GE T60 Instruction Manual page 394

5.6 GROUPED ELEMENTS
The following examples explain how the restraining signal is created for maximum sensitivity and security. These examples
clarify the operating principle and provide guidance for testing of the element.
5
EXAMPLE 1: EXTERNAL SINGLE-LINE-TO-GROUND FAULT
Given the following inputs: IA = 1 pu ∠0°, IB = 0, IC = 0, and IG = 1 pu ∠180°
The relay calculates the following values:
Igd = 0, IR0 abs 3
=
The restraining signal is twice the fault current. This gives extra margin should the phase or neutral CT saturate.
EXAMPLE 2: EXTERNAL HIGH-CURRENT SLG FAULT
Given the following inputs: IA = 10 pu ∠0°, IB = 0, IC = 0, and IG = 10 pu ∠–180°
The relay calculates the following values:
Igd = 0, IR0 abs 3
=
EXAMPLE 3: EXTERNAL HIGH-CURRENT THREE-PHASE SYMMETRICAL FAULT
Given the following inputs: IA = 10 pu ∠0°, IB = 10 pu ∠–120°, IC = 10 pu ∠120°, and IG = 0 pu
The relay calculates the following values:
Igd = 0, IR0 abs 3 0
=
EXAMPLE 4: INTERNAL LOW-CURRENT SINGLE-LINE-TO-GROUND FAULT UNDER FULL LOAD
Given the following inputs: IA = 1.10 pu ∠0°, IB = 1.0 pu ∠–120°, IC = 1.0 pu ∠120°, and IG = 0.05 pu ∠0°
The relay calculates the following values:
I_0 = 0.033 pu ∠0°, I_2 = 0.033 pu ∠0°, and I_1 = 1.033 pu ∠0°
Igd = abs(3 × 0.0333 + 0.05) = 0.15 pu, IR0 = abs(3 × 0.033 – (0.05)) = 0.05 pu, IR2 = 3 × 0.033 = 0.10 pu,
IR1 = 1.033 / 8 = 0.1292 pu, and Igr = 0.1292 pu
Despite very low fault current level the differential current is above 100% of the restraining current.
5-248
Figure 5–118: RESTRICTED GROUND FAULT SCHEME LOGIC
1
 
× -- - – ( ) 1 2 pu , IR2
– = =
 
3
1
 
× -- - – ( )
10 20 pu , IR2
– =
 
3
( × 0 ( ) )
0 pu , IR2
– = =
T60 Transformer Protection System
1 1 3 ⁄
3 × -- - 1 pu , IR1 --------- -
= = =
3
8
10
× ----- -
3 10 pu , IR1 3
= = =
3
10

× × ----- - 0
3 0 0 pu , IR1 3
= =

3
5 SETTINGS
0.042 pu , and Igr = 2 pu
10 10
 
× ----- - ----- -
0 , and Igr = 20 pu.
– =
 
3 3

10 pu , and Igr = 10 pu.
– =

GE Multilin