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Paralleling Generators
Field excitation and fuel consumption variations causes different behaviours in power generation when generators are in parallel or in single

Hello

Why when generators are in parallel increasing the fuel will produce more active power without affecting the speed of the prime mover and consequently the frequency? A "single" generator instead will increase the speed/frequency with more fuel injection.

The same for the field excitation that, if increased, will increase, in parallel, the reactive power instead of the voltage when in single?

Regards

Since you are looking at generator fundamentals you would be better thinking of generators in parallel as 'the grid'. Once a single generator is synchronised to the grid, it cannot slow down or speed up - it is synchronised until such time it is electrically disconnected from the grid.

Thus increasing the fuel will produce more active power without affecting the speed.

Ok, but what will keep the frequency of one generator "linked" to that of the other one or as you say of the grid?

As an example, we know that reverse power can occur. I have seen that in order to test the reverse power somebody forces one generator to slowdown until reverse power happen. But the forcing action is performed by increasing or decreasing the speed with the usual button on the main switchboards.

This operate on the fuel injection and so on the speed of the prime mover.

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

JeffersonAirplane,

>Ok, but what will keep the frequency of one generator
>"linked" to that of the other one or as you say of the grid?

Magnetism. The forces of magnetism keep the generator rotor spinning at the speed that is proportional to the frequency of the AC current flowing in the generator stator windings when the generator breaker is closed. And, the magnetic field associated with the flow of current in the generator stator windings "appears" to rotate around the stator, at a speed that is proportional to the frequency of the grid.

I would suggest you read some references on basic AC synchronous generator fundamentals. And, remember F=(P*N)/120, where "F" is frequency in Hertz, "P" is the number of poles of the generator rotor (which is always fixed), and "N" is the speed of the generator rotor (in RPM). Synchronism and synchronous speed are VERY important words, hence the importance attached to the act of synchronizing a generator to a grid with other generators.

And, it's all really about magnetism. Unlike poles attract each other--with great force, as we all know. And like poles repel each other--with great force, as we all know. And because there are two magnetic fields at work inside a generator (the one associated with the rotor and the one associated with the stator), magnetic forces of attraction are responsible for keeping ALL generators SYNCHRONIZED together on a grid running at the same frequency--which is proportional to speed, based on the above formula (which can be solved for speed or frequency if the other variable is known, along with the number of poles of the generator (which doesn't ever change); N=(120*F)/P is the same as F=(P*N/120)).

As for reverse power, if the torque being applied to the generator rotor (which is a function of the energy flow-rate into the prime mover) is exactly equal to that required to keep the generator rotor spinning at synchronous speed (100% speed) while the generator breaker is closed, the power "output" of the generator will be zero watts (or zero kW, or zero MW). If the energy flow-rate into the prime mover is increased when the generator breaker is closed, the generator speed DOES NOT change (because the generator rotor is magnetically locked into a speed that is proportional to the frequency of the grid the generator is synchronized to (which is 100% speed when the grid frequency is at rated!)), but the additional torque which is trying to accelerate the rotor to a higher speed is converted by the synchronous generator to amperes--and the power output of the generator increases above 0 watts/kW/MW. (Generators convert torque to amperes. Electric motors convert amperes to torque. There is no real difference between generators and motors, except the direction of current flowing in the windings, which determines whether the electric machine is a generator or a motor. The traction motors or most electric vehicles actually becomes generators during braking, to charge the batteries by using the traction motors as generators.)

Conversely, if the energy flow-rate into the synchronous generator prime mover is reduced below that required to maintain 100% speed when the generator breaker is closed then the power "output" of the generator will be negative. Actually, current from other generators and their prime movers on the grid will keep the generator rotor--and the prime mover--spinning at synchronous speed (again, because the generator is always magnetically locked into synchronous speed proportional to the grid frequency). When this happens, the synchronous generator actually becomes a synchronous motor, and a load on the grid (because real power, amperes, are flowing into the generator to keep it spinning at synchronous speed.

As you know, it takes a certain amount of energy flowing into the prime mover of a generator just to get it to and to maintain rated speed (synchronous speed) when synchronizing a generator to the grid--or to produce rated frequency if a single generator is going to power a load. That power is always required to keep the generator running at rated speed, even if it is connected to a grid with other generators. But, if that amount of energy flowing into the prime mover is not sufficient to keep the generator (and the prime mover) running at synchronous speed when the generator breaker is closed and the unit is synchronized to a grid then the generator will become a motor because it MUST spin at synchronous speed due to the magnetic forces at work between the generator rotor and the generator stator in the generator when the generator breaker is closed.

