Chapter 3

Transients on Integrated PowerSystem

3.1 Line Dropping and Load Rejection

3.1.1 Line Dropping

• In three phase circuit capacitance switching, the determination of the voltage trapped afterswitching off is based on the ratio of the positive sequence capacitance C1 (total capacitanceper phase) to the zero sequence capacitance C0(capacitance from line-to-ground).

• The ratio is very large when the capacitor bank has ungrounded neutral. In this case,the voltage across the first phase approaches three times the line-to-ground voltage. Fortransmission lines and cables, the ratio is typically ranges from 1.6 to 2.0. This means thatthe voltage across fist pole to clear is in the range of 2.2 to 2.4 times of line-to-neutralvoltage. It reduces the chances of a restrike and also results in a significant reduction inthe transient voltage causes a restrike.

• During the restrike, the voltage across the switch will be divided into source side and tothe line based on their respective surge impedances. The wave travels up and down theline corresponding to the oscillatory swing of the bank in the case of lumped capacitance.

• Resistance switching method is used in this case whereby a resistor with a series switch isincluded in parallel with the breaker. The resistor and the switch included in the circuitwhile switching and the resistance value chosen to effectively damp the transient oscillationsdue to transient. Later the resistor removed after switching.

• While switching with the resistance switching method, the load is a combination of R andC. Hence the system voltage is no longer at a peak at current zero when the interruptionoccur. Therefore, less voltage trapped on the line and reduces the chance of a restrike.

• The line voltage of an energized line will not be uniform throughout the line. When the lineis removed from the source, a non-uniform charge will be trapped on the line. It will resultwaves with amplitudes depending upon the degree of nonuniformity. The waves travel upand down until losses damp them out.

• Sometimes a line is switched with transformer at line end. In this case, the trappedcharge dissipate through the magnetizing impedance of the transformer. Until staurationof transformer core, transformer offers high impedance and the discharge is very slow.But, the impedance drops abruptly when the core runs into saturation. This results sharpincrease in current and a rapid discharge of the line.Anyhow, the restriking of the circuitbreaker connected with a transformer is less than for unloaded lines.

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Figure 3.1: Equivalent circuit during load rejection

• The secondary side switching of transformers connected to high voltage lines can itselfcreate overvoltages, even without restriking. This comes about again as a result of thenonlinear properties of the transformer core.

• It is well known that magnetizing current of a transformer contains 3rd hormanics. Thereis a possibility for considerable harmonic voltage generation due to higher impedance ofboth line and transformer.

3.1.2 Load Rejection

• Where the line is connected to a major load on a generating station, sudden load rejectionwill result in over speeding of the machines and a rise in voltage until such times as thiscan be checked by the governers and exciters.

• The amplitude of the over voltage can be evaluated approximately, as illustrated by Fig.3.1,by,

V = EXc

Xc −Xs

where ‘E’ is the voltage behind the transient ractance, which is assumed to be constantover the subtransient period and its value before the incident, Xs the transient ractance ofthe generator in series with the thransformer reactance, and Xc the equivalent capacitiveinput reactance of the system.

3.2 Voltage Transients on Closing and Reclosing Lines

The transient phenomena will vary according to the system configuration, the source type, linelength, and terminations. For a line open at the far end, three types of transients on switchingare shown in Fig.3.2.

Figure 3.2(a) shows a transient which can be expected when the source is mainly inductive,for example, a single line connected to a transformer. It is single-frequency transient, and ifswitching takes place at maximum voltage in a phase, the transient oscillates to almost twice thevalue of the system voltage across the entire line length.

Figure 3.2(b) shows a high-frequency transient which can be expected with infinite sourceimpedance. This means that the system from which the line originates has a number of cablesor lines connected to it, and the line being switched is not longer than the incoming lines. Manyline terminations and connections at the point of origination of the line being switched meansthat these terminations present an overall low characteristic impedance compared to the linebeing switched. As a result the transient occurs at its natural frequency.

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Figure 3.2: Profiles of switching surges on closing a line, dependent upon power system configu-ration

Figure 3.3: (a).Switching a open ended line (b) and (c) Voltage profiles on bus and line side,respectively

Figure 3.2(c) shows the pattern of switching transient with complex source impedance, con-sisting of inductance of transformers and the surge impedance of other lines and cables feedingthe system. The transient overvoltage occurs at a number of frequencies.

