Mo d u le 6. 6.0 0
Fault Current
Calculation
By: Dr. Hamid Jaffari
Power system Review
Fau l t C u r r e n t s
Symmetrical Fault
Asymmetrical fault fault
Power System Review
F a u l t A n a ly ly s i s
Analysis Type Power Flow: normal operating conditions Faults: abnormal operating conditions
Fault Types a l Fa Fa u l t Balanced or S y m m e t r i c al
t r i c al a l F au au l t s Unbalanced or U n s y m m e tr
Three Phase Short Circuit 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
V ar i o u s T y p es o f Fau l t s a)Symmetrical Fault a
a
b
b
c
c
ISymmetrical-fault (3 )
VF Z1 Zfault
b)Unsymmetrical Fault
line - to - line Fault
double line - to - ground Fault
line - to - ground Fault
a
a
a
b
b
b
c
c
c
j 3VF Ifault(line - to - line) Z1 Z2 Zfault
Ifault(Line - to - ground)
3VF Z1 Z2 ( Z0 3 Z n) 3Zfault
Asymmetrical Fault Calculation
Power System Review
R -L C ir c u i t T r an s i en t s R
e(t )
2 V sin( wt )
+ -
L
SW Closed @ t 0
di(t )
Ri(t ) 2V sin( t ) t 0 dt t V T Solution : i (t ) iac(t ) idc(t ) 2 [sin( t ) sin( )e ] Z forced Solution natural Solution
Equation : L
Symmetrical Fault / Steady State Fault Current ( forced ) :
iac (t )
2
V sin( t ) amp Z
dc Offset Current (transient ) : t V idc (t ) 2 sin( )e T Z
Z R 2 X 2 R 2 ( l ) 2
X wl tg 1 R R
tg 1
T
L X X R R 2 fR
A s y m m et r i c al f au l t
i(t ) iac(t ) idc(t ) 2
V
t
[sin( t ) sin( )e T ]
Z •Dc offset Magnitude depends on angle α: (
2
)
0 dc offset
•In order to get the largest fault current:
2 I ac
where : I ac(rms ac fault current )
Set : (
2
V Z
)
i (t ) iac (t ) idc (t ) 2 I ac[sin( t
2
)e
t T
]
A s y m m et r ic al f au l t Note: i(t) is not completely periodic. So, how do we get the rms value of i(t) ?
Assume :
irms(t )
e
t T
C (constant)
Now calculate the RMS Asymmetrical Fault Current:
( I ac) 2 ( I dc) 2 [ I ac]2 [ 2 I dce
t
] I ac 1 2e
T 2
2 t T
Amp
X Note : T
L X X R R 2 fR R
&
t
f
; where is time in cycles
irms (t ) I ac 1 2e I rms( ) k ( ) I ac
2 t T
I ac 1 2e
2
X fR 2
f
I ac 1 2e
where : k ( ) asymmetrical factor 1 2e
4 ( X / R )
Amp
4 ( X / R )
Per Unit
A s y m m et r ic a l Fau l t C al c u l at i o n
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
A s y m m et r i c a l Fau l t C al c u l at i o n Solution:
a ) Z R jX R j ( L) 0.4 j ( 2 60)(8 x10 3 ) 0.4 j 3.016 3.04282.4 I ac
Z Z 3.04282.4
V Z
2400volts 3.042
788.95A
b) @ t 0; I rms (0) I ack (0) 788.95 1 2 1366.46 A
c)
X R
( Ratio )
3.016 0. 4
e(t ) 2 2,400sin(wt )
7.54
R 4
+ -
L 20mH
SW Closed @ t 0
4 ( 0.1)
k ( 0.1cycle ) 1 2e
7.54
4 ( 3)
k ( 3cycle ) 1 2e
7.54
1 1.693 1.641
1 6.739 x10 3 1.00
I rms ( 0.1cycle ) I ac k (0.1cycle ) x1.641 1,294.69 A I rms ( 3cycle ) I ac k (3cycle ) 788.95 A
