Symmet
S
trical Fa
ault Anaalysis
Fault:
F
A fau
ult in a circuuit is any faiilure that intterferes with
h the normall flow of currrent to the load. In
most
m faults, a current paath forms beetween two oor more phasses, or betweeen one or more
m
phases and the
neutral
n
(grouund). Since the
t impedannce of a new path is usuaally low, an eexcessive cuurrent may flow.
Most
M of the fault on the power systeem leads to a short circuit condition.. When such
h a conditionn occurs,
a heavy cu
urrent (calledd short-circuuit current) flows throough the eqquipment, caausing consiiderable
damage
d
to th
he equipmennt and interruuption of serrvice to conssumers.
The
T choice of apparatuss and the design and arrrangement of
o practicallyy every equip
pment in thee power
hort-circuit current
c
consiiderations.
system depeends upon sh
Types of Faults
F
The
power
system
b
behaves
mally undeer conditioons of
abnorm
symmeetrical
short
circuit
(symm
metrical threee-phase faultt).
Symm
metrical fault may be a solid
three-pphase shorrt-circuit or
o may
involvve arc impedaance.
metrical fau
ult conditioons are
Symm
causedd in the system acccidently
througgh
insulattion
failure
of
equipm
ments or flashover
f
of lines
initiateed by a liightning strroke or
througgh accidental faulty operration.
The syystem must be
b protectedd against
flow oof heavy sh
hort-circuit currents
c
(whichh can cause permanent damage
to major equipmennt) by disconnnecting
the fauulty part of thhe system byy means
of
o circuit breeakers operaated by proteective relayinng.
For
F proper choice of ciircuit breakeers and prottective relayying, the maagnitude of currents
c
thaat would
flow under short-circuit
s
conditions must
m be estim
mated – this is the scopee of fault anaalysis.
High-voltag
H
ve strings off insulators supporting
s
eeach phase. The
T insulatoors must
e transmissiion lines hav
be
b large enoough to prev
vent flashoveer – a condiition when the
t voltage ddifference beetween the line
l
and
the
t ground is
i large enou
ugh to ionizee the air arouund insulatorrs and thus pprovide a currrent path beetween a
phase
p
and a tower.
If
I flashover occurs on a single phasse of the linee, an arc will be produceed. Such fauults are calledd single
line-to-grou
l
und faults. Since the short-circuit path has a low impeddance, very high currennts flow
through
t
the faulted linee into the ground
g
and bback into thhe power syystem. Faultss involving ionized
current
c
pathhs are also caalled transieent faults. Thhey usually clear if pow
wer is removed from the line for
a short time and then resstored.
o occur if onne phase of the
t line breaaks and com
mes into contaact with
Single line-tto-ground faaults can also
the
t ground or if insulattors break. This
T
fault iss called a peermanent faault since it will remainn after a
quick
q
powerr removing.
Approximat
A
tely 75% off all faults inn power sysstems are eiither transiennt or permaanent single line-toground
g
faultts.
Sometimes, all three ph
hases of a transmission
t
n line are shhorted togethher – symm
metrical threee-phase
faults.
f
Two
T phases of a line maay touch, or flashover maay occur bettween two phhases – a lin
ne-to-line fau
ult.
When
W
two liines touch eaach other annd also touchh the groundd, the fault is called a dou
uble line-to--ground
fault
f
Lighting
L
stro
okes cause most
m faults on
o high-volttage transmiission lines pproducing a very high transient
t
that
t
greatly exceeds thee rated voltaage of the line. This volltage usuallyy causes flashover betw
ween the
phase
p
and thhe ground of
o the line creating an arc.
a Once th
he current staarts flowingg through the arc, it
remains
r
even after the liighting disapppears.
High
H
curren
nts due to a fault must be
b detected by protectivve circuitry and the circcuit breakerss on the
affected
a
trannsmission lin
ne should au
utomaticallyy open for a brief periodd (about 1/33 second). T
This will
allow
a
ionizeed air to deeionize. If thhe fault wass transient, normal operration shoulld be restored after
reclosing
r
thee breaker. Thherefore, maany transientt faults are cleared
c
autom
matically. Otherwise, thee circuit
breaker
b
shouuld open agaain isolating the transmisssion line.
(A) Phase-to
o-earth fault
(C) Phase-too-phase-to-eaarth fault
(E) Three phhase-to-earthh fault
(G) Pilot-earrth fault *
*In undergrround mining
g applicationn only
(B)) Phase-to-phase fault
(D)) Three phasse fault
(F)) Phase-to-piilot fault *
Symmetrica
al Fault: Thhe fault on th
he power syystem which
h gives rise o symmetriccal fault currrent (i.e.
o
equal
e
fault current
c
in thee line with 120 displaceement) is called symmetrrical fault.
Protectio
P
on againsst fault
High
H
curren
nts due to a fault must be
b detected by protectivve circuitry and the circcuit breakerss on the
affected
a
trannsmission lin
ne should au
utomaticallyy open for a brief periodd (about 1/33 second). T
This will
allow
a
ionizeed air to deeionize. If thhe fault wass transient, normal operration shoulld be restored after
reclosing
r
thee breaker. Thherefore, maany transientt faults are cleared
c
autom
matically. Otherwise, thee circuit
breaker
b
shouuld open agaain isolating the transmisssion line.
Selecting an
n appropriatee circuit breaaker (type, siize, etc.) is im
mportant
Transien
nt on a Transmisssion Linee
Let
L us consiider the shorrt circuit traansient on a transmission
n line. Certaain simplifyiing assumptiions are
made
m
at this stage.
1. The line is fed frrom a constaant voltage soource.
2. Shorrt-circuit takees place wheen the line iff unloaded
3. Line capacitancee is negligiblle and the linne can be reppresented byy a lumped RL
R series circcuit.
4. The impedance is zero in short-circuit.
5. The short circuit is assumed to take place at t=0. The parameter α controls the instant on the
voltage wave when short circuit is occurs.
6. The instantaneous voltage equation for this circuit is:
di(t )
Ri(t ) + L
= v = 2V sin(ωt + α )
dt
7.
It is known from circuit theory that the current after short-circuit is composed of two parts, i.e.
i (t ) = i (t ) + i (t ) = i (t ) + i (t )
s
t
ac
dc
2V
i (t ) = i (t ) =
sin(ωt + α − θ ) [Symmetrical short circuit current]
s
ac
Z
V
The rms value of symmetrical or steady-state fault current is: I =
ac
Z
i (t ) = i (t ) = 2 I sin(ωt + α − θ )
s
ac
ac
2V
i (t ) = i (t ) = −
sin(α − θ )e −t / τ L
t
dc
Z
[DC off - set
current]
i (t ) = i (t ) = − 2 I sin(α − θ )e − t / τ L
t
dc
ac
is (or iac) is called steady state or ac current, it (or idc is called transient or dc current [it is such that i(0)
= is(0) + it(0) =0 being an inductive circuit, it decays corresponding to the time constant L/R.]
L
X
X
⎛ ωL ⎞
Z = R 2 + (ωL) 2 = R 2 + X 2 ; X = ωL ; θ = tan − 1⎜
=
s
⎟; τL = =
R ωR 2πfR
⎝ R ⎠
The sinusoidal steady state current is called the symmetrical short-circuit (steady-state) fault current
and the unidirectional transient current is called the dc off-set current.
Since the maximum value of dc off-set current depends on α, this current causes the total short-circuit
current to be unsymmetrical till the transient decays.
dc offset current + Symmetrical fault current = Asymmetrical Fault Current (Total)
2V
2V
i(t ) = i (t ) + i (t ) = i (t ) + i (t ) =
sin(ωt + α − θ ) −
sin(α − θ )e −t / τ L
s
t
ac
dc
Z
Z
[
]
i(t ) = i (t ) + i (t ) = i (t ) + i (t ) = 2 I sin(ωt + α − θ ) − sin(α − θ )e − t / τ L
s
t
ac
dc
ac
The total current is called the asymmetrical current.
At the instant of applying the voltage, the dc and steady-state components always has the same
magnitude but are the opposite sign in order to express the zero value of current then existing.
If the value of the steady-state term is not zero at t = 0, the dc component appears in the solution in
order to satisfy the physical conditions of zero current at the instant of closing the switch.
DMAM (3/22)
At t = 0;
2V
i (t ) = i (t ) = −
sin(α − θ ) = − 2 I sin(α − θ )
t
dc
ac
Z
when α - θ = 0 or α - θ = π, i (t = 0) = i (t = 0) = 0
t
dc
The dc term does not exist, if the circuit is closed at point on the voltage wave such that α - θ = 0 or α
- θ = π.
2V
= m 2I
when α - θ = ±π /2, i (t = 0) = i (t = 0) = m
t
dc
ac
Z
The dc component has its maximum initial value, which is equal to maximum value of the
sinusoidal component, if the circuit is closed at point on the voltage wave such that α - θ = ±π /2.
Fig.9.2 Waveform of a short-circuit current on a transient line [5].
DMAM (4/22)
Problem: The source voltage v = 151sin(377t+α) V is applied in the following circuit where R = 0.125
Ω, L = 10 mH. Find the current response after closing the switch S for the following cases: (i) no dcoffset, and (ii) maximum dc-offset. Sketch the current waveform up to 3 cycles corresponding to case
(i) and (ii).
Solution:
Here, ω = 377, XL = ωL = 377×0.01 = 3.77 Ω
Z = 0.125 + j[377 × 0.01] = 0.125 + j 3.77 = 3.772∠88.1°
V 151
1
40
151
I =
=
×
=
A
V =
Z
2 ; ac
2 3.772
2
L
0.01
=
= 0.08 s
R 0.125
The current response is then given by
i (t ) = 40 sin(377t + α − 88.1°) − sin(α − 88.1°)e − t / 0.08
(i)a No dc offset, if switch is closed when α = θ = 88.1o.
τL =
[
]
(ii) Maximum dc offset, if switch is closed when α = θ - 90o = -1.9o.
The first peak is called the maximum momentary short-circuit current, Imm.
2V
2V
I
=
−
sin(α − θ ) = 2 I [1 − sin(α − θ )]
ac
mm
Z
Z
Since the transmission line resistance is small, θ ≈ 90o.
2V
2V
=
+
cos α = 2 I [1 + cos α ]
I
ac
mm
Z
Z
This is the maximum possible value for α = 0, i.e. short-circuit occurring when the voltage wave is
going through zero. Thus
2V
I
=2
= 2 2 I = twice the maximum of symmetrical short-circuit current
mm(max possiibe)
ac
Z
(doubling effect).
For the selection of circuit breakers, momentary circuit is taken corresponding to its maximum
possible value (a safe choice).
when α - θ =-π /2,
π
π
π
⎡
⎤
⎡
⎤
i (t ) = 2 I ⎢sin(ωt − ) − sin( − )e −t / τ L ⎥ = 2 I ⎢sin(ωt − ) + e −t / τ L ⎥
ac ⎣
ac ⎣
2
2
2
⎦
⎦
− t /τ 2
− 2t / τ
L] =I
L
= [ I ]2 + [ I (t )]2 = [ I ]2 + [ 2 I e
1 + 2e
I
rms
ac
dc
ac
ac
ac
τ is time in cycle
t =τ / f ;
DMAM (5/22)
2t
τ
=
L
2τ
4πτ
=
f ( X / 2πfR) ( X / R)
−
I
rms
=I
ac
1 + 2e
−
4πτ
( X / R)
4πτ
( X / R)
;
K (τ ) = 1 + 2e
K (τ = 0) = 3
= K (τ ) I
I
rms
ac
Irms decreases from √3Iac when τ=0 to Iac when is τ very large.
