01F 03 Symmetrical Components Lectures

Dr.

Theof e Cabanabscirccon

Mohamm

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Dr.

Cacan

Mohamm

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age 66 of

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zero-se

Zero-seq

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quence

equence

quence

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current

Dr.

Cacurcirc

Mohamm

se 5: -rrent, noculate w

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Zersepas s

Mohamm

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system network

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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

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20 Oa small

and ng twork.

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One-line l power

the zero-

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

Dr.

ExanetwG1:TraT1:T2:M1M2Sel

Mohamm

ample 1work, an: 300 MVansmiss 350 MV 300 MV: 200 M: 100 M

lect the g

mad Abd

11.4: Drnd (iii) zVA, 20 sion lineVA, 230VA, 220

MVA, 13MVA, 13

generato

dul Mann

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

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uence nek for the %, Xn= 0X2=0.5

5%, Xn= 5% generato

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age 71 of

etwork, system .4 /km, X0=

0.4 or circui

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(ii) negshown

=1.5 /

it.

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gative-sein the fi

/km.

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equence igure:



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



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

Dr.

Mohammmad Abddul Mann

Symm

nan

metrical C

Componen

Pa

nts

age 74 off 90

AIUB/EEEE



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.

Dr.

Mohammmad Abddul Mann

Symm

nan

metrical C

Componen

Pa

nts

age 76 off 90 AIUB/E

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



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



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



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



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

Dr.

For

TX

Mohamm

r T3:

13 = XT

mad Abd

23 =TX

dul Mann

3= TX

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nan

1.00 =

metrical C

72.2022

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252

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0.16150 =

f 90

pu1

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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

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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.

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n the fold as foll, Xn = 0, Xn = 0

u, Xn = 0

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llowing ows: .2. .5.

0.32.

Dr.

Theseq(5 +per Profigu

Mohamm

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mission lmpedanhms. Dr

2: The one genera

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ne line dators, the

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diagrame motors

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metrical C

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m of a pos and the

re for P

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nd negat and th

network

ower syse transfo

Problem

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age 85 of

tive seqhe zero sk with al

stem is sormers a

m 2

f 90

quence imsequencell the va

shown inare rated

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mpedane impedalues ma

n the fold as foll

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nce zero dance of arked in

llowing ows:

f



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

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.

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n the folncluded

ked in p

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llowing d in one

per unit

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

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age 88 of

nd zero s

f 90

sequenc

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ce netwo

EEE

orks for


Symm

nan

metrical C

Componen

Pa

nts


EEE

Documents

01F 03 Symmetrical Components Lectures