A s y m m et r i c a l Fau l t -U n l o ad e d Sy n c h r o n o u s M ac h i n e
Three Stages: Subtransient, Transient, and Steady State
i (t ) iac (t ) idc(t ) iac(t ) idc(t )
Instantaneous Current
t
t
1 ' 2 E g [( " ' )e T d ( ' )e T d ] sin( t ) X d X d X d X d X d 2 d-axis E g 2 " e t / T A 2 I "e t / T A Maximum dc offset N X d 1
1
"
1
1
Where :
Stator winding Uniform air-gap Stator
"
"
"
X d direct axix Subtransient Reactance
I
X ' d direct axix Transient Reactance
I ' E g / X ' d
X
I E g / X d
d direct axix Synchronous Reactance/SteadyStateReactance
E g / X d
q-axis
Rotor
S d axis direct axis q axis quadrature axis
TA armature time constant Note : Manufactur eres provide : Machine Reactances Time Constants
"
'
X d , X d , X T"d , T ' d , T
Rotor winding
d A
&
Sy n c h r o n o u s M ac h i n e A s y m m et r ic a l Fau l t En v el o p e s
Asymmetry Sources: (1) Open Phase and (2) SLG Fault iac(t ) Subtransient fault Current
I "
E g X "d
Transient fault Current I '
"
2 I
2 I '
AC current envelopes idc - MAX(t)
2 E g X " d
e
t T A
2 I "e
t T A
E g '
X d
I
S.S fault Current E g X
d
2 I t
I "
I '
I
E g
2 I "
X "d
E g
Stages of Asymmetrical Fault near Generator Subtransient
2 I '
'
X d
Transient
E g X d
"
Steady State
dc offset
2 I
2 I '
Asymmetrical Fault
Fault Current Calculation
Power System Review
Fau l t C u r r e n t A n al y s i s 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) N o t e: Th i s c o u r s e w i l l f o c u s o n P U & M V A M et h o d s
Power System Review
Fau l t C u r r en t A n al y s i s
Ohmic Method
Power System Review
Oh m ic M et h o d This Method Requires: Transferring all impedances to high/low
voltage side of transformer using square N N of XFMR turn ratio N OR N 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
Fau l t C u r r en t A n al y s i s
Infinite Bus method
Power System Review
In f i n i t e B u s C al c u l at i o n •Infinite Bus calculation is a convenient way to estimate the maximum 3 ᶲ fault current flow on the sec side of the transformer •The following steps are necessary to calculate the ISC Step1: Calculate Ztotal ( pu ) Z utility Z transformer Step2 : Calculate ISC
1.0 pu
Step3 : Calculate IBase
3 x kVLL
Step4 : ISC actual I Bsae x I SC
Z total ( pu ) Z utility Z transformer where;
Z pu KVA 3
Note1: If Utility Short Circuit is Known
Z utility
Ztotal
MVAbase MVASC
& Z transformer
Z % 100
Note2 : If Utility Short Circuit is Unknown Z total Z transformer where; Z transformer
Z % 100
pu & Z utility 0
In f i n i t e B u s C al c u l at i o n U n k n o w n U t i l it y S C Dat a Example1: Calculate the maximum 3 ᶲ fault current on 5000 KVA Transformer’s secondary bus.
