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8/3/2019 BOWest Library - Electrical System Formulae
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OWest Library - Electrical System Formulae
Library -Electrical System Formulae
ontentsNotation Impedance Admittance Reactance Resonance Reactive Loads and Power Factor Complex Power Three Phase Power Per-unit System
Symmetrical Components Fault Calculations Three Phase Fault Level Thermal Short-time Rating Instrument Transformers Power Factor Correction Reactors Harmonic Resonance
otation
he library uses the symbol font for some of the notation and formulae. If the symbols for the letterspha beta delta' do not appear here [ a b d] then the symbol font needs to be installed before all notation
nd formulae will be displayed correctly.
susceptancecapacitancevoltage sourcefrequencyconductanceh-operatorcurrent
j-operatorinductanceactive powerreactivepower
[siemens, S][farads, F][volts, V][hertz, Hz][siemens, S][1120][amps, A]
[190][henrys, H][watts, W][VAreactive,VArs]
Q R S t V W X Y Z f w
quality factorresistanceapparent powertimevoltage dropenergyreactance
admittanceimpedancephase angleangularfrequency
[number][ohms, ][volt-amps, VA][seconds, s][volts, V][joules, J][ohms, ][siemens, S][ohms, ][degrees, ][rad/sec]
mpedance
he impedance Z of a resistance R in series with a reactance X is:= R + jX
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ectangular and polar forms of impedance Z:= R + jX = (R 2 + X2)tan -1(X / R) = |Z| f = |Z|cos f + j|Z|sin f
ddition of impedances Z1 and Z2:
1 + Z2 = (R 1 + jX1) + (R 2 + jX2) = (R 1 + R 2) + j(X1 + X2)
ubtraction of impedances Z1 and Z2:
1 - Z2 = (R 1 + jX1) - (R 2 + jX2) = (R 1 - R 2) + j(X 1 - X2)
ultiplication of impedances Z1 and Z2:
1 * Z2 = |Z 1 |f 1 * |Z2|f 2 = ( |Z 1| * |Z 2| )(f 1 + f 2 )
ivision of impedances Z1 and Z2:
1 / Z2 = |Z 1 |f 1 / |Z 2|f 2 = ( |Z 1| / |Z 2 | )(f 1 - f 2 )
summary:use the rectangular form for addition and subtraction,use the polar form for multiplication and division.
dmittance
n impedance Z comprising a resistance R in series with a reactance X can be converted to an admittancecomprising a conductance G in parallel with a susceptance B:= Z -1 = 1 / (R + jX) = (R - jX) / (R 2 + X2) = R / (R 2 + X2) - jX / (R 2 + X2) = G - jB = R / (R 2 + X2) = R / |Z| 2 = X / (R 2 + X2) = X / |Z| 2 sing the polar form of impedance Z:= 1 / |Z| f = |Z| -1- f = |Y| - f = |Y|cos f - j|Y|sin f
onversely, an admittance Y comprising a conductance G in parallel with a susceptance B can beonverted to an impedance Z comprising a resistance R in series with a reactance X:= Y -1 = 1 / (G - jB) = (G + jB) / (G 2 + B 2) = G / (G 2 + B 2) + jB / (G 2 + B 2) = R + jX = G / (G 2 + B 2) = G / |Y| 2 = B / (G 2 + B 2) = B / |Y| 2 sing the polar form of admittance Y:= 1 / |Y| -f = |Y| -1f = |Z| f = |Z|cos f + j|Z|sin f
he total impedance ZS of impedances Z1, Z2, Z3,... connected in series is:
S = Z1 + Z1 + Z1 +...
he total admittance YP of admittances Y1, Y2, Y3,... connected in parallel is:
P = Y1 + Y1 + Y1 +...
summary:use impedances when operating on series circuits,use admittances when operating on parallel circuits.
eactance
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ductive Reactance he inductive reactance XL of an inductance L at angular frequency w and frequency f is:
L = wL = 2 pfL
or a sinusoidal current i of amplitude I and angular frequency w:= I sin wt sinusoidal current i is passed through an inductance L, the voltage e across the inductance is:= L di/dt = wLI cos wt = X LI cos wt
he current through an inductance lags the voltage across it by 90.
