Siemens SIPROTEC 7SJ62 Instruction Manual page 230

Functions
6-90
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:
1
------ - I ⋅ 0.58 I ⋅
= =
I
2
3
Examples:
Motor: I
N Motor
I
2 long-term prim
I
2 short-term prim
Current CT
Transformers
Set Value 46-1 Address 4002
Set Value 46-2 Address 4004
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:
1
------ - I ⋅ 0.58 I ⋅
I = =
2
3
− 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
Transformer Base Rating
Nominal High Side Voltage
Nominal Low Side Voltage
Transformer Connection
High Side CT Ratio
The following faults may be detected at the lower-voltage side:
= 545A
/ I = 0.11 long-term
N Motor
/I = 0.55 for T
N Motor
= 600A/1A
= 0.11 · 545A · (1/600A) = 0.10A
= 0.55 · 545A · (1/600A) = 0.50A
1
-- - I ⋅ 0.33 I ⋅
I = =
2
3
:
16 MVA
V = 110 kV
HS
V = 20 kV
LS
Delta-Grounded Wye
CTR = 100 A / 1 A
= 1s
max
7SJ62 Manual
C53000-G1140-C121-1