E
I would like to know what happens to the power produced by a synchronous generator, if the load is of power factor 1 (resistive), and the excitation is increased above that required to produce the voltage of the grid?
This is a difficult question to answer, even though it seems it should be very straightforward. But, here goes, anyway, ("Once more, unto the breach!").
Are you familiar with the power triangle? The kW side of the power triangle (the real power side) is a function of the amount of power being developed by the synchronous generator's prime mover and applied to the synchronous generator. And when the power factor of a synchronous generator is 1.0 (unity) then the hypotenuse of the power triangle (the kVA side, or the apparent power side) is exactly equal to the kW side of the triangle, and the kVAr side (the reactive power side) of the power triangle is 0 (zero). So, if the amount of power being developed by the prime mover and being applied to the generator remains constant, then the kW side of the power triangle will remain constant. And if the power factor is unity (1.0) then virtually all of the power going into the synchronous generator from the prime mover is going out as real power (Watts; kW; MW).
When the excitation being applied to the field of the synchronous generator is increased above the amount required to maintain unity power factor (1.0) while holding the kW (real power) side of the triangle fixed the kVAr (reactive power) side of the triangle increases. Now, there are mixed thoughts about what happens to the kW (real power) side of the power triangle when this happens--mostly because most textbooks, references and prognosticators forget about where the power for the excitation comes from. It doesn't come from a completely separate source--in which case in my understanding of how a synchronous generator works would mean that the kW (real power) output of the synchronous generator would remain fixed.
But, most synchronous generators obtain the power for the excitation from the synchronous generator (either from the generator shaft or from the generator terminals)--which means the power for the excitation system ultimately comes from the generator prime mover. In other words, if the power being applied by the prime mover to the synchronous generator remained unchanged while the excitation was increased, a portion of the power being applied to the synchronous generator from the prime mover would go into the excitation system--meaning that the kW output from the synchronous generator would decrease slightly. How much? Well, that depends on how much real power (Watts; kW; MW) the generator was producing at the time of the excitation change--and how much the excitation was increased. The decrease in real power when changing excitation can be seen more easily when the real power output is low, and not so easily when the real power output is high.
You don't get something for nothing in this world. In my understanding of how synchronous generators work, and how the power triangle can be used to explain how synchronous generators work (which does NOT take into account where the excitation power comes from!), if one holds the total power out of the synchronous generator constant (so the hypotenuse of the power triangle, the kVA (apparent power) side of the power triangle is fixed) while also holding the kW (real power) side of the power triangle fixed while increasing the kVAr side (the reactive power) of the power triangle, then something has to give way--either the kVA decreases or the kW decreases.
Again, I want to point out that most synchronous generators (every one I've ever worked for more than 30 years) gets the power for the excitation system from the synchronous generator in some fashion--either from the generator rotor (rotating, or "brushless" exciter), or from the generator output terminals (a potential (voltage) source system, or a current-transformer system), but from the generator. Which means that the prime mover is ultimately providing the power for the excitation system--and when one increases the excitation power, while holding the power being applied by the prime mover to the synchronous generator fixed, then the power output of the generator has to decrease.
Don't ask me for the maths of all this--I've spent decades arguing with theory types about how generators REALLY work, not how they theoretically work in textbooks and references. But, these are the observations I have repeatedly seen over more than three decades of working on power generation equipment. The changes in real power are always more apparent at low real power outputs, and are much more difficult to observe/quantify at larger real power outputs.
Hope this helps!
Are you familiar with the power triangle? The kW side of the power triangle (the real power side) is a function of the amount of power being developed by the synchronous generator's prime mover and applied to the synchronous generator. And when the power factor of a synchronous generator is 1.0 (unity) then the hypotenuse of the power triangle (the kVA side, or the apparent power side) is exactly equal to the kW side of the triangle, and the kVAr side (the reactive power side) of the power triangle is 0 (zero). So, if the amount of power being developed by the prime mover and being applied to the generator remains constant, then the kW side of the power triangle will remain constant. And if the power factor is unity (1.0) then virtually all of the power going into the synchronous generator from the prime mover is going out as real power (Watts; kW; MW).
