H7 SHORT-CIRCUIT EMERSON EDUARDO RODRIGUES.pdf

5
SHORT-CIRCUIT
CURRENTS CALCULATION
Nouredine Hadjsaid, Ion Tri�tiu, and Lucian Toma
5.1 INTRODUCTION
A short circuit is defined as any accidental contact-direct or through an impedance
between two points of a power system that normally have different voltage,
for example, between the conductors of two phases of a line or transformer, or
between a conductor and earth, or between a conductor and permanently grounded
components, such as the ground wires of overhead lines and the frames of electrical
machines.
The analysis of short-circuit currents and related voltages as well as the power flows is
useful for
• adequately setting up protection devices, which detect the fault and actuate the
circuit breakers of the faulted component (line, transformer, generator, motor)
thereby minimizing short-circuit consequences;
• checkup (determining) the breaking capacity of circuit breakers;
• changing the network structure, where possible, so as to make the short-circuit
currents less severe.
HandbookofElectricalPower SystemDynamics: Modeling, Stability, and Control. Edited by Mircea Eremia and
Mohammad Shahidehpour.
© 2013 by The Institute of Electrical and Electronics Engineers, Inc. Published 2013 by John Wiley & Sons, Inc.
229

230 SHORT-CIRCUIT CURRENTS CALCULATION
5.1 .1 The Main Types of Short Circuits
There are various types of short circuits that can occur in electrical installations [1]:
• Phase-to-earth (80% of faults).
• Phase-to-phase (15% of faults). This type of fault often degenerates into a three
phase fault.
• Three phase (only 5% of initial faults).
These different short circuits (abnormal connections) and their corresponding currents
are presented in Figure 5.1 [2]. The primary characteristics of a short circuit are as follows:
• Duration (self-extinguishing, transient, and steady state).
• Origin:
- lightning or switching overvoltages;
- insulation breakdown due to heat, humidity, or a corrosive environment;
- mechanical (break in a conductor, accidental electrical contact between two
conductors via a foreign conducting body such as a tool or an animal).
• Location (inside or outside a machine or an electrical switchboard).
In Figure 5.1 the following notations have been used:
r�l is the initial phase-to-earth short-circuit current;
r�2 is the initial phase-to-phase short-circuit current;
r�2Ea' r�2Eb are the initial phase-to-phase short-circuit currents, flowing to earth from
phase a or b, respectively;
r�E2E is the initial phase-to-phase short-circuit current flowing to earth;
r�3 is the initial three-phase short-circuit current.
Note: It is necessary to distinguish between the short-circuit current at the short-circuit
location and the partial short-circuit currents in the branches at any point of the network.
E! ffi� := �=!
§� �
c
(a) (b) Figure 5 . 1 . Different types of short cir-
cu its and their currents [2):
a a (a) symmetrical th ree-phase short
b b circu it; (b) phase-to-phase short circuit
c c--;:,
141
<J-- clear of earth; (c) phase-to-phase-to-
42Ea to.Fb earth short circu it; (d) phase-to-earth
short circu it. � short-circu it
<I-- ---I>
<I--
cu rrent
---I>
--t>-- partial short-circuit currents i n
(c) (d) conductors and earth retu rn

INTRODUCTION
5.1 .2 Consequences of Short Circuits
The consequences are variable depending on the type and the duration ofthe fault, the point
in the installation where the fault occurs, and the short-circuit power. According to the
work [2] of Schneider Electric, the consequences include
• at the fault location, the presence of electrical arcs, resulting in:
- damage of insulation;
- welding of conductors;
- fire and danger to life;
• on the faulty circuit:
- electrodynamic forces, resulting in: deformation of the busbars, disconnection of
cables;
- excessive temperature rise due to an increase in Joule losses, with the risk of
damage to insulation;
• on other circuits in the network or in nearby by networks:
- voltage dips during the time required to clear the fault, ranging from a few
milliseconds to a few hundred milliseconds;
- shutdown of a part of the network, the extent of that part depending on the
design of the network and the discrimination levels offered by protection
devices;
- dynamic instability and/or the loss of machine synchronization;
- disturbances in control/monitoring circuits, and so on.
As a consequences of these abnormal connections, the structure of the power system
impedance (or admittance) network suddenly changes, entailing the passage of abnormal
currents both through the accidental connection represented by a short circuit itself and
through the different components of the power system; such currents are called short
circuit currents [3].
Usually, the maximum currents that may flow in the various components ofthe power
system are the short circuit ones. These currents, which may reach very high values, have
particularly adverse effects on the different components of the system (i.e., not only
those directly involved by the fault) and on the operation ofthe system itself. Among these
effects, the main ones (which explain why short-circuit currents should be known) are the
following:
• Thermal effects, that is, increased temperature of conductors due to the increase of
losses by the Joule effects; this effect obviously depends on both intensity and
duration of the current.
• Electrodynamic stresses between the conductors of the same components (line,
transformer, motors, generators) also depending on their spacing.
• Variations ofpowerflows on links and, thus, ofthe powers generated by alternators,
jeopardizing the stability ofthe system. In the effect, a short circuit represents a large
perturbation and can cause loss a synchronism of one or more generators opening
of links, separation of the network in two or more parts, nonsupply of all or part of
the load.
231

5.2 CHARACTERISTICS OF SHORT-CIRCUIT CURRENTS
As it was shown in subsection 2. 1.3.7, the short-circuit current contains a subtransient
period, a transient, and a steady state one. The maximum value of the short-circuit current
during the transient period occurs when the fault happens as the voltage passes through zero
(Figure 2.1Sa), the minimal value occurs when the fault happens at the peak value of the
voltage, in which case the aperiodical component disappears.
The transient and subtransient periods of short-circuit current are not identical on the
three phases of the network. The three-phase stabilized short-circuit currents are, however,
identical in their effective values on all the three phases.
A complete calculation of short-circuit currents should give the currents as a function
of time at the short-circuit location from the initiation of short circuit up to its end,
corresponding to the instantaneous value ofthe voltage at the beginning ofshortcircuit [1].
If three-phase generators are short-circuited, either directly or through system
impedances, this results in short-circuit currents that normally start with a high peak
value but which decay to a steady-state value in accordance with the time constants of the
circuits. The curve showing this process is represented in Figures S.2a and b.
The transient conditions, prevailing while the short-circuit current develops, differ
depending on the distance between the fault location and the generator. This distance is not
necessarily physical, but means that the generator impedances are less than the link
impedance, between the generator and the fault location [2].
• Far-from-generator short circuit. Short circuit during which the magnitude of the
symmetrical AC component of the prospective (available) short-circuit current
remains essentially constant.
• Near-to-generator short circuit. Short circuit to which at least one synchronous
machine delivers an initial symmetrical short-circuit current, which is more than
twice the machine's rated current, or a short circuit to which synchronous or
asynchronous machines contribute more than S% of the initial symmetrical
short-circuit current (l�) without motors.
In the case of a far-from-generator short circuit (Figure S.2a), the short-circuit current
can be considered as the sum of two components:
• The AC component with constant amplitude during the whole short circuit.
• The aperiodic DC component beginning with an initial value A decaying to zero.
In case of a near-to-generator short circuit (Figure S.2b), the short-circuit current can
be considered as the sum of the following two components:
• The AC component with decaying amplitude during the short circuit.
• The aperiodic DC component beginning with an initial value A decaying to zero.
In practical applications, the following specific quantities present a major interest:
• The initial symmetrical short-circuit current, l�, is the effective value of the AC
symmetrical component and serves as a basis for the calculation of the peak
asymmetrical short-circuit current, ip, as well as the breaking current and capacity.

CHARACTERISTICS OF SHORT-CIRCUIT CURRENTS
Current
Top envelope
a----+'
..,.
' "/ Decaying (aperiodic) component ido
'){...
.---- '.. .......... ... .......... ..........
" /
'.
.:� Time
........
"y....... .......... ...
Bottom envelope
�
N
(a)
Current
Top envelope
",'/'1'��:��
.
��e
��OdIC) component 'do
" .. ....... ... 2-/2 'k
, '
"
(b)
Time
Figure 5.2. Typical wave of a short-circuit cu rrent: (a) far-from-generator short circu it with
constant AC component; (b) near-to-generator short circu it with decaying AC component.
(Repri nted with permission from lEe 60909. [1])
• Thepeak asymmetrical short-circuit current, ip, is the maximum instantaneous value
of this current during initial short-circuit conditions and is indicated as a peak value.
In addition to the AC component, it includes the DC component id.c., which occurs
during a sudden short circuit. The value ip determines the dynamic stresses in the
components of the installation and also the load carried by the switchgear.
• The breaking current, h, is the effective value of the AC component of the short
circuit current at the instant of contact separation in the circuit breaker;
• Initial symmetrical short-circuitpower S�. Fictive value determined as a product of
the initial symmetrical short-circuit current I� and the rated system voltage
Ur(S� = V3Ur/�).
• DC component id.c. of the short-circuit current and its initial value A.
• The steady-state short-circuit current, h, is rms value of a symmetrical current,
which remains after the decay ofthe transient phenomena; it is quoted as an effective
value.
In most practical cases, a determination like this is not necessary. Depending on the
application of the results, it is of interest to know the rms value of the symmetrical AC
component and the peak value ip of the short-circuit current following the occurrence of the
short circuit. The highest value ip depends on the time constant of the decaying aperiodic
233

component andthe frequencyf, thatis, on theratioRIX orXIRofthe short-circuitimpedance
�k> and is reached ifthe short circuit starts at zero voltage. The current ip also depends on the
decay of the symmetrical AC component of the short-circuit current (see 5.4.2).
In meshed networks, there are several direct current time constants. That is why it is
possible to give an easy method for calculating ip and id.c.. Special methods to calculate ip
with sufficient accuracy are given in Standard IEC 60909 [1].
In the case of a near-to-generator short circuit (Figure 5.2b), the variation in the
impedance of the generator, in this case, the dominant impedance damps the short-circuit
current.
The transient current-development conditions are complicated by the variation in the
source voltage (electromotive force) resulting from the short circuit. For simplicity, the
electromotive force is assumed to be constant and the internal reactance of the machine is
variable [2].
The reactance develops in three stages:
• Subtransient (the first 10-20 milliseconds of the fault).
• Transient (up to 500 milliseconds).
• Steady state (or synchronous reactance).
Note that in the indicated order, the reactance acquires a higher value at each stage,
that is, the subtransient reactance is less than the transient reactance, itself less than the
steady-state reactance.
The successive effect of the three reactances leads to a gradual reduction in the short
circuit current, which is the sum of four components (Figure 5.3):
• The three alternating components (subtransient, transient, and steady state).
• The aperiodic component resulting from the development ofthe current in the circuit
(inductive).
Note that the decrease in the generator reactance is faster than that of the aperiodic
component. This is a rare situation that can cause saturation of the magnetic circuits and
interruption problems because several periods occur before the current passes through zero
[2].
Practically speaking, information on the development of the short circuit is not
essential:
• In a LV installation, due to the speed of the breaking devices, the value of the
subtransient short-circuit current, denoted I�, and of the maximum asymmetrical
peak amplitude ip is sufficient when determining the breaking capacities of the
protection devices and the electrodynamic forces.
• In LV power distribution and in HVapplications, however, the transient short-circuit
current is often used ifbreaking occurs before the steady-state stage, in which case it
becomes useful to use the short-circuit breaking current, denoted h, which deter
mines the breaking capacity ofthe time-delayed circuit breakers. Ib is the value ofthe
short-circuit current at the instant at which interruption is effective, that is, following
a time t after the development of the short circuit, where t = tmin. Time tmjn
(minimum time delay) is the sum of the minimum operating time of a protection
relay and the shortest opening time of the associated circuit breaker, that is, the

