from the power plant department...
 Posted by rahul on 15 August, 2010 - 7:52 pm
Two turbines each with load of 15 mw and having capacity of 30 mw were running.

One was in isochronous mode and other in the droop mode. Station load was 28 mw and 2 mw export. I observed every time islanding occurred in droop mode turbine load decreases and in frequency mode turbine load increases if we do not change speed setpoint. Why is that in droop mode always load decreases during islanding and in frequency mode it increases and to what point it can go. Our droop is 5%. Please help me

 Posted by CSA on 16 August, 2010 - 1:24 pm
This is a good example, with only two turbines of relatively equal capacity supplying an "island". I'm not exactly clear on the conditions at your plant, but if you will just read on you should be able to relate the explanation and examples to your situation.

Imagine one tandem bicycle with two riders each pedaling at about 50% of capacity while riding at a constant speed and desiring to maintain speed constant during their journey to deliver some packages to another town on a relatively flat and smooth road. The cranks of the tandem bicycle are the same size (number of teeth) are coupled together with a chain, meaning that the rate of pedal speed for both riders is exactly the same (just like two isochronous generators (alternators) would be locked into the same frequency and speed during operation, islanded or when connected to a grid).

Poorly attached to racks on the bicycle are several packages, and a couple of the heavy ones on the rear rack of the bike fall off without the two riders noticing. This means the load has decreased and if the two riders keep pedaling at the same rate the bicycle speed will increase above desired. So, both riders immediately notice the increase in speed and both decrease their torque output, but the speed now falls below desired and they both increase their torque output, and it takes some time before they get the speed back to desired and stable.

Now, if they had made some agreement before they started their journey that one rider or the other would adjust his torque output to maintain the speed and the other rider would keep pedaling at a constant torque output, then it's likely they would not have had the disturbance in speed when they load decreased. But, because both riders were attempting to respond to the change in speed (because of the change in load) it took some time for them to return to the desired speed.

In the example above, when both riders are attempting to adjust their torque output to respond to changes in load (which caused the speed to change) they could be compared to the governors of both turbines being in Isochronous ("frequency") control mode. Both riders (turbines) saw a change in speed (analogous to a change in frequency) because of a change in load and both riders (turbines) tried to change their torque outputs to have the speed (frequency) return to normal.

If when the packages fell off neither rider changed their torque output and the bicycle speed increased, then that could be compared to both turbine governors being in Droop control mode, with nothing adjusting their outputs to maintain frequency because neither is attempting to control speed (frequency) and they are just keeping their torque output relatively constant regardless of speed.

If one rider had agreed to keep his torque output relatively constant during the ride (let's identify him as Rider D) with the other rider adjusting his torque output as necessary to maintain the speed constant during the journey (let's identify him as Rider I) then it's likely that the speed would remain relatively constant during the trek. This is an example of how properly tuned governors will work during island operation. One turbine governor should be in Isochronous control mode and the other should be in Droop control mode. As load changes, which will tend to cause a change in frequency, the Isochronous governor will automatically change load to keep the frequency constant and the other governor will just maintain its output relatively constant (presuming frequency remains relatively constant).

The catch here is that if the load of the machine operating in Isochronous mode reaches its maximum output rating then it will not be able to respond to any further increases in load by increasing its output; it's already at maximum and so it can't do any more! In this situation, the operators must increase the output of the Droop machine, which would tend to increase frequency (presuming load is relatively stable), and the Isochronous machine will reduce its output to maintain frequency and then be able to respond to any further increases in load automatically.

A lot of operators believe that island operation can be totally automatic, and that's just NOT the case for most situations. In your case, if I understand the situation correctly, you had a load of 28 MW with a capacity of 30 MW. If either unit was at rated power output and load continued to increase the machine whose output was already at rated couldn't contribute any more power to the island. If that machine is the Isochronous machine, then the frequency will decrease of load is increased. So, it's the operator's responsibility to change the load of the Droop machine to unload the Isochronous machine so that it can respond to any further increases in load.