When there is one single generator supplying a load, the amount of energy flowing into the prime mover must be equal to the amount required to spin the generator at the proper speed to achieve the rated frequency (based on the formula above) AND to supply the electrical load(s) the generator is powering (motors, lights, televisions, computers, computer monitors, tea kettles, coffee makers, etc.). If the energy flow-rate into the prime mover is not sufficient to do BOTH (maintain rated speed/frequency) AND power the load(s), then the frequency of the generator (and the loads connected to it) will decrease. There is no other source of current to keep the generator rotor locked into synchronous speed--only the amount of torque being provided by the prime mover. So, if that torque is insufficient to do BOTH (maintain rated speed/frequency AND power the load(s)) then the frequency will decrease.

If the energy flow-rate into the prime mover is greater than that required for a single generator to maintain rated speed/frequency AND power the electrical load(s) connected to the generator then the frequency of the generator (and the loads connected to it) will increase above rated.

Read up on basic AC electrical generator (and motor) fundamentals, and always remember that there are two magnetic fields at work inside a generator when the generator is supplying power to a grid (or load(s)). There are also many excellent (and some not-so-excellent) videos on YouTube and other websites about basic AC electrical generating fundamentals. When the prime mover and generator is synchronized to a grid with other generators and their prime movers, it MUST run at the speed that is proportional to the frequency of the grid (from the formula above) because of those magnetic forces at work inside the generator. The second magnetic field--the one associated with the generator stator) is only present when the generator breaker is closed and current is flowing in the generator stator windings.

For a single generator supplying a load(s), there is still that second magnetic field when current is flowing in the generator stator windings, there's just no other source of current (than the torque being provided by the generator's prime mover). The generator can't become a motor because of reverse power--because there's no other source of current from other generators and their prime movers. So, if the energy flowing into the prime mover is more or less than that required to maintain rated speed/frequency AND power the electrical load(s) connected to the generator, then the frequency of the generator and the load(s) connected to the generator will be higher or lower, respectively, than rated.

Hope this helps!(By the way, do you work in the oil patch, possibly on a supply boat or an oil rig?)

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

How could this reply have been more helpful (to the person who gave it a thumbs-down)? I have struggled with trying to explain this before (more than once) and I'm open to any suggestions or constructive criticisms.

If I've made any errors with physics, I'm open to corrections. But, it's about as simple as I can make it for someone who's asking a very basic question with, seemingly, little knowledge of AC power fundamentals (which aren't that hard, really).

If there's a better way to explain why machines all spin at their synchronous speeds when synchronized to a grid with other machines and their prime movers, help me find it. 'Cause I've tried. And I don't seem to know how to explain it without emf's and counter-emf's and load angles and vectors--none of which can be measured on site with the instrumentation which is typically supplied with most power generating equipment. And a lot of maths--none of which most people are interested in or care to learn about. One simple formula (the basic formula for frequency and speed)--that's all.

Yes, there's a lot more that can be done--if we could use diagrams and pictures on control.com. But, we can't. And the information that gets posted to web-sharing sites usually disappears after a few months, so that's not a long-term option, either. And, maths can often just serve to add to the confusion, if the basic fundamentals are better understood. (I remember my Electrical 101, 102, 201 and 202 courses in university--they were brutal with all the maths, and no real clear explanations in the texts or from the lectures. The books and the instructor just seem to think the basic "operation" is intuitive and can be derived from the formulae and maths. And, for many people (I speak for many of my fellow students at the time) it just wasn't. We were just memorizing formulae and diagrams--but what was really happening, what was trying to be described by the formulae and diagrams--just wasn't clear. Not until many years later, with lots of hands-on experience, re-reading the texts (some of which were seriously lacking, like on droop speed control!), and piecing it all together in my mind.

Which might be the problem.?.?.? I don't always tend to grasp things the same way other do--most things, yes. But some things, not. Is this one of those things?

I'm looking for some help with explaining this concept as simply and succinctly as possible.

