Scilab Companion Electric Machinery and Transformers_B. S. Guru and H. R. Hiziroglu

7/25/2019 Scilab Companion Electric Machinery and Transformers_B. S. Guru and H. R. Hiziroglu

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Book Description

Title: Electric Machinery And Transformers

Author: B. S. Guru And H. R. Hiziroglu

Publisher: Oxford University Press, New York

Edition: 3

Year: 2004

ISBN: 9780195138900

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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 Review of electric circuit theory 9

2 Review of basic laws of electromagnetism 18

3 Principles of Electromechanical Energy Conversion 25

4 Transformers 33

5 Direct Current Generators 46

6 Direct Current Motors 54

7 Synchronous Generators 63

8 Synchronous motors 72

9 Polyphase Induction Motor 79

10 Analysis of a single phase induction Motor 89

11 Synchronous Generator Dynamics 97

12 Permanent magnet motors 100

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List of Scilab Codes

Exa 1.1 finding the max power delivered . . . . . . . . . . . . 9Exa 1.2 Finding the current in the circuit and plot V vs T and

I vs T curve . . . . . . . . . . . . . . . . . . . . . . . 10Exa 1.3 Finding the value of capacitor. . . . . . . . . . . . . . 12Exa 1.4 Determine the line current and phase currents and power

absorbed by the load and power dessipated by transmis-sion line . . . . . . . . . . . . . . . . . . . . . . . . . . 13

Exa 1.6 Determine load current load voltage load power andpower factor . . . . . . . . . . . . . . . . . . . . . . . 15

Exa 1.7 Determine the reading of two wattmeters total powerand power factor . . . . . . . . . . . . . . . . . . . . . 16

Exa 2.1 Find the induced emf in coil. . . . . . . . . . . . . . . 18Exa 2.6 Find the magnetic flux density . . . . . . . . . . . . . 19

Exa 2.10 Find the percentage of flux setup by coil 1 links coil 2 19Exa 2.11 Find the Inductance of each coil mutual inductance and

coefficient of coupling . . . . . . . . . . . . . . . . . . 20Exa 2.12 Find effective inductance when connected in parallel

aiding and parallel opposing. . . . . . . . . . . . . . . 21Exa 2.13 Find hysteresis loss and eddy current loss . . . . . . . 22Exa 2.14 Find the minimum length of magnet for maintaining

max energy in air gap . . . . . . . . . . . . . . . . . . 23Exa 3.1 Find the mass of object and energy stored in the feild 25Exa 3.3 Find the energy stored in the magnetic feild . . . . . . 26Exa 3.4 Find the current in the coil and energy stored in the

system . . . . . . . . . . . . . . . . . . . . . . . . . . 26Exa 3.5 Find the current in the coil . . . . . . . . . . . . . . . 28

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Exa 3.6 Find the frequency of induced emf max value of induced

emf rms value of induced emf average value of inducedemf . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29Exa 3.7 Find the synchronous speed and percent slip of the motor 30Exa 3.8 Find the rotor speed and average torque developed by

motor . . . . . . . . . . . . . . . . . . . . . . . . . . . 30Exa 3.9 Find the restraining force of the spring. . . . . . . . . 31Exa 4.2 Find the a ratio and current in primary and the power

supplied to load and the flux in the core . . . . . . . . 33Exa 4.3 Find the efficiency of transformer . . . . . . . . . . . . 34Exa 4.4 Find the efficiency of transformer . . . . . . . . . . . . 35Exa 4.6 Find efficiency and voltage regulation of transformer . 36

Exa 4.7 Find the KVA rating at max efficiency . . . . . . . . . 37Exa 4.9 Find the generator voltage generator current and effi-

ciency . . . . . . . . . . . . . . . . . . . . . . . . . . . 38Exa 4.10 Find the primary winding voltage secondary winding

voltage ratio of transformation and nominal rating oftransformer . . . . . . . . . . . . . . . . . . . . . . . . 40

Exa 4.11 Find the efficiency and voltage regulation . . . . . . . 42Exa 4.13 Find the line voltages and line currents and efficiency of

the transformer. . . . . . . . . . . . . . . . . . . . . . 43Exa 4.14 Find the line current line voltage and power . . . . . . 45

Exa 5.1 Find the coil pitch for 2 pole winding and 4 pole winding 46Exa 5.3 Find the induced emf in the armature winding inducedemf per coil induced emf per turn induced emf per con-ductor . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

Exa 5.4 Find the current in each conductor the torque developedthe power developed . . . . . . . . . . . . . . . . . . . 48

Exa 5.5 Find induced emf at full load power developed torquedeveloped applied torque efficiency external resistancein feild winding voltage regulation . . . . . . . . . . . 49

Exa 5.6 Find Rfx and terminal voltage voltage regulation Effi-ciency . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

Exa 5.7 Find the voltage between far end of feeder and bus bar 52Exa 5.9 Find maximum efficiency of generator . . . . . . . . . 53Exa 6.1 Find armature current at rated load efficiency at full

load no of turns per pole new speed of motor and drivingtorque when armature current reduces . . . . . . . . . 54

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Exa 6.3 Find power developed and speed for cumulative com-

pound motor differential compound motor . . . . . . . 55Exa 6.4 Find the motor speed power loss in external resistanceefficiency . . . . . . . . . . . . . . . . . . . . . . . . . 56

Exa 6.5 Find the new motor speed power loss in external resis-tance efficiency . . . . . . . . . . . . . . . . . . . . . . 57

Exa 6.6 Find the value of external resistance when motor devel-ops torque of 30 Nm at 2000rpm torque of 30Nm at 715rpm . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

Exa 6.7 Find the torque and efficiency of the motor . . . . . . 60Exa 6.8 Find the reading on the scale . . . . . . . . . . . . . . 61Exa 6.9 Find the external resistance breaking torque at the in-

stant of plugging when the speed of motor approacheszero . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

Exa 7.2 Find the pitch factor. . . . . . . . . . . . . . . . . . . 63Exa 7.3 Find the distribution factor . . . . . . . . . . . . . . . 64Exa 7.5 Find the frequency of induced voltage phase voltage line

voltage . . . . . . . . . . . . . . . . . . . . . . . . . . 64Exa 7.6 Find the voltage regulation . . . . . . . . . . . . . . . 65Exa 7.7 Find the voltage regulation efficiency torque developed 66Exa 7.8 Find synchronous reactance per phase and voltage reg-

ulation . . . . . . . . . . . . . . . . . . . . . . . . . . 67

Exa 7.9 Find the voltage regulation and power developed by thegenerator . . . . . . . . . . . . . . . . . . . . . . . . . 68Exa 7.10 Find per phase terminal voltage armature current power

supplied total power output . . . . . . . . . . . . . . . 69Exa 8.1 Find the generated voltage and efficiency of motor . . 72Exa 8.2 Find the excitation voltage and power developed . . . 73Exa 8.3 Find power factor power angle line to line excitation

voltage torque developed . . . . . . . . . . . . . . . . 74Exa 8.4 Find the excitation voltage and other parameters . . . 75Exa 8.6 Find the new armature current and new power factor. 76Exa 8.7 Find the overall power factor and power factor of motor

to improve overall power factor . . . . . . . . . . . . . 77Exa 9.1 Find the synchronous speed and slip and rotor frequency 79Exa 9.2 Find the efficiency . . . . . . . . . . . . . . . . . . . . 80Exa 9.3 Find the efficiency of the motor. . . . . . . . . . . . . 81

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Exa 9.4 Find the max power developed and slip and the torque

developed . . . . . . . . . . . . . . . . . . . . . . . . . 82Exa 9.5 Find the breakdown slip and the breakdown torque andpower developed by the motor . . . . . . . . . . . . . 83

Exa 9.6 Find the breakdown slip and the breakdown torque andstarting torque and the value of external resistance . . 84

Exa 9.7 Find the torque range and current range . . . . . . . . 85Exa 9.8 Find Eqv circuit parameters. . . . . . . . . . . . . . . 86Exa 9.10 Find the equivalent rotor impedance as reffered to stator 87Exa 10.1 Find the per unit slip in the direction of rotation and in

opposite direction and effective rotor resistance in eachbranch . . . . . . . . . . . . . . . . . . . . . . . . . . 89

Exa 10.2 Find the shaft torque and the efficiency of the motor . 90Exa 10.3 Find the line current . . . . . . . . . . . . . . . . . . . 91Exa 10.4 Find the equivalent circuit parameters . . . . . . . . . 94Exa 10.5 Find the induced emf in the armature . . . . . . . . . 95Exa 11.7 Find the rms value of symmetric subtransient and tran-

sient currents . . . . . . . . . . . . . . . . . . . . . . . 97Exa 11.8 Find per unit power and critical fault clearing time . . 98Exa 12.1 Find the speed of motor and torque under blocked rotor

condition . . . . . . . . . . . . . . . . . . . . . . . . . 100Exa 12.2 Find the magnetic flux. . . . . . . . . . . . . . . . . . 101

Exa 12.3 Find the developed power and copper loss in the sec-ondary side . . . . . . . . . . . . . . . . . . . . . . . . 102

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List of Figures

1.1 Finding the current in the circuit and plot V vs T and I vs Tcurve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

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Chapter 1

Review of electric circuit

theory

Scilab code Exa 1.1 finding the max power delivered

1 / / C ap ti on : f i n d i n g t h e max p ower d e l i v e r e d2 / / Exa : 1 . 13 close ;

4 clc ;

5 clear ;

6 / / on a p p l y i n g KVL we g e t7 i = 7 5 / 5 0 ; // i n A mpe re s8 v _ t h = ( 3 0 * i ) + 2 5 ; / / E q u i v al e n t T he ve ni n v o l t a g e ( i n

V o l t s )9 r _ t h = ( 2 0 * 3 0 ) / ( 2 0 + 3 0 ) ; / / E q u i v a l e n t t h e v e n i n

r e s i s t a n c e ( i n Ohms )10 R _ l o a d = r _ t h ; // Load r e s i s t a n c e =t he v e ni n r e s i s t a n c e (in Ohms)

11 disp ( R _ l o a d , l o ad r e s i s t a n c e ( i n ohms )= ) // i n ohms12 i _ l o a d = v _ t h / ( r _ t h + R _ l o a d ) ; // i n A mpe re s

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Figure 1.1: Finding the current in the circuit and plot V vs T and I vs Tcurve

13 p _ m a x = ( i _ l o a d ^ 2 ) * r _ t h ; // i n Watt s14 disp ( p _ m a x , max p o we r ( i n w a t t s )= ) // maximum po wer

d i s s i p i a t e d

Scilab code Exa 1.2 Finding the current in the circuit and plot V vs Tand I vs T curve

1 / / C ap ti on : F in d in g t he c u r r e nt i n t he c i r c u i t andp l ot V v s T and I v s T c ur ve

2 / / Exa : 1 . 23 clc ;

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4 clear ;

5 close ;6 // R ef er t o f i g u r e 1 . 5 a7 L = 1 * 1 0 ^ - 3 ; / / h e n e r y8 R = 3 ; //ohms9 C = 2 0 0 * 1 0 ^ - 6 ; / / f a r a d a y10 disp ( v ( t ) = 14. 142 c os 10 00 t )11 V _ m = 1 4 . 1 4 2 ; // Peak v al ue o f a p p l i e d v o l t a g e ( i n V o lt s

)12 V = V _ m / sqrt ( 2 ) ; //RMS v a l u e o f a p p l i e d v o l t a g e ( i n

V o l t s )13 / /On c o m pa r i ng w i t h s t a n d a r d e q u a t i o n v ( t )=a c o s w t

14 w = 1 0 0 0 ; // i n r a d i a n / s e c o nd15 / / I n d u c t i v e i m p e d a nc e=j wL16 Z _ L = % i * w * L ; // i n ohms17 / / c a p a c i t i v e i mp ed an ce=j /wC18 Z _ c = - % i / ( w * C ) ; // in ohms19 / / Impedance o f t he c i r c u i t i s g iv en by20 Z = Z _ L + Z _ c + R ; // i n ohms21 I = V / Z / / C ur re nt i n t h e c i r c u i t // i n Amperes22 r = real ( I ) ;

23 i = imag ( I ) ;

24 m a g n _ I = sqrt ( ( r ^ 2 ) + ( i ^ 2 ) ) ;/ / m ag ni tu de o f c u r r e n t ( i nAmperes)

25 p h a s e _ I = a t a n d ( i / r ) ; // p h as e o f c u r r e nt ( i n d e gr e e )26 disp ( m a g n _ I , m ag n it ud e o f c u r r e n t ( i n Am peres ) ) ;27 disp (pha se_I , p ha s e o f c u r r e n t ( i n D e gr e es ) ) ;28 xset ( window ,1);29 xtitle ( c u r r e n t t im e p l o t , t i m e ( i n S e co n d s ) ,

c u r r e n t ( i n Amp er es ) ) ;30 z = linspace ( 0 , 2 0 , 1 0 ) ;

31 x = linspace (0,%pi ,100);

32 z = 2 . 8 2 8 * cos ( ( 1 0 0 0 * x ) + ( atan ( i / r ) ) ) ;