When the excitation being applied to the field of the synchronous generator is increased above the amount required to maintain unity power factor (1.0) while holding the kW (real power) side of the triangle fixed the kVAr (reactive power) side of the triangle increases. Now, there are mixed thoughts about what happens to the kW (real power) side of the power triangle when this happens--mostly because most textbooks, references and prognosticators forget about where the power for the excitation comes from. It doesn't come from a completely separate source--in which case in my understanding of how a synchronous generator works would mean that the kW (real power) output of the synchronous generator would remain fixed.
But, most synchronous generators obtain the power for the excitation from the synchronous generator (either from the generator shaft or from the generator terminals)--which means the power for the excitation system ultimately comes from the generator prime mover. In other words, if the power being applied by the prime mover to the synchronous generator remained unchanged while the excitation was increased, a portion of the power being applied to the synchronous generator from the prime mover would go into the excitation system--meaning that the kW output from the synchronous generator would decrease slightly. How much? Well, that depends on how much real power (Watts; kW; MW) the generator was producing at the time of the excitation change--and how much the excitation was increased. The decrease in real power when changing excitation can be seen more easily when the real power output is low, and not so easily when the real power output is high.
You don't get something for nothing in this world. In my understanding of how synchronous generators work, and how the power triangle can be used to explain how synchronous generators work (which does NOT take into account where the excitation power comes from!), if one holds the total power out of the synchronous generator constant (so the hypotenuse of the power triangle, the kVA (apparent power) side of the power triangle is fixed) while also holding the kW (real power) side of the power triangle fixed while increasing the kVAr side (the reactive power) of the power triangle, then something has to give way--either the kVA decreases or the kW decreases.
Again, I want to point out that most synchronous generators (every one I've ever worked for more than 30 years) gets the power for the excitation system from the synchronous generator in some fashion--either from the generator rotor (rotating, or "brushless" exciter), or from the generator output terminals (a potential (voltage) source system, or a current-transformer system), but from the generator. Which means that the prime mover is ultimately providing the power for the excitation system--and when one increases the excitation power, while holding the power being applied by the prime mover to the synchronous generator fixed, then the power output of the generator has to decrease.
Don't ask me for the maths of all this--I've spent decades arguing with theory types about how generators REALLY work, not how they theoretically work in textbooks and references. But, these are the observations I have repeatedly seen over more than three decades of working on power generation equipment. The changes in real power are always more apparent at low real power outputs, and are much more difficult to observe/quantify at larger real power outputs.
Hope this helps!
E
EGACC
Thank you Mr. CSA, your explanation is very clear and I understand all your comment. I know this is a hypothetical (not real) situation, but only to better understand the behavior of the synchronous generators connected to the grid, and I would like you to help me, and I apologize for insisting on this non-real case.
If the driver torque could be increased, the real power could increase, and if the excitation increases, then, the VARs increase, but since the load is totally resistive (FP = 1), where are the VARs going? Thank you again and i appreciate so much your opinion.
If the driver torque could be increased, the real power could increase, and if the excitation increases, then, the VARs increase, but since the load is totally resistive (FP = 1), where are the VARs going? Thank you again and i appreciate so much your opinion.
EGACC,
Generators are devices for converting torque into amperes; motors are devices for converting amperes into torque. Light bulbs are devices for converting amperes into heat. And computers, ..., well, computers are devices for converting amperes into virtual torque. ;-)
Per your first post, we are talking about generators connected-<i>synchronized to</i>--a grid, so this applies only to synchronous generators synchronized to a large AC power transmission/distribution grid. When the generator breaker is closed if the real power being produced by the generator is zero--that is, the amount of torque being applied by the prime mover to the generator rotor is exactly equal to what's required to keep the generator spinning at synchronous speed--then the real power (Watts, kW, MW) are zero. Increasing the torque applied by the prime mover increases the real power output. In effect, what is happening when the generator is being "loaded" is that the generator is taking on some of the total load on the grid--all of the motors and lights and computers and televisions. If one could see the total load on the grid and the grid frequency as a generator is loaded <i>if no other generators are unloaded</i> the load will not change (the total number of motors and lights and computers and televisions doesn't change) BUT the frequency will tend to increase. So, the grid operators have to reduce the load on some other generator(s) in order to keep the grid frequency constant or near desired. This is one reason why grid operators usually like to know when synchronous generators are being synchronized and loaded, and when they are being unloaded (when possible!) so they can balance the generation on the grid to maintain grid frequency.
