9-2 L90 Line Differential Relay GE Multilin9.1 CT REQUIREMENTS 9 APPLICATION OF SETTINGS99.1.2 CALCULATION EXAMPLE 1To check performance of a class C400 ANSI/IEEE CT, ratios 2000/1800/1600/1500 : 5 A connected at 1500:5, and where:ï maximum Ifp = 14 000 Aï maximum Ifg = 12 000 Aï impedance angle of source and line = 78∞ï CT secondary leads are 75 m of AWG No. 10.BURDEN CHECK:ANSI/IEEE class C400 requires that the CT can deliver 1 to 20 times the rated secondary current to a standard B-4 burden(4 Ω or lower) without exceeding a maximum ratio error of 10%.The maximum allowed burden at the 1500/5 tap is . Now,Therefore, the . This is less than the allowed 3 Ω,which is OK.KNEEPOINT VOLTAGE CHECK:The maximum voltage available from the .The system X/R ratio .The CT Voltage for maximum phase fault is:The CT Voltage for maximum ground fault is:The CT will provide acceptable performance in this application.9.1.3 CALCULATION EXAMPLE 2To check the performance of an IEC CT of class 5P20, 15 VA, ratio 1500:5 A, assume identical parameters as for ExampleNumber 1.BURDEN CHECK:The IEC rating requires the CT deliver up to 20 times the rated secondary current without exceeding a maximum ratio errorof 5%, to a burden of:The total Burden = Rr + Rl = 0.008 + 0.52 = 0.528 Ω, which is less than the allowed 0.6 Ω, which is OK.KNEEPOINT VOLTAGE CHECK:Use the procedure shown for Example Number 1 above.1500 2000⁄( ) 4× 3 Ω=R CT 0.75 Ω=R r0.2 VA5 A( )2------------------ 0.008 Ω= =R L 2 75 m 3.75 Ω1000 m--------------------×× 2 0.26 Ω× 0.528 Ω= = =Total Burden R CT R r R L+ + 0.75 Ω 0.008 Ω 0.52 Ω+ + 1.28 Ω= = =CT 1500 2000⁄( ) 400× 300 V= =78°tan= 4.71=V 14000 Aratio of 300:1----------------------------------- 4.71 1+( ) 0.75 0.26 0.008 Ω+ +( )×× 271.26 V (< 300 V, which is OK)= =V 12000 Aratio of 300:1----------------------------------- 4.71 1+( ) 0.75 0.52 0.008 Ω+ +( )×× 291.89 V (< 300 V, which is OK)= =Burden 15 VA5 A( )2---------------- 0.6 Ω at the 5 A rated current= =
