Synchronous generator on an infinite bus

[ffp]

I'm reviewing some subjects that I long forgot and now I'm studying synchronous machines.
So, when you change the field current (of the rotor) of s synchronous generator, the result is a change in the magnetic flux, which change the internal voltage (Ea). That will result in a change in the terminal voltage (Vt).

Now, when the generator is connected to an infinite bus, it is said that a change in the field current changes the reactive power provided by the generator.
I know that an infinite bus has constant frequency and voltage.
I also know that what determines real power output is the speed and torque of the axis of the generator.

And I could understand the logic used by books to explain that with phasorial diagrams. However, I can't understand how this physically works. What's behind that.

For example, consider a synchronous generator connected to an infinite bus that has its field current increased (by decreasing rotor resistance). That increase will increase the flux, that will increase the internal and terminal voltages. If the bus has a fixed voltage, and the generator terminals were initially in that voltage, wouldn't that increase just create a difference in voltage between the generator terminals and the bus?

 

[anorlunda]

Re read this PF Insights Article.

https://www.physicsforums.com/insights/ac-power-analysis-part-2-network-analysis/

I also know that what determines real power output is the speed and torque of the axis of the generator.

No. Not speed and torque, but angle. Then ask yourself, how is angle related to speed?

 

[ffp]

Could you provide some explanation about the two mantras? The link doesn't seem to show.

In a synchronous generator, the real power is dependent only on the torque and speed of rotation. Pin= T×ω

 

[anorlunda]

Torque times speed is the definition of mechanical power for any rotating machine to for example s diesel engine. But if you are treating power electrically to another place such as the infinite bus place, we're express it in terms of voltage and current. With AC, every voltage and current has a magnitude and a phase angle. The phase is the important part for real power.

The equation to ship power to the next node on the grid has nothing to do with a generator. To ship power from node 1 through inductance X to node 2 the equation for real power is $P=\frac{V1•V2}{X} \sin{\theta}$ where $\theta$ is the phase angle difference between V1 and V2.

Edit: I finished the last sentence in italics.

 

[anorlunda]

@ffp Did you get the answer you need?

 

[ffp]

@ffp Did you get the answer you need?

No, you didn't even finish theblast dentence of your last post.
And you didn't adress field current at all. I think we are going past the question...

 

[berkeman]

No, you didn't even finish theblast dentence of your last post.

Well, what do you think $\theta$ represents?

And you didn't adress field current at all. I think we are going past the question...

Could you maybe rephrase your question to help you get the focused replies that you want? Thank you.

BTW -- @anorlunda is one of the subject matter experts here at PF in this topic. Please keep that in mind in your responses going forward. Thank you,

 

[anorlunda]

So, when you change the field current (of the rotor) of s synchronous generator, the result is a change in the magnetic flux, which change the internal voltage (Ea). That will result in a change in the terminal voltage (Vt).

That's essentially correct.

I also know that what determines real power output is the speed and torque of the axis of the generator.

That's misleading. Since we are neglecting losses and since energy is conserved we can say P_mechanical=P_generator=P₁=P₂ where bus 1 is the generator terminals and bus 2 is the infinite bus. We can write 4 equations for those 4 powers, but when we put numbers in, all 4 must produce the same value for P. Torque*speed is the equation for P_mechanical, real(V1*I1^*) is the formula for power sent from bus 1.

So, when you increase field current, the primary result is to increase the magnitude at the generator terminals which is V₁. That doesn't tell us what happens to the phase angle of V₁. So how does that change the real and imaginary powers, P and Q?

$P=\frac{V1•V2}{X} \sin{\theta}$
$Q=\frac{V1(V1-V2)}{X}\cos{\theta}$
where $\theta$ is the phase angle difference between V1 and V2.

Now, let's increase field voltage which caused the voltage V1 to increase. As you can see, both P and Q increase with V1 increasing. However, since P is constrained to be equal to P_mechanical the angle $\theta$ must also change to keep P constant. That is the part missing in your thinking, voltage has a magnitude in volts that you can measure with a voltmeter, but it also has a phase angle that is critically important. If you had an oscilloscope instead of a voltmeter, you could see the whole story.

