So i got this while i was searching on how load is controlled on Mark V. The logic is understandable but making sense to field equipments needs further digging. however this information is quite useful. Please refer to Control.com for further information.
What makes a generator increase load
Posted by kumarash on 10 August, 2008 – 3:49 pm
If a generator is running at 100MW of a gas turbine rated 120MW and a command is issued from Mark V to increase load to 120MW. What happens to generator that it increases load to 120MW?
Posted by CSA on 12 August, 2008 – 1:01 am
Isn’t it obvious? It does so because it is commanded to do so, you said so yourself. All good generators do as commanded, don’t they?
Seriously, though, and in as few words as possible, when an operator clicks on Raise Spd/Ld or increases the Pre-Selected Load Control setpoint when the unit is being operated in Pre-Selected Load Control mode the fuel flow-rate into the turbine is increased, which increases the torque being produced by the turbine, which the generator converts to amps, which results in an increase in load.
For the exact details of what’s happening, read on.
If one wants to make something spin, one needs to supply some force to it. That force is usually referred to as torque. The more torque applied to something, the faster it will usually spin. Decrease the torque applied to something, and it will usually slow down.
A gas turbine is a device that produces torque, and the amount of torque being produced can be varied, and is in direct proportion (usually) to the amount of fuel being burned in the combustor(s) of the gas turbine. Increase the fuel flow-rate, and the amount of torque being produced by the turbine will increase. Decrease the fuel flow-rate, and the amount of torque being produced by the turbine will decrease.
The generator (more correctly called an alternator) used with GE-design heavy duty gas turbines is a synchronous generator. It is the nature of synchronous generators being operated in parallel with other synchronous generators on an AC grid to supply a load which is much greater than any single generator that they will all be running at the same speed, and no single generator and its prime mover (the gas turbine in your case) can operate at speeds higher or lower than the other generators with which is it connected, because they are synchronous generators (alternators).
There is a formula that describes the relationship between the frequency of an AC system and the speed and type of synchronous generators connected to the system: F = (P * N)/120, where F is the frequency (in Hertz) of the grid, P is the number of poles of the generators (an even number never less than two, and the number of poles of any generator is usually fixed and can’t be changed while the generator is running), and N is the speed of the generator rotor, usually the field (in RPM). The (synchronous) generators used with Frame 9E GE-design heavy duty gas turbines are two-pole generators and are directly connected to the turbines (i.e., there is no reduction gear between the turbine and the generator, and the generator is not connected to a “free” turbine which spins independently of the axial compressor of the turbine), so that’s why the 100% speed rating of the turbine is 3000 RPM because the grid the generator is connected to is a 50 Hz system (nominally). (Note there are no approximations in this formula, and note that speed and frequency are directly proportional. Increase the speed of a synchronous generator and it’s frequency will increase; decrease the frequency of a grid to which a generator is connected and the generator speed will decrease proportionally; speed and frequency are directly related.
So, we’ve established that when a two-pole (synchronous) generator is connected to a 50 Hz grid with other generators that it will be operating at 3000 RPM (N = (120 * 50)/2 = 3000 RPM), and when the prime mover (the turbine) is directly coupled to the generator that it, too, will spin at 3000 RPM. Again, this is because the generator rotor is locked into the same speed (in synchronism with) the other generators at a speed that is dictated by the frequency of the grid and the number of poles of the generator (which never changes), and the turbine shaft is directly connected/coupled to the generator rotor.
Further, it is the nature of most AC grids that there are so many generators, of various number of poles, so many all supplying such a large load that any one single generator and its prime mover can’t usually have any appreciable effect on the frequency of all the other generators and the speeds of their prime movers (unless the prime mover being loaded or unloaded has a very large power rating/capacity in relation to all the other generators and the load).
When a generator is accelerated to rated speed during starting prior to synchronization, the prime mover is producing just enough torque to keep the generator (and turbine and compressor, in this case) speed equal to the frequency of the grid (in this case, 3000 RPM). When the unit is synchronized, the power “output” of the generator is very low, usually, because during synchronization the turbine speed (and generator frequency) is adjusted to be just slightly higher than a speed which would be equal to grid frequency. This means that there is slightly more torque being produced by the turbine than is required to keep the generator rotor spinning at a frequency equal to grid frequency. When the generator breaker is closed, the speed of the generator rotor, and the turbine and axial compressor, actually slows down. Yes, that’s right; the speed decreases because once the generator breaker closes the generator is then locked into synchronous speed which is a speed that is proportional to the grid frequency of all the other generators it’s now operating in parallel with.
