1. A three-phase fault occurred at Bus 5 in the power system. The fault current (IF) was calculated to be 5 pu based on the total fault impedance of 16.0j ohms as seen from Bus 5. 2. The fault current supplied by Generator 1 (IG1F) was calculated to be 2.1 pu, and by Generator 2 (IG2F) was calculated to be 8.0 pu. 3. The voltages during fault condition were calculated to be 0.68 pu at Bus 5 and 0.24 pu at Bus 3, indicating a voltage drop from the pre-fault voltage of 1.0 pu at all buses.

1. ELE B7 Power SystemsELE B7 Power Systems
EngineeringEngineering
Balanced (Symmetrical) FaultsBalanced (Symmetrical) Faults

2. Fault Analysis
• Fault?
When the insulation of the system breaks down or a
conducting object comes in touch with a live point, a short
circuit or a fault occurs.
• Sources?
• This breakdown can be due to a variety of different factors
• lightning
• wires blowing together in the wind
• animals or plants coming in contact with the wires
• salt spray or pollution on insulators

3. Fault Types
• There are two classes of faults
– Balanced faults (symmetric faults): A Fault involving all the three
phases
• system remains balanced;
• these faults are relatively rare, but are the easiest to analyze so
we’ll consider them first.
– Unbalanced faults (unsymmetric faults): A fault involving only one
or two phases
• The majority of the faults are unsymmetrical.
• system is no longer balanced;
• very common, but more difficult to analyze
• The most common type of fault on a three phase system is
the single line-to-ground (SLG), followed by the line-to-
line faults (LL), double line-to-ground (DLG) faults, and
balanced three phase faults

4. Fault Types
IFb
Transformer
IFa
IFc
IF
Transformer
IF
Single Line to Ground Fault
IF
IF
IF
Transformer IF
Line to Line Fault
IF
IFb
Transformer
IF
IF
IF
IFc
Double line to ground Fault
Three Phases Fault (Balanced Fault)

5. Fault Analysis
• Fault Analysis involve finding the voltage and current
distribution throughout the system during fault condition.
• Fault currents cause equipment damage due to both
thermal and mechanical processes
• Why do we need that?
– To adjust and set the protective devices so we can
detect any fault and isolate the faulty portion of the
system.
– To protect the human being and the equipment during
the abnormal operating conditions.
– need to determine the maximum current to insure
devices can survive the fault

6. Symmetrical Faults
(Balanced Faults)
• It is the most severe fault but can be easily calculated
• It is an important type of fault
• The circuit breaker rated MVA breaking capacity is
selected based on the three phase short circuit MVA.
A Fault involving
all the three phases.
Transformer
Three Phase
Short Circuit

7. Symmetrical Faults
(Balanced Faults)
• It is the most severe fault but can be easily calculated
• It is an important type of fault
• The circuit breaker rated MVA breaking capacity is
selected based on the three phase short circuit MVA.
A Fault involving
all the three phases.
Transformer
Three Phase
Short Circuit

8. 1. Synchronous machines are represented by a constant
voltage sources behind subtransient reactances. The EMF
of all generators are 1 per unit making zero angle.
o
V 01
The following assumptions are made in three phase fault
calculation :
This means that the system voltage is at its nominal
value and the system is operating at no-load conditions
at the time of fault.
Therefore, All the generators can be replaced by a single
generator since all EMFs are equal and are in phase.

9. 2. Transformers are represented by their leakage reactances
Winding resistances, shunt admittances and Δ-Y phase
shifts are neglected
3. The shunt capacitances of the transmission
Lines are neglected.
4. The system resistances are neglected and only
the inductive reactance of different elements are
taken into account.

10. Symmetrical
Fault Calculation
Transformer
Three Phase
Short Circuit
Simple Circuits
Load Ignored
Use Thevenin's Equivalent
Simple Circuits
Load not Ignored
Use Thevenin's Equivalent
Find Pre-fault voltage
Large Circuits
Construct
The Bus Impedance Matrix
Three Phase Fault
Calculation
2
3
1

11. 1. Simple Circuits and Load is ignored
1. Draw a single line diagram for
the system.
The Calculations for the three phase fault are easy because the
circuit is completely symmetrical and calculations can be done
for only one phase.
T1
T2
G1
G2
1
2
3 phase
fault e
Steps For Calculating Symmetrical Faults:
2. Select a common base and find
out the per unit reactances of all
generators, transformers,
transmission lines, etc.

