GE Multilin g60, Инструкция по эксплуатации

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5-202 G60 Generator Protection System GE Multilin
5.6 GROUPED ELEMENTS 5 SETTINGS
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, , , , and Igr = 2 pu
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, , , , and Igr = 20 pu.
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, , , , and Igr = 10 pu.
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.
EXAMPLE 5: INTERNAL LOW-CURRENT, HIGH-LOAD SINGLE-LINE-TO-GROUND FAULT WITH NO FEED FROM
THE GROUND
Given the following inputs: IA = 1.10 pu ∠ 0°, IB = 1.0 pu ∠ –120°, IC = 1.0 pu ∠ 120°, and IG = 0.0 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 + 0.0) = 0.10 pu, IR0 = abs(3 × 0.033 – (0.0)) = 0.10 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 75% of the restraining current.
EXAMPLE 6: INTERNAL HIGH-CURRENT SINGLE-LINE-TO-GROUND FAULT WITH NO FEED FROM THE GROUND
Given the following inputs: IA = 10 pu
∠
0°, IB = 0 pu, IC = 0 pu, and IG = 0 pu
The relay calculates the following values:
I_0 = 3.3 pu
∠
0°, I_2 = 3.3 pu
∠
0°, and I_1 = 3.3 pu
∠
0°
Igd = abs(3
×
3.3 + 0.0) = 10 pu, IR0 = abs(3
×
3.3 – (0.0)) = 10 pu, IR2 = 3
×
3.3 = 10 pu, IR1 = 3
×
(3.33 – 3.33) = 0
pu, and Igr = 10 pu
The differential current is 100% of the restraining current.
IR0 abs 3
1
3
--- ×
1
–
( )
–
 
 
2 pu
= =
IR2 3
1
3
--- ×
1 pu
= =
IR1
1 3 ⁄
8
----------
0.042 pu
= =
IR0 abs 3
1
3
--- ×
10
–
( )
–
 
 
20 pu
= =
IR2 3
10
3
------ ×
10 pu
= =
IR1 3
10
3
------
10
3
------
–
 
 
×
0
= =
IR0 abs 3 0 × 0 ( )
–
( ) 0 pu
= =
IR2 3 0 × 0 pu
= =
IR1 3
10
3
------ 0
–
 
 
× 10 pu
= =