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Scilab Textbook Companion for

Transmission & Distribution Of Electrical

Power

by P. Jain1

Created byHarpreet Singh

B TECHElectrical Engineering

Vyas Institute of Engineering & Technology

College TeacherNA

June 1, 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: Transmission & Distribution Of Electrical Power

Author: P. Jain

Edition: 1

Year: 2012

ISBN: 978-93-80343-73-0

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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 Supply System 5

2 Distribution System 9

3 Mechanical Features of Overhead Line 15

4 Transmission Line Parameters 27

5 Performance of Short and Medium Transmission Lines 40

6 Performance of Long Transmission Lines 57

7 Corona 79

8 Insulators 85

9 Underground Cables 91

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List of Scilab Codes

Exa 1.1 Find the copper saving . . . . . . . . . . . . . . . . . 5Exa 1.4 Calculate volume of conductor required in 1 phase 2 wire

and 3 phase 3 wire system. . . . . . . . . . . . . . . . 5Exa 1.5 Calculate DC supply voltage . . . . . . . . . . . . . . 7Exa 2.1 Calculate the most economical cross sectional area . . 9Exa 2.2 Calculate the most economical current density . . . . 10Exa 2.3 Calculate the most economical current density and di-

ameter of conductor . . . . . . . . . . . . . . . . . . . 10Exa 2.4 Calculate the most economical cross sectional area . . 11Exa 2.5 Calculate the most economical cross sectional area . . 12Exa 2.6 Calculate the most economical cross sectional area . . 13Exa 3.1 calculate the weight of the conductor required . . . . . 15Exa 3.2 Calculate the max sag . . . . . . . . . . . . . . . . . . 16

Exa 3.3 Calculate the hieght above ground at which conductorshould be supported . . . . . . . . . . . . . . . . . . . 16

Exa 3.4 Calculate horizontal component of tension and max sag 17Exa 3.5 Calculate the max sag in still air and wind pressure. . 18Exa 3.6 Calculate the max sag . . . . . . . . . . . . . . . . . . 19Exa 3.7 Calculate the vertical sag . . . . . . . . . . . . . . . . 19Exa 3.8 Calculate the minimum clearance of conductor and water 20Exa 3.9 Calculate sag from taller of the two supports . . . . . 21Exa 3.10 find the clearance of conductor from ground . . . . . . 22Exa 3.11 Find stringing tension in the conductor . . . . . . . . 23Exa 3.12 find the clearance of conductor from water level at mid

point . . . . . . . . . . . . . . . . . . . . . . . . . . . 23Exa 3.13 find the clearance of conductor from ground 1 At its

lowest elevation 2 the min clearance of the line . . . . 24Exa 3.14 Determine Sag and Tension under erection conditions 25

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Exa 4.1 Find the loop inductance and reactance . . . . . . . . 27

Exa 4.2 Find the loop inductance . . . . . . . . . . . . . . . . 27Exa 4.3 Calculate GMR pf ACSR conductor . . . . . . . . . . 28Exa 4.4 Find the total inductance of the line . . . . . . . . . . 29Exa 4.5 Find the loop inductance . . . . . . . . . . . . . . . . 29Exa 4.6 Find the inductance per phase of 30 km line. . . . . . 30Exa 4.7 Find the inductance of a 3 phase line situated at cornes

of a triangle. . . . . . . . . . . . . . . . . . . . . . . . 30Exa 4.8 Find the inductance of a 3 phase line arranged in hori-

zontal plane. . . . . . . . . . . . . . . . . . . . . . . . 31Exa 4.9 Find the loop inductance per phase. . . . . . . . . . . 31Exa 4.10 Find the loop inductance per phase. . . . . . . . . . . 32

Exa 4.11 Find the inductance of an ASCR 3 phase line . . . . . 33Exa 4.12 Find inductive reactance of 3 phase bundled conductor 34Exa 4.13 Find the capacitance of 1 phase line . . . . . . . . . . 35Exa 4.14 Find the capacitance of 2 wire 1 phase line . . . . . . 35Exa 4.15 Find the capacitance of 3 phase line . . . . . . . . . . 36Exa 4.16 Find the capacitance of 3 phase 3 wire line . . . . . . 36Exa 4.17 Find the capacitance and charging current . . . . . . . 37Exa 4.18 find capacitive reactance to neutral and charging current 37Exa 4.19 Calculate the capacitance per phase . . . . . . . . . . 38Exa 5.1 Find voltage at sending end and percentage regulation

and transmission efficiency . . . . . . . . . . . . . . . 40Exa 5.2 voltage at sending end and percentage regulation andtotal line losses and transmission efficiency. . . . . . . 41

Exa 5.3 find sending end voltage and regulation . . . . . . . . 41Exa 5.4 Find sending end voltage and power factor and efficieny

and regulation . . . . . . . . . . . . . . . . . . . . . . 42Exa 5.5 Find load end voltage and efficieny . . . . . . . . . . . 43Exa 5.6 Find current and voltage of sending end and percentage

regulation and line losses and sending end power factorand transmission efficiency . . . . . . . . . . . . . . . 44

Exa 5.7 Find current and voltage of sending end and percentage

regulation and transmission efficiency . . . . . . . . . 45Exa 5.8 Find current and voltage of sending end and percentage

regulation . . . . . . . . . . . . . . . . . . . . . . . . . 46Exa 5.9 Find current and voltage of sending end . . . . . . . . 47

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Exa 5.10 Find regulation and charging current using nominal T

method . . . . . . . . . . . . . . . . . . . . . . . . . . 48Exa 5.11 find sending end voltage and current and power andefficiency . . . . . . . . . . . . . . . . . . . . . . . . . 49

Exa 5.12 Find ABCD parameters and sending end voltage andcurrent and power factor and transmission eficiency. . 51

Exa 5.13 find sending end voltage and current and power andefficiency . . . . . . . . . . . . . . . . . . . . . . . . . 53

Exa 5.14 Determine ABCD constant and sending end power factor 54Exa 6.1 Determine auxiliary constants. . . . . . . . . . . . . . 57Exa 6.2 Determine sending end voltage and current . . . . . . 58Exa 6.3 Determine percentage rise in voltage . . . . . . . . . . 60

Exa 6.4 alculate constants of equivalent circuit of line . . . . . 61Exa 6.5 calculate constants of equivalent circuit of line. . . . . 62Exa 6.6 calculate Ao and Bo and Co and Do constants . . . . 63Exa 6.7 calculate equivalent T and pi constants . . . . . . . . 65Exa 6.8 find sending end reactive and active power. . . . . . . 66Exa 6.9 find sending end voltage and regulation and recieving

end rective and synchornous power . . . . . . . . . . . 68Exa 6.10 find sending end voltage and charging current and power 69Exa 6.11 Determine sending end voltage and current and power

factor and MVA and power angle . . . . . . . . . . . . 71

Exa 6.12 Find sending end voltage and current and power factor 72Exa 6.13 Find characteristics impedance and propogation con-stant and ABCD constants . . . . . . . . . . . . . . . 74

Exa 6.14 Determine recieving end voltage and current. . . . . . 75Exa 6.15 Determine the induced voltage in the telephone line . 77Exa 7.1 Determine line voltage for commencing of corona . . . 79Exa 7.2 Determine whether corona will be there or not . . . . 79Exa 7.3 Determine critical discruptive voltage for line . . . . . 80Exa 7.4 Find spacing between the conductor . . . . . . . . . . 81Exa 7.5 Determine critical discruptive voltage for line . . . . . 81Exa 7.6 Determine critical discruptive voltage for line and corona

loss . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82Exa 7.7 Determine critical disruptive voltage and Visual critical

voltage and Corona loss . . . . . . . . . . . . . . . . . 82Exa 7.8 Find corona characteristics . . . . . . . . . . . . . . . 83

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Exa 8.1 find voltage distribution across each insulator and string

efficiency . . . . . . . . . . . . . . . . . . . . . . . . . 85Exa 8.2 find max safe working voltage and string efficiency . . 86Exa 8.3 find ratio of ground to mutual capacitance and system

line voltage and string efficiency . . . . . . . . . . . . 86Exa 8.4 find system line voltage and string efficiency. . . . . . 87Exa 8.5 find max safe working voltage . . . . . . . . . . . . . . 87Exa 8.7 Find the values of line to pin capacitance . . . . . . . 88Exa 8.8 find string efficiency . . . . . . . . . . . . . . . . . . . 88Exa 8.9 Calculate voltage on line end unit and capacitance Cx

required . . . . . . . . . . . . . . . . . . . . . . . . . . 89Exa 9.1 Determine insulation resistance . . . . . . . . . . . . . 91

Exa 9.2 Determine resistivity of dielectric in a cable . . . . . . 91Exa 9.3 Find max and min electrostatic stresses and capacitance

and charging current . . . . . . . . . . . . . . . . . . . 92Exa 9.4 Find max electrostatic stresses and charging kVA . . . 93Exa 9.5 Determine internal diameter of shealth D and diameter

of conductor d . . . . . . . . . . . . . . . . . . . . . . 93Exa 9.6 Determine most economical value of diameter and over-

all diameter of insulation . . . . . . . . . . . . . . . . 94Exa 9.7 Determine most economical value of diameter of single

core cable . . . . . . . . . . . . . . . . . . . . . . . . . 94

Exa 9.8 find safe working voltage of cable . . . . . . . . . . . . 95Exa 9.9 find radial thickness and safe working voltage of cable 96Exa 9.10 find the voltage on the intersheaths. . . . . . . . . . . 96Exa 9.11 find the position and voltage on the intersheaths and

max and min stress . . . . . . . . . . . . . . . . . . . 97Exa 9.12 Calculate the charging current . . . . . . . . . . . . . 99Exa 9.13 Calculate the kVA taken. . . . . . . . . . . . . . . . . 99Exa 9.14 Calculate the charging current . . . . . . . . . . . . . 100

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Chapter 1

Supply System

Scilab code Exa 1.1 Find the copper saving

1 / / Fin d t he c op pe r s a v i ng2 clear ;

3 clc ;

4 // s o l t i o n5 / / g i v e n6 v 1 = 2 4 0 ; / / v o l t / / i n i t i a l v o l t a g e o f t h e s y s te m

7 v 2 = 5 0 0 ; // v o l t // f i n a l v o l t a g e o f t he s ys te m8 printf ( Volume a t 2 40 v o l t s ( v o l 1 ) = ( 4P2dl ) / (%d

W) \n , v 1 ^ 2 / 4 ) ;9 printf ( Volume a t 5 00 v o l t s ( v o l 2 ) = ( 4P2dl ) / (%d

W) \n , v 2 ^ 2 / 4 ) ;10 printf ( P e r ce n ta g e s a v in g i n c op pe r = ( ( v ol 1v o l 2 )

1 0 0 ) / v o l 1 \n ) ;11 s = ( ( (1 / v 1 ^ 2 ) - (1 / v 2 ^ 2) ) / ( 1 / v 1 ^ 2) ) * 1 0 0 ;

12 printf ( The p e r ce n t ag e s a v i ng o f t he c op pe r i s , % . 2 f p e r c e n t , s ) ;

Scilab code Exa 1.4 Calculate volume of conductor required in 1 phase 2wire and 3 phase 3 wire system

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1 / / C a l c u l a te volume o f c o nd u ct o r r e q u i r e d i n 1 p ha se

2 w ir e and 3 p ha se 3 w ir e sy st e m2 clear ;3 clc ;

4 / / s o l u t i o n5 / / g i v e n6 p f = 0 . 8 ; // p ow er f a c t o r7 p M V A = ( 2 . 5 * 1 0 ^ 6 ) ; / / v o l t a mp er e8 v = ( 3 3 * 1 0 ^ 3 ) ; / / v o l t s9 l = 5 0 * 1 0 ^ 3 ; / /m/ / l e n g t h o f t h e l i n e

10 p = p M V A * p f ; / / w a t t s / / p o we r t r a s m i t t e d = p ow er i n MVAp . f .

