GE D60 Instruction Manual page 499

9 THEORY OF OPERATION
All four comparators and the overcurrent supervision are satisfied.
The MHO phase A ground element will operate for this fault.
c) MHO PHASE A TO GROUND ELEMENT (AFTER MEMORY EXPIRES)
After the memory expires, the relay checks the actual positive-sequence voltage and compares it with 10% of the nominal
voltage:
_1 | = 58.83 V > 0.1 × 69.28 V
| V
A
After the memory expires the relay will use the actual voltage for polarization.
I × Z + I_0 × K0 × Z + I
A G
V _1
A
I _2 × Z
A D
I_0 × Z
I_0 × Z
D
Overcurrent supervision: | 3 × I_0 | = 4.09 A > 3 A
•
Mho difference angle = | –3.9° – (–2.1°) | = 1.8° < 75°
•
Reactance difference angle = | –3.9° – 19.8° | = 23.7° < 75°
•
Zero-sequence directional difference angle = | 19.8° – (–2.1°) | = 21.9° < 75°
•
Negative-sequence directional difference angle = | 19.8° – (–2.1°) | = 21.9° < 75°
•
Fault-type comparator difference angle = | 19.8° – 19.8° | = 0.0° < 50°
•
All four comparators and the overcurrent supervision are satisfied.
The Zone 1 MHO phase A ground element will operate for this fault.
• Zero-sequence directional difference angle for zones 2 and higher (phase A) = | 19.8° – 8.4° | = 11.4° < 90°.
Zones 2 and hihger phase A ground elements will pick-up, time-out and operate.
d) MHO AB PHASE ELEMENT
(I – I ) × Z – (V – V ) = 88.65 V ∠–78.7°
A B A B
(V – V )_1M = 112.08 V ∠30.0°
A B
) × Z = 103.50 V ∠–21.2°
(I – I
A B
) × Z = 7.39 V ∠–21.2°
(I – I
A B D
• Overcurrent supervision: | (I
Mho difference angle = | –78.7° – 30.0° | = 108.7° > 75°
•
Reactance difference angle = | –78.7° – (–21.2°) | = 57.5° < 75°
•
Directional difference angle = | –21.2° – 30.0° | = 51.2° < 75°
•
The mho comparator is not satisfied.
The MHO AB phase element will not operate for this fault.
Repeating the above analysis one concludes that out of the six distance elements only the ground element in phase A will
operate for this fault.
e) QUAD PHASE A TO GROUND ELEMENT (BEFORE MEMORY EXPIRES)
I × Z + I_0 × K0 × Z + I × K0M × Z – V
A G
V _1M
A
j3
j × I_0 × e
_2 × Z
I
A D
I_0 × Z
D
I × Z + I_0 × K0 × Z + I × K0M × Z
A R R G
× Z + I_0 × K0 × Z
I
A R R
GE Multilin
× K0M × Z – V = 103.33 V ∠–3.9°
A
= 58.83 V ∠–2.1°
= 1.37 V ∠19.8°
= 19.11 V ∠19.8°
= 1.37 V ∠19.8°
| = 4.27 A > 3 A
– I ) / 3
A B
= 103.33 V ∠–3.9°
A
= 64.71 V ∠0.0°
= 1.37 A ∠24.8°
= 1.37 V ∠19.8°
= 1.37 V ∠19.8°
– V = 87.6 V ∠–109.2°
R A
= 91.5 V ∠–93.0°
D60 Line Distance Protection System
9.1 DISTANCE ELEMENTS
9-9
9