Transcript
Ontoseno Penangsang1)
1) Electrical Department, Sepuluh Nopember of Institute Technology
Surabaya, 60111, Indonesia
Short Circuit Objectives
Transient Phenomenon During Short Circuit
Thevenin Equivalent
Short Cicuit Capacity
Bus Admittance Matrix (Ybus)
Bus Impedance Matrix (Zbus)
Conclusions
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Calculate system fault current duties and compare them with :
- the first cycle momentary or close-
and-latch ratings
- the interrupting ratings
of circuit- interrupting devices, such as circuit breakers and fuses
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Calculate system fault current duties to compare with :
- short-time ratings
- withstand ratings
of system components, such as busway,cables, transformers, disconnect switches,etc
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Selection ratings or settings of short-circuit protective devices, such as molded case breakers, solid-state trip units, fuses, relays.
Evaluate short-circuit current flow and voltage levels in the overall system for short-circuits in specific areas
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Sources of Short Circuit Currents
Short Circuit Current from Induction Motor is
usually neglected
Synchronous generators
Synchronous motors and
condensers
Induction machines
Electric utility system
(Note : Power capacitors can also produce extremely high
transient fault or switching currents, but usually of short duration
and of natural frequency much higher than power frequency)
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Assumptions in Short Circuit Current
Calculation
• Normal Loads, Line Charging Cpacitancs, Shunt Elements connected to ground are heglected.
• All System Internal Voltages have the same Magnitude and Phase Angle (1.0 0o)
• Series Resistance of Transmission Line and Transformer Impedance is neglected.
• All Transformers have Nominal Tap Position.
• Generators, Motors are represented by a constant Voltage Source connected in series with :
– Sub-Transient Reactance (The system is in a sub-transient condition)
– OR, Transient Reactance (The system is in a transient condition)
– OR, Synchronous Reactance (The system in a steady-state condition)
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Basic Assumptions
To simplify the short-circuit calculations, a number of assumptions
are required.
The short-circuit current, during a three phase short-circuit, is assumed
to occur simultaneously on all three phases;
During the short-circuit, the number of phases involved does not
change, i.e. a three-phase fault remains three-phase and a phase-
to-earth fault remains phase-to-earth
For the entire duration of the shortcircuit, the voltages responsible
for the flow of the current and the short-circuit impedance do not
change significantly;
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Transformer regulators or tap-changers are assumed to be set to a
medium position (if the short-circuit occurs away from the generator,
the actual position of the transformer regulator or tap-changers does
not need to be taken into account;
Arc resistances are not taken into account;
Basic Assumptions
All line capacitances are neglected;
Load currents are neglected;
All zero-sequence impedances are taken into account.
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Transient Phenomenon During Short Circuit
Waktu
arus sinusoidal
arus
Short circuit current wave in a steady state condition is the
same as the current wave before the short circuit occurs
(sinusoidal), only the magnitude is different.
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dt
diLRitEm )sin(
)sin()( tEte m
Electrical Power System Network
Ideal Voltage Source
tL
R
m etZ
Eti )sin()sin()(
21
222 LRZ
R
L 1tan
tL
R
m
Z
E
)sin(
Differential Equations :
Short Circuit Current :
DC Component :
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Waktu
Arus asimetris total
komponen DC
arus AC simetrisa
rus
The magnitude of a short circuit current is
maximum at the time when the short circuit occurs,
and decreasing exponentially to a steady state
value.
Total Asymmetrical Current
DC ComponentSymmetrical AC Current
Cu
rre
nt
Time
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Short Circuit Current
from a Generator
Short Circuit Current
without DC Component
Reactance for Short Circuit
Current Calculation
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Short Circuit Currents
in phase a, b and c with
DC Component ≠ 0
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Notes :
• The occurance of a short circuit can not be predicted, therefore α is notknown.
• DC component vanishes very fast, usually in 8 to 10 cycles.
• The reactance of synchronous machines changes with time
• To calculate the short circuit current, the following synchronous reactancesare used:
"
dx Sub-Transient Reactance, to calculate the short circuit current in the
first cycle after the occurrance of a short circuit, i.e. within 0.05 - 0.1
seconds,
Transient Reactance, to calculate the short circuit current in a few
cycles after the occurrance of a short circuit, i.e. within 0.2 – 2
seconds,
Synchronous Reactance, to calculate the short circuit current after the
steady state condition is reached.
