View
217
Download
0
Category
Preview:
Citation preview
1
SESSION 3
Short Circuit Calculations, Unsymmetrical Faults
Leonard Bohmann, Michigan State University
Elham Makram, Clemson University
2
Short Circuit CalculationsLeonard Bohmann
Michigan State University
3
Line to Ground Short
2003 IEEE T & D CONFERENCE
42003 IEEE T & D CONFERENCE
-3
-2
-1
0
1
2
3
4
5
6
7
0 0.05 0.1 0.15 0.2tA
mp
s(k
A)
Short Circuit Fault Currents
high frequency transient
dc offset decaying sinusoid
52003 IEEE T & D CONFERENCE
Short Circuit Fault Currents (continued)
v(t)
X(t) R
]e)tcos(
)t[cos()t(Z
V)t(i
then
)t(jXR)t(Z
)tcos(V)t(v
if
)t(XtR
0
m
m
ω−θ−α+ω−
θ−α+ω=
+=θ∠α+ω=
-3
-2
-1
0
1
2
3
4
5
0 0.05 0.1 0.15 0.2
6
dc offset
-3
-2
-1
0
1
2
3
4
5
0 0.05 0.1 0.15 0.2
θ−α+ω−θ−α+ω=
ω− XtR
0m e)tcos()tcos(
Z
V)t(i
2003 IEEE T & D CONFERENCE
7
Variable magnitude
-6
-4
-2
0
2
4
6
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
t
2003 IEEE T & D CONFERENCE
)tcos()t(Z
V)t(i m θ−α+ω=
8
Reactance of Synchronous Machine
00.1
0.20.3
0.40.5
0.60.7
0.80.9
1
0 0.2 0.4 0.6 0.8 1 1.2 1.4
t
reac
tan
ce(p
u)
X'd
X"d
Xd
2003 IEEE T & D CONFERENCE
9
• Use sub-transient reactance to calculatefault currents– worst case for when CB is opening
• Do not calculate dc offset directly– use X/R to approximate offset
(asymmetry ratio) at opening time– X/R is the time constant of the decay in
cycles
2003 IEEE T & D CONFERENCE
10
How do youcalculate the
fault current (If)at bus 3?
ZT12
ZT35
1 2
3
4
ZG1
ZG4
ZT15
ZL5
5
ZL2
If
ZT23
ZT34
2003 IEEE T & D CONFERENCE
11
Use Thevenin equivalent
VTh
ZTh
3
If
• VTh: open circuitvoltage
– also called the pre-fault voltage, Vf
– often use 1∠ 0°
2003 IEEE T & D CONFERENCE
122003 IEEE T & D CONFERENCE
Use for:
• symmetrical faults
– single phase networks
– three phase faults in three phasenetworks (per phase equivalent)
• concept for solving all faults
132003 IEEE T & D CONFERENCE
• This method calculates the currentgoing into the fault
• Often want the current elsewherein the network (through a circuitbreaker, in bus bars, in other lines,etc.)
14
A current source, with the magnitude ofthe fault current, placed at the fault, hasthe same effect on the rest of the circuitas the short circuit.
Vf
ZTh
3
If
Vf
ZTh
3
If
2003 IEEE T & D CONFERENCE
15
Adding a current source tothe original circuit hasthe same electricaleffect as adding it tothe equivalent circuit
2003 IEEE T & D CONFERENCE
Vf
ZTh
3
If
ZT12
ZT35
1 2
3
4
ZG1
ZG4
ZT23ZT15
ZL5
5
ZL2
ZT34
If
16
All the voltages and allthe currents can be foundusing the new circuit.
