32.Power System Study-sc, Rc & Dynamic Testing

8/9/2019 32.Power System Study-sc, Rc & Dynamic Testing

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POWER SYSTEM STUDY-SHORT CIRCUIT STUDY, RELAY CO-ORDINATION & Dynamic Testing TechniqueBy

KAMIN DAVE(DOBLE ENGINEERING PVT.LTD), KEIL SHAH

Power System Study:SHORT CIRCUIT Study, Relay Co-ordination

&Dynamic Testing Technique of Relay

KAMIN DAVE (DOBLE ENGINEERING PVT.LTD), KEIL SHAH (EXPELPROSYS.)

“KNOWLEDGE IS POWER”

“MULTIPLICATION OF THIS DOCUMENT IS STRICTLY PROHIBITED”

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INTRODUCTION:

Successful Operation of a power system depends largely on the engineer’s ability toprovide reliable and uninterrupted service to loads. The reliability of the power supplyimplies much more than merely being available. Ideally, the loads must be fed atconstant voltage and frequency at all times. In practical terms this means thatconsumer’s equipment may operate satisfactorily. For example, a drop in voltage of 10-15% or a reduction of the system frequency of only a few hertz may lead to stalling ofthe motor loads on the system.

As electrical utilities have grown in size, and the number of interconnections hasincreased, planning for future expansion has become increasingly complex. Theincreasing cost of additions and modifications has made it imperative that utilitiesconsider a range of design options, and perform detailed studies of the effects on the

system of each option, based on the number of assumptions: like normal and abnormaloperating conditions, peak and off-peak loadings, and present and future years ofhandled.

Future transmission and distribution systems will be far more complex than those oftoday. This means that the power system planner’s task will be more complex. If thesystems being planned are to be optional with respect to construction cost, performance,and operating efficiency, better planning tools are required.

Parameter Conversion:Power transmission lines are operated at voltage levels where kilovolts is the mostconvenient unit to express voltage. The amount of power transmitted is in terms of

kilowatts or megawatts and kilo amperes or mega amperes. However the quantities,current and Ohms are often expressed as a percent or per unit of base value. The perunit value of any quantity is defined as the ratio of the quantity to its base valueexpressed as a decimal. Both the per unit (p.u.) and percent methods of calculation aresimpler than the use of actual amperes, Ohms, and voltage values.

The per unit method has an advantage over the percent method because the product oftwo quantities expressed in per unit is expressed in per unit itself, but the product of twoquantities expressed in percent must be divided by 100 to obtain the result in percent.

The per unit value of a line to neutral voltage on the line to neutral voltage base is equalto the per unit value of the line to line voltage at the same point on the line to line voltage

base if the system is balanced. Similarly, the three-phase kVA is three times thekVA/phase and the three-phase kVA base is three times the base kVA per phase.Therefore the per unit value of the three-phase kVA on the three-phase kVA base isidentical to the per unit value of the kVA per phase on the kVA per phase base.

Base impedance and base current value can be computed directly from three-phasevalues of base kilovolts and base kilo-amperes.

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Base Current = {Base kVA (3-ph) / 1.7325*Base kV}

Where, Base kV is the line-to-line voltage.Base Z = { (Base kV / 1.7325)2*1000 / (Base kVA) / 1.7325 }

Base Z = { (Base kV)2 / (Base MVA) }

Sometimes the per unit impedance of a component of a system is expressed on a baseother than the one selected as base for the part of the system in which the component islocated. Since all impedances in any one part of a system must be expressed on thesame impedance base when making computations, it is necessary to have a means ofconverting per unit impedances from one base to another. The per unit impedance isgiven by following equation;

Per unit Z = { ( Actual Z in Ohms*Base MVA ) / ( Base kV )2 }

Which shows that per unit impedance is directly proportional to “base MVA” andinversely proportional to the square of the base voltage. Therefore, to change from perunit impedance on a given base to per unit impedance on a new base, the followingequation is used,

Per unit Znew = ( Per unit Zgiven )*( Base kVgiven / Base kVnew )2*( Base MVAnew / Base MKVAgiven )

The Ohmic values of resistance and leakage reactance of a transformer depends onwhether they are measured from the LT side or HT side of a transformer. If they areexpressed in “p.u.”, the base MVA rating of the transformer which is same as referredfrom HT side or LT side. The base kV is selected as the voltage of LT winding, if theohmic values are referring to LT side, else it is selected as voltage of HT winding, if theohmic values are referring to HT side of transformer. Whereas the “PU” values remainssame regardless of whether they are determined from HT side or LT side.

The advantages of the “PU” method are;

• The “PU” impedance of machines of same type and widely different ratingsusually lie within a narrow range, although the ohmic values differ for machinesof different ratings. For this reason, when the impedance is not known definitely,it is generally possible to select from the tabulated values a “PU” impedancewhich will be reasonably correct.

• When impedance in ohms is specified in an equivalent circuit, each impedancemust be referred to the same circuit by multiplying it by the square of the ratio ofrated voltages of the two sides of a transformer. The “PU” impedance, onceexpressed in proper base, remains same either referring from HT side or LT side.

• The way in which transformers are connected in three phase circuits does notaffect the “PU” impedances of the equivalent circuit, although the transformerconnection does determine the relation between the voltage bases on the twosides of the transformer.

Two winding Transformer Parameter Conversions:

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Manufacturers usually specify the impedance of a piece of apparatus in percent or peron the base of the nameplate rating. It is converted to common base using MVA ratingand the voltage rating of transformer. Sometimes the voltage ratings of the transformer does not match exactly with the base voltage on their respective sides, in case thetransformer parameters are converted to the base values of voltage and MVA. To beginwith, assuming that transformer tap is on primary side ( HV side ), the given impedance

is converted to common base as;

Znewpu = { (Zoldpu)*(MVAnew / MVAold)*(Rated kVsec / Base kVsec)2 }

If the transformer parameters are given in actual units (ohms). Then the values areconverted to common base as;

Zpu = (Zohms)*(Base MVA / BasekV2)

Base kV is the voltage referred to the side at which measurements are made.

The transformer R/X ratio is used to separate the transformer resistance and reactancevalues from the impedance. If number of units are in parallel then the effectiveequivalent impedance is computed by dividing the impedance by units.

X = [ {√Z2 / { 1 + ( R / X )2 }} / Units ]

R = (X)*(R / X)

The minimum tap value is computed as;

Tapmin(pu) = [ { VTap min kV / Base kVpri }*{ Base kVsec / Rated kVsec } ]

The maximum tap value is computed as;

Tapmax(pu) = [ { VTap max kV / Base kVpri }*{ Base kVsec / Rated kVsec } ]

The tap step value (pu) is computed as;

Tap step(pu) = [ { VTap max kV – VTap min kV } / { NTap max – NTap min } ]

Nominal tap value (pu) is computed as;

Tap nom = [ { VTap min kV + { ( NTap nom – NTap min )*Tap step kV } } ]*[ Base kV sec / Rated kV sec ]

The vector groups shows the connection of phases of two windings of a transformer andthe numerical index for the phase displacement of the vectors of the two star-voltages.The numerical index shows by what multiples of 30o the low voltage vector lags ( anti-clockwise rotation of vectors ) behind the high voltage vector with the correspondingterminal designation. For example the groups are interpreted as;

• Vector Group Dy5 – High voltage in Delta and low voltage in star connection.

• Vector Group Yz11 – High voltage in Star and low voltage in zigzag connection.

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The transformer vector group information is required for 3-phase load flow andunbalanced fault studies. The different vector groups used are;

• Star with neutral isolated.• Star with neutral grounded.• Star with neutral impedance.

• Delta connected.

The zero sequence impedances differ greatly depending on the type of connection andthe construction of the transformers. Conductors connected to transformer windings withdelta connection or with star with an insulated neutral point cannot carry a zerosequence current. The zero sequence impedance is therefore infinite. When the neutralpoint of star winding is earthed or connected, zero sequence can flow in the associatedsystem. If the transformer is star connected on primary side and delta connected onsecondary side, then shunt impedance will exists from primary node to ground and vice-versa.

