View
671
Download
5
Category
Preview:
Citation preview
Learning OutlineLearning Outline
Symmetrical Components TheorySymmetrical Components Theory
Sequence Impedance: Load, Line, Generator, T fTransformer
Fault Analysis: Line‐Ground, Line‐Line, Line‐Line‐Ground.
Examples and Class Exercisesp
1
Fault Analysisy ____________________
• Fault types:– balanced faults (<5%)
• three‐phase to ground
• Three‐phase
– unbalanced faults• single‐line to ground (60%‐75%)
d bl li t d (15% 25%)• double‐line to ground (15%‐25%)
• line‐to‐line faults (5%‐15%)line to line faults (5% 15%)
2
Example impact of faultExample impact of faultThe second largest blackout in the history of TEPCO(The Tokyo Electric Power Company Inc ) hit central(The Tokyo Electric Power Company, Inc.) hit centralTokyo area at about 7:38 a.m. on August 14, 2006. Itwas caused by a floating crane on a barge goingupstream on a river on the eastern edge of the city.
The workers on the boat did not realize that the 33meter crane was raised too high, so it hit TEPCO's275 kV double circuit transmission lines that runacross the riveracross the river.
As a result of the accident the transmission lines were short‐circuited and the wires damaged. The relay protection operated and tripped both lines
3
Symmetrical Componentsy p __________
• Three phase voltage or current is in a balance condition if it has the following characteristic:the following characteristic:
– Magnitude of phase a,b, and c is all the same
– The system has sequence of a,b,cThe system has sequence of a,b,c
– The angle between phase is displace by 120 degree
• If one of the above is character is not satisfied, unbalanced occur. Example:
4
Symmetrical Componentsy p __________
• For unbalanced system, power system analysis cannot be analyzed using per phase as in Load Flow analysis oranalyzed using per phase as in Load Flow analysis or Symmetrical fault ‐>Symmetrical components need to be used.
• Symmetrical component allow unbalanced phase quantities such as current and voltages to be replaced by three separate balanced symmetrical componentsbalanced symmetrical components.
5
Symmetrical Componentsy p __________
6
Symmetrical Componentsy p __________
By convention, the direction of rotation of the phasors is taken to be counterclock wiseto be counterclock‐wise.
Positive sequence:111 0 aaa III =°∠=
1211 240 aab IaII =°∠=
111 120 III °∠
(10.1)
111 120 aac aIII =°∠=
Where we defined an operator a that causes a counterclockwise rotation of 120 degree such that:degree, such that:
866.05.0120sin120cos1201 jja +−=°+°=°∠=
866.05.02401)1201()1201(2 ja −−=°∠=°∠×°∠= 01 2 =++ aa(10.2)
0136013 ja +=°∠= (10.3)
7
Symmetrical Componentsy p __________
Negative sequence:
°∠= 022aa II
222 120 aab aIII =°∠=2222 240 III °∠
(10.4)
2222 240 aac IaII =°∠=
000cba III ==
Zero sequence:
(10.5)
8
Symmetrical Componentsy p __________
Consider the three‐phase unbalanced current of cba III ,,
210
210
bbbb
aaaa
IIII
IIII
++=
++=(10.6)
210cccc IIII ++=
Based on (10.1), (10.4) and (10.5), (10.6) can be rewrite all in terms of phase acomponents
2120
210aaaa IIII ++=
⎥⎤
⎢⎡
⎥⎤
⎢⎡
⎥⎤
⎢⎡
1
0
2
111 aa II
2210
2120
aaac
aaab
IaaIII
aIIaII
++=
++= (10.7)
⎥⎥⎥
⎦⎢⎢⎢
⎣⎥⎥⎥
⎦⎢⎢⎢
⎣
=⎥⎥⎥
⎦⎢⎢⎢
⎣2
1
2
2
11
a
a
c
b
II
aaaa
II (10.8)
9
Symmetrical Componentsy p __________
Equation 10.8 can be written as:
012a
abc AII =
Where A is known as symmetrical components transformation matrix,
(10.9)
which transforms phasor currents into components currents and
abcI 012aI
⎥⎤
⎢⎡
2
111(10 10)
⎥⎥⎥
⎦⎢⎢⎢
⎣
=2
2
11
aaaaA (10.10)
Solving (10.9) for the symmetrical components of currents:
abca IAI −=012 (10.11)
The inverse of A is given by:⎥⎤
⎢⎡ 111
1 (10 12)
⎥⎥⎥
⎦⎢⎢⎢
⎣
=−
aaaa
2
2
11
31A (10.12)
10
Symmetrical Componentsy p __________
From (10.10) and (10.12), we conclude that
(10.13)*
31 AA =−
Substituting for A‐1 in (10.11), we have:Substituting for A in (10.11), we have:
⎥⎥⎥⎤
⎢⎢⎢⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
⎥⎥⎥⎤
⎢⎢⎢⎡
b
a
a
a
III
aaIII
2
2
2
1
0
11
111
31 (10.14)
⎥⎦⎢⎣⎥⎦⎢⎣⎥⎦⎢⎣ ca IaaI 22 1
or in component form, the symmetrical components are:
)(31
)(31
21
0
cbaa
cbaa
IaaIII
IIII
++=
++=
(10.15)
)(31 22
cbaa aIIaII ++=
11
Symmetrical Componentsy p __________
Similar expressions exists for voltage:
(10.16)2210
2120
210
aaac
aaab
aaaa
VaaVVV
aVVaVV
VVVV
++=
++=
++=012a
abc AVV = (10.17)
aaac
1
The symmetrical components in terms of unbalanced voltages are:
)(31
)(31
21
0
cbaa
cbaa
VaaVVV
VVVV
++=
++=
(10.18) abca VAV −=012 (10.19)
)(31 22
cbaa aVVaVV ++=
12
Symmetrical Componentsy p __________
The apparent power may also be expressed in terms of the symmetrical componentscomponents.
*
)3(abcabcT
S IV=φ (10.20)
*
*012012)3( )()( a
Ta
T
S AIAV=φ
Substituting (10.9) and (10.17) in (10.20), we obtain:
(10.21)*012*012a
Ta
T
IAAV=
3, * == AAAA TTSince complex power becomes
(10.21)
*012012T
*** 221100
012012)3(
333
)(3
aaaaaa
T
S
IVIVIV
IV
++=
=φ
Total power for unbalance 3 phase system can be obtained from the sum of
(10.22)
Total power for unbalance 3‐phase system can be obtained from the sum of symmetrical components powers.
13
Example 1p _______________________
One conductor of a three‐phase line is open. The current flowing to delta‐connected load through line a is 10 A With the current in line a asconnected load through line a is 10 A. With the current in line a asreference and assuming that line c is open, find the symmetricalcomponents of the line currents.
a AIa °∠= 010
)(31
)(31
21
0
cbaa
cbaa
IaaIII
IIII
++=
++=
b AIb °∠= 18010
AI 0
)(31
)(3
22cbaa
cbaa
aIIaII ++=
c AIc 0=
14
Solution________________________
The line current are:
°∠= 010aI °∠= 18010bI 0=cI
1From (10.15): )(
31
)(31
21
0
cbaa
cbaa
IaaIII
IIII
++=
++=
0)018010010(31)0( =+°∠+°∠=aI
Ia +°+°∠+°∠= )0)120180(10010(31)1(
0 Sequence
+ Sequence
)(313
22cbaa aIIaII ++=
Aj °−∠=−= 3078.589.25
Aj
Ia
°∠=+=
+°+°∠+°∠=
307858925
)0)240180(10010(31)2(
‐ SequenceAj ∠=+= 3078.589.25
From (10.4) 0)0( =bI 0)0( =cI
AIb °−∠= 15078.5)1( AIc °∠= 9078.5)1(
AIb °∠= 15078.5)2( AIc °−∠= 9078.5)2(
15
Example 2p _______________________
16
Exercise 1_______________________
: that Show
°∠=++
)(1
1201)a(1
a)(1 (a)
2
2
°−∠=+− 1803
a)(1a)(1(b) 2
2
17
Exercise 2_______________________Obtain the symmetrical components for the set of unbalanced voltages
°−∠=°∠=°−∠= 30100,90200,120300 cba VVV
)(31
)(31
21
0
cbaa
cbaa
VaaVVV
VVVV
++=
++=
°−∠=
1202650.42012V
The symmetrical components of a set of unbalanced three‐phase currents are
)(313
22cbaa aVVaVV ++=°−∠
°−∠8961.849473.86
1351852.193
The symmetrical components of a set of unbalanced three phase currents are
Obtain the original unbalanced phasors.
