I am a non-electrical engineer and have hard time understanding some electrical aspects of the generators.

Background: Steam driven turbine, AC, three phase generator.

Excitation current determines the strength of the magnetic field of the coil (along with speed of the rotor). Got it. This magnetic field then induces voltage at the generator terminals. If the generator is not connected to the load (or anything), there is no power flow, but voltage is established with this magnetic field. The voltage will be strictly a function of rotor speed and excitation current (and voltage). Got it.

Connected to the load: When connected to the load, the generator voltage drops with the increasing load. Voltage regulator however will increase (change) excitation current/voltage to maintain constant generator voltage.

Connected to the grid: When connected to the grid, the voltage is set by the system (grid), but the excitation current determines VARS coming in/out of the generator. (I am getting little lost here, but so far so good).

My information from operators is as follows and it may be wrong:

Automatic Voltage Regulator (AVR) controls the generator voltage. Grid transformer lowers the voltage from the grid. Grid transformer has a low-tab changer (terminology?) that can adjust the stepped-down voltage from the grid. BUS is being fed from grid and generator. BUS feeds users. When low-tab-changer is decreased, and thus stepped-down grid voltage decreases, the AVR of the generator will adjust the excitation to 'make-up' for the decreased voltage. The result will be that the BUS will end up at the same voltage as before, but now the generator is 'pushing' more to maintain this voltage. This all happens while the power (MW) is the same coming from the grid and from the generator to the users.

My question:
My basic electrical understanding is that power flow can happen only across a voltage drop. (From higher voltage to lower voltage). So, how is it possible that the generator increases its voltage (to maintain BUS voltage) while the power going out of the generator to the system remains the same? At the end, the BUS voltage is the same, the generator voltage has to be higher, thus the delta Voltage is different, yet the power flow is the same. What am I missing, assuming wrong, or where is the conflicting information I have?

Thanks.

 

There is a phenomenon in electrical systems called reactance. This applies to AC-only systems, and involves elements that store energy during one half of the AC cycle, then release it again during the other half. In elements containing reactance, the current is not directly related to the applied voltage as it is in resistance, but is 90 degrees out of phase with the voltage. In particular, generators are dominated by inductive reactance which involves energy stored in a magnetic field proportional to the square of the current. VARs effectively represent the energy needed to build up and discharge the energy stored in the magnetic field.

The voltage across an inductive reactance is proportional to the rate of change of current, and current lags voltage by 90 degrees. The winding of an alternator is primarily inductive, with a very small resistive volt drop. The power transmitted through an inductive element is proportional to the product of the magnitudes of the voltages at either end and the sine of the phase angle between them, and inversely proportional to the reactance of the winding.

So we can get more power out of an alternator by keeping the terminal or bus voltage constant, and the internal or excitation voltage constant, if the angle between the two voltages increases. In practice, if you were somehow able to watch the relative position of the rotor using a stroboscope synchronised to the AC bus voltage peak, you would see the rotor position move from in phase with the peak at no load to about 60 degrees behind it at rated power - all without any change in excitation.

There is no convenient mechanical analogue I can propose that might help you see what is happening - the nearest I can suggest is a reciprocating piston drive with springs instead of a solid connecting rod.

 

Rudy,

Good questions, and some good statements.

Where to start...? I want to preface the below by stating most of it is theoretical and doesn't take into account a lot of real-world factors, but that doesn't change the facts or what happens on AC power systems ("cause and effect").

I would like to correct one term: Transformers can have tap changers, which allow the transformer ratio to be varied by changing "taps" to short or un-short transformer windings. The taps are, in essence, just large contacts. If the taps can be changed while current is flowing through the transformer, they are "special" taps and have a special sequence and are called "load tap changers"--meaning the taps can be changed under load (again, when current is flowing through the transformers between the generator and the load(s)).

Tap changers are typically used when large voltage swings are anticipated on transmission/distribution lines or when large reactive current changes/flows are anticipated on transmission/distribution lines. Most excitation systems are limited to +/- 5% of rated generator terminal voltage, and if the expected grid voltage excursions and/or reactive current flows are larger than this range then many times tap changers are employed to assist with maintaining grid voltage with a limited range of excitation change.

I have even been to sites that operate the excitation system at a constant generator field current/voltage and use the load tap changers to do virtually all voltage and reactive current changes. It seemed odd at the time, but it suited their purposes. They made tap changer adjustments during synchronization, and only made small incremental excitation changes during synchronization to "match" generator terminal voltage to the voltage on the low side of the step-up transformer. (This is always done ("voltage matching") to prevent excessive VArs when the generator breaker is closed during synchronization.)

You also made a distinction that I found a little unusual. You talked about the generator being connected to a load, and then about the generator being connected to a grid. The grid can be considered the load from a generator's perspective. I don't see the need for a distinction. A grid is the network of generating stations (generators and their prime mover) and the load (homes, offices, factories, water treatment plants, water pumping plants, etc.) and the wires that connect the generators and their prime movers to the loads, or "the" load. All of the motors and lights and computers and computer monitors connected to a grid can be considered as one large load. And, all the generators (and the prime movers driving them) can be considered as one generator.

I also sense some confusion about the function of the transformer on the generator output. It's purpose is to step up the generator voltage to a higher voltage which thereby reduces the current flowing in the high voltage lines <b>for the same amount of real power</b> (watts) which is one of the main reasons AC power systems are used: to minimize I^2*R heating losses in long transmission lines by reducing the current flow by increasing the voltage. The voltage is then reduced at substations and local transformers to required levels for consumers, which increases the current, but over much shorter distances--all of which increases the overall efficiency of the electric generation, transmission and distribution system.

Changing transformer taps (in effect changing the transformer ratio) is sort of a "coarse" excitation change, in relative terms (see below). At the same time the high-side voltage is changing, the low-side voltage changes. We're not usually talking about large voltage changes, I think on the order of less than 0.5 kV per tap (on the low side) in most cases, but I haven't seen a lot of transformers with tap changers.

Now, I have jumped a little ahead of my explanations by associating reactive current and excitation--but that's how one should relate the two in very simple terms. Let's ignore, for the moment, the variable step-up transformer and presume the transformer taps are in the right position to allow optimal operation of the excitation system (meaning the excitation system can be increased or decreased by almost +/-5% without reaching upper or lower limits).

On to the heart of the story. Let's say the generator breaker connecting the generator to the step-up transformer is closed. Further, let's say that the amount of excitation is exactly equal to the amount required to make the generator terminal voltage equal to the voltage on the low side of the step-up transformer. Finally, let's say the steam flow-rate is exactly equal to the amount required to keep the turbine-generator rotor spinning at synchronous speed (that speed which is proportional to grid frequency).

In this case the power meter will read zero watts (or KW or MW), and the VAr meter will read zero VArs (of KVArs or MVArs) . There is NO real- or reactive power flowing in the generator windings.

