appa-module 6-fault current analysis
TRANSCRIPT
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Module 6.0
Fault Current Calculation
By: Dr. Hamid Jaffari
Power system Review
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Fault Currents
Symmetrical Fault
Asymmetrical fault
Power System Review
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Faul t Analys is
Analysis Type
Power Flow: normal operating conditions Faults: abnormal operating conditions
Fault Types
Balanced or Symmetr ical Fault
Three Phase Short Circuit
Unbalanced or Unsymmetr ical Faul ts
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
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Various Types o f Fau lts
FaultlSymmetrica)a
FaultcalUnsymmetri)b
Faultline-to-line Faultground-to-linedouble Faultground-to-line
fault1
F
ZZ
V)(3I
l-faultSymmetrica
fault021
Ffault
3Z)3Z(ZZ
3Vground)-to-(LineI
nZ
fault21
Ffault
ZZZ
V3line)-to-(lineI
j
a
b
c
a
b
c
a
b
c
a
b
c
a
b
c
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Asymmetrical
Fault Calculation
Power System Review
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R-L Circu i t Trans ien tsR
+
-0
@
t
ClosedSW
L
)sin(2)( wtVte
0)sin(2)()(: ttVtRidt
tdiLEquation
])sin()[sin(2)()()(: Tt
etZ
VtititiSolution dcac
amptZ
V
tiac )sin(2)(
T
t
eZ
Vtidc
)sin(2)(
2222 )( lRXRZ
R
wltg
R
Xtg 11
fR
X
R
X
R
LT
2
:)(/ forcedCurrentFaultStateSteadyFaultlSymmetrica :)(transientCurrentOffsetdc
Solutionforced Solutionnatural
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Asymmetr ical faul t
])sin()[sin(2)()()( Tt
etZ
Vtititi dcac
Dc offset Magnitude depends on angle :
acIoffsetdc 20 )2(
Z
VcurrentfaultacrmsIwhere ac )(:
])2
[sin(2)()()(
)2
(:
T
t
etItititi
Set
acdcac
In order to get the largest fault current:
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Asymmetr ical faul t
Note: i(t) is not completely periodic. So, how do weget the rms value of i(t) ?
Assume :
Now calculate the RMS Asymmetrical Fault Current:
)constant(Ce Tt
AmpeIeIIIIti Tt
acT
t
dcacdcacrms
2
222221]2[][)()()(
cyclesintimeiswhere
f
t
fR
X
R
X
R
X
R
LTNote
;&
2
:
AmpeIeIeIti RXacfR
Xf
acT
t
acrms)/(
4
2
2
2
212121)(
UnitPer21factoralasymmetric)(:)()(
)/(
4
RX
acrms ekwhereIkI
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Asymmetr ical Faul t Calculat ion
Example:In the following Circuit, V=2.4kV, L=8mH,R=0.4, and =260 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 maximumdc offset.
+
-0@ tClosedSW
mHL 20
)sin(24002)( wtte
4R
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Asymmetr ical Fau lt Calcu lat ion
Solution:
4.82042.3
4.82042.3016.34.0)108)(602(4.0)() 3
ZZ
jxjLjRjXRZa
A95.788042.3
2400
volts
Z
VIac
A46.13662195.788)0()0(;0@) kIItb acrms
00.110739.6121)3(
641.1693.1121)1.0(
54.74.0
016.3)()
354.7
)3(4
54.7
)1.0(4
xecyclek
ecyclek
RatioR
Xc
A95.788)3()3(
A69.294,1641.1)1.0()1.0(
cyclekIcycleI
xcyclekIcycleI
acrms
acrms
+
- 0@
t
ClosedSW
mHL 20
)sin(400,22)( wtte
4R
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Asymmetr ical Fau lt-Unloaded
Synch ronous Mach ine
Three Stages: Subtransient, Transient, and Steady State
constanttimearmatureT
/I
/I
/I
:
offsetdcMaximum22)(
)2
sin(]1
)11
()11