33 plot ( x , z ) ;34 xset ( window ,2);35 xtitle ( v o l t a g e t im e p l o t , t i m e ( i n S e co n d s ) ,

v o l t a g e ( i n V o lt s ) ) ;36 x = linspace (0,%pi ,100);

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37 y = linspace ( 0 , 2 0 , 1 0 ) ;

38 y = 1 4 . 1 4 2 * cos ( 1 0 0 0 * x ) ;39 plot ( x , y ) ;

Scilab code Exa 1.3 Finding the value of capacitor

1 // C ap ti on : F in d in g t he v a l ue o f c a p a c i t o r2 / / Ex n o . 1 . 33 clc ;

4 clear ;

5 close ;

6 I = 1 0 ; / / C u r re n t drawn by t h e l o a d ( i n Amp eres )7 p f 1 = 0 . 5 ; // l a g g i n g power f a c t o r8 p f 2 = 0 . 8 ;

9 Q 1 = a c o s d ( p f 1 ) ;

10 Q 2 = a c o s d ( p f 2 ) ;

11 I _L = 1 0 * ( c o s d ( - Q1 ) + % i * s i n d ( - Q1 ) ) ; // i n A mpe re s12 V = 1 2 0 ; // s o u r ce v o l t a g e ( i n V o l ts )13 f = 6 0 ; / / f r e q u e n c y o f s o u r c e ( i n H e rt z )14 // R ef er t o f i g 1 . 6 ( b )15 / / I L c =I L+ I c16 S = V * conj ( I _ L ) ; / / c om pl ex p ow er a b s o r be d by l o a d ( i n

Watts)17 //On c o n n ec t i ng c a p a c i t o r a c r o s s l o ad c u r r e n t ( I )

ha ve 0 . 8 p f l a g g i n g18 I _ L c o = real ( S ) / ( V * p f 2 ) ; // c u r re n t s u p pl i e d by l oa d

a f t e r c o n ne c t in g c a p a c i t o r ( i n Amperes )19 I _ L c = I _ L c o * ( c o s d ( - Q 2 ) + % i * ( s i n d ( - Q 2 ) ) ) ; / / i n A mp er es20 I _ c = I _ L c - I _ L ; // i n A mpe re s21 Z _ c = V / I _ c ; / / c a p a c i t i v e i m pe d an c e ( i n Ohms )22 / / Z c=j X c

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23 X _ c = Z _ c / ( - % i ) ; // C a p a c i t i v e r e a c t a n c e

24 C = 1 / ( 2 * % p i * f * X _ c ) ;25 disp ( real ( C ) , V al ue o f c a p a c i t a n c e ( i n Fa ra d ) i s = )

Scilab code Exa 1.4 Determine the line current and phase currents and

power absorbed by the load and power dessipated by transmission line

1 / / C ap ti on : D et er mi ne t h e l i n e c u r r e n t and p ha sec u r r e n t s , p ow er a b s o r be d by t h e l o a d and p ow erd e s s i p a t e d by t r an s m i s s i o n l i n e

2 / / Ex n o : 1 . 43 clc ;

4 clear ;

5 close ;

6 / / Make d e l t a s t a r c o n v e r s i on o f l oa d7 Z _ L = 1 + % i * 2 ; / / I mp ed an ce o f e a ch w i r e ( i n Ohms )

8 Z _ p = ( 1 7 7 - % i * 2 4 6 ) ; / / p e rp h a s e i m p e d a nc e ( i n Ohms )9 Z _ p Y = ( 1 7 7 - % i * 2 4 6 ) / 3 ; / / p e rp ha se i mp ed an ce i n Y

c o n n e c t i o n ( i n Ohms )10 Z = Z _ L + Z _ p Y ; / / T o t a l p e r p h a s e i m p ed a n ce ( i n Ohms )11 V = 8 6 6 / sqrt ( 3 ) ; //Perp ha se v o l t a g e ( i n V o lt s )12 V _ p h a s e = 0 ;

13 I = V / Z ; // C u rr en t i n t he c i r c u i t ( i n Ampere )14 r = real ( I ) ;

15 i = imag ( I ) ;

16 I _ m a g = sqrt ( ( r ^ 2 ) + ( i ^ 2 ) ) ; // m a gn it ud e o f c u r r e n t ( i n

Amperes)17 I _ p h a s e = a t a n d ( i / r ) ; // p h as e o f c u r r e nt ( i n D eg re es )18 p f = c o s d ( I _ p h a s e ) ; // p ower f a c t o r19 // R e fe r t o f i g : 1 . 1 3 ( b )20 // S ou rc e a r e c o nn e ct ed i n s t ar , s o p ha se c u r r e n t s =

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l i n e c u r r e nt s

21 I _ n a _ m a g = I _ m a g ; / / Ma gn it ud e o f S o ur c e c u r r e n t t hr o ug hna ( i n A mp er es )22 I _ n b _ m a g = I _ m a g ; / / Ma gn it ud e o f S o ur c e c u r r e n t t hr o ug h

nb ( i n A mp er es )23 I _ n c _ m a g = I _ m a g ; / / Ma gn it ud e o f S o ur c e c u r r e n t t hr o ug h

nc ( i n A mp er es )24 I _ n a _ p h a s e = I _ p h a s e + ( 0 ) ; // p ha se a n gl e o f c u r r e nt

t h ro u g h na ( i n D eg r ee )25 I _ n b _ p h a s e = I _ p h a s e + ( - 1 2 0 ) ; // p ha se a n gl e o f c u r r e n t

t h ro u g h nb ( i n D eg r ee )26 I _ n c _ p h a s e = I _ p h a s e + ( 1 2 0 ) ; // p ha se a n gl e o f c u r r e nt

t h ro u g h nc ( i n D eg r ee )27 disp (I_ na_mag , I n a m a g ( i n A mp er es )= ) ;28 disp (I_na_phase , I n a p h a s e ( i n D e gr e es )= ) ;29 disp (I_ nb_mag , I n b m a g ( i n A mp er es )= ) ;30 disp (I_nb_phase , I n b p h a s e ( i n D e gr e es )= ) ;31 disp (I_ nc_mag , I n c m a g ( i n A mp er es )= ) ;32 disp (I_nc_phase , I n c p h a s e ( i n D e gr e es )= ) ;33 // Load i s c o nn e ct e d i n d e l t a n et wo rk34 I _ A B _ m a g = I _ m a g / sqrt ( 3 ) ; / / m ag ni tu de o f c u r r e n t

t h r o u g h AB ( i n A mp er es )35 I _ B C _ m a g = I _ m a g / sqrt ( 3 ) ;

/ / m ag ni tu de o f c u r r e n tt h r o u g h BC ( i n A mp er es )36 I _ C A _ m a g = I _ m a g / sqrt ( 3 ) ; / / m ag ni tu de o f c u r r e n t

t h r o u g h CA ( i n A mp er es )37 I _ A B _ p h a s e = I _ n a _ p h a s e + 3 0 ; // p ha se a n gl e o f c u r r e nt

t h r o u g h AB ( i n D e g r e e s )38 I _ B C _ p h a s e = I _ n b _ p h a s e + 3 0 ; // p ha se a n gl e o f c u r r e nt

t h r o u g h BC ( i n D e g r e e s )39 I_CA_phase =I_nb_phase -90; // p ha se a n gl e o f c u r r e nt

t h r o u g h CA ( i n D e g r e e s )40 disp (I_ AB_mag , I A B ma g ( i n A mp er es )= ) ;

41 disp (I_AB_phase , I A B p h a se ( i n D e g r ee s )= ) ;42 disp (I_ BC_mag , I B C m ag ( i n A mp er es )= ) ;43 disp (I_BC_phase , I B C p h a se ( i n D e g r ee s )= ) ;44 disp (I_ CA_mag , I C A m ag ( i n A mp er es )= ) ;45 disp (I_CA_phase , I C A p h as e ( i n D e g r ee s )= ) ;

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46 I _ A B = I _ A B _ m a g * ( c o s d ( I _ A B _ p h a s e ) + % i * s i n d ( I _ A B _ p h a s e ) )

; / / ( i n A mp er es )47 P _ l o a d = 3 * I _ A B _ m a g ^ 2 * real ( Z _ p ) ; / / i n w a t t s48 disp ( real ( P _ l o a d ) , P ower d i s s i p a t e d ( i n Watts ) = ) ;49 P _ l i n e = 3 * I _ m a g ^ 2 * real ( Z _ L ) ; / / i n w a t ts50 disp ( P _ l i n e , Power d i s s i p a t e d by t r an s m i s s i o n l i n e (

i n W at ts )= )

Scilab code Exa 1.6 Determine load current load voltage load power andpower factor

1 / / C a pt i on : D e te r mi n e l o a d c u r r e n t , l o a d v o l t a g e , l o a dp ower and p ower f a c t o r

2 / / Exa : 1 . 63 clc ;

4 clear ;

5 close ;6 // R ef er t o t he f i g : 1 . 1 67 R = 4 0 ; // in ohms8 L = % i * 3 0 ; // i n ohms9 V = 1 1 7 * ( ( c o s d ( 0 ) + % i * s i n d ( 0 ) ) ) ; / / i n V o l t s10 / / E q ui v al e nt l o ad i mp ed an ce i s o b ta i n ed by p a r a l l e l

c om bi na ti o n o f R e s i s t a n ce R and I n du c ta n ce L11 Z _ L = ( R * L ) / ( R + L ) ; / / l o a d i m p e d an c e ( i n Ohms )12 Z 1 = 0 . 6 + % i * 1 6 . 8 ; / / i n Ohms13 Z = Z _ L + Z 1 ; // E q u i v a le n t i mp ed an ce o f c i r c u i t ( i n Ohms )14 I = V / Z ;

/ / c u r r e n t t h r ou g h l o a d ( i n Am peres )15 r1 = real ( I ) ;16 i1 = imag ( I ) ;

17 I _ m a g = sqrt ( r 1 ^ 2 + i 1 ^ 2 ) ; // m ag ni tu de o f c u r r e nt f l o w i n gt h r ou g h l o a d ( i n Am peres )

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18 disp ( I _ m a g , R e a d i ng o f a mm et er ( i n A mp er es )= ) ;

19 V _ L = I * Z _ L ; // v o l t a g e a c r o s s l o ad ( i n V o lt s )20 r2 = real ( V _ L ) ;21 i2 = imag ( V _ L ) ;

22 V _ L _ m a g = sqrt ( r 2 ^ 2 + i 2 ^ 2 ) ; // m a gn it ud e o f v o l t a g ea c r o s s l oa d ( i n V ol t s )

23 disp (V_L _mag , R ea di ng o f v o l t m e t e r ( i n V o l t s )= ) ;24 P = real ( V _ L * conj ( I ) ) ; / / Po wer d e v e l o p e d ( i n Wa tts )25 disp (P , R e ad i ng o f w a t t me t e r ( i n W at ts )= ) ;26 p f = P / ( V _ L _ m a g * I _ m a g ) ; / / Po wer f a c t o r27 disp ( p f , p o wer f a c t o r = )

Scilab code Exa 1.7 Determine the reading of two wattmeters total powerand power factor

1 / / C a pt i on : D e te r mi n e t h e r e a d i n g o f two w a tt m et e rs ,

t o t a l power and power f a c t o r2 / / Exa : 1 . 73 clc ;

4 clear ;

5 close ;

6 / / t r a n s fo r m i n g d e l t a c o nn ec te d s o u r ce i n t o ane q u i v a l e n t S ta rc o n ne c te d s o u r c e

7 V _ s = 1 3 5 1 ; // s o u r c e v o l t a g e ( i n V o l t s )8 V = 1 3 5 1 / sqrt ( 3 ) ; // i n v o l t s9 V _ p h a s e = 0 ;

10 Z = 3 6 0 + % i * 1 5 0 ;/ / p e r

p h a s e i m p e d an c e ( i n ohms )11 I = V / Z ; // c u r r e nt i n t he c i r c u i t ( i n Amperes )12 r = real ( I ) ;

13 i = imag ( I ) ;

14 I _ m a g = sqrt ( r ^ 2 + i ^ 2 ) ; / / i n a m pe re

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15 I _ p h a s e = a t a n d ( i / r ) ; / / d e g r e e

16 // R ef er t o f i g 1 . 1 9 ( a )17 V _ a b = 1 3 5 1 * ( c o s d ( - 3 0 ) + % i * s i n d ( - 3 0 ) ) ; / / i n V o l t s18 I _ a A = 2 * ( c o s d ( I _ p h a s e ) + % i * s i n d ( I _ p h a s e ) ) ; // i n A mpe re s19 V _ c b = 1 3 5 1 * ( c o s d ( - 9 0 ) + % i * s i n d ( - 9 0 ) ) ; / / i n V o l t s20 I _ c C = 2 * ( c o s d ( I _ p h a s e - 1 2 0 ) + % i * s i n d ( I _ p h a s e - 1 2 0 ) ) ; // in

Amperes21 P1 = real ( V _ a b * conj ( I _ a A ) ) ; // r e a d in g o f w at tm et er 1 (

i n Wa tts )22 disp ( P 1 , R e ad i n g o f w a t t me t e r W1 ( i n W at ts ) = ) ;23 P2 = real ( V _ c b * conj ( I _ c C ) ) ; // r e a d in g o f w at tm et er 2 (

i n Wa tts )

