Module 6.0
Fault Current
Calculation
By: Dr. Hamid Jaffari
Power system Review
Fault Currents
 Symmetrical Fault
 Asymmetrical fault
Power System Review
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
Various Types of Faults
a)Symmetrical Fault
a
a
b
b
c
c
VF
ISymmetrical-fault(3 ) 
Z1  Zfault
b)Unsymmetrical Fault
line - to - line Fault
a
double line - to - ground Fault
b
c
Ifault(line - to - line) 
a
b
 j 3VF
Z1  Z2  Zfault
c
line - to - ground Fault
a
b
c
Ifault(Line - to - ground) 
3VF
Z1  Z2  ( Z0  3Zn)  3Zfault
Asymmetrical
Fault Calculation
Power System Review
R-L Circuit Transients
R
e(t )  2 V sin( wt   )
+
-
L
SW Closed @
t 0
di(t )
 Ri (t )  2V sin(t   ) t  0
dt
t

V
Solution : i(t )  iac(t )  idc(t )  2 [sin(t     )  sin(   )e T ]
Z
forced Solution natural Solution
Equation : L
Symmetrical Fault / Steady State Fault Current( forced ) :
iac(t )  2
V
sin(t     ) amp
Z
Z
R2  X 2 
R 2  (l ) 2 
X
1  wl 
  tg 

R
 R
L
X
X
T 

R R 2fR
  tg 1 
dc Offset Current(transient ) :
t

V
T
idc(t )  2 sin(   )e
Z
Asymmetrical fault
t

V
i(t )  iac(t )  idc(t )  2 [sin(t     )  sin(   )e T ]
Z
•Dc offset Magnitude depends on angle α:
    ( 

2
)
0  dc offset  2 Iac
where : Iac(rms ac fault current ) 
V
Z

•In order to get the largest fault current:
Set :   (  )
2
t


i (t )  i (t )  i (t )  2 I [sin(t  )  e T ]
2
ac
dc
ac
Asymmetrical fault
 Note: i(t) is not completely periodic. So, how do we
get the rms value of i(t) ?
 Assume : e  C (constant)
 Now calculate the RMS Asymmetrical Fault Current:

t
T

t
T 2
irms(t )  ( Iac) 2  ( Idc) 2  [ Iac]2  [ 2 Idce ]  Iac 1  2e
X
Note : T 
L 
X
X



R
R R 2fR
&
t

f
irms(t )  Iac 1  2e
2t
T
2t
T
Amp
; where  is time in cycles



 Iac 1  2e
2
 X 

f 
 2fR 
 Iac 1  2e
Irms( )  k ( ) Iac where : k ( )  asymmetrical factor  1  2e


4
( X / R)
4
( X / R)
Amp
Per Unit
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.
L  20mH
R  4
e(t )  2 2400 sin( wt   )
+
-
SW Closed @ t  0
Asymmetrical Fault Calculation
 Solution:
a) Z  R  jX  R  j (L)  0.4  j (2 60)(8 x10 3 )  0.4  j 3.016  3.04282.4
Z  Z  3.04282.4
Iac 
V 2400volts

 788.95A
Z
3.042
b) @ t  0; Irms(0)  Iack (0)  788.95 1  2  1366.46 A
X
3.016
c) ( Ratio ) 
 7.54
R
0.4
k (  0.1cycle )  1  2e
k (  3cycle )  1  2e
e(t )  2 2,400 sin( wt   )
 4 ( 0.1)
7.54
 4 ( 3 )
7.54
R  4
+
-
L  20mH
SW Closed @
t 0
 1  1.693  1.641
 1  6.739 x10 3  1.00
Irms  (  0.1cycle )  Iac k (0.1cycle ) x1.641  1,294.69 A
Irms  (  3cycle )  Iac k (3cycle )  788.95 A
Asymmetrical Fault-Unloaded
Synchronous Machine
 Three Stages: Subtransient, Transient, and Steady State
i (t )  iac(t )  idc(t )
 Instantaneous Current
 '
1
1  T "d
1
1
1

iac(t )  2 Eg[( "  ' )e
( ' 
)e T d  ] sin(t    )
X d Xd
Xd X d
Xd
2
d-axis
Eg
idc(t )  2 " e t / TA  2 I "e t / TA M aximum dc offset
N
X d
Where :
t
X
"
I
d  direct axix Subtransient Reactance
X 'd
t
"