11 w = 0 . 2 * p ; / / w a tt s / / l i n e l o s s e s = 2 0% o f po wert r a n s m i t t e d

12 d = 2 . 8 5 / 1 0 ^ 8 ; //ohm meter // r e s i s t i v i t y o f aluminiu m13 printf ( 1 p ha se 2 w ir e s ys te m\n ) ;14 i 1 = p M V A / v ;

15 a 1 = ( 2 * i 1 ^ 2 * d * l ) / w ;

16 printf ( Load c u r r en t i n 1 p ha se 2 w i re s ys te m= %f ampere \n , i 1 ) ;

17 printf ( C r o s s s e c t i o n a l a re a o f 1 p h a s e 2 w i resy st e m= %f m2\n , a 1 ) ;

18 v o l 1 = 2 * a 1 * l ;

19 printf ( Volume o f a lu min iu m c on d uc to r r e q u i r e d i n 1p ha se 2 w i r e s ys te m = %f m et er c ub e \n\n , v o l 1 ) ;

20 printf ( 3 p ha se 3 w ir e s ys te m\n ) ;21 i 2 = p M V A / ( 3 ^ 0 . 5 * v ) ;

22 a 2 = ( 3 * i 2 ^ 2 * d * l ) / w ;

23 printf ( Load c u r r en t i n 3 p ha se 3 w i re s ys te m= %f ampere \n , i 2 ) ;

24 printf ( C r o s s s e c t i o n a l a re a o f 3 p h a s e 3 w i resy st e m= %f m2\n , a 2 ) ;

25 v o l 2 = 3 * a 2 * l ;

26 printf ( Volume o f a lu mi ni um c o n du c to r r e q u i r e d i n 3p ha se 3 w i re s ys te m = %f m et er cu be , v o l 2 ) ;

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Scilab code Exa 1.5 Calculate DC supply voltage

1 / / f i n d t h e DC s u p p l y v o l t a g e2 clear ;

3 clc ;

4 // s o l t i o n5 / / g i v e n6 / / c o n s i d e r 1 p h as e AC s y s t em7 p f = 0 . 8 ;

8 v = ( 3 3 * 1 0 ^ 3 ) ; / / v o l t s9 r 1 = 0 . 1 5 ; //ohm// t o t a l r e s i s t a n c e o f t he 1 p ha se l i n e

10 P D 1 = 0 . 2 ; // p e r ce n t ag e v o l t a g e d ro p i n 1 p ha se ACs y s t e m

11 V d = P D 1 * v ; // v o l t // v o l t a g e dr op i n t he l i n e12 I 1 = V d / r 1 ; / / a mp er e / / l o a d c u r r e n t13 p = v * I 1 * p f ; / / w a t ts / / p ow er r e c i e v e d by t h e c on su me r14 P = p / 1 0 ^ 8 ;

15 printf ( 1 p h a s e AC s y s t em \n ) ;16 printf ( V o l t a g e d ro p= %d v o l t s\n , V d ) ;

17 printf ( L oa d c u r r e n t = %d a m pe re\n , I 1 ) ;18 printf ( Power r e c i e v e d by c on su me r= %d w a t ts o r= %f

105 kW \n\n , p , P ) ;19 / / c o n s i d e r DC 2 w i r e s ys te m20 r 2 = 0 . 1 ; //ohm// t o t a l r e s i s t a n c e o f t he DC 2 w ir e l i n e21 P D 2 = 0 . 2 5 ; // p e r ce n t a ge v o l t a g e d ro p i n DC 2 w i re

s y s t e m22 printf ( DC 2 w i r e s y s te m\n ) ;23 printf ( L oa d c u r r e n t i n DC s y s t e m= %f /V \n , p ) ;24 printf ( V o l t a ge d ro p= L oad c u r r e tl i n e r e s i s t a n c e=

I 2 R2= (%d/V) %f \n , p , r2 );25 printf ( Given v o l t a g e dr op i s 25 p e rc e n t ag e o f max

v o l t a g e= . 2 5V \n ) ;26 V = sqrt ( ( p * r 2 ) / P D 2 ) ;

27 printf ( E q u a t i n g a b ov e e q u a t i o n we g e t V= %f KV ,V

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/ 1 0 0 0 ) ;

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Chapter 2

Distribution System

Scilab code Exa 2.1 Calculate the most economical cross sectional area

1 // C a l cu l a t e t he most e co no mi ca l c r o s s s e c t i o n a l a re a2 clear ;

3 clc ;

4 // s o l t i o n5 / / g i v e n6 i d = 0 . 1 5 ; // i n t e r e s t & d e p r e c i a t i o n c h ar g es

7 i = 2 6 0 ; / / a m p er e / / max c u r r e n t8 d = 0 . 1 7 3 ; / /ohm/ / r e s i s t a n c e o f c o n du c t or9 c s t = . 0 3 ; // r s // c o s t o f e ne rg y p er u n it

10 t = ( 3 6 5 * 2 4 ) / 2 ; // t im e o f e ne rg y l o s s11 printf ( Annual c o st o f 2 c or e f e e de r c a b l e i s Rs ( 9 0 a

+ 10 ) p e r m e te r\n ) ;12 P 3 = ( 2 * i ^ 2 * d * t * c s t ) / 1 0 0 0 / /kWh/ / a n n u al c o s t o f e n e r g y

l o s s13 printf ( E n er g y l o s s p e r annum= P3 / a= %f / a \n , P 3 ) ;14 P 2 = 9 0 * 1 0 0 0 * i d ; / / e n e r g y l o s t p e r annum

15 printf ( C a p i t a l c o s t = P2a= %da \n , P2 );16 a = sqrt ( P 3 / P 2 ) ;17 printf ( Economic c r o s s s e c t i o n o f c o n du ct o r i s= (

P3 / P2 ) = %f s q u a r e cm , a ) ;

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Scilab code Exa 2.2 Calculate the most economical current density

1 // C a l c u l a te t he most e c on o mi c al c u r r e nt d e n s i t y2 clear ;

3 clc ;

4 // s o l t i o n5 / / g i v e n6 i d = 0 . 1 ; // i n t e r e s t & d e p r e c i a t i o n c ha r ge s7 d = 1 . 7 8 * 1 0 ^ - 8 ; // ohm m// r e s i s t i v i t y8 R = ( d * 1 0 0 0 ) / 1 0 ^ - 4 ; // ohm // r e s i s t a n c e o f c o n du c t or9 c s t = . 5 0 ; // r s // c o s t o f e ne rg y p er u n it

10 t = ( 3 6 5 * 2 4 ) ; // t im e o f e ne rg y l o s s11 l f = . 7 ; // l o ad f a c t o r o f l o s s e s12 printf ( Annua l c o s t o f c a b l e i s Rs ( 2 8 00 a + 130 0) p e r km

\n\n ) ;13 printf ( R e s i s t a n c e o f e ac h c o n du c to r= %f /a \n , R ) ;14 P 3 = ( R * t * c s t * l f ) / 1 0 0 0 ; // I 2 / /kWh/ / a n n u a l c o s t o f

e ne rg y l o s s15 printf ( Annua l c o s t o f e n er g y l o s s = P3 /a= ( %f I 2 ) / a

\n , P3 );16 P 2 = 2 8 0 0 * i d ; / / e n e r g y l o s t p e r annum17 printf ( Annual c h ar g e o n a c co un t o f i n t r e s t and

d e p r e c i a t i o n on v a r i a b l e c o st o f l i n e = P2a= %da\n , P 2 ) ;

18 J = sqrt ( P 2 / P 3 ) ; // c u r r e n t d e n s i t y I / a19 printf ( Economic c u r r e nt d e n s i t y o f c o nd u ct or i s %f

A/ cm s q u a r e , J ) ;

Scilab code Exa 2.3 Calculate the most economical current density anddiameter of conductor

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1 // C a l c u l a te t he most e c on o mi c al c u r r e nt d e n s i t y and

d i am et er o f c on d uc to r2 clear ;3 clc ;

4 // s o l t i o n5 / / g i v e n6 i d = 0 . 1 ; // i n t e r e s t & d e p r e c i a t i o n c ha r ge s7 c s t = . 0 2 ; // r s // c o s t o f e ne rg y p er u n it8 d = 0 . 1 7 3 ; / /ohm/ / r e s i s t a n c e o f c o n du c t or9 p f = . 8 ; / / l a g g i n g

10 P = 1 5 0 0 * 1 0 ^ 3 ; //Wat ts// l oa d11 V = 1 1 0 0 0 ; / / v o l t s / / s u p p l y v o l t a g e

12 t = 2 0 0 * 8 ; // hou r s13 printf ( an n u a l c o s t o f 3 c or e f e e d e r c ab l e i s Rs

( 8 0 0 0 + 2 0 00 0 a ) p e r km\n ) ;14 printf ( R e s i s t a n c e o f e a ch c o n d u c t o r= %. 3 f / a \n , d )

;

15 i = P / ( sqrt ( 3 ) * V * p f ) ; //ampere16 printf ( C u rr en t i n e ac h c o n du c t or= %. 3 f A\n , i ) ;17 P 2 = 2 0 0 0 0 * i d ; / / e n e r g y l o s t p e r annum18 printf ( C a p i t a l c o s t = P2a= %da \n , P2 );19 P 3 = ( 3 * i ^ 2 * d * t * c s t ) / 1 0 0 0 ; / /kWh/ / a n n u al c o s t o f e n e r g y

l o s s20 printf ( E n er g y l o s s p e r annum= P3 / a= %f / a \n , P 3 ) ;21 a = sqrt ( P 3 / P 2 ) ;

22 printf ( Economic c r o s s s e c t i o n o f c o n du ct o r i s= (P3 / P2 ) = %f s q u a r e cm \n , a ) ;

23 printf ( D i a me te r o f c o n d u c t o r= %. 1 f cm \n , sqrt ( 4 * a/ % p i ) ) ;

24 printf ( C u r r en t d e n s i t y = %f A/cm s q u a r e , i / a) ;

Scilab code Exa 2.4 Calculate the most economical cross sectional area

1 // C a l cu l a t e t he most e co no mi ca l c r o s s s e c t i o n a l a re a2 clear ;

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3 clc ;

4 // s o l t i o n5 / / g i v e n6 i d = 0 . 1 ; // i n t e r e s t & d e p r e c i a t i o n c ha r ge s7 p f = . 8 ; / / l a g g i n g8 P = 1 0 ^ 6 ; //Wat ts// l oa d9 V = 1 1 0 0 0 ; / / v o l t s / / s u p p l y v o l t a g e

10 c s t = . 1 5 ; // r s // c o s t o f e ne rg y p er u n it11 d = 1 . 7 5 * 1 0 ^ - 6 ; / / ohm cm / / s p e c i f i c r e s i s t a n c e12 l = 1 0 0 0 / /m// l e n g t h o f t h e c a b l e13 t = 3 0 0 0 ; // hou r s14 printf ( Annual c o st o f 2 c or e f e e de r c a b l e i s Rs ( 3 0

+ 5 00 a ) p e r m et er \n ) ;15 R = ( d * 1 0 0 0 * 1 0 0 ) ; / /ohm/ / r e s i s t a n c e o f c o n du c t or16 printf ( R e s i s t a n c e o f e ac h c o n du c to r= %f /a \n , R ) ;17 i = P / ( V * p f ) ; //ampe r e18 printf ( C u rr en t i n e ac h c o n du c t or= %f A\n , i ) ;19 P 2 = 5 0 0 * 1 0 ^ 3 * i d ; / / e n e r g y l o s t p e r annum20 printf ( C a p i t a l c o s t = P2a= %da \n , P2 );21 P 3 = ( 2 * i ^ 2 * R * t * c s t ) / 1 0 0 0 ; / /kWh/ / a n n u al c o s t o f e n e r g y

l o s s22 printf ( E n er g y l o s s p e r annum= P3 / a= %f / a \n , P 3 ) ;23 a = sqrt ( P 3 / P 2 ) ;

24 printf ( Economic c r o s s s e c t i o n o f c o n du ct o r i s= (P3 / P2 ) = %f s q u a r e cm \n , a ) ;

25 printf ( D i a me te r o f c o n d u c t o r= %f cm \n , sqrt ( 4 * a /% p i ) ) ;

Scilab code Exa 2.5 Calculate the most economical cross sectional area

12 // C a l cu l a t e t he most e co no mi ca l c r o s s s e c t i o n a l a re a3 clear ;

4 clc ;

5 // s o l t i o n

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6 / / g i v e n

7 i d = 0 . 1 ; // i n t e r e s t & d e p r e c i a t i o n c ha r ge s8 p f = . 8 5 ; / / l a g g i n g9 P m = 1 0 ^ 3 ; // Watts //Max Demand

10 P t = 5 * 1 0 ^ 6 / /kWh/ / T o a t a l e n e r g y c o n s u m p t i o n11 V = 1 1 0 0 0 ; / / v o l t s / / s u p p l y v o l t a g e12 c s t = . 0 5 ; // r s // c o s t o f e ne rg y p er u n it13 d = 1 . 7 2 * 1 0 ^ - 6 ; / / ohm cm / / s p e c i f i c r e s i s t a n c e14 t = ( 3 6 5 * 2 4 ) ; // t im e o f e ne rg y l o s s15 printf ( Annual c o s t o f c a b le i s Rs ( 80 00 0 a + 2 00 00 )

p e r km\n ) ;16 l f = P t / ( P m * t ) // Annual l o ad f a c t o r

17 printf ( Annua l l o a d f a c t o r = %f \n , lf );18 l l f = . 2 5 * l f + . 7 5 * l f ^ 2 ; // L os s l oa d f a c t o r19 printf ( L os s l oa d f a c t o r = %f \n , l l f ) ;20 i = P m * 1 0 0 0 / ( sqrt ( 3 ) * V * p f ) ; //ampe r e21 printf ( C u rr en t i n e ac h c o n du c t or= %. 1 f A\n , i ) ;22 P 2 = 8 0 0 0 0 * i d ; / / e n e r g y l o s t p e r annum23 printf ( C a p i t a l c o s t = P2a= %dal \n , P2 );24 R = d * 1 0 0 * 1 0 0 0 ; //ohm25 P 3 = ( 3 * i ^ 2 * R * t * c s t * l l f ) / 1 0 0 0 ; / /kWh/ / a n n u a l c o s t o f

e ne rg y l o s s26 printf (

E n er gy l o s s p e r annum= ( P3l ) / a= ( % f l ) / a \n , P 3 ) ;27 a = sqrt ( P 3 / P 2 ) ;

28 printf ( Economic c r o s s s e c t i o n o f c o n du ct o r i s= (P3 / P2 ) = %f s q u a r e cm \n , a ) ;

29 //THERE I S TYPOGRAPHICAL ERROR IN THE ANS IN BOOK ITI S 0 . 2 4 0 4 cm 2

Scilab code Exa 2.6 Calculate the most economical cross sectional area

1 // C a l cu l a t e t he most e co no mi ca l c r o s s s e c t i o n a l a re a2 clear ;

3 clc ;

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4 // s o l t i o n

5 / / g i v e n6 i d = 0 . 1 ; // i n t e r e s t & d e p r e c i a t i o n c ha r ge s7 V = 2 0 0 0 0 ; / / v o l t s / / s u p p l y v o l t a g e8 d = 1 . 7 2 * 1 0 ^ - 6 ; / / ohm cm / / s p e c i f i c r e s i s t a n c e9 c s t = . 6 ; // r s // c o s t o f e ne rg y p er u n it

10 p 1 = 1 5 0 0 // k i l o w a t t s11 t 1 = 8 // hour s12 p f 1 = . 8 / / po wer f a c t o r13 p 2 = 1 0 0 0 // k i l o w a t t s14 t 2 = 1 0 // hou r s15 p f 2 = . 9 / / po wer f a c t o r

