BSE-515-Test-2016

BSE515In-ClassTestNov.16,2016Showninthefigureisafeedercircuitwiththedesigncurrentof500Amadebyone150mlong4/c300mm2PVC/SWA/PVCcable.Itisprotectedbya630AMCCBwithVIIDMTrelaysatamainLVswitchboardagainstoverloadcurrent.ThemainLVswitchboardisprotectedbya2500ACBwithVIIDNTrelays.CTRatiosfortheserelaysaregiveninthefigure.The1500kVAtransformerhasashort-circuitimpedanceof6%inaper-unitsystem,andtheshort-circuitcapacityat11kVisequalto150MVA.Theswitchboardimpedanceisassumedtobej0.5ohm.Cableimpedance(IETvalues)isgivenbyrtab=0.135(mV/m),xtab=0.130(mV/m)andztab=0.185(mV/m)a)Determinethe3-phaseprospectivefaultcurrentsatthebeginandtheremoteendsofthiscircuit.(4marks)b)Determinetheappropriatecurrentsetting(PS)andtimemultipliersetting(TS)oftheIDMTrelaysforMCCMsothatthemaximumoperationtimeislessthan0.5sec.underthese3-phasefaultcurrent.(3marks)c)AssumetherelaysforACBhavethesettingofPS=1005,andTM=0.1,ifthegradingtimeis0.2s.determinetheMCCBrelayssettingagainsothatACBwillnottripundertheminimumfaultcurrentinthefeedercircuit.(3marks)TheVIIDMTrelaysaredescribedwith𝑡𝑡=TM×13.5𝑃𝑃𝑃𝑃𝑃𝑃−1and𝑃𝑃𝑃𝑃𝑃𝑃=×𝐼𝐼𝑓𝑓𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶×𝑃𝑃𝑃𝑃whereTSisintherangefrom0.1to1.0instepsof0.1,andSisintherageof50%-200%inthestepof10%.ChillerPlant630AMCCB4/c300mm2PVC/SWA/PVCcableO/CO/C2500AACB2250A/5A1500kVA,11kV/380V500A/5ASolution(a)(4marks)Zhv=j380^2/150MVA=j0.962mohmZt=6%*380^380/1500kVA=j5.776mohmZswb=j0.5ohmZcab=(0.135*1.15+j0.130)/1,732*150=13.445+j11.259mohmIf1=220/(Zhv+Zt+Zswb)=220/(0.962+5.776+0.5)=30.4kAIf2=220/(Zhv+Zt+Zswb+Zcab)=220/(1i*(0.962+5.776+0.5+11.259)+13.455)=9.6kA(b)(3marks)MCCBPS=100%(Ib=Is,p)ConsideringtheworstcaseIf2PSM=If2/CTratingxPS=9.6kA/0.5*100%=19.2T=TM*13.5/(19.2-1)=0.742TM0.5TM0.67TM=0.6(c)(3marks)ACBPS=100%andTM=0.1PSM=If2/CTratingxPS=9.6kA/2.25*100%=4.27Tacb=TM*13.5/(4.27-1)=4.128*0.1=0.413sTmccb=0.742TMTacb-0.2=0.413-0.2=0.213sTM=0.213/0.472=0.45--.TM=0.4