symmetrical fault analysis

39
F m n M a d T o F H b t p I l t c a t q A g f T W f Fault: A fau most faults, neutral (grou Most of the a heavy cu damage to th The choice system depe Types of F of circuit bre For proper flow under s High-voltag be large eno the ground i phase and a If flashover line-to-grou through the current path a short time Single line-t the ground quick power Approximat ground fault Sometimes, faults. Two phases When two li fault ult in a circu a current pa und). Since t fault on the urrent (called he equipmen of apparatus ends upon sh Faults eakers opera choice of ci short-circuit e transmissi ough to prev is large enou tower. occurs on a und faults. faulted line hs are also ca and then res to-ground fa or if insulat r removing. tely 75% of ts. all three ph of a line ma ines touch ea uit is any fai ath forms be the impedan power syste d short-circu nt and interru s and the de hort-circuit c ated by prote ircuit breake conditions m ion lines hav vent flashove ugh to ionize a single phas Since the s e into the g alled transie stored. aults can also tors break. T all faults in hases of a t ay touch, or ach other an ilure that int etween two o nce of a new em leads to a uit current) uption of ser sign and arr current consi ective relayin ers and prot must be estim ve strings of er – a condi e the air arou se of the line hort-circuit ground and b ent faults. Th o occur if on This fault is n power sys transmission flashover ma nd also touch S terferes with or more phas path is usua a short circu flows thro rvice to cons rangement o iderations. ng. tective relay mated – this f insulators s ition when t und insulator e, an arc wil path has a back into th hey usually ne phase of t s called a pe stems are ei n line are sh ay occur bet h the ground Symmet h the normal ses, or betwe ally low, an e it condition. ough the eq sumers. of practically The abnorm symme (symm Symm three-p involv Symm caused throug equipm initiate throug The sy flow o (which to maj the fau ying, the ma is the scope supporting e the voltage d rs and thus p l be produce low imped he power sy clear if pow the line brea ermanent fa ither transien horted togeth tween two ph d, the fault is trical Fa l flow of cur een one or m excessive cu . When such quipment, ca y every equip power mally unde etrical metrical three metrical faul phase shor ve arc impeda metrical fau d in the gh insulat ments or f ed by a li gh accidenta ystem must b of heavy sh h can cause or equipmen ulty part of th agnitude of c e of fault ana each phase. T difference be provide a cur ed. Such fau dance, very ystem. Faults wer is remov aks and com ault since it nt or perma her – symm hases – a lin called a dou ault Ana rrent to the more phases urrent may fl h a condition ausing consi pment in the system b er conditio short e-phase fault t may be rt-circuit o ance. ult conditio system ac tion failu flashover o ightning str l faulty oper be protected hort-circuit c permanent nt) by discon he system by currents tha alysis. The insulato etween the l rrent path be ults are called high curren s involving ed from the mes into conta will remain anent single metrical thre ne-to-line fau uble line-to- alysis load. In and the ow. n occurs, iderable e power behaves ons of circuit t). a solid or may ons are ccidently re of of lines roke or ration. d against currents damage nnecting y means at would ors must line and etween a d single nts flow ionized line for act with n after a line-to- ee-phase ult. -ground

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Fault analysis, power systems,

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    Fault: A faumost faults, neutral (grou

    Most of the a heavy cudamage to thThe choice system depe

    Types of F

    of circuit breFor proper flow under sHigh-voltagbe large enothe ground iphase and a If flashover line-to-grouthrough the current patha short time Single line-tthe ground quick powerApproximatground faultSometimes, faults. Two phases When two lifault

    ult in a circua current pa

    und). Since t

    fault on the urrent (calledhe equipmenof apparatus

    ends upon sh

    Faults

    eakers operachoice of cishort-circuit e transmissi

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    hs are also caand then res

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    power systed short-circunt and interrus and the dehort-circuit c

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    a single phasSince the se into the galled transiestored. aults can alsotors break. T

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    ground and bent faults. Th

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    Symmeth the normalses, or betweally low, an e

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    anent single

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  • LtprHaarb

    e PHaarb

    Lm

    Lighting strothat greatly phase and thremains eveHigh currenaffected tranallow ionizereclosing thebreaker shou

    (A) Phase-to(C) Phase-to(E) Three ph(G) Pilot-ear*In undergr

    Symmetricaequal fault c

    ProtectioHigh currenaffected tranallow ionizereclosing thebreaker shouSelecting an

    TransienLet us consimade at this

    1. The 2. Shor3. Line

    okes cause mexceeds the

    he ground on after the li

    nts due to a nsmission lined air to dee breaker. Thuld open aga

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    al Fault: Thcurrent in the

    on againsnts due to a nsmission lined air to dee breaker. Thuld open agan appropriate

    nt on a Tider the shor stage. line is fed frrt-circuit take capacitance

    most faults oe rated voltaof the line cighting disapfault must bne should au

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    arth fault h fault

    g application

    he fault on the line with 1

    st fault fault must bne should au

    eionize. If thherefore, maain isolating e circuit brea

    ransmissrt circuit tra

    rom a constaes place whee is negligibl

    on high-voltage of the lireating an a

    ppears. be detected utomaticallyhe fault wasany transientthe transmis

    (B) (D) (F)

    n only

    he power sy20o displace

    be detected utomaticallyhe fault wasany transientthe transmis

    aker (type, si

    sion Lineansient on a

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    arc. Once th

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  • DMAM (3/22)

    4. The impedance is zero in short-circuit. 5. The short circuit is assumed to take place at t=0. The parameter controls the instant on the

    voltage wave when short circuit is occurs. 6. The instantaneous voltage equation for this circuit is:

    7. )sin(2)()( +==+ tVv

    dttdiLtRi

    It is known from circuit theory that the current after short-circuit is composed of two parts, i.e.

    )()()()()( tdcitacittitsiti +=+= [ ]currentcircuitshortlSymmetrica)sin(2)()( +== t

    ZVtacitsi

    The rms value of symmetrical or steady-state fault current is: ZV

    acI =

    )sin(2)()( +== tacItacitsi

    [ ]currentset-off DCLteZVtdcitti

    /)sin(2)()( == LteacItdcitti

    /)sin(2)()( == is (or iac) is called steady state or ac current, it (or idc is called transient or dc current [it is such that i(0) = is(0) + it(0) =0 being an inductive circuit, it decays corresponding to the time constant L/R.]

    222)(2 XRLRZ +=+= ; LX = ;

    =RL 1tan ; s

    2 fRX

    RX

    RL

    L === The sinusoidal steady state current is called the symmetrical short-circuit (steady-state) fault current and the unidirectional transient current is called the dc off-set current. Since the maximum value of dc off-set current depends on , this current causes the total short-circuit current to be unsymmetrical till the transient decays. dc offset current + Symmetrical fault current = Asymmetrical Fault Current (Total)

    LteZVt

    ZVtdcitacittitsiti

    /)sin(2)sin(2)()()()()( +=+=+= [ ]LtetacItdcitacittitsiti /)sin()sin(2)()()()()( +=+=+=

    The total current is called the asymmetrical current. At the instant of applying the voltage, the dc and steady-state components always has the same magnitude but are the opposite sign in order to express the zero value of current then existing. If the value of the steady-state term is not zero at t = 0, the dc component appears in the solution in order to satisfy the physical conditions of zero current at the instant of closing the switch.

  • DMAM (4/22)

    At t = 0; )sin(2)sin(2)()( === acIZVtdcitti

    when - = 0 or - = , 0)0()0( ==== tdcitti The dc term does not exist, if the circuit is closed at point on the voltage wave such that - = 0 or - = .

    when - = /2, acIZVtdcitti 2

    2)0()0( mm ===== The dc component has its maximum initial value, which is equal to maximum value of the sinusoidal component, if the circuit is closed at point on the voltage wave such that - = /2.

    Fig.9.2 Waveform of a short-circuit current on a transient line [5].

  • DMAM (5/22)

    Problem: The source voltage v = 151sin(377t+) V is applied in the following circuit where R = 0.125 , L = 10 mH. Find the current response after closing the switch S for the following cases: (i) no dc-offset, and (ii) maximum dc-offset. Sketch the current waveform up to 3 cycles corresponding to case (i) and (ii). Solution: Here, = 377, XL = L = 3770.01 = 3.77

    =+=+= 1.88772.377.3125.0]01.0377[125.0 jjZ

    2151=V

    ;A

    ZV

    acI 240

    772.31

    2151 ===

    sRL

    L 08.0125.001.0 ===

    The current response is then given by [ ]08.0/)1.88sin()1.88377sin(40)( tetti += (i)a No dc offset, if switch is closed when = = 88.1o.

    (ii) Maximum dc offset, if switch is closed when = - 90o = -1.9o.

    The first peak is called the maximum momentary short-circuit current, Imm.