12. 4. Reduce the single line reactance diagram by using
series, parallel and Delta-Why transformations
keeping the identity of the fault point intact. Find the
total reactance of the system as seen from the fault
point (Using Thevenin’s Theorem)
3. From the single line
diagram of the system
draw a single line
reactance diagram showing
one phase and neutral.
Indicate all the reactances,
etc. on the single line
reactance diagram.
GG
15.0j
20.0j 16.0j
10.0j
23.0j
GG
01
01
23.0j

13. 5. Find the fault current and the fault MVA in
per unit.
Convert the per unit values to actual values.
6. Retrace the steps to calculate the voltages
and the currents throughout different parts
of the power system

14. T1
T2
G1
G2
X=0.2 ohm/phase/km
1
2
50 MVA,
11 kV
10% Reactance
100 MVA,
11 kV
15% Reactance
100 MVA, 11/132 kV
10% Reactance
50 MVA, 11/132 kV
8% Reactance
200 km
3 phase
fault e
Example:
A three phase fault occurs in the system as shown in the Figure.
Find the total fault current, the fault level and fault current
supplied by each generator.

15. Step 1: Draw a single line diagram for the system.
The single line diagram for the system is given in the
example as shown.
Solution:
T1
T2
G1
G2
X=0.2 ohm/phase/km
1
2
50 MVA,
11 kV
10% Reactance
100 MVA,
11 kV
15% Reactance
100 MVA, 11/132 kV
10% Reactance
50 MVA, 11/132 kV
8% Reactance
200 km
3 phase
fault e

16. Step 2: Select a common base and find the per unit reactances of all
generators, transformers, etc.
Select the common base as:
100 MVA (100,000 kVA)
11 kV for Transformer low voltage side (LV)
132 kV for Transformer high voltage side (HV)
T1
T2
G1
G2
X=0.2 ohm/phase/km
1
2
50 MVA,
11 kV
10% Reactance
100 MVA,
11 kV
15% Reactance
100 MVA, 11/132 kV
10% Reactance
50 MVA, 11/132 kV
8% Reactance
200 km
3 phase
fault e
kVVBase 11
kVVBase 132

17. T1
T2
G1
G2
X=0.2 ohm/phase/km
1
2
50 MVA,
11 kV
10% Reactance
100 MVA,
11 kV
15% Reactance
100 MVA, 11/132 kV
10% Reactance
50 MVA, 11/132 kV
8% Reactance
200 km
3 phase
fault e
kVVBase 11
kVVBase 132
G1
G2
15.01 jXG The per unit reactance
2.0
50
100
*1.02 jjXG The per unit reactance
1.01 jXT The per unit reactance
16.0
50
100
*08.02 jjXT The per unit reactance
23.0
)132(
100
)200*2.0(
2
j
kV
MVA
j
Z
X
X
Base
LINE
LINE The per unit reactance
T1
T2
TL

18. From the single line diagram of the system draw a single line reactance diagram
showing one phase and neutral. Indicate all the reactances, etc. on the single line
reactance diagram.
T1
T2
G1
G2
1
2
3 phase
fault e
G
15.0j
20.0j 16.0j
10.0j
23.0j
G
01
01
23.0j

19. G
25.0j
36.0j
G
01
01
115.0j
Find the total impedance (reactance) of the system as seen from the
fault side.
25.0j
36.0j
115.0j
pujX
j
jj
jj
X
Total
Total
2625.0
115.0
36.025.0
36.0*25.0





20. puj
j
IF 8095.3
2625.0
01



AI sidekVBase 4.437
132*3
1000*100
)( 132 
AjI o
ActualF 9027.16664.437*8095.3)( 
01
2625.0j
LLBase
Base
Lbase
V
S
I
||3
||
||
3

BasepuFActualF III *)()( 

21. The Fault Level is:
100)( BaseMVA
95.380100*8095.3)( LevelActualMVA
puMVA Level 8095.3)( 
T1
T2
G1
G2
X=0.2 ohm/phase/km
1
2
50 MVA,
11 kV
10% Reactance
100 MVA,
11 kV
15% Reactance
100 MVA, 11/132 kV
10% Reactance
50 MVA, 11/132 kV
8% Reactance
200 km
3 phase
fault e
1)( GFI
2)( GFI
FI
kVVBase 132kVVBase 11

22. The Fault Current at 11 kV side supplied by the two generators is:
AI sidekVBase 8.5248
11*3
1000*100
)( 11 
AjI o
ActualF 90199958.5248*8095.3)( 
At 11 kV Side:
G
15.0j
20.0j 16.0j
10.0j
23.0j
G
01
01
23.0j
TotalFI )(
1)( GFI
2)( GFI
kVVBase 132kVVBase 11

