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AC generator on grid: Governor and Amp

  1. Jun 24, 2014 #1
    Hi Forum, once again I seek your help

    I have a very good understanding of rotating magnetic field due to stator, armature reaction, and the phasor diagram.

    from energy point of view it is very clear that if Gov increases, while output voltage remains locked by the gird, Amp will increase
    From phasor diagram, it is very clear power angle between Ea and Vout will increase, thus the Amp needs to be longer to prop Ea up.
    and from torque's point of view, Amps increases in magnitude and its resulting Armature flux will have a reduced angle with respect to Field Flux (as in closer to 90 degrees than before), thus explaining the torque increase.

    HOWEVER, all these explanations only explain the necessity of Amp's increase, it's like saying "due to conservation of energy, heat must be generated somewhere" I need an explanation like "due to impact and molecule interactions, heat was transferred."
    I CAN NOT think of a reason Amp increases for the first place. When Gov increases, field flux does not change in magnitude nor the change rate, how is the torque of water translated into current increase?

  2. jcsd
  3. Jun 27, 2014 #2
    How does the increase of governor translate into current increase?

    Hi Forum, I seek your help.

    when a synchronous generator is running on a large grid, How does the increase of water flow/push onto the turbine translate into stator current increase? (assuming generator producing positive vars, exciter remains constant)

    Thank you for your kindness of helping or intention to help, however, I am not looking for answers from "conservation of energy" point of view, I am not looking for a "phasor diagram" explanation either. I my opinion, those are very good explanations to explain the end result: voltage induced by flux of rotor alone (most literatures refer to that voltage as Ea) will lead output voltage even more (torque angle has increased), Var will decrease a bit, output power will increase as the portion of amp that is in phase with output voltage has increased.

    My real puzzle is: with excitation be constant, speed of the rotor constant, you will have a constant induced voltage on stator. which may lead or lag, but nonetheless constant in value. what caused the current in stator to start to increase?

    My guess is that: at the transient of increasing water flow onto the turbine, rotor actually speeds up a bit, and Ea increases in magnitude (so that means governor sometimes DOES have an effect on the voltage and vars when on the grid). The increase of magnitude of Ea will cause current in the stator to increase, which causes more magnetic counter torque, which will slow down the the "rotor speeding up", reducing the torque angle between rotor flux and stator flux, thus increaseing the counter torque even more. And eventually rotor will come back to synchronous speed ending up with the same speed and same Ea (the magnitude of which is again solely determined by the excitor).

    assuming no wire will burn no matter the current: hyperthetically, no matter how huge the sudden increase of water will be, rotor will NEVER EVER fly out of control at the transient. because the change of amp's phase angle is always bigger than Ea's phase angle, so the counter torque will always compenstate the torque from water.
  4. Jun 27, 2014 #3

    jim hardy

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    Removed for editing... back later

    Found my mistake - my alleged brain reversed something... apologies hereby extended.


    No takers yet ?

    You do realize that the reason we resort to phasor diagrams is because "word pictures" do not paint the same image in different peoples' minds. So you wind up with 'chicken or egg' dilemmas and confusion.

    So what'll make sense to you depends on how you've envisioned the system.
    I'll assume you're speaking of a machine connected to a system with which it can exchange vars. Else your stator voltage wouldn't be constant.

    What works for me is :
    to consider the field(rotor) and armature as two rotating mmf's, one dragging the other along.
    The sum of those two mmf's divided by magnetic reluctance is flux. Let us assume flux takes a sine shape, for simplicity. Real generators come pretty close to that.

    Voltage generated is 90 degrees out of phase with flux as we know because d(sine) = cosine.

    I'll take terminal voltage as my reference point for phase angles.

    Also for simplicity,
    i'll use for my mental model a simple two pole machine with only one armature winding of just a single turn.
    It helps to draw a sketch of that .

    Now imagine the machine excited but no armature current, ie connected to an open circuit..
    That means the only MMF present is that of the rotor. No armature current means no armature MMF.

    here was my mistake - jhFreeze frame your machine at the instant of terminal voltage maximum (not zero as i had earlier), the sinewave peak. That's when the rotor poles are directly aligned with the plane of the winding.

    Since that may seem counter-intuitive:
    If you are a believer in "Flux Linking",
    at that instant rotor flux lies in the plane of the winding so the flux linking (enclosed by) the single armature turn is zero. Recall that a sinewave has greatest slope right at its zero crossing, so d[phi]/dt for our sine-shaped flux is maximum at that instant... and so is voltage.
    If you are instead a believer in "Flux Cutting",
    at that instant rotor flux lies in the plane of the winding so the lines of flux are cutting the armature conductors at their greatest rate, so d[phi]/dt is maximum at that instant... and so is voltage.