    [ ])sin(12)sin(22 == acIZV

    ZV

    mmI

    Since the transmission line resistance is small, 90o. [ ] cos12cos22 +=+= acIZ

    VZV

    mmI

    This is the maximum possible value for = 0, i.e. short-circuit occurring when the voltage wave is going through zero. Thus

    acIZV

    mmI 2222)possiibe(max == = twice the maximum of symmetrical short-circuit current

    (doubling effect). For the selection of circuit breakers, momentary circuit is taken corresponding to its maximum possible value (a safe choice). when - =- /2,

    +=

    = LtLt etacIetacIti // )

    2sin(2)

    2sin()

    2sin(2)(

    LteacILteacIacItdcIacIrmsI

    /2212]

    /2[2][2)]([2][

    +=+=+= ft /= ; is time in cycle

  • DMAM (6/22)

    )/(4

    )2/(22

    RXfRXfL

    t

    ==

    )/(4

    21 RXeacIrmsI

    +=

    )/(4

    21)( RXeK

    += ; 3)0( ==K

    acIKrmsI )(= Irms decreases from 3Iac when =0 to Iac when is very large. Example 7.1 [Ref. 3, p. 360] A bolted circuit occurs in the series RL circuit with V = 20 kV, X = 8 , R = 0.8 , and with maximum dc offset. The circuit breaker opens 3 cycles after fault inception. Determine (i) the rms ac fault current, (ii) the rms momentary current at =0.5 cycle, which pass through the breaker before it opens, and (iii) the rms asymmetrical fault current that the breaker interrups. Solution:

    (i) kA488.28.08

    2022

    =+

    ==ZV

    acI

    (ii) X/R= 8/0.8=10, =0.5 cycle

    438.1105.04

    21)/(4

    21)( =

    +=

    +=

    eRXeK A576.3488.2438.1)( ==== acIKmomentaryIrmsI

    (iii) =0.5 cycle

    023.11034

    21)/(4

    21)( =

    +=

    +=

    eRXeK A544.2488.2023.1)( === acIKrmsI

    Short Circuit of a Synchronous Machine (on No Load) Armature resistance being small can be neglected. Under steady-state short-circuit conditions; the armature reaction of a synchronous machine produces a demagnetizing flux which is modeled as a reactance Xa (known as fictitious reactance) in series with the induced emf. The combination of fictitious (Xa) and leakage (Xl) reactances is called synchronous reactance, Xd (direct axis synchronous reactance in the case of salient pole machine). The steady-state short-circuit model of a synchronous machine is shown in Fig. 9.3(a) on per phase basis. Immediately upon short-circuit, the DC off-set currents appear in all three-phases, each with a different magnitudes since the point on the voltage wave at which short-circuit occurs is different for each phase. These DC off-set current are accounted for separately on an empirical bass and, therefore, for short circuit studies, we need to concentrate our attention on symmetrical (sinusoidal) short-circuit current only. Immediately in the event of short circuit, the symmetrical short-circuit current is limited by leakage reactance. Since the air-gap flux cannot be changed instantaneously to counter the demagnetization of the armature short-circuit current, currents appear in the field winding as well as damper winding in a

  • DMAM (7/22)

    direction to help the mail flux. The time constant of damper winding (which has low leakage inductance) is much less than that of the field winding (which has high leakage inductance). Thus the initial part of the short-circuit, the damper winding and field windings have transformer currents induced in them so that in the circuit model their reactances Xf of field winding and Xdw of damper winding appear in parallel with Xa as shown in Fig. 9.3(b). As the damper winding currents are first to die out, Xdw effectively becomes open-circuited and at a later stage Xf becomes open-circuited. The machine reactances thus changes from the parallel combination of Xf, Xa, and Xdw during the initial period of the short-circuit to Xf, and Xa in parallel [Fig. 9.3(b)] in the middle period of the short-circuit, and finally to Xa in steady-state [Fig. 9.3(a)]. The reactance, which is called subtransient reactance, presented by the machine in the initial period of the short-circuit, i.e.

    )]/1()/1()/1[(1////''

    dwXfXaXlXdwXfXaXlXdX +++=+=

    The reactance, which is called transient reactance, after the damper winding currents have died out is:

    )]/1()/1[(1//'

    fXaXlXfXaXlXdX ++=+=

    The reactance, which is called synchronous reactance, after the field winding currents have died out is:

    aXlXdX += Obviously, dXdXdX

  • DMAM (8/22)

    (c) Approximate circuit model during transient period of a short-circuit

    Fig. 9.3 [Ref. 5] The currents and reactances discussed above are defined by the following equations, which apply to an alternator operating at no-load before the occurrence of a three-phase (or symmetrical) fault at its terminals:

    The rms value of steady-state current: dXgEoaI ==

    2

    The rms value of transient current excluding dc off-set component: '2'

    dX

    gEobI ==

    The rms value of subtransiente current excluding dc off-set componnet: ''2''

    dX

    gEocI ==

    Xd = direct-axis synchronous reactance Xd = direct-axis transient reactance Xd = direct-axis subtransient reactance |Eg| = rms voltage from one terminal to neutral at no load oa, ob, and oc = intercepts shown in Fig. 9.4.

    Obviously, III >> ''' There are similar quadrature axis reactances. Xq = quadrature-axis synchronous reactance Xq = quadrature -axis transient reactance Xq = quadrature -axis subtransient reactance If the armature resistance is small, the quadrature axis reactances do not significantly affect the short-circuit current. Using the above direct-axis reactances, the instantaneous ac current can be written as:

    )2

    sin(1)'/(1

    '1)

    ''/('

    1''

    12)( +

    +

    +

    = t

    dXdTte

    dXdXdTte

    dXdXgEtiac

  • DMAM (9/22)

    (a) Symmetrical short-circuit armature current in synchronous machine

    (b) Envelope of synchronous machine symmetrical short circuit current.

    Fig. 9.4 [Ref. 1 and 5] At t=0, when the fault occurs, the rms value of i(t) i.e. subtransient fault current is:

    '')0(''

    dX

    gEII ac ==

    The duration of I is determined by time constant Td, called the direct-axis short-circuit subtransient time constant. When t> Td but t < Td (direct-axis short-circuit transient time constant), the first exponential term of above equation has decayed almost zero, but the second exponential has not decayed significantly. The rms fault current then equal the rams transient fault current, given by

    ''

    dX

    gEI =

    When t is much larger than Td, the rms fault current approaches its steady state value, given by

    dac X

    gEIII == )(

  • DMAM (10/22)

    Since the three-phase no-load voltages are displaced 120o from each other, the three-phase ac fault currents are displaced 120o fro each other. In addition to that ac fault current, each phase has a different dc offset. The maximum dc offset in any oe phase, which occurs when =0 is

    )/(2)/(''

    2)( ''max AAdc

    TteITte

    dX

    gEti ==

    Where, TA is called the armature time constant. Example 7.2 [Ref. 3, p. 363] A 500 MVA, 20 kV, 60 Hz synchronous generator with Xd=0.15, Xd=0.24, Xd=1.1 pu and time constant Td=0.035, Td= 2.0, TA=0.2 s is connected to the circuit breaker. The generator is operating at 5% above rated voltage and at no-load when a bolted three-phase short-circuit occurs on the load side of the breaker. The breaker interrupts the fault 3 cycles after fault inception. Determine (i) the subtransient fault current in pu and kA rms, (ii) the maximum dc-offset as a function of time; and (iii) rms asymmetrical fault current, which the breaker interrupts, assuming maximum dc-offset. Solution: (i) The generator is operating at 5% above rated voltage and at no-load when a bolted three-phase short-circuit occurs, thus Eg = 1.05 pu. The subtransient fault current that occurs in each of the three phases is

    pu0.715.005.1

    '''' ===

    dX

    gEI

    The generator base current is: Ibase= 500/(3)(20) = 14.43 kA The rms subtransient fault current in kA is: kA0.10143.140.7'' ==I (ii) The maximum dc offset that may occur in any phase is:

    kA)2.0/(9.142)2.0/(1012)/(2)( ''maxteteTteIti Adc

    === (iii) The rms ac fault current at t= 3 cycle =0.05 s is

    pu920.41.1

    1)0.2/05.0(1.1

    124.01)035.0/05.0(

    24.01

    15.0105.1)s05.0( =

    +

    +

    = eeI ac

    kA01.7143.14920.4)s05.0( ==acI 2])/(2[2][2)]([2][)( '' ATteIacItdcIacItrmsI

    +=+=

    kA1322])2.0/05.0(101[22]01.71[)s05.0( =+= ermsI Example 10.1 [Ref. 1, p. 253] Two generators are connected in parallel to the low-voltage side of three-phase -Y transformer as shown in Fig. 10.5. Generator 1 is rated 50,000 kVA, 13.8 kV. Generator 2 is rated 25,000 kVA, 13.8 kV. Each generator has a subtransient reactance of 25%. The transformer is rated 75,000 kVA, 13.8/69Y kV, with a reactance of 10%. Before the fault occurs, the voltage on the high-tension side of the transformer is 66 kV. The transformer is unloaded, and there is no circulating current between generators. Find the subtransient current in each generator when the three-phase short-circuit occurs on the high-tension side of the transformer.

  • DMAM (11/22)

    Fig. 10.5 One-line diagram of Example 10.1.

    Solution: Select as base in the high-tension circuit 69 kV, 75 MVA. Then the base voltage on the low-tension side is 13.8 kV.