24 disp ( P 2 , R e ad i n g o f w a t t me t e r W2 ( i n W at ts )= ) ;25 P = P 1 + P 2 ; / / t o t a l po we r d e v el o p ed ( i n Watts )26 disp (P , T o t a l p ow er d e v e l o p e d ( i n Wat ts ) = ) ;27 p f = c o s d ( I _ p h a s e ) ; // p ower f a c t o r28 disp ( p f , p o wer f a c t o r = )

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Chapter 2

Review of basic laws of

electromagnetism

Scilab code Exa 2.1 Find the induced emf in coil

1 / / C ap ti on : Fi nd t h e i n du c ed emf i n c o i l2 / / Exa : 2 . 13 clc ;

4 clear ;

5 close ;

6 N = 1 0 0 0 ; / / Number o f t u r n s7 p h y _ 1 = 1 0 0 * 1 0 ^ - 3 ; / / i n i t i a l m a g n e t i c f l u x ( i n w e b e rs )8 p h y _ 2 = 2 0 * 1 0 ^ - 3 ; // f i n a l m ag ne ti c f l u x ( i n w eb er s )9 p h y = p h y _ 2 - p h y _ 1 ; // c ha ng e i n m ag ne ti c f l u x10 t = 5 ; / / ( i n s e c o n d s )11 e = ( - 1 ) * N * ( p h y / t ) ; / / i n d u c ed em f ( i n v o l t s )12 disp (e , I n du c ed emf ( i n v o l t s )= )

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Scilab code Exa 2.6 Find the magnetic flux density

1 / / C ap ti on : Fi nd t h e m a gn e ti c f l u x d e n s i t y2 / / Exa : 2 . 63 clc ;

4 clear ;

5 close ;

6 u _ o = 4 * % p i * 1 0 ^ - 7 ; // p e r me a b li t y o f a i r7 u _ r = 1 2 0 0 ; // p e r m e a b l it y o f m a gn e ti c m a t e r i a l8 N = 1 5 0 0 ; //No . o f t u r ns9 I = 4 ; // c u r r e n t i n t he c o i l ( i n Amperes )10 r _ i = 1 0 * 1 0 ^ - 2 ; // i n n er r a d i i o f m ag ne ti c c o re ( i n

me t e r s )11 r _ o = 1 2 * 1 0 ^ - 2 ; // o u te r r a d i i o f m ag ne ti c c o re ( i n

me t e r s )12 r _ m = ( r _ i + r _ o ) / 2 ; // mean r a d i i o f m ag ne ti c c o r e ( i n

me t e r s )13 l _ g = 1 * 1 0 ^ - 2 ; // l e n g th o f a i r gap ( i n m et er s )14 l _ m = 2 * % p i * ( r _ m - l _ g ) ; / / i n m e t er s15 // R e fe r t o f i g : 2 . 1 416 A _ m = ( r _ o - r _ i ) ^ 2 ; / / c r o s s s e c t i o n a l a re a o f m a g n e t i c

p a th ( i n m e te r 2 )17 R _ m = l _ m / ( u _ o * u _ r * A _ m ) ;

// r e l u c t a n c e o f m ag ne ti cm a t e r i a l18 R _ g = l _ g / ( u _ o * A _ m ) ; // r e l u c t a n c e o f a i r gap19 //R m a nd R g i n s e r e i s20 R = R _ m + R _ g ;

21 B _ m = N * I / ( R * A _ m ) ; // m a gn et ic f l u x d e n s i t y ( i n T es l a )22 disp ( B _ m , m a gn et i c f l u x d e n s i t y ( i n T e sl a )= )

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Scilab code Exa 2.10 Find the percentage of flux setup by coil 1 links coil

2

1 // C ap ti on : Fi nd t he p e r ce n t ag e o f f l u x s e tu p by c o i l1 l i n k s c o i l 2

2 / / Exa : 2 . 1 03 clc ;

4 clear ;

5 close ;

6 // R e fe r t o eqn 2 . 2 67 e _ 2 1 = 2 0 ; // v o l t a g e i nd uc ed i n c o i l 2 ( i n v o l t s )8 I 1 = 2 0 0 0 ; // r a t e o f c h a ng e o f c u r r e n t i n c o i l 1 ( i n

A mpe re s/ se c o nd )9 M = e _ 2 1 / I 1 ; // i n he nr y10 L 1 = 2 5 * 1 0 ^ - 3 ; / / i n h e n ry11 L 2 = 2 5 * 1 0 ^ - 3 ; / / i n h e n ry12 // R e fe r t o eqn 2 . 3 213 k = ( M / L 1 ) * 1 0 0 ; // c o e f f i c i e n t o f co up li ng14 disp (k , p e r c e n t a g e (%)= )

Scilab code Exa 2.11 Find the Inductance of each coil mutual inductanceand coefficient of coupling

1 / / C ap ti on : Fi nd ( a ) I n d u ct a n ce o f e ac h c o i l ( b ) m ut ua li n du c t an c e ( c ) c o e f f i c i e n t of c o up l i ng

2 / / Exa : 2 . 1 13 clc ;

4 clear ;

5 close ;

6 / /L1 , L2=i n d u c t a n c e s o f c o i l 1&27 / /M=m u tu al i n d u c t a n c e b /w c o i l 1&2

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8 L _ a i d = 2 . 3 8 ; // e f f e c t i v e i n d uc t an c e when c on n ec te d i n

s e r e i s a i d i n g9 L _ o p p = 1 . 0 2 ; // e f f e c t i v e i n d uc t an c e when c on n ec te d i ns e r e i s o pp os in g

10 // L1+L2+2M=L a i d11 //L1+L22M=L opp12 M = ( L _ a i d - L _ o p p ) / 4 ; / / i n h e n ry13 disp (M , m ut ua l i n d u c t a n c e ( i n h e n ry )= )14 //L1=16L215 L1=( L_aid -2*M) /17; / / i n h e n ry16 disp ( L 1 , i n d uc t an c e o f c o i l 1 ( i n h e n ry )= )17 L2=L_aid -(2*M) -L1; / / i n h e n ry

18 disp ( L 2 , i n d uc t an c e o f c o i l 2 ( i n h e n ry )= )19 k = M / ( sqrt ( L 1 * L 2 ) ) ;

20 disp (k , c o e f f i c i e n t of co up li n g= )

Scilab code Exa 2.12 Find effective inductance when connected in parallelaiding and parallel opposing

1 / / C ap ti on : Fi nd e f f e c t i v e i n d u c t a nc e when c o n ne c te di n ( a ) p a r a l l e l a i d i n g ( b) p a r a l l e l o pp os in g

2 / / Exa : 2 . 1 23 clc ;

4 clear ;

5 close ;

6 L 1 = 1 . 6 ; // s e l f i n d uc t an c e o f c o i l 1 ( i n Henry )7 L 2 = 0 . 1 ;

// s e l f i n d uc t an c e o f c o i l 2 ( i n Henry )8 M = 0 . 3 4 ; / / m ut ua l i n d u c t a n c e ( i n H en ry )9 / / R e f e r t o eqn 2. 4510 L _ a i d = ( ( L 1 * L 2 ) - M ^ 2 ) * 1 0 ^ 3 / ( L 1 + L 2 - ( 2 * M ) ) ; // i n m i l i

Henry

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11 disp ( L _ a i d , e f f e c t i v e i nd uc ta nc e i n p a r a l l e l a i d i n g

( i n m i li

H e nr y ) = )12 / / R e f e r t o eqn 2. 4613 L _ o p p = ( ( L 1 * L 2 ) - M ^ 2 ) * 1 0 ^ 3 / ( L 1 + L 2 + ( 2 * M ) ) ; // i n m i l i

h e n r y14 disp ( L _ o p p , e f f e c t i v e i nd uc ta nc e i n p a r a l l e l

o pp os in g ( i n mi niH e nr y ) = )

Scilab code Exa 2.13 Find hysteresis loss and eddy current loss

1 // C ap ti on : Fi nd h y s t e r e s i s l o s s and eddyc u r r e n t l o s s2 / / Exa : 2 . 1 33 clc ;

4 clear ;

5 close ;

6 // r e f e r t o eqn 2. 50

7 // eqn : 2 . 5 1 , 2 . 52 & 2 . 5 3 a r e o b ta i ne d8 f = [ 25 25 6 0] ; / / i n h e r t z9 disp (f , f r e q u e n c y ( i n h e r t z )= ) ;10 B _m = [ 1 .1 1 .5 1 .1 ];

11 P _m = [ 0 .4 0 .8 1 .2 ];

12 / /On s o l v i n g e qn : 2 . 5 1 & e q n : 2 . 5 313 k _ e = ( 0 . 0 1 6 - 0 . 0 2 ) / ( 3 0 . 2 5 - 7 2 . 6 ) ;

14 / /on s o l v i n g eqn : 2 . 5 1 & e q n : 2 . 5 215 n =( log ( ( 0 . 0 1 6 - ( 3 0 . 2 5 * k _ e ) ) / ( 0 . 0 3 2 - ( 5 6 . 2 5 * k _ e ) ) ) ) / (

log ( 1 . 1 / 1 . 5 ) ) ;

16 k _ h = ( 0 . 0 1 6 - ( 3 0 . 2 5 * k _ e ) ) / 1 . 1 ^ n ;

17 P _ h = k _ h * f . * B _ m ^ n // h y s t e r e s i s l o s s18 disp ( P _ h , H y s t e r e s i s l o s s ( i n Watts )= ) ;19 P _ e d d y = k _ e * ( f ^ 2 ) . * B _ m ^ 2 // eddy c u r r e nt l o s s20 disp ( P _ e d d y , e ddy c u r r e n t l o s s ( i n Watts )= ) ;

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Scilab code Exa 2.14 Find the minimum length of magnet for maintainingmax energy in air gap

1 / / C a p t i o n : F in d t h e minimum l e n g t h o f m ag ne t f o rm a i nt a in i n g max e ne rg y i n a i r gap

2 / / Exa : 2 . 1 43 clc ;

4 clear ;

5 close ;

6 u _ o = 4 * % p i * 1 0 ^ - 7 ; // p e r me a b li t y o f a i r7 u _ r = 5 0 0 ; // p e r m ea b li t y o f s t e e l8 l _ g = 1 * 1 0 ^ - 2 ; // l e ng t h o f a i r gap s e c t i o n ( i n m ete r )9 A _ g = 1 0 * 1 0 ^ - 4 ; / / c r o s s s e c t i o n a l a re a o f a i r gap

s e c t i o n ( i n m et er 2 )10 A _ m = 1 0 * 1 0 ^ - 4 ; / / c r o s s s e c t i o n a l a re a o f magnet

s e c t i o n ( i n m et er 2 )11 A _ s = 1 0 * 1 0 ^ - 4 ; / / c r o s s s e c t i o n a l a r e a o f s t e e l

s e c t i o n s ( i n m et er 2 )12 l _ s = 5 0 * 1 0 ^ - 2 ; // l e ng t h o f s t e e l s e c t i o n ( i n me t e r )13 // R e fe r t o f i g : 2 . 29 ( D e m a g ne t i z a ti o n and e n er g y

p ro du ct c u r ve s o f a magnet )14 H _ m = - 1 4 4 * 1 0 ^ 3 ; // ( i n A mper e /me t e r )15 B _ m = 0 . 2 3 ; // M ag ne ti c f l u x d e n s i t y ( i n T es l a )16 // r e f e r t o eqn : 2 . 5 517 l _ m = ( - 1 * 1 0 0 ) * ( ( ( l _ g * A _ m ) / ( u _ o * A _ g ) ) + ( ( 2 * l _ s * A _ m ) / (

u _ o * u _ r * A _ s ) ) ) * ( B _ m / H _ m ) ; // ( i n c e n t i m e t e r )

18 disp ( l _ m , minimum l e n g t h o f ma gne t ( i n c e n t i m e t e r )= )

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Chapter 3

Principles of Electromechanical

Energy Conversion

Scilab code Exa 3.1 Find the mass of object and energy stored in the feild

1 / / C ap ti on : Fi nd t h e mass o f o b j e c t and e n er g y s t o r e di n t h e f e i l d

2 / / Exa : 3 . 13 clc ;

4 clear ;

5 close ;

6 A = 2 0 * 1 0 ^ - 4 ; // s u r f a c e a re a o f e a c h c ap a ci t o r s p l a t e7 d = 5 * 1 0 ^ - 3 ; // s e p a r a t i o n b et we en t he p l a t e s8 e = ( 1 0 ^ - 9 ) / ( 3 6 * % p i ) ; // p e r me t i vi t y o f a i r9 V = 1 0 * 1 0 ^ 3 ; / / p o t e n t i a l d i f f . b et we en t h e p l a t e s

10 F _ e = ( e * A * V ^ 2 ) / ( 2 * d ^ 2 ) ; // e l e c t r i c f o r c e11 g = 9 . 8 1 ; // a c c e l e r a t i o n due t o g r a v i ty ( i n m et e r /s e c o n d 2 )

12 / / Fo r c o nd t o f b a l a n c i n g e l e c t r i c f o r c e =w e i gh t o f o b j e c t

