Siemens SIPROTEC 7UT613 series Manual page 214

2 Functions
214
for 1-pole faults: I2> = 0.1 A, i.e. earth fault current as from approx. 0.3 A.
I = 5 A results in 5 times the secondary value. Consider the current transformer ratios
N
when setting the device with primary values.
For a power transformer, unbalanced load protection may be used as sensitive pro-
tection for low magnitude phase-to-earth and phase-to-phase faults. In particular, this
application is well suited for delta-wye transformers where low side phase-to-ground
faults do not generate a high side zero sequence current.
Since transformers transform symmetrical currents according to the transformation
ratio "TR", the relationship between negative sequence currents and total fault current
for phase-to-phase faults and phase-to-earth faults are also valid for the transformer
as long as the turns ratio "TR" is taken into consideration.
Considering a power transformer with the following data:
Rated apparent power S
primary nominal voltage U
secondary nominal voltage U
Vector Group
Primary CT set
The following faults may be detected at the low side:
If the pickup setting of the device on the high side is set to I2> = 0.1 A, then a phase-
to-earth fault current of
for single-phase,
for two-pole faults can be detected. This corresponds to 36 % and 20 % of the trans-
former nominal current respectively.
To prevent false operation for faults in other zones of protection, the delay time T I2>
must be coordinated with the time grading of other relays in the system.
For generators and motors, the setting depends on the permissible unbalanced load
of the protected object. If the I
sequence current, it can be used as an alarm stage with a long time delay. The I
stage is then set to a short-term negative sequence current with the delay time permit-
ted here.
Example:
I
Motor
N Motor
I
2 dd prim
I
2 max prim
Current trans- TR
former
I2>
Setting
= 16 MVA
NT
= 110 kV
N
= 20 kV
N
Dyn5
100 A/1 A
> stage is set to the continuously permissible negative
2
= 545 A
/ I
= 0.11 continuous
N Motor
/ I
= 0.55 for T
N Motor
= 600 A / 1 A
= 0.11 · 545 A = 60 A primary or
0.11· 545 A · (1/600) = 0.10 A secondary
= 1 s
max
7UT613/63x Manual
C53000-G1176-C160-2
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