°∠=°∠=°−∠= 304,905,303 210aaa III
210aaaa IIII ++=
2210
2120
aaac
aaab
aaaa
IaaIII
aIIaII
++=
++=
°−∠°∠
=
3042163.421854.8
Iabc
°−∠ 2163.1021854.8
18
Exercise 3_______________________The line‐to‐line voltages in an unbalanced three‐phase supply are
°∠=°−∠=°∠= 120500,1500254.866,01000 cabcab VVV
Determine the symmetrical components for line and phase voltages, then find the phase voltages Van, Vbn, and Vcn.
°∠°∠
=
8934107626763300.0
VL 012
°∠=
00.0Va 012
°−∠=
1066.199586.440Vabc
°∠°−∠
306751.2888934.107626.763
°∠°−∠
606667.1668934.409586.440
°∠°−∠
603333.3331021.1669252.600
19
Sequence Impedanceq p ______________
• The impedance of an equipment or component to theThe impedance of an equipment or component to thecurrent of different sequences.
iti i d (Z1) I d th t• positive‐sequence impedance (Z1): Impedance thatcauses a positive‐sequence current to flow
• negative‐sequence impedance (Z2): Impedance thatcauses a negative‐sequence current to flow
• zero‐sequence impedance (Z0): Impedance that causesa zero‐sequence current to flow
20
Sequence Impedance of Y‐Connected Loadq p
Line to ground voltages are:
(10.23)
nncsbmamc
nncmbsamb
nncmbmasa
IZIZIZIZVIZIZIZIZVIZIZIZIZV
++++=+++=+++=
(10.24)cban IIII ++=
Kirchhoff’ current law:
(10 25)
Substituting In into (10.23):
⎥⎤
⎢⎡
⎥⎤
⎢⎡ +++
⎥⎤
⎢⎡ anmnmnsa IZZZZZZV )(
(10.25)
(10.26)
⎥⎥⎥
⎦⎢⎢⎢
⎣⎥⎥⎥
⎦⎢⎢⎢
⎣ ++++++=
⎥⎥⎥
⎦⎢⎢⎢
⎣ c
b
nsnmnm
nmnsnm
c
b
II
ZZZZZZZZZZZZ
VV
)()(
abcabcabc IZV = (10.26)IZV =
21
Sequence Impedance of Y‐Connected Loadq p
(10 27)⎥⎥⎤
⎢⎢⎡
++++++
)()( nmnmns
abc ZZZZZZZZZZZZ
Z (10.27)⎥⎥⎥
⎦⎢⎢⎢
⎣ ++++++=
)()(
nsnmnm
nmnsnm
ZZZZZZZZZZZZZ
Writing Vabc and Iabc in terms of their symmetrical components:
(10.28)012012a
abca AIZAV =
Multiplying (10.28) by A‐1 :
(10.29)012012
012012 )(
a
aabc
a
IZ
IAZAV
=
= −
AZAZ abc−=012where (10.30)
(10.31)⎥⎥⎤
⎢⎢⎡
⎥⎥⎤
⎢⎢⎡
++++++
⎥⎥⎤
⎢⎢⎡
22012 1111
)()(
1111
1 aaZZZZZZZZZZZZ
aaZnmnmns
Substituting for Zabc, A and A‐1 from (10.27), (10.10) and (10.12):
⎥⎥⎦⎢
⎢⎣⎥⎥⎦⎢
⎢⎣ +++
+++⎥⎥⎦⎢
⎢⎣
=22 1
1)(
)(11
3aaaa
ZZZZZZZZZZZZ
aaaaZ
nsnmnm
nmnsnm
22
Sequence Impedance of Y‐Connected Loadq p
⎤⎡
Performing the multiplication in (10.31):
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−
++=
)(000)(000)23(
012
ms
ms
mns
ZZZZ
ZZZZ (10.32)
⎥⎤
⎢⎡ +
0)(000)3(
012ns
ZZZ
Z
When there is no mutual coupling, Zm = 0, and the impedance matrix becomes
(10.33)
⎥⎥⎥
⎦⎢⎢⎢
⎣
=)(00
0)(0012
s
s
ZZZ (10.33)
23
Sequence Impedance of Transmission Linesq p
For sequence impedance transmission line, Z1 = Z2, whereas Z0 is different and larger approximately 3 times than positive and negative sequence.
S I d f S h M hiSequence Impedance of Synchronous MachineThe positive-sequence generator impedance is the value found when positive-sequence current flows from the action of an imposed positive-sequence set of
"2
q p p qvoltages.
