scheider electric MiCOM P746, Техническое руководство

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Application Notes P746/EN AP/G31
MiCOM P746 (AP) 6-
35
Treating this as an open bus coupler circuit breaker as before the topology algorithm will
have extended zone 1 with the area located between the CT and the circuit breaker. This
then fully replicates the scheme up to the open bus coupler CB. Remember that in this
example zone 2’s limit extended up to the circuit breaker but this zone has been tripped
already.
The circuit breaker is now open and the fault current would flow to feed the fault. The
differential current in the main zone 2 would equal zero, as the current is flowing into zone 1
whereas the current measured will be equal to the fault current i
fault
.
Zone 2
I
diff
=
I
5
+ I
6
=
i
diff
Z2 = 0
Zone 1
I
diff
=
I
1
+ I
2
=
i
diff
Z1 = i
fault
> (I
D
>2 and k2 x
I
Bias
)
Check zone I
diff
=
I
1
+ I
2
+ I
5
+ I
6
=
i
diff
Z1 = i
fault
> (I
D
CZ>2 and kCZ x
I
Bias
)
Hence, the system reacts to the continuing presence of the fault and trips the zone 1 as the
check zone I
diff
> (I
D
CZ>2 and kCZ x
I
Bias
) and the zone I
diff
> (I
D
>2 and k2 x
I
Bias
).
In this example it can be seen that the opposite zone is tripped first but the dynamic topology
reacts to the changed scheme and subsequently trips the adjacent main zone.
7.3.7 CTs on both sides of coupler, CB closed and fault evolves between CT and CB.
2 CT Coupler with the CB closed and Fault between a CT and the CB
P0846ENa
AP
FIGURE 21: CT’ ON BOTH SIDES OF BUS COUPLER,
CB CLOSED FAULT OCCURS BETWEEN A CT & THE CB
Treating this as a closed bus section circuit breaker the topology algorithm will have created
an overlapped zone that surrounds the circuit breaker with the bus coupler CTs as its limits
made by zone 1 and 2. This then fully replicates the scheme.
Under normal operating conditions when the circuit breaker is closed load current would flow
through the circuit breaker and hence both zones. The differential current in the two main
zones would equal zero, as the current flowing into the zones would still equal the current
flowing out.
However, if a fault was to occur in the overlapped zone, current would flow into both zones
and feed the fault. The differential current in the two main zones will be equal to that of the
fault current.
The main zones would operate. When the check zone element is calculated, the differential
current which results from the presence of the fault in the coupler, will confirm the presence
of a fault and initiate a simultaneous trip of both main.
(1) Hence, the system reacts to a fault occurring between the CT and the CB simultaneously
tripping both zones.
Zone 1
I
diff
=
I
1
+ I
2
+ I
4
=
i
diff
Z1 = i
fault
> (I
D
>2 and k2 x
I
Bias
)
Zone 2
I
diff
=
I
3
+ I
5
+ I
6
=
i
diff
Z2 = i
fault
> (I
D
>2 and k2 x
I
Bias
)
Check zone
I
diff
=
I
1
+ I
2
+ I
5
+ I
6
=
i
diff
Zx = i
fault