Highest overvoltages occur when unloaded high-voltage transmission lines are energized andreenergized and this imposes voltage stresses on circuit breakers. Figure 3.3(a) shows closing ofa line of surge impedance Z0 and length l, open at the far end. Before the breaker is closed, thevoltage on the supply side of the breaker terminal is equal to the power system voltage, while theline voltage is zero. At the moment of closing, the voltage at the sending end must rise from zeroto the power frequency voltage. This takes place in the form of a traveling wave on the line withits peak at um interacting with supply system parameters. As an unloaded line has capacitiveimpedance, the steady-state voltage at the supply end is higher than the system voltage, and dueto Ferranti effect, the receiving-end voltage is higher than the sending end. Overvoltage factorcan be defined as follows:

OVtotal =umun

where, um is the highest peak voltage at a given point and un is the power frequency voltage onthe supply side of the breaker before switching.

The increase in power-frequency overvoltage depends considerably on the line length. Thetransient voltage is not so simple to determine and depends upon the phase angle at the closinginstant (Fig. 3.3). At instant t = t1, maximum superposition of transient and power-frequency

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voltage occurs.

3.3 Overvoltages by faults

Overvoltage may be produced by certain types of asymmetrical fault such as a ground faulton one of the phase conductors of a three-phase transmission line. The situation is somewhatanalogous to the switching transients except that here instead ofinjecting a current, a voltageequal and opposite to the prefault voltage at the fault point is applied. The losses in the systemwill dampen the transient by some percentage.

Figure 3.4: Faulted Network

• Figure 3.4 shows the faulted system in which a ground fault has occured on the “a” phaseat point F.

• Figure 3.5 shows the prefault steady-state voltage VfSa at the fault point.

• If it is assumed that a fault occurs at the peak of the prefault voltage and t = 0 at thisinstant, then the voltage injected at the fault point is as shown in Figure 3.5 (b).

• Figure 3.6 shows the de-energised network to which is applied the voltage Vffa. This isalso known as the superimposed voltage.

We can study the problem of overvoltage by asking how the de-energised network behaves inresponse to the application of this voltage. Complex analysis based on travelling wave theory(such as Bewley lattice diagrams) can be used to depict the level of overvoltage likely to beinduced, particularly on the unfaulted phases due to the injection of the suddenly applied voltageat the fault point. A line-ground fault of the type considered here can produce an overvoltageon an unfaulted phase as high as approximately twice the normal line-to-neutral voltage.

Figure 3.7 gives an example of such an overvoltage, in particular when the fault occurs at themidpoint of the line; then maximum voltage is at the midpoint ofthe unfaulted conductor.

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Figure 3.5: Faulted Network Voltage

Figure 3.6: Superimposed Network

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Figure 3.7: Overvoltage phenomena on a faulted three phase system

3.4 Switching Surges on Integrated System

The disturbance produced by the switching operation as modified by the interconnected systemspreads through the system, setting up waves that travel along the lines and reflect to and froat open ends.

Figure 3.8(a) shows two buses with voltages of 138kV and 345kV interconnected by auto-transformer. The 138kV bus is fed through the generator transformer. There are lines connectedto both the buses. Figure 3.8 (b) shows a single phase representation. A fault at short distancedown the line of 354kV bus is considered for the analysis. The switching operation due to thisfault will evoke a response from both the line and the system.

Figure 3.8 (c) shows the equivalent circuit of the system with a common voltage base. Thelines are represented by the resistances R1 and R2 under transient conditions until it is modifiedby reflection from points down the line. The surge impedance of the circuit is given by,√

L1

C1

The source circuit can be replaced by a parallel RL circuit in which L = L1 + L2 + L3 andR = R1. The response of a parallel RL circuit to a ramp of current I0t is given by,

v(s) =RI0

s[s+ (R/L)]

in operational form or

V (t) = LI0(1− e−RtL )

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Figure 3.8: Various representation of a generating and swtiching station.

as a time function. The fault current I0 is given by,

I0 =VpL

where, Vp is the peak system voltage. Substituting the value of I0,

V (t) = Vp(1− e−RtL ) = Vp(1− e−αt)

This shows that the fault will rise exponentially with a time constant. The approximate responseof the equivalent circuit to a ramp of current is shown in Figure 3.9. Furthermore, a voltage ofthis form will travel down each of the connected lines. But, while considering the capacitance,it introduce a delay ′τ ′ as indicated in Fig.3.9(b)

The effect of the many waves on the bus voltage of a station will depend on the impedance

Figure 3.9: Approximate repsonse of Fig.3.8(c) to a ramp of current. (a) Capacitance neglected.(b) Capacitance included.

