Electrical Machinery By Dr. P S Bimbhra t3b45

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Scilab Textbook Companion for Electrical Machinery by Dr. P S Bimbhra1 Created by Tejash Ajay Sharma B.Tech Electrical Engineering Visvesvaraya National Institute of Technology College Teacher None Cross-Checked by None April 4, 2017

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 ed from the ”Textbook Companion Project” section at the website http://scilab.in


Book Description Title: Electrical Machinery Author: Dr. P S Bimbhra Publisher: Khanna Publishers, New Delhi Edition: 7 Year: 1995 ISBN: 81-7409-173-4

1


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.

2


Contents List of Scilab Codes

4

1 Transformers

5

2 Electromechanical Energy Conversion Principle

78

3 Basic Concepts of Rotating Electrical Machines

93

4 DC machines

118

5 Polyphase synchronous machine

193

6 Polyphase Induction Motors

237

7 Armature Windings

296

8 Appendix A

307

9 Appendix B

317

3


List of Scilab Codes Exa 1.1 Exa 1.2 Exa 1.3 Exa 1.4 Exa 1.5 Exa 1.6 Exa 1.8

Exa 1.9 Exa 1.10 Exa 1.11 Exa 1.12 Exa 1.14 Exa 1.15 Exa 1.16

finding number of primary and secondary turns and net cross section area of core . . . . . . . . . . . . . . . . Determining number of turns in all three windings of transformer . . . . . . . . . . . . . . . . . . . . . . . . the components of exciting current primary current and power factor . . . . . . . . . . . . . . . . . . . . . . . finding saving in weight of core and wire material . . . determination of no load voltage and frequency of secondary side . . . . . . . . . . . . . . . . . . . . . . . . finding primary winding current and power factor . . . Determining 1 primary resistance and reactance referred to secondary 2 secondary resistance and reactance referred to primary 3 equivalent resistance and leakage reactance referred to primary and secondary 4 total ohmic losses 5 voltage applied to hv side . . . . . . . . . . . finding secondary terminal voltage . . . . . . . . . . . finding primary current secondary terminal voltage power factor power output and efficiency . . . . . . . . . . . determining parameters of equivalent circuit referred to l v and h v sides . . . . . . . . . . . . . . . . . . . . . determining equivalent circuit parameters referred to h v side and its efficiency . . . . . . . . . . . . . . . . . p u value of leakage impedance exciting current referred to l v and h v sides . . . . . . . . . . . . . . . . . . . . determining p u values of circuit parameters referred to both sides . . . . . . . . . . . . . . . . . . . . . . . . Determining parameters of equivalent circuit diagram

4


5 6 6 7 8 10

10 12 13 14 15 16 17 19

Exa 1.18 Exa 1.19 Exa 1.20 Exa 1.21 Exa Exa Exa Exa Exa Exa

1.22 1.24 1.25 1.26 1.27 1.28

Exa 1.29 Exa 1.30 Exa 1.31 Exa 1.32 Exa 1.33


Exa 1.39 Exa 1.40 Exa 1.41

Determining voltage regulation and terminal voltage at different power factor . . . . . . . . . . . . . . . . . . Determining voltage regulation and secondary terminal voltage . . . . . . . . . . . . . . . . . . . . . . . . . . Determining voltage applied to high voltage side . . . voltage at sending end of feeder primary terminals active and reactive power power factor . . . . . . . . . . . . input power and power factor . . . . . . . . . . . . . . hysteresis and eddy current losses . . . . . . . . . . . . total core loss . . . . . . . . . . . . . . . . . . . . . . determining losses and output at 60 hz . . . . . . . . . finding efficiency and losses . . . . . . . . . . . . . . . determining efficiency load voltage regulation and secondary terminal voltage . . . . . . . . . . . . . . . . . Determining transformer core loss ohmic loss and pu value of equivalent resistance . . . . . . . . . . . . . . determining voltage regulation at given power factor . Determining efficiency on full load full load power factor and load power factor for zero voltage drop . . . . . . Determine input power and power factor at a certain load . . . . . . . . . . . . . . . . . . . . . . . . . . . . determining load power factor at minimum secondary terminal voltage and minimum secondary terminal voltage . . . . . . . . . . . . . . . . . . . . . . . . . . . . Determining full day efficiency . . . . . . . . . . . . . Determining hysteresis and eddy current losses at given frequencies . . . . . . . . . . . . . . . . . . . . . . . . Determining reading of wattmeter connected on l v side Determine leakage impedance per phase in ohm and p u system . . . . . . . . . . . . . . . . . . . . . . . . . Determine power transformed power conducted power output and auto transformer efficiency at same power factor . . . . . . . . . . . . . . . . . . . . . . . . . . . Determine current in various parts of circuit . . . . . . Determine KVA output KVA transformed and KVA conducted of auto transformer . . . . . . . . . . . . . . . determining currents in various branch of autotransformer . . . . . . . . . . . . . . . . . . . . . . . . . . . 5


20 21 21 22 23 24 25 26 27 28 30 31 31 32

33 34 35 37 38

38 40 41 41

Exa 1.42 Exa 1.43 Exa 1.44

Exa 1.45 Exa 1.46 Exa 1.48

Exa 1.49

Exa 1.50 Exa 1.51 Exa 1.52 Exa 1.53


Finding primary current and primary input . . . . . . Finding current in three section of autotransformer . . voltage and current rating KVA rating efficiency percentage impedance regulation and short circuit current on each side . . . . . . . . . . . . . . . . . . . . . . . . Determining load voltage at given power factors . . . . Finding ratio of full load KVA delivered to sum of individual KVA ratings . . . . . . . . . . . . . . . . . . . Determining greatest load that can be put across parallel combination of transformers and secondary terminal voltage . . . . . . . . . . . . . . . . . . . . . . . . . . 1 How transformers share a given load 2 Determine the greatest load that can be supplied by parallel combination of transformers and 3 Determining magnitude of the equivalent leakage impedance under normal loading 1 How transformers share a given load 2 Determine the secondary terminal voltage for part1 . . . . . . . . . . Determining largest KVA that can be shared by parallel combination of transformer without overloading . . . . Determining no load circulating current and ohmic loss caused by it and no load terminal voltage . . . . . . . Finding no load circulating current current supplied by each transformer and KVA KW and power factor of each transformer . . . . . . . . . . . . . . . . . . . . . . . . Determining value of reactance connected in series with transformer B so that load current is equally shared . Determining tap setting to maintain rated voltage on secondary side for given loads . . . . . . . . . . . . . . Determining tap setting to maintain given voltage at load terminal . . . . . . . . . . . . . . . . . . . . . . . Calculating primary winding current and line current Calculating KVA required to maintain the given voltage Calculating the rating of the primary and secondary winding . . . . . . . . . . . . . . . . . . . . . . . . . . Determining the magnitude of maximum induced emf and For supply voltage finding the limit of output voltage and finding the rating of regulator for given load current . . . . . . . . . . . . . . . . . . . . . . . . . . 6


42 43

44 45 46

47

48 49 50 51

52 54 54 55 56 57 57

58

Exa 1.61

Exa 1.62

Exa 1.63

Exa 1.64

Exa 1.65 Exa 1.66 Exa 1.67 Exa 1.68


Exa 1.76

1 Determining self impedances of primary and secondary windings 2 Finding the value of equivalent circuit parameter referred to both sides 3 Determining secondary terminal voltage for given load . . . . . . . . . . . . . 1 Determining self inductances of hv and lv windings 2 Calculating mutual inductance between h v and l v windings 3 Finding coupling factors k1 and k2 for both windings and coefficient of coupling . . . . . . . . . . Calculating the current that would flow in the winding 1 when the winding is connected to given supply and given load . . . . . . . . . . . . . . . . . . . . . . . . 1 Determining the transformers turns ratio for maximum transfer of power 2 Determining load current voltage and power under maximum transfer of power . . . Finding turns ratio for maximum transfer of power 2 Computing the load voltage at given frequencies . . . Computing secondary line voltage line current and output KVA for the given connection . . . . . . . . . . . Determining transformer rating and phase and line current on both high and low voltage sides . . . . . . . . 1 Finding power consumed by load 2 KVA rating of transformer 3 Phase and line currents on both h v and l v side . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Determining number of primary turns 2 current power factor and power on primary side under no load . . . . Determining magnetizing current for the given figs . . Determining load voltage load power factor and load power . . . . . . . . . . . . . . . . . . . . . . . . . . . Calculating power delivered by each source and power dissipated in given resistor . . . . . . . . . . . . . . . . Calculating efficiency and regulation under given condition . . . . . . . . . . . . . . . . . . . . . . . . . . . . Determining current taken from source primary input impedance and input power . . . . . . . . . . . . . . . Finding currents drawn by primary and secondary of transformer A and open circuit voltage across secondary of transformer B . . . . . . . . . . . . . . . . . . . . . Determining the reading of ideal voltmeter . . . . . . 7


59

60

61

62 63 64 65

66 67 68 69 70 71 72

73 73

Exa 1.77 Exa 1.78 Exa 1.79 Exa 2.4 Exa 2.6 Exa 2.7

Exa 2.8 Exa 2.9 Exa 2.13 Exa 2.15 Exa 2.16 Exa 2.17 Exa 2.19 Exa 2.21 Exa 2.22 Exa 2.23 Exa 2.25

Exa 2.26 Exa 2.28 Exa 2.29 Exa 2.30

Finding current drawn by transformer under no load test Determining the efficiency and voltage regulation of transformer . . . . . . . . . . . . . . . . . . . . . . . . . . . Determining the KVA rating of different section of autotransformer . . . . . . . . . . . . . . . . . . . . . . . Determining magnitude of average magnetic force . . . Determining force required for axial alignment of electromagnets . . . . . . . . . . . . . . . . . . . . . . . . Determining 1 air gap flux densities and coil inductance 2 energy stored in magnetic field 3 electromagnetic force 4 mechanical work done 5 electrical energy supplied by load . . . . . . . . . . . . . . . . . . . . . . . . . . . . Determining torque . . . . . . . . . . . . . . . . . . . Determining torque . . . . . . . . . . . . . . . . . . . Determining magnitude and direction of torque for different cases . . . . . . . . . . . . . . . . . . . . . . . Determining magnitude of exciting current for closing relay and keeping it closed . . . . . . . . . . . . . . . . Determining magnitude of unbalanced magnetic pull on armature . . . . . . . . . . . . . . . . . . . . . . . . . Determining magnitude and direction of magnetic force Determining 1 mechanical work done 2 energy supplied by each electrical source 3 change in field energy . . . Determining force between two magnetic surfaces . . . Determining force between two plates . . . . . . . . . Determining magnitude of voltage applied between two plates . . . . . . . . . . . . . . . . . . . . . . . . . . . Determining 1 coil inductance and magnetic energy 2 electrical energy supplied by source 3 mechanical work done . . . . . . . . . . . . . . . . . . . . . . . . . . . . Determining force required for armature alignment with field structure . . . . . . . . . . . . . . . . . . . . . . Determining energy stored in inductor . . . . . . . . . Determining and comparing field energy stored and field energy density in iron and air gap . . . . . . . . . . . Determining 1 coil inductance field energy and force on armature 2 mechanical work done 3 magnetic force 4 mechanical work done for constant flux linkage . . . . 8


75 75 76 78 79

79 81 82 82 83 85 85 86 87 87 87

88 89 89 90

91

Exa 3.2 Exa 3.3 Exa 3.4 Exa 3.5 Exa 3.6 Exa 3.8 Exa 3.10 Exa 3.11

Exa 3.12 Exa 3.13 Exa 3.14


Exa 3.33 Exa 3.35

Determining rating of generator for given parallel paths Calculating generated armature voltage for different types of machine . . . . . . . . . . . . . . . . . . . . . . . . Determining 1 frequency of EMF 2 number of poles 3 number of synchronous motor poles . . . . . . . . . . Calculating frequency and magnitude of per phase EMF for different speed of rotor . . . . . . . . . . . . . . . . Calculating fundamental third and fifth harmonic belt factors for stator . . . . . . . . . . . . . . . . . . . . . Determining distribution and winding factor . . . . . . Finding ratio of outputs and amount of copper required for different combinations . . . . . . . . . . . . . . . . 1 Estimating resultant line and phase voltage for given configurations 2 Determining circulating current for delta connected machine . . . . . . . . . . . . . . . . . . . . Calculating percentage increase in per phase rms emf due to harmonic . . . . . . . . . . . . . . . . . . . . . Determining phase and line EMF for given windings . Calculating third and fifth phase EMF in of fundamental phase emfs and the ratio of resultant line emf to resultant phase emf . . . . . . . . . . . . . . . . . . Finding rms value of voltage in single turn coil . . . . Calculating rms value of fundamental emf per phase . Determining resultant emf for given combination of series coil . . . . . . . . . . . . . . . . . . . . . . . . . . Determining maximum and rms value of peak of fundamental mmf wave . . . . . . . . . . . . . . . . . . . . Calculating peak amplitude of mmf wave peak and rms value of fundamental mmf wave . . . . . . . . . . . . . Finding linear velocity of travelling mmf wave . . . . . Calculating resultant peak gap mmf peak gap flux density total gap energy electromagnetic torque and electromagnetic power . . . . . . . . . . . . . . . . . . . . Determining torque for given pole fields . . . . . . . . Finding peak value of fundamental mmf peak value of fundamental air gap flux density wave fundamental value of flux per pole and rms value of phase and line emfs at no load and rated speed . . . . . . . . . . . . . . . . . 9


93 94 95 96 97 98 99

100 102 103

104 105 106 107 108 108 109

109 110

111

Exa 3.36

Exa Exa Exa Exa Exa Exa Exa Exa

3.37 3.39 3.40 3.41 3.42 3.43 4.2 4.3

Exa 4.4 Exa 4.5 Exa 4.6 Exa 4.9 Exa Exa Exa Exa

4.11 4.12 4.13 4.14


Calculating 1 peak value of fundamental air gap flux density 2 peak value of fundamental mmf wave 3 peak value of armature mmf wave and resultant mmf wave per pole 4 rms value of armature current and its power factor . . . . . . . . . . . . . . . . . . . . . . . . . . . Determining full load efficiency of given transformer . Determining KW rating of motor . . . . . . . . . . . . Determining continuous KW rating of motor . . . . . Determining final steady temperature rise and new KVA rating of transformer . . . . . . . . . . . . . . . . . . . Determining one hour rating of induction motor . . . . Calculating ratio of core loss to ohmic loss . . . . . . . Determining electromagnetic power and internal torque Determining total current emf power developed in armature and electromagnetic torque . . . . . . . . . . . Determining 1 generated emf at no load 2 terminal voltage at full load . . . . . . . . . . . . . . . . . . . . . Determining terminal voltage of generator . . . . . . . Determining ratio of speed as a generator to speed as a motor . . . . . . . . . . . . . . . . . . . . . . . . . . . Determining demagnetizing and cross magnetizing ampere turns per pole for different positions of brushes . Determining number of turns required on each pole . Determining time of commutation . . . . . . . . . . . Determining value of commutating field . . . . . . . . Determining number of pole face conductors of compensating winding in each pole . . . . . . . . . . . . . . . Determining 1 compensating winding conductors per pole 2 number of turns on each interpole . . . . . . . . Determining number of series field turns per pole . . . Determining demagnetizing effect of armature reaction at rated load and speed . . . . . . . . . . . . . . . . . Determining terminal voltage at given speed . . . . . . Determining 1 open circuit voltage 2 critical value of shunt field resistance 3 critical speed 4 open circuit voltage for given field resistance 5 terminal voltage . . . .

10


112 114 114 115 116 116 117 118 119 119 120 121 122 123 123 124 124 125 126 126 127

128

Exa 4.20

Exa 4.21 Exa 4.22 Exa 4.23 Exa 4.24 Exa 4.25 Exa 4.26 Exa Exa Exa Exa

4.27 4.28 4.29 4.30


Exa 4.37 Exa 4.38

Exa 4.39

Determining 1 no load emf 2 output current and shunt field current 3 maximum output current and terminal voltage 4 steady state short circuit current 5 additional resistance that must be inserted in field circuit . . . . Determining range of external rheostat and dissipating power . . . . . . . . . . . . . . . . . . . . . . . . . . . Determining field circuit resistance and flux per pole . Determining 1 value of regulator resistance 2 no load terminal voltage for given speeds . . . . . . . . . . . . Determining terminal voltage and number of series field turns per pole . . . . . . . . . . . . . . . . . . . . . . Determining series field turns and resistance of diverter for achieving desired performance . . . . . . . . . . . . Determining emf generated in armature and percentage change in series field ampere turns due to diverter . . Determining speed and developed torque at full load . Determining motor speed and rated shaft torque . . . Determining external resistance inserted in field circuit Determining 1 speed and internal torque developed 2 shaft power shaft torque and efficiency . . . . . . . . Determining shaft power operating speed armature current and motor efficiency . . . . . . . . . . . . . . . . Determining speed of motor . . . . . . . . . . . . . . Determining 1 speed in rpm 2 shaft power output . . . Determining operating speed and current drawn from source . . . . . . . . . . . . . . . . . . . . . . . . . . . Determining number of series field turns . . . . . . . Determining 1 shunt field current 2 effective armature reaction 3 number of series field turns 4 speed at rated armature current and at rated voltage 5 internal starting torque . . . . . . . . . . . . . . . . . . . . . . . . . . . Determining steady state speed and armature current drawn from source . . . . . . . . . . . . . . . . . . . . Determining 1 external resistance 2 value of first resistance element 3 external resistance cut out in second step 4 total number of steps . . . . . . . . . . . . . . Calculating 1 resistance of each step 2 voltage at which ors should be close . . . . . . . . . . . . . . . 11


130 132 133 134 136 137 138 139 140 141 142 143 145 145 147 148

149 152

153 154

Exa Exa Exa Exa Exa

4.40 4.41 4.42 4.43 4.44

Exa Exa Exa Exa Exa Exa

4.45 4.46 4.48 4.49 4.50 4.51

Exa 4.52 Exa 4.53 Exa 4.54 Exa Exa Exa Exa

4.55 4.56 4.57 4.58

Exa 4.59 Exa 4.60 Exa 4.61


Determining load torque for different cases . . . . . . . Finding motor speed . . . . . . . . . . . . . . . . . . . Finding motor speed . . . . . . . . . . . . . . . . . . . Finding speed regulation . . . . . . . . . . . . . . . . Determining 1 maximum value of current and corresponding torque 2 ultimate speed and armature current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Determining armature current . . . . . . . . . . . . . Determining new speed and armature current . . . . . Determining resistance inserted in shunt field . . . . . Determining new speed of motor . . . . . . . . . . . . Determining percentage change in field flux . . . . . . Determining number of series turns per pole to reduce speed . . . . . . . . . . . . . . . . . . . . . . . . . . . Determining motor speed . . . . . . . . . . . . . . . . Determining motor speed and current . . . . . . . . . Determining 1 armature current and speed of motor 2 value of external resistance inserted in field winding . Determining current and percentage in flux . . . . . . Determining additional resistance to obtain rated torque Determining voltage and current in magnetic circuit . Calculating armature current for different control methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . Determining speed range for full load and no load 2 minimum motor field current . . . . . . . . . . . . . . Determining motor torque and motor speed . . . . . . Determining 1 speed and current for torque 2 speed and torque for given armature current 3 speed and torque for given armature current . . . . . . . . . . . . . . . . Determining size of DC motor and hoist speed . . . . Determining additional resistance inserted in motor circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . Determining speed of motor for given condition . . . . Determining motor speed and armature current . . . . Determining shaft power input efficiency at rated load maximum efficiency and power output . . . . . . . . . Determining efficiency and speed of motor . . . . . . .

12


156 157 158 159

160 160 161 162 163 163 164 165 166 167 168 169 169 170 171 173

174 176 177 177 178 178 179

Exa 4.69 Exa Exa Exa Exa

4.70 4.71 4.72 4.73

Exa 4.74 Exa Exa Exa Exa

4.75 4.77 4.78 4.81

Exa 4.82 Exa 4.83 Exa 4.84 Exa 4.85 Exa 4.86 Exa 4.87 Exa 5.1 Exa 5.3 Exa Exa Exa Exa

5.4 5.5 5.6 5.7


Determining 1 no load current 2 speed 3 armature current . . . . . . . . . . . . . . . . . . . . . . . . . . . . Determining efficiency of motor . . . . . . . . . . . . . Determining armature voltage drop . . . . . . . . . . . Determining shaft torque and efficiency of motor . . . Determining shaft torque shaft power and motor efficiency . . . . . . . . . . . . . . . . . . . . . . . . . . Determining KW output efficiency and percentage change in speed from no load to full load . . . . . . . . . . . . Determining full load efficiency of motor . . . . . . . . Determining efficiency of both machines . . . . . . . . Determining efficiency of both machines . . . . . . . . Determining field current and power gain at rated output and when compensation is zero . . . . . . . . . . To plot the external characteristics for given field current Determining field winding current . . . . . . . . . . . Determining output voltage of generator and reference voltage of potentiometer . . . . . . . . . . . . . . . . . Determining reference voltage of potentiometer . . . . Determining amplifier gain . . . . . . . . . . . . . . . Determining 1 no load rotational losses 2 motor output 3 stall torque . . . . . . . . . . . . . . . . . . . . . . Determining percentage voltage regulation of alternator by different methods . . . . . . . . . . . . . . . . . . . Determining shaft power line current pf and efficiency for maximum power output and maximum power input Determining line current and pf for synchronous motor Determining torque developed and new pf . . . . . . . Determining pf for increased input power . . . . . . . Determining new value of armature current and power factor . . . . . . . . . . . . . . . . . . . . . . . . . . . Determining power factor at which machine is operating Calculating 1 pf efficiency excitation emfs input current and 2 load angle and maximum output . . . . . . . . . Determining 1 excitation emfs 2 mechanical power developed and pf 3 minimum excitation voltage . . . . . Calculating current drawn from supply and pf . . . . Determining synchronous reactance . . . . . . . . . . . 13


180 181 182 183 183 184 185 186 186 187 188 189 190 190 191 192 193 196 197 198 199 199 200 201 202 203 204


Exa 5.20 Exa 5.21 Exa 5.22 Exa 5.23 Exa 5.24 Exa 5.27 Exa 5.29 Exa 5.31 Exa 5.32 Exa 5.34 Exa 5.35

Exa 5.36

Finding mechanical power delivered by motor and input KVA . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204 Determining 1 load angle pu armature current and pf 2 excitation voltage load angle and pf . . . . . . . . . . 205 Determining 1 new value of armature current load angle and pf 2 armature current load angle and power delivered 206 Determining operating pf and load angle . . . . . . . 207 Determining armature current power factor and excitation emf . . . . . . . . . . . . . . . . . . . . . . . . . . 208 Determining load angle power factor and armature current . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208 Determining 1 excitation emf 2 power transfer and 3 maximum power transfer armature current terminal voltage and power factor . . . . . . . . . . . . . . . . . . 209 Determining 1 maximum power output 2 armature current and power factor . . . . . . . . . . . . . . . . . . 211 Calculating data needed to plot v curves and pf variation with field current . . . . . . . . . . . . . . . . . . 211 Determining data to plot v curves and pf variation with field current neglecting armature resistance . . . . . . 212 Determining armature current power factor and load angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 Determining pu excitation voltage . . . . . . . . . . . 214 Calculating 1 power angle armature current and pf 2 maximum power output and power angle . . . . . . . 215 Determining change in synchronous power and reactive power . . . . . . . . . . . . . . . . . . . . . . . . . . . 216 Determining maximum load armature current and pf . 217 Determining minimum excitation voltage and maximum stable load angle . . . . . . . . . . . . . . . . . . . . . 217 Determining load angle and excitation voltage . . . . . 218 Determining excitation voltage power synchronizing power per electrical and mechanical degrees and corresponding torque maximum value of load angle and corresponding power . . . . . . . . . . . . . . . . . . . . . . . . . . . 219 Calculate synchronizing power and torque per mechanical degrees 1 at no load and 2 at full load . . . . . . . 220

14


Exa 5.37 Exa 5.38

Exa 5.41 Exa 5.43 Exa 5.44 Exa 5.45 Exa 5.46 Exa 5.47 Exa 5.48 Exa 5.49 Exa Exa Exa Exa

5.53 5.54 5.55 5.56


Exa 6.3 Exa 6.4

Determining 1 synchronizing current power and torque 2 armature current and angle by which rotor slips back Determining 1 load angle armature current and pf 2 maximum load load angle and armature current 3 minimum excitation voltage . . . . . . . . . . . . . . . . . Determining 1 excitation voltage 2 reluctance power developed . . . . . . . . . . . . . . . . . . . . . . . . . . Determining efficiency at 1 half load 2 full load . . . . Determining effective resistance in pu and ohms and ratio of ac to dc resistance . . . . . . . . . . . . . . . Determining 1 efficiency and 2 maximum efficiency . . Determining KVA rating of synchronous condenser and KVA of factory . . . . . . . . . . . . . . . . . . . . . . Determining 1 total load KVA 2 KVA capacity of motor 3 power factor of motor . . . . . . . . . . . . . . . . . Calculating permissible additional load and rating of synchronous condenser . . . . . . . . . . . . . . . . . . Determining new pf and percentage reduction in line current . . . . . . . . . . . . . . . . . . . . . . . . . . Determining voltage regulation at full load . . . . . . Determining terminal voltage and current . . . . . . . Determining load angle and power factor . . . . . . . . Determining 1 terminal voltage and armature current 2 load angle and excitation voltage . . . . . . . . . . . . Calculating maximum power output and minimum pu excitation . . . . . . . . . . . . . . . . . . . . . . . . . Determining voltage regulation . . . . . . . . . . . . . Determining synchronous reactance . . . . . . . . . . . Determining the mechanical angle through which rotor moves . . . . . . . . . . . . . . . . . . . . . . . . . . Determining 1 full load slip and rotor frequency 2 relative speed of stator with stator structure and rotor structure 3 relative speed of rotor with stator structure rotor structure and stator field . . . . . . . . . . . . . . . . Determining rotor speed . . . . . . . . . . . . . . . . . Determining 1 speed of rotor 2 ratio of voltages at slip rings . . . . . . . . . . . . . . . . . . . . . . . . . . .

15


221

222 224 225 225 226 227 228 229 229 230 232 233 233 234 235 236 237

238 239 240

Exa 6.5

Exa 6.6 Exa 6.7

Exa 6.8 Exa 6.9 Exa 6.10

Exa 6.11 Exa 6.13

Exa 6.14

Exa 6.15 Exa 6.16 Exa 6.17 Exa 6.18 Exa 6.19 Exa 6.20 Exa 6.21 Exa 6.22

Determining 1 rotor current rotor power factor and torque 2 rotor current rotor power factor and torque after insertion of external resistance . . . . . . . . . . . . . . Comparing rotor ohmic losses . . . . . . . . . . . . . Determining 1 stator core loss 2 total rotor losses at full load 3 total rotor ohmic loss at full load 4 full load speed 5 internal torque shaft torque and efficiency . . . . . . Determining slip of motor . . . . . . . . . . . . . . . . Determining stator current rotor speed output torque and efficiency . . . . . . . . . . . . . . . . . . . . . . . Determining 1 slip for maximum torque and maximum torque 2 rotor current and starting torque 3 external resistance inserted in rotor circuit 4 internal power developed 5 maximum internal power developed and corresponding slip . . . . . . . . . . . . . . . . . . . . . Determining maximum internal torque for different cases determining 1 slip and rotor speed 2 rotor ohmic loss 3 starting torque 4 starting current 5 stator current 6 full load efficiency 7 slip at maximum torque for new resistance 8 full load slip 9 starting torque 10 starting current 11 rotor losses 12 power . . . . . . . . . . . . Determining 1 maximum torque 2 full load rotor ohmic loss 3 slip at maximum torque 4 full load slip 5 full load torque . . . . . . . . . . . . . . . . . . . . . . . . . . . Determining percentage reduction in rotor circuit resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . Determining external resistance inserted in rotor circuit and percentage change in current and power factor . . Determining starting torque . . . . . . . . . . . . . . . Determining 1 slip at maximum torque 2 full load slip 3 rotor current in of full load current . . . . . . Determining starting and maximum torque . . . . . . Determining 1 slip 2 rotor current rotor ohmic loss and rotor power factor 3 power output . . . . . . . . . . . Determining 1 rotor and stator input 2 starting torque Determining maximum torque corresponding slip and starting torque . . . . . . . . . . . . . . . . . . . . . .

16


241 242

242 244 244

246 248

250

252 253 254 255 256 257 257 258 259

Exa 6.23

Exa 6.25

Exa Exa Exa Exa

6.26 6.27 6.28 6.29

Exa Exa Exa Exa

6.30 6.31 6.32 6.33

Exa 6.34

Exa 6.35 Exa 6.36 Exa 6.37 Exa 6.41 Exa 6.43 Exa 6.44 Exa 6.45 Exa 6.46 Exa 6.47

Determining 1 maximum torque 2 starting torque 3 full load rotor ohmic loss 4 slip at full load 5 full load torque 6 slip at maximum torque . . . . . . . . . . . . . . . . Determining 1 speed range of DC motor 2 KVA rating of stator 3 DC motor rating and maximum torque 4 number of poles 5 new speed range . . . . . . . . . . Determining starting torque . . . . . . . . . . . . . . . Determining slip frequency and slip at maximum torque Determining frequency of motor . . . . . . . . . . . . Determining power factor input current equivalent rotor current and torque developed maximum torque and corresponding speed . . . . . . . . . . . . . . . . . . . Determining motor speed and power output . . . . . Determining percentage change in motor speed and losses Determining no load speed of motor . . . . . . . . . . Determining minimum voltage impressed on motor and additional resistance inserted in rotor circuit . . . . . . Comparing starting current starting torque and maximum torque at given frequencies and comparing voltage for given frequencies . . . . . . . . . . . . . . . . . . . Determining 1 slip at rated load 2 starting torque . . . Finding ratio of starting current starting torque and maximum torque at given frequencies . . . . . . . . . Determining 1 maximum internal torque and internal starting torque 2 slip at maximum torque . . . . . . . Determining rotational losses and equivalent circuit parameters . . . . . . . . . . . . . . . . . . . . . . . . . Determining starting torque . . . . . . . . . . . . . . . Determining starting torque . . . . . . . . . . . . . . Determining 1 mechanical power output 2 net torque 3 efficiency of motor . . . . . . . . . . . . . . . . . . . . Determining new operating speed . . . . . . . . . . . . Determining 1 line current pf slip torque and efficiency 2 maximum possible pf and corresponding line current 3 maximum power output and maximum power input 4 slip at maximum torque and maximum torque 5 starting torque . . . . . . . . . . . . . . . . . . . . . . . . . . .

17


260

262 263 264 264

265 267 267 268 269

270 270 271 272 273 274 275 276 277

278

Exa 6.48



Exa 6.65 Exa 7.1 Exa 7.2 Exa 7.3

Determining 1 external resistance inserted in rotor circuit 2 stator currents power factor 3 power output operating power factor and efficiency . . . . . . . . . . . Determining 1 rotational and core loss 2 electric power output power factor and efficiency . . . . . . . . . . . Determining per phase value of capacitance and total KVA rating of capacitor bank . . . . . . . . . . . . . . Determining capacitance of bank and each unit and percentage saving in energy lost . . . . . . . . . . . . . . Determining tapping on autotransformer and line current at starting . . . . . . . . . . . . . . . . . . . . . . Determining starting torque in of full load torque Determining maximum permissible KW rating of motor for different cases . . . . . . . . . . . . . . . . . . . . Determining 1 voltage applied to motor terminals 2 current drawn by motor 3 line current drawn from supply mains . . . . . . . . . . . . . . . . . . . . . . . . . . . Determining ratio of starting torque to full load torque for different starters . . . . . . . . . . . . . . . . . . . Determining resistance of feeder and percentage increase in starting torque . . . . . . . . . . . . . . . . . . . . Determining minimum allowable cross section area of each conductor of feeder . . . . . . . . . . . . . . . . . Determining starting torque for different connections Determining value of resistance elements for a 4 step starter . . . . . . . . . . . . . . . . . . . . . . . . . . . Deg 5 sections of a 6 stud starter . . . . . . . . Determining 1 starting current and starting torque 2 external resistance to limit starting current and starting torque under this condition . . . . . . . . . . . . . . . Determining 1 line current 2 power returned to three phase supply 3 efficiency of motor . . . . . . . . . . . Determining the number of commutator segments back pitch and front pitch and commutator pitch . . . . . . Deg the progressive simplex lap winding with two coil sides per slot . . . . . . . . . . . . . . . . . . . . . Determining winding table and position of brushes on commutator . . . . . . . . . . . . . . . . . . . . . . . . 18


280 282 283 285 286 287 287

288 289 290 291 291 292 292

293 294 296 297 297

Exa Exa Exa Exa Exa

7.5 7.6 7.7 7.8 7.9

Exa 7.10 Exa 7.11 Exa 7.12 Exa 7.13 Exa 7.15 Exa 8.1 Exa 8.2 Exa 8.3 Exa 8.4 Exa Exa Exa Exa

8.5 8.6 8.7 8.8


Exa 9.4

Deg a simplex lap winding with the given details Deg a simplex wave winding with given details . Deg the simplex wave winding . . . . . . . . . . Deg the winding and determining the speed . . . Determining the suitable arrangement of equilizer ring for given details . . . . . . . . . . . . . . . . . . . . . Determining the resistance measured between two adjacent commutator segments . . . . . . . . . . . . . . . Determining the details for winding diagram and distribution factor . . . . . . . . . . . . . . . . . . . . . . . Determining the details for winding diagram and distribution factor . . . . . . . . . . . . . . . . . . . . . . . Determining the details for winding table and effective turns per phase . . . . . . . . . . . . . . . . . . . . . . Determining and deg the details for winding of 3 phase machine . . . . . . . . . . . . . . . . . . . . . . Determining the reluctance of ring and current required to establish the required flux . . . . . . . . . . . . . . Determining the exciting current in coil . . . . . . . . Determining the exciting current with and without magnetic leakage and fringing . . . . . . . . . . . . . . . . Determining the coil current for different values of relative permeability . . . . . . . . . . . . . . . . . . . . . Calculating the exciting current to set up required flux Calculating the coil current to establish required flux . Finding the exciting current in given coil . . . . . . . . Determining the emf induced in conductor for different angles of conductor with field flux . . . . . . . . . . . Determining the emf induced in square coil for different angles of coil with field flux . . . . . . . . . . . . . . . Determining the emf induced in conductor . . . . . . . Determining the inductance for given circuit by using the given expression . . . . . . . . . . . . . . . . . . . Determining the inductance of coils for given circuit . Determining phase and line current power factor total active and reactive power for star and delta connected load . . . . . . . . . . . . . . . . . . . . . . . . . . . . Determining per phase circuit parameters . . . . . . . 19


298 299 300 301 302 302 303 304 304 305 307 307 308 309 310 311 312 313 314 314 315 316

317 318

Exa 9.5 Exa 9.6 Exa 9.7 Exa 9.8

Finding total line current power factor total and reactive power . . . . . . . . . . . . . . . . . . . . . . . . . . . Determining input power line current power factor and shaft power . . . . . . . . . . . . . . . . . . . . . . . . Determining no load losses and no load power factor . Calculating reading of wattmeters and input power to load . . . . . . . . . . . . . . . . . . . . . . . . . . . .

20


319 319 320 320

Chapter 1 Transformers

Scilab code Exa 1.1 finding number of primary and secondary turns and net cross section area of core 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

clc ; f =50; // f r e q u e n c y i n Hz Et =13; // emf p e r t u r n i n v o l t s E1 =2310; // p r i m a r y v o l t a g e i n v o l t s E2 = 220; // s e c o n d a r y v o l t a g e i n v o l t s B =1.4; // maximum f l u x d e n s i t y i n T e s l a // c a l c u l a t i n g t h e number o f t u r n s i n p r i m a r y and secondary side N2 = round ( E2 / Et ) ; // s e c o n d a r y s i d e t u r n s printf ( ’ Number o f s e c o n d a r y t u r n s i s %f\n ’ , N2 ) ; N1 = round ( N2 *( E1 / E2 ) ) ; // p r i m a r y s i d e t u r n s printf ( ’ Number o f p r i m a r y t u r n s i s %f\n ’ , N1 ) ; disp ( ’ The v a l u e o f p r i m a r y t u r n s d o e s n o t s a t i s f y with the ’ ); disp ( ’ v a l u e o f s e c o n d a r y t u r n s s o t a k i n g v a l u e o f N2 =18( n e x t n e a r e s t i n t e g e r ) ’ ) ; N2 =18; // new v a l u e o f s e c o n d a r y t u r n s N1 =18*( E1 / E2 ) ; printf ( ’ Number o f p r i m a r y t u r n s i s %f\n ’ , N1 ) ; 21


18 printf ( ’ Number o f s e c o n d a r y t u r n s i s %f\n ’ , N2 ) ; 19 // c a l c u l a t i n g n e t c o r e a r e a 20 A =(220/(18* sqrt (2) * %pi * B *50) ) *10^4; // where N2=18 21 printf ( ’ Net a r e a o f c o r e i s %f cmˆ2 ’ ,A ) ;

Scilab code Exa 1.2 Determining number of turns in all three windings of transformer 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

clc ; f =50; // f r e q u e n c y i n h e r t z B =1.2; // maximum f l u x d e n s i t y i n T e s l a A =75*10^ -4; // n e t c o r e a r e a i n mˆ2 E1 =220; // p r i m a r y s i d e v o l t a g e i n v o l t s E2 =600; // s e c o n d a r y s i d e v o l t a g e i n v o l t s E3 =11; // t e r t i a r y s i d e v o l t a g e i n v o l t s n3 = round ( E3 /2) ; // number o f t u r n s i n h a l f o f t h e t e r t i a r y winding Et = round ( sqrt (2) * %pi *50* B * A ) ; // c a l c u l a t i n g emf p e r turn N3 = Et * n3 ; // t o t a l number o f t u r n s i n t e r t i a r y winding printf ( ’ t o t a l number o f t u r n s i n t e r t i a r y winding i s %f\n ’ , N3 ) ; N2 = round ( E2 *( n3 / E3 ) ) ; // t o t a l number o f t u r n s i n secondary winding printf ( ’ t o t a l number o f t u r n s i n secondary winding i s %f\n ’ , N2 ) ; N1 = round ( E1 *( n3 / E3 ) ) ; // t o t a l number o f t u r n s i n secondary winding printf ( ’ t o t a l number o f t u r n s i n p r i m a r y w i n d i n g i s %f ’ , N1 ) ;

22


Scilab code Exa 1.3 the components of exciting current primary current and power factor 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

16 17 18 19

clc ; f =50; // f r e q u e n c y i n h e r t z E1 =2200; // s u p p l y v o l t a g e i n v o l t s E2 =220; // s e c o n d a r y s i d e v o l t a g e i n v o l t s P =361; // c o r e l o s s i n w a t t s Io =0.6; // e x c i t i n g c u r r e n t i n Ampere Is =60; // s e c o n d a r y l o a d c u r r e n t i n Ampere pf =0.8; // power f a c t o r Ic = P / E1 ; // c o r e l o s s component o f c u r r e n t printf ( ’ c o r e l o s s component o f e x c i t i n g c u r r e n t i s %f A\n ’ , Ic ) ; Im = sqrt ( Io ^2 - Ic ^2) ; // m a g n e t i s i n g component o f current printf ( ’ m a g n e t i s i n g component o f e x c i t i n g c u r r e n t i s %f A\n ’ , Im ) ; ip = Is *( E2 / E1 ) ; // p r i m a r y c u r r e n t r e q u i r e d t o n e u t r a l i s e the secondary current Iv = ip * pf + Ic ; // t o t a l v e r t i c a l compartment o f primary c u r r e n t Ih = ip *0.6+ Im ; // t o t a l h o r i z o n t a l compartment o f p r i m a r y c u r r e n t , p f c o s ( t h e t a ) =0.8 s o s i n ( t h e t a ) =0.6 Ip = sqrt ( Iv ^2+ Ih ^2) ; // t o a t l p r i m a r y c u r r e n t printf ( ’ T o t a l p r i m a r y c u r r e n t i s %f A\n ’ , Ip ) ; ppf = Iv / Ip ; // p r i m a r y power f a c t o r printf ( ’ p r i m a r y power f a c t o r i s %f ( l a g g i n g ) ’ , ppf ) ;

Scilab code Exa 1.4 finding saving in weight of core and wire material 1 clc ; 2 disp ( ’ w e i g h t o f

laminations is directly proportion t o c o r e volume d e n s i t y , which i s d i r e c t l y 23


3 4 5 6 7 8 9

10 11 12 13

p r o p o r t i o n a l t o p r o d u c t o f a r e a and h e i g h t o f l i m b s and w h i l e t a k i n g t h e r a t i o o f w e i g h t o f CRGO l a m i n a t i o n s and h o t r o l l e d laminations , h e i g h t o f limbs g e t s c a n c e l l e d out ( h e i g h t o f l i m b s a r e assumed t o be e q u a l ) . So , i n t h e end r a t i o of weights of laminations i s equal to r a t i o o f a r e a o f c o r e . Now a r e a o f c o r e i s g i v e n by maximum f l u x / f l u x d e n s i t y . A c c o r d i n g t o q u e s t i o n maximum f l u x r e m a i n same s o , w h i l e t a k i n g r a t i o o f a r e a s t h e maximum f l u x g e t s c a n c e l l e d ’ ) B1 =1.2; // f l u x d e n s i t y i n h o t r o l l e d s t e e l laminations B2 =1.6; // f l u x d e n s i t y i n CRGO s t e e l l a m i n a t i o n s W1 =100; // w e i g h t o f H . R c o r e i n kg W2 = W1 *( B1 / B2 ) ; // c a l c u l a t i n g w e i g h t o f CRGO l a m i n a t i o n s i n kg s =(( W1 - W2 ) / W1 ) *100; // c a l c u l a t i n g s a v i n g i n c o r e material printf ( ’ p e r c e n t a g e s a v i n g i n c o r e m a t e r i a l i s %f p e r c e n t \n ’ ,s ) ; disp ( ’ w e i g h t o f w i r e i s d i r e c t l y p r o p o r t i o n a l t o p r o d u c t o f l e n g t h o f t u r n a r o u n d c o r e and c r o s s s e c t i o n o f w i r e . ( Wire c r o s s s e c t i o n i s assumed t o be same i n CRGO and HR l a m i n a t i o n s s o g e t s c a n c e l l e d out w h i l e t a k i n g r a t i o ) a l s o the length of turn i s i n v e r s e l y p r o p o r t i o n a l to square root of flux density ’) w1 = 80 // w e i g h t o f Hot r o l l e d w i r e w2 = w1 *( sqrt (1.2/1.6) ) ; // w e i g h t o f CRGO w i r e s =(( w1 - w2 ) / w1 ) *100; // s a v i n g i n w e i g h t o f w i r e printf ( ’ P e r c e n t a g e s a v i n g i n w e i g h t o f w i r e i s %f p e r c e n t ’ ,s ) ;

Scilab code Exa 1.5 determination of no load voltage and frequency of secondary side 24


1 2 3 4 5

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

clc ; v1 =240; // h i g h v o l t a g e s i d e v o l t a g e v2 =120; // low v o l t a g e s i d e v o l t a g e f1 =50; // f r e q u e n c y i n Hz disp ( ’ v1 i s d i r e c t l y p r o p o r t i o n a l t o p r o d u c t o f f r e q u e n c y and maximum f l u x . c o n s i d e r i n g q1 be maximum f l u x f o r v1 and q2 be maximum f l u x f o r v11 t h e n Q=q2 / q1 can be c a l c u l a t e d a s f o l l o w ’ ) disp ( ’ c a s e a ’ ) v11 =240; // new s u p p l y v o l t a g e f2 =40; // new s u p p l y f r e q u e n c y Q =( v11 * f1 ) /( v1 * f2 ) ; v22 =( v2 * f2 * Q ) / f1 ; printf ( ’ s e c o n d a r y v o l t a g e f o r c a s e a i s %f v \n ’ , v22 ) ; disp ( ’ c a s e b ’ ) v11 =120; // new s u p p l y v o l t a g e f2 =25; // new s u p p l y f r e q u e n c y Q =( v11 * f1 ) /( v1 * f2 ) ; v22 =( v2 * f2 * Q ) / f1 ; printf ( ’ s e c o n d a r y v o l t a g e f o r c a s e a i s %f v \n ’ , v22 ) ; disp ( ’ c a s e c ’ ) v11 =120; // new s u p p l y v o l t a g e f2 =50; // new s u p p l y f r e q u e n c y Q =( v11 * f1 ) /( v1 * f2 ) ; v22 =( v2 * f2 * Q ) / f1 ; printf ( ’ s e c o n d a r y v o l t a g e f o r c a s e a i s %f v \n ’ , v22 ) ; disp ( ’ c a s e d ’ ) v11 =480; // new s u p p l y v o l t a g e f2 =50; // new s u p p l y f r e q u e n c y Q =( v11 * f1 ) /( v1 * f2 ) ; v22 =( v2 * f2 * Q ) / f1 ; printf ( ’ s e c o n d a r y v o l t a g e f o r c a s e a i s %f v \n ’ , v22 ) ; disp ( ’ c a s e e ’ ) v11 =240; // new s u p p l y v o l t a g e 25


32 f2 =0; // new s u p p l y f r e q u e n c y 33 disp ( ’ s i n c e f r e q u e n c y i s z e r o .

S o u r c e i s a DC s o u r c e so a very high c u r r e n t w i l l flow in primary s i d e which w i l l damage t h e t r a n s f o r m e r and t h e s e c o n d a r y i n d u c e d emf i s z e r o ’ )

Scilab code Exa 1.6 finding primary winding current and power factor 1 2 3 4 5 6 7 8 9

10 11 12 13 14 15

clc ; N1 =100; // no . o f p r i m a r y t u r n s N2 =160; // No . o f s e c o n d a r y t u r n s N3 =60; // No . o f t e r t i a r y t u r n s I2 =10; // s e c o n d a r y s i d e c u r r e n t I3 =20; // t e r t i a r y s i d e c u r r e n t F2 = N2 * I2 ; // mmf o f s e c o n d a r y w i n d i n g F3 = N3 * I3 ; // mmf o f t e r t i a r y w i n d i n g disp ( ’ l o a d c o n n e c t e d t o s e c o n d a r y i s p u r e l y r e s i s t i v e and l o a d c o n n e c t e d t o t e r t i a r y i s p u r e l y c a p a c i t i v e ’ ); F23 = sqrt ( F2 ^2+ F3 ^2) ; // r e s u l t a n t l o a d mmf F1 = F23 ; // p r i m a r y w i n d i n g mmf b a l a n c e s t h i s l o a d mmf I1 = F1 / N1 ; printf ( ’ p r i m a r y c u r r e n t i s %f A\n ’ , I1 ) ; pf = F2 / F1 ; printf ( ’ p r i m a r y s i d e power f a c t o r i s %f l e a d i n g ’ , pf ) ;

Scilab code Exa 1.8 Determining 1 primary resistance and reactance referred to secondary 2 secondary resistance and reactance referred to primary 3 equivalent resistance and leakage reactance referred to primary and secondary 4 total ohmic losses 5 voltage applied to hv side 26


1 2 3 4 5 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

clc ; P =33000; // r a t e d power o f t r a n s f o r m e r E1 =2200; // p r i m a r y v o l t a g e E2 =220; // s e c o n d a r y v o l t a g e k = E2 / E1 ; // t u r n ’ s r a t i o r1 =2.4; // p r i m a r y w i n d i n g r e s i s t a n c e i n ohm x1 =6; // p r i m a r y w i n d i n g r e a c t a n c e i n ohm r2 =0.03; // s e c o n d a r y w i n d i n g r e s i s t a n c e i n ohm x2 =0.07; // s e c o n d a r y w i n d i n g r e a c t a n c e i n ohm r12 = r1 * k ^2; // p r i m a r y r e s i s t a n c e r e f e r r e d t o secondary x12 = x1 * k ^2; // p r i m a r y r e a c t a n c e r e f e r r e d t o secondary printf ( ’ p r i m a r y r e s i s t a n c e and r e a c t a n c e r e f e r r e d t o s e c o n d a r y a r e %f ohm and %f ohm\n ’ ,r12 , x12 ) ; r21 = r2 / k ^2; // 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 primary x21 = x2 / k ^2; // 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 primary printf ( ’ 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 r e f e r r e d t o p r i m a r y a r e %f ohm and %f ohm\n ’ ,r21 , x21 ) ; re1 = r1 + r21 ; xe1 = x1 + x21 ; printf ( ’ e q u i v a l e n t r e s i s t a n c e and r e a c t a n c e r e f e r r e d t o p r i m a r y a r e %f ohm and %f ohm\n ’ ,re1 , xe1 ) ; re2 = r2 + r12 ; xe2 = x2 + x12 ; printf ( ’ e q u i v a l e n t r e s i s t a n c e and r e a c t a n c e r e f e r r e d t o s e c o n d a r y a r e %f ohm and %f ohm\n ’ ,re2 , xe2 ) ; Ip = P / E1 ; printf ( ’ p r i m a r y f u l l l o a d c u r r e n t i s %f A\n ’ , Ip ) ; Is = P / E2 ; printf ( ’ s e c o n d a r y f u l l l o a d c u r r e n t i s %f A\n ’ , Is ) ; O = Ip ^2* re1 ; printf ( ’ ohmic l o s s e s a t f u l l l o a d i s %f W\n ’ ,O ) ; Ils =160; // s e c o n d a r y s i d e l o a d c u r r e n t Ilp = Ils * k ; // p r i m a r y s i d e l o a d c u r r e n t Ze1 = sqrt ( re1 ^2+ xe1 ^2) ; 27


31 V = Ilp * Ze1 ; 32 printf ( ’ V o l t a g e a p p l i e d t o h . v s i d e

in order to o b t a i n 160A s h o r t c i r c u i t c i r r e n t i n low v o l t a g e w i n d i n g i s %f V\n ’ ,V ) ; 33 Pi = Ilp ^2* re1 ; 34 printf ( ’ power i n p u t i s %f W’ , Pi ) ;

Scilab code Exa 1.9 finding secondary terminal voltage 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

clc ; P =10000; // r a t e d power o f t r a n s f o r m e r E1 =2500; // r a t e d p r i m a r y s i d e v o l t a g e E2 =250; // r a t e d s e c o n d a r y s i d e v o l t a g e // i n i t i a l i s i n g p r i m a r y s i d e p a r a m e t e r s r1 =4.8; // p r i m a r y r e s i s t a n c e i n ohm x1 =11.2; // p r i m a r y l e a k a g e r e a c t a n c e i n ohm // i n i t i a l i s i n g s e c o n d a r y s i d e p a r a m e t e r s r2 =0.048; // s e c o n d a r y r e s i s t a n c e i n ohm x2 =0.112; // 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 n ohm k = E2 / E1 ; // t u r n ’ s r a t i o 1 z =5+ %i *3.5; re2 = r2 + r1 * k ^2; // r e s i s t a n c e r e f e r r e d t o s e c o n d a r y xe2 = x2 + x1 * k ^2; // r e a c t a n c e r e f e r r e d t o s e c o n d a r y ze2 = re2 + %i * xe2 ; zt = z + ze2 ; // t o t a l l o a d on s e c o n d a r y Z = abs ( zt ) ; I2 = E2 / Z ; // l o a d c u r r e n t on s e c o n d a r y disp ( ’ c a s e a ’ ) ; V2 = round ( I2 * abs ( z ) ) ; 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 i s %f V\n ’ , V2 ) ; disp ( ’ c a s e b ’ ) ; pf =0.8; // power f a c t o r I2l = P / E2 ; // r a t e d c u r r e n t o f s e c o n d a r y s i d e VD = I2l *( re2 * pf + xe2 * sqrt (1 - pf ^2) ) ; // v o l t a g e d r o p i n t r a n s f o r m e r l e a k a g e impedance 28


26 Vt = E2 - VD ; 27 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 i s %f V ’ , Vt )

Scilab code Exa 1.10 finding primary current secondary terminal voltage power factor power output and efficiency 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28

clc ; E1 =250; // v o l t a g e on low t e n s i o n s i d e E2 =2500; // v o l t a g e on h i g h t e n s i o n s i d e k = E2 / E1 ; // t u r n s r a t i o Z =380+230* %i ; // g i v e n l o a d c o n n e c t e d t o h i g h tension side Zl = Z / k ^2; // l o a d r e f e r r e d t o low t e n s i o n s i d e zl =0.2+0.7* %i ; // l e a k a g e i m p e d a n c e o f t r a n s f o r m e r zt = Zl + zl ; // t o t a l s e r i e s i m p e d a n c e ztm = abs ( zt ) ; // m a g n i t u d e o f t o t a l s e r i e s i m p e d a n c e I1 = E1 / zt ; I1m = abs ( I1 ) ; // m a g n i t u d e o f p r i m a r y l o a d c u r r e n t I2 = I1m / k ; // s e c o n d a r y l o a d c u r r e n t vt =5* abs ( Z ) ; 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 i s %f V\n ’ , vt ) ; R =500; // s h u n t b r a n c h r e s i s t a n c e X =250; // s h u n t b r a n c h l e a k a g e r e a c t a n c e Ic = E1 / R ; // c o r e l e s s component o f c u r r e n t Im = E1 /( %i * X ) ; // m a g n e t i z i n g component o f c u r r e n t Ie = Ic + Im ; // t o t a l e x c i t i n g c u r r e n t It = I1 + Ie ; // t o t a l c u r r e n t on low t e n s i o n s i d e Itm = abs ( It ) ; printf ( ’ p r i m a r y c u r r e n t i s %f A\n ’ , Itm ) ; pf = cos ( atan ( imag ( It ) , real ( It ) ) ) ; printf ( ’ power f a c t o r i s %f l a g g i n g \n ’ , pf ) , lpf = real ( Z ) / abs ( Z ) ; op = vt * I2 * lpf ; printf ( ’ o u t p u t power i s %f W\n ’ , op ) ; 29


29 30 31 32 33

pc = Ic ^2* R ; // c o r e l e s s power poh = I1m ^2* real ( zl ) ; // ohmic l o s s e s pin = E1 * Itm * pf ; // i n p u t power n =( op / pin ) *100; // e f f i c i e n c y printf ( ’ e f f i c i e n c y o f t r a n s f o r m e r i s %f p e r c e n t ’ ,n ) ;

Scilab code Exa 1.11 determining parameters of equivalent circuit referred to l v and h v sides 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

clc ; p =10000; // r a t e d o u t p u t o f t r a n s f o r m e r E1 =2500; // p r i m a r y s i d e r a t e d v o l t a g e E2 =250; // s e c o n d a r y s i d e r a t e d v o l t a g e k = E2 / E1 ; // t u r n ’ s r a t i o // i n i t i a l i s i n g r e s u l t s o f open c i r c u i t r e s u l t s on l . v side Vo =250; // open c i r c u i t v o l t a g e Io =1.4; // no l o a d c u r r e n t Po =105; // open c i r c u i t v o l t a g e // i n i t i a l i s i n g t h e r e s u l t s o f s h o r t c i r c u i t r e s u l t s on h . v s i d e Vsc =104; // s h o r t c i r c u i t v o l t a g e Isc =8; // s h o r t c i r c u i t c u r r e n t Psc =320; // power d i s s i p a t e d theta = Po /( Vo * Io ) ; // no l o a d power f a c t o r Ic = Io * theta ; // c o r e l e s s component o f c u r r e n t Im = Io * sqrt (1 - theta ^2) ; // m a g n e t i s i n g component o f current Ro = round ( Vo / Ic ) ; // s h u n t b r a n c h r e s i s t a n c e Xo = round ( Vo / Im ) ; // s h u n t b r a n c h i m p e d a n c e Zsc = Vsc / Isc ; // s h o r t c i r c u i t i m p e d a n c e reh = Psc / Isc ^2; // t o t a l t r a n s f o r m e r r e s i s t a n c e xeh = sqrt ( Zsc ^2 - reh ^2) ; // t o t a l t r a n s f o r m e r l e a k a g e impedance // e q u i v a l e n t c i r c u i t r e f e r r e d t o l . v s i d e 30


23 rel = reh * k ^2; 24 xml = xeh * k ^2; 25 printf ( ’ s h u n t b r a n c h r e s i s t a n c e and r e a c t a n c e 26

27 28 29 30

31

i s %f

ohm and %f ohm\n ’ ,Ro , Xo ) ; printf ( ’ v a l u e o f t r a n s f o r m e r r e s i s t a n c e and l e a k a g e r e a c t a n c e r e f e r r e d t o l . v s i d e i s %f ohm and %f ohm\n ’ ,rel , xml ) ; // e q u i v a l e n t c i r c u i t r e f e r r e d t o h . v s i d e Rch = Ro / k ^2; Xmh = Xo / k ^2; printf ( ’ s h u n t b r a n c h r e s i s t a n c e and r e a c t a n c e r e f e r r e d t o h . v s i d e i s %f ohm and %f ohm\n ’ ,Rch , Xmh ) ; printf ( ’ v a l u e o f t r a n s f o r m e r r e s i s t a n c e and l e a k a g e r e a c t a n c e r e f e r r e d t o h . v s i d e i s %f ohm and %f ohm\n ’ ,reh , xeh ) ;

Scilab code Exa 1.12 determining equivalent circuit parameters referred to h v side and its efficiency 1 2 3 4 5 6 7 8 9 10 11 12 13

clc ; P =200000; // r a t e d power o u t p u t o f t r a n s f o r m e r E1 =11000; // p r i m a r y s i d e v o l t a g e E2 =400; // s e c o n d a r y s i d e v o l t a g e // i n i t i a l i s i n g t h e r e s u l t s o f t h e open c i r c u i t t e s t p e r f o r m e d on l v s i d e Vo =400; // open c i r c u i t v o l t a g e i n V Io =9; // no l o a d c u r r e n t i n A Po =1500; // c o r e l o s s i n W // i n i t i a l i s i n g t h e r e s u l t s o f s h o r t c i r c u i t t e s t p e r f o r m e d on h v s i d e Vsc =350; // v o l t a g e a p p l i e d i n s h o r t c i r c u i t t e s t Isc = P /(3* E1 ) ; // s h o r t c i r c u i t c u r r e n t Psc =2100; // power d i s s i p a t e d i n s h o r t c i r c u i t t e s t E2p = E2 / sqrt (3) ; // p e r p h a s e v o l t a g e 31


14 pop = Po /3; // p e r p h a s e c o r e l o s s 15 Ic = pop / E2p ; // c o r e l o s s c u r r e n t 16 Im = sqrt ( Io ^2 - Ic ^2) ; // m a g n e t i s i n g component o f

current 17 R = E2p / Ic ; // c o r e l o s s r e s i s t a n c e i n ohm 18 X = E2p / Im ; // m a g n e t i z i n g r e a c t a n c e 19 Rh = R *( E1 / E2p ) ^2; // c o r e l o s s r e s i s t a n c e r e f e r r e d 20 21

22 23 24 25 26

27 28

to h v side Xh = floor ( X *( E1 / E2p ) ^2) ; // m a g n e t i z i n g component r e f e r r e d to h v s i d e printf ( ’ c o r e l o s s r e s i s t a n c e and m a g n e t i z i n g r e a c t a n c e r e f e r r e d t o h v s i d e i s %f ohm and %f ohm\n ’ ,Rh , Xh ) ; Ps = Psc /3; // ohmic l o s s p e r p h a s e Z = Vsc / Isc ; // t o t a l i m p e d a n c e o f t r a n s f o r m e r Re = Ps / Isc ^2; // T o t a l r e s i s t a n c e o f t r a n s f o r m e r r e f r r e d to high voltage s i d e Xe = sqrt ( Z ^2 - Re ^2) ; // t o t a l l e a k a g e i m p e d a n c e o f transformer r e f e r r e d to h v s i d e printf ( ’ t r a n s f o r m e r r e s i s t a n c e and l e a k a g e i m p e d a n c e r e f e r r e d t o h v s i d e a r e %f ohm and %f ohm\n ’ ,Re , Xe ) ; n =(1 -( pop + Ps /2^2) /( P /6+ pop + Ps /2^2) ) *100; // e f f i c i e n c y at h a l f load printf ( ’ e f f i c i e n c y a t h a l f l o a d i s %f p e r c e n t ’ ,n ) ;

Scilab code Exa 1.14 p u value of leakage impedance exciting current referred to l v and h v sides 1 2 3 4 5 6

clc ; p =20000; // r a t e d power o f t r a n s f o r m e r vbh =2500; // b a s e v o l t a g e i n v o l t s f o r h . v . s i d e vbl =250; // b a s e v o l t a g e i n v o l t s f o r l . v . s i d e ibh = p / vbh ; // b a s e c u r r e n t i n Ampere f o r h . v . s i d e zbh = vbh / ibh ; // b a s e i m p e d a n c e i n ohm 32


7 ze =2.6+4.3* %i ; // e q u i v a l e n t 8 9 10 11 12 13 14 15 16 17 18 19 20 21

22 23 24 25 26 27

l e a k a g e impedance r e f e r r e d t o h . v . s i d e i n ohm zepu = ze / zbh ; // p e r u n i t v a l u e i n ohm disp ( ’ Per u n i t v a l u e o f e q u i v a l e n t l e a k a g e i m p e d a n c e r e f e r r e d to h . v . s i d e i s ’ ); disp ( zepu ) ; k = vbl / vbh ; // t u r n ’ s r a t i o zel = ze * k ^2; // e q u i v a l e n t l e a k a g e i m p e d a n c e r e f e r r e d t o l . v . s i d e i n ohm ibl = p / vbl ; // b a s e c u r r e n t i n Ampere f o r l . v . s i d e zbl = vbl / ibl ; // b a s e i m p e d a n c e f o r l . v . s i d e zelpu = zel / zbl ; // p e r u n i t v a l u e i n ohm disp ( ’ Per u n i t v a l u e o f e q u i v a l e n t l e a k a g e i m p e d a n c e r e f e r r e d to l . v . s i d e i s ’ ); disp ( zelpu ) ; zepum = abs ( zepu ) ; // m a g n i t u d e o f p e r u n i t i m p e d a n c e vhl = zepum * vbh ; // t o t a l l e a k a g e i m p e d a c e d r o p on h . v . side vbl = zepum * vbl ; // t o t a l l e a k a g e i m p e d a c e d r o p on l . v . side printf ( ’ T o t a l l e a k a g e i m p e d a n c e d r o p on h . v . s i d e and l . v . s i d e a r e %f V and %f V r e s p e c t i v e l y \n ’ , vhl , vbl ) ; Ieh =4.8; // e x c i t i n g c u r r e n t i n Ampere Iepu = Ieh / ibh ; // p u v a l u e o f e x c i t i n g c u r r e n t r e f e r r e d to h . v . s i d e printf ( ’ Per u n i t v a l u e o f e x c i t i n g c u r r e n t r e f e r r e d t o h . v . s i d e i s %f p . u . \n ’ , Iepu ) ; Iel = Ieh / k ; // e x c i t i n g c u r r e n t r e f e r r e d t o l . v . side Ielpu = Iel / ibl ; // p u v a l u e o f e x c i t i n g c u r r e n t r e f e r r e d to l . v . s i d e printf ( ’ Per u n i t v a l u e o f e x c i t i n g c u r r e n t r e f e r r e d t o l . v . s i d e i s %f p . u . \n ’ , Ielpu ) ;

33


Scilab code Exa 1.15 determining p u values of circuit parameters referred to both sides 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

22 23 24 25

clc ; P =10000; // r a t e d power o f t r a n s f o r m e r vbh =2000; // b a s e v o l t a g e f o r h v s i d e i n v o l t s ibh = P / vbh ; // b a s e c u r r e n t f o r h v s i d e i n Ampere vbl =200; // b a s e v o l t a g e f o r l v s i d e i n v o l t s ibl = P / vbl ; // b a s e c u r r e n t f o r l v s i d e i n Ampere k = vbl / vbh ; // t u r n s r a t i o r1 =3.6; // r e s i s t a n c e o f h v s i d e o f t r a n s f o r m e r i n ohm x1 =5.2; // l e a k a g e r e a c t a c e h v s i d e o f t r a n s f o r m e r i n ohm z = vbh / ibh ; // b a s e i m p e d a n c e f o r h v s i d e ’ r1pu = r1 / z ; // p u v a l u e f o r r e s i s t a n c e o f h v s i d e o f t r a n s f o r m e r i n ohm x1pu = x1 / z ; // p u v a l u e f o r l e a k a g e r e a c t a n c e o f h v s i d e o f t r a n s f o r m e r i n ohm r2 =0.04; // r e s i s t a n c e o f l v s i d e o f t r a n s f o r m e r i n ohm x2 =0.056; // l e a k a g e r e a c t a c e l v s i d e o f t r a n s f o r m e r i n ohm // t o t a l r e s i s t a n c e r e f e r r e d t o h v s i d e re = r1 + r2 / k ^2; repu = re / z ; // t o t a l l e a k a g e i m p e d a n c e r e f e r r e d t o h v s i d e xe = x1 + x2 / k ^2; xepu = xe / z ; printf ( ’ t o t a l p e r u n i t r e s i s t a n c e and p e r u n i t l e a k a g e i m p e d a n c e r e f e r r e d t o h v s i d e a r e %f and %f\n ’ , repu , xepu ) ; Z = vbl / ibl ; // b a s e i m p e d a n c e f o r l v s i d e Re = r2 + r1 * k ^2; // t o t a l r e s i s t a n c e r e f e r r e d t o l v side Repu = Re / Z ; Xe = x2 + x1 * k ^2; // t o t a l l e a k a g e i m p e d a n c e r e f e r r e d t o l v side 34


26 27

Xepu = Xe / Z ; printf ( ’ t o t a l p e r u n i t r e s i s t a n c e and p e r u n i t l e a k a g e impedance r e f e r r e d to l v s i d e a r e %f and %f ’ , Repu , Xepu ) ;

Scilab code Exa 1.16 Determining parameters of equivalent circuit diagram 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

clc ; P =200000; // r a t e d power o f t r a n s f o r m e r E1 =4000; // p r i m a r y s i d e r a t e d v o l t a g e E2 =1000; // s e c o n d a r y s i d e r a t e d v o l t a g e n =0.97; // e f f i c i e n c y pfn =0.25; // power f a c t o r a t no l o a d pff =0.8; // power f a c t o r a t f u l l l o a d vr =5; // p e r c e n t a g e v o l t a g e r e g u l a t i o n Pl =((1/ n ) -1) *200000; // t o t a l l o s s e s a t f u l l l o a d Pf = Pl *0.6; // t o t a l l o s s e s a t 60% o f f u l l l o a d Po =( Pl - Pf ) /(1 -0.36) ; // ohmic l o s s e s Pc = Pl - Po ; // c o r e l o s s e s re2 =( Po / P ) *100; // P U t o t a l r e s i s t a n c e r e f e r r e d t o l . v . side xe2 =( vr - re2 * pff ) / sqrt (1 - pff ^2) ; // P U t o t a l l e a k a g e reactance r e f e r r e d to l . v . s i d e re2 =( re2 * E2 ^2) /(100* P ) ; // t o t a l r e s i s t a n c e i n ohms disp ( ’ T o t a l r e s i s t a n c e r e f e r r e d t o l . v . s i d e i s ’ ) ; printf ( ’ %f ohm ’ , re2 ) ; xe2 =( xe2 * E2 ^2) /(100* P ) ; // t o t a l l e a k a g e r e a c t a n c e i n ohms disp ( ’ T o t a l l e a k a g e r e a c t a n c e r e f e r r e d t o l . v . s i d e i s ’ ); printf ( ’ %f ohm ’ , xe2 ) ; Rc = E2 ^2/ Pc ; disp ( ’ C o r e l o s s r e s i s t a n c e i s ’ ) ; printf ( ’ %f ohm ’ , Rc ) ; 35


24 Ie2 = Pc /( E2 * pfn ) ; // e x c i t i n g c u r r e n t i n Ampere 25 Ic = Pc / E2 ; // c o r e l o s s c u r r e n t 26 Im = sqrt ( Ie2 ^2 - Ic ^2) ; // m a g n e t i z i n g component o f

exciting current 27 Xm = E2 / Im ; // m a g n e t i z i n g r e a c t a n c e 28 disp ( ’ M a g n e t i z i n g r e a c t a n c e i s ’ ) ; 29 printf ( ’ %f ohm ’ , Xm ) ; 30 disp ( ’ A l l p a r a m e t e r s a r e known . So ,

equivalent c i r c u i t d i a g r a m r e f e r r e d t o l . v . s i d e can be drawn . ’ ) ;

Scilab code Exa 1.18 Determining voltage regulation and terminal voltage at different power factor 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

clc ; P =20000; // r a t e d power o f t r a n s f o r m e r E1 =2500; // p r i m a r y s i d e v o l t a g e E2 =500; // s e c o n d a r y s i d e v o l t a g e r1 =8; // p r i m a r y r e s i s t a n c e i n ohm x1 =17; // p r i m a r y l e a k a g e r e a c t a n c e i n ohm r2 =0.3; // s e c o n d a r y r e s i s t a n c e i n ohm x2 =0.7; // s e c o n d a r y l e a k a g e r e a c t a n e i n ohm k = E2 / E1 ; // t u r n s r a t i o re2 = r2 + r1 * k ^2; // 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 secondary winding xe2 = x2 + x1 * k ^2; // e q u i v a l e n t l e a k a g e r e a c t a n c e r e f e r r e d to secondary winding Il = P / E2 ; // f u l l l o a d s e c o n d a r y c u r r e n t disp ( ’ c a s e a ’ ) ; pf =0.8; // l a g g i n g power f a c t o r vd = Il *( re2 * pf + xe2 * sqrt (1 - pf ^2) ) ; // V o l t a g e d r o p i n impedance i n v o l t s vt = E2 - vd ; // s e c o n d a r y t e r m i n a l v o l t a g e 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 f o r a l a g g i n g power f a c t o r i s %f v \n ’ , vt ) ; 36


18 vr =(( E2 - vt ) / E2 ) *100; // v o l t a g e r e g u l a t i o n 19 printf ( ’ v o l t a g e r e g u l a t i o n f o r a l a g g i n g power 20 21 22 23 24 25 26

f a c t o r i s %f p e r c e n t \n ’ , vr ) ; disp ( ’ c a s e b ’ ) ; pf =0.8; // l e a d i n g power f a c t o r vd = Il *( re2 * pf - xe2 * sqrt (1 - pf ^2) ) ; // V o l t a g e d r o p i n impedance i n v o l t s vt = E2 - vd ; // s e c o n d a r y t e r m i n a l v o l t a g e 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 f o r a l e a d i n g power f a c t o r i s %f v \n ’ , vt ) ; vr =(( E2 - vt ) / E2 ) *100; // v o l t a g e r e g u l a t i o n printf ( ’ v o l t a g e r e g u l a t i o n f o r a l e a d i n g power f a c t o r i s %f p e r c e n t \n ’ , vr ) ;

Scilab code Exa 1.19 Determining voltage regulation and secondary terminal voltage 1 2 3 4 5 6 7 8 9 10

clc ; rpu =0.02; // P U e q u i v a l e n t r e s i s t a n c e xpu =0.05; // P U e q u i v a l e n t l e a k a g e r e a c t a n c e E2 =440; // S e c o n d a r y f u l l l o a d v o l t a g e pf =0.8; // l a g g i n g power f a c t o r vr = rpu * pf + xpu * sqrt (1 - pf ^2) ; // P U v o l t a g e regulation printf ( ’ F u l l l o a d p . u . v o l t a g e r e g u l a t i o n i s %f o r %f p e r c e n t \n ’ ,vr , vr *100) ; dv = E2 * vr ; // c h a n g e i n t e r m i n a l v o l t a g e V2 = E2 - dv ; // s e c o n d a r y t e r m i n a l v o l t a g e 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 i s %f V ’ , V2 ) ;

Scilab code Exa 1.20 Determining voltage applied to high voltage side 1 clc ;

37


2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

P =10000; // r a t e d power o f t r a n s f o r m e r i n VA E1 =2000; // f u l l l o a d p r i m a r y v o l t a g e E2 =400; // f u l l l o a d s e c o n d a r y v o l t a g e k = E2 / E1 ; // t u r n s r a t i o pf =0.8; // l a g g i n g power f a c t o r // i n i t i a l i s i n g r e s u l t s o f s h o r t c i r c u i t t e s t v =60; // v o l t a g e a p p l i e d f o r s h o r t c i r c u i t t e s t i =4; // s h o r t c i r c u i t c u r r e n t p =100; // power d i s s i p a t e d i n s h o r t c i r c u i t ; reh = p / i ^2; // t o t a l r e s i s t a n c e r e f e r r e d t o h v s i d e zeh = v / i ; // t o t a l i m p e d a n c e r e f e r r e d t o h v s i d e xeh = sqrt ( zeh ^2 - reh ^2) ; // t o t a l l e a k a g e r e a c t a n c e r e f e r r e d to h v s i d e rel = reh * k ^2; // r e s i s t a n c e r e f e r r e d t o l v s i d e xel = xeh * k ^2; // r e a c t a n c e r e f e r r e d t o l v s i d e i2l = P / E2 ; // f u l l l o a d s e c o n d a r y c u r r e n t vr = i2l *( rel * pf + xel * sqrt (1 - pf ^2) ) ; // v o l t a g e regulation v2 = E2 + vr ; // t o t a l v o l t a g e o f s e c o n d a r y when t r a n s f o r m e r i s o p e r a t i n g on f u l l l o a d v1 = v2 / k ; // v o l t a g e a p p l i e d t o p r i m a r y t o d e l i v e r f u l l load printf ( ’ v o l t a g e a p p l i e d t o p r i m a r y t o d e l i v e r f u l l l o a d i s %f v ’ , v1 ) ;

Scilab code Exa 1.21 voltage at sending end of feeder primary terminals active and reactive power power factor 1 2 3 4 5 6 7

clc ; zf =30+120* %i ; // f e e d e r i m p e d a n c e E1 =33000; // p r i m a r y s i d e v o l t a g e E2 =3300; // s e c o n d a r y s i d e v o l t a g e k = E2 / E1 ; // t u r n s r a t i o P =100000; // l o a d power pf =0.8; // l e a d i n g power f a c t o r o f l o a d 38


8 zl =0.3+1.4* %i ; // l e a k a g e i m p e d a n c e r e f e r r e d 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27

to l v

side zfl = zf * k ^2; // f e e d e r i m p e d a n c e r e f e r r e d t o l v s i d e vt =3300; // t e r m i n a l v o l t a g e il = P /( vt * pf ) ; // l o a d c u r r e n t R = real ( zfl ) + real ( zl ) ; // t o t a l r e s i s t a n c e r e f e r r e d to l v s i d e X = imag ( zfl ) + imag ( zl ) ; // t o t a l i m p e d a n c e r e f e r r e d t o l v side vfl = vt + il *( R * pf - X * sqrt (1 - pf ^2) ) ; // v o l t a g e a t t h e s e n d i n g end o f f e e d e r r e f e r r e d t o l v s i d e vf = vfl / k ; // v o l t a g e a t t h e s e n d i n g end o f f e e d e r printf ( ’ V o l t a g e a t t h e s e n d i n g end o f f e e d e r i s %f v \n ’ , vf ) ; v2 = vt + il *( real ( zl ) * pf - imag ( zl ) * sqrt (1 - pf ^2) ) ; // voltage induced in secondary windings v1 = round ( v2 / k ) ; printf ( ’ v o l t a g e a t t h e p r i m a r y t e r m i n a l s o f t r a n s f o r m e r i s %f v \n ’ , v1 ) ; ap = il ^2* R ; printf ( ’ a c t i v e power l o s s i s %f W\n ’ , ap ) ; ar = il ^2* X ; printf ( ’ r e a c t i v e power l o s s i s %f W\n ’ , ar ) ; =P - P * %i * tan ( acosd ( pf ) *( %pi /180) ) ; // c o m p l e x power a t l o a d end i n VA s = +(( ap + ar * %i ) ) ; // c o m p l e x power a t f e e d e r end i n VA pfs = cos ( atan ( imag ( s ) , real ( s ) ) ) ; printf ( ’ power f a c t o r a t t h e s e n d i n g end i s %f l e a d i n g ’ , pfs ) ;

Scilab code Exa 1.22 input power and power factor 1 clc ; 2 P =10000; // r a t e d power o f

transformer

39


3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28

E1 =2000; // p r i m a r y s i d e v o l t a g e E2 =200; // s e c o n d a r y s i d e v o l t a g e f =50; // f r e q u e n c y i n h e r t z po =125; // no l o a d power pfo =0.15; // no l o a d power f a c t o r zbh = E1 ^2/ P ; // b a s e i m p e d a n c e on h v s i d e k = E2 / E1 ; // t u r n s r a t i o zl =0.5+1* %i ; // p e r c e n t l e a k a g e i m p e d a n c e zlh = zl *( zbh * k ^2) ; // p e r c e n t l e a k a g e i m p e d a n c e r e f e r r e d to h v s i d e Rc = E1 ^2/ po ; // c o r e l o s s r e s i s t a n c e Io = po /( E1 * pfo ) ; // No l o a d c u r r e n t Xm = E1 /( Io * sqrt (1 - pfo ^2) ) ; // m a g n e t i z i n g r e a c t a n c e p =10000; // l o a d power pf =0.8; // power f a c t o r o f l o a d il = p /( E2 * pf ) ; // s e c o n d a r y l o a d c u r r e n t ilp = il * k ; // p r i m a r y l o a d c u r r e n t vp = E1 + ilp *( real ( zlh ) * pf + imag ( zlh ) * sqrt (1 - pf ^2) ) ; ap = ilp ^2* real ( zlh ) ; // a c t i v e power l o s s i n s e r i e s resistance ar = ilp ^2* imag ( zlh ) ; // r e a c t i v e power l o s s i n s e r i e s reactance Ap = vp ^2/ Rc ; // a c t i v e power l o s s i n c o r e l o s s resistance Ar = vp ^2/ Xm ; // r e a c t i v e power l o s s i n m a g n e t i z i n g reactance l = p *(1+ %i * tan ( acos (0.8) ) ) ; // c o m p l e x power a t l o a d end i n VA i =( real ( l ) + ap + Ap ) + %i *( imag ( l ) + ar + Ar ) ; // c o m p l e x power i n p u t t o t r a n s f o r m e r VA printf ( ’ r e a l power i n p u t t o t r a n s f o r m e r i s %f W\n ’ , real ( i ) ) ; ipf = cos ( atan ( imag ( i ) , real ( i ) ) ) ; printf ( ’ i n p u t power f a c t o r i s %f l a g g i n g ’ , ipf ) ;

40


Scilab code Exa 1.24 hysteresis and eddy current losses 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

clc ; pc1 =52; // c o r e l o s s a t f =40 f1 =40; // f r e q u e n c y i n h e r t z pc2 =90; // c o r e l o s s a t f =60 f2 =60; // f r e q u e n c y i n h e r t z f =[ f1 f1 ^2; f2 f2 ^2]; pc =[ pc1 ; pc2 ]; k = inv ( f ) * pc ; // p r o p o r t i o n a l i t y c o n s t a n t s f o r h y s t e r e s i s and eddy current l o s s e s are kh = k (1) ; disp ( kh ) // p r o p o r t i o n a l i t y c o n s t a n t s f o r hysteresis losses ke = k (2) ; disp ( ke ) // p r o p o r t i o n a l i t y c o n s t a n t s f o r eddy c u r r e n t l o s s e s // d e t e r m i n i n g b o t h l o s s e s a t 50 h e r t z f =50; ph = kh * f ; printf ( ’ h y s t e r e s i s l o s s e s a t 50 h e r t z i s %f W\n ’ , ph ) ; pe = ke * f ^2; printf ( ’ eddy c u r r e n t l o s s e s a t 50 h e r t z i s %f W’ , pe ) ; // a n s w e r f o r eddy c u r r e n t l o s s e s i s m i s p r i n t e d i n book

Scilab code Exa 1.25 total core loss 1 2 clc ; 3 // s u b s c r i p t s 1 and 2 a r e u s e d t h e

quantities r e f e r r e d t o 60 hz and 50 hz f r e q u e n c y respectively 4 v1 =220; // r a t e d v o l t a g e a t 60 hz 41


5 6 7 8 9 10 11 12 13 14 15

f1 =60; // o p e r a t i n g f r e q u e n c y ph1 =340; // h y s t e r e s i s l o s s a t 60 hz pe1 =120; // eddy c u r r e n t l o s s a t 60 hz v2 =230; // r a t e d v o l t a g e a t 50 hz f2 =50; // o p e r a t i n g f r e q u e n c y s =1.6; // S t e i n m e t z ’ s c o n s t a n t B =( f1 / f2 ) *( v2 / v1 ) ; // r a t i o o f f l u x d e n s i t i e s Bm2/ Bm1 ph2 = ceil ( ph1 *(50/60) * B ^ s ) ; // h y s t e r e s i s l o s s a t 50 hz pe2 = pe1 *( f2 / f1 ) ^2*( B ) ^2; // eddy c u r r e n t l o s s a t 50 hz pc = ph2 + pe2 ; printf ( ’ T o t a l c o r e l o s s a t 50 hz i s %f W’ , pc ) ;

Scilab code Exa 1.26 determining losses and output at 60 hz 1 clc ; 2 // s u b s c r i p t s 1 and 2 a r e u s e d t o 3 4 5 6 7 8 9 10 11 12 13 14

r e f e r 50 hz and 60 hz q u a n t i t y r e s p e c t i v e l y // v o l t a g e and c u r r e n t i s same f o r b o t h t h e c a s e s s =1.6; // S t e i n m e t z ’ s c o e f f i c i e n t poh1 =1.6; // p e r c e n t a g e ohmic l o s s e s ph1 =0.9; // p e r c e n t a g e h y s t e r e s i s l o s s e s pe1 =0.6; // p e r c e n t a g e eddy c u r r e n t l o s s e s f1 =50; // f r e q u e n c y i n h e r t z f2 =60; // f r e q u e n c y i n h e r t z B = f1 / f2 // s i n c e v o l t a g e l e v e l a r e same f o r b o t h c a s e s r a t i o o f f l u x d e n s i t i e s i . e Bm2/Bm1=f 1 / f 2 ph2 = ph1 *( f2 / f1 ) * B ^ s ; // p e r c e n t a g e h y s t e r e s i s l o s s e s pe2 = pe1 *( f2 / f1 ) ^2* B ^2; // p e r c e n t a g e eddy c u r r e n t losses poh2 = poh1 ; // s i n c e t h e v o l t a g e and c u r r e n t l e v e l s a r e same t h e r e f o r e ohmic l o s s e s a r e same // f o r t h e t o t a l l o s s e s t o be r e m a i n same a t b o t h 42


15 16 17 18

19 20 21 22

t h e f r e q u e n c i e s o n l y ohmic l o s s e s can be v a r i e d p = poh1 + ph1 + pe1 ; // t o t a l l o s s e s a t 50 hz pc = ph2 + pe2 ; // t o t a l c o r e l o s s e s a t 60 hz pnoh =p - pc ; // p e r m i s s i b l e v a l u e f o r new ohmic l o s s e s x = sqrt ( pnoh / poh1 ) ; // f a c t o r by which o u t p u t a t 50 hz s h o u l d be m u l t i p l i e d t o g e t t h e same o u t p u t a t 60 hz printf ( ’ ohmic l o s s e s a t 60 hz i s %f p e r c e n t \n ’ , poh2 ) ; printf ( ’ h y s t e r e s i s l o s s e s a t 60 hz i s %f p e r c e n t \n ’ , ph2 ) ; printf ( ’ eddy c u r r e n t l o s s e s a t 60 hz i s %f p e r c e n t \n ’ , pe2 ) ; printf ( ’ f a c t o r by which o u t p u t a t 50 hz s h o u l d be m u l t i p l i e d t o g e t t h e same o u t p u t a t 60 hz i s %f ’ ,x ) ;

Scilab code Exa 1.27 finding efficiency and losses 1 clc ; 2 // s u b s c r i p t s 1 and 2 a r e u s e d t o i n d i c a t e

3 4 5 6 7

8

t r a n s f o r m e r o f 11 kv a t 25 hz and 22 kv a t 50 hz respectively // f o r same c u r r e n t power i s d o u b l e d t h e r e f o r e P2=2 P1 poh1 =1.8; // ohmic l o s s e s a s a p e r c e n t a g e o f t o t a l power P1 ph1 =0.8; // h y s t e r e s i s l o s s e s a s a p e r c e n t a g e o f t o t a l power P1 pe1 =0.3; // eddy c u r r e n t l o s s e s a s a p e r c e n t a g e o f t o t a l power P1 poh2 = poh1 /2; // ohmic l o s s e s do n o t c h a n g e w i t h f r e q u e n c y but c h a n g e s w i t h v o l t a g e s i n c e p1=2p1 we g e t t h e r e s u l t shown // s i n c e f r e q u e n c y a l s o g e t s d o u b l e d whwn v o l t a g e 43


9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

l e v e l s d o u b l e t h e r e f o r e t h e r e i s no c h a n g e i n f l u x d e n s i t y i . e i s Bm1=Bm2 f1 =25; // f r e q u e n c y i n h e r t z f2 =50; // f r e q u e n c y i n h e r t z ph2 =( f2 / f1 ) * ph1 ; // h y s t e r e s i s l o s s e s a r e d i r e c t l y p r o p o r t i o n a l to frequency pe2 =( f2 / f1 ) ^2* pe1 ; // eddy c u r r e n t l o s s e s a r e d i r e c t l y p r o p o r t i o n a l to frequency // we know p2=2p1 ph2p = ph2 /2; // h y s t e r e s i s l o s s e s as a percentage of t o t a l power P2 pe2p = pe2 /2; // eddy c u r r e n t l o s s e s a s a p e r c e n t a g e o f t o t a l power P2 printf ( ’ ohmic l o s s e s a s a p e r c e n t a g e o f t o t a l power a t 50 hz i s %f p e r c e n t \n ’ , poh2 ) ; printf ( ’ h y s t e r e s i s l o s s e s a s a p e r c e n t a g e o f t o t a l power a t 50 hz i s %f p e r c e n t \n ’ , ph2p ) ; printf ( ’ eddy c u r r e n t l o s s e s a s a p e r c e n t a g e o f t o t a l power a t 50 hz i s %f p e r c e n t \n ’ , pe2p ) ; // e f f i c i e n c y a t f 1 , v1 n1 =(1 -(( poh1 + ph1 + pe1 ) /100) /(1+(( poh1 + ph1 + pe1 ) /100) ) ) *100; printf ( ’ e f f i c i e n c y a t 25 hz i s %f p e r c e n t \n ’ , n1 ) ; // e f f i c i e n c y a t f 2 , v2 n2 =(1 -(( poh2 + ph2p + pe2p ) /100) /(1+(( poh2 + ph2p + pe2p ) /100) ) ) *100; printf ( ’ e f f i c i e n c y a t 50 hz i s %f p e r c e n t ’ , n2 ) ;

