Exercise 3
Electric Power System 2014
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Exercise 3 Sequence components, unsymmetrical faults
V phase  AV sequence
1

where A  1

1
1 1 
2
j

2
a a  and a  e 3  1120
a a 2 

3.1 Generating positive sequence voltages
The phase voltages of a synchronous generator is 2200, 220-120°, 220120° V. It
is connected to a balanced three phase load. Compute the sequence representation of
the generator voltages.
3.2 Generating negative sequence voltages
The rotating direction of the generator is reversed. The phase voltages are now
2200, 220120°, 220120°. Compute the sequence representation of the
generator voltages.
3.3 Generating zero sequence voltages
A single phase AC voltage source is connected to a balanced three phase load, so that
all three phase voltages are the same, 2200. Compute the sequence representation of
the phase voltages.
3.4 Unsymmetrical load (Exam 010108, Problem 5)
A 0.4kV single-phase load is fed from a 10kV system through a 10MVA, 10/0.4kV
/Y transformer. The load is a 0.016 resistance and by mistake it has been
connected between phase a and ground.
The transformer has Xeq=0.1p.u. and the neutral point of the Y winding is grounded
through a reactance Xn=0.05p.u.
The network feeding the transformer can be described as Thévenin equivalents for the
positive sequence with VTH1=10kV, ZTH1=j0.02p.u., for the negative sequence with
VTH2=0, ZTH2=ZTH1, and for the zero sequence with VTH0=0, ZTH0=j0.01p.u.
All per unit values are given using the bases 10MVA and rated transformer voltages.
The transformer phase shift can be ignored.
a)
Draw sequence diagrams of the system (including network and transformer) as
seen from the 0.4kV bus before the load is connected.
b)
Determine the steady state sequence currents at the 0.4kV side of the
transformer when the load has been connected.
c)
Determine the steady state zero sequence current at the 10kV side of the
transformer.
3.5 Unsymmetrical fault (Exam 001019, Problem 2)
x12
1
2
3
x13
x14
x23
Pm
x24
x34
4
Figure 1. Single line diagram with positive Figure 2. Power plant with synchronous
sequence line reactances.
generator and Y- transformer.
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Exercise 3
Electric Power System 2014
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The system of Figure 1 is used with a power plant (Figure 2) at bus 1 and an infinite
bus at bus 2. The cable connecting buses 1 and 3 is damaged near bus 1 during
excavations (digging). It is important that a protective relay quickly detects this and
that circuit breakers can interrupt the fault current to deenergize the cable.
Prefault voltage is 55 kV and prefault current and DC offset can be neglected.
Positive, negative and zero sequence Zbus are the same. With Sbase=100MVA and
Vbase=50kV, and buses 3 and 4 eliminated Zbus is:
0.2 0.1
Zbus = j 
 p.u.
0.1 0.3
The generator has a solidly grounded neutral point. P.u. reactances are: Xd=1.0,
X’d=0.25, X”d=0.1, X2=0.1, X0=0.05, with ratings 25MVA and 20kV as bases.
The Y- transformer has a solidly grounded neutral point and is rated 50 MVA and
20kV/50kV. Xeq is 0.1 p.u. using rated values as bases.
a) Suggest the minimum current rating of the circuit breaker by determining phase
values in A of the maximum fault current due to a zero-ohm short-circuit.
b) Determine phase values in A of the initial fault current for the least severe fault
type where the short-circuit current goes through the excavator (digging machine) to
ground. The excavator has a resistive impedance of 10.
3.6 Currents during unsymmetrical fault in five-bus system
Example 7.5 (pp 396) and Example 9.8 (pp 424) describe symmetrical and
unsymmetrical faults in a five-bus power system shown on p 399. The per unit zero,
positive and negative sequence bus impedance matrices Zbus0, Zbus1 and Zbus2 are:
Zbus0=0.0125
0
0
0
0
Zbus1=0.0280
0.0177
0.0085
0.0123
0.0204
0
0.1089
0
0.0044
0.0112
0.0177
0.0570
0.0136
0.0197
0.0256
0
0
0.0125
0
0
0.0085
0.0136
0.0182
0.0164
0.0123
0
0.0044
0
0.0089
0.0021
0
0.0112
0
0.0021
0.0158
0.0123
0.0197
0.0164
0.0236
0.0178
0.0204
0.0256
0.0123
0.0178
0.0295
Zbus2=Zbus1
Calculate the fault current for an unsymmetrical fault at bus two. Use the bus
impedance matrices and a prefault voltage VF=1.05 p.u. to determine Thévenin
equivalents of the sequence networks as seen from bus two. Then use the fault
currents in the formula sheet to determine the sequence currents at the faulted bus. Do
this for:
a) Zero impedance single-line to ground fault in phase a. As a comparison, the phase
A current is 14.14 p.u. according to Table 9.4 on p 499.
b) Zero impedance line to line fault between phases b and c.
c) Zero impedance double line to ground fault on phases b and c.
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Exercise 3
Electric Power System 2014
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3.7 Unsymmetrical faults (Exam 011025 Problem 3)
Sequence fault currents (a-c) and phase fault currents (A-C) are shown below as
phasors for three different fault types (I-III) all with fault impedance ZF=0.
I2
I2
I0
I0=0
a)
I1
I1
b)
Ib
Ib=Ic=0