    =

    oldbase,

    newbase,

    newbase,

    oldbase,newpu, oldpu, S

    S

    V

    VZZ

    If newbase,oldbase, VV = then

    =

    oldbase,

    newbase,newpu, oldpu, S

    SZZ

    If newbase,oldbase, SS = then

    =

    newbase,

    oldbase,newpu, oldpu, V

    VZZ

    Generator 1:

    375.0507525.0'' 1 =

    =dX pu

    957.06966

    1 ==gE pu

    Base current, 8.133

    75000

    base33base

    base1 == LLVS

    I Generator 2:

    750.0257525.0'' 2 =

    =dX pu

    957.06966

    2 ==gE pu

    8.13375000

    base33base

    base2 == LLVS

    I Transformer 1:

    10.0=X pu

  • DMAM (12/22)

    Figure 10.6 shows the reactance diagram before the fault. A three-phase fault at P is simulated by closing switch S. The internal voltages of the two machines may be considered to be in parallel since they must be identical in magnitude and phase if no circulating current flows between them. The equivalent parallel subtransient reactance is:

    ''2

    ''1

    ''2

    ''1''

    2//''1

    dXdXdXdX

    dXdXgpX +== pu

    Fig. 10.6 Reactance diagram for Example 10.1.

    25.0750.0375.0750.0375.0 =+

    ==gpX pu Therefore, as a phasor with Eg=0.957 as reference, the subtransient current in the short-circuit is:

    735.210.025.0

    957.0'' jjjjXgpjX

    gEI =+=+= pu

    The voltage on the delta side is 2735.0)10.0)(735.2('' === jjXIV pu

    The subtransient current in generator 1 is:

    823.1375.0

    275.0957.0''1

    1''1 jj

    djX

    VgEI ==

    = pu

    A57208.133

    75000823.1base1'',1

    ''1 === IpuII

    The subtransient current in generator 2 is:

    912.0750.0

    275.0957.0''2

    2''2 jj

    djX

    VgEI ==

    = pu

    A28608.133

    75000912.0base2'',2

    ''2 === IpuII

    Example 9.1 [Ref. 5] For the radial network shown in Fig. 9.6, a three-phase fault occurs at F. Determine the fault current and the line voltage at 11 kV bus under fault conditions.

  • DMAM (13/22)

    Fig. 9.6

    Fig. 9.7

    Example 9.2 [Ref. 5] A 25 MVA, 11 kV generator with Xd = 20% is connected through a transformer, line and a transformer to a bus that supplies three identical motors as shown in Fig. 9.8. Each motor has Xd =25% and Xd = 30% on a base of 5 MVA, 6.6 kV. The three-phase rating of the step-up transformer is 25 MVA, 11/66 kV with a leakage reactance of 10% and that of the step-down transformer is 25 MVA, 6/6. kV with a leakage reactance of 10%. The bus voltage at motors is 6. kV when a three-phase fault occurs at the point F. For specified fault, calculate:

    (i) the subtransient current in the fault, (ii) the subtransient current in the breaker B, (iii) the momentary current in braker B, and (iv) the current to be interrupted by breaker B in five cycles.

    Given, Reactance of the transmission line = 15% o a base of 25 MVA, 66 kV. Assume that the system is operating on no load when the fault occurs.

    Fig. 9.8

  • DMAM (14/22)

    (a)

    (b)

    (c)

    (d)

    Fig. 9.9 Solution: Choose a system base of 25 MVA. For generator base voltage: 11 kV For line base voltage: 66 kV For Motor base voltage: 6.6 Kv (i) For each motor

    pu25.152525.0'' jjdmX ==

    Line, transformer and generator reactance are already given on proper base reactances. The circuit model of the system for fault calculations is given in Fig. 9.9(a). The system being initially on no-load, the generator and motor induced emfs are identical. The circuit can therefore be reduced to that of Fig. 9.9(b) and the to Fig. 9.9(c). Now

    pu22.455.1

    125.1

    13 jjjSC

    I ==

    Base current in 6. kV circuit = A21876.63

    100025 ==

  • DMAM (15/22)

    A9229218722.4 ==SCI (ii) From Fig. 9.9(c), current through circuit breaker B is

    pu42.355.1

    125.1

    12)( jjjBSCI ==

    A5.7479218742.3)( ==BSCI (iii) For finding momentary current through the breaker, we must add the dc off-set current to the symmetrical subtransient current obtained in part (i). Rather than calculating the dc off-set current, allowance is made for it on an empirical basis. Momentary current through breaker B = 1.67479.5 = 11967 A (iv) To compute the current to interrupt by the breaker, motor subtransient reactance (Xd = j0.25) is now replaced by transient reactance (Xd = j0.30).

    pu5.1

    5253.0' jjdmX ==

    The reactances of the circuit of Fig. 9.9(c) now modify to that of Fig. 9.9(d). Current (symmetrical) to

    be interrupted by the breaker (as shown by arrow) pu1515.355.1

    15.1

    12 jjj

    == Allowance made of the dc off-set value by multiplying with a factor of 1.1. Therefore, the current to be interrupted is

    A758121871515.31.1)int( ==BI Short Circuit of Loaded Synchronous Machine (Ref. 5, p. 313) [Internal Voltages of Loaded Machines Under Transient Conditions (Ref. 1, p. 254)]

    (a) Usually steady-state equivalent circuit (Xs=Xd) of synchronous generator with

    load.

    (b) Equivalent circuit of synchronous generator for calculation of I.

    Zext: external impedance between generator terminal and fault point (F or P) where the fault occurs IL: current flowing before fault occurs Vf: voltage at the fault point Vt: terminal voltage of the generator Eg: subtransient internal voltage or voltage behind the subtransient reactance Eg: transient internal voltage or voltage behind the transient reactance

    (c) Equivalent circuit of synchronous generator for calculation of I.

    Fig. 10.7 [Ref. 1] Equivalent circuit for a synchronous generator supplying balanced three-phase load.

  • DMAM (16/22)

    Consider a generator that is loaded when the fault occurs. Fig. 10.7(a) shows the equivalent circuit of a synchronous generator that has a balance three-phase load. The equivalent circuit of the synchronous generator is its no-load voltage Eg in series with its synchronous impedance Xs or Xd. If a three-phase fault occurs at point F or P in the system, a short circuit from P to neutral in the equivalent circuit does not satisfy the conditions for calculating subtransient current since the reactance of the generator must be Xd if subtransient current I is calculated or Xd if transient current I is calculated. As shown in Fig. 10.7 the current relation of voltage can be written as:

    dXLjItVgE += Here IL is called prefault current. ''''

    dXLjItVgE += Here IL is called postfault subtransient current. ''dXLjItVgE += Here IL is called postfault transient current.

    Similarly, the subtarnsient internal voltage and transient internal voltage for a shynchronous motor are given by:

    ''''dXLjItVmE = ''dXLjItVmE =

    Example 10.2 [Ref.1, p.256] A synchronous generator and motor are rated 30 MVA, 13.2 kV, and both have subtransient reactances 20%. The line connecting them has a reactance of 10% on the basis of the machine ratings. The motor is drawing 20 MW at 0.8 power factor leading and a terminal voltage of 12.8 kV when a symmetrical three-phase fault occurs at the motor terminals. Find the subtransient current in the generator, motor, and fault by using the internal voltage of the machine.

    Fig. One-line diagram for Example 10.2.

    Solution: Chose a base 30 MVA, 13.2 kV. Fig. 10.8 shows the equivalent circuit of the system as described.

    Neutral bus

    Neutral bus

    (a) Before the fault (b) After the fault Fig. 10.8 Equivalent circuit for Example 10.2.

    Base current, A13123102.133

    61030 =

    =baseI

  • DMAM (17/22)

    Given, the voltage at the fault is 12.8 kV which we choose a reference.

    pu097.02.138.12 ==fV

    A9.3611288.03108.123

    61020 =

    =LI

    pu52.069.09.3686.01312

    9.361128 jLI +===

    For the generator: Vt =0.97+j0.1(0.69+j0.52)=0.918+j0.069 pu Eg=0.918+j0.069+j0.2(0.69+j0.52)=0.814+j0.207 pu Ig=(0.814+j0.207)/(j0.3)=0.69-j2.71 pu = 1312(0.69-j2.71) A = (905 j3550) A For the motor: Vt = Vt =0.970o. Em=0.918+j0.0-j0.2(0.69+j0.52)=1.074-j0.138 pu Im=(1.074-j0.138)/(j0.2)=-0.69-j5.37 pu = 1312(-0.69-j5.37) A = (-905 j7050) A In the fault: If= Ig+ Im=(0.69-j2.71)+( -0.69-j5.37)= -j8.08 pu = 1312(-j8.08) A = -j10,600 A. Example 9.3 [Ref.5, p.314] A synchronous generator and a synchronous motor each rated 25 MVA, 11 KV having 15% subtransient reactance are connected through transformers and a line as shown in Fig. 9.12(a). The transformers are rated 25 MVA, 11/66 kV with leakage reactance of 10% each. The line has a reactance of 10% o a base of 25 MVA, 66 kV. The motor is drawing 15 MW at 0.8 power factor leading and a terminal voltage of 10.6 kV when a symmetrical three-phase fault occurs at the motor terminals. Find the subtransient current in the generator, motor and fault. [Hint. same as Example 10.2 [Ref.1, p.256]]

    (a) One-line diagram for the system of Example 9.3.

