The question
A plant maintenance team records the transformer rating plate before a shutdown. The plate gives one rating in kVA and two voltages. The team needs three numbers from that plate. It needs the current on the high voltage side. It needs the current on the low voltage side. It needs the fault current the low voltage board must withstand.
The inputs
- Rating: 1000 kVA
- Phases: 3
- Primary voltage: 11 000 V
- Secondary voltage: 400 V
- Impedance: 5 %
The impedance also sits on the rating plate. It is the percentage of rated voltage that drives rated current through a short-circuited winding.
The clause
AS/NZS 3000:2018 Cl 2.5.4.1 covers the prospective fault current at a supply point. A transformer fed installation takes its fault level from the transformer rating and impedance.
The steps
The tool divides the rating by the voltage and by the square root of three.
- Primary full load current: 52.49 A
- Secondary full load current: 1,443.38 A
It then divides the secondary current by the impedance in per unit. A 5 % impedance is 0.05 per unit, so the fault current is twenty times the full load current.
- Secondary fault current: 28,867.51 A
The tool reports one assumption with that figure. The fault current assumes an infinite upstream source. A real 11 kV network has its own impedance, so the true figure at the board is lower. Ask the network company for the fault level at the point of supply before a board is specified.
The result
The secondary full load current is 1,443.38 A. A main switchboard on this transformer needs a rating above that figure, and the network company sets the final number.
The primary current of 52.49 A sets the high voltage protection. It also sets the cable from the ring main unit. The 11 kV side carries far less current than the 400 V side, because the voltage is far higher.
The fault current of 28,867.51 A is the withstand figure to check against. A board rated at 25 kA does not cover it. A board rated at 36 kA does, with the upstream network impedance still to be taken off.
Reading the impedance
A lower impedance gives a stiffer transformer and a higher fault current. At 4 % the same 1000 kVA unit would give a fault current a quarter higher again. At 6 % it would drop by a sixth. The impedance moves the fault level more than any other single number on the plate.
Voltage regulation runs the other way. A low impedance holds the secondary voltage steadier under load. Transformer selection trades the two against each other. The switchboard withstand rating follows the choice.
Using the figures on site
The secondary full load current sets three things at once. It sets the main switch rating. It sets the busbar rating. It sets the size of the cable or busway from the transformer to the board.
The primary full load current sets the high voltage fuse or relay. A fuse on the 11 kV side has to carry 52.49 A without ageing. It also has to survive the transformer inrush at switch-on. Inrush runs several times the full load current for a few cycles.
The fault current matters twice. It sets the withstand rating of the board. It also sets how fast the protection has to clear. Let-through energy rises with the square of the current and with the time. A board that is fed at 28,867.51 A needs both checks recorded in the design file.
Record the rating plate data before the shutdown ends. The plate carries the rating, the voltages, the impedance and the vector group. Those four values drive every figure on this page.
This page is a design aid. Verify every value against the current edition of the standard.
See also: Transformer secondary fault current for 500 kVA, Generator fault current calculation at 400 V and Short-circuit withstand and the k factor.