Experiment No
: 02
FAULT STUDY
Instructed By:
Name
Group Members:
: Weerasinghe W.M.S.C.
Weranga K.S.K.
- 060529
Index No
: 060526V
Wickramasinghe V.B.
- 060531
Group
: 12
Yasaranga H.B.D.
- 060562
Dept
: EE
Abeygoonasekara
-060567
Date
: 14/06/2007

CALCULATION
For this experiment following simplified power system was used. Fault calculations
were done with respect to this system.
x1= 0.3
x2= 0.2
x0= 0.05
x = 0.09
x1= 0.1
POLPITIYA
Z=0
.
x0= 0 002+j0.00
5
.02
ANURADHAPURA
4
0
0.
.8
7+j0
0.34
Z = 2.5
x0=
j
7+
01
0. 15
= 0.
Z 0=
x
Z = 0.058+j0.102
x0= 0.4
Z=
x0= 0.156
1.1 +j0.3
4
1
Z = 0.19+j0.44
x0= 2.0
Z = 0.057+j0.13
x0= 0.45
BOLAWATTA
LAXAPANA
x = 0.02
x1= 0.05
x2= 0.04
x0= 0.01
xp = 0.051
xt = 0.055
xs = 0.045
KOLONNAWA
x = 0.08
 Practical Calculation
1. Single Line to Earth Fault (L-G Fault)
Supply
Side
Ia
Va = 0
Ib
Vb
Ic
Vc
The fault impedance is zero. So,
Va  0
Ib  0
Ic  0
(a) the fault currents could be calculated as follows,
I a  1
I   1
 b 
I c  1
1  I a0 

  I a1 
 2  I a2 
1
2

1
I a0 
1
I   1
 a1  3 
1
I a2 

1

2
 I a0  I a1  I a2 
1  I a

2 
  I b  0
  I c  0 
Ia
3
Therefore Fault current, I f = I a
a
b
c
I f  3I a0
I f  3  46mA
I f  138mA
Actual current in the circuit,
I f  138  10 3  4000  2640 A
I f  1457.284kA
(b) fault voltages can be calculated as follows,
Va  1
V   1
 b 
Vc  1
1  Va0 

  Va1 
 2  Va2 
1
2

Va  0 1
V
  1
 b
 
Vc
 1
1
2

1   - 18 

   32
 2   - 14 
Vb  Va0   2 Va1  Va2
Vb  -18  32240 0  14120 0
Vb  48.1248  124.1278 0 kV
Actual voltage,
Vb  48.1248  2640 - 124.1278
Vb  127.0495 - 124.1278 kV
Vc  Va0  Va1   2 Va2
Vc  -18  32120 0  14240 0
Vc  48.1248124.1278 0 V
Actual voltage,
Vc  48.1248  26404.1278
Vc  127.0495124.1278 kV
2. Double Line to Earth Fault (LL-G Fault)
Ia
Va
Ib
Vb=0
Ic
Vc=0
Supply
Side
Fault impedance is zero. So
Ia  0
Vb  0
Vc  0
(a) the fault currents could be calculated as follows,
I a  1
I   1
 b 
I c  1
1
2

I a  0 1
I
  1
b

 
I c
 1
1  I a0 

  I a1 
  I a2 
1
2

1  - 39 

   92
  - 53 
a
b
c
I b  I a0   2 I a1  I a2
I b  39  92240 0  53120 0
I b  138.5316  114.979 0 mA
Actual current in the circuit
I b  138.5316  10 3  4000  2640  114.979 0
I b  1462.8937  114.979 0 kA
I c  I a0  I a1   2 I a2
I c  39  92120 0  53240 0
I c  138.5316114.979 0 mA
Actual current in the circuit
I c  138.5316  10 3  4000  2640114.979 A
I c  1462.8936114.979 kA
(c) fault voltages can be calculated as follows,
Va  1
V   1
 b 
Vc  1
1  Va0 

  Va1 
  Va2 
1
2

Va
 1
V  0  1
 b
 
Vc  0  1
1
2

Va  Va0  Va1  Va2
Va  15  15  15
Va  45 V
Actual voltage in the circuit
Va  45  2640
Va  118.8 kV
1  15

  15
  15
3.
Line to Line Fault (L-L Fault)
Supply
Side
Ia
Va
Ib
Vb
Ic
Vc
Fault impedance is zero. So
Ia  0
Vb  Vc
I b  I c
(a) Calculating fault currents
1
1  I a0 
I a  1