CHARACTERISTICS OF SHORT-CIRCUIT CURRENTS
(b) 0 AAAAnAAAoAAAA_l • tIs]
ik� I I
�VVV�VllVi/ViVN'�
I I I
(c) ik��1i'Y:Y':S • tIs]
I�
"
(d) Of--�====--
...
�
--�" tIs]
-+ ,. +
(e)
QI Q3 Q5
Figure 5.3. Tota l short-circu it current ik and
contribution of its components [2]: (a) subtran-
o "-----''-:II-AAAAAAI-Wrl-Wrl-Wrl-Wrl-Wrltr--. t[s] sient reactance; (b) transient reactance; (c)
Subtrans.ent Transient Steady state
steady-state reactance; (d) aperiodic component.
shortest time between the appearance of the short-circuit current and the initial
separation of the pole contacts on the switching device.
Figure 5.4 presents the various currents of the short circuits defined above.
In the calculation ofthe short-circuitcurrents in systems supplied by generators, power
station units and motors (near-to-generator and/or near-to-motor short circuits), it is of
Symmetrical t--'f---+-+--+-+---1L----f-----+---:--f----l.--,-
Asymmetrical
Subtransient Transient Steady state
r-----�-----------------------r----�--�t
Figure 5.4. Short-ci rcu its cu rrent near a generator [2].
235

interest not only to know the initial symmetrical short-circuit current l� and the peak short
circuit current ip, but also the symmetrical short-circuit breaking current lb and the steady
state short-circuit current h. In this case, the symmetrical short-circuit breaking current lb
is smaller than the initial symmetrical short-circuit current l�. Normally, the steady-state
short-circuit current lk is smaller than the symmetrical short-circuit breaking current lb.
For the calculation of the initial symmetrical short-circuit current l�, the symmetrical
short-circuit breaking current h, and the steady-state short-circuit current h at the short
circuit location, the system may be converted by network reduction into an equivalent
short-circuit impedance �k seen from the short-circuit location. This procedure is not
allowed when calculating the peak short-circuit current ip; in this case, it is necessary to
distinguish between networks with and without parallel branches.
While using fuses or current-limiting circuit breakers to protect substations, the initial
symmetrical short-circuit current is first calculated as if these devices were not available.
From the calculated initial symmetrical short-circuit current and characteristic curves of
the fuses or current-limiting circuitbreakers, the cut-offcurrent is determined, which is the
peak short-circuit current of the downstream substation.
5.3 METHODS OF SHORT-CIRCUIT CURRENTS CALCULATION
5.3.1 Basic Assumptions
To simplify the short-circuit currents calculations, a number of assumptions are required.
These impose limits for which the calculations are valid but usually provide good
approximations, facilitating comprehension of the physical phenomena, and consequently
the short-circuit current calculations. They nevertheless maintain a fully acceptable level of
accuracy, "erring" systematically on the conservative side.
IEC 60909 is applicable for calculation of short-circuit currents [1]
• in low voltage three-phase AC systems;
• in high voltage three-phase AC systems.
Systems at highest of 550kV and above with long transmission lines need special
considerations.
The assumptions used are as follows [1,2] :
• The short-circuit current, during a three-phase short-circuit, is assumed to occur
simultaneously on all three phases.
• During the short circuit, the number of phases involved does not change, that is, a
three-phase fault remains three-phase and a phase-to-earth fault remains phase-to
earth.
• For the entire duration ofthe short circuit, the voltages responsible for the flow ofthe
current and the short-circuit impedance do not change significantly.
• Transformerregulators or tap changers are assumed to be set to a medium position (if
the short circuit occurs away from the generator, the actual position of the
transformer regulator or tap changers does not need to be taken into account).
• Arc resistances are not taken into account.
• All line capacitances are neglected.

METHODS OF SHORT-CIRCUIT CURRENTS CALCULATION
• Load currents are neglected.
• All zero sequence are taken into account.
When calculating short-circuit currents in systems with different voltage levels, it is
necessary to transfer impedance values from one voltage level to another, usually to that
voltage level at which the short-circuit current is to be calculated. For per unit or other
similar unit systems no transformation is necessary, if these systems are coherent, that is,
UrTHV/UrTLV = UnHV/UnLV for each transformer in the system with partial short-circuit
currents. UrTHv/UrTLvl is normally not equal to UnHv/UnLv2 [1].
• UrTHV and UrTLV are the rated voltages ofthe transformer on the high voltage or low
voltage side;
• UnTHV and UnTLV are the nominal system voltages of the transformer on the high
voltage or low voltage side.
The impedances of the equipment in superimposed or subordinated networks are to be
divided or multiplied by the square of the rated transformation ratio Np Voltages and
currents are to be converted by the rated transformation ratio Np
5.3.2 Method of Equivalent Voltage Source
One way of calculating the short-circuit current assumes the introduction of an equivalent
voltage source at the short-circuit location. The equivalent voltage source is the only active
voltage of the system. All network feeders, synchronous and asynchronous machines are
replaced by their internal impedance.
In all cases, it is possible to determinate the short-circuit current at the short-circuit
location F with the help of an equivalent voltage source. Operational data and the load of
consumers, tap changer position of transformers, excitation of generators, and so on, are
dispensable; additional calculations about all the different possible load flows at the instant
of short circuit are superfluous.
Figure 5.5 shows an example of the equivalent voltage source at the short
circuit location F as the only active voltage of the system fed by a transformer without
Figure 5 . 5 . System diagram and equ iva lent circuit d iagram for network feeders: (a) without
transformer; (b) with transformer. (Repri nted with permission from lEe 60909 [11.)
I Ur denotes the rated phase-to-phase voltage;
2 The nominal system voltage (U,,) is the voltage (phase-to-phase) at which a system is designated, and to which
certain operating characteristics are referred.
237

or with on-load tap changer. All other active voltages in the system are assumed to be
zero.
• If a three-phase short circuit is fed from a network in which only the initial
symmetrical short-circuit current l�s at feeder connection point S is known
(Figure 5.5a), then the equivalent impedance �s of the network (positive-sequence
short-circuit impedance) at the feeder connection point S should be determined by
cUns
Zs = 131"
V.J kS
If Rs/Xs is known, then Xs shall be calculated by
Zs
Xs = ----r'=�==
VI+ (RS/XS)2
(5. 1)
(5.2)
• Ifthe short circuit is fed by a transformer (Figure 5.5b) from a medium- or high-voltage
network in which only the initial symmetrical short-circuit current l�s at the feeder
connection point S is known, then the positive-sequence equivalent short-circuit imped
ance ZST referred to the low-voltage side of the transformer is to be determined by
where
cUns 1
ZST = 131" .
N2
V.J kS r
Uns is the nominal system voltage at the feeder connection point S;
(5.3)
l�s is the initial symmetrical short-circuit current at the feeder connection point S;
c is the voltage factor (see Table 5.1) for the voltage Un S;
Nr is the rated transformation ratio at which on-load tap changer is in the main position.
• In the case of high-voltage feeders with nominal voltage above 35 kV fed by
overhead lines (Figure 5.6), the equivalent impedance �s may be considered in
many cases as a reactance, that is, �s � jXs. In other cases, if no accurate value is
known for the resistance Rs of network feeders, we may substitute Rs = 0. 1 Xs,
where Xs = 0.995 Zs.
T A B L E 5.1. Voltage Factor c (Reprinted from IEC 60909-0 (1))
Voltage Factor c for the calculation of
Nominal Voltage VII
Low voltage:
100-1000V
Medium voltage: 1-35 kV
High voltage
d
: >35 kV
Maximum Short-Circuit
Currents Cmaxa
1.05"
1.10c
1.10
Minimum Short-Circuit
Currents Cmin
0.95
1.00
aCmaxUn should not exceed the highest voltage Um for equipment of power systems.
b
For low-voltage systems with a tolerance of +6%. for example systems renamed from 380V to 400V.
cFor low-voltage systems with a tolerance of +10%.
d
If no nominal voltage is defined. cmaxUn = Um or CminUn = 0.9Um should be applied.

S T A Bfk3
mt t$
L
(])
I
/J"s HV
N,:I
LV Un
(s (a)
(c)
Figure 5.6. Il lustrations for ca lcu lating the i n itial symmetrical short-circuit cu rrent f� in compli
ance with the procedu re for the eq u ivalent voltage sou rce: (a) system d iagram; (b) equ iva lent
d iagram of the positive-sequence system; (c) eq u ivalent d iagram with short-circuit impedance.
(Repri nted with permission from IEC 60909 [1])
The initial symmetrical short-circuit currents l�s max and l�s min on the high
voltage side of the transformer shall be given by the supply company or by an
adequate calculation according to the lEe 60909 standard.
Shunt admittances (e.g., line capacitances and passive loads) are not to be
considered when calculating short-circuit currents (Figure 5.6b).
Note: If there are no national standards, it seems adequate to choose a voltage
factor c according to Table 5.1, considering that the highest voltage in a normal
system does not differ, on average, by more than approximately +5% (some LV
systems) or +10% (some HV systems) from the nominal system voltage Un [1].
5.3.3 Method of Symmetrical Components
5.3.3.1 General Principles. The majority of faults in a power system are not
symmetrical three-phase short circuits, but occur between one phase and earth or less
commonly occur between phases where they may also involve earth. In addition, there is
usually a finite fault impedance rather than a short circuit. Thus, instead of a completely
balanced three-phase circuit, which can be solved by calculating the current flowing in one
of the three phases, an unbalanced three-phase circuit must now be solved. This can most
conveniently be done by using the symmetrical components of an unbalanced system of
currents or voltages [4].
In 1918, one of the most powerful tools for dealing with unbalanced polyphase circuit
was discussed by c .L. Fortescue at a meeting of the American Institute of Electrical
Engineers [5]. The method of symmetrical components can be applied to any polyphase
system containing any number ofphases, but the three-phases system is the only ofinterest
here.
According the Fortescue's theorem, three unbalanced phasors ofa three-phases system
can be resolved into three fictitious balanced systems of phasors called symmetrical
components of the original phasors (Figure 5.7).
239

240
v+
-b
Positive sequence
J{,
Negative sequence
SHORT-CIRCUIT CURRENTS CALCULATION
Zero sequence
Figure 5.7. Th ree sets of ba lanced phasors that form the symmetrical components of three
u n balanced phasors.
The three balanced systems of phasors are as follows [6]:
(i) A positive-sequence system consisting of three components (voltages or cur
rents) equal in magnitude, shifted from each other by 2lT/3, with the phase
sequence same to that of the original phasors a, b, c and denoted by I�, U;, I�
or V+ V+ V+
- a ' - b ' -c ·
(ii) A negative-sequence system consisting of three components (voltages or cur
rents) equal in magnitude, shifted from each other by 2lT/3, with the phase
sequence opposite to that of the original phasors a, b, c and denoted by I�, Ib,
r V- V- V-
-c or _ a ' - b ' -c .
(iii) A zero-sequence system consisting of three components (voltages or currents)
equal in magnitude and with zero phase shifting from each other, and denoted by
10 10 10 VO VO VO
-a' -b' - C
or - a' - b' -c·
In order to represent the phase shifting of 2lT/3 between the components of the
positive- and negative-sequence systems, it is convenient to use a rotation operator denoted
by �. This operator causes a rotation of 2lT/3 in the trigonometrical direction. It is a
complex number of unit magnitude with an angle of 2lT/3 and is defined as follows:
Ifthe operator � is applied to rotate a phasor twice in succession, the phasor is rotated
by 2 . 2lT/3, which is equivalent to -2lT/3, and
Three successive applications of � rotate the phasor by 2lT, thus
�3 = 1 ilK = 1 � = 1
Therefore,
1 + � + �2 = 0

5.3.3.2 The Symmetrical Components of Unsymmetrical Phasors. Since each
of the original unbalanced phasors is the sum of its components, the original phasors
expressed in terms of theirs components are [6]
v = VO + V+ + v- (5.4a)
- a - a - a - a
Yb= VO +
-b V+ +
-b v-
-b (5.4b)
V = VO + V+ + v- (5.4c)
-c -c -c -c
In order to resolve the three unsymmetrical phasors ( Ya, Y b, and Y J into their
symmetrical components, we reduce the number of unknown quantities by expressing each
component of Yb and Yc in terms of the operator � and the components of Ya.
Inspection of Figure 5.7 allows us to write:
VO= VO V+= V+ v-= v-
- a - - a - - a -
y�= VO Y�= a2 V Yb"= a V
VO= VO V+= a V+ v-= a2 v-
-c - -c -c
Substituting the equations (5.5) into the equations (5.4) gives:
or expressing in matrix form
V = VO + V+ + V-
- a - - -
Y b= yO + �2 y+ + a V-
Yc = yO + � y+ + a2 V-
(5.5)
(5.6a)
(5.6b)
(5.6c)
(5.7)
Figure 5.8 illustrates the construction of the phase voltages Ya, Yb' and Yc in terms
of the sequence components yO, y+, and Y-, based on equations (5.5) and (5.6).
Let us denote by
(5.8)
a matrix with constant (complex or real) and independent time coefficients. It allows
expressing the unbalanced phasors in terms of balanced components. In this way each of
241

c
a
b (a) b (b)
Figure 5.8. Representation of the voltage phasors i n terms of seq uence voltages.
the unbalanced phase components (current or voltage) can be expressed as a linear
combination of positive-, negative-, and zero-sequence components.
Equation (5.7) can thus be written as follows:
[ Yab c]= [T] [ yo+-] (5.7')
The balanced components can be also expressed in terms of the unbalanced compo
nents. It may be verified easily that
1
(5.8')
Premultiplying both sides of equation (5.7) by [Tr1 achieve
(5.9)
This matrix equation can be written as separate equations in ordinary fashion:
(5. lOa)
(5.lOb)
(5. lOc)
Th t Vo v+ v- VO V+ d V- b .c d fr
e sequence componen s _
b' _
b ' _
b ' _
c' _
c ' an _
c can e loun . om
equation (5.5).

