Toshiba GR-200-5 Instruction Manual

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6F2T0207 (0.01)GRE200 (5,6)- 1124 -INF 144 145 146 147 148CorrespondedFunctionIDCorrespondedData ID inprimary valuesCoefficients MEASemantics I I, V I, V,P, QIN,VENIa,Ib,Ic,I1,I2,I0,Ie,IseVa,Vb,Vc,V1,V2,V0,Ve,Vs,P, Q, f(V2 secondary)V0 OK 0x7110817110814302061056(V0 secondary)53.74606142 3413.333333Ve OK OK 0x7110817110814302c31056(V4 primary)53.74606142 3413.333333Vs OK 0x7110817110814302081056(Vs primary)53.74606142 3413.333333P OK OK 0x71108171108143030b1098(P)2.23942E-06 1706.666667Q OK OK 0x71108171108143040c1098(Q)2.23942E-06 1706.666667f OK 0x71108171108143060f105C(f)68.26666667 3413.333333Note: The above coefficients are derived with the following equations:(1) Currents in three-phase (IL1, IL2, IL3) and in symmetrical component (I1, I2, I0):Coefficient= 4096In×2.4where, In: Secondary current ratingFor example, in In=1A , we can have coefficient 1706.666667.(2) Earth fault current (Ie) and Ie used for SEF (Ise) :Coefficient= 4096In×1.2where, In: Secondary current rating (or Ise rating current)For example, in In=1A , we can have coefficient 3413.333333.(3) Voltages in three-phase (VL1, VL2, VL3), in symmetrical component (V1, V2, V0), in earth-fault (Ve) , and in reference (Vs):Coefficient= 4096Vn√3⁄ ×1.2where, Vn: Secondary phase-phase voltage ratingFor example, in Vn=110V , we can have coefficient 53.74606142.