Solutions Manual for Electric Machinery Fundamentals, 5th Edition

i Solutions Manual to accompany Chapman Electric Machinery Fundamentals Fifth Edition Stephen J Chapman BAE Systems Australia ii Solutions Manual to accompany Electric Machinery Fundamentals Fifth Edition Copyright 2012 McGrawHill Inc All rights reserved Printed in the United States of America No part of this book may be used or reproduced in any manner whatsoever without written permission with the following exception homework solutions may be copied for classroom use iii TABLE OF CONTENTS Preface iv 1 Introduction to Machinery Principles 1 2 Transformers 23 3 AC Machine Fundamentals 73 4 Synchronous Generators 81 5 Synchronous Motors 123 6 Induction Motors 152 7 DC Machinery Fundamentals 202 8 DC Motors and Generators 214 9 SinglePhase and Special Purpose Motors 276 A Review of ThreePhase Circuits 287 B Coil Pitch and Distributed Windings 295 C SalientPole Theory of Synchronous Machines 302 S1 Introduction to Power Electronics 308 E Errata 348 iv PREFACE TO THE INSTRUCTOR This Instructors Manual is intended to accompany the fifth edition of Electric Machinery Fundamentals To make this manual easier to use it has been made selfcontained Both the original problem statement and the problem solution are given for each problem in the book This structure should make it easier to copy pages from the manual for posting after problems have been assigned Many of the problems in Chapters 2 4 5 and 8 require that a student read one or more values from a magnetization curve The required curves are given within the textbook but they are shown with relatively few vertical and horizontal lines so that they will not appear too cluttered Electronic copies of the corresponding opencircuit characteristics shortcircuit characteristics and magnetization curves as also supplied with the book They are supplied in as ASCII text files Students can use these files for electronic solutions to homework problems The ASCII files can be read into MATLAB and used to interpolate points along the curve Each curve is given in ASCII format with comments at the beginning For example the magnetization curve in Figure P81 is contained in file p81magdat Its contents are shown below This is the magnetization curve shown in Figure P81 The first column is the field current in amps and the second column is the internal generated voltage in volts at a speed of 1200 rmin To use this file in MATLAB type load p81magdat The data will be loaded into an N x 2 array named p81mag with the first column containing If and the second column containing the opencircuit voltage MATLAB function interp1 can be used to recover a value from this curve 0 0 00132 667 003 1333 0033 16 0067 3130 01 4546 0133 6026 0167 7506 02 8974 0233 1044 0267 11886 03 13286 0333 14646 0367 15978 04 17218 0433 18398 0467 19504 v 05 20518 0533 21452 0567 22306 06 2312 0633 238 0667 24414 07 24974 0733 25508 0767 2592 08 26374 0833 2676 0867 2708 09 2736 0933 27614 0966 278 1 27974 1033 28148 1067 28294 11 28428 1133 28548 1167 28654 12 2873 1233 28786 1267 28836 13 28882 1333 2892 1367 289375 14 289567 1433 289689 1466 289811 15 289950 To use this curve in a MATLAB program the user would include the following statements in the program Get the magnetization curve Note that this curve is defined for a speed of 1200 rmin load p81magdat ifvalues p81mag1 eavalues p81mag2 n0 1200 The solutions in this manual have been checked twice but inevitably some errors will have slipped through If you locate errors which you would like to see corrected please feel free to contact me at the address shown below or at my email address schapmantpgicomau I greatly appreciate your input My physical and email addresses may change from time to time but my contact details will always be available at the books Web site which is httpwwwmhhecomchapman Thank you Stephen J Chapman Melbourne Australia March 31 2011 Chapter 1 Introduction to Machinery Principles 11 A motors shaft is spinning at a speed of 1800 rmin What is the shaft speed in radians per second SOLUTION The speed in radians per second is 1 min 2 rad 1800 rmin 1885 rads 60 s 1 r 12 A flywheel with a moment of inertia of 4 kg m2 is initially at rest If a torque of 6 N m counterclockwise is suddenly applied to the flywheel what will be the speed of the flywheel after 5 s Express that speed in both radians per second and revolutions per minute SOLUTION The speed in radians per second is 2 6 N m 5 s 75 rads 4 kg m t t J The speed in revolutions per minute is 1 r 60 s 75 rads 716 rmin 2 rad 1 min n 13 A force of 10 N is applied to a cylinder as shown in Figure P11 What are the magnitude and direction of the torque produced on the cylinder What is the angular acceleration of the cylinder SOLUTION The magnitude and the direction of the torque on this cylinder is CCW sin ind rF ind 015 m 10 N sin 30 075 N m CCW The resulting angular acceleration is 2 2 075 N m 0188 rads 4 kg m J 14 A motor is supplying 50 N m of torque to its load If the motors shaft is turning at 1500 rmin what is the mechanical power supplied to the load in watts In horsepower SOLUTION The mechanical power supplied to the load is 1 1 min 2 rad 50 N m 1500 rmin 7854 W 60 s 1 r P 1 hp 7854 W 105 hp 746 W P 15 A ferromagnetic core is shown in Figure P12 The depth of the core is 5 cm The other dimensions of the core are as shown in the figure Find the value of the current that will produce a flux of 0005 Wb With this current what is the flux density at the top of the core What is the flux density at the right side of the core Assume that the relative permeability of the core is 800 SOLUTION There are three regions in this core The top and bottom form one region the left side forms a second region and the right side forms a third region If we assume that the mean path length of the flux is in the center of each leg of the core and if we ignore spreading at the corners of the core then the path lengths are 2275 cm 55 cm 30 cm and 30 cm The reluctances of these regions are 1l 2l 3l 1 7 055 m 729 kA tWb 800 4 10 Hm 005 m 015 m r o l l A A R 2 7 030 m 597 kA tWb 800 4 10 Hm 005 m 010 m r o l l A A R 3 7 030 m 1194 kA tWb 800 4 10 Hm 005 m 005 m r o l l A A R The total reluctance is thus TOT 1 2 3 729 597 1194 252 kA tWb R R R R and the magnetomotive force required to produce a flux of 0005 Wb is 0005 Wb 252 kA tWb 1260 A t F R and the required current is 1260 A t 25 A 500 t i N F The flux density on the top of the core is 2 0005 Wb 067 T 015 m 005 m B A The flux density on the right side of the core is 0005 Wb 20 T 005 m 005 m B A 16 A ferromagnetic core with a relative permeability of 1500 is shown in Figure P13 The dimensions are as shown in the diagram and the depth of the core is 5 cm The air gaps on the left and right sides of the core are 0050 and 0070 cm respectively Because of fringing effects the effective area of the air gaps is 5 percent larger than their physical size If there are 300 turns in the coil wrapped around the center leg of the core and if the current in the coil is 10 A what is the flux in each of the left center and right legs of the core What is the flux density in each air gap SOLUTION This core can be divided up into five regions Let be the reluctance of the lefthand portion of the core be the reluctance of the lefthand air gap R be the reluctance of the righthand portion of the core be the reluctance of the righthand air gap and be the reluctance of the center leg of the core Then the total reluctance of the core is 1 R 3 2 R R4 5 R 1 2 3 4 TOT 5 1 2 3 4 R R R R R R R R R R 1 1 7 0 1 111 m 168 kA tWb 1500 4 10 Hm 007 m 005 m r l A R 2 2 7 0 2 00007 m 152 kA tWb 4 10 Hm 007 m 005 m 105 l A R 3 3 7 0 3 111 m 168 kA tWb 1500 4 10 Hm 007 m 005 m r l A R 4 4 7 0 4 00005 m 108 kA tWb 4 10 Hm 007 m 005 m 105 l A R 3 5 5 7 0 5 037 m 561 kA tWb 1500 4 10 Hm 007 m 005 m r l A R The total reluctance is 1 2 3 4 TOT 5 1 2 3 4 168 152 168 108 561 204 kA tWb 168 152 168 108 R R R R R R R R R R The total flux in the core is equal to the flux in the center leg center TOT TOT 300 t 10 A 000147 Wb 204 kA tWb F R The fluxes in the left and right legs can be found by the flux divider rule which is analogous to the current divider rule 3 4 left TOT 1 2 3 4 168 108 000147 Wb 000068 Wb 168 152 168 108 R R R R R R 1 2 right TOT 1 2 3 4 168 152 000147 Wb 000079 Wb 168 152 168 108 R R R R R R The flux density in the air gaps can be determined from the equation BA left left eff 000068 Wb 0185 T 007 cm 005 cm 105 B A right right eff 000079 Wb 0215 T 007 cm 005 cm 105 B A 17 A twolegged core is shown in Figure P14 The winding on the left leg of the core N1 has 600 turns and the winding on the right N2 has 200 turns The coils are wound in the directions shown in the figure If the dimensions are as shown then what flux would be produced by currents i1 05 A and i2 10 A Assume r 1200 and constant 4 SOLUTION The two coils on this core are would so that their magnetomotive forces are additive so the total magnetomotive force on this core is TOT 1 1 2 2 600 t 05 A 200 t 100 A 500 A t N i N i F The total reluctance in the core is TOT 7 0 260 m 766 kA tWb 1200 4 10 Hm 015 m 015 m r l A R and the flux in the core is TOT TOT 500 A t 000653 Wb 766 kA tWb F R 18 A core with three legs is shown in Figure P15 Its depth is 5 cm and there are 100 turns on the leftmost leg The relative permeability of the core can be assumed to be 2000 and constant What flux exists in each of the three legs of the core What is the flux density in each of the legs Assume a 5 increase in the effective area of the air gap due to fringing effects 5 SOLUTION This core can be divided up into four regions Let be the reluctance of the lefthand portion of the core be the reluctance of the center leg of the core be the reluctance of the center air gap and be the reluctance of the righthand portion of the core Then the total reluctance of the core is the reluctance of the lefthand leg plot the parallel combination of the reluctances of the righthand and center legs 1 R R2 3 R 4 R 2 3 TOT 1 2 3 R R R R R R R R 4 4 1 1 7 0 1 108 m 955 kA tWb 2000 4 10 Hm 009 m 005 m r l A R 2 2 7 0 2 034 m 180 kA tWb 2000 4 10 Hm 015 m 005 m r l A R 3 3 7 0 3 00005 m 510 kA tWb 4 10 Hm 015 m 005 m 104 l A R 4 4 7 0 4 108 m 955 kA tWb 2000 4 10 Hm 009 m 005 m r l A R The total reluctance is 2 3 4 TOT 1 2 3 4 180 510 955 955 1355 kA tWb 180 510 950 R R R R R R R R The total flux in the core is equal to the flux in the left leg left TOT TOT 100 t 20 A 000148 Wb 1355 kA tWb F R The fluxes in the center and right legs can be found by the flux divider rule which is analogous to the current divider rule 4 center TOT 2 3 4 955 000148 Wb 000086 Wb 180 510 955 R R R R 6 2 3 right TOT 2 3 4 180 510 000235 Wb 000062 Wb 180 510 955 R R R R R The flux density in the legs can be determined from the equation BA left left 000148 Wb 0329 T 009 m 005 m B A center center 000086 Wb 0115 T 015 m 005 m B A left right 000062 Wb 0138 T 009 cm 005 cm B A 19 A wire is shown in Figure P16 which is carrying 20 A in the presence of a magnetic field Calculate the magnitude and direction of the force induced on the wire SOLUTION The force on this wire can be calculated from the equation 2 A 1 m 05 T 100 N into the page i ilB F l B 110 The wire is shown in Figure P17 is moving in the presence of a magnetic field With the information given in the figure determine the magnitude and direction of the induced voltage in the wire SOLUTION The induced voltage on this wire can be calculated from the equation shown below The voltage on the wire is positive downward because the vector quantity v B points downward 7 ind cos 45 10 ms 02 T 025 m cos 45 0354 V positive down e vBl v B l 111 Repeat Problem 110 for the wire in Figure P18 SOLUTION The induced voltage on this wire can be calculated from the equation shown below The total voltage is zero because the vector quantity v B points into the page while the wire runs in the plane of the page ind cos 90 1 ms 05 T 05 m cos 90 0 V e vBl v B l 112 The core shown in Figure P14 is made of a steel whose magnetization curve is shown in Figure P19 Repeat Problem 17 but this time do not assume a constant value of µr How much flux is produced in the core by the currents specified What is the relative permeability of this core under these conditions Was the assumption in Problem 17 that the relative permeability was equal to 1200 a good assumption for these conditions Is it a good assumption in general 8 SOLUTION The magnetization curve for this core is shown below 192 9 The two coils on this core are wound so that their magnetomotive forces are additive so the total magnetomotive force on this core is TOT 1 1 2 2 600 t 05 A 200 t 100 A 500 A t N i N i F Therefore the magnetizing intensity H is 500 A t 192 A tm 260 m c H l F From the magnetization curve 017 T B and the total flux in the core is TOT 017 T 015 m 015 m 000383 Wb BA The relative permeability of the core can be found from the reluctance as follows A l r 0 TOT TOT R F Solving for µr yields TOT 7 TOT 0 000383 Wb 26 m 704 500 A t 4 10 Hm 015 m 015 m r l A F The assumption that r 1200 is not very good here It is not very good in general 113 A core with three legs is shown in Figure P110 Its depth is 5 cm and there are 400 turns on the center leg The remaining dimensions are shown in the figure The core is composed of a steel having the magnetization curve shown in Figure 110c Answer the following questions about this core a What current is required to produce a flux density of 05 T in the central leg of the core b What current is required to produce a flux density of 10 T in the central leg of the core Is it twice the current in part a c What are the reluctances of the central and right legs of the core under the conditions in part a d What are the reluctances of the central and right legs of the core under the conditions in part b e What conclusion can you make about reluctances in real magnetic cores 10 SOLUTION The magnetization curve for this core is shown below a A flux density of 05 T in the central core corresponds to a total flux of TOT 05 T 005 m 005 m 000125 Wb BA By symmetry the flux in each of the two outer legs must be 1 2 0000625 Wb and the flux density in the other legs must be 1 2 0000625 Wb 025 T 005 m 005 m B B The magnetizing intensity H required to produce a flux density of 025 T can be found from Figure 110c It is 50 Atm Similarly the magnetizing intensity H required to produce a flux density of 050 T is 75 Atm The mean length of the center leg is 21 cm and the mean length of each outer leg is 63 dm so the total MMF needed is TOT center center outer outer H l H l F TOT 75 A tm 021 m 50 A tm 063 m 473 A t F and the required current is TOT 473 A t 012 A 400 t i N F b A flux density of 10 T in the central core corresponds to a total flux of TOT 10 T 005 m 005 m 00025 Wb BA By symmetry the flux in each of the two outer legs must be 1 2 000125 Wb and the flux density in the other legs must be 1 2 000125 Wb 050 T 005 m 005 m B B 11 The magnetizing intensity H required to produce a flux density of 050 T can be found from Figure 110c It is 75 Atm Similarly the magnetizing intensity H required to produce a flux density of 100 T is about 160 Atm Therefore the total MMF needed is TOT center center outer outer H I H I F TOT 160 A tm 021 m 75 A tm 063 m 808 A t F and the required current is TOT 808 A t 0202 A 400 t i N This current is not twice the current in part a c The reluctance of the central leg of the core under the conditions of part a is TOT cent TOT 75 A tm 021 m 126 kA tWb 000125 Wb F R The reluctance of the right leg of the core under the conditions of part a is TOT right TOT 50 A tm 063 m 504 kA tWb 0000625 Wb F R d The reluctance of the central leg of the core under the conditions of part b is TOT cent TOT 160 A tm 021 m 134 kA tWb 00025 Wb F R The reluctance of the right leg of the core under the conditions of part b is TOT right TOT 75 A tm 063 m 378 kA tWb 000125 Wb F R e The reluctances in real magnetic cores are not constant 114 A twolegged magnetic core with an air gap is shown in Figure P111 The depth of the core is 5 cm the length of the air gap in the core is 005 cm and the number of turns on the coil is 1000 The magnetization curve of the core material is shown in Figure P19 Assume a 5 percent increase in effective airgap area to account for fringing How much current is required to produce an airgap flux density of 05 T What are the flux densities of the four sides of the core at that current What is the total flux present in the air gap 12 SOLUTION The magnetization curve for this core is shown below An airgap flux density of 05 T requires a total flux of eff 05 T 005 m 005 m 105 000131 Wb BA This flux requires a flux density in the righthand leg of 13 right 000131 Wb 0524 T 005 m 005 m B A The flux density in the other three legs of the core is top left bottom 000131 Wb 0262 T 010 m 005 m B B B A The magnetizing intensity required to produce a flux density of 05 T in the air gap can be found from the equation ag ag o B H ag ag 7 0 05 T 398 kA tm 4 10 Hm B H The magnetizing intensity required to produce a flux density of 0524 T in the righthand leg of the core can be found from Figure P19 to be right 410 A tm H The magnetizing intensity required to produce a flux density of 0262 T in the top left and bottom legs of the core can be found from Figure P19 to be top left bottom 240 A tm H H H The total MMF required to produce the flux is TOT ag ag right right top top left left bottom bottom H l H l H l H l H l F TOT 398 kA tm 00005 m 410 A tm 040 m 3 240 A tm 040 m F TOT 2786 164 288 651 A t F and the required current is TOT 651 A t 0651 A 1000 t i N F The flux densities in the four sides of the core and the total flux present in the air gap were calculated above 115 A transformer core with an effective mean path length of 6 in has a 200turn coil wrapped around one leg Its crosssectional area is 025 in2 and its magnetization curve is shown in Figure 110c If current of 03 A is flowing in the coil what is the total flux in the core What is the flux density 14 SOLUTION The magnetizing intensity applied to this core is 200 t 03 A 394 A tm 6 in 00254 min c c Ni H l l F From the magnetization curve the flux density in the core is B 135 T The total flux in the core is 2 2 00254 m 135 T 025 in 0000218 Wb 1 in BA 116 The core shown in Figure P12 has the flux shown in Figure P112 Sketch the voltage present at the terminals of the coil 15 SOLUTION By Lenz Law an increasing flux in the direction shown on the core will produce a voltage that tends to oppose the increase This voltage will be the same polarity as the direction shown on the core so it will be positive The induced voltage in the core is given by the equation ind d e N dt so the voltage in the windings will be Time dt N d eind 0 t 2 s 500 t 0010 Wb 2 s 250 V 2 t 5 s 500 t 0020 Wb 3 s 333 V 5 t 7 s 500 t 0010 Wb 2 s 250 V 7 t 8 s 500 t 0010 Wb 1 s 500 V The resulting voltage is plotted below 16 117 Figure P113 shows the core of a simple dc motor The magnetization curve for the metal in this core is given by Figure 110c and d Assume that the crosssectional area of each air gap is 18 cm2 and that the width of each air gap is 005 cm The effective diameter of the rotor core is 5 cm a We wish to build a machine with as great a flux density as possible while avoiding excessive saturation in the core What would be a reasonable maximum flux density for this core b What would be the total flux in the core at the flux density of part a c The maximum possible field current for this machine is 1 A Select a reasonable number of turns of wire to provide the desired flux density while not exceeding the maximum available current 17 SOLUTION The magnetization curve for this core is shown below The relative permeability of this core is shown below Note This is a design problem and the answer presented here is not unique Other values could be selected for the flux density in part a and other numbers of turns could be selected in part c These other answers are also correct if the proper steps were followed and if the choices were reasonable a From Figure 110c a reasonable maximum flux density would be about 12 T Notice that the saturation effects become significant for higher flux densities b At a flux density of 12 T the total flux in the core would be 12 T005 m005 m 00030 Wb BA c The total reluctance of the core is 18 TOT stator air gap 1 rotor air gap 2 R R R R R At a flux density of 12 T the relative permeability r of the stator is about 3800 so the stator reluctance is stator stator 7 stator stator 060 m 503 kA tWb 3800 4 10 Hm 005 m 005 m l A R At a flux density of 12 T the relative permeability r of the rotor is 3800 so the rotor reluctance is rotor rotor 7 stator rotor 005 m 42 kA tWb 3800 4 10 Hm 005 m 005 m l A R The reluctance of both air gap 1 and air gap 2 is air gap air gap 1 air gap 2 7 2 air gap air gap 00005 m 221 kA tWb 4 10 Hm 00018 m l A R R Therefore the total reluctance of the core is TOT stator air gap 1 rotor air gap 2 R R R R R TOT 503 221 42 221 496 kA tWb R The required MMF is TOT TOT 0003 Wb 496 kA tWb 1488 A t F R Since F Ni and the current is limited to 1 A one possible choice for the number of turns is N 2000 This would allow the desired flux density to be achieved with a current of about 074 A 118 Assume that the voltage applied to a load is 208 30 V V and the current flowing through the load is 2 20 A I a Calculate the complex power S consumed by this load b Is this load inductive or capacitive c Calculate the power factor of this load d Calculate the reactive power consumed or supplied by this load Does the load consume reactive power from the source or supply it to the source SOLUTION a The complex power S consumed by this load is 208 30 V 2 20 A 208 30 V 2 20 A S VI 416 50 VA S b This is a capacitive load c The power factor of this load is PF cos 50 0643 leading d This load supplies reactive power to the source The reactive power of the load is sin 208 V 2 A sin 50 319 var Q VI 119 Figure P114 shows a simple singlephase ac power system with three loads The voltage source is 240 0 V impedances of these three loads are V 1 10 30 Z 2 10 45 Z 3 10 90 Z 19 Answer the following questions about this power system a Assume that the switch shown in the figure is initially open and calculate the current I the power factor and the real reactive and apparent power being supplied by the source b How much real reactive and apparent power is being consumed by each load with the switch open c Assume that the switch shown in the figure is now closed and calculate the current I the power factor and the real reactive and apparent power being supplied by the source d How much real reactive and apparent power is being consumed by each load with the switch closed e What happened to the current flowing from the source when the switch closed Why I V Z1 Z2 3 Z 120 0 