Short-circuit current at the secondary terminals of a MV/LV distribution transformer
The case of one transformer
- In a simplified approach, the impedance of the MV system is assumed to be negligibly small, so that:
[math]\displaystyle{ Isc = \frac {In \times 100}{Usc} }[/math]
where
[math]\displaystyle{ In = \frac{S \times 10^3}{U_{20}\sqrt 3} }[/math]
and:
S = kVA rating of the transformer
U20 = phase-to-phase secondary no-load voltage
In = rated current of the transformer
Isc = short-circuit current in amps
Usc = short-circuit impedance voltage of the transformer in %.
Typical values of Usc for distribution transformers are given in Figure G33
| Transformer rating (kVA) | Usc in % | |
|---|---|---|
| Oil-immersed | Cast-resin dry type | |
| 50 to 750 | 4 | 6 |
| 800 to 3,200 | 6 | 6 |
Example
400 kVA transformer, 420 V at no load
Usc = 4%
[math]\displaystyle{ In =\frac{400 \times 10^3}{420\times \sqrt 3} =550A }[/math]
[math]\displaystyle{ Isc =\frac{550 \times 100}{4} = 13.7 kA }[/math]
The case of several transformers in parallel feeding a busbar
The value of fault current on an outgoing circuit immediately downstream of the busbars (see Fig. G34) can be estimated as the sum of the Isc from each transformer calculated separately.
It is assumed that all transformers are supplied from the same MV network, in which case the values obtained from Figure G33 when added together will give a slightly higher fault-level value than would actually occur.
Other factors which have not been taken into account are the impedance of the busbars and of the cable between transformers and circuit breakers.
The conservative fault-current value obtained however, is sufficiently accurate for basic installation design purposes. The choice of circuit breakers and incorporated protective devices against short-circuit and fault currents is described in Selection of a circuit-breaker .