Eg / X
"
q-axis
X d  direct axix Synchronous Reactance/SteadyStateReactance
TA  armature time constant
S
d  axis  direct axis
q  axis  quadrature axis
Note : M anufactureres provide :
X "d , X ' d , X
d
Time Constants
T"d , T ' d , T
A
Stator
Rotor
I  Eg / Xd
M achine Reactances
Uniform air-gap
Rotor winding
d
I '  Eg / X ' d
 direct axix Transient Reactance
Stator winding
&
Synchronous Machine
Asymmetrical Fault Envelopes
 Asymmetry Sources: (1) Open Phase and (2) SLG Fault
iac(t )
Subtransient fault Current
2I
"
'
Eg Transient fault Current 2I
I  "
Xd
Eg
I'  '
S.S fault Current
Xd
Eg
I
Xd
"
AC current envelopes
t

2 Eg  TtA
"
TA
idc - MAX(t) 
e

2
I
e
X "d
2I
t
I" 
I' 
I
Eg
X "d
Eg
X 'd
Eg
Xd
2I
"
2I "
Stages of Asymmetrical Fault near Generator
Subtransient
2I '
Transient
Steady State
dc offset
2I '
Asymmetrical Fault
Fault Current
Calculation
Power System Review
Fault Current Analysis
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
Power System Review
Fault Current Analysis
Ohmic Method
Power System Review
Ohmic Method
This Method Requires:
 Transferring all impedances to high/low
voltage side of transformer using square
of XFMR turn ratio  NN  OR  NN 
2
2
1
2
2
1
Using your AC circuit theory knowledge
Voltage & Current dividers
Thevenin & Norton equivalents
Kramer’s Rule, etc
Power System Review
Fault Current Analysis
Infinite Bus method
Power System Review
Infinite Bus Calculation
•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
Step1: Calculate Ztotal( pu)  Zutility  Ztransformer
1.0 pu
Step2 : Calculate ISC 
Zpu
KVA 3
Step3 : Calculate IBase 
Ztotal
3 x kVLL
Step4 : ISC actual  IBsae x ISC
Note1: If Utility Short Circuit is Known
Ztotal( pu)  Zutility  Ztransformer
where;
MVAbase
Z%
Zutility 
& Ztransformer 
MVASC
100
Note2 : If Utility Short Circuit is Unknown
Ztotal  Ztransformer
where;
Z%
Ztransformer 
pu & Zutility  0
100
Infinite Bus Calculation
Unknown Utility SC Data
Example1: Calculate the maximum 3ᶲ fault current on 5000 KVA
Transformer’s secondary bus.
VS
Z% 7.5
Step1: Calculate Zpu 

 0.075 pu No Source Data 5000KVA
100 100
13.8kV/4.16kV
Z  7.5%
1.0 pu 1.0
Step2 : Calculate ISC 

 13.333
Zpu .075
KVA 3
5000
Step3 : Calculate IBase 

 693.95 A
3 x kVLL
3 x 4.16kV
Step4 : ISC actual  IBsae x ISC  13.333 x 693.95  9252.4 A
Infinite Bus Calculation
with Known Utility SC Data
Example2: Calculate the maximum 3ᶲ fault current on 5000 KVA
Transformer’s secondary bus.
VS
Calculate
Zutility
Ztransformer 
Ztoal  Zutility  Ztransformer
M BAbase 150
Zutility 

 1 pu
M VASC 150
2
2
 kVold   SbaseNew 
 4.16   5 
ZUtility New  ZpuOld 
 1x

 
  .033 pu
 kVnew   SbaseOld 
 4.16   150 
SC  150MVA
5000KVA
13.8kV/4.16kV
Z  7.5%
Ztotal  0.075  0.033  0.108 pu
Z % 7.5

 0.075 pu
100 100
Calculatio n Steps : Step1: Calculate Ztotal  Zutility  Ztransformer  0.033  .075  0.108 pu
1.0 pu
1.0