apacitive Reactance he capacitive reactance XC of a capacitance C at angular frequency w and frequency f is:
C = 1 / wC = 1 / 2 pfC
or a sinusoidal voltage v of amplitude V and angular frequency w:= V sin wt sinusoidal voltage v is applied across a capacitance C, the current i through the capacitance is:
= C dv/dt = wCV cos wt = V cos wt / X C
he current through a capacitance leads the voltage across it by 90.
esonance
eries Resonance series circuit comprising an inductance L, a resistance R and a capacitance C has an impedance ZS of:
S = R + j(X L - XC)
here XL = wL and XC = 1 / wC
t resonance, the imaginary part of ZS is zero:
C = XL
Sr = R
r = (1 / LC) = 2 pfr he quality factor at resonance Q r is:
r = wrL / R = (L / CR 2) = (1 / R )(L / C) = 1 / wrCR
arallel resonance
parallel circuit comprising an inductance L with a series resistance R, connected in parallel withcapacitance C, has an admittance YP of:
P = 1 / (R + jX L) + 1 / (- jX C) = (R / (R 2 + XL2)) - j(XL / (R 2 + XL2) - 1 / X C)
here XL = wL and XC = 1 / wC
t resonance, the imaginary part of YP is zero:
C = (R 2 + XL2) / XL = XL + R 2 / XL = XL(1 + R 2 / XL2)
Pr = YPr -1 = (R 2 + XL2) / R = X LXC / R = L / CR
r = (1 / LC - R 2 / L2) = 2 pfr
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he quality factor at resonance Qr is:
r = wrL / R = (L / CR 2 - 1) = (1 / R )(L / C - R 2)
ote that for the same values of L, R and C, the parallel resonance frequency is lower than theries resonance frequency, but if the ratio R / L is small then the parallel resonance frequency is close toe series resonance frequency.
eactive Loads and Power Factor
esistance and Series Reactance he impedance Z of a reactive load comprising resistance R and series reactance X is:= R + jX = |Z| f onverting to the equivalent admittance Y:= 1 / Z = 1 / (R + jX) = (R - jX) / (R 2 + X2) = R / |Z| 2 - jX / |Z| 2
When a voltage V (taken as reference) is applied across the reactive load Z, the current I is:= VY = V(R / |Z| 2 - jX / |Z| 2) = VR / |Z| 2 - jVX / |Z| 2 = IP - jIQ
he active current IP and the reactive current IQ are:
= VR / |Z| 2 = |I|cos f = VX / |Z| 2 = |I|sin f
he apparent power S , active power P and reactive power Q are:= V|I| = V 2 / |Z| = |I| 2 |Z| = VI P = IP 2|Z| 2 / R = V 2R / |Z| 2 = |I| 2R
= VI Q = IQ2 |Z| 2 / X = V 2X / |Z| 2 = |I| 2X
he power factor cos f and reactive factor sin f are:os f = IP / |I| = P / S = R / |Z| n f = IQ / |I| = Q / S = X / |Z|
esistance and Shunt Reactance he impedance Z of a reactive load comprising resistance R and shunt reactance X is found from:/ Z = 1 / R + 1 / jX onverting to the equivalent admittance Y comprising conductance G and shunt susceptance B:= 1 / Z = 1 / R - j / X = G - jB = |Y| -f
When a voltage V (taken as reference) is applied across the reactive load Y, the current I is:= VY = V(G - jB) = VG - jVB = I P - jIQ
he active current IP and the reactive current IQ are:
= VG = V / R = |I|cos f = VB = V / X = |I|sin f
he apparent power S , active power P and reactive power Q are:= V|I| = |I| 2 / |Y| = V 2|Y| = VI P = IP 2 / G = |I| 2G / |Y| 2 = V2G
= VI Q = IQ2 / B = |I| 2B / |Y| 2 = V2B
he power factor cos f and reactive factor sin f are:
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os f = IP / |I| = P / S = G / |Y| n f = IQ / |I| = Q / S = B / |Y|
omplex Power
When a voltage V causes a current I to flow through a reactive load Z, the complex power S is:= VI* where I* is the conjugate of the complex current I.
ductive Load = R + jX L
= IP - jIQ
os f = R / |Z| (lagging)= IP + jIQ
= P + jQ n inductive load is a sink of lagging VArs (a source of leading VArs).
