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Scilab Textbook Companion for

Switchgear Protection And Power Systems

by S. S. Rao1

Created byMayank Gupta

BEElectrical Engineering

Thapar UniversityCollege Teacher

Dr. Sunil Kumar SinglaCross-Checked by

Lavitha Pereira

November 13, 2013

1Funded by a grant from the National Mission on Education through ICT,http://spoken-tutorial.org/NMEICT-Intro. This Textbook Companion and Scilabcodes written in it can be downloaded from the Textbook Companion Projectsection at the website http://scilab.in


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Book Description

Title: Switchgear Protection And Power Systems

Author: S. S. Rao

Publisher: Khanna Publisher, New Delhi

Edition: 13

Year: 2012

ISBN: 8174092323

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Scilab numbering policy used in this document and the relation to theabove book.

Exa Example (Solved example)

Eqn Equation (Particular equation of the above book)

AP Appendix to Example(Scilab Code that is an Appednix to a particularExample of the above book)

For example, Exa 3.51 means solved example 3.51 of this book. Sec 2.3 meansa scilab code whose theory is explained in Section 2.3 of the book.

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Contents

List of Scilab Codes 5

3 Fundamentals of Fault Clearing and Switching Phenomena 10

17 Electrical Substations and Equipments and Busbar Layouts 17

18 Neutral Grounding or Earthing 18

19 Introduction to Fault Calculations 21

20 Symmetric Faults and Current Limiting Reactors 26

21 Symmetric Components 48

22 Unsymmetrical Faults on Unloaded Generator 57

23 Faults On Power Systems 69

32 Protection of transformers 78

33 Protection of Generators 80

35 Current Transformers and their Applications 83

36 Voltage Transformer and their Application 86

44 Power System Stability and Auto Reclosing Schemes 87

45 Voltage Control and Compensation of ReacTve Power 93

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46 Economic operation of Power Systems 97

57 Power Flow Calculations 100

58 Applications of switchgear 108

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List of Scilab Codes

Exa 3.1 To find the transient current of RL circuit . . . . . . . 10Exa 3.2 to find the DC component and instantaneous value of

currents and voltages . . . . . . . . . . . . . . . . . . 10Exa 3.3 To find Max Rate of restriking voltage and time for

RRRV and the frequency . . . . . . . . . . . . . . . . 12Exa 3.4 To find the peak striking voltage and its frequency and

the avg of RRRV and its max rate . . . . . . . . . . . 13Exa 3.5 The average rate of rise of restriking voltage. . . . . . 13Exa 3.6 To estimate the average rate of restriking voltage . . . 14Exa 3.7 to find the peak striking voltage and the time to reach it 15Exa 3.8 To find the value of resistance to be used across the

contact space . . . . . . . . . . . . . . . . . . . . . . . 15Exa 17.1 to find the min force on the conductors . . . . . . . . 17

Exa 18.1 To calculate the ohmic value of impedence. . . . . . . 18Exa 18.2 to find the value of reactance . . . . . . . . . . . . . . 18Exa 18.3 calculate the reactance to neutralize different value of

line capacitance . . . . . . . . . . . . . . . . . . . . . 19Exa 18.4 To find the inductance and the KVA rating . . . . . . 19Exa 19.1 expressing the quantities in per unit form . . . . . . . 21Exa 19.2 conversion in per unit . . . . . . . . . . . . . . . . . . 21Exa 19.3 to find the new pu reactance . . . . . . . . . . . . . . 22Exa 19.4 drawing the reactance diagram of the system . . . . . 22Exa 19.5 to find the fault current . . . . . . . . . . . . . . . . . 23Exa 19.6 The reactance calculations. . . . . . . . . . . . . . . . 23

Exa 19.7 to find the pu impedences . . . . . . . . . . . . . . . . 24Exa 19.9 To calculate the new fault level . . . . . . . . . . . . . 25Exa 20.1 Calculate Fault MVA and current . . . . . . . . . . . 26Exa 20.2 To find the steady state fault current. . . . . . . . . . 27

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Exa 35.02 Calculate the effective burden of the current transformer 83

Exa 35.03 To find out the flux density of core . . . . . . . . . . . 84Exa 35.04 To calculate the ratio error of CT. . . . . . . . . . . . 84Exa 36.03 To calculate the VA of the output of voltage transformer 86Exa 44.01 To calculate max possible power transfer through the

transmission line . . . . . . . . . . . . . . . . . . . . . 87Exa 44.02 To calculate max possible power transfer through the

transmission line . . . . . . . . . . . . . . . . . . . . . 87Exa 44.03 To calculate the steady state limit . . . . . . . . . . . 88Exa 44.04.aTo determine the Inertia Constants and Angular Mo-

mentum . . . . . . . . . . . . . . . . . . . . . . . . . . 88Exa 44.04 To calculate the kinetic energy of rotor. . . . . . . . . 89

Exa 44.05 To find the stored energy and angular acceleration . . 89Exa 44.06 To calculate the Angular momentum and acceleration

of rotor . . . . . . . . . . . . . . . . . . . . . . . . . . 90Exa 44.07 To calculate the power and increase in the shaft power 90Exa 44.08 To calculate the critical clearing angle . . . . . . . . . 91Exa 45.B.2 To find the overall power factor of the sub station. . . 93Exa 45.B.3 Calculate the KVAr required of capacitor . . . . . . . 94Exa 45.B.4 Calculate the economical pf . . . . . . . . . . . . . . . 94Exa 45.B.5 Calculate the most economical pf . . . . . . . . . . . . 94Exa 45.B.6 Calculate the kW and power factor of substation . . . 95

Exa 45.01 To find the power factor and KVA . . . . . . . . . . . 96Exa 46.01 To determine the load allocation of various units . . . 97Exa 46.02 To calculate the load distribution on basis of economic

loading . . . . . . . . . . . . . . . . . . . . . . . . . . 98Exa 46.03 Comparison of Economic and Equal loading . . . . . . 99Exa 57.01 To find the branch current and branch admittance . . 100Exa 57.02 To find the admittance of the circuit . . . . . . . . . . 100Exa 57.04 To find the Voltage of the circuit . . . . . . . . . . . . 101Exa 57.05 To calculate power angle between source and load voltage 101Exa 57.06 Reactive and complex power flow . . . . . . . . . . . . 102Exa 57.07 To calculate the pu active power flow. . . . . . . . . . 102

Exa 57.08 sending end voltage and average reactive power flow . 102Exa 57.09 To calculate the complex and real power of the system 103Exa 57.11 Determine the voltage and phase angle at bus 2 by gauss

seidal method. . . . . . . . . . . . . . . . . . . . . . . 103Exa 57.12 to determine the modified bus voltage . . . . . . . . . 104

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Exa 57.13 To calculate the voltage of bus 2 by NR method . . . 104

Exa 57.14 to calculate the power flows and line losses . . . . . . 105Exa 57.15 To find the sending end power and DC voltage . . . . 106Exa 57.16 to calculate the power flow of given line . . . . . . . . 107Exa 57.17 To calculate the power flow through the lines . . . . . 107Exa 58.02 To find the over current factor . . . . . . . . . . . . . 108

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Chapter 3

Fundamentals of Fault Clearing

and Switching Phenomena

Scilab code Exa 3.1 To find the transient current of RL circuit

1

2 clear ;

3 close ;

4 clc ;

5 R = 1 0 ;

6 L = 0 . 1 ;

7 f = 5 0 ;

8 w = 2 * % p i * f ;

9 k = sqrt ( ( R ^ 2 ) + ( ( w * L ) ^ 2 ) ) ;

10 a n g l e = atan ( w * L / R ) ;

11 E = 4 0 0

12 A = E * sin ( a n g l e ) / k ;

13 i = A * exp ( ( - R ) * . 0 2 / L ) ;

14 i = round ( i * 1 0 0 ) / 1 0 0 ;

15 mprintf ( t h e t r a n s i e n t c u r r e n t =%fA ,i )

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Scilab code Exa 3.2 to find the DC component and instantaneous value

of currents and voltages

1

2 clear ;

3 close ;

4 clc ;

5

6 R = 1 0 ;

7 L = 0 . 1 ;

8 f = 5 0 ;

9 w = 2 * % p i * f ;

10 k = sqrt ( ( R ^ 2 ) + ( ( w * L ) ^ 2 ) ) ;11 a n g l e = atan ( w * L / R ) ;

12 E = 1 0 0 ;

13 Em = sqrt ( 2 ) * E ;

14 A = E m * sin ( a n g l e ) / k ;

15 i 1 = A ;

16 Em = round ( E m * 1 0 ) / 1 0 ;

17 i1 = round ( i 1 * 1 0 ) / 1 0 ;

18 mprintf ( c u r r e n t i n a mp er es f o r p a r t 1=%fA\n , i 1 ) ;19 mprintf ( c u r r en t i n p a rt 2& p a rt 3= 0\n ) ;20 mprintf (

t h e DC c om po ne nt v a n i s h e s i f e=%fV, E m ) ;

//t h e e r r o r i s due t o t h e e rr on e o us v a l ue s i n t h et e x t b o o k

21

22 t 1 = 0 . 5 * . 0 2 ;

23 i 2 = A * exp ( ( - R ) * t 1 / L ) ;

24 mprintf ( \ n c ur r en t a t . 5 c y c l e s f o r t 1=% fs ec \n c u r r e n t i n t h e p ro bl em = %fA , t 1 , i 2 ) ;

25 t 2 = 1 . 5 * . 0 2 ;

26 i 3 = A * exp ( ( - R ) * t 2 / L ) ;

27 mprintf ( \ n c ur r en t a t 1 . 5 c y c l e s f o r t 2=% fs ec \

n c u r r e n t i n t h e p ro bl em = %fA , t 2 , i 3 ) ;28 t 3 = 5 . 5 * . 0 2 ;

29 i 4 = A * exp ( ( - R ) * t 3 / L ) ;

30 mprintf ( \ n c ur r en t a t 5 . 5 c y c l e s f o r t 3=% fs ec \n c u r r e n t i n t h e p ro bl em = %fA , t 3 , i 4 ) ;

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31

3233 disp ( t h e d i f f e r e n c e i n r e s u l t i s due t o e rr on e o us

v a l u e i n t e x tb o o k . )

Scilab code Exa 3.3 To find Max Rate of restriking voltage and time forRRRV and the frequency

1 clear ;

2 close ;3 clc ;

4 C = . 0 0 3 e - 6

5 L = 1 . 6 e - 3

6 y = sqrt ( L * C ) ;

7 y = round ( y * 1 e 7 ) / 1 e 7 ;

8 f = ( 2 * 3 . 1 4 * y ) ^ - 1 ;

9 f = round ( f / 1 0 0 ) * 1 0 0 ;

10 i = 7 5 0 0 ;

11 E = i * 2 * 3 . 1 5 * L * 5 0 ;

12 E m = 1 . 4 1 4 * E ;

13 Em = round ( E m / 1 0 ) * 1 014 t = y * % p i / 2 ;

15 t = t * 1 e 6 ;

16 t = round ( t * 1 0 0 ) / 1 0 0 ;

17 e = E m / y ;

18 e = round ( ( e ) / 1 e 6 ) * 1 e 6 ;

19 e = fix ( e / 1 e 7 ) * 1 e 7

20 mprintf ( fr eq ue nc y of o s c i l l a t i o n s=%fc/ s , f ) ;21 mprintf ( \ n ti me o f maximum r e s t r i k i n g v o l t a g e =

% f m i c r o s e c , t ) ;

22 mprintf ( \nmaximum r e s t r i k i n g v o l t a g e =%dV/ m i c r o s e c s , e / 1 e 6 ) ;

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Scilab code Exa 3.4 To find the peak striking voltage and its frequency

and the avg of RRRV and its max rate

1

2 clear ;

3 close ;

4 clc ;

5 R =5

6 f = 5 0

7 L = R / ( 2 * % p i * f ) ;

8 V = 1 1 e 3 ;

9 V p h = 1 1 / sqrt ( 3 ) ;

10 C = 0 . 0 1 d - 6 ;11 y = sqrt ( L * C ) ;

12 Em = sqrt ( 2 ) * V p h ;

13 e p = 2 * E m ;

14 ep = round ( e p * 1 0 ) / 1 0 ;

15 y = round ( y * 1 e 7 ) / 1 e 7 ;

16 t = y * % p i ;

17 t = fix ( t * 1 e 7 ) / 1 e 7

18 e a = e p / t ;

19 ea = round ( e a / 1 e 3 ) * 1 e 3

20 f n = ( 2 * 3 . 1 4 * y ) ^ - 1 ;

21 Em = round ( E m )

22 E m a x = E m / y ;

23 E m a x = round ( E m a x / 1 0 0 0 ) * 1 e 3 ;

24 mprintf ( p e ak r e s t r i k i n g v o l t a g e =%dkV , e p ) ;25 printf ( \ nf re qu en cy of o s c i l l a t i o n s=%dc/ s , f n ) ;26 printf ( \ n a ve ra g e r a t e o f r e s t r i k i n g v o l t a g e=%fkV/

m i c r o s e c s , e a / 1 e 6 ) ;27 printf ( \nmax r e s t r i k i n g v o l t a g e =%dV/ m i c r o s e c s , E m a x

/ 1 e 3 ) ;

Scilab code Exa 3.5 The average rate of rise of restriking voltage

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1

2 clear ;3 close ;

4 clc ;

5 E = 1 9 . 1 * 1 e 3 ;

6 L = 1 0 * 1 e - 3 ;

7 C = . 0 2 * 1 e - 6 ;

8 Em = sqrt ( 2 ) * E ;

9 y = sqrt ( L * C ) ;

10 t = % p i * y * 1 e 6 ;

11 e m a x = 2 * E m ;

12 e a v g = e m a x / t ;

13 e a v g = round ( e a v g / 1 0 ) * 1 014 printf ( a v e r a g e r e s t r i k i n g v o l t a g e =%dV/ m i c r o s e c s ,

e a v g ) ;

Scilab code Exa 3.6 To estimate the average rate of restriking voltage

1 clear ;

2 close ;

3 clc ;4 V = 7 8 e 3 ;

5 V p h = V / sqrt ( 3 ) ;

6 E m = 2 * V p h ;

7 p f = 0 . 4 ;

8 a n g l e = acos ( p f ) ;

9 k1 = sin ( a n g l e ) ;

10 k1 = round ( k 1 * 1 0 0 ) / 1 0 0 ;

11 k 2 = . 9 5 1 ;

12 k 3 = 1 ;

13 k = k 1 * k 2 * k 3 ;14 k = round ( k * 1 0 0 0 ) / 1 e 3 ;

15 E = k * E m ;

16 f = 1 5 0 0 0 ;

17 t = 1 / ( 2 * f ) ;

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18 t = round ( t * 1 e 6 ) ;

19 e a v g = 2 * E / t ;20 e a v g = round ( e a v g / 1 0 0 ) * 1 0 0 ;

