Theory of Alternating Current Machinery_A. S. Langsdorf

Scilab Textbook Companion for Theory of Alternating Current Machinery by A. S. Langsdorf1 Created by Ajay Kumar G Bachelor of Engineering (B.E.) Electrical Engineering Sri Jayachamarajendra College of Engineering College Teacher R S Ananda Murthy Cross-Checked by

August 10, 2013

1 Funded

by a grant from the National Mission on Education through ICT, http://spoken-tutorial.org/NMEICT-Intro. This Textbook Companion and Scilab codes written in it can be downloaded from the ”Textbook Companion Project” section at the website http://scilab.in

Book Description Title: Theory of Alternating Current Machinery Author: A. S. Langsdorf Publisher: Tata McGraw - Hill Education Edition: 1 Year: 1999 ISBN: 978-0-07-099423-2

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Scilab numbering policy used in this document and the relation to the above 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 particular Example of the above book) For example, Exa 3.51 means solved example 3.51 of this book. Sec 2.3 means a scilab code whose theory is explained in Section 2.3 of the book.

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Contents List of Scilab Codes

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1 Fundamental Principles of Transformer

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2 Transformer Connections and Operation

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3 Transformer structure Insulation Heating and Load Stresses

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10 The Synchronous Generator

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16 The Mercury Arc Rectifier

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List of Scilab Codes Exa 1.6.14 To find secondary resistance and reactance . . Exa 1.9.18 To find the secondary terminal voltage . . . . . Exa 1.13.28To find the regulation of transformer . . . . . . Exa 1.14.29To find regulation by percent method . . . . . Exa 1.14.31To find the per unit regulation . . . . . . . . . Exa 1.14.33To find the load loss of transformer . . . . . . . Exa 1.16.37To measure the core loss of transformer . . . . Exa 1.17.41To find the efficiency at different loads . . . . . Exa 2.3.69 To find primary voltage and current supplied . Exa 2.6.76 To find branch currents and voltages . . . . . . Exa 2.22.111Conductively and Inductively transferred power Exa 2.29.130Positive and negative sequence voltages . . . . Exa 2.29.131Positive Negative and Zero sequence voltages . Exa 3.16.161To find radial force due to current . . . . . . . Exa 10.9.407To find the field excitation required . . . . . . Exa 10.10.413 Regulation by emf method . . . . . . . . . . . Exa 10.12.416 Regulation by mmf method . . . . . . . . . . . Exa 16.9.617Effect of phase control . . . . . . . . . . . . . .

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Chapter 1 Fundamental Principles of Transformer

Scilab code Exa 1.6.14 To find secondary resistance and reactance // E x a m p l e 1 6 p g 1 4 . s c e // To f i n d s e c o n d a r y r e s i s t a n c e and r e a c t a n c e // Theory o f A l t e r n a t i n g C u r r e n t M a c h i n e r y by Alexander Langsdorf 4 // F i r s t E d i t i o n 1 9 9 9 , T h i r t y S e c o n d r e p r i n t 5 // Tata McGraw H i l l P u b l i s h i n g Company 6 // Example i n Page 14 1 2 3

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clear ; clc ; close ; // Given d a t a volt_amp = 10 e +3; // V o l t Ampere r a t i n g o f t r a n s f o r m e r i s 10kA volt_ratio = 440/110; // T r a n s f o r m e r v o l t a g e r a t i o freq_tr = 60; // F r e q u e n c y o f t r a n s f o r m e r u s a g e i s 60 c p s o r 60 Hz pri_res = 0.50; // Primary r e s i s t a n c e i s 0 . 5 0 Ohm sec_res = 0.032; // S e c o n d a r y r e s i s t a n c e i s 0 . 0 3 2 5

Ohm 17 pri_reac = 0.90; // Primary l e a k a g e r e a c t a n c e i s 0 . 9 0 Ohm 18 sec_reac = 0.06; // S e c o n d a r y l e a k a g e r e a c t a n c e i s 0 . 0 6 Ohm 19 20 21

// C a l c u l a t i o n s printf ( ” The r a t i o o f t r a n s f o r m a t i o n i s %d” , volt_ratio ) ; 22 sec_res_ref_pri = sec_res *( volt_ratio ^2) ; // Ohms 23 sec_reac_ref_pri = sec_reac *( volt_ratio ^2) ; // Ohms 24 25 26

disp ( ’ Hence , ’ ) ; printf ( ” S e c o n d a r y r e s i s t a n c e r e f e r r e d = %0 . 3 f Ohm \n ” , sec_res_ref_pri ) ; 27 printf ( ” S e c o n d a r y r e a c t a n c e r e f e r r e d = %0 . 2 f Ohm” , sec_reac_ref_pri ) ; //

to the primary // Ohms to the primary Ohms

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// R e s u l t // The r a t i o o f t r a n s f o r m a t i o n i s 4 // S e c o n d a r y r e s i s t a n c e r e f e r r e d t o t h e p r i m a r y i s 0 . 5 1 2 Ohm 32 // S e c o n d a r y r e a c t a n c e r e f e r r e d t o t h e p r i m a r y i s 0 . 9 6 Ohm

Scilab code Exa 1.9.18 To find the secondary terminal voltage // E x a m p l e 1 9 p g 1 8 . s c e // To f i n d t h e s e c o n d a r y t e r m i n a l v o l t a g e // Theory o f A l t e r n a t i n g C u r r e n t M a c h i n e r y by Alexander Langsdorf 4 // F i r s t E d i t i o n 1 9 9 9 , T h i r t y S e c o n d r e p r i n t 5 // Tata McGraw H i l l P u b l i s h i n g Company 1 2 3

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6 7 8 9 10 11 12 13 14 15 16 17 18 19

// Example i n Page 18

clear ; clc ; close ;

// Given d a t a v1 = 2000; // Primary v o l t a g e , v o l t s v2 = 400; // S e c o n d a r y Open V o l t a g e , v o l t s pf = +0.8; // Power f a c t o r l a g g i n g 80% r1 = 5.5; // R e s i s t a n c e R1 , Ohms r2 = 0.2; // R e s i s t a n c e R2 , Ohms x1 = 12; // R e a c t a n c e X1 , Ohms x2 = 0.45; // R e a c t a n c e X2 , Ohms va_rating = 10 e +3 // v o l t −ampere r a t i n g o f t r a n s f o r m e r , VA 20 voltage1 = v1 ; // S u p p l y i n p u t v o l t a g e , V o l t s 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

// C a l c u l a t i o n s current1 = va_rating / voltage1 ; // Amperes current2 = current1 ; // Amperes turns_ratio = v1 / v2 ; r2dash = turns_ratio ^2 * r2 ; // r 2 a s r e f e r r e d t o p r i m a r y s i d e , Ohms sum_ofr = r1 + r2dash ; // t o t a l e q u i v a l e n t r e s i s t a n c e r e f e r r e d t o p r i m a r y , Ohms x2dash = turns_ratio ^2 * x2 ; // x2 a s r e f e r r e d t o p r i m a r y s i d e , Ohms sum_ofx = x1 + x2dash ; // Sum o f r e a c t a n c e s , Ohms // Taking c u r r e n t a x i s a s t h e r e f e r e n c e a s p e r t h e problem vec_current1 = 5 + 0* %i ; // V e c t o r C u r r e n t 1 , Amperes vec_current2 = vec_current1 ; // V e c t o r C u r r e n t 2 , Amperes theta = acos (0.8) ; // l a g g i n g p h a s e a n g l e i n r a d i a n s vector_volt1 = voltage1 ; // V o l t s function y = ff ( voltage2 ) // To s o l v e f o r s e c o n d a r y v o l t a g e from t h e 7