During synchronization, the fuel flow-rate, on the other hand, is held constant once the desired speed is achieved, which means that the torque is being held constant. During synchronization, when the turbine (and generator) speed is increased to be just slightly above grid frequency there is more torque being produced than is required to keep the rotor spinning at a speed that is equal to grid frequency. When the generator breaker closes and the generator speed slows down, which slows down the turbine, the torque remains the same because the fuel was not changed. That extra torque is converted into amps by the generator, and more amps equals more load. That’s what a generator is: a device for converting torque into amps. (A motor is a device for converting amps into torque. Actually, the only difference between a motor and a generator is the “direction” of current- and torque flow into or out of the machine.)
The basic formula for power (Watts) is: P = V * I, where P is power (in Watts), V is generator terminal voltage (in Volts), and I is armature current (the alternating current flowing in the stator of the generator). (For a three-phase generator the entire formula is P = V* I * (3^(0.5)) * PF, where, 3^(0.5) is the square root of three (a fixed value, 1.732, I think), and PF is the power factor of the generator (which is a number never greater than 1.0, and which we will presume to be 1.0 for the purposes of our discussion). Coincidentally, the terminal voltage of most synchronous generators is almost a fixed value, as well, and doesn’t usually vary by more than approximately +/- 5.0%, which on an 11,0000 Volt generator is only about 550 Volts (out of 11,000).
So, since one of the terms of the three-phase power formula *is* a fixed value (the square root of three), and we are presuming one of the terms (PF) to be fixed and equal to 1.0, and the generator terminal voltage is, for all intents and purposes, a fixed value, the way that a generator produces more power is to increase the number of amps flowing in the stator. The way that amps are increased in the generator stator is by providing more torque from the turbine into the generator; more torque equals more amps. Less torque equals less amps. (We are presuming that the prime mover is always producing at least sufficient torque to keep the generator rotor spinning at synchronous speed. When it doesn’t, the generator actually becomes a motor and keeps spinning at synchronous speed and draws current from other generators on the grid. This is what’s known as “reverse power” or, “motorizing the generator.” It’s *very* bad for steam turbines and reciprocating engines even for very short periods of time; not so injurious for gas turbines for short periods of times, especially single-shaft gas turbines like GE-design heavy duty Frame 9E gas turbines.)
Now some people are going to say that when a unit with a GE Speedtronic turbine control system is automatically synchronized to the grid that it loads up to “Spinning Reserve”, and they are correct. There is some logic that, once the generator breaker is closed during auto synch’ing, increases the amount of fuel being put into the turbine which increases the torque being produced by the turbine and since the turbine speed can’t increase (because it’s directly coupled to the (synchronous) generator which is now connected to the grid and it’s speed is being controlled by the frequency of the grid to which it’s connected) the extra torque that’s being produced by the addition of the fuel gets converted into amps.
So, it should be clear that torque, amps, and load are proportional to each other. An increase in fuel results in an increase in torque which results in an increase in load; a decrease in torque results in a decrease in amps which results in a decrease in load.
Now. for a GE-design heavy duty gas turbine to be operated in parallel with other generators on a grid, it is operated in Droop speed control mode. This is one of two governing modes for most prime movers (Isochronous being the other one) and is the mode that allows the turbine and generator to smoothly and stably participate in powering a large load while paralleled with other generators. (Some people refer to this as “sharing load”, and while that’s technically correct the same term is also used in another description of Droop speed control and this double usage causes lots of problems for most people. So, we’re going to refer Droop speed control mode as the governor (control system) mode that allows the stable and smooth production of power by a prime mover and generator when connected to a grid with other generators.)
We discussed what happens during synchronization, when the turbine speed is increased to make the generator frequency just slightly higher than the grid frequency (this is generally referred to as speed matching). And to increase the speed requires an increase in torque, which comes from increasing the fuel flow-rate to the turbine. When the generator breaker is closed, the turbine speed can’t change and any attempt to increase the turbine speed will just cause additional torque to be produced, and the generator converts the torque to amps, which becomes load.