The neutral impedance given in ohms, converted to common base as;

Base Zpri = ( Base kVpri2 / Base MVA ) in Ohm

Base Zsec = ( Base kVsec2 / Base MVA ) in Ohm

Rpu neutral pri = ( ROhm neutral pri / Base Z pri )

Rpu neutral sec = ( ROhms neutral sec / Base Zsec )

EXAMPLE:

Rated MVA = 315Primary Voltage = 420 kVSecondary Voltage = 240 kVPositive sequence impedance = 0.125PU or 12.5%Zero sequence impedance = 0.100PU or 10%TAPmin = 1.0TAPmax = 17.0TAPnormal = 12.0Minimum TAP Voltage = 360 kVMaximum TAP Voltage = 440 kVNeutral Rpri = Rsec = 2.0 Ohm.Connection = YnYn0

The transformer is connected to a bus on HT side with voltage 400 kV and on LT side isconnected to a bus with voltage 220 kV. Hence primary base voltage = 400 kV and thesecondary base voltage is 220 kV.The common base MVA = 100

Zpositive seq. in PU = (%Z / 100)*(Base MVA / Rated MVA)*(Rated kV / Base kV)2

= (12.5 / 100)*(100 / 315)*(240/220)2

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= 0.047225 PU

Xpositive seq. in PU = [ {√Z2 / { 1 + ( R / X )

2 }} / Units ]

= [ {√(0.047225)2 } / { 1 + (0.05)

2 } ]

= 0.047166 PU

Rpositive seq. in PU = (0.04766 * 0.05)= 0.002358 PU

Zzero seq. in PU = (%Z / 100)*(Base MVA / Rated MVA)*(Rated kV / Base kV)2

= (10 / 100)*(100 / 315)*(240/220)2 = 0.0377804 PU

Xzero seq. in PU = [ {√Z2 / { 1 + ( R / X )

2 }} / Units ]

= [ {√(0.0377804)2 } / { 1 + (0.05)

2 } ]

= 0.0377332 PU

Rzero seq. in PU = (0.0377332 * 0.05)

= 0.00188666 PU

Tapmin(pu) = [ { VTap min kV / Base kVpri }*{ Base kVsec / Rated kVsec } ]= [ { 360 / 400 }*{ 220 / 240 } ]= 0.82500 PU

Tapmax(pu) = [ { VTap max kV / Base kVpri }*{ Base kVsec / Rated kVsec } ]= [ { 440 / 400 }*{ 220 / 240 } ]= 1.00833 PU

Tap step(pu) = [ { VTap max kV – VTap min kV } / { NTap max – NTap min } ] = [ { 440 – 360 } / { (17 – 1)*400 } ]= 0.0125 PU

Tap = [ { VTap min kV + { ( NTap nom – NTap min )*Tap step kV } } ]*[ Base kV sec / Rated kV sec ]nom= [ { 360 + { (12 – 1)*5 } } / 400 ]*[ 220 / 240 ] = 0.95104 PU

The neutral impedance values are computed as;


= ( 400 )2 / 100= 1600 Ohm.


= ( 220 )2 / 100= 484 Ohm.

Rpu neutral pri = ( ROhm neutral pri / Base Z pri )= ( 2 / 1600 )= 0.00125 PU

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Rpu neutral sec = ( ROhms neutral sec / Base Zsec )= ( 2 / 484 )= 0.004132 PU

Three Winding Transformer Parameter Conversions:The MVA rating of a two-winding transformer is same on primary and secondary side,whereas all the three windings of a three winding transformer may have different MVAratings. The impedance of each winding of a three-winding transformer may be given inpercent or PU based on the rating of its own winding. The transformer impedancevalues, which are measured by short circuit test, are

• Impedance measured in primary with secondary short circuited and tertiary open(Z ps).

• Impedance measured in primary with tertiary short circuited and secondary open(Z pt).

• Impedance measured in secondary with tertiary short circuited and primary open

(Z st).

If the three impedances measured in Ohms are referred to the voltage of one of thewindings, the impedance of each separate winding referred to that same winding arerelated to the measured impedances as;

Z ps = Zp + Zs Z pt = Zp + Zt Zst = Zs + Zt

Where Zp, Zs, and Zt are the impedance of primary, secondary and tertiary windingsreferred to primary circuit. If Zps, Zpt, Zst are the measured impedances refer to primary

circuit, the real and reactive parts are separated as;

X ps = [ { √Z ps2 } / {( R / X ps )

2 + 1} ] / Units

X pt = [ { √Z pt2 } / {( R / X pt )

2 + 1} ] / Units

X st = [ { √Z st2 } / {( R / X st )

2 + 1} ] / Units

Solving the above impedance simultaneous equations for Rp, Rs, & Rt, Xp, Xs & Xtyields,

Rp = [ R ps + R pt – R st ] / 2

Rs = [ R ps + R st – R pt ] / 2

Rt = [ R pt + R st – R ps ] / 2

Xp = [ X ps + X pt – X st ] / 2

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Xs = [ X ps + X st – X pt ] / 2

Xt = [ X pt + X st – X ps ] / 2

The impedance of the three windings are connected in star (Y) to represent the singlephase equivalent circuit of the three winding transformer. Since the ohmic values of theimpedances must be referred to the same voltage, it follows that conversion to PUrequires the same MVA base for all the three circuits and requires voltage bases in threecircuits of the transformer.

The neutral impedances, if present is converted to per unit values on common base as;

Base Z pri = ( Base kV pri )2 / Base MVA

Base Z sec = ( Base kV sec )2 / Base MVA

Base Z ter = ( Base kV ter )2 / Base MVA

Rpu neutral pri = Rohm neural pri / Base Z pri

Rpu neutral sec = Rohm neutral sec / Base Z sec

Rpu neutral ter = Rohm neutral ter / Base Z ter

EXAMPLE:Rated Primary MVA = 15Primary Voltage = 66 kVRated Secondary MVA = 10Secondary Voltage = 13.2 kVRated tertiary MVA = 5.0Tertiary Voltage = 2.3 kVZps = 7% on 15 MVA, 66 kVRX-Ratio-ps = 0.05Zpt = 9% on 15 MVA, 66 kVRX-Ratio-pt = 0.05Zst = 8% on 15 MVA, 66 kVRX-Ratio-st = 0.05TAPmin = 1.0TAPmax = 17.0TAPnormal = 12.0Minimum TAP Voltage = 59.4 kVMaximum TAP Voltage = 72.6 kVRpri = Rsec = Rtertiary = 2.0 Ohm.Connection = YYnYn0

Assuming the common base values as 15MVA and 66 kV. First step is to convert all theimpedance to common base on primary side. Zps & Zpt are measured at primaryratings, need no conversion while the Zst measured at different ratings it is converted tocommon base values as;

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Zst in PU = (%Z / 100)*(Base MVA / Rated MVA)*(Rated kV / Base kV)2

= (8.0 / 100)*(15 / 10)*(13.2 / 13.2 )2 = 0.12 PU

Zps in PU = (%Z / 100)*(Base MVA / Rated MVA)*(Rated kV / Base kV)2

= (7.0 / 100)*(15 / 15)*(13.2 / 13.2 )2 = 0.07 PU

Zpt in PU = (%Z / 100)*(Base MVA / Rated MVA)*(Rated kV / Base kV)2

= (9.0 / 100)*(15 / 15)*(13.2 / 13.2 )2 = 0.09 PU

X ps = [ { √Z ps2 } / {( R / X ps )

2 + 1} ] / Units= [ { √(0.07)2 } / {( 0.05 )2 + 1} ] / Units= 0.06991 PU

X pt = [ { √Z pt2

} / {( R / X pt )2

+ 1} ] / Units= [ { √(0.09)2 } / {(0.05)2 + 1} ] / Units= 0.0898877 PU

X st = [ { √Z st2 } / {( R / X st )

2 + 1} ] / Units= [ { √(0.12)2 } / {(0.05)2 + 1} ] / Units= 0.1198503 PU

Rst = Xst * RXRatio-st = 0.1198503 * 0.05= 0.0059925 PU

Rps = Xps * RXRatio-ps = 0.06991 * 0.05= 0.0034955 PU

Rpt = Xpt * RXRatio-pt = 0.0898877 * 0.05= 0.00449438 PU

Rp = [ R ps + R pt – R st ] / 2= [ 0.0034955 + 0.00449438 – 0.0059925 ] / 2

= 0.00099865 PU

Rs = [ R ps + R st – R pt ] / 2= [ 0.0034955 + 0.0059925 – 0.00449438 ] / 2= 0.00249685 PU

Rt = [ R pt + R st – R ps ] / 2= [ 0.00449438 + 0.0059925 – 0.0034955 ] / 2

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= 0.00349565 PU

Xp = [ X ps + X pt – X st ] / 2= [ 0.06991 + 0.0898877 – 0.1198503 ] / 2= 0.0199737 PU

Xs = [ X ps + X st – X pt ] / 2= [ 0.06991 + 0.1198503 – 0.0898877 ] / 2= 0.0499363 PU

Xt = [ X pt + X st – X ps ] / 2= [ 0.0898877 + 0.1198503 – 0.06991 ] / 2= 0.069914 PU

Tapmin(pu) = [ { VTap min kV / Base kVpri }*{ Base kVsec / Rated kVsec } ]= [ { 59.4 / 66 }*{ 13.2 / 13.2 } ]

= 0.9 PU

Tapmax(pu) = [ { VTap max kV / Base kVpri }*{ Base kVsec / Rated kVsec } ]= [ { 72.6 / 66 }*{ 13.2 / 13.2 } ]= 1.10 PU

Tap step(pu) = [ { VTap max kV – VTap min kV } / { NTap max – NTap min } ] = [ { 72.6 – 59.4 } / { (17 – 1)*66 } ]= 0.0125 PU

Tap = [ { VTap min kV + { ( NTap nom – NTap min )*Tap ste kV } } ]*[ Base kV sec / Rated kV sec ]nom p = [ { 59.4 + { (12 – 1)*0.825 } } / 66 ]*[ 13.2 / 13.2 ]

= 1.0375 PU

The neutral impedance values are computed as;


= ( 66 )2 / 15= 290.4 Ohm.


= ( 13.2 )2 / 15= 11.616 Ohm.

Base Zter = ( Base kVter 2 / Base MVA ) in Ohm

= ( 2.3 )2 / 15= 0.35267 Ohm.