Step1: Calculate Z pu Step2 : Calculate ISC
Z%
7 .5
VS
0.075 pu
100 100 1.0 pu 1.0
Step3 : Calculate IBase
Z pu .075 KVA 3 3 x kVLL
No Source Data
5000KVA 13.8kV/4.16kV Z 7.5%
13.333
5000 3 x 4.16kV
693.95 A
Step4 : ISC actual I Bsae x I SC 13.333 x 693.95 9252.4 A
In f i n i t e B u s C al c u l at i o n w i th K n o w n U ti li t y SC Dat a Example2: Calculate the maximum 3 ᶲ fault current on 5000 KVA Transformer’s secondary bus. VS Calculate
Ztoal Zutility Ztransformer Zutility
Zutility
M BAbase M VASC
Z Utility New Z puOld Z transformer
SC 150MVA
Z % 100
7.5 100
13.8kV/4.16kV
150 150
kV old kV new
5000KVA
1 pu
2
S base New 4.16 1 x 4.16 S baseOld
Z 7.5% 2
Ztotal 0.075 0.033 0.108 pu
5 .033 pu 150
0.075 pu
Calculatio n Steps : Step1: Calculate Ztotal Z utility Z transformer 0.033 .075 0.108 pu Step2 : Calculate ISC
1.0 pu Z total ( pu )
Step3 : Calculate IBase
1.0
0.108
KVA 3 3 x kVLL
9.26 5000 3 x 4.16kV
693.95 A
Step4 : ISC actual I Bsae x I SC 9.26 x 693.95 6426 A
Fau l t C u r r en t A n al y s i s
Per-Unit Method
Power System Review
Fau l t C u r r e n t A n al y s i s : Per-Unit Meth o d 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
Fau l t Cu r r en t A n al y s i s : Per-Unit Meth o d
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 3 ᶲ) Power System Review
Sy m m et r ic al Co m p o n e n t s
Steps involved: 1. 2. 3. 4.
Draw a single-line diagram of the desired power system(equivalent per phase) Define zones using transformation point as a point of demarcation Select a common MVAbase for all zones Select a kVBase for one zone & Calculate a. kVBase for other zones b. Zbase, and Ibase for all zones Power System Review
Sy m m et r i c al C o m p o n e n t s ..c o n t
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
3 Φ Sy m m et r i c al Fau l t A n a ly s i s (PU Meth o d )
Symmetrical Fault refers to a balanced 3 Φ fault, in a balanced 3 Φ 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
S y m m et r i c a l Fau l t M o d el in g f o r a B o l t ed Fau l t (PU Meth o d ) I1
Z1 eq
I 0 0
+
+ VF _
Ia
SEQ
I 2 0
V1=0 _
Ic
I 1 fault ( PU )
() SEQ
+
Ib
V f ( PU )
+
+ Vb Vc _ _
Z 1eq( PU )
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
Prac tice Exam p le (PU Metho d ):
In the following power system Calculate(a)3ᶲ Symmetrical fault current @ Bus3 and select an appropriate Breaker Size @ Bus 3 500MVA
7 50MVA
13.8kVΔ / 115kVΥ
115kV / 13.8 kV
XT1"
50 0MVA
Bus1
0.15PU
G1
XT13 2
13.8kV X
"
0.15 PU
Bus 2
XT1 6
XT23 4
XT2 "
0.18PU 75 0MVA G2
13.2kV
Sbase 75 0MVA
Sbase 75 0MVA
Kvbase 13.8kV
Kvbase 115kV
Kvbase 13.8kV
Zbase .254
Zbase 17.63
Zbase .254
Bus 3
SBase 750 MVA
Sbase 75 0MVA
X
"
0.20 PU
B reaker Selec tion
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:
X/R <15 →Use ISymmetrical →It means the dc offset has not decayed 2. If X/R>15 to an acceptable level. Thus, calculate I Asymmetrical. 1. If
3. Calculate I Asymmetrical at calculated fault location. 4. Breaker Interrupting Capability should be 20% greater
than the calculated fault current.