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13 // F e=mg

14 m = F _ e / g ;15 disp ( m * 1 0 0 0 , m ass o f o b j e c t ( i n grams )= ) ;16 W _ f = ( e * A * V ^ 2 ) / ( 2 * d ) ;

17 disp ( W _ f * 1 0 0 0 0 0 0 , e n er g y s t o r e d i n t h e f e i l d ( i nmi c r oj o u l e s )= )

Scilab code Exa 3.3 Find the energy stored in the magnetic feild

1 / / C ap ti on : Fi nd t h e e n er g y s t o r e d i n t h e m a gn et i cf e i l d

2 / / Exa : 3 . 33 clc ;

4 clear ;

5 close ;

6 / / i =c u r r e n t i n t h e c k t ( i n Amperes )

7 // x= t o t a l f l u x l i n k a g e8 function i = f ( x ) , i = x / ( 6 - ( 2 * x ) ) , e n d f u n c t i o n ;

9 / / R e f er t o eqn : 3 . 1 810 W _ m = intg ( 0 , 2 , f ) ; / / E ne rg y s t o r e d i n m a g ne t i c f e i l d11 disp ( W _ m , E ne rg y s t o r e d i n m a gn et i c f e i l d ( i n J o u l e s

)= )

Scilab code Exa 3.4 Find the current in the coil and energy stored in thesystem

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1 // C ap ti on : Fi nd t he c u r r en t i n t he c o i l and e ne rg y

s t o r e d i n t he s ys te m2 / / Exa : 3 . 43 clc ;

4 clear ;

5 close ;

6 N = 1 0 0 ; // no . o f t ur ns o f c o i l7 A = 1 0 ^ - 4 ; // ar e a8 x = 1 * 1 0 ^ - 2 ; // l e ng t h o f a i r gap9 u _ o = 4 * % p i * 1 0 ^ - 7 ; // p e r me a b li t y o f a i r10 u _ r = 2 0 0 0 ; // p e r m e a b l it y o f m a gn e ti c m a t e r i a l11 D = 7 . 8 5 * 1 0 ^ 3 ; // d e n s i t y o f m a t e r i a l ( i n kg /m 3 )

12 V = 1 1 * 1 0 ^ - 6 ; // v olume o f m a t e r i a l13 m = D * V ; // mass o f m a t e r i a l14 g = 9 . 8 1 ; // a c c e l e r a t i o n due t o g r a v i ty15 // R ef er t o f i g : 3 . 716 R _ o = ( 1 5 . 5 * 1 0 ^ - 2 ) / ( u _ o * u _ r * A ) ; // r e l u c t a n c e o f o u te r

l e g s17 R _ c = ( 5 . 5 * 1 0 ^ - 2 ) / ( u _ o * u _ r * A ) ; // r e l u c t a n c e o f c e n t r a l

l e g18 function y = L ( x ) ; / / i n d u c t a n c e19 y = ( N ^2) / R ( x ) ;

20 e n d f u n c t i o n ;

21 function y = R ( x ) ; // t o t a l r e l u c t a nc e22 y = R _c + R _ g ( x ) + ( 0 .5 * ( R _ o + R _g ( x ) ) ) ;


24 function y = R_g ( x ) ; // r e l u c t a n c e o f a i r gap25 y = x /( u _o * A );


27 x = [0.01 ] ; // P oi nt s o f i n t e r e s t28 t =[ diag ( d e r i v a t i v e ( L , x ) ) ] ; // t =dL/dx ( at x = 0. 01m)29 // s i n c e t


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35 disp ( W _ f * 1 0 ^ 3 , e n er g y s t o r e d i n t h e m a gn e ti c f e i l d

( i n m i l i

J o u l e s )= )

Scilab code Exa 3.5 Find the current in the coil

1 // C ap ti on : Fi nd t he c u r r en t i n t he c o i l2 / / Exa : 3 . 53 clc ;

4 clear ;

5 T = 2 0 ; / / t o r q ue e x e r t e d by s p r i n g ( i n Newtonmet er )6 r = 0 . 2 ; // r a d i u s o f s p r i n g ( i n m et er )7 F _ s = T / r ; // f o r c e e x er t ed by s p r i n g on m ag ne ti c p l a t e8 N = 1 0 0 0 ; // no . o f t ur ns i n c o i l9 u _ o = 4 * % p i * 1 0 ^ - 7 ; // p e r m a b li t y o f a i r10 A = 9 * 1 0 ^ - 4 ; / / a r e a ( i n m e te r 2 )11 function y = L ( x ) ; / / i n d u c t a n c e

12 y = ( N ^2) / R ( x ) ;13 e n d f u n c t i o n ;

14 function y = R ( x ) ; // r e l u c t a n c e o f a i r gap15 y = ( 2* x ) /( u _ o *A ) ;


17 x = [ 0. 00 1 ] ; // P oi nt s o f i n t e r e s t18 t =[ diag ( d e r i v a t i v e ( L , x ) ) ] ; // t =dL/dx ( at x = 0. 001m)19 // s i n c e t


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Scilab code Exa 3.6 Find the frequency of induced emf max value of in-duced emf rms value of induced emf average value of induced emf

1 // C ap ti on : Fi nd t he ( a ) f r e qu e n cy o f i nd uc ed emf ( b )max v a lu e o f i nd uc ed emf ( c ) rms v a lu e o f i nd uc ed

emf ( d ) a ve ra ge v al u e o f i nd uc ed emf 2 / / Exa : 3 . 63 clc ;

4 clear ;

5 close ;

6 N = 1 0 0 ; // no . o f t ur ns i n c o i l7 P = 4 ; / / number o f p o l e s8 N _ m = 1 8 0 0 ; / / r o t o r s p e ed ( i n rpm )9 f l u x _ p = 4 . 5 * 1 0 ^ - 3 ; // f l u x p e r p o l e ( i n Wb)10 f = ( P * N _ m ) / 1 2 0 ; / / R e f e r t o e qn : 3 . 3 0 a

11 disp (f , ( a ) f r e q u e n c y o f i n du c ed emf ( i n H er tz )= ) ;12 // r e f e r t o eqn : 3 . 3 113 E _ m = ( 2 * % p i * P * f l u x _ p * N _ m ) / 1 2 0 ; // max v a l u e o f i n du c ed

emf p e r t u rn14 E _ m c = N * E _ m ;

15 disp ( E _ m c , ( b ) max v al u e o f i nd uc ed emf i n c o i l ( i nV o l t s )= ) ;

16 E _ r m s = E _ m c / sqrt ( 2 ) ;

17 disp ( E _ r m s , ( c ) rms v a l u e o f i n du c ed emf ( i n V o l t s )= ) ;

18 E _ a v g = ( 2 * E _ m c ) / % p i ;

19 disp ( E _ a v g , ( d ) a v er a ge v a lu e o f i nd uc ed emf ( i nV o l t s )= )

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Scilab code Exa 3.7 Find the synchronous speed and percent slip of themotor

1 / / C a pt i on : F in d t h e s y n c h ro n o u s s p e ed a nd p e r c e n ts l i p o f t he motor

2 / / Exa : 3 . 73 clc ;

4 clear ;

5 close ;

6 P = 4 ; // no . o f p o l e7 f = 5 0 ; / / f r e q u e n c y ( i n Hz )8 N _ r = 1 2 0 0 ; / / s p e ed o f r o t o r ( i n rpm )9 N _ s = ( 1 2 0 * f ) / P ;

10 disp ( N _ s , s y n c h r o n o u s s p e e d ( i n rpm ) = ) ;11 s = ( N _ s - N _ r ) / N _ s ; / / s l i p

12 s _ p = s * 1 0 0 ;13 disp ( s _ p , p e r c e n t s l i p o f t h e m oto r (%)= )

Scilab code Exa 3.8 Find the rotor speed and average torque developed

by motor

1 / / C ap ti on : Fi nd t h e r o t o r s p ee d and a v e r ag e t o r q ued e v e l o p e d by m ot or

2 / / Exa : 3 . 8

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3 clc ;

4 clear ;5 close ;

6 N = 2 ; // no . o f p o l e s7 f = 6 0 ; / / f r e q u e n c y i n Hz8 I _ r m s = 1 0 ; // c u r r e n t i n t a k e9 L _ q = 1 ; / / min i n d u c t an c e ( i n H)10 L _ d = 2 ; / / max i n d u c t a n c e ( i nH )11 w = 2 * % p i * f ;

12 disp (w , r o t o r s p e e d ( i n r a d / s e c )= ) ;13 / / R e f er t o eqn : 3 . 5 214 T _ a v g = ( - 1 ) * 0 . 1 2 5 * ( L _ d - L _ q ) * ( ( I _ r m s * sqrt ( 2 ) ) ^ 2 ) * s i n d

( 2 * 4 5 ) ;15 if ( T_avg


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9 A = 2 . 2 5 * 1 0 ^ - 4 ; // a r ea o f a i r gap ( i n m 2)

10 x = 0 . 0 0 2 ; // l e n g th o f a i r gap ( i n m)11 i = I * 1 . 5 0 ; // min c u r r e nt n ee d ed f o r a c t i v a t i n g r e l a y12 F _ m = u _ o * A * 0 . 5 * ( ( N * i ) / x ) ^ 2 ; // R e fe r t o eqn 3 . 5 313 disp ( F _ m , r e s t r a i n i n g f o r c e o f t he s p r i n g ( i n Newton )

= )

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Chapter 4

Transformers

Scilab code Exa 4.2 Find the a ratio and current in primary and the powersupplied to load and the flux in the core

1 / / C a pt i on : F in d t h e ( a ) ar a t i o ( b ) c ur r e nt i n

p ri ma ry ( c ) t he power s u p p l i e d t o l o ad ( d ) andt h e f l ux i n t h e c or e

2 / / Exa : 4 . 23 clc ;

4 clear ;

5 close ;

6 N _ p = 1 5 0 ; // no . o f t u rn s i n p ri ma ry w in di ng7 N _ s = 7 5 0 ; // no . o f t u rn s i n s e co n da r y w in di ng8 f = 5 0 ; / / f r e q u e n c y i n Hz9 I _ 2 = 4 ; / / l o a d c u r r e n t ( i n Am peres )

10 V _ 1 = 2 4 0 ; // v o l t a g e on p ri ma ry s i d e ( i n V o lt s )11 p f = 0 . 8 ; // p ow er f a c t o r12 a = N _ p / N _ s ;

13 disp (a , ( a ) ar a t i o = ) ;14 I _ 1 = I _ 2 / a ;

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15 disp ( I _ 1 , ( b ) c u r r e n t i n p ri m ar y ( i n Amperes )= ) ;

16 V _ 2 = V _ 1 / a ;17 disp ( V _ 2 , ( c ) v o l t a g e on s e c on d a r y s i d e ( i n V o l t s )= ) ;

18 P _ L = V _ 2 * I _ 2 * p f ;

19 disp ( P _ L , ( d ) power s u p p l i e d t o t he l o ad ( i n Watts )= ) ;

20 f l u x = V _ 1 / ( 4 . 4 4 * f * N _ p ) ;

21 disp ( f l u x * 1 0 ^ 3 , ( e ) f l u x i n t he c o re ( i n m i li Weber)= ) ;

Scilab code Exa 4.3 Find the efficiency of transformer

1 // C ap ti on : Fi nd t he e f f i c i e n c y o f t r a n s fo r m e r2 / / Exa : 4 . 33 clc ;

4 clear ;5 close ;

6 R _ 1 = 4 ; // i n ohms7 R _ 2 = 0 . 0 4 ; // in ohms8 X _ 1 = 1 2 ; // i n ohms9 X _ 2 = 0 . 1 2 ; // in ohms10 p f = 0 . 8 6 6 ; // p owe r f a c t o r11 V _ p = 2 3 0 0 ; // p ri ma ry v o l t a g e i n v o l t s12 V _ s = 2 3 0 ; // S ec on da ry v o l t a g e i n v o l t s13 S = 2 3 0 0 0 ; //VA14 t h e t a = a c o s d ( p f ) ;

15 I _ 2 = ( S * 0 . 7 5 / V _ s ) * ( c o s d ( t h e t a ) + % i * s i n d ( t h e t a ) ) ; //s e c on d a r y c u r r e n t ( i n Amperes )

16 Z _ 2 = R _ 2 + % i * X _ 2 ; / / s e c o n d a r y w i nd i n g i mp ed an ce ( i nohms)

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17 E _ 2 = V _ s + I _ 2 * Z _ 2 ; / / i n du c ed emf i n s e c on d a r y w in d in g (

i n V o l t s )18 a = V _ p / V _ s ; // t r a n s f o rm a t i o n r a t i o19 E _ 1 = a * E _ 2 ; / / i n du c ed emf i n p ri ma ry w in d in g ( i n V o l t s

)20 I _ 1 = I _ 2 / a ; / / c u r r e n t i n p ri ma ry w i nd i ng21 Z _ 1 = R _ 1 + % i * X _ 1 ; / / p r i ma r y w i nd i n g i mp ed an ce ( i n ohms )22 V _ 1 = E _ 1 + I _ 1 * Z _ 1 ; / / s o u r c e v o l t a g e23 P _ o = real ( V _ s * conj ( I _ 2 ) ) ; / / o u t p u t p o we r ( i n W at ts )24 P _ i n = real ( V _ 1 * conj ( I _ 1 ) ) ; / / i n p u t p o we r25 E f f = P _ o / P _ i n ;