Example 7.1 [Ref. 3, p. 360] A bolted circuit occurs in the series RL circuit with V = 20 kV, X = 8 Ω,
R = 0.8 Ω, and with maximum dc offset. The circuit breaker opens 3 cycles after fault inception.
Determine (i) the rms ac fault current, (ii) the rms momentary current at τ=0.5 cycle, which pass
through the breaker before it opens, and (iii) the rms asymmetrical fault current that the breaker
interrups.
Solution:
V
20
(i) I =
=
= 2.488 kA
2
ac
Z
8 + 0.8 2
(ii) X/R= 8/0.8=10, τ=0.5 cycle
4πτ
4π × 0.5
−
−
( X / R)
10
K (τ ) = 1 + 2e
= 1 + 2e
= 1.438
I
=I
= K (τ ) I = 1.438 × 2.488 = 3.576
rms
momentary
ac
(iii) τ=0.5 cycle
4πτ
4π × 3
−
−
( X / R)
10 = 1.023
K (τ ) = 1 + 2e
= 1 + 2e
I
= K (τ ) I = 1.023 × 2.488 = 2.544 A
rms
ac
A
Short Circuit of a Synchronous Machine (on No Load)
Armature resistance being small can be neglected.
Under steady-state short-circuit conditions; the armature reaction of a synchronous machine produces
a demagnetizing flux which is modeled as a reactance Xa (known as fictitious reactance) in series with
the induced emf. The combination of fictitious (Xa) and leakage (Xl) reactances is called synchronous
reactance, Xd (direct axis synchronous reactance in the case of salient pole machine). The steady-state
short-circuit model of a synchronous machine is shown in Fig. 9.3(a) on per phase basis.
Immediately upon short-circuit, the DC off-set currents appear in all three-phases, each with a different
magnitudes since the point on the voltage wave at which short-circuit occurs is different for each
phase. These DC off-set current are accounted for separately on an empirical bass and, therefore, for
short circuit studies, we need to concentrate our attention on symmetrical (sinusoidal) short-circuit
current only.
Immediately in the event of short circuit, the symmetrical short-circuit current is limited by leakage
reactance. Since the air-gap flux cannot be changed instantaneously to counter the demagnetization of
the armature short-circuit current, currents appear in the field winding as well as damper winding in a
DMAM (6/22)
direction to help the mail flux. The time constant of damper winding (which has low leakage
inductance) is much less than that of the field winding (which has high leakage inductance).
Thus the initial part of the short-circuit, the damper winding and field windings have transformer
currents induced in them so that in the circuit model their reactances ⎯ Xf of field winding and Xdw of
damper winding ⎯ appear in parallel with Xa as shown in Fig. 9.3(b).
As the damper winding currents are first to die out, Xdw effectively becomes open-circuited and at a
later stage Xf becomes open-circuited. The machine reactances thus changes from the parallel
combination of Xf, Xa, and Xdw during the initial period of the short-circuit to Xf, and Xa in parallel [Fig.
9.3(b)] in the middle period of the short-circuit, and finally to Xa in steady-state [Fig. 9.3(a)].
The reactance, which is called subtransient reactance, presented by the machine in the initial period
of the short-circuit, i.e.
1
X ' ' = X + X // X // X
=X +
d
l
a
f
dw
l [(1 / X ) + (1 / X ) + (1 / X )]
a
f
dw
The reactance, which is called transient reactance, after the damper winding currents have died out is:
1
X ' = X + X // X = X +
d
l
a
f
l [(1 / X ) + (1 / X )]
a
f
The reactance, which is called synchronous reactance, after the field winding currents have died out
is:
X =X +X
d
l
a
Obviously, X ' ' < X ' < X . The machine thus offers a time-varying reactance which changes from
d
d
d
Xd″ to Xd′ and finally to Xd.
If the short-circuit current of a synchronous machine is examined after the dc off-set current have been
removed from it, the current wave shape is found as shown in Fig. 9.4(a). The envelope of the current
wave shape is plotted in Fig. 9.4 (b).
The short-circuit current is divided into three periods: (i) initial subtransient period hen the current is
large as the machine offers subtransient reactance, Xd″, (ii) The middle transient period hen the current
is large as the machine offers transient reactance, Xd′, and (iii) finally the steady-state period when the
machine offers synchronous reactance, Xd.
IF the transient envelop is extrapolated backwards in time, the difference between the transient and
subtransient envelopes is the current ∆i″ (corresponding to the damper winding current) which decays
fast according to the damper winding time constant. Similarly, the difference ∆i′ between the steadystate and transient envelopes decays in accordance with the field time constant.
(a) Steady-state short circuit model of a
synchronous machine
DMAM (7/22)
(b) Approximate circuit model during subtransient
period of a short-circuit
(c) Approximate circuit model during transient period of a short-circuit
Fig. 9.3 [Ref. 5]
The currents and reactances discussed above are defined by the following equations, which apply to an
alternator operating at no-load before the occurrence of a three-phase (or symmetrical) fault at its
terminals:
E
oa
g
The rms value of steady-state current: I =
=
2 Xd
E
ob
g
=
The rms value of transient current excluding dc off-set component: I ' =
2 X'
d
E
oc
g
=
The rms value of subtransiente current excluding dc off-set componnet: I ' ' =
2 X ''
d
Xd = direct-axis synchronous reactance
Xd = direct-axis transient reactance
Xd′′ = direct-axis subtransient reactance
|Eg| = rms voltage from one terminal to neutral at no load
oa, ob, and oc = intercepts shown in Fig. 9.4.
Obviously, I ' ' > I ' > I
There are similar quadrature axis reactances.
Xq = quadrature-axis synchronous reactance
Xq = quadrature -axis transient reactance
Xq′′ = quadrature -axis subtransient reactance
If the armature resistance is small, the quadrature axis reactances do not significantly affect the shortcircuit current. Using the above direct-axis reactances, the instantaneous ac current can be written as:
⎡⎛
⎜ 1
1
iac (t ) = 2 E ⎢⎢⎜
−
g ⎜ ''
'
⎢⎣⎝ X d X d
DMAM (8/22)
⎤
⎞ − (t / T ' ' ) ⎛
⎞
'
⎟
⎜ 1
1 ⎟ − (t / Td )
1 ⎥
π
d
e
+⎜
−
+
sin(ωt + α − )
⎟e
⎟
⎥
X ⎟
X
2
⎟
⎜ X'
d ⎠
d ⎥⎦
⎠
⎝ d
(a) Symmetrical short-circuit armature current in synchronous machine
(b) Envelope of synchronous machine symmetrical short circuit current.
Fig. 9.4 [Ref. 1 and 5]
At t=0, when the fault occurs, the rms value of i(t) i.e. subtransient fault current is:
E
g
I '' = I ac (0) =
X ''
d
The duration of I″ is determined by time constant Td′′, called the direct-axis short-circuit subtransient
time constant.
When t> Td′′ but t < Td′ (direct-axis short-circuit transient time constant), the first exponential term of
above equation has decayed almost zero, but the second exponential has not decayed significantly. The
rms fault current then equal the rams transient fault current, given by
E
g
'
I =
X'
d
When t is much larger than Td′, the rms fault current approaches its steady state value, given by
E
g
I = I ac I (∞ ) =
Xd
DMAM (9/22)
Since the three-phase no-load voltages are displaced 120o from each other, the three-phase ac fault
currents are displaced 120o fro each other. In addition to that ac fault current, each phase has a
different dc offset. The maximum dc offset in any oe phase, which occurs when α=0 is
idc max (t ) =
2E
g − (t / T A )
− (t / T A )
e
= 2 I '' e
X ''
d
Where, TA is called the armature time constant.
Example 7.2 [Ref. 3, p. 363] A 500 MVA, 20 kV, 60 Hz synchronous generator with Xd′′=0.15,
Xd′=0.24, Xd=1.1 pu and time constant Td′′=0.035, Td′= 2.0, TA=0.2 s is connected to the circuit breaker.
The generator is operating at 5% above rated voltage and at no-load when a bolted three-phase shortcircuit occurs on the load side of the breaker. The breaker interrupts the fault 3 cycles after fault
inception. Determine (i) the subtransient fault current in pu and kA rms, (ii) the maximum dc-offset as
a function of time; and (iii) rms asymmetrical fault current, which the breaker interrupts, assuming
maximum dc-offset.
Solution:
(i) The generator is operating at 5% above rated voltage and at no-load when a bolted three-phase
short-circuit occurs, thus Eg = 1.05 pu.
The subtransient fault current that occurs in each of the three phases is
E
g 1.05
''
=
= 7.0 pu
I =
X ' ' 0.15
d
The generator base current is: Ibase= 500/(√3)(20) = 14.43 kA
The rms subtransient fault current in kA is: I '' = 7.0 × 14.43 = 101.0 kA
(ii) The maximum dc offset that may occur in any phase is:
− (t / T A )
− (t / 0.2)
− (t / 0.2)
idc max (t ) = 2 I '' e
= 2 × 101e
= 142.9e
kA
(iii) The rms ac fault current at t= 3 cycle =0.05 s is
⎡⎛ 1
1 ⎞ − (0.05 / 0.035) ⎛ 1
1 ⎞ − (0.05 / 2.0) 1 ⎤
I ac (0.05 s) = 1.05⎢⎜
−
+⎜
−
+ ⎥ = 4.920 pu
⎟e
⎟e
1.1⎦
⎝ 0.24 1.1 ⎠
⎣⎝ 0.15 0.24 ⎠
I ac (0.05 s) = 4.920 × 14.43 = 71.01 kA
I
I
rms
rms
(t ) = [ I
ac
]2 + [ I
dc
(t )]2 = [ I
(0.05 s) = [71.01] 2 + 2[101e
ac
] 2 + [ 2 I '' e
− (t / T A ) 2
]
− (0.05 / 0.2) 2
] = 132 kA
Example 10.1 [Ref. 1, p. 253] Two generators are connected in parallel to the low-voltage side of
three-phase ∆-Y transformer as shown in Fig. 10.5. Generator 1 is rated 50,000 kVA, 13.8 kV.
Generator 2 is rated 25,000 kVA, 13.8 kV. Each generator has a subtransient reactance of 25%. The
transformer is rated 75,000 kVA, 13.8∆/69Y kV, with a reactance of 10%. Before the fault occurs, the
voltage on the high-tension side of the transformer is 66 kV. The transformer is unloaded, and there is
no circulating current between generators. Find the subtransient current in each generator when the
three-phase short-circuit occurs on the high-tension side of the transformer.
DMAM (10/22)
Fig. 10.5 One-line diagram of Example 10.1.
Solution: Select as base in the high-tension circuit 69 kV, 75 MVA. Then the base voltage on the lowtension side is 13.8 kV.