(And if the thumbs-down is about asking if the OP (Original Poster) worked in the oil patch or on a supply boat or rig, that was because they often use multiple generators in parallel (synchronized to each other), but which sometimes operate singly, and the loads can, and do, change very quickly and can lead to some "unusual" operating scenarios, especially if prime mover governors are not properly set up. I have witnessed some drilling rig operations with local generators, and it's pretty impressive, and scary, at the same time. Those generators and their prime movers take a lot of abuse--mechanically, electrically, and otherwise. Some handle it better than others, I'm told. And I wouldn't be surprised to find some technician or mechanic asking this kind of question on control.com trying to understand what is or might be happening; someone without a lot of technical background or electrical education/knowledge. Hence, the question. No other reason.)

Absolutely great.

I provide control system for equipment in the industrial field and at the moment in the naval. We generically call it automation for on-board equipment and this includes also power management. So i deal with hw, sw, and control principles, but I normally do not need to go in deep with some matters. In the case of the PMS i deal with tresholds of power and control the insertion / disinsertions of generators on the net, monitoring the interlocks and so on. But even if i do not need to get the true principles behind, I need to understand that for myself.

One doubt remain. Reading your explanation it seems to me that if the grid is very powerful (big generators synchronized all together that almost nothing could perturb from their constant revolution) the grid will force the generator newly connected to the speed proportional to the frequency of the grid itself. In the case of only two generators, each one forces the other to turn at the speed proportional of the freequency? Alternatively sometimes one dirves the other and the other drives the one?

Thank you,
Roberto

I am totally agreed with you. A simple answer but very powerful, then the math and formulas can follow. Magnetism. And then all the other detail explained in a very intuitive way.

Again, a great post.

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

JeffersonAirplane,

Thank you very much for the kind words. I'm working on another post at the moment, but will respond to your latest questions soon.

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

Roberto,

>One doubt remain. Reading your explanation it seems to me
>that if the grid is very powerful (big generators
>synchronized all together that almost nothing could perturb
>from their constant revolution) the grid will force the
>generator newly connected to the speed proportional to the
>frequency of the grid itself.

Doubt is not my favorite word (look it up). I can provide clarification, but I'm not good at assuaging doubts.

Anyway, you are 100% correct--a large grid (often referred to as an infinite grid) will force a generator being synchronized to its synchronous speed with the grid frequency. As you may know, when synchronizing a generator to the grid it is customary for the speed/frequency of the generator to be just a little higher than the frequency of the grid it is being synchronized to. And, there is usually either a synchroscope which rotates to indicate fastness or slowness and the rate of fastness or slowness, or a set of synchronizing lights which also indicate the rate of fastness or slowness (sometimes there are both).

Basically what the synchroscope indicates when it is approaching 12 o'clock (and what the synch lights indicate when they are going dim (usually!)) is that the North pole(s) of the generator rotor are coming into alignment with what will be the South poles of the generator stator field. And usually just before the synch'scope reaches 12 o'clock (or the synch lights go dim) the generator breaker is told to close. There is a slight time lag for the generator breaker to actually close (hence, why the signal is sent a little "early") but the idea is that when the generator breaker actually closes the magnetic poles will be in alignment--and the rotor is "locked" into the speed which is proportional to the grid frequency with very little mechanical force required to "capture" and "lock" the rotor into synchronism and synchronous speed.

Now, the governor is still supplying slightly more fuel than is required to maintain rated speed, and that did, when the generator breaker was open, keep the generator rotor spinning slightly faster than synchronous speed. But, when the generator breaker closes and the generator rotor gets locked into synchronism that extra torque from the generator prime mover gets converted into a small amount of amperes flowing in the generator stator windings, which means the power output of the generator is positive and above zero watts/kW/MW.

If the generator was not synchronized before the generator breaker was closed when the generator breaker was closed one of two things could happen. Either the magnetic forces of the generator rotor and the generator stator would be out of alignment--and force them apart, possibly in the wrong direction!--or the magnetic forces would cause the generator rotor to spin extremely fast as the forces of unlike poles attract each other. In either case, as soon as the unlike magnetic poles attract each other the generator would effectively stop (for a very brief instant in time)--which would transmit forces trying to stop the prime mover back to the prime mover through the coupling connecting the generator rotor to the prime mover output shaft. This causes very great mechanical forces for very short periods of time--but these forces can be very destructive, even catastrophic.

That's why synchronizing a generator to other generators is so important--and so critical--to protect the generator being synchronized, and also to protect the generator breaker and the grid components. So, the generator breaker is closed when the unlike poles of the generator rotor and stator would be in alignment thereby reducing the forces required to lock the rotor into synchronism with the stator. (There is some force as the rotor is "captured" and "locked" into synchronism, but if the synch-scope is going relatively slowly then the forces arent't too great.)

