10-2 L90 Line Current Differential System GE Multilin10.1 CT REQUIREMENTS 10 APPLICATION OF SETTINGS10(EQ 10.3)3. This gives a total burden of:. (EQ 10.4)4. This is less than the allowed 3 Ω, which is OK.The following procedure verifies the kneepoint voltage.1. The maximum voltage available from the .2. The system X/R ratio .3. The CT voltage for maximum phase fault is:(EQ 10.5)4. The CT voltage for maximum ground fault is:(EQ 10.6)5. The CT will provide acceptable performance in this application.10.1.3 CALCULATION EXAMPLE 2To check the performance of an IEC CT of class 5P20, 15 VA, ratio 1500:5 A, assume the following values:• maximum Ifp = 14 000 A• maximum Ifg = 12 000 A• impedance angle of source and line = 78°• CT secondary leads are 75 m of AWG 10.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:(EQ 10.7)The total Burden = R r + Rl = 0.008 + 0.52 = 0.528 Ω, which is less than the allowed 0.6 Ω, which is OK.The following procedure verifies the kneepoint voltage.1. The maximum voltage available from the .2. The system X/R ratio .3. The CT voltage for maximum phase fault is:(EQ 10.8)4. The CT voltage for maximum ground fault is:(EQ 10.9)5. The CT will provide acceptable performance in this application.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= =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)= =
