MODULE-2

SYMMETRICAL FAULT ANALYSIS: Introduction, Transient on a Transmission Line, Short

Circuit of a Synchronous Machine(On No Load), Short Circuit of a Loaded Synchronous
Machine, Selection of circuit breakers.

Introduction:

Normally, a power system operates under balanced conditions. When the system becomes
unbalanced due to the failures of insulation at any point or due to the contact of live wires,
a short–circuit or fault, is said to occur in the line. Faults may occur in the power system
due to the number of reasons like natural disturbances (lightning, high-speed winds,
earthquakes), insulation breakdown, falling of a tree, bird shorting, etc.

Faults that occurs in transmission lines are broadly classified as

• Symmetrical faults
• Unsymmetrical faults

Symmetrical faults

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In such types of faults, all the phases are short-circuited to each other and often to earth.
Such fault is balanced in the sense that the systems remain symmetrical, or we can say the
lines displaced by an equal angle (i.e. 120° in three phase line). It is the most severe type of
fault involving largest current, but it occurs rarely. For this reason balanced short- circuit
calculation is performed to determine these large currents.

Symmetrical short circuit of a synchronous generator (on no load &

constant Excitation):

Short Circuit of a Synchronous Machine – Under steady state short circuit conditions, the
armature reaction of a synchronous generator produces a demagnetizing flux. In terms of a
circuit this effect is modelled as a reactance Xa in series with the induced emf. This
reactance when combined with the leakage reactance Xl of the machine is
called synchronous reactance Xd (direct axis synchronous reactance in the case of salient
pole machines). Armature resistance being small can be neglected. The steady state short
circuit model of a synchronous machine is shown in Fig. 9.3a on per phase basis
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Consider now the sudden short circuit (three-phase) of a synchronous generator initially
operating under open circuit conditions. The machine undergoes a transient in all the three
phase finally ending up in steady state conditions described above. The circuit breaker
must, of course, interrupt the current much before steady conditions are reached.
Immediately upon short circuit, the DC off-set currents appear in all the three phases, each
with a different magnitude since the point on the voltage wave at which short circuit occurs
is different for each phase. These DC off-set currents are accounted for separately on an
empirical basis and, therefore, for short circuit studies, we need to concentrate our
attention on symmetrical (sinusoidal) short circuit current only. Immediately in the

event of a short circuit, the symmetrical short circuit current is limited only by the leakage
reactance of the machine. Since the air gap flux cannot change instantaneously (theorem of
constant flux linkages), to counter the demagnetization of the armature short circuit
current, currents appear in the field winding as well as in the damper winding in a
direction to help the main flux. These currents decay in accordance with the winding time
constants. The time constant of the damper winding which has low leakage inductance is
much less than that of the field winding, which has high leakage inductance. Thus during
the initial part of the short circuit, the damper and field windings have transformer
currents induced in them so that in the circuit model their reactances—Xf of field winding
and Xdw of damper winding—appear in parallel* with Xa as shown in Fig. 9.3b. As the
damper winding currents are first to die out, Xdw effectively becomes open circuited and at
a later stage Xf becomes open circuited. The machine reactance thus changes from the
parallel combination of Xa, Xf and Xdw during the initial period of the short circuit to Xa and
Xf in parallel (Fig. 9.3c) in the middle period of the short circuit, and finally to X a in steady
state (Fig. 9.3a). The reactance presented by the machine in the initial period of the short
circuit, i.e.

is called the subtransient reactance of the machine. While the reactance effective after the
damper winding currents have died out, i.e.

is called the transient reactance of the machine. Of course, the reactance under steady
conditions is the synchronous reactance of the machine. Obviously X″d<X′d<Xd. The machine

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thus offers a time-varying reactance which changes from X″d to X′d and finally to Xd.
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If we examine the oscillogram of the short circuit current of a synchronous machine after
the DC off-set currents have been removed from it, we will find the current wave shape as
given in Fig. 9.4a. The envelope of the current wave shape is plotted in Fig. 9.4b. The short
circuit current can be divided into three periods—initial subtransient period when the
current is large as the machine offers subtransient reactance, the middle transient period
where the machine offers transient reactance, and finally the steady state period when the
machine offers synchronous reactance.