apacitive Load = R - jX C
= IP + jIQ
os f = R / |Z| (leading)= IP - jIQ
= P - jQ capacitive load is a source of lagging VArs (a sink of leading VArs).
hree Phase Power
or a balanced star connected load with line voltage Vline
and line current Iline
:
star = Vline / 3
ar = Iline
star = Vstar / Istar = Vline / 3Iline
star = 3V star Istar = 3Vline Iline = Vline 2 / Zstar = 3I line 2Zstar
or a balanced delta connected load with line voltage Vline and line current Iline :
delta = Vline
elta = Iline / 3
delta = Vdelta / Idelta = 3Vline / Iline delta = 3V delta Idelta = 3Vline Iline = 3V line 2 / Zdelta = Iline 2Zdelta
he apparent power S , active power P and reactive power Q are related by:2 = P 2 + Q 2 = Scos f = Ssin f here cos f is the power factor and sin f is the reactive factor
ote that for equivalence between balanced star and delta connected loads:delta = 3Z star
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er-unit System
or each system parameter, per-unit value is equal to the actual value divided by a base value:pu = E / E base
u = I / Ibase
pu = Z / Z base
elect rated values as base values, usually rated power in MVA and rated phase voltage in kV:base = S rated = 3E line Iline
base = E phase = E line / 3
he base values for line current in kA and per-phase star impedance in ohms/phase are:ase = S base / 3E base ( = S base / 3E line )
base = E base / Ibase = 3E base 2 / S base ( = E line 2 / S base )
ote that selecting the base values for any two of S base , Ebase , Ibase or Zbase fixes the base values of all
ur. Note also that Ohm's Law is satisfied by each of the sets of actual, base and per-unit values foroltage, current and impedance.
ransformers he primary and secondary MVA ratings of a transformer are equal, but the voltages and currents ine primary (subscript 1) and the secondary (subscript 2) are usually different:
3E 1line I1line = S = 3E 2line I2line
onverting to base (per-phase star) values:E 1base I1base = S base = 3E 2base I2base
1base / E 2base = I2base / I1base 1base / Z2base = (E 1base / E 2base )2
he impedance Z21pu referred to the primary side, equivalent to an impedance Z2pu on the secondary side, is:
21pu = Z2pu (E1base / E 2base )2
he impedance Z12pu referred to the secondary side, equivalent to an impedance Z1pu on the primary side, is:
12pu = Z1pu (E2base / E 1base )2
ote that per-unit and percentage values are related by:pu = Z% / 100
ymmetrical Components
any three phase system, the line currents Ia , Ib and Ic may be expressed as the phasor sum of:
a set of balanced positive phase sequence currents Ia1 , Ib1 and Ic1 (phase sequence a-b-c),
a set of balanced negative phase sequence currents Ia2 , Ib2 and Ic2 (phase sequence a-c-b),
a set of identical zero phase sequence currents Ia0 , Ib0 and Ic0 (cophasal, no phase sequence).
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he positive, negative and zero sequence currents are calculated from the line currents using:1 = (Ia + hI b + h 2Ic) / 3
2 = (Ia + h 2Ib + hI c) / 3
0 = (Ia + Ib + Ic ) / 3
he positive, negative and zero sequence currents are combined to give the line currents using:= Ia1 + Ia2 + Ia0
= Ib1 + Ib2 + Ib0 = h 2Ia1 + hI a2 + Ia0 = Ic1 + Ic2 + Ic0 = hI a1 + h 2Ia2 + Ia0
he residual current Ir is equal to the total zero sequence current:
= Ia0 + Ib0 + Ic0 = 3I a0 = Ia + Ib + Ic = Ie
hich is measured using three current transformers with parallel connected secondaries.is the earth fault current of the system.
milarly, for phase-to-earth voltages Vae , Vbe and Vce , the residual voltage Vr is equal to the total
ro sequence voltage:r = Va0 + Vb0 + Vc0 = 3V a0 = Vae + Vbe + Vce = 3V ne hich is measured using an earthed-star / open-delta connected voltage transformer.ne is the neutral displacement voltage of the system.