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equation // vector volt1 = vector volt2 + vec current2 ∗ ( ( s u m o f r ) +( s u m o f x ) ∗ %i ) ; // v e c t o r v o l t 2 = v o l t a g e 2 ∗ ( c o s ( t h e t a )+s i n ( t h e t a ) ∗ %i ) ; // v e c t o r v o l t 1 = v o l t a g e 2 ∗ ( c o s ( t h e t a )+s i n ( t h e t a ) ∗ %i ) + v e c c u r r e n t 2 ∗ ( ( s u m o f r ) +( s u m o f x ) ∗ %i ) ; // S e p a r a t i n g r e a l and i m a g i n a r y p a r t s and c a l c u l a t i n g t h e a b s o l u t e v a l u e s , and e q u a t i n g i t t o z e r o ( o r h e r e y ( 1 ) ) , t h e e x p r e s s i o n would l o o k l i k e below // y ( 1 ) = −( v e c t o r v o l t 1 ˆ 2 ) + ( c o s ( t h e t a ) ∗ v o l t a g e 2 ( 1 ) + a b s ( v e c c u r r e n t 2 ) ∗ s u m o f r ) ˆ2 + ( v o l t a g e 2 ( 1 ) ∗ s i n ( t h e t a ) + abs ( v e c c u r r e n t 2 ) ∗ sum ofx ) ˆ 2 ; y (1) = -( vector_volt1 ^2) + ( cos ( theta ) * voltage2 (1) + abs ( vec_current2 ) *( sum_ofr ) ) ^2 + ( sin ( theta ) * voltage2 (1) + abs ( vec_current2 ) *( sum_ofx ) ) ^2; endfunction sec_volt_in_terms_of_pri = fsolve ([0.1] , ff ) ; // i n Volts sec_voltage = sec_volt_in_terms_of_pri / turns_ratio ; // i n V o l t s printf ( ” \ n S e c o n d a r y V o l t a g e a s r e f e r r e d t o p r i m a r y i s %. 2 f v o l t s \n ” , sec_volt_in_terms_of_pri ) ; printf ( ” S e c o n d a r y T e r m i n a l V o l t a g e a t f u l l l o a d i s % . 2 f v o l t s \n ” , sec_voltage ) ;

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// R e s u l t // S e c o n d a r y V o l t a g e a s r e f e r r e d t o p r i m a r y i s 1887.30 volts 52 // S e c o n d a r y T e r m i n a l V o l t a g e a t f u l l l o a d i s 3 7 7 . 4 6 volts

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Scilab code Exa 1.13.28 To find the regulation of transformer // E x a m p l e 1 1 3 p g 2 8 . s c e // To f i n d t h e r e g u l a t i o n o f t r a n s f o r m e r // Theory o f A l t e r n a t i n g C u r r e n t M a c h i n e r y by Alexander Langsdorf 4 // F i r s t E d i t i o n 1 9 9 9 , T h i r t y S e c o n d r e p r i n t 5 // Tata McGraw H i l l P u b l i s h i n g Company 6 // Example i n Page 28

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7 8 9 clear ; clc ; close ; 10 11 // Given d a t a 12 v1 = 1100; // Primary v o l t a g e , V o l t s 13 v2 = 110; // S e c o n d a r y Open V o l t a g e , V o l t s 14 volt_sc = 33; // V o l t a g e f o r S h o r t C i r c u i t

f u l l load

current , Volts 15 pow_sc_in = 85; // S h o r t C i r c u i t i n p u t Power , Watts 16 pf = +0.8; // Power f a c t o r l a g g i n g 80% 17 va_rating = 5 e +3 // v o l t −ampere r a t i n g o f

t r a n s f o r m e r , VA 18 19 20 21 22 23 24 25 26

// C a l c u l a t i o n s // Method b a s e d on Eq . 1−35 // v1 ˆ2 = ( v2 + v o l t s c ∗ c o s ( t h e t a e − t h e t a 2 ) ) ˆ2 + ( v o l t s c ∗ sin ( thetae − theta2 ) ) ˆ2; current1 = va_rating / v1 ; // C u r r e n t i n Amperes thetae = acos ( pow_sc_in /( volt_sc * current1 ) ) ; theta2 = acos ( pf ) ; function y = ff1 ( v2 ) 9

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y (1) = -( v1 ^2) + ( v2 + volt_sc * cos ( thetae - theta2 ) ) ^2 + ( volt_sc * sin ( thetae - theta2 ) ) ^2; endfunction volt2 = fsolve ([0.1] , ff1 ) ; // v o l t a g e i n v o l t s // R e g u l a t i o n = ( ( v1 − v o l t 2 ) / v1 ) ∗ 100 Regulation1 = (( v1 - volt2 ) / v1 ) *100; printf ( ” \ n R e g u l a t i o n o f t h e T r a n s f o r m e r by method 1 i s %. 2 f %% \n ” , Regulation1 ) ; // Method b a s e d on Eq . 1−36 // v1 ˆ2 = ( v2 + c u r r e n t 1 ∗ r e ∗ c o s ( t h e t a 2 ) + c u r r e n t 1 ∗ xe ∗ s i n ( t h e t a 2 ) ) ˆ2 + ( c u r r e n t 1 ∗ xe ∗ c o s ( t h e t a 2 ) − current1 ∗ re ∗ sin ( theta2 ) ) ˆ2; current1 = va_rating / v1 ; // C u r r e n t i n Amperes thetae = acos ( pow_sc_in /( volt_sc * current1 ) ) ; theta2 = acos ( pf ) ; ze = volt_sc / current1 ; // i m p e d a n c e i n Ohms re = pow_sc_in /( current1 ^2) ; // R e s i s t a n c e i n Ohms xe = ( ze ^2 - re ^2) ^0.5; // R e a c t a n c e i n Ohms function y = ff2 ( v2 ) y (1) = -( v1 ^2) + ( v2 + current1 * re * cos ( theta2 ) + current1 * xe * sin ( theta2 ) ) ^2 + ( current1 * xe * cos ( theta2 ) - current1 * re * sin ( theta2 ) ) ^2; endfunction volt2 = fsolve ([0.1] , ff2 ) ; // R e g u l a t i o n = ( ( v1 − v o l t 2 ) / v1 ) ∗ 100 Regulation2 = (( v1 - volt2 ) / v1 ) *100; printf ( ” R e g u l a t i o n o f t h e T r a n s f o r m e r by method 2 i s %. 2 f %% \n ” , Regulation2 ) ;

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// R e s u l t // R e g u l a t i o n o f t h e T r a n s f o r m e r by method 1 i s 2 . 8 5 % 52 // R e g u l a t i o n o f t h e T r a n s f o r m e r by method 2 i s 2 . 8 5 %

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Scilab code Exa 1.14.29 To find regulation by percent method // E x a m p l e 1 1 4 p g 2 9 . s c e // To f i n d r e g u l a t i o n by p e r c e n t method // Theory o f A l t e r n a t i n g C u r r e n t M a c h i n e r y by Alexander Langsdorf 4 // F i r s t E d i t i o n 1 9 9 9 , T h i r t y S e c o n d r e p r i n t 5 // Tata McGraw H i l l P u b l i s h i n g Company 6 // Example i n Page 29

