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Symmetrical Faults
Chapter 7
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Faults
Shunt faults:
Three phaseabc
Line to line
Line to ground
2 Line to ground
ba
c
ab
c
ab
c
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Faults
Series faults
One open phase:a
bc
2 open phasesa
bc
Increased phase
impedance
Z a
b
c
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Why Study Faults?
Determine currents and voltages in thesystem under fault conditions
Use information to set protective devices
Determine withstand capability thatsystem equipment must have:
Insulating level
Fault current capability of circuit breakers: Maximum momentary current
Interrupting current
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Symmetrical Faults
t=0
2 V
i(t)
Fault at t = 0AC
R L
)sin(2)( tVte
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Symmetrical Faults
For a short circuit at generator terminals at t=0
and generator initially open circuited:
dt
di
LRite )(
dt
diLRitVSin )(2
by using Laplace transforms i(t) can be found
(L is considered constant)
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Symmetrical Faults
]/
)()([2
)(Tt
eSintSin
Z
Vti
2222)( XRLRZ
R
XTan
R
LTan
11
Where:
R
X
R
L
T Time Constant
]/
)()([2)(Tt
eSintSinac
Iti
Where: Iac = acRMS fault current at t=0 (Examples)
Note that for a 3-
phase system will be
different for each
phase. Therefore, DC
offset will be different
for each phase
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t = 0
acI2
iac
Idc = 0
]/
)()([2)(Tt
eSintSinac
Iti
o
90
V2 e(t)
o90
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]/
)()([2)(Tt
eSintSinac
Iti
0
o90
V2 e(t)
t = 0
iac02 acI
02 acIidc
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iac
02 acI
02 acI id
c
0
22ac
I
t
0
o90
]/
)()([2)(Tt
eSintSinac
Iti
)(ti
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Symmetrical Faults
Iac and Idc are independent after t = 0
22
dcI
acI
RMSI
Tt
eacoIdcI
2
Substituting:
Tte
acI
Tt
eac
Iac
IRMSI
221)222((max)
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Asymmetry Factor
IRMS(max) = K() Iac
Asymmetry Factor = K()
rx
eK
4
21)(
Where:
= number of cycles
(Example 7.1)
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Example 7.1
Fault at a time to produce maximum DC offset
Circuit Breaker opens 3 cycles after fault inception
IFault at t = 0AC
R = 0.8 XL = 8
V = 20 kVLN-
+
CB
Find:
1. Iac at t = 0
2. IRMS Momentary at = 0.5 cycles
3. IRMS Interrupting Current
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Example 7.1
a. RMSACkAI 488.2
88.0
20)0(
22
b.
438.121)5.0(
)10
5.(4
eKKAImomentary 577.3)488.2)(438.1(
c.023.121)3(
)10
3(4 eKKAI ngInterrupti 545.2)488.2)(023.1(
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AC DecrementIn the previous analysis we treated the
generator as a constant voltage behind aconstant impedance for each phase. The
constant inductance is valid for steady state
conditions but for transient conditions, thegenerator inductance is not constant.
Recall that for steady state conditions, the
equivalent machine reactance is made upof 2 parts: a) Armature leakage reactance
b) Armature reaction
(See Phasor Diagram)
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AC Decrement
Steadystate model of generatorXL is leakage reactance
XAR is a fictitious reactance and XAR>> XL
XARis due to flux linkages of armature current with the fieldcircuit. Flux linkages can not change instantaneously.Therefore, if the generator is initially unloaded when afaultoccurs the effective reactance is XL which is referred to asSubtransient Reactance, x.
AC
EI
R XL XAR
Load
I I
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jILXL
jILXAR(t)
EIField
Flux
Armature Reaction
Resultant
Field
ET
XL XAR
-
+
EI
I=IL
Load
Loaded Generator
I 0
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EField
Flux
Armature Reaction = 0
Resultant
Field
ET0
t = 0 -
XL XAR=0ET0
-
+
E = E = E = ET0
I=0
Unloaded Generator
I 0
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XL XAR
-
+
E = E = E = ET0
I=0
t=0
EField
Flux
Armature Reaction = 0
Resultant
Field
ET0 = 0
Faulted Generator
I I
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XL XAR=0
-
+
E = E = E = ET0
I = I
E = jIXL
t=0+
Field
Flux
Resultant
Field
ET = 0
I
Armature Reaction = 0
I I
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XL XAR
-
+
E = E = E = ET0
I = I
E = jI(XL + XAR)
t 3Cyc.
Field
Flux
Resultant
Field
ET = 0
I
Armature Reaction = 0
I I
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XL XAR
-
+
E = E = E = ET0
I = I
E = jI(XL + XAR)
t =
Field
Flux
Resultant
Field
ET = 0
I
Armature Reaction = 0
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AC Decrement
As fault current begins to flow, armature reaction willincrease with time thereby increasing the apparent
reactance. Therefore, the ac component of the fault
current will decrease with time to a steady state
condition as shown in the figure below.