Rpu neutral pri = ( ROhm neutral pri / Base Z pri )= ( 2 / 290.4 )= 0.00688 PU

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Rpu neutral sec = ( ROhms neutral sec / Base Zsec )= ( 2 / 11.616 )= 0.172176 PU

Rpu neutral ter = ( ROhms neutral ter / Base Zter )= ( 2 / 0.35267 )= 5.671 PU

Transmission Line Parameter Conversions:For a transmission line, line resistance, reactance and the susceptance of the line aregiven in actual units (Ohms) per km length of the line per circuit. Zero sequenceimpedance denotes the impedance of zero sequence system of a three phase line perphase in which equal and in-phase currents are flowing through the three phaseconductors of the system. The operative zero sequence impedance is affected by,among other things, the electrical conductivity of the earth and the presence of earthwires. The same formule are used for both positive and zero sequence parameter

conversion. The line positive sequence parameters are converted to per unit on commonbase values as;

Base Z = ( Base kV )2 / Base MVA

R1pu = ( R1ohm / Base Z )*( Length / Circuits )

X1pu = ( X1ohm / Base Z )*( Length / Circuits )

B1pu = ( B1mho * Base Z )*( Length * Circuits )

EXAMPLE: Number of Circuits = 1.0Length of Line = 181km.Positive sequence R = 0.0288864 Ohm/km.Zero sequence R = 0.072216 Ohm/km.Positive sequence X = 0.32704 Ohm/km.Zero sequence X = 0.8176 Ohm/km.Positive sequence B = 1.78087E-06 mho/km.Zero sequence B = 1.52374E-06 mho/km.Common base MVA = 100 and Base voltage = 400.0 kV.

Base Z = ( Base kV )2 / Base MVA

= ( 400 )2

/ 100= 1600 Ohm.

R1pos. pu = ( R1ohm / Base Z )*( Length / Circuits )= ( 0.0288864 / 1600 )*(181 / 1)= 0.003267 PU

X1pos.pu = ( X1ohm / Base Z )*( Length / Circuits )

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= ( 0.32704 / 1600 )*(181 / 1)= 0.03699 PU

B1pos.pu = ( B1mho * Base Z )*( Length * Circuits )= (1.78087E-06 * 1600)*(181*1)

= 0.51574 PU

R1zero. pu = ( R0ohm / Base Z )*( Length / Circuits )= ( 0.072216 / 1600 )*(181 / 1)= 0.008169 PU

X1zero.pu = ( X0ohm / Base Z )*( Length / Circuits )= ( 0.8176 / 1600 )*(181 / 1)= 0.09249 PU

B1zero.pu = ( B0mho * Base Z )*( Length * Circuits )= (1.52374E-06 * 1600)*(181*1)

= 0.44127 PU

Motor Parameter Conversions:The induction motor parameters are given in PU on its own rating. The motor isrepresented as a shunt impedance between node to which motor is connected andground. The shunt impedance value is obtained after simplifying the exact equivalentcircuit as shown in fig below. The motor parameters are converted to common base byusing formula.

Rmotor = Rold ( Base kV old / Base kV new )2*( MVA new / MVA old )

Rmotor = Rmotor / Units

Xmotor = Xold ( Base kV old / Base kV new )2*(MVA new / MVA old )

Xmotor = Xmotor / Units

By using above formula, stator resistance, rotor resistance, stator reactance, rotorreactance and magnetizing reactance are converted to common base. On simplifying theequivalent circuit, we obtain equations for the equivalent resistance and reactance asdenoted below;

Rmotorq = Rstator + [ { ( Slip * Rrotor * ( Xmag )2 ) } / { Rrotor

2 + ( Slip2 ( Xrotor + Xmag )2 } ]

Xmotorq = Xstator + [ ( Rrotor 2*Xmag ) + ( Slip

2*Xrotor *Xmag ( Xrotor + Xmag ) ) ] / [ Rrotor

2+ ( Slip

2 ( Xrotor + Xmag )

2 ) ]

Type of motor winding connection is used for zero sequence network calculations only.The motor neutral impedance, value is converted to common base and 3-times of it isadded to the zero sequence impedance value of the motor to get the effective zerosequence impedance.

EXAMPLE:

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Number of Units = 1.0Voltage Rating = 2.3 kVRating = 1.6785 MW, 0.8 P.F.Stator resistance = 0.029 Ohm.Stator reactance = 0.226 Ohm.Rotor resistance = 0.022 Ohm.

Rotor reactance = 0.226 Ohm.Magnetizing reactance = 13.04 Ohm.Winding type = Delta.Common base = 100 MVA and base volt = 2.3 kV.

Motor Zbase = (Base kV)2 / (Base MVA)

= (2.3)2 / (2.098125)= 2.5213 Ohm.

Rstator = Rold ( Base kV old / Base kV new )2*( MVA new / MVA old )

= (0.029 / 2.5213)*( 2.3 / 2.3 )2 *(100 / 2.098125)= 0.5482 PU.

Xstator = Xold ( Base kV old / Base kV new )2*(MVA new / MVA old )

= (0.226 / 2.5213)*( 2.3 / 2.3 )2 *(100 / 2.098125)= 4.2722 PU.

Rrotor = Rold ( Base kV old / Base kV new )2*( MVA new / MVA old )

= (0.022 / 2.5213)*( 2.3 / 2.3 )2 *(100 / 2.098125)= 0.41588 PU.

Xrotor = Xold ( Base kV old / Base kV new )2*(MVA new / MVA old )

= (0.226 / 2.5213)*( 2.3 / 2.3 )2 *(100 / 2.098125)= 4.2722 PU.

Xmagnetizing = Xold ( Base kV old / Base kV new )2*(MVA new / MVA old )

= (13.04 / 2.5213)*( 2.3 / 2.3 )2 *(100 / 2.098125)= 246.504 PU.

Rmotorq = Rstator + [ { ( Slip * Rrotor * ( Xmag )2 ) } / { Rrotor

2 + ( Slip2 ( Xrotor + Xmag )2 } ]

= 0.5482 + [{ (0.0077*0.41588*(246.504)2 )} / {(0.41588)2 + ( (0.0077)2 ( 4.2722 + 246.504)2 }]= 50.010 PU.

Xmotorq = Xstator + [ ( Rrotor 2*Xmag ) + ( Slip

2*Xrotor *Xmag ( Xrotor + Xmag ) ) ] / [ Rrotor

2+ ( Slip

2 ( Xrotor + Xmag )

2 ) ]

= 4.2722+[ (0.41 882*246.504)+(0.0077

2*4.2722*246.504 (4.2722+246.504))] / [ 0.41588

2(4.2722+246.504)

2) ]5

= 19.215899 PU.

Series Reactor Parameter Conversions: The reactors are used as branch elements to limit the current during fault conditions, tocyclic voltage fluctuations caused by repetitive loads (Cyclic loads) in conjunction withcondensers and to limit the motor starting currents. The series reactor resistance andreactance values are given in PU on its own rating. The values are converted tocommon base as:

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R new = (R old)* (Base kV old / Base kV new)2* (MVA new / MVA old)

X new = (X old)* (Base kV old / Base kV new)2* (MVA new / MVA old)

EXAMPLE:Rated Voltage = 11 kV.Rated Current = 262 Amp.kVar = 256.

Reactor MVA = 1.7325 * 11*262 = 5 MVA

X in Ohm = (kVar) / (3* I2)= (256*1000) / (3*2622) = 1.2431 Ohm/Phase

X in PU = (1.2431 *5) / (11)2

= 0.0514 PU

Common base MVA = 100, Base Voltage = 11kV.

Rreactor = 0.0 PU.

Xreactor = (0.514)*(100/5)*(11/11)2

= 1.028 PU

Shunt Capacitor Parameter Conversions: The shunt capacitor banks are used extensively to correct power factors and as results,improve voltage regulation at the point of connection. The shunt capacitor conductance

and susceptance values are given in PU on its own rating. The values are converted tocommon base as:

G new = (G old)* (Base kV new / Base kV old)2 * (MVA old / MVA new)

B new = (B old)* (Base kV new / Base kV old)2 * (MVA old / MVA new)

Series Capacitor Parameter Conversions:The series capacitors are sometimes used on transmission and distribution lines tocompensate for the inductive reactance drop or to improve the system stability byincreasing the amount of power that can be transmitted on tie lines. The series capacitorconductance and susceptance values are given in PU on its own rating. The values areconverted to common base as:

G new = (G old)* (Base kV new / Base kV old)2 * (MVA old / MVA new)

All series elements are molded in impedance form, hence the parameters are convertedto impedance form by inverting the admittance.

R pu = (G pu) / (G2 pu + B2 pu)

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V gen pu = (V gen kV) / (Base kV)

Type of generator winding connection is used for zero sequence network calculationsonly. The neutral impedance is given in actual units (Ohms), it is converted to PU oncommon base as:

Base Z = (Base kV)2 / (Base MVA)

R neutral pu = (R neutral Ohms) / (Base Z)

The generator neutral impedance value is converted to common base and 3 times of it isadded to the zero sequence impedance value of the generator to get the effective zerosequence impedance.

Tdo’ = No load time constant determines the excitation and de-magnetisation with Openstator circuit.

Td” = Sub-transient time constant determines the excitation and de-magnetization withthree phase short circuit.Td’ = Transient time constant determines the excitation and de-magnetization with threephase short circuit.

EXAMPLE:Number of units = 1.0Voltage Rating = 11.0 kVRated MVA = 260 MVAPositive sequence resistance = 0.00154 PU.Steady state direct axis reactance = 2.22 PU.Transient direct axis reactance = 0.265 PU.