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7 8 9 clear ; clc ; close ; 10 11 // Given d a t a 12 v1 = 1100; // Primary v o l t a g e , v o l t s 13 v2 = 110; // S e c o n d a r y Open V o l t a g e , v o l t s 14 volt_sc = 33; // V o l t a g e f o r S h o r t C i r c u i t

f u l l load

current , volts 15 pow_sc_in = 85; // S h o r t C i r c u i t i n p u t Power , 16 pf = +0.8; // Power f a c t o r l a g g i n g 80% 17 va_rating = 5 e +3 // v o l t −ampere r a t i n g o f

watts

t r a n s f o r m e r , VA 18 19 20 21 22 23 24 25 26

// C a l c u l a t i o n s // Method b a s e d on Eq . 1−38 // % r e g u l a t i o n = r p c ∗ c o s ( t h e t a 2 ) + xpc ∗ s i n ( t h e t a 2 ) + ( ( xpc ∗ c o s ( t h e t a 2 ) − r p c ∗ s i n ( t h e t a 2 ) ) ˆ 2 ) / 2 0 0 ; current1 = va_rating / v1 ; // C u r r e n t i n Amperes thetae = acos ( pow_sc_in /( volt_sc * current1 ) ) ; theta2 = acos ( pf ) ; ze = volt_sc / current1 ; // Impedance i n Ohms 11

re = pow_sc_in /( current1 ^2) ; // R e s i s t a n c e i n Ohms xe = ( ze ^2 - re ^2) ^0.5; // Impedance i n Ohms rpc = ( current1 * re / v1 ) *100; xpc = ( current1 * xe / v1 ) *100; percent_regulation = rpc * cos ( theta2 ) + xpc * sin ( theta2 ) + (( xpc * cos ( theta2 ) - rpc * sin ( theta2 ) ) ^2) /200; 32 printf ( ” R e g u l a t i o n o f t h e T r a n s f o r m e r by per −c e n t method i s %. 2 f %% \n ” , percent_regulation ) ;

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// R e s u l t // R e g u l a t i o n o f t h e T r a n s f o r m e r by per −c e n t method i s 2.85 %

Scilab code Exa 1.14.31 To find the per unit regulation // E x a m p l e 1 1 4 p g 3 1 . s c e // To f i n d t h e p e r u n i t r e g u l a t i o n // Theory o f A l t e r n a t i n g C u r r e n t M a c h i n e r y by Alexander Langsdorf 4 // F i r s t E d i t i o n 1 9 9 9 , T h i r t y S e c o n d r e p r i n t 5 // Tata McGraw H i l l P u b l i s h i n g Company 6 // Example i n Page 31

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clear ; clc ; close ; // Given d a t a r_pu = 0.017; // Per−u n i t r e s i s t a n c e x_pu = 0.0247; // Per−u n i t r e a c t a n c e power_factor = 1; // U n i t y Power F a c t o r overload = 0.25; // 25% o v e r l o a d

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17 // C a l c u l a t i o n s 18 phi = acos ( power_factor ) ; 19 OL_factor = 1.00 + overload ; 20 r_pu = r_pu * OL_factor ; // Base v a l u e h a s t o be

changed f o r 0 . 2 5 o v e r l o a d 21 x_pu = x_pu * OL_factor ; // Base v a l u e h a s t o be changed f o r 0 . 2 5 o v e r l o a d 22 // Formula f o r r e g u l a t i o n i s , Per−u n i t −r e g u l a t i o n = r pu ∗ c o s ( phi ) + x pu ∗ s i n ( phi ) + 0 . 5 ∗ ( x pu ∗ c o s ( phi ) − r p u ∗ s i n ( p h i ) ) ˆ2 23 perunit_regulation = r_pu * cos ( phi ) + x_pu * sin ( phi ) + 0.5*( x_pu * cos ( phi ) - r_pu * sin ( phi ) ) ^2; 24 25 26 27 28 29

// d i s p ( ’ Hence , ’ ) ; printf ( ” Per−u n i t r e g u l a t i o n = %0 . 4 f ” , perunit_regulation ) ; // R e s u l t // Per−u n i t r e g u l a t i o n = 0 . 0 2 1 7

Scilab code Exa 1.14.33 To find the load loss of transformer // E x a m p l e 1 1 5 p g 3 3 . s c e // To f i n d t h e l o a d l o s s o f t r a n s f o r m e r // Theory o f A l t e r n a t i n g C u r r e n t M a c h i n e r y by Alexander Langsdorf 4 // F i r s t E d i t i o n 1 9 9 9 , T h i r t y S e c o n d r e p r i n t 5 // Tata McGraw H i l l P u b l i s h i n g Company 6 // Example i n Page 33

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clear ; clc ; close ;

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// Given d a t a Total_Culoss1 = 630; // T o t a l Copper L o s s a t 20 degree c e l c i u s , watts 13 TrueCopper_loss1 = 504; // Copper l o s s due t o True Ohmic r e s i s t a n c e a t 20 d e g r e e c e l c i u s , w a t t s 14 temp1 = 20; // Temperature , d e g r e e c e l c i u s 15 temp2 = 75; // Temperature , d e g r e e c e l c i u s 11 12

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// C a l c u l a t i o n s eddy_loss1 = Total_Culoss1 - TrueCopper_loss1 ; // Eddy C u r r e n t l o s s a t 20 d e g r e e c e l s i u s , w a t t s 19 TrueCopper_loss2 = TrueCopper_loss1 * ( temp2 + 234.5) / ( temp1 + 234.5) ; // True Copper l o s s a t 75 d e g r e e c e l c i u s , w a t t s 20 eddy_loss2 = eddy_loss1 * ( temp1 + 234.5) / ( temp2 + 234.5) ; // Eddy C u r r e n t l o s s a t 75 d e g r e e c e l s i u s , watts 21 load_loss = TrueCopper_loss2 + eddy_loss2 ; // Load l o s s a t 75 d e g r e e c e l s i u s , w a t t s 22 23

printf ( ” Eddy C u r r e n t l o s s a t 20 d e g r e e c e l c i u s = %. 0 f w a t t s \n ” , eddy_loss1 ) ; 24 printf ( ” True Copper l o s s a t 75 d e g r e e c e l c i u s = %. 0 f w a t t s \n ” , TrueCopper_loss2 ) ; 25 printf ( ” Load l o s s a t 75 d e g r e e c e l c i u s = %. 0 f w a t t s ” , load_loss ) ; 26 27 28

// R e s u l t // Eddy C u r r e n t l o s s a t 20 d e g r e e c e l c i u s = 126 watts 29 // True Copper l o s s a t 75 d e g r e e c e l c i u s = 613 w a t t s 30 // Load l o s s a t 75 d e g r e e c e l c i u s = 717 w a t t s

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Scilab code Exa 1.16.37 To measure the core loss of transformer // E x a m p l e 1 1 6 p g 3 7 . s c e // To m e a s u r e t h e c o r e l o s s o f t r a n s f o r m e r // Theory o f A l t e r n a t i n g C u r r e n t M a c h i n e r y by Alexander Langsdorf 4 // F i r s t E d i t i o n 1 9 9 9 , T h i r t y S e c o n d r e p r i n t 5 // Tata McGraw H i l l P u b l i s h i n g Company 6 // Example i n Page 37

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clear ; clc ; close ; // f1 B1 P1 f2 B2 P2 P3

Given d a t a = 30; // F r e q u e n c y , Hz = 8; // Flux D e n s i t y , k i l o g a u s s = 0.135; // Core l o s s , w a t t s p e r l b = 60; // F r e q u e n c y , Hz = 12; // Flux D e n s i t y , k i l o g a u s s = 0.75; // Core l o s s , w a t t s p e r l b = 0.31; // Core l o s s , w a t t s p e r l b

