Electrical Generation Questions - gen excitation, gen pf

You don't mention whether the plant is connected to a grid. First consider a single machine connected to a large (i.e. "infinite") grid which presents a constant voltage and negligable impedance. You try to raise the voltage of the generator but the grid says "no way". the result is that as the internal generator voltage is increased and the power is held constant, the reactive output changes. The machine generates more reactive and for a generator, this implies going more lagging. The grid obliges by reducing reactive generation elsewhere as the load reactive is unchanged. Draw phasor diagrams- that is the easiest, assuming Vt and P are constant. and the internal generator voltage varies (are you familiar with V curves for a synchronous motor?- similar to this). Overexcited generator goes lagging and underexcited goes leading. I can give the equations for this. Let me know if you need them or any examples. I do have a DOS based Turbo Basic program which allows jockeying between two machines feeding a common load.

Now if your two machines are the only machines on the system, they must supply the power and reactive of the load. If the load is 10MW at 0.8 pf lag, the load requires 6MVAR reactive (note that loads generally absorb reactive- hence the need to generate it). This can be split between the machines and in your situation it is good to split the power (8MW) and the reactive equally between the machines. If you try to reduce the voltage on one machine, the terminal voltage drops a bit (because the "grid" is finite) but the current distribution changes. The power is determined by the governor settings so the main thing that changes is the reactive. The result is that one machine can take 1MVAR (at 4MW and o.97 pf lag) while the other picks up the slack and carries 5MVAR , 4MW at 0.62 pf lag. Go further and one could get one machine at -1MVAR (leading) while the other is at

Transmission line reactive can be lagging or leading depending on line length and load. Your boosting of voltage at your system connection is sending more reactive into the system to compensate for reactive losses. You have to supply this. Adjusting the substation autotransformer can try to boost voltage and so increase reactive flow. Your generators will have to supply this reactive.

What is your background as knowing this can help considerably in attempts to explain?.

Don,

Thanks for the great response. To answer a few unclear portions of my original post - yes the machines are in parallel and connected to the grid. I am still very wet behind the ears regarding power plant engineering (I'm an EE with less than 5 years experience). I've not seen any V curves but that is most likely something I can google for the theory behind it. But yes - I would love the equations for the over/under excited generators yielding lagging/leading pf. An example probably wouldn't hurt either. Thanks!

I follow your logic regarding the loading and reactive power sharing. Regarding the transmission lines - the overall length is about 7 miles. Fundamentally, it would make sense that the transmission lines act as big capacitors (with the air between the lines being the dielectric) hence the need to transmit at a lagging power factor to compensate. It is still not clear to me why an inductive power factor is being injected into a grid which likely (cant say for sure) has massive inductive loads on it.

I guess I am also getting stuck with the concept of voltage increasing yielding more reactive power. Is this voltage increase basically elongating the hypotenuse of the power triangle thus increasing the pf?

Your response was great and I appreciate you taking the time to help me. Thank you.

I will pump this book:

Don did a good explanation, but one other thing you should understand is the load capability curve. This is a curve of the generators VAR and W output capability. Obviously the turbine sets a limit on the power output. But when the load is reduced, the VAR output can go up because the limit near full load is usually the stator winding current. With lower power out, the VAR output can increase without overloading the stator. But two more limits occur. The leading VAR is limited by how much current can be pumped through the rotor winding - either the winding limit, or the limit of the exciter supply. The lagging VAR limit is from two things - one is stability, the other is rotor iron heating. With lower rotor field, the generator coupling to the grid is reduced. It can fall out of sync easier. The rotor iron also heats at low field current because the 60Hz current now starts going through the edge of the rotor, heating it up.

You may be confusing 'lagging pf' for a generator and 'lagging pf' for a load. As Don mentioned, when a load (such as an induction motor) has a lagging pf, the current is lagging behind the voltage waveform, literally. But in a generator, the 'real' component of current is the opposite of voltage waveform (power is flowing *out* of the generator, where in a motor power is flowing *in*). So current that is lagging the voltage waveform in a generator has reactive power flowing *out* of it, while in a motor with current lagging the voltage waveform, the reactive power is flowing *into* it. (where, by long convention, lagging reactive power is called 'positive' and leading reactive power is called 'negative').

In an isolated system, the total load pf determines the amount of reactive power being consumed, and the generator(s) must supply it. If it does not, then voltage on the terminals will change, changing the generator current until the reactive power flowing out of the generator does match the load.

