CLIENT -

CONSULTANT -

CONTRACTOR -

PROJECT:

PACKAGE-II :

DOC. NAME : DESIGN CALCULATION SHEET 51NS CORE FOR NEUTRAL CT OF Converter duty TRANSFORMER OF SS2

DOCUMENT NO. - REV.-00 Dated:03/07/2014

SystembasicInputDataTransformerrating = 630KVA

CTratios = 600/1ADesign = 3PH4WIREVoltage = 690 VACFrequency = 50 HzBasicInputDataofCurrentTransformer:Ratio = 300 / 1 AAccuracyclassforCoreII(51NS) = 5P 15AccuracyclassforCoreI = PSSelectedBurden(VA)forcoreI(PS) = 10 VA

BasicInputDataofDistributionTransformer:PrimaryVoltage(v) = 11 kVSecondaryVoltage(V) = 690 VTransformerrating = 0.630 MVAImpedenceoftransformer = 4 %%impedancevariationonnegativeside = 0 (nonegativetolerance

forcalculationpurpose)DistributionSystemFaultCurrent:

Fullloadcurrent,IFL =

6= (0.63x10)

(1.7322x690)= A

MinTransformerImpedance,ZM = (%impedancex(100%tolerance))/100= 4x(1000))/100= 4 %= (IFx100)/ZM= (527.17x100)/4)

IF = 13179.3 Amp

== 43.94 Amp

1.2

1.1

1.4

IF/CTRMax.throughFaultcurrentatCTSecondarySideIfs

MaximumthroughfaultcurrentonLVsideoftransfromer,MVA,IF

527.17

(3XSec.Voltage)(MVAX106)

1.3

1 of 4

CTsizingcalculationforIDMTEarthFaultprotectionRelay(51NS)NumberofIdenticalCTS = 1 NOSLocationofCTS = Core_1ofNCTonTrafoorseperateCT

IDMTEarthFaultprotectionRelay(51NS)InputdataRelayModule =E/FSetting = 10 %CurrentI/Pburden,VA = 0.1 VARelatyoperatingCurrent,IR = 0.1 AmpCoreICTInputdataRatio = 300 /1ASelectedBurden(VA) = 10 VAAccuracyClass = 5P 15CalculatedofCTBurdens:CableLengthl(m) = mLeadresistanceat70Cfor2.5mm2copper(/km) = /kmLeadResistance(RL)() = ()LeadBurden(VA) = VATotalMeterBurden(VA) = VATotalBurdenonCT(VA) = VACTSecondaryresistanceat750C(Rct)(DuetoDesignlimitationofCTmanufacturer,Rct) = 4

Selectedburdenismorethanrequired. HENCESAFE.WhenCT,CoreIisusedwithratio600/1A,Incaseofphasetophasefaultmaximumprimarycurrenttobemeasuredbytherelayforoperation = 20%IfFaultcurrentatSecondary If = 13.2 kA

VoltagedevelopedbyCTVct = If X(Rct+Rl)Pri.CTRatio

= 13.2X1000X(4+0.887)/300)

CTVct = 215.03 V

AccordinglyminimumVAburdenofconsideredCTRatingwithALF15

ALF = 15= ((Vct/ALF)xCTseccurrent)(RctxCTsecCurrent)

= ((215.03/15)X1)(4X1)= 14.34 4= 10.34

Hence10VACThaving5P15accuracyisadequateHenceNCTcore1orseperateCTon0.63MVAConvertertrafoCT300/1A,5P15,10VA,RCT=

3 of 4

4 of 4