Yes; you read that correctly. Change the load of the Droop machine to allow the Isochronous machine to respond to any further increases in load. That's because the setpoint for the Isochronous governor is frequency (speed)!!! Remember: The Isochronous machine is trying to control frequency (speed), so if an operator changes the setpoint of the Isochronous machine he is changing the frequency (speed) setpoint! And, if the Isochronous machine is already at rated output, changing it's frequency setpoint will have no effect!!!

When a governor is in Droop control mode, it's not trying to control frequency. It's primary purpose is just to be a stable contributor to the power required to supply the load while some other machine (or entity) is controlling frequency. 'Stable contributor' means that if the frequency remains constant that even if the total load on the system (the island in this example) changes the Droop machine output will not change. (Note the caveat: if the frequency remains stable.)

If the system frequency is stable the speed of all the machines will be stable. Because most large AC generators (alternators) are synchronous machines, they are all locked into a speed which is directly proportional to frequency, so that if the frequency of the system changes the speed of all the machines operating in parallel (be it two, or twenty, or two hundred, or two thousand) will change. And that's an extremely important concept to remember as we continue this discussion.

In an island situation, the Droop machine is just there to help with the load that the Isochronous machine cannot handle. So, the Isochronous machine does the work of responding to load changes, which would tend to cause frequency changes, and the Droop machine--presuming the frequency remains relatively constant (that the the Isochronous machine's governor is doing it's job)--is just there to contribute to the load, and that the contribution remains stable as long as the frequency remains stable or unless the operator changes the setpoint of the Droop machine (which would have the opposite effect on the Isochronous machine's load).

If the torque being provided by bicycle Rider I (who had agreed to adjust his output) was near his maximum, he could ask the other rider to increase his torque output for a while. This would allow Rider I to reduce his torque output as Rider D increased his output while the bicycle maintained a constant speed.

In our bicycle example if Rider I's torque output was was about at his maximum ability and the cousin of the other rider (Rider D) who was running alongside the bicycle suddenly jumped on the handlebars of the bicycle then it's likely that the speed of the bicycle would drop below desired because Rider I's torque output might be unable to provide the necessary torque to keep the bicycle traveling at the desired rate of speed. In this case, Rider D would have to consciously increase his torque output to return the bicycle to the desired speed to carry the additional load of his cousin.

Just so with our island example. If the Isochronous machine had reached its maximum output and the load increased, then the frequency would start to decrease. The only choice the operators would have would be to manually increase the load of the Droop machine in order to return the frequency to desired. Changing the setpoint of the Isochronous machine would have virtually no effect because the Isochronous machine is already at rated output!

That's very basically what happens on a small power island with a variable load with two machines operating in parallel and with a relatively stable frequency. The governor of one machine should be in Isochronous control mode and the governor of the other machine should be in Droop control mode. The operators should be constantly watching the load on the Isochronous machine, and when it gets very close to the rated output of the Isochronous machine they should increase the load on the Droop machine in order to give the Isochronous machine the ability to automatically respond to any further changes in load. They should not be responding to changes in load by changing the setpoint or setting of the Isochronous machine, except to keep the Isochronous machine's output below rated by adjusting the load of the Droop machine.

Conversely, if the load of the Isochronous machine is near zero, and if the load of the island (system) decreases which would tend to increase frequency, then the Isochronous machine will likely not be able to keep the frequency from increasing by very much, and the generator breaker of the Isochronous machine might be tripped on reverse power. So, the operators should be reducing the load of the Droop machine to make sure the load on the Isochronous machine stays above zero so that it can respond to any further decreases in load. Note, the operators do not change the setpoint of the Isochronous machine, because that setpoint, and the governor of that machine when operating in Isochronous mode, should be adjusted to and capable of responding to the majority of expected changes in load (very large and sudden load swings might be difficult for some prime movers and their governors to respond to, while other types of prime movers and/or governors might be better able to handle such load changes).

