bowest library - electrical system formulae

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



10/13


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:



11/13


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 :



12/13


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



13/13


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

Page AccessCount since 18

Updated 22 December 2001Copyright 1998-2001 BOWest Pty Ltd

Please refer to the disclaimer and copyright notices before using this information

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