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Library - Electrical System Formulae

Contents

Notation 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

Notation

Thesymbol fontis used for some notation and formulae. If the Greek symbols foralpha beta delta do not

appear here [] the symbol font needs to be installed for correct display ofnotation and formulae.

BCEfGh

IjLPQ

susceptancecapacitancevoltage sourcefrequencyconductanceh-operator

currentj-operatorinductanceactive powerreactive power

[siemens, S][farads, F][volts, V][hertz, Hz][siemens, S]

[1120]

[amps, A][190][henrys, H][watts, W][VAreactive,VArs]

QRStVW

XYZ

quality factorresistanceapparent powertimevoltage dropenergy

reactanceadmittanceimpedancephase angleangularfrequency

[number]

[ohms, ][volt-amps,VA][seconds, s][volts, V]

[joules, J][ohms, ][siemens, S]

[ohms, ][degrees, ][rad/sec]
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The total impedance ZS of impedances Z1, Z2, Z3,... connected in series is:ZS = Z1 + Z1 + Z1 +...The total admittanceYP of admittancesY1,Y2,Y3,... connected in parallel is:

YP = Y1 + Y1 + Y1 +...

In summary:- use impedances when operating on series circuits,- use admittances when operating on parallel circuits.

Reactance

Indu ct ive Reactance

The inductive reactance XL of an inductance L at angular frequency andfrequency fis:

XL = L = 2fLFor a sinusoidal current i of amplitude I and angular frequency :i = I sintIf sinusoidal current i is passed through an inductance L, the voltage e across theinductance is:

e = L di/dt = LI cost = XLI costThe current through an inductance lags the voltage across it by 90.

Capacit ive Reactanc e

The capacitive reactance XC of a capacitance C at angular frequency andfrequency fis:

XC = 1 / C = 1 / 2fCFor a sinusoidal voltage v of amplitude V and angular frequency :v = V sintIf sinusoidal voltage v is applied across a capacitance C, the current i throughthe capacitance is:

i = C dv/dt = CV cost = V cost / XCThe current through a capacitance leads the voltage across it by 90.

Resonance

Ser ies ResonanceA series circuit comprising an inductance L, a resistance R and a capacitance Chas an impedance ZS of:


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ZS = R + j(XL - XC)

where XL = L and XC = 1 / CAt resonance, the imaginary part ofZS is zero:XC = XL

ZSr= Rr= (1 / LC) = 2frThe quality factor at resonance Qr is:

Qr= rL / R = (L / CR2) = (1 / R )(L / C) = 1 / rCRParal lel resonanc e

A parallel circuit comprising an inductance L with a series resistance R,connected in parallel with a capacitance C, has an admittanceYP of:

YP = 1 / (R + jXL) + 1 / (- jXC) = (R / (R2 + XL

2)) - j(XL / (R2 + XL

2) - 1 / XC)

where XL = L and XC = 1 / CAt resonance, the imaginary part ofYP is zero:XC = (R

2 + XL2) / XL = XL + R

2 / XL = XL(1 + R2 / XL

2)ZPr= YPr

-1 = (R2 + XL2) / R = XLXC / R = L / CRr= (1 / LC - R2 / L2) = 2fr

The quality factor at resonance Qr is:

Qr= rL / R = (L / CR2 - 1) = (1 / R )(L / C - R2)Note that for the same values ofL, R and C, the parallel resonance frequency islower than the series resonance frequency, but if the ratio R / L is small then theparallel resonance frequency is close to the series resonance frequency.

