We have Brush AVR A-30 for our GT based power plant. There are few things which are not much clear to me like:

there are two controls available
1) Main control
2)Standby control.

i)So how these two controls are different as far as protection and control of alternator is concerned?

ii) Do the limiters like O/E, U/E or V/F remain in line during Standby control?

iii) Is Standby control in AVR A-30 same as Manual control in other types of AVRS?

Thanks n regards,
ashish

 

ashish,

What does the Brush manual say about these two modes?

 

CSA,

What i have learned from the Brush manual for AVR A-30 is that in Main control mode we can select Voltage control, PF control and VAR control along with var shed whereas in Standby control mode we can select either Generator output voltage control or Excitation current control. If it is voltage control then output voltage will be kept constant (depending upon the set value) irrespective of the load whereas in excitation current control the output voltage will not remain constant with varying load (this is what i believe).

Again the shifting of control from Main to Standby will occur whenever there is U/V, O/V, U/E limiter or O/E limiter operated alarm etc appear in the MONITOR card. So during Standby mode can we reset those limiter alarms etc which appeared during Main to Standby mode shifting?

So i request the domain experts to give their views on AVR system (spec. Brush AVR A-30) which i believe has not been discussed much in this forum.

How the AVR system will be different in these two modes of operation?

Thanks n regards,
ashish

 

Ashish,

That's excellent; you've read the fine manual.

A synchronous generator's terminal voltage is a function of the amount of excitation applied to the rotor's windings. At synchronous speed (which is what most synchronous generators should be operated at) which should be stable and for all intents and purposes fixed, when the excitation is increased the generator terminal voltage will tend to increase, and when the excitation is decreased the generator terminal voltage will tend to decrease.

When the generator is NOT connected in parallel with other generators, the generator terminal voltage will absolutely increase or decrease in direct proportion to the amount of excitation applied. When the generator is being operated in parallel with other generators the terminal voltage doesn't usually appear to change very much because of the effect of the system voltage which appears to keep the generator terminal voltage relatively constant. (Depending on many external factors, the generator terminal voltage may actually be seen to increase or decrease, but not by the same amount as if the generator were NOT operating in parallel with other generators.)

It's generally intended to keep the generator terminal voltage relatively constant when the machine is operating in parallel with other generators. So, the amount of excitation is varied as necessary to maintain the generator terminal voltage setpoint. Usually, there are some PTs (Potential Transformers) used to provide the feedback to the circuit that's used to vary the excitation as required to maintain the actual generator terminal voltage equal to the setpoint.

When a synchronous generator is being operated in parallel with other generators on a large or infinite grid and one increases the excitation above the amount required to keep the generator terminal voltage exactly equal to the system voltage, then reactive current will begin to "flow" through the generator's stator windings. As the excitation is increased the Power Factor of the generator will begin to decrease from 1.0 in the lagging direction, and the VAr meter will increase above 0 in the lagging direction.

If the excitation is decreased below that required to keep the generator terminal voltage exactly equal to the system voltage the Power Factor of the generator will decrease from 1.0 in the leading direction, and the VAr meter will increase above 0 in the leading direction.

So, power factor and VArs are related to excitation. Different control loops can be created, using Power Factor sensors or VAr sensors, to control the Power Factor or VAr "flow" of the generator by varying excitation as required, "automatically." The operator sets a Power Factor setpoint, or a VAr setpoint, and the excitation control system automatically adjusts the excitation as required to maintain the desired Power Factor or VAr "flow."

When this is happening (excitation is being increased or decreased when the generator is being operated in parallel with other generators on a large or infinite grid), the generator terminal voltage may or may not appear to change by very much--again, that's a function of many factors. But when the excitation is exactly equal to the amount required to keep the generator terminal voltage equal to the voltage of the system which which it is synchronized then the Power Factor of the generator will be equal to 1.0 (Unity) and the VAr meter will be at 0.0 VArs.

Since the amount of excitation being applied to the generator rotor is being varied, there is an "inner loop" to this generator terminal voltage control mode, commonly referred to by most other manufacturers as Automatic Generator Voltage Control Mode, or Auto Mode, or AVR (Automatic Voltage Regulator) mode. When the excitation control system is in "Auto" mode, or what Brush appears to be referring to as Main control mode, the setpoint is generator terminal voltage and the feedback is generator terminal voltage.

The "inner loop" is the excitation current, which must be stable for the generator terminal voltage to be stable. In this loop, the setpoint might be either excitation voltage or excitation current, and the feedback would be either excitation voltage or excitation current, respectively.

Most excitation control systems have a method for only controlling the excitation terminal voltage or excitation current--the "inner loop." And most manufacturers call this the "Manual" control mode--what Brush appears to be calling Standby control mode.