When a GE-design heavy duty gas turbine with a Speedtronic turbine control panel is being operated in Droop speed control mode, and the operator wants to raise or lower the load, what happens is that the turbine speed reference is increased or decreased, which causes the fuel flow-rate to be increased or decreased, in an attempt to make the actual turbine speed increase or decrease by increasing or decreasing the amount of torque being produced by the turbine. But, since the actual turbine speed can’t increase or decrease, any torque increase results in an increase in amps which results in an increase in load, and any torque decrease results in a decrease in amps which results in a decrease in load.
So, when an operator clicks on Raise Spd/Ld or increases the Pre-Selected Load Setpoint when the unit is being operated in Pre-Selected Load Control, what’s really happening is that the turbine speed reference is increasing, which results in more fuel being admitted to the turbine, which results in extra torque which can’t result in increased speed so the generator converts the torque to amps, which results in an increase in load.
And all good generators do as commanded, especially when it’s at the behest of a Mark V Speedtronic turbine control system.
Posted by Mikas on 12 August, 2008 – 10:36 pm
That is great explanation CSA.
I’d like to extend this story on steam turbines. I believe that basic logic is the same: what is fuel with gas turbines, that is steam with steam turbines. If generator’s circuit breaker is open, then more steam supplied to the turbine will cause turbine to rotate faster, but after immediately after synchronization, more steam will produce more torque and because, speed is constant, there will be more amps and hence, more power output to the grid.
However, there are some things that are not clear to me. Generator’s output power can be increased by adding more steam to the turbine, say, by opening turbine’s control valves (I assume turbine has only high pressure and low pressure stage). I think that turbine can increase its torque also by not opening control valves, but increasing steam pressure which is equivalent.
I wonder what approach is used more frequently.
I also believe it has a lot of in common with so called “exploatation concept” (turbine or boiler leading).
In turbine leading mode, turbine’s controller is controlling power output by adjusting control valves. That, inevitably, leads to steam’s pressure changing. In such case, boiler controller is increasing or decreasing coal feeders speed. More speed means more coal in the boiler. More coal in the boiler means more heat, more heat means higher pressure (or more steam!?!).
Am I on the right track?
Posted by nameless on 29 June, 2012 – 12:15 am
hey, adding more heat doesn’t raise the pressure in this case because theoretically the steam generation is isobaric (in steam turbines). The pressure only comes from the pumps that pump the condensed water before it is being heated up into steam.
Posted by electrics [Control.com Member] on 12 June, 2009 – 2:05 pm
I ponder on this issue for a long time, there are a few questions if you allow me to ask sir.
now, if we increase the load of a syn. generator while opeating under normal conditions and doesn’t increase the fuel so what happens? i think if we don’t increase the fuel any so the generator will begin to slow and it will stop till the kinetic energy of the primemover and alternator is consumed totally and the generator will have so big currents true? but this is just a amateurish guess pls tell me what happens if frequency controller doesnt intervene with the load incrase? in theory now the load is not equal the power being produced so what happens? this is my first question.
my second question is, they always say that the generator speed gets slower if load increases, so this is a matter reagrding frequency controller or the nature of a sync. generator?
“The mechanical source of power for the generator is a prime mover such as diesel engines or steam, gas, water, and wind turbines. All prime movers behave in a similar fashion. As the power drawn from them increases, the rotational speed decreases. In general, this decrease in speed is nonlinear. However, the governor makes this decrease in speed linear with increasing power demand.” for example this is a passage from a book, so pls tell me how can a sync. gen can get slower as load increases? what makes it to slow down?
I also wanna ask you as third question, a diesel generator has a slightly higher frequency than 50 hz while working no-load true? so does it make difference between working alone and working in synchronisation with an infinite grid? if true why need to be a bit higher than 50 ?
pls englihten me sir…
Posted by klarg on 1 May, 2011 – 12:30 pm
Many people have a hard time grasping “why more torque results in more armature (stator) amps?” That is, “what is happening on an electrical basis in the armature that results in greater current flow?”
For some reason there are more explanations written on the synchronous motor dynamics than the synchronous generator dynamics. (If I goof in re-stating the case for the generator please correct the narrative or terminology).
When more torque is applied to the rotor the torque angle (angular displacement between the center of the rotor and stator poles) increases.
The rotor speed does not change, but the torque angle increases (technically some momentary delta V must exist to cause the angle to increase).