    (b) Prefault equivalent circuit (c) Equivalent circuit during fault

    Fig. 9.12

  • DMAM (18/22)

    Fault Current Calculation Using Thevenins Theorem [(Ref. 1, p. 258)] The Thevenin impedance at fault point P of Fig. 10.7(b) is given as:

    ''

    )''(//)''(

    djXextZLZdjXextZLZ

    extZdjXextZthZ +++=+=

    The Thevenin voltage equal to Vf, the voltage at the fault point before the fault occurs. Upon the occurance of a three phase short circuit at P, the subtransient current in the fault is

    )''(

    )''(''

    djXextZLZ

    djXextZLZfV

    ZfVIth +

    ++==

    Fault Current Calculation Using Superposition Theorem [(Ref. 1, p. 258)] The fault is divided between the parallel circuits of the generators inversely as their impedances, the resulting values are the currents from each machine due to nly the change in voltage at the fault point. The fault currents thus attributed to the two machines must be added the current flowing in each before the fault occurred to find the total current in the machines after the fault. The supper position theorem supplies the reason for adding the current flowing before the fault to the current computed by Theveins theorem. Fig. 10.10(a) shows the a generator having a voltage Vf conncetd at the fault and equal to the voltage at the fault before the fault occurs. This generator has no effect on the current flowing before the fault occurs, and the circuit corresponding to that of Fig. 10.8 (a). Adding in series with Vf another generator having the circuit of Fig. 10.10(b), which corresponds to that of Fig. 10.8(b). The principle of superposition, applied by first shorting Eg, Em, and Vf, gives the currents found by distributing the fault current between the two generators inversely as the impedances of their circuits. Then shorting the remaining generator Vf with Eg, Em, and Vf, in the circuit gives the current flowing before the fault. Adding the two values of current in each branch gives the current in the branch after the fault. If Vf equals the prefault voltage at the fault, then the current calculation with Vf can be consider zero i.e. If2=0 and Vf, has no effect, since it represents the system before fault occurs.

    (a) Before the fault (b) During the fault

    Fig. 10.10 Circuits illustrating the application of the superposition theorm to determine the proportion of the fault current in each branch of the system.

  • DMAM (19/22)

    Example 10.3 [Ref.1, p.259] Solve Example 10.2 by the using of Thevenins theorem. Solution:

    12.02.03.02.03.0//)''( j

    jjjj

    LZdjXextZthZ =+=+=

    pu097.02.138.12 ==fV

    08.812.0

    097.0'' jjZ

    fVIth

    f === = 1312(-j8.08) A = -j10,600 A. Example 10.3.1 [Ref.1, p.260] Apply the above principle in Example 10.3 by the using of superposition theorem. Solution: Using the current division Subtransient generator current neglecting prefault current:

    pu23.308.83.02.0

    2.0''3.02.0

    2.0''1 jjfIgI =+=+=

    Subtransient motor current neglecting prefault current:

    pu85.408.83.02.0

    3.0''3.02.0

    3.0''1 jjfImI =+=+=

    To these current must be added the prefault current IL to obtain the total subtransient currents in the machines: Subtransient generator current including prefault current:

    pu71.269.052.069.023.3'' 1'' jjjIgIgI L =++=+=

    Subtransient motor current including prefault current:

    pu37.569.052.069.085.4'' 1'' jjjImImI L ===

    Example 7.3 [Ref.3, p.366] The synchronous generator in the following figure is operated at rated MVA, 0.95 power factor lagging and at 5% above rated voltage when a bolted three-phase short circuit occurs at bus 1. Calculate the per-unit values of (i) subtransient current; (ii) Subtransient generator and motor current, neglecting prefault current, and (iii) subtransient generator and motor currents including prefault current.

    Fig. 7.3 Single-line diagram of a synchronous generator feeding a synchronous motor.

    Solution: (i) Using a 100 MVA base, the base impedance in the zone of the transmission line is

    == 44.190100

    )138(,

    2

    linebaseZ

    And pu1050.044.190

    20 ==lineX The per-unit reactances are shown in Fig.7.4. From the first circuit in Fig. 7.4(d), the Thevenin

  • DMAM (20/22)

    impedance as viewed from the fault is

    pu11565.0505.015.0505.015.0 jjTHXTHZ =+

    == And the prefault voltage at the generator terminal is

    pu01 =fV The subtransient fault current is then

    pu079.911565.0

    01'' jjTHZ

    fV

    fI ===

    (a) Three-phase short-circuit (b) Short-circuit represented by two opposing voltage sources

    (c) Application of superposition

    (d) Vf set equal to prefault voltage at fault.

    Fig. 7.4 Application of superposition to a power system three-phase circuit. (ii) Using the current division in the first circuit of Fig. 7.4(d)

    pu000.7)079.9)(7710.0(''15.0505.0

    505.0''1 jjfIgI ==+=

    pu079.2)079.9)(2290.0(''15.0505.0

    15.0''1 jjfImI ==+=

    If Vf equals the prefault voltage at the fault, then the current calculation with Vf can be consider zero i.e. If2=0 and Vf, has no effect, since it represents the system before fault occurs. (iii) The generator base current is

  • DMAM (21/22)

    kA1837.48.133

    100, ==genbaseI

    And the prefault generator current is

    kA18199845.395.0cos)05.18.13(3

    100 1 ==

    LI

    pu2974.09048.095.0cos18199524.01837.4

    18199845.3 1 jLI ===

    The subtarnsient generator and motor currents, including prefault current and then,

    pu9.82353.7297.79048.02974.09048.0000.7'' 1'' ==+=+= jjjIgIgI L

    pu1.243999.1782.19048.02974.09048.0000.7'' 1'' ==+== jjjImImI L

    The Bus Impedance Matrix in Fault Calculations [Ref. 1, p. 261]

    (a) One line diagram of a power system. (b) Reactance diagram at steady-state condition. Fig. 7.3

    Let us proceed to the general equations by starting with a specific network with which we are already familiar. If we change the reactance in series with the generated voltages of the circuit shown in Fig. 7.3 become subtransient internal voltages, we have the network shown in Fig. 10.11. A three-phase fault at bus 4 is simulated by the network of Fig. 10.12 where the impedance values of Fig. 10.11 have been changed to admittances. The generated voltages Vf and Vf in series constitute the short circuit. Generated voltage Vf alone in this branch would cause no current in the branch. With Vf and Vf in series the branch is a short circuit and the branch current If. If Ea, Eb, Ec, and Vf are short-circuited, the voltages and current are those only to Vf. Then the only current entering a node from a source is that from Vf and is - If into node 4 (If from node 4) since there is no current in the branch until the insertion of -Vf. The node equations in matrix form for the network with Vf the only source ar

    =

    =

    f

    bus

    ff VVVV

    VVVV

    j

    I3

    2

    1

    3

    2

    1

    '' 0.180.80.50.50.883.175.20.40.55.283.100.00.50.40.033.12

    000

    Y

    Where subscript indicates that the change in voltage due to the fault and the due only to Vf.

  • DMAM (22/22)

    Fig. 10.11 Reactance diagram obtained from Fig. 7.3 by subtatituteing subtransient for synchronous reactances of the machines and subtarnsient internal voltages for no-load generated voltages. Reactance values are marked as per unit.

    Fig. 10.12 Circuit of Fig. 10.11 with admittances marhed in per unit and a three phase fault on bus 4 of the system simulated by Vf and Vf in seies.

    The bus voltages due to Vf are given by

    =

    =

    =

    ''

    1

    ''

    1

    3

    2

    1

    000

    000

    000

    0.180.80.50.50.883.175.20.40.55.283.100.00.50.40.033.12

    f

    bus

    f

    bus

    ff III

    j

    VVVV

    ZY

    =

    =

    ''44434241

    34333231

    24232221

    14131211

    ''3

    2

    1

    000

    000

    ff

    bus

    f IZZZZZZZZZZZZZZZZ

    IVVVV

    Z

    If total number of bus is N, and fault is occurred in bus k, then considering fk VV = ; ffk III == ''''

    =

    +

    +

    +

    ++

    ++

    +

    +

    +

    +

    ++

    ++

    +

    +

    +

    +

    +

    +

    +

    +

    +

    +

    +

    0....00

    ....00

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....''

    )2(

    )1(

    2

    1

    1)2(

    )2)(2(

    )2)(1(

    )2(

    )2(2

    )2(1

    )1(

    )1)(2(

    )1)(1(

    )1(

    )1(2

    )1(1

    )2(

    )1(

    2

    1

    12

    2)2(

    2)1(

    2

    22

    12

    1

    1)2(

    1)1(

    1

    21

    11

    2

    1

    2

    1

    fk

    NN

    Nk

    Nk

    kN

    N

    N

    NkN

    kk

    kk

    kk

    k

    k

    kN

    kk

    kk

    kk

    k

    k

    Nk

    kk

    kk

    kk

    k

    k

    NN

    k

    k

    k

    N

    k

    k

    k

    N

    k

    k

    fk I

    Z

    ZZ

    Z

    ZZ

    ZZ

    ZZZ

    ZZ

    Z

    ZZZ

    ZZ

    Z

    ZZ

    Z

    ZZ

    ZZ

    ZZ

    Z

    ZZ

    Z

    ZZ

    Z

    ZZ

    V

    VV

    V

    VV

    The diagonal elements of Zbus matrix are called self-impedance. And the off-diagonal impedances are called mutual-impedance.