Negative sequence reactance = 0.225 PU.Zero sequence reactance = 0.125 PU.Winding connection = Star connection.Neutral resistance = 2.0 Ohms.Neutral reactance = 0.0 Ohms.

Common base = 100 MVA and base voltage = 11 kV.

R gen = (R old)* (Base kV old / Base kV new)2* (MVA new / MVA old)

= (0.00154)*(11 / 11)2* (100 / 260)= 0.00059 PU.

X gen pos. = (X old)* (Base kV old / Base kV new)2

* (MVA new / MVA old) = (2.22)*(11 / 11)2* (100 / 260)= 0.8538 PU.

X gen Neg. = (X old)* (Base kV old / Base kV new)2* (MVA new / MVA old)

= (0.225)*(11 / 11)2* (100 / 260)= 0.08654 PU

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X d’ = (X old)* (Base kV old / Base kV new)2* (MVA new / MVA old)

= (0.265)*(11 / 11)2* (100 / 260)= 0.10192 PU.

Base Impedance = (Base Voltage)2 / Base MVA= (11)2 / 100= 3.24 Ohm.

Generator technical data-sheet of Brush Electric Machine, USA are as shown belowwhich is installed in Search-Chem industries ltd, Bharuch;

Rated MVA = 35Terminal Voltage = 11.0kVFrequency = 50HzSpeed = 3000RPMPower Factor = 0.85

Applicable national standard = IEC 34-3Rated Air inlet temperature = 15 deg.Unsaturated Direct axis Synchronous Reactance (Xd) = 243.9%Saturated Direct axis Synchronous Reactance (Xd) = 243.9%Saturated Direct axis Transient Reactance (Xd’) = 25.1%Unsaturated Direct axis Transient Reactance (Xd’) = 33.9%Saturated Direct axis Sub-transient Reactance (Xd”) = 18%Unsaturated Direct axis Sub-transient Reactance (Xd”) = 24.1%Unsaturated Quadrature axis Synchronous Reactance (Xq) = 224.1%Saturated Quadrature axis Synchronous Reactance (Xq) = 224.1%Unsaturated Quadrature axis Transient Reactance (Xq’) = 48.2%Saturated Quadrature axis Sub-transient Reactance (Xq”) = 17.7%Unsaturated Quadrature axis Sub-transient Reactance (Xq”) = 24.1%Saturated Negative sequence Reactance (X2) = 17.3%

Unsaturated Negative sequence Reactance (X2) = 23.2%Saturated Zero Sequence Reactance (X0) = 10.2%Unsaturated Zero Sequence Reactance (X0) = 12.9%Leakage Reactance, Overexcited (XLM,OEX) = 21.2%Leakage Reactance, Underexcited (XLM,UEX) = 21.2%Short Circuit Ratio = 0.43Open Circuit Time constant (Td0’) = 6.016second3-ph Short-circuit transient time constant (Td3’) = 0.620secondLine to Line short-circuit transient time constant (Td2’) = 0.975secondLine to Neutral short-circuit transient time constant (Td1’) = 1.165secondShort-circuit sub-transient time constant (Td”) = 0.015secondOpen Circuit Sub-transient time constant (Td0”) = 0.021second

Open Circuit Time constant (Tq0’) = 0.522second 3-ph Short-circuit transient time constant (Tq’) = 0.522second Short-circuit sub-transient time constant (Tq”) = 0.015secondOpen Circuit Sub-transient time constant (Tq0”) = 0.041second

Generator technical data-sheet of Brush Electric Machine, USA are as shown belowwhich is installed in Search-Chem industries ltd, Bharuch;

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Rated MVA = 60Terminal Voltage = 11.0kVFrequency = 50HzSpeed = 3000RPMPower Factor = 0.85

Applicable national standard = IEC 34-3

Rated Air inlet temperature = 15 deg.Unsaturated Direct axis Synchronous Reactance (Xd) = 257%Saturated Direct axis Transient Reactance (Xd’) = 22.7%Saturated Direct axis Sub-transient Reactance (Xd”) = 15.9%Unsaturated Zero Sequence Reactance (X0) = 10.4%Unsaturated Negative sequence Reactance (X2) = 19.4%Short Circuit Ratio = 0.41

Generator technical data-sheet of ALSTOM Electric Machine, UK are as shown belowwhich is installed in Sanghi industries ltd, Kuttch;

Rated MVA = 15

Terminal Voltage = 11.0kVFrequency = 50HzSpeed = 600RPMPower Factor = 0.80

Applicable national standard = IEC 60034 Ambient temperature = 50 deg.Unsaturated Direct axis Synchronous Reactance (Xd) = 169.4%Saturated Direct axis Synchronous Reactance (Xd) = -Saturated Direct axis Transient Reactance (Xd’) = 28.1%Unsaturated Direct axis Transient Reactance (Xd’) = 34.7%Saturated Direct axis Sub-transient Reactance (Xd”) = 18.7%Unsaturated Direct axis Sub-transient Reactance (Xd”) = 23.1%

Unsaturated Quadrature axis Synchronous Reactance (Xq) = 100.2%Saturated Quadrature axis Synchronous Reactance (Xq) = -Saturated Quadrature axis Sub-transient Reactance (Xq”) = 21.5%Unsaturated Quadrature axis Sub-transient Reactance (Xq”)= 25.6%

Unsaturated Negative sequence Reactance (X2) = 24.5%Unsaturated Zero Sequence Reactance (X0) = 6.5%Unsaturated Zero Sequence Resistance (R0) = 0.914%

Short Circuit Ratio = 0.637Open Circuit Time constant (Td0’) = 3.63second3-ph Short-circuit transient time constant (Td’) = 0.743second

3-ph Short-circuit saturated transient time constant (Td’) = 0.590secondShort-circuit sub-transient time constant (Td”) = 0.05secondShort-circuit saturated sub-transient time constant (Td”) = 0.035secondMaximum kVAr available at zero p.f. under-excited = 0.51 PUMaximum kVAr available at zero p.f. over-excited = 0.76 PUOver speed = 720RPM

Typical Parameters Considered for fault study:

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Generator:

• Transient (Xd’) or sub-transient (Xd”) reactance is considered for positivesequence.

• Negative sequence reactance which is approximately equal to (Xd”).

• Zero sequence reactance, which is comparatively small around 0.1 to 0.7 timesof (Xd”).

• Assume Short Circuit Ratio = 1 / Xd while value of Xd is not given in data-sheet

Transformer:* Positive sequence, Negative sequence, & Zero sequence impedances are equal.

Per Unit (PU) and Percentage Quantity (%):

Per Unit Quantity = Percentage Quantity / 100

• Quantity => Voltage, Current, MVA, Impedance

• E.g. Z = 10% => Z = 0.10 PU; V = 105% => V = 1.05PU• Per Unit Computation is slightly advantageous over percentage computation.

• Product of two quantities expressed in PU. Result also in PU.• Product of two quantities expressed in percent. Result shall be divided by 100 to

get percent.

• Fault level calculations are generally performed using PU only.

Per Unit Quantity:

Q(PU) = Q (ACTUAL) / Q (BASE)

e.g. Vbase = 6.6kV; Vactual = 3.3kV; => V = 0.5 PU

e.g. Pbase = 100 MVA; Pactual = 200 MW; => P = 2 PU

Choosing Base:

In general, MVA (3-ph) & Voltage (L to L) chosen as base

Base Current = Base MVA / 1.7325 * Base Voltage

Base Impedance = Base Voltage / 1.7325 * Base Current

= {Base Voltage / (1.7325 * Base MVA / 1.7325 * Base Voltage) }

= (Base Voltage)2 / Base MVA

Base Voltage changes on either side of Transformer:

• Choose Base Voltage as 11 kV and Base Power as 100 MVA• Let there be 11 / 132 kV Transformer.

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• On the HT side of Transformer, base voltage is Automatically 132 kV.• You can not independently choose another base voltage on other side of

Transformer.

• Base Power is 100 MVA on either side of Transformer.

On Low voltage side:

Base voltage = 11 kV (Always L to L )

Base MVA = 100 (Always 3-ph Power)

Base Current = 100 / 1.7325 * 11 = 5.2486 kA

Base Impedance = (11)2 / 100 = 1.21 Ohms

On High voltage side:

Base Voltage = 132 kV (Always L to L)

Base MVA = 100 (Always 3-ph Power)

Base Current = 100 / 1.7325 * 132 = 0.4374 kA

Base Impedance = (132)2 / 100 = 174.24 Ohms

Advantages of Calculations in Per Unit System:

Per Unit Impedance of Transformer is same whether referred to Primary orSecondary.

e.g. 11 / 33 kV, 50 MVA, Z = 10% (0.1PU)

• In PU, Z = 0.1 on either 11 kV or 33 kV side

• In Ohms, on 11 kV side:

Z base = (Base Voltage)2 / Base MVA = (11)2 / 50 = 2.42 Ohms

Z 11 = (Z base * Z pu) = (2.42)* (0.1) = 0.242 Ohms.

On 33 kV side:


Z 33 = (Z base* Z pu) = (21.78)* (0.1) = 2.178 Ohms

• Per Unit Impedance lie within a Narrow Band while Ohmic values can bewidely different.