Because of the torque angle increase the phase angle between the voltage impressed on the stator from the grid, and the voltage impressed on the stator from the (rotor) field is also increased. The resolved sum of these voltage vectors results in a higher net voltage. This higher ‘net’ voltage is what draws a greater current in the stator.
Posted by CSA [Control.com Most Valuable Participant – 247 helpful votes.] on 2 May, 2011 – 11:33 am
I always find it interesting that no one questions how current is converted to torque in a motor, but they have great difficulty in understanding how torque is converted to current in a generator.
I always love to ask the question: Why do we generate electricity? The responses range from “the deer in the headlights” look with silence, to a wrinkled brow with silence–but always silence.
The exact physical principles are difficult to describe, except to theoretical in-their-head types who like maths and vector diagrams. Most power plant operators and technicians just can’t get their arms (or heads) around such concepts, and yet neither can they understand that the generator actually drives the motors and lights and computers that they switch on and off in the power plant, and many more motors and lights and computers in far-flung locations.
Electricity is what allows work to be done in many different places by producing work in a centralized places (power plants). The torque produced by the prime mover is converted to amps in the generator, and wires connect motors (and lights and computers) to the generator(s) and work is done at the remote locations connected to the generator(s) by wires. It’s very, very similar to hydraulics. Work is used to produce high pressure in the pump and that pressure is used to do work at some remote location(s) by connecting the devices at the remote locations to the pump with pipes or hoses.
Electricity is just a way of producing power in one place (the power plant) and transmitting it to many different places (via the electrical grid) to accomplish work. Otherwise, we would have bazillions of generators at every place we needed work to be done. The prime mover driving the generator at the power plant(s) actually does the work at the various remote locations, connected to the devices at the remote locations by wires.
Posted by Enric_Tgn on 12 August, 2008 – 2:34 am
In essence, you are increasing the torque applied on the generator shaft (power, but speed is constant) And this increases the angle between Eo (induced voltage in stator) and the voltage at output terminals (theoretical U and f, constants, and connected to an infinite power net). This angle is torque-dependant (called torque angle) more angle > more power delivered > more power drained from driving machine. If the generator is working as a motor, this angle is negative. If the machine is under no load condition (open circuit) this angle is 0.
Posted by Kushwaha_r_k [Control.com Member] on 27 October, 2008 – 4:12 pm
It’s true that more fuel to prime mover increases the torque, which increases the load angle in the generator. It’s also true that magnetic field strength of rotor increases as load angle increases, which causes to increase the current through stator windings thus the MW. But what causes the rotor current to increase, i.e. excitation current to increase?
Posted by CSA [Author has 11 helpful votes.] on 28 October, 2008 – 5:21 am
Rotor field strength is a function of excitation current from the synchronous generator exciter, sometimes referred to as the “AVR” (Automatic Voltage Regulator). The AVR, when operating in automatic mode, is attempting to keep the generator terminal voltage equal to the generator terminal voltage setpoint by varying the excitation current applied to the generator rotor which is directly proportional to the rotor field strength.
As the current in the stator windings increases, the tendency is for the generator terminal voltage to decrease (because of what’s called “armature reaction”), but the AVR will increase the excitation current which increases rotor field strength which overcomes armature reaction to keep the generator terminal voltage constant.
So, I take exception to your statement that it’s true “that magnetic field strength of rotor increases as load angle increases, which causes to increase the current through stator windings thus the MW”. One cannot appreciably change the load (MW) of a synchronous generator by changing the rotor field strength; it’s done by changing the torque being applied to the generator.
The increase in rotor field strength is the *result* of increased stator amps which are tending to decrease generator terminal voltage, not the cause of increased stator amps. If the stator amps are increased (by increasing the torque applied to the generator rotor) and nothing was done to change the rotor field strength the generator terminal voltage would decrease and the reactive power output of the generator would change.
Remember the formula for the power produced by a three-phase synchronous generator: P = V * I * (3^(0.5) * PF. Look at the rating of the generator’s terminal voltage; it’s usually plus or minus 5% of some nominal value, say 11 KV, or 13.8 KV. So, about the maximum effect one can have on the output of the generator with the rotor field strength is to change it by plus or minus 5%, which is very little, and that’s only under theoretically ideal conditions.
