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Generator VAr vs PF control
AVR VAr control VS Voltage control

Hello all,

Can any one give a explanation on the difference between VAr control or Voltage control on a synchronous generator automatic voltage regulator (AVR)?

I know by lowering the field current on one generator (two generators in parallel) you increase the VAr consumption going in to that generator which will read a lag Power Factor PF.

Also by raising the field current to one generator (two generators in parallel) you increase the VAr production going out of that generator witch will read a lead PF.

So if an AVR was in VAr control mode will the system voltage drop?

When is VAr control used?

When is voltage used?

How does it work?

Thanks

Hope someone can give a reply. I interested to know it as well.

1 out of 2 members thought this post was helpful...

gustavo_marcelo,

You seem to have an islanded system at your site, that can be used to gather some data when it separates from the grid. AND, you seem to have some modeling capability, as well?

Why don't you provide some data from the plant? Or from the model?

Let's say, after the unit separates and is stabilized, and the system voltage is at rated, what is the power factor of the load?

If you have two units operating in parallel in island mode, what happens when you then start increasing the excitation on one unit? What happens to the power factor of the unit you are increasing the excitation on? What happens to the MVAr output of the unit you are increasing the excitation on? What happens to the power factor of the other unit? What happens to the MVAr output of the other unit? During this process, what happens to the system (island) voltage when you are increasing the excitation on one generator? And, what is happening to the terminal voltage of the other generator?

Let's say you increased the excitation on one generator until the power factor of that generator is at 1.0, and the power factor of the other generator reaches the same level at which you started this experiment--what happens if you continue to increase to the excitation of the same generator you have been increasing the excitation of? What happens to the power factors of both generators? What happens to the MVAr outputs of both generator? What happens to the island system voltage? What happens to the generator terminal voltage of both units?

Or, if you only have one unit operating in island mode, what happens when the system voltage is at rated? What is the power factor at the generator output? What is the MVAr output of the generator? Now, let's say you start raising the excitation of the generator, what happens to the power factor at the generator? What happens to the MVAr output of the generator? What happens to the generator terminal voltage? What happens to the island system voltage?

In my experience, when operating as an island, the power factor (VAr requirement; Q)) of the island load is what it is (the sum total of all the reactive loads on the system), and can't be changed by changing the power factor of the generator(s). Neither can the real power (MW; P) required by the load be changed by increasing or decreasing the power being produced by the generator(s)--it is the sum total of all the real power loads on the system.

So, the question is: What happens when one or more generators on an island system (because that seems to be what the original poster was asking about!) increase (or decrease) their excitation? What happens to the generator outputs--power factor, MVAr, and voltage? What happens to the grid values--power factor, MVAr, and voltage--when changing the generator excitation? I'm assuming one would start this test from a condition where the island voltage was at rated, and it would be excellent if it was possible to know what the actual island reactive power "consumption" was (MVAr; power factor)--but I think they would be the same values as when the island system voltage was at rated.

But, it would be excellent if you could help to answer your own question with data from your plant. You could make up a table to fill out during your experiment, or from the archived data recorder. Or from the model you have built or have access to. We don't need to speculate or quibble about what might happen--you can provide real data and feedback on what does happen, or at least what your model says might happen. You shouldn't have to change excitation by very much to see some results, especially on an islanded power system (unless it's very large and has several synchronous generators synchronized together powering the island load). And, you could convince the Operations Supervisors that this would be an excellent method for gathering operating data the operators could use to improve their operating capability--and response to unusual situations. A win-win for everyone, wouldn't you say? (I would!) One unit powering a power island, or even two units powering an island, it shouldn't take too much deviation from rated island system voltage to see effects pretty quickly. And, again, it would give the operators some experience with what happens and what to expect when operating in island mode.

I'm really looking forward to your test results!!! Even from the model (and it would GREAT if you could tell us a little about the model--who built (programmed) it; what it's capabilities are; etc.). Yes, I'm VERY excited to hear back on this topic.

>You seem to have an islanded system at your site, that can
>be used to gather some data when it separates from the grid.
>AND, you seem to have some modeling capability, as well?

Yes CSA, we can operate both.Yes, I did some modelling, but for this case I only used a generic model and parameters to see the behaviour.I always do it,asking many people so that I can gather and compare various feedback and opinion.

>Why don't you provide some data from the plant? Or from the
>model?

It is a generic one. I can share if I got the opportunity to perform it for recording purpose ( perhaps can see it from trends).