This is enough for now; in the next response, we'll talk about the "frequency" response of Droop control mode on a small island.

 Posted by Manit on 16 August, 2010 - 7:10 pm
Good Explanation

 Posted by CSA on 17 August, 2010 - 11:07 am
Well, it's time to correct a mistake in this posting....

First, at the end of the second paragraph it was written:

>...(just like two isochronous generators >(alternators) would be locked into the same >frequency and speed...

>...(just like two synchronous generators >(alternators) would be locked into the same >frequency and speed...

('Isochronous generator' is kind of a bad name, since generators aren't Droop or isochronous "machines", rather it's the prime mover governor that is in either Droop or Isochronous control. A generator with a prime mover operating in Isochronous control mode can be (usually is) a synchronous generator, but there shouldn't normally be two machines both with their governors operating in Isochronous control mode at the same time on any grid.)

I believe that's the extent of the glaring errors in this posting, and I apologize for any confusion it might have caused. Technically, it's correct, but it's not what was intended and could convey an incorrect message that was contradictory to other information in the posting.

If there are any other errors, just post them and we'll try to get them fixed!

 Posted by Timus on 18 July, 2012 - 10:08 am
Dear CSA,

My best wishes.

I am a novice to this site, though I have been going through the various threads for few months.I came through this old thread and I need to ask something:

I read through your classic bicycle example for understanding Droop & Isochronous governor modes.
I need to understand about droop percentage i.e.
Earlier we had a droop characteristics of 4% at our site which we have recently revised to 5% as per our local state grid requirement.

Ours is a unit with average loading of 125-130 Mw power requirement with 3 GTs & 1 STG. All our units operate in synchronised mode with grid in droop mode.

Kindly give some insights.

Best Regards

 Posted by CSA on 18 July, 2012 - 1:05 pm
It's the feedback that makes the information on control.com most valuable. Don't forget that.

I'm not really seeing a question here.

Increasing the droop from 4% to 5% means that for every 1% error in speed (difference between reference and actual when not at full load) the unit will change load by 20% (for a 5% droop characteristic) instead of 25% (for a 4% droop characteristic). Otherwise, for all intents and purposes (see below, though), the unit will operate similar to the way it operated before.

If the unit at your site has a GE Speedtronic control system, if you changed the droop characteristic without changing the loading rates appropriately, there will be difference in loading (and unloading) times. Depending on the vintage of control system, there might also be some interaction between primary- and back-up exhaust temperature control curves, too.

Hope this helps!

 Posted by Bruce Durdle on 16 August, 2010 - 4:02 pm
If you are exporting 2 MW while connected to the grid, and then the plant is islanded, the exported power has nowhere to go. In island mode, the turbine will need to supply only the 28 MW station load. So as long as you are exporting you will always reduce load when going from synchronised to islanded.

Now if you are importing when you island, you might have a problem as your equipment will try to pick up the balance of load ...

Bruce.

 Posted by WAC on 17 August, 2010 - 8:41 am
IF YOU ARE EXPORTING 2 MW, SO YOU ARE NOT IN ISLAND MODE.

 Posted by CSA on 17 August, 2010 - 12:40 pm
While this is a simple concept it's very much a function of many parameters which can appear to complicate matters, especially when one can't use graphs and charts and sketches to make points. And I've used as much maths as necessary because we're just dealing with concepts here, not complete power systems analyses and studies. Every attempt has been made to try to make it as "simple" as possible, and I believe that if one relates the concepts to the bicycle examples it should become very clear, perhaps after a few readings.

I apologize profusely in advance for any typographical errors and any mistakes, which I will correct as soon as I become aware of them.

