GE L90 Instruction Manual page 527

9 THEORY OF OPERATION
• 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 cal-
culated 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 9–7: COMPOSITE SIGNALS AT ALL THREE RELAYS
VALUE RELAY 1
V 0.38781 pu 0.26811°
LOC(X)
V 0.30072 pu –12.468°
REM1(X)
V 0.37827 pu 8.9388°
REM2(X)
I 1.3839 pu –84.504°
LOC(X)
I 4.5704 pu –85.236°
REM1(X)
I 1.7033 pu –56.917°
REM2(X)
The line impedances entered in secondary ohms are recalculated as follows (refer to the previous sub-section for equa-
tions).
Table 9–8: PER-UNIT LINE IMPEDANCE
VALUE RELAY 1
Local to tap 0.088509 pu 80.5°
Remote 1 to tap 0.15174 pu
Remote 2 to tap 0.069551 pu
Using the data in the previous two tables, the tap voltages are calculated as follows (refer to the previous sub-section for
equations).
Table 9–9: CALCULATED TAP VOLTAGES USING TERMINAL DATA
VALUE RELAY 1
V 0.26581 pu 2.2352°
T(LOC)
V 0.39758 pu –178.9°
T(REM1)
V 0.26535 pu 2.4583°
T(REM2)
From the above table, it is already visible that:
• Looking from relay 1 there is no fault between the tap and the local terminal and between the tap and remote 2 termi-
nal. 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 ter-
minal. 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 sub-section for equations).
GE Multilin
RELAY 2
0.30072 pu –12.468°
0.38781 pu 0.26811°
0.37827 pu 8.9388°
5.4844 pu –85.236°
1.6607 pu –84.504°
2.0439 pu –56.917°
RELAY 2
0.12644 pu 80.5°
80.5° 0.073754 pu 80.5°
80.5° 0.057957 pu 80.5°
RELAY 2
0.39755 pu –178.9°
0.26582 pu 2.2351°
0.26535 pu 2.4587°
L90 Line Current Differential System
9.4 FAULT LOCATOR
RELAY 3
0.37827 pu 8.9388°
0.38781 pu 0.26811°
0.30072 pu –12.468°
1.2775 pu –56.917°
1.0379 pu –84.504°
3.4278 pu –85.236°
RELAY 3
0.092735 pu 80.5°
0.11801 pu 80.5°
0.20232 pu 80.5°
RELAY 3
0.26535 pu 2.4583°
0.26581 pu 2.2352°
0.39758 pu –178.9°
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