⎛V
⎞⎛ S
⎞
⎜ base, old ⎟⎜ base, new ⎟
= Z pu,old ⎜
Z
pu, new
⎜ Vbase, new ⎟⎟⎜⎜ S base, old ⎟⎟
⎝
⎠⎝
⎠
If Vbase,old = Vbase,new
If S base,old = S base,new
⎞
⎟
⎟⎟
⎠
⎛V
⎞
⎜ base, old ⎟
= Z pu,old ⎜
then Z
pu, new
⎜ Vbase, new ⎟⎟
⎝
⎠
⎛S
⎜ base, new
= Z pu,old ⎜
then Z
pu, new
⎜ S base, old
⎝
Generator 1:
⎛ 75 ⎞
X ' ' = 0.25⎜ ⎟ = 0.375 pu
d1
⎝ 50 ⎠
66
E =
= 0.957 pu
g1 69
S
75000
base3φ
=
Base current, I base1 =
3V
3 × 13.8
baseLL
Generator 2:
⎛ 75 ⎞
X ' ' = 0.25⎜ ⎟ = 0.750 pu
d2
⎝ 25 ⎠
66
E
=
= 0.957 pu
g 2 69
S
75000
base3φ
I
=
=
base2
3V
3 × 13.8
baseLL
Transformer 1:
X = 0.10 pu
DMAM (11/22)
Figure 10.6 shows the reactance diagram
before the fault. A three-phase fault at P
is simulated by closing switch S.
The internal voltages of the two machines
may be considered to be in parallel since
they must be identical in magnitude and
phase if no circulating current flows
between them. The equivalent parallel
subtransient reactance is:
X '' X ''
'
'
'
'
d1 d 2 pu
Fig. 10.6 Reactance diagram for Example 10.1.
X
= X // X
=
gp
d1
d2
'
'
'
'
X +X
d1
d2
0.375 × 0.750
X
==
= 0.25 pu
gp
0.375 + 0.750
Therefore, as a phasor with Eg=0.957 as reference, the subtransient current in the short-circuit is:
E
0.957
g
I ''=
=
= − j 2.735 pu
jX + jX
j 0.25 + j 0.10
gp
The voltage on the delta side is
V∆ = I ' ' X = (− j 2.735)( j 0.10) = 0.2735 pu
The subtransient current in generator 1 is:
E −V
g1 ∆ 0.957 − 0.275
I '' =
=
= − j1.823 pu
1
j 0.375
jX ' '
d1
75000
I '' = I '' I
= 1.823 ×
= 5720 A
1
1, pu base1
3 × 13.8
The subtransient current in generator 2 is:
E −V
g2
∆ 0.957 − 0.275
I '' =
=
= − j 0.912 pu
2
j 0.750
jX ' '
d2
75000
I '' = I ''
I
= 0.912 ×
= 2860 A
base2
2
2, pu
3 × 13.8
Example 9.1 [Ref. 5] For the radial network shown in Fig. 9.6, a three-phase fault occurs at F.
Determine the fault current and the line voltage at 11 kV bus under fault conditions.
DMAM (12/22)
Fig. 9.6
Fig. 9.7
Example 9.2 [Ref. 5] A 25 MVA, 11 kV generator with Xd″ = 20% is connected through a transformer,
line and a transformer to a bus that supplies three identical motors as shown in Fig. 9.8. Each motor
has Xd″ =25% and Xd′ = 30% on a base of 5 MVA, 6.6 kV. The three-phase rating of the step-up
transformer is 25 MVA, 11/66 kV with a leakage reactance of 10% and that of the step-down
transformer is 25 MVA, 6/6. kV with a leakage reactance of 10%. The bus voltage at motors is 6. kV
when a three-phase fault occurs at the point F. For specified fault, calculate:
(i)
the subtransient current in the fault,
(ii)
the subtransient current in the breaker B,
(iii)
the momentary current in braker B, and
(iv)
the current to be interrupted by breaker B in five cycles.
Given, Reactance of the transmission line = 15% o a base of 25 MVA, 66 kV. Assume that the system
is operating on no load when the fault occurs.
Fig. 9.8
DMAM (13/22)
(a)
(b)
(d)
(c)
Fig. 9.9
Solution: Choose a system base of 25 MVA.
For generator base voltage: 11 kV
For line base voltage: 66 kV
For Motor base voltage: 6.6 Kv
(i) For each motor
25
X ' ' = j 0.25 ×
= j1.25 pu
dm
5
Line, transformer and generator reactance are already given on proper base reactances.
The circuit model of the system for fault calculations is given in Fig. 9.9(a). The system being initially
on no-load, the generator and motor induced emfs are identical. The circuit can therefore be reduced to
that of Fig. 9.9(b) and the to Fig. 9.9(c). Now
1
1
I
= 3×
×
= − j 4.22 pu
SC
j1.25 j1.55
25 × 1000
= 2187 A
Base current in 6. kV circuit = =
3 × 6.6
DMAM (14/22)
= 4.22 × 2187 = 9229 A
SC
(ii) From Fig. 9.9(c), current through circuit breaker B is
1
1
I
= 2×
×
= − j3.42 pu
SC ( B)
j1.25 j1.55
I
= 3.42 × 2187 = 7479.5 A
SC( B)
(iii) For finding momentary current through the breaker, we must add the dc off-set current to the
symmetrical subtransient current obtained in part (i). Rather than calculating the dc –off-set current,
allowance is made for it on an empirical basis.
Momentary current through breaker B = 1.6×7479.5 = 11967 A
(iv) To compute the current to interrupt by the breaker, motor subtransient reactance (Xd″ = j0.25) is
now replaced by transient reactance (Xd′ = j0.30).
25
X ' = j 0.3 ×
= j1.5 pu
dm
5
The reactances of the circuit of Fig. 9.9(c) now modify to that of Fig. 9.9(d). Current (symmetrical) to
1
1
×
= − j3.1515 pu
be interrupted by the breaker (as shown by arrow) = 2 ×
j1.5 j1.55
Allowance made of the dc off-set value by multiplying with a factor of 1.1. Therefore, the current to be
interrupted is
I
= 1.1 × 3.1515 × 2187 = 7581 A
int(B)
I
[Internal
Voltages of Loaded Machines Under Transient Conditions (Ref. 1,
p. 254)]
Short Circuit of Loaded Synchronous Machine
(a) Usually steady-state equivalent circuit
(Xs=Xd) of synchronous generator with
load.
(Ref. 5, p. 313)
(b) Equivalent circuit of synchronous
generator for calculation of I″.
Zext: external impedance between generator
terminal and fault point (F or P) where the fault
occurs
IL: current flowing before fault occurs
Vf: voltage at the fault point
Vt: terminal voltage of the generator
Eg″: subtransient internal voltage or voltage
behind the subtransient reactance
Eg′: transient internal voltage or voltage behind
the transient reactance
(c) Equivalent circuit of synchronous
generator for calculation of I′.
Fig. 10.7 [Ref. 1] Equivalent circuit for a synchronous generator supplying balanced three-phase load.
DMAM (15/22)
Consider a generator that is loaded when the fault occurs. Fig. 10.7(a) shows the equivalent circuit of a
synchronous generator that has a balance three-phase load.
The equivalent circuit of the synchronous generator is its no-load voltage Eg in series with its
synchronous impedance Xs or Xd. If a three-phase fault occurs at point F or P in the system, a short
circuit from P to neutral in the equivalent circuit does not satisfy the conditions for calculating
subtransient current since the reactance of the generator must be Xd″ if subtransient current I″ is
calculated or Xd′ if transient current I′ is calculated.
As shown in Fig. 10.7 the current relation of voltage can be written as:
E = V + jI X
Here IL is called prefault current.
g
t
L d
E ' ' = V + jI X ' '
Here IL is called postfault subtransient current.
g
t
L d
E ' = V + jI X '
Here IL is called postfault transient current.
g
t
L d
Similarly, the subtarnsient internal voltage and transient internal voltage for a shynchronous motor are
given by:
E ' ' = V − jI X ' '
m
t
L d
'
E = V − jI X '
m
t
L d
Example 10.2 [Ref.1, p.256] A synchronous generator and motor are rated 30 MVA, 13.2 kV, and
both have subtransient reactances 20%. The line connecting them has a reactance of 10% on the basis
of the machine ratings. The motor is drawing 20 MW at 0.8 power factor leading and a terminal
voltage of 12.8 kV when a symmetrical three-phase fault occurs at the motor terminals. Find the
subtransient current in the generator, motor, and fault by using the internal voltage of the machine.
Fig. One-line diagram for Example 10.2.
Solution: Chose a base 30 MVA, 13.2 kV. Fig. 10.8 shows the equivalent circuit of the system as
described.
Neutral bus
Neutral bus
(a) Before the fault
(b) After the fault
Fig. 10.8 Equivalent circuit for Example 10.2.
Base current, I
DMAM (16/22)
base
=
30 × 106
3 × 13.2 × 103
= 1312 A
Given, the voltage at the fault is 12.8 kV which we choose a reference.
12.8
Vf =
= 0.97∠0° pu
13.2
20 × 106
I =
= 1128∠36.9° A
L
3 × 12.8 × 103 × 0.8
1128∠36.9°
I =
= 0.86∠36.9° = 0.69 + j 0.52 pu
L
1312
For the generator:
Vt =0.97+j0.1(0.69+j0.52)=0.918+j0.069 pu
Eg″=0.918+j0.069+j0.2(0.69+j0.52)=0.814+j0.207 pu
Ig″=(0.814+j0.207)/(j0.3)=0.69-j2.71 pu = 1312(0.69-j2.71) A = (905 –j3550) A
For the motor:
Vt = Vt =0.97∠0o.
Em″=0.918+j0.0-j0.2(0.69+j0.52)=1.074-j0.138 pu
Im″=(1.074-j0.138)/(j0.2)=-0.69-j5.37 pu = 1312(-0.69-j5.37) A = (-905 –j7050) A
In the fault:
If″= Ig″+ Im″=(0.69-j2.71)+( -0.69-j5.37)= -j8.08 pu = 1312(-j8.08) A = -j10,600 A.
Example 9.3 [Ref.5, p.314] A synchronous generator and a synchronous motor each rated 25 MVA,
11 KV having 15% subtransient reactance are connected through transformers and a line as shown in
Fig. 9.12(a). The transformers are rated 25 MVA, 11/66 kV with leakage reactance of 10% each. The
line has a reactance of 10% o a base of 25 MVA, 66 kV. The motor is drawing 15 MW at 0.8 power
factor leading and a terminal voltage of 10.6 kV when a symmetrical three-phase fault occurs at the
motor terminals. Find the subtransient current in the generator, motor and fault. [Hint. same as
Example 10.2 [Ref.1, p.256]]
(a) One-line diagram for the system of Example 9.3.
(b) Prefault equivalent circuit
(c) Equivalent circuit during fault
Fig. 9.12
DMAM (17/22)
Fault Current Calculation Using Thevenin’s Theorem [(Ref. 1, p. 258)]
The Thevenin impedance at fault point P of Fig. 10.7(b) is given as:
Z (Z
+ jX ' ' )
'
'
L
ext
d
Z = (Z
+ jX ) // Z
=
th
ext
d
ext
Z +Z
+ jX ' '
L
ext
d
The Thevenin voltage equal to Vf, the voltage at the fault point before the fault occurs.