Now, lets' say the excitation remains equal to the amount required to keep the generator terminal voltage equal to the voltage on the low side of the step-up transformer (what the generator "sees" as "grid" voltage) but the steam flow-rate is increased. The tendency of the increased steam flow-rate is to try to increase the speed of rotation of the turbine-generator rotor but the speed is, for all intents and purposes, being held constant by the AC grid (we're talking about a well-regulated grid of some size with many other turbines and generators all synchronized together). The increase in torque caused by the increase in steam flow-rate can't increase the turbine-generator rotor speed but the generator converts that "excess" torque into amperes (watts, KW or MW) flowing in the generator stator windings and "out" through the step-up transformer. (And that torque, converted to amperes, flows through motors at, sometimes, long distances away from the generator, and is converted back to torque in the motor(s) and lights and computers and computer monitors. In reality, the steam turbine is doing the work of many motors and lights and computer and computer monitors from a remote location (where the boiler and the source of fuel for the boiler is located) connected to the loads by wires and using amperes as the torque transmission medium.)

Conversely, if the steam flow-rate were decreased from the exact amount required just to keep the turbine-generator rotor spinning at synchronous speed then amperes would flow "into" the generator stator ("motorizing" the generator as it's called; or, reverse power) in order to keep the turbine-generator rotor spinning at synchronous speed. (This is NOT good for steam turbines, by the way, to be spun by the generator acting as a motor.)

So the real power output (watts, or KW or MW) of an AC generator is a function of the steam flow-rate into the steam turbine (in your case) driving the generator. An increase in steam flow-rate would try to increase the speed of the turbine-generator rotor, but the speed is fixed by the frequency of the grid to which it is connected, so the generator converts the extra torque into amperes. (And motors convert the amperes back into torque to produce work at the other end of the wires. Electricity is just a way of transmitting torque--just like a hydraulic system uses a pump to create pressure and flow and a cylinder or ram or hydraulic motor at the other end of the pipe or hose converts that pressure and flow back into torque and useful work.)

Now, let's talk about excitation and its effect on generator output. Starting from the same conditions as above (zero watts and zero vars with the generator breaker closed--meaning the steam flow-rate is exactly what's required to keep the turbine-generator rotor spinning at synchronous speed, and the excitation is exactly what's required to keep the generator terminal voltage equal to the voltage on the low side of the step-up side of the transformer), let's say the excitation was increased. The tendency would be for the generator terminal voltage to increase, but since the generator is connected to a grid with many other generators the generator terminal voltage can't increase by very much (in some places on a grid and under some conditions it might increase more than others, and it might not seem to increase at all in other places on the same grid; the amount of the increase is a function of many factors). Since the generator terminal voltage can't change by very much (in most cases!) the generator converts the "excess" excitation into reactive current "flowing" in the generator stator--Lagging reactive current in this case.

So, just as when the steam flow-rate increased which would have tended to increase the turbine-generator speed above synchronous speed and the generator converted the "excess" torque to amperes (watts, or KW or MW), the "excess" excitation which would tend to increase the generator terminal voltage above "grid" voltage (the voltage it is connected to on the low side of the step-up transformer) is converted by the generator into reactive current flowing in the generator stator windings. The VAr meter would increase in the Lagging direction (and the Power Factor meter would decrease from 1.0, in the Lagging direction.)

If the excitation were reduced from that required to keep the generator terminal voltage exactly equal to "grid" voltage, then leading reactive current would start to flow in the generator stator windings. The VAr meter would increase in the Leading direction (and the Power Factor meter would decrease from 1.0, in the Leading direction).

So, the generator converts torque to amperes (real power) flowing in the generator stator because the speed can't change appreciably. And when excitation is increased or decreased the reactive power flowing in the generator stator varies because the generator terminal voltage can't change by very much.

Now, let's say the steam flow-rate is stable and the machine is producing 100 MW. And the excitation is exactly equal to the amount required to make the generator terminal voltage equal to the voltage on the low side of the step-up transformer, so the reactive power in the generator stator windings is zero. BUT, now the step-up transformer taps are changed such that the voltage on the low side of the step-up transformer decreases. This is, effectively, the same as increasing the excitation because now the generator terminal voltage is suddenly "higher" than the voltage on the low side of the step-up transformer. Lagging reactive current will begin to "flow" in the generator stator windings when this happens if the excitation remains unchanged.

Torque equals real power; excitation equals reactive power.

Now, let's do some of the real-world stuff. When the steam flow-rate of the steam turbine increases (or decreases) the turbine-generator rotor does actually change speed, if only for a split second (acceleration, or de-acceleration, as the case may be). The magnitude of the steady-state real-world change is directly proportional to the magnitude of the change with respect to all of the other machines providing torque to their generators. If there are a LOT of other machines on the grid, some very large, and the amount of the change on your machine is very, very small in relation then the steayd-state speed change will be imperceptible. If the grid is small and/or the steam flow-rate (torque) change is large then the speed change may be more noticeable.

On a well-regulated grid of just about any size the speed change (frequency change) caused by one machine increasing or decreasing the amount of torque being applied to the generator rotor will be virtually imperceptible because the grid operators will be monitoring grid frequency and responding accordingly. The increased torque is converted to watts, or KW, or MW, which is real power since the turbine-generator speed can't be appreciably changed.

The same is basically true of generator terminal voltage versus grid voltage and reactive power flow. Bucking (trying to reduce grid voltage by reducing generator terminal voltage) and boosting (trying to increase grid voltage by increasing generator terminal voltage--by varying excitation--causes changes in reactive power flow.

And whether generator excitation is changed with respect to the voltage on the low side of the step-up transformer, or the voltage on the low side of the step-up transformer is changed with respect to the generator terminal voltage (assuming generator excitation is constant), reactive power flow will change.

If the excitation were held constant as the steam flow-rate was increased then, yes, the generator terminal voltage would decrease. This occurs for several reasons, one of which is that exciter regulators generally have a "droop" component (PLEASE--let's NOT go there!), and there are some internal reactions inside the generator (emf's and counter-emf's and back-emf's and all of that stuff) that also cause a decrease in the generator terminal voltage with respect to the voltage on the low side of the step-up transformer. So, operators generally need to increase excitation as real power is increased to keep the VAr flow constant and to keep the power factor meter close to unity or as desired.

When talking to the operators, who have a real hard time explaining what they do, other than to say, "That's the way we've always done it!", or, "It's never done that before!" and even, as in your case, have a difficult time communicating the names of equipment they are entrusted with operating, try to temper everything they say. And, never, ever ask too many questions; that's a sure way to get their ire up higher than it usually is when someone is watching them. It is a true testament to engineers and the early designers of electrical systems that they operate without too much human intervention and that today's control systems don't require a high level of knowledge or expertise. Most operators can only do what they've been told by others, or do what they've done in response to situations they have encountered (usually several times). They are not required by their supervisors to really understand how the equipment they operate works--they just have to operate it, and, again it's a credit to the early designers and control system engineers that it works as well as it does.