 9.26
Ztotal( pu) 0.108
KVA 3
5000
Step3 : Calculate IBase 

 693.95 A
3 x kVLL
3 x 4.16kV
Step4 : ISC actual  IBsae x ISC  9.26 x 693.95  6426 A
Step2 : Calculate ISC 
Fault Current Analysis
Per-Unit Method
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
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)
Power System Review
Symmetrical Components
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
Power System Review
Symmetrical Components..cont
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
Power System Review
Symmetrical
Fault Calculation
Power System Review
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
Symmetrical Fault Modeling
for a Bolted Fault (PU Method)
I1
Z1 eq
I0 0
+
+
VF
_
Ia
SEQ
I2 0
V1=0
_
Ic
Vf ( PU )
I 1 fault( PU ) 
Z 1eq ( PU )
() SEQ
+
Ib
+
+
Vb
Vc
_ _
Va
_
g
Z0 eq
I0=0
Z2 eq
I2=0
+
Vo=0
(0) SEQ
_
+
Phase
Ib = -Ia = Ic = ISC
_
Vbg = Vag = Vcg =0
V2=0
() SEQ
Note: VF=Pre Fault Voltage
Practice Example (PU Method):
 In the following power system Calculate(a)3ᶲ Symmetrical
fault current @ Bus3 and select an appropriate Breaker
Size @ Bus 3
500MVA
750MVA
13.8kVΔ / 115kVΥ
115kV / 13.8 kV 
XT1"
500MVA
Bus1
 0.15PU
G1
XT13  2
13.8kV
Bus 2
XT1  6
XT23  4
XT2 "
 0.18PU
750MVA
G2
13.2kV
"
X  0.15 PU
"
Sbase  750MVA
Sbase  750MVA
Kvbase  13.8kV
Kvbase  115kV
Kvbase  13.8kV
Zbase  .254
Zbase  17.63
Zbase  .254
Bus3
SBase  750 MVA
Sbase  750MVA
X  0.20 PU
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.
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:
X
ISymmetrical
 Unknown  IBreaker Interrupting Capability 
R
0.8
Practice Example (PU Method):
500MVA
750MVA
13.8kVΔ / 115kVΥ
115kV / 13.8 kV 
Bus1
XT1"  0.15PU
500MVA
G1
Bus 2
XT1  6
XT13  2 XT23  4
13.8kV
XT2 "  0.18PU
G2
13.2kV
"
X  0.15 PU
Breaker Selection :
750MVA
"
Sbase  750MVA
Sbase  750MVA
Kvbase  13.8kV
Kvbase  115kV
Kvbase  13.8kV
Zbase  .254
Zbase  17.63
Zbase  .254
Bus3
Sbase  750MVA
X  0.20 PU
SBase  750 MVA
Breaker Voltage Class :115 kV
Breaker Cycle :3 cycle
ISymmetrical  13,291.2 A
13,291.2
IBreaker Interrupting Capability 
 16,614.2 A
0.8
Symmetrical Fault Current
Analysis…MVA-Method
MVA Method
Power System Review
Fault Current Calculation-MVA Method
 This method follows a four steps process:
1. Calculate the Admittance of every component in its own
infinite bus.
Y (Admittance) 
100
Z%
2. Multiply the calculated admittances in step(1) by the
MVA rating of each component to get MVASC.
MVAsc  MVA x Y (Admittance)
3. Combine short-circuit MVAs & follow the Admittance
series & parallel rules:
b) Series M VAs :
a) Parallel M VAs :
1
1
1
1
MVAtotal  MVA1  MVA2  ........MVAn


 ........
MVAtotal MVA1 MVA2
MVAn
4. Convert MVAs to Symmetrical fault current
MVAsc (Total )
Isymmetrical 
3 x kVll
Power System Review
MVA Equivalent Network
Series M VAs :
1
1
1
1


 ........
MVAtotal MVA1 MVA2
MVAn
MVA1
MVA2
MVA3
MVATotal
Parallel M VAs :
MVAtotal  MVA1  MVA2  ........MVAn
MVA1
MVA2
1
1
1
1