,
dx
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Fault Types
A. SHUNT FAULT (SHORT CIRCUIT)
1. SYMMETRICAL SHORT CIRCUIT
a. THREE PHASE (L-L-L)
b. THREE PHASE TO GROUND (L-L-L-G)
2. UNSYMMETRICAL SHORT CIRCUIT
a. ONE PHASE TO GROUND (1L-G)
b. PHASE TO PHASE (L-L)
c. TWOPHASA TO GROUND (2L-G)
B. SERIES FAULT (OPEN LINE)
1. ONE LINE OPEN (1L-O)
2. TWO LINES OPEN (2L-O)
3. UNBALANCED SERIES IMPEDANCE
C. SIMULTANEOUS FAULT
1. SHUNT - SHUNT
2. SHUNT - SERIES
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a. One Phase to Ground
b. Phase to Phase
c. Two Phase to Ground
d. Three Phase to Ground
e. Three Phase
f. Three Phase to Ground
through Impedance
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Thévenin equivalent
Represents passive network
• No-load voltage VTH
• Short-circuit impedance ZTH
• All sources zero for ZTH
Also for entire power system
+/0/- seq. equivalents
VTH
ZTH
~
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Short-circuit current
– Z=0 connected at terminals
– Short-circuit current
• Limited by ZTH (good!)
• ISC=VTH/ZTH≈1/ZTH p.u. (VTH≈1)
• Determines breaker rating
VTH
ZTH
~ ISC
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Short-circuit power
– Short-circuit power
• Short-circuit capacity
• Fault level
– SSC=VTHISC≈ISC≈1/ZTH p.u.
– SSC not useful power
VTH
ZTH
~ ISC
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Network strength
SLOAD relative to SSC
• SLOAD << SSC : strong
• SLOAD ≈ SSC/2 : weak
• SLOAD > SSC/2 : impossible VTH
ZTH
~ SLOAD
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Bus admittance matrix Ybus
Admittance representation
Nodal current balances
I= Ybus Vbus
Reference bus removed
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Example: Ybus
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Ybus properties
• With reference
– Row and column sums zero
• Reference removed
– Dimensions N-1 x N-1
• One Ybus for each sequence
• Sparse and symmetric
• Compact network model
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Bus impedance matrix Zbus
Vbus = Zbus I
If Ybus is invertible:
Zbus = Ybus-1
Zbus by inspection difficult
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ZTH from Zbus
Element ii of Zbus
– Short-circuit impedance ZTH at bus i
• Conditions
– Zbus has neutral as reference
– Generators have internal impedance
– Loads can be included in Zbus
• Practical for large systemsOntoseno Penangsang
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Simplified Machine Model
No Load Generator under Fault Condition
Internal Voltages of Loaded Machines under Fault Condition
Zbus Method
The Selection of Circuit BreakersOntoseno Penangsang
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Simplified Machine Model
For steady-state operation, generators are represented
with a constant emf behind a synchronous reactance, XS
For salient-pole rotors, there is a direct axis and quadrature axis
Reactances
Under transient conditions, the machine reactance
changes due to the effect of the armature (transformer)
reaction and eddy currents in the damping circuits
For analysis it is useful to imagine the synchronous
reactance as three components
direct axis sub-transient reactance
direct axis transient reactance
direct axis steady-state reactance
these transient reactances have an associated time-constant
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max
max
'
max
max'
"
max
max"
I
EX
I
EX
I
EX
d
d
d
NO LOAD GENERATORUNDER FAULT CONDITIONS
Emax : maximum line-neutral voltage of a Generator
Imax : maximum symmetrical short circuit current
Eg : rms line-neutral voltage of a Generator
I˝ : sub-transient current (rms, no DC component)
I΄ : transient current (rms, no DC component)
I : steady-state current ( rms)
d
g
d
g
d
g
X
EI
X
EI
X
EI
'
'
"
"
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INTERNAL VOLTAGES OF LOADEDMACHINES UNDER FAULT CONDITIONS
Before a SC occurs After a SC occurs
After a SC occurs, sub-transient internal voltage ( E˝ ) OR transient internal
voltage ( E΄ ) is used
Generator : Eg˝ = Vt + j IL Xd˝
Eg΄ = Vt + j IL Xd΄
Motor : Em˝ = Vt – j IL Xd˝
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Example 1 :
(sub-transient internal voltage ( E˝ ) is used)
MOTOR, GENERATOR : 30.000 kVA, 13.2 kV, X˝= 20 %
LINE : X = 10 % (BASE : MACHINE RATING)
THE MOTOR IS DRAWING 20.000 kW, p.f. : 0.8 LEADING
MOTOR TERMINAL VOLTAGE : 12.8 kV WHWN A
SYMMETRICAL THREE PHASE FAULT OCCURS AT THE
MOTOR TERMINALS.