• Transformer 415 V to 400 kV and 500 kVA to 500 MVA.Z lies between 5% (0.05PU) to 15% (0.15PU)

• Generator 1 MVA to 500 MVA,X’d lies between 20% (0.2PU) to 30% (0.3PU)

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PU = only realistic way to solve big and practical problems.

Basic Considerations of Short-Circuit Calculations:

Why Short-Circuit Calculations?

Several sections of the National Electrical Code relate to proper over-current protection.Safe and reliable application of over-current protective devices based on these sectionsmandate that a short circuit study and a selective coordination study be conducted.

The protection for an electrical system should not only be safe under all serviceconditions but, to insure continuity of service, it should be selectively coordinated aswell. A coordinated system is one where only the faulted circuit is isolated withoutdisturbing any other part of the system. Over-current protection devices should alsoprovide short-circuit as well as overload protection for system components, such as bus,wire, motor controllers, etc.

To obtain reliable, coordinated operation and assure that system components are

protected from damage, it is necessary to first calculate the available fault current atvarious critical points in the electrical system.

Once the short-circuit levels are determined, the engineer can specify proper interruptingrating requirements, selectively coordinate the system and provide componentprotection.

General Comments on Short-Circuit Calculations:Short Circuit Calculations should be done at all critical points in the system.These would include:- Service Entrance- Panel Boards

- Motor Control Centers- Motor Starters- Transfer Switches- Load CentersNormally, short circuit studies involve calculating a bolted 3-phase fault condition. Thiscan be characterized as all three phases “bolted” together to create a zero impedanceconnection. This establishes a “worst case” condition that results in maximum thermaland mechanical stress in the system. From this calculation, other types of faultconditions can be obtained.

Sources of short circuit current that are normally taken under consideration include:- Utility Generation

- Local Generation- Synchronous Motors and- Induction MotorsCapacitor discharge currents can normally be neglected due to their short time duration.Certain IEEE (Institute of Electrical and Electronic Engineers) publications detail how tocalculate these currents if they are substantial.

Asymmetrical Components:

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Short circuit current normally takes on an asymmetrical characteristic during the first fewcycles of duration. That is, it is offset about the zero axis, as indicated in Figure 1.

Figure 1.

In Figure 2, note that the total short-circuit current Ia is the summation of two

components - the symmetrical RMS current IS, and the DC component, IDC. The DCcomponent is a function of the stored energy within the system at the initiation of theshort circuit. It decays to zero after a few cycles due to I2R losses in the system, at whichpoint the short circuit current is symmetrical about the zero axis. The RMS value of thesymmetrical component may be determined using Ohm’s Law. To determine theasymmetrical component, it is necessary to know the X/R ratio of the system. To obtainthe X/R ratio, the total resistance and total reactance of the circuit to the point of faultmust be determined. Maximum thermal and mechanical stress on the equipment occursduring these first few cycles. It is important to concentrate on what happens during thefirst half cycle after the initiation of the fault.

Figure 2.

Where,Ia = Asymmetrical RMS Fault Current in kA.IDC = DC component depend on X/R in kA

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Is = Asymmetrical RMS Fault Current in kA.Ip = Instantaneous Peak Current in kA.

Figure 2 illustrates a worst case waveform that 1 phase of the 3 phase system willassume during the first few cycles after the fault initiation.

The key portions are:- Symmetrical RMS Short Circuit Current = Is- Instantaneous Peak Current = Ip- Asymmetrical RMS Short Circuit Current(worst case single phase) = Ia

Interrupting Rating, Interrupting Capacity and Short-Circuit Currents:Interrupting Rating can be defined as “the maximum short-circuit current that a protectivedevice can safely clear, under specified test conditions.”Interrupting Capacity can be defined as “the actual short circuit current that a protectivedevice has been tested to interrupt.”

Interrupting Rating:Equipment intended to break current at fault levels shall have an interrupting ratingsufficient for the system voltage and the current, which is available at the line terminalsof the equipment.

Available Short-Circuit Current:Service Equipment shall be suitable for the short circuit current available at its supplyterminals.

Low voltage fuses have their interrupting rating expressed in terms of the symmetricalcomponent of short-circuit current, IS. They are given an RMS symmetrical interruptingrating at a specific power factor. This means that the fuse can interrupt any asymmetrical

current associated with this rating. Thus only the symmetrical component of short-circuitcurrent need be considered to determine the necessary interrupting rating of a lowvoltage fuse. For U.L. listed low voltage fuses, interrupting rating equals its interruptingcapacity.

Low voltage molded case circuit breakers also have their interrupting rating expressed interms of RMS symmetrical amperes at a specific power factor. However, it is necessaryto determine a molded case circuit breaker’s interrupting capacity in order to safely applyit. The reader is directed to Buss bulletin PMCB II for an understanding of this concept.

Three Phase fault through Impedance:

Fault MVA = Base MVA / (Z1 + Zf )Ia1 = Fault MVA / 1.7325*(Rated kV), Ia2 = 0, Ia0 = 0

Let, Ia1 = Positive sequence Fault Current in kA,Ia2 = Negative sequence Fault Current in kAIa0 = Zero sequence Fault Current in kAZ1 = Positive sequence impedance in PU.Zf = Fault impedance in PU.

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Example: (Z1 + Zf ) = 0.2 PU., Base MVA = 15, Rated kV = 11, Base kV = 11.

PU method:

Step:-1

Fault MVA = 15 / 0.2 = 75 MVA

Step:-2Ia1 = 75 / 1.7325*(11) = 3.93kA.

Ohm method:

Step:-1Find Base Impedance value in Ohm


Step:-2Find Z value in OhmZ in Ohm = (Z base * Z pu) = (8.066)* (0.2) = 1.613 Ohms.

Step:-3Find Fault current

Ia1 = Rated kV / 1.7325*(Z in Ohm) = 3.93kA.

Single-line to ground fault:

(Z1’ + Z2’ + Z0’) = (Z1 + Z2 + Z0) / 3 Fault MVA = Base MVA / (Z1’ + Z2’ + Z0’)Ia0 = Fault MVA / 1.7325*(Rated kV)

Let, Ia1 = Positive sequence Fault Current in kA,Ia2 = Negative sequence Fault Current in kAIa0 = Zero sequence Fault Current in kAZ1 = Positive sequence impedance in PU.Zf = Fault impedance in PU.

Example: Z1 = 1.35 PU., Z2 = 1.35 PU, Z0 = 1.0 PU, Base MVA = 100, Rated kV = 6.6,Base kV = 6.6.

PU method:

Step:-1(Z1’ + Z2’ + Z0’) = (Z1 + Z2 + Z0) / 3 = (1.35 + 1.35 + 1.0) / 3 = 1.23 PU

Step:-1Fault MVA = 100 / 1.23 = 81.30 MVA

Step:-2Ia0 = 81.30 / 1.7325*(6.6) = 7.11kA.

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Ohm method:

Step:-1

Find Base Impedance value in OhmZ base = (Base Voltage)

2 / Base MVA = (6.6)2 / 100 = 0.4356 Ohms

Step:-2Find Z value in Ohm

Z in Ohm = (Z base * Z pu) = (0.4356)* (1.23) = 0.5357 Ohms.

Step:-3Find Fault current

Ia1 = Rated kV / 1.7325*(Z in Ohm) = 7.11kA.

Sudden three-phase short circuit on an unloaded synchronous generator :

A sudden three-phase short circuit on an initially unloaded synchronous generator willhave several types of current components: a unidirectional (or DC) transient current, an

AC transient current, and an AC steady state current. Upon close examination, the ACtransient current is seen to decay first rapidly, then more slowly. The initial rapid decay iscalled the sub-transient (Xd”) part and the slower decay is called the transient part. The

AC component is symmetrical, since all three phases will have essentially the sameRMS current values at any time, but the DC transients are asymmetrical. Thus thesymmetrical short-circuit current excludes the DC terms and the asymmetrical currentincludes both the AC and the DC terms.

Figure 2.1.1 shows plots that resemble actual short-circuit currents in the three armaturephases. The phase a current is completely offset in the negative direction, while the band c phase currents are offset half as much in the positive direction. The exact offsetswill depend on the instant at which the fault occurs relative to the pre-fault voltagewaveform.

DC offsets:

The physical reason for the existence of the DC offsets is that the current and fluxwaveforms of an inductive circuit cannot change instantly. Since the pre-fault current iszero, the DC offset in any given phase must take on a value equal in magnitude andopposite in sign to the value of the AC waveform just after the fault. Since the ACcurrents are shifted by 120o from each other, the DC components of the three phaseswill always add to zero.

The maximum DC offset in a phase current is the peak value of the AC initialcomponent. This occurs in one phase at most, since the three DC offsets must add tozero as discussed above. That means that the maximum DC offset is 1.414 times theinitial RMS AC waveform.

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View of DOBLE “DATC” software which is shown a DC OFFSET with L/R=300ms

Symmetrical short-circuit current:

In most short-circuit current calculations, the AC (or symmetrical) current is needed. Weconsider a sudden three-phase short circuit on an unloaded synchronous generator.

The sub-transient component of this current decays with a time constant that isdetermined by the time constant of the damper windings, which are windings on the polefaces of salient-pole synchronous motors and generators. Cylindrical-rotor machinesusually have no damper windings, but eddy currents in their solid-iron rotors have much

the same effect. Just after the short circuit, the RMS value of the current is I" = E / Xd"where E is the pre-fault voltage and Xd" is the generator sub-transient reactance. If E isin volts (line- neutral) and Xd" is in ohms, then I" will be in amps. If E and Xd" are in per-unit, then I" will be in per unit. Since Xd" is usually tabulated in percent or per unit, andsince E is approximately 1.00 per unit, we will use per-unit equations for our calculations.