Reactive Loads and Power Factor

Resistance and Series ReactanceThe impedance Z of a reactive load comprising resistance R and seriesreactance X is:

Z = R + jX = |Z|Converting to the equivalent admittanceY:

Y = 1 / Z = 1 / (R + jX) = (R - jX) / (R2 + X2) = R / |Z|2 - jX / |Z|2

When a voltage V (taken as reference) is applied across the reactive load Z, thecurrent I is:I = VY = V(R / |Z|2 - jX / |Z|2) = VR / |Z|2 - jVX / |Z|2 = IP - jIQThe active current IP and the reactive current IQ are:

IP = VR / |Z|2 = |I|cos

IQ = VX / |Z|2 = |I|sin


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The apparent powerS, active powerP and reactive powerQ are:S = V|I| = V2 / |Z| = |I|2|Z|P = VIP = IP

2|Z|2 / R = V2R / |Z|2 = |I|2RQ = VIQ = IQ

2|Z|2 / X = V2X / |Z|2 = |I|2X

The power factorcos and reactive factorsin are:cos = IP / |I| = P / S = R / |Z|sin = IQ / |I| = Q / S = X / |Z|Resistance and Shunt ReactanceThe impedance Z of a reactive load comprising resistance R and shunt reactanceX is found from:1 / Z = 1 / R + 1 / jXConverting to the equivalent admittanceY comprising conductance G and shuntsusceptance B:

Y = 1 / Z = 1 / R - j / X = G - jB = |Y|When a voltage V (taken as reference) is applied across the reactive loadY, thecurrent I is:I = VY = V(G - jB) = VG - jVB = IP - jIQThe active current IP and the reactive current IQ are:

IP = VG = V / R = |I|cosIQ = VB = V / X = |I|sinThe apparent powerS, active powerP and reactive powerQ are:S = V|I| = |I|2 / |Y| = V2|Y|P = VIP = IP

2 / G = |I|2G / |Y|2 = V2G

Q = VIQ = IQ2

/ B = |I|2

B / |Y|2

= V2

B

The power factorcos and reactive factorsin are:cos = IP / |I| = P / S = G / |Y|sin = IQ / |I| = Q / S = B / |Y|

Complex Power

When a voltage V causes a current I to flow through a reactive load Z, the

complex powerS is:S = VI* where I* is the conjugate of the complex current I.

Indu ct ive LoadZ = R + jXLI = IP - jIQ

cos = R / |Z| (lagging)I* = IP + jIQ


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S = P + jQAn inductive load is a sink of lagging VArs (a source of leading VArs).

Capacit ive LoadZ = R - jXC

I = IP + jIQcos = R / |Z| (leading)I* = IP - jIQS = P - jQ

A capacitive load is a source of lagging VArs (a sink of leading VArs).

Three Phase Power

For a balanced starconnected load with line voltage Vline and line current Iline:

Vstar= Vline / 3Istar= IlineZstar= Vstar/ Istar= Vline / 3IlineSstar= 3VstarIstar= 3VlineIline = Vline2 / Zstar= 3Iline2ZstarFor a balanced delta connected load with line voltage Vline and line current Iline:Vdelta = Vline

Idelta = Iline / 3Zdelta = Vdelta / Idelta = 3Vline / IlineSdelta = 3VdeltaIdelta = 3VlineIline = 3Vline2 / Zdelta = Iline2ZdeltaThe apparent powerS, active powerP and reactive powerQ are related by:S2 = P2 + Q2

P = ScosQ = Ssinwhere cos is the power factor and sin is the reactive factorNote that for equivalence between balanced star and delta connected loads:Zdelta = 3Zstar

Per-unit System

For each system parameter, per-unit value is equal to the actual value divided bya base value:Epu = E / EbaseIpu = I / IbaseZpu = Z / Zbase


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Select rated values as base values, usually rated power in MVA and rated phasevoltage in kV:

Sbase = Srated = 3ElineIlineEbase = Ephase = Eline/ 3The base values for line current in kA and per-phase star impedance inohms/phase are:

Ibase = Sbase / 3Ebase ( = Sbase / 3Eline)Zbase = Ebase / Ibase = 3Ebase

2 / Sbase ( = Eline2 / Sbase)

Note that selecting the base values for any two ofSbase, Ebase, Ibase orZbase fixesthe base values of all four. Note also that Ohm's Law is satisfied by each of thesets of actual, base and per-unit values for voltage, current and impedance.

