Project:Customer:EndUser:Consultant:MECONLIMITED6.6kVSwitchboardatElectricalSubstation
Bus1ofSS1SwitchgearSwitchboardLoad = 1438.2 kWSystemVoltage =
6.6 kVPFImprovement
From = 0.90 LaggingTo = 0.98 Leading
[email protected] =
(kWRating)x(PFCorrectionFactor)=
(kWRating)x{[tan(arccos0.8)][tan(arccos0.95)]}= tan(across0.80)
0.490= tan(across0.95) 0.200= 1438.2x(0.490.2)= 417.08 kVAR
Voltageat6.6kVBus1 = VVoltagedropatReactor =
VLVoltagedropatCapacitor = VcInductiveReactance(Ohms/ph) =
XLCapacitiveReactance(Ohms/ph = XcReactanceofEqcircuit(Ohms/ph) =
XeqCurrentthroughtheEquivalentcircuit = ICLVoltageequationofcircuit
= V= VcVL
= Xeq= XcXL= Xeq= Xc0.06Xc
ReactanceofEqcircuit(Ohms/ph) = Xeq= 0.94XcorXeq/0.96
whereXeq(@6.6kV'Basevoltage.) = Xeq=
kV^2x1000/(kVARofCapacitor)(6.6^2x1000)/417.078)104.45
CapacitiveReactance(Ohms/ph) = Xc= Xeq/0.998(104.45/0.998)104.66
Henry
InductiveReactance(Ohms/ph) = XL= Xcx6%(104.66x6%)0.210 Ohms
CurrentthroughtheEquivalentcircuit = ICL(If)=
(Basevoltageperphase)/XeqBaseVoltagePerPhase = Iph= (6.6/3)
3.82 VoltReactanceofEqcircuit(Ohms/ph) = Xeq= 104.45
= ICL(If)= (3.82x1000/104.45)36.58 A
ReactorVoltagedropforCurrent = VL= ICLxXL36.58x0.21
= 7.682 V= 0.0100 kV
VoltageacrossEquivalentcircuit(BusVoltage) = V=
ICLxXeq/1000(36.58x104.45)/(1000)3.830 kV(PhN)
CapacitorVoltage = Vc= V+VL(3.83+0.01)
3.84 kV(PhN)6.660 kV(PhPh)
CapacitorshallberatedforabovevoltagetomeetthereactorVoltagedrops
kVARfromCapacitoratabovevoltage = VcxICL= VcxVcx1000/XC
(6.66x6.66x1000)/104.66Hencerequiredcapacitorrating 423.81
kVARCapacitorselectedaspercontract kVAR
SeriesreactorisRated0.2%ofvalueoftheCapacitorFinallyselected
HenceReactorvalue =
[([email protected])/((7.2/6.6)2x0.940]x0.06(423.81/(POWER((7.2/6.6),2)x0.998)x0.002)0.720
kVARat6.6kV
NewXc = VxV/MVAR= (3.83x3.83)/(423.81x1000)= 34.610
InductiveReactanceXL(Ohms/ph) = 6%ofnewXc(34.61x6%)2.077
Ohms
Hence,NetImpedanceX'c = XcXL(34.612.0766)46.410 Ohms
Currentat7.2kV(I) = V/(1.732xnewXc)3.83/(1.732x34.61)63.900
Amps
SeriesReactorRating = (IxIxXL)/1000
19.44 kVAR
ReactorVoltageDropforCurrent = IxXL = 63.9x0.21) V
13.419 kV
Conclusion:
CapacitorvalueforBus1in6.6kVSwitchboardatElectricalSubstation750KVAR
Thus1050kVARcapacitorissufficientforBus1in6.6kVSwitchboardatSS1.ThisisatypicalcalculationandforBus2in6.6
