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Section 11
Testing functionality by secondary injection
The previous calculations are in primary values. They are transferred to secondary values to perform
injections by a test set. Primary values are transferred to secondary values by taking into account the
CT ratio and the VT ratio (respectively 9000/1 A and 13.8/0.1 kV in the example).
The magnitude of the secondary voltages, that are related to the points RE and SE of the R-X plane,
needs to be checked.
RE (R
FwdR
V
=
t FwdZs
,
EQUATION14054 V1 EN-US
SE (R
RvsR
V
V
=
t RvsZs
,
EQUATION14055 V1 EN-US
The tests, which are described in this section, may require voltages that have magnitude equal to
110% of the previous values. The continuous permissive overload voltage of the protection voltage
channels of the TRM module is 420 V; so the previous voltages may be applied to the analog
channels of the IED continuously. Limitations may be related to the available test set; the current I
was calculated by using a factor 2.5 (instead of the maximum value 4) in order to reduce the
magnitude of the test voltage for the points RE and SE.
Test sets usually do not have a feature to simulate a real network during a power swing and apply the
related analog quantities at the terminal of the generator. The scope of the present test is not a
simulation of a real network. Voltages and currents are supplied in order to measure an impedance
that changes in the time and traverses the plane R-X and, in particular, the area inside the lens
characteristic. The test may be performed by applying:
•
Symmetric three-phase voltage at 50 Hz. The magnitude depends on the point of the
characteristic that needs to be verified. The following three main points of the line segment SE-
RE need to be checked:
•
•
•
The phase angle of the test voltages is equal to:
•
arctan (ForwardX/ForwardR) for tests in the quadrant 1 and 2 of the R-X plane
•
arctan (ReverseX/ReverseR) -180° for tests in the quadrant 3 and 4 of the R-X plane
•
Symmetric three-phase current, where the current is the summation of two currents that have
the same magnitude, but different frequencies.
I
50
GUID-F02E8D18-FF87-45BE-8142-E8FA19F6966B V1 EN-US
The first current I
phase angle 0º.
The second current I
(at the starting time of the test) and frequency:
•
•
140
, X
):
FwdX
V
0 1
.
VT s
,
11931
V
×
=
×
t FwdZ
,
V
13 8
.
,
VT p
, X
):
RvsX
V
0 1
.
,
VT s
5899
×
=
×
t RvsZ
,
13 8
.
V
,
VT p
the point RE (R
, X
FwdR
a point which is related to the parameter ReachZ1 (boundary between zone 1 and zone 2)
the point SE (R
, X
RvsR
I
20918
t
I
10459
=
=
=
=
tf
2
2
has frequency 50 Hz, magnitude 10459 A (that is, 1.162 A secondary) and
50
has magnitude 10459 A (that is, 1.162 A secondary), phase angle 180º
tf
49.5 Hz for the test as generator in the quadrant 1 and 2 of the R-X plane
50.5 Hz for the test as generator in the quadrant 3 and 4 of the R-X plane
© 2017 - 2022 Hitachi Energy. All rights reserved
86 45
.
V
=
42 75
.
V
=
)
FwdX
)
RvsX
A
1MRK 504 165-UEN Rev. K
(Equation 18)
(Equation 19)
(Equation 20)
Transformer protection RET670
Commissioning manual
t
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