Does synchronous Reactance vary with frequency?

[I_am_learning]

Synchronous generator synchronous reactance Xs is composed of Xs = Xa + Xl
where Xl is the armature leakage ractance can Xa is there to model the armature reactance.
Although Xl is frequency dependent (Xl = Lw, where L is the leakage inductance which is constant), is Xa also frequency dependent ? How ?

Xa is there to model the drop in terminal voltage due to demagnetizing nature of field produced by armature conductor? Unlike the Induced internal EMF (E) of the generator which is proportional to frequency, this demagnetizing field seems to be proportional to the armature current only. So, it appears to me that Xa should be independent of frequency.
But I'm not sure, so here I am to ask.

 

[jim hardy]

You sure are learning!

let's think for a second here.
i have to because this is forty year old memories
and i remember how things work better than i remember formulas

thought experiment -
fixed field current
short circuit on machine
so armature current is internal EMF divided by Xa (assume Xleakage is much smaller)

shut off prime mover and machine coasts down

when it reaches half speed
armature current should still be same ,
because we have same effect of armature amp-turns cancelling field amp-turns,
and neither field amps nor turns has changed

but internal EMF is now halved as you observed

half voltage / same current = half Xa

and that jibes with my vivid memory of us doing that experiment in machinery lab about 1966

professor said to us "You see boys, synchronous impedance is a function of frequency"
which i had to believe though didn't really quite understand why.

until years later when i was shown how direct and quadrature currents interact with field.


i hope this helps you.

if i am wrong please point it out - i am a plodder who values progress more than delusions of perfection.


you know, come to think of it don't recall seeing a formula for Xa,

for the big generators they give instead Short Circuit Ratio which is related to Xa.
Short Circuit Ratio is this:
[ Apply excitation to produce nominal rated voltage with machine unloaded,
apply short circuit,
measure fraction of nominal rated current that flows and that's short circuit ratio ]

our main generator (894 mva) was i think 0.58

our system guys used the inverse of that , called "per unit impedance", our was 171%.

if there's a genuine power engineer here who sees an error please correct me.

old jim

 

[I_am_learning]

You sure are learning!

let's think for a second here.
i have to because this is forty year old memories
and i remember how things work better than i remember formulas

thought experiment -
fixed field current
short circuit on machine
so armature current is internal EMF divided by Xa (assume Xleakage is much smaller)

shut off prime mover and machine coasts down

when it reaches half speed
armature current should still be same ,
because we have same effect of armature amp-turns cancelling field amp-turns,
and neither field amps nor turns has changed

but internal EMF is now halved as you observed

half voltage / same current = half Xa

and that jibes with my vivid memory of us doing that experiment in machinery lab about 1966

professor said to us "You see boys, synchronous impedance is a function of frequency"
which i had to believe though didn't really quite understand why.

until years later when i was shown how direct and quadrature currents interact with field.


i hope this helps you.

if i am wrong please point it out - i am a plodder who values progress more than delusions of perfection.


you know, come to think of it don't recall seeing a formula for Xa,

for the big generators they give instead Short Circuit Ratio which is related to Xa.
Short Circuit Ratio is this:
[ Apply excitation to produce nominal rated voltage with machine unloaded,
apply short circuit,
measure fraction of nominal rated current that flows and that's short circuit ratio ]

our main generator (894 mva) was i think 0.58

our system guys used the inverse of that , called "per unit impedance", our was 171%.

if there's a genuine power engineer here who sees an error please correct me.

old jim

Thanks jim.
I didn't get the bold part. I thought Armature current = Enternal EMF / X.
(X being whatever is present in the closed path, in our case, X = Xa)

 

[jim hardy]

remember Xa is a number calculated to represent a circuit element that would fit observed behavior of the alternator's armature reaction.

We are accustomed to X being determined by number of turns and area and permeability and of course frequency.

instead here it's ratio of calculated internal volts/observed current
where calculated internal volts is basically field current X speed X some constants.



"" this demagnetizing field seems to be proportional to the armature current only.""
indeed. and how much armature current flows?
Under short circuit, just enough to cancel out the field current else terminal voltage would not be zero.

since we locked field current it still takes same armature current to cancel field(amp-turns hasn't changed either place),
and we halved internal voltage by halving speed,
so result of X calculation is halved.

-------------------------------------

Maybe this is a better way to think about it..

Xa = (2 X pi X f) X La
where La is result of physical construction of armature and field.

in power side of EE we're not really accustomed to operating alternators off frequency
so we think of Xa as a machine constant.
that's how it's treated in basic machinery courses.
but really it's correct only at nameplate RPM.
La is what's the machine constant, function of area, turns, mu, airgap, etc.

sound plausible?

old jim

 

[I_am_learning]

remember Xa is a number calculated to represent a circuit element that would fit observed behavior of the alternator's armature reaction.