B reaker Selec tion Criterio n
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 R
Unknown IBreaker Interrupting Capability
ISymmetrical 0.8
Prac tice Exam p le (PU Metho d ): 50 0MVA
75 0MVA
13.8kVΔ / 115kVΥ
115kV / 13.8 kV
Bus1
"
XT1
0.15PU
500MVA
G1
XT13 2
13.8kV
Bus 2
XT1 6
XT2
"
0.18PU
750MVA
G2
XT23 4
13.2 kV
"
X
"
0.15 PU
Breaker Selection :
Sbase
750MVA
Sbase
750MVA
Bus3
Sbase
750MVA
Kvbase 13.8kV
Kvbase 115kV
Kvbase 13.8kV
Zbase .254
Zbase 17.63
Zbase .254
X
0.20 PU
SBase 750 MVA
Breaker Voltage Class :115 kV Breaker Cycle :3 cycle ISymmetrical 13,291.2 A
IBreaker Interrupti ng Capability
13,291.2 08
16,614.2 A
S y m m et r i c a l Fau l t C u r r e n t Analysis…MVA-Method
MVA Method
Power System Review
M VA M et h o d - Fau lt Curren t Calcu lation
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. MVA sc MVA x Y (Admittance) 3. Combine short-circuit MVAs & follow the Admittance
series & parallel rules: a) Parallel MVAs : MVAtotal MVA 1 MVA 2 ........MVAn
b) Series MVAs : 1
MVAtotal
1
MVA1
1
MVA2
4. Convert MVAs to Symmetrical fault current MVAsc (Total ) Isymmetrical 3 x kV ll Power System Review
........
1
MVAn
M VA Eq u i v al en t N et w o r k Series MVAs : 1
MVAtotal
1
MVA1
MVA1
1
MVA2
MVA2
........
1
MVAn
MVA3
1
1
MVAtotal MVA1 MVA2 MVA3
MVAtotal MVA1 MVA2 MVA3
MVAtotal MVA 1 MVA2 ........MVAn
MVA2
1
MVATotal
Parallel MVAs :
MVA1
1
MVA3
MVATotal
W h y U s e t h e MVA M et h o d ?
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
M VA M et h o d A s s u m p t io n s
Two Conditions must be satisfied:
1.
X R
10
2. Steady StateOperation
Power System Review
S y m m et r i c a l Fau l t Cu r r e n t Analysis... M V A - M e t h o d
Formulas:
Utility : MVA fau lt MVAsc 3 x kV ll x I sc( KA) kV ll 2 Cable : MVA fa u lt Z ( ) Generator / Sycnhroonous Motor : MVA fault MVA x Transformer : MVA fault MVA x
100 Z xfmr %
Note: Impedances (Z) are steady state values
Power System Review
100 X d "Gen%
S y m m et r i c a l Fau l t Cu r r e n t Analysis... M V A - M e t h o d Motor :
Motor : MVA fault MVAmotor x
100 X d "Gen%
Induction Motor : MVA fau lt MVAmotor x
I locked rotor I full load amp
Where: X”d=direct-axis Subtransient Reactance
X”d= I Full-load amp/I Locked Rotor amp
Power System Review
S y m m et r i c a l Fau l t Cu r r en t Analysis...MVA-Method
Summary:
MVA parallel total MVA1 MVA2 MVAn MVA series total
1 [(1/ MVA1) (1/ MVA2) (1/ MVAn)]
I fault ( KA)
MVA total 3 x kV LL
Power System Review
Example1:Fault C al cu l at io n (M VA m et h o d ) 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
Motor 2MVA Y 4.16kV X”d=0.25pu
M
4.16kV
( M VA M et ho d ) S tep 1: N et w o r k M o d el i n g Utility Source 13.8kV, 15KA fault current
Bus1 13.8kV
MVA sou rce 3 x (13.8kv) x(15kA) 358.5 MVA
Transformer 7MVA 13.8kV/4.16kV Z=9%
MVAtransformer MVA x
Generator
MVAGenerator MVA x
100 Z xfmr % 1 "
X d
7 x100
1.5 x
3-500McM cables, 2000 ft Z=0.2Ω