// C a l c u l a t i o n s a = f2 / f1 ; x =( log ( B2 ^2*( P2 - a ^2 * P3 ) /(( P2 - a * P3 ) * B1 ^2 - a *( a -1) * P1 * B2 ^2) ) ) /( log ( B2 / B1 ) ) ; kh = ( P2 - a ^2 * P3 ) /( f2 *(1 - a ) *( B2 ^ x ) ) ; ke = (( P2 - a * P3 ) * a ) /(( a -1) * f2 ^2* B2 ^2) ; Ph1 = kh * f1 * B1 ^ x ; Pe1 = ke * f1 ^2* B1 ^2; // H y s t e r e s i s Power l o s s , w a t t s Ph2 = kh * f2 * B2 ^ x ; Pe2 = ke * f2 ^2* B2 ^2; // H y s t e r e s i s Power l o s s , w a t t s Ph3 = kh * f1 * B2 ^ x ; Pe3 = ke * f1 ^2* B2 ^2; // H y s t e r e s i s Power l o s s , w a t t s Pt1 = Ph1 + Pe1 ; // T o t a l Power l o s s , w a t t s Pt2 = Ph2 + Pe2 ; // T o t a l Power l o s s , w a t t s Pt3 = Ph3 + Pe3 ; // T o t a l Power l o s s , w a t t s disp ( ’ V a l u e o f x i s ’ ) ; disp ( x ) ; 15

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36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51

disp ( ’ V a l u e o f kh i s ’ ) ; disp ( kh ) ; disp ( ’ V a l u e o f ke i s ’ ) ; disp ( ke ) ; printf ( ” \n −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−\ n f | B , k i l o g a u s s | Ph , w a t t s p e r l b | Pe , w a t t s per lb \n −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−\ n %d | %d | %. 3 f | %. 3 f \n %d | %d | %. 3 f | %. 3 f \n %d | %d | %. 3 f | %. 3 f \n −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−\ n ” , f1 , B1 , Ph1 , Pe1 , f2 , B2 , Ph2 , Pe2 , f1 , B2 , Ph3 , Pe3 ) ; // // // // // // // // // // // // // // //

Result Value o f x i s 2.0637323 V a l u e o f kh i s 0.0000484 V a l u e o f ke i s 0.0000005

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− 52

//

53

//

f | B , k i l o g a u s s | Ph , w a t t s p e r l b | Pe , w a t t s per lb −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

16

54 55 56 57

// // // //

30 | 60 | 30 |

8 12 12

| | |

0.106 0.490 0.245

| | |

0.029 0.260 0.065

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

Scilab code Exa 1.17.41 To find the efficiency at different loads // E x a m p l e 1 1 7 p g 4 1 . s c e // To f i n d t h e e f f i c i e n c y a t d i f f e r e n t l o a d s // Theory o f A l t e r n a t i n g C u r r e n t M a c h i n e r y by Alexander Langsdorf 4 // F i r s t E d i t i o n 1 9 9 9 , T h i r t y S e c o n d r e p r i n t 5 // Tata McGraw H i l l P u b l i s h i n g Company 6 // Example i n Page 41

1 2 3

7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

clear ; clc ; close ; // Given d a t a va = 50 e +3; // VA r a t i n g o f t r a n s f o r m e r , VA v1 = 2200; // V o l t s v2 = 220; // V o l t s f = 60; // F r e q u e n c y , Hz core_loss = 350; // Power l o s s , w a t t s cu_loss = 630; // Power l o s s , w a t t s pf0 = 1; pf1 = 0.8; // C a l c u l a t i o n s turns_ratio = v1 / v2 ; upf_full_load_eff = ( va * pf0 /( va * pf0 + core_loss + 17

cu_loss ) ) *100; // F u l l Load E f f i c i e n c y a t u p f 24 upf_three_fourth_eff = ((0.75* va * pf0 ) /(0.75* va * pf0 + core_loss + (0.75^2) * cu_loss ) ) *100; // E f f i c i e n c y a t t h r e e −f o u r t h l o a d a t u p f 25 full_load_eff = (( va * pf1 ) /( va * pf1 + core_loss + cu_loss ) ) *100; // E f f i c i e n c y a t f u l l l o a d a t 0 . 8 pf 26 three_fourth_eff = ((0.75* va * pf1 ) /(0.75* va * pf1 + core_loss + (0.75^2) * cu_loss ) ) *100; // E f f i c i e n c y a t t h r e e −f o u r t h l o a d a t 0 . 8 p f 27 28

printf ( ’ E f f i c i e n c y a t F u l l l o a d & u n i t y power f a c t o r = %. 1 f %% \n ’ , upf_full_load_eff ) ; 29 printf ( ’ E f f i c i e n c y a t Three−f o u r t h t h e f u l l l o a d & u n i t y power f a c t o r = %. 1 f %%\n ’ , upf_three_fourth_eff ) ; 30 printf ( ’ E f f i c i e n c y a t F u l l l o a d e f f i c i e n c y a t 80%% power f a c t o r = %. 1 f %%\n ’ , full_load_eff ) ; 31 printf ( ’ E f f i c i e n c y a t t h r e e −f o u r t h l o a d e f f i c i e n c y a t 80%% power f a c t o r = %. 1 f %%\n ’ , three_fourth_eff ) ; 32 33 34

// R e s u l t // E f f i c i e n c y a t F u l l l o a d & u n i t y power f a c t o r = 98.1 % 35 // E f f i c i e n c y a t Three−f o u r t h t h e f u l l l o a d & u n i t y power f a c t o r = 9 8 . 2 % 36 // E f f i c i e n c y a t F u l l l o a d e f f i c i e n c y a t 80% power factor = 97.6 % 37 // E f f i c i e n c y a t t h r e e −f o u r t h l o a d e f f i c i e n c y a t 80% power f a c t o r = 9 7 . 7 %

18

Chapter 2 Transformer Connections and Operation

Scilab code Exa 2.3.69 To find primary voltage and current supplied // E x a m p l e 2 3 p g 6 9 . s c e // To f i n d p r i m a r y v o l t a g e and c u r r e n t s u p p l i e d // Theory o f A l t e r n a t i n g C u r r e n t M a c h i n e r y by Alexander Langsdorf 4 // F i r s t E d i t i o n 1 9 9 9 , T h i r t y S e c o n d r e p r i n t 5 // Tata McGraw H i l l P u b l i s h i n g Company 6 // Example i n Page 69 1 2 3

7 8 9 clear ; clc ; close ; 10 11 // Given d a t a 12 13 // T r a n s f o r m e r A d a t a 14 va_A = 100 e +3; // VA r a t i n g o f T r a n s f o r m e r 15 v1_A = 4600; // V o l t a g e i n v o l t s 16 v2_A = 230; // V o l t a g e i n v o l t s 17 x_A = 0.027; // R e a c t a n c e i n Ohms 18 r_A = 0.008; // R e s i s t a n c e i n Ohms

19

19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54

// T r a n s f o r m e r B d a t a va_B = 200 e +3; // VA r a t i n g o f T r a n s f o r m e r v1_B = 4610; // V o l t a g e i n v o l t s v2_B = 225; // V o l t a g e i n v o l t s x_B = 0.013; // R e a c t a n c e i n ohms r_B = 0.003; // R e s i s t a n c e i n ohms // Common Data P_load = 150 e +3; // Power i n Watts pf = +0.85; // + d e n o t e s l a g g i n g power f a c t o r vg = 225; // V o l t a g e i n v o l t s

// C a l c u l a t i o n s // T r a n s f o r m e r A a_1 = v1_A / v2_A ; z_1 = r_A + x_A * %i ; y_1 = 1 / z_1 ; y_1_HVside = y_1 / a_1 ; // T r a n s f o r m e r B a_2 = v1_B / v2_B ; z_2 = r_B + x_B * %i ; y_2 = 1 / z_2 ; y_2_HVside = y_2 / a_2 ; y_K = y_1 + y_2 ; y_K_HVside = y_1_HVside + y_2_HVside ; // To f i n d t h e c u r r e n t I = P_load / ( vg * pf ) ; V2_vec = vg ; theta = acos (0.85) ; I_vec = I *( cos ( theta ) - sin ( theta ) * %i ) ; // − s i g n indicates I lags V