I   1  2   I 
  a1 
 b 

I c  1 
  I a2 
1
1  0 
I a  0  1

I
  1  2    75
b


 


I c  I b  1 
  - 75
I b  I a0   2 I a1  I a2
I b  0  75240 0  75120 0
I b  129.9038  90 0 mA
a
b
c
Actual current in the circuit
I b  129.9038  10 3  4000  2640  90 0 A
I b  1371.7842  90 0 kA
I c  I a0  I a1   2 I a2
I c  0  75120 0  75240 0
I c  129.903890 0 mA
Actual current in the circuit
I c  129.9038  10 3  4000  264090 0 A
I c  1371.784290 0 kA
(d) fault voltages can be calculated as follows,
Va  1
V   1
 b 
Vc  1
1


2
Va  0  1
V
  1
 b
 
Vc  Vb  1
1  Va0 

  Va1 
  Va2 
1


2
1   0

  22
  22
Vb  Va0   2 Va1  Va2
Vb  0  22240 0  22120 0
Vb  22180 0 V
Actual voltage in the circuit
Vb  22  2640180 0
Vb  58.08180 0 kV
 Vc  58.08180 kV
 Theoretical Calculation

Single Line to Earth Fault (L-G Fault)
Va  0 , I b  0 , Ic  0
Since Z1 =0.208, Z2 = 0.1915, Z3 = 0.4495
Fault Current can be given as,
3E f
If 
Z1  Z2  Z0
If 
3  132
0.208  0.1915  0.4495
I f  466.43 kA
Fault Voltages from the diagram
Va0   Z 0 I a0  - 0.4495  155.48 kA  -69.89 kV
Va1  E f  Z1 I a1  132 kV - 0.208  155.48 kA  99.46 kV
Va2   Z 2 I a2  - 0.1915  155.48 kA  - 29.77kV
Vb  Vb0  Vb1  Vb2
Vb  -69.89   2  99.46    29.77
Vb  -69.89  99.46240  29.77120
Vb  153.28 - 133.10 0 kV
Vc  Vc0  Vc1  Vc2
Vc  -69.89    99.46   2  29.77
Vc  -69.89  99.46120  29.77240
Vc  153.28133.10 0 kV

Double Line to Earth Fault (LL-G Fault)
Ia  0, Vb  0, Vc  0
Since Z1 =0.208, Z2 = 0.1915, Z3 = 0.4495
Z1 =832, Z2 = 766, Z3 = 1796
I a1  132 /( 0.208  0.1915 // 0.4495)
I a1  385.64kA
I a 2  (132  385.64  0.208) / 0.1915
I a 2  270.42kA
I a 0  (132  385.64  0.208) / 0.4495
I a 0  115.21kA
Va  3  Va1
Va  3  132  0.208  385.64 
Va  155.36 kV
I b  I b0  I b1  I b2
I b  I a0   2 I a1  I a2
I b  115.21  385.64240  115.21120
I b  500.85  120 0 kA
I c  I c0  I c1  I c2
I c  I a0  I a1   2 I a2
I c  115.21  385.64120  115.21240
I c  500.85120 0 kA

Line to Line Fault (L-L Fault)
I a  0, Vb  Vc , I b  Ic
Since Z1 =0.208, Z2 = 0.1915, Z3 = 0.4495
I a1 
Ef
Z1  Z 2
I a1 
132 kV
0.208  0.1915
I a1 
132 kV
0.461
I a1  330.41 kA
I a2   330.41 kA
Va1  Va2  63.556kV
Vb  Vb0  Vb1  Vb2
Vb  Va0   2 Va1  Va2
Vb  0  63.556240  63.556120
Vb  63.556180 0 kV
 Vc  63.556180 0 kV
I b  I b0  I b1  I b2
I b  I a0   2 I a1  I a2
I b  0  330.41240  330.41120
I b  572.29  90 0 kA
I c  I b
I c  572.2990
0
Practical Values
Faul
t
Typ
e
Fault Currents
kA
Fault Voltages
kV
Va
Ia
Ib
Ic
L-G
1457.2
84
0
0
0
L-L
0
1371.7842  90 0
1371.784290 0
0
LL-G
0
Vc
Vb
48.1248 - 124.1278 48.1248124.1278
58.08180 0
58.08180
0
0
1462.8937  114.979 0 1462.8937  114.979 0 45
Theoretical Values
Fault
Type
Fault Currents
kA
Fault Voltages
kV
Ia
Ib
Ic
Va
Vb
Vc
L-G
466.43
0
0
0
153.28  133.10 0
153.28133.10 0
L-L
0
572.29  90 0
572.2990 0
0
63.5561800
63.5561800
L-LG
0
500.85  120 0
500.85120 0
155.36
0
0