So, if I read what you wrote correctly, there was some overlapping of concepts that aren't necessarily related, and some incorrect terms that might have also lead to some confusion. My explanation comes from my university training (decades ago), and from operational experience on small power islands (ships, specifically) up to and including large, infinite grids. I have very little experience with poorly-regulated grids and with uncontrollable grid frequency excursions, and transient conditions--which by their nature don't occur very often but seem to get held up as typical of all grids by some contributors to similar threads.

Yes, there are all manner of physical and electrical principles and vectors and maths involved in AC electrical power generation, most of which aren't important to operators and to stable grid operation. I choose to concentrate on "cause and effect" to describe synchronous generator operation, instead of emf-this and emf-that. My explanation above is pretty simple and theoretical, while still trying to deal with some real-world circumstances and avoiding all of the maths and vectors and emf's (which do exist, but which just serve to confuse the general explanation which I believed you were asking for).

Remember the basic electrical power formula: P = V * I. Remember also that the generator voltage is usually pretty stable and constant, and most excitation systems can't change the terminal voltage by more or less than approximately 5% of rated (which is about 700 volts for a 13.8 KV generator). So, changing voltage has little to do with changing the power of a generator; it's MUCH more about the current--which is directly proportional to the torque being applied to the generator rotor by the prime mover (steam turbine in your case).

Reactive power, which only occurs in AC electrical systems, is a function of the generator terminal voltage which is a function of the excitation (which controls the field strength). Remember also that the speed of the generator is basically fixed (doesn't change by very much, so as you correctly noted since the speed is constant the way to change voltage is to change excitation. And that affects the reactive current flowing in the stator windings.

Many people will try to talk about load angle, which is a reflection of the amount of "twist" applied to the generator rotor by the prime mover. More torque equals more twist, and more twist also means more amperes. It's just another way of describing how torque is related to amperes, or is converted to amperes. There are multiple ways to describe what happens; trying to keep them all straight can be a problem!

Hope this helps! (Sorry for the length; but I was trying to address all of the issues in your original post.) I'm sure others will also be able to clarify some of the concepts if you have more questions.

Good on you, though, for making the effort to understand! And for trying to keep the operators' explanations in perspective.

 

Bruce has a great answer, but try and start with this concept:

Apparent power (we'll label it S) has two components, an active part measured in watts, often abbreviate as P, and a reactive (some people call it 'imaginary', but that kind of clouds the picture) measured in Volt Amp Reactance, or VARs, often abbreviated as Q.

So, S=P+jQ, where j=sqrt(-1)... don't worry about j right now, it comes into play because the voltage and current are sine waves.

When you increase voltage, you are increasing Q, not P... that is why you don't see a change is watts flowing out of the system. If you were measuring S, you would see an increase (S is measured in Volt Amps).

If that makes sense, the next step is to dive into Bruce's comments if you're looking for a more detailed understanding.

cheers,
nic

 

I can feel the heat of the flames already!

> It is a true testament to engineers and the early designers of electrical systems that they operate
> without too much human intervention and that today's control systems don't require a high level of knowledge or expertise.

Control systems (of any vintage) DO require a higher level of knowledge and expertise. What I meant to type was:

"... and that today's control system don't require a high level of knowledge or expertise <b><i>to operate</b></i>....."

I'm not slighting power plant operators. Their management and supervision has fostered and allowed this situation to occur.

And, it's really no different than driving an automobile. Who among us really know exactly how every system on a car works? We only care that it starts reliably and gets us back and forth between our destinations. And, considering how little "driver intervention" automobiles require today, versus even fifty years ago, it's a testament to the engineers and designers that we have such reliable and long-lasting conveyances today.

The difference, for me, is that electricity is so vital to our lives today. There's so much that's done with electricity today, and more every day. Our lives depend on a reliable supply of electricity to the point of criticality these days.

 

Thank you Bruce for the answer.

I think my mental block was that I was thinking about AC current like a DC current or an energy flow in fluid networks - as you pointed out.

Now, if I was forced to make a mechanical analogue, I would say that the best comparison is with a metal bar that moves horizontally back and forth. The movement is done by an infinite force (grid). You can clip (connect) a user to the bar and transform the back-and-forth movement of the bar into whatever usable energy the user needs. We attached a load, and work is being done. The bar moves 60 times a second.

We can also attach a 'generator' by connecting to the bar a device that helps with the pushing and pulling. The amount of pushing and pulling is based on power flow (thus the prime mover).

This analogue shows that the energy flow is dependent on users and sources (generators, grid) connected to the bus.

Now, voltage and current. I think Voltage is like force. If you push on the bar (voltage), but the bar does not move (no current), then no work is being done. The force applied on the bar has amplitude and is cyclical, just like voltage. The current is really speed of the bar. Displacement per second, just like electrons per second in AMPS. If you multiple the force F by displacement per second, you get work in Watts. Just like when you multiple Current and Voltage.
The displacement is also cyclical and has amplitude.

Now, in an in deal world, you would want the force of the generator to be in the sync with the movement. In other words, you don't want to push the bar when it is already at its maximum displacement, or already going back when you should be pulling. So even if you apply force, you may not do full or any work if you do it out of synch. If you pull when you are supposed to be pushing and vice-versa, you would be doing no work. If you don't push/pull enough, the bar is going to push/pull you and you are motorized.

You gave me the above idea with the piston suggestion, and may be this is exactly what you meant.

Now, attaching a spring to the bar, I can see that you now have some residual left-over forces on the bar outside of its natural back and forth movement. Measuring these forces (and work) would give you the VARS.

I can actually imagine this and make the connection to the AC current.

Can you please comment whether this is close, or I stilled missed something. I am sure the real world has many more factors, but is the above analogue close, or is there something fundamentally wrong in comparison with the AC current generation.

 

CSA,

I can't thank you enough for the time and effort you took to address my question. I specifically like that you did not go into details that would not matter (at this point) but did not omit any that do matter. Thanks for clarifying terminology and explaining simple stuff and correcting my assumptions.

I also hope that you expected that your answer will make me ask more questions .

Please have a look at the follow up post I made to Bruce above. Let me know whether you agree with the mechanical analogue I made there.

However your detail explanation about the excitation makes me ask more questions.

The system I am involved in is 4 turbines (biggest one is 50 MW) connected to 3 BUSes with heavy industrial machines in a pulp mill. The three BUSes, each having load tap changer, is connected to the US grid. They can run connected to the grid, off the grid supplying power to the local machines with multiple generators, and off the grid with one generator (running part of the mill).

I think I understand the dynamics of the RPM of the above three scenarios. Where the grid locks the speed of all the machines to 60 Hz (through magnetic field of the AC current). When one generator is supplying power to the users, then the governor is responsible to compensate for the load swings and maintain 3600 RPM. When multiple generators (off the grid) are connected then all are spinning at the same speed, and if load swings, the RPM change for all of the generators. I think I understand this and we do not have to go here unless you feel I am missing something.

