Electrical Power Systems_C. L. Wadhwa.pdf

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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

May 23, 2016

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 4

1 FUNDAMENTALS OF POWER SYSTEMS 5

2 LINE CONSTANT CALCULATIONS 7

3 CAPACITANCE OF TRANSMISSION LINES 11

4 PERFORMANCE OF LINES 13

5 HIGH VOLTAGE DC TRANSMISSION 23

6 CORONA 26

7 MECHANICAL DESIGN OF TRANSMISSION LINES 30

8 OVERHEAD LINE INSULATORS 34

9 INSULATED CABLES 35

10 VOLTAGE CONTROL 40

11 NEUTRAL GROUNDING 44

12 TRANSIENTS IN POWER SYSTEMS 46

13 SYMMETRICAL COMPONENTS AND FAULT CALCU-

LATIONS 51

14 PROTECTIVE RELAYS 65

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15 CIRCUIT BREAKERS 71

17 POWER SYSTEM SYNCHRONOUS STABILITY 75

18 LOAD FLOWS 85

19 ECONOMIC LOAD DISPATCH 91

20 LOAD FREQUENCY CONTROL 94

21 COMPENSATION IN POWER SYSTEMS 96

22 POWER SYSTEM VOLTAGE STABILITY 98

23 STATE ESTIMATION IN POWER SYSTEMS 102

24 UNIT COMMITMENT 107

25 ECONOMIC SCHEDULING OF HYDROTHERMAL PLANTS

AND OPTIMAL POWER FLOWS 110

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List of Scilab Codes

Exa 1.1 To determine the Base values and pu values . . . . . . 5Exa 2.2 To dtermine inductance of a 3 phase line . . . . . . . 7Exa 2.3 Determine the equivalent radius of bundle conductor

having its part conductors r on the periphery of circleof dia d . . . . . . . . . . . . . . . . . . . . . . . . . . 7

Exa 2.4 To determine the inductance of single phase Transmis-sion line . . . . . . . . . . . . . . . . . . . . . . . . . . 8

Exa 2.5 To determine the inductance per Km of 3 phase line . 9Exa 2.6 To determine the inductance of double circuit line . . 10Exa 2.7 To determine the inductance per Km per phase of single

circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . 10Exa 3.1 To determine the capacitance and charging current . 11Exa 3.2 To determine the capacitance and charging current . 11

Exa 3.3 To determine the capacitance and charging current . 12Exa 4.1 To determine the sending end voltage and current power

and power factor Evaluate A B C D parameters . . . . 13Exa 4.2 To determine power input and output i star connected

ii delta connected . . . . . . . . . . . . . . . . . . . . 14Exa 4.3 To determine efficiency and regulation of line . . . . . 15Exa 4.4 To calculate the voltage across each load impedence and

current in the nuetral . . . . . . . . . . . . . . . . . . 15Exa 4.5 To determine efficiency and regulation of 3 phase line. 16Exa 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 . . . . . . . . . . . . . . . . . . . . . . . . 17

Exa 4.7 To determine of efficiency of line . . . . . . . . . . . . 19

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Exa 4.8 To determine the ABCD parameters of Line . . . . . . 19

Exa 4.9 To determine the sending end voltage and efficiency us-ing Nominal pi and Nominal T method . . . . . . . . 20Exa 4.10 To determine the sending end voltage and current power

and power factor Evaluate A B C D parameters . . . . 21Exa 5.1 To determine the dc output voltage when delay anglw

a0 b30 c45 . . . . . . . . . . . . . . . . . . . . . . . . 23Exa 5.2 To determine the necessary line secondary voltage and

tap ratio required . . . . . . . . . . . . . . . . . . . . 23Exa 5.3 To determine the effective reactance per phase . . . . 24Exa 5.4 Calculate the direct current delivered. . . . . . . . . . 24Exa 6.1 To determine the critical disruptive voltage and critical

voltage for local and general corona . . . . . . . . . . 26Exa 6.2 To determine whether corona will be present in the air

space round the conductor. . . . . . . . . . . . . . . . 27Exa 6.3 To determine the critical disruptive voltage and corona

loss . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27Exa 6.4 To determine the voltage for which corona will com-

mence on the line . . . . . . . . . . . . . . . . . . . . 28Exa 6.5 To determine the corona characterstics . . . . . . . . . 28Exa 7.1 Calculate the sag . . . . . . . . . . . . . . . . . . . . 30Exa 7.2 To calculate the maximum Sag . . . . . . . . . . . . . 30

Exa 7.3 To determine the Sag . . . . . . . . . . . . . . . . . . 31Exa 7.4 To determine the clearence between the conductor andwater level . . . . . . . . . . . . . . . . . . . . . . . . 32

Exa 8.1 To determine the maximum voltage that the string ofthe suspension insulators can withstand . . . . . . . . 34

Exa 9.1 To determine the economic overall diameter of a 1corecable metal sheathead . . . . . . . . . . . . . . . . . . 35

Exa 9.2 To determine the minimum internal diameter of the leadsheath . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

Exa 9.3 To determine the maximum safe working voltage . . . 36Exa 9.4 To determine the maximum stresses in each of the three

layers . . . . . . . . . . . . . . . . . . . . . . . . . . . 37Exa 9.5 o dtermine the equivalent star connected capacity and

the kVA required . . . . . . . . . . . . . . . . . . . . . 37

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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. . . . . . . . . . . . . . . . . . . . . 38