In Manual control mode (or Brush's Standby control mode), the unit operator (a person) manually controls generator terminal voltage, or generator Power Factor, or VAr "flow" by adjusting the excitation terminal voltage setpoint or the excitation current setpoint.

Now, here's where it gets really dicey--especially on control.com. If the generator terminal voltage setpoint is not changed as a generator's load (Watts; KW; MW) are increased the Power Factor of the generator will "decrease" in the leading direction, and the VAr "flow" will increase in the leading direction. So, as a generator's real power increases, it's generally necessary to increase the excitation just to maintain a constant Power Factor or VAr "flow". There is some disagreement about the actual mechanics of how this happens amongst some posters here on control.com, but it is agreed that unless some action is taken as the generator is loaded the Power Factor and VAr meter will move to the leading directions.

Conversely, if a generator is operating at some load and the Power Factor is at 1.0 and the VAr meter is at 0.0 VArs, and the generator is unloaded, then the generator Power Factor will decrease in the lagging direction and the VAr meter will increase in the lagging direction. Again, the actual mechanics of how this happens in the generator is in contention, but it is agreed that this will happen.

Now, if a generator is being operated strictly on generator terminal voltage control mode (no Power Factor control or VAr control) and the system voltage changes, then the generator's Power Factor and VArs will change. And, system voltages do usually change during the course of a day, sometimes dramatically during the day depending on conditions and how the grid is "operated" or "controlled." So, this is another reason plants typically use either Power Factor control or VAr control, to make sure the generator's Power Factor or VAr "flow" stays relatively constant regardless of external factors (like changing system voltage).

Finally, some generator excitation systems have a method of reducing excitation automatically when a machine is being unloaded to prevent "excessive" lagging VArs from flowing, and thereby preventing the generator's Power Factor from becoming low in the lagging direction. This is usually called "VAr Shedding".

Hope this helps. It seems the Brush's Main Control Mode is the same as most other manufacturer's Auto excitation control mode (sometimes called the AC control mode), and the Brush's Standby control mode is the same most other manufacturer's Manual control mode (sometimes also called the DC Excitation control mode).

Warning: Be VERY careful about your electrical terms if you reply. (For example, don't refer to reactive "power". Reactive current is fine; reactive power is not--for one person on control.com.)

 

Dear CSA,

I am confused with the terminology you have used, it would be nice if you give a little explanation . If the generator excitation is lower than required for a terminal voltage as grid, the machine will behave as inductive (receiving reactive power from the grid). The inductive load has lagging power factor (if I am not wrong! :S ). Whereas you have used terminology of leading power factor in this case . Please explain if there is any difference in conventions!

Cheers!!

 

When one is dealing with VArs, one has to take the flow in the context of the discussion. From a <b>generator</b> perspective, lagging VArs feed a lagging load. Lagging VArs flowing "out" of a generator are considered to be positive, while VArs flowing "into" a generator are considered to be negative, and "leading".

The same is true of substations: one always has to consider the context of the discussion and the flow of VArs into/out of the substation when referring to leading or lagging, positive or negative.

It is opposite to the direct definition of inductive and capacitive--but it is still how VAr flow is viewed from a generator perspective.

I believe that if you refer to some power system manuals and technical reference documents you will find a more detailed explanation, complete with maths if you like, to support the usage.

 

Thanks!!!

 

My question is related use of transformer tap changer during high system voltage.

As system voltage increase, gen. terminal voltage also increases, generator VARS increases in leading direction and AVR decreases excitation current also to keep machine terminal voltage within limits.
SO gen. can trip on loss of excitation protection.

So what should be done with generator load (either increase or decrease)and either transformer tap changer position (suppose it was at 14 position) to increased or decreased to avoid generator trip on loss of field. Also justify transformer tap changer position w.r.t system voltage.

 

If generator-A is operating at 40MW with 0 MVAR and generator-B is operating at 40MW with 20MVAR, is there any difference between fuel inputs of two generators? although there will be difference between stator currents.

 

engr,

You didn't say if the MVars of Generator B were leading or lagging....

I would be very interested to hear what you believe or have observed in regards to your question. Do you think that the fuel flow-rates for the two prime movers driving the generators in your scenario are equal, or do they differ? If they differ, what is the magnitude of the difference--1%, 5%, 10%, 20%?

I've considered this many times myself, and had many discussions with others--and there is no shortage of opinions, most of it without the benefit of maths. I welcome discussion and correction of my opinions and theory, but the analysis below is my personal opinion--without the benefit of maths.

Referring to the AC power system power triangle, the apparent power of Generator A is 40.0 MVA, and the apparent power of Generator B is 44.7 MVA. MVA is called "apparent power" and represents the total power in an AC circuit: resistive (real) and reactive--from every definition I've read. I interpret 'total power' to mean ALL the power being produced by a generator--real (resistive) and reactive.

