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Examples:
Motor with the following data:
Nominal current
Continuously permissible negative sequence current I
Briefly permissible negative sequence current
Current transformer
Setting value
Setting value
When protecting feeder or cable systems, unbalanced load protection may serve to identify low magnitude un-
symmetrical faults below the pickup values of the directional and non-directional overcurrent elements.
Here, the following must be observed:
A phase-to-ground fault with current I corresponds to the following negative sequence current:
On the other hand, with more than 60% of unbalanced load, a phase-to-phase fault can be assumed. The delay
time 46-2 DELAY must be coordinated with the system grading of phase-to-phase faults.
For a power transformer, unbalanced load protection may be used as sensitive protection for low magnitude
phase-to-ground and phase-to-phase faults. In particular, this application is well suited for delta-wye transform-
ers where low side phase-to-ground faults do not generate high side zero sequence currents (e.g. vector group
Dy).
Since transformers transform symmetrical currents according to the transformation ratio "CTR", 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 "CTR" is taken into consideration.
Consider a transformer with the following data:
Base Transformer Rating
Primary Nominal Voltage
Secondary Nominal Voltage
Vector Groups
High Side CT
The following fault currents may be detected at the low side:
If 46-1 PICKUP on the high side of the devices is set to = 0.1, then a fault current of I = 3 · TR
PICKUP = 3 · 110/20 · 100 · 0.1 A = 165 A for single-phase faults and √3 · TR
can be detected for two-phase faults at the low side. This corresponds to 36% and 20% of the transformer
nominal current respectively. It is important to note that load current is not taken into account in this simplified
example.
SIPROTEC, 7SK80, Manual
E50417-G1140-C344-A5, Release date 11.2012
= 545 A
I
Nom Motor
/I
2 dd prim
Nom Motor
/I
I
2 long-term prim
Nom Motor
I
/I
Nomprim
Nomsec
46-1 Pickup = 0.11 · 545 A · (1/600 A) = 0.10 A
46-2 Pickup = 0.55 · 545 A · (1/600 A) = 0.50 A
S
= 16 MVA
NomT
V
= 110 kV
Nom
(TR
V
= 20 kV
V
Nom
Dy5
100 A / 1 A
(CT
I
2.6 Negative Sequence Protection 46
= 0.11 continuous
= 0.55 for T max = 1 s
= 600 A/1 A
= 110/20)
= 100)
V
·46-1 PICKUP = 95 A
· TR
V
I
Functions
· 46-1
· TR
I
125
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