engr,

I've been doing some research regarding load tap changers, or on-load tap changers. It seems that different manufacturers make different types of transformers/taps depending on the application.

It's really difficult to say what should be done at your site without understanding more about the type of load tap changer at your site.

To my way of thinking about the problem, if the "grid" voltage the generator terminals are seeing (from the low voltage side of the step-up transformer) is really high versus the amount of excitation being supplied to the generator then a lot of leading reactive current will be flowing in the generator stator windings and it's entirely likely that in an effort to reduce the voltage at the generator terminals the excitation system and/or the operators will reduce the excitation which can lead to a very dangerous situation (loss of excitation, or, worse, loss of synchronism (pole slipping)).

So, if it's possible to reduce the voltage on the low voltage side of the step-up transformer with the load tap changer without having a negative effect on the voltage on the high voltage side of the step-up transformer so that the generator excitation can be increased that would seem to be the thing to do <b>based on the information provided.</b>

But, we don't know what will happen to the voltage on the high side of the step-up transformer--which is important to understand. There may be something happening on the high side of the step-up transformer that we don't know about or understand. I've read about some transformers that can change taps and only affect the voltage on one side of the transformer (high or low), and then I've read about some transformers that, when the taps are changed, there is a difference on both sides of the transformer.

So, there's not enough information for us to be of much help. I think you should be working with the grid operator/regulators to understand why the voltage on the low side of the transformer is so high. And, if this only happens on occasion (it's not a typical operating circumstance), why is it happening and what can be done to correct it?

I wish I could be more help. From the lack of other replies, I would guess that there isn't anyone else with any experience in a situation similar to yours that can comment. And, perhaps that's just because there's too much we don't know.

Please write back to let us know how this progresses.

 

Phil Corso,

Why is it so hard to just answer a question with an answer?

Even if you have to provide more information than requested because the original poster didn't provide enough information.

You seem so eager to use unexplained terms, inconsistent significant figures, and now semantics.

 

Sorry, CSA...

it's all your getting. I must admit, however, that we are somewhat equal... I know as much about gas-turbines as you know about electricity.

How long have you been involved with your avocation?

Regarding electricity, my student is willing to tutor you.

Best to you,
Phil

 

Sorry, CSA...

The poster provided enough information for the study, except for alternator efficiency!

If you believe my student’s guesstimate of efficiency is in error, then please be good enough to provide a value you may have learned during your career.

Phil

 

Engr,

Was the answer adequate, or not, for your purposes?

Phil

 

Phil Corso,

I'm sure I would flunk your class, or more likely be expelled.

Thank your student for the tutoring option--if he/she actually offered. My understanding of electricity is adequate for my needs--and those of my Customers/colleagues, even though it doesn't meet your standards or conventions.

I don't' quibble much with the estimate; I was only asking for an explanation of terms. I endeavour always to try to explain terms when I introduce them to a conversation/thread (even if I've explained them in twenty-three other threads) because I like to have new terms explained to me, and if I stumble on a thread/conversation it's nice to know what is being talked about. I find my understanding is enhanced--but I know that is opposition to your propensity to obfuscate.

Again, so we are in agreement--there <b>IS</b> a difference in fuel flow, even if it's not of any great magnitude. My electrical instructors didn't allow us to ignore losses, as they can, and do, add up. I don't necessarily agree with the exact numbers, but I have already put forth my theory (~45 MVA vs. 40 MVA), and I'll leave it at that. I'm satisfied that there is a difference in fuel flow-rate, and I did state that since VArs are not relatable (as Watts are, to say, horsepower) that it would be difficult to quantify. It is, after all, heat--whether it be in the stator windings, and/or the field windings, or the stator end-turns. I guess if we knew reactance/resistance we could estimate further, to even more significant figures that could be indiscriminately rounded off.

Thanks, again, Phil, and ESaD.

 

CSA,

Oh dear, oh dear, oh dear! I thought we had dealt with the subject of "reactive power" a year or so ago!