Exa 9.7 To calculate the induced emf in each sheath . . . . . 38Exa 9.8 To determine the ratio of sheath loss to core loss of the

cable . . . . . . . . . . . . . . . . . . . . . . . . . . . 39Exa 10.1 To determine the total power active and reactive sup-

plied by the generator and the pf at which the generatormust operate . . . . . . . . . . . . . . . . . . . . . . . 40

Exa 10.2 Determine the settings of the tap changers required tomaintain the voltage of load bus bar . . . . . . . . . . 41

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 . . . . . . . 41

Exa 10.4 Determine the KV Ar of the Modifier and the maximumload that can be transmitted . . . . . . . . . . . . . . 42

Exa 11.1 To find the inductance and KVA rating of the arc sup-pressor coil in the system . . . . . . . . . . . . . . . . 44

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 . . . . 44Exa 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 . . . . . . 46

Exa 12.2 Find the voltage rise at the junction due to surge . . 47Exa 12.3 To find the surge voltages and currents transmitted into

branch line . . . . . . . . . . . . . . . . . . . . . . . . 48Exa 12.4 Determine the maximum value of transmitted wave. . 48Exa 12.5 Determine the maximum value of transmitted surge . 49Exa 12.6 Determine i the value of the Voltage wave when it has

travelled through a distance 50 Km ii Power loss andHeat loss . . . . . . . . . . . . . . . . . . . . . . . . . 49

Exa 13.1 Determine the symmetrical components of voltages . . 51Exa 13.2 Find the symmetrical component of currents . . . . . 51Exa 13.3 Determine the fault current and line to line voltages . 52

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

the fault . . . . . . . . . . . . . . . . . . . . . . . . . 53Exa 13.5 determine the fault current and line to line voltages atthe fault . . . . . . . . . . . . . . . . . . . . . . . . . 54

Exa 13.6 Determine the fault current when i LG ii LL iii LLGfault takes place at P . . . . . . . . . . . . . . . . . . 55

Exa 13.8 Determine the percent increase of busbar voltage . . . 57Exa 13.9 Determine the short circuit capacity of the breaker . . 57Exa 13.10 To determine the short circuit capacity of each station 57Exa 13.11 Determine the Fault MVA . . . . . . . . . . . . . . . 58Exa 13.12 To Determine the subtransient current in the alternator

motor and the fault . . . . . . . . . . . . . . . . . . . 58

Exa 13.13 To Determine the reactance of the reactor to preventthe brakers being overloaded . . . . . . . . . . . . . . 59

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 . . . . . . . . . . . . . . 60

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 . . . . . . . . . . . . . 61

Exa 13.16 Determine the shorrt circuit MVA of the transformer 62Exa 13.17 To determine the line voltages and currents in per unit

on delta side of the transformer . . . . . . . . . . . . . 63Exa 14.1 To determine the time of operation of relay . . . . . . 65Exa 14.2 To determine the phase shifting network to be used. . 65Exa 14.3 To provide time current grading . . . . . . . . . . . . 66Exa 14.4 To determine the proportion of the winding which re-

mains unprotected against earth fault . . . . . . . . . 66Exa 14.5 To determine i percent winding which remains unpro-

tected ii min value of earthing resistance required toprotect 80 percent of winding . . . . . . . . . . . . . 67

Exa 14.6 To determine whether relay will operate or not . . . . 68Exa 14.7 To determine the ratio of CT on HV side . . . . . . . 68

Exa 14.8 To determine the number of turns each current trans-former should have . . . . . . . . . . . . . . . . . . . . 68

Exa 14.9 To determine the R1 R2 and C also The potential acrossrelays . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

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Exa 14.10 To determine the kneepoint voltage and cross section of

core . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69Exa 14.11 To determine the VA output of CT . . . . . . . . . . 70Exa 15.1 To determine the voltage appearing across the pole of

CB also determine the value of resistance to be usedacross contacts . . . . . . . . . . . . . . . . . . . . . . 71

Exa 15.2 To determine the rate of rise of restriking voltage . . . 71Exa 15.3 To Determine the average rate of rise of restriking voltage 72Exa 15.4 To determine the rated normal current breaking current

making current and short time rating current . . . . . 73Exa 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 . . . . . . . . . . . . . . . . 73

Exa 17.1 To determine the acceleration Also determine the changein torque angle and rpmat the end of 15 cycles . . . . 75

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. . . . . . . . . . 76

Exa 17.3 To calculate the maximum value of d during the swing-ing of the rotor around its new equilibrium position. . 77

Exa 17.4 To calculate the critical clearing angle for the conditiondescribed . . . . . . . . . . . . . . . . . . . . . . . . . 77Exa 17.5 To calculate the critical clearing angle for the generator

for a 3phase fault. . . . . . . . . . . . . . . . . . . . . 78Exa 17.6 determine the critical clearing angle . . . . . . . . . . 79Exa 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. . . . . . . . . . . . . . . . . . . 80

Exa 17.8 Compute the prefault faulted and post fault reduced Ymatrices. . . . . . . . . . . . . . . . . . . . . . . . . . 80

Exa 17.9 Determine the reduced admittance matrices for prefault

fault and post fault conditions and determine the powerangle characterstics for three conditions . . . . . . . . 81

Exa 17.10 To Determine the rotor angle and angular frequency us-ing runga kutta and eulers modified method . . . . . . 82