If the transient envelope is extrapolated backwards in time, the difference between the
transient and subtransient envelopes is the current Δi″ (corresponding to the damper
winding current) which decays fast according to the damper winding time constant.
Similarly, the difference Δi′ between the steady state and transient envelopes decays in
accordance with the field time constant.

In terms of the oscillogram, the currents and reactances discussed above, we can write

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The intercept Ob for finding transient reactance can be determined accurately by means of
a logarithmic plot. Both Δi″ and Δi′ decay exponentially as
The plot of log (Δi″+Δi′) versus time for t≫Tdw therefore, becomes a straight line with a
slope of (-Δi′0/Tf) as shown in Fig. 9.5. As the straight line portion of the plot is extrapolated
(straight line extrapolation is much more accurate than, the exponential extrapolation of
Fig. 9.4), the intercept corresponding to t = 0 is

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ra = AC resistance of the armature winding per phase.

Though the machine reactance’s are dependent upon magnetic saturation (corresponding
to excitation), the values of reactance’s normally lie within certain predictable limits for
different types of machines. Table 9.1 gives typical values of machine reactance are which
can be used in fault calculations and in stability studies.

Normally both generator and motor subtransient reactance’s are used to determine the
momentary current flowing on occurrence of a short circuit. To decide the interrupting
capacity of circuit breakers, except those which open
Instantaneously, subtransient reactance is used for generators and transient reactance for
synchronous motors

Short circuit of a loaded synchronous generator:

Short Circuit of a Loaded Synchronous Machine – In the previous article on the short circuit
of a synchronous machine, it was assumed that the machine was operating at no load prior
to the occurrence of short circuit. The analysis of short circuit on a loaded synchronous
machine is complicated and is beyond the scope of this book. We shall, however, present
here the methods of computing short circuit current when short circuit occurs under
loaded conditions.

Figure 9.10 shows the circuit model of a synchronous generator operating under steady
conditions supplying a load current I° to the bus at a terminal voltage of V°. Eg is the

induced emf under loaded condition and Xd is the direct axis synchronous reactance of the
machine. When short circuit occurs at the terminals of this machine, the circuit model to be
used for computing short circuit current is given in Fig. 9.11a for subtransient current, and
in Fig. 9.11b for transient current. The induced emfs to be used in these models are given

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The voltage E″g is known as the voltage behind the subtransient reactance and the
voltage E′g is known as the voltage behind the transient reactance. In fact, if I° is zero
(no load case), E″g=E′g=Eg, the no load voltage, in which case the circuit model
reduces
Synchronous motors have internal emfs and reactances similar to that of a generator
except that the current direction is reversed. During short circuit conditions these can be
replaced by similar circuit models except that the voltage behind subtransient/transient
reactance is given by

Whenever we are dealing with short circuit of an interconnected system, the synchronous

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machines (generators and motors) are replaced by their corresponding circuit models
having voltage behind subtransient (transient) reactance in series with subtransient
(transient) reactance. The rest of the network being passive remains unchanged.

Selection of circuit breakers:

There are a few different criteria to consider when selecting a circuit breaker including
voltage, frequency, interrupting capacity, continuous current rating, unusual operating
conditions and product testing.

The overall voltage rating is calculated by the highest voltage that can be applied across all
end ports, the distribution type and how the circuit breaker is directly integrated into the
system. It is important to select a circuit breaker with enough voltage capacity to meet the
end application.

Frequency
Circuit breakers up to 600 amps can be applied to frequencies of 50-120 Hz. Higher than 120 Hz
frequencies will end up with the breaker having to derate. During higher frequency projects, the
eddy currents and iron losses causes greater heating within the thermal trip components thus
requiring the breaker to be derated or specifically calibrated. The total quantity of deration
depends on the ampere rating, frame size as well as the current frequency. A general rule of thumb
is the higher the ampere rating in a specific frame size the greater the derating needed.

All higher rated breakers over 600 amps contain a transformer-heated bimetal and are suitable for
60 Hz AC maximum. For 50 Hz AC minimum applications special calibration is generally available.
Solid state trip breakers are pre-calibrated for 50 Hz or 60 Hz applications. If doing a diesel
generator project the frequency will either be 50 Hz or 60 Hz. It is best to check ahead of time with
an electrical contractor to make sure calibration measures are in place before moving forward with
a 50 Hz project.

