4
\$\begingroup\$

I understand that reactive power injection can be used by generators to support the grid voltage when sagging, but I’ve never understood why this works. I thought it could be charging transmission line capacitance or something, but I can’t make sense of it when I try to follow through with this starting point.

I’ve seen similar questions on here with great answers. In Why is it desirable to inject reactive power into a transmission system? the answer explains how capacitive compensation on the transmission line can reduce the reactive power draw from a generator to power a partially-reactive load, ultimately explaining why reactive power is relevant in a power system. In Why does reactive power affect voltage? the answer from Olin explains how a reactive load can draw additional current from a generator and result in real power loss and voltage drop across the transmission line. I am guessing there is enough info from these answers that I could be able to come to an answer to this question, but I think I need some spoon-feeding.

The assumption in my question is that reactive power injection from a generator supports grid voltage. Is this generally true? If so, could someone help me understand how reactive power injection can be used to raise a sagging grid voltage, or the voltage at a local bus? Are there truths or assumptions made about the grid’s loads and lines in doing this?

\$\endgroup\$
5
  • \$\begingroup\$ Forget the grid. Too big, Too distributed. Look at power factor correction for an induction motor. This can be extrapolated for the grid. \$\endgroup\$
    – StainlessSteelRat
    Nov 26, 2021 at 20:06
  • \$\begingroup\$ Tip: when you paste a Stack Exchange question URL into a question it is automatically rendered as the question title with a hyperlink. \$\endgroup\$
    – Transistor
    Nov 26, 2021 at 20:10
  • \$\begingroup\$ Thank you Transistor, I’ll edit. StainlessSteelRat, i can make sense of simplifying this that way. I think you are saying if we apply power factor correction with reactive power injection, it will reduce the apparent power of loads that are burdening other generators, which would mitigate a sag? \$\endgroup\$
    – Mitchell Easley
    Nov 26, 2021 at 20:16
  • 1
    \$\begingroup\$ It decreases reactive power source has to supply, which decreases apparent power the source has to supply, which decreases current the source has to supply. Injecting leading reactive power decreases load reactive power, improving power factor and decreasing current. Same thing as power factor correction with a capacitor at a load. \$\endgroup\$
    – StainlessSteelRat
    Nov 26, 2021 at 20:46
  • \$\begingroup\$ Easiest way to see it is to sketch out a simple system and assign some power flow (P, Q). Stick a generator in place providing P, but no Q. Then work it again with the generator providing Q to bus. Recall \$I=\frac{S^*}{V^*}\$ \$\endgroup\$
    – relayman357
    Nov 26, 2021 at 21:51

1 Answer 1

Reset to default
8
\$\begingroup\$

The primary assumption about a grid is that a high percentage of the load energy is used by induction motors. Induction motors have a lagging power factor and thus require reactive volt-amperes, aka reactive power. Reactive VA is a measure of energy stored during one half-cycle and released during the following half-cycle of the current waveform.

If the generator must supply the reactive VA, the current from the generator to the induction motor loads is higher than is necessary to supply the energy taken by the motor and converted to mechanical energy. If the reactive VA can be supplied from capacitors located closer to the motor, the voltage drop and lost energy between the generator and motor is reduced.

Locating capacitors or "injecting reactive power" close to the loads having lagging power factor is essential. Doing that at the generator helps with losses and voltage drop inside the generator, but does nothing to help with the transmission system component losses. Reactive power can be "injected" anyplace between the generator and the reactive load, but the benefit increases as the injection point gets closer to the load.

The word "injected" is really not very appropriate. It implies forcing something into a place against some opposing force. It is really more like providing an outlet. It is more like providing a temporary storage place or overflow reservoir

What is reactive energy? may be of interest.

\$\endgroup\$
2
  • 1
    \$\begingroup\$ I myself am not well versed in this, but from what I know, and from your answer: it is not really "injecting reactive power", but rather "injecting reactive power mitigation" that helps (which is what the capacitor banks do). \$\endgroup\$
    – anrieff
    Nov 26, 2021 at 23:10
  • \$\begingroup\$ @anrieff Thank you. I added a paragraph to my answer. \$\endgroup\$
    – user80875
    Nov 27, 2021 at 0:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.