Cahier technique no 195 - Schneider Electric

account instead of the maximum through current.The aim is to protect a motor, the CTs have aratio of 100/1.v Result obtained with the through current(7 I n of CT):V k u 14 (R ct + 2 R L ).v Result obtained with switchboard I sc(I sc = 40 kA):V k u 800 (R ct + 2 R L )It is not necessary to go into too much detail tounderstand the importance of choosing the rightparameter!The table in figure 9 gives the through currentvalues to be taken into account when thethrough current is the CT calculation base (seeCahier Technique no. 194).c With line differential protections, taking intoaccount the pilot wires in the calculation of R wiring .In point of fact, R L is given by the wiring linkingthe CTs to the relay located on the same side(end) of the line (see fig. 10 ).You must not take into account the length of thepilot wires which run from one end to the other ofthe protected line.RemindersWith respect to high impedance differentialprotections:c For the calculation of min. V k , take account ofthe through current (see fig. 9).c Calculation of the stabilising resistance R st is afunction of min. V k and of the relay settingcurrent.c Calculation of peak voltage (V p ) is based onthe internal I sc of the protected zone and on thereal V k of the CT.Applications Through Imax Excess through CommentsImaxBB differential Switchboard real I scSwitchboard I thTake real I scif noprotectionincrease possible. Elsetake I thMotor differential Motor starting I 7 I n(motor) If you know neither theprotection otherwise starting I nor the motor I n,7 x I n(CT) take 7 x I n(CT)Generator differential Generator I sccontribution 7 I n(generator) X’’ = generatorprotection only, i.e. I n(100 / X’’) otherwise subtransient reactance7 x I n(CT) as a %. If unknown, weassume X’’ % u 15i.e. 100/15 = 6.67(7 is taken by excess)Restricted earth fault I scseen at the CT If P aunknown, we P a= upstream shortdifferentialprotection primary for a fault at the take circuit power and P t=transformer secondary, P sc= P tpower limited byi.e. I sc= P sc/Ue P t= P n(100 / Z sc) transformerP sc= (P t.P a) / (P t+ P a)Z sc% = transformershort-circuit impedanceLine differential I scat 80 % of line Switchboard I thby Switchboard I thbyprotection default defaultFig. 9 : determining the through current properly.R LPilot wires (several kilometers)R LFig. 10 : R L is given by the wiring between the CT and the relays located on the same side of the line.Cahier Technique Schneider Electric no. 195 / p.11