ABB ACS580-04 - Calculating the Short-Circuit Current of the Installation

Note: See also section Protecting the drive and input power cable in
short-circuits (page 92).
In multicable installations, install only one fuse per phase (not one fuse per conductor).
Fuses with higher current rating than the recommended ones must not be used. Fuses
with lower current rating can be used.
Fuses from other manufacturers can be used if they meet the ratings and the melting
curve of the fuse does not exceed the melting curve of the fuse mentioned in the table.
￭ Calculating the short-circuit current of the installation
Make sure that the short-circuit current of the installation is at least the value given
in the fuse table.
The short-circuit current of the installation can be calculated as follows:
I
k2-ph
=
2 ·
R
c
2
+ (Z
k
+ X
c
)
2
U
where
Short-circuit current in symmetrical two-phase short-circuit I
k2-ph
Network line-to-line voltage (V) U
Cable resistance (ohm) R
c
Z
k
= z
k
· U
n
2
/ S
n
= transformer impedance (ohm) Z
k
Transformer impedance (%) z
k
Transformer rated voltage (V) U
n
Nominal apparent power of the transformer (kVA) S
n
Cable reactance (ohm) X
c
Example:
The drive type is ACS580-04-725A-4 with a nominal voltage (U) of 400 V.
The supply transformer specifications are:
• rated power S
n
= 1000 kVA
• rated voltage (drive supply voltage) U
n
= 410 V
• impedance z
k
= 5%.
The supply cable specifications are:
• length = 100 m
• resistance/length = 0.125 ohm/km
• reactance/length = 0.074 ohm/km.
Calculate the short-circuit current as follows:
Technical data 183