This document discusses fault current calculation methods. It covers symmetrical and asymmetrical faults, and describes analyzing power systems under both normal and abnormal operating conditions. The infinite bus method and per unit methods for calculating fault current are introduced. Synchronous machine response to asymmetrical faults is examined, including the subtransient, transient, and steady state stages. Fault current envelopes are presented.

1. Module 6.0
Fault Current Calculation
By: Dr. Hamid Jaffari
Power system Review

2. Fault Currents
 Symmetrical Fault
 Asymmetrical fault
Power System Review

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


5. Asymmetrical
Fault Calculation
Power System Review

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

13. nt
Subtransie
Transient
State
Steady
offset
dc
Fault
al
Asymmetric
Generator
near
Fault
al
Asymmetric
of
Stages
d
g
X
E
I "
"

"
2I
d
g
X
E
I '
'

'
2I
d
g
X
E
I 
"
2I
'
2I

14. Fault Current
Calculation
Power System Review

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

16. Fault Current Analysis
Power System Review
Ohmic Method

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

18. Infinite Bus method
Power System Review
Fault Current Analysis

19. Infinite Bus Calculation
SC
Bsae
pu
pu
SC
r
transforme
utility
pu
total
I
x
I
Z
Z
Z





actual
I
:
Step4
kV
x
3
3
KVA
I
Calculate
:
Step3
0
.
1
I
Calculate
:
Step2
Z
Calculate
:
Step1
SC
LL
Base
)
(

•Infinite Bus calculation is a convenient way to
estimate the maximum 3ᶲ fault current flow on the sec
side of the transformer
•The following steps are necessary to calculate the ISC
100
%
&
;
Known
is
Circuit
Short
Utility
If
:
Note1
)
(
Z
Z
MVA
MVA
Z
where
Z
Z
Z
r
transforme
SC
base
utility
r
transforme
utility
pu
total




0
&
100
%
;
Unknown
is
Circuit
Short
Utility
If
:
Note2



utility
r
transforme
r
transforme
total
Z
pu
Z
Z
where
Z
Z
total
Z

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

22. Fault Current Analysis
Power System Review
Per-Unit Method

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

27. Symmetrical
Fault Calculation
Power System Review

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 

34. Symmetrical Fault Current
Analysis…MVA-Method
MVA Method
Power System Review

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
)
(


36. MVA Equivalent Network
n
total MVA
MVA
MVA
MVA ........
:
MVAs
Parallel
2
1 


n
total MVA
MVA
MVA
MVA
1
........
1
1
1
:
MVAs
Series
2
1



1
MVA 2
MVA 3
MVA
1
MVA 2
MVA 3
MVA
3
2
1 MVA
MVA
MVA
MVAtotal 


3
2
1
1
1
1
1
MVA
MVA
MVA
MVAtotal



Total
MVA
Total
MVA

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

44. Step 2: Network Reduction(MVA Method)
52
.
358
77
.
77
10
8
53
.
86 10
8
76
.
36
53
.
86
1
77
.
77
1
52
.
358
1
1
:
MVAs
Series



total
MVA
76
.
36
53
.
86
1
77
.
77
1
52
.
358
1
1




MVAtotal
3
2
1
:
MVA
MVA
MVA
MVA
MVAs
Parallel
total 


76
.
54
8
76
.
36
10 



total
MVA
76
.
54
MVA
Fault

45. Step 3:Fault MVA Conversion to Ifault
76
.
54

fault
MVA
6003
.
7
)
16
.
4
(
3
76
.
54
)
16
.
4
(
3
)
3
(
)
( 


LL
LL
fault
fault
kV
x
kV
x
MVA
kA
I

Amp
l
Symmetrica
Ifault 3
.
600
,
7
)
( 
kV
kVll 16
.
4

:
2 Quantities
Bus
Bus2 Fault Current:

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 

48. 76
.
36
8
10
A
kV
x
MVA
I
on
Contributi
Generator
Gen
fault 9
.
387
,
1
16
.
4
3
10
:



Gen
MVA
Motor
MVA
)
( Line
Xfmr
Utility
MVA 

A
kV
x
MVA
I
on
Contributi
Motor
motor
fault 3
.
110
,
1
16
.
4
3
8
:



Ex1: Motor/Gen Fault Contribution
(MVA Method)
A
kV
x
MVA
I
on
Contributi
Utility
fault 102
,
5
205
.
7
76
.
36
16
.
4
3
76
.
36
:



A
2
.
600
,
7
3
.
110
,
1
9
.
387
,
1
102
,
5
:






 

 Gen
f
utility
f
motor
f
fault I
I
I
I
Current
Fault
Total

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
(



















51.  22.86 kV Utility Source:
 Line:
 Transformers:
Power System Review
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
Solution to Example2 (MVA method):

52. Solution to Example2 (MVA method):
 Generators:
 Motors:
Power System Review
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
(



















53. Example 2:Symmetrical Fault Current
Calculation (MVA-method)
Power System Review
593.903
MVA
2903.220
MVA
222.222
MVA
41.667
MVA
50 MVA 13.333
MVA
9.375
MVA
14.286
MVA
BUS 1
BUS 2
Modeling
Network
:
Step1

54. Symmetrical Fault Current
Analysis…MVA-Method
 Series MVAs:
 Parallel MVAs:
Power System Review
)]
/
1
(
)
/
1
(
)
/
1
[(
1
2
1 n
MVA
MVA
MVA
total
series
MVA








n
MVA
MVA
MVA
total
parallel
MVA 






 2
1
Reduction
MVA
Network
:
Step2

55. Example2: Symmetrical Fault Current
Analysis…MVA-Method
 MVA series:
MVA=1/[(1/593.903)+(1/2,903.220)+(1/222.222)]
MVA=1/[(.0017)+(.0003)+(.0045)]=153.846
 Bus1 (parallel)=153.846+41.667+13.333=208.846
 MVA series @Bus2:
MVA=1/[(1/208.846)+(1/50)]
MVA=1/[(.0048)+(.0200)]=40.323
Power System Review
Reduction
MVA
Network
:
Step2
50 MVA
9.375
MVA
14.286
MVA
BUS 2
208.846MVA

56. Ex2: Short Circuit MVA Calculation
@ Bus 2(MVA method)
153.846
MVA
41.667
MVA
50
MVA
13.333
MVA
9.375
MVA
14.286
MVA
BUS 1
BUS 2
846
.
153
)]
22
.
222
/
1
(
)
22
.
903
,
2
/
1
(
)
903
.
593
/
1
[(
1




total
series
MVA
50
MVA
9.375
MVA
14.286
MVA
BUS 2
MVA
846
.
208
333
.
13
667
.
41
846
.
153 



parallel
MVA
208.846
MVA
n
Calculatio
MVA
Fault
:
Step3

57. 40.323
MVA
9.375
MVA
14.286
MVA
BUS 2
MVA
984
.
63
375
.
9
286
.
14
323
.
40
2
@ 



Bus
MVA
323
.
40
)]
50
/
1
(
)
846
.
208
/
1
[(
1



series
MVA
MVA
984
.
63
2
@ 
Bus
MVA fault
Ex2: Short Circuit MVA Calculation
@ Bus 2(MVA method)

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 

62. Example3: Symmetrical Fault
Analysis(MVA-method)
Power System Review
1700.36
MVA
1500
MVA
117.65
MVA
27.32
MVA
5 MVA_____
(13.2/13.8)²x0.2
=27.32
102.52
MVA
27.32
MVA
MVA Fault= 102.52+27.32
= 129.84
Ifault= 129.84/(1.732x13.8)
= 5,432.3 Amps
65
.
117
1
36
.
1700
1
1500
1
1



MVA
r
Transforme
Line
Source
Motor

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).

65. Solution
 Answer: 37.5 KVA
Break Time !!!!!