> 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.

It is incorrect to refer to MVA or "apparent power" as "total power". The apparent power is found by simply multiplying the voltage as indicated on a voltmeter by the current indicated by an ammeter - ignoring the fact that in the AC systems we are talking about the two are generally out of phase, with the phase angle affecting the power factor.

> 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.

Reactive power does not represent heat - it refers to the power needed to transfer energy from one stored or useful form to another. In an electrical system, energy can be stored in a capacitor (as anyone who has got themselves across a charged capacitor can attest). This energy is given by 1/2 C v^2. (C is capacitance, v is voltage: I use lower-case letters to indicate instantaneous values rather than RMS or other averages).

Similarly, energy can be stored in an inductor, and is proportional to the square of current ( 1/2 L i^2, where L is the inductance and i is the instantaneous current). This inductive energy is what causes the spark when a current through an inductive element is interrupted.

To change the voltage across a capacitor, energy must flow. In an AC situation, the energy is 0 at a voltage zero, and will increase as the voltage rises to its peak value. The flow of energy into the capacitor gives rise to a power which can be considered as "positive". When the voltage has reached its peak value and is decreasing, the energy stored in the capacitor must reduce. The flow of energy out of the capacitor can be thought of as a "negative" power. When capacitor voltage is zero, there is no energy stored. When the voltage commences to charge up to the opposite peak on the next half-cycle, energy must again flow into the capacitor - the "v^2" term in the energy relationship means that the polarity of the voltage has no effect.

So over one complete cycle of voltage applied to a capacitor we have power into the capacitor for 1/4 cycle; power out for 1/4 cycle; power in again for 1/4 cycle; and out again for 1/4 cycle. Using basic trigonometrical identities, it can be shown that if the voltage is (V sin f), the energy is V^2 sin^2 f = V^2 /2 (1 - cos 2f). The power flowing to inject and remove this energy is given by V^2 sin 2f. So the average value of power over one complete cycle is zero. However, there MUST be an alternating component of power flow to allow for the energy changes involved with the charge and discharge of the capacitor.

Similarly, maintaining a sinusoidal current flow through an inductor requires energy to be added to and removed from the inductor twice per cycle, with energy flowing in while current is increasing and out again while it is decreasing.

Neither of these forms of power involve any dissipation or heating in the inductor or capacitor. There will be some small power losses in the capacitor due to atomic-level effects in the dielectric, and slightly higher losses in the inductor because of iron and copper losses dependent on the current. However, these are generally mush smaller than the reactive power flowing in and out again.

The situation is parallel to the use of regenerative or reactive (the same word) braking in an electric vehicle. In a conventionally-powered vehicle, all the kinetic energy (1/2 m V^2) of the vehicle is converted to heat when stopping at traffic lights. To regain the original speed, fuel must be burnt to restore the kinetic energy to the original level. With regenerative braking, the kinetic energy is converted to electrical energy via generator action of the motors: this energy is stored in a battery which is to all intents and purposes just a very large capacitor. On re-starting, this stored energy can be re-used to accelerate the vehicle: the only fuel needed is that used to cover losses.

The amount of heating in a circuit containing capacitors or inductors does not depend on the reactive elements at all, but on how they are connected to the rest of the electrical circuit. The interesting trick comes about because, in each of these devices, the voltage and current are 90 degrees out of phase; but the phase relationship is reversed - CIVIL - in a capacitor, current leads voltage by 90 degrees; in an inductor, voltage leads current by 90 degrees (or current lags voltage by 90 degrees). So if a capacitor and inductor are connected in parallel to an AC voltage, the current into the capacitor is 180 degrees out of phase with the current into the inductor, and the total current seen by an ammeter connected into the common supply will be less than the current in the L-C circuit. The power needed to charge the capacitor while its voltage is increasing can be partly supplied from the energy released because the inductor current is reducing.