26 disp ( E f f * 1 0 0 , E f f i c i e n c y (%)= ) ;

Scilab code Exa 4.4 Find the efficiency of transformer

1 // C ap ti on : Fi nd t he e f f i c i e n c y

2 / / Exa : 4 . 43 clc ;

4 clear ;

5 close ;

6 //From E xa : 4 . 37 V _ 2 = 2 3 0 ; / / i n V o l t s8 Z _ 1 = 4 + % i * 1 2 ;

9 I _ s = 7 5 * ( c o s d ( 3 0 ) + % i * s i n d ( 3 0 ) ) ; // i n A mpe re s10 a = 1 0 ; // t r a n s f o rm a t i o n r a t i o11 E _ 1 = 2 2 8 2 . 8 7 * ( c o s d ( 2 . 3 3 ) + % i * s i n d ( 2 . 3 3 ) ) ; / / i n V o l t s12 E _ 2 = 2 2 8 . 2 8 7 * ( c o s d ( 2 . 3 3 ) + % i * s i n d ( 2 . 3 3 ) ) ;

/ / i n V o l t s13 I _ p = 7 . 5 * ( c o s d ( 3 0 ) + % i * s i n d ( 3 0 ) ) ; // i n A mpe re s14 P _ o = 1 4 9 3 8 . 9 4 ; // i n Watt s15 R _ c 1 = 2 0 0 0 0 ; // c o r e l o s s r e s i s t a n c e on p ri m a ry s i d e16 X _ m 1 = 1 5 0 0 0 ; / / m a g n et i z in g r e a c t a n c e on p ri ma ry s i d e

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17 I _ c = E _ 1 / R _ c 1 ; // i n A mpe re s

18 I _ m = E _ 1 / ( % i * X _ m 1 ) ; // i n A mpe re s19 I _ p h y = I _ c + I _ m ; // in Amperes20 I _ 1 = I _ p + I _ p h y ; // in Amperes21 V _ 1 = E _ 1 + Z _ 1 * I _ 1 ; / / i n V o l t s22 P _ i n = real ( V _ 1 * conj ( I _ 1 ) ) ; // i n Watt s23 E f f = P _ o / P _ i n ;

24 disp ( E f f * 1 0 0 , E f f i c i e n c y (%)= )

Scilab code Exa 4.6 Find efficiency and voltage regulation of transformer

1 // C ap ti on : Fi nd e f f i c i e n c y and v o l t a g e r e g u l a t i o n o f t r a n s f o r m e r

2 / / Exa : 4 . 63 clc ;

4 clear ;

5 close ;6 S = 2 2 0 0 ; / / V o l tAmpere7 V _ s = 2 2 0 ; // s e co n da r y s i d e v o l t a g e ( i n V o l ts )8 V _ 2 = V _ s ;

9 V _ p = 4 4 0 ; // p r im ar y s i d e v o l t a g e ( i n V o lt s )10 R _ e 1 = 3 ; // i n ohms11 X _ e 1 = 4 ; // i n ohms12 R _ c 1 = 2 . 5 * 1 0 0 0 ; // in ohms13 X _ m 1 = 2 0 0 0 ; // in ohms14 a = V _ p / V _ 2 ; // t r a n s f o rm a t i o n r a t i o15 p f = 0 . 7 0 7 ;

// l a g g i n g power f a c t o r16 t h e t a = - a c o s d ( p f ) ;17 I _ 2 = ( S / V _ 2 ) * ( c o s d ( t h e t a ) + % i * s i n d ( t h e t a ) ) ; / / ( i n

Amperes)18 // R ef er t o e q ui v a l e n t c i r c u i t ( f i g : 4 . 1 6 )

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19 I _ p = I _ 2 / a ; // i n A mpe re s

20 V _ 2 p = a * V _ 2 ;21 V _ 1 = V _ 2 p + I _ p * ( R _ e 1 + % i * X _ e 1 ) ;

22 I _ c = V _ 1 / R _ c 1 ; // c o r e l o s s c u r r e nt ( i n Amperes )23 I _ m = V _ 1 / ( % i * X _ m 1 ) ;

24 I _ 1 = I _ p + I _ c + I _ m ; // c u r r e n t s u p p l i e d by s o u r ce ( i nAmperes)

25 P _ o = real ( V _ p * conj ( I _ p ) ) ; / / o u t pu t p ow er ( i n Wa tts )26 P _ i n = real ( V _ 1 * conj ( I _ 1 ) ) ; / / i n p u t p ow er ( i n W at ts )27 E f f = P _ o / P _ i n ; // E f f i c i e n c y28 disp ( E f f * 1 0 0 , E f f i c i e n c y (%)= ) ;29 V R = ( abs ( V _ 1 ) -abs ( V _ p ) ) / V _ p ;

30 disp ( V R * 1 0 0 , v o l t a g e r e g u l a t i o n (%)= )

Scilab code Exa 4.7 Find the KVA rating at max efficiency

1 / / C ap ti on : Fi nd ( a )KVA r a t i n g a t max e f f i c i e n c y ( b )max e f f i c i e n c y ( c ) E f f i c i e n c y a t f u l l l oa d and0 . 8 p f l a g g i n g ( d ) e q u i v a l e nt c o r e r e s i s t a n c e

2 / / Exa : 4 . 73 clc ;

4 clear ;

5 close ;

6 S = 1 2 0 0 0 0 ; / / V o l tAmpere7 V _ p = 2 4 0 0 ; // i n v o l t s8 V _ s = 2 4 0 ; // i n v o l t s9 R _ 1 = 0 . 7 5 ;

// in ohms10 R _ 2 = 0 . 0 1 ; // in ohms11 X _ 1 = 0 . 8 ; // i n ohms12 X _ 2 = 0 . 0 2 ; // in ohms13 p f = 0 . 8 ; / / l a g g i n g

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14 t h e t a = - a c o s d ( p f ) ;

15 a = V _ p / V _ s ; // t r a n s f o rm a t i o n r a t i o16 I _ p = S / V _ p ; / / r a t e d l o a d c u r r e n t ( i n Amperes )17 I _ p _ e t a = 0 . 7 * I _ p ; // l oa d c u r r e n t a t max e f f i c i e n c y18 K V A = I _ p _ e t a * V _ p / 1 0 0 0 ;

19 disp ( K V A , ( a ) KVA r a t i n g a t max e f f i c i e n c y = ) ;20 P _ c u _ e t a = I _ p _ e t a ^ 2 * ( R _ 1 + a ^ 2 * R _ 2 ) ; / / c o pp e r l o s s ( i n

Watts)21 P _ m = P _ c u _ e t a ; // c o r e l o s s22 P _ o = V _ p * I _ p _ e t a * p f ; // p ower o ut pu t a t max e f f i c i e n c y23 P _ i n = P _ o + P _ m + P _ c u _ e t a ; / / p ow er i n p u t a t max

e f f i c i e n c y

24 e t a = P _ o / P _ i n ;25 disp ( e t a * 1 0 0 , ( b ) max e f f i c i e n c y (%)= ) ;26 P _ o _ F L = V _ p * I _ p * p f ; / / po wer o u tp ut a t f u l l l o a d27 P _ c u _ F L = I _ p ^ 2 * ( R _ 1 + a ^ 2 * R _ 2 ) ; / / c o pp e r l o s s a t f u l l

l o a d28 P _ i n _ F L = P _ c u _ F L + P _ o _ F L + P _ m ;

29 E f f = P _ o _ F L / P _ i n _ F L ;

30 disp ( E f f * 1 0 0 , ( c ) E f f i c i e n c y a t f u l l l o a d (%)= ) ;31 R _ c 1 = V _ p ^ 2 / P _ c u _ e t a ;

32 disp ( R _ c 1 , ( d ) e q u i v a l e n t c o re r e s i s t a n c e ( i n ohms )=

) ;

Scilab code Exa 4.9 Find the generator voltage generator current and ef-ficiency

1 // C ap ti on : Fi nd t he ( a ) g e n e r a t or v o l t a g e ( b )g e n e r a to r c u r r e n t ( c ) e f f i c i e n c y

2 / / Exa : 4 . 93 clc ;

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4 clear ;

5 close ;6 // R ef er t o f i g : 4 . 2 97 / / For r e g i o n A8 V _ b A = 2 3 0 ; / / i n V o l t s9 S _ b A = . 4 6 0 0 0 ; / / V o l tAmpere10 I _ b A = S _ b A / V _ b A ; // i n A mpe re s11 Z _ b A = V _ b A / I _ b A ; // in ohms12 Z _ g _ p u = ( 0 . 0 2 3 + % i * 0 . 0 9 2 ) / Z _ b A ;

13 R _ L _ p u = 0 . 0 2 3 / Z _ b A ;

14 X _ L _ p u = 0 . 0 6 9 / Z _ b A ;

15 / / For r e g i o n B

16 / / Pe r u n i t p a r am e te r s on h ig hv o l t a g e s i d e o f t h es t e p up t r a n s f o r m e r

17 V _ b B = 2 3 0 0 ; / / i n V o l t s18 S _ b B = 4 6 0 0 0 ; / / V o l tAmpere19 I _ b B = S _ b B / V _ b B ; // i n A mpe re s20 Z _ b B = V _ b B / I _ b B ; // in ohms21 R _ H _ p u = 2 . 3 / Z _ b B ;

22 X _ H _ p u = 6 . 9 / Z _ b B ;

23 R _ c H _ p u 1 = 1 3 8 0 0 / Z _ b B ;

24 X _ m H _ p u 1 = 6 9 0 0 / Z _ b B ;

25 Z _ l _ p u = ( 2 . 0 7 + % i * 4 . 1 4 ) / Z _ b B ;//P e r

u n i t i mp ed an ce o f t r an s m i s s i o n l i n e26 / / Pe r u n i t p a r am e te r s on h ig hv o l t a g e s i d e o f t h e

s t e p down t r a n s f o r m e r27 X _ m H _ p u 2 = 9 2 0 0 / Z _ b B ;

28 R _ c H _ p u 2 = 1 1 5 0 0 / Z _ b B ;

29 / / For r e g i o n C30 V _ b C = 1 1 5 ; / / i n V o l t s31 S _ b C = 4 6 0 0 0 ; / / V o l tAmpere32 I _ b C = S _ b C / V _ b C ; // i n A mpe re s33 Z _ b C = V _ b C / I _ b C ; // in ohms

34 R _ L _ p u = 0 . 0 0 5 7 5 / Z _ b C ;35 X _ L _ p u = 0 . 0 1 7 2 5 / Z _ b C ;

36 V _ L _ p u = 1 * ( c o s d ( 0 ) + % i * s i n d ( 0 ) ) ;

37 I _ L _ p u = 1 * ( c o s d ( - 3 0 ) + % i * s i n d ( - 3 0 ) ) ;

38 E _ l _ p u = V _ L _ p u + ( R _ L _ p u + % i * X _ L _ p u ) * I _ L _ p u ;

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39 I _ l _ p u = I _ L _ p u + E _ l _ p u * ( 0 . 0 1 - % i * ( 1 / 8 0 ) ) ;

40 E _ g _ p u = E _ l _ p u + I _ l _ p u * ( 0 . 0 2 + % i * 0 . 0 6 + 0 . 0 1 8 + % i* 0 . 0 3 6 + 0 . 0 2 + % i * 0 . 0 6 ) ;

41 I _ g _ p u = I _ l _ p u + E _ g _ p u * ( ( 1 / 1 2 0 ) - % i * ( 1 / 6 0 ) ) ;

42 V _ g _ p u = E _ g _ p u + I _ g _ p u * ( 0 . 0 2 + 0 . 0 2 + % i * 0 . 0 8 + % i * 0 . 0 6 ) ;

43 V _ g = V _ b A * V _ g _ p u ;

44 disp ( abs ( V _ g ) , ( a ) G e ne r at o r V o l ta g e ( i n V o l t s )= ) ;45 disp ( a t a n d ( imag ( V _ g ) / real ( V _ g ) ) , P ha se o f g e n e r a t ed

v o l t a g e ( i n d e g r e e )= ) ;46 I _ g = I _ b A * I _ g _ p u ;

47 disp ( abs ( I _ g ) , ( b ) G e n e ra t o r c u r r e n t ( i n A mperes ) = );

48 disp ( a t a n d ( imag ( I _ g ) / real ( I _ g ) ) , P ha se o f g e n e r a t o rc u r r e n t ( i n d e g r e e )= ) ;

49 P _ o _ p u = 0 . 8 6 6 ; / / r a t e d p ow er o u tp u t a t p f = 0. 86 6l a g g i n g

50 P _ i n _ p u = real ( V _ g _ p u * conj ( I _ g _ p u ) ) ;

51 E f f = P _ o _ p u / P _ i n _ p u ;

52 disp ( E f f * 1 0 0 , ( c ) E f f i c i e n c y (%)= ) ;

Scilab code Exa 4.10 Find the primary winding voltage secondary wind-ing voltage ratio of transformation and nominal rating of transformer