Anyway, synchronous generators have a second capability--to "produce" or "consume" VArs, reactive power. And as you have noted, that is done by increasing or decreasing the amount of excitation above or below the amount required to make the generator terminal voltage exactly equal to the grid voltage. (When the amount of excitation being applied to the synchronous generator rotor is exactly equal to that required to make the terminal voltage equal to the grid voltage, the power factor is 1.0, and the reactive current is zero (no leading and no lagging reactive current).
On an AC grid, the inductive and capacitive loads affect the power factor of the load, the reactive power of the load. When the excitation of a synchronous generator is increased above that required to maintain the generator's terminal voltage equal to the grid voltage what happens is that the generator "produces" VArs, which "feed" the reactive load of the grid. If one could see the power factor of the entire load one would see the power factor change as synchronous generator excitation is changed. Again, the nature of the loads isn't changing--the inductances and capacitances aren't changing when synchronous generator excitation is changing--but the relationship of the voltage- and current sine waves change. And, that is what changes the power factor of the load.
If the majority of loads on an AC system are inductive then the grid requires some lagging VArs from synchronous generators to keep the voltage and current sine waves from getting too far out of phase with each other (which will lead to brown-outs and black-outs). Power factor correction capacitors can also be used--but they are typically not smoothly variable--they can only be added or subtracted in "chunks" (which can literally be felt as "bumps" when in power stations or sub-stations if the capacitor banks are nearby). (The same is true for a highly capacitive system load, the sine waves just shift in the opposite direction.)
So, when excitation is increased on a synchronous generator above that required to maintain unity power factor (1.0) (which is the same as keeping the generator terminal voltage equal to grid voltage) the generator "produces" reactive current (lagging VArs, from a generator's perspective) to overcome some of the inductive loads on the grid causing the grid power factors to be less than 1.0. And because most AC systems have reactive loads (inductive and capacitive) there is a "need" for some VArs. IN other words, most systems are easily capable of accepting some VArs to help with maintaining power factor (which is another way of expressing the phase shift between the voltage- and current sine waves--isn't this electrical stuff fun?!?!?!).
Again, we're talking about AC power transmission and distribution grids--not islanded operation. And, we're talking about larger grids/systems, as opposed to smaller systems. And islanded systems present whole 'nother set of challenges and opportunities, and since they're not very common, let's not get into that area. Please. This is as much theory as I can handle.
And, grid operators also have to pay very close attention to grid power factor and reactive power. They can also ask or require power plants to change their excitation to support grid stability and maintain a proper relationship between the voltage- and current sine waves on the grid. Which is as important as the frequency, which is a function of load. (If there isn't enough generation on the grid (meaning the total amount of generation doesn't match the total load (motors, lights, computers, televisions, etc.) then the grid frequency will decrease--because some of the torque that goes into maintaining frequency (speed) has to go into supporting the load (Watts, kW, MW).)
Lastly, motors and even transformers "require" VArs to operate. Induction motors, in particular, cause the shift between the voltage- and current sine waves, and if there are a LOT of induction motors operating on a grid (water pumps; air conditioner compressor motors and fan motors; electric fans; refrigerator compressor motors; etc.) without some source of VArs to "counter" them then the voltage- and current sine waves will get too far out of phase, and first brown-outs and in some extreme cases black-outs can occur. So, grid operator have to make sure that there is some "source" of VArs to overcome the effect of the motors and inductive loads so keep the sine waves in phase. And, the easiest way to do this is to change synchronous generator excitation.
