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Armature reaction drop

  1. Aug 6, 2015 #1


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    In synchronous generator, how does armature reaction cause a 'drop' ? How is it different from the leakage reactance of stator windings? I can't visualize it..Please help..
    Last edited by a moderator: Aug 6, 2015
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  3. Aug 6, 2015 #2

    jim hardy

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    The way to visualize it is to freeze-frame your thinking. Stop the rotor with its pole centered underneath a phase winding and think of it as DC.
    Armature amp-turns will either aid or oppose field amp-turns.
    So they'll add to or subtract from field, affecting terminal volts just as if they were an internal impedance
    Look up "Synchronous Impedance"
  4. Aug 7, 2015 #3


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    In case of unity pf, armature amp-turns neither add to nor subtract from the field amp turns. They just distort the field amp turns. And if they enhance or weaken field flux at other power factors, this will reduce the 'generated emf'. What I read in book was 'part of generated emf is used to overcome the armature reaction reactance'. That means generated emf is constant. But with reactive loads, its not true because some part of armature amp turns enhance or weaken the field amp turns. Is there anything similar to 'back emf' that is induced in the armature windings due to armature reaction? What does the term 'drop' mean in this context? Is it reduction in the generated voltage or a part of it being used to overcome synchronous reactance? If it is later, how does that reactance work??
  5. Aug 7, 2015 #4

    jim hardy

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    Armature reaction and synchronous reactance are two names for the same thing, and reflect two ways of thinking about it.
    The math works out nicely to treat it as an internal reactance, you'll probably see it treated that way in your books and on phasor diagrams.

    Let's really simplify our thinking down to a synchronous machine with a one turn armature and a permanent magnet rotor.
    We'll allow the complication that the rotor is machined to produce sine shaped flux, though.
    Recall that voltage and flux have a derivative/integral relationship meaning our sine shaped flux gives a cosine shaped voltage
    because e = - n*dΦ/dt
    and cosine is 90 degrees out of phase with sine. That's a quarter cycle.

    so--- peak voltage appears at the flux zero crossing
    and zero voltage appears at the flux peak
    because of that sine-cosine relation.

    Do we all accept that unity pf current is in phase with terminal voltage ?
    and zero pf current is 90 degrees ahead or behind it ?

    We'll make our simple one turn machine with only one pole pair so electrical and mechanical degrees are the same, 90 deg = ¼ turn.

    please -
    pardon my crude sketch i just dont do well with computers.

    The circles are my one-turn armature windings. I've freeze-framed my thinking as mentioned earlier, so these would be like high speed photographs.
    Left sketch:
    Flux from armature zero pf(real) current is perpendicular to flux from field
    so there is little interaction beyond a phase shift, and that's why there is a power angle.
    At this instant, zero pf current is zero so it does nothing.

    Right sketch:
    A quarter turn later (or earlier) when unity pf (real?) current is zero
    and reactive(zero pf) current is maximum,
    look where the rotor is !
    That's why zero pf current in the armature adds or subtracts directly to field flux. It peaks when rotor pole is right underneath the armature turns.

    You should stick with the notation in your textbook

    but i find this mental image useful to understand why there's such a thing as synchronous reactance. It's not same as leakage reactance.
    It does affect terminal voltage by reducing flux in the machine(,or raising it )
    but the external characteristics are same as if it were another circuit element
    and that's how most authors present it.

    Is that any help ?

    old jim
    Last edited: Aug 7, 2015
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