MVAtotal MVA1 MVA2 MVA3
MVAtotal  MVA1  MVA2  MVA3
MVA3
MVATotal
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
MVA Method Assumptions
Two Conditions must be satisfied:
X
1.  10
R
2. Steady StateOperation
Power System Review
Symmetrical Fault Current
Analysis...MVA-Method
 Formulas:
Utility : MVAfault  MVAsc  3x kVll x Isc( KA)
kVll 2
Cable : MVAfault 
Z ()
Generator / Sycnhroonous Motor : MVAfault  MVA x
100
Transformer : MVAfault  MVA x
Zxfmr %
Note: Impedances (Z) are steady state values
Power System Review
100
Xd "Gen%
Symmetrical Fault Current
Analysis...MVA-Method
Motor :
Motor : MVAfault  MVAmotor x
100
Xd "Gen%
Ilocked rotor
Induction Motor : MVAfault  MVAmotor x
Ifull  load amp
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
 Summary:
MVA parallel  total  MVA1  MVA2      MVAn
1
MVA series  total 
[(1/ MVA1)  (1/ MVA2)      (1/ MVAn)]
MVA total
I fault ( KA) 
3 x kVLL
Power System Review
Example1:Fault Calculation(MVA method)
In the following Power System, Calculate the fault current @ Bus2 & fault current
contributions from both Gen & Motor?
Utility Source
13.8kV, 15KA fault current
Bus 1 13.8kV
Transformer
7MVA
13.8kV/4.16kV
Z=9%
Generator
1.5MVA Y
4.16kV
X”d=0.15pu
3-500McM cables, 2000 ft
Z=0.2Ω
Bus 2 4.16kV
Motor
2MVA Y
4.16kV
X”d=0.25pu
M
Step1:Network Modeling(MVA Method)
Utility Source
13.8kV, 15KA fault current
MVAsource  3 x (13.8kv) x(15kA)  358.5MVA
Bus1 13.8kV
Transformer
7MVA
13.8kV/4.16kV
Z=9%
MVAtransformer  MVA x
Generator
MVAGenerator  MVA x
100
7 x100

 77.77 MVA
Zxfmr %
9
MVALine 
2
2
86.53
10
kV
(4.16)

 86.53MVA
Zline
0.2
Bus2 4.16kV
Motor
2MVA Y
4.16kV
X”d=0.25
MVAMotor  MVA x
M
77.77
1
1
 1.5 x
 10MVA
"
Xd
0.15
3-500McM cables, 2000 ft
Z=0.2Ω
1.5MVA Y
4.16kV
X”d=0.15
358.52
1
1

2
x
 8MVA
Xd "
0.25
8
Step 2: Network Reduction(MVA Method)
358.52
77.77
86.53
10
Series M VAs :
1
1
1
1



MVAtotal 358.52 77.77 86.53
MVAtotal 
1
 36.76
1
1
1


358.52 77.77 86.53
36.76
10
Fault MVA
54.76
8
8
Parallel MVAs :
MVAtotal  MVA1  MVA2  MVA3
MVAtotal  10  36.76  8  54.76
Step 3:Fault MVA Conversion to Ifault
Bus 2 Quantities :
MVAfault  54.76
kVll  4.16kV
Bus2 Fault Current:
MVAfault(3 )
54.76
Ifault(kA) 

 7.6003
3x(4.16kVLL)
3x(4.16kVLL)
Ifault(Symmetrical )  7,600.3 Amp
Example1:Fault Analysis(PU Method)
In the following Power System, Calculate the fault current @ Bus2 & fault current
contributions from both Gen & Motor using PU Method?
Utility Source
13.8kV, 15KA fault current
Bus1 13.8kV
Transformer
7MVA
13.8kV/4.16kV
Z=9%
Generator
1.5MVA
Y
4.16kV
X”d=0.15
Zutility
Vf  1.0 pu
3-500McM cables, 2000 ft
Z=0.2Ω
Bus2 4.16kV
Motor
2MVA Y
4.16kV
M
X”d=0.25
ZXfmr
ZGen
Zmotor
ZLine
()SEQ Network for Bus 2
Example 1: Symmetrical Fault Current
Calculation Comparison between
PU & MVA Methods
MVA method calculatio n :
Ifault @ Bus2  Ifault( pu) xIbase  0.548x13,879 A  7,605.7 A
Per  Unit Method calculatio n :
Ifault @ Bus 2  7,600.3 Amp
Ex1: Motor/Gen Fault Contribution
(MVA Method)
Utility Contributi on :
Ifault 
36.76MVA 36.76