Equation (S.lOa) shows that no zero-sequence component exists if the sum of
unbalanced phasors is zero. In a perfectly symmetrical and balanced three-phase system,
the sum of the voltage phasors is always zero, and contains no zero-sequence compo
nents [6] .
Similar to voltages, the currents phasors can be resolved using the theory of
symmetrical components. They may be represented either analytically or graphically.
The phase currents can thus be written in terms of their sequence components and the
rotation operator as
or
1 = 1° + 1+ + r
----a - - - (S. l la)
(S. l lb)
(S. l lc)
(S. 12a)
(S. 12b)
(S. 12c)
Ifa neutral wire is added to the three-phase system, the currentthatreturns through the
neutral !.., is equal the sum of the phase currents, that is
Substituting equation (S. 13) in equation (S.12a) yield:
1 = 31°
-n -
(S. 13)
Ifno return path through neutral exists in the three-phase system, then !.., = 0, and the
phase currents contain no zero-sequence components. A Ll-connected load or transformer
provides no return path, and thus the phase currents can contain no zero-sequence
components.
DECOUPLING THESEQUENCEIMPEDANCES. The relationship between voltages and currents
on a three-phase electrical line can be written as [7, 13]
(S. 14)
243

a L I
y,
b Ib
Yb
C L
y,
(a)
,
0
1+21M
o �
Y Zero sequence
o 0
,
+
I-IM
+ �
Y
0 0
Positive sequence
, I-IM
�
Y Negative sequence
o 0
(b)
Figure 5.9. Representation of an electrical l i ne: (a) phase components model with mutual
coup l i ng; (b) sequence components model.
where, [?-abel is the impedance matrix of the line (Figure 5.9a), given as:
[?-abcl= [f
-ca
�ab
�b
�cb
;ae 1
-be
Z
-c
(5.15)
where, �a' �b' and �e are the self-impedances of the phases a, b, and c, respectively,
whereas �ab= �ba' �ac= �ea' and �bc= �cb are the mutual impedance between
corresponding phases.
Assuming that the three-phase line is constructed perfectly symmetrical, the self
impedances are equal, that is, �a= �b= �e= �, and all the mutual impedances are
equal between them and equal to �M' Under these conditions, the impedance matrix
becomes
(5. 15/)
The above matrix is now fully symmetrical. In order to simplify the calculations, the
phase components impedance matrix from (5.15/) can be decoupled using the theory of
symmetrical components. Therefore, by appropriate transformation, and taking into
account the expression 1 + � + �2= 0, the sequence components impedance matrix
is achieved [7]:
(5.16)

The sequence components impedance matrix is thereby diagonal, the nondiagonal
terms being equal to zero, which allow decoupling the sequence components. Figure 5.9a
shows the representation of an electrical line in the three-phase components, whereas
Figure 5.9b illustrates the three sequence networks, with no coupling between them.
Assuming now that the original three-phase system in not symmetrical, that is, the
impedances �a' �b' and �c are not equal, and neglecting the mutual impedances between
phases, the impedance matrix from (5. 15) becomes [7]
(5. 17)
Under these conditions, transformation into sequence components give:
1
[ Z+ �b + �c �a + �2 �b + � �c
= � �a-: � �b + �2 �c �a + �b + �c
�a + �2 �b + � �c �a + � �b + �2 �c
�a + � �b + �2 �c 1
�a + �2 �b + � �c
�a + �b + �c
(5.18)
The resulting matrix shows that under asymmetrical construction of the three-phase
system, the sequence matrix cannot be decoupled. Due to the network asymmetry,
unbalanced currents on the three phases occur. The degree of asymmetry is, in general,
small, and thereby in practical calculations the original three-phase system is considered
symmetrical. When afaultoccurs in the power system, except forthe case ofa symmetrical
fault, the network impedances are no longer equal and the voltages and currents are
unbalanced; the greatest unbalance occurs at the fault point [7].
CHARACTERISTICS OF THE SYMMETRICAL VOLTAGES. Taking into account the representa
tions from Figures 5.7 and 5.8, the characteristics of the sequence voltages are as
follows [7] :
• The phasors yO are called zero-sequence voltage components, are equal in magni
tude in all phases (a, b, and c) and are cophasial, that is, all components are varying
together.
• The phasors y+ are called positive-sequence voltage components, are equal in
magnitude in all phases and form a system with the same phase sequence as the
original three-phase system, that is, phase b lags phase a by 2n/3, phase c lags phase
b, and phase a lags phase c; the positive-sequence is also referred to as abc sequence.
• The phasors Y- are called negative-sequence voltage components, are equal in
magnitude in each phase and are shifted also by 2n/3; the sequence of voltages is
changed as compared to the positive sequence, that is, phase b leads phase a rather
than lagging, while phase c lags phase a; the order of sequence it thus acba.
245

(a) (b)
(c)
Figure 5 . 1 0. Three-phase AC system representation for calculation of the short-circuit imped
ances at the short-circuit location F: (a) positive-seq uence short-circuit i mpedance; (b) negative
seq uence short-circu it impedance; (c) zero-seq uence short-circu it impedance. (Adopted from IEC
60909[1])
• The order of sequence of the voltage components corresponds to the order in which
each voltage attain the maximum value; the rotation of phasors is counterclockwise
for all three sets of sequence components, as was assumed for the original
unbalanced phasors (Figure 5.7).
These sequence currents are related to the sequence voltages by the corresponding
sequence impedances (�+, �-, and �o).
For the purpose of Standard IEC 60909, one has to make a distinction between short
circuit impedances at the short-circuit location F and the short-circuit impedances of
individual electrical equipment [1].
• The positive-sequence short-circuit impedance �+ at the short-circuit location F is
obtained when a symmetrical system of voltages ofpositive-sequence phase order is
applied to the short-circuit location F, and all synchronous and asynchronous
machines are replaced by their internal impedances (Figure 5.lOa):
v+
z+ =-=
- /+
(5.19')
• The negative-sequence short-circuit impedance �- at the short-circuit location F is
obtained when a symmetrical system of voltages ofnegative-sequence phase order is
applied to the short-circuit location F (Figure 5. lOb):
v-
z- =-=-
- 1- (5.19/1)
Note: The value of �+ and �- impedances can differ from each other only in the
case ofrotating machines. When far-from-generator short circuits are calculated, it is
generally allowed to take �- = �+.
• The zero-sequence short-circuit impedance �o at the short-circuit location F is
obtained if an AC voltage is applied between the three short-circuited phase

conductors and the joint return (e.g., earthing systems, neutral conductor, earth
wires, cable sheaths, cable armoring) (Figure 5.lOc):
(5.19"')
Note: The capacitances ofoverhead lines and cables of low-voltage networks may
be neglected.
5.3.3.3 Sequence Impedance of Network Components. Decoupling the
sequence impedances, as shown by equation (5.16), makes possible constructing indepen
dent sequence networks. A sequence networkis a single-phase equivalent circuit consisting
of a sequence impedance and eventually a voltage source. The voltage drop caused by the
flow of the sequence current through the corresponding sequence network depends on
the equivalent impedance of that network [6]. The impedance of a sequence network may
be different from the impedances of the other sequence networks. The value of an
impedance depends on the type of equipments from the analyzed part of the electrical
network, for example, transmission lines, transformers, generators, and so on.
In the case of nonrotating equipment like transformers, the negative-sequence
impedance is equal to the positive-sequence impedance, whereas in the case of rotating
equipments, the two impedances are different. In general, for all type of equipments the
zero-sequence impedance is different from both positive- and negative-sequence imped
ance (see Table 5.3). The sequence impedances are calculated using either equipment data
from manufacturers or estimated data [7].
(i) Construction of sequence networks Three types of sequence networks are
defined: positive, negative, and zero. The sequence networks are designed as
viewed from the point of fault or unbalance, denoted by F, or F+, F- and pO,
respectively [7]. The sequence voltages are measured from the point of fault to
the neutral point, marked by N, or N+, N- and NO, respectively for each sequence
network (Figure 5.1 1). Each network is designed in the form of a dipole through
which sequence current flows.
(a)
The voltages applied to the network terminals are taken as phase-to-neutral
voltages. The electrical generators are designed to supply balanced voltages and
1
F
N
(b) (c)
Figure 5 . 1 1 . Sequence networks rep
resentation: (a) positive seq uence; (b)
negative sequence; (c) zero seq uence.
247

thus an emf §. is represented on the positive-sequence network only. Phase "a"
voltage is taken as reference voltage and the voltages of phases b and c are
expressed in terms of phase a voltage. The outward direction of the current flow
in sequence networks is the positive direction (see Figure 5.1 1). Zero-sequence
currents will flow through the network only ifreturn path exists through which a
completed circuit is provided.
The sequence networks carrying the sequence currents are interconnected to
represents unbalance fault conditions [6].
A voltage source, §., which is the Thevenin equivalent (Figure 5.11a), is
represented on the positive-sequence network only. Applying the voltage
Kirchhoff's law on each sequence network we may write:
v+ = E - /+ Z+
v- = r Z- (5.20)
or in matrix form
(5.21)
Equation (5.21), together with the equation that describes the conditions at the
fault location, is used to derive L+ in terms of §, �+, �-, and �o, for each
type of fault.
(ii) Sequence impedance of electrical lines
• Positive- and negative-sequence impedances An electrical line is a passive
element. Assuming that the three phases of the line are symmetrical, the
sequence of the phase voltages applied, that is, abc (positive sequence) or acb
(negative sequence) makes no difference as regards the voltage drops
produced. Therefore, the positive- and negative-sequence impedances of a
symmetrical three-phase electrical line are identical. Their values are equal
to the electrical line impedance used for steady-state calculations [8]. Note
that for long transmission lines without phase transposition, the phase
impedances and thus the sequence impedances may have slightly different
values.
• Zero-sequence impedance The zero-sequence impedance of an electrical line
is different by the positive-sequence and negative-sequence impedances
because it also involves the paths to/through the ground and thus it accounts
for the ground and the protection wires(s) impedances. The zero-sequence
currents flowing through the phases of a three-phase electrical line are equal
and cophasial, and their sum returns through the paths to/through the ground.