>In the case of only two
>generators, each one forces the other to turn at the speed
>proportional of the freequency? Alternatively sometimes one
>dirves the other and the other drives the one?

In the case of only two generators, usually the one with the more powerful prime mover controls the frequency--not always, but usually. But, again, you are right; they are still synchronized with each other and neither can go faster or slower than the other. There is only one frequency coming out of the receptacle on the wall, remember! There can't be multiple generators all running independently at different frequencies synchronized together on a grid producing a single frequency for the user. It's just not physically possible.

So, I hope this clarifies things for you.

That's very ok thanks again. Now I go back for a while to my first post where I was speaking about field excitation.

Why I hear speaking of increase (decrease) of reactive power when field excitation increases (decreases) in paralleled generators (while in single that variation modify the voltage output)? Isn't it that the load determines the type of power? If I had only a resistance to drive isn't it a active power?

So many questions I know. Hope this does not disturb.

Regards

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

Roberto,

Excitation is very much like watts (kW; MW). When a generator and its prime mover are connected to a large ("infinite") grid, there is an unlimited amount of load (watts/kW/MW), AND an unlimited amount of reactive load (VArs; kVArs; MVArs). So, when the generator is at rated speed and the generator breaker is OPEN, increasing excitation will increase generator terminal voltage.

But, when the generator is synchronized to a large ("infinite") grid (the generator breaker is closed) and the excitation is exactly equal to what is required to make the generator terminal voltage equal to the grid voltage the generator will not be "producing" (or "consuming") and VArs. The power factor will be unity, 1.0.

If the excitation is increased, the generator is trying to increase the grid voltage--but it can't increase the grid voltage by any appreciable amount, but what happens is that the generator begins to produce VArs--called lagging VArs at the generator terminal.

So, it's very similar to when the energy flow-rate into the generator prime mover is increased which would make the speed increase if the generator breaker were open, but when the generator is synchronized to the grid and generator breaker is closed the speed can't increase and the watts/kW/MW increase.

If the excitation is decreased below the amount required to make the generator terminal voltage equal to the grid voltage then what happens is that VArs (reactive power) flows "into" the generator and the generator becomes a reactive load on the grid. These VArs are called "leading" VArs at the generator terminal.

When there is just one generator supplying a load, you are correct: the reactive load is what it is and the generator can't do anything about that. BUT, changing the excitation does change the generator terminal voltage and the grid voltage at this point. And when there are two generators synchronized together supplying a load, real and reactive, the amount of reactive load is still fixed but the split of reactive load being "carried" by each of the generators can be affected by changing the excitation of one of them (similar to what happens to the real load when two generators are synchronized together).

Finally, if the load on any system is purely resistive then there won't be any reactive power required by the system and changing excitation will have an appreciable effect on system voltage. But, since the majority of electric motors are induction motors the likelihood of that happening is very low. It does happen at some plants (arc furnaces, for example), but even those, I'm told, have some reactive load.

Hope this helps!

Then, can we say that when two generator are paralleled and load is purely resistive, if excitation field increases in one of the two it could be considered like if the other was decreasing its excitation (this because the reference is the output voltage)?. And so is it becoming a reactive load?. In this way even with a pure resistive load, anyway the generator with the increased field excitation will produce reactive power while the other will absorb it being a reactive load.

But as you say, this case is very peculiar (very low probability) and the system will not be working correctly and generating reverse power it seems to me. So control should avoid this.

Then when a generator is generating alone the power for the load and the load is both active and reactive normally, it seems to me, it will not need to modify the excitation field if the voltage remains stable at the output terminals. And if load change there will be more fuel injected whatever it is the type of load. Am I right with this?

If yes in the case of 2 paralleled generators I do not see very clearly the why for modifying excitation field unless that of transferring the reactive load from one to the other as you were describing (when you were speaking of the split).

Regards

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

When two generators are synchronized to each other and not to a larger grid, if any part of the load is reactive then the excitation systems of the two generators can be used to make the two generators split the reactive load, or have one generator take the reactive load.

In my experience, if one excitation system continues to increase its excitation then the system voltage will increase AND the other generator will eventually start "consuming" reactive power. Regardless of how much of the load the two are supplying is resistive.