16 p 3 = 1 0 0 / / k i l o w a t t s17 t 3 = 6 // hour s18 p f 3 = 1 / / po wer f a c t o r19 t = 3 6 5 / / no . o f d ay s20 i 1 = p 1 * 1 0 0 0 / ( sqrt ( 3 ) * V * p f 1 ) ; / / am pe re / / c u r r e n t a t t i me

t 121 i 2 = p 2 * 1 0 0 0 / ( sqrt ( 3 ) * V * p f 2 ) ; / / am pe re / / c u r r e n t a t t i me

t 222 i 3 = p 3 * 1 0 0 0 / ( sqrt ( 3 ) * V * p f 3 ) ; / / am pe re / / c u r r e n t a t t i me

t 323 R = d * 1 0 0 * 1 0 0 0 ;

//ohm24 P 2 = 8 0 0 0 * i d ; // L os s l oa d f a c t o r25 printf ( Annual c o s t o f c a b le i s Rs ( 80 00 0 a + 2 00 00 )

p e r km\n ) ;26 printf ( C a p i t a l c o s t = P2a= %dal \n , P2 );27 P 3 = ( 3 * ( ( i 1 ^ 2 * t 1 ) + ( i 2 ^ 2 * t 2 ) + ( i 3 ^ 2 * t 3 ) ) * R * t * c s t ) / 1 0 0 0 ;

//kWh// a n n u al c o s t o f e n er g y l o s s28 printf ( E n er gy l o s s p e r annum= ( P3l ) / a= ( % f l ) / a \n

, P 3 ) ;29 a = sqrt ( P 3 / P 2 ) ;

30 printf ( Economic c r o s s s e c t i o n o f c o n du ct o r i s= (

P3 / P2 ) = %f s q u a r e cm \n , a ) ;

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Chapter 3

Mechanical Features of

Scilab code Exa 3.1 calculate the weight of the conductor required

1 / / c a l c u l a t e t he w ei gh t o f t he c on du ct or r e q ui r e d2 clear ;

3 clc ;

4 // s o l t i o n5 / / g i v e n6 p = 3 0 * 1 0 ^ 6 ; / / w a t ts / / p ow er t o b e t r a n s m i t t e d7 v = 1 3 2 * 1 0 ^ 3 ; / / v o l t s / / L i n e v o l t a g e8 l = 1 2 0 * 1 0 ^ 3 ; //m// l e n g th o f 3 p ha se 3 w i re l i n e9 n = 0 . 9 ; // e f f i c i e n y o f t h e t r a n sm i s s i o n l i n e

10 p f = . 8 ; / / po wer f a c t o r11 d 1 = 1 . 7 8 * 1 0 ^ - 8 ; //ohm m// r e s i s t i v i t y o f co pp er12 d 2 = 2 . 6 * 1 0 ^ - 8 ; //ohm m// r e s i s t i v i t y o f alum inui m13 D 1 = 8 . 9 * 1 0 ^ 3 ; / / ( k g /m 3 ) / / s p e c i f i c g r a v i t y o f t h e

c o p p e r

14 D 2 = 2 * 1 0 ^ 3 ; / / ( k g /m 3 ) / / s p e c i f i c g r a v i t y o f t h eal umi ni um15 printf ( Weight o f t he c on d uc to r r e q u i r e d \n\n ) ;16 printf ( W=(3dl 2PD) /( ( 1n ) V 2p f 2 ) k g\n\n\n ) ;17 W 1 = ( 3 * d 1 * l ^ 2 * p * D 1 ) / ( ( 1 - n ) * v ^ 2 * p f ^ 2 ) ;

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18 printf ( W ei gh t o f c o p pp e r r e q u i r e d = %d k g\n\n , round

( W 1 / 1 0 0 0 ) * 1 0 0 0 ) ;19 W 2 = ( 3 * d 2 * l ^ 2 * p * D 2 ) / ( ( 1 - n ) * v ^ 2 * p f ^ 2 ) ;

20 printf ( W ei gh t o f a lu mi nu im r e q u i r e d = %d k g\n\n\n ,round ( W 2 / 1 0 0 ) * 1 0 0 ) ;

Scilab code Exa 3.2 Calculate the max sag

1 / / C a l c u l a t e t h e max s a g

2 clear ;3 clc ;

4 // s o l t i o n5 / / g i v e n6 W = . 6 ; / / k g /m/ / L i n e c o n d u c t o r w i e g h t7 L = 3 0 0 ; // m et er / / sp an o f t he l i n e8 T = 1 2 0 0 ; // k g // max a l l o w a b l e t e n s i o n9 printf ( Max sa g= (WL2 ) /( 8 T) \n ) ;

10 s a g = ( W * L ^ 2) / ( 8 * T ) ;

11 printf ( Sag= %.3 f m , s a g ) ;

Scilab code Exa 3.3 Calculate the hieght above ground at which conduc-tor should be supported

1 / / C a l c u l a t e t h e h i e g h t a bo ve g ro un d a t w hi chc o n du c to r s h o ul d be s u p po r t ed

2 clear ;

3 clc ;

4 // s o l t i o n

5 / / g i v e n6 W = 6 8 0 ; / / k g /km/ / L i n e c o n d u c t o r w e i g h t7 L = 2 4 0 ; // m et er / / sp an o f t he l i n e8 U = 3 2 0 0 ; / / kg / / U l t i m a te s t r e n g t h9 s f = 2 ; // s a f e t y f a c t o r

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10 T = U / s f ; // k g // max a l l o w a b l e t e n s i o n

11 g c = 8 ; / /m/ / g r o un d c l e a r a n c e12 w = W / 1 0 0 0 ; / / kg /m/ / We ig ht o f c o n d u c t o r i n m et er13 printf ( Max sa g= (WL2 ) /( 8 T) \n ) ;14 s a g = ( w * L ^ 2) / ( 8 * T ) ;

15 printf ( Sag= %. 2 f m\n , s a g ) ;16 H = g c + s a g ;

17 printf ( H e i gh t a bo ve w hi ch c o n du c t or s h o u ld bes u p p o r t e d \n= g r ou n d c l e a r a n c e + s a g= %. 2 f m , H ) ;

Scilab code Exa 3.4 Calculate horizontal component of tension and maxsag

1 / / C a l c u l a t e h o r i z o n t a l c om po nen t o f t e n s i o n and maxs a g

2 clear ;

3 clc ;

4 // s o l t i o n5 / / g i v e n6 W = 7 5 0 ; / / k g /km/ / L i n e c o n d u c t o r w e i g h t

7 L = 3 0 0 ; // m et er / / sp an o f t he l i n e8 T = 3 4 0 0 ; // k g // max a l l o w a b l e t e n s i o n9 w = W / 1 0 0 0 ; / / kg /m/ / We ig ht o f c o n d u c t o r i n m et er

10 printf ( Max sa g= (WL2 ) /( 8 Th) \n ) ;11 x = ( w * L ^2 ) / (8 ) ;

12 printf ( Sag= %. 1 f /Th\n\n , x ) ;13 printf ( Max t e n s i o n = Th + wS\n ) ;14 T h = ( T + sqrt ( T ^ 2 + 4 * w * x ) ) / 2 ;

15 printf ( H o r i z o n t a l c om po ne nt o f t e n s i o n ( Th )= %. 3 f kg\n , T h ) ;

16 s a g = ( w * L ^ 2) / ( 8 * T h ) ;17 printf ( Sag= %. 3 f m\n , s a g ) ;18 y = s a g / 2 ;

19 z = sqrt ( ( 2 * T h * y ) / w ) ;

20 printf ( P oi nt a t whi ch s ag w i l l be h a l f = %. 3 f m\n ,

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z ) ;

21 //THERE I S TYPOGRAPHICAL ERROR IN BOOK DUE TO THATTHERE I S A VARIATION22 // IN BOOK Th=3 44 8.1 91 kg23 //MAX SAG=2 .4 46 m24 // P oi nt a t which s ag w i l l be h a l f = 1 0 6 .0 4 5 m

Scilab code Exa 3.5 Calculate the max sag in still air and wind pressure

1 / / C a l c u l a t e t h e max s a g i n s t i l l a i r and win dp r e s s u r e2 clear ;

3 clc ;

4 // s o l t i o n5 / / g i v e n6 W c = 1 . 1 3 ; / / k g /m/ / L i n e c o n d u c t o r w e i g h t7 P = 3 3 . 7 / / k g /m 2 / / Wind p r e s s u r e8 L = 1 8 0 ; // m et er / / sp an o f t he l i n e9 f u = 4 2 2 0 ; / / kg / / U l t im a t e s t r e s s

10 s f = 5 ; // s a f e t y f a c t o r

11 f = f u / s f ; // kg // w o rk in g s t r e s s12 D = 1 . 2 7 ; / /cm // d i a o f c o p p e r13 r = 1 . 2 5 ; //cm// R a di a l t h i c k n e s s o f i c e14 a = ( % p i * D ^ 2 ) / 4 ; //cm 2 // a r ea o f c r o s s s e c t i o n15 printf ( Area o f c r o s s s e c t i o n = %3f cm2\n , a ) ;16 T = f * a ; / / kg / /max a l l o w a b l e t e n s i o n17 printf ( W or ki ng t e n s i o n = %. 2 f k g\n , T ) ;18 s a g 1 = ( W c * L ^ 2) / ( 8 * T ) ; / / s a g i n s t i l l a i r19 printf ( Sag i n s i l l a i r = %. 2 f m\n , s a g 1 ) ;20 W i = 2 8 9 0 . 3 * r * ( D + r ) * 1 0 ^ - 4 ;

21 printf ( Wei ght o f i c e c o a t i n g = %. 2 f k g \n , W i ) ;22 W w = P * ( D + 2 * r ) * 1 0 ^ - 2 ;23 printf ( Wind f o r c e = %. 5 f k g \n , W w ) ;24 Wr = sqrt ( ( W c + W i ) ^ 2 + W w ^ 2 ) ;

25 s a g 2 = ( W r * L ^ 2) / ( 8 * T ) ; // s ag i n wind + i c e

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26 printf ( Max Sag= %.3 f m\n , s a g 2 ) ;

Scilab code Exa 3.6 Calculate the max sag

1 / / C a l c u l a t e t h e max s a g2 clear ;

3 clc ;

4 W c = . 8 5 ; / / k g /m/ / L i n e c o n d u c t o r w i e g h t5 L = 2 7 5 ; // m et er / / sp an o f t he l i n e

6 U = 8 0 0 0 ; / / kg / / U l t i m a te s t r e n g t h7 s f = 2 ; // s a f e t y f a c t o r8 P = 3 9 ; / / k g /m 2 / / Wind p r e s s u r e9 T = U / s f ; // k g // max a l l o w a b l e t e n s i o n

10 D = 1 9 . 5 ; //mm// di a of c opp e r11 r = 1 3 ; //cm// R a di a l t h i c k n e s s o f i c e12 W i = 9 1 0 * % p i * r * ( D + r ) * 1 0 ^ - 6 ;

13 W w = P * ( D + 2 * r ) * 1 0 ^ - 3 ; / / Wind f o r c e /m l e n g h t14 Wr = sqrt ( ( W c + W i ) ^ 2 + W w ^ 2 ) ; // r e s u l t a n t s ag15 s a g = ( W r * L ^ 2) / ( 8 * T ) ; // s ag i n wind + i c e16 printf ( Max Sag= %.3 f m\n , s a g ) ;

Scilab code Exa 3.7 Calculate the vertical sag

1 / / C a l cu l a t e t he v e r t i c a l s ag2 clear ;

3 clc ;

4 // s o l t i o n5 / / g i v e n

6 W = 1 1 7 0 ; / / k g /km/ / L i n e c o n d u c t o r w i e g h t7 P = 1 2 2 ; / / k g /m 2 / / Wind p r e s s u r e8 L = 2 0 0 ; // m et er / / sp an o f t he l i n e9 A = 1 . 2 9 ; //cm 2 // c r o s s s e c t i o n a l a r ea

10 U = 4 2 1 8 * A ; / / kg / / B r e a ki n g s t r e n g t h

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11 s f = 5 ; // s a f e t y f a c t o r

12 T = U / s f ; // k g // max a l l o w a b l e t e n s i o n13 W c = W / 1 0 0 0 ; / / kg /m/ / Wei gh t o f c o n d u c t o r i n m et er14 D = sqrt ( ( 4 * A ) / % p i ) ; / /cm // d i a m e t e r o f t h e c o n d u c t o r15 printf ( D i a me te r o f t h e c o n d u c t o r= %. 2 f cm\n , D ) ;16 W w = P * ( D ) * 1 0 ^ - 2 ; / / Wind f o r c e /m l e n g h t17 printf ( Wind f o r c e = %. 2 f k g \n , W w ) ;18 Wr = sqrt ( W c ^ 2 + W w ^ 2 ) ; // r e s u l t a n t w ei gh t19 printf ( R e s u l t a n t s a g= %. 2 f k g \n , W r ) ;20 s a g = ( W r * L ^ 2) / ( 8 * T ) ; / /m/ / S l a n t s a g21 printf ( S l a n t S ag= %. 2 f m\n , s a g ) ;22 T h = a t a n d ( W w / W c ) ; / / d e g r e e / / a n g l e b et we en s l a n t s a g

and v e r t i c a l s ag23 V s a g = s a g * c o s d ( T h ) ; / /m/ / V e r t i c a l s a g24 printf ( V e r t i c a l s a g= %. 3 fm , V s a g ) ;

Scilab code Exa 3.8 Calculate the minimum clearance of conductor andwater

1 / / C a l c u l a t e t h e minimum c l e a r a n c e o f c o n du c to r and

w a t e r2 clear ;

3 clc ;

4 // s o l t i o n5 / / g i v e n6 W = 1 . 5 ; / / k g /m/ / L i n e c o n d u c t o r w i e g h t7 L = 5 0 0 ; // m et er / / sp an o f t he l i n e8 T = 1 6 0 0 ; // k g // max a l l o w a b l e t e n s i o n9 T 1 = 3 0 ; / /m// h e i g h t o f t h e t ow er 1

