ABB RES670 2.1 Commissioning Manual page 110

Section 11
Testing functionality by secondary injection
104
IEC10000141 V2 EN
Figure 25: 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 of the protection current channels of the transformer module.
ovrl
If the rated secondary current I
test I is
ts
I I 4 I 4 A
≤ = × =
ts ovrl rs
EQUATION14041 V1 EN
If the CT of the generator has ratio 9000/1 A, then in primary values
I
9000
rp
I I I 4
≤ = × = ×
t ovrl p , ovrl
I
rs
EQUATION14042 V1 EN
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 . I 2 5 8367 20918 .
= × = ×
t Base
EQUATION14043 V1 EN
I = I(50 Hz) + I(49.5 Hz)
of the analog channel is 1 A, then the maximum current
rs
36000 A
=
1
A
=
1MRK 511 366-UUS -
(Equation 1)
(Equation 2)
(Equation 3)
Commissioning manual