power system analysis lab 5

8/16/2019 Power System Analysis Lab 5

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David Mubayiwa12596724

Power System Analysis

(301/603) Name: David Mubayiwa

Student number: 12596724

Title of the e!eriment: "ault #evel $al%ulation& '&in(

)MTD$*+S$,D

#aboratory (rou!: Monday 1-:../15:..

#aboratory &u!ervi&or:

#aboratory !artner&: 0ia& handari

Date !erformed: 13th May 2.15

Due date: 26th

May 2.15

Date &ubmitted: 26th May 2.15

I hereby declare that this report is entirely my own work and has not been copied from any

other student or past student. Font: Times New Roman 12 Italic

Student &i(nature: //////////////////////////////////////////////////////////


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Introduction This experiment is about to investiate how to use a commercia! so"tware

proram to ca!cu!ate "au!t !eve!s and extended app!ications#

#

$ims and %b&ectives

The aim o" this !aboratory was to demonstrate student how to use o"

'()$D so"tware proram to ca!cu!ate "au!t !eve!s and extended

app!ications#


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$pparatus '()$D so"tware

'rinter

Method1# *sin the system data iven in Tab!e 1 and the impedance diarams

shown in +iures 2,4- compute the positive- neative and .ero se/uence 0,

bus matrices#

2# Invert the 0,bus matrices to obtain the ,bus matrices and the Thevenin

impedances "or each bus#

# 3un the '()$D proram#

4# )hoose the "au!t !ocation- type- start time and duration as per theinstructor iven va!ues#

5# )a!cu!ate the rms "au!t current se/uence components#

Theory and )ircuits

Figure 1 4-bus power system

Table 1 4 Bus System Data


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Figure 2 Positive Sequence Diagram

Figure 3 egative Sequence Diagram

Figure 4 !ero Sequence Diagram


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Figure " Fault location# $ault type# $ault start time an% $ault %uration selectors in PS&'D

3esu!ts

o +au!t 3esu!ts

Figure ( o Fault &urrent

Figure ) o Fault *oltages


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'hase $ to round 3esu!ts

Figure + P,ase ' to roun% &urrent

Figure . P,ase ' to roun% *oltages

'hase $ to round 3esu!ts

Figure 1/ P,ase 'B to roun% &urrent


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Figure 11 P,ase 'B to roun% *oltage

'hase $) to round

Figure 12 P,ase 'B& to roun% &urrents

Figure 13 P,ase 'B& to roun% *oltages

'hase $ 3esu!ts


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Figure 14 P,ase 'B &urrents

Figure 1" P,ase 'B *oltages

'hase $) 3esu!ts

Figure 1( P,ase 'B& &urrents


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Figure 1) P,ase 'B& *oltages

Discussion,%%ordin( to the data li&ted in lab &heet when bu& - i& fault we %an obtain:

Z TH (1)=Z TH (2)= j0.15 pu

Z TH (0)= j 0.1 pu

"or &in(le line to (round fault

I a1= I a2= I a0= Ea

Z TH (1 )+Z TH (2 )+Z TH (0 )=

1

j (0.15+0.15+0.1)=− j 2.5 pu

[

I a I b

I c

]=

[

1 1 1

1 a2

a

1 a a

2

] [

I a 0 I a 1

I a 2

]=

[

1 1 1

1 a2

a

1 a a

2

].

[

− j 2.5− j 2.5

− j2.5

]=

[

− j 7.50

0

] pu


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V a 0V a 1V a 2

=[0

1

0]−[

j 0.1 0 0

0 j 0.15 0

0 0 j 0.15] I a 0 I a 1 I a 2

=[ −0.25

0.625

−0.375] puSo

V agV bgV cg

=[1 1 1

1 a2

a

1 a a2][−0.250.6250.375 ]=[

0

0.944∠−113.4 °0.944∠113.4 ° ] pu

"or double line to (round fault

I a1= Ea

Z TH (1 )+Z TH (2) ∙ Z TH ( 0 )

Z TH (2 )+Z TH (0 )

= 1

j 0.15+ j 0.015

j0.25

=− j 4.76 pu

V a1=V a2=V a0=V a

3= Ea− I a1 ∙ Z TH (1)=1−(− j4.76) ( j 0.15 )=0.286 pu

I a 0=−V a 0

Z TH (0 )= j 2.86 pu I a2=

−V a 2

Z TH (2 )=

0.286

j 0.15= j 1.9 pu

[ I a I b I c ]=

[1 1 1

1 a2

a

1 a a2] [

I a 0 I a 1 I a 2]

=

[1 1 1

1 a2

a

1 a a2]

.