V V SOLUTION a With the switch open only loads 1 and 2 are connected to the source The current in Load 1 is 1I 1 240 0 V 24 30 A 10 30 A I The current in Load 2 is 2I 2 240 0 V 24 45 A 10 45 A I Therefore the total current from the source is 1 2 24 30 A 24 45 A 4759 375 A I I I The power factor supplied by the source is PF cos cos 375 0793 lagging Note that the angle used in the power factor and power calculations is the impedance angle which is the negative of the current angle as long as voltage is at 0 The real reactive and apparent power supplied by the source are cos 240 V 4759 A cos 375 9061 W P VI cos 240 V 4759 A sin 375 6953 var Q VI 240 V 4759 A 11420 VA S VI b The real reactive and apparent power consumed by Load 1 are cos 240 V 24 A cos 30 4988 W P VI cos 240 V 24 A sin 30 2880 var Q VI cos 240 V 24 A 5760 VA S VI The real reactive and apparent power consumed by Load 2 are cos 240 V 24 A cos 45 4073 W P VI 20 cos 240 V 24 A sin 45 4073 var Q VI cos 240 V 24 A 5760 VA S VI As expected the real and reactive power supplied by the source are equal to the sum of the real and reactive powers consumed by the loads c With the switch closed all three loads are connected to the source The current in Loads 1 and 2 is the same as before The current in Load 3 is 3I 3 240 0 V 24 90 A 10 90 A I Therefore the total current from the source is 1 2 3 24 30 A 24 45 A 24 90 A 3808 75 A I I I I The power factor supplied by the source is PF cos cos 75 0991 lagging The real reactive and apparent power supplied by the source are cos 240 V 3808 A cos 75 9061 W P VI cos 240 V 3808 A sin 75 1193 var Q VI 240 V 3808 A 9140 VA S VI d The real reactive and apparent power consumed by Load 1 are cos 240 V 24 A cos 30 4988 W P VI cos 240 V 24 A sin 30 2880 var Q VI cos 240 V 24 A 5760 VA S VI The real reactive and apparent power consumed by Load 2 are cos 240 V 24 A cos 45 4073 W P VI cos 240 V 24 A sin 45 4073 var Q VI cos 240 V 24 A 5760 VA S VI The real reactive and apparent power consumed by Load 3 are cos 240 V 24 A cos 90 0 W P VI cos 240 V 24 A sin 90 5760 var Q VI cos 240 V 24 A 5760 VA S VI As expected the real and reactive power supplied by the source are equal to the sum of the real and reactive powers consumed by the loads e The current flowing decreased when the switch closed because most of the reactive power being consumed by Loads 1 and 2 is being supplied by Load 3 Since less reactive power has to be supplied by the source the total current flow decreases 120 Demonstrate that Equation 159 can be derived from Equation 158 using simple trigonometric identities 2 cos cos p t v t i t VI t t 158 cos 1 cos2 sin sin2 p t VI t VI t 159 21 SOLUTION The first step is to apply the following identity 1 cos cos cos cos 2 The result is 2 cos cos p t v t i t VI t t 1 2 cos cos 2 p t VI t t t t cos cos 2 p t VI t Now we must apply the angle addition identity to the second term cos cos cos sin sin The result is cos cos2 cos sin2 sin p t VI t t Collecting terms yields the final result cos 1 cos2 sin sin2 p t VI t VI t 121 A linear machine has a magnetic flux density of 05 T directed into the page a resistance of 025 a bar length l 10 m and a battery voltage of 100 V a What is the initial force on the bar at starting What is the initial current flow b What is the noload steadystate speed of the bar c If the bar is loaded with a force of 25 N opposite to the direction of motion what is the new steady state speed What is the efficiency of the machine under these circumstances SOLUTION a The current in the bar at starting is 100 V 400 A 025 VB i R Therefore the force on the bar at starting is 400 A 1 m 05 T 200 N to the right i F l B b The noload steadystate speed of this bar can be found from the equation vBl e VB ind 22 100 V 200 ms 05 T 1 m VB v Bl c With a load of 25 N opposite to the direction of motion the steadystate current flow in the bar will be given by ilB F F ind app app 25 N 50 A 05 T 1 m F i Bl The induced voltage in the bar will be ind 100 V 50 A 025 875 V B e V iR and the velocity of the bar will be 875 V 175 ms 05 T 1 m VB v Bl The input power to the linear machine under these conditions is in 100 V 50 A 5000 W B P V i The output power from the linear machine under these conditions is out 875 V 50 A 4375 W B P V i Therefore the efficiency of the machine under these conditions is out in 4375 W 100 100 875 5000 W P P 122 A linear machine has the following characteristics 0 B 5 T into page R 025 l 05 m 120 V B V a If this bar has a load of 20 N attached to it opposite to the direction of motion what is the steadystate speed of the bar b If the bar runs off into a region where the flux density falls to 045 T what happens to the bar What is its final steadystate speed c Suppose B V is now decreased to 100 V with everything else remaining as in part b What is the new steadystate speed of the bar d From the results for parts b and c what are two methods of controlling the speed of a linear machine or a real dc motor SOLUTION a With a load of 20 N opposite to the direction of motion the steadystate current flow in the bar will be given by ilB F F ind app app 20 N 80 A 05 T 05 m F i Bl The induced voltage in the bar will be 23 24 ind 120 V 80 A 025 100 V B e V iR and the velocity of the bar will be ind 100 V 400 ms 05 T 05 m e v Bl b If the flux density drops to 045 T while the load on the bar remains the same there will be a speed transient until again The new steady state current will be app ind 20 N F F app ind F F il B app 20 N 889 A 045 T 05 m F i Bl The induced voltage in the bar will be ind 120 V 889 A 025 978 V B e V iR and the velocity of the bar will be ind 978 V 433 ms 045 T 05 m e v Bl c If the battery voltage is decreased to 100 V while the load on the bar remains the same there will be a speed transient until again The new steady state current will be app ind 20 N F F app ind F F il B app 20 N 889 A 045 T 05 m F i Bl The induced voltage in the bar will be ind 100 V 889 A 025 778 V B e V iR and the velocity of the bar will be ind 778 V 344 ms 045 T 05 m e v Bl d From the results of the two previous parts we can see that there are two ways to control the speed of a linear dc machine Reducing the flux density B of the machine increases the steadystate speed and reducing the battery voltage VB decreases the steadstate speed of the machine Both of these speed control methods work for real dc machines as well as for linear machines Chapter 2 Transformers 21 A 100kVA 8000277V distribution transformer has the following resistances and reactances 5 P R 0005 S R 6 XP 0006 S X 25 50 k C R 10 k XM The excitation branch impedances are given referred to the highvoltage side of the transformer a Find the equivalent circuit of this transformer referred to the lowvoltage side b Find the perunit equivalent circuit of this transformer c Assume that this transformer is supplying rated load at 277 V and 085 PF lagging What is this transformers input voltage What is its voltage regulation d What are the copper losses and core losses in this transformer under the conditions of part c e What is the transformers efficiency under the conditions of part c SOLUTION a The turns ratio of this transformer is a 8000277 2888 Therefore the primary impedances referred to the low voltage secondary side are 2 2 5 0006 2888 P P R R a 2 2 6 00072 2888 P P R X a and the excitation branch elements referred to the secondary side are 2 2 50 k 60 2888 C C R R a 2 2 10 k 12 2888 M M X X a The resulting equivalent circuit is 0006 j00072 0005 j0006 60 j12 b The rated kVA of the transformer is 100 kVA and the rated voltage on the secondary side is 277 V so the rated current in the secondary side is 100 kVA277 V 361 A Therefore the base impedance on the primary side is base base base 277 V 0767 361 A V Z I Since base the resulting perunit equivalent circuit is as shown below actual pu Z Z Z 00078 j00094 00065 j00078 782 j156 c To simplify the calculations use the simplified equivalent circuit referred to the secondary side of the transformer 00005 j12 60 j00132 0011 The secondary current in this transformer is 100 kVA 318 A 361 318 A 277 V S I Therefore the primary voltage on this transformer referred to the secondary side is EQ EQ P S S R jX V V I 277 0 V 0011 00132 361 318 A 283 04 V P j V The voltage regulation of the transformer under these conditions is 283 277 VR 100 22 277 d Under the conditions of part c the transformers output power copper losses and core losses are OUT cos 100 kVA 085 85 kW P S 2 2 CU EQ 361 011 1430 W S P I R 26 2 2 core 283 1602 W 50 P C V P R e The efficiency of this transformer is OUT OUT CU core 85000 100 100 966 85000 1430 1602 P P P P 22 A singlephase power system is shown in Figure P21 The power source feeds a 100kVA 1424kV transformer through a feeder impedance of 382 j140 The transformers equivalent series impedance referred to its lowvoltage side is 010 j04 The load on the transformer is 90 kW at 08 PF lagging and 2300 V a What is the voltage at the power source of the system b What is the voltage regulation of the transformer c How efficient is the overall power system SOLUTION To solve this problem we will refer the circuit to the secondary lowvoltage side The feeders impedance referred to the secondary side is 2 line 24 kV 382 140 112 411 14 kV Z j j The secondary current is given by SI 90 kW 4688 A 2400 V 08 SI The power factor is 080 lagging so the impedance angle cos 1 08 3687 and the phasor current is 4688 3687 A S I a The voltage at the power source of this system referred to the secondary side is EQ line source Z Z S S S I I V V source 2400 0 V 4688 3687 A 112 411 4688 3687 A 010 040 j j V 27 source 2576 30 V V Therefore the voltage at the power source is source 2576 30 V 14 kV 155 30 kV 24 kV V b To find the voltage regulation of the transformer we must find the voltage at the primary side of the transformer referred to the secondary side under these conditions S ZEQ S P I V V 2400 0 V 4688 3687 A 010 040 2415 03 V P j V There is a voltage drop of 15 V under these load conditions Therefore the voltage regulation of the transformer is 2415 2400 VR 100 063 2400 c The overall efficiency of the power system will be the ratio of the output power to the input power The output power supplied to the load is POUT 90 kW The input power supplied by the source is 2 2 IN OUT LOSS OUT 90 kW 4688 A 122 9268 kW P P P P I R IN source cos 2415 V 4688 A cos 3657 9093 kW S P V I Therefore the efficiency of the power system is OUT IN 90 kW 100 100 971 9268 kW P P Note Problem 23 was printed incorrectly in the first edition of this text By accident a portion of Problem 24 was printed here instead of the appropriate text This should be fixed by the second printing of the book 23 Consider a simple power system consisting of an ideal voltage source an ideal stepup transformer a transmission line an ideal stepdown transformer and a load The voltage of the source is S 480 0 V The impedance of the transmission line is V line 3 4 Z j and the impedance of the load is load 30 40 Z j a Assume that the transformers are not present in the circuit What is the load voltage and efficiency of the system b Assume that transformer 1 is a 15 stepup transformer and transformer 2 is a 51 stepdown transformer What is the load voltage and efficiency of the system c What transformer turns ratio would be required to reduce the transmission line losses to 1 of the total power produced by the generator SOLUTION a The equivalent circuit of this power system is shown below 28 S 480 0 V V line 3 4 Z j load 30 40 Z j The load current in this system is load 480 0 V 8727 5313 A 3 4 30 40 j j I The load voltage is load 8727 5313 A 30 40 4364 0 A j V The power consumed by the load is 2 load 8727 A 30 2285 W P The power consumed by the transmission line is 2 line 8727 A 3 2285 W P The efficiency of the power system is OUT load IN load line 2285 W 100 100 100 909 2285 W 2285 W P P P P P b The equivalent circuit of this power system is shown below S 480 0 V V line 3 4 Z j load 30 40 Z j 15 51 The line impedance referred to primary side of T1 is 2 2 line line 02 3 4 012 016 Z a Z j j The load impedance referred to primary side of T1 is the same as the actual impedance since the turns ratios of the stepup and stepdown transformers undo each others changes load 30 40 Z j The resulting equivalent circuit referred to the primary side of T1 is 29 S 480 0 V V line 012 016 Z j load 30 40 Z j The load current in this system is load 480 0 V 9562 5313 A 012 016 30 40 j j I The load voltage is load 9562 5313 A 30 40 478 34 A j V The power consumed by the load is 2 load 9562 A 30 2743 W P The power consumed by the transmission line is 2 line 8727 A 012 11 W P The efficiency of the power system is OUT load IN load line 2743 W 100 100 100 996 2743 W 11 W P P P P P c Since the power in a resistor is given by 2 P I R the total power consumed in the line resistance will be directly proportional to the ratio of the line resistance to the total resistance in the circuit The load resistance is 30 and that must be 99 of the total resistance in order for the efficient to be 1 Therefore the referred line resistance must be line load 001 099 R R line load 001 001 30 0303 099 099 R R Since the referred line resistance is 0303 and the actual line resistance is 3 the turns ration must be 2 line line Z a Z 2 0303 3 a 2 0303 3 a 0318 a in order for 1 of the power to be consumed in the transmission line 24 The secondary winding of a real transformer has a terminal voltage of The turns ratio of the transformer is 100200 a 050 If the secondary current of the transformer is what is the primary current of this transformer What are its voltage regulation and efficiency The impedances of this transformer referred to the primary side are v t t s 2828 sin 377 V i t t s 7 07 3687 sin 377 A 30 eq 020 R 300 C R eq 080 X 100 XM SOLUTION The equivalent circuit of this transformer is shown below Since no particular equivalent circuit was specified we are using the approximate equivalent circuit referred to the primary side The secondary voltage and current are 2828 0 V 200 0 V 2 S V 707 3687 A 5 3687 A 2 S I The secondary voltage referred to the primary side is 100 0 V S a S V V The secondary current referred to the primary side is 10 3687 A S S a I I The primary circuit voltage is given by eq eq P S S R jX V V I 100 0 V 10 3687 A 020 080 1065 28 V P j V The excitation current of this transformer is EX 1065 28 V 1065 28 V 0355 28 1065 872 300 100 C M j I I I EX 112 688 A I Therefore the total primary current of this transformer is EX 10 3687 112 688 110 400 A P S I I I The voltage regulation of the transformer at this load is 1065 100 VR 100 100 65 100 P S S V aV aV The input power to this transformer is 31 IN cos 1065 V 110 A cos 28 400 P P P V I IN 1065 V 110 A cos 428 860 W P The output power from this transformer is OUT cos 200 V 5 A cos 3687 800 W S S P V I Therefore the transformers efficiency is OUT IN 800 W 100 100 930 860 W P P 25 When travelers from the USA and Canada visit Europe they encounter a different power distribution system Wall voltages in North America are 120 V rms at 60 Hz while typical wall voltages in Europe are 230 V at 50 Hz Many travelers carry small stepup stepdown transformers so that they can use their appliances in the countries that they are visiting A typical transformer might be rated at 1kVA and 115230 V It has 500 turns of wire on the 115V side and 1000 turns of wire on the 230V side The magnetization curve for this transformer is shown in Figure P22 and can be found in file p22mag at this books Web site a Suppose that this transformer is connected to a 120V 60 Hz power source with no load connected to the 240V side Sketch the magnetization current that would flow in the transformer Use MATLAB to plot the current accurately if it is available What is the rms amplitude of the magnetization current What percentage of fullload current is the magnetization current b Now suppose that this transformer is connected to a 240V 50 Hz power source with no load connected to the 120V side Sketch the magnetization current that would flow in the transformer Use MATLAB to plot the current accurately if it is available What is the rms amplitude of the magnetization current What percentage of fullload current is the magnetization current c In which case is the magnetization current a higher percentage of fullload current Why 32 Note An electronic version of this magnetization curve can be found in file p22magdat which can be used with MATLAB programs Column 1 contains the MMF in A turns and column 2 contains the resulting flux in webers SOLUTION a When this transformer is connected to a 120V 60 Hz source the flux in the core will be given by the equation cos M P V t N t 2101 The magnetization current required for any given flux level can be found from Figure P22 or alternately from the equivalent table in file p22magdat The MATLAB program shown below calculates the flux level at each time the corresponding magnetization current and the rms value of the magnetization current Mfile prob25am Mfile to calculate and plot the magnetization current of a 120240 transformer operating at 120 volts and 60 Hz This program also calculates the rms value of the mag current Load the magnetization curve It is in two columns with the first column being mmf and the second column being flux load p22magdat mmfdata p221 fluxdata p222 Initialize values S 1000 Apparent power VA Vrms 120 Rms voltage V VM Vrms sqrt2 Max voltage V NP 500 Primary turns Calculat freq 60 Freq Hz e angular velocity for 60 Hz w 2 pi freq Calculate flux versus time time 013000130 0 to 130 sec flux VMwNP cosw time Calculate the mmf corresponding to a given flux using the MATLAB interpolation functio mmf interp1fluxdatammfdataflux n Calculate the im mmf NP magnetization current Calculate the rms value of the current irms sqrtsumim2lengthim dispThe rms current at 120 V and 60 Hz is num2strirms 33 Calculate the fullload current ifl S Vrms Calculate the percentage of fullload current percnt irms ifl 100 dispThe magnetization current is num2strpercnt of fullload current Plot the magnetization current figure1 plottimeim title bfMagnetization Current at 120 V and 60 Hz xlabel bfTime s ylabel bfitIm rmA axis0 004 05 05 grid on When this program is executed the results are prob25a The rms current at 120 V and 60 Hz is 031863 The magnetization current is 38236 of fullload current The rms magnetization current is 0318 A Since the fullload current is 1000 VA 120 V 833 A the magnetization current is 382 of the fullload current The resulting plot is b When this transformer is connected to a 240V 50 Hz source the flux in the core will be given by the equation cos t N V t S M The magnetization current required for any given flux level can be found from Figure P22 or alternately from the equivalent table in file p22magdat The MATLAB program shown below calculates the flux level at each time the corresponding magnetization current and the rms value of the magnetization current Mfile prob25bm Mfile to calculate and plot the magnetization 34 current of a 120240 transformer operating at 35 240 volts and 50 Hz This program also calculates the rms value of the mag current Load the magnetization curve It is in two columns with the first column being mmf and the second column being flux load p22magdat mmfdata p221 fluxdata p222 Initialize values S 1000 Apparent power VA Vrms 240 Rms voltage V VM Vrms sqrt2 Max voltage V NP 1000 Primary turns Calculate angular velocity for 50 Hz freq 50 Freq Hz w 2 pi freq Calculate flux versus time time 012500125 0 to 125 sec flux VMwNP cosw time Calculate the mmf corresponding to a given flux using the MATLAB interpolation functio mmf interp1fluxdatammfdataflux n Calculate the magnetization current im mmf NP Calculate the rms value of the current irms sqrtsumim2lengthim dispThe rms current at 50 Hz is num2strirms Calculate the fullload current ifl S Vrms Calculate the percentage of fullload current percnt irms ifl 100 dispThe magnetization current is num2strpercnt of fullload current Plot the magnetization current figure1 plottimeim title bfMagnetization Current at 240 V and 50 Hz xlabel bfTime s ylabel bfitIm rmA axis0 004 05 05 grid on When this program is executed the results are prob25b The rms current at 50 Hz is 022973 The magnetization current is 55134 of fullload current The rms magnetization current is 0230 A Since the fullload current is 1000 VA 240 V 417 A the magnetization current is 551 of the fullload current The resulting plot is c The magnetization current is a higher percentage of the fullload current for the 50 Hz case than for the 60 Hz case This is true because the peak flux is higher for the 50 Hz waveform driving the core further into saturation 26 A 1000VA 230115V transformer has been tested to determine its equivalent circuit The results of the tests are shown below Opencircuit test on secondary side Shortcircuit test on primary side VOC 115 V VSC 171 V IOC 011 A ISC 87 A POC 39 W PSC 381 W a Find the equivalent circuit of this transformer referred to the lowvoltage side of the transformer b Find the transformers voltage regulation at rated conditions and 1 08 PF lagging 2 10 PF 3 08 PF leading c Determine the transformers efficiency at rated conditions and 08 PF lagging SOLUTION a OPEN CIRCUIT TEST referred to the lowvoltage or secondary side EX 011 A 00009565 S 115 V C M Y G jB 1 1 OC OC OC 39 W cos cos 720 115 V 011 A P V I EX 00009565 72 S 00002956 00009096 S C M Y G jB j 1 3383 C C R G 1 1099 M M X B 36 SHORT CIRCUIT TEST referred to the highvoltage or primary side EQ EQ EQ 171 V 197 87 A jX Z R 1 1 SC SC SC 381 W cos cos 752 171 V 87 A P V I EQ EQ EQ 197 752 0503 1905 Z R jX j EQ 0503 R EQ 1905 X j To convert the equivalent circuit to the secondary side divide each series impedance by the square of the turns ratio a 230115 2 Note that the excitation branch elements are already on the secondary side The resulting equivalent circuit is shown below EQS 0126 R EQS 0476 X j 3383 RC S 1099 XM S b To find the required voltage regulation we will use the equivalent circuit of the transformer referred to the secondary side The rated secondary current is 8 70 A 115 V 1000 VA SI We will now calculate the primary voltage referred to the secondary side and use the voltage regulation equation for each power factor 1 08 PF Lagging EQ 115 0 V 0126 0476 87 36 87 A P S S Z j V V I P 1184 13 V V 1184115 VR 100 296 115 2 10 PF EQ 115 0 V 0126 0476 87 00 A P S S Z j V V I P 1162 204 V V 37 1162115 VR 100 104 115 3 08 PF Leading EQ 115 0 V 0126 0476 87 36 87 A P S S Z j V V I P 1135 20 V V 1135115 VR 100 13 115 c At rated conditions and 08 PF lagging the output power of this transformer is OUT cos 115 V 87 A 08 800 W S S P V I The copper and core losses of this transformer are 2 2 CU EQ 87 A 0126 95 W S S P I R 2 2 core 1184 V 41 W 3383 P C V P R Therefore the efficiency of this transformer at these conditions