So, if I understood your question correctly when a synchronous generator is connected to a grid and the excitation is increased above a unity power factor the generator then "produces" some lagging VArs, which are "consumed" by lagging loads. Really, the generator VArs are overcoming the effect of the lagging VArs from the inductive loads on the system. It's not exactly technically correct to say VArs are "produced" and "consumed" but it's a convenient way to visualize what's happening. (Lagging VArs are considered to be "pushed" out of the synchronous generator--produced; and leading VArs are considered to be "sucked" into the synchronous generator--consumed.) Too much inductive load without any lagging VArs from some source will, again, cause the sine waves to shift too far out of phase with each other. And that's the effect of VArs. So, adding VArs to a system (powering more inductive motors) will shift the sine waves further out of phase--unless some source of reactive current is used to return the sine waves to more of an in-phase relationship. So, VArs can also be thought of as being consumed by reactive loads, and as being produced by synchronous generators (or power factor correction capacitors).
Hope this helps! (Remember, I advised this was not an easy question, and I hope my attempts at making this easier to visualize and understand don't offend readers who think and speak in terms of vectors and maths and emfs and counter-emfs and formulae.)
Generators are devices for converting torque into amperes; motors are devices for converting amperes into torque. Light bulbs are devices for converting amperes into heat. And computers, ..., well, computers are devices for converting amperes into virtual torque. ;-)
Per your first post, we are talking about generators connected-<i>synchronized to</i>--a grid, so this applies only to synchronous generators synchronized to a large AC power transmission/distribution grid. When the generator breaker is closed if the real power being produced by the generator is zero--that is, the amount of torque being applied by the prime mover to the generator rotor is exactly equal to what's required to keep the generator spinning at synchronous speed--then the real power (Watts, kW, MW) are zero. Increasing the torque applied by the prime mover increases the real power output. In effect, what is happening when the generator is being "loaded" is that the generator is taking on some of the total load on the grid--all of the motors and lights and computers and televisions. If one could see the total load on the grid and the grid frequency as a generator is loaded <i>if no other generators are unloaded</i> the load will not change (the total number of motors and lights and computers and televisions doesn't change) BUT the frequency will tend to increase. So, the grid operators have to reduce the load on some other generator(s) in order to keep the grid frequency constant or near desired. This is one reason why grid operators usually like to know when synchronous generators are being synchronized and loaded, and when they are being unloaded (when possible!) so they can balance the generation on the grid to maintain grid frequency.
Anyway, synchronous generators have a second capability--to "produce" or "consume" VArs, reactive power. And as you have noted, that is done by increasing or decreasing the amount of excitation above or below the amount required to make the generator terminal voltage exactly equal to the grid voltage. (When the amount of excitation being applied to the synchronous generator rotor is exactly equal to that required to make the terminal voltage equal to the grid voltage, the power factor is 1.0, and the reactive current is zero (no leading and no lagging reactive current).
On an AC grid, the inductive and capacitive loads affect the power factor of the load, the reactive power of the load. When the excitation of a synchronous generator is increased above that required to maintain the generator's terminal voltage equal to the grid voltage what happens is that the generator "produces" VArs, which "feed" the reactive load of the grid. If one could see the power factor of the entire load one would see the power factor change as synchronous generator excitation is changed. Again, the nature of the loads isn't changing--the inductances and capacitances aren't changing when synchronous generator excitation is changing--but the relationship of the voltage- and current sine waves change. And, that is what changes the power factor of the load.
If the majority of loads on an AC system are inductive then the grid requires some lagging VArs from synchronous generators to keep the voltage and current sine waves from getting too far out of phase with each other (which will lead to brown-outs and black-outs). Power factor correction capacitors can also be used--but they are typically not smoothly variable--they can only be added or subtracted in "chunks" (which can literally be felt as "bumps" when in power stations or sub-stations if the capacitor banks are nearby). (The same is true for a highly capacitive system load, the sine waves just shift in the opposite direction.)