Scilab code Exa 1.28 determining efficiency load voltage regulation and secondary terminal voltage 1 2 3 4

clc ; P =10000; // r a t e d power o f t r a n s f o r m e r i n VA E1 =2500; // p r i m a r y s i d e v o l t a g e E2 =250; // s e c o n d a r y s i d e v o l t a g e 44


5 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

pf =0.8; // power f a c t o r // i n i t i a l i s i n g t h e r e s u l t s o f open c i r c u i t t e s t vo =250; // open c i r c u i t v o l t a g e io =0.8; // no l o a d c u r r e n t po =50; // open c i r c u i t v o l t a g e // i n i t i a l i s i n g t h e r e s u l t s o f open c i r c u i t t e s t vsc =60; // s h o r t c i r c u i t v o l t a g e isc =3; // s h o r t c i r c u i t c u r r e n t psc =45; // power d i s s i p a t e d i n t e s t ifl = P / E1 ; // f u l l l o a d c u r r e n t on p r i m a r y s i d e poh = psc *( ifl / isc ) ^2; // ohmic l o s s e s a t f u l l l o a d current disp ( ’ c a s e a ( 1 ) ’ ) ; n =(1 -( po +( poh /4^2) ) /( po +( poh /4^2) +( P * pf ) /4) ) *100; // e f f i c i e n c y a t 1/4 l o a d printf ( ’ e f f i c i e n c y a t 1/4 l o a d i s %f p e r c e n t \n ’ ,n ) ; disp ( ’ c a s e a ( 2 ) ’ ) ; n =(1 -( po +( poh /2^2) ) /( po +( poh /2^2) +( P * pf ) /2) ) *100; // e f f i c i e n c y a t 1/2 l o a d printf ( ’ e f f i c i e n c y a t 1/2 l o a d i s %f p e r c e n t \n ’ ,n ) ; disp ( ’ c a s e a ( 3 ) ’ ) ; n =(1 -( po +( poh /1^2) ) /( po +( poh /1^2) +( P * pf ) /1) ) *100; // e f f i c i e n c y at f u l l load printf ( ’ e f f i c i e n c y a t f u l l l o a d i s %f p e r c e n t \n ’ ,n ) ; disp ( ’ c a s e a ( 4 ) ’ ) ; n =(1 -( po +(( poh *5^2) /4^2) ) /( po +(( poh *5^2) /4^2) +( P * pf *5) /4) ) *100; // e f f i c i e n c y a t 1 ∗ 1 / 4 l o a d printf ( ’ e f f i c i e n c y a t 5/4 l o a d i s %f p e r c e n t \n ’ ,n ) ; // l e t maximum e f f i c i e n c y o c c u r s a t x t i m e s t h e r a t e d KVA // maximum e f f i c i e n c y o c c u r s when c o r e l o s s becomes e q u a l t o ohmic l o s s e s x = sqrt ( po / poh ) ; nm =( x * P ) /1000; // VA o u t p u t a t maximum nmax =(1 -(2* po ) /( nm *1000* pf +2* po ) ) *100; printf ( ’KVA l o a d a t which maximum e f f i c i e n c y o c c u r s i s %f KVA\n ’ , nm ) ; printf ( ’ Maximum e f f i c i e n c y i s %f p e r c e n t \n ’ , nmax ) ; 45


35 // from s h o r t c i r c u i t t e s t 36 reh = psc / isc ^2; // t o t a l r e s i s t a n c e 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51

r e f e r r e d to h v side zeh = vsc / isc ; // t o t a l i m p e d a n c e r e f e r r e d t o h v s i d e xeh = sqrt ( zeh ^2 - reh ^2) ; // t o t a l l e a k a g e r e a c t a n c e r e f e r r e d to h v s i d e er =( ifl * reh ) / E1 ; // p u r e s i s t a n c e ex =( ifl * xeh ) / E1 ; // p u r e a c t a n c e vr =( er * pf + ex * sqrt (1 - pf ^2) ) *100; // p u v o l t a g e regulation printf ( ’ p u v o l t a g e r e g u l a t i o n f o r l a g g i n g power f a c t o r i s %f p e r c e n t \n ’ , vr ) ; dv = E2 *( vr /100) ; // v o l t a g e d r o p i n s e r i e s i m p e d a n c e v2 = E2 - dv ; 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 f o r l a g g i n g power f a c t o r o f 0 . 8 i s %f v \n ’ , v2 ) ; // v o l t a a g e r e g u l a t i o n f o r l e a d i n g power f a c t o r vr =( er * pf - ex * sqrt (1 - pf ^2) ) *100; // p u v o l t a g e regulation printf ( ’ p u v o l t a g e r e g u l a t i o n f o r l e a d i n g power f a c t o r i s %f p e r c e n t \n ’ , vr ) ; dv = E2 *( vr /100) ; // v o l t a g e d r o p i n s e r i e s i m p e d a n c e v2 = E2 - dv ; 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 f o r l e a d i n g power f a c t o r o f 0 . 8 i s %f v \n ’ , v2 ) ;

Scilab code Exa 1.29 Determining transformer core loss ohmic loss and pu value of equivalent resistance 1 clc ; 2 p =20000; // r a t e d c a p a c i t y o f t r a n s f o r m e r 3 n =0.98; // e f f i c i e n c y o f t r a n s f o r m e r a t f u l l

load

and h a l f l o a d 4 c =[ 1 1; 1 1/4]; 5 o =[ ((1/ n ) -1) * p ; ((1/ n ) -1) *( p /2) ]; 46


6 l = inv ( c ) * o ; 7 printf ( ’ Core l o s s e s a r e %f W\n ’ ,l (1) ) ; 8 printf ( ’ Ohmic l o s s e s a r e %f W\n ’ ,l (2) ) ; 9 re = l (2) / p ; 10 printf ( ’ p . u . v a l u e o f e q u i v a l e n t r e s i s t a n c e

i s %f ’ ,

re ) ;

Scilab code Exa 1.30 determining voltage regulation at given power factor 1 clc ; 2 P =100000; // VA o f t r a n s f o r m e r 3 nmax =0.98; // maximum e f f i c i e n c y o f t r a n s f o r m e r 4 pf =0.8; // power f a c t o r a t which maximum e f f i c i e n c y 5 6 7 8 9 10 11 12 13 14 15

occurs l =80; // p e r c e n t a g e o f f u l l l o a d a t which maximum e f f i c i e n c y occurs po = P * pf *( l /100) ; // o u t p u t a t maximum e f f i c i e n c y pl =((1/ nmax ) -1) * po ; // t o t a l l o s s e s pc = pl /2; // c o r e l o s s poh = pc ; // a t maximum e f f i c i e n c y v a r i a b l e l o s s e s a r e equal to constant l o s s e s pohl = poh *(100/ l ) ^2; // ohmic l o s s e s a t f u l l l o a d z =0.05; // p u l e a k a g e i m p e d a n c e r = pohl / P ; // p u r e s i s t a n c e x = sqrt ( z ^2 - r ^2) ; // p u l e a k a g e r e a c t a n c e vr =( r * pf + x * sqrt (1 - pf ^2) ) *100; // v o l t a g e r e g u l a t i o n printf ( ’ V o l t a g e r e g u l a t i o n a t 0 . 8 p . f . l a g g i n g i s %f p e r c e n t ’ , vr ) ;

Scilab code Exa 1.31 Determining efficiency on full load full load power factor and load power factor for zero voltage drop 47


1 clc ; 2 vdr =2; // p e r c e n t a g e 3 4

5 6 7 8 9 10 11

f u l l l o a d v o l t a g e drop i n resistance vdx =4; // p e r c e n t a g e f u l l l o a d v o l t a g e d r o p i n leakage reactance // f u l l l o a d ohmic l o s s e s a r e e q u a l t o 0 . 0 2 ∗VA r a t i n g o f t r a n s f o r m e r which i s e q u a l t o i r o n losses n =100/(1+( vdr /100) +( vdr /100) ) ; printf ( ’ E f f i c i e n c y on f u l l l o a d a t u n i t y p . f i s %f p e r c e n t \n ’ ,n ) ; // maximum v o l t a g e d r o p means v o l t a g e r e g u l a t i o n i s a l s o maximum pf = vdr / sqrt ( vdr ^2+ vdx ^2) ; printf ( ’ F u l l l o a d power f a c t o r a t which v o l t a g e r e g u l a t i o n i s maximum i s %f l a g g i n g \n ’ , pf ) ; pf = vdx / sqrt ( vdr ^2+ vdx ^2) ; printf ( ’ l o a d power f a c t o r a t which v o l t a g e r e g u l a t i o n i s z e r o i s %f l e a d i n g ’ , pf ) ;

Scilab code Exa 1.32 Determine input power and power factor at a certain load 1 2 3 4 5 6 7 8 9 10 11

clc ; P =20000; // r a t e d VA o f t r a n s f o r m e r E1 =3300; // r a t e d v o l t a g e o f p r i m a r y E2 =220; // r a t e d v o l t a g e o f s e c o n d a r y v2 =220; // v o l t a g e a t which l o a d i s g e t t i n g delivered p =14960; // l o a d power i n Watts pf =0.8; // power f a c t o r a t on l o a d pc =160; // c o r e l o s s pfo =0.15; // power f a c t o r a t no l o a d il = p /( v2 * pf ) ; // l o a d c u r r e n t is = P / E2 ; // r a t e d c u r r e n t o f s e c o n d a r y 48


12 vr =1 ; // p e r c e n t a g e 13 14 15 16 17 18 19 20 21 22 23 24

25 26 27

v o l t a g e drop o f r a t e d v o l t a g e in total r e s i s t a n c e vx =3 ; // p e r c e n t a g e v o l t a g e d r o p o f r a t e d v o l t a g e in total leakage reactance re2 =( E2 * vr ) /( is *100) ; // t o t a l r e s i s t a n c e r e f e r r e d to secondary xe2 =( E2 * vx ) /( is *100) ; // t o t a l l e a k a g e r e a c t a n c e r e f e r r e d to secondary poh = il ^2* re2 ; // ohmic l o s s e s pi = poh + pc + p ; // t o t a l i n p u t power // E2 a s a r e f e r e n c e i2 = il *( pf - %i * sqrt (1 - pf ^2) ) ; E2n = v2 + i2 *( re2 + %i * xe2 ) ; // s e c o n d a r y w i n d i n g v o l t a g e io = pc /( pfo * E2 ) ; // no l o a d c u r r e n t ic = pc / E2 ; // c o r e l o s s c u r r e n t im = sqrt ( io ^2 - ic ^2) ; // m a g n e t i z i n g c u r r e n t I = i2 +( ic - im * %i ) ; // t o t a l i n p u t c u r r e n t , n e g a t i v e s i g n b e f o r e im i n d i c a t e s t h a t i t l a g s b e h i n d E2 by 90 d e g r e e pfi = cos ( atan ( imag ( I ) , real ( I ) ) ) ; // i n p u t power factor printf ( ’ T o t a l i n p u t power i s %f W \n ’ , pi ) ; printf ( ’ I n p u t power f a c t o r i s %f l a g g i n g ’ , pfi ) ;

Scilab code Exa 1.33 determining load power factor at minimum secondary terminal voltage and minimum secondary terminal voltage 1 2 3 4 5

clc ; P =500000; // VA r a t i n g o f t r a n s f o r m e r E2 =400; // r a t e d s e c o n d a r y v o l t a g e nmax =0.98; // maximum e f f i c i e n c y o f t r a n s f o r m e r l =80; // p e r c e n t a g e o f f u l l l o a d a t which maximum e f f i c i e n c y occurs 6 ze2 =4.5; // p e r c e n t a g e i m p e d a n c e 7 pt =((1/ nmax ) -1) * P *( l /100) ; // t o t a l l o s s e s 49


8 pc = pt /2; // c o r e

l o s s = ohmic l o s s a t maximum

efficiency 9 poh = pc ; // ohmic l o s s 10 pohl = poh *(100/ l ) ^2; // f u l l l o a d ohmic l o s s e s 11 re2 =( pohl / P ) *100; // p e r c e n t a g e r e s i s t a n c e 12 xe2 = sqrt ( ze2 ^2 - re2 ^2) ; // p e r c e n t a g e l e a k a g e

reactance pfl = re2 / ze2 ; // l o a d power f a c t o r vr = re2 * pfl + xe2 * sqrt (1 - pfl ^2) ; // v o l t a g e r e g u l a t i o n dv =( E2 * vr ) /100; // c h a n g e i n t e r m i n a l v o l t a g e V2 = E2 - dv ; // S e c o n d a r y t e r m i n a l v o l t a g e printf ( ’ Load power f a c t o r a t which s e c o n d a r y t e r m i n a l v o l t a g e i s minimum i s %f\n ’ , pfl ) ; 18 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 i s %f v ’ , V2 ) ; 19 // a n s w e r f o r t o t a l l o s s e s i s g i v e n wrong i n t h e book

13 14 15 16 17

Scilab code Exa 1.34 Determining full day efficiency 1 2 3 4 5 6 7 8 9 10 11 12 13 14

clc ; P =5000; // r a t e d VA o f t r a n s f o r m e r pc =40; // c o r e l o s s , i t r e m a i n s f i x e d f o r w h o l e day poh =100; // ohmic l o s s e s // d a t a f o r d u r a t i o n 7 A .M t o 1 P .M p1 =3000; // power consumed pf1 =0.6 // power f a c t o r o f l o a d pk1 = p1 / pf1 ; // VA l o a d poh1 = poh *( pk1 / P ) ^2; // ohmic l o s s e s f o r g i v e n duration // d a t a f o r d u r a t i o n 1 P .M t o 6 P .M p2 =2000; // power consumed pf2 =0.8 // power f a c t o r o f l o a d pk2 = p2 / pf2 ; // VA l o a d poh2 = poh *( pk2 / P ) ^2; // ohmic l o s s e s f o r g i v e n duration 50


15 // d a t a f o r d u r a t i o n 6 P .M t o 1 A .M 16 p3 =6000; // power consumed 17 pf3 =0.9 // power f a c t o r o f l o a d 18 pk3 = p3 / pf3 ; // VA l o a d 19 poh3 = poh *( pk3 / P ) ^2; // ohmic l o s s e s f o r 20 21 22 23 24 25 26

given

duration // d a t a f o r d u r a t i o n 1 A .M t o 7 A .m =no l o a d poht = poh1 *6+ poh2 *5+ poh3 *7; // e n e r g y l o s t i n ohmic losses pct =( pc *24) ; // d a i l y e n e r g y l o s t a s c o r e l o s s ptl = poht + pct ; // t o t a l e n e r g y l o s t po = p1 *6+ p2 *5+ p3 *7; // o u t p u t n =(1 -( ptl /( ptl + po ) ) ) *100; printf ( ’ A l l day e f f i c i e n c y i s %f p e r c e n t ’ ,n ) ;

Scilab code Exa 1.35 Determining hysteresis and eddy current losses at given frequencies 1 clc ; 2 3 //V/ f

4 5 6 7 8 9 10 11 12 13 14 15

r a t i o i s same f o r e v e r y c a s e h e n c e h y s t e r e s i s l o s s e s and eddy c u r r e n t l o s s e s can be c a l c u l a t e d separately // d a t a f o r column 1 vt1 =214; // t e r m i n a l v o l t a g e f1 =50; // f r e q u e n c y i n hz p1 =100; // power i n p u t i n Watts vp1 = vt1 ; // p e r p h a s e v o l t a g e pv1 = p1 /3; // p e r p h a s e power pc1 = pv1 / f1 ; // c o r e l o s s p e r c y c l e // d a t a f o r column 2 vt2 =171; // t e r m i n a l v o l t a g e f2 =40; // f r e q u e n c y i n hz p2 =72.5; // power i n p u t i n Watts vp2 = vt2 ; // p e r p h a s e v o l t a g e 51


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 44 45 46

pv2 = p2 /3; // p e r p h a s e power pc2 = pv2 / f2 ; // c o r e l o s s p e r c y c l e // d a t a f o r column 3 vt3 =128; // t e r m i n a l v o l t a g e f3 =30; // f r e q u e n c y i n hz p3 =50; // power i n p u t i n Watts vp3 = vt3 ; // p e r p h a s e v o l t a g e pv3 = p3 /3; // p e r p h a s e power pc3 = pv3 / f3 ; // c o r e l o s s p e r c y c l e // d a t a f o r column 4 vt4 =85.6; // t e r m i n a l v o l t a g e f4 =20; // f r e q u e n c y i n hz p4 =30; // power i n p u t i n Watts vp4 = vt4 ; // p e r p h a s e v o l t a g e pv4 = p4 /3; // p e r p h a s e power pc4 = pv4 / f4 ; // c o r e l o s s p e r c y c l e // V a l u e s o f k1 and k2 have b e e n o b t a i n e d from g r a p h k1 =0.39; k2 =( pc1 - k1 ) /50; F1 =60; // f r e q u e n c y a t which l o s s e s h a s t o be calculated ph1 = k1 * F1 ; // p e r p h a s e h y s t e r e s i s l o s s a t 60 hz pe1 = k2 * F1 ^2; // p e r p h a s e eddy c u r e n t l o s s a t 60 hz pht =3* ph1 ; // t o t a l h y s t e r e s i s l o s s pet =3* pe1 ; // t o t a l eddy c u r r e n t l o s s printf ( ’ T o t a l h y s t e r e s i s and eddy c u r r e n t l o s s e s a t 60 hz a r e %f W and %f W r e s p e c t i v e l y \n ’ ,pht , pet ) ; F2 =40; // f r e q u e n c y a t which l o s s e s h a s t o be calculated ph2 = k1 * F2 ; // p e r p h a s e h y s t e r e s i s l o s s a t 40 hz pe2 = k2 * F2 ^2; // p e r p h a s e eddy c u r e n t l o s s a t 40 hz pht =3* ph2 ; // t o t a l h y s t e r e s i s l o s s pet =3* pe2 ; // t o t a l eddy c u r r e n t l o s s printf ( ’ T o t a l h y s t e r e s i s and eddy c u r r e n t l o s s e s a t 40 hz a r e %f W and %f W r e s p e c t i v e l y ’ ,pht , pet ) ;

52


Scilab code Exa 1.36 Determining reading of wattmeter connected on l v side 1 clc ; 2 E1 =230; // p r i m a r y r a t i n g 3 4 5 6 7 8 9 10 11 12

13 14 15 16

17 18 19

o f t r a n s f o r m e r 1 and transformer 2 E2 =400; // s e c o n d a r y r a t i n g o f t r a n s f o r m e r 1 e2 =410; // s e c o n d a r y r a t i n g o f t r a n s f o r m e r 2 iv =25; // c u r r e n t f e e d e d by v o l t a g e r e g u l a t o r t o h v s e r i e s winding pc =200; // c o r e l o s s i n e a c h t r a n s f o r m e r r =1 // a s s u m i n g r e s i s t a n c e o f t r a n s f o r m e r t o be 1 x =3* r // a s p e r q u e s t i o n l e a k a g e r e a c t a n c e i s t h r i c e of resistance il1 =( iv * E2 ) / E1 ; // p r i m a r y c u r r e n t o f t r a n s f o r m e r 1 il2 =( iv * e2 ) / E1 ; // p r i m a r y c u r r e n t o f t r a n s f o r m e r 2 pf = r / sqrt ( r ^2+ x ^2) ; // power f a c t o r d u r i n g s h o r t circuit // As p e r t h e c i r c u i t d i a g r a m g i v e n i n q u e s t i o n , by K i r c h o f f s c u r r e n t law c u r r e n t t h r o u g h c u r r e n t c o i l o f w a t t m e t e r W1 i s g i v e n by I = il2 - il1 ; // 2∗ c o r e l o s s i s t h e r e a d i n g o f w a t t m e t e r 2 W = E1 * I * pf ; // r e a d i n g o f w a t t m e t e r 1 c o n n e c t e d on l v side // i n c i r c u i t d i a g r a m i f t e r m i n a l a i s c o n n e c t e d t o c and t e r m i n a l b i s c o n n e c t e d t o d t h e c u r r e n t I and I o ( no l o a d c u r r e n t ) f l o w i n t h e same d i r e c t i o n o f c u r r e n t c o i l o f Wattmeter . Hence i t s reading i s i n c r e a s e d to Wt =2* pc + W ; printf ( ’ r e a d i n g o f w a t t m e t e r a s p e r t h e c o n n e c t i o n d e s c r i b e d i s %f W\n ’ , Wt ) ; // i n c i r c u i t d i a g r a m i f t e r m i n a l c i s c o n n e c t e d t o 53


b and t e r m i n a l d i s c o n n e c t e d t o a t h e c u r r e n t I and I o ( no l o a d c u r r e n t ) f l o w i n t h e o p p o s i t e d i r e c t i o n t h r o u g h c u r r e n t c o i l o f Wattmeter . Hence i t s reading i s decreased to 20 Wt =2* pc - W ; 21 printf ( ’ r e a d i n g o f w a t t m e t e r a s p e r t h e c o n n e c t i o n d e s c r i b e d i s %f W’ , Wt ) ;

Scilab code Exa 1.37 Determine leakage impedance per phase in ohm and p u system 1 clc ; 2 E1 =3300; // r a t e d p h a s e v o l t a g e 3 4 5 6 7 8 9 10 11 12

of primary of a

t h r e e phase t r a n s f o r m e r v =360; // v o l t a g e i n j e c t e d i n open d e l t a h v w i n d i n g to c i r c u l a t e f u l l load current vph = v /3; // v o l t a g e a c r o s s e a c h p h a s e P =300; // r a t e d KVA o f t r a n s f o r m e r Pph = P /3; // KVA p e r p h a s e Iph =( Pph *1000) / E1 ; // p e r p h a s e c u r r e n t z = vph / Iph ; printf ( ’ Per Phase l e a k a g e i m p e d a n c e i s %f ohms \n ’ ,z ) ; zb = E1 / Iph ; // b a s e i m p e d a n c e zpu = z / zb ; printf ( ’ l e a k a g e i m p e d a n c e p e r p h a s e i n p e r u n i t s y s t e m i s %f p . u ’ , zpu ) ;

Scilab code Exa 1.38 Determine power transformed power conducted power output and auto transformer efficiency at same power factor 1 clc ; 2 P =20000; // r a t e d VA o f

transformer 54


3 4 5 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

E1 =2300; // r a t e d v o l t a g e o f p r i m a r y E2 =230; // r a t e d v o l t a g e o f s e c o n d a r y pf =0.6; // power f a c t o r n =0.96; // e f f i c i e n c y ih = P / E1 ; // r a t e d c u r r e n t o f h v w i n d i n g il = P / E2 ; // r a t e d c u r r e n t o f l v w i n d i n g // As p e r t h e c o n n e c t i o n s g i v e n i n f i g 1 4 . 1 ( a ) , two voltages are in s e r i e s aiding Et = E1 + E2 ; // o u t p u t v o l t a g e o f a u t o t r a n s f o r m e r disp ( ’ c a s e a ’ ) ; // By K i r c h o f f s law a t p o i n t b , s u p p l y c u r r e n t i s g i v e n by I = il + ih ; Pa1 = Et * il ; // VA r a t i n g o f a u t o t r a n s f o r m e r Po1 =( Pa1 /1000) ; // power o u t p u t a t f u l l l o a d u n i t y power f a c t o r Pt1 =( E2 * il ) /1000; // power t r a n s f o r m e d Pc1 =( Po1 - Pt1 ) ; // power c o n d u c t e d printf ( ’ For t h e g i v e n c o n n e c t i o n , o u t p u t power i s %f kW\n ’ , Po1 ) ; printf ( ’ For t h e g i v e n c o n n e c t i o n , t r a n s f o r m e d power i s %f kW\n ’ , Pt1 ) ; printf ( ’ For t h e g i v e n c o n n e c t i o n , c o n d u c t e d power i s %f kW\n ’ , Pc1 ) ; disp ( ’ c a s e b ’ ) ; // As p e r t h e c o n n e c t i o n s g i v e n i n f i g 1 4 . 1 ( b ) , two voltages are in s e r i e s opposition Et = E1 - E2 ; // o u t p u t v o l t a g e o f a u t o t r a n s f o r m e r // By K i r c h o f f s law a t p o i n t b , s u p p l y c u r r e n t i s g i v e n by I = il - ih ; Pa2 = E1 * I ; // VA r a t i n g o f a u t o t r a n s f o r m e r Po2 = Pa2 /1000; // power o u t p u t a t f u l l l o a d u n i t y power f a c t o r Pt2 =( E2 * il ) /1000; // power t r a n s f o r m e d Pc2 =( Po2 - Pt2 ) ; // power c o n d u c t e d printf ( ’ For t h e g i v e n c o n n e c t i o n , o u t p u t power i s %f kW\n ’ , Po2 ) ; 55


31 32 33 34

35 36 37 38 39 40

printf ( ’ For t h e g i v e n c o n n e c t i o n , t r a n s f o r m e d power i s %f kW\n ’ , Pt2 ) ; printf ( ’ For t h e g i v e n c o n n e c t i o n , c o n d u c t e d power i s %f kW\n ’ , Pc2 ) ; pl =((1/ n ) -1) * P * pf ; // l o s s e s i n 2− w i n d i n g transformer // a u t o t r a n s f o r m e r o p e r a t e s a t r a t e d c u r r e n t and r a t e d v o l t a g e s o e f f i c i e n c y and l o s s e s r e m a i n constant disp ( ’ E f f i c i e n c y f o r c a s e a ’ ) ; n1 =(1 -( pl /( Po1 *1000* pf + pl ) ) ) *100; printf ( ’ E f f i c i e n c y o f a u t o t r a n s f o r m e r f o r %f VA i s %f p e r c e n t \n ’ ,Po1 , n1 ) ; disp ( ’ E f f i c i e n c y f o r c a s e b ’ ) ; n2 =(1 -( pl /( Po2 *1000* pf + pl ) ) ) *100; printf ( ’ E f f i c i e n c y o f a u t o t r a n s f o r m e r f o r %f VA i s %f p e r c e n t ’ ,Po2 , n2 ) ;

Scilab code Exa 1.39 Determine current in various parts of circuit 1 clc ; 2 // c o n n e c t i o n s have b e e n made i n

3 4 5 6 7 8 9 10 11

f i g 1 . 4 2 i n book t o s u i t v o l t a g e r e q u i r e m e n t o f 3 0 0 0V, 3 5 0 0V and 1 0 0 0V . E1 =1000; // p r i m a r y w i n d i n g of transformer E2 =2000; // s e c o n d a r y w i n d i n g o f t r a n s f o r m e r E3 =500; // t e r t i a r y w i n d i n g o f t r a n s f o r m e r l1 =1050; // l o a d i n KVA a c r o s s 3 5 0 0 V l2 =180; // l o a d i n KVA a c r o s s 1 0 0 0 V i1 =( l1 *1000) /( E1 + E2 + E3 ) ; // c u r r e n t t h r o u g h l o a d o f 1 0 5 0 KVA i2 =( l2 *1000) /( E1 ) ; // c u r r e n t t h r o u g h l o a d o f 180 KVA kt = l1 + l2 ; // T o t a l KVA l o a d s u p p l i e d I =( kt *1000) /( E1 + E2 ) ; 56


printf ( ’ c u r r e n t t h r o u g h %f KVA l o a d i s %f A\n ’ ,l1 , i1 ); 13 printf ( ’ c u r r e n t t h r o u g h %f KVA l o a d i s %f A\n ’ ,l2 , i2 ); 14 printf ( ’ c u r r e n t drawn from s u p p l y i s %f A ’ ,I ) ; 12

Scilab code Exa 1.40 Determine KVA output KVA transformed and KVA conducted of auto transformer 1 2 3 4 5

6 7 8 9 10 11 12 13 14 15 16

clc ; E1 =2500; // p r i m a r y s i d e v o l t a g e E2 =250; // s e c o n d a r y s i d e v o l t a g e P =10000; // r a t e d VA o f t r a n s f o r m e r // t o a c h i e v e a v o l t a g e l e v e l o f 2 6 2 5 , two e q u a l p a r t s o f 125 V e a c h o f s e c o n d a r y w i n d i n g a r e c o n n e c t e d i n p a r a l l e l w i t h e a c h o t h e r and i n s e r i e s with primary winding Eo = E1 + E2 /2; // d e s i r e d o u t p u t o f a u t o t r a n s f o r m e r il = P / E2 ; // r a t e d c u r r e n t o f l v w i n d i n g i =2* il ; // T o t a l o u t p u t c u r r e n t K =( i * Eo ) /1000; // Auto t r a n s s f o r m e r KVA r a t i n g ip = P / E1 ; // r a t e d c u r r e n t o f h v w i n d i n g I = i + ip ; // c u r r e n t drawn from s u p p l y Pt =( i *( E2 /2) ) /1000; // KVA t r a n s f o r m e d Pc =K - Pt ; // KVA c o n d u c t e d printf ( ’KVA o u t p u t o f a u t o t r a n s f o r m e r i s %f KVA\n ’ ,K ); printf ( ’KVA t r a n s f o r m e d i s %f KVA\n ’ , Pt ) ; printf ( ’KVA c o n d u c t e d i s %f KVA ’ , Pc ) ;

Scilab code Exa 1.41 determining currents in various branch of autotransformer 57


1 clc ; 2 E1 =440; // p r i m a r y s u p p l y v o l t a g e 3 E2 =380; // v o l t a g e a t which l o a d a t s e c o n d a r y 4 5 6 7 8 9 10 11 12

terminal i s being supplied l1 =40000; // power r a t i n g o f l o a d i n w a t t s pf =0.8; // l a g g i n g power f a c t o r I2 = l1 /( sqrt (3) * E2 * pf ) ; // p e r p h a s e KVA i n p u t=p e r p h a s e KVA o u t p u t I1 =( E2 / E1 ) * I2 ; In = I2 - I1 ; printf ( ’ C u r r e n t i n p r i m a r y b r a n c h i s %f A\n ’ , I1 ) ; printf ( ’ c u r r e n t i n s e c o n d a r y b r a n c h i s %f A\n ’ , I2 ) ; printf ( ’ c u r r e n t b e t w e e n n e u t r a l and t a p p i n g p o i n t s i s %f A ’ , In ) ;

Scilab code Exa 1.42 Finding primary current and primary input 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

clc ; // From f i g 1 . 4 5 N1 =1000; // no o f t u r n s on p r i m a r y N2 =400; // no . o f t u r n s on s e c o n d a r y n2 =300; // no . o f t u r n s a c r o s s p o i n t s A and B l1 =600; // a l o a d o f 600 KW c o n n e c t e d b e t w e e n p o i n t s A and C l2 =60+60* %i ; // l o a d c o n n e c t e d b e t w e e n p o i n t s A and B E =30000; // p r i m a r y s u p p l y v o l t a g e vac = E *( N2 / N1 ) ; // s e c o n d a r y s i d e v o l t a g e I1 =( l1 *1000) / vac ; // c u r r e n t t h r o u g h l o a d o f 600 KW vab =( vac / N2 ) * n2 ; // v o l a t g e a c r o s s p i n t s A and B I2 = vab / l2 ; // l o a d c u r r e n t t h r o u g h l o a d o f 60+60 i iba = I1 + I2 ; // c u r r e n t t h r o u g h s e c t i o n Ab o f w i n d i n g mfs = iba * n2 + I1 *( N2 - n2 ) ; // s e c o n a d r y mmf ip = mfs / N1 ; printf ( ’ p r i m a r y c u r r e n t i s %f%fi A\n ’ , real ( ip ) , imag ( 58


17 18 19 20

ip ) ) ; Pi =( E * abs ( ip ) * cos ( atan ( imag ( ip ) , real ( ip ) ) ) ) /1000; printf ( ’ p r i m a r y power i n p u t i s %f KW\n ’ , Pi ) ; pf = cos ( atan ( imag ( ip ) , real ( ip ) ) ) printf ( ’ power f a c t o r a t p r i m a r y t e r m i n a l i s %f l a g g i n g ’ , pf )

Scilab code Exa 1.43 Finding current in three section of autotransformer 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

clc ; E =400; // s u p p l y v o l t a g e l1 =200; // l o a d c o n n e c t e d a c r o s s 75% t a p p i n g l2 =400; // l o a d c o n n e c t e d b e t w e e n 25% and 100% tapping t1 =25; // 25% t a p p i n g p o i n t t2 =50; // 50% t a p p i n g p o i n t t3 =75; // 75% t a p p i n g p o i n t V2 =( t3 /100) * E ; // v o l t a g e a c r o s s 200 ohm l o a d I2 = V2 / l1 ; // c u r r e n t t h r o u g h 200 ohm l o a d I1 =( V2 * I2 ) / E ; // from f i g . ( 1 . 4 6 b ) , KCL a t p o i n t d g i v e s idb = I2 - I1 ; // same s e c o n d a r y v o l t a g e i s a p p l i e d a g a i n s t l o a d o f 400 ohm I2 = V2 / l2 ; // c u r r e n t t h r o u g h 400 ohm l o a d I1 =( V2 * I2 ’) / E ; // from f i g ( 1 . 4 6 c ) , KCL a t p o i n t c g i v e s ica = I2 - I1 ; // s u p e r i m p o s i n g t h e c u r r e n t s o f a b o v e r e s u l t s c u r r e n t i n t h r e e p o r t i o n o f w i n d i n g can be known icd = ica ; disp ( ’ c u r r e n t i n s e c t i o n cd o f w i n d i n g i s ’ ) printf ( ’ %f A\n ’ , icd ) ; ibc = I1 ; disp ( ’ c u r r e n t i n s e c t i o n bc o f w i n d i n g i s ’ ) 59


24 printf ( ’ %f A\n ’ , ibc ) ; 25 iab = idb + I1 ; 26 disp ( ’ c u r r e n t i n s e c t i o n ab o f w i n d i n g 27 printf ( ’ %f A\n ’ , iab ) ;

is ’)

Scilab code Exa 1.44 voltage and current rating KVA rating efficiency percentage impedance regulation and short circuit current on each side 1 2 3 4 5 6 7 8 9 10 11 12 13

14

15 16 17

clc ; P =100000; // VA r a t i n g o f two w i n d i n g t r a n s f o r m e r E1 =2000; // r a t e d v o l t a g e o f h v s i d e E2 =200; // r a t e d v o l t a g e o f l v s i d e l =2.5; // p e r c e n t a g e o f l o s s i n two w i n d i n g transformer vr =3; // p e r c e n t a g e o f v o l t a g e r e g u l a t i o n i n two winding transformer z =4; // p e r c e n t a g e o f l e a k a g e i m p e d a n c e i n two winding transformer ih = P / E1 ; // f u l l l o a d c u r r e n t o f h v s i d e il = P / E2 ; // f u l l l o a d c u r r e n t o f l v s i d e V1 = E1 ; // r a t e d v o l t a g e on l v s i d e o f autotransformer V2 = E1 + E2 ; // r a t e d v o l t a g e on h v s i d e o f autotransformer Il = il + ih ; // r a t e d c u r r e n t on l v s i d e o f autotransformer printf ( ’ Rated v o l t a g e on l v and h v s i d e o f a u t o t r a n s f o r m e r a r e %f v and %f v r e s p e c t i v e l y \n , ’ ,V1 , V2 ) ; printf ( ’ Rated c u r r e n t on h v and l v s i d e o f a u t o t r a n s f o r m e r a r e %f A and %f A r e s p e c t i v e l y \n , ’ ,il , Il ) ; k = E1 / V2 ; // t u r n s r a t i o f o r a u t o t r a n s f o r m e r K =((1/(1 - k ) ) * P ) /1000; printf ( ’ Rated KVA o f a u t o t r a n s f o r m e r i s %f KVA\n ’ ,K ) 60


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

; pl =(1 - k ) * l ; // p e r c e n t f u l l l o a d l o s s e s i n autotransformer n =100 - pl ; printf ( ’ E f f i c i e n c y o f a u t o t r a n s f o r m e r i s %f p e r c e n t \n ’ ,n ) ; Z =(1 - k ) * z ; printf ( ’ P e r c e n t a g e i m p e d a n c e a s an a u t o t r a n s f o r m e r i s %f \n ’ ,Z ) ; VR =(1 - k ) * vr ; printf ( ’ p e r c e n t a g e v o l t a g e r e g u l a t i o n a s an a u t o t r a n s f o r m e r i s %f \n ’ , VR ) ; Is =(1/(1 - k ) ) *(100/ z ) ; // s h o r t c i r c u i t p u c u r r e n t Ish =( Is * il ) /1000; printf ( ’ S h o r t c i r c u i t o f a u t o t r a n s f o r m e r on h v s i d e i s %f KA \n ’ , Ish ) ; Isl =( Is * Il ) /1000; printf ( ’ S h o r t c i r c u i t o f a u t o t r a n s f o r m e r on l v s i d e i s %f KA \n ’ , Isl ) ;

Scilab code Exa 1.45 Determining load voltage at given power factors 1 clc ; 2 v1 =10; // v o l t a g e 3 4 5 6 7 8 9

a p p l i e d t o p r i m a r y when s e c o n d a r y is short circuited ip =60; // p r i m a r y c u r r e n t when s e c o n d a r y i s s h o r t circuited k =0.8; // t u r n s r a t i o E1 =250; // i n p u t v o l t a g e f o r l o a d v o l t a g e h a s t o be calculated E2 =200; // r a t e d v o l t a g e o f s e c o n d a r y il =100; // l o a d c u r r e n t pfo =0.24; // power f a c t o r d u r i n g s h o r t c i r c u i t t e s t f =(1 - k ) ^2/ k ^2; // f a c t o r by which s e c o n d a r y i m p e d a n c e h a s t o be m u l t i p l i e d f o r r e f e r r i n g i t 61


10 11 12 13 14 15 16 17 18 19 20 21 22

to primary // z e 1=z 1+f ∗ z 2 t h e r e f o r e by ohm s law ze1 = v1 / ip ; // t o t a l i m p e d a n c e r e f e r r e d t o p r i m a r y re1 = ze1 * pfo ; // t o t a l 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 xe1 = ze1 * sqrt (1 - pfo ^2) ; // t o t a l l e a k a g e r e a c t a n c e r e f e r r e d to primary disp ( ’ c a s e a ’ ) ; pf =0.8; // l a g g i n g power f a c t o r o f l o a d Ip =( E2 * il ) / E1 ; // c u r r e n t i n p r i m a r y due t o l o a d current v2 =( E1 - Ip *( re1 * pf + xe1 * sqrt (1 - pf ^2) ) ) * k ; 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 l a g g i n g power f a c t o r i s %f v \n ’ ,pf , v2 ) ; disp ( ’ c a s e b ’ ) pf =1; // u n i t y power f a c t o r v2 =( E1 - Ip *( re1 * pf + xe1 * sqrt (1 - pf ^2) ) ) * k ; 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 u n i t y power f a c t o r i s %f v ’ , v2 ) ;

Scilab code Exa 1.46 Finding ratio of full load KVA delivered to sum of individual KVA ratings 1 clc ; 2 r1 =9 ; // r a t i o 3 4 5

6 7

of reactance to r e s i s t a n c e f o r transformer 1 r2 =3 ; // r a t i o o f r e a c t a n c e t o r e s i s t a n c e f o r transformer 2 d = atand ( r1 ) - atand ( r2 ) ; // d i f f e r e n e b e t w e e n a n g l e s o f t r a n s f o r m e r ’ s l e a k a g e impedance // l e a k a g e i m p e d a n c e o f b o t h t r a n s f o r m e r s a r e e q u a l z 1=z2 , t h r e e f o r e c u r r e n t s i n b o t h t r a n s f o r m e r s a r e e q u a l t h a t i s i 1=i 2 ; I =1/ cos (( d /2) *( %pi /180) ) ; // r a t i o o f n u m e r i c a l sum o f i 1 and i 2 t o p h a s o r sum o f i 1 and i 2 k = cos (( d /2) *( %pi /180) ) ; 62


8

printf ( ’ r a t i o o f f u l l l o a d KVA d e l i v e r e d t o sum o f b o t h t r a n s f o r m e r s KVA r a t i n g s i s %f ’ ,k ) ;

Scilab code Exa 1.48 Determining greatest load that can be put across parallel combination of transformers and secondary terminal voltage 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

clc ; P =400000; // r a t e d KVA o f t r a n s f o r m e r P1 =11000; // r a t e d p r i m a r y v o l t a g e S2 =6600; // r a t e d s e c o n d a r y v o l t a g e v1 =360; // v o l t a g e r e c o r d e d d u r i n g s h o r t c i r c u i t o f l v winding f o r f i r s t transformer p1 =3025; // power d i s s i p a t e d d u r i n g s h o r t c i r c u i t o f l v winding f o r f i r s t transformer v2 =400; // v o l t a g e r e c o r d e d d u r i n g s h o r t c i r c u i t o f l v winding f o r second transformer p2 =3200; // power d i s s i p a t e d d u r i n g s h o r t c i r c u i t o f l v winding f o r second transformer v3 =480; // v o l t a g e r e c o r d e d d u r i n g s h o r t c i r c u i t t e s t of l v winding t h i r d transformer p3 =3250; // power d i s s i p a t e d d u r i n g s h o r t c i r c u i t o f l v winding f o r t h i r d transformer l1 =( P +( v1 / v2 ) * P +( v1 / v3 ) * P ) /1000; printf ( ’ The g r e a t e s t l o a d t h a t can be put on t h e t r a n s f o r m e r s i s %f KVA\n ’ , l1 ) ; is = P / S2 ; // s e c o n d a r y r a t e d c u r r e n t // t r a n s f o r m e r 1 i s f u l l y l o a d e d , i t s c a r r i e s f u l l load current re2 = p1 / is ^2; // t o t a l r e s i s t a n c e r e f e r r e d t o secondary side vd = is * re2 ; // v o l t a g e d r o p f o r t r a n s f o r m e r 1 E2 = S2 - vd ; 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 i s %f v ’ , E2 ) ;

63


Scilab code Exa 1.49 1 How transformers share a given load 2 Determine the greatest load that can be supplied by parallel combination of transformers and 3 Determining magnitude of the equivalent leakage impedance under normal loading 1 clc ; 2 disp ( ’ c a s e b ’ ) ; 3 // KVA r a t i n g s and l e a k a g e i m p e d a n c e s f o r t h e 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

transformers are k1 =100; // KVA r a t i n g f o r t r a n s f o r m e r 1 ; z1 =0.02; // p u i m p e d a n c e f o r t r a n s f o r m e r 1 ; k2 =75; // KVA r a t i n g f o r t r a n s f o r m e r 2 ; z2 =0.03; // p u i m p e d a n c e f o r t r a n s f o r m e r 2 ; k3 =50; // KVA r a t i n g f o r t r a n s f o r m e r 3 ; z3 =0.025; // p u i m p e d a n c e f o r t r a n s f o r m e r 3 ; disp ( ’ c a s e b ( 1 ) ’ ) ; // assumng k1 a s a b a s e KVA S =225; // l o a d which h a s t o be s h a r e d by t h r e e transformers ze1 = z1 *100; // p e r c e n t a g e i m p e d a n c e f o r t r a n s f o r m e r 1 ze2 =( k1 / k2 ) * z2 *100; // p e r c e n t a g e i m p e d a n c e f o r transformer 2 ze3 =( k1 / k3 ) * z3 *100; // p e r c e n t a g e i m p e d a n c e f o r transformer 3 zt =(1/ ze1 ) +(1/ ze2 ) +1/( ze3 ) ; // t o t a l p e r c e n t a g e l e a k a g e impedance s1 = S /( ze1 * zt ) ; s2 = S /( ze2 * zt ) ; s3 = S /( ze3 * zt ) ; printf ( ’ l o a d s h a r e d by t r a n s f o r m e r 1 , 2 and 3 a r e %f KVA, %f KVA and %f KVA r e s p e c t i v e l y \n ’ ,s1 , s2 , s3 ) ; disp ( ’ c a s e b ( 2 ) ’ ) ; // s i n c e t r a n s f o r m e r 1 h a s l o w e s t l e a k a g e i m p e d a n c e 64


23 24 25 26

27 28 29 30 31

among t h r e e , i t w i l l be l o a d e d t o i t s r a t e d capacity S = k1 * ze1 * zt ; // t o t a l KVA s h a r e d printf ( ’ g r e a t e s t l o a d t h a t can be s h a r e d by t r a n s f o r m e r s i s %f KVA\n ’ ,S ) ; disp ( ’ c a s e b ( 3 ) ’ ) ; // f o r s u c c e s s f u l p a r a l l e l o p e r a t i o n o f t r a n s f o r m e r a l l t h e t h r e e l e a k a g e i m p e d a n c e s b a s e d on t h e i r KVA r a t i n g s h o u l d be e q u a l . S i n c e m a g n i t u d e o f l e a k a g e impedance o f t r a n s f o r m e r 1 i s f i x e d that i s 2 p e r c e n t z 2=z 3=2 p e r c e n t ze1 =2; ze2 = ze1 *( k1 / k2 ) ; ze3 = ze1 *( k1 / k3 ) ; zt =(1/ ze1 ) +(1/ ze2 ) +(1/ ze3 ) ; // T o t a l l e a k a g e impedance printf ( ’ m a g n i t u d e o f e q u i v a l e n t l e a k a g e i m p e d a n c e i s %f p e r c e n t \n ’ , zt ) ;

Scilab code Exa 1.50 1 How transformers share a given load 2 Determine the secondary terminal voltage for part1 1 clc ; 2 // shorts 3 4 5 6 7 8 9 10 11

c i r c u i t s t e s t on two t r a n s f o r m e r s g a v e the f o l l o w i n g r e s u l t s P1 =200000; // KVA o f t r a n s f o r m e r 1 V1 =3; // p e r c e n t a g e r a t e d v o l t a g e pf1 =0.25; // l a g g i n g power f a c t o r f o r Xmer1 P2 =500000; // KVA o f t r a n s f o r m e r 2 V2 =4; // p e r c e n t a g e r a t e d v o l t a g e pf2 =0.3 // l a g g i n g power f a c t o r f o r Xmer2 l =560000; // l o a d c o n n e c t e d a c r o s s p a r a l l e l c o m b i n a t i o n o f t r a n s f o r m e r s i n KW pf =0.8; // power f a c t o r o f l o a d E1 =11000; // Rated p r i m a r y v o l t a g e 65


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

E2 =400; // Rated s e c o n d a r y v o l t a g e ib =1; // b a s e c u r r e n t vb =1; // b a s e v o l t a g e z1 =( V1 /100) *1; // b a s e i m p e d a n c e Ze1 = z1 *( pf1 + %i * sqrt (1 - pf1 ^2) ) ; // p u i m p e d a n c e z2 =( V2 /100) *1; // b a s e i m e d a n c e Ze2 = z2 *( pf2 + %i * sqrt (1 - pf2 ^2) ) ; // p u i m p e d a n c e pb = P2 ; // b a s e f o r p u c o n v e r s i o n ze1 =( pb / P1 ) * Ze1 ; ze2 =( pb / P2 ) * Ze2 ; zt = ze1 + ze2 ; // t o t a l i m p e d a n c e s = l / pf ; // KVA r a t i n g o f t r a n s f o r m e r S = s *( pf - %i * sqrt (1 - pf ^2) ) ; // c o m p l e x form o f KVA rating s1 =( S * ze2 ) /( zt ) ; // KVA s h a r e d by f i r s t t r a n s f o r m e r PF1 = cos ( atand ( imag ( s1 ) , real ( s1 ) ) *( %pi /180) ) ; s1w = round (( abs ( s1 ) * PF1 ) /1000) ; printf ( ’KW l o a d s h a r e d by t r a n s f o r m e r 1 i s %f a t %f power f a c t o r l a g g i n g \n ’ ,s1w , PF1 ) ; s2 =( S * ze1 ) /( zt ) ; // KVA s h a r e d by f i r s t t r a n s f o r m e r PF2 = cos ( atand ( imag ( s2 ) , real ( s2 ) ) *( %pi /180) ) ; s2w = ceil (( abs ( s2 ) * PF2 ) /1000) ; printf ( ’KW l o a d s h a r e d by t r a n s f o r m e r 2 i s %f a t %f power f a c t o r l a g g i n g \n ’ ,s2w , PF2 ) ; i1 = abs ( s1 ) / P1 ; // p u c u r r e n t s h a r e d by t r a n s f o r m e r 1 i2 = abs ( s2 ) / P2 ; // p u c u r r e n t s h a r e d by t r a n s f o r m e r 2 vr = i1 *( real ( Ze1 ) * PF1 + imag ( Ze1 ) * sqrt (1 - PF1 ^2) ) ; // voltag regulation dv = E2 * vr ; // c h a n g e i n t e r m i n a l v o l t a g e Vt = E2 - dv ; // t e r m i n a l v o l t a g e 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 i s %f v ’ , Vt ) ;

66


Scilab code Exa 1.51 Determining largest KVA that can be shared by parallel combination of transformer without overloading 1 2 3 4 5 6 7 8 9 10

11 12

clc ; k1 =1000; // Rated KVA o f t r a n s f o r m e r 1 k2 =500; // Rated KVA o f t r a n s f o r m e r 2 ze1 =0.02+ %i *0.06; // p u l e a k a g e i m p e d a n c e o f transformer 1 ze2 =0.025+ %i *0.08; // p u l e a k a g e i m p e d a n c e o f transformer 2 zb =1000; // b a s e i m p e d a n c e Z1 =( zb / k1 ) * ze1 ; // i m p e d a n c e o f t r a n s f o r m e r 1 Z2 =( zb / k2 ) * ze2 ; // i m p e d a n c e o f t r a n s f o r m e r 2 zt = Z1 + Z2 ; // t o t a l i m p e d a n c e S = k1 *( abs ( zt ) / abs ( Z2 ) ) ; // s i n c e ze1
Scilab code Exa 1.52 Determining no load circulating current and ohmic loss caused by it and no load terminal voltage 1 clc ; 2 // two t r a n s f o r m e r s a r e c o n n e c t e d i n 3 4 5 6 7

p a r a l l e l and

has f o l l o w i n g data P1 =100; // r a t e d KVA o f t r a n s f o r m e r 1 E11 =6600; // r a t e d p r i m a r y v o l t a g e f o r t r a n s f o r m e r 1 E21 =230; // r a t e d s e c o n d a r y v o l t a g e f o r t r a n s f o r m e r 1 z1 =1.5+4* %i // p e r c e n t a g e l e a k a g e i m p e d a n c e f o r transformer 1 P2 =200; // r a t e d KVA o f t r a n s f o r m e r 2 67


8 E12 =6600; // r a t e d p r i m a r y v o l t a g e f o r t r a n s f o r m e r 2 9 E22 =220; // r a t e d s e c o n d a r y v o l t a g e f o r t r a n s f o r m e r 10 11 12 13 14 15 16 17 18 19 20 21

2 z2 =1+5* %i // p e r c e n t a g e l e a k a g e i m p e d a n c e f o r transformer 2 i1 =( P1 *1000) / E21 ; // f u l l l o a d c u r r e n t f o r transformer 1 ze1 =( z1 /100) *( E21 ^2/( P1 *1000) ) ; // l e a k a g e i m p e d a n c e f o r t r a n s f o r m e r 1 i n ohm i2 =( P2 *1000) / E22 ; // f u l l l o a d c u r r e n t f o r transformer 2 ze2 =( z2 /100) *( E22 ^2/( P2 *1000) ) ; // l e a k a g e i m p e d a n c e f o r t r a n s f o r m e r 2 i n ohm io =( E21 - E22 ) / abs ( ze1 + ze2 ) ; // c i r c u l a t i n g c u r r e n t a t no l o a d printf ( ’ No l o a d c i r c u l a t i n g c u r r e n t i s %f A\n ’ , real ( io ) ) ; poh = abs ( io ) ^2*( real ( ze1 ) + real ( ze2 ) ) ; printf ( ’ Ohmic l o s s e s due t o c i r c u l t i n g c u r r e n t i s %f W\n ’ , poh ) ; vd = abs ( io ) * abs ( ze1 ) ; // v o l t a g e d r o p i n l e a k a g e impedance vt = E21 - vd ; // s e c o n d a r y t e r m i n a l v o l t a g e 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 i s %f v ’ , vt ) ;

Scilab code Exa 1.53 Finding no load circulating current current supplied by each transformer and KVA KW and power factor of each transformer 1 2 3 4 5

clc ; k1 =500000; // r a t e d VA o f t r a n s f o r m e r 1 r1 =0.015; // p u r e s i s t a n c e o f t r a n s f o r m e r 1 x1 =0.05; // p u r e a c t a n c e o f t r a n s f o r m e r 1 s1 =405; // s e c o n d a r y no l o a d v o l t a g e f o r t r a n s f o r m e r 1 6 k2 =250000; // r a t e d VA o f t r a n s f o r m e r 2 68


7 r2 =0.01; // p u r e s i s t a n c e o f t r a n s f o r m e r 2 8 x2 =0.05; // p u r e a c t a n c e o f t r a n s f o r m e r 2 9 s2 =415; // s e c o n d a r y no l o a d v o l t a g e f o r t r a n s f o r m e r 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33

2 l =750000; // KVA r a t i n g o f l o a d pf =0.8; // power f a c t o r o f l o a d v =400; // v o l t a g e a t which l o a d i s b e i n g d e l i v e r e d z1 =( r1 + %i * x1 ) *( v ^2/ k1 ) ; // i m p e d a n c e f o r t r a n s f o r m e r 1 i n ohms z2 = ( r2 + %i * x2 ) *( v ^2/ k2 ) ; // i m p e d a n c e f o r t r a n s f o r m e r 2 i n ohms il = l / v ; // l o a d c u r r e n t zl = v / il ; // l o a d i m p e d a n c e zl = zl *( pf + %i * sqrt (1 - pf ^2) ) ; // c o m p l e x form o f l o a d impedance zt = z1 + z2 ; // e q u i v a l e n t i m p e d a n c e o f b o t h transformer io =( s2 - s1 ) / abs ( zt ) ; // c i r c u l a t i n g c u r r e n t a t no load aio = cos ( atand ( imag ( zt ) , real ( zt ) ) *( %pi /180) ) ; // power f a c t o r printf ( ’ C i r c u l a t i n g c u r r e n t a t no l o a d i s %f A a t a power f a c t o r o f %f l a g \n ’ ,io , aio ) ; Ia =(( s1 * z2 ) +( s1 - s2 ) * zl ) /(( z1 * z2 ) +( zl * zt ) ) ; ia = abs ( Ia ) ; printf ( ’ C u r r e n t s h a r e d by t r a n s f o r m e r 1 i s %f A\n ’ , ia ) ; Ib =(( s2 * z1 ) -( s1 - s2 ) * zl ) /(( z1 * z2 ) +( zl * zt ) ) ; ib = abs ( Ib ) ; printf ( ’ C u r r e n t s h a r e d by t r a n s f o r m e r 2 i s %f A\n ’ , ib ) ; kv1 =( ia * v ) /1000; pf1 = cos ( atand ( imag ( Ia ) , real ( Ia ) ) *( %pi /180) ) ; kw1 = kv1 * pf1 ; printf ( ’KVA s h a r e d by t r a n s f o r m e r 1 i s %f KVA a t %f l a g g i n g power f a c t o r \n ’ ,kv1 , pf1 ) ; printf ( ’KW s h a r e d by t r a n s f o r m e r 1 i s %f KW\n ’ , kw1 ) ; kv2 =( ib * v ) /1000; 69


34 pf2 = cos ( atand ( imag ( Ib ) , real ( Ib ) ) *( %pi /180) ) ; 35 kw2 = kv2 * pf2 ; 36 printf ( ’KVA s h a r e d by t r a n s f o r m e r 2 i s %f KVA a t %f 37

l a g g i n g power f a c t o r \n ’ ,kv2 , pf2 ) ; printf ( ’KW s h a r e d by t r a n s f o r m e r 2 i s %f KW\n ’ , kw2 ) ;

Scilab code Exa 1.54 Determining value of reactance connected in series with transformer B so that load current is equally shared 1 clc ; 2 z1 =0.4+2.2* %i ; // l e a k a g e i m p e d a n c e r e f e r r e d 3 4 5 6 7 8 9 10 11

to secondary for transformer 1 s1 =510; // s e c o n d a r y no l o a d v o l t a g e f o r t r a n s f o r m e r 1 z2 =0.6+1.7* %i ; // l e a k a g e i m p e d a n c e r e f e r r e d t o secondary for transformer 2 s2 =500; // s e c o n d a r y no l o a d v o l t a g e f o r t r a n s f o r m e r 2 zl =5+3* %i ; // l o a d i m p e d a n c e // c u r r e n t s h a r e d by b o t h t r a n s f o r m e r s s h o u l d be e q u a l i . e . I 1=I 2 t =( s2 *( z1 ) -( s1 - s2 ) * zl -( s1 - s2 ) * zl ) / s1 ; xn = sqrt ( abs ( t ) ^2 - real ( z2 ) ^2) ; xe = xn - imag ( z2 ) ; // e x t e r n a l r e a c t a n c e t o be added i n series printf ( ’ E x t e r n a l r e a c t a n c e c o n n e c t e d i n s e r i e s w i t h l e a k a g e impedance o f t r a n s f o r m e r 2 so that the c u r r e n t s h a r e d by b o t h t r a n s f o r m e r s a r e e q u a l i s %f ohm ’ , xe ) ;

Scilab code Exa 1.55 Determining tap setting to maintain rated voltage on secondary side for given loads 70


1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

clc ; k =100; // Kva r a t i n g o f t r a n s f o r m e r disp ( ’ c a s e ( a ) ’ ) ; l1 =90; // KVA r a t i n g o f l o a d pf =0.8; // l a g g i n g p f o f l o a d z =0.0075+0.09* %i ; // p u l e a k a g e i m p e d a n c e o f transformer l1 = l1 / k ; // l o a d e x p r e s s e d i n p u w i t h b a s e KVA=100 vd = l1 *( real ( z ) * pf + imag ( z ) * sqrt (1 - pf ^2) ) ; // p u voltage regulation ts = vd *100; // t a p s e t t i n g printf ( ’ Number o f t u r n s i n t h e p r i m a r y w i n d i n g s h o u l d be t a p p e d down by %f p e r c e n t \n ’ , ts ) ; disp ( ’ c a s e ( b ) ’ ) ; lk =100; // KW r a t i n g o f l o a d pf =0.8; // l a g g i n g p f o f l o a d l1 = lk / pf ; // KVA r a t i n g o f l o a d z =0.0075+0.09* %i ; // p u l e a k a g e i m p e d a n c e o f transformer l1 = l1 / k ; // l o a d e x p r e s s e d i n p u w i t h b a s e KVA=100 vd = l1 *( real ( z ) * pf + imag ( z ) * sqrt (1 - pf ^2) ) ; // p u voltage regulation ts = vd *100; // t a p s e t t i n g printf ( ’ Number o f t u r n s i n t h e p r i m a r y w i n d i n g s h o u l d be t a p p e d down by %f p e r c e n t \n ’ , ts ) ;

Scilab code Exa 1.56 Determining tap setting to maintain given voltage at load terminal 1 2 3 4 5 6

clc ; p =11000; // p e r p h a s e h v s i d e v o l t a g e s =433; // l v s i d e v o l t a g e sp = s / sqrt (3) ; // p e r p h a s e l v s i d e v o l t a g e K =10000; // VA r a t i n g o f s t a r c o n n e c t e d l o a d pf =0.8; // power f a c t o r o f l o a d 71


7 z =0.5+1* %i ; // Impedance p e r l e a d 8 zh =300+1500* %i ; // p e r p h a s e h v s i d e 9 10 11 12 13 14 15 16 17 18 19 20

leakage impedance zl =0.2+1* %i ; // p e r p h a s e l v s i d e l e a k a g e i m p e d a n c e vl =400; // l o a d v o l t a g e k = p / sp ; // t u r n s r a t i o zhl = zh / k ^2; // h v s i d e l e a k a g e i m p e d a n c e r e f e r r e d to l v s i d e zt = zhl + zl + z ; // t o t a l i m p e d a n c e p e r p h a s e b e t w e e n t r a n s f o r m e r and s e c o n d a y l o a d il = K /( sqrt (3) * vl ) ; // p e r p h a s e l o a d c u r r e n t vd = round ( il *( real ( zt ) * pf + imag ( zt ) * sqrt (1 - pf ^2) ) ) ; // v o l t a g e drop per phase vlp = vl / sqrt (3) ; // p e r p h a s e l o a d t e r m i n a l v o l t a g e E2 = vlp + vd ; // p e r p h a s e v o l t a g e t h a t must m a i n t a i n e d l v terminals vb = E2 - sp ; // v o l t a g e b o o s t t h a t t a p c h a n g e r must provide ts = round (( vb / sp ) *100) ; printf ( ’ t a p down i f t h e t a p p e d c o i l s a r e on h v s i d e o r t a p up i f t h e t a p p e d c o i l s a r e on l v s i d e by %f p e r c e n t ’ , ts ) ;

Scilab code Exa 1.57 Calculating primary winding current and line current 1 2 3 4 5 6

clc ; k =5; // E f f e c t i v e t u r n s r a t i o E1 =400; // s u p p l y v o l t a g e f o r p r i m a r y il =10; // l o a d c u r r e n t E2 = E1 / k ; // m a g n i t u d e o f maximum s e c o n d a r y i n d u c e d E1o = E1 + E2 ; // maximum p o s s i b e v a l u e o f o u t p u t voltage 7 P =( il * E2 ) /1000; // R a t i n g o f s e c o n d a r y w i n d i n g 8 ip =( il * E2 ) / E1 ; // n e g l e c t i n g n o l o a d c u r r e n t , p r i m a r y 72


winding current 9 I = ip + il ; 10 printf ( ’ T o t a l p r i m a r y l i n e c u r r e n t i s %f A ’ ,I ) ;

Scilab code Exa 1.58 Calculating KVA required to maintain the given voltage 1 2 3 4 5

clc ; E1 =430; // s u p p l y v o l t a g e k =100000; // VA r a t i n g o f l o a d il = k /( sqrt (3) * E1 ) ; // l o a d c u r r e n t v1 =380; v2 =460; // r a n g e i n which v o l t a g e a t f e e d e r end v a r i e s 6 vr = E1 - v1 ; // maximum v a r i a t i o n i n o u t p u t s i d e o f regulator 7 P =( sqrt (3) * vr * il ) /1000; 8 printf ( ’KVA r a t i n g o f i n d u c t i o n r e g u l a t o r i s %f KVA ’ ,P ) ;

Scilab code Exa 1.59 Calculating the rating of the primary and secondary winding 1 2 3 4 5 6 7 8 9

clc ; V1 =400; // s u p p l y v o l t a g e E2 =50; // i n d u c e d s e c o n d a r y v o l t a g e l =8; // KW r a t i n g o f l o a d pf =0.8; // power f a c t o r o f l o a d n =0.85; // e f f i c i e n c y o f i n d u c t i o n r e g u l a t o r ip =( l *1000) /( sqrt (3) * pf * n * V1 ) ; // i n p u t c u r r e n t vmo = V1 - E2 ; // minimum o u t p u t v o l t a g e imo =( l *1000) /( sqrt (3) * pf * vmo ) ; // maximum o u t p u t current 73


10 ks =( sqrt (3) * E2 * imo ) /1000; // KVA r a t i n g

of secondary

winding 11 Vmo = V1 + E2 ; // maximum o u t p u t v o l t a g e 12 Imo =( l *1000) /( sqrt (3) * pf * Vmo ) ; // minimum o u t p u t 13

14 15 16 17

current // p r i m a r y w i n d i n g h a s t o c a r r y t h e c u r r e n t i n d u c e d i n i t by s e c o n d a r y c u r r e n t due t o t r a n s f o r m e r a c t i o n and t h e d i f f e r e n c e o f i n p u t c u r r e n t and output cu r r en t ipm =(( Imo * E2 ) / V1 ) + ip - Imo ; // maximum p r i m a r y w i n d i n g current kp =( sqrt (3) * V1 * ipm ) /1000; printf ( ’ R a t i n g o f p r i m a r y w i n d i n g i s %f KVA\n ’ , kp ) ; printf ( ’ R a t i n g o f s e c o n d a r y w i n d i n g i s %f KVA\n ’ , ks ) ;

Scilab code Exa 1.60 Determining the magnitude of maximum induced emf and For supply voltage finding the limit of output voltage and finding the rating of regulator for given load current 1 2 3 4 5 6 7 8 9 10

clc ; p =10; // number o f f u l l p i t c h c o i l i n s t a t o r t =120; // p h a s e s p r e a d o f c o i l q =10.5*10^ -3; // f l u x p e r p o l e i n Weber f =50; // f r e q u e n c y of induction regulator n =2; // number o f t u r n s i n e a c h c o i l ; E1 =230; // s u p p l y v o l t a g e il =30; // l o a d c u r r e n t N = p * n ; // T o t a l number o f t u r n s kw = sin (( t /2) *( %pi /180) ) /(( t /2) *( %pi /180) ) ; // distribution factor 11 E2 = sqrt (2) * %pi * f * N * q * kw ; // maximum s e c o n d a r y i n d u c e d EMF 12 Vo1 = E1 + E2 ; Vo2 = E1 - E2 ; // r a n g e o f o u t p u t v o l t a g e 13 k =( il * E2 ) /1000; 74


printf ( ’ Maximum i n d u c e d v o l t a g e i s %f v \n ’ , E2 ) ; printf ( ’ r a n g e o f o u t p u t v o l t a g e i s %f v−%f v \n ’ ,Vo2 , Vo1 ) ; 16 printf ( ’KVA r a t i n g o f i n d u c t i o n r e g u l a t o r i s %f KVA ’ ,k ) ; 14 15

Scilab code Exa 1.61 1 Determining self impedances of primary and secondary windings 2 Finding the value of equivalent circuit parameter referred to both sides 3 Determining secondary terminal voltage for given load 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

clc ; // s u b s c r i p t 1 and 2 i n d i c a t e s h v and l v w i n d i n g P =10000; // r a t e d VA o f t r a n s f o r m e r E1 =2300; // r a t e d v o l t a g e E2 =230; // r a t e d v o l t a g e r1 =10; // t o t a l r e s i s t a n c e r2 =0.10; // t o t a l r e s i s t a n c e l1 =40*10^ -3 ; l2 =4*10^ -4; // s e l f −i n d u c t a n c e s m =10; // mutual i n d u c t a n c e k = E2 / E1 ; // t u r n s r a t i o f =50; // f r e q u e n c y o f s u p p l y ; disp ( ’ c a s e 1 ’ ) ; L1 =( m / k ) + l1 ; printf ( ’ Primary s e l f i n d u c t a n c e i s %f H\n ’ , L1 ) ; L2 =( m * k ) + l2 ; printf ( ’ S e c o n d a r y s e l f i n d u c t a n c e i s %f H\n ’ , L2 ) ; disp ( ’ c a s e b ’ ) ; r21 = r2 / k ^2; // l v s i d e r e s i s t a n c e r e f e r r e d t o h v side 19 l21 = l2 / k ^2; // l v s i d e s e l f i n d u c t a n c e r e f e r r e d t o h v side 20 M1 = m / k ; // mutual i n d u c t a n c e r e f e r r e d t o h v s i d e 21 printf ( ’ c i r c u i t p a r a m e t e r s r e f e r r e d t o p r i m a r y w i n d i n g a r e R1=%f ohm , R2=%f ohm , L1=%f H, L2=%f H and Lm1=%f H\n ’ ,r1 , r21 , l1 , l21 , M1 ) ; 75


22 r12 = r1 * k ^2; // h v s i d e

r e s i s t a n c e r e f e r r e d to l v

side 23 l12 = l1 * k ^2; // h v s i d e

s e l f inductance r e f e r r e d to

l v side 24 M2 = m * k ; // mutual i n d u c t a n c e r e f e r r e d t o l v s i d e 25 printf ( ’ c i r c u i t p a r a m e t e r s r e f e r r e d t o s e c o n d a r y

26 27 28 29 30 31 32 33 34

w i n d i n g a r e R1=%f ohm , R2=%f ohm , L1=%f H, L2=%f H and Lm2=%f H\n ’ ,r12 , r2 , l12 , l2 , M2 ) ; disp ( ’ c a s e c ’ ) ; lo =5+5* %i ; // l o a d c o n n e c t e d t o s e c o n d a r y x1 =2* %pi * f * l12 ; // l e a k a g e r e a c t a n c e x2 =2* %pi * f * l2 ; // l e a k a g e r e a c t a n c e re2 = real ( lo ) + r2 + r12 ; // t o t a l r e s i s t a n c e a f t e r r e f e r r i n g to l v s i d e xe2 = imag ( lo ) + x1 + x2 ; // t o t a l r e a c t a n c e a f t e r r e f e r r i n g to l v s i d e Z = re2 + %i * xe2 ; // t o t a l i m p e d a n c e vt =( E2 / abs ( Z ) ) * abs ( lo ) ; 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 i s %f v ’ , vt ) ;

Scilab code Exa 1.62 1 Determining self inductances of hv and lv windings 2 Calculating mutual inductance between h v and l v windings 3 Finding coupling factors k1 and k2 for both windings and coefficient of coupling 1 2 3 4

clc ; n1 =590; // p r i m a r y s i d e t u r n s n2 =295; // s e c o n d a r y s i d e t u r n s V1 =230; // v o l t a g e s o u r c e from which h v s i d e was energised during t e s t 5 io1 =0.35; // no l o a d c u r r e n t f o r when h v s i d e i s energised 6 V2 =110; // i n d u c e d v o l t a g e a c r o s s open c i r c u i t e d l v w i n d i n g due e n e r g i s e d h v s i d e 7 v2 =115; // v o l t a g e s o u r c e from which l v s i d e was energised during t e s t 76


8 io2 =0.72; // no l o a d c u r r e n t

f o r when l v s i d e i s

energised 9 v1 =226; // i n d u c e d v o l t a g e 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

a c r o s s open c i r c u i t e d h v w i n d i n g due e n e r g i s e d l v s i d e f =50; // f r e q u e n c y o f s u p p l y ; w1 = V1 /( sqrt (2) * %pi *50) ; // Maximum v a l u e o f f l u x l i n k a g e with h v winding L1 = w1 *(1/( sqrt (2) * io1 ) ) ; printf ( ’ s e l f i n d u c t a n c e o f h v w i n d i n g i %f H\n ’ , L1 ) ; w2 = v2 /( sqrt (2) * %pi *50) ; // Maximum v a l u e o f f l u x l i n k a g e with l v winding L2 = w2 *(1/( sqrt (2) * io2 ) ) ; printf ( ’ s e l f i n d u c t a n c e o f l v w i n d i n g i %f H\n ’ , L2 ) ; M =( V2 /( sqrt (2) * %pi * f ) ) *(1/( sqrt (2) * io1 ) ) ; printf ( ’ mutual i n d u c t a n c e b e t w e e n h v and l v w i n d i n g i s %f H\n ’ ,M ) ; k1 =( n1 / n2 ) *( M / L1 ) ; // c o u p l i n g f a c t o r f o r h v s i d e k2 =( n2 / n1 ) *( M / L2 ) ; // c o u p l i n g f a c t o r f o r l v s i d e k = sqrt ( k1 * k2 ) ; // c o e f f i c i e n t o f c o u p l i n g printf ( ’ c o u p l i n g f a c t o r f o r h v s i d e i s %f\n ’ , k1 ) ; printf ( ’ c o u p l i n g f a c t o r f o r l v s i d e i s %f\n ’ , k2 ) ; printf ( ’ c o e f f i c i e n t o f c o u p l i n g i s %f\n ’ ,k ) ;

Scilab code Exa 1.63 Calculating the current that would flow in the winding 1 when the winding is connected to given supply and given load 1 2 3 4 5 6 7

clc ; L1 =4*10^ -3; // s e l f i n d u t a n c e o f w i n d i n g 1 L2 =6*10^ -3; // s e l f i n d u t a n c e o f w i n d i n g 2 M =1.8*10^ -3; // mutual i n d u c t a n c e o f two w i n d i n g s E1 =130; // s u p p l y v o l t a g e f o r w i n d i n g 1 f =500/ %pi ; // f r e q u e n c y o f s u p p l y l =0.2*10^ -3; // l o a d c o n n e c t e d t o w i n d i n g 2 77


8

9 10 11 12 13 14 15 16 17

// w r i t i n g v o l t a g e i n rms form i n m a t r i x form V1=r 1 ∗ I 1+jw ∗ L1 ∗ I1 −jw ∗m∗ I2 , V2=−r 2 ∗ I2 −jw ∗ L2 ∗ I 2+jw ∗m∗ I 1 t o d e t e r m i n e I 1 and I 2 t1 = %i *2* %pi * f * L1 ; t2 = - %i *2* %pi * f * M ; t3 = %i *2* %pi * f * M ; t4 = - %i *2* %pi * f *l - %i *2* %pi * f * L2 ; // w r i t i n g d i f f e r e n t o f matrix s =[ t1 t2 ; t3 t4 ]; v =[130;0]; i = inv ( s ) * v ; // c a l c u l a t i n g c u r r e n t i n two w i n d i n g s p = - imag ( i (1) ) ; printf ( ’ m a g n i t u d e o f c u r r e n t i n p r i m a r y w i n d i n g i s %f A ’ ,p ) ;

Scilab code Exa 1.64 1 Determining the transformers turns ratio for maximum transfer of power 2 Determining load current voltage and power under maximum transfer of power 1 clc ; 2 r =60; // r e s i s t i v e 3

4

5 6 7 8 9 10

l o a d which i s c o u p l e d t o electronic circuit v =5; R =3000; // e l e c t r o n i c c i r c u i t r e p r e s e n t e d by a v o l t a g e s o u r c e in s e r i e s with a i n t e r n a l resistance // f o r maximum t r a n s f e r o f power , l o a d 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 must be e q u a l t o i n t e r n a l resitance of source k = sqrt ( R / r ) ; printf ( ’ t u r n s r a t i o f o r maximum t r a n s f e r o f power i s %f\n ’ ,k ) ; // r e f e r r i g a l l q u a n t i t i e s t o l o a d s i d e vl = v / k ; // s o u r c e v o l t a g e r e f e r r e d t o l o a d s i d e Rl = R / k ^2; // s o u r c e r e f e r r e d t o l o a d s i d e il =( vl /( Rl + r ) ) *1000 ; // l o a d c u r r e n t 78


11 vld =( il * r ) /1000; // l o a d v o l t a g e 12 p =( il ^2* r ) /1000; // l o a d power 13 printf ( ’ l o a d v o l t a g e i s %f v \n ’ , vld ) ; 14 printf ( ’ l o a d c u r r e n t i s %f mA\n ’ , il ) ; 15 printf ( ’ l o a d power i s %f mW\n ’ ,p ) ;

Scilab code Exa 1.65 Finding turns ratio for maximum transfer of power 2 Computing the load voltage at given frequencies 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

clc ; r1 =20; // r e s i s t a n c e o f p r i m a r y s i d e l1 =1*10^ -3; // l e a k a g e i n d u c t a n c e o f p r i m a r y s i d e r2 =0.5; // r e s i s t a n c e o f s e c o n d a r y s i d e l2 =0.025*10^ -3; // l e a k a g e i n d u c t a n c e o f s e c o n d a r y side m =0.2; // mutual i n d u c t a n c e l =50; // l o a d i n ohm c o n n e c t e d t o t r a n s f o r m e r v =5; // v o l t a g e s o u r c e R =2000; // i n t e r n a l r e s i s t a n c e o f s o u r c e k = sqrt ( R / l ) ; // t u r n s r a t i o f o r maximum power transfer printf ( ’ Turns r a t i o i s %f\n ’ ,k ) ; r21 =0.5* k ^2; // 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 primary l21 = l2 * k ^2; // s e c o n d a r y i n d u c t a n c e r e f e r r e d t o primary lp = l * k ^2; // l o a d 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 rs = r1 + r21 + lp + R ; // t o t a l s e r i e s r e s i s t a n c e rp =(( R + r1 ) *( R + r1 ) ) / rs ; // e q u i v a l e n t r e s i s t a n c e leq = l1 + l21 ; // e q u i v a l e n t i n d u c t a n c e f1 =100; // f r e q u e n c y i n h e r t z a t which l o a d v o l t a g e h a s t o be c a l c u l a t e d Vl =(1/ k ) *( R / rs ) * v *(1/( sqrt (1+( rp /(2* %pi * f1 * m ) ) ^2) ) ) ; printf ( ’ l o a d v o l t a g e a t %f hz i s %f v \n ’ ,f1 , Vl ) ; f2 =5000; // f r e q u e n c y i n h e r t z a t which l o a d v o l t a g e 79


22 23 24 25 26

h a s t o be c a l c u l a t e d Vl =(1/ k ) *( R / rs ) * v ; printf ( ’ l o a d v o l t a g e a t %f hz i s %f v \n ’ ,f2 , Vl ) ; f3 =15000; // f r e q u e n c y i n h e r t z a t which l o a d v o l t a g e h a s t o be c a l c u l a t e d Vl =(1/ k ) *( R / rs ) *(1/( sqrt (1+((2* %pi * f3 * l1 ) / rs ) ^2) ) ) * v ; printf ( ’ l o a d v o l t a g e a t %f hz i s %f v \n ’ ,f3 , Vl ) ;

Scilab code Exa 1.66 Computing secondary line voltage line current and output KVA for the given connection 1 clc ; 2 // Answer f o r 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

case c , secondary l i n e voltage i s g i v e n wrong i n book k =12; // p e r p h a s e t u r n s r a t i o E1 =11000; // s u p p l y v o l t a g e from f e e d e r l i n e ip =20; // p r i m a r y l i n e c u r r e n t disp ( ’ c a s e a : s t a r −d e l t a ’ ) ; vph = E1 / sqrt (3) ; // p r i m a r y p h a s e v o l t a g e iph = ip ; // p h a s e c u r r e n t on p r i m a r y sph = vph / k ; // s e c o n d a r y p h a s e v o l t a g e vls = sph ; printf ( ’ L i n e v o l t a g e on s e c o n d a r y i s %f v \n ’ , vls ) ; isph = k * iph ; // p h a s e c u r r e n t on s e c o n d a r y isl = sqrt (3) * isph ; printf ( ’ l i n e c u r r e n t on s e c o n d a r y i s %f A\n ’ , isl ) ; Kv =(3* sph * isph ) /1000; printf ( ’ Output KVA i s %f KVA\n ’ , Kv ) ; disp ( ’ c a s e b : d e l t a −s t a r ’ ) ; vph = E1 ; // p r i m a r y p h a s e v o l t a g e iph = ip / sqrt (3) ; // p h a s e c u r r e n t on p r i m a r y sph = vph / k ; // s e c o n d a r y p h a s e v o l t a g e vls = sqrt (3) * sph ; printf ( ’ L i n e v o l t a g e on s e c o n d a r y i s %f v \n ’ , vls ) ; 80


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

isph = k * iph ; // p h a s e c u r r e n t on s e c o n d a r y isl = isph ; printf ( ’ l i n e c u r r e n t on s e c o n d a r y i s %f A\n ’ , isl ) ; Kv =(3* sph * isph ) /1000; printf ( ’ Output KVA i s %f KVA\n ’ , Kv ) ; disp ( ’ c a s e c : d e l t a −d e l t a ’ ) ; vph = E1 ; // p r i m a r y p h a s e v o l t a g e iph = ip / sqrt (3) ; // p h a s e c u r r e n t on p r i m a r y sph = vph / k ; // s e c o n d a r y p h a s e v o l t a g e vls = sph ; printf ( ’ L i n e v o l t a g e on s e c o n d a r y i s %f v \n ’ , vls ) ; isph = k * iph ; // p h a s e c u r r e n t on s e c o n d a r y isl = sqrt (3) * isph ; printf ( ’ l i n e c u r r e n t on s e c o n d a r y i s %f A\n ’ , isl ) ; Kv =(3* sph * isph ) /1000; printf ( ’ Output KVA i s %f KVA\n ’ , Kv ) ; disp ( ’ c a s e d : s t a r −s t a r ’ ) ; vph = E1 / sqrt (3) ; // p r i m a r y p h a s e v o l t a g e iph = ip ; // p h a s e c u r r e n t on p r i m a r y sph = vph / k ; // s e c o n d a r y p h a s e v o l t a g e vls = sqrt (3) * sph ; printf ( ’ L i n e v o l t a g e on s e c o n d a r y i s %f v \n ’ , vls ) ; isph = k * iph ; // p h a s e c u r r e n t on s e c o n d a r y isl = isph ; printf ( ’ l i n e c u r r e n t on s e c o n d a r y i s %f A\n ’ , isl ) ; Kv =(3* sph * isph ) /1000; printf ( ’ Output KVA i s %f KVA\n ’ , Kv ) ;

Scilab code Exa 1.67 Determining transformer rating and phase and line current on both high and low voltage sides 1 2 3 4

clc ; E1 =11000; // l i n e v o l t a g e o f p r i m a r y E2 =415; // l i n e v o l t a g e o f s e c o n d a r y kw =30; // KW r a t i n g o f 3 p h a s e i n d u c t i o n motor 81


5 6 7 8 9 10 11 12 13 14 15 16

n =0.9; // e f f i c i e n y pf =0.833; // power f a c t o r a t which l o a d i s o p e r a t i n g Kv = kw /( n * pf ) ; // KVA r a t i n g o f t r a n s f o r m e r printf ( ’KVA r a t i n g o f t r a n s f o r m e r i s %f KVA\n ’ , Kv ) ; ilv =( Kv *1000) /( sqrt (3) * E2 ) ; // l i n e c u r r e n t on l v side // s e c o n d a r y i s s t a r c o n n e c t e d t h e r e f o r e l i n e c u r r e n t =p h a s e c u r r e n t printf ( ’ L i n e c u r r e n t on l v s i d e i s %f A\n ’ , ilv ) ; printf ( ’ Phase c u r r e n t on l v s i d e i s %f A\n ’ , ilv ) ; ilp =( Kv *1000) /( sqrt (3) * E1 ) ; // l i n e c u r r e n t on h v side ipp = ilp / sqrt (3) ; printf ( ’ L i n e c u r r e n t on h v s i d e i s %f A\n ’ , ilp ) ; printf ( ’ Phase c u r r e n t on h v s i d e i s %f A\n ’ , ipp ) ;

Scilab code Exa 1.68 1 Finding power consumed by load 2 KVA rating of transformer 3 Phase and line currents on both h v and l v side 1 2 3 4 5 6 7 8 9 10 11 12 13 14

clc ; il =100; // l o a d c u r r e n t pf =0.8; E1 =11000; // p r i m a r y l i n e v o l t a g e E2 =400; // s e c o n d a r y l i n e v o l t a g e p =( sqrt (3) * E2 * il * pf ) /1000; printf ( ’ power consumed by l o a d i s %f KW\n ’ ,p ) ; k =( sqrt (3) * E2 * il ) /1000; printf ( ’KVA r a t i n g o f l o a d i s %f KVA\n ’ ,k ) ; iph =( k *1000) /( sqrt (3) *11000) ; // p h a s e c u r r e n t on h v side // p r i m a r y i s s t a r c o n n e c t e d t h e r e f o r e l i n e c u r r e n t= phase c u r r e n t printf ( ’ L i n e c u r r e n t on h v s i d e i s %f A\n ’ , iph ) ; printf ( ’ Phase c u r r e n t on h v s i d e i s %f A\n ’ , iph ) ; ipl = il / sqrt (3) ; 82


15 16

printf ( ’ L i n e c u r r e n t on l v s i d e i s %f A\n ’ , il ) ; printf ( ’ Phase c u r r e n t on l v s i d e i s %f A\n ’ , ipl ) ;

Scilab code Exa 1.69 1 Determining number of primary turns 2 current power factor and power on primary side under no load 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

clc ; B =1.2; // maximum f l u x d e n s i t y i n c o r e H =600; // m a g n e t i c f i e l d i n t e n s i t y i n A/m d =7.8; // gram d e n s i t y i n g /cmˆ3 cl =2; // c o r e l o s s p e r kg e =1200; // s u p p l y v o l t a g e f =50; // f r e q u e n c y o f s u p p l y v o l t a g e t =10; // t h i c k n e s s o f c o r e i n cm w =40; L =30; // d i m e n s i o n s o f c o r e s =0.9; // s t a c k i n g f a c t o r ga = t ^2*10^ -4; // g r o s s c o r e a r e a i n mˆ2 nga = ga * s ; // n e t g r o s s c o r e a r e a i n mˆ2 q = nga * B ; // maximum f l u x i n c o r e N =( e ) /( sqrt (2) * %pi * f * q ) ; printf ( ’ Number o f t u r n s i n p r i m a r y i s %f\n ’ ,N ) ; ml =(( w + t ) +( L + t ) ) *2; // mean f l u x p a t h i n cm mmf = H *( ml /100) ; // mmf o f t h e c o r e mi = mmf / N ; // maximum v a l u e o f m a g n e t i z i n g c u r r e n t irm = mi / sqrt (2) ; // rms v a l u e o f m a g n e t i z i n g c u r r e n t cv =( ml /100) * nga ; // c o r e volume wc = cv * d *10^3; // w e i g h t o f c o r e pc = wc * cl ; // t o t a l c o r e l o s s Ic = pc / e ; // c o r e l o s s c u r r e n t io = sqrt ( Ic ^2+ irm ^2) ; printf ( ’ No l o a d c u r r e n t on p r i m a r y s i d e i s %f A\n ’ , io ) ; 27 pf = Ic / io ; 28 printf ( ’ No l o a d power f a c t o r i s %f\n ’ , pf ) ; 83