So, one does not usually change the load (MW) of a synchronous generator by changing rotor field strength; that’s done by changing the amount of torque being applied to the generator rotor. If the torque is increased, the amperes flowing in the generator stator windings increase, which increases the armature reaction which tends to decrease the generator terminal voltage. So, to maintain generator terminal voltage, the exciter regulator output is increased which causes more rotor field strength.
If you really want to understand armature reaction, you can send your name, the name of the company you are employed by (or he’ll probably take the name of your school if you’re a student), and your email address to email@example.com and request a copy of his armature reaction document.
Posted by ssv on 31 March, 2010 – 10:11 am
While increasing fuel flow, torque will increase, which in turn will increase torque angle and which in turn will increase more current at the stator and so MW which in turn will increase armature reaction and which in turn will increase rotor field current to maintain the terminal voltage. All this is clear.
Now my confusion is as follows:
As we know that Torque = Force * angular displacement. Now for increasing load we just increase fuel to the turbine so as to get more torque. But my question is we don’t have the angular displacement of the rotor because the rotor speed is locked once it is on grid, so as per above equation the Torque would be zero because again the speed is fixed. So How we are increasing the Torque by increasing fuel ? to get our desired MW. What is the mechanism happening which is causing Torque to become more ? even while angular displacement is zero.
Please don’t tell me that torque is increased due to increase in fuel. My query is WHY as per above explanation.
Posted by CSA [Control.com Most Valuable Participant – 247 helpful votes.] on 31 March, 2010 – 1:29 pm
When the generator rotor is locked in synchronous speed with other generators on a large grid it is effectively a “brake” preventing any increase in speed which would normally be the result of an increase in fuel.
When a generator is being synchronized with a grid, it is usually done so with the generator rotor spinning slightly faster than synchronous speed (the synch scope is rotating clock-wise in the Fast direction). When the generator breaker is closed the speed of the generator rotor (and the turbine) slow down to synchronous speed, even though the fuel is held constant. This results in a “positive” power output of the generator. The “extra” speed (torque) that was causing the generator rotor to spin faster than synchronous speed before the generator breaker was closed is converted to amperes in the generator when the rotor speed is reduced to synchronous speed.
(If the synchroscope were held stationary at the 12 o’clock position and the generator breaker were closed, there would be zero power output from the generator, because the turbine is spinning the generator rotor at synchronous speed.)
Another way to think of it is that the turbine buckets are at a fixed distance from the turbine shaft. Hot gases impinging on the turbine buckets develop torque (force x distance, right?). Increasing the temperature of the hot gases increases the force and therefore the (force x distance) and therefore the torque.
But, the generator rotor, because it’s locked into synchronous speed with the grid, can’t be spun any faster even though the turbine rotor is “twisting” the load coupling shaft harder because of the increased torque. That’s the torque angle, the increased twist being applied to try to spin the generator rotor faster than synchronous speed.
Again, torque is force x distance. If the distance is fixed, then increasing the force (by increasing the fuel) will increase the torque.
If a turbine-generator is producing power at say, 25% of rated output, and suddenly the generator breaker were opened but the fuel wasn’t reduced, then the turbine shaft and generator rotor speed would increase very fast, and the unit would probably overspeed. That’s because the amount of fuel being burned to produce the force on the turbine buckets is much more than is required to keep the generator rotor spinning at synchronous speed.
Please have a look at the http://www.wikipedia.org definition of Torque, specifically the section titled ‘Relationship between torque, power and energy’. Note, that the physical and mechanical engineering definitions of torque can be different.
When additional torque is applied through the load coupling between a turbine and a generator, the coupling is “twisted” and the change in angular displacement can actually be measured. Also, there is a change in the relationships of the magnetic fields in the generator (the rotor field and the stator field(s)), an angular displacement.
So, there is angular displacement.
Posted by Bruce Durdle [Control.com Most Valuable Participant – 70 helpful votes.] on 31 March, 2010 – 3:05 pm
My basic mechanics text says that torque =- Force x radius of action. And increasing drive torque will increase the angular displacement by which the rotor field leads the stator field – up to 90 degrees or so.
Posted by zahid on 17 June, 2012 – 4:50 am
Kindly explain in detail, why increase in prime mover output causes increase in load angle and with increase in load angle, how gen. load increases? Is load angle is proportional to gen. load?
