3. Fault Analysis
Analysis Type
Power Flow: normal operating conditions
Faults: abnormal operating conditions
Fault Types
Balanced or Symmetrical Fault
Three Phase Short Circuit
Unbalanced or Unsymmetrical Faults
Single line-to-ground
Double line-to-ground
Line-to-line
What are the results used for?
o Determining the circuit breaker rating
o Protective Relaying settings
4. Various Types of Faults
Fault
l
Symmetrica
)
a
Fault
cal
Unsymmetri
)
b
Fault
line
-
to
-
line Fault
ground
-
to
-
line
double Fault
ground
-
to
-
line
fault
1
F
Z
Z
V
)
(3
I
l-fault
Symmetrica
fault
0
2
1
F
fault
3Z
)
3
Z
(
Z
Z
3V
ground)
-
to
-
(Line
I
n
Z
fault
2
1
F
fault
Z
Z
Z
V
3
line)
-
to
-
(line
I
j
a
b
c
a
b
c
a
b
c
a
b
c
a
b
c
6. R-L Circuit Transients
R
+
-
0
@
t
Closed
SW
L
)
sin(
2
)
(
wt
V
t
e
0
)
sin(
2
)
(
)
(
:
t
t
V
t
Ri
dt
t
di
L
Equation
]
)
sin(
)
[sin(
2
)
(
)
(
)
(
: T
t
e
t
Z
V
t
i
t
i
t
i
Solution dc
ac
amp
t
Z
V
t
iac )
sin(
2
)
(
T
t
e
Z
V
t
idc
)
sin(
2
)
(
2
2
2
2
)
( l
R
X
R
Z
R
wl
tg
R
X
tg 1
1
fR
X
R
X
R
L
T
2
:
)
(
/ forced
Current
Fault
State
Steady
Fault
l
Symmetrica :
)
(transient
Current
Offset
dc
Solution
forced Solution
natural
7. Asymmetrical fault
]
)
sin(
)
[sin(
2
)
(
)
(
)
( T
t
e
t
Z
V
t
i
t
i
t
i dc
ac
•Dc offset Magnitude depends on angle α:
ac
I
offset
dc 2
0
)
2
(
Z
V
current
fault
ac
rms
I
where ac
)
(
:
]
)
2
[sin(
2
)
(
)
(
)
(
)
2
(
:
T
t
e
t
I
t
i
t
i
t
i
Set
ac
dc
ac
•In order to get the largest fault current:
8. Asymmetrical fault
Note: i(t) is not completely periodic. So, how do we
get the rms value of i(t) ?
Assume :
Now calculate the RMS Asymmetrical Fault Current:
)
constant
(
C
e T
t
Amp
e
I
e
I
I
I
I
t
i T
t
ac
T
t
dc
ac
dc
ac
rms
2
2
2
2
2
2
1
]
2
[
]
[
)
(
)
(
)
(
cycles
in
time
is
where
f
t
fR
X
R
X
R
X
R
L
T
Note
;
&
2
:
Amp
e
I
e
I
e
I
t
i R
X
ac
fR
X
f
ac
T
t
ac
rms
)
/
(
4
2
2
2
2
1
2
1
2
1
)
(
Unit
Per
2
1
factor
al
asymmetric
)
(
:
)
(
)
( )
/
(
4
R
X
ac
rms e
k
where
I
k
I
9. Asymmetrical Fault Calculation
Example: In the following Circuit, V=2.4kV, L=8mH,
R=0.4Ω, and ω=2π60 rad/sec. Determine (a) the rms
symmetrical fault current; (b) the rms asymmetrical fault
current; (c) the rms asymmetrical fault current for .1 cycle
& 3 cycle after the switch closes, assuming the maximum
dc offset.
+
- 0
@
t
Closed
SW
mH
L 20
)
sin(
2400
2
)
(
wt
t
e
4
R
10. Asymmetrical Fault Calculation
Solution:
4
.
82
042
.
3
4
.
82
042
.
3
016
.
3
4
.
0
)
10
8
)(
60
2
(
4
.
0
)
(
) 3
Z
Z
j
x
j
L
j
R
jX
R
Z
a
A
95
.
788
042
.
3
2400
volts
Z
V
Iac
A
46
.
