GE L90 Instruction Manual page 799

CHAPTER 10: 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
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
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 section for equations).
Table 10-44: Per-unit line impedance
Value Relay 1
Local to tap 0.088509 pu ∠80.5°
Remote 1 to tap 0.15174 pu ∠80.5°
Remote 2 to tap 0.069551 pu ∠80.5°
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
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 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
LOC-REM1 0.66337 pu
L90 LINE CURRENT DIFFERENTIAL SYSTEM – INSTRUCTION MANUAL
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°
0.073754 pu ∠80.5°
0.057957 pu ∠80.5°
Relay 2 Relay 3
0.39755 pu ∠–178.9° 0.26535 pu ∠2.4583°
0.26582 pu ∠2.2351° 0.26581 pu ∠2.2352°
0.26535 pu ∠2.4587° 0.39758 pu ∠–178.9°
Relay 2 Relay 3
0.66334 pu 0.0011344 pu
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°
FAULT LOCATOR
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