I keep re-reading the original post and now I'm even more confused about the site conditions. It seems others (Bruce Durdle and WAC) have the same concerns; I don't understand how there can be an export of 2 MW when operating in island mode. Island mode presumes completely disconnected from any external load, so no export. Hmmm....

These terms can be very misleading, can't they? That's why it's SO important to properly state all of the conditions when posting to a forum like this for help. You might be operating independently of a grid with your island, but consider some portion of the power output to be "exported" to some other area of the plant or geographical location that is also independent of the grid and other generators and can be disconnected from the generators and the prime movers driving the generators. But, we don't know that unless you tell us. Export almost always means "to an external grid" and when you are connected to an external grid unless you have some special kind of devices that prevent any power from "back-flowing" into your island (which I've never seen), then you are operating in parallel with other generators and their prime movers, and under such conditions neither of your machines should be operating in frequency control (unless there's something else about your situation and control scheme that we don't know because you haven't told us).

If they operate primarily in export mode while connected to a larger grid, and then separate or get separated from the grid and supply their own, independent load, then the total load on the units will decrease by an amount equal to the amount that was being exported. The load is the sum total of the lights and motors and computers and such, and is not a function of how much is being generated or the capacity of the generators which are on line. On an AC (alternating current) system, generation (the amount of power being produced by any and all generators connected to the grid must be exactly equal to the load or the grid frequency will not be at rated, regardless of the capacity of the generators connected to the grid. If more power is being produced than is required, then that will translate into a higher than desired frequency, in the same way that any less power being produced than is required will result in a lower frequency. (Actually, the power will be whatever is required, it's just that if there isn't sufficient power that the frequency won't be at rated, just as if too much power is being produced the power won't actually increase only the frequency will increase.)

It's just like our riders on the bicycle. If the load is constant and the riders don't produce enough torque, the bicycle will keep moving, it just won't move as fast. The load hasn't changed, just the torque being applied to the bicycle cranks to move the load. And if the riders produce too much torque when the load isn't changing, the load won't increase just the speed. At the same time, if the load is changing, then the riders need to coordinate their efforts so as to not be working against each other and causing the speed to be unstable. It's the same with electrical grids, of any size; it's a balancing act since the amount of generation must exactly match the load in order for the frequency to be at desired. Too much generation (torque) for the load and the frequency will be higher than desired; too little generation (torque) for the load and the frequency will be lower than desired. That's what system-, grid- and transmission and distribution system operators all have to do every day: Balance the generation to match the load so the frequency is stable.

At any rate, let's tackle the dreaded frequency response of Droop speed control. "Once more unto the breach, dear friends. Once more ... "

Here's our example for this discussion (on a completely isolated "power island" with no power export to any external grid which has other generators and load): two generators each capable of 15 MW; one generator's prime mover operating in Isochronous control mode and one generator's prime mover operating in Droop control mode, with a droop setting of 5%, let's say this is a 50 Hz system. Let's also say that both prime movers are in a new and clean condition and running at nearly rated conditions, so their outputs are are nearly 100% efficient and capable of achieving rated load. Further, we're not going to consider any reactive "loads" here, and consider the loads to be almost purely resistive (pretty nice conditions, eh?). And, lastly, we're going to presume there is no "external" system controlling load and/or frequency by sending signals to the prime mover governors (i.e., no load-sharing or frequency control schemes from a local or remote control system).

The important concept to understand about the frequency response aspect of Droop control mode is that a change in electrical load that results in a change in frequency (speed) will result in a change in percent of rated output that is inversely proportional to the droop setting, and must take into account the power output of the unit before the frequency excursion began. So, a droop setting of 5% means that for every 1% change in frequency (speed) the prime mover's output will change by 20% (1/5 = 0.20, or 20%) of rated up to rated load. This means that for every 1% change in frequency (speed) that the governor will change the prime mover's output by 20% of rated, up to rated load. In our example, 20% of rated load would be (15 MW * 0.20 =) 3.0 MW, making each 1% change in frequency equivalent to a 3 MW change in load. So, theoretically if the Droop unit was connected to a load or system but not putting out any power when the frequency was at 50 Hz and if the frequency went down to 95% of rated (47.5 Hz) because the load increased then the unit's power output would increase to rated output, or 15 MW for each of the units in our example.