21 printf ( a v e r a g e r e s t r i k i n g v o l t a g e =%fkV/ m i c r o s e c s ,e a v g / 1 e 3 ) ;

Scilab code Exa 3.7 to find the peak striking voltage and the time to reachit

1 clear ;2 clc ;

3 E m = 1 0 0 e 3

4 t = 7 0 e - 6

5 E a = E m / t / 1 e 6

6 f = 1 / ( 2 * t ) ;

7 Ea = round ( E a / 1 0 ) * 1 0 ;

8 f = round ( f ) ;

9 printf ( a v e r a g e v o l t a g e i n v o l t s =%dV/ m i c r o s e c s\n , Ea) ;

10 printf ( f r e q ue nc y of o s c i l l a t i o n =%dc /s , f ) ;

Scilab code Exa 3.8 To find the value of resistance to be used across thecontact space

1

2 clc ;

3 L = 6 ;

4 C = 0 . 0 1 e - 6 ;

5 i = 1 0 ;6 v = i * sqrt ( L / C ) ;

7 R = . 5 * v / i ;

8 R = round ( R / 1 0 ) * 1 0 ;

9 printf ( d am pi ng r e s i s t a n c e i n ohms=%fkohms , R / 1 e 3 ) ;

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Chapter 17

Electrical Substations and

Equipments and Busbar

Layouts

Scilab code Exa 17.1 to find the min force on the conductors

1 clear ;

2 clc ;

3 I s c = 2 5 e3 ;

4 i = 2 . 5 5 * I s c ;

5 L = 1 ;

6 r = 0 . 2 4 ;

7 F = 2 . 0 4 6 * ( i ^ 2 ) * 1 0 ^ - 5 / r ;

8 mprintf ( t h e f o r c e on b us ba r p er m et er l e n g t h =%dk g f p e r m et er , F / 1 e 3 ) ;

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Chapter 18

Neutral Grounding or Earthing

Scilab code Exa 18.1 To calculate the ohmic value of impedence

1 clc ;

2 clear ;

3 P = 2 0 0 0 e 3 ;

4 V = 4 0 0 ;

5 r = . 4 ;

6 z = V ^ 2 / ( r * P ) ;

7 mprintf ( t h e v a l u e o f z=%f ohm , z ) ;

Scilab code Exa 18.2 to find the value of reactance

1 clc ;

2 clear ;

3 w = 3 1 4 ;

4 c = . 0 1 5 e - 6 ;

5 l = 1 / ( 3 * w ^ 2 * c ) ; // t h e d i f f e r e n c e i n r e s u l t i s due t oe r ro n eo u s c a l c u l a t i o n i n t ex tb oo k .

6 l = round ( l * 1 0 ) / 1 0 ;

7 mprintf ( i n d u c t a n c e =%f H e n r i es , l ) ;

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8 disp ( t h e d i f f e r e n c e i n r e s u l t i s due t o e rr on e o us

c a l c u l a t i o n i n t ex tb o ok . )

Scilab code Exa 18.3 calculate the reactance to neutralize different valueof line capacitance

1 clc ;

2 clear ;

3 c 1 = 1 . 5 e - 6 ;

4 w = 2 * % p i * 5 0 ;5 L 1 = 1 / ( 3 * c 1 * ( w ^ 2 ) ) ;

6 c 2 = . 9 * c 1 ;

7 L 2 = 1 / ( 3 * c 2 * ( w ^ 2 ) ) ;

8 c 3 = . 9 5 * c 1 ;

9 L 3 = 1 / ( 3 * c 3 * ( w ^ 2 ) ) ;

10 L1 = round ( L 1 * 1 0 0 ) / 1 0 0 ;

11 L2 = round ( L 2 * 1 0 ) / 1 0 ;

12 L3 = round ( L 3 * 1 0 0 ) / 1 0 0 ;

13 mprintf ( t he i nd uc ta nc e f o r 100 p e r c e nt l i n ec a p a c i t a n c e =%f h e n r i e s \n , L 1 ) ;

14 mprintf ( f o r 90 p e r ce n t l i n e c a p ac i ta n c e , t hei n d u c t a n c e =%f h e n r i e s\n , L 2 ) ;

15 mprintf ( f o r 9 5 p e r c e n t l i n e c a p a c i t a n e i n d u c t an c e=%f h e n r i e s , L 3 ) ;

Scilab code Exa 18.4 To find the inductance and the KVA rating

1 clc ;

2 clear ;3 c = . 0 1 e - 6 * 5 0 ;

4 w = 2 * % p i * 5 0 ;

5 L = 1 / ( 3 * c * ( w ^ 2 ) ) ;

6 L = round ( L * 1 0 0 ) / 1 0 0 ;

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7 V = 3 3 e 3 / sqrt ( 3 ) ;

8 I = V / ( w * L ) ;9 I = round ( I * 1 0 0 0 ) / 1 0 0 0 ;

10 I = round ( I * 1 0 0 ) / 1 0 0 ;

11 R = V * I / 1 e 3 ; // t h e d i f f e r e n c e i n r e s u l t i s due t oe r ro n eo u s c a l c u l a t i o n i n t ex tb oo k .

12 mprintf ( t h e v a l u e o f L=%fH a n d r a t i n g =%fkVA , L , R ) ;13 disp ( t h e d i f f e r e n c e i n r e s u l t i s due t o e rr on e o us

c a l c u l a t i o n i n t ex tb o ok . ) ;

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Chapter 19

Introduction to Fault

Calculations

Scilab code Exa 19.1 expressing the quantities in per unit form

1 clc ;

2 clear ;

3 i = 1 0 ;

4 v = 2 0 0 ;

5 z = v / i ;

6 I 1 = 2 0 / i ;

7 I 2 = . 2 / i ;

8 v 1 = 5 0 / v ;

9 r = 2 / z ;

10 mprintf ( t he bas e i mpe de nc e =%dohm\n , z ) ;11 mprintf ( t h e b a s e v a l u e s f o r 2 0A=%dp . u . \ n . t h e b a s e

v a l u e s f o r 2A=%fp . u .\n t he b a s e v a l u e s f o r 5 0V=%fp. u .\ n t h e b a s e v a l u e s f o r 2 ohm=%fp . u , I 1 , I 2 , v 1 , r );

Scilab code Exa 19.2 conversion in per unit

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1 clc ;

2 clear ;3 z = 2 ;

4 v = 1 1 e 3 ;

5 r = 1 0 0 0 e 3 ;

6 z b = v ^ 2 / r ;

7 y = z / z b ;

8 y = round ( y * 1 0 0 0 0 ) / 1 0 0 0 0 ;

9 mprintf ( t h e p e r u n i t r e s i s t a n c e =%fp . u , y ) ;

Scilab code Exa 19.3 to find the new pu reactance

1 clc ;

2 clear ;

3 v = 1 1 e 3 ;

4 r = 1 5 0 0 0 e 3 ;

5 z p = . 1 5 ;

6 v n e w = 1 1 0 e 3 ;

7 r n e w = 3 0 0 0 0 e 3 ;

8 z b = v ^ 2 / r ;

9 Z = z p * z b ;10 z b n e w = v n e w ^ 2 / r n e w ;

11 Z p = Z / z b n e w ;

12 mprintf ( t h e new p e r u n i t r e a c t a n c e =%fp . u , Z p / 1 0 ) ;

Scilab code Exa 19.4 drawing the reactance diagram of the system

1 clc ;

2 clear ;3 v 1 = 1 1 e 3 ;

4 v 2 = 2 2 e 3 ;

5 v 3 = 3 . 3 e 3 ;

6 r = 1 0 0 0 0 e 3 ;

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7 z b 1 = v 1 ^ 2 / r ;

8 z b 2 = v 2 ^ 2 / r ;9 z b 3 = v 3 ^ 2 / r ;

10 z p 1 = 3 0 0 / z b 3 ;

11 z p 2 = 3 0 0 * ( z b 2 / z b 3 ) / z b 2 ;

12 z p 3 = 3 0 0 * ( z b 1 / z b 3 ) / z b 1 ;

13 z p 1 = round ( z p 1 * 1 0 ) / 1 0 ;

14 z p 1 = round ( z p 1 ) ;

15 z p 2 = round ( z p 2 * 1 0 ) / 1 0 ;

16 z p 2 = round ( z p 2 ) ;

17 z p 3 = round ( z p 3 * 1 0 ) / 1 0 ;

18 z p 3 = round ( z p 3 ) ;

19 mprintf ( t h e p e r u n i t v a l u e s =%dp . u . ; %dp . u . ; %dp . u. , z p 1 , z p 2 , z p 3 ) ;

Scilab code Exa 19.5 to find the fault current

1 clc ;

2 clear ;

3 z = 0 . 2 * % i * 0 . 1 5 5 / ( 0 . 2 + 0 . 1 5 5 ) ;

4 v = 1 ;5 i = v / z ;

6 ir = real ( i ) ;

7 im = imag ( i ) ;

8 im = round ( i m * 1 0 0 ) / 1 0 0 ;

9 mprintf ( t h e f a u l t c u r r e n t i s =%d+( % f j )A , i r , i m ) ;

Scilab code Exa 19.6 The reactance calculations

1 clc ;

2 clear ;

3 r = 3 0 0 0 0 e 3 ;

4 v 1 = 1 1 e 3 ;

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5 v 2 = 1 1 0 e 3 ;

6 z b 1 = v 1 ^ 2 / r ;7 z b 2 = v 2 ^ 2 / r ;

8 z p 1 = 8 0 / z b 2 ;

9 z p 2 = . 1 * % i * 3 0 0 0 0 / 3 5 0 0 0 ;

10 z p 3 = . 2 * % i * 3 0 0 0 0 / 1 0 0 0 0 ;

11 z p 3 r = real ( z p 3 ) ;

12 z p 2 r = real ( z p 2 ) ;

13 z p 3 i = imag ( z p 3 ) ;

14 z p 2 i = imag ( z p 2 ) ;

15 z b 2 = round ( z b 2 * 1 0 ) / 1 0 ;

16 z p 1 = round ( z p 1 * 1 0 0 0 ) / 1 0 0 0 ;

17 z p 2 i = round ( z p 2 i * 1 0 0 0 0 ) / 1 0 0 0 0 ;18 z p 3 i = round ( z p 3 i * 1 0 ) / 1 0 ;

19 mprintf ( t he b as e i mpe d enc e o f t r a n sm i s s i on l i n ec i r c u t i = % f o h m\ n pe r u n it r e a ct a n ce o f t r a n sm i s s i on

l i n e=%fp . u . \ n , z b 2 , z p 1 ) ;20 mprintf ( p er u n it r e a ct a n ce o f t r an s f o r m er t o new

ba se=%f+(% fj )p . u. \nPer u n i t r e a ct a n ce o f motor t onew ba se=%f+( %f j ) p . u . , z p 2 r , z p 2 i , z p 3 r , z p 3 i ) ;

Scilab code Exa 19.7 to find the pu impedences

1 clc ;

2 clear ;

3 r 1 = 1 0 e 6 ;

4 r 2 = 7 . 5 e 6 ;

5 r 3 = 5 e 6 ;

6 v 1 = 6 6 e 3 ;

7 v 2 = 1 1 e 3 ;

8 v 3 = 3 . 3 e 3 ;9 z s t = . 0 6 * r 1 * % i / r 2 ;

10 z p s = . 0 7 * % i ;

11 z p t = . 0 9 * % i ;

12 Z p = ( z s t + z p s - z s t ) / 2 ;

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13 Z s = ( z p s + z s t - z p t ) / 2 ;

14 Z t = ( z p t + z s t - z p s ) / 2 ;15 Z p i = imag ( Z p ) ;

16 Z s i = imag ( Z s ) ;

17 Z t i = imag ( Z t ) ;

18 Z p i = round ( Z p i * 1 0 0 ) / 1 0 0 ;

19 mprintf ( t he p er u n i t i m p e d e n cc e o f c i r c u i t \nZp=%f j p . u ;\ n Zs=%fjp . u ;\ n Zt=%fjp . u , Z p i , Z s i , Z t i ) ;

Scilab code Exa 19.9 To calculate the new fault level

1 clc ;

2 clear ;

3 o l d = 5 0 0 0 ;

4 b a n k = 2 0 0 ;

5 n e w = o l d - b a n k ;

6 mprintf ( new f a u l t =%dMVA , n e w ) ;

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Chapter 20

Symmetric Faults and Current

Limiting Reactors

Scilab code Exa 20.1 Calculate Fault MVA and current

1 clear ;

2 clc ;

3 V = 6 . 6 e 3 ;

4 r = 5 e 6 ;

5 X = . 1 2 ;

6 F = r / X ;

7 I = ( F / V ) / ( % i * sqrt ( 3 ) ) ;

8 Ir = real ( I ) ;

9 Ii = imag ( I ) ;

10 I m o d = sqrt ( ( I r ^ 2 ) + ( I i ^ 2 ) ) ;

11 I a n g l e = a t a n d ( I r / I i ) - 9 0 ;

12 F = fix ( F / 1 e 5 ) * 1 e 5 ;

13 I m o d = fix ( I m o d ) ;

14 mprintf ( Method 1 \ nthe va lu e of f a u l t MVA=%fMVA \n

t h e f a u l t c u r r e n t i s = %d / %d A\n , ( F / 1 e 6 ) , I m o d ,I a n g l e ) ;15 / / m et ho d 216 V b a s e = V / sqrt ( 3 ) ;

17 I f a u l t p u = 1 / ( X * % i ) ;

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18 I b a s e = r / ( V b a s e * 3 ) ;

19 I f a u l t = I f a u l t p u * I b a s e ;20 P = sqrt ( 3 ) * I f a u l t * V ;

21 Ir = real ( I f a u l t ) ;

22 Ii = imag ( I f a u l t ) ;

23 I m o d = sqrt ( ( I r ^ 2 ) + ( I i ^ 2 ) ) ;

24 Pr = real ( P ) ;

25 Pi = imag ( P ) ;

26 P m o d = sqrt ( ( P r ^ 2 ) + ( P i ^ 2 ) ) ;

27 P a n g l e = a t a n d ( P r / P i ) - 9 0 ;

28 P m o d = fix ( P m o d / 1 e 5 ) * 1 e 5 ;

29 I m o d = fix ( I m o d ) ;

30 mprintf ( From method 2\n t h e v a l u e o f f a u l t MVA=%f /%d MVA \n t h e f a u l t c u r r e n t i s = %d A , ( P m o d / 1 e 6

) , P a n g l e , I m o d ) ;

31 / / m et ho d 332 v 1 = 6 . 4 e 3 ;

33 I = ( v 1 / V ) / X ;

34 I f a u l t = I b a s e * I ;

35 p = sqrt ( 3 ) * I f a u l t * v 1 ; // t h e d i f f e r e n c e i n r e s u l t i sdue t o e r ro n e o u s c a l c u l a t i o n i n t ex tb oo k .