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

gauss seidal method . . . . . . . . . . . . . . . . . . . 85Exa 18.2 Determine the voltages starting with a flat voltage profile 86Exa 18.3 Solve the prevous problem for for voltages at the end of

first iteration . . . . . . . . . . . . . . . . . . . . . . . 87Exa 18.4 Determine the set of load flow equations at the end of

first iteration by using Newton Raphson method . . . 88Exa 18.5 Determine the equations at the end of first iteration

after applying given constraints . . . . . . . . . . . . . 90Exa 19.1 To Determine the economic operating schedule and the

corresponding cost of generation b Determine the sav-ings obtained by loading the units . . . . . . . . . . . 91

Exa 19.2 Determine the incremental cost of recieved power andpenalty factor of the plant. . . . . . . . . . . . . . . . 92

Exa 19.4 Determine the minimum cost of generation . . . . . . 92Exa 20.1 Determine the load taken by the set C and indicate the

direction in which the energy is flowing . . . . . . . . 94Exa 20.2 Determine the load shared by each machine . . . . . . 94Exa 20.3 Determine the frequency to which the generated voltage

drops before the steam flow commences to increase tomeet the new load . . . . . . . . . . . . . . . . . . . . 95

Exa 21.1 Determine the load bus voltage . . . . . . . . . . . . . 96

Exa 22.2 To Determine the source voltage when the load is dis-connected to load pf i unity ii 8 lag . . . . . . . . . . . 98Exa 22.3 To determine thee Ac system voltage when the dc sys-

tem is disconnected or shutdown . . . . . . . . . . . . 99Exa 22.4 To Calculate the new on and off times for constant energy 99Exa 22.6 To discuss the effect of tap changing . . . . . . . . . . 100Exa 22.7 To determine the effect of tapping to raise the secondary

voltage by 10percent . . . . . . . . . . . . . . . . . . . 100Exa 22.8 Calculate the additional reactive power capability at full

load . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101Exa 23.1 To determine the state vector at the end of first iteration 102

Exa 23.2 Determine The States of the systems at the end of firstiteration. . . . . . . . . . . . . . . . . . . . . . . . . . 104

Exa 23.3 Problem on State Estimator Linear Model . . . . . . . 105Exa 23.4 Determine theta1 Theta2 . . . . . . . . . . . . . . . . 105Exa 24.3 Priority List Method. . . . . . . . . . . . . . . . . . . 107

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Exa 24.4 illustrate the dynamic programming for preparing an

optimal unit commitment . . . . . . . . . . . . . . . . 108Exa 25.1 illustrating the procedure for economic scheduling clearall . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

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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 ue o f v o l t a ge 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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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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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 ) F or 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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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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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 ce 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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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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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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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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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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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 ) ;

7 mprintf ( s h o r t c i r c u i t c a p a c i t y =%. 2 f MVA\n , S c c ) ;

Scilab code Exa 13.10 To determine the short circuit capacity of each sta-

tion

1 / / To d e te rm in e t h e s h o r t c i r c u i t c a p a ci t y o f e a c hs t a t i o n

2 clear

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3 clc ;

4 X = 1 2 0 0 * 1 0 0 / 8 0 0 ; // p e rc e n t r e a ct a n ce o f o t he rg e n e r a t i n g s t a t i o n5 X c = . 5 * 1 2 0 0 / ( 1 1 * 1 1 ) ;

6 S c = 1 2 0 0 * 1 0 0 / 8 6 . 5 9 ; // s h o rt c i r c u i t MVA o f t he bus7 X f = 1 1 9 . 8 4 ; // e q u i v a l e n t f a u l t i mp ed en ce b etw een F

and n e u t r a l b us8 M V A = 1 2 0 0 * 1 0 0 / X f ;

9 mprintf ( s h o rt c i r c u i t c a p ac i t y o f e a ch s t a t i o n=%. 0 f MVA\n , M V A ) ;

Scilab code Exa 13.11 Determine the Fault MVA

1 / / D e t er m i ne t h e F a u l t MVA2 clear

3 clc ;

4 S b = 1 0 0 ; // ba se power (MVA)5 S C = S b / . 1 4 ;

6 mprintf ( S . C . MVA =%. 2 f MVA\n , S C ) ;

Scilab code Exa 13.12 To Determine the subtransient current in the al-ternator motor and the fault

1 / / To De t er m ine t he s u b t r a n si e n t c u r r e n t i n t hea l t e r n a t o r , motor and t he f a u l t

2 clear

3 clc ;

4 I b = 5 0 * 1 0 0 0 / ( sqrt ( 3 ) * 1 3 . 2 ) ; / / b a se c u r r e n t ( amps . )

5 V f = 1 2 . 5 / 1 3 . 5 ; // t he P r e f a u l t V o lt a ge ( p . u )6 X f = ( % i * . 3 ) * ( % i * . 2 ) / ( % i * . 5 ) ; / / F a u l t i m p e d e n c e ( p . u )7 I f = . 9 4 6 9 / ( X f ) ; / / F a u l t c u r r e n t ( p . u )8 I f l = 3 0 * 1 0 0 0 / ( ( sqrt ( 3 ) * 1 2 . 5 * . 8 ) ) ; / / f u l l l o a d c u r r e n t

(amps)

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27 V a b = . 2 5 6 4 - V b ;

