Trip Decision Example; Trip Decision Test - GE L90 Instruction Manual
Line differential relay
ur series
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8.2 OPERATING CONDITION CHARACTERISTICS
Settings: S1 = 10%, S2 = 10%, BP = 5 pu secondary, P = 0.5 pu
Assumed Current: I_L= 4.0 pu ∠0°, I_R= 0.8 pu ∠0°
The assumed condition is a radial line with a high resistance fault, source at the local end only, and through resistive load
current.
2
– (
I
I_L
I_R
=
+
op
As the current at both ends is less than the breakpoint of 5.0, equation (1), for 2-terminal mode, is used to calculate
restraint.
2
2
(
⋅
I
2 S
=
Rest
1
(
⋅
(
2
0.1
=
0.8328
=
σ
where
= 0, assuming a pure sine wave.
2
I
23.04
Op
>
⇒
----------- -
----------------- -
1
2
0.8328
I
Rest
The use of the
CURRENT DIFF PICKUP
are discussed in the Current Differential section of Chapter 5.
The following figure shows how the relay's main settings are affecting the restraint characteristics. Remote and local cur-
rents are 180° apart which represent an external fault. The breakpoint between two slopes indicates the point where the
restraint area is becoming wider to override uncertainties coming from CT saturation, fault noise, harmonics etc. Increasing
the slope percentage makes the restraint area wider.
8
8-18
2
2
)
∠
∠
4.0 0°
0.8 0°
23.04
=
+
=
2
2
2
⋅
)
(
⋅
⋅
)
I_L
2 S
I_R
2P
+
+
1
2
2
2
2
)
⋅
)
(
⋅
(
)
⋅
)
4
2
0.1
0.8
+
>
⇒
27.67
1
Trip
=
,
CURRENT DIFF RESTRAINT 1
loc pu
20
OPERATE
16
10
BP=4, P=1, S1=20%, S2=40%
8
4
0
4
8
0
Figure 8–8: SETTINGS IMPACT ON RESTRAINT CHARACTERISTIC
L90 Line Differential Relay
2
σ
+
2
⋅
(
)
2
0.5
0
+
+
,
CURRENT DIFF RESTRAINT 2
RESTRAINT
BP=8, P=2, S1=30%, S2=50%
BP=4, P=1, S1=30%, S2=50%
OPERATE
12
16
8 THEORY OF OPERATION
8.2.2 TRIP DECISION EXAMPLE
8.2.3 TRIP DECISION TEST
, and
CURRENT DIFF BREAK PT
rem pu
20
831725A1.CDR
GE Multilin
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