The amount of the resistive load shared by the two generators is controlled by the amount of energy flowing into the prime movers of the two generators. And, if either energy flow-rate causes the total energy flowing into the two generators to exceed the amount required to maintain the load AND maintain rated speed/frequency, then frequency/speed of the units will increase.

When two generators and their prime movers are synchronized together and supplying a load (resistive; resistive-reactive) it's customary for one of the to be operated in Isochronous speed control mode and the other to be operated in Droop speed control mode. Sometimes, a separate PMS is used to control the loads AND the frequency of two such generators and both are operated in Droop speed control mode. It can get VERY complicated VERY fast.

>Then when a generator is generating alone the power for the
>load and the load is both active and reactive normally, it
>seems to me, it will not need to modify the excitation field
>if the voltage remains stable at the output terminals. And
>if load change there will be more fuel injected whatever it
>is the type of load. Am I right with this?

Yes.

>If yes in the case of 2 paralleled generators I do not see
>very clearly the why for modifying excitation field unless
>that of transferring the reactive load from one to the other
>as you were describing (when you were speaking of the
>split).

Yes.

Very well. Thank you very much.

Best regards,
Roberto

If the energy flow-rate into the prime mover is
>increased when the generator breaker is closed, the
>generator speed DOES NOT change (because the generator rotor
>is magnetically locked into a speed that is proportional to
>the frequency of the grid the generator is synchronized to
>(which is 100% speed when the grid frequency is at rated!)),
>but the additional torque which is trying to accelerate the
>rotor to a higher speed is converted by the synchronous
>generator to amperes--and the power output of the generator
>increases above 0 watts/kW/MW.

If the grid is small,and has a fix load request, and I increase the fuel only to one diesel generator out of 3 that are already on net sharing this load, What happens? the the DG that receive more fuel will take the load from others, keeping the frequency stable?

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

If the diesel engines are all connected to a grid and one engine had been operated with increase fuel, that engine will be exporting power to the grid. The other engines more or less maintain the same power.

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

>If the diesel engines are all connected to a grid and one
>engine had been operated with increase fuel, that engine
>will be exporting power to the grid. The other engines more
>or less maintain the same power.

That's impossible to say with any degree of certainty without fully understanding how large the "grid" is, and what modes the three DG sets are operating in, and if there is any kind of external load and/or frequency control active and controlling the DG sets.

And, because the cac86 did not provide any of that information it's not possible to respond to his question. There are simply too many variables to be able to say exactly how the three DG sets would respond to his scenario. Are the units all in Droop Speed Control mode?

Is any unit in Isochronous Speed Control mode?

Is any unit in Isochronous Load Sharing control mode?

Is there any "Power Management" or "Load Sharing" control system monitoring the grid frequency and sending signals to any of the three DG sets?

Is there any "Power Management" or "Load Sharing" control system monitoring the loads of any or all of the three DG sets and sending signals to control the loads of any or all of the three DG sets in response to some operator-selected matrix or formula?

We could speculate (as one responder already has without clearly stating the conditions of his response), and we could try to cover all the possible variables and conditions--but, why not ask cac86 to provide the necessary information about the site conditions, and then respond more concisely?

There is not one answer to cac86's question--without fully understanding the conditions. He may not even fully understand the conditions, but, we should be clear and concise--and at least give him a chance to respond with more information.

Guys, first of all is a pleasure to discuss with you.

The grid is small, for example i will say that the power of the generator is 6MW (all 3 are the same) at 100% and the design load is at 75% of 2 generator

There is isochronous load sharing active (and backup with droop mode) there is PMS with fast trip of not essential users and over and lower frequency protection for breakears together with all standard safeties.

I can say that for the above scenario, all generators are running at 30% load, and due to a malfunction the governor of the prime mover DG1 go to 100%. This DG1 will take all the load from the other DG's? and this because the generator start to feed extra current than before (as i read on previous answer in this page) or something else will happens?

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cac86,

The plot thickeneth. It's not an operator action causing the increase in load of one of the three units supplying a small grid (also known commonly as an "island load" because it is operating independently of a larger grid with more generators); rather, it's some kind of failure of the governor of one of the DG sets.

AND, there is Isochronous load sharing and a PMS (seemingly just meant to control load in the event of loss (tripping) of generator(s) (to "shed" load to try to prevent a total black-out).