36 p = round ( p / 1 e 5 ) * 1 e 5 ;

37 mprintf (\ n th e new f a u l t c u r r e nt a t 6 . 4 kV i s = %fA \n t h e n e w f a u lt p ower a t s e r v i c e v o l t a g e i s =%fMVA

, I f a u l t , p / 1 e 6 ) ;38 disp ( t h e d i f f e r e n c e i n r e s u l t i s due t o e rr on e o us


Scilab code Exa 20.2 To find the steady state fault current

1 clear ;2 clc ;

3 V = 3 0 0 0 e 3 ;

4 r 1 = 3 0 ;

5 r = 5 0 0 0 e 3 ;

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6 v b 2 = 1 1 e 3 ;

7 v b 3 = 3 3 e 3 ;8 x = . 2 ;

9 X t = . 0 5 * r / V ;

10 X l = r 1 * r / ( v b 3 ^ 2 ) ;

11 x t o t a l = ( x + X t + X l ) * % i ;

12 M V A = r * % i * 1 e - 6 / x t o t a l ;

13 I f a u l t = M V A * 1 e 6 / ( sqrt ( 3 ) * v b 3 * % i ) ;



16 I m o d = sqrt ( ( I r ^ 2 ) + ( I i ^ 2 ) ) ;


18 I m o d = round ( I m o d ) ;19 M V A = round ( M V A * 1 0 ) / 1 0 ;

20 mprintf ( t h e v a lu e o f f a l u t c u r r e n t = %d/ %d Amp \nf a u l t MVA =%f MVA , I m o d , I a n g l e , M V A ) ;

Scilab code Exa 20.03 to find the fault MVA

1 clear ;

2 clc ;3 r a t i n g = 2 5 e 6 ;

4 v b = 1 1 e 3 ;

5 x = . 1 6 / 4 ;

6 f a u l t M V A = r a t i n g * 1 e - 6 / x ;

7 mprintf ( th e f a u l t MVA f rom method 1=%dMVA , f a u l t M V A) ;

8 / / m et ho d 29 I f a u l t = 1 / ( x * % i ) ;

10 I b = r a t i n g / ( sqrt ( 3 ) * v b ) ;

11 I s c = I b * 2 5 ;12 M V A = sqrt ( 3 ) * v b * I s c / 1 e 6 ;

13 mprintf ( \n th e f a u l t MVA fro m method 2=%dMVA , M V A ) ;

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Scilab code Exa 20.04 calculate the fault current and MVA

1 clear ;

2 clc ;

3 R = 3 e 6 ;

4 R b = 6 0 0 0 e 3 ;

5 v b 1 = 1 1 e 3 ;

6 v b 2 = 2 2 e 3 ;

7 X = . 1 5 ;

8 x = . 1 5 * R b / R ;

9 x e q = x / 2 ;

10 M V A = R b / x e q ;

11 I f a u l t = M V A / ( sqrt ( 3 ) * v b 1 * % i ) ;



14 I m o d = sqrt ( ( I r ^ 2 ) + ( I i ^ 2 ) ) ;


16 I m o d = round ( I m o d / 1 0 ) * 1 0 ;

17 mprintf ( f o r f a u l t on g e n e ra t o r s i d e \n Fa u l t MVA=

%dMVA \n Fa ul t cu rr en t=%d/ %dAmp , M V A / 1 e 6 , I m o d ,I a n g l e ) ;

18 x 2 = . 0 5 ;

19 X e q = x 2 + x e q ;

20 M V A = R b / X e q ;

21 I f a u l t = M V A / ( 1 . 7 3 4 * v b 2 * % i ) ;



24 I m o d = sqrt ( ( I r ^ 2 ) + ( I i ^ 2 ) ) ;


26 mprintf ( \n f o r f a u l t on t r a n s m i s s i o n s i d e \n F a u l tMVA=%dMVA \n F au l t c u r r e n t =%d/ %dAmp( l a g ) , M V A / 1e 6 , I m o d , I a n g l e ) ;

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Scilab code Exa 20.05.a Calculate the Fault MVA and current

1 clear ;

2 clc ;

3 R = 3 e 6 ;

4 R b = 6 e 6 ;

5 v b 2 = 1 1 e 3 ;

6 v b 3 = 6 6 e 3 ;

7 x = . 2 ;

8 X g = x * R b / R ;

9 x t = . 0 5 ;

10 x l = v b 3 ^ 2 / R b ;

11 x l 1 = 2 0 * . 1 / x l ;

12 x l 2 = x l 1 * 4 ;

13

14 X 1 = X g + x t + x l 2 ;

15 X 2 = X g + x t + x l 1 ;

16 X = inv ( inv ( X 1 ) + inv ( X 2 ) ) ;

17 I f a u l t p u = 1 / ( X * % i ) ;

18 I f a u l t = I f a u l t p u * R b / ( sqrt ( 3 ) * v b 3 ) ;19 M V A = sqrt ( 3 ) * v b 3 * I f a u l t * % i ;



22 I m o d = sqrt ( ( I r ^ 2 ) + ( I i ^ 2 ) ) ;


24 M V A = fix ( M V A / 1 e 5 ) * 1 e 5 ;

25 I m o d = fix ( I m o d ) ;

26 mprintf ( \n F a u l t MVA=%fMVA \n F a u l t c u r r e n t =%d/%dAmp , M V A / 1 e 6 , I m o d , I a n g l e ) ;

27 / / a n o t h e r m et ho d28 M V A = R b / X ;

29 I f a u l t = M V A / ( sqrt ( 3 ) * v b 3 * % i ) ;



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32 I m o d = sqrt ( ( I r ^ 2 ) + ( I i ^ 2 ) ) ;

33 I a n g l e = a t a n d ( I r / I i ) - 9 0 ;34 M V A = fix ( M V A / 1 e 5 ) * 1 e 5 ;

35 I m o d = fix ( I m o d ) ;

36 mprintf ( \n \n f ro m s e c o nd metho d\ n F a u l t MVA=%fMVA \n Fa ul t cu rr en t=%d/ %dAmp, M V A / 1 e 6 , I m o d , I a n g l e ) ;

Scilab code Exa 20.05.b calculating the fault current

1 clear ;2 clc ;

3 v 1 = 6 6 e 3 ;

4 v 2 = 1 1 e 3 ;

5 x 2 = . 4 6 1 ;

6 x 1 = . 4 5 2 7 ;

7 I f = 2 2 9 ;

8 I 1 = I f * x 2 / ( x 1 + x 2 ) ;

9 I 2 = I f * x 1 / ( x 1 + x 2 ) ;

10 I = I 1 + I 2 ;

11 I g 1 = I 1 * v 1 / v 2 ;

12 I g 1 = fix ( I g 1 ) ;13 I1 = round ( I 1 * 1 0 ) / 1 0 ;

14 I2 = round ( I 2 * 1 0 ) / 1 0 ;

15 mprintf ( t he f a u l t c u r r e n t s u pp l i e d by e a cht r an s f o r me r i s\n I1=%fA\ nI2=%fA\ nI3=I1+I2=%dA\n ,I 1 , I 2 , I ) ;

16 I2 = fix ( I 2 ) ;

17 I g 2 = I 2 * v 1 / v 2 ;

18 mprintf ( t he f a u l t c u r r e n t s u pp l i e d by e a chg e n e r a t o r \n I g 1=%dA\n I g 2=%dA\n , I g 1 , I g 2 ) ;

Scilab code Exa 20.06 To calculate the current supplied by alternator

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1 clear ;

2 clc ;3 r = 6 e 6 ;

4 v 1 = 1 1 e 3 ;

5 v 2 = 6 6 e 3 ;

6 x g = . 1 ;

7 x t = . 0 9 ;

8 z = 4 + ( 1 * % i ) ;

9 z b = v 2 ^ 2 / r ;

10 z p u = z / z b ;

11 E = 1 ;

12 I f a u l t = E / ( z p u + ( ( x g + x t ) * % i ) ) ;

13 Ir = real ( I f a u l t ) ;14 Ii = imag ( I f a u l t ) ;

15 I m o d = sqrt ( ( I r ^ 2 ) + ( I i ^ 2 ) ) ;

16 I b = r / ( sqrt ( 3 ) * v 2 ) ;

17 i = I m o d * I b ;

18 i g b = r / ( sqrt ( 3 ) * v 1 ) ;

19 i g = i g b * I m o d ;

20 i = fix ( i ) ;

21 ig = fix ( i g ) ;

22 mprintf ( t h e b a se c u r r e n t on HT s i d e = %dA\n t h e

c u r r e n t f ro m g e n e r a t o r = %dA, i , i g ) ;

Scilab code Exa 20.07 finding the current supplied by generator

1 clear ;

2 clc ;

3 r 1 = 2 0 e 6 ;

4 r b = 3 0 e 6 ;

5 v 1 = 1 1 e 3 ;6 v 2 = 1 1 0 e 3 ;

7 x 1 g = . 2 * r b / r 1 ;

8 x 1 t = . 0 8 * r b / r 1 ;

9 x 2 g = . 2 ;

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10 x 2 t = . 1 ;

11 x l = . 5 1 6 ;12 x 0 = x l / 2 ;

13 x 1 = x 1 g + x 1 t ;

14 x 2 = x 2 g + x 2 t ;

15 x = inv ( inv ( x 2 ) + inv ( x 1 ) ) ;

16 z = x + x 0 ;

17 E = 1 ;

18 i s c = E / z ;

19 i g 1 = i s c * x 2 / ( x 1 + x 2 ) ;

20 i g 2 = i s c * x 1 / ( x 1 + x 2 ) ;

21 i = i g 1 + i g 2 ;

22 i b = r b / ( 1 . 7 3 5 5 * v 1 ) ;23 i g 1 = fix ( i g 1 * 1 0 0 0 ) / 1 0 0 0 ;

24 I g 1 = i g 1 * i b ;

25 ib = fix ( i b ) ;

26 i g 2 = fix ( i g 2 * 1 0 0 ) / 1 0 0 ;

27 I g 2 = i g 2 * i b ;

28 I g 2 = fix ( I g 2 ) ;

29 mprintf ( t h e c u r r e n t t a k e n f r om G1=%dA ( l a g g i n g ) \nt h e c u r r e n t t a k e n f r o m G2=%dA( l a g g i n g ) , I g 1 , I g 2 ) ;

Scilab code Exa 20.08 to calulate the subtransient fault current and breakercurrent rating

1 clear ;

2 clc ;

3 r = 2 5 e 6 ;

4 r b = 5 e 6 ;

5 v 1 = 6 . 6 e 3 ;

6 v 2 = 2 5 e 3 ;7 x s = . 2 ;

8 x t = . 3 ;

9 X s = x s * r / r b ;

10 X t = x t * r / r b ;

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11 Z = . 1 2 5 ;

12 v = 1 ;13 I = v / ( Z ) ;

14 i b = r / ( 1 . 7 3 5 5 * v 1 ) ;

15 ib = fix ( i b ) ;

16 i = i b * 8 ;

17 i g = I * . 2 5 / . 5 ;

18 i m = I - i g ;

19 i t = 3 * 1 + i m ;

20 I a = i b * i t ;

21 I m o m = 1 . 6 * I a ;

22 x t = . 1 5 ;

23 Z t h = . 3 7 5 * . 2 5 / ( . 3 7 5 + . 2 5 ) ;24 I = v / x t ;

25 i g e n = I * . 3 7 5 / . 6 2 5 ;

26 i m o t = . 2 5 * I * . 2 5 / . 6 2 5 ;

27 i t o t = i g e n + ( 3 * i m o t ) ; / /symm b r e a k i n g c u r r e n t28 i b r = i t o t * 1 . 1 ; / /asymm b r e a k i n g c u r r e n t29 i s = i t o t * i b ;

30 i a = i b r * i b * 1 . 0 1 ;

31 ia = fix ( i a / 1 0 0 ) * 1 0 0 ;

32 r b r e a k i n g = 1 . 7 3 9 * v 1 * i a ;

33 r b r e a k i n g = fix ( r b r e a k i n g / 1 e 6 ) * 1 e 6 ;

34 I m o m = round ( I m o m / 1 0 ) * 1 0 ;

35 ia = round ( i a ) ;

36 is = fix ( i s / 1 0 0 ) * 1 0 0 ;

37 mprintf ( t he s u b t r a n si e n t f a u l t c u r r e nt I f = %d/ 90A\n s u b t a n s i e n t c u r r e n t i n b r e a k e r A=%dA\n t h emome nt ar y c u r r e n t = %dA\n , t he c u r r e nt t o bei n t e r r u p t e d a s y m m e t r i c=%dA \n s y m m et r i ci n t e r r u p t i n g c u r r e n t =%dA\n t h e r a t i n g o f t h e CBi n kv a=%dkVA , i , I a , I m o m , i a , i s , r b r e a k i n g / 1 e 3 ) ;

Scilab code Exa 20.09 to calculate the fault level

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1 clc ;

2 clear ;3 r b = 1 0 0 e 6 ;

4 r f = 1 e 6 ;

5 v = 3 . 3 e 3 ;

6 x = r f / r b ;

7 x p u = . 6 ;

8 x t o t = x + x p u ;

9 r f 2 = r f / x t o t ;

10 r f 2 = round ( r f 2 / 1 e 4 ) * 1 e 4 ;

11 I f = r f 2 / ( 1 . 7 2 * v ) ;

12 If = fix ( I f ) ;

13 mprintf ( t h e f a u l t l e v e l i s =%fMVA\n t h e f a u l tc u r r e n t=%dA , r f 2 / 1 e 6 , I f ) ; ;

Scilab code Exa 20.10 to calculate the max possible fault level

1 clear ;

2 clc ;

3 r = 5 0 0 e 3 ;

4 x = 4 . 7 5 / 1 0 0 ;5 f a u l t = r / x ;

6 f a u l t = fix ( f a u l t / 1 e 5 ) * 1 e 5 ;

7 mprintf ( t h e f a u l t l e v e l on LT s i d e =%dkVA , f a u l t / 1 e 3) ;

Scilab code Exa 20.11 to calculate the fault level

1 clc ;2 clear ;

3 r 1 = 7 5 e 6 ;

4 r 2 = 1 5 0 e 6 ;

5 r b = r 1 + r 2 ;

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6 r f = r b ;

7 x = . 0 5 ;8 x n = x * r b / 1 e 6 ;

9 x e q = r b / r f ;

10 X = x n + x e q ;

11 f a u l t = r b / X ;

12 f = r b / x n ;

13 f a u l t = round ( f a u l t / 1 e 4 ) * 1 e 4

14 mprintf ( f a u l t l e v e l on LT s i d e o f t r a n s f o r m e r =%fMVA\n f a u l t l e v e l when s ou rc e o f r e a c t an c e i s

n e g l e c t e d=%fMVA, f a u l t / 1 e 6 , f / 1 e 6 ) ;

Scilab code Exa 20.12 To calculate the fault level at any point of line

1 clear ;

2 clc ;

3 r b = 1 0 0 e 6 ;