[(2)(
CurrentousInstantane)()()(
A
ReactanceStateSteadyReactance/sSynchronouaxixdirect
''ReactanceTransientaxixdirect
'
ReactancentSubtransieaxixdirect"
/"/
"
''"
""
'"
dgd
dgd
dgd
TtTt
d
g
dc
d
T
t
dd
T
t
dd
gac
XEX
XEX
XEX
Where
eIeX
E
ti
tX
eXX
eXX
Eti
tititi
AA
dd
dcac
Add
ddd XXX
T,T,TConstantsTime
&ReactancesMachine
:provideeresManufactur:Note
'"
,'
,"
Stator
Uniform air-gap
Stator winding
Rotor
Rotor winding
N
S
d-axis
q-axis
axisquadratureaxisqaxisdirectaxisd
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Synch ronous Mach ine
Asymmetr ical Faul t Envelopes
Asymmetry Sources: (1) Open Phase and (2) SLG Fault
d
g
X
EI
"
"
d
g
X
EI
'
'
d
g
X
EI
CurrentfaultntSubtransie
CurrentfaultTransient
CurrentfaultS.S
)(tiac
t
envelopescurrentAC
AA T
t
T
t
d
geIe
XE
"
"MAX-dc 22(t)i
"2I
I2
'2I
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ntSubtransie
Transient
StateSteady
offsetdc
FaultalAsymmetric
GeneratornearFaultalAsymmetricofStages
d
g
X
EI
"
"
"2I
d
g
X
EI'
'
'
2I
d
g
X
EI
"2I
'2I
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Fault Current
Calculation
Power System Review
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Fault Current Analys is
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 co urse wi l l focus on PU & MVA Methods
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Fault Curren t Analys is
Power System Review
Ohmic Method
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Ohm ic 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 equivalentsKramers Rule, etc
2
1
22
2
1
N
NOR
N
N
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Infinite Bus method
Power System Review
Fault Curren t Analys is
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In f in i te Bus Calcu lat ion
SCBsae
pu
puSC
rtransformeutilityputotal
IxI
Z
ZZ
actualI:Step4kVx3
3KVAICalculate:Step3
0.1ICalculate:Step2
ZCalculate:Step1
SC
LL
Base
)(
Infinite Bus calculation is a convenient way to
estimate the maximum 3fault current flow on the secside of the transformer
The following steps are necessary to calculate the ISC
100
%&
;
KnownisCircuitShortUtilityIf:Note1)(
ZZ
MVA
MVAZ
where
ZZZ
rtransforme
SC
baseutility
rtransformeutilityputotal
0&100
%
;
UnknownisCircuitShortUtilityIf:Note2
utilityrtransforme
rtransformetotal
ZpuZ
Z
where
ZZ
totalZ
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7.5%Z
kVkV/4.1613.8
KVA5000
VS
In f in i te Bus Calcu lat ion
Unknown Uti l ity SC Data
A4.925295.693333.13actualI:Step4
A95.693
16.43
5000
kVx3
3KVAICalculate:Step3
333.13075.
0.10.1ICalculate:Step2
075.0100
5.7
100
Z%ZCalculate:Step1
SC
LL
Base
pu
xIxIkVx
Z
pu
SCBsae
pu
puSC
Example1:Calculate the maximum 3fault current on 5000 KVATransformers secondary bus.
DataSourceNo
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7.5%Z
kVkV/4.1613.8
KVA5000
VS
Inf in i te Bus Calcu lat ion
w i th Known Uti li ty SC Data
A642695.69326.9actualI:Step4
A95.69316.43
5000
kVx3
3KVAICalculate:Step3
26.9108.0
0.10.1ICalculate:Step2
108.0075.033.0ZCalculate:Step1
SC
LL
Base
)(
total
xIxI
kVx
Z
puZZ
SCBsae
putotal
puSC
rtransformeutility
Example2:Calculate the maximum 3fault current on 5000 KVA
Transformers secondary bus.