DISCUSSION
1. Importance of Fault Study
Fault of a power system is required in order to provide information for the selection of
switchgear, setting of relays and stability of system operation. The power system which is given is not
a static one. So that system changes during operation (Switching on or off of generators and
transmission lines) and during planning (addition of generators and transmission lines). So fault study
need to be routinely performed by utility engineers.
2. Assumptions made in Fault Study
To simplify the calculations following assumptions are usually made in fault study of
the power systems. They are as follows,

All sources are balanced and equal in magnitude and phase.

Sources represented by the Thevenin’s voltage prior to fault at the fault point.

Large systems may be represented by infinite bus-bars.

Transformers are on nominal tap position.

Resistances are negligible compared to reactance.

Transmission lines are assumed fully transposed and all three phase have same Z.

Loads currents are negligible compared to the fault currents.

Line charging currents can be completely neglected.
3. DC Network Analyzer
The DC network analyzer is design to analyze the fault currents in power system. The symmetrical
andunsymmetrical fault analysis can be done with help of analyzer. The unit comprises with variable
power supply sources, variable resistance, Milliammeter and ohmmeter. The unit is design with
number of sections in power system. The sections are considered for general case. The unit will have
two sections of alternators, two sections of sending end transformers with busbars, four sections of
transmission lines, four sections of receiving end transformers with busbars and four load sections.
The fault impedance diagram can be prepared on per phase basis for symmetrical faults and
currents are calculated. The same can be analyzed with dc network analyzer panel. For the
unsymmetrical faults, the sequence reactance diagrams are prepared and the positive, negative & zero
sequence reactance diagrams are connected in series or parallel according to type of fault for fault
current analysis. The DC network analyzer will give the simulated fault current in agnitude but the
exact phase can not be found. The assumption is made that fault resistance is very small than the
reactance. (i.e. Rf <<< XF) So the fault current is pure reactive & logging behind the phase voltage by
90. For analyzing currents with its phase and magnitude the AC network analyzer is required. The DC
network analyzer is simple to operate, easy to understand and full proof with protections. The power
supply units in analyzer can be independently used for other applications also. The power supply
manual is enclosed separately.
The DC network analyzer unit is totally enclosed, free standing, dustproof. The unit powder coated
and screen printed for different section parameters.
4. Importance of using Sequence Components
Considering the power system the analyzing procedure is easy if the power system is balanced.
But the systems are not balance in every time. To analyze the unbalanced systems the method of
sequence components is using. In this method the unbalanced system splits in to three balanced
components. They are, Positive sequence networks (balanced & same phase sequence as unbalanced
system), Negative sequence (balanced & opposite phase sequence to the unbalanced system) networks
and Zero sequence (balanced & having same phase) networks. Because of this decomposition of the
unbalanced system removes the complex parts of the system properties. So the analyzing matrix
calculation will be easy.
5. Relationships between the sequence impedance for generators, transformers and transmission
lines
Depending on the component used in the power system impedance of the components
used for fault calculation may changed for various sequences. Some detail about the relationship
between the sequential impedance for generators, transformers and transmission lines are given below.

Generator
The generator has an inherent direction of rotation and the sequence considered may either
have the same direction (no relative motion) or the opposite direction (relative motion at twice
the speed). Thus the rotational emf generated for the positive sequence and the negative
sequence would also be different. So that the generator has different values for positive
sequence, negative sequence and zero sequence impedances.

Transformer
Transformer is also passive and stationary. So it does not have an inherent direction of
rotation. So it always has the same positive sequence impedance and negative sequence
impedance and even zero sequence impedances. But the zero sequence paths across the
windings of a transformer depend on the winding connection and even grounding impedance.

Transmission Lines
Likes a transformer, the conductors of the transmission lines are passive and stationary. So
they do not have an inherent direction. Thus, they have the same positive sequence and
negative sequence impedances. But the zero sequence paths involve the earth wire and or the
earth return path. The result is higher the zero sequence impedance.

References
o EE 423 – Power System Analysis study notes