You would see that after the field voltage increase, V1 increased in magnitude, but it also changed phase to become closer to the phase of V2.

Here's the moral of the story: In AC analysis, you can't discuss voltages or currents without including the phase angles of both. Unfortunately, phase angles can't be seen on voltmeters or ammeters.

 

[ffp]

That's essentially correct.

That's misleading. Since we are neglecting losses and since energy is conserved we can say P_mechanical=P_generator=P₁=P₂ where bus 1 is the generator terminals and bus 2 is the infinite bus. We can write 4 equations for those 4 powers, but when we put numbers in, all 4 must produce the same value for P. Torque*speed is the equation for P_mechanical, real(V1*I1^*) is the formula for power sent from bus 1.

So, when you increase field current, the primary result is to increase the magnitude at the generator terminals which is V₁. That doesn't tell us what happens to the phase angle of V₁. So how does that change the real and imaginary powers, P and Q?

$P=\frac{V1•V2}{X} \sin{\theta}$
$Q=\frac{V1(V1-V2)}{X}\cos{\theta}$
where $\theta$ is the phase angle difference between V1 and V2.

Now, let's increase field voltage which caused the voltage V1 to increase. As you can see, both P and Q increase with V1 increasing. However, since P is constrained to be equal to P_mechanical the angle $\theta$ must also change to keep P constant. That is the part missing in your thinking, voltage has a magnitude in volts that you can measure with a voltmeter, but it also has a phase angle that is critically important. If you had an oscilloscope instead of a voltmeter, you could see the whole story.

View attachment 319549
You would see that after the field voltage increase, V1 increased in magnitude, but it also changed phase to become closer to the phase of V2.

Here's the moral of the story: In AC analysis, you can't discuss voltages or currents without including the phase angles of both. Unfortunately, phase angles can't be seen on voltmeters or ammeters.

Ok, now we seem to be adressing the question, thanks so much.

A change in field current is caused by a change in the rotor resistence, right? Why would that affect the angle of the current or even the angle in the generator voltage terminals?

As I uderstand If( field current) generates Ea(internal voltage) that, after the impedances, results in Vt (terminals voltage). Changing the magnitude of If would change the angle of Vt??

I feel that the answer is, as you said, P is fixed. And thus, for the equation to make sense, or in a phasor diagram analysis, the angle between Ea and Vt. But I can't understand that physically.

 

[ffp]

Well, what do you think $\theta$ represents? Could you maybe rephrase your question to help you get the focused replies that you want? Thank you.

BTW -- @anorlunda is one of the subject matter experts here at PF in this topic. Please keep that in mind in your responses going forward. Thank you,

Theta is the angle between the voltage of bus 1 and bus 2. I'm don't know how this help, though.

The question is: In a synchronous generator, why a change in field current changes terminal voltage when it is alone but changes reactive power (not changing terminals voltage) when connected to an infinite bus?

 

[anorlunda]

Changing the magnitude of If would change the angle of Vt??

I feel that the answer is, as you said, P is fixed. And thus, for the equation to make sense, or in a phasor diagram analysis, the angle between Ea and Vt.

Yes, correct again. You're getting there.

The question is: In a synchronous generator, why a change in field current changes terminal voltage when it is alone but changes reactive power (not changing terminals voltage) when connected to an infinite bus?

It does change terminal voltage in both cases. But reactive power is very sensitive to changes (V1-V2) so you get a big change in Q for only a small change in (V1-V2). In real life, V2 is not infinite, so if V1 increases, the neighboring bus voltage V2 will also increase some.

By the way, there are many ways to change field voltage and current, not just changing a resistance. It doesn't matter how you did it. It matters only that it changes.

feel that the answer is, as you said, P is fixed. And thus, for the equation to make sense, or in a phasor diagram analysis, the angle between Ea and Vt. But I can't understand that physically.

I wish you had read that Insights article more closely. The answer is there.

