CHAPTER 10: THEORY OF OPERATION
FAULT LOCATOR
L90 LINE CURRENT DIFFERENTIAL SYSTEM – INSTRUCTION MANUAL
10-57
10
•
Relay 1: 0.38781 pu
∠
0.26811°
•
Relay 2: 0.30072 pu
∠
–12.468°
•
Relay 3: 0.37827 pu
∠
8.9388°
Since they have a common per-unit base, the composite voltages are used at all locations. The currents are ratio matched
using the tap settings.
For example, the composite current at relay 1 is 1.3839 pu of its local CT; that is, 1.3839 × 1200 A = 1.6607 kA. When
calculated at relay 2 from the data sent from relay 1 to relay 2, this value is 1.6607 kA / 1000 A = 1.6607 pu of the relay 2 CT.
This is due to the procedure of applying tap settings to the received phase currents before calculating the composite
signal.
As a result, the three relays work with the following signals.
Table 10-43: Composite signals at all three relays
The line impedances entered in secondary ohms are recalculated as follows (refer to the previous section for equations).
Table 10-44: Per-unit line impedance
Using the data in the previous two tables, the tap voltages are calculated as follows (refer to the previous section for
equations).
Table 10-45: Calculated tap voltages using terminal data
From this table, it is visible that
•
Looking from relay 1, there is no fault between the tap and the local terminal and between the tap and remote 2
terminal. Therefore, the fault must be between the remote 1 terminal = relay 2 and the tap.
•
Looking from relay 2, there is no fault between the tap and the remote 1 terminal, and between the tap and remote 2
terminal. Therefore, the fault must be between the local terminal = relay 2 and the tap.
•
Looking from relay 3, there is no fault between the tap and the remote 1 terminal, and between the tap and the local
terminal. Therefore, the fault must be between the remote 2 terminal = relay 2 and the tap.
Note that the correct value of the tap voltage is equal for all three relays. This is expected since the per-unit base for the
composite voltages is equal for all three relays.
The three relays calculate the differences as follows (refer to the previous section for equations).
Table 10-46: Tab voltage differences using terminal data
Value
Relay 1
Relay 2
Relay 3
V
LOC(X)
0.38781 pu
∠
0.26811°
0.30072 pu
∠
–12.468°
0.37827 pu
∠
8.9388°
V
REM1(X)
0.30072 pu
∠
–12.468°
0.38781 pu
∠
0.26811°
0.38781 pu
∠
0.26811°
V
REM2(X)
0.37827 pu
∠
8.9388°
0.37827 pu
∠
8.9388°
0.30072 pu
∠
–12.468°
I
LOC(X)
1.3839 pu
∠
–84.504°
5.4844 pu
∠
–85.236°
1.2775 pu
∠
–56.917°
I
REM1(X)
4.5704 pu
∠
–85.236°
1.6607 pu
∠
–84.504°
1.0379 pu
∠
–84.504°
I
REM2(X)
1.7033 pu
∠
–56.917°
2.0439 pu
∠
–56.917°
3.4278 pu
∠
–85.236°
Value
Relay 1
Relay 2
Relay 3
Local to tap
0.088509 pu
∠
80.5°
0.12644 pu
∠
80.5°
0.092735 pu
∠
80.5°
Remote 1 to tap
0.15174 pu
∠
80.5°
0.073754 pu
∠
80.5°
0.11801 pu
∠
80.5°
Remote 2 to tap
0.069551 pu
∠
80.5°
0.057957 pu
∠
80.5°
0.20232 pu
∠
80.5°
Value
Relay 1
Relay 2
Relay 3
V
T(LOC)
0.26581 pu
∠
2.2352°
0.39755 pu
∠
–178.9°
0.26535 pu
∠
2.4583°
V
T(REM1)
0.39758 pu
∠
–178.9°
0.26582 pu
∠
2.2351°
0.26581 pu
∠
2.2352°
V
T(REM2)
0.26535 pu
∠
2.4583°
0.26535 pu
∠
2.4587°
0.39758 pu
∠
–178.9°
Value
Relay 1
Relay 2
Relay 3
LOC-REM1
0.66337 pu
0.66334 pu
0.0011344 pu