55

20

56 57 58 59 60 61 62

V1_vec = (( V2_vec * y_K ) + I_vec ) / ( y_K_HVside ) ; I1_vec = ( I_vec + V1_vec *(( y_K / a_1 ) - y_K_HVside ) ) / ( z_1 * y_K ) ; I2_vec = I_vec - I1_vec ;

printf ( ’ Primary V o l t a g e o f t r a n s f o r m e r = %f / %f V o l t s \n ’ , abs ( V1_vec ) , ( atan (( imag ( V1_vec ) ) /( real ( V1_vec ) ) ) ) *180/ %pi ) ; 63 printf ( ’ C u r r e n t S u p p l i e d by t r a n s f o r m e r A = %f / %f V o l t s \n ’ , abs ( I1_vec ) , ( atan (( imag ( I1_vec ) ) /( real ( I1_vec ) ) ) ) *180/ %pi ) ; 64 printf ( ’ C u r r e n t S u p p l i e d by t r a n s f o r m e r B = %f / %f V o l t s \n ’ , abs ( I2_vec ) , ( atan (( imag ( I2_vec ) ) /( real ( I2_vec ) ) ) ) *180/ %pi ) ; 65 66 67

// R e s u l t // Primary V o l t a g e o f t r a n s f o r m e r = 4 6 7 8 . 8 6 7 6 9 8 / 1.211839 Volts 68 // C u r r e n t S u p p l i e d by t r a n s f o r m e r A = 3 6 1 . 3 2 4 4 0 3 / − 4 4. 4 00 7 1 5 V o l t s 69 // C u r r e n t S u p p l i e d by t r a n s f o r m e r B = 4 3 8 . 8 5 8 3 8 6 / − 2 1. 4 31 5 5 3 V o l t s

Scilab code Exa 2.6.76 To find branch currents and voltages // E x a m p l e 2 6 p g 7 6 . s c e // To f i n d b r a n c h c u r r e n t s and v o l t a g e s // Theory o f A l t e r n a t i n g C u r r e n t M a c h i n e r y by Alexander Langsdorf 4 // F i r s t E d i t i o n 1 9 9 9 , T h i r t y S e c o n d r e p r i n t 5 // Tata McGraw H i l l P u b l i s h i n g Company 1 2 3

21

6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43

// Example i n Page 76

clear ; clc ; close ; // Given d a t a // T r a n s f o r m e r d a t a va = 100 e +3; // VA r a t i n g o f T r a n s f o r m e r v1 = 11500; // V o l t a g e i n v o l t s v2 = 230; // V o l t a g e i n v o l t s f = 60; // F r e q u e n c y i n Hz OC_pow = 560; // Power i n w a t t s pf = +0.155; sc_volt = 217.5; // V o l t s sc_curr = 8.7; // Amperes sc_pow = 1135; // Power i n w a t t s ll_volt = 15000; // L i n e t o l i n e v o l t a g e z_1 = 0.6; // Impedance pf2 = +0.866; pf3 = -0.5; // C a l c u l a t i o n s power_factor = sc_pow / ( sc_volt * sc_curr ) ; theta_e = acos ( power_factor ) ; transformation_ratio = v1 / v2 ; // HT v a l u e s z = sc_volt / sc_curr ; r = z * cos ( theta_e ) ; x = z * sin ( theta_e ) ; // LT v a l u e s z_lt = z /( transformation_ratio ^2) ; r_lt = r /( transformation_ratio ^2) ; 22

44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65

x_lt = x /( transformation_ratio ^2) ; zz = r_lt + %i * x_lt ; // R e f e r r i n g t o f i g u r e 2 . 1 6 ( b ) i n p a g e 77 z1 = z_1 + zz ; z_2 = z_1 *( pf2 + %i * abs ( pf3 ) ) ; z2 = z_2 + zz ; z_3 = z_1 *( abs ( pf3 ) - %i * pf2 ) ; z3 = z_3 + zz ; disp ( ’ z 1 = ’ ) disp ( z1 ) ; disp ( ’ z 2 = ’ ) disp ( z2 ) ; disp ( ’ z 3 = ’ ) disp ( z3 ) ; disp ( ’ By r e f e r r i n g t o F i g u r e 2 . 1 6 ( b ) i n p a g e 7 7 , E A , E B , E C can be w r i t t e n i n t e r m s o f t h e unknowns x and y . ’ ) ;

66 67

printf ( ” \nE A = −(x − 1 5 0 ) + j ( 2 5 9 . 8 − y ) \ nE B = −x − j y \nE C = ( 3 0 0 − x ) − j y ” ) ; 68 printf ( ” \n\ nI A = E A / z 1 \ n I B = E B / z 2 \ n I C = E C / z 3 \n ” ) ; 69 70

71 72 73 74

printf ( ” \ nI A = −1.649 x −0.0218 y + 2 5 3 . 0 1 + j ( 4 2 5 . 1 4 −1.649 y + 0 . 0 2 1 8 x ) \ n I B = −1.415 x −0.829 y + j ( 0 . 8 2 9 x − 1 . 4 1 5 y ) \ n I C = −0.860 x + 1 . 4 3 9 y +258 + j ( − 1 . 4 3 9 x −0.860 y + 4 3 1 . 7 ) \n ” ) ; // I A + I B + I C = 0 ; disp ( ’On s i m p l i f i c a t i o n and by s e p a r a t i n g t h e r e a l 23

and i m a g i n a r y p a r t s , we g e t two e q u a t i o n s c o n s i s t i n g o f x and y a s v a r i a b l e s a s shown ’ ) ; 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102

printf ( ” \n −3.924 x + 0 . 5 8 8 y + 5 1 1 . 0 1 = 0\ n −0.588 x −3.924 y + 8 5 6 . 8 4 = 0\ n ” ) ; function y = ff ( x ) ; y (1) = -3.924* x (1) +0.588* x (2) +511.01; y (2) = -0.588* x (1) -3.924* x (2) +856.84; endfunction answer = fsolve ([100;100] , ff ) ; // Answers g i v e n i n p r o b i s s u p p o s e d t o have some m i s t a k e i n v a l u e s o f x and y x = answer ([1]) ; y = answer ([2]) ; E_A = -( x - 150) + %i *(259.8 - y ) ; E_B = -x - %i * y ; E_C = (300 - x ) - %i * y ; I_A = E_A / z1 ; I_B = E_B / z2 ; I_C = E_C / z3 ; printf ( ” \n\ nI A = %0 . 2 f / %0 . 2 f Amps” , abs ( I_A ) , atan ( imag ( I_A ) / real ( I_A ) ) *180/ %pi ) ; printf ( ” \n\ n I B = %0 . 2 f / %0 . 2 f Amps” , abs ( I_B ) , atan ( imag ( I_B ) / real ( I_B ) ) *180/ %pi ) ; printf ( ” \n\ n I C = %0 . 2 f / %0 . 2 f Amps” , abs ( I_C ) , atan ( imag ( I_C ) / real ( I_C ) ) *180/ %pi ) ; printf ( ” \n\nE A = %0 . 2 f / %0 . 2 f V o l t s ” , abs ( E_A ) , atan ( imag ( E_A ) / real ( E_A ) ) *180/ %pi ) ; printf ( ” \n\ nE B = %0 . 2 f / %0 . 2 f V o l t s ” , abs ( E_B ) , atan ( imag ( E_B ) / real ( E_B ) ) *180/ %pi ) ; printf ( ” \n\nE C = %0 . 2 f / %0 . 2 f V o l t s ” , abs ( E_C ) , atan ( imag ( E_C ) / real ( E_C ) ) *180/ %pi ) ; 24

103 104 105 106 107 108 109 110 111 112 113 114 115 116 117

// // // // // // // // // // // // // //

Result z1 = 0.6059982 + 0.0080014 i z2 = 0.5255982 + 0.3080014 i z3 = 0.3059982 − 0.5115986 i