• Start with a steady state P_mechanical=P_generator.
• Increase field voltage and/or current by any means. P_generator increases.
• Now we have an imbalance P_mechanical < P_generator. The generator starts slowing down.
• Since the generator frequency is no longer identical to infinite bus frequency, the relative angle $\theta$ starts changing. It changes in the direction to return P_generator to the original value even if the voltage is higher.
• When P_mechanical=P_generator returns to balanced, generator speed returns to synchronous speed, and the frequency matches infinite bus frequency. But $\theta$ does not return to the original value.

In the non-infinite case, any change on the grid causes temporary changes in frequency which changes all those angles until things are balanced again. It's fun to watch that happen on a continental scale as in this video. In the video, the disturbance was huge, but even tiny changes such as 1% change in field current, or some customer turning on a light bulb, causes analogous propagation of angle changes everywhere on the the grid.

 

[ffp]

It does change terminal voltage in both cases. But reactive power is very sensitive to changes (V1-V2) so you get a big change in Q for only a small change in (V1-V2). In real life, V2 is not infinite, so if V1 increases, the neighboring bus voltage V2 will also increase some.

You are talking about changes in the angle between V1 and V2, right? Because a change in module would not affect reactive power.

By the way, there are many ways to change field voltage and current, not just changing a resistance. It doesn't matter how you did it. It matters only that it changes.

It doesn't? It doesn't matter if the angle of the field current is cnahged or not?

• Now we have an imbalance P_mechanical < P_generator. The generator starts slowing down.

Why would the generator slow down? The speed/frequency of the generator isn't dependent only on the prime mover speed?

 

[anorlunda]

You are talking about changes in the angle between V1 and V2, right?

Right.

It doesn't? It doesn't matter if the angle of the field current is cnahged or not?

The field is DC. It doesn't have an angle.

Why would the generator slow down? The speed/frequency of the generator isn't dependent only on the prime mover speed?

Yes it is. The prime mover and the generator are bolted together on the same shaft.

Another way to look at it is conservation of energy. Energy is conserved both instantaneously and on the average. So any imbalance between mechanical and electrical power must go someplace. It goes into kinetic energy stored in the rotating masses. Speed must change when there is an imbalance to preserve conservation of energy.By the way, are you a power plant operator?

 

[anorlunda]

@ffp, synchronous generators can also use permanent magnets instead of adjustable field exciters. Call them PMGs.

Ask yourself, how would a PMG work connected to an infinite bus? Think of your questions in post #1.

Ask yourself why we don't use PMGs on the power grid?

 

[ffp]

Right.The field is DC. It doesn't have an angle.

Oh, certainly. Can't believe I forgot field current is DC.
Well, then why a change in the field current would change the angle of the terminal voltage?

Yes it is. The prime mover and the generator are bolted together on the same shaft.

Another way to look at it is conservation of energy. Energy is conserved both instantaneously and on the average. So any imbalance between mechanical and electrical power must go someplace. It goes into kinetic energy stored in the rotating masses. Speed must change when there is an imbalance to preserve conservation of energy.By the way, are you a power plant operator?

No, I'm not. I'm just studying that subject again after a long time. Why?

I know that when you increase the load (resistive), for example, the terminal voltage goes down. Not the speed of the shaft. The angular speed is determined by the turbines or whatever is spinning to create the energy. They might have to directly increase the speed to make up for the lower Vt, though.

Oh, and the permanent magnet generator woul not change its field current. We don't use them in the power grid because we have to conatantly adjust for the reactive power imbalance, by changing the field current, if I'm not mistaken. This still doesn't explain why changing the DC field current affect reactive power or the angle of Vt.

 

[cnh1995]

Well, then why a change in the field current would change the angle of the terminal voltage?

A change in the field current changes the rotor angle w.r.t. the angle of the terminal voltage. The magnitude and phase angle of the terminal voltage are fixed (by the grid).
You must have seen the phasor diagram of this scenario in your notes/textbook.
Consider the following two cases:
1) The generator is supplying active power P1 to the grid, and prime mover torque is increased:
Increased torque tends to increase the angle between E and V, and after a transient, the generator rotor settles down at a greater power angle such that P2=EVsinδ_new/X_s is greater than P1.
The terminal voltage remains constant (as it is established by the grid).