4 r 1 = 5 0 e 6 ;

5 r 2 = r b ;

6 x 1 = r b / r 1 ;

7 x 2 = r b / r 2 ;8 x e q = inv ( inv ( x 1 ) + inv ( x 2 ) ) ;

9 f = r b / x e q ;

10 mprintf ( t h e f a u l t l e v e l on t h e l i n e =%dMVA , f / 1 e 6 ) ;

Scilab code Exa 20.13 to find initial short circuit current and peak SCcurrent

1 clear ;2 clc ;

3 x = . 2 3 ;

4 r = 3 7 5 0 e 3 ;

5 v = 6 6 0 0 ;

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6 r e s = . 8 6 6 ;

7 x 1 = x * ( v ^ 2 ) / r ;8 z = sqrt ( ( r e s ^ 2 ) + ( x 1 ^ 2 ) ) ;

9 i = 1 . 1 * v / ( sqrt ( 3 ) * z ) ;

10 f = r e s / x 1 ;

11 x = 1 . 3 8 ;

12 i = round ( i / 1 0 0 ) * 1 0 0

13 is = sqrt ( 2 ) * x * i ;

14 is = round ( i s / 1 0 ) * 1 0 ;

15 mprintf ( i n i t i a l s h o r t c i r c u i t c u r r e n t =%dA \n p ea ks h o r t c i r c u i t c u r r e n t=%dA , i , i s ) ;

Scilab code Exa 20.14 to find the subtransient currents

1 clear ;

2 clc ;

3 r b = 7 5 0 0 0 e 3 ;

4 r o = 5 0 e 6 ;

5 v 1 = 1 1 e 3 ;

6 v 2 = 6 6 e 3 ;

7 x a = . 2 5 * r b / r o ;8 x b = . 7 5 ;

9 x t = . 1 ;

10 v = 1 ;

11 x e q = inv ( inv ( x a ) + inv ( x b ) ) + x t ;

12 i = v / x e q ;

13 i = round ( i * 1 0 0 ) / 1 0 0 ;

14 i a = i * x b / ( x a + x b ) ;

15 i b = i * x a / ( x a + x b ) ;

16 ia = round ( i a * 1 0 0 ) / 1 0 0 ;

17 i l t = r b / ( sqrt ( 3 ) * v 1 ) ;18 i h t = r b / ( sqrt ( 3 ) * v 2 ) ;

19 i = i * i h t ;

20 i = fix ( i )

21 i a = i a * i l t ;

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22 i l t = r b / ( 1 . 7 3 * v 1 ) ;

23 i b = i b * i l t ;24 ia = round ( i a ) ;

25 ib = round ( i b / 1 0 ) * 1 0 ;

26 mprintf ( s ub t r a n s i e n t c u r r e n t g e n e r a t o r A=%dA \ng e n er a t o r B=%dA \n HT s i d e=%dA, i a , i b , i ) ;

Scilab code Exa 20.15 to find SC current and rms current and makingand breaking capacity required

1 clear ;

2 clc ;

3 x = 1 ;

4 e = 1 ;

5 i = e / x ;

6 r = 7 . 5 e 6 ;

7 v = 6 . 6 e 3 ;

8 i = r / ( sqrt ( 3 ) * v ) ;

9 i = fix ( i ) ;

10 x 2 = . 0 9 ;

11 i 2 = e / x 2 ;12 I 2 = i 2 * i ;

13 I2 = fix ( I 2 / 1 0 ) * 1 0

14 i d c = sqrt ( 2 ) * I 2 ;

15 m c = i d c * 2 ;

16 x 3 = . 1 5 ;

17 i 3 = e / x 3 ;

18 I 3 = i 3 * i ;

19 i b = I 3 * 1 . 4 ;

20 M v a = sqrt ( 3 ) * v * i b ;

21 i d c = round ( i d c / 1 e 2 ) * 1 e 2 ;22 mc = round ( m c / 1 e 2 ) * 1 e 2 ;

23 I3 = round ( I 3 / 1 0 ) * 1 0 ;

24 M v a = fix ( M v a / 1 e 4 ) * 1 e 4

25 mprintf ( s u s t a i n e d s h o r t c i r c u i t c u r r e n t=%dA\

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n i n i t i a l symm etri c SC cu rr en t=%fkA\nmaximum dc

component=%fkA\n ma ki ng c a p a c i t y r e q u i r e d =%fkA\n t r a n s i e n t s h o r t c i r c u i t c u r r e nt=%fkA\ni n t e r r u p t i n g c a p a c i t y r e q u i r e d =%fMVA, A sy mm et ri c ,i , I 2 / 1 e 3 , i d c / 1 e 3 , m c / 1 e 3 , I 3 / 1 e 3 , M v a / 1 e 6 ) ;

Scilab code Exa 20.16.a to find the short circuit current

1 clear ;

2 clc ;3 r b = 2 e 6 ;

4 r = 1 . 2 e 6 ;

5 x = 7 * r b / r ;

6 v = 6 . 6 e 3 ;

7 i = r b / v ;

8 z b = v / i ;

9 r = 1 2 0 0 e 3 ;

10 r b = 2 0 0 0 e 3 ;

11 v = 6 . 6 e 3 ;

12 i = r b / v ;

13 x = . 1 ;14 z 0 = v * x / i ;

15 x 1 = 7 * r b / r ;

16 z 1 = v * x 1 / ( 1 0 0 * i ) ;

17 z 2 = 2 ;

18 z = z 0 + z 1 + z 2 ;

19 i s h = v / z ;

20 zb = round ( z b * 1 0 ) / 1 0 ;

21 i s h = round ( i s h / 1 0 ) * 1 0 ;

22 mprintf ( t he s h o r t c i r c u i t c u r r en t b y d i r e c t ohmic

method=%fA\n , i s h ) ;23 mprintf ( t he bas e i mpe de nc e =%fohm , z b ) ;

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Scilab code Exa 20.16.b to find SC current by ohmic method

1 clear ;

2 clc ;

3 r b = 2 e 6 ;

4 r = 1 . 2 e 6 ;

5 x = 7 * r b / r ;

6 x 1 = 1 0 ;

7 x 2 = 1 1 . 7 ;

8 v = 6 . 6 e 3 ;

9 i = r b / v ;

10 z b = v / i ;

11 r = 1 2 0 0 e 3 ;12 r b = 2 0 0 0 e 3 ;

13 v = 6 . 6 e 3 ;

14 x t = . 1 1 7 ;

15 x f = 2 / z b * 1 0 0 ;

16 x t o t = x f + x 1 + x 2 ;

17 i s h = i * 1 0 0 / x t o t ;

18 i s h = round ( i s h / 1 0 ) * 1 0 ;

19 mprintf ( t he s h or t c i r c u i t c u r r e n t by p e r ce nt a g ere ac ta nc e method=%fA , i s h ) ;

Scilab code Exa 20.16.c To find the new SC current

1 clear ;

2 clc ;

3 x 1 = 5 ;

4 x 2 = 1 0 ;

5 x 3 = 1 1 . 7 ;

6 x 4 = 9 . 1 ;7 i = 3 0 3 ;

8 x t = x 1 + x 2 + x 3 + x 4 ;

9 i s h = 3 0 3 * 1 0 0 / x t ;

10 mprintf ( t h e SHORT CIRCUIT CURRENT=%dA , i s h )

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Scilab code Exa 20.17.a To find the SC current of the circuit

1 clear ;

2 clc ;

3 v = 3 . 3 e 3 ;

4 r b = 3 e 6 ;

5 r 1 = 1 e 6 ;

6 r 2 = 1 . 5 e 6 ;

7 x 1 = 1 0 ;8 x 2 = 2 0 ;

9 X 1 = x 1 * r b / r 1 ;

10 X 2 = x 2 * r b / r 2 ;

11 x = inv ( inv ( X 1 ) + inv ( X 2 ) ) ;

12 k v a = r b * 1 0 0 / x ;

13 i s h = k v a / ( 1 . 7 3 8 8 * v ) ;

14 i s h = round ( i s h ) ;

15 printf ( t h e v a l ue o f s h o r t c i r c u i t c u r r e nt=%dA , i s h );

Scilab code Exa 20.17.b to find the reactance of the reactor

1 clear ;

2 clc ;

3 v = 3 . 3 e 3 ;

4 r b = 3 e 6 ;

5 r 1 = 1 e 6 ;

6 r 2 = 1 . 5 e 6 ;

7 x 1 = 1 0 ;

8 x 2 = 2 0 ;

9 X 1 = x 1 * r b / r 1 ;

10 X 2 = x 2 * r b / r 2 ;

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11 x = inv ( inv ( X 1 ) + inv ( X 2 ) ) ;

12 k v a = r b * 1 0 0 / x ;13 i s h = k v a / ( sqrt ( 3 ) * v ) ;

14 r x = 1 0 e 6 ;

15 x 2 = r b * 1 0 0 / r x ;

16 r = inv ( inv ( X 1 ) - inv ( X 2 ) ) - 3 0 ;

17 printf ( t he r e ac t a nc e o f g e n er a t o r t o b e c on v e r t ed=% d p e r c e n t , r ) ;

Scilab code Exa 20.18.a To calculate the reactance of the reactor to limitSC MVA

1 clear ;

2 clc ;

3 r 1 = 3 e 6 ;

4 x = 1 0 ;

5 r = 1 5 0 e 6 ;

6 r b = 9 e 6 ;

7 x 1 = x * r b / r 1 ;

8 xc = inv (2* inv ( x 1 ) ) ;

9 x t = r b * 1 0 0 / r ;10 x =( inv ( inv ( x t ) - inv ( x c ) ) ) - 5 ;

11 printf ( t h e r e a c t a n c e t h a t s h o u ld be a dd ed= %dp e r c e n t , x ) ;

Scilab code Exa 20.18.b fault level at generator bus

1 clear ;

2 clc ;3 z = 4 0 0 0 ;

4 z b = 9 ;

5 x 1 = z b / z * 1 0 0 ;

6 x 2 = 5 ;

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7 x 3 = 3 0 ;

8 x 4 = 3 0 ;9 x = inv ( inv ( x 1 + x 2 ) + inv ( x 3 ) + inv ( x 4 ) ) ;

10 x = round ( x * 1 0 0 ) / 1 0 0 ;

11 f a u l t = z b * 1 e 3 / x * 1 0 0 ;

12 f a u l t = fix ( f a u l t / 1 e 3 ) * 1 e 3 ;

13 mprintf ( t h e new f a u l t l e v e l o f g e n e r a t o r b us=%dMVA, f a u l t / 1 e 3 ) ;

Scilab code Exa 20.19 to calculate the current fed to the faults

1 clear ;

2 clc ;

3 r b = 2 0 e 6 ;

4 r = 1 0 e 6 ;

5 v 1 = 1 1 e 3 ;

6 v 2 = 6 6 e 3 ;

7 x 1 = 5 ;

8 X 1 = x 1 * r b / r ;

9 x a = 2 0 ;

10 x b = 2 0 ;11 x c = 2 0 ;

12 x d = 2 0 ;

13 x b u s = 2 5 ;

14 x t r = X 1 ;

15 x c d = inv ( inv ( x c ) + inv ( x d ) ) ;

16 x a b = inv ( inv ( x a ) + inv ( x b ) ) ;

17 x c d b u s = x c d + x b u s ;

18 xn = inv ( inv ( x a b ) + inv ( x c d b u s ) ) ;

19 x t h = x t r + x n ;

20 m v a = r b / x t h * 1 0 0 ;21 i = m v a / ( 1 . 7 4 5 * v 2 ) ;

22 i = round ( i ) ;

23 printf ( t h e SC MVA=%fMVA \n t he SC c u r r e n t =%dA ,mva/ 1 e 6 , i ) ;

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Scilab code Exa 20.20.b to calculate the percentage change of reactors R

1 clear ;

2 clc ;

3 g = 2 0 ;

4 v = 1 1 e 3 ;

5 r = 2 0 e 6 ;

6 n = 4 ;

7 x = . 4 ;8 x 1 = g / ( n - 1 ) ;

9 z = ( ( x 1 / x ) - ( x 1 ) ) / 1 . 3 3 ;

10 R = ( z / 1 0 0 ) * ( v ^ 2 ) / r ;

11 R = round ( R * 1 0 0 0 ) / 1 0 0 0 ;

12 printf ( t h e v a l u e o f r e a c t a n c e =%fohms , R ) ;

Scilab code Exa 20.21 calculate the MVA and current by both generator

and transformer side

1 clear ;

2 clc ;

3 x s t = 2 0 ;

4 x t r = 2 8 ;

5 x s = 2 5 0 ;

6 x t = 1 5 ;

7 v 1 = 2 5 e 3 ;

8 r 1 = 5 0 0 e 6 / . 8 ;

9 v 2 = 2 2 0 e 3 ;

10 r b = 6 0 0 e 6 ;

11 v b = 2 5 e 3 ;

12 x f = r b / r 1 ;

13 x s t = x s t * x f / 1 0 0 ;

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14 x t r = x t r * x f / 1 0 0 ;

15 x s = x s * x f / 1 0 0 ;16 x t = x t / 1 0 0 ;

17 x e q s = inv ( inv ( x s t ) + inv ( x t ) ) ;

18 x e q t = inv ( inv ( x t r ) + inv ( x t ) ) ;

19 x e g = inv ( inv ( x s ) + inv ( x t ) ) ;

20 e = 1 ;

21 x e q s = round ( x e q s * 1 0 0 0 ) / 1 e 3 ;

22 i s = e / x e q s ;

23 is = round ( i s ) ;

24 i t = e / x e q t ;

25 i g = e / x e g ;

26 i 1 = i s * x t / ( x t + x s t ) ;27 i 2 = i s * x s t / ( x s t + x t ) ;

28 i b = r b / ( 1 . 7 2 6 * 2 2 . 2 * 1 e 3 ) ;

29 I s = i s * i b ;

30 i1 = round ( i 1 * 1 0 ) / 1 0 ;

31 Is = round ( I s / 1 e 3 ) * 1 e 3 ;

32 i2 = fix ( i 2 * 1 0 0 ) / 0 1 0 0 ;

33 I 1 = i 1 * i b ;

34 I 2 = i 2 * i b ;

35 I1 = fix ( I 1 / 1 e 2 ) * 1 e 2 ;

36 I2 = fix ( I 2 / 1 e 2 ) * 1 e 2 ;

37 mprintf ( t o t a l s u b t r a n si e n t c u r r e n t Toff=%fkA\n s u b t r a n s i e n t c u r r e n t on g e n e r a t o r s i d e =%fkA\ns u b t r a n s i e n t c u r r e n t on t r a n s f o r m e r s i d e =%fkA , Is/ 1 e 3 , I 1 / 1 e 3 , I 2 / 1 e 3 ) ;

Scilab code Exa 20.22 calculate the short circuit level and normal andeffective fault current

1 clc ;

2 clear ;

3 m v a n = 6 8 0 0 e 6 ;