The negative-sequence reactance is close to the positive-sequencesubstransient reactance i e : "2
dXX ≈substransient reactance, i.e :
Zero-sequence reactance is approximated to the leakage reactance, i.e :
lXX ≈0
24
Sequence Impedances of Transformerq p
• Series Leakage Impedance.– the magnetization current and core losses represented by the shunt branch t e ag et at o cu e t a d co e osses ep ese ted by t e s u t b a c
are neglected (they represent only 1% of the total load current)– the transformer is modeled with the equivalent series leakage impedance
• Since transformer is a static device, the leakage impedance will not change if the phase sequence is changedif the phase sequence is changed.
• Therefore, the positive and negative sequence impedance are the same; Z0 = Z1 = Z2 = Zl
• Wiring connection always cause a phase shift. In Y‐Delta or Delta‐Y transformer:– Positive Sequence rotates by a +30 degrees from HV to LV side – Negative Sequence rotates by a ‐30 degrees from HV to LV side – Zero Sequence does not rotateZero Sequence does not rotate
• The equivalent circuit for zero‐sequence impedance depends on the winding connections and also upon whether or not the neutrals are grounded.
25
Sequence Impedances of Transformerq pConnection diagram Zero‐sequence circuit
Figure (a)
Figure (b)g ( )
Figure (c)Figure (c)
Figure (d)Figure (d)
Figure (e)
26
Figure (e)
Sequence Impedances of Transformerq p
Description of Zero sequence Equivalent Circuit
(a) Y‐Y connections with both neutrals grounded – We know that the zero sequence currentequals the sum of phase currents. Since both neutrals are grounded, there is a path for the zerosequence current to flow in the primary and secondary, and the transformer exhibits theequivalent leakage impedance per phase as shown in Fig. (a).
(b) Y‐Y connections with primary the neutral grounded – The primary neutral is grounded, but( ) p y g p y g ,since the secondary neutral is isolated, the secondary phase current must sum up to zero. Thismeans that the zero‐sequence current in the secondary is zero. Consequently, the zerosequence current in the primary is zero, reflecting infinite impedance or an open circuit asshown in Fig. (b).g ( )
27
Sequence Impedances of Transformerq pc) Y‐Δ with grounded neutral – in this configuration, the primary currents canflow because the zero‐sequence circulating current in the Δ‐connectedsecondary and a ground return path for the Y‐connected primary. Note that noy g p p yzero‐sequence current can leave the Δ terminals, thus there is an isolationbetween the primary and secondary sides as shown in figure (c)
d) Y Δ ti ith i l t d t l i thi fi ti b thd) Y‐Δ connection with isolated neutral – in this configuration, because theneutral is isolated, zero sequence current cannot flow and the equivalentcircuit reflects an infinite impedance or an open as shown in figure (d)
e) Δ‐Δ connection – in this configuration, zero‐sequence currents circulate inthe Δ‐connected windings, but no currents can leave the Δ terminals, and theequivalent circuit is as shown in figure (e)
Notice that the neutral impedance plays an important part in the equivalentcircuit. When the neutral is grounded through an impedance Zn, becauseIn=3Io, in the equivalent circuit, the neutral impedance appears as 3Zn in the
28
path of Io.
Sequence Impedances of a Loaded Generatorq p
A synchronous machine generates balanced three‐phase internal voltages and is represented as a positive‐sequence set of phasors
aabc E
aa⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡= 2
1E
(10.44)
a ⎥⎦⎢⎣
29
Sequence Impedances of a Loaded Generatorq p
The machine is supplying a three‐phase balanced load. Applying kirchhoff’s voltage law to each phase we obtain:
(10.45)nnbsbb
nnasaa
IZIZEV
IZIZEV
−−=
−−=
nncscc IZIZEV −−=
Substituting for In = Ia + Ib + Ic into (10.45):