> 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>

If the fuel flow remains unchanged while the MVAR is increased, the active power available to the load will be reduced slightly but only because of the increased iron and resistive losses due to the increased current. 40 MW + 20 MVAr will give 44.72 MVA (to 2 decimal points) which is an increase in current of 1.18 %. Typical current-related losses will be about 2 % of full-load rating at full load, so the change in active power will be 2 % of 1.18% of full-load - about 0.23 %. Changes in power required for the excitation system are similarly of very small magnitude.

From my observations, a relatively small change in ambient temperature or relative humidity on a gas turbine (or cooling water temperature for a steam turbine) can change the efficiency by much more than this. So, to me, the correct answer to the original question is:
" Yes -it will have a small effect - but one that is insignificant compared with other factors that can change fuel consumption".

 

Bruce Durdle,

Thanks; once again I was hoping for your input and clarification.

It's greatly appreciated--and respected.

 

Bruce Durdle,

Dear, oh dear, oh dear, my mistakes were:

1) I did (do) consider MVA to be "power"--specifically "total power", since that's the amount of "power" being produced (both real and reactive). I keep forgetting about the "apparent power" thing. In my mind if I operate a synchronous generator at unity power factor (0 VArs) and stable, constant energy flow-rate into the prime mover and change excitation, the watts will also change. Which has little or no effect on the hypotenuse of the power triangle (which I visualize as "total" power: watts plus VArs). I know this is especially true at very low real power outputs, but also true at higher power outputs (more easily visible with digital meters and transducers than with analog meters).

So to my mind the "total power" (being output by the generator) doesn't change when energy flow-rate into the prime mover is held constant but excitation is varied to change the "split" of power between watts and VArs. In other words, if the hypotenuse of the power triangle is held constant (by not changing the energy flow-rate into the prime mover) but the excitation changes then MVA won't change but the real power (watts) will change as the VArs change. Again, starting from 0 VArs and some watts if I hold energy flow-rate into the prime mover constant and change excitation the real power will decrease as the VArs increase--but, for all intents and purposes the MVA won't change.

I haven't done a lot of maths for this (I'm told I have an aversion to maths.), but I know I've seen watts go down when energy flow-rate is held constant and VArs are changed from 0. To get watts back to the same value as before the excitation was changed it's necessary to increase energy flow-rate into the generator prime mover.

If I want to increase VArs (lower power factor)--and maintain real power (watts) I have to also increase energy flow-rate into the prime mover as I change excitation. I explained it very poorly, and I definitely used the wrong term ("total power" as opposed to "apparent power"--but I have a dislike for the term apparent power because while one component of it doesn't do any useful work that doesn't diminish it's reality because some work is lost when the reactive power component is increased. Again, since it's not tangible (like horsepower) it's difficult to grasp (mathematically it can be quantified, but relating that to real-world work is difficult at best).

2) I equated the increased current (real plus reactive) flowing through conductors to increased heat--instead of considering the losses due to inductive effects. My bad here, too.

I do agree--the correct answer is that the energy flow-rate into the prime is not the same for both conditions stated by the poster (40 MW, 0 MVArs vs. 40 MW, 20 MVArs). It's not "free"--on more than one account--choosing to split the energy produced by a generator between watts and VArs if one wants to maintain a constant or stable real power component.

Thanks again for your patient explanation. I really do need to get over my dislike of the term "apparent power" and quit confusing it with "total power." I know what I see happen to meters (and digital feedback) when energy flow-rate is held constant and excitation is varied: watts change, a lot a low watts, and by not so much at higher watts. But, the watts do change. And, to get back to the same watts it's necessary to increase energy flow-rate into the prime mover.

How about some help with engr's question about on-load tap changers? I just don't have enough experience with on-load tap changers to be of much help. I haven't been very successful on line, nor had much time to go looking for written documentation, on on-load tap changers.

Thanks again--and thanks in advance for helping with engr's on-load tap changer question if you can.