  • DMAM (23/22)

    =

    1566.01058.00967.00978.01058.01341.00799.00864.00967.00799.01554.00651.00978.00864.00651.01488.0

    jbusZ

    And so

    ''44 ff IZV = ;

    44

    ''

    ZV

    I ff =

    ''

    44

    14''141 ff IZ

    ZIZV ==;

    ''

    44

    24''242 ff IZ

    ZIZV ==;

    ''

    44

    34''343 ff IZ

    ZIZV ==

    The faulted network is assumed to have been without loads before the fault. In such a case no current is flowing before the fault, and all voltages throughout the network are the same and equal to Vf. This assumption simplifies our work considerably , and applying the principle of superposition gives:

    ===+=

    44

    14

    44

    1414

    ''11 1 Z

    ZVZZVVZIVVVV ffffff

    ===+=

    44

    24

    44

    2424

    ''22 1 Z

    ZVZZVVZIVVVV ffffff

    ===+=

    44

    34

    44

    3434

    ''33 1 Z

    ZVZZVVZIVVVV ffffff

    04 == ff VVV These voltages exist when subtransient current flows and Zbus has been formed for a network having subtransient values for generator reactances. In general terms for a fault on bus k, and neglecting prefault currents,

    )16.10(kk

    ff Z

    VI = And the postfault voltage at bus n is

    ===+=

    kk

    nkf

    kk

    nkffnkffnfn Z

    ZV

    ZZ

    VVZIVVVV 1

    Usually, Vf is assumed to be 1.00o per-unit and with this assumption for our faulted network puj

    jI f 386.61566.0

    1'' ==

    pujjV 3755.0

    1566.00978.0111 =

    =

    pujjV 3825.0

    1566.00967.0112 =

    =

    pujjV 3244.0

    1566.01058.0113 =

    =

    Current flow from bus 1 to bus 3

  • DMAM (24/22)

    pujj

    VVyz

    VVI 2044.025.0

    3244.03755.0)( 311313

    31''13 ====

    Current flow from generator Ga to bus 1

    pujj

    VEyz

    VEI g

    g 0817.23.0

    3755.01)( 1''

    1111

    1''

    ''13 ===

    = Other currents can be found in a similar manner, and voltages and currents with the fault on any other bus are calculated just as easily from the impedance matrix. Example 10.3.2 A synchronous generator and motor are connected through a transmission line as shown in the following figure. The motor is drawing 20 MW at 0.8 power factor leading and a terminal voltage of 12.8 kV when a symmetrical three-phase fault occurs at the motor terminals. (i) Determine the bus impedance matrix. (ii) For a bolted three-phase short-circuit at bus 2, use Zbus to calculate the subtransient fault current and the contribution to the fault current from the transmission line. (iii) the subtransient generator and motor current including prefault current.

    Solution: Given, the voltage at the fault is 12.8 kV which we choose a reference.

    pu097.02.138.12 ==fV

    A13123102.133

    61030 =

    =baseI

    A9.3611288.03108.123

    61020 =

    =LI

    pu52.069.09.3686.01312

    9.361128 jLI +===

    y11 = y22 =1/ j0.2 = - j5.0; y12 = y21 = 1/ j0.1= - j10.0

    puj

    jjj

    bus

    =0.15

    0.100.100.15

    Y

    pujj

    jj

    bus

    =12.008.0

    08.012.0

    Z

    The subtransient fault current at bus 2 is

    pujjZ

    VZV

    I fkk

    ff 0833.812.0

    097.0

    22

    ====

    =

    =

    = 03233.0

    12.008.01097.01

    22

    211 j

    jZZVV f

    0122

    222 =

    =

    ZZVV f

    pujjz

    VVI 041.408.03233.00

    21

    1221 ===

    Subtransient generator current neglecting prefault current:

  • DMAM (25/22)

    pu23.308.83.02.0

    2.03.02.0

    2.0'' jjIgfI f =+=+= Subtransient motor current neglecting prefault current:

    pu85.408.83.02.0

    3.03.02.0

    3.0'' jjImfI f =+=+= To these current must be added the prefault current IL to obtain the total subtransient currents in the machines: Subtransient generator current including prefault current:

    pu71.269.052.069.023.3'''' jjjIgfIgI L =++=+= Subtransient motor current including prefault current:

    pu37.569.052.069.085.4'''' jjjImfImI L === Example 7.4 [Ref. 3, p. 369] Faults at bus 1 and 2 in Fig. 7.3 are of interest. The prefault voltage is 1.05 pu and prefault load current is neglected. (i) Determine the bus impedance matrix. (ii) For a bolted three-phase short-circuit at bus 1, use Zbus to calculate the subtransient fault current and the contribution to the fault current from the transmission line. (iii) Repeat part (b) for bolted three-phase short-circuit at bus 2.

    Fig. 7.3 Single-line diagram of a synchronous generator feeding a synchronous motor.

    (a) Reactance diagram obtained from Fig. 7.3. Reactance values are marked as per unit.

    (b) Circuit of Fig. (a) with admittances marked in per unit.

    Fig. 7.4 Solution: (i) The circuit of Fig. 7.4 (a) is redrawn in Fig. 7.4(b) showing per-unit admittance rather than per-unit impedance values. Neglectring prefault load current . Eg= Em = Vf =1.050o pu. From Fig. 7.4 (b) the bus admittance matrix is

    pujbus

    = 2787.8

    2787.32787.3

    9454.9Y

    puj-bus

    +==13893.004580.0

    04580.011565.01

    busYZ

    (ii) The subtransient fault current at bus 1 is

  • DMAM (26/22)

    pujjZ

    VZV

    I fkk

    ff 079.911565.0

    005.1

    11

    ====

    0111

    111 =

    =

    ZZVV f

    =

    =

    = 06342.0

    11565.004580.01005.11

    11

    212 j

    jZZVV f

    pujjz

    VVI 079.23050.0

    06343.0

    21

    1221 ===

    (iii) The subtransient fault current at bus 2 is

    pujjZ

    VZV

    I fkk

    ff 558.713893.0

    005.1

    22

    ====

    =

    =

    = 07039.0

    13893.004580.01005.11

    22

    121 j

    jZZVV f

    0122

    222 =

    =

    ZZVV f

    pujjz

    VVI 308.23050.0

    039.7.0

    12

    2112 ===

    Example 10.3 [Ref. 2, p. 394] A three phase fault occurs at bus 2 of the network of Fig. 10.5. Determine the initial symmetrical fault current (that is subtransient current) in the fault, the voltages at buses 1, 3, 4 during fault, the current flow in the line from bus 3 to 1 and the current contribution to the fault from lines 3 to 2 and 1 to 2, 4 to 2. Take the prefault voltage Vf at bus equal to 1.00o pu and neglect all prefault currents.

    Fig. 10.5 Solution: [[[y11=0; z12=0.125i; z21=z12; y12=1/z12; y21=y12;

  • DMAM (27/22)

    z13=0.25i; z31=z13; y13=1/z13; y31=y13;z14=0.4i; z41=z14; y14=1/z14; y41=y14; y22=0; z23=0.25i; z32=z23; y23=1/z23; y32=y23; z24=0.2i; z42=z24; y24=1/z24; y42=y24; z33=0.3i; y33=1/z33; y34=0; y43=0; z44=0.3i; y44=1/z44; Y11=y11+y12+y13+y14; Y22=y21+y22+y23+y24; Y33=y31+y32+y33+y34; Y44=y41+y42+y43+y44; Y12=-y12; Y21=Y12; Y13=-y13; Y31=Y13; Y14=-y14; Y41=Y14; Y23=-y23; Y32=Y23; Y24=-y24; Y42=Y24; Y34=-y34; Y43=Y34; Ybus=[Y11 Y12 Y13 Y14 Y21 Y22 Y23 Y24 Y31 Y32 Y33 Y34 Y41 Y42 Y43 Y44]; Zbus=inv(Ybus);]]]

    =833.10

    00.55.2

    033.11

    0.40.4

    0.50.4

    0.170.8

    5.20.40.8

    5.14

    j

    jj

    jjj

    jjjj

    jjjj

    Ybus

    =

    1954.01046.01506.01456.0

    1046.02954.01494.01544.0

    1506.01494.02295.01938.0

    1456.01544.01938.02436.0

    jjjj

    jjjj

    jjjj

    jjjj

    Zbus

    pujjZ

    VZV

    I fkk

    ff 3573.42295.0

    0.1

    22

    ====

    ===+=

    kk

    nkf

    kk

    nkffnkffnfn Z

    ZVZZVVZIVVVV 1

    1556.02295.01938.010.11

    22

    121 =

    =

    =

    jj

    ZZVV f

    0122

    222 =

    =

    ZZVV f

    3490.02295.01494.010.11

    22

    323 =

    =

    =

    jj

    ZZVV f

  • DMAM (28/22)

    3438.02295.01506.010.11

    22

    424 =

    =

    =

    jj

    ZZVV f

    =

    1954.01046.01506.01456.0

    1046.02954.01494.01544.0

    1506.01494.02295.01938.0

    1456.01544.01938.02436.0

    jjjj

    jjjj

    jjjj

    jjjj

    Zbus

    The current flow in the line from bus 3 to 1

    7736.025.0

    1556.03490.0

    31

    1331 jjz

    VVI === The current contribution to the fault from lines 1 to 2:

    pujjz

    VI 2448.1125.0

    1556.0

    12

    112 ===

    The current contribution to the fault from lines 3 to 2.

    pujjz

    VI 3960.125.0

    3490.0

    32

    332 ===

    The current contribution to the fault from lines 4 to 2:

    pujjz

    VI 7190.120.0

    3490.0

    42

    442 ===

    A Bus Impedance Matrix Equivalent Netwrok [Ref. 1, p. 264]

    Fig. 7.6 [R. 3, p. 317] Bus equivalent circuit (rake equivalent).