If the Droop unit were putting out 50% of rated power (7.5 MW for our example prime movers) and the load increased causing the frequency to go down by 1% (to 49.5 Hz) then the power output of the prime mover would be increased by 20%, up to 70% of rated output, or 10.5 MW. If the load continued to increase causing the frequency to continue to decrease by another 1% (to 49.0 Hz), then the output would increase to 90% of rated, or 13.5 MW. And, if the load continued to increase even further causing the frequency to continue to decrease to 48.75 Hz (another 0.5% decrease in frequency) the machine output would be at rated, 15.0 MW. (But the frequency would still be less than rated; it's just that the governor operating in droop speed control mode would respond to a change in frequency up to the rated output of the prime mover!)

Let's define some more conditions for our totally isolated power island (two generators and their prime movers operating in parallel, and not in parallel with any other generators across any tie line, with no external signals coming from any kind of load-sharing or power control system, with one prime mover's governor in Isochronous control mode and the other prime mover's governor in Droop control mode.) Let's say the total island load at the present time is a stable 25 MW at a stable 50.028 Hz frequency, and the load on the Droop machine is 11 MW, which means the load on the Isochronous machine is 14 MW.

Now, let's say some motors and lights in some part of the plant are being brought on-line, increasing the load to a total of 26 MW. As the motors and lights are being energized, the frequency will tend to decrease but the governor of the Isochronous machine will increase the load of the unit up to 15 MW to keep the frequency stable at approximately 50 Hz (plus or minus a few hundredths of Hz). The load of the Droop machine will remain constant at 11 MW, running along just fine, thank you, very much! It's output is stable as long as the frequency remains relatively stable.

It's late in the day now, and let's say that some more motors are being started and it's dusk and the automatic lighting circuits have illuminated some perimeter lighting, and the total power island load has increased to 29 MW. Well, the Isochronous machine is already at its limit at 15 MW (rated power output) and the setpoint of the Droop machine has not changed, so as the load increases and the Isochronous machine can't increase it's load any further the frequency of the power island will start to decrease as the load increases. As the frequency decreases, the load of the Droop machine will increase, up to the maximum output of the Droop machine.

We know that for every 1% change in frequency that the load will change by 20% of rated, and vice versa, if the load changes by 20% then the Droop machine's frequency will change by 1% of rated. In this example, the load has changed by 3 MW, or approximately 20% of rated, and this means that the frequency will decrease by 1% (each 1% change in frequency is equal to a 0.5 Hz change) to 49.5 Hz. Now the output of the Droop machine is at 14 MW, but the frequency of BOTH machines is down to 49.5 Hz (synchronous machines are locked together at speeds that are proportional to the frequency).

There are a lot of other considerations that come into play to say exactly how fast the frequency will decrease in such a situation, but that's a lot of maths and physics and we're just trying to understand concepts here, not details (of which there aren't enough known about this site, and which would take a smarter technician than I to solve). And, we're not considering how fast the dusk-to-dawn lights came on or how fast the motors were started or how slowly the operators responded to the load increase and the frequency change. We're just talking fundamental concepts here.

At this point the operators (who have just awoken from their post carbohydrate binge-induced slumber and are probably freaking out!) can start increasing the Droop speed setpoint of the Droop machine, and the frequency of the power island will increase back to rated, and the output of the Droop machine will remain at 14 MW, the output of the Isochronous machine will be at 15 MW and the frequency will be at approximately 50.0 Hz. If they increased the output of the Droop machine to 15 MW, that would reduce the output of the Isochronous machine to 14 MW, meaning it would be able to respond to any more load increases--as long as they didn't exceed 1 MW (making the total plant load greater than 30 MW, the combined rated output of the two prime movers).

