Hi,

I've always used this rule of thumb:

- lagging generator (over-excited): exporting Watt and exporting VAR

- leading generator (under-excited): exporting Watt and importing VAR

- lagging motor (under-excited): importing Watt and importing VAR

- leading motor (over-excited): importing Watt and exporting VAR

so along this line of thinking, lagging generator θ is positive, leading generator θ is negative, lagging motor θ is negative, leading motor θ is positive. where θ is the impedance phase angle in the "power triangle."

Is this correct?

I'm getting confused with the phase difference between voltage and current (see attached photos from graffeo book). in his lagging generator explanation, the θ is negative. is the θ between voltage and current different than the θ in the power triangle? I thought these were the same. Or is the rule of thumb i've been using wrong from the start?

can someone please shed some light here for me. thank you.

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Your thumb rule is correct. PF angle is same as power triangle when we have one voltage across the load. These two concepts and the note below should not be mixed till one is thorough in the all the concepts. So it will be a good idea to understand them separately and thoroughly.

Note -This concept is only for the machines connected to Grid (Vt is constant) else for stand alone cases the scenario will be different.

Let me give a try for Gen?Motor? connected to grid. In generator and motor case the pf angle theta, if not specified otherwise, is the angle between terminal voltage, repeat terminal voltage, repeat terminal voltage and the current. There is no change in the notation of this angle (negative when behind the voltage and positive when ahead) leading implies ahead of Vt and lagging means behind Vt. Now build your phasor diagram from here; E=Vt+IX for generator and E=Vt-IX for motor. If E obtained like this is more than Vt, the machine (whether G or M) is over excited and if less it is under excited.

Let us come to whether VARs are consumed or supplied. For this, one has to compare reactive power with real power and decide. Say a generator delivering a power to a lagging pf load. By Vt x I* method, we see that both P and jQ are positive ,so both, real and reactive powers are supplied by Generator and consumed by load and from phasor we get Gen is over excited. This makes sense as the excitation current in the generator is responsible for generation of reactive power and turbine input for generation of active power.

Similarly for leading pf, Vt x I* will give us reactive power as negative and real power positive means load is consuming and hence Gen is supplying the real power and load is supplying and Gen is consuming the reactive power. From phasor Gen is under excited. This also makes sense when we say Gen is under excited the E will be less than Vt, at the same time the flux required to produce power will be less (due to less excitation), so some one from out side has to strengthen the Gen flux else terminal voltage will drop which is not possible because the thing is connected to grid.

For motor, as it is a load, the explanation is a bit simple. When P and Q are both positive, it is consuming both, real and reactive power and again from phaser we will see that it is under excited and I is lagging. When the motor is over excited, it will have I as leading resulting into negative real power implying it is consuming real power and supplying reactive power.

One has to go on thinking about it till one gets it, because the stuff involves mixing of many concepts. If it has not helped, you can frame your question pin pointed at one point (instead of understanding the whole lot of concepts in one go) so that we can discuss that part in detail.