he h-operator he h -operator (1 120) is the complex cube root of unity:= - 1 / 2 + j 3 / 2 = 1 120 = 1 -240 2 = - 1 / 2 - j 3 / 2 = 1 240 = 1 -120
ome useful properties of h are:+ h + h 2 = 0 + h 2 = - 1 = 1 180 - h 2 = j3 = 390 2 - h = - j 3 = 3-90
ault Calculations
he different types of short-circuit fault which occur on a power system are:single phase to earth,double phase,double phase to earth,hree phase,hree phase to earth.
or each type of short-circuit fault occurring on an unloaded system:he first column states the phase voltage and line current conditions at the fault,he second column states the phase 'a' sequence current and voltage conditions at the fault,he third column provides formulae for the phase 'a' sequence currents at the fault,he fourth column provides formulae for the fault current and the resulting line currents.y convention, the faulted phases are selected for fault symmetry with respect to reference phase 'a'.
= fault current
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= earth fault current
a = normal phase voltage at the fault location
1 = positive phase sequence network impedance to the fault
2 = negative phase sequence network impedance to the fault
0 = zero phase sequence network impedance to the fault
ngle phase to earth - fault from phase 'a' to earth:
Va = 0
b = Ic = 0
f = Ia = Ie
Ia1 = Ia2 = Ia0 = Ia / 3
Va1 + Va2 + Va0 = 0
Ia1 = E a / (Z1 + Z2 + Z0)
Ia2 = Ia1
Ia0 = Ia1
I f = 3I a0 = 3E a / (Z1 + Z2 + Z0) = Ie
Ia = I f = 3E a / (Z1 + Z2 + Z0)
ouble phase - fault from phase 'b' to phase 'c':
Vb = Vc
a = 0
f = Ib = - Ic
Ia1 + Ia2 = 0
Ia0 = 0
Va1 = Va2
Ia1 = E a / (Z1 + Z2)
Ia2 = - Ia1
Ia0 = 0
I f = - j 3Ia1 = - j 3E a / (Z1 + Z2) Ib = I f = - j3E a / (Z1 + Z2) Ic = - I f = j3E a / (Z1 + Z2)
ouble phase to earth - fault from phase 'b' to phase 'c' to earth:
Vb = Vc = 0
a = 0
f = Ib + Ic = Ie
Ia1 + Ia2 + Ia0 = 0
Va1 = Va2 = Va0
Ia1 = E a / Znet
Ia2 = - Ia1 Z0 / (Z2 + Z0)
Ia0 = - Ia1 Z2 / (Z2 + Z0)
I f = 3I a0 = - 3E aZ2 / Szz = Ie Ib = I f / 2 - j 3E a (Z2 / 2 + Z 0) / Szz Ic = I f / 2 + j 3E a (Z2 / 2 + Z 0) / Szz
Znet = Z1 + Z2Z0 / (Z2 + Z0) and Szz = Z1Z2 + Z2Z0 + Z0Z1 = (Z2 + Z0)Znet
hree phase (and three phase to earth) - fault from phase 'a' to phase 'b' to phase 'c' (to earth):
Va = Vb = Vc (= 0)
a + Ib + Ic = 0 (= I e )
f = Ia = hI b = h 2Ic
Va0 = Va (= 0)
Va1 = Va2 = 0
Ia1 = E a / Z1
Ia2 = 0
Ia0 = 0
I f = Ia1 = E a / Z1 = Ia
Ib = E b / Z1
Ic = E c / Z1
he values of Z1, Z2 and Z0 are each determined from the respective positive, negative and zero
quence impedance networks by network reduction to a single impedance.
ote that the single phase fault current is greater than the three phase fault current if Z0 is less than (2Z1
Z2).
ote also that if the system is earthed through an impedance Zn (carrying current 3I0) then an impedance
Zn (carrying current I0) must be included in the zero sequence impedance network.
hree Phase Fault Level
he symmetrical three phase short-circuit current Isc of a power system with no-load line and phase
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oltages E line and Ephase and source impedance ZS per-phase star is:
c = E phase / ZS = E line / 3ZS
he three phase fault level S sc of the power system is:
sc = 3I sc 2ZS = 3E phase Isc = 3E phase 2 / ZS = E line 2 / ZS
ote that if the X / R ratio of the source impedance ZS (comprising resistance RS and reactance XS )
sufficiently large, then ZS
XS
.