By r e f e r r i n g t o F i g u r e 2 . 1 6 ( b ) i n p a g e 7 7 , E A , E B , E C can be w r i t t e n i n t e r m s o f t h e unknowns x and y . 118 // E A = −(x − 1 5 0 ) + j ( 2 5 9 . 8 − y ) 119 // E B = −x − j y 120 // E C = ( 3 0 0 − x ) − j y 121 // 122 // I A = E A / z 1 123 // I B = E B / z 2 124 // I C = E C / z 3 125 // 126 // I A = −1.649 x −0.0218 y + 2 5 3 . 0 1 + j ( 4 2 5 . 1 4 −1.649 y +0.0218 x ) 127 // I B = −1.415 x −0.829 y + j ( 0 . 8 2 9 x − 1 . 4 1 5 y ) 128 // I C = −0.860 x + 1 . 4 3 9 y +258 + j ( − 1 . 4 3 9 x −0.860 y +431.7) 129 // 130 // On s i m p l i f i c a t i o n and by s e p a r a t i n g t h e r e a l and i m a g i n a r y p a r t s , we g e t two e q u a t i o n s c o n s i s t i n g o f x and y a s v a r i a b l e s a s shown 131 // 132 // −3.924 x + 0 . 5 8 8 y + 5 1 1 . 0 1 = 0 133 // −0.588 x −3.924 y + 8 5 6 . 8 4 = 0 134 // 25

135 136 137 138 139 140 141 142 143 144 145 146

// // // // // // // // // // // //

I A = 108.89 /

−82.59 Amps

I B = 412.73 /

2 0 . 3 0 Amps

I C = 402.59 /

4 . 9 9 Amps

E A = 65.99 /

−81.84 V o l t s

E B = 251.44 /

50.67 Volts

E C = 240.00 /

−54.13 V o l t s

Scilab code Exa 2.22.111 Conductively and Inductively transferred power // E x a m p l e 2 2 2 p g 1 1 1 . s c e // C o n d u c t i v e l y and I n d u c t i v e l y t r a n s f e r r e d power // Theory o f A l t e r n a t i n g C u r r e n t M a c h i n e r y by Alexander Langsdorf 4 // F i r s t E d i t i o n 1 9 9 9 , T h i r t y S e c o n d r e p r i n t 5 // Tata McGraw H i l l P u b l i s h i n g Company 6 // Example i n Page 111

1 2 3

7 8 9 clear ; clc ; close ; 10 11 // Given d a t a 12 13 // T r a n s f o r m e r d a t a 14 va = 10 e +3; // VA r a t i n g o f T r a n s f o r m e r , VA 15 v1 = 2300; // V o l t a g e i n v o l t s 16 v2 = 230; // V o l t a g e i n v o l t s 17 disp ( ’ R e f e r r i n g t o F i g 2 . 5 7 , we have ’ ) ;

26

18 19 20 21 22 23 24 25 26 27 28 29

// C a l c u l a t i o n s V_1 = v1 + v2 ; // V o l t a g e i n v o l t s I_1 = va / v2 ; // V o l t a g e i n v o l t s I_3 = va / v1 ; // V o l t a g e i n v o l t s I_2 = I_1 + I_3 ; // C u r r e n t i n Amperes a = V_1 / v1 ; P = V_1 * I_1 ; // Power i n w a t t s P_i = P * ( a - 1) / a ; // Power i n w a t t s P_c = round ( P / a ) ; // Power i n w a t t s printf ( ” \n\ n T o t a l v o l t −a m p e r e s s u p p l i e d from t h e s o u r c e i s = %d VA \ nVolt −Amperes s u p p l i e d i n d u c t i v e l y i s = %d VA\ nPower s u p p l i e d c o n d u c t i v e l y i s %d VA\n ” , P , P_i , P_c ) ;

30 31 32 33 34 35

// R e s u l t // R e f e r r i n g t o F i g 2 . 5 7 , we have // // // T o t a l v o l t −a m p e r e s s u p p l i e d from t h e s o u r c e i s = 1 1 0 0 0 0 VA 36 // Volt −Amperes s u p p l i e d i n d u c t i v e l y i s = 1 0 0 0 0 VA 37 // Power s u p p l i e d c o n d u c t i v e l y i s 1 0 0 0 0 0 VA

Scilab code Exa 2.29.130 Positive and negative sequence voltages // E x a m p l e 2 2 9 p g 1 3 0 . s c e // P o s i t i v e and n e g a t i v e s e q u e n c e v o l t a g e s // Theory o f A l t e r n a t i n g C u r r e n t M a c h i n e r y by Alexander Langsdorf 4 // F i r s t E d i t i o n 1 9 9 9 , T h i r t y S e c o n d r e p r i n t 5 // Tata McGraw H i l l P u b l i s h i n g Company

1 2 3

27

6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

// Example i n Page 130

clear ; clc ; close ; // Given d a t a V_1 V_2 V_3 a =

= 1000 + %i *50; = -800 + %i *100; = -200 - %i *150; cos (2* %pi /3) + %i * sin (2* %pi /3) ;

// C a l c u l a t i o n s

disp ( ’ A c c o r d i n g t o E q u a t i o n s 2−88 and 2−89 i n p a g e 130 ’ ) ; 21 V_1p = ( V_1 + V_2 * a + V_3 * a ^2) /3; 22 V_1n = ( V_1 + V_2 * a ^( -1) + V_3 * a ^( -2) ) /3; 23 24

printf ( ” \n\ n P o s i t i v e s e q u e n c e v o l t a g e i s = %0 . 4 f / %0 . 2 f V o l t s \ n N e g a t i v e s e q u e n c e v o l t a g e i s = %0 . 4 f / %0 . 2 f V o l t s \n ” , abs ( V_1p ) , atan ( imag ( V_1p ) / real ( V_1p ) ) *180/ %pi , abs ( V_1n ) , atan ( imag ( V_1n ) / real ( V_1n ) ) *180/ %pi ) ;

25 26 27 28 29 30

// R e s u l t // A c c o r d i n g t o E q u a t i o n s 2−88 and 2−89 i n p a g e 130 // // // P o s i t i v e s e q u e n c e v o l t a g e i s = 4 5 2 . 7 7 4 0 / −19.11 Volts 31 // N e g a t i v e s e q u e n c e v o l t a g e i s = 6 0 5 . 5 2 6 5 / 19.11 Volts

28

Scilab code Exa 2.29.131 Positive Negative and Zero sequence voltages // E x a m p l e 2 2 9 p g 1 3 1 . s c e // P o s i t i v e N e g a t i v e and Z e r o s e q u e n c e v o l t a g e s // Theory o f A l t e r n a t i n g C u r r e n t M a c h i n e r y by Alexander Langsdorf 4 // F i r s t E d i t i o n 1 9 9 9 , T h i r t y S e c o n d r e p r i n t 5 // Tata McGraw H i l l P u b l i s h i n g Company 6 // Example i n Page 131

1 2 3

7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

24 25 26 27 28 29

clear ; clc ; close ; // Given d a t a V_1 = 1000 + 50* %i ; V_2 = -800 + 100* %i ; V_3 = -1100 - 270* %i ; a = cos (2* %pi /3) + %i * sin (2* %pi /3) ; // C a l c u l a t i o n s disp ( ’ A c c o r d i n g t o E q u a t i o n s 2 −90 , 2−88 and 2−89 ’ ) ; V_0 = ( V_1 + V_2 + V_3 ) /3; V_1p = ( V_1 + V_2 * a + V_3 * a ^2) /3; V_1n = ( V_1 + V_2 * a ^( -1) + V_3 * a ^( -2) ) /3; printf ( ” \n\ n Z e r o s e q u e n c e v o l t a g e i s = %0 . 4 f / %0 . 2 f V o l t s \ n P o s i t i v e s e q u e n c e v o l t a g e i s = %0 . 4 f / %0 . 2 f V o l t s \ n N e g a t i v e s e q u e n c e v o l t a g e i s = %0 . 4 f / %0 . 2 f V o l t s \n ” , abs ( V_0 ) , atan ( imag ( V_0 ) / real ( V_0 ) ) *180/ %pi , abs ( V_1p ) , atan ( imag ( V_1p ) / real ( V_1p ) ) *180/ %pi , abs ( V_1n ) , atan ( imag ( V_1n ) / real ( V_1n ) ) *180/ %pi ) ; // R e s u l t // A c c o r d i n g t o E q u a t i o n s 2 −90 , 2−88 and 2−89 // // // Z e r o s e q u e n c e v o l t a g e i s = 3 0 2 . 6 5 4 9 / 7 . 5 9 V o l t s 29

