Download - 01F 03 Symmetrical Components Lectures
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Dr.
Theof e Cabanabscirccon
Mohamm
e positiveach typ
se 1: Y-nk is ungsence ofcuit exisnnected
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negativnnection
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Neutral -sequencone winquence cmer.
onnectio
metrical C
ence of entical.
Groundce currending prcurrent
on Diag
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ded If eitent cannrevents between
grams
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age 64 of
phase tra
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Dr.
Cathe seq
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Mohamm
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k, Both Nxists forcan flow
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Neutralr zero-s
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age 65 of
this connnt in botansform
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Dr.
Cacan
Mohamm
Sym
se 4: Y-nnot flow
Sym
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mbols
- bankw in the
mbols
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k, Ungrotransfor
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onnectio
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on Diag
Y An uinding.
on Diag
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grams
ungroun
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nts
age 66 of
Z
nded Y
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f 90
Zero-seq
zero-se
Zero-seq
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quence
equence
quence
EEE
current
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Dr.
Cacurcirc
Mohamm
se 5: -rrent, noculate w
Sym
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- banko zero-se
within the
mbols
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k Since aequencee win
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Symm
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a circue currendings.
onnectio
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uit provnt can fl
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grams
nts
age 67 of
return po a -
Z
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path forbank, a
Zero-seq
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r zero-sealthough
quence
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equence h it can
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Dr.
Zersepas s
Mohamm
ro-sequeparately shown in
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ence equare readn Figs. 1
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uivalentdily com11.19 an
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t circuitsmbined tnd 11.20
metrical C
s determto form 0.
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age 68 of
or variomplete z
ig. 11.19a small pcorrespo
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9 One-lipower syonding z
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s of the quence n
ine diagystem anzero-sequork.
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system network
gram of nd the uence
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Dr. Mohammmad Abddul Mann
Symm
nan
metrical C
Componen
Pa
nts
age 69 of
Fig. diagrsystecorreseque
f 90
11.2ram of m
espondinence net
AIUB/E
20 Oa small
and ng twork.
EEE
One-line l power
the zero-
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Symmetrical Components
Dr. Mohammad Abdul Mannan Page 70 of 90 AIUB/EEE
FOR SYNCHROMOUS MACHINES Negative Sequence Impedance < Positive Sequence Impedance Zero Sequence Impedance = Variable Items = May be taken equal to positive sequence impedance if its value is not given
FOR TRANSFORMER Negative Sequence Impedance = Positive Sequence Impedance Zero Sequence Impedance = Positive Sequence Impedance, if there is circuit for earth current = Infinite, if there is no through circuit for earth current
FOR TRANSMISSION LINE Negative Sequence Impedance = Positive Sequence Impedance Zero Sequence Impedance = Variable Item = May be taken as three times the positive sequence impedance if its value is not given
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Dr.
ExanetwG1:TraT1:T2:M1M2Sel
Mohamm
ample 1work, an: 300 MVansmiss 350 MV 300 MV: 200 M: 100 M
lect the g
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11.4: Drnd (iii) zVA, 20 sion lineVA, 230VA, 220
MVA, 13MVA, 13
generato
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raw (i) thzero-seqkV, X
e: 64 km0/20 kV0/13.2 k.2 kV, X.2 kV, X
or rating
Symm
nan
he positquence n= 20%,
m (4- mi, X = 1
kV, X =X = 20%X = 20%g as base
metrical C
tive-sequnetwork X0= 5%), X1=X20%
= 10% %, X0=5%, X0= 5e in the
Componen
Pa
uence nek for the %, Xn= 0X2=0.5
5%, Xn= 5% generato
nts
age 71 of
etwork, system .4 /km, X0=
0.4 or circui
f 90
(ii) negshown
=1.5 /
it.
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gative-sein the fi
/km.
EEE
equence igure:
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Symmetrical Components
Dr. Mohammad Abdul Mannan Page 72 of 90 AIUB/EEE
Generator Side
Vbase = 20 kV X1 = X2 =X =0.2(20/20) (300/300)=0.2 pu X0= 0.05(20/20) (300/300)=0.05 pu Xbase = (20)2/300 = 1.333 3Xn= (30.4)/1.333 = 0.9 pu
Transmission Line side
Vbase = (230/20)20 = 230 kV X1=X2 = 640.5 = 32 X0= 641.5 = 96 Zbase = (230)2/300 = 176.3 X1=X2 = 32/176.3 = 0.1815 pu X0= 96/176.3 = 0.5445 pu
Motor side
Vbase = (13.2/220)230 = 13.8 kV For M1: X1 = X2 =X = 0.2(13.2/13.8) (300/100)=0.2745 pu
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Symmetrical Components
Dr. Mohammad Abdul Mannan Page 73 of 90 AIUB/EEE
X0=0.05(13.2/13.8)(300/100)=0.0686 pu, Zbase = (13.2)2/300 = 0.635 3Xn=(3 0.4)/0.635= 1.89 pu For M2: X1 = X2 =X = 0.2(13.2/13.8) (300/200)=0.549 pu X0=0.05(13.2/13.8)(300/200)=0.1372 pu,
For T1: 0857.03503002
2302301.0'' ==X pu
For T2: 0915.03003002
13.813.21.0'' =
=X pu
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Dr.