1 / / C ap ti on : Fi nd ( a ) p ri ma ry w in d in g v o l t a g e ( b )s ec on da r y w i nd in g v o l t a g e ( c ) r a t i o o f t r a ns f o rm a t i o n ( d ) n omi na l r a t i n g o f t r a ns f or m e r

2 / / Exa : 4 . 1 03 clc ;

4 clear ;

5 close ;

6 V _ 1 = 2 4 0 0 ; / / i n V o l t s

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7 V _ 2 = 2 4 0 ; / / i n V o l t s

8 S _ o = 2 4 * 1 0 0 0 ; / / V o l t

Ampere9 I _ 1 = 1 0 ; // i n A mpe re s10 I _ 2 = 1 0 0 ; // i n A mpe re s11 // R ef er t o f i g : 4 . 3 1 ( a )12 V _ 1 a = V _ 1 + V _ 2 ;

13 V _ 2 a = V _ 2 ;

14 a _ T 1 = V _ 1 a / V _ 2 a ;

15 a _ T 2 = V _ 2 a / V _ 1 a ;

16 a _ T 3 = V _ 1 a / V _ 1 ;

17 a _ T 4 = V _ 1 / V _ 1 a ;

18 S _ o a _ 1 = V _ 1 a * I _ 1 ;

19 S _ o a _ 2 = V _ 1 a * I _ 1 ;20 S _ o a _ 3 = V _ 1 a * I _ 2 ;

21 S _ o a _ 4 = V _ 1 a * I _ 2 ;

22 disp ( R e fe r t o f i g : 4 . 3 1 a ) ;23 disp ( V _ 1 a , ( a ) p ri m ar y w in d in g v o l t a g e ( i n V o l t s )= )

;

24 disp ( V _ 2 a , ( b ) s e c on d a r y w in d in g v o l t a g e ( i n V o l t s )= ) ;

25 disp ( a _ T 1 , ( c ) r a t i o o f t r a n s f o rm a t i o n= ) ;26 disp ( S _ o a _ 1 / 1 0 0 0 , ( d ) n om in al r a t i n g o f t r an s f o r m er

(KVA)= ) ;

27 disp ( R e fe r t o f i g : 4 . 3 1 b ) ;28 disp ( V _ 2 a , ( a ) p ri m ar y w in d in g v o l t a g e ( i n V o l t s )= )

;

29 disp ( V _ 1 a , ( b ) s e c on d a r y w in d in g v o l t a g e ( i n V o l t s )= ) ;


(KVA)= ) ;32 disp ( R e fe r t o f i g : 4 . 3 1 c ) ;33 disp ( V _ 1 a , ( a ) p ri m ar y w in d in g v o l t a g e ( i n V o l t s )= )

;34 disp ( V _ 1 , ( b ) s e c on d a ry w in d in g v o l t a g e ( i n V o l t s )=

) ;


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(KVA)= ) ;

37 disp ( R e fe r t o f i g : 4 . 3 1 d ) ;38 disp ( V _ 1 , ( a ) p ri ma ry w in d in g v o l t a g e ( i n V o l t s )= ) ;39 disp ( V _ 1 a , ( b ) s e c on d a r y w in d in g v o l t a g e ( i n V o l t s )=

) ;40 disp ( a _ T 4 , ( c ) r a t i o o f t r a n s f o rm a t i o n= ) ;41 disp ( S _ o a _ 4 / 1 0 0 0 , ( d ) n om in al r a t i n g o f t r an s f o r m er

(KVA)= ) ;

Scilab code Exa 4.11 Find the efficiency and voltage regulation

1 // C ap ti on : Fi nd t he e f f i c i e n c y and v o l t a g e r e g u l a t i o n2 / / Exa : 4 . 1 13 clc ;

4 clear ;

5 close ;

6 V _ 2 a = 4 8 0 ; // i n v o l t s7 p f = 0 . 7 0 7 ; / / l e a d i n g8 t h e t a = a c o s d ( p f ) ;

9 a _ T = 1 2 0 / 4 8 0 ; // r a t i o o f t r a ns f o r m a t io n o f s te pupt r a n s f o r m e r

10 a = 3 6 0 / 1 2 0 ; // r a t i o o f t r a ns f o rm a t io n o f twow i n d i n gt r a n s f o r m e r

11 R _ c H = 8 . 6 4 * 1 0 0 0 ; // in ohms12 R _ H = 1 8 . 9 ; // in ohms13 X _ H = 2 1 . 6 ; // in ohms14 X _ L = 2 . 4 ;

// i n ohms15 R _ L = 2 . 1 ; // i n ohms16 X _ m H = 6 . 8 4 * 1 0 0 0 ; // in ohms17 R _ c L = R _ c H / a ^ 2 ; // e q ui v a l e n t c or e l o s s r e s i s t a n c e i n

ohms

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18 X _ m L = X _ m H / a ^ 2 ; / / m a g n e t i z i n g r e a c t a n c e

19 I _ 2 a = ( 7 2 0 / 3 6 0 ) * ( c o s d ( t h e t a ) + % i * s i n d ( t h e t a ) ) ;20 I _ H = I _ 2 a ;

21 I _ p a = I _ 2 a / a _ T ;

22 I _ c o m = I _ p a - I _ 2 a ; / / c u r r e n t t h r o u g h common w i n d i n g ( i nAmperes)

23 / / on a p p l y i n g KVL t o t h e o u tp u t l o o p24 E _ L = ( I _ 2 a * ( R _ H + % i * X _ H ) + V _ 2 a - I _ c o m * ( R _ L + % i * X _ L ) ) / 4 ;

25 V _ 1 a = E _ L + I _ c o m * ( R _ L + % i * X _ L ) ;

26 I _ c a = V _ 1 a / R _ c L ; // c o r e l o s s c u r r e nt i n Amperes27 I _ m a = - % i * V _ 1 a / X _ m L ; / / m a g n e t i z i n g c u r r e n t i n Amp eres28 I _ p h y _ a = I _ c a + I _ m a ; // e x c i t a t i o n c u r re n t

29 I _ 1 a = I _ p a + I _ p h y _ a ;30 P _ o = real ( V _ 2 a * conj ( I _ 2 a ) ) ;

31 P _ i n = real ( V _ 1 a * conj ( I _ 1 a ) ) ;

32 E f f = P _ o / P _ i n ;

33 disp ( E f f * 1 0 0 , E f f i c i e n c y (%)= ) ;34 V _ 2 a n L = V _ 1 a / a _ T ; // no l oa d v o l t a g e35 V R = ( abs ( V _ 2 a n L ) - V _ 2 a ) / V _ 2 a ;

36 disp ( V R * 1 0 0 , V o l ta g e r e g u l a t i o n (%)= ) ;

Scilab code Exa 4.13 Find the line voltages and line currents and effi-ciency of the transformer

1 // C ap ti on : Fi nd t he l i n e v o l t a ge s , t he l i n e c u r r e n t sand e f f i c i e n c y o f t he t r a n sf o r m e r

2 / / Exa : 4 . 1 33 clc ;

4 clear ;

5 close ;

6 R _ H = 1 3 3 . 5 * 1 0 ^ - 3 ; // i n ohms

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7 X _ H = 2 0 1 * 1 0 ^ - 3 ; // in ohms

8 R _ L = 3 9 . 5 * 1 0 ^ - 3 ; // in ohms9 X _ L = 6 1 . 5 * 1 0 ^ - 3 ; // in ohms10 R _ c L = 2 4 0 ; // in ohms11 X _ m L = 2 9 0 ; // in ohms12 p f = 0 . 8 ; / / l a g g i n g13 t h e t a = - a c o s d ( p f ) ;

14 V _ 2 n = 1 3 8 . 5 6 4 * ( c o s d ( 0 ) + % i * s i n d ( 0 ) ) ; // r a t e d l o a dv o l t a g e f o r Y/Y c o n ne c t io n

15 I _ 2 A = 8 6 . 6 * ( c o s d ( t h e t a ) + % i * s i n d ( t h e t a ) ) ; // l o adc u r r e n t

16 a = 1 2 0 / 1 3 8 . 5 6 4 ; // t r a n s f o r ma t i o n r a t i o

17 I _ p A = ( I _ 2 A / a ) * ( c o s d ( 3 0 ) + % i * s i n d ( 3 0 ) ) ; / / p e r p h a s ec u r r e nt i n p ri ma ry w in di ng

18 E _ 2 n = V _ 2 n + I _ 2 A * ( 0 . 0 4 4 5 + % i * 0 . 0 6 7 ) ; / / v o l t a g e i n d u ce di n s e c on d a r y w in d in g

19 E _ 2 L = sqrt ( 3 ) * E _ 2 n * ( c o s d ( 3 0 ) + % i * s i n d ( 3 0 ) ) ;

20 E _ 1 n = a * E _ 2 n * ( c o s d ( 3 0 ) + % i * s i n d ( 3 0 ) ) ; / / v o l t a g e i n d u ce di n p ri ma ry w in di n g

21 I _ 1 A = I _ p A + E _ 1 n * ( ( 1 / 2 4 0 ) - % i * ( 1 / 2 9 0 ) ) ;

22 disp ( abs ( I _ 2 A ) , L in e c u r re n t i n s ec on da r y s i d e ( i nAmperes)= ) ;

23 disp ( a t a n d ( imag ( I _ 2 A ) / real ( I _ 2 A ) ) , p ha se a n g le o f i nd uc ed l i n e c u r r e nt i n s e co n da r y ( i n D eg re e )= ) ;

24 disp ( abs ( I _ 1 A ) , L in e c u r re n t i n p ri ma ry s i d e ( i nAmperes)= ) ;

25 disp ( a t a n d ( imag ( I _ 1 A ) / real ( I _ 1 A ) ) , p ha se a n g le o f i nd uc ed l i n e c u r r e nt i n p ri ma ry ( i n D eg re e ) = ) ;

26 disp ( abs ( E _ 2 L ) , L in e v o l t a g e i nd uc ed i n s ec on da rys i d e ( i n V o l t s )= ) ;

27 disp ( a t a n d ( imag ( E _ 2 L ) / real ( E _ 2 L ) ) , p ha se a n g le o f i nd uc ed l i n e v o l t a g e i n s e co n da r y ( i n D eg re e )= ) ;

28 V _ 1 n = E _ 1 n + I _ 1 A * ( R _ L + % i * X _ L ) ;

29 V _ 1 L = sqrt ( 3 ) * V _ 1 n * ( c o s d ( 3 0 ) + % i * s i n d ( 3 0 ) ) ;30 disp ( abs ( V _ 1 L ) , L in e v o l t a g e i nd uc ed i n p ri ma ry

s i d e ( i n V o l t s )= ) ;31 disp ( a t a n d ( imag ( V _ 1 L ) / real ( V _ 1 L ) ) , p ha se a n g le o f

i nd uc ed l i n e v o l t a g e i n p ri ma ry ( i n D eg re e )= ) ;

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32 P _ o = 3 * real ( 1 3 8 . 5 6 4 * conj ( I _ 2 A ) ) ;

33 P _ i n = 3 * real ( V _ 1 n * conj ( I _ 1 A ) ) ;34 E f f = P _ o / P _ i n ;

35 disp ( E f f * 1 0 0 , E f f i c i e n c y (%)= ) ;

Scilab code Exa 4.14 Find the line current line voltage and power

1 // C ap ti on : Fi nd t he l i n e c u rr e nt , l i n e v o l t a g e andpower

2 / / Exa : 4 . 1 43 clc ;

4 clear ;

5 close ;

6 I _ L = 4 * 8 0 / 5 ;

7 disp ( I _ L , L i n e c u r r e n t ( i n Amp eres ) = ) ;8 V _ L = 1 1 0 * 1 0 0 / 1 ;

9 disp ( V _ L , L i ne v o l t a g e ( i n V o l t s )= ) ;10 P = ( 1 0 0 / 1 ) * ( 8 0 / 5 ) * 3 5 2 ;

11 disp (P , Power on t he t r a n s m i s s i o n l i n e ( i n Watts )= );

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Chapter 5

Direct Current Generators

Scilab code Exa 5.1 Find the coil pitch for 2 pole winding and 4 polewinding

1 // C ap ti on : Fi nd t he c o i l p i t c h f o r ( a ) 2p o l e w in d in g

( b ) 4p o l e w in d in g2 / / Exa : 5 . 13 clc ;

4 clear ;

5 close ;

6 P 1 = 2 ;

7 P 2 = 4 ;

8 S = 1 0 ; // no . o f s l o t s9 S _ p 1 = S / P 1 ; // s l o t s p er p o l e10 y1 = int ( S _ p 1 ) ; // c o i l p i t ch i n s l o t s

11 S _ s 1 = 1 8 0 / S _ p 1 ; // s l o t s pa n12 C _ p 1 = S _ s 1 * y 1 ; // co i l pi tc h ( e l e c t r i c a l )13 disp ( C _ p 1 , c o i l p i t ch f o r 2po le win din g ( e l e c t r i c a l

)= ) ;14 S _ p 2 = S / P 2 ; // s l o t s p er p o l e

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15 S _ s 2 = 1 8 0 / S _ p 2 ; // s l o t s pa n