ransformers a transformer of rating S T (taken as base) and per-unit impedance ZTpu is fed from a source with
nlimited fault level (infinite busbars), then the per-unit secondary short-circuit current I2pu and fault level
2pu are:
pu = E 2pu / ZTpu = 1.0 / Z Tpu
2pu = I2pu = 1.0 / Z Tpu
the source fault level is limited to S S by per-unit source impedance ZSpu (to the same base as ZTpu ),
en the secondary short-circuit current I2pu and fault level S 2pu are reduced to:pu = E 2pu / (ZTpu + ZSpu ) = 1.0 / (Z Tpu + ZSpu )
2pu = I2pu = 1.0 / (Z Tpu + ZSpu )
here ZSpu = S T / S S
hermal Short-time Rating
a conductor which is rated to carry full load current Iload continuously is rated to carry a maximum
ult current Ilimit for a time tlimit , then a lower fault current Ifault can be carried for a longer timeault according to:
limit - Iload )2 tlimit = ( Ifault - Iload )2 tfault
earranging for Ifault and tfault :
ault = ( I limit - Iload ) ( t limit / t fault ) + Iload
ault = t limit ( Ilimit - Iload )2 / ( Ifault - Iload )2
Iload is small compared with Ilimit and Ifault , then:
mit 2 t limit Ifault 2 tfault ault Ilimit ( t limit / t fault )
ault tlimit ( Ilimit / Ifault )2
ote that if the current Ifault is reduced by a factor of two, then the time tfault is increased by a factor of four.
nstrument Transformers
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oltage Transformer or a voltage transformer of voltampere rating S , rated primary voltage VP and rated secondary voltage
S , the maximum secondary current ISmax , maximum secondary burden conductance GBmax and
aximum primary current IPmax are:
max = S / V S
Bmax = ISmax / VS = S / V S 2
max = S / V P = ISmax VS / VP
urrent Transformer or a measurement current transformer of voltampere rating S , rated primary current IP and rated
condary current IS , the maximum secondary voltage VSmax , maximum secondary burden resistance
Bmax and maximum primary voltage VPmax are:
Smax = S / I S
Bmax = VSmax / IS = S / I S 2
Pmax = S / I P = VSmax IS / IP
or a protection current transformer of voltampere rating S , rated primary current IP , rated secondary
urrent IS and rated accuracy limit factor F, the rated secondary reference voltage VSF , maximumcondary burden resistance RBmax and equivalent primary reference voltage VPF are:
SF = SF / I S
Bmax = VSF / ISF = S / I S 2
PF = SF / I P = VSF IS / IP
mpedance Measurement the primary voltage Vpri and the primary current Ipri are measured at a point in a system, then the
imary impedance Zpri at that point is:
pri = Vpri / Ipri
the measured voltage is the secondary voltage Vsec of a voltage transformer of primary/secondary ratio
V and the measured current is the secondary current Isec of a current transformer of primary/secondary
tio NI, then the primary impedance Zpri is related to the secondary impedance Zsec by:
pri = Vpri / Ipri = Vsec NV / Isec NI = Z sec NV / NI = Z sec NZ
here NZ = NV / NI
the no-load (source) voltage Epri is also measured at the point, then the source impedance ZTpri to
e point is:Tpri = (E pri - Vpri ) / Ipri = (E sec - Vsec )NV / Isec NI = ZTsec NV / NI = ZTsec NZ
ow er Factor Correction
an inductive load with an active power demand P has an uncorrected power factor of cos f 1 lagging, andrequired to have a corrected power factor of cos f 2 lagging, the uncorrected and corrected reactive
ower demands, Q1 and Q2, are:
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1 = P tan f 1
2 = P tan f 2 here tan f n = (1 / cos 2f n - 1)
he leading (capacitive) reactive power demand QC which must be connected across the load is:
C = Q 1 - Q 2 = P (tan f 1 - tan f 2)
he uncorrected and corrected apparent power demands, S1
and S2, are related by:
1cos f 1 = P = S 2cos f 2 omparing corrected and uncorrected load currents and apparent power demands:/ I1 = S 2 / S 1 = cos f 1 / cos f 2
the load is required to have a corrected power factor of unity, Q2 is zero and:
C = Q 1 = P tan f 1 / I1 = S 2 / S 1 = cos f 1 = P / S 1
hunt Capacitors
