Electrical Power Systems_C. L. Wadhwa

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Scilab Textbook Companion for

Electrical Power Systems

by C. L. Wadhwa1

Created byAnuj Bansal

B.EElectrical Engineering

Thapar University, Patiala(Punjab)College Teacher

Dr. Sunil Kumar SinglaCross-Checked by

Lavitha Pereira

July 4, 2014

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: Electrical Power Systems

Author: C. L. Wadhwa

Publisher: New Age International Pvt. Ltd., New Delhi

Edition: 6

Year: 2010

ISBN: 9788122428391

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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

1 FUNDAMENTALS OF POWER SYSTEMS 12

2 LINE CONSTANT CALCULATIONS 14

3 CAPACITANCE OF TRANSMISSION LINES 18

4 PERFORMANCE OF LINES 20

5 HIGH VOLTAGE DC TRANSMISSION 30

6 CORONA 33

7 MECHANICAL DESIGN OF TRANSMISSION LINES 37

8 OVERHEAD LINE INSULATORS 41

9 INSULATED CABLES 42

10 VOLTAGE CONTROL 47

11 NEUTRAL GROUNDING 51

12 TRANSIENTS IN POWER SYSTEMS 53

13 SYMMETRICAL COMPONENTS AND FAULT CALCU-

LATIONS 58

14 PROTECTIVE RELAYS 72

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15 CIRCUIT BREAKERS 78

17 POWER SYSTEM SYNCHRONOUS STABILITY 82

18 LOAD FLOWS 92

19 ECONOMIC LOAD DISPATCH 98

20 LOAD FREQUENCY CONTROL 101

21 COMPENSATION IN POWER SYSTEMS 103

22 POWER SYSTEM VOLTAGE STABILITY 105

23 STATE ESTIMATION IN POWER SYSTEMS 109

24 UNIT COMMITMENT 114

25 ECONOMIC SCHEDULING OF HYDROTHERMAL PLANTS

AND OPTIMAL POWER FLOWS 117

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List of Scilab Codes

Exa 1.1 To determine the Base values and pu values . . . . . . 12Exa 2.2 To dtermine inductance of a 3 phase line . . . . . . . 14Exa 2.3 Determine the equivalent radius of bundle conductor

having its part conductors r on the periphery of circleof dia d . . . . . . . . . . . . . . . . . . . . . . . . . . 14

Exa 2.4 To determine the inductance of single phase Transmis-sion line . . . . . . . . . . . . . . . . . . . . . . . . . . 15

Exa 2.5 To determine the inductance per Km of 3 phase line . 16Exa 2.6 To determine the inductance of double circuit line . . 17Exa 2.7 To determine the inductance per Km per phase of single

circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . 17Exa 3.1 To determine the capacitance and charging current . 18Exa 3.2 To determine the capacitance and charging current . 18

Exa 3.3 To determine the capacitance and charging current . 19Exa 4.1 To determine the sending end voltage and current power

and power factor Evaluate A B C D parameters . . . . 20Exa 4.2 To determine power input and output i star connected

ii delta connected . . . . . . . . . . . . . . . . . . . . 21Exa 4.3 To determine efficiency and regulation of line . . . . . 22Exa 4.4 To calculate the voltage across each load impedence and

current in the nuetral . . . . . . . . . . . . . . . . . . 22Exa 4.5 To determine efficiency and regulation of 3 phase line. 23Exa 4.6 To find the rms value and phase values i The incident

voltage to neutral at the recieving end ii The reflected

voltage to neutral at the recieving end iii The incidentand reflected voltage to neutral at 120 km from the re-cieving end . . . . . . . . . . . . . . . . . . . . . . . . 24

Exa 4.7 To determine of efficiency of line . . . . . . . . . . . . 26

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Exa 4.8 To determine the ABCD parameters of Line . . . . . . 26

Exa 4.9 To determine the sending end voltage and efficiency us-ing Nominal pi and Nominal T method . . . . . . . . 27Exa 4.10 To determine the sending end voltage and current power

and power factor Evaluate A B C D parameters . . . . 28Exa 5.1 To determine the dc output voltage when delay anglw

a0 b30 c45 . . . . . . . . . . . . . . . . . . . . . . . . 30Exa 5.2 To determine the necessary line secondary voltage and

tap ratio required . . . . . . . . . . . . . . . . . . . . 30Exa 5.3 To determine the effective reactance per phase . . . . 31Exa 5.4 Calculate the direct current delivered. . . . . . . . . . 31Exa 6.1 To determine the critical disruptive voltage and critical

voltage for local and general corona . . . . . . . . . . 33Exa 6.2 To determine whether corona will be present in the air

space round the conductor. . . . . . . . . . . . . . . . 34Exa 6.3 To determine the critical disruptive voltage and corona

loss . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34Exa 6.4 To determine the voltage for which corona will com-

mence on the line . . . . . . . . . . . . . . . . . . . . 35Exa 6.5 To determine the corona characterstics . . . . . . . . . 35Exa 7.1 Calculate the sag . . . . . . . . . . . . . . . . . . . . 37Exa 7.2 To calculate the maximum Sag . . . . . . . . . . . . . 37

Exa 7.3 To determine the Sag . . . . . . . . . . . . . . . . . . 38Exa 7.4 To determine the clearence between the conductor andwater level . . . . . . . . . . . . . . . . . . . . . . . . 39

Exa 8.1 To determine the maximum voltage that the string ofthe suspension insulators can withstand . . . . . . . . 41

Exa 9.1 To determine the economic overall diameter of a 1corecable metal sheathead . . . . . . . . . . . . . . . . . . 42

Exa 9.2 To determine the minimum internal diameter of the leadsheath . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

Exa 9.3 To determine the maximum safe working voltage . . . 43Exa 9.4 To determine the maximum stresses in each of the three

layers . . . . . . . . . . . . . . . . . . . . . . . . . . . 44Exa 9.5 o dtermine the equivalent star connected capacity and

the kVA required . . . . . . . . . . . . . . . . . . . . . 44

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Exa 9.6 Determine the capacitance a between any two conduc-

tors b between any two bunched conductors and thethird conductor c Also calculate the charging currentper phase per km. . . . . . . . . . . . . . . . . . . . . 45

Exa 9.7 To calculate the induced emf in each sheath . . . . . 45Exa 9.8 To determine the ratio of sheath loss to core loss of the

cable . . . . . . . . . . . . . . . . . . . . . . . . . . . 46Exa 10.1 To determine the total power active and reactive sup-

plied by the generator and the pf at which the generatormust operate . . . . . . . . . . . . . . . . . . . . . . . 47

Exa 10.2 Determine the settings of the tap changers required tomaintain the voltage of load bus bar . . . . . . . . . . 48

Exa 10.3 i Find the sending end Voltage and the regulation ofline ii Determine the reactance power supplied by theline and by synchronous capacotor and pf of line iii De-termine the maximum power transmitted . . . . . . . 48

Exa 10.4 Determine the KV Ar of the Modifier and the maximumload that can be transmitted . . . . . . . . . . . . . . 49

Exa 11.1 To find the inductance and KVA rating of the arc sup-pressor coil in the system . . . . . . . . . . . . . . . . 51

Exa 11.2 Determine the reactance to neutralize the capacitanceof i 100 percent of the length of line ii 90 percent of the

length of line iii 80 percent of the length of line . . . . 51Exa 12.1 To determine the i the neutral impedence of line ii linecurrent iii rate of energy absorption rate of reflectionand state form of reflection iv terminating resistance vamount of reflected and transmitted power . . . . . . 53

Exa 12.2 Find the voltage rise at the junction due to surge . . 54Exa 12.3 To find the surge voltages and currents transmitted into

branch line . . . . . . . . . . . . . . . . . . . . . . . . 55Exa 12.4 Determine the maximum value of transmitted wave. . 55Exa 12.5 Determine the maximum value of transmitted surge . 56Exa 12.6 Determine i the value of the Voltage wave when it has

travelled through a distance 50 Km ii Power loss andHeat loss . . . . . . . . . . . . . . . . . . . . . . . . . 56

Exa 13.1 Determine the symmetrical components of voltages . . 58Exa 13.2 Find the symmetrical component of currents . . . . . 58Exa 13.3 Determine the fault current and line to line voltages . 59

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Exa 13.4 determine the fault current and line to line voltages at

the fault . . . . . . . . . . . . . . . . . . . . . . . . . 60Exa 13.5 determine the fault current and line to line voltages atthe fault . . . . . . . . . . . . . . . . . . . . . . . . . 61

Exa 13.6 Determine the fault current when i LG ii LL iii LLGfault takes place at P . . . . . . . . . . . . . . . . . . 62

Exa 13.8 Determine the percent increase of busbar voltage . . . 64Exa 13.9 Determine the short circuit capacity of the breaker . . 64Exa 13.10 To determine the short circuit capacity of each station 64Exa 13.11 Determine the Fault MVA . . . . . . . . . . . . . . . 65Exa 13.12 To Determine the subtransient current in the alternator

motor and the fault . . . . . . . . . . . . . . . . . . . 65

Exa 13.13 To Determine the reactance of the reactor to preventthe brakers being overloaded . . . . . . . . . . . . . . 66

Exa 13.14 Determine the subtransient currents in all phases of ma-chine1 the fault current and the voltages of machine 1and voltage at the fault point . . . . . . . . . . . . . . 67

Exa 13.15 To determine the i pre fault current in line a ii the sub-transient current in pu iii the subtransient current ineach phase of generator in pu . . . . . . . . . . . . . 68

Exa 13.16 Determine the shorrt circuit MVA of the transformer 69Exa 13.17 To determine the line voltages and currents in per unit

on delta side of the transformer . . . . . . . . . . . . . 70Exa 14.1 To determine the time of operation of relay . . . . . . 72Exa 14.2 To determine the phase shifting network to be used. . 72Exa 14.3 To provide time current grading . . . . . . . . . . . . 73Exa 14.4 To determine the proportion of the winding which re-

mains unprotected against earth fault . . . . . . . . . 73Exa 14.5 To determine i percent winding which remains unpro-

tected ii min value of earthing resistance required toprotect 80 percent of winding . . . . . . . . . . . . . 74

Exa 14.6 To determine whether relay will operate or not . . . . 75Exa 14.7 To determine the ratio of CT on HV side . . . . . . . 75

Exa 14.8 To determine the number of turns each current trans-former should have . . . . . . . . . . . . . . . . . . . . 75

Exa 14.9 To determine the R1 R2 and C also The potential acrossrelays . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

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Exa 14.10 To determine the kneepoint voltage and cross section of

core . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76Exa 14.11 To determine the VA output of CT . . . . . . . . . . 77Exa 15.1 To determine the voltage appearing across the pole of

CB also determine the value of resistance to be usedacross contacts . . . . . . . . . . . . . . . . . . . . . . 78

Exa 15.2 To determine the rate of rise of restriking voltage . . . 78Exa 15.3 To Determine the average rate of rise of restriking voltage 79Exa 15.4 To determine the rated normal current breaking current

making current and short time rating current . . . . . 80Exa 15.5 TO Determine i sustained short circuit current in the

breaker ii initial symmetrical rms current in the breaker

iii maximum possible dc component of the short circuitcurrent in the breaker iv momentary current rating ofthe breaker v the current . . . . . . . . . . . . . . . . 80

Exa 17.1 To determine the acceleration Also determine the changein torque angle and rpmat the end of 15 cycles . . . . 82

Exa 17.2 To determine the frequency of natural oscillations if thegenrator is loaded to i 60 Percent and ii 75 percent ofits maximum power transfer capacity. . . . . . . . . . 83

Exa 17.3 To calculate the maximum value of d during the swing-ing of the rotor around its new equilibrium position. . 84

Exa 17.4 To calculate the critical clearing angle for the conditiondescribed . . . . . . . . . . . . . . . . . . . . . . . . . 84Exa 17.5 To calculate the critical clearing angle for the generator

for a 3phase fault. . . . . . . . . . . . . . . . . . . . . 85Exa 17.6 determine the critical clearing angle . . . . . . . . . . 86Exa 17.7 To determine the centre and radius for the pull out curve

ans also minimum output vars when the output powersare i 0 ii 25pu iii 5pu. . . . . . . . . . . . . . . . . . . 87

Exa 17.8 Compute the prefault faulted and post fault reduced Ymatrices. . . . . . . . . . . . . . . . . . . . . . . . . . 87

Exa 17.9 Determine the reduced admittance matrices for prefault

fault and post fault conditions and determine the powerangle characterstics for three conditions . . . . . . . . 88

Exa 17.10 To Determine the rotor angle and angular frequency us-ing runga kutta and eulers modified method . . . . . . 89

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Exa 18.1 Determine the voltages at the end of first iteration using

gauss seidal method . . . . . . . . . . . . . . . . . . . 92Exa 18.2 Determine the voltages starting with a flat voltage profile 93Exa 18.3 Solve the prevous problem for for voltages at the end of

first iteration . . . . . . . . . . . . . . . . . . . . . . . 94Exa 18.4 Determine the set of load flow equations at the end of

first iteration by using Newton Raphson method . . . 95Exa 18.5 Determine the equations at the end of first iteration

after applying given constraints . . . . . . . . . . . . . 97Exa 19.1 To Determine the economic operating schedule and the

corresponding cost of generation b Determine the sav-ings obtained by loading the units . . . . . . . . . . . 98

Exa 19.2 Determine the incremental cost of recieved power andpenalty factor of the plant. . . . . . . . . . . . . . . . 99

