I = I(50 Hz) + I(49.5 Hz)
IEC10000141 V2 EN
Figure 40: 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 is 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 (Equation 1)
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 (Equation 2)
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 I A
t Base
= × = × =2 5 2 5 8367 20918. .
EQUATION14043 V1 EN (Equation 3)
Section 11 1MRK 504 165-UUS -
Testing functionality by secondary injection
152 Transformer protection RET670 2.2 ANSI
Commissioning manual
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