    Fig. 7.6 shows a bus impedance equivalent circuit that illustrates the short-circuit current in an N-bus system. This circuit is given the name rake equivalent due to its shape, which is similar to a garden rake. Neglecting prefault load currents, the internal vpltage sources of all synchronous machines are equal both in magnitude and phase. As such, they can be connected, as shown in in Fig. 7.7, and replaced by one equivalent source Vf from neutral bus 0 to a references bus, denoted k. This equivalent source is also shon in the rake equivalent of Fig. 7.6.

    Using Zbus, the fault currents in Fig. 7.6 are givn by

  • DMAM (29/22)

    =

    +

    +

    +

    +

    +

    ++

    ++

    +

    +

    +

    +

    ++

    ++

    +

    +

    +

    +

    +

    +

    +

    +

    ++

    +

    N

    k

    k

    k

    NN

    Nk

    Nk

    kN

    N

    N

    NkN

    kk

    kk

    kk

    k

    k

    kN

    kk

    kk

    kk

    k

    k

    Nk

    kk

    kk

    kk

    k

    k

    NN

    k

    k

    k

    N

    k

    k

    k

    Nf

    kf

    kf

    kf

    f

    f

    I

    II

    I

    II

    Z

    ZZ

    Z

    ZZ

    ZZ

    ZZZ

    ZZ

    Z

    ZZZ

    ZZ

    Z

    ZZ

    Z

    ZZ

    ZZ

    ZZ

    Z

    ZZ

    Z

    ZZ

    Z

    ZZ

    VV

    VVVV

    VV

    VVVV

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    ....

    )2(

    )1(

    2

    1

    )2(

    )1(

    2

    1

    1)2(

    )2)(2(

    )2)(1(

    )2(

    )2(2

    )2(1

    )1(

    )1)(2(

    )1)(1(

    )1(

    )1(2

    )1(1

    )2(

    )1(

    2

    1

    12

    2)2(

    2)1(

    2

    22

    12

    1

    1)2(

    1)1(

    1

    21

    11

    )2(

    )1(

    2

    1

    Where, I1, I2, . Are the branch currents and (Vf V1), (Vf V1), . Are the voltages across the branches. If switch S in Fig. 7.6 is open, all currents are zero and the voltage at each bus with respect to the neutral equals Vf. This corresponds to prefault conditions, neglecting prefault load currents. If switch S is closed, corresponding to a short-circuit at bus k, Vk = 0 and all currents except Ik remain zero. The fault current If= Ik= Vf/Zkk, which agree with the above driven equation. This fault current also induces a voltage drop ZnkIk = (Znk/Zkk)Vf across each branch n. The voltage at bus n with respect to the neutral then equals Vf minus this voltage drop, which agree with the derived equation. As shown in Fig. 7.6, subtransient fault currents throughout an N-bus system can be determined from the bus impedance matrix and the prefault voltage. Zbus can be computed by first constructing Ybus, via nodal equations, and then inverting Ybus. Once Zbus has been obtained, these fault currents are easily computed.

    Fig. 7.7 [Ref. 3, p. 372] Parallel connecion of unloaded synchronous machine internal voltage sources. Example 10.4 [Ref.1, p. 265] Determine the bus matrix for the following network. Generator at buses 1 and 3 are rated 270 and 225 MVA, respectively. The generator subtransient reactances plus the reactances of the transformers conncting them to the buses are each 0.30 per-unit on the generator rating as base. The turns ratios of transformers are such that the voltage base in each generator circuit is equal to the voltage rating of the generator. Include the generator and transformer reactances in the matrix. Find the subtransient current in a three-phase fault at bus 4 and the current coming to the faulted bus over each line. Prefault current is to be neglected and all voltages are assumed to be 1.0 per-unit before the fault occurs. System base is 100 MVA. Neglect all resistances.

  • DMAM (30/22)

    One-line diagram for example 10.4.

    Solution: Converted to the 100 MVA base, the combined generator and transformer reactances are:

    Generator at bus 1: puX 1111.02701003.0 ==

    Generator at bus 3: puX 1333.02251003.0 ==

    The network with admittances marked in per-unit is shown in Fig. 10.14 from which the node admittance matrix is

    =

    667.16968.3762.40.0

    937.7

    968.3944.6

    976.20.00.0

    762.4976.2

    175.23937.7

    0.0

    0.00.0

    937.7889.13

    952.5

    937.70.00.0

    952.5889.22

    jbusY

    Fig. 10.14 Reactance diagram for Example 10.4.

    This 55 bus is inverted on a digital computer to yield the short-circuit matrix

  • DMAM (31/22)

    =

    1301.01002.00603.00605.00608.0

    1002.02321.00720.00630.00511.0

    0603.00720.00875.00664.00382.0

    0605.00630.00664.01338.00558.0

    0608.00511.00382.00558.00793.0

    jbusZ

    The subtransient current in a three-phase fault on bus 4 is

    pujjZ

    VI ff 308.42321.0

    0.1

    44

    '' === At buses 3 and 5 the voltages are:

    nkffn ZIVV =

    pujjZIVV ff 6898.0)0720.0)(308.4(0.134''

    3 === pujjZIVV ff 5683.0)1002.0)(308.4(0.154

    ''5 ===

    Currents to the fault = Current from bus 3 to bus 4 + Current from bus 5 to bus 4 pujjjjjI f 308.4255.2053.2)968.3(5683.0)976.2(6898.0 ==+=

    From the same short-circuit matrix we can find similar information for fault on any other buses. The Selection of Circuit Breakers [CBs] [Ref. 1, p. 267] Circuit Breaker (CB): A circuit breaker is a mechanical switch capable of interrupting fault current and of reclosing. When a CB contacts separate while carrying current, an arc forms. The CB is designed to extinguish the arc by elongating and cooling it. Classification of CB: CBs are classified as power circuit breakers when they are intended for service in ac circuits above 1.5 kV, and as low-voltage circuit breakers in ac circuits up to 1.5 kV. Depending on the medium: air, oil, SH6 gas, vacuum - in which the arc is elongated. The arc can be elongated either by a magnetic force or by a blast of air. Reclosing the CB: Some CBs are equipped with a high speed automatic reclosing capability. Voltage Ratings Rated Maxium Voltage: Designates the maximum rms line-to-line operating voltage. The breaker should be used in system with an operating voltage less than or equal to this rating. Rated Voltage Range Factor: The range of voltage for which the symmetrical interrupting capability times the operating voltage is constant. Rated Low Frequency Withstand Voltage: The maximum 60 Hz rms line-to-line voltage that the circuit breaker can withstand insulation damage. Rated Impluse Withstand Voltage: The maximum crest voltage of a voltage pulse with standard rise and delay times that the breaker insulation can withstand. Current Ratings Rated Continuous Current: The maximum 60 Hz rms current that the breaker can carry continuously while it is the closed position without overheating. RatedShort-circuit Current: The maximum rms symmetrical current that the breaker can safely interrupt at rated maximum voltage. Rated momentary Current: The maximum rms asymmetrical current that the breaker can withstand while in the closed position without damage. Rated momentary current for stabdard breaker is 1.6 time the symmetrical interrupting capability. Rated Interuupting Time: The time in cycles on a 60 Hz basis from the instant the trip coil is energized to the instat the fault current is cleared.

  • DMAM (32/22)

    Modern Circuit breakers standards are based on symmetrical interrupting current. It is usually, necessary to calculate only symmetrical fault current at a system location, and then select a breaker with a symmetrical interrupting capability equal to or above the calculated current. The CB has the additional capability to interrupt the asymmetrical (or total) fault current if the dc off-set is not too large. The maximum asymmetrical factor K(=0) =3, which occurs at fault inception (=0). After fault inception, the dc off-set current decays exponentially with time constant L= (L/R) = (X/R), and the asymmetrical factor decreases. Power circuit breaker with a 2 cycle rated interruption time are designed for an asymmetrical interruption capability up to 1.4 times their symmetrical interruption capability, whereas slower CBs have a lower asymmetrical interrupting capability. [[Inclusion of dc component results in a rms value of current immediately after the fault, which is higher than the subtransient current. For oil circuit breakers above 5 kV the subtransient current multiplied by 1.6 is considered to be the rms value of the current whose disruptive forces the breaker must withstand during the first half-cycle after the fault occurs. This current is called momentary current, and for many years circuit breakers were rated in terms of their momentary current. The interrupting rating of a circuit breaker was specified in kVAs or MVAs. The interrupting kVA equal 3 times the kV of the bus to which circuit breaker is connected times the current which the breaker must be capable of interrupting when its contact part. This current is lower than the momentary current and depends on the speed of the breaker, such as 8, 5, 3, or 1.5 cycles, which is a measure of the time from the occurrence of the fault to the extinction of the arc.

    nginterrupti of capable bemust breaker the whichcurrent connected isbreaker circuit which to bus the ofkV 3kVA ngInterrupti =

    The current which a breaker must interrupt is usually asymmetrical since it still contains some of decaying dc component. A schedule of preferred ratings for as high-voltage oil circuit breakers specifies the interrupting current rating of breakers in terms of the component of the asymmetrical current which is symmetrically about the zero axis. This current is properly called the required symmetrical interrupting capability or simply the rated symmetrical short-circuit current. Selection of circuit breakers may also be made on the basis of total current (dc component included).]] What is the current to be interrupted? Modern day circuit breakers are designed to interrupt in first few cycles (five cycles or less). The dc off-set has not been yet died out and so contributes to the current to be interrupted. Rather than computing the value of the dc off-set at the time of interruption (since it is very complex for a large network), the symmetrical short-circuit current alone is calculated. The figure is then increased by an empirical multiplying factor to account for the dc off-set current as shown in the following Table [Ref.5, p. 318].