c)
I0=I1=I2
Ia=0
Ic
B)
Ib
Ia
Ia=0
Ic
C)
A)
I. Double line-to-ground II. Single line-to-ground III. Line-to-line
Combine phasor diagrams with the appropriate fault type. Do this by using the
characteristics of the fault types in terms of sequence and phase currents. Mention this
characteristic in each case as motivation to your choice.
Show how the phase currents and sequence currents are related by drawing how each
phase current is composed in the three cases. Prefault currents are zero.
3.8 Earth fault and reactance earthing
Somewhere in Småland, a zero  earth fault occurs in the a-phase on the 20 kV side
of a 20 MVA, 145/20kV /Y transformer fed from a 145kV, 50 Hz system.
The network feeding the transformer can be described as Thévenin equivalents for
positive sequence (VTH1=145/3 kV, ZTH1), negative sequence (VTH2=0, ZTH2=ZTH1)
and zero sequence (VTH0=0, ZTH0).
The transformer has series reactance Xeq and the neutral point of the Y winding can be
connected to ground through a reactance Xn=Ln – a Petersen coil.
The transformer feeds a 40 km long 20 kV overhead line that can be modeled as a
link with only shunt susceptance jB=jC. C1=C2=9.95 nF/km and C0=4.4 nF/km.
a. Draw the three sequence diagrams of the system (including network, transformer
and cable) as seen from the 20 kV bus (where VTH1=20/3 kV). Connect Xn to
ground through a switch.
b. With the switch open and transformer ungrounded, consider the fault situation. Set
Xeq, ZTH1, ZTH2 and ZTH0 to zero as they are << 1/B. Redraw the diagrams that are
now connected by the fault.
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Exercise 3
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c. Determine the capacitive earth fault current in the a-phase in Amps.
Note: B1 should not influence the fault current.
d. Close the switch and show that the fault current is eliminated by a Petersen coil with Xn =
6 k.
e. The standard way to state the reactance or inductance value of a Petersen coil is to state
the current in Amps through it at nominal line to ground voltage, corresponding to a zero
 earth fault. What value in Amps should the Petersen coil have here?
3.9 Capacitive earth fault current when cables replace overhead lines
The overhead line in Exercise 3.8 is replaced by underground cable. The 40 km long
cable has C0=0.33 F/km. Determine the capacitive earth fault current Ic at a zero 
earth fault at the busbar. With what factor did Ic change as compared to Exercise 3.8?
3.10 Neutral point resistor
In reality, the impedance of a Petersen coil also has a resistive component, which
gives a resistive or active component of the fault current.
a. Add a resistance RN of 3 k in parallel with the Petersen coil in Exercise 3.8
but using a cable feeder as in Exercise 3.9 and determine the active component
of the fault current.
b. Draw the capacitive earth fault current, the current through the (purely
inductive) Petersen coil and the current through RN in the complex plane.
This current component cannot be eliminated by the coil and thus sets a lower limit of
the fault current. On the other hand, if there are many feeders this current component
goes only through the faulted feeder. This is the basis for selective clearing of earth
faults with (active component) residual1 overcurrent protection. Since a resistive
component in the earthing is desirable, explicit neutral point resistors are often used in
parallel with the earthing reactance (Petersen coil).
An earth fault changes the neutral point potential or zero sequence voltage from zero
in an intact network to another value. This can be used by protection to detect earth
faults. The zero sequence voltage is typically measured in the transformer station.
With this method it is not possible to say on which feeder the earth fault is, so the
protection disconnects all feeders. The residual overvoltage protection thus clears the
fault non-selectively and is used as backup protection.
c. With RN in place, determine the zero sequence voltage during a 5k earth
fault.
1
Zero sequence voltage is often termed residual voltage – what is left when you add the phase
voltages. The same applies for current.
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Exercise 3
Electric Power System 2014
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Answers
3.1: V0=0, V1=2200 V, V2=0
3.2: V0=0, V1=0, V2=2200 V
3.3: V0=2200 V, V1=0, V2=0
3.4
I1
ZTH1
Pos
seq
X eq
VF
+
V1
–
I2
ZTH2
+
X eq
Neg
seq
3R
V2
–
I0
ZTH0
Zero
seq
X eq
3Xn
+
V0
–
a)
b) I0=I1=I2=0.329–9.3° p.u.
c) I0=0 @ 10 kV
3.5
a) The breaker must be able to interrupt a current of 7.33p.u. or 8468A.
b) A phase current of 2.54p.u. or 2930A should be classified as a fault.
3.6
a) I0=I1=I2=–j4.71 p.u.
b) I0=0, I1=–j9.21 p.u., I2=j9.21 p.u
c) I0=j3.82 p.u., I1=–j11.12 p.u., I2=j7.30 p.u
3.7 LLG: I+b+C
3.8 a
Pos.
seq.
LL: III+a+B SLG: II+c+A
ZTH1
Xeq
B1
VTH
ZTH2
B1
B2
ZTH0
B2
Xeq
3Xn
d. Z0>106  If=0
c. Ic=1.9 A
3.9 144 A and 75 times
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VTH
Xeq
Neg.
seq.
Zero
seq.
3.8 b.
B0
3Xn
e. Reactance current = 1.9 A = Ic!
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B0