Restricted Ground Fault Scheme Logic; External Single-Line-To-Ground Fault; External High-Current Slg Fault; External High-Current Three-Phase Symmetrical Fault - GE T60 Instruction Manual

5.6 GROUPED ELEMENTS
5
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.
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
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
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
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 5–119: RESTRICTED GROUND FAULT SCHEME LOGIC
1
 
 –  
abs 3 -- - 1 2 pu ,
= – =
 
3
1
 
 -- - –  
abs 3 10 20 pu
= – =
 
3
  0   
abs 3 0 0 pu , IR2
= – =
T60 Transformer Protection System
1

IR2 3 -- - 1 pu , IR1
= = =
3
10
 ----- -
, IR2 3 10 pu , IR1
= =
3

3 0 0 pu , IR1 3
= = =
1 3 
--------- - 0.042 pu , and Igr = 2 pu
=
8
10 10
 
 ----- - ----- -
3 0 , and Igr = 20 pu.
= – =
 
3 3
10
 
 ----- - 0
10 pu , and Igr = 10 pu.
– =
 
3
5 SETTINGS
GE Multilin