The magnetic field produced by the flow of the zero-sequence currents
differs from that produced by the flow of the positive- or negative-sequence
currents, which are dephased by 2n/3, and the return current is zero. The
resulting effect of the flow of the zero-sequence current is a zero-sequence
reactance several times greater than the positive-sequence reactance, approx
imately by a factor [9] :
• Single-circuit overhead lines without ground wires or with steel ground wires
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5
• Single-circuit overhead lines with copper or aluminum ground wires . . . . 2
• Double-circuit overhead lines without ground wires or with steel ground wires
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5
• Double-circuit overhead lines with copper or aluminum ground wires . . . 3
• Three-phase cables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 to 5
• Single-phase cables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
The zero-sequence mutual impedance between parallel circuits of the same
towers or even on different towers placed on the same right-of-way can also be
important. Since the grounding impedance depends on the type of soil simplify
ing assumptions should be made.
(iii) Sequence impedances of transformers
• Positive- and negative-sequence impedances Similar to transmission lines,
the transformers are passive elements, having no moving parts. Assuming that
the transformer is symmetrical, the phase sequence has no effect on the
winding reactances. Therefore, the equivalent circuits and the associated
impedances are identical for both the positive-sequence and the negative
sequence networks. For fault analysis, only the series leakage impedance is
considered.
• Zero-sequence impedance The zero-sequence impedance and the equivalent
circuit depend upon the winding connection, method ofneutral grounding, and
so on.
Table 5.2 illustrates the sequence circuits of the two-winding and three
winding transformers for different winding connections.
The grounding impedance. The neutral point of Y-connected winding of the
transformer may be grounded either solidly or through a neutral impedance �n'
Under symmetrical and balanced conditions, the potential of the neutral point is
zero and no current will flow through the grounding impedance. However, if
asymmetry or unbalance occurs, the potential of the neutral point will be
different by zero and a current L, = 3 !_? will flow through the impedance
�n causing a voltage drop 3 �n !_?, which is added to all the phase-to-ground
voltages and hence to the zero-sequence voltages [10]. In order to determine the
zero-sequence current LO, which flows through the zero-sequence network, an
impedance 3 �n must be added in this network in series with the transformer
impedance [9].
Let us define the transformer impedance by �� as the sum of the perphase
linkage reactances and resistances of the high (H) and low (L) voltage windings,
that is, �� = �� + ��, which may be about equal to the positive-sequence
249

T A B l E 5.2. The Sequence Equivalent Circuits of Transformers for Different Winding Connections
Case Winding Connections
2
3
4
5
6
7
8
9
lO
Zero-Sequence Circuit
o o
N° Ref.
0
M
H
N° rc o-+L
Ref.
o-+M
H 3�
o-+L
N°
Ref.
�
-M
�
o-+M
W
-H
k'
o-+L
Ref.
�
M
�
-M
H
-H
N0
k'
L
Q Q Ref.
0
Positive- or Negative- Sequence Circuit
o
Ref.
o
Ref.
Q
Ref.
o
Ref.
o
Ref.
o
(>-0 --------00 Ref.

impedance. If the neutral of the Y-connected primary winding is grounded
through an impedance �n' the total zero-sequence impedance as seen from the
primary side is �o + 3 �n. Since a Ll-connected winding provides no path for
zero-sequence currents flowing in the line, the impedance seen from the
secondary side is infinite.
Zero-sequence currents will flow through windings on one side of the
transformer and into the connected lines only if a return path exists through
which a completed circuit is provided. Also, a path for the corresponding current
in the other windings must exist [1 1].
In the case 1 presented in Table 5.2, on the primary windings side a return path
for the zero-sequence currents is provided by the grounding system of the Y
connection, whereas on the secondary windings side a closed circuit is provide by
the Ll-connection that serves to balance the currents in the primary windings. Of
course, no zero-sequence currents will flow in the lines connected to the
secondary winding. In the zero-sequence network an open circuit must be
represented between the line and the reference bus on the Ll side.
If the neutral point of the Y-connected windings on one side is ungrounded, as
in the cases 3 and6 given in Table 5.2, an open circuit exists in the zero-sequence
network and no zero-sequence currents can flow between the two parts of the
system connected by the transformer. Thus, the zero-sequence impedance seen
from either the primary side or the secondary side is infinite. Since a Ll
connection provides no return path for zero-sequence currents, no zero-sequence
current can flow into a Ll-Ll bank, as shown in the case 4, although it can circulate
within the Ll windings. The Y-Y transformer with both neutrals grounded (case 5)
is a case in which zero-sequence currents can flow in both windings and outside
the transformer on both sides. In this case, the zero-sequence network is similar
to the positive-sequence network.
The cases 7-10 given in Table 5.2 illustrate the sequence networks for three
winding transformers. The effects ofthe leakage reactances and resistances ofthe
three-windings are represented by a star-connected equivalent. Connection of
this equivalent to the primary, secondary, andtertiary terminals, and the reference
bus (Ref.) depends on the winding connections.
(iv) Sequence impedances ofsynchronous machines
• Positive-sequence impedances Detailed representation of the synchronous
machines in the positive-sequence network applicable for stability studies
was discussed in Chapter 2. This representation is also valid for short-circuit
currents calculation for the entire study period, including the subtransient,
transient, and steady-state conditions.
Figure 5.12 illustrates the simplified positive-sequence network of a synchro
nous machine. Thus, following the occurrence ofafault at the machine terminals,
R+ x+
��II�-1••••'---"� J(
t
Reference
1--------0bus
Figure 5 . 1 2. positive-sequence network of a synchronous
mach ine.
251

the armature fluxes and the currents change. Three positive-sequence reactances
X+ are defined depending on the time frame of interest: subtransient, transient,
and steady state. The positive-sequence resistance R+ is the armature resistance.
The sequence network from Figure 5.12 was represented as a single-phase
circuit under the assumption of symmetrical construction of the electrical
machine. The reference bus is the neutral of the generator.
Neglecting the saliency effects, the positive-sequence reactance is: X+ = X�,
for the subtransient conditions; X+ = X�, for the transient conditions; X
+ = Xd,
for steady-state conditions.
• Negative- and zero-sequence impedances The negative-sequence and zero
sequence impedances are specified by the manufacturers and are determined
by tests.
A synchronous machine is designed to normally produce positive-sequence
voltages and currents. When an unbalance is created, for example, due to a short
circuit occurring near the generator, to single-phase loads or to open conductors,
negative-sequence currents flow in the armature winding that produce a field
rotating in opposite direction to the positive-sequence field created by the rotor
winding and at double synchronous speed with respect to the rotor speed. The
rotor is thus subjected to double frequency flux pulsation [12]. The existence of
the negative-sequence currents results in rotor losses in the form of heating as
well as small armature losses. The field produced by the negative-sequence
currents passes alternatively through the direct axis and the quadrature axis, and
thus the negative-sequence reactance is:
where X� andX� are the direct-axis and quadrature-axis subtransient reactances.
The negative-sequence resistance R- is greater than the positive-sequence
resistance and depends on the damper winding, since the double frequency
currents are induced in the rotor field and damper winding.
Figure 5.13 illustrates the equivalent circuit of a synchronous machine
appropriate for use in negative-sequence networks. No emf is represented, since
the synchronous machine produces positive-sequence voltages only. The refer
ence bus of the equivalent circuit is the neutral of the generator.
When zero-sequence currents flow through the armature, their instantaneous
values in the three phases are equal and the resultant air gap flux is zero.
Therefore, the zero-sequence reactance is approximately equal to the leakage
reactance.
The flow of the zero-sequence currents in the three phases and the neutral of
the synchronous machine is shown in Figure 5. 14a. The zero-sequence equiv
alent circuit is illustrated in Figure 5. 14b.
R X
1
t-_
_
_
_
_
_
-o Reference
bus
Figure 5. 1 3 . Negative-sequence network of a synchronous
mach ine.

a
l�
l'
� t yO
1°
-b
b Reference
l� bus
c -
(a) (b)
Figure 5 . 1 4. Representation of the synchronous mach ine for zero seq uence: (a) cu rrents paths; (b)
equ iva lent circuit (si ngle phase).
The zero-sequence currents flowing through the three phases return through
the grounding impedance �n as a sum !..., = r:. + £i: + � = 3r:.. This is why
a term 3�n is introduced in the equivalent circuit of Figure 5.14b. Ifthe neutral is
isolated from the ground, then Zn = 00 and !..., = O. The grounding impedance
of the synchronous machine is not a part of either the positive- or negative
sequence network because neither positive- nor negative-sequence current can
flow through this impedance [6]. The reference bus ofthe equivalent circuit is the
ground at the generator.
Since all the neutral points of a symmetrical three-phase system through which
balanced currents flow are at the same potential, all the neutral points have the same
potential when either positive-sequence or negative-sequence currents flow [9].
Table 5.3 shows the zero-sequence characteristics of the various elements in
an electrical network.
5.3.3.4 Unsymmetrical Fault Calculations. In the sections 5.3.3.2 and 5.3.3.3,
construction of the three sequence networks as seen from the fault point in terms of
the characteristics of the power system elements has been presented. For unsymmetrical
faultcurrent calculations, the three separate networks can be connected in a certain manner,
depending on the type of the fault [7].
T A B l E 5.3. Zero-Sequence Characteristic of the Various Elements in an Electrical N etwork
Elements
Transformer (seen from secondary winding)
No neutral
Yyn or Zyn free flux
forced flux
Dyn or YNyn
primary D or Y + Zn
Machine
Synchronous
Asynchronous
Electrical Line
00
00
10 to lSx+
x+
0.1 to O.2 X+
253

SHUNT FAULTS OF ZERO IMPEDANCE. The following cases of faults to ground can be
defined, assuming zero fault impedance [12] :
(i) Single phase-to-earth-on the a phase
Ib = 0
Ie = 0
Ya = 0
(ii) Phase-to-phase-between b and c phases
I = 0
- a
Ib = - Ie
Yb = Ye
(iii) Phase-to-phase-to-earth-between b and c phases
(iv) Three phase
I = 0
-a
Ia + Ib + Ie = 0
Ya = Yb
Yb = Ye
(v) Single-phase open-circuit fault
I = 0
-a
�b = 0
�e = 0
(S.22)
(S.23)
(S.24)
(S.2S)
(S.26)
Notice from the above that three equations that define the fault conditions can be written.
(i) Single phase-to-earthfault Consider a fault defined by equations (S.22) and by
Figure S. ISa.
The conditions at the fault are expressed by the following equations:
I = 0
-e v = 0
-a

F
a --�--------------
b --�--�----------
(al
II II
1------,
J' r=J'. tOJ't I
- F' - - n - - F
O
I
I I I
I I I
I I I
I I I
I I I
_
Z' I _
T I i' I
-
[1
I I T
I I I
I I I
I I I
I I I
I I I
N' I N- I N
° I
I I....J I_J I
L______________________ J
(bl
Figure. 5 . 1 5 Single phase-to-earth fault at F: (a) defin ition of fau lt; (b) equ iva lent circu it.
With b, = 0 and !..c = 0, the symmetrical components of the current from
(5. 12 a,b, and c), are given by
(5.27)
so that !....O, !....+, and C each equal !....a/3 and
(5.28)
Substituting !....+ for C and !....O in equation (5.21), we obtain:
(5.29)
Performing the indicated matrix multiplication and subtraction an equality of
two column matrices is obtained. Premultiplying both column matrices by the
row matrix [ 1 1 1 ] gives
(5.30)
Since Ya = yO + y+ + y- = 0, we solve equation (5.30) for !....+ and obtain:
E
1+ = -
.,--
---'
=--
----;;c
�+ + �- + ZO
(5.31)
Equations (5.28) and (5.31) are particular equations for a single phase-to-earth
fault. They are used with equation (5.21) and the symmetrical-components
relations to determine all the voltages and currents at the fault.
255

If the three sequence networks are connected in series (Figure 5.15b), we see
that the currents and voltages resulting therefrom satisfy the equations above,
and the three sequence impedances are in series with the voltage §. With the
sequence networks so connected, thevoltage across each sequence network is the
symmetrical component of § of that sequence.
If the neutral of the generator is not grounded, the zero-sequence network is
open circuited and �o is infinite. Since equation (5.31) shows that !..
+ is zero
when �o is infinite, C and 1....0 must be zero. Thus no current flows in phase a
since � is the sum of its components, all of which are zero.
The fault current � is:
3 §
1 = 3 10 = -:------,;-
=..a -
�+ + �- + ZO
The voltage of phase b to ground under fault conditions is:
(5.32)
Z- (a2 - a) + ZO(a2 - 1)
V = a2 V+ + a V- + VO = V - - - - - (5.33)
- b - - - - - - a
�+ + �- + ZO
Similarly, the voltage of phase c can be calculated.
(ii) Phase-phase fault Consider a fault defined by equation (5.23) and by
Figure 5.16.
� = O � = - !..c Yb = Yc
With Yb = Yc' the symmetrical components voltage from equation (5.9) are
given by:
from which we find
F
a --�-------------
b --�--�----------
c --�--�---------
(a)
V + = V-
,------1
I I
F + F
t r
I I
L_______J
(b)
(5.34)
Figure 5. 1 6. Phase-phase-earth fau lt at F: (a) defi n ition of fau lt; (b). equ iva lent circu it.