28 V b c = V b - V c ;29 V c a = V c - . 2 5 6 4 ;

30 V A 1 = E a - I A 1 * ( % i * . 0 5 ) ;

31 V A 2 = - I A 2 * ( % i * . 0 5 ) ;

32 V A = V A 1 + V A 2 ;

33 V B = (( ( - .5 - % i * .8 66 ) * V A1 ) + (( - . 5 + % i * .8 6 6) * V A2 ) ) ;

34 V C= V A1 * ( -. 5 + %i * . 86 6) + V A2 * ( -. 5 - %i * . 86 6) ;

35 V A B = V A - V B ;

36 V B C = V B - V C ;

37 V C A = V C - V A ;

38 / / An swe rs don t match due t o d i f f e r e n c e i n r o un d in g

o f f o f d i g i t s39 disp ( I a , f a u l t c u r r e n t s , I a= ) ;40 disp ( I b , Ib= ) ;41 disp ( I c , Ic = ) ; / / C a l c u l a t i o n i n bo ok i s wrong .42 disp ( I A , IA= ) ;43 disp ( I B , IB ) ;44 disp ( I C , IC ) ;45 disp ( V o l ta g es a t f a u l t p o in t ) ;46 disp ( V a b , Vab( p . u )= ) ;47 disp ( V b c , Vbc( p . u )= ) ;48 disp ( V c a ,

Vca ( p . u )=) ;

49 disp ( V A B , VAB= ) ;50 disp ( V B C , VBC=) ;51 disp ( V C A , VCA= ) ;

Scilab code Exa 13.15 To determine the i pre fault current in line a iithe subtransient current in pu iii the subtransient current in each phase ofgenerator in pu

1 / / To d e t er m i ne t h e ( i ) p re f a u l t c ur re nt i n l i n e a( i i ) t he s u b t r a n s i e n t c u r r e nt i n p . u ( i i i ) t he

s u b t r a ns i e n t c u r r e nt i n e a ch p ha se o f g e n er a t o ri n p . u

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2 clear

3 clc ;4 I a 1 = - .8 - %i * 2 .6 + . 8 - %i * . 4;

5 I a 2 = - % i * 3 ;

6 I a o = - % i * 3 ;

7 A = - .8 - %i * 2. 6 + .8 + %i * 2;

8 a = . 8 ;

9 b = . 6 ;

10 I p f =a + %i * b;

11 I s f c = 3 * I a 1 ;

12 i A 1 = .8 - % i * . 4;

13 i A 2 = - % i * 1 ;

14 i A o = 0 ;15 I A 1 = % i * i A 1 ;

16 I A 2 = - % i * i A 2 ;

17 I A = IA 1 + I A2 ;

18 L = c o sd ( 1 2 0 ) + % i * s in d ( 1 2 0 ) ;

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

20 I C = ( L ^2 ) * IA 2 + I A1 * L ;

21 disp ( I p f , ( i ) p r e f a u l t c u r r e n t i n l i n e a= ) ;22 disp ( I s f c , ( i i ) t h e s u b t r a n si e n t f a u l t c u r r e n t i n p .

u=) ;23 disp ( I A ,

IA=) ;

24 disp ( I B , IB= ) ;25 disp ( I C , IC=) ;

Scilab code Exa 13.16 Determine the shorrt circuit MVA of the trans-former

1 / / D et er mi ne t he s h o r r t c i r c u i t MVA o f t he

t r a n s f o r m e r2 clear3 clc ;

4 S . C . M V A = . 5 / . 0 5 ;

5 mprintf ( S .C .MVA=%. 0 f MVA , S . C . M V A ) ;

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Chapter 14

PROTECTIVE RELAYS

Scilab code Exa 14.1 To determine the time of operation of relay

1 / / To d e te rm in e t h e t i m e o f o p er a ti o n o f r e l a y .2 clear

3 clc ;

4 I f = 4 0 0 0 ; // f a u l t c u r r e n t5 I = 5 * 1 . 2 5 ; // o p er a t i ng c u r r e n t o f r e l a y6 C T = 4 0 0 / 5 ; // CT r a t i o

7 P S M = I f / ( I * C T ) ; // p l u g s e t t i n g m u l t i p l i e r8 mprintf ( PSM=%. 3 f\n , P S M ) ;9 mprintf ( o p e r a t i n g t i me f o r PSM=8 i s 3 . 2 s e c . \ n ) ;

10 mprintf ( a c t u a l o p e r a t i n g t im e = 1 . 9 2 s e c . ) ;

Scilab code Exa 14.2 To determine the phase shifting network to be used

1 / / To d et er mi ne t he p ha se s h i f t i n g n et wo rk t o be

used .2 clear

3 clc ;

4 Z = 1 0 0 0 *( c o sd ( 6 0 ) + % i * s i nd ( 6 0 ) ) ; / / i m p e d e n c e

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5 X = t a n d ( 5 0 ) * 1 0 0 0 * c o s d ( 6 0 ) ;

6 X l = 1 0 0 0 * s i n d ( 6 0 ) ;7 X c = X l - X ;

8 C = 1 0 0 0 0 0 0 / ( 3 1 4 * X c ) ;

9 / / An swe rs don t match due t o d i f f e r e n c e i n r o un d in go f f o f d i g i t s

10 disp (X , X= ) ;11 disp ( X c , Xc= ) ;12 disp (C , C ( m i c r o f a r a d s )= ) ;

Scilab code Exa 14.3 To provide time current grading

1 //To p r o v i d e t im e c u r r e nt g r a di n g .2 clear

3 clc ;

