GE Multilin G60 Generator Management Relay 5-1535 SETTINGS 5.5 GROUPED ELEMENTS5Figure 5–91: 100% STATOR GROUND APPLICATION EXAMPLEWe have the magnitude of neutral voltage VN as:(EQ 5.36)and the magnitude of the neutral and zero-sequence voltages as:(EQ 5.37)Therefore, under the normal conditions described above, we set the operating quantities as follows:(EQ 5.38)In actual practice, the pickup ratio may vary from 0.4 to 0.85.Example 2: Operating quantities for a fault at a fraction k from the neutral grounding point.For analysis, consider the above figure and assume that .In this case, we have the magnitude of the neutral voltage at:(EQ 5.39)and the magnitude of the neutral and zero-sequence voltages as:(EQ 5.40)830737A1.CDRSVn (3rd)Va (3rd)Vb (3rd)Vc (3rd)Pickup3rd)()rd3()rd3(0<+ VVVnnandPickup13rd)()rd3()rd3(00 ->+ VVVnandnSupervisio3rd)()rd3( 0 >+ VVnVa Vb VcE3E3E3Vnk3V0 (3rd)V NR E3×R jX c–------------------ 10 5×5 j5–---------------- V 101 j–---------- V= = =V N⇒ 7.07 V=V N V0+V0jX c– E3×R jX c–------------------------ j50–5 j5–-------------- j10–1 j–-----------= = = V N V0+⇒ 10 j10–1 j–---------------------=V N V0+⇒ 10 pu=Pickup V NV N V0+-----------------------> 0.70710--------------- 0.707 pu= =Supervision V N V0+< 10 pu=E3 10 V= , R 5Ω, X c 5Ω, and k 0.15= = =V N k E3× 0.15 10× 1.5= = =V N V0+V01 k–( )E3 1 k–( )E3 1 k–( )E3+ +3------------------------------------------------------------------------------------- 3 0.85 10××3---------------------------------- 8.5= = = V N V0+⇒ 8.5 1.5+ 10= =V N V0+⇒ 10 pu=