1366
2
1
95
.
788
)
0
(
)
0
(
;
0
@
)
k
I
I
t
b ac
rms
00
.
1
10
739
.
6
1
2
1
)
3
(
641
.
1
693
.
1
1
2
1
)
1
.
0
(
54
.
7
4
.
0
016
.
3
)
(
)
3
54
.
7
)
3
(
4
54
.
7
)
1
.
0
(
4
x
e
cycle
k
e
cycle
k
Ratio
R
X
c
A
95
.
788
)
3
(
)
3
(
A
69
.
294
,
1
641
.
1
)
1
.
0
(
)
1
.
0
(
cycle
k
I
cycle
I
x
cycle
k
I
cycle
I
ac
rms
ac
rms
+
- 0
@
t
Closed
SW
mH
L 20
)
sin(
400
,
2
2
)
(
wt
t
e
4
R
11. Asymmetrical Fault-Unloaded
Synchronous Machine
Three Stages: Subtransient, Transient, and Steady State
constant
time
armature
T
/
I
/
I
/
I
:
offset
dc
Maximum
2
2
)
(
)
2
sin(
]
1
)
1
1
(
)
1
1
[(
2
)
(
Current
ous
Instantane
)
(
)
(
)
(
A
Reactance
State
Steady
Reactance/
s
Synchronou
axix
direct
'
'
Reactance
Transient
axix
direct
'
Reactance
nt
Subtransie
axix
direct
"
/
"
/
"
'
'
"
"
"
'
"
d
g
d
d
g
d
d
g
d
T
t
T
t
d
g
dc
d
T
t
d
d
T
t
d
d
g
ac
X
E
X
X
E
X
X
E
X
Where
e
I
e
X
E
t
i
t
X
e
X
X
e
X
X
E
t
i
t
i
t
i
t
i
A
A
d
d
dc
ac
A
d
d
d
d
d X
X
X
T
,
T
,
T
Constants
Time
&
Reactances
Machine
:
provide
eres
Manufactur
:
Note
'
"
,
'
,
"
Stator
Uniform air-gap
Stator winding
Rotor
Rotor winding
N
S
d-axis
q-axis
axis
quadrature
axis
q
axis
direct
axis
d
12. Synchronous Machine
Asymmetrical Fault Envelopes
Asymmetry Sources: (1) Open Phase and (2) SLG Fault
d
g
X
E
I "
"
d
g
X
E
I '
'
d
g
X
E
I
Current
fault
nt
Subtransie
Current
fault
Transient
Current
fault
S.S
)
(t
iac
t
envelopes
current
AC
A
A T
t
T
t
d
g
e
I
e
X
E
"
"
MAX
-
dc 2
2
(t)
i
"
2I
I
2
'
2I
15. Fault Current Analysis
Power System Review
Four methods to calculate the fault current:
1.Ohmic Method (not preferred)
2.Infinite Bus Method (Convenient & Easy)
3.Per Unit Method (Most Common)
4.MVA Method (Quick & Easy)
Note: This course will focus on PU & MVA Methods
17. Ohmic Method
Power System Review
This Method Requires:
Transferring all impedances to high/low
voltage side of transformer using square
of XFMR turn ratio
Using your AC circuit theory knowledge
Voltage & Current dividers
Thevenin & Norton equivalents
Kramer’s Rule, etc
2
1
2
2
2
1
N
N
OR
N
N
20. 7.5%
Z
kV
kV/4.16
13.8
KVA
5000
VS
Infinite Bus Calculation
Unknown Utility SC Data
A
4
.
9252
95
.
693
333
.
13
actual
I
:
Step4
A
95
.
693
16
.
4
3
5000
kV
x
3
3
KVA
I
Calculate
:
Step3
333
.
13
075
.
0
.
1
0
.
1
I
Calculate
:
Step2
075
.
0
100
5
.
7
100
Z%
Z
Calculate
:
Step1
SC
LL
Base
pu
x
I
x
I
kV
x
Z
pu
SC
Bsae
pu
pu
SC
Example1: Calculate the maximum 3ᶲ fault current on 5000 KVA
Transformer’s secondary bus.
Data
Source
No
21. 7.5%
Z
kV
kV/4.16
13.8
KVA
5000
VS
Infinite Bus Calculation
with Known Utility SC Data
A
6426
95
.