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puj
pu
Axx
I
Ax
I
puV
kVkVABASE
L
BASE
f
52,069,0
9,3686,0
9,3611288,1238,0
000.20
13122,133
000.30
097,02,13
8,12
2,13 , 000.30:
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Aj
pujj
jI
puj
jjjE
pujjjV
Generator
g
g
t
3550905
71,269,03,0
207,0814,0
207,0814,0
)52,069,0(2,0069,0918,0
069,0918,0)52,069,0(1,097,0
:
"
"
Aj
pujj
jI
pujjjE
puVV
Motor
m
m
ft
7050905
37,569,02,0
138,0074,1
138,0074,1)52,069,0(2,097,0
097,0
:
"
"
Aj
puj
III
LocationFaultAt
mgf
10600
08,8
:
"""
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puV
pujjj
jjZ
f
th
097,0
12,02,03,0
)2.0)(3,0(
Example 2 :
(Thevenin’s theorem is used)
08,8
12.0
097.0
:
0"
puj
jI
LocationFaultAt
f
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pujjj
jI g 23,308,8
5,0
2,0"
pujjj
jIm 85,408,8
5,0
3,0"
From Generator :
From Motor :
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Example 1
IL≠0
Example 2
IL=0
Example 2
+ IL
Ig” 0,69-j2,71 pu -j3,23 pu 0,69-j2,71 pu
Im” 0,69-j5,37 pu -j4,85 pu 0,69-j5,37 pu
If” -j8,08 pu -j8,08 pu -j8,08 pu
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Zbus METHOD
Single Line Diagram
Impedance Diagram
Vf: Prefault
Voltage at
Bus 2
Three Phase Short
Circuit occurs on bus 2.
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Admittance Diagram
If” : Short Circuit current
Vf – Vf = 0
Short Circuit
occurs on Bus 2
I = Ybus . VOntoseno Penangsang
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4
3
1
"
67,20 33,3 0,10 33,3
33,3 67,11 33,3 0,0
0,10 33,3 67,16 33,3
33,3 0,0 33,3 0,10
0
0
0
V
V
V
V
I ff
Ybus MatrixOntoseno Penangsang
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I = Ybus . V
V = [Ybus]-1 . I
V = Zbus . I
Zbus = [Ybus]-1
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0
0
0
"
44434241
34333231
24232221
14131211
4
3
1
ff I
zzzz
zzzz
zzzz
zzzz
V
V
V
V
V = Zbus . I
22
"
z
VI
f
f Short Circuit
current from
Bus 2
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ff
ff
ff
Vz
zzIV
Vz
zzIV
Vz
zzIV
22
4242
"
4
22
3232
"
3
22
1212
"
1
42
"
44
32
"
33
2
12
"
11
0
zIVVVV
zIVVVV
VVV
zIVVVV
fff
fff
ff
fff
The changes of voltages
at Bus 1, 3 and 4
(Total) Voltage at
each bus.
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kk
f
fz
VI
f
kk
nkfn V
z
zVV
Vf is assumed
(prefault current is neglected)
Three Phase Short Circuit
occurs on Bus k
Short Circuit
current from
Bus k
Voltage on
Bus n
pu00,1
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1155,0 0602,0 0953,0 0702,0
0602,0 1226,0 0692,0 0431,0
0953,0 0692,0 1471,0 0807,0
0702,0 0431,0 0807,0 1502,0
jZbus
Three Phase Short Circuit
occurs on bus 2
Calculate Short
Circuit Current
flowing in each line
and from Generator
Example 3 :
Vf: Prefault
Voltage at
Bus 2
Impedance
Diagram
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THE SELECTION OF CIRCUIT BREAKERS
The electric utility company
furnishes data to a customer who
must determine the fault current in
order to specify circuit breakers
properly for an industrial plant or
industrial power distribution system
connected to the utility system at a
certain point.
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c
b
a
If
0
Time
In the short circuit calculation, we calculate the subtransient
current called the initial symmetrical current, which does not
include the component.
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Devices
Types of
Short Circuit
current
Reactance used for SC calcMultplying
Factor
Synchr.
Generator
Synchr.
Motor
Induction
Motor
none
Molded
Case CB
Symmetrical
Ampere
Sub-
transient
(X”)
Sub-
transient
(X”)
Sub-
transient
(X”)
CB diatas
600 volt
8 cycles
Symmetrical
Interrupting
Ampere
Sub-
transient
(X”)
Transient
(X)neglected
Genera
l Case
Special
Case
1.0 1.1*)
Asymmetrical
Momentary
Ampere
Sub-
transient
(X”)
Sub-
transient
(X”)
Sub-
transient
(X”)
1.6 1.5**)
Fuse diatas
1500 volt
Asymmetrical
Interrupting
Ampere
Sub-
transient
(X”)
Sub-
transient
(X”)
Sub-
transient
(X”)
1.6 1.2***)
Multiplying Factor (standar ANSI/IEEE C37.5-1953)
WHERE :
Vpf : VOLTAGE (L-L) BEFORE THE SHORT CIRCUIT
OCCURS (VOLT)
I˝ : RMS VALUE OF THE INITIAL SYMMETRICAL
SHORT CIRCUIT CURRENT (AMPERE)
ξ : MULTIPLYING FACTOR
THE RATING OF CIRCUIT BREAKERS
I˝. ξ .10-3 kA
√3 Vpf . I˝ . ξ . 10-6 MVA
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