Upon the occurance of a three phase short circuit at P, the subtransient current in the fault is
+ jX ' ' )
V
V (Z + Z
f
f
L
ext
d
=
I '' =
Z th
Z (Z
+ jX ' ' )
L ext
d
Fault Current Calculation Using Superposition Theorem [(Ref. 1, p.
258)]
The fault is divided between the parallel circuits of the generators inversely as their impedances, the
resulting values are the currents from each machine due to nly the change in voltage at the fault point.
The fault currents thus attributed to the two machines must be added the current flowing in each before
the fault occurred to find the total current in the machines after the fault.
The supper position theorem supplies the reason for adding the current flowing before the fault to the
current computed by Thevein’s theorem.
Fig. 10.10(a) shows the a generator having a voltage Vf conncetd at the fault and equal to the voltage at
the fault before the fault occurs. This generator has no effect on the current flowing before the fault
occurs, and the circuit corresponding to that of Fig. 10.8 (a).
Adding in series with Vf another generator having the circuit of Fig. 10.10(b), which corresponds to
that of Fig. 10.8(b).
The principle of superposition, applied by first shorting Eg′′, Em′′, and Vf, gives the currents found by
distributing the fault current between the two generators inversely as the impedances of their circuits.
Then shorting the remaining generator – Vf with Eg′′, Em′′, and Vf, in the circuit gives the current
flowing before the fault.
Adding the two values of current in each branch gives the current in the branch after the fault.
If Vf equals the prefault voltage at the fault, then the current calculation with Vf can be consider zero
i.e. If2=0 and Vf, has no effect, since it represents the system before fault occurs.
(a) Before the fault
(b) During the fault
Fig. 10.10 Circuits illustrating the application of the superposition theorm to determine the proportion
of the fault current in each branch of the system.
DMAM (18/22)
Example 10.3 [Ref.1, p.259] Solve Example 10.2 by the using of Thevenin’s theorem.
Solution:
j 0.3 × j 0.2
Z = (Z
+ jX ' ' ) // Z =
= j 0.12
th
ext
d
L j 0.3 + j 0.2
12.8
Vf =
= 0.97∠0° pu
13.2
V
0.97∠0°
f
I 'f' =
=
= − j8.08 = 1312(-j8.08) A = -j10,600 A.
Z th
j 0.12
Example 10.3.1 [Ref.1, p.260] Apply the above principle in Example 10.3 by the using of
superposition theorem.
Solution:
Using the current division
Subtransient generator current neglecting prefault current:
0.2
0.2
I '' =
I '' =
× − j8.08 = − j3.23 pu
g1 0.2 + 0.3 f 0.2 + 0.3
Subtransient motor current neglecting prefault current:
0.3
0.3
I '' =
I '' =
× − j8.08 = − j 4.85 pu
m1 0.2 + 0.3 f 0.2 + 0.3
To these current must be added the prefault current IL to obtain the total subtransient currents in the
machines:
Subtransient generator current including prefault current:
I ' ' = I ' ' + I L = − j3.23 + 0.69 + j 0.52 = 0.69 − j 2.71 pu
g
g1
Subtransient motor current including prefault current:
I ' ' = I ' ' − I L = − j 4.85 − 0.69 − j 0.52 = −0.69 − j5.37 pu
m
m1
Example 7.3 [Ref.3, p.366] The synchronous generator in the following figure is operated at rated
MVA, 0.95 power factor lagging and at 5% above rated voltage when a bolted three-phase short circuit
occurs at bus 1. Calculate the per-unit values of (i) subtransient current; (ii) Subtransient generator and
motor current, neglecting prefault current, and (iii) subtransient generator and motor currents including
prefault current.
Fig. 7.3 Single-line diagram of a synchronous generator feeding a synchronous motor.
Solution:
(i) Using a 100 MVA base, the base impedance in the zone of the transmission line is
(138) 2
=
= 190.44 Ω
Z
base, line
100
20
=
= 0.1050 pu
And X
line 190.44
The per-unit reactances are shown in Fig.7.4. From the first circuit in Fig. 7.4(d), the Thevenin
DMAM (19/22)
impedance as viewed from the fault is
0.15 × 0.505
Z
=X
= j
= j 0.11565 pu
TH
TH
0.15 + 0.505
And the prefault voltage at the generator terminal is
V = 1∠0° pu
f
The subtransient fault current is then
V
1∠0°
f
I '' =
=
= − j 9.079 pu
f Z
j 0.11565
TH
(a) Three-phase short-circuit
(b) Short-circuit represented by two
opposing voltage sources
(c) Application of superposition
(d) Vf set equal to prefault voltage at fault.
Fig. 7.4 Application of superposition to a power system three-phase circuit.
(ii) Using the current division in the first circuit of Fig. 7.4(d)
0.505
I '' =
I ' ' = (0.7710)(− j9.079) = − j 7.000 pu
g1 0.505 + 0.15 f
0.15
I '' =
I ' ' = (0.2290)(− j 9.079) = − j 2.079 pu
m1 0.505 + 0.15 f
If Vf equals the prefault voltage at the fault, then the current calculation with Vf can be consider zero
i.e. If2=0 and Vf, has no effect, since it represents the system before fault occurs.
(iii) The generator base current is
DMAM (20/22)
100
=
= 4.1837 kA
base, gen
3 × 13.8
And the prefault generator current is
100
I =
∠ − cos −1 0.95 = 3.9845∠ − 1819° kA
L
3 × (13.8 × 1.05)
I
3.9845∠ − 1819°
= 0.9524∠ − 1819°∠ − cos −1 0.95 = 0.9048 − j 0.2974 pu
L
4.1837
The subtarnsient generator and motor currents, including prefault current and then,
I ' ' = I ' ' + I L = − j 7.000 + 0.9048 − j 0.2974 = −0.9048 − j 7.297 = 7.353∠ − 82.9° pu
g
g1
I ' ' = I ' ' − I L = − j 7.000 − 0.9048 + j 0.2974 = −0.9048 − j1.782 = 1.999∠243.1° pu
m
m1
I
=
The Bus Impedance Matrix in Fault Calculations [Ref. 1, p. 261]
(a) One line diagram of a power system.
(b) Reactance diagram at steady-state condition.
Fig. 7.3
Let us proceed to the general equations by starting with a specific network with which we are already
familiar. If we change the reactance in series with the generated voltages of the circuit shown in Fig.
7.3 become subtransient internal voltages, we have the network shown in Fig. 10.11.
A three-phase fault at bus 4 is simulated by the network of Fig. 10.12 where the impedance values of
Fig. 10.11 have been changed to admittances.
The generated voltages Vf and –Vf in series constitute the short circuit. Generated voltage Vf alone in
this branch would cause no current in the branch. With Vf and –Vf in series the branch is a short circuit
and the branch current If′′.
If Ea′′, Eb′′, Ec′′, and Vf are short-circuited, the voltages and current are those only to –Vf. Then the only
current entering a node from a source is that from –Vf and is - If′′ into node 4 (If′′ from node 4) since
there is no current in the branch until the insertion of -Vf. The node equations in matrix form for the
network with –Vf the only source ar
⎡ V1∆ ⎤
⎡ 0 ⎤
0.0
4.0
5.0 ⎤ ⎡ V1∆ ⎤
⎡− 12.33
⎢ ∆ ⎥
⎢
⎥
⎢ 0 ⎥
⎢
V
− 10.83
2.5
5.0 ⎥⎥ ⎢ V2∆ ⎥
⎢
⎥ = j ⎢ 0.0
= Ybus ⎢ 2∆ ⎥
∆ ⎥
⎢
⎢
⎢ 0 ⎥
⎢ 4.0
V3 ⎥
− 17.83 8.0 ⎥ V3
2.5
⎢
⎥
⎢
⎥
⎢
⎥
⎢
⎥
''
− 18.0⎦ ⎣⎢− V f ⎦⎥
5.0
8.0
⎣ 5.0
⎣⎢− − I f ⎦⎥
⎣⎢− V f ⎦⎥
Where subscript ∆ indicates that the change in voltage due to the fault and the due only to –Vf.
DMAM (21/22)
Fig. 10.11 Reactance diagram obtained from Fig. 7.3
by subtatituteing subtransient for synchronous
reactances of the machines and subtarnsient internal
voltages for no-load generated voltages. Reactance
values are marked as per unit.
Fig. 10.12 Circuit of Fig. 10.11 with
admittances marhed in per unit and a three
phase fault on bus 4 of the system simulated
by –Vf and Vf in seies.
The bus voltages due to –Vf are given by
−1
⎡ V1∆ ⎤
0.0
4.0
5.0 ⎤ ⎡ 0 ⎤
⎡0⎤
⎡ 0 ⎤
⎡− 12.33
⎢ ∆ ⎥
⎢
⎥
⎢
⎥
⎢ 0 ⎥
⎢
0⎥
2.5
5.0 ⎥⎥ ⎢ 0 ⎥
− 10.83
−1 ⎢
⎢ V2 ⎥ = j ⎢ 0.0
⎥
= Ybus
= Z bus ⎢
⎢ V3∆ ⎥
⎢0⎥
⎢ 0 ⎥
⎢ 4.0
2.5
− 17.83 8.0 ⎥ ⎢ 0 ⎥
⎢
⎥
⎢ ⎥
⎢ '' ⎥
⎢
⎥ ⎢ '' ⎥
5.0
8.0
− 18.0⎦ ⎣⎢− I f ⎦⎥
⎢⎣− V f ⎥⎦
⎣ 5.0
⎣⎢ I f ⎦⎥
⎣⎢− I f ⎦⎥
⎡ V1∆ ⎤
⎡ 0 ⎤ ⎡ Z11 Z12 Z13 Z14 ⎤ ⎡ 0 ⎤
⎢ ∆ ⎥
⎢
⎥ ⎢
⎥
⎥⎢
⎢ V2 ⎥ = Z ⎢ 0 ⎥ = ⎢ Z 21 Z 22 Z 23 Z 24 ⎥ ⎢ 0 ⎥
bus
⎢ V3∆ ⎥
⎢ 0 ⎥ ⎢ Z 31 Z 32 Z 33 Z 34 ⎥ ⎢ 0 ⎥
⎢
⎥
⎢ '' ⎥ ⎢
⎥ ⎢ '' ⎥
⎢⎣− V f ⎥⎦
⎣⎢− I f ⎦⎥ ⎣ Z 41 Z 42 Z 43 Z 44 ⎦ ⎣⎢− I f ⎦⎥
If total number of bus is N, and fault is occurred in bus k, then considering
Vk∆ = −V f ;
I k''= − I f''= − I f
Z 1( k +1)
Z 1( k + 2 )
.... Z 1k
.... Z1N ⎤ ⎡ 0 ⎤
Z 12
⎡ V1∆ ⎤ ⎡ Z 11
⎢ ∆ ⎥ ⎢
Z 2 ( k +1)
Z 2 ( k + 2) .... Z 2 N ⎥⎥ ⎢ 0 ⎥
Z 22 .... Z 2 k
⎢ V2 ⎥ ⎢ Z 21
⎢
⎥
⎢ .... ⎥ ⎢ ....
....
....
....
....
....
....
