Circuit Response
every model has an edge, and the figure carries it
Concept

Power factor correction — where it appears

A capacitor across a supply, sized to cancel a load's reactive power by one division rather than found by trial. It is exact at the load and the frequency it was computed from and at no other, and at the point where it is exact the energy stored in the installation is exactly twice what it was.

Named by 4 essays across one field — each of them below, with the objects they name alongside it.

One capacitor of 77.3 µF, against every load it was not sized for. computed by solving, not by drawing. Sized from the 20 Ω load, the capacitor takes the power factor to 1.000000 there and leaves 0.0e+0 var of 1636 VA. At 178 Ω the same installation sits at 0.23 leading. The correction is exact at one point on this axis and nowhere else on it.

The capacitor that was right once

Cancelling a load's reactive power needs one division and no iteration, and the answer is exact. It is exact for the load it was computed from, at the frequency it was computed at, and the figure here is about what it does everywhere else — which includes making the installation worse than it was before anything was fitted.

power · Power factor correction Correcting a 20 Ω, 50 mH load to unity, and what it stores. computed by solving, not by drawing. A capacitor across a 230 V, 50 Hz supply is swept from nothing to 300 µF against a load drawing 1636 W and 1285 var. The reactive power falls through zero at 77.31 µF and keeps going; the energy stored in the installation rises from 2044.9 mJ to 4089.7 mJ at that point — exactly twice, because unity power factor means the two stores are equal — and goes on rising afterwards. Only the cable current has a least value, 7.113 A against 9.044 A.

The energy a unity power factor doubles

Reactive power can be computed three ways, and their agreement is easily called a verification. Two of the three are one theorem written twice and cannot disagree about anything; only the third is independent, and what it computes is a difference. Correcting a 20 Ω, 50 mH load to a power factor of 1.000000 takes its reactive power from 1,285 var to nothing and takes the energy stored in the installation from 2,044.9 mJ to 4,089.7 mJ — exactly twice, at every load and every frequency.

power · Reactive power A capacitor of 67% of the load's reactive power, moved along a feeder whose load is spread evenly over 10 km. computed by solving, not by drawing at each of 30 positions: one phase of an 11 kV feeder of 0.3 + j0.35 Ω/km, carrying 2 MW at 0.8 lagging spread over 30 equal loads. Uncorrected it loses 16.8 kW a phase; with every load corrected to unity at its own terminals, 11.1 kW — so 5.66 kW is the reactive current's share. This capacitor does best at 67% of the way out, where the loss is 11.8 kW and 87.6% of the reactive share is gone. At the substation end it removes nothing; at the far end it overshoots.

The capacitor two-thirds of the way out

A feeder's reactive current costs loss in every section it passes through, and on a feeder whose load is spread along its length that current is largest at the substation and falls to nothing at the far end. Correcting it at the substation removes no loss at all. One phase of an 11 kV feeder carrying 2 MW at 0.8 lagging over ten kilometres loses 16.8 kW, of which 5.66 kW is the reactive current's; the best single capacitor is about two-thirds of the load's reactive power placed about two-thirds of the way out, and it removes 87.6% of that share — against the eight-ninths the textbook rule gives for an idealised line. Two capacitors at 2/5 and 4/5 remove 95.6%. The same capacitor left in at light load costs more than it saves below half load, and the rule for a continuous load is a statement about many taps.

power · Reactive power A 640 kvar bank on a 16 MVA 400 V bus: alone its resonance with the supply is at h = 5.00, and with a 7% reactor in series at 3.01, with the bank's own null at 3.78 between that and everything above it. computed by solving, not by drawing: the impedance per phase at a 400 V bus, with a short-circuit level of 16 MVA and a 640 kvar correction bank, against harmonic order, in closed form and at the fifth by a netlist. The bank alone (dashed) resonates with the supply at h = 5.00, where the bus impedance peaks. With a reactor of 7% of the capacitor's fundamental reactance in series (solid), the bank's own series resonance is at h = 1/√0.07 = 3.78, a deep null, and the parallel peak has moved down to h = 3.01. Above the null the bank is an inductor, in parallel with the supply's inductance, so there is nothing for any higher harmonic to resonate with.

The reactor that keeps a bank out of it

A correction bank on a stiff bus resonates with the supply at a harmonic set by the two sizes, and at 640 kvar on 16 MVA that harmonic is the fifth, which it multiplies by 34. A reactor of 7 per cent of the capacitor's reactance in series puts a null at h = 3.78, and above that null the bank is an inductor, in parallel with another inductor, with nothing left to resonate. So the fifth is never raised at any size of bank. It was not free. The capacitor now stands 7.5 per cent above the bus, the resonance has gone below the null and can land on the third, 22.7 times at 654 kvar, and the number that looks like a detail, 7 rather than the 4 that would tune the bank to the fifth, is what keeps an ageing bank on the right side of its null.

power · Power factor correction

Named alongside it

The objects these essays reach for when they reach for this one.

Power factorSeries resonanceHarmonic amplificationOvercorrectionReactive powerClosed formComplex powerDesign tradeoffEnergy-storageGyratorNullorQuality factor

All concepts