693
26
.
9
actual
I
:
Step4
A
95
.
693
16
.
4
3
5000
kV
x
3
3
KVA
I
Calculate
:
Step3
26
.
9
108
.
0
0
.
1
0
.
1
I
Calculate
:
Step2
108
.
0
075
.
033
.
0
Z
Calculate
:
Step1
SC
LL
Base
)
(
total
x
I
x
I
kV
x
Z
pu
Z
Z
SC
Bsae
pu
total
pu
SC
r
transforme
utility
Example2: Calculate the maximum 3ᶲ fault current on 5000 KVA
Transformer’s secondary bus.
150MVA
SC
pu
Z
Z
pu
x
S
S
kV
kV
Z
Z
pu
r
transforme
Old
base
New
base
new
old
pu
Utility
SC
base
utility
Old
New
075
.
0
100
5
.
7
100
%
033
.
150
5
16
.
4
16
.
4
1
1
150
150
MVA
MBA
Z
Z
Z
Z
Calculate
2
2
r
transforme
utility
toal
pu
108
.
0
33
0
.
0
0.075
Ztotal
utility
Z
:
Steps
n
Calculatio
23. Power System Review
Fault Current Analysis:
Per-Unit Method
PU analysis is used for both symmetrical &
unsymmetrical fault calculations.
•All components are defined in PU system.
•Analysis is performed using equivalent per phase
circuit modeling.
•Requires knowledge of symmetrical components
•Requires selecting two system bases for
calculating all base & PU quantities:
kVBase & MVAbase
24. Power System Review
Fault Current Analysis:
Per-Unit Method
This Method requires:
•Knowledge of symmetrical components
Positive sequence (+ SEQ)
Negative sequence(-SEQ)
Zero sequence (0 SEQ)
•Interconnecting positive, negative, and
zero networks for calculating the various
unsymmetrical faults(LG, LL/LLG, and 3ᶲ)
25. Symmetrical Components
Power System Review
Steps involved:
1. Draw a single-line diagram of the desired
power system(equivalent per phase)
2. Define zones using transformation point as
a point of demarcation
3. Select a common MVAbase for all zones
4. Select a kVBase for one zone & Calculate
a. kVBase for other zones
b. Zbase, and Ibase for all zones
26. Symmetrical Components..cont
Power System Review
6. Replace each component with its
equivalent reactance in per-unit
7. Draw sequence networks(+, -, 0)
8. Use (+)SEQ network for Symmetrical
Fault analysis
9. Combine appropriate networks for
calculating various Unsymmetrical
Fault analysis
28. 3Φ Symmetrical Fault Analysis
(PU Method)
Symmetrical Fault refers to a balanced 3Φ
fault, in a balanced 3Φ system operating in
steady state, which is either :
Bolted fault: LLLG fault with Zfault=0
Non-Bolted fault: LLLG fault with Zfault≠0
Only the (+)SEQ network exists.
(0)SEQ & (-)SEQ currents are equal to “Zero”.
Power System Review
29. Symmetrical Fault Modeling
for a Bolted Fault (PU Method)
Z0 eq
Note: VF=Pre Fault Voltage
+
_
Vo=0
Z2 eq
VF
Z1 eq
+
_
+
_
V1=0
I0=0
I1 Ia
Ib
Ic
Vc
Vb
Va
+
+
+
_ _ _
Ib = -Ia = Ic = ISC
Vbg = Vag = Vcg =0
Phase
g
+
_
V2=0
I2=0
)
(
1
)
(
)
(
1
PU
eq
PU
f
Z
V
I PU
fault
0
2
I
0
0
I
SEQ
SEQ
)
(
SEQ
)
(
SEQ
)
0
(
30. Practice Example (PU Method):
In the following power system Calculate(a)3ᶲ Symmetrical
fault current @ Bus3 and select an appropriate Breaker
Size @ Bus 3
G1
G2
PU
.15
0
X
kV
13.8
MVA
500
"
.15PU
0
T1
Υ
115kV
/
Δ
13.8kV
MVA
500
"
X
PU
.20
0
X
kV
13.2
MVA
750
"
.18PU
0
T2
kV
8
.