.... ⎥ ⎢ .... ⎥
⎢
⎥ ⎢
⎥⎢
⎥
Z k ( k +1)
Z k ( k + 2) .... Z kN ⎥ ⎢− I fk'' ⎥
Z k 2 .... Z kk
⎢− V fk ⎥ = ⎢ Z k1
⎢ V ∆ ⎥ ⎢ Z ( k +1)1 Z ( k +1) 2 .... Z ( k +1) k Z ( k +1)( k +1) Z ( k +1)( k + 2 ) .... Z ( k +1) N ⎥ ⎢ 0 ⎥
⎢ k∆+1 ⎥ ⎢
⎥⎢
⎥
⎢ Vk + 2 ⎥ ⎢ Z ( k + 2 )1 Z ( k + 2 ) 2 .... Z ( k + 2) k Z ( k + 2)( k +1) Z ( k + 2)( k + 2 ) .... Z ( k + 2) N ⎥ ⎢ 0 ⎥
⎢ .... ⎥ ⎢ ....
....
....
....
....
....
....
.... ⎥ ⎢ .... ⎥
⎢
⎥ ⎢
⎥⎢
⎥
Z N ( k +1)
Z N ( k + 2) Z N 1 Z NN ⎥⎦ ⎣⎢ 0 ⎦⎥
Z N 2 Z N 1 Z Nk
⎢⎣ V N∆ ⎥⎦ ⎢⎣ Z N 1
The diagonal elements of Zbus matrix are called self-impedance. And the off-diagonal impedances are
called mutual-impedance.
DMAM (22/22)
⎡0.1488
⎢ 0.0651
Z bus = j ⎢
⎢0.0864
⎢
⎣0.0978
And so
− V f = − Z 44 I 'f'
;
V
f
I 'f' =
Z 44
0.0651 0.0864 0.0978⎤
0.1554 0.0799 0.0967⎥⎥
0.0799 0.1341 0.1058⎥
⎥
0.0967 0.1058 0.1566⎦
Z14 ''
If
Z 44 ;
Z
V2∆ = − Z 24 I 'f' = − 24 I 'f'
Z 44 ;
V1∆ = − Z14 I 'f' = −
Z 34 ''
If
Z 44
The faulted network is assumed to have been without loads before the fault. In such a case no current
is flowing before the fault, and all voltages throughout the network are the same and equal to Vf. This
assumption simplifies our work considerably , and applying the principle of superposition gives:
⎛ Z ⎞
Z
V1 = V f + V1∆ = V f − I 'f' Z 14 = V f − V f 14 = V f ⎜⎜1 − 14 ⎟⎟
Z 44
⎝ Z 44 ⎠
V3∆ = − Z 34 I 'f' = −
V2 = V f + V2∆ = V f − I 'f' Z 24 = V f − V f
⎛ Z ⎞
Z 24
= V f ⎜⎜1 − 24 ⎟⎟
Z 44
⎝ Z 44 ⎠
V3 = V f + V3∆ = V f − I 'f' Z 34 = V f − V f
⎛ Z ⎞
Z 34
= V f ⎜⎜1 − 34 ⎟⎟
Z 44
⎝ Z 44 ⎠
V4 = V f − V f = 0
These voltages exist when subtransient current flows and Zbus has been formed for a network having
subtransient values for generator reactances.
In general terms for a fault on bus k, and neglecting prefault currents,
Vf
(10.16)
If =
Z kk
And the postfault voltage at bus n is
⎛ Z ⎞
Z
Vn = V f + Vn∆ = V f − I f Z nk = V f − V f nk = V f ⎜⎜1 − nk ⎟⎟
Z kk
⎝ Z kk ⎠
Usually, Vf is assumed to be 1.0∠0o per-unit and with this assumption for our faulted network
1
I 'f' =
= − j 6.386 pu
j 0.1566
⎛
j 0.0978 ⎞
⎟ = 0.3755 pu
V1 = 1⎜⎜1 −
j 0.1566 ⎟⎠
⎝
⎛
j 0.0967 ⎞
⎟ = 0.3825 pu
V2 = 1⎜⎜1 −
j 0.1566 ⎟⎠
⎝
⎛
j 0.1058 ⎞
⎟ = 0.3244 pu
V3 = 1⎜⎜1 −
j 0.1566 ⎟⎠
⎝
Current flow from bus 1 to bus 3
DMAM (23/22)
V1 − V3
0.3755 − 0.3244
= y13 (V1 − V3 ) =
= − j 0.2044 pu
z13
j 0.25
Current flow from generator Ga to bus 1
E g'' − V1
1 − 0.3755
''
I 13 =
= y11 ( E g'' − V1 ) =
= − j 2.0817 pu
z11
j 0.3
Other currents can be found in a similar manner, and voltages and currents with the fault on any other
bus are calculated just as easily from the impedance matrix.
I 13'' =
Example 10.3.2 A synchronous generator and motor are connected through a transmission line as
shown in the following figure. The motor is drawing 20 MW at 0.8 power factor leading and a terminal
voltage of 12.8 kV when a symmetrical three-phase fault occurs at the motor terminals. (i) Determine
the bus impedance matrix. (ii) For a bolted three-phase short-circuit at bus 2, use Zbus to calculate the
subtransient fault current and the contribution to the fault current from the transmission line. (iii) the
subtransient generator and motor current including prefault current.
Solution: Given, the voltage at the fault is 12.8 kV which we choose a reference.
12.8
Vf =
= 0.97∠0° pu
13.2
30 × 106
I
=
= 1312 A
base
3
3 × 13.2 × 10
20 × 106
= 1128∠36.9° A
I =
L
3 × 12.8 × 103 × 0.8
1128∠36.9°
I =
= 0.86∠36.9° = 0.69 + j 0.52 pu
L
1312
y11 = y22 =1/ j0.2 = - j5.0; y12 = y21 = 1/ j0.1= - j10.0
⎡− j15.0 j10.0 ⎤
Ybus = ⎢
⎥ pu
⎣ j10.0 − j15.0⎦
⎡ j 0.12 j 0.08⎤
Z bus = ⎢
⎥ pu
⎣ j 0.08 j 0.12⎦
The subtransient fault current at bus 2 is
Vf
Vf
0.97∠0°
If =
=
=
= − j8.0833 pu
Z kk Z 22
j 0.12
⎛ Z ⎞
⎛
j 0.08 ⎞
⎟ = 0.3233∠0°
V1 = V f ⎜⎜1 − 21 ⎟⎟ = 0.97∠0°⎜⎜1 −
j 0.12 ⎟⎠
⎝
⎝ Z 22 ⎠
⎛ Z ⎞
V2 = V f ⎜⎜1 − 22 ⎟⎟ = 0
⎝ Z 22 ⎠
V2 − V1 0 − 0.3233
=
= j 4.041 pu
z 21
j 0.08
Subtransient generator current neglecting prefault current:
I 21 =
DMAM (24/22)
0.2
0.2
I '' =
I =
× − j8.08 = − j3.23 pu
gf 0.2 + 0.3 f 0.2 + 0.3
Subtransient motor current neglecting prefault current:
0.3
0.3
I '' =
If =
× − j8.08 = − j 4.85 pu
mf 0.2 + 0.3
0.2 + 0.3
To these current must be added the prefault current IL to obtain the total subtransient currents in the
machines:
Subtransient generator current including prefault current:
I ' ' = I ' ' + I L = − j3.23 + 0.69 + j 0.52 = 0.69 − j 2.71 pu
g
gf
Subtransient motor current including prefault current:
I ' ' = I ' ' − I L = − j 4.85 − 0.69 − j 0.52 = −0.69 − j5.37 pu
m
mf
Example 7.4 [Ref. 3, p. 369] Faults at bus 1 and 2 in Fig. 7.3 are of interest. The prefault voltage is
1.05 pu and prefault load current is neglected. (i) Determine the bus impedance matrix. (ii) For a
bolted three-phase short-circuit at bus 1, use Zbus to calculate the subtransient fault current and the
contribution to the fault current from the transmission line. (iii) Repeat part (b) for bolted three-phase
short-circuit at bus 2.
Fig. 7.3 Single-line diagram of a synchronous generator feeding a synchronous motor.
(a) Reactance diagram obtained from Fig. 7.3.
(b) Circuit of Fig. (a) with admittances marked
Reactance values are marked as per unit.
in per unit.
Fig. 7.4
Solution:
(i) The circuit of Fig. 7.4 (a) is redrawn in Fig. 7.4(b) showing per-unit admittance rather than per-unit
impedance values. Neglectring prefault load current . Eg′′= Em′ = Vf =1.05∠0o pu. From Fig. 7.4 (b) the
bus admittance matrix is
⎡ 9.9454 − 3.2787⎤
Ybus = − j ⎢
⎥ pu
⎣− 3.2787 8.2787 ⎦
⎡0.11565 0.04580⎤
-1
Z bus = Ybus
= + j⎢
⎥ pu
⎣0.04580 0.13893⎦
(ii) The subtransient fault current at bus 1 is
DMAM (25/22)
If =
Vf
Z kk
=
Vf
Z 11
=
1.05∠0°
= − j 9.079 pu
j 0.11565
⎛ Z ⎞
V1 = V f ⎜⎜1 − 11 ⎟⎟ = 0
⎝ Z11 ⎠
⎛ Z ⎞
⎛
j 0.04580 ⎞
⎟ = 0.6342∠0°
V2 = V f ⎜⎜1 − 21 ⎟⎟ = 1.05∠0°⎜⎜1 −
j 0.11565 ⎟⎠
⎝
⎝ Z11 ⎠
V2 − V1 0.6343 − 0
=
= − j 2.079 pu
z 21
j 0.3050
(iii) The subtransient fault current at bus 2 is
Vf
Vf
1.05∠0°
If =
=
=
= − j 7.558 pu
Z kk Z 22
j 0.13893
I 21 =
⎛ Z ⎞
⎛
j 0.04580 ⎞
⎟ = 0.7039∠0°
V1 = V f ⎜⎜1 − 12 ⎟⎟ = 1.05∠0°⎜⎜1 −
j 0.13893 ⎟⎠
⎝
⎝ Z 22 ⎠
⎛ Z ⎞
V2 = V f ⎜⎜1 − 22 ⎟⎟ = 0
⎝ Z 22 ⎠
I 12 =
V1 − V2 0.7.39 − 0
=
= − j 2.308 pu
z12
j 0.3050
Example 10.3 [Ref. 2, p. 394] A three phase fault occurs at bus 2 of the network of Fig. 10.5.
Determine the initial symmetrical fault current (that is subtransient current) in the fault, the voltages at
buses 1, 3, 4 during fault, the current flow in the line from bus 3 to 1 and the current contribution to the
fault from lines 3 to 2 and 1 to 2, 4 to 2. Take the prefault voltage Vf at bus equal to 1.0∠0o pu and
neglect all prefault currents.
