The reactor that keeps a bank out of it

The capacitor that was right once measured what a correction capacitor does everywhere other than where it was sized, and it found the worst of those places at a harmonic. A capacitor that exactly cancels a load’s reactive power at 50 Hz forms a parallel resonance with the supply’s inductance, and a nonlinear load is a current source at exactly the frequencies that could reach it. It ended that discussion with a sentence: the engineered answer is a detuning reactor, a small inductance deliberately put in series with each capacitor to move the resonance below the lowest harmonic present, typically 7 per cent of the capacitor’s reactance, and it costs real money and real losses.

That sentence holds three claims. A series reactor moves the resonance; it moves it below the lowest harmonic present; and 7 per cent is the right number. Taken literally, the second is not generally achievable. Followed through on one real bus, the first is more interesting than it sounds, and the third is why this reactor is 7 per cent rather than something that looks more efficient. The reason, when it comes, is about how capacitors age rather than how they resonate.

The bus #

The bus is a 400 V bus at 50 Hz with a short-circuit level of 16 MVA, which is about what a 1 MVA transformer at 6 per cent gives. The whole of the supply’s inductance is 32 µH per phase at this level, with a Q of 40 applied so that its resonance has a finite height. The correction bank is star-equivalent per phase, sized by its reactive power at the bus’s voltage. Its reactor is stated in the conventional way as a percentage $p$ of the capacitor’s fundamental reactance, with a small series resistance from the reactor’s own Q of 30. The fifth harmonic’s bus impedance is checked against a netlist of those elements, and it agrees to a part in a thousand with the closed form.

Without a reactor, the parallel resonance of the bank and the supply is at $h = \sqrt{S_{sc}/Q_c}$. That is the classic formula and it is what the figure finds: $h = 8.95$ for 200 kvar, 6.32 for 400 kvar, 5.00 for 640 kvar and 4.47 for 800 kvar. A plant that adds capacitance a step at a time therefore walks its resonance down the harmonic series, and at each step something different is amplified. On this bus, 640 kvar puts it exactly on the fifth.

An inductor above its null #

A capacitor with a series inductor is a series resonator with a null in its impedance at $h_t = 1/\sqrt p$, which is 3.78 for 7 per cent, or 189 Hz. This is the essential fact, and a simple sentence states what follows from it. Below the null the bank is a capacitor and above the null it is an inductor. So above the null the bus has two inductors in parallel, the supply’s and the bank’s, and two inductors have no resonance. Every harmonic above $h_t$ sees a bus impedance lower than the supply’s alone, and it does so at every size of bank. The bank has been removed from the question, for all of those harmonics at once.

Below the null the bank is still a capacitor, and the supply is still an inductor, and they still have a resonance. It is now below the null, because the inductance in series with $C$ has risen. On the figure it has gone from 5.00 to 3.01. No reactor gets rid of that resonance: it can only be placed. The sentence that says it moves below the lowest harmonic present is about choosing $p$ so that no harmonic is there. That works for a 3-wire 400 V bus where the third has no path, and that condition is doing a lot of work, as the next figure shows.

What it does to the fifth, and to the third #

Swept against the size of the bank, the difference is striking.

Without a reactor the fifth in the supply is multiplied by 1.45 at 200 kvar, by 34.2 at 642 kvar and by 1.77 at 1,000 kvar. The peak is a real event and its size is set only by the circuit’s Q. Thirty-four times a 20 per cent fifth is more than the fundamental, which is why a correction bank on a harmonic-rich bus can fail rather than merely be ineffective, as the capacitor that was right once noted in general terms. With the reactor the same fifth is 0.71, 0.43 and 0.32 of what was injected. At every size of bank, the bank now helps. That is the first half of what the reactor buys: it does not need to be sized to avoid anything, and a plant can add steps to its bank without any of them landing on the fifth.

And it has done something else, shown by the dots. At 654 kvar with a 7 per cent reactor, the third in the supply is multiplied by 22.7. The resonance has not gone. It has been moved onto the next harmonic down. This is one reason a 7 per cent bank is specified only on buses where the third is known to be small: a 3-wire delta-connected bank on a balanced load has no third-harmonic path, since triple-n currents in a balanced 3-wire system have nowhere to flow. The sequence a harmonic belongs to says why. A 4-wire bus with single-phase rectifier loads has a very large third in its neutral, and a star-connected bank with a neutral on that bus would see this resonance. The standard answer is a much bigger reactor, 14 per cent, which places the null at $h = 2.67$, below the third.