2003 IEEE T & D CONFERENCE
ZT12
ZT35
1 2
3
4
ZG1
ZG4
ZT23ZT15
ZL5
5
ZL2
ZT34
If
Use superposition:– Use the original sources
(or assume currents are 0and voltages are 1∠ 0°)
– Use just the currentsource, gives changesdue to the fault
17
Calculation detailsBus Admittance Matrix: Ybus
2003 IEEE T & D CONFERENCE
I = Ybus V
vector of currentinjections
vector of busvoltages
ZT12
ZT35
1 2
3
4
ZG1
ZG4
ZT23ZT15
ZL5
5
ZL2
ZT34
If
182003 IEEE T & D CONFERENCE
ZT12
ZT35
1 2
3
4
ZG1
ZG4
ZT23ZT15
ZL5
5
ZL2
ZT34
If
=
555351
4443
35343332
232221
151211
bus
00
000
0
00
00
Y
YYY
YY
YYYY
YYY
YYY
Diagonals:Σ of Y connected at nodeY22 = 1/ZT12 + 1/ZL2 + 1/ZT23
Off diagonals:- of Y connected between nodesY53 = -1/ZT35
192003 IEEE T & D CONFERENCE
Ybus is sparse (lots of zero)
– small size
– easy to build
– quick algorithms to manipulate
20
Bus Impedance Matrix: Zbus
2003 IEEE T & D CONFERENCE
V = Zbus I
vector of currentinjections
vector of busvoltages ZT12
ZT35
1 2
3
4
ZG1
ZG4
ZT23ZT15
ZL5
5
ZL2
ZT34
If
Zbus = Ybus-1
21
=
5554535251
4544434241
3534333231
2524232221
1514131211
busZ
ZZZZZ
ZZZZZ
ZZZZZ
ZZZZZ
ZZZZZ
2003 IEEE T & D CONFERENCE
Zbus is a full matrix– slow to invert– slow to manipulate– avoid using if
possible
Diagonal element is the TheveninImpedance at that bus– Z33 is the Thevenin Impedance at bus 3
222003 IEEE T & D CONFERENCE
To find the change in voltages due tothe fault current at bus 3:
⋅
=
∆∆∆∆∆
0
0
I-
0
0
V
V
V
V
V
f
5554535251
4544434241
3534333231
2524232221
1514131211
5
4
3
2
1
ZZZZZ
ZZZZZ
ZZZZZ
ZZZZZ
ZZZZZ
23
To calculate current in line 2-3:
2003 IEEE T & D CONFERENCE
23T
32f23 Z
VVI
∆−∆=∆
1 2
3
4
ZT23
5 If
∆I23f
24
Asymmetry ratio: S
• Method calculates the RMS worst caseof the 60 Hz component– this is needed to set protective relays
• For circuit breaker selection, need toknow the total RMS current (60 Hzand dc offset)– Circuit breakers are rated on symmetrical
fault current, assuming an X/R ratio of 152003 IEEE T & D CONFERENCE
252003 IEEE T & D CONFERENCE
( ) RXt2
ac
RMS
RXt2
acRMS
2RX
t
ac2acRMS
2dc
2acRMS
21isS(t)wheretSI
I
21II
I2II
III
ω−
ω−
ω−
⋅+=
⋅+=
+=
+=
e
e
e
Asymmetry ratio
26
Calculating X/R• Equations presented are
for a single order system• For higher order systems,
– ignore R and findequivalent X,
– ignore X, and findequivalent R
• Actual decay is typicallyquicker than thisapproximation
ZT12
ZT35
1 2
3
4
ZG1
ZG4
ZT23ZT15
ZL5
5
ZL2
ZT34
If
2003 IEEE T & D CONFERENCE
27
Algorithm
2003 IEEE T & D CONFERENCE
• Calculate Thevenin Impedance• Calculate current into fault• Find the needed elements of Zbus (not all
of them)• Find the needed changes in bus voltages• Find the line currents of of interest• Find the dc offset
28
Example• The oneline drawing of a system, along with Zbus
are shown below.
j 0.08 pu
j0.03 pu
1 2
3
4
j0.1pu
j0.1 pu
j0.025 puj0.05 pu
j15pu
5
j10 pu
If
j0.05 pu
2003 IEEE T & D CONFERENCE
puj
=
0799004290064400623005590
0429006580048700450003340
0644004870073100675005020
0623004500067500831005380
0559003340050200538006560
Zbus
.....
.....
.....
.....
.....
i) Calculate the fault current for athree phase solid fault at bus 3.
ii) Calculate the voltages at bus 1,2, and 3 during this fault.
iii) Calculate the current flowingfrom bus 2 to bus 3.
29
Part (i): Calculate the fault current for a threephase solid fault at bus 3.
• Assume all initial busvoltages are 1∠ 0°– therefore all initial
currents are 0
• Use the 3,3 elementof Zbus for theThevenin impedance.
1∠ 0°
j0.731
3
If
2003 IEEE T & D CONFERENCE
puj
j
6813I
07310
1I
f
f
.
.
−=
=
30
Multiply the negative of the fault current by elements inthe 3rd column to get the change in voltages
2003 IEEE T & D CONFERENCE
Part (ii): Calculate the voltages at bus 2, 3, and4 during this fault.