 

engr

Interesting debate and I am happy that it seems to have come to the right conclusion despite some distractions along the way.
I undertake many performance tests of CCGT plant using ASME PTC test codes and in doing so have to make a correction if the unit is not operating on the specified power factor.
The correction is quite complex and takes into account the difference in generator losses (friction, core, field, brushes etc. etc.) between rated and actual conditions but the results of a recent test may be of interest and compare reasonably well with the comments from Bruce above.

Generator Actual Values:
Measured Gross Power = 359942.3kW
Power Factor = 0.982
Calculated Losses = 2696kW

Generator Rated Values at same power:
Power Factor = 0.85
Nominal losses = 3609kW

Difference in losses caused by generator not on rated power factor:
3609 – 2696 = 913 kW or 0.25%
Small in percentage terms but still a significant amount of power.

 

The use of pu (per unit) is widespread in electrical calculations as it allows the voltage variations through a transmission system to be ignored - 1 pu voltage is the rated voltage for whatever part of the system we are interested in at the time. So with a generating station with nominal output voltage of 14.4 kV, connected through a step-up transformer to a 220 kV transmission system, with additional transformers at the far end stepping down to 33 kV and then to 11 kV, we can treat the whole thing as a single series circuit with a source voltage of 1 pu.

To fully define the parameters of a system in pu, we need to define a base MVA rating. For a generator this is usually the nominal MVA rating of the set, but the choice of an MVA base is quite arbitrary. The base current is then defined as base MVA divided by base voltage, and pu impedance (reactance or resistance) is the value that would drop 1 pu voltage at 1 pu current.

For a typical alternator, if the nameplate voltage and MVA rating are used as bases, the internal reactances and resistances are generally very similar. So one of my references (very dated) gives synchronous reactance of 1.45 pu (Kimbark: "Power System Stability", 1956.) without specifying the MVA rating of the machine involved.

 

One thing I want to add is suppose the the current Ia is composed of Ir and Ix. When such a current flows through a conductor, the active power dissipation in that conductor is proportional to the sum of the squares of Ir and Ix. But the reactive power, proportional to square of Ix, is not dissipated!

 

I am with CSA. In a sense it is correct to say apparent power as total power: Consider a lifting electromagnet. The useful work it does in lifting an object is from conversion of its magnetic field energy i.e reactive energy to mechanical energy. Another example: a conventional relay.

> It is incorrect to refer to MVA or "apparent power" as "total power".
>
>> Reactive power doesn't perform any useful work.

 

<b><i>MOST EXCELLENT!</i></b> As an instructor of power plant operations and plant simulator training, the topic of excitation, power factor, and vars is probably the most difficult for operators to grasp. Your article is one of the best I have seen, in over 40 years.

<b><i>Thank you and well done!</i></b>

 

I have few questions regarding AVR. We have two power plants using Unitrol AVR.

1. There are three modes(in our plant only auto mode is using)
Auto/manual mode
Var mode
Pf mode

Can we select two or all modes in one time because for operation of the plant.

2. Our two plants connected to same system, one providing 300Mw and 80Mvar while other plant 300Mw and -10Mvars. What's the reason.

 

We are combined cycle power plant with 2 GE 7FA gas turbine and 1 D11 steam turbine with Mark6e Conversion (upgraded Mark5.) In the last few days we have just resolved similar questions here. Mostly the questions arose because the steam turbine and the gas turbine control system had different labels on the control buttons, and in different locations. We also did not fully understand the functionality of each mode of the generator control.

After fully understanding what the modes of operation are, we re-labeled the buttons and made them match- gas turbine to steam turbine (something GE should do.)

If you have VAR or PF selected, the control system will try to maintain VAR or PF set point by varying the voltage of the generator (as compared to the bus or grid) so the generator is not holding the generator voltage steady which is what AVR is meant to do, and typically what the grid operator (or grid controller) wants. The grid operator also needs to have generators raise or lower their respective voltage (or adjust their VAR setting) on demand to deal with localized high or low voltage situations on the grid but wants our generators in AVR following any requested VAR adjustment.

