PARALLEL AND BANK OPERATIONS TRANSFORMERS MATCHING
How To Match Transformers For Banking and Parallel Operation?
The following rules must be obeyed in order to successfully connect two or more transformers in parallel with each other:
1. The turns ratios of all of the transformers must be nearly equal.
2. The phase angle displacements of all of the transformers must be identical.
3. The series impedances of all transformers must be nearly equal, when expressed as ‘‘%Z’’ using the transformer impedance base.
The first two rules are required so that the open-circuit secondary voltages of the transformers are closely matched in order to avoid excessive circulating currents when the parallel connections are made.
The last rule is based on the fact that for a given voltage rating and %Z, the ohmic impedance of a transformer is inversely proportional to its KVA rating. When transformers having the same %Z are connected in parallel, the load currents will split in proportion to the KVA ratings of the units.
Therefore, transformers with different KVA ratings can be successfully operated in parallel as long as their %Z values are all approximately the same.
Example
(This example is based on an actual event.)
Two three-phase 10,000 KVA 66,000Δ-12,470Y volt transformers were in parallel operation in a substation. The primaries of the two transformers are connected to a 66 kV transmission line through a single air break switch.
This switch is designed to interrupt magnetizing current only, which is less than 1 A. The transformers were being removed from service and the secondary loads had been removed. A switchman then started to open the air break switch, expecting to see a small arc as the magnetizing current was interrupted.
Instead, there was a loud ‘‘bang’’ and there was a ball of flame where the air break switch contacts had vaporized. Something was obviously wrong.
Upon closer inspection, it was revealed that the two transformers had been set on widely different taps: The first transformer was on the 62,700 V primary tap and the second transformer was on the 69,300 V primary tap.
Both transformers had a 7% impedance. Because the turns ratios were unequal, a circulating current was set up even without any secondary load. The opencircuit secondary voltage difference, assuming 66 kV at the transformer primaries, is calculated below.
ΔEs = 66,000 x ( 12,470/ 62,700 - 12,470/69,300 ) V = 1250 V = 0.10 per unit
The per-unit circulating current in the secondary loop is equal to ΔEs divided by the sum of the per-unit impedances of the two transformers:
Ic = 0.10/ 0.14 = 0.714 per unit
Converting Ic into amperes:
Ic = 0.714 x 10,000 KVA/(12.47 kV 1.732) = 331 A per phase
Since Ic flows in a loop in the secondary circuit, the current out of the secondary of the first transformer equals the current into the secondary of the second transformer. But since the turns ratios are not equal, Ic does not get transformed into equal and opposite currents at the primaries.
Primary current of first transformer
65.8 A per phase
Primary current of second transformer
59.6 A per phase
The net current through the air break switch, IAB, is the difference in the primary currents:
IAB 65.8 A per phase 59.6 A per phase 6.2 A per phase
The current through the air break switch supplies the I2c Xs reactive losses of both transformers and therefore lags the primary voltage by 90°. The resulting current exceeded the interrupting rating of the switch, causing it to fail.
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