10 T 2 = 9 0 ; / /m// h e i g h t o f t h e t ow er 2

11 h = T 2 - T 1 ; / /m/ / d i f f e r e n c e i n t h e b et we en s u p po r t12 printf ( D i s t a n c e o f s u p po r t T1 f ro m O ( L ow es t p o i n t )be x1\n ) ;

13 printf ( D i s t a n c e o f s u p po r t T2 f ro m O ( L ow es t p o i n t )be x2\n ) ;

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14 printf ( x1+x2= %dm\n , L ) ;

15 d i f = ( ( h * 2 * T ) / ( W * L ) ) ; //x2x116 printf ( x2x1= %dm\n , d i f ) ;17 x 2 = ( L + d i f ) / 2 ; //m18 x 1 = L - x 2 ; //m19 printf ( x1= %dm, x2= %dm\n , x 1 , x 2 ) ;20 s a g = ( ( W * x1 ^ 2 ) / ( 2 * T ) ) ;//m21 printf ( S a g ( From t o w e r 1 ) = %d m\n , round ( s a g ) ) ;22 C = T 1 - s a g ; / / C l e a r a n c e23 printf ( C l ea r a nc e o f t he l o we s t p o in t from w a te r

l e v e l = %dm\n , C ) ;

Scilab code Exa 3.9 Calculate sag from taller of the two supports

1 / / C a l c u l a t e s a g f ro m t a l l e r o f t h e two s u p p o r t s2 clear ;

3 clc ;

4 // s o l t i o n5 / / g i v e n6 W c = 1 . 9 2 5 ; / / k g /m/ / L i n e c o n d u c t o r w i e g h t

7 L = 6 0 0 ; // m et er / / sp an o f t he l i n e8 h = 1 5 //m//T1T29 W i = 1 // k g // W ieg ht o f t h e i c e

10 W r = W i + W c ; // r e s u l t a n t w ei gh t11 A = 2 . 2 //cm212 U = 8 0 0 0 * A ; / / kg / / B r e a ki n g s t r e n g t h13 s f = 5 ; // s a f e t y f a c t o r14 T = U / s f ; // k g // max a l l o w a b l e t e n s i o n15 printf ( x1+x2= %dm\n , L ) ;16 d i f = ( ( h * 2 * T ) / ( W r * L ) ) ; //x2x1

17 printf ( x2x1= %dm\n , d i f ) ;18 x 2 = ( L + d i f ) / 2 ; //m19 x 1 = L - x 2 ; //m20 printf ( x1= %dm, x2= %dm\n , round ( x 1 ) , round ( x 2 ) ) ;21 s a g = ( ( W r *( round ( x 2 ) ) ^ 2 ) / ( 2 * T ) ) ; //m

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22 printf ( S ag ( f ro m t a l l e r o f t h e t wo s u p p o r t s )= %. 3 f m

\n , s a g ) ;

Scilab code Exa 3.10 find the clearance of conductor from ground

1 / / f i n d t he c l e a r a n c e o f c on d uc to r from g ro un d2 clear ;

3 clc ;

4 // s o l t i o n

5 / / g i v e n6 W = 1 ; / / k g /m/ / L i n e c o n d u c t o r w i e g h t7 L = 3 0 0 ; // m et er / / sp an o f t he l i n e8 T = 1 5 0 0 ; // k g // max a l l o w a b l e t e n s i o n9 T 1 = 2 2 - 2 ; //m// e f f e c t i v e h e i g ht o f t he t o we rs

10 g = 1 / 2 0 ; // s i n // g r ad i e n t11 h = L * g / /m/ / v e r t i c a l d i s t a n c e b et we en two t o w er s12 printf ( x 1+ x 2 %dm\n , L ) ;13 d i f = ( ( h * 2 * T ) / ( W * L ) ) ; //x2x114 printf ( x2x1= %dm\n , d i f ) ;15 x 2 = ( L + d i f ) / 2 ; //m

16 x 1 = L - x 2 ; //m17 printf ( x1= %dm, x2= %dm\n , round ( x 1 ) , round ( x 2 ) ) ;18 s a g = ( ( W * x2 ^ 2 ) / ( 2 * T ) ) ;//m19 printf ( Sag= %. 3 f m\n , s a g ) ;20 T 2 = T 1 + h ; / /m/ / h i e g h t o f t h e s e c o nd t o we r21 g f = x 1 * t a n d ( a s i n d ( 1 / 2 0 ) ) ; //m// e l e v a t i o n o f t h e

g r o u n d a t max s a g22 O G = T 2 - s a g - g f ; / /m/ / g r o un d c l e a r a n c e23 printf ( C l e a ra n ce o f t he l o w es t p o i nt O fro m g ro un d

i s % . 2 f m , O G ) ;

24 //SINCE THERE I S NO REFRENCE OF WATERLEVEL IN THEQUESTION THEREFORE THE EXTRA SOLUTION I S ANTYPOGRAPHICAL ERROR

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Scilab code Exa 3.11 Find stringing tension in the conductor

1 // Find s t r i n g i n g t e n s i o n i n t he c on du ct or2 clear ;

3 clc ;

4 // s o l t i o n5 / / g i v e n6 W = 0 . 7 ; / / k g /m/ / L i n e c o n d u c t o r w i e g h t7 L = 2 5 0 ; // m et er / / sp an o f t he l i n e8 T 1 = 2 5 ; / /m// h e i g h t o f t h e t ow er 19 T 2 = 7 5 ; / /m// h e i g h t o f t h e t ow er 2

10 h = T 2 - T 1 ; / /m/ / d i f f e r e n c e i n t h e b et we en s u p po r t11 T m = 4 5 ; / /m/ / h i e g h t o f midway b e tw e en t h e t o w e r s12 h m = T m - T 1 ; / /m/ / midway p o i n t b e tw e en t h e t wo t o w e r s13 L m = L / 2 ; / /m/ / h a l f o f t h e s pa n14 printf ( We know th at \nh=(WL( L2x ) ) /( 2T) \n ) ;15 printf ( Fo r t h e two t o w er s \n%d=(%. 1 f%d(%d2x ) ) /( 2

T) \n , h , W , L , L ) ;16 printf ( Fo r t h e mid p o i n t \n%d=(%. 1 f%d(%d2x ) ) /( 2

T) \n , h m , W , L m , L m ) ;17 x = - ( 2 * L ) + ( 2 . 5 * L m ) ;

18 printf ( By a b ov e e q u a t i o n x= %d m\n , x ) ;19 T = ( W * L * ( L - 2 * x ) ) / ( 2 * h ) ;

20 printf ( S t r i n g i n g T e n s i o n ( T)=% . 2 f k g ,T )

Scilab code Exa 3.12 find the clearance of conductor from water level atmid point

1 / / f i n d t he c l e a r a n c e o f c on du ct or from w at er l e v e la t mid p o i nt

2 clear ;

3 clc ;

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4 // s o l t i o n

5 / / g i v e n6 W = . 8 4 4 ; / / k g /m/ / L i n e c o n d u c t o r w i e g h t7 L = 3 0 0 ; // m et er / / sp an o f t he l i n e8 T = 1 8 0 0 ; // k g // max a l l o w a b l e t e n s i o n9 T 1 = 4 0 ; / /m// h e i g h t o f t h e t ow er 1

10 T 2 = 8 0 ; / /m// h e i g h t o f t h e t ow er 211 h = T 2 - T 1 ; / /m/ / d i f f e r e n c e i n t h e b et we en s u p po r t12 x = L / 2 - ( T * h ) / ( W * L ) ;

13 printf ( D i s t a n c e b et we en m i dp o in t and l o w e s t p o i n t=% . 2 f m\n , ( L / 2 ) - x ) ;

14 S m i d = ( W * ( L / 2 - x ) ^ 2 ) / ( 2 * T ) ;

15 printf ( H e i gh t b et we en m i dp o in t and l o w e s t p o i n t= %. 3 f m\n , S m i d ) ;

16 S 2 = ( W * ( L - x ) ^ 2 ) / ( 2 * T ) ;

17 printf ( H e i g h t b et we en t a l l e r t o we r and l o w e s t p o i n t= % . 3 fm\n , S 2 ) ;

18 C = T 2 - ( S 2 - S m i d ) ;

19 printf ( C l ea r a nc e o f c on du ct or from w at er l e v e l a tmid p o i n t = %. 3 fm ,C )

Scilab code Exa 3.13 find the clearance of conductor from ground 1 Atits lowest elevation 2 the min clearance of the line

1 // f i n d t he c l e a r a n c e o f c on d uc to r from g ro un d i ) Ati t s l o w e s t e l e v a t i o n i i ) t h e min c l e a r a nc e o f t hel i n e

2 clear ;

3 clc ;

4 // s o l t i o n

5 / / g i v e n6 W = . 8 ; / / k g /m/ / L i n e c o n d u c t o r w i e g h t7 L = 3 0 0 ; // m et er / / sp an o f t he l i n e8 T = 1 5 0 0 ; // k g // max a l l o w a b l e t e n s i o n9 T 1 = 3 0 ; / /m/ / h e i g h t o f t h e t o w e r s

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10 g = 1 / 2 0 ; // t a n // gr ound s l o p e

11 h = L * g / /m/ / v e r t i c a l d i s t a n c e b et we en two t o w er s12 T 2 = T 1 + h ; //m// h e i g ht o f t he t ow er a l on g t he s l o p e13 x 1 = L / 2 - ( T * h ) / ( W * L ) ;

14 printf ( D i s t a n c e b et we en t o we r on g ro un d and s a g=x1=% . 2 f m\n , x 1 ) ;

15 S 1 = ( W * x 1 ^ 2 ) / ( 2 * T ) ;

16 printf ( S ag f o r t o we r on g ro un d ( S 1 )= %. 5 fm\n , S 1 ) ;17 S 2 = ( W * ( L - x 1 ) ^ 2 ) / ( 2 * T ) ;

18 printf ( S ag f o r t o we r on h i l l ( S 2 )= %. 5 fm\n , S 2 ) ;19 C = T 1 - S 1 - x 1 * g ;

20 printf ( C l ea r a nc e o f c on du ct or from l o we s t e l e v a t i o n

= % . 5 fm\n , C ) ;21 x = poly (0 , x ) ;22 C1 = poly ( [ C - g W / (2 * T ) ], x , c ) ;23 d = derivat ( C 1 ) ;

24 xa = roots ( d ) ;

25 C a = C - g * x a + W / ( 2 * T ) * x a ^ 2 ;

26 printf ( Minimum c l e a r a n c e f r om gr ound= %dm , C a ) ;

Scilab code Exa 3.14 Determine Sag and Tension under erection condi-tions

1 // D et er mi ne Sag & T en si on u nd er e r e c t i o n c o n d i t i o n s2 clear ;

3 clc ;

4 // s o l t i o n5 / / g i v e n6 W = . 9 ; / / k g /m/ / L i n e c o n d u c t o r w i e g h t7 L = 3 0 0 ; // m et er / / sp an o f t he l i n e

8 a = 2 . 4 0 * 1 0 ^ - 4 //m2/ / ar ea9 D = 1 9 . 5 //mm// di a me te r10 U = 8 0 0 0 ; / / kg / / U l t i m a te s t r e n g t h11 s f = 2 ; // s a f e t y f a c t o r12 P = 3 8 . 5 ; / / k g /m 2 / / Wind p r e s s u r e

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13 T 1 = U / s f ; / / kg / /max a l l o w a b l e t e n s i o n

14 E = 9 3 2 0 * 1 0 ^ 6 ; // kg/m2/ / Young s Modulus15 a l p = 1 8 . 4 4 * 1 0 ^ - 6 ; / / 1/ C / / L i n e a r e x p an s i o n16 t 1 = 5 // C // t e m pe r a tu r e u nd er n or ma l c o n d i t i o n17 t 2 = 3 5 / / C / / t e mp e r at u r e u nd er w o rs t c o n d i t i o n18 d t = t 2 - t 1 ; // C // d i f f e r e n c e i n t em p er at u re19 f 1 = T 1 / a ;

20 W w = P * ( D ) * 1 0 ^ - 3 ; / / w e i g ht d ue t o wi nd21 printf ( Wind f o r c e = %. 2 f k g \n , W w ) ;22 Wr = sqrt ( W ^ 2 + W w ^ 2 ) ; // r e s u l t a n t w ei gh t23 C 1 = W ^ 2 * L ^ 2 * E / ( 2 4 * a ^ 2 ) ;

24 C 2 = - f 1 + W r ^ 2 * L ^ 2 * E / ( 2 4 * f 1 ^ 2 * a ^ 2 ) + d t * a l p * E ;

25 p = poly ([ - C1 0 C2 1] , f2 , c ) ;26 r = roots ( p ) ;

27 f 2 = 1 1 9 51 2 9 2; // a c ce p te d v a lu e o f f 228 s a g = ( W * L ^ 2 ) / ( 8 * f 2 * a ) ;

29 printf ( Sag a t e r e c t i o n = %. 3 f m , s a g ) ;30 / /The b ook h as u s e d i n c o r r e c t v al ue o f f 2 and i n i t

t h e s a g i s 2 . 1 21m

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Chapter 4

Transmission Line Parameters

Scilab code Exa 4.1 Find the loop inductance and reactance

1 / / Fin d t he l o op i n d uc t an c e and r e a c t a n ce2 clear ;

3 clc ;

4 // s o l t i o n5 / / g i v e n6 r = ( 1 . 2 1 3 * 1 0 ^ - 2 ) / 2 ; / /m/ / r a d i u s o f t h e c o n d u c t o r

7 d = 1 . 2 5 ; //m// sp ac in g8 f = 5 0 ; //Hz // f r e q9 r e = r * exp ( - 1 / 4 ) ;