2) The generator is supplying active power P1 to the grid, and its field excitation is increased:
An increase in the field excitation increases the electromagnetic torque on the rotor. As the primer mover torque is unchanged, the increased EM torque tends to decelerate the rotor. But since the generator speed is locked by the infinite bus (or the grid), the rotor settles down at a lower power angle such that E_oldsinδ_old=E_newsinδ_new. Here, the active power supplied by the generator is unchanged (since the speed and prime mover torque are unchanged), but at a lower power angle.

When a generator is connected to the infinite bus:
1. Its terminal voltage (magnitude and phase angle) is constant (often denoted by V∠0°).
2. The frequency of the infinite bus is fixed. Hence, in steady state, the generator always rotates at the synchronous speed.

When excitation/prime mover torque is changed in the generator, the generator responds by changing its power angle which changes P and Q accordingly.

Hope this helps.

 

[anorlunda]

I know that when you increase the load (resistive),

When you run a generator with a resistive load across the terminals, it is not synchronous with a grid. It could run at any frequency, and there is no angle difference because there's only one node.

Oh, and the permanent magnet generator woul not change its field current. We don't use them in the power grid because we have to conatantly adjust for the reactive power imbalance, by changing the field current,

Yup. That's right.

This still doesn't explain why changing the DC field current affect reactive power or the angle of Vt.

Sigh. I answered those questions in the earlier posts. If you still don't understand I'm at a loss.

 

[anorlunda]

The magnitude and phase angle of the terminal voltage are fixed (by the grid).

Ay ay ay. Although you are trying to be helpful, you mixed up the terminal bus with the infinite bus so badly that several of your statements (like the one above) are counter productive.

The OP question was about a synchronous generator connected to an infinite bus, as in the one line diagram below. It is the infinite bus voltage (U1 in the diagram) that is normally taken as 1∠0°. Since there is power flow, there must voltage drop across the transmission reactance Z. Therefore the terminal voltage (U2 in the diagram) can not be 1∠0°. Also, it's magnitude and phase must vary when power varies.

The terminal voltage remains constant (as it is established by the grid).

Same mistake. If terminal voltage and angle were constant, and infinite bus voltage and angle were constant, there could never be any change in the current or power sent from one to the other. (See U1 and U2 in the diagram above)

When a generator is connected to the infinite bus:
1. Its terminal voltage (magnitude and phase angle) is constant (often denoted by V∠0°).

Same mistake. It is not the terminal bus, but rather the infinite bus denoted by V∠0°

I am trying to get @ffp to understand synchronous operation of a generator when connected to the grid. It can not be understood by thinking of the generator in isolation. The impedance Z in that diagram is just as much a part of the system as internals of the generator. A generator serving a local load is not interconnected and it is not synchronous with anything external. This case of a generator connected to an infinite bus through an impedance Z, is the simplest possible case of interconnected operation. That system can only be analyzed by simultaneous consideration of all the components including Z.

Edit: One more thing. Terminal voltage is not fixed by the grid. Quite the opposite. There is a voltage regulator at the power plant. It is a closed loop controller that varies excitation (field) voltage to hold terminal voltage at the set point. If terminal voltage was fixed by the grid, there would be no need for the voltage regulator.

 

[cnh1995]

I was explaining this scenario.

The generator is connected to the infinite bus through its synchronous reactance alone. There is no transmission line. So the terminal voltage is same as the infinite bus voltage.

 

[anorlunda]

I was explaining this scenario.
View attachment 319587
The generator is connected to the infinite bus through its synchronous reactance alone. There is no transmission line. So the terminal voltage is same as the infinite bus voltage.

That's better. But it is not what the OP in this thread asks about, so it is misleading to quote it here.

That E in your diagram is not physically accessible. It is like the voltage in a Thevenin's equivalent; a useful analogy but not anything real.

 

[ffp]

Sigh. I answered those questions in the earlier posts. If you still don't understand I'm at a loss.

I'm sorry, but i couldn't find the answer.