4 I s e c 1 = 4 0 0 0 / 4 0 ; / / s e c o n d a r y c u r r e n t ( amps )5 P S M = 1 0 0 / 5 ; // PSM i f 1 00% s e t t i n g i s u se d6 I s e c 2 = 4 0 0 0 / 4 0 ;

7 P S M 2 = 1 0 0 / 6 . 2 5 ; //PSM i f s e t t i n g u se d i s 1 25%8 T M S b = . 7 2 / 2 . 5 ;

9 P S M 1 = 5 0 0 0 / ( 6 . 2 5 * 4 0 ) ;10 t o = 2 . 2 ;

11 t b = t o * T M S b ;

12 P S M a = 5 0 0 0 / ( 6 . 2 5 * 8 0 ) ;

13 T M S = 1 . 1 3 8 / 3 ;

14 P S M a 1 = 6 0 0 0 / ( 6 . 2 5 * 8 0 ) ;

15 t a = ( 2 . 6 * . 3 7 9 ) ;

16 mprintf ( A c tu al o p e r a t i n g t im e o f r e a l y a t b=%. 3 f s e c . \n , t b ) ;

17 mprintf ( A c tu al o p e r a t i n g t im e o f r e a l y a t a=%. 3 f

s e c . \n , t a ) ;

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Scilab code Exa 14.4 To determine the proportion of the winding which

remains unprotected against earth fault

1 / / To d e te rm i ne t he p r o p o r ti o n o f t he w in di ng wh ichr em ai ns u np r o t ec t e d a g a i n s t e a rt h f a u l t .

2 clear

3 clc ;

4 V p h = 6 6 0 0 / ( sqrt ( 3 ) ) ;

5 I f u l l = 5 0 0 0 / ( sqrt ( 3 ) * 6 . 6 ) ;

6 I b = I f u l l * . 2 5 ;

7 x = I b * 8 0 0 / V p h ;

8 mprintf ( p e r c e n t o f t h e w in di n g r e ma i ns u n p r o t ec t e d=

%.2 f \n , x ) ;

Scilab code Exa 14.5 To determine i percent winding which remains un-protected ii min value of earthing resistance required to protect 80 percentof winding

1 / / To d e t e r m i n e ( i ) % w i nd i n g w hi ch r e m a in su n pr o t ec t ed ( i i ) min . v a lu e o f e a r t h i n g r e s i s t a n c e

r e q ui r e d t o p r o t ec t 80% o f w in di ng2 clear

3 clc ;

4 I p h = 1 0 0 0 0 / sqrt ( 3 ) ; // p ha se v o l t a g e o f a l t e r n a t o r (V)5 x = 1 . 8 * 1 0 0 * 1 0 * 1 0 0 0 / ( 5 * I p h ) ;

6 mprintf ( ( i ) p e r c e n t w in d in g w hi ch r em a in sun pr ot e c t e d=%. 2 f \n , x ) ;

7 I p = I p h * . 2 ;

8 R = 1 . 8 * 1 0 0 0 / ( 5 * I p ) ;

9 mprintf ( ( i i ) minimum v a l ue o f e a r t h i n g r e s i s t a n c e

r e q u i r e d t o p r o t e c t 80 p e r ce n t o f w in di ng =%. 4 f ohms \n ,R )

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2 clear

3 clc ;4 I c = 5 * . 2 5 ; // o p e r a t i n g c u r r e n t ( amp )5 V s e c = 5 / 1 . 2 5 ; // s e c on d a r y v o l t a g e (V)6 B m = 1 . 4 ;

7 f = 5 0 ;

8 N = 5 0 ;

9 V = 1 5 * V s e c ;

10 A = 6 0 / ( 4 . 4 4 * B m * f * N ) ;

11 mprintf ( t he k ne e p oi nt must b e s l i g h t l y h i g h ert han =%. 3 f V\n , V ) ;

12 mprintf ( a r ea o f c r o s s s e c t i o n=%. 6 f m 2\n , A ) ;

Scilab code Exa 14.11 To determine the VA output of CT

1 / / To d e t er m i ne t h e VA o u t pu t o f CT .2 clear

3 clc ;

4 o . p = 5 * 5 * ( . 1 +. 1 ) + 5;

5 mprintf ( VA out pu t of CT =%. 0 f VA\n , o . p ) ;

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1 / / To d e t e r m in e t h e r a t e o f r i s e o f r e s t r i k i n g

v o l t a g e2 clear3 clc ;

4 V n l = 1 3 2 * sqrt ( 2 ) / sqrt ( 3 ) ; // p eak v al ue o f peak t on e u t r a l v o l t a g e ( kV )

5 V r 1 = V n l * . 9 5 ; // r e c o v e r y v o l t a g e ( kV )6 V r = 1 0 2 . 4 * . 9 1 6 ; // a c t i v e r e c o ve r y v o l t a g e ( kV )7 V r m a x = 2 * V r ;

8 f n = 1 6 * ( 1 0 ^ 3 ) ;

9 t = 1 / ( 2 * f n ) ;

10 R R R V = V r m a x * ( 1 0 ^ - 6 ) / t ;

11 mprintf ( r a t e o f r i s e o f r e s t r i k i n g v o l t a g e , RRRV=%. 0 f kV / m i c rose c , R R R V ) ;

Scilab code Exa 15.3 To Determine the average rate of rise of restrikingvoltage

1 // To D et er m ine t he a v e ra ge r a t e o f r i s e o f r e s t r i k i n g v o l t a g e

