Functions
6-90 7SJ62 Manual
C53000-G1140-C121-1
interpreted as a phase-to-phase fault, the time delay of this element should be coor-
dinated with fault protection relays. The magnitude of the negative sequence current
with respect to the phase current when one phase is out of service is given as follows:
Examples:
This ensures
When protecting a feeder
, negative sequence protection may serve to identify low
magnitude unsymmetrical faults below the pickup values of the directional and non-
directional overcurrent elements. To detect load magnitude faults, the pickup value of
the negative sequence time-overcurrent elements must be set below the following:
− a phase-to-phase fault (I) results in the following negative sequence current:
− a phase-to-ground fault (I) corresponds to the following negative sequence current:
To prevent false operations for fault in other zones of protection, the time delay should
be coordinated with other fault protection relays in the system.
For a transformer
, negative sequence protection may be used as sensitive protection
for low magnitude phase-to-ground and phase-to-phase faults. In particular, this ap-
plication is well suited for delta-wye transformers where low side phase-to-ground
faults do not generate high side zero sequence currents.
The relationship between negative sequence currents and total fault current for phase-
to-phase faults and phase-to-ground faults are valid for the transformer as long as the
turns ratio is taken into consideration.
Consider a transformer with the following data:
The following faults may be detected at the lower-voltage side:
Motor: I
N Motor
= 545A
I
2 long-term prim
/ I
N Motor
= 0.11 long-term
I
2 short-term prim
/I
N Motor
= 0.55 for T
max
= 1s
Current
Transformers
CT = 600A/1A
Set Value 46-1 Address 4002 = 0.11 · 545A · (1/600A) = 0.10A
Set Value 46-2 Address 4004 = 0.55 · 545A · (1/600A) = 0.50A
Transformer Base Rating 16 MVA
Nominal High Side Voltage V
HS
= 110 kV
Nominal Low Side Voltage V
LS
= 20 kV
Transformer Connection Delta-Grounded Wye
High Side CT Ratio CTR = 100 A / 1 A
I
2
1
3
-------
I⋅ 0.58 I⋅
==
I
2
1
3
-------
I⋅ 0.58 I⋅
==
I
2
1
3
---
I⋅ 0.33 I⋅
==
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