite Bus" ?
it's just an electrical node with so much capacity that no single machine connected to it can affect its voltage.
an "Infinite Bus" load will accept real and reactive power an any amount your machine can deliver,
The Grid is almost an infinite bus, and for learning theory we consider how would a machine behave if connected to a genuine infinite bus.When you say " constant load and variable excitation" it leaves me wondering...
because in utility parlance "Load" means Megawatts being passed into the grid,
reactive power is called "Megavars"
If you ask operator "What load do you have right now? " he'll tell you what his megawatt meter reads
to find out reactive power you'll have to ask him "How much reactive?" or "How many megavars"
He adjusts megavars with excitation
and megawatts with the steam valveso to answer your question it is necessary to define "constant load"
are you asking about a machine connected to infinite bus, or connected to some fixed impedance ?

I'm talking about a fixed impedance like a lamp bank,which so far we've been working on in labs. The concept of infinite bus has just been introduced to us, just a week ago. So all the characteristics I've mentioned are true only for a machine connected to an infinite bus??

 

[anorlunda]

Wait! Constant-Increase-Decrease-grid caused-machine caused-field caused; that is a wrong way to think about things, and those are the wrong questions to ask.

You should be using the equivalent circuit for a synchronous machine linked in #11. The circuit tells you how things behave, just as in any circuit.

Is the mathematical approach a problem for your students?

 

[jim hardy]

So all the characteristics I've mentioned are true only for a machine connected to an infinite bus??

well "all the characteristics" invites me to say "yes" but that's not the best answer. Generalizations cause trouble, generally .

Lamp loads , unless they're inductively ballasted lamps or SMPS like CFL's , are a pure resistive loadCan current be anything but perfectly in phase with terminal volts ?
so can power factor be anything other than 1.0 ?

For constant load and variable excitation, how does the power factor change? Icosø remains the same but both I and cosø change with excitation.

there's your trouble. Which angle do you mean by ø ?

Angle between terminal volts and amps doesn't change with excitation, for power factor is determined by your load.

Angle between rotor position and terminal volts does change with excitation, for excitation affects the strength of the magnetic field that's coupling energy from the rotor into the stator windings.
Teacher should have you observe and plot that with a stroboscope, it's really fun to watch the machine respond.
Angle between terminal volts zero crossing and rotor position is called "power angle"

and it's same as angle between field mmf and total mmf

observe if you increase field mmf by raising excitation, everything grows in proportion
armature amps stay in phase with terminal volts, and increase,
terminal volts increase as they must to drive more current through lamps,
but power angle stays about the same unlike when providing constant mw to infinite bus...


Could it be you've confused "power factor angle" with "power angle"?

Learn to work that simple one-turn machine from post 9 in your head. It makes lot of formulas intuitive and will help you with D-Q as mentioned by Anorlunda. D-Q is the coin of the realm, you'll need to be fluent in it, so start now building mental models to facilitate becoming fluent with the math.

old jim