In an 'infinite bus' situation, the voltage on the bus is fixed. So the infinite bus itself can supply the loads with whatever reactive power they need. The infinite bus can also absorb any amount of reactive power the generator wants to generate. So the reactive power flow out of the generator is determined solely by the excitation applied, the machine's characturistics (synchronous impedance for steady-state stuff) and the real power being carried by the machine.

If you draw a single phase circuit with an 'ideal' voltage source, an internal inductance and then the generator terminals, finally connected to the 'infinite bus' (another ideal voltage source that is fixed voltage and frequency), you'll probably get the idea. The difference between 'ideal' voltage source inside the generator and the infinite bus voltage is simply the voltage drop across the internal inductance. The current through the machine is a function of this voltage difference and the phase difference between the 'ideal' source and the infinite bus. Changing excitation is equivalent to varying the 'ideal' voltage source inside the generator..

Seven miles for the transmission line isn't at all unusual. Our plants have lines that run more than 20 miles to the next station, and even that is not out of the ordinary. Our lines are slightly capacitive when unloaded, but the substation load at the other end is dominant. But the regional operator gets most of the voltage support he needs (MVAR generation) from other equipment (capacitor banks, peaking units, etc [we're a base load unit]) and we only carry enough MVAR to stay lagging by about 0.98. This is for stability reasions mostly.

Yet another factor in deciding just how much reactive load to carry is system stability. If the voltage at your end of the transmission line is too low, then the power you're generating won't actually flow down the line. So your generator doesn't apply as much counter-torque to the prime-mover and the shaft speed picks up. You 'slip poles', drop out of synchronism with the grid and "bad things" happen. Many generator unit voltage regulators have "URAL" (Under Reactive Ampere Limit) circuits that cut in automatically to prevent excitation from being reduced too far. Others have some procedural curve the operators must follow, something like MW on the horizontal axis and MVAR on the vertical, with shaded areas saying "Always operate above this line..."

Such limits also consider the possibility of a line tripping (if feeding multiple lines). When one line trips, the generating end will increase its torque angle (higher impedance in the transmission line). If this gets too close to 90 degrees, "bad things" happen (generator could pull out of sync and start slipping). Some installations even consider the effects of whether a line has 'fast reclosers' or not. If the line trips but recloses quickly enough, the generator end will not have advanced its torque angle very far before reclosure. This is common if the lines are subject to lightning strikes. Lightning causes flashover, the line trips, the generating unit starts to increase its torque angle, the line recloses, generator returns back to original torque angle (well, there are a few 'oscillations' before it damps out).

This is the case for us, if certain transmission line protective equipment is out of service, we must use a different curve to determine minimum MVAR (have to carry more in case of no 'fast reclosers').

Welcome to the world of *POWER* electrical engineering :-)

daestrom P.S. Another good book on this stuff is "Electric Power Transmission Systems", J.Robert Eaton / Edwin Cohen. Mines getting pretty old, but still a lot of good info.

I appreciate alot of the feedback but some of the information appears to be 2 steps ahead of where I currently am regarding theory digestion. Plus, some of the nomenclature (and theory) you folks are using is passing right over my head. Advanced rotating machinery and the associated equipment typically used in power generation are not areas covered in college. It also doesnt help that I do not have a mentor or any senior engineering type person to keep me on the straight and narrow. I am going to have to teach this info to myself.

I will attempt to get ahold of the recommended reading material since seeing diagrams supported by mathematical equations have always solidified things more in my brain. Hope it works out.

Thanks for all the help.

Well, start with AC circuit analysis of the simple scheme I mentioned. A variable AC source to represent the 'internal voltage' of the generator, a series inductance that represents the machine's 'synchronous impedance' (don't worry about the name, just treat it as an inductor). Zero resistance in the circuit (it's several orders of magnitude less than the inductance and can be ignored with little error). And a fixed AC source (the 'infinite bus').

Another thing to look up would be 'four-terminal networks'. Studying these can give you insight into how real and reactive power flow between an AC source and load over a line.

Yes, the phasor diagrams can help a lot.

Also, if you're not familiar with P.U. (per unit) calculations/terms, take a moment with those. They're not real hard or anything, but a lot of graphs and nomographs are based on P.U. terms (some xfmr and gen ratings such as synch. impedance are also listed on a P.U. basis).

