## Generator VAr Import and Reverse Power

Hi guys

I am not sure if this is the right forum, so please feel free to move this topic around to where it belongs. This is not homework help, but something I have been wondering about lately.

Anyway.
I am wondering about som differences between VAr Import and reverse power for generators.

In an example, let's say that we have 2 equally sized generator connected to the same bus in an island.
G1 is running at 690 Volts
G2 is running at 700 Volts.
Prime mover is to diesel engines.

Now G1 would import reactive effect because it is seen as an inductive load from the Net.
Does that mean that G2 is putting 10 volts on the stator windings of G1?

Now, let's say that G1 completely loses excitation and is left with remanent voltage of about 100 volts. This would lead to a much greater VAr import from G2, but would it be the same as putting 600 volts on the stator windings? And if so, would it be some kind of reverse power where G2 is "helping" G1's diesel engine to turn?

(In the above examples all generator protections are disabled)
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 Recognitions: Homework Help Hey Twinflower! Haven't seen many questions by you lately... Well, I'm not really an expert, but hopefully what I'll say is of some use. The voltage of a generator will depend on the speed of the engine. If you connect G1 to G2, their voltages will have to be the same. Note that if you run the engine a current is induced. But you also have that if you send a current through, the engine will start turning, like an electric motor. It means that some of the current of G1 is send through G2, slowing or speeding its engine, in such a way that any voltage differences cancel out. However, conservation of energy still makes sure that all the fuel burnt in the engines will find its way to useful utilization (other than standard losses). I mean that no energy will be lost in slowing or speeding an engine.
 Hi I Like Serena! It's been awhile, that's true.. (But it's kind off a good thing that I figure out my questions before coming here to ask for help) Anyway. It's correct that a generator's voltage is dependent of the speed of the generator (or the prime mover per se). But it is not correct that two generators would need the same voltage in order to be synchronized and connected. The same speed - yes, but not the same voltage as you can change the voltage by another mean than changing the speed: namely changing the excitation current used to produce the magnetic field in the rotor. You say that if you send a current through a generator, it will start to turn like an electric motor, does that mean that any kind of VAr Import would result in a reverse power scenario where the generator that is importing VAr will "help" the prime mover turn? This is basicly my initial question, because i want to know the difference of reverse power (when losing prime mover input and the generator is turned as a large motor) and VAr import (where current is _in_troduced to the generator and the generator is seen as a large inductor from the net) I hope this was understandable to some degree, it's kinda hard to explain some times, and I have an awful head ache after last night

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## Generator VAr Import and Reverse Power

 Quote by Twinflower Hi I Like Serena! It's been awhile, that's true.. (But it's kind off a good thing that I figure out my questions before coming here to ask for help)
Aww, that just means you've learned how to easily understand the material.
Shouldn't you be studying more challenging material then?

 Anyway. It's correct that a generator's voltage is dependent of the speed of the generator (or the prime mover per se). But it is not correct that two generators would need the same voltage in order to be synchronized and connected. The same speed - yes, but not the same voltage as you can change the voltage by another mean than changing the speed: namely changing the excitation current used to produce the magnetic field in the rotor.
In a single (ideal) wire you can not have 2 different voltages - they are short circuited.
So if you connect the poles of the 2 generators, with or without load, their voltages will forcibly be the same.

 You say that if you send a current through a generator, it will start to turn like an electric motor, does that mean that any kind of VAr Import would result in a reverse power scenario where the generator that is importing VAr will "help" the prime mover turn?
Yes.

 I hope this was understandable to some degree, it's kinda hard to explain some times, and I have an awful head ache after last night
Hope you're feeling better then.
What did you do?

 Quote by I like Serena Aww, that just means you've learned how to easily understand the material. Shouldn't you be studying more challenging material then? In a single (ideal) wire you can not have 2 different voltages - they are short circuited. So if you connect the poles of the 2 generators, with or without load, their voltages will forcibly be the same. Yes. Hope you're feeling better then. What did you do?
I again I Like Serena :)
Sorry for my late reply, I've been busy at work lately.
I'm currently in China (which is a long way from home for me) doing commissioning on a power plant on a major drill ship. We have 8 generators, 7,2 MW each at 11 kV.
So this being my work, I often lay wake at night philosophizing about this and that when it comes to generators.

However, I am a bit unsure about what you say that two generators cannot have a different voltage. On a single conductor, yes. Directly connected, yes. But when connected to a switchboard with loads of different loads (bad sentence) between them, I have seen many times that their voltage differs by a few volts - and that this may cause unbalanced reactive power sharing.