29 pf = e * io * pf ; 30 printf ( ’ power i n p u t a t no l o a d

i s %f W’ , pf ) ;

Scilab code Exa 1.70 Determining magnetizing current for the given figs 1 clc ; 2 // Three c o r e t y p e t r a n s f o r m e r s a r e g i v e n i n

fig

1.80 3 // For f i r s t c o r e t y p e t r a n s f o r m e r 4 im1 =4; // m a g n e t i z i n g c o r e 5 e2 =100; // emf i n d u c e d i n s e c o n d a r y w i n d i n g 6 B =1; // maximum f l u x d e n s i t y i n T e s l a 7 // mmf i s d i r e c t l y p r o p o r t i o n a l maximum f l u x

8

9 10 11 12 13 14 15 16

17

in core i . e im ∗N( no . o f t u r n s )=kq ( f l u x ) , k i s proportionality constant // f o r f i g ( 8 0 ( b ) ) , qm2 ( f l u x f o r c o r e t r a n s f o r m e r 1 )= qm1 ( f l u x f o r c o r e t r a n s f o r m e r 2 ) , t h a t i s f l u x i n b o t h c o i l s i n c o r e t r a n s f o r m e r 2 i s qm1 / 2 ; // f o r u p p e r c o i l im2 ∗N i s d i r e c t l y p r o p o r t i o n a l t o qm1/2 // f o r l o w e r c o i l Im2 ∗N i s d i r e c t l y p r o p o r t i o n a l t o qm2/2 // a d d i n g a b o v e r e l a t i o n we g e t im2+Im2 =4( m a g n e t i z i n g current ) Im2 = im1 /2; im2 = Im2 ; // m a g n e t i z i n g c u r r e n t o f e a c h c o i l i s 2 A imt = Im2 + im2 ; // t o t a l m a g n e t i z i n g c u r r e n t f o r transformer 2 // s i n c e f l u x i s same f o r b o t h t r a n s f o r m e r , emf i n d u c e d i s a l s o same // s i n c e f l u x i s same f o r b o t h t r a n s f o r m e r , a r e a i s same , t h e r e f o r e m a g n e t i c f l u x d e n s i t y i s a l s o same printf ( ’ M a g n e t i z i n g c u r r e n t f o r t r a n s f o r m e r 2 i s %f A\n ’ , imt ) ; 84


18 19 20 21 22 23 24 25 26 27 28

printf ( ’ emf i n d u c e d i n s e c o n d a r y f o r t r a n s f o r m e r 2 i s %f v \n ’ , e2 ) ; printf ( ’ M a g n e t i c f l u x d e n s i t y i n t r a n s f o r m e r 2 i s %f T\n ’ ,B ) ; // f o r f i g ( 8 0 ( c ) ) , qm3=qm1/2 // qm1˜ im1 ∗N, qm3˜ im3 ∗N ; ˜− s i g n o f d i r e c t l y p r o p o r t i o n a l ( assumption ) // from a b o v e two r e l a t i o n s , we g e t im3 = im1 /4; B3 = B /2; E2 = e2 /2; printf ( ’ M a g n e t i z i n g c u r r e n t f o r t r a n s f o r m e r 3 i s %f A\n ’ , im3 ) ; printf ( ’ emf i n d u c e d i n s e c o n d a r y f o r t r a n s f o r m e r 3 i s %f v \n ’ , E2 ) ; printf ( ’ M a g n e t i c f l u x d e n s i t y i n t r a n s f o r m e r 3 i s %f T\n ’ , B3 ) ;

Scilab code Exa 1.71 Determining load voltage load power factor and load power 1 2 3 4 5 6 7 8 9 10 11 12 13 14

clc ; P =20; // Rated KVA o f t r a n s f o r m e r E1 =250; // r a t e d p r i m a r y v o l t a g e E2 =125; // r a t e d s e c o n d a r y v o l t a g e r1 =0.15; // r e s i s t a n c e o f p r i m a r y s i d e x1 =0.25; // l e a k a g e r e a c t a n c e o f p r i m a r y s i d e r2 =0.03; // r e s i s t a n c e o f s e c o n d a r y s i d e x2 =0.04; // l e a k a g e r e a c t a n c e o f s e c o n d a r y s i d e // g i v e n E1=V1 ( p r i m a r y t e r m i n a l v o l t a g e ) k = E2 / E1 ; // t u r n s r a t i o ip =( P *1000) / E1 ; // f u l l l o a d p r i m a r y c u r r e n t // v o l t a g e r e g u l a t i o n =0 , b e c a u s e E1=V1 t h e r e f o r e pf = - atand ( r1 , x1 ) ; // p h a s e a n g l e b e t w e e n E1 and i p // n e g a t i v e s i g n i n d i c a t e s t h a t c u r r e n t l e a d s 85


15 16 17 18 19

20 21 22 23 24 25 26

voltage re2 = r2 / k ^2; // 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 primary xe2 = x2 / k ^2; // 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 r e f e r r e d to primary ip = ip *( cos ( pf *( %pi /180) ) - %i * sin ( pf *( %pi /180) ) ) ; // c o m p l e x form o f p r i m a r y c u r r e n t V2 = E1 - ip *( re2 + xe2 * %i ) ; pfl = - atand ( imag ( V2 ) , real ( V2 ) ) - pf ; // p h a s e a n g l e by which p r i m a r y c u r r e n t l e a d s s e c o n d a r y t e r m i n a l v o l t a g e r e f e r r e d to primary PF = cos ( pfl *( %pi /180) ) ; vl = abs ( V2 ) /2; isl =( P *1000) / E2 ; pl = vl * isl * PF ; printf ( ’ Load v o l t a g e i s %f v \n ’ , vl ) ; printf ( ’ Load power f a c t o r i s %f l e a d i n g \n ’ , PF ) ; printf ( ’ Load power i s %f W’ , pl ) ;

Scilab code Exa 1.72 Calculating power delivered by each source and power dissipated in given resistor 1 2 3 4 5 6 7 8

clc ; // from f i g 1 . 8 2 E1 =5; // s u p p l y v o l t a g e E2 =20; // i n d u c e d s e c o n d a r y v o l t a g e k = E2 / E1 ; // t u r n s r a t i o r1 =6; // p r i m a r y p a r a m e t e r s r2 =16; // s e c o n d a r y p a r a m e t e r s r21 = r2 / k ; // s e c o n d a r y p a r a m e t e r s r e f e r r e d t o primary 9 E21 =( E2 *2) / k ; // s e c o n d a r y v o l t a g e r e f e r r e d t o primary 10 theta =60*( %pi /180) ; // p h a s e a n g l e a s s o c o a t e d w i t h E2 86


11 12 13 14 15 16 17 18 19 20

21 22 23 24 25

// a f t e r r e f e r r i n g t o p r i m a r y s i d e , w i t h E1 a s a reference V = E21 *( cos ( theta ) - %i * sin ( theta ) ) - E1 ; // r e s u l t a n t voltage I = abs ( V ) /( r1 + r21 ) ; // m a g n i t u d e o f c u r r e n t pd1 = I ^2* r1 ; pd2 = I ^2* r21 ; printf ( ’ power d i s s i p a t e d i n %f ohm r e s i s t o r i s %f W\ n ’ ,r1 , pd1 ) ; printf ( ’ power d i s s i p a t e d i n %f ohm r e s i s t o r i s %f W\ n ’ ,r21 , pd2 ) ; // C u r r e n t l a g s E1 by 90 d e g r e e teta1 =90*( %pi /180) ; // S i n c e E2 l a g s E1 by 60 d e g r e e and C u r r e n t due t o r e s u l t a n t v o l t a g e l a g s E1 by 9 0 , t h e r e f o r e p h a s e d i f f e r e n c e C u r r e n t I and E2 i s teta2 =(90 -60) *( %pi /180) ; Pd1 = E1 * I * cos ( teta1 ) ; Pd2 = E21 * I * cos ( teta2 ) ; printf ( ’ power d e l i v e r e d by %f v s o u r c e i s %f W\n ’ ,E1 , Pd1 ) ; printf ( ’ power d e l i v e r e d by %f v s o u r c e i s %f W\n ’ ,E2 , Pd2 ) ;

Scilab code Exa 1.73 Calculating efficiency and regulation under given condition 1 2 3 4 5

clc ; P =1200; // r a t e d VA o f t r a n s f o r m e r E1 =240; // r a t e d p r i m a r y v o l t a g e E2 =110; // r a t e d s e c o n d a r y v o l t a g e xe =5; // t o t a l l e a k a g e r e a c t a n c e o f t r a n s f o r m e r r e f e r r e d to primary 6 re =1; // t o t a l r e s i s t a n c e o f t r a n s f o r m e r r e f e r r e d t o primary 87


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

vl =240; // r a t e d v o l t a g e o f l o a d R =2500; // c o r e r e s i s t a n c e pf =1; // power f a c t o r ip = P / E1 ; // f u l l l o a d p r i m a r y c u r r e n t // a s s u m i n g E1 a s a r e f e r e n c e Vin = sqrt (( E1 + ip * re ) ^2+( ip * xe ) ^2) ; // i n p u t v o l t a g e when l o a d i s c o n n e c t e d pi = ip ^2* re ; // ohmic l o s s e s pc = Vin ^2/ R ; // c o r e l o s s n =( P * pf *100) /( P * pf + pi + pc ) ; printf ( ’ E f f i c i e n c y a t f u l l l o a d i s %f p e r c e n t \n ’ ,n ) ; vr =(( Vin - E1 ) / Vin ) *100; printf ( ’ V o l t a g e r e g u l a t i o n f o r f u l l l o a d i s %f p e r c e n t ’ , vr ) ;

Scilab code Exa 1.74 Determining current taken from source primary input impedance and input power 1 2 3 4 5 6 7 8

9 10 11 12 13 14

clc ; k1 =4; // t u r n s r a t i o f o r t r a n s f o r m e r 1 k2 =3; // t u r n s r a t i o f o r t r a n s f o r m e r 2 E1 =120; // s u p p l y a c r o s s which two t r a n s f o r m e r s a r e connected in p a r a l l e l E21 = E1 / k1 ; // s e c o n d a r y v o l t a g e f o r t r a n s f o r m e r 1 E22 = E1 / k2 ; // s e c o n d a r y v o l t a g e f o r t r a n s f o r m e r 2 R =10; // l o a d r e s i s t a n c e // S e c o n d a r y w i n d i n g s a r e c o n n e c t e d i n s e r i e s w i t h t h e p o l a r i t y s u c h t h a t t h e v o l t a g e s E21 and E22 aid each other E2 = E21 + E22 ; // r e s u l t a n t o u t p u t v o l t a g e il = E2 / R ; // Load c u r r e n t // f o r mmf b a l a n c e , p r i m a r y c u r r e n t f o r T1 ip1 = il / k1 ; // f o r mmf b a l a n c e , p r i m a r y c u r r e n t f o r T2 ip2 = il / k2 ; 88


15 ip = ip1 + ip2 ; // 16 printf ( ’ C u r r e n t 17 zi = E1 / ip ; 18 printf ( ’ Primary 19 pi = E2 * il ; 20 printf ( ’ Primary

t o t a l c u r r e n t drawn from s o u r c e drawn from s o u r c e i s %f A\n ’ , ip ) ; i n p u t i m p e d a n c e i s %f ohm\n ’ , zi ) ; i n p u t power i s %f W’ , pi ) ;

Scilab code Exa 1.75 Finding currents drawn by primary and secondary of transformer A and open circuit voltage across secondary of transformer B 1 clc ; 2 // A and B a r e two t r a n s f o r m e r shown i n f i g 1 . 8 5 3 im =0.1; // m a g n e t i z i n g c u r r e n t 4 // S i n c e SA ( s e c o n d a r y w i n d i n g f o r t r a n s f o r m e r A)

5 6 7 8 9

is

c o n n e c t e d d i r e c t l y a c r o s s 200 v s u p p l y , s o t h e magnetizing current required to e s t a b l i s h the f l u x must f l o w t h r o u g h SA , t h e r e f o r e c u r r e n t t a k e n by w i n d i n g PA( p r i m a r y w i n d i n g f o r t r a n s f o r m e r A) i s z e r o // S i n c e v o l t a g e a c r o s s SA i s 200 v , v o l t a g e a c r o s s PA h a s t o be 200 v by t r a n s f o r m e r a c t i o n // As a r e s u l t o f i t , v o l t a g e a c r o s s PB i s z e r o , t h e r e f o r e i n d u c e d emf i n SB i s z e r o printf ( ’ C u r r e n t i n s e c o n d a r y w i n d i n g o f t r a n s f o r m e r A i s %f A ’ , im ) ; disp ( ’ C u r r e n t i n p r i m a r y w i n d i n g o f t r a n s f o r m e r A i s z e r o ’ ); disp ( ’ V o l t a g e a c r o s s s e c o n d a r y w i n d i n g o f t r a n s f o r m e r B i s z e r o ’ );

Scilab code Exa 1.76 Determining the reading of ideal voltmeter 1 clc ;

89


2 3 4 5 6 7

8 9 10 11

12 13 14 15

16 17

// 1 and 2 s u b s c r i p t s a r e u s e d f o r t r a n s f o r m e r 1 and transformer 2 // P and S a r e u s e d f o r p r i m a r y and s e c o n d a r y winding E =230; // Rated p r i m a r y v o l t a g e o f b o t h t r a n s f o r m e r disp ( ’ c a s e 1 ’ ) ; R =0; // l o a d c o n n e t e d a c r o s s s e c o n d a r y o f transformer 1 // R=0 means w i n d i n g i s s h o r t c i r c u i t e d , t h e r e f o r e v o l t a g e a c r o s s S1 i s z e r o , t h e r e f o r e w h o l e v o l t a g e i s applied a c r o s s s2 t h e r e f o r e printf ( ’ R e a d i n g o f v o l t m e t e r f o r R=%f i s %f v \n ’ ,R , E ); disp ( ’ c a s e 2 ’ ) ; R =115; // l o a d c o n n e t e d a c r o s s s e c o n d a r y o f transformer 1 // A r e s i s t a n c e o f 115 ohm s h o u l d c a u s e a c u r r e n t o f 1 A b e c a u s e v o l t a g e a c r o s s P1 i s 115 v but t h i s means t h a t P1 and P2 c a r r y i n g 1A c u r r e n t . A c c o r d i n g t o t h i s v o l t a g e a c r o s s P2 i s ma gn eti zin g impedance o f t r a n s f o r m e r 2 times m a g n e t i z i n g c u r r e n t . But m a g n e t i z i n g i m p e d a n c e i s v e r y l a r g e , due t o t h i s v o l t a g e a c r o s s P2 r i s e s much a b o v e 115 v and v o l t a g e a c r o s s P1 f a l l s , due t o which m a g n e t i z i n g c u r r e n t t h r o u g h S1 decreases . disp ( ’ V o l t m e t e r r e a d i n g i s much more t h a n 115 v but l e s s t h a n 230 v . L e t i t be Vb ’ ) ; disp ( ’ c a s e 3 ’ ) ; R =1000; // l o a d c o n n e t e d a c r o s s s e c o n d a r y o f transformer 1 // For R=1000 ohm , c u r r e n t i n P1 and S1 a r e r e d u c e d . Hence c u r r e n t i n P2 i s a l s o r e d u c e d . T h e r e f o r e v o l t a g e a c r o s s P2 i s a l s o r e d u c e d but s t i l l i t i s more t h a n 115 v disp ( ’ V o l t m e t e r r e a d i n g i s more t h a n 115 v but l e s s t h a n Vb ’ ) ; disp ( ’ c a s e 4 ’ ) ; 90


// f o r R= i n f i n i t y S1 i s open c i r c u i t e d , t h e r e f o r e v o l t a g e a c r o s s P1=115 v and P2=115 v 19 disp ( ’ R e a d i n g o f v o l t m e t e r i s 115 v ’ ) ; 18

Scilab code Exa 1.77 Finding current drawn by transformer under no load test 1 2 3 4 5 6 7 8

clc ; E1 =200; // r a t e d v o l t a g e o f l v s i d e E2 =400; // r a t e d v o l t a g e o f h v s i d e f =50; // f r e q u e n c y o f s u p p l y W =80; // open c i r c u i t i n p u t w a t t a g e=c o r e l o s s m =1.91; // mutual i n d u c t i o n b e t w e e n h v and l v s i d e Ic = W / E2 ; // c o r e l o s s c u r r e n t Qmax = E1 /( sqrt (2) * %pi * f ) ; // maximum v a l u e o f f l u x l i n k a g e with l v winding 9 Im = Qmax /( sqrt (2) * m ) ; 10 Io = sqrt ( Ic ^2+ Im ^2) ; 11 printf ( ’ C u r r e n t t a k e n by t r a n s f o r m e r when no l o a d t e s t i s c o n d u c t e d on h v s i d e i s %f A ’ , Io ) ;

Scilab code Exa 1.78 Determining the efficiency and voltage regulation of transformer 1 2 3 4 5

clc ; P =100000; // r a t e d VA o f t r a n s f o r m e r n =0.98; // maximum p o s s i b l e e f f i c i e n c y l =80000; // r a t e d KVA o f l o a d vrm =0.04; // maximum p o s s i b l e v o l t a g e r e g u l a t i o n i s equal to ze2 in p u 6 pf =0.8; // power f a c t o r a t which e f f i c i e n c y a n f v o l t a g e r e g u l a t i o n h a s t o be d e t e r m i n e d 7 pl =((1/ n ) -1) * l ; // t o t a l l o s s e s i n t r a n s f o r m e r 91


8 pc = pl /2; // c o r e 9 10 11 12 13 14 15

l o s s e s ; a t maximum e f f i c i e n c y ohmic l o s s e s = core l o s s e s po =(1/ pf ) ^2* pc ; // ohmic l o s s e s a t g i v e n power factor N =( P * pf *100) /( l + po + pc ) ; printf ( ’ E f f i c i e n c y a t %f l a g g i n g power f a c t o r i s %f p e r c e n t \n ’ ,pf , N ) ; re2 = po / P ; // r e s i s t a n c e i n p u xe2 = sqrt ( vrm ^2 - re2 ^2) ; vr =( re2 * pf + xe2 * sqrt (1 - pf ^2) ) *100; printf ( ’ V o l t a g e r e g u l a t i o n a t %f l a g g i n g power f a c t o r i s %f p e r c e n t ’ ,pf , vr ) ;

Scilab code Exa 1.79 Determining the KVA rating of different section of autotransformer 1 clc ; 2 // a f t e r 3 4 5 6 7 8 9 10 11 12 13 14 15

d e r i v i n g the e x p r e s s i o n f o r both the t r a n s f o r m e r and a u t o t r a n s f o r m e r k =0.8; // r a t i o o f V2/V1 ( t u r n s r a t i o ) W =100; // Power t o be d e l i v e r e d i n KW pf =1; // u n i t y power f a c t o r n =0.96; // g i v e n e f f i c i e n c y disp ( ’ f o r two w i n d i n g t r a n s f o r m e r ’ ) ; ks = W ; printf ( ’KVA r a t i n g f o r s e c o n d a r y o f two w i n d i n g t r a n s f o r m e r i s %f KVA\n ’ , ks ) ; kp = ks / n ; printf ( ’KVA r a t i n g f o r p r i m a r y o f two w i n d i n g t r a n s f o r m e r i s %f KVA\n ’ , kp ) ; disp ( ’ f o r a u t o t r a n s f o r m e r ’ ) ; kab =( W / n ) *(1 - k ) ; printf ( ’KVA r a t i n g f o r s e c t i o n AB o f a u t o t r a n s f o r m e r i s %f KVA\n ’ , kab ) ; kbc = W *(1 - k / n ) ; 92


16

printf ( ’KVA r a t i n g f o r s e c t i o n BC o f a u t o t r a n s f o r m e r i s %f KVA\n ’ , kbc ) ;

93


Chapter 2 Electromechanical Energy Conversion Principle

Scilab code Exa 2.4 Determining magnitude of average magnetic force 1 2 3 4 5 6 7 8 9 10 11 12 13

clc ; l =0.02; // a i r gap l e n g t h i1 =20; // i n t e r m e d i a t e c u r r e n t i2 =40; // c u r r e n t d u r i n g a r m a t u r e movement from open to c l o s e p o s i t i o n // from f i g 2 . 1 1 f1 =0.04* i1 ; // f l u x l i n k a g e d u r i n g open p o s i t i o n a t A f2 =1.2+0.03*( i1 -20) ; // f l u x l i n k a g e d u r i n g c l o s e p o s i t i o n at D f3 =0.04* i2 ; // f l u x l i n k a g e d u r i n g open p o s i t i o n a t B f4 =1.2+0.03*( i2 -20) ; // f l u x l i n k a g e d u r i n g c l o s e p o s i t i o n at C // M e c h a n i c a l work done=a r e a ODCFEO−a r e a OABFEO W =(( i1 * f2 ) /2) +((( f2 + f4 ) * i1 ) /2) -(( i2 * f3 ) /2) ; fe = W / l ; printf ( ’ A v e r a g e e l e c t r o m a g n e t i c f o r c e i s %d N ’ , fe ) ;

94


Scilab code Exa 2.6 Determining force required for axial alignment of electromagnets 1 2 3 4 5 6 7 8

clc ; g =0.003; // gap l e n g t h wp =0.006; // p o l e w i d t h B =0.8; // f l u x d e n s i t y i n a i r gap uo =4* %pi *10^ -7; // f r e e s p a c e p e r m e a b i l i t y // a f t e r t h e d e r i v a t i o n o f e x p r e s s i o n fe =( B ^2* wp * g ) /(2* uo ) ; printf ( ’ F o r c e t e n d i n g t o b r i n g e l e c t r o m a g n e t s i n t o a x i a l a l i g n m e n t i s %f N ’ , fe ) ;

Scilab code Exa 2.7 Determining 1 air gap flux densities and coil inductance 2 energy stored in magnetic field 3 electromagnetic force 4 mechanical work done 5 electrical energy supplied by load 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

clc ; N =1500; // number o f t u r n s i n c o i l i =3; // c u r r e n t c a r r i e d by c o i l uo =4* %pi *10^ -7; // f r e e s p a c e p e r m e a b i l i t y l =0.04; // s i d e o f p l u n g e r A = %pi *( l /2) ^2; // c r o s s s e c t i o n a r e a o f p l u n g e r disp ( ’ c a s e a ’ ) ; disp ( ’ f o r gap l e n g t h G=2 cm ’ ) ; g =0.02; // a i r gap l e n g t h i n cm x =0; // d i s p l a c e m e n t o f p l u n g e r G1 =g - x ; // gap l e n g t h B =( uo * i * N ) / G1 ; // a i r gap f l u x d e n s i t y printf ( ’ A i r gap f l u x d e n s i t y i s %f Wb/mˆ2\ n ’ ,B ) ; L1 =( N ^2* uo * A ) / G1 ; printf ( ’ C o i l i n d u c t a n c e i s %f H ’ , L1 ) ; 95


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 44 45 46 47 48 49 50 51 52 53

disp ( ’ f o r gap l e n g t h G=1.5 cm ’ ) ; g =0.02; // a i r gap l e n g t h i n cm x =0.005; // d i s p l a c e m e n t o f p l u n g e r G2 =g - x ; // gap l e n g t h B =( uo * i * N ) / G2 ; // a i r gap f l u x d e n s i t y printf ( ’ A i r gap f l u x d e n s i t y i s %f Wb/mˆ2\ n ’ ,B ) ; L2 =( N ^2* uo * A ) / G2 ; printf ( ’ C o i l i n d u c t a n c e i s %f H ’ , L2 ) ; disp ( ’ f o r gap l e n g t h G=1 cm ’ ) ; g =0.02; // a i r gap l e n g t h i n cm x =0.01; // d i s p l a c e m e n t o f p l u n g e r G3 =g - x ; // gap l e n g t h B =( uo * i * N ) / G3 ; // a i r gap f l u x d e n s i t y printf ( ’ A i r gap f l u x d e n s i t y i s %f Wb/mˆ2\ n ’ ,B ) ; L3 =( N ^2* uo * A ) / G3 ; printf ( ’ C o i l i n d u c t a n c e i s %f H ’ , L3 ) ; disp ( ’ f o r gap l e n g t h G=0.5 cm ’ ) ; g =0.02; // a i r gap l e n g t h i n cm x =0.015; // d i s p l a c e m e n t o f p l u n g e r G4 =g - x ; // gap l e n g t h B =( uo * i * N ) / G4 ; // a i r gap f l u x d e n s i t y printf ( ’ A i r gap f l u x d e n s i t y i s %f Wb/mˆ2\ n ’ ,B ) ; L4 =( N ^2* uo * A ) / G4 ; printf ( ’ C o i l i n d u c t a n c e i s %f H ’ , L4 ) ; disp ( ’ c a s e b ’ ) ; disp ( ’ f o r gap l e n g t h G=2 cm ’ ) ; W =( i ^2* L1 ) /2; printf ( ’ Energy s t o r e d i s %f watt−s e c \n ’ ,W ) ; disp ( ’ f o r gap l e n g t h G=1.5 cm ’ ) ; W =( i ^2* L2 ) /2; printf ( ’ Energy s t o r e d i s %f watt−s e c \n ’ ,W ) ; disp ( ’ f o r gap l e n g t h G=1 cm ’ ) ; W =( i ^2* L3 ) /2; printf ( ’ Energy s t o r e d i s %f watt−s e c \n ’ ,W ) ; disp ( ’ f o r gap l e n g t h G=0.5 cm ’ ) ; W =( i ^2* L4 ) /2; printf ( ’ Energy s t o r e d i s %f watt−s e c \n ’ ,W ) ; disp ( ’ c a s e c ’ ) ; 96


54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74

disp ( ’ f o r gap l e n g t h G=2 cm ’ ) ; fe = round (( i ^2* g * L1 ) /(2* G1 ^2) ) ; printf ( ’ E l e c t r o m a g n e t i c f o r c e i s %f N\n ’ , fe ) ; disp ( ’ f o r gap l e n g t h G=1.5 cm ’ ) ; fe =( i ^2* g * L1 ) /(2* G2 ^2) ; printf ( ’ E l e c t r o m a g n e t i c f o r c e i s %f N\n ’ , fe ) ; disp ( ’ f o r gap l e n g t h G=1 cm ’ ) ; fe = round (( i ^2* g * L1 ) /(2* G3 ^2) ) ; printf ( ’ E l e c t r o m a g n e t i c f o r c e i s %f N\n ’ , fe ) ; disp ( ’ f o r gap l e n g t h G=0.5 cm ’ ) ; fe = round (( i ^2* g * L1 ) /(2* G4 ^2) ) ; printf ( ’ E l e c t r o m a g n e t i c f o r c e i s %f N\n ’ , fe ) ; disp ( ’ c a s e 4 ’ ) ; // f o r g=2 cm and g =0.5cm , d i s p l a c e m e n t i s g i v e n by xi =0; xf =0.015; Wm = integrate ( ’ ( i ˆ2∗ g ∗ L1 ) / ( 2 ∗ ( g−x ) ˆ 2 ) ’ , ’ x ’ ,xi , xf ) ; printf ( ’ M e c h a n i c a l work done i s %f watt−s e c \n ’ , Wm ) ; disp ( ’ c a s e e ’ ) ; We = integrate ( ’ ( i ˆ2∗ g ∗ L1 ) / ( g−x ) ˆ2 ’ , ’ x ’ ,xi , xf ) ; printf ( ’ E l e c t r i c a l e n e r g y s u p p l i e d by s o u r c e i s %f watt−s e c \n ’ , We ) ;

Scilab code Exa 2.8 Determining torque 1 2 3 4 5 6 7 8 9

clc ; r =50*10^ -3; // r a d i u s o f r o t o r g =2*10^ -3; // a i r gap l e n g t h l =10*10^ -3; // l e n g t h n o r m a l t o r a d i u s r B =2.2; // maximum a i r gap f l u x d e n s i t y uo =4* %pi *10^ -7; // f r e e s p a c e p e r m e a b i l i t y // a f t e r t h e d e r i v a t i o n o f t h e e x p r e s s i o n T =( B ^2* g * r * l ) / uo ; printf ( ’ Magnitude o f t o r q u e i s %f N−m ’ ,T ) ;

97


Scilab code Exa 2.9 Determining torque 1 2 3 4 5 6 7 8 9

clc ; r =50*10^ -3; // r a d i u s o f r o t o r g =2*10^ -3; // a i r gap l e n g t h l =10*10^ -3; // l e n g t h n o r m a l t o r a d i u s r B =2.2; // maximum a i r gap f l u x d e n s i t y uo =4* %pi *10^ -7; // f r e e s p a c e p e r m e a b i l i t y // a f t e r t h e d e r i v a t i o n o f t h e e x p r e s s i o n T =( B ^2* g *( r +( g /2) ) * l ) / uo ; printf ( ’ Magnitude o f t o r q u e i s %f N−m ’ ,T ) ;

Scilab code Exa 2.13 Determining magnitude and direction of torque for different cases 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

clc ; f =50; // f r e q u e n c y w =2* %pi * f ; // a n g u l a r s p e e d y =60; // y=a n g u l a r p o s i t i o n o f r o t o r Ls =0.6+0.2* cosd (2* y ) // s e l f i n d u c t a n c e o f s t a t o r dLs = -2*0.2* sind (2* y ) ; // d e r i v a t i v e o f Ls w i t h y Lr =0.75+0.3* cosd (2* y ) // s e l f i n d u c t a n c e o f r o t o r dLr = -2*0.3* sind (2* y ) ; // d e r i v a t i v e o f Lr w i t h y Ms =0.8* cosd ( y ) // mutual i n d u c t a n c e o f s t a t o r dMs = -0.8* sind ( y ) ; // d e r i v a t i v e o f Ms w i t h y disp ( ’ c a s e a ’ ) ; is =20; // s t a t o r c u r r e n t ir =10; // r o t o r c u r r e n t te =( is ^2* dLs ) /2+( ir ^2* dLr ) /2+( is * ir ) * dMs ; printf ( ’ Magnitude o f t o r q u e i s %f Nm and s i n c e i t i s n e g a t i v e i t a c t s in such a d i r e c t i o n so as to d e c r e a s e a n g u l a r p o s i t i o n \n ’ ,- te ) ; 98


16 is =20; // s t a t o r c u r r e n t 17 ir = -10; // r o t o r c u r r e n t 18 te =(( is ^2* dLs ) /2) +(( ir ^2* dLr ) /2) +(( is * ir ) * dMs ) ; 19 printf ( ’ Magnitude o f t o r q u e i s %f Nm and i t a c t s i n 20 21 22 23 24 25

26 27 28 29 30 31

32 33 34 35 36 37

c l o c k w i s e d i r e c t i o n \n ’ , te ) ; is =20; // s t a t o r c u r r e n t ir =0; // r o t o r c u r r e n t te =(( is ^2* dLs ) /2) +(( ir ^2* dLr ) /2) +(( is * ir ) * dMs ) ; printf ( ’ Magnitude o f t o r q u e i s %f Nm and i t a c t s i n c o u n t e r −c l o c k w i s e d i r e c t i o n \n ’ ,- te ) ; disp ( ’ c a s e b ’ ) ; // r o t o r w i n d i n g i s s h o r t c i r c u i t e d r o t o r v o l t a g e =0 and i s =s q r t ( 2 ) ∗20∗ s i n ( wt ) h e r e a v e r a g e t o r q u e i s n e e d e d s o f o r c a l c u l a t i o n we n e e d n o t t o w o r r y a b o u t s i n ( wt ) is = sqrt (2) *20; // s t a t o r c u r r e n t ir =( - Ms / Lr ) * is ; // r o t o r c u r r e n t te =(( is ^2* dLs ) /2) +(( ir ^2* dLr ) /2) +(( is * ir ) * dMs ) ; printf ( ’ Magnitude o f t o r q u e i s %f Nm and i t a c t s i n c l o c k w i s e d i r e c t i o n \n ’ , te /2) ; disp ( ’ c a s e c ’ ) ; // v s=s q r t ( 2 ) ∗ 3 1 4 ∗ s i n ( wt ) a g a i n h e r e a v e r a g e t o r q u e i s n e e d e d s o f o r c a l c u l a t i o n we n e e d n o t t o w o r r y a b o u t s i n ( wt ) vs = sqrt (2) *314; // s t a t o r w i n d i n g v o l t a g e ls1 =( Ls -( Ms ^2/ Lr ) ) ; // s h o r t c i r c u i t i n d u c t a n c e o f s t a t o r winding is = vs /( w * ls1 ) ; // s t a t o r c u r r e n t ir =( - Ms / Lr ) * is ; // r o t o r c u r r e n t te =(( is ^2* dLs ) /2) +(( ir ^2* dLr ) /2) +(( is * ir ) * dMs ) ; printf ( ’ Magnitude o f t o r q u e i s %f Nm and i t a c t s i n c l o c k w i s e d i r e c t i o n \n ’ , te /2) ;

Scilab code Exa 2.15 Determining magnitude of exciting current for closing relay and keeping it closed 99


1 2 3 4 5 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

clc ; B =[ 0.2 0.4 0.6 0.8 1 1.2]; H =[ 50 100 160 225 300 400]; plot (H , B ) ; xlabel ( ’ m a g n e t i c f i e l d i n t e n s i t y ’ ) ; ylabel ( ’ m a g n e t i c f l u x d e n s i t y ’ ) ; title ( ’B−H c u r v e ’ ) ; p =0.2; // f o r c e e x e r t e d by s p r i n g g1 =0.5*10^ -2; // a i r gap l e n g t h g2 =0.1*10^ -2; // r e d u c e d gap l e n g t h a f t e r c o i l i s energised n =2000; // c o i l t u r n s l =0.2; // mean l e n g t h o f m a g n e t i c i r o n p a t h A =0.2*10^ -4; // a r e a o f c r o s s −s e c t i o n g =9.81; // a c c e l e r a t i o n due t o g r a v i t y l1 =6; // gap l e n g t h l2 =3; // gap l e n g t h b e t w e e n s p r i n g and c o r e uo =4* %pi *10^ -7; // f r e e s p a c e p e r m e a b i l i t y disp ( ’ c a s e a ’ ) ; fe =( p * g * l2 ) / l1 ; // e l e c t r o m a g n e t i c t o r q u e Bg = sqrt ((2* fe * uo ) / A ) ; // a i r gap f l u x d e n s i t y printf ( ’ A i r gap f l u x d e n s i t y i s %f T\n ’ , Bg ) ; // c o r r e s p o n d i n g t o v a l u e o f Bg from B−H c u r v e H i s Hg =87.7; // m a g n e t i c f l u x i n t e n s i t y ATi = l * Hg ; // t o t a l ampere t u r n f o r i r o n p a t h ATg =( Bg * g1 ) / uo ; // ampere t u r n f o r a i r gap AT = ATi + ATg ; // t o t a l ampere t u r n s ie = AT / n ; printf ( ’ E x c i t i n g c u r r e n t r e q u i r e d t o c l o s e t h e a r m a t u r e r e l a y i s %f A\n ’ , ie ) ; disp ( ’ c a s e b ’ ) ; ATg =( Bg * g2 ) / uo ; // ampere t u r n f o r a i r gap AT = ATi + ATg ; // t o t a l ampere t u r n s ie = AT / n ; printf ( ’ E x c i t i n g c u r r e n t r e q u i r e d t o k e e p t h e a r m a t u r e c l o s e d i s %f A\n ’ , ie ) ;

100


Scilab code Exa 2.16 Determining magnitude of unbalanced magnetic pull on armature 1 2 3 4 5 6 7 8 9

clc ; x =0.1*10^ -3; // d i s p l a c e m e n t o f a r m a t u r e B =0.8; // a i r gap f l u x d e n s i t y A =200*10^ -4; // a r e a u n d e r p o l e g =0.6*10^ -2; // a i r gap l e n g t h uo =4* %pi *10^ -7; // f r e e s p a c e p e r m e a b i l i t y // a f t e r d e r i v e d e x p r e s s i o n p =( B ^2* A * g * x ) /( uo *( g ^2 - x ^2) ) ; printf ( ’ U n b a l a n c e d m a g n e t i c p u l l on a r m a t u r e i s %f N ’ ,p ) ;

Scilab code Exa 2.17 Determining magnitude and direction of magnetic force 1 clc ; 2 i = sqrt (2) *1000; // maximum c u r r e n t 3 4 5 6 7 8 9

c a r r i e d by

conductors z =2; // number o f c o n d u c t o r s i n s l o t l =1; // embedded l e n g t h w =0.05; // s l o t w i d t h uo =4* %pi *10^ -7; // f r e e s p a c e p e r m e a b i l i t y // a f t e r d e r i v e d e x p r e s s i o n fe =( uo * z ^2* l * i ^2) /(2* w ) ; printf ( ’ Magnitude o f m a g n e t i c f o r c e i s %f N and i t s d i r e c t i o n i s t o w a r d s t h e bottom o f t h e s l o t \n ’ , fe /2) ;

101


Scilab code Exa 2.19 Determining 1 mechanical work done 2 energy supplied by each electrical source 3 change in field energy 1 clc ; 2 // L1 =3+(1/(2∗X) ) − s e l f i n d u c t a n c e o f c o i l 1 3 // L1 =2+(1/(2∗X) ) − s e l f i n d u c t a n c e o f c o i l 2 4 // M=1/(2∗X) −mutual i n d u c t a n c e 5 // X i s d i s p l a c e m e n t 6 i1 =10; // c u r r e n t o f c o i l 1 7 i2 = -5; // c u r r e n t o f c o i l 2 8 // from e x p r e s s i o n W=(L1 ∗ i 1 ˆ 2 ) /2 + ( L2 ∗ i 2 ˆ 2 ) /2 + ( i 1

9

10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

∗ i 2 ∗M) where W i s work done and i s e q u a l t o 1 75+ 25 /( 4 ∗ x ) ; // f e = −25/(4∗ x ˆ 2 ) where f e i s e l e c t r o m a g n e t i c f o r c e and i s e q u a l t o r a t e o f c h a n g e o f work done w i t h r e s p e c t to x disp ( ’ c a s e a ’ ) ; x1 =0.5; x2 =1; Wm = integrate ( ’ −25/(4∗ x ˆ 2 ) ’ , ’ x ’ ,x1 , x2 ) ; printf ( ’ M e c h a n i c a l work done f o r g i v e n i n c r e m e n t i n d i s p l a c e m e n t i s %f watt−s e c \n ’ , Wm ) ; disp ( ’ c a s e b ’ ) // p s i 1 =(3+(1/(2∗ x ) ) ) ∗ i 1 + ( 1 / ( 2 ∗ x ) ) ∗ i 2 p s i 1 i s f l u x l i n k a g e with c o i l 1 // p s i 2 =(2+(1/(2∗ x ) ) ) ∗ i 2 + ( 1 / ( 2 ∗ x ) ) ∗ i 1 p s i 2 i s f l u x l i n k a g e with c o i l 2 We1 =(((3+(1/(2* x2 ) ) ) * i1 +(1/(2* x2 ) ) * i2 ) -((3+(1/(2* x1 ) ) ) * i1 +(1/(2* x1 ) ) * i2 ) ) * i1 ; We2 =(((2+(1/(2* x2 ) ) ) * i2 +(1/(2* x2 ) ) * i1 ) -((2+(1/(2* x1 ) ) ) * i2 +(1/(2* x1 ) ) * i1 ) ) * i2 ; printf ( ’ E l e c t r i c a l e n e r g y s u p p l i e d by s o u r c e 1 i s %f watt−s e c \n ’ , We1 ) ; printf ( ’ E l e c t r i c a l e n e r g y s u p p l i e d by s o u r c e 2 i s %f watt−s e c \n ’ , We2 ) ; disp ( ’ c a s e c ’ ) ; // W=175+(25/(4∗ x ) ) ; dw =175+(25/(4* x2 ) ) -(175+(25/(4* x1 ) ) ) ; 102


25

printf ( ’ Change i n f i e l d e n e r g y i s %f Watt−s e c \n ’ , dw ) ;

Scilab code Exa 2.21 Determining force between two magnetic surfaces 1 2 3 4 5 6

clc ; B =1.6; // f l u x d e n s i t y b e t w e e n m a g n e t i c s u r f a c e s a =1; // a r e a o f m a g n e t i c s u r f a c e s uo =4* %pi *10^ -7; // f r e e s p a c e p e r m e a b i l i t y f =( B ^2* a ) /(2* uo ) ; printf ( ’ F o r c e b e t w e e n two m a g n e t i c s u r f a c e s i s %f N ’ ,f ) ;

Scilab code Exa 2.22 Determining force between two plates 1 2 3 4 5

clc ; A =1; // a r e a o f p l a t e s E =3*10^6; // e l e c t r i c f i e l d i n t e n s i t y b e t w e e n p l a t e s Eo =(10^ -9/(36* %pi ) ) ; // vaccum p e r m i t t i v i t y // a f t e r u s i n g b o t h e n e r g y and c o e n e r g y method expression for force is derived 6 f =( E ^2* Eo * A ) /2; 7 printf ( ’ F o r c e b e t w e e n two p l a t e s i s %f N ’ ,f ) ;

Scilab code Exa 2.23 Determining magnitude of voltage applied between two plates 1 clc ; 2 w =5*10^ -3; // c o u n t e r w e i g h t 3 a =30*10^ -4; // a r e a o f c r o s s

section of plates

103


4 5 6 7 8 9 10 11 12

d =1*10^ -2; // d i s t a n c e b e t w e e n two p l a t e s g =9.81; // a c c e l e r a t i o n due t o g r a v i t y Eo =(10^ -9/(36* %pi ) ) ; // vaccum p e r m i t t i v i t y l1 =8*10^ -2; // h o r i z o n t a l d i s t a n c e b e t w e e n p i v o t and plates l2 =10*10^ -2; // h o r i z o n t a l d i s t a n c e b e t w e e n p i v o t and c o u n t e r w e i g h t // f o r t h e b a r t o r e m a i n h o r i z o n t a l e l e c t r o s t a t i c f o r c e i s g i v e n by fe =( w * l2 * g ) / l1 ; v = sqrt (( fe *2* d ^2) /( Eo * a ) ) ; printf ( ’ V o l t a g e a p p l i e d b e t w e e n t h e p l a t e s i s %f KV ’ ,v /1000) ;

Scilab code Exa 2.25 Determining 1 coil inductance and magnetic energy 2 electrical energy supplied by source 3 mechanical work done 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

clc ; n =1000; // number o f t u r n s i n e x c i t i n g c o i l a =5*5*10^ -4; // c r o s s s e c t i o n a r e a o f c o r e g =1*10^ -2; // l e n g t h o f a i r gap uo =4* %pi *10^ -7; // f r e e s p a c e p e r m e a b i l i t y disp ( ’ c a s e a ’ ) ; i =5; // c o i l c u r r e n t l =( n ^2* uo * a ) /(2* g ) ; printf ( ’ I n d u c t a n c e o f c o i l i s %f H\n ’ ,l ) ; E =( l * i ^2) /2; printf ( ’ F i e l d e n e r g y s t o r e d i n i n d u c t o r i s %f Watt− s e c \n ’ ,E ) ; fe =( i ^2* a * n ^2* uo ) /(4* g ^2) ; printf ( ’ F o r c e on t h e a r m a t u r e i s %f N\n ’ , fe ) ; disp ( ’ c a s e b ’ ) ; g1 =0.5*10^ -2; // r e d u c e d l e n g t h o f a i r gap We =(( n ^2* uo * a ) /(2* g1 ) -( n ^2* uo * a ) /(2* g ) ) * i ^2; printf ( ’ E l e c t r i c a l e n e r g y s u p p l i e d by s o u r c e i s %f 104


Watt−s e c \n ’ , We ) ; 18 disp ( ’ c a s e c ’ ) ; 19 w = integrate ( ’ ( n ˆ2∗ uo ∗ a ∗ i ˆ 2 ) / ( 4 ∗ ( g−x ) ˆ 2 ) ’ , ’ x ’ ,0 , g1 ) ; 20 printf ( ’ M e c h a n i c a l work done i s %f Watt−s e c ’ ,w ) ;

Scilab code Exa 2.26 Determining force required for armature alignment with field structure 1 2 3 4 5 6 7 8

clc ; de =110*( %pi /180) ; // p o l e p i t c h g =0.4*10^ -2; // a i r gap l e n g t h B =0.5; // a i r gap f l u x d e n s i t y d =0.3; // a r m a t u r e d i a m e t e r uo =4* %pi *10^ -7; // f r e e s p a c e p e r m e a b i l i t y fe =( B ^2* d * de * g ) /(2* uo ) ; printf ( ’ F o r c e t h a t t e n d s t o p u l l t h e a r m a t u r e i n t o a l i g n m e n t i s %f N ’ , fe )

Scilab code Exa 2.28 Determining energy stored in inductor 1 2 3 4 5 6 7

8 9 10 11

clc ; r =4; // r e s i s t a n c e o f i n d u c t o r v =8; // maximum v o l t a g e L =2; // i n d u c t a n c e t =2; // t i m e r e q u i r e d t o r e a c h maximum v o l t a g e v a l u e disp ( ’ c a s e a ’ ) ; // a f t e r s o l v i n g l a p l a c e e q u a t i o n we g e t e x p r e s s i o n for i ( t ) ( transient current ) i . e i ( t ) =(( exp ˆ−2∗ t ) +(2∗ t −1) ) /2 // f o r t =2 s e c i ( t ) i s g i v e n by it =(( exp ( -2* t ) ) +2* t -1) /2; E =( L * it ^2) /2; printf ( ’ Energy s t o r e d d u r i n g %d s e c i s %f J \n ’ ,t , E ) ; 105


12 disp ( ’ c a s e b ’ ) ; 13 i = v / r ; // c u r r e n t when t r a n s i e n t s a r e o v e r 14 E =( L * i ^2) /2; 15 printf ( ’ Energy s t o r e d a f t e r t r a n s i e n t s a r e o v e r

is

%d J \n ’ ,E ) ;

Scilab code Exa 2.29 Determining and comparing field energy stored and field energy density in iron and air gap 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

clc ; l =1.2; // l e n g t h o f i r o n p a t h a =5*5*10^ -4; // a r e a o f c r o s s s e c t i o n uo =4* %pi *10^ -7; // p e r m e a b i l i t y f o r f r e e s p a c e ur =1500; // r e l a t i v e p e r m e a b i l i t y f o r i r o n i =2; // e x c i t i n g c u r r e n t n =1000; // number o f t u r n s o f c o i l g =0.5*10^ -2; // a i r gap l e n g t h r =( l /( uo * ur * a ) ) +( g /( uo * a ) ) ; // n e t r e l u c t a n c e f =( n * i ) / r ; // f l u x i n c o i l fe1 =(( f ^2* l ) /( uo * ur * a ) ) /2; printf ( ’ F i e l d e n e r g y s t o r e d i n i r o n i s %f J \n ’ , fe1 ) ; fe2 =(( f ^2* g ) /( uo * a ) ) /2; printf ( ’ F i e l d e n e r g y s t o r e d i n a i r gap i s %f J \n ’ , fe2 ) ; r1 = fe2 / fe1 ; printf ( ’ R a t i o o f f i e l d e n e r g y s t o r e d i n a i r gap t o f i e l d e n e r g y s t o r e d i n i r o n i s %f \n ’ , r1 ) ; d1 = fe1 /( l * a ) ; printf ( ’ Energy d e n s i t y i n i r o n i s %f J /mˆ3\ n ’ , d1 ) ; d2 = fe2 /( g * a ) ; printf ( ’ Energy d e n s i t y i n a i r gap i s %f J /mˆ3\ n ’ , d2 ) ; r2 = d2 / d1 ; printf ( ’ R a t i o o f e n e r g y d e n s i t y i n a i r gap t o e n e r g y d e n s i t y i n i r o n i s %f \n ’ , r2 ) ; 106


Scilab code Exa 2.30 Determining 1 coil inductance field energy and force on armature 2 mechanical work done 3 magnetic force 4 mechanical work done for constant flux linkage 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27

clc ; n =1200; // number o f t u r n s i n e x c i t i n g c o i l a =6*5*10^ -4; // a r e a o f c r o s s s e c t i o n o f c o r e disp ( ’ c a s e a ’ ) ; x =0.01; // d i s p l a c e m e n t o f c o i l i =2; // e x c i t i n g c u r r e n t uo =4* %pi *10^ -7; // f r e e s p a c e p e r m e a b i l i t y Wf =( n ^2* uo * a * i ^2) /(4* x ) ; printf ( ’ F i e l d e n e r g y s t o r e d i s %f J \n ’ , Wf ) ; F =( - n ^2* uo * a * i ^2) /(4* x ^2) ; printf ( ’ F o r c e on a r m a t u r e i s %f N\n ’ ,F ) ; disp ( ’ c a s e b ’ ) ; x1 =0.005; // f u r t h e r d i s p l a c e m e n t o f c o i l v1 =( n ^2* uo * a * i ) /(2* x ) ; // f l u x l i n k a g e c o r r e s p o n d i n g t o d i s p l a c e m e n t 1 cm v2 =( n ^2* uo * a * i ) /(2* x1 ) ; // f l u x l i n k a g e corresponding to displacement 0 . 5 cm M =(( v2 - v1 ) * i ) /2; printf ( ’ M e c h a n i c a l e n e r g y o u t p u t i s %f J \n ’ ,M ) ; disp ( ’ c a s e c ’ ) ; // a f t e r d e r i v i n g e x p r e s s i o n Wm = integrate ( ’ (−n ˆ2∗ uo ∗ a ) / x ˆ2 ’ , ’ x ’ ,x , x1 ) ; printf ( ’ M e c h a n i c a l work done i s %f J \n ’ , Wm ) ; disp ( ’ c a s e d ’ ) ; // f o r x = 0 . 0 0 5 f l u x l i n k a g e i s c o n s t a n t t h a n c u r r e n t w i l l change i2 = v1 /((2* n ^2* uo * a ) /(2* x ) ) ; Wm =(( i - i2 ) * v1 ) /2; 107


28 29 30 31 32

printf ( ’ M e c h a n i c a l work done i f f l u x l i n k a g e a r e m a i n t a i n e d c o n s t a n t i s %f J \n ’ , Wm ) ; disp ( ’ c a s e e ’ ) ; // a f t e r t h e e x p r e s s i o n i s d e r i v e d Wm = integrate ( ’ (−v1 ˆ 2 ) / ( n ˆ2∗ uo ∗ a ) ’ , ’ x ’ ,x , x1 ) ; printf ( ’ M e c h a n i c a l work done i f f l u x l i n k a g e a r e m a i n t a i n e d c o n s t a n t i s %f J \n ’ , Wm ) ;

108


Chapter 3 Basic Concepts of Rotating Electrical Machines

Scilab code Exa 3.2 Determining rating of generator for given parallel paths 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

clc ; n =24; // Number o f a r m a t u r e c o n d u c t o r v =2; // a v e r a g e v o l t a g e p e r c o n d u c t o r i =5; // c u r r e n t c a r r y i n g c a p a c i t y o f e a c h c o n d u c t o r disp ( ’ c a s e a ’ ) ; a =2; // number o f p a r a l l e l p a t h sc = n / a ; // s e r i e s c o n n e c t e d c o n d u c t o r i n e a c h p a t h Ea = sc * v ; // o u t p u t v o l t a g e Ia = i * a ; // o u t p u t c u r r e n t p = Ea * Ia ; // power r a t i n g printf ( ’ G e n e r a t o r r a t i n g i s %f W\n ’ ,p ) ; disp ( ’ c a s e b ’ ) ; a =4; // number o f p a r a l l e l p a t h sc = n / a ; // s e r i e s c o n n e c t e d c o n d u c t o r i n e a c h p a t h Ea = sc * v ; // o u t p u t v o l t a g e Ia = i * a ; // o u t p u t c u r r e n t p = Ea * Ia ; // power r a t i n g printf ( ’ G e n e r a t o r r a t i n g i s %f W\n ’ ,p ) ; 109


19 20 21 22 23 24 25

disp ( ’ c a s e c ’ ) ; a =6; // number o f p a r a l l e l p a t h sc = n / a ; // s e r i e s c o n n e c t e d c o n d u c t o r i n e a c h p a t h Ea = sc * v ; // o u t p u t v o l t a g e Ia = i * a ; // o u t p u t c u r r e n t p = Ea * Ia ; // power r a t i n g printf ( ’ G e n e r a t o r r a t i n g i s %f W\n ’ ,p ) ;

Scilab code Exa 3.3 Calculating generated armature voltage for different types of machine 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

clc ; p =4; // number o f p o l e s s =60; // number o f s l o t s c =8; // number o f c o n d u c t o r s p e r s l o t f =20*10^ -3; // f l u x p e r p o l e nr =1500; // r e l a t i v e s p e e d i n rpm b e t w e e n f i e l d f l u x and a r m a t u r e w i n d i n g disp ( ’ c a s e a ’ ) ; // w i n d i n g i s l a p c o n n e c t e d a = p ; // f o r l a p c o n n e c t e d w i n d i n g , number o f p a r a l l e l p a t h=number o f p o l e z = s * c ; // t o t a l number o f c o n d u c t o r s n = nr /60; // s p e e d i n r p s E =( f * z * n * p ) / a ; printf ( ’ G e n e r a t e d EMF i s %f V\n ’ ,E ) ; disp ( ’ c a s e b ’ ) ; kw =0.96; // w i n d i n g f a c t o r nt = z /2; // T o t a l number o f t u r n s nph = nt /3; // number o f s e r i e s t u r n s p e r p h a s e fg =( p * n ) /2; // g e n e r a t e d EMF f r e q u e n c y E = sqrt (2) * %pi * fg * nph * kw * f ; printf ( ’ G e n e r a t e d EMF p e r p h a s e i s %f V\n ’ ,E ) ; e = round ( sqrt (3) * E ) ; printf ( ’ G e n e r a t e d EMF b e t w e e n l i n e t e r m i n a l i s %f V\ 110


n ’ ,e ) ;

Scilab code Exa 3.4 Determining 1 frequency of EMF 2 number of poles 3 number of synchronous motor poles 1 clc ; 2 p1 =4; // number o f 3 4 5 6 7 8

9 10 11 12 13

14 15 16 17 18 19 20 21

poles in s l i p ring induction

motor p2 =6; // number o f p o l e s i n s y n c h r o n o u s motor f =50; // f r e q u e n c y o f s u p p l y ns =(120* f ) / p2 ; // s y n c h r o n o u s motor s p e e d ni =(120* f ) / p1 ; // i n d u c t i o n motor s p e e d disp ( ’ c a s e a ( 1 ) ’ ) ; // when s y n c h r o n o u s motor i s d r i v e n i n d i r e c t i o n o p p o s i t e to the r o t a t i n g f i e l d produced the i n d u c t i o n motor s t a t o r nr = ns + ni ; // r e l a t i v e s p e e d F =( p1 * nr ) /120; printf ( ’ F r e q u e n c y o f EMF a t r o t o r s l i p r i n g t e r m i n a l s i s %f Hz\n ’ ,F ) ; disp ( ’ c a s e a ( 2 ) ’ ) ; // when s y n c h r o n o u s motor i s d r i v e n i n d i r e c t i o n o f t h e r o t a t i n g f i e l d p r o d u c e d t h e i n d u c t i o n motor stator nr = ni - ns ; // r e l a t i v e s p e e d F =( p1 * nr ) /120; printf ( ’ F r e q u e n c y o f EMF a t r o t o r s l i p r i n g t e r m i n a l s i s %f Hz\n ’ ,F ) ; disp ( ’ c a s e b ’ ) ; fn =150; // f r e q u e n c y o f r o t o r t e r m i n a l v o l t a g e required // l e t new number o f p o l e be pn t h e n r e l a t i v e s p e e d i s n r=n s + ( 1 2 0 ∗ 5 0 ) / pn ; pn =(( fn *120) -(120* f ) ) / ns ; printf ( ’ Number o f p o l e s t h a t t h e i n d u c t i o n motor 111


22 23 24 25

must have i s %f \n ’ , pn ) ; disp ( ’ c a s e c ’ ) ; pi =8; // number o f p o l e s i n i n d u c t i o n motor ps =(120* f * pi ) /(( fn *120) -(120* f ) ) ; printf ( ’ Number o f s y n c h r o n o u s motor p o l e s r e q u i r e d i s %f ’ , ps ) ;

Scilab code Exa 3.5 Calculating frequency and magnitude of per phase EMF for different speed of rotor 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

clc ; p =4; // number o f p o l e f =50; // f r e q u e n c y o f s u p p l y ns =420; // s t a t o r t u r n s nr =240; // r o t o r t u r n s F =30*10^ -3; // f l u x p e r p o l e kw =0.96; // w i n d i n g f a c t o r f o r b o t h s t a t o r and r o t o r nsph = ns /3; // s t a t o r t u r n p e r p h a s e nrph = nr /3; // r o t o r t u r n p e r p h a s e es = sqrt (2) * %pi * f * kw * nsph * F ; // s t a t o r t u r n p e r p h a s e disp ( ’ c a s e a ’ ) ; // r o t o r i s s t a t i o n a r y s =1; // a t s t a n d s t i l l s l i p =1 er = sqrt (2) * %pi * f * kw * nrph * F ; printf ( ’ f r e q u e n c y o f EMF i n s t a t o r i s %f Hz\n ’ ,f ) ; printf ( ’ f r e q u e n c y o f EMF i n r o t o r i s %f Hz\n ’ ,f ) ; printf ( ’ Per p h a s e s t a t o r EMF i s %f V\n ’ , es ) ; printf ( ’ Per p h a s e r o t o r EMF i s %f V\n ’ , er ) ; disp ( ’ c a s e b ’ ) ; sr =1440; // s p e e d o f r o t o r i n rpm i n d i r e c t i o n o f rotating flux Ns =(120* f ) / p ; // s p e e d o f r o t a t i n g f l u x s =( Ns - sr ) / Ns ; // s l i p fr = s * f ; // f r e q u e n c y o f EMF i n r o t o r er = sqrt (2) * %pi * fr * kw * nrph * F ; 112


25 printf ( ’ f r e q u e n c y 26 printf ( ’ f r e q u e n c y 27 printf ( ’ Per p h a s e 28 printf ( ’ Per p h a s e 29 disp ( ’ c a s e c ’ ) ; 30 sr =1440; // s p e e d 31 32 33 34 35 36 37

o f EMF i n s t a t o r i s %f Hz\n ’ ,f ) ; o f EMF i n r o t o r i s %f Hz\n ’ , fr ) ; s t a t o r EMF i s %f V\n ’ , es ) ; r o t o r EMF i s %f V\n ’ , er ) ;

o f r o t o r i n rpm o p p o s i t e t o t h e direction of rotating flux s =( Ns + sr ) / Ns ; // s l i p fr = s * f ; // f r e q u e n c y o f EMF i n r o t o r er = sqrt (2) * %pi * fr * kw * nrph * F ; printf ( ’ f r e q u e n c y o f EMF i n s t a t o r i s %f Hz\n ’ ,f ) ; printf ( ’ f r e q u e n c y o f EMF i n r o t o r i s %f Hz\n ’ , fr ) ; printf ( ’ Per p h a s e s t a t o r EMF i s %f V\n ’ , es ) ; printf ( ’ Per p h a s e r o t o r EMF i s %f V\n ’ , er ) ;

Scilab code Exa 3.6 Calculating fundamental third and fifth harmonic belt factors for stator 1 2 3 4 5 6 7 8 9 10 11 12 13 14

clc ; disp ( ’ c a s e a ’ ) ; s =54; // number o f s l o t s i n s t a t o r , 3 p h a s e p =6; // number o f p o l e s spp = s /(3* p ) ; // s l o t s p e r p o l e p e r p h a s e v =( p *180) / s ; // s l o t a n g u l a r p i t c h k1 = sin ((( spp * v ) /2) *( %pi /180) ) /( spp * sin (( v /2) *( %pi /180) ) ) ; // f u n d a m e n t a l h a r m o n i c s k3 = sin (((3* spp * v ) /2) *( %pi /180) ) /( spp * sin (((3* v ) /2) *( %pi /180) ) ) ; // t h i r d h a r m o n i c k5 = sin (((5* spp * v ) /2) *( %pi /180) ) /( spp * sin (((5* v ) /2) *( %pi /180) ) ) ; // f i f t h h a r m o n i c printf ( ’ F i r s t h a r m o n i c component i s %f\n ’ , k1 ) ; printf ( ’ T h i r d h a r m o n i c component i s %f\n ’ , k3 ) ; printf ( ’ F i f t h h a r m o n i c component i s %f\n ’ , k5 ) ; disp ( ’ c a s e b ’ ) ; s =48; // number o f s l o t s i n s t a t o r , 3 p h a s e 113


15 16 17 18 19 20 21 22 23 24 25

p =6; // number o f p o l e s spp = s /(3* p ) ; // s l o t s p e r p o l e p e r p h a s e sk = spp *3; v =( p *180) / s ; // s l o t a n g u l a r p i t c h ps = spp * v ; // p h a s e s p r e a d k1 = sin ((( ps ) /2) *( %pi /180) ) /( sk * sin ((( ps ) /( sk *2) ) *( %pi /180) ) ) ; // f u n d a m e n t a l h a r m o n i c s k3 = sin (((3* ps ) /2) *( %pi /180) ) /( sk * sin (((3* ps ) /( sk *2) ) *( %pi /180) ) ) ; // t h i r d h a r m o n i c k5 = sin (((5* ps ) /2) *( %pi /180) ) /( sk * sin (((5* ps ) /( sk *2) ) *( %pi /180) ) ) ; // f i f t h h a r m o n i c printf ( ’ F i r s t h a r m o n i c component i s %f\n ’ , k1 ) ; printf ( ’ T h i r d h a r m o n i c component i s %f\n ’ , k3 ) ; printf ( ’ F i f t h h a r m o n i c component i s %f\n ’ , k5 ) ;

Scilab code Exa 3.8 Determining distribution and winding factor 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

clc ; disp ( ’ c a s e a ’ ) ; cs =160*( %pi /180) ; // c o i l s p a n i n r a d i a n ps =120*( %pi /180) ; // p h a s e s p r e a d kd = sin ( ps ) /( ps /2) ; // d i s t r i b u t i o n f a c t o r f o r uniformly d i s t r i b u t e d winding e =180 -( cs *(180/ %pi ) ) ; // c h o r d i n g a n g l e kp = cos (( e /2) *( %pi /180) ) ; // C o i l s p a n f a c t o r wf = kd * kp ; // w i n d i n g f a c t o r disp ( ’ D i s t r i b u t i o n f a c t o r i s ’ ) ; disp ( kd ) ; disp ( ’ Winding f a c t o r i s ’ ) ; disp ( wf ) ; disp ( ’ c a s e b ’ ) ; s =9; // number o f s l o t s p e r p o l e sa =180/ s ; // s l o t a n g u l a r p i t c h // f o r a p h a s e s p r e a d o f 120 , 6 ∗20=120 , 6 a d j a c e n t s l o t s must b e l o n g t o t h e same phase , t h e r e f o r e 114


17 p =6; // p o l e s b e l o n g i n g t o same p h a s e 18 kd = sin ( ps /2) /( p * sin ( ps /(2*6) ) ) ; 19 wf = kd * kp ; 20 disp ( ’ D i s t r i b u t i o n f a c t o r i s ’ ) ; 21 disp ( kd ) ; 22 disp ( ’ Winding f a c t o r i s ’ ) ; 23 disp ( wf ) ;

Scilab code Exa 3.10 Finding ratio of outputs and amount of copper required for different combinations 1 clc ; 2 disp ( ’ c a s e a ’ ) ; 3 f1 =2/3; // f r a c t i o n o f s l o t wound 4 p1 = f1 *180; // p h a s e s p r e a d , 2/3 o f t h e

s l o t s are

wound 5 kd1 = sin (( p1 /2) *( %pi /180) ) /(( p1 /2) *( %pi /180) ) ; // 6 7 8 9 10

11 12 13 14 15 16

distribution factor p2 =180; // p h a s e s p r e a d , A l l t h e s l o t s a r e wound f2 =1; // f r a c t i o n o f s l o t wound kd2 = sin (( p2 /2) *( %pi /180) ) /(( p2 /2) *( %pi /180) ) ; // distribution factor // o u t p u t i s d i r e c t l y p r o p o r t i o n a l t o t h e p r o d u c t o f f r a c t i o n o f s l o t s wound and d i s t r i b u t i o n f a c t o r ro =( f2 * kd2 ) /( f1 * kd1 ) ; // I t i s assumed t h a t f r e q u e n c y , f l u x p e r p o l e and t h e c o n d u c t o r c r o s s s e c t i o n i s same printf ( ’ R a t i o o f o u t p u t s i s %f \n ’ , ro ) ; rc = f2 / f1 ; printf ( ’ r a t i o o f c o p p e r r e q u i r e d i s %f\n ’ , rc ) ; disp ( ’ c a s e b ’ ) ; p3 =60; // f o r 3− p h a s e w i n d i n g , p h a s e s p r e a d i s 60 degrees kd3 = sin (( p3 /2) *( %pi /180) ) /(( p3 /2) *( %pi /180) ) ; // distribution factor 115


17 18 19 20 21 22 23 24 25 26 27 28 29 30

// s i n c e a l l t h e s l o t s a r e wound f o r b o t h 1− p h a s e and 3−phase , f r a c t i o n o f t h e s l o t s wound i s 1 f3 =1; // f r a c t i o n o f t h e s l o t s wound ro = kd3 / kd2 ; printf ( ’ R a t i o o f o u t p u t s i s %f \n ’ , ro ) ; rc = f2 / f3 ; printf ( ’ r a t i o o f c o p p e r r e q u i r e d i s %f\n ’ , rc ) ; disp ( ’ c a s e c ’ ) ; f4 =1; // f r a c t i o n o f t h e s l o t s wound p4 =90; // f o r 2− p h a s e w i n d i n g , p h a s e s p r e a d i s 90 degrees kd4 = sin (( p4 /2) *( %pi /180) ) /(( p4 /2) *( %pi /180) ) ; // distribution factor ro = kd3 / kd4 ; printf ( ’ R a t i o o f o u t p u t s i s %f \n ’ , ro ) ; rc = f3 / f4 ; printf ( ’ r a t i o o f c o p p e r r e q u i r e d i s %f\n ’ , rc ) ;

Scilab code Exa 3.11 1 Estimating resultant line and phase voltage for given configurations 2 Determining circulating current for delta connected machine 1 2 3 4 5 6 7 8 9 10

clc ; d =0.28; // a i r gap d i a m e t e r l =0.23; // c o r e l e n g t h o f a l t e r n a t o r spp =4; // s l o t s p e r p o l e p e r p h a s e b1 =0.87; // a m p l i t u d e o f f l u x d e n s i t y i n f u n d a m e n t a l harmonic i n Tesla b3 =0.24; // a m p l i t u d e o f f l u x d e n s i t y i n t h i r d harmonic i n Tesla b5 =0.14; // a m p l i t u d e o f f l u x d e n s i t y i n f i f t h harmonic i n Tesla p =6; // number o f p o l e s i n a l t e r n a t o r np =3; // number o f p h a s e s c =8; // number o f c o n d u c t o r p e r s l o t 116


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

f =50; // f r e q u e n c y o f s u p p l y f1 =(2* d * l * b1 ) / p ; // f l u x f o r f u n d a m e n t a l h a r m o n i c f3 =(2* d * l * b3 ) /( p *3) ; // f l u x f o r t h i r d h a r m o n i c f5 =(2* d * l * b5 ) /( p *5) ; // f l u x f o r f i f t h h a r m o n i c ap =180/( spp * np ) ; // s l o t a n g u l a r p i t c h kd1 = sin ((( spp * ap ) /2) *( %pi /180) ) /( spp * sin (( ap /2) *( %pi /180) ) ) ; // d i s t r i b u t i o n f a c t o r f o r f u n d a m e n t a l harmonic kd3 = sin (((3* spp * ap ) /2) *( %pi /180) ) /( spp * sin (((3* ap ) /2) *( %pi /180) ) ) ; // d i s t r i b u t i o n f a c t o r f o r t h i r d harmonic kd5 = sin (((5* spp * ap ) /2) *( %pi /180) ) /( spp * sin (((5* ap ) /2) *( %pi /180) ) ) ; // d i s t r i b u t i o n f a c t o r f o r f i f t h harmonic // c o i l i s s h o r t p i t c h e d by one s l o t , t h e r e f o r e e =180/( spp * np ) ; // c h o r d i n g a n g l e cs1 = cos (( e /2) *( %pi /180) ) ; // c o i l s p a n f a c t o r f o r fundamental harmonic cs3 = cos (((3* e ) /2) *( %pi /180) ) ; // c o i l s p a n f a c t o r f o r t h i r d harmonic cs5 = cos (((5* e ) /2) *( %pi /180) ) ; // c o i l s p a n f a c t o r f o r f i f t h harmonic kw1 = cs1 * kd1 ; // w i n d i n g f a c t o r f o r f u n d a m e n t a l harmonic kw3 = cs3 * kd3 ; // w i n d i n g f a c t o r f o r t h i r d h a r m o n i c kw5 = cs5 * kd5 ; // w i n d i n g f a c t o r f o r f i f t h h a r m o n i c ts = spp * np * p ; // t o t a l number o f s l o t s tt =( ts * c ) /2; // t o t a l number o f t u r n s nph = tt / np ; // s e r i e s t u r n p e r p h a s e ep1 = sqrt (2) * %pi * f * kw1 * nph * f1 ; // emf p e r p h a s e f o r fundamental harmonics ep3 =( ep1 * kw3 *3* f3 ) /( kw1 * f1 ) ; // emf p e r p h a s e f o r t h i r d harmonics ep5 =( ep1 * kw5 *5* f5 ) /( kw1 * f1 ) ; // emf p e r p h a s e f o r f i f t h harmonics disp ( ’ c a s e a ( 1 ) : s t a r c o n n e c t e d a l t e r n a t o r ’ ) ; ep = sqrt ( ep1 ^2+ ep3 ^2+ ep5 ^2) ; printf ( ’ R e s u l t a n t EMF p e r p h a s e i s %f V\n ’ , ep ) ; 117


36 37 38 39 40

41 42 43 44 45

46 47

// t h i r d f r e q u e n c y l i n e emf d o e s n o t a p p e a r i n l i n e voltage el = sqrt (3) * sqrt ( ep1 ^2+ ep5 ^2) ; printf ( ’ R e s u l t a n t l i n e v o l t a g e i s %f V\n ’ , el ) ; disp ( ’ c a s e a ( 2 ) : D e l t a c o n n e c t e d a l t e r n a t o r ’ ) ; // t h i r d f r e q u e n c y l i n e emf d o e s n o t a p p e a r i n l i n e and p h a s e v o l t a g e a s t h e y a r e s h o r t c i r c u i t e d by closed delta ep = sqrt ( ep1 ^2+ ep5 ^2) ; printf ( ’ R e s u l t a n t EMF p e r p h a s e ( a l s o l i n e v o l t a g e ) i s %f V\n ’ , ep ) ; disp ( ’ c a s e b : d e l t a c o n n e c t e d a l t e r n a t o r ’ ) ; rpp =10; // r e a c t a n c e p e r p h a s e // emf t o due f i r s t and t h i r d h a r m o n i c c a n c e l s e a c h o t h e r but t h i r d h a r m o n i c g i v e s r i s e t o circulating current I =(3* ep3 ) /(3* np * rpp ) ; printf ( ’ C i r c u l a t i n g c u r r e n t i s %f A ’ ,I ) ;

Scilab code Exa 3.12 Calculating percentage increase in per phase rms emf due to harmonic 1 2 3 4 5 6 7 8 9 10

clc ; spp =3; // s l o t s p e r p o l e p e r p h a s e np =3; // number o f p h a s e s cs =8; // c o i l s p a n fp =0.20; // f r a c t i o n o f t h i r d h a r m o n i c i n f l u x d e n s i t y wave i n a i r gap sp = spp * np ; // s l o t s p e r p o l e v =180/ sp ; // s l o t a n g u l a r p i t c h kd1 = sin ((( spp * v ) /2) *( %pi /180) ) /( spp * sin (( v /2) *( %pi /180) ) ) ; // d i s t r i b u t i o n f a c t o r // f o r a c o i l s p a n o f 8 s l o t s t h e c o i l i s s h o r t p i t c h e d by one s l o t e = v ; // c h o r d i n g a n g l e 118


11 kp1 = cos (( e /2) *( %pi /180) ) ; // c o i l s p a n f a c t o r 12 kw1 = kp1 * kd1 ; // w i n d i n g f a c t o r 13 kd3 = sin (((3* spp * v ) /2) *( %pi /180) ) /( spp * sin ((( v *3) /2)

14 15 16 17 18 19

*( %pi /180) ) ) ; // d i s t r i b u t i o n f a c t o r f o r t h i r d harmonic kp3 = cos (((3* e ) /2) *( %pi /180) ) ; // c o i l s p a n f a c t o r f o r t h i r d harmonic kw3 = kd3 * kp3 ; // w i n d i n g f a c t o r f o r t h i r d h a r m o n i c er =( kw3 / kw1 ) * fp ; // r a t i o o f t h i r d h a r m o n i c emf t o f u n d a m e n t a l emf ep = sqrt (1+ er ^2) ; // r a t i o o f n e t emf t o f u n d a m e n t a l emf pi =(( ep -1) /1) *100; printf ( ’ P e r c e n t a g e i n c r e a s e i n p e r p h a s e rms emf i s %f p e r c e n t ’ , pi ) ;

Scilab code Exa 3.13 Determining phase and line EMF for given windings 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

clc ; p =6; // number o f p o l e s i n a l t e r n a t o r s =42; // number o f s l o t s i n a l t e r n a t o r f =0.012; // f l u x p e r p o l e t =8; // number o f t u r n s i n f u l l p i t c h c o i l F =50; // f r e q u e n c y o f a l t e r n a t o r disp ( ’ c a s e a ’ ) ; np =2; // number o f p h a s e s spp =42/( p * np ) ; // s l o t s p e r p o l e p e r p h a s e // spp i s n o t an i n t e g e r , t h e 2− p h a s e w i n d i n g i s a f r a c t i o n a l s l o t w i n d i n g , t h e r e f o r e Sk i s g i v e n by Sk = spp *2; v =90; // p h a s e s p r e a d f o r 2− p h a s e w i n d i n g kd = sind ( v /2) /( Sk * sind ( v /(2* Sk ) ) ) ; // d i s t r i b u t i o n factor kw = kd ; // w i n d i n g f a c t o r a s kp=1 nph =( s * t ) / np ; // p e r p h a s e s e r i e s t u r n 119


16 17 18 19 20 21 22 23 24 25 26 27 28 29

eph = sqrt (2) * F * %pi * kw * nph * f ; el = sqrt (2) * eph ; printf ( ’ Phase emf i s %f V\n ’ , eph ) ; printf ( ’ L i n e emf i s %f V\n ’ , el ) ; disp ( ’ c a s e b ’ ) ; np =3; // number o f p h a s e s v =60; // p h a s e s p r a e d f o r 3− p h a s e w i n d i n g kd = sind ( v /2) /( Sk * sind ( v /(2* Sk ) ) ) ; // d i s t r i b u t i o n factor kw = kd ; // w i n d i n g f a c t o r a s kp=1 nph =( s * t ) / np ; // p e r p h a s e s e r i e s t u r n eph = sqrt (2) * F * %pi * kw * nph * f ; el = sqrt (3) * eph ; printf ( ’ Phase emf i s %f V\n ’ , eph ) ; printf ( ’ L i n e emf i s %f V\n ’ , el ) ;

Scilab code Exa 3.14 Calculating third and fifth phase EMF in of fundamental phase emfs and the ratio of resultant line emf to resultant phase emf 1 2 3 4 5 6 7 8 9

10 11

clc ; s =81; // number o f s l o t s p =6; // number o f p o l e s np =3; // number o f p h a s e s cs =13; // c o i l s p a n i n t e r m s o f s l o t p i t c h e s v =60; // p h a s e s p r e a d f o r t h r e e p h a s e w i n d i n g f3 =0.4; // r a t i o o f t h i r d h a r m o n i c f l u x t o f i r s t harmonic f l u x f5 =0.25; // r a t i o o f f i f t h h a r m o n i c f l u x t o f i r s t harmonic f l u x spp = s /( p * np ) ; // // spp i s n o t an i n t e g e r , t h e 2− phase winding i s a f r a c t i o n a l s l o t winding , t h e r e f o r e Sk i s g i v e n by Sk = spp *2; ap =( p *180) / s ; 120


12 Cs = cs * ap ; // c o i l s p a n 13 e =180 - Cs ; // c h o r d i n g a n g l e 14 kd1 = sind ( v /2) /( Sk * sind ( v /(2* Sk ) ) ) ; // d i s t r i b u t i o n

f a c t o r f o r fundamental harmonic 15 kp1 = cosd ( e /2) ; // c o i l s p a n f a c t o r 16 kd3 = sind ((3* v ) /2) /( Sk * sind ((3* v ) /(2* Sk ) ) ) ; //

d i s t r i b u t i o n f a c t o r f o r t h i r d harmonic span f a c t o r f o r t h i r d harmonic kd5 = sind ((5* v ) /2) /( Sk * sind ((5* v ) /(2* Sk ) ) ) ; // d i s t r i b u t i o n f a c t o r f o r f i f t h harmonic kp5 = cosd ((5* e ) /2) ; // c o i l s p a n f a c t o r f o r f i f t h harmonic kw1 = kd1 * kp1 ; // w i n d i n g f a c t o r f o r f u n d a m e n t a l harmonics kw3 = kd3 * kp3 ; // w i n d i n g f a c t o r f o r t h i r d h a r m o n i c kw5 = kd5 * kp5 ; // w i n d i n g f a c t o r f o r f i f t h h a r m o n i c ep3 =( kw3 * f3 ) / kw1 ; printf ( ’ rms v a l u e o f t h i r d h a r m o n i c emf i s %f t i m e s t h e f u n d a m e n t a l h a r m o n i c emf \n ’ , ep3 ) ; ep5 =( kw5 * f5 ) / kw1 ; printf ( ’ rms v a l u e o f f i f t h h a r m o n i c emf i s %f t i m e s t h e f u n d a m e n t a l h a r m o n i c emf \n ’ , ep5 ) ; ep = sqrt (1+ ep3 ^2+ ep5 ^2) ; // r e s u l t a n t p h a s e emf el = sqrt (3) * sqrt (1+ ep5 ^2) ; // r e s u l t a n t l i n e emf r = el / ep ; printf ( ’ R a t i o o f r e s u l t a n t l i n e emf t o r e s u l t a n t p h a s e emf i s %f ’ ,r ) ;

17 kp3 = cosd ((3* e ) /2) ; // c o i l 18 19 20 21 22 23 24 25 26 27 28 29 30

Scilab code Exa 3.15 Finding rms value of voltage in single turn coil 1 2 3 4

clc ; B =1; // peak f l u x d e n s i t y i n T e s l a l =0.8; // l e n g t h o f a r m a t u r e c o n d u c t o r v =20; // v e l o c i t y o f c o i l 121


5 6

7 8 9 10 11

// f o r 0< t h e t a <30 c o i l aa ’ i s moving i n z e r o B− wave , emf f o r t h i s r a n g e i s z e r o // f o r 30< t h e t a < 60 c o i l s i d e a i s c u t t i n g t h r o u g h B−wave and c o i l s i d e a ’ i s c u t t i n g z e r o B−wave , therefore e1 = B * l * v ; // emf a t g i v e n p o s i t i o n o f c o i l // f o r 60< t h e t a < 150 b o t h c o i l s i d e s a r e c u t t i n g t h r o u g h B−wave e2 =2* B * l * v ; // n e t emf a t g i v e n p o s i t i o n o f c o i l rms = sqrt ((1/ %pi ) *((( e1 ^2* %pi *2) /6) +(( e2 ^2* %pi ) /2) ) ) ; printf ( ’RMS v a l u e o f g e n e r a t e d emf i n one s i n g l e t u r n c o i l i s %f V ’ , rms ) ;

Scilab code Exa 3.16 Calculating rms value of fundamental emf per phase 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

clc ; f =50; // f r e q u e n c y o f a l t e r n a t o r B =1; // peak f l u x d e n s i t y t =360; // t o t a l t u r n s v =60; // p h a s e s p r e a d pi =0.6; // p o l e p i t c h l =0.8; // s t a t o r l e n g t h cs =180; // c o i l s p a n i n e l e c t r i c a l d e g r e e s nph = t /3; // s e r i e s t u r n p e r p h a s e Bp =(4* B * cosd ( v /2) ) / %pi ; // f u n d a m e n t a l v a l u e o f peak flux density F =(2* l * pi * Bp ) / %pi ; // Fundamental a i r −gap f l u x p e r pole kd = sind ( v /2) /(( v /2) *( %pi /180) ) ; // d i s t r i b u t i o n factor kw = kd ; // w i n d i n g f a c t o r , a s kp=1 eph = sqrt (2) * %pi * f * F * kw * nph ; printf ( ’RMS v a l u e o f f u n d a m e n t a l emf p e r p h a s e i s %f V ’ , eph ) ;

122


Scilab code Exa 3.18 Determining resultant emf for given combination of series coil 1 2 3 4 5 6 7 8 9 10 11 12

13 14 15 16 17 18 19 20 21

clc ; p =6; // number o f p o l e s s =54; // number o f s l o t s n =1000; // s p e e d o f a l t e r n a t o r i n rpm t =80; // number o f t u r n i n c o i l s A and B f =0.015; // f l u x p e r p o l e F =50; // g i v e n f r e q u e n c y o f a l t e r n a t o r // C o i l A i s o v e r p i t c h e d by one s l o t and c o i l B i s s h o r t p i t c h e d by one s l o t pp = s / p ; // p o l e p i t c h sap =( p *180) / s ; // s l o t a n g u l a r p i t c h e1 =( %pi * F * f * t ) / sqrt (2) ; // EMF g e n e r a t e d i n one c o i l s i d e of c o i l A or B // same EMF i s g e n e r a t e d i n c o l s i d e 11 but w i t h a p h a s e o f (180+ s a p ) d e g r e e s . R e s u l t a n t o f emf i n c o i l s i d e 1 and 11 i s g i v e n by Ea =2* e1 * cosd ( sap /2) ; // n e t emf i n c o i l s i d e 1 Eb = Ea ; // n e t emf i n c o i l s i d e 2 // Ea and Eb a r e i n p h a s e w i t h e a c h o t h e r from p h a s o r diagram ( f i g . 3 . 2 6 ) disp ( ’ c a s e a ’ ) ; en = Ea + Eb ; printf ( ’ R e s u l t a n t e .m. f when c o i l s A and B a r e c o n n e c t e d i n s e r i e s a i d i n g i s %f V\n ’ , en ) ; disp ( ’ c a s e b ’ ) ; en = Ea - Eb ; printf ( ’ R e s u l t a n t e .m. f when c o i l s A and B a r e c o n n e c t e d i n s e r i e s o p p o s i n g i s %f V\n ’ , en ) ;

123


Scilab code Exa 3.19 Determining maximum and rms value of peak of fundamental mmf wave 1 2 3 4 5 6 7 8 9 10 11 12 13 14

clc ; np =3; // number o f p h a s e s p =2; // number o f p o l e s spp =5; // s l o t s p e r p o l e p e r p h a s e n =4; // number o f t u r n s i n c o i l i =20; // p e r p h a s e c u r r e n t v =( spp *180) /( spp * np ) ; // p h a s e s p r e a d imax = sqrt (2) * i ; // maximum v a l u e o f c u r r e n t mmf = spp * n * imax ; // r e s u l t a n t a m p l i t u d e o f mmf kd = sind ( v /2) /(( v /2) *( %pi /180) ) ; // d i s t r i b u t i o n factor fp =(4* mmf * kd ) / %pi ; // peak v a l u e o f f u n d a m e n t a l component fr =(4*3* spp * n * i ) / %pi ^2; // rms v a l u e o f f u n d a m e n t a l component printf ( ’ Maximum v a l u e o f t h e peak o f f u n d a m e n t a l m.m . f wave i s %f AT/ p o l e \n ’ , fp ) ; printf ( ’RMS v a l u e o f t h e peak o f f u n d a m e n t a l m.m. f wave i s %f AT/ p o l e \n ’ , fr ) ;

Scilab code Exa 3.20 Calculating peak amplitude of mmf wave peak and rms value of fundamental mmf wave 1 2 3 4 5 6 7 8

clc ; p =2; // number o f p o l e i =24; // p h a s e c u r r e n t t =300; // f u l l p i t c h e d t u r n s v =60; // p h a s e s p r e a d np =3; // number o f p h a s e s nph = t / np ; // s e r i e s t u r n p e r p h a s e j =( nph * sqrt (2) * i *180) /( v * %pi ) ; // peak v a l u e o f uniform current density 124


9 disp ( ’ c a s e a ’ ) ; 10 A =( j * v * %pi ) /(2*180) ; // peak a m p l i t u d e o f 11 12 13 14 15 16 17

t r a p e z o i d a l m.m. f wave printf ( ’ Peak a m p l i t u d e o f t r a p e z o i d a l m.m. f wave i s %f ATs/ p o l e \n ’ ,A ) ; disp ( ’ c a s e b ’ ) ; kd = sind ( v /2) /(( v /2) *( %pi /180) ) ; // d i s t r i b u t i o n factor fp =(4* kd * A ) / %pi ; printf ( ’ Peak v a l u e o f f u n d a m e n t a l mmf wave i s %f AT/ p o l e \n ’ , fp ) ; fr =(4*3* A ) /( %pi ^2* sqrt (2) ) ; printf ( ’RMS v a l u e o f f u n d a m e n t a l mmf wave i s %f AT/ p o l e \n ’ , fr ) ;

Scilab code Exa 3.30 Finding linear velocity of travelling mmf wave 1 2 3 4 5

clc ; p =6; // number o f p o l e s i n i n d u c t i o n motor f =50; // f r e q u e n c y o f motor d =1.2; // s t a t o r b o r e d i a m e t e r // i n one r e v o l u t i o n p e r i p h e r a l d i s t a n c e o f Pi ∗ diameter i s transversed 6 v =(2* f * %pi * d ) / p ; 7 printf ( ’ L i n e a r v e l o c i t y o f t r a v e l l i n g mmf wave i s %f m/ s e c ’ ,v ) ;