Based on the information provided, and because Isochronous load sharing can be programmed in many different methods, that it's still very difficult to say precisely what would happen in this case. I would expect the grid (island) frequency would suddenly increase.

I don't know how the Isochronous load control scheme at your site would respond to that. I suspect it would then start to reduce the loads of the DG2 and DG3 to try to reduce the frequency.

In my experience with Isochronous load control schemes, one or more units are set to maintain a certain load (as if they were operating in Droop Speed Control!) and those units are considered "secondary" units. And one or more units are set to respond to load changes in order to maintain frequency, and those units are considered "primary" units. And, there is usually some kind of "matrix" that is used in the event of loss (tripping) of one of the primary units to select one of the secondary units to automatically become a primary unit and adjust its load as required to maintain frequency.

But, again--there are many flavors of Isochronous load sharing; there is no one standard. And, in my mind Isoch load sharing is just a de-tuned Isoch mode, somewhere between Isoch and Droop. Isoch load sharing either requires the governors of multiple units to share load information with each other via some kind of analog signal, or some kind of external system which monitors the load of all the units AND the frequency of the grid (island) and then decides which units to load first, and second, and third, and so on. And, which units to unload first, and second, and third, and so on. It actually gets very complicated very quickly.

In a very simple system with one unit (DG1) running in Isochronous speed control mode and all other units (DG2 & DG3) synchronized together with the Isoch unit running in Droop speed control mode, if the Isoch unit's governor (DG1) suddenly decided to feed a LOT of fuel to it's diesel what would happen would be that the grid frequency would increase. The other units (DG2 & DG3), operating in Droop speed control mode, would decrease their load because of the frequency increase (because the grid frequency caused the actual speed to increase while the speed reference remained the same, the speed error would then decrease (I'm using the formula Speed Error=(Speed Reference - Actual Speed)) which would cause the fuel flow-rates to DG2 and DG3 to drop. There may come a point at which DG2 & DG3 trip on reverse power because DG1 is putting out so much power and has caused the grid (island) frequency to increase so much.

>I can say that for the above scenario, all generators are
>running at 30% load, and due to a malfunction the governor
>of the prime mover DG1 go to 100%. This DG1 will take all
>the load from the other DG's? and this because the generator
>start to feed extra current than before (as i read on
>previous answer in this page) or something else will
>happens?

So, cac86, it sounds like this was a real event at your site. What happened when DG1 suddenly went to 100% power? Did DG2 and DG3 hold their power outputs? Did they trip--if they tripped, what was the alarm given for the cause of the trips?

If all the generators were running at 30% of their capacity and suddenly one unit puts out 100% of its capacity, without understanding exactly how the Isoch load sharing scheme at your site is configured to run it's next to impossible to say what would happen. My guess is that the system frequency would be excessive at some point. 30% of 2 MW is 0.6 MW, and 3 x 0.6 MW equals 1.8 MW--which is less than the capacity of any one of the machines. So, if one generator tried to go to 2.0 MW, the frequency would definitely go high--especially if the other units didn't reduce their loads (which they would do if they were running in Droop Speed Control--but we don't know how the "Isoch load sharing with Droop Speed Control back-up" is configured to work!).

But, if you told us what actually happened we might be able to help you understand why it happened (not why DG1's governor went awry!, but why the other two units and the system did what it did). But, there's always going to be the unknown: How the Isoch load sharing scheme with Droop Speed Control back-up is configured to work.

The "power" of the grid (island) is the sum of all the loads on the system (the electric motors, and electric lights, and televisions, and computers and computer monitors, and tea kettles, etc.) that are on/running at any instant in time. All of the generators synchronized together have to act as one generator to supply the load at any instant in time. If the generator(s) try to put out more power than the total of the loads (the electric motors, electric lights, televisions, computers and computer monitors, and tea kettles) for that instant in time the grid (island) frequency will increase. Or, if the generator(s) do not put out enough power to supply the total of the loads (the electric motors, electric lights, televisions, computers and computer monitors, and tea kettles) for that instant in time the grid (island) frequency will decrease. The power of the system (the generators AND the loads!) can never be more than required by the loads--if the generators try to put out more power than the loads require, the grid frequency will increase. And, if the generators don't put out enough power required by the loads then the grid (island) frequency will decrease. The load in either case (the total of all the electric motors, electric lights, televisions, computers and computer monitors, and tea kettles) doesn't change--only the frequency changes if the generators can't keep amount of power being produced equal to the amount of power required while maintaining frequency.