2 clear3 clc ;

4 V m = 1 3 2 * sqrt ( 2 ) / sqrt ( 3 ) ;

5 K 1 = . 9 ;

6 K 2 = 1 . 5

7 K = K 1 * K 2 ;

8 s i n q = . 9 2 ;

9 V r = K * V m * s i n q ;

10 f n = 1 6 * ( 1 0 ^ 3 ) ;

11 R R R V = 2 * V r * ( 1 0 ^ - 6 ) * f n * 2 ;

12 mprintf ( a ve r a ge r a t e o f r i s e o f r e s t r i k i n g v ol ta ge ,RRRV=%. 3 f kV/ mi cr os e c\n , R R R V ) ;

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Chapter 17

POWER SYSTEM

SYNCHRONOUS STABILITY

Scilab code Exa 17.1 To determine the acceleration Also determine thechange in torque angle and rpmat the end of 15 cycles

1 / / To d et er mi ne t he a c c e l e r a t i o n . A ls o d et er mi net h e c ha ng e i n t o r q ue a n g l e and r . p . mat t h e en d o f

15 c y c l e s2 clear

3 clc ;

4 H = 9 ;

5 G = 2 0 ; // mach ine Ra ti ng (MVA)6 K E = H * G ;

7 mprintf ( ( a )K .E s t o r e d i n t he r o t o r =%. 0 f MJ\n , K E ) ;8 P i = 2 5 0 0 0 * . 7 3 5 ;

9 P G = 1 5 0 0 0 ;

10 P a = ( P i - P G ) / ( 1 0 0 0 ) ;

11 f = 5 0 ;

12 M = G * H / ( % p i * f ) ;13 a = P a / M ;

14 mprintf ( ( b ) The a c c e l e r a t i n g p ow er =%. 3 f MW\n , P a ) ;15 mprintf ( A c c e l e r a t i o n =%. 3 f r a d / s e c 2\n , a ) ;16 t = 1 5 / 5 0 ;

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Scilab code Exa 17.3 To calculate the maximum value of d during theswinging of the rotor around its new equilibrium position

1 / /To c a l c u l a t e t h e maximum v a l u e o f d d u r i ng t h es w i ng i n g o f t he r o t o r ar ound i t s new e q u i l i b r i u mp o s i t i o n

2 clc

3 clear

4 a = . 2 5 ; // s i nd o =.255 d o = a s i n d ( a ) ; //6 b = . 5 // s i nd c =. 57 d c = a s i n d ( b ) ;

8 c = c o s d ( d o ) + . 5 * d o * % p i / 1 8 0 ;

9 d m = d c ;

10 e = 1 ;

11 while ( e > . 0 0 0 1 )

12 d m = d m + . 1 ;

13 e = abs ( c - ( ( ( . 5 * d m * % p i ) / 1 8 0 ) + c o s d ( d m ) ) ) ;

14 end

15 printf ( dm a p p r ox i m a te l y f ou nd t o be %d d e g r e e , d m ) ;

Scilab code Exa 17.4 To calculate the critical clearing angle for the con-dition described

1 // To c a l c u l a t e t he c r i t i c a l c l e a r i n g a n gl e f o r t hec o n d i t i o n d e s c r i b e d .

2 clear

3 clc ;4 s i n d o = . 5 ;

5 d 0 = a s i n d ( s i n d o ) * % p i / 1 8 0 ;

6 r 1 = . 2 ;

7 r 2 = . 7 5 ;

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13 r 1 = P m a x 1 / P m a x o ;

14 r 2 = P m a x 2 / P m a x o ;15 P s = 1 ;

16 s i n d o = P s / P m a x o ;

17 d o = a s i n d ( s i n d o ) ;

18 d 0 = a s i n d ( s i n d o ) * % p i / 1 8 0 ;

19 s i n d m = 1 / P m a x 2 ;

20 c o s d m = c o s d ( a s i n d ( s i n d m ) ) ;

21 D m = % p i * ( 1 8 0 - ( a s i n d ( s i n d m ) ) ) / 1 8 0 ;

22 D c = ( ( ( s i n d o * ( D m - d 0 ) ) - ( r 2 * c o s d m ) ) - ( r 1 * c o s d ( d o ) ) ) / ( r 2 -

r 1 ) ;

23 d c = a c o s d ( D c ) ; // c r i t i c a l an gl e

24 mprintf ( The c r i t i c a l c l e a r i n g a n g l e i s g i v e n by= %. 1 f , d c ) ;

Scilab code Exa 17.6 determine the critical clearing angle

1 / / (A) d e t er m i ne t h e c r i t i c a l c l e a r i n g a n g l e2 clear

3 clc ;

4 P m = %i * . 12 + % i * .0 35 + ( ( %i * . 25 * % i * .3 ) / %i * . 55 ) ;5 P m 1 = 0 ;

6 P m 2 = 1 . 1 * 1 / . 4 0 5 ;

7 r 1 = 0 ;

8 r 2 = 2 . 7 1 6 / 3 . 7 7 5 ;

9 d 0 = ( a s i n d ( 1 / 3 . 7 7 5 ) ) ;

10 d M = ( 1 8 0 - a s i n d ( 1 / 2 . 7 1 6 ) ) ;

11 d o = d 0 * % p i / 1 8 0 ;

12 d m = d M * % p i / 1 8 0 ;