Scilab code Exa 3.32 Calculating resultant peak gap mmf peak gap flux density total gap energy electromagnetic torque and electromagnetic power 1 clc ; 2 p =2; // number o f p o l e s 3 f =50; // f r e q u e n c y o f machine

125


4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

D =1.6; // d i a m e t e r o f c y l i n d r i c a l r o t o r l =1.8; // l e n g t h o f c y l i n d r i c a l r o t o r g =0.012; // a i r gap l e n g t h rm =4000; // peak v a l u e o f r o t o r mmf rs =6000; // peak v a l u e o f s t a t o r mmf ph =140; // p h a s e d i f f e r e n c e b e t w e e n s t a t o r mmf and r o t o r mmf uo =4* %pi *10^ -7; // f r e e s p a c e p e r m e a b i l i t y 0 disp ( ’ a ’ ) ; rp = sqrt ( rm ^2+ rs ^2+2* rm * rs * cosd ( ph ) ) ; printf ( ’ R e s u l t a n t peak gap mmf i s %f AT/ p o l e \n ’ , rp ) ; disp ( ’ b ’ ) ; Bp =( uo * rp ) / g ; printf ( ’ Peak gap f l u x d e n s i t y i s %f T\n ’ , Bp ) ; disp ( ’ c ’ ) ; ge =( uo * %pi * D * l * rp ^2) /(4* g ) ; printf ( ’ T o t a l gap e n e r g y i s %f J o u l e s \n ’ , ge ) ; disp ( ’ d ’ ) ; T =( p * uo * %pi * D * l * rs * rm * sind ( ph ) ) /(4* g ) ; printf ( ’ E l e c t r o m a g n e t i c t o r q u e i s %f Nm\n ’ ,T ) ; disp ( ’ e ’ ) ; wm =(4* %pi * f ) /2; // s y n c h r o n o u s s p e e d P =( T * wm ) /1000 printf ( ’ E l e c t r o m a g n e t i c power i s %f KW’ ,P ) ;

Scilab code Exa 3.33 Determining torque for given pole fields 1 2 3 4 5 6 7 8

clc ; d =0.8; // d i a m e t e r o f r o t o r machine l =0.5; // l e n g t h o f r o t o r machine g =0.005; // a i r gap l e n g t h as =10000; // peak c u r r e n t d e n s i t y f o r s t a t o r ar =6000; // peak c u r r e n t d e n s i t y f o r r o t o r t =60; // t o r q u e a n g l e disp ( ’ c a s e a ’ ) ; 126


9 10 11 12 13 14 15 16 17 18 19 20 21

p =2; // number o f p o l e uo =4* %pi *10^ -7; // f r e e s p a c e p e r m e a b i l i t y Fs =( as * d ) / p ; // peak s t a t o r mmf p e r p o l e Fr =( ar * d ) / p ; // peak r o t o r mmf p e r p o l e Te =( p * uo * %pi * d * l * Fs * Fr * sind ( t ) ) /(4* g ) ; printf ( ’ Torque f o r g i v e n number o f p o l e s i s %f Nm\n ’ , Te ) ; disp ( ’ c a s e b ’ ) ; p =6; // number o f p o l e uo =4* %pi *10^ -7; // f r e e s p a c e p e r m e a b i l i t y Fs =( as * d ) / p ; // peak s t a t o r mmf p e r p o l e Fr =( ar * d ) / p ; // peak r o t o r mmf p e r p o l e Te =( p * uo * %pi * d * l * Fs * Fr * sind ( t ) ) /(4* g ) ; printf ( ’ Torque f o r g i v e n number o f p o l e s i s %f Nm\n ’ , Te ) ;

Scilab code Exa 3.35 Finding peak value of fundamental mmf peak value of fundamental air gap flux density wave fundamental value of flux per pole and rms value of phase and line emfs at no load and rated speed 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

clc ; p =4; // number o f p o l e s np =3; // number o f p h a s e s f =50; // f r e q u e n c y o f a l t e r n a t o r sap =8; // s l o t a n g u l a r p i t c h c =12; // number o f c o n c e n t r i c c o i l s i n f i e l d w i n d i n g tf =6; // t u r n s p e r f i e l d c o i l ta =28; // s e r i e s a r m a t u r e t u r n p e r p h a s e ar =0.6; // a r m a t u r e r a d i u s la =4; // a r m a t u r e l e n g t h g =0.06; // gap l e n g t h w =0.96; // w i n d i n g f a c t o r f o r a r m a t u r e w i n d i n g fc =1000; // f i e l d c u r r e n t disp ( ’ c a s e a ’ ) ; kd = sind (( np * sap ) /2) /( np * sind ( sap /2) ) ; // 127


16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32

distribution factor kp =1; // c o i l s p a n f a c t o r kf = kd * kp ; // w i n d i n g f a c t o r f o r f i e l d w i n d i n g nf = tf * c ; // number o f f i e l d t u r n F =(4* kf * nf * fc ) /( %pi * p ) ; printf ( ’ Peak v a l u e o f f u n d a m e n t a l mmf p r o d u c e d by f i e l d w i n d i n g i s %f AT/ p o l e \n ’ ,F ) ; disp ( ’ c a s e b ’ ) ; uo =4* %pi *10^ -7; // f r e e s p a c e p e r m e a b i l i t y B =( uo * F ) / g ; printf ( ’ Peak v a l u e o f f u n d a m e n t a l f l u x d e n s i t y wave i s %f T\n ’ ,B ) ; disp ( ’ c a s e c ’ ) ; v =(4* B * la * ar ) / p ; printf ( ’ Fundamental v a l u e o f a i r gap f l u x p e r p o l e i s %f W\n ’ ,v ) ; disp ( ’ c a s e d ’ ) ; eph = sqrt (2) * %pi * f * v * ta * w ; printf ( ’EMF p e r p h a s e i s %f V\n ’ , eph ) ; el = sqrt (3) * round ( eph ) ; printf ( ’ L i n e EMF i s %f V ’ , el ) ;

Scilab code Exa 3.36 Calculating 1 peak value of fundamental air gap flux density 2 peak value of fundamental mmf wave 3 peak value of armature mmf wave and resultant mmf wave per pole 4 rms value of armature current and its power factor 1 2 3 4 5 6 7 8

clc ; np =3; // number o f p h a s e s p =6; // number o f p o l e s f =50; // f r e q u e n c y o f a l t e r n a t o r e =415; // open c i r c u i t emf ; s =36; // number o f s l o t s i n a r m a t u r e t =4; // number o f t u r n s p e r c o i l g =0.18; // a i r gap d i a m e t e r 128


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

l =0.4; // c o r e l e n g t h G =0.002; // gap l e n g t h T =42; // number o f t u r n s i n f i e l d w i n d i n g kf =0.96; // w i n d i n g f a c t o r uo =4* %pi *10^ -7; // f r e e s p a c e p e r m e a b i l i t y disp ( ’ c a s e a ’ ) ; nph =( s * t ) / np ; // s e r i e s t u r n p e r p h a s e spp = s /( p * np ) ; // s l o t s p e r p o l e p e r p h a s e v =180/ p ; // s l o t a n g u l a r p i t c h kd = sind (( spp * v ) /2) /( spp * sind ( v /2) ) ; // d i s t r i b u t i o n factor Flu = e /( sqrt (2) * sqrt (3) * %pi * f * nph * kd ) ; // f l u x p e r pole B =( p * Flu *2) /(4* l * g ) ; printf ( ’ Peak v a l u e o f f u n d a m e n t a l f l u x d e n s i t y wave i s %f T\n ’ ,B ) ; disp ( ’ c a s e b ’ ) ; Fl =( G * B ) / uo ; // peak f u n d a m e n t a l f i e l d mmf wave printf ( ’ Peak v a l u e o f f u n d a m e n t a l mmf wave i s %f AT/ p o l e \n ’ , Fl ) ; If =( %pi * Fl * p ) /(4* kf * T ) ; printf ( ’DC f i e l d c u r r e n t i s %f A\n ’ , If ) ; disp ( ’ c a s e c ’ ) ; Te =114; // g i v e n t o r q u e Ta =146; // t o r q u e a n g l e Fm = floor (( Te *4* G ) /( p * uo * %pi * g * l * Fl * sind ( Ta ) ) ) ; printf ( ’ Peak v a l u e o f f u n d a m e n t a l a r m a t u r e mmf i s %f AT/ p o l e \n ’ , Fm ) ; Fr = sqrt ( Fl ^2+ Fm ^2+2* Fl * Fm * cosd ( Ta ) ) ; printf ( ’ R e s u l t a n t mmf p e r p o l e i s %f AT/ p o l e \n ’ , Fr ) ; disp ( ’ c a s e d ’ ) ia =( Fm *2* %pi * p ) /(12* kd * nph * sqrt (2) ) ; printf ( ’RMS v a l u e o f a r m a t u r e c u r r e n t i s %f A\n ’ , ia ) ; ns =1000; // s p e e d i n rpm wm =(2* %pi * ns ) /60; // a n g u l a r s p e e d i n r p s pf =( Te * wm ) /( sqrt (3) * e * ia ) ; printf ( ’ Power f a c t o r i s %f l a g g i n g ’ , pf ) ; 129


Scilab code Exa 3.37 Determining full load efficiency of given transformer 1 2 3 4

clc ; n1 =0.95; // e f f i c i e n c y o f t r a n s f o r m e r 1 lo =((1/ n1 ) -1) ; // f r a c t i o n o f o u t p u t l o s t d =2; // L i n e a r d i m e n s i o n o f t r a n s f o r m e r B i s two times the Linear dimension of transformer A 5 nb =(1/(1+((1* lo ) / d ) ) ) *100; 6 printf ( ’ F u l l l o a d e f f i c i e n c y o f t r a n s f o r m e r B i s %f p e r c e n t ’ , nb ) ;

Scilab code Exa 3.39 Determining KW rating of motor 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

clc ; t0 =0; // a c c e l e r a t i n g p e r i o d t1 =30; // d e c e l e r a t i n g p e r i o d l1 =2000; // maximum l o a d d u r i n g a c c e l e r a t i n g p e r i o d lf =600; // maximum l o a d d u r i n g d e c e l e r a t i n g p e r i o d l =1000; // l o a d d u r i n g f u l l l o a d tf =60; // f u l l l o a d d u r a t i o n td =10; // d e c e l e r a t i n g d u r a t i o n tde =20; // d e c t i n g p e r i o d sa = l1 / t1 ; // s l o p e d u r i n g a c c e l e r a t i n g sd = lf / td ; // s l o p e d u r i n g d e c e l e r a t i n g e1 = integrate ( ’ ( s a ∗ t ) ˆ2 ’ , ’ t ’ ,t0 , t1 ) ; // term 1 f o r f i n d i n g motor r a t i n g e2 = l ^2* tf ; // term 2 f o r f i n d i n g motor r a t i n g e3 = integrate ( ’ ( s d ∗ t ) ˆ2 ’ , ’ t ’ ,t0 , td ) ; // term 3 f o r f i n d i n g motor r a t i n g T = t1 + tf + td + tde ; // t o t a l d u r a t i o n R = sqrt ((1/120) *( e1 + e2 + e3 ) ) ; 130


17 18

printf ( ’KW r a t i n g o f motor i s %f KW’ ,R ) ; disp ( ’ Choose a motor o f r a t i n g a b o v e t h e c a l c u l a t e d r a t i n g ’ );

Scilab code Exa 3.40 Determining continuous KW rating of motor 1 clc ; 2 T =80; // t o t a l d u r a t i o n 3 t1 =5 -0; // d u r a t i o n o f f i r s t 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

increasing loading

period t2 =36 -5; // d u r a t i o n o f s e c o n d i n c r e a s i n g l o a d i n g period t3 =39 -36; // d u r a t i o n o f f i r s t d e c r e a s i n g l o a d i n g period t4 =55 -39; // d u r a t i o n o f s e c o n d d e c r e a s i n g l o a d i n g period t5 =80 -55; // d u r a t i o n o f u n i f o r m l o a d i n g l1 =150; // i n i t i a l l o a d l2 =1000; // l o a d a t 5 t h s e c l3 =1400; // l o a d a t 36 t h s e c l4 =300; // l o a d a t 39 t h s e c l5 =150; // l o a d d u r i n g u n i f o r m l o a d i n g T1 =( t1 /3) *( l1 ^2+ l2 ^2+ l1 * l2 ) ; // term 1 f o r e v a l u a t i n g rms power T2 =( t2 /3) *( l2 ^2+ l3 ^2+ l2 * l3 ) ; // term 2 f o r e v a l u a t i n g rms power T3 =( t3 /3) *( l3 ^2+ l4 ^2+ l3 * l4 ) ; // term 3 f o r e v a l u a t i n g rms power T4 =( t4 /3) *( l4 ^2+ l5 ^2+ l4 * l5 ) ; // term 4 f o r e v a l u a t i n g rms power T5 = t5 * l5 ^2; // term 5 f o r e v a l u a t i n g rms power R = sqrt ((1/ T ) *( T1 + T2 + T3 + T4 + T5 ) ) ; printf ( ’ As p e r t h e l o a d t i m e g r a p h r a t i n g i s %f KW’ , R); disp ( ’ Choose a motor o f r a t i n g a b o v e t h e c a l c u l a t e d 131


r a t i n g ’ );

Scilab code Exa 3.41 Determining final steady temperature rise and new KVA rating of transformer 1 2 3 4 5 6 7 8 9 10 11

12 13 14 15 16 17 18 19 20

clc ; p =200; // r a t e d KVA o f t r a n s f o r m e r n =0.98; // e f f i c i e n c y t1 =20; // t e m p e r a t u r e a f t e r one h o u r o f o p e r a t i o n t2 =34; // t e m p e r a t u r e a f t e r two h o u r o f o p e r a t i o n r =1/3; // r a t i o o f f u l l l o a d c o r e l o s s e s t o ohmic loss disp ( ’ c a s e a ’ ) ; t =[( t2 / t1 ) -1]; th = -1/ log ( t ) ; // h e a t i n g t i m e c o n s t a n t i n h o u r s theta = t1 /(1 - exp ( -1/ th ) ) ; printf ( ’ F i n a l s t e a d y t e m p e r a t u r e r i s e o f t h e t r a n s f o r m e r on r a t e d l o a d i s %f d e g r e e c e l s i u s \n ’ , theta ) ; disp ( ’ c a s e b ’ ) ; f =1.2; // w i t h i n c r e a s e d h e a t d i s s i p a t i o n , r a t i o o f new l o s s t o o l d l o s s Pn = sqrt (( f *(1+ r ) ) -r ) * p ; printf ( ’ New KVA r a t i n g o f t r a n s f o r m e r i s %f KVA\n ’ , Pn ) ; // f o r a t e m p e r a t u r e r i s e o f 78 d e g r e e t3 =78; f =( t3 / theta ) * f ; // r a t i o o f new l o s s t o o l d l o s s Pn = sqrt (( f *(1+ r ) ) -r ) * p ; printf ( ’ New KVA r a t i n g o f t r a n s f o r m e r i s %f KVA\n ’ , Pn ) ;

Scilab code Exa 3.42 Determining one hour rating of induction motor 132


1 2 3 4 5 6 7 8 9 10 11 12 13 14

clc ; p =100; // KW r a t i n g o f t r a n s f o r m e r al =1; // r a t i o o f c o r e l o s s t o ohmic l o s s th =3; // h e a t i n g t i m e c o n s t a n t i n h o u r s h =1; // d u r a t i o n i n h o u r f o r which KVA r a t i n g h a s t o be d e t e r m i n e d disp ( ’ c a s e a ’ ) ; // c o n s t a n t l o s s e s a r e e q u a l t o v a r i a b l e l o s s e s pn = p * sqrt (((1+ al ) /(1 - exp ( - h / th ) ) ) - al ) ; printf ( ’ One h o u r r a t i n g i s %f KW\n ’ , pn ) ; disp ( ’ c a s e b ’ ) ; // c o n s a t a n t l o s s e s a r e n e g l e c t e d al =0; // r a t i o o f c o r e l o s s t o ohmic l o s s pn = p * sqrt (((1+ al ) /(1 - exp ( - h / th ) ) ) - al ) ; printf ( ’ One h o u r r a t i n g i s %f KW\n ’ , pn ) ;

Scilab code Exa 3.43 Calculating ratio of core loss to ohmic loss 1 clc ; 2 t =1/2; // r a t i o

o f c o n t i n u o u s r a t i n g t o one h o u r

rating 3 p =2; // r a t i o o f new KVA r a t i n g t o o l d KVA r a t i n g 4 al =2*( p * t ) ; 5 printf ( ’ R a t i o o f c o r e l o s s t o ohmic l o s s i s %f ’ , al )

;

133


Chapter 4 DC machines

Scilab code Exa 4.2 Determining electromagnetic power and internal torque 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

clc ; l =0.3; // c o r e l e n g t h r =0.2; // r a d i u s n =20; // s p e e d i n r . p . s . Ia =20; // a r m a t u r e c u r r e n t Z =500; // t o t a l c o n d u c t o r s Bav =0.5; // avg . f l u x d e n s i t y a =4; // l a p −wound P =4; // no o f p o l e s Wm =2* %pi * n phi =((0.5*2* %pi *0.2*0.3) /4) ; Ea =(( P * n * Z * phi ) / a ) ; // g e n e r a t e d emf Pm = Ea * Ia ; // g r o s s m e c h a n i c a l power d e v e l o p e d Te =(( Ea * Ia ) / Wm ) ; // i n t e r n a l t o r q u e printf ( ’ G e n e r a t e d E .M. F . i s %f V. \ n ’ , Ea ) ; printf ( ’ G r o s s m e c h a n i c a l power d e v e l o p e d i s %f W. \ n ’ , Pm ) ; 17 printf ( ’ I n t e r n a l Torque i s %f Nm. ’ , Te ) ;

134


Scilab code Exa 4.3 Determining total current emf power developed in armature and electromagnetic torque 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27

clc ; P =6; // no o f p o l e s Z =300; // no o f c o n d u c t o r s phi =0.015; // f l u x p e r p o l e i n w e b e r s n =30; // s p e e d i n r . p . s . Ic =80; // c u r r e n t p e r c o n d u c t o r Wm =2* %pi * n ; Eav = P * n * phi ; // avg . emf p e r c o n d u c t o r // when c o n d u c t o r s a r e wave c o n n e c t e d disp ( ’ Wave C o n n e c t e d ’ ) a1 =2; // no o f p a r a l l e l p a t h s Ia = Ic * a1 ; // t o t a l c u r r e n t Ea = Eav *( Z / a1 ) ; //E .M. F . Pa = Ea * Ia ; // power d e v e l o p e d i n a r m a t u r e Te = Ea * Ia / Wm ; // E l e c t r o m a g n e t i c t o r q u e printf ( ’ G e n e r a t e d E .M. F . i s %f V. \ n ’ , Ea ) ; printf ( ’ Power d e v e l o p e d i n a r m a t u r e i s %f W. \ n ’ , Pa ) ; printf ( ’ E l e c t r o m a g n e t i c Torque i s %f Nm. \ n ’ , Te ) ; // when c o n d u c t o r s a r e l a p c o n n e c t e d disp ( ’ Lap C o n n e c t e d ’ ) a2 =4; // no o f p a r a l l e l p a t h s Ia2 = Ic * a2 ; // t o t a l c u r r e n t Ea2 = Eav *( Z / a2 ) ; //E .M. F . Pa2 = Ea2 * Ia2 ; // power d e v e l o p e d i n a r m a t u r e Te2 = Ea2 * Ia2 / Wm ; // E l e c t r o m a g n e t i c t o r q u e printf ( ’ G e n e r a t e d E .M. F . i s %f V. \ n ’ , Ea2 ) ; printf ( ’ Power d e v e l o p e d i n a r m a t u r e i s %f W. \ n ’ , Pa2 ) ; 28 printf ( ’ E l e c t r o m a g n e t i c Torque i s %f Nm. ’ , Te2 ) ;

Scilab code Exa 4.4 Determining 1 generated emf at no load 2 terminal voltage at full load 135


1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

clc ; p =6; // number o f p o l e s c =240; // number o f c o i l s t =2; // number o f t u r n s p e r c o i l rt =0.03; // r e s i s t a n c e o f one t u r n l =0.5; // l e n g t h o f a r m a t u r e d =0.4; // d i a m e t e r o f a r m a t u r e B =0.6; // a i r gap f l u x d e n s i t y a = p ; // number o f p a r a l l e l p a t h s i s same a s number of p o l e s f o e lap winding an =40; // m e c h a n i c a l a n g l e s u b t e n d e d by p o l e n =1200; // a r m a t u r e s p e e d th = an *( p /2) ; // e l e c t r i c a l a n g l e s u b t e n d e d by p o l e f =(2* %pi *( d /2) * l * th * B ) /( p *180) ; // f l u x p e r p o l e Z =2* c * t ; // t o t a l c o n d u c t o r s disp ( ’ c a s e a ’ ) ; Ea =( f * Z * n * p ) /(60* a ) ; printf ( ’ G e n e r a t e d EMF a t no l o a d i s %f V\n ’ , ceil ( Ea ) ); disp ( ’ c a s e b ’ ) ; ia =40; // a r m a t u r e c u r r e n t at =( c * t ) / a ; // number o f a r m a t u r e t u r n s p e r p a t h r = at * rt ; // r e s i s t a n c e o f one p a t h Ra = r / a ; // r e s i s t a n c e o f a r m a t u r e c i r c u i t vt = Ea - ia * Ra ; printf ( ’ 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 %f V\n ’ , ceil ( vt ) ) ;

Scilab code Exa 4.5 Determining terminal voltage of generator 1 2 3 4 5

clc Pout =24000; // r a t e d o u t p u t power i n w a t t s Et =250; // r a t e d t e r m i n a l v o l t a g e Ra =0.1; // a r m a t u r e r e s i s t a n c e N =1600; // s p e e d i n rpm 136


6 7 8 9 10 11 12 13 14

// Ea ( t e r m i n a l v o l t a g e )= k ∗ (N∗ p h i ) , where k i s constant & phi i s f l u x per pole // At no l o a d , 260=k ∗ 1 6 0 0 ∗ p h i . . . . ( 1 ) Ia = Pout / Et ; // i f t h e g e n e r a t e d v o l t a g e u n d e r r a t e d l o a d i s Ea1 , then // Ea=k ∗ 1 5 0 0 ∗ p h i . . . . ( 2 ) // From e q u a t i o n ( 1 ) &(2) , ( Ea1 / 2 6 0 ) = ( ( 1 5 0 0 ∗ p h i ) /(1600∗ phi ) ) Ea1 =(1500*260) /1600; Vt = Ea1 - Ia * Ra // t e r m i n a l v o l t a g e a t r a t e d l o a d printf ( ’ The t e r m i n a l v o l t a g e o f g e n e r a t o r u n d e r g i v e n c o n d i t i o n s i s %f V . ’ , Vt )

Scilab code Exa 4.6 Determining ratio of speed as a generator to speed as a motor 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

clc ; Vt =230; // t e r m i n a l v o l t a g e o f dc s h u n t machine Il =40; // L i n e c u r r e n t Ra =0.5; // Armature c i r c u i t r e s i s t a n c e Rf =115; // F i e l d c i r c u i t r e s i s t a n c e //GENERATOR OPERATION If = Vt / Rf ; // f i e l d c u r r e n t Ia1 = If + Il ; // a r m a t u r e c u r r e n t Ea1 = Vt + Ia1 * Ra ; // g e n e r a t e d emf // Ea1=k ∗Ng∗ p h i . . . . ( 1 ) , where Ng i s t h e g e n e r a t o r speed & phi i s f l u x per pole p r o p o r t i o n a l to I f //MOTOR OPERATION Ia2 = Il - If ; // a r m a t u r e c u r r e n t Ea2 = Vt - Ia2 * Ra ; // g e n e r a t e d emf // Ea2=k ∗Nm∗ p h i . . . . ( 2 ) , where Nm i s t h e motor s p e e d & phi i s f l u x per pole p r o p o r t i o n a l to I f // From e q u a t i o n ( 1 ) &(2) , ( Ea1 / Ea2 ) =((Ng∗ p h i ) / (Nm∗ p h i )) 137


16 N = Ea1 / Ea2 ; // r a t i o 17

o f s p e e d o f g e n e r a t o r t o motor , N=

Ng/Nm printf ( ’ The r a t i o o f s p e e d a s a g e n e r a t o r t o t h e s p e e d a s a motor i s %f . ’ ,N )

Scilab code Exa 4.9 Determining demagnetizing and cross magnetizing ampere turns per pole for different positions of brushes 1 2 3 4 5 6 7 8 9 10 11

12 13 14 15 16

17

clc ; A =2; //No o f p a r a l l e l p a t h s f o r a r m a t u r e c o n d u c t o r s P =6; //No . o f p o l e s If =2; // F i e l d c u r r e n t Il =148; // L i n e c u r r e n t Ia = If + Il ; // Armature c u r r e n t Z =480; //No o f c o n d u c t o r s // b r u s h e s on GNA, t h e t a =0 ATd1 =0 // d e m a g n e t i z i n g ampere t u r n s ATc1 =(( Ia * Z ) /(2* A * P ) ) // C r o s s m a g n e t i z i n g ampere turns printf ( ’ When b r u s h e s a r e on GNA t h e d e m a g n e t i z i n g ampere t u r n s & C r o s s m a g n e t i z i n g ampere t u r n s a r e e q u a l t o %f & %f ATs/ p o l e r e s p e c t i v e l y . \ n ’ , ATd1 , ATc1 ) ; // b r u s h e s a r e s h i f t e d from GNA by 5 d e g r e e s e l e c t r i c a l , t h e t a =5 theta =5; ATd2 =((2* theta * Ia * Z ) /(180*2* A * P ) ) // d e m a g n e t i z i n g ampere t u r n s ATc2 =3000 - ATd2 ; // C r o s s m a g n e t i z i n g ampere t u r n s printf ( ’ When t h e b r u s h e s a r e s h i f t e d from GNA by 5 d e g r e e s e l e c t r i c a l t h e d e m a g n e t i z i n g ampere t u r n s & C r o s s m a g n e t i z i n g ampere t u r n s a r e e q u a l t o %f & %f ATs/ p o l e r e s p e c t i v e l y . \ n ’ , ATd2 , ATc2 ) ; // b r u s h e s a r e s h i f t e d from GNA by 5 d e g r e e s m e c h a n i c a l , t h e t a m =5 138


theta_m =5; // m e c h a n i c a l a n g l e theta_e =( P /2) * theta_m ; // e l e c t r i c a l a n g l e ATd3 =((2* theta_e * Ia * Z ) /(180*2* A * P ) ) // d e m a g n e t i z i n g ampere t u r n s 21 ATc3 =3000 - ATd3 ; // C r o s s m a g n e t i z i n g ampere t u r n s 22 printf ( ’ When t h e b r u s h e s a r e s h i f t e d from GNA by 5 d e g r e e s m e c h n i c a l t h e d e m a g n e t i z i n g ampere t u r n s & C r o s s m a g n e t i z i n g ampere t u r n s a r e e q u a l t o %f & %f ATs/ p o l e r e s p e c t i v e l y ’ , ATd3 , ATc3 ) ; 18 19 20

Scilab code Exa 4.11 Determining number of turns required on each pole 1 2 3 4 5 6 7 8 9 10 11

clc ; P =4; //No o f p o l e s Pout =100000; // Output power i n w a t t s Vt =200; // t e r m i n a l v o l t a g e Z =256; //No o f c o n d u c t o r s A =4; // no o f p a r a l l e l p a t h s o f a r m a t u r e c o n d u c t o r s Ia = Pout / Vt ; // a r m a t u r e c u r r e n t B =0.25; // i n t e r p o l a r f l u x d e n s i t y i n t e s l a g =0.01; // i n t e r p o l a r a i r gap l e n g t h U =4* %pi *0.0000001; // p e r m e a b i l i t y o f a i r F =(( Ia * Z ) /(2* A * P ) ) +(( B / U ) *( g ) ) ; // The i n t e r p o l a r m.m. f . p e r p o l e 12 N = F / Ia ; 13 printf ( ’ The t u r n s on e a c h i n t e r p o l e s s h o u l d be e q u a l t o %f . ’ , round ( N ) ) ;

Scilab code Exa 4.12 Determining time of commutation 1 clc ; 2 D =50; // d i a m e t e r o f commutator 3 N =1000; // s p e e d o f r o t a t i o n o f commutator i n rpm

139


4 Wb =1.5; // b r u s h w i d t h 5 V = %pi * D * N /60; // p e r i p h e r a l v e l o c i t y o f commutator 6 Tc =( Wb *1000) / V ; // t i m e o f commutation i n ms 7 printf ( ’ Time o f commutation i s %f ms . ’ , Tc ) ;

Scilab code Exa 4.13 Determining value of commutating field 1

// The a n s w e r g i v e n i n book f o r t h i s q u e s t i o n i s wrong .

2 3 4 5 6 7 8 9 10 11 12 13

clc ; P =4; //No o f p o l e s Ia =120; // a r m a t u r e c u r r e n t A =4; //No o f p a r a l l e l p a t h s f o r a r m a t u r e c o n d u c t o r L =0.02 // i n d u c t a n c e i n mH // Et=L ∗ ( d i / d t ) , T r a n s f o r m e r emf i n c o i l // d i =2∗ I a /A, c h a n g e o f c u r r e n t d u r i n g commutation // d t=Tc , t i m e o f commutation // Et = 0 . 0 2 ∗ 0 . 0 0 1 ∗ ( 6 0 / Tc ) . . . . ( 1 ) // Er =2∗( Bav∗ l ∗ v ) , r o t a t i o n a l emf i n s i n g l e t u r n c o i l // Er =2∗( p h i c /Tc ) . . . . ( 2 ) , p h i c i s t h e avg v a l u e o f f l u x i n t h e commutating z o n e 14 // For l i n e a r commutation , Er=Et , from e q u a t i o n ( 1 ) &(2) 15 phi_c =60*0.02*0.001/2; // p h i c i s t h e avg v a l u e o f f l u x i n t h e commutating z o n e 16 printf ( ’THE AVG. VALUE OF FLUX IN THE COMMUTATING ZONE I S %f Wb. ’ , phi_c )

Scilab code Exa 4.14 Determining number of pole face conductors of compensating winding in each pole 1 clc ;

140


2 3 4 5 6 7 8 9 10 11 12 13

Pout =2000000; // o u t p u t power i n w a t t s Vt =400; // o u t p u t v o l t a g e P =14; //No o f p o l e s A =14; //No o f p a r a l l e l p a t h s o f c o n d u c t o r Pr =0.7; // p o l e a r c t o p o l e p i t c h r a t i o Z =1100; // t o t a l a r m a t u r e c o n d u c t o r s Ia = Pout / Vt ; // a r m a t u r e c u r r e n t A_z =( Ia * Z ) /( A * P ) ; // a r m a t u r e ampere c o n d u c t o r s p e r pole A_z1 = Pr * A_z ; // a r m a t u r e ampere c o n d u c t o r s p e r p o l e t o be c o m p e n s a t e d by p o l e f a c e w i n d i n g // The c o m p e n s a t i n g w i n d i n g c a r r i e s t h e e n t i r e armature c u r r e n t o f 5000 A. Wc = A_z1 / Ia ; // c o m p e n s a t i n g w i n d i n g c o n d u c t o r s p e r pole printf ( ’ T o t a l c o m p e n s a t i n g w i n d i n g c o n d u c t o r s p e r p o l e a r e %f . ’ , round ( Wc ) ) ;

Scilab code Exa 4.15 Determining 1 compensating winding conductors per pole 2 number of turns on each interpole 1 2 3 4 5 6 7 8

clc ; ATp =15000; // a r m t u r e ampere t u r n s p e r p o l e Pr =0.68; // r a t i o o f p o l e a r c t o p o l e p i t c h Ia =850; // r a t e d a r m a t u r e c u r r e n t B =0.25; // i n t e r p o l a r f l u x d e n s i t y i n t e s l a g =0.01; // i n t e r p o l a r a i r gap l e n g t h U =4* %pi *0.0000001; // p e r m e a b i l i t y o f a i r ATc = Pr * ATp ; // c o m p e n s a t i n g w i n d i n g ampere t u r n s p e r pole 9 C =2*( ATc / Ia ) ; // c o m p e n s a t i n g w i n d i n g c o n d u c t o r s p e r pole 10 MMF_ag =( B / U ) * g ; //M.M. F . r e q u i r e d f o r t h e a i r gap under the i n t e r p o l e 11 MMF = MMF_ag + ATp ; // i n t e r p o l e M.M. F . w i t h o u t 141


compensating winding 12 MMF_c = MMF - ATc ; // ampere t u r n s f u r n i s h e d by e a c h interpole 13 N = MMF_c / Ia ; //No . o f t u r n s on e a c h i n t e r p o l e 14 printf ( ’ Number o f t u r n s on e a c h i n t e r p o l e i s %f . ’ , round ( N ) ) ;

Scilab code Exa 4.16 Determining number of series field turns per pole 1 2 3 4 5 6 7 8 9 10

clc ; Pout =10000; // o u t p u t power o f dc g e n e r a t o r i n w a t t s Vt =250; // t e r m i n a l v o l t a g e i n v o l t s If =2; // f i e l d c u r r e n t i n ampere a t no l o a d If1 =2.2; // f i e l d c u r r e n t i n ampere a t r a t e d l o a d Tp =1400; // t u r n s on e a c h p o l e Ia = Pout / Vt ; // a r m a t u r e c u r r e n t MMF_rl = If1 * Tp ; //M.M. F . r e q u i r e d a t r a t e d l o a d MMF_nl = If * Tp ; //M.M. F . r e q u i r e d a t no l o a d MMF_s = MMF_rl - MMF_nl ; //M.M. F . s u p p l i e d by s e r i e s winding 11 Is = Ia ; // s e r i e s c u r r e n t a t f u l l l o a d 12 Ts = MMF_s / Is ; // s e r i e s f i e l d t u r n s 13 printf ( ’ S e r i e s f i e l d t u r n s a r e e q u a l t o %f . ’ , Ts ) ;

Scilab code Exa 4.17 Determining demagnetizing effect of armature reaction at rated load and speed 1 clc ; 2 // t a b l e

i s given in question for plotting magnetising curve 3 if1 =[ 0 0.2 0.4 0.6 1 1.4 1.8 2 ]; 4 Ea =[ 6 40 80 120 194 246 269 274]; 5 plot ( if1 , Ea ) ; 142


xlabel ( ’ I f ’ ) ; ylabel ( ’ Ea ’ ) ; title ( ’ m a g n e t i s i n g c u r v e ’ ) v =230; // r a t e d v o l t a g e o f g e n e r a t o r p =10000; // r a t e d power o f g e n e r a t o r n =1500; // r a t e d s p e e d o f g e n e r a t o r rf =184; // s h u n t f i e l d r e s i s t a n c e ra =0.443; // a r m a t u r e r e s i s t a n c e ifl =1.7; // r a t e d f i e l d c u r r e n t il = p / v ; // f u l l l o a d c u r r e n t printf ( ’ T o t a l a r m a t u r e c u r r e n t i s %f A\n ’ , il + ifl ) ; printf ( ’ Armature r e s i s t a n c e d r o p i s %f ohms \n ’ ,( il + ifl ) * ra ) ; 18 disp ( ’ I n f i g 4 . 1 7 ( t e x t b o o k ) ,AB i s made e q u a l t o armature r e s i s t a n c e drop then through B a h o r i z o n t a l l i n e i s made m e e t i n g c u r v e a t c ’ ) ; 19 disp ( ’ D e m a g n e t i s i n g e f f e c t i s g i v e n by BC which i s e q u a l to 0 . 2 5 A ’ ); 6 7 8 9 10 11 12 13 14 15 16 17

Scilab code Exa 4.18 Determining terminal voltage at given speed 1 2 3 4 5 6 7 8 9 10 11 12 13

clc ; vt1 =50; // t e r m i n a l v o l t a g e rf =100; // r e s i s t a n c e o f f i e l d c i r c u i t n1 =1000; // s p e e d c o r r e s p o n d i n g t o v t 1 =50 vt2 =225; // t e r m i n a l v o l t a g e n2 =2000; // s p e e d c o r r e s p o n d i n g t o v t 2 =225 vt3 =405; // t e r m i n a l v o l t a g e n3 =3000; // s p e e d c o r r e s p o n d i n g t o v t 3 =405 disp ( ’ c a s e a ’ ) ; ifl1 = vt1 / rf ; // f i e l d c u r r e n t f o r n=1000 rpm ifl2 = vt2 / rf ; // f i e l d c u r r e n t f o r n=2000 rpm ifl3 = vt3 / rf ; // f i e l d c u r r e n t f o r n=3000 rpm printf ( ’ F i e l d c u r r e n t f o r s p e e d=%f rpm i s %f A\n ’ ,n1 , ifl1 ) ; 143


14 15 16 17 18 19 20 21 22 23 24 25

printf ( ’ F i e l d c u r r e n t f o r s p e e d=%f rpm i s %f A\n ’ ,n2 , ifl2 ) ; printf ( ’ F i e l d c u r r e n t f o r s p e e d=%f rpm i s %f A\n ’ ,n3 , ifl3 ) ; vt11 = vt1 *( n2 / n1 ) ; printf ( ’ T e r m i n a l v o l t a g e=%f V a t %f rpm i s e q u i v a l e n t t o %f V a t %f rpm\n ’ ,vt1 , n1 , vt11 , n2 ) ; vt33 = vt3 *( n2 / n3 ) ; printf ( ’ T e r m i n a l v o l t a g e=%f V a t %f rpm i s e q u i v a l e n t t o %f V a t %f rpm\n ’ ,vt3 , n3 , vt33 , n2 ) ; disp ( ’ U s i n g a b o v e data , m a g n e t i s i n g c u r v e i s drawn f o r n=2000 rpm ’ ) ; // from f i g 4 . 3 7 disp ( ’ For f i e l d r e s i s t a n c e =80 ohms t e r m i n a l v o l t a g e i s g i v e n by BC ’ ) ; disp ( ’BC=253 , h e n c e t e r m i n a l v o l t a g e c o r r e s p o n d i n g t o f i e l d r e s i s t a n c e o f 80 ohms i s 253 V ’ ) ; disp ( ’ For f i e l d r e s i s t a n c e =70 ohms t e r m i n a l v o l t a g e i s g i v e n by QP ’ ) ; disp ( ’QP=268 , h e n c e t e r m i n a l v o l t a g e c o r r e s p o n d i n g t o f i e l d r e s i s t a n c e o f 70 ohms i s 268 V ’ ) ;

Scilab code Exa 4.19 Determining 1 open circuit voltage 2 critical value of shunt field resistance 3 critical speed 4 open circuit voltage for given field resistance 5 terminal voltage 1 clc ; 2 n =1500; // s p e e d o f g e n e r a t o r 3 // d a t a i s g i v e n i n q u e s t i o n f o r 4 5 6 7 8

magnetising curve a t n=1500 rpm If =[ 0 0.4 0.8 1.2 1.6 2 2.4 2.8 3]; Ea =[6 60 120 172.5 202.5 221 231 237 240]; subplot (221) ; plot ( If , Ea ) ; xlabel ( ’ f i e l d c u r r e n t ’ ) ; 144


9 ylabel ( ’ g e n e r a t e d EMF ’ ) ; 10 title ( ’ M a g n e t i s i n g c u r v e f o r n=1500 ’ ) ; 11 disp ( ’ c a s e a ’ ) 12 rf =100; // f i e l d r e s i s t a n c e 13 // r f =100 l e t s s a y v o l t a g e =240 and f i e l d

14 15 16 17 18 19 20 21 22 23

24 25 26 27 28 29 30

31

c u r r e n t =2.4 which i s shown by p o i n t A, s t r a i g h t l i n e p a s s i n g t h r o u g h A and o r i g i n m e e t s m a g n e t i s i n g c u r r e n t a t B which i s no l o a d v o l t a g e Eo =230; printf ( ’ No l o a d v o l t a g e i s %f V\n ’ , Eo ) ; disp ( ’ c a s e b ’ ) ; // a l i n e OF i s drawn p a s s i n g t h r o u g h o r i g i n s l o p e of this line gives c r i t i c a l resistance vt =180; // t e r m i n a l v o l t a g e ifl =1.2; // f i e l d c u r r e n t c o r r e s p o n d i n g t o t e r m i n a l voltage rfl = vt / ifl ; printf ( ’ C r i t i c a l v a l u e o f s h u n t f i e l d r e s i s t a n c e i s %f ohms \n ’ , rfl ) ; disp ( ’ c a s e c ’ ) ; // Choose S ( any p o i n t ) on l i n e a r p a r t o f m a g n e t i s i n g c u r v e . A v e r t i c a l l i n e from S m e e t s f i e l d r e s i s t a n c e l i n e a t t and h o r i z o n t a l l i n e a t y . Now e1 =90; // t e r m i n a l v o l t a g e c o r r e s p o n d i n g t o p o i n t s e2 =60; // t e r m i n a l v o l t a g e c o r r e s p o n d i n g t o p o i n t t n2 =( e2 / e1 ) * n ; printf ( ’ C r i t i c a l s p e e d f o r g i v e n s h u n t f i e l d r e s i s t a n c e i s %f rpm\n ’ , n2 ) ; disp ( ’ c a s e d ’ ) ; n3 =1200; // s p e e d a t which m a g n e t i s i n g c u r v e i s drawn // d a t a f o r m a g n e t i s i n g c u r v e a t n=1200 can be o b t a i n e d by m u l t i p l y i n g v o l t a g e o f m a g n e t i s i n g c u r v e a t n=1500 by f a c t o r 1 2 0 0 / 1 5 0 0 and a t p o i n t C f i e l d r e s i s t a n c e l i n e f o r 100 ohms meet a t m a g n e t i s i n g c u r v e . T h i s p o i n t g i v e s no l o a d EMF EAn = Ea *( n3 / n ) ; 145


32 33 34 35 36 37 38 39 40 41 42 43

subplot (222) ; plot ( If , EAn ) ; xlabel ( ’ f i e l d c u r r e n t ’ ) ; ylabel ( ’ g e n e r a t e d EMF ’ ) ; title ( ’ M a g n e t i s i n g c u r v e f o r n=1200 ’ ) ; Eo =165; printf ( ’ No l o a d EMF i s %f V\n ’ , Eo ) ; disp ( ’ c a s e e ’ ) ; ia =50; // a r m a t u r e c u r r e n t ra =0.3; // a r m a t u r e r e s i s t a n c e vd = ia * ra ; // a r m a t u r e r e s i s t a n c e d r o p // To o b t a i n t e r m i n a l v o l t a g e c u t OD e q u a l t o vd and draw DG p a r a l l e l t o f i e l d r e s i s t a n c e l i n e . From G draw v e r t i c a l l i n e meeting f i e l d r e s i s t a n c e l i n e at H. Point c o r r e s p o n d i n g to H g i v e s t e r m i n a l v o l t a g e which i s 44 vt =207; 45 printf ( ’ T e r m i n a l v o l t a g e i s %f V\n ’ , vt ) ;

Scilab code Exa 4.20 Determining 1 no load emf 2 output current and shunt field current 3 maximum output current and terminal voltage 4 steady state short circuit current 5 additional resistance that must be inserted in field circuit 1 2 3 4 5 6 7

clc ; ra =0.5; // a r m a t u r e r e s i s t a n c e rf =180; // s h u n t f i e l d r e s i s t a n c e n =1100; // s p e e d a t which g e n e r a t o r i s b e i n g d r i v e n n1 =1000; // s p e e d f o r which d a t a i s g i v e n disp ( ’ c a s e a ’ ) ; // from t h e d a t a g i v e n i n q u e s t i o n m a g n e t i s i n g c u r v e i s drawn ( f i g 4 . 4 6 ) 8 If =[ 0 0.2 0.4 0.6 0.8 1 1.2 1.4 ]; 9 Ea =[5 50 100 140 170 190 200 205]; 10 Ean =( n / n1 ) * Ea 146


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

plot ( If , Ean ) ; xlabel ( ’ f i e l d c u r r e n t ’ ) ; ylabel ( ’ g e n e r a t e d EMF ’ ) ; title ( ’ M a g n e t i s i n g c u r v e f o r n=1100 ’ ) ; // l i n e c o r r e s p o n d i n g t o r f =180 ohms t o meet s a t u r a t i o n c u r v e a t 221 V which i s no l o a d EMF Eo =221; printf ( ’ No l o a d EMF i s %f V\n ’ , Eo ) ; disp ( ’ c a s e b ’ ) ; vt =190; // t e r m i n a l v o l t a g e // from c u r v e a r m a t u r e r e s i s t a n c e d r o p i s g i v e n by l i n e BC vd =22.5; // a r m a t u r e r e s i s t a n c e d r o p ia = vd / ra ; // a r m a t u r e c u r r e n t ifl = vt / rf ; // f i e l d c u r r e n t printf ( ’ Shunt f i e l d c u r r e n t i s %f A\n ’ , ifl ) ; printf ( ’ Output c u r r e n t i s %f A\n ’ ,ia - ifl ) ; disp ( ’ c a s e c ’ ) ; // OP r e p r e s e n t s maximum a r m a t u r e r e s i s t a n c e d r o p i . e OP=46.5 V vd =46.5; ia = vd / ra ; // a r m a t u r e r e s i s t a n c e // t a n g e n t p o i n t a t R g i v e s f i e l d c u r r e n t which i s ifl =0.635; printf ( ’ Maximum o u t p u t c u r r e n t i s %f A ’ ,ia - ifl ) ; disp ( ’ c a s e d ’ ) ; // u n d e r s t e a d y s t a t e s h o r t c i r c u i t t e r m i n a l v o l t a g e =0 V and r e s i d u a l f l u x EMF i s E =5.5; // r e s i d u a l f l u x EMF printf ( ’ S t e a d y s t a t e s h o r t c i r c u i t c u r r e n t i s %f A\n ’ ,E / ra ) ; disp ( ’ c a s e e ’ ) ; Eo =210; // no l o a d v o l t a g e // f o r Eo OD r e p r e s e n t s f i e l d r e s i s t a n c e f i e l d current i s 1.015 ifl =1.015; // f i e l d c u r r e n t rfn = Eo / ifl ; // f i e l d r e s i s t a n c e printf ( ’ A d d i t i o n a l r e s i s t a n c e r e q u i r e d i s %f ohms \n ’ 147


43 44 45 46 47 48 49 50 51

,rfn - rf ) ; disp ( ’ c a s e f ’ ) ; rf =150; // s h u n t f i e l d r e s i s t a n c e vt =180; // t e r m i n a l v o l t a g e p =0.04; // r e d u c t i o n i n f l u x due t o a r m a t u r e reaction ifl = vt / rf ; // f i e l d c u r r e n t Ea =220*(1 - p ) ; // g e n e r a t e d v o l t a g e ia =( Ea - vt ) / ra ; // a r m a t u r e c u r r e n t il = ia - ifl ; // l o a d c u r r e n t printf ( ’ Load power i s %f KW’ ,( vt * il ) /1000) ;

Scilab code Exa 4.21 Determining range of external rheostat and dissipating power 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

clc ; Vrated =30; // r a t e d o u t p u t v o l t a g e o f g e n e r a t e r Irated =200; // r a t e d o u t p u t c u r r e n t o f g e n e r a t o r Ra =0.03; // a r m a t u r e r e s i s t a n c e ( i n c l u d i n g b r u s h e s ) Rf =2.4; // f i e l d w i n d i n g r e s i s t a n c e //No−l o a d s a t u r a t i o n c u r v e a t 2 2 0 0 rpm If =[2 4 6 8 10 12]; Ea =[15 27 35 40 43 45]; plot ( If , Ea ) ; // m a g n e t i z a t i o n c u r v e a t 2 2 0 0 rpm

// ( 1 )AT 2 2 0 0 rpm // a t no l o a d Ea1 =28; // i n d u c e d v o l t a g e i n a r m a t u r e // f o r t h i s v o l t a g e , t h e f i e l d c u r r e n t r e q u i r e d , from magnetization curve is − 16 If1 =4.23; // f i e l d c u r r e n t i n ampere 17 Rt1 = Ea1 / If1 ; // t o t a l s h u n t f i e l d r e s i s t a n c e 18 Re = Rt1 - Rf ; // e x t e r n a l r e s i s t a n c e 19

148


// a t f u l l l o a d Ea1_ =28+ Irated * Ra ; // i n d u c e d v o l t a g e i n a r m a t u r e // f o r t h i s v o l t a g e , t h e f i e l d c u r r e n t r e q u i r e d , from magnetization curve is − 23 If1_ =5.67; // f i e l d c u r r e n t 24 Rt1_ = Ea1 / If1_ ; // t o t a l s h u n t f i e l d r e s i s t a n c e 25 Re_ = Rt1_ - Rf ; // e x t e r n a l r e s i s t a n c e 20 21 22

26 27 28 // ( 2 )AT 4 5 0 0 rpm 29 // a t no l o a d 30 Ea2__ =28; // i n d u c e d v o l t a g e i n a r m a t u r e a t 4 5 0 0 rpm 31 Ea2 =28*(2200/4500) ; // Ea a t 2 2 0 0 rpm 32 // f o r t h i s v o l t a g e , t h e f i e l d c u r r e n t r e q u i r e d , from

magnetization curve is − 33 If2 =1.833; // f i e l d c u r r e n t i n ampere 34 Rt2 = Ea2__ / If2 ; // t o t a l s h u n t f i e l d r e s i s t a n c e 35 Re__ = Rt2 - Rf ; // e x t e r n a l r e s i s t a n c e 36 37 38 39 40 41 42 43 44 45

// a t f u l l l o a d Ea2___ =28+ Irated * Ra ; // i n d u c e d v o l t a g e i n a r m a t u r e a t 4 5 0 0 rpm Ea2_ =34*(2200/4500) ; // Ea a t 4 5 0 0 rpm // f o r t h i s v o l t a g e , t h e f i e l d c u r r e n t r e q u i r e d , from magnetization curve is − If2_ =2.17; // f i e l d c u r r e n t Rt2_ = Ea2__ / If2_ ; // t o t a l s h u n t f i e l d r e s i s t a n c e Re___ = Rt2_ - Rf ; // e x t e r n a l r e s i s t a n c e Pmax = If1_ * If1_ * min ( Re , Re_ , Re__ , Re___ ) ; printf ( ’ The minimum & maximum v a l u e o f e x t e r n a l r e s i s t a n c e i s %f & %fohm r e s p e c t i v e l y . \ nMaximum power d i s s i p a t e d t h r o u g h r h e o s t a t i s %f ohm . ’ , min ( Re , Re_ , Re__ , Re___ ) ,max ( Re , Re_ , Re__ , Re___ ) , Pmax ) ;

Scilab code Exa 4.22 Determining field circuit resistance and flux per pole 149


1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

clc ; Vt =100; // t e r m i n n a l v o l t a g e P =2; // no o f p o l e s Z =1000; // no o f c o n d u c t o r s A =2; // no o f p a r a l l e l p a t h s f o r a r m a t u r e c o n d u c t o r s Ra_ =2*10 e -3; // r e s i s t a n c e o f e a c h a r m a t u r e Ra =500* Ra_ *(1/2) ; // t o t a l a r m a t u r e r e s i s t a n c e // L e t I f be f i e l d c u r r e n t // Ea=Vt+( I l + I f ) ∗ 0 . 5 // Ea1=100+(10+ I f ) ∗ 0 . 5 , b e c a u s e a t 1 0 5 5 rpm I l =10. // Ea2=100+(20+ I f ) ∗ 0 . 5 , b e c a u s e a t 1 1 0 5 rpm I l =20. // But , Ea=k1 ∗ I f ∗ s p e e d // T h e r e f o r e , ( ( I f ∗ 1 0 5 5 ) / ( I f ∗ 1 1 0 5 ) ) =((100+(10+ I f ) ∗ 0 . 5 ) /(100+(20+ I f ) ∗ 0 . 5 ) ) , which g i v e s − If =1; // f i e l d c u r r e n t Ea1 =100+(10+1) *0.5; // a t 1 0 5 5 rpm N =1055; // s p e e d o f r o t o r phi =( Ea1 *60* A ) /( Z * N * P ) ; Rf = Vt / If ; // f i e l d c i r c u i t r e s i s t a n c e printf ( ’ F i e l d c i r c u i t r e s i s t a n c e i s %f ohm . \ n ’ , Rf ) ; printf ( ’ Flux p e r p o l e i s %f Wb. ’ , phi ) ;

Scilab code Exa 4.23 Determining 1 value of regulator resistance 2 no load terminal voltage for given speeds 1 2 3 4 5 6 7 8 9 10

clc ; disp ( ’ c a s e a ’ ) ; v =90; // v o l t a g e b u i l d up by r e g u l a t i n g r e s i s t a n c e rs =( v * v ) /(2* v ) ; // s h u n t w i n d i n g r e s i s t a n c e IF =[500 1000 1500 2000 2500 3000]; VA =[154 302 396 458 505 538 ]; subplot (313) ; plot ( IF , VA ) ; xlabel ( ’ F i e l d ATs/ p o l e ’ ) ; ylabel ( ’ G e n e r a t e d e .m. f E , ( V) ’ ) ; 150


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

title ( ’ M a g n e t i z i n g c u r v e f o r n=500 r . p .m. ’ ) ; // from m a g n e t i z i n g c u r v e f o r v =90 , f i e l d c u r r e n t i s 0.89 A ifl =0.89; // f i e l d c u r r e n t re =( v /(2* ifl ) ) - rs ; printf ( ’ V a l u e o f t h e r e s i s t a n c e i n t h e r e g u l a t o r i s %f ohms \n ’ , re ) ; disp ( ’ c a s e b ’ ) ; t1 =800; // t u r n s p e r p o l e o f s e p a r a t e l y e x c i t e d winding r1 =160; // r e s i s t a n c e o f w i n d i n g vc =220; // c o n s t a n t s u p p l y v o l t a g e t2 =500; // t u r n s p e r p o l e o f s h u n t w i n d i n g r2 =200; // r e s i s t a n c e o f w i n d i n g AT1 =( vc * t1 ) / r1 ; // Ampere t u r n s o f s e p a r a t e l y e x c i t e d winding // AT2=( t 2 / r 2 ) ∗E i s Ampere t u r n s o f s h u n t w i n d i n g and E i s g e n e r a t e d EMF AT3 = AT1 +( t2 / r2 ) * VA n1 =500; n2 =600; // g i v e n s p e e d s VA2 =( n2 / n1 ) * VA ; // g e n e r a t e d EMF f o r n=600 rpm subplot (323) ; plot ( IF , VA2 ) ; xlabel ( ’ F i e l d ATs/ p o l e ’ ) ; ylabel ( ’ G e n e r a t e d e .m. f E , ( V) ’ ) ; title ( ’ M a g n e t i z i n g c u r v e f o r n=600 r . p .m. ’ ) ; subplot (333) ; plot ( AT3 , VA ) ; ylabel ( ’ G e n e r a t e d e .m. f E , ( V) ’ ) ; xlabel ( ’ T o t a l ATs/ p o l e due t o b o t h f i e l d w i n d i n g ’ ) ; title ( ’ G e n e r a t e d e .m. f E , ( V) v s t o t a l Ampere t u r n s / p o l e ’ ); // p l o t o f v a r i a t i o n o f g e n e r a t e d e .m. f w i t h t o t a l Ampere t u r n s p e r p o l e i n t e r s e c t s m a g n e t i z i n g c u r v e f o r n=500 rpm a t P and m a g n e t i z i n g c u r v e f o r n=600 rpm a t Q. ( r e f e r f i g . 4 . 4 8 ) // P o i n t P g i v e s no−l o a d t e r m i n a l v o l t a g e a t 500 rpm 151


and Q g i v e s no−l o a d t e r m i n a l v o l t a g e a t 600 rpm 40 disp ( ’ No l o a d v o l t a g e a t 500 rpm i s 490 V and a t 600 rpm i s 621 V ’ ) ;

Scilab code Exa 4.24 Determining terminal voltage and number of series field turns per pole 1 clc ; 2 // m a g n e t i z a t i o n c u r v e a t 1 2 0 0 rpm 3 If =[0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8]; // f i e l d

c u r r e n t i n rpm 4 Ea =[6 53 106 160 209 241 258 272 282 288]; // i n d u c e d 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

20

v o l t a g e in armature plot ( If , Ea ) ; xlabel ( ’ I f ’ ) ; ylabel ( ’ Ea ’ ) ; // m a g n e t i z a t i o n c u r v e a t 1 2 0 0 rpm Pout =10000; // g e n e r t o r o u t p u t i n w a t t s Vt =230; // T e r m i n a l V o l t a g e Ra =0.5; // a r m a t u r e r e s i s t a n c e w i t h b r u s h e s Ns =1000; // t u r n s o f s h u n t w i n d i n g Nf =4; // t u r n s o f f i e l d w i n d i n g Z =1000; //No o f c o n d u c t o r s //PART (A) // At r a t e d o u t p u t c u r r e n t t h e s p e e d i s 1 1 5 0 rpm & shunt f i e l d c u r r e n t i s 1 A If_ =1; // f i e l d c u r r e n t a t r a t e d o / p c u r r e n t Il = Pout / Vt ; // r a t e d o u t p u t c u r r e n t Ia = Il + If ; // a r m a t u r e c u r r e n t a t r a t e d l o a d Is = Ia ; // f o r l o n g s h u n t compound g e n e r a t o r t h e s e r i e s f i e l d c u r r e n t i s equal to armature c u r r e n t // S i n c e t h e compound g e e n e r a t o r i s c u m u l a t i v e l y compounded , t h e t o t a l p o l e p e r m.m. f . i s ( Nf ∗ I f + Ns∗ I s ) ampere t u r n s // Thus t h e e q u i v a l e n t s h u n t f i e l d c u r r e n t i s g i v e n by 1/ Nf ∗ ( Nf ∗ I f +Ns∗ I s ) = 1 + ( 4 ∗ 4 4 . 5 / 1 0 0 0 ) =1.18 A . The 152


g e n e r a t e d emf f o r t h i s f i e l d c u r r e n r from t h e m a g n e t i z a t i o n c u r v e i s 257 v o l t s . 21 // For s p e e d o f 1 1 5 0 rpm t h e g e n e r a t e d emf i s − 22 Ea_ =257*(1150/1200) ; 23 Vt_ = Ea_ - Ia * Ra ; // t e r m i n a l v o l t a g e 24 25 //PART(B) 26 Eg = Vt + Ia * Ra ; // g e n e r a t e d emf i n t h e a r m a t u r e a t 1 1 5 0 27

28 29 30 31 32 33

rpm //By u s i n g t h e m a g n e t i z a t i o n c u r v e , t h e g e n e r a t e d emf a t 1 2 0 0 rpm w i l l be 2 5 2 . 2 5 ∗ ( 1 2 0 0 / 1 5 0 0 ) = 2 6 3 . 3 volts . // From t h e open c i r c u i t c h a r a c t e r i s t i c s , t h e f i e l d c u r r e n t c o r r e s p o n d i n g to 263.3 v o l t s i s 1.26 A. MMFt =1.26*1000; // T o t a l MMF // T o t a l MMF must be p r o d u c e d by t h e combined a c t i o n o f shunt & s e r i e s windings . // 1 . 2 6 ∗ 1 0 0 0 = 1 . 0 0 ∗ 1 0 0 0 + Ns ∗ ( 4 4 . 5 ) ; Ns_ =(0.26*1000) /44.5; // s e r i e s f i e l d t u r n s printf ( ’ The number o f s e r i e s f i e l d t u r n s s h o u l d be %f . ’ , round ( Ns_ ) ) ;

Scilab code Exa 4.25 Determining series field turns and resistance of diverter for achieving desired performance 1 clc ; 2 // r e p e a t p a r t ( b ) o f e x a m p l e 4 . 2 1 3 4 //PART( a )− 5 //When t h e d e m a g n e t i z i n g e f f e c t i s

ed for , t h e n from e q u a t i o n :− Net mmf = Nf ∗ I f +Ns∗ I s −ATd ....(1) 6 // 1 . 2 6 ∗ 1 0 0 0 = 1 . 0 0 ∗ 1 0 0 0 + 1 0 I s −0.022 I s ∗ 1 0 0 0 7 Ns = round (0.3578*1000/44.5) ; // no o f t u r n s i n s e r i e s f i e l d winding 153


8 9 10 11 12 13 14 15 16 17 18

//PART( b )− // I f t h e r e a r e 10 s e r i e s f i e l d t u r n s , t h e n from equation (1) , // 1 . 2 6 ∗ 1 0 0 0 = 1 . 0 0 ∗ 1 0 0 0 + 1 0 I s − 0 . 0 0 2 2 I s ∗ 1 0 0 0 Is =0.26/0.0078 // Out o f t h e t o t a l a r m a t u r e c u r r e n t o f 4 4 . 5 A, o n l y I s ( 3 3 . 3 ) should flow through the s e r i e s f i e l d . // T h i s can be a c h i e v e d by p u t t i n g a r e s i s t o r i n p a r a l l e l with the s e r i e s f i e l d winding . // 3 3 . 3 = ( 4 4 . 5 ∗ Rdi ) / ( 0 . 0 5 + Rdi ) Rdi =0.05/0.3363; printf ( ’NO OF TURNS IN SERIES FIELD WINDING ARE %f . ’ , Ns ) ; printf ( ’ \nTHE RESISTANCE OF DIVERTER Rdi SHOULD BE %f OHMS. ’ , Rdi ) ;

Scilab code Exa 4.26 Determining emf generated in armature and percentage change in series field ampere turns due to diverter 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

clc ; Vt =250; // r a t e d o / p v o l t a g e o f g e n e r a t o r Pout =10000; // o / p o f g e n e r a t o r i n w a t t s Ra =0.4; // a r m a t u r e r e s i s t a n c e Rse =0.2; // s e r i e s f i e l d r e s i s t a n c e Rs =125; // s h u n t f i e l d r e e s i s t a n c e Vb =2; // t o t a l b r u s h c o t a c t d r o p Il = Pout / Vt ; // l o a d c u r r e n t //PART( a )−LONG SHUNT CONNECTION If = Vt / Rs ; // s h u n t f i e l d c u r r e n t Ia = Il + If ; // a r m a t u r e c u r r e n t // s e r i e s f i e l d w i n d i n g a l s o c a r r i e s I a . Eal = Vt + Ia *( Rse + Ra ) + Vb ; // g e n e r a t e d emf i n a r m a t u r e printf ( ’ The g e n e r a t e d EMF i n a r m a t u r e when t h e 154


g e n e r a t e d i s c o n n e c t e d a s l o n g s h u n t machine i s %f . \ n ’ , Eal ) ; 16 17 //PART( b )−SHORT SHUNT CONNECTION 18 V = Vt + Il * Rse ; // v o l t a g e a c r o s s s h u n t 19 20 21 22

23 24 25 26 27 28 29

f i e l d and

armature t e r m i n a l s If_ = V / Rs ; // s h u n t f i e l d c u r r e n t Ia_ = Il + If_ ; // a r m a t u r e c u r r e n t Eas = V + Ia * Ra + Vb ; // g e n e r a t e d emf i n a r m a t u r e printf ( ’ The g e n e r a t e d EMF i n a r m a t u r e when t h e g e n e r a t e d i s c o n n e c t e d a s s h o r t s h u n t machine i s %f . ’ , Eas ) ; //PART( c )− // S e r i e s f i e l d ampere t u r n s a r e p r o p o r t i o n a l t o s e r i e s −f i e l d current I // I s = 0 . 3 / 0 . 5 ∗ I , where , I s i s s e r i e s f i e l d c u r r e n t with d i v e r t e r . // s e r i e s f i e l d ampere−t u r n s w i t h d v i d e r = K∗ 0 . 6 ∗ I s , where K i s a c o n s t a n t . // p e r c e n t a g e r e d u c t i o n i n s e r i e s f i e l d ampere t u r n s i s − ( ( I−O. 6 I ) / I ) ∗ 1 0 0 . disp ( ’ P e r c e n t a g e r e d u c t i o n i n s e r i e s f i e l d ampere t u r n s i s 40%. ’ ) ;

Scilab code Exa 4.27 Determining speed and developed torque at full load

1 2 3 4 5 6

clc ; Vt =230; // o u t p u t v o l t a g e Ra =0.3; // a r m a t u r e c i r c u i t r e s i s t a n c e Rf =160; // f i e l d c i r c u i t r e s i s t a n c e Il =40; // l i n e c u r r e n t a t f u l l l o a d & r a t e d v o l t a g e Ia1 =3.33 // a r m a t u r e c u r r e n t a t r a t e d v o l t a g e & no l o a d s p e e d o f 1 0 0 0 rpm 155


7 8 9 10 11 12 13 14 15 16

17 18 19 20 21 22 23

//No l o a d c o u n t e r emf i s − Ea1 = Vt - Ia1 * Ra ; If = Vt / Rf ; // f i e l d c u r r e n t // At f u l l l o a d a r m a t u r e c u r r e n t i s − Ia2 = Il - If ; Ea2 = Vt - Ia2 * Ra ; // C o u n t e r emf a t f u l l l o a d // At f u l l l o a d , t h e f i e l d f l u x i s − // P h i 2 =0. 96 ∗ p h i 1 // The c o n t e r emf Ea , i s g i v e n by− Ea=Ka∗ p h i ∗Wm // Ea1 / Ea2=(Ka∗ p h i 1 ∗Wm1) / ( Ka∗ p h i 2 ∗Wm2) =( p h i 1 ∗ n1 ) / ( p h i 2 ∗ n2 ) o r 2 2 9 / 2 1 8 . 4 3 = ( 1 0 0 0 ∗ p h i 1 ) / n2 ∗ ( 0 . 9 6 ∗ phi 1 ) . . . ( 1 ) // from e q u a t i o n ( 1 ) n2 =995; // f u l l l o a d s p e e d // At f u l l l o a d , Ea2=Ka∗ p h i 2 ∗Wm //Ka∗ p h i 2=Ea2 /Wm // E l e c t r o m a g n e t i c o r d e v e l o p e d , t o r q u e a t f u l l l o a d i s , Te=Ka∗ p h i 2 ∗ I a 2 Te =( Ea2 *60) /(2* %pi * n2 ) * Ia2 ; // E l e c t r o m a g n e t i c t o r q u e developed . printf ( ’ E l e c t r o m a g n e t i c o r d e v e l o p e d , t o r q u e a t f u l l l o a d i s %f . ’ , Te ) ;

Scilab code Exa 4.28 Determining motor speed and rated shaft torque 1 2 3 4 5 6 7

clc ; // Armature r e a c t i o n i s n e g l e c t e d Vt =220; // o u t p u t v o l t a g e Ra =0.2; // a r m a t u r e c i r c u i t r e s i s t a n c e Rf =110; // f i e l d c i r c u i t r e s i s t a n c e n1 =1500; // s p e e d o f r o t o r a t no l o a d Il1 =5; // c u r e n t drawn by motor i n ampere a t no l o a d & 1 5 0 0 rpm 8 Il2 =52; // c u r e n t drawn by motor i n ampere a t r a t e d load & rated voltage 156


9 10 11 12 13 14 15 16 17

18

19 20 21 22 23 24

If = Vt / Rf ; // s h u n t f i e l d c u r r e n t Ia1 = Il1 - If ; // a r m a t u r e c u r r e n t a t no l o a d Ea1 = Vt - Ia1 * Ra ; // c o u n t e r emf a t no l o a d Pr = Ea1 * Ia1 ; // r o t a t i o n a l l o s s e s a t no l o a d // R o t a t i o n a l l o s s e s a t f u l l l o a d & no l o a d a r e same Ia2 = Il2 - If ; // a r m a t u r e c u r r e n t a t f u l l l o a d Ea2 = Vt - Ia2 * Ra ; // c o u n t e r emf a t f u l l l o a d Pem = Ia2 * Ea2 ; // e l e c t r o m a g n e t i c power // Here p h i 1 ( no l o a d f l u x )=p h i 2 ( f u l l l o a d f l u x ) , because the f i e l d current i s constant & e f f e c t of AR i s n e g l e c t e d . // Ea1 / Ea2=(n1 ∗ p h i 1 ) / ( n2 ∗ p h i 2 ) ; where n1 & n2 a r e s p e e d o f r o t o r a t no l o a d & f u l l l o a d respectively . n2 = fix (( n1 * Ea2 ) /( Ea1 ) ) ; Psh = Pem - Pr ; // s h a f t power Wm =(2* %pi * n2 ) /60; // a n g u l a r v e l o c i t y o f s h a f t a t f u l l load Tsh = Psh / Wm // s h a f t t o r q u e printf ( ’ The motor s p e e d i s %f rpm . \ n ’ , n2 ) ; printf ( ’ Rated s h a f t t o r q u e i s %f Nm. ’ , Tsh ) ;

Scilab code Exa 4.29 Determining external resistance inserted in field circuit 1 2 3 4 5 6

clc ; Ra =0.4; // a r m a t u r e r e s i s t a n c e i n ohm Rf =200; // f i e l d c i r c u i t r e s i s t a n c e i n ohm Vt =230; // t e r m i n a l v o l t a g e f o r dc motor If_1 =1.1; // f i e l d c u r r e n t f o r dc g e n e r a t o r a t open c i r c u i t v o l t a g e o f 210 V . 7 If_2 =0.9; // f i e l d c u r r e n t f o r dc g e n e r a t o r a t open c i r c u i t v o l t a g e o f 230 V . 8 Ia =24; // a r m a t u r e c u r r e n t f o r dc s h u n t motor a t 1 5 0 0 157


9 10 11 12 13 14 15 16

rpm Ea = Vt - Ia * Ra ; // c o u n t e r e .m. f . f o r dc motor a t 1 5 0 0 rpm and f u l l l o a d // For g e n e r a t e d e .m. f . , Ea=230 V, f i e l d c u r r e n t i s 1 . 1 A & f o r Ea=210 V, f i e l d c u r r e n t i s 0 . 9 A // The c h a n g e i n g e n e r a t e d e .m. f . i s 20 V f o r f i e l d v a r i a t i o n o f 0 . 2 A & t h i s change i s l i n e a r . // T h e r e f o r e f o r a g e n e r a t e d e .m. f . o f Ea = 2 2 0 . 4 V a t 1 5 0 0 rpm , t h e f i e l d c u r r e n t would be− If =0.9+(0.2/20) *10.4; // 0 . 9 A f o r 210 V & ( 0 . 2 / 2 0 ) ∗10.4 f o r remaining 10.4 V. Rsh = Vt / If ; // Shunt f i e l d r e s i s t a n c e r e q u i r e d f o r a f i e l d c u r r e n t ( I f ) w i t h t e r m i n a l v o l t a g e ( Vt ) . Rext = Rsh - Rf ; // E x t e r n a l r e s i s t a n c e t h a t must be i n s e r t e d in shunt f i e l d c i r c u i t printf ( ’ The e x t e r n a l r e s i s t a n c e t h a t must be i n s e r t e d i n s h u n t f i e l d c i r c u i t = %f ohm . ’ , Rext ) ;

Scilab code Exa 4.30 Determining 1 speed and internal torque developed 2 shaft power shaft torque and efficiency 1 2 3 4 5 6 7 8 9 10 11 12 13 14

clc ; // Armature r e a c t i o n i s n e g l e c t e d . Vt =250; // S u p p l y v o l t a g e P =4; //No o f p o l e s A =2; //No o f p a r a l l e l p a t h s f o r a r m a t u r e c o n d u c t o r s Z =500; //No o f a r m a t u r e c o n d u c t o r s Ra =0.25; // a r m a t u r e c i r c u i t r e s i s t a n c e i n ohm Rf =125; // f i e l d r e s i s t a n c e i n ohm phi =0.02; // f l u x p e r p o l e i n weber Il =14; // c u r r e n t drawn by motor from s u p p l y mains Ish = Vt / Rf ; // c o n s t a n t s h u n t f i e l d c u r r e n t Pr =300; // r o t a t i o n a l l o s s e s i n w a t t s Pi = Vt * Il ; // power i n p u t i n w a t t s

158


15 //PART( a )− 16 Ia = Il - Ish ; // a r m a t u r e c u r r e n t 17 Ea = Vt - Ia * Ra ; // c o u n t e r / back emf 18 // Ea=(P∗ p h i ∗Z∗N) / ( 6 0 ∗A) 19 N =(60* A * Ea ) /( P * phi * Z ) ; // s p e e d o f

r o t a t i o n o f motor

i n rpm Wm =(2* %pi * N ) /60; // a n g u l a r v e l o c i t y o f motor Pe = Ea * Ia ; // e l e c t r o m a g n e t i c power Ti = Pe / Wm ; // I n t e r n a l t o r q u e d e v e l o p e d i n Nm. printf ( ’ Speed o f r o t a t i o n o f motor i s %f rpm . ’ ,N ) ; printf ( ’ \ n I n t e r n a l t o r q u e d e v e l o p e d = %f Nm. ’ , Ti ) ;

20 21 22 23 24 25 26 //PART( b )− 27 Psh = Pe - Pr ; // s h a f t power 28 Tsh = Psh / Wm ; // s h a f t t o r q u e 29 %n =( Psh / Pi ) *100; // p e r c e n t a g e e f i c i e n c y 30 printf ( ’ \ n S h a f t power = %f w a t t s . ’ , Psh ) ; 31 printf ( ’ \ n S h a f t t o r q u e = %f Nm. ’ , Tsh ) ; 32 printf ( ’ \ n E f f i c i e n c y o f motor i s %f p e r c e n t . ’ , %n ) ;

Scilab code Exa 4.31 Determining shaft power operating speed armature current and motor efficiency 1 2 3 4 5 6 7 8 9

clc ; Vt =230; // S u p p l y v o l t a g e P =4; //No o f p o l e s A =2; //No o f p a r a l l e l p a t h s f o r a r m a t u r e c o n d u c t o r s Z =600; //No o f a r m a t u r e c o n d u c t o r s Ra =0.25; // a r m a t u r e c i r c u i t r e s i s t a n c e i n ohm phi =0.01; // f l u x p e r p o l e i n weber Pr =500; // r o t a t i o n a l l o s s e s i n w a t t s // g e n e r a t e d emf i n a r m a t u r e , Ea=( p h i ∗Z∗P∗n ) / ( 6 0 ∗A) i f n i s speed in armature 10 // C o u n t e r emf i s Ea = ( 0 . 0 1 ∗ 6 0 0 ∗ 4 ∗ n ) / ( 6 0 ∗ 2 ) =0.2 n v o l t s 11 // Vt=Ea+I a ∗Ra 159


12 13 14

15 16 17 18

19 20 21 22 23

24 25 26

27 28 29 30 31 32 33 34 35

// I a =(Vt−Ea ) /Ra/ // S h a f t o / p i n w a t t s , Psh=Ea∗ Ia −Pr , Psh = ( 0 . 2 n ) ∗ ( 9 2 0 − 0 . 8 n ) −500 . . . . ( 1 ) n =[700 800 900 1000 1100]; // d i f f e r e n t s p e e d s o f motor f o r which t h e s h a f t o / p power i s t o be measured // Psh = ( 0 . 1 8 4 ∗ n ) − ( 1 . 6 ∗ ( 1 0 e −4) ∗n∗n ) −0.5 , S h a f t o / p power i n KW. Psh1 =49.1; // S h a f t o /p power i n KW a t n=700 rpm Psh2 =44.3; // S h a f t o /p power i n KW a t n=800 rpm Psh3 =33.5; // S h a f t o /p power i n KW a t n=900 rpm ∗ Psh1 , Psh2 , Psh3 , Psh4 , Psh5 a r e c a l c l u l a t e d from e q u a t i o n ( 1 ) ∗ Psh4 =23.5; // S h a f t o /p power i n KW a t n=1000 rpm Psh5 =8.3; // S h a f t o / p power i n KW a t n=1100 rpm Psh =[ Psh1 Psh2 Psh3 Psh4 Psh5 ]; Pi =[4.5 8.5 14 21.1 30]; // i / p power s u p p l i e d t o f a n i n KW. plot (n , Psh ,n , Pi ) ; xlabel ( ’RPM ’ ) ; ylabel ( ’KW’ ) ; // S h a f t o / p power v e r s u s s p e e d o f motor and power i / p v e r s u s s p e e d o f f a n a r e p l o t t e d on t h e same g r a p h // For p l o t : −∗ b l u e l i n e − motor c h a r a c t e r i s t i c , g r e e n line − fan c h a r a c t e s t i c ∗ // The i n t e s e c t o n o f t h e s e two c u r v e s i s c a l l e d a s OPERATING POINT . // At o p e r a t i n g p o i n t t h e s p e e d i s 1 0 1 2 rpm & pwer o / p o f motor o r power i / p t o t h e f a n i s 22 KW . . . . ( from t h e i n t e r s e c t i o n p o i n t o f two c u r v e s ) n_ =1012; // s p e e d a t o p e r a t i n g p o i n t P_o =22000; // Power o u t p u t l o s s Ia =920 -(0.8* n_ ) ; // a r m a t u r e c u r r e n t Parm = Ia * Ia * Ra ; // a r m a t u r e l o s s P_ip = P_o + Pr + Parm ; // Power i n p u t %n =( P_o / P_ip ) *100; // motor e f f i c i e n c y printf ( ’ Armature c u r r e n t i s %f A\n . ’ , Ia ) ; printf ( ’ O p e r a t i n g s p e e d i s %f rpm\n . ’ , n_ ) printf ( ’ Motor e f f i c i e n c y i s %f p e r c e n t . ’ , %n ) ;

160


Scilab code Exa 4.32 Determining speed of motor 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

clc ; Vt =230; // S u p p l y v o l t a g e P =4; //No o f p o l e s A =2; //No o f p a r a l l e l p a t h s f o r a r m a t u r e c o n d u c t o r s Z =500; //No o f a r m a t u r e c o n d u c t o r s Ra =0.2; // a r m a t u r e c i r c u i t r e s i s t a n c e i n ohm Rs =0.1; // f i e l d r e s i s t a n c e i n ohm Il =40; // l i n e c u r r e n t N =1000; // r a t e d s p e e d i n rpm Ia1 =40; // a r m a t u r e c u r r e n t f o r dc s e r i e s motor a t 40 A line current Ia2 =20; // a r m a t u r e c u r r e n t f o r dc s e r i e s motor a t 20 A line current // For 40A l i n e c u r r e n t Ea1 = Vt - Ia1 *( Ra + Rs ) ; // c o u n t e r emf // For 20A l i n e c u r r e n t Ea2 = Vt - Ia2 *( Ra + Rs ) ; // c o u n t e r emf // Let , p h i 1=f l u x a t 40 A, p h i 2=f l u x a t 20 A l i n e current // p h i 2 = 0 . 6 ∗ ( p h i 1 ) // ( Ea1 / Ea2 ) =(n1 ∗ p h i 1 ) / ( n2 ∗ p h i 2 ) // ( 2 1 8 / 2 2 4 ) = ( 1 0 0 0 ∗ ( p h i 1 ) ) / ( n2 ∗ ( 0 . 6 ∗ ( p h i 1 ) ) n2 =(1000*224) /(218*0.6) ; // s p e e d o f motor a t l i n e c u r r e n t o f 20 A a t 230 V printf ( ’ Speed o f motor a t l i n e c u r r e n t o f 20 A a t 230 V i s %f rpm . ’ , round ( n2 ) ) ;

Scilab code Exa 4.33 Determining 1 speed in rpm 2 shaft power output

161


1

//ANSWER GIVEN IN THE BOOK FOR THIS QUESTION I S INCORRECT .

2 3 clc ; 4 // N e g l e c t i n g a r m a t u r e 5 // Assuming r o t a t i o n a l 6 V =230; // S u p p l y v o l t a g e 7 P =15000; // power r a t i n g 8 Il_1 =80; // l i n e c u r r e n t 9 Il_2 =40; // l i n e c u r r e n t 10 11 12 13 14 15 16 17 18 19

r e a c t i o n & magnetic s a t u r a t i o n l o s s e s to remain c o n s t a n t

o f dc s e r i e s motor i n w a t t s rated a s s u m i n g t h a t motor draws h a l f the rated current at rated voltage Ia_1 = Il_1 ; // a r m a t u r e c u r r e n t a t l i n e c u r r e n t e q u a l t o 80 A . Ia_2 = Il_2 ; // a r m a t u r e c u r r e n t a t l i n e c u r r e n t e q u a l t o 40 A . n1 =1000; // r a t e d s p e e d i n rpm // F u l l l o a d l o s s e s e x p r e s s e d a s p e r c e n t a g e o f motor i n p u t :− // Armature ohmic l o s s =2.8%( i n c l u d i n g b r u s h l o s s ) // F i e l d ohmic l o s s =2.2% // R o t a t i o n a l l o s s =2.2% P_ip = V * Il_1 ; // f u l l l o a d i n p u t P_ohmic = P_ip *(5.4/100) // As p e r c e n t o f t o t a l ohmic l o s s e s = 2 . 2 + 2 . 8 = 5 . 4% // But P ohmic= I l ∗ I l ∗ ( Ra+Rs ) ; where ( Ra+Rs ) =( a r m a t u r e + series field ) resistance // ( Ra+Rs )=P ohmic / ( I l ∗ I l ) = 0 . 1 1 5 ohms // Let , r =(Ra+Rs ) r =0.115;

20 21 22 23 24 //PART( a )− 25 Ea1 =V -( Ia_1 * r ) ; // c o u n t e r emf a t 26 Ea2 =V -( Ia_2 * r ) ; // c o u n t e r emf a t 27 // S i n c e t h e m a g n e t i c s a t u r a t i o n

l i n e c u r r e n t = 80 A l i n e c u r r e n t = 40 A i s n e g l e c t e d , p h i 1= k ∗80 & p h i 2=k ∗ 4 0 ; where k=c o n s t a n t & p h i 1 & p h i 2 a r e f l u x p e r p o l e a t l i n e c u r r e n t s 80 & 40 A respectively . 28 // ( Ea1 / Ea2 ) =(n1 ∗ p h i 1 ) / ( n2 ∗ p h i 2 ) o r ( 2 2 0 . 8 / 2 2 5 . 4 ) 162


= ( 1 0 0 0 ∗ 8 0 ) / ( n2 ∗ 4 0 ) ; where Ea1 = 2 2 0 . 8 V Ea2 = 2 2 5 . 4 V . 29 n2 =(1000*80*225.4) /(40*220.8) ; // s p e e d i n rpm 30 printf ( ’ The s p e e d o f r o t a t i o n o f motor when t h e motor draws h a l f t h e r a t e d c u r r e n t a t r a t e d v o l t a g e i s %f rpm . ’ , round ( n2 ) ) ; 31 32 //PART( b )− 33 Pr = P_ip *(2.2/100) ; // r o t a t i o n a l l o s s e s 34 Psh = Ea2 * Ia_2 - Pr ; 35 printf ( ’ \ nThe s h a f t o u t p u t power i s %f W. ’ , Psh ) ;

Scilab code Exa 4.34 Determining operating speed and current drawn from source 1 2 clc ; 3 Pout =40000; // o u t p u t power 4 V =250; // s u p p l y v o l t a g e i n v o l t s 5 r =0.2; // sum o f a r m a t u r e c i r c u i t 6 7 8 9 10 11 12 13

resistance & series

field circuit resistance n =1500; // s p e e d o f dc s e r i e s motor a t r a t e d c u r r e n t //PART( a )−

Ir = Pout / V ; // r a t e d c u r r e n t Ea =V - Ir * r ; // c o u n t e r emf a t r a t e d l o a d Wm =(2* %pi * n ) /60; // a n g u l a r s p e e d o f r o t a t i o n o f motor Te = round (( Ea * Ir ) / Wm ) ; // r a t e d e l e c t r o m a g n e t i c t o r q u e i n Nm. 14 //Now t h e r e l a t i o n g i v i n g t h e t o r q u e s p e e d c h a r a c t e r i s t i c s o f s e r i e s motor must be d e v e l o p e d . 15 // S i n c e t h e m a g n e t i c s a t u r a t i o n i s n e g l e c t e d 16 // Te=Ka∗ p h i ∗ I 163


17 // Te=K1∗ I a ˆ2 18 K1 = Te /( Ir ^2) ; // I a=r a t e d c u r r e n t 19 // Ea=K2∗ p h i ∗n=K2∗ I a ∗n 20 K2 = Ea /( Ir * n ) ; // v a l u e s o f c o n s t a n t o f 21 22 23 24 25 26 27 28 29 30 31 32 33

proportionality // The v a l u e s o f c o n s t a n t s k1 & K2 a r e o b t a i n e d from rated conditions // I a =(V−Ea ) / r & Ea=K2∗ I a ∗n ; I a =1250 −0.00454∗ I a ∗n // I a = 1 2 5 0 / ( 1 + 0 . 0 0 4 5 4 n ) . . . . ( 1 ) ; I a=a r m a t u r e c u r r e n t a t any s p e e d // Te=K1∗ I a ˆ2 n =[1400 1450 1500 1550 1600 1650 1700]; Te = K1 .*((( V / r ) ^2) ./(1+( K2 / r ) .* n ) .^2) ; // Electromagnetic torque Tl =5.* sqrt ( n ) ; // l o a d t o r q u e plot ( Te ,n , Tl , n ) ; xlabel ( ’T(Nm) ’ ) ; ylabel ( ’ Speed ( rpm ) ’ ) ; title ( ’ Speed−t o r q u e c h a r a c t e r i s t i c s f o r S e r i e s −motor and f o r l o a d ’ ) ; //THE INTERSECTION OF SERIES MOTOR & LOAD CHARACTERISTICS GIVES THE OPERATING POINT

34 35

// From t h e c u r v e t h e o p e r a t i n g p o i n t i s o b t a i n e d a t 1 5 9 1 rpm & t o r q u e i s 1 9 9 . 5 Nm 36 disp ( ’ The o p e r a t i n g s p e e d o f motor i s 1 5 9 1 rpm . ’ ) ; 37 38 //PART( b )− 39 // C u r r e n t drawn from t h e s o u r c e i s − 40 Ia =( V / r ) /(1+( K2 / r ) *1591) ; // From e q u a t i o n ( 1 ) 41 printf ( ’ C u r r e n t drawn from t h e s o u r c e i s %f A . ’ , Ia ) ;

Scilab code Exa 4.35 Determining number of series field turns

164


1 2 3 4 5 6 7 8 9

10 11

clc ; V =230; // s u p p l y c u r r e n t Pout =15000; // o u t p u t v o l t a g e Ra =0.2; // a r m a t u r e c i r c u i t r e s i s t a n c e Rse =0.1; // r e s i s t a n c e o f s e r i e s f i e l d w i n d i n g Nf =1000; // s h u n t f i e l d t u r n s p e r p o l e // Data f o r m a g n e t i z a t i o n c u r v e a t 1 5 0 0 rpm If_ =[0 0.2 0.4 0.6 0.8 1.02 1.15 1.32 1.56 1.92 2.4]; // f i e l d c u r r e n t a t 1 5 0 0 rpm Ea =[6 40 80 120 160 200 220 240 260 280 300]; // c o u n t e r emf a t 1 5 0 0 rpm and r e s p e c t i v e f i e l d current plot ( If_ , Ea ) ; // M a g n e t i z a t i o n c u r v e a t 1 5 0 0 rpm Ia_1 =4; // a r m a t u r e c u r r e n t a t r a t e d v o l t a g e and r a t e d load Ia_2 =70; // a r m a t u r e c u r r e n t a t 1 2 0 0 rpm

12 13 14 // At no l o a d 15 Ea_ =V - Ia_1 * Ra ; // c o u n t e r emf 16 // f i e l d c u r r e n t r e q u i r e d f o r Ea ( i e . 2 2 9 . 2 V) , from O .