Onto questions:

> You also made a distinction that I found a little unusual. You talked about the generator being connected to a load,
> and then about the generator being connected to a grid. The grid can be considered the load from a generator's
> perspective. I don't see the need for a distinction. A grid is the network of generating stations (generators and
> their prime mover) and the load (homes, offices, factories, water treatment plants, water pumping plants, etc.) and
> the wires that connect the generators and their prime movers to the loads, or "the" load. All of the motors and lights
> and computers and computer monitors connected to a grid can be considered as one large load. And, all the generators
> (and the prime movers driving them) can be considered as one generator.

Understood, but for my understanding I needed to make the distinction to understand that the speed of the turbine is 'locked' when connected to large number of generators (grid). In order to increase speed, one would have to do so for all of the grid connected generators, which is impossible - for our purpose. Whereas if you are off the grid, you do have capability to affect speeds of the local generators (with loads connected). Also it seems that the excitation affects VARS more so in the first case and Voltage in the second - to that later.

In mechanical world, if you were to measure how much does the speed (RPM) change with constant load, by changing torque of the prime mover, I would call it elasticity of the system. If you are not connected to the grid, only supplying power to local loads, then by increasing steam flow to the turbine, the speed will change instantly. In my words I'd say the system is elastic. If more generators are connected to the bus, but not the grid, then under the same situation, increasing the steam flow will have to change the speed of all of the generators at the same time, even though the steam flow changed only for one turbine. In this case, the system would be also elastic but less so, because you have more shafts and turbines to change speed. If you are connected to infinite number of generators (grid) then increasing the steam flow will not change the speed (or may be only so little that it is not observed for practical purpose), and I'd say that this system has very little elasticity, or even no elasticity.

So, yes I agree that the three situations above are the same from the generator point of view, no distinction is required, but the 'elasticity' changes significantly with those situations and I wanted to distinguish between them.

----

I am omitting it here as I have no further question, but thanks for the real life scenarios you went over with increasing / decreasing steam and what it does to the system. Very helpful.

----

> Now, let's talk about excitation and ....

> So, just as when the steam flow-rate increased which would have tended to increase the turbine-generator speed
> above synchronous speed and the generator converted the "excess" torque to amperes (watts, or KW or MW), the
> "excess" excitation which would tend to increase the generator terminal voltage above "grid" voltage (the voltage it is
> connected to on the low side of the step-up transformer) is converted by the generator into reactive current flowing
> in the generator stator windings. The VAr meter would increase in the Lagging direction (and the Power Factor meter
> would decrease from 1.0, in the Lagging direction.)

But this makes me think that the VARS out of the generator could be controlled by the excitation. Why not then, adjust the excitation to the perfect VARS = 0 condition and run all generators at power factor of 1.0? If I can go from Leading to Lagging condition with the excitation adjustments, I have to cross 1.0 power factor, so why not stop and hold it there?

----

Let's talk little bit more about VARS. In our case, the generators are connected to the individual BUSes, which in turn are connected to individual loads. The BUSes are also connected to each other through the step-down-transformer and then (for the most of the time) connected to the grid.

The local loads are big 3 phase coil motors for the most part, so heavy inductive loads (I think). I understand that VARS are generated here and thus exist in the BUS. Reactive power, as I understand it now, is the residual current flow caused by coils as the real current is fluctuating. The coils generate magnetic field, that then has to collapse, which causes current flow back to the the system = reactive power (I am getting little lost here, but so far so good). Capacitors do the opposite, but we don't deal with this type of reactive power, so for my purpose, I don't need to know much about them, other than they cancel the inductive reactive power (leading vs. lagging).

So, the VARS have been generated and they exist on the BUS. If, e.g. other interconnected BUSes in this internal mill network have only ideal load users (power factor 1.0), but a BUS # 2 has loads with power factor 0.9, would you see the reactive power in the entire network or only in BUS #2? Would BUS #1 see reactive power(VARS, power factor 0.9)?

Second, how does the generator handle this? Are these VARS effecting the generator? If the VARS from the user are e.g. power factor 0.9 leading, would you want to run the generator at power factor 0.9 lagging to compensate?

Are VARS eventually consumed and dissipated as heat in the system or sent to the grid?

I heard of a situation where power flows one way through a transmission line between generator and the bus, but the VARS flow in the opposite direction. How is that possible?

Operator also said that the utility company penalizes us if we send VARS to them. If we however have low power factor loads in the mill (given), how do we 'consume' these VARS so we don't send them to the grid? I guess excitation current of the generator has something to this with this. I heard of adding capacitance to the system, but these are not used in our industry (to my knowledge), so let's no go that route.

I also heard that VARS and thus power factor can be 'pushed' out of the generator, but it cannot be destroyed. Or in other words, the power factor of the whole system stays the same, one can only change the power factor of a local generator. As if VARS load shifting... Now I really do not know what I am talking about, just writing remnants of what I remember.

Thanks again for the help. I read books, but they don't help. They either assume reader knowledge that one does not have, go into details that one can be easily confused by on the 2nd page, or don't cover the topics in depth needed for proper understanding. Books also don't correct misunderstandings and mental blocks.

 

Thanks nic. Combination of all the posts helps to clarify things, as everyone explains it from a different angle.

 

Now you've got me thinking ...

You're on the right track with your mental model. Unfortunately, while there are a lot of direct mechanical analogues of electrical effects, they do tend to break down if we examine them too closely. I've tried to come up with a relatively straightforward system that can be directly translated from electrical to mechanical or vice versa, but its not easy!

But if you have your your reciprocating bar as a bus, with its travel thought of as 'voltage" then we can couple a reciprocating cylinder to it through a spring. At no load, the bar and spring will be in step, but if there is some load applied to the cylinder (using it as a pump for example) the spring has to stretch a bit to pull the cylinder out, and compress a bit to push it in and there will be a phase difference between the two with the cylinder lagging. However, if the cylinder is used to add power to the system, it will compress the spring on the extension stroke and extend it on the retraction part of the cycle.

Now if we were to drive the "cylinder" end by a doohickey having adjustable stroke, we can supply a given amount of power by having a low cylinder stroke with large angular difference, or a large stroke with a smaller angular difference. By adjusting the stroke at the doohickey we can regulate the angle difference which is effectively the power factor. The controller setting the stroke of the doohickey is the equivalent of a voltage regulator.

(This is about where the analogues start to break down as the current in an electrical system is the same at both ends of the reactance, whereas our forces at the two ends of the linkage will be different).

If there is only one drive cylinder acting on the bar, you have the "islanded" case. With two or three, there is a degree of interaction between them similar to that in a small installation such as you have described - and every additional driver reduces the effects of any one.

