Parallel Operation of Transformers

Parallel operation means two or more transformers are connected to the same supply bus bars on the primary side and to a common bus bar or load on the secondary side.

The reliability argument is usually decisive. A substation with two half-size units can be maintained without interrupting supply, whereas one full-size unit cannot — and the capital cost penalty is modest.

Everything is referred to the secondary side, since that is where the two machines meet. The no-load secondary EMFs are then \(E_A = V_1/a_1\) and \(E_B = V_1/a_2\), both driven by the same primary voltage.

For transformers sharing a common load:

1. The voltage ratio must be the same.
2. The per-unit impedance of each machine on its own base must be the same.
3. The polarity must be the same, so that there is no circulating current between the transformers.
4. The phase sequence must be the same, and no phase difference must exist between the voltages of the two transformers.

Condition 3 is the same check as Chapter 47's polarity test, and for the same reason. Condition 4 concerns three-phase transformers and their vector groups, which Part 3 takes up later; for single-phase units it reduces to getting the polarity right.

Three equations describe the arrangement completely: one node equation and two loop equations.

Solving (2) and (3) for \(I_A\): equating the two expressions for \(V_L\) gives

Setting the load to zero leaves only the second component, which therefore exists whether or not any load is connected.

Note how small a difference suffices. The impedances are only a few percent, so their sum is small, and a ratio mismatch of even 1 % can drive a circulating current of several percent of rated. This is why the taps of paralleled transformers must be matched, not merely their nameplate ratios.

With \(a_1 = a_2\) the circulating term vanishes and only the sharing term remains:

The condition \(a_1 = a_2\) is easier to state than to achieve, since the ratio depends on the tap position as well as the nameplate. In practice both machines are set to the same tap, and the residual mismatch is accepted.

Multiplying each current by the common load voltage converts the sharing rule into kVA.

If per-unit equivalent impedances are not equal, transformers will not share the load in proportion to their kVA ratings, and the overall rating of the transformer bank will be reduced.

• Transformers are paralleled because a single large unit may be unavailable, demand grows, reliability improves, and transport is easier.

• Four conditions: same voltage ratio, same per-unit impedance on own base, same polarity, same phase sequence. Polarity and phase sequence are absolute.

• Each current has two components: a share of the load, and a circulating current.

• The circulating current \(I_c = (E_A - E_B)/(Z_A + Z_B)\) flows even at no load, is nearly purely reactive, wastes copper loss and reduces the permissible kVA.

• With equal ratios, \(I_A/I_B = Z_B/Z_A\) — currents divide inversely as the impedances.

• Likewise \(S_A/S_B = Z_B/Z_A\) — the kVA load is inversely proportional to impedance.

• For sharing in proportion to rating, impedances must be inversely proportional to ratings, i.e. equal percentage impedance on each machine's own base, equivalently \(I_AZ_A = I_BZ_B\).

• If not, the bank loses capacity: the lower-impedance machine overloads while the other still has headroom.

• Equalising percentage impedance on a common base. The condition is equal on each machine's own base — which means unequal on a common one.

• Expecting equal machines to share equally regardless of impedance. The impedance decides, not the rating.

• Getting the sharing ratio upside down. \(S_A/S_B = Z_B/Z_A\) — the lower impedance takes more.

• Thinking the circulating current appears only on load. It flows at no load too.

• Adding \(\left|I_A\right|\) and \(\left|I_B\right|\) arithmetically. They must be added as phasors.

• Using primary-referred impedances with secondary currents. Refer everything to one side first.

• Ignoring tap position. Equal nameplate ratios on different taps are not equal ratios.

• Paralleling without a polarity check. Reversed polarity is a dead short through both machines.

• Assuming a lower percentage impedance is better. It overloads that machine and raises the fault level.

• Forgetting phase sequence and vector group for three-phase banks.

For sharing problems, convert both impedances to a common base first — that is what makes them proportional to the ohmic values the formula needs.

1. P49.1 Two transformers with equal ratios have \(Z_A = 0.15 + j0.45~\Omega\) and \(Z_B = 0.10 + j0.30~\Omega\) referred to the secondary. Find the ratio in which they share a load.

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   \[Z_A = 0.4743~\Omega, \qquad Z_B = 0.3162~\Omega\]

   \[\frac{I_A}{I_B} = \frac{Z_B}{Z_A} = \frac{0.3162}{0.4743} = 0.6667\]

   B takes 1.5 times as much as A. Note both impedances have the same angle, so the currents are in phase and the magnitudes add directly.

2. P49.2 A 100 kVA transformer at 3 % and a 200 kVA at 4 % share 250 kVA. Find each share.

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   \[Z_A = (3)\frac{100}{100} = 3.000\,\%, \qquad Z_B = (4)\frac{100}{200} = 2.000\,\%\]

   \[S_A = (250)\frac{2}{5} = 100.0~\mathrm{kVA}, \qquad S_B = (250)\frac{3}{5} = 150.0~\mathrm{kVA}\]

   A is at 100 % of rating, B at 75 % — the bank is already limited.

