Removed for editing... back later
Found my mistake - my alleged brain reversed something... apologies hereby extended.
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No takers yet ?
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?
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