To change the reactive power, the Ea (and thus Vt) have to be lagung or leading in relation to the voltage of the infinite bus, right?

Ea is equal to K(constructive constant) x Φ (flux) x ω (angular velocity if shaft).Since the If (field current) is DC, and is practically linear in relation to flux I don't know how it could change the angle of Ea.
The angle of Ea is determined by... ω?

 

[anorlunda]

I'm sorry, but i couldn't find the answer.

I repeat from #11.

• Start with a steady state Pmechanical=Pgenerator.
• Increase field voltage and/or current by any means. Pgenerator increases.
• Now we have an imbalance Pmechanical < Pgenerator. The generator starts slowing down.
• Since the generator frequency is no longer identical to infinite bus frequency, the relative angle θ starts changing. It changes in the direction to return Pgenerator to the original value even if the voltage is higher.
• When Pmechanical=Pgenerator returns to balanced, generator speed returns to synchronous speed, and the frequency matches infinite bus frequency. But θ does not return to the original value.

In #16, @cnh1995 said basically the same thing in different words.

A change in the field current changes the rotor angle w.r.t. the angle of the terminal voltage. The magnitude and phase angle of the terminal voltage are fixed (by the grid).
You must have seen the phasor diagram of this scenario in your notes/textbook.
Consider the following two cases:
1) The generator is supplying active power P1 to the grid, and prime mover torque is increased:
Increased torque tends to increase the angle between E and V, and after a transient, the generator rotor settles down at a greater power angle such that P2=EVsinδnew/Xs is greater than P1.
The terminal voltage remains constant (as it is established by the grid).

2) The generator is supplying active power P1 to the grid, and its field excitation is increased:
An increase in the field excitation increases the electromagnetic torque on the rotor. As the primer mover torque is unchanged, the increased EM torque tends to decelerate the rotor. But since the generator speed is locked by the infinite bus (or the grid), the rotor settles down at a lower power angle such that Eoldsinδold=Enewsinδnew. Here, the active power supplied by the generator is unchanged (since the speed and prime mover torque are unchanged), but at a lower power angle.

 

[ffp]

I repeat from #11.

Icreasing field current increases Φ (flux), which increases Ea (internal voltage). Increase in voltage increases real power, and I believe that's what you mean by Pgenerator, right?

Ok, in this case Pin, which is equal to τ.ω (torque x angular velocity) is also increased by the change in If, since the torque is proportional to the fields of rotor and stator. So, wouldn't the Pin and Pgenerator both increase and stay equal to each other?
The only way I can see the speed of the generator shaft going slower is by litetally tirning the turbines slower.

In #16, @cnh1995 said basically the same thing in different words.

Whats is wrt?
1) isnt angle between Ea and Vt determined by load?

2) The power angle δ is the angle between Ea and Vt, ad well as Brotor and Bstator (magnetic fields).
How is it related to power factor (reactive power) that is the angle difference between Vt and Ia?

Also, i have trouble understanding the concept of EM torque in generator mode. The torque is supplied by the turbines, right? Which torque is the EM?

 

[anorlunda]

The only way I can see the speed of the generator shaft going slower is by litetally tirning the turbines slower.

Bingo. Yes, the turbine and the generator are locked together in RPM.

Ok, in this case Pin, which is equal to τ.ω (torque x angular velocity) is also increased by the change in If, since the torque is proportional to the fields of rotor and stator.

Wrong. Turbine torque is proportional to steam pressure and steam flow rate. You can't change it directly by doing stuff in the generator.

 

[anorlunda]

In addition, you just gave me a chance to use my favorite graphic.

That is the James Watt flyball governor. It was the classical way of adjusting the power of power plant generators for half a century. It was attached to the turbine, and the throttle controlled steam flow. So, if you do anything on the electric loads, the grid, or the generator that changes real power distribution, it eventually causes this governor (and every other governor in all the plants on the grid) to react.

 

[cnh1995]

Also, i have trouble understanding the concept of EM torque in generator mode. The torque is supplied by the turbines, right? Which torque is the EM?

