vintageplayer said:
Pretty much any electrical textbook defines reactive power pretty precisely... Q = Im{VI*}
Not having a textbook available, I looked at Wiki and other sources and noticed that they all seem to rely on a diagram that implies the energy is a vector quantity. That is pretty nonensical as the basis for strict treatment of the topic. Reactive Power may be a useful number to use when describing 'good' or 'bad' loads but it seems to be based on a very dodgy first step in its derivation. Reactive components do not dissipate power so they cannot, in themselves, consume any power from the supply.
I found a very revealing paragraph http://www.allaboutcircuits.com/textbook/alternating-current/chpt-11/true-reactive-and-apparent-power/:
"We know that reactive loads such as inductors and capacitors dissipate zero power, yet the fact that they drop voltage and draw current gives the deceptive impression that they actually
do dissipate power. This “phantom power” is called
reactive power, and it is measured in a unit called
Volt-Amps-Reactive (VAR), rather than watts".
He actually comes clean about it. What's being described is not Power - it's all in a name that's been (I would say mis-)applied to make the practicalities a bit more approachable. It's an analogy, in fact.
When you bring the idea of supplying coal to a power station you are begging a lot of questions and assuming that the term is fully justified. If you take a simple voltage source / load model and change the impedance of the load from resistive to partly reactive then the power dissipated by the load will be affected and the power supplied will be affected by the same amount. No Loss would be involved; the power delivered by the Voltage source will change by the same amount as the load power changes. So what's the difference between an ideal voltage source and a power station? An alternator winding has resistance and so will the supply cable. If a load is partly reactive then, in order to get the same Power transferred to the load, it resistance would have to be reduced (or the alternator voltage increased). That would increase the peak current values in the supply, which would increase the dissipation in the
Resistive Parts of the supply circuit. The economics of supplying Electricity are crucial and supply circuit resistance is highly relevant. If the peak current increases by 5%, the lost power will be 10% more. That's not a 'Reactive Loss'; it's a Resistive Loss, brought about by the presence of reactive elements.
The increased stress on components, when PF is not unity can also be relevant but the "phantom power" is not relevant - it's the increased Voltage or Current.