13
/
115kV
MVA
750
"
X
6
XT1
2
X 13
T
17.63
Zbase
115kV
Kvbase
MVA
750
Sbase
1
Bus 2
Bus
3
Bus
4
X 23
T
.254
Zbase
13.8kV
Kvbase
MVA
750
Sbase
.254
Zbase
13.8kV
Kvbase
MVA
750
Sbase
MVA
750
SBase
31. Breaker Selection
Modern Circuit Breaker standards are designed based on
ISymmetrical. The following steps are required to determine an
appropriate breaker size:
1. Use “E/X” method to calculate the minimum ISymmetrical.
2. Calculate X/R ratio:
1. If X/R <15 →Use ISymmetrical
2. If X/R>15 →It means the dc offset has not decayed
to an acceptable level. Thus, calculate IAsymmetrical.
3. Calculate IAsymmetrical at calculated fault location.
4. Breaker Interrupting Capability should be 20% greater
than the calculated fault current.
32. Breaker Selection Criterion
Generator/ Synchronous Motor/Large Induction motors
Breakers:
Use subtransient Reactance X”d to calculate ISymmetrical.
Use 2 cycle Breaker
Transmission Breakers:
Use 3 cycle Breakers if X/R>15
Use 5 cycle Breaker if X/R<15
Distribution Breakers:
Use 3 cycle or 5 cycle Breakers
If X/R ratio is unknown Use:
8
.
0
I
I
l
Symmetrica
Capability
ng
Interrupti
Breaker
Unknown
R
X
33. A
2
.
614
,
16
8
.
0
13,291.2
I Capability
ng
Interrupti
Breaker
G1 G2
PU
.15
0
X
kV
13.8
MVA
500
"
.15PU
0
T1
Υ
115kV
/
Δ
13.8kV
MVA
500
"
X
PU
.20
0
X
kV
13.2
MVA
750
"
.18PU
0
T2
kV
8
.
13
/
115kV
MVA
750
"
X
6
XT1
2
X 13
T
17.63
Zbase
115kV
Kvbase
MVA
750
Sbase
1
Bus 2
Bus
3
Bus
4
X 23
T
.254
Zbase
13.8kV
Kvbase
MVA
750
Sbase
.254
Zbase
13.8kV
Kvbase
MVA
750
Sbase
MVA
750
SBase
kV
115
:
Class
Voltage
Breaker
:
Selection
Breaker
cycle
3
:
Cycle
Breaker
Practice Example (PU Method):
A
2
.
291
,
13
I l
Symmetrica
35. Fault Current Calculation-MVA Method
This method follows a four steps process:
1. Calculate the Admittance of every component in its own
infinite bus.
2. Multiply the calculated admittances in step(1) by the
MVA rating of each component to get MVASC.
3. Combine short-circuit MVAs & follow the Admittance
series & parallel rules:
4. Convert MVAs to Symmetrical fault current
Power System Review
%
100
)
Admittance
(
Z
Y
)
Admittance
(
Y
x
MVA
MVAsc
n
total MVA
MVA
MVA
MVA ........
:
MVAs
Parallel
a)
2
1
n
total MVA
MVA
MVA
MVA
1
........
1
1
1
:
MVAs
Series
b)
2
1
ll
kV
x
Total
MVAsc
al
Isymmetric
3
)
(
37. Why Use the MVA Method?
This method is internationally used and accepted by most
protection engineers.
The network set up is easier than Ohmic or PU method.
You can calculate Ifault in a shorter time period.
This method makes it easier to see the fault contributions
@ every point in the system.
Calculation accuracy is within 3% to 5% compared to PU &
Ohmic method.
Power System Review
38. MVA Method Assumptions
Power System Review
10
.
1
R
X
Two Conditions must be satisfied:
Operation
State
Steady
.