Electromagnetic torque is the torque produced by the interaction of the armature flux and the rotor flux. This torque opposes the rotor rotation in a generator. In steady state, the electromagnetic torque is equal and opposite to the prime mover torque (electromechanical energy conversion principle) and the generator frequency matches the grid frequency.

This and my earlier post #16 are machine-centric.
I believe thinking simultaneously about the internal working of the generator (stator, rotor, fields etc) and changes on the grid (P, Q, phase angles ,voltages etc) would be confusing and counter-productive for you in the beginning.

You can upload the circuit diagram and relevant text from the book you are studying, and see if you can relate the answers in this thread to that.
Your confusion could be due differences in the nomenclature used here and that in the book.

 

[ffp]

Bingo. Yes, the turbine and the generator are locked together in RPM.Wrong. Turbine torque is proportional to steam pressure and steam flow rate. You can't change it directly by doing stuff in the generator.

Then the torque that is changed by If is the electromagnetic torque, which is opposed to the prime mover?? Wouldn't that mean that increasing If ( and thus EM torque) reduce shaft speed?

 

[anorlunda]

Then the torque that is changed by If is the electromagnetic torque, which is opposed to the prime mover?? Wouldn't that mean that increasing If ( and thus EM torque) reduce shaft speed?

Yes.

Conservation of energy applies to the turbine also. If the power into the turbine is less than the power out, even momentarily, it slows down. The only way it speeds up again is if power in is greater than power out, even momentarily.

You can think through all these scenarios solely on the basis of energy. Flux, torque, voltage, current, only add smoke and confusion. You would be better off ignoring those things. Energy is never created or destroyed here, it is only transported and/or stored. That is an extremely powerful analytical tool and I recommend that you make more use of it.

 

[ffp]

Electromagnetic torque is the torque produced by the interaction of the armature flux and the rotor flux. This torque opposes the rotor rotation in a generator. In steady state, the electromagnetic torque is equal and opposite to the prime mover torque (electromechanical energy conversion principle) and the generator frequency matches the grid frequency.

This and my earlier post #16 are machine-centric.
I believe thinking simultaneously about the internal working of the generator (stator, rotor, fields etc) and changes on the grid (P, Q, phase angles ,voltages etc) would be confusing and counter-productive for you in the beginning.

You can upload the circuit diagram and relevant text from the book you are studying, and see if you can relate the answers in this thread to that.
Your confusion could be due differences in the nomenclature used here and that in the book.

Thanks for the torque explanation. I have forgoten that in generator mode, there is a torque opposing the rotation and it is proportional to the load, right?

About the book, it's Electric Machinery Fundamentals by Chapman. I stopped in this part:

I can understand the explanation using the phasor diagram, but not how it would work inside the machine. Also, why an generator operating alone have it's Vt changed by If, but when connected to an infinite bus, it doesn't?

EDIT: Also, the book says that a generator operating alone has the power dictated by the load. While connected with an infinite bus, the power is constant (hes considering ω unchanged). Wouldn't the power be ALWAYS according to τ and ω?

EDIT2: Curiosity question: it says that when connected to infinite bus, speed *cannot* change. It's physically impossible? What happens if you speed up the shaft (by letting more steam go, for example)? Would the oposing torque make that physically impossible, or that would just mess up electrical frequency and is not wanted (but possible)?

 

[cnh1995]

Also, why an generator operating alone have it's Vt changed by If, but when connected to an infinite bus, it doesn't?

See the diagram attached in #19.

When analysing things from the machine side, the machine is assumed to be connected to an infinite bus. The voltage, phase angle and frequency of the infinte bus are fixed. It is an ideal bus whose voltage, frequency and phase angles cannot be changed by connecting/disconnecting a single generator. In practice, there are several systems and mechanisms in action that make the real-life grid "almost" an infinite bus.
This is the simplest model to analyse things from the generator side.

The E in the diagram (in #19) is the excitation emf and V is the terminal voltage of the generator, which is also the infinite bus voltage.