⎤⎡⎤⎡⎤⎡⎤⎡ IZZZZEV )(
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
++
+−
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
c
b
a
nsnn
nnsn
nnns
c
b
a
c
b
a
III
ZZZZZZZZZZZZ
EEE
VVV
)()(
)(
(10.46)
In compact form: abcabcabcabc IZEV −= (10.47)
30
Sequence Impedances of a Loaded Generatorq pTransforming the terminal voltages and currents phasors into their symmetrical components:
(10.48)012012012a
abcaa AIZAEAV −=
Multiplying (10.48) by A‐1:
(10.49)012012012
012012012 )(
aa
aabc
aa
IZE
IAZAEV
−=
−= −
h ⎤⎡⎤⎡ +⎤⎡ 111)(111 ZZZZ(10.50)
Where:
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
++
+
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
2
2
2
2012
11
111
)()(
)(
11
111
31
aaaa
ZZZZZZZZZZZZ
aaaa
nsnn
nnsn
nnns
Z
Performing the above multiplication:
⎥⎥⎤
⎢⎢⎡
=⎥⎥⎤
⎢⎢⎡ +
= 1
0
012 0000
0000)3(
ZZ
ZZZ
s
ns
Z (10.51)
⎥⎥
⎦⎢⎢
⎣⎥⎥⎦⎢
⎢⎣
20000 ZZs
s
31
Sequence Impedances of a Loaded Generatorq pSince the generated emf is balanced, there is only positive‐sequence voltage, i.e:
(10 52)⎥⎤
⎢⎡ 0
(10.52)⎥⎥⎥
⎦⎢⎢⎢
⎣
=0
012aa EE
( )⎥⎤
⎢⎡
⎥⎤
⎢⎡
⎥⎤
⎢⎡
⎥⎤
⎢⎡ 000 000 aa IZV
Substituting for and in (10.49):012aE 012Z
111
000 0 aa IZV −=(10.54)
⎥⎥⎥
⎦⎢⎢⎢
⎣⎥⎥⎥
⎦⎢⎢⎢
⎣
−⎥⎥⎥
⎦⎢⎢⎢
⎣
=⎥⎥⎥
⎦⎢⎢⎢
⎣2
1
2
1
2
1
0000
0 a
aa
a
a
II
ZZE
VV (10.53) or
222
111
0 aa
aaa
IZV
IZEV
−=
−=
32
Sequence Impedances of a Loaded Generatorq pThe three equations in (10.54) can be represented by the three equivalent sequence networks:
• Important observations:– The three sequences are independent.
Th i i k i h h li di d i– The positive‐sequence network is the same as the one‐line diagram used in studying balance three‐phase currents and voltages.
– Only the positive‐sequence network has a source and no voltage source for other sequences.h l f h h f f d– The neutral of the system is the reference for positive‐ and negative‐sequence networks, but ground is the reference for zero‐sequence networks. Thus, zero sequence current can only flow if the circuit from the system neutrals to ground is complete.The grounding impedance is reflected in the zero sequence network as 3Zn– The grounding impedance is reflected in the zero sequence network as 3Zn
– The three‐sequence systems can be solved separately on a per phase basis. The phase currents and voltages can then be determined by superposing their symmetrical components of current and voltage respectively. 33
Single Line‐To‐Ground Faultg
Three‐phase generator with neutral grounded through impedance Zn and SLGF occurs at phase a through impedance Zf .
Assuming the generator is initially on no‐load, the boundary conditions at the fault point are:
afa IZV = (10.55)
0== cb II (10.56)34
Single Line‐To‐Ground Faultg
Substituting for Ib = Ic = 0, the symmetrical components of currents from (10.14) are:are:
⎥⎥⎥⎤
⎢⎢⎢⎡
⎥⎥⎥⎤
⎢⎢⎢⎡
=⎥⎥⎥⎤
⎢⎢⎢⎡
01111
31
2
2
2
1
0a
a
a IaaI
I(10.57)
⎥⎦⎢⎣⎥⎦⎢⎣⎥⎦⎢⎣ 013 22
a aaI
From the above equation, we find that:
1 (10 58) 111
000 0 aa IZV −=
aaaa IIII31210 ===
Phase a voltage in terms of symmetrical components is :
(10.58)
222
111
0 aa
aaa
IZV
IZEV
−=
−=
210aaaa VVVV ++=
: noting and (10.54) from and, ngSubstituti 210210aaaaaa IIIVVV ==
(10.59)
0210 )( aaa IZZZEV ++−= (10. 60)
35
Single Line‐To‐Ground Faultg
:get we,3 noting and (10.55), fromfor ngSubstituti .3 Where 0a
0aans IIVZZZ =+=
(10 61)02100 )(3 aaaf IZZZEIZ ++−=
or
(10.61)
f
aa ZZZZ
EI
32100
+++=
h f l
(10.62)
The fault current is
f
aaa ZZZZ
EII
33
3 2100
+++== (10.63)
f
In order to obtain symmetrical voltage at the point of fault Equation, (10.63) is substituted into Eq. (10.54)
36
Single Line‐To‐Ground Faultg
Eq. (10.58) and (10.62) can be represented by connecting the sequence networks in series as shown in the following figure.
aaaa IIII31210 ===
f
aa ZZZZ
EI
32100
+++=(10.58) (10.62)
37
Line‐To‐Line Fault
Three‐phase generator with a fault through an impedance Zf between phase band c.