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Figure 2.1.1 Synchronous machine armature short-circuit currents for sudden 3-ph Fault

The transient component decays with a much longer time constant (about 1 secondcompared to several cycles for the sub-transient) that is due to the time constant of thefield winding. A few cycles after the fault occurs, the RMS value of the current is I' = E /Xd' where Xd' is the generator transient reactance.

Figure 2.1.2 shows the rms ac (or symmetrical) short-circuit current for the same casethat is illustrated in the previous figure.

If the short circuit is left on until the transients die out, the steady-state short-circuitcurrent may be determined. We subtract the steady-state current from the accomponent, and plot the result on semi-log paper, as shown in Figure 2.1.3, todetermine transient and sub-transient time constants and reactances.

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Figure 2.1.3 The transient time constant Td' is the slope of the line fitted by ignoring theinitial part of this curve. The excess current for the first few cycles may be plotted on asimilar graph to obtain the sub-transient time constant Td".

Calculation: According to ANSI/IEEE Std. 242-1996, the total AC component of armature currentconsists of the steady-state (Id) value and two components that decaying at a rate

according to their respective time constant.Iac = (Id” – Id’)*(e(2*3.14*f*t)/(Td”) ) + (Id’ – Id)*(e(2*3.14*f*t)/(Td’) ) + Id-----------(1)

(1) Subtransient Component, Id”

Id” = (e”/Xd”) p.u. where e” = (et + Xd” Sin(theta) )----------------(2)When machine is at no-load, e”=et

(2) Transient component, Id’

Id’ = (e’/Xd’) p.u.(3) Steady-state component, Id

Id = (et/Xd)* (If/Ifg)----------------------------------------------------------(3)(4) The DC component of the armature current is controlled by the sub-transient

reactance and the armature time constant:

Idc = (1.414)* (Id”)* (e(2*3.14*f*t)/(X/R) )= (1.414)* (Id”)* (e(t/(L/R)) )---------------------------------------------(4)

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General Description of Calculation Methodology:

LOW ZERO-SEQUENCE IMPEDANCES ON GENERATORS:

Good design practice dictates that short circuit calculations be performed to ensure that

the IEC rating specified for equipment is higher than the anticipated fault currents.Properly rated equipment is not only a code requirement but can impact the cost of aproject. Equipment that is underrated must be replaced, and overrated equipment costsmore than properly rated equipment.

Short circuit calculations during the design phase require some assumptions andgeneralizations, which are validated or corrected when a detailed study is completed.The detailed study is usually completed prior to the review of equipment submittals.Underrated equipment identified in the study can be changed before orders are releasedor factories prepare the equipment for shipment.

In the case study mentioned in the introduction, the detailed short circuit study was not

completed until after the equipment had been installed. Since the engineer had used thepositive sequence impedance of the generator for his calculations (or used generatordecrement curve data, which is based on the positive sequence impedance), some ofthe equipment was underrated. As will be discussed below, the lower zero sequenceimpedance of a generator can result in line-to-ground fault currents that can be as highas one and a half times the phase fault currents.

ANALYSISTo graphically illustrate the variation in magnitude between line-to-ground faults andphase faults, we will begin by looking at the one-line diagram for a typical installation.This is shown in Figure below. The service transformers are 2500 kVA and thegenerators are 1500 kW.

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Table 1 summarizes the impedance values for the transformers and generators shown. All values are shown in per unit for each phase. The kVA base is 10,000 and the kVbase for the primary is 13.8 and 0.48 for the secondary. Note that the values for thegenerator reactance are sub-transient values.

As can be seen by observation, the generator zero-sequence X(0) impedance issomewhat lower than the positive-sequence X(+). This contrasts markedly with thetransformers and the utility source impedance, where the zero-sequence impedanceX(0) is essentially the same as the positive sequence impedance X(+).

TABLE:-1: SOURCE & TRANSFORMER DATA

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For the purpose of this analysis, we will assume that the utility and generators will neverbe paralleled. As such, there is no need to consider the implications of the lowergenerator zero-sequence impedance being in parallel with the utility transformer zero-sequence impedance.

To determine the available three-phase fault currents, the pre-fault voltage (usuallytaken to be the nominal voltage) would be divided by the equivalent positive sequenceimpedance of the network, as seen from the faulted bus.

Using the impedance values for the conductors shown in Table 2, for a fault at the ATS,

the positive sequence network, with the generator as the source, would be as shown inFigure. (Since the line-to-ground fault current magnitudes for the transformer source arethe same as the phase fault current magnitudes, we will confine our analysis to thegenerator source.)

TABLE:-2: CONDUCTOR IMPEDANCE VALUE

POSITIVE SEQUENCE DIAGRAM OF NETWORK

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The equivalent impedance of the positive sequence network would then be the parallelcombination of the series combination of generators and their conductors in series withthe conductor to the ATS.

Neglecting the impedance of the conductors, neglecting the resistance of the generators,

and using only the positive sequence impedance would yield the following results:

We now turn our attention to the available line-to-ground fault currents. For this

calculation, the pre-fault voltage would be divided by the equivalent zero sequenceimpedance of the network, as seen from the faulted bus. Looking at the impedance seenat the ATS for a line-to-ground fault, the following sequence network is generated.

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Sequence Network for Line-to-Ground Faults

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As before, we will neglect the impedance of the conductors and the resistance of thegenerators. This would yield the following results:

As can be seen from this calculation, the line-to-ground fault current is almost 125% ofthe three-phase fault current. If the model were expanded to include five generators, thethree-phase fault current would be approximately 27.75 kA and the line-to-ground faultcurrent would be 34.5 kA. If the AIC rating of the equipment to be used were based onthe three phase currents, the equipment would be underrated. Of course, in a case with

fewer and/or smaller machines, the disparity between the phase fault currents and theground fault currents may not be a problem, that is, 125% of the phase fault currentsmay still be well below the minimum IEC rating of equipment in a given voltage rating.However, it is always important to check before the equipment is approved formanufacture and shipment.

CONCLUSIONSThis simple, two machine model demonstrates the higher magnitudes that can beanticipated for line-to-ground faults when a facility is fed from a generator source. Whenspecifying the IEC rating of equipment, it is important to consider the source of the faultand the nature of that source.

Although simple installations with small machines may not be affected, good designpractice would dictate that consideration be given to the calculation of line-to-groundfault currents. Failure to compare the equipment short circuit ratings to the calculatedline-to-ground fault currents can cause construction delays and expensive equipmentreplacement.

ANSI/IEEE CALCULATION METHODS:

In ANSI/IEEE short-circuit calculations, an equivalent voltage source at the fault location,which equals the pre-fault voltage at the location, replaces all external voltage sources

and machine internal voltage sources.

All machines are represented by their internal impedances. Line capacitances and staticloads are neglected. Transformer taps can be set at either the nominal position or at thetapped position, and different schemes are available to correct transformer impedanceand system voltages if off-nominal tap setting exists. It is assumed the fault is bolted,therefore, arc resistances are not considered. System impedances are assumed to bebalanced three-phase, and the method of symmetrical components is used forunbalanced fault calculations.

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Three different impedance networks are formed to calculate momentary, interrupting,and steady-state short-circuit currents, and corresponding duties for various protectivedevices. These networks are: ½ cycle network (sub-transient network), 1.5-4.0 cyclenetwork (transient network), and 30-cycle network (steady-state network).

ANSI/IEEE standards recommend the use of separate “R” and “X” networks to calculate

X/R values. An X/R ratio is obtained for each individual faulted bus and short-circuitcurrent. This X/R ratio is then used to determine the multiplying factor to account for thesystem DC OFFSET.

½ Cycle Fault:

It is used to calculate momentary short-circuit and protective device duties at the ½Cycle after the fault. The following table shows the type of device and its associatedduties using the ½ cycle fault.

Type of Device Duty

High Voltage Circuit breaker Closing & Latching capability

Low Voltage Circuit breaker Interrupting capability

Fuse Interrupting capability

Switch gear & MCC Bus bracing

Relay Instantaneous settings

½ cycle fault is also consider as a sub-transient period fault, primarily because allrotating machines are represented by their sub-transient reactances, as shown in table:

Type of Machine Xsc

Utility X”

Turbo generator Xd”

Hydro generator with amortisseur winding Xd”

Hydro generator without amortisseurwinding

0.75*Xd”

Condenser Xd”

Synchronous motor Xd”

Induction machine >1000hp @ 1800rpm orless

Xd” = 1/LRC (LRC = Locked rotor current)

Induction machine >250hp @ 3600rpm Xd” = 1/LRC

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Induction machine >50hp 1.2*Xd”

Induction machine 1000hp @ 1800rpm orless 1.5*Xd” = 1/LRC (LRC = Locked rotorcurrent)

Induction machine >250hp @ 3600rpm 1.5*Xd” = 1/LRC

Induction machine >50hp 3.0*Xd”

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30 Cycle Fault:

It is used to calculate the steady state short-circuit current and duties for some of theprotective devices 30 Cycle after the fault. The following table shows the type of device

and its associated duties using the 1.5 – 4.0 cycle fault.