 5,102 A
3 x 4.16kV 7.205
MVA(Utility  Xfmr  Line)
MVAGen
Generator Contributi on :
10
36.76
Ifault  Gen 
8
Motor Contributi on :
MVAMotor
Ifault  motor 
Total Fault Current :
Ifault  If
 motor
10MVA
 1,387.9 A
3 x 4.16kV
 If
 utility
 If
 Gen
8MVA
 1,110.3 A
3x 4.16kV
 5,102  1,387.9  1,110.3  7,600.2 A
Ex1:Symmetrical Fault Current Analysis
PU & MVA Methods Comparison
MVA method calculatio n :
If
 motor
 1,110.3 Amp
Per  Unit Method calculatio n :
If-motor  1,110 A
Symmetrical Fault Current Calculation
MVA Method
Example2: Calculate the Symmetrical fault current @ Bus2 using the MVA Method
MVAfault  3 x 22.86 kVLLx15kA  593.903
kV 2 (22.86kV ) 2
MVAfault 
 2,903.22
Zline
0.18
MVAXfmr
20

 222.222
Z
%
0
.09




 100 
MVAXfmr 3.5
MVAfault 

 50
 Z %  0.07


 100 
Utility Source
22.86kV, 15KA fault current
Generator
3-500McM cables, 2000 ft
Z=.18 Ω
MVAfault 
MVA
5

 41.667 MVA
 Z %  0.12


 100 
MVA
2
MVAfault(G 2) 

 14.286 MVA
 Z %  0.14


 100 
Transformer
20MVA Delta-Yn
22.86/4.16kV
Z=9%
Generator
Y
5MVA
4.16kV
Z=12%
MVAfault(G1) 
MVA
2

 13.333 MVA
Z
%

 0.15


 100 
MVA
1.5
MVAfault(G 2) 

 9.375 MVA
 Z %  0.16


 100 
MVAfault( M 1) 
BUS 1
Transformer
3.5MVA Delta-Yn
4.16kV/480V
Z=7%
Y
Motor
2MVA Y
4.16kV
Z=15%
M
BUS 2
Generator
2MVA
480 V
Z=14%
M
Bolted Fault
Motor
1.5MVA Y
480V
Z=16%
Solution to Example2 (MVA method):
 22.86 kV Utility Source:
MVAfault  3 x 22.86 kVLLx15kA  593.903
 Line:
kV 2 (22.86kV ) 2
MVAfault 
 2,903.22
Zline
0.18
 Transformers:
MVAXfmr
20
MVAfault 

 222.222
 Z %  0.09


100


MVAXfmr 3.5
MVAfault 

 50
 Z %  0.07


 100 
Power System Review
Solution to Example2 (MVA method):
 Generators:
MVA
5

 41.667 MVA
 Z %  0.12


 100 
MVA
2
MVAfault(G 2) 

 14.286 MVA
 Z %  0.14


100


MVAfault(G1) 
 Motors:
MVA
2
MVAfault( M 1) 