Posted by CSA [Control.com Most Valuable Participant – 247 helpful votes.] on 17 June, 2012 – 4:22 pm
You can use your preferred Internet search engine to search for “generator load angle” and “synchronous generator load angle” and find MANY relevant search results. I liked this one best (not because I like maths, or because the grammar and spelling are like mine lately) but because it was quite concise in the use of maths:
Imagine you are twisting a shaft to try to make a rotating mass spin at a constant speed and the speed of that rotating mass can’t change. Imagine there is a brightly painted straight line along the shaft and that you have a strobe light fixed at the same frequency as the rotating shaft allowing you to observe the painted line.
If you applied more torque to the shaft to try to increase (accelerate) the speed of the rotating mass (whose speed is fixed and can’t change) the painted line will begin to “bend” because the additional torque you are applying to the shaft is causing it to deform slightly. The more torque you apply the larger the angular deformation; the less torque applied, the smaller the angular deformation.
When the rotating mass is a synchronous generator connected in parallel with other synchronous generators on a grid whose frequency is being properly maintained, the increased torque from the prime mover tries to accelerate the generator rotor–but the magnetic forces of the grid voltage and current flowing in the generator stator winding and the magnetic forces of the DC current flowing in the rotor windings keep the generator rotor speed fixed. The generator converts the increased torque from the prime mover into amps, which are transmitted over wires to motors and other devices that convert the amps back into torque to produce useful work (or light or heat).
The more torque that is applied to the generator rotor shaft the more twist (the greater the load angle) there will be on the shaft because the prime mover is trying to increase the generator rotor speed but the magnetic forces inside the generator rotor will not let the speed increase and the generator converts the torque that would otherwise have increased the rotor speed into amperes.
There is generally a coupling shaft between most prime movers and generator rotors. These couplings, called load couplings, have to be strong enough to withstand the torsional forces developed during the generation of electrical power when torque is transmitted to generator rotors. They also have to have enough “give” in them yet not break so that rated torque (load) can be applied to the generator rotor so the generator can produce rated power.
So, load angle is directly proportional to the amount of torque being applied to the generator rotor (which is almost never measured or monitored) and also directly proportional to the amount of load current flowing in the generator armature. By the way, load angle is almost never measured or monitored in real life, either. It’s just one of the ways people can make other people think they are smarter by talking about terms that most people never see or have use for.
What you want to remember (for operating prime movers and generators–not necessarily for electrical coursework) is that load is proportional to torque, and torque is a function of the energy being admitted to the prime mover driving the generator. For a steam turbine, it’s the amount and pressure of the steam flowing into and through the turbine. For a gas turbine, it’s the amount of fuel being burned and the amount of air flowing through the turbine. For a hydro turbine, it’s the amount and pressure of the water flowing through the turbine. For a wind turbine, it’s the wind speed and flow-rate. More torque means more load; less torque means less load.
Load angle takes care of itself. Most power plant operators only look at the power output of the plant–which is directly proportional to the amount of current flowing since voltage is maintained relatively constant. When the power plant operator wants more power, they increase the energy input into the prime mover. But, they almost never know what the load angle is.
I’ve only seen one hydro plant that had a torque meter which was used to calculate a load angle–but it was only for demonstration purposes and the operators never bothered with the information from it. It was a “pet project” of some engineer who had long since moved on to another project. It was interesting to watch, but it provided no useful information to the power plant operators.
Load angle is simply another means of describing the amount of torque transmitted from the prime mover to the generator–just as electrical load is also a means of measuring the torque transmitted from the prime mover to the generator. Because load is directly proportional to torque, and vice versa.
It’s a wonderful thing to understand and to be able to describe, but unless you are designing primer movers, generators and/or load couplings, or are involved in the study of electrical power system transmission and stability it’s pretty, well, useless. (Also, electrical coursework where textbooks and professors like to talk about these things as if they were real and were actually part of some control scheme for most generators or their prime movers.)