150MVASC
puZ
Z
puxS
S
kV
kVZZ
pu
rtransforme
Oldbase
Newbase
new
oldpuUtility
SC
baseutility
OldNew
075.0100
5.7
100
%
033.150
5
16.4
16.41
1
150
150
MVA
MBAZ
ZZZCalculate
22
rtransformeutilitytoal
pu108.0330.00.075Ztotal utilityZ
:StepsnCalculatio
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Fault Curren t Analys is
Power System Review
Per-Unit Method
F lt C t A l i
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Power System Review
Fault Current Analys is:
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
F lt C t A l i
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Power System Review
Fau lt Curren t Analys is:
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 variousunsymmetrical faults(LG, LL/LLG, and 3)
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Symmetr ical 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 asa point of demarcation
3. Select a common MVAbase for all zones
4. Select a kVBasefor one zone & Calculatea. kVBase for other zones
b. Zbase, and Ibasefor all zones
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Symmetr ical Components..con t
Power System Review
6. Replace each component with itsequivalent reactance in per-unit
7. Draw sequence networks(+, -, 0)
8. Use (+)SEQ network for SymmetricalFault analysis
9. Combine appropriate networks for
calculating various UnsymmetricalFault analysis
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Symmetrical
Fault Calculation
Power System Review
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3Symmetr ical Faul t Analys is(PU Method)
Symmetrical Fault refers to a balanced 3fault, in a balanced 3system operating in
steady state, which is either :
Bolted fault: LLLG fault with Zfault=0Non-Bolted fault: LLLG fault with Zfault0
Only the (+)SEQ network exists. (0)SEQ & (-)SEQ currents are equal to Zero.
Power System Review
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Symmetr ical Fau lt Model ing
for a Bo lted Fau lt (PU Method)
Z0eq
Note:VF=Pre Fault Voltage
+
_Vo=0
Z2 eq
VF
Z1eq
+
_
+
_V1=0
I0=0
I1 Ia
Ib
Ic
VcVb
Va+
+
+
_ _ _
Ib = -Ia = Ic = ISCVbg = Vag = Vcg =0
Phase
g
+
_V2=0
I2=0
)(1
)()(1PUeq
PUf
ZVI PUfault
02I
00I
SEQ
SEQ)(
SEQ)(SEQ)0(
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Prac t ice Example (PU Method):
In the following power system Calculate(a)3Symmetrical
fault current @ Bus3 and select an appropriate BreakerSize @ Bus 3
G1G2
PU.150X
kV13.8
MVA500
"
.15PU0T1
115kV/13.8kV
MVA500
"X
PU.200X
kV13.2
MVA750
"
.18PU0T2
kV8.13/115kV
MVA750
"X
6XT1
2X 13T
17.63Zbase
115kVKvbase
MVA750
Sbase
1Bus 2Bus
3Bus
4X 23T
.254Zbase
13.8kVKvbase
MVA750
Sbase
.254Zbase
13.8kVKvbase
MVA750
Sbase
MVA750SBase
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Breaker Selec t ion
Modern Circuit Breaker standards are designed based on
ISymmetrical. The following steps are required to determine anappropriate breaker size:
1. Use E/X method to calculate the minimum ISymmetrical.
2. Calculate X/R ratio:
1. If X/R 15 It means the dc offset has not decayed
to an acceptable level. Thus, calculate IAsymmetrical.
3. Calculate IAsymmetricalat calculated fault location.
4. Breaker Interrupting Capability should be 20%greater
than the calculated fault current.
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Breaker Selec t ion Cri ter ion
Generator/ Synchronous Motor/Large Induction motors
Breakers: Use subtransient Reactance Xdto calculate ISymmetrical.
Use 2 cycle Breaker
TransmissionBreakers:
Use 3 cycle Breakers if X/R>15
Use 5 cycle Breaker if X/R
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A2.614,16
8.0
13,291.2I CapabilityngInterruptiBreaker
G1 G2
PU.150X
kV13.8
MVA500
"
.15PU0T1
115kV/13.8kV
MVA500
"
X
PU.200X
kV13.2
MVA750
"
.18PU0T2
kV8.13/115kV
MVA750
"
X
6XT1
2X 13T
17.63Zbase
115kVKvbase
MVA750
Sbase
1Bus 2Bus
3Bus
4X 23T
.254Zbase
13.8kVKvbase
MVA750
Sbase
.254Zbase
13.8kVKvbase
MVA750
Sbase
MVA750SBase
kV115:ClassVoltageBreaker:SelectionBreaker
cycle3:CycleBreaker
Prac t ice Example (PU Method):
A2.291,13I lSymmetrica
S t i l F l t C t
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Symmetr ical Fault Current
AnalysisMVA-Method
MVA Method
Power System Review
F lt C t C l l t i
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Fau lt Curren t Calcu lat ion-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(
ZY
)Admittance(YxMVAMVAsc
ntotal MVAMVAMVAMVA ........