I 0
Zs
Ia=0
NZsZs
EaEb
Ec
Z
Ib
Va
ZfIcVb
Vc
bfb IZVV =− I = 0
Assuming the generator is initially on no‐load, the boundary conditions at the fault point are:
(10.64) (10.66)bfcb IZVV Ia 00=+ cb II
(10.64)
(10.65)
(10.66)
38
Line‐To‐Line Fault
Substituting for Ia = 0, and Ic = ‐Ib, the symmetrical components of the currents from (10.14) are:from (10.14) are:
⎥⎥⎤
⎢⎢⎡
⎥⎥⎤
⎢⎢⎡
=⎥⎥⎤
⎢⎢⎡
ba
a
IaaII 0
1111
1 21
0
(10.67)
⎥⎥⎦⎢
⎢⎣−⎥⎥⎦⎢
⎢⎣⎥
⎥
⎦⎢⎢
⎣ b
b
a
a
IaaI 13 22
From the above equation we find that:From the above equation, we find that:
00 =aI
1
(10.68)
(10 69)ba IaaI )(
31 21 −=
ba IaaI )(31 22 −=
(10.69)
(10.70)
ba 3
39
Line‐To‐Line Fault
Also, from (10.69) and (10.70), we note that:21 (10 71)21aa II −=
From (10.16), we have:
VVVVVVVV )()( 22102120
(10.71)
2120
210
aaab
aaaa
aVVaVV
VVVV
++=
++=
(10.16)
bf
aa
aaaaaacb
IZVVaa
VaaVVaVVaVVV
=−−=
++−++=−
))((
)()(212
22102120
(10.72)
2210aaac VaaVVV ++=
000 0 aa IZV −=:get we, noting and (10.54) from and for ngSubstituti 1221
a aaa IIVV −=
bfaa IZIZZEaa =+−− ])()[( 1212 (10.73)(10.54)
222
111
0
0
aa
aaa
aa
IZV
IZEV
V
−=
−=
:get we(10.69), from for ngSubstituti bI
))((3
)( 22
1121
aaaaI
ZIZZE afaa −−
=+− (10.74) ba IaaI )(31 21 −= (10.69)
))((
40
Line‐To‐Line Fault
:in results for solving,3 Since 1aIa))(aa(a 22 =−−
)( 211
f
aa ZZZ
EI
++= (10.75)
The phase currents are
⎥⎤
⎢⎡⎥⎤
⎢⎡
⎥⎤
⎢⎡ 0111aI
(10 76)
⎥⎥⎥
⎦⎢⎢⎢
⎣−⎥⎥⎥
⎦⎢⎢⎢
⎣
=⎥⎥⎥
⎦⎢⎢⎢
⎣1
1
2
2
11
a
a
c
b
II
aaaa
II
The fault current is
(10.76)
The fault current is
12 )( acb IaaII −=−= or 13 ab IjI −=(10.77) (10.78)
41
Line‐To‐Line Fault
Eq. (10.71) and (10.75) can be represented by connecting the positive and negative –sequence networks as shown in the following figure.