TransformersThe primary and secondary MVA ratings of a transformer are equal, but the

voltages and currents in the primary (subscript 1) and the secondary (subscript 2)are usually different:3E1lineI1line = S = 3E2lineI2lineConverting to base (per-phase star) values:3E1baseI1base = Sbase = 3E2baseI2baseE1base / E2base = I2base / I1baseZ1base / Z2base = (E1base / E2base)

2

The impedance Z21pu referred to the primary side, equivalent to an impedanceZ2pu on the secondary side, is:

Z21pu = Z2pu(E1base / E2base)2

The impedance Z12pu referred to the secondary side, equivalent to an impedanceZ1pu on the primary side, is:Z12pu = Z1pu(E2base / E1base)

2

Note that per-unit and percentage values are related by:Zpu = Z% / 100

Symmetrical Components

In any three phase system, the line currents Ia, Ib and Ic may be expressed as thephasor sum of:- a set of balanced positive phase sequence currents Ia1, Ib1 and Ic1 (phasesequence a-b-c),- a set of balanced negative phase sequence currents Ia2, Ib2 and Ic2 (phasesequence a-c-b),


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- a set of identical zero phase sequence currents Ia0, Ib0 and Ic0 (cophasal, nophase sequence).

The positive, negative and zero sequence currents are calculated from the linecurrents using:

Ia1 = (Ia + hIb + h

2

Ic) / 3Ia2 = (Ia + h2Ib + hIc) / 3

Ia0 = (Ia + Ib + Ic) / 3

The positive, negative and zero sequence currents are combined to give the linecurrents using:Ia = Ia1 + Ia2 + Ia0Ib = Ib1 + Ib2 + Ib0 = h

2Ia1 + hIa2 + Ia0Ic = Ic1 + Ic2 + Ic0= hIa1 + h

2Ia2 + Ia0

The residual current Ir is equal to the total zero sequence current:

Ir= Ia0 + Ib0 + Ic0 = 3Ia0 = Ia + Ib + Ic = Iewhich is measured using three current transformers with parallel connectedsecondaries.Ie is the earth fault current of the system.

Similarly, for phase-to-earth voltages Vae, Vbe and Vce, the residual voltage Vr isequal to the total zero sequence voltage:Vr= Va0 + Vb0 + Vc0 = 3Va0 = Vae + Vbe + Vce = 3Vnewhich is measured using an earthed-star / open-delta connected voltagetransformer.Vne is the neutral displacement voltage of the system.

The h-operator

The h-operator (1120) is the complex cube root of unity:

h = - 1 / 2 + j3 / 2 = 1120 = 1-240h2 = - 1 / 2 - j3 / 2 = 1240 = 1-120Some useful properties ofh are:1 + h + h2 = 0

h + h2 = - 1 = 1180h - h2 = j3 = 390h2 - h = - j3 = 3-90

Fault Calculations

The different types of short-circuit fault which occur on a power system are:- single phase to earth,- double phase,


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- double phase to earth,- three phase,- three phase to earth.

For each type of short-circuit fault occurring on an unloaded system:

- the first column states the phase voltage and line current conditions at the fault,- the second column states the phase 'a' sequence current and voltageconditions at the fault,- the third column provides formulae for the phase 'a' sequence currents at thefault,- the fourth column provides formulae for the fault current and the resulting linecurrents.By convention, the faulted phases are selected for fault symmetry with respect toreference phase 'a'.