So the phrase “below the lowest harmonic present” conceals a real decision. Below the null there is still a peak somewhere, and the choice of $p$ is a choice of where to put it.

The price at the fundamental #

The reactor has a cost at 50 Hz as well, and it is exact.

The inductor’s voltage opposes the capacitor’s, so the capacitor stands above the bus by exactly what the inductor drops, and its fundamental voltage is $1/(1 - p)$ of the bus. At 7 per cent that is 1.075 times, or 430 V on a 400 V bus, before any harmonic has arrived. It is the first thing a bank’s specification has to get right, because a capacitor’s life falls steeply with its voltage: a bank that is built with 400 V capacitors and given a reactor is overvoltaged by 7.5 per cent all day. Detuned banks are built with higher-rated capacitors for that reason, typically 440 V or 480 V, and a higher-rated capacitor’s kvar is rated at its rated voltage. So what is written on the capacitor and what the bank delivers are two quite different numbers.

The same factor raises what the bank delivers, for a given capacitance: 1.075 times the reactive power of the capacitor alone, since the current has risen in the same ratio. That is the reactor giving something back, and it shows up in the correction calculation. A bank sized with a calculation that ignores the reactor will come out 7.5 per cent too large. It would be a small error if the correction’s own exactness had not been the subject of the previous essay: the answer to the exact calculation now has a reactor in it.

A filter is a different job #

A reader looking at the null will have an obvious question. Why not put it on the fifth? A reactor of 4 per cent places the null at exactly $h = 5$, where the bank has almost no impedance, and that takes the fifth out of the supply entirely. It is called a tuned filter and it is a thing people build.

At 4 per cent, a 500 kvar bank takes 1.00 of the fifth and leaves 0.01 in the supply. That is what a filter does, and the figure shows what it costs. Every bit of the fifth that was going into the supply from every load on the bus is now going through this capacitor, and for a bank that has to carry 20 or 30 per cent of its fundamental current as fifth harmonic that is a large rise in its current and a large rise in its peak voltage, since the voltage is the sum. A filter has to be rated for that and it has to be designed with the rest of the bus’s harmonic spectrum in view. The reactance cancelled, and the resonance it buys described the same exchange in series compensation: a resonance placed deliberately has to be paid for in that branch.

The 7 per cent bank does something else. It takes 0.51 of the fifth and leaves 0.49, a fairly even division, and it is not trying to clean the bus. It carries a bearable fifth, and it does not raise anything anywhere above its null. Those are modest aims, and it meets them for almost any bus. The two designs differ by three per cent of a capacitor’s reactance, and they are different products doing different jobs.

Now look to the left of 4 per cent. At 3 per cent, the bank has 1.47 of the fifth in it and the supply 0.47; at 2 per cent, 2.77 and 1.77, so that the supply is carrying more than was injected; and at 1 per cent, 20.7 and 19.9. The null is now above the fifth. So at the fifth the bank is a capacitor, the supply’s resonance is back, and it takes only a small change in the reactor’s share to get from “a filter” to “an amplifier”. The null’s position is what decides which of those it is, and it can move.

The wrong side of its own null #

And that is the reason for 7. A real capacitor does not hold its value. Self-healing metallised capacitors lose capacitance through their life as their film clears local faults. A bank’s fuses can blow one element at a time. So over a bank’s life its capacitance falls, and the null at $h_t = 1/\sqrt{p\,C/C_0}$ rises.

A bank tuned at 4 per cent has the fifth on the wrong side of its null at the first per cent lost, because it was on the null to begin with. From that moment the bank is capacitive at the fifth, and as it ages its capacitive reactance grows until it matches the supply’s. For this 200 kvar bank that is at about a quarter of the capacitance lost, where the resonance has come back on exactly the harmonic the filter was meant to remove. Real tuned filters are therefore tuned a little below their harmonic, at about 4.7 rather than 5, and their aging is monitored, with a lost-capacitance alarm. That margin is the difference between what the formula says and what will be true in ten years.