4 v = 1 3 2 e 3 ;

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5 m v a c = 2 0 0 e 6 ;

6 m v a e = m v a n - m v a c ;7 n = m v a n / ( sqrt ( 3 ) * v ) ;

8 e = m v a e / ( 1 . 6 8 1 * v ) ;

9 e = fix ( e / 1 0 ) * 1 0 ;

10 n = fix ( n / 1 0 ) * 1 0 ;

11 printf ( n or ma l f a u l t c u r r e n t=%f / 90 kA\n E f f e c t i v ef a u l t c u r r e n t=%f / 90 kA , n / 1 e 3 , e / 1 e 3 ) ;

Scilab code Exa 20.23 calculate the SC ratio and effective SC ratio ofHVDC current

1 clear ;

2 clc ;

3 v = 4 0 0 e 3 ;

4 m v a n = 3 0 0 0 0 e 6 ;

5 m w = 1 5 0 0 e 6 ;

6 m v a c = 6 0 0 e 6 ;

7 n = m v a n / m w ;

8 m v a e = m v a n - m v a c ; // // t h e d i f f e r e n c e i n r e s u l t i s due

t o e r ro n eo u s c a l c u l a t i o n i n t ex tb oo k .9 e = m v a e / m w ;

10 mprintf ( t h e SC r a i o =%d\n e f f e c t i v e f a u l t l e v e l =%fMVA\n e f f e c t i v e c i r c u i t l e v e l o f HVDC s y s t em ( ESCR )=%f , n , m v a e / 1 e 6 , e ) ;

11 disp ( t h e d i f f e r e n c e i n r e s u l t i s due t o e rr on e o usc a l c u l a t i o n i n t ex tb oo k . ) ;

Scilab code Exa 20.24 to calculate the fault levels on secondary sides oftransformer

1 clear ;

2 clc ;

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3 s = 1 ;

4 x t = 5 ;5 m = s / x t * 1 0 0 ;

6 n = 2 * s / x t * 1 0 0 ;

7 mprintf ( f a u l t l e v e l on l t s i d e =%dMVA\n f a u l t l e v e lon HT s i d e=%dMVA , m , n ) ;

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Chapter 21

Symmetric Components

Scilab code Exa 21.01 Calculate the symmetric components of unbalancedlines

1 clear ;

2 clc ;

3 v a = 1 0 0 * ( % e ^ ( % p i * % i / 2 ) ) ;

4 v b = 1 1 6 * ( % e ^ ( % i * 0 ) ) ;

5 v c = 7 1 * ( % e ^ ( % i * ( 2 2 4 . 8 * % p i / 1 8 0 ) ) ) ;

6 a = 1 * % e ^ ( % i * ( 1 2 0 * % p i / 1 8 0 ) ) ;7 b = a ^ 2 ;

8 v a 0 = 1 / 3 * ( v a + v b + v c ) ;

9 v a 1 = 1 / 3 * ( v a + ( a * v b ) + ( b * v c ) ) ;

10 v a 2 = 1 / 3 * ( v a + ( b * v b ) + ( a * v c ) ) ;

11 v a 0 r = real ( v a 0 ) ;

12 v a 0 i = imag ( v a 0 ) ;

13 v a 0 m = sqrt ( ( v a 0 r ^ 2 ) + ( v a 0 i ^ 2 ) ) ;

14 v a 0 a = a t a n d ( v a 0 i / v a 0 r ) ;

15 v a 1 r = real ( v a 1 ) ;

16 v a 1 i = imag ( v a 1 ) ;17 v a 1 m = sqrt ( ( v a 1 r ^ 2 ) + ( v a 1 i ^ 2 ) ) ; // t h e d i f f e r e n c e i nr e s u l t i s due t o e rr on e o us c a l c u l a t i o n i nt e x t b o o k .

18 v a 1 a = a t a n d ( v a 1 i / v a 1 r ) ;

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19 v a 2 r = real ( v a 2 ) ;

20 v a 2 i = imag ( v a 2 ) ;21 v a 2 m = sqrt ( ( v a 2 r ^ 2 ) + ( v a 2 i ^ 2 ) ) ;

22 v a 2 a = a t a n d ( v a 2 i / v a 2 r ) ;

23 mprintf ( t h e s y mm e tr i c c om po ne nt s a r e \n va0=%f+j%fV \t o r\ t %f / %d V , v a 0 r , v a 0 i , v a 0 m , v a 0 a ) ;

24 mprintf ( \n va1=%f+j%f V \t o r\ t %f / %d V , v a 1 r , v a 1 i ,v a 1 m , v a 1 a ) ;

25 mprintf ( \n va2=%f+j (%f ) V \t o r\ t %f / %d V , v a 2 r ,v a 2 i , v a 2 m , v a 2 a ) ;

26 disp ( t h e d i f f e r e n c e i n r e s u l t i s due t o e rr on e o usc a l c u l a t i o n i n t ex tb oo k . )

Scilab code Exa 21.02 to calculate the line voltages

1 clear ;

2 clc ;

3 v a = 2 2 + ( 1 6 . 6 6 * % i ) ;

4 v b = - 2 5 . 3 3 + ( % i * 8 9 . 3 4 ) ;

5 v c = 3 . 3 3 - ( % i * 6 ) ;

6 a = 1 * % e ^ ( % i * ( 1 2 0 * % p i / 1 8 0 ) ) ;7 b = a ^ 2 ;

8 v a 0 = ( v a + v b + v c ) ;

9 v a 1 = ( v a + ( b * v b ) + ( a * v c ) ) ;

10 v a 2 = ( v a + ( a * v b ) + ( b * v c ) ) ;

11 v a 0 r = real ( v a 0 ) ;

12 v a 0 i = imag ( v a 0 ) ;

13 v a 0 m = sqrt ( ( v a 0 r ^ 2 ) + ( v a 0 i ^ 2 ) ) ;

14 v a 0 a = a t a n d ( v a 0 i / v a 0 r ) ;

15 v a 1 r = real ( v a 1 ) ;

16 v a 1 i = imag ( v a 1 ) ;17 v a 1 m = round ( sqrt ( ( v a 1 r ^ 2 ) + ( v a 1 i ^ 2 ) ) * 1 0 ) / 1 0 ;

18 v a 1 a = a t a n d ( v a 1 i / v a 1 r ) ;

19 v a 2 r = round ( real ( v a 2 ) ) ;

20 v a 2 i = round ( imag ( v a 2 ) ) ;

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21 v a 2 m = round ( sqrt ( ( v a 2 r ^ 2 ) + ( v a 2 i ^ 2 ) ) ) ;

22 v a 2 a = a t a n d ( v a 2 i / v a 2 r ) ;23 mprintf ( t h e v o l t a g e l e v e l s a re \n va=%f+j%f V \t o r\t %f / %d V , v a 0 r , v a 0 i , v a 0 m , v a 0 a ) ;

24 mprintf ( \n vb=%f+j ( %f) V \t o r\ t %f / %d V ,va1r,va1i, v a 1 m , v a 1 a ) ;

25 mprintf ( \n vc=%f+j ( %f) V \t o r\ t %f / %d V ,va2r,va2i, v a 2 m , v a 2 a ) ;

Scilab code Exa 21.03 To determine the line currents

1 clear ;

2 clc ;

3 i b = 5 0 ;

4 i c = 1 0 * % e ^ ( % i * % p i / 2 ) ;

5 i a = 1 0 * % e ^ ( % i * % p i ) ;

6 a = 1 * % e ^ ( % i * ( 1 2 0 * % p i / 1 8 0 ) ) ;

7 b = a ^ 2 ;

8 i a 0 = ( i a + i b + i c ) ;

9 i a 1 = ( i a + ( b * i b ) + ( a * i c ) ) ;

10 i a 2 = ( i a + ( a * i b ) + ( b * i c ) ) ;11 i a 0 r = real ( i a 0 ) ;

12 i a 0 i = imag ( i a 0 ) ;

13 i a 0 m = sqrt ( ( i a 0 r ^ 2 ) + ( i a 0 i ^ 2 ) ) ;

14 i a 0 a = a t a n d ( i a 0 i / i a 0 r ) ;

15 i a 1 r = real ( i a 1 ) ;

16 i a 1 i = imag ( i a 1 ) ;

17 i a 1 m = sqrt ( ( i a 1 r ^ 2 ) + ( i a 1 i ^ 2 ) ) ;

18 i a 1 a = a t a n d ( i a 1 i / i a 1 r ) ;

19 i a 2 r = real ( i a 2 ) ;

20 i a 2 i = imag ( i a 2 ) ;21 i a 2 m = sqrt ( ( i a 2 r ^ 2 ) + ( i a 2 i ^ 2 ) ) ;

22 i a 2 a = a t a n d ( i a 2 i / i a 2 r ) ;

23 mprintf ( t h e c ur r e nt l e v e l s a re \n i a=%f+j %f A \t o r\t %f / %d A , i a 0 r , i a 0 i , i a 0 m , i a 0 a ) ;

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24 mprintf ( \n ib=%f+j ( %f) A \t o r\ t %f / %d A ,ia1r,ia1i

, i a 1 m , i a 1 a ) ;25 mprintf ( \n i c =%f+ j ( %f ) A \t o r\ t %f / %d A ,ia2r,ia2i, i a 2 m , i a 2 a ) ;

Scilab code Exa 21.04 to find the symmetric components of line currents

1 clear ;

2 clc ;

3 i a = 2 0 ;4 i b = 2 0 * ( % e ^ ( % i * % p i ) ) ;

5 i c = 0 ;

6 a = 1 * % e ^ ( % i * ( 1 2 0 * % p i / 1 8 0 ) ) ;

7 b = a ^ 2 ;

8 i a 0 = 1 / 3 * ( i a + i b + i c ) ;

9 i a 1 = 1 / 3 * ( i a + ( a * i b ) + ( b * i c ) ) ;

10 i a 2 = 1 / 3 * ( i a + ( b * i b ) + ( a * i c ) ) ;

11 i a 0 r = real ( i a 0 ) ;

12 i a 0 i = imag ( i a 0 ) ;

13 i a 0 m = sqrt ( ( i a 0 r ^ 2 ) + ( i a 0 i ^ 2 ) ) ;

14 i a 0 a = 0 - a t a n d ( i a 0 r / i a 0 i ) ;15 i a 1 r = real ( i a 1 ) ;

16 i a 1 i = imag ( i a 1 ) ;

17 i a 1 m = sqrt ( ( i a 1 r ^ 2 ) + ( i a 1 i ^ 2 ) ) ;

18 i a 1 a = a t a n d ( i a 1 i / i a 1 r ) ;

19 i a 2 r = real ( i a 2 ) ;

20 i a 2 i = imag ( i a 2 ) ;

21 i a 2 m = sqrt ( ( i a 2 r ^ 2 ) + ( i a 2 i ^ 2 ) ) ;

22 i a 2 a = a t a n d ( i a 2 i / i a 2 r ) ;

23 mprintf ( t h e s y mm et ri c c om po ne nt s a r e \n ia 0=%f+j%f

A \t o r\ t %f / %d A , i a 0 r , i a 0 i , i a 0 m , i a 0 a ) ;24 mprintf ( \n i a 1=%f+j %f A \t o r\ t %f / %d A , i a 1 r , i a 1 i ,i a 1 m , i a 1 a ) ;

25 mprintf ( \n i a 2=%f+ j ( %f ) A \t o r\ t %f / %d A , i a 2 r ,i a 2 i , i a 2 m , i a 2 a ) ;

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26 i b 1 = b * i a 1 ;

27 i b 2 = a * i a 2 ;28 i c 1 = a * i a 1 ;

29 i c 2 = b * i a 2 ;

30 i b 0 = i a 0 ;

31 i c 0 = i a 0 ;

32 i b 1 r = real ( i b 1 ) ;

33 i b 1 i = imag ( i b 1 ) ;

34 i b 1 m = sqrt ( ( i b 1 r ^ 2 ) + ( i b 1 i ^ 2 ) ) ;

35 i b 1 a = a t a n d ( i b 1 i / i b 1 r ) ;

36 i b 2 r = real ( i b 2 ) ;

37 i b 2 i = imag ( i b 2 ) ;

38 i b 2 m = sqrt ( ( i b 2 r ^ 2 ) + ( i b 2 i ^ 2 ) ) ;39 i b 2 a = a t a n d ( i b 2 i / i b 2 r ) ;

40 i c 1 r = real ( i c 1 ) ;

41 i c 1 i = imag ( i c 1 ) ;

42 i c 1 m = sqrt ( ( i c 1 r ^ 2 ) + ( i c 1 i ^ 2 ) ) ;

43 i c 1 a = a t a n d ( i c 1 i / i c 1 r ) ;

44 i c 2 r = real ( i c 2 ) ;

45 i c 2 i = imag ( i c 2 ) ;

46 i c 2 m = sqrt ( ( i c 2 r ^ 2 ) + ( i c 2 i ^ 2 ) ) ;

47 i c 2 a = a t a n d ( i c 2 i / i c 2 r ) ;

48 mprintf (\n \n i b0=%fA

, i b 0 ) ;

49 mprintf ( \n ib 1=%f+j%f A \t o r\ t %f / %d A , i b 1 r , i b 1 i ,i b 1 m , i b 1 a ) ;

50 mprintf ( \n i b 2=%f+ j ( %f ) A \t o r\ t %f / %d A , i b 2 r ,i b 2 i , i b 2 m , i b 2 a ) ;

51 mprintf ( \n \n i c 0 =%f A , i c 0 ) ;52 mprintf ( \n i c 1=%f+j %f A \t o r\ t %f / %d A , i c 1 r , i c 1 i ,

i c 1 m , i c 1 a ) ;

53 mprintf ( \n i c 2=%f+ j ( %f ) A \t o r\ t %f / %d A , i c 2 r ,i c 2 i , i c 2 m , i c 2 a ) ;

Scilab code Exa 21.05 to calculate the voltages of phase and line voltages

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1 clear ;

2 clc ;3 v b = . 5 8 4 + ( 0 * % i ) ;

4 v c = . 5 8 4 + ( 0 * % i ) ;

5 v a = 0 ;

6 a = 1 * % e ^ ( % i * ( 1 2 0 * % p i / 1 8 0 ) ) ;

7 b = a ^ 2 ;

8 v a e = ( v a + v b + v c ) ;

9 v b e = ( v a + ( b * v b ) + ( a * v c ) ) ;

10 v c e = ( v a + ( a * v b ) + ( b * v c ) ) ;

11 v a 0 = v a e - v b e ;

12 v a 1 = v b e - v c e ;

13 v a 2 = v c e - v a e ;14 v a 0 r = real ( v a 0 ) ;

15 v a 0 i = imag ( v a 0 ) ;

16 v a 0 m = sqrt ( ( v a 0 r ^ 2 ) + ( v a 0 i ^ 2 ) ) ;

17 v a 0 a = a t a n d ( v a 0 i / v a 0 r ) ;