10 L = 4 * 1 0 ^ - 7 * log ( d / r e ) ;

11 L k m = L * 1 0 0 0 ;

12 printf ( I n d u c t a n c e p e r km ( L )=% . 2 f 10 4 H/Km\n ,Lkm* 1 0 ^ 4 ) ;

13 X = 2 * % p i * f * L k m ;

14 printf ( Re ac ta nc e (X)= %. 1 f ohm/km, X ) ;

Scilab code Exa 4.2 Find the loop inductance

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1 / / Fin d t he l o op i n d uc t an c e

2 clear ;3 clc ;

4 // s o l t i o n5 / / g i v e n6 r = ( 1 * 1 0 ^ - 2 ) / 2 ; / /m/ / r a d i u s o f t h e c o n d u c t o r7 d = 2 ; //m// sp ac in g8 u = 5 0 // r e l a t i v e p e r m e ab i l i t y o f s t e e l and c o pp er9 L = ( 1 + 4 * log ( d / r ) ) * 1 0 ^ - 7 * 1 0 0 0 ;

10 L m H = L * 1 0 0 0 ;

11 printf ( I n d u c t a n c e p e r km ( L ) c o p p e r c o n d u c t o r =%. 3 f mH\n , L m H ) ;

12 L r = ( u + 4 * log ( d / r ) ) * 1 0 ^ - 7 * 1 0 0 0 ;13 printf ( I n d u c t a n c e p e r km (L ) s t e e l c o n d u c t o r=%. 3 f mH

\n , L r * 1 0 0 0 ) ;

Scilab code Exa 4.3 Calculate GMR pf ACSR conductor

1 // C a l c ul at e GMR pf ACSR c on duc t or2 clear ;

3 clc ;4 // s o l t i o n5 / / g i v e n6 r = 3 ; / /mm/ / r a d i u s o f t h e c o n d u c t o r7 r e = r * exp ( - 1 / 4 ) ;

8 d 1 1 = r e ;

9 d 1 2 = 2 * r //=d17=d16 ;10 d 1 4 = 4 * r ;

11 d 1 3 = sqrt ( d 1 4 ^ 2 - d 1 2 ^ 2 ) ; //=d1512 D s 1 = ( d 1 1 * d 1 2 * d 1 3 * d 1 4 * d 1 3 * d 1 2 * d 1 2 ) ;

13 D s 1 _ = D s 1 / ( r ^ 7 ) ;14 printf ( D s1= ( % f ) ( 1 / 7 ) r \n , D s 1 _ ) ;15 d 7 1 = 2 * r ; //=d72=d73=d74=d75=d7616 D s 7 = ( d 7 1 ^ 6 * r e ) ;

17 D s 7 _ = D s 7 / ( r ^ 7 ) ;

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18 printf ( D s7= ( % f ) ( 1 / 7 ) r \n , D s 7 _ ) ;

19 D s = ( D s 1 ^ 6 * D s 7 ) ^ ( 1 / 4 9 ) ;20 printf ( GMR= %. 4 fmm , D s ) ;

Scilab code Exa 4.4 Find the total inductance of the line

1 // Find t h e t o t a l i nd uc ta nc e o f t h e l i n e2 clear ;

3 clc ;

4 // s o l t i o n5 / / g i v e n6 r = 1 . 4 ; / /cm/ / r a d i u s o f t h e c o n du c t or7 r e = r * exp ( - 1 / 4 ) ;

8 d 1 2 = 2 0 ; //cm// sp ac i n g b/w 1&29 d 1 1 _ = 2 0 + 1 2 0 ; / / cm / / s p a c i n g b /w 1 &1

10 d 1 2 _ = 2 0 + 1 2 0 + 2 0 ; / / cm // s p a c i n g b /w 1 &2 11 d 2 1 _ = 1 2 0 ; / /cm / / s p a c i n g b /w 2 &1 12 d 2 2 _ = 2 0 + 1 2 0 ; / / cm / / s p a c i n g b /w 2 &2 13 D m = ( d 1 1 _ * d 1 2 _ * d 2 1 _ * d 2 2 _ ) ^ ( 1 / 4 ) ;

14 printf ( Mut ua l GMD= %. 2 fcm \n , D m ) ;

15 Ds = floor ( ( r e * d 1 2 * r e * d 1 2 ) ^ ( 1 / 4 ) * 1 0 0 ) / 1 0 0 ;16 printf ( S e l f GMD= %. 2 fcm\n , D s ) ;17 L = 0 . 4 * log ( D m / D s ) ;

18 printf ( L oo p I n d u c t a n c e o f l i n e = %. 5 f mH/km , L ) ;

Scilab code Exa 4.5 Find the loop inductance

1 / / Fin d t he l o op i n d uc t an c e

2 clear ;3 clc ;

4 // s o l t i o n5 / / g i v e n6 r = 1 / 2 ; / /cm/ / r a d i u s o f t h e c o n du c t or

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7 r e = r * exp ( - 1 / 4 ) ;

8 d 1 2 = 2 0 0 ; //cm// sp ac i n g b/w 1&29 d 1 1 _ = 3 0 0 ; / /cm / / s p a c i n g b /w 1 &1 10 d 1 2 _ = sqrt ( ( 3 0 0 ) ^ 2 + ( 2 0 0 ) ^ 2 ) ; / / cm // s p a c i n g b /w 1 &2 11 d 2 1 _ = d 1 2 _ ; / / cm // s p a c i n g b /w 2 &1 12 d 2 2 _ = 3 0 0 ; / /cm / / s p a c i n g b /w 2 &2 13 D m = ( d 1 1 _ * d 1 2 _ * d 2 1 _ * d 2 2 _ ) ^ ( 1 / 4 ) ;

14 printf ( Mut ua l GMD= %. 3 fcm \n , D m ) ;15 D s = ( r e * d 1 2 * r e * d 1 2 ) ^ ( 1 / 4 ) ;

16 printf ( S e l f GMD= %. 3 fcm\n , D s ) ;17 L = 0 . 4 * log ( D m / D s ) ;

18 printf ( Loop I n d u c t a n c e o f l i n e = %. 3 fmH/km\n , L ) ;

Scilab code Exa 4.6 Find the inductance per phase of 30 km line

1 / / Find t he i n du c t a nc e p er p ha se o f 30 km l i n e2 clear ;

3 clc ;

4 // s o l t i o n5 / / g i v e n

6 r = ( 1 5 ) / 2 ; / /mm/ / r a d i u s o f t h e c o n d u c t o r7 r e = r * exp ( - 1 / 4 ) ;

8 d = 1 . 5 * 1 0 0 0 ; //mm// sp a c i n g9 L = 0 . 2 * log ( d / r e ) ;

10 printf ( L oo p I n d u c t a n c e o f l i n e = %. 2 f mH/km\n , L ) ;11 L l = L * 3 0 / 1 0 0 0 ;

12 printf ( I n d uc t an c e p er p ha se o f 30 km l o ng l i n e = %. 4f H , L l ) ;

Scilab code Exa 4.7 Find the inductance of a 3 phase line situated atcornes of a triangle

1 // Find t he i n du c t a nc e o f a 3 p ha se l i n e ( t r i a n g l e )

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2 clear ;

3 clc ;4 // s o l t i o n5 / / g i v e n6 r = 1 ; //cm/ / r a d i u s o f t h e c o n du c t or7 r e = r * exp ( - 1 / 4 ) ;

8 d 1 = 6 0 0 ; // cm/ / s p a c i n g o f t h e t r i a n g u l a r s ha pe d s ys te m9 d 2 = 7 0 0 ; // cm/ / s p a c i n g o f t h e t r i a n g u l a r s ha pe d s ys te m

10 d 3 = 8 0 0 ; // cm/ / s p a c i n g o f t h e t r i a n g u l a r s ha pe d s ys te m11 L = 0 . 2 * log ( ( ( d 1 * d 2 * d 3 ) ^ ( 1 / 3 ) ) / r e ) ;

12 printf ( L oo p I n d u c t a n c e o f l i n e = %. 4 f mH/km\n , L ) ;

Scilab code Exa 4.8 Find the inductance of a 3 phase line arranged inhorizontal plane

1 // F in d t he i n d uc t an c e o f a 3 p ha se l i n e ( p l an e )2 clear ;

3 clc ;

4 // s o l t i o n5 / / g i v e n

6 r = 1 ; //cm/ / r a d i u s o f t h e c o n du c t or7 r e = r * exp ( - 1 / 4 ) ;

8 d = 3 0 0 ; / /cm // s p a c i n g b /w c o n d u c t o r s9 C 1 = 0 . 2 * [ log ( d / r e ) + 0 . 5 * log ( 2 ) ] ;

10 C 2 = 0 . 2 * ( ( sqrt ( 3 ) ) / 2 ) * log ( 2 ) ;

11 L a = c o m p l e x ( C 1 , - C 2 ) ;

12 L b = 0 . 2 * log ( d / r e ) ;

13 L c = c o m p l e x ( C 1 , C 2 ) ;

14 printf ( La= (%. 2 f %. 2 f j )mH\n , real ( L a ) , imag ( L a ) ) ;15 printf ( Lb= %. 4 fmH\n , L b ) ;16 printf ( Lc= (%.2 f +%. 2 f j )mH\n , real ( L c ) , imag ( L c ) ) ;

Scilab code Exa 4.9 Find the loop inductance per phase

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1 / / Fin d t he l o op i n d uc t an c e p er p ha se

2 clear ;3 clc ;

4 // s o l t i o n5 / / g i v e n6 r = 5 ; / /mm/ / r a d i u s o f t h e c o n d u c t o r7 r e = r * exp ( - 1 / 4 ) ;

8 d = 3 5 0 0 ; //mm// sp a c i n g9 L = 2 * 1 0 ^ ( - 7 ) * log ( d / r e ) ;

10 L _ = L * 1 0 ^ 6 ;

11 printf ( I n d u c t a n c e p e r km ( L )=% . 4 f 10 6 H\n , L _ ) ;12 printf ( Lav=210 7{ l o g ( dp/r ) +1/3l o g ( 2 )}\n ) ;

13 printf ( Lav= L\n ) ;14 Z=(L /(2*10^ -7) -1/3* log ( 2 ) ) ;

15 d p = r e * exp ( Z ) ;

16 d p _ = d p / 1 0 0 0 ;

17 printf ( A f t e r s o v i n g a bo ve e q u a t i on \n ) ;18 printf ( S p ac in g b et we en t he c o n du c to r s i n t he p l an e (

dp) = %. 3f m , d p _ ) ;

Scilab code Exa 4.10 Find the loop inductance per phase

1 / / Fin d t he l o op i n d uc t an c e p er p ha se2 clear ;

3 clc ;

4 // s o l t i o n5 / / g i v e n6 r = 2 0 ; / /mm/ / r a d i u s o f t h e c o n d u c t o r7 r e = r * exp ( - 1 / 4 ) ;

8 d = 7 0 0 0 ; //mm// sp a c i n g

9 L = 0 . 1 * log (( sqrt ( 3 ) ) * d / ( 2 * r e ) ) ;10 printf ( I n d u c t a n c e p e r km ( L )=% . 4 f mH\n , L ) ;

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Scilab code Exa 4.11 Find the inductance of an ASCR 3 phase line

1 // F in d t he i n d uc t an c e o f an ASCR 3 p ha se l i n e2 clear ;

3 clc ;

4 // s o l t i o n5 / / g i v e n6 r = 5 / 2 ; / /mm/ / r a d i u s o f t h e c o n d u c t o r7 r e = r * 2 . 1 7 7 * 1 0 ^ - 3 ; //m8 d x = 6 ; / /m/ / s p a c i n g i n X d i r e c t i o n9 d y = 8 ; / /m/ / s p a c i n g i n Y d i r e c t i o n

10 d a a _ = sqrt ( d x ^ 2 + ( 2 * d y ) ^ 2 ) ;

11 d b b _ = 6 ;12 d c c _ = d a a _ ;

13 d a b = 8 ;

14 d a b _ = sqrt ( d x ^ 2 + d y ^ 2 ) ;

15 d b c = 8 ;

16 d b c _ = sqrt ( d x ^ 2 + d y ^ 2 ) ;

17 d c a _ = 6 ;

18 d c a = 1 6 ;

19 D s a = sqrt ( r e * d a a _ ) ;

20 D s b = sqrt ( r e * d b b _ ) ;

21 D s c = sqrt ( r e * d c c _ ) ;

22 D s = ( D s a * D s b * D s c ) ^ ( 1 / 3 ) ;

23 printf ( S e l f GMD or GMR, Ds= %.4 fm\n , D s ) ;24 D a b = sqrt ( d a b * d a b _ ) ;

25 D b c = sqrt ( d b c * d b c _ ) ;

26 D c a = sqrt ( d c a * d c a _ ) ;

27 D m = ( D a b * D b c * D c a ) ^ ( 1 / 3 ) ;

28 printf ( GMD, Dm= %. 2 fm\n , D m ) ;29 L = 0 . 2 * log ( D m / D s ) ;

30 printf ( I n d u c t a nc e o f 1 00 km l i n e ( L)=%. 4 f H\n ,L* 0 . 1 ) ;

31 L _ = 0 . 1 * log ( ( 2 ^ ( 1 / 3 ) ) * ( d y / r e ) * ( ( d x ^ 2 + d y ^ 2 ) / ( 4 * d y ^ 2 + d x^ 2 ) ) ^ ( 1 / 3 ) ) ;

32 printf ( I n d u c t a n c e ( By a n o t h e r metho d ) p e r p h a se p e rkm( L)=%. 4 f H\n , L _ * . 1 ) ;

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Scilab code Exa 4.12 Find inductive reactance of 3 phase bundled con-ductor

1 / / Find i n d u c t i v e r e ac t a nc e o f 3 p ha se bu nd le dc o n d u c t o r

2 clear ;

3 clc ;

4 // s o l t i o n5 / / g i v e n6 r = 1 . 7 5 * 1 0 ^ - 2 ; / /m/ / r a d i u s o f t h e c o n d u c t o r7 r e = r * exp ( - 1 / 4 ) ;