// P o s i t i v e s e q u e n c e v o l t a g e i s = 5 5 8 . 9 0 5 0 / Volts 31 // N e g a t i v e s e q u e n c e v o l t a g e i s = 7 5 7 . 9 5 2 4 / Volts

30

30

13.62 −3.15

Chapter 3 Transformer structure Insulation Heating and Load Stresses

Scilab code Exa 3.16.161 To find radial force due to current // E x a m p l e 3 1 6 p g 1 6 1 . s c e // To f i n d r a d i a l f o r c e due t o c u r r e n t // Theory o f A l t e r n a t i n g C u r r e n t M a c h i n e r y by Alexander Langsdorf 4 // F i r s t E d i t i o n 1 9 9 9 , T h i r t y S e c o n d r e p r i n t 5 // Tata McGraw H i l l P u b l i s h i n g Company 6 // Example i n Page 161

1 2 3

7 8 9 10 11 12 13 14 15 16

clear ; clc ; close ; // Given d a t a va = 200 e +3; // V o l t Amperes o f t r a n s f o r m e r , VA v1 = 11000; // V o l t a g e i n v o l t s v2 = 2300; // V o l t a g e i n v o l t s T = 46.3; // Mean l e n g t h o f t h e t u r n , i n c h e s n = 455; // Number o f t u r n s 31

17 I = 1320; // C u r r e n t i n Amperes 18 l = 35; // l e n g t h i n i n c h e s 19 k = 1.8; 20 zeq_ht = 8.33; 21 22 // C a l c u l a t i o n s 23 24 F_av = (0.45/1 e +7) *(( T * n ^2* I ^2) /( k * l ) ) ; 25 printf ( ” \n The r a d i a l f o r c e due t o t h e c u r r e n t

o f %d Amps f o r g i v e n d a t a i s %d l b \n ” , I , round ( F_av ) )

; 26 27 28

// R e s u l t // The r a d i a l f o r c e due t o t h e c u r r e n t o f 1 3 2 0 Amps f o r g i v e n data i s 11930 l b

32

Chapter 10 The Synchronous Generator

Scilab code Exa 10.9.407 To find the field excitation required // E x a m p l e 1 0 9 p g 4 0 7 . s c e // To f i n d t h e f i e l d e x c i t a t i o n r e q u i r e d // Theory o f A l t e r n a t i n g C u r r e n t M a c h i n e r y by Alexander Langsdorf 4 // F i r s t E d i t i o n 1 9 9 9 , T h i r t y S e c o n d r e p r i n t 5 // Tata McGraw H i l l P u b l i s h i n g Company 6 // Example i n Page 407

1 2 3

7 8 9 10 11 12 13 14 15 16 17 18 19 20

clear ; clc ; close ; // Given d a t a va = 2500 e +3; // V o l t Ampere r a t i n g o f machine , VA vll = 6600; // L i n e t o L i n e v o l t a g e i n v o l t s N = 3000; // Number o f t u r n s f = 50; // F r e q u e n c y i n Hz slots = 60; n = 4; poles =2; r = 0.073; x = 0.87; 33

21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54

55

pf1 = pf2 = pf3 = phase

0.8; 1; 0; = 3;

// C a l c u l a t i o n s // For 80% power f a c t o r phi = acos ( pf1 ) ; V = vll / sqrt (3) ; I = round ( va / ( phase * V ) ) ; IR_a = I * r ; IX_a = I * x ; V_vec = V *( cos ( phi ) + %i * sin ( phi ) ) ; E = V_vec + I *( r + %i * x ) ; E_mag = sqrt ( real ( E ) ^2 + imag ( E ) ^2) ; conductors = slots * n ; turns = conductors /2; N_p = turns / ( poles * phase ) ; q = slots / ( poles * phase ) ; gama = 360 / slots ; gama = gama * %pi /2; k_b1 = ( sin ( q * gama /2) ) /( q * sin ( gama /2) ) ; k_p1 = 1; A = (2* sqrt (2) / %pi ) * phase * k_b1 * k_p1 * N_p * I ; cos_alpha = ( real ( E ) / E_mag ) ; sin_alpha = ( imag ( E ) / E_mag ) ; alpha = acos ( cos_alpha ) ; F_r_mag = 17500; F_r = F_r_mag *( cos ( alpha + %pi /2) + %i * sin ( alpha + %pi /2) ) ; F = F_r - A ; F_mag = sqrt ( real ( F ) ^2 + imag ( F ) ^2) ; disp ( ’ The open− c i r c u i t v o l t a g e c o r r e s p o n d i n g t o t h i s e x c i t a t i o n , d e t e r m i n e d from F i g . 10 −12 , i s 4 4 5 0 v o l t s ; ’ ); oc_volt = 4450; 34

56 57

regulation80 = (( oc_volt - V ) / V ) *100; printf ( ” \n\ nThe r e g u l a t i o n f o r 80%% power f a c t o r i s %0 . 1 f %% ” , regulation80 ) ;

58 59 // For power f a c t o r 1 . 0 60 61 phi = acos ( pf2 ) ; 62 V_vec = V *( cos ( phi ) + %i * sin ( phi ) ) ; 63 E = V_vec + I *( r + %i * x ) ; 64 E_mag = sqrt ( real ( E ) ^2 + imag ( E ) ^2) ; 65 cos_alpha = ( real ( E ) / E_mag ) ; 66 sin_alpha = ( imag ( E ) / E_mag ) ; 67 alpha = acos ( cos_alpha ) ; 68 F_r_mag = 16500; 69 F_r = F_r_mag *( cos ( alpha + %pi /2) + %i * sin ( alpha + 70 71 72

73 74 75

%pi /2) ) ; F = F_r - A ; F_mag = sqrt ( real ( F ) ^2 + imag ( F ) ^2) ; disp ( ’ The open− c i r c u i t v o l t a g e c o r r e s p o n d i n g t o t h i s e x c i t a t i o n , d e t e r m i n e d from F i g . 10 −12 , i s 4 1 5 0 v o l t s ; ’ ); oc_volt = 4150; regulation100 = (( oc_volt - V ) / V ) *100; printf ( ” \n\ nThe r e g u l a t i o n f o r 100%% power f a c t o r i s %0 . 1 f %% ” , regulation100 ) ;

76 77 // For power f a c t o r 0 78 79 phi = acos ( pf3 ) ; 80 E = V + I *( x ) ; 81 F_r_mag = 18000; 82 F_r = F_r_mag + 11300; 83 printf ( ” \ nThe v a l u e F R c o r r e s p o n d i n g t o F i g 10−12

i s %d V o l t s \n ” , F_r ) ; 84 disp ( ’ The open− c i r c u i t v o l t a g e c o r r e s p o n d i n g t o t h i s e x c i t a t i o n , d e t e r m i n e d from F i g . 10 −12 , i s 4 5 0 0 v o l t s ; ’ ); 85 oc_volt = 4500; 35