2
39. Symmetrical Fault Current
Analysis...MVA-Method
Power System Review
)
(
3
: KA
I
x
kV
x
MVAsc
MVA
Utility sc
ll
fault
)
(
:
2
Z
kV
MVA
Cable
ll
fault
%
100
:
/ "
Gen
d
fault
X
x
MVA
MVA
Motor
us
Sycnhroono
Generator
%
100
:
xfmr
fault
Z
x
MVA
MVA
r
Transforme
Formulas:
Note: Impedances (Z) are steady state values
40. Where: X”d=direct-axis Subtransient Reactance
X”d= I Full-load amp/I Locked Rotor amp
Power System Review
Symmetrical Fault Current
Analysis...MVA-Method
%
100
: "
Gen
d
motor
fault
X
x
MVA
MVA
Motor
amp
load
full
rotor
locked
motor
fault
I
I
x
MVA
MVA
Motor
Induction
:
:
Motor
41. Summary:
Power System Review
Symmetrical Fault Current
Analysis...MVA-Method
LL
kV
x
MVA
I
total
KA
fault
3
)
(
)]
/
1
(
)
/
1
(
)
/
1
[(
1
2
1 n
MVA
MVA
MVA
total
series
MVA
n
MVA
MVA
MVA
total
parallel
MVA
2
1
42. Example1:Fault Calculation(MVA method)
Generator
M
Utility Source
13.8kV, 15KA fault current
Motor
2MVA Y
4.16kV
X”d=0.25pu
3-500McM cables, 2000 ft
Z=0.2Ω
Transformer
7MVA
13.8kV/4.16kV
Z=9%
1.5MVA Y
4.16kV
X”d=0.15pu
Bus 1 13.8kV
Bus 2 4.16kV
In the following Power System, Calculate the fault current @ Bus2 & fault current
contributions from both Gen & Motor?
43. Step1:Network Modeling(MVA Method)
Generator
M
Utility Source
13.8kV, 15KA fault current
Motor
2MVA Y
4.16kV
X”d=0.25
3-500McM cables, 2000 ft
Z=0.2Ω
Transformer
7MVA
13.8kV/4.16kV
Z=9%
1.5MVA Y
4.16kV
X”d=0.15
MVA
x
x
MVAsource 5
.
358
)
kA
15
(
)
kv
8
.
13
(
3
MVA
x
Z
x
MVA
MVA
xfmr
r
transforme 77
.
77
9
100
7
%
100
MVA
x
X
x
MVA
MVA
d
Generator 10
15
.
0
1
5
.
1
1
"
MVA
x
X
x
MVA
MVA
d
Motor 8
25
.
0
1
2
1
"
52
.
358
77
.
77
10
8
MVA
Z
kV
MVA
line
Line 53
.
86
2
.
0
)
16
.
4
( 2
2
53
.
86
Bus1 13.8kV
Bus2 4.16kV
46. Generator
M
Utility Source
13.8kV, 15KA fault current
Motor
2MVA Y
4.16kV
X”d=0.25
3-500McM cables, 2000 ft
Z=0.2Ω
Transformer
7MVA
13.8kV/4.16kV
Z=9%
1.5MVA
Y
4.16kV
X”d=0.15
Bus1 13.8kV
Bus2 4.16kV
pu
Vf 0
.
1
2
)
( Bus
for
Network
SEQ
utility
Z
Xfmr
Z
Line
Z
Gen
Z motor
Z
In the following Power System, Calculate the fault current @ Bus2 & fault current
contributions from both Gen & Motor using PU Method?
Example1:Fault Analysis(PU Method)
47. Example 1: Symmetrical Fault Current
Calculation Comparison between
PU & MVA Methods
Amp
I Bus
fault 3
.
600
,
7
2
@
A
7
.
605
,
7
A
879
,
13
548
.
0
)
(
2
@
x
xI
pu
I
I base
fault
Bus
fault
:
n
calculatio
method
MVA
:
n
calculatio
Method
Unit
Per
49. Ex1:Symmetrical Fault Current Analysis
PU & MVA Methods Comparison
Amp
I motor
f 3
.
110
,
1
A
110
,
1
f-motor
I
:
n
calculatio
method
MVA
:
n
calculatio
Method
Unit
Per
50. Symmetrical Fault Current Calculation
MVA Method
Example2: Calculate the Symmetrical fault current @ Bus2 using the MVA Method
Generator
M
M
Generator
Utility Source
22.86kV, 15KA fault current
Transformer
20MVA Delta-Yn
22.86/4.16kV
Z=9% 5MVA
4.16kV
Z=12%
Transformer
3.5MVA Delta-Yn
4.16kV/480V
Z=7%
Motor
2MVA Y
4.16kV
Z=15%
Motor
1.5MVA Y
480V
Z=16%
Y
Y
3-500McM cables, 2000 ft
Z=.18 Ω
Bolted Fault
Generator
2MVA
480 V
Z=14%
BUS 1
BUS 2
903
.