What I'm trying to say is this: If the power required by the load(s) on a grid (island) is 1.8 MW, if the generators synchronized together supplying the grid (island) load(s) is 1.8 MW and the frequency is at rated (let's say 50.0 Hz), then all is good. But, if one generator decides to put out more power all of a sudden, it's individual power output might change--BUT the power required by the grid (island) load(s) isn't going to change. And, the difference, if it can't be compensated for by reducing the power output of the other generator(s) synchronized to the grid (island), is going to result in a higher frequency (greater than 50.0 Hz). The TOTAL of the three power outputs of the three generators is still going to be 1.8 MW--but it's NOT going to be 1.8 MW at 50.0 Hz. The load(s) on the grid (island) aren't going to change--they're still going to require 1.8 MW. It's just that if one or more of the generators synchronized to the grid tries to cause the total power of the generators to be more than 1.8 MW what's going to happen is the grid frequency is going to increase. Generators can't make more power (amperes) than is required by the load(s) on the grid (island). If the generator(s) synchronized to the grid (island) try to make more power (amperes) than required by the load the net result will be an increase in frequency--not power.

Conversely, if one of the three generators suddenly trips off line, the power required by the load(s) on the grid (island) doesn't suddenly change (unless the PMS suddenly sheds loads!), it's still 1.8 MW--but the frequency of the grid (island) is going to decrease UNTIL the remaining two generators can be adjusted to provide 1.8 MW at 50.0 Hz.

Hope this helps! Again, if you can tell us what happened when this scenario occurred we might be able to help understand why it happened--but probably not a full and complete understanding unless we know how the Isoch load sharing with Droop Speed Control back-up scheme is configured.

Unfortunately i don't have much more information. I'am describing a real accident reported from a friend on board of a cargo ship.

My friend told me that, after the governor of the engine went to 100% they have a blackout,they didn't realize exactly the sequence of events, because everything was too fast.

Anyway they found reverse power on the other 2 running DG. They found over frequency on users breakers, and they found overspeed on the faulty DG.

cac86,

>My friend told me that, after the governor of the engine
>went to 100% they have a blackout,they didn't realize
>exactly the sequence of events, because everything was too
>fast.

>Anyway they found reverse power on the other 2 running DG.
>They found over frequency on users breakers, and they found
>overspeed on the faulty DG.

All of this is consistent with what I would expect would happen for a very simple system consisting of one DG operating in Isochronous Speed Control Mode, and two DG sets operating in Droop Speed Control Mode at a low(er) load. The Isoch unit burning excess fuel for the conditions--and going to, essentially, max fuel flow-rate--would cause the frequency of the grid to increase. The Droop machines would then start to unload as the frequency increased (REMEMBER: Droop Speed Control is all about the error between the speed reference (which isn't changing in this case!) and the actual speed (which is increasing in this case)--which, in this case, would be driving the fuel flow-rate, and the power outputs of the Droop units, down--to reverse power.) The Isoch unit would go to overspeed at some point, and the overall result would be a black-out. All of which is consistent with what your shipboard friend reported happened.

The Droop units are not going to contribute to frequency control--that's the job of the Isoch unit, and if it's governor has malfunctioned resulting in 100% fuel flow-rate then the frequency (speed) of the grid is going to increase (since the load on the system isn't very high). The decreasing speed error between the speed reference (let's say it was at 101.33% speed) and the actual speed (which was increasing and approaching 101.33%) would be driving the speed error to zero as the actual speed got closer and closer to the speed reference, and that would be driving the fuel flow-rate lower and lower.

Fuel flow-rate--when the unit is synchronized and running at rated speed (frequency)--is directly proportional to the amperes being produced by the generator. In the same ways motor convert amperes to mechanical torque, generators convert mechanical power (torque) to amperes. The torque the generators convert to amperes comes from the prime mover driving the generators--diesel engines in this case. If the amount of fuel is not sufficient to make the generator have positive amperes (flowing out of the generator on to the grid) the reverse power protective relay is going to trip the generator breaker to protect the diesel(s). Now all of the power being produced is coming from the Isoch unit--whose fuel flow-rate is "uncontrolled" and much more than is required to supply the loads AND maintain rated speed (frequency). And, so the protection functions are going to shut down the Isoch unit. Probably initially because of a high frequency (speed) which may have been close or at some point exceeded the overspeed setting of the governor.

Hope this helps!