So my next question is that if the above is true, and if one generator is outputting 10 more volts than another generator on the same bus - would it then mean that the +10V-generator is applying 10 volts on the other one? Because, 10 volts difference would mean a hell of a lot of VAr Import to the one with lower voltage.
 Recognitions: Gold Member Science Advisor Reactive current in generators either aids or opposes the DC field. So the underexcited machine will accept reactive current from the other one, 90 degrees out of phase, attempting to make their terminal voltages equal. Circulating current will be (Δterminal volts)/(impedance connecting them). A total loss of excitation would make that machine into an induction generator and its speed would change. Its engine would try to accelerate it until the current induced in its rotor makes enough field to support terminal voltage. Presumably a loss of field detector circuit would trip it well before that happened. Circulating current gets very high in that circumstance. old jim

 Quote by jim hardy Reactive current in generators either aids or opposes the DC field. So the underexcited machine will accept reactive current from the other one, 90 degrees out of phase, attempting to make their terminal voltages equal. Circulating current will be (Δterminal volts)/(impedance connecting them). A total loss of excitation would make that machine into an induction generator and its speed would change. Its engine would try to accelerate it until the current induced in its rotor makes enough field to support terminal voltage. Presumably a loss of field detector circuit would trip it well before that happened. Circulating current gets very high in that circumstance. old jim
Thanks a bunch for a really understandable reply.

New question:
When exactly does the generator turn into a motor?

When there is noe more reactive export - only import?
Or when the reactive import is "so and so" large compared to rated power?
Or when the reactive imported current is so large that it opposed the field enough to make it zero (or negative?)

(just a few more details left to grasp)

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 When exactly does the generator turn into a motor?
When its prime mover stops applying torque to the shaft.
The only difference betwen a generator and a motor is direction of power flow. If you shut off the fuel to one of the diesels the electrical machine will commence being a motor and spin the diesel at synchronous speed, drawing power from the other machine.. At least until some protective relay trips it.
The old British term "Dynamo" was to me a more intuitive name than motor or generator.

 When there is noe more reactive export - only import?
There's a naming convention.
For the scenaio you described , Reactive current circulates between the machines.
Have you ever played with the voltage adjust? There's a point where current meter indicates a minimum, either raising or lowering the voltage adjust from there causes current to rise. At that point your machines are matched and any reactive power is going out to the external load.

So let's pretend for a moment you have no external load, just two machines in parallel.
When they are matched the current will be zero(assuming engine throttles are also matched).
Raise the voltage on either machine and curent increases - reactive current is circulating between the machines.
By convention, the overexcited machine is said to be exporting and the underexcited one importing. Also by convention, the overexcited one is said to be "in the lag" and the underexcited one "in the lead".
Lead and lag refers to the phase of current WRT terminal voltage. It'll be 90 deg ahead or behind.
Leading current AIDS the machine's internal field, raising terminal volts
Lagging current OPPOSES it lowering terminal volts
For our simple case of two machines circulating only reactive current,
one is in the lead and one is in the lag
leading machine(underexcited) is accepting(importing) reactive power and lagging machine(overexcited) is supplying(exporting) it.

The point at which the machine changes from exporting to importing would be the point where circulating current falls to zero, ie machines matched.
So the best answer to your question would be "When reactive current is zero" edit er, make that minimum but i thought it needed that lead-in. Sorry if i'm being too long winded.

Does that help any?
It's best to start with minimum system, just two machines and grasp how reactive current helps or hinders the internal field.
Then add an external load , it just offsets things by however much reactive current is drawn by the load. So instead of current going through zero as you change voltage it'd go through a minimum.

Since you're on a boat, I encourage you to experiment with your machines.
I had the luxury of two 450kw generators and freedom to do these things, there's nothing like grabbing that knob and getting the feel of how much knob gives how much circulating current. I could easily cause 300 amps of circulating current, therer wasn't much impedance between my machines.

So, you asked initially to effect "is one generator applying 10 volts to the other one'

well almost. It's applying enough current to raise the other one's voltage by ten volts(?)

I hope i answered the right questions - sometimes i wander.

old jim
 Recognitions: Homework Help Yep. I've got nothing more to add (with my limited knowledge). I'm just left wondering why old jim did not answer sooner.
 Hi again Jim First off, I love your wandering! Every old piece of information presented in a new way makes another piece of the puzzle fall to its place. So, secondly i think you misunderstood my question. I often "play around" with the voltage setpoint to find the sweetspot where reactive export for all generators are equal - knowing that this is the generator's "best" voltage even though it might be 691 V on one generator and 690 on another. (I think this is due to transformers and some measurement inaccuracies of the AVR sensing voltage - correct me if i'm wrong) Anyways, my first initial question in this thread is that when a generator is importing VAR, is that current doing any work that makes the prime mover's work on the shaft easier? Also I think that even though the current is supposed to 90 degrees leading the voltage, it still must be somewhat lower (89? 89.9?)due to heat losses in the windings providing som real power (?) And, IF a generator is importing huge amounts of reactive current from another generator (all protection functions are disabled, machine will eventually burn or pole slip will occur etc), still with prime mover input -- will that current make it easier for the prime mover to turn the shaft? another question: when reactive imported current is opposing the excitation current (in the rotor, right?), does that mean that the excitation current is canceled out? Or does the AVR somehow turning down the excitation by some kind of detection of something? so many questions.. so many pieces laying on the table which needs to make my picture whole :)
 Recognitions: Homework Help Oh good you didn't address me, since I have no clue! Old jim?

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 I'm just left wondering why old jim did not answer sooner.
I feel like i'm maybe making a nuisance of myself here by overposting, so i stay on EE and Nuke threads mostly. And i wait for others to have their say usually somebody else gives a better answer than i could have.

 (I think this is due to transformers and some measurement inaccuracies of the AVR sensing voltage - correct me if i'm wrong)
i think you're exactly right on that one.

 So, secondly i think you misunderstood my question.
Quite likely, i do that a lot. It's called 'Asperger's Syndrome".

 Anyways, my first initial question in this thread is that when a generator is importing VAR, is that current doing any work that makes the prime mover's work on the shaft easier?
No. VARS transmit no work.
There's two different ways to show that.
1. . Recall VAR current is 90 deg out of phase.
Watts is Volts X Amps X cosine of angle between them, and since cos(90deg) = zero that product is also zero - no watts so no torque. That's a math solution.
2. . If you draw two sinewaves 90 deg out of phase, label one Volts and the other Amps,
now quadrant by quadrant multiply them to get 'instantaneous' watts. You'll see by symmetry that there's equal positive and negative products so the area under the resulting curve is zero. That's a graphic solution.
Now shift them back in phase and repeat - all area is now positive (because negative volts times negative amps = positive watts).
Observe if you shift them 180 out all area is negative, which means you swapped from generator to motor.
Now it's true that VARS exchange energy between source and load for fractions of a cycle but they average out to zero. That's the positive and negative areas you saw in the graphical solution.. In a single phase machine that would make a pulsating torque but it would average zero.

 Also I think that even though the current is supposed to 90 degrees leading the voltage, it still must be somewhat lower (89? 89.9?)due to heat losses in the windings providing som real power (?)
I believe you could say that. But don't get hung up on that thought. Practically, there'll always be a resistive and a reactive component to load current so it'll have some angle between -90 and +90 deg.. For learning we break load current into ideal components at zero and 90 degrees and study them separately.

 And, IF a generator is importing huge amounts of reactive current from another generator (all protection functions are disabled, machine will eventually burn or pole slip will occur etc), still with prime mover input -- will that current make it easier for the prime mover to turn the shaft?
Hmmmm there's a possibliliy we're not on same track here. Try this thought:
That imported reactive current does not transmit any torque to the shaft so it niether helps nor hinders the prime mover.
What it does do is this: it adds to the flux that's made by the rotating field of the underexcited machine. Now - if that underexcited machine has so little excitation that its field can no longer handle the torque of its prime mover, it will accelerate above synchronous speed and try to run as an induction generator. That can be mechanically pretty violent so we dont want it to happen.
But the reason the machine sped up is not because the other one helped it, instead because its own field got too weak to hold on to the torque from its engine.

Let me lift a phrase fom an old post last January

If as a kid you ever played with magnets, put one on top of a table and one underneath and moved them from below, you know magnets can transmit force.
Of course so can electromagnets. It shouldn't be much of astretch to realize they can transmit torque, too.
Rotor of a synchronous machine is an electromagnet that physically rotates. A stator with external voltage applied is also an electromagnet and its field rotates because of the precession of three phase currents around windings. In a motor the stator field drags the rotor along, in a ngenerator the rotor drags the stator field around.

We had a long discussion with a cool guy named Bassalisk about this last January, you might check these two threads. He had some good drawings and some very good thoughts.
http://www.physicsforums.com/showthread.php?t=538388
http://www.physicsforums.com/showthread.php?t=571174

 another question: when reactive imported current is opposing the excitation current (in the rotor, right?), does that mean that the excitation current is canceled out?
Well - it's flux we should be thinking of, and magneto-motive force MMF.
Flux is MMF divided by (properties of the iron magnetiic path ~ reluctance)
and MMF is amp-turns.
The amp-turns of the reactive armature current oppose the amp-turns of the field
so it's the MMFs that cancel each other out.
Both currents still flow but the end result is not unlike they cancelled one another. That's called Armature Reaction and is a big part of those two links i gave.

 Or does the AVR somehow turning down the excitation by some kind of detection of something?
The AVR measures reactive current and modifies the excitation to make the machines share vars gracefully. Often they measure phase B current and add that to phase AC voltage, note they're already 90 degrees out of phase so it's a dirt simple way to get var correction. It'll be called "Reactive Droop". Typical setting is 5% meaning full reactive amps would lower voltage by 5%. That way a machine that's hogging vars will be cut back by its AVR. Or a slacker will be kicked up a little.

If machines are tied together by transformers you'd subtract transformer impedance from that 5% setting. Note that with a high impedance transformer , > the droop you want, you might have to reverse the AVR droop setting to make it look like there's 5% impedance between machine and bus. Then it'll be callaed a "line loss compensator" or "reactive rise" or something the manufacturer found to his liking. Be careful with that adjustment - it chenges the reactive compensation from negative into positive feedback so the AVR gets real touchy, might need more damping.

Sigh it's endless, but really fun. I envy you that setup - have a good time !!

I sense you want an intuitive feel for these machines. Until tha sinks in that i cant make myself believe the formulas, afterward they come easily.. I hope i have helped you

Wow - an engineer on a boat in China. Just like "Sand Pebbles" , one of my favorite movies.

old jim

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