C.C. i s 1.23 A. 17 18 19 20 21 22 23 24 25 26 27

// At l o a d Ea__ =V - Ia_2 *( Ra + Rse ) ; // c o u n t e r emf a t 1 2 0 0 rpm Ea___ =209*(1500/1200) ; // Ea a t s p e e d o f 1 5 0 0 rpm // F i e l d c u r r e n t c o r r e s p o n d i n g t o E a ( i e . 2 6 1 . 2 5 V) , from O . C . C . i s 1 . 5 7 5 A . // T o t a l D−a x i s mmf p e r p o l e=Nf ∗ I f +Ns∗ I s If =1.23; // f i e l d c u r r e n t a t 2 2 9 . 2 V i s 1 . 2 3 A If1 =1.23; // f i e l d c u r r e n t a t 2 6 1 . 2 5 V i s 1 . 5 7 5 A // 1 . 5 7 5 ∗ 1 0 0 0 = 1 . 2 3 ∗ 1 0 0 0 + Ns ∗ ( 7 0 ) Ns =(0.345*1000) /70; // s e r i e s f i e l d t u r n s printf ( ’ For l o n g −s h u n t c o n n e c t i o n , s e r i e s f i e l d t u r n s i s e q u a l t o %f . ’ , round ( Ns ) ) ;

165


Scilab code Exa 4.36 Determining 1 shunt field current 2 effective armature reaction 3 number of series field turns 4 speed at rated armature current and at rated voltage 5 internal starting torque 1 // Answer f o r p a r t ( e ) i n book i s i n c o r r e c t . 2 clc ; 3 // M a g n e t i z a t i o n c u r v e i s same a s t h a t o f e x a m p l e 4 5 6 7

4.33 Ra =0.2; // a r m a t u r e r e s i s t a n c e ( i n c l u d i n g b r u s h e s ) Nf =2000; // s h u n t f i e l d t u r n s N =1500; // motor s p e e d i n rpm a t no l o a d a s w e l l a s rated load . Ia =36; // motor a r m a t u r e c u r r e n t i n Amperes a t r a t e d load

8 9 //PART( a )−At no l o a d , 10 Vt =230; // s u p p l y mains i n v o l t s 11 Ea_a = Vt ; // c o u n t e r emf , n e g l e c t i n g a r m a t u r e

circuit

resistance 12 If_a =1.23; // f i e l d c u r r e n t i n a m p e r e s 13 printf ( ’ ( a ) Shunt f i e l d c u r r e n t i s %f A. \ n ’ , If_a ) ; 14 // Thus c o n s t a n t s h u n t f i e l d c u r r e n t I f from O . C . C . i s 1 . 2 3 A c o r r e s p o n d i n g t o 230V . 15 16 17 18

//PART( b )−At f u l l l o a d , Ea_b = Vt - Ia * Ra ; // c o u n t e r emf //A p o i n t i s drawn on m a g n e t i z a t i o n c u r v e w i t h c o o r d i n a t e s A( I f , Ea b ) . 19 // The h o r i z o n t a l d i s t a n c e b e t w e e n Pt . A & t h e magnetization curve , g i v e s the e f f e c t i v e armature r e a c t i o n in o f shunt f i e l d current , i t s value i s 0.06 A. 20 AT_arm = Nf *0.06; // a r m a t u r e r e a c t i o n i n amopere t u r n s per pole 21 printf ( ’ ( b ) E f f e c t i v e a r m a t u r e r e a c t i o n i s %f ampere t u r n s p e r p o l e . \ n ’ , AT_arm ) ; 22 23

//PART( c )−At r a t e d l o a d , w i t h s e r i e s w i n d i n g i n 166


24 25 26 27 28 29 30 31

c i r c u i t & motor i s c u m u l a t i v e l y compounded , Rse =0.05; // s e r i e s f i e l d r e s i s t a n c e i n ohm Ea__c = Vt - Ia *( Ra + Rse ) ; // c o u n t e r emf a t 1 3 5 0 rpm Ea_c = Ea__c *(1500/1350) ; // Ea a t 1 5 0 0 rpm // From m a g n e t i z a t i o n c u r v e , Ea = 2 4 5 . 5 V r e q u i r e s I f c =1.365 A. // From e q u a t i o n − Net MMF =Nf ∗ I f +Ns∗ I s −ATd . . . . ( 1 ) // 1 . 3 6 5 ∗ 2 0 0 0 = 1 . 2 3 ∗ 2 0 0 0 + Ns ∗ ( 3 6 ) −120 Ns = round (65/6) ; // S e r i e s f i e l d c u r r e n t printf ( ’ ( c ) R e q u i r e d no o f s e r i e s f i e l d t u r n s a r e %f . \ n ’ , Ns ) ;

32 33 //PART( d )− I f t h e s e r i e s f i e l d w i n d i n g h a s 20 t u r n s − 34 Ns_ =20; // no o f t u r n s o f s e r i e s f i e l d w i n d i n g 35 // Net mmf = Nf ∗ I f +Ns ∗ I s −AT arm . . . . ( f o r m u l a ) 36 mmf_net = If_a * Nf + Ns_ * Ia - AT_arm ; // Net f i e l d mmf i n 37 38 39 40 41 42 43 44 45 46 47 48 49

t e r m s o f ATs If_d =( If_a * Nf + Ns_ * Ia - AT_arm ) / Nf ; // Net f i e l d mmf i n t e r m s o f t h e e q u i v a l e n t s h u n t f i e l d c u r r e n t (A) . // From t h e m a g n e t i z a t i o n c u r v e , t h e v a l u e o f Ea c o r r e s p o n d i n g t o I f =1.53A i s 258V a t 1 5 0 0 rpm . // But t h e c o u n t e r emf , Ea c o r r e s p o n d i n g t o r a t e d c u r r e n t i s 2 3 0 − 3 6 ( 0 . 2 + 0 . 0 5 ) =221 V . // T h e r e f o r e t h e motor s p e e d n c o r r e s p o n d i n g t o Ea =221V i s − // ( 2 2 1 / 2 5 8 ) =(n / 1 5 0 0 ) n =(221/258) *1500; printf ( ’ ( d ) Speed a t r a t e d v o l t a g e r a t e d a r m a t u r e c u r r e n t i s %f rpm . \ n ’ , round ( n ) ) ; //PART( e )−Assuming d e m a g n e t i z i n g e f f e c t o f a r m a t u r e r e a c t i o n t o be 200 ampere t u r n s p e r p o l e . ATarm =200; // d e m a g n e t i z i n g e f f e c t o f a r m a t u r e r e a c t i o n i n ampere t u r n s p e r p o l e . Ia_e =50; // a r m a t u r e c u r r e n t i n a m p e r e s . . . . ( g i v e n ) mmfnet = If_a * Nf + Ns_ * Ia_e - ATarm ; // from e q u a t i o n no ....(1) If_e = mmfnet / Nf ; // Net f i e l d mmf i n t e r m s o f t h e 167


50 51 52 53 54 55

e q u i v a l e n t s h u n t f i e l d c u r r e n t (A) . // From t h e m a g n e t i z i n g c u r v e , c o r r e s p o n d i n g t o f i e l d c u r r e n t I f e ( 1 . 6 3 A) , Ea a t 1 5 0 0 rpm i s 264 V . // But , Ea=Ka∗ p h i ∗Wm ; where , p h i = f l u x p e r p o l e // Thus , Ka∗ p h i = ( 2 6 4 ∗ 6 0 ) / ( 2 ∗ %pi ∗ 1 5 0 0 ) Kaphi =264/(50* %pi ) ; //Ka∗ p h i Test = Kaphi * Ia_e ; // s t a r t i n g t o r q u e printf ( ’ ( e ) When t h e a r m a t u r e c u r r e n t i s l i m i t e d t o 50 A t h e s t a r t i n g t o r q u e i s %f Nm. ’ , Test ) ;

Scilab code Exa 4.37 Determining steady state speed and armature current drawn from source 1 2 3 4 5 6 7

8 9 10 11 12 13 14 15 16 17

clc ; V =230; // s u p p l y v o l t a g e i n v o l t s Ra =0.5; // a r m a t u r e r e s i s t a n c e i n ohm N =250; // r a t e d s p e e d o f motor I =100; // r a t e d c u r r e n t i n ampere // For t h e s e p a r a t e l y e x c i t e d dc motor t o r q u e −s p e e d c h a r a c t e r i s t i c s i s g i v e n by Tl=500−W, where W i s r o t a t i o n a l s p e e d i n r a d / s e c & Tl i s l o a d t o r q u e i n Nm. // At r a t e d l o a d , motor c o u n t e r emf i s − Ea =V - I * Ra ; // Ea=Km∗Wr ; Km = motor c o n s t a n t , Wr = r a t e d motor speed in rad / s e c Wr =(2* %pi *250) /60; // r a t e d motor s p e e d i n r a d / s e c Km = Ea / Wr ; // motor c o n s t a n t i n V−s / r a d // Armature c u r r e n t a t any s p e e d W i s g i v e n by− Ia =( V - Ea ) / Ra ; // i e . I a =(230−Km∗W) / 0 . 5 // Motor t o r q u e , Te=Km∗ I a =(Km/ 0 . 5 ) ∗(230 −Km∗W) // Under s t e a d y s t a t e , motor t o r q u e , Te=l o a d t o r q u e , Tl // Thus , (Km/ 0 . 5 ) ∗(230 −Km∗W) =500−10∗W 168


18 W =((( V * Km ) / Ra ) -500) /(( Km ^2/ Ra ) -10) ; // a n g u l a r s p e e d

i n rpm 19 N_ =( W *60) /(2* %pi ) ; // Speed i n rpm 20 Ia_ =(230 - Km * W ) /0.5 // a r m a t u r e c u r r e n t 21 printf ( ’ S t e a d y s t a t e s p e e d o f motor i s %f rpm\n . ’ , N_

); 22 printf ( ’ Armature c u r r e n t drawn by motor a t s t e a d y s t a t e i s %f A . ’ , Ia_ ) ;

Scilab code Exa 4.38 Determining 1 external resistance 2 value of first resistance element 3 external resistance cut out in second step 4 total number of steps 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

clc ; v =230; // r a t e d v o l t a g e o f dc motor p =10000; // r a t e d power o f dc motor rf =115; // f i e l d r e s i s t a n c e ra =0.348; // n e t a r m a t u r e r e s i s t a n c e ifs = v / rf ; // s h u n t f i e l d c u r r e n t ia =( p / v ) - ifs ; // r a t e d a r m a t u r e c u r r e n t disp ( ’ c a s e a ’ ) ; rx1 =( v /(2* ia ) ) - ra ; printf ( ’ E x t e r n a l r e s i s t a n c e r e q u i r e d a t t h e t i m e o f s t a r t i n g i s %f ohms \n ’ , rx1 ) ; disp ( ’ c a s e b ’ ) ; Ea1 =v - ia *( rx1 + ra ) ; // c o u n t e r emf a t s t u d 1 r2 =( v - Ea1 ) /(2* ia ) ; // r e s i s t a n c e when h a n d l e i s moved t o 2 nd s t u d rx2 = r2 - ra ; // e x t e r n a l r e s i s t a n c e rc = rx1 - rx2 ; printf ( ’ R e s i s t a n c e t h a t must be c u t o u t i n f i r s t s t e p i s %f ohms \n ’ , rc ) ; disp ( ’ c a s e c ’ ) ; Ea2 =v - ia * rc ; // c o u n t e r emf a t s t u d 2 r3 =( v - Ea2 ) /(2* ia ) ; // r e s i s t a n c e when h a n d l e i s 169


20 21 22 23 24 25 26 27

moved t o 3 r d s t u d rc = rc - r3 ; printf ( ’ R e s i s t a n c e t h a t must be c u t o u t i n s e c o n d s t e p i s %f ohms \n ’ , rc ) ; disp ( ’ c a s e d ’ ) ; Ea3 =v - ia * rc ; // c o u n t e r emf a t s t u d 3 r4 =( v - Ea3 ) /(2* ia ) ; // r e s i s t a n c e when h a n d l e i s moved t o 4 t h s t u d rc = rc - r4 ; printf ( ’ R e s i s t a n c e t h a t must be c u t o u t i n t h i r d s t e p i s %f ohms \n ’ , rc ) ; disp ( ’ T o t a l number o f s t e p s i s 3 ’ ) ;

Scilab code Exa 4.39 Calculating 1 resistance of each step 2 voltage at which ors should be close 1 2 3 4 5 6 7 8 9 10 11 12 13 14

clc ; v =240; // r a t e d v o l t a g e o f dc s h u n t motor i =50; // r a t e d c u r r e n t o f dc s h u n t motor ra =0.2; // a r m a t u r e r e s i s t a n c e n =4; // number o f r e s i s t a n c e e l e m e n t N =1500; // r a t e d s p e e d o f motor vb =1; // pu b a s e v o l t a g e ia =1; // pu b a s e c u r r e n t rb = v / i ; // pu b a s e r e s i s t a n c e ra = ra / rb ; // p e r u n i t a r m a t u r e r e s i s t a n c e disp ( ’ c a s e a ’ ) ; ia1 =1.4; // pu maximum a l l o w a b l e a r m a t u r e c u r r e n t R = vb / ia1 ; // n e t r e s i s t a n c e al =( ra / R ) ^(1/ n ) ; // r a t i o o f t o t a l r e s i s t a n c e s on two a d j a c e n t s t u d s 15 r1 = R *(1 - al ) ; 16 printf ( ’ R e s i s t a n c e c u t o u t when h a n d l e i s a t s t u d 2 i s %f pu o r %f ohms \n ’ ,r1 , r1 * rb ) ; 17 r2 = al * r1 ; 170


18 19 20 21 22 23 24 25

26 27 28 29

30 31 32 33

34 35 36 37

38

printf ( ’ R e s i s t a n c e c u t o u t when h a n d l e i s a t s t u d 3 i s %f pu o r %f ohms \n ’ ,r2 , r2 * rb ) ; r3 = al * r2 ; printf ( ’ R e s i s t a n c e c u t o u t when h a n d l e i s a t s t u d 4 i s %f pu o r %f ohms \n ’ ,r3 , r3 * rb ) ; r4 = al * r3 ; printf ( ’ R e s i s t a n c e c u t o u t when h a n d l e i s a t s t u d 5 i s %f pu o r %f ohms \n ’ ,r4 , r4 * rb ) ; disp ( ’ c a s e b ’ ) ; ia2 = al * ia1 ; // pu minimum a r m a t u r e c u r r e n t // a t s t u d 1 a r m a t u r e c u r r e n t=i a 2 a f t e r t 1 where t i s t i m e r e c k o n e d from t h e i n s t a n t motor i s s w i t c h e d on Ea1 = vb - ia2 * R ; // c o u n t e r EMF a t s t u d 1 Va1 = Ea1 + ia2 * ra ; // v o l t a g e a c r o s s a t a r m a t u r e terminal at i n s t a n t t1 printf ( ’ The f i r s t c o n t a c t o r s h o u l d c l o s e a t %f pu o r %f V\n ’ ,Va1 , Va1 * v ) ; // a t s t u d 2 a r m a t u r e c u r r e n t=i a 2 a f t e r t 2 where t i s t i m e r e c k o n e d from t h e i n s t a n t motor i s s w i t c h e d on Ea2 = vb - ia2 *( R - r1 ) ; // c o u n t e r EMF a t s t u d 2 Va2 = Ea2 + ia2 * ra ; // v o l t a g e a c r o s s a t a r m a t u r e terminal at i n s t a n t t2 printf ( ’ The s e c o n d c o n t a c t o r s h o u l d c l o s e a t %f pu o r %f V\n ’ ,Va2 , Va2 * v ) ; // a t s t u d 3 a r m a t u r e c u r r e n t=i a 2 a f t e r t 3 where t i s t i m e r e c k o n e d from t h e i n s t a n t motor i s s w i t c h e d on Ea3 = vb - ia2 *( R - r1 - r2 ) ; // c o u n t e r EMF a t s t u d 3 Va3 = Ea3 + ia2 * ra ; // v o l t a g e a c r o s s a t a r m a t u r e terminal at i n s t a n t t3 printf ( ’ The t h i r d c o n t a c t o r s h o u l d c l o s e a t %f pu o r %f V\n ’ ,Va3 , Va3 * v ) ; // a t s t u d 4 a r m a t u r e c u r r e n t=i a 2 a f t e r t 4 where t i s t i m e r e c k o n e d from t h e i n s t a n t motor i s s w i t c h e d on Ea4 = vb - ia2 *( R - r1 - r2 - r3 ) ; // c o u n t e r EMF a t s t u d 4 171


39 Va4 = Ea4 + ia2 * ra ; // v o l t a g e 40 41 42 43 44 45 46 47 48 49 50 51

a c r o s s at armature

terminal at i n s t a n t t4 printf ( ’ The f o u r t h c o n t a c t o r s h o u l d c l o s e a t %f pu o r %f V\n ’ ,Va4 , Va4 * v ) ; disp ( ’ c a s e c ’ ) ; Ea = vb - ia * ra ; // pu f u l l l o a d c o u n t e r EMF n1 = Ea1 / Ea ; // pu s p e e d when h a n d l e i s a t s t u d 1 printf ( ’ Speed o f dc s h u n t motor when h a n d l e i s a t s t u d 1 i s %f pu o r %f rpm\n ’ ,n1 , n1 * N ) ; n2 = Ea2 / Ea ; // pu s p e e d when h a n d l e i s a t s t u d 2 printf ( ’ Speed o f dc s h u n t motor when h a n d l e i s a t s t u d 2 i s %f pu o r %f rpm\n ’ ,n2 , n2 * N ) ; n3 = Ea3 / Ea ; // pu s p e e d when h a n d l e i s a t s t u d 3 printf ( ’ Speed o f dc s h u n t motor when h a n d l e i s a t s t u d 3 i s %f pu o r %f rpm\n ’ ,n3 , n3 * N ) ; n4 = Ea4 / Ea ; // pu s p e e d when h a n d l e i s a t s t u d 4 printf ( ’ Speed o f dc s h u n t motor when h a n d l e i s a t s t u d 4 i s %f pu o r %f rpm\n ’ ,n4 , n4 * N ) ; disp ( ’ U s i n g a b o v e d a t a s k e t c h o f v a r i a t i o n o f a r m a t u r e c u r r e n t and s p e e d can be o b t a i n e d w i t h time ’ );

Scilab code Exa 4.40 Determining load torque for different cases 1 2 3 4 5 6 7 8 9 10 11

clc ; v =200; // r a t e d v o l t a g e o f s h u n t motor i =22; // r a t e d c u r r e n t o f dc s h u n t motor n1 =1000; // s p e e d a t which motor i s r u n n i n g rf =100; // f i e l d r e s i s t a n c e ra =0.1; // a r m a t u r e r e s i s t a n c e n2 =800; // r e d u c e d s p e e d a t which motor i s t o run iF = v / rf ; // f i e l d c u r r e n t ia =i - iF ; // a r m a t u r e c u r r e n t disp ( ’ c a s e a ’ ) ; // l o a d t o r q u e i s i n d e p e n d e n t o f s p e e d 172


12 Ea1 =v - ia * ra ; // c o u n t e r EMF a t 1 0 0 0 rpm 13 rg =( v - ia * ra -( n2 * Ea1 ) / n1 ) / ia ; 14 printf ( ’ A d d i t i o n a l r e s i s t a n c e i n s e r t e d i n a r m a t u r e 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33

c i r c u i t i s %f ohms \n ’ , rg ) ; printf ( ’ L o s s i n a d d i t i o n a l r e s i s t a n c e i s %f W\n ’ , ia ^2* rg ) ; disp ( ’ c a s e b ’ ) ; // l o a d t o r q u e i s d i r e c t l y p r o p o r t i o n a l t o s p e e d ia2 =( n2 / n1 ) * ia ; // a r m a t u r e c u r r e n t a t 800 rpm rg =( v - ia2 * ra -( n2 * Ea1 ) / n1 ) / ia2 ; printf ( ’ A d d i t i o n a l r e s i s t a n c e i n s e r t e d i n a r m a t u r e c i r c u i t i s %f ohms \n ’ , rg ) ; printf ( ’ L o s s i n a d d i t i o n a l r e s i s t a n c e i s %f W\n ’ , ia2 ^2* rg ) ; disp ( ’ c a s e c ’ ) ; // l o a d t o r q u e v a r i e s a s t h e s q u a r e o f s p e e d ia2 =( n2 / n1 ) ^2* ia ; // a r m a t u r e c u r r e n t a t 800 rpm rg =( v - ia2 * ra -( n2 * Ea1 ) / n1 ) / ia2 ; printf ( ’ A d d i t i o n a l r e s i s t a n c e i n s e r t e d i n a r m a t u r e c i r c u i t i s %f ohms \n ’ , rg ) ; printf ( ’ L o s s i n a d d i t i o n a l r e s i s t a n c e i s %f W\n ’ , ia2 ^2* rg ) ; disp ( ’ c a s e d ’ ) ; // l o a d t o r q u e v a r i e s a s t h e c u b e o f s p e e d ia2 =( n2 / n1 ) ^3* ia ; // a r m a t u r e c u r r e n t a t 800 rpm rg =( v - ia2 * ra -( n2 * Ea1 ) / n1 ) / ia2 ; printf ( ’ A d d i t i o n a l r e s i s t a n c e i n s e r t e d i n a r m a t u r e c i r c u i t i s %f ohms \n ’ , rg ) ; printf ( ’ L o s s i n a d d i t i o n a l r e s i s t a n c e i s %f W\n ’ , ia2 ^2* rg ) ;

Scilab code Exa 4.41 Finding motor speed 1 clc ; 2 V =240; // r a t e d

v o l t a g e o f dc s h u n t motor 173


3 4 5 6 7 8 9 10 11 12 13 14

n =800; // r a t e d s p e e d o f dc s h u n t motor i =50; // r a t e d c u r r e n t o f dc s h u n t motor ra =0.2; // a r m a t u r e r e s i s t a n c e pr =0.6; // r e d u c t i o n i n l o a d t o r q u e a s a f r a c t i o n o f f u l l load torque rg =2; // s e r i e s r e s i s t a n c e i n a r m a t u r e c i r c u i t fr1 =0.04; // w e a k e n i n g o f f i e l d f l u x a t f u l l l o a d fr2 =0.02; // w e a k e n i n g o f f i e l d f l u x a t 60% o f f u l l load Ea1 =V -( i * ra ) ; // c o u n t e r EMF a t r a t e d l o a d ia2 =( i * pr ) *((1 - fr1 ) /(1 - fr2 ) ) ; // a r m a t u r e c u r r e n t a t reduced load torque Ea2 =V - ia2 *( rg + ra ) ; // c o u n t e r EMF a t r e d u c e d l o a d torque n2 =( n * Ea2 *(1 - fr1 ) ) /( Ea1 *(1 - fr2 ) ) ; printf ( ’ Motor s p e e d a t r e d u c e d l o a d t o r q u e i s %f rpm ’ , n2 ) ;

Scilab code Exa 4.42 Finding motor speed 1 2 3 4 5 6 7 8 9

10 11 12 13

clc ; N =1000; // s p e e d o f dc s e r i e s motor v =250; // s u p p l y from mains i =50; // c u r r e n t drawn from mains r =0.6; // a r m a t u r e + f i e l d r e s i s t a n c e rg =4.4; // a d d i t i o n a l r e s i s t a n c e // f i e l d f l u x i s p r o p o r t i o n a l t o a r m a t u r e c u r r e n t Ea1 =v - i * r ; // c o u n t e r EMF a t 1 0 0 0 rpm // Ea2=v−(n2 / 2 0 ) ∗ ( r+r g ) where Ea2 i s c o u n t e r EMf a t s p e e d n2 . t a k i n g r a t i o o f Ea2 / Ea1 we o b t a i n a q u a d r a t i c e q u a t i o n i n n2 whose t e r m s a r e g i v e n by t1 =( Ea1 * i ) / N ; t2 =( N * i ) *(( r + rg ) /( N / i ) ) ; t3 = -( N * i * v ) ; p =[ t1 t2 t3 ]; 174


14 n = roots ( p ) ; 15 printf ( ’ New s p e e d o f motor i s %f rpm ’ , ceil ( n (2) ) ) ;

Scilab code Exa 4.43 Finding speed regulation 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

clc ; v =230; // r a t e d v o l t a g e o f dc s h u n t motor n1 =900; // s p e e d a t which motor i s r u n n i n g ia1 =2; // a r m a t u r e c u r r e n t a t n=900 rpm ra =0.5; // a r m a t u r e r e s i s t a n c e ia2 =20; // a r m a t u r e c u r r e n t a t r a t e d l o a d and r a t e d voltage Ea =v - ia1 * ra ; // c o u n t e r EMF a t no l o a d k =( Ea *60) /(2* %pi * n1 ) ; // c o n s t a n t term u s e d f o r c a l c u l a t i n g back EMF disp ( ’ c a s e a ’ ) ; rs =2; // r e s i s t a n c e i n s e r i e s w i t h a r m a t u r e rp =3; // r e s i s t a c e i n p a r a l l e l w i t h s e r i e s c o m b i n a t i o n o f r s and r a A = rp /( rp + rs ) ; wmo =(1/ k ) *( A *v - ia1 *( A * rs + ra ) ) ; // no−l o a d s p e e d wml =(1/ k ) *( A *v - ia2 *( A * rs + ra ) ) ; // f u l l −l o a d s p e e d sr =(( wmo - wml ) / wml ) *100; // p e r c e n t s p e e d r e g u l a t i o n printf ( ’ Speed r e g u l a t i o n f o r f i r s t c a s e i s %f p e r c e n t \n ’ , sr ) ; disp ( ’ c a s e b ’ ) ; rs =3; // r e s i s t a n c e i n s e r i e s w i t h a r m a t u r e wmo =(1/ k ) *( v - ia1 *( rs + ra ) ) ; // no−l o a d s p e e d wml =(1/ k ) *( v - ia2 *( rs + ra ) ) ; // f u l l −l o a d s p e e d sr =(( wmo - wml ) / wml ) *100; // p e r c e n t s p e e d r e g u l a t i o n printf ( ’ Speed r e g u l a t i o n f o r s e c o n d c a s e i s %f p e r c e n t \n ’ , sr ) ;

175


Scilab code Exa 4.44 Determining 1 maximum value of current and corresponding torque 2 ultimate speed and armature current 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

clc ; v =200; // r a t e d v o l t a g e o f dc s h u n t motor ra =0.1; // a r m a t u r e r e s i s t a n c e n =1000; // r u n n i n g s p e e d o f motor ia =50; // a r m a t u r e c u r r e n t a t n=1000 rpm re =0.1; // r e d u c t i o n i n f i e l d f l u x disp ( ’ c a s e a ’ ) ; Ea1 =v - ia * ra ; // i n i t i a l c o u n t e r EMF Ea2 = Ea1 *(1 - re ) ; // c o u n t e r EMF a f t e r r e d u c e d f i e l d flux iam =( v - Ea2 ) / ra ; printf ( ’ Maximum v a l u e o f a r m a t u r e c u r r e n t i s %f A\n ’ , iam ) ; T =( iam / ia ) *(1 - re ) ; printf ( ’ Torque c o r r e s p o n d i n g t o maximum a r m a t u r e c u r r e n t i s %f t i m e s i n i t i a l t o r q u e \n ’ ,T ) ; disp ( ’ c a s e b ’ ) ; ia2 =(1/(1 - re ) ) * ia ; printf ( ’ Armature c u r r e n t when t r a n s i e n t s a r e o v e r i s %f A\n ’ , ia2 ) ; Ea2 =v - ia2 * ra ; // c o u n t e r EMF when t r a n s i e n t s a r e over n2 =( Ea2 * n ) /( Ea1 *(1 - re ) ) ; printf ( ’ U l t i m a t e s p e e d a f t e r t r a n s i e n t s a r e o v e r i s %f rpm ’ , ceil ( n2 ) ) ;

Scilab code Exa 4.45 Determining armature current 1 2 3 4

clc ; clc ; v =200; // r a t e d v o l t a g e o f dc s h u n t motor ra =0.1; // a r m a t u r e r e s i s t a n c e 176


5 6 7 8 9 10 11 12 13 14

n =1000; // r u n n i n g s p e e d o f motor ia =50; // a r m a t u r e c u r r e n t a t n=1000 rpm re =0.1; // r e d u c t i o n i n f i e l d f l u x ia2 =(1/(1 - re ) ) * ia ; // a r m a t u r e c u r r e n t when t r a n s i e n t s are over Ea1 =v - ia2 * ra ; // c o u n t e r EMF when t r a n s i e n t s a r e over // w i t h s u d d e n i n c r e a s e from 0 . 9 ∗ f t o f ( f=f l u x ) , c o u n t e r EMF r i s e s t o Ea2 = Ea1 *(1/(1 - re ) ) ; i =( v - Ea2 ) / ra ; printf ( ’ Armature c u r r e n t i s %f A ’ ,i ) ; disp ( ’ S i n c e a r m a t u r e c u r r e n t i s n e g a t i v e , machine a c t s a s a g e n e r a t o r . Speed r e d u c e s t i l l c o u n t e r EMF becomes l e s s t h a n s u p p l y v o l t a g e , s o t h a t motor a c t i o n t a k e s p l a c e and t o r q u e b a l a n c e i s obtained ’)

Scilab code Exa 4.46 Determining new speed and armature current 1 2 3 4 5

clc ; v =220; // s u p p l y v o l t a g e n1 =2000; // s p e e d o f f a n motor ia1 =60; // c u r r e n t c o r r e s p o n d i n g t o n=2000 rpm // f l u x i s d i r e c t l y p r o p o r t i o n a l t o e x c i t i n g c u r r e n t and l o a d t o r q u e i n c r e a s e a s s q u a r e o f s p e e d 6 // f o u r f i e l d c o i l s a r e c o n n e c t e d i n two p a r a l l e l g r o u p s a l s o n ˆ2 i s d i r e c t l y p r o p o r t i o n a l t o armature c u r r e n t t h e r e f o r e 7 r = sqrt ((2* ia1 ^2) / n1 ^2) ; // r a t i o o f a r m a t u r e c u r r e n t c o r r e s p o n d i n g t o n2 and n2 where n2=new s p e e d 8 // c o u n t e r EMF a r e d i r e c t l y p r o p o r t i o n a l t o p r o d u c t n∗ i a and r a ( a r m a t u r e r e s i s t a n c e ) and r s ( s e r i e s ) r e s i s t a n c e are not given , t h e r e f o r e t a k i g r a t i o o f n1 ∗ i a 1 and n2 ∗ i a 2 we can d e t e r m i n e v a l u e o f n2 177


9 n2 = sqrt (( ia1 * n1 *2) / r ) ; 10 printf ( ’ New s p e e d i s %f rpm\n ’ , n2 ) ; 11 ia2 = n2 * r ; 12 printf ( ’ New a r m a t u r e c u r r e n t i s %f A\n ’ , ia2 ) ;

Scilab code Exa 4.48 Determining resistance inserted in shunt field 1 2 3 4 5 6 7 8 9 10

11 12 13 14 15 16 17 18 19 20

clc ; v =230; // r a t e d v o l t a g e o f dc s h u n t motor ra =0.4; // a r m a t u r e c i r c u i t r e s i s t a n c e rf =115; // f i e l d r e s i s t a n c e n1 =800; // i n i t i a l s p e e d n2 =1000; // f i n a l s p e e d ia1 =20; // a r m a t u r e c u r r e n t a t n=800 rpm // t o r q u e a t b o t h s p e e d i s same t h e r e f o r e f 1 ∗ i a 1=f 2 ∗ i a 2 where f= f i e l d f l u x t h e r e f o r e Ea1 =v - ia1 * ra ; // c o u n t e r EMF a t 800 rpm // i a 2=i a 1 ∗ k where k=f 1 / f 2 now w r i t i n g Ea2 ( c o u n t e r EMF a t 1 0 0 0 rpm ) i n t e r m s o f k and f i n d i n g v a l u e o f k by s o l v i n g q u a d r a t i c e q u a t i o n i n k whose are t1 = ia1 * ra * n1 ; t2 = - v * n1 ; t3 = Ea1 * n2 ; p =[ t1 t2 t3 ]; k = roots ( p ) ; if1 = v / rf ; // i n i t i a l f i e l d c u r r e n t if2 = if1 / k (2) ; // f i n a l f i e l d c u r r e n t c o r r e p o n d i n g t o n=1000rpm rs = v / if2 ; // new s h u n t f i e l d c i r c u i t r e s i s t a n c e re = rs - rf ; printf ( ’ R e s i s t a n c e t h a t must be i n s e r t e d i n s h u n t f i e l d c i r c u i t i s %f ohms \n ’ , floor ( re ) ) ;

178


Scilab code Exa 4.49 Determining new speed of motor 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

clc ; v =250; // r a t e d v o l t a g e o f dc s h u n t motor ra =0.5; // a r m a t u r e r e s i s t a n c e rf =250; // f i e l d r e s i s t a n c e n1 =600; // s p e e d o f motor i =21; // c u r r e n t drawn by motor when n=600 rpm re =250; // a d d i t i o n a l r e s i s t a n c e i n f i e l d c i r c u i t if1 = v / rf ; // f i e l d c u r r e n t ia =i - if1 ; // a r m a t u r e c u r r e n t Ea1 =v - ia * ra ; // c o u n t e r EMF a t n=600 rpm if2 = v /( rf + re ) ; // f i e l d c u r r e n t a f t e r a d d i t i o n o f external resistance ia2 = ia *( if1 / if2 ) ; // a r m a t u r e c u r r e n t a f t e r a d d i t i o n of external resistance Ea2 =v - ia2 * ra ; // c o u n t e r EMF a t new s p e e d n2 =( n1 * Ea2 * if1 ) /( Ea1 * if2 ) ; printf ( ’ New s p e e d o f motor i s %f rpm ’ , n2 ) ;

Scilab code Exa 4.50 Determining percentage change in field flux 1 2 3 4 5 6 7 8 9

clc ; v =250; // s u p p l y v o l t a g e i =50; // c u r r e n t drawn from s u p p l y ic =0.4; // p e r c e n t a g e i n c r e a s e i n s p e e d T =1.2; // r a t i o o f f i n a l and i n i t i a l t o r q u e n =1.4; // r a t i o o f f i n a l and i n i t i a l s p e e d ra =0.5; // a r m a t u r e r e s i s t a n c e Ea1 =v - i * ra ; // c o u n t e r EMF a t i n i t i a l s p e e d // i a 2 =(T2/T1 ) ∗ i a 1 ∗ k where k=f 1 / f 2 and T1 i s i n i t i a l t o r q u e and T2 i s f i n a l t o r q u e now w r i t i n g Ea2 ( 179


10 11 12 13 14 15 16

c o u n t e r EMF a t 1 0 0 0 rpm ) i n t e r m s o f k and f i n d i n g v a l u e o f k by s o l v i n g q u a d r a t i c e q u a t i o n i n k whose t e r m s a r e t1 = T * ra * i ; t2 = - v ; t3 = n * Ea1 ; p =[ t1 t2 t3 ]; k = roots ( p ) ; fr =(1 -(1/ k (2) ) ) *100; printf ( ’ P e r c e n t a g e r e d u c t i o n i n f i e l d f l u x i s %f p e r c e n t ’ , fr ) ;

Scilab code Exa 4.51 Determining number of series turns per pole to reduce speed 1 2 3 4 5 6 7 8 9 10

11 12 13 14 15 16 17

clc ; v =250; // r a t e d v o l t a g e o f dc s h u n t motor n1 =1200; // no l o a d s p e e d N =1000; // t u r n s p e r p o l e i n s h u n t f i e l d w i n d i n g n2 =900; // r e d u c e d s p e e d ia =100; // f u l l l o a d a r m a t u r e c u r r e n t rf =0.2; // s e r i e s f i e l d r e s i s t a n c e ra =0.1; // a r m a t u r e r e s i s t a n c e ar =0.04; // a r m a t u r e r e a c t i o n a s a f r a c t i o n o f main f i e l d m.m. f // At no l o a d c o u n t e r EMF=v t h e r e f o r e from magnetization curve given in f i g 4.76 f i e l d current is IF =[ 0.38 0.58 0.8 1.1 1.36 1.76]; EA =[125 180 215 250 275 300]; plot ( IF , EA ) ; xlabel ( ’ f i e l d c u r r e n t ’ ) ; ylabel ( ’ c o u n t e r EMF ’ ) ; title ( ’ M a g n e t i s i n g c u r v e ’ ) ; ifs1 =1.1; // f i e l d c u r r e n t 180


18 Ats1 = ifs1 * N ; // ampere t u r n s f o r f i e l d c u r r e n t 19 Ea2 =v - ia *( ra + rf ) ; // c o u n t e r EMF a t f u l l l o a d 20 // m a g n e t i z a t i o n c u r v e i s f o r 1 2 0 0 rpm , t h e r e f o r e

f u l l l o a d c o u n t e r EMF c o r r e s p o n d i n g t o i t 21 Ea2 = Ea2 *( n1 / n2 ) ; 22 // c o r r e s p o n d i n g t o a b o v e c o u n t e r EMF f i e l d

is current

from m a g n e t i z a t i o n c u r v e i s ifs2 =1.62; Ats2 =( ifs2 * N ) /(1 - ar ) ; // ampere t u r n s f o r f i e l d current at f u l l load 25 at = Ats2 - Ats1 ; // s e r i e s f i e l d 26 t = at / ia ; 27 printf ( ’ Number o f s e r i e s t u r n s p e r p o l e t o r e d u c e s p e e d t o %f rpm i s %f ’ ,n2 , ceil ( t ) ) ; 23 24

Scilab code Exa 4.52 Determining motor speed 1 2 3 4 5 6 7 8 9

10 11 12 13 14

clc ; v =240; // s u p p l y v o l t a g e n =1000; // s p e e d o f motor i =40; // c u r r e n t drawn from s u p p l y rf =0.2; // f i e l d r e s i s t a n c e ra =0.25; // a r m t u r e r e s i s t a n c e rd =0.3; // d i v e r t e r r e s i s t a n c e // t o r q u e i s c o n s t a n t f o r d i f f e r e n t s p e e d s // when d i v e r t e r i s put i n p a r a l l e l w i t h s e r i e s r e s i s t a n c e t h e n some f r a c t i o n o f a r m a t u r e c u r r e n t f l o w s through s e r i e s c i r c u i t t h i s current f o r c o n s t a n t t o r q u e i s g i v e n by ia2 = sqrt ( i ^2/( rd /( rf + rd ) ) ) ; Ea1 =v - i *( ra + rf ) ; // c o u n t e r EMf a t n=1000 rpm Ea2 =v - ia2 *( ra +(( rf * rd ) /( rf + rd ) ) ) ; // c o u n t e r EMF a t new s p e e d n2 =( Ea2 * n * i ) /( Ea1 *( rd /( rf + rd ) ) * ia2 ) ; printf ( ’ Motor s p e e d a f t e r d i v e r t e r i s put i n 181


p a r a l l e l with s e r i e s ceil ( n2 ) ) ;

f i e l d w i n d i n g i s %f rpm ’ ,

Scilab code Exa 4.53 Determining motor speed and current 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

clc ; v =230; // r a t e d v o l t a g e o f motor p =6; // number o f p o l e s f =4*10^ -3; // f l u x p e r p o l e i n Wb/A T =20; // l o a d t o r q u e n =800; // s p e e d a t T=20 N−m a =2; // f o r wave c o n n e c t e d c o n d u c t o r s number o f p a r a l l e l path z =432; // number o f c o n d u c t o r s r =1; // t o t a l r e s i s t a n c e o f motor // i t i s g i v e n t h a t T=kn ˆ 2 , t h e r e f o r e k = T / n ^2; // p r o p o r t i o n a l i t y o f c o n s t a n t r1 =( f * z * p ) /(60* a ) ; // r a t i o o f back EMF t o p r o d u c t o f s p e e d and a r m a t u r e c u r r e n t t =( r1 *60) /(2* %pi ) ; // r a t i o o f f u l l l o a d t o r q u e t o square o f armature c u r r e n t // a l s o Ea ( back EMf )=v−i a ∗ r a and r 1=Ea / ( n ∗ i a ) c o m p a i r i n g b o t h we g e t i a=v /(1+ r 1 ∗n ) ; // A l s o k ∗ n2ˆ2= t ∗ i a ˆ2 , i n t h i s e x p r e s s i o n p u t t i n g v a l u e o f i a and s o l v i n g q u a d r a t i c e q u a t i o n i n n2 t1 = sqrt ( k ) * r1 ; t2 = sqrt ( k ) ; t3 = - sqrt ( t ) * v ; p =[ t1 t2 t3 ]; n2 = roots ( p ) ; printf ( ’ Speed o f motor i s %f rpm\n ’ , n2 (2) ) ; ia = v /(1+ r1 * n2 (2) ) ; printf ( ’ Armature c u r r e n t when motor i s c o n n e c t e d t o r a t e d s u p p l y i s %f A ’ , ia ) ;

182


Scilab code Exa 4.54 Determining 1 armature current and speed of motor 2 value of external resistance inserted in field winding 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

19 20 21 22 23 24

clc ; v =230; // r a t e d v o l t a g e o f dc s h u n t motor n =1000; // r a t e d s p e e d o f motor rf =115; // f i e l d r e s i s t a n c e ra =0.5; // a r m a t u r e r e s i s t a n c e ia =4; // no l o a d a r m a t u r e c u r r e n t k =( v - ia * ra ) /(2* %pi * n /60) ; // c o n s t a n t term i n f o r m u l a o f back EMF disp ( ’ c a s e a ’ ) ; t =80; // l o a d t o r q u e ia2 = t / k ; // a r m a t u r e l o a d c u r r e n t Ea2 =v - ia2 * ra ; // c o u n t e r EMF c o r r e s p o n d i n g t o l o a d armature c u r r e n t printf ( ’ Armature c u r r e n t f o r g i v e n l o a d i s %f A\n ’ , ia2 ) ; n2 =( Ea2 *60) /( k *2* %pi ) ; printf ( ’ Speed o f motor a t g i v e n l o a d i s %f rpm\n ’ , n2 ); disp ( ’ c a s e b ’ ) ; pd =8000; // power d e v e l o p e d by motor n3 =1250; // s p e e d a t power i s d e v e l o p e d // d e t e r m i n i n g v a l u e o f a r m a t u r e c u r r e n t c o r r e s p o n d i n g t o power by s o l v i n g q u a d r a t i c e q u a t i o n whose t e r m s a r e t1 = ra ; t2 = - v ; t3 = pd ; p =[ t1 t2 t3 ]; ia3 = roots ( p ) ; Ea3 =v - ia3 (2) * ra ; // c o u n t e r EMF f o r l o a d a r m a t u r e current 183


25 k1 = k /( v / rf ) ; // c o n s t a n t term i n f o r m u l a o f back EMF

for f i e l d current = 1 A 26 ifn =( Ea3 *60) /(2* %pi * n3 * k1 ) ; 27 rfn = v / ifn ; 28 printf ( ’ E x t e r n a l r e s i s t a n c e t h a t must be i n s e r t e d

in

s e r i e s w i t h f i e l d w i n d i n g i s %f ohms ’ ,rfn - rf ) ;

Scilab code Exa 4.55 Determining current and percentage in flux 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

clc ; v =250; // r a t e d v o l t a g e o f dc s e r i e s motor ra =0.25; // a r m a t u r e r e s i s t a n c e rf =0.15; // s e r i e s f i e l d r e s i s t a n c e disp ( ’ c a s e a ’ ) ; t =80; // d e v e l o p e d t o r q u e n1 =1200; // s p e e d a t d e v e l o p e d t o r q u e // s o l v i n g q u a d r a t i c e q u a t i o n i n i a t1 = ra + rf ; t2 = - v ; t3 =( t *2* %pi * n1 ) /60; p =[ t1 t2 t3 ]; ia = roots ( p ) ; printf ( ’ C u r r e n t f o r d e v e l o p i n g g i v e n t o r q u e i s %f A\ n ’ , ia (2) ) ; disp ( ’ c a s e b ’ ) ; n2 =1800; ia2 = ia (2) /2; Ea1 =v - ia (2) *( ra + rf ) ; // c o u n t e r EMF c o r r e s p o n d i n g t o armature c u r r e n t o f case 1 Ea2 =v - ia2 *( ra + rf ) ; // c o u n t e r EMF c o r r e s p o n d i n g t o armature c u r r e n t i a 2 fr =( n1 * Ea2 ) /( n2 * Ea1 ) ; // r a t i o o f f l u x e s f o r two armatures current pr =(1 - fr ) *100; printf ( ’ P e r c e n t a g e r e d u c t i o n i n f l u x i s %f p e c e n t ’ , 184


pr ) ;

Scilab code Exa 4.56 Determining additional resistance to obtain rated torque 1 2 3 4 5 6 7 8 9 10

11 12 13 14 15

clc ; v =230; // r a t e d v o l t a g e o f dc s e r i e s motor n =1500; // s p e e d a t r a t e d o u t p u t i =20; // c u r r e n t drawn a t r a t e d o u t p u t ra =0.3; // a r m a t u r e r e s i s t a n c e rf =0.2; // f i e l d r e s i s t a n c e disp ( ’ c a s e a ’ ) ; // At s t a r t i n g Ea=0 , t h e r e f o r e re =( v / i ) -( ra + rf ) ; printf ( ’ E x t e r n a l r e s i s t a n c e t o be added i n motor a r m a t u r e c i r c u i t t o d e v e l o p r a t e d t o r q u e i s %f ohms \n ’ , re ) ; disp ( ’ c a s e b ’ ) ; n2 =1000; // s p e e d a t r a t e d t o r q u e h a s t o be developed Ea2 =( n2 / n ) *( v - i *( ra + rf ) ) ; // c o u n t e r EMF a t n=1000 rpm re =( v - Ea2 - i *( ra + rf ) ) / i ; printf ( ’ E x t e r n a l r e s i s t a n c e t o be added i n motor a r m a t u r e c i r c u i t t o d e v e l o p r a t e d t o r q u e i s %f ohms \n ’ , re ) ;

Scilab code Exa 4.57 Determining voltage and current in magnetic circuit 1 clc ; 2 // i n book v o l t a g e s

a r e c a l c u l a t e d f o r r =0.5 n o t f o r r = 0 . 4 ( a s a s k e d i n q u e s t i o n ) t h a t i s why a n s w e r is differing 185


3 4 5 6 7 8 9 10

11 12 13 14 15 16 17

18 19 20 21 22

v =230; // s u p p l y v o l t a g e n1 =800; // s p e e d a t s u p p l y v o l t a g e i =20; // c u r r e n t drawn from s u p p l y r =0.4; // dc s e r i e s motor r e s i s t a n c e n2 =1000; // r a i s e d s p e e d Ea1 =v - i * r ; // c o u n t e r EMF a t 800 rpm disp ( ’ c a s e a ’ ) ; // when m a g n e t i c c i r c u i t i s s a t u r a t e d f l u x i s c o n s t a n t . Under s t e a d y s t a t e c o n d i t i o n f u l l l o a d t o r q u e=t o r q u e a t any l o a d t h e r e f o r e i2 = i *( n2 / n1 ) ^2; // new c u r r e n t drawn from s u p p l y Ea2 = Ea1 *( n2 / n1 ) ; // c o u n t e r EMF a t 1 0 0 0 rpm vt = Ea2 + i2 * r ; printf ( ’ C u r r e n t f o r s a t u r a t e d m a g n e t i c c i r c u i t i s %f A\n ’ , i2 ) ; printf ( ’ V o l t a g e f o r s a t u r a t e d m a g n e t i c c i r c u i t i s %f V\n ’ , vt ) ; disp ( ’ c a s e b ’ ) ; // when m a g n e t i c c i r c u i t i s n o t s a t u r a t e d f l u x i s d i r e c t l y p r o p o r t i o n a l t o a r m a t u r e c u r r e n t and torque i s d i r e c t l y p r o p o r t i o n a l to square of armature c u r r e n t i3 =( n2 / n1 ) * i ; Ea3 =( n2 * Ea1 * i3 ) /( n1 * i ) ; // c o u n t e r EMF a t 1 0 0 0 rpm vt = Ea3 + i3 * r ; printf ( ’ C u r r e n t f o r u n s a t u r a t e d m a g n e t i c c i r c u i t i s %f A\n ’ , i3 ) ; printf ( ’ V o l t a g e f o r u n s a t u r a t e d m a g n e t i c c i r c u i t i s %f V\n ’ , vt ) ;

Scilab code Exa 4.58 Calculating armature current for different control methods 1 clc ; 2 s =4; // s p e e d r a n g e

186


3 ia =60; // a r m a t u r e c u r r e n t a t s p e e d n 4 disp ( ’ F i e l d f l u x c o n t r o l ’ ) ; 5 // For c o n s t a n t power l o a d Ea∗ I a i s c o n s t a n t 6 7

8 9 10

11 12

13 14

t h e r e f o r e i a i s c o n a t a n t a t 4∗ n printf ( ’ The a r m a t u r e c u r r e n t a t r e q u i r e d s p e e d i s %f A\n ’ , ia ) ; // For c o n s t a n t t o r q u e l o a d , s p e e d i s 4 t i m e s o f i n i t i a l s p e e d t h e r e f o r e f l u x c h a n g e s by 1/4 t i m e s and h e n c e t o m a i n t a i n t o r q u e c o n s t a n t a r m a t u r e c u r r e n t s h o u l d be f o u r t i m e s printf ( ’ The a r m a t u r e c u r r e n t a t r e q u i r e d s p e e d i s %f A\n ’ ,4* ia ) ; disp ( ’ Armature v o l t a g e c o n t r o l ’ ) ; // For c o n s t a n t power l o a d Ea∗ I a i s c o n s t a n t t h e r e f o r e at 4 times speed armature v o l t a g e i s 4 t i m e s and t h e a r m a t u r e c u r r e n t g e t s r e d u c e d by 1/4 t i m e s printf ( ’ The a r m a t u r e c u r r e n t a t r e q u i r e d s p e e d i s %f A\n ’ , ia /4) ; // For c o n s t a n t power l o a d Ea∗ I a i s c o n s t a n t t h e r e f o r e at 4 times speed , f l u x i s constant therefore // a r m a t u r e c u r r e n t i s c o n s t a n t printf ( ’ The a r m a t u r e c u r r e n t a t r e q u i r e d s p e e d i s %f A\n ’ , ia ) ;

Scilab code Exa 4.59 Determining speed range for full load and no load 2 minimum motor field current 1 2 3 4 5

clc ; v =220; // r a t e d v o l t a g e o f motor i =15; // r a t e d c u r r e n t o f motor ra =0.4; // n e t a r m a t u r e r e s i s t a n c e n =1500; // s p e e d f o r which m a g n e t i z a t i o n c u r v e i s given 187


6 IF =[ 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.2 1.45]; 7 EA =[120 160 197 210 220 228 232 236 243 248]; 8 plot ( IF , EA ) ; 9 xlabel ( ’ f i e l d c u r r e n t ’ ) ; 10 ylabel ( ’ v o l t a g e ’ ) ; 11 title ( ’ m a g n e t i s i n g c u r v e ’ ) ; 12 disp ( ’ c a s e a ’ ) ; 13 ifg1 =0.15; // i n i t i a l g e n e r a t o r f i e l d c u r r e n t 14 ifg2 =1.4; // f i n a l generator f i e l d current 15 ifm =0.6; // motor f i e l d c u r r e n t 16 // c o r r e s p o n d i n g i f g 1 c o u n t e r EMF o f g e n e r a t o r from

magnetization curve i s 17 Ea1 =60; 18 vd = i *( ra + ra ) ; // v o l t a g e d r o p i n two a r m a t u r e 19 20 21 22 23 24 25 26 27 28 29 30 31 32

33 34

resistance Ea2 = Ea1 - vd ; // c o u n t e r EMF o f motor // but motor c o u n t e r EMF f o r 0 . 6 A f i e l d c u r r e n t a t 1 5 0 0 rpm i s Ea3 =210; nmin1 =( Ea2 / Ea3 ) * n ; // minimum motor s p e e d // c o r r e s p o n d i n g i f g 2 c o u n t e r EMF o f g e n e r a t o r from magnetization curve i s Ea4 =247; Ea5 = Ea4 - vd ; // c o u n t e r EMF o f motor nmax1 =( Ea5 / Ea3 ) * n ; // maximum motor s p e e d sr = nmax1 / nmin1 ; // s p e e d r a n g e printf ( ’ Speed r a n g e f o r f u l l l o a d a r m a t u r e c u r r e n t i s %f : 1 \ n ’ , sr ) ; // f o r no l o a d g e n e r a t o r c o u n t e r EMF= motor c o u n t e r EMf nmin2 =( Ea1 / Ea3 ) * n ; // minimum motor s p e e d pr =(( nmin2 - nmin1 ) / nmin2 ) *100; printf ( ’ P e r c e n t s p e e d d r o p from no l o a d t o f u l l l o a d f o r c o n d i t i o n o f minimum s p e e d i s %f p e r c e n t \n ’ , pr ) ; // f o r maximum g e n e r a t o r f i e l d c u r r e n t g e n e r a t o r c o u n t e r EMF= motor c o u n t e r EMf a t no l o a d nmax2 =( Ea4 / Ea3 ) * n ; // maximum motor s p e e d 188


35 sr = nmax2 / nmin2 ; // s p e e d r a n g e 36 printf ( ’ Speed r a n g e f o r f u l l l o a d a r m a t u r e c u r r e n t 37 38

39 40 41 42 43 44 45 46 47

i s %f : 1 \ n ’ , sr ) ; pr =(( nmax2 - nmax1 ) / nmax2 ) *100; printf ( ’ P e r c e n t s p e e d d r o p from no l o a d t o f u l l l o a d f o r c o n d i t i o n o f maximum s p e e d i s %f p e r c e n t \n ’ , pr ) ; disp ( ’ c a s e b ’ ) // f o r g e n e r a t o r f i e l d c u r r e n t =1 A c o u n t e r EMF from magnetization curve i s Ea6 =236; Ea7 = Ea6 - vd ; // motor c o u n t e r EMF a t f u l l l o a d Ea8 =( Ea7 /(2* nmax1 ) ) * n ; printf ( ’ Motor c o u n t e r EMF f o r %f rpm i s %f V\n ’ ,n , Ea8 ) ; // C o r r e s p o n d i n g t o Ea8 , f i e l d c u r r e n t i s ifmi =0.25; printf ( ’ Minimum motor f i e l d c u r r e n t i s %f A\n ’ , ifmi ) ;

Scilab code Exa 4.60 Determining motor torque and motor speed 1 clc ; 2 p =4000; // r a t e d power o f 3 4 5 6 7 8 9 10 11

s e p a r a t e l y e x c i t e d dc s e r i e s motor v =230; // r a t e d v o l t a g e o f motor n =1000; // r a t e d s p e e d o f motor e =260; // a c v o l t a g e s u p p l i e d t o motor t h r o u g h converter i =2; // c u r r e n t drawn a t no l o a d no =1100; // s p e e d a t no l o a d ra =0.5; // a r m a t u r e r e s i s t a n c e vd =2; // v o l t a g e d r o p i n t h y r i s t o r de =30; // f i r i n g a n g l e d e l a y ia =20; // r a t e d a r m a t u r e c u r r e n t 189


12 k =((((2* sqrt (2) * e ) / %pi ) -( i * ra ) - vd ) *60) /(2* %pi * no ) ;

// c o n a t a n t term i n f o r m u l a o f back EMf 13 disp ( ’ c a s e a ’ ) ; 14 printf ( ’ Motor t o r q u e i s %f Nm\n ’ ,k * ia ) ; 15 disp ( ’ c a s e b ’ ) ; 16 wm =((((2* sqrt (2) * e * cosd ( de ) ) / %pi ) - vd ) / k ) -(( ra * ia ) / k ) 17

; printf ( ’ Motor s p e e d i s %f rpm ’ , ceil (( wm *60) /(2* %pi ) ) );

Scilab code Exa 4.61 Determining 1 speed and current for torque 2 speed and torque for given armature current 3 speed and torque for given armature current 1 clc ; 2 n =800; // s p e e d a t which m a g n e t i z a t i o n c u r v e 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

is given // c a s e a v =600; // dc v o l t a g e s o u r c e ra =0.3; // a r m a t u r e r e s i s t a n c e rf =0.25; // f i e l d r e s i s t a n c e T =300; // g i v e n t o r q u e VT =[ 200 375 443 485 510 518]; // t e r m i n a l v o l t a g e IF =[ 15 30 45 60 75 90 ]; // f i e l d c u r r e n t EA1 = VT + IF * ra ; // g e n e r a t e d EMF EA2 =v - IF *( ra + rf ) ; // g e n e r a t e d EMF f o r v=600 V N2 = n *( EA2 ./ EA1 ) ; // s p e e d f o r v= 600 V TE =( EA2 .* IF .* EA1 *60) ./(2* %pi * n * EA2 ) ; // t o r q u e subplot (221) ; plot ( TE , N2 ) ; xlabel ( ’ Torque (Nm) ’ ) ; ylabel ( ’ s p e e d ( rpm ) ’ ) ; title ( ’ s p e e d −t o r q u e ’ ) ; subplot (222) ; plot ( TE , IF ) ; 190


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

xlabel ( ’ Torque (Nm) ’ ) ; ylabel ( ’ c u r r e n t (A) ’ ) ; title ( ’ c u r r e n t −t o r q u e ’ ) ; disp ( ’ from c u r v e s , f o r a t o r q u e o f 300 Nm, s p e e d i s 940 rpm and c u r r e n t i s 5 2 . 5 A ’ ) ; disp ( ’ c a s e b ’ ) ; rd =0.25; // d i v e r t e r r e s i s t a n c e put i n p a r a l l e l w i t h s e r i e s c o m b i n a t i o n o f a r m a t u r e and f i e l d resistance ia1 =30; ia2 =60; // a r m a t u r e c u r r e n t s if1 = ia1 *( rd /( rd + rf ) ) ; // f i e l d c u r r e n t c o r r e s p o n d i n g to ia1 Ea1 =204.5; // g i v e n c o u n t e r EMF f o r f i e l d c u r r e n t Ea2 =v - ia1 *( ra +(( rd * rf ) /( rd + rf ) ) ) ; // c o u n t e r EMF f o r v o l t a g e s u p p l y o f 600 V n2 = n *( Ea2 / Ea1 ) ; printf ( ’ Speed a t %f A a r m a t u r e c u r r e n t i s %f rpm\n ’ , ia1 , n2 ) ; T =( Ea1 *60* ia1 ) /(2* %pi * n ) ; printf ( ’ Torque a t %f A a r m a t u r e c u r r e n t i s %f Nm\n ’ , ia1 , T ) ; if1 = ia2 *( rd /( rd + rf ) ) ; // f i e l d c u r r e n t c o r r e s p o n d i n g to ia1 Ea1 =384; // g i v e n c o u n t e r EMF f o r f i e l d c u r r e n t Ea2 =v - ia2 *( ra +(( rd * rf ) /( rd + rf ) ) ) ; // c o u n t e r EMF f o r v o l t a g e s u p p l y o f 600 V n2 = n *( Ea2 / Ea1 ) ; printf ( ’ Speed a t %f A a r m a t u r e c u r r e n t i s %f rpm\n ’ , ia2 , n2 ) ; T =( Ea1 *60* ia2 ) /(2* %pi * n ) ; printf ( ’ Torque a t %f A a r m a t u r e c u r r e n t i s %f Nm\n ’ , ia2 , T ) ; disp ( ’ c a s e c ’ ) ; ia3 =75; // a r m a t u r e c u r r e n t t =0.8; // t a p p i n g p e r c e n t a g e o f f i e l d w i n d i n g a s a f r a t i o n of f u l l s e r i e s turn ifl = t * ia3 ; // c o r r e s p o n d i n g f i e l d c u r r e n t 191


47 Ea =503; // g i v e n c o u n t e r EMF f o r f i e l d c u r r e n t 48 Ea2 =v - ia3 *( ra + t * rf ) ; // c o u n t e r EMF f o r v o l t a g e 49 50 51 52

s u p p l y o f 600 V n2 = n *( Ea2 / Ea ) ; printf ( ’ Speed a t %f A a r m a t u r e c u r r e n t i s %f rpm\n ’ , ia3 , n2 ) ; T =( Ea *60* ia3 ) /(2* %pi * n ) ; printf ( ’ Torque a t %f A a r m a t u r e c u r r e n t i s %f Nm\n ’ , ia3 , T ) ;

Scilab code Exa 4.62 Determining size of DC motor and hoist speed 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

clc ; ra =0.3; // a r m a t u r e r e s i s t a n c e n =0.7; // e f f i c i e n c y o f dc s h u n t motor l =800; // w e i g h t o f l o a d v =3; // s p e e d o f r a i s i n g a l o a d vi =230; // i n i t i a l v a l u e o f s u p p l y v o l t a g e vf =190; // f i n a l v a l u e o f s u p p l y v o l t a g e g =9.81; // a c c e l e r a t i o n due t o g r a v i t y f = l * g ; // r e s i s t i n g f o r c e due t o g r a v i t a t i o n a l p u l l p = f * v ; // power r e q u i r e d f o r l i f t i n g t h e l o a d P = p / n ; // power r a t i n g o f dc machine printf ( ’ R e q u i r e d r a t i n g o f power i s %f KW\n ’ ,P /1000) ; // f o r s u p p l y v o l t a g e o f 230 V Ea=230− i a ∗ r a f i n d i n g q u a d r a t i c e q u a t i o n i n i a whose t e r m s a r e t1 = ra ; t2 = - vi ; t3 = P ; p =[ t1 t2 t3 ] ; ia = roots ( p ) ; Ea = vf - ia (2) * ra ; // c o u n t e r EMF f o r s u p p l y v o l t a g e o f 190 V v =( Ea * ia (2) * n ) /( l * g ) ; 192


21

printf ( ’ New h o i s t s p e e d i s %f m/ s ’ ,v ) ;

Scilab code Exa 4.63 Determining additional resistance inserted in motor circuit 1 2 3 4 5 6 7 8 9 10

clc ; v1 =6; // h o i s t s p e e d i =60; // s e r i e s c u r r e n t v =600; // s u p p l y v o l t a g e r =0.5; // n e t r e s i s t a n c e g =9.81; // a c c e l e r a t i o n due t o g r a v i t y v2 =4; // r e d u c e d h o i s t s p e e d Ea1 =v - i * r ; // c o u n t e r EMf c o r r e s p o n d i n g t o v1 rx =(( v -(( v2 / v1 ) * Ea1 ) ) / i ) -r ; printf ( ’ E x t e r n a l r e s i s t a n c e t o be added i s %f ohms ’ , rx ) ;

Scilab code Exa 4.64 Determining speed of motor for given condition 1 2 3 4 5 6 7 8

clc ; v =450; // s u p p l y v o l t a g e i =25; // c u r r e n t drawn from s u p p l y n =600; // f u l l l o a d s p e e d z =500; // number o f c o n d u c t o r s f =1.7*10^ -2* sqrt ( i ) ; // f l u x p e r p o l e p =4; // number o f p o l e s a = p ; // number o f p a r a l l e l p a t h s f o r wave wound w i n d i n g i s same a s number o f p o l e s 9 Ea1 =( f * n * z * p ) /(60* a ) ; // c o u n t e r EMF 10 ra =( v - Ea1 ) / i ; // a r m a t u r e r e s i s t a n c e 11 // T=k ∗ f ∗ i a where f i s f l u x and i a i s a r m a t u r e c u r r e n t As p e r q u e s t i o n new t o r q u e i s h a l f o f i n i t i a l torque 193


12 i2 =(( i ^1.5) /2) ^(1/1.5) ; // new a r m a t u r e c u r r e n t 13 Ea2 =( v /2) - i2 * ra ; // c o u n t e r EMF f o r new a r m a t u r e

current 14 n2 =( Ea2 * f * sqrt ( i ) * n ) /( Ea1 * f * sqrt ( i2 ) ) ; 15 printf ( ’ New s p e e d a t which motor w i l l run i s %f rpm ’ , floor ( n2 ) ) ;

Scilab code Exa 4.66 Determining motor speed and armature current 1 clc ; 2 // a n s w e r 3 4 5 6 7 8 9 10 11 12 13 14 15 16

i s c a l c u l a t e d f o r t o r q u e =30 but i t i s a s k e d f o r t o r q u e =40 i . e why a n s w e r v a r i e s p =4; // number o f dc s e r i e s motor f =4*10^ -3; // r a t i o o f f l u x p e r p o l e t o a r m a t u r e current T =40; // t o r q u e o f f a n n =1000; // s p e e d o f motor a =2; // number o f p a r a l l e l p a t h f o r waave w i n d i n g z =480; // number o f c o n d u c t o r s ra =1; // a r m a t u r e r e s i s t a n c e v =230; // s u p p l y v o l t a g e re = sqrt (( T *2* %pi * a ) /( p * z * f * n ^2) ) ; // r a t i o o f a r m a t u r e c u r r e n t and new s p e e d // Ea=vt−i a ∗ r a w r i t i n g i a i n t e r m s o f n s o l v i n g f o r n ( n i s new s p e e d ) n2 = v /( re +(( p * f * z ) /(60* a ) ) ) ; printf ( ’ Motor s p e e d i s %f rpm\n ’ , n2 ) ; ia = re * n2 ; printf ( ’ Armature c u r r e n t i s %f A ’ , ia ) ;

Scilab code Exa 4.67 Determining shaft power input efficiency at rated load maximum efficiency and power output 194


1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

clc ; p =10000; // r a t e d power o f g e n e r a t o r v =250; // r a t e d v o l t a g e o f g e n e r a t o r l1 =400; // r o t a t i o n l l o s s e s ra =0.5; // a r m a t u r e r e s i s t a n c e rf =250; // s h u n t f i e l d r e s i s t a n c e ifl = v / rf ; // c o n s t a n t f i e l d c u r r e n t lc = ifl * rf + l1 ; // c o n s t a n t l o s s e s io = p / v ; // o u t p u t c u r r e n t o f g e n e r a t o r ia = io + ifl ; // a r m a t u r e c u r r e n t la = ia ^2* ra ; // a r m a t u r e c i r c u i t l o s s ps = p + lc + la ; // g e n e r a t o r s h a f t power i n p u t printf ( ’ G e n e r a t o r s h a f t power i n p u t i s %f W\n ’ , ps ) ; ng =(1 -(( lc + la ) / ps ) ) *100; printf ( ’ E f f i c i e n c y a t r a t e d l o a d i s %f p e r c e n t \n ’ , ng ); // a t maximum e f f i c i e n c y v a r i a b l e l o s s e s = c o n s t a n t losses ia = sqrt ( lc / ra ) ; // a r m a t u r e c u r r e n t a t maximum efficiency io = floor ( ia ) - ifl ; // o u t p u t c u r r e n t o f g e n e r a t o r po = v * io ; // o u t p u t power printf ( ’ G e n e r a t o r o u t p u t a t maximum e f f i c i e n c y i s %f W\n ’ , po ) ; pi = po +2* lc ; nm =(1 -(( lc + lc ) / pi ) ) *100; printf ( ’ Maximum e f f i c i e n c y i s %f p e r c e n t \n ’ , nm ) ;

Scilab code Exa 4.68 Determining efficiency and speed of motor 1 2 3 4 5

clc ; v =250; // r a t e d v o l t a g e o f s h u n t motor p =15000; // r a t e d power o f motor nm =0.88; // maximumu e f f i c i e n c y o f motor n =700; // s p e e d o f motor 195


6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

rf =100; // r e s i s t a n c e o f s h u n t f i e l d i =78; // c u r r e n t drawn by mains f =0.8; // f r a c t i o n o f r a t e d o u t p u t b e i n g d e l i v e r e d l =((1/ nm ) -1) * f * p ; // t o t a l l o s s e s // a t maximum l o s s e s c o n s t a n t l o s s e s = v a r i a b l e losses lc = l /2; // c o n s t a n t l o s s e s pi = f * p + l ; // i n p u t t o motor a t maximum e f f i c i e n c y il = pi / v ; // i n p u t l i n e c u r r e n t ia = il -( v / rf ) ; // a r m a t u r e c u r r e n t ra = lc / ia ^2; // a r m a t u r e r e s i s t a n c e ia2 =i -( v / rf ) ; // a r m a t u r e c u r r e n t a t g i v e n l o a d pi = i * v ; // t o t a l power i n p u t tl = ia2 ^2* ra + lc ; // t o t a l l o s s e s n1 =(1 -( tl / pi ) ) *100; // e f f i c i e n c y a t l i n e c u r r e n t o f 75 A Ea1 =v - ia * ra ; // c o u n t e r EMF Ea2 =v - ia2 * ra ; // c o u n t e r EMF c o r r e s p o n d i n g t o l i n e c u r r e n t o f 75 A // f i e l d c u r r e n t i s c o n s t a n t s o f l u x i s c o n s t a n t n2 =( Ea2 / Ea1 ) * n ; printf ( ’ E f f i c i e n c y a t l i n e c u r r e n t o f %d A i s %f p e r c e n t \n ’ ,i , ceil ( n1 ) ) ; printf ( ’ Speed a t l i n e c u r r e n t o f %d A i s %f rpm ’ ,i , floor ( n2 ) ) ;

Scilab code Exa 4.69 Determining 1 no load current 2 speed 3 armature current 1 2 3 4 5 6

clc ; p =10000; // r a t e d power o f t r a n s f o r m e r n =900; // s p e e d o f motor v =400; // r a t e d v o l t a g e o f motor ra =1; // a r m a t y r e r e s i s t a n c e rf =400; // f i e l d r e s i s t a n c e 196


7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29

ne =0.85; // e f f i c i e n c y a t r a t e d l o a d l =((1/ ne ) -1) * p ; // t o t a l l o s s e s disp ( ’ c a s e a ’ ) ; pi = p + l ; // power i n p u t il = pi / v ; // l i n e c u r r e n t ia = il -( v / rf ) ; // a r m a t u r e c u r r e n t la = ia ^2* ra ; // a r m a t u r e c i r c u i t l o s s e s lf = v *( v / rf ) ; // s h u n t f i e l d l o s s e s wo =l - la - lf ; // no l o a d l o s s e s iao = wo / v ; // no l o a d c u r r e n t n e g l e c t i n g a r m a t u r e l o s s e s a t no l o a d printf ( ’ No l o a d a r m a t u r e c u r r e n t i s %f A\n ’ , iao ) ; disp ( ’ c a s e b ’ ) ; Ea1 =v - ia * ra ; // c o u n t e r EMF a t r a t e d l o a d il =20; // c u r r e n t drawn by motor ia = il -( v / rf ) ; // a r m a t u r e c u r r e n t Ea2 =v - ia * ra ; // c o u n t e r EMF a t l i n e c u r r e n t o f 20 A n2 =( Ea2 / Ea1 ) * n ; printf ( ’ Speed o f motor w h i l e d r a w i n g c u r r e n t o f %d A from mains i s %f rpm\n ’ ,il , ceil ( n2 ) ) ; disp ( ’ c a s e c ’ ) ; k =( Ea1 *60) /(2* %pi * n ) ; // c o n s t a n t term i n c o u n t e r EMF f o r m u l a T =98.5; // e l e c t r o m a g n e t i c t o r q u e ia = T / k ; printf ( ’ Armature c u r r e n t a t g i v e n t o r q u e i s %f A ’ , ceil ( ia ) ) ;

Scilab code Exa 4.70 Determining efficiency of motor 1 2 3 4 5

clc ; v =240; // r a t e d v o l t a g e o f motor and s u p p l y v o l t a g e i =5.2; // l i n e c u r r e n t p =10000; // r a t e d power o f motor no =1200; // no l o a d s p e e d 197


6 7 8 9 10 11

12 13 14 15 16 17 18 19

ra =0.25; // a r m a t u r e r e s i s t a n c e rf =160; // f i e l d r e s i s t a n c e ifl = v / rf ; // c o n s t a n t f i e l d c u r r e n t iao =i - ifl ; // no l o a d a r m a t u r e c u r r e n t wo = v * iao - iao ^2* ra ; // no l o a d r o t a t i o n a l l o s s e s // by u s i n g e q u a t i o n o f e l e c t r o m a g n e t i c power s o l v i n g q u a d r a t i c equation in armature c u r r e n t whose t e r m s a r e t1 = ra ; t2 = - v ; t3 = p + wo ; P =[ t1 t2 t3 ]; ia = roots ( P ) ; pi =( v - ia (2) * ra ) * ia (2) + ia (2) ^2* ra + ifl * v ; // motor input nm =( p / pi ) *100; printf ( ’ Motor e f f i c i e n c y a t r a t e d l o a d i s %f p e r c e n t ’ , nm ) ;

Scilab code Exa 4.71 Determining armature voltage drop 1 2 3 4 5 6 7 8

clc ; v =440; // r a t e d v o l t a g e o f mootor no =2000; // no l o a d s p e e d n1 =1000; // s p e e d a t f u l l l o a d t o r q u e Tl =0.5; // l o a d t o r q u e a s a f r a c t i o n o f r a t e d t o r q u e n2 =1050; // i n c r e a s e d s p e e d due t o r e d u e d t o r q u e // f i e l d c u r r e n t i s c o n s t a n t s o f l u x i s c o n s t a n t // s i n c e t o r q u g e t s r e d u c e d by h a l f new a r m a t u r e c u r r e n t a l s o g e t s r e d u c e d h a l f i . e i a 2=i a 1 / 2 ; 9 vd =( v *( n2 - n1 ) ) /( n2 -( n1 /2) ) ; 10 printf ( ’ Armature v o l t a g e d r o p a t f u l l l o a d i s %d V ’ , vd ) ;

198


Scilab code Exa 4.72 Determining shaft torque and efficiency of motor 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

clc ; v =230; // s o u r c e v o l t a g e ra =0.1; // r e s i s t a n c e o f a r m a t u r e ia =100; // a r m a t u r e c u r r e n t n =1600; // s p e e d o f dc s h u n t motor wl =300; // f r i c t i o n and w i n d a g e l o s s e s lo =1200; // no l o a d c o r e l o s s lc =2500; // c o p p e r l o s s e s Ls =0.01; // s t r a y l o s s e s a s a f r a c t i o n o f o u t p u t Ea =v - ia * ra ; // c o u n t e r EMF pe = Ea * ia ; // e l e c t r o m a g n e t i c power wo = wl + lo ; // no l o a d r o t a t i o n a l l o s s e s po = pe - wo ; // s h a f t power + s t r a y l o a d l o s s e s psh = po /(1+ Ls ) ; Tsh =( psh *60) /(2* %pi * n ) ; printf ( ’ S h a f t t o r q u e i s %f Nm\n ’ , Tsh ) ; pi = pe + lc ; // power i n p u t t o motor nm =( psh / pi ) *100; printf ( ’ Motor e f f i c i e n c y i s %f p e r c e n t ’ , nm ) ;

Scilab code Exa 4.73 Determining shaft torque shaft power and motor efficiency 1 clc ; 2 // s h a f t power i s 3 4 5 6

g i v e n l i t t l e b i t more t h a n a c t u a l value in question w1 =25; w2 =9; // s p r i n g b a l a n c e r e a d i n g s i n kg d =19.5*10^ -2; // o u t s i d e p u l l e y d i a m e t e r t =0.5*10^ -2; // b e l t t h i c k n e s s 199


7 8 9 10 11 12 13 14 15 16 17

g =9.81; // a c c e l e r a t i o n due t o g r a v i t y n =1500; // motor s p e e d v =230; // a p p l i e d v o l t a g e il =12.5; // l i n e c u r r e n t Ts =( w1 - w2 ) *(( d /2) +( t /2) ) * g ; printf ( ’ S h a f t t o r q u e i s %f Nm\n ’ , Ts ) ; psh =(2* %pi * n * Ts ) /60; printf ( ’ S h a f t power i s %f W\n ’ , psh ) ; pi = v * il ; // motor i n p u t nm =( psh / pi ) *100; printf ( ’ Motor e f f i c i e n c y a t r a t e d l o a d i s %f p e r c e n t ’ , nm ) ;

Scilab code Exa 4.74 Determining KW output efficiency and percentage change in speed from no load to full load 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

clc ; v =400; // r a t e d v o l t a g e o f dc s h u n t motor io =5; // c u r r e n t a t no l o a d ra =0.5; // a r m a t u r e r e s i s t a n c e rf =200; // f i e l d r e s i s t a n c e i =50; // c u r r e n t a t f u l l l o a d ifl = v / rf ; // c o n s t a n t s h u n t f i e l d c u r r e n t iao = io - ifl ; // no l o a d a r m a t u r e c u r r e n t wo = v * iao - iao ^2* ra ; // no l o a d r o t a t i o n a l l o s s e s ia =i - ifl ; // f u l l l o a d a r m a t u r e c u r r e n t la = ia ^2* ra ; // f u l l l o a d a r m a t u r e c i r c u i t l o s s e s lf = v * ifl ; // c o n s t a n t s h u n t f e l d l o s s e s tl = la + lf + wo ; // t o t a l f i e l d l o s s e s pi = i * v ; // motor i n p u t a t f u l l l o a d nm =(1 -( tl / pi ) ) *100; printf ( ’ Output power i s %f KW\n ’ ,( pi - tl ) /1000) ; printf ( ’ E f f i c i e n c y on f u l l l o a d i s %f p e r c e n t \n ’ , nm ) ; 18 Ea1 =v - iao * ra ; // no l o a d c o u n t e r EMF 200


19 Ea2 =v - ia * ra ; // f u l l l o a d c o u n t e r EMF 20 pr =(( Ea1 - Ea2 ) / Ea1 ) *100; // Ea i s d i r e c t l y

p r o p o r t i o a l t o s p e e d s o p e r c e n t a g e c h a n g e i n Ea i s same a s p e r c e n t a g e i n s p e e d ; 21 22

printf ( ’ P e r c e n t a g e c h a n g e i n s p e e d from no l o a d t o f u l l l o a d i s %f p e r c e n t ’ , pr ) ;

Scilab code Exa 4.75 Determining full load efficiency of motor 1 2 3 4 5 6 7 8 9 10 11 12

13 14 15 16 17 18 19 20 21 22 23

clc ; v =400; // r a t e d v o l t a g e o f dc s h u n t motor p =20000; // r a t e d power o f motor i =2.5; // no l o a d c u r r e n t ra =0.5; // a r m a t u r e r e s i s t a n c e rf =800; // f i e l d c u r r e n t vb =2; // v o l t a g e d r o p i n b r u s h ifl = v / rf ; // c o n s t a n t s h u n t f i e l d c u r r e n t iao =i - ifl ; // no l o a d a r m a t u r e c u r r e n t wo = v * iao - iao ^2* ra ; // no l o a d r o t a t i o n a l l o s s e s tl = wo + v * ifl ; // t o t a l l o s s e s // by u s i n g e q u a t i o n o f power i n p u t= o u t p u t power + l o s s e s , s o l v i n g q u a d r a t i c equation in armature c u r r e n t whose t e r m s a r e t1 = ra ; t2 = vb - v ; t3 = p + tl - v *( v / rf ) ; P =[ t1 t2 t3 ]; ia = roots ( P ) ; lo = ia (2) ^2* ra ; // a r m a t u r e ohmic l o s s e s lb = ia (2) * vb ; // b r u s h d r o p l o s s tl = tl + lo + lb ; // t o t a l l o s s e s a t r a t e d l o a d pi = p + tl ; // i n p u t power nm =( p / pi ) *100; printf ( ’ F u l l l o a d e f f i c i e n c y i s %f p e r c e n t ’ , nm ) ; 201


Scilab code Exa 4.77 Determining efficiency of both machines 1 clc ; 2 // Hopkinson ’ s method g a v e f o l l o w i n g 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

r e s u l t f o r two

i d e n t i c a l dc s h u n t m a c h i n e s v =230; // l i n e v o l t a g e il =30; // l i n e c u r r e n t e x c l u d i n g b o t h f i e l d c u r r e n t s ia =230; // motor a r m a t u r e c u r r e n t ifl1 =4; ifl2 =5; // f i e l d c u r r e n t s ra =0.025; // a r m a t u r e c u r r e n t // from f i g 4 . 8 5 ig = ia - il ; // g e n e r a t o r a r m a t u r e c u r r e n t la1 = ig ^2* ra ; // a r m a t u r e c i r c u i t l o s s e s i n g e n e r a t o r la2 = ia ^2* ra ; // a r m a t u r e c i r c u i t l o s s e s i n motor pd = v * il ; // power drawn from s u p p l y ( e x c l u d i n g f i e l d loss ) wo = pd - la1 - la2 ; // no l o a d r o t a t i o n a l l o s s e s f o r b o t h machines pg = v * ig ; // g e n e r a t o r o u t p u t k tl =( wo /2) + v * ifl2 + la1 ; // t o t a l l o s s e s f o r g e n e r a t o r ng =(1 -( tl /( tl + pg ) ) ) *100; pi = v *( ia + ifl1 ) ; // i n p u t power f o r motor tl =( wo /2) + v * ifl1 + la2 ; // t o t a l l o s s e s f o r motor nm =(1 -( tl / pi ) ) *100; printf ( ’ Motor e f f i c i e n c y i s %f p e r c e n t \n ’ , nm ) ; printf ( ’ G e n e r a t o r e f f i c i e n c y i s %f p e r c e n t \n ’ , ng ) ; // I f b o t h machine a r e assumed t o have same e f f i c i e n c y then n = sqrt ( ig / ia ) *100; printf ( ’ E f f i c i e n c y o f machine i s %d p e r c e n t ’ ,n ) ;

Scilab code Exa 4.78 Determining efficiency of both machines 202


1 clc ; 2 // f i e l d s 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

t e s t on two s i m i l a r machine g a v e f o l l o w i n g test iam =60; // motor a r m a t u r e c u r r e n t vam =500; // v o l t a g e a c r o s s a r m a t u r e vfm =40; // v o l t a g e a c r o s s f i e l d vt =450; // t e r m i n a l v o l t a g e f o r g e n e r a t o r io =46; // o u t p u t c u r r e n t f o r g e n e r a t o r vfg =40; // v o l t a g e a c r o s s f i e l d ra =0.25; // a r m t u r e r e s i s t a n c e pi =( vam + vfm + vfg ) * iam ; // power i n p u t t o w h o l e s e t pog = vt * io ; // g e n e r a t o r o u t p u t tl = pi - pog ; // t o t a l l o s s i n w h o l e s e t poh = iam ^2* ra + iam *( vfm + vfg ) + io ^2* ra ; // t o t a l ohmic losses wo =( tl - poh ) /2; // no l o a d r o a t i o n a l l o s s e s f o r e a c h machines pim =( vam + vfm ) * iam ; // motor power i n p u t plm = iam ^2* ra + iam * vfm + wo ; // t o t a l motor l o s s nm =(1 -( plm / pim ) ) *100; printf ( ’ Motor e f f i c i e n c y i s %f p e r c e n t \n ’ , nm ) ; plg = io ^2* ra + iam * vfm + wo ; // t o t a l motor l o s s pgm = pog + plg ; // g e n e r a t o r i n p u t ng =(1 -( plg / pgm ) ) *100; printf ( ’ G e n e r a t o r e f f i c i e n c y i s %f p e r c e n t ’ , ng ) ;

Scilab code Exa 4.81 Determining field current and power gain at rated output and when compensation is zero 1 2 3 4 5 6

clc ; p =3000; // power o f a m p l i d y n e v =300; // v o l t a g e o f a m p l i d y n e w =200; // a n g u l a r v e l o c i t y o f a m p l i d y n e rf =50; // f i e l d r e s i s t a n c e ra =5; // a r m a t u r e r e s i s t a n c e 203


rc =1; // c o m p e n s a t i n g w i n d i n g r e s i s t a n c e kqf =250; kdq =100; kqd =80; // v o l t a g e c o n s t a n t s A =( kdq * kqf ) /( ra * rf ) ; // v o l t a g e a m p l i f i c a t i o n f a c t o r id = p / v ; // r a t e d c u r r e n t vf =( v + id *( ra + rc ) ) / A ; // f i e l d v o l t a g e ifl = vf / rf ; // f i e l d c u r r e n t pk =( v * id ) /( vf * ifl ) ; // power g a i n printf ( ’ F i e l d c u r r e n t i s %f A\n ’ , ifl ) ; printf ( ’ Power g a i n a t r a t e d o u t p u t i s %f \n ’ , pk ) ; // when c o m p e n s a t i o n i s z e r o vf =( v + id *((( kdq * kqd ) / ra ) + ra ) ) / A ; // f i e l d v o l t a g e ifl = vf / rf ; printf ( ’ F i e l d c u r r e n t a t z e r o c o m p e n s a t i o n i s %f A\n ’ , ifl ) ; 22 pk =( v * id ) /( vf * ifl ) ; // power g a i n 23 printf ( ’ Power g a i n a t r a t e d o u t p u t a t z e r o c o m p e n s a t i o n i s %f \n ’ , pk ) ; 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