When the generator is operating in stand-alone mode, the infinite bus is replaced by an impedance and it is just like a normal ac circuit.
But when the generator is operating on the inifinte bus, its steady state speed, frequency and terminal voltage are locked by the infinite bus.
Before you connect the generator to the infinite bus (or "the grid" in real life), you must match generator's voltage, phase angle and frequency with the grid voltage, phase angle and frequency .
There must be another chapter in your book about how this synchronization is done in real life.

: Also, the book says that a generator operating alone has the power dictated by the load. While connected with an infinite bus, the power is constant (hes considering ω unchanged). Wouldn't the power be ALWAYS according to τ and ω?

The output mechanical power of the generator is always the product of torque and angular speed. But the angular speed is locked by the infinite bus. So, to change the power output, you only need to change the prime mover torque. If the steam input is increased, the rotor would accelerate -- more "active" armature current will flow -- this will increase the electromagnetic torque and decelerate the rotor -- the rotor will oscillate about a new angle δ -- finally settle down at δ and resume synchronous speed.
The infinite bus locks the generator speed. Hence, when you increase/decrease the prime mover torque, the generator undergoes a transient of oscillations and settles down at an advanced angular position w.r.t the infinite bus voltage, and supplies the active power proportional to the product of T and w.
This active power is proportional to the sine of the phase angle difference between E and V.

What happens if you speed up the shaft (by letting more steam go, for example)? Would the oposing torque make that physically impossible, or that would just mess up electrical frequency and is not wanted (but possible)?

For a machine connected to an infinite bus, the speed of the generator will never change and the generator would continue to deliver active power to the bus until its steady state power limit is reached.

In the real life grid, this affects all the other connected generators and they respond accordingly to maintain the grid frequency constant.
To understand how it works, one will require understanding of the concepts such as load-frequency control, automatic voltage regulators etc (which are already mentioned/linked in some earlier posts).

 

[ffp]

See the diagram attached in #19.

When analysing things from the machine side, the machine is assumed to be connected to an infinite bus. The voltage, phase angle and frequency of the infinte bus are fixed. It is an ideal bus whose voltage, frequency and phase angles cannot be changed by connecting/disconnecting a single generator. In practice, there are several systems and mechanisms in action that make the real-life grid "almost" an infinite bus.
This is the simplest model to analyse things from the generator side.

The E in the diagram (in #19) is the excitation emf and V is the terminal voltage of the generator, which is also the infinite bus voltage.

When the generator is operating in stand-alone mode, the infinite bus is replaced by an impedance and it is just like a normal ac circuit.
But when the generator is operating on the inifinte bus, its steady state speed, frequency and terminal voltage are locked by the infinite bus.
Before you connect the generator to the infinite bus (or "the grid" in real life), you must match generator's voltage, phase angle and frequency with the grid voltage, phase angle and frequency .
There must be another chapter in your book about how this synchronization is done in real life.

Yes, I have already seen synchronization of parallel generators.
How exactly is frequency (speed) and voltage fixed in the grid? I know that, as you said, to connect a new generator in parallel, thyçey have to match frequency (a bit higher before connecting) and voltage. But that doesn't mean you can't change any of them after connecting. So, my Edit2 question remains. It is physically impossible to accelerate the prine mover after connecting to the grid (opposing torque makes it impossible, or something like that), or it isbpossible but it's something undesirable and always avoided because of some problems this could cause in the grid (like messingbwithvother generators connected)?

The output mechanical power of the generator is always the product of torque and angular speed. But the angular speed is locked by the infinite bus. So, to change the power output, you only need to change the prime mover torque. If the steam input is increased, the rotor would accelerate -- more "active" armature current will flow -- this will increase the electromagnetic torque and decelerate the rotor -- the rotor will oscillate about a new angle δ -- finally settle down at δ and resume synchronous speed.
The infinite bus locks the generator speed. Hence, when you increase/decrease the prime mover torque, the generator undergoes a transient of oscillations and settles down at an advanced angular position w.r.t the infinite bus voltage, and supplies the active power proportional to the product of T and w.
This active power is proportional to the sine of the phase angle difference between E and V.

Well, then the power of generators operating alone isn't determined by the load, but the torque and speed?