Since h = - !..c and L,. = 0, the symmetrical components of current are
given by equation:
and therefore
Since [0 is zero,
r = - /+
- -
(5.35')
(5.35")
(5.36)
Equation (5.29), with the substitutions allowed by equations (5.34), (5.35'),
(5.35/1), (5.36), becomes:
(5.37)
Performing the indicated matrix operations and premultiplying the resulting
matrix equation by the row matrix [ 1 1 - 1 ] gives
o = E - /+ z+ - /+ z-
or
and solving for [+ yields
E
/+ = -
.,----
-
�+ + �-
(5.38)
Equations (5.34), (5.35'), (5.35/1), and (5.38) are the special equations for a
phase-to-phase fault. They are used with equation (5.29) and the symmetrical
component relations to determine all voltages and currents at the fault.
From equation (5. 1 1b) we have
(5.39)
257

F
a -+-------
b -*-�-----
c -
.......
-
-".r-
-
-
-
-
(a)
1----------1
I ___ .L ___ _
I I I
F + F F
O
t r t
(b)
Figure 5. 1 7 . Phase-phase-earth fau lt at F: (a) defi n ition of fau lt; (b). equ iva lent circu it.
The fault current is
I = - I =
-
jV3 §
�
-c z+ + �
(5.40)
The constraints imposed by equations (5.35') and (5.34) indicate that there is
no zero-sequence network connection in the equivalent circuit and that the
positive- and negative-sequence networks are connected in parallel. Figure 5.16b
shows the equivalent circuit satisfying the above equations.
(iii) Phase-phase-earth fault Consider a fault defined by equation (5.24) and by
Figure 5.17.
The condition at the fault are expressed by the following equations:
v = 0
-
c
With Yb = 0 and Yc = 0, the symmetrical components of voltage are given
by equation
Therefore yO, Y+, and Y- equal Ya/3, and
v+ = v- = VO
- - -
(5.41)
Substituting (§ -
I+ �+) for Y+, Y-, and yOin equation (5.29) and
premultiplying both sides by [�rl, where
1
ZO 0 0
[ZO 0
�
r
-
[ �rl =
�
Z+
1
0
Z+
0
0 Z- 1
0 0
Z-
-

give
1
ZO
0
0
1
z+
0 0
0
0
1
z-
[
§ - L+ �+
l
E - /+ z+
- - -
E - /+ z+
- - -
1
ZO
0 0
[!1
0
1
0
z+
1
0 0
z-
Premultiplying both sides of equation (S.42) by the row matrix [ 1
recognizing that
we have
E
ZO
Z+ E E Z+ E
/+ -=- + -=- - /+ + -=- - /+ -=- = -=
- ZO Z+ - Z- - Z- Z+
and upon collecting terms we obtain
and
§
(�- + �O)
/+ = ------�----��----�
�+ �- + �+ �o
+ �- �o
E
[n
(S.42)
1 ] and
(S.43)
(S.44)
(S.4S)
Equations (S.41) and (S.4S) are the special equations for a double phase-to
earth fault. They are used with equation (S.29) and the symmetrical-component
relations to determine all the voltages and currents at the fault.
Equations (S.41) indicates that the sequence networks should be connected in
parallel (Figure S.17b), since the positive-, negative-, and zero-sequence voltages
are equal at the fault. Examination of Figure S. 17b shows that all the conditions
derived above forthe double-phase-to-earth fault are satisfied by this connection.
The diagram of network connections shows that the positive-sequence current
!..+ is determined by the voltage § impressed on �+in series with the parallel
combination of �- and �o
. The same relation is given by equation (S.4S).
(iv) Three-phase fault Consider a symmetrical three-phase fault defined by equa
tion (S.2S) and by Figure S.18.
!..a + � + & = O
Ya = Yb = Yc
259

F
a --�--------------
b --�--�---------- F
c --�--�--�------
(a) (b)
Figure 5 . 1 8. Three-phase fault at F: (a) defin ition of fau lt; (b) equ iva lent circuit.
In terms of symmetrical components
(5.46)
v+ = y- = yO = 0 (5.47)
There are no zero-sequence or negative-sequence emfvalues, andthe voltages
yO and Y- at the fault point F are each equal to zero. Therefore, zero- and
negative-sequence currents do not flow anywhere in the system. As expected,
only the positive-sequence network is involved. The fault is simulated by
shorting the fault point F+ and the reference bus N+of the positive-sequence
network (Figure 5.1Sb).
(v) Single-phase open circuitfault The single-phase open circuit fault is shown in
Figure 5. 19. Let �a' �b' and �c be series voltages drop in phases between FJ and
F2 in phases a, b, c, respectively.
I � I
(a)
(b)
Figure 5 . 1 9. Open circuit on phase "a": (a) circu it d iagram; (b) eq u iva lent circu it.

At the fault point, the boundary conditions are (Figure S.19a) [12]
!..a = 0
Hence, from equation (S.9):
and therefore
1
vO= - V
-a 3 -a
1
v+ = - V
-a 3 -a
1
v- = - V
-a 3 -a
1
v+ = v- = vO= - V
-a -a -a 3 -a
From equation (S. l 1a), for !..a = 0
!..a = L+ + C + LO= 0
(S.48)
(S.49)
(S.SO)
From equation (S.49), it can be concluded that the sequence networks are
connected in parallel, as shown in Figure S.19b.
FAULTSTHROUGH IMPEDANCE. All the faults discussed in the preceding sections consisted of
direct short circuits between phases and from one or two phases to earth. Most faults are the
result ofinsulator flashovers, where the impedance between the phase and earth depends on
the resistance of the arc, ofthe tower itself, and of the tower footing if ground wires are not
used. Tower-footing resistances form the major part of the resistance between phase and
earth and depend on the soil conditions [6].
The effect ofimpedance in the fault is found by deriving equations similar to those for
faults through zero impedance. Connections of the hypothetical stubs for faults through
impedance are shown in Figure S.20.
a ----�-----------------
b ----��------------
c ----��---------
(a)
�-------
I
I
I
I I
F ' II
8I
I
I
N' I
I I
L.. _______ J
(b)
Figure 5.20. Connection of sequence network to simulate a th ree-phase fault through imped
ance at point F.
261

A balanced system remains symmetrical after the occurrence of a three-phase fault
having the same impedance between each phase and a common point. Only positive
sequence currents flow.
With the fault impedance �f equal in all phases (Figure 5.20a), the voltage at the fault
locations is
and since only positive-sequence currents flow,
and
(5.51)
The sequence-network connection is shown in Figure 5.20b.
A formal derivation can be made for the single phase-to-earth and double phase-to
earth faults through impedance shown in Figures 5.21a and b [6].
• For a single phase-to-earthfault through �f (Figure 5.21a):
/+ = /- = /0
• For a double phase-to-earth fault through impedance �f (Figure 5.21b):
v+ = v-
1- 1-
I I
I I
F'
F-
J'

-

N ' N
°
- -
1------------1
: F> ,�---L--�,
�O
I
N 03Zr
l______ �---�---�
t
I
----r---J
!____________ J
(a) (b)
(5.52)
(5.53)
Figure 5.2 1 . Connections of the seq uence networks for: (a) single phase-to-earth fau lt through
�f; (b) double phase-to-earth fault through �f.

Figure 5.22. Con nection of the
seq uence networks to simu late a
phase-to-phase fault through
impedance.
• A phase-to-phase (phases b and c) fault through impedance �f is shown in
Figure 5.22.
The conditions at the fault are
�, � and L: bear the same relations to each other as in the phase-to-phase fault without
impedance. Therefore L+ = - C
The sequence components of voltage are given by
or
therefore
Since L+ = - C
and, upon substituting L b in (5.56), we obtain
v+ - v- = /+ Z
- - - - /
(5.54)
(5.55')
(5.55/1)
(5.56)
(5.57)
Equation (5.57) requires the insertion of �f between the fault points in the positive
and negative-sequence networks to fulfill the required conditions for the fault.
263

5.4 CALCULATION OF SHORT-CIRCUIT CURRENT COMPONENTS
3
In the short-circuit currents calculation in a node, the maximum and minimum values are of
interest ifthe thermal and mechanical stresses and the power quality are involved, respectively.
5.4.1 Initial Symmetrical Short-Circuit Current II:
Considering two cases [1]:
For the common case when �o > �+ = �-, the highest initial short-circuit current
will occur for the three-phase short circuit.
For short circuits near transformers with low zero-sequence impedance, �o may be
smaller than �+. In that case, the highest initial short-circuit current I�E2E will occur for a
phase-to-phase short circuit with earth connection (see Figure 5.23) for �-/ �+ = 1 and
�-/ �o > 1, where �- = �+.
5.4.1.1 Three-Phase Short Circuit. In general, the initial symmetrical short
circuit current l� shall be calculated using equation (5.58) with the equivalent voltage
source (cUn/v'3) at the short-circuit location and the short-circuit impedance
( �k = Rk +jXk):
(5.58)
The equivalent voltage source shall be introduced at the short-circuit location (see
Figure 5.6) with the factor c according to Table 5.1.
(i) Single-fedshort circuits For a far-from-generator short circuit fed from a single
source (see Figure 5.23a) the short-circuit current is calculated with equation
D
G
0)
G
0)
S A
T CD tdF
Uns
U T /"
'"
U",
LV T ..... HV
T � t /
«
;
)"
S
�k3
L
(a)
Un F
�k3
L
Un F
(b)
L
�k3 (c)
Un F
Figure 5.23. Examples of single-fed short circu its: (a) short circuit fed from a network feeder via a
transformer; (b) short circu it fed from one generator (without u n it transformer); (c) short circuit
fed from one power station u n it generator and u n it transformer with or without on-load tap
changer, . (Reprinted with perm ission from IEC 60909[1))
3 This section is partially reproduced from the IEC 60909 Standard with permission by International Electro
technical Commission [IJ .

