Identification; Application; Setting Guidelines - ABB RELION 670 Series Applications Manual

1MRK 504 152-UEN B
8.3.1

Identification

Function description
Instantaneous residual overcurrent
protection
8.3.2

Application

In many applications, when fault current is limited to a defined value by the object impedance,
an instantaneous earth-fault protection can provide fast and selective tripping.
The Instantaneous residual overcurrent EFPIOC, which can operate in 15 ms (50 Hz nominal
system frequency) for faults characterized by very high currents, is included in the IED.
8.3.3

Setting guidelines

The parameters for the Instantaneous residual overcurrent protection EFPIOC are set via the
local HMI or PCM600.
Some guidelines for the choice of setting parameter for EFPIOC is given.
GlobalBaseSel : Selects the global base value group used by the function to define ( IBase ),
UBase ) and ( SBase ).
(
The setting of the function is limited to the operate residual current to the protection (
The basic requirement is to assure selectivity, that is EFPIOC shall not be allowed to operate
for faults at other objects than the protected object (line).
For a normal line in a meshed system single phase-to-earth faults and phase-to-phase-to-earth
faults shall be calculated as shown in figure
the protection are calculated. For a fault at the remote line end this fault current is I
calculation the operational state with high source impedance Z
should be used. For the fault at the home busbar this fault current is I
operational state with low source impedance Z
used.
~
IEC09000022 V1 EN-US
Figure 248: Through fault current from A to B: I
Application manual
IEC 61850
identification
EFPIOC
I
fB
A
Z Z
A L
IED
IEC 60617 ANSI/IEEE C37.2
identification device number
IN>>
IEF V1 EN-US
248 and figure 249. The residual currents (3I
and low source impedance Z
A
and high source impedance Z
A
B
Z
~
B
Fault
IEC09000022-1-en.vsd
fB
Section 8
Current protection
M14887-1 v4
50N
M12699-3 v5
IP14985-1 v1
M12762-44 v2 M12762-4 v4
M12762-6 v7
IN>> ).
) to
0
. In this
fB
. In this calculation the
fA
should be
B
B
403