Real power is the result of torque from the prime mover being converted to amperes in the generator stator at a relatively stable voltage and those amperes are converted back into useful work at the other end of the wires connected to the generator--work that is quantifiable and tangible.

Reactive power doesn't perform any useful work, and mostly results in the "generation" of unwanted heat. Reactive "power" (we have to be careful here at control.com lest we upset a curmudgeonly MVP) does no useful, quantifiable, tangible work--other than "generating" unwanted heat in the conductors in which it is flowing.

I keep returning to the heat "generated" by reactive power because heat is a form of energy which requires real power to produce (think of an electric space heater, or an incandescent lamp--both purely resistive devices that produce heat from current flowing through the elements). So, in my opinion, this is one reason the fuel flow-rate of Generator B would be higher. How much higher is a subject of some debate.

Returning to the power triangle, the amount of real power is directly related to the fuel flow-rate of the generator prime mover. In other words, the length of the real power side of the power triangle is a function of fuel flow-rate to the generator prime mover. To be more specific, still, for all intents and purposes the length of the real power side of the power triangle is most directly related to the fuel flow-rate of the generator prime mover.

If both generators in your example had their fuel control valves fixed (clamped in one position) when operating at 40 MW and 0 MVar then the apparent power of both generators would be the same: 40.0 MVA, because the total power being produced by the generator is equal to the real power being produced since there is no reactive power.

Now, if the excitation of Generator B was changed to "produce" 20 MVar (the length of the reactive side of the power triangle was increased from 0 MVar to 20 MVar) <i>while the fuel flow-rate remained unchanged</i> in my personal opinion the apparent power side of the power triangle of the generator would remain unchanged at 40.0 MVA, but the real power would decrease to 34.6 MW--because the total energy going into the generator would be split between real power and reactive power <i>while the fuel flow-rate remained unchanged.</i>

So, to make Generator B produce 40 MW at 20 MVar it would be necessary to increase the fuel flow-rate to Generator B. But, I don't believe the change in fuel flow-rate is significant (more than a few percent) because the "work" done by reactive power is relatively low (the heat "generated" by reactive current flowing in the AC circuit of the generator).

To muddy matters even further, for the overwhelming majority of generators the power (real power, Watts) for the excitation system comes from the generator (static excitation) or from the prime mover driving the generator (brushless excitation). So, that means when excitation is increased to produce lagging VArs that some power is taken away from the total power being produced by the prime mover and generator, so the fuel flow-rate must be increased to maintain the same net real power output of the generator.

The amount of excitation applied to a synchronous generator rotor is directly proportional to the amount of reactive current flowing in the generator stator windings. If the fuel flow-rate is held constant and the excitation is varied, the reactive current varies and so does the power factor--the efficiency of the generator at producing real power output for the total amount of energy being provided to the generator. So, if the fuel flow-rate is held constant and the excitation varies, the amount of real power being produced by the generator will change--by a lot if the real power output is low, by not so much if the real power output is high.

This business of holding the fuel flow-rate constant while varying reactive power (current) is my way of analyzing your scenario. If both generators start from 40 MVA (40 MW, 0 MVar) and then one generator's reactive current is changed while it's fuel flow-rate is held constant I believe the real power output will decrease--because the power factor will decrease as the reactive power increases for the same fuel flow-rate, the same amount of torque being applied by the prime mover to the generator rotor. Changing the excitation changes the reactive power flow--and to an extent, the real power flow if the fuel flow-rate doesn't change to maintain the same real power output.

So, that's my opinion and theory. If I'm wrong, then I'll have learned from the ensuing discussion. I don't want to know about armature reaction or counter-emf or back-emf or load angle. I just want to be able to relate everything to the power triangle--very simple maths that most everyone can visualize and understand. So, if we have a discussion about this--let's keep it simple, please.

 

Eng... the simple reply to your query is :

If one ignores losses, then neither lag nor lead MVAr wii influence fuel rate!

Regards,
Phil Corso

 

Phil Corso,

Let's not ignore "the losses." What would be the difference in fuel flow-rates?

And, let's start from an equal MVA, 0 MVar, holding fuel flow constant and then increase the MVar to 20 lagging MVar. What would happen to the fuel flow-rate? And the real power (MW)?

And, then, if one increased the real power to 40 MW while maintaining 20 MVar, lagging, what would happen to the fuel flow-rate?

Inquiring minds would like to know.

Thank you in advance. Short, concise answers are all that's required for short, simple scenarios.

 

CSA... I take it then, you won't object to my using formlae?

Regards,
Phil Corso

 

CSA...

I believe you will agree that if load MW is constant so is fuel-rate. Paraphrasing the poster's query, "while maintaining a 40MW load, will a change in the load's MVAr component affect fuel-rate?"