23. 111,2 )()()( GFkVTFGF III 
A
jj
j
I GF 903.11800
25.036.0
36.0
)9019995()( 1 


AI GF 907.8194)( 2 
G
25.0j
36.0j
G
01
01
115.0j
T1
T2
G1
G2
1
2
1)( GFI
2)( GFI
1)( GFI
2)( GFI
TotalFI )(
A
jj
j
II kVTFGF 903.11800
25.036.0
36.0
)()( 11,1 



24. Example:
Consider the single-line diagram of a power system shown below.
The transient reactance of each part of the system is as shown
and expressed in pu on a common 100 MVA base.
G1 G22.0j
2.0j
4.0j
8.0j
4.0j
1.0j
1.0j
Bus 1
Bus 3 Bus 4
Bus 2
Bus 5
Assuming that all generators are
working on the rated voltages, when a
three-phase fault with impedance of
j0.16 pu occurs at bus 5. Find: The fault
currents and buses voltages.

25. G1 G22.0j
2.0j
4.0j
8.0j
4.0j
1.0j
1.0j
Bus 1
Bus 3 Bus 4
Bus 2
Bus 5
2.0j
2.0j
4.0j
8.0j
4.0j
1.0j
1.0j
Bus 1
Bus 3 Bus 4
Bus 2
Bus 5
16.0j
2.0j
4.0j
4.0j
8.0j
4.0j
Bus 3 Bus 4
Bus 5
16.0j

26. 5.016.034.0 jjjZTH 
pu0.2j
5.0j
1
)I(
0
5BusF 

puj
jj
j
II FGF 2.1
4.06.0
6.0
)( 1 


puj
jj
j
II FGF 8.0
4.06.0
4.0
)( 2 


16.0j
Bus 5
34.0j
FI
16.0j
Bus 5
34.0j
FI
2.0j 4.0j
16.0j
Bus 3 Bus 4
Bus 5
2.0j 2.0j
1.0j
FI
1FGI 2FGI
16.0j
Bus 5
6.0j
1.0j
4.0j
FI
2FGI1FGI

27. 0V&01V
:VoltagesPrefault
Ground
o
BUS 
pu
VprefaultV
VoltageFault
o
76.024.001
:
33


pu
jj
jIV
isFaultDuringVariationVoltage
G
24.0
)2.1()2.0(0
)2.0(0
:
13



68.0
)0.2)(16.0(01
:
5

 jjV
FaultduringVariationVoltage
o
pu
VprefaultV
VoltageFault
o
32.068.001
:
55


NOTE: If the fault impedance is zero.
0.1)I)(0.0j(01V
:FaultduringVariationVoltage
F
o
5  pu0.00.101VprefaultV o
55  and
1GI 2GI
FI
o
01
2.0j 4.0j
16.0j
Bus 3 Bus 4
Bus 5
2.0j 2.0j
1.0j
+_



5V



3V

28. G1 G22.0j
2.0j
4.0j
8.0j
4.0j
1.0j
1.0j
Bus 1
Bus 3 Bus 4
Bus 2
Bus 5
Problem:
For the network shown in the figure, find the current flowing between
buses 3 and 4 during symmetrical three phase fault at bus 5 as in the
previous example.
16.0j
2.0j
2.0j
4.0j
8.0j
4.0j
1.0j
1.0j
Bus 1
Bus 3 Bus 4
Bus 2
Bus 5

29. pu
jjprefault
VprefaultV
pu
VprefaultV
VoltageFaultDuring
o
o
68.032.001
)8.0)(4.0(0
76.024.001
:
44
33





puj
jj
j
II FGF 2.1
4.06.0
6.0
)( 1 

 puj
jj
j
II FGF 8.0
4.06.0
4.0
)( 2 


puj
j
VV
IF 1.0
8.0
)( 43
34 


4.0
)( 53
35
j
VV
IF


4.0
)( 54
45
j
VV
IF

and
2.0j
2.0j
4.0j
8.0j
4.0j
1.0j
1.0j
Bus 1
Bus 3 Bus 4
Bus 2
Bus 5
16.0j
1GI
2GI
34F )I(
45F )I(35F )I(


4V


5V

30. Symmetrical
Fault Calculation
Transformer
Three Phase
Short Circuit
Three Phase Fault
Calculation
Simple Circuits
Load Ignored
Use Thevenin's Equivalent
Simple Circuits
Load not Ignored
Use Thevenin's Equivalent
Find Pre-fault voltage
Large Circuits
Construct
The Bus Impedance Matrix
2

31. 2. Symmetrical Faults Considering the load current
Generally, the fault currents are much larger than the load currents. Therefore, the
load current can be neglected during fault calculations.
The pre-fault voltage and the ZTH are used in calculating the fault current.
The fault current represents two components:
1. The Load Current
2. The Short Circuit Current
There are some cases where considering the load current is an essential factor in
fault calculations. Superposition technique is proposed for such cases to compute the
fault current.
For such case, it is necessary to compute the terminal voltage at fault location
before fault takes place. This terminal voltage is known as the pre-fault voltage.
Connecting the load to the system causes the current to flow in the network. Voltage
drop due to system impedance cause the voltage magnitude at different buses to be
deviated from 1.0 pu.