    Now, connect the machine to a system and allow some real load current to flow precisely in phase with the voltage, that is power factor is exactly 1.0.

    By right hand rule the resulting mmf from armature current is at a right angle to that of the rotor.
    So the sum of the two mmf's has shifted off the axis of the rotor. Flux of course follows.
    The rotor must come to a new position restore the angle of flux to match phase of terminal voltage.
    And that's how power angle is affected by load.

    Now here's the trick to answering your question:
    With no armature current there was only one mmf
    but now there are two
    so the summation of mmf's has now become a triangle instead of just a single line
    a triangle with field amp-turns as adjacent side
    with armature amp-turns(from real current not reactive) as opposite side
    and their vector sum as hypotenuse,
    and hypotenuse is same length as the single line of field amp-turns was beforehand, else terminal volts would change,,,

    which means the adjacent side must have shrunk, mustn't it ? In order to satisfy Pythagoras with constant excitation AND constant terminal volts(flux) ? Sketch it.

    How'd it shrink? We didnt change excitation.

    Aha ! Back to alignment of rotor poles and our one turn winding.
    Armature current that's in phase with terminal volts makes a mmf that adds at 90 degrees to field mmf
    BUT ! armature current that's at 90 degrees out of phase to terminal volts makes mmf that adds or subtracts DIRECTLY to/from field mmf, not perpendicular.

    Now that's cool... Draw that one turn machine and convince yourself of it, for it's basic to really understanding alternators..
    So - as we come to the new rotor angle, terminal volts which is determined by the system to which we are tied will push a little reactive current into the machine to shrink that adjacent side for us.
    It's a beautifully self balancing system , a machine on an infinite bus.

    And that's why, as you said, vars will change slightly when you change load at constant excitation. The other machines on the system provide the amp-turns necessary to keep flux constant, by injecting reactive amps into the armature.. that's what caused stator amps to change.

    Work this in your head in your morning shower and when driving home from work.
    When you miss your turnpike exit because you were thinking about synchronous machines you are approaching my level of nerdiness - I'm certified potentate.

    Certainly you'll arrive at a different word picture than mine. Only advice i have is this - it won't impress the girls. Stick to tales of dragon slaying.

    have fun

    and i hope i addressed the right question , sans phasors as requested

    old jim
    Last edited: Jun 27, 2014
  5. Jun 27, 2014 #4

    jim hardy

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  6. Jun 27, 2014 #5

    jim hardy

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    fixed my mistake in post #3 jh
  7. Jun 28, 2014 #6
    Hi Mr Jim, it is good to see you flexing your grey cells again =)

    I agree to your diagnose about my problem, it is a chicken and egg problem. I also felt that we were using the same mental picture. but, i think my 'fundamental driving force' symptom is acting up again.
    I followed your post until the part "now here is the trick to your question"

    something here I would like to confirm:
    during the power factor 1.0 operation, field flux should be the hypotenuse, or the longest element, shouldn't it?
    you were explaining the increase of load In your post, and I am perfect ok with that scenario. however, increasing governor is different from load increase.

    but, regardless, i think the key info in your post is the 'current flowing back into the generator'. my new understanding is as follows:
    so, at the instance of gov increase, assuming power factor 1 operation at that instance, the original stator current's torque is not sufficient enough to maintain the relative power angle. field flux will advance more in angle and total flux will reduce, output voltage drops below grid voltage, grid current rushes in to aid voltage (thru some phasor math, i can see, but still not very intuitively sure that, the current rushing in is mostly reactive, and the real power component of current somehow increases. this maybe the reason the grid loves positive reactive power. during a pinch you can always turn reative power into real power)

    so my original reasoning of Ea (voltage induced by field flux alone) increase is wrong, Ea may increase by miniscule amount, but the main reason is really the reactive current rushing in to aid voltage drop of the generator (or the generator was pushing large amount of reactive current into the grid, but now it is not pushing as much as before)
  8. Jun 28, 2014 #7

    jim hardy

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    Will take me a while to digest all your points...

    a machine connected to a grid has its terminal voltage pretty well fixed by the system to which it is connected.
    So the key point is i think as you stated reactive current aids or opposes field current , which became clear to me once i saw the relative physical alignment of poles to armature winding.
    They tend to teach that as "Synchronous Impedance" which fits the math equations well but didnt help me with the physical picture.

    If you increase torque applied to the generator the rotor will pull ahead because rotor and stator are just two electromagnets in a tug of war,,, as it pulls ahead the mmf summing triangle widens out and pythagoras determines the result.

    Thank you for the kind words..
    I'll continue exercising the few gray cells i have left.

