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Chapter 4
Fault Analysis
4.1. Introduction Fault is the unintentional connecting together of two or more conductors which ordinary operatesbetween conductors may be physical metallic contact or through an arc. with a difference of potential between them. Connection There are two categorized of fault in power system: (i) Three phase balanced fault Three phase fault Three phase to earth fault (ii) Three phase unbalanced fault Line to line fault Line to earth fault Double line to earth fault / line to line to earth fault When fault occurred in a system, it will produced large current to flow through the transmission line and will caused damage to the equipment located at the end of the network. Actually these faults can be isolated using relays and circuit breakers which are used to separate the network from the high current. Nowadays, FACTS (Flexible AC Transmission System) and Custom Power are use to solve the fault problem. In fault analysis, values of these currents are calculated for the differences types of faults at various locations in the system. The data/ result obtained are then serve to determine the settings of relays whichcontrol the circuit breakers
2.2. Types of Fault
a) one line to earth fault
b) line to line fault
c) double line to earth fault
d) three phase to earth fault
e) three phase fault
Objectives of fault analysis. i) ii) To determine maximum and minimum 3-phase short circuit current.To determine the unsymmetrical fault current for single or double line to earth, line to line fault and other faults. iii) To determine fault current distribution & busbar voltage levels during faults. iv) Investigating of the operation of protection relays.
X x pu
X x pu
X x pu
G
G
G
' 3
' 2
' 1
Step 1
G
F
0.24pu 0.27pu 1
2
G1 G
1
2 0.1pu 1
2
N
Step 2 : Short circuit voltage source
N
1
2 0.1pu 0.27pu 1
2 0.24pu 1
2
F
Step 3 : Reduction of impedance network.
F N
0.054pu 1
2
Step 4 : Put back voltage sources and short circuit FN, than findIF
V (^) pu 1. 0 pu 0 o
F N
0.054pu 1
2 G
I I xI x MVAkV A
I pu Fact Fpu Base
o Fpu
- (^5360) ( 118 ) 5431
, ,
,
Load MVA / Fault MVA
MVA
MVA
X
MVA
OR
MVA
kV
xV xI
eq
Base
B F
Example 2 An interconnected generator-reactor system is shown in Figure. The base values for the given percent reactances are the ratings of the individual pieces of equipment. A 3θ short circuit occurs at point F. determine the fault current and the fault kVA if the busbar line o line voltage is 11kV. Choose 50MVA as base values.
I x kVMVA A
MVA M MVA kVA f
f
- (^2521110666)
050. 246 203.^252203252
Example 3:
Find the fault current in the following. Take base of 10MVA, 6.6kV.
6.6/33kV,x=6%, 5000kVA
T
R 20ohm
6.6/33kV,x=5%, 7500kVA
G
6.6kV,x=10%,6000kVA
G1 T
6.6,x=12%,4000kVA
Solution:
X x pu
X x pu
X x pu
X x pu
T
T
G
G
2
1
2
1
Z pu
X x MVAk pu
Z X X X X X
T
r
T G T G T r
2
1 1 2 2
z
B pu F
base Base base
I
I I
x k
M
xkV
I MVA
3 ^10333 174. 95
I (^) F IpuxIbase XVpupuxIB 0. 3341 pupux 174. 95 523. 82 A
Fault MVA 3 I (^) F VB 3 x 523. 82 x 33 k 30 MVA
Example 4 In the network shown in figure 1, each line has a reactance of 0.4Ω/km. using a base of 100MVA, calculate the 3θ symmetrical fault current in amperes and the corresponding fault level (in MVA) at busbars A and B.
100MVA,x=0.15pu
9 km
75MVA ,x=0.14pu
T2 10 km
100MVA ,x=0.17pu
66kV
40MVA ,x=0.12pu
60MVA,x=0.15pu
A
33kV
C
G2 D 10 km
G1 T
(^9) km^ B
66kV
T
12 km
Figure 1 Solution:
Line CA 12km(0.4Ω/km) = 4.8Ω Line CD 9km(0.4Ω/km) = 3.6Ω Line DA 10km(0.4Ω/km) = 4.0Ω Choose 100MVA as base value. Then:
L
1 2
1 2
L
1^ L3 2
1 2
j0.17 j0.
1 2
j0.
1 2
j0.
1 2
j0.083^1 j0.
j0.
j0.
j0.
j0.
1 2
j0.
j21.
j0. 1 2
L
L
L
j0.