⋅
=
∆∆−
−−
0
0
13.68--
0
0
0799004290064400623005590
0429006580048700450003340
0644004870073100675005020
0623004500067500831005380
0559003340050200538006560
V
V
1
9230
6870
5
4
jj
.....
.....
.....
.....
.....
.
.
312003 IEEE T & D CONFERENCE
Part (ii): Calculate the voltages at bus 2, 3, and4 during this fault. (continued)
pu
pu
pu
0101VVV
07660923001VVV
3130687001VVV
33i3f
22i2f
11i1f
=−°∠=∆−==−°∠=∆−==−°∠=∆−=
..
..
The voltages during the fault are the initial voltagesminus the changes in the voltages
322003 IEEE T & D CONFERENCE
Part (iii): Calculate the current flowing from bus 2to bus 3.
The current is the differencebetween the voltagesdivided by the impedance
pujj
8630250
007660I
Z
VVI
23f
T23
3f2f23f
..
. −=−=
−=
1 2
3
4
5-j13.68 pu
-j3.86 pu
33
Problem
1 2
3
4
j0.1 puj0.1 pu
j0.1 pu
j0.025 puj0.05 pu
j0.05 pu
j 0.08
2003 IEEE T & D CONFERENCE
A small power system is sketched below along with Zbus.
a) What is the fault current for a three-phase fault toground at bus 3?
b) What are the voltages at busses 2, 3, and 4?c) What is the current due to the fault from bus 4 to 3 and
from 2 to 3?
puj 2bus 10
4045106342221742
1063658463432613
4222634353440443
1742261304437834
Z −×
=
....
....
....
....
12003 IEEE T & D CONFERENCE
Unsymmetrical FaultsElham Makram
Clemson University
1
SINGLE LINE TO GROUND FAULT
Fig.1. Single line to ground fault on phase A
2003 IEEE T & D CONFERENCE
Zf
Ibf IcfIaf
a
b
c
n
0f210
affa
210
12
0
21
22
12
22
10212
0
afff
cfbf
3I*ZVVVOr,
I*ZVNow
III
0Ia)a(IThus
II
a)I–(aa)I–(a
0IaaIIaIIaI
equations,currenttheFrom
I*ZV
0II
locationfaultatConditionsBoundary
=++=
===++
==
=++=++
===
22003 IEEE T & D CONFERENCE
SINGLE LINE TO GROUND FAULT (CONTD.)
Fig.2. Sequence connection for A-G fault
I0
To satisfy the voltage and current boundary conditions, the sequence networks should beconnected in series as shown in Fig. 2.
3Zf
ZeroSequenceNetwork
ZeroSequenceNetwork
ZeroSequenceNetwork
I1
I2
V0
+
-
+V1-
V2-
+
32003 IEEE T & D CONFERENCE
LINE TO LINE FAULT
Fig.3. Line to line fault
a
Zf
Ibf IcfIaf
b
c
n
Vaf
+
Vbf Vcf
1f21
22
f12
12
22
10
212
0f212
0
bffcfbf
21
22
12
22
1212
0
cfbf
af
IZVV
a)V-(aZa)I-(aa)V-(a
toleadsThis
VaaVV
)aIIa(IZaVVaV
IZVV
conditionsboundaryVoltage
I-I
a)I(a-a)I(a
0IaaIaIIa
0I
Thus,
I-I
0I
ConditionsboundaryCurrent
+=+=
+++
++=++
+=
=+=+
=+++
=
==
42003 IEEE T & D CONFERENCE
Fig.4. Connection of sequence network for BC fault
To satisfy the voltage and current boundary conditions, the sequence networks shouldbe connected as shown in Fig. 4.
LINE TO LINE FAULT (CONTD.)
Zf
PositiveSequenceNetwork
NegativeSequenceNetwork
I1
I2
+V1-
V2-
+
52003 IEEE T & D CONFERENCE
DOUBLE LINE TO GROUND FAULT
Fig.5. Double line to ground fault
0f10
f22
12
12
0212
0
21
22
10212
0
cfbf
fcfbfbf
210
af
I3ZV-VThus
Z)IaaIaI
Ia2I(aVVaV
Also,
VVThus
VaaVVaVVaV
Thus,
VV
and,
)ZI(IV
conditionsboundaryVoltage
0III
Thus,
0I
location,faultat the
ConditionsboundaryCurrent
=++
++=++
=++=++
=
+=
=++
=
Zf
a
b
c
n
+
Ibf IcfIaf
Vaf Vbf Vcf
62003 IEEE T & D CONFERENCE
DOUBLE LINE TO GROUND FAULT (CONTD.)
