9-28 L90 Line Current Differential System GE Multilin9.3 DISTANCE ELEMENTS 9 THEORY OF OPERATION9All 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 nominalvoltage:| VA_1 | = 58.83 V 0.1 69.28 VAfter the memory expires the relay will use the actual voltage for polarization.IA Z + I_0 K0 Z + I G K0M Z – VA = 103.33 V –3.9°VA_1 = 58.83 V –2.1°IA_2 Z D = 1.37 V 19.8°I_0 Z = 19.11 V 19.8°I_0 ZD = 1.37 V 19.8°• 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(IA – IB) Z – (VA – VB) = 88.65 V –78.7°(VA – VB)_1M = 112.08 V 30.0°(IA – IB) Z = 103.50 V –21.2°(IA – IB) ZD = 7.39 V –21.2°• Overcurrent supervision: | (IA – IB) / | = 4.27 A 3 A• 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 willoperate for this fault.e) QUAD PHASE A TO GROUND ELEMENT (BEFORE MEMORY EXPIRES)IA Z + I_0 K0 Z + I G K0M Z – VA = 103.33 V –3.9°VA_1M = 64.71 V 0.0°j I_0 e j3 = 1.37 A 24.8°IA_2 Z D = 1.37 V 19.8°I_0 ZD = 1.37 V 19.8°IA ZR + I_0 K0 Z R + I G K0M Z R – VA= 87.6 V –109.2°IA ZR + I_0 K0 Z R = 91.5 V –93.0°3