We are accustomed to X being determined by number of turns and area and permeability and of course frequency.

instead here it's ratio of calculated internal volts/observed current
where calculated internal volts is basically field current X speed X some constants.



"" this demagnetizing field seems to be proportional to the armature current only.""
indeed. and how much armature current flows?
Under short circuit, just enough to cancel out the field current else terminal voltage would not be zero.

since we locked field current it still takes same armature current to cancel field(amp-turns hasn't changed either place),
and we halved internal voltage by halving speed,
so result of X calculation is halved.

-------------------------------------

Maybe this is a better way to think about it..

Xa = (2 X pi X f) X La
where La is result of physical construction of armature and field.

in power side of EE we're not really accustomed to operating alternators off frequency
so we think of Xa as a machine constant.
that's how it's treated in basic machinery courses.
but really it's correct only at nameplate RPM.
La is what's the machine constant, function of area, turns, mu, airgap, etc.

sound plausible?

old jim

So, in fact, the known quantity is the armature current (which can be logically calculated by realizing that Armature mmf should cancel the field mmf, if the terminal voltage is to be 0), and from that we calculate, Xa, a quantity we invented to model the observed behavior of the Generator, And it happens that it comes to be dependent on frequency.
Oh! Got it.
Thanks a ton!

Now, I got an answer to my previous question,
When synchronous generator is run in isolated mode with fixed excitation, It won't get indefinite increase in terminal voltage with increase in Generator RPM, because with the generated emf, also the Xa gets increased. So, actually the terminal voltage and armature current remains pretty much very near the rated value even when the generator RPM shoots very high.

Here is a little calc I performed.
Enternal Emf = (1+2j) pu
X = 2j pu
Load R = 1 pu
So, current (I) = Emf / (X + R) = (1+2j)/(1+2j) = 1 pu
So, terminal Voltage Vt = Emf - I*X = (1+2j) - 1*2j = 1pu.
So, in this initial condition terminal voltage is exactly 1 pu.

Now, let's increase generator RPM by 5 times
So,
Emf = 5*(1+2j) pu
X = 5*2j (just seen that X is function of frequency, i.e. RPM)
So, current I = Emf / (X+R) = 5*(1+2j)/(10j+1) = 1.0396 - 0.39604j = 1.11249<-20.8545
So, Terminal Voltage = current*R = 1.11249<-20.8545 ;

Even when, RPM increases by 20 times,
Terminal voltages becomes only 1.1178
In fact, the maximum possible is only 1.1180 pu.

So, MATLAB wasn't wrong in showing the terminal voltage not increasing with RPM, but I was. :)

 

[jim hardy]

Sure looks to me like you got it.

""So, in fact, the known quantity is the armature current (which can be logically calculated by realizing that Armature mmf should cancel the field mmf, if the terminal voltage is to be 0),.."

known quantity is Ia?

yes, in my lab so many years ago we indeed shorted the terminals of an alternator (not a real big one probably 20KW or so) and measured it with a meter. then let it coast down and observed nearly constant indication..

fifteen years later had reason to operate a farily big one (about 300mva) into a short circuit, though at only 3600RPM.
Armature current was very well behaved and tracked field current.
we actually had a flux probe in that machine and watched every rotor slot go by on an oscilloscope. that let us locate shorted rotor turns.
flux is driven to near zero by armature reaction but right adjacent the rotor conductors there's field leakage flux and we detected that

[ Not my idea, it was thought up by a GE guy named Don Allbright.
i think you'll like his site, here
http://www.generatortech.com/C-Page3-Our Company-History.html

photo courtesy of generatortech, Don Allbright ]


......

i quickly went over your algebra,
at first glance i expected higher terminal volts
but i was remembering how our huge generator behaved open circuit.
hard to get one's thinking turned around at my age.
your math looks right to me
and your example certainly shows how an alternator with lots of armature reaction would be well behaved in an automobile application - the voltage regulator has an easy job.

further, my intuition says you're right
because your results look a lot like the graphs in that MIT paper.

Now - what would happen if your load suddenly became open circuit?
terminal volts would leap to your "Enternal Emf"...

and this is observed to happen in automobiles when a battery cable gets loose.
that was a painful learning curve for electronics industry. National claims in an appnote that 200 volt spikes are not unlikely and their line of automotive semiconductors is (nowadays) protected against them.
so when you see a friend's battery cables getting corroded and loose, implore him to fix it. it's more important these days with all the electronics in cars.

honestly, i really learned from your questions and your analysis..
a blend of math and mechanical 'working it in your head' can be very powerful.

thank you!

.........