is OUT OUT CU core 800 W 100 983 800 W 95 W 41 W P P P P 27 A 30kVA 8000230V distribution transformer has an impedance referred to the primary of 20 j100 The components of the excitation branch referred to the primary side are and 100 k C R 20 k XM a If the primary voltage is 7967 V and the load impedance is L Z 20 j07 what is the secondary voltage of the transformer What is the voltage regulation of the transformer b If the load is disconnected and a capacitor of j30 is connected in its place what is the secondary voltage of the transformer What is its voltage regulation under these conditions SOLUTION a The easiest way to solve this problem is to refer all components to the primary side of the transformer The turns ratio is a 8000230 3478 Thus the load impedance referred to the primary side is 3478 2 20 07 2419 847 ZL j j The referred secondary current is 7967 0 V 7967 0 V 3045 212 A 20 100 2419 847 2616 212 S j j I and the referred secondary voltage is 3045 212 A 2419 847 7804 19 V S S ZL j V I The actual secondary voltage is thus 38 7804 19 V 2244 19 V 3478 S S a V V The voltage regulation is 79677804 VR 100 209 7804 b The easiest way to solve this problem is to refer all components to the primary side of the transformer The turns ratio is again a 3478 Thus the load impedance referred to the primary side is 3478 2 30 3629 ZL j j The referred secondary current is 7967 0 V 7967 0 V 2258 897 A 20 100 3629 3529 897 S j j I and the referred secondary voltage is 2258 897 A 3629 8194 03 V S S ZL j V I The actual secondary voltage is thus 8914 03 V 2563 03 V 3478 S S a V V The voltage regulation is 7967 8914 VR 100 106 8194 28 A 150MVA 15200kV singlephase power transformer has a perunit resistance of 12 percent and a per unit reactance of 5 percent data taken from the transformers nameplate The magnetizing impedance is j80 per unit a Find the equivalent circuit referred to the lowvoltage side of this transformer b Calculate the voltage regulation of this transformer for a fullload current at power factor of 08 lagging c Calculate the copper and core losses in transformer at the conditions in b d Assume that the primary voltage of this transformer is a constant 15 kV and plot the secondary voltage as a function of load current for currents from noload to fullload Repeat this process for power factors of 08 lagging 10 and 08 leading a The base impedance of this transformer referred to the primary lowvoltage side is 2 2 base base base 15 kV 15 150 MVA V Z S so EQ 0012 15 0018 R EQ 005 15 0075 X 80 15 120 XM 39 The equivalent circuit is EQ 0018 P R EQ 0075 P X j not specified C R 120 XM b If the load on the secondary side of the transformer is 150 MVA at 08 PF lagging and the referred secondary voltage is 15 kV then the referred secondary current is LOAD 150 MVA 12500 A PF 15 kV 08 S S P I V 12500 3687 A S I P The voltage on the primary side of the transformer is EQ P S S Z V V I 15000 0 V 12500 3687 A 0018 0075 15755 223 V P j V Therefore the voltage regulation of the transformer is 1575515000 VR 100 503 15000 c This problem is repetitive in nature and is ideally suited for MATLAB A program to calculate the secondary voltage of the transformer as a function of load is shown below Mfile prob28m Mfile to calculate and plot the secondary voltage of a transformer as a function of load for power factors of 08 lagging 10 and 08 leading These calculations are done using an equivalent circuit referred to the primary side Define values for this transformer VP 15000 Primary voltage V amps 012512500 Current values A Req 0018 Equivalent R ohms Xeq 0075 Equivalent X ohms Calculate the current values for the three power factors The first row of I contains the lagging currents the second row contains the unity currents and the third row contains the leading currents I zeros3lengthamps 40 I1 amps 08 j06 Lagging I2 amps 10 Unity I3 amps 08 j06 Leading Calculate VS referred to the primary side for each current and power factor aVS VP ReqI jXeqI Refer the secondary voltages back to the secondary side using the turns ratio VS aVS 20015 Plot the secondary voltage in kV versus load plotampsabsVS11000bLineWidth20 hold on plotampsabsVS21000kLineWidth20 plotampsabsVS31000rLineWidth20 title bfSecondary Voltage Versus Load xlabel bfLoad A ylabel bfSecondary Voltage kV legend08 PF lagging10 PF08 PF leading grid on hold off The resulting plot of secondary voltage versus load is shown below 29 A 5000kVA 230138kV singlephase power transformer has a perunit resistance of 1 percent and a perunit reactance of 5 percent data taken from the transformers nameplate The opencircuit test performed on the lowvoltage side of the transformer yielded the following data V OC 138 kV OC 211 A I OC 908 kW P a Find the equivalent circuit referred to the lowvoltage side of this transformer 41 b If the voltage on the secondary side is 138 kV and the power supplied is 4000 kW at 08 PF lagging find the voltage regulation of the transformer Find its efficiency SOLUTION a The opencircuit test was performed on the lowvoltage side of the transformer so it can be used to directly find the components of the excitation branch relative to the lowvoltage side EX 211 A 0001529 138 kV C M Y G jB 1 1 OC OC OC 908 kW cos cos 7183 138 kV 211 A P V I EX 0001529 7183 S 00004456 00013577 S C M Y G jB j 1 2244 C C R G 1 737 M M X B The base impedance of this transformer referred to the secondary side is 2 2 base base base 138 kV 3809 5000 kVA V Z S so and EQ 001 3809 038 R EQ 005 3809 19 X The resulting equivalent circuit is shown below 0 38 REQs 91 EQs j X 2244 RC s 737 XM s b If the load on the secondary side of the transformer is 4000 kW at 08 PF lagging and the secondary voltage is 138 kV the secondary current is LOAD 4000 kW 3623 A PF 138 kV 08 S S P I V 3623 3687 A S I The voltage on the primary side of the transformer referred to the secondary side is EQ P S SZ V V I 42 13800 0 V 3623 3687 A 038 19 14330 19 V P j V There is a voltage drop of 14 V under these load conditions Therefore the voltage regulation of the transformer is 14330 13800 VR 100 384 13800 The transformer copper losses and core losses are 2 2 CU EQ 3623 A 038 499 kW S S P I R 2 2 core 14330 V 915 kW 2244 P C V P R Therefore the efficiency of this transformer at these conditions is OUT OUT CU core 4000 kW 100 966 4000 kW 499 kW 915 kW P P P P 210 A threephase transformer bank is to handle 500 kVA and have a 34511kV voltage ratio Find the rating of each individual transformer in the bank high voltage low voltage turns ratio and apparent power if the transformer bank is connected to a YY b Y c Y d e open f open Yopen SOLUTION For the first four connections the apparent power rating of each transformer is 13 of the total apparent power rating of the threephase transformer For the open and openYopen connections the apparent power rating is a bit more complicated The 500 kVA must be 866 of the total apparent power rating of the two transformers so 250 kVA must be 866 of the apparent power rating of a single transformer Therefore the apparent power rating of each transformer must be 288 kVA The ratings for each transformer in the bank for each connection are given below Connection Primary Voltage Secondary Voltage Apparent Power Turns Ratio YY 199 kV 635 kV 167 kVA 2501 Y 199 kV 110 kV 167 kVA 1441 Y 345 kV 635 kV 167 kVA 4331 345 kV 110 kV 167 kVA 2501 open 345 kV 110 kV 288 kVA 2501 openYopen 199 kV 110 kV 288 kVA 1441 Note The openYopen answer assumes that the Y is on the highvoltage side if the Y is on the low voltage side the turns ratio would be 4331 and the apparent power rating would be unchanged 211 A 100MVA 230115kV Y threephase power transformer has a perunit resistance of 0015 pu and a perunit reactance of 006 pu The excitation branch elements are 100 pu C R and M 20 pu X a If this transformer supplies a load of 80 MVA at 08 PF lagging draw the phasor diagram of one phase of the transformer b What is the voltage regulation of the transformer bank under these conditions c Sketch the equivalent circuit referred to the lowvoltage side of one phase of this transformer Calculate all the transformer impedances referred to the lowvoltage side d Determine the losses in the transformer and the efficiency of the transformer under the conditions of part b 43 SOLUTION a The transformer supplies a load of 80 MVA at 08 PF lagging Therefore the secondary line current of the transformer is 80000000 VA 402 A 3 3 115000 V LS LS S I V The base apparent power is and the base line voltage on the secondary side is so the base value of the secondary line current is base 100 MVA S base 115 kV VLS base base base 100000000 VA 502 A 3 3 115000 V LS LS S I V so the perunit secondary current is 1 pu pu 402 A cos 08 08 3687 502 A LS LS LS I I I The perunit phasor diagram is shown below I 08318 V 100 S VP 08 3687 I P 1039 17 V b The perunit primary voltage of this transformer is EQ 10 0 08 3687 0015 006 1039 17 P S Z j V V I and the voltage regulation is 1039 10 VR 100 39 10 c The secondary side of this transfer is Yconnected so the base phase voltage of the low voltage secondary side of this transformer is base base 115 kV 664 kV 3 3 LS S V V The base impedance of the transformer referred to the lowvoltage side is 2 2 base base base 3 3 664 kV 133 100 MVA V S Z S Each perunit impedance is converted to actual ohms referred to the lowvoltage side by multiplying it by this base impedance The resulting equivalent circuit is shown below 44 200 j798 133 k j266 k EQ 0015 133 200 S R EQ 006 133 798 S X 100 133 133 k C R 20 133 266 k XM d The perunit losses in the series resistance are 2 2 08 0015 00096 pu EQ EQ P I R and the actual losses in the series resistance are base pu 00096 pu 100 MVA 096 MW EQ EQ P S P The perunit losses in the excitation branch are 2 2 pu 1039 00108 pu 100 EX EX V P R and the actual losses in the excitation branch are base pu 00108 pu 100 MVA 108 MW EX EX P S P The perunit power supplied to the load load loadpu base 80 MW 080 pu 100 MVA P P S Therefore the transformers efficiency is OUT IN 080 100 100 975 080 00096 00108 P P 212 Three 20kVA 24000277V distribution transformers are connected in Y The opencircuit test was performed on the lowvoltage side of this transformer bank and the following data were recorded V lineOC 480 V IlineOC 4 10 A P3 OC W 945 The shortcircuit test was performed on the highvoltage side of this transformer bank and the following data were recorded 1400 V lineSC V 1 80 A lineSC I P3 SC W 912 a Find the perunit equivalent circuit of this transformer bank 45 b Find the voltage regulation of this transformer bank at the rated load and 090 PF lagging c What is the transformer banks efficiency under these conditions SOLUTION a The equivalent of this threephase transformer bank can be found just like the equivalent circuit of a singlephase transformer if we work on a perphase bases The opencircuit test data on the lowvoltage side can be used to find the excitation branch impedances referred to the secondary side of the transformer bank Since the lowvoltage side of the transformer is Yconnected the perphase open circuit measurements are OC 277 V V OC 410 A I OC 315 W P The excitation admittance is given by 410 A 001480 S 277 V OC EX OC I Y V The admittance angle is 1 1 315 W cos cos 739 277 V 410 A OC OC OC P V I Therefore 001483 739 000410 001422 EX C M Y G jB j 1 244 C C R G 1 703 M M X B The base impedance for a single transformer referred to the lowvoltage side is 2 2 base base base 277 V 3836 20 kVA S S V Z S so the excitation branch elements can be expressed in perunit as 244 636 pu 3836 RC 703 183 pu 3836 XM The shortcircuit test data taken in the highvoltage side can be used to find the series impedances referred to the highvoltage side Note that the highvoltage is connected so SC SC 1400 V L V V SC SC 3 1039 A L I I and SC SC 3 304 W P P 1400 V 1347 1039 A SC EQ SC V Z I 1 1 304 W cos cos 779 1400 V 1039 A SC SC SC P V I 1347 779 282 1317 EQ P EQ P EQ P Z R jX j The base impedance referred to the highvoltage side is 2 2 base base 24000 V 23040 25 kVA S P V Z S The resulting perunit impedances are 46 282 00122 pu 23040 REQ 1317 00572 pu 23040 XEQ The perunit perphase equivalent circuit of the transformer bank is shown below VS VP IS IP RC jXM REQ jXEQ 00122 j00572 636 j183 b If this transformer is operating at rated load and 090 PF lagging then current flow will be at an angle of or 258 The perunit voltage at the primary side of the transformer will be cos 1 90 EQ 10 0 10 258 00125 00588 1038 262 P S SZ j V V I The voltage regulation of this transformer bank is 1038 10 VR 100 38 10 c The output power of this transformer bank is OUT cos 10 10 09 09 pu S S P V I The copper losses are 2 2 CU EQ 10 00122 00122 pu S P I R The core losses are 2 2 core 1038 00169 pu 636 P C V P R Therefore the total input power to the transformer bank is IN OUT CU core 09 00122 00169 0929 P P P P and the efficiency of the transformer bank is OUT IN 09 100 100 969 0929 P P 213 A 14400480V threephase Yconnected transformer bank consists of three identical 100kVA 8314480V transformers It is supplied with power directly from a large constantvoltage bus In the shortcircuit test the recorded values on the highvoltage side for one of these transformers are 47 SC 510 V V 6 A ISC 12 SC 3000 W P a If this bank delivers a rated load at 08 PF lagging and rated voltage what is the linetoline voltage on the primary of the transformer bank b What is the voltage regulation under these conditions c Assume that the primary phase voltage of this transformer is a constant 8314 V and plot the secondary voltage as a function of load current for currents from noload to fullload Repeat this process for power factors of 08 lagging 10 and 08 leading d Plot the voltage regulation of this transformer as a function of load current for currents from noload to fullload Repeat this process for power factors of 08 lagging 10 and 08 leading e Sketch the perunit equivalent circuit of this transformer SOLUTION From the shortcircuit information it is possible to determine the perphase impedance of the transformer bank referred to the highvoltage primary side Note that the shortcircuit information is given for one transformer of the three in the bank The voltage across this transformer is SC 510 V V the shortcircuit phase current is SC 126 A I and the power per phase is SC 3000 W P Thus the perphase impedance is EQ EQ EQ 510 V 4048 126 A Z R jX 1 1 SC SC SC 3000 W cos cos 621 510 V 126 A P V I EQ EQ EQ 4048 621 1894 3577 Z R jX j EQ 1894 R EQ 3577 X j a If this Y transformer bank delivers rated kVA 300 kVA at 08 power factor lagging while the secondary voltage is at rated value then each transformer delivers 100 kVA at a voltage of 480 V and 08 PF lagging Referred to the primary side of one of the transformers the load on each transformer is equivalent to 100 kVA at 8314 V and 08 PF lagging The equivalent current flowing in the secondary of one transformer referred to the primary side is 100 kVA 1203 A 8314 V S I 1203 3687 A S I The voltage on the primary side of a single transformer is thus P S S P Z EQ I V V 8413 0 V 1203 3687 A 1894 3577 8856 134 V P j V The linetoline voltage on the primary of the transformer is 48 LL 3 3 8856 V 1534 kV P P V V b The voltage regulation of the transformer is 88568314 VR 100 652 8314 Note It is much easier to solve problems of this sort in the perunit system For example compare this solution to the simpler solution of Problem 29 c The base values of this transformer bank on the primary side are base 300 KVA S base base 144 kVA VLL V base base base 300 KVA 1203 A 3 3 144 kV L LL S I V base base 1237 A 714 A 3 3 LI I This sort of repetitive operation is best performed with MATLAB Note that in this case the problem is specifying a fixed primary phase voltage of 8314 V and asking what the secondary voltage will be as a function of load Therefore we must subtract the voltage drop inside the transformer at each load and convert the resulting voltage from the primary side to the secondary low voltage side A suitable MATLAB program is shown below Mfile prob213cm Mfile to calculate and plot the secondary voltage of a threephase Ydelta transformer bank as a function of load for power factors of 085 lagging 10 and 085 leading These calculations are done using an equivalent circuit referred to the primary side Define values for this transformer VL 14400 Primary line voltage V VPP VL sqrt3 Primary phase voltage V amps 00012031203 Phase current values A Req 1894 Equivalent R ohms Xeq 3577 Equivalent X ohms Calculate the current values for the three power factors The first row of I contains the lagging currents the second row contains the unity currents and the third row contains the leading currents re 085 im sinacosre I zeros3lengthamps I1 amps re jim Lagging I2 amps 10 Unity I3 amps re jim Leading 49 Calculate secondary phase voltage referred to the primary side for each current and power factor aVSP VPP ReqI jXeqI Refer the secondary phase voltages back to the secondary side using the turns ratio Because this is a deltaconnected secondary this is also the line voltage VSP aVSP 4808314 Plot the secondary voltage versus load plotampsabsVSP1bLineWidth20 hold on plotampsabsVSP2kLineWidth20 plotampsabsVSP3rLineWidth20 title bfSecondary Voltage Versus Load xlabel bfLoad A ylabel bfSecondary Voltage V legend085 PF lagging10 PF085 PF leading grid on hold off The resulting plot is shown below d This sort of repetitive operation is best performed with MATLAB A suitable MATLAB program is shown below Mfile prob213dm Mfile to calculate and plot the voltage regulation of a threephase Ydelta transformer bank as a function of load for power factors of 085 lagging 10 and 085 leading These calculations are done using an equivalent circuit referred to the primary side 50 51 Define values for this transformer VL 14400 Primary line voltage V VPP VL sqrt3 Primary phase voltage V amps 00012031203 Phase current values A Req 1894 Equivalent R ohms Xeq 3577 Equivalent X ohms Calculate the current values for the three power factors The first row of I contains the lagging currents the second row contains the unity currents and the third row contains the leading currents re 085 im sinacosre I zeros3lengthamps I1 amps re jim Lagging I2 amps 10 Unity I3 amps re jim Leading Calculate secondary phase voltage referred to the primary side for each current and power factor aVSP VPP ReqI jXeqI Calculate the voltage regulation VR VPP absaVSP absaVSP 100 Plot the voltage regulation versus load plotampsVR1bLineWidth20 hold on plotampsVR2kLineWidth20 plotampsVR3rLineWidth20 title bfVoltage Regulation Versus Load xlabel bfLoad A ylabel bfVoltage Regulation legend085 PF lagging10 PF085 PF leading grid on hold off The resulting plot is shown below e The base phase voltage on the primary side is given by base 144 kV 8314 kV 3 3 LP P V V The base impedance on the primary side is given by 2 2 base base base 8314 kV 230 300 kVA S S V Z S The perphase impedance on the primary side is EQ 1894 R EQ 3577 X The perunit impedance is EQ EQpu base 1894 0082 pu 230 R R Z EQ EQpu base 3577 0156 pu 230 X X Z The excitation branch information was not given for the transformer so the perunit perphase equivalent circuit of the transformer bank is shown below 52 VS VP IS IP RC jXM REQ jXEQ 0082 j0156 214 A 138kV singlephase generator supplies power to a load through a transmission line The loads impedance is and the transmission lines impedance is Zload 500 3687 60 60 Zline a If the generator is directly connected to the load Figure P23a what is the ratio of the load voltage to the generated voltage What are the transmission losses of the system b What percentage of the power supplied by the source reached the load what is the efficiency of the transmission system c If a 110 stepup transformer is placed at the output of the generator and a 101 transformer is placed at the load end of the transmission line what is the new ratio of the load voltage to the generated voltage What are the transmission losses of the system now Note The transformers may be assumed to be ideal d What percentage of the power supplied by the source reached the load now 53 e Compare the efficiencies of the transmission system with and without transformers SOLUTION a In the case of the directlyconnected load the line current is line load 138 0 kV 2483 393 A 60 60 500 3687 I I The load voltage is load load load 2483 393 A 500 3687 1242 243 kV Z V I The resistance in the load is load load cos 500cos 60 250 R Z The power supplied to the load is 2 2 load line load 2483 A 250 154 kW P I R The ratio of the load voltage to the generated voltage is 1242138 0900 The resistance in the transmission line is line line cos 60cos 60 30 R Z so the transmission losses in the system are 2 2 loss line line 2483 A 30 185 kW P I R b The efficiency of this power system is out out in out loss 154 kW 100 100 100 893 154 kW 185 kW P P P P P c In this case a 110 stepup transformer precedes the transmission line and a 101 stepdown transformer follows the transmission line If the transformers are removed by referring the transmission line to the voltage levels found on either end then the impedance of the transmission line becomes 2 2 line line 1 1 60 60 060 60 10 10 Z Z The current in the referred transmission line and in the load becomes line load 138 0 kV 2757 369 A 060 60 500 3687 I I The load voltage is load load load 2757 369 A 500 3687 13785 003 kV Z V I The resistance in the load is load load cos 500cos 60 250 R Z The power supplied to the load is 2 2 load line load 2757 A 250 190 kW P I R The ratio of the load voltage to the generated voltage is 13785138 09989 Also the transmission losses in the system are reduced The current in the transmission line is 54 line load 1 1 2757 A 2757 A 10 10 I I and the losses in the transmission line are 2 2 loss line line 2757 A 30 228 W P I R d The efficiency of this power system is out out in out loss 190 kW 100 100 100 999 190 kW 0228 kW P P P P P e Transmission losses have decreased by a factor of more than 80 when the transformers were added to the system 215 An autotransformer is used to connect a 126kV distribution line to a 138kV distribution line It must be capable of handling 2000 kVA There are three phases connected YY with their neutrals solidly grounded a What must the turns ratio be to accomplish this connection N C N SE b How much apparent power must the windings of each autotransformer handle c What is the power advantage of this autotransformer system d If one of the autotransformers were reconnected as an ordinary transformer what would its ratings be SOLUTION a The transformer is connected YY so the primary and secondary phase voltages are the line voltages divided by 3 The turns ratio of each autotransformer is given by SE 138 kV 3 126 kV 3 C H L C N N V V N SE 126 126 138 C C N N N 126 SE 12 C N N Therefore SE 126 12 105 NC N b The power advantage of this autotransformer