In a lot of cases, inertia is taken as an analogue of inductance, with current equivalent to velocity and voltage to force. But this doesn't translate well to a system where voltage is held more or less constant.

I just hope this hasn't added to the confusion - I've been trying to come up with a simple explanation of power factor/reactance to mechanically-oriented guys for years. Now I usually reply that I'll explain power factor after they've explained entropy!

 

> But this makes me think that the VARS out of the generator could be controlled by the excitation. Why not then, adjust
> the excitation to the perfect VARS = 0 condition and run all generators at power factor of 1.0? If I can go from
> Leading to Lagging condition with the excitation adjustments, I have to cross 1.0 power factor, so why not stop and hold it there?
>
>----

An induction motor acts electrically as if it had two basic components. First, the magnetic field it needs is set up by current flowing through the stator coils of the motor. Because this is inductive, it depends on the applied voltage but lags the voltage by 90 degrees. The amount of current needed to set up the magnetic field is pretty well fixed and depends only on the voltage applied to the motor. The energy needed to set up the magnetic field is returned to the system when the field collapses (twice per cycle) and this current is purely reactive, lagging the voltage by 90 degrees. As long as the voltage remains fixed, the VArs drawn by the motor will be fixed.

If the motor is loaded, it has to develop some torque - this comes from the interaction between the rotor current and the magnetic field. The rotor current is converted through transformer action into an additional current in the stator coils - but this one is in phase with voltage as it represents actual power converted from electrical to mechanical, and effectively dissipated by the motor. The total current is the vector sum of the reactive current needed to set up the magnetic field and the active current converted to watts at the shaft. A typical induction motor has an inherent power factor which is usually around 0.85 at rated load, falling to 0.2 or less if unloaded.

Now, while the reactive current does not require any power to generate, it does add to the energy lost in the resistances of the system. This energy has to be generated and that costs. If the power is coming from an external supplier, they may well have additional charges based on power factor falling below a specified limit (maybe 0.85 or 0.9).

So your load will have an inbuilt VAr requirement based on the energy flowing into and out of the motor windings twice per cycle. Even though this averages out at 0 over a complete cycle, it will affect the instantaneous balance, and the moment-by-moment VAr consumption of the load must be matched by VArs supplied from sources. If you run your generators at a power factor of 1.0 (ie, 0 VAr being generated) the total VAr requirement must be imported and this will greatly reduce the power factor seen by an incoming supply.

Another secondary point is related to the internal behaviour of the alternator. While the main excitation is developed by current flowing in the main field windings on the rotor, the load current flowing in the stator windings also sets up a magnetic field. If the machine is generating at a lagging power factor, this field is out of phase or opposes the main field and the main field has to be strengthened in response. However, the result is a very "stiff" control system and a fairly large change in stator current is needed to give a change in power factor. If the power factor developed by the alternator is 1.0 or leading, the stator field tends to support the main field which is reduced in consequence - as a result, a small change in stator or load current can have a very large effect on the overall machine behaviour. In extreme cases, the voltage regulator becomes totally ineffective and field instability results.

It happens (very nicely!) that the optimum value for alternator power factor for good stability is about 0.85 - about the same as that of the typical industrial load.

A useful topic that might help with all of this relates to the capability diagram or circuit diagram of an alternator. Google "alternator capability curve" - the paper by Cummins Power seems to give a good explanation.

As well, there is another whole topic related to capacitors and capacitive reactance - a capacitor also stores energy but the current leads the voltage by 90 degrees rather than lags as in an inductor, so capacitors can also be used to adjust the VAr consumption in an installation.

Hope you don't regret asking the question!

 

CSA,

I am happy that you can feel the heat of the flames! As a curious and studious operator I thank you for the in depth explanation and easily understood concepts of basic power generation. I am looking forward to reading more posts about this subject.

Max.

 

Rudy,

You are welcome, and, yes, I anticipated there would be more questions.

I have yet to find a really good mechanical analogy to AC electrical systems. I think Bruce Durdle's reciprocating engine with springs for connecting rods is as good as I've heard. Coming from primarily a mechanical background myself, I've tried for years to come up with something but nothing really fits the bill entirely. So, I don't want to spend a lot of time on this because I think it detracts from understanding AC electrical systems by spending time trying to make analogies that, to date, no one has successfully made. And AC electrical systems have been around for a LONG time. They are what they are, and they require understanding what they are, and what they aren't, too.

When a plant like yours is separated from the utility and supplying its own electrical power, that's typically referred to as "island" operation. You are operating as your own power "island". In that case, what typically happens is one of the governors is put into what's called Isochronous Speed Control and it automatically responds to changes in frequency as motors and lights and computers and computer monitors are started and stopped by adjusting the steam flow-rate VERY quickly to keep the turbine-generator speed and frequency at rated. The other steam turbine governors should be in Droop Speed Control (and let's NOT go there in this thread, please!).

This means that all the other turbines and generators synchronized with that one machine will all be running at the same frequency (that's the definition of synchronism) and if the Isochronous governor doesn't maintain speed/frequency then all the other turbine-generators will also run at the same speed/frequency as the Isochronous machine.

I prefer to think of VArs as being produced and consumed. FROM A GENERATOR'S PERSPECTIVE, Lagging VArs "feed" a lagging VAr "load." I like to think of VArs as something that are necessary for induction motors to work--and since the majority of loads on most grids (small power islands or even large infinite grids) are induction motors the power factor of most grids is "lagging" (again from the generator's perspective). So, the "consumption" of VArs occurs in loads with inductive characteristics (induction motors being the most common).

The effect of VARs on an AC electrical system is to shift the voltage- and current sine waves out of phase with each other. (When the load is purely resistive, the voltage- and current sine waves are in phase with each other.) There are capacitive loads, too, such as flourescent lights, which serve to "counter" inductive loads; but the majority of loads are inductive in nature.

Allowed to continue "unchecked", if the voltage- and current sine waves get too far out of phase with each other then bad things start happening to the grid, beginning with brown-outs and excessive heating of conductors (wires; bus bars; etc.). So, something must "produce" VArs to counter the effect of VArs on the system, to supply the "consumers" of VArs. It doesn't have to be a one-for-one production/consumption, but if no VArs were "produced" then bad things would happen, including poor effieciency of the entire electrical system.

One way that "producing" VArs can be done is by over-exciting the field of a synchronous generator, as described in the previous response (increasing the excitation above that which is required to maintain generator terminal voltage exactly equal to grid voltage). The VAr load of any AC electrical system can't be "controlled" as much as it can be "responded" to or "dealt" with. The number of inductive loads (again, primarily induction motors) and the efficiency of those loads (the power factor, which is a measure of efficiency) dictates the "amount" of the VAr load on the system. All that can be done is to "produce" VArs to counter the "consumption" to try to keep the voltage- and current sine waves in phase with each other to keep the AC electrical system operating at optimal efficiency.