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Figure 3.10: First-generation encounter between wave and station

of that station and connected lines, and also other stations that the wave has encountered.As a first approximation, each station can be represented by a parallel RL circuit. The

inducteance will depend upon the source impedance of the station and the resistance upon thetransmission lines connected to its bus. Figure 3.10 shows the first generation of encounters tobe evaluated. This requires the application of the reflection and refraction coefficients.

The reflected wave will be,

Vr = aVi = Vi

(Z2 − Z1

Z2 + Z1

)where, Vi is the incident wave, Z2 is the impedance of the station being encountered, and Z1 isthe impedance of the line along which the incident wave is advancing.

The transmitted wave or refracted wave is given by,

Vt = bVi = Vi

(2Z2

Z2 + Z1

)It describes the voltage waves that travel down the several lines that R represents.

After encounters of this type, the reflected and refracted waves return from where they cameor proceed to more distant stations; normally termed as second generation encounters. Thereflected wave and the transmitted wave are modified by the reflection and refraction coefficientsappropriate to the next station or discontinuity. The coefficents will have the same general formbut differ numerically because of different station and line characteristics.

3.5 EMTP (Electromagnetic Transients Program) for tran-sient application:

• The EMTP is a computer program designed to solve electrical transient problems in lumpedcircuits, distributed circuits, or both. It was introduced by H.W.Dommel in the early 1960s.

• Transient analyses can be carried out in circuits with any arbitrary configutation of lumpedparameters (R,L and C). Transmission lines, with distributed parameters, transposed oruntransposed, can be included in the network. Losses in lines are approximated by lumpedresistance. Nonlinear resistaors (for surge arresters) and nonlinear inductors (for saturabledevices) can be represented. It is also possible to open or close switches to simulate breakeroperations, flashovers etc. Both voltage and current sources are available with sine, ramp,or step functions. Trapped charges can be recognized.

The EMTP is based on the application of the trapezoidal rule to convert the differentialequations of the network components to algebraic equations. This approach is demonstrated inthe following text for the inductance, capacitance, and lossless line.

For the inductance L of a branch between the nodes k and m, it holds

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Figure 3.11: (a) Inductance (b) Equivalent Impedance Network

vk − vm = L

(dik,mdt

)thus,

ik,m(t) = ik,m(t−∆t) +1

L

∫ t

t−∆t

(vk − vm)dt

Integration by means of the trapezoidal rule gives the following equations,

ik,m(t) =∆t

2L

(vk(t)− vm(t)

)+ Ik,m(t−∆t)

Ik,m(t−∆t) = ik,m(t−∆t) +∆t

2L

(vk(t−∆t)− vm(t−∆t)

)For the capacitance C of a branch between the nodes k and m, it holds

Figure 3.12: (a) Capacitance (b) Equivalent Impedance Network

vk(t)− vm(t) = vk(t−∆t)− vm(t−∆t) +1

C

∫ t

t−∆t

ik,m(t)dt

Integration by means of the trapezoidal rule gives the following equations

ik,m(t) =2C

∆t

(vk(t)− vm(t)

)+ Ik,m(t−∆t)

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Ik,m(t−∆t) = −ik,m(t−∆t)− 2C

∆t

(vk(t−∆t)− vm(t−∆t)

)For a single-phase lossless line between the terminals k and m, the following equation must

be true.

um(t− τ) + Zim,k(t− τ) = uk(t)− Zim,k(t)

where, τ is the travel time(s).Resistance is simply represented by Ohm’s law,

Figure 3.13: Representation of resistance

ik,m(t) =1

R

(vk(t)− vm(t)

)Considering one each of the components just described joining at node 1 to adjacent nodes

2,3,4 and 5 as shown in Fig.3.14.

Figure 3.14: Assembly of components connected to node 1 of a network

The model equation for current at node 1 can be written as,

i12(t) + i13(t) + i14(t) + i15(t) = i1(t)

Substituting the values of currents from the above analysis,[1

Z0+

∆t

2L+

2C

∆t+

1

R

]v1(t)−∆t

2Lv3(t) =

2C

∆tv4(t)− 1

Rv5(t) = i1(t)−I12(t−τ)−I13(t−τ)−I14(t−τ)−I15(t−τ)

This equation is a linear, algebric equation in unknown voltages, with the right hand side

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known from the values of preceding steps.Each of the n nodes of the network can be trated in the same way to give similar equations.

For easy calculation, these are assembled in matrix form:

[G][v(t)] = [i(t)]− [I]

where, [G] - Nodal conductance matrix, [v(t)]-Vector of n node voltages, [i(t)]-Vector of currentsources, and [I] - Vector of past history terms.