1.5MVA Y 4.16kV X”d=0.15
2
MVA Line
kV
Z line
(4.16) 0.2
9 1 0.15
358.52
77.77 MVA
10 MVA
10
2
86.53
86.53 MVA
Bus2 4.16kV Motor 2MVA Y 4.16kV X”d=0.25
MVA Motor MVA x M
1 X d "
2 x
1 0.25
77.77
8 MVA
8
( M VA M eth o d ) S tep 2 : Net w o r k R ed u c t i o n Series MVAs :
358.52
1 MVAtotal
77.77
MVAtotal
1 358.52
10
86.53
10
1 358.52
1 1 77.77
36.76
1 77.77
1
1 86.53
36.76
86.53
Fault MVA
54.76 8
8 Parallel MVAs : MVAtotal MVA1 MVA 2 MVA3
MVAtotal 10 36.76 8 54.76
f ault S tep 3 : Fau l t MV A C o n v er s i o n t o I
Bus 2 Quantities :
MVA fault 54.76
kV ll 4.16kV Bus2 Fault Current:
I fault (kA)
MVA fault (3 ) 3 x(4.16kV LL )
54.76 3 x(4.16kV LL)
7.6003
I fault (Symmetrical ) 7,600.3 Amp
( PU Metho d ) E x am p l e1: Fau l t A n a l y s i s 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
Z utility V f 1.0 pu
3-500McM cables, 2000 ft Z=0.2Ω
Z Xfmr
Z Gen
Z motor
Z Line
Bus2 4.16kV
()SEQ Network for Bus 2 Motor 2MVA Y 4.16kV M X”d=0.25
E xa m p l e 1 : S ym m et ri ca l Fa ul t Cu r re nt C al c u l a t io n C o m p a r i s o n b et w een PU & M VA M et h o d s MVA method calculatio n : fault ( pu) xI base 0.548x13,879 A 7,605.7 A I fault @ Bus 2 I
Per Unit Method calculatio n :
I fault @ Bus2 7,600.3 Amp
E x 1: M o t o r /G en F au l t Co n t r i b u t i o n (M VA M eth o d ) Utility Contributi on : I fau lt
36.76 MVA
3 x 4.16kV
36.76 7.205
5,102 A
MVAGen
MVA(Utility Xfmr Line)
Generator Contributi on :
10
36.76
I fau lt Gen
8
I fau lt motor
Total Fault Current : motor
3 x 4.16kV
1,387.9 A
Motor Contributi on :
MVA Moto r
I fau lt I f
10 MVA
I f
utility
I f
Gen
8 MVA 3 x 4.16kV
1,110.3 A
5 102 1 387 9 1 110 3 7 600 2 A
E x 1: S y m m et r i c a l Fau l t Cu r r e n t A n al y s i s PU & M VA M et h o d s C o m p a ri s o n MVA method calculatio n :
I f motor 1,110.3 Amp Per Unit Method calculatio n :
I f-motor 1,110 A
S y m m e t r i c a l F a u l t Cu C u r r en e n t C al cu la t io n M VA V A M et et h o d Example2: Calculate the Symmetrical fault current @ Bus2 using the MVA MVA Method MVA MVA fault 3 x 22.86 kV LL x15kA 593 593.903 903 MVA MVA fault
MVA fau lt
kV 2 (22.86kV ) 2 Z line
0.18
2,903.22
MVA Xfmr 20 222.222 Z % 0.09 100
MVA Xfmr 3.5 50 MVA fau lt Z % 0.07 100 MVA fault (G1)
Utility Source 22.86kV, 22.86kV, 15KA fault current
Generator
3-500McM cables, 2000 ft Z=.18 Ω Transformer 20MVA Delta-Yn 22.86/4.16kV Z=9%
Generator
Y
5MVA 4.16kV Z=12%
MVA 5 41.667 MVA Z % 0.12 100
MVA 2 14.286 MVA MVA fault (G 2) Z % 0.14 100 MVA fau lt ( M 1)
MVA fau lt (G 2)
MVA 2 13.333 MVA Z % 0.15 100 MVA
Z % 100
1 .5 0.16
BUS 1 Transformer 3.5MVA Delta-Yn 4.16kV/480V Z=7%
Y
Motor 2MVA Y 4.16kV Z=15%
M
9.375 MVA
BUS 2 Generator 2MVA 480 V Z=14%
M
Bolted Fault
Motor 1.5MVA Y 480V
): S o l u t i o n t o E x a m p l e2 e 2 ( M V A m e t h o d ):
22.86 kV Utility Source: MV MVA fault 3 x 22.86 kV LL x15kA 59 593 3.90 903 3
Line:
2
MV MVA fau lt