Type of Device Duty

High Voltage Circuit breaker N/A

Low Voltage Circuit breaker N/A

Fuse N/A

Switch gear & MCC N/A

Relay IDMtL Over current relay settings

The type of rotating machine and its representation in the 30-cycle fault is shown infollowing table. Note that induction machines, synchronous motors, and condensers arenot considered in the 30-cycle fault calculation.

Type of Machine Xsc

Utility X”

Turbo generator Xd’

Hydro generator with amortisseur winding Xd’

Hydro generator without amortisseurwinding

Xd’

Condenser Infinite

Synchronous motor Infinite

Induction machine >1000hp @ 1800rpm orless

Infinite

Induction machine >250hp @ 3600rpm Infinite

Induction machine >50hp Infinite

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The ANSI (American National Standards & Institute) multiplying factor is determined bythe equivalent system X/R ratio at a particular fault location. The X/R ratio is calculatedby the separate “R” and “X” networks.

Momentary (1/2 Cycle) short-circuit current calculation (Buses & HV CB):

The momentary short-circuit current at the ½ cycle represents the highest or maximumvalue of the short-circuit current (before its AC & DC components decay toward thesteady state value). Although, in reality, the highest or maximum short-circuit currentactually occurs slightly before the ½ cycle, the ½ cycle network is used for thiscalculation.

The following procedure is used to calculate momentary short-circuit current:

Step:-1

Calculate the symmetrical rms value of momentary short-circuit current using thefollowing formula:

Imom,rms,symm = ( Vpre-fault ) / ((1.7325)*Zeq)

Where Zeq is the equivalent impedance at the faulted bus from the ½ cycle fault.

Step:-2

Calculate the Asymmetrical rms value of momentary short-circuit current using thefollowing formula:

Imom,rms,Asymm = (MFm)*( Imom,rms,symm )

Step:-3

Calculate the Multiplying or Asymmetrical factor using the following formula:

(MFm) = Under root( 1 + (2e)-2*3.14/(X/R)

)

Step:-4

Calculate the Peak value of momentary short-circuit current using the following formula:

Imom,Peak = (MFP)*( Imom,rms,Asymm )

Step:-5

Calculate the Peak Multiplying factor using the following formula:

(MFP) = 1.414*(Under root( 1 + (e)-3.14/(X/R)

))

In both equations for MFm and MFP calculation, X/R is the ratio of X to R at the faultlocation obtained from separate X and R networks at ½ cycle. The value of the faultcurrent calculated by this method can be used for following purpose:

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• Check closing & latching capabilities of high voltage circuit breakers.

• Check bus-bar capacity.

• Adjust relay instantaneous settings.

• Check interrupting capabilities of fuses and low voltage circuit breakers.

High voltage Circuit breaker Interrupting short-circuit current calculation:

The interrupting fault currents for high voltage circuit breakers correspond to the 1.5 –4.0 cycle short-circuit for high voltage circuit breakers:

The following procedure is used to calculate the interrupting short-circuit current:

Step:-1

Calculate the symmetrical rms value of the interrupting short-circuit current using the

following formula:

Iint,rms,symm = ( Vpre-fault ) / ((1.7325)*Zeq)

Where Zeq is the equivalent impedance at the faulted bus from the 1.5 – 4.0 cycle fault.

Step:-2

Calculate the short-circuit current contributions to the fault location from the surroundingbuses.

Step:-3

If contribution is from a Remote bus, the symmetrical value is corrected by the factor ofMFr, calculated from

MFr = Under root( 1 + (2e)-4*3.14*t/(X/R)

)

Where t is the circuit breaker contact parting time in cycles, as given in the followingtable:Circuit breaker rating in Cycles Contact Parting time in Cycles8 4.05 3.03 2.02 1.5

Low voltage Circuit breaker Interrupting short-circuit current calculation:

Due to the instantaneous action of low voltage circuit breakers at maximum short-circuitvalues, the ½ cycle network is used for calculating the interrupting short-circuit current.

The following procedure is used to calculate the interrupting short-circuit current:

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Step:-1

Calculate the symmetrical rms value of the interrupting short-circuit current using thefollowing formula:

Iint,rms,symm = ( Vpre-fault ) / ((1.7325)*Zeq)

Where Zeq is the equivalent impedance at the faulted bus from the ½ cycle fault.

Step:-2

Calculate the adjusted asymmetrical rms value of the interrupting short-circuit currentduty using the following formula:

Iint,rms,adj = MF * Iint,rms,symm

Step:-3(MF) = {1.414*(Under root( 1 + (e)

-3.14/(X/R)))} / {1.414*(Under root( 1 + (e)

-3.14/(X/R)TEST))} for

unfused power breakers.

(MF) = {1.414*(Under root( 1 + (2e)-2*3.14/(X/R)

))} / {1.414*(Under root( 1 + (2e)-2*3.14/(X/R)TEST

))} forfused power or molded case breakers.

Where (X/R)test is calculated based on the test power factor. The manufacturermaximum testing power factors given in the following table are used as the defaultvalues:Circuit breaker type Max Design %PF (X/R)testPower breaker (Unfused) 15 6.59Power breaker (fused) 20 4.90Molded case (>20000A) 20 4.90Molded case (10001-20000A) 30 3.18

Molded case (


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Sub-transient Voltage (E”) of a Synchronous machine:This is the rms value of the symmetrical internal voltage of a synchronous machinewhich is active behind the sub-transient reactance Xd” at the moment of short-circuit.

Sub-transient Reactance (Xd”) of a Synchronous machine:

This is the effective reactance at the moment of short-circuit. For the calculation of short-circuit currents, the saturated value of (Xd”) is taken.

According to IEC Standard 909, the synchronous motor impedance used in IEC short-circuit calculations is calculated in the same way as the synchronous generator.

Zk = KG(R+Xd”)

KG = (kVn * Cmax) / kVr * {1 + (Xd”)*(SIN(pHi))}

Where kVn and kVr are the nominal voltage of the terminal bus and the motor ratedvoltage respectively, Cmax is determine based on machine rated voltage, Xd” is

machine sub-transient reactance (Per unit in motor base), and “pHi” is the machine ratedpower factor angle.

Minimum time delay (Tmin) of a Circuit breaker:This is the shortest time between the beginning of the short-circuit current and the firstcontact separation of one pole of the switching device.

Voltage factor C:This is the factor used to adjust the value of the equivalent voltage source for minimumand maximum current calculation according to the following table:

Nominal Voltage Vn Cmax Cmin

Low voltage; 100V to1000V

1.05 1.00

Medium voltage:>1kV to35kV

1.05 1.00

230V/400V 1.00 0.95High voltage:>35kV to230kV

1.10 1.00

Initial symmetrical short-circuit current calculation:I”k = (C * Vn) / {(1.7325)*Zk}

Where Zk is the equivalent impedance at the fault location.

Peak short-circuit current calculation:IP = 1.414* (I”k)* (R/X)

DC component of short-circuit current calculation:

IDC = 1.414* (I”k)* (e(2*3.14*f*t)/(X/R) )

Asymmetrical Short-circuit current calculation for MV CB:

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IK,ASYMM. = (I”k)*{1 + (2e(-4*3.14*f*t)/(X/R) )}

DC component of short-circuit current calculation for MC CB:

IDC = 1.414* (I”k)* (e(2*3.14*f*t)/(X/R) )

Asymmetrical Short-circuit current calculation for LV CB:

IK,ASYMM. = (I”k)*{1 + (2e(-4*3.14*f*t)/(X/R) )}

Asymmetrical Short-circuit current calculation for FUSE:

IK,ASYMM. = (I”k)*{1 + (2e(-4*3.14*f*t)/(X/R) )}

The Importance of the X/R Ratio in Low-Voltage Short Circuit Studies:

Introduction:

In some short circuit studies, the X/R ratio is ignored when comparing the short circuitrating of the equipment to the available fault current at the equipment. What is notalways realized is that when low-voltage gear is tested, it is tested at a certain X/R ratio.The X/R ratio is important because it determines the peak asymmetrical fault current.The asymmetrical fault current can be much larger than the symmetrical fault current.The purpose of this article is to introduce such terms as the X/R ratio and asymmetricalfault current and to relate the importance of the X/R ratio to the rating of low-voltageequipment.

X/R Ratio and Asymmetrical Fault Current:

In AC electrical systems, impedance has two components. The first is called reactance(X). Reactance depends on two things: (1) the inductance and (2) the frequency.Inductance reflects how hard it is to change the current. All conductors have someinductance, but a more useful example of a component having inductance is a coil ofwire. Frequency is fixed at either 60 or 50Hz, depending upon where in the world theelectrical system is, so the reactance is solely dependent upon the inductance.

The second component of impedance is the familiar resistance (R). Resistance is ameasure of how hard it is for current to flow. When current flows through a materialhaving resistance, heat is transferred from the material to the surroundings.

The resistance and reactance of a circuit establishes a power factor. The power factor

(p.f.) is given by the following equation:

p.f. = cos(tan-1(X/R))

If the power factor is unity (1), then the impedance only has resistance. If the powerfactor is zero, then the impedance only has reactance.

The power factor also determines how much the voltage and current waveforms (sinewaves) are out of phase. Remember that both voltage and current are sine waves in

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linear AC electrical systems. For purely resistive systems, the voltage and current are inphase. For purely reactive systems, the voltage and current are 90-degress (one-quarterof a cycle) out of phase, with the voltage leading the current. Figure 2 below illustratesthis.