I need to say that in reality neither watts nor VArs are really produced or consumed. They just "flow", like hydraulic fluid in a hydraulic system, to a cylinder or hydraulic motor under pressure from a hydraulic pump and then back to the sump to repeat the cycle. However, watts are generally considered to be produced and consumed, and VArs can also be thought of in the same way. For me, it simplifies the explanations. (Negative VArs, from a generator's perspective, are Leading VArs. Positive VArs, from a generator's perspective, are Lagging VArs. By most conventions I've ever seen.) And, further, I need to say that my explanations are what I like to call "cause and effect" explanations, not so much technically precise explanations. I leave that to the mathematicians and textbook authors, who don't usually have a lot of real-world experience and tend to think in physical and mathematical terms instead of "when this happens, that happens." So, there might be responses which will use equations, calculus, vectors and "ivory tower" explanations--all of which may be technically precise. My observations and explanations are my own, and if they aren't technically 100% precise they are still real-world observations that many have found useful over many years.

I'm a little confused about the "step down" transformer thing. Are you saying all of the transformers connected to the generators step the voltage down to a lower voltage on the BUS side of the transformer (lower than the generator side)? Is the transformer that connects the plant to the utility also a step-down transformer (meaning that the utility side of the transformer is at a lower voltage than the plant side)? Sorry; but I'm very picky about my terms (it's the cause of more engineering problems and misunderstandings than just about anything else; two people can be (have been!) talking about the same thing using completely different terms and never even know it--the English language is such a wonderful thing, technically, anyway).

I'm out of time for tonight; I'll try to write more later after I read more of your reply and questions. I do caution you to not spend too much time trying to find a mechanical analogy to AC electrical systems. Just try to get a grasp of the concepts, and then mull them over in your mind. They will become more and more apparent and obvious the more you review them--just like many of the mechanical concepts you have learned.

I like to quote, "Learning is finding out what you already know. Doing is demonstrating that you know it. Teaching is reminding others that they know it as well as you do." (Richard Bach, 'Illusions') Remember what it's like when someone explains something to you so that you finally understand it? A lot of people say to themselves, or even out loud, "I knew that!" Because we all do know this stuff; it's just that it takes something a little different for it to "click" for everyone.

It's very enjoyable seeing people discovering what they already knew.

 

Rudy,

I'll do my little bit to try and fill in the gaps, Bruce and CSA have done a fantastic job explaining the concepts so far.

> But this makes me think that the VARS out of the generator could be controlled by the excitation. Why not then, adjust
> the excitation to the perfect VARS = 0 condition and run all generators at power factor of 1.0? If I can go from
> Leading to Lagging condition with the excitation adjustments, I have to cross 1.0 power factor, so why not stop and hold it there?

Yes, the generator can control the VArs and the excitation can be set to maintain a power factor of 1.0, or 0kVAr. Lets say there is a large amount of inductive load on you site (many induction motors). If the load and generator is connected to the utility supply, the excitation can be increased so the generator starts producing more kVAr, making the power factor more lagging on the generator and the magnetic coupling between the rotor and the stator is strengthened. The effect of this on the mains supply is that it now starts supplying less kVAr which is moving the power factor closer to 1. The limitation here is the heating effect of the excitation current. Typically a generator rating states the kVA and the power factor. The maximum kW output of the generator is kVA * power factor. If the power factor becomes more lagging than the maximum power factor for inductive load, then the excitation windings will overheat and get damaged.

What needs to be considered are the penalties that utilities would charge when the kVAr demand reaches a certain level from the site load. This can be mitigated by using one of the following:

1. In the above scenario, if there is enough capacity in the generator, over excite the generator so that it now supplies the kVAr demand on the site.

2. Power factor correction loads such as a bank of capacitors can be used.

3. Change the induction motors to synchronous motors where the power factor can be controlled in a similar manner as described above, however I imagine that this would be an expensive option.

The risk of using power factor correction equipment in point 2 is that to ensure the site load power factor does not become too leading for the generator when there is a load shift (large banks of motors switching off is one example). When the power factor becomes leading, the generator absorbs kVArs and the magnetic coupling between the rotor and the stator is weakened. While generators are designed to absorb a small amount, if it starts absorbing too much kVArs, the risk here is that and could get weakened to a point the magnetic coupling is broken briefly which causes a very dangerous phenomenon called pole slipping (can cause serious damage to the drive train such as the coupling and/or the prime mover).

I guess it is important to know what the limitations of the generator package are to ensure the correct controls measures (excitation control or other) are used to minimize the cost of the site load.

> Let's talk little bit more about VARS. In our case, the generators are connected to the individual BUSes, which
> in turn are connected to individual loads. The BUSes are also connected to each other through the
> step-down-transformer and then (for the most of the time) connected to the grid.
>
> The local loads are big 3 phase coil motors for the most part, so heavy inductive loads (I think). I understand
> that VARS are generated here and thus exist in the BUS. Reactive power, as I understand it now, is the residual
> current flow caused by coils as the real current is fluctuating. The coils generate magnetic field, that then has
> to collapse, which causes current flow back to the the system = reactive power (I am getting little lost here, but so
> far so good). Capacitors do the opposite, but we don't deal with this type of reactive power, so for my
> purpose, I don't need to know much about them, other than they cancel the inductive reactive power (leading vs. lagging).
>
> So, the VARS have been generated and they exist on the BUS. If, e.g. other interconnected BUSes in this internal
> mill network have only ideal load users (power factor 1.0), but a BUS # 2 has loads with power factor 0.9, would you
> see the reactive power in the entire network or only in BUS #2? Would BUS #1 see reactive power(VARS, power factor 0.9)?
>
> Second, how does the generator handle this? Are these VARS effecting the generator? If the VARS from the user
> are e.g. power factor 0.9 leading, would you want to run the generator at power factor 0.9 lagging to compensate?

Imagine that you have monitoring equipment on four different points, the first being the local generator source, the second and third being bus#1 and bus#2, lastly would be the utility source. When the generator is operating connected to the grid, if it is set to "base load" then you will set what the kW and kVAr outputs on the generator be setting the fuel and excitation settings accordingly, lets assume in this case you set the kVAr output for the generator to 1 (or supplying 0 kVAr). Any shift in the load will be taken up by the mains supply, however note that there is always a small delay in the controller for each generator and as a load changes, ther will be a slight "bump" in the kW and kVAr outputs on the generator itself until it reaches its steady state.

I assume that bus#1 and bus#2 only have the respective loads connected. Then the only monitors that will see the steady state reactive load would be the one on bus#2 and the one on the mains supply. Using the description above on controlling the kVAr output from a generator by biasing the excitation, the generator can take on the reactive load from bus#2 causing a reading of 0.9pf.

If you take the next step, there are other methods of control where the load is monitored at the utility side and the generator fuel and excitation is controlled to ensure the utility kW and kVAr levels are maintained and the generator takes the swing.

> Are VARS eventually consumed and dissipated as heat in the system or sent to the grid?