593
15
86
.
22
3
kA
x
kV
x
MVA LL
fault
22
.
903
,
2
18
.
0
)
86
.
22
( 2
2
kV
Z
kV
MVA
line
fault
50
07
.
0
5
.
3
100
%
222
.
222
09
.
0
20
100
%
Z
MVA
MVA
Z
MVA
MVA
Xfmr
fault
Xfmr
fault
MVA
Z
MVA
G
MVA
MVA
Z
MVA
G
MVA
fault
fault
286
.
14
14
.
0
2
100
%
)
2
(
667
.
41
12
.
0
5
100
%
)
1
(
MVA
Z
MVA
G
MVA
MVA
Z
MVA
M
MVA
fault
fault
375
.
9
16
.
0
5
.
1
100
%
)
2
(
333
.
13
15
.
0
2
100
%
)
1
(
58. Example2: Symmetrical Fault
Current Analysis…MVA-Method
Bus2 (total) = 40.323+14.286+9.375=63.984 MVA
Now, Calculate the Short Circuit MVA @Bus1?
Power System Review
Available Fault Current @Bus 2:
Ifault=63.984 MVA/[ x 0.48kV]=76,963 A
3
59. Ex2:Calculate Short Circuit MVA@ Bus1
(MVA method)
153.864
MVA
41.667
MVA
50 MVA
9.375
MVA
14.286
MVA
13.333
MVA
195.531
MVA
13.333
MVA
50
MVA
9.375+14.286=23.661
MVA
208.864+16.051=224.915 MVA
208.864= 195.531+13.333
MVA
1/[(1/50)+(1/23.661)]=1/.0623=16.051
MVA
Power System Review
BUS 1 BUS 1
BUS 1
BUS 1
BUS 2
MVA
531
.
195
667
.
41
846
.
153
parallel
MVA
MVA
915
.
224
1
@
Bus
fault
MVA
60. S.C or Fault MVA @ Bus1:
S.C or Fault MVA= 224.915
I fault @Bus1= 224.915 MVA/( x4.16kV)
Power System Review
Ex2: Calculate Short Circuit MVA
@ Bus 1 (MVA method)
3
Available Fault Current at Bus 1:
I fault @Bus1=31,216 A
61. Example 3: Symmetrical Fault Analysis
Power System Review
Source M
1500 MVA
Fault
69 kV
X=2.8Ω
10 MVA
X=8.5%
69kV Δ/Υ-n 13.8kV
13.2 kV
X=0.2
Calculate the symmetrical fault current at the secondary terminals of a 10 MVA XFMR
using both the PU-Method & the MVA Method. Use 15 MVA & 69 kV base values for
the transmission line.
5 MVA Υ-n
Zone 1 Zone 2
kV
V 69
1
Base
-
lL
A
kV
x
S
IBase 57
.
627
3 1
Base
Base
2
4
.
317
15
69
S
2
Base
Base
2
Base
1
1
1
kV
Z
MVA
S 15
Base
MVA
S 15
Base
7
.
12
15
8
.
13 2
Base2
Z
kV
V 8
.
13
2
Base
-
lL
63. Symmetrical Fault Calculation
Comparison Between PU & MVA
Methods
I fault= 5,410.3 Amp
I fault = 5,432.3 Amp
:
method
PU
:
method
MVA
Example 3:
Power System Review
64. References
1. J.D. Golver, M.S. Sarma, Power System Analysis and design,
4th ed., (Thomson Crop, 2008).
2. M.S. Sarma, Electric Machines, 2nd ed., (West Publishing Company,
1985).
3. A.E. Fitzgerald, C. Kingsley, and S. Umans, Electric
Machinery, 4th ed. (New York: McGraw-Hill, 1983).
4. P.M. Anderson, Analysis of Faulted Power systems(Ames, IA: Iowa
Satate university Press, 1973).
5.W.D. Stevenson, Jr., Elements of Power System Analysis, 4th
ed. (New York: McGraw-Hill, 1982).