Parallel Operation and Load Sharing

A percentage impedance is only meaningful against its own base, so before the two can be compared they must be referred to one common kVA base. Take 500 kVA:

The 250 kVA machine is electrically the stiffer of the two by a factor of 2.5 — it is 5% impedant at 250 kVA, which is 10% at 500 kVA.

Equal ratios mean the two windings sit across the same terminal voltage, so on the common base the current, and hence the kVA, divides inversely as the impedance:

Divide the 600 kVA in that ratio. The base cancels if the shares are written straight from the nameplates as rating divided by percentage impedance:

The two add to 600 kVA exactly, which they may do here only because the impedance angles are equal — Problem 2 shows what happens when they are not.

Express each share as a fraction of its own rating:

The unit with the lower percentage impedance is the harder worked of the two, whatever their ratings. With 600 kVA drawn from a bank rated 750 kVA in total, one machine is at 86% and the other loafing at 69%.

The power factor is the same for both and equal to the load's, because the two impedances have the same angle. Each machine therefore delivers 0.8 of its share as kW:

Put the impedances in polar form, because it is the angles that make this problem different from Problem 1:

A ratio \(X/R\) of 4.0 for the larger unit and 1.2 for the smaller is entirely normal: leakage reactance grows with size faster than winding resistance falls.

The load current, with the busbar voltage as reference:

Divide it inversely as the impedances, keeping the angles:

Transformer A, the more reactive of the two, carries a current lagging by nearly 49°; transformer B carries one lagging by only 23°. Neither is at the load's 36.87°.

Convert to volt-amperes at the busbar voltage. Each machine's power factor is the cosine of its own current angle:

Now the point of the problem. Add the two kVA figures arithmetically and compare with the load:

Nothing has been created. The two currents are not in phase, so their magnitudes cannot be added; only their components can.

Check by resolving into kW and kvar, which must add:

240 kW and 180 kvar are exactly the load's \(300\times0.8\) and \(300\times0.6\). The kVA column is the only one that does not balance.

What the nameplate rule of Problem 1 would have predicted. The percentage impedances are \(0.04123/(400^2/200000) = 5.15\%\) and \(0.04690/(400^2/150000) = 4.40\%\), giving shares of 159.7 kVA and 140.3 kVA.

The ratio it predicts, \(|Z_B|/|Z_A| = 1.138\), is exactly right — current division by magnitude is unaffected by the angles. What is wrong is the assumption that the two shares add to 300.

Form the sharing weights as rating over percentage impedance, exactly as in Problem 1 but for three units:

Ask which unit reaches its rating first. Divide each share fraction by the machine's own rating — or, more directly, note that the fractional loading of a unit is proportional to \(1/Z\%\), so the smallest percentage impedance is always the first to saturate. Here that is P, at 4.0%.

Scale the total until P is exactly full. P carries 0.2174 of whatever the bank carries, so

Check the other two at that load:

Nearly 350 kVA of perfectly good transformer is unusable, and the reason is entirely a mismatch of percentage impedance.

The single change that recovers it. Every unit reaches 100% together only if the loading fractions are equal, and since loading \(\propto 1/Z\%\) that requires

Making all three 6% (the largest present) would divide 1500 kVA as 250 : 500 : 750 — exactly in proportion to the ratings. In practice the 4% and 5% units would be re-specified, or a series reactor fitted, to bring them up to 6%; nothing can be done to the 6% machine to help.

On open circuit the two secondaries form a closed loop. The 4 V difference between the induced emfs is impressed on the two impedances in series — there is nowhere else for it to act:

Judge it against the rating. Rated secondary current is

A ratio error of 4 V in 220 V — 1.8% — produces a permanent 16% current. The reason is visible in per unit: the driving voltage is 1.8% while the impedance opposing it is only \(5.03\% + 6.14\% = 11.2\%\), so \(0.0182/0.1117 = 0.163\) pu.

This current does no useful work at all. It leaves transformer A, passes through transformer B against its emf, and returns — heating both windings and consuming part of the thermal rating that the load should have had. It is present whenever the transformers are energised, load or no load.

Now connect the load. Its current, referred to the 220 V busbar:

Superpose the two effects. Writing \(E_A = I_AZ_A+V\) and \(E_B = I_BZ_B+V\) with \(I_A+I_B = I\) and eliminating \(V\):

Note the two impedances are almost parallel in angle here (70.82° and 70.35°), so the load-division term barely rotates the current; it is the circulating term that skews everything.

Convert to volt-amperes at the 220 V busbar:

A 150 kVA load on a 200 kVA bank, and transformer A is already at 96.5% of rating. Had the ratios matched, the division term alone would have given 82.5 kVA and 67.5 kVA — still unequal, because \(|Z_A| < |Z_B|\), but nothing like as lopsided. The extra 14 kVA on A and the 13 kVA missing from B are the circulating current at work.

Being three-phase changes nothing about the sharing rule. Both nameplate kVA figures are three-phase totals and both percentage impedances are already per-unit quantities, so the weights are formed exactly as before:

Divide the 2000 kVA:

The smaller machine, being the stiffer one, is carrying almost its full rating while the larger sits at two-thirds.

Secondary line currents from the three-phase kVA at 433 V:

The currents add arithmetically here only because the impedance angles were taken equal, so all three phasors are collinear.

Maximum bank load. T2 saturates first, at 45.45% of whatever the board carries:

2200 kVA out of an installed 2600 kVA, or 84.6%. The 400 kVA shortfall is the price of the 1.5-point impedance mismatch.

The matched impedance. Proportional sharing needs equal loading fractions, hence equal percentage impedances:

With that value the shares at 2600 kVA would be 1600 kVA and 1000 kVA, both units exactly full. It is worth noting that raising T2's impedance increases the bank's capacity — the opposite of what intuition suggests about impedance.

The conditions, in order of severity. The first three cannot be violated at all; the last two govern only how well the bank performs:

The four displacement groups. Every standard three-phase connection puts the secondary line voltage at one of four angles relative to the primary, and the group number is that angle in units of 30° measured clockwise:

The reason groups 3 and 4 can be bridged is that interchanging two phases on the primary and the secondary reverses the phase sequence through the transformer, which negates the displacement: \(-30^\circ\) becomes \(+30^\circ\). Relabelling terminals can only rotate a set of phasors by \(\pm120^\circ\), so it can never turn 0° into 180°, and groups 1 and 2 stay apart.

The circulating voltage produced by a displacement. Two secondary phasors of equal magnitude \(E\) separated by an angle \(\theta\) leave a difference

Unlike the ratio error of Problem 4, this difference is not a small trim — it is a fixed fraction of the whole secondary voltage.

Thirty degrees. In per unit, with \(E = 1.0\) and the two 6% impedances in series:

Over four times full-load current, permanently, with no load connected. The bank would trip on overcurrent within seconds, and would burn out if it did not.

Sixty degrees — a group 3 unit paralleled with a group 4 unit without the phase interchange:

The full secondary voltage now appears across the loop impedance. For comparison, 180° would give \(2.0/0.12 = 16.7\) pu, which is simply a short circuit fed from both ends.

Put the three cases side by side against the 16.3% of Problem 4:

A tap mismatch wastes capacity; a group mismatch destroys the transformers. The two faults differ by two orders of magnitude.