Whether you realize it or not, when you ride a bicycle the same things are happening all the time. In this case, the load coupling is the chain. When you want to increase the speed of the bicycle you apply more torque to the pedals and the instantaneous effect of that increase in torque is to stretch the chain a little bit. As the speed of the bicycle catches up to the increased torque the chain stretch decreases a little bit. If you were riding up a hill and trying to maintain the same speed on the bicycle you would need, at some point, to increase the torque to maintain the same speed as when you were riding on flat ground, and the chain would stretch as the torque increased. Unless you changed gears on the bicycle, the crank speed would need to remain unchanged relative to wheel speed but more torque would be required to maintain the same speed. That increase in torque causes the chain between the crank sprocket and the driven sprocket to stretch. If the chain is strong enough, it won’t break. And, if you could measure an angular difference between the crank at steady speed on flat ground and at the same steady speed going uphill there would be a slight difference–the effect of trying to increase (or maintain) the speed as the load (hill) increases.
“Learning is finding out what you already knew.” (Richard Bach, ‘Illusions’) Learning should be fun, and when you grasp the concept you are learning, it should cause the feeling, “Yeah! I knew that–I just never thought of it in that way before!”
Hope this helps!
Posted by zahid on 20 June, 2012 – 3:39 pm
kindly explain if a generator is operating with 50MW, 05 MVARS and with 50MW, 25 MVARS, is there any difference of fuel input in two cases?
Posted by CSA [Control.com Most Valuable Participant – 247 helpful votes.] on 21 June, 2012 – 12:22 am
I believe there are two reasons why the fuel flow “to the generator” is higher when the generator is producing 50 MW with 25 MVAr than when it is producing 50 MW with 5 MVAr.
First, the VA is higher at 50 MW and 25 MVAr. This is a measure of the total energy (reactive …, er, … uh, … power plus real power). The power factor (a measure of the efficiency) says that at 50 MW and 25 MVAr the amount of real work versus the total work is much less than at 50 MW and 5 MVAr.
Have a look at wikipedia for “power triangle” or “apparent power” for more explanation. Although no real “work” is done by the reactive component, there is heat generated in conductors which requires work.
Second, to “increase” MVAr one has to increase the excitation. The power for the exciter for every synchronous generator I’ve ever worked on came from the prime mover, either by using generator terminal power through a diode bridge or by a rotating “brushless” exciter which is powered by the prime mover. So, to produce more excitation for the same real power there has to be more torque from the prime mover which means that more fuel must be flowing at the higher MVAr than at the lower MVAr for the same MW.
Maths and vectors can come from some other reply.
Posted by Phil Corso [Control.com Most Valuable Participant – 83 helpful votes.] on 21 June, 2012 – 11:50 am
The answer, for the parameters you cited, is the fact that for the 50MW/5MVAr case the generator’s output is 50.2MVA! And for the 50MW/25MVAr case the generator’s output is 55.9MVA.
Increased MVA means a corresponding 1ncrease in armature-current, thus higher generator-losses (I^2 x R) requiring more fuel!
Finally, generator losses are not reflected in the 50MW measurement!
Regards, Phil Corso
Posted by Jesus Santos on 12 August, 2008 – 2:36 am
What is rated to 120MW, the turbine or generator?
Posted by Phil Corso [Control.com Most Valuable Participant – 83 helpful votes.] on 6 April, 2010 – 11:24 pm
Kumarash… All answers thus far presume your generator is connected to a Grid. Is that the case?
Regards, Phil Corso
Posted by Maint [Control.com Member] on 18 April, 2010 – 9:58 am
need to clarify one thing with Generator mode Isochronous and Droop?
the Raise/lower speed after synchronizing which change with mode of turbine , that mean FSR how control in Isoch/Droop ? which change actual in both mode ?
Posted by CSA [Control.com Most Valuable Participant – 247 helpful votes.] on 18 April, 2010 – 11:03 pm
The Raise- and Lower Speed/Load switches change the Turbine Speed Reference (TNR) regardless of whether or not the unit is operating in Isochronous or Droop Speed Control modes.
When in Isochronous mode, changing the turbine speed reference will change the frequency at which the unit is operating. Not the load, the frequency.
When in Droop mode, changing the turbine speed reference will change the power output of the unit. Not the frequency, the load.
Isochronous mode is proportional plus integral control. Droop mode is proportional control.
Other than this, I don’t understand the question.
Posted by Greg N on 29 June, 2010 – 8:43 pm
I have the same question, but what if the generator is only rated for 100MW?
My question is: If the prime mover can produce 120MW of power, but the generator is rated at 100MW, how will the generator react when if the prime mover is putting out 120MW of power?