Scilab code Exa 4.82 To plot the external characteristics for given field current 1 clc ; 2 // p l o t 3 4 5 6 7 8 9 10 11

f o r open c i r c u i t c h a r a c t e r i s t i c s i s g i v e n i n f i g 4.10 IF =[ 0 11.5 23 36.5 59.5 79 110 160]; EA =[0 40 80 120 160 180 200 220 ]; subplot (221) ; plot ( IF , EA ) ; xlabel ( ’ f i e l d ATs ’ ) ; ylabel ( ’ v o l t a g e ’ ) ; title ( ’ m a g n e t i s i n g c u r v e ’ ) ; nf =800; // f i e l d w i n d i n g t u r n s rd =0.5; // t o t a l a r m a t u r e r e s i s t a n c e a l o n g d−a x i s 204


12 ifl =0.2; // f i e l d w i n d i n g c u r r e n t 13 d =10; // p r o d u c t o f ( d i f f e r e n c e b e t w e e n mmf o f

14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31

c o m p e n s a t i n g w i n d i n g and a r m a t u r e mmf a l o n g d− c i r c u i t ) and l o a d c u r r e n t nf1 = nf * ifl ; // f i e l d w i n d i n g t u r n s f o r f i e l d c u r r e n t o f 200mA il = nf1 / d ; // maximum l o a d c u r r e n t printf ( ’ Maximum f i e l d c u r r e n t i s %d A\n ’ , il ) ; IL =[0 2 4 6 8 10 12 14 16]; // l o a d c u r r e n t s ATD = nf1 - d * IL ; // n e t d−a x i s ATs disp ( ’ Net d−a x i s ATs i s ’ ) ; disp ( ATD ) ; // c o r r e s p o n d i n g t o e a c h ATD open c i r c u i t EMF i s o b t a i n e d from m a g n e t i s i n g c u r v e EO =[220 213 204.7 194 180.5 161.4 128 70 0 ]; // open c i r c u i t EMF VRD = rd * IL ; // d−a x i s r e s i s t a n c e d r o p VO = EO - VRD ; disp ( ’ Output v o l t a g e (V) i s ’ ) ; disp ( VO ) ; subplot (222) ; plot ( IL , VO ) ; xlabel ( ’ l o a d c u r r e n t (A) ’ ) ; ylabel ( ’ Output v o l t a g e ( v ) ’ ) ; title ( ’ Output v o l t a g e v s Load c u r r e n t ’ ) ;

Scilab code Exa 4.83 Determining field winding current 1 2 3 4 5 6

clc ; // from f i g 4 . 7 9 vo =206; // o u t p u t v o l t a g e il =8; // l o a d c u r r e n t ifl =0.2; // f i e l d c u r r e n t Eo =280; // open c i r c u i t v o l t a g e f o r which f i e l d w i n d i n g c u r r e n t i s t o be d e t e r m i n e d 205


7 r =0.5; // n e t r e s i s t a n c e 8 n =800; // d−a x i s ampere t u r n s 9 // w i t h s a t u r a t i o n i g n o r e d o u t p u t v o l t a g e vd i s

g i v e n by vd=(n ∗ i f −10∗ i l )K− i l ∗ r 10 K =( vo + il * r ) /(800* ifl -10* il ) ; // s l o p e

of straight

l i n e in curve 11 ifl = Eo /( K * n ) ; 12 printf ( ’ For g i v e n open c i r c u i t v o l t a g e f i e l d c u r r e n t i s %f mA ’ , ifl *1000) ;

Scilab code Exa 4.84 Determining output voltage of generator and reference voltage of potentiometer 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

clc ; A =100; // a m p l i d y n e v o l t a g e a m p l i f i c a t i o n vo =200; // DC g e n e r a t o r o u t p u t v o l t a g e rf =125; // f i e l d w i n d i n g r e s i s t a n c e vfb =0.1; // r a t i o o f f e e d b a c k v o l t a g e t o o u t p u t voltage of generator vr =50; // r e f e r e n c e v o l t a g e // a m p l i d y n e o u t p u t v o l t a g e , Va =( vr−v f b ∗ v t ) ∗A // i g=va / r f i g i s g e n e r a t o r f i e l d c u r r e n t // vog=i g ∗ vo vog i s g e n e r a t o r o u t p u t v o l t a g e −1 // s i m p l i f y i n g 1 we g e t vt =( vr * A ) /(( vfb * A ) +( rf / vo ) ) ; printf ( ’ Output v o l t a g e o f g e n e r a t o r i s %f V\n ’ , vt ) ; // now f e e d b a c k v o l t a g e i s r e d u c e d t o z e r o vr =( vt * rf ) /( A * vo ) ; printf ( ’ R e f e r e n c e v o l t a g e t o o b t a i n r e q u i r e d o u t p u t g e n e r a t o r v o l t a g e i s %f V ’ , vr ) ;

Scilab code Exa 4.85 Determining reference voltage of potentiometer 206


1 2 3 4 5 6 7 8 9 10 11 12

clc ; g1 =1.5; // g a i n f a c t o r o f a m p l i f i e r g2 =80; // g a i n f a c t o r o f g e n e r a t o r vo =250; // o u t p u t v o l t a g e a t no l o a d s =0.2; // f e e d b a c k p o t e n t i o m e t e r s e t t i n g // f o r g e n e r a t e d v o l t a g e= 80V f i e l d c u r r e n t i s 1 A ifl = vo / g2 ; // f i e l d c u r r e n t f o r g e n e r a t e d v o l t a g e= 250V vi = ifl / g1 ; // a m p l i f i e r i n p u t v o l t a g e f o r f i e l d c u r r e n t c o r r e s p o n d i n g t o g e n e r a t e d v o l t a g e= 250V vfb = s * vo ; // f e e d b a c k v o l t a g e vr = vfb + vi ; printf ( ’ R e f e r e n c e v o l t a g e f o r g i v e n p o t e n t i o m e t e r s e t t i n g i s %f V\n ’ , vr ) ; printf ( ’ When f e e d b a c k s e t t i n g i s z e r o , r e f e r e n c e v o l t a g e i s %f V ’ , vi ) ;

Scilab code Exa 4.86 Determining amplifier gain 1 2 3 4 5 6 7 8 9 10 11 12 13 14

clc ; p =4000; // r a t e d power o f g e n e r a t o r v =250; // r a t e d v o l t a g e o f g e n e r a t o r ra =0.25; // a r m a t u r e r e s i s t a n c e rf =100; // f i e l r e s i s t a n c e vr =20; // i m p r o v i n g f a c t o r f o r v o l t a g e r e g u l a t i o n g1 =120; // g e n e r a t o r g a i n // a f t e r d e r i v i n g r e q u i r e d e x p r e s s i o n il = p / v ; // l o a d c u r r e n t vgr =(( il * ra ) / v ) *(1/ vr ) ; // pu f u l l l o a d g e n e r a t o r regulation dvt = - vgr * v ; // d e c r e a s e i n t e r m i n a l v o l t a g e o f g e n e r a t o r from no l o a d t o f u l l l o a d disp ( ’ c a s e a ’ ) ; s =0.1; // f e e d b a c k p o t e n t i o m e t e r s e t t i n g A =( - dvt * rf - il * ra * rf ) /( dvt * s * g1 ) ; 207


15 printf ( ’ A m p l i f i e r g a i n i s %f\n ’ ,A ) ; 16 disp ( ’ c a s e b ’ ) ; 17 s =1; // f e e d b a c k p o t e n t i o m e t e r s e t t i n g 18 A =( - dvt * rf - il * ra * rf ) /( dvt * s * g1 ) ; 19 printf ( ’ A m p l i f i e r g a i n i s %f\n ’ ,A ) ;

Scilab code Exa 4.87 Determining 1 no load rotational losses 2 motor output 3 stall torque 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

clc ; v =48; // s u p p l y v o l t a g e n =2400; // s p e e d o f permanent magnet DC motor i =0.8; // c u r r e n t drawn by motor ra =1; // a r m a t u r e r e s i s t a n c e o f motor disp ( ’ c a s e a ’ ) ; Ea =v - i * ra ; // g e n e r a t e d EMF l = Ea * i ; printf ( ’ No l o a d r o t a t i o n a l l o s s e s i s %f W\n ’ ,l ) ; disp ( ’ c a s e b ’ ) ; km =( Ea *60) /(2* %pi * n ) ; // s p e e d v o l t a g e c o n s t a n t v =40; // s u p p l y v o l t a g e n1 =1600; // s p e e d a t s u p p l y v o l t a g e Ea =( km *2* %pi * n1 ) /60; // g e n e r a t e d EMF ia =( v - Ea ) / ra ; // new a r m a t u r e c u r r e n t pe = Ea * ia ; // E l e c t r o m a g n e t i c power d e v e l o p e d po = pe - l ; printf ( ’ Output power i s %f W\n ’ , po ) ; disp ( ’ c a s e c ’ ) ; v =20; // s u p p l y v o l t a g e // when motor s t a l l s Ea=0 ia = v / ra ; // s t a l l c u r r e n t T = km * ia ; printf ( ’ S t a l l t o r q u e i s %f Nm ’ ,T ) ;

208


Chapter 5 Polyphase synchronous machine

Scilab code Exa 5.1 Determining percentage voltage regulation of alternator by different methods 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

clc ; v =220; // r a t e d v o l t a g e o f a l t e r n a t o r f =50; // f r e q u e n c y o f s u p p l y r =0.06; // r e s i s t a n c e p e r p h a s e p =6; // number o f p o l e s i =40; // f u l l l o a d c u r r e n t pf =0.8; // l a g g i n g power f a c t o r vt = v / sqrt (3) ; // r a t e d p e r p h a s e v o l t a g e IF =[ 0.2 0.4 0.6 0.8 1 1.2 1.4 1.8 2.2 2.6 3 3.4]; EA =[ 29 58 87 116 146 172 194 232 261.5 284 300 310]; subplot (313) ; plot ( IF , EA / sqrt (3) ) ; xlabel ( ’ F i e l d c u r r e n t ’ ) ; ylabel ( ’ open c i r c u i t v o l t a g e ’ ) ; title ( ’ open c i r c u i t c h a r a c t e r i s t i c s ’ ) ; IF1 =[ 0.2 0.4 0.6 0.8 1 1.2 1.4 1.8 ]; ISC =[ 6.6 13.2 20 26.5 32.4 40 46.3 59 ]; 209


18 subplot (323) ; 19 plot ( IF1 , ISC ) ; 20 xlabel ( ’ F i e l d c u r r e n t ’ ) ; 21 ylabel ( ’ s h o r t c i r c u i t c u r r e n t ’ ) ; 22 title ( ’ s h o r t c i r c u i t c h a r a c t e r i s t i c s ’ ) ; 23 ZPF =[ 0 0 0 0 0 0 29 88 140 177 208 230]; 24 subplot (333) ; 25 plot ( IF , ZPF ) ; 26 xlabel ( ’ F i e l d c u r r e n t ’ ) ; 27 ylabel ( ’ t e r m i n a l v o l t a g e ’ ) ; 28 title ( ’ f u l l l o a d z e r o power f a c t o r c h a r a c t e r i s t i c s ’ ) 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

; disp ( ’EMF method ’ ) ; // v a l u e o f s y n c h r o n o u s r e a c t a n c e i s t a k e n from given table EA1 =[ 29 58 87 116 146 172 194 232] ZS = EA1 ./( ISC * sqrt (3) ) ; disp ( ’ s y n c h r o n o u s i m p e d a n c e ( ohms ) i s ’ ) ; disp ( ZS ) ; XS = ZS ; // RSˆ2 i s n e g l i g i b l e disp ( ’ s y n c h r o n o u s r e a c t a n c e ( ohms ) i s ’ ) ; disp ( XS ) ; xs =2.27; ia = i *( pf - %i * sqrt (1 - pf ^2) ) ; // f u l l l o a d c u r r e n t i n c o m p l e x form E = vt + ia *( r + %i * xs ) ; // E x c i t a t i o n v o l t a g e vr = floor ((( abs ( E ) - vt ) / vt ) *100) ; printf ( ’ V o l t a g e r e g u l a t i o n i s %f p e r c e n t \n ’ , vr ) ; disp ( ’Mmf method ’ ) ; // w i t h i a a s r e f e r e n c e E = vt *( pf + %i * sqrt (1 - pf ^2) ) + i * r ; // E x c i t a t i o n v o l t a g e // from f i g 5 . 3 0 ,E=127 V oc =1.69; // c u r r e n t f o r g i v e n e x c i t a t i o n v o l t a g e o b t a i n e d from open c i r c u i t c h a r a c t e r i s t i c s sc =1.2; // f i e l d c u r r e n t r e q u i r e d t o c i r c u l a t e f u l l load short c i r c u i t current al = atand ( imag ( E ) , real ( E ) ) ; // a n g l e b e t w e e n i a and E Ff =( oc *( - sind ( al ) + %i * cosd ( al ) ) ) - sc ; // f i e l d mmf 210


51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81

printf ( ’ f i e l d mmf i s %f A\n ’ , abs ( Ff ) ) ; // c o r r e s p o n d i n g t o Ff , E= 1 6 3 . 5 v from O . C . C Ef =163.5; vr =(( Ef - vt ) / vt ) *100; printf ( ’ V o l t a g e r e g u l a t i o n i s %f p e r c e n t \n ’ , vr ) ; disp ( ’ Z e r o power f a c t o r method ’ ) ; // As p e r t h e d e s c r i p t i o n g i v e n i n method vd =30; // v o l t a g e d r o p a r m a t u r e l e a k a g e r e a c t a n c e xa = vd / i ; // a r m a t u r e l e a k a g e r e a c t a n c e // w i t h i a a s r e f e r e n c e Er = vt *( pf + %i * sqrt (1 - pf ^2) ) + i *( r + %i * xa ) ; // Excitation voltage // from f i g 5 . 3 0 ,E= 1 4 8 . 6 V oc =2.134; // c u r r e n t f o r g i v e n e x c i t a t i o n v o l t a g e o b t a i n e d from open c i r c u i t c h a r a c t e r i s t i c s Fa =0.84; // a r m a t u r e mmf from p o t i e r t r i a n g l e be = atand ( imag ( Er ) , real ( Er ) ) ; // a n g l e b e t w e e n i a and E Ff =( oc *( - sind ( be ) + %i * cosd ( be ) ) ) - Fa ; // f i e l d mmf printf ( ’ f i e l d mmf i s %f A\n ’ , abs ( Ff ) ) ; // c o r r e s p o n d i n g t o Ff = 2 . 7 9 7 A, E=169 v from O . C . C Ef =169; vr =(( Ef - vt ) / vt ) *100; printf ( ’ V o l t a g e r e g u l a t i o n i s %f p e r c e n t \n ’ , vr ) ; disp ( ’ New A . S . A method ’ ) ; // p a r a m e t e r s n e e d e d i n t h i s method a r e c a l c u l a t e d in part c id =0.366; // d i f f e r e n c e i n f i e l d c u r r e n t b e t w e e n OCC and a i r gap l i n e from f i g 5 . 3 0 th = acosd ( pf ) ; ig =1.507; // f i e l d c u r r e n t c o r r e s p o n d i n g t o r a t e d rated per phase v o l t a g e Ff = ig + sc *( %i * pf + sqrt (1 - pf ^2) ) ; // f i e l d mmf w i t h o u t saturation Ff = abs ( Ff ) + id ; // i e l d mmf w i t h s a t u r a t i o n printf ( ’ f i e l d mmf i s %f A\n ’ , Ff ) ; // c o r r e s p o n d i n g t o Ff = 2 . 7 9 1 A, E=169 v from O . C . C Ef =169; 211


82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99

vr =(( Ef - vt ) / vt ) *100; printf ( ’ V o l t a g e r e g u l a t i o n i s %f p e r c e n t \n ’ , vr ) ; disp ( ’ S a t u r a t e d s y n c h r o n o u s r e a c t a n c e method ’ ) ; // f o r E= 1 4 8 . 5 v ( from p a r t c ) , Era =179.5; // a i r l i n e gap v o l t a g e k = Era / abs ( Er ) ; // s a t u r a t i o n f a c t o r vdg =100.5; // v o l t a g e d r o p i n u n s a t u r a t e d synchronous reactance xag = vdg / i ; // u n s a t u r a t e d s y n c h r o n o u s r e a c t a n c e xas = xa +(( xag - xa ) / k ) ; // s a t u r a t e d s y n c h r o n o u s reactance // w i t h v t a s r e f e r e n c e Ef = vt + ia *( r + %i * xas ) ; ok =2.15; // r e s u l t a n t mmf from f i g 5 . 3 0 Ff =( abs ( Ef ) / abs ( Er ) ) * ok ; printf ( ’ f i e l d mmf i s %f A\n ’ , Ff ) ; // c o r r e s p o n d i n g t o Ff =2.78 A, E=169 v from O . C . C Ef =169; vr =(( Ef - vt ) / vt ) *100; printf ( ’ V o l t a g e r e g u l a t i o n i s %f p e r c e n t \n ’ , vr ) ;

Scilab code Exa 5.3 Determining shaft power line current pf and efficiency for maximum power output and maximum power input 1 2 3 4 5 6 7 8

clc ; e =400; // r a t e d v o l t a g e o f motor E =510; // e x c i t a t i o n emf zs =0.5+ %i *4; // s y n c h r o n o u s i m p e d a n c e p e r p h a s e l =900; // n e t l o s s al =90 - atand ( imag ( zs ) , real ( zs ) ) ; disp ( ’ c a s e a ’ ) ; Pmax =(( e * E ) / abs ( zs ) ) -(( E ^2* real ( zs ) ) / abs ( zs ) ^2) ; // maximum o u t p u t power 9 sp = Pmax *3 - l ; 10 printf ( ’ S h a f t power i s %f W\n ’ , sp ) ; 212


11 ia =( sqrt ( e ^2+ E ^2 -2* e * E * cosd ( atand ( imag ( zs ) , real ( zs ) )

) ) ) / abs ( zs ) ; 12 il = sqrt (3) * ia ; // l i n e c u r r e n t 13 printf ( ’ L i n e c u r r e n t i s %f A\n ’ , il ) ; 14 di = acosd (( e * abs ( zs ) -E * real ( zs ) ) /( ia * abs ( zs ) ^2) ) ; 15 pf = cosd ( atand ( imag ( zs ) , real ( zs ) ) - di ) ; 16 printf ( ’ Power f a c t o r i s %f l a g g i n g \n ’ , pf ) ; 17 disp ( ’ c a s e b ’ ) ; 18 Pmax =(( e * E ) / abs ( zs ) ) +(( e ^2* real ( zs ) ) / abs ( zs ) ^2) ; // 19 20 21 22 23 24 25 26

maximum i n p u t power ia =( sqrt ( e ^2+ E ^2 -2* e * E * cosd (90+ al ) ) ) / abs ( zs ) ; sp = floor ( Pmax *3 - ia ^2* real ( zs ) *3 -900) ; printf ( ’ S h a f t power i s %f W\n ’ , sp ) ; il = sqrt (3) * ia ; // l i n e c u r r e n t printf ( ’ L i n e c u r r e n t i s %f A\n ’ , il ) ; di = acosd (( e + E * cosd ( atand ( imag ( zs ) , real ( zs ) ) ) ) /( ia * abs ( zs ) ) ) ; pf = cosd ( atand ( imag ( zs ) , real ( zs ) ) - di ) ; printf ( ’ Power f a c t o r i s %f l a g g i n g \n ’ , pf ) ;

Scilab code Exa 5.4 Determining line current and pf for synchronous motor 1 2 3 4 5 6 7 8 9 10 11

clc ; v =3300; // r a t e d v o l t a g e o f motor zs =0.4+ %i *5; // s y n c h r o n o u s i m p e d a n c e p e r p h a s e E =4000; // e x c i t a t i o n EMF pi =1000; // i n p u t power vp = v / sqrt (3) ; // p e r p h a s e r a t e d v o l t a g e ep = E / sqrt (3) ; // p e r p h a s e e x c i t a t i o n EMF al = atand ( real ( zs ) , imag ( zs ) ) ; t1 =( pi *1000) /3; t2 =( vp ^2/ abs ( zs ) ^2) * real ( zs ) ; t3 = abs ( zs ) /( vp * ep ) ; // t e r m s n e e d e d t o e v a l u a t e l o a d angle 213


12 di = asind (( t1 - t2 ) * t3 ) + al ; // l o a d a n g l e 13 ia =( sqrt ( vp ^2+ ep ^2 -2* ep * vp * cosd ( di ) ) ) / abs ( zs ) ; 14 pf =( pi *1000) /(3* ia * vp ) ; 15 // h e r e ep ∗ c o s ( d i )+i a ∗ r a ∗ pf > vp ; h e n c e l e a d i n g power

factor 16 printf ( ’ L i n e c u r r e n t i s %f A\n ’ , ia ) ; 17 printf ( ’ Power f a c t o r i s %f l e a d i n g ’ , pf ) ;

Scilab code Exa 5.5 Determining torque developed and new pf 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

clc ; v =230; // r a t e d v o l t a g e o f motor f =50; // f r e q u e n c y p =4; // number o f p o l e s zs =0.6+3* %i ; // s y n c h r o n o u s i m p e d a n c e ia1 =10; // c u r r e n t drawn by motor a t u p f ia2 =40; // f i n a l c u r r e n t a f t e r l o a d i s i n c e a s e d t o certain value vt = v / sqrt (3) ; // p e r p h a s e v o l t a g e al = atand ( real ( zs ) , imag ( zs ) ) ; Ef = sqrt (( vt - ia1 * real ( zs ) ) ^2+( ia1 * imag ( zs ) ) ^2) ; // e x c i t a t i o n EMF t1 =( ia2 * abs ( zs ) ) ^2; t2 = Ef ^2+ vt ^2; t3 = -2* Ef * vt ; // t e r m s n e e d e d t o e v a l u a t e l o a d a n g l e de = acosd (( t1 - t2 ) / t3 ) ; // l o a d a n g l e pi =( Ef * vt * sind ( de - al ) ) / abs ( zs ) +( vt ^2* real ( zs ) ) / abs ( zs ) ^2; // i n p u t power pf = pi /( vt * ia2 ) ; printf ( ’ Power f a c t o r i s %f l a g g i n g \n ’ , pf ) ; pd =3*( pi - ia2 ^2* real ( zs ) ) ; // d e v e l o p e d power ns =(120* f ) / p ; // s y n c h r o n o u s s p e e d T =( pd *60) /(2* %pi * ns ) ; printf ( ’ Torque d e v e l o p e d i s %f N−m ’ ,T ) ;

214


Scilab code Exa 5.6 Determining pf for increased input power 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

clc ; v =6600; // r a t e d v o l t a g e o f motor zs =1.5+12* %i ; // p e r p h a s e s y n c h r o n o u s i m p e d a n c e pi1 =1000; // i n p u t power pf =0.8; // power f a c t o r pi2 =1500; // power a t which power f a c t o r i s t o be found out vt = v / sqrt (3) ; // p e r p h a s e v o l t a g e al = atand ( real ( zs ) , imag ( zs ) ) ; ia =( pi1 *1000) /( sqrt (3) * v * pf ) ; Ef = sqrt (( vt * pf - ia * real ( zs ) ) ^2+( vt * sqrt (1 - pf ^2) + ia * imag ( zs ) ) ^2) ; // e x c i t a t i o n EMF t1 =( pi2 *1000) /3; t2 =( vt ^2/ abs ( zs ) ^2) * real ( zs ) ; t3 = abs ( zs ) /( vt * Ef ) ; // t e r m s n e e d e d t o e v a l u a t e l o a d angle di = asind (( t1 - t2 ) * t3 ) + al ; // l o a d a n g l e ia =( sqrt ( vt ^2+ Ef ^2 -2* Ef * vt * cosd ( di ) ) ) / abs ( zs ) ; // new a r m a t u r e c u r r e n t pfn =(( pi2 - pi1 ) *1000) /( ia * vt ) ; // a s Ef ∗ c o s ( d i )+i a ∗ ra > v t h e n c e l e a d i n g power factor printf ( ’ New power f a c t o r i s %f l e a d i n g ’ , pfn ) ;

Scilab code Exa 5.7 Determining new value of armature current and power factor 1 clc ; 2 v =2300; // r a t e d v o l t a g e o f motor 3 xs =12 ; // p e r p h a s e s y n c h r o n o u s r e a c t a n c e

215


4 5 6 7 8 9 10 11 12 13 14 15 16

p =200000; // VA r a t i n g o f motor l1 =120000; // i n i t i a l l o a d l2 =60000; // f i n a l l o a d vt = v / sqrt (3) ; // r a t e d p e r p h a s e v o l t a g e ia = l1 /(3* vt ) ; // minimum a r m a t u r e c u r r e n t ia1 =1.5* ia ; // a r m a t u r e c u r r e n t a t r e d u c e d l o a d ( 5 0% increment ) pf =1/1.5; // power f a c t o r Ef = sqrt (( vt * pf ) ^2+( vt * sqrt (1 - pf ^2) + ia1 * xs ) ^2) ; // e x c i t a t i o n EMF de = asind (( l2 * xs ) /(3* vt * Ef ) ) ; // new l o a d a n g l e ia2 =( sqrt ( vt ^2+ Ef ^2 -2* Ef * vt * cosd ( de ) ) ) / xs ; // new armature c u r r e n t printf ( ’ New v a l u e o f a r m a t u r e c u r r e n t i s %f A\n ’ , ia2 ); pfn = l2 /(3* vt * ia2 ) ; printf ( ’ Power f a c t o r a t new a r m a t u r e c u r r e n t i s %f l e a d i n g ’ , pfn ) ;

Scilab code Exa 5.8 Determining power factor at which machine is operating 1 clc ; 2 ef =1.2; // r a t i o 3 4 5 6 7 8 9 10 11

of e x c i t a t i o n voltage to rated per phase v o l t a g e i =0.7; // r a t i o o f a r m a t u r e c u r r e n t t o r a t e d c u r r e n t r =0.01; // p e r c e n t a g e r e s i s t a n c e o f motor x =0.5; // p e r c e n t a g e r e a c t a n c e o f motor // a s p e r t h e e x p r e s s i o n g i v e n i n book t1 = ef ^2 -( r * i ) ^2 -( x * i ) ^2 -1; t2 = sqrt ((2* i * r ) ^2+(2* i * x ) ^2) ; t3 = atand ((2* i * r ) /(2* i * x ) ) ; // t e r m s n e e d e d t o f i n d o u t power f a c t o r pf = cosd ( asind ( t1 / t2 ) - t3 ) ; printf ( ’ Power f a c t o r i s %f l a g g i n g ’ , pf ) ; 216


Scilab code Exa 5.9 Calculating 1 pf efficiency excitation emfs input current and 2 load angle and maximum output 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

clc ; v =400; // r a t e d v o l t a g e o f motor zs =0.13+ %i *1.3 ; // p e r p h a s e s y n c h r o n o u s i m p e d a n c e p =100000; // VA r a t i n g o f motor l =4000; // s t r a y l o s s e s pl =75000; // power d e l i v e r e d t o l o a d disp ( ’ c a s e a ’ ) ; il = p /( sqrt (3) * v ) ; // l i n e c u r r e n t vt = v / sqrt (3) ; // p e r p h a s e r a t e d v o l t a g e pd = pl + l ; // power d e v e l o p e d poh =3* il ^2* real ( zs ) ; lt = poh + l ; // t o t a l l o s s e s pi = pl + lt ; // i n p u t power pf = pi / p ; // power f a c t o r n =(1 -( lt / pi ) ) *100; // e f f i c i e n c y printf ( ’ Power f a c t o r i s %f\n ’ , pf ) ; printf ( ’ E f f i c i e n c y i s %f p e r c e n t \n ’ ,n ) ; Ef1 = round ( sqrt (( vt * pf - il * real ( zs ) ) ^2+( - vt * sqrt (1 - pf ^2) + il * imag ( zs ) ) ^2) ) ; // e x c i t a t i o n EMF de = atand (( - vt * sqrt (1 - pf ^2) + il * imag ( zs ) ) /( vt * pf - il * real ( zs ) ) ) + acosd ( pf ) ; // l o a d a n g l e printf ( ’ E x c i t a t i o n EMf a t u n d e r e x c i t a t i o n i s %f v \n ’ , Ef1 ) ; printf ( ’ Load a n g l e a t u n d e r e x c i t a t i o n i s %f d e g r e e s \n ’ , de ) ; Ef2 = round ( sqrt (( vt * pf - il * real ( zs ) ) ^2+( vt * sqrt (1 - pf ^2) + il * imag ( zs ) ) ^2) ) ; // e x c i t a t i o n EMF de = atand (( vt * sqrt (1 - pf ^2) + il * imag ( zs ) ) /( vt * pf - il * real ( zs ) ) ) - acosd ( pf ) ; // l o a d a n g l e printf ( ’ E x c i t a t i o n EMf a t o v e r e x c i t a t i o n i s %f v \n ’ , Ef2 ) ; 217


25 26 27 28 29 30 31 32 33 34

printf ( ’ Load a n g l e a t o v e r e x c i t a t i o n i s %f d e g r e e s \ n ’ , de ) ; i = pi /( sqrt (3) * v ) ; printf ( ’ I n p u t c u r r e n t i s %f A\n ’ ,i ) ; disp ( ’ c a e s b ’ ) ; de = acosd ( real ( zs ) / abs ( zs ) ) ; // l o a d a n g l e pmax =(( vt * Ef1 ) / abs ( zs ) ) -(( Ef1 ^2* real ( zs ) ) / abs ( zs ) ^2) ; pt = pmax *3; printf ( ’ Load a n g l e f o r maximum power o u t p u t i s %f d e g r e e s \n ’ , de ) ; printf ( ’ Maximum o u t p u t p e r p h a s e i s %f W\n ’ , pmax ) ; printf ( ’ T o t a l maximum o u t p u t i s %f W\n ’ , pt ) ;

Scilab code Exa 5.10 Determining 1 excitation emfs 2 mechanical power developed and pf 3 minimum excitation voltage 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

clc ; v =6600; // r a t e d v o l t a g e o f motor xs =20 ; // p e r p h a s e s y n c h r o n o u s r e a c t a n c e p =500000; // VA r a t i n g o f motor il = p /( sqrt (3) * v ) ; // r a t e d a r m a t u r e c u r r e n t vt = v / sqrt (3) ; // p e r p h a s e r a t e d v o l t a g e disp ( ’ c a s e a ’ ) ; de =10; // l o a d a n g l e c1 =1; c2 = -2* vt * cosd ( de ) ; c3 = vt ^2 -( il * xs ) ^2; // c o e f f i c i e n t s o f q u a d r a t i c e q u a t i o n i n Ef p = [ c1 c2 c3 ]; Ef = roots ( p ) ; printf ( ’ Per p h a s e e x c i t a t i o n EMF a t l a g g i n g p f i s %f v \n ’ , Ef (2) ) ; printf ( ’ E x c i t a t i o n l i n e EMF a t l a g g i n g p f i s %f v \n ’ , sqrt (3) * Ef (2) ) ; 218


16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34

printf ( ’ Per p h a s e e x c i t a t i o n EMF a t l e a d i n g p f i s %f v \n ’ , Ef (1) ) ; printf ( ’ E x c i t a t i o n l i n e EMF a t l e a d i n g p f i s %f v \n ’ , sqrt (3) * Ef (1) ) ; disp ( ’ c a s e b ’ ) ; disp ( ’ For l a g g i n g p f ’ ) ; pd =(3* vt * Ef (2) * sind ( de ) ) / xs ; pf = pd /( sqrt (3) * v * il ) ; printf ( ’ M e c h a n i c a l power d e v e l o p e d i s %f W\n ’ , pd ) ; printf ( ’ Power f a c t o r i s %f l a g g i n g \n ’ , pf ) ; disp ( ’ For l e a d i n g p f ’ ) ; pd =(3* vt * Ef (1) * sind ( de ) ) / xs ; pf = pd /( sqrt (3) * v * il ) ; printf ( ’ M e c h a n i c a l power d e v e l o p e d i s %f W\n ’ , pd ) ; printf ( ’ Power f a c t o r i s %f l e a d i n g \n ’ , pf ) ; disp ( ’ c a s e c ’ ) ; p =200000; // d e l i v e r e d power de =90; // l o a d a n g l e f o r f a l l i n g o u t o f s t e p // motor f a l l s o u t o f s t e p a t de= 90 d e g r e e s Ef =( p * xs ) /(3* sind ( de ) * vt ) ; printf ( ’ Minimum e x c i t a t i o n v o l t a g e p e r p h a s e i s %f v ’ , Ef ) ;

Scilab code Exa 5.11 Calculating current drawn from supply and pf 1 2 3 4 5 6 7 8 9 10

clc ; v =415; // r a t e d v o l t a g e o f motor f =50; // f r e q u e n c y o f motor ef =520; // l i n e t o l i n e e x c i t a t i o n emf p =6; // number o f p o l e s xs =2; // p e r p h a s e s y n c h r o n o u s r e a c t a n c e t =220; // t o r q u e d e v e l o p e d vt = v / sqrt (3) ; // r a t e d p e r p h a s e v o l t a g e eft = ef / sqrt (3) ; // p e r p h a s e e x c i t a t i o n emf ws =(4* %pi * f ) / p ; // s y n c h r o n o u s s p e e d 219


11 de = asind (( t * ws * xs ) /(3* vt * eft ) ) ; // load angle 12 ia =( sqrt ( vt ^2+ eft ^2 -2* eft * vt * cosd ( de ) ) ) / xs ; //

from phasor diagram ( f i g 5 . 4 8 ) , armature c u r r e n t 13 pf =( ef * sind ( de ) ) /( xs * ia * sqrt (3) ) ; 14 printf ( ’ C u r r e n t drawn from s u p p l y i s %f A\n ’ , ia ) ; 15 printf ( ’ Power f a c t o r i s %f l e a d i n g ’ , pf ) ;

Scilab code Exa 5.12 Determining synchronous reactance 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

clc ; v =415; // r a t e d v o l a t g e o f motor pf =0.9; // l e a d i n g power f a c t o r ps =15000; // s h a f t power E =400; // e x c i t a t i o n emf r =0.5; // p e r p h a s e r e s i s t a n c e vt = v / sqrt (3) ; // r a t e d p e r p h a s e v o l t a g e e = E / sqrt (3) ; // p e r p h a s e e x c i t a t i o n emf c1 =1.5; c2 = - sqrt (3) * v * pf ; c3 = ps ; // c o e f f i c i e n t s o f q u a d r a t i c e q u a t i o n i n armature c u r r e n t p = [ c1 c2 c3 ]; ia = roots ( p ) ; // h i g h e r v a l u e o f a r m a t u r e c u r r e n t i s n e g l e c t e d xs =(( vt * sqrt (1 - pf ^2) ) -( sqrt ( e ^2 -( vt * pf + ia (2) * r ) ^2) ) ) / ia (2) ; printf ( ’ S y n c h r o n o u s r e a c t a n c e i s %f ohm ’ , xs ) ;

Scilab code Exa 5.13 Finding mechanical power delivered by motor and input KVA 1 clc ; 2 e =1.2; // pu e x c i t a t i o n emf

220


3 xs =0.8; // pu s y n c h r o n o u s r e a c t a n c e 4 vt =1; // pu r a t e d v o l t a g e 5 ia =1; // pu a r m a t u r e c u r r e n t f o r KVA=100 %, v t ∗ i a =1

t h e r e f o r e i a =1; 6 pf = cosd ( asind (( e ^2 - xs ^2 -1) /( -2* xs ) ) ) ; // l e a d i n g 7 8 9 10 11 12 13

power f a c t o r pd = vt * ia * pf ; printf ( ’ M e c h a n i c a l power d e v e l o p e d by motor i s %f pu \n ’ , pd ) ; e =1; // pu e x c i t a t i o n emf r e d u c e d t o g e n e r a t e 100% emf de = asind (( pf * xs ) /( vt * e ) ) ; // l o a d a n g l e ia =( sqrt ( vt ^2+ e ^2 -2* e * vt * cosd ( de ) ) ) / xs ; // new armature c u r r e n t p =( vt * ia ) *100; printf ( ’ New KVA r a t i n g i s %f p e r c e n t ’ ,p ) ;

Scilab code Exa 5.14 Determining 1 load angle pu armature current and pf 2 excitation voltage load angle and pf 1 2 3 4 5 6 7 8 9 10 11 12 13 14

clc ; xs =0.8; // pu r e a c t a n c e Ef =1.3; // pu e x c i t a t i o n EMF p =0.5; // pu o u t p u t power vt =1; // r a t e d p e r p h a s e v o l t a g e disp ( ’ c a s e a ’ ) ; de = asind (( p * xs ) /( vt * Ef ) ) ; printf ( ’ Load a n g l e i s %f d e g r e e s \n ’ , de ) ; ia =( sqrt ( vt ^2+ Ef ^2 -2* vt * Ef * cosd ( de ) ) ) / xs ; // from phasor diagram ( f i g 5 . 4 9 ) printf ( ’ Armature c u r r e n t i s %f p . u . \ n ’ , ia ) ; pf = p /( vt * ia ) ; printf ( ’ Power f a c t o r i s %f l a g g i n g \n ’ , pf ) ; disp ( ’ c a s e b ’ ) ; // Under g i v e n c o n d i t i o n m a g n i t u d e o f power f a c t o r 221


15 16 17 18

r e m a i n s same but i t becomes l e a d i n g Ef = sqrt (( vt * pf ) ^2+( vt * sqrt (1 - pf ^2) - ia * xs ) ^2) ; // e x c i t a t i o n EMF printf ( ’ E x c i t a t i o n EMF i s %f p . u . \ n ’ , Ef ) ; de = asind (( p * xs ) /( vt * Ef ) ) ; printf ( ’ Load a n g l e i s %f d e g r e e s \n ’ , de ) ;

Scilab code Exa 5.15 Determining 1 new value of armature current load angle and pf 2 armature current load angle and power delivered 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

clc ; v =11000; // r a t e d v o l t a g e o f motor zs =1+10* %i ; // p e r p h a s e s y n c h r o n o u s i m p e d a n c e ia =100; // a r m a t u r e c u r r e n t a t u n i t y power f a c t o r vt = v / sqrt (3) ; // p e r p h a s e v o l t a g e Ef = sqrt (( vt + ia * real ( zs ) ) ^2+( ia * imag ( zs ) ) ^2) ; // e x c i t a t i o n EMF from p h a s o r d i a g r a m al = atand ( real ( zs ) , imag ( zs ) ) ; de = atand (( ia * imag ( zs ) ) /( vt + ia ) ) ; // l o a d a n g l e p =( Ef * vt * sind ( de + al ) / abs ( zs ) ) -(( vt ^2* real ( zs ) ) / abs ( zs ) ^2) ; // p e r p h a s e power d e l i v e r e d disp ( ’ c a s e a ’ ) ; Ef1 =1.15* Ef ; // E x c i t a t i o n EMF a f t e r an i n c r e m e n t o f 15% t1 = p ; t2 =( vt ^2/ abs ( zs ) ^2) * real ( zs ) ; t3 = abs ( zs ) /( vt * Ef1 ) ; // t e r m s n e e d e d t o e v a l u a t e load angle di = asind (( t1 + t2 ) * t3 ) - al ; // l o a d a n g l e ia1 =( sqrt ( vt ^2+ Ef1 ^2 -2* Ef1 * vt * cosd ( di ) ) ) / abs ( zs ) ; // armature c u r r e n t pf = p /( vt * ia1 ) ; printf ( ’ New v a l u e o f a r m a t u r e c u r r e n t i s %f A\n ’ , ia1 ); printf ( ’ New v a l u e o f l o a d a n g l e i s %f d e g r e e s \n ’ , di ) 222


20 21 22 23 24 25 26 27 28 29 30 31 32 33

; printf ( ’ New power f a c t o r i s %f l a g g i n g \n ’ , pf ) ; disp ( ’ c a s e b ’ ) ; // a t u n i t y p f pf =1; c1 =1+ imag ( zs ) ^2; c2 =2* vt ; c3 = vt ^2 - Ef1 ^2; // c o e f f i c i e n t s o f q u a d r a t i c e q u a t i o n in armature c u r r e n t p = [ c1 c2 c3 ]; ia = roots ( p ) ; printf ( ’ Armature c u r r e n t u n d e r g i v e n c o n d i t i o n i s %f A\n ’ , ia (2) ) ; P =( vt * ia (2) * pf ) /1000; Pt = P *3; printf ( ’ Per p h a s e power d e l i v e r e d i s %f KW\n ’ ,P ) ; printf ( ’ Net power d e l i v e r e d i s %f KW\n ’ , Pt ) ;

Scilab code Exa 5.16 Determining operating pf and load angle 1 clc ; 2 vt =1; // r a t e d p e r p h a s e v o l t a g e 3 zs =0.02+0.8* %i ; // p e r p h a s e p . u s y n c h r o n o u s 4 5 6 7 8 9 10 11 12

impedance // At t h e t i m e o f s y n c h r o n i z a t i o n e x c i t a t i o n EMF= r a t e d p e r p h a s e v o l t a g e and l o a d a n g l e =0 ef =1; // pu e x c i t a t i o n EMF ia =1; // pu a r m a t u r e c u r r e n t // from p h a s o r d i a g r a m 5 . 5 1 t1 = ef ^2 - real ( zs ) ^2 - imag ( zs ) ^2 -1; t2 = sqrt ((2* real ( zs ) ) ^2+(2* imag ( zs ) ) ^2) ; t3 = atand ( - real ( zs ) / imag ( zs ) ) ; // t e r m s n e e d e d t o f i n d o u t power f a c t o r pf = cosd ( asind ( t1 / t2 ) + t3 ) ; printf ( ’ O p e r a t i n g power f a c t o r i s %f l e a d i n g \n ’ , pf ) ; 223


13 14 15 16

al = atand ( real ( zs ) , imag ( zs ) ) ; t1 = vt * ia * pf ; t2 =( vt ^2/ abs ( zs ) ^2) * real ( zs ) ; t3 = abs ( zs ) /( vt * ef ) ; // t e r m s n e e d e d t o e v a l u a t e l o a d angle 17 de = floor ( asind (( t1 + t2 ) * t3 ) ) - al ; // l o a d a n g l e 18 printf ( ’ Load a n g l e o f g e n e r a t o r i s %f d e g r e e s ’ , de ) ;

Scilab code Exa 5.17 Determining armature current power factor and excitation emf 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

clc ; p =40*10^6; // r a t e d power o f t u r b o g e n e r a t o r v =11000; // r a t e d v o l t a g e o f g e n e r a t o r xs =0.8; // p . u . s y n c h r o n o u s r e a c t a n c e ra =0.5; // s e r i e s r e a c t a n c e o f i n f i n i t e bus vt = v / sqrt (3) ; // r a t e d p e r p h a s e v o l t a g e disp ( ’ c a s e b ’ ) ; ia = p /( v * sqrt (3) ) ; // a r m a t u r e c u r r e n t printf ( ’ Armature c u r r e n t i s %f A\n ’ , ia ) ; xs = xs *( vt / ia ) ; // x s i n ohms vd = ia * ra ; // v o l t a g e d r o p i n s e r i e s r e s i s t a n c e pf = cosd ( asind (( vd /2) / vt ) ) ; printf ( ’ A l t e r n a t e power f a c t o r i s %f l a g g i n g \n ’ , pf ) ; disp ( ’ c a s e c ’ ) ; Ef = sqrt (( vt * pf ) ^2+(( vd /2) +( ia * xs ) ) ^2) ; printf ( ’ E x c i t a t i o n EMF l i n e t o n e u t r a l i s %f V\n ’ , Ef ); 17 printf ( ’ E x c i t a t i o n EMF l i n e t o l i n e i s %f V\n ’ , sqrt (3) * Ef ) ;

Scilab code Exa 5.18 Determining load angle power factor and armature current 224


1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

clc ; v =400; // r a t e d v o l t a g e o f motor pi =5472; // i n p u t power np =3; // number o f p h a s e s xs =10; // s y n c h r o n o u s r e a c t a n c e ef = v ; // e x c i t a t i o n v o l t a g e vt = v / sqrt (3) ; // r a t e d p e r p h a s e v o l t a g e de = round ( asind (( pi * xs * np ) /( np * v ^2) ) ) ; printf ( ’ Load a n g l e i s %f d e g r e e s \n ’ , de ) ; // from f i g . 5 . 5 3 , v t=e f ( e x c i t a t i o n v o l t a g e p e r p h a s e ) a r m a t u r e r e s i s t a n c e =0 pf = cosd ( de /2) ; printf ( ’ Power f a c t o r i s %f l a g g i n g \n ’ , pf ) ; // from f i g . 5 . 5 3 ia = floor ((2* vt * sind ( de /2) ) / xs ) ; printf ( ’ Armature c u r r e n t i s %f A ’ , ia ) ;

Scilab code Exa 5.19 Determining 1 excitation emf 2 power transfer and 3 maximum power transfer armature current terminal voltage and power factor 1 2 3 4 5 6 7 8 9 10 11 12 13

clc ; v =2000; // r a t e d v o l t a g e o f motor xsm =2; // s y n c h r o n o u s r e a c t a n c e o f motor xsg =3; // s y n c h r o n o u s r e a c t a n c e o f g e n e r a t o r xt =1.5; // t r a n s m i s s i o n l i n e r e a c t a n c e ia =100; // c u r r e n t drawn by motor pf =1; // power f a c t o r disp ( ’ c a s e a ’ ) ; vt = v / sqrt (3) ; // r a t e d p e r p h a s e v o l t a g e Efm = floor ( sqrt ( vt ^2+( ia * xsm ) ^2) ) ; // e x c i t a t i o n EMF printf ( ’ E x c i t a t i o n EMF f o r motor i s %f V\n ’ , Efm ) ; Efg = sqrt ( vt ^2+( ia *( xsg + xt ) ) ^2) ; // e x c i t a t i o n EMF printf ( ’ E x c i t a t i o n EMF f o r a l t e r n a t o r i s %f V\n ’ , Efg ); 225


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

disp ( ’ c a s e b ’ ) ; de1 = acosd ( vt / Efm ) ; // l o a d a n g l e f o r motor de2 = acosd ( vt / Efg ) ; // l o a d a n g l e f o r a l t e r n a t o r de = de1 + de2 ; // power a n g l e b e t w e e n Efm and Efg pt =( Efg * Efm * sind ( de ) ) /( xsm + xsg + xt ) ; P = pt *3; printf ( ’ Per p h a s e power t r a n s f e r b e t w e e n a l t e r n a t o r and motor i s %f KW\n ’ , pt /1000) ; printf ( ’ Net power t r a n s f e r b e t w e e n a l t e r n a t o r and motor i s %f KW\n ’ ,P /1000) ; disp ( ’ c a s e c ’ ) ; // from p h a s o r d i a g r a m f i g 5 . 5 4 ia = sqrt ( Efm ^2+ Efg ^2) /( xsm + xsg + xt ) ; // f o r maximum t r a n s f e r o f power , power a n g l e =90 degrees de =90 pmax =( Efg * Efm * sind ( de ) ) /( xsm + xsg + xt ) ; P = pmax *3; printf ( ’ Per p h a s e maximum power t r a n s f e r b e t w e e n a l t e r n a t o r and motor i s %f KW\n ’ , pmax /1000) ; printf ( ’ Net maximum power t r a n s f e r b e t w e e n a l t e r n a t o r and motor i s %f KW\n ’ ,P /1000) ; // from p h a s o r d i a g r a m s d e t e r m i n i n g v a r i o u s p a r a m e t e r s n e e d e d t o f i n d power f a c t o r be = acosd ( Efm /( ia *( xsm + xsg + xt ) ) ) ; Vp = sqrt (( Efm - ia * xsm * cosd ( be ) ) ^2+( ia * xsm * sind ( be ) ) ^2) ; // p h a s e v o l t a g e Vl = sqrt (3) * Vp ; // l i n e v o l t a g e printf ( ’ Armature c u r r e n t f o r g i v e n c o n d i t i o n i s %f A \n ’ , ia ) ; printf ( ’ T e r m i n a l v o l t a g e o f s y n c h r o n o u s motor i s %f V\n ’ , Vp ) ; // from p h a s o r d i a g r a m aoc = asind (( ia * xsm * sind ( be ) ) / Vp ) ; pf = cosd (90 - be - aoc ) ; printf ( ’ Power f a c t o r a n g l e o f motor i s %f l e a d i n g ’ , pf ) ;

226


Scilab code Exa 5.20 Determining 1 maximum power output 2 armature current and power factor 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

clc ; p =20*10^6; // VA r a t i n g o f a l t e r n a t o r z =5; // i m p e d a n c e o f a l t e r n a t o r r =0.5; // r e s i s t a n c e o f a l t e r n a t o r v =11000; // v o l t a g e r a t i n g o f bus b a r s e =12000; // e x c i t a t i o n v o l t a g e vt = v / sqrt (3) ; // a l t e r n a t o r p e r p h a s e v o l t a g e Ef = e / sqrt (3) ; // a l t e r n a t o r p e r p h a s e e x c i t a t i o n voltage pmax = round (((( Ef * vt ) / z ) -(( vt ^2* r ) / z ^2) ) /10^6) ; P = round ( pmax *3) ; printf ( ’ Per p h a s e maximum o u t p u t power from a l t e r n a t o r i s %f MW\n ’ , pmax ) ; printf ( ’ T o t a l maximum o u t p u t power from a l t e r n a t o r i s %f MW\n ’ ,P ) ; disp ( ’ c a s e b ’ ) ; pf = r / z ; // power f a c t o r ia = round (( sqrt ( vt ^2+ Ef ^2 -2* Ef * vt * pf ) ) / z ) ; // armature c u r r e n t printf ( ’ Armature c u r r e n t i s %f A\n ’ , ia ) ; pf =( Ef *z - vt * r ) /( ia * z ^2) ; printf ( ’ Power f a c t o r u n d e r maximum power c o n d i t i o n i s %f l e a d i n g ’ , pf ) ;

Scilab code Exa 5.21 Calculating data needed to plot v curves and pf variation with field current 1 clc ; 2 v =2200; // r a t e d

v o l t a g e o f motor 227


3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27

r =0.32; // p e r p h a s e a r m a t u r e r e s i s t a n c e p =1500; // KW r a t i n g o f motor ie =15; // e x c i t i n g c u r r e n t is =750; // s h o r t c i r c u i t c u r r e n t cl =60; // c o r e l o s s i n KW fl =40; // f r i c t i o n a l and w i n d a g e l o s s i n KW IF =[5 10 15 20 25 30]; EFO =[ 760 1500 2140 2650 3040 3340]; // e x c i t a t i o n EMF p e r p h a s e EFP = EFO / sqrt (3) ; disp ( ’ E x c i t a t i o n EMF p e r p h a s e (V) i s ’ ) ; disp ( EFP ) ; // from t a b l e g i v e n f o r i e =15 , Ef =2140; // E x c i t a t i o n EMF np =3; // number o f p h a s e s ef = Ef / sqrt (3) ; // p e r p h a s e open c i r c u i t v o l t a g e zs = ef / is ; // s y n c h r o n o u s i m p e d a n c e vt = floor ( v / sqrt (3) ) ; // p e r p h a s e t e r m i n a l v o l t a g e i = floor ( vt / zs ) ; // c u r r e n t p h a s o r l a g g i n g v t ia = vt /(2* r ) ; // a r m a t u r e c u r r e n t pd =(( p /2) + cl + fl ) / np ; // m e c h a n i c a l power d e v e l o p e d per phase at half −f u l l load output R = ceil ( sqrt (( ia ^2 -(( pd *1000) / r ) ) ) ) ; printf ( ’ R a d i u s o f power c i r c l e i s %f A\n ’ ,R ) ; printf ( ’ C u r r e n t p h a s o r i s %f A\n ’ ,i ) ; printf ( ’ S y n c h r o n o u s i m p e d a n c e i s %f ohm\n ’ , zs ) ; disp ( ’ u s i n g a b o v e d a t a and t a b l e g i v e n i n s o l u t i o n , V−c u r v e s and v a r i a t i o n o f p . f . w i t h f i e l d c u r r e n t s can be p l o t t e d ’ ) ;

Scilab code Exa 5.22 Determining data to plot v curves and pf variation with field current neglecting armature resistance 1 clc ; 2 v =2200; // r a t e d

v o l t a g e o f motor 228


3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

18 19 20 21

p =1500; // KW r a t i n g o f motor ie =15; // e x c i t i n g c u r r e n t is =750; // s h o r t c i r c u i t c u r r e n t cl =60; // c o r e l o s s i n KW fl =40; // f r i c t i o n a l and w i n d a g e l o s s i n KW // from t a b l e g i v e n i n q u e s t i o n f o r i e =15 , Ef =2140; // E x c i t a t i o n EMF np =3; // number o f p h a s e s ef = Ef / sqrt (3) ; // p e r p h a s e open c i r c u i t v o l t a g e xs = ef / is ; // s y n c h r o n o u s r e a c t a n c e vt = floor ( v / sqrt (3) ) ; // p e r p h a s e t e r m i n a l v o l t a g e i = floor ( vt / xs ) ; // c u r r e n t p h a s o r l a g g i n g v t pd =(( p /2) + cl + fl ) / np ; // m e c h a n i c a l power d e v e l o p e d per phase at half −f u l l load output ia = pd *1000/ vt ; // w o r k i n g component o f a r m a t u r e current // a s r e s i s t a n c e =0 , a r m a t u r e c u r r e n t =0 t h e r e f o r e r a d i u s o f power c i r c l e =0 t h a t i s i t becomes l i n e with c e n t r e at zero printf ( ’ Working component o f a r m a t u r e c u r r e n t i s %f A\n ’ , ia ) ; printf ( ’ T e r m i n a l v o l t a g e i s %f A\n ’ , vt ) ; printf ( ’ S y n c h r o n o u s r e a c t a n c e i s %f ohm\n ’ , xs ) ; disp ( ’ u s i n g a b o v e d a t a and t a b l e g i v e n i n s o l u t i o n , V−c u r v e s and v a r i a t i o n o f p . f . w i t h f i e l d c u r r e n t s can be p l o t t e d ’ ) ;

Scilab code Exa 5.23 Determining armature current power factor and load angle 1 2 3 4 5

clc ; v =1100; // r a t e d v o l t a g e o f motor ef =1650; // emf p =350; // i n p u t power i n KW zs =0.7+3.2* %i ; // s y n c h r o n o u s i m p e d a n c e p e r p h a s e 229


6 vt = v / sqrt (3) ; // r a t e d p e r p h a s e v o l t a g e 7 eft = ef / sqrt (3) ; // p e r p h a s e emf 8 i1 = vt / abs ( zs ) ; 9 printf ( ’ C u r r e n t p h a s o r l a g g i n g t e r m i n a l v o l t a g e 10 11 12 13 14 15 16 17 18 19 20

is %f A\n ’ , i1 ) ; i2 = eft / abs ( zs ) ; printf ( ’ C u r r e n t p h a s o r l a g g i n g e x c i t a t i o n EMF i s %f A\n ’ , i2 ) ; ia =( p *1000) /(3* vt ) ; printf ( ’ Working component o f a r m a t u r e c u r r e n t i s %f A ’ , ia ) ; disp ( ’ u s i n g t h i s d a t a v e c t o r d i a g r a m i s drawn and v a l u e o f i a and power f a c t o r i s o b t a i n e d ’ ) ; ia =194.5; pf =19.5; printf ( ’ Power f a c t o r i s %f l e a d i n g \n ’ , pf ) ; printf ( ’ Armature c u r r e n t i s %f A\n ’ , ia ) ; de = acosd (( ia ^2 - i1 ^2 - i2 ^2) /( -2* i1 * i2 ) ) ; printf ( ’ Load a n g l e i s %f d e g r e e s \n ’ , de ) ;

Scilab code Exa 5.24 Determining pu excitation voltage 1 2 3 4 5 6 7 8 9 10 11

clc ; xd =1.2; // d a x i s s y n c h r o n o u s r e a c t a n c e xq =0.8; // q a x i s s y n c h r o n o u s r e a c t a n c e ra =0.025; // a r m a t u r e r e s i s t a n c e vt =1; // pu r a t e d p e r p h a s e v o l t a g e disp ( ’ c a s e a ’ ) ; disp ( ’ For l a g g i n g power f a c t o r ’ ) pf =0.8; // power f a c t o r ia =1*( pf - sqrt (1 - pf ^2) * %i ) ; // a r m a t u r e c u r r e n t Ef1 = vt + %i *( ia * xq ) + ia * ra ; // e x c i t a t i o n EMF id =1* sind ( atand ( imag ( Ef1 ) , real ( Ef1 ) ) + acosd ( pf ) ) ; // d component o f a r m a t u r e c u r r e n t 12 iq =1* cosd ( atand ( imag ( Ef1 ) , real ( Ef1 ) ) + acosd ( pf ) ) ; // 230


13 14

15 16 17 18 19 20 21 22 23

q component o f a r m a t u r e c u r r e n t Ef = abs ( Ef1 ) + id *( xd - xq ) ; // e x c i t a t i o n EMF printf ( ’ E x c i t a t i o n EMF i s %f p . u . a t a l o a d a n g l e o f %f d e g r e e s \n ’ , abs ( Ef ) , atand ( imag ( Ef1 ) , real ( Ef1 ) ) ); disp ( ’ c a s e b ’ ) ; disp ( ’ For l e a d i n g power f a c t o r ’ ) pf =0.8; // power f a c t o r ia =1*( pf + sqrt (1 - pf ^2) * %i ) ; // a r m a t u r e c u r r e n t Ef1 = vt + %i *( ia * xq ) + ia * ra ; // e x c i t a t i o n EMF id =1* sind ( atand ( imag ( Ef1 ) , real ( Ef1 ) ) - acosd ( pf ) ) ; // d component o f a r m a t u r e c u r r e n t iq =1* cosd ( atand ( imag ( Ef1 ) , real ( Ef1 ) ) - acosd ( pf ) ) ; // q component o f a r m a t u r e c u r r e n t Ef = abs ( Ef1 ) + id *( xd - xq ) ; // e x c i t a t i o n EMF printf ( ’ E x c i t a t i o n EMF i s %f p . u . a t a l o a d a n g l e o f %f d e g r e e s \n ’ , abs ( Ef ) , atand ( imag ( Ef1 ) , real ( Ef1 ) ) ) ;

Scilab code Exa 5.27 Calculating 1 power angle armature current and pf 2 maximum power output and power angle 1 2 3 4 5 6 7 8 9 10 11 12 13

clc ; v =400; // r a t e d v o l t a g e o f s y n c h r o n o u s machine vt = v / sqrt (3) ; // p e r p h a s e r a t e d v o l t a g e ps =9500; // s h a f t l o a d xd =5; // p e r p h a s e d−a x i s s y n c h r o n o u s r e a c t a n c e xq =3.2; // p e r p h a s e q−a x i s s y n c h r o n o u s r e a c t a n c e l =500; // f r i c t i o n w i n d a g e and c o r e l o s s e s np =3; // number o f p h a s e s // a t r a t e d v o l t a g e e x c i t a t i o n EMF Ef = v / sqrt (3) ; // e x c i t a t i o n EMF disp ( ’ c a s e a ’ ) ; pt = ps + l ; // power d e v e l o p e d // by u s i n g f o r m u l a pd=np ∗ ( ( ( Ef ∗ Ef ∗ s i n d ( de ) ) / xd ) +(( 231


14 15 16 17 18 19 20 21 22 23 24 25

Ef ˆ2∗ s i n d ( 2 ∗ de ) ∗ 0 . 5 ∗ ( xd−xq ) ) / ( xd ∗ xq ) ) ) , and h i t and t r i a l method we o b t a i n v a l u e o f l o a d a n g l e de =11.623; // l o a d a n g l e id =( Ef - Ef * cosd ( de ) ) / xd ; // d−a x i s component o f armature c u r r e n t iq =( Ef * sind ( de ) ) / xq ; // q−a x i s component o f a r m a t u r e current ia = sqrt ( id ^2+ iq ^2) ; printf ( ’ Armature c u r r e n t i s %f A\n ’ , ia ) ; pf = cosd ( acosd ( iq / ia ) - de ) ; printf ( ’ Power f a c t o r i s %f l a g g i n g \n ’ , pf ) ; disp ( ’ c a s e b ’ ) ; de = acosd (( - Ef * xq ) /(4* Ef *( xd - xq ) ) +( sqrt (0.5+(( Ef * xq ) /(4* vt *( xd - xq ) ) ) ^2) ) ) ; pd = np *((( Ef * Ef * sind ( de ) ) / xd ) +(( Ef ^2* sind (2* de ) *0.5*( xd - xq ) ) /( xd * xq ) ) ) ; // maximum power d e v e l o p e d po = pd - l ; printf ( ’ Maximum power o u t p u t i s %f W’ , po ) ;

Scilab code Exa 5.29 Determining change in synchronous power and reactive power 1 2 3 4 5 6 7 8 9 10

clc ; Ef =1.4; // p . u e x c i t a t i o n EMF xs =1.2; // p . u s y n c h r o n o u s r e a c t a n c e p =0.5; // p . u s y n c h r o n o u s power b e i n g d e l i v e r e d i =1; // p e r c e n t a g e i n c r e m e n t i n p r i m e mover t o r q u e vt =1; // r a t e d p e r p h a s e v o l t a g e de = asind (( p * xs ) /( Ef * vt ) ) ; // l o a d a n g l e dp =( i * p ) /100; // i n c r e a s e i n p . u r e a l power ip =( dp / p ) *100; printf ( ’ P e r c e n t a g e i n c r e a s e i n r e a l power i s %d p e r c e n t o f i t s p r e v i o u s v a l u e \n ’ , ip ) ; 11 iq = - tand ( de ) * ip ; 12 printf ( ’ P e r c e n t a g e d e c r e a s e i n r e a c t i v e power i s %f 232


p e r c e n t o f i t s p r e v i o u s v a l u e \n ’ ,- iq ) ;

Scilab code Exa 5.31 Determining maximum load armature current and pf 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

clc ; v =400; // r a t e d v o l t a g e o f motor xd =6; // d−a x i s s y n c h r o n o u s r e a c t a n c e xq =4; // q−a x i s s y n c h r o n o u s r e a c t a n c e vt =400/ sqrt (3) ; // r a t e d p e r p h a s e v o l t a g e // As p e r t h e t h e o r y g i v e n i n q u e s t i o n , p h a s o r d i a g r a m i s drawn and f o r m u l a i s d e r i v e d // A f t e r t h e d e r i v e d e x p r e s s i o n , f o r maximum power l o a d a n g l e =45 de =45; P =( vt ^2/2) *((1/ xq ) -(1/ xd ) ) * sind (2* de ) ; printf ( ’ Maximum l o a d t h a t can be put on s y n c h r o n o u s motor i s %f W p e r p h a s e \n ’ ,P ) ; printf ( ’ Maximum l o a d t h a t can be put on s y n c h r o n o u s motor i s %f W f o r 3− p h a s e \n ’ ,3* P ) ; iq =( vt * sind ( de ) ) / xq ; // q−a x i s component o f a r m a t u r e current id =( vt * sind ( de ) ) / xd ; // d−a x i s component o f a r m a t u r e current ia = sqrt ( iq ^2+ id ^2) ; printf ( ’ Armature c u r r e n t i s %f A\n ’ , ia ) ; pf =(3* P ) /( sqrt (3) * v * ia ) ; printf ( ’ Power f a c t o r a t maximum power i s %f l a g g i n g ’ , pf ) ;

Scilab code Exa 5.32 Determining minimum excitation voltage and maximum stable load angle 233


1 clc ; 2 // Answer f o r minimum e x c i t a t i o n 3 4 5 6 7 8 9 10 11

12 13 14 15

voltage i s given wrong i n book v =400; // r a t e d v o l t a g e o f motor xd =6; // d−a x i s s y n c h r o n o u s r e a c t a n c e xq =4; // q−a x i s s y n c h r o n o u s r e a c t a n c e vt =400/ sqrt (3) ; // r a t e d p e r p h a s e v o l t a g e p =21; // l o a d c a r r i e d by motor pph =( p /3) *1000; // p e r p h a s e l o a d c a r r i e d by motor // As p e r t h e t h e o r y g i v e n i n q u e s t i o n , e x p r e s s i o n is derived // A f t e r t h e d e r i v e d e x p r e s s i o n , // c o s ( de ) =( v t ˆ 2 ∗ ( xd−xq ) ∗ s i n ( de ) ˆ 3 ) / ( pph ∗ xd ∗ xq ) , v a l u e o f de ( l o a d a n g l e i s o b t a i n e d by t r i a l and e r r o r method and v a l u e o f l o a d a n g l e i s ) de =63.2; Ef =( pph -( vt ^2/2) *(( xd - xq ) /( xd * xq ) ) * sind (2* de ) ) /(( vt / xd ) * sind ( de ) ) ; printf ( ’ Maximum s t a b l e l o a d a n g l e i s %f d e g r e e s \n ’ , de ) ; printf ( ’ Minimum e x c i t a t i o n v o l t a g e i s %f v \n ’ , Ef ) ;

Scilab code Exa 5.34 Determining load angle and excitation voltage 1 2 3 4 5 6 7 8

clc ; vt =1; // pu r a t e d v o l t a g e xd =1; // pu d−a x i s s y n c h r o n o u s r e a c t a n c e xq =0.6; // pu q−a x i s s y n c h r o n o u s r e a c t a n c e p =0.9; // pu power b e i n g d e l i v e r e d pf =0.8; // power f a c t o r ia = p /( vt * pf ) ; // pu a r m a t u r e c u r r e n t de = atand (( ia * xq + vt * sind ( acosd ( pf ) ) ) /( vt * pf ) ) - acosd ( pf ) ; 9 printf ( ’ Load a n g l e i s %f d e g r e e s \n ’ , de ) ; 10 Ef = vt * cosd ( de ) + ia * sind ( de + acosd ( pf ) ) * xd ; 234


11 12 13 14 15 16

printf ( ’ E x c i t a t i o n v o l t a g e i s %f p . u . \ n ’ , Ef ) ; // when l o s s o f e x c i t a t i o n t a k e s p l a c e Ef =0 , f o r maximum power l o a d a n g l e =45 de =45; // l o a d a n g l e pmax =( vt ^2*( xd - xq ) * sind (2* de ) ) /(2* xd * xq ) ; printf ( ’ Maximum power i s %f p . u . \ n ’ , pmax ) ; disp ( ’ As maximum power i s l e s s t h a n t h e power b e i n g d e l i v e r e d g e n e r a t o r w i l l l o s e synchronism ’ );

Scilab code Exa 5.35 Determining excitation voltage power synchronizing power per electrical and mechanical degrees and corresponding torque maximum value of load angle and corresponding power 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

clc ; v =11000; // r a t e d v o l t a g e o f motor P =20*10^6; // r a t e d power o f motor p =12; // number o f p o l e s f =50; // f r e q u e n c y xd =5; // d−a x i s s y n c h r o n o u s r e a c t a n c e xq =3; // q−a x i s s y n c h r o n o u s r e a c t a n c e vt = v / sqrt (3) ; // p e r p h a s e r a t e d v o l t a g e ia = P /( sqrt (3) * v ) ; // p e r p h a s e a r m a t u r e c u r r e n t disp ( ’ c a s e a ’ ) ; // from p h a s o r d i a g r a m Ef = vt - %i *( ia * xq ) ; de = atand (( ia * xq ) ,( vt ) ) ; // l o a d a n g l e id = ia * sind ( de ) ; // d−a x i s component o f a r m a t u r e current Ef = ceil ( abs ( Ef ) + id *( xd - xq ) ) ; printf ( ’ E x c i t a t i o n v o l t a g e i s %f V t o n e u t r a l \n ’ , Ef ) ; disp ( ’ c a s e b ’ ) ; po =((( Ef * vt * sind ( de ) ) / xd ) +(( vt ^2*( xd - xq ) * sind (2* de ) ) /(2* xd * xq ) ) ) /1000; printf ( ’ Power i s %f KW p e r p h a s e \n ’ , po ) ; 235


20 po =3* po *1000; 21 printf ( ’ C a l c u l a t e d power %f W i s 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37

almost equal to g i v e n r a t e d power i . e %f W\n ’ ,po , P ) ; disp ( ’ c a s e c ’ ) ; ps =(( Ef * vt * cosd ( de ) ) / xd ) +(( vt ^2*( xd - xq ) * cosd (2* de ) ) /( xd * xq ) ) ; printf ( ’ S y n c h r o n o z i n g power p e r e l e c t r i c a l d e g r e e i s %f KW\n ’ ,(3* ps * %pi ) /(180*1000) ) ; ws =(4* %pi * f ) / p ; // s y n c h r o n o u s s p e e d t =(3* ps * %pi ) /(180* ws ) ; printf ( ’ Torque c o r r e s p o n d i n g t o s y n c h r o n o u s power i s %f N−m\n ’ ,t ) ; disp ( ’ c a s e d ’ ) ; printf ( ’ S y n c h r o n o z i n g power p e r m e c h a n i c a l d e g r e e i s %f KW\n ’ ,(3* ps * %pi * p ) /(2*180*1000) ) ; t =(3* ps * %pi * p ) /(2*180) ; printf ( ’ Torque c o r r e s p o n d i n g t o s y n c h r o n o u s power i s %f N−m\n ’ ,t ) ; disp ( ’ c a s e e ’ ) ; de = acosd (( - Ef * xq ) /(4* Ef *( xd - xq ) ) +( sqrt (0.5+(( Ef * xq ) /(4* vt *( xd - xq ) ) ) ^2) ) ) ; printf ( ’ Maximum v a l u e o f power a n g l e i s %f d e g r e e s \n ’ , de ) ; pmax =((( Ef * vt * sind ( de ) ) / xd ) +(( vt ^2*( xd - xq ) * sind (2* de ) ) /(2* xd * xq ) ) ) /1000; printf ( ’ Maximum power i s %f KW p e r p h a s e \n ’ , pmax ) ; printf ( ’ Maximum power i s %f KW f o r 3 p h a s e \n ’ , pmax *3) ;

Scilab code Exa 5.36 Calculate synchronizing power and torque per mechanical degrees 1 at no load and 2 at full load 1 clc ; 2 v =6600; // r a t e d v o l t a g e 3 p =8; // number o f p o l e s

o f motor

236


4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

f =50; // f r e q u e n c y o f motor xs =20; // p e r c e n t a g e s y n c h r o n o u s r e a c t a n c e P =3000; // r a t e d power o f motor m =3; // number o f p h a s e s vt = v / sqrt (3) ; // p e r p h a s e r a t e d v o l t a g e ia =( P *1000) /( sqrt (3) * v ) ; // p e r p h a s e a r m a t u r e current xs =( xs * vt ) /(100* ia ) ; // s y n c h r o n o u s r e a c t a n c e i n ohm disp ( ’ c a s e a ’ ) ; de =0; // a t no l o a d l o a d a n g l e =0 and e x c i t a t i o n v o l t a g e=p e r p h a s e r a t e d v o l t a g e ps = ceil ((( m * vt ^2* cosd ( de ) * %pi * p ) /( xs *360) ) /1000) ; printf ( ’ S y n c h r o n o z i n g power p e r m e c h a n i c a l d e g r e e i s %f KW\n ’ , ps ) ; T =(( ps *1000* p *60) /(2* %pi *120* f ) ) /1000; printf ( ’ C o r r e s p o n d i n g s y n c h r o n i z i n g t o r q u e i s %f KN− m\n ’ ,T ) ; disp ( ’ c a s e b ’ ) ; pf =0.8; // l a g g i n g power f a c t o r Ef = vt + %i * ia *( pf - sqrt (1 - pf ^2) * %i ) * xs ; // E x c i t a t i o n EMF de = atand ( imag ( Ef ) , real ( Ef ) ) ; ps =(( m * vt * abs ( Ef ) * cosd ( de ) * %pi * p ) /( xs *360) ) /1000; printf ( ’ S y n c h r o n o z i n g power p e r m e c h a n i c a l d e g r e e i s %f KW\n ’ , ceil ( ps ) ) ; T =(( ps *1000* p *60) /(2* %pi *120* f ) ) /1000; printf ( ’ C o r r e s p o n d i n g s y n c h r o n i z i n g t o r q u e i s %f KN− m\n ’ ,T ) ;

Scilab code Exa 5.37 Determining 1 synchronizing current power and torque 2 armature current and angle by which rotor slips back 1 clc ; 2 vt =3300; // t e r m i n a l v o l t a g e 3 xs =11; // s y n c h r o n o u s r e a c t a n c e p e r p h a s e

237


4 p =8; // number o f p o l e s 5 f =50; // f r e q u e n c y o f motor 6 m =3; // number o f p h a s e s 7 // from f i g 5 . 8 2 8 // a t no l o a d l o a d a n g l e =0 and e x c i t a t i o n 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

v o l t a g e=

terminal voltage de =0; s = p /2; // e l e c t r i c a l d e g r e e e q u i v a l e n t o f one mechanical degree in space es =2* vt * sind ( s /2) ; // s y n c h r o n i z i n g v o l t a g e is = es / xs ; printf ( ’ S y n c h r o n i z i n g c u r r e n t i n t h e a r m a t u r e i s %f A\n ’ , is ) ; ps = m * vt * is * cosd ( de + s /2) ; printf ( ’ s y n c h o n i z i n g power i s %f KW\n ’ , floor ( ps /1000) ) ; ws =(2* %pi *120* f ) /(60* p ) ; T = ps / ws ; printf ( ’ S y n c h r o n i z i n g t o r q u e t e n d i n g t o r e s t o r e r o t o r t o i t s p r e v i o u s p o s i t i o n i s %f Nm\n ’ ,T ) ; disp ( ’ c a s e b ’ ) ; ia =30; // a r m a t u r e c u r r e n t dde =2*( asind (( ia * xs ) /(2* vt ) ) ) ; // c h a n g e i n l o a d angle in e l e c t r i c a l degrees s = dde *(2/ p ) ; printf ( ’ R o t o r s l i p s back by %f m e c h a n i c a l d e g r e e s ’ ,s );

Scilab code Exa 5.38 Determining 1 load angle armature current and pf 2 maximum load load angle and armature current 3 minimum excitation voltage 1 clc ; 2 v =400; // r a t e d v o l t a g e o f motor 3 f =50; // f r e q u e n c y o f motor

238


4 5 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

r =1; // p e r p h a s e r e s i s t a n c e x =5; // p e r p h a s e r e a c t a n c e m =3; // number o f p h a s e s p =15000; // r a t e d power o f motor disp ( ’ c a s e a ’ ) ; EF =480; // E x c i t a t i o n v o l t a g e ph = p / m ; // p e r p h a s e power vt = v / sqrt (3) ; // t e r m i n a l v o l t a g e R0 = vt /(2* r ) ; printf ( ’ R a d i u s o f z e r o power c i r c l e i s %f A\n ’ , R0 ) ; R1 = sqrt ( R0 ^2 -( ph / r ) ) ; printf ( ’ R a d i u s o f p e r p h a s e power c i r c l e i s %f A\n ’ , R1 ) ; Ef = EF / sqrt (3) ; // p e r p h a s e e x c i t a t i o n v o l t a g e i1 = vt / sqrt ( r ^2+ x ^2) ; i2 = Ef / sqrt ( r ^2+ x ^2) ; // c u r r e n t p h a s o r s l a g g i n g t e r m i n a l and e x c i t a t i o n v o l t a g e printf ( ’ C u r r e n t p h a s o r s l a g g i n g t e r m i n a l v o l t a g e i s %f A\n ’ , i1 ) ; printf ( ’ C u r r e n t p h a s o r s l a g g i n g e x c i a t i o n v o l t a g e i s %f A\n ’ , i2 ) ; disp ( ’ u s i n g t h e a b o v e d a t a power c i r c l e d i a g r a m i s drawn and v a l u e o f a r m a t u r e c u r r e n t and power f a c t o r i s o b t a i n e d ’ ); ia =26; pf =0.955; printf ( ’ Armature c u r r e n t i s %f A\n ’ , ia ) ; printf ( ’ Power f a c t o r i s %f l e a d i n g \n ’ , pf ) ; disp ( ’ c a s e b ’ ) ; // from power c i r c l e diagram , r a d i u s f o r maximum power i s 61 A R2 =61; // r a d i u s f o r maximum power pmax =( R0 ^2 - R2 ^2) * r ; printf ( ’ Maximum power p e r p h a s e i s %f KW\n ’ , pmax /1000) ; printf ( ’ Maximum power f o r 3− p h a s e i s %f KW\n ’ ,(3* pmax ) /1000) ; disp ( ’ c a s e c ’ ) ; 239


33 l =12; // l o a d on motor i n KW 34 lp =( l /3) *1000; // p e r p h a s e l o a d 35 ef =( sqrt ( r ^2+ x ^2) ) *( R0 - sqrt ( R0 ^2 - lp / r ) ) ; 36 printf ( ’ Minimum e x c i t a t i o n v o l t a g e i s %f V ’ , ef ) ;

Scilab code Exa 5.41 Determining 1 excitation voltage 2 reluctance power developed 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

clc ; n =1490; // s p e e d o f machine i n rpm p =4; // number o f p o l e s f =50; // f r e q u e n c y v =11000; // r a t e d v o l t a g e o f machine p =20*10^6; // r a t e d power o f machine v1 =30; v2 =25; // p e r p h a s e v o l t a g e f o r p h a s e A o f machine i1 =10; i2 =6.5; // p e r p h a s e c u r r e n t f o r p h a s e A o f machine ns =(120* f ) / p ; // s y n c h r o n o u s s p e e d o f machine xd = v1 / i2 ; // d−a x i s s y n c h r o n o u s r e a c t a n c e xq = v2 / i1 ; // q−a x i s s y n c h r o n o u s r e a c t a n c e disp ( ’ c a s e a ’ ) ; ia = p /( sqrt (3) * v ) ; // a r m a t u r e c u r r e n t vt = v / sqrt (3) ; // p e r p h a s e a r m a t u r e v o l t a g e Ef = vt + ia * xq * %i ; de = atand ( imag ( Ef ) , real ( Ef ) ) ; // l o a d a n g l e id = ia * sind ( de ) ; // d−a x i s c u r r e n t Ef1 = abs ( Ef ) + id *( xd - xq ) ; printf ( ’ Per p h a s e e x c i t a t i o n v a l u e i s %f V\n ’ , ceil ( Ef1 ) ) ; printf ( ’ L i n e v a l u e o f e x c i t a t i o n EMf i s %f V\n ’ , ceil ( sqrt (3) * Ef1 ) ) ; disp ( ’ c a s e 2 ’ ) ; pr =( vt ^2*( xd - xq ) * sind (2* de ) ) /(2* xd * xq ) ; printf ( ’ R e l u c t a n c e power d e v e l o p e d by machine i s %f 240


KW p e r p h a s e \n ’ , pr /1000) ; 26 printf ( ’ T o t a l r e l u c t a n c e power d e v e l o p e d by machine i s %f KW’ ,(3* pr ) /1000) ;

Scilab code Exa 5.43 Determining efficiency at 1 half load 2 full load 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

clc ; p =100*10^3; // r a t e d power o f a l t e r n a t o r v =440; // r a t e d v o l t a g e o f a l t e r n a t o r m =3; // number o f p h a s e s l1 =340; // f r i c t i o n and w i n d a g e l o s s e s l2 =480; // open c i r c u i t c o r e l o s s e s rf =180; // f i e l d w i n d i n g r e s i s t a n c e a t 75 d e g r e e c e l . ra =0.02; // a r m a t u r e r e s i s t a n c e p e r p h a s e vf =220; // v o l t a g e a p p l i e d t o f i e l d w i n d i n g pf =0.8; // power f a c t o r disp ( ’ At h a l f l o a d ’ ) ; ia = p /( sqrt (3) * v ) ; // f u l l l o a d a r m a t u r e c u r r e n t l3 =( m * ia ^2* ra ) /4; // s h o r t c i r c u i t l o a d l o s s a t h a l f load l4 = vf ^2/ rf ; // f i e l d c i r c u i t l o s s lt = l1 + l2 + l3 + l4 ; // n e t l o s s n =(1 -( lt /(( p /2) * pf + lt ) ) ) *100; printf ( ’ E f f i c i e n c y i s %f p e r c e n t \n ’ ,n ) ; disp ( ’ At f u l l l o a d ’ ) ; l3 = m * ia ^2* ra ; // s h o r t c i r c u i t l o a d l o s s a t f u l l load lt = l1 + l2 + l3 + l4 ; // n e t l o s s n =(1 -( lt /( p * pf + lt ) ) ) *100; printf ( ’ E f f i c i e n c y i s %f p e r c e n t \n ’ ,n ) ;

241


Scilab code Exa 5.44 Determining effective resistance in pu and ohms and ratio of ac to dc resistance 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

clc ; p =40000; // r a t e d power o f machine v =400; // r a t e d v o l t a g e o f machine l =1500; // s h o r t c i r c u i t l o a d l o s s m =3; // number o f p h a s e s ia1 =1; // a r m a t u r e c u r r e n t i n p . u . ra =0.118; // dc a r m a t u r e r e s i s t a n c e a t 30 d e g r e e c e l . ia2 = p /( sqrt (3) * v ) ; // r a t e d a r m a t u r e c u r r e n t l1 = l / p ; // s h o r t c i r c u i t l o s s i n p . u . ra1 = l1 / ia1 ^2; printf ( ’ E f f e c t i v e a r m a t u r e r e s i s t a n c e i s %f p . u . \ n ’ , ra1 ) ; l2 = l / m ; // s h o r t c i r c u i t l o a d l o s s p e r p h a s e ra2 = l2 / ia2 ^2; printf ( ’ E f f e c t i v e a c a r m a t u r e r e s i s t a n c e i s %f ohm p e r p h a s e \n ’ , ra2 ) ; r = ra2 / ra ; printf ( ’ R a t i o o f a c t o dc a r m a t u r e r e s i s t a n c e i s g i v e n a s %f ’ ,r ) ;

Scilab code Exa 5.45 Determining 1 efficiency and 2 maximum efficiency 1 2 3 4 5 6 7 8 9

clc ; p =500*10^3; // r a t e d power o f a l t e r n a t o r v =11000; // r a t e d v o l t a g e o f a l t e r n a t o r m =3; // number o f p h a s e s l1 =1500; // f r i c t i o n and w i n d a g e l o s s e s l2 =2500; // open c i r c u i t c o r e l o s s e s ra =4; // a r m a t u r e r e s i s t a n c e p e r p h a s e l3 =1000; // f i e l d c o p p e r l o s s pf =0.8; // power f a c t o r 242


10 disp ( ’ c a s e a : H a l f l o a d ’ ) ; 11 ia = p /( sqrt (3) * v ) ; // a r m a t u r e c u r r e n t 12 l4 =( m * ia ^2* ra ) /4; // s h o r t c i r c u i t l o a d

l o s s at h a l f

load 13 lt = l1 + l2 + l3 + l4 ; // n e t l o s s 14 n =(1 -( lt /(( p /2) * pf + lt ) ) ) *100; 15 printf ( ’ E f f i c i e n c y i s %f p e r c e n t \n ’ ,n ) ; 16 disp ( ’ c a s e b ’ ) ; 17 // f o r maximum e f f i c i e n c y v a r i a b l e l o s s e s =c o n s t a n t 18 19 20 21 22

losses iam = sqrt (( l1 + l2 + l3 ) /( m * ra ) ) ; // a r m a t u r e c u r r e n t a t maximum e f f i c i e n c y pout = m *( v / sqrt (3) ) * iam * pf ; // o u t p u t power t a maximum e f f i c i e n c y lt =2*( l1 + l2 + l3 ) ; // n e t l o s s e s nm =(1 -( lt /( pout + lt ) ) ) *100; printf ( ’ Maximum e f f i c i e n c y i s %f p e r c e n t \n ’ , nm ) ;

Scilab code Exa 5.46 Determining KVA rating of synchronous condenser and KVA of factory 1 2 3 4 5 6 7 8 9 10 11 12

clc ; l =1800; // t o t a l l o a d on f a c t o r y pf =0.6; // power f a c t o r pfn =0.95; // d e s i r e d power f a c t o r // from p h a s o r d i a g r a m 5 . 1 0 7 l1 = l / pf ; // l o a d i n VA a1 = acosd ( pf ) ; // p h a s e a n g l e b e t w e e n t e r m i n a l v o l t a g e and a r m a t u r e c u r r e n t a2 = acosd ( pfn ) ; // p h a s e a n g l e c o r r e s p o n d i n g t o d e s i r e d power f a c t o r k1 = l1 * sind ( a1 ) ; // KVAr o f l o a d k2 = l * tand ( a2 ) ; // combined KVAr disp ( ’ c a s e a ’ ) ; s = k1 - k2 ; 243


printf ( ’ S y n c h r o n o u s c o n d e n s e r r a t i n g i s %f KVA\n ’ , floor ( s ) ) ; 14 disp ( ’ c a s e b ’ ) ; 15 r = sqrt ( l ^2+ k2 ^2) ; 16 printf ( ’ T o t a l KVA o f f a c t o r y i s %f KVA ’ ,r ) ; 13