Maximum Interrupting Capacity

The interrupting rating is generally accepted as the highest amount of fault current the
breaker can interrupt without causing system failure to itself. Determining the maximum
amount of fault current supplied by a system can be calculated at any given time. The one
infallible rule that must be followed when applying the correct circuit breaker is that the
interrupting capacity of the breaker must be equal or greater than the amount of fault
current that can be delivered at the point in the system where the breaker is applied.
Failure to apply the correct amount of interrupting capacity will result in damage to the

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breaker for specific application.

Analysis of three phase symmetrical faults

The symmetrical fault can be analysed on single phase basis using reactance diagram or by
using per unit reactance diagram. The symmetrical fault analysis has to be performed
separately for subtransient, transient and steady state conditions of the fault, because the
reactances and internal emfs of the synchronous machines will be different in each state.
Once the per unit reactance diagram of the power system is formed for a particular state
(subtransient/transient/steady state) of fault condition, then the currents and voltages in
the various parts of the system can be determined by any one of the following method:

i)Using Kirchoff's laws

ii)Using Thevenin's theorem.

iii)By forming the bus impedance matrix.

The following procedure can be followed to directly calculate voltages and currents
during symmetrical fault condition in a power system, using Kirchoff's laws,

i) Select appropriate base values and determine the prefault condition reactance diagram
of the given power system. (The prefault condition reactance diagram is separately formed
for subtransient, transient and steady state condition of the fault).

ii) Calculate the internal emfs of the synchronous machines and the prefault voltages at the
fault point using prefault current. (load current). Note: If the power system is unloaded (i.e
if there is no prefault current), then the prefault voltage at the fault point is 1 p.u. Also the
internal emfs for subtransient and transient state are the same as steady state induced emf.

iii) Draw the fault condition reactance diagram of the system. This diagram is same as
prefault reactance diagram except that the fault is represented by a short circuit or by the
specified fault impedance. The currents in this reactance diagram are fault condition
currents.

iv)Calculate the p.u value of the fault currents in various parts of the system and at the fault
point.

v) The actual values of the fault currents are obtained by multiplying the p.uvalues by the

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respective base currents.

Symmetrical fault analysis using Thevenin's theorem.

The following procedure can be followed to calculate the voltages and currents during
symmetrical fault using Thevenin's theorem.

1) Select appropriate base values and determine the prefault condition reactance diagram
of the given power system.

2) Calculate the prefault Thevenin's voltage at the fault point using the prefault
current(load current). If the system is unloaded, then the prefault voltage is 1p.u.

3) Determine the Thevenin's impedance of the system at the fault point by shorting all
voltage sources.

4) Draw the Thevenin's equivalent at the fault point. Then the p.u value of fault current is
given by If=VTH/(ZTH+Zf). Multiplying the p.u value by the base value gives the actual
value of the fault current. Here, Zf is the fault impedance of the system. For a solid three
phase short circuit, Zf=0
5) The fault current in other parts of the network are determined from the knowledge of
change in current due to fault and prefault current. The fault current (i.e post fault current)
in any part of the system is given by sum of prefault current and change in current due to
fault. The change in current due to fault can be estimated by connecting the Thevenin's
source with reversed polarity at the fault. Replace all other sources by zero values sources.
Now the currents in various part of the system are the change in currents due to fault.
These currents are calculated using any conventional technique.

Example 1: A synchronous generator and motor are rated for 30,000 KVA, 13.2 KV and
both have subtransients reactance of 20%. The line connecting them has a reactance of 1%
on the base of machine ratings. The motor is drawing 20,000KW at 0.8 p.f. leading. The
terminal voltage of the motor is 12.8KV. When a symmetrical three- phase fault occurs at
motor is 12.8KV. When a symmetrical three phase fault occurs at motor terminals, find the
subtransient current in generator, motor and at th fault point.

Example 2.1:

For the radial network shown in fig. 2.6, a three phase fault occurs at F. Determine

the fault current and the line voltage at 11kV bus under fault conditions.

