ABB Relion REG650 Commissioning Manual page 52

Section 7
Testing functionality
1. Calculate the magnitude of a fundamental frequency current I that results in the measured
reactance
[ ]
I A
[ ]
I A
[ ]
I A
IECEQUATION2321 V1 EN-US
2. Three symmetrical fundamental frequency phase voltages must be fed to IED, with
magnitudes proportional to U
in the IED. If three symmetrical fundamental frequency currents, proportional to I = 14107
A, that is, taking into the account the CT ratio, are fed to the IED, and they all lag their
phase voltages by 90 electrical degrees, OOSPPAM function must measure a reactance
equal to
reactance is constant and stationary, as shown in point A in
the currents must be programmed for a signal generator, for example, Omicron.
Voltages (instantaneous values) of the fundamental frequency, that is, 50 Hz or 60 Hz,
must be as follows. Example for fundamental frequency 50 Hz:
u
L 1
IECEQUATION2328 V1 EN-US
u
L 2
IECEQUATION2329 V1 EN-US
u
L 3
IECEQUATION2330 V1 EN-US
Where U
applied to this primary value of the phase voltage, so that the actual value of the
secondary input voltage applied to IED can be calculated.
The three symmetrical currents of the fundamental frequency, 50 Hz (or 60 Hz), must be
as follows:
i
L1
IECEQUATION2325 V1 EN-US
i
L2
IECEQUATION2326 V1 EN-US
i
L3
IECEQUATION2327 V1 EN-US
Where I = U
primary current so that correct value of the secondary injected current can be calculated.
Expected result: The function must measure a reactance X =
Zbase, which corresponds to 0.565 Ohm. The TRIP signal must remain FALSE (0).
3. In the last step of the commissioning procedure, apart from the fundamental frequency
current described above, another current component must be injected, with a frequency
equal to the nominal frequency ± 1.000 Hz. For nominal frequency 50 Hz, first 49. 000 Hz
and then 51.000 Hz. The magnitude of this additional current component must be 0.9
times current I. If the symmetrical additional current component in all three phases has
49.000 Hz, the trajectory of the complex impedance has to be exactly as illustrated in
Figure
as shown in
shown in
transformer HV-side terminals, as described in
zone 2, and this can be observed as the Boolean output TRIPZ2, which must be periodically
(with 1 Hz) set to TRUE (1), and then reset to FALSE (0). The output trip must be set to
46
ForwardX , as shown in point A in
= U / 3 / ForwardX
gen
= 13 8 . kV / 3 0 565 / . Ohm
14107
=
ForwardX , that is, 58.38 percent of Zbase, corresponding to 0.565 Ohm. This
(
= 2 ⋅ U ⋅ sin 2 ⋅ ⋅ 50 0
π
phase
(
= 2 ⋅ U ⋅ sin 2 ⋅ ⋅ 50 0
π
phase
(
= 2 ⋅ U ⋅ sin 2 ⋅ 50 0 .
π
phase
= U / √(3) = 13800 V / √(3) = 7967.4 Volts. The actual PT ratio must be
phase gen
(
2 sin 2 50 0 .
= ⋅ ⋅ I ⋅ ⋅ ⋅ − t
π
(
= 2 ⋅ ⋅ I sin 2 ⋅ ⋅ 50 0 . ⋅ − t
π
(
= 2 ⋅ ⋅ sin 2 ⋅ ⋅ 50 0 . ⋅ −
I t
π
/ √(3) / ForwardX = 14107 A. The actual CT ratio must be applied to this
gen
11. The complex impedance Z(R, X) travels from the right side towards the left side,
Figure 11. The Boolean output signal GENMODE must be set to TRUE (1) as
Figure 12. If the X-line is set in a way, so that the zone 2 begins on the power
© Copyright 2011 ABB. All rights reserved
Figure
/ √(3), that is, taking into the account the actual PT ratio
gen
)
. ⋅ t
)
. ⋅ − ⋅ t 2 / 3
π
)
⋅ − ⋅ t 4 / 3
π
)
/ 2
π
)
/ 2 2 − ⋅ / 3
π π
)
/ 2 4 − ⋅ / 3
π π
Figure
11.
Figure 11. The voltages and
ForwardX = 58.38 percent of
11, then all the pole slips occur in
Generator protection REG650
Commissioning manual
1MRK 502 035-UEN A
(Equation 1)
(Equation 2)
(Equation 3)
(Equation 4)
(Equation 5)
(Equation 6)
(Equation 7)