is IO SE 105 115 C C C W C C S N N N N S N N so 1115 of the power in each transformer goes through the windings Since 13 of the total power is associated with each phase the windings in each autotransformer must handle 2000 kVA 635 kVA 3 105 SW c As determined in b the power advantage of this autotransformer system is 115 d The voltages across each phase of the autotransformer are 138 3 7967 V and 126 3 7275 V The voltage across the common winding is 7275 kV and the voltage across the series winding is 7967 kV 7275 kV 692 V Therefore a single phase of the autotransformer connected as an ordinary transformer would be rated at 7275692 V and 635 kVA C N SE N 55 216 Prove the following statement If a transformer having a series impedance is connected as an autotransformer its perunit series impedance Zeq Zeq as an autotransformer will be Z N N N Z C eq SE SE eq Note that this expression is the reciprocal of the autotransformer power advantage SOLUTION The impedance of a transformer can be found by shorting the secondary winding and determining the ratio of the voltage to the current of its primary winding For the transformer connected as an ordinary transformer the impedance referred to the primary C N is NC NSE V1 V2 Z1 Z2 2 eq 1 2 C SE N Z Z Z N The corresponding equivalent circuit is NC NSE V1 V2 Zeq When this transformer is connected as an autotransformer the circuit is as shown below If the output windings of the autotransformer are shorted out the voltages will be zero and the voltage will be VH L V NC NSE VL VH Zeq VC VSE ISE IL IC C Zeq L I V where eq is the impedance of the ordinary transformer However Z 56 C SE C SE C SE C C SE C L N N N N N I I I I I I or L C C N N N I I SE SE so the input voltage can be expressed in terms of the input current as eq eq Z N N N Z L C SE SE C L I I V The input impedance of the autotransformer is defined as L L Z eq V I so eq eq Z N N N Z C SE SE L L I V This is the expression that we were trying to prove 217 A 10kVA 480120V conventional transformer is to be used to supply power from a 600V source to a 120V load Consider the transformer to be ideal and assume that all insulation can handle 600 V a Sketch the transformer connection that will do the required job b Find the kilovoltampere rating of the transformer in the configuration c Find the maximum primary and secondary currents under these conditions SOLUTION a For this configuration the common winding must be the smaller of the two windings and SE 4 C N N The transformer connection is shown below 120 V 600 V NC NSE b The kVA rating of the autotransformer can be found from the equation SE IO SE 4 10 kVA 125 kVA 4 C C C W C N N N N S S N N c The maximum primary current for this configuration will be 12500 VA 2083 A 600 V P P S I V and the maximum secondary current is 12500 VA 104 A 120 V S S S I V 218 A 10kVA 480120V conventional transformer is to be used to supply power from a 600V source to a 480V load Consider the transformer to be ideal and assume that all insulation can handle 600 V 57 a Sketch the transformer connection that will do the required job b Find the kilovoltampere rating of the transformer in the configuration c Find the maximum primary and secondary currents under these conditions d The transformer in Problem 218 is identical to the transformer in Problem 217 but there is a significant difference in the apparent power capability of the transformer in the two situations Why What does that say about the best circumstances in which to use an autotransformer SOLUTION a For this configuration the common winding must be the larger of the two windings and SE 4 NC N The transformer connection is shown below 480 V 600 V NC NSE b The kVA rating of the autotransformer can be found from the equation SE SE SE IO SE SE 4 10 kVA 50 kVA C W N N N N S S N N c The maximum primary current for this configuration will be 50000 VA 833 A 600 V P P S I V and the maximum secondary current is 50000 VA 104 A 480 V S S S I V d Note that the apparent power handling capability of the autotransformer is much higher when there is only a small difference between primary and secondary voltages Autotransformers are normally only used when there is a small difference between the two voltage levels 219 Two phases of a 144kV threephase distribution line serve a remote rural road the neutral is also available A farmer along the road has a 480 V feeder supplying 200 kW at 085 PF lagging of three phase loads plus 60 kW at 09 PF lagging of singlephase loads The singlephase loads are distributed evenly among the three phases Assuming that the openYopen connection is used to supply power to his farm find the voltages and currents in each of the two transformers Also find the real and reactive powers supplied by each transformer Assume that the transformers are ideal What is the minimum required kVA rating of each transformer SOLUTION The farmers power system is illustrated below 58 Load 1 Load 2 VLLP VLLS ILP ILS The loads on each phase are balanced and the total load is found as 1 200 kW P 1 1 1 tan 200 kW tan cos 085 124 kvar Q P 2 60 kW P 1 2 2 tan 60 kW tan cos 09 29 kvar Q P TOT 260 kW P TOT 153 kvar Q 1 1 TOT TOT 153 kvar PF cos tan cos tan 0862 lagging 260 kW Q P The line current on the secondary side of the transformer bank is TOT 260 kW 363 A 3 PF 3 480 V 0862 LS LS P I V The openYopen connection is shown below From the figure it is obvious that the secondary voltage across the transformer is 480 V and the secondary current in each transformer is 246 A The primary voltages and currents are given by the transformer turns ratios to be 7967 V and 148 A respectively If the voltage of phase A of the primary side is arbitrarily taken as an angle of 0 then the voltage of phase B will be at an angle of 120 and the voltages of phases A and B on the secondary side will be 480 0 V and VAS 120 V 480 VBS respectively Note that line currents are shifted by 30 due to the difference between line and phase quantities and by a further 305 due to the power factor of the load 59 VA 83140 V VB 8314 120 V VAS 4800 V VBS 480120 V IAS 363605 A IBS 3631805 A ICS 363595 A IB 3631205 A A B n A B C IAP 2096605 A IBP 20961805 A In 2096595 A The real and reactive powers supplied by each transformer are calculated below cos 480 V 363 A cos 0 605 858 kW A AS A P V I cos 480 V 363 A sin 0 605 1517 kvar A AS A Q V I cos 480 V 363 A cos 120 1205 1742 kW B BS B P V I sin 480 V 363 A sin 120 1205 15 kvar B BS B Q V I Notice that the real and reactive powers supplied by the two transformers are radically different put the apparent power supplied by each transformer is the same Also notice that the total power A B P P supplied by the transformers is equal to the power consumed by the loads within roundoff error while the total reactive power supplied by the transformers is equal to the reactive power consumed by the loads QA Q B 220 A 50kVA 20000480V 60Hz singlephase distribution transformer is tested with the following results Opencircuit test measured from secondary side Shortcircuit test measured from primary side VOC 480 V VSC 1130 V IOC 41 A ISC 130 A POC 620 W PSC 550 W a Find the perunit equivalent circuit for this transformer at 60 Hz b What is the efficiency of the transformer at rated conditions and unity power factor What is the voltage regulation at those conditions c What would the ratings of this transformer be if it were operated on a 50Hz power system d Sketch the equivalent circuit of this transformer referred to the primary side if it is operating at 50 Hz e What is the efficiency of the transformer at rated conditions on a 50 Hz power system with unity power factor What is the voltage regulation at those conditions f How does the efficiency of a transformer at rated conditions and 60 Hz compare to the same physical device running a 50 Hz SOLUTION a The base impedance of this transformer referred to the primary side is 60 2 2 base 20000 V 8 k 50 kVA P P V Z S The base impedance of this transformer referred to the secondary side is 2 2 base 480 V 4608 50 kVA S S V Z S The open circuit test yields the values for the excitation branch referred to the secondary side 410 A 000854 S 480 V OC EX OC I Y V 1 1 620 W cos cos 716 480 V 41 A OC OC OC P V I 000854 716 000270 000810 EX C M Y G jB j 1 370 C C R G 1 123 M M X B The excitation branch elements can be expressed in perunit as 370 803 pu 4608 RC 123 267 pu 4608 XM The short circuit test yields the values for the series impedances referred to the primary side 1130 V 869 130 A SC EQ SC V Z I 1 1 550 W cos cos 680 1130 V 130 A SC SC SC P V I 869 68 326 806 EQ EQ EQ Z R jX j The resulting perunit impedances are 326 0041 pu 8000 REQ 806 0101 pu 8000 XEQ The perunit equivalent circuit is 61 VS VP IS IP RC jXM REQ jXEQ 0041 j0101 803 j267 b The perunit primary voltage at rated conditions and unity power factor is EQ P S SZ V V I 1 0 V 1 0 0041 0101 1046 554 pu P j V The perunit power consumed by REQ is 2 2 EQ 1 pu 0041 pu 0041 pu P I R The perunit power consumed by C R is 2 2 C 1046 00136 pu 803 P C V P R Therefore the efficiency of this transformer at rated load and unity power factor is out out in out EQ 100 100 100 100 948 100 0041 00136 C P P P P P P and the voltage regulation is 1046 100 VR 100 46 100 c The voltage and apparent power ratings of this transformer must be reduced in direct proportion to the decrease in frequency in order to avoid flux saturation effects in the core At 50 Hz the ratings are rated 50 Hz 50 kVA 417 kVA 60 Hz S rated 50 Hz 20000 V 16667 kV 60 Hz VP rated 50 Hz 480 V 400 V 60 Hz VS d The transformer parameters referred to the primary side at 60 Hz are base pu 8 k 803 642 k C C R Z R base pu 8 k 267 214 k M M X Z X EQ base E pu 8 k 0041 328 Q R Z R EQ base E pu 8 k 0101 808 Q X Z X At 50 Hz the resistance will be unaffected but the reactances are reduced in direct proportion to the decrease in frequency At 50 Hz the reactances are 50 Hz 214 k 178 k 60 Hz XM EQ 50 Hz 808 673 60 Hz X 62 The resulting equivalent circuit referred to the primary at 50 Hz is shown below VS VP IS IP RC jXM REQ jXEQ 642 k j178 k 328 j673 e The base impedance of this transformer at 50 Hz referred to the primary side is 2 2 base 16667 V 666 k 417 kVA P P V Z S The base impedance of this transformer at 50 Hz referred to the secondary side is 2 2 base 400 V 3837 417 kVA S S V Z S The excitation branch elements can be expressed in perunit as 642 k 964 pu 666 k RC 178 k 267 pu 666 k XM The series impedances can be expressed in perunit as 328 00492 pu 666 k REQ 673 0101 pu 666 k XEQ The perunit primary voltage at rated conditions and unity power factor is EQ P S SZ V V I 1 0 V 1 0 00492 0101 1054 549 pu P j V The perunit power consumed by REQ is 2 2 EQ 1 pu 00492 pu 00492 pu P I R The perunit power consumed by C R is 2 2 C 1054 00115 pu 964 P C V P R Therefore the efficiency of this transformer at rated load and unity power factor is out out in out EQ 100 100 100 100 943 100 00492 00115 C P P P P P P 63 and the voltage regulation is 1054 100 VR 100 54 100 f The efficiency of the transformer at 50 Hz is almost the same as the efficiency at 60 Hz just slightly less but the total apparent power rating of the transformer at 50 Hz must be less than the apparent power rating at 60 Hz by the ratio 5060 In other words the efficiencies are similar but the power handling capability is reduced 221 Prove that the threephase system of voltages on the secondary of the Y transformer shown in Figure 2 37b lags the threephase system of voltages on the primary of the transformer by 30 SOLUTION The figure is reproduced below VC VB VA VC VA VB VA VB VC VA VB VC Assume that the phase voltages on the primary side are given by 0 P A V V 120 P B V V 120 P C V V Then the phase voltages on the secondary side are given by 0 S A V V 120 S B V V 120 S C V V 64 where Since this is a Y transformer bank the line voltage on the primary side is a V V P S ab V 30 3 120 0 P P P B A ab V V V V V V and the voltage Note that the line voltage on the secondary side lags the line voltage on the primary side by 30 0 S A a b V V V 222 Prove that the threephase system of voltages on the secondary of the Y transformer shown in Figure 2 37c lags the threephase system of voltages on the primary of the transformer by 30 SOLUTION The figure is reproduced below VA VA VB VB VC VC Assume that the phase voltages on the primary side are given by 0 P A V V 120 P B V V 120 P C V V Then the phase voltages on the secondary side are given by 0 S A V V 120 S B V V 120 S C V V where Since this is a Y transformer bank the line voltage on the primary side is just equal to a V V P S ab V A V V 0 P The line voltage on the secondary side is given by 65 30 3 120 0 P P P C A a b V V V V V V Note that the line voltage on the secondary side lags the line voltage on the primary side by 30 223 A singlephase 10kVA 480120V transformer is to be used as an autotransformer tying a 600V distribution line to a 480V load When it is tested as a conventional transformer the following values are measured on the primary 480V side of the transformer Opencircuit test measured on secondary side Shortcircuit test measured on primary side VOC 120 V VSC 100 V IOC 160 A ISC 106 A VOC 38 W PSC 25 W a Find the perunit equivalent circuit of this transformer when it is connected in the conventional manner What is the efficiency of the transformer at rated conditions and unity power factor What is the voltage regulation at those conditions b Sketch the transformer connections when it is used as a 600480V stepdown autotransformer c What is the kilovoltampere rating of this transformer when it is used in the autotransformer connection d Answer the questions in a for the autotransformer connection SOLUTION a The base impedance of this transformer referred to the primary side is 2 2 base 480 V 2304 10 kVA P P V Z S The base impedance of this transformer referred to the secondary side is 2 2 base 120 V 144 10 kVA P P V Z S The open circuit test yields the values for the excitation branch referred to the secondary side 160 A 001333 S 120 V OC EX OC I Y V 1 1 38 W cos cos 786 120 V 160 A OC OC OC P V I 001333 786 0002635 001307 EX C M Y G jB j 1 380 C C R G 1 765 M M X B The excitation branch elements can be expressed in perunit as 380 263 pu 144 RC 765 531 pu 144 XM The short circuit test yields the values for the series impedances referred to the primary side 100 V 0943 106 A SC EQ SC V Z I 66 1 1 25 W cos cos 764 100 V 106 A SC SC SC P V I 0943 764 0222 0917 EQ EQ EQ Z R jX j The resulting perunit impedances are 0222 000963 pu 2304 REQ 0917 00398 pu 2304 XEQ The perunit equivalent circuit is VS VP IS IP RC jXM REQ jXEQ 000963 j00398 263 j531 At rated conditions and unity power factor the output power to this transformer would be 10 pu The input voltage would be IN P EQ P S SZ V V I 1 0 V 1 0 000963 00398 101 223 pu P j V The core losses in resistor C would be R 2 2 core 101 000388 pu 263 C V P R The copper losses in resistor EQ would be R 2 2 CU EQ 10 000963 000963 pu P I R The input power of the transformer would be IN OUT CU core 10 000963 000388 10135 P P P P and the transformer efficiency would be OUT IN 10 100 100 987 10135 P P The voltage regulation of the transformer is 101 100 VR 100 10 100 67 b The autotransformer connection for 600480 V stepdown operation is 480 V 600 V NC NSE VSE VC c When used as an autotransformer the kVA rating of this transformer becomes SE IO SE 4 1 10 kVA 50 kVA 1 C W N N S S N d As an autotransformer the perunit series impedance is decreased by the reciprocal of the power advantage so the series impedance becomes EQ Z 000963 000193 pu 5 REQ 00398 000796 pu 5 XEQ while the magnetization branch elements are basically unchanged At rated conditions and unity power factor the output power to this transformer would be OUT 10 pu The input voltage would be P EQ P S SZ V V I 1 0 V 1 0 000193 000796 1002 05 pu P j V The core losses in resistor C would be R 2 2 core 1002 000382 pu 263 C V P R The copper losses in resistor EQ would be R 2 2 CU EQ 10 000193 00019 pu P I R The input power of the transformer would be IN OUT CU core 10 00019 000382 10057 P P P P and the transformer efficiency would be OUT IN 10 100 100 994 10057 P P The voltage regulation of the transformer is 10057 100 VR 100 06 100 68 224 Figure P24 shows a oneline diagram of a power system consisting of a threephase 480V 60Hz generator supplying two loads through a transmission line with a pair of transformers at either end NOTE Oneline diagrams are described in Appendix A the discussion of threephase power circuits a Sketch the perphase equivalent circuit of this power system b With the switch opened find the real power P reactive power Q and apparent power S supplied by the generator What is the power factor of the generator c With the switch closed find the real power P reactive power Q and apparent power S supplied by the generator What is the power factor of the generator d What are the transmission losses transformer plus transmission line losses in this system with the switch open With the switch closed What is the effect of adding Load 2 to the system SOLUTION This problem can best be solved using the perunit system of measurements The power system can be divided into three regions by the two transformers If the perunit base quantities in Region 1 left of transformer 1 are chosen to be 1000 kVA and 480 V then the base quantities in Regions 2 between the transformers and 3 right or transformer 2 will be as shown below Sbase1 Lbase1 V Region 1 Region 2 Region 3 Sbase1 Sbase2 Sbase3 1000 kVA 1000 kVA 1000 kVA VLbase2 480 V VLbase2 14400 V VLbase3 480 V The base impedances of each region will be 2 2 1 base1 base1 3 3 277 V 0238 1000 kVA V Z S 2 2 2 base2 base2 3 3 8314 V 2074 1000 kVA V Z S 2 2 3 base3 base3 3 3 277 V 0238 1000 kVA V Z S a To get the perunit perphase equivalent circuit we must convert each impedance in the system to perunit on the base of the region in which it is located The impedance of transformer is already in perunit to the proper base so we dont have to do anything to it 1T 0 010 1 pu R 0 040 1 pu X The impedance of transformer is already in perunit but it is perunit to the base of transformer so it must be converted to the base of the power system T2 2 T 69 2 base 1 base 2 pu on base 2 pu on base 1 2 base 2 base 1 V S R X Z R X Z V S 260 2 2pu 2 8314 V 1000 kVA 0020 0040 8314 V 500 kVA R 2 2pu 2 8314 V 1000 kVA 0085 0170 8314 V 500 kVA X The perunit impedance of the transmission line is line linepu base2 15 10 000723 00482 2074 Z j Z j Z The perunit impedance of Load 1 is load1 load1pu base3 045 3687 1513 1134 0238 Z Z j Z The perunit impedance of Load 2 is load2 load2pu base3 08 336 0238 Z j Z j Z The resulting perunit perphase equivalent circuit is shown below 10 T1 T2 Line L1 L2 0010 j0040 000723 j00482 0040 j0170 1513 j1134 j336 b With the switch opened the equivalent impedance of this circuit is EQ 0010 0040 000723 00482 0040 0170 1513 1134 Z j j j j EQ 15702 13922 2099 416 Z j The resulting current is 1 0 04765 416 2099 416 I The load voltage under these conditions would be Loadpu Load 04765 416 1513 1134 0901 47 Z j V I Load Loadpu base3 0901 480 V 432 V V V V The power supplied to the load is 2 2 Loadpu Load 04765 1513 0344 P I R Load Loadpu base 0344 1000 kVA 344 kW P P S The power supplied by the generator is pu cos 1 04765 cos416 0356 G P VI 70 pu sin 1 04765 sin 416 0316 QG VI pu 1 04765 04765 SG VI pu base 0356 1000 kVA 356 kW G G P P S pu base 0316 1000 kVA 316 kVAR G G Q Q S pu base 04765 1000 kVA 4765 kVA G G S S S The power factor of the generator is PF cos 416 0748 lagging c With the switch closed the equivalent impedance of this circuit is EQ 1513 1134 336 0010 0040 000723 00482 0040 0170 1513 1134 336 j j Z j j j j j EQ 0010 0040 000788 00525 0040 0170 2358 0109 Z j j j j EQ 2415 0367 2443 865 Z j The resulting current is 1 0 0409 865 2443 865 I The load voltage under these conditions would be Loadpu Load 0409 865 2358 0109 0966 60 Z j V I Load Loadpu base3 0966 480 V 464 V V V V The power supplied to the two loads is the power supplied to the resistive component of the parallel combination of the two loads 2358 pu 2 2 Loadpu Load 0409 2358 0394 P I R Load Loadpu base 0394 1000 kVA 394 kW P P S The power supplied by the generator is pu cos 1 0409 cos60 0407 G P VI pu sin 1 0409 sin60 00428 QG VI pu 1 0409 0409 SG VI pu base 0407 1000 kVA 407 kW G G P P S pu base 00428 1000 kVA 428 kVAR G G Q Q S pu base 0409 1000 kVA 409 kVA G G S S S The power factor of the generator is PF cos 60 0995 lagging d The transmission losses with the switch open are 2 2 linepu line 04765 000723 000164 P I R line l pu base 000164 1000 kVA 164 kW ine P P S The transmission losses with the switch closed are 2 2 linepu line 0409 000723 000121 P I R 71 72 line l pu base 000121 1000 kVA 121 kW ine P P S Load 2 improved the power factor of the system increasing the load voltage and the total power supplied to the loads while simultaneously decreasing the current in the transmission line and the transmission line losses This problem is a good example of the advantages of power factor correction in power systems Chapter 3 AC Machinery Fundamentals 31 The simple loop is rotating in a uniform magnetic field shown in Figure 31 has the following characteristics 10 T to the right B r 01 m l 03 m m 377 rads a Calculate the voltage induced in this rotating loop e t tot b What is the frequency of the voltage produced in this loop c Suppose that a 10 resistor is connected as a load across the terminals of the loop Calculate the current that would flow through the resistor d Calculate the magnitude and direction of the induced torque on the loop for the conditions in c e Calculate the instantaneous and average electric power being generated by the loop for the conditions in c f Calculate the mechanical power being consumed by the loop for the conditions in c How does this number compare to the amount of electric power being generated by the loop m r vab vcd B N S B is a uniform magnetic field aligned as shown a b c d SOLUTION a The induced voltage on a simple rotating loop is given by ind 2 sin e t r Bl t 38 ind 2 01 m 377 rads 10 T 03 m sin377 e t t ind 226 sin377 V e t t b The angular velocity of the voltage produced in the loop is 377 rads Frequency is related to angular velocity by the equation 2 f so 377 60 Hz 2 2 f c If a 10 resistor is connected as a load across the terminals of the loop the current flow would be ind 226 sin377 V 226 sin 377 A 10 e t i t t R 73 d The induced torque would be ind 2 sin t rilΒ 317 ind 2 01 m 226 sin A 03 m 10 T sin t t t 2 ind 0136 sin 377 N m counterclockwise t t e The instantaneous power generated by the loop is 2 ind 226 sin377 V 226 sin 377 A 511 sin 377 W P t e i t t t The average power generated by the loop is 2 ave 1 511 sin 2555 W T P t dt T f The mechanical power being consumed by the loop is 2 2 ind 0136 sin 377 N m 377 rads 513 sin W P t t Note that