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relation of PF, MVAR, voltage, kV

relation of PF, MVAR, voltage, kV

relation of PF, MVAR, voltage, kV

(OP)
good afternoon all..

can somebody explain why:
- inceasing voltage of generator makes power factor decreases (lagging condition), MVAR decreases, and kV from generatot decreases.
and vice versa.


any idea would be appreciated... thank you

RE: relation of PF, MVAR, voltage, kV

(OP)
what is the best for generator to have PF? in lagging or leading condition.

- what is PF in capacitive and inductive, or lagging and leading.

- we have one turbine is running on PF positive, and other is running on PF negative. why this system can work?

-what makes generator to have PF negative and positive.

thank you very much

RE: relation of PF, MVAR, voltage, kV

There are a lot of misconceptions embodied in your questions.  

I'd suggest finding a basic text on generator concepts  - the Electrical Engineers Handbook would be a good starting point.

Increasing excitation to generator increases the output voltage of the generator and causes it to create more VARs.  A generator that is producing vars is operating at a lagging pf.  A generator that is consuming vars is operating at a leading power factor.  

Generator voltage and power factor is controlled by the amount of field excitation current.   

RE: relation of PF, MVAR, voltage, kV

Deja Vu all over again!
Check this thread out;
thread238-217958: Reactive power

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: relation of PF, MVAR, voltage, kV

Generator is the source as well as sink of VArs. When sourcing VArs, it operates at lagging power factor and when consuming VArs it operates at leading power factor. This is done by excitation current level. Low field current means leading PF and consumes VArs. High field current meand lagging pf and hence delivers VArs.

RE: relation of PF, MVAR, voltage, kV

The underlying assumption is that the generator is connected to a (large) network. This means that the voltage at the terminals of the generator is fixed, that is, both magnitude and phase are fixed. Let's call this voltage U. The phase angle can be selected to be zero.

In addition, the power of the generator is fixed, determined by the machine turning the generator (the "prime mover"). Only the magnetization current can be controlled. It determines the  magnitude of the voltage induced in the stator windings. The phase of the induced voltage cannot be independently controlled. The induced voltage is E cos(fii) + jE sin(fii), where E is the magnitude of the induced voltage and fii is the phase angle.

The current in the stator winding is I = (E cos(fii) + jE sin(fii) -U) / jX , where X is the generator reactance. It is assumed that the resistance is so small that it can be neglected, and that the windings are connected in a wye. The complex power of the generator is equal to the product of the voltage times the complex conjugate of the current, multiplied by three, because of the three phases, S = 3UI*. The real power is the real part of the complex power, P = Re(S) = 3 U E sin(fii)/X.

When the magnetization current is increased, the magnitude of the induced voltage E increases. But because the power is fixed, the phase angle fii must change so that the product E sin(fii) stays constant. Because the induced voltage changes, the current (phase and magnitude) will also change. The reactive power and the power factor can thus be adjusted in this way by changing the magnetization current.

The interesting part is that sin(180deg - fii) = sin(fii), so that the same power can be obtained with two different phase angles, in principle, at least. I do not know, how this is achieved in practice. Maybe someone can explain this?

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