Good Luck, come back anytime if you get 'stuck'. Myself, or others (Don and Charles are two that come to mind that are quite well versed in this area) will be glad to help you get through any 'rough spots'.

daestrom

For me, when you say 'brush' generators I think of Charles F. Brush. Do you really mean synchronous generator? (With slip rings and carbon brushes.)

What is an aeroderivative turbine?

Consider a generator (reactive) capability curve - a plot of watts vs vars showing the capability of a synchronous machine based on allowed temperature. Generally it shows 2 quadrants out of 4. The horizontal axis starts at 0 and represents watts out of the generator. The vertical axis is along the left side and represents vars.

The power system considers inductive reactance as being positive.* (The * is a footnote)

In power system speak, inductors consume vars and capacitors supply (make) vars. A synchronous machine can do either. A more complete but slightly confusing statement is that inductors consume vars supplied by capacitors and capacitors consume vars supplied by inductors. Since vars represent energy stored in magnetic or electric fields, vars are not actually 'consumed' but shuttle back and forth between reactive elements.

Expand the capability curve to 4 quadrants. To the right of the var axis is positive watts (watts 'made') and to the left is negative watts (watts 'consumed'). In a similar fashion above the watt axis is positive vars (vars 'made') and below is negative vars (vars 'consumed').

_Generators take on the pf of the load they serve._

I assumed that meant slip rings and brushes vs. a brushless excitation system.

A turbine derived from a airplane engine. Usually the power section (compressor, combustor & compressor drive turbine) is nearly identical to a turbojet power section. The turbine that spins the generator is downstream and is customized for shaft power delivery vs. the original design that was designed for aircraft.

-snipped-

Yep, absolutely. The plant I worked at years ago had frequency, MW, MVAR, stator kA, stator kV, and in the substation kV and A for each line. I don't remember a PF meter. Commands from dispatch were in MW and MVAR, with voltage limit bands.

------ A seven mile T-line has negligable capacitance as far as its effect on reactive is concerned. If it did, the generator would have to become more leading, not lagging. The basic reason why the generator has to supply reactive is that loads generally suck reactive. An induction motor, for example, will have a very low no- load pf and at rated load may be in the order of 0.8 pf lag. The generator has to supply this.

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Simple case: load drawing 1 per unit power at 0.8 pf lag (S=0.8+j0.6) at 1 pu voltage The generator is supplying this load through a line of impedance 0+j0.1 pu the load current is S*/V =0.8-j0.6 pu The generator voltage is 1 +(0.8-j0.6)(0+j0.1)=1.06 +j 0.08 pu or 1.063 @4 .3 degrees ahead of the load voltage. The complex power generated is 0.8 +j 0.6 +(I^2) (0+j0.1) =0.8+j0.7 P=0.8 and Q = 0.7 so the pf is 0.75 lag

Basic equations. Load at voltage Vl and generator at voltage Vg. Assume that the load voltage is constant at all times and we can take it as reference (angle 0) All V and I below are phasor quantities Current Il = Vg-Vl/Zline =|Il| @ angle theta To simplify take Z=jX Sl=Pl+jQl (load complex power =(Vl)Il* (* implying conjugate) generator Vg =Vl +jX(Il) =|Vg| @ angle delta Sg =VgIl* =( Vg)Il* =Vg(Vg-Vl)/jX do the manipulations and you will end up with Sg =((|Vg||Vl|)/X )sin delta +j[ ((|Vg|^2) /X) -((|Vg|^2)/X) cos delta]

Your machines are not affecting the grid so frequency and voltage at the load are constant try to increase voltage and there is a local voltage increase but the grid voltage is unchanged so that Vg in the above changes magnitude and delta changes to compensate. This affects the reactive. Try to change speed and the generator power increases so that there is a change in delta (and some change in Vg resulting in a change in reactive.

If you have two machines feeding a load over separate lines and they are the only generation sources, then you have two racoons in a bag.

I can send you a program for the latter situation but what limits are on files that you can get?

Again, I apologize for the confusion. The generator has a brushless, perm magnet exciter utilizing a single shaft. Brush is the manufacturer. No slip ring. No burshes.

A gas combustion turbine of the construction typically used in aricraft engines. Very simple in desgin when compared to the massive frame engines. Not quite as robust an engine as the frame machines.

Thanks again all. I will come back and respond when I get a chance to review all this wonderful information. I just happen to glance and see the above questions - which may or may not be causing problems.