Fig.6. Sequence network connection for doubleline to ground fault
Considering the voltage and current boundary conditions, the sequence networks areconnected as shown in Fig. 6.
3Zf
I0
ZeroSequenceNetwork
PositiveSequenceNetwork
NegativeSequenceNetwork
I1
I2
V0
+
-
+V1-
V2-
+
72003 IEEE T & D CONFERENCE
Example 1
Consider the power system shown in Fig. 7.
Ratings
G1 and G2: 100 MVA, 20 kV
X1 = X2 = 0.3 pu
X0 = 0.04 pu
T1 and T2: 100 MVA, 345/20 kV
XT1 = XT2 =0.08 pu
T. L. X1 = X2 = 0.15 pu
X0 = 0.5 pu
Find the voltages at Bus 1 and thecurrents from T1 to P due to
(i) Single line to ground fault at P
(ii) Line to line fault at P
(iii) Double line to ground fault at P
Fig. 7. Example 1
G1 G2
T1 T21 2P T. L.
82003 IEEE T & D CONFERENCE
Positive sequence network
Negative sequence network Zero sequence network
Z0th = j0.08
j0.2213
Pj0.08 j0.5Z1th = j0.2213
P
j0.3 j0.3
j0.08 j0.15 j0.08
1
P
2 j0.53j0.38
P
j0.2213
P
j0.2213j0.15)j0.08(j0.3||j0.8)(0.3Z
01VEEIf
th
21
=+++=°∠===
92003 IEEE T & D CONFERENCE
(i) Single line to ground fault
P--T1fromcomponentsphasecurrentThe
0.15304-(j0.8)j1.913-ZI-V
0.42332-(j0.38)(j1.114)-ZI-V
0.57668j0.38)(-j1.114)(–1.0ZI-EV
:arelocationfaultat thecomponentsvoltagesequenceThe
j1.9134-I
j1.114-I
114.10.38)(0.53
0.53*II
P--T1fromcomponentscurrentsequencethearefollowingThe
j1.9134-j0.08)j0.2213(j0.2213
01
)ZZ(Z
EIII
000
222
1111
0
2
11
2th1th0th021
>===
======
==
−=+
=
>
=++
°∠++
===
j
=
a2
a1
a0
2
2
c
b
a
I
I
I
*
1
1
111
I
I
I
aa
aa
102003 IEEE T & D CONFERENCE
-j0.799c
-j0.799b
-j3.027
c-j5.739= 3 I0
-j2.712
j0.799-j1.913–(-j1.114)aa(-j1.114)I
j0.799-j1.913–a(-j1.114)(-j1.114)aI
j3.027-j1.913-j1.114--j1.114I
2c
2b
a
=+=
=+=
==
j0.866-0.2297
))(-0.42332240(1)(0.57668)120(10.15304-VaaVVV
j0.866--0.2297
))(-0.42332120(1)(0.57668)240(10.15304-aVVaVV
00.42332–0.576680.15304-VVVV
:arelocationfaultat thevoltagesphaseThe
22
10c
212
0b
210a
+=°∠+°∠+=++=
=°∠+°∠+=++=
=+=++=
112003 IEEE T & D CONFERENCE
(ii) Line to line fault
°∠=°∠=
°∠===
==−=
=+=
=+=
==>
−=+
−=>
−=+
°∠=+
=−=
00.5V
0-0.5V
01V
5.0ZI-V
0.5j0.38)(-j1.316)(-1.0ZIEV
:Voltages
2.279(j1.316)a(-j1.316)aI
2.279a(j1.316)(-j1.316)aI
0I
316.1jP)--1T(I
316.1j)0.53(0.38
0.5326.2jP)--1T(I
26.2j0.2213)(j0.2213
01
)Z(Z
EII
:Currents
c
b
a
2a2a2
1a1a1
2c
2b
a
2
1
2th1th21 j
122003 IEEE T & D CONFERENCE
SEQUENCE PHASE SHIFT THROUGH WYE/DELTA TRANSFORMER BANK
•Positive and negative sequence impedance is independent of connection
•These networks ignore phase shift
•When current and voltages are transformed from one side to the other phase shift must
be considered.
•Standard ANSI connection has been shown in figs. 8 (a) and 8 (b).