8 d = 7 ; / / s p a c i n g9 S = 0 . 4 ; / / s p a c i n g b et we en s u b c o n d u c t o r s

10 Ds = sqrt ( r e * S ) ; //GMR11 d a b = 7 ;

12 d a b _ = 7 . 4 ;

13 d a _ b = 6 . 6 ;

14 d a _ b _ = 7 ;

15 D a b = ( d a b * d a b _ * d a _ b * d a _ b _ ) ^ . 2 5 ;

16 D b c = D a b ;17 d c a = 1 4 ;

18 d c a _ = 1 3 . 6 ;

19 d c _ a = 1 4 . 4 ;

20 d c _ a _ = 1 4 ;

21 D c a = ( d c a * d c a _ * d c _ a * d c _ a _ ) ^ . 2 5 ;

22 D m = ( D a b * D c a * D b c ) ^ ( 1 / 3 ) ; //GMD23 L = 0 . 2 * log ( D m / D s ) ;

24 printf ( In du ct an ce (L )=%.4 f mH/km\n , L ) ;25 X l = 2 * % p i * 5 0 * L * 1 0 ^ - 3 ;

26 printf ( I n d u c t i ve r e a c t a n ce= %. 1 f /km\n , X l ) ;27 r1 = sqrt ( 2 * ( ( r * 1 0 ^ 2 ) ^ 2 ) ) ;

28 r e 1 = r 1 * exp ( - 1 / 4 ) ;

29 D a b 1 = 7 ;

30 D b c 1 = 7 ;

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31 D c a 1 = 1 4 ;

32 D m 1 = ( D a b 1 * D b c 1 * D c a 1 ) ^ ( 1 / 3 ) ; / /GMD o f s i n g l e c o n d u c t o rl i n e33 L 1 = 0 . 2 * log ( D m 1 / ( r e 1 * 1 0 ^ - 2 ) ) ;

34 printf ( In du ct an ce (L )=%.3 f mH/km\n , L 1 ) ;35 X l 1 = 2 * % p i * 5 0 * L 1 * 1 0 ^ - 3 ;

36 printf ( I n d u c t i ve r e a c t a n ce= %. 3 f /km , X l 1 ) ;

Scilab code Exa 4.13 Find the capacitance of 1 phase line

1 // Find t he c a p ac i t a n c e o f 1 p ha se l i n e2 clear ;

3 clc ;

4 // s o l t i o n5 / / g i v e n6 r = 1 5 / 2 ; / /mm/ / r a d i u s o f t h e c o n d u c t o r7 d = 1 5 0 0 ; //mm// sp a c i n g8 L = 3 0 0 0 0 ; / /m/ / l e n g t h o f t h e l i n e9 E o = 8 . 8 5 * 1 0 ^ - 1 2 // p e r mi t i v i t y o f t h e a i r

10 C = % p i * E o * L / ( log ( d / r ) ) ;

11 C _ = C * 1 0 ^ 6 ;12 printf ( C a pa ci ta nc e o f 30km l i n e = %f F , C _ ) ;

Scilab code Exa 4.14 Find the capacitance of 2 wire 1 phase line

1 / / Find t h e c a pa c it a nc e o f 2 w ir e 1 p h a s e l i n e2 clear ;

3 clc ;

4 // s o l t i o n5 / / g i v e n6 r = 0 . 2 5 ; // cm/ / r a d i u s o f t h e c o n du c to r7 d = 1 5 0 ; //cm// sp ac i n g8 L = 5 0 0 0 0 ; / /m/ / l e n g t h o f t h e l i n e

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9 h = 7 0 0 / /cm// h e i g h t o f c o n du c t or a bo ve e a r t h

10 E o = 8 . 8 5 4 * 1 0 ^ - 1 2 // p e r m i t i v i t y o f t he a i r11 C = % p i * E o * L / ( log (120/( sqrt ( 1 + ( d ^ 2 / ( 2 * h ) ^ 2 ) ) * r ) ) ) ;12 C _ = C * 1 0 ^ 6 ;

13 printf ( C a p a c it a n ce o f 5 0km l i n e = %. 3 f F , C _ ) ;

Scilab code Exa 4.15 Find the capacitance of 3 phase line

1 // Find t he c a p ac i t a n c e o f 3 p ha se l i n e

2 clear ;3 clc ;

4 // s o l t i o n5 / / g i v e n6 r = 1 ; //cm/ / r a d i u s o f t h e c o n du c t or7 d = 2 5 0 ; //cm// sp ac i n g8 L = 1 0 0 0 0 0 ; //m// l e n g t h o f t h e l i n e9 E o = 8 . 8 5 4 * 1 0 ^ - 1 2 // p e r m i t i v i t y o f t he a i r

10 C = 2 * % p i * E o * L / ( log ( d / r ) ) ;

11 C _ = C * 1 0 ^ 6 ;

12 printf ( C a p a c it a n ce o f 1 00km l i n e = %. 4 f F , C _ ) ;

Scilab code Exa 4.16 Find the capacitance of 3 phase 3 wire line

1 / / Find t h e c a pa c it a nc e o f 3 p h a s e 3 w ir e l i n e2 clear ;

3 clc ;

4 // s o l t i o n5 / / g i v e n

6 r = 0 . 0 1 ; / /m/ / r a d i u s o f t h e c o n d u c t o r7 d 1 = 3 . 5 ; //m// sp ac in g8 d 2 = 5 ; //m// sp ac in g9 d 3 = 8 ; //m// sp ac in g

10 L = 1 0 0 0 ; //m/ / l e n g t h o f t h e l i n e

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11 E o = 8 . 8 5 4 * 1 0 ^ - 1 2 // p e r m i t i v i t y o f t he a i r

12 d e = ( d 1 * d 2 * d 3 ) ^ ( 1 / 3 )13 C = 2 * % p i * E o * L / ( log ( d e / r ) ) ;

14 C _ = C * 1 0 ^ 6 ;

15 printf ( C ap ac it an ce o f l i n e = %. 4 f F , C _ ) ;

Scilab code Exa 4.17 Find the capacitance and charging current

1 / / Fin d t he c a p a c i t a n ce and c h a rg i n g c u r r e n t

2 clear ;3 clc ;

4 // s o l t i o n5 / / g i v e n6 f = 5 0 ; / / f r e q u e n c y7 V p h = 2 2 0 * 1 0 0 0 / sqrt ( 3 ) ; // p h as e v o l t a g e8 r = 0 . 0 1 ; / /m/ / r a d i u s o f t h e c o n d u c t o r9 d 1 = 3 ; //m// sp ac in g

10 d 2 = 3 ; //m// sp ac in g11 d 3 = 6 ; //m// sp ac in g12 L = 1 0 0 0 ; //m/ / l e n g t h o f t h e l i n e

13 E o = 8 . 8 5 4 * 1 0 ^ - 1 2 // p e r m i t i v i t y o f t he a i r14 d e = ( d 1 * d 2 * d 3 ) ^ ( 1 / 3 )

15 C = 2 * % p i * E o * L / ( log ( d e / r ) ) ;

16 C _ = C * 1 0 ^ 9 ;

17 printf ( C a p ac it a nc e o f l i n e = %. 4 f 1012F\n , C _ ) ;18 I c = 2 * % p i * f * C * V p h ;

19 printf ( C h ar gi ng c u r r e n t p e r p ha s e i s = %. 3 fmA , I c ) ;

Scilab code Exa 4.18 find capacitive reactance to neutral and chargingcurrent

1 / / f i n d c a p a c i t i v e r e a ct a n ce t o n e u t r a l and c h ar g in gc u r r e n t

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2 clear ;

3 clc ;4 // s o l t i o n5 / / g i v e n6 r = 1 . 2 5 * 1 0 ^ - 2 ; / /m/ / r a d i u s o f t h e c o n d u c t o r7 f = 5 0 / / f r e q u e n c y8 V p h = 1 3 2 * 1 0 0 0 / sqrt ( 3 ) ; // p h as e v o l t a g e9 E o = 8 . 8 5 * 1 0 ^ - 1 2 // p e r mi t i v i t y o f t h e a i r

10 d r r _ = sqrt ( 7 ^ 2 + ( 4 + 4 ) ^ 2 ) ;

11 d b b _ = d r r _ ;

12 d y y _ = 9 ;

13 D s r = sqrt ( r * d r r _ ) ;

14 D s y = sqrt ( r * d y y _ ) ;15 D s b = sqrt ( r * d b b _ ) ;

16 D s = ( D s r * D s y * D s b ) ^ ( 1 / 3 ) ;

17 d r y = sqrt ( 4 ^ 2 + ( 4 . 5 - 3 . 5 ) ^ 2 ) ;

18 d r y _ = sqrt ( ( 9 - 1 ) ^ 2 + 4 ^ 2 ) ;

19 D r y = sqrt ( d r y * d r y _ ) ;

20 D y b = D r y ;

21 D b r = sqrt ( 8 * 7 ) ;

22 D m = ( D y b * D b r * D r y ) ^ ( 1 / 3 ) ;

23 C = 2 * % p i * E o / ( log ( D m / D s ) ) ;

24 printf ( C a p a c i t a n c e p e r p h a se= %. 2 f 10 9 F/km\n

,C

* 1 0 ^ 1 2 ) ;

25 C r = 1 / ( 2 * % p i * f * C * 1 0 0 0 ) ;

26 printf ( C a p a c it a n ce p e r p ha s e= %. 2 f k \n , C r / 1 0 0 0 ) ;27 I c = ( 2 * % p i * f * C * 1 0 0 0 ) * V p h ;

28 printf ( C h a r g i n g c u r r e n t = % . 4 f A/km , I c ) ;

Scilab code Exa 4.19 Calculate the capacitance per phase

1 // C a l c u l a te t he c a p a c i t a n ce p er p ha se2 clear ;

3 clc ;

4 // s o l t i o n

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5 / / g i v e n

6 E o = 8 . 8 5 * 1 0 ^ - 1 2 // p e r mi t i v i t y o f t h e a i r7 V p h = 1 3 2 * 1 0 0 0 / sqrt ( 3 ) ; // p h as e v o l t a g e8 d 1 = 8 ; / /m/ / d i s t a n c e s9 d 2 = 6 ; //m

10 r = 3 * 2 . 5 * 1 0 ^ - 3 ; //m// r a d i u s o f c o n du c t or i n m11 C = 4 * % p i * E o / log ( ( 2 ^ ( 1 / 3 ) ) * ( d 1 / r ) * ( ( d 2 ^ 2 + d 1 ^ 2 ) / ( 4 * d 1

^ 2 + d 2 ^ 2 ) ) ^ ( 1 / 3 ) ) ;

12 C _ = C * 1 0 0 * 1 0 0 0 ;

13 printf ( C a p ac it a nc e o f 1 00 km l i n e = %. 3 f f , C_* 1 0 ^ 6 ) ;

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Chapter 5

Performance of Short and

Medium Transmission Lines

Scilab code Exa 5.1 Find voltage at sending end and percentage regula-tion and transmission efficiency

1 // Find v o l t a g e a t s e n di ng end , p e rc e nt a ge r e g u l a t i o nand t r an s m i s s i o n e f f i c i e n c y

2 clear ;

3 clc ;

4 // s o l t i o n5 / / g i v e n6 P = 3 3 0 0 ; //kW//power7 V r = 3 3 0 0 0 ; //kV/ / r e c i e v i n g v o l t a g e8 p f = 0 . 8 ; // p ea k f a c t o r9 R = 2 ; / /ohm // r e s i s t a n c e

10 X = 3 ; / /ohm // l o o p r e a c t a n c e11 I = P * 1 0 0 0 / ( V r * p f ) ;

12 Vs = sqrt ( ( V r * p f + I * R ) ^ 2 + ( ( V r * s i n d ( a c o s d ( p f ) ) ) + I * X ) ^ 2 ) ;

13 printf ( V o l t a g e a t s e n d i n g en d ( Vs )= %. 3 fV\n , V s ) ;14 P r = ( ( V s - V r ) * 1 0 0 ) / V r ;15 printf ( P e r ce n ta g e r e g u l a t i o n = %f p e r ce n t \n , P r ) ;16 L l = I * I * R / 1 0 0 0 ; // l i n e l o s s e s17 n t = P * 1 0 0 / ( P + L l ) ;

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18 printf ( T r a ns m is s io n e f f i c i e n c y = %. 2 f p e r ce n t , n t )

Scilab code Exa 5.2 voltage at sending end and percentage regulation andtotal line losses and transmission efficiency

1 // v o l t a g e a t s e nd i ng end , p e r ce n t a ge r e g u l a t i o n ,t o t a l l i n e l o s s e s and t r a n s m i s s i o n e f f i c i e n c y

2 clear ;

3 clc ;

4 // s o l t i o n5 / / g i v e n6 P = 5 0 0 0 ; //kW//power7 V = 2 2 0 0 0 ; / /kV// r e c i e v i n g v o l t a g e8 p f = 0 . 8 ; // p ea k f a c t o r9 R = 4 ; / /ohm // r e s i s t a n c e

10 X = 6 ; / /ohm // l o o p r e a c t a n c e11 V r = V / sqrt ( 3 ) ;

12 I = P * 1 0 0 0 / ( 3 * round ( V r ) * p f ) ;

13 Vs = round ( V r ) + ( I * R * p f ) + ( I * X * s i n d ( a c o s d ( p f ) ) ) ;

14 V s l = sqrt ( 3 ) * V s ;

15 printf ( S e n di n g end l i n e v o l t a g e = %. 3 fkV\n ,Vsl/1000)

16 P r = ( ( V s l - V ) * 1 0 0 ) / V ;

17 printf ( P e r ce n ta g e r e g u l a t i o n = %. 2 f p e r ce n t \n , P r ) ;18 L l = 3 * ( round ( I ) ) ^ 2 * R / 1 0 0 0 ; // l i n e l o s s e s19 printf ( T o t a l L i n e L o s s e s = %. 3 fkW\n , L l ) ;20 n t = P * 1 0 0 / ( P + L l ) ;