13 d c = a c o s d ( ( ( ( d m - d o ) * s i n d ( d 0 ) ) - ( r 1 * c o s d ( d 0 ) ) + ( r 2 * c o s d (

d M ) ) ) / ( r 2 - r 1 ) ) ;14 mprintf ( dc=%. 2 f , d c ) ;

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Scilab code Exa 17.7 To determine the centre and radius for the pull out

curve ans also minimum output vars when the output powers are i 0 ii 25puiii 5pu

1 // To d e te rm in e t he c e n tr e and r a d i u s f o r t he p u l lo ut c u rv e a ns a l s o minimum o u tp u t v a r s when t h eo u t p u t p o w er s a r e ( i ) 0 ( i i ) . 2 5 p . u ( i i i ) . 5 p . u

2 clear

3 clc ;

4 P c = 0 ;

5 V = . 9 8 ;

6 Qc=V ^2*((1/.4) -(1/1.1))/2;

7 R = V ^ 2 * ( ( 1 / . 4 ) + ( 1 / 1 . 1 ) ) / 2 ;8 Q = - ( . 9 8 ^2 * (( 1 . 1 - . 4 ) / . 44 ) / 2 ) + ( . 9 8^ 2 ) * 1 . 5 / ( 2* . 4 4) ;

9 mprintf ( ( i )Q=%. 2 f MVAr\n , abs ( Q ) * 1 0 0 ) ;10 P = . 2 5 ;

11 Q 2 = - ( ( 1 . 6 37 ^ 2 ) - ( . 25 ^ 2 ) ) ^ . 5 + . 7 6 3 9 ;

12 mprintf ( ( i i )Q=%. 4 f p . u\n , Q 2 ) ;13 Q 3 = - ( (1 . 6 37 ^ 2) - ( .5 ^ 2) ) ^ . 5 + . 7 6 39 ;

14 mprintf ( ( i i i )Q=%. 4 f p . u , Q 3 ) ;

Scilab code Exa 17.8 Compute the prefault faulted and post fault reducedY matrices

1 / / Compute t he p r e f a u l t , f a u l t e d and p o st f a u l tr e d u c e d Y m a t r i c e s

2 clear

3 clc ;

4 y = [ - %i * 5 0 %i * 5 ; 0 - %i * 5 %i * 5; % i *5 %i * 5 - %i * 10 ];

5 Y AA = [ - %i * 5 0 ;0 - %i * 5 ];

6 Y A B = [ % i * 5 ; % i * 5 ] ;7 Y B A = [ % i * 5 % i * 5 ];

8 Y B B = [ % i * 1 0 ] ;

9 Y = Y A A - Y A B * ( inv ( Y B B ) ) * Y B A ;

10 Y fu ll = [ - %i * 5 0 % i *5 ;0 - %i * 7 .5 % i * 2. 5; % i * 5 % i * 2. 5 - %i

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Chapter 18

LOAD FLOWS

Scilab code Exa 18.1 Determine the voltages at the end of first iterationusing gauss seidal method

1 // D et er mi ne t he v o l t a g e s a t t he end o f f i r s ti t e r a t i o n u si ng g au ss s e i d a l method

2 clear

3 clc ;

4 Y = [ 3 - % i * 12 - 2+ % i * 8 - 1+ % i * 4 0 ; - 2+ % i * 8 3 .6 6 6 - % i * 1 4 . 66 4

- .6 6 6+ % i * 2 . 6 6 64 - 1+ % i * 4 ; - 1+ % i * 4 - .6 6 6+ % i * 2 . 6 6 643 .6 6 6 - % i * 1 4 . 66 4 - 2+ % i * 8 ; 0 - 1+ % i * 4 - 2+ % i * 8 3 - % i

*12];

5 P 2 = - . 5 ;

6 P 3 = - . 4 ;

7 P 4 = - . 3 ;

8 Q 4 = - . 1 ;

9 Q 3 = - . 3 ;

10 Q 2 = - . 2 ;

11 V 2 = 1 ;

12 V 3 = 1 ;13 V 4 = 1 ;

14 V 1 0 = 1 . 0 6 ;

15 V 3 0 = 1 ;

16 V 4 0 = 1 ;

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17 V 2 1 = ( ( ( P 2 - % i * Q 2 ) / V 2 ) - Y ( 2 , 1 ) * V 1 0 - Y ( 2 , 3 ) * V 3 0 - Y ( 2 , 4 ) *

V 4 0 ) / ( Y ( 2 , 2 ) ) ;18 V 2 1 a c c = 1 + 1 . 6 * ( V 2 1 - 1 ) ;

19 disp (V21acc , V21acc=) ;20 V 3 1 = ( ( ( P 3 - % i * Q 3 ) / V 3 ) - Y ( 3 , 1 ) * V 1 0 - Y ( 3 , 2 ) * V 2 1 a cc - Y ( 3 , 4 )

* V 4 0 ) / ( Y ( 3 , 3 ) ) ;

21 V 3 1 a c c = 1 + 1 . 6 * ( V 3 1 - 1 ) ;

22 disp (V31acc , V31acc=) ;23 V 4 1 = ( ( ( P 4 - % i * Q 4 ) / V 4 ) - Y ( 4 , 2 ) * V 2 1 a c c - Y ( 4 , 3 ) * V 3 1 a c c ) / ( Y

( 4 , 4 ) ) ;

24 V 4 1 a c c = 1 + 1 . 6 * ( V 4 1 - 1 ) ;

25 disp (V41acc , V41acc=) ;

Scilab code Exa 18.2 Determine the voltages starting with a flat voltageprofile

1 // D et e r mi ne t he v o l t a g e s s t a r t i n g w i th a f l a tv o l t a ge p r o f i l e .