I f= fault current

Ie = earth fault currentEa = normal phase voltage at the fault locationZ1 = positive phase sequence network impedance to the faultZ2 = negative phase sequence network impedance to the faultZ0 = zero phase sequence network impedance to the fault

Single phase to earth- fault from phase 'a' to earth:

Va = 0Ib = Ic = 0I f= Ia = Ie

Ia1 = Ia2 = Ia0 = Ia / 3Va1 + Va2 + Va0 = 0

Ia1 = Ea / (Z1 + Z2 + Z0)Ia2 = Ia1Ia0 = Ia1

I f= 3Ia0 = 3Ea / (Z1 + Z2 + Z0) = IeIa = I f= 3Ea / (Z1 + Z2 + Z0)

Double phase- fault from phase 'b' to phase 'c':

Vb = VcIa = 0I f= Ib = - Ic

Ia1 + Ia2 = 0Ia0 = 0Va1 = Va2

Ia1 = Ea / (Z1 + Z2)Ia2 = - Ia1Ia0 = 0

I f= - j3Ia1 = - j3Ea / (Z1 + Z2)Ib = I f= - j3Ea / (Z1 + Z2)Ic = - I f= j3Ea / (Z1 + Z2)

Double ph ase to earth- fault from phase 'b' to phase 'c' to earth:

Vb = Vc = 0Ia = 0I f= Ib + Ic = Ie

Ia1 + Ia2 + Ia0 = 0

Va1 = Va2 = Va0

Ia1 = Ea / ZnetIa2 = - Ia1Z0 / (Z2 + Z0)Ia0 = - Ia1Z2 / (Z2 + Z0)

I f= 3Ia0 = - 3EaZ2 / zz = IeIb = I f / 2 - j3Ea(Z2 / 2 + Z0) / zzIc = I f / 2 + j3Ea(Z2 / 2 + Z0) / zz

Znet = Z1 + Z2Z0 / (Z2 + Z0) and zz = Z1Z2 + Z2Z0 + Z0Z1 = (Z2 + Z0)ZnetThree phase (and th ree phase to earth)- fault from phase 'a' to phase 'b' tophase 'c' (to earth):


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Va = Vb = Vc (= 0)Ia + Ib + Ic = 0 (= Ie)I f= Ia = hIb = h

2Ic

Va0 = Va (= 0)Va1 = Va2 = 0

Ia1 = Ea / Z1Ia2 = 0Ia0 = 0

I f= Ia1 = Ea / Z1 = IaIb = Eb / Z1Ic = Ec / Z1

The values ofZ1, Z2 and Z0 are each determined from the respective positive,

negative and zero sequence impedance networks by network reduction to asingle impedance.

Note that the single phase fault current is greater than the three phase faultcurrent ifZ0 is less than (2Z1 - Z2).

Note also that if the system is earthed through an impedance Zn (carrying current3I0) then an impedance 3Zn (carrying current I0) must be included in the zerosequence impedance network.

Three Phase Fault Level

The symmetrical three phase short-circuit current Isc of a power system with no-load line and phase voltages Eline and Ephase and source impedance ZS per-phase star is:

Isc = Ephase / ZS = Eline / 3ZSThe three phase fault level Ssc of the power system is:Ssc = 3Isc

2ZS = 3EphaseIsc = 3Ephase2 / ZS = Eline

2 / ZS

Note that if the X / R ratio of the source impedance ZS (comprising resistance RSand reactance XS) is sufficiently large, then ZS XS.TransformersIf a transformer of rating ST (taken as base) and per-unit impedance ZTpu is fedfrom a source with unlimited fault level (infinite busbars), then the per-unitsecondary short-circuit current I2pu and fault level S2pu are:I2pu = E2pu / ZTpu = 1.0 / ZTpuS2pu = I2pu = 1.0 / ZTpu

If the source fault level is limited to SS by per-unit source impedance ZSpu (to the

same base as ZTpu), then the secondary short-circuit current I2pu and fault levelS2pu are reduced to:I2pu = E2pu / (ZTpu + ZSpu) = 1.0 / (ZTpu + ZSpu)S2pu = I2pu = 1.0 / (ZTpu + ZSpu)where ZSpu = ST / SS