At our plant, a single generator could not select 2 modes as per your question. As a matter of fact, the control modes were labelled on the steam turbine Off/VAR/ or PF. So the Automatic Voltage Regulator (AVR) is truly only active when Excitation System is in Automatic and the control mode is Off (not in VAR or PF control.) This did seem counter-intuitive to have the control mode Off in order to be in AVR, furthermore, our gas turbines had the same (Off) button labelled AUTO. That added to our confusion until we dug into the logic and discovered they were the same point. We Re-labelled this Off button (control mode on the steam turbine) and the Auto button (control mode of the gas turbine) to "AVR On", because that is truly when the AVR is active. We also relocated the Raise and Lower Voltage buttons close to these control mode buttons, as well as the Bus and Generator Voltage and MVAR reading to allow for easier setting.

We now ensure that AVR On is always selected and VAR adjustments can be made by simply selecting Raise or Lower on the voltage until the desired VAR setting is reached or the operator can select VAR mode and enter a set point but must return the control mode to AVR On as soon as the unit reaches the desired VAR setting.

As to why your units operate at 80MVARS and -10MVARS; When multiple generators are connected on a bus with multiple loads, the grid operator may have different voltage (VAR setting) requirements for each generator depending on the types of load (more capacitive or more inductive) and depending on the voltage at other points of the grid, and also depending what other generation is available. There are lots of variables and depending on what country you live in and where exactly you live in that country (or where the generator is located on the grid in the country.)

Sometimes it is just the grid operator does not know that one generator is set at a higher voltage than the other and all they see is the net voltage or VAR of -70MVAR. It is worth investigating further because you are expending more energy in unbalanced/wasted heating of the generator that you do not necessarily get compensated for in MWhr payment. If that is the case, you may be able to adjust the voltage or VARS to level the system out, but that needs to be coordinated with the grid operator.

For clarification (at our plant): -VAR means the Generator Voltage is less than the Bus Voltage, +VAR means Generator Voltage is Higher than Bus Voltage. We set AVR on and we are at about 18KV. The MVAR will float +/- 10MVAR throughout the week as bus voltage goes above and below 18KV. Occasionally the grid operator will have us adjust MVAR on a unit...Good Luck

 

Ayaan Alam,

Most excitation systems ("AVRs"--Automatic Voltage Regulators) are pretty similar in how they operate. An excitation system provides (ultimately) DC current/voltage to the rotating generator field windings. There are "brushed" and "brushless" types of excitation systems. Brushed systems use carbon brushes riding on rings called "slip rings" to apply DC from some source (these days usually the output of a rectifier bridge that converts AC to DC at a controlled and variable rate). A brushless system has no slip rings or brushes, but rather applies a small DC variable voltage/current to a stationary field that produces AC on a rotating armature attached to the generator rotor (inside the stationary field) and that AC is then rectified using diodes on the generator rotor to convert the AC to DC.

Most larger generators (above 50-80 MW or so) use brushed exciters with brushes and slip rings, and not knowing anything about Unitrol AVRs I will describe this type (brushed).

The exciter regulator--the part of the exciter that controls the amount of DC applied to the brushes/slip rings has two basic elements, or two basic control loops. One controls the amount of DC current/voltage applied to the generator field and doesn't care about the generator terminal voltage. This is called the DC regulator, or Manual Regulator, or Manual mode. In this type of operation, the operator has to monitor the generator terminal voltage and adjust the regulator manually to achieve the desired voltage. (There is usually a DC current/voltage setpoint that is adjusted by the operator when in Manual mode, but, the setpoint value usually isn't visible to the operator. But, when the operator is clicking on RAISE or LOWER the DC voltage/current setpoint is changing, and the actual DC voltage/current changes to match the setpoint. This causes the generator terminal voltage to change, and that's what the operator usually sees.)