Exa 19.4 Determine the minimum cost of generation . . . . . . 99Exa 20.1 Determine the load taken by the set C and indicate the

direction in which the energy is flowing . . . . . . . . 101Exa 20.2 Determine the load shared by each machine . . . . . . 101Exa 20.3 Determine the frequency to which the generated voltage

drops before the steam flow commences to increase tomeet the new load . . . . . . . . . . . . . . . . . . . . 102

Exa 21.1 Determine the load bus voltage . . . . . . . . . . . . . 103

Exa 22.2 To Determine the source voltage when the load is dis-connected to load pf i unity ii 8 lag . . . . . . . . . . . 105Exa 22.3 To determine thee Ac system voltage when the dc sys-

tem is disconnected or shutdown . . . . . . . . . . . . 106Exa 22.4 To Calculate the new on and off times for constant energy 106Exa 22.6 To discuss the effect of tap changing . . . . . . . . . . 107Exa 22.7 To determine the effect of tapping to raise the secondary

voltage by 10percent . . . . . . . . . . . . . . . . . . . 107Exa 22.8 Calculate the additional reactive power capability at full

load . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108Exa 23.1 To determine the state vector at the end of first iteration 109

Exa 23.2 Determine The States of the systems at the end of firstiteration. . . . . . . . . . . . . . . . . . . . . . . . . . 111

Exa 23.3 Problem on State Estimator Linear Model . . . . . . . 112Exa 23.4 Determine theta1 Theta2 . . . . . . . . . . . . . . . . 112Exa 24.3 Priority List Method. . . . . . . . . . . . . . . . . . . 114

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Exa 24.4 illustrate the dynamic programming for preparing an

optimal unit commitment . . . . . . . . . . . . . . . . 115Exa 25.1 illustrating the procedure for economic scheduling clearall . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

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Chapter 1

FUNDAMENTALS OF

POWER SYSTEMS

Scilab code Exa 1.1 To determine the Base values and pu values

1 / / To d e t er m i ne t h e B as e v a l u e s and p . u v a l u e s2 clear

3 clc ;

4 S b = 1 0 0 ; / / b a s e v a l u e o f p o we r (MVA)5 V b = 3 3 ; // b as e v al u e o f v o l t ag e ( Kv )6 V b l = V b * 1 1 0 / 3 2 ;

7 V b m = V b l * 3 2 / 1 1 0 ;

8 Z p . u t = 0 . 0 8 * 1 0 0 * 3 2 * 3 2 / ( 1 1 0 * 3 3 * 3 3 ) ;

9 Z p . u . l = 5 0 * 1 0 0 / ( V b l ^ 2 ) ;

10 Z p . u m 1 = . 2 * 1 0 0 * 3 0 * 3 0 / ( 3 0 * 3 3 * 3 3 ) ;

11 Z p . u m 2 = . 2 * 1 0 0 * 3 0 * 3 0 / ( 2 0 * 3 3 * 3 3 ) ;

12 Z p . u m 3 = . 2 * 1 0 0 * 3 0 * 3 0 / ( 5 0 * 3 3 * 3 3 ) ;

13 mprintf ( Base v al u e o f v o l t ag e i n l i n e = %. 2 f kV\n ,V b l ) ;

14 mprintf ( B ase v a lu e o f v o l t a g e i n motor c i r c u i t =%. 0 f kV\n , V b m ) ;15 mprintf ( p . u v a l u e o f r e a c t a n c e t r a n s f o r m e r =%. 5 f p .

u\n , Z p . u t ) ;16 mprintf ( p . u v a l u e o f i m pe de n ce o f l i n e =%. 4 f p . u\n ,

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Z p . u . l ) ;

17 mprintf ( p . u v a l u e o f r e a c t a n c e o f m oto r 1 =%. 4 f p . u\n , Z p . u m 1 ) ;18 mprintf ( p . u v a l u e o f r e a c t a n c e o f m oto r 2 =%. 3 f p . u

\n , Z p . u m 2 ) ;19 mprintf ( p . u v a l u e o f r e a c t a n c e o f m oto r 3 =%. 4 f p . u

\n , Z p . u m 3 ) ;

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Chapter 2

LINE CONSTANT

CALCULATIONS

Scilab code Exa 2.2 To dtermine inductance of a 3 phase line

1 / /To d te rm in e i n du c t an c e o f a 3 p ha se l i n e2 clear

3 clc ;

4 G M D = 0 . 7 7 8 8 * 0 . 8 / ( 2 * 1 0 0 ) ;

5 M g m d = ( ( 1 . 6 * 3 . 2 * 1 . 6 ) ^ ( 1 / 3 ) ) ;

6 Z = 2 * ( 1 0 ^ - 4 ) * 1 0 0 0 * log ( 2 . 0 1 5 / . 0 0 3 1 1 5 ) ;

7 mprintf ( The s e l f GMD o f t h e c o n d u c t o r =% . 6 f m e t r e s \n , G M D ) ;

8 mprintf ( The m u tu a l GMD o f t h e c o n d u c t o r =% . 3 f m e t r e s \n , M g m d ) ;

9 mprintf ( In du ct an ce =%.3 f mH/km\n , Z ) ;

Scilab code Exa 2.3 Determine the equivalent radius of bundle conductorhaving its part conductors r on the periphery of circle of dia d

1 / /What w i l l be t he e q u i v a l e n t r a d i us o f b un dl ec on du ct or h av in g i t s p ar t c on d uc t o r s r on t he

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p e r i ph e r y o f c i r c l e o f d ia d i f t he number o f

c on du c to r s i s 2 , 3 , 4 , 6 ?23 clear

4 clc ;

5 r = poly (0 , r ) ;6 D 1 1 = r ^ 1 ;

7 D 1 2 = 2 * r ;

8 D 1 4 = 4 * r

9 D 1 3 = sqrt ( 1 6 - 4 ) * r ;

10 D s 1 = ( ( 1 * 2 * 2 * sqrt ( 3 ) * 4 * 2 * sqrt ( 3 ) * 2 * 2 ) ^ ( 1 / 7 ) ) * r ;

11 D s 7 = ( ( 2 * 1 * 2 * 2 * * 2 * 2 * 2 ) ^ ( 1 / 7 ) ) * r ; // we g e t t h i s a f t e r

Taking r o u t s i d e t he 1/7 t h r o ot12 D s = ( ( ( ( 1 * 2 * 2 * sqrt ( 3 ) * 4 * 2 * sqrt ( 3 ) * 2 * 2 ) ^ ( 1 / 7 ) ) ^ 6 )

* ( ( 2 * 1 * 2 * 2 * * 2 * 2 * 2 ) ^ ( 1 / 7 ) ) ) ^ ( 1 / 7 ) * r ;

13 D s e q = ( ( . 7 7 8 8 ) ^ ( 1 / 7 ) ) * D s ;

14 disp ( D s e q , D se q .= ) ;

Scilab code Exa 2.4 To determine the inductance of single phase Trans-mission line

1 / / To d et er mi ne t he i n du c t a nc e o f s i n g l e p ha seT ra n sm i ss i on l i n e

2 clear

3 clc ;

4 G M D a = 0 . 0 0 1 9 4 7 ; / / GMD o f c o n d u c t o r i n g ro u p A5 D S A = ( ( . 0 0 1 9 4 7 * 6 * 1 2 * . 0 0 1 9 4 7 * 6 * 6 * 0 . 0 0 1 9 4 7 * 6 * 1 2 ) ^ ( 1 / 9 ) )

;

6 D S B = sqrt ( 5 * ( 1 0 ^ - 3 ) * . 7 7 8 8 * 6 ) ;

7 D a e = sqrt ( ( 9 ^ 2 ) + 6 ^ 2 ) ;

8 D c d = sqrt ( ( 1 2 ^ 2 ) + 9 ^ 2 ) ;9 D M A = ( ( 9 * 1 0 . 8 1 * 1 0 . 8 1 * 9 * 1 5 * 1 0 . 8 1 ) ^ ( 1 / 6 ) ) ;

10 L A = 2 * ( 1 0 ^ - 7 ) * ( 1 0 ^ 6 ) * log ( D M A / D S A ) ;

11 L B = 2 * ( 1 0 ^ - 7 ) * ( 1 0 ^ 6 ) * log ( D M A / D S B ) ;

12 T o t = L A + L B ;

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13 mprintf ( i n d u c t a n c e o f l i n e A , LA=% . 3 f mH/km\n , L A ) ;

// A ns we rs don t match d ue t o d i f f e r e n c e i nr o u n d i n g o f f o f d i g i t s14 mprintf ( i n d u c t a n c e o f l i n e B , LB=% . 1 f mH/km\n , L B ) ;

// A ns we rs don t match d ue t o d i f f e r e n c e i nr o u n d i n g o f f o f d i g i t s

15 mprintf ( t o t a l i n d u c t a n c e o f l i n e =%. 2 f mH/km\n ,Tot) ; // A ns we rs don t match du e t o d i f f e r e n c e i nr o u n d i n g o f f o f d i g i t s

Scilab code Exa 2.5 To determine the inductance per Km of 3 phase line

1 / / To d e t er m in e t he i n d uc t a nc e p er Km o f 3p h a s el i n e

2 clear

3 clc ;

4 G M D c = 1 . 2 6 6 * 0 . 7 7 8 8 * ( 1 0 ^ - 2 ) ; // s e l f GMD of e ac hc o n d u c t o r

5 D b c = sqrt ( ( 4 ^ 2 ) + ( . 7 5 ^ 2 ) ) ;

6 D a b = D b c ;

7 D a b = sqrt ( ( 4 ^ 2 ) + ( 8 . 2 5 ^ 2 ) ) ;8 D a a = sqrt ( ( 8 ^ 2 ) + ( 7 . 5 ^ 2 ) ) ;

9 D m 1 = ( D b c * 8 * 7 . 5 * 9 . 1 6 8 5 ) ^ ( 1 / 4 ) ;

10 D m 2 = ( D b c * D b c * 9 . 1 6 8 5 * 9 . 1 6 8 5 ) ^ ( 1 / 4 ) ;

11 D m 3 = D m 1 ;

12 D m = ( ( D m 1 * D m 2 * D m 3 ) ^ ( 1 / 3 ) ) ;

13 D s 1 = sqrt ( G M D c * D a a ) ; / / s e l f GMD o f e a c h p h a s e14 D s 3 = D s 1 ;

15 D s 2 = sqrt ( G M D c * 9 ) ;

16 D s = ( ( D s 1 * D s 2 * D s 3 ) ^ ( 1 / 3 ) ) ;

17 Z = 2 * ( 1 0 ^ - 4 ) * ( 1 0 0 0 ) * log ( D m / D s ) ;18 mprintf ( in du ct an ce=%.3 f mH/km/ph ase \n , Z ) ;

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Scilab code Exa 2.6 To determine the inductance of double circuit line

1 / / To d et er mi ne t he i n du c t a nc e o f d ou bl e c i r c u i tl i n e

2 clear

3 clc ;

4 G M D s = . 0 0 6 9 ; / / s e l f GMD o f t h e c o n d u c t o r5 D a b = sqrt ( ( 3 ^ 2 ) + . 5 ^ 2 ) ;

6 D b c = D a b ;

7 D a c = 6 ;

8 D a b = sqrt ( ( 3 ^ 2 ) + 6 ^ 2 ) ;

9 D a a = sqrt ( ( 6 ^ 2 ) + 5 . 5 ^ 2 ) ;

10 D m 1 = ( ( 3 . 0 4 * 6 * 5 . 5 * 6 . 7 0 8 ) ^ . 2 5 ) ;11 D m 2 = ( ( 3 . 0 4 * 3 . 0 4 * 6 . 7 0 8 * 6 . 7 0 8 ) ^ . 2 5 ) ;

12 D m = 4 . 8 9 ;

13 D s 1 = sqrt ( G M D s * D a a ) ;

14 D s 2 = 0 . 2 2 1 7 ;

15 D s = . 2 2 8 ;

16 Z = 2 * ( 1 0 ^ - 7 ) * ( 1 0 ^ 6 ) * log ( D m / D s ) ;

17 mprintf ( in du ct an ce =%.3 f mH/km, Z ) ;

Scilab code Exa 2.7 To determine the inductance per Km per phase ofsingle circuit

1 / / / / To d e te r m in e t h e i n d u c t a nc e p e r Km p er p ha seo f s i n g l e c i r c u i t

2 clear

3 clc ;

4 Ds = sqrt ( 0 . 0 2 5 * . 4 * . 7 7 8 8 ) ;

5 D m = ( ( 6 . 5 * 1 3 . 0 * 6 . 5 ) ^ ( 1 / 3 ) ) ;

6 Z = 2 * ( 1 0 ^ - 4 ) * 1 0 0 0 * log ( D m / D s ) ;7 mprintf ( i n du c t an c e =%. 3 f mH/km/ phase , Z ) ;

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Chapter 3

CAPACITANCE OF

TRANSMISSION LINES

Scilab code Exa 3.1 To determine the capacitance and charging current

1 / /To d et er m in e t he c a p a c i t an c e and c h a rg i n g c u r r e nt2 clear

3 clc ;

4 D m = 2 . 0 1 5 ; / / m u tu a l GMD o f c o n d u c t o r s (m)5 r = . 4 ; / / r a d i u s o f c o n du c to r ( cm )6 C = 1 0 ^ - 9 * 1 0 0 0 / ( 1 8 * log ( 2 0 1 . 5 / . 4 ) ) ;

7 I c = 1 3 2 * 1 0 0 0 * 8 . 9 2 8 * 3 1 4 * ( 1 0 ^ - 9 ) / sqrt ( 3 ) ;

8 mprintf ( c a p a c i t a n c e =%. 1 3 f F/km\n , C ) ; / / A n s w er s d on t match due t o d i f f e r e n t r e p r e n t a t i o n