>Let's say, after the unit separates and is stabilized, and
>the system voltage is at rated, what is the power factor of
>the load?
Around 0.9 is the normal value.

>If you have two units operating in parallel in island mode,
>what happens when you then start increasing the excitation
>on one unit?
Field current will increase, the Q also increase wirh a reduction in PF.

>What happens to the power factor of the unit
>you are increasing the excitation on? What happens to the
>MVAr output of the unit you are increasing the excitation
>on?
Reduced in lagging direction and increased of Q

>What happens to the power factor of the other unit? What
>happens to the MVAr output of the other unit?

Moving towards leading and reduced Q ( absorbing it from other unit).

>During this process, what happens to the system (island) voltage when
>you are increasing the excitation on one generator? Increase

>And,
>what is happening to the terminal voltage of the other
>generator?
Reduced with the increase of Q

>Let's say you increased the excitation on one generator
>until the power factor of that generator is at 1.0, and the
>power factor of the other generator reaches the same level
>at which you started this experiment--what happens if you
>continue to increase to the excitation of the same generator
>you have been increasing the excitation of?

>What happens to the power factors of both generators? What happens to the
>MVAr outputs of both generator? What happens to the island
>system voltage? What happens to the generator terminal
>voltage of both units?
>
>Or, if you only have one unit operating in island mode, what
>happens when the system voltage is at rated? What is the
>power factor at the generator output? What is the MVAr
>output of the generator? Now, let's say you start raising
>the excitation of the generator, what happens to the power
>factor at the generator? What happens to the MVAr output of
>the generator? What happens to the generator terminal
>voltage? What happens to the island system voltage?
>
>In my experience, when operating as an island, the power
>factor (VAr requirement; Q)) of the island load is what it
>is (the sum total of all the reactive loads on the system),
>and can't be changed by changing the power factor of the
>generator(s). Neither can the real power (MW; P) required by
>the load be changed by increasing or decreasing the power
>being produced by the generator(s)--it is the sum total of
>all the real power loads on the system.

Yes correct

I will come back to this post with some info from the modelling result.

2 out of 2 members thought this post was helpful...

gustavo_marcelo,

>>If you have two units operating in parallel in island
>mode,
>>what happens when you then start increasing the excitation
>>on one unit?
>Field current will increase, the Q also increase with a
>reduction in PF.

I agree with this; it should happen.

>>What happens to the power factor of the unit
>>you are increasing the excitation on? What happens to the
>>MVAr output of the unit you are increasing the excitation
>>on?
>Reduced in lagging direction and increased of Q

Maybe.

>>What happens to the power factor of the other unit? What
>>happens to the MVAr output of the other unit?

>Moving towards leading and reduced Q ( absorbing it from
>other unit).

Maybe.

>>During this process, what happens to the system (island)
>voltage when
>>you are increasing the excitation on one generator?
>Increase

I agree.

>>And,
>>what is happening to the terminal voltage of the other
>>generator?
> Reduced with the increase of Q

I can agree with this if the increase of Q (MVAr) is in the leading direction.

Here's what I've seen in power plants. When two units are synchronized to an island load and the reactive power "drawn" by the load is relatively constant, if the voltage of one generator is equal to the system voltage and its power factor is 1.0 (and its Q (MVAr) is 0 (no leading or lagging MVArs)), the power factor of the other generator is equal to the power factor of the load (something less than 1.0)--it is providing the reactive power drawn by the load. (I'm used to islanded loads having an inductive reactive load (mostly from induction motors).)

If one starts increasing the excitation of the generator which has the
low power factor and is providing the reactive current required by the island load, the island can't "absorb" any more reactive power. The number of motors and transformers and inductive loads doesn't change--isn't changing--just because one tries to increase the reactive current being provided by one generator). So, the other generator starts to "absorb" the excess reactive current and its power factor decreases in the leading direction (again, I'm presuming the island load is inductive), and the MVArs of the second generator starts increasing in the leading direction. The excess lagging MVArs being produced by the first generator (the one whose excitation is being increased) has to go somewhere--and the island loads can't do anything with it. Sometimes it's hard to see, but if one has a good generator terminal voltmeter with enough resolution one can see the voltage actually decreasing slightly. And, the island voltage is actually increasing--because of the increase from the excitation of the first generator.

