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Hello everyone..

The subject question is related to over excited generator. The answer in the book is (B). 

But I when I tried to answer this question, my choice was (C) because it is over excited generator which means induced voltage is higher than terminal voltage. My understanding under this case the power factor for this system is lagging and I got more confident when I saw the the phasor diagram for synchronous generator supplying a lagging power factor in the NCEES reference book page#51. Therefore, I chose (C) because in lagging power factor VARs will be absorbed from the system and when the system is leading it means VARs will be supplied to the system, is it correct? 

Appreciate your help to clarify this question...

WhatsApp Image 2020-11-20 at 10.43.20 PM.jpeg

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When you have lagging power factor, the generator is sending VARs to the grid.

Here's an example:

Suppose you're at 1pu Voltage and 1pu Current in a lagging system, say by 30 degrees

that means V = 1pu and I = 1pu -30°

Now what are VARs?

VARS = VIsin(θ) = VIsin(θv - θi) = 1*1sin(0- (-30)) = 1sin(30) = 0.5pu

Now what if you were leading by 30°?

then VARS = VIsin(θv - θi) = 1*1sin(0-30) = -0.5pu

You may also just want to memorize that from a generator's perspective, lagging means supplying VARs and leading means absorbing VARs with the math above as a sanity check.

Edited by jd5191
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One thing I should note.

When a synchronous generator is over-excited, it has a lagging power factor.

When a synchronous motor is over-excited, it has a leading power factor.

But either way, when a synchronous generator or synchronous motor is over-excited, it delivers reactive power (Q, vars).

Synch Gen.png

Synch Motor.png

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Also a few phasor diagrams I drew out (not the most artistic) to try to explain over-excited synchronous machines.

I learned these from the Engineering Pro Guides book.

Basically, just remember:

  • Any synchronous machine (synch gen or synch motor) that is over-excited always delivers reactive power (Q, vars).
  • Overexcitation occurs when the real component of internal induced voltage Ea is greater than the magnitude of terminal voltage Vt: Re(Ea) > |Vt|

Overexcited synch gen phasor.png

Overexcited synch motor phasor.png

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1 hour ago, Mohammed Ahmed said:

Thank you all for explanation and clarification, but why synchronous motor delivers reactive power while in equation (shown above) it is in negative which means in my understanding (when Q or P negative) it is absorbing instead of supplying power? 

The synchronous motor is a load. This is shown by the current I entering the positive terminal of the load. And when a load's reactive power value is negative, that means the load is delivering reactive power.

When a source's reactive power Q value is negative, the source is absorbing reactive power.

But when a load's reactive power Q value is negative, the load is delivering reactive power.

Edited by akyip
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2 hours ago, akyip said:

The synchronous motor is a load. This is shown by the current I entering the positive terminal of the load. And when a load's reactive power value is negative, that means the load is delivering reactive power.

When a source's reactive power Q value is negative, the source is absorbing reactive power.

But when a load's reactive power Q value is negative, the load is delivering reactive power.

I got it.. thank you so much 

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