9 mprintf ( c h ar gi ng c u r r e n t =%. 4 f amp/km , I c ) ;



3 clc ;

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4 G M D m = 6 . 6 1 ; // mu tu al GMD(m)

5 D s 1 = sqrt ( 1 . 2 5 * ( 1 0 ^ - 2 ) * 1 0 . 9 6 5 ) ;6 D s 3 = D s 1 ;

7 D s 2 = sqrt ( 1 . 2 5 * ( 1 0 ^ - 2 ) * 9 ) ;

8 D s = ( ( D s 1 * D s 2 * D s 3 ) ^ . 3 3 3 3 3 3 ) ;

9 C = 1 / ( 1 8 * log ( G M D m / D s ) ) ;

10 I c = 2 2 0 * 1 0 0 0 * 3 1 4 * . 0 1 9 0 5 * ( 1 0 ^ - 6 ) / sqrt ( 3 ) ;

11 mprintf ( c a p a c i t a n c e =%. 6 f m i cr oFarad/km\n , C ) ;12 mprintf ( c h a r g i n g c u r r e n t =% . 2 f amp /km , I c ) ;



3 clc ;

4 G M D = 8 . 1 9 ;

5 Ds = sqrt ( 2 . 2 5 * ( 1 0 ^ - 2 ) * . 4 ) ;

6 C = 1 / ( 1 8 * log ( G M D / D s ) ) ;

7 I c = 2 2 0 * 1 0 0 0 * 3 1 4 * C * ( 1 0 ^ - 6 ) / sqrt ( 3 ) ;

8 mprintf ( c a p a c i t a n c e p e r km =%. 5 f m i cr oFarad\n , C ) ;

9 mprintf ( c h a r g i n g c u r r e n t =%. 3 f amp , I c ) ;

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Chapter 4

PERFORMANCE OF LINES

Scilab code Exa 4.1 To determine the sending end voltage and currentpower and power factor Evaluate A B C D parameters

1 / /To d et re mi ne t he t he v o l t ag e a t t he g e n e r a t i n gs t a t i o n and e f f i c i e n c y o f t r a n s m i s s i o n

2 clear

3 clc ;

4 R = 0 . 4 9 6 ; // r e s i s t a n c e

5 X = 1 . 5 3 6 ;6 V r = 2 0 0 0 ;

7 Z = ( 1 0 * 2* 2 / ( 11 * 1 1 ) ) + % i * 3 0 * 2 *2 / ( 1 1* 1 1 ) ;

8 Z t = ( .0 4 +( 1 .3 * 2* 2 /( 1 1* 1 1) ) ) + % i * (. 12 5 +

( 4 . 5 * 2 * 2 / ( 1 1 * 1 1 ) ) ) ; / / T r a n s f o r m e r i m p e d e n c e9 I l = 2 5 0 * 1 0 0 0 / 2 0 0 0 ; // l i n e c u r r e n t ( amps . )

10 P l = I l * I l * R ; / / l i n e l o s s (kW)11 P o = 2 5 0 * 0 . 8 ; // ou tp ut (kW)12 c o s r = 0 . 8 ; // power f a c t o r13 s i n r = . 6 ;

14 % n = 2 0 0 * 1 0 0 / ( 2 0 0 + 7 . 7 ) ;15 V s = ( V r * c o s r + I l * R ) + % i * ( V r * s i n r + I l * X ) ;

16 V = sqrt ( ( 1 66 2 ^ 2) + ( 1 3 92 ^ 2 ) ) ;

17 mprintf ( e f f i c i e n c y = %. 1 f p e r ce n t \n , % n ) ;18 mprintf ( S e n di ng end v o l t a g e ,|Vs |=%. 0 f v o l t s , V ) ;

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Scilab code Exa 4.2 To determine power input and output i star con-nected ii delta connected

1 / /To d e t er m i ne p ower i n p u t and o u tp ut ( i ) s t a rc o n n e c te d ( i i ) d e l t a c o n n e c t ed

2 clear

3 clc ;

4 mprintf ( when l o a d i s s t a r c on n ec te d \n ) ;

5 V l n = 4 0 0 / sqrt ( 3 ) ; // L in e t o n e u t r a l v o l t a g e (V)6 Z = 7+ % i * 11 ; / / I mp ed en ce p e r p h a se7 I l = 2 3 1 / Z ; // l i n e c u r r e nt ( amp . )8 I = abs ( 2 3 1 / Z ) ;

9 P i = 3 * I * I * 7 ;

10 P o = 3 * I * I * 6 ;

11 mprintf ( p o we r i n p u t =%. 0 f w a t t s \n , P i ) ; //A nsw e r sdon t match due t o d i f f e r e n c e i n r ou nd in g o f f o f d i g i t s

12 mprintf ( p o w er o u t p u t =% . 0 f w a t t s \n , P o ) ; //A nsw e r sdon t match due t o d i f f e r e n c e i n r ou nd in g o f f o f d i g i t s

13 mprintf ( when l o ad i s d e l t a c o nn e ct ed \n ) ;14 Z e = 2 + % i * 3; / / e q u i v a l e n t i m pe de n ce ( ohm )15 Z p = 3 + % i * 5 ; / / i mp ed en ce p e r p ha s e16 i l = 2 3 1 / Z p ; / / L i n e c u r r e n t ( a mps . )17 IL = abs ( i l ) ;

18 p i = 3 * I L * I L * 3 ;

19 p o = 3 * I L * I L * 2 ;

20 mprintf ( p o w er i n p u t =% . 1 f w a t t s \n , p i ) ; / / A n s w er s d on t match due t o d i f f e r e n c e i n r ou nd in g o f f o f

d i g i t s21 mprintf ( p ow er o u tp u t = %. 0 f w a t ts \n , p o ) ; //A nsw e r s

don t match due t o d i f f e r e n c e i n r ou nd in g o f f o f d i g i t s

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Scilab code Exa 4.3 To determine efficiency and regulation of line

1 // To d e t er mi ne e f f i c i e n c y and r e g u l a t i o n o f l i n e2 clear

3 clc ;

4 a = 1 0 0 / . 5

5 X l = 2 * ( 1 0 ^ - 7 ) * log ( 1 0 0 / . 5 ) ; // i n du c t an c e (H/ me t e r )6 X L = 2 0 * ( 1 0 0 0 ) * X l ; // i n d uc t an c e o f 20 km l e n g t h7 R = 6 . 6 5 ; // r e s i s t a n c e ( ohm )8 R c = 2 0 * 1 0 0 0 / ( 5 8 * 9 0 ) ; // r e s i s t a n c e o f c op pe r ( ohm )9 I = 1 0 * 1 0 0 0 / ( 3 3 * . 8 * sqrt ( 3 ) ) ; / / t h e c u r r e n t ( amps . )

10 P l = 3 * I * I * R c / ( 1 0 ^ 6 ) ; // l o s s (MW)11 n = 1 0 / ( 1 0 + P l ) ;

12 mprintf ( e f f i c i e n c y =%. 4 f p e r c e n t \n , n ) ;13 V r = 1 9 0 5 2 ;

14 c o s r = . 8 ; / / po wer f a c t o r15 s i n r = . 6 ;

16 Vs = abs ( (( V r * c o s r + I * Rc ) + % i * ( V r * s in r + I * R ) ) ) ;

17 mprintf ( Vs =%. 0 f v o l t s\n , V s ) ; //A nswe r don t matc h

due t o d i f f e r e n c e i n r o u n d i n g o f f o f d i g i t s18 R e g = ( V s - V r ) * 1 0 0 / V r ;

19 mprintf ( r e g u l a t i o n =%. 2 f p e r c e n t , R e g )

Scilab code Exa 4.4 To calculate the voltage across each load impedenceand current in the nuetral

1 / /To c a l c u l a t e t he v o l t a ge a c r o s s e a ch l oa d

i mp ed en ce and c u r r e n t i n t he n u e t r a l2 clear

3 clc ;

4 I R = ( 4 0 0 ) / ( ( sqrt ( 3 ) * ( 6 . 3 + % i * 9 ) ) ) ;

5 I Y = 2 3 1 * ( c o sd ( - 12 0) + % i * s in d ( - 1 20 ) ) / 8 . 3;

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6 I B = 2 3 1 * ( c o sd ( 1 2 0 ) + % i * s i nd ( 1 2 0 ) ) / ( 6. 3 - % i * 8 ) ;

7 In = abs (( I R + IY + IB ) ); / / N e u tr a l c u r r e n t8 mprintf ( N e u t r a l c u r r e n t =%. 2 f amps\n , I n ) ;9 VR = abs ( I R * ( 6+ % i * 9) ) ;

10 VY = abs ( I Y * ( 8 ) ) ;

11 VB = abs ( I B * ( 6 - % i * 8 ) ) ;

12 mprintf ( V o l t a ge a c r o s s P ha se R =%. 1 f v o l t s \n , V R ) ;13 mprintf ( V o l t a ge a c r o s s P ha se Y =%. 2 f v o l t s \n , V Y ) ;14 mprintf ( V o l t a ge a c r o s s P ha se B =%. 0 f v o l t s \n , V B ) ;

Scilab code Exa 4.5 To determine efficiency and regulation of 3 phase line

1 / / To d et er mi ne e f f i c i e n c y and r e g u l a t i o n o f 3 p h as el i n e

2 clear

3 clc ;

4 R = 1 0 0 * . 1 ; // R e s i s t a n ce o f l i n e ( ohm )5 X l = 2 * ( 1 0 ^ - 7 ) * 1 0 0 * 1 0 0 0 * log ( 2 0 0 / . 7 5 ) ; / / i n d u c t a n c e o f

l i n e6 X 2 = X l * 3 1 4 ; // i n d u c t i v e r e a c t a nc e

7 C = 2 * ( % p i * 1 0 0 ) * 8 . 8 5 4 * ( 1 0 ^ - 1 2 ) * 1 0 0 * 1 0 0 0 * ( 1 0 ^ 6 ) / ( log( 2 0 0 / . 7 5 ) ) ; // c a p a c i t a n ce p er p ha se ( m ic ro f a r a d )

8 mprintf ( U s i n g N om in alT method\n ) ;9 I r = 2 0 * 1 0 0 0 / ( sqrt ( 3 ) * 6 6 * . 8 ) ;

10 V r = 6 6 * 1 0 0 0 / sqrt ( 3 ) ;

11 V c = ( 3 8 1 04 * . 8 + I r * 5) + % i * ( 3 8 1 04 * . 6+ I r * 1 7 .5 5 ) ; //v o l t a g e a c r o s s c on de ns er

12 I c = % i * 3 1 4 * ( V c ) * . 9 9 5 4 * ( 1 0 ^ - 6 ) ;

13 i s = I r + I c ;

14 Is = abs ( I r + I c ) ;

15 Vs = abs ( Vc + ( is * (5 + %i * 1 7. 53 ) )) ;16 VR = abs ( V s * ( - % i * 3 1 9 9 ) / ( 5 - % i * 3 1 8 1 ) ) ; // no l o adr e c i e v i n g end v o l t a g e

17 R e g = ( V R - V r ) * 1 0 0 / V r ;

18 P l = 3 * ( I r * Ir * 5 + I s * Is * 5 ) / 1 0 0 0 00 0 ;

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19 % n = 2 0 * 1 0 0 / ( 2 0 + P l ) ;

20 mprintf ( p e r c e n t r e g u l a t i o n =%. 1 f \n , R e g ) ;21 mprintf ( p e r c e n t e f f i c i e n c y =%. 1 f \n\n , % n ) ;22 mprintf ( U s i n g N om in alp i m et ho d\n ) ;23 I r 1 = 2 1 8 . 6 8 * ( . 8 - % i * . 6 ) ;

24 I c 1 = % i * 3 1 4 * . 4 9 7 7 * ( 1 0 ^ - 6 ) * V r ;

25 I l = I r 1 + I c 1 ;

26 v s 1 = V r + I l * ( 1 0 + % i * 3 5 . 1 ) ;

27 V s 1 = abs ( v s 1 ) ;

28 V r 1 = V s 1 * ( - % i * 6 3 9 8 ) / ( 1 0 - % i * 6 3 6 3 ) ;

29 V R 1 = abs ( V r 1 ) ; // no l oa d r e c i e v i n g end v o l t a ge30 R e g 2 = ( V R 1 - V r ) * 1 0 0 / V r ;

31 IL = abs ( I r 1 + I c 1 ) ;32 L o s s = 3 * I L * I L * 1 0 ;

33 % n = 2 0 * 1 0 0 / 2 1 . 3 8 8 ;

34 mprintf ( p e r c e n t r e g u l a t i o n =%. 2 f \n , R e g 2 ) ;35 mprintf ( p e r c e n t e f f i c i e n c y =%. 1 f \n , % n ) ;

Scilab code Exa 4.6 To find the rms value and phase values i The incidentvoltage to neutral at the recieving end ii The reflected voltage to neutral at

the recieving end iii The incident and reflected voltage to neutral at 120 kmfrom the recieving end

1 / /To f i n d t h e rms v a l u e and p ha se v a l u e s ( i ) Thei n c i d e n t v o l t a ge t o n e u t r al a t t he r e c i e v i n g end( i i ) The r e f l e c t e d v o l t a g e t o n e u t r a l a t t her e c i e v i n g end ( i i i ) The i n c i d e n t and r e f l e c t e dv o l t a g e t o n e u t ra l a t 120 km from t he r e c i e v i n gend .