21aa II −= 1 aE
I =
42
aa)( 21
fa ZZZ
I++
=
Double Line‐To‐Ground FaultFigure 10.14 shows a three‐phase generator with a fault on phases b and c through an impedance Zf to ground. Assuming the generator is initially on no‐load the boundary conditions at the fault point areload, the boundary conditions at the fault point are
0
)(210 =++=
+==
aaaa
cbfcb
IIII
IIZVV (10.79)(10.80)
From (10.16), the phase voltages Vb and Vc are
Figure 10 14Figure 10.14Double line‐to‐ground fault
43
Double Line‐To‐Ground Fault
(10.81)2210
2120aaab
VaaVVV
aVVaVV
++=
++=
(10 82)
21 ( )
aaac VaaVVV ++ (10.82)
thatnote weabove from ,VSinceVb c=
21aa VV =
Substituting for the symmetrical components of current in (10.79), we get
(10.83)
)2(
)(210
22102120)(
aaaf
aaaaaafb
IIIZ
IaaIIaIIaIZV
−−=
+++++=
03 af IZ= (10.84)
44
:havewe(10.81), into (10.83)fromfor and(10.84)from for ngSubstituti 2b aVV
10
1200 )(3
aa
aaaf
VV
VaaVIZ
−=
++=
( ),( )( )g b a
(10.85)
110 aa IZE
I−
:get we,for solving and(10.85)into(10.54)from voltageofcomponentslsymmetrica for the ngSubstituti
0aI
(10 86))3( 0
0
f
aaa ZZ
I+
−=
Also, substituting for the symmetrical components of voltage in (10.83), we obtain11IZE − (10 87)
(10.86)
22
ZIZE
I aaa
−−=
E
:get we,for solving and (10.80) into I and for ngSubstituti 120a aa II
(10.87)
f
f
aa
ZZZZZZ
Z
EI
3)3(
02
021
1
++
++
= (10.88)
45
sequence-negative theofn combinatio paralel with theseriesin impedancesequence-positive theconnectingby drepresentebecan (10.88)-(10.86)Equation
(10 8)fromfoundthenarecurrentphaseThefoundareIandIand (10.87), and (10.86)in dsubstitute is (10.86) from found I of valueThe
10.15. figure ofcircuit equivalent in theshown as networks sequence-zero and
20
1a
from obtained iscurrent fault theFinally,(10.8).fromfoundthen arecurrent phaseThefound. are I and I and aa
03bf IIII =+= (10.89)3 acbf IIII + ( )
Figure 10.15 Sequence network connection for double line‐to‐ground fault
46
The one-line diagram of a simple power system is show in Figure 10 16 The neutral of each
EXAMPLE
The one-line diagram of a simple power system is show in Figure 10.16. The neutral of each generator is grounded through a current-limiting reactor of 0.25/3 per unit on a 100-MVA base. The system data expressed in per unit on a common 100-mva base tabulated below. The generators are running on no-load at their rated voltage and rated frequency with their emfs in phasephase.Determine the fault current for the following faultsa. A balaced three-phase fault at bus 3 through a fault impedance = j 0.1 per unitb. A single line-to-ground fault at bus 3 through a fault impedance = j 0.1 per unit c A line-to-line fault at bus 3 through a fault impedance = j 0 1 per unit
fZfZ
fZc. A line-to-line fault at bus 3 through a fault impedance j 0.1 per unit d. A double line-to-ground fault at bus 3 through a fault impedance = j 0.1 per unit
fZfZ
Item Base Rated X1 X2 X0
MVA VoltageG1 100 20-kV 0.15 0.15 0.05G2 100 20 kV 0.15 0.15 0.05T1 100 20/220 kV 0.10 0.10 0.10T1 100 20/220 kV 0.10 0.10 0.10T2 100 20/220 kV 0.10 0.10 0.10L12 100 220 kV 0.125 0.125 0.30L13 100 220 kV 0.15 0,15 0.35L23 100 220 kV 0 25 0 25 0 7125L23 100 220 kV 0.25 0.25 0.7125
47
Figure 10 16Figure 10.16
Fault
Item BaseMVA
RatedVoltage
X1 X2 X0
G1 100 20-kV 0.15 0.15 0.05G2 100 20 kV 0 15 0 15 0 05G2 100 20 kV 0.15 0.15 0.05T1 100 20/220 kV 0.10 0.10 0.10T2 100 20/220 kV 0.10 0.10 0.10L12 100 220 kV 0.125 0.125 0.30
48
L13 100 220 kV 0.15 0,15 0.35L23 100 220 kV 0.25 0.25 0.7125
To find Thevenin impedance viewed from the faulted bus (bus 3), we convert the delta formed by buses 123 to an equivalent Y as shown belowformed by buses 123 to an equivalent Y as shown below
