INFORME DE LABORATORIO FISICA IIDescripción completa
Informe Capacitor
nmFull description
Project : Customer : End User : Consultant : MECON LIMITED 6.6 kV Switchboard at Electrical Substation Bus1 of SS1 Switchgear Switchboard Load System Voltage PF Improvement
= =
From = To =
1438.2
kW
6.6
kV
0.90 0.98
Lagging Leading
kVAR required for PF Improvement @ 6.6kV
= = = = = =
(kW Rating) x (PF Correction Factor) (kW Rating) x { [tan (arccos 0.8)] ‐ [tan (arccos 0.95)] } tan (across 0.80) 0.490 tan (across 0.95) 0.200 1438.2 x (0.49 ‐ 0.2) 417.08 kVAR
Voltage at 6.6 kV Bus‐1 Voltage drop at Reactor Voltage drop at Capacitor Inductive Reactance (Ohms/ph) Capacitive Reactance (Ohms/ph Reactance of Eq circuit(Ohms/ph) Current through the Equivalent circuit
= = = = = = =
V VL Vc XL Xc Xeq ICL
Voltage equation of circuit
Reactance of Eq circuit(Ohms/ph)
= = = =
V= Xeq = Xeq = Xeq =
where Xeq (@ 6.6 kV' Base voltage.)
= Xeq=
Capacitive Reactance (Ohms/ph)
= Xc =
Inductive Reactance (Ohms/ph)
= XL =
Current through the Equivalent circuit Base Voltage Per Phase
= ICL(If) =
Reactance of Eq circuit(Ohms/ph)
= Xeq = = ICL(If) =
Reactor Voltage drop for Current
= VL =
= Iph =
= =
Voltage across Equivalent circuit(Bus Voltage)
= V =
Capacitor Voltage
= Vc =
Capacitor shall be rated for above voltage to meet the reactor Voltage drops
Vc – VL Xc –XL Xc ‐ 0.06 Xc 0.94 Xc or Xeq / 0.96 kV ^2 x 1000 / (kVAR of Capacitor) (6.6^2 x 1000 ) / 417.078) 104.45 Xeq / 0.998 (104.45 / 0.998 ) Henry 104.66 Xc x 6 % (104.66 x 6 % ) Ohms 0.210 (Base voltage per phase) / Xeq (6.6/ √ 3 ) Volt 3.82 104.45 (3.82 x 1000/104.45) A 36.58 ICL x XL 36.58 x 0.21 V 7.682 kV 0.0100 ICL x Xeq/1000 (36.58 x 104.45)/(1000) 3.830 kV(Ph‐N) V + VL (3.83+0.01) 3.84 kV(Ph‐N) 6.660 kV(Ph‐Ph)
kVAR from Capacitor at above voltage
Hence required capacitor rating Capacitor selected as per contract
= Vc x ICL = Vc x Vc x 1000 / XC
(6.66 x 6.66 x 1000)/104.66 423.81
kVAR kVAR
Series reactor is Rated 0.2% of value of the Capacitor Finally selected Hence Reactor value
(423.81/(POWER((7.2/6.6),2) x 0.998) x 0.002 ) 0.720 New Xc
= V x V / MVAR = (3.83 x 3.83 )/( 423.81 x 1000) = 34.610
Inductive Reactance XL (Ohms/ph)
= 6% of new Xc (34.61 x 6%) 2.077
Hence, Net Impedance X'c
Amps
= (I x I x XL) / 1000
19.44 Reactor Voltage Drop for Current
Ohms
= V / (1.732 x new Xc)
3.83/ (1.732 x 34.61) 63.900 Series Reactor Rating
Ohms
= Xc – XL
(34.61 ‐ 2.0766) 46.410 Current at 7.2 kV (I)
kVAR at 6.6kV
kVAR
= I x XL = 63.9 x 0.21)
13.419
V kV
Conclusion : Thus 1050 kVAR capacitor is sufficient for Bus‐1 in 6.6 kV Switchboard at SS‐1. This is a typical calculation and for Bus‐2 in 6.6 Capacitor value for Bus‐1 in 6.6 kV Switchboard at Electrical Substation 750KVAR
