GE Multilin G30 Generator Protection System 5-1815 SETTINGS 5.6 GROUPED ELEMENTS5Figure 5–71: RESTRICTED GROUND FAULT SCHEME LOGICThe following examples explain how the restraining signal is created for maximum sensitivity and security. These examplesclarify the operating principle and provide guidance for testing of the element.EXAMPLE 1: EXTERNAL SINGLE-LINE-TO-GROUND FAULTGiven 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 puThe restraining signal is twice the fault current. This gives extra margin should the phase or neutral CT saturate.EXAMPLE 2: EXTERNAL HIGH-CURRENT SLG FAULTGiven 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 FAULTGiven the following inputs: IA = 10 pu ∠0°, IB = 10 pu ∠–120°, IC = 10 pu ∠120°, and IG = 0 puThe relay calculates the following values:Igd = 0, , , , and Igr = 10 pu.EXAMPLE 4: INTERNAL LOW-CURRENT SINGLE-LINE-TO-GROUND FAULT UNDER FULL LOADGiven 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 puDespite very low fault current level the differential current is above 100% of the restraining current.SETTINGSETTINGSETTINGSETTINGSETTINGSSETTINGFLEXLOGIC OPERANDSACTUAL VALUESRESTD GND FT1FUNCTION:RESTD GND FT1BLOCK:RESTD GND FT1SOURCE:RESTD GND FT1PICKUP:RESTD GND FT1 RESETDELAY:RESTD GND FT1 PICKUPDELAY:RESTD GND FT1SLOPE:RESTD GND FT1 OPRESTD GND FT1 DPORESTD GND FT1 PKPRGF 1 Igd MagRGF 1 Igr MagOff=0Enabled=1AND828002A3.CDRRUNRUNIgd > PICKUPINIGI_0I_1I_2AND> SLOPE *Igd IgrDifferentialandRestrainingCurrentstPKP tRSTIR0 abs 3 13---× 1–( )– 2 pu= = IR2 3 13---× 1 pu= = IR1 1 3⁄8---------- 0.042 pu= =IR0 abs 3 13---× 10–( )– 20 pu= = IR2 3 103------× 10 pu= = IR1 3 103------ 103------– × 0= =IR0 abs 3 0× 0( )–( ) 0 pu= = IR2 3 0× 0 pu= = IR1 3 103------ 0– × 10 pu= =