:MVAsParallela)
21 ntotal MVAMVAMVAMVA
1........111
:MVAsSeriesb)
21
llkVx
TotalMVAscalIsymmetric
3
)(
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MVA Equ ivalent Network
ntotal MVAMVAMVAMVA ........
:MVAsParallel
21
ntotal MVAMVAMVAMVA
1........
111
:MVAsSeries
21
1MVA 2MVA 3MVA
1MVA 2MVA 3MVA
321 MVAMVAMVAMVAtotal
321
1111
MVAMVAMVAMVAtotal
TotalMVA
TotalMVA
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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 Ifaultin 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
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MVA Method Assumpt ions
Power System Review
10.1 R
X
Two Conditions must be satisfied:
OperationStateSteady.2
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Symmetr ical Fau lt Current
Analysis.. .MVA-Method
Power System Review
)(3: KAIxkVxMVAscMVAUtility scllfault
)(:
2
Z
kV
MVACable
ll
fault
%
100:/
"Gend
fault
XxMVAMVAMotorusSycnhroonoGenerator
%
100:
xfmr
fault
ZxMVAMVArTransforme
Formulas:
Note:Impedances (Z) are steady state values
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Where: Xd=direct-axis Subtransient Reactance
Xd
= IFull-load amp
/ILocked Rotor amp
Power System Review
Symmetr ical Fau lt Current
Analysis.. .MVA-Method
%
100:
"Gend
motorfault
XxMVAMVAMotor
amploadfull
rotorlockedmotorfault
IIxMVAMVAMotorInduction
:
:Motor
Symmetr ical Fau l t Curren t
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Summary:
Power System Review
Symmetr ical Fau lt Curren t
Analysis.. .MVA-Method
LLkVxMVAI
totalKAfault
3)(
)]/1()/1()/1[(
1
21 nMVAMVAMVAtotalseriesMVA
nMVAMVAMVAtotalparallelMVA 21
E l 1 F lt C l l t i (MVA th d )
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Example1:Fault Calcu lat ion (MVA method )
Generator
M
Utility Source
13.8kV, 15KA fault current
Motor
2MVA Y
4.16kV
Xd=0.25pu
3-500McM cables, 2000 ft
Z=0.2
Transformer
7MVA
13.8kV/4.16kV
Z=9%
1.5MVA Y
4.16kV
Xd=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?
S
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Step1:Network Model ing(MVA Method)
Generator
M
Utility Source
13.8kV, 15KA fault current
Motor2MVA Y
4.16kV
Xd=0.25
3-500McM cables, 2000 ft
Z=0.2
Transformer
7MVA
13.8kV/4.16kV
Z=9%
1.5MVA Y
4.16kV
Xd=0.15
MVAxxMVAsource 5.358)kA15()kv8.13(3
MVAx
ZxMVAMVA
xfmr
rtransforme 77.779
1007
%
100
MVAx
X
xMVAMVAd
Generator 10
15.0
15.1
1"
MVAxXxMVAMVA dMotor 825.0
12
1"
52.358
77.77
10
8
MVAZ
kVMVA
line
Line 53.862.0
)16.4(22
53.86
Bus1 13.8kV
Bus2 4.16kV
St 2 N t k R d t i
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Step 2: Network Reduc t ion(MVA Method)
52.358
77.77
10
8
53.86 10
8
76.36
53.86
1
77.77
1
52.358
11
:MVAsSeries
totalMVA
76.36
53.86
1
77.77
1
52.358
1
1
MVAtotal
321
:
MVAMVAMVAMVA
MVAsParallel
total
76.54876.3610 totalMVA
76.54
MVAFault
St 3 F lt MVA C i t I
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Step 3:Fau l t MVA Conversion to Ifault
76.54faultMVA
6003.7)16.4(3
76.54
)16.4(3
)3()(
LLLL
faultfault
kVxkVx
MVAkAI
AmplSymmetricaIfault 3.600,7)(
kVkVll 16.4
:2QuantitiesBus
Bus2 Fault Current:
Example1:Faul t Analysis (PU Method)
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Generator
M
Utility Source
13.8kV, 15KA fault current
Motor2MVA Y
4.16kV
Xd=0.25
3-500McM cables, 2000 ft
Z=0.2
Transformer
7MVA
13.8kV/4.16kV
Z=9%
1.5MVA
Y
4.16kV
Xd=0.15
Bus1 13.8kV
Bus2 4.16kV
puVf 0.1
2)( BusforNetworkSEQ
utilityZ
XfmrZ
LineZ
GenZmotorZ
In the following Power System, Calculate the fault current @ Bus2 & fault current
contributions from both Gen & Motor using PU Method?