2 clear

3 clc ;4 R = 0 . 2 ;

5 L = 1 . 3 ;

6 C = 0 . 0 1 * ( 1 0 ^ - 6 ) ;

7 z = R + % i * L * 3 1 4 * ( 1 0 ^ - 3 ) ; // s e r i e i mp ed en ce

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8 y = % i * 3 1 4 * C ; / / s hu nt a d mi t ta n ce

9 Zc = sqrt ( z / y ) ; // c h a r a c t e r s t i c i mp ed en ce10 Y = sqrt ( y * z ) ;11 V r = 1 3 2 * 1 0 0 0 / sqrt ( 3 ) ;

12 I r = 0 ;

13 V in = ( V r + I r * Zc ) / 2; // i n c i d e n t v o l t a g e t o n e u t r a l a tt h e r e c i e v i n g end

14 mprintf ( Vr =%. 3 f v o l t s \n , V r ) ; // Answer don t matchdue t o d i f f e r e n c e i n r o u n d i n g o f f o f d i g i t s

15 mprintf ( ( i ) The i n c i d e n t v o l t a g e t o n e u t ra l a t t her e c i e v i n g end =%. 3 f v o l t s \n , V i n ) ; // Answer don t

match d ue t o d i f f e r e n c e i n r o un di ng o f f o f

d i g i t s16 V in 2 =( V r - I r * Zc ) / 2; // The r e f l e c t e d v o l t a ge t o

n e u t r al a t t h e r e c i e v i n g end17 mprintf ( ( i i ) The r e f l e c t e d v o l t a g e t o n e u t ra l a t t he

r e c i e v i n g end=%. 3 f v o l t s \n , V i n 2 ) ; // Answer don t match d ue t o d i f f e r e n c e i nr ou nd i n g o f f o f d i g i t s

18 V r p = V r * exp ( . 2 7 1 4 * 1 2 0 * ( 1 0 ^ - 3 ) ) * exp ( %i

* 1 . 1 6 9 * 1 2 0 * ( 1 0 ^ - 3 ) ) / 1 0 0 0 ; // Takin g Vrp=Vr+19 V r m = V r * exp ( - 0 . 0 3 2 5 ) * exp ( - % i * . 1 4 0 ) / 1 0 0 0 ; // Takin g Vrm=

Vr20 v 1 = V r m / 2 ; // r e f l e c t e d v o l t a ge t o n e ut r a l a t 1 20 kmfrom t he r e c i e v i n g end

21 p h a s e _ v 1 = a t a n d ( imag ( v 1 ) / real ( v 1 ) ) ; // P hase a n g le o f v1

22 v 2 = V r p / 2 ; // i n c i d e n t v o l t a g e t o n e u t r a l a t 120 kmfrom t he r e c i e v i n g end

23 p h a s e _ v 2 = a t a n d ( imag ( v 2 ) / real ( v 2 ) ) ; // P hase a n g le o f v2

24 mprintf ( ( i i i ) r e f l e c t e d v o l t a ge t o n e ut r a l a t 120km fro m t he r e c i e v i n g end =%. 2 f a t a n g le o f % . 2 f

\n , abs ( v 1 ) , p h a s e _ v 1 ) ;25 mprintf ( i n c i d e n t v o l t ag e t o n e u t r a l a t 120 km f rom

t he r e c i e v i n g end = %. 2 f a t a n gl e o f %. 2 f \n , abs (v 2 ) , p h a s e _ v 2 ) ;

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Scilab code Exa 4.7 To determine of efficiency of line

1 //To d et er mi ne o f e f f i c i e n c y o f l i n e2 clear

3 clc ;

4 I r = 4 0 * 1 0 0 0 / ( sqrt ( 3 ) * 1 3 2 * . 8 ) ;

5 V r = 1 3 2 * 1 0 0 0 / sqrt ( 3 ) ;

6 Z c = 3 8 0* ( c o s d ( - 1 3. 0 6) + % i * s i nd ( - 1 3. 0 6) ) ;

7 I R = I r *( c o s d ( - 3 6 . 8) + % i * s in d ( - 3 6. 8 ) ) ;

8 V sp = ( V r + I R * Z c ) * ( 1 .0 3 3 *( c o sd ( 8 . 0 2 ) + % i * s in d ( 8 . 0 2) ) )

/2;

9 V sm = ( V r - I R * Z c ) * ( .9 6 8 *( c o sd ( - 8 .0 2) + % i * s in d ( - 8 .0 2 ) ) )

/2;

10 v s = V sp + V sm ;

11 Vs = abs ( v s ) ;

12 i s = ( V s p - V s m ) / Z c ;

13 Is = abs ( i s )

14 P = 3 * V s * I s * c o s d ( 3 3 . 7 2 ) / 1 0 ^ 6 ;

15 n = 4 0 * 1 0 0 / P ;

16 mprintf ( e f f i c i e n c y =%. 1 f , n ) ; //Answer don t matchdue t o d i f f e r e n c e i n r o u n d i n g o f f o f d i g i t s

Scilab code Exa 4.8 To determine the ABCD parameters of Line

1 / /To d e t e r m i ne t h e ABCD p a r a m e t e r s o f L i n e2 clear

3 clc ;

4 y l = ( 0 . 27 1 4 + % i * 1 . 16 9 ) * 1 2 0 *( 1 0 ^ - 3 ) ;5 I r = 4 0 * 1 0 0 0 / ( sqrt ( 3 ) * 1 3 2 * . 8 )

6 A = cosh ( y l ) ;

7 p h a s e _ A = a t a n d ( imag ( A ) / real ( A ) ) ; // P ha se a n g le o f A8 I R = I r *( c o s d ( - 3 6 . 8) + % i * s in d ( - 3 6. 8 ) )

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9 V r = 1 3 2 * 1 0 0 0 / sqrt ( 3 ) ;

10 Z c = 3 8 0* ( c o s d ( - 1 3. 0 6) + % i * s i nd ( - 1 3. 0 6) ) ;11 B = Z c * sinh ( y l ) ;

12 p h a s e _ B = a t a n d ( imag ( B ) / real ( B ) ) ; // P ha se a n g le o f B13 V s = ( A * V r + B * I R ) ;

14 f = abs ( B ) ;

15 d = abs ( V s ) ;

16 C = sinh ( y l ) / Z c ;

17 p h a s e _ C = a t a n d ( imag ( C ) / real ( C ) ) ; // P ha se a n g le o f C18 D = cosh ( y l ) ;

19 p h a s e _ D = a t a n d ( imag ( D ) / real ( D ) ) ; // P ha se a n g le o f D20 mprintf ( A=%. 2 f a t a n a n g l e o f %. 2 f \n , abs ( A ) ,

p h a s e _ A )21 mprintf ( B=%. 1 f a t a n a n g l e o f %. 0 f \n , abs ( B ) ,

p h a s e _ B )

22 mprintf ( C=%. 2 f a t a n a n g l e o f %. 2 f \n , abs ( C ) ,p h a s e _ C )

23 mprintf ( D=%. 2 f a t a n a n g l e o f %. 2 f \n , abs ( D ) ,p h a s e _ D )

Scilab code Exa 4.9 To determine the sending end voltage and efficiencyusing Nominal pi and Nominal T method

1 / /To d et er m in e t he s e nd i n g end v o l t a g e ande f f i c i e n c y u s i ng N om in a l p i and NominalT method

2 clear

3 clc ;

4 I r = 2 1 8 . 7 * ( . 8 - % i * . 6 ) ;

5 I c 1 = % i * 3 1 4 * . 6 * ( 1 0 ^ - 6 ) * 7 6 2 0 0 ;

6 I l = I c 1 + I r ;

7 V s = 7 6 20 0 + I l * (2 4+ % i * 48 . 38 ) ;8 p h a s e _ V s = a t a n d ( imag ( V s ) / real ( V s ) ) ; // p h as e a n g le o f VS

9 P l = 3 * 2 4 * abs ( I l ) * abs ( I l ) / 1 0 0 0 0 0 0 ; // The L o s s (MW)10 n = 4 0 * 1 0 0 / ( 4 0 + P l ) ;

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11 mprintf ( U s i n g N om in al p i m et ho d\n ) ;

12 mprintf ( Vs=%. 0 f v o l t s a t an a n g l e o f % . 2 f \n , abs (V s ) , p h a s e _ V s )13 mprintf ( e f f i c i e n c y =%. 2 f p e r c e n t \n , n ) ;14 mprintf ( \ nU si ng N omi nalT method\n ) ;15 V c = 7 6 2 0 0* ( . 8 + % i * . 6) + 2 1 8 .7 * ( 12 + % i * 2 4 .4 9 ) ;

16 I c = % i * 3 1 4* 1 . 2* ( 1 0^ - 6 ) * ( 6 3 5 84 + % i * 5 1 07 6 ) ;

17 I s = 1 9 9 . 4 6+ % i * 2 3 .9 5 ;

18 V s =( V c + I s * (1 2+ % i * 24 .4 9 ) ) / 10 0 0;

19 p h a s e _ V s = a t a n d ( imag ( V s ) / real ( V s ) ) ; // P hase a n g le o f Vs

20 P l1 = 3 * 1 2 *( ( 2 0 0. 8 9 ^ 2 ) + 2 1 8 .7 ^ 2 ) / 1 0 0 00 0 0 ; //The l o s s (MW

)21 n 1 = 4 0 * 1 0 0 / ( 4 0 + P l 1 ) ;

22 mprintf ( Vs=%. 2 f a t an a n g l e o f % . 2 f \n , abs ( V s ) ,p h a s e _ V s )

23 mprintf ( e f f i c i e n c y =%. 2 f p e r c e n t \n , n 1 ) ;

Scilab code Exa 4.10 To determine the sending end voltage and currentpower and power factor Evaluate A B C D parameters

1 / / To d e te rm i ne t he s e nd i ng end v o l t a g e and c u r r e nt, power and power f a c t o r . E va lu a te A , B , C , Dp a r a m e t e r s .

2 clear

3 clc ;

4 R = . 1 5 5 7 * 1 6 0

5 G M D = ( 3 . 7 * 6 . 4 7 5 * 7 . 4 ) ^ ( 1 / 3 ) ;

6 Z 1 = 2 * ( 1 0 ^ - 7 ) * log ( 5 6 0 / . 9 7 8 ) * 1 6 0 * 1 0 0 0 ;

7 X L = 6 3 . 8 ;

8 C = ( 1 0 ^ - 9 ) * 2 * ( 1 0 ^ 6 ) * % p i * 1 6 0 * 1 0 0 0 / ( 3 6 * % p i * log( 5 6 0 / . 9 7 8 ) ) ;

9 Z = sqrt ( ( . 15 5 7 ^ 2) + . 3 9 87 5 ^ 2) * ( c o s d ( 6 8 . 67 ) + % i * s in d

( 6 8 . 6 7 ) ) ;

10 j w C = % i * 3 1 4 * 1 . 3 9 9 * ( 1 0 ^ - 6 ) / 1 6 0 ;

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11 Zc = sqrt ( Z / j w C ) ;

12 y = sqrt ( Z * j w C ) ;13 y l = y * 1 6 0 ;

14 A = cosh ( y l ) ;

15 B = Z c * sinh ( y l )

16 C = sinh ( y l ) / Z c ;

17 I r = 5 0 0 0 0 / ( sqrt ( 3 ) * 1 3 2 ) ;

18 V s = ( A * 7 6 .2 0 8 ) + ( B * ( 10 ^ - 3 ) * I r * ( c o sd ( - 3 6. 8 7) + % i * s i n d

( - 3 6 . 8 7 ) ) ) ;

19 V S = 1 5 2 . 3 4 ;

20 I s = C * 7 6 . 20 8 * ( 10 ^ 3 ) + ( A * Ir * ( c o s d ( - 3 6 . 87 ) + % i * s i nd

( - 3 6 . 8 7 ) ) ) ;

21 P s = 3 * abs ( V s ) * abs ( I s ) * c o s d ( 3 3 . 9 6 ) ;22 p f = c o s d ( 3 3 . 9 6 ) ;

23 V n l = abs ( V s ) / abs ( A ) ;

24 r e g = ( V nl - 7 6 . 2 0 8 ) * 1 0 0 / 7 6 . 2 0 8 ;

25 n = 5 0 0 0 0 * . 8 * 1 0 0 / abs ( P s ) ;

26 mprintf ( Vs l i n e t o l i n e =%. 2 f kV\n , V S ) ;27 disp ( I s , s e n d i n g end c u r r e n t I s (A)= ) ; // Answer don t

match d ue t o d i f f e r e n c e i n r o un di ng o f f o f d i g i t s

28 mprintf ( s e n d i n g e nd p o we r=% . 0 f kW\n , P s ) ;29 mprintf (

s e n d i n g e nd p . f =%. 3 f \n, p f ) ;

30 mprintf ( p e r c e n t r e g u l a t i o n =%. 1 f \n , r e g ) ;31 mprintf ( p e r c e n t e f f i c e n c y =%. 1 f , n ) ;

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Chapter 5

HIGH VOLTAGE DC

TRANSMISSION

Scilab code Exa 5.1 To determine the dc output voltage when delay anglwa0 b30 c45

1 / /To d e t er m i ne t h e d . c . o u tp u t v o l t a g e when d e l a ya ng lw ( a ) 0 ( b ) 3 0 ( c ) 4 5

2 clear

3 clc ;