the amount of mechanical power consumed by the loop is equal to the amount of electrical power created by the loop within roundoff error This machine is acting as a generator converting mechanical power into electrical power 32 Develop a table showing the speed of magnetic field rotation in ac machines of 2 4 6 8 10 12 and 14 poles operating at frequencies of 50 60 and 400 Hz SOLUTION The equation relating the speed of magnetic field rotation to the number of poles and electrical frequency is 120 se sm f n P The resulting table is Number of Poles ef 50 Hz ef 60 Hz ef 400 Hz 2 3000 rmin 3600 rmin 24000 rmin 4 1500 rmin 1800 rmin 12000 rmin 6 1000 rmin 1200 rmin 8000 rmin 8 750 rmin 900 rmin 6000 rmin 10 600 rmin 720 rmin 4800 rmin 12 500 rmin 600 rmin 4000 rmin 14 4286 rmin 5143 rmin 3429 rmin 33 The first ac power system in the USA ran at a frequency of 133 Hz If the ac power for this system were produced by a 4pole generator how fast would the shaft of the generator have to rotate SOLUTION The equation for the speed of the shaft is 120 133 Hz 120 3990 rmin 4 se sm f n P 34 A threephase Yconnected fourpole winding is installed in 24 slots on a stator There are 40 turns of wire in each slot of the windings All coils in each phase are connected in series The flux per pole in the machine is 0060 Wb and the speed of rotation of the magnetic field is 1800 rmin a What is the frequency of the voltage produced in this winding b What are the resulting phase and terminal voltages of this stator SOLUTION 74 a The frequency of the voltage produced in this winding is 1800 rmin 4 poles 60 Hz 120 120 sm se n P f b There are 24 slots on this stator with 40 turns of wire per slot Since this is a fourpole machine there are two sets of coils in 8 slots associated with each phase The voltage in the coils in one pair of slots is 2 2 40 t 0060 Wb 60 Hz 640 V A C E N f There are eight slots associated with each phase and the all of the coils in a slot are connected in series so the total phase voltage is 8 640 V 5120 V V Since the machine is Yconnected 3 8868 V VL V 35 A threephase Δconnected sixpole winding is installed in 36 slots on a stator There are 150 turns of wire in each slot of the windings All coils in each phase are connected in series The flux per pole in the machine is 0060 Wb and the speed of rotation of the magnetic field is 1000 rmin a What is the frequency of the voltage produced in this winding b What are the resulting phase and terminal voltages of this stator SOLUTION a The frequency of the voltage produced in this winding is 1000 rmin 6 poles 50 Hz 120 120 sm se n P f b There are 36 slots on this stator with 150 turns of wire per slot Since this is a sixpole machine there are three sets of coils in 12 slots associated with each phase The voltage in the coils in one pair of slots is 2 2 150 t 0060 Wb 50 Hz 2000 V A C E N f There are 12 slots associated with each phase and the all of the coils in a slot are connected in series so the total phase voltage is 12 2000 V 24000 V V Since the machine is Δ connected 24000 V VL V 36 A threephase Yconnected 60Hz twopole synchronous machine has a stator with 5000 turns of wire per phase What rotor flux would be required to produce a terminal linetoline voltage of 132 kV SOLUTION The phase voltage of this machine should be 3 7621 V L V V The induced voltage per phase in this machine which is equal to at noload conditions is given by the equation V 2 A C E N f so 7621 V 00057 Wb 2 2 5000 t 60 Hz A C E N f 75 76 37 Modify the MATLAB in Example 31 by swapping the currents flowing in any two phases What happens to the resulting net magnetic field SOLUTION This modification is very simplejust swap the currents supplied to two of the three phases Mfile magfield2m Mfile to calculate the net magetic field produced by a threephase stator Set up the basic conditions bmax 1 Normalize bmax to 1 freq 60 60 Hz w 2pifreq angluar velocity rads First generate the three component magnetic fields t 016000160 Baa sinwt cos0 jsin0 Bbb sinwt2pi3 cos2pi3 jsin2pi3 Bcc sinwt2pi3 cos2pi3 jsin2pi3 Calculate Bnet Bnet Baa Bbb Bcc Calculate a circle representing the expected maximum value of Bnet circle 15 coswt jsinwt Plot the magnitude and direction of the resulting magnetic fields Note that Baa is black Bbb is blue Bcc is magneta and Bnet is red for ii 1lengtht Plot the reference circle plotcirclek hold on Plot the four magnetic fields plot0 realBaaii0 imagBaaiikLineWidth2 plot0 realBbbii0 imagBbbiibLineWidth2 plot0 realBccii0 imagBcciimLineWidth2 plot0 realBnetii0 imagBnetiirLineWidth3 axis square axis2 2 2 2 drawnow hold off end When this program executes the net magnetic field rotates clockwise instead of counterclockwise 38 If an ac machine has the rotor and stator magnetic fields shown in Figure P31 what is the direction of the induced torque in the machine Is the machine acting as a motor or generator SOLUTION Since ind net k R τ B B the induced torque is clockwise opposite the direction of motion The machine is acting as a generator 39 The flux density distribution over the surface of a twopole stator of radius r and length l is given by B B t M m cos 337b Prove that the total flux under each pole face is 2rlBM SOLUTION The total flux under a pole face is given by the equation 77 d B A Under a pole face the flux density B is always parallel to the vector dA since the flux density is always perpendicular to the surface of the rotor and stator in the air gap Therefore B dA A differential area on the surface of a cylinder is given by the differential length along the cylinder dl times the differential width around the radius of the cylinder rd dA dl rd where r is the radius of the cylinder Therefore the flux under the pole face is B dl r d Since r is constant and B is constant with respect to l this equation reduces to rl B d Now cos cos M B B t BM when we substitute t so rl B d 2 2 2 2 cos sin 1 1 M M M rl B d rlB rlB 2 rlBM 310 In the early days of ac motor development machine designers had great difficulty controlling the core losses hysteresis and eddy currents in machines They had not yet developed steels with low hysteresis and were not making laminations as thin as the ones used today To help control these losses early ac motors in the USA were run from a 25 Hz ac power supply while lighting systems were run from a separate 60 Hz ac power supply a Develop a table showing the speed of magnetic field rotation in ac machines of 2 4 6 8 10 12 and 14 poles operating at 25 Hz What was the fastest rotational speed available to these early motors b For a given motor operating at a constant flux density B how would the core losses of the motor running at 25 Hz compare to the core losses of the motor running at 60 Hz c Why did the early engineers provide a separate 60 Hz power system for lighting SOLUTION a The equation relating the speed of magnetic field rotation to the number of poles and electrical frequency is 120 e m f n P 78 The resulting table is Number of Poles ef 25 Hz 2 1500 rmin 4 750 rmin 6 500 rmin 8 375 rmin 10 300 rmin 12 250 rmin 14 2143 rmin The highest possible rotational speed was 1500 rmin b Core losses scale according to the 15th power of the speed of rotation so the ratio of the core losses at 25 Hz to the core losses at 60 Hz for a given machine would be 1500 15 ratio 0269 3600 or 269 c At 25 Hz the light from incandescent lamps would visibly flicker in a very annoying way 311 In later years motors improved and could be run directly from a 60 Hz power supply As a result 25 Hz power systems shrank and disappeared However there were many perfectlygood working 25 Hz motors in factories around the country that owners were not ready to discard To keep them running some users created their own 25 Hz power in the plant using motorgenerator sets A motor generator set consists of two machines connected on a common shaft one acting as a motor and the other acting as a generator If the two machines have different numbers of poles but exactly the same shaft speed then the electrical frequency of the two machines will be different due to Equation 334 What combination of poles on the two machines could convert 60 Hz power to 25 Hz power 120 sm se n P f 334 SOLUTION From Equation 334 the speed of rotation of the 60 Hz machines would be 1 1 1 1 120 60 Hz 120 7200 se sm f n P P 1P and the speed of rotation of the 25 Hz machines would be 2 2 2 2 120 25 Hz 120 3000 se sm f n P P 2P If the machines are tied together in a motorgenerator set the shaft speed must be the same for both machines so 1 2 7200 3000 P P 1 2 7200 72 3000 30 P P This result says that a 72pole machine at 60 Hz would rotate at the same speed as a 30pole machine at 25 Hz Each machine would rotate at 100 rmin 1 1 1 120 60 Hz 120 100 rmin 72 se sm f n P 79 80 2 2 2 120 25 Hz 120 100 rmin 30 se sm f n P Chapter 4 Synchronous Generators 41 At a location in Europe it is necessary to supply 1000 kW of 60Hz power The only power sources available operate at 50 Hz It is decided to generate the power by means of a motorgenerator set consisting of a synchronous motor driving a synchronous generator How many poles should each of the two machines have in order to convert 50Hz power to 60Hz power SOLUTION The speed of a synchronous machine is related to its frequency by the equation 120 sm se n P f 334 To make a 50 Hz and a 60 Hz machine have the same mechanical speed so that they can be coupled together we see that sync 1 2 120 50 Hz 120 60 Hz n P P 2 1 6 12 5 10 P P Therefore a 10pole synchronous motor must be coupled to a 12pole synchronous generator to accomplish this frequency conversion 42 A 138kV 50MVA 09powerfactorlagging 60Hz fourpole Yconnected synchronous generator has a synchronous reactance of 25 and an armature resistance of 02 At 60 Hz its friction and windage losses are 1 MW and its core losses are 15 MW The field circuit has a dc voltage of 120 V and the maximum is 10 A The current of the field circuit is adjustable over the range from 0 to 10 A The OCC of this generator is shown in Figure P41 IF 81 a How much field current is required to make the terminal voltage V or line voltage equal to 138 kV when the generator is running at no load T L V b What is the internal generated voltage A of this machine at rated conditions E c What is the phase voltage V of this generator at rated conditions d How much field current is required to make the terminal voltage V equal to 138 kV when the generator is running at rated conditions T e Suppose that this generator is running at rated conditions and then the load is removed without changing the field current What would the terminal voltage of the generator be f How much steadystate power and torque must the generators prime mover be capable of supplying to handle the rated conditions g Construct a capability curve for this generator Note An electronic version of this open circuit characteristic can be found in file Figp41occdat which can be used with MATLAB programs Column 1 contains field current in amps and column 2 contains opencircuit terminal voltage in volts a If the noload terminal voltage is 138 kV the required field current can be read directly from the opencircuit characteristic It is 350 A b This generator is Yconnected so A L I I At rated conditions the line and phase current in this generator is 82 50 MVA 2092 A 3 3 13800 V A L L P I I V at an angle of 258 The phase voltage of this machine is 3 7967 V T V V The internal generated voltage of the machine is A A A R jX E V I SI A 7967 0 020 2092 258 A 25 2092 258 A A j E A 11544 231 V E c The phase voltage of the machine at rated conditions is V 7967 V From the OCC the required field current is 10 A d The equivalent opencircuit terminal voltage corresponding to an A E of 11544 volts is oc 3 11544 V 20 kV VT From the OCC the required field current is 10 A e If the load is removed without changing the field current 11544 V A V E The corresponding terminal voltage would be 20 kV f The input power to this generator is equal to the output power plus losses The rated output power is OUT 50 MVA 09 45 MW P 2 2 CU 3 3 2092 A 02 26 MW A A P I R FW 1 MW P core 15 MW P stray assumed 0 P IN OUT CU FW core stray 501 MW P P P P P P Therefore the prime mover must be capable of supplying 501 MW Since the generator is a fourpole 60 Hz machine to must be turning at 1800 rmin The required torque is IN APP 501 MW 265800 N m 1 min 2 rad 1800 rmin 60 s 1 r m P e The rotor current limit of the capability curve would be drawn from an origin of 2 2 3 3 7967 V 7617 MVAR 25 S V Q X The radius of the rotor current limit is 3 3 7967 V 11544 V 110 MVA 25 A E S V E D X The stator current limit is a circle at the origin of radius 83 3 3 7967 V 2092 A 50 MVA A S V I A MATLAB program that plots this capability diagram is shown below Mfile prob42m Mfile to display a capability curve for a synchronous generator Calculate the waveforms for times from 0 to 130 s Q 7617 DE 110 S 50 Get points for stator current limit theta 95195 Angle in degrees rad theta pi 180 Angle in radians scurve S cosrad jsinrad Get points for rotor current limit orig jQ theta 651115 Angle in degrees rad theta pi 180 Angle in radians rcurve orig DE cosrad jsinrad Plot the capability diagram figure1 plotrealscurveimagscurvebLineWidth20 hold on plotrealrcurveimagrcurverLineWidth20 Add x and y axes plot 75 750 0k plot 0075 75k Set titles and axes title bfSynchronous Generator Capability Diagram xlabelbfPower MW ylabelbfReactive Power MVAR axis 75 75 75 75 axis square hold off The resulting capability diagram is shown below 84 43 Assume that the field current of the generator in Problem 42 has been adjusted to a value of 5 A a What will the terminal voltage of this generator be if it is connected to a connected load with an impedance of 24 25 b Sketch the phasor diagram of this generator c What is the efficiency of the generator at these conditions d Now assume that another identical connected load is to be paralleled with the first one What happens to the phasor diagram for the generator e What is the new terminal voltage after the load has been added f What must be done to restore the terminal voltage to its original value SOLUTION a If the field current is 50 A the opencircuit terminal voltage will be about 16500 V and the open circuit phase voltage in the generator and hence EA will be 16500 3 9526 V The load is connected with three impedances of 24 25 8 25 From the Y transform this load is equivalent to a Yconnected load with three impedances of The resulting perphase equivalent circuit is shown below EA 020 j25 Z V IA 8 25 85 The magnitude of the phase current flowing in this generator is 9526 V 9526 V 1004 A 02 25 8 25 949 A A A S E I R jX Z j Therefore the magnitude of the phase voltage is 1004 A 8 8032 V A V I Z and the terminal voltage is 3 3 8032 V 13910 V VT V b Armature current is and the phase voltage is 1004 25 A A I 8032 0 V V Therefore the internal generated voltage is A A A R jX E V I SI A 8032 0 020 1004 25 A 25 1004 25 A A j E A 9530 133 V E The resulting phasor diagram is shown below not to scale 8032 0 V V A 9530 133 V E 1004 25 A A I S A jX I A RA I c The efficiency of the generator under these conditions can be found as follows OUT 3 cos 3 8032 V 1004 A cos 25 219 MW A P V I 2 2 CU 3 3 1004 A 02 605 kW A A P I R FW 1 MW P core 15 MW P stray assumed 0 P IN OUT CU FW core stray 25 MW P P P P P P OUT IN 219 MW 100 100 876 25 MW P P d To get the basic idea of what happens we will ignore the armature resistance for the moment If the field current and the rotational speed of the generator are constant then the magnitude of A K E A E S jX I is constant The quantity increases in length at the same angle while the magnitude of must remain constant Therefore swings out along the arc of constant magnitude until the new fits exactly between and S A jX I A E A E S V 86 V A E A I S A jX I AI A E V e The new impedance per phase will be half of the old value so The magnitude of the phase current flowing in this generator is 4 25 Z W 9526 V 9526 V 1680 A 02 25 4 25 567 A A A S E I R jX Z j Therefore the magnitude of the phase voltage is 1680 A 4 6720 V A V I Z and the terminal voltage is 3 3 6720 V 11640 V VT V f To restore the terminal voltage to its original value increase the field current FI 44 Assume that the field current of the generator in Problem 42 is adjusted to achieve rated voltage 138 kV at full load conditions in each of the questions below a What is the efficiency of the generator at rated load b What is the voltage regulation of the generator if it is loaded to rated kilovoltamperes with 09 PFlagging loads c What is the voltage regulation of the generator if it is loaded to rated kilovoltamperes with 09 PFleading loads d What is the voltage regulation of the generator if it is loaded to rated kilovoltamperes with unity powerfactor loads e Use MATLAB to plot the terminal voltage of the generator as a function of load for all three power factors SOLUTION a This generator is Yconnected so At rated conditions the line and phase current in this generator is LI I A 50 MVA 2092 A 3 3 13800 V A L L P I I V at an angle of 258 The phase voltage of this machine is 3 7967 V T V V The internal generated voltage of the machine is A A A R jX E V I SI A 7967 0 020 251 3687 A 25 2092 258 A A j E 87 88 A 11544 231 V E The input power to this generator is equal to the output power plus losses The rated output power is OUT 50 MVA 09 45 MW P 2 2 CU 3 3 2092 A 02 26 MW A A P I R FW 1 MW P core 15 MW P stray assumed 0 P IN OUT CU FW core stray 501 MW P P P P P P OUT IN 45 MW 100 100 898 501 MW P h P b If the generator is loaded to rated MVA with lagging loads the phase voltage is Vf 7967 0 V and the internal generated voltage is A 11544 231 V E Therefore the phase voltage at noload would be 11544 0 V The voltage regulation would be V 11544 7967 VR 100 449 7967 c If the generator is loaded to rated kVA with leading loads the phase voltage is and the internal generated voltage is Vf 7967 0 V A A A R jX E V I SI A 7967 0 020 2092 258 A 25 2092 258 A A j E A 7793 388 V E The voltage regulation would be 7793 7967 VR 100 22 7967 d If the generator is loaded to rated kVA at unity power factor the phase voltage is Vf 7967 0 V and the internal generated voltage is A A A R jX E V I SI A 7967 0 020 2092 0 A 25 2092 0 A A j E A 9883 32 V E The voltage regulation would be 9883 7967 VR 100 24 7967 e For this problem we will assume that the terminal voltage is adjusted to 138 kV at no load conditions and see what happens to the voltage as load increases at 09 lagging unity and 09 leading power factors Note that the maximum current will be 2092 A in any case A phasor diagram representing the situation at lagging power factor is shown below I A V EA I R A A jX SI A By the Pythagorean Theorem 2 2 2 cos sin cos sin A A A S A S A A S E V R I X I X I R I 2 2 cos sin cos sin A S A A S A A S A V E X I R I R I X I In this case 2584 so cos 09 and sin 06512 A phasor diagram representing the situation at leading power factor is shown below I A V EA I R A A jX SI A By the Pythagorean Theorem 2 2 2 cos sin cos sin A A A S A S A A S E V R I X I X I R I 2 2 cos sin cos sin A S A A S A A S A V E X I R I R I X I In this case 2584 so cos 09 and sin 06512 A phasor diagram representing the situation at unity power factor is shown below I A V EA I R A A jX SI A By the Pythagorean Theorem 2 2 2 A S A E V X I 2 2 A S A V E X I In this case 0 so cos 10 and sin 0 The MATLAB program is shown below takes advantage of this fact 89 90 Mfile prob44em Mfile to calculate and plot the terminal voltage of a synchronous generator as a function of load for power factors of 08 lagging 10 and 08 leading Define values for this generator EA 7967 Internal gen voltage I 020922092 Current values A R 020 R ohms X 250 XS ohms Calculate the voltage for the lagging PF case VPlag sqrt EA2 XI09 RI 065122 RI09 XI 06512 VTlag VPlag sqrt3 Calculate the voltage for the leading PF case VPlead sqrt EA2 XI09 RI 065122 RI09 XI 06512 VTlead VPlead sqrt3 Calculate the voltage for the unity PF case VPunity sqrt EA2 XI2 VTunity VPunity sqrt3 Plot the terminal voltage versus load plotIabsVTlag1000bLineWidth20 hold on plotIabsVTunity1000kLineWidth20 plotIabsVTlead1000rLineWidth20 title bfTerminal Voltage Versus Load xlabel bfLoad A ylabel bfTerminal Voltage kV legend09 PF lagging10 PF09 PF leading axis0 2200 0 20 grid on hold off The resulting plot is shown below 45 Assume that the field current of the generator in Problem 42 has been adjusted so that it supplies rated voltage when loaded with rated current at unity power factor a What is the torque angle of the generator when supplying rated current at unity power factor b What is the maximum power that this generator can deliver to a unity power factor load when the field current is adjusted to the current value c When this generator is running at full load with unity power factor how close is it to the static stability limit of the machine SOLUTION a The torque angle can be found by calculating A E A A A R jX E V I SI A 7967 0 020 2092 0 A 25 2092 0 A A j E A 9883 32 V E Thus the torque angle 32 b The static stability limit occurs at 90 This generator is a approaching that limit If we ignore the internal resistance of the generator the output power will be given by 3 sin A S V E P X and the output power is proportional to sin The maximum possible power will occur when sin 90 That power is 3 3 7967 V 9883 V sin sin90 945 MW 25 A S V E P X c The static stability limit is about twice the rated output power of the generator 91 46 The internal generated voltage of a 2pole connected 60 Hz three phase synchronous generator is 144 kV and the terminal voltage is 128 kV The synchronous reactance of this machine is 4 and the armature resistance can be ignored A E T V a If the torque angle of the generator δ 18 how much power is being supplied by this generator at the current time b What is the power factor of the generator at this time c Sketch the phasor diagram under these circumstances d Ignoring losses in this generator what torque must be applied to its shaft by the prime mover at these conditions SOLUTION a If resistance is ignored the output power from this generator is given by 3 3 128 kV 144 kV sin sin18 427 MW 4 A S V E P X b The phase current flowing in this generator can be calculated from A jXS E V I A A A S jX E V I 144 18 kV 128 0 kV 1135 114 A 4 A j I and the power factor is cos 114 098 lagging Therefore the impedance angle 114 c The phasor diagram is 128 0 kV V A 144 18 kV E 1035 114 A A I S A jX I d The induced torque is given by the equation conv ind m P With no losses conv app ind 427 MW 113300 N m 2 60 hz m P 47 A 100MVA 144kV 08PFlagging 50Hz twopole Yconnected synchronous generator has a per unit synchronous reactance of 11 and a perunit armature resistance of 0011 a What are its synchronous reactance and armature resistance in ohms 92 b What is the magnitude of the internal generated