Example1:Faul t Analysis(PU Method)
Examp le 1: Symmetr ical Fault Current
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Examp le 1: Symmetr ical Fault Current
Calculat ion Comparison between
PU & MVA Methods
AmpI Busfault 3.600,72@
A7.605,7A879,13548.0)(2@ xxIpuII basefaultBusfault
:ncalculatiomethodMVA
:ncalculatioMethodUnitPer
E 1 M t /G F lt C t ib t i
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76.36
8
10
AkVx
MVAI
onContributiGenerator
Genfault 9.387,116.43
10
:
GenMVA
MotorMVA
)( LineXfmrUtilityMVA
AkVx
MVAI
onContributiMotor
motorfault 3.110,116.43
8
:
Ex1: Moto r/Gen Fau lt Con tr ibu t ion
(MVA Method)
AkVx
MVAI
onContributiUtility
fault 102,5205.7
76.36
16.43
76.36
:
A2.600,73.110,19.387,1102,5
:
Genfutilityfmotorffau lt IIII
CurrentFaultTotal
Ex1:Symmetr ical Fau l t Current Analys is
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Ex1:Symmetr ical Fau lt Current Analys is
PU & MVA Methods Comparison
AmpI motorf 3.110,1
A110,1f-motorI
:ncalculatiomethodMVA
:ncalculatioMethodUnitPer
Symmetr ical Fau l t Curren t Calculat ion
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Symmetr ical Fau lt Curren t Calculat ion
MVA MethodExample2: 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
22.86/4.16kVZ=9% 5MVA
4.16kV
Z=12%
Transformer
3.5MVA Delta-Yn
4.16kV/480V
Z=7%
Motor
2MVA Y
4.16kVZ=15%
Motor
1.5MVA Y
480V
Z=16%
Y
Y
3-500McM cables, 2000 ft
Z=.18
Bolted Fault
Generator
2MVA
480 V
Z=14%
BUS 1
BUS 2
903.5931586.223 kAxkVxMVA LLfault
22.903,218.0
)86.22( 22
kV
Z
kVMVA
line
fault
50
07.0
5.3
100%
222.22209.0
20
100
%
Z
MVAMVA
Z
MVAMVA
Xfmrfau lt
Xfmrfau lt
MVAZ
MVAGMVA
MVAZ
MVAGMVA
fault
fault
286.1414.0
2
100
%)2(
667.4112.0
5
100
%)1(
MVAZ
MVAGMVA
MVAZ
MVAMMVA
fau lt
fau lt
375.916.0
5.1
100
%)2(
333.1315.0
2
100
%)1(
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22.86 kV Utility Source:
Line:
Transformers:
Power System Review
903.5931586.223 kAxkVxMVA LLfault
22.903,218.0
)86.22(22
kV
Z
kVMVA
line
fault
5007.0
5.3
100
%
222.22209.0
20
100
%
Z
MVAMVA
Z
MVAMVA
Xfmrfault
Xfmrfault
Solut ion to Example2 (MVA method):
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Solut ion to Example2 (MVA method):
Generators:
Motors:
Power System Review
MVA
Z
MVAGMVA
MVAZMVAGMVA
fault
fault
286.14
14.0
2
100%
)2(
667.4112.05
100
%)1(
MVAZ
MVAGMVA
MVAZ
MVAMMVA
fault
fault
375.916.0
5.1
100
%)2(
333.1315.0
2
100
%)1(
Example 2:Symmetr ical Fau l t Current
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Example 2:Symmetr ical Fau lt Current
Calculat ion (MVA-method)
Power System Review
593.903MVA
2903.220MVA
222.222MVA
41.667
MVA
50 MVA 13.333MVA
9.375MVA
14.286MVA
BUS 1
BUS 2
ModelingNetwork:Step1
Symmetr ical Faul t Curren t
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Symmetr ical Fault Curren t
AnalysisMVA-Method
Series MVAs:
Parallel MVAs:
Power System Review
)]/1()/1()/1[(
1
21 nMVAMVAMVAtotalseriesMVA
nMVAMVAMVAtotalparallelMVA 21
ReductionMVANetwork:Step2
E l 2 S t i l F l t C t
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Example2: Symmetr ical Fau lt Current
AnalysisMVA-Method
MVAseries:
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
ReductionMVANetwork:Step2
50 MVA
9.375MVA
14.286MVA
BUS 2
208.846MVA
Ex2: Short Circu i t MVA Calcu lat ion
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Ex2: Short Circu i t MVA Calcu lat ion
@ Bus 2(MVA method)
153.846MVA
41.667MVA
50MVA
13.333MVA
9.375MVA
14.286MVA
BUS 1
BUS 2
846.153)]22.222/1()22.903,2/1()903.593/1[(
1
totalseriesMVA
50MVA
9.375MVA
14.286MVA
BUS 2
MVA846.208333.13667.41846.153 parallelMVA
208.846MVA
nCalculatioMVAFault:Step3
Ex2: Short Circu i t MVA Calcu lat ion
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40.323MVA
9.375MVA
14.286MVA
BUS 2
MVA984.63375.9286.14323.402@ BusMVA
323.40)]50/1()846.208/1[(
1
seriesMVA
MVA984.632@ BusMVA fault
Ex2: Short Circu i t MVA Calcu lat ion
@ Bus 2(MVA method)
Example2: Symmetr ical Fau l t
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Example2: Symmetr ical Fau lt
Current AnalysisMVA-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 A3
Ex2:Calcu late Short Circu it MVA@ Bus1
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Ex2:Calcu late Short Circu it MVA@ Bus1
(MVA method)
153.864MVA
41.667
MVA
50 MVA
9.375MVA
14.286MVA
13.333MVA
195.531MVA
13.333MVA
50MVA
9.375+14.286=23.661MVA
208.864+16.051=224.915MVA
208.864= 195.531+13.333MVA
1/[(1/50)+(1/23.661)]=1/.0623=16.051MVA
Power System Review
BUS 1 BUS 1
BUS 1
BUS 1
BUS 2
MVA531.195667.41846.153 parallelMVA
MVA915.2241@ BusfaultMVA
Ex2: Calcu late Sho rt Circu it MVA
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S.C or Fault MVA @ Bus1:
S.C or Fault MVA= 224.915
Ifault @Bus1= 224.915 MVA/( x4.16kV)
Power System Review
@Bus 1 (MVA method)
3
Available Fault Current at Bus 1:
I fault @Bus1=31,216 A
E l 3 S t i l F l t A l i
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Example 3: Symmetr ical Faul t Analys is
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Source M
1500 MVA
Fault69 kV
X=2.8
10 MVA
X=8.5%69kV/-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 -n
Zone 1 Zone 2
kVV 691Base-lL
AkVx
SIBase 57.6273
1Base
Base2 4.31715
69
S
2
Base
Base2
Base1
1
1
kV
Z
MVAS 15Base MVAS 15Base
7.1215
8.132
Base2Z
kVV 8.132Base-lL
Example3: Symmetr ical Faul t
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Example3: Symmetr ical Fault
Analys is(MVA-method)
Power System Review
1700.36MVA
1500MVA
117.65MVA
27.32MVA
5 MVA_____
(13.2/13.8)x0.2=27.32
102.52MVA
27.32MVA
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
11
MVA
rTransforme
Line
Source
Motor
Example 3:
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Symmetr ical Fau lt Calcu lat ion
Compar ison Between PU & MVA
Methods
I fault=5,410.3 Amp
I fault = 5,432.3 Amp
:methodPU
:methodMVA
Example 3:
Power System Review
R f
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References
1. J.D. Golver, M.S. Sarma, Power System Analysis and design,
4thed., (Thomson Crop, 2008).2. M.S. Sarma, Electric Machines, 2nded., (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).
Solut ion
Break Time !!!!!
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Solut ion
Answer: 37.5 KVA
Break Time !!!!!