CALCULATION OF SHORT-CIRCUIT CURRENT COMPONENTS
(5.58), with
where
Rk = RSI + RTK + RL
Xk = XSI + XTK + XL
Rk and Xk are the sum of series-connected resistances and reactances of the positive
sequence system, respectively (Figure 5.23a);
RL is the line resistance for a conductor temperature of 20°C, when calculating the
maximum short-circuit currents;
�TK = RTK +jXTK = KT(RT +jXT) is the corrected transformer impedance;
KT is the impedance correction factor, KT = 0.95 .
. XT
Cmax
o 6
' where XT is the
1 + . . XT
relative reactance of the transformer XT =
(
2 ).
VrT/SrT
Resistances Rk less than 0.3 . Xk may be neglected. The impedance of the
network feeder �Sk = RSI +jXsl is referred to the voltage ofthe transformer side
connected to the short-circuit location (F).
For the examples in Figures 5.23b and c, the initial symmetrical short circuit is
calculated with the corrected impedances of the generator and the power station
unit in series with a line impedance �L = RL +jXL. The impedances of
generator (G), network transformer (T), and power station units (S) shall be
multiplied with the impedance correction factors KG, KT,and Ks, respectively,
Kso when calculating short-circuit currents with the equivalent voltage source at
the short-circuit location according to this Standard (See IEC 60909-4).
The short-circuit impedances for these examples are given by the following
equations:
�k = �Gk + �L = KG (RG +jX�
)
+ �L for Figure 5.23b (5.59')
�k = �s + �L = Ks (N; �G + �THv
)
+ �L for Figure 5.23c (5.59")
where �Gk shall be determined from equation
with
�s from equation
Vn Cmax
KG = - . -:--
---::c
c;;-
:--
VrG 1 + X� sin !PrG
'
(5.60)
(5.61')
(5.62')
265

266
with
where
Uns UrTLV
Kso = ----,----,- . -- .
UrG(1 + PG) UrTHV
(5.61/1)
(5.62/1)
PG is the range of generator voltage regulation andPT is the range oftransformer voltage
adjustment;
Cmax is the factor voltage for the calculation of maximum short-circuit currents.
The generator impedance shall be transferred to the high-voltage side
using the rated transformation ratio Nr• The unit transformer impedance
�THV = RTHV +jXTHV according to the equations (5.63') and (5.63/1) without
KT is referred to the high-voltage side:
where
ZT = � '
U;T
100% SrT
RT = � '
U;T
100% SrT
XT = Jz� - R�
(5.63')
(5.63/1)
(5.63/11)
UKr is the short-circuit voltage at rated current in percent;
URr is the rated resistive component of the short-circuit voltage in percent;
UrT is the rated voltage of the transformer on the high-voltage or low-voltage side;
SrT is the rated apparent power of the transformer.
(ii) Short circuits fed from nonmeshed networks When there is more than one
source contributing to the short-circuit current, and the sources are unmeshed,
as shown in Figure 5.24, the initial symmetrical short-circuit current I� at the
G
S
1;T J'
4M
LpT LpM
IbT ibM
LkT IkM
M
3-
k3
Figure 5.24. Example of a non-
F meshed network.

---<I>---_-�_ s
(a)
Figure 5.25. A meshed network fed from several sou rces: (a) system d iagram; (b) equ iva lent
circuit d iagram for the ca lcu l ation with the equ iva lent voltage sou rce cUn/V? at the short-circuit
location (Repri nted with perm ission from IEC 60909[1]).
short-circuit location F is the sum of the individual branch short-circuit currents.
Each branch short-circuit current can be calculated as an independent single
source three-phase short-circuit current in accordance with equation (5.58).
The initial short-circuit current at the short-circuit location F is the phasor,
which is sum of the individual partial short-circuit currents [1]:
/" - """"' /"
k - L-- ki (5.64)
The short-circuit current at the short-circuit location F is considered, with
good accuracy, as being the sum of the absolute values of the individual partial
short-circuit currents.
(iii) Short circuits in meshed networks In meshed networks, such as those shown in
Figure 5.25, it is generally necessary to determine the short-circuit impedance
�k = �+ by network reduction (series connection, parallel connection, and
delta-star transformation, for example) using the positive-sequence short-circuit
impedances of electrical equipment.
The impedances in systems connected through transformers to the system, in
which short circuit occurs, have to be transferred by the square of the rated
transformation ratio. Ifthere are several transformers with slightly differing rated
transformation ratios (NrTI, NrT2, NrTn), in between two systems, the arithmetic
mean value can be used.
The initial symmetrical short-circuit current shall be calculated with the equiv
alent voltage source cUn/V3 at the short-circuit location using equation (5.58).
5.4.1.2 Phase-to-Phase Short Circuit. In the case of a phase-to-phase short
circuit, according to Figure 5.1b the initial short-circuit current shall be calculated by
(5.65)
267

During the initial stage of the short circuit, the negative impedance is approximately
equal to the positive-sequence impedance, independent of whether the short circuit is a
near-to-generator or a far-from-generator short circuit. Therefore in equation (S.6S), it is
possible to introduce �- =
�+.
Only during the transient of the steady-state stage, the short-circuit impedance �- is
different from �+ if the short circuit is a near-to-generator short circuit [1].
5.4.1.3 Phase-to-Phase Short Circuit with Earth Connection. To calculate the
initial symmetrical short-circuit currents it is necessary to distinguish between the currents
r�2Ea' r�2Eb and r�E2E (see Figure S. lc).
For far-from-generator short circuits, �- is approximately equal to �+. Ifin this case
�o is less than �- the current r�E2E in the phase-to-phase short circuit with earth
connection generally is the largest ofall initial symmetrical short-circuit currents r�, r�2'
I" I"
-k2E' and -kl·
The equations (S.66/) and (S.66") are given for the calculation of r�2Ea and r�2Eb in
Figure S. lc:
ZO _ a Z-
['k2Ea == -jcUn - - -
�+ �- + �+ �o +
Z - ZO
ZO _ a2 Z-
r'k2Eb = -jcUn - - -
�+ �- + �+ �o + �- �o
(S.66/)
(S.66")
The initial short-circuit current r�2E flowing to earth and/or earthed wires,
(Figure S. lc), is calculated by
I"
-kE2E = - �+ �- + �+ �o +
Z- ZO
(S.66///)
For a far-from-generator short circuit with Z- =
�+, these equations lead to the
absolute values:
(S.67/)
(S.67///)
I" _ .,--_
V3
_
3_
c_
U
_
n�
kE2E -
I
�+ + 2 �ol (S.67///)
5.4.1.4 Phase-to-Earth Short Circuit. The initial phase-to-earth short-circuit
current Gl in Figure S.la shall be calculated by
/"
V3cUn
-kl - �+ + �- +
ZO
(S.68)

For a far-from-generator short circuit with Z+ = �-, the absolute value is
calculated by
(S.68')
If �o is less than �- = �+, the initial phase-to-earth short-circuit current 1�1 is
larger than the three-phase short-circuit current l�, but smaller than I�E2E' However, 1�1 will
be the highest current to be interrupted by a circuit breaker if 1 .0 > �oI �+ > 0.23 [1].
5.4.2 Peak Short-Circuit Current ip
5.4.2.1 Three-Phase Short Circuit.
(i) Short circuits in nonmeshed networks For three-phase short circuits fed from
nonmeshed networks as in Figures S.23 and S.24 the contribution to the peak
short-circuit current from each branch can be expressed by
(S.69)
The factor K for the RIX or XIR ratio shall be obtained from Figure S.26 or
calculated by the following expression [1]:
K = 1 .02 + 0.98 e-
3R/X (S.70)
Equations (S.69) and (S.70) presume that the short circuit starts at zero
voltage, and that ip is reached approximately after one half-cycle. For a
synchronous generator use RGf (fictions resistance RGf = (O.OS...O.IS)x� in
function of UrG and SrG).
The peak short-circuit current ip at a short-circuit location F, fed from sources
that are not meshed with one another, (Figure S.24), is the sum ofthe branch peak
short-circuit currents:
2.0
i 18
I 16
J(
1.4
1.2
10
1

'
'"r-....
......r-------
o 0.2 0.4 0.6 0.8
---
1.2
RlX
2.0
1 18
J(
16
1.4
/
1.2
10
V
0.5 1 2
V
/""e--
.-/
V
10 20 50 100 200
XIR
Figure 5.26. Factor K for series circuit as a function of ratio R/X or X/R.
(S.71)
269

For the example from Figure S.24 results:
(S.72)
(ii) Short circuits in meshed networks When calculating the peak short-circuit
current ip in meshed networks, equation (S.69) shall be used with K determined
using one of the following methods (a, b, or c) [1].
(a) Uniform ratio R/X or X/R
Forthismethodthe factor K is determined from Figure S.26 taking the smallest ratio of
R/X or the largest ratio of X/R of all branches of the network, having values between 1.6
and 1 .9 in high voltage networks and 1 .3 and 1.S in low voltage networks.
It is only necessary to choose the branches, which carrypartial short-circuit currents at
the nominal voltage corresponding to the short-circuit location and branches with
transformers adjacent to the short-circuit location. Any branch may be a series combination
of several impedances.
(b) Ratio R/X or X/R at the short-circuit location
For this method the factor K is multiplied by a factor LIS to cover inaccuracies caused
by using the ratio R/X from a network reduction with complex impedances:
(S.73)
As long as R/X remains smaller than 0.3 in all branches, it is not necessary to use the
factor LIS. It is not necessary for the product LIS · Kb to exceed 1 .8 in low-voltage
networks or to exceed 2.0 in medium- and high-voltage networks.
The factor Kb is found from Figure S.26 for the ratio Rk/Xk given by the short-circuit
impedance �k = Rk +jXk at the short-circuit location F, calculated for frequency
f = SO Hz (or 60 Hz).
(c) Equivalentfrequency fe
An equivalent impedance �c of the system as seen from the short-circuit
location is calculated assuming a frequencyfe = 20 Hz (for a nominal frequency
off = SO Hz) orfc = 24 Hz (for a nominal frequency off = 60 Hz). The R/X
or X/R ratio is then determined according to equation (S.74):
R
X
X
R
where
Re fe
Xc f
Xc f
_ . -
Re fe
(S.74 a)
(S.74 b)
�e = Re + jXe is the equivalent impedance of the system as seen from the short-circuit
location for the assumed frequency fe;
Re is the real part of �e (Re is generally not equal to the R at nominal frequency);
Xc is the imaginary part of �e (Xc is generally not equal to the X at nominal frequency).
The factor K is found from Figure S.26 using the R/X or X/R ratio from
equation (S.74a), or with equation (S.74b).

Method (c) is recommended in meshed networks. When using this method in
meshed networks with transformers, generators, and power station units, the
impedance correction factors KT, KG, and Ks, respectively, Kso shall be
introduced with the same values as for the 50 Hz (or 60 Hz) calculations.
5.4.2.2 Phase-to-Phase Short Circuit. For a phase-to-phase short circuit the peak
short-circuit current can be expressed by
(5.75)
The factor K shall be calculated according to 5.4.2. 1, case (i) or to 5.4.2.2, case (ii),
depending on the system configuration. For simplification, it is permitted to use the same
value of K as for the three-phase short circuit.
When �+ = �-, the phase-to-phase peak short-circuit current ip2 is smaller than the
three-phase peak short-circuit current ip as shown in equation:
. J3 .
Ip2 =
T Ip (5.76)
5.4.2.3 Phase-to-Phase Short Circuit with Earth Connection. For a phase-to
phase short circuit with earth connection, the peak short-circuit current can be expressed by
(5.77)
The factor K shall be calculated according to 5.4.2. 1, case (i) or to 5.4.2. 1, case (ii),
value of K as for the three-phase short circuit.
It is only necessary to calculate ip2E, when �o is much less than �+ (less than about
1/4 of �+) [1].
5.4.2.4 Phase-to-Earth Short Circuit. For a phase-to-earth short circuit, the peak
short-circuit current can be expressed by
(5.78)
The factor K shall be calculated according to 5.4.2. 1, case (i) or to 5.4.2. 1, case (ii),
value of K as for the three-phase short circuit [1].
5.4.3 DC Component of the Short-Circuit Current
The maximum decaying aperiodic component id.c. of the short-circuit current as shown in
Figures 5.2a and 5.2b may be calculated with sufficient accuracy by equation
where
.
_ (;;.
21" -2nft R/X
{d.c. - V L k e
l� is the initial symmetrical short-circuit current;
f is the nominal frequency;
t is the time;
(5.79)
R/X is the ratio according to 5.4.2. 1 orthe ratios according to the methods (a) and (c).
271