If two generators are synchronized together and powering an island load together independent of a grid and the island load is fairly stable and isn't changing by a great amount and the two units are both operating in Droop Speed Control mode and the grid frequency is stable, if the load of one unit is increased the load of the other unit will decrease. The island load isn't changing--the number of motors and lights and televisions and tea kettles and hot plates and computers and computer monitors doesn't change because someone changed the load of one of the generators. The extra power doesn't have anywhere to go--but to cause the load of the other unit to decrease. Now, the relative sizes of the two units will have an effect on the split of power between the two as will the Droop setpoints of the two units. So, in addition to this, it's very possible the grid frequency may also deviate from rated.

Again, changing the reactive power output of a generator doesn't change the amount of reactive load being powered; that's determined by the nature of the loads. And, one can't change the amount of real power on the grid by increasing the amount of power being produced by one generator--it's fixed by the loads on the grid. So, excess (or a deficiency of) power being supplied to a load will affect the frequency of the loads (and of the generators synchronized together powering the grid). And, an excess (or deficiency) of MVArs will cause the voltage of the loads to suffer (and have an effect on the voltage of the other generators on the grid, also).

Even on a very large (or, "infinite") grid: An excess of power being produced (more than the load requires) will result in an increase in the frequency of the grid. And, similarly, a deficiency of MVArs being produced (less than the load requires) will result in a decrease in grid voltage.

It's not possible to produce more power than the load requires and still maintain the grid frequency at rated. And, if one wants the voltage of the grid to remain at rated one has to produce the amount of reactive power required by the grid, or the grid voltage suffers. (The difference here is that while grids don't want to deviate from rated frequency by very much, if at all, it's acceptable for grid voltage to fluctuate much more (as a percentage of rated) than the frequency is allowed to fluctuate.)

It's all a balancing act. And, if thought of that way, it can be reasoned through relatively easily.

This is my experience, and understanding. For stable and steady operating conditions it has been pretty consistent (my experience, that is). The load is what it is--real and reactive. If you want to keep the frequency of the system constant, you have to provide the amount of real power required by the load. If you want to keep the voltage of the system constant, you have to provide the reactive power required by the load. Changes to the amount of real or reactive power being generated will not change the real or reactive power required by the load being supplied--but it will change the frequency or voltage of the system. I tend to think of watts (kw; MW) as being produced and consumed. And, in the same way I tend to think of VArs (kVArs; MVArs) as being produced and consumed; it's not technically correct, but it can be useful in understanding and anticipating changes and effects. Small power islands are seemingly a little more difficult to understand, but only because on a large ("infinite") grid the same effects happen--they are just "spread out" over a much wider group of units and therefore affect a much smaller individual effect on each of the units.

Any number of generators synchronized together powering any number of load(s) are operating like one single generator power a single load (because all of the loads appear as one to the power system--just as all the generators appear as one to the system). The load can have a real and reactive component. The torque being provided to the generator(s) provides the real power, and the excitation being applied to the generator(s) provides the reactive power. And, the total amount of real power being provided to the system must equal the real power being required by the load(s) for the system frequency to remain at rated. And, the total amount of reactive power being provided to the system must equal the reactive power required by the load(s) for the system frequency to be equal to rated. No matter the number of generators. Cause an imbalance of either power or VArs and frequency or voltage is going to suffer.

Trying to answer questions like this can be really difficult--because to properly answer the question the conditions have to be fully stated and understood. And, when trying to understand a small, island power system when we're only really experienced with large power grids can be very difficult. The original poster titled the thread as a question about VAr vs PF control. But the description hinted about AVR VAr control versus voltage control. And the original post was kind of all over the place. That's probably why nobody responded to it before now.

It's like trying to understand Droop Speed Control when the textbook or reference book says it's the change in speed when the load changes. That just doesn't happen on a well-regulated grid. The text or reference doesn't state that this only happens when the generator set is operating completely independently of any other system in Droop Speed Control (instead of Isochronous Speed Control), and load is ADDED to or REMOVED from the system being powered by the generator set--and the operator isn't taking ANY action to maintain rated frequency.

All of the conditions have to be stated and understood--not implied and neglected. And sometimes stating all of the conditions can be very difficult, as can be understanding the conditions being implied when few are stated.

This is a technical industry, and details matter.

Thanks a lot CSA. Your reply is very informative and useful.

2 out of 3 members thought this post was helpful...

Mohammeibr752,

Thank you for the kind words, but, as usual there is an error....

>And, the total amount of
>reactive power being provided to the system must equal the
>reactive power required by the load(s) for the system
>voltage to be equal to rated. No matter the number of
>generators. BEcause an imbalance of either power or VArs and
>frequency or voltage is going to suffer.

Sincere apologies for any confusion.