MV design guide - Schneider Electric

Design rules
Short-circuit currents
The complexity in calculating
the three-phase short-circuit current
basically lies in determining the
impedance value in the network
upstream of the fault location.
Example of a three-phase calculation
Impedance method
All the components of a network (supply network, transformer, alternator,
motors, cables, bars, etc) are characterised by an impedance (Z)
comprising a resistive component (R) and an inductive component (X) or
so-called reactance. X, R and Z are expressed in ohms.
c The relation between these different values is given by:
(cf. example 1 opposite)
Z = (R 2 + X 2 )
Example 1:
Network layout
Tr1 Tr2
c The method involves:
v breaking down the network into sections
v calculating the values of R and X for each component
v calculating for the network:
- the equivalent value of R or X
- the equivalent value of impedance
- the short-circuit current.
c The three-phase short-circuit current is:
A
Equivalent layouts
Zr
Isc =
U
e • Zsc
Za
Zt1
Zt2
Isc : short-circuit current (in kA)
U : phase to phase voltage at the point in question
before the appearance of the fault, in kV.
Zsc : short-circuit impedance (in ohms)
Z = Zr + Zt1//Zt2
Z = Zr + Zt1 • Zt2
Zt1 + Zt2
(cf. example 2 below)
Za
Zsc = Z//Za
Zsc = Z • Za
Z + Za
Example 2:
c Zsc = 0.72 ohm
c U = 10 kV
Isc =
10
= 21.38 kA
e • 0,27
Schneider Electric
Merlin Gerin MV design guide
17