18 v a 1 r = real ( v a 1 ) ;

19 v a 1 i = imag ( v a 1 ) ;

20 v a 1 m = sqrt ( ( v a 1 r ^ 2 ) + ( v a 1 i ^ 2 ) ) ;

21 v a 1 a = 0 ;

22 v a 2 r = real ( v a 2 ) ;

23 v a 2 i = imag ( v a 2 ) ;

24 v a 2 m = sqrt ( ( v a 2 r ^ 2 ) + ( v a 2 i ^ 2 ) ) ;

25 v a 2 a = a t a n d ( v a 2 i / v a 2 r ) + 1 8 0 ;

26 mprintf ( t h e v o l t a g e l e v e l s a re \n vab=%f+j%f V \t o r\ t %f / %d V , v a 0 r , v a 0 i , v a 0 m , v a 0 a ) ;

27 mprintf ( \n vbc=%f+j ( %f) V \t o r\ t %f / %d V , v a 1 r ,v a 1 i , v a 1 m , v a 1 a ) ;

28 mprintf ( \n vca=%f+j (%f ) V \t o r\ t %f / %d V , v a 2 r ,v a 2 i , v a 2 m , v a 2 a ) ;

Scilab code Exa 21.06 to calculate the value of Ia

1 clear ;

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2 clc ;

3 e = 1 ;4 x 1 = . 2 5 * % i ;

5 x 2 = . 3 5 * % i ;

6 x 0 = . 1 * % i ;

7 i a 0 = e / ( x 1 + x 2 + x 0 ) ;

8 i a 1 = i a 0 ;

9 i a 2 = i a 0 ;

10 i a = i a 0 + i a 1 + i a 2 ;

11 i a r = real ( i a ) ;

12 i a i = imag ( i a ) ;

13 i a m = round ( sqrt ( ( i a r ^ 2 ) + ( i a i ^ 2 ) ) * 1 0 0 ) / 1 0 0 ;

14 i a a = 0 ;15 mprintf ( t h e c ur r e nt l e v e l s a re \n i a=%f+ j ( %f ) A \

t o r\ t %f / %d A , i a r , i a i , i a m , i a a ) ;

Scilab code Exa 21.07 to find the line and phase voltage of phase a

1 clear ;

2 clc ;

3 z 1 = . 2 5 * % i ;4 z 2 = . 3 5 * % i ;

5 z 0 = . 1 * % i ;

6 e a = 1 ;

7 i a 1 = inv ( z 1 + inv ( inv ( z 2 ) + inv ( z 0 ) ) ) * e a ;

8 v a 1 = e a - ( i a 1 * z 1 ) ;

9 v a 0 = v a 1 ;

10 v a 2 = v a 0 ;

11 i a 0 = - v a 0 / z 0 ;

12 i a 2 = - v a 2 / z 2 ;

13 i a = i a 1 + i a 2 + i a 0 ;14 v a = v a 1 + v a 2 + v a 0 ;

15 va = fix ( v a * 1 0 0 0 ) / 1 e 3 ;

16 mprintf ( t h e c u r r e n t i a =%dA\ tVa=%fV , i a , v a ) ;

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Scilab code Exa 21.08 to find positive sequence component of fault cur-rent

1 clear ;

2 clc ;

3 r 0 = . 1 ;

4 v = 1 ;

5 r 1 = . 0 5 ;

6 r 2 = . 0 5 ;

7 r 3 = . 2 ;

8 r 4 = . 2 ;

9 r 3 4 = inv ( inv ( r 3 ) + inv ( r 4 ) ) ;

10 r 2 3 4 = r 2 + r 3 4 ;

11 r 1 0 = r 1 + r 0 ;

12 r = inv ( inv ( r 2 3 4 ) + inv ( r 1 0 ) ) ;

13 i p = v / r ;

14 mprintf ( t h e p o s i t i v e s e q ue n c e c u r r e n t=%fpu , i p ) ;

Scilab code Exa 21.09 calculate the symmetric components of the fault

1 clear ;

2 clc ;

3 i a = 8 6 . 6 + ( % i * 5 0 ) ;

4 i b = 2 5 - ( 4 3 . 3 * % i ) ;

5 i c = - 3 0 ;

6 a = 1 * % e ^ ( % i * ( 1 2 0 * % p i / 1 8 0 ) ) ;

7 b = a ^ 2 ;

8 i a 0 = 1 / 3 * ( i a + i b + i c ) ;9 i a 1 = 1 / 3 * ( i a + ( a * i b ) + ( b * i c ) ) ;

10 i a 2 = 1 / 3 * ( i a + ( b * i b ) + ( a * i c ) ) ;

11 i a 0 r = real ( i a 0 ) ;

12 i a 0 i = imag ( i a 0 ) ;

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13 i a 0 m = sqrt ( ( i a 0 r ^ 2 ) + ( i a 0 i ^ 2 ) ) ;

14 i a 0 a = a t a n d ( i a 0 r / i a 0 i ) ;15 i a 1 r = real ( i a 1 ) ;

16 i a 1 i = imag ( i a 1 ) ;

17 i a 1 m = sqrt ( ( i a 1 r ^ 2 ) + ( i a 1 i ^ 2 ) ) ;

18 i a 1 a = a t a n d ( i a 1 i / i a 1 r ) ;

19 i a 2 r = real ( i a 2 ) ;

20 i a 2 i = imag ( i a 2 ) ;

21 i a 2 m = sqrt ( ( i a 2 r ^ 2 ) + ( i a 2 i ^ 2 ) ) ;

22 i a 2 a = a t a n d ( i a 2 i / i a 2 r ) ;

23 i n = i a + i b + i c ;

24 mprintf ( t h e s y mm et ri c c om po ne nt s a r e \n i r 0=%f+j %f

A \t o r\ t %f / %d A , i a 0 r , i a 0 i , i a 0 m , i a 0 a ) ;25 mprintf ( \n i r 1=%f+j %f A \t o r\ t %f / %d A , i a 1 r , i a 1 i ,

i a 1 m , i a 1 a ) ;

26 mprintf ( \n i r 2 =%f+ j ( % f ) A \t o r\ t %f / %d A\n n e u t r a lc u r r e n t i n = %fA , i a 2 r , i a 2 i , i a 2 m , i a 2 a , i n ) ;

Scilab code Exa 21.10 to calculate the zero components of currents

1 clear ;2 clc ;

3 i n = 9 ;

4 i a = i n / 3 ;

5 i b = i a ;

6 i c = i b ;

7 mprintf ( t h e z e r o s e q u e n c e c om po ne nt s a r e i a 0 =%dA \ tib0=%dA \ t i c 0=%d , i a , i b , i c ) ;

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Chapter 22

Unsymmetrical Faults on

Unloaded Generator

Scilab code Exa 22.01 to calculate the sub transient currents for differenttypes of fault

1 clear ;

2 clc ;

3 v = 1 1 e 3 / sqrt ( 3 ) ;

4 r = 2 5 e 6 ;

5 x 2 = . 3 5 * % i ;

6 x 0 = . 1 * % i ;

7 x 1 = . 2 5 * % i ;

8 e = 1 ;

9 i a 0 = e / ( x 0 + x 1 + x 2 ) ;

10 i a 0 = round ( i a 0 * 1 0 0 ) / 1 0 0 ;

11 i a 1 = i a 0 ;

12 i a 2 = i a 0 ;

13 i a = 3 * i a 0 ;

14 i b a s e = r / ( ( 3 ) * v ) ;15 I f a u l t = 3 * i a 0 * i b a s e ;

16 I f a u l t = round ( I f a u l t / 1 0 ) * 1 0 ;

17 v a 1 = e - ( i a 1 * x 1 ) ;

18 v a 2 = - i a 2 * x 2 ;

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19 v a 0 = - i a 0 * x 0 ;

20 a = 1 * % e ^ ( % i * ( 1 2 0 * % p i / 1 8 0 ) ) ;21 b = a ^ 2 ;

22 v a = ( v a 1 + v a 2 + v a 0 ) ;

23 v b = ( v a 0 + ( b * v a 1 ) + ( a * v a 2 ) ) ;

24 v c = ( v a 0 + ( a * v a 1 ) + ( b * v a 2 ) ) ;

25 v a b = v a - v b ;

26 v b c = v b - v c ;

27 v c a = v c - v a ;

28 v a b = v a b * v ;

29 v b c = v b c * v ;

30 v c a = v c a * v ;

31 v a 0 r = real ( v a b ) ;32 v a 0 i = imag ( v a b ) ;

33 v a 0 m = sqrt ( ( v a 0 r ^ 2 ) + ( v a 0 i ^ 2 ) ) ;

34 v a 0 a = a t a n d ( v a 0 i / v a 0 r ) ;

35 v a 1 r = real ( v b c ) ;

36 v a 1 i = imag ( v b c ) ;

37 v a 1 m = sqrt ( ( v a 1 r ^ 2 ) + ( v a 1 i ^ 2 ) ) ;

38 v a 1 a = a t a n d ( v a 1 i / v a 1 r ) ;

39 v a 2 r = real ( v c a ) ;

40 v a 2 i = imag ( v c a ) ;

41 v a 2 m = sqrt ( ( v a 2 r ^ 2 ) + ( v a 2 i ^ 2 ) ) ;

42 v a 2 a = a t a n d ( v a 2 i / v a 2 r ) ;

43 mprintf ( t h e s u bt r a n si e n t v o l t a g e l e v e l s a re \n v ab=%f+j%f V \t o r\ t %f / %d kV , round ( v a 0 r * 1 0 0 / 1 e 3 )/100, round ( v a 0 i * 1 0 0 / 1 e 3 ) / 1 0 0 , round ( v a 0 m * 1 0 0 / 1 e 3 )

/ 1 0 0 , v a 0 a ) ;

44 mprintf ( \n vbc=%f+j ( %f) kV \t o r\ t %f / %d V , round (v a 1 r * 1 0 0 / 1 e 3 ) / 1 0 0 , round ( v a 1 i * 1 0 0 / 1 e 3 ) / 1 0 0 , round (

v a 1 m * 1 0 0 / 1 e 3 ) / 1 0 0 , round ( v a 1 a ) + 1 8 0 ) ;

45 mprintf ( \n vca=%f+j ( %f) kV \t o r\ t %f / %d V , round (v a 2 r * 1 0 0 / 1 e 3 ) / 1 0 0 , round ( v a 2 i * 1 0 0 / 1 e 3 ) / 1 0 0 , round (

v a 2 m * 1 0 0 / 1 e 3 ) / 1 0 0 , 1 8 0 + v a 2 a ) ;46

47 I a r = real ( I f a u l t ) ;

48 I a i = imag ( I f a u l t ) ;

49 I a m o d = sqrt ( ( I a r ^ 2 ) + ( I a i ^ 2 ) ) ;

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50 i a a = a t a n d ( I a r / I a i ) - 9 0 ;

51 mprintf ( \n t he s u b t r a n si e n t l i n e c u r r e n t \n Ia=%f+j( % f ) A \t o r\ t %f / %d A , I a r , I a i , I a m o d , i a a ) ;

Scilab code Exa 22.02 To find ratio of line currents to single line to groundfaults

1 clear ;

2 clc ;

3 v = 1 1 e 3 ;4 r = 1 0 e 6 ;

5 x 1 = . 0 5 * % i ;

6 x 2 = . 1 5 * % i ;

7 x 0 = . 1 5 * % i ;

8 e = 1 ;

9 i a 1 = e / ( x 0 + x 1 + x 2 ) ;

10 i a = 3 * i a 1 ;

11 i c = e / x 0 ;

12 c = i a / i c ;

13 mprintf ( t h e r a t i o o f l i n e t o g round f a u l t t o 3 ph a s e

f a u l t = % f , c ) ;

Scilab code Exa 22.03 to calculate line current for single line to groundfault

1 clear ;

2 clc ;

3 v = 1 1 e 3 ;

4 r = 2 5 e 6 ;5 e = 1 ;

6 x g 0 = . 0 5 * % i ;

7 x 1 = . 1 5 * % i ;

8 x 2 = . 1 5 * % i ;

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9 z b a s e = v ^ 2 / r ;

10 r e s = . 3 ;11 x d = r e s / z b a s e ;

12 x 0 = x g 0 + ( 3 * x d * % i ) ;

13 x = x 1 + x 2 + x 0 ;

14 i a 0 = e / x ;

15 i a = 3 * i a 0 ;

16 i a b a s e = r / ( 1 . 7 3 9 8 * v ) ;

17 i a = i a * i a b a s e ;

18 ia = fix ( i a ) ;

19 printf ( t h e l i n e c ur r e nt f o r a l i n e t o gr ound f a u l t=%dA ,- imag ( i a ) ) ;

Scilab code Exa 22.04.a To calculate subtransient voltage between dou-ble line to ground fault

1 clear ;

2 clc ;

3 v = 1 1 e 3 / sqrt ( 3 ) ;

4 r = 2 5 e 6 ;

5 x 1 = . 2 5 * % i ;6 x 2 = . 3 5 * % i ;

7 x 0 = . 1 * % i ;

8 x n = 0 ;

9 e = 1 ;

10 i a 1 = e / ( x 1 + ( x 0 * x 2 / ( x 0 + x 2 ) ) ) ;

11 v a 1 = e - ( i a 1 * x 1 ) ;

12 v a 2 = v a 1 ;

13 v a 0 = v a 2 ;

14 i a 2 = - v a 2 / x 2 ;

15 i a 0 = - v a 0 / x 0 ;16 a = 1 * % e ^ ( % i * ( 1 2 0 * % p i / 1 8 0 ) ) ;

17 b = a ^ 2 ;

18 i a = ( i a 0 + i a 1 + i a 2 ) ;

19 i b = ( i a 0 + ( b * i a 1 ) + ( a * i a 2 ) ) ;

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20 i c = ( i a 0 + ( a * i a 1 ) + ( b * i a 2 ) ) ;

21 i n = 3 * i a 0 ;22 v a = 3 * v a 1 ;

23 v b = 0 ;

24 v c = v b ;

25 v a b = v a ;

26 v b c = v b - v c ;

27 v c a = - v a ;

28 v a b = v * v a b ;

29 v c a = v * v c a ;

30 i = r / ( 3 * v ) ;

31 i a 0 r = real ( i a ) ;

32 i a 0 i = imag ( i a ) ;33 i a m = sqrt ( ( i a 0 r ^ 2 ) + ( i a 0 i ^ 2 ) ) ;

34 i a 1 r = real ( i b ) ;

35 i a 1 i = imag ( i b ) ;

36 i b m = sqrt ( ( i a 1 r ^ 2 ) + ( i a 1 i ^ 2 ) ) ;

37 i a 2 r = real ( i c ) ;

38 i a 2 i = imag ( i c ) ;

39 i c m = sqrt ( ( i a 2 r ^ 2 ) + ( i a 2 i ^ 2 ) ) ;

40 i c = i c m * i ;

41 i b = i b m * i ;