1 2 j0. j0.083^1
1 2
1 2
C 1 2
1 2 F
j0. B
j0. 1 2
j0.32 N
A
1 2
1 2 1 2
1puV V N
1 2 F
Example 5 Refer to figure below; Generator G 1 (^) , G 2 25 MVA , x 0. 1 pu Motor M 20 MVA , x 0. 1 pu Transmission lines AB, BC,CD,AC,BD. Impedance 10Ω each. Take base of 100MVA, 33kV. Assume machines produces rated voltage of 33kV at no load. By network reduction , find fault current and fault MVA level at busbar D, if the fault at busbar C what will be the fault MVA at busbar C. G
M
D
G
B C
A
Solution Base 100MVA , 33kV (a) Fault occur at busbar D
0.1pu
10ohm
D
C 10 ohm
A
M
10ohm 10 ohm
0.1pu
10 ohm
B
(b) Fault occur at busbar C
I I xI X x xMVAk x A kA
X
F pu B T
T (^11003330). 773751 1749. 55 22. 61
Fault MVA 3 V (^) B IF 3 x 33 kx 22. 61 k 129. 24 MVA
4.4. UNBALANCE FAULT USING SYMMETRICAL COMPONENTS Three unbalanced phasors of a three-phase system can be resolved into three balancedsystems of phasors:-
a) Positive-sequence components - consisting of three phasors equal in magnitude, displaced from each other by 120º in phase, and having the same phase sequence as the original phasors.(Subscript 1 show the positive sequence) b) Negative-sequence components - consisting of three phasors equal in magnitude, displaced from each other by 120º in phase, and having the phase sequence opposite to that of the original phasors. .(Subscript 2 show the negative sequence) c) Zero-sequence components – consisting of three phasors equal in magnitude and with zero phase displacement from each other. .(Subscript 0 show the zero sequence)
In matrix form, the voltages are:-
2
1
0 2
2 1
a
a
a c
b
a V
V
V
a a
a a V
V
V
So,
Vb 1
Va 1
Vb Va 1
Vb Vc
Va 2 Vbo
Vc 1
Vco Vao
c
b
a a
a
a V
V
V
a a
a a V
V
V
2
2 2
1
0 1
In matrix form, the currents are:-
2
1
0 2
2 1
a
a
a c
b
a I
I
I
a a
a a I
I
I
So,
c
b
a a
a
a I
I
I
a a
a a I
I
I
2
2 2
1
0 1
The unloaded symmetrical voltage component can be written as:
2
1
0 2
1
0 2
1
0 0 0
a
a
a a a
a
a I
I
I
Z
Z
Z
E
V
V
V
a)Line to earth fault
Ec c Ib
Eb
Fault occur at line a. 0
Io = Ia
Ia
Ea
0
a
Zn
Ic
b
b c
a I I
V
(^1120)
1 1 1 2 1 0
1 1 1 2 1 0 0
Z Z Z
I E
I Z I Z I Z E
V
V E I Z I Z I Z
a^ a
a a a a
a
a a a a a
Ia
Z2^ -
Vao Zg
Z
Va
Z
Zg
+^ Ea
3Zn Iao
3Zn
Ia
Ia
Ea
Z
Va
b) Line to line fault
Eb
Zn
Ic
Io = In
c
Ib
Ia
b
a
0
Ea
Ec
b c
a
b c I I
I
V V
c
b
a a
a
a V
V
V
a a
a a V
V
V
2
2 2
1
0 1
b
b
a a
a
a V
V
V
a a
a a V
V
V
2
2 2
1
0 1
Va (^) 1 Va (^2) IIba ^ ^0 Ic
c
c a
a
a I
I
a a
a a I
I
I 0
2
2 2
1
0
2 1
a a
a I I
I
because 3
2
1 a^ c
a c I^ I
I I
1
1 2
1
0 1
1
a
a a a
a
a I
I
Z
Z
Z
E
V
V
V
Va 0 0 because Ia 0 0
(^112)
1 2 1 1
1 1 2
1 1 1
Z Z I E
I Z E I Z
V I Z
V E I Z
a^ a
a a a
a a
a a a
2
1
(^10)
2
1 1 0 0 0 1
Z
Z
Z
Z
Z
Z
Z
2
1
0 1
1
1
1 1
1 1
1 1
2
1
0 0
2
1
0 a
a
a a Z
Z
Z a a
a a
a a I
I
I
E
E I Z
E I Z
E I Z
Z
Z
Z
So,
21 1 2
11 1 1 1
01 1 0
a a a
a a a a
a a a
E ZI Z I
E ZI Z ZE I
E ZI Z I
Knowing that Ia Ia 1 Ia 2 Ia 0 0 ,
^
2
1 0 1 1 0 2
0 2
2
1 1 1
1 1 0
1 1 0 2
2
1 1 1 1 2
1 1 0 1
1 1 0
2
1 1 1 1
1 1 0
1 1
( ) 1
0
0
Z
Z Z I Z Z Z
E Z Z
Z
I Z Z
I Z Z
I Z Z
E Z
E
Z
I Z Z
E Z
E Z
I Z Z
E Z
I Z Z
E
Z
E I Z Z
E Z
E I Z Z
E I Z
a a
a a a a a
a a a a a a a
a a a a a a a
0 2
(^1102)
0 2
(^110202)
0 2 1 2 1 0
0 2 2
1 0 2 0 0 2 1 0 2
1 0 2
1 2
1 0 0 2 1
0 2
Z Z
Z Z Z
I E
Z Z
Z Z Z Z Z
I E
Z Z Z Z ZZ
E Z Z
Z
ZZ Z
Z
Z Z ZZ Z
I E Z Z
I
Z
Z
Z
Z Z Z
E Z Z
a^ a
a a
a a a
a a
Tutorial 4 FAULT ANALYSIS
- Refer to the figures 1, 2 ,3 ,4 and 5 below, draw an impedance diagram. Calculate the fault MVA if fault occur at : (a) Lets take 60MVA as new base.
G
118kV
60MVA0.1pu 20MVA0.09pu G
20MVA0.08pu G
Figure 1 (b). Choose 50MVA as base values.