Fig. 10.5
Solution:
[[[y11=0;
z12=0.125i;
DMAM (26/22)
z21=z12;
y12=1/z12;
y21=y12;
z13=0.25i;
z14=0.4i;
z31=z13;
z41=z14;
y13=1/z13;
y14=1/z14;
y31=y13;
y41=y14;
y22=0;
z23=0.25i;
z24=0.2i;
z32=z23;
z42=z24;
y23=1/z23;
y24=1/z24;
y32=y23;
y42=y24;
z33=0.3i;
y33=1/z33;
y34=0;
y43=0;
z44=0.3i;
y44=1/z44;
Y11=y11+y12+y13+y14;
Y33=y31+y32+y33+y34;
Y12=-y12;
Y13=-y13;
Y14=-y14;
Y21=Y12;
Y31=Y13;
Y41=Y14;
Y23=-y23;
Y24=-y24;
Y32=Y23;
Y42=Y24;
Y34=-y34;
Y43=Y34;
Ybus=[Y11
Y21
Y31
Y41
Y12
Y22
Y32
Y42
Y13
Y23
Y33
Y43
Y22=y21+y22+y23+y24;
Y44=y41+y42+y43+y44;
Y14
Y24
Y34
Y44];
Zbus=inv(Ybus);]]]
Ybus
j 4.0
j 2.5 ⎤
j8.0
⎡− j14.5
⎢ j8.0 − j17.0
j 4.0
j5.0 ⎥⎥
⎢
=
⎥
⎢ j 4.0
0
j 4.0 − j11.33
⎥
⎢
− j10.833⎦
0
j5.0
⎣ j 2.5
⎡ j 0.2436
⎢ j 0.1938
Z bus = ⎢
⎢ j 0.1544
⎢
⎣ j 0.1456
Vf
Vf
If =
=
=
Z kk Z 22
j 0.1938
j 0.2295
j 0.1494
j 0.1506
j 0.1544
j 0.1494
j 0.2954
j 0.1046
j 0.1456⎤
j 0.1506⎥⎥
j 0.1046⎥
⎥
j 0.1954⎦
1.0
= − j 4.3573 pu
j 0.2295
Vn = V f + Vn∆ = V f − I f Z nk = V f − V f
⎛ Z
Z nk
= V f ⎜⎜1 − nk
Z kk
⎝ Z kk
⎛ Z ⎞
⎛
j 0.1938 ⎞
⎟ = 0.1556
V1 = V f ⎜⎜1 − 12 ⎟⎟ = 1.0⎜⎜1 −
j 0.2295 ⎟⎠
⎝
⎝ Z 22 ⎠
⎛ Z ⎞
V2 = V f ⎜⎜1 − 22 ⎟⎟ = 0
⎝ Z 22 ⎠
⎛ Z ⎞
⎛
j 0.1494 ⎞
⎟ = 0.3490
V3 = V f ⎜⎜1 − 32 ⎟⎟ = 1.0⎜⎜1 −
j 0.2295 ⎟⎠
⎝
⎝ Z 22 ⎠
DMAM (27/22)
⎞
⎟⎟
⎠
⎛ Z ⎞
⎛
j 0.1506 ⎞
⎟ = 0.3438
V4 = V f ⎜⎜1 − 42 ⎟⎟ = 1.0⎜⎜1 −
j 0.2295 ⎟⎠
⎝
⎝ Z 22 ⎠
Z bus
⎡ j 0.2436
⎢ j 0.1938
=⎢
⎢ j 0.1544
⎢
⎣ j 0.1456
j 0.1938
j 0.2295
j 0.1494
j 0.1506
j 0.1544
j 0.1494
j 0.2954
j 0.1046
j 0.1456⎤
j 0.1506⎥⎥
j 0.1046⎥
⎥
j 0.1954⎦
The current flow in the line from bus 3 to 1
V − V1 0.3490 − 0.1556
I 31 = 3
=
= − j 0.7736
z 31
j 0.25
The current contribution to the fault from lines 1 to 2:
V
0.1556
I 12 = 1 =
= − j1.2448 pu
z12
j 0.125
The current contribution to the fault from lines 3 to 2.
V
0.3490
I 32 = 3 =
= − j1.3960 pu
z 32
j 0.25
The current contribution to the fault from lines 4 to 2:
V
0.3490
I 42 = 4 =
= − j1.7190 pu
z 42
j 0.20
A Bus Impedance Matrix Equivalent Netwrok [Ref. 1, p. 264]
Fig. 7.6 [R. 3, p. 317] Bus equivalent circuit (rake equivalent).
Using Zbus, the fault currents in Fig. 7.6 are givn by
DMAM (28/22)
Fig. 7.6 shows a bus impedance
equivalent circuit that illustrates
the short-circuit current in an Nbus system. This circuit is given
the name rake equivalent due to
its shape, which is similar to a
garden rake.
Neglecting prefault load currents,
the internal vpltage sources of all
synchronous machines are equal
both in magnitude and phase. As
such, they can be connected, as
shown in in Fig. 7.7, and replaced
by one equivalent source Vf from
neutral bus 0 to a references bus,
denoted k. This equivalent source
is also shon in the rake equivalent
of Fig. 7.6.
Z 1( k +1)
Z 1( k + 2 )
Z 12
.... Z 1N ⎤ ⎡ I 1 ⎤
.... Z 1k
⎡ V f − V1 ⎤ ⎡ Z 11
⎢ V −V ⎥ ⎢
Z 2 ( k +1)
Z 2 ( k + 2) .... Z 2 N ⎥⎥ ⎢ I 2 ⎥
Z 22 .... Z 2 k
2 ⎥
⎢ Z 21
⎢ f
⎢
⎥
⎥ ⎢ ....
⎢
....
....
....
.... ⎥ ⎢ .... ⎥
....
....
....
....
⎥⎢
⎥ ⎢
⎢
⎥
Z k ( k +1)
Z k ( k + 2 ) .... Z kN ⎥ ⎢ I k ⎥
Z k 2 .... Z kk
⎢ V f − Vk ⎥ = ⎢ Z k 1
⎢V f − V( k +1) ⎥ ⎢ Z ( k +1)1 Z ( k +1) 2 .... Z ( k +1) k Z ( k +1)( k +1) Z ( k +1)( k + 2) .... Z ( k +1) N ⎥ ⎢ I ( k +1) ⎥
⎥⎢
⎥ ⎢
⎢
⎥
⎢V f − V( k + 2 ) ⎥ ⎢ Z ( k + 2 )1 Z ( k + 2 ) 2 .... Z ( k + 2 ) k Z ( k + 2)( k +1) Z ( k + 2 )( k + 2) .... Z ( k + 2 ) N ⎥ ⎢ I ( k + 2 ) ⎥
⎥ ⎢ ....
⎢
....
....
....
....
.... ⎥ ⎢ .... ⎥
....
....
....
⎥⎢
⎥ ⎢
⎢
⎥
Z N ( k +1)
Z N ( k + 2 ) Z N 1 Z NN ⎥⎦ ⎢⎣ I N ⎥⎦
Z N 2 Z N 1 Z Nk
⎢⎣ V f − V N ⎥⎦ ⎢⎣ Z N 1
Where, I1, I2, …. Are the branch currents and (Vf – V1), (Vf – V1), …. Are the voltages across the
branches.
If switch S in Fig. 7.6 is open, all currents are zero and the voltage at each bus with respect to the
neutral equals Vf. This corresponds to prefault conditions, neglecting prefault load currents.
If switch S is closed, corresponding to a short-circuit at bus k, Vk = 0 and all currents except Ik remain
zero.
The fault current If′′= Ik= Vf/Zkk, which agree with the above driven equation. This fault current also
induces a voltage drop ZnkIk = (Znk/Zkk)Vf across each branch n. The voltage at bus n with respect to the
neutral then equals Vf minus this voltage drop, which agree with the derived equation.
As shown in Fig. 7.6, subtransient fault currents throughout an N-bus system can be determined from
the bus impedance matrix and the prefault voltage. Zbus can be computed by first constructing Ybus, via
nodal equations, and then inverting Ybus. Once Zbus has been obtained, these fault currents are easily
computed.
Fig. 7.7 [Ref. 3, p. 372] Parallel connecion of unloaded synchronous machine internal voltage sources.
Example 10.4 [Ref.1, p. 265] Determine the bus matrix for the following network. Generator at buses
1 and 3 are rated 270 and 225 MVA, respectively. The generator subtransient reactances plus the
reactances of the transformers conncting them to the buses are each 0.30 per-unit on the generator
rating as base. The turns ratios of transformers are such that the voltage base in each generator circuit
is equal to the voltage rating of the generator. Include the generator and transformer reactances in the
matrix. Find the subtransient current in a three-phase fault at bus 4 and the current coming to the
faulted bus over each line. Prefault current is to be neglected and all voltages are assumed to be 1.0
per-unit before the fault occurs. System base is 100 MVA. Neglect all resistances.
DMAM (29/22)
One-line diagram for example 10.4.
Solution:
Converted to the 100 MVA base, the combined generator and transformer reactances are:
100
= 0.1111 pu
Generator at bus 1: X = 0.3 ×
270
100
= 0.1333 pu
Generator at bus 3: X = 0.3 ×
225
The network with admittances marked in per-unit is shown in Fig. 10.14 from which the node
admittance matrix is
7.937 ⎤
0.0
0.0
⎡− 22.889 5.952
⎢ 5.952 − 13.889 7.937
0.0 ⎥⎥
0.0
⎢
Ybus = j ⎢ 0.0
7.937 − 23.175 2.976 4.762 ⎥
⎢
⎥
2.976 − 6.944 3.968 ⎥
0.0
⎢ 0.0
⎢⎣ 7.937
4.762 3.968 − 16.667 ⎥⎦
0.0
Fig. 10.14 Reactance diagram for Example 10.4.
This 5×5 bus is inverted on a digital computer to yield the short-circuit matrix
DMAM (30/22)
⎡0.0793 0.0558 0.0382 0.0511 0.0608⎤
⎢0.0558 0.1338 0.0664 0.0630 0.0605⎥
⎢
⎥
Z bus = j ⎢0.0382 0.0664 0.0875 0.0720 0.0603⎥
⎢
⎥
⎢ 0.0511 0.0630 0.0720 0.2321 0.1002⎥
⎢⎣0.0608 0.0605 0.0603 0.1002 0.1301⎥⎦
The subtransient current in a three-phase fault on bus 4 is
Vf
1.0
I 'f' =
=
= − j 4.308 pu
Z 44
j 0.2321
At buses 3 and 5 the voltages are: Vn = V f − I f Z nk
V3 = V f − I 'f' Z 34 = 1.0 − ( j 4.308)( j 0.0720) = 0.6898 pu
V5 = V f − I 'f' Z 54 = 1.0 − ( j 4.308)( j 0.1002) = 0.5683 pu
Currents to the fault = Current from bus 3 to bus 4 + Current from bus 5 to bus 4
I f = 0.6898(− j 2.976) + 0.5683(− j3.968) = − j 2.053 − j 2.255 = − j 4.308 pu
From the same short-circuit matrix we can find similar information for fault on any other buses.
The Selection of Circuit Breakers [CBs] [Ref. 1, p. 267]
Circuit Breaker (CB): A circuit breaker is a mechanical switch capable of interrupting fault current
and of reclosing.
When a CB contacts separate while carrying current, an arc forms. The CB is designed to extinguish
the arc by elongating and cooling it.
Classification of CB: CBs are classified as power circuit breakers when they are intended for service
in ac circuits above 1.5 kV, and as low-voltage circuit breakers in ac circuits up to 1.5 kV.