So, when excitation is increased on a synchronous generator above that required to maintain unity power factor (1.0) (which is the same as keeping the generator terminal voltage equal to grid voltage) the generator "produces" reactive current (lagging VArs, from a generator's perspective) to overcome some of the inductive loads on the grid causing the grid power factors to be less than 1.0. And because most AC systems have reactive loads (inductive and capacitive) there is a "need" for some VArs. IN other words, most systems are easily capable of accepting some VArs to help with maintaining power factor (which is another way of expressing the phase shift between the voltage- and current sine waves--isn't this electrical stuff fun?!?!?!).
Again, we're talking about AC power transmission and distribution grids--not islanded operation. And, we're talking about larger grids/systems, as opposed to smaller systems. And islanded systems present whole 'nother set of challenges and opportunities, and since they're not very common, let's not get into that area. Please. This is as much theory as I can handle.
And, grid operators also have to pay very close attention to grid power factor and reactive power. They can also ask or require power plants to change their excitation to support grid stability and maintain a proper relationship between the voltage- and current sine waves on the grid. Which is as important as the frequency, which is a function of load. (If there isn't enough generation on the grid (meaning the total amount of generation doesn't match the total load (motors, lights, computers, televisions, etc.) then the grid frequency will decrease--because some of the torque that goes into maintaining frequency (speed) has to go into supporting the load (Watts, kW, MW).)
Lastly, motors and even transformers "require" VArs to operate. Induction motors, in particular, cause the shift between the voltage- and current sine waves, and if there are a LOT of induction motors operating on a grid (water pumps; air conditioner compressor motors and fan motors; electric fans; refrigerator compressor motors; etc.) without some source of VArs to "counter" them then the voltage- and current sine waves will get too far out of phase, and first brown-outs and in some extreme cases black-outs can occur. So, grid operator have to make sure that there is some "source" of VArs to overcome the effect of the motors and inductive loads so keep the sine waves in phase. And, the easiest way to do this is to change synchronous generator excitation.
So, if I understood your question correctly when a synchronous generator is connected to a grid and the excitation is increased above a unity power factor the generator then "produces" some lagging VArs, which are "consumed" by lagging loads. Really, the generator VArs are overcoming the effect of the lagging VArs from the inductive loads on the system. It's not exactly technically correct to say VArs are "produced" and "consumed" but it's a convenient way to visualize what's happening. (Lagging VArs are considered to be "pushed" out of the synchronous generator--produced; and leading VArs are considered to be "sucked" into the synchronous generator--consumed.) Too much inductive load without any lagging VArs from some source will, again, cause the sine waves to shift too far out of phase with each other. And that's the effect of VArs. So, adding VArs to a system (powering more inductive motors) will shift the sine waves further out of phase--unless some source of reactive current is used to return the sine waves to more of an in-phase relationship. So, VArs can also be thought of as being consumed by reactive loads, and as being produced by synchronous generators (or power factor correction capacitors).
Hope this helps! (Remember, I advised this was not an easy question, and I hope my attempts at making this easier to visualize and understand don't offend readers who think and speak in terms of vectors and maths and emfs and counter-emfs and formulae.)
E
EGACC
CSA I appreciate so much your excellent answer. it's very clear and very explanatory! As I said before I know it is not a real case, but it helps me better understand the operation of the synchronous generator connected to the Grid.
Thank you
Thank you
| Thread starter | Similar threads | Forum | Replies | Date |
|---|---|---|---|---|
|
Generator Power Factor drops to 0.3 On Inductive Load. | Power Generation | 2 | |
| L | Power Factor and Load Angle in Synchronous Generator | Power Generation | 0 | |
| T | Load Angle and Power Factor | General Automation Chat | 3 | |
| I | Power factor, frequency and load | Power Generation | 4 | |
| B | Power Factor During Load Sharing | General Automation Chat | 0 |