3. P49.3 For P49.2, find the maximum load the bank can supply and the capacity lost.

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   A carries the fraction \(2/5 = 0.400\) of the load, so it reaches 100 kVA at

   \[S_L = \frac{100}{0.400} = 250~\mathrm{kVA}\]

   \[\text{lost} = 300 - 250 = 50~\mathrm{kVA} = 16.67\,\%\]

4. P49.4 Two transformers have secondary EMFs of 420 V and 415 V, with \(Z_A + Z_B = 0.25 + j0.90~\Omega\). Find the circulating current at no load.

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   \[\left|Z_A + Z_B\right| = \sqrt{0.0625 + 0.81} = 0.9341~\Omega\]

   \[I_c = \frac{420 - 415}{0.9341} = \frac{5}{0.9341} = 5.353~\mathrm{A}\]

   at an angle of \(-74.48^{\circ}\) — nearly purely reactive.

5. P49.5 A 400 kVA transformer at 5 % is to be paralleled with a 600 kVA unit. What percentage impedance should the new unit have?

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   \[\%Z = 5\,\%\]

   Equal percentage impedance on their own bases. The ohmic impedances are then in the ratio \(600 : 400\), i.e. inversely as the ratings.

6. P49.6 For P49.5, if the new unit arrives at 4 %, what fraction of a 1000 kVA load does each carry?

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   \[Z_A = (5)\frac{100}{400} = 1.250\,\%, \qquad Z_B = (4)\frac{100}{600} = 0.6667\,\%\]

   \[S_A = (1000)\frac{0.6667}{1.9167} = 347.8~\mathrm{kVA}, \qquad S_B = 652.2~\mathrm{kVA}\]

   A is at 87.0 % and B at 108.7 % — B is overloaded at a total load well within the installed 1000 kVA.

7. P49.7 Derive the two components of \(I_A\) from the circuit equations.

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   From the two loop equations, both equal to \(V_L\):

   \[E_A - I_AZ_A = E_B - \left(I_L - I_A\right)Z_B\]

   \[E_A - E_B = I_AZ_A + I_AZ_B - I_LZ_B\]

   \[I_A\left(Z_A + Z_B\right) = \left(E_A - E_B\right) + I_LZ_B\]

   \[I_A = \frac{I_LZ_B}{Z_A + Z_B} + \frac{E_A - E_B}{Z_A + Z_B}\]

   The first term is A's share of the load current, set by the current-divider rule. The second is the circulating current, driven by the difference of the two EMFs round the secondary loop.

   By symmetry \(I_B\) has the same sharing form with \(Z_A\) in the numerator, but the circulating term carries a minus sign. The circulating component therefore cancels in the sum, leaving \(I_A + I_B = I_L\) — it circulates between the machines and never reaches the load.

8. P49.8 Why is the circulating current almost purely reactive, and why does that matter?

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   Because it is driven by \((E_A - E_B)\) through the impedance \((Z_A + Z_B)\), and a transformer's impedance is mostly reactance — typically \(X\) is three or four times \(R\). The current therefore lags the driving voltage by close to 90°, as Example 49.1's 73.8° shows.

   Why it matters, two ways. It transfers almost no real power between the machines, so it is not doing useful work — it merely occupies winding capacity. And because it is nearly in quadrature with the load current, it adds to the load current almost at right angles, so its effect on the total current magnitude is smaller than its own size would suggest.

   What it does do is consume rated current in one machine, which is the real cost: the bank's capacity is reduced even though little energy is wasted.

9. P49.9 Explain why equal percentage impedance on a common base is the wrong condition.

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   Because equal impedance on a common base means equal ohmic impedance, and the sharing rule \(S_A/S_B = Z_B/Z_A\) then gives an equal split of kVA — regardless of how the ratings differ.

   Example. A 200 kVA machine at 4 % and a 300 kVA machine at 6 % both come to 2 % on a 100 kVA base. They therefore divide a load equally: 200 kVA each. A is then at 100 % of its rating while B is at only 67 %, and the 500 kVA bank is limited to 400 kVA.

   The correct condition is equal percentage impedance on each machine's own base, because \(\%Z_{\text{own}} = Z_{\Omega}S_{\text{rated}}/V^{2}\), so equal \(\%Z_{\text{own}}\) makes \(Z_{\Omega} \propto 1/S_{\text{rated}}\) — exactly what the sharing rule needs to divide load in proportion to rating.

   Equal on their own bases means unequal on a common base, in the ratio of the ratings.

10. P49.10 Two identical transformers are paralleled, but one is on a tap 2.5 % higher. Discuss the consequences.

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    The two secondary EMFs now differ by 2.5 %, so a circulating current flows even at no load, given by \(I_c = (E_A - E_B)/(Z_A + Z_B)\).

    Magnitude. With two identical machines of, say, 5 % impedance each, the sum is 10 % on the machine base, so \(I_c \approx (2.5/10) = 25\,\%\) of rated current — ten times the tap mismatch, because the opposing impedance is itself only a few percent.

    Consequences. The higher-tap machine carries 25 % of rated current circulating at no load, and on load that current adds to its share. It therefore reaches rated current when the bank is at roughly 75 % of installed capacity, while the other machine is well below rating.

    There is also a continuous copper loss \(I_c^{2}(R_A + R_B)\) for no return, and the current is nearly reactive, so it does not even transfer useful power.

    The remedy is trivial: put both machines on the same tap. This is why tap positions are checked as a matter of routine before paralleling.