21 printf ( T r a ns m is s io n e f f i c i e n c y = %. 3 f p e r ce n t , n t )

Scilab code Exa 5.3 find sending end voltage and regulation

1 // f i n d s e n di n g end v o l t a g e and r e g u l a t i o n2 clear ;

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3 clc ;

4 // s o l t i o n5 / / g i v e n6 P = 5 0 0 0 ; //kW//power7 V = 1 1 0 0 0 ; / /kV// r e c i e v i n g v o l t a g e8 p f = 0 . 8 ; // p ea k f a c t o r9 L = 1 . 1 * 1 0 ^ - 3 / /H p e r km p e r p h as e / / L i n e i n d u c t a n c e

10 L l = 0 . 1 2 * P * 1 0 0 0 ;

11 V r = V / sqrt ( 3 ) ;

12 I = P * 1 0 0 0 / ( 3 * round ( V r ) * p f ) ;

13 R = L l / ( 3 * I ^ 2 ) ;

14 X = 5 . 1 8 3 6 ;

15 Vs = round ( V r ) + ( round ( I ) * R * p f ) + ( I * X * s i n d ( a c o s d ( p f ) ) ) ;16 V s l = sqrt ( 3 ) * V s ;

17 printf ( L i n e v o l t a g e a t s e n d i n g end= %. 3 f kV\n ,Vsl/ 1 0 0 0 ) ;

18 P r = ( ( V s l - V ) * 1 0 0 ) / V ;

19 printf ( P e r ce n ta g e r e g u l a t i o n = %. 3 f p e r ce n t \n , P r ) ;

Scilab code Exa 5.4 Find sending end voltage and power factor and effi-

cieny and regulation

1 // Find s e n di n g end v ol t a g e , power f a c t o r , e f f i c i e n yand r e g u l a t i o n

2 clear ;

3 clc ;

4 // s o l t i o n5 / / g i v e n6 S = 1 2 0 0 0 ; / /kVA/ / p o w er s u p p l i e d7 p f = 0 . 8 ; // p ow er f a c t o r

8 d e l = 1 . 7 3 * 1 0 ^ - 6 ;9 d = 1 4 0 / /cm// d i s t a n c e o f t h e c o n du c to r10 l = 5 0 * 1 0 ^ 3 ;

11 V r l = 3 3 0 0 0 ; / /V/ / r e c i e v i n g en d v o l t a g e12 I = S * 1 0 0 0 / ( sqrt ( 3 ) * V r l ) ;

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13 L l = 0 . 1 5 * S * 1 0 0 0 * p f ;

14 R = L l / ( 3 * I * I ) ;15 a = d e l * l * 1 0 0 / ( R ) ;

16 r = sqrt ( a / % p i ) ;

17 r e = r * exp ( - 1 / 4 ) ;

18 L = 0 . 2 * 5 0 * ( 1 0 ^ - 3 ) * log ( d / r e ) ;

19 X = 2 * % p i * 5 0 * L ;

20 X_ = floor ( X * 1 0 0 ) / 1 0 0 ;

21 V s = V r l / sqrt ( 3 ) + ( I * R * p f ) + ( I * X _ * s i n d ( a c o s d ( p f ) ) ) ;

22 V s l = sqrt ( 3 ) * V s ;

23 printf ( S e n di n g end l i n e v o l t a g e = %. 4 fkV\n ,Vsl/1000)

24 s p f = ( V r l * p f / sqrt ( 3 ) + I * R ) / V s ;25 printf ( S e n di n g end p ower f a c t o r = %. 3 f l a g g i n g\n ,

s p f ) ;

26 n t = S * p f * 1 0 0 / ( S * p f + ( L l / 1 0 0 0 ) ) ;

27 printf ( T r a ns m is s io n e f f i c i e n c y = %. 3 f p e r ce n t \n , n t )28 P r = ( ( V s l - V r l ) * 1 0 0 ) / V r l ;

29 printf ( P e r ce n ta g e r e g u l a t i o n = %. 3 f p e r ce n t \n , P r ) ;

Scilab code Exa 5.5 Find load end voltage and efficieny

1 // Find l oa d end v o l t a g e and e f f i c i e n y2 clear ;

3 clc ;

4 // s o l t i o n5 / / g i v e n6 P = 3 0 0 0 //kW//output7 V s l = 1 1 0 0 0 / / v o l t s8 p f = 0 . 8 / / l a g g i n g / / p ow er f a c t o r

9 R = 3 * 0 . 4 ; / /ohm/ / r e s i s t a n c e o f e ac h c o n du c t or10 X = 3 * 0 . 8 ; / /ohm // r e a c t a n c e o f e a ch c o n d u c t o r11 V s = V s l / sqrt ( 3 ) ;

12 Z = ( R * p f + X * s i n d ( a c o s d ( p f ) ) ) ;

13 V s _ = round ( V s ) ;

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14 printf ( Vr=%d % . 1 f I\n , V s _ , Z ) ;

15 I _ = P * 1 0 0 0 / ( 3 * p f )16 Vr = poly (0 , Vr ) ;17 printf ( I=%. 0 f /Vr\n , I _ ) ;18 A = 2 . 4 * I _ - V s _ * V r + V r ^ 2

19 a n s w = roots ( A ) ;

20 V r = 5 8 3 7 . 0 4 1 ;

21 V r l = sqrt ( 3 ) * V r ;

22 printf ( L i n e v o l t a g e a t t h e end ( V rl )= %d V\n , V r l ) ;23 I = I _ / V r ;

24 L l = 3 * I * I * R ;

25 n t = P * 1 0 0 0 * 1 0 0 / ( P * 1 0 0 0 + L l ) ;

26 printf ( T r a ns m is s io n e f f i c i e n c y = %. 1 f p e r ce n t , n t )

Scilab code Exa 5.6 Find current and voltage of sending end and percent-age regulation and line losses and sending end power factor and transmissionefficiency

1 / / Fin d c u r r e nt and v o l t a g e o f s e nd i ng end ,p e rc e n t ag e r e g u la t i o n , l i n e l o s s e s , s en d in g end

power f a c t o r and t r an s m i s s i o n e f f i c i e n c y2 clear ;

3 clc ;

4 // s o l t i o n5 / / g i v e n6 R = 0 . 6 1 2 5 * 1 0 0 ; / /ohm/ / t o t a l r e s i s t a n c e7 X = 1 * 1 0 0 ; //ohm// r e ac t a nc e8 Y = 1 7 . 5 * 1 0 ^ - 4 ; / /S // t o t a l s u s e c p t a n c e9 V r = 6 6 * 1 0 0 0 ; //V

10 p f = 0 . 8 ; // p ow er f a c t o r

11 P = 2 0 * 1 0 ^ 6 ; // wat ts12 I r = ( P / ( V r * p f ) ) * c o m p l e x ( p f , - 0 . 6 ) ;13 I c = c o m p l e x ( 0 , Y * V r ) ;

14 I s = I r + I c ;

15 t h e t a 1 = a t a n d ( ( imag ( I s ) / real ( I s ) ) ) ;

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16 printf ( S en di ng end c u r r e nt= %. 2 f % . 3 f A \n , abs (

I s ) , t h e t a 1 ) ;17 V s = V r + I s * ( c o m p l e x ( R , X ) ) ;

18 t h e t a 2 = a t a n d ( ( imag ( V s ) / real ( V s ) ) ) ;

19 printf ( S e n di n g end v o l t a g e = %. 3 f % . 2 f V o l t s\n ,abs ( V s ) , t h e t a 2 ) ;

20 p h i = t h e t a 2 - t h e t a 1 ;

21 printf ( s e n d i n g e nd p ow er f a c t o r = %. 3 f ( l a g ) \n , c o s d (p h i ) ) ;

22 P r = ( ( abs ( V s ) - V r ) * 1 0 0 ) / V r ;

23 printf ( P e r ce n ta g e r e g u l a t i o n = %. 1 f p e r ce n t \n , P r ) ;24 L l = ( abs ( I s ) ) ^ 2 * R / 1 0 0 0 ; // l i n e l o s s e s

25 printf ( T o t a l L i n e L o s s e s = %. 3 fkW\n , L l ) ;26 n t = P * 1 0 0 / ( P + L l * 1 0 0 0 ) ;

27 printf ( T r a ns m is s io n e f f i c i e n c y = %. 2 f p e r ce n t , n t )

Scilab code Exa 5.7 Find current and voltage of sending end and percent-age regulation and transmission efficiency

1 / / Fin d c u r r e nt and v o l t a g e o f s e nd i ng end a nd

p e r c e nt ag e r e g u l a t i o n and t r an s m i s s i o n e f f i c i e n c y2 clear ;

3 clc ;

4 // s o l t i o n5 / / g i v e n6 R = 0 . 2 * 1 5 0 ; / /ohm/ / t o t a l r e s i s t a n c e7 X = 0 . 5 * 1 5 0 ; //ohm// r e ac t a nc e8 Y = 1 5 0 * 3 * 1 0 ^ - 6 ; // S // t o t a l s u s e c p t a n c e9 V r l = 1 3 2 * 1 0 0 0 ; //V

10 p f = 0 . 8 ; // p ow er f a c t o r

11 P = 4 0 * 1 0 ^ 6 ; //MVA12 V r = V r l / sqrt ( 3 ) ;13 I r _ = ( P / ( sqrt ( 3 ) * V r l ) ) ;

14 I r = I r _ * c o m p l e x ( p f , - 0 . 6 ) ;

15 Z = c o m p l e x ( R , X ) ; //ohm//Load impeda nce

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16 V _ = V r + I r * ( Z / 2 ) ;

17 I c = V _ * ( % i ) * Y ;18 I s = I r + I c ;

19 t h e t a 1 = a t a n d ( ( imag ( I s ) / real ( I s ) ) ) ;

20 printf ( S e n di n g end c u r r e n t= %. 3 f % . 2 f A\n , abs (I s ) , t h e t a 1 ) ;

21 V s = V _ + I s * ( Z / 2 ) ;

22 t h e t a 2 = a t a n d ( ( imag ( V s ) / real ( V s ) ) ) ;

23 V l s = sqrt ( 3 ) * abs ( V s ) / 1 0 0 0 ;

24 printf ( S e n di n g end l i n e v o l t a g e = %. 2 fkV\n , V l s ) ;25 P r = ( ( abs ( V s ) - V r ) * 1 0 0 ) / V r ;

26 printf ( P e r ce n ta g e v o l t a g e r e g u l a t i o n = %. 1 f p e r ce n t \

n , P r ) ;27 p h i = t h e t a 2 - t h e t a 1 ;

28 n t = ( V r l * I r _ * p f * 1 0 0 ) / ( V l s * 1 0 0 0 * abs ( I s ) * c o s d ( p h i ) ) ;

29 printf ( T r a ns m is s io n e f f i c i e n c y = %. 2 f p e r ce n t , n t ) ;

Scilab code Exa 5.8 Find current and voltage of sending end and percent-age regulation

1 / / Fin d c u r r e nt and v o l t a g e o f s e nd i ng end a ndp e rc e n t ag e r e g u l a t i o n

2 clear ;

3 clc ;

4 // s o l t i o n5 / / g i v e n6 R = 0 . 1 4 2 5 * 2 0 0 ; / /ohm/ / t o t a l r e s i s t a n c e7 X = 0 . 4 9 * 2 0 0 ; //ohm// r e ac t a nc e8 Y = 8 * 1 0 ^ - 4 ; / /S // t o t a l s u s e c p t a n c e9 V r l = 1 3 2 * 1 0 0 0 ; //V

10 p f = 0 . 8 ; // p ow er f a c t o r11 P = 5 0 * 1 0 ^ 6 ; //MVA12 Vr = round ( V r l / sqrt ( 3 ) ) ;

13 I r _ = ( P / ( sqrt ( 3 ) * V r l ) ) ;

14 I r = I r _ * c o m p l e x ( p f , - 0 . 6 ) ;

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15 I c r = 0 . 5 * ( % i * Y ) * V r ;

16 I l = I r + I c r ;17 Z = c o m p l e x ( R , X ) ; //ohm//Load impeda nce18 V s = V r + I l * ( Z ) ;

19 t h e t a = a t a n d ( ( imag ( V s ) / real ( V s ) ) ) ;

20 printf ( S en di ng end v o l t a g e= %. 3 f % . 3 f \n , abs ( Vs) , t h e t a ) ;

21 V l s = sqrt ( 3 ) * abs ( V s ) / 1 0 0 0 ;

22 printf ( S e n di n g end l i n e v o l t a g e = %. 2 fkV\n , V l s ) ;23 M = 1 + 0 . 5 * ( % i * Y ) * Z ; //THE BOOK HAS USED 0 . 9 9 6 2 BUT IT

I S 0 . 96 224 V r l o = V l s / abs ( M ) ;

25 P r = ( ( V r l o * 1 0 0 0 - V r l ) * 1 0 0 ) / V r l ;26 printf ( P e r ce n ta g e v o l t a g e r e g u l a t i o n = %. 1 f p e r ce n t \

n , P r ) ;27 //THE ANS OF THE REGULATION I S 2 1 . 4% BECAUSE OF

TYPOLOGICAL ERROR

Scilab code Exa 5.9 Find current and voltage of sending end

1 / / Find c u r r e nt and v o l t a g e o f s en d in g end2 clear ;

3 clc ;

4 // s o l t i o n5 / / g i v e n6 R = 0 . 1 * 1 5 0 ; / /ohm/ / t o t a l r e s i s t a n c e7 X = 0 . 5 * 1 5 0 ; //ohm// r e ac t a nc e8 Y = 3 * 1 5 0 * 1 0 ^ - 6 ; // S // t o t a l s u s e c p t a n c e9 V r l = 1 1 0 * 1 0 0 0 ; //V