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Solution:

Base values:

Let us choose,

base MVA=100
base kV in the overhead line=33

we calculate,

base kV on the generator side=33×11/33= 11

base kV on the cable side=33×6.6/33= 6.6

Reactance of generator G1:

XG1, new = XG1, old × ( (MVA)B, new / (MVA)B, old ) × ( (kV)2B, old / (kV)2B, new )

= j0.15 × (100 / 10 ) × (112/ 112)

= j 1.5 p.u

Reactance of generator G2:

XG2, new = XG2, old × ( (MVA)B, new / (MVA)B, old ) × ( (kV)2 B, old / (kV)2B, new )

= j0.125 × (100 / 10 ) × (112/ 112)

= j 1.25 p.u

Reactance of transformer T1: (calculated secondary side it)HV or HT

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XT1, new = XT1, old × ( (MVA)B, new / (MVA)B, old ) × ( (kV)2B, old / (kV)2B, new

= j0.1 × (100 / 10 ) × (332/ 332)

= j 1.0 p.u

Reactance of transformer T2: (calculated primary side of it)HV or HT

XT2, new = XT2, old × ( (MVA)B, new / (MVA)B, old ) × ( (kV)2B, old / (kV)2B, new )

= j0.08 × (100 / 5 ) × (332/ )

= j 1.6 p.u

Impedance of O.H line:

ZO.H,p.u= ZO.H( Ω ) ×(MVA)B,new / (kV)2B

= (30 ×(0.27+j0.36)) × 100 / 332

= 0.744+j0.99 p.u
Impedance of cable:

Zc= Zc( Ω ) ×(MVA)B,new / (kV)2B

= (3 ×(0.135+j0.08)) × 100 / 6.62

= 0.93+j0.55 p.u

the prefault impedance diagram of the given system is as shown in fig. 2.7

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Since the system is unloaded prior to occurrence of fault, Vpf is assumed as 1p.u.
Thevenin's theorem is employed here to find the fault current.

VTH=Vpf=1p.u

To find ZTH:

Shorting the generated voltages, we obtain the equivalent circuit of the system

prior to the fault as in fig. 2.8

ZTH=((j1.5×j1.25) /

(j1.5+j1.25))+(j1.0+0.744+j0.99+j1.6+0.93+j0.55)=1.674+j4.82=5.1∠70.8° p.u

Thus, the Thevenin's equivalent circuit of the system with respect to fault
point is as shown if fig 2.9

Now short circuiting the terminals of the Thevenin's equivalent circuit as shown in fig. 2.10
is equivalent to the fault condition. The current flowing through the short circuit is the fault
current.

The p.u value of fault current, If=VTH/ZTH=(1.0∠0°) / (5.1∠70.8°) = 0.196 ∠-70.8°p.u

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The base current, Ib= (1000 × base power) / (√3 × base voltage)= (1000×100) / (√3 × 6.6)
= 8747A therefore, absolute value of fault current , If= 0.196 ∠-70.8° × 8747 = 1714 ∠-70.8°
Amperes.

To find voltage at 11 kV bus during fault:

From fig 2.7, it can be observed that the total impedance between point F and

11kV bus is,

=(0.93+j0.55)+j1.6+(0.744+j0.99)+j1.0

=(1.674+j4.14)p.u

=4.466∠67.98°

Voltage at 11kV bus=4.466∠67.98°×0.196 ∠-70.8°=0.875∠-2.82° p.u

The absolute value of the voltage at 11kV bus=0.875∠-2.82°×11=0.9625∠-2.82°kV

Example 3. A 25MVA, 13.8kV generator with Xd”=15% is connected through a transformer
to a bus that supplies four identical motors as shown in fig. 2.18. Each motor has Xd”=20%
& Xd'=30% on a base of 5MVA, 6.9kV. The three phase rating of the transformer is 25MVA,
13.8-6.9 kV. With a leakage reactance of 10%. The bus voltage at the motors is 6.9kV when
a three-phase fault occurs at the point P. For the fault specified determine:

a)The subtransient current in the fault.

b) The subtransient current in the breaker A.

c) The momentary current in breaker A.

d)The current to be interrupted by breaker A in 5 cycles.