Scilab code Exa 5.47 Determining 1 total load KVA 2 KVA capacity of motor 3 power factor of motor 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

clc ; l0 =300; // t o t a l l o a d on f a c t o r y pf =0.6; // l a g g i n g power f a c t o r n =88; // p e r c e n t a g e e f f i c i e n c y o f motor pfn =0.9; // d e s i r e d power f a c t o r l1 =60; // m e c h a n i c a l l o a d t o be s u p p l i e d // from p h a s o r d i a g r a m 5 . 1 0 8 pi = l1 /( n /100) ; // s y n c h r o n o u s motor i n p u t lt = pi + l0 ; // combined l o a d disp ( ’ c a s e a ’ ) ; k = lt / pfn ; printf ( ’ T o t a l l o a d i s %f KVA\n ’ ,k ) ; disp ( ’ c a s e b ’ ) ; a1 = acosd ( pf ) ; // p h a s e a n g l e b e t w e e n t e r m i n a l v o l t a g e and a r m a t u r e c u r r e n t a2 = acosd ( pfn ) ; // p h a s e a n g l e c o r r e s p o n d i n g t o d e s i r e d power f a c t o r k1 = l0 * tand ( a1 ) ; // KVAr o f l o a d k2 = lt * tand ( a2 ) ; // combined KVAr s = k1 - k2 ; // l e a d i n g KVAr s u p p l i e d by s y n c h r o n o u s motor r = sqrt ( s ^2+ pi ^2) ; printf ( ’ C a p a c i t y o f s y n c h r o n o u s motor i s %f KVA\n ’ ,r ); disp ( ’ c a s e c ’ ) ; pfs = pi / r ; 244


23

printf ( ’ S y n c h r o n o u s motor o p e r a t i n g power f a c t o r i s %f l e a d i n g ’ , pfs ) ;

Scilab code Exa 5.48 Calculating permissible additional load and rating of synchronous condenser 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

clc ; p0 =1000; // f u l l l o a d power r a t i n g o f s u b s t a t i o n pf =0.71; // l a g g i n g power f a c t o r pfn =0.87; // d e s i r e d power f a c t o r // from p h a s o r dagram 5 . 1 0 9 p1 = p0 * pf ; // l o a d KW p2 = sqrt ( p0 ^2 - p1 ^2) ; // l o a d KVAr pn = p0 * pfn ; // new power d e l i v e r e d t o l o a d p0n = pn / pf ; // new l o a d KVA pl = p0n - p0 ; printf ( ’ P e r m i s s i b l e a d d i t i o n a l l o a d a t %f l a g g i n g power f a c t o r i s %f KVA\n ’ ,pf , pl ) ; p2n = sqrt ( p0n ^2 - pn ^2) ; // new l o a d KVAr p2ns = sqrt ( p0 ^2 - pn ^2) ; // new l o a d KVAr w i t h t h e u s e of synchronous condensor R = p2n - p2ns ; printf ( ’ R a t i n g o f s y n c h r o n o u s c o n d e n s o r i s %f KVA ’ ,R );

Scilab code Exa 5.49 Determining new pf and percentage reduction in line current 1 2 3 4

clc ; p =4000; // l o a d t a k e n by i n d u s t r i a l p l a n t i n KW pf =0.8; // l a g g i n g power f a c t o r l =400; // r a t i n g o f i n d u c t i o n motor t o be r e p l a c e d by s y n c h r o n o u s motor 245


5 n =0.9; // e f f i c i e n c y

o f i n d u c t i o n motor and

s y n c h r o n o u s motor 6 pf1 =0.9; // l a g g i n g power f a c t o r a t which i n d u c t i o n

motor o p e r a t e s 7 pf2 =0.8; // l e a d i n g power f a c t o r a t which 8 9 10 11 12 13 14 15 16

s y n c h r o n o u s motor o p e r a t e s A =p - %i * p * tand ( acosd ( pf ) ) ; // c o m p l e x power o f p l a n t pi = l / pf1 ; // power i n p u t t o i n d u c t i o n motor B = pi - %i * pi * tand ( acosd ( pf1 ) ) ; // c o m p l e x power r e q u i r e m e n t o f i n d u c t i o n motor C = pi + %i * pi * tand ( acosd ( pf2 ) ) ; // c o m p l e x power r e q u i r e m e n t o f s y n c h r o n o u s motor pfn = cosd ( atand ( imag (A - B + C ) , real (A - B + C ) ) ) ; printf ( ’ New power f a c t o r o f t h e p l a n t i s %f l a g g i n g \ n ’ , pfn ) ; r =( abs (A - B + C ) / sqrt (3) ) /( p /( sqrt (3) * pf ) ) ; // r a t i o o f new l i n e c u r r e n t t o o r i g i n a l l i n e c u r r e n t pr =(1 - r ) *100; printf ( ’ P e r c e n t a g e r e d u c t i o n i n l i n e c u r r e n t i s %f p e r c e n t ’ , pr ) ;

Scilab code Exa 5.53 Determining voltage regulation at full load 1 2 3 4 5 6 7 8 9 10 11 12

clc ; m =3; // number o f p h a s e s p =2; // number o f p o l e s P =4*10^6; // r a t e d power o f g e n e r a t o r v =11000; // r a t e d v o l t a g e o f g e n e r a t o r as =72; // a r m a t u r e s l o t s cs =4; // c o n d u c t o r p e r a r m a t u r e s l o t rs =24; // r o t o r s l o t s rp =10; // r o t o r s l o t a n g u l a r p i t c h cr =20; // c o n d u c t o r s p e r r o t o r s l o t z =0.1+2* %i ; // a r m a t u r e l e a k a g e i m p e d a n c e p e r p h a s e pf =0.8; // l a g g i n g power f a c t o r 246


13 vt = v / sqrt (3) ; // t e r m i n a l v o l t a g e 14 ia = P /( sqrt (3) * v ) ; // f u l l l o a d a r m a t u r e c u r r e n t 15 // Open c i r c u i t c h a r a c t e r i s t i c s have b e e n p l o t t e d

16 17 18 19 20 21 22 23 24

25 26 27 28 29 30 31 32 33 34 35 36 37 38

u s i n g t a b l e g i v e n i n q u e s t i o n . Per p h a s e v a l u e o f excita tion voltage i s obtained fron table IF =[ 40 80 120 160 200 240 280 320 360]; EA =[ 2490 4980 7470 9390 10620 11460 12030 12450 12660 ]; plot ( IF , EA / sqrt (3) ) ; xlabel ( ’ F i e l d c u r r e n t ’ ) ; ylabel ( ’ open c i r c u i t v o l t a g e ’ ) ; title ( ’ open c i r c u i t c h a r a c t e r i s t i c s ’ ) ; Er = vt + ia *( pf - %i * sqrt (1 - pf ^2) ) * z ; // a i r gap v o l t a g e printf ( ’ A i r gap v o l t a g e i s %f V\n ’ , abs ( Er ) ) ; disp ( ’ C o r r e s p o n d i n g t o m a g n i t u d e o f a i r gap v o l t a g e v a l u e o f f i e l d MMF i n t e r m s o f f i e l d c u r r e n t i s o b t a i n e d from O . C . C ( f o r t e x t b o o k r e f e r f i g . 5 . 1 1 4 ) ’ ); Fr =242; // f i e l d MMF i n t e r m s o f f i e l d c u r r e n t q = rs / p ; // r o t o r s l o t s p e r p o l e kd = sind ( q * rp /2) /( q * sind ( rp /2) ) ; // d i s t r i b u t i o n factor kp =1 ; // c o i l s p a n f a c t o r f o r f u l l p i t c h f i e l d c o i l kw = kd * kp ; // w i n d i n g f a c t o r Nf =( rs * cr ) / p ; // t o t a l f i e l d t u r n s F1f =(4* kw * Nf ) /( %pi * p ) ; // r a t i o o f f u n d a m e n t a l f i e l d mmf p e r p o l e t o f i e l d c u r r e n t Nph =( as * cs ) /( m * p ) ; // s e r i e s t u r n p e r p h a s e q1 = as /( m * p ) ; // a r m a t u r e s l o t s p e r p o l e p e r p h a s e v1 =( p *180) / as ; // a r m a t u r e s l o t a n g u l a r p i t c h kd =( sind ( q1 * v1 /2) ) /( q1 * sind ( v1 /2) ) ; // d i s t r i b u t i o n factor kw = kd * kp ; // w i n d i n g f a c t o r Fa =(0.9* m * Nph * ia * kw ) /( p * F1f ) ; // a r m a t u r e mmf i n o f f i e l d c u r r e n t B = acosd ( pf ) + atand ( imag ( Er ) , real ( Er ) ) ; // power f a c t o r a n g l e + a n g l e by which a i r gap v o l t a g e leads terminal voltage 247


39 Ff = sqrt ( Fr ^2+ Fa ^2 -2* Fr * Fa * cosd (90+ B ) ) ; // e q u i v a l e n t

f i e l d current printf ( ’ E q u i v a l e n t f i e l d c u r r e n t i s %f A\n ’ , Ff ) ; // c o r r e s p o n d i n g t o e q u i v a l e n t f i e l d c u r r e n t , e x c i t a t i o n v o l t a g e i s o b t a i n e d from O . C . C 42 Ef =7168; // e x c i t a t i o n EMF p e r p h a s e 43 vr =(( Ef - vt ) / vt ) *100; 44 printf ( ’ V o l t a g e r e g u l a t i o n a t f u l l l o a d %f l a g g i n g power f a c t o r i s %f p e r c e n t ’ ,pf , vr ) ; 40 41

Scilab code Exa 5.54 Determining terminal voltage and current 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

clc ; p =2*10^6; // r a t e d power o f a l t e r n a t o r v =11000; // r a t e d v o l t a g e o f a l t e r n a t o r zs =0.3+5* %i ; // s y n c h r o n o u s i m p e d a n c e p e r p h a s e pf =0.8; // l a g g i n g power f a c t o r vt = v / sqrt (3) ; // t e r m i n a l v o l t a g e ia = p /( sqrt (3) * v ) ; // f u l l l o a d a r m a t u r e c u r r e n t // w i t h v t a s r e f e r e n c e p h a s o r Ef = vt + ia *( pf - sqrt (1 - pf ^2) * %i ) * zs ; // now e x c i t a t i o n l e v e l i s same but l o a d power fcator is leading printf ( ’ Load c u r r e n t i s %f A\n ’ , ia ) ; de = cosd ( atand ( imag ( Ef ) , real ( Ef ) ) ) ; // a n g l e b e t w e e n e x c i t a t i o n and t e r m i n a l v o l t a g e vt = abs ( Ef ) *( de + sqrt (1 - de ^2) * %i ) - ia *( pf + sqrt (1 - pf ^2) * %i ) * zs ; printf ( ’ T e r m i n a l v o l t a g e a t %f l e a d i n g power f a c t o r i s %f V p e r p h a s e \n ’ ,pf , abs ( vt ) ) ; printf ( ’ L i n e t e r m i n a l v o l t a g e i s %f KV ’ ,( sqrt (3) * abs ( vt ) ) /1000) ;

248


Scilab code Exa 5.55 Determining load angle and power factor 1 2 3 4 5 6 7 8 9 10 11 12 13 14

clc ; p =30*10^6; // r a t e d power o f a l t e r n a t o r v =11000; // r a t e d v o l t a g e o f a l t e r n a t o r zs =0.005+0.70* %i ; // p . u s y n c h r o n o u s r e a c t a n c e Ef =1.5; // p . u . e x c i t a t i o n EMF ia =1; // p . u . a r m a t u r e c u r r e n t vt =1; // p . u . t e r m i n a l v o l t a g e t1 = Ef ^2 -( real ( zs ) * ia ) ^2 -( imag ( zs ) * ia ) ^2 -1; t2 = sqrt ((2* ia * real ( zs ) ) ^2+(2* ia * imag ( zs ) ) ^2) ; t3 = atand ((2* ia * real ( zs ) ) /(2* ia * imag ( zs ) ) ) ; // t e r m s n e e d e d t o f i n d o u t power f a c t o r pf = cosd ( asind ( t1 / t2 ) - t3 ) ; printf ( ’ Power f a c t o r i s %f l a g g i n g \n ’ , pf ) ; de = acosd (( ia * abs ( zs ) ^2 - Ef ^2 - vt ^2) /( -2* Ef * vt ) ) ; printf ( ’ Load a n g l e i s %f d e g r e e s ’ , de ) ;

Scilab code Exa 5.56 Determining 1 terminal voltage and armature current 2 load angle and excitation voltage 1 2 3 4 5 6 7 8 9

clc ; xd =1.2; // pu d−a x i s s y n c h r o n o u s r e a c t a n c e xq =0.8; // pu q−a x i s s y n c h r o n o u s r e a c t a n c e xe =0.1; // pu e x t e r n a l r e a c t a n c e vb =1; // v o l t a g e o f i n f i n i t e bus pf =0.9; // l a g g i n g power f a c t o r disp ( ’ c a s e a ’ ) ; // vb=vt−j ∗ i a ∗ xe −(1) where i a i s a r m a t u r e c u r r e n t // i a=i a ∗ ( pf −%i ∗ s q r t (1− p f ˆ 2 ) ) ; // c o m p l e x form o f armature c u r r e n t 10 // v t ∗ i a =1 t h e r e f o r e i a =1/ v t s o l v i n g e q u a t i o n 1 we g e t a q u a d r a t i c e q u a t i o n i n v t whose t e r m s a r e 11 t1 =1; 12 t3 = -2* xe * sqrt (1 - pf ^2) - vb ; 249


13 t5 =( xe * sqrt (1 - pf ^2) ) ^2 -( pf * xe ) ^2; // t e r m s o f

quadratic equation in terminal voltage 14 p = [ t1 0 t3 0 t5 ]; 15 vt = roots ( p ) ; 16 ia =1/ vt (2) ; // pu a r m a t u r e c u r r e n t 17 printf ( ’ G e n e r a t o r t e r m i n a l v o l t a g e

i s %f pu \n ’ , vt (2)

); printf ( ’ Armature c u r r e n t i s %f pu \n ’ , ia ) ; disp ( ’ c a s e b ’ ) ; Ef = vt (2) + ia *( pf - %i * sqrt (1 - pf ^2) ) * %i * xq ; de = atand ( imag ( Ef ) , real ( Ef ) ) ; pa = acosd ( pf ) ; // power f a c t o r a n g l e id = ia * sind ( de + pa ) ; // d−a x i s component o f a r m a t u r e current 24 Ef = abs ( Ef ) + id *( xd - xq ) ; 25 printf ( ’ Load a n g l e i s %f d e g r e e s \n ’ , de ) ; 26 printf ( ’ E x c i t a t i o n v o l t a g e i s %f pu ’ , Ef ) ; 18 19 20 21 22 23

Scilab code Exa 5.57 Calculating maximum power output and minimum pu excitation 1 2 3 4 5 6 7

8 9 10

clc ; xd =0.85; // r e a c t a n c e a l o n g d−a x i s xq =0.55; // r e a c t a n c e a l o n g q−a x i s vt =1; // pu bus v o l t a g e Ef =1.2; // pu e x c i t a t i o n EMF // P=( Ef ∗ v t ∗ s i n ( de ) ) / xd + ( v t ˆ 2 / 2 ) ∗ ( ( 1 / xq ) −(1/ xd ) ) ∗ s i n ( 2 ∗ de ) where p i s power and de i s l o a d a n g l e // f o r maximum power dp / dde ( d e r i v a t i v e w i t h r e s p e c t t o l o a d a n g l e ) i s z e r o . S o l v i n g we g e t a q u a d r a t i c e q u a t i o n whose t e r m s a r e p =[ ( vt ^2/2) *((1/ xq ) -(1/ xd ) ) *4 ( Ef * vt ) / xd -( vt ^2/2) *((1/ xq ) -(1/ xd ) ) *2 ]; l = roots ( p ) ; an = l (2) ; 250


11 de = acos ( an ) *(180/ %pi ) ; // l o a d a n g l e 12 13 pmax =( Ef * vt * sin ( de *( %pi /180) ) ) / xd + ( vt ^2/2) *((1/ xq ) 14

15 16 17 18 19 20

-(1/ xd ) ) * sin (2* de *( %pi /180) ) ; printf ( ’ Maximum power o u t p u t t h a t motor can s u p p l y w i t h o u t l o s s o f s y n c h r o n i z a t i o n i s %f pu \n ’ , pmax ) ; // c o s ( de ) =( v t ˆ2/ p ) ∗ ( ( xd−xq ) / ( xd+xq ) ) ∗ s i n ( de ) ˆ3 where de i s l o a d a n g l e f o r minimum e x c i t a t i o n EMF // by t r i a l and e r r o r v a l u e o f de i s de =63; P =1; // pu power Ef =( P -(( vt ^2/2) *(( xd - xq ) /( xd * xq ) ) * sind (2* de ) ) ) /(( vt / xd ) * sind ( de ) ) ; printf ( ’ Minimum e x c i t a t i o n EMF f o r machine t o s t a y i n s y n c h r o n i s m i s %f pu \n ’ , Ef ) ;

Scilab code Exa 5.58 Determining voltage regulation 1 2 3 4 5 6 7 8 9 10 11 12 13 14

clc ; p =3*10^6; // r a t e d power o f a l t e r n a t o r v =11000; // r a t e d v o l t a g e o f a l t e r n a t o r r =0.4; // p e r p h a s e e f f e c t i v e r e s i s t a n c e vl =12370; // l i n e t o l i n e v o l t a g e a t z e r o l e a d i n g power f a c t o r i =100; // l o a d c u r r e n t a t z e r o power f a c t o r pf =0.8; // l a g g i n g power f a c t o r a t which v o l t a g e r e g u l a t i o n h a s t o be d e t e r m i n e d vt = vl / sqrt (3) ; // p e r p h a s e t e r m i n a l v o l t a g e Ef = v / sqrt (3) ; // p e r p h a s e e x c i t a t i o n EMF ia = p /( sqrt (3) * v ) ; // f u l l l o a d p h a s e c u r r e n t // f o r z e r o power f a c t o r l o a d a n g l e =0 zs =( vt - Ef ) / i ; // s y n c h r o n o u s i m p e d a n c e xs = sqrt ( zs ^2 - r ^2) ; // s y n c h r o n o u s r e a c t a n c e // From p h a s o r d i a g r a m 251


15 Ef1 = sqrt (( Ef * pf + ia * r ) ^2+( Ef * sqrt (1 - pf ^2) + ia * xs ) ^2) ;

// e x c i t a t i o n EMF a t 0 . 8 power f a c t o r 16 vr =(( Ef1 - Ef ) / Ef ) *100; 17 printf ( ’ V o l t a g e r e g u l a t i o n a t %f l a g g i n g power

f a c t o r i s %f p e r c e n t ’ ,pf , vr ) ;

Scilab code Exa 5.59 Determining synchronous reactance 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

clc ; v =11000; // v o l t a g e o f i n f i n i t e bus po =15*10^6; // o u t p u t power o f a l t e r n a t o r pf =0.8; // o p e r a t i n g power f a c t o r o f s y n c h r o n o u s machine p =130; // p e r c e n t a g e i n c r e a s e i n e x c i t a t i o n EMF m =3; // number o f p h a s e s ia = po /( sqrt (3) * pf * v ) ; // p e r p h a s e a r m a t u r e c u r r e n t vb = v / sqrt (3) ; // p e r p h a s e bus v o l t a g e // from p h a s o r d i a g r a m s i n f i g 5 . 1 1 7 ( a ) and 5 . 1 1 7 ( b ) xs =( sqrt ((( p /100) * vb ) ^2 -( vb * pf ) ^2) -( vb * sqrt (1 - pf ^2) ) ) / ia ; // s y n c h r o n o u s r e a c t a n c e printf ( ’ S y n c h r o n o u s r e a c t a n c e o f machine i s %f ohms \ n ’ , xs ) ; de = asind (( po * xs ) /( m * vb ^2) ) ; printf ( ’ Load a n g l e o f machine b e f o r e e x c i t a t i o n EMF i s i n c r e a s e d i s %f d e g r e e s \n ’ , de ) ; pf = cosd ( de /2) ; printf ( ’ Power f a c t o r o f t h e machine b e f o r e e x c i t a t i o n EMF i s i n c r e a s e d i s %f l e a d i n g \n ’ , pf ) ; ia =(2* vb * sind ( de /2) ) / xs ; printf ( ’ Armature c u r r e n t o f t h e machine b e f o r e e x c i t a t i o n EMF i s i n c r e a s e d i s %f A ’ , ia ) ;

252


Chapter 6 Polyphase Induction Motors

Scilab code Exa 6.1 Determining the mechanical angle through which rotor moves 1 clc ; 2 // a f t e r 3 4 5 6 7 8 9 10 11 12 13 14

changing to phase b disp ( ’ c a s e a ’ ) ; P =2; // number o f te =(2/ P ) *120; printf ( ’ Number o f r o t o r moves i s disp ( ’ c a s e b ’ ) ; P =4; // number o f te =(2/ P ) *120; printf ( ’ Number o f r o t o r moves i s disp ( ’ c a s e c ’ ) ; P =6; // number o f te =(2/ P ) *120; printf ( ’ Number o f r o t o r moves i s

dc s u p p l y t e r m i n a l s from p h a s e a

poles m e c h a n i c a l d e g r e e s t h r o u g h which %d d e g r e e s \n ’ , te ) ; poles m e c h a n i c a l d e g r e e s t h r o u g h which %d d e g r e e s \n ’ , te ) ; poles m e c h a n i c a l d e g r e e s t h r o u g h which %d d e g r e e s \n ’ , te ) ;

253


Scilab code Exa 6.2 Determining 1 full load slip and rotor frequency 2 relative speed of stator with stator structure and rotor structure 3 relative speed of rotor with stator structure rotor structure and stator field 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

clc ; Nf =1440; // f u l l l o a d s p e e d f =50; // f r e q u e n c y disp ( ’ c a s e a ’ ) ; P = fix ((120* f ) / Nf ) ; // f o r m u l a f o r f i n d i n g p o l e s mprintf ( ’ The number o f P o l e s i s %d\n ’ ,P ) ; disp ( ’ c a s e b ’ ) ;

Ns =(120* f ) / P ; // f i n d i n g s y n c h r o n o u s s p e e d s =( Ns - Nf ) / Ns ; // f i n d i n g s l i p a t f u l l l o a d f2 = s * f ; // r o t o r f r e q u e n c y mprintf ( ’ The f u l l l o a d s l i p i s %f and t h e r o t o r f r e q u e n c y i s %f Hz\n ’ ,s , f2 ) ;

17 18 disp ( ’ c a s e c ’ ) ; 19 20 21 // s p e e d o f s t a t o r f i e l d w . r . t s t a t o r s t r u c t u r e i s Ns 22 Nss = Ns ; 23 // a n s w e r f o r s p e e d o f s t a t o r f i e l d w i t h r e s p e c t t o 24 25 26 27

s t a t o r s t r u c t u r e i s g i v e n wrong i n book Wss =(2* %pi * Ns ) /60; Nsr = Ns - Nf ; // s p e e d o f s t a t o r f i e l d w . r . t r o t o r structure Wsr =(2* %pi * Nsr ) /60; printf ( ’ The s p e e d o f s t a t o r f i e l d w . r . t s t a t o r i s %f 254


r a d / s e c , %f rpm\n and w . r . t r o t o r i s %f r a d / s e c , %f rpm\n ’ ,Wss , Nss , Wsr , Nsr ) ; 28 29 30 31 32 33 34 35

disp ( ’ c a s e d ’ ) ;

// s p e e d o f r o t o r f i e l d w . r . t s t a t o r s t r u c t u r e i s Nf+ Ns Nrr =(120* f2 ) / P ; // s p e e d o f r o t o r f i e l d w . r . t r o t o r structure Nrs = Nf + Nrr ; // a n s w e r f o r s p e e d o f r o t o r f i e l d w i t h r e s p e c t t o r o t o r s t r u c t u r e i s g i v e n wrong i n book Wrs =(2* %pi * Nrs ) /60;

36 37 38 Wrr =(2* %pi * Nrr ) /60; 39 // The s t a t o r and r o t o r

f i e l d s are s t a t i o n a r y w. r . t to each other 40 printf ( ’ The s p e e d o f r o t o r f i e l d w . r . t s t a t o r s t r u c t u r e i s %f r a d / s e c , %f rpm\n and w . r . t r o t o r s t r u c t u r e i s %f r a d / s e c , %f rpm and s p e e d o f r o t o r f i e l d w . r . t s t a t o r f i e l d i s 0 ’ ,Wrs , Nrs , Wrr , Nrr ) ;

Scilab code Exa 6.3 Determining rotor speed 1 2 3 4 5 6 7 8 9

clc ; f =50; // f r e q u e n c y o f s t a t o r P =6; NofO =90; // number o f o s c i l l a t i o n f2 = NofO /60; // r o t o r f r e q u e n c y s = f2 / f ; // s l i p Ns =(120* f ) / P ; // s y n c h r o n o u s s p e e d Nr = Ns *(1 - s ) ; // r o t o r s p e e d 255


10 11

mprintf ( ’ The r o t o r s p e e d i s %f rpm ’ , Nr ) ;

Scilab code Exa 6.4 Determining 1 speed of rotor 2 ratio of voltages at slip rings 1 2 3 4 5 6 7 8 9 10

clc ; f1 =50; // f r e q u e n c y o f s u p p l y f2 =20; // f r e q u e n c y r e q u i r e d by t h e l o a d P =4; // f o r p a r t a Nrf_ss =(120* f1 ) / P ; // Speed o f r o t o r f i e l d w . r . t stator structure Nrf_rs =(120* f2 ) / P ; // Speed o f s t a t o r f i e l d w . r . t rotor structure // Nr (+ or −) s p e e d o f r o t o r f i e l d w . r . t r o t o r = s p e e d of s t a t o r f i e l d w. r . t s t a t o r // f o r +ve s i g n r o t o r must be d r i v e n i n t h e d i r e c t i o n of s t a t o r f i e l d at a speed Nr1 = Nrf_ss - Nrf_rs ; Nr2 = Nrf_ss + Nrf_rs ; mprintf ( ’ The two s p e e d s a r e %d and %d \n ’ ,Nr1 , Nr2 ) ;

11 12 13 14 15 16 // f o r p a r t b 17 18 // f o r r o t o r s p e e d Nr1 19 s1 =( Nrf_ss - Nr1 ) / Nrf_ss ; 20 // f o r r o t o r s p e e d Nr2 21 s2 =( Nrf_ss - Nr2 ) / Nrf_ss ; 22 //On e v a l u a t i o n t h e r a t i o o f v o l t a g e s i s f o u n d a s 23 R = s1 / s2 ; //R=E1/E2 24 mprintf ( ’ The r a t i o o f two v o l t a g e s a v a i l a b l e a t t h e

s l i p r i n g s a t t h e two s p e e d s i s %d ’ ,R ) ; 256


25 26

// f o r p a r t c

Scilab code Exa 6.5 Determining 1 rotor current rotor power factor and torque 2 rotor current rotor power factor and torque after insertion of external resistance 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

clc ; P =4; N =1440; f =50; r2 =0.2; x2 =1; E2 =120; // m i s t a k e i n T e f l

// f o r p a r t a disp ( ’ For p a r t a ’ ) ; Ns =(120* f ) / P ; I2_st =120/( sqrt (( r2 * r2 ) +( x2 * x2 ) ) ) ; Rpf =( r2 ) /( sqrt (( r2 * r2 ) +( x2 * x2 ) ) ) ; Ws =(2*3.14* Ns ) /60; Te_st =(3/ Ws ) *( I2_st ) *( I2_st ) *( r2 /1) ; s_fl =( Ns - N ) / Ns ; I2_fl =( s_fl * E2 ) /( sqrt ( r2 * r2 +( s_fl * x2 * s_fl * x2 ) ) ) ; Rpf_fl =( r2 ) /( sqrt ( r2 * r2 +( s_fl * x2 * s_fl * x2 ) ) ) ; Te_fl =((3) *( I2_fl ) *( I2_fl ) *( r2 ) ) /( Ws * s_fl ) ; RATIOst_fl = I2_st / I2_fl ; RATIOtst_tfl = Te_st / Te_fl ; mprintf ( ’ At s t a r t i n g \n t h e r o t o r c u r r e n t i s %f amp \n R o t o r power f a c t o r i s %f \n Torque i s %f r a d / s e c \n ’ , I2_st , Rpf , Te_st ) ; 26 mprintf ( ’ At f u l l l o a d \n t h e r o t o r c u r r e n t i s %f amp 257


\n R o t o r power f a c t o r i s %f \n Torque i s %f r a d / s e c \n ’ , I2_fl , Rpf_fl , Te_fl ) ; 27 28 29 30 31 32 33 34 35

// f o r p a r t b disp ( ’ For p a r t b ’ ) ; r2_n = r2 +1; I2_stn = E2 /( sqrt (( r2_n * r2_n ) +( x2 * x2 ) ) ) ; Rpf_stn =( r2_n ) /( sqrt ((( r2_n ) *( r2_n ) ) +(( x2 ) *( x2 ) ) ) ) ; Te_stn =(3/ Ws ) *( I2_stn ) *( I2_stn ) *( r2_n /1) ; mprintf ( ’ At s t a r t i n g \n t h e r o t o r c u r r e n t i s %f amp \n R o t o r power f a c t o r i s %f \n Torque i s %f r a d / s e c \n ’ , I2_stn , Rpf_stn , Te_stn ) ;

Scilab code Exa 6.6 Comparing rotor ohmic losses 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

clc ; P =6; f =50; N_f =960; Ns =(120* f ) / P ; n1 =800; n2 =400; s_fl =( Ns - N_f ) / Ns ; s_1 =( Ns - n1 ) / Ns ; s_2 =( Ns - n2 ) / Ns ; Ratio_1 = s_1 / s_fl ; Ratio_2 = s_2 / s_fl ; mprintf ( ’ The R a t i o a t %d rpm i s %f \n ’ ,n1 , Ratio_1 ) ; mprintf ( ’ The R a t i o a t %d rpm i s %f \n ’ ,n2 , Ratio_2 ) ;

258


Scilab code Exa 6.7 Determining 1 stator core loss 2 total rotor losses at full load 3 total rotor ohmic loss at full load 4 full load speed 5 internal torque shaft torque and efficiency 1 2 3 4 5 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

clc ; Psh =10000; P =4; f =50; Pi =660; Pw =420; I_l =8; rs =1.2; Pi_fl =11200; I_fl =18; Ns =(120* f ) / P ; Ws =(2*3.14* Ns ) /60;

// f o r p a r t a disp ( ’ f o r p a r t a ’ ) ; Pstl = Pi - Pw -((3* I_l * I_l * rs ) /(3) ) ; mprintf ( ’ The s t a t o r c o r e l o s s i s \n %f W \n ’ , Pstl ) ; // f o r p a r t b disp ( ’ f o r p a r t b ’ ) ; Pg = Pi_fl - Pstl -(3*( I_fl / sqrt (3) ) *( I_fl / sqrt (3) ) * rs ) ; Prl = Pg - Psh ; mprintf ( ’ The r o t o r l o s s i s %f W \n ’ , Prl ) ; // f o r p a r t c disp ( ’ f o r p a r t c ’ ) ; Prol = Prl - Pw ; mprintf ( ’ The r o t o r ohmic l o s s i s %f W \n ’ , Prol ) ; // f o r p a r t d 259


35 36 37 38 39 40 41 42 43 44 45 46 47

disp ( ’ f o r p a r t d ’ ) ; s_fl = Prol / Pg ; Nr = Ns *(1 - s_fl ) ; mprintf ( ’ F u l l Load s p e e d o f r o t o r i s %f rpm \n ’ , Nr ) ; // f o r p a r t e disp ( ’ f o r p a r t e ’ ) ; Te = Pg / Ws ; Tsh = Psh /(( Ws ) *(1 - s_fl ) ) ; E =( Psh / Pi_fl ) *100; mprintf ( ’ The i n t e r n a l t o r q u e i s %f Nm \n The s h a f t t o r q u e i s %f Nm \n The motor E f f i c i e n c y i s %f p e r c e n t ’ ,Te , Tsh , E ) ;

Scilab code Exa 6.8 Determining slip of motor 1 2 3 4 5 6 7 8 9 10 11

clc ; E =0.9; L =45000; Tl =((1/0.9) -1) * L ; Rl =( Tl *2) /7; // A c c o r d i n g t o t h e g i v e n c o n d i t o i n s Pg = L + Rl +( Rl /2) ; s = Rl / Pg ; mprintf ( ’ S l i p i s %f ’ ,s ) ;

Scilab code Exa 6.9 Determining stator current rotor speed output torque and efficiency 260


1 2 3 4 5 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

clc ; P =4; r1 =0.15; x1 =0.45; r2 =0.12; x2 =0.45; Xm =28.5; s =0.04; V =400; f =50; Pfixed =400; t1 = complex (( r2 / s ) , x2 ) ; t2 = complex (0 , Xm ) ; t3 = complex (( r2 / s ) ,( x2 + Xm ) ) ; T =( t1 * t2 ) / t3 ; Zab = complex ( r1 , x1 ) + T ; Rf = real ( T ) ; I1 = V /( sqrt (3) * abs ( Zab ) ) ; ian = atand ( imag ( Zab ) , real ( Zab ) ) ; Pf = cosd ( ian ) ; I1_mag = sqrt ( real ( I1 ) * real ( I1 ) + imag ( I1 ) * imag ( I1 ) ) ; Psti = sqrt (3) * I1_mag * V * Pf ; Pg =3* I1 * I1 * Rf ; ns =(2* f ) / P ; nr =(1 - s ) * ns ; Ws =2*3.14* ns ; Pm =(1 - s ) * Pg ; Psh = Pm - Pfixed ; To =( Psh ) /((1 - s ) * Ws ) ; Psto =3* I1_mag * I1_mag * r1 ; Prto = s * Pg ; Tls = Psto + Prto + Pfixed ; Pi = Psh + Tls ; E =(1 -( Tls / Pi ) ) *100; mprintf ( ’ s t a r o r c u r r e n t = %f amp a t l a g g i n g p h a s e 261


a n g l e o f %f w . r . t i n p u t v o l t a g e \n r o t o r s p e e d = %f r p s o r %f rpm o u t p u t t o r q u e = %f Nm \n E f f i c i e n c y = %f p e r c e n t ’ ,I1 , ian , nr , nr *60 , To , E ) ;

Scilab code Exa 6.10 Determining 1 slip for maximum torque and maximum torque 2 rotor current and starting torque 3 external resistance inserted in rotor circuit 4 internal power developed 5 maximum internal power developed and corresponding slip 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

clc ; // from 6 . 9 p r o b l e m P =4; r1 =0.15; x1 =0.45; r2 =0.12; x2 =0.45; Xm =28.5; s =0.04; V =400; f =50; Pfixed =400; t =1.2; // r o t o r e f f e c t i v e t u r n s r a t i o // f o r p a r t a // A c c o r d i n g t o t h e c o n d i t i o n s and d i a g r a m t1 = complex ( r1 , x1 ) ; t2 = complex (0 , Xm ) ; t3 = complex ( r1 , x2 + Xm ) ; Ze =( t1 * t2 ) /( t3 ) ; Re = real ( Ze ) ; Xe = imag ( Ze ) ; t4 = complex ( Re ,( x2 + Xe ) ) ; 262


27 28 29 30 31 32 33 34

SmT =( r2 ) /( sqrt (( Re * Re ) +(( x2 + Xe ) *( x2 + Xe ) ) ) ) ; Ve =( V / sqrt (3) ) *( Xm /( x2 + Xm ) ) ; Ws =(4* %pi * f ) / P ; Tem =(3/ Ws ) * Ve ^2*(1/2) *(1/( Re + sqrt ( Re ^2+( x2 + Xe ) ^2) ) ) ; Pm = Tem *(1 - SmT ) * Ws ; Psh = Pm - Pfixed ; Tsh = Psh /( Ws *(1 - SmT ) ) ; mprintf ( ’ f o r p a r t a \n s l i p = %f \n maximun t o r q u e = %f Nm \n power o u t p u t = %f KW \n ’ ,SmT , Tem , Psh /1000) ;

35 36 37 // f o r p a r t b 38 s =1; 39 I2st =( Ve ) /( sqrt (( r2 + Re ) *( r2 + Re ) +( x2 + Xe ) *( x2 + Xe ) ) ) ; 40 Test =(3/ Ws ) * I2st * I2st *( r2 ) ; 41 mprintf ( ’ f o r p a r t b r o t o r c u r r e n t = %f A \n t o r q u e

= %f Nm \n ’ , I2st , Test ) ; 42 43 44 // f o r p a r t c 45 R = sqrt ( Re ^2+( x2 + Xe ) ^2) - r2 ; 46 Ra = R /( t ^2) ; 47 mprintf ( ’ f o r p a r t c \n e x t e r n a l

r e s i s i t a n c e value i s

= %f Ohm \n ’ , Ra ) ; 48 49 // f o r p a r t d 50 s1 =0.04; 51 Pm =((3*( Ve ) *( Ve ) ) * r2 *((1 - s1 ) / s1 ) ) /((( Re + r2 +(( r2 *(1 -

s1 ) / s1 ) ) ) ) *(( Re + r2 +(( r2 *(1 - s1 ) / s1 ) ) ) ) +(( x2 + Xe ) *( x2 + Xe ) ) ) ; 52 mprintf ( ’ f o r p a r t d \n power d e v e l o p e d i s %f KW \n ’ , Pm /1000) ; 53 54 // f o r p a r t e 55 SmP =( r2 ) /( sqrt ((( Re + r2 ) *( Re + r2 ) ) +(( x2 + Xe ) *( x2 + Xe ) ) ) +

r2 ) ; 56 Pmn =((3/2) * Ve * Ve ) /( Re + r2 + sqrt (( r2 + Re ) *( r2 + Re ) +( x2 + Xe 263


) *( x2 + Xe ) ) ) ; 57 mprintf ( ’ f o r p a r t e \n s l i p = %f \n power d e v e l o p e d = %f KW’ ,SmP , Pmn /1000) ;

Scilab code Exa 6.11 Determining maximum internal torque for different cases 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

clc ; P =4; r1 =0.15; x1 =0.45; r2 =0.12; x2 =0.45; Xm =28.5; s =0.04; V =400; f =50; Pfixed =400; // from p r o b l e m 6 . 1 0 Re =0.1476; Xe =0.443; r2 =0.12; x2 =0.45;

a = Xm /( x2 + Xm ) ; // Ve=a ∗V1 ; Wr =(4* %pi * f ) / P ; b =(3/ Wr ) *(1/2) *(1/(( Re ) +( sqrt (( Re * Re ) +(( x2 + Xe ) *( x2 + Xe ) ) ) ) ) ) ; 23 //Tem=b∗Ve∗Ve 24 25 // f o r p a r t a 26 V1 =230; 27 Ve1 = a * V1 ;

264


28 29

Tem1 = b * Ve1 * Ve1 ; mprintf ( ’ f o r p a r t a \n maximum i n t e r n a l t o r q u e d e v e l o p e d i s %f Nm \n ’ , Tem1 ) ;

30 31 // f o r p a r t b 32 V2 =115; 33 Ve2 = a * V2 ; 34 Tem2 = b * Ve2 * Ve2 ; 35 mprintf ( ’ f o r p a r t b \n maximum i n t e r n a l

torque

d e v e l o p e d i s %f Nm \n ’ , Tem2 ) ; 36 37 38 39 40 41 42 43 44 45 46

// f o r f =25 Hz Xe1 =(1/2) * Xe ; x21 =(1/2) * x2 ; Ws1 =(1/2) * Wr ;

// f o r p a r t c V3 =115; Ve3 = a * V3 ; Tem3 =(3/ Ws1 ) * Ve3 * Ve3 *(1/2) *(1/(( Re ) +( sqrt (( Re * Re ) +(( x21 + Xe1 ) *( x21 + Xe1 ) ) ) ) ) ) 47 mprintf ( ’ f o r p a r t c \n maximum i n t e r n a l t o r q u e d e v e l o p e d i s %f Nm \n ’ , Tem3 ) ; 48 49 50 51 52 53 54 55 56 57 58 59

// f o r f =5 Hz Xe2 =(1/10) * Xe ; x22 =(1/10) * x2 ; Ws2 =(1/10) * Wr ;

// f o r p a r t d f3 =5; // f 3 = ( 1 / 1 0 ) ∗ f V4 =23; Ve4 = a * V4 ; Tem4 =(3/ Ws2 ) * Ve4 * Ve4 *(1/2) *(1/(( Re ) +( sqrt (( Re * Re ) +((( x22 + Xe2 ) *( x22 + Xe2 ) ) ) ) ) ) ) 60 mprintf ( ’ f o r p a r t d \n maximum i n t e r n a l t o r q u e 265


d e v e l o p e d i s %f Nm \n ’ , Tem4 ) ;

Scilab code Exa 6.13 determining 1 slip and rotor speed 2 rotor ohmic loss 3 starting torque 4 starting current 5 stator current 6 full load efficiency 7 slip at maximum torque for new resistance 8 full load slip 9 starting torque 10 starting current 11 rotor losses 12 power 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

// a n s w e r match + r o o t s clc ; Pm =10000; V =400; f =50; smT =0.1; P =4; Ns =(120* f ) / P ; // f o r ( i ) disp ( ’ ( i ) ’ ) ; // As p e r g i v e n c o n d i t i o n s t h e s l i p i s g i v e n by e q u a t i o n S f l 2 −0.4 S f l +0.01=0 V =[1 -0.4 0.01]; R = roots ( V ) ; Sfl = R (2) ; Nr = Ns *(1 - Sfl ) ; mprintf ( ’ The s l i p i s %f \n The r o t o r s p e e d i s %f r . p .m ’ ,Sfl , ceil ( Nr ) ) ;

19 20 // f o r ( i i ) 21 disp ( ’ ( i i ) ’ ) ; 22 Pg = Pm /(1 - Sfl ) ; 23 Prot = Sfl * Pg ; 24 mprintf ( ’ The r o t o r ohmic l o s s 25 26 // f o r ( i i i )

i s %f W \n ’ , Prot ) ;

266


27 disp ( ’ ( i i i ) ’ ) ; 28 Tefl = Pg /(2*3.14*( Ns /60) ) ; 29 Test =(4* Tefl ) /(( smT ) +(1/ smT ) ) ; 30 mprintf ( ’ s t a r t i n g t o r q u e i s %f Nm \n ’ , Test ) ; 31 32 // f o r ( i v ) 33 disp ( ’ ( i v ) ’ ) ; 34 a = sqrt ((( Sfl * Sfl ) +( smT * smT ) ) /(( Sfl ) *( Sfl ) *(1+( smT ) *( 35

smT ) ) ) ) ; mprintf ( ’ s t a r t i n g c u r r e n t = %f f u l l l o a d c u r r e n t \n ’ , a);

36 37 // f o r ( v ) 38 disp ( ’ ( v ) ’ ) ; 39 // a n s w e r i s s l i g h t l y d i f f e r e n t i n book 40 b = sqrt ((1/2) *(1+( smT / Sfl ) ^2) ) ; 41 mprintf ( ’ s t a t o r c u r r e n t a t maximun t o r q u e = %f f u l l

l o a d c u r r e n t \n ’ ,b ) ; 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61

// f o r ( v i ) disp ( ’ ( v i ) ’ ) ; E =( Pm / Pg ) *100; mprintf ( ’ f u l l l o a d e f f i c i e n c y

i s = %f p e r c e n t \n ’ ,E ) ;

// f o r ( v i i ) disp ( ’ ( v i i ) ’ ) ; // As p e r g i v e n c o n d i t i o n s smT1 =3* smT ; mprintf ( ’ New s l i p v a l u e i s %f \n ’ , smT1 ) ; // f o r ( v i i i ) disp ( ’ ( v i i i ) ’ ) ; // A c c o r d i n g t o t h e g i v e n c o n d i t i o n s s 1 ( 2 ) −1.2 s +0.09 VV =[1 -1.2 0.09]; RR = roots ( VV ) ; s1 = RR (2) ; Nr1 = Ns *(1 - s1 ) ; mprintf ( ’ f u l l l o a d s l i p i s %f r o t o r s p e e d i s %f r . p . 267


m ’ ,s1 , Nr1 ) ; 62 63 // f o r ( i x ) 64 disp ( ’ ( i x ) ’ ) ; 65 Test1 =((2) /((1/0.3) +(0.3) ) ) *(2* Tefl ) ; 66 mprintf ( ’ s t a r t i n g t o r q u e i s %f Nm \n ’ , Test1 ) ; 67 68 // f o r ( x ) 69 disp ( ’ ( x ) ’ ) ; 70 c = sqrt (( s1 ^2+ smT1 ^2) /( s1 ^2*(1+ smT1 ^2) ) ) ; 71 mprintf ( ’ s t a r t i n g c u r r e n t = %f f u l l l o a d c u r r e n t \n ’

,c ) ; 72 73 74 75 76

// f o r ( x i ) disp ( ’ ( x i ) ’ ) ; Protfl = s1 * Pg ; mprintf ( ’ R o t o r ohmic l o s s a t f u l l l o a d t o r q u e i s %f W \n ’ , Protfl ) ;

77 78 // f o r ( x i i ) 79 disp ( ’ ( x i i ) ’ ) ; 80 Pm1 =(1 - s1 ) * Pg ; 81 E = Pm1 / Pg ; 82 mprintf ( ’ E f f i c i e n c y

i s %f p e r c e n t ’ ,E *100) ;

Scilab code Exa 6.14 Determining 1 maximum torque 2 full load rotor ohmic loss 3 slip at maximum torque 4 full load slip 5 full load torque 1 2 3 4 5 6 7

clc ; Pm =60000; P =6; s =0.04; V =400; smT =0.2; 268


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

f =50; Ns =(120* f ) / P ; Ws =(2* %pi * Ns ) /60; Wr = Ws *(1 - s ) ; Tefl = Pm / Wr ; // f o r p a r t a Tem =((( smT / s ) +( s / smT ) ) /2) * Tefl ; mprintf ( ’ f o r p a r t a \n t h e maximun t o r q u e i s %f Nm\n ’ , Tem ) ; // f o r p a r t b Prot =( s /(1 - s ) ) *( Pm ) ; mprintf ( ’ f o r p a r t b \n t h e r o t o r ohmic l o s s i s \n ’ , Prot ) ;

22 23 // f o r p a r t c 24 smT1 =2* smT ; 25 mprintf ( ’ f o r p a r t c \n 26 27 // f o r p a r t d 28 //On a n a l y s i s t h e s l i p 29 s2 =0.084; 30 mprintf ( ’ f o r p a r t d \n 31 32 // f o r p a r t e 33 T2 = Pm /(( Ws ) *(1 - s2 ) ) ; 34 mprintf ( ’ f o r p a r t e \n

THe new s l i p

%f W

i s %f \n ’ , smT1 ) ;

i s g i v e n by f u l l load s l i p

i s %f \n ’ , s2 ) ;

t h e f u l l l o a d t o r q u e i s %f Nm

\n ’ , T2 ) ;

Scilab code Exa 6.15 Determining percentage reduction in rotor circuit resistance 1

269


2 3 4 5 6 7 8

clc ; sfl =0.05; // F u l l l o a d s l i p // T e s t /Tem=a // T f l /Tem=b a =1/2; b =1/1.6; // As p e r t h e g i v e n e q u a t i o n we g e t smT1 ˆ2 −2.5 smT1 +1=0 9 Q =[1 -2.5 1]; 10 R = roots ( Q ) ; 11 smT1 = R (2) ; 12 13

// For f u l l l o a d s l i p o f 0 . 0 5 we g e t t h e e q u a t i o n smT2 ˆ2 −0.20 smT2 + 0 . 0 0 2 5 14 Q1 =[1 -0.20 0.0025]; 15 R1 = roots ( Q1 ) ; 16 smT2 = R1 (1) ; 17 18 P =(( smT1 - smT2 ) / smT1 ) *100; 19 mprintf ( ’ P e r c e n t a g e r e d u c t i o n i n

rotor circuit

r e s i s t a n c e i s %f p e r c e n t ’ ,P ) ;

Scilab code Exa 6.16 Determining external resistance inserted in rotor circuit and percentage change in current and power factor 1 2 clc ; 3 r2 =0.04; 4 x2 =0.2; 5 6 // As p e r g i v e n

c o n d i t i o n s we g e t a q u a d r a t i c e q u a t i o n i n smT which i s smTˆ2 −4∗smT+1 7 t1 =1; t2 = -4; t3 =1; 8 p =[ t1 t2 t3 ]; 9 smT = roots ( p ) ; 270


10 11 r22 = x2 * smT (2) ; 12 R = r22 - r2 ; 13 mprintf ( ’ The e x t e r n a l

r e s i s t a n e n e e d e d t o be i n s e r t e d i s %f Ohm \n ’ ,R ) ;

14 15 16 17 18 19 20 21 22 23 24 25 26 27 28

// s a y V=400( I n p u t v o l t a g e ) V =400; // w i t h o u t e x t e r n a l r e s i s t a n c e Ist = V /( sqrt (( r2 ) *( r2 ) +( x2 ) *( x2 ) ) ) ; pf = r2 /( sqrt (( r2 ) *( r2 ) +( x2 ) *( x2 ) ) ) ; // w i t h e x t e r n a l r e s i s t a n c e Ist1 = V /( sqrt (( r22 ) *( r22 ) +( x2 ) *( x2 ) ) ) ; pf1 = r22 /( sqrt (( r22 ) *( r22 ) +( x2 ) *( x2 ) ) ) ;

a =(( Ist - Ist1 ) / Ist ) *100; b =(( pf1 - pf ) / pf ) *100; mprintf ( ’ P e r c e n t a g e i n s t a r t i n g c u r r e n t i s %f \n ’ ,a ) ; 29 mprintf ( ’ P e r c e n t a g e i n power f a c t o r i s %f \n ’ ,b ) ;

Scilab code Exa 6.17 Determining starting torque 1 clc ; 2 // r 2 / x2=a 3 a =.5; 4 Test =25; 5 6 // f o r p a r t a 7 disp ( ’ For p a r t a ’ ) ; 8 // b=3(V1 ) 2/ r2Ws 9 // As p e r g i v e n c o n d i t i o n s 10 b = Test *5;

271


11 12 13 14 15 16 17 18 19 20

//When r o t o r r e s i s t a c e i s d o u b l e d Test1 = b *(1/4) ; mprintf ( ’ The s t a r t i n g t o r q u e i s %f Nm\n ’ , Test1 ) ; // f o r p a r t b disp ( ’ For p a r t b ’ ) ; // r e s i s a n c e i s h a l f Test2 = b *(2/17) ;

mprintf ( ’ The s t a r t i n g t o r q u e i s %f Nm ’ , Test2 ) ;

Scilab code Exa 6.18 Determining 1 slip at maximum torque 2 full load slip 3 rotor current in of full load current 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

// e q u a t i o n clc ; // T e s t / T e f l = 1 . 5 ; d =1.5; //Tem/ T e f l = 2 . 5 ; e =2.5; // f o r p a r t a // d=T e s t / T e f l ; // e q u a t i o n o f t o r q u e g i v e s f o l l o w i n g e q u a t i o n Q =[1 -3.33 1]; R = roots ( Q ) ; smT = R (2) ; mprintf ( ’ The s l i p a t maximun t o r q u e i s %f \n ’ , smT ) // f o r p a r t b // e q u a t i o n o f t o r q u e g i v e s Q =[1 -1.665 0.111]; R = roots ( Q ) ; sfl = R (2) ; 272


22 mprintf ( ’ The s l i p a t f u l l l o a d i s %f \n ’ , sfl ) 23 24 // f o r p a r t c 25 // I 2 s t=c ∗ I s f l As p e r t o r q u e e q u a t i o n 26 c = sqrt (( d ) *(1/ sfl ) ) ; 27 mprintf ( ’ The r o t o r c u r r e n t = %f t i m e s f u l l l o a d

c u r r e n t \n ’ ,c )

Scilab code Exa 6.19 Determining starting and maximum torque 1 2 3 4 5 6 7 8 9 10

clc ; Te =200; s =0.04; c =4; // g i v e n m u l t i p l y i n g f a c t o r o f l e a k a g e r e a c t a n c e // 3V∗V=a ∗WS a = Te * s *(((1+(1/ s ) ) *(1+(1/ s ) ) ) +(( c + c ) *( c + c ) ) ) ; Test = a *(1/((1+1) *(1+1) +( c + c ) *( c + c ) ) ) ; Tem = a *(1/2) *(1/(1+ sqrt ((1) *(1) +( c + c ) *( c + c ) ) ) ) ; mprintf ( ’ The s t a r t i n g t o r q u e i s %f Nm \n The maximun Torque i s %f Nm ’ , Test , Tem ) ;

Scilab code Exa 6.20 Determining 1 slip 2 rotor current rotor ohmic loss and rotor power factor 3 power output 1 clc ; 2 sA =0.05; // s l i p 3 4 // f o r p a r t a 5 disp ( ’ f o r p a r t a ’ ) ; 6 // Torque i s p r o p o r t i o n a l t o s / r 2 7 // As p e r g i v e n c o n d i t i o n s sB=a ∗sA 8 a =4;

273


9 sB = a * sA ; 10 mprintf ( ’ The s l i p

i s %d t i m e s p r e v i o u s s l i p and \n ’ ,

a); 11 12 // f o r p a r t b 13 disp ( ’ f o r p a r t b ’ ) ; 14 // I 2 i s d i r e c t l y p r o p o r t i o n a l t o s / r 2 15 // As p e r g i v e n c o n d i t i o n s I2B=b ∗ I2A 16 b = sB /( a * sA ) ; 17 // R o t o r ohmic l o s s e s i s d i r e c t l y p r o p o r t i o n a l t o I ∗ I 18 19 20 21 22

∗ r2 // As p e r g i v e n c o n d i t i o n s P2=c ∗P1 c=a*b; // As p e r g i v e n c o n d i t i o n s Pf2=d ∗ Pf1 d=b; mprintf ( ’ r o t o r c u r r e n t f o r new r o t o r r e s i s t a n c e i s e q u a l t o i n i t i a l r o t o r c u r r e n t \n R o t o r ohmic l o s s e s f o r new r o t o r r e s i s t a n c e =%f t i m e s i n i t i a l ohmic l o s s e s \n power f a c t o r f o r new r o t o r r e s i s t a n c e i s e q u a l t o i n i t i a l power f a c t o r ’ ,c ) ;

23 24 // f o r p a r t c 25 disp ( ’ f o r p a r t c ’ ) ; 26 // As p e r g i v e n c o n d i t i o n s Wa=e ∗Ws 27 e =1 - sA ; 28 //Wb=f ∗Ws 29 b =1 - sB ; 30 //PB=g ∗PA 31 g = b / e ; 32 mprintf ( ’ The power o u t p u t i s r e d u c e d t o %f t i m e s

p r e v i o u s v a l u e ’ ,g ) ;

Scilab code Exa 6.21 Determining 1 rotor and stator input 2 starting torque 1 clc ;

274


2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

f =50; P =6; Pmsh =10000; // S h a f t Output N =930; Pw =600; Pf =0.01* Pmsh ; // F r i c t i o n and Windage l o s s e s Ns =(120* f ) / P ; NmT =800; // Speed a t maximum t o r q u e

// f o r p a r t a disp ( ’ f o r p a r t a ’ ) ; sfl =( Ns - N ) / Ns ; Pm = Pmsh + Pf ; Pg = Pm /(1 - sfl ) ; Pst = Pg + Pw ; mprintf ( ’ T o t a l R o t o r i n p u t i s %f W \n T o t a l S t a t o r i n p u t i s %f W \n ’ ,Pg , Pst ) ;

19 20 // f o r p a r t b 21 disp ( ’ f o r p a r t b ’ ) ; 22 smT =( Ns - NmT ) / Ns ; 23 Ws =(2* %pi * Ns ) /60; 24 Tefl = Pg / Ws ; 25 Test =((( smT / sfl ) +( sfl / smT ) ) /2) *(2/(( smT ) +(1/ smT ) ) ) *

Tefl ; 26 mprintf ( ’ Maximun Torque i s %f Nm ’ , Test ) ;

Scilab code Exa 6.22 Determining maximum torque corresponding slip and starting torque 1 2 3 4

clc ; Pm =7500; V =420; f =50; 275


5 6 7 8 9 10 11 12 13 14 15 16 17 18

P =4; s =0.04; r1 =1.2; x1 =1.4; x2 =1.4; Xm =38.6;

// As p e r Thevenin ’ s E q u i v a l e n t c i r c u i t Re =( r1 * Xm ) /( Xm + x2 ) ; Xe =( x1 * Xm ) /( x2 + Xm ) ; Ve =( V / sqrt (3) ) *( Xm /( x2 + Xm ) ) ; r2 =(3) *(1 - s ) * s * Ve * Ve *(1/ Pm ) ; smT = r2 /( sqrt (( Re * Re ) +(( Xe + x2 ) *( Xe + x2 ) ) ) ) ; Tem =((3* Ve * Ve ) /((((120* f ) / P ) /60) *2* %pi ) ) *(1/2) *(1/( Re +( sqrt (( Re * Re ) +(( Xe + x2 ) *( Xe + x2 ) ) ) ) ) ) ; 19 Test =((3* Ve * Ve ) /((((120* f ) / P ) /60) *2* %pi ) ) *( r2 /((( Re + r2 ) *( Re + r2 ) ) +(( Xe + x2 ) *( Xe + x2 ) ) ) ) ; 20 mprintf ( ’ maximum t o r q u e i s %f Nm \n s l i p i s %f \n s t a r t i n g t o r q u e i s %f Nm ’ ,Tem , smT , Test ) ;

Scilab code Exa 6.23 Determining 1 maximum torque 2 starting torque 3 full load rotor ohmic loss 4 slip at full load 5 full load torque 6 slip at maximum torque 1 2 3 4 5 6 7 8 9 10 11

clc ; Pm =100000; V =420; P =6; f =50; sfl =0.04; smT =0.2; // f o r p a r t a disp ( ’ f o r p a r t a ’ ) ; Pg = Pm /(1 - sfl ) ; 276


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 44 45 46 47 48

Ws =(4* %pi * f ) / P ; Tefl = Pg / Ws ; // a=T e f l /Tem a =(1/(2/(( sfl / smT ) +( smT / sfl ) ) ) ) ; Tem = a * Tefl ; mprintf ( ’ Maximum Torque i s %f Nm \n ’ , Tem ) ; // f o r p a r t b disp ( ’ f o r p a r t b ’ ) ; // b=T e s t /Tem b =2/((1/ smT ) +( smT ) ) ; Test = b * Tem ; mprintf ( ’ The s t a r t i n g Torque i s %f Nm \n ’ , Test ) // f o r p a r t c disp ( ’ f o r p a r t c ’ ) ; Prot = sfl * Pg ; mprintf ( ’ R o t o r Ohmic l o s s e s a r e %f W \n ’ , Prot ) // f o r p a r t d disp ( ’ f o r p a r t d ’ ) ; // Output i s p r o p o r t i o n a l t o ( s (1− s ) ) / r 2 // Given c o n d i t i o n s g i v e s t h e e q u a t i o n a s s 1 ∗ s1−s 1 +0.0768 Q =[1 -1 0.0768]; R = roots ( Q ) ; s1 = R (2) ; mprintf ( ’ S l i p i s %f \n ’ , s1 ) // f o r p a r t e disp ( ’ f o r p a r t e ’ ) ; Tefl =( Pm /(1 - s1 ) ) / Ws ; mprintf ( ’ f u l l −l o a d t o r q u e i s %f Nm \n ’ , Tefl ) // f o r p a r t f disp ( ’ f o r p a r t f ’ ) ; smT1 =2* smT ; mprintf ( ’ s l i p a t maximum t o r q u e i s %f ’ , smT1 ) ; 277


Scilab code Exa 6.25 Determining 1 speed range of DC motor 2 KVA rating of stator 3 DC motor rating and maximum torque 4 number of poles 5 new speed range 1 2 3 4 5 6 7 8 9 10 11 12 13 14

clc ; P =10; f =50; Pm =48000; pf =0.8; f21 =120; // min f r e q u e n c y r a n g e f22 =300; //max f r e q u e n c y r a n g e Ns =(120* f ) / P ; // f o r f 2 =300 Nr1 =((120* f21 ) / P ) - Ns ; // f o r f 2 =600 Nr2 =((120* f22 ) / P ) - Ns ; mprintf ( ’ Thus t h e dc motor c h a n g e s s p e e d from %f t o %f rpm \n ’ ,Nr1 , Nr2 )

15 16 17 18 19 20 21 22 23 24 25 26 27

// f o r p a r t b and c s1 =( Nr1 + Ns ) / Ns ; s2 =( Nr2 + Ns ) / Ns ; Pr = Pm / pf ; Pr1 = Pr / s1 ; Pr2 = Pr / s2 ; R1 =( s1 -1) * Pr1 * pf ; R2 =( s2 -1) * Pr2 * pf ; T1 =( R1 *60) /(2* %pi * Nr1 ) ; T2 =( R2 *60) /(2* %pi * Nr2 ) ; // s t a t o r s h o u l d be a b l e t o h a n d l e h i g h e r KVA mprintf ( ’KVA r a t i n g o f i n d u c t i o n motor s t a t o r i s %f KVA\n ’ , Pr1 /1000) 28 mprintf ( ’DC motor r a t i n g i s %f KW \n Maximum t o r q u e 278


o u t p u t from DC motor i s %f Nm \n ’ , R2 /1000 , T1 ) ; 29 30 // f o r p a r t d 31 //When s p e e d i s l i m i t e d t o 2 7 0 0 rpm 32 P1 =((120* f22 ) -(120* f ) ) /2700; 33 P1 = ceil ( P1 ) ; 34 mprintf ( ’ Number o f P o l e s i s %d \n ’ , P1 ) ; 35 36 // f o r p a r t e 37 Nr11 =(( f22 *120) / P1 ) -((120* f ) / P1 ) ; 38 Nr22 =(( f21 *120) / P1 ) -((120* f ) / P1 ) ; 39 mprintf ( ’ Thus t h e new s p e e d r a n g e o f dc motor i s

from %f t o %f rpm \n ’ , Nr22 , Nr11 ) ;

Scilab code Exa 6.26 Determining starting torque 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

clc ; f =50; P =4; Pm =10000; // Rated o u t p u t N =1425; Nm =1200; // Speed a t which maximun t o r q u e i s developed Ns =(120* f ) / P ; s =( Ns - N ) / Ns ; Ws =(2* %pi * Ns ) /60; Tefl =( Pm / Ws ) *(1/(1 - s ) ) ; smT =( Ns - Nm ) / Ns ; Tem = Tefl *(( s / smT ) +( smT / s ) ) *(1/2) ; Test = Tem *(2) *(1/((1/ smT ) +( smT /1) ) ) ; mprintf ( ’ The s t a r t i n g t o r q u e i s %f Nm ’ , Test ) ;

279


Scilab code Exa 6.27 Determining slip frequency and slip at maximum torque 1 2 3 4 5 6 7 8 9

clc ; fs =2; // s l i p f r e q u e n c y V =400; f =50; V2 =340; //New v o l t a g e f2 =40; //New f r e q u e n c y smT =0.1; // s l i p a t which i t d e v e l o p s maximum t o r q u e

// maximun t o r q u e ’ s s l i p (1/ f ) 10 smT1 =( f / f2 ) * smT ; 11 12 13 14 15 16 17 18 19 20

i s d i r e c t l y p r o p o r t i o n a l to

// Maximun Torque i s d i r e c t l y p r o p o r t i o n a l t o ( (V∗V) /( f ∗ f ) ) s = fs / f ; // Ted ( D e v e l o p e d Torque ) i s p r o p o r t i o n a l t o (Tem/smT ) ∗ ( s /smT ) // Ted1 ( 4 0 0V, 5 0 Hz ) p r o p o r t i o n a l a a =(( V * V ) /( f * f ) ) *( s / smT ) ; // e q u a t i n g t h e d e v e l o p e d t o r q u e e q u a t i o n s1 = a *((( f2 ) *( f2 ) ) /(( V2 ) *( V2 ) ) ) *( smT1 ) ; fs1 = s1 * f2 ; mprintf ( ’ The new s l i p f r e q u e n c y i s %f Hz ’ , fs1 ) ;

Scilab code Exa 6.28 Determining frequency of motor 1 2 3 4 5 6

clc ; f =50; V =440; P =4; N =1490; // Rated s p e e d N1 =1600; //New Speed 280


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

Ns =(120* f ) / P ; s =( Ns - N ) / Ns ; // With n e g l e c t i n g r e s i s t a n c e s and l e a k a g e r e a c t a n c e s // Torque i s d i r e c t l y p r o p o r t i o n a l t o s / ( f r 2 ) // A p p l l y i n g t h e c o n d i t i o n f o r same t o r q u e we g e t // a=s / f a =( s / f ) ; // Ns / s=b b =120/ P ; // s =(Ns−N1 ) / Ns // U s i n g a b o v e e q u a t i o n we g e t e q u a t i o n ( f ∗ f ) −7500 f −400000 Q =[1 -7500 400000] R = roots ( Q ) ; f1 = R (2) ; mprintf ( ’ V a l u e o f new F r e q u e n c y i s %f Hz ’ , f1 ) ;

Scilab code Exa 6.29 Determining power factor input current equivalent rotor current and torque developed maximum torque and corresponding speed 1 2 3 4 5 6 7 8 9 10 11 12 13

// debug clc ; V1 =420; // s u p p l y v o l t a g e r1 =2.95; x1 =6.82; r2 =2.08; x2 =4.11; Iml =6.7; // m a g n e t i z i n g l i n e c u r r e n t Pw =269; // c o r e l o s s s =0.03; // s l i p P =12; f =50; N =(120* f ) / P ; 281


14 Ns =(120* f ) / P ; 15 16 Im = Iml / sqrt (3) ; 17 //V1=E1+Im ( r 1+j x 1 ) 18 // Above e q u a t i o n on s o l v i n g 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

g i v e s t h e s o l u t i o n a s E1

∗E1 +52.8 E1 − 1 7 5 5 7 2 . 6 5 Q =[1 52.8 -175572.62]; R = roots ( Q ) ; E1 = R (2) ; Xm = E1 / Im ; // As p e r t h e c i r c u i t d i a g r a m a = r2 / s ; Zf =((( r2 / s ) + x2 * %i ) * Xm * %i ) /(( r2 / s ) +(( x2 + Xm ) * %i ) ) ; Rf = real ( Zf ) ; Zab = complex (( real ( Zf ) + r1 ) ,( imag ( Zf ) + x1 ) ) ; I1 =420/ Zab ; I1M = sqrt (( real ( I1 ) * real ( I1 ) ) +( imag ( I1 ) * imag ( I1 ) ) ) ; an1 = atand ( imag ( I1 ) , real ( I1 ) ) ; pf = cosd ( atand ( imag ( I1 ) / real ( I1 ) ) ) ; I2 = I1 *( Xm * %i ) *(1/(( r2 / s ) +(( x2 + Xm ) * %i ) ) ) ; an2 = atand ( imag ( I2 ) , real ( I2 ) ) ; I2M = sqrt (( real ( I2 ) * real ( I2 ) ) +( imag ( I2 ) * imag ( I2 ) ) ) ; T =3*(60/(2* %pi * N ) ) * I1M * I1M * Rf ; mprintf ( ’ The power f a c t o r i s %f Lag \n The i n p u t c u r r e n t i s %f A l a g g i n g by an a n g l e o f %f d e g r e e s \n The o u t p u t r o t o r c u r r e n t i s %f A l a g g i n g by an a n g l e o f %f d e g r e e s \n The Torque d e v e l o p e d i s %f Nm \n ’ ,pf , I1M , - an1 , I2M , - an2 , T ) ;

// For maximun Torque X1 = x1 + Xm ; Re =( r1 * Xm ) / X1 ; Xe =( x1 * Xm ) / X1 ; smT = r2 /( sqrt (( Re ) *( Re ) +( x2 + Xe ) *( x2 + Xe ) ) ) ; Nm = Ns *(1 - smT ) ; Tem =3*( E1 ) *( E1 ) *(1/( Re +( sqrt (( Re ) *( Re ) +( x2 + Xe ) *( x2 + 282


Xe ) ) ) ) ) *(1/2) *(1/(2* %pi *( N /60) ) ) ; 47 mprintf ( ’ maximum t o r q u e d e v e l o p e d i s %f Nm \n c o r r e s p o n d i n g s p e e d i s %f rpm ’ ,Tem , Nm ) ;

Scilab code Exa 6.30 Determining motor speed and power output 1 2

3 4 5 6 7 8 9 10 11 12 13 14 15 16

// I n s o l u t i o n t h e y have t a k e n d i f f e r e n t v a l u e o f s p e e d a t r a t e d t o r q u e from what i s g i v e n i n q u e s t i o n t h a t i s why a n s w e r i s v a r y i n g clc ; P =4; Pm =10000; //OUTPUT POWER f =50; //FREQUENCY N =1440; //SPEED AT WHICH RATED TORQUE I S OBTAINED Ns =(120* f ) / P ; //SYNCHRONOUS SPEED s =( Ns - N ) / Ns ; // Torque i s d i r e c t l y p r o p o r t i o n a l t o t h e s l i p // As p e r g i v e n c o n d i t i o n s s1 =(1/2) * s ; Nr = Ns *(1 - s1 ) ; Pm1 =(1/2) *((( Pm *60) /(2* %pi * N ) ) ) *((2* %pi * Nr ) /(60) ) ; mprintf ( ’ The motor s p e e d i s %f rpm \n The power o u t p u t i s %f W’ ,Nr , Pm1 ) ;

Scilab code Exa 6.31 Determining percentage change in motor speed and losses 1 2 clc ; 3 N =1455; 4 Ns =1500; // G e n e r a l c a s e

c o n s i d e r e d i n the problem 283


5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

s1 =( Ns - N ) / Ns ; // f o r V1=0.9V //V1/V=a a =0.9; //T=(3VVs ) / ( Wsr2 ) // As t o r q u e i s c o n s t a n t s2 =( s1 ) /( a * a ) ; Nr = Ns *(1 - s2 ) ; // I=s1V / r 2 // I 2 2 / I 2 1=b b =( s2 * a ) / s1 ; // L o s s e s R a t i o=c R=b*b;

d =(( N - Nr ) / N ) *100; e =(( R -1) /1) *100; mprintf ( ’ P e r c e n t a g e r e d u c t i o n i n s p e e d i s %f p e r c e n t \n ’ ,d ) ; 23 mprintf ( ’ P e r c e n t a g e r e d u c t i o n i n ohmic l o s s e s i s %f p e r c e n t \n ’ ,e ) ;

Scilab code Exa 6.32 Determining no load speed of motor 1 2 3 4 5 6 7 8 9 10 11

clc ; P =7500; // r a t e d power o f i n d u c t i o n motor v =400; // r a t e d v o l t a g e o f motor To =6; // no l o a d t o r q u e fs =0.04; // f u l l l o a d s l i p p =6; // number o f p o l e s f =50; // f r e q u e n c y ns =(120* f ) / p ; // s y n c h r o n o u s s p e e d Tl =( P *60) /(2* %pi * ns *(1 - fs ) ) ; // f u l l l o a d t o r q u e s =( To * fs * v ^2) /( Tl *( v /2) ^2) ; // s l i p a t no l o a d no = ns *(1 - s ) ; 284


12

printf ( ’ No l o a d s p e e d o f motor i s %f rpm\n ’ , no ) ;

Scilab code Exa 6.33 Determining minimum voltage impressed on motor and additional resistance inserted in rotor circuit 1 clc ; 2 tr =2.5; // r a t i o

o f maximum t o r q u e t o f u l l l o a d

torque 3 sm =0.18; // maximum s l i p 4 r =1; // p e r p h a s e r o t o r r e s i s t a n c e 5 x2 = r / sm ; // r o t o r r e a c t a n c e 6 // u s i n g e x p r e s s i o n f o r t r we o b t a i n a q u a d r a t i c 7 8 9 10 11 12 13

14

15 16 17 18 19 20

e q u a t i o n i s s ( f u l l l o a d s l i p ) whose t e r m s a r e t1 =1; t2 = - tr *2* sm ; t3 = sm ^2; t =[ t1 t2 t3 ]; s = roots ( t ) ; x = sqrt ((2* x2 ) /((( r / s (2) ) ^2+ x2 ^2) * s (2) ) ) ; printf ( ’ Minimum v o l t a g e t h a t c o u l d be i m p r e s s e d s o t h a t motor can s u p p l y r a t e d t o r q u e i s %f t i m e s r a t e d v o l t a g e o r %f p e r c e n t o f r a t e d v o l t a g e \n ’ ,x , x *100) ; // from e x p r e s s i o n f o r maximum t o r q u e and f u l l l o a d t o r q u e we g e t a q u a d r a t i c e q u a t i o n i n R( e x t e r n a l l r e s i s t a n c e ) whose t e r m s a r e t1 =1; t2 =2 -2* x2 ; t3 =1+ x2 ^2 -2* x2 ; t =[ t1 t2 t3 ]; R = roots ( t ) ; printf ( ’ E x t e r n a l r e s i s t a n c e i n s e r t e d i n r o t o r c i r c u i t i s %f ohms \n ’ ,R (2) ) ;

285


Scilab code Exa 6.34 Comparing starting current starting torque and maximum torque at given frequencies and comparing voltage for given frequencies 1 clc ; 2 f1 =50; // r a t e d f r e q u e n c y o f 3− p h a s e i n d u c t i o n 3 4 5 6 7 8 9 10 11 12 13 14

15 16

17

motor f2 =40; // a p p l i e d f r e q u e n c y vr =0.9; // r a t i o o f a p p l i e d v o l t a g e t o r a t e d v o l t a g e m =3; // number o p h a s e s fr = f2 / f1 ; // r a t i o o f f r e q u e n c i e s ir = fr / vr ; printf ( ’ R a t i o o f s t a r t i n g c u r r e n t a t %d Hz t o s t a r t i n g c u r r e n t a t %d Hz i s %f \n ’ ,f1 , f2 , ir ) ; tr =( m / f1 ) *( f2 / m ) *( fr / vr ) ^2; printf ( ’ R a t i o o f s t a r t i n g t o r q u e a t %d Hz t o s t a r t i n g t o r q u e a t %d Hz i s %f \n ’ ,f1 , f2 , tr ) ; tmr =( m / f1 ) *( f2 / m ) *( fr /( vr ) ^2) ; printf ( ’ R a t i o o f maximum t o r q u e a t %d Hz t o maximum t o r q u e a t %d Hz i s %f \n ’ ,f1 , f2 , tmr ) ; vr1 = sqrt (( m / f1 ) *( f2 / m ) * fr ^2) ; printf ( ’ For t h e same s t a r t i n g t o r q u e r a t i o o f v o l t a g e a t %d Hz t o r a t i o o f v o l t a g e a t %d Hz i s %f\n ’ ,f2 , f1 , vr1 ) ; vr2 = sqrt (( m / f1 ) *( f2 / m ) * fr ) ; printf ( ’ For t h e same maximum t o r q u e r a t i o o f v o l t a g e a t %d Hz t o r a t i o o f v o l t a g e a t %d Hz i s %f\n ’ , f2 , f1 , vr2 ) ; // a n s w e r f o r r a t i o o f v2 / v1 f o r same s t a r t i n g t o r q u e i s s l i g h t l y d i f f e r e n t from what i s g i v e n i n book

Scilab code Exa 6.35 Determining 1 slip at rated load 2 starting torque 286


1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

clc ; P =60000; // r a t e d power o f 3− p h a s e i n d u c t i o n motor p =4; // number o f p o l e s f =50; // f r e q u e n c y po =3000; // no l o a d l o s s e s i =0.3; // r a t i o o f r a t e d c u r r e n t t o r a t e d v o l t a g e when motor i s p r e v e n t e d from r o t a t i n g pi =4000; // power i n p u t when motor i s p r e v e n t e d from rotating pr =0.3; // r a t i o o f m e c h a n i c a l l o s s e s t o no l o a d losses pm = pr * po ; // m e c h a n i c a l l o s s e s lsc1 = po - pm ; // s t a t o r c o r e l o s s lsc2 = pi /2; // s t a t o r c o p p e r l o s s =r o t o r c o p p e r l o s s disp ( ’ c a s e a ’ ) ; pg = P + pm + lsc2 ; // a i r gap power s = lsc2 / pg ; printf ( ’ S l i p a t r a t e d l o a d i s %f\n ’ ,s ) ; disp ( ’ c a s e b ’ ) ; pim = pi / i ^2; // power i n p u t t o motor d u r i n g b l o c k e d rotor test pg = pim - lsc1 - lsc2 ; // a i r gap power ws =(4* %pi * f ) / p ; // s y n c h r o n o u s s p e e d T = pg / ws ; printf ( ’ S t a r t i n g t o r q u e a t r a t e d a p p l i e d v o l t a g e i s %f Nm\n ’ ,T ) ;

Scilab code Exa 6.36 Finding ratio of starting current starting torque and maximum torque at given frequencies 1 clc ; 2 sm =0.2; // s l i p 3 f1 =50; // r a t e d f r e q u e n c y o f 3− p h a s e i n d u c t i o n

motor 4 f2 =45; // a p p l i e d

frequency 287


5 fr = f2 / f1 ; // r a t i o o f f r e q u e n c i e s i r = f r / v r ; 6 ir = sqrt (( sm ^2+1) /( sm ^2+ fr ^2) ) ; 7 printf ( ’ R a t i o o f s t a r t i n g c u r r e n t a t %d Hz t o

s t a r t i n g c u r r e n t a t %d Hz i s %f \n ’ ,f2 , f1 , ir ) ; 8 tr =( sm ^2+1) /( sm ^2+ fr ^2) ; 9 printf ( ’ R a t i o o f s t a r t i n g

t o r q u e a t %d Hz t o s t a r t i n g t o r q u e a t %d Hz i s %f \n ’ ,f2 , f1 , tr ) ; 10 tmr =1/ fr ; 11 printf ( ’ R a t i o o f maximum t o r q u e a t %d Hz t o maximum t o r q u e a t %d Hz i s %f \n ’ ,f2 , f1 , tmr ) ;

Scilab code Exa 6.37 Determining 1 maximum internal torque and internal starting torque 2 slip at maximum torque 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

clc ; P =20000; // r a t e d power o f i n d u c t i o n motor v =400; // r a t e d v o l t a g e o f motor f =50; // f r e q u e n c y m =3; // number o f p h a s e s p =4; // number o f p o l e s r1 =0.2; // s t a t o r r e s i s t a n c e x =0.45; // s t a t o r / r o t o r l e a k a g e r e a c t a n c e xm =18; // m a g n e t i s i n g r e a c t a n c e s =0.04; // s l i p pg = P /(1 - s ) ; // a i r gap power pr = s * pg ; // r o t o r c o p p e r l o s s vp = v / sqrt (3) ; // p e r p h a s e v o l t a g e ve =( vp * xm ) /( x + xm ) ; // T h e v e n i n v o l t a g e re =( r1 * xm ) /( x + xm ) ; // T h e v e n i n r e s i s t a n c e xe =( x * xm ) /( x + xm ) ; // T h e v e n i n r e a c t a n c e // u s i n g Thevenin ’ s t h e o r r m and r o t o r c o p p e r l o s s e x p r e s s i o n we g e t a q u a d r a t i c e q u a t i o n i n r 2 ( r o t o r r e s i s t a n c e ) whose t e r m s a r e 18 t1 = pr / s ^2; 19 t2 =((2* pr * re ) / s ) -( m * ve ^2) ; 288


20 21 22 23 24 25 26 27 28 29 30 31

t3 = pr *(( xe + x ) ^2+ re ^2) ; t =[ t1 t2 t3 ]; r2 = roots ( t ) ; disp ( ’ c a s e a ’ ) ; ws =(4* %pi * f ) / p ; // s y n c h r o n o u s s p e e d Tm =( m * ve ^2) /( ws *2*( re + sqrt ( re ^2+( x + xe ) ^2) ) ) ; printf ( ’ Maximum i n t e r n a l t o r q u e i s %f Nm\n ’ , Tm ) ; Ti =( m * ve ^2* r2 (1) ) /( ws *(( re + r2 (1) ) ^2+( x + xe ) ^2) ) ; printf ( ’ I n i t i a l s t a r t i n g t o r q u e i s %f Nm\n ’ , Ti ) ; disp ( ’ c a s e b ’ ) ; sm = r2 (1) /( sqrt ( re ^2+( xe + x ) ^2) ) ; printf ( ’ S l i p a t maximum t o r q u e i s %f ’ , sm ) ;

Scilab code Exa 6.41 Determining rotational losses and equivalent circuit parameters 1 clc ; 2 P =10000; // r a t e d power o f

s q u i r r e l cage induction

motor V =400; // r a t e d v o l t a g e o f motor m =3; // number o f p h a s e s // no l o a d t e s t r e s u l t s Vo =400; // a p p l i e d v o l t a g e io =8; // no l o a d c u r r e n t Po =250; // no l o a d power // b l o c k e d r o t o r t e s t vb =90; // a p p l i e d v o l t a g e ib =35; // c u r r e n t pb =1350; // i n p u t power // a c r e s i s t a n c e i s 1 . 2 t i m e s dc r e s i s t a n c e rs =0.6; // p e r p h a s e dc r e s i s t a n c e o f s t a t o r w i n d i n g pr = Po - m *( io / sqrt (3) ) ^2*(1.2* rs ) ; // no l o a d rotational losses 16 znl = Vo /( io / sqrt (3) ) ; // no l o a d i m p e d a n c e 17 rnl = Po /( m *( io / sqrt (3) ) ^2) ; // no l o a d r e s i s t a n c e 3 4 5 6 7 8 9 10 11 12 13 14 15

289


18 19 20 21 22 23 24 25 26 27 28 29 30 31

xnl = sqrt ( znl ^2 - rnl ^2) ; // no l o a d r e a c t a n c e zbr = vb /( ib / sqrt (3) ) ; // b l o c k r o t o r t e s t i m p e d a n c e Rbr = pb /( m *( ib / sqrt (3) ) ^2) ; // b l o c k r o t o r r e s i s t a n c e xbr = sqrt ( zbr ^2 - Rbr ^2) ; // b l o c k r o t o r r e a c t a n c e x1 = xbr /2; xm = xnl - x1 ; X2 = xm + x1 ; r2 =( Rbr -1.2* rs ) *( X2 / xm ) ^2; printf ( ’ R o t a t i o n a l l o s s e s a r e %f w a t t s \n ’ , pr ) ; printf ( ’ S t a t o r r e s i s t a n c e i s %f ohms \n ’ ,1.2* rs ) ; printf ( ’ R o t o r r e s i s t a n c e i s %f ohms \n ’ , r2 ) ; printf ( ’ M a g n e t i s i n g r e a c t a n c e i s %f ohms \n ’ , xm ) ; printf ( ’ S t a t o r r e a c t a n c e i s %f ohms \n ’ , x1 ) ; printf ( ’ R o t o r r e a c t a n c e i s %f ohms ’ , x1 ) ;

Scilab code Exa 6.43 Determining starting torque 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

clc ; p =10000; // r a t e d power o f SCIM v =420; // r a t e d v o l t a g e o f SCIM p =4; // number o f p o l e s f =50; // f r e q u e n c y o f SCIM // r e s u l t s o f b l o c k e d rotor test vb =210; // a p p l i e d v o l t a g e ib =20; // a p p l i e d c u r r e n t pb =5000; // power d i s s i p a t e d l =300; // s t a t o r c o r e l o s s rs =0.6; // dc s t a t o r r e s i s t a n c e m =3; // number o f p h a s e s R =( rs *3) /2; // p e r p h a s e s t a t o r r e s i s t a n c e Rs =1.2* R ; // E f f e c t i v e s t a t o r r e s i s t a n c e p e r p h a s e pi = pb *( v / vb ) ^2; // power i n p u t a t r a t e d v o l t a g e during block rotor t e s t 16 is = ib *( v / vb ) ; // s t a t o r c u r r e n t a t r a t e d v o l t a g e during block rotor t e s t 290


17 pg = pi - m *( is / sqrt (3) ) ^2* Rs - l ; // a i r gap power 18 ws =(4* %pi * f ) / p ; 19 printf ( ’ s y n c h r o n o u s s p e e d i s %f r a d / s e c \n ’ , ws ) ; 20 T = pg / ws ; 21 printf ( ’ S t a r t i n g t o r q u e i s %f Nm ’ ,T ) ;

Scilab code Exa 6.44 Determining starting torque 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

clc ; p =6; // number o f p o l e s m =3; // number o f p h a s e s f =50; // f r e q u e n c y o f motor P =40000; // r a t e d power o f i n d u c t i o n motor v =400; // r a t e d v o l t a g e o f i n d u c t i o n motor // r e s u l t s o f b l o c k e d rotor test vb =200; // a p p l i e d v o l t a g e ib =110; // a p p l i e d c u r r e n t pf =0.4; // power f a c t o r f1 =45; // f r e q u e n c y a t s t a r t i n g t o r q u e i s t o be determined e =380; // v o l t a g e a t s t a r t i n g t o r q u e i s t o be determined vbp = vb / sqrt (3) ; // p e r p h a s e v o l t a g e d u r i n g b l o c k e d rotor test zb = vbp / ib ; // t o t a l i m p e d a n c e r e f e r r e d t o s t a t o r R = zb * pf ; // n e t r e s i s t a n c e r e f e r r e d t o s t a t o r X = zb *( sqrt (1 - pf ^2) ) ; // n e t r e a c t a n c e r e f e r r e d t o stator X = X *( f1 / f ) ; // n e t r e a c t a n c e a t f r e q u e n c y =45 Z = R + X * %i ; // i m p e d a n c e a t f r e q u e n c y =45 v1 = e / sqrt (3) ; // p e r p h a s e s t a t o r is = v1 /( Z ) ; // s t a r t i n g c u r r e n t ws =(4* %pi * f ) / p ; // s y n c h r o n o u s s p e e d T =(3/ ws ) * abs ( is ) ^2*( R /2) ; printf ( ’ S t a r t i n g t o r q u e i s %f Nm ’ ,T ) ; 291