or star-connected shunt capacitors each of capacitance Cstar on a three phase system of line voltageline and frequency f, the leading reactive power demand QCstar and the leading reactive line current Iline are:
Cstar = Vline 2 / XCstar = 2 pfCstar Vline 2
ne = Q Cstar / 3Vline = Vline / 3XCstar
star = Q Cstar / 2pfVline 2
or delta-connected shunt capacitors each of capacitance Cdelta on a three phase system of line voltage
line and frequency f, the leading reactive power demand QCdelta and the leading reactive line current
ne are:
Cdelta = 3V line 2 / XCdelta = 6 pfCdelta Vline 2 ne = Q Cdelta / 3Vline = 3Vline / XCdelta
delta = Q Cdelta / 6pfVline 2
ote that for the same leading reactive power QC:
Cdelta = 3X Cstar
delta = C star / 3
eries Capacitors or series line capacitors each of capacitance C
seriescarrying line current I
lineon a three phase system
frequency f, the voltage drop Vdrop across each line capacitor and the total leading reactive power
emand QCseries of the set of three line capacitors are:
drop = Iline XCseries = Iline / 2pfCseries
Cseries = 3V drop 2 / XCseries = 3V drop Iline = 3I line 2XCseries = 3I line 2 / 2pfC series
series = 3I line 2 / 2pfQCseries
ote that the apparent power rating S rating of the set of three series line capacitors is based on the
ne voltage Vline and not the voltage drop Vdrop :
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rating = 3Vline Iline
eactors
hunt Reactors or star-connected shunt reactors each of inductance Lstar on a three phase system of line voltage Vline
nd frequency f, the lagging reactive power demand QLstar and the lagging reactive line current Iline are:
Lstar = Vline 2 / XLstar = Vline 2 / 2pfLstar
ne = Q Lstar / 3Vline = Vline / 3XLstar
star = Vline 2 / 2pfQLstar
or delta-connected shunt reactors each of inductance Ldelta on a three phase system of line voltage
line and frequency f, the lagging reactive power demand QLdelta and the lagging reactive line current
ne are:
Ldelta = 3V line 2 / XLdelta = 3V line 2 / 2pfLdelta
ne= Q
Ldelta/ 3V
line= 3V
line/ X
Ldelta
delta = 3V line 2 / 2pfQLdelta
ote that for the same lagging reactive power QL:
Ldelta = 3X Lstar
delta = 3L star
eries Reactors or series line reactors each of inductance Lseries carrying line current Iline on a three phase system
frequency f, the voltage drop Vdrop across each line reactor and the total lagging reactive power
emand QLseries of the set of three line reactors are:
drop = Iline XLseries = 2 pfLseries Iline
Lseries = 3V drop 2 / XLseries = 3V drop Iline = 3I line 2XLseries = 6 pfLseries Iline 2
series = Q Lseries / 6pfIline 2
ote that the apparent power rating S rating of the set of three series line reactors is based on the line
oltage Vline and not the voltage drop Vdrop :
rating = 3Vline Iline
armonic Resonance
a node in a power system operating at frequency f has a inductive source reactance XL per phase and
as power factor correction with a capacitive reactance XC per phase, the source inductance L and
e correction capacitance C are:= XL / w = 1 / wXC here w = 2 pf
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he series resonance angular frequency wr of an inductance L with a capacitance C is:
r = (1 / LC) = w(XC / XL)
he three phase fault level S sc at the node for no-load phase voltage E and source impedance Z per-
hase star is:sc = 3E 2 / |Z| = 3E 2 / |R + jX L|
the ratio XL / R of the source impedance Z is sufficiently large, |Z| XL so that:
sc 3E 2 / XL
he reactive power rating QC of the power factor correction capacitors for a capacitive reactance XC
er phase at phase voltage E is:C = 3E 2 / XC
he harmonic number fr / f of the series resonance of XL with XC is:
/ f = w r / w = (XC / XL) (S sc / Q C)
ote that the ratio XL / XC which results in a harmonic number fr / f is:
L / XC = 1 / ( f r / f )2
for fr / f to be equal to the geometric mean of the third and fifth harmonics:
/ f = 15 = 3.873
L / XC = 1 / 15 = 0.067
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