voltage at the rated conditions What is its torque angle at these conditions EA c Ignoring losses in this generator what torque must be applied to its shaft by the prime mover at full load SOLUTION SOLUTION The base phase voltage of this generator is base 14400 3 8314 V V Therefore the base impedance of the generator is 2 2 base base base 3 3 8314 V 2074 100000000 VA V Z S a The generator impedance in ohms are 0011 2074 00228 A R 11 2074 2281 S X b The rated armature current is 100 MVA 4009 A 3 3 144 kV A L T S I I V The power factor is 08 lagging so 4009 3687 A A I Therefore the internal generated voltage is A A A R jX E V I SI A 8314 0 00228 4009 3687 A 2281 4009 3687 A A j E A 15660 276 V E Therefore the magnitude of the internal generated voltage A E 15660 V and the torque angle 276 c Ignoring losses the input power would equal the output power Since OUT 08 100 MVA 80 MW P and sync 120 50 Hz 120 3000 rmin 2 fse n P the applied torque would be app ind 80000000 W 254700 N m 3000 rmin 2 radr 1 min60 s 48 A 200MVA 12kV 085PFlagging 50Hz 20pole Yconnected water turbine generator has a per unit synchronous reactance of 09 and a perunit armature resistance of 01 This generator is operating in parallel with a large power system infinite bus a What is the speed of rotation of this generators shaft b What is the magnitude of the internal generated voltage at rated conditions EA c What is the torque angle of the generator at rated conditions d What are the values of the generators synchronous reactance and armature resistance in ohms 93 e If the field current is held constant what is the maximum power possible out of this generator How much reserve power or torque does this generator have at full load f At the absolute maximum power possible how much reactive power will this generator be supplying or consuming Sketch the corresponding phasor diagram Assume is still unchanged IF SOLUTION a The speed of rotation of this generators shaft is sync 120 50 Hz 120 300 rmin 20 fse n P b The perunit phase voltage at rated conditions is 10 0 V and the perunit phase current at rated conditions is since the power factor is 09 lagging so the perunit internal generated voltage is 10 258 A I S A A A A R jX E V I I 1 0 01 1 258 09 1 258 A j E A 169 274 pu E The base phase voltage is base 12 kV 3 6928 V V so the internal generated voltage is 169 274 pu 6928 V 11710 274 V A E c The torque angle of the generator is 274 d The base impedance of the generator is 2 2 base base base 3 3 6928 V 072 200000000 VA V Z S Therefore the synchronous reactance is 09 072 0648 S X and the armature resistance is 01 072 0072 A R e If the field current is held constant and the armature resistance is ignored the power out of this generator is given by 3 sin A S V E P X The max power is given by max 3 3 6928 V 11710 V sin90 376 MW 0648 A S V E P X Since the full load power is 200 MVA 085 170 MW P this generator is supplying 45 of the maximum possible power at full load conditions 94 f At the maximum power possible the torque angle 90 so the phasor will be at an angle of 90 and the current flowing will be A E A A A R jX E V I S I A A A A S R jX E V I 11710 90 kV 6298 0 kV 20400 346 A 0072 0648 A j I The impedance angle 346 and the reactive power supplied by the generator is 3 sin 3 6298 V 20400 A sin 346 219 Mvar A Q V I 6298 0 V V A 11170 90 V E 20 400 346 A A I S A jX I A RA I 49 A 480V 250kVA 08PFlagging twopole threephase 60Hz synchronous generators prime mover has a noload speed of 3650 rmin and a fullload speed of 3570 rmin It is operating in parallel with a 480V 250kVA 085PFlagging fourpole 60Hz synchronous generator whose prime mover has a no load speed of 1800 rmin and a fullload speed of 1780 rmin The loads supplied by the two generators consist of 300 kW at 08 PF lagging a Calculate the speed droops of generator 1 and generator 2 b Find the operating frequency of the power system c Find the power being supplied by each of the generators in this system d What must the generators operators do to adjust the operating frequency to 60 Hz e If the current line voltage is 460 V what must the generators operators do to correct for the low terminal voltage SOLUTION The noload frequency of generator 1 corresponds to a frequency of nl1 3650 rmin 2 60833 Hz 120 120 nm P f The fullload frequency of generator 1 corresponds to a frequency of 95 fl1 3570 rmin 2 595 Hz 120 120 nm P f The noload frequency of generator 2 corresponds to a frequency of nl2 1800 rmin 4 6000 Hz 120 120 nm P f The fullload frequency of generator 2 corresponds to a frequency of fl2 1780 rmin 4 59333 Hz 120 120 nm P f a The speed droop of generator 1 is given by nl fl 1 fl 3650 rmin 3570 rmin SD 100 100 224 3570 rmin n n n The speed droop of generator 2 is given by nl fl 2 fl 1800 rmin 1780 rmin SD 100 100 112 1780 rmin n n n b The full load power supplied by generator 1 is 1full load cos 250 kVA 08 200 kW P S The maximum power supplied by generator 2 is 2full load cos 250 kVA 085 2125 kW P S The power supplied by generator 1 as a function of frequency is given by 1 1 nl1 sys P P s f f and the power supplied by generator 2 as a function of frequency is given by 2 2 nl2 sy P P s f f s The power curves slope for generator 1 is 1 nl fl 02 MW 015 MWHz 60833 Hz 595 Hz P P s f f The power curves slope for generator 1 is 2 nl fl 02125 MW 0319 MWHz 6000 Hz 59333 Hz P P s f f The noload frequency of generator 1 is 60833 Hz and the noload frequency of generator 2 is 60 Hz The total power that they must supply is 300 kW so the system frequency can be found from the equations LOAD 1 2 P P P LOAD 1 nl1 sys 2 nl2 sys P P P s f f s f f sys sys 300 kW 015 MWHz 60833 Hz 0319 MWHz 600 Hz f f 96 sys sys 300 kW 9125 kW 015 MWHz 19140 kW 0319 MWHz f f sys 0469 MWHz 9125 kW 19140 kW 300 kW f sys 27965 kW 59627 Hz 0469 MWHz f c The power supplied by generator 1 is 1 1 nl1 sys 015 MWHz 60833 Hz 59627 Hz 181 kW P P s f f The power supplied by generator 2 is 2 2 nl2 sys 0319 MWHz 600 Hz 59627 Hz 119 kW P P s f f d The get the system frequency to 60 Hz the operators of the generators must increase the noload frequency setpoints of both generators simultaneously That action will increase thefrequency of the system without changing the power sharing between the generators e If the terminal voltage is 460 V the operators of the generators must increase the field currents on both generators simultaneously That action will increase the terminal voltages of the system without changing the reactive power sharing between the generators 410 Three physically identical synchronous generators are operating in parallel They are all rated for a full load of 100 MW at 08 PF lagging The noload frequency of generator A is 61 Hz and its speed droop is 3 percent The noload frequency of generator B is 615 Hz and its speed droop is 34 percent The no load frequency of generator C is 605 Hz and its speed droop is 26 percent a If a total load consisting of 230 MW is being supplied by this power system what will the system frequency be and how will the power be shared among the three generators b Create a plot showing the power supplied by each generator as a function of the total power supplied to all loads you may use MATLAB to create this plot At what load does one of the generators exceed its ratings Which generator exceeds its ratings first c Is this power sharing in a acceptable Why or why not d What actions could an operator take to improve the real power sharing among these generators SOLUTION a Speed droop is defined as nl fl nl fl fl fl SD 100 100 n n f f n f so nl fl SD 1 100 f f Thus the fullload frequencies of generators A B and C are nlA flA A 61 Hz 59223 Hz SD 30 1 1 100 100 f f nlB flB B 615 Hz 59478 Hz SD 34 1 1 100 100 f f 97 nlC flC C 605 Hz 58967 Hz SD 26 1 1 100 100 f f and the slopes of the powerfrequency curves are 100 MW 5627 MWHz 61 Hz 59223 Hz SPA 100 MW 4946 MWHz 615 Hz 59478 Hz SPB 100 MW 6523 MWHz 605 Hz 58967 Hz SPC a The total load is 230 MW so the system frequency is LOAD nlA sys nlB sys nlC sys PA PB PC P s f f s f f s f f sys sys sys 230 MW 5627 610 4946 615 6523 605 f f f sys sys sys 230 MW 3433 5627 3042 4946 3946 6523 f f f 17096 sys 10191 f sys 5961 Hz f The power supplied by each generator will be nlA sys 5627 MWHz 610 Hz 5961 Hz 782 MW A PA P s f f nlB sys 4946 MWHz 615 Hz 5961 Hz 935 MW B PB P s f f nlC sys 6523 MWHz 605 Hz 5961 Hz 581 MW C PC P s f f b The equation in part a can be rewritten slightly to express system frequency as a function of load LOAD sys sys sys 5627 610 4946 615 6523 605 P f f f f LOAD sys sys sys 3433 5627 3042 4946 3946 6523 P f f sys LOAD 17096 10421 f P LOAD sys 10421 17096 P f A MATLAB program that uses this equation to determine the power sharing among the generators as a function of load is shown below Mfile prob410bm Mfile to calculate and plot the power sharing among three generators as a function of load Define values for this generator fnlA 610 Noload freq of Gen A fnlB 615 Noload freq of Gen B fnlC 605 Noload freq of Gen C spA 5627 Slope of Gen A MWHz spB 4946 Slope of Gen B MWHz 98 spC 6523 Slope of Gen C MWHz Pload 010300 Load in MW Calculate the system frequency fsys 10421 Pload 17096 Calculate the power of each generator PA spA fnlA fsys PB spB fnlB fsys PC spC fnlC fsys Plot the power sharing versus load plotPloadPAbLineWidth20 hold on plotPloadPBkLineWidth20 plotPloadPCrLineWidth20 plot0 300100 100kLineWidth10 plot0 3000 0k title bfPower Sharing Versus Total Load xlabel bfTotal Load MW ylabel bfGenerator Power MW legendGenerator AGenerator BGenerator CPower Limit grid on hold off The resulting plot is shown below This plot reveals that there are power sharing problems both for high loads and for low loads Generator B is the first to exceed its ratings as load increases Its rated power is reached at a total load of 253 MW On the other hand Generator C gets into trouble as the total load is reduced When the total load drops to 78 MW the direction of power flow reverses in Generator C c The power sharing in a is acceptable because all generators are within their power limits d To improve the power sharing among the three generators in a without affecting the operating frequency of the system the operator should decrease the governor setpoints on Generator B while simultaneously increasing them in Generator C 99 411 A paper mill has installed three steam generators boilers to provide process steam and also to use some its waste products as an energy source Since there is extra capacity the mill has installed three 10MW turbine generators to take advantage of the situation Each generator is a 4160V 125 MVA 60 Hz 08 PFlagging twopole Yconnected synchronous generator with a synchronous reactance of 110 and an armature resistance of 003 Generators 1 and 2 have a characteristic powerfrequency slope of 5 MWHz and generator 3 has a slope of 6 MWHz sP a If the noload frequency of each of the three generators is adjusted to 61 Hz how much power will the three machines be supplying when actual system frequency is 60 Hz b What is the maximum power the three generators can supply in this condition without the ratings of one of them being exceeded At what frequency does this limit occur How much power does each generator supply at that point c What would have to be done to get all three generators to supply their rated real and reactive powers at an overall operating frequency of 60 Hz d What would the internal generated voltages of the three generators be under this condition SOLUTION a If the system frequency is 60 Hz and the noload frequencies of the generators are 61 Hz then the power supplied by the generators will be 1 1 nl1 sys 5 MWHz 61 Hz 60 Hz 50 MW P P s f f 2 2 nl2 sys 5 MWHz 61 Hz 60 Hz 50 MW P P s f f 3 3 nl3 sys 6 MWHz 61 Hz 60 Hz 60 MW P P s f f Therefore the total power supplied by the generators is 16 MW b The maximum power supplied by any one generator is 125 MVA08 10 MW Generator 3 will be the first machine to reach that limit Generator 3 will supply this power at a frequency of sys 10 MW 6 MWHz 61 Hz f sys 5933 Hz f At this point the power supplied by Generators 1 and 2 is 1 2 1 nl1 sys 5 MWHz 61 Hz 5933 Hz 835 MW P P P s f f The total power supplied by the generators at this condition is 267 MW c To get each of the generators to supply 10 MW at 60 Hz the noload frequencies of Generator 1 and Generator 2 would have to be adjusted to 62 Hz and the noload frequency of Generator 3 would have to be adjusted to 6167 Hz The field currents of the three generators must then be adjusted to get them supplying a power factor of 080 lagging At that point each generator will be supplying its rated real and reactive power d Under the conditions of part c which are the rated conditions of the generators the internal generated voltage would be given by A A A R jX E V I I S A The phase voltage of the generators is 4160 V 3 2402 V and since the generators are Yconnected their rated current is 100 125 MVA 1735 A 3 3 4160 V A L T S I I V The power factor is 080 lagging so 1735 3687 A A I Therefore A A A R jX E V I I S A 2402 0 003 1735 3687 A 11 1735 3687 A A j E A 3888 226 V E 412 Suppose that you were an engineer planning a new electric cogeneration facility for a plant with excess process steam You have a choice of either two 10 MW turbinegenerators or a single 20 MW turbine generator What would be the advantages and disadvantages of each choice SOLUTION A single 20 MW generator will probably be cheaper and more efficient than two 10 MW generators but if the 20 MW generator goes down all 20 MW of generation would be lost at once If two 10 MW generators are chosen one of them could go down for maintenance and some power could still be generated 413 A 25MVA 122kV 09PFlagging threephase twopole Yconnected 60Hz synchronous generator was tested by the opencircuit test and its airgap voltage was extrapolated with the following results Opencircuit test Field current A 275 320 365 380 475 570 Line voltage kV 122 130 138 141 152 160 Extrapolated airgap voltage kV 133 154 175 183 228 274 The shortcircuit test was then performed with the following results Shortcircuit test Field current A 275 320 365 380 475 570 Armature current A 890 1040 1190 1240 1550 1885 The armature resistance is 06 per phase a Find the unsaturated synchronous reactance of this generator in ohms per phase and in perunit b Find the approximate saturated synchronous reactance at a field current of 380 A Express the answer both in ohms per phase and in perunit X S c Find the approximate saturated synchronous reactance at a field current of 475 A Express the answer both in ohms per phase and in perunit d Find the shortcircuit ratio for this generator e What is the internal generated voltage of this generator at rated conditions f What field current is required to achieve rated voltage at rated load SOLUTION a The unsaturated synchronous reactance of this generator is the same at any field current so we will look at it at a field current of 380 A The extrapolated airgap voltage at this point is 183 kV and the shortcircuit current is 1240 A Since this generator is Yconnected the phase voltage is 183 kV 3 10566 V V and the armature current is AI 1240 A Therefore the unsaturated synchronous reactance is 101 10566 V 852 1240 A XSu The base impedance of this generator is 2 2 base base base 3 3 7044 V 595 25000000 VA V Z S Therefore the perunit unsaturated synchronous reactance is pu 852 143 595 XSu b The saturated synchronous reactance at a field current of 380 A can be found from the OCC and the SCC The OCC voltage at FI 380 A is 141 kV and the shortcircuit current is 1240 A Since this generator is Yconnected the corresponding phase voltage is 141 kV 3 8141 V V and the armature current is 1240 IA A Therefore the saturated synchronous reactance is 8141 V 657 1240 A S X and the perunit unsaturated synchronous reactance is pu 852 110 595 XS c The saturated synchronous reactance at a field current of 475 A can be found from the OCC and the SCC The OCC voltage at FI 475 A is 152 kV and the shortcircuit current is 1550 A Since this generator is Yconnected the corresponding phase voltage is 152 kV 3 8776 V V and the armature current is 1550 AI A Therefore the saturated synchronous reactance is 8776 V 566 1550 A S X and the perunit unsaturated synchronous reactance is pu 566 0951 595 XS d The rated voltage of this generator is 122 kV which requires a field current of 275 A The rated line and armature current of this generator is 25 MVA 1183 A 3 122 kV LI The field current required to produce a shortcircuit current of 1183 A is about 365 A Therefore the shortcircuit ratio of this generator is 275 A SCR 075 365 A e The internal generated voltage of this generator at rated conditions would be calculated using the saturated synchronous reactance which is about 657 S X 1183 A if the field current is 380 A Since the power factor is 09 lagging the armature current is 258 A I The phase voltage is 12200 3 0 V 7044 0 V V Therefore 102 103 S A A A A R jX E V I I 7044 0 060 1183 258 A 657 1183 258 A A j E A 12930 312 V E f If the internal generated voltage is 12930 V per phase the corresponding line value would be 12930 V 3 22400 V This would require a field current of about 470 A 414 During a shortcircuit test a Yconnected synchronous generator produces 100 A of shortcircuit armature current per phase at a field current of 25 A At the same field current the opencircuit line voltage is measured to be 440 V a Calculate the saturated synchronous reactance under these conditions b If the armature resistance is 03 Ω per phase and the generator supplies 60 A to a purely resistive Yconnected load of 3 Ω per phase at this field current setting determine the voltage regulation under these load conditions SOLUTION a The saturated synchronous reactance at a field current of 25 A can be found from the information supplied in the problem The open circuit line voltage at FI 25 A is 440 V and the shortcircuit current is 100 A Since this generator is Yconnected the corresponding phase voltage is 440 V 3 254 V V and the armature current is AI 100 A Therefore the saturated synchronous reactance is 254 V 254 100 A S X S A b Assume that the desired line voltage is 440 V which means that the phase voltage The armature current is so the internal generated voltage is 254 0 V V A 60 0 A I A A A R jX E V I I 254 0 030 60 0 A 254 60 0 A A j E A 312 293 V E This is also the phase voltage at no load conditions The corresponding line voltage at no load conditions would be nl 312 V 3 540 V VL The voltage regulation is nl fl fl 540 440 VR 100 100 227 440 T T T V V V 415 A threephase Yconnected synchronous generator is rated 120 MVA 138 kV 08PFlagging and 60 Hz Its synchronous reactance is 12 per phase and its armature resistance is 01 per phase a What is its voltage regulation b What would the voltage and apparent power rating of this generator be if it were operated at 50 Hz with the same armature and field losses as it had at 60 Hz c What would the voltage regulation of the generator be at 50 Hz SOLUTION a The rated armature current is 120 MVA 5020 A 3 3 138 kV A L T S I I V The power factor is 08 lagging so 5020 3687 A A I The phase voltage is 138 kV 3 7967 V Therefore the internal generated voltage is A A A R jX E V I SI A 7967 0 01 5020 3687 A 12 5020 3687 A A j E A 12800 207 V E The resulting voltage regulation is 12800 7967 VR 100 607 7967 b If the generator is to be operated at 50 Hz with the same armature and field losses as at 60 Hz so that the windings do not overheat then its armature and field currents must not change Since the voltage of the generator is directly proportional to the speed of the generator the voltage rating and hence the apparent power rating of the generator will be reduced by a factor of 56 rated 5 138 kV 115 kV 6 VT rated 5 120 MVA 100 MVA 6 S Also the synchronous reactance will be reduced by a factor of 56 5 12 100 6 S X c At 50 Hz rated conditions the armature current would be 100 MVA 5020 A 3 3 115 kV A L T S I I V The power factor is 08 lagging so 5020 3687 A A I The phase voltage is 115 kV 3 6640 V Therefore the internal generated voltage is A A A R jX E V I SI A 6640 0 01 5020 3687 A 10 5020 3687 A A j E A 10300 188 V E The resulting voltage regulation is 10300 6640 VR 100 551 6640 Problems 416 to 426 refer to a sixpole Yconnected synchronous generator rated at 500 kVA 32 kV 09 PF lagging and 60 Hz Its armature resistance is 07 The core losses of this generator at rated conditions are 8 kW and the friction and windage losses are 10 kW The opencircuit and shortcircuit characteristics are shown in Figure P42 RA 104 Note An electronic version of the saturated open circuit characteristic can be found in file p42occdat and an electronic version of the airgap characteristic can be found in file p42agoccdat These files can be used with MATLAB programs Column 1 contains field current in amps and column 2 contains opencircuit terminal voltage in volts An electronic version of the short circuit characteristic can be found in file p42sccdat Column 1 contains field current in amps and column 2 contains shortcircuit terminal current in amps 416 a What is the saturated synchronous reactance of this generator at the rated conditions b What is the unsaturated synchronous reactance of this generator c Plot the saturated synchronous reactance of this generator as a function of load SOLUTION a The rated armature current for this generator is 500 kVA 902 A 3 3 3200 V A L T S I I V The field current required to produce this much shortcircuit current may be read from the SCC It is 070 A1 The open circuit voltage at 070 A is 2447 V2 so the opencircuit phase voltage A E is 2447 3 1313 V The approximate saturated synchronous reactance S X is 1413 V 157 902 A S X b The unsaturated synchronous reactance XSu is the ratio of the airgap line to the SCC This is a straight line so we can determine its value by comparing the ratio of the airgap voltage to the short circuit current at any given field current For example at FI 15 A the airgap line voltage is 4470 V and the SCC is 175 A 4470 V 3 1475 175 A XSu c This task can best be performed with MATLAB The opencircuit characteristic is available in a file called p42occdat and the shortcircuit characteristic is available in a file called p42sccdat Each of these files are organized in two columns where the first column is field current and the second 1 If you have MATLAB available you can use the file p42sccdat and the interp1 function to look up this value as shown below Note that column 1 of p42scc contains field current and column 2 contains shortcircuit terminal current load p42sccdat if1 interp1p42scc2p42scc1902 if1 07032 2 If you have MATLAB available you can use the file p42occdat and the interp1 function to look up this value as shown below Note that column 1 of p42occ contains field current and column 2 contains opencircuit terminal voltage load p42occdat vt interp1p42occ1p42occ2070 vt 