kV (22.86kV ) Z line
0.18
2
2,90 903 3.22
Transformers:
MVA Xfmr 20 fa u lt 222.222 MVA fau Z % 0.09 100 MVA Xfmr 3.5 fa u lt 50 MVA fau Z % 0.07 100 Power System Review
): S o l u t i o n t o E x a m p l e2 e 2 ( M V A m e t h o d ):
Generators: MVA fau fa u lt (G1)
MVA 5 41.667 MVA Z % 0.12 100
MVA 2 14.286 MVA MVA fau fa u lt (G 2) Z % 0.14 100
Motors:
MVA 2 13.333 MVA MVA faul fa ult t ( M 1) Z % 0.15 100 MVA faul fa ult t (G 2)
MVA 1.5 9.375 MVA Z % 0.16 100 Power System Review
E x am p l e 2: S y m m et r i c a l Fau l t C u r r e n t C a l c u l a t i o n ( MV A- m et ho d ) 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
S y m m et r i c a l Fau l t C u r r en t
Analysis…M VA -M e th od Step2 : Network MVA Reduction
Series MVAs:
MVA series total
1 [(1/ MVA1) (1/ MVA2) (1/ MVAn)]
Parallel MVAs:
MVA parallel total MVA1 MVA2 MVAn
Power System Review
E x am p l e2: S y m m et r i c a l Fau l t C u r r e n t 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)]
BUS 2
MVA=1/[(.0048)+(.0200)]=40.323 14.286 MVA
Power System Review
9.375 MVA
E x 2: S h o r t C i r c u i t M VA C al c u l at i o n
@ B u s 2 ( M V A m et h o d ) Step3 : Fault MVA Calculatio n MVA series total
153.846 MVA
1 [(1 / 593.903) (1 / 2,903.22) (1 / 222.22)] 41.667 MVA
153.846 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
E x 2: S h o r t C i r c u i t M VA C al c u l at i o n
@ B u s 2 ( M V A m et h o d ) MVA series
1 [(1 / 208.846) (1 / 50)]
40.323
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
E x am p l e2: S y m m et r i c a l Fau l t 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
E x2 :Cal cu l at e Sh o r t C ir c u i t M VA @ B u s 1 (M V A m et h o d ) MVA parallel 153.846 41.667 195.531MVA 41.667 MVA
153.864 MVA
195.531
BUS 1
MVA
BUS 1
50 MVA
13.333 MVA
BUS 2
14.286 MVA
13.333 MVA
50 MVA
9.375+14.286=23.661 MVA
9.375 MVA 208.864 =
208.864+16.051=224.915 MVA
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: Calcu late Sho rt Circu it MVA @ B u s 1 (M V A m et h o d )
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
Exam p le 3: Sy m m et r i c al Fau l t A n a ly s i s 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
5 MVA Υ-n
M
Source
V lL-Base2 13.8 kV
V lL-Base1 69 kV Z Base1
2
kV
Base1
SBase1
13.2 kV X=0.2
2
69
15
S Base 15 MVA
Z Base2
317.4
Zone 1
I Base 2
2
13.8
12.7 15 S Base 3 x kV Base1
Zone 2 S Base 15 MVA
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627.57 A
E x am p l e 3: S y m m et r i c a l Fau l t Analysis ( MV A- m et ho d ) Source
Line
Transformer
1
1500 MVA
MVA
1700.36 MVA
1
1
1
1500 1700.36 117.65
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: S y m m et r i c al Fau l t Cal c u l at i o n C o m p a r is o n B et w een P U & M VA 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, 2 nd 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, 4 th ed. (New York: McGraw-Hill, 1982 ).
