Table of Contents
Advertisement
1MRK 511 409-UUS Rev. K
•
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)
IEC10000141 V2 EN-US
Figure 25:
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
current channels of the transformer module.
If the rated secondary current I
I
I
≤
ts
ovrl
EQUATION14041 V1 EN-US
If the CT of the generator has ratio 9000/1 A, then in primary values
I
I
≤
t
ovrl p
,
EQUATION14042 V1 EN-US
Reference is made to the numerical values of the example, explained in the "Setting guidelines" of the
Application Manual. A test current equal to 2.5 time the base current of the generator is chosen; this
choice is related to the selected test voltage that is applied while testing the point SE and RE.
I
2 5
.
=
×
t
EQUATION14043 V1 EN-US
The parameter ReachZ1 defines the boundary between zone 1 and zone 2; it is expressed in percent of
the parameter ForwardX. If the setting of ReachZ1 = 12%, then corresponding primary value of the
reactance is
ReachZ
X
=
RZ 1
EQUATION14044 V1 EN-US
Phasor measurement unit RES670
Commissioning manual
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
of the analog channel is 1 A, then the maximum current test I
rs
4
I
4
A
= ×
=
rs
I
9000
rp
I
4
36000
=
×
= ×
=
ovrl
I
1
rs
I
2 5 8367 20918
.
A
=
×
=
Base
1
ForwardX
×
×
ZBase
100
100
© 2017 - 2022 Hitachi Energy. All rights reserved
I = I(50 Hz) + I(49.5 Hz)
A
12
59 33
.
=
×
×
0 9522 0
.
=
100
100
Testing functionality by secondary injection
of the protection
ovrl
.0 0 68Ω
Section 11
is
ts
(Equation 1)
(Equation 2)
(Equation 3)
(Equation 4)
91
Advertisement
Table of Contents