The second part of the exciter regulator looks at generator terminal voltage and automatically adjusts the DC current/voltage as necessary to keep the generator terminal voltage equal to a generator terminal voltage setpoint which the operator adjusts, and which may or may not be visible to the operator. Usually the generator terminal voltage setpoint isn't visible to the operator; the operator just looks at the generator terminal voltmeter/display value and adjusts the setpoint until the voltage is what is required. This is called the AC regulator, or the Automatic regulator.

Most excitation systems these days start and operate in Automatic mode. If there is a failure of the generator terminal voltage feedback or some other problem with the voltage monitoring function, the excitation system will change (automatically) to Manual mode. And, an alarm will alert the operator to the change. Some corrective action must be taken, and usually, the operator must re-select Automatic mode to return to Automatic mode of operation.

VAr mode is another control loop that tells the excitation system to adjust the DC current/voltage to whatever level is required to maintain a VAr setpoint. The VAr setpoint is usually visible to the operator, or can be manually set using a keypad or a mouse.

PF or Power Factor control is yet another control loop that tells the excitation system to adjust the DC voltage/current to whatever level is required to maintain a Power Factor setpoint.

It is NOT possible to operate a single generator in more than one mode--it can only be in either Manual, Automatic, VAr or PF (Power Factor) control. If you have more than one generator on the site, each generator's excitation system can usually be operated in any one of the four modes, also. Each generator can even be operated in two different modes, though that can lead to problems.

As for why one of the generators has a positive VAr flow and the other a negative VAr flow, well, that's how the two are being operated. Again, the generator outputs are independent of each other (for the most part; if the two generators share a single transformer connection to the grid (unless the transformer has two independent sets of windings) there can be some ugly interaction between the two generators if not properly tuned/adjusted). It may also be a requirement of the grid operators/regulators to have the two operate at separate VAr levels, though that would seem unusual unless there are other factors we can't know about (such as one generator feeds one transmission line, and the other a separate, different transmission line, for example).

I will offer that many grid regulators are now requiring power plants to ONLY use Automatic or Manual mode of operation, and NOT to use VAr or Power Factor control. This is because of the way the modes respond to grid disturbances; Automatic mode is more "predictable" than VAr or PF control according to most grid operators/regulators. But, it varies around the world, though, most regions seem to be adopting this requirement. VAr and PF control were meant to be convenient means for operators to set and maintain VAr or PF setpoints automatically (meaning the operators didn't have to do anything....). But, in today's electricity generating environment that's changing (much to the displeasure of operators).

Hope this helps!

 

Michael Borrelli,

(Many of the better gas turbine controls people are named Michael! But not me....)

Excellent description of the various modes, and congratulations for fixing GE's convoluted buttons and descriptions. Isn't it amazing that two different groups in GE (the steam turbine group and the gas turbine group) using the same excitation system can be so different in their nomenclature--and be so for so many years???

Anyway, thanks for the information and description--and interesting to hear about the grid regulator involvement. Personally, I don't believe it will be long before grid regulators have control of generator terminal voltage (VArs; PF) in the same way many have control of load (using something called AGC--Automatic Generation Control).

 

Thanks for your reply.

Also we have a diesel generator of 3.2 MW having AVR operating in PF control but facing some issues.

1. The pf set point is o.9 but it fluctuates with time and not automatically controlled, every time operator controlled the pf.

2. In power factor mode, the active and reactive power both will vary to keep the pf constant. My question is AVR only control the reactive power so by changing only the reactive power how it will keep the power factor constant.

waiting for your reply

 

Hi There.

I'm also a bit confuse on the difference of cosphi reg and Var reg.
As far as I know, cosphi/power factor will be change if var was changed, so why cosphi mode is needed as a different option?

Additionally, we have situation here at site where we used OLTC for 13.8/230kV Xformer. when we set up the OLTC to auto mode, it automatically changes to 8L tap, we cannot find out the reason why.