42 i a = i a m * i ;

43 ib = round ( i b / 0 1 e 2 ) * 1 e 2 ;

44 ic = round ( i c / 0 1 e 2 ) * 1 e 2 ;

45 i n = i n * i * % i ;

46 mprintf ( t h e l i n e v o l t a g e s a re \nvab=%fV \ t vbc=%fkV\ t v c a=%f / 180k V\ n t h e l i n e c u r r e nt s a re \ nia=%fA \t i b=%dA \ t i c=%dA \ t i n=%dA , v a b / 1 e 3 , v b c / 1 e 3 , -v c a / 1 e 3 , i a , - i b , i c , - real ( i n ) ) ;

Scilab code Exa 22.04.b To calculate fault current following through theneutral reactor

1 clear ;

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2 clc ;

3 v = 1 1 e 3 / sqrt ( 3 ) ;4 r = 2 5 e 6 ;

5 x 1 = . 2 5 * % i ;

6 x 2 = . 3 5 * % i ;

7 x g 0 = . 1 * % i ;

8 x n = 0 . 1 * % i ;

9 e = 1 ;

10 x 0 = x g 0 + ( 3 * x n ) ;

11 i a 1 = e / ( x 1 + ( x 0 * x 2 / ( x 0 + x 2 ) ) ) ;

12 v a 1 = e - ( i a 1 * x 1 ) ;

13 v a 2 = v a 1 ;

14 v a 0 = v a 2 ;15 i a 2 = - v a 2 / x 2 ;

16 i a 0 = - v a 0 / x 0 ;

17 a = 1 * % e ^ ( % i * ( 1 2 0 * % p i / 1 8 0 ) ) ;

18 b = a ^ 2 ;

19 i a = ( i a 0 + i a 1 + i a 2 ) ;

20 i b = ( i a 0 + ( b * i a 1 ) + ( a * i a 2 ) ) ;

21 i c = ( i a 0 + ( a * i a 1 ) + ( b * i a 2 ) ) ;

22 i a 0 r = real ( i a ) ;

23 i a 0 i = imag ( i a ) ;

24 i a m = sqrt ( ( i a 0 r ^ 2 ) + ( i a 0 i ^ 2 ) ) ;

25 i a 1 r = real ( i b ) ;

26 i a 1 i = imag ( i b ) ;

27 i b m = sqrt ( ( i a 1 r ^ 2 ) + ( i a 1 i ^ 2 ) ) ;

28 i a 2 r = real ( i c ) ;

29 i a 2 i = imag ( i c ) ;

30 i c m = sqrt ( ( i a 2 r ^ 2 ) + ( i a 2 i ^ 2 ) ) ; // t he d i f f e r e n c e i nr e s u l t i s due t o e rr on e o us c a l c u l a t i o n i nt e x t b o o k .

31 i a a = 0 ;

32 i b a = a t a n d ( i a 1 i / i a 1 r ) ;

33 i c a = a t a n d ( i a 2 i / i a 2 r ) ;34 mprintf ( t h e s y mm et ri c c om po ne nt s a r e \n ia 0=%f+j%f

A \t o r\ t %f / %d A , i a 0 r , i a 0 i , i a m , i a a ) ;35 mprintf ( \n ib=%f+j%f A \t o r\ t %f / %d A , i a 1 r , i a 1 i ,

i b m , i b a ) ;

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36 mprintf ( \n i c =%f+ j ( %f ) A \t o r\ t %f / %d A ,ia2r,ia2i

, i c m , i c a ) ;37 i n = i b + i c ;

38 mprintf ( \ n n e u t a l c u r r e n t I n=%fA ,( imag ( i n ) * 1 3 1 0 ) ) ;39 disp ( // t h e d i f f e r e n c e i n r e s u l t i s due t o e rr on eo us

c a l c u l a t i o n i n t ex t bo o k . )

Scilab code Exa 22.05 TO find fault current and line to neutral voltagesat generator terminals

1 clear ;

2 clc ;

3 r = 1 0 e 6 ;

4 v = 1 1 e 3 ;

5 e = 1 ;

6 x 1 = . 2 6 * % i ;

7 x 2 = . 1 8 * % i ;

8 x 0 = . 3 6 * % i ;

9 i a 1 = e / ( x 1 + ( x 0 * x 2 / ( x 0 + x 2 ) ) ) ;

10 v a 1 = e - ( i a 1 * x 1 ) ;

11 v a 2 = v a 1 ;12 v a 0 = v a 2 ;

13 i a 2 = - v a 2 / x 2 ;

14 i a 0 = - v a 0 / x 0 ;

15 a = 1 * % e ^ ( % i * ( 1 2 0 * % p i / 1 8 0 ) ) ;

16 b = a ^ 2 ;

17 i a = ( i a 0 + i a 1 + i a 2 ) ;

18 i b = ( i a 0 + ( b * i a 1 ) + ( a * i a 2 ) ) ;

19 i c = ( i a 0 + ( a * i a 1 ) + ( b * i a 2 ) ) ;

20 i = r / ( sqrt ( 3 ) * v ) ;

21 i a = i a * i ;22 i b = i b * i ;

23 i c = i c * i ;

24 i a 0 r = real ( i a ) ;

25 i a 0 i = imag ( i a ) ;

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26 i a m = sqrt ( ( i a 0 r ^ 2 ) + ( i a 0 i ^ 2 ) ) ;

27 i a 1 r = real ( i b ) ;28 i a 1 i = imag ( i b ) ;

29 i b m = sqrt ( ( i a 1 r ^ 2 ) + ( i a 1 i ^ 2 ) ) ;

30 i a 2 r = real ( i c ) ;

31 i a 2 i = imag ( i c ) ;

32 i c m = sqrt ( ( i a 2 r ^ 2 ) + ( i a 2 i ^ 2 ) ) ;

33 i c m = round ( i c m ) ;

34 i b m = round ( i b m ) ;

35 i a a = 0 ;

36 i b a = 1 8 0 + a t a n d ( i a 1 i / i a 1 r ) ;

37 i c a = a t a n d ( i a 2 i / i a 2 r ) ;

38 mprintf ( t h e s y mm et ri c c om po ne nt s a r e \n ia 0=%f+j%fA \t o r\ t %f / %d A , i a 0 r , i a 0 i , i a m , i a a ) ;

39 mprintf ( \n ib=%f+j%f A \t o r\ t %f / %d A , i a 1 r , i a 1 i ,i b m , i b a ) ;

40 mprintf ( \n i c =%f+ j ( %f ) A \t o r\ t %f / %d A ,ia2r,ia2i, i c m , i c a ) ;

41 i n = i b + i c ;

42 mprintf ( \ n n e u t a l c u r r e n t I n=%fA ,( imag ( i n ) * 1 3 1 0 ) ) ;43 // a t g e n e r a to r44 x 1 = . 1 6 * % i ;

45 x 2 = . 0 8 * % i ;

46 x 0 = . 0 6 * % i ;

47 v a 1 = 1 - ( i a 1 * x 1 ) ;

48 v a 2 = - i a 2 * x 2 ;

49 v a 0 = i a 0 * x 0 ;

50 v a = ( v a 0 + v a 1 + v a 2 ) ;

51 v b = ( v a 0 + ( b * v a 1 ) + ( a * v a 2 ) ) ; // t h e d i f f e r e n c e i n r e s u l ti s due t o e r r o n e o u s c a l c u l a t i o n i n t ex tb oo k .

52

53 v c = ( v a 0 + ( a * v a 1 ) + ( b * v a 2 ) ) ;

54 v = v / sqrt ( 3 ) ;

55 v a = v * v a / 1 e 3 ;56 v b = v * v b / 1 e 3 ;

57 v c = v * v c / 1 e 3 ;

58 v a 0 r = real ( v a ) ;

59 v a 0 i = imag ( v a ) ;

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60 v a 0 m = sqrt ( ( v a 0 r ^ 2 ) + ( v a 0 i ^ 2 ) ) ;

61 v a 0 a = a t a n d ( v a 0 i / v a 0 r ) ;62 v a 1 r = real ( v b ) ;

63 v a 1 i = imag ( v b ) ;

64 v a 1 m = sqrt ( ( v a 1 r ^ 2 ) + ( v a 1 i ^ 2 ) ) ;

65 v a 1 a = a t a n d ( v a 1 i / v a 1 r ) ;

66 v a 2 r = real ( v c ) ;

67 v a 2 i = imag ( v c ) ;

68 v a 2 m = sqrt ( ( v a 2 r ^ 2 ) + ( v a 2 i ^ 2 ) ) ;

69 v a 2 a = a t a n d ( v a 2 i / v a 2 r ) ;

70 mprintf ( \ n t h e v o l t a g e l e v e l s a re \n va=%f+j%f kV \t o r\ t %f / %d kV , v a 0 r , v a 0 i , v a 0 m , v a 0 a ) ;

71 mprintf ( \n vb=%f+j ( %f) kV \t o r\ t %f / %d kV , v a 1 r ,v a 1 i , v a 1 m , v a 1 a ) ; // t h e d i f f e r e n c e i n r e s u l t i s due

t o e r ro n eo u s c a l c u l a t i o n i n t ex tb oo k .72 mprintf ( \n vc=%f+j ( %f) kV \t o r\ t %f / %d kV , v a 2 r ,

v a 2 i , v a 2 m , v a 2 a ) ;

73 disp ( t h e d i f f e r e n c e i n r e s u l t i s due t o e rr on e o usc a l c u l a t i o n i n t ex tb o ok . ) ;

Scilab code Exa 22.06 To calculate subtransient voltage between line toline fault

1 clear ;

2 clc ;

3 r = 1 2 5 0 e 3 ;

4 v = 6 0 0 ;

5 z 1 = . 1 5 * % i ;

6 z 2 = . 3 * % i ;

7 z 3 = . 0 5 * % i ;

8 z 4 = . 5 5 * % i ;9 x1 = inv ( inv ( z 2 ) + inv ( z 1 ) ) ;

10 x 2 = x 1 ;

11 x0 = inv ( inv ( z 3 ) + inv ( z 4 ) ) ;

12 e = 1 ;

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13 i a 1 = e / ( x 1 + x 2 + x 0 ) ;

14 i a 2 = i a 1 ;15 i a 0 = i a 2 ;

16 i a = 3 * i a 1 ; // t h e d i f f e r e n c e i n r e s u l t i s due t oe r ro n eo u s c a l c u l a t i o n i n t ex tb oo k .

17 b a s e = r / ( sqrt ( 3 ) * v ) ;

18 i t a = i a * b a s e ;

19 mprintf ( t h e f a u l t c u r r e n t =%fA ,- imag ( i t a ) ) ;20 disp ( t h e d i f f e r e n c e i n r e s u l t i s due t o e rr on e o us


Scilab code Exa 22.07 ratio of line currents for line to line to three phasefaults

1 clc ;

2 clear ;

3 e = 1 ;

4 x 1 = . 1 5 * % i ;

5 x 2 = . 1 5 * % i ;

6 i a 1 = e / ( x 1 + x 2 ) ;

7 a = 1 * % e ^ ( % i * ( 1 2 0 * % p i / 1 8 0 ) ) ;8 b = a ^ 2 ;

9 i a 2 = - i a 1 ;

10 i a = ( b - a ) * i a 1 ;

11 i a p = e / x 1 ;

12 c = real ( i a ) / imag ( i a p ) ;

13 mprintf ( th e r a t i o t o l i n e t o l i n e f a u l t t o t h re ep h a se f a u l t =%f , c ) ;

Scilab code Exa 22.08 To calculate the percentage reactance and resis-tance

1 clear ;

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2 clc ;

3 e = 1 ;4 x 1 = . 6 ;

5 x 2 = . 2 5 ;

6 x 0 = . 1 5 ;

7 i a = 1 ;

8 x n = ( 3 * e / 3 * i a ) - ( ( x 1 + x 2 + x 0 ) / 3 ) ;

9 i f a u l t = 1 ;

10 r = sqrt ( 8 / 9 ) ;

11 mprintf ( t he p e rc e n t ag e r e ac t a nc e t ha t s ho ul d beadded i n t he g e n e r a to r n e u t r a l =% fp er ce nt \n , xn* 1 0 0 ) ;

12 mprintf ( r e s i s t a n c e t o be added i n n e u t ra l t o gr oundc i r c u i t t o a c h i e v e t he same p ur p os e i s %f , r ) ;

Scilab code Exa 22.09 To find the SC current and ratio of generator con-tribution

1 clear ;

2 clc ;

3 x 1 = . 0 7 * % i ;4 x 2 = . 0 4 * % i ;

5 x 0 = . 1 * % i ;

6 e = 1 ;

7 i a = 3 * e / ( x 1 + x 2 + x 0 ) ;

8 i a = - imag ( i a ) ;

9 i a 0 = i a / 3 ;

10 i a 1 = i a / 3 ;

11 i a 2 = i a 1 ;

12 i a 1 = i a 1 / 3 ;

13 i a 2 = i a 1 ;14 i g 1 = i a 0 + i a 2 + i a 1 ;

15 i g 2 = i a 1 + i a 2 ;

16 i g 3 = i g 2 ;

17 c = i g 1 / i g 2 ;

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18 ia = round ( i a * 1 0 ) / 1 0 ;

19 c = 4 . 0 5 * c ;20 d = 4 . 0 5 ;

21 mprintf ( f o r s i n g l e l i n e t o ground f a u l t I a=j% fA ,i a ) ;

22 mprintf ( \ n t h e r a t i o o f c o n t r i bu t i o n o f g e n e ra t o r I ,I I and I I I i s %d : %d : %d , c , d , d ) ;

23 i 3 = e / ( x 1 ) ;

24 i l = 3 * e / ( x 1 + x 2 + x 0 ) ;

25 y = i 3 / i l ;

26 mprintf ( \ n t h e r a t i o o f 3p ha se t o l i n e t o groundf a u l t = % f , y ) ;

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Chapter 23

Faults On Power Systems

Scilab code Exa 23.03 To calculate the fault current

1 clear ;

2 clc ;

3 v f = 1 ;

4 r = 1 2 5 0 e 3 ;

5 V = 6 0 0 ;

6 x 1 = . 5 ;

7 x 2 = . 5 ;8 x 3 = . 0 2 ;

9 i a 2 = v f / ( x 1 + x 2 + x 3 ) ;

10 i a = 3 * i a 2 ;

11 i a 1 = i a 2 ;

12 i a 0 = i a 1 ;

13 i a b = r / ( sqrt ( 3 ) * V ) ;

14 i a b = round ( i a b / 1 0 ) * 1 0 ;

15 ia = round ( i a * 1 0 0 ) / 1 0 0 ;

16 I f = i a * i a b ; // t h e d i f f e r e n c e i n r e s u l t i s due t o

e r ro n eo u s c a l c u l a t i o n i n t ex tb oo k .17 printf ( f a u l t c u r r e n t I f =%fA , I f ) ;18 disp ( t h e d i f f e r e n c e i n r e s u l t i s due t o e rr on e o us


69


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1 clear ;

2 clc ;

3 v = 1 ;

4 r = 1 2 5 0 e 3 ;

5 V = 6 0 0 ;

6 x 1 = . 0 5 * % i ;

7 x 2 = . 0 5 * % i ;

8 x 0 = . 0 2 * % i ;

9 a = 1 * % e ^ ( % i * ( 1 2 0 * % p i / 1 8 0 ) ) ;

10 b = a ^ 2 ;

11 i a 1 = v / ( x 1 + inv ( inv ( x 2 ) + inv ( x 0 ) ) ) ;

12 i b a s e = 1 2 0 0 ;

13 v a 1 = v - ( i a 1 * x 1 ) ;

14 i a 2 = - v a 1 / x 2 ;

15 i a 0 = - v a 1 / x 0 ;

16 i a = ( i a 0 + i a 1 + i a 2 ) ;

17 i b = ( i a 0 + ( b * i a 1 ) + ( a * i a 2 ) ) ;

18 i c = ( i a 0 + ( a * i a 1 ) + ( b * i a 2 ) ) ;19 i a 0 r = real ( i a ) ;

20 i a 0 i = imag ( i a ) ;

21 i a m = sqrt ( ( i a 0 r ^ 2 ) + ( i a 0 i ^ 2 ) ) ;

22 i a 1 r = real ( i b ) ;

23 i a 1 i = imag ( i b ) ;

24 i b m = sqrt ( ( i a 1 r ^ 2 ) + ( i a 1 i ^ 2 ) ) ; // t he d i f f e r e n c e i nr e s u l t i s due t o e rr on e o us c a l c u l a t i o n i nt e x t b o o k .