"2I '2II2
"2I
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AC Decrement
For a round rotor machine we only need toconsider the direct axis reactance.
dX
EI
"
"2"2 Subtransient
dX
EI
'
'2'2
dX
EI 22
Transient
Synchronous(steadystate)
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AC Decrement
Can write the ac decrement equation )])'()'"(2)( '" tSinIeIIeIItaci dT
tdTt
For an unloaded generator
(special case):TEEEE '"
Td: Subtransient time constant
(function of amortisseur winding X/R)Td: Transient time constant
(function of field winding X/R)
Look at equation for t=0 and t=infinity
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AC Decrement
For t = 0
)])'()'"(2)( '"
tSinIeIIeIItaci dT
tdTt
For t =
IIiac 2]00[2(max)
"2])'()'"[(2(max) IIIIIIiac
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ac and dc Decrement
Transform ac decrement equation to phasor form
/')'(")'"(
_
IdT
t
eIIdT
t
eIIacI
dc decrement equation:
A
T
t
eSinIdc
I
)("2
Where TA = Armature circuit time constant
(Example 7.2)
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Network Equivalent circuit
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Network Equivalent circuit
Lecture 9 Short circuit calculation
Electrical netw ork can be represented by a Thvenin equivalent circuit, voltage source and
series connected inductive reactance and a small resistance. The resistance is neglected
most of the time.
The voltage is the rated netw ork line to neutral voltage. The reactance is calculated f rom the
short circuit current. The pow er company calculates the short circuit current at every
substation.
Example: A 500 kV netw ork short circuit current is 30kA. Calculate the Thvenin equivalent
circuit and the short circuit current time function.
Netw ork Thvenin Equivalent circuit is:
AC VlnZnet
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Network Equivalent circuit
Vnet 500kV Ishort 30kA 2 60 Hz
The network line to line voltage and the equivalent netw ork reactance and estimated small
resistance
Vln
Vnet
3288.675kV X
net
Vln
Ishort
9.623
Rnet
Xnet
20 Estimated value Lnet
Xnet
25.524mH
The supply voltage is:
Vsup t ( ) 2 Vln sin t ( )
Laplace transformation of the supply voltage is:
2 Vln cos ( ) s sin ( )( )
2
s2
2 Vln sin t ( ) has Laplace transform
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Network Equivalent circuitLaplace transformation of the supply voltage is:
2 Vln cos ( ) s sin ( )( )
2
s2
2 Vln sin t ( ) has Laplace transform
The system equation w hen the load current is neglected
2 Vln cos ( ) s sin ( )( )
2
s2
s Lnet Rnet inet
Short c ircuit current in s domain is: inet s ( )2 Vln cos ( ) s sin ( )( )
2
s2
s Lnet Rnet
Inverse Laplace transform is:
2 Vln cos ( )( )
2
s2
s Lnet Rnet has inverse Laplace transform
2 Vln cos ( ) Rnet sin t 2
Lnet e
Rnet t
Lnet
2
Lnet cos t 2
2
2 Lnet2 2 Rnet2
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Network Equivalent circuit
Ish1 t ( )2 Vln cos ( ) Rnet sin t
2 Lnet e
R
net
t
Lnet
2
Lnet cos t 2
2
2
Lnet2
2
Rnet2
2 Vln s sin ( )( )
2 s2 s Lnet Rnet has inverse Laplace transform
2 Vln sin ( ) Rnet cos t 2
Rnet e
Rnet t
Lnet
Lnet sin t 2
2
Lnet
2
2 R
net
2
Ish2 t ( )2 Vln sin ( ) Rnet cos t
2 Rnet e
Rnet t
Lnet
Lnet sin t 2
2
Lnet2
2
Rnet2
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Network Equivalent circuit
The short circuit current time function
IShort t ( ) Ish1 t ( ) Ish2 t ( )
DC component of the current is
IDC t ( )
2 Vln cos ( ) Lnet e
Rnet t
Lnet
Lnet2
2
Rnet2
2 Vln sin ( ) Rnet e
Rnet t
Lnet
Lnet2
2
Rnet2
IDC t ( )2 Vln cos ( ) Lnet e
Rnet t
Lnet
Lnet
2
2 Rnet
2
2 Vln sin ( ) Rnet e
Rnet t
Lnet
Lnet
2
2 Rnet
2
IDC 0s 0deg ( ) 42.321kA IDC 0s 90deg ( ) 2.116 kA
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Network Equivalent circuit
t 0s 0.01ms 150ms Ishort_max 78.67kA IDC_max 42.321kA
0 50 100 15050
0
50
100
IShort t 0.0deg ( )
kA
IDC t 0deg( )
kA
t
ms
Current and DC component maximum at 0deg
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Network Equivalent circuit
t 0s 0.01ms 150ms IShort_max 42.215kA
0 50 100 150
60
40
20
020
40
60
IShort t 90.0deg ( )
kA
t
ms
No DC component at 90deg