Scilab code Exa 6.45 Determining 1 mechanical power output 2 net torque 3 efficiency of motor 1 2 3 4 5 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

clc ; v =400; // r a t e d v o l t a g e o f motor m =3; // number o f p h a s e s r =2; // r a t i o o f l e a k a g e r e a c t a n c e o f s t a t o r t o leakage reactance of rotor ns =1000; // s y n c h r o n o u s s p e e d n =960; // s p e e d o f motor f =50; // f r e q u e n c y // no l o a d t e s t r e s u l t s Vo =400; // a p p l i e d v o l t a g e io =7.5; // no l o a d c u r r e n t pfo =0.135; // power f a c t o r // b l o c k e d r o t o r t e s t vb =150; // a p p l i e d v o l t a g e ib =35; // c u r r e n t pfb =0.44; // power f a c t o r znl = Vo /( io * sqrt (3) ) ; // no l o a d i m p e d a n c e rnl = znl * pfo ; // no l o a d r e s i s t a n c e xnl = sqrt ( znl ^2 - rnl ^2) ; // no l o a d r e a c t a n c e zbr = vb /( ib * sqrt (3) ) ; // b l o c k r o t o r t e s t i m p e d a n c e Rbr = zbr * pfb ; // b l o c k r o t o r r e s i s t a n c e xbr = sqrt ( zbr ^2 - Rbr ^2) ; // b l o c k r o t o r r e a c t a n c e x2 = xbr /3; // l e a k a g e r e a c t a n c e o f r o t o r x1 = x2 *2; // l e a k a g e r e a c t a n c e o f s t a t o r xm = xnl - x1 ; // m a g n e t i s i n g r e a c t a n c e r1 = Rbr /2; // s t a t o r r e s i s t a n c e / r o t o r r e s i s t a n c e V1 = v / sqrt (3) ; // p e r p h a s e s t a t o r v o l t a g e Ve =( V1 * xm ) /( x1 + xm ) ; // t h e v e n i n v o l t a g e Re =( r1 * xm ) /( x1 + xm ) ; // t h e v e n i n r e s i s t a n c e Xe =( x1 * xm ) /( x1 + xm ) ; // t h e v e n i n r e s i s t a n c e lr = sqrt (3) * v * io * pfo - m * io ^2* r1 ; // r o t a t i o n a l l o s s e s 292


31 s =( ns - n ) / ns ; // s l i p 32 ir = Ve /( Re +( r1 / s ) + %i *( Xe + x2 ) ) ; // r o t o r 33 34 35 36 37 38 39 40 41 42 43 44 45 46

current at slip Pm = m * abs ( ir ) ^2* r1 *((1 - s ) / s ) ; disp ( ’ c a s e a ’ ) ; Psh = Pm - lr ; printf ( ’ M e c h a n i c a l power o u t p u t i s %f KW\n ’ , Psh /1000) ; disp ( ’ c a s e b ’ ) ; wr =((2* %pi * f ) *(1 - s ) ) / m ; // s p e e d a t which motor i s running T = Psh / wr ; printf ( ’ Net t o r q u e i s %f Nm\n ’ ,T ) ; disp ( ’ c a s e c ’ ) ; lor =( Pm * s ) /(1 - s ) ; // r o t o r / s t a t o r ohmic l o s s e s Tl = lor *2+ lr ; // t o t a l l o s s e s pi = Tl + Psh ; // i n p u t power ne = Psh / pi ; printf ( ’ E f f i c i e n c y o f motor i s %f p e r c e n t ’ , ne *100) ;

Scilab code Exa 6.46 Determining new operating speed 1 2 3 4 5 6 7 8 9 10 11

clc ; f =60; // f r e q u e n c y p =6; // number o f p o l e s n =1175; // s p e e d o f i n d u c t i o n motor re =0.06; // r e d u c t i o n i n f r e q u e n c y dv =0.1; // r e d u c t i o n i n v o l t a g e ws1 =(120* f ) / p ; // s y n c h r o n o u s s p e e d s1 =( ws1 - n ) / ws1 ; // s l i p s2 =((1 - re ) /((1 - dv ) ^2) ) * s1 ; // new s l i p ws2 = ws1 *(1 - s2 ) *(1 - re ) ; printf ( ’ New o p e r a t i n g s p e e d i s %f rpm ’ , ws2 ) ;

293


Scilab code Exa 6.47 Determining 1 line current pf slip torque and efficiency 2 maximum possible pf and corresponding line current 3 maximum power output and maximum power input 4 slip at maximum torque and maximum torque 5 starting torque 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

clc ; P =15000; // r a t e d power o f i n d u c t i o n motor V =400; // r a t e d v o l t a g e o f motor f =50; // f r e q u e n c y m =3; // number o f p h a s e s po =4; // number o f p o l e s // no l o a d t e s t r e s u l t s Vo =400; // a p p l i e d l i n e v o l t a g e io =9; // no l o a d l i n e c u r r e n t Po =1310; // power i n p u t // b l o c k e d r o t o r t e s t vb =200; // l i n e v o l t a g e ib =50; // l i n e c u r r e n t pb =7100; // i n p u t power pfo = po /( sqrt (3) * io * Vo ) ; // no l o a d power f a c t o r pfb = pb /( sqrt (3) * ib * vb ) ; // s h o r t c i r c u i t power factor isc =( V / vb ) * ib ; // s h o r t c i r c u i t c u r r e n t printf ( ’ S h o r t c i r c u i t c u r r e n t i s %d A\n ’ , isc ) ; // c i r c l e d i a g r a m i s drawn i n f i g 6 . 3 7 w i t h s c a l e 6 A= 1 cm disp ( ’ c a s e a ’ ) ; x =6; // s c a l e pps =( V / sqrt (3) ) * x ; // p e r p h a s e power s c a l e fp = P /3; // f u l l l o a d power p e r p h a s e // a s p e r t h e c o n s t r u c t i o n we o b t a i n OP=6.05 which corresponds to f u l l load current ifl = x *6.05; printf ( ’ F u l l l o a d l i n e c u r r e n t i s %f A\n ’ , ifl ) ; 294


27 // from f i g a n g l e POV1= 2 9 . 5 ; 28 fpf = cosd (29.5) ; 29 printf ( ’ F u l l l o a d power f a c t o r 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44

45 46 47

48 49 50 51

52 53

i s %f l a g g i n g \n ’ , fpf )

; // f u l l l o a d s l i p i s g i v e n by r a t i o ba /bP where ba = 2 . 5 , bP =38.5 fs =2.5/38.5; printf ( ’ F u l l l o a d s l i p i s %f \n ’ , fs ) ; ws =(2* %pi * f *120) /( po *60) ; // s y n c h r o n o u s s p e e d Ft =(3.85* pps * m ) / ws ; printf ( ’ F u l l l o a d t o r q u e i s %f Nm\n ’ , Ft ) ; // e f f i c i e n c y i s g i v e n by r a t i o aP/dP where aP = 3 . 6 , dP =4.45 ne =3.6/4.45; printf ( ’ F u l l l o a d e f f i c i e n c y i s %f p e r c e n t \n ’ , ne *100) ; disp ( ’ c a s e b ’ ) ; // OP t u r n s o u t t o be t a n g e n t t o c i r c u l a r l o c u s , therefore disp ( ’ Maximum power f a c t o r i s 0 . 8 7 l a g g i n g ’ ) ; disp ( ’ Maximum l i n e c u r r e n t i s 3 6 . 3 A ’ ) ; disp ( ’ c a s e c ’ ) ; // a c c o r d i n g t o c o n s t r u c t i o n s g i v e n i n s o l u t i o n we o b t a i n AA’ = 5 . 3 from which maximum power o u t p u t can be c a l c u l a t e d mpo =5.3* m * pps ; printf ( ’ Maximum o u t p u t power i s %f KW\n ’ , mpo /1000) ; // a c c o r d i n g t o c o n s t r u c t i o n s g i v e n i n s o l u t i o n we o b t a i n CC’=8.45= r a d i u s o f c i r c l e from which maximum power i n p u t can be c a l c u l a t e d mpi =8.45* m * pps + po ; printf ( ’ Maximum i n p u t power i s %f KW\n ’ , mpi /1000) ; disp ( ’ c a s e d ’ ) ; // a c c o r d i n g t o c o n s t r u c t i o n s g i v e n i n s o l u t i o n we o b t a i n BB’ = 6 . 6 5 from which maximum t o r q u e can be calculated Mt =(6.65* m * pps ) / ws ; printf ( ’ Maximum t o r q u e i s %f Nm\n ’ , Mt ) ; 295


54 55 56 57 58

59 60

// maximum s l i p i s g i v e n by r a t i o fb ’ / BB’ where fb ’ = 1 . 5 8 , BB’ = 6 . 6 5 s =1.58/6.65; printf ( ’ Maximum s l i p i s %f \n ’ ,s ) ; disp ( ’ c a s e e ’ ) ; // a c c o r d i n g t o c o n s t r u c t i o n s g i v e n i n s o l u t i o n we o b t a i n DG=3.3 from which s t a r t i n g t o r q u e can be calculated St =(3.3* m * pps ) / ws ; printf ( ’ S t a r t i n g t o r q u e i s %f Nm\n ’ , St ) ;

Scilab code Exa 6.48 Determining 1 external resistance inserted in rotor circuit 2 stator currents power factor 3 power output operating power factor and efficiency 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

clc ; P =4500; // r a t e d power o f i n d u c t i o n motor V =400; // r a t e d v o l t a g e o f motor f =50; // f r e q u e n c y m =3; // number o f p h a s e s // no l o a d t e s t r e s u l t s Vo =400; // a p p l i e d l i n e v o l t a g e io =4.2; // no l o a d l i n e c u r r e n t Po =480; // power i n p u t // b l o c k e d r o t o r t e s t vb =215; // l i n e v o l t a g e ib =15; // l i n e c u r r e n t pb =1080; // i n p u t power rs =1.2; // r o t o r r e s i s t a n c e r e f e r r e d t o s t a t o r p e r phase nt =2; // s t a t o r t o r o t o r t u r n s r a t i o pfo = Po /( sqrt (3) * io * Vo ) ; // no l o a d power f a c t o r pfb = pb /( sqrt (3) * ib * vb ) ; // s h o r t c i r c u i t power factor isc =( V / vb ) *( ib * sqrt (3) ) ; // p e r p h a s e s h o r t c i r c u i t 296


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

current iop = io / sqrt (3) ; // p e r p h a s e no l o a d c u r r e n t x =1; // s c a l e 1 A= 1 cm // c i r c l e d i a g r a m i s drawn i n f i g 6 . 3 8 disp ( ’ c a s e a ’ ) ; // v a l u e o f maximum t o r q u e a t s t a r t i n g i s n o t g i v e n // now we n o t e Bf =4.6 and B ’ f =1.25 u s i n g t h e s e v a l u e s e x t e r n a l r e s i s t a n c e t o be i n s e r t e d i s calculated re =(4.6/1.25) *1.2; // e x t e r n a l r e s i s t a n c e printf ( ’ E x t e r n a l r e s i s t a n c e r e f e r r e d t o r o t o r i s %f ohms \n ’ , re / nt ^2) ; // a s p e r t h e c o n s t r u c t i o n we o b t a i n OB= 1 1 . 2 4 which i s needed to c a l c u l a t e s t a r t i n g l i n e c u r r e n t is =11.24* sqrt (3) ; printf ( ’ S t a r t i n g c u r r e n t i s %f A\n ’ , is ) ; // a n g l e OBB’ = 4 5 . 5 which i s n e e d e d t o c a l c u l a t e power f a c t o r pf = cosd (45.5) ; printf ( ’ power f a c t o r i s %f l a g g i n g \n ’ , pf ) ; pps = x * V ; // p e r p h a s e power s c a l e fp = P / m ; // f u l l l o a d power p e r p h a s e disp ( ’ c a s e b ’ ) ; // now t o r q u e i s 1 . 2 5 t i m e s f u l l l o a d t o r q u e // now we n o t e NK=2.9 and N’ K=2.1 u s i n g t h e s e v a l u e s e x t e r n a l r e s i s t a n c e t o be i n s e r t e d i s c a l c u l a t e d re =(2.9/2.1) *1.2; // e x t e r n a l r e s i s t a n c e printf ( ’ E x t e r n a l r e s i s t a n c e r e f e r r e d t o r o t o r i s %f ohms \n ’ , re / nt ^2) ; // a s p e r t h e c o n s t r u c t i o n we o b t a i n ON= 1 4 . 3 5 which i s needed to c a l c u l a t e s t a r t i n g l i n e c u r r e n t is =14.35* sqrt (3) ; printf ( ’ S t a r t i n g c u r r e n t i s %f A\n ’ , is ) ; // a n g l e ONN’ = 5 8 . 3 which i s n e e d e d t o c a l c u l a t e power f a c t o r pf = cosd (58.3) ; printf ( ’ power f a c t o r i s %f l a g g i n g \n ’ , pf ) ; disp ( ’ c a s e c ’ ) ; 297


47 48 49 50 51 52 53 54 55

// we o b t a i n OH=5.35 which i s p e r p h a s e o u t p u t current // t h e t a g =41.3 opf = cosd (41.3) ; printf ( ’ O p e r a t i n g power f a c t o r i s %f l e a d i n g \n ’ , opf ) ; po = m *5.35* V * opf ; printf ( ’ Output power i s %f KW\n ’ , po /1000) ; // we n o t e HL=3.95 and Ha =4.90 which i s n e e d e d f o r efficiency ne =3.95/4.9; printf ( ’ I n d u c t i o n g e n e r a t o r e f f i c i e n c y i s %f p e r c e n t ’ , ne *100) ;

Scilab code Exa 6.49 Determining 1 rotational and core loss 2 electric power output power factor and efficiency 1 2 3 4 5 6 7 8 9 10 11 12

13 14 15 16

clc ; p =150000; // r a t e d power o f i n d u c t i o n motor v =400; // r a t e d v o l t a g e o f i n d u c t i o n motor m =3; // number o f p h a s e s r1 =0.02; // s t a t o r r e s i s t a n c e r2 =0.04; // r o t o r r e s i s t a n c e xm =9.8; // m a g n e t i s i n g r e a c t a n c e x1 =0.2; // l e a k a g e r e a c t a n c e o f s t a t o r o r r o t o r s =0.04; // s l i p n =0.93; // e f f i c i e n c y disp ( ’ c a s e a ’ ) ; Zf =((( r2 / s ) + %i * x1 ) * %i * xm ) /(( r2 / s ) + %i *( xm + x1 ) ) ; // p e r p h a s e i m p e d a n c e o f f e r e d t o s t a t o r by r o r a t i n g a i r gap f i e l d z = r1 + %i * x1 ; // i m p e d a n c e o f s t a t o r Z = Zf + z ; // t o t a l i m p e d a n c e is = v /( sqrt (3) * abs ( Z ) ) ; // s t a t o r c u r r e n t pg = m * is ^2* real ( Zf ) ; // a i r gap power 298


17 18 19 20 21 22 23 24

25 26 27 28 29 30 31 32 33 34 35 36 37

l1 = m * is ^2* r1 ; // s t a t o r c o p p e r l o s s l2 = s * pg ; // r o t o r c o p p e r l o s s Tl =((1/ n ) -1) * p ; // t o t a l l o s s e s lr = Tl -( l1 + l2 ) ; // r o t a t i o n a l and c o r e l o s s e s printf ( ’ R o t a t i o n a l and c o r e l o s s e s a r e %f W\n ’ , lr ) ; disp ( ’ c a s e b ’ ) ; s = -0.04; // s l i p Zf =((( r2 / s ) + %i * x1 ) * %i * xm ) /(( r2 / s ) + %i *( xm + x1 ) ) ; // p e r p h a s e i m p e d a n c e o f f e r e d t o s t a t o r by r o r a t i n g a i r gap f i e l d Z = Zf + z ; // t o t a l i m p e d a n c e is = v /( sqrt (3) * abs ( Z ) ) ; // s t a t o r c u r r e n t pf = cosd (180 - atand ( imag ( Z ) , real ( Z ) ) ) ; // power f a c t o r printf ( ’ Power f a c t o r a t t h e g e n e r a t o r t e r m i n a l i s %f l e a d i n g \n ’ , pf ) ; po = sqrt (3) * is * v * pf ; // e l e c t r i c a l o u t p u t printf ( ’ E l e c t r i c a l o u t p u t i s %f KW\n ’ , po /1000) ; pg = - m * is ^2* real ( Zf ) ; // a i r gap power l1 = m * is ^2* r1 ; // s t a t o r c o p p e r l o s s l2 = - s * pg ; // r o t o r c o p p e r l o s s Tl = l1 + l2 + lr ; // t o t a l l o s s e s pi = Tl + po ; // m e c h a n i c a l power i n p u t ne = po / pi ; printf ( ’ E f f i c i e n c y i s %f p e r c e n t ’ , ne *100) ;

Scilab code Exa 6.50 Determining per phase value of capacitance and total KVA rating of capacitor bank 1 2 3 4 5 6 7

clc ; v =400; // b a l a n c e d s u p p l y v o l t a g e i =10; // l i n e c u r r e n t f =50; // f r e q u e n c y o f s u p p l y m =3; // number o f p h a s e s pf =0.8; // l a g g i n g power f a c t o r pfn =0.9; // i m p r o v e d power f a c t o r 299


8 disp ( ’ s t a o r i n s t a r ’ ) ; 9 i = i *( pf - %i * sqrt (1 - pf ^2) ) ; // c o m p l e x form o f 10 11 12 13 14 15 16 17 18

19 20 21 22 23 24

25 26 27 28 29 30 31

line current il = real ( i ) / pfn ; // l i n e c u r r e n t a t i m p r o v e d power factor il = il *( pfn - %i * sqrt (1 - pfn ^2) ) ; // c o m p l e x form o f new line current // from f i g . 6 . 3 9 ic = -( imag ( i ) - imag ( il ) ) ; // r e a c t i v e component o f c u r r e n t t o be n e u t r a l i s e d // c a p a c i t o r bank i s s t a r c o n n e c t e d xcs = v /( ic * sqrt (3) ) ; // c a p a c i t a n c e r e a c t a n c e Cs =1/(2* %pi * f * xcs ) ; // c a p a c i t a n c e K = m * ic * v / sqrt (3) ; printf ( ’ Per p h a s e v a l u e o f c a p a c i t a n c e f o r s t a r c o n n e c t e d c a p a c i t o r bank i s %f m i c r o f a r a d \n ’ , Cs *10^6) ; printf ( ’ T o t a l KVA r a t i n g f o r s t a r c o n n e c t e d c a p a c i t o r bank i s %f KVA\n ’ ,K /1000) ; // d e l t a c o n n e c t e d c a p a c i t o r bank // c a p a c i t o r bank i s d e l t a c o n n e c t e d , converting i n t o e q u i v a l e n t s t a r X s t a r=X d e l t a /3 xcd = v /( ic * sqrt (3) ) ; // c a p a c i t a n c e r e a c t a n c e Cd =1/(2* %pi * f * xcd * m ) ; // c a p a c i t a n c e printf ( ’ Per p h a s e v a l u e o f c a p a c i t a n c e f o r d e l t a c o n n e c t e d c a p a c i t o r bank i s %f m i c r o f a r a d \n ’ , Cd *10^6) ; printf ( ’ T o t a l KVA r a t i n g f o r d e l t a c o n n e c t e d c a p a c i t o r bank i s %f KVA\n ’ ,K /1000) ; disp ( ’ S t a t o r i n d e l t a ’ ) ; i =( abs ( i ) / sqrt (3) ) *( pf - %i * sqrt (1 - pf ^2) ) ; // c o m p l e x form o f l i n e c u r r e n t il = real ( i ) / pfn ; // l i n e c u r r e n t a t i m p r o v e d power factor il = il *( pfn - %i * sqrt (1 - pfn ^2) ) ; // c o m p l e x form o f new line current // from f i g . 6 . 3 9 ic = -( imag ( i ) - imag ( il ) ) ; // r e a c t i v e component o f 300


32 33 34 35 36 37

38 39 40 41 42

43

c u r r e n t t o be n e u t r a l i s e d // c a p a c i t o r bank i s s t a r c o n n e c t e d // c a p a c i t o r bank i s s t a r c o n n e c t e d , converting i n t o e q u i v a l e n t d e l t a X d e l t a =3∗ X s t a r xcs = v / ic ; // c a p a c i t a n c e r e a c t a n c e Cs = m /(2* %pi * f * xcs ) ; // c a p a c i t a n c e K = m * ic * v ; printf ( ’ Per p h a s e v a l u e o f c a p a c i t a n c e f o r s t a r c o n n e c t e d c a p a c i t o r bank i s %f m i c r o f a r a d \n ’ , Cs *10^6) ; printf ( ’ T o t a l KVA r a t i n g f o r s t a r c o n n e c t e d c a p a c i t o r bank i s %f KVA\n ’ ,K /1000) ; // d e l t a c o n n e c t e d c a p a c i t o r bank xcd = v / ic ; // c a p a c i t a n c e r e a c t a n c e Cd =1/(2* %pi * f * xcd ) ; // c a p a c i t a n c e printf ( ’ Per p h a s e v a l u e o f c a p a c i t a n c e f o r d e l t a c o n n e c t e d c a p a c i t o r bank i s %f m i c r o f a r a d \n ’ , Cd *10^6) ; printf ( ’ T o t a l KVA r a t i n g f o r d e l t a c o n n e c t e d c a p a c i t o r bank i s %f KVA\n ’ ,K /1000) ;

Scilab code Exa 6.51 Determining capacitance of bank and each unit and percentage saving in energy lost 1 2 3 4 5 6 7 8 9 10 11

clc ; v =3300; // b a l a n c e d s u p p l y v o l t a g e p =500000; // r a t e d power o f i n d u c t i o n motor f =50; // f r e q u e n c y o f s u p p l y m =3; // number o f p h a s e s pf =0.7; // l a g g i n g power f a c t o r pfn =0.9; // i m p r o v e d power f a c t o r vc =420; // r a t e d v o l t a g e o f c a p a c i t o r n =0.86; // motor e f f i c i e n c y i = p /( sqrt (3) * v * pf * n ) ; // l i n e c u r r e n t i = i *( pf - %i * sqrt (1 - pf ^2) ) ; // c o m p l e x form o f l i n e 301


12 13 14 15 16 17 18 19 20 21 22 23

24 25 26 27 28

current il = real ( i ) / pfn ; // l i n e c u r r e n t a t i m p r o v e d power factor il = il *( pfn - %i * sqrt (1 - pfn ^2) ) ; // c o m p l e x form o f new line current // from f i g . 6 . 3 9 ic = -( imag ( i ) - imag ( il ) ) ; // r e a c t i v e component o f c u r r e n t t o be n e u t r a l i s e d // c a p a c i t o r bank i s d e l t a c o n n e c t e d // c a p a c i t o r bank i s d e l t a c o n n e c t e d , converting i n t o e q u i v a l e n t s t a r X s t a r=X d e l t a /3 xcd = v /( ic * sqrt (3) ) ; // c a p a c i t a n c e r e a c t a n c e Cd =1/(2* %pi * f * xcd * m ) ; // c a p a c i t a n c e // now e a c h c a p a c i t o r i s r a t e d a t 420 V, number o f capacitor connected in s e r i e s i s n = ceil ( v / vc ) ; C = Cd * n ; printf ( ’ Per p h a s e v a l u e o f e a c h c a p a c i t a n c e f o r d e l t a c o n n e c t e d c a p a c i t o r bank i s %f m i c r o f a r a d \n ’ ,C *10^6) ; // l e t R be r e s i s t a n c e o f d i s t r i b u t i o n c i r c u i t // power l o s t w i t h o u t c a p a c i t o r bank i s m∗ a b s ( i ) ˆ2∗R // power l o s t w i t h c a p a c i t o r bank i s m∗ a b s ( i l ) ˆ2∗R therefore ps =( abs ( i ) ^2 - abs ( il ) ^2) / abs ( i ) ^2 printf ( ’ P e r c e n t a g e s a v i n g i n l o s s e s i s %f p e r c e n t ’ , ps *100) ;

Scilab code Exa 6.53 Determining tapping on autotransformer and line current at starting 1 clc ; 2 fs =0.05; // f u l l l o a d s l i p 3 ir =6; // r a t i o o f s t a r t i n g

c u r r e n t and f u l l l o a d

current 302


4 t =1; // r a t i o o f s t a r t i n g t o r q u e t o f u l l l o a d t o r q u e 5 x = sqrt ( t /(( ir ^2) * fs ) ) ; 6 printf ( ’ Tapping p o i n t i s a t %f p e r c e n t \n ’ ,x *100) ; 7 is = x ^2* ir ; 8 printf ( ’ S t a r t i n g c u r r e n t i s %f t i m e s f u l l l o a d

c u r r e n t \n ’ , is ) ;

Scilab code Exa 6.54 Determining starting torque in of full load torque 1 clc ; 2 vr =0.4; // v o l t a g e 3 4 5 6 7 8 9

applied during blocked rotor t e s t as a f r a c t i o n of rated v o l t a g e ir =2.5; // l i n e c u r r e n t d u r i n g b l o c k e d r o t o r t e s t a s a f r a c t i o n of f u l l load current tr =0.3; // s t a r t i n g t o r q u e a s a f r a c t i o n o f r a t e d torque is =1.5; // s t a r t i n g c u r r e n t a s a f r a c t i o n o f f u l l load current isc = ir / vr ; // s h o r t c i r c u i t c u r r e n t a t r a t e d l o a d x = sqrt ( is / isc ) ; // s t a r t i n g c u r r e n t a s a f r a c t i o n o f short c i r c u i t current at rated load T =( x / vr ) ^2* tr ; printf ( ’ S t a r t i n g t o r q u e i s %f p e r c e n t o f f u l l l o a d t o r q u e ’ ,T *100) ;

Scilab code Exa 6.55 Determining maximum permissible KW rating of motor for different cases 1 2 3 4

clc ; v =440; // r a t e d v o l t a g e o f d i s t r i b u t i o n c i r c u i t im =1200; // maximum c u r r e n t t h a t can be s u p p l i e d n =0.85; // e f f i c i e n c y o f i n d u c t i o n motor 303


5 pf =0.8; // power f a c t o r o f motor 6 ir =5; // r a t i o o f s t a r t i n g c u r r e n t t o 7 8 9 10 11 12 13 14 15 16 17 18 19 20

f u l l load current disp ( ’ c a s e a ’ ) ; il = im / ir ; // r a t e d l i n e c u r r e n t p = sqrt (3) * v * il * n * pf ; printf ( ’ Maximum KW r a t i n g i s %f KW\n ’ ,p /1000) ; disp ( ’ c a s e b ’ ) ; x =0.8; // r a t e d o f a p p l i e d v o l t a g e and s t e p p e d down voltage il = im /( x ^2* ir ) ; // r a t e d l i n e c u r r e n t p = sqrt (3) * v * il * n * pf ; printf ( ’ Maximum KW r a t i n g i s %f KW\n ’ ,p /1000) ; disp ( ’ c a s e c ’ ) ; // s t a r −d e l t a c o n v e r t e r i s same a s a u t o t r a n s f o r m e r s t a r t e r with 57.8 % tapping t h e r e f o r e il = im /(0.578^2* ir ) ; // r a t e d l i n e c u r r e n t p = sqrt (3) * v * il * n * pf ; printf ( ’ Maximum KW r a t i n g i s %f KW\n ’ ,p /1000) ;

Scilab code Exa 6.56 Determining 1 voltage applied to motor terminals 2 current drawn by motor 3 line current drawn from supply mains 1 2 3 4 5 6 7 8 9 10

clc ; p =10000; // r a t e d power o f motor v =400; // r a t e d v o l t a g e o f motor n =0.87; // f u l l l o a d e f f i c i e n c y pf =0.85; // power f a c t o r ir =5; // r a t i o o f s t a r t i n g c u r r e n t t o f u l l l o a d current tr =1.5; // r a t i o o f s t a r t i n g t o r q u e t o f u l l l o a d torque disp ( ’ c a s e a ’ ) ; vt = v / sqrt ( tr ) ; printf ( ’ V o l t a g e a p p l i e d t o motor t e r m i n a l i s %f V\n ’ 304


11 12 13 14 15 16 17

, vt ) ; disp ( ’ c a s e b ’ ) ; ifl = p /( sqrt (3) * v * pf * n ) ; // f u l l l o a d c u r r e n t il =( ir * vt * ifl ) / v ; printf ( ’ C u r r e n t drawn by motor i s %f A\n ’ , il ) ; disp ( ’ c a s e c ’ ) ; i =( vt / v ) * il ; printf ( ’ L i n e c u r r e n t drawn from s u p p l y mains i s %f A ’ ,i ) ;

Scilab code Exa 6.57 Determining ratio of starting torque to full load torque for different starters 1 clc ; 2 tm =2; // r a t i o o f maximum t o r q u e t o f u l l l o a d t o r q u e 3 r =0.2; // p e r p h a s e r o t o r r e s i s t a n c e r e f e r r e d t o 4 5 6 7 8 9 10 11 12 13 14 15

stator x =2; // p e r p h a s e r e a c t a n c e r e f e r r e d t o s t a t o r s = r / x ; // s l i p a t maximum t o r q u e disp ( ’ c a s e a ’ ) ; ts1 =(2* s * tm ) /( s ^2+1) ; printf ( ’ R a t i o o f s t a r t i n g t o r q u e t o f u l l l o a d t o r q u e i s %f\n ’ , ts1 ) ; disp ( ’ c a s e b ’ ) ; ts2 = ts1 /3; printf ( ’ R a t i o o f s t a r t i n g t o r q u e t o f u l l l o a d t o r q u e w i t h s t a r −d e l t a s t a r t e r i s %f\n ’ , ts2 ) ; disp ( ’ c a s e c ’ ) ; t =0.7; // t a p p i n g p o i n t ts3 = ts1 * t ^2; printf ( ’ R a t i o o f s t a r t i n g t o r q u e t o f u l l l o a d t o r q u e w i t h a u t o t r a n s f o r m e r s t a r t e r i s %f\n ’ , ts3 ) ;

305


Scilab code Exa 6.58 Determining resistance of feeder and percentage increase in starting torque 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

17 18 19 20

21 22 23

24

clc ; v =400; // s u p p l y v o l t a g e f =50; // f r e q u e n c y o f s u p p l y // r e s u l t s o f s h o r t c i r c u i t t e s t V =200; // a p p l i e d v o l t a g e i =100; // s h o r t c i r c u i t c u r r e n t pf =0.4; // power f a c t o r zsc =( V * sqrt (3) ) / i ; // s h o r t c i r c u i t i m p e d a n c e rsc = zsc * pf ; // s h o r t c i r c u i t r e s i s t a n c e xsc = sqrt ( zsc ^2 - rsc ^2) ; // s h o r t c i r c u i t r e a c t a n c e R = sqrt ((( xsc ^2+ rsc ^2) -3*(( rsc /3) ^2+( xsc /3) ^2) ) /2) ; // r e s i s t a n c e o f f e e d e r disp ( ’ w i t h s t a r c o n n e c t i o n ’ ) ; ts1 =(3*( v / sqrt (3) ) ^2* rsc ) /(( R + rsc ) ^2+ xsc ^2) ; // p r o d u c t o f s t a r t i n g t o r q u e and s y n c h r o n o u s s p e e d // now two f e e d e r s a r e c o n n e c t e d i n p a r a l l e l t h e r e f o r e net r e s i s t a c e of f e e d e r i s rp = R ^2/( R + R ) ; ts2 =(3*( v / sqrt (3) ) ^2* rsc ) /(( rp + rsc ) ^2+ xsc ^2) ; // p r o d u c t o f new s t a r t i n g t o r q u e and s y n c h r o n o u s speed pr =( ts2 - ts1 ) / ts1 ; printf ( ’ P e r c e n t a g e i n c r e a s e i n s t a r t i n g t o r q u e w i t h s t a r c o n n e c t i o n i s %f p e r c e n t \n ’ , pr *100) ; disp ( ’ With d e l t a c o n n e c t i o n ’ ) ; ts1 =(3*( v / sqrt (3) ) ^2*( rsc /3) ) /(( R +( rsc /3) ) ^2+( xsc /3) ^2) ; // p r o d u c t o f s t a r t i n g t o r q u e and synchronous speed // now two f e e d e r s a r e c o n n e c t e d i n p a r a l l e l t h e r e f o r e net r e s i s t a c e of f e e d e r i s rp = R ^2/( R + R ) ; ts2 =(3*( v / sqrt (3) ) ^2*( rsc /3) ) /(( rp +( rsc /3) ) ^2+( xsc /3) ^2) ; // p r o d u c t o f new s t a r t i n g t o r q u e and synchronous speed pr =( ts2 - ts1 ) / ts1 ; 306


25

printf ( ’ P e r c e n t a g e i n c r e a s e i n s t a r t i n g t o r q u e w i t h d e l t a c o n n e c t i o n i s %f p e r c e n t \n ’ , pr *100) ;

Scilab code Exa 6.59 Determining minimum allowable cross section area of each conductor of feeder 1 2 3 4 5 6 7

8 9 10 11 12 13 14

clc ; z =1.2+3* %i ; // p e r p h a s e s t a n d s t i l l i m p e d a n c e v =400; // s u p p l y v o l t a g e l =500; // l e n g t h o f f e e d e r l i n e tr =30; // maximum p e r c e n t a g e r e d u c t i o n p o s s i b l e i n s t a r t i n g torque ro =0.02; // r e s i s t i v i t y o f f e e d e r m a t e r i a l // e q u a t i n g e x p r e s s i o n o f s t a r t i n g t o r q u e w i t h and w i t h o u t f e e d e r we g e t a q u a d r a t i c e q u a t i o n i n R ( f e e d e r r e s i s t a n c e ) whose t e r m s a r e t1 =(1 -( tr /100) ) ; t2 =2* real ( z ) * t1 ; t3 = t1 * abs ( z ) ^2 - abs ( z ) ^2; p =[ t1 t2 t3 ]; R = roots ( p ) ; A =( ro * l ) / R (2) ; printf ( ’ Minimum a l l o w a b l e c r o s s s e c t i o n i s %f mmˆ2 ’ , A);

Scilab code Exa 6.60 Determining starting torque for different connections 1 2 3 4 5

clc ; f =50; // f r e q u e n c y p =10; // number o f p o l e s pb =120000; // power d i s s i p a t e d i n b l o c k r o t o r t e s t // s t a t o r ohmic l o s s e s = r o t o r ohmic l o s s e s 307


6 7 8 9 10 11 12 13 14

pr = pb /2; // t o t a l r o t o r l o s s disp ( ’ c a s e a ’ ) ; ws =(4* %pi * f ) / p ; // s y n c h r o n o u s s p e e d Ts = pr / ws ; printf ( ’ S t a r t i n g t o r q u e i s %f Nm\n ’ , Ts ) ; disp ( ’ c a s e b ’ ) ; pr = pr /3; // t o t a l r o t o r ohmic l o s s Ts = pr / ws ; printf ( ’ S t a r t i n g t o r q u e i s %f Nm\n ’ , Ts ) ;

Scilab code Exa 6.62 Determining value of resistance elements for a 4 step starter 1 2 3 4 5 6 7 8 9 10 11 12 13 14

clc ; s =0.03; // f u l l l o a d s l i p R =0.015; // r o t o r r e s i s t a n c e p e r p h a s e n =4; // number o f s t e p i n s t a r t e r al = s ^(1/ n ) ; R1 = R / s ; // r e s i s t a n c e o f w h o l e s e c t i o n r1 = R1 *(1 - al ) ; printf ( ’ R e s i s t a n c e o f f i r s t e l e m e n t i s %f ohms \n ’ , r1 ); r2 = r1 * al ; printf ( ’ R e s i s t a n c e o f s e c o n d e l e m e n t i s %f ohms \n ’ , r2 ) ; r3 = r1 * al ^2; printf ( ’ R e s i s t a n c e o f t h i r d e l e m e n t i s %f ohms \n ’ , r3 ); r4 = r1 * al ^3; printf ( ’ R e s i s t a n c e o f f o u r t h e l e m e n t i s %f ohms \n ’ , r4 ) ;

Scilab code Exa 6.63 Deg 5 sections of a 6 stud starter 308


1 clc ; 2 fs =0.02; // f u l l l o a d s l i p 3 ir =2; // r a t i o o f s t a r t i n g

current to f u l l load

current 4 n =5; // number o f s e c t i o n 5 R =0.03; // r o t o r r e s i s t a n c e 6 // i r ∗ i f l =(E2/R) ∗sm where i f l 7 8 9 10 11 12 13 14 15 16 17 18 19

i s f u l l load current and E2 i s i n d u c e d v o l t a g e i n r o t o r t h e r e f o r e sm = fs * ir ; // maximum s l i p al = sm ^(1/ n ) ; R1 = R / sm ; // r e s i s t a n c e o f w h o l e s e c t i o n r1 = R1 *(1 - al ) ; printf ( ’ R e s i s t a n c e o f f i r s t e l e m e n t i s %f ohms \n ’ , r1 ); r2 = r1 * al ; printf ( ’ R e s i s t a n c e o f s e c o n d e l e m e n t i s %f ohms \n ’ , r2 ) ; r3 = r1 * al ^2; printf ( ’ R e s i s t a n c e o f t h i r d e l e m e n t i s %f ohms \n ’ , r3 ); r4 = r1 * al ^3; printf ( ’ R e s i s t a n c e o f f o u r t h e l e m e n t i s %f ohms \n ’ , r4 ) ; r5 = r1 * al ^4; printf ( ’ R e s i s t a n c e o f f i f t h e l e m e n t i s %f ohms \n ’ , r5 );

Scilab code Exa 6.64 Determining 1 starting current and starting torque 2 external resistance to limit starting current and starting torque under this condition 1 2 3 4

clc ; v =3300; // r a t e d v o l t a g e o f i n d u c t i o n motor p =6; // number o f p o l e s f =50; // f r e q u e n c y 309


5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27

t =3.2; // s t a t o r t o r o t o r t u r n s r =0.1; // r o t o r r e s i s t a n c e x =1; // r o t o r l e a k a g e r e a c t a n c e R = t ^2* r ; // r o t o r r e s i s t a n c e r e f e r r e d t o s t a t o r X = t ^2* x ; // r o t o r r e a c t a n c e r e f e r r e d t o s t a t o r ws =(4* %pi * f ) / p ; // s y n c h r o n o u s s p e e d disp ( ’ c a s e a ’ ) ; is =( v / sqrt (3) ) /( sqrt ( R ^2+ X ^2) ) ; printf ( ’ S t a r t i n g c u r r e n t a t r a t e d v o l t a g e i s %f A\n ’ , is ) ; Ts =(3* is ^2* R ) / ws ; printf ( ’ S t a r t i n g t o r q u e a t r a t e d v o l t a g e i s %f Nm\n ’ , Ts ) ; disp ( ’ c a s e b ’ ) ; is =50; // s t a r t i n g c u r r e n t // i s =Vp / ( s q r t ( ( R+r e x ) ˆ2+Xˆ 2 ) where r e x i s e x t e r n a l r e s i s t a n c e and Vp i s p h a s e v o l t a g e // s o l v i n g a b o v e e q u a t i o n we g e t a q u a d r a t i c e q u a t i o n i n r e x whose t e r m s a r e t1 =1; t2 =2* R ; t3 =( R ^2+ X ^2) -(( v / sqrt (3) ) / is ) ^2; p =[ t1 t2 t3 ]; rex = roots ( p ) ; printf ( ’ E x t e r n a l r e s i s t a n c e r e f e r r e d t o r o t o r i s %f ohms \n ’ , rex (2) / t ^2) ; Ts =(3* is ^2*( R + rex (2) ) ) / ws ; printf ( ’ S t a r t i n g t o r q u e u n d e r new c o n d i t i o n i s %f Nm \n ’ , Ts ) ;

Scilab code Exa 6.65 Determining 1 line current 2 power returned to three phase supply 3 efficiency of motor 1 clc ; 2 p =6; // number o f

poles 310


3 4 5 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

v =400; // r a t e d v o l t a g e o f i n d u c t i o n motor m =3; // number o f p h a s e s f =50; // f r e q u e n c y r1 =0.2; // s t a t o r r e s i s t a n c e r2 =0.5; // r o t o r r e s i s t a n c e xm =48; // m a g n e t i s i n g r e a c t a n c e x1 =2; // l e a k a g e r e a c t a n c e o f s t a t o r o r r o t o r n =1050; // s p e e d o f motor ns =(120* f ) / p ; // s y n c h r o n o u s s p e e d s =( ns - n ) / ns ; // o p e r a t i n g s l i p disp ( ’ c a s e a ’ ) ; Zf =((( r2 / s ) + %i * x1 ) * %i * xm ) /(( r2 / s ) + %i *( xm + x1 ) ) ; // p e r p h a s e i m p e d a n c e o f f e r e d t o s t a t o r by r o r a t i n g a i r gap f i e l d z = r1 + %i * x1 ; // i m p e d a n c e o f s t a t o r Z = Zf + z ; // t o t a l i m p e d a n c e is = v /( sqrt (3) * abs ( Z ) ) ; // s t a t o r c u r r e n t pf = cosd ( atand ( imag ( Z ) , real ( Z ) ) ) ; printf ( ’ S t a t o r l i n e c u r r e n t i s %f A\n ’ , is ) ; disp ( ’ c a s e b ’ ) ; Po = m *( v / sqrt (3) ) * is * pf ; // n e g a t i v e power i n d i c a t e s i n d u c t i o n machine i s ac ting as generator printf ( ’ Power f e d back t o 3 p h a s e s u p p l y s y s t e m i s %f W\n ’ ,- Po ) ; disp ( ’ c a s e c ’ ) ; lr =600; // r o t a t i o n a l and c o r e l o s s e s pg = m * is ^2* real ( Zf ) ; // a i r gap power l1 = m * is ^2* r1 ; // s t a t o r c o p p e r l o s s l2 = s * pg ; // r o t o r c o p p e r l o s s Tl = lr + l1 + l2 ; // t o t a l l o s s e s pi = - Po + Tl ; // m e c h a n i c a l power i n p u t ne = - Po / pi ; printf ( ’ E f f i c i e n c y o f i n d u c t i o n motor i s %f p e r c e n t \ n ’ , ne *100) ;

311


Chapter 7 Armature Windings

Scilab code Exa 7.1 Determining the number of commutator segments back pitch and front pitch and commutator pitch 1 2 3 4 5 6 7 8 9 10 11 12 13 14

clc ; p =6; // number o f p o l e s c =40; // number o f c o i l s w =2; // w i n d i n g p i t c h f o r s i m p l e x l a p w i n d i n g printf ( ’ Number o f commutator s e g m e n t s i s e q u a l t o number o f c o i l s =%f\n ’ ,c ) ; k =1/3; // i n t e g e r added ( o r s u b t r a c t e d ) t o c a l c u l a t e back p i t c h t o make i t an odd i n t e g e r yb =((2* c ) / p ) -k ; printf ( ’ Back p i t c h i s %f \n ’ , yb ) ; yf = yb - w ; printf ( ’ F r o n t p i t c h f o r p r o g r e s s i v e w i n d i n g i s %f\n ’ , yf ) ; yf = yb + w ; printf ( ’ F r o n t p i t c h f o r r e t r o g r e s s i v e w i n d i n g i s %f\ n ’ , yf ) yc =1; printf ( ’ For s i m p l e x l a p w i n d i n g , commutator p i t c h i s e q u a l t o %f ’ , yc ) ;

312


Scilab code Exa 7.2 Deg the progressive simplex lap winding with two coil sides per slot 1 clc ; 2 p =4; // number o f p o l e s 3 c =12; // number o f c o i l s 4 // Number o f commutator s e g m e n t s 5 6 7 8 9 10 11 12 13 14 15 16 17 18

19 20

i s e q u a l t o number o f c o i l s =12 // Each c o i l h a s two c o i l s i d e t h e r e f o r e t o t a l c o i l s i d e s a r e 24 s =(2* c ) /2 ; // t o t a l number o f s l o t s r e q u i r e d k =1; // i n t e g e r added ( o r s u b t r a c t e d ) t o c a l c u l a t e back p i t c h t o make i t an odd i n t e g e r w =2; // w i n d i n g p i t c h yb1 =((2* c ) / p ) -k ; // back p i t c h // o r yb2 =((2* c ) / p ) + k ; // back p i t c h disp ( ’ Back p i t c h i s ’ ) ; disp ( yb1 , ’ o r ’ , yb2 ) ; yf1 = yb1 -2; // f r o n t p i t c h f o r yb=5 yf2 = yb2 -2; // f r o n t p i t c h f o r yb=7 disp ( ’ f r o n t p i t c h f o r p r o g r e s s i v e w i n d i n g i s ’ ) ; disp ( yf1 , ’ o r ’ , yf2 ) ; disp ( ’ I t i s d e s i r a b l e t h a t ( yb+y f ) /2 s h o u l d be e q u a l to pole p i t c h that i s 6( in o f c o i l s i d e s p e r p o l e ) . So c h o o s e yb=7 and y f =5 ’ ) ; disp ( ’ Commutator p i t c h f o r p r o g r e s s i v e l a p w i n d i n g i s ’ ); disp (1) ;

Scilab code Exa 7.3 Determining winding table and position of brushes on commutator 313


1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

clc ; p =4; // number o f p o l e s s =14; // number o f s l o t s =2; // c o i l s i d e s p e r s l o t s w =2; // w i n d i n g p i t c h C =( s * ) /2; // number o f c o i l s yb =(2* C ) / p ; disp ( ’ Back p i t c h i s ’ ) ; disp ( yb ) ; yf = yb - w ; disp ( ’ F r o n t p i t c h i s ’ ) ; disp ( yf ) ; disp ( ’ w i n d i n g t a b l e f o r p r o g r e s s i v e l a p w i n d i n g i s ’ ) ; disp ( ’ (1 −8) −(3 −10) −(5 −12) −(7 −14) −(9 −16) −(11 −18) −(13 −20) −(15 −22) −(17 −24) −(19 −26) ’ ) ; disp ( ’ −(21 −28) −(23 −2) −(25 −4) −(27 −6) ’ ) ; disp ( ’ from w i n d i n g d i a g r a m ’ ) disp ( ’ Brush A i s t o u c h i n g s e g m e n t s 1 and 2 p a r t l y ’ ) ; disp ( ’ Brush B i s a t s e g m e n t 5 ’ ) ; disp ( ’ Brush C i s a t s e g m e n t 8 ’ ) ;

Scilab code Exa 7.5 Deg a simplex lap winding with the given details 1 2 3 4 5 6

clc ; disp ( ’ c a s e a ’ ) ; s =30; // number o f s l o t s c =60; // number o f c o i l s p =4; // number o f p o l e s k =1; // i n t e g e r added ( o r s u b t r a c t e d ) t o c a l c u l a t e back p i t c h t o make i t an odd i n t e g e r 7 tc = c *2; // total coil sides 8 u = tc / s ; // c o i l s i d e s p e r s l o t s 9 yb1 =( tc / p ) + k ; 314


10 yb2 =( tc / p ) -k ; 11 disp ( ’ Back p i t c h i s ’ ) ; 12 disp ( yb1 ) ; 13 disp ( ’ o r ’ ) ; 14 disp ( yb2 ) ; 15 disp ( ’ f o r back p i t c h =29 , t o p

16 17 18 19 20 21 22 23 24 25 26 27 28 29

c o i l s i d e s 1 and 3 i n s l o t 1 a r e c o n n e c t e d t o bottom c o i l 30 and 32 i n s l o t 8 . Due t o t h i s a r r a n g e m e n t s p l i t c o i l s can be a v o i d e d . But f o r back p i t c h= 3 1 , c o i l s i d e s 34 which i s i n s l o t 9 h a s t o be used , s o s p l i t c o i l s are needed ’ ) disp ( ’ c a s e b ’ ) ; s =20; // number o f s l o t s c =60; // number o f c o i l s p =4; // number o f p o l e s k =1; // i n t e g e r added ( o r s u b t r a c t e d ) t o c a l c u l a t e back p i t c h t o make i t an odd i n t e g e r tc = c *2; // t o t a l c o i l s i d e s u = tc / s ; // c o i l s i d e s p e r s l o t s yb1 =( tc / p ) + k ; yb2 =( tc / p ) -k ; disp ( ’ Back p i t c h i s ’ ) ; disp ( yb1 ) ; disp ( ’ o r ’ ) ; disp ( yb2 ) ; disp ( ’ f o r back p i t c h =29 , t o p c o i l s i d e s 1 , 3 and 5 a r e c o n n e c t e d t o bottom c o i l 3 0 , 32 and 3 4 . Due t o t h i s a r r a n g e m e n t s p l i t c o i l s c a n n o t be a v o i d e d . But f o r back p i t c h= 3 1 , c o i l s i d e s 1 , 3 and 5 a r e c o n n e c t e d t o bottom c o i l s i d e s 3 2 , 34 and 36 which a r e i n s l o t 6 , s o s p l i t c o i l s a r e n o t n e e d e d ’ );

Scilab code Exa 7.6 Deg a simplex wave winding with given details

315


1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

clc ; p =4; // number o f p o l e s s =11; // number o f s l o t s ts =2; // c o i l s i d e s p e r s l o t C =( s * ts ) /2; // t o t a l c o i l s w =((2* C ) +2) /( p /2) ; // w i n d i n g p i t c h // s i n c e b o t h back and f r o n t p i t c h s h o u l d be odd choose Yb =7; Yf =5; disp ( ’ Back p i t c h i s ’ ) disp ( Yb ) ; disp ( ’ F r o n t p i t c h i s ’ ) disp ( Yf ) ; yc =( C +1) /( p /2) ; disp ( ’ commutator p i t c h ’ ) ; disp ( yc ) ; disp ( ’ U s i n g t h i s d a t a w i n d i n g d i a g r a m can be drawn ’ ) ; disp ( ’ Winding t a b l e i s ’ ) ; disp ( ’ (1 −8) −(13 −20) −(3 −10) −(15 −22) −(5 −12) −(17 −2) −(7 −14) −(19 −4) −(9 −16) −(21 −6) −(11 −18)−1 ’ ) ;

Scilab code Exa 7.7 Deg the simplex wave winding 1 2 3 4 5 6

clc ; p =6; // number o f p o l e s s =72; // number o f s l o t s ts =4; // number o f c o i l s i d e s p e r s l o t C =( s * ts ) /2; // t o t a l number o f c o i l s // To make commutator p i t c h an i n t e g e r one c o i l made dummy c o i l t h e r e f o r e 7 C =C -1; 8 yc =( C +1) /( p /2) ; 9 disp ( ’ commutator p i t c h ’ ) ; 316


is

10 disp ( yc ) ; 11 yw =((2* C ) +2) /( p /2) ; 12 disp ( ’ Winding p i t c h i s ’ ) ; 13 disp ( yw ) ; 14 // s i n c e back and f r o n t p i t c h s h o u l d be odd c h o o s e 15 yb =49; 16 disp ( ’ Back p i t c h i s ’ ) ; 17 disp ( yb ) ; 18 yf =47; 19 disp ( ’ F r o n t p i t c h i s ’ ) ; 20 disp ( yf ) ;

Scilab code Exa 7.8 Deg the winding and determining the speed 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

clc ; p =4; // number o f p o l e s z =2540; // number o f c o n d u c t o r s s =32; // number o f s l o t s c =127; // number o f commutator s e c t o r s = t o t a l number of c o i l s v =500; // i n d u c e d v o l t a g e r e q u i r e d f =5*10^ -3; // f i e l d f l u x p e r p o l e a =2; // number o f p a r a l l e l p a t h s zs = ceil ( z / s ) ; // c o n d u c t o r s p e r s l o t // f o r z s =80 Z = zs * s ; // t o t a l c o n d u c t o r s t = floor ( Z /(2* c ) ) ; // t u r n p e r c o i l C = Z /(2* t ) ; // a c t u a l number o f c o i l s // I t i s n e c e s s a r y t h a t a c t u a l c o i l s s h o u l d be same a s commutator s e g m e n t s s o one c o i l i s made dummy disp ( ’ commutor p i t c h i s ’ ) disp (( c +1) /( p /2) ) ; disp ( ’ o r ’ ) ; disp (( c -1) /( p /2) ) ; disp ( ’ Winding p i t c h i s ’ ) 317


20 21 22 23 24 25 26 27 28

disp (((2* c ) +2) /( p /2) ) ; disp ( ’ o r ’ ) ; disp (((2* c ) -2) /( p /2) ) ; disp ( ’ For p r o g r e s s i v e w i n d i n g , back p i t c h =65 and f r o n t p i t c h =63 ’ ) ; disp ( ’ For r e t r o g r e s s i v e w i n d i n g , back p i t c h =63 and f r o n t p i t c h =63 ’ ) ; // s i n c e dumy c o i l i s n o t i n c i r c u i t , number o f active conductor i s Z = c * t *2; n =( v * a *60) /( f * Z * p ) ; printf ( ’ Speed f o r r e q u i r e d i n d u c e d v o l t a g e i s %f rpm ’ ,n ) ;

Scilab code Exa 7.9 Determining the suitable arrangement of equilizer ring for given details 1 2 3 4 5 6 7 8 9

clc ; p =8; // number o f p o l e s c =240; // number o f c o i l s r =10; // number o f e q u i l i z e r r i n g Yeq =(2* c ) / p ; printf ( ’ E q u i p o t e n t i a l p i t c h i s %f c o i l s \n ’ , Yeq ) ; Ytp =(2* c ) /( r * p ) ; printf ( ’ Tapping p o i n t p i t c h i s %f c o i l s ’ , Ytp ) ; disp ( ’ Arrangement i s shown i n t a b u l a r form i n example 7 . 9 ’ );

Scilab code Exa 7.10 Determining the resistance measured between two adjacent commutator segments 1 clc ; 2 disp ( ’ b ( 1 ) ’ ) ;

318


3 c =12; // number o f c o i l s 4 r =0.1; // r e s i s t a n c e o f e a c h c o i l 5 // any one c o i l c o n n e c t e d t o commutator s e g m e n t i s

i n p a r a l l e l w i t h o t h e r 11 therefore 6 R =11* r ; // r e s i s t a n c e o f 11 7 req =( r * R ) /( r + R ) ; 8 printf ( ’ R e s i s t a n c e m e a s u r e d commutator s e g m e n t s i s %f

s e r i e s connected c o i l s coil b e t w e e n two a d j a c e n t ohm\n ’ , req ) ;

Scilab code Exa 7.11 Determining the details for winding diagram and distribution factor 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

clc ; disp ( ’ a ’ ) ; s =24; // t o t a l number o f s l o t s p =4; // number o f p o l e s np =3; // number o f p h a s e s ph =60; // p h a s e s p r e a d // g i v e n a r m a t u r e h a s d o u b l e l a y e r w i n d i n g and f u l l p i t c h c o i l span v =( p *180) / s ; printf ( ’ S l o t a n g u l a r p i t c h i s %d d e g r e e s \n ’ ,v ) ; disp ( ’ Number o f a d j a c e n t s l o t s i n one p h a s e b e l t i s ’ ); disp ( ph / v ) ; cs = s / p ; printf ( ’ C o i l s p a n i s %d s l o t s \n ’ , cs ) ; disp ( ’ U s i n g t h i s d a t a w i n d i n g t a b l e f o r t h e t h r e e p h a s e s i s shown i n Ex7 . 1 1 ’ ) disp ( ’ d ’ ) ; sp = s /( p * np ) ; // s l o t s p e r p o l e p e r p h a s e disp ( ’ D i s t r i b u t i o n f a c t o r i s ’ ) ; disp ( sind ( ph /2) /( sp * sind ( v /2) ) ) ;

319


Scilab code Exa 7.12 Determining the details for winding diagram and distribution factor 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

clc ; disp ( ’ a ’ ) ; s =24; // t o t a l number o f s l o t s p =4; // number o f p o l e s np =3; // number o f p h a s e s ph =120; // p h a s e s p r e a d // g i v e n a r m a t u r e h a s d o u b l e l a y e r w i n d i n g and f u l l p i t c h c o i l span v =( p *180) / s ; printf ( ’ S l o t a n g u l a r p i t c h i s %d d e g r e e s \n ’ ,v ) ; disp ( ’ Number o f a d j a c e n t s l o t s i n one p h a s e b e l t i s ’ ); disp ( ph / v ) ; cs = s / p ; printf ( ’ C o i l s p a n i s %d s l o t s \n ’ , cs ) ; disp ( ’ U s i n g t h i s d a t a w i n d i n g t a b l e f o r t h e t h r e e p h a s e s i s shown i n Ex7 . 1 2 ’ ) disp ( ’ d ’ ) ; sp = s /( p * np ) ; // s l o t s p e r p o l e p e r p h a s e disp ( ’ D i s t r i b u t i o n f a c t o r i s ’ ) ; disp ( sind ( ph /2) /( sp * sind ( ph /(2* sp ) ) ) ) ;

Scilab code Exa 7.13 Determining the details for winding table and effective turns per phase 1 2 3 4

clc ; np =3; // number o f p h a s e sp =9; // s l o t s p e r p o l e zs =4; // c o n d u c t o r s p e r s l o t 320


5 f =0.8; // c o i l s p a n a s a f r a c t i o n o f p o l e p i t c h 6 ph =60; // p h a s e s p r e a d 7 v =180/ sp ; // s l o t a n g u l a r p i t c h 8 disp ( ’ Number o f a d j a c e n t s l o t s b e l o n g i n g t o any 9 10 11 12 13 14 15 16 17 18 19 20 21 22

phase i s ’ ); disp ( ph / v ) ; printf ( ’ P o l e p i t c h i s %f s l o t s \n ’ , sp ) ; c = floor (0.8* sp ) ; printf ( ’ C o i l s p a n i s o f %f s l o t s \n ’ ,c ) ; disp ( ’ U s i n g t h i s data , w i n d i n g t a b l e i s shown i n Ex7 . 1 3 ’ ); t =( sp * zs *4) /2; // t o t a l t u r n s i n machine spp = sp / np ; // s l o t s p e r p o l e p e r p h a s e kd = sind ( ph /2) /( spp * sind ( v /2) ) ; // d i s t r i b u t i o n factor = c * v ; // c o i l s p a n i n d e g r e e s e =180 - ; // c h o r d i n g a n g l e kp = cosd ( e /2) ; // c o i l s p a n f a c t o r kw = kd * kp ; // w i n d i n g f a c t o r tp =( t * kw ) / np ; printf ( ’ Number o f e f f e c t i v e t u r n s p e r p h a s e i s %f ’ , tp ) ;

Scilab code Exa 7.15 Determining and deg the details for winding of 3 phase machine 1 2 3 4 5 6 7 8 9

clc ; s =24; // number o f s l o t s p =4; // number o f p o l e s ph =60; // p h a s e s p r e a d ap =( p *180) / s ; // s l o t a n g u l a r p i t c h pp = s / p ; // p o l e p i t c h printf ( ’ P o l e p i t c h i s %d s l o t s \n ’ , pp ) ; printf ( ’ s l o t a n g u l a r p i t c h i s %d d e g r e e s ’ , ap ) ; disp ( ’ u s i n g t h e s e data , h a l f c o i l and w h o l e c o i l 321


s i n g l e l a y e r c o n c e n t r i c windings diagram a re drawn ’ ) ;

322


Chapter 8 Appendix A

Scilab code Exa 8.1 Determining the reluctance of ring and current required to establish the required flux 1 2 3 4 5 6 7 8 9 10 11

clc ; // a n s w e r i s g i v e n wrong i n t h e book d =0.2; // mean d i a m e t e r o f m i l d s t e e l r i n g ac =50*10^ -4; // c r o s s s e c t i o n a l a r e a o f c o r e uo =4* %pi *10^ -7; // f r e e s p a c e p e r m e a b i l i t y ur =800; // r e l a t i v e p e r m e a b i l i t y f =1*10^ -3; // r e q u i r e d f l u x N =200; // Number o f t u r n s l = %pi * d // l e n g t h o f c o r e R = l /( uo * ur * ac ) ; // r e l u c t a n c e o f r i n g printf ( ’ r e l u c t a n c e o f f e r e d by r i n g i s %f AT/Wb\n ’ ,R ) ; 12 mmf = f * R ; // mmf p r o d u c e d i n r i n g 13 i = mmf / N ; 14 printf ( ’ c u r r e n t r e q u i r e d t o p r o d u c e t h e d e s i r e d f l u x i s %f A ’ ,i ) ;

Scilab code Exa 8.2 Determining the exciting current in coil 323


1 2 3 4 5 6 7 8 9 10 11 12 13

clc ; ur =10000; // r e l a t i v e p e r m e a b i l i t y o f i r o n lc =0.5; // c o r e l e n g t h lg =4*10^ -3; // a i r gap l e n g t h N =600; // number o f t u r n s B =1.2; // d e s i r e d f l u x d e n s i t y uo =4* %pi *10^ -7; // f r e e s p a c e p e r m e a b i l i t y Ac =25*10^ -4; // c o r e a r e a mfc =( B * lc ) /( uo * ur ) ; // mmf f o r c o r e mfg =( B * lg ) / uo ; // mmf f o r a i r gap mft = mfc + mfg ; // n e t mmf i = mft / N ; printf ( ’ e x c i t i n g c u r r e n t r e q u i r e d t o e s t a b l i s h t h e d e s i r e d f l u x i s %f A ’ ,i ) ;

Scilab code Exa 8.3 Determining the exciting current with and without magnetic leakage and fringing 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

clc ; lc =0.5; // c o r e l e n g t h i n m e t r e dc =2.85*10^ -2; // d i a m e t e r o f c r o s s s e c t i o n o f c o r e lg =2*10^ -3; // l e n g t h o f a i r gap N =500; // Number o o f t u r n s o f c o i l f =0.8*10^ -3; // a i r gap f l u x uo =4* %pi *10^ -7; // p e r m e a b i l i t y o f f r e e s p a c e HATM =[1500 2210 2720 3500 4100]; BT =[0.9 1.1 1.2 1.275 1.3]; plot ( HATM , BT ) ; xlabel ( ’ m a g n e t i c f i e l d i n t e n s i t y ’ ) ; ylabel ( ’ f l u x d e n s i t y ’ ) ; disp ( ’ c a s e a ’ ) ; ur =500; // r e l a t i v e p e r m e a b i l i t y Ac =( %pi /4) * dc ^2; // Area o f c o r e Rlg = lg /( uo * Ac ) ; // r e l u c t a n c e o f a i r gap Rlc = lc /( uo * ur * Ac ) ; // r e l u c t a n c e o f i r o n c o r e 324


18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35

Rt = Rlg + Rlc ; // T o t a l r e l u c t a n c e I =( f * Rt ) / N ; // E x c i t i n g c u r r e n t printf ( ’ E x c i t i n g c u r r e n t i n c o i l i s %f A\n ’ ,I ) ; disp ( ’ c a s e b ’ ) ; Ag =( %pi /4) *( dc +2* lg ) ^2; // a i r gap a r e a Rlg = lg /( uo * Ag ) ; // r e l u c t a n c e o f a i r gap I =( f *( Rlc + Rlg ) ) / N ; // E x c i t i n g c u r r e n t printf ( ’ E x c i t i n g c u r r e n t a f t e r a c c o u n t i n g f o r f l u x f r i n g i n g i s %f A\n ’ ,I ) ; disp ( ’ c a s e c ’ ) ; Bg = f / Ac ; // A i r gap f l u x d e n s i t y Atg =( Bg * lg ) / uo ; // a i r gap mmf // from t h e p l o t we can g e t t h e v a l u e s o f c o r e f l u x d e n s i t y and m a g n e t i c f i e l d i n t e n s i t y Bc =1.245; // c o r e f l u x d e n s i t y i n T e s l a H =3200; // m a g n e t i c f i e l d i n t e n s i t y i n Ats /m Atc = H * lc ; // c o r e mmf mt = Atg + Atc ; // t o t a l mmf I = mt / N ; // E x c i t i n g c u r r e n t printf ( ’ E x c i t i n g c u r r e n t f o r t h i r d c a s e i s %f A ’ ,I ) ;

Scilab code Exa 8.4 Determining the coil current for different values of relative permeability 1 2 3 4 5 6 7 8 9 10 11

clc ; N =1000; // Number o f t u r n s f =1*10^ -3; // f l u x i n c e n t r a l l i m b Ac =8*10^ -4; // Area o f c e n t r a l l i m b Ao =4*10^ -4; // Area o f o u t e r l i m b lg =2*10^ -3; // l e n g t h o f a i r gap lc =0.15; // l e n g t h o f c e n t r a l l i m b i n m e t r e lo =0.25; // l e n g t h o f o u t e r l i m b i n m e t r e uo =4* %pi *10^ -7; // p e r m e a b i l i t y o f f r e e s p a c e disp ( ’ c a s e a ’ ) ; // f o r u r= i n f i n i t y , r e l u c t a n c e o f f e r e d by c a s t s t e e l 325


12 13 14 15 16 17 18 19 20 21 22 23 24

i s zero Rl1 = lg /( uo * Ao ) ; // r e l u c t a n c e o f f e r e d by o u t e r l i m b Rl2 = lg /( uo * Ac ) ; // r e l u c t a n c e o f f e r e d by c e n t r a l limb // Assuming m a g n e t i c c i r c u i t a s a c l o s e c i r c u i t , a p p l y i n g KVl i n one o f l o o p g i v e s I =( f *( Rl2 +( Rl1 /2) ) ) / N ; printf ( ’ C o i l c u r r e n t f o r f i r s t c a s e i s %f A\n ’ ,I ) ; disp ( ’ c a s e b ’ ) ; ur =6000; // r e l a t i v e p e r m a b i l i t y Rlc1 =( lc + lo ) /( uo * ur * Ao ) ; // r e l u c t a n c e o f o u t e r s t e e l c o r e ( i n c l u d i n g the top ) Rlc2 =( lc ) /( uo * ur * Ac ) ; // r e l u c t a n c e o f f e r e d by central s t e e l core r =( Rlc1 + Rl1 ) /2; // r e s u l t a n t o f o u t e r r e l u c t a n c e // By kVL we g e t I =( f *( Rlc2 + Rl2 + r ) ) / N ; printf ( ’ C o i l c u r r e n t f o r s e c o n d c a s e i s %f A\n ’ ,I ) ;

Scilab code Exa 8.5 Calculating the exciting current to set up required flux 1 clc ; 2 N =500; // number o f t u r n s i n c e n t r a l l i m b 3 ac =600*10^ -6; // c r o s s s e c t i o n a l a r e a o f c e n t r a l 4 5 6 7 8 9 10 11 12

limb ao =375*10^ -6; // c r o s s s e c t i o n a l a r e a o f o u t e r l i m b f =0.9*10^ -3; // r e q u i r e d f l u x i n Weber lg =0.8*10^ -3; // l e n g t h o f a i r gap lc =180*10^ -3; // l e n g t h o f c e n t r a l l i m b lo =400*10^ -3; // l e n g t h o f o u t e r l i m b uo =4* %pi *10^ -7; // f r e e s p a c e p e r m e a b i l i t y Bg = f / ac ; // a i r gap f l u x d e n s i t y Hg = Bg / uo ; // m a g n e t i c f i e l d i n t e n s i t y i n a i r gap mg = Hg * lg ; // mmf r e q u i r e d f o r a i r gap 326


13 14 15 16 17 18 19 20 21 22

// from f i g A . 7 , f o r B=1.5T , H f o r c a s t s t e e l i s 3 0 0 0 Ats /m H =3000; // m a g n e t i c f i e l d i n t e n s i t y f o r c a s t s t e e l mc = H * lc ; // mmf i n c e n t r a l l i m b Bo = f /(2* ao ) ; // f l u x d e n s i t y i n e a c h o u t e r l i m b // f o r B= 1 . 2 , H=1400 H =1400; // m a g n e t i c f i e l d i n t e n s i t y f o r c a s t s t e e l for given flux density mo = H * lo ; // mmf f o r o u t e r l i m b // By KVL I =( mg + mo + mc ) / N ; printf ( ’ The e x c i t i n g c u r r e n t r e q u i r e d t o e s t a b l i s h t h e d e s i r e d f l u x i s %f A ’ ,I ) ;

Scilab code Exa 8.6 Calculating the coil current to establish required flux 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

clc ; N =400; // number o f t u r n s i n c o i l ac =20*10^ -4; // a r e a o f c e m n t r a l l i m b ao =15*10^ -4; // a r e a o f o u t e r i i m b lg =1*10^ -3; // l e n g t h o f a i r gap lc =40*10^ -2; // l e n g t h o f c e n t r a l l i m b lo =60*10^ -2; // l e n g t h o f e a c h o u t e r l i m b f =0.9*10^ -3; // r e q u i r e d f l u x uo =4* %pi *10^ -7; // f r e e s p a c e p e r m e a b i l i t y Bg = f / ao ; // a i r gap f l u x d e n s i t y mg =( Bg * lg ) / uo ; // mmf o r a i r gap // f o r B= 0 . 6 ,H=575 AT/m from f i g A . 7 H =575; // m a g n e t i c f l u x i n t e n s i t y f o r g i v e n f l u x density mo = H * lo ; // mmf o f o u t e r l i m b which c o n t a i n a i r gap mt = mo + mg ; // combined mmf o f a i r gap and o u t e r l i m b // t h i s mmf a c t s a c r o s s t h e o t h e r o u t e r l i m b haeb = mt / lo ; // m a g n e t i c f i e l d i n t e n s i t y i n o u t e r l i m b which d o e s n o t c o n t a i n a i r gap 327


18 // f o r H= 1 3 7 0 . 7 7 , B=1.19 T from f i g A . 7 19 Bo =1.19; // f l u x d e n s i t y f o r g i v e n m a g n e t i c 20 21 22 23 24 25 26 27 28

field intensity faeb = Bo * ao ; // f l u x i n o u t e r l i m b fnet = f + faeb ; // n e t f l u x t h r o u g h c e n t r a l l i m b Bc = fnet / ac ; // f l u x d e n s i t y i n c e n t r a l l i m b // from f i g A. 7 H =1900; // m a g n e t i c f i e l d i n t e n s i t y f o r g i v e n f l u x density mc = H * lc ; // mmf i n c e n t r a l l i m b // by KVL i n one o f t h e l o o p I =( mc + mt ) / N ; printf ( ’ E x c i t i n g c u r r e n t r e q u i r e d t o e s t a b l i s h t h e g i v e n f l u x i s %f A ’ ,I )

Scilab code Exa 8.7 Finding the exciting current in given coil 1 clc ; 2 a =30*10^ -4; // c r o s s 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

s e c t i o n a l area of ferromagnetic core uo =4* %pi *10^ -7; // f r e e s p a c e p e r m e a b i l i t y ur =4000; // r e l a t i v e p e r m e a b i l i t y f o r c o r e f =10*10^ -3; // f l u x i n c e n t r a l l i m b n1 =200; // number o f t u r n s i n c o i l 1 m1 =5000; // mmf f o r c o i l 1 n2 =100; // number o f t u r n s i n c o i l 2 lc =0.3; // l e n g t h o f c e n t r a l l i m b lo =0.6; // l e n g t h o f o u t e r l i m b lg =1*10^ -3; // l e n g t h o f a i r gap rc = lc /( uo * ur * a ) ; // r e l u c t a n c e f o r c e n t r a l l i m b ro = lo /( uo * ur * a ) ; // r e l u c t a n c e f o r o u t e r l i m b rg = lg /( uo * a ) ; // r e l u c t a n c e f o r a i r gap mc = f *( rc + rg ) ; // mmf i n c e n t r a l l i m b // by KML, f l u x i n o u t e r l i m b c o n t a i n i n g c o i l 1 i s f1 =( m1 - mc ) / ro ; 328


18 19 20 21 22 23 24

// By f l u x law a t node a i n f i g A. 1 7 , f l u x i n o u t e r limb contaning c o i l 2 i s f2 = f1 - f ; // by mmf law , mmf i n c o i l 2 i s m2 = mc - f2 * ro ; I2 = m2 / n2 ; // c u r r e n t i n c o i l 2 , u p p e r p o l a r i t y i s assumed p o s i t i v e printf ( ’ C u r r e n t i n c o i l 2 i s %f A ’ , I2 ) ; disp ( ’ As t h e mmf o f c o i l 2 i s p o s i t i v e , assumed polarity i s correct . Therefore terminal A i s p o s i t i v e b e c a u s e c u r r e n t e n t e r s t h r o u g h i t and t e r m i n a l B i s n e g a t i v e ’ );

Scilab code Exa 8.8 Determining the emf induced in conductor for different angles of conductor with field flux 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

clc ; l =0.8; // l e n g t h o f c o n d u c t o r B =1.2; // f l u x d e n s i t y o f u n i f o r m m a g n e t i c f i e l d v =30; // s p e e d o f c o n d u c t o r disp ( ’ c a s e a ’ ) ; // c o n d u c t o r m o t i o n i s n o r m a l t o f i e l d f l u x theta =90; // a n g l e b e t w e e n d i r e c t i o n o f m o t i o n and field flux e = B * l * v * sin ( theta *( %pi /180) ) ; printf ( ’EMF i n d u c e d i s %f V\n ’ ,e ) ; disp ( ’ c a s e b ’ ) ; // c o n d u c t o r m o t i o n i s a t an a n g l e o f 30 d e g r e e s from d i r e c t i o n o f f i e l d theta =30; // a n g l e b e t w e e n d i r e c t i o n o f m o t i o n and field flux e = B * l * v * sin ( theta *( %pi /180) ) ; printf ( ’EMF i n d u c e d i s %f V\n ’ ,e ) ; disp ( ’ c a s e c ’ ) ; // c o n d u c t o r m o t i o n i s p a r l l e l t o f i e l d f l u x 329


theta =0; // a n g l e b e t w e e n d i r e c t i o n o f m o t i o n and field flux 18 e = B * l * v * sin ( theta *( %pi /180) ) ; 19 printf ( ’EMF i n d u c e d i s %f V\n ’ ,e ) ; 17

Scilab code Exa 8.9 Determining the emf induced in square coil for different angles of coil with field flux 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

clc ; // A f t e r d e r i v i n g t h e e x p r e s s i o n a =0.1; // s i d e o f s q u a r e c o i l N =100; // number o f t u r n s n =1000; // s p e e d o f r o t a t i o n on rpm B =1; // f l u x d e n s i t y o f a u n i f o r m m a g n e t i c disp ( ’ c a s e a ’ ) ; theta =90; // a n g l e o f c o i l w i t h t h e f i e l d w =(2* %pi * n ) /60; // a n g u l a r s p e e d o f c o i l i n e = N * B * a ^2* w * cos ( theta *( %pi /180) ) ; printf ( ’ Emf i n d u c e d i n c o i l i s %f V\n ’ ,e ) ; disp ( ’ c a s e b ’ ) ; theta =30; // a n g l e o f c o i l w i t h t h e f i e l d w =(2* %pi * n ) /60; // a n g u l a r s p e e d o f c o i l i n e = N * B * a ^2* w * cos ( theta *( %pi /180) ) ; printf ( ’ Emf i n d u c e d i n c o i l i s %f V\n ’ ,e ) ; disp ( ’ c a s e c ’ ) ; theta =0; // a n g l e o f c o i l w i t h t h e f i e l d w =(2* %pi * n ) /60; // a n g u l a r s p e e d o f c o i l i n e = N * B * a ^2* w * cos ( theta *( %pi /180) ) ; printf ( ’ Emf i n d u c e d i n c o i l i s %f V\n ’ ,e ) ;

field

rad / s

rad / s

rad / s

Scilab code Exa 8.10 Determining the emf induced in conductor 1 clc ;

330


2 l =0.5; // l e n g t h o f c o n d u c t o r l y i n g a l o n g Y−a x i s 3 B =1.2; // Flux d e n s i t y a l o n g t h e X−a x i s 4 v =2; // v e l o c i t y o f c o n d u c t o r 5 // e=Blv ; f o r maximum i n d u c e d emf a l l t h e t h r e e

q u a n t i t i e s s h o u l d be p e r p e n d i c u l a r t o e a c h o t h e r 6 e=B*l*v; 7 printf ( ’ Maximum i n d u c e d EMF i n c o n d u c t o r i s %f V ’ ,e ) ;

Scilab code Exa 8.12 Determining the inductance for given circuit by using the given expression 1 clc ; 2 disp ( ’ c a s e a ’ ) ; 3 // a s p e r t h e d a t a t a k e n from Ex 1 3 4 rlg =24.948*10^5; // a i r gap r e l u c t a n c e 5 6 7 8 9 10 11 12 13

14 15 16 17 18 19

f o r example 1 3 (a) rlc =12.474*10^5; // i r o n c o r e r e l u c t a n c e f o r e x a m p l e 1 3 (a) rl = rlg + rlc ; // n e t r e l u c t a n c e N =500; // Number o f t u r n s L =( N ^2/ rl ) *1000; printf ( ’ I n d u c t a n c e f o r c a s e a i s %f mH\n ’ ,L ) ; disp ( ’ c a s e b ’ ) ; // a s p e r t h e d a t a t a k e n from Ex 1 3 p a r t ( c ) B =1.254; // c a l c u l a t e d f l u x d e n s i t y H =3200; // m a g n e t i c f i e l d i n t e n s i t y o b t a i n e d from magnetisation curve corresponding to the f l u x density calculated uo =4* %pi *10^ -7; // f r e e s p a c e p e r m e a b i l i t y ur = B /( H * uo ) ; // r e l a t i v e p e r m e a b i l i t y o f i r o n c o r e d =2.85*10^ -2; // d i a m e t e r o f c r o s s s e c t i o n A =( %pi * d ^2) /4; // a r e a o f c o r e l =0.5; // c o r e l e n g t h rlc = l /( ur * uo * A ) ; // r e l u c t a n c e o f i r o n c o r e f o r p a r t 331


C 20 rt = rlg + rlc ; // n e t r e l u c t a n c e 21 L =( N ^2/ rt ) *1000; 22 printf ( ’ I n d u c t a n c e f o r c a s e b i s %f mH\n ’ ,L ) ;

Scilab code Exa 8.13 Determining the inductance of coils for given circuit 1 clc ; 2 // d a t a t a k e n from Ex A . 7 , f i g A. 1 6 3 N1 =200; // number o f t u r n s i n c o i l 1 4 f1 =53.97*10^ -3; // f l u x i n o u t e r l i m b c o n t a i n i n g 5 6 7 8 9 10 11 12 13 14

coil 1 m1 =5000; // mmf f o r c o i l 1 I1 = m1 / N1 ; // c u r r e n t i n c o i l 1 N2 =100; // number o f t u r n s i n c o i l 2 f2 =43.97*10^ -3; // f l u x i n o u t e r l i m b c o n t a i n i n g coil 2 m2 =1102; // mmf f o r c o i l 2 I2 = m2 / N2 ; // c u r r e n t i n c o i l 2 L1 =( N1 * f1 ) / I1 ; printf ( ’ I n d u c t a n c e f o r c o i l 1 i s %f H\n ’ , L1 ) ; L2 =( N2 * f2 ) / I2 ; printf ( ’ I n d u c t a n c e f o r c o i l 2 i s %f H\n ’ , L2 ) ;

332


Chapter 9 Appendix B

Scilab code Exa 9.3 Determining phase and line current power factor total active and reactive power for star and delta connected load 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

clc ; vl =400; // l i n e v o l t a g e z =10+7.5* %i ; // l o a d i m p e d a n c e p e r p h a s e disp ( ’ For s t a r c o n n e c t e d l o a d ’ ) ; vp = vl / sqrt (3) ; // p h a s e v o l t a g e ip = vp / abs ( z ) ; // p h a s e and l i n e c u r r e n t a r e same i n the case of s t a r connected load an = atand ( - imag ( z ) , real ( z ) ) ; pf = cosd ( an ) ; P = sqrt (3) * vl * ip ; pa = sqrt (3) * vl * ip * pf ; pr = - sqrt (3) * vl * ip * sind ( an ) ; printf ( ’ Phase and l i n e c u r r e n t s a r e %f A\n ’ , ip ) ; printf ( ’ Power f a c t o r i s %f l a g g i n g \n ’ , pf ) ; printf ( ’ T o t a l v o l t ampere i s %f VA\n ’ ,P ) ; printf ( ’ T o t a l a c t i v e power i s %f W\n ’ , pa ) ; printf ( ’ T o t a l r e a c t i v e power i s %f VAr\n ’ , pr ) ; disp ( ’ For d e l t a c o n n e c t e d l o a d ’ ) ; vp = vl // p h a s e v o l t a g e and l i n e v o l t a g e a r e same i n the case of s t a r connected load 333


19 20 21 22 23 24 25 26 27 28 29 30 31

ip = vp / abs ( z ) ; il = ip * sqrt (3) ; an = atand ( - imag ( z ) , real ( z ) ) ; pf = cosd ( an ) ; P = sqrt (3) * vl * il ; pa = sqrt (3) * vl * il * pf ; pr = - sqrt (3) * vl * il * sind ( an ) ; printf ( ’ Phase c u r r e n t i s %f A\n ’ , ip ) ; printf ( ’ L i n e c u r r e n t i s %f A\n ’ , il ) ; printf ( ’ Power f a c t o r i s %f l a g g i n g \n ’ , pf ) ; printf ( ’ T o t a l v o l t ampere i s %f VA\n ’ ,P ) ; printf ( ’ T o t a l a c t i v e power i s %f W\n ’ , pa ) ; printf ( ’ T o t a l r e a c t i v e power i s %f VAr\n ’ , pr ) ;

Scilab code Exa 9.4 Determining per phase circuit parameters 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

clc ; il =48; // l o a d c u r r e n t ( l e a d i n g ) p =30; // l o a d power i n KW vl =500; // l i n e v o l t a g e f =50; // s u p p l y f r e q u e n c y pf =( p *1000) /( sqrt (3) * vl * il ) ; vp = vl / sqrt (3) ; // p h a s e v o l t a g e zp = vp / il ; // m a g n i t u d e o f p h a s e i m p e d a n c e rp = zp * pf ; // s i n c e c u r r e n t i s l e a d i n g o t h e r p a r a m e t e r must be a capacitor xc = zp * sqrt (1 - pf ^2) ; // r e a c t a n c e c =(10^6) /(2* %pi * f * xc ) ; disp ( ’ c i r c u i t p a r a m e t e r s a r e ’ ) ; printf ( ’ Load r e s i s t a n c e i s %f ohm\n ’ , rp ) ; printf ( ’ Load c a p a c i t a n c e i s %f m i c r o f a r a d ’ ,c ) ;

334


Scilab code Exa 9.5 Finding total line current power factor total and reactive power 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

clc ; zs =10+15* %i ; // s t a r c o n n e c t e d l o a d p e r p h a s e zd =12 -15* %i ; // d e l t a c o n n e c t e d l o a d p e r p h a s e vl =400; // s u p p l y l i n e v o l t a g e disp ( ’ c a s e a ’ ) ; // c o n v e r t i n g d e l t a c o n n e c t e d l o a d t o s t a r c o n n e c t e d load zd = zd /3; vp = vl / sqrt (3) ; i1 = vp / zs ; // l i n e c u r r e n t i n s t a r c o n n e c t e d l o a d i2 = vp / zd ; // l i n e c u r r e n t i n d e l t a c o n n e c t e d l o a d i = abs ( i1 + i2 ) ; printf ( ’ T o t a l l i n e c u r r e n t i s %f A\n ’ ,i ) ; an = atand ( imag ( i1 + i2 ) , real ( i1 + i2 ) ) ; pf = cosd ( an ) ; P =( sqrt (3) * vl * i * pf ) ; pr = sqrt (3) * vl * i * sqrt (1 - pf ^2) ; printf ( ’ Power f a c t o r i s %f l e a d i n g \n ’ , pf ) ; printf ( ’ T o t a l power i s %f W\n ’ ,P ) ; printf ( ’ T o t a l r e a c t v e power i s %f VAr ’ , pr ) ;

Scilab code Exa 9.6 Determining input power line current power factor and shaft power 1 2 3 4 5 6 7 8

clc ; w1 =85; // r e a d i n g o f w a t t m e t e r 1 ; w2 =35; // r e a d i n g o f w a t t m e t e r 2 ; P = w1 + w2 ; // t o t a l i n p u t power n =0.85; // e f f i c i e n c y o f motor vl =1100; // s u p p l y v o l t a g e pf = cosd ( atand (( sqrt (3) *( w1 - w2 ) ) /( w1 + w2 ) ) ) ; il =( P *1000) /( sqrt (3) * vl * pf ) ; // l i n e c u r r e n t 335


9 ps = n * P ; 10 printf ( 11 printf ( 12 printf ( 13 printf (

’ I n p u t power ’ Line current ’ power f a c t o r ’ s h a f t power

i s %f KW\n ’ ,P ) ; i s %f A\n ’ , il ) ; i s %f l a g g i n g \n ’ , pf ) ; i s %f KW’ , ps ) ;

Scilab code Exa 9.7 Determining no load losses and no load power factor 1 clc ; 2 w1 =2000; // r e a d i n g o f w a t t m e t e r 1 u n d e r no l o a d 3 w2 = -400; // r e a d i n g o f w a t t m e t e r 2 u n d e r no l o a d ,

4 5 6 7 8

s i n c e t h e c o n n e c t i o n s a r e r e v e r s e d t h a t i s why negative sign theta = atand (( sqrt (3) *( w1 - w2 ) ) /( w1 + w2 ) ) ; pl = w1 + w2 ; pf = cosd ( theta ) ; printf ( ’ No l o a d l o s s e s a r e %f W\n ’ , pl ) ; printf ( ’ No l o a d power f a c t o r i s %f l a g g i n g ’ , pf ) ;

Scilab code Exa 9.8 Calculating reading of wattmeters and input power to load 1 2 3 4 5 6 7 8 9 10 11

clc ; vl =230; // l i n e v o l t a g e f =50; // f r e q u e n c y o f s u p p l y c =100*10^ -6; // v a l u e o f c a p a c i t a n c e i n e a c h p h a s e vp =230/ sqrt (3) ; // p h a s e v o l t a g e zp =1/(2* %pi * f * c ) ; // p h a s e i m p e d a n c e il = vp / zp ; // l i n e c u r r e n t // v a l u e o f c o s ( t h e t a ) i s t a k e n from f i g B . 1 5 w1 = vl * il * cosd (120) ; w2 = vl * il * cosd (60) ; printf ( ’ R e a d i n g o f w a t t m e t e r 1 i s %f W\n ’ , w1 ) ; 336


12 printf ( ’ R e a d i n g o f w a t t m e t e r 2 i s %f W\n ’ , w2 ) ; 13 p = w1 + w2 ; 14 printf ( ’ T o t a l i n p u t power i s %f W’ ,p ) ;

337


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