16 y2 = int ( S _ p 2 ) ; // c o i l p i t ch i n s l o t s17 C _ p 2 = S _ s 2 * y 2 ; // co i l pi tc h ( e l e c t r i c a l )18 disp ( C _ p 2 , c o i l p i t ch f o r 4po l e wi nd in g ( e l e c t r i c a l )

= ) ;

Scilab code Exa 5.3 Find the induced emf in the armature winding in-duced emf per coil induced emf per turn induced emf per conductor

1 / / C a pt i on : F in d t h e ( a ) i n d u c ed em f i n t h e a r ma t ur ew in d in g ( b ) i n du c ed emf p e r c o i l ( c ) i n du c ed emf p e r t u rn ( d ) i n du c ed emf p e r c o n du c to r

2 / / Ex : 5 . 33 clc ;

4 clear ;

5 close ;

6 C = 2 4 ; // no . o f c o i l s7 N _ c = 1 8 ; // no . o f t ur ns p er c o i l8 P = 2 ; // no . o f p o l e9 W _ m = 1 8 3 . 2 ; / / a n g u l a r v e l o c i t y ( i n r a d / s e c )10 Z = 2 * C * N _ c ; / / t o t a l a rm at ur e c o n d u c to r s11 a = 2 ; // no . o f p a r a l l e l p at h s12 L = 0 . 2 ; / / e f f e c t i v e l e n g t h o f m ac hi ne ( i n m et er )13 r = 0 . 1 ; / / r a d i u s o f a r ma t ur e ( i n m et er )14 A _ p = ( 2 * % p i * r * L ) / P ; // a c t u a l p o l e a r ea15 A _ e = A _ p * 0 . 8 ; // e f f e c t i v e p o l e a re a16 B = 1 ;

// f l u x d e n s i t y p er p o l e ( i n T es l a )17 P h y = B * A _ e ; // e f f e c t i v e f l u x p e r p ol e18 K _ a = ( Z * P ) / ( 2 * % p i * a ) ; / / m ac hi ne c o n s t a n t19 E _ a = K _ a * P h y * W _ m ;

20 disp ( E _ a , ( a ) i nd uc ed emf i n a rm at ur e w in di ng ( i n

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V o l t s )= ) ;

21 E _ c o i l = E _ a / ( C / a ) ;22 disp ( E _ c o i l , ( b ) i nd uc ed emf p er c o i l ( i n V ol t s )= );

23 E _ t u r n = E _ c o i l / N _ c ;

24 disp ( E _ t u r n , ( c ) i n du c ed emf p e r t u rn ( i n V o l t s )= );

25 E _ c o n d = E _ t u r n / 2 ;

26 disp ( E _ c o n d , ( d ) i nd uc ed emf p er c o nd u ct o r ( i nV o l t s )= ) ;

Scilab code Exa 5.4 Find the current in each conductor the torque devel-oped the power developed

1 // C ap ti on : Fi nd t he ( a ) c u r r e n t i n e ac h c on d uc to r ( b )t h e t o r q u e d e v e l o p e d ( c ) t h e p ow er d e v e l o p e d

2 / / Exa : 5 . 43 clc ;

4 clear ;

5 close ;

6 K _ a = 1 3 7 . 5 1 ; / / R e f er t o e xa : 5 . 37 P h y = 0 . 0 5 ; // f l u x p er p o l e ( R ef er t o ex a : 5 . 3 )8 E _ a = 1 2 5 9 . 6 ; / / i n du c ed emf ( R e f er t o e xa : 5 . 3 )9 I = 2 5 ; / / c u r r e n t i n t h e m ac hi ne ( i n Amp er es )10 a = 2 ; // no . o f p a r a l l e l p at h s11 I _ c o n d = I / a ;

12 disp ( I _ c o n d , ( a ) c u r r e nt i n e a ch c on du ct or ( i nAmperes)= ) ;

13 T _ d = K _ a * P h y * I ;

14 disp ( T _ d , ( b ) t o r q u e d e v e l o p e d by m ac hi ne ( i n Newtonmeter )= ) ;

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15 P _ d = E _ a * I ;

16 disp ( P _ d , ( c ) P owe r d e v e l o p e d ( i n W at ts )= ) ;

Scilab code Exa 5.5 Find induced emf at full load power developed torquedeveloped applied torque efficiency external resistance in feild winding volt-

age regulation

1 / / C a pt i on : F in d ( a ) i n d u ce d emf a t f u l l l o a d ( b ) p ow erd e v el o p ed ( c ) t o r q ue d e v el o p ed ( d ) a p p l i e d t o r q ue (e ) e f f i c i e n c y ( f ) e x t e r n a l r e s i s t a n c e i n f e i l dw in di ng ( g ) v o l t a g e r e g u l a t i o n

2 / / Exa : 5 . 53 clc ;

4 clear ;

5 close ;

6 N _ m = 6 0 0 ; // s pe ed o f r o t o r ( i n rpm )

7 R _ a = 0 . 0 1 ; // a r ma tu re r e s i s t a n c e ( i n ohms )8 R _ f w = 3 0 ; / / f e i l d w i n d i n g r e s i s t a n c e ( i n ohms )9 V _ f = 1 2 0 ; // v o lt a g e o f e x t e r n a l s o u r c e ( i n v o l t s )10 N _ f = 5 0 0 ; // no . o f t ur n s p er p o l e11 P _ r = 1 0 0 0 0 ; / / i n w a t t s12 V _ t = 2 4 0 ; // t e r m in a l v o l t a g e ( i n v o l t s )13 P _ o = 2 4 0 * 1 0 ^ 3 ; / / r a t e d p ow er ( i n w a t ts )14 I _ L = P _ o / V _ t ; // l o a d c u r r e n t15 I _ a = I _ L ; / / a rm at ur e c u r r e n t16 E _ a f l = V _ t + ( I _ a * R _ a ) ; // r e f e r t o eqn : 5 . 2 717 disp ( E _ a f l ,

( a ) i n du c ed emf a t f u l l l o a d ( i n V o l t s )= ) ;18 P _ d = E _ a f l * I _ a ;

19 disp ( P _ d , ( b ) p ow er d e v e l o p e d ( i n w a t ts )= ) ;20 W _ m = ( 2 * % p i * N _ m ) / 6 0 ; // a n g ul a r v e l o c i t y ( R e fe r t o Eqn

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: 5 . 5 & 5 . 6 )

21 T _ d = P _ d / W _ m ;22 disp ( T _ d , ( c ) t o r q u e d e v e l o p e d ( i n Newtonmet er )= ) ;23 P _ i n m = P _ d + P _ r ; / / m e c h a n i c al p ow er i n p u t24 T _ s = P _ i n m / W _ m ;

25 disp ( T _ s , ( d ) A p p li e d t o r q u e ( i n Newtonmeter )= ) ;26 / / R e f er f i g : 5 . 2 1 ( m a g n e t i z at i o n c u rv e )27 I _ f = 2 . 5 ; // e f f e c t i v e f e i l d c u r r e n t28 m m f = ( 2 . 5 * N _ f ) + ( 0 . 2 5 * I _ a ) ; // t o t a l mmf 29 I _ f a = m m f / N _ f ; / / a c t u a l f e i l d c u r r e n t30 P _ i n = P _ i n m + ( V _ f * I _ f a ) ; // t o t a l p ower i n p u t31 E f f = ( P _ o / P _ i n ) * 1 0 0 ;

32 disp ( E f f , ( e ) e f f i c i e n c y (%)= ) ;33 R _ f = V _ f / I _ f a ;

34 R _ f x = R _ f - R _ f w ;

35 disp ( R _ f x , ( f ) e x t e r n a l r e s i s t a n c e i n f e i l d w in di ng( i n ohms) = ) ;

36 V R = ( ( 2 6 6 - V _ t ) / V _ t ) * 1 0 0 ; // R ef e r t o f i g : 5 . 2 137 disp ( V R , ( g ) v o l t a g e r e g u l a t i o n (%)= ) ;

Scilab code Exa 5.6 Find Rfx and terminal voltage voltage regulation Ef-ficiency

1 / / C ap ti on : Fi nd R f x and ( a ) t e r m i n a l v o l t a g e ( b )v o l t a ge r e g u l a t i o n ( c ) E f f i c i e n c y

2 / / Exa : 5 . 63 clc ;

4 clear ;

5 close ;

6 R _ f w = 3 0 ; // i n ohms7 R _ a = 0 . 2 ; // i n ohms

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8 N _ f = 2 0 0 ; / / t u r n s / p o l e

9 P _ r = 1 2 0 0 ; // i n Wat t s10 I _ L = 1 0 0 ;11 D _ m m f = 0 . 5 * I _ L ; / / d e m a g n e t i z i n g mmf 12 V _ n L = 1 7 0 ; // no l o ad v o l t a g e ( i n V o lt s )13 // R e fe r t o f i g : 5 . 2 6 ( m a g n et i z at i o n c ur ve )14 I _ f = 3 . 5 ; / / f i e l d c u r r e n t i n A mp er es15 R _ f = V _ n L / I _ f ;

16 R _ f x = R _ f - R _ f w ;

17 disp ( R _ f x , R f x ( i n ohms ) = ) ;18 // F i r s t i t e r a t i o n :19 //Assume

20 E _ a = 1 7 0 ;21 V_t1=E_a -103.5* R_a;

22 // S ec on d i t e r a t i o n :23 I _ f 2 = V _ t 1 / R _ f ; / / a c t u a l f i e l d c u r r e n t24 I _ f e 2 = ( N _ f * I _ f 2 - D _ m m f ) / N _ f ;

25 // R ef er t o f i g : 5 . 2 626 E _ a 2 = 1 6 5 ;

27 V_t2=E_a2 -103.07* R_a;

28 // t h i r d i t e r a t i o n29 I _ f 3 = V _ t 2 / R _ f ; / / a c t u a l f i e l d c u r r e n t30 I _ f e = ( N _ f * I _ f - D _ m m f ) / N _ f ;

31 // R e fe r t o f i g :32 E _ a 3 = 1 6 3 ;

33 V_t3=E_a3 -102.97* R_a;

34 V _ t = V _ t 3 ;

35 disp ( V _ t , ( a ) T er mi na l v o l t a g e ( i n V o l t s )= ) ;36 I _ f = V _ t / R _ f ;

37 E _ a = E _ a 3 ;

38 V R = ( V _ n L - V _ t ) * 1 0 0 / V _ t ;

39 disp ( V R , ( b ) V o l t a g e R e g u l a t i o n (%)= ) ;40 P _ o = V _ t * I _ L ; / / p o we r o u t p u t

41 P _ c u = R _ a * ( I _ L + I _ f ) ^ 2 + R _ f * I _ f ^ 2 ; / / c o pp e r l o s s42 P _ d = P _ o + P _ c u ; / / p ow er d e v e l o p e d43 P _ i n = P _ d + P _ r ; / / p ow er i n p u t44 E f f = P _ o * 1 0 0 / P _ i n ;

45 disp ( E f f , ( c ) E f f i c i e n c y (%)= ) ;

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Scilab code Exa 5.7 Find the voltage between far end of feeder and busbar

1 // C ap ti on : Fi nd t he v o l t a g e b et wee n f a r end o f f e e d e r

and b us b a r2 / / Exa : 5 . 73 clc ;

4 clear ;

5 close ;

6 V _ o = 2 4 0 ; // b us b ar v o l t a g e ( i n V o lt s )7 I _ d = 0 ;

8 I _ s = 3 0 0 ; // c u r r e nt i n s e r i e s w in di ng ( i n Amperes )9 R _ s = 0 . 0 3 ; // r e s i s t a n c e o f s e r i e s f e i l d w in di ng ( i n

ohms)10 R _ a = 0 . 0 2 ; // r e s i s t a n c e o f a rm at ur e w in d in g ( i n ohms )11 R _ f e = 0 . 2 5 ; // r e s i s t a n c e o f f e e d e r ( i n ohms )12 / / R e f er t o eqn : 5 . 3 313 I _ a = I _ s ;

14 E _ a = 0 . 4 * I _ s ; / / i n d u c e d e mf 15 V _ d = I _ s * ( R _ s + R _ a + R _ f e ) ; / / v o l t a g e d ro p ( i n V o l t s )16 V _ t = V _ o + E _ a - V _ d ;

17 disp ( V _ t , v o l t a g e be t w ee n f a r end o f f e e d e r and busb ar ( i n V o l ts )= )

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Scilab code Exa 5.9 Find maximum efficiency of generator

1 / / C a pt i on : F in d maximum e f f i c i e n c y o f g e n e r a t o r2 / / Exa : 5 . 93 clc ;

4 clear ;

5 close ;

6 R _ a = 5 0 * 1 0 ^ - 3 ; // a r ma tu re r e s i s t a n c e ( i n ohms )7 R _ s = 2 0 * 1 0 ^ - 3 ; // s e r i e s f i e l d r e s i s t a n c e8 R _ s h = 4 0 ; / / s hu nt f i e l d r e s i s t a n c e9 P _ r o t = 2 0 0 0 ; // r o t a t i o n a l l o s s ( i n w at ts )10 V = 1 2 0 ; // v o l t ag e ( i n v o l l t s )