86 87 88 89 90

91 92 93

94 95 96 97 98

99 100

regulation0 = (( oc_volt - V ) / V ) *100; printf ( ” \ nThe r e g u l a t i o n f o r 0%% power f a c t o r i s %0 . 1 f %% \n ” , regulation0 ) ; // R e s u l t // The open− c i r c u i t v o l t a g e c o r r e s p o n d i n g t o t h i s e x c i t a t i o n , d e t e r m i n e d from F i g . 10 −12 , i s 4 4 5 0 volts ; // // The r e g u l a t i o n f o r 80% power f a c t o r i s 1 6 . 8 % // The open− c i r c u i t v o l t a g e c o r r e s p o n d i n g t o t h i s e x c i t a t i o n , d e t e r m i n e d from F i g . 10 −12 , i s 4 1 5 0 volts ; // // The r e g u l a t i o n f o r 100% power f a c t o r i s 8 . 9 % // The v a l u e F R c o r r e s p o n d i n g t o F i g 10−12 i s 2 9 3 0 0 Volts // // The open− c i r c u i t v o l t a g e c o r r e s p o n d i n g t o t h i s e x c i t a t i o n , d e t e r m i n e d from F i g . 10 −12 , i s 4 5 0 0 volts ; // // The r e g u l a t i o n f o r 0% power f a c t o r i s 1 8 . 1 %

Scilab code Exa 10.10.413 Regulation by emf method // E x a m p l e 1 0 1 0 p g 4 1 3 . s c e // R e g u l a t i o n by emf method // Theory o f A l t e r n a t i n g C u r r e n t M a c h i n e r y by Alexander Langsdorf 4 // F i r s t E d i t i o n 1 9 9 9 , T h i r t y S e c o n d r e p r i n t 5 // Tata McGraw H i l l P u b l i s h i n g Company 6 // Example i n Page 413 1 2 3

36

7 8 9 clear ; clc ; close ; 10 11 // Given d a t a 12 va = 2500 e +3; // Volt −Ampere r a t i n g 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38

of the

t r a n s f o r m e r , VA vll = 6600; // L i n e t o L i n e v o l t a g e i n v o l t s r = 0.073; // R e s i s t a n c e i n Ohms pf1 = 0.8; phase = 3; vref = 3640; // R e f e r e n c e f o r v o l t a g e i n v o l t s iref = 340; // R e f e r e n c e f o r c u r r e n t i n Amperes // C a l c u l a t i o n s z_s = vref / iref ; x_s = sqrt ( z_s ^2 - r ^2) ; disp ( ’ By R e f e r r i n g t o F i g . 10−19 ’ ) ; phi = acos ( pf1 ) ; V = vll / sqrt (3) ; I = round ( va / ( phase * V ) ) ; V_vec = V *( cos ( phi ) + %i * sin ( phi ) ) ; E = V_vec + I *( r + %i * x_s ) ; E_mag = sqrt ( real ( E ) ^2 + imag ( E ) ^2) ; Regulation = (( E_mag - V ) / V ) *100; printf ( ” R e g u l a t i o n i s f o u n d t o be %. 2 f %%” , Regulation ) ;

// R e s u l t // By R e f e r r i n g t o F i g . 10 −19 // R e g u l a t i o n i s f o u n d t o be 4 5 . 7 3 %

37

Scilab code Exa 10.12.416 Regulation by mmf method // E x a m p l e 1 0 1 2 p g 4 1 6 . s c e // R e g u l a t i o n by mmf method // Theory o f A l t e r n a t i n g C u r r e n t M a c h i n e r y by Alexander Langsdorf 4 // F i r s t E d i t i o n 1 9 9 9 , T h i r t y S e c o n d r e p r i n t 5 // Tata McGraw H i l l P u b l i s h i n g Company 6 // Example i n Page 416 1 2 3

7 8 clear ; clc ; close ; 9 10 // Given d a t a 11 va = 2500 e +3; // V o l t Ampere r a t i n g 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28

of the

t r a n s f o r m e r , VA vll = 6600; // L i n e t o L i n e v o l t a g e , V o l t s r = 0.073; // R e s i s t a n c e i n Ohms x = 0.87; // R e a c t a n c e i n Ohms pf1 = 0.8; phase = 3; // C a l c u l a t i o n s phi = acos ( pf1 ) ; V = vll / sqrt (3) ; I = round ( va / ( phase * V ) ) ; IR_a = I * r ; IX_a = I * x ; V_vec = V *( cos ( phi ) + %i * sin ( phi ) ) ; E = V_vec + IR_a ; E_mag = sqrt ( real ( E ) ^2 + imag ( E ) ^2) ; F_r1_mag = 16500; 38

29 30 31 32 33 34 35 36 37 38 39 40

cos_alpha = ( real ( E ) / E_mag ) ; sin_alpha = ( imag ( E ) / E_mag ) ; alpha = acos ( cos_alpha ) ; F_r1 = F_r1_mag *( cos ( %pi /2 + alpha ) + %i * sin ( %pi /2 + alpha ) ) ; A_plus_Ax = 10000; F = F_r1 - ( A_plus_Ax ) ; F_mag = sqrt ( real ( F ) ^2 + imag ( F ) ^2) ; printf ( ” \n Magnitude o f F i s %0 . 2 f amp−t u r n s p e r p o l e ” , F_mag ) ; disp ( ’ T h i s m a g n i t u d e o f F c o r r e s p o n d s t o Open− c i r c u i t v o l t a g e o f 4330 V o l t s ’ ); oc_volt = 4330; regulation = (( oc_volt - V ) / V ) *100; printf ( ” \ n R e g u l a t i o n i s f o u n d t o be %0 . 1 f %% \n ” , regulation ) ;

41 42 43 44

// R e s u l t // Magnitude o f F i s 2 3 8 6 6 . 0 2 amp−t u r n s p e r p o l e // T h i s m a g n i t u d e o f F c o r r e s p o n d s t o Open− c i r c u i t v o l t a g e o f 4330 V o l t s 45 // 46 // R e g u l a t i o n i s f o u n d t o be 1 3 . 6 %

39

Chapter 16 The Mercury Arc Rectifier

Scilab code Exa 16.9.617 Effect of phase control // E x a m p l e 1 6 9 p g 6 1 7 . s c e // E f f e c t o f p h a s e c o n t r o l // Theory o f A l t e r n a t i n g C u r r e n t M a c h i n e r y by Alexander Langsdorf 4 // F i r s t E d i t i o n 1 9 9 9 , T h i r t y S e c o n d r e p r i n t 5 // Tata McGraw H i l l P u b l i s h i n g Company 6 // Example i n Page 617

1 2 3

7 8 9 clear ; clc ; close ; 10 11 // Given d a t a 12 13 phi = 20; 14 alpha1 = 30; 15 alpha2 = 0; 16 17 // C a l c u l a t i o n s 18 19 ans1 = ( cos ( phi * %pi /(180*2) ) * cos ( phi * %pi /(180*2) +

alpha1 * %pi /180) *100) ; 40

20

ans2 = round ( cos ( phi * %pi /(180*2) ) * cos ( phi * %pi /(180*2) + alpha2 * %pi /180) *100) ; 21 Effect = ( ans1 / ans2 ) *100; 22 23

24 25 26

printf ( ” \n\ n E f f e c t o f p h a s e c o n t r o l h e r e i s t o r e d u c e t h e dc v o l t a g e t o %0 . 2 f %% o f t h e v a l u e i t would have i n t h e a b s e n c e o f p h a s e c o n t r o l \n ” , Effect ) ; // R e s u l t // E f f e c t o f p h a s e c o n t r o l h e r e i s t o r e d u c e t h e dc v o l t a g e t o 7 7 . 7 7 % o f t h e v a l u e i t would have i n the absence o f phase c o n t r o l

41