Solution:

base values:

let we choose,

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base MVA=25

base kV in the generator circuit=13.8

we calculate, base voltage on the motor side=13.8×6.9/13.8=6.9

Reactance of generator G:

XdG” =j0.15(same as old p.u value in given because base values have been chosen

on the same machine ratings)

XdG' =j0.15(same as subtransient reactance as it is not specified in data).

Reactance of transformer T:

XT=j0.1(same as old p.u value in given because base values have been chosen on

the same machine ratings)

Reactances of motors:
XdM,p.u,new” = XdM,p.u,old” × ( (MVA)B, new / (MVA)B, old ) × ( (kV)2B, old / (kV)2B, new )

= j0.2 × (25 / 5 ) × (6.92/ 6.92)

= j 1.0 p.u

XdM,p.u,new' = XdM,p.u,old' × ( (MVA)B, new / (MVA)B, old ) × ( (kV)2B, old / (kV)2B, new )

= j0.3 × (25 / 5 ) × (6.92/ 6.92)

= j 1.5 p.u

The prefault voltage at the point P is 6.9kV=6.9/6.9=1p.u and the base current in the 6.9kV
circuit is,

Ib= (1000 × base power) / (√3 × base voltage)= (1000 × 25) / (√3 ×6.9)=2091.8A
The reactance diagram with subtransient values of the reactance marked is shown in fig
2.19.

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a) therefore subtransient fault current, If”=4×(1/j1.0)+(1/j0.25)= -j8p.u The absolute value
of the current is If”= -j8 × 2091.8 = - j16734.4 A

b) therefore subtransient current in breaker A, I”= 3×(1/j1.0)+(1/j0.25)= -j7p.u .The

absolute value of the current is I”= -j7 × 2091.8 = - j14642.6 A

c)To find the momentary current in the breaker A, we must account for the d.c. Offset
current. This is done empirically as follows:

Momentary current through breaker A= 1.6×14642.6=23428.16 A

d)To compute the current to be interrupted by the breaker A, it is required to obtain the
transient reactance model of the system. This is shown in fig 2.20
The current to be interrupted by the breaker A now is =3×(1/j1.5)+(1/j0.25)= -j6p.u
Allowance is made for the d.c. Offset current by multiplying with a factor 1.1(seetable 2.1).
therefore the current to be interrupted is, 1.1×6×2091=13805.88A

Example 4. Two synchronous motors are connected to the bus of a large system through a
short transmission line as shown in fig 2.33. The ratings of the various components
are:Motors (each): 1MVA, 440V, 0.1p.u, transient reactance line: 0.05 ohm reactance large
system: short circuit MVA at its bus at 440V is 8. when the motors are operating at 440V,
calculate the short circuit current fed into a three phase fault at the motor bus.

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Solution:

base values: let us choose the motor ratings as base values

therefore, base MVA=1 ,base kV on the motor side=0.44

The large system can be considered as a source of constant voltage feeding the

line through an “infinite bus”. The voltage rating of the bus is 440V(as given)

Reactance of Motors: XM=j0.1 p.u

Reactance of line: XTL,p.u= XTL( Ω ) ×(MVA)B,new / (kV)2B

= j0.05 × 1 / 0.442

=j0.258 p.u

Hence the p.u reactance diagram is as shown in fig 2.34.

The prefault voltage at the motor bus, Vpf=400V p.u value of prefault voltage,
Vpf=400/440=0.909p.u p.u value of voltage at infinite bus, VL=440/440=1p.u

The fault condition of the system is shown in fig 2.35. The total fault currentIf' is sum of the
fault current If1' and If2'

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From the fig 2.35, it can be observed that,

If1'=VL/j0.2583=1∠0°/j0.2583= -j3.87 p.u

If2'=Vpf/(j0.1/2)=0.909∠0°/j0.5= -j18.18 p.u

If'= If1'+If2'= -j3.87 -j 18.18 = -j22.05 p.u = 22.05∠ -90° p.u

Actual value of fault current = p.u value of fault current × base current

=22.05∠ -90° × (1000 ×1) / (√3 × 0.44)

=28933.12 ∠ -90° A.