2 clear

3 clc ;

45 Y = [ 3 - % i * 12 - 2+ % i * 8 - 1+ % i * 4 0 ; - 2+ % i * 8 3 .6 6 6 - % i * 1 4 . 66 4

- .6 6 6+ % i * 2 . 6 6 64 - 1+ % i * 4 ; - 1+ % i * 4 - .6 6 6+ % i * 2 . 6 6 64

3 .6 6 6 - % i * 1 4 . 66 4 - 2+ % i * 8 ; 0 - 1+ % i * 4 - 2+ % i * 8 3 - % i

*12];

6 P 2 = . 5 ;

7 P 3 = - . 4 ;

8 P 4 = - . 3 ;

9 Q 4 = - . 1 ;

10 Q 3 = - . 3 ;

11 V 3 = 1 ;12 V 4 = 1 ;

13 V 1 = 1 . 0 6 ;

14 V 2 = 1 . 0 4 ;

15 V 3 0 = 1 ;

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16 V 4 0 = 1 ;

17 Q 2 = - imag ( [ V 2 * [ Y ( 2 , 1 ) * V 1 + Y ( 2 , 2 ) * V 2 + Y ( 2 , 3 ) * V 3 + Y ( 2 , 4 ) *V 4 ] ] ) ;

18 V 2 1 = ( ( ( P 2 - % i * Q 2 ) / V 2 ) - Y ( 2 , 1 ) * V 1 - Y ( 2 , 3 ) * V 3 0 - Y ( 2 , 4 ) * V 4 0

) / ( Y ( 2 , 2 ) ) ;

19 d = a t a n d ( 0 . 0 2 9 1 4 7 3 / 1 . 0 4 7 2 8 6 8 ) ;

20 V 2 1 = 1 . 0 4 * ( c o s d ( d ) + % i * s i n d ( d ) ) ;

21 disp ( V 2 1 , V21= ) ;22 V 3 1 = ( ( ( P 3 - % i * Q 3 ) / V 3 ) - Y ( 3 , 1 ) * V 1 - Y ( 3 , 2 ) * V 2 1 - Y ( 3 , 4 ) * V 4 0

) / ( Y ( 3 , 3 ) ) ;

23 disp ( V 3 1 , V31= ) ;24 V 4 1 = ( ( ( P 4 - % i * Q 4 ) / V 4 ) - Y ( 4 , 2 ) * V 2 1 - Y ( 4 , 3 ) * V 3 1 ) / ( Y ( 4 , 4 ) )

;25 disp ( V 4 1 , V41= ) ;

Scilab code Exa 18.3 Solve the prevous problem for for voltages at theend of first iteration

1 / / S o lv e t he p re vo us pr obl em f o r f o r v o l t a g e s a t t heend o f f i r s t i t e r a t i o n . f o r .2


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14 V 2 = 1 ;

15 V 3 0 = 1 ;16 V 4 0 = 1 ;

17 Q 2 = . 2 ;

18 V 3 = 1 ;

19 V 2 1 = ( ( ( P 2 - % i * Q 2 ) / V 2 ) - Y ( 2 , 1 ) * V 1 - Y ( 2 , 3 ) * V 3 0 - Y ( 2 , 4 ) * V 4 0

) / ( Y ( 2 , 2 ) ) ;

20 V 3 1 = ( ( ( P 3 - % i * Q 3 ) / V 3 ) - Y ( 3 , 1 ) * V 1 - Y ( 3 , 2 ) * V 2 1 - Y ( 3 , 4 ) * V 4 0

) / ( Y ( 3 , 3 ) ) ;

21 V 4 1 = ( ( ( P 4 - % i * Q 4 ) / V 4 ) - Y ( 4 , 2 ) * V 2 1 - Y ( 4 , 3 ) * V 3 1 ) / ( Y ( 4 , 4 ) )

;

22 disp ( V 2 1 , V21= ) ;

23 disp ( V 3 1 , V31= ) ;24 disp ( V 4 1 , V41= ) ;

Scilab code Exa 18.4 Determine the set of load flow equations at the endof first iteration by using Newton Raphson method

1 // D et e r mi ne t he s e t o f l oa d f l o w e q ua t i on s a t t heend o f f i r s t i t e r a t i o n by u s i n g Newton Raphson

metho d .2 clear

3 clc ;

4 Y = [ 6. 2 5 - % i * 1 8 . 75 - 1. 25 + % i * 3 . 75 - 5+ % i * 1 5 ; - 1 . 25 + % i

* 3 .7 5 2 . 91 6 - % i * 8 . 7 5 - 1 .6 66 + % i * 5 ; - 5+ % i * 1 5 - 1 .6 6 6+

% i * 5 6 .6 6 6 - % i * 2 0 ];

5 V 1 = 1 . 0 6 ;

6 G 1 1 = 6 . 2 5 ;

7 G 1 2 = - 1 . 2 5 ;

8 G 2 1 = G 1 2 ;

9 G 1 3 = - 5 ;10 G 3 1 = G 1 3 ;

11 G 2 2 = 2 . 9 1 6 ;

12 G

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