Fig. 8 (a) Wye/Delta Fig. 8 (b) Delta/Wye
High side Low sideIA
IB
IC
Ia
Ib
Ic
A
B
C
a
b
c
nIA
nIB
nIC
IA
IB
IC
n:1
BC
A
bc
a
High side Low side
Ia
Ib
Ic
IA
IB
IC
a
b
c
A
B
C
Ia/n
Ib/n
Ic/n
n:1
BC
A
bc
a
132003 IEEE T & D CONFERENCE
30VV
unitperInor
30V3na)-(1nV)aV-n(V)V-n(VV
:voltagesequenceNegative
30II
unitperInor
30I3n)a-(1nI)Ia-n(I)I-n(II
:currentsequenceNegative
30VV
unit,perinor
30V3n)a-(1nV)Va-n(VV
:volatgesequencePositive
30II
unit,perIn
30-I3na)-(1nI)aI-n(I)I-n(II
:currentsequencePositive
8(a)fig.Refer to
a2A2
a2a2a2a2b2a2A2
a2A2
A22
A2A22
A2C2A22a
a1A1
a12
aa12
a1A1
a1A1
A1A1A1A1C1A11a
°−∠=
°−∠====
°−∠=
°∠====
°∠=
°∠===
°∠=
°∠====
142003 IEEE T & D CONFERENCE
°−∠=
°−∠=
°−∠==
°∠=
°−∠==
°∠=
°∠===
30VV
voltagesequenceNegative
30II
unitperInor
30In
3)aI-(I
n
1I
:currentsequenceNegative
30VV
unit,perinor,
30Vn
3)aV-(V
n
1V
:voltagesequencePostitive
30II
unit,perinor,
30In
3)Ia-(I
n
1)I-(I
n
1I
:currentsequencePostitive
(b)8fig.Refer to
a2A2
a2A2
a2a2a2A2
a1A1
A1A1A1a1
a1A1
a1a12
a1b1a1A1
152003 IEEE T & D CONFERENCE
°∠=°+∠=
∆∆
30-voltageLowsequenceNegativegeHigh voltasequenceNegative(ii)
30voltageLowsequencePositivegeHigh voltasequencePositive(i)
:sconnectionY-or-Yingeneral,In
162003 IEEE T & D CONFERENCE
Example 2
Find the currents and voltages at G1 of Example 1, due to single line to ground fault at P.
0I
j1.9295)60(1.114)120(1.114
)60-(1.1141201)120-(1.1142401aIIaI
j1.9295-60-j1.114120-j1.114-III
0I
60-j1.11430j1.114-I
120-j1.114-30-j1.114-I
:Currents(i)
cg
ag2ag12
bg
ag21agag
ag0
ag2
1ag
==°∠+°∠=
°∠°∠+°∠°∠=+=
=°∠+°∠=+=
=
°∠=°+∠=
°∠=°∠=
172003 IEEE T & D CONFERENCE
°∠°∠
°−∠=
=
°∠=+=
°∠=
°∠°∠=+=
°−∠=°∠°∠=+=
°∠=
°∠==
==
==
901
12.240577.0
60577.0
V
V
V
*
1
1
111
V
V
V
901.0VaaVV
240.120.577
1500.334-2100.666aVVaV
60577.0300.334-30-0.666VVV
300.334-V
30-0.666V
shift)phaseneglectingNetworksequence(from0.3342-
ZI–V
shift)phaseneglectingNetworksequence(from0.666
j0.3)(-j1.114)(-1.0ZI–EV
:Voltages(ii)
ag2
ag1
ag0
2
2
cg
bg
ag
ag22
ag1cg
ag2ag12
bg
ag2ag1ag
ag2
ag1
g2a2ag2
g1a1ag1
aa
aa
Example 2 ( contd.)
182003 IEEE T & D CONFERENCE
PROBLEM #1
For the system shown below, find the currents in the transformer and generator windings inper unit and amperes for a single-line-to-ground fault at
(i) P
(ii) R
G1
15/115 kVP
G1R
G: 100 MVA, 15 kV
X1=X2= 0.1,
X0 = 0.05
Transformer:
100 MVA, 15/115 kV
X = 0.1
PROBLEMS
19
PROBLEM #2
Solve examples 1 and 2 for a double-line-to-ground fault at P.
PROBLEMS
2003 IEEE T & D CONFERENCE
202003 IEEE T & D CONFERENCE
PROBLEM #3
Solve examples 1 and 2 for a line-line fault at P.
PROBLEMS
Recommended