A bank at 7 per cent has its null at 3.78, far below the fifth. Losing capacitance moves the null upward and the bank’s reactance at the fifth falls, so the supply’s share falls, and the fifth is actually reduced a little as the bank ages. A loss of 43 per cent would be needed to bring the null to the fifth, by which time the bank would have been replaced for having lost most of its correction. So the 7 per cent number is not a fine optimisation but the largest margin on the fifth that is reasonable to pay for. And that is why the standard numbers are 7 and 14, rather than the 5 or 6 that would absorb more: the capacitor that was right once said the sizing was exact and could not be installed alone, and a reactor can’t be chosen as if the capacitor it is paired with will stay the same.

Sizing, with a reactor in it #

Together, these change how a correction bank’s specification reads. The bank’s target reactive power, at the bus’s voltage, gives the capacitance after dividing by $1/(1 - p)$. The capacitor’s rating is set by $V/(1 - p)$ plus the harmonic voltage it carries, with a margin for the bus’s own tolerance. The reactor has to carry the fundamental plus the fifth and seventh without saturating, because a saturating reactor’s inductance falls, which raises the null towards the harmonic it was meant to be below. It has its own loss, usually about 0.5 per cent of the kvar, which is real heat in a real cabinet. All of that makes a detuned bank perhaps half again as expensive as a plain one per kvar.

That cost is what a plant is buying a guarantee with. A plain bank may be fine at its first step, and it becomes a hazard when a variable-speed drive is added or the plant’s transformer changes, both of which move the bus’s resonance. The plain bank is a calculation that is right once. A detuned bank is a design that stays within its limits as the plant changes, at every size of bank and through the bank’s own ageing, and it is the same thing being asked of a correction system that the gain a tolerance demands asks of an oscillator: a margin that is paid for in advance, because the value that would be exact will not stay there.

What the bus leaves out #

The supply here is a pure inductance with a Q, and the load is a current source at each harmonic with no impedance of its own. A real load’s resistance damps the resonance, often heavily, so the 34 is an upper bound; a bus with a large resistive load has its peak well below that. The fifth is treated on its own, where a real drive injects a fifth and a seventh at a stated phase relative to each other and to the voltage. The bank has no switching transient in this model, although switching a capacitor onto a stiff bus is its own large problem, of the kind the first cycle no steady state contains measures, and it is what the reactor helps with in another way, by limiting inrush. And a real bank’s parallel resonance on a stiff transformer interacts with other banks on the same bus: two banks with different reactors have two nulls and a peak between them, and that can land anywhere.

The number to keep #

A series reactor of share $p$ puts a null at $h = 1/\sqrt p$, and above it a bank is an inductor with nothing left to resonate with: 34.2 times the fifth becomes 0.43 at the bank size that was worst. The resonance is not removed. It is below the null, 22.7 times the third at 654 kvar on this bus, which is why a 7 per cent bank belongs where the third has no path and a 14 per cent one where it does. The capacitor stands $1/(1 - p)$ above the bus, 430 V on 400, and delivers 7.5 per cent more reactive power for it. At 4 per cent the bank is a filter that carries the whole fifth, and it is on the wrong side of its null at the first per cent of capacitance lost. Seven per cent keeps it well clear of the fifth, and ageing moves it the safe way.

Still open: a bank of steps, a saturating reactor, and two banks on one bus #

A bank that switches in steps. A real correction bank has steps, with reactive power switched in and out by a controller, and each step has its own reactor. Whether all steps should have one share of reactance, and what happens to the bus’s peak between two steps with unequal capacitance, is the same calculation with several branches.

A reactor that saturates. The reactor has to carry the fundamental, the fifth and any inrush, and an iron-core reactor’s inductance falls at high current. Since the null rises as the inductance falls, a saturating reactor moves the bank towards the harmonic at the moment it is carrying most of it. How much margin a reactor’s saturation current has to have over the worst combination is a nonlinear calculation on this same bus.

Two banks on one bus. Two banks with different reactors have two nulls, with a parallel peak between them that neither has alone. Where that peak falls depends on the ratio of their sizes, and it can land on a harmonic neither bank was near. Whether two plants sharing one transformer can each be correct alone and wrong together is a direct extension of the bus impedance here.