windings split into two groups per phase, when each group is equipped with current transformers. The
protection function performs a simple summation of the currents of the two channels I3P1 and I3P2.
11.4.12.1 Verifying the settings
GUID-708C4033-3111-481F-9868-FF207C18C9F9 v2
The test of the out-of-step protection function is made to verify that the trip is issued if the following
events happen.
• the impedance, seen by the function, enters the lens characteristic from one side and leaves it from
the opposite side
• the trip is issued according to the settings TripAngle and tBreaker
The tripping zone needs to be detected and confirmed. The test may be performed by taking into account
the following key points that are shown in Figure
43:
• the point RE (RE = Receiving End)
• the intersection between the line segment SE-RE and the X-line, which is defined through the
setting ReachZ1
• the point SE (SE = Sending End)
I = I(50 Hz) + I(49.5 Hz)
IEC10000141 V2 EN-US
Figure 43: Trajectory of the impedance Z(R, X) for the injected current with two components: a 50
Hz component and a 49.5 Hz current component
The test of the out-of-step protection function requires the injection of the analog quantities for a quite
long time. The rating of the analogue channels must be considered in order to avoid any hardware
damage. The test current is lower than the continuous permissive overload current I
ovrl
of the protection
current channels of the transformer module.
If the rated secondary current I
rs
of the analog channel is 1 A, then the maximum current test I
ts
is
I I I A
ts ovrl rs
≤ = × =4 4
EQUATION14041 V1 EN-US (Equation 5)
If the CT of the generator has ratio 9000/1 A, then in primary values
I I I
I
I
A
t ovrl p ovrl
rp
rs
≤ = × = × =
,
4
9000
1
36000
EQUATION14042 V1 EN-US
(Equation 6)
1MRK 504 165-UUS Rev. J Section 11
Testing functionality by secondary injection
Transformer protection RET670 145
Commissioning manual
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