32. Consider a load (synchronous Motor) connected to a synchronous generator through
a transmission line. The circuit model of the network could be represented as shown
in the Figure
MIGI
jXG jXM
G M
jXL
FI
G M
FMIFGI
jXL
jXG jXM
Normal Operation Condition:
LMG III 
Fault condition at Load terminal:
LGFG III  )(
LMFM III  )(
MFGFF III )()( 
During fault condition in the system, the generator as well as the load (synchronous
motor) will supply the faulted terminals with power from the energy stored in their
windings.

33. The Motor is drawing 40 MW at 0.8 pf. Leading with terminal
voltage of 10.95 kV.
Calculate the total current in the generator and motor during 3
phase s.c.
Example
100 MVA, 11 kV
Generator
0.25 pu
sub-transient
reactance
50 MVA, Motor
0.2 pu
sub-transient
reactance
0.05 pu. Reactance
On 100 MVA Base
3 phase
fault
G M

34. A
kV
MVA
CurretBase
kVkVBase
MVAMVABase
88.5248
)11(3
100
11
100



Solution:
The network could be represented as
shown in the Fig.
A4.2636
)8.0)(kV95.10(3
MW40
CurretLoadThe 
pujI
jI
I
L
L
L
)301.0402.0(
)6.08.0(502.0
)8.0(cos502.0 1


 
puI puL 502.0
88.5248
4.2636
)( 
Under Normal Operation Condition (Prefault condition)
leading0.8isfactorpowerThe
25.0j 4.0j
G M
05.0j MIGI

35. Consider the motor terminal voltage as a reference.
puvoltage)terminal(Motor pu 995.0
11
95.10

pu)0201.0j9799.0(V
995.0)050.0j)(301.0j402.0(V
V)050.0j(IV
G
G
MLG



pu)(V o
puM 0995.0 
Using KVL, calculate the pre-fault voltage
at the generator terminal.
Find the equivalent reactance as seen
from the fault terminals.
pujX
jj
jj
X
TH
TH
1607.0
)45.0()25.0(
)45.0)(25.0(



25.0j 4.0j
05.0j
THX
25.0j 4.0j
G M
05.0j
MV

36. Calculate the fault current
puj)(I
j
j
)(I
X
Voltaegefaultpre
)(I
puF
puF
TH
puF
)09.6125.0(
1607.0
0201.09799.0





pu(I)
puj(I)
jj(I)
I)(I(I)
o
G
G
G
LGFG
4.8264.3
)609.3482.0(
)301.0402.0()91.308.0(




Fault Current from Generator Side
puj)(I
pujj)(I
)(I)(I)(I
MF
MF
GFpuFMF
)18.2045.0(
)91.308.0()09.6125.0(



Fault Current from Motor Side
Current from Generator Side:
puj)(I
j0.25j0.45
j0.45
)(I)(I
GF
puFGF
)91.308.0( 


pu(I)
puj(I)
jj(I)
I)(I(I)
o
M
M
M
LMFM
8.261505.2
)48.2357.0(
)301.0402.0()18.2045.0(




Current from Motor Side:
25.0j 4.0j
G M
05.0j
FI
G(I) M(I)

37. Consider the system shown
in the Figure.
The two generators are
supplying the buses 1, 2, 3
and 4.
Let the pre-fault voltage
between bus 2 and ref. Bus
be (Vf ).
3 - Symmetrical Fault Calculation Using Bus Impedance
Matrix
3
T1
T2
G1
G2
1
2
4
fV

38. If a 3 phase short circuit
occurs a bus 2.
A huge current
will run in the system
during the fault
condition.
"
fI
3
2
1
4
"
fI
3
T1
T2
G1
G2
1 2
4
fV
3 phase
fault e
If the pre-fault voltage
between bus 2 and ref. bus is
Vf , then the voltage during
fault is zero.
Then, the system can be
analyzed using the reactance
circuit.