25 i a 2 r = real ( i c ) ;

26 i a 2 i = imag ( i c ) ;

27 i c m = sqrt ( ( i a 2 r ^ 2 ) + ( i a 2 i ^ 2 ) ) ;

28 i a a = 0 ;

29 i b a = a t a n d ( i a 1 i / i a 1 r ) ;

30 i c a = a t a n d ( i a 2 i / i a 2 r ) ;

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31 i m = i b m * i b a s e ;

32 mprintf ( f a u l t c u r r e n t f o r d ou bl e l i n e t o gr oundfault=%fA , i m )33 disp ( t h e d i f f e r e n c e i n r e s u l t i s due t o e rr on e o us



1 clear ;

2 clc ;3 v = 1 ;

4 r = 1 2 5 0 e 3 ;

5 V = 6 0 0 ;

6 x 1 = . 0 5 * % i ;

7 x 2 = . 0 5 * % i ;

8 x 0 = . 0 2 * % i ;

9 i a 1 = v / ( x 1 + x 2 ) ;

10 i a 2 = - i a 1 ;

11 i a = i a 1 + i a 2 ;

12 i a 0 = 0 ;

13 a = 1 * % e ^ ( % i * ( 1 2 0 * % p i / 1 8 0 ) ) ;14 b = a ^ 2 ;

15 i a = ( i a 0 + i a 1 + i a 2 ) ;

16 i b = ( i a 0 + ( b * i a 1 ) + ( a * i a 2 ) ) ;

17 i c = ( i a 0 + ( a * i a 1 ) + ( b * i a 2 ) ) ;

18 i a 0 r = real ( i a ) ;

19 i a 0 i = imag ( i a ) ;

20 i a m = sqrt ( ( i a 0 r ^ 2 ) + ( i a 0 i ^ 2 ) ) ;

21 i a 1 r = real ( i b ) ;

22 i a 1 i = imag ( i b ) ;

23 i b m = sqrt ( ( i a 1 r ^ 2 ) + ( i a 1 i ^ 2 ) ) ;24 i a 2 r = real ( i c ) ;

25 i a 2 i = imag ( i c ) ;

26 i c m = sqrt ( ( i a 2 r ^ 2 ) + ( i a 2 i ^ 2 ) ) ;

27 i a a = 0 ;

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28 i b a = a t a n d ( i a 1 i / i a 1 r ) ;

29 i c a = a t a n d ( i a 2 i / i a 2 r ) ;30 i b a s e = r / ( sqrt ( 3 ) * V ) ;

31 i b m = i b m * i b a s e ;

32 i b m = round ( i b m / 1 0 0 ) * 1 0 0 ;

33 mprintf ( f a u l t c u r r e n t f o r d ou bl e l i n e t o gr oundf a u l t=%dA , i b m ) ;

Scilab code Exa 23.06 to find the subtransient fault currents

1 clear ;

2 clc ;

3 r = 1 2 5 0 e 3 ;

4 v = 6 0 0 ;

5 z 1 = . 1 5 * % i ;

6 z 2 = . 3 * % i ;

7 z 3 = . 0 5 * % i ;

8 z 4 = . 5 5 * % i ;

9 x1 = inv ( inv ( z 2 ) + inv ( z 1 ) ) ;

10 x 2 = x 1 ;

11 x0 = inv ( inv ( z 3 ) + inv ( z 4 ) ) ;12 e = 1 ;

13 i a 1 = e / ( x 1 + x 2 + x 0 ) ;

14 i a 2 = i a 1 ;

15 i a 0 = i a 2 ;

16 i a = 3 * i a 1 ; // t h e d i f f e r e n c e i n r e s u l t i s due t oe r ro n eo u s c a l c u l a t i o n i n t ex tb oo k .

17 b a s e = r / ( sqrt ( 3 ) * v ) ;

18 i t a = i a * b a s e ;

19 mprintf ( t h e f a u l t c u r r e n t =%fA ,- imag ( i t a ) ) ;

20 disp ( t h e d i f f e r e n c e i n r e s u l t i s due t o e rr on e o usc a l c u l a t i o n i n t ex tb o ok . ) ;

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Scilab code Exa 23.07 To calculate the fault current for different cases

1 clear ;

2 clc ;

3 e = 1 ;

4 r = 1 5 0 0 e 3 ;

5 v = 1 1 e 3 ;

6 x 1 = . 1 ;

7 i a = 3 * e / ( x 1 * 3 ) ;

8 i b a s e = r / ( sqrt ( 3 ) * v ) ;

9 i = i a * i b a s e ;

10 mprintf ( t he s i n g l e l i n e t o ground f a u l t = %dA , i ) ;

11 i a 1 = e / ( 2 * x 1 ) ;12 ib = sqrt ( 3 ) * i a 1 ;

13 i b = i b a s e * i b ;

14 mprintf ( \n l i n e t o l i n e f a u l t c u r r e nt=%dA , i b ) ;

Scilab code Exa 23.08 To calculate fault current and phase voltages

1 clear ;

2 clc ;3 X 1 = 6 . 6 * % i ;

4 X 2 = 6 . 3 * % i ;

5 X 0 = 1 2 . 6 * % i ;

6 r = 3 7 . 5 e 6 ;

7 v = 3 3 e 3 ;

8 e = 1 ;

9 z b = v ^ 2 / r ;

10 x 1 = X 1 / z b ;

11 x 2 = X 2 / z b ;

12 x 0 = X 0 / z b ;13 x 1 g = . 1 8 * % i ;

14 x 2 g = . 1 2 * % i ;

15 x 0 g = . 1 * % i ;

16 x 1 = x 1 + x 1 g ;

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17 x 2 = x 2 + x 2 g ;

18 x 0 = x 0 + x 0 g ;19 i a = 3 * e / ( x 1 + x 2 + x 0 ) ;

20 i a 1 = i a / 3 ;

21 a = 1 * % e ^ ( % i * ( 1 2 0 * % p i / 1 8 0 ) ) ;

22 b = a ^ 2 ;

23 i b a s e = r / ( sqrt ( 3 ) * v ) ;

24 i a n = i a * i b a s e ;

25 printf ( f a u l t c u r r e n t =%djAmp, imag ( i a n ) ) ;26 v a = e - ( i a 1 * x 1 g ) ;

27 v b = - i a 1 * x 2 g ;

28 v c = - i a 1 * x 0 g ;

29 v a 0 = ( v a + v b + v c ) ;30 v a 1 = ( v a + ( b * v b ) + ( a * v c ) ) ;

31 v a 2 = ( v a + ( a * v b ) + ( b * v c ) ) ;

32 v = v / sqrt ( 3 ) ;

33 v a 0 = v a 0 * v ;

34 v a 1 = v a 1 * v ;

35 v a 2 = v a 2 * v ;

36 v a 0 r = real ( v a 0 ) ;

37 v a 0 i = imag ( v a 0 ) ;

38 v a 0 m = sqrt ( ( v a 0 r ^ 2 ) + ( v a 0 i ^ 2 ) ) ;

39 v a 0 a = a t a n d ( v a 0 i / v a 0 r ) ;

40 v a 1 r = real ( v a 1 ) ;

41 v a 1 i = imag ( v a 1 ) ;

42 v a 1 m = sqrt ( ( v a 1 r ^ 2 ) + ( v a 1 i ^ 2 ) ) ;

43 v a 1 a = a t a n d ( v a 1 i / v a 1 r ) - 1 2 0 ;

44 v a 2 r = real ( v a 2 ) ;

45 v a 2 i = imag ( v a 2 ) ;

46 v a 2 m = sqrt ( ( v a 2 r ^ 2 ) + ( v a 2 i ^ 2 ) ) ;

47 v a 2 a = a t a n d ( v a 2 i / v a 2 r ) + 1 2 0 ;

48 mprintf ( \ n t h e v o l t a g e l e v e l s a re \n va=%f+j%f V \t o r\ t %d/ %d kV , v a 0 r / 1 e 3 , v a 0 i / 1 e 3 , v a 0 m / 1 e 3 , v a 0 a )

;49 mprintf ( \n vb=%f+j ( %f) kV \t o r\ t %d/ %d kV , v a 1 r / 1

e 3 , v a 1 i / 1 e 3 , v a 1 m / 1 e 3 , v a 1 a ) ;

50 mprintf ( \n vc=%f+j ( %f) kV \t o r\ t %d/ %d kV , v a 2 r / 1e 3 , v a 2 i / 1 e 3 , v a 2 m / 1 e 3 , v a 2 a ) ;

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32 i a 1 r = real ( i b ) ;

33 i a 1 i = imag ( i b ) ;34 i b m = sqrt ( ( i a 1 r ^ 2 ) + ( i a 1 i ^ 2 ) ) ;

35 i a 2 r = real ( i c ) ;

36 i a 2 i = imag ( i c ) ;

37 i c m = sqrt ( ( i a 2 r ^ 2 ) + ( i a 2 i ^ 2 ) ) ;

38 i a a = - 9 0 ;

39 i b a = 1 8 0 + a t a n d ( i a 1 i / i a 1 r ) ;

40 i c a = a t a n d ( i a 2 i / i a 2 r ) ;

41 mprintf ( t h e s ym me tr ic c om po ne nt s f o r t h r e e p ha s ef a u l t a re \n i a 0=%f+j %f A \t o r\ t %f / %d A , i a 0 r ,i a 0 i , i a m , i a a ) ;

42 mprintf ( \n ib=%f+j%f A \t o r\ t %f / %d A , i a 1 r , i a 1 i ,i b m , i b a ) ;

43 mprintf ( \n i c =%f+ j ( %f ) A \t o r\ t %f / %d A ,ia2r,ia2i, i c m , i c a ) ;

44 i a 1 = e / ( x 1 + x 2 ) ;

45 i a 2 = - i a 1 ;

46 i a 0 = 0 ;

47 i a = ( i a 0 + i a 1 + i a 2 ) ;

48 i b = ( i a 0 + ( b * i a 1 ) + ( a * i a 2 ) ) ;

49 i c = ( i a 0 + ( a * i a 1 ) + ( b * i a 2 ) ) ;

50 i = r / ( sqrt ( 3 ) * v ) ;

51 i a = i a * i ;

52 i b = i b * i ;

53 i c = i c * i ;

54 i a 0 r = real ( i a ) ;

55 i a 0 i = imag ( i a ) ;

56 i a m = sqrt ( ( i a 0 r ^ 2 ) + ( i a 0 i ^ 2 ) ) ;

57 i a 1 r = real ( i b ) ;

58 i a 1 i = imag ( i b ) ;

59 i b m = sqrt ( ( i a 1 r ^ 2 ) + ( i a 1 i ^ 2 ) ) ;

60 i a 2 r = real ( i c ) ;

61 i a 2 i = imag ( i c ) ;62 i c m = sqrt ( ( i a 2 r ^ 2 ) + ( i a 2 i ^ 2 ) ) ;

63 i a a = 0 ;

64 i b a = 1 8 0 + a t a n d ( i a 1 i / i a 1 r ) ;

65 i c a = a t a n d ( i a 2 i / i a 2 r ) ;

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66 i c m = round ( i c m / 1 0 ) * 1 0 ;

67 i b m = round ( i b m / 1 0 ) * 1 0 ;68 mprintf ( \ n th e s ymm et ri c c omponent s f o r l i n e t o l i n ef a u l t a re \n i a 0=%f+j %f A \t o r\ t %f / % f A , i a 0 r ,

i a 0 i , i a m , i a a ) ;

69 mprintf ( \n ib=%f+j%f A \t o r\ t %f / % f A , i a 1 r , i a 1 i ,i b m , i b a ) ;

70 mprintf ( \n i c =%f+ j ( %f ) A \t o r\ t %f / % f A ,ia2r,ia2i, i c m , i c a ) ;

71 i a 1 = e / ( x 1 + x 2 + z 0 ) ;

72 i a 2 = i a 1 ;

73 i a 0 = i a 2 ;

74 i a = ( i a 0 + i a 1 + i a 2 ) ;75 i b = ( i a 0 + ( b * i a 1 ) + ( a * i a 2 ) ) ;

76 i c = ( i a 0 + ( a * i a 1 ) + ( b * i a 2 ) ) ;

77 i = r / ( sqrt ( 3 ) * v ) ;

78 i a = i a * 8 7 4 ;

79 i a 0 r = real ( i a ) ;

80 i a 0 i = imag ( i a ) ;

81 i a m = sqrt ( ( i a 0 r ^ 2 ) + ( i a 0 i ^ 2 ) ) ;

82 i a 1 r = real ( i b ) ;

83 i a 1 i = imag ( i b ) ;

84 i b m = sqrt ( ( i a 1 r ^ 2 ) + ( i a 1 i ^ 2 ) ) ;

85 i a 2 r = real ( i c ) ;

86 i a 2 i = imag ( i c ) ;

87 i c m = sqrt ( ( i a 2 r ^ 2 ) + ( i a 2 i ^ 2 ) ) ;

88 i a a = a t a n d ( i a 0 i / i a 0 r ) ;

89 i b a = 0 ;

90 i c a = 0 ;

91 mprintf ( \ n th e s y mm et ri c c omp on en ts f o r s i n g l e l i n et o g round f a u l t a re \n i a 0=%f+j %f A \t o r\ t %f / %f

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