The VArs are a function of the load itself and will always be dissipated as heat by the load, the *source* for the VArs depends on the scenario described above.

> I heard of a situation where power flows one way through a transmission line between generator and the bus, but
> the VARS flow in the opposite direction. How is that possible?

This is a good question, the best way to think of this is that for an inductive load, the generator needs to supply a magnetising current as well as the load required for the work. As the AC voltage is cycling, the current is also doing the same. The problem here is that the magnetic load presents some inertia and there is a slight delay (lag)in the current compared to the voltage. The effect this has on the generator is that it needs to work harder to maintain the voltage so it requires more excitation.

For a capacitive load, wikipedia (http://en.wikipedia.org/wiki/Capacitor) provides an excellent example in the Hydraulic analogy section as follows:

______________________________________

" In the hydraulic analogy, charge carriers flowing through a wire are analogous to water flowing through a pipe. A capacitor is like a rubber membrane sealed inside a pipe. Water molecules cannot pass through the membrane, but some water can move by stretching the membrane. The analogy clarifies a few aspects of capacitors:

- The current alters the charge on a capacitor, just as the flow of water changes the position of the membrane. More specifically, the effect of an electric current is to increase the charge of one plate of the capacitor, and decrease the charge of the other plate by an equal amount. This is just like how, when water flow moves the rubber membrane, it increases the amount of water on one side of the membrane, and decreases the amount of water on the other side.

- The more a capacitor is charged, the larger its voltage drop; i.e., the more it "pushes back" against the charging current. This is analogous to the fact that the more a membrane is stretched, the more it pushes back on the water.

- Charge can flow "through" a capacitor even though no individual electron can get from one side to the other. This is analogous to the fact that water can flow through the pipe even though no water molecule can pass through the rubber membrane. Of course, the flow cannot continue the same direction forever; the capacitor will experience dielectric breakdown, and analogously the membrane will eventually break.

- The capacitance describes how much charge can be stored on one plate of a capacitor for a given "push" (voltage drop). A very stretchy, flexible membrane corresponds to a higher capacitance than a stiff membrane.

- A charged-up capacitor is storing potential energy, analogously to a stretched membrane."

______________________________________

Now the "push" described in the above analogy is the absorption of the kVArs by the generator.

Lastly, the effects caused by reactive load is independent to the real load required for actual work.

One example is an induction motor driving a pump, a magnetisation current is required for the magnetic coupling (kVAr) and the pump load on the motor (kW) draws current from the source to ensure the required speed is maintained.

The effect of a capacitive load bank is better described in the above hydraulic analogy.

> Operator also said that the utility company penalizes us if we send VARS to them. If we however have low power
> factor loads in the mill (given), how do we 'consume' these VARS so we don't send them to the grid? I guess excitation
> current of the generator has something to this with this. I heard of adding capacitance to the system, but these are
> not used in our industry (to my knowledge), so let's no go that route.
>
> I also heard that VARS and thus power factor can be 'pushed' out of the generator, but it cannot be destroyed.
> Or in other words, the power factor of the whole system stays the same, one can only change the power factor of a local
> generator. As if VARS load shifting... Now I really do not know what I am talking about, just writing remnants of what I remember.

It is easier to understand if you think in terms of kW and kVAr. Power factor is calculated out using the following formula:

Power Factor = kW / kVA

Using the original formula above:

(kVA)^2 = (kW)^2 + (kVAr)^2

If you are given a combination of 2 of the above figures you can determine the remainder.

Just remember, the load requires a certain amount of kW and kVAr and whatever one source doesn't supply, the other will.

Sorry if I have just added to the confusion, but I hope this explanation helps you somewhat...

 

I did electrical engineering in high school a while ago and for some reason i cannot remember, How does one calculate the relations between input voltage, turns ratio and speed in either induction or brush motors? This would help me out a lot with the project i am running..

 

Electrical engineering was not available as a course in high school when I attended decades ago.

But, the comment attracted my attention to a amazing thread that will assist me in taking complex topics and presenting these topics in a less than complex way.

Thanks!

 

i'm working on a project and am looking to find out what method for power generation will give me the best outcome, and what would be the best way to control and manipulate the power created so I can use it as I need it.

 

Hi, Semos,

Congratulations on your project! (Sounds like a course assignment designed to make one investigate and think about electrical power generation....)

Electrical power generation is all about producing torque in one location, converting that torque to amperes (at some voltage), transmitting those amperes using wires to another, or many, remote locations, and then converting those amperes back into torque to do work at the remote locations. This "work" includes lights and computers and computer monitors, as well as pumps (water pumping is the biggest user of electrical power in the world--all kinds of water: fresh water, sewage, etc.).

So, you need a way to produce torque--called a prime mover. That may be a reciprocating engine using gasoline or diesel fuel or natural gas, or a steam turbine, or a combustion turbine, or a hydro turbine, or a wind turbine.

The prime mover is coupled to a synchronous generator (usually, at least for AC (Alternating Current) power--which is the most common type of power available in the world) and supplies the torque that the generator converts into amperes.

Wires and circuit breakers (of various types) and transformers are used to transmit and distribute the amperes produced the generator to various locations. Devices at the locations then convert the amperes back into torque (primarily using electric motors--which are generators "in reverse" in that they convert amperes back into torque) to drive pumps and fans, as well as producing light and "virtual torque" (computers and computer monitors, etc.).

So, when designing a power generation system one needs to know what sources are available to produce torque: hydro power (water), steam (boilers), combustion turbines (liquid or gas fuels), wind, etc. And, then one needs to find the proper-sized prime mover and generator (called a generator set) and the appropriate devices (circuit breakers and transformers) to connect it to "loads (electric motors, lights, computers and computer monitors) using wires and switches and such.

Your question is very broad, and as such, it's very difficult to provide much more of an answer. I suggest you use your preferred Internet search engine to look up "electrical power systems" or "AC power systems" (using the double quotes as shown) for more information about various systems and how they work.

The "method" of power generation depends on what's available where you are going to produce the electrical power. You can't build a hydro turbine-generator in the desert, and you wouldn't build a combustion turbine in a rain forest--unless you had a pipeline to supply fuel (liquid or gaseous) to the turbine. And you wouldn't build a wind turbine-generator inside a building.

The nature and amount of your "loads" (motors, lights, computers and computer monitors) will also play a large factor in the type and size of power generation method. You may be synchronizing to (connecting to) other generators--or you may not. All of this affects the choices one makes.

There are also DC (Direct Current) power systems, but they are less common and used for special applications. And you didn't specify, so I presumed AC.

Hope this helps!

 

CSA,

"The increase in torque caused by the increase in steam flow-rate can't increase the turbine-generator rotor speed but the generator converts that "excess" torque into amperes (watts, KW or MW) flowing in the generator stator windings and "out" through the step-up transformer."