 13.333 MVA
 Z %  0.15


 100 
MVA
1.5
MVAfault(G 2) 

 9.375 MVA
 Z %  0.16


 100 
Power System Review
Example 2:Symmetrical Fault Current
Calculation (MVA-method)
Step1: Network Modeling
593.903
MVA
2903.220
MVA
41.667
MVA
222.222
MVA
BUS 1
50 MVA
13.333
MVA
BUS 2
14.286
MVA
9.375
MVA
Power System Review
Symmetrical Fault Current
Analysis…MVA-Method
Step2 : Network MVA Reduction
 Series MVAs:
1
MVA series  total 
[(1/ MVA1)  (1/ MVA2)      (1/ MVAn)]
 Parallel MVAs:
MVA parallel  total  MVA1  MVA2      MVAn
Power System Review
Example2: Symmetrical Fault Current
Analysis…MVA-Method
Step2 : Network MVA Reduction
 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
208.846MVA
 MVA series @Bus2:
50 MVA
MVA=1/[(1/208.846)+(1/50)]
MVA=1/[(.0048)+(.0200)]=40.323
BUS 2
14.286
MVA
Power System Review
9.375
MVA
Ex2: Short Circuit MVA Calculation
@ Bus 2(MVA method)
Step3 : Fault MVA Calculatio n
MVA series  total 
153.846
MVA
1
 153.846
[(1 / 593.903)  (1 / 2,903.22)  (1 / 222.22)]
41.667
MVA
MVA parallel  153.846  41.667  13.333  208.846MVA
208.846
MVA
BUS 1
50
MVA
13.333
MVA
50
MVA
BUS 2
BUS 2
14.286
MVA
9.375
MVA
14.286
MVA
9.375
MVA
Ex2: Short Circuit MVA Calculation
@ Bus 2(MVA method)
MVA series 
1
 40.323
[(1 / 208.846)  (1 / 50)]
40.323
MVA
BUS 2
14.286
MVA
9.375
MVA
MVA @ Bus 2  40.323  14.286  9.375  63.984 MVA
MVA
fault
@ Bus 2  63.984 MVA
Example2: Symmetrical Fault
Current Analysis…MVA-Method
 Bus2 (total) = 40.323+14.286+9.375=63.984 MVA
Available Fault Current @Bus 2:
Ifault=63.984 MVA/[
3 x 0.48kV]=76,963 A
 Now, Calculate the Short Circuit MVA @Bus1?
Power System Review
Ex2:Calculate Short Circuit MVA@ Bus1
(MVA method)
MVA parallel  153.846  41.667  195.531MVA
41.667
MVA
153.864
MVA
BUS 1
50 MVA

13.333
MVA
13.333
MVA
50
MVA
BUS 2
14.286
MVA
BUS 1
195.531
MVA
9.375+14.286=23.661
MVA
9.375
MVA
208.864+16.051=224.915 MVA
208.864= 195.531+13.333
MVA

BUS 1
1/[(1/50)+(1/23.661)]=1/.0623=16.051
MVA
Power System Review
BUS 1
MVA fault @ Bus1  224.915 MVA
Ex2: Calculate Short Circuit MVA
@ Bus 1 (MVA method)
 S.C or Fault MVA @ Bus1:
 S.C or Fault MVA= 224.915
I fault @Bus1= 224.915 MVA/( 3x4.16kV)
Available Fault Current at Bus 1:
I fault @Bus1=31,216 A
Power System Review
Example 3: Symmetrical Fault Analysis
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.
1500 MVA
Fault
10 MVA
69kV Δ/Υ-n 13.8kV
X=8.5%
69 kV
X=2.8Ω
M
Source
13.82
Z Base2 
 12.7
15
S Base
IBase2 
 627.57 A
3 x kVBase1
kV 2 Base1 692


 317.4
SBase1
15
SBase  15 MVA
13.2 kV
X=0.2
VlL-Base2  13.8 kV
VlL-Base1  69 kV
Z Base1
5 MVA Υ-n
Zone 1
Zone 2 SBase  15 MVA
Power System Review
Example3: Symmetrical Fault
Analysis(MVA-method)
Source
Line
Transformer
1500
MVA
1
1
1
1



MVA 1500 1700.36 117.65
1700.36
MVA
102.52
MVA
MVA Fault= 102.52+27.32
= 129.84
117.65
MVA
27.32
MVA
Motor
27.32
MVA
Ifault= 129.84/(1.732x13.8)
= 5,432.3 Amps
5 MVA_____ =27.32
(13.2/13.8)²x0.2
Power System Review
Example 3:
Symmetrical Fault Calculation
Comparison Between PU & MVA
Methods
PU method :
I fault= 5,410.3 Amp
MVA method : I
fault =
5,432.3 Amp
Power System Review
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).
Solution
Break
Time !!!!!
 Answer: 37.5 KVA