Reactive power compensation
0% of Q
shunt bank
noneexact130%
A shunt branch across the load supplying the reactive power the load
wants, so the supply no longer has to. Sized as a share of
the load's Q rather than in kvar, so it stays put while you drag φ.
Past 100% it overshoots and the supply current starts leading again.
It is a capacitor only for a lagging load. Drag φ negative and
the same control fits a reactor instead, because a capacitor
on an already-leading load makes it worse. The readout names whichever
it is — they are different hardware.
Where it goes matters. The load still draws its full current.
Only the circuit upstream of the bank sees the reduction, so
the run between bank and load saves nothing. That is why a bank goes
as close to the load as the money allows — at the transformer it
protects only the grid connection, at the cabinet it protects the
feeder too.
Reading the picture
The wheel. Both arrows turn together at ω and the angle between
them never changes — that fixed angle is the power factor.
Each arrow is drawn at the full radius of its own axis, so their
lengths are not comparable; the angle is what carries meaning.
The dotted lines. Each arrow's height above the centre is
exactly the value plotted on the curve beside it. The sine wave is the
shadow of the rotating arrow, not an analogy for it.
The dashed triangle. The current splits into a part along the
voltage — I cos φ, which does all the work — and a
part across it, I sin φ, which does none. The green
leg is the current this load would draw at pf 1.00. Everything
beyond it is heating cable for nothing.
The power curve. p = v × i pulses at
twice line frequency, around a mean equal to the active
power. The red lobes are energy flowing back out of the load, having
done nothing at all. At pf 1.00 there are none; at pf 0 the
mean is zero and nothing but sloshing is left.
Compensation. Fit a bank and the current arrow swings back
towards the voltage and gets shorter. The green band does not move —
the useful current never changes. What shrinks is the red leg, and the
violet segment beyond it is exactly the part the bank is now supplying
instead of the grid. All three tips sit on one line at right angles to
the voltage, which is the whole of power factor correction in one
picture.
Where the cost actually is. Watch the two numbers disagree: at
pf 0.80 only about 3% of the energy comes back each
cycle, but the current is 25% higher — and heating goes with
the square of current, so the run wastes about 56% more. The
returned energy is not the expense. The extra current is.
Pure sinusoids throughout, so cos φ here is the true power
factor. A real rectifier front end also draws distorted current, whose
distortion power is a third axis this tool does not have.