4 V o = 3 * sqrt ( 2 ) * 1 1 0 / % p i ;

5 V d = V o * ( c o sd ( 0 ) + c o sd ( 1 5 ) ) / 2 ;

6 V d 1 = V o * ( c os d ( 3 0 ) + c o sd ( 4 5 ) ) / 2 ;

7 V d 2 = V o * ( c os d ( 4 5 ) + c o sd ( 6 0 ) ) / 2 ;

8 mprintf ( ( a ) F or a= 0, Vd=%. 2 f kV\n , V d ) ;9 mprintf ( ( b) For a=30 ,Vd=%.2 f kV\n , V d 1 ) ;

10 mprintf ( ( c ) For a =45,Vd=%.2 f kV\n , V d 2 ) ;

Scilab code Exa 5.2 To determine the necessary line secondary voltageand tap ratio required

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1 / / To d et er mi ne t he n e c es s a r y l i n e s ec on da ry v o l t a g e

and t a p r a t i o r e q u i r e d .2 clear3 clc ;

4 V L = 1 0 0 * 2 * % p i / ( 3 * sqrt ( 2 ) *( c o s d ( 3 0) + c o sd ( 4 5 ) ) ) ;

5 mprintf ( VL=%. 2 f kV\n , V L ) ; / / A n s w er s d on t m at ch d u et o d i f f e r e n c e i n r o u n d i n g o f f o f d i g i t s

6 T r = V L / 1 1 0 ;

7 mprintf ( t a p r a t i o =%. 2 f \n , T r ) ;

Scilab code Exa 5.3 To determine the effective reactance per phase

1 // To d e te rm in e t he e f f e c t i v e r e a ct a n ce p er p ha se2 clear

3 clc ;

4 V d = 1 0 0 0 0 0 ;

5 I d = 8 0 0 ; // c u r r e nt6 X = ( ( 3 * sqrt ( 2 ) * 9 4 . 1 1 5 * . 8 6 6 * 1 0 0 0 / % p i ) - V d ) * % p i / ( 3 * I d ) ;

7 mprintf ( e f f e c t i v e r e a c t a n c e p e r p ha se , X=%. 2 f ohm\n , X ) ; / / Answer don t match d ue t o d i f f e r e n c e i n

r o u n d i n g o f f o f d i g i t s

Scilab code Exa 5.4 Calculate the direct current delivered

1 / / C a l cu l a t e t he d i r e c t c u r r e n t d e l i v e r e d2 clear

3 clc ;

4 a = 1 5 ;

5 d 0 = 1 0 ;6 y = 1 5 ;

7 X = 1 5 ;

8 R = 1 0 ;

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9 I d = ( 3 * sqrt ( 2 ) * 1 20 * ( c o sd ( a ) - c o s d ( d 0 + y) ) * 1 0 0 0) / ( ( R +

( 3 * 2 * X ) / % p i ) * % p i ) ;10 mprintf ( Id=%.2 f amp . \ n , I d ) ;

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Chapter 6

CORONA

Scilab code Exa 6.1 To determine the critical disruptive voltage and crit-ical voltage for local and general corona

1 / /To d e t er m i ne t h e c r i t i c a l d i s r u p t i v e v o l t a g e andc r i t i c a l v o l t a g e f o r l o c a l and g e n e r a l c o ro n a .

2 clear

3 clc ;

4 t = 2 1 ; // a i r t em p er a tu r e

5 b = 7 3 . 6 ; // a i r p r e s s u r e6 d o = 3 . 9 2 * 7 3 . 6 / ( 2 7 3 + t ) ;

7 m = . 8 5 ;

8 r = . 5 2 ;

9 d = 2 5 0 ;

10 V d = 2 1 .1 * m * d o * r * log ( 2 5 0 / . 5 2 ) ;

11 vd = sqrt ( 3 ) * V d ;

12 m = . 7 ;

13 v v = 2 1. 1 * m * d o * r * (1 + ( . 3/ sqrt ( r * d o ) ) ) * log ( 2 5 0 / . 5 2 ) ;

14 V v = v v * sqrt ( 3 ) ;

15 V v g = V v * . 8 / . 7 ;16 mprintf ( c r i t i c a l d i s r u p t i v e l i n e t o l i n e v o l t a g e =%. 2 f kV \n , v d ) ;

17 mprintf ( v i s u a l c r i t i c a l v o l t a g e f o r l o c a l c o ro n a=%. 2 f kV \n , v v ) ;

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18 mprintf ( v i s u a l c r i t i c a l v o l t a g e f o r g e n e r a l c o ro n a=

% . 2 f kV \n , V v g ) ;

Scilab code Exa 6.2 To determine whether corona will be present in theair space round the conductor

1 / / To d et er mi ne w h et h er c or on a w i l l be p r e s e nt i nt he a i r s pa ce r ound t he c on du ct or

2 clear

3 clc ;4 d = 2 . 5 ;

5 d i = 3 ; // i n t e r n a l d ia me te r6 d o = 9 ; // e x t e r n al d ia me te r7 r i = d i / 2 ; // i n t e r n a l r a d i u s8 r o = d o / 2 ; // e x t e r n a l d ia me te r9 g 1 m a x = 2 0 / ( 1 . 2 5 * log ( r i / ( d / 2) ) + . 2 0 8* 1 . 5* log ( r o / r i ) ) ;

10 mprintf ( g1max=%. 0 f kV/cm \n , g 1 m a x ) ;11 mprintf ( S i n c e t he g r a d i e n t e x ce e ds 2 1 . 1 /kV/cm ,

c or on a w i l l be p r e s e nt . )

Scilab code Exa 6.3 To determine the critical disruptive voltage and coronaloss

1 / / To d e t er m i ne t h e c r i t i c a l d i s r u p t i v e v o l t a g e andc or on a l o s s

2 clear

3 clc ;

4 m = 1 . 0 7 ;

5 r = . 6 2 56 V = 2 1* m * r *log ( 3 0 5 / . 6 2 5 ) ;

7 V l = V * sqrt ( 3 ) ;

8 mprintf ( c r i t i c a l d i s r u p t i v e v o l t a g e =% . 0 f kV\n , V ) ;

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9 mprintf ( s i n c e o p er a ti n g v o l t a ge i s 110 kV , c o ro na

l o s s= 0 ) ;

Scilab code Exa 6.4 To determine the voltage for which corona will com-mence on the line

1 / /To d et er mi ne t he v o l t a g e f o r which c or on a w i l lcommence on t h e l i n e

2 clear

3 clc ;4 r = . 5 ;

5 V = 2 1 * r * log ( 1 0 0 / . 5 ) ;

6 mprintf ( c r i t i c a l d i s r u p t i v e v o l t a g e =% . 1 f kV , V ) ;

Scilab code Exa 6.5 To determine the corona characterstics

1 //To d et er mi ne t he c or on a c h a r a c t e r s t i c s2 clear

3 clc ;

4 D = 1 . 0 3 6 ; / / c o n d u c t o r d i a m e t e r ( cm )5 d = 2 . 4 4 ; / / d e l t a s p a c i n g (m)6 r = D / 2 ; // r a di us ( cm)7 R a t i o = d * 1 0 0 / r ;

8 j = r / ( d * 1 0 0 ) ;

9 R a t 2 = sqrt ( j ) ;

10 t = 2 6 . 6 7 ; / / t e m p e r a t u r e11 b = 7 3 . 1 5 ; // b a r om e tr i c p r e s s u r e12 m v = . 7 2 ;

13 V = 6 3 . 5 ;14 f = 5 0 ; / / f r e q u e n c y15 d o = 3 . 9 2 * b / ( 2 7 3 + t ) ; / / d o = d e l l16 v d = 2 1 . 1 * . 8 5 * d o * r * log ( R a t i o ) ;

17 mprintf ( c r i t i c a l d i s r u p t i v e v o l t a g e =% . 2 f kV\n , v d ) ;

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18 V v = 2 1. 1 * m v * d o * r * (1 + ( . 3/ sqrt ( r * d o ) ) ) * log ( R a t i o ) ;

19 P l = 2 4 1 * ( 1 0 ^ - 5 ) * ( f + 2 5 ) * R a t 2 * ( ( V - v d ) ^ 2 ) / d o ; //pow e rl o s s20 V d = . 8 * v d ;

21 P l 2 = 2 4 1 * ( 1 0 ^ - 5 ) * ( f + 2 5 ) * R a t 2 * ( ( V - V d ) ^ 2 ) * 1 6 0 / d o ; / / l o s sp e r p h a se /km

22 T o t al = 3 * P l2 ;

23 mprintf ( v i s u a l c r i t i c a l v o l t a g e =% . 0 f kV\n , V v ) ;24 mprintf ( Power l o s s=%.3 f kW/ pha se /km\n , P l ) ;25 mprintf ( u nde r f o u l w ea th er c o n d i t i o n ,\n ) ;26 mprintf ( c r i t i c a l d i s r u p t i v e v o l t a g e =% . 2 f kV\n , V d ) ;27 mprintf ( T o t a l l o s s =%. 0 f kW\n , T o t a l ) ;

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Chapter 7

MECHANICAL DESIGN OF

TRANSMISSION LINES

Scilab code Exa 7.1 Calculate the sag

1 // C a l c u l a te t he s ag2 clear

3 clc ;

4 s f = 5 ; // F ac to r o f s a f e t y5 d = . 9 5 ; / / c o n d u c t o r d i a ( cm )6 W s = 4 2 5 0 / s f ; // w or ki ng s t r e s s ( k g/ cm 2 )7 A = % p i * ( d ^ 2 ) / 4 ; // a r e a ( cm 2 )8 W p = 4 0 * d * ( 1 0 ^ - 2 ) ; / / wi nd p r e s s u r e ( k g /cm )9 W = sqrt ( ( . 65 ^ 2 ) + ( . 38 ^ 2 ) ) ; // T ot al e f f e c t i v e w ei gh t (

kg/m)10 T = 8 5 0 * A ; // w or k in g t e n s i o n ( kg )11 c = T / W ;

12 l = 1 6 0 ;

13 d = l ^ 2 / ( 8 * 8 0 0 ) ;

14 mprintf ( sag , d=%.0 f me tre s \n , d ) ;

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Scilab code Exa 7.2 To calculate the maximum Sag

1 / / To c a l c u l a t e t h e maximum S ag2 clear

3 clc ;

4 D = 1 .9 5 + 2 .6 ; / / o v e r a l l d i a me t er ( cm )5 A = 4 . 5 5 * ( 1 0 ^ - 2 ) ; / / a r e a ( m 2 )6 d = 1 9 . 5 ; / / d i a m e t e r o f c o n d u c t o r (mm)7 r = d / 2 ; / / r a d i u s o f c o n d u c t o r (mm)8 W p = A * 3 9 ; / / wi nd p r e s s u r e ( k g / m 2 )9 t = 1 3 ; // i c e c o at i n g (mm)

10 U S = 8 0 0 0 ; // u l t i m a te s t r e n g t h ( k g )

11 A i c e = % p i * (1 0 ^ - 6 ) * (( r + t ) ^ 2 - r ^ 2 ) ; // a r e a s e c t i o n o f i c e ( m 2 )

12 W i c e = A i c e * 9 1 0 ;

13 W =( sqrt ( (. 85 + W i ce ) ^ 2 + W p ^2 ) ) ;// t o t a l w e i g h t o f i c e(kg/m)

14 T = U S / 2 ; // w o rk in g t e a ns i o n ( kg )15 c = T / W ;

16 l = 2 7 5 ; / / l e n g t h o f s p an (m)17 S m a x = l * l / ( 8 * c ) ;

18 mprintf ( Maximum sa g=%.1 f me tr es \n , S m a x ) ;

Scilab code Exa 7.3 To determine the Sag

1 / /To d e t e r m i ne t h e S ag2 clear

3 clc ;

4 A = 1 3 . 2 ; // c r o s s s e c t i o n o f c on du ct or ( mm 2 )5 A r = 4 . 1 * ( 1 0 ^ - 3 ) ; // p r o j e c t e d a re a

6 W p = A r * 4 8 . 8 2 ; / / w in d l o a d i n d /m( k g /m)7 w = . 1 1 5 ;8 W = sqrt ( ( .1 1 57 ^ 2) + ( W p ^2 ) ) ;// e f f e c t i v e l o a d i n g p e r

m e t r e ( k g )9 q 1 = W / . 1 1 5 ;

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10 b = w / A ;

11 f 1 = 2 1 ; // w o rk in g s t r e s s12 T 1 = f 1 * A ;13 c = T 1 / W ;

14 l = 4 5 . 7 ;

15 S = l * l / ( 8 * c ) ;

16 d T = 3 2 . 2 - 4 . 5 ; // d i f f e r e n c e i n t em pe ra tu re17 E = 1 . 2 6 * ( 1 0 0 0 0 ) ;

18 a = 1 6 . 6 * ( 1 0 ^ - 6 ) ;

19 d = 8 . 7 6 5 * ( 1 0 ^ - 3 ) ;

20 K = f 1 - ( ( l * d * q 1 ) ^ 2 ) * E / ( 2 4 * f 1 * f 1 ) ;

21 p = poly ( [ - 84 .2 3 0 - 14 .4 4 1] , f2 , c ) ;

22 r = roots ( p ) ;23 f 2 = 1 4 . 8 23 3 3 2; // a cc ep te d v al ue o f f 224 T = f 2 * A ;

25 c = T / w ;

26 d 1 = l * l / ( 8 * c ) ;

27 mprintf ( s a g a t 3 2 . 2 C e l s i u s , d=%. 4 f m et re s , d 1 ) ;

Scilab code Exa 7.4 To determine the clearence between the conductor

and water level

1 / / To d e te rm i ne t he c l e a r e n c e b etw een t he c o nd u ct o rand w at er l e v e l

2 clear

3 clc ;

4 T = 2 0 0 0 ; // w o rk in g t e n si o n ( kg )5 w = 1 ;

6 c = T / w ;

7 h = 9 0 - 3 0 ;

8 l = 3 0 0 ; // span (m)9 a = ( l / 2 ) - ( c * h / l ) ;10 b = 5 5 0 ;

11 d 1 = a * a / ( 2 * c ) ;

12 d 2 = ( 4 0 0 ^ 2 ) / ( 2 * c ) ; // s a g a t 4 00 m e tr es (m)

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13 H m = d 2 - d 1 ; // h e ig h t o f mid p o in t w i th r e s p e c t t o A

14 C l = 3 0 + H m ;15 mprintf ( t h e c l e a r e n c e b et we en t he c o nd u ct o r andw a te r l e v e l midway b et we en t h e t o w e r s= %. 3 f m e t r e s \n , C l ) ;

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Chapter 8

OVERHEAD LINE

INSULATORS

Scilab code Exa 8.1 To determine the maximum voltage that the stringof the suspension insulators can withstand

1 / / To d e t er m i ne t h e maximum v o l t a g e t h a t t h e s t r i n go f t he s u s pe n si o n i n s u l a t o r s can w it hs ta nd .