Depending on the medium: air, oil, SH6 gas, vacuum - in which the arc is elongated. The arc can be
elongated either by a magnetic force or by a blast of air.
Reclosing the CB: Some CBs are equipped with a high speed automatic reclosing capability.
Voltage Ratings
Rated Maxium Voltage: Designates the maximum rms line-to-line operating voltage. The breaker
should be used in system with an operating voltage less than or equal to this rating.
Rated Voltage Range Factor: The range of voltage for which the symmetrical interrupting capability
times the operating voltage is constant.
Rated Low Frequency Withstand Voltage: The maximum 60 Hz rms line-to-line voltage that the
circuit breaker can withstand insulation damage.
Rated Impluse Withstand Voltage: The maximum crest voltage of a voltage pulse with standard rise
and delay times that the breaker insulation can withstand.
Current Ratings
Rated Continuous Current: The maximum 60 Hz rms current that the breaker can carry continuously
while it is the closed position without overheating.
RatedShort-circuit Current: The maximum rms symmetrical current that the breaker can safely
interrupt at rated maximum voltage.
Rated momentary Current: The maximum rms asymmetrical current that the breaker can withstand
while in the closed position without damage. Rated momentary current for stabdard breaker is 1.6 time
the symmetrical interrupting capability.
Rated Interuupting Time: The time in cycles on a 60 Hz basis from the instant the trip coil is
energized to the instat the fault current is cleared.
DMAM (31/22)
Modern Circuit breakers standards are based on symmetrical interrupting current. It is usually,
necessary to calculate only symmetrical fault current at a system location, and then select a breaker
with a symmetrical interrupting capability equal to or above the calculated current. The CB has the
additional capability to interrupt the asymmetrical (or total) fault current if the dc off-set is not too
large.
The maximum asymmetrical factor K(τ=0) =√3, which occurs at fault inception (τ=0). After fault
inception, the dc off-set current decays exponentially with time constant τL= (L/R) = (Xω/R), and the
asymmetrical factor decreases. Power circuit breaker with a 2 cycle rated interruption time are
designed for an asymmetrical interruption capability up to 1.4 times their symmetrical interruption
capability, whereas slower CBs have a lower asymmetrical interrupting capability.
[[Inclusion of dc component results in a rms value of current immediately after the fault, which is
higher than the subtransient current.
For oil circuit breakers above 5 kV the subtransient current multiplied by 1.6 is considered to be the
rms value of the current whose disruptive forces the breaker must withstand during the first half-cycle
after the fault occurs. This current is called momentary current, and for many years circuit breakers
were rated in terms of their momentary current.
The interrupting rating of a circuit breaker was specified in kVAs or MVAs. The interrupting kVA
equal √3 times the kV of the bus to which circuit breaker is connected times the current which the
breaker must be capable of interrupting when its contact part. This current is lower than the
momentary current and depends on the speed of the breaker, such as 8, 5, 3, or 1.5 cycles, which is a
measure of the time from the occurrence of the fault to the extinction of the arc.
Interrupting kVA = 3 × kV of the bus to which circuit breaker is connected× current
which the breaker must be capable of interrupting
The current which a breaker must interrupt is usually asymmetrical since it still contains some of
decaying dc component. A schedule of preferred ratings for as high-voltage oil circuit breakers
specifies the interrupting current rating of breakers in terms of the component of the asymmetrical
current which is symmetrically about the zero axis. This current is properly called the required
symmetrical interrupting capability or simply the rated symmetrical short-circuit current.
Selection of circuit breakers may also be made on the basis of total current (dc component included).]]
What is the current to be interrupted?
Modern day circuit breakers are designed to interrupt in first few cycles (five cycles or less). The dc
off-set has not been yet died out and so contributes to the current to be interrupted. Rather than
computing the value of the dc off-set at the time of interruption (since it is very complex for a large
network), the symmetrical short-circuit current alone is calculated. The figure is then increased by an
empirical multiplying factor to account for the dc off-set current as shown in the following Table
[Ref.5, p. 318].
Circuit Breaker Speed
8 cycles or slower
5 cycles
3 cycles
2 cycles
Multiplying Factor
1.0
1.1
1.2
1.4
Breakers are identified by nominal-voltage class, such as 69 kV.
DMAM (32/22)
Breakers are identified by nominalvoltage class, such as 69 kV. Among
other factors specified are rated
continuous current 1200 A, rated
short-circuit current at rated max
voltage I=19 kA, rated maximum
voltage Vmax=72.5, voltage range
factor K=1.21, and rated short circuit
current times operating voltage
(72.5×19000) is constant.
Imax=KI = 1.21×19=23 A at operating
voltage Vmin = Vmax/K = 72.5/1.21 =
60 kV.
At operating voltages V between
Vmin and Vmax, the symmetrical
interrupting
capability
is Fig. 7.8 [Ref.3, p. 381] Symmetrical interrupting capability of
a 69 kV class breaker.
I×Vmax/V=1378/V kA.
At operating voltages below Vmin, the symmetrical interrupting capability remains at Imax= 23 kA.
Breakers of the 115 kV class and higher have a K of 1.0.
Example 10.6 [Ref. 3, p.405] A 69-kV circuit breaker having a voltage range factor K of 1.21 and a
continuous current rating of 1200 A has a rated short-circuit current of 19,000 A at the maximum
rated voltage of 72 .5 kV. Determine the maximum symmetrical interrupting capability of the
breaker and explain its significance at lower operating voltages.
Solution: The maximum symmetrical interrupting capability is given by
Rated short-circuit current = 1.21×19,000 = 22,990 A
This value of symmetrical interrupting current must not be exceeded. From the definition, we have
Lower limit of operating voltage = 72.5/1.21 = 60 kV.
Hence, in the operating voltage range 72.5 to 60 kV, the symmetrical interrupting current may exceed
the rated short-circuit current of 19,000 A, but it is limited to 22,990 A. For example, at 66 kV the
interrupting current can be = 72.5/66×19,000 = 20, 871 A.
E/X method:
A simplified procedure for calculating the symmetrical short-circuit current, called the E/X method,
disregards all resistance, all static loads, and all prefault current. Subtransient reactance is used for
generators in the E/X method, and for synchronous motors the recommended reactance is the Xd′′ of the
motor times 1.5, which is the approximate value of the transient reactance of the motor.
The impedance by which the voltage Vf at the fault is divided to find short-circuit current must
examined when E/X method is used. In specifying a breaker for bus k this impedance is Zkk of the bus
impedance matrix with the proper machine reactances since the short-circuit current is expressed by
Eq. (0.16).
If the ratio of X/R of this impedance is 15 or less, a breaker of the correct voltage and kVA may be
used if its interrupting current rating is equal to or exceeds the calculation current.
If X/R ratio is unkown, the calculated current should be no more than 80% of the allowed value for the
breaker at the existing bus voltage.
The ANSI (American National Standards Institute) application guide specifies a corrected method to
account for ac and dc times constants for the decay of the current amplitude if the X/R ratio exceeds
15. The corrected method also considers breaker speed.
DMAM (33/22)
[[Induction motor
m
less thhan 50 hp are neglectedd, and variou
us multiplyinng factors arre applied too the Xd′′
of
o lager indu
uction motorrs dependingg on their sizze. If no moors are preseented, symm
metrical shorrt-circuit
current
c
equaals subtransieent current.]]
Table 7.10
0 Preferred rating for outtdoor CBs (A
ANSI)
V
Voltage
C
Current
Ratedd short
Rated
Rated voltage
Rated Maxx
Nominal
circuit current
continuus
Voltage (kV
V,
rangee factor
Threee-Phase
(at rateed max
current at 600
MVA
A Class
rms)
(K
K)
Hz (a, rms))
kV) (kA
A, rms)
500
5
25.8
2.15
1200
11
1500
48.3
1.21
1200
17
2500
72.5
1.21
1200
19
Ideentification
Nominal
voltage claass
(kV, rms))
23
46
69
Example
E
7..7 [Ref.3, p.. 381] The calculated
c
syymmetrical fault currennt is 17 kA at a three-ppase bus
where
w
the opperating volltage is 64 kV.
k The X/R
R ratio is unkknown. Select a CB from
m Table 7.100 for tis
bus.
b
T
69 kV
V class brreaker has a symmeetrical interrrupting cappability I×Vmax/V=
Soluion: The
19×72.5/64=
=21.5 kA at the operatinng voltage 644 kV.
The
T calculatted symmetrrical faul currrent, 17 kA,, is less than 80% of thiss capability, wich is requuirement
when
w
X/R is unknown. Therefore,
T
w select the 69 kV class breaker from
we
m Table 7.10
0.
Example
E
100.5 [Ref.1, p.
p 268] A 255 MVA 13.88 kV generaator with Xd′′ =15% is connected
c
thhrough a
transformer
t
to a bus whhich suppliees four identtical motors, as shown in Fig. 10.15. The subttransient
reactance
r
Xd′′ of each motor
m
is 20% on a basse of 5 MV
VA, 6.9 kV. The three-pphase ratingg of the
transformer
t
is 25 MVA
A, 13.8/6.9 kV
V with a leaakage reactaance 10%. T
The bus voltaage at the m
motors is
6.9 kV whenn a three-phaase fault occcurs at point P. For the faault specifiedd, determinee (i) the subttransient
current
c
in thhe fault, (ii)) the subtrannsient currennt in breaker A, and (iiii) the symm
metrical shorrt-circuit
interrupting
i
curent (as defined
d
for ciircuit breakeer applicationn) in the fauult and in breeaker A.
Fig. 10.115 One-line diagram
d
for Example
10.16 Reactance diagram
d
for Example
E
10..5.
100.5.
Solution:
(i) For a basse of 25 MVA
A, 13.8 kV in
i the generaator circuit, the base for the motors is
i 25 MVA, 6.9 kV.
The
T subtranssient reactan
nce of each motor
m
is
25
X d'' = 0.20 ×
= 1.0 pu
p
5
Fig.
F 10.16 iss the diagram
m with subtraansient values of reactannce marked. For a fault at
a P.
V f = 1.0 puu
Z th = j 0.125 pu
1.0
I 'f' =
= − j8.0 pu
j 0.125
The base current in the 6.9 kV circuit is
25 × 1000
I base =
= 2090 A
3 × 6.9
I 'f' = 8 × 2090 = 16720 A
(ii) The generator contributes a current of
j 0.25
I g = − j8.0 ×
= − j 4.0 pu
j 0.50
The four motors contributes a current of = -j8.0 – (-j4.0) = -j4.0 pu
Each motor contributes a current of = -j4.0/4 = -j1.0 pu
Through breaker A comes the contribution from the generator and three of the four motor.
The current which pass through the breaker A is
I '' = − j 4.0 + 3(− j1.0) = − j 7.0 pu = 7.0 × 2090 = 14630 A
(iii) To compute the current through breaker A to be interrupted, replace the subtransient reactance of
j1.0 by the transient reactance of j1.5 in the motor circuits of Fig. 10.16. Then
(1.5 / 4) × 0.25
= j 0.15 pu
(1.5 / 4) + 0.25
The total current is: (1/j0.15)
The generator contributes a current of: (1/j0.15)×[0.375/(0.375+0.25)]= -j4.0 pu
Each motor contributes a current of: (1/4)×(1/j0.15)×[0.25/(0.375+0.25)]= -j0.67 pu
The symmetrical short-circuit current to be interrupted is: (4.0+3×0.67) ×2090 = 12560 A
Z th = j
The usual procedure is to rate all the breakers connected to a bus on the basis of the current into a fault
on the bus. In that case the short-circuit current interrupting rating of the breakers connected to the 6.9
kV must be at least: 4+4×0.67=6.67 pu or 6.67 ×2090 = 13940 A.