It is usual for known voltages to be driving some of the nodes. These are handled by parti-tioning the nodes in above equation into the A nodes with unknown voltages and the B nodeswith known voltages to give,[

[GAA] [GAB ][GBA] [GBB ]

] [[vA(t)][vB(t)]

]=

[[iA(t)][iB(t)]

]−[[IA][IB ]

]from which the unknown voltage [vA(t)] is found by solving,

[GAA][vA(t)] = [Itotal]− [GAB ][vB(t)]

where,[Itotal] = [iA(t)][IA]

This equation helps to obtain the solution of linear equations in each time step with a constantcoefficient matrix [GAA], provided ∆t is not changed. But, the right side of the equation mustbe recalculated in each time step.

The equation[GAA][vA(t)] = [Itotal]− [GAB ][vB(t)]

is best solved by traingular factorization of the augmented matrix [GAA], [GAB ] once for all,before entering the time loop. The same process is then extended to the vector [Itotal] in eachtime step in the so-called forward solution, following by a back substitution to obtain [vA(t)] asindicated in Fig. 3.15.

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Figure 3.15: Flowchart of transients program

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3.6 Short Line or Kilometric Fault

Faults occuring between a few kilometers to some hundreds of kilometers from the breaker aretermed ‘short-line faults (SLF)’ or ‘kilometric faults’.

For such faults, the line impedance limits the current and, consequently, supports some ofthe system voltage. The generated voltage is divided on either side of the breaker in proportionto the impedance of the line and the source.

Figure 3.16: Distribution of voltage on a faulted system

Fig.3.16 shows the distribution of voltage. When the circuit breaker interrupts at currentzero the generated voltage will be near the peak value.

Let L be the source inductance, L1 be the line inductance to the fault and C1 be the linecapacitance to the fault. E is the peak value of the e.m.f driving the fault.

Peak voltage at the breaking point is given by,

V =L1

L+ L1E

Opening of the breaker is simulated by injecting a cancellation current,

− E

ω(L+ L1)ωt = − E

(L+ L1)t

This current wave, travelling on the line, produces a voltage wave,

− E

(L+ L1)t.Z

where Z is the surge impedance of the line.

v =1√L1C1

and

Z =

√L1

C1

Time taken by the wave to reach faulted end, T is given by,√L1C1

When the wave reaches the faulted end of the line in time T , the voltage due to the wave at

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the line side circuit terminal is,

− E

(L+ L1)

√L1

C1

√L1C1 = − E

(L+ L1)L1 = −V

The addition of initial voltage and the voltage due to transient wave at the circuit breakerterminal is zero. The voltage wave is reflected with opposite sign at the short circuit point andwith the same sign at the circuit breaker terminal.

Figure 3.17: Voltage at line terminal of circuit breaker during kilometric fault

The voltage at the circuit line terminal is given by Fig.3.17. It can be seen that the lineside terminal of the circuit breaker can make a considerable swing in potential, +V to −V , ina comparatively short time 2T . In this time, the other terminal of the circuit breaker wouldbe still having a potential approximately equal to V. It is this fast rate of rise of the transientrecovery voltage that may prejudice the successful operation of the circuit breaker.

If the source can be represented by a parallel L and C circuit, the source side response willbe (1-cosine) recovery wave with a period 2π

√LC and an amplitute of

E − E

(L+ L1)L1 =

E

(L+ L1)L

Therefore the potential if the circuit breaker terminal on the source side is given by,

E

(L+ L1)L

(1− cos

t√LC

)The breaker sees the difference of potential on the two terminals as shown in Fig.3.18.

Figure 3.18: Circuit breaker recovery voltage following a short-line fault

Contents

1 Standing Waves and Natural Frequencies . . . . . . . . . . . . . . . . . . . . . 11.1 Case:1 (r=g=0; Losses Neglected) . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1.1 Location of Poles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1.2 Residues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2 General Case (Considering r,l,g,c) . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2 Refelction and Refraction Coefficient . . . . . . . . . . . . . . . . . . . . . . . . 7

3 Transients on Integrated Power System . . . . . . . . . . . . . . . . . . . . . . 93.1 Line Dropping and Load Rejection . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3.1.1 Line Dropping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93.1.2 Load Rejection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

3.2 Voltage Transients on Closing and Reclosing Lines . . . . . . . . . . . . . . . . . 103.3 Overvoltages by faults . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123.4 Switching Surges on Integrated System . . . . . . . . . . . . . . . . . . . . . . . . 143.5 EMTP (Electromagnetic Transients Program) for transient application: . . . . . 163.6 Short Line or Kilometric Fault . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

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