2 clear

3 clc ;

4 E 3 = 1 7 . 5 ;

5 E 1 = 6 4 * E 3 / 8 9 ;

6 E 2 = 9 * E 1 / 8 ;

7 E = E 1 + E 2 + E 3 ;

8 mprintf ( t h e maximum v o l t a g e t h at t he s t r i n g o f t hes u s p e n s i o n i n s u l a t o r s ca n w i t hs t a nd=%. 2 f kV\n ,E );

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2 clear

3 clc ;4 e 1 = 4 ;

5 e 2 = 4 ;

6 e 3 = 2 . 5 ;

7 g 1 m a x = 5 0 ;

8 g 2 m a x = 4 0 ;

9 g 3 m a x = 3 0 ;

10 r = . 5 ; // r a d i u s ( cm )11 r 1 = r * e 1 * g 1 m a x / ( e 2 * g 2 m a x ) ;

12 r 2 = r 1 * e 2 * g 2 m a x / ( e 3 * g 3 m a x ) ;

13 V = 6 6 ;

14 l n c = ( V - ( ( r * g 1 m a x * log ( r 1 / r ) ) + ( r 1 * g 2 m a x * log ( r 2 / r 1 ) ) ) ) ;15 m = l n c / ( r 2 * g 3 m a x ) ;

16 R = r 2 * ( % e ^ m ) ;

17 D = 2 * R ;

18 mprintf ( minimum i n t e r n a l d i am e te r o f t he l e a dsh ea th ,D=%.2 f cms\n , D ) ;

Scilab code Exa 9.3 To determine the maximum safe working voltage

1 / / To d e t e r m i n e t h e maximum s a f e w o rk i ng v o l t a g e2 clear

3 clc ;

4 r = . 5 ; / / r a d i u s o f c o n d u c t o r ( cm )5 g 1 m a x = 3 4 ;

6 e r = 5 ;

7 r 1 = 1 ;

8 R = 7 / 2 ; / / e x t e r n a l d i a ( cm )9 g 2 m a x = ( r * g 1 m a x ) / ( e r * r 1 ) ;

10 V = ( ( r * g 1 m a x * log ( r 1 / r ) ) + ( r 1 * g 2 m a x * log ( R / r 1 ) ) ) ;11 V = V / ( sqrt ( 2 ) ) ;

12 mprintf ( Maximum s a f e w o r k i ng v o l l t a g e ,V =%. 2 f kV r.m. s \n , V ) ;

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Scilab code Exa 9.4 To determine the maximum stresses in each of thethree layers

1 / /To d e t er m i ne t h e maximum s t r e s s e s i n e ac h o f t h et hr ee l a y e r s .

2 clear

3 clc ;

4 r = . 9 ;

5 r 1 = 1 . 2 5

6 r 2 = r 1 + . 3 5 ;

7 r 3 = r 2 + . 3 5 ; // r a d i us o f o ut er mo st l a y e r8 V d = 2 0 ; // v o l t a ge d i f f e r e n c e ( kV )9 g 1 m a x = V d / ( r * log ( r 1 / r ) ) ;

10 g 2 m a x = V d / ( r 1 * log ( r 2 / r 1 ) ) ;

11 g 3 m a x = ( 6 6 - 4 0 ) / ( r 2 * log ( r 3 / r 2 ) ) ;

12 mprintf ( g1max =%. 1 f kV/cm\n , g 1 m a x ) ;13 mprintf ( g2max =%. 2 f kV/cm\n , g 2 m a x ) ;14 mprintf ( g3max =%. 0 f kV/cm\n , g 3 m a x ) ;

Scilab code Exa 9.5 o dtermine the equivalent star connected capacityand the kVA required

1 //To d t er mi ne t he e q u i v a l e n t s t a r c on n ec te d c a p a c i tyand t h e kVA r e q u i r e d .

2 clear

3 clc ;

4 V = 2 0 ; / / v o l t a g e ( kV )

5 w = 3 1 4 ;6 C = 2 * 3 . 0 4 * 1 0 ^ - 6 ; / / c a p a c i t a n c e p e r p h a s e ( m i cr of a r a d )7 K V A = V * V * w * C * 1 0 0 0 ;

8 mprintf ( 3phase kVA r e q u i r e d =%. 0 f kVA , K V A ) ; //Answer don t match du e t o d i f f e r e n c e i n r o un d in g

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o f f o f d i g i t s

Scilab code Exa 9.6 Determine the capacitance a between any two con-ductors b between any two bunched conductors and the third conductor cAlso calculate the charging current per phase per km

1 / / D et er mi ne t h e c a p a c i t a n c e ( a ) b et we en any twoc o n d u c t o r s ( b ) b et we en a ny two b un ch ed c o n d u c t o r sand t he t h i r d c on d uc to r ( c ) A ls o c a l c u l a t e t he

c h a rg i n g c u r r e n t p er p ha se p er km2 clear3 clc ;

4 C 1 = . 2 0 8 ;

5 C 2 = . 0 9 6 ;

6 C x = 3 * C 1 ;

7 w = 3 1 4 ;

8 V = 1 0 ;

9 C y = ( C 1 + 2 * C 2 ) ;

10 C o = ( ( 1 . 5 * C y ) - ( C x / 6 ) ) ;

11 C = C o / 2 ;

12 mprintf ( ( i ) C a p a c i t a n c e b e tw e en a ny t wo c o n d u c t o r s =%. 3 f m ic roFarad/km\n , C ) ;

13 c = ( (2 * C 2 + ( (2 /3 ) * C1 ) ) ) ;

14 mprintf ( ( i i ) C a p a c i t a n c e b e tw e en a ny t wo b un ch edc o n d u c t o r s and t h e t h i r d c o n d u c t o r=%. 2 f m ic roFarad/km\n , c ) ;

15 I = V * w * C o * 1 0 0 0 * ( 1 0 ^ - 6 ) / sqrt ( 3 ) ;

16 mprintf ( ( i i i ) t h e c h a r g i n g c u r r e n t p e r p ha s e p e r km=%. 3 f A\n , I ) ;

Scilab code Exa 9.7 To calculate the induced emf in each sheath

1 / / To c a l c u l a t e t he i nd uc ed emf i n e ac h s he at h .

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2 clear

3 clc ;4 rm= (2.28/2) -(.152/2); // mean r a d i u s o f s h ea t h ( cm )5 d = 5 . 0 8 ;

6 a = d / r m ;

7 w = 3 1 4 ;

8 X m = 2 * ( 1 0 ^ - 7 ) * log ( a ) ; / / m ut ua l i n d u c t a n c e (H/m)9 X m 2 = 2 0 0 0 * X m ;

10 V = w * X m 2 * 4 0 0 ;

11 mprintf ( V o l t a ge i n du c ed =%. 2 f v o l t s \n , V ) ; //Answerdon t match e x a c t l y due t o d i f f e r e n c e i n

r o u n d i n g o f f o f d i g i t s i bet w ee n c a l c u l a t i o n s

Scilab code Exa 9.8 To determine the ratio of sheath loss to core loss ofthe cable

1 //To d et er mi ne t he r a t i o o f s he at h l o s s t o c or e l o s so f t h e c a b l e

2 clear

3 clc ;

4 R = 2 * . 1 6 2 5 ;5 R s = 2 * 2 . 1 4 ;

6 M = 3 1 4 ;

7 w = 6 . 2 6 8 * 1 0 ^ - 4 ;

8 r = R s * M * M * w * w / ( R * ( ( R s ^ 2 ) + ( M * M * w * w ) ) ) ;

9 mprintf ( r a t i o=%.4 f \n , r ) ;

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Chapter 10

VOLTAGE CONTROL

Scilab code Exa 10.1 To determine the total power active and reactivesupplied by the generator and the pf at which the generator must operate

1 / / To d et er mi ne t he t o t a l power , a c t i v e andr e a c t i v e , s u p p l i e d by t he g e ne r a to r and t he p . f a t whi ch t he g e n er a t o r must o p er a te .

2 clear

3 clc ;

4 V = 1 ; / / v o l t a g e ( p . u )5 P a = . 5 ; / / a c t i v e p ow er a t A ( p . u )6 P r = . 3 7 5 ; / / r e a c t i v e p ower a t A( p . u )7 X c a = 0 . 0 7 5 + 0 . 0 4 ; / / r e a c t a n c e b et we en C and A8 P l = ( ( P a ^ 2 ) + ( P r ^ 2 ) ) * X c a / ( V ^ 2 ) ;

9 p a c = 1 . 5 ;

10 p r c = 2 ;

11 P t a = . 5 + 1 . 5 ; // t o t a l a c t i v e power b et we en E and C12 P t r = P r + P l + 2 ; // r e a c t i v e power b etw een E a nd C13 X t = . 0 5 + . 0 2 5 ; // t o t a l r e a c t a nc e b et ew ee n E an C

14 P l2 = ( ( 2* 2) + ( 2 .4 1 99 ^ 2) ) ; // l o s s ( p . u )15 P a t = 2 0 0 ;16 P r t = 3 1 5 . 9 ;

17 p f = . 5 3 4 9 ;

18 mprintf ( T o ta l a c t i v e power s u p p l i e d by g e n e r a to r =%

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. 0 f MW\n , P a t ) ;

19 mprintf ( T ot al r e a c t i v e power s u p p li e d by g e n er a t o r=%. 1 f MW \n , P r t ) ;20 mprintf ( p . f o f t h e g e n e r a t o r =%. 4 f \n , p f ) ;

Scilab code Exa 10.2 Determine the settings of the tap changers requiredto maintain the voltage of load bus bar

1 // D et er m in e t he s e t t i n g s o f t he t ap c ha ng e r s

r e q u i r e d t o m ai nt ai n t h e v o l t a ge o f l oa d bus b ar2 clear

3 clc ;

4 l 1 = 1 5 0 ;

5 t s t r = 1 ;

6 l o a d 2 = 7 2 . 6 5 ;

7 R = 3 0 ;

8 P = ( l 1 * ( 1 0 ^ 6 ) ) / 3 ;

9 X = 8 0 ;

10 Q = ( l o a d 2 * ( 1 0 ^ 6 ) ) / 3 ;

11 V s = ( 2 3 0 * ( 1 0 ^ 3 ) ) / sqrt ( 3 ) ;

12 V r = V s ;13 t s 2 = 1 / ( 1 - ( ( ( R * P ) + ( X * Q ) ) / ( V s * V r ) ) ) ;

14 ts = sqrt ( t s 2 ) ;

15 mprintf ( t s=%. 2 f p . u\n , t s ) ;

Scilab code Exa 10.3 i Find the sending end Voltage and the regulation ofline ii Determine the reactance power supplied by the line and by synchronouscapacotor and pf of line iii Determine the maximum power transmitted

1 // ( i ) Fin d t he s e nd i ng end V o lt ag e and t her e g u l a t i o n o f l i n e ( i i ) De t er m ine t he r e a ct a n cepower s u p p l i e d by t he l i n e and by s yn ch ro n ou s

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c a p a c o t o r and p . f o f l i n e ( i i i ) D et er mi ne t h e

maximum p o we r t r a n s m i t t e d2 clear3 clc ;

4 A = . 8 9 5 ;

5 V r = 2 1 5 ;

6 B = 1 8 2 . 5 ;

7 x = A * ( V r ^ 2 ) / B ;

8 y = 7 8 . 6 - 1 . 4 ; // ba9 p = a c o s d ( . 9 ) ;

10 X 1 = x / 5 0 ;

11 V s = 2 6 5 * 1 8 2 . 5 / 2 1 5 ;

12 V r 1 = V s / A ;13 R e g = 1 0 0 * ( V r 1 - V r ) / V r ;

14 mprintf ( ( i ) s e n d i n g e nd v o l t a g e ( kV )=%. 1 f kV\n , V s );

15 mprintf ( r e c i e v i n g end v o l t a g e =%. 0 f kV\n , V r 1 ) ;16 mprintf ( R e g u l a t i o n = %. 2 f p e r c e n t \n , R e g ) ;17 V s 1 = 2 3 6 ;

18 Q = V s 1 * V r / B ;

19 Q P = . 2 5 * 5 0 ;

20 P R = . 5 0 * 5 0 ;

21 c o s Q = . 9 5 8 ;