2447 106 column is either opencircuit terminal voltage or shortcircuit current A program to read these files and calculate and plot S is shown below X Mfile prob416cm Mfile to calculate and plot the saturated synchronous reactance of a synchronous generator Load the opencircuit characteristic It is in two columns with the first column being field current and the second column being terminal voltage load p42occdat ifocc p42occ1 vtocc p42occ2 Load the shortcircuit characteristic It is in two columns with the first column being field current and the second column being line current armature current load p42sccdat ifscc p42scc1 iascc p42scc2 Calculate Xs if1 0010014 Current steps vt interp1ifoccvtoccif1 Terminal voltage ia interp1ifscciasccif1 Current Xs vt sqrt3 ia Plot the synchronous reactance figure1 plotif1XsLineWidth20 title bfSaturated Synchronous Reactance itXs rm xlabel bfField Current A ylabel bfitXs rmbfOmega grid on 107 The resulting plot is 417 a What are the rated current and internal generated voltage of this generator b What field current does this generator require to operate at the rated voltage current and power factor SOLUTION a The rated line and armature current for this generator is 500 kVA 902 A 3 3 3200 V A L T S I I V The power factor is 09 lagging so 902 258 A A I The rated phase voltage is V 32 kV 3 1850 V The saturated synchronous reactance at rated conditions was found to be 157 in the previous problem Therefore the internal generated voltage is A A A R jX E V I I S A 1850 0 07 902 258 A 157 902 258 A A j E A 2814 263 V E b This internal generated voltage corresponds to a noload terminal voltage of 3 2814 4874 V From the opencircuit characteristic the required field current would be 32 A 418 What is the voltage regulation of this generator at the rated current and power factor SOLUTION The voltage regulation is nl fl fl 4874 3200 VR 100 100 523 3200 T T T V V V 419 If this generator is operating at the rated conditions and the load is suddenly removed what will the terminal voltage be 108 SOLUTION If the load is suddenly removed V The internal generated voltage at full load was 2814 V so V EA A E 2814 V after the load is removed Since the generator is Yconnected 3 4874 V VT V when the load is removed 420 What are the electrical losses in this generator at rated conditions SOLUTION The current flow at rated conditions is AI 902 A so the electrical losses are 2 2 CU 3 3 902 A 07 171 kW A A P I R 421 If this machine is operating at rated conditions what input torque must be applied to the shaft of this generator Express your answer both in newtonmeters and in poundfeet SOLUTION To get the applied torque we must know the input power The input power to this generator is equal to the output power plus losses The rated output power and the losses are OUT 1 MVA 09 900 kW P 2 2 CU 3 3 902 A 07 171 kW A A P I R FW 10 kW P core 8 kW P stray assumed 0 P IN OUT CU FW core stray 9351 kW P P P P P P Therefore the applied torque is IN APP 9351 kW 7441 N m 2 rad 1 min 1200 rmin 1 r 60 s m P or APP 704 9351 kW 704 5486 lb ft 1200 rmin m P n 422 What is the torque angle of this generator at rated conditions SOLUTION From the calculations in Problem 417 263 423 Assume that the generator field current is adjusted to supply 3200 V under rated conditions What is the static stability limit of this generator Note You may ignore to make this calculation easier How close is the fullload condition of this generator to the static stability limit A R SOLUTION From Problem 417 the phase voltage of this generator at rated conditions is 1850 0 V and the internal generated voltage at rated conditions is A 2814 263 V E Therefore the static stability limit is MAX 3 3 1850 V 2814 V 995 kW 157 A S V E P X The maximum possible power of this generator is about twice the fullload rated power 109 424 Assume that the generator field current is adjusted to supply 3200 V under rated conditions Plot the power supplied by the generator as a function of the torque angle SOLUTION The power supplied by the generator as a function of the torque angle is given by the equation 3 sin A S V E P X A MATLAB program that calculates terminal voltage as function of impedance angle is shown below Mfile prob424m Mfile to calculate and plot the power supplied by the generator as a function of torque angle assuming that the field current has been adjusted to supply rated voltage at full load conditions Define values for this generator EA 2814 Internal gen voltage V VP 1850 Phase voltage V XS 157 XS ohms Calculate power vs torque angle delta 00190 power 3 VP EA XS sindelta pi180 Plot the power vs torque angle figure1 plotdeltapower1000bLineWidth20 title bfGenerator Power vs Torque Angle delta xlabel bfTorque Angle deg ylabel bfOutput Power kW grid on hold off The resulting plot of output power vs torque angle is 110 425 Assume that the generators field current is adjusted so that the generator supplies rated voltage at the rated load current and power factor If the field current and the magnitude of the load current are held constant how will the terminal voltage change as the load power factor varies from 09 PF lagging to 09 PF leading Make a plot of the terminal voltage versus the load impedance angle SOLUTION If the field current is held constant then the magnitude of will be constant although its angle will vary Also the magnitude of the armature current is constant Since we also know and the current angle we know enough to find the phase voltage and therefore the terminal voltage At lagging power factors can be found from the following relationships A E E RA S X V VT V I A V A I R A A jX SI A By the Pythagorean Theorem 2 2 2 sin cos sin cos S A A S A S A A A R I X I X I R I V E sin cos sin cos 2 2 A S A A S A A S A X I R I R I X I E V At unity power factor can be found from the following relationships V I A V EA I R A A jX SI A By the Pythagorean Theorem 2 2 2 A S E V X I A 2 2 A S A X I E V At leading power factors can be found from the following relationships V I A V EA I R A A jX SI A 111 By the Pythagorean Theorem 2 2 2 sin cos sin cos S A A S A S A A A R I X I X I R I V E sin cos sin cos 2 2 A S A A S A A S A X I R I R I X I E V If we examine these three cases we can see that the only difference among them is the sign of the term sin If is taken as positive for lagging power factors and negative for leading power factors in other words if is the impedance angle then all three cases can be represented by the single equation sin cos sin cos 2 2 A S A A S A A S A X I R I R I X I E V A MATLAB program that calculates terminal voltage as function of impedance angle is shown below Mfile prob425m Mfile to calculate and plot the terminal voltage of a synchronous generator as a function of impedance angle as PF changes from 090 lagging to 090 leading Define values for this generator EA 2814 Internal gen voltage V VP 1850 Phase voltage V RA 07 Armature resistance ohms XS 157 XS ohms IA 902 Current A Calculate impedance angle theta in degrees for the specified range of power factors theta 2580258258 In degrees th theta pi180 In radians Calculate the phase voltage and terminal voltage VP sqrt EA2 XSIAcosth RAIAsinth2 RAIAcosth XSIAsinth VT VP sqrt3 Plot the terminal voltage versus power factor figure1 plotthetaabsVT1000bLineWidth20 title bfTerminal Voltage Versus Impedance Angle xlabel bfImpedance Angle deg ylabel bfTerminal Voltage kV grid on hold off The resulting plot of terminal voltage versus impedance angle with field and armature currents held constant is shown below 112 426 Assume that the generator is connected to a 3200V infinite bus and that its field current has been adjusted so that it is supplying rated power and power factor to the bus You may ignore the armature resistance A when answering the following questions R a What would happen to the real and reactive power supplied by this generator if the field flux and therefore A is reduced by 5 percent E b Plot the real power supplied by this generator as a function of the flux as the flux is varied from 80 to 100 of the flux at rated conditions c Plot the reactive power supplied by this generator as a function of the flux as the flux is varied from 80 to 100 of the flux at rated conditions d Plot the line current supplied by this generator as a function of the flux as the flux is varied from 80 to 100 of the flux at rated conditions SOLUTION a If the field flux is reduced by 5 nothing would happen to the real power The reactive power supplied would decrease as shown below V EA1 jX SI A Qsys QG Q 2 Q 1 EA2 IA2 IA1 VT Q I sin A b If armature resistance is ignored the power supplied to the bus will not change as flux is varied Therefore the plot of real power versus flux is 113 c If armature resistance is ignored the internal generated voltage will increase as flux increases but the quantity A E EAsin will remain constant Therefore the voltage for any flux can be found from the expression Ar r A E E and the angle for any A can be found from the expression E r A Ar E E sin sin 1 where is the flux in the machine r is the flux at rated conditions is the magnitude of the internal generated voltage at rated conditions and Ar E r is the angle of the internal generated voltage at rated conditions From this information we can calculate A for any given load from equation I S A A jX V E I and the resulting reactive power from the equation sin 3 Q V IA where is the impedance angle which is the negative of the current angle The rated line and armature current for this generator is 500 kVA 902 A 3 3 3200 V A L T S I I V 114 The power factor is 09 lagging so 902 258 A A I The rated phase voltage is V 32 kV 3 1850 V The saturated synchronous reactance at rated conditions was found to be 157 in the previous problem Therefore the internal generated voltage ignoring A is R A jXS E V I A 1850 0 157 902 258 A A j E A 2776 273 V E so 2776 V and EAr r 273 A MATLAB program that calculates the reactive power supplied voltage as a function of flux is shown below Mfile prob426cm Mfile to calculate and plot the reactive power supplied to an infinite bus as flux is varied from 80 to 100 of the flux at rated conditions Define values for this generator fluxratio 080001100 Flux ratio Ear 2776 Ea at full flux dr 273 pi180 Torque ang at full flux Vp 1850 Phase voltage Xs 157 Xs ohms Calculate Ea for each flux Ea fluxratio Ear Calculate delta for each flux d asin Ear Ea sindr Calculate Ia for each flux Ea Ea cosd jsind Ia Ea Vp jXs Calculate reactive power for each flux theta atan2imagIarealIa Q 3 Vp absIa sintheta Plot the power supplied versus flux figure1 plotfluxratio100Q1000bLineWidth20 title bfReactive power versus flux xlabel bfFlux of fullload flux ylabel bfitQrmbf kVAR grid on hold off 115 When this program is executed the plot of reactive power versus flux is d The program in part c of this program calculated as a function of flux A MATLAB program that plots the magnitude of this current as a function of flux is shown below A I Mfile prob426dm Mfile to calculate and plot the armature current supplied to an infinite bus as flux is varied from 80 to 100 of the flux at rated conditions Define values for this generator fluxratio 080001100 Flux ratio Ear 2776 Ea at full flux dr 273 pi180 Torque ang at full flux Vp 1850 Phase voltage Xs 157 Xs ohms Calculate Ea for each flux Ea fluxratio Ear Calculate delta for each flux d asin Ear Ea sindr Calculate Ia for each flux Ea Ea cosd jsind Ia Ea Vp jXs Plot the armature current versus flux figure1 plotfluxratio100absIabLineWidth20 title bfArmature current versus flux xlabel bfFlux of fullload flux ylabel bfitIArmbf A grid on hold off 116 When this program is executed the plot of armature current versus flux is Note that if the flux variation started at less than 80 of full load flux we would see the armature current both fall and rise in a classic V curve For example if the flux varied from 65 to 100 of fullload flux the curve would be as shown below 427 Two identical 25MVA 1200V 08PFlagging 60 Hz threephase synchronous generators are connected in parallel to supply a load The prime movers of the two generators happen to have different speed droop characteristics When the field currents of the two generators are equal one delivers 1200 A at 09 PF lagging while the other delivers 900 A at 075 PF lagging a What are the real power and the reactive power supplied by each generator to the load b What is the overall power factor of the load 117 c In what direction must the field current on each generator be adjusted in order for them to operate at the same power factor SOLUTION a The real and reactive powers are 1 3 cos 3 1200 V 1200 A 09 2245 kW T L P V I 1 1 3 sin 3 1200 V 1200 A sin cos 09 1087 kVAR T L Q V I 2 3 cos 3 1200 V 900 A 075 1403 kW T L P V I 1 2 3 sin 3 1200 V 900 A sin cos 075 1237 kVAR T L Q V I b The overall power factor can be found from the total real and reactive power supplied to the load TOT 1 2 2245 kW 1403 kW 3648 kW P P P TOT 1 2 1087 kVAR 1237 kVAR 2324 kVAR Q Q Q The overall power factor is 1 TOT TOT PF cos tan 0843 lagging Q P c The field current of generator 1 should be increased and the field current of generator 2 should be simultaneously decreased 428 A generating station for a power system consists of four 300MVA 15kV 085PFlagging synchronous generators with identical speed droop characteristics operating in parallel The governors on the generators prime movers are adjusted to produce a 3Hz drop from no load to full load Three of these generators are each supplying a steady 200 MW at a frequency of 60 Hz while the fourth generator called the swing generator handles all incremental load changes on the system while maintaining the systems frequency at 60 Hz a At a given instant the total system loads are 650 MW at a frequency of 60 Hz What are the no load frequencies of each of the systems generators b If the system load rises to 725 MW and the generators governor set points do not change what will the new system frequency be c To what frequency must the noload frequency of the swing generator be adjusted in order to restore the system frequency to 60 Hz d If the system is operating at the conditions described in part c what would happen if the swing generator were tripped off the line disconnected from the power line SOLUTION a The fullload power of these generators is 300 MVA 085 255 MW and the droop from no load to fullload is 3 Hz Therefore the slope of the powerfrequency curve for these four generators is 255 MW 85 MWHz 3 Hz Ps If generators 1 2 and 3 are supplying 200 MW each then generator 4 must be supplying 50 MW The noload frequency of the first three generators is 118 1 1 nl1 sys P P s f f nl1 200 MW 85 MWHz 60 Hz f nl1 6235 Hz f The noload frequency of the fourth generator is 4 4 nl4 sy P P s f f s nl1 50 MW 85 MWHz 60 Hz f nl1 6059 Hz f b The setpoints of generators 1 2 3 and 4 do not change so the new system frequency will be LOAD 1 nl1 sys 2 nl2 sys 3 nl3 sys 4 nl4 sys P P P P P s f f s f f s f f s f f sys sys sys sys 725 MW 85 6235 85 6235 85 6235 85 6059 f f f f sys 725 2104940 340 f sys 2032440 340 f sys 5978 Hz f c The governor setpoints of the swing generator must be increased until the system frequency rises back to 60 Hz At 60 Hz the other three generators will be supplying 200 MW each so the swing generator must supply 725 MW 3200 MW 125 MW at 60 Hz Therefore the swing generators setpoints must be set to 4 4 nl4 sy P P s f f s nl1 125 MW 85 MWHz 60 Hz f nl1 6147 Hz f d If the swing generator trips off the line the other three generators would have to supply all 725 MW of the load Therefore the system frequency will become LOAD 1 nl1 sys 2 nl2 sys 3 nl3 sys P P P P s f f s f f s f f sys sys sys 725 MW 85 6235 85 6235 85 6235 f f f sys 725 1589925 255 f sys 1517425 255 f sys 5951 Hz f Each generator will supply 2417 MW to the loads 429 A 100MVA 144kV 08PFlagging Yconnected synchronous generator has a negligible armature resistance and a synchronous reactance of 10 perunit The generator is connected in parallel with a 60 Hz 144kV infinite bus that is capable of supplying or consuming any amount of real or reactive power with no change in frequency or terminal voltage 119 a What is the synchronous reactance of the generator in ohms b What is the internal generated voltage E of this generator under rated conditions A c What is the armature current in this machine at rated conditions I A d Suppose that the generator is initially operating at rated conditions If the internal generated voltage E A is decreased by 5 percent what will the new armature current I be A e Repeat part d for 10 15 20 and 25 percent reductions in E A f Plot the magnitude of the armature current as a function of You may wish to use MATLAB to create this plot I A EA SOLUTION a The rated phase voltage of this generator is 144 kV 3 8313 V The base impedance of this generator is 2 2 base base base 3 3 8313 V 207 100000000 VA V Z S Therefore 0 negligible A R 10 207 207 S X b The rated armature current is 100 MVA 4009 A 3 3 144 kV A L T S I I V The power factor is 08 lagging so 4009 3687 A A I Therefore the internal generated voltage is A A A R jX E V I SI A 8313 0 207 4009 3687 A A j E A 14858 2654 V E c From the above calculations 4009 3687 A A I d If is decreased by 5 the armature current will change as shown below Note that the infinite bus will keep and A E V m constant Also since the prime mover hasnt changed the power supplied by the generator will be constant V EA1 jX SI A EA2 I A2 I A1 Q I sin A 120 3 sin constant A S V E P X so 1 1 2 sin sin A A E E 2 With a 5 decrease and 2 14115 V EA 1 1 1 2 2 2 14858 V sin sin sin sin 2654 280 14115 V A A E E Therefore the new armature current is 2 14115 280 8313 0 3777 321 A 207 A A jXS j E V I e Repeating part d With a 10 decrease and 2 13372 V EA 1 1 1 2 2 2 14858 V sin sin sin sin 2654 298 13372 V A A E E Therefore the new armature current is 2 13372 298 8313 0 3582 263 A 207 A A jXS j E V I With a 15 decrease and 2 12629 V EA 1 1 1 2 2 2 14858 V sin sin sin sin 2654 317 12629 V A A E E Therefore the new armature current is 2 12629 317 8313 0 3414 201 A 207 A A jXS j E V I With a 20 decrease and 2 11886 V EA 1 1 1 2 2 2 14858 V sin sin sin sin 2654 340 11886 V A A E E Therefore the new armature current is 2 11886 340 8313 0 3296 131 A 207 A A jXS j E V I With a 25 decrease and 2 11144 V EA 1 1 1 2 2 2 14858 V sin sin sin sin 2654 366 11144 V A A E E Therefore the new armature current is 2 11144 366 8313 0 3224 54 A 207 A A jXS j E V I 121 122 f A MATLAB program to plot the magnitude of the armature current as a function of is shown below IA A E Mfile prob429fm Mfile to calculate and plot the armature current supplied to an infinite bus as Ea is varied Define values for this generator Ea 05500110014858 Ea Vp 8313 Phase voltage d1 2654pi180 Torque angle at full Ea Xs 818 Xs ohms Calculate delta for each Ea d asin 14858 Ea sind1 Calculate Ia for each flux Ea Ea cosd jsind Ia Ea Vp jXs Plot the armature current versus Ea figure1 plotabsEa1000absIabLineWidth20 title bfArmature current versus itEArm xlabel bfitEArmbf kV ylabel bfitIArmbf A grid on hold off The resulting plot is shown below Chapter 5 Synchronous Motors 51 A 480V 60 Hz 400hp 08PFleading eightpole connected synchronous motor has a synchronous reactance of 06 and negligible armature resistance Ignore its friction windage and core losses for the purposes of this problem Assume that E A is directly proportional to the field current FI in other words assume that the motor operates in the linear part of the magnetization curve and that E A 480 V when FI 4 A a What is the speed of this motor b If this motor is initially supplying 400 hp at 08 PF lagging what are the magnitudes and angles of and E A I A c How much torque is this motor producing What is the torque angle How near is this value to the maximum possible induced torque of the motor for this field current setting d If E A is increased by 30 percent what is the new magnitude of the armature current What is the motors new power factor e Calculate and plot the motors Vcurve for this load condition SOLUTION a The speed of this motor is given by 120 60 Hz 120 900 rmin 8 se m f n P b If losses are being ignored the output power is equal to the input power so the input power will be IN 400 hp 746 Whp 2984 kW P This situation is shown in the phasor diagram below V EA jX SI A I A The line current flow under these circumstances is 2984 kW 449 A 3 PF 3 480 V 08 L T P I V Because the motor is connected the corresponding phase current is 449 3 259 A A I The angle of the current is so cos 1 080 3687 259 3687 A A I The internal generated voltage is A E A jXS E V IA 480 0 V 06 259 3687 A 406 178 V A j E 123 c This motor has 6 poles and an electrical frequency of 60 Hz so its rotation speed is 1200 rmin The induced torque is m n OUT ind 2984 kW 3166 N m 1 min 2 rad 900 rmin 60 s 1 r m P The maximum possible induced torque for the motor at this field setting is the maximum possible power divided by m indmax 3 3 480 V 406 V 10340 N m 1 min 2 rad 900 rmin 06 60 s 1 r A m S V E X The current operating torque is about 13 of the maximum possible torque d If the magnitude of the internal generated voltage is increased by 30 the new torque angle can be found from the fact that A E constant sin P E A 2 1 130 130 406 V 4872 V A A E E 1 1 1 2 1 2 406 V sin sin sin sin 178 148 4872 V A A E E The new armature current is 2 2 480 0 V 4872 148 V 208 41 A 06 A A jXS j V E I The magnitude of the armature current is 208 A and the power factor is cos 241 0913 lagging e A MATLAB program to calculate and plot the motors Vcurve is shown below Mfile prob51em Mfile create a plot of armature current versus Ea for the synchronous motor of Problem 51 Initialize values Ea 090001170406 Magnitude of Ea volts Ear 406 Reference Ea deltar 178 pi180 Reference torque angle Xs 06 Synchronous reactance ohms Vp 480 Phase voltage at 0 degrees Ear Ear cosdeltar j sindeltar Calculate delta2 delta2 asin absEar absEa sindeltar Calculate the phasor Ea Ea Ea cosdelta2 j sindelta2 Calculate Ia Ia Vp Ea j Xs Plot the vcurve 124 figure1 plotabsEaabsIabLinewidth20 xlabelbfitEArmbf V ylabelbfitIArmbf A title bfSynchronous Motor VCurve grid on The resulting plot is shown below 52 Assume that the motor of Problem 51 is operating at rated conditions a What are the magnitudes and angles of E and and A I A FI b Suppose the load is removed from the motor What are the magnitudes and angles of and I now E A A SOLUTION a The line current flow at rated conditions is 2984 kW 449 A 3 PF 3 480 V 08 L T P I V Because the motor is connected the corresponding phase current is 449 3 259 A AI The angle of the current is cos 1 080 3687 so A 259 3687 A I The internal generated voltage A is E A jXS E V IA 480 0 V 06 259 3687 A 587 122 V A j E The field current is directly proportional to E A with 480 V when FI 4 A Since the real E A is 587 V the required field current is 2 2 