Synchronous generator on an infinite bus

[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.

 

[ffp]

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

Yeah, I'll try to not mix things up. I too feel I'm going in circles, but mostly because I'm not focusing where matters.

I was re-reading the thread and I'll try to put what I think as simply as I can.

There are 4 scenarios:

A.1) generator alone with load gets turbine speed increased
A.2)generator alone with load gets field current increased

B.1) generator in grid gets turbine speed increased
B.2) generator in grid gets field current increased

I will make affirmations about the scenarios and would like to be corrected when wrong.

A.1) increasing turbine speed increases Ea (=KΦω), which increases Vt (=Ea-IajXs). Both terminal voltage (Vt) and active power (P) increases. Since the angle between Ia and Vt didn't change, reactive power (Q) remains unchanged.

A.2) Increasing If will increase Φ which increases Ea(=KΦω), which increases Vt(=Ea-IajXs). However, P (=τω) remains unchanged. For that to happen, the angle between Ea and Vt, ad well as Vt and Ia must change, so reactive power (Q) changes.

B.1) increasing ω (τω) governor means increasing P (=τω). Since the grid fixates ω and Vt, there is an increase in the opposing EM torque, which reduces ω, but makes τ increase, after a transitory. So the turbines speed is still synchronous speed, but it's torque is higher. Result is higher P and no change in Q.

B.2) increasing If increases Φ, which increases Ea(=KΦω). P(=τω) remains constant and Vt tries to increase. However Vt is fixated by the grid. So, there is a bigger stator current (Ia), and changes in angles between Ea-Vt and Ia-Vt: Ea-Vt reduces and Ia-Vt increases.This way, P remains constant, but Q increases (angle between Ia and Vt increases). What exactly causes this changes in angles, I'm not sure.There must be several wrong afirmatives there. If you could point them and correct then, or maybe do the same 4 scenarios from scratch, that'd be nice. I know that thinking in terms of all variables (Ea, If, τ,ω,Φ,P,etc) might not be the best way, but that's exactly what i need to understand.

 

[Tom.G]

Hi @ffp, I have been sometimes reading this thread, but not in detail. Also I am NOT at all an expert in this field!

Anyhow, here are a few observations:
a) there seemed to be a confusion about the definition of an Infinite Bus
b) there was maybe a confusion between an AC generator and a DC generator

I think in your assertion A2, that the reasoning and conclusion should be identical to assertion A1.

Assertions B1 and B2 are so far out of my knowledge that I can not make any statements about them.

If I have made an error in this post, I hope that the more knowledgable people here will set us both straight!

Cheers,
Tom

p.s. I am impressed by your persistence and amount of effort you have put in to understanding this subject. I sincerely hope that you can get a complete and correct understanding of the subject here.

 

[cnh1995]

A.1) increasing turbine speed increases Ea (=KΦω), which increases Vt (=Ea-IajXs). Both terminal voltage (Vt) and active power (P) increases. Since the angle between Ia and Vt didn't change, reactive power (Q) remains unchanged.

A.2) Increasing If will increase Φ which increases Ea(=KΦω), which increases Vt(=Ea-IajXs). However, P (=τω) remains unchanged. For that to happen, the angle between Ea and Vt, ad well as Vt and Ia must change, so reactive power (Q) changes.

In a stand alone generator, increasing field current or generator torque will change the terminal voltage and speed. Any changes in load impedance and load power factor will also affect terminal voltage and frequency. But I do not have much information on how the terminal voltage and frequency is regulated in a stand alone generator.

Result is higher P and no change in Q.

Higher P and lower Q. With an increase in the power angle, the reactive power is reduced. Look up the formula for Q. It depends on the cosine of the power angle.

B.2) increasing If increases Φ, which increases Ea(=KΦω). P(=τω) remains constant and Vt tries to increase. However Vt is fixated by the grid. So, there is a bigger stator current (Ia), and changes in angles between Ea-Vt and Ia-Vt: Ea-Vt reduces and Ia-Vt increases.This way, P remains constant, but Q increases (angle between Ia and Vt increases).

Yes.

What exactly causes this changes in angles, I'm not sure.

Read #16.