22 mprintf ( \n ( i i )QP(MVAr)=%. 1 f MV Ar\n , Q P ) ;23 mprintf ( PR(MVAr)=%. 0 f MV Ar\n , P R ) ;24 mprintf ( CosQ=%. 3 f \n , c o s Q ) ;25 M N = 4 . 5 5 ;

26 S b m a x = M N * 5 0 ;

27 mprintf ( maximum power tr an sm it te d =%.1 f MW\n ,Sbmax) ;

Scilab code Exa 10.4 Determine the KV Ar of the Modifier and the max-imum load that can be transmitted

1 // D et er mi ne t he KV Ar o f t he M o d i f i e r and t he

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maximum l o a d t h a t c an b e t r a n s m i t t e d

2 clear3 clc ;

4 a = 0 ;

5 b = 7 3 . 3

6 A = 1 ;

7 B = 2 0 . 8 8 ;

8 V s = 6 6 ;

9 V r = 6 6 ;

10 L o a d = 7 5 ;

11 p = poly ( [1 4 62 4 4 00 1 ] , Qr , c ) ;12 r = roots ( p ) ;

13 Q r = - 4 0 . 70 1 5 38 ;14 C = - Q r + ( 7 5* . 6/ . 8) ;

15 S m a x = ( V r ^ 2 ) * ( 1 - c o s d ( b ) ) / B ;

16 mprintf ( The p h as e m o d i f i e r c a p a c i t y =%. 2 f MV Ar\n ,C ) ;

17 mprintf ( Maximum power tr a n s m i t t ed ,Pmax =%.2 f MW ,S m a x ) ;

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Chapter 11

NEUTRAL GROUNDING

Scilab code Exa 11.1 To find the inductance and KVA rating of the arcsuppressor coil in the system

1 / / To f i n d t he i n d uc t a nc e and KVA r a t i n g o f t he a r cs u pp r e s so r c o i l i n t h e s y s t e m

2 clear

3 clc ;

4 C 1 = 2 * % p i * ( 1 0 ^ - 9 ) / ( 3 6 * % p i * log ( ( 4 * 4 * 8 ) ^ ( 1 / 3 )

/ ( 1 0 * ( 1 0 ^ - 3 ) ) ) ) ;5 C = C 1 * 1 9 2 * ( 1 0 ^ 9 ) ; // c a p a c i t an c e p er p ha se ( m ic ro

f a r a d )6 L = ( 1 0 ) ^ 6 / ( 3 * 3 1 4 * 3 1 4 * C ) ;

7 V = 1 3 2 ; / / v o l t a g e ( kV )8 M V A = V * V / ( 3 * 3 1 4 * L ) ;

9 mprintf ( i n d u c t a n c e , L=%. 2 f H\n , L ) ;10 mprintf ( MVA r a t i n g o f s u p p r e s s o r c o i l =%. 3 f MVA

p er c o i l , M V A ) ;

Scilab code Exa 11.2 Determine the reactance to neutralize the capaci-tance of i 100 percent of the length of line ii 90 percent of the length of lineiii 80 percent of the length of line

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1 // D et er m in e t he r e ac t a nc e t o n e u t r a l i z e t he

c a p ac i t an c e o f ( i ) 100% o f t he l e ng t h o f l i n e ( i i )90% o f t he l e ng t h o f l i n e ( i i i ) 8 0% o f t he l e ng t ho f l i n e

2 clear

3 clc ;

4 w L = 1 / ( 3 * 3 1 4 * ( 1 0 ) ^ - 6 ) ;

5 mprintf ( ( i ) i n d u c t i v e r e ac t a nc e f o r 100 p e rc e n t o f t h e l e n g t h o f l i n e =%. 1 f ohms\n , w L ) ;

6 w L = 1 0 ^ 6 / ( 3 * 3 1 4 * . 9 ) ;

7 mprintf ( ( i i ) i n d u c t i v e r e a c t a n ce f o r 90 p e r ce n t o f t h e l e n g t h o f l i n e =%. 1 f ohms\n , w L ) ;

8 w L = 1 / ( 3 * 3 1 4 * ( 1 0 ) ^ - 6 ) / . 8 ;9 mprintf ( ( i i i ) i n d u c t i v e r e a c t a nc e f o r 80 p e r ce n t o f

t h e l e n g t h o f l i n e =%. 1 f ohms\n , w L ) ;

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Chapter 12

TRANSIENTS IN POWER

SYSTEMS

Scilab code Exa 12.1 To determine the i the neutral impedence of line iiline current iii rate of energy absorption rate of reflection and state form ofreflection iv terminating resistance v amount of reflected and transmittedpower

1 // To d e te rm i ne t he ( i ) t he n e u t r a l i mp ed en ce o f l i n e

( i i ) l i n e c u r r e n t ( i i i ) r a t e o f e n er g y a b s o r p t i o n, r at e o f r e f l e c t i o n and s t a t e form o f r e f l e c t i o n

( i v ) t e r m in a t i n g r e s i s t a n c e ( v ) amount o f r e f l e c t e d and t r a n s mi t t ed power

2 clear

3 clc ;

4 L = 2 * ( 1 0 ^ - 7 ) * log ( 1 0 0 / . 7 5 ) ; / / i n d u c t an c e p er u n i tl e n g t h

5 C = 2 * % p i * ( 1 0 ^ - 9 ) / ( 3 6 * % p i * log ( 1 0 0 / . 7 5 ) ) ; / / C a p a c i t a n c ep er p ha se p er u n i t l e n g th ( F/m)

6 Z1 = sqrt ( L / C ) ;7 E = 1 1 0 0 0 ;

8 mprintf ( ( i ) t h e n a t u r a l i m pe de n ce o f l i n e =%. 0 f ohms\n , Z 1 ) ;

9 I l = E / ( sqrt ( 3 ) * Z 1 ) ; / / l i n e c u r r e n t ( amps )

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10 mprintf ( ( i i ) l i n e c u r r e n t =%. 1 f amps\n , I l ) ;

11 R = 1 0 0 0 ;12 Z 2 = R ;

13 E 1 = 2 * Z 2 * E / ( ( Z 1 + Z 2 ) * sqrt ( 3 ) ) ;

14 P r = 3 * E 1 * E 1 / ( R * 1 0 0 0 ) ; / / Ra te o f p ow er c o ns u mp t io n15 V r = ( Z 2 - Z 1 ) * E / ( sqrt ( 3 ) * ( Z 2 + Z 1 ) * 1 0 0 0 ) ; / / R e f l e c t e d

v o l t a g e16 E r = 3 * V r * V r * 1 0 0 0 / Z 1 // r a t e o f r e f l e c t e d v o l t a ge17 mprintf ( ( i i i ) r a t e o f e n e r g y a b s o r p t i o n =%. 1 f kW\n ,

P r ) ;

18 mprintf ( r a t e o f r e f l e c t e d e ne rg y =%. 1 f kW\n , E r ) ;19 mprintf ( ( i v ) T er mi na ti ng r e s i s t a n c e s h ou l d be e q ua l

t o s u r ge i mp ed en ce o f l i n e =%. 0 f ohms\n , Z 1 ) ;20 L = . 5 * ( 1 0 ^ - 8 ) ;

21 C = 1 0 ^ - 1 2 ;

22 Z = sqrt ( L / C ) ; // s u r g e i mp ed en ce23 V R = 2 * Z * 1 1 / ( ( Z 1 + Z ) * sqrt ( 3 ) ) ;

24 V r l = ( Z - Z 1 ) * 1 1 / ( ( Z 1 + Z ) * sqrt ( 3 ) ) ;

25 P R 1 = 3 * V R * V R * 1 0 0 0 / ( Z ) ;

26 d = abs ( V r l ) ;

27 P r l = 3 * d * d * 1 0 0 0 / Z 1 ;

28 mprintf ( ( v ) R e f r a c t e d p o we r =% . 1 f kW\n , P R 1 ) ;29 mprintf (

R e f l e c t e d p o we r =% . 1 f kW\n, P r l ) ;

30 // // Answer don t match e x a c t l y due t o d i f f e r e n c e i nr o u n d i n g o f f o f d i g i t s i bet w ee n c a l c u l a t i o n s

Scilab code Exa 12.2 Find the voltage rise at the junction due to surge

1 // Find t h e v o l t a ge r i s e a t t he j u nc t i o n due t o s ur ge2 clear

3 clc ;4 X l c = . 3 * ( 1 0 ^ - 3 ) ; // i n d uc t an c e o f c a b l e (H)5 X c c = . 4 * ( 1 0 ^ - 6 ) ; // c a pa c it a nc e o f c a b l e ( F)6 X l o = 1 . 5 * ( 1 0 ^ - 3 ) ; / / i n d u c t a nc e o f o v er h ea d l i n e (H)7 X c o = . 0 1 2 * ( 1 0 ^ - 6 ) ; // c a p ac i t a n c e o f o ve rh ea d l i n e ( F)

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8 Z n c = sqrt ( ( X l c / X c c ) ) ;

9 Z n l = sqrt ( ( X l o / X c o ) ) ;10 mprintf ( N a t u r a l i m pe de n ce o f c a b l e =%. 2 f ohms \n ,Z n c ) ;

11 mprintf ( N a t u r a l i m pe de n ce o f o v e rh e a d l i n e =%. 1 f ohms \n , Z n l ) ;

12 E = 2 * Z n l * 1 5 / ( 3 5 3 + 2 7 ) ;

13 mprintf ( v o l t a ge r i s e a t t he j u nc t i o n due t o s ur ge =% . 2 f kV \n , E ) ;

Scilab code Exa 12.3 To find the surge voltages and currents transmittedinto branch line

1 // To f i n d t he s u rg e v o l t a g e s and c u r r e n t st r an s mi t te d i n t o br an ch l i n e

2 clear

3 clc ;

4 Z 1 = 6 0 0 ;

5 Z 2 = 8 0 0 ;

6 Z 3 = 2 0 0 ;

7 E = 1 0 0 ;8 E 1 = 2 * E / ( Z 1 * ( ( 1 / Z 1 ) + ( 1 / Z 2 ) + ( 1 / Z 3 ) ) ) ;

9 I z 2 = E 1 * 1 0 0 0 / Z 2 ;

10 I z 3 = E 1 * 1 0 0 0 / Z 3 ;

11 mprintf ( T r a n s m i t t ed v o l t a g e =%. 2 f kV \n , E 1 ) ;12 mprintf ( The t r a n s m i t t e d c u r r e n t i n l i n e Z2=%. 2 f

amps \n , I z 2 ) ;13 mprintf ( The t r a n s m i t t e d c u r r e n t i n l i n e Z3=%. 1 f

amps \n , I z 3 ) ;14 // // Answer don t match e x a c t l y due t o d i f f e r e n c e i n

r o u n d i n g o f f o f d i g i t s i bet w ee n c a l c u l a t i o n s

Scilab code Exa 12.4 Determine the maximum value of transmitted wave

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1 / / D e te r mi n e t h e maximum v a l u e o f t r a n s m i t t e d wave

2 clear3 clc ;

4 Z = 3 5 0 ; / / s u r g e i m p e d en c r ( o hms )5 C = 3 0 0 0 * ( 1 0 ^ - 1 2 ) ; // e a rt h c a p a c i t a nc e ( F)6 t = 2 * ( 1 0 ^ - 6 ) ;

7 E = 5 0 0 ;

8 E 1 = 2 * E * ( 1 - exp ( ( - 1 * t / ( Z * C ) ) ) ) ;

9 mprintf ( t h e maximum v a l u e o f t r a n s m i t t e d v o l t a g e =%. 0 f kV \n , E 1 ) ;

Scilab code Exa 12.5 Determine the maximum value of transmitted surge

1 / / D e te r mi n e t h e maximum v a l u e o f t r a n s m i t t e d s u r g e2 clear

3 clc ;

4 Z = 3 5 0 ; / / s u r g e i m p e d en c r ( o hms )5 L = 8 0 0 * ( 1 0 ^ - 6 ) ;

6 t = 2 * ( 1 0 ^ - 6 ) ;

7 E = 5 0 0 ;

8 E 1 = E * ( 1 - exp ( ( - 1 * t * 2 * Z / L ) ) ) ;9 mprintf ( The maximum v a l u e o f t r a n s m i t t e d v o l t a g e =%

. 1 f kV \n , E 1 ) ;

Scilab code Exa 12.6 Determine i the value of the Voltage wave when ithas travelled through a distance 50 Km ii Power loss and Heat loss

1 / / D e te rm i ne ( i ) t h e v a l u e o f t h e V o l t a g e wave when

i t ha s t r a v e l l e d t hr ou gh a d i s t a n c e 50 Km. ( i i )Power l o s s and Heat l o s s .

2

3 clear

4 clc ;

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5 e o = 5 0 ;

6 x = 5 0 ;7 R = 6 ;

8 Z = 4 0 0 ;

9 G = 0 ;

10 v = 3 * ( 1 0 ^ 5 ) ;

11 e = 2 . 6 8 ;

12 e 1 = ( e o * ( e ^ ( ( - 1 / 2 ) * R * x / Z ) ) ) ;

13 / / a ns we ss d oe s no t match due t o t he d i f f e r e n c e i nr o u n d i n g o f f o f d i g i t s .