[ j

2.857

− j 4.762

j 1.905 ]=

[ 0

7.190∠−143.4 °

7.190∠36.6 ° ] pu

V agV bgV cg

=[1 1 1

1 a2

a

1 a a2] .

V a 0V a 1V a 2

=[1 1 1

1 a2

a

1 a a2] .[0.2860.2860.286]=[

0.858

0

0 ] pu

"or line to line fault


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I a1= Ea

Z th1 +Z th

2 =

1

j 0.15+ j 0.15=− j 3.333 pu

I a 0=0 I a 2=− I a 1= j 3.33 pu

[ I a I b I c

]=[1 1 1

1 a2

a

1 a a2] [

I a 0 I a 1 I a 2

]=[1 1 1

1 a2

a

1 a a2] [ 0−3.333.33 ]=[

0

−5.775.77

] pu

V a0V a1

V a2

=[0

1

0]−[

j 0.1 0 0

0 j0.15 0

0 0 j 0.15] I a0 I a1

I a2

=[ 0

0.5005

0.4995] pu

,nd

V agV bgV cg

=[1 1 1

1 a2

a

1 a a2][ 00.50050.4995 ]=[

1

−0.5

−0.5] pu

n thi& e!erien%e the fault analy&i& i& analy&ed by +S$,D &oftware in numeri%al analy&i&

a!!lied to !ower &y&tem8 +S$,D &how& the waveform& for variou& level& of fault ty!e in a

&u((e&ted bu& and !ha&e8 Different fault ty!e !rodu%e different waveform&8

Q1

The %al%ulated re&ult& &how that the mo&t &evere fault ty!e i& the &in(le line to (round fault8

The %urrent throu(h !ha&e , i& 785!u whi%h i& 19869, %om!arin( to other two ty!e& of

fault& line to line fault and double line to (round fault8

Q2


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f for &in(le line to (round fault i& 785!u8

f for line to line fault i& 7819!u8

f for double line to (round fault i& .835!u8

Q3

I a1= Ea

Z TH (1)+Z TH (0 )∥Z TH (2 )

I a 0=−V a 0

Z TH (0 )

I a

2= −V a2

Z TH (2 )

Z TH (1 )=Z TH (1 )∥Z 441= j 0.15 ∥( j 0.4 × 10050 + j 0.12)= j 0.1289

Z TH (2 )=Z TH (1 )∥ Z 442= j 0.15 ∥( j 0.45× 10050 + j 0.12)= j 0.132

Z TH (0 )= j 0.07

V a1=V a2=V a0= Ea− I a1×Z TH (1 )=0.259 pu

{ I a 1=− j 5.7

I a 2= j 2

I a 0= j 3.71

Thu&

[ I a I b I c

]=[1 1 1

1 a2

a

1 a a2] [

I a 0 I a 1 I a 2

]=1 1 1

1 a2

a

1 a a2[

j 3.71

− j 5.7

j 2 ]=[ 0

8.7∠140.28 °

8.7∠39.72° ] pu

[V agV bgV cg

]=[1 1 1

1 a2

a

1 a a2] .[

V a 0V a 1V a 2

]=[1 1 1

1 a2

a

1 a a2] .[0.2590.2590.259]=[

0.778

0

0 ] pu


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)onc!usions To conc!ude- the "au!t ana!ysis is an important too! app!ied to power

system by comparin the vo!tae and current wave"orm when the

dierence "au!ts happened at dierence bus# (tudents in this experiment

ana!yse the power 8ow and ca!cu!ate "au!t by usin '()$D proram- thusthis !aboratory is success"u!#

power system analysis lab 5

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