    Circuit Breaker Speed Multiplying Factor 8 cycles or slower 1.0 5 cycles 1.1 3 cycles 1.2 2 cycles 1.4

    Breakers are identified by nominal-voltage class, such as 69 kV.

  • DMAM (33/22)

    Breakers are identified by nominal-voltage class, such as 69 kV. Among other factors specified are rated continuous current 1200 A, rated short-circuit current at rated max voltage I=19 kA, rated maximum voltage Vmax=72.5, voltage range factor K=1.21, and rated short circuit current times operating voltage (72.519000) is constant. Imax=KI = 1.2119=23 A at operating voltage Vmin = Vmax/K = 72.5/1.21 = 60 kV. At operating voltages V between Vmin and Vmax, the symmetrical interrupting capability is IVmax/V=1378/V kA.

    Fig. 7.8 [Ref.3, p. 381] Symmetrical interrupting capability of a 69 kV class breaker.

    At operating voltages below Vmin, the symmetrical interrupting capability remains at Imax= 23 kA. Breakers of the 115 kV class and higher have a K of 1.0. Example 10.6 [Ref. 3, p.405] A 69-kV circuit breaker having a voltage range factor K of 1.21 and a continuous current rating of 1200 A has a rated short-circuit current of 19,000 A at the maximum rated voltage of 72 .5 kV. Determine the maximum symmetrical interrupting capability of the breaker and explain its significance at lower operating voltages. Solution: The maximum symmetrical interrupting capability is given by Rated short-circuit current = 1.2119,000 = 22,990 A This value of symmetrical interrupting current must not be exceeded. From the definition, we have Lower limit of operating voltage = 72.5/1.21 = 60 kV. Hence, in the operating voltage range 72.5 to 60 kV, the symmetrical interrupting current may exceed the rated short-circuit current of 19,000 A, but it is limited to 22,990 A. For example, at 66 kV the interrupting current can be = 72.5/6619,000 = 20, 871 A. E/X method: A simplified procedure for calculating the symmetrical short-circuit current, called the E/X method, disregards all resistance, all static loads, and all prefault current. Subtransient reactance is used for generators in the E/X method, and for synchronous motors the recommended reactance is the Xd of the motor times 1.5, which is the approximate value of the transient reactance of the motor. The impedance by which the voltage Vf at the fault is divided to find short-circuit current must examined when E/X method is used. In specifying a breaker for bus k this impedance is Zkk of the bus impedance matrix with the proper machine reactances since the short-circuit current is expressed by Eq. (0.16). If the ratio of X/R of this impedance is 15 or less, a breaker of the correct voltage and kVA may be used if its interrupting current rating is equal to or exceeds the calculation current. If X/R ratio is unkown, the calculated current should be no more than 80% of the allowed value for the breaker at the existing bus voltage. The ANSI (American National Standards Institute) application guide specifies a corrected method to account for ac and dc times constants for the decay of the current amplitude if the X/R ratio exceeds 15. The corrected method also considers breaker speed.

  • oc

    Ewb

    Tw Etrt

    ci

    T

    F

    [[Induction mof lager inducurrent equa

    Ide

    Nominalvoltage cla

    (kV, rms)

    23 46 69

    Example 7.where the opbus. Soluion: T1972.5/64=The calculatwhen X/R is

    Example 10transformer reactance Xtransformer 6.9 kV whencurrent in thinterrupting

    Fig. 10.1

    Solution: (i) For a basThe subtrans

    X d 20.0'' =

    Fig. 10.16 ispuV f 0.1=

    motor less thuction motorals subtransie

    entification

    ass )

    NoThreeMVA

    512

    .7 [Ref.3, p.perating vol

    The 69 kV=21.5 kA at ted symmetr unknown. T

    0.5 [Ref.1, pto a bus wh

    Xd of each mis 25 MVA

    n a three-phahe fault, (ii)curent (as d

    15 One-line d10

    se of 25 MVAsient reactan

    p0.1525 =

    s the diagramu

    han 50 hp ars dependingent current.]

    Table 7.10

    minal e-Phase A Class

    500 500 500

    . 381] The cltage is 64 k

    V class brthe operatin

    rical faul curTherefore, w

    p. 268] A 25hich suppliemotor is 20

    A, 13.8/6.9 kVase fault occ) the subtrandefined for ci

    diagram for 0.5.

    A, 13.8 kV ince of each m

    pu

    m with subtra

    re neglectedg on their siz]

    0 Preferred rV

    Rated MaxVoltage (kV

    rms)

    25.8 48.3 72.5

    calculated sykV. The X/R

    reaker has ng voltage 64rrent, 17 kA,

    we select the

    5 MVA 13.8es four ident% on a basV with a lea

    curs at point nsient currenircuit breake

    Example

    in the generamotor is

    ansient valu

    d, and variouze. If no mo

    ating for outVoltage

    x V,

    Rated range

    (K

    211

    ymmetrical R ratio is unk

    a symme4 kV. , is less than 69 kV class

    8 kV generatical motorsse of 5 MVakage reactaP. For the fa

    nt in breakeer application

    10.16

    ator circuit,

    es of reactan

    us multiplyinors are prese

    tdoor CBs (A

    voltage e factor K)

    .15

    .21

    .21

    fault currenknown. Sele

    etrical inter

    80% of thisbreaker from

    ator with Xd, as shown

    VA, 6.9 kV. ance 10%. Tault specifiedr A, and (iiin) in the fau

    Reactance d

    the base for

    nce marked.

    ng factors arented, symm

    ANSI) C

    Rated continuus

    current at 60Hz (a, rms)

    1200 1200 1200

    nt is 17 kA ct a CB from

    rrupting cap

    s capability, m Table 7.10

    d =15% is cin Fig. 10.1The three-p

    The bus voltad, determinei) the symm

    ult and in bre

    diagram for E

    the motors i

    For a fault a

    re applied tometrical shor

    Current

    0 )

    Ratedcircuit (at ratekV) (kA

    111

    at a three-pm Table 7.10

    pability I

    wich is requ0.

    connected th5. The subtphase ratingage at the me (i) the subt

    metrical shoreaker A.

    Example 10.

    is 25 MVA,

    at P.

    o the Xd rt-circuit

    d short current

    ed max A, rms) 1 7 9

    pase bus 0 for tis

    Vmax/V=

    uirement

    hrough a transient g of the

    motors is transient rt-circuit

    .5.

    6.9 kV.

  • DMAM (35/22)

    pujZth 125.0= puj

    jI f 0.8125.0

    0.1'' == The base current in the 6.9 kV circuit is

    AIbase 20909.63100025 =

    = AI f 1672020908

    '' == (ii) The generator contributes a current of

    pujjjjI g 0.450.0

    25.00.8 == The four motors contributes a current of = -j8.0 (-j4.0) = -j4.0 pu Each motor contributes a current of = -j4.0/4 = -j1.0 pu Through breaker A comes the contribution from the generator and three of the four motor. The current which pass through the breaker A is

    ApujjjI 1463020900.70.7)0.1(30.4'' ===+= (iii) To compute the current through breaker A to be interrupted, replace the subtransient reactance of j1.0 by the transient reactance of j1.5 in the motor circuits of Fig. 10.16. Then

    pujjZth 15.025.0)4/5.1(25.0)4/5.1( =+

    = The total current is: (1/j0.15) The generator contributes a current of: (1/j0.15)[0.375/(0.375+0.25)]= -j4.0 pu Each motor contributes a current of: (1/4)(1/j0.15)[0.25/(0.375+0.25)]= -j0.67 pu The symmetrical short-circuit current to be interrupted is: (4.0+30.67) 2090 = 12560 A The usual procedure is to rate all the breakers connected to a bus on the basis of the current into a fault on the bus. In that case the short-circuit current interrupting rating of the breakers connected to the 6.9 kV must be at least: 4+40.67=6.67 pu or 6.67 2090 = 13940 A. A 14.4 kV circuit breaker has a rated maximum voltage 15.5 kV and a K of 2.67. At 15.5 kV its rated short-circuit interrupting current is 8900 A. The constant value of rated = 15.58900 This CB is rated for a symmetrical short-circuit interrupting current of: 2.678900 = 23760 A (maximum voltage) at a voltage of 15.5/2.76 =5.8 kV (minimum voltage). This current is the maximum current that can be interrupted even thoug the breaker may be in circuit of lower voltage. The short circuit interrupting current rating at 6.9 kV is: (15.58900)/6.9 = 20,000 A The required capability of 13,940 A is well below 80% of 20,000 A, and the breaker is suitable with respect to short-circuit current. Calculation of Short-Circuit Current Using Bus Impedance Matrix Equivalent Network

  • DMAM (36/22)

    10.16 Reactance diagram for Example 10.5. 10.17 Bus impedance equivalent network for the bus impedance matrix of Example 10.5. Two nodes have been identified in Fig. 10.16. Node 1 is the bus on the low-tension side of the transformer, and node 2 is on the high-tension side. For motor reactance of 1.5 per unit.