Why can't the flow rate increase the turbine-generator rotor speed? I understand that it must be at grid frequency but, if the flow rate increase speed must be more than 3000 rpm. what does stop the rotor at 3000rpm?

it really confuses me. Hope my question is clear enough.

 

serpenson,

The formula which describes the frequency produced by a synchronous generator when it is NOT synchronized to a grid is:<pre>F = (P * N) / 120

where
F = Frequency (in Hertz)

P = Number of poles of synchronous generator rotor (always an even number)

N = Speed of rotation of generator rotor (in RPM)</pre>
When synchronized with other generators and their prime movers the formula that describes the speed of the generator rotor is:<pre>
N = (F * 120) / P
where N = Speed of rotation of generator rotor (in RPM)
F = Frequency (in Hertz)
P = Number of poles of synchronous generator rotor (always an even number)</pre>

When synchronized to a grid with other synchronous generators and their prime movers there are two (2) magnetic fields which interact with each other. One is that of the rotating synchronous generator rotor created by DC that is applied to the rotor windings (this is typically called the generator "field"). This is a very strong magnetic field with North and South poles.

The other magnetic field is created by the flow of AC (alternating current) in the synchronous generator stator, or armature, windings. Because of the alternating nature of the current, the magnetic field created in the stator windings appears to rotate around the stator. This synchronous generator stator magnetic field also has North and South poles--which attract--and are attracted to--the South and North poles, respectively, of the synchronous generator rotor.

Now, we've all handled (played with) two magnets. We know that the North pole of one magnet is very attracted to--and attracts--the South pole of the other magnet, pulling them together with magnetic forces relative to the strength of the magnets, and requiring some physical force to pull them apart. We also know that when we try to force the two South (or North) poles of the magnets together they DO NOT want to be anywhere near each other. It's said that opposite poles attract each other, and similar poles repel each other.

Inside a synchronous generator the strength of magnetic fields is a function of the amp-turn relationship of the currents flowing in the winding conductors (the "turns"). The number of turns (conductors) in the windings don't change (they're fixed by the design and construction of the generator and rotor) but as the amperes flowing in each set of windings--the generator rotor and the generator stator--the strengths of the two magnetic fields change. And they are VERY strong; <b>VERY</b> strong.

So, what happens is that when AC (alternating current) is flowing in the synchronous generator stator windings (when it's synchronized to a grid and producing power (watts)) the strength of the generator stator magnetic field, as it appears to rotate around the stator due to its alternating nature, <b>grabs (attracts) and holds the opposite pole of the generator rotor</b> and keeps it locked with the rotating magnetic field of the stator windings--the speed of which is proportional to the frequency of the AC flowing in the generator stator windings (which is the same as the frequency of the grid to which it is synchronized with).

It's magnetic forces that keep the generator rotor locked into a speed that's proportional to the frequency of the alternating current flowing in the generator stator which is the same as the grid frequency. So, magnetic forces won't allow the generator rotor to spin any faster or slower than the speed proportional to the grid frequency.

Applying more torque<b>which would <i>tend</i> to make the speed increase--</b>gets converted to more amperes (current) flowing in the generator stator windings, which means more power (watts) is being produced by the generator.

It's all about magnetism and the two formulas at the top of this response--the second of which is really just the first rearranged to solve for speed instead of frequency. There's only one formula relating speed and frequency (and the number of poles--which doesn't change during synchronous generator operation).

When a synchronous generator is unsynchronized, there is no alternating current flowing in the generator stator windings--so there is no resultant magnetic field associated with the stator windings (just a voltage waveform produced by the rotor magnetic fields "cutting" the stator windings). So, an increase in torque WILL result in an increase in speed--when the generator IS NOT synchronized to a grid.

(To be clear, this discussion is about synchronous generators synchronized to a grid with other synchronous generators and their prime movers and operating in Droop speed control governor mode. A single generator supplying a load which is less than the rating of the prime mover driving the generator will be operating in Isochronous speed control mode and will change the energy flow-rate into the prime mover to change the speed of the synchronous generator rotor to control--and maintain--the speed <b>and frequency</b> of the generator output (and of the power being applied to the loads powered by the generator and its prime mover).

Most people understand that (induction) AC electric motors operate at a fairly constant speed regardless of load (presuming they are not powered by a variable frequency drive, that is). And, the same formula above applies to AC electric motors: F=(P*N)/120, or N=(120*F)/P. Changing the load of the an induction AC electric motor does not appreciably change the speed of the motor. So, throttling the discharge valve of a centrifugal pump being driven by an induction AC electric motor to change the flow-rate through the pump which changes the torque required by the pump being supplied by the electric motor--which changes the load on the pump motor--does not appreciably change the speed of the pump motor (or the pump). Nobody seems to have a problem with this. As the flow-rate through the centrifugal pump varies so will the torque required by the pump, which also changed the alternating current being drawn by the electric motor driving the pump--and current and torque are directly related: an increase in amperes (current) results in an increase in torque, and an increase in torque results in an increase in amperes (current) flowing in the motor windings.

An AC electric generator being driven by a prime mover does the opposite. The generator converts torque to amperes--at a constant frequency (on a well-regulated grid); and wires transmit the current (torque) to motors (and other loads) which then convert the amperes back to torque (and other "useful" work--though surfing the World Wide Web cannot generally be considered useful work, and a lot of watts are consumed doing that!). One can't have motors without generators--and amperes flowing through wires connect the two. The prime movers driving the generators actually do the work of the motors at the other end of the wires. The amperes (current) flowing in the wires just connect the generators to the motors (and other loads)--it's just a tranmission medium (method) for getting to be done in many remote locations by electric motors (and other loads) to be done by prime movers located in different places (where the ability to produce torque is more easily possible). Without electric generation, transmission and distribution, and electric motors (and other loads) we'd have to have bazillions of fossil fuel fired, or water (hydro) -driven "engines" to power pumps and fans and elevator motors and air conditions and refrigerators (both of which use electric motors) and lights and such. Computers probably wouldn't even be possible--or they'd be VERY different from what we know today.)

And, all of this is done at a constant frequency. No single generator synchronized to a grid with other generators can operate at a frequency that is even 0.1 Hz different than the speed proportional to the grid frequency for that generator (the generator's synchronous speed). And motors have synchronous speeds, too. And it's the synchronous speed that keeps motor speeds relatively constant, and synchronous speed is proportional to (related to) frequency.

And it's all about magnetism and magnetic fields. Have a look--there are many videos and presentations available on the World Wide Web about synchronous generators and the magnetic fields of the stators (armatures) that "rotate" because of the alternating nature of the current flowing in the stators (armatures). And, this captures and holds (locks) the magnetic fields of the generator rotors into synchronism with the stator magnetic field "rotation."

Hope this helps!

 

Thank you CSA for the very understandable answers. There is one question from my side, you stated that in case of grid system the speed and voltage will be kept constant by the grid, but what will keep the voltage and speed in case of single generator?

Thanks in advance.