Mohammmad Abddul Mann
Symm
nan
metrical C
Componen
Pa
nts
age 74 off 90
AIUB/EEEE
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Symmetrical Components
Dr. Mohammad Abdul Mannan Page 75 of 90 AIUB/EEE
Example 11.4.1: The one line diagram of a power system with the reactances of the two sections of the transmission line is shown in the figure. The generators and the transformers are rated as follows: G1: 50 MVA, 13.8 kV, X1 = 0.2 pu, X2 = 0.15 pu, X0 = 0.05 pu, Xn = 0.2pu G2: 30 MVA, 18 kV, X1 = 0.2 pu, X2 = 0.15 pu, X0 = 0.08 pu. G3: 30 MVA, 20 kV, X1 = 0.2 pu, X2 = 0.15 pu, X0 = 0.1 pu, Xn = 0.04 pu. T1: 25 MVA, 220-Y/13.8- kV, X=10% T2: Single phase units each rated 10 MVA, 127/18 kV, X=10% T3: 35 MVA, 220-Y/22-Y kV, X=10% Draw the positive, negative and zero sequence network with all the values marked in per unit. Choose a base of 50 MVA, 13.0 kV in the circuit of generator 1.
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Dr.
Mohammmad Abddul Mann
Symm
nan
metrical C
Componen
Pa
nts
age 76 off 90 AIUB/E
EEE
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Symmetrical Components
Dr. Mohammad Abdul Mannan Page 77 of 90 AIUB/EEE
New Base: 50 MVA, 13.0 kV in the circuit of generator 1 Transformer T2: Given single phase rating: 10 MVA, 127/18 kV, since T2 is Y- connected the three phase rating: 310=30 MVA and (3127)/18 kV or 220/18 kV, Region # 1: Given, Vbase = 13.0 kV; Generator 1 Region # 2: Vbase =(220/13.8) 13 = 207.25 kV; Transmission Lines Line 1 and Line 2 Region # 3: Vbase =(18/220) 207.25 = 16.96 kV; Generator 2 Region # 4: Vbase =(22/220) 207.25 = 20.725 kV; Generator 3 For G1:
pu062.113
8.13)(
11 ===
nbaseVgE
gE
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Symmetrical Components
Dr. Mohammad Abdul Mannan Page 78 of 90 AIUB/EEE
pu0.563420502
138.132.01_1 =
=gX
pu0.442620502
138.1315.021 =
=gX
pu0.140920502
138.1305.001 =
=gX
pu0.563420502
138.132.01 =
=ngX
pu1.69025634.0313 ==ngX For Line 1 and Line 2:
=== 859.0512 50
2)25.207(2
baseSbaseVbaseZ
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Symmetrical Components
Dr. Mohammad Abdul Mannan Page 79 of 90 AIUB/EEE
pu 0.0466859.0512
401)1(2)1(1 ====baseZX
LXLX
pu0.1397859.0512
1200)1(0 ===baseZX
LX
pu 0.0582859.0512
501)2(2)2(1 ====baseZX
LXLX
pu0.1746859.0512
1500)2(0 ===baseZX
LX
For G2:
pu062.196.16
1822 ===
baseVgE
gE
pu0.375530502
96.16182.012 =
=gX
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Symmetrical Components
Dr. Mohammad Abdul Mannan Page 80 of 90 AIUB/EEE
pu0.281630502
96.161815.022 =
=gX
pu0.150230502
96.161808.002 =
=gX
For G3:
pu962.0725.20
20)(
33 ===
nbaseVgE
gE
pu0.310430502
725.20202.013 =
=gX
pu0.232830502
725.202015.023 =
=gX
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Symmetrical Components
Dr. Mohammad Abdul Mannan Page 81 of 90 AIUB/EEE
pu0.155230502
725.20201.003 =
=gX
pu0.022430502
725.202004.03 =
=ngX
pu0.06720224.0333 ==ngX For T1:
pu0.225425502
138.131.0012111 =
=== TXTXTX For T2:
pu0.187730502
96.16181.0022212 =
=== TXTXTX
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Dr.