Effect of P.F. upon Voltage (------) and Current (------) waveform

The above equation means that the power factor and X/R ratio are related. Therefore,power factor and X/R ratio are different ways of saying the same thing. Please note thatas power factor decreases, the X/R ratio increases.

Right after a fault occurs, the current waveform is no longer a sine wave. Instead, it canbe represented by the sum of a sine wave and a decaying exponential. Figure 3 belowillustrates this phenomenon.

Sine-wave (-----) , Decaying Exponential (-----), and their Sum (-----)

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Please note that the decaying exponential added to the sine wave causes the current toreach a much larger value than that of the sine wave alone. The waveform that equalsthe sum of the sine wave and the decaying exponential is called the asymmetricalcurrent because the waveform does not have symmetry above and below the time axis.The sine wave alone is called the symmetrical current because it does have symmetryabove and below the time axis.

The actual waveform of the asymmetrical fault current is hard to predict because itdepends on what time in the voltage cycle waveform the fault occurs. However, thelargest asymmetrical fault current occurs when a fault happens at a point when thevoltage is zero. Then, the asymmetrical fault current depends only on the X/R ratio, orpower factor, and the magnitude of the symmetrical fault current.

Figure 4 below shows how the ratio of the peak asymmetrical current to the RMSsymmetrical current varies with the X/R ratio. (RMS symmetrical current equals the peaksymmetrical current divided by the square root of 2.) What Figure 4 shows is that thepeak asymmetrical current increases with the X/R ratio.

Figure 4. Peak asymmetrical current as a function of symmetrical RMS current. (Datataken from notes on the GE Electrical Distribution & Control Low-voltage Protector

Application Seminar.)

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Role of X/R Ratio when Comparing Short Circuit Ratings:

Low voltage devices have one rating, as opposed to medium-voltage gear, which haveboth a momentary and interrupting rating. This rating is reported in terms of symmetricalcurrent. Therefore, the rating must be compared to the calculated symmetrical current.

But the story does not end here. All low voltage protective devices are tested at an X/Rratio. The X/R ratio at which a device is tested depends upon the device type. Table 1below summarizes the device types and the X/R ratios at which they are tested.

Table 1. X/R ratios at which low voltage protective devices are tested.

Although low voltage devices do not have asymmetrical ratings, if we know thesymmetrical current rating and the test X/R ratio, Figure 4 gives us the maximumasymmetrical fault current. So, in a way, there is an asymmetrical fault current rating, butit is not explicit. Therefore, in any short circuit study, both the X/R ratio and thesymmetrical fault current must be taken into account.

Remember that, for a calculated value of RMS symmetrical current, as X/R ratioincrease, the maximum asymmetrical current (peak or RMS) also increases.

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If the calculated symmetrical fault current is larger than the device short circuit rating, thedevice in underrated, regardless of X/R ratio. However, it is possible for the device to beunderrated even if the short circuit rating exceeds the calculated symmetrical faultcurrent. How is this possible? We will discuss this next.

Consider some equipment whose calculated symmetrical fault current is less than the

short circuit rating of the equipment. Also, the calculated X/R ratio is less than or equalto the test X/R ratio. The maximum calculated asymmetrical fault current will be lessthan the maximum asymmetrical current that corresponds to the short circuit rating andthe test X/R ratio. The device will be properly rated.

Now consider another possibility. What if the symmetrical fault current is the same as theequipment’s rated current, but the actual X/R ratio is larger than the tested X/R ratio?Now, the maximum asymmetrical fault current will be larger than the maximumasymmetrical current corresponding to the short circuit rating and the test X/R ratio.

Although the available symmetrical fault current is equal to the rating, the asymmetricalfault current is higher than that when the device was tested. The device is not ratedproperly.

The above two paragraphs motivate a de-rating factor, or multiplying factor (MF), that isdefined by the following formula:

If the calculated X/R ratio at a device is larger than the test X/R ratio of the device, thenthe calculated symmetrical fault current must be multiplied by the multiplying factor. Or,equivalently, the short circuit rating must be divided by the multiplying factor. Themultiplying factor is equal to the ratio of the calculated asymmetrical fault current to the

asymmetrical fault current at the test X/R ratio and the rated symmetrical current.

Here is an example of the process. After running a fault analysis, the symmetrical faultcurrent at some low voltage switchgear is found to be 62kA during the first half-cycle.The switchgear contains power circuit breakers rated at 65kA. The asymmetrical peakfault current was found to be 149kA. The X/R ratio was calculated to be 11.1.

The test X/R ratio of low voltage power circuit breakers is 6.6. Although the symmetricalfault current is lower than the rating of the circuit breakers, the fact that the X/R ratio ishigher than the test value means that we must use the multiplying factor.

Therefore, the effective symmetrical fault current is 1.07 X 62kA = 66kA. Because 66kA> 65kA, the switchgear is underrated. We can also de-rate the switchgear. Then, theeffective rating of the gear is 65kA / 1.07 = 61kA. Now, because 62kA > 61kA, theswitchgear is under-rated.

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Summary of X/R Ratio:

When performing short circuit calculations, it is important to consider the X/R ratio. Thehigher the X/R ratio, the higher the asymmetrical peak fault current. Therefore, whenverifying the ratings of electrical equipment, both the symmetrical short circuit rating andthe X/R ratio must be taken into consideration.

If the calculated X/R ratio is larger than the test X/R ratio, then the equipment shortcircuit rating must be de-rated by a multiplying factor. This multiplying factor equals theratio of the calculated peak asymmetrical fault current divided by the peak asymmetricalcurrent corresponding to the rated symmetrical current and the test X/R ratio.

Procedures and Methods:

Short-Circuit Current Calculations & Relay-Coordination:

To determine the fault current at any point in the system, first draw a one-line diagram

showing all of the sources of short-circuit current feeding into the fault, as well as theimpedances of the circuit components.

To begin the study, the system components, including those of the utility system, arerepresented as impedances in the diagram.

It must be understood that short circuit calculations are performed without currentlimiting devices in the system. Calculations are done as though these devices areReplaced with copper bars, to determine the maximum “available” short circuit current.This is necessary to project how the system and the current limiting devices will perform.

Also, current limiting devices do not operate in series to produce a “compounding”

current limiting effect. The downstream, or load side, fuse will operate alone under ashort circuit condition if properly coordinated.

Three-phase Short-Circuit Calculation by Per-Unit Method:The Per-Unit method is generally used for calculating short-circuit currents when theelectrical system is more complex.

After establishing a one-line diagram of the system, proceed to the followingcalculations:

Step-1:

Infinite Source (Power Grid) Impedance:Source Impedance Zs = MVA Base / MVA Fault

Step-2:Transmission Line Impedance:

Step-3:Power/Distribution Transformer Impedance:

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X PU = (%X)* (KVA Base) / (100)* (KVA X’mer )

R PU = (%R)* (KVA Base) / (100)* (KVA X’mer )

-------------------- --------------------------------- --------------- ---

Z2 PU = √(X PU)2 + (R PU)2 OrZ2 PU = (%Z)* (KVA Base) / (100)* (KVA X’mer )

Step-4:Cable:

X PU = {(X in Ohms)* (KVA Base)} / {(1000)* (kV Base)2}

R PU = {(R in Ohms)* (KVA Base)} / {(1000)* (kV Base)2}

-------------------- --------------------------------- --------------- ---

Z3 PU = √(X PU)2 + (R PU)2

Step-5:Determine the total Single Circuit line Impedance of Power Grid:

Z’ Total PU = (Zs PU + Z1 PU + Z2 PU + Z3 PU)

Step-6:Determine the total Short Circuit MVA from Power Grid:

MVA SC1 = (MVA Base) / (Z’ Total PU)

Step-7:Determine Short Circuit Current Contribution from Power Grid:

I kA1 = (MVA SC1) / (1.7325* Rated KV)

Step-8:Determine the Short Circuit Contribution from Generator:

Z gen PU = {(%X’d)* (MVA Base)} / {(100)* (MVA Rated)}

Step-9:Determine the total Short Circuit MVA from Generator:

MVA SC2 = (MVA Base) / (Z gen PU)

Step-10:Determine Short Circuit Current Contribution from Generator:

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Step-11:Determine the Short Circuit Contribution from Motors or Dynamic loads:

Z loads PU = {(%Z)* (MVA Base)} / {(100)* (MVA Rated)}

Generally, %Z = 25% consider for LT motors or loads and %Z = 15-20% for HT motorsor loads.

Step-12:Determine the total Short Circuit MVA from Dynamic loads or Motors:

MVA SC3 = (MVA Base) / (Z loads PU)

Step-13:Determine Short Circuit Current Contribution from Dynamic loads or Motors:


Step-14:Determine Total Short Circuit Current at Fault:

I kA symm RMS = (I kA1 + I kA2 + I kA3)

Step-15:Determine X/R ratio of the system to the point of fault:

X/Rratio = (X PU total) / (R PU total)

Step-16:Determine the Asymmetrical RMS short-circuit current to the point of fault:

I kA asymm RMS = (I kA symm RMS)* (Asym Factor)

Step-17:Determine the Asymmetrical Factor:

Asym Factor or Momentary multiplying factor (MFm) = Under root( 1 + (

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