    Hmmm i had assumed the hypotenuse to be total flux, sum of field and armature reaction.
    But that was from intuition not rigorous analysis. And memories from a 1965 machinery class, complete with cobwebs.
    Which does your textbook author use in his phasor analysis?
    I'll look into my old motor book tonight.

    old jim
  9. Jun 28, 2014 #8
    Hi Mr Jim

    totally flux is the sum of field and armature. I agree.
    but, since field flux almost always needs to be bigger in magnitude to counter the armature flux to form the total flux, total flux is typically smaller. or, in this way: armature flux is always lagging field flux by more than 90 degrees.

    of course my assumption is always that generator is producing positive vars, or current is lagging.

    thank you again, Mr Jim.
    if I may, could I know your background?
  10. Jun 29, 2014 #9

    jim hardy

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    It's heartwarming to see somebody this interested in machinery. I loved my machinery classes, especially the labs where we had modest size motors and generators (~10 hp) to work on.

    Okay i braved MSPaint and will now show my abysmal skill with it.

    You were right about the hypotenuse at unity power factor. Dawing the pictures really cleared it up for me, i shoulda done it yesterday before posting. But it was way past bedtime as it is again now.

    Hmm computer just typed something from days ago - -something going wrong here. Well lets see if i can finish this post..

    The circle just represents locus of constant mmf or voltage .. helped me draw .

    Machine unloaded, some base excitation and voltage. No armature current, hence no power and theoretically no torque (windage and bearings and i suppose magnetic losses in core that we'll ignore).

    Machine with same excitation and same terminal volts, as we apply torque and let armature current flow however it sees fit.
    Observe machine went into lead - the system delivered the excitation shortfall by supplying reactive current . Green current is ahead of voltage.
    Hah ! There's my blooper ! That triangle is isoscles not right.. What was i thinking? Belay my claim about hypotenuses !

    This time we applied torque but adjusted excitation to maintain vars at exactly zero.
    http://[URL=http://s232.photobucket.com/user/oldjimh/media/ugenetic3_zps0b47bca3.gif.html][PLAIN]http://i232.photobucket.com/albums/ee289/oldjimh/ugenetic3_zps0b47bca3.gif [Broken]

    Okay, this time there IS a right triangle because of unity power factor , amps and volts both perpendicular to total flux

    I'm more than a little embarassed at my floundering around with this.
    But i prefer to go back to basics and figure it out, for you see what happens when i try to remember.
    Thanks for the refresher, "I needed that".

    I hope this simplistic thinking is of help . With a good mental picture one can figure out the equations instead of memorizing them.

    You asked my background - i majored in EE so long ago it doesn't matter, suffice it to say " before pocket calculators. "
    I went to work for the local electric utility as an engineer in the maintenance department of a nuclear plant. The utility found that having engineers assigned to maintenance improved the skill sets of both groups, so they rotated a lot of central office engineers through the plant.
    I preferred plant work to office so stayed there.

    let me know if i made any more bloopers.

    old jim
    Last edited by a moderator: May 6, 2017
  11. Jun 29, 2014 #10
    ahha, i can see a steady hand behind all these drawings. i have done so much photoshop, i know how hard it is to get those little arrows right with a mouse =)

    thank your sir for spending so much time on me. I am happy to let you know that, with your knowledge, i have Finally obtained a a very very robut and simple mental model for sync generators, that can explain most steady state and transient conditions. And i am well assured of its accuracy, because an on site Engineer with at least half of a century experience has approved of it =).

    all my previous posts about fundamental driver or d(phi)/dt were all really for this moment.

    i am a volunteer right now, working in a small town called Gillam in Canada. I m trying to gain some EE experience since I was Educated in it a while ago, but had been working in other fields. Most of the things I do involves writing code for PLC and design HMIs. I felt duty and morally bounded to understand how generators work.

    Thank you very much again, Mr Old Jim.
  12. Jun 29, 2014 #11

    jim hardy

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    Thanks for the kind words, they mean more than you know.
    If i lent a small helping hand, well - you sure earned it..

    I found Gillam on googlemaps and it looks like a delightful place. Lake trout in that reservoir?
    If i ever finish my Volksplane will come try out that little lakeside airport. Fly rod or deep trolling rig?

    I guess you work at that hydro plant? Impressive switchyard - how much generation there?

    old jim
  13. Jun 29, 2014 #12
    we have 5 and buildig 2 new ones, we have 4000MW now, eventually there will be more than 5000 MW here. half of the Canadian rivers drain to us.
    I work at the town head quarter, while the stations report problems to us. we have equipments spanning 4 decades.

    you are building a plane =)?
  14. Jun 30, 2014 #13

    jim hardy

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    Honestly it's still in the dream stage.
    I have a set of plans for Evans VP-2 and a Porsche type4 engine .
    Acquiring parts slowly.

    A fellow has to have dreams....
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