T A B L E 5.4. Ratio of equivalent and nominal frequencies for various fxt values
f x t
fclf
<1
0.27
<2.5
0.15
<5
0.092
<12.5
0.055
The correct resistance RG of the generator armature should be used and not RGf.
For meshed networks, the ratio RjX or XjR is to be determined by the method (c) in
5.4.2. 1, case (ii). Depending on the productfx t wheref is the frequency and t is the time,
the equivalent frequency fe should be used (Table 5.4) [l].
5.4.4 Symmetrical Short-Circuit Breaking Current lb
The breaking current at the short-circuit location consists in general of symmetrical current
h and an aperiodic current id.c. at the time tmin according to equation (5.79).
Note: for some near-to-generator short circuits the value of id.e. at tmin may exceed the
peak value of h and this can lead to missing current zeros [1].
5.4.4.1 Far-trom-Generator Short Circuit. For far-from-generator short
circuits, the short-circuit breaking currents are equal to the initial short-circuit
currents [1]:
h = I� (5.80a)
h2 = I�2 (5.80b)
(5.80c)
(5.80d)
5.4.4.2 Near-to-Generator Short Circuit.
SINGLE-FED THREE-PHASE SHORT CIRCUIT. For a near-to-generator short circuit, in the case
of a single fed short circuit as in Figure 5.23b and 5.23c or from nonmeshed networks as in
Figure 5.24, the decay to the symmetrical short-circuit breaking current is taken into
account by the factor f1 (according to equation (5.82)):
(5.81)
The factor f1 depends on the minimum time delay tmin and the ratio I�GjIrG, where IrG
is the rated generator current. The values of f1 in equation (5.82) apply if synchronous
machines are excited by rotating exciters or by static converterexciters (provided, for static
exciters, the minimum time delay tmin is less than 0.25 s and the maximum excitation
voltage is less than 1.6 times rated load excitation voltage). For all other cases take f1 = 1,
if the exact value is unknown [1].
When there is a unit transformer between the generator and the short-circuit location,
the partial short-circuit current I�s at the high-voltage side of the unit transformer (in
Figure 5.23c) shall be transferred by the rated transformation ratio to the terminal of the

l0.
9 ��t------+------t
11
0.8 1--+--+-+-+�:-+-�::--+--1---1
o.7 1--+--+-+--I---'liF;..;;p�+--I---1
0.6 1--+--+-+--1--+--+-+--=:::'1-:::--1
Figure 5.27. Factor J.L for calculation
of short-circu it breaking cu rrent lb (1).
generator I�G = Nrl�s before calculating f..L, using the following equations [1]:
f..L = 0.84 + 0.26e-O.26 · /�G/lru for tmin = 0.02 s
f..L = 0.71 + 0.51e-0.3o . /�G/lru for tmin = 0.05 s
f..L = 0.62 + 0.72e-0.32 . /�G/lru for tmin = O. lO s
f..L = 0.56 + 0.94e-O.38 · /�G/lru for tmin :2: 0.25 s
(5.82)
If l�dIrG is not greater than 2, apply f..L = 1 for all values ofthe minimum time delay
tmin' The factor f..L may also be obtained from Figure 5.27. For other values of minimum
time delay, linear interpolation between curves is acceptable.
5.4.5 Steady-State Short-Circuit Current lk
The calculation of the steady-state short-circuit current lk is less accurate than the
calculation of the initial short-circuit current I�.
5.4.5.1 Three-Phase Short Circuit of One Generator or One Power Station
Unit. For near-to-generator three-phase short circuits fed directly from one synchronous
generator or one power station unit only, according to Figures 5.23b or 5.23c the steady
state short-circuit currentlk depends on the excitation system, the voltage regulator action,
and saturation influences.
Synchronous machines (generators, motors, or compensators) with terminal-fed static
exciters do not contribute to Ik in the case of a short circuit at the terminals of the machine,
but they contribute to lk ifthere is an impedance between the terminals andthe short-circuit
location. A contribution is also given if, in case of a power station unit, the short circuit
occurs on the high-voltage side of the unit transformer (Figure 5.23c).
• Maximum steady-state short-circuit current. For the calculation of the maximum
steady-state short-circuit current, the synchronous generator may be set at the
maximum excitation:
hmax = A.maxlrG (5.83)
273

274
If..
2.8
2.6
2.4
2.2
2.0
1 .8
1 .6
1 .4
1 .2
1 .0
0.8
0.6
0.4
0.2
�
I'
-J-
I
I--
,
f..,,=
�;...--
/-,
p-
f.."""
--
X
d Sal
1.2
--I-- 1.4
- �
1.6
1.8
"""- 2.0
2.2
0
0 1 2 3 4 5 6 7 8 9
Three-phase short circuit 4c/I,c
(a)
If..
2.8
2.6
2.4
2.2
2.0
1 .8
X
d Sal - 1.2
f..m� ./i-"'" -- 1.4
��
1.6
-- 1.8
�l:;::::; 2.0
2.2
�
1 .6
1 .4 I
1 .2
I
,
1 .0
0.8
0.6
0.4
r-!-- f..m,,,
I --
,
0.2
0
0 1 2 3 4 5 6 7 8 9
Three-phase short circuit I;c/I,c
(b)
Figure 5.28. Factors "min and "max for cyl indrical rotor generators: (a) series 1; (b) series 2.
For static excitation systems fed from the generator terminals and a short circuit at
the terminals, the field voltage collapses as the terminal voltage collapses, therefore
take "min = "max = 0 in this case.
- "max may be obtained from Figures 5.28 or 5.29 for cylindrical rotor generators or
salient-pole generators. The saturated reactance Xdsat is the reciprocal ofthe saturated
no-load short-circuit ratio;
- "max-curves of series 1 are based on the highest possible excitation voltage
according to either 1 .3 times the rated excitation at rated apparent power and power
factor for cylindrical rotor generators (Figure 5.28a) or 1 .6 times the rated excitation
voltage for salient-pole generators (Figure 5.29a);
If..
5.5
5.0
4.5
4.0
Xd sat
f..",," _ o.6
3.5
3.0
2.5
2.0
1.5
1 .0
0.5
....-
/ -- 0.8
- 1.0
L��-::- 1.2
Ie::�
1.7
2.0
I �
I f..m",
---,�
�
I
,
0
0 1 2 3 4 5 6 7 8 9
(a)
If..
5.5
5.0
4.5
4.0
3.5
3.0
2.5
2.0
�'0 6
f..",," ./'
V -- 0.8
l-V - 1.0
�r- - 1.2
- 1.7
}'r"" 2.0
r
1 .5
1 .0
I
L f..m",
---, ---
I
0,5 ,
0
0 1 2 3 4 5 6 7 8 9
(b)
Figure 5.29. Factors "min and "max for sa l ient-pole rotor generators: (a) series 1; (b) series 2.

- �max--curves of series 2 are based on the highest possible excitation voltage
according to either 1 .6 times the rated excitation at rated apparent power and power
factorfor cylindrical rotor generators (Figure 5.28b), or 2.0 times the rated excitation
voltage for salient pole generators (Figure 5.29b).
Note: When calculating maximum short-circuit currents, it is necessary to
introduce the following conditions:
- Voltage factor Cmax according to Table 5.1 shall be applied for the calculation of
maximum short-circuit currents in the absence of a national standard.
- Choose the system configuration and the maximum contribution from power plants
and network feeders, which lead to the maximum value of short-circuit current at the
short-circuit location, or for accepted sectioning of the network to control the short
circuit current.
- When equivalent impedances �s are used to represent external networks, the
minimum equivalent short-circuit impedance shall be used, which corresponds to
the maximum short-circuit current contribution from the network feeders.
- Motors shall be included.
- Resistance RLof overhead lines and cables are to be introduced at a temperature of
20°C.
• Minimum steady-state short-circuit current. For the minimum steady-state short
circuit current in the case of a single-fed short circuit from one generator or one
power station unit according to Figures 5.23b and 5.23c, constant no-load excitation
(voltage regulator not being effective) of the synchronous machine is assumed:
Ik min = �minIrG (5.84)
�min may be obtained from Figures 5.28 and 5.29. In the case ofminimum steady-state
short circuit introduce c = Cmin according to Table 5.l.
The calculation of the minimum steady-state short-circuit current in the case of a
near-to-generator short circuit, fed by one or several similar and parallel working
generators with compound excitation, is made as follows:
CminUn
Ik min = ----'7======
V3JR�+X�
For the effective reactance of the generators, introduce
UrG
XdP = ---
V3hp
(5.85)
(5.86)
where IkP is the steady-state short-circuit current of a generator at a three-phase
terminal short circuit. The value should be obtained from the manufacturer.
Note: When calculating minimum short-circuit currents, it is necessary to
introduce the following conditions:
- Voltage factor Cmax for the calculation of minimum short-circuit currents has to be
applied according to Table 5.1.
275

- Choose the system configuration and the minimum contribution from power stations
and network feeders, which lead to a minimum value of short-circuit current at the
short-circuit location.
- Motors are to be neglected.
- Resistances RL of overhead lines and cables, line conductors, and neutral conductors,
are to be introduced at a higher temperature.
where
- RL20 is the resistance at a temperature of 20°C.
- 8e is the conductor temperature in degree Celsius at the end of the short-circuit
duration.
- a is a factor equal to 0.004/K, valid with sufficient accuracy for most practical
purposes for copper, aluminum, and aluminum alloy.
- Amax is curves of series 1 or 2 may also be applied in the case of terminal-fed static
exciters, if the short circuit is at the high-voltage side of the unit transformer of a
power station unit or in the system, and if the maximum excitation voltage is chosen
with respect to the partial breakdown of the terminal voltage of the generator during
the short circuit.
5.4.5.2 Three-Phase Short Circuit in Nonmeshed Networks. In the case of a
three-phase short-circuit in nonmeshed networks, as in Figure 5.24, the steady-state short
circuit current at the short-circuit location can be calculated by the summation of the
individual steady-state short-circuit current contributions:
(5.87)
For example from Figure 5.24 we obtained:
(5.88)
where A (Amax or Amin) is found from Figures 5.28 and 5.29. IrG! is the rated current of the
generator transferred to the high voltage side of the unit transformer in Figure 5.24.
In the case of network feeders or network feeders in series with transformers (see
Figure 5.24), h = l� is valid (far-from-generator short circuit).
5.4.5.3 Three-Phase Short Circuit in Meshed Networks. In meshed networks
with several sources, the steady-state short-circuit current may be calculated approxi
mately by
hmax = l�max M
hmin = l�min
l�max = l� is found according to (5.89) and l�min according to (5.90).
(5.89)
(5.90)

Equations (S.89) and (S.90) are valid in the case of far-from-generator and in the case
of near-to-generator short circuits.
5.4.5.4 Unbalanced Short Circuits. In all cases for steady-state unbalanced short
circuits, the flux decay in the generator is not taken into account and the following
equations should be used:
h2 = 1�2 (S.91a)
h2E = 1�2E (S.91b)
IkE2E = I�2E (S.91c)
hi = 1�1 (S.91d)
In the case of minimum steady-state short circuits introduce C = Cmin according to
Table S.l.
5.4.6 Applications
Application 1. Single-fed short circuit
Given the radial electrical network from Figure S.30, fed from only one source, determine
the rms and the peak values of the short-circuit current caused by the symmetrical three
phase fault indicated by k3. The equipment parameters are also presented in Figure S.30.
Solution
Since a three-phase fault is concerned, the pOSItIve-sequence network is used. The
equivalent positive-sequence network of the network from Figure S.30 is presented in
Figure S.31.
�
T1A 1 l 0 kV k3
�
/ = 30 km
. L..l
1--+-
-
-
-
-
-
-
-
--1 T,
rt= 0.034 Q/km - /::,.
Uns = 400 kV
S" = 4000 MVA
RsIXs = 0. 1
xt= 0.33 Q/km SnT = 250 MVA
!!.p".n = 600 kW
u" = 14%
Figure 5.30. Rad ial electric network fed from one sou rce.
(a) (b)
Figure 5.3 1 . Equ iva lent circu it: (a) positive-sequence network; (b) red uced positive-sequence
network.
277