11 I _ f = V / R _ s h ; / / s h u nt f i e l d c u r r e n t12 / / R e f er t o eq n 5 . 4 913 I _ L m = sqrt ( ( P _ r o t + ( R _ a + R _ s + R _ s h ) * ( I _ f ^ 2 ) ) / ( R _ a + R _ s ) ) ;

14 P _ o = I _ L m * V ; // p ower o ut pu t a t max e f f i c i e n c y15 P _ c u = ( ( ( I _ L m ^ 2 ) * ( R _ a + R _ s ) ) + ( ( I _ f ^ 2 ) * R _ s h ) ) ; / / t o t a l

c op pe r l o s s16 P _ d = P _ o + P _ c u ; // Power d e ve l op e d a t max e f f i c i e n c y17 P _ i n = P _ d + P _ r o t ;

18 E f f = ( P _ o / P _ i n ) * 1 0 0 ;

19 disp ( E f f , Max e f f i c i e n c y o f g e n er a t or (%)= ) ;

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Chapter 6

Direct Current Motors

Scilab code Exa 6.1 Find armature current at rated load efficiency at fullload no of turns per pole new speed of motor and driving torque when ar-mature current reduces

1 / / C ap ti on : Fi nd ( a ) a rm at ur e c u r r e n t a t r a t e d l o a d ( b )e f f i c i e n c y a t f u l l l oa d ( c ) no . o f t u rn s p er p o le( d ) new s pe ed o f motor and d r i v i n g t o rq u e whena rm at ur e c u r r e n t r e d uc e s t o 1 6 . 6 7A

2 / / Exa : 6 . 13 clc ;

4 clear ;

5 close ;

6 P _ o = 1 0 * 7 4 6 ; / / o u t p ut p ow er ( i n Wa tts )7 V _ s = 2 2 0 ;

8 P _ r o t = 1 0 4 0 ; // r o t a t i o n a l l o s s ( i n Watts )9 R _ a = 0 . 7 5 ; // a r ma tu re r e s i s t a n c e ( i n ohms )10 R _ s = 0 . 2 5 ; // s e r i e s w in di ng r e s i s t a n c e ( i n ohms )11 N _m = 1 20 0; / / ( i n rpm )12 P _ d = P _ o + P _ r o t ;

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13 function y = r oo t ( a , b , c ) ;

14 y = ( ( - b ) - sqrt ( ( b ^ 2 ) - ( 4* a * c ) ) ) / ( 2 * a ) ;15 e n d f u n c t i o n ;

16 I _ a = r o o t ( 1 , - 2 2 0 , 8 5 0 0 ) ;

17 disp ( I _ a , ( a ) a rm at u re c u r r e n t a t r a te d l oa d ( i nAmperes)= ) ;

18 P _ i n = V _ s * I _ a ;

19 disp ( ( P _ o / P _ i n ) * 1 0 0 , ( b ) E f f i c i e n c y a t f u l l l o ad (%)= ) ;

20 N _ s = 1 5 0 / I _ a ;

21 disp ( N _ s , ( c ) no . o f t u r ns p er p o l e= ) ;22 I _ a n = 1 6 . 6 7 ;

23 E _ a n = V _ s - ( I _ a n * ( R _ a + R _ s ) ) ;24 N _ m n = ( E _ a n * N _ m ) / 9 0 ;

25 disp ( int ( N _ m n ) , ( d ) new s p e e d o f m ot or ( i n rpm )= ) ;26 T _ d n = ( E _ a n * I _ a n ) / 2 8 3 . 9 ;

27 disp ( T _ d n , d r i v i n g t o r q u e ( i n Newtonmet er )= ) ;

Scilab code Exa 6.3 Find power developed and speed for cumulative com-pound motor differential compound motor

1 / / C ap ti on : Fi nd p ower d e v el o p e d and s p ee d f o r ( a )cu mu la ti ve compound motor (b ) d i f f e r e n t i a lcompound motor

2 / / Exa : 6 . 33 clc ;

4 clear ;

5 close ;

6 disp ( ( a ) F or C u m ul a t iv e co mpoun d m ot or ) ;7 V = 2 4 0 ; // i n v o l t s ( R e fe r t o ex a : 6 . 2 )8 R _ a = 0 . 4 ; // a rm at ur e r e s i s t a n c e ( R ef er t o ex a : 6 . 2 )

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9 T = 2 0 . 6 8 ; / / t o r q u e ( R e f er t o e xa : 6 . 2 )

10 R _ x = 0 . 1 ; // i n ohms11 I _ a = 2 2 . 5 ; // a r ma tu re c u r r e nt o f s hu nt motor ( R e fe r t oe xa : 6 . 2 )

12 I _ a c = I _ a / ( 1 + 0 . 1 2 5 ) ; // a r m at ur e c u r r e n t o f c um mu la ti vecompound motor

13 E _ a c = V - ( I _ a c * ( R _ a + R _ x ) ) ;

14 P _ d c = E _ a c * I _ a c ;

15 disp ( P _ d c , P ower d e v e l o p e d ( i n W atts ) = ) ;16 N _ m c = ( P _ d c * 6 0 ) / ( T * 2 * % p i ) ;

17 disp ( int ( N _ m c ) , s p e e d ( i n rpm ) = ) ;18 disp ( (b ) For d i f f e r e n t i a l compound motor ) ;

19 I _ a d = I _ a / ( 1 - 0 . 1 2 5 ) ; // a r m at ur e c u r r e n t o f c um mu la ti vecompound motor

20 E _ a d = V - ( I _ a d * ( R _ a + R _ x ) ) ;

21 P _ d d = E _ a d * I _ a d ;

22 disp ( P _ d d , P ower d e v e l o p e d ( i n W atts ) = ) ;23 N _ m d = ( P _ d d * 6 0 ) / ( T * 2 * % p i ) ;

24 disp ( int ( N _ m d ) , s p e e d ( i n rpm ) = ) ;

Scilab code Exa 6.4 Find the motor speed power loss in external resis-tance efficiency

1 // C ap ti on : Fi nd t he ( a ) motor s pe ed ( b ) power l o s si n e x t e r n a l r e s i s t a n c e ( c ) e f f i c i e n c y

2 / / Exa : 6 . 43 clc ;

4 clear ;

5 close ;

6 V = 1 2 0

7 N _ m f L = 2 4 0 0 ; / / f u l l l o a d s p e ed o f m ot or

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8 R _ a = 0 . 4 ; // a r ma tu re r e s i s t a n c e ( i n ohms )

9 R _ s h = 1 6 0 ; // s h u nt f i e l d w in d in g r e s i s t a n c e10 I _ f L = 1 4 . 7 5 ; / / c u r r e n t drawn a t f u l l l o a d ( i n Amp eres )11 I _ n L = 2 ; / / c u r r e n t d ra wn a t no l o a d ( i n Am peres )12 R _ x = 3 . 6 ; // e x t e r na l r e s i s t a n c e13 I _ f = V / R _ s h ; / / f e i l d c u r r e n t14 I _ a n L = I _ n L - I _ f ; // a r ma tu re c u r r e n t a t no l o ad15 E _ a n L = V - ( I _ a n L * R _ a ) ; / / no l o a d b ac k emf 16 P _ d n L = E _ a n L * I _ a n L ; / / po wer d e v el o p e d a t no l o a d17 I _ a f L = I _ f L - I _ f ; / / a rm at ur e c u r r e n t a t f u l l l o a d18 E _ a f L = V - ( I _ a f L * R _ a ) ; / / f u l l l o a d b a ck e mf 19 P _ d f L = E _ a f L * I _ a f L ; / / po wer d e v el o p e d a t f u l l l o a d

20 N _ m n L = ( E _ a n L / E _ a f L ) * N _ m f L ; // no l o ad s pe ed21 P _ i n _ f L = V * I _ f L ; / / po wer i n p ut a t f u l l l o a d22 E _ a _ n = V - ( I _ a f L * ( R _ a + R _ x ) ) ; / / new b a c k e mf 23 P _ d _ n = E _ a _ n * I _ a f L ; / / new p ow er d e v e l o p e d24 N _ m _ n = ceil ( ( E _ a _ n / E _ a f L ) * N _ m f L ) ;

25 disp ( A f t er i n s e r t i o n o f e x t e r n al r e s i s t a n c e i n t hea r ma t ur e c k t ) ;

26 disp ( N _ m _ n , ( a ) m ot or s p e ed ( i n rpm )= ) ;27 P _ r o t _ n = ( N _ m _ n / N _ m n L ) * P _ d n L ;

28 P _ o _ n = P _ d _ n - P _ r o t _ n ;

29 P _ x = ( I _ a f L ^ 2 ) * R _ x ;

30 disp ( P _ x , ( b ) power l o s s i n e x t e r n al r e s i s t a n c e ( i nWatts)= ) ;

31 E f f = P _ o _ n / P _ i n _ f L ;

32 disp ( E f f * 1 0 0 , ( c ) e f f i c i e n c y (%)= ) ;

Scilab code Exa 6.5 Find the new motor speed power loss in external re-sistance efficiency

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1 / / C a pt i on : F in d t h e new ( a ) m ot or s p e ed ( b ) p ow er

l o s s i n e x t e r n a l r e s i s t a n c e ( c ) e f f i c i e n c y2 / / Exa : 6 . 53 clc ;

4 clear ;

5 close ;

6 R _ x = 8 0 ; // e x t e r na l r e s i s t a n c e7 / / R e f er t o Exa 6 . 48 R _ s h = 1 6 0 ; // s hu nt r e s i s t a n c e9 V = 1 2 0 ; // i n v o l t s10 E _ a = 1 1 4 . 4 ; / / ba ck emf a t f u l l l o a d11 N _ m = 2 4 0 0 ; / / s p e ed o f m ot or

12 P _ r o t = 1 4 3 ; // r o t a t i o n a l l o s s e s13 I _ f n = V / ( R _ x + R _ s h ) ; / /new f i e l d w in d in g c u r r e n t14 I _ f = 0 . 7 5 ; / / f i e l d c u r r e n t a t f u l l l o a d15 c = sqrt ( I _ f / I _ f n ) ; // r a t i o o f new f l u x t o o ld f l u x16 R _ a = 0 . 4 ; // a r ma tu re r e s i s t a n c e17 I _ a = 1 4 ; // a r ma tu re r e s i s t a n c e18 I _ a n = I _ a * c ;

19 E _ a n = V - ( I _ a n * R _ a ) ;

20 N _ m n = c * ( E _ a n / E _ a ) * N _ m ;

21 disp ( int ( N _ m n ) , ( a ) new m ot or s p e e d ( i n rpm )= ) ;22 P _ x = ( I _ f n ^ 2 ) * R _ x ;

23 disp ( P _ x , ( b ) Power l o s s i n e x t e r n al r e s i s t a n c e ( i nWatts)= ) ;

24 P _ i n = V * ( I _ f n + I _ a n ) ;

25 P _ d n = E _ a n * I _ a n ;

26 P _ o = P _ d n - P _ r o t ;

27 E f f = P _ o / P _ i n ;

28 disp ( E f f * 1 0 0 , ( c ) E f f i c i e n c y (%)= ) ;

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Scilab code Exa 6.6 Find the value of external resistance when motor de-

velops torque of 30 Nm at 2000rpm torque of 30Nm at 715 rpm

1 // C ap ti on : Fi nd t he v a lu e o f e x t e r n a l r e s i s t a n c e whenmotor d e ve l op s ( a ) t or qu e o f 30 Nm a t 2 00 0rpm (

b ) t or qu e o f 30Nm a t 7 15 rpm2 / / Exa : 6 . 63 clc ;

4 clear ;

5 close ;

6 V _ s = 1 2 0 ; / / i n V o l t s7 R _ f e = 3 0 ; / / r e s i s t a n c e o f f e i l d w in d in g

8 I _ a = 5 0 ; / / a r m at u re c u r r e n t ( i n Am peres )9 R _ a g = 0 . 2 ; // a r ma tu re r e s i s t a n c e o f g e n e r a to r ( i n ohms

)10 R _ a m = 0 . 3 ; // a r ma tu re r e s i s t a n c e o f motor ( i n ohms )11 N _ m 1 = 2 0 0 0 ;

12 N _ m 2 = 7 1 5 ;

13 T = 3 0 ; / / t o r q u e ( i n Newto nmet er )14 w _ m = ( N _ m 1 * 2 * % p i ) / 6 0 ;

15 P _ d = T * w _ m ; / / p ow er d e v e l o p e d16 E _ a m = P _ d / I _ a ; / / ba ck emf o f m oto r17 E _ a m n = E _ a m * N _ m 2 / N _ m 1 ;

/ / new b a c k e mf 18 V _ t = E _ a m + ( I _ a * R _ a m ) ;19 V _ t n = E _ a m n + ( I _ a * R _ a m ) ;

20 E _ a g = V _ t + ( I _ a * R _ a g ) ; / / i n du c ed emf o f g e n e r a t o r21 E _ a g n = V _ t n + ( I _ a * R _ a g ) ; // new i n du c ed emf o f g e n e r a t o r22 I _ f = 1 . 7 5 ; // R e f er t o m

Scilab Companion Electric Machinery and Transformers_B. S. Guru and H. R. Hiziroglu

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Transcript of Scilab Companion Electric Machinery and Transformers_B. S. Guru and H. R. Hiziroglu