Posted by CSA [Control.com Most Valuable Participant – 247 helpful votes.] on 29 June, 2010 – 11:51 pm
If the generator rating is lower than that of the prime mover, then there should be some kind of load limiting function to prevent the prime mover from putting out more torque than the generator can convert into amps.
When current flows in a conductor, heat is generated. The ability of the generator stator (and rotor) to be cooled to prevent damage to insulation and from expansion (things usually expand when they get hot, and they get hot when current flows) is one of the limits of generator operation.
If the prime mover is producing more torque than the generator can safely convert to amps, then there will be too much heat in the generator and eventually something will fail, usually the insulation.
There’s also the coupling between the prime mover and the generator which must be considered. There are limits to the amount of torque couplings can transmit, and the load coupling, as it’s called, must be capable of transmitting “rated” torque to the generator. Usually the load coupling (and the generator) are rated slightly higher than the prime mover.
But, if there are no other extenuating circumstances or mitigating factors which we are unaware of, strictly speaking the prime mover output should somehow be limited from producing more power than the generator is rated for in this question.
If the prime mover output is not limited and it exceeds the rating of the generator, the generator will continue to produce power until it overheats and fails (presuming the load coupling is capable of transmitting the high torque, also).
Posted by M.B. on 20 October, 2010 – 2:53 pm
I have a gas genset (Caterpillar engine + kato gen) output 570kW (858A) but the average load in the platform is 300 to 350A, means approx. 41% load. Now the engine is giving many problems and require frequently maintenance, so the Client has required to increase the load in the gen in order to has at least 50% loaded the generator and let the engine to have a good performance. Honesty I do not know too much about that units but there are other way to increase the load than increasing the fuel and torque, so we can implement a feasible solution in site.
Thanks so much for your reply
Posted by Ampao on 20 July, 2011 – 8:23 am
I have a question that may require your expertise CSA.. How can generators supplying a considerable number of loads be able to keep up with the continuous change or variation of load levels.
Its Like we consider a prime mover of a generator to be an engine of a car where the varying loads as the off road terrain. What part/device of the generator that assumes the role of the driver in a car. And how can that part/device be able to keep with the abrupt and unpredictable changes of the loads.
A realistic example would be say i have a 2KVA portable generator supplying a residential load of 1kW (resistive). Then suddenly I switch on five 100w light bulbs altogether as part of the loads in the house. Will the generator be able to provide 100% power to the light bulb right away? Or it will take sometime? or worse will it trip? Thanks in advance for you reply.
Posted by Joe E. [Author has 1 helpful vote.] on 20 July, 2011 – 4:16 pm
This would be a function of the governor. To extend somewhat your analogy of the car, imagine setting the cruise control. As the speed drops a little, it opens the throttle to compensate. A standalone generator sets the “grid” frequency and voltage, so a sudden increase in load may well cause the speed to drop for a short period of time. How much it drops and for how long are functions of the governor’s response time and the generator/prime mover capacity.
Posted by John on 13 July, 2012 – 2:08 am
Who knows the Dynamic relation between Mload and output power of LP turbine? How are they connected to the speeds of LP and GG?
Posted by ncs [Author has 9 helpful votes.] on 13 July, 2012 – 4:55 pm
Could you define Mload and GG
Posted by CSA [Control.com Most Valuable Participant – 247 helpful votes.] on 16 July, 2012 – 11:10 am
The question is not clear, and the abbreviations are unclear, as well.
I’m pretty sure ‘GG’ stands for gas generator, the High Pressure shaft of a multi-shaft machine. I’m not sure what ‘Mload’ means, but possibly Megawatt load?
Since the LP turbine shaft typically is directly connected to the load (generator or compressor, etc.), the amount of torque produced by the LP turbine is directly transferred to the load being driven by the LP turbine.
The GG includes the High Pressure turbine and axial compressor. So, a lot of the energy developed by the High Pressure turbine is used to power the axial compressor. The remainder of the energy is directed to the Low Pressure turbine which drives the load (generator or compressor, etc.).
Generally, the speed of the GG is not directly controlled by the turbine control system. The speed of the LP turbine and its load is the usually the controlled variable, meaning that the speed of the GG is allowed to vary as required in order to control the speed of the LP turbine and its load. In other words, the design of the machine is such that the GG speed can vary as required in order to control the LP speed.
Does this answer the question?