39. During the three-phase fault, the voltage between bus 2 and the
reference becomes zero. Therefore, the 3 phase fault at bus 2 can be
simulated by making the voltage between bus 2 and Ref. is equal to
zero during the fault condition.
3
2
1
4
"
fI
fVfV
This is simulated by inserting
a source with a magnitude
equal to
Therefore, the voltage
between bus 2 and reference
is zero and fault current
runs in the system.
)( fV
This means that, the Fault current during the three-phase short
circuit results from the voltage source
"
fI
)( fV

40. Therefore, to compute the
short circuit current, we have
to find the current entering
bus 2 due to only )( fV












4
3
2
1
V
V
V
V

















4
3
1
V
V
Vf
V



  
BUSZ
44434241
34333231
24232221
14131211
ZZZZ
ZZZZ
ZZZZ
ZZZZ












=













0
0
I
0
''
f=
The Nodal Impedance Equation
during the fault is:
Note:
is the voltage difference between node i and the reference node
due to the current .
iV
"
fI
Ignoring all other currents entering
other buses (ignore sources).
3
2
1
4
"
fI
fV

41. Or












4
3
2
1
V
V
V
V

















4
3
1
V
V
Vf
V





















''
f42
''
f32
''
f22
''
f12
IZ
IZ
IZ
IZ
= =
And from the second row, the fault current is:
22
''
Z
V
I
f
f 
Substituting in the previous equation:












4
3
2
1
V
V
V
V

















4
3
1
V
V
Vf
V



























f
22
42
f
22
32
f
f
22
12
V
Z
Z
V
Z
Z
V
V
Z
Z
= =

42. Assuming no load is connected in the system, no current will flow
before fault and all bus voltages are the same and equal to the pre-
fault voltage )V( f
Using Superposition, then during the fault:












4
3
2
1
V
V
V
V




fV























22
42
22
32
22
12
Z
Z
1
Z
Z
1
0
Z
Z
1
= +
i)voltagefaultPre(f)voltageiBus(i VVV  
iV is the voltage change in bus i due to fault current in bus 2.












4
3
2
1
V
V
V
V














f
f
f
f
V
V
V
V
=

43. Example:
A three-phase fault occurs at bus
2 of the network shown in the
Figure.
Determine:
1. the sub-transient current
during the fault,
2.the voltages at all the buses
3.The current flow during the
fault from buses 3-2, 1-2 and 4-2.
Consider the pre-fault voltage at
bus 2 equal to 1 pu. and neglect
the pre-fault currents.
3
T1
T2
G1
G2
1 2
4
fV
3 phase
fault e

44. Solution:
Construct the reactance circuit
under normal operation condition.
pu2.0j pu1.0j
pu125.0j
pu25.0j
pu4.0j
pu25.0j
up2.0j
3
2
1
4
pu2.0j pu1.0j
Construct the bus impedance
Matrix (ZBUS)















1954.0j1046.0j1506.0j1456.0j
1046.0j1954.0j1494.0j1544.0j
1506.0j1494.0j2295.0j1938.0j
1456.0j1544.0j1938.0j2436.0j
ZBUS
NOTE:
Constructing ZBus will be discussed later.

45. Since there are no-load currents, the pre-fault voltage at all the buses is equal to
1.0 pu. When the fault occurs at bus 2,
pu357.4j
2295.0j
0.1
Z
V
I
22
f''
f 
fV























22
42
22
32
22
12
Z
Z
1
Z
Z
1
0
Z
Z
1
=












4
3
2
1
V
V
V
V
fV























22
42
22
32
22
12
Z
Z
1
Z
Z
1
0
Z
Z
1
=












4
3
2
1
V
V
V
V
=























2295.0j
1506.0j
1
2295.0j
1494.0j
1
0
2295.0j
1938.0j
1












343.0
349.0
0
155.0
=















1954.0j1046.0j1506.0j1456.0j
1046.0j1954.0j1494.0j1544.0j
1506.0j1494.0j2295.0j1938.0j
1456.0j1544.0j1938.0j2436.0j
ZBUS
3
T1
T2
G1
G2
1 2
4
fV
3 phase
fault e

46. The current flows from bus 1 to bus 2 is:
125.0j
VV
I 21
12


pu2448.1jI12 
125.0j
1556.0
I12 
pu2.0j pu1.0j
pu125.0j
pu25.0j
pu4.0j
pu25.0j
pu2.0j
3
2
1
4
"
fI
pu2.0j pu1.0j
The current flows from bus
3 to bus 2 is:
25.0j
V3V
I 2
32


pu396.1jI32 
pu719.1jI42 
The current flows
from bus 4 to bus 2 is:
The current flows from
bus 3 to bus 1 is:
25.0j
VV
I 13
31

 pu7736.0j