Calculation can be performed either in absolute units or in per units.
Calculation using absolute units
All parameters are referred to the nominal voltage of the network in which the fault
occurs, that is UnK= 110kV.
(a) Calculation of the network parameters
- network feeder parameters
xt = U�s . (UnK)2 = U�K = 1102 =3.0250
Sse UnS Sse 4000
+ Rs +
Rs =- . Xs =0.1 . 3.025 =0.30250
Xs
- electrical line parameters
Rt =r6 . I .
xt =X6 . I .
- transformers parameters
(UnK)2 =0.034 . 30 .
UnL
(UnK)2 =0.33 . 30 .
UnL
G��Y=0.07710
(110)2
400 =0.74870
10-3 =0.11620
z+ =� . U�K =� . 1102 =6.7760
Tl,2 100 SnT 100 250
X+ -
Tl,2 - ( +
)2 ( +
)2 _
vi
2 2 _
ZTl,2 - RTl,2 - 6.776 -0.1162 -6.7750
(b) Calculation of the short-circuit impedance
z+ . Z+
�: = Z++ �t+ - r1 -� = (0.3025+j3.025) +(0.0771 +jO.7487)+
- 5
�Tl + �T2
(0.1162+j6.775) . (0.1162+j6.775) =(0.4377 '7.1612)0
+(0.1162+j6.775)+(0.1162+j6.775) +J
zt = J(Rt)2+ (Xt)
2 = V0.43772+7.16122 =7.17460
(c) Calculation of the initial symmetrical short-circuit current I� =/+ (5.58)
I
"
= CUnK = 1.1 · 110 =9.737kA
k v'3zt v'3 . 7.1746
(d) Calculation of the peak short-circuit current ip

- factor K (5.70)
3 0.4377
K = 1 .02 + 0.98e-
3Rk/Xk = 1 .02 + 0.98e- 7.1612 = 1 .836
- peak short-circuit current (5.69)
ip = KhI� = 1 .836 . h . 9.737 = 25.28 kA
Calculation in per units
For calculation in per-unit, definition of the base quantities is necessary:
- base voltage (nominal voltage of the network at the fault location) Ub = UnK =
1 1O kV;
- base apparent power: Sb = 100MVA.
The base current and base impedance are therefore achieved:
Sb 100
h = -- = = 0.525 kA
V3Ub V3 . 1 10
U2 1 102
Zb =
_
b = -- = 121 0
Sb 100
Sb 100
xt = - = -- = 0.025 p.u.
Sse 4000
rt =
Rs . xt = 0. 1 . 0.025 = 0.0025 p.u.
Xs
- electrical line parameters
+ + Sb 100
rL = ro . I . -2
- = 0.034 . 30 . --
2 = O.00064 p.u.
UnL 400
Sb 100
xt = x6 . I . -2
- = 0.33 . 30 . --
2 = 0.00619 p.u.
UnL 400
- transformers parameters
+ Sb -
3 100 _3
rT1,2 = D.Pse,n . -
S
2 . 10 = 600 . --
2 . 10 = 0.00096 p.u.
nT 250
+ = � . � = � .
100
= 0 056 U
ZTI,2 100 SnT 100 250
. p. .
41,2 = (Zfl,2)
2
- (rfl ,2)
2
=
VO.0562 - 0.000962
= 0.05599 0
279

(b) Calculation of the short-circuit impedance
z+ . z+
It = I� + It + -
II -I2 = (0.0025+jO.025) +(0.00064+jO.00619)+
ITI + IT2
(0.00096+jO.05599) . (0.00096+jO.05599) = (0.00362 "0.05919) .u.
+(0.00096+jO.05599) +(0.00096+jO.05599) +) p
zt = V(rt)2+ (xt)2 = VO.003622+0.059192 = 0.05929n
(c) Calculation of the initial symmetrical short-circuit current i� = i+ (5.58)
.11
c 1.1
lk =
zt
=
0.05929 = 18.552p.u.
l� = i� . h = 18.552 . 0.525 = 9.737kA
(d) Calculation of the peak short-circuit current ip
- factor K (5.70)
K = 1.02+0.98e-3rk/Xk = 1.02+0.98e-3 ggg�1§ = 1.836
- peak short-circuit current (5.69)
ip = K . J2 . l� = 1.836 . J2 . 9.737 = 25.28kA
Application 2. Short circuit fed from nonmeshed network
Given the nonmeshed electrical network from Figure 5.32, fed from two sources,
determine the rms and the peak values of the short-circuit current caused by the
symmetrical three-phase fault indicated by k3. The equipment parameters are also
presented in Figure 5.32
Solution
Similar to the previous case, the positive-sequence network will be used (Figure 5.33).
110 kV
1, = IO km
rt, = 0. 193 Q/km
Uns = 1 1 0 kV xt, = 0.386 Q/km
S", = 2000 MVA
RslXs = 0. 1
k3
I, = 20 km
rt, = 0. 122 Q/km
xt, = 0.372 Q/km SnT = 25 MVA
IlP",n = 140 kW
U
",
= 1 1%
G
SnO = 21 MVA
� UnO = 10.5 kV
x "
d = 22%
RnO = 0.0809 Q
Figure 5.32. Non meshed electrica l network fed from two sou rces.

Figure 5.33. Equ ivalent circu it of the
nonmeshed electric network from Figure
5.32: (a) positive-seq uence network; (b)
red uced positive-sequence network.
The calculation is performed in absolute units. Therefore, all parameters are calculated
referred to the nominal voltage of the network in which the fault occurs, that is
UnK = 110kY.
xt = U�s . (UnK)2 = U�K =
1102 = 6.05f!
Sse UnS Sse 2000
+ Rs +
Rs = - . Xs = 0.1 · 6.05 = 0.605f!
Xs
- generator parameters
Rt; = RnG . (�:�)2 = 0.0809 . (11�.�)2 = 8.8788f!
X" U2 22 1102
x
+
= _
d
.
�
= - . -- = 126.7619f!
G 100 SnG 100 21
- electrical lines parameters
- transformer parameters
Rtl = r61 . II = 0.193 . 10 = 1.93f!
Xtl = X61 . II = 0.386 . 10= 3.86f!
Rt2 = r62 . 12 = 0.122 . 20= 2.44f!
xt2 = xt2 . 12 = 0.372 . 20= 7.44f!
140 . ;102 . 10-3 = 2.7104f!
25
z+
=
�
. U�K=� . 1102 = 53.24f!
T 100 SnT 100 25
Xf = J(Zf)2- (Rf)2 = V53.242-2.71042 = 53.171f!
281

(b) Calculation of the short-circuit impedances
�tt = �t + �tl = (0.605 +)6.05) + (1.93 +)3.86) = (2.535 +)9.91) n
ztt = J(Rtl)2 + (xtt)2 = )2.5352 + 9.912 = 10.2291 n
+ �6 + �� + �t2 = (8.8788 +)126.7619) + (2.7104 +)53.171)+
�k2 =
+(2.44 +)7.44) = (14.0292 +)187.3729) n
ztz = J(Rt2
)2 + (xtz)2 = )14.02922 + 187.37292 = 187.8973 n
z+ =
�tt . �0.
=
(2.535 +)9.91) . (14.0292 +)187.3729)
= (2.3209 ·9.4268) n
-k
�tt + �t2
(2.535 +)9.91) + (14.0292 +)187.3729)
+}
zt = J(Rt)2 + (Xt)2 = )2.32092 + 9.42682 = 9.7083 n
Calculation of the initial symmetrical short-circuit currents (5.58)
I" =
1+ =
CUnK
=
1.1 · 110 =
6.829kA
kl 1
v'3Ztl v'3 . 10.2291
I" =
1+ =
CUnK
=
1.1 · 110 =
0.372kA
k2 2 v'3Zt2 v'3 . 187.8973
I" - /+ - CUnK _
1.1 · 110 -7 196kA
k - -
v'3zt
-
v'3 . 9.7083 - .
Calculation of the peak short-circuit current
- factor K (5.70)
2.535
-3·
KI =
1.02+ 0.98c3Rkl/Xkl =
1.02+ 0.98e 9.91 =
1.475
14.0292
-3·
K2 =
1.02+ 0.98e-3Rk,jXk2 =
1.02+ 0.98e 187.3729 =
1.803
- peak short-circuit currents (5.69)
ipl =
KI . y'2 . I�l =
1.475 . y'2 . 6.829 =
14.245kA
ip2 =
K2 . y'2 . I�2 =
1.803 . y'2 . 0.372=
0.948kA
ip =
ipl + ip2 =
9.616+ 0.67 =
15.193kA
Application 3. Short circuit in meshed network
Given the meshed electrical network from Figure 5.34, determine the rms and the peak
values of the short-circuit current caused by a symmetrical three-phase fault indicated by

Uns = 1 10 kV
S" = 2000 MVA
Rs/Xs = O. 1
z
O
/z+ = 1.5
r;'
.3 = 0. 122 nlkm
X;2,J = 0.402 nlkm
rg'
.3 = 0.325 nlkm
XS,.3 = 1 .306 nlkm
r;, = 0.049 Q/km
x;, = 0.194 nlkm
rg, = 0.254 Q/km
x8, = 1 . 132 nlkm
I, = 20 km
110 kV
110 kV
SnT' = 16 MVA
!1P",n = 94 kW
u" = 1 1%
Un.nJUnMrn = 110/22 kV
SnT2� = 63 MVA
= 265 kW
20 kV Usc = 12%
k3, k1
G
SnO = 15 MVA
UnO = 6.3 kV
� X "d = 12%
X- = 1 8%
RnO = 0.0617 Q
Figure 5.34. Short circuit fed from a meshed electrica l network.
k3 and a single-phase fault indicated by kl, respectively. The equipment parameters are
also shown in Figure 5.34
Solution
Figure 5.35 shows the positive-, negative-, and zero-sequence network of the meshed
electrical network, used for three-phase and single-phase short-circuit current calculation.
W (b)
z+
-G
(e)
Figure 5.35. Equ iva lent circu it of the meshed electrical network: (a) positive-sequence network;
(b) negative-seq uence network; (c) zero-seq uence network.
283

F+
z:
-5 Z; F+
��1+
Ii
N+ N+
Figure 5.36. Red uced eq u ivalent positive-sequence network.
The calculation is performed in absolute units. Therefore, all parameters are referred to
the nominal voltage of the network in which the fault occurs, that is UnK = 1 10 kY.
(i) Reducing the positive-sequence network
Figure 5.36 illustrates the reduced equivalent positive-sequence network.
xt =
U�s . (UnK)2 =
U�K =
202
= 0.2 f!
Sse UnS Sse 2000
+ Rs +
Rs = - . Xs = 0. 1 · 0.2 = 0.02 f!
Xs
- generator parameters
+ _ (Unk)2 _ ( 20 )2 _
RG -RnG · - - 0.0617 · - - 0.2239 f!
UnG 10.5
X+ -
X�
. U�K =
� .
202 = 3.2 f!
G -
100 SnG 100 15
- electrical lines parameters
(UnK)2 = 0.194 . 10 . (�)2 = 0.0641 f!
UnL1 1 10
R+ - r+ I
L2,3 - 02,3 . 2,3 .
xt2,3 = X62,3 . 12,3 .
( UnK )2 = 0. 122 . 15
UnL2,3
( UnK )2 = 0.402 . 15
UnL2,3
(20 )2
1 10
= 0.0605 f!
C2
1
0
0
r= 0. 1993 f!

H7 SHORT-CIRCUIT EMERSON EDUARDO RODRIGUES.pdf

H7 SHORT-CIRCUIT EMERSON EDUARDO RODRIGUES.pdf

Recommended

Recommended

More Related Content

Similar to H7 SHORT-CIRCUIT EMERSON EDUARDO RODRIGUES.pdf

Similar to H7 SHORT-CIRCUIT EMERSON EDUARDO RODRIGUES.pdf (20)

More from EMERSON EDUARDO RODRIGUES

More from EMERSON EDUARDO RODRIGUES (20)

Recently uploaded

Recently uploaded (20)

H7 SHORT-CIRCUIT EMERSON EDUARDO RODRIGUES.pdf