03571430)15.0)(125.0( jjjZ )250)(1250( jj
Fig. 10.17Positive-sequence impedance
0357143.0525.01 j
jZ S == 0595238.0
525.0)25.0)(125.0(
2 jj
jjZ S ==
0714286.05250
)25.0)(15.0(3 j
jjjZ S ==
525.0j
49
0714286.05952381.0
)3095238.0)(2857143.0(133 j
jjjZ +=
22.0jj
=
50
To find thevenin impedance viewed from the faulted bus (bus 3), we convert the delta formed by buses 123 to an equivalent Y as shown in figure 10.19(b)
)350)(300( jj 0770642.03625.1
)35.0)(30.0(1 j
jjjZ S ==
15688070)7125.0)(30.0( jjjZ ==
j0.077064
1568807.03625.12 j
jZ S ==
1830257.03625.1
)7125.0)(35.0(3 j
jjjZ S ==
51Fig. 10.19Zero-sequence impedance
j
Combining the parallel branches, the zero‐sequence thevenin impedance isg p q p
1830275.073394490
)2568807.0)(4770642.0(033 j
jjjZ +=
35.07339449.033
j
jj
=
j0.077064
So, the zero‐sequence impedance diagram is show in fig. 10.20
52
Fig. 10.20 Zero‐sequence network
(a) Balanced three‐phase fault at bus 3Assuming the no‐load generated emfs are equal to 1.0 per unit, the fault current is
A90820 1j3 1250.1(F)I )0(3 °∠V a
A90-820.1pu-j3.1250.1 j0.22Z Z
(F)If
133
)0(33 °∠==
+=
+=
j
(b) Single line‐to ground fault at bus 3F (10 62) th t f th f lt tFrom (10.62), the sequence component of the fault current are
puj
V a
-j0.91743(j0.1)0.35 j0.22 j0.22
0.1 3ZZZ Z
II If
033
233
133
)0(323
13
03 =
+++=
+++===
The fault current is :
jIII a ⎤⎡−⎤⎡⎤⎡⎤⎡⎤⎡ 752323111 03
033
pujI
III
aaaa
III
c
b
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
=⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
007523.2
00
3
11
111 3
03
03
3
2
2
3
3
3
53
(c) Line‐to Line fault at bus 3The zero‐sequence component of current is zero, i.e.,
0 I03 =
The positive‐and negative‐sequence components of the fault current are
V a
puV a
-j1.8519j0.1 j0.22 j0.22
0.1 ZZZ Z
I If
033
233
133
)0(323
13 =
++=
+++=−=
The fault current is
pujj
aaaa
III
c
b
a
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
2075.32075.30
8519.18519.10
11
111
2
2
3
3
3
j ⎦⎣⎦⎣⎦⎣⎦⎣ 3
(d) Double Line‐to Line‐fault at bus 3The positive‐sequence component of the fault current is
puV a
-j2.6017
j0.3j0.35j0.22j0.3)(j0.35 j0.22 j0.22
0.1
)Z3ZZ)Z3Z(Z
Z I
f033
233
f033
2331
33
)0(313 =
+++
+=
+++
+=
)f3333
54
The negative‐sequence component of current is :
pujjV a
j1.9438 j0.22
)6017.2)(22.0(0.1 Z
IZ I 2
33
133
133)0(32
3 =−−
−=−
−=
Th t f t iThe zero‐sequence component of current is:
pujjV a
j0.6579j0 3j0 35
)6017.2)(22.0(0.1 3ZZ
IZ I 0
133
133)0(30
3 =+−−
−=+
−−=
j0.3j0.353ZZ f33 ++
And the phase currents are :
jI a ⎤⎡⎤⎡⎤⎡⎥⎤
⎢⎡ 06579.01113
pujj
j
aaaa
II
c
b
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
°∠°∠=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡−
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
07.14058.493.165058.4
0
9438.16017.2
6579.0
11
2
2
3
3
3
And the fault currents is:
°∠=+= 909732.1)( 333cb IIFI
55
UNBALANCED FAULT ANALYSIS USING BUS IMPEDANCE MATRIX
Single Line‐to‐Ground Fault Using Zbus
kkkk
VIII )0(021
210 === (10.90)
fkkkkkkkkk ZZZZ 3021 +++
k. busat oltageprefault v theis)0(V andmatrix impedance bus
ingcorrespond theof axisk in the elements diagonal theare Zand Z, ZWhere 2kk
1kk
k
okk
: iscurrent phasefault The
=abckI A 012
kI (10.91)
Line‐to‐Line Fault Using Zbus
00 =kIV )0(
(10.92)
fkkkk
kkk ZZZ
VII++
=−= 2121 )0(
(10.93)
56
Double Line‐to‐Ground Fault Using Zbus
fkkkk
fkkkkkk
kk
ZZZZZZ
Z
VI
3)3(
)0(
02
021
1
++
++
= (10.94)
fkkkk
2
112 )0(
kk
kkkkk Z
IZVI −−= (10.95)
kk
kkkkk ZZ
IZVI3
)0(0
110
+−
−= (10.96)fkk ZZ 3+
iscurrent result theand(10.91), fromobtainedarecurrentsphaseThe matrix. impedance bus ingcorrespond theof axisk in the elements diagonal theare Zand , Zand , ZWhere 2
kk1kk
okk
( ),pp
Ck
bkk IIFI +=)( (10.97)
57
Recommended