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Thermal Short-time Rating

If a conductor which is rated to carry full load current Iload continuously is rated tocarry a maximum fault current Ilimit for a time tlimit, then a lower fault current Ifaultcan be carried for a longer time tfault according to:

( Ilimit - Iload )

2

tlimit = ( Ifault - Iload )

2

tfault

Rearranging forIfault and tfault:Ifault = ( Ilimit - Iload ) ( tlimit / tfault )

+ Iloadtfault = tlimit ( Ilimit - Iload )

2 / ( Ifault - Iload )2

IfIload is small compared with Ilimit and Ifault, then:

Ilimit2 tlimitIfault2 tfault

Ifault Ilimit ( tlimit / tfault )tfault tlimit ( Ilimit / Ifault )2Note that if the current Ifault is reduced by a factor of two, then the time tfault isincreased by a factor of four.

Instrument Transformers

Voltage TransformerFor a voltage transformer of voltampere rating S, rated primary voltage VP andrated secondary voltage VS, the maximum secondary current ISmax, maximumsecondary burden conductance GBmax and maximum primary current IPmax are:

ISmax = S / VSGBmax = ISmax / VS = S / VS

2IPmax = S / VP = ISmaxVS / VP

Current TransformerFor a measurement current transformer of voltampere rating S, rated primarycurrent IP and rated secondary current IS, the maximum secondary voltage VSmax,maximum secondary burden resistance RBmax and maximum primary voltageVPmax are:VSmax = S / ISRBmax = VSmax / IS = S / IS

2

VPmax = S / IP = VSmaxIS / IP

For a protection current transformer of voltampere rating S, rated primary currentIP, rated secondary current IS and rated accuracy limit factorF, the ratedsecondary reference voltage VSF, maximum secondary burden resistance RBmaxand equivalent primary reference voltage VPF are:VSF = SF / IS


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RBmax = VSF / ISF = S / IS2

VPF = SF / IP = VSFIS / IP

Imp edance MeasurementIf the primary voltage Vpri and the primary current Ipri are measured at a point in a

system, then the primary impedance Zpri at that point is:Zpri = Vpri / Ipri

If the measured voltage is the secondary voltage Vsec of a voltage transformer ofprimary/secondary ratio NV and the measured current is the secondary currentIsec of a current transformer of primary/secondary ratio NI, then the primaryimpedance Zpri is related to the secondary impedance Zsec by:Zpri = Vpri / Ipri = VsecNV / IsecNI = ZsecNV / NI = ZsecNZwhere NZ = NV / NI

If the no-load (source) voltage Epri is also measured at the point, then the source

impedance ZTpri to the point is:ZTpri = (Epri - Vpri) / Ipri = (Esec - Vsec)NV / IsecNI = ZTsecNV / NI = ZTsecNZ

Power Factor Correction

If an inductive load with an active power demand P has an uncorrected power

factor ofcos1 lagging, and is required to have a corrected power factor ofcos2lagging, the uncorrected and corrected reactive power demands, Q1 and Q2, are:

Q1 = P tan1Q2 = P tan2where tann = (1 / cos2n - 1)The leading (capacitive) reactive power demand QC which must be connectedacross the load is:

QC = Q1 - Q2 = P (tan1 - tan2)The uncorrected and corrected apparent power demands, S1 and S2, are relatedby:

S1cos1 = P = S2cos2Comparing corrected and uncorrected load currents and apparent power

demands:I2 / I1 = S2 / S1 = cos1 / cos2If the load is required to have a corrected power factor of unity, Q2 is zero and:

QC = Q1 = P tan1I2 / I1 = S2 / S1 = cos1 = P / S1


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Shunt Capaci torsFor star-connected shunt capacitors each of capacitance Cstaron a three phasesystem of line voltage Vline and frequency f, the leading reactive power demandQCstarand the leading reactive line current Iline are:

QCstar= Vline2 / XCstar= 2fCstarVline2

Iline = QCstar/ 3Vline = Vline / 3XCstarCstar= QCstar/ 2fVline2For delta-connected shunt capacitors each of capacitance Cdelta on a three phasesystem of line voltage Vline and frequency f, the leading reactive power demandQCdelta and the leading reactive line current Iline are:

QCdelta = 3Vline2 / XCdelta = 6fCdeltaVline2

Iline = QCdelta / 3Vline = 3Vline / XCdeltaCdelta = QCdelta / 6fVline2Note that for the same leading reactive powerQC:

XCdelta = 3XCstarCdelta = Cstar / 3

Series Capacito rsFor series line capacitors each of capacitance Cseries carrying line current Iline ona three phase system of frequency f, the voltage drop Vdrop across each linecapacitor and the total leading reactive power demand QCseries of the set of threeline capacitors are:

Vdrop = IlineXCseries = Iline / 2fCseriesQCseries = 3Vdrop

2 / XCseries = 3VdropIline = 3Iline2XCseries = 3Iline

2 / 2fCseriesCseries = 3Iline

2 / 2fQCseriesNote that the apparent power rating Srating of the set of three series linecapacitors is based on the line voltage Vline and not the voltage drop Vdrop:

Srating = 3VlineIline

Reactors

Shunt ReactorsFor star-connected shunt reactors each of inductance Lstaron a three phase

system of line voltage Vline and frequency f, the lagging reactive power demandQLstarand the lagging reactive line current Iline are:

QLstar= Vline2 / XLstar= Vline

2 / 2fLstarIline = QLstar / 3Vline = Vline / 3XLstarLstar= Vline

2 / 2fQLstar


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For delta-connected shunt reactors each of inductance Ldelta on a three phasesystem of line voltage Vline and frequency f, the lagging reactive power demandQLdelta and the lagging reactive line current Iline are:

QLdelta = 3Vline2 / XLdelta = 3Vline

2 / 2fLdeltaIline = QLdelta / 3Vline = 3Vline / XLdeltaLdelta = 3Vline

2

/ 2fQLdeltaNote that for the same lagging reactive powerQL:XLdelta = 3XLstarLdelta = 3Lstar

Series Reacto rsFor series line reactors each of inductance Lseries carrying line current Iline on athree phase system of frequency f, the voltage drop Vdrop across each linereactor and the total lagging reactive power demand QLseries of the set of threeline reactors are:

Vdrop = IlineXLseries = 2fLseriesIlineQLseries = 3Vdrop

2 / XLseries = 3VdropIline = 3Iline2XLseries = 6fLseriesIline2

Lseries = QLseries / 6fIline2Note that the apparent power rating Srating of the set of three series line reactorsis based on the line voltage Vline and not the voltage drop Vdrop:

Srating = 3VlineIline

Harmonic Resonance

If a node in a power system operating at frequency fhas a inductive sourcereactance XL per phase and has power factor correction with a capacitivereactance XC per phase, the source inductance L and the correction capacitanceC are:

L = XL / C = 1 / XCwhere fThe series resonance angular frequency rof an inductance L with acapacitance C is:r= (1 / LC) = (XC / XL)The three phase fault level Ssc at the node for no-load phase voltage E andsource impedance Z per-phase star is:Ssc = 3E

2 / |Z| = 3E2 / |R + jXL|

If the ratio XL / R of the source impedance Z is sufficiently large, |Z| XL so that:Ssc 3E2 / XL


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The reactive power rating QC of the power factor correction capacitors for acapacitive reactance XC per phase at phase voltage E is:QC = 3E

2 / XC

The harmonic numberfr / fof the series resonance ofXL with XC is:

fr / f = r / = (XC / XL) (Ssc / QC)Note that the ratio XL / XC which results in a harmonic numberfr / fis:XL / XC = 1 / ( fr / f )

2so forfr / fto be equal to the geometric mean of the third and fifth harmonics:

fr / f = 15 = 3.873XL / XC = 1 / 15 = 0.067

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