1 1 A F A F I I E E 125 2 2 1 1 587 V 4 A 489 A 480 V A F F A I I E E b When the load is removed from the motor the magnitude of E A remains unchanged but the torque angle goes to 0 The resulting armature current is 480 0 V 587 0 1783 90 A 06 A A jXS j V E I 53 A 230V 50 Hz twopole synchronous motor draws 40 A from the line at unity power factor and full load Assuming that the motor is lossless answer the following questions a What is the output torque of this motor Express the answer both in newtonmeters and in pound feet b What must be done to change the power factor to 085 leading Explain your answer using phasor diagrams c What will the magnitude of the line current be if the power factor is adjusted to 085 leading SOLUTION a If this motor is assumed lossless then the input power is equal to the output power The input power to this motor is IN 3 cos 3 230 V 40 A 10 1593 kW T L P V I The rotational speed of the motor is 120 50 Hz 120 1500 rmin 4 se m f n P The output torque would be OUT LOAD 1593 kW 1014 N m 1 min 2 rad 1500 rmin 60 s 1 r m P In English units OUT LOAD 704 1593 kW 704 748 lb ft 1500 rmin m P n b To change the motors power factor to 08 leading its field current must be increased Since the power supplied to the load is independent of the field current level an increase in field current increases EA while keeping the distance EA sin constant This increase in changes the angle of the current eventually causing it to reach a power factor of 08 leading A E A I 126 V EA1 jX SI A EA2 IA2 IA1 Q I sin A P P c The magnitude of the line current will be 1593 kW 500 A 3 PF 3 230 V 08 L T P I V 54 A 2300V 1000hp 08PF leading 60Hz twopole Yconnected synchronous motor has a synchronous reactance of 25 and an armature resistance of 03 At 60 Hz its friction and windage losses are 30 kW and its core losses are 20 kW The field circuit has a dc voltage of 200 V and the maximum is 10 A The opencircuit characteristic of this motor is shown in Figure P51 Answer the following questions about the motor assuming that it is being supplied by an infinite bus IF a How much field current would be required to make this machine operate at unity power factor when supplying full load b What is the motors efficiency at full load and unity power factor 127 c If the field current were increased by 5 percent what would the new value of the armature current be What would the new power factor be How much reactive power is being consumed or supplied by the motor d What is the maximum torque this machine is theoretically capable of supplying at unity power factor At 08 PF leading Note An electronic version of this open circuit characteristic can be found in file p51occdat which can be used with MATLAB programs Column 1 contains field current in amps and column 2 contains opencircuit terminal voltage in volts SOLUTION a At full load the input power to the motor is CU core mech OUT IN P P P P P We cant know the copper losses until the armature current is known so we will find the input power and armature current ignoring that term and then correct the input power after we know it IN 1000 hp 746 Whp 30 kW 20 kW 796 kW P Therefore the line and phase current at unity power factor is 796 kW 200 A 3 PF 3 2300 V 10 A L T P I I V The copper losses due to a current of 200 A are 2 2 CU 3 3 200 A 03 360 kW A A P I R Therefore a better estimate of the input power at full load is IN 1000 hp 746 Whp 30 kW 20 kW 36 kW 832 kW P and a better estimate of the line and phase current at unity power factor is 832 kW 209 A 3 PF 3 2300 V 10 A L T P I I V The phasor diagram of this motor operating a unity power factor is shown below jX SI A V I A EA I R A A The phase voltage of this motor is 2300 3 1328 V The required internal generated voltage is A A A R jX E V I SI A 1328 0 V 03 209 0 A 25 209 0 A A j E 1370 2244 V A E 128 This internal generated voltage corresponds to a terminal voltage of 3 1370 2371 V This voltage would require a field current of 454 A b The motors efficiency at full load and unity power factor is OUT IN 746 kW 100 100 897 832 kW P P c To solve this problem we will temporarily ignore the effects of the armature resistance If is ignored then RA A R EA sin is directly proportional to the power supplied by the motor Since the power supplied by the motor does not change when is changed this quantity will be a constant FI If the field current is increased by 5 then the new field current will be 477 A and the new value of the opencircuit terminal voltage will be 2450 V The new value of EA will be 2435 V 3 1406 V Therefore the new torque angle will be 1 1 1 2 1 2 1370 V sin sin sin sin 2244 239 1406 V A A E E Therefore the new armature current will be 1328 0 V 1406 239 V 227 26 A 03 25 A A A S R jX j V E I The new current is about the same as before but the phase angle has become positive The new power factor is cos 26 0999 leading and the reactive power supplied by the motor is 3 sin 3 2300 V 227 A sin 26 410 kVAR T L Q V I d The maximum torque possible at unity power factor ignoring the effects of A is R indmax 3 3 1328 V 1370 V 5790 N m 1 min 2 rad 3600 rmin 25 60 s 1 r A m S V E X If we are ignoring the resistance of the motor then the input power would be 7968 kW note that copper losses are ignored At a power factor of 08 leading the current flow will be 796 kW 250 A 3 PF 3 2300 V 08 A L T P I I V so A 250 3687 A The internal generated voltage at 08 PF leading ignoring copper losses is I A A A R jX E V I SI A 1328 0 V 25 250 3687 A A j E 1775 164 V A E Therefore the maximum torque at a power factor of 08 leading is indmax 3 3 1328 V 1775 V 7503 N m 1 min 2 rad 3600 rmin 25 60 s 1 r A m S V E X 129 55 Plot the Vcurves versus for the synchronous motor of Problem 54 at noload halfload and fullload conditions Note that an electronic version of the opencircuit characteristics in Figure P51 is available at the books Web site It may simplify the calculations required by this problem I A IF Note This problem can be greatly simplified if the armature resistance and copper losses are ignored This solution does not ignore them Instead it tries to estimate the copper losses by first getting an estimate of the armature current SOLUTION The input power at noload halfload and fullload conditions is given below Note that we are assuming that A is negligible in each case R INnl 30 kW 20 kW 50 kW P INhalf 500 hp 746 Whp 30 kW 20 kW 423 kW P INfull 1000 hp 746 Whp 30 kW 20 kW 796 kW P If the power factor is adjusted to unity then armature currents will be nl 50 kW 126 A 3 PF 3 2300 V 10 A T P I V half 423 kW 106 A 3 PF 3 2300 V 10 A T P I V full 796 kW 200 A 3 PF 3 2300 V 10 A T P I V The copper losses due to the armature currents are 2 2 CUnl 3 3 126 A 03 014 kW A A P I R 2 2 CUhalf 3 3 106 A 03 101 kW A A P I R 2 2 CUfull 3 3 200 A 03 360 kW A A P I R Therefore a better estimate of the input power at the various loads is INnl 30 kW 20 kW 014 kW 501 kW P INhalf 500 hp 746 Whp 30 kW 20 kW 101 kW 433 kW P INfull 1000 hp 746 Whp 30 kW 20 kW 36 kW 832 kW P and a better estimate of the line and phase current at unity power factor is nl 501 kW 126 A 3 PF 3 2300 V 10 A T P I V half 433 kW 109 A 3 PF 3 2300 V 10 A T P I V full 832 kW 209 A 3 PF 3 2300 V 10 A L T P I I V 130 The corresponding internal generated voltages at unity power factor are 131 S A A A A R jX E V I I nl 1328 0 V 03 126 0 A 25 126 0 A 1325 13 V A j E half 1328 0 V 03 109 0 A 25 109 0 A 1324 119 V A j E full 1328 0 V 03 209 0 A 25 209 0 A 1369 224 V A j E These values of and at unity power factor can serve as reference points in calculating the synchronous motor Vcurves The MATLAB program to solve this problem is shown below A E Mfile prob55m Mfile create a plot of armature current versus field current for the synchronous motor of Problem 64 at noload halfload and fullload First initialize the field current values 21 values in the range 3858 A if1 25018 Get the OCC load p51occdat ifvalues p51occ1 vtvalues p51occ2 Now initialize all other values Xs 25 Synchronous reactance Vp 1328 Phase voltage The following values of Ea and delta are for unity power factor They will serve as reference values when calculating the Vcurves dnl 13 pi180 delta at noload dhalf 119 pi180 delta at halfload dfull 224 pi180 delta at fullload Eanl 1325 Ea at noload Eahalf 1324 Ea at halfload Eafull 1369 Ea at fullload Calculate the actual Ea corresponding to each level of field current Ea interp1ifvaluesvtvaluesif1 sqrt3 Calculate the armature currents associated with each value of Ea for the noload case First calculate delta delta asin Eanl Ea sindnl Calculate the phasor Ea Ea2 Ea cosdelta j sindelta Now calculate Ia Ianl Vp Ea2 j Xs 132 Calculate the armature currents associated with each value of Ea for the halfload case First calculate delta delta asin Eahalf Ea sindhalf Calculate the phasor Ea Ea2 Ea cosdelta j sindelta Now calculate Ia Iahalf Vp Ea2 j Xs Calculate the armature currents associated with each value of Ea for the fullload case First calculate delta delta asin Eafull Ea sindfull Calculate the phasor Ea Ea2 Ea cosdelta j sindelta Now calculate Ia Iafull Vp Ea2 j Xs Plot the vcurves hold off plotif1absIanlkLinewidth20 hold on plotif1absIahalfbLinewidth20 plotif1absIafullrLinewidth20 xlabelbfField Current A ylabelbfArmature Current A title bfSynchronous Motor VCurve legendNo loadHalf loadFull load grid on The resulting plot is shown below The flattening visible to the right of the Vcurves is due to magnetic saturation in the machine 56 If a 60Hz synchronous motor is to be operated at 50 Hz will its synchronous reactance be the same as at 60 Hz or will it change Hint Think about the derivation of X S SOLUTION The synchronous reactance represents the effects of the armature reaction voltage and the armature selfinductance The armature reaction voltage is caused by the armature magnetic field and the amount of voltage is directly proportional to the speed with which the magnetic field sweeps over the stator surface The higher the frequency the faster sweeps over the stator and the higher the armature reaction voltage is Therefore the armature reaction voltage is directly proportional to frequency Similarly the reactance of the armature selfinductance is directly proportional to frequency so the total synchronous reactance is directly proportional to frequency If the frequency is changed from 60 Hz to 50 Hz the synchronous reactance will be decreased by a factor of 56 stat E S B S B stat E X S 57 A 208V Yconnected synchronous motor is drawing 50 A at unity power factor from a 208V power system The field current flowing under these conditions is 27 A Its synchronous reactance is 16 Assume a linear opencircuit characteristic a Find V and A for these conditions E b Find the torque angle c What is the static stability power limit under these conditions d How much field current would be required to make the motor operate at 080 PF leading e What is the new torque angle in part d SOLUTION a The phase voltage of this motor is V 120 V and the armature current is A 50 0 A I Therefore the internal generated voltage is A A A R jX E V I SI A 120 0 V 16 50 0 A A j E 144 337 V A E 133 134 que angle of this machine is 337 b The tor c The static stability power limit is given by max W 16 XS of the motor operating at a power factor of 078 leading is shown below 3 3 120 V 144 V 324 k V EA P d A phasor diagram V EA1 jX SI A EA2 IA2 IA1 P P Since the power supplied by the motor is constant the quantity IA cos which is directly proportional generated voltag required to produce this current would be is directly proportional to the field flux and we have assumed in this proportional to the fiel to power must be constant Therefore 2 08 50 A 100 AI A2 625 3687 A I The internal e 2 2 2 A A A S A R jX E V I I 2 120 0 V 16 6250 3687 A A j E 2 A 197 239 V E The internal generated voltage A E problem that the flux is directly d current Therefore the required field current is 2 2 1 197 V 27 A 370 A A F F E I I 1 144 V A E new torque angle of this machine is 239 8 ected threephase synchronous motor has a sistance of 01 per unit If this motor is running power of the motor is e The 5 A 412 kV 60 Hz 3000hp 08PFleading Δconn synchronous reactance of 11 per unit and an armature re at rated voltage with a line current of 300 A at 085 PF leading what is the internal generated voltage per phase inside this motor What is the torque angle δ SOLUTION 3000 hp 746 Whp 2238 kW The output If we take this as rated power the ratings of this machine are base PF 2238 k S P W 08 2798 kVA base 4120 V VL base 4120 V V base base base 2798 kVA 392 A 3 3 4120 V L L S I V base base 392 A 226 A 3 3 LI I Therefore the line current of 300 A in perunit is pu base 300 A 0765 pu 392 A L L L I I I and the final perunit current is pu 0765 318 pu I The internal generated voltage in perunit is A A A R jX E V I SI A 1 0 01 0765 318 11 0765 318 A j E 1572 287 pu A E In volts the internal generated voltage is 1572 287 pu 4120 V 6477 287 V A E And the torque angle δ is 287 59 Figure P52 shows a synchronous motor phasor diagram for a motor operating at a leading power factor with no For this motor the torque angle is given by RA cos tan sin S A S A X I V X I 1 cos tan sin S A S A X I V X I Derive an equation for the torque angle of the synchronous motor if the armature resistance is included SOLUTION The phasor diagram with the armature resistance considered is shown below 135 jX SI A V I A EA I R A A I R A A cos X SI A sin X SI A cos Therefore cos sin tan sin cos S A A A S A A A X I R I V X I R I 1 cos sin tan sin cos S A A A S A A A X I R I V X I R I 510 A synchronous machine has a synchronous reactance of 10 per phase and an armature resistance of 01 per phase If E 46010 V and 4800 V is this machine a motor or a generator How much power P is this machine consuming from or supplying to the electrical system How much reactive power Q is this machine consuming from or supplying to the electrical system A V SOLUTION This machine is a motor consuming power from the power system because EA is lagging V It is also consuming reactive power because EA cos V The current flowing in this machine is 480 0 V 460 10 V 839 13 A 01 10 A A A S R jX j V E I Therefore the real power consumed by this motor is 3 cos 3 480 V 893 A cos 13 1253 kW A P V I and the reactive power consumed by this motor is 3 sin 3 480 V 893 A sin 13 289 kVAR A Q V I 511 A 500 kVA 600 V 08PFleading Yconnected synchronous motor has a synchronous reactance of 10 per unit and an armature resistance of 01 per unit At the current time E 10012 pu and 10 pu A V a Is this machine currently acting as a motor or a generator b How much power P is this machine consuming from or supplying to the electrical system c How much reactive power Q is this machine consuming from or supplying to the electrical system d Is this machine operating within its rated limits SOLUTION a This machine is a generator supplying power to the power system because EA is leading V 136 It is also consuming reactive power because EA cos V b The perunit current flowing in this machine assuming that it is a generator is A A A R jX E V I S I A 1 12 pu 1 0 pu 0208 117 A 01 10 A A A S R jX j E V I The current angle in this generator is 117 so the impedance angle is 117 Therefore the real power supplied to the power system by this machine is 3 cos 3 10 0208 cos 117 0611 pu A P V I Converting from perunit to real power in watts we get base pu 500 kVA 0611 pu 3055 kW P S P c The reactive power consumed by this motor is 3 sin 3 10 0208 sin 117 0127 pu A Q V I Converting from perunit to reactive power in var we get base pu 500 kVA 0127 pu 635 kvar Q S Q d The total apparent power of this machine is 2 2 2 2 3055 kW 635 kvar 312 kVA S P Q The machine is operating within limits 512 Figure P53 shows a small industrial plant supplied by an external 480 V threephase power supply The plant includes three main loads as shown in the figure Answer the following questions about the plant The synchronous motor is rated at 100 hp 460 V and 08PFleading The synchronous reactance is 11 pu and armature resistance is 001 pu The OCC for this motor is shown in Figure P54 a If the switch on the synchronous motor is open how much real reactive and apparent power is being supplied to the plant What is the current LI in the transmission line The switch is now closed and the synchronous motor is supplying rated power at rated power factor b What is the field current in the motor c What is the torque angle of the motor c What is the power factor of the motor d How much real reactive and apparent power is being supplied to the plant now What is the current LI in the transmission line Now suppose that the field current is increased to 30 A e What is the real and reactive power supplied to the motor f What is the torque angle of the motor g What is the power factor of the motor 137 h How much real reactive and apparent power is being supplied to the plant now What is the current LI in the transmission line i How does the line current in part d compare to the line current in part h Why SOLUTION a The real and reactive power supplied by Load 1 is 1 100 kW P 1 1 1 1 tan cos PF 100 kW tan cos 09 484 kvar Q P The real and reactive power supplied by Load 2 is 2 2 cos 80 kVA 08 64 kW P S 138 1 1 2 2 sin cos PF 80 kW sin cos 08 48 kvar Q S The real reactive and apparent power supplied to the plant is TOT 1 2 100 kW 64 kW 164 kW P P P TOT 1 2 484 kvar 48 kvar 964 kvar Q Q Q 2 2 2 2 TOT TOT TOT 164 kW 964 kVA 190 kVA S P Q The line current is given by TOT 190 kVA 2285 A 3 3 480 V L L S I V Note In the following sections we will treat the synchronous motor as though it were Y connected The problem doesnt specify the connection and the answers are the same whether we assume a or a Y connection so we will just choose one for convenience b The rated power of the motor is 100 hp 746 Whp 746 kW motor P Assuming that the motor is Yconnected the base quantities for the synchronous motor are base PF 746 kW 08 932 kVA S P base 460 V VL base Lbase 3 460 V 3 266 V V V base base base 932 kVA 117 A 3 3 460 V L L S I V base base 117 A L I I 2 2 base base base 3 3 266 V 2278 93200 VA V Z S Therefore the impedances can be expressed in ohms per phase as base pu 2278 001 0023 A A R Z R base pu 2278 11 251 S S X Z X The rated real and reactive power of the motor is 100 hp 746 Whp 746 kW motor P 1 1 tan cos PF 746 kW tan cos 08 56 kvar motor motor Q P Note that the motor is supplying reactive power not consuming it The phase voltage would be 3 480 V 3 277 V L V V and magnitude of the armature current would be 139 746 kW 112 A 3 cos 3 277 V 08 motor A P I V Since the power factor is 08 leading the current A 112 3687 A Ι and the internal generated voltage is A A A R jX E V I S I A 277 0 V 0023 112 3687 A 251 112 3687 A A j E 498 270 pu A E The field current must corresponding to 498 V is 186 A c The torque angle of the motor is 27 d The real reactive and apparent power supplied to the plant is TOT 1 2 100 kW 64 kW746 kW 2386 kW motor P P P P TOT 1 2 484 kvar 48 kvar 56 kvar 404 kvar motor Q Q Q Q 2 2 2 2 TOT TOT TOT 2386 kW 404 kVA 242 kVA S P Q The line current is given by TOT 242 kVA 291 A 3 3 480 V L L S I V e If the field current is increased to 2 A the magnitude of the internal generated voltage will rise from 498 V to 517 V see the OCC curve The power supplied to the load will remain constant because the load torque and rotational speed are unchanged so the distance EA sin P will remain constant Therefore the torque angle becomes 1 1 1 2 1 2 498 V sin sin sin sin 27 259 517 V A A E E The new armature current is 2 2 277 0 V 517 259 V 117 398 A 251 A A jXS j V E I The current angle is 398 so the impedance angle 398 The real power supplied by the motor is 3 cos 3 277 V 117 A cos 398 746 kW motor P V I 3 cos 3 277 V 117 A sin 398 622 kvar Qmotor V I f The torque angle of the motor is 398 g The power factor of the motor is cos 398 0768 leading h The real reactive and apparent power supplied to the plant is TOT 1 2 100 kW 64 kW746 kW 2386 kW motor P P P P 140 TOT 1 2 484 kvar 48 kvar 622 kvar 262 kvar motor Q Q Q Q 2 2 2 2 TOT TOT TOT 2386 kW 262 kVA 240 kVA S P Q The line current is given by TOT 240 kVA 2887 A 3 3 480 V L L S I V i The line current in part d is greater than the line current in part h because the synchronous motor is supplying more of the reactive power being consumed by the other two loads required less reactive power from the infinite bus 513 A 480V 100kW 08PF leading 50Hz fourpole Yconnected synchronous motor has a synchronous reactance of 18 and a negligible armature resistance The rotational losses are also to be ignored This motor is to be operated over a continuous range of speeds from 300 to 1500 rmin where the speed changes are to be accomplished by controlling the system frequency with a solidstate drive a Over what range must the input frequency be varied to provide this speed control range b How large is at the motors rated conditions EA c What is the maximum power the motor can produce at rated speed with the calculated in part b EA d What is the largest value that could be at 300 rmin EA e Assuming that the applied voltage V is derated by the same amount as what is the maximum power the motor could supply at 300 rmin EA f How does the power capability of a synchronous motor relate to its speed SOLUTION a A speed of 300 rmin corresponds to a frequency of 300 rmin 4 10 Hz 120 120 m se n P f A speed of 1500 rmin corresponds to a frequency of 1500 rmin 4 50 Hz 120 120 m se n P f The frequency must be controlled in the range 10 to 50 Hz b The armature current at rated conditions is 100 kW 1503 A 3 PF 3 480 V 08 A L T P I I V so A 1503 3687 A This machine is Yconnected so the phase voltage is V I 480 3 277 V The internal generated voltage is A A A R jX E V I SI A 277 0 V 18 1503 3687 A A j E 489 262 V A E 141 So A E 489 V at rated conditions c The maximum power that the motor can produce at rated speed with the value of from part b is A E max 3 3 277 V 489 V 226 kW 18 A S V E P X d Since A must be decreased linearly with frequency the maximum value at 300 rmin would be E 300 10 Hz 489 V 978 V 50 Hz EA e If the applied voltage V is derated by the same amount as A E then V 1050277 554 V Also note that S X 105018 036 The maximum power that the motor could supply would be max 3 3 978 V 554 V 451 kW 036 A S V E P X f As we can see by comparing the results of c and e the powerhandling capability of the synchronous motor varies linearly with the speed of the motor 514 A 2300V 400hp 60Hz eightpole Yconnected synchronous motor has a rated power factor of 085 leading At full load the efficiency is 90 percent The armature resistance is 08 and the synchronous reactance is 11 Find the following quantities for this machine when it is operating at full load a Output torque b Input power c m n d E A e I A f conv P g P P P mech core stray SOLUTION a Since this machine has 8 poles it rotates at a speed of 120 