For
TX
Mohamm
r T3:
13 = XT
mad Abd
23 =TX
dul Mann
3= TX
Symm
nan
1.00 =
metrical C
72.2022
Componen
Pa
35502
252
nts
age 82 of
0.16150 =
f 90
pu1
AIUB/EEEE
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Dr. Mohammmad Abddul Mann
Symm
nan
metrical C
Componen
Pa
nts
age 83 off 90 AIUB/E
EEE
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Dr.
ProfiguG1G2M1M2T1: T2:
Mohamm
oblem 1ure. The: 200 M: 100 M
1: 100 M2: 80 MV
350 MV300 MV
mad Abd
1: The one genera
MVA, 16MVA, 16MVA, 25VA, 25 KVA, 20/VA, 25
dul Mann
ne line dators, the KV, X1 KV, X1
5 KV, XKV, X1/275 KV/275 KV
Symm
nan
diagrame motors1 = 0.2 p1 = 0.2 p
X1 = 0.2 p= 0.2 pu
V, X=10%V, X=10
Figur
metrical C
m of a pos and thepu, X2 =pu, X2 =pu, X2 =u, X2 = % 0%
re for P
Componen
Pa
ower syse transfo
= 0.15 pu= 0.15 pu= 0.15 pu0.15 pu
Problem
nts
age 84 of
stem is sormers au, X0 = 0u, X0 = 0u, X0 = , X0 = 0
m 1
f 90
shown inare rated0.05 pu,0.15 pu,0.25 pu
0.05 pu.
AIUB/E
n the fold as foll, Xn = 0, Xn = 0
u, Xn = 0
EEE
llowing ows: .2. .5.
0.32.
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Dr.
Theseq(5 +per Profigu
Mohamm
e transmquence im+ j50) or unit.
oblem 2ure. The
mad Abd
mission lmpedanhms. Dr
2: The one genera
dul Mann
line hasnce of (2raw the
ne line dators, the
Symm
nan
s the pos2.5 + j25zero seq
diagrame motors
Figur
metrical C
sitive an5) ohms quence n
m of a pos and the
re for P
Componen
Pa
nd negat and th
network
ower syse transfo
Problem
nts
age 85 of
tive seqhe zero sk with al
stem is sormers a
m 2
f 90
quence imsequencell the va
shown inare rated
AIUB/E
mpedane impedalues ma
n the fold as foll
EEE
nce zero dance of arked in
llowing ows:
f
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Symmetrical Components
Dr. Mohammad Abdul Mannan Page 86 of 90 AIUB/EEE
Draw the zero sequence network with all the values marked in per unit.
Gen MVA Rating
Voltage Rating (kV)
Xd pu
X2 pu
X0 pu
G1 100 15 0.18 0.18 0.07 G2 100 15 0.20 0.20 0.10 G3 100 15 0.15 0.15 0.05 G4 100 15 0.3 0.4 0.10 T1 100
(base) 15/765 0.1 0.1 0.1
T2 100 (base)
15/765 0.1 0.1 0.1
T3 500 15/765 0.12 0.12 0.12 T4 750 15/765 0.11 0.11 0.11
Line-Line
X1=X2 Ohm
X0 Ohm
1-2 50 150 1-3 40 100 2-3 40 100
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Dr.
Profiguline
Dta
Mohamm
oblem 3ure. Thee diagram
Draw theaking th
mad Abd
3: The one generam.
e zero se base 2
dul Mann
ne line dators, th
sequenc200MAV
Symm
nan
diagramhe moto
Figu
ce netwoV and 26
metrical C
m of a poors and
ure for P
orks wit68 kV o
Componen
Pa
ower systhe tran
Problem
th all thon transm
nts
age 87 of
stem is snsformer
m 3
he valuemission
f 90
shown inrs are in
es markline.
AIUB/E
n the folncluded
ked in p
EEE
llowing d in one
per unit
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Dr.
Pth
Mohamm
Problemhe follow
mad Abd
m 4: Drawing on
dul Mann
aw the pne-line d
Symm
nan
positive-diagram.
metrical C
, negati
Componen
Pa
ve -, an
nts
age 88 of
nd zero s
f 90
sequenc
AIUB/E
ce netwo
EEE
orks for
-
Dr. Mohammmad Abddul Mann
Symm
nan
metrical C
Componen
Pa
nts
age 89 off 90 AIUB/E
EEE
-
Dr. Mohammmad Abddul Mann
Symm
nan
metrical C
Componen
Pa
nts
age 90 off 90 AIUB/E
EEE