 

[cnh1995]

But that doesn't mean you can't change any of them after connecting. So, my Edit2 question remains. It is physically impossible to accelerate the prine mover after connecting to the grid (opposing torque makes it impossible, or something like that), or it isbpossible but it's something undesirable and always avoided because of some problems this could cause in the grid (like messingbwithvother generators connected)?

If you increase the steam input with the intention of running the generator faster, the rotor will accelerate momantarily, but the infinite bus will force it to resume the synchronous speed after a transient period of damped oscillations (which depends on many factors like machine inertia, damping arrangement etc).

If the steam input is increased, the rotor would accelerate -- more "active" armature current will flow -- this will increase the electromagnetic torque and decelerate the rotor -- the rotor will oscillate about a new angle δ -- finally settle down at δ and resume synchronous speed.
The infinite bus locks the generator speed. Hence, when you increase/decrease the prime mover torque, the generator undergoes a transient of oscillations and settles down at an advanced angular position w.r.t the infinite bus voltage, and supplies the active power proportional to the product of T and w.

So to answer your question, yes it is impossible to increase the steady state speed and frequency of the generator connected to infinite bus. If you try, you will end up changing the active power output of the generator. The steady state speed and frequency will remain unchanged.
If you attempt to change the terminal voltage of the generator (which is same as the infinte bus voltage) by changing its field excitation, you will end up changing its reactive power output. The terminal voltage will remain unchanged.

Well, then the power of generators operating alone isn't determined by the load, but the torque and speed?

It is determined by the output voltage and load current. Output voltage depends on the speed, load current depends on the load impedance and torque depends on the "real component" of the load current.It is no different from any other ac circuit.
More foucus on stand-alone operation of the generator will surely confuse you.

 

[anorlunda]

How exactly is frequency (speed) and voltage fixed in the grid?

Review the video posted in #11. You can see disturbances in frequency propagating across the entire grid.

There are no infinite buses in real life. So grid frequency is actually a variable. It is held nearly constant by the action of the modern equivalents of the flyball governors as shown in post #25.

This is my last post in this thread. We're going in circles because you ignore or don't understand things said earlier in the thread.

 

[ffp]

@anorlunda and @cnh1995 h Thanks a lot for your answers. The explanations about how the grid fix frequency and voltage were useful for understanding.

If you increase the steam input with the intention of running the generator faster, the rotor will accelerate momantarily, but the infinite bus will force it to resume the synchronous speed after a transient period of damped oscillations (which depends on many factors like machine inertia, damping arrangement etc).

So to answer your question, yes it is impossible to increase the steady state speed and frequency of the generator connected to infinite bus. If you try, you will end up changing the active power output of the generator. The steady state speed and frequency will remain unchanged.
If you attempt to change the terminal voltage of the generator (which is same as the infinte bus voltage) by changing its field excitation, you will end up changing its reactive power output. The terminal voltage will remain unchanged.

My issue was with the notion that increasing If would change reactive power. That's not intuitive at all.
As I was studying induction motors and DC machines before, the increase in the magnetic flux always meant more active power: the stronger the flux, stronger the torque.

 

[cnh1995]

the stronger the flux, stronger the torque.

Not necessarily.
Torque in a rotating machine is produced by the interaction of the stator and rotor magnetic fields.
If you look up the torque equation in terms of the fields, the electromagnetic torque produced is proportional to the product of field strengths but more importantly, it is also proportional to the sine of the angle between the two field vectors. This means, only that component of the armature field which is perpendicular to the rotor field is responsible for torque production. Look up armature reaction in a synchronous machine.

As I was studying induction motors and DC machines before, the increase in the magnetic flux always meant more active power

The construction of a dc machine ensures that the stator and rotor magnetic fields are almost always perpendicular to each other. So more flux could mean more torque.
But in case of an induction motor: If you reduce the rotor resistance keeping the supply voltage unchanged, the starting rotor current increases. This means same stator flux but increased rotor flux at starting. Yet, the starting torque decreases because though the magnitude of the rotor flux has increased, its perpendicular component to the stator flux has decreased.

I belive you are drawing incorrect conclusions by mixing up many things at a time.