    67.121.0

    14/5.1

    111 jjj

    Y =+= 101.0

    112 jj

    Y == 67.161.0

    115.0

    122 jjj

    Y =+=

    The node admittance matrix is

    =67.160.10

    0.1067.12

    jbusY and its inverse is

    =114.0

    900.0900.0

    150.0jbusZ

    Fig. 10.17 is the network corresponding to the bus impedance matrix closing S1 with S2 open represents a fault on bus 1. The symmetrical short-circuit interrupting current in a three-phase fault at node 1 is

    pujj

    I SC 67.615.00.1 ==

    The bus impedance matrix also gives us the voltage at bus 2 with the fault on bus 1. pujjZIV SC 4.0)09.0)(67.6(0.10.1 212 ===

    And since the admittance between nodes 1 and 2 is -j10.0, the current into the fault from the transformer is

    (0.4-0.0)(-j10)= - j4.0 pu We also know immediately the short-circuit current in a three-phase fault at node 2 which, by referring to Fig.10.17 with S1 open and S2 closed is

    pujj

    I SC 77.8114.00.1 ==

    Short Circuit Capacity (SCC)/ Fault Level of Bus SCC is defined by the product of the magnitude of pre-fault voltage (Vf) and the post fault current (If). For a solid fault, Zf= 0 Say, prefault voltage, Vf Fault current, |If |= |Vf |/ XT ; XT = Total impedance in the faulted circuit SCC= prefault voltage postfault current SCC= |Vf | |If |= |Vf | |Vf |/ XT = |Vf |2/ XT For, Vf 1 p.u.; SCC= 1/ XT p.u. (If all the values are given in p.u) SCC= 1/ XT MVA (If all the values are given in actual values) SCC= Base MVA/ Xs. Power companies furnish data to a customer who must determine the fault current to specify circuit breakers for an industrial plant or distribution system ay any point. Usually, the data supplied lists the short-circuit MVA instead of Thevenin impedance, where

  • DMAM (37/22)

    310)kVnominal(3MVACircuitShort = SCI Where Isc in amperes is the rms magnitude of the short-circuit current in a three-phase fault at the connection point. Base megavoltampere are related to base kilovolt anad base amperes by Short-circuit MVA = 3 (nominal kV) Isc10-3 If base kilovolt equal nominal kilovolt then Short-circuit MVA = Isc pu At nominal voltage the Thevenin equivalent circuit looking back into the system from the point of connection is an emf of 1.00o pu in series with the pu impedance Zth. Therefore, under short-circuit conditions,

    puMVAcircuitShort

    puI

    ZSC

    th0.10.1 ==

    With resistance and shunt capacitance neglected Zth =Xth, the single-phase Thevenin equivalent circuit which represents the system is an emf equal to the nominal line voltage divided by 3 in series with an inductive reactance of

    =SC

    th IX 1000)3kV/nominal(

    = MVACircuitShort)kVnominal( 2

    thX

    If base kV is equal to nominal kV, converting to per unit yields

    puX thSC

    base

    II

    MVACircuitShortMVABase ==

    Strength of a Bus A symmetrical fault occurs at bus-3. Say, the pre-fault voltage at bus-3 is 1.0 p.u. and just after the fault this voltage will be reduced to almost zero. The reduction in voltage of various buses indicates the strength of the network. So, the strength of a bus is the ability of the bus to maintain its voltage when a fault takes place at other bus:

    Strength of a bus related to SCC. The higher the SCC of the bus, the more it will be able to maintain its voltage in the case of

    fault on any other bus If XT =0 then SCC = . For this case the bus is known as Infinite Strong Bus

    Example: A small generating station has a bus-bar divided into three sections. Each section is through a reactor rated at 5MVA, 0.1 p.u. A generator of 8MVA. 0.15p.u. is connected to each section of the bas bar. Determine the SCC of the bus if a three-phase fault takes place on one of the section of the sections of bus-bar. Solution: Let the base: 8MVA in the generator circuit. Xg = 0.15 pu

  • Ea

    XBC

    C

    C

    P0ctWr

    So the reacXP=j(0.15+

    5.15.1

    jjX eq =

    j0.15 and jSCC = 1/XConsiderin

    Example: Tare linked wstation. [

    Solution: LeXg1= 0.1 p.uBase impedaCable = 0.5/

    Say fault at

    SCC= 1/Xeq ConsideringSCC = 10.2

    Say fault at

    SCC= 1/Xeq ConsideringSCC = 11.73

    Problems 10.25 per unconnected bthe short-cirWhen the vrms current

    Solution: Le

    ctors reactan+0.16)/2=j0.

    016.0(5016.0(5

    jj

    +++

    (0.16+0.155Xeq =1/j0.1016ng only the m

    Two generatiwith an inter-[Xg1=Xg2= 0.

    et the base: u. ance= (11)2 /[(11)2 /1200

    t the termin

    =1/j0.098 = g only the ma

    p.u. = 10.2

    t the termin

    =1/j0.085 = g only the ma3 p.u. = 11.7

    10.6 [Ref. 1,nit, respectivy a line havircuit megavovoltage at tt in a three-

    et base powe

    nce at 8MVA.155

    .0)155.0)155.0 j=

    5) is parallel6 = -j9.86 p.

    magnitude SC

    ing station h-connected c.1 p.u]

    1200MVA, /1200

    0] = 4.96 p.u

    nal of genera

    -j10.2 agnitude; 1200= 12,2

    nal of genera

    -j11.73 agnitude; 731200= 14

    p. 273] Twvely, on a bing a reactanoltamperes othe motor bu-phase fault

    er 2000 kVA

    A, XT = 0.18

    1016.

    l to each with.u. CC = 9.86 p.

    having 1200Mcable having

    11kV in the X

    u

    ator-1: X eq

    234MVA

    ator-2: X eq

    4,072 MVA

    wo synchronobase of 480 nce of 0.023 of the powerus is 440 V,

    at the motor

    A, base volta

    8/5 = 0.16 p

    h respect to

    .u. = 9.86 pu

    MVA and 80g reactance o

    generator ciXg2= 0.1(12

    0(1.00(1.0

    jjjj

    +=

    (1.015.0

    jjjj

    +=

    ous motors V, 2000 kV to a bus o

    r system are neglect loa

    r bus.

    age 480 V

    pu

    the fault bus

    u 8MVA =

    00MVA respof 0.5 per

    ircuit. 200/800)=0.1

    )96.415.0)96.415.0

    ++

    )96.41.0()96.41.0(j

    ++

    having subtrVA are connof a power sy

    9.6 MVA fad current a

    s bar

    = 78.73MVA

    pectively andphase. Dete

    15 p.u.

    098.0j=

    085.0j=

    ransient reacnected to a ystem. At thfor a nominaand find the

    A

    d operating aermine SCC

    ctances of 0bus. This m

    he power sysal voltage ofinitial symm

    at 11kV of each

    0.80 and motor is stem bus f 480 V. metrical

  • DMAM (39/22)

    When fault occurs, the bus voltage is: Vf = 440/480 = 0.92 pu Base impedance of line = (0.480)21000/20000 = 0.1152 pu Reactance of transmission line, Xline = 0.023/0.1152 = 0.2 pu At the power system bus, ISC=9.6106/(3480)= 11547 A Ibase = 2106/(3480)= 2406 A

    Reactance of the power system as a source

    puX s 21.0115472406

    II

    9.62

    MVACircuitShortMVABase

    SC

    base ===== Thevenin reactance at fault point F

    puX T j0.13j0.41)(j0.19)//(j0.2)125)//(j0.2(j0.8//j0. ==+= kApu

    XI

    Tf 17240607.707.70.13

    1Vf ===== References [1] Willaim D. Stevenson, Elements of Power System Analysis, Fouth Edition, McGraw-Hill International Editions, Civil Engineering Series, McGraw-Hill Inc. [2] John J. Grainger, William D. Steevnson, Jr., Power System Analysis, McGraw-Hill Series in Electrical and Conputer Engineering, McGraw-Hill Inc. [3] J. Duncan Glover, Mulukutla S. Sharma, Thomas J. Overbye, Power System Analysis and Design, Fouth Edition (India Edition), Course Technology Cengage Learning [4] Hadi Saadat, Power System Analysis, Tata McGraw-Hill Publishing Company Limited [5] I J Nagrath, D P Lothari, Modern Power System Analysis, Second Edition, Tata McGraw-Hill Publishing Company Liited [6] V. K. Mehta, Rohit Mehta, Principles of Power System, Multicolor Illustrative Edition, S. Chand and Company Limited Shortcut for 22 matrices For

    =

    dcba

    A , the inverse can be found using this formula:

    =

    =acbd

    bcadacbd

    A1

    ]det[11A

    Example:

    =4321

    A ;

    =

    =

    2/12/3

    121324

    211A