A 14.4 kV circuit breaker has a rated maximum voltage 15.5 kV and a K of 2.67.
At 15.5 kV its rated short-circuit interrupting current is 8900 A.
The constant value of rated = 15.5×8900
This CB is rated for a symmetrical short-circuit interrupting current of: 2.67×8900 = 23760 A
(maximum voltage) at a voltage of 15.5/2.76 =5.8 kV (minimum voltage).
This current is the maximum current that can be interrupted even thoug the breaker may be in circuit of
lower voltage.
The short circuit interrupting current rating at 6.9 kV is: (15.5×8900)/6.9 = 20,000 A
The required capability of 13,940 A is well below 80% of 20,000 A, and the breaker is suitable with
respect to short-circuit current.
Calculation of Short-Circuit Current Using Bus Impedance Matrix Equivalent Network
DMAM (35/22)
10.16 Reactance diagram for Example 10.5.
10.17 Bus impedance equivalent network for
the bus impedance matrix of Example 10.5.
Two nodes have been identified in Fig. 10.16. Node 1 is the bus on the low-tension side of the
transformer, and node 2 is on the high-tension side. For motor reactance of 1.5 per unit.
1
1
1
= − j10
Y22 =
+
= − j16.67
j 0.1
j 0.15 j 0.1
⎡0.150 0.090⎤
⎡ − 12.67 10.0 ⎤
and its inverse is Z bus = j ⎢
The node admittance matrix is Ybus = j ⎢
⎥
⎥
⎣ 0.090 0.114⎦
⎣ 10.0 − 16.67 ⎦
Fig. 10.17 is the network corresponding to the bus impedance matrix closing S1 with S2 open
represents a fault on bus 1.
The symmetrical short-circuit interrupting current in a three-phase fault at node 1 is
1.0
I SC =
= − j 6.67 pu
j 0.15
The bus impedance matrix also gives us the voltage at bus 2 with the fault on bus 1.
V2 = 1.0 − I SC Z 21 = 1.0 − (− j 6.67)( j 0.09) = 0.4 pu
And since the admittance between nodes 1 and 2 is -j10.0, the current into the fault from the
transformer is
(0.4-0.0)(-j10)= - j4.0 pu
Y11 =
1
1
+
= − j12.67
j1.5 / 4 j 0.1
Y12 =
We also know immediately the short-circuit current in a three-phase fault at node 2 which, by referring
to Fig.10.17 with S1 open and S2 closed is
1.0
I SC =
= − j8.77 pu
j 0.114
Short Circuit Capacity (SCC)/ Fault Level of Bus
SCC is defined by the product of the magnitude of pre-fault voltage (Vf) and the post fault current (If).
For a solid fault, Zf= 0
Say, prefault voltage, Vf
Fault current, |If |= |Vf |/ XT ; XT = Total impedance in the faulted circuit
SCC= prefault voltage × postfault current
SCC= |Vf | ×|If |= |Vf |× |Vf |/ XT = |Vf |2/ XT
For, Vf ≈ 1 p.u.; SCC= 1/ XT p.u. (If all the values are given in p.u)
SCC= 1/ XT MVA (If all the values are given in actual values)
SCC= Base MVA/ Xs.
Power companies furnish data to a customer who must determine the fault current to specify circuit
breakers for an industrial plant or distribution system ay any point. Usually, the data supplied lists the
short-circuit MVA instead of Thevenin impedance, where
DMAM (36/22)
Short − Circuit MVA = 3 × (nominal kV) × I SC × 10 −3
Where ⎪Isc⎪ in amperes is the rms magnitude of the short-circuit current in a three-phase fault at the
connection point. Base megavoltampere are related to base kilovolt anad base amperes by
Short-circuit MVA = √3 × (nominal kV) × ⎪Isc⎪×10-3
If base kilovolt equal nominal kilovolt then
Short-circuit MVA = ⎪Isc⎪ pu
At nominal voltage the Thevenin equivalent circuit looking back into the system from the point of
connection is an emf of 1.0∠0o pu in series with the pu impedance Zth. Therefore, under short-circuit
conditions,
1. 0
1. 0
Z th =
pu =
pu
I SC
Short circuit MVA
With resistance and shunt capacitance neglected Zth =Xth, the single-phase Thevenin equivalent circuit
which represents the system is an emf equal to the nominal line voltage divided by √3 in series with an
inductive reactance of
(nominal kV/ 3 ) × 1000
X th =
Ω
I SC
(nominal kV) 2
Ω
Short − Circuit MVA
If base kV is equal to nominal kV, converting to per unit yields
I
Base MVA
= base
X th =
pu
Short − Circuit MVA I SC
X th =
Strength of a Bus
A symmetrical fault occurs at
bus-3.
Say, the pre-fault voltage at
bus-3 is 1.0 p.u. and just after
the fault this voltage will be
reduced to almost zero. The
reduction in voltage of
various buses indicates the
strength of the network.
So, the strength of a bus is the ability of the bus to maintain its voltage when a fault takes place at
other bus:
• Strength of a bus related to SCC.
• The higher the SCC of the bus, the more it will be able to maintain its voltage in the case of
fault on any other bus
• If XT =0 then SCC = ∞. For this case the bus is known as Infinite Strong Bus
Example: A small generating station has a busbar divided into three sections. Each section is
through a reactor rated at 5MVA, 0.1 p.u. A
generator of 8MVA. 0.15p.u. is connected to
each section of the bas bar. Determine the SCC
of the bus if a three-phase fault takes place on
one of the section of the sections of bus-bar.
Solution: Let the base: 8MVA in the generator circuit. Xg = 0.15 pu
DMAM (37/22)
So the reacctors reactannce at 8MVA
A, XT = 0.1×88/5 = 0.16 pu
p
XP=j(0.15+
+0.16)/2=j0..155
j1.5 × j (0.16 + 0.155)
X eq =
= j 0.1016
j1.5 + j (0.16 + 0.155)
j0.15 and j(0.16+0.155
5) is parallell to each withh respect to the fault buss bar
Xeq =1/j0.1016
6 = -j9.86 p..u.
SCC = 1/X
Considerinng only the magnitude
m
SC
CC = 9.86 p..u. = 9.86 puu × 8MVA = 78.73MVA
A
Example:
E
T generatiing station having
Two
h
1200M
MVA and 8000MVA resppectively andd operating at
a 11kV
are
a linked with
w an inter--connected cable
c
havingg reactance of
o 0.5Ω per phase. Deteermine SCC of each
station.
[X
[ g1=Xg2= 0..1 p.u]
Solution: Leet the base: 1200MVA, 11kV in the generator ciircuit.
Xg1= 0.1 p.uu.
Xg2= 0.1×(12
200/800)=0.115 p.u.
Base
B
impedaance= (11)2 /1200
Cable
C
= 0.5//[(11)2 /12000] = 4.96 p.uu
j 0.1 × j (0.15 + 4.96)
Say fault att the termin
nal of generaator-1: X eq =
= j 0.098
j 0.1 + j (0.15 + 4.96)
SCC= 1/Xeq =1/j0.098 = -j10.2
Considering
C
g only the maagnitude;
SCC = 10.2 p.u. = 10.2×
×1200= 12,2234MVA
j 0.15 × j (0.1 + 4.96)
nal of generaator-2: X eq =
= j 0.085
Say fault att the termin
j 0.1 + j (0.1 + 4.96)
SCC= 1/Xeq =1/j0.085 = -j11.73
Considering
C
g only the maagnitude;
SCC = 11.733 p.u. = 11.7
73×1200= 144,072 MVA
Problems
P
1
10.6
[Ref. 1, p. 273] Tw
wo synchronoous motors having subtrransient reacctances of 0.80
0
and
0.25
0
per unnit, respectivvely, on a base
b
of 480 V, 2000 kV
VA are connnected to a bus. This motor
m
is
connected
c
by a line haviing a reactan
nce of 0.023 Ω to a bus of
o a power syystem. At thhe power sysstem bus
the
t short-cirrcuit megavooltamperes of
o the powerr system are 9.6 MVA ffor a nominaal voltage off 480 V.
When
W
the voltage
v
at the
t motor bu
us is 440 V, neglect loaad current aand find the initial symm
metrical
rms
r
currentt in a three--phase fault at the motorr bus.
Solution: Leet base poweer 2000 kVA
A, base voltaage 480 V
When fault occurs, the bus voltage is: Vf = 440/480 = 0.92 pu
Base impedance of line = (0.480)2×1000/20000
= 0.1152 pu
Reactance of transmission line, Xline =
0.023/0.1152 = 0.2 pu
At
the
power
system
bus,
ISC=9.6×106/(√3×480)= 11547 A
Ibase = 2×106/(√3×480)= 2406 A
Reactance of the power system as a source
I
Base MVA
2
2406
Xs =
=
= base =
= 0.21
Short − Circuit MVA 9.6 I SC 11547
Thevenin reactance at fault point F
X T = (j0.8//j0.25)//(j0.21 + j0.2) = (j0.19)//(j0.41) = j0.13
V
1
If = f =
= 7.07 pu = 7.07 × 2406 = 17 kA
X T 0.13
pu
pu
References
[1] Willaim D. Stevenson, Elements of Power System Analysis, Fouth Edition, McGraw-Hill
International Editions, Civil Engineering Series, McGraw-Hill Inc.
[2] John J. Grainger, William D. Steevnson, Jr., Power System Analysis, McGraw-Hill Series in
Electrical and Conputer Engineering, McGraw-Hill Inc.
[3] J. Duncan Glover, Mulukutla S. Sharma, Thomas J. Overbye, Power System Analysis and Design,
Fouth Edition (India Edition), Course Technology Cengage Learning
[4] Hadi Saadat, Power System Analysis, Tata McGraw-Hill Publishing Company Limited
[5] I J Nagrath, D P Lothari, Modern Power System Analysis, Second Edition, Tata McGraw-Hill
Publishing Company Liited
[6] V. K. Mehta, Rohit Mehta, Principles of Power System, Multicolor Illustrative Edition, S. Chand
and Company Limited
Shortcut for 2×2 matrices
⎡a b ⎤
For A = ⎢
⎥ , the inverse can be found using this formula:
⎣c d ⎦
1 ⎡ d − b⎤
1 ⎡ d − b⎤
=
A −1 =
⎢
⎥
det[ A] ⎣− c a ⎦ ad − bc ⎢⎣− c a ⎥⎦
⎡1 2⎤
Example: A = ⎢
⎥;
⎣3 4⎦
DMAM (39/22)
A−1 =
1 ⎤
1 ⎡ 4 − 2⎤ ⎡ − 2
=⎢
⎢
⎥
− 2 ⎣− 3 1 ⎦ ⎣3 / 2 − 1 / 2⎥⎦