14 mprintf ( ( i ) t h e v a l u e o f t h e V o l ta g e wave when i th as t r a v e l l e d t h ro u gh a d i s t a n c e 5 0 Km=%. 1 f kV \n

, e 1 ) ;15 P l = e 1 * e 1 * 1 0 0 0 / 4 0 0 ;

16 i o = e o * 1 0 0 0 / Z ;

17 t = x / v ;

18 H = - ( 5 0 * 1 2 5 * 4 0 0 * ( ( e ^ - . 7 5 ) - 1 ) ) / ( 6 * 3 * 1 0 ^ 5 )

19 mprintf ( ( i i ) P owe r l o s s=%. 3 fkW \n h e at l o s s =%. 3 f kJ , P l , H ) ;

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Chapter 13

SYMMETRICAL

COMPONENTS AND FAULT

CALCULATIONS

Scilab code Exa 13.1 Determine the symmetrical components of voltages

1 / / D et er mi ne t h e s y m me t r ic a l c om po ne nt s o f v o l t a g e s .2 clear

3 clc ;

4 V a = 1 0 0* ( c o s d ( 0 ) + % i * s in d ( 0 ) ) ;

5 V b = 3 3 *( c o s d ( - 10 0) + % i * s i nd ( - 10 0 ) ) ;

6 V c = 3 8 * ( c o sd ( 1 7 6. 5 ) + % i * s in d ( 1 7 6 . 5) ) ;

7 L = 1 * ( c o sd ( 1 20 ) + % i * s i nd ( 1 20 ) ) ;

8 V a 1 = (( V a + L * Vb + ( L ^ 2) * V c )) / 3;

9 V a 2 = (( V a + L * Vc + ( L ^ 2) * V b )) / 3;

10 V c o =( ( Va + Vb + Vc ) ) /3 ;

11 disp ( V a 1 , Va1= ) ;12 disp ( V a 2 , Va2= ) ;

13 disp ( V c o , Vco= ) ;

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Scilab code Exa 13.2 Find the symmetrical component of currents

1 / / F i nd t h e s y m me t r ic a l co mp on ent o f c u r r e n t s2 clear

3 clc ;

4 I a = 5 0 0+ % i * 1 50 ; // L in e c u r r e n t i n p h a se a5 I b = 1 00 - % i * 6 00 ; // L in e c u r r e n t i n p h a se b6 I c = - 30 0+ % i * 6 00 ; // L i ne c u r r e n t i n p h a s e c7 L = ( c o s d ( 1 2 0) + % i * s i nd ( 1 2 0 ) ) ;

8 I a o = ( I a + I b + I c ) / 3 ;

9 I a 1 = ( I a + I b * L + ( L ^ 2 ) * I c ) / 3 ;

10 I a2 = ( I a + ( L ^ 2) * I b + ( L* I c )) / 3;

11 disp ( I a o , Ia o (amps )= ) ;12 disp ( I a 1 , Ia 1 (amps )= ) ;13 disp ( I a 2 , Ia 2 (amps )= ) ; // Answer i n t he book i s n ot

c o r r e c t . wrong c a l c u l a t i o n i n t he book

Scilab code Exa 13.3 Determine the fault current and line to line voltages

1 // D e t e r m i n e t he f a u l t c u r r e n t and l i n e t o l i n e

v o l t a g e s2 clear

3 clc ;

4 E a = 1 ;

5 Z 1 = . 2 5 * % i ;

6 Z 2 = . 3 5 * % i ;

7 Z o = . 1 * % i ;

8 I a 1 = E a / ( Z 1 + Z 2 + Z o ) ;

9 L = - . 5 + % i * . 8 6 6 ;

10 I a 2 = I a 1 ;

11 I a o = I a 2 ;12 I a = I a 1 + I a 2 + I a o ;

13 I b = 2 5 * 1 0 0 0 / ( ( sqrt ( 3 ) * 1 3 . 2 ) ) ;

14 I f = I b * abs ( I a ) ;

15 V a 1 = E a - ( I a 1 * Z 1 ) ;

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16 V a 2 = - I a 2 * Z 2 ;

17 V a 0 = - I a o * Z o ;18 V a = V a 1 + V a 2 + V a 0 ;

19 V b 1 = ( L ^ 2 ) * V a 1 ;

20 V b 2 = L * V a 2 ;

21 V b o = V a 0 ;

22 V c o = V a 0 ;

23 V c 1 = L * V a 1 ;

24 V c 2 = ( L ^ 2 ) * V a 2 ;

25 V b = V b1 + V b2 + V bo ;

26 V c = V c o + V c 1 + V c 2 ;

27 V a b = V a - V b ;

28 V a c = V a - V c ;29 V b c = V b - V c ;

30 v a b = ( 1 3 . 2 * abs ( V a b ) ) / sqrt ( 3 ) ;

31 v a c = ( 1 3 . 2 * abs ( V a c ) ) / sqrt ( 3 ) ;

32 v b c = ( 1 3 . 2 * abs ( V b c ) ) / sqrt ( 3 ) ;

33 disp ( I f , f a u l t c u r r e n t ( amps )= ) ; // Answer don tmatch d ue t o d i f f e r e n c e i n r o u n d i n g o f f o f d i g i t s

34 disp ( V a b , Vab(kV)= ) ; / / A n sw er d on t m at ch d ue t od i f f e r e n c e i n r o u n d i n g o f f o f d i g i t s

35 disp ( V a c , Vac(kV)= ) ; / / A n sw er d on t m at ch d ue t o

d i f f e r e n c e i n r o u n d i n g o f f o f d i g i t s36 disp ( V b c , Vbc(kV)= ) ; / / A n sw er d on t m at ch d ue t od i f f e r e n c e i n r o u n d i n g o f f o f d i g i t s

Scilab code Exa 13.4 determine the fault current and line to line voltagesat the fault

1 / / D et e r mi n e t he f a u l t c u r r e n t and l i n e t o l i n e

v ol t a g e a t t h e f a u l t .2 clear3 clc ;

4 E a = 1 ;

5 L = ( c o s d ( 1 2 0) + % i * s i nd ( 1 2 0 ) ) ;

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6 Z 1 = % i * . 2 5 ;

7 Z 2 = % i * . 3 5 ;8 I a 1 = E a / ( Z 1 + Z 2 ) ;

9 I a 2 = - I a 1 ;

10 I a o = 0 ;

11 I b 1 = ( L ^ 2 ) * I a 1 ;

12 I b 2 = L * I a 2 ;

13 I b o = 0 ;

14 I b = I b 1 + I b2 + I b o ;

15 I b a = 1 0 9 3 ;

16 I f = I b a * abs ( I b ) ;

17 V a 1 = E a - ( I a 1 * Z 1 ) ;

18 V a 2 = - I a 2 * Z 2 ;19 V a o = 0 ;

20 V a = V a 1 + V a 2 + V a o ;

21 V b = ( L ^2 ) * Va 1 + L * V a2 ;

22 V c = V b ;

23 V a b = V a - V b ;

24 V a c = V a - V c ;

25 V b c = V b - V c ;

26 mprintf ( F a u l t c u r r e n t =%. 2 f amps\n , I f ) ; //Answerdon t match due t o d i f f e r e n c e i n r ou nd in g o f f o f

d i g i t s27 v a b = ( abs ( V a b ) * 1 3 . 2 ) / sqrt ( 3 ) ;28 v b c = ( abs ( V b c ) * 1 3 . 2 ) / sqrt ( 3 ) ;

29 v a c = ( abs ( V a c ) * 1 3 . 2 ) / sqrt ( 3 ) ;

30 mprintf ( Vab=%. 2 f kV\n , v a b ) ;31 mprintf ( Vac=%. 2 f kV\n , v a c ) ;32 mprintf ( Vbc=%. 2 f kV\n , v b c ) ;

Scilab code Exa 13.5 determine the fault current and line to line voltagesat the fault

1 // d et er mi ne t h e f a u l t c u r r e n t and l i n e t o l i n ev o l t ag e s a t t h e f a u l t

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2 clear

3 clc ;4 E a = 1+ 0 * %i ;

5 Z o = % i * . 1 ;

6 Z 1 = % i * . 2 5 ;

7 Z 2 = % i * . 3 5 ;

8 I a 1 = E a / ( Z 1 + ( Z o * Z 2 / ( Z o + Z 2 ) ) ) ;

9 V a 1 = E a - I a 1 * Z 1 ;

10 V a 2 = V a 1 ;

11 V a o = V a 2 ;

12 I a 2 = - V a 2 / Z 2 ;

13 I a o = - V a o / Z o ;

14 I = I a 2 + I a o ;15 I f = 3 * I a o ; // f a u l t c u r r e n t16 I b = 1 0 9 3 ; // b as e c u r r e n t17 I f l = abs ( I f * I b ) ;

18 disp ( I f l , F a u l t c u r r e n t ( amps ) = ) ; //Answer don tmatch d ue t o d i f f e r e n c e i n r o u n d i n g o f f o f d i g i t s

19 V a = 3 * V a 1

20 V b = 0 ;

21 V c = 0 ;

22 V a b = abs ( V a ) * 1 3 . 2 / sqrt ( 3 ) ;

23 V a c = abs ( V a ) * 1 3 . 2 / sqrt ( 3 ) ;

24 V b c = abs ( V b ) * 1 3 . 2 / sqrt ( 3 ) ;

25 mprintf ( Vab=%. 3 f kV\n , V a b ) ;26 mprintf ( Vac=%. 3 f kV\n , V a c ) ;27 mprintf ( Vbc=%. 3 f kV\n , V b c ) ;

Scilab code Exa 13.6 Determine the fault current when i LG ii LL iii LLGfault takes place at P

1 / / D et er mi ne t h e f a u l t c u r r e n t when ( i ) LG ( i i ) LL (i i i )LLG f a u l t t ak es p l a c e a t P .

2 clear

3 clc ;

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4 V b l = 1 3 . 8 * 1 1 5 / 1 3 . 2 ; // b a s e v o l t a g e on t h e l i n e s i d e

o f t r a n s f o r m e r ( kV )5 V b m = 1 2 0 * 1 3 . 2 / 1 1 5 ; // b as e v o l t a g e on t he motor s i d eo f t r a n s f o r m e r ( kV )

6 X t = 1 0 * ( ( 1 3 . 2 / 1 3 . 8 ) ^ 2 ) * 3 0 / 3 5 ; // p e rc e n t r e ac t a nc e o f t r a n s f o r m e r

7 X m = 2 0 * ( ( 1 2 . 5 / 1 3 . 8 ) ^ 2 ) * 3 0 / 2 0 ; // p e rc e n t r e ac t a nc e o f motor

8 X l = 8 0 * 3 0 * 1 0 0 / ( 1 2 0 * 1 2 0 ) ; // p e rc e n t r e a ct a n ce o f l i n e9 X n = 2 * 3 * 3 0 * 1 0 0 / ( 1 3 . 8 * 1 3 . 8 ) ; // n e u t r a l r e ac t a nc e

10 X z = 2 0 0 * 3 0 * 1 0 0 / ( 1 2 0 * 1 2 0 ) ;

11 Z n = % i * . 1 4 6 ; / / n e g a t i v e s e q ue n c e i mp ed en ce

12 Z o = . 0 6 7 6 7 ; / / z e r o s e q u en c e i mp ed en ce13 Z = % i * . 3 5 9 6 ; / / t o t a l i m pe d en c e14 I a 1 = 1 / Z ;

15 I a 2 = I a 1 ;

16 I a o = I a 2 ;

17 I f 1 = 3 * I a 1 ;

18 I b = 3 0 * 1 0 0 0 / ( sqrt ( 3 ) * 1 3 . 8 ) ;

19 I b l = 3 0 * 1 0 0 0 / ( sqrt ( 3 ) * 1 2 0 ) ;

20 I f c = I b l * abs ( I f 1 ) ;

21 Z 1 = % i * . 1 4 6 ;

22 Z 2 = Z 1 ;

23 I A 1 = 1 / ( Z 1 + Z 2 )

24 I A 2 = - I A 1

25 L = ( c o s d ( 1 2 0) + % i * s i nd ( 1 2 0 ) ) ;

26 I A o = 0 ;

27 I B = ( L ^2 ) * IA 1 + L * I A2 ;

28 I C = - I B ;

29 IF = abs ( I B ) * I b l ;

30 Z o = % i * . 0 6 7 6 7 ;

31 i a 1 = 1 / ( Z 1 + ( Z o * Z 2 / ( Z o + Z 2 ) ) ) ;

32 i a 2 = i a 1 * Z o / ( Z 2 + Z o ) ;

33 i a o = % i * 3 . 5 5 3 ;34 I f 2 = 3 * i a o ;

35 I F 2 = abs ( I f 2 * I b l ) ;

36 mprintf ( F a u l t C u rr e nt ( i ) LG f a u l t , I f =%. 0 f amps\n , I f c ) ;

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37 mprintf ( ( i i )LL f a u l t , I f =%. 1 f amps\n , I F ) ;

38 mprintf ( ( i i i )LLG , I f =%. 0 f amps\n , I F 2 ) ;

Scilab code Exa 13.8 Determine the percent increase of busbar voltage

1 / / De t e r mi ne t he p e rc e n t i n c r e a s e o f b us ba r v o l t a g e2 clear

3 clc ;

4 v x = 3 ; // p er c e nt r e a c ta n ce o f t h e s e r i e s e l em en t

5 s i n r = . 6 ;6 V = v x * s i n r ;

7 mprintf ( P e r c en t d ro p o f v o l t s =%. 1 f p e r c e n t \n , V ) ;

Scilab code Exa 13.9 Determine the short circuit capacity of the breaker

1 / / De t e r mi n e t he s h or t c i r c u i t c a p a ci t y o f t heb r e a k e r

2 clear3 clc ;

4 S b = 8 ; // Base MVA5 Z e q = ( % i * . 1 5 ) * ( % i * . 3 1 5 ) / ( % i * . 4 6 5 ) ;

6 S c c = abs ( S b / Z e q ) ;

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Electrical Power Systems_C. L. Wadhwa