I now realize my mistake. Of course it will "influence" fuel-rate, but, to what degree?

The change is so small that it won't "impact" fuel-rate.

Do you want me to provide additional discussion proving my thesis?

Regards,
Phil

 

Phil Corso,

I don't generally object to the use of mathematical formulae (when it's spelled correctly, of course). I do object to allusions, incomplete, in-concise--albeit short--replies.

Have at it--but please just be clear, concise, and complete. While being short and simple. We're not in an organized class with textbooks and reference materials required or suggested. So, it just needs to be complete and concise--recognizing that this isn't a Q&A session after a lecture after required reading and homework. It needs to be concise and complete and clear.

 

Phil Corso,

Yes, please.

 

CSA...

considering your admitted disdain for academia, I thought you might accept a university student’s answer. So I assigned the task to my students. Following is the winner’s solution:

Start of Student Study

1) This study will show that changing an alternator’s load-PF (hence its armature-current, Ia) will cause a change of the prime-mover’s fuel-flow rate! Additionally, it will quantify the magnitude of the change!

2) The prime-mover’s mechanical output must match only the real-component of the alternator’s connected electrical load plus alternator losses!

3) When Load PF = 1.0, armature-current is Ia, and losses are proportional to Ia^2!

4) When Load PF &#8800; 1.0, armature-current is Ia’, and losses are proportional to Ia’ = SQRT(Ir^2+Ix^2) which can also be presented as Ia’ = (Ia+&#8710;Ia), where &#8710;Ia represents the change due to Load-PF!

5) An example using Engr’s parameters: 45 MVA; 40 MW; 20 MVAr; PF, 0.8; Est. Eff, 98.3%; Losses, 0.7 MW:

o Convert Ir and Ix to per-unit; Ir = 2.0 pu, Ix = 1.0 pu
o For case, PF = 1.0, Ia = Ir = 2.0 pu.
o For case, PF &#8800; 1,0, Ia’= (Ia+&#8710;Ia) = 2.24 pu, and &#8710;Ia = 2.24-2.0 = 0.24 pu.

8) Let MWr represent losses at rated MW and PF, and MWo represent losses at PF = 1.0, then MWo = [(Ia/Ia’)^2]*MWr = 0.56 MW!

9) The magnitude of the Fuel-flow rate change, in this case, is 0.14 MW!

End of Student Study

CSA… QED (although it may not be as short and concise as your answers are!

Regards,
Phil

 

Phil Corso,

Were it not for education I would not be where I am today. It's just that some academics forget what it's like to have to learn a concept--which is much more than maths. People operating and maintaining equipment would certainly be better off if they understood all of the maths, and it does little good to start an explanation with maths for those who really don't need that level of understanding. Most every one of us learned simple y=mx+b in primary grades, or even in secondary school, but it's truly amazing how few people can calculate slope (gain) and offset.

Maths are a tool to enhance understanding--not the way to start explaining a complex subject. Again, especially for someone who's not in a classroom environment, and who probably didn't do well in a classroom environment as a student (except maybe socially, but not academically).

As for this exclamation pointer-riddled "answer".... (Are you teaching grammar and punctuation now, too?!) I'm not quite sure it hits the mark. I'll keep trying to make sense of it, but something doesn't seem quite right. I think it's with the first statement (bound):

> 2) The prime-mover’s mechanical output must match only the real-component of
> the alternator’s connected electrical load plus alternator losses!

But then I lose it here because I don't know what Ir and Ix are:

> o Convert Ir and Ix to per-unit; Ir = 2.0 pu, Ix = 1.0 pu
> o For case, PF = 1.0, Ia = Ir = 2.0 pu.
> o For case, PF &#8800; 1,0, Ia'= (Ia+&#8710;Ia) = 2.24 pu, and &#8710;Ia = 2.24-2.0 = 0.24 pu.

And why is Ir 2.0 pu (what units?), and why is only Ir used (Ia = Ir = 2.0 pu) in any calculations?

And, how did we get to (Ia+&#8710;Ia) = 2.24 pu?

Why did 0.68 MW get rounded to 0.7 MW, and then from (8) we have two decimal places (0.56 MW). What kind of math is this? Shouldn't students be taught to use consistent decimal places to avoid rounding errors, or is this a Marketing class?

Let's sort the above out, then we can deal with the formula in (8), because I'm missing something there, too.

 

In case of high leading VARS due to grid demand, generator is under excited due to AVR reaction. Then what should be done with transformer tap position (suppose it is at 14 position) and generator MW load (either increase or decrease MW load) to prevent tripping of generator on loss of field protection?

 

CSA...

so is it your position that the solution to Engr's query boils down to application of "Significant Figures?"

Phil