Siemens siprotec SJ62, User Manual

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2 Functions
75
SIPROTEC 4, 7SJ62/63/64 Handbuch
C53000-G1140-C147-A, Edition 07.2015
50-2 Element The pickup value of the relay element 50-2 is set at address
1202
, the assigned time
delay 50-2 DELAY at address
1203
. This stage is often used for current grading in
view of impedances such as transformers, motors or generators. It is specified such
that it picks up for faults up to this impedance.
Example: Transformer used to distribution bus supply with the following data:
Based on the data above, the following fault currents are calculated:
The nominal current of the transformer is:
Due to the following definition
the following setting applies to the protection device: The 50-2 relay element must be
set higher than the maximum fault current, which is detected during a low side fault on
the high side. To reduce fault probability as much as possible even when fault power
varies, the following setting is selected in primary values:
I >>/ I
Nom
= 10, i.e. I >> =
1000 A.
Increased inrush currents, if the fundamental component exceeds the setting value,
are rendered harmless by delay times (address
1203 50-2 DELAY
).
For motor protection, the 50-2 relay element must be set smaller than the smallest
phase-to-phase fault current and larger than the largest motor starting current. Since
the maximum appearing startup current is usually below 1.6 x the rated startup current
(even with unfavorable conditions), the following setting is adequate for fault current
stage 50-2:
1.6 x I
Startup
< 50-2 Pickup < I
ϕϕ
–Min
The potential increase in starting current caused by overvoltage conditions is already
accounted for by the 1.6 factor. The 50-2 element may be set with no delay (
50-2
DELAY
= 0.00 s) since, unlike with e.g. the transformer, no saturation of the shunt re-
actance occurs in a motor.
The principle of the "reverse interlocking" utilizes the multi-element function of the time
overcurrent protection: element 50–2 is used as accelerated busbar protection with a
Rated Power of the Transformer S
NomT
= 16 MVA
Transformer Impedance ZTX = 10 %
Primary Nominal Voltage V
Nom1
= 110 kV
Secondary Nominal Voltage V
Nom2
= 20 kV
Vector Groups Dy 5
Starpoint Grounded
Fault power on 110 kV–side 1 GVA
3-Phase High Side Fault Current at 110 kV = 5250 A
3-Phase Low Side Fault Current at 20 kV = 3928 A
Current flowing on the High Side at 110 kV = 714 A
I
NomT, 110
= 84 A (High side)
I
NomT, 20
= 462 A (Low side)
Current Transformer (High Side) 100 A / 1 A
Current Transformer (Low Side) 500 A / 1 A