10 p f = 0 . 8 ; // p ow er f a c t o r

11 P = 5 0 * 1 0 ^ 6 ; //M w at t s12 Vr = floor ( V r l / sqrt ( 3 ) ) ;13 I r _ = ( P / ( sqrt ( 3 ) * V r l * p f ) ) ;

14 I r = I r _ * c o m p l e x ( p f , - 0 . 6 ) ;

15 I c 1 = V r * ( % i * Y / 2 ) ;

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16 I l = I r + I c 1 ;

17 Z = c o m p l e x ( R , X ) ;18 V s = V r + I l * Z ;

19 t h e t a = a t a n d ( ( imag ( V s ) / real ( V s ) ) ) ;

20 V l s = sqrt ( 3 ) * abs ( V s ) / 1 0 0 0 ;

21 printf ( S en di ng end l i n e v o l t a g e= %. 2 f kV\n , V l s ) ;22 I c 2 = V s * ( % i * Y / 2 ) ;

23 I s = I l + I c 2 ;

24 printf ( S e n d i n g e nd c u r r e n t ( I s )= %. 1 f A , abs ( I s ) ) ;

Scilab code Exa 5.10 Find regulation and charging current using nominalT method

1 // F in d r e g u l a t i o n and c h a rg i n g c u r r e n t u s i ng n om in alT method

2 clear ;

3 clc ;

4 // s o l t i o n5 //FUNCTIONS6 function [ z ] = r x r ( A , B ) / / F u nc ti o n f o r t h e

m u l t i p l i c a t i o n o f r e c t a n gu l a r7 z ( 1 ) = A ( 1 ) * B ( 1 )

8 z ( 2 ) = A ( 2 ) + B ( 2 )

9 e n d f u n c t i o n

10

11 function [ a ] = r 2 p ( z ) // F un ct io n f o r r e c t a n g u l a r t op o l a r

12 a = z ( 1 ) * c o m p l e x ( c o s d ( z ( 2 ) ) , s i n d ( z ( 2 ) ) )

13 e n d f u n c t i o n

14

15 / / g i v e n16 P = 5 0 * 1 0 ^ 6 ; //M w at t s17 V r l = 1 3 2 * 1 0 0 0 ; //V18 p f = 0 . 8 //pow e r f a c t i r19 V r = [ floor ( V r l / sqrt ( 3) ) 0 ];

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20 I r = [ floor ( P / ( sqrt ( 3) * V r l * p f ) ) - a c os d ( p f ) ] ;

21 A = [ 0 . 95 1 . 4] ;22 B = [ 9 6 7 8] ;

23 C = [ 0 . 0 0 15 9 0 ];

24 D = A ;

25 Z 1 = r x r ( A , V r ) ;

26 Z 2 = r x r ( B , I r ) ;

27 A V = r 2 p ( Z 1 ) ;

28 B I = r 2 p ( Z 2 ) ;

29 V s = A V + B I ;

30 t h e t a 1 = a t a n d ( ( imag ( V s ) / real ( V s ) ) ) ;

31 printf ( S e n di n g e nd v o l t a ge= %. 0 f % . 2 f V ol ts \n ,

abs ( V s ) , t h e t a 1 ) ;32 Y 1 = r x r ( C , V r ) ;

33 Y 2 = r x r ( D , I r ) ;

34 C V = r 2 p ( Y 1 ) ;

35 D I = r 2 p ( Y 2 ) ;

36 I s = C V + D I ;

37 I r a = r 2 p ( I r ) ;

38 I c = I s - I r a ;

39 t h e t a 2 = a t a n d ( imag ( I c ) / real ( I c ) ) ;

40 I c _ = sqrt ( round ( imag ( I c ) ) ^ 2 + round ( real ( I c ) ) ^ 2 ) ;

41 printf ( Ch a r g i n g c ur r e nt= %. 1 f % f A\n

, I c _ ,

t h e t a 2 ) ;

42 P r = ( ( abs ( V s ) / A ( 1 ) - V r ) * 1 0 0 ) / V r ;

43 printf ( P e r ce n ta g e r e g u l a t i o n = %. 0 f p e r ce n t \n , P r ) ;44 / / 1 . The M ag ni tu de o f S en d in g en d v o l t a g e i s 9 4 06 6 ,

i t i s due t o r o un di ng some o f t he v a lu e s45 // 2 . The a n g le i n t h e b ook i s 9 3. 1 i n c ha rg in g

c u r r e n t

Scilab code Exa 5.11 find sending end voltage and current and power andefficiency

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1 // f i n d s e nd i n g end v o l t a g e and c u r r e n t and power and

e f f i c i e n c y2 clear ;3 clc ;

4 // s o l t i o n5 //FUNCTIONS6 function [ z ] = r x r ( A , B ) / / F u nc ti o n f o r t h e

m u l t i p l i c a t i o n o f r e c t a n gu l a r7 z ( 1 ) = A ( 1 ) * B ( 1 )

8 z ( 2 ) = A ( 2 ) + B ( 2 )

9 e n d f u n c t i o n

10

11 function [ a ] = r 2 p ( z ) // F un ct io n f o r r e c t a n g u l a r t op o l a r

12 a = z ( 1 ) * c o m p l e x ( c o s d ( z ( 2 ) ) , s i n d ( z ( 2 ) ) )

13 e n d f u n c t i o n

14 / / g i v e n15 P = 5 0 * 1 0 ^ 6 ; //VA16 V r l = 1 1 0 * 1 0 0 0 ; //V17 p f = 0 . 8 //pow e r f a c t i r18 V r = [ V r l / sqrt ( 3) 0 ];

19 I r = [ P / ( sqrt ( 3) * V r l ) - a c os d ( p f ) ] ;

20 A = [0 .9 8 3 ];

21 B = [ 1 1 0 7 5 ];

22 C = [ 0 . 0 0 05 8 0 ];

23 D = [0 .9 8 3 ];

24 Z 1 = r x r ( A , V r ) ;

25 Z 2 = r x r ( B , I r ) ;

26 A V = r 2 p ( Z 1 ) ;

27 B I = r 2 p ( Z 2 ) ;

28 V s = A V + B I ;

29 t h e t a 1 = a t a n d ( ( imag ( V s ) / real ( V s ) ) ) ;

30 printf ( S e n di n g end v o l t a g e = %. 0 f V\n , abs ( V s ) ) ;

31 Y 1 = r x r ( C , V r ) ;32 Y 2 = r x r ( D , I r ) ;

33 C V = r 2 p ( Y 1 ) ;

34 D I = r 2 p ( Y 2 ) ;

35 I s = C V + D I ;

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36 t h e t a 2 = a t a n d ( imag ( I s ) / real ( I s ) ) ;

37 printf ( M ag ni tu de o f s e n d i ng end c u r r e n t= %d A\n ,abs ( I s ) ) ;38 phis= theta2 -theta 1;

39 P s = 3 * abs ( V s ) * abs ( I s ) * c o s d ( p h i s ) ;

40 printf ( Sen din g end power=%.1fMW\n , floor ( P s / 1 0 ^ 5 )/ 1 0 ) ;

41 P r = P * p f ;

42 n = P r * 1 0 0 / ( floor ( P s / 1 0 ^ 5 ) * 1 0 ^ 5 ) ;

43 printf ( T r a n s m i ss i o n E f f i c i e n c y = %. 1 f p e r c e n t , n ) ;44 // The v al ue o f v o l t a g e i s 8 74 27 V

Scilab code Exa 5.12 Find ABCD parameters and sending end voltageand current and power factor and transmission eficiency

1 / / F in d ABCD p a r a m e t e r s and s e n d i n g en d v o l t a g e a ndc u r r e nt and power f a c t o r and t r a n s m i s s i o ne f i c i e n c y

2 clear ;

3 clc ;

4 // s o l t i o n5 //FUNCTIONS6 function [ z ] = r x r ( A , B ) / / F u nc ti o n f o r t h e

m u l t i p l i c a t i o n o f r e c t a n gu l a r7 z ( 1 ) = A ( 1 ) * B ( 1 )

8 z ( 2 ) = A ( 2 ) + B ( 2 )

9 e n d f u n c t i o n

10

11 function [ a ] = r 2 p ( z ) // F un ct io n f o r r e c t a n g u l a r t op o l a r

12 a = z ( 1 ) * c o m p l e x ( c o s d ( z ( 2 ) ) , s i n d ( z ( 2 ) ) )13 e n d f u n c t i o n

14

15 / / g i v e n16 P = 8 0 * 1 0 ^ 6 ; // wat ts

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17 V r l = 2 2 0 * 1 0 0 0 ; //V

18 p f = 0 . 8 //pow e r f a c t i r19 V r = [ V r l / sqrt ( 3) 0 ];20 I r _ = [ P / ( sqrt ( 3) * V r l * p f ) - a c o sd ( p f ) ] ;

21 I r = r 2 p ( I r _ ) ;

22 Z = [ 2 0 0 8 0 ];

23 Y = [ 0 . 0 0 13 9 0 ];

24 a = r x r ( Z , Y ) ;

25 A c = 1 + r 2 p ( a ) / 2 ;

26 A =[ abs ( A c ) a ta nd ( ( imag ( A c ) / real ( A c ) ) ) ] ;

27 D = A ;

28 printf ( A=D= %.3 f % .1 f \n , A ( 1 ) , A ( 2 ) ) ;

29 b = r x r ( Z , Y ) ;30 B c = 1 + r 2 p ( b ) / 4 ;

31 B =[ abs ( B c ) a ta nd ( ( imag ( B c ) / real ( B c ) ) ) ] ;

32 B = r x r ( Z , B ) ;

33 printf ( B= %. 1 f % . 2 f ohm\n , B ( 1 ) , B ( 2 ) ) ;34 C = Y ;

35 printf ( C=%. 4 f % d s i em en s \n , C ( 1 ) , C ( 2 ) ) ;36 Z 1 = r x r ( A , V r ) ;

37 Z 2 = r x r ( B , I r _ ) ;

38 A V = r 2 p ( Z 1 ) ;

39 B I = r 2 p ( Z 2 ) ;

40 V s = A V + B I ;

41 t h e t a 1 = a t a n d ( ( imag ( V s ) / real ( V s ) ) ) ;

42 V s l = sqrt ( 3 ) * abs ( V s ) ;

43 printf ( S e n d i n g e nd v o l t a g e = %dkV\n , round ( V s l / 1 0 0 0 )) ;

44 Y 1 = r x r ( C , V r ) ;

45 Y 2 = r x r ( D , I r _ ) ;

46 C V = r 2 p ( Y 1 ) ;

47 D I = r 2 p ( Y 2 ) ;

48 I s = C V + D I ;

49 t h e t a 2 = a t a n d ( imag ( I s ) / real ( I s ) ) ;50 printf ( S en di ng end c u r r e nt= %. 1 f % . 1 f A \n , abs (

I s ) , t h e t a 2 ) ;

51 phis= theta2 -theta 1;

52 P s = 3 * abs ( V s ) * abs ( I s ) * c o s d ( p h i s ) ;

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53 printf ( Sen din g end power=%.2fMW\n , P s / 1 0 ^ 6 ) ;

54 n = P * 1 0 0 / P s ;55 printf ( T r a n s m i ss i o n E f f i c i e n c y = %. 1 f p e r c e n t , n ) ;

Scilab code Exa 5.13 find sending end voltage and current and power andefficiency

1 // f i n d s e nd i n g end v o l t a g e and c u r r e n t and power ande f f i c i e n c y

2 clear ;3 clc ;

4 // s o l t i o n5 //FUNCTIONS6 function [ z ] = r x r ( A , B ) / / F u nc ti o n f o r t h e

m u l t i p l i c a t i o n o f r e c t a n gu l a r7 z ( 1 ) = A ( 1 ) * B ( 1 )

8 z ( 2 ) = A ( 2 ) + B ( 2 )

9 e n d f u n c t i o n

10

11 function [ a ] = r 2 p ( z ) // F un ct io n f o r r e c t a n g u l a r t o

p o l a r12 a = z ( 1 ) * c o m p l e x ( c o s d ( z ( 2 ) ) , s i n d ( z ( 2 ) ) )

13 e n d f u n c t i o n

14 / / g i v e n15 P = 5 0 * 1 0 ^ 6 ; //VA16 V r l = 1 1 0 * 1 0 0 0 ; //V17 p f = 0 . 8 //pow e r f a c t i r18 V r = [ V r l / sqrt ( 3) 0 ];

19 I r = [ P / ( sqrt ( 3) * V r l ) - a c os d ( p f ) ] ;

20 A = [0 .9 8 3 ];

21 B = [ 1 1 0 7 5 ];22 C = [ 0 . 0 0 05 8 0 ];

23 D = [0 .9 8 3 ];

24 Z 1 = r x r ( A , V r ) ;

25 Z 2 = r x r ( B , I r ) ;

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26 A V = r 2 p ( Z 1 ) ;

27 B I = r 2 p ( Z 2 ) ;28 V s = A V + B I ;

29 t h e t a 1 = a t a n d ( ( imag ( V s ) / real ( V s ) ) ) ;

30 printf ( S e n di n g end v o l t a g e = %. 0 f V\n , abs ( V s ) ) ;31 Y 1 = r x r ( C , V r ) ;

32 Y 2 = r x r ( D , I r ) ;

33 C V = r 2 p ( Y 1 ) ;

34 D I = r 2 p ( Y 2 ) ;

35 I s = C V + D I ;

36 t h e t a 2 = a t a n d ( imag ( I s ) / real ( I s ) ) ;

37 printf ( M ag ni tu de o f s e n d i ng end c u r r e n t= %d A\n ,

abs ( I s ) ) ;38 phis= theta2 -theta 1;

39 P s = 3 * abs ( V s ) * abs ( I s ) * c o s d ( p h i s ) ;

40 printf ( Sen din g end power=%.1fMW\n , floor ( P s / 1 0 ^