Which is the correct diagram (synchronous alternator varying PF)

[jaus tail]

Hi,

Sorry this question may be silly but I have an exam on 11th Feb on machines, maths and electronics.

Book had two diagrams for alternator external characteristics. But then which one of 2 diagrams is right?

Book says:
For zero pf loads, it is seen V(t) = E(f) +/- I(a)X(s)
It's minus for lagging loads, and plus for leading loads.
For lagging loads as load increases terminal voltage drops.
And for leading loads as load increases terminal voltage will rise.

But then book says that left diagram is for no load, then why is I(a) varying as in (on a scale). For no load, I(a) = 0. So terminal voltage must be constant horizontal line.

Then for right diagram book says:
With alternator running at rated V(t) and current, what happens if I(a) is reduced to zero(This means open circuit right?)
Book says that no load voltage may rise or fall depending on the load pf. But why?
If I(a) is zero then load pf must have no effect on excitation voltage.

Book says that E(f) > V(t) at lagging load. And E(f) < V(t) for leading loads. But I'm not understanding why the above two graphs are different?

Could this be correct theory:

Like for left graph. I have three alternators. One has lagging pf, one has unity pf, one has leading pf.
I run them on no load and adjust excitation (field current, same field current for three machines) such that terminal voltage is constant.
Now I add load to three machines (lagging, unity and leading) to and see the change in current.

For lagging loads, the addition of load reduces terminal voltage, whereas for leading load the addition of load increases terminal voltage.

But how does this experiment give right graph results?

 

[anorlunda]

But then book says that left diagram is for no load, then why is I(a) varying as in (on a scale). For no load, I(a) = 0. So terminal voltage must be constant horizontal line.

No load means real power is zero. Imaginary power is nonzero.

You are confusing no load with open circuit.

 

[jaus tail]

Thanks.

For right diagram/graphs, like suppose I have three machines
One has lagging load, second has leading load and third is at unity power factor.
Now the rated current is at rated voltage. How did this come at same point for all machines?
Is the excitation same or different for three loads?

I mean for leading V(t) = E + I(a) times X(s) and for lagging V(t) = E - I(a) times X(s).
So either the V(t) or the E must be different.

 

[anorlunda]

You can't consider the generator in isolation .The load plays a part too.

You can have 1.0 V and 1.0 I with different PF with different loads.

You should read this.
https://www.physicsforums.com/insights/ac-power-analysis-part-2-network-analysis/

 

[jim hardy]

I think you're right about the typo, i'd expect that line to be for 0pf leading.

That aside,
Why cannot both curves be correct ? They describe the machine at two different values of field current. Right hand family of curves just doesn't show the zero PF lines. But observe all curves cross at value of armature current they stated, zero or 1 pu.

Maybe this thread will help you understand why armature and field currents interact , and what is Xs.
https://www.physicsforums.com/threads/synchronous-generator-excitation.829603/

 

[jaus tail]

I think you're right about the typo, i'd expect that line to be for 0pf leading.

That aside,
Why cannot both curves be correct ? They describe the machine at two different values of field current. Right hand family of curves just doesn't show the zero PF lines. But observe all curves cross at value of armature current they stated, zero or 1 pu.

In the right hand family of curves how am I reducing I(a)? Is it done by reducing the KW load or by KVAR load or by both? Cause if I'm reducing only one load then the pf won't remain constant.

Like for leading pf, if I reduce the KW load, the load will reduce in magnitude but it will become more leading. So the curve should go upwards instead of down (for leading pf, in right side curve)

I'm not able to understand how are they reducing or increasing the currents keeping the pf constant as pf depends on load and excitation and excitation is constant.

No load means real power is zero. Imaginary power is nonzero.

You are confusing no load with open circuit.

So by no load and non-zero armature current, does it mean I have shorted the armature terminals? Or is current flowing through stray capacitance. In that case how will the power factor change as power factor depends on load and excitation and book says excitation is held constant?

 

[anorlunda]

The diagram is contradictory. It says that the entire curve on the left (a) is no-load. But on the curve, one of the lines is labeled "zero load" Clearly, the other lines in (a) are not zero load.

Just think back on the definition of $PF=\frac{P}{\sqrt{P^2+Q^2}}$ Forget generator properties. We know that both P and Q can be plus, minus, or zero independently.

You say the book says E is constant, but we can't see the entire context. Same for Fig 5.7, we can't see the entire context of what the book says about the assumptions behind that figure.

Once again, when you say stray capacitance, that suggests that you are thinking only of the generator and ignoring the load. The system includes the generator, the load, and transmission to connect the two. You must start thinking of the entire system.

 

[jim hardy]

I'm not able to understand how are they reducing or increasing the currents keeping the pf constant as pf depends on load and excitation and excitation is constant.

Think in small steps.

The fact terminal volts can go all the way to zero says this machine is not tied to the grid,
it's being tested by itself as in your school's machinery lab

spun at constant RPM (we had a big DC motor to drive ours)

and with a load bank consisting of adjustable resistors and capacitors.

So you can lock excitation, adjust load impedance to get desired amps and pf in armature
and plot results.
They made two such plots
left one with number of field amps that make 1PU volts open circuit
right one with number of field amps that make 1PU volts at rated armature current and pf=1.0

that's all it is.

 

[jim hardy]

I mean for leading V(t) = E + I(a) times X(s) and for lagging V(t) = E - I(a) times X(s).
So either the V(t) or the E must be different.

Isn't Et what's different, and plotted on the vertical axis?

Like for leading pf, if I reduce the KW load, the load will reduce in magnitude but it will become more leading. So the curve should go upwards instead of down (for leading pf, in right side curve)

You still haven't accepted that reactive current either aids or opposes field and that's what "Synchronous Reactance" Xs really is.

Thought experiment:
spin a machine at constant RPM, nothing connected to terminals except a voltmeter
apply field amps until terminal voltage is 1 pu
hold field amps at that value and apply a short circuit to the terminals

How much current flows ? E divided by Xs, as you said in first post

V(t) = E(f) +/- I(a)X(s)

what becomes of terminal volts V(t)? Falls to zero of course. so I(a) = -/+ E(f) / X(s)
What then is flux in the machine ? Basically zero else there'd be terminal volts.
Where did the flux go ? MMF from field amp-turns got canceled out by armature amp-turns so no flux flows.
That's the phenomenon X(s) describes. I tried to explain that in the link i gave in post 5 above.

X(s) could equally well be described by ratio of numbers of field turns to armature turns.
But most authors prefer to describe it as a separate inductive reactance because it behaves as if it were one.

Drawing the phasors will help you.

Understand what's going on , then work toward your formulas. Saves a lot of memorization.

 

[jaus tail]

Isn't Et what's different, and plotted on the vertical axis?
Thanks a lot for the replies.
The E here is the emf due to resultant flux right?
Like there is one flux due to field mmf, and one flux due to armature current.
So the phasor addition of that flux will be F(r) and that gives this E(t), right?

 

[jim hardy]

Thanks a lot for the replies.
The E here is the emf due to resultant flux right?
Like there is one flux due to field mmf, and one flux due to armature current. mmf

So the phasor addition of that flux will be F(r) and that gives this E(t), right?

That sounds right.

 

[jaus tail]

F(r) is the resultant flux. Thanks.