For a leading generator (absorbing vars), how is its MVA represented using rectangular coordinates? is it P + jQ or P - jQ? Appreciate a response and an explanation of the sign for the reactive term, if possible.

A leading generator would absorb vars from a leading load, correct? The apparent power for the load would be P - jQ (leading), correct?

 

VJP...

An AC generator (more aptly, an Alternator) produces two time-varying electrical outputs. one is terminal volts (V), the other, armature-current (A). Both, sinusoidal in nature, are usually separated in time. That time-difference is expressed as a third variable, eg., theta, in degrees.

The products of the three variables yield the so-called "Power Triangle." It's a right-triangle, having, of course, two sides, and an hypotenuse, identified as follows:

Hypotenuse = product of V x A. Units in Volt-Amperes.

Horiz-side = product V x A x Cos(theta). Units in Watts.

Vrt'l-side = product V x A x Sin(theta). Units in Volt-Amps reactive.


Using Vector-notation, the Horz-side represents the 'Real' component (P) and the Vrt'l-leg represents the 'Reactive' component (jQ).

o If the reactive component is inductive the equation is P + jQ.

o If the reactive component is capacitive the equation is P - jQ.

o If both components are present the equation is P + j(Qi-Qc)

For a more detailed discussion, I suggest you search Control.Com Archive for:

http://www.control.com/thread/1026236550

Regards,
Phil Corso

 

VJP...

Please note that for simplification purposes the Sqrt(3) multiplier, related to the 3-ph case, has been intentionally omitted.

Phil