The important thing to note here is that a Droop machine being operated at rated power output cannot and will not respond by increasing its output if the frequency decreases. It's already at its maximum output, and it can't give any more to "the cause;" it can only contribute power output up to its rated output, regardless of the change in frequency.

There have been thousands of words written on control.com about Droop speed control, and we're not going to go into that in this thread. Use the 'Search' function of control.com to find other threads if you want to understand Droop speed control and references and errors and such. We're just talking about the frequency response aspect of Droop speed control in this thread.

[For those of you operating combustion turbines for electrical power generation, take note: When a combustion turbine operating at Base Load for the given ambient and machine conditions experiences a frequency decrease, it's power output actually DECREASES because the axial compressor speed decreases which decreases the amount of air entering the machine which decreases the amount of fuel that can be burned which decreases the amount of power than can be produced. Regardless of the droop setting of the governor--because the machine is already making as much power as it can make without burning itself up and with the reduced airflow it has to further reduce power output! Isn't this stuff fun?]

But, the more I read the original post, I'm not convinced the originator was asking about a power island. I believe when he was referring to "frequency control" he was referring to some kind of external signal coming into the prime mover(s) to try to help control the frequency of a larger grid with which the plant is connected. Or, that there is some kind of plant load sharing scheme that is sending signals to the prime mover governors to try to respond to load changes and frequency excursions. I'm honestly very confused about what the originator was describing, but I'm not confused about the concept of droop control on a power island.

The load on any power system (islanded or large, "infinite" grid) is the sum total of all the lights and motors and computers and heaters out there. And, the amount of generation must exactly match the load for the frequency to be at rated.

Getting back to our bicyclists, if neither rider is paying attention to the speed of the bicycle, just pedaling along applying the same amount of torque to their pedals, and the load changes then the speed will change. If the load increases the speed will decrease, and if the load decreases the speed will increase.

Same with our two generators. Let's say the governors of both prime movers are set to operate in Droop control mode and the load at that time is relatively constant. Let's say the droop setting of both governors is the same, 5%. And let's say that one machine is carrying 10 MW of load and the other machine is carrying 5 MW, making the total load equal to 15 MW. If the load on the power island increases to 18 MW, but because both machines are being operated in Droop control mode, so they are not going to increase their fuel to try to maintain the frequency; they will increase the fuel in order to maintain the load, and since the droop setting of both machines is the same they will split the increase in load between them with each machine's load increasing by 1.5 MW (half the 3 MW increase). However, the frequency of both machines will decrease to 49.75 Hz.

Now, the operators could "increase" the Droop speed/load setpoint of one or both machines until the frequency returned to 50 Hz, but that's not "automatic" control, and the frequency changes any time the load changes until the operators respond to the load change to return the frequency to normal.

Again, all of the above presumes a completely isolated power island, with no ties to any external (local or remote) load-sharing or frequency control scheme. It's pretty "ideal" and probably won't describe exactly what happens in a real situation with reactive loads and machines with different inertias and electrical characteristics and different prime movers, but it's a decent explanation of what is intended to happen under ideal conditions for an isolated power island being supplied by two prime movers and generators of roughly equal rating and similar droop setpoints responding to changes in load (relatively slow and incremental changes, not fast, step-changes) with well-tuned governors.

To the originator, rahul, please provide some clarification and some indication of whether or not this is the kind or reply you were looking for. I hope it was helpful for others who may have similar curiosities and questions, but, we're trying to answer your question(s) and we need to know if we've done so. So, we're awaiting some kind of clarification and response.

 Posted by S.N.BHATTACHARYYA on 26 May, 2012 - 7:37 am

i did not understand one point. you mentioned an export of 2MW. that means you were connected to the grid. then how could you set one generator in "speed" mode. the grid controls the speed. is the plant load steady or fluctuating?

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