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Excitation on a rotor

  1. Mar 16, 2010 #1
    I know as a fact that excitation (amount of current on the generator rotor) determines the amount of reactive power output from the generator.

    But why? How does it turning the excitation up or down influence the reactive power output?

    Thanks.
     
  2. jcsd
  3. Mar 16, 2010 #2
    I am confused. If a customer at the end of a transmission line has a purely resistive load, he will not have any reactive voltage on the load. Neglecting transmission line effects, the generator will not be generating any reactive voltage (power) either. Are you discussing switchgear to reduce an overvoltage condition at the generator? If a customer has a large amount of induction motor load, the power factor improves (increases) slightly if the generator output voltage decreases.

    Bob S
     
  4. Mar 17, 2010 #3
    you have completely confused me!

    im just curious as to how the excitation has an effect on reactive power output.
     
  5. Mar 17, 2010 #4
    According to wikipedia, re: large electric power generators:

    "Because power transferred into the field circuit is much less than in the armature circuit, AC generators nearly always have the field winding on the rotor and the stator as the armature winding."

    So I'll talk about this arrangement.

    The "rotor" rotates, and the "stator" is stationary. The "field circuit" supplies a magnetic field to the generator and the "armature" developes electic current for external power usage.

    In any case the windings called "field windings" are feed what is called "excitation current."
    These windings establish a magnetic field through which the stator windings pass. The stator windings feed power into the grid. (A lot of lingo, hu?)

    The greater the excitation current, the greater the magnetic field developed by the rotor and presented to the stator windings. The greater the magnetic field, the greater the voltage across the stator. Does this answer your question?
     
    Last edited: Mar 17, 2010
  6. Mar 17, 2010 #5
    I understand what that all means, but how does the amount of current on the rotor (field) influence the reactive power output?
     
  7. Mar 18, 2010 #6
    The output voltage is proportional to the field current. The output power depends upon the load. As you can imagine, if there is nothing connected to the generator, it will put out zero power.
     
  8. Mar 18, 2010 #7

    sophiecentaur

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    I can't see why people are discussing the 'power into' the field winding. An alternator will work perfectly well with a permanent field magnet, into which no power is 'fed' and a permanent magnet is not a source of power. The power delivered from an alternator is surely provided by the motor that drives the alternator shaft.
     
  9. Mar 18, 2010 #8
    No one is discussing power into the field windings.
     
  10. Mar 18, 2010 #9

    sophiecentaur

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    " "Because power transferred into the field circuit is much less than in the armature circuit, AC generators nearly always have the field winding on the rotor and the stator as the armature winding."
    "
    One of your own quotes, Phrak, from Wikkers.
     
  11. Mar 18, 2010 #10
    I am so lost. :(
     
  12. Mar 18, 2010 #11

    sophiecentaur

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    Have you a reference for this, please?
    I can't help thinking that the amount of 'resonant' energy in the exciter circuit would depend on the source impedance of the exciter supply compared with any coupling with the load - via the rotor.
    Hence my previous difficulty with the idea that the field winding has to provide anything but a field, (which could be supplied with a permanent magnet).

    A power generator is usually run as a low impedance source, in order to reduce losses, rather than to be a matched source, to maximise power transfer. Is there some suggestion that imaginary currents flowing to the load are affected by the exciter field where the real currents are not?

    I have only a nodding aquaintance with these things but I should like to know if this has a firm basis or is it more apocryphal? I could imagine that the Maths could get quite heavy quite quickly.
     
  13. Mar 18, 2010 #12
    I don't have a reference for that, but iam 99% sure that is correct. Is it not right?

    I always thought that if you wanted to increase power output from a generator, give the turbine more steam. If you want to increase reactive power output from a generator, put more excitation current on the rotor (field winding).
     
  14. Mar 18, 2010 #13
    The best way for a utility to control reactive power output is to encourage all customers to correct low power factors, either by using more efficient electric motors (with less reactive impedance) or by using capacitor banks to correct the power factor at the load. There are probably things the utility can do to match the generator to a reactive customer load, but this does not reduce the impact of the reactive load on the grid, power transformers, etc.

    Bob S
     
  15. Mar 18, 2010 #14

    sophiecentaur

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    But how can it work that way?
    If you work into a resistive load you can't have any out of phase current flowing to it, so exciter current can't make any difference to that. If the load has a reactive component then the phase angle is defined by the Load R and X. Is there some second-order effect, here, involving phase lags inside the alternator?
    There must be a reason for your belief - you seem so convinced - but I wouldn't know where to start looking for the info (not as fast as you could:smile:).

    A specific field current and turbine power must be required for a given alternator at a given speed to produce a given Power and, if your (pure resistance) load is a certain, given distance away, the line capacitance will be independent of load R. This would imply that the power factor could go down as the load resistance decreases (and the load current increases) - and that could give a some sort of relationship between exciter current and PF - but it seems to go the wrong way (!?)
     
  16. Mar 18, 2010 #15

    stewartcs

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    Yes that is correct for a synchronous generator.

    Correct again for a synchronous generator.

    It influences the reactive power by adjusting the power factor. If you overexcite the field windings more reactive power is generated on the line. Draw a phasor diagram to see exactly how the reactive power changes.

    This is why synchronous condensers are used in many industrial applications that have high inductive loads. By correcting the power factor the total line current is reduced to a minimum.

    CS
     
  17. Mar 18, 2010 #16
    It is surprising that a synchronous capacitor is another name for an unloaded salient-pole synchronous motor. This is why industries with a large amount of induction motors also used synchronous motors to correct a low power factor.
    From http://en.wikipedia.org/wiki/Power_factor
    "Instead of using a set of switched capacitors, an unloaded synchronous motor can supply reactive power. The reactive power drawn by the synchronous motor is a function of its field excitation. This is referred to as a synchronous condenser. It is started and connected to the electrical network. It operates at full leading power factor and puts vars onto the network as required to support a system’s voltage or to maintain the system power factor at a specified level.
    The condenser’s installation and operation are identical to large electric motors. Its principal advantage is the ease with which the amount of correction can be adjusted; it behaves like an electrically variable capacitor. Unlike capacitors, the amount of reactive power supplied is proportional to voltage, not the square of voltage; this improves voltage stability on large networks. Synchronous condensors are often used in connection with high voltage direct current transmission projects or in large industrial plants such as steel mills."


    Bob S
     
  18. Mar 18, 2010 #17

    sophiecentaur

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    "It influences the reactive power by adjusting the power factor. If you overexcite the field windings more reactive power is generated on the line. Draw a phasor diagram to see exactly how the reactive power changes"

    I just don't understand how you can say that "reactive power is generated on the line". If the load is resistive, then how will the current into it not be in phase with the PD across it? What phasor diagram are you thinking of?
     
  19. Mar 18, 2010 #18

    stewartcs

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    If this is a theoretically purely resistive load then no, it won't. However, in real life the transmission line will have some inductive and capacitive reactance even if the load is purely resistive. I presumed the OP was referring to real life.

    CS
     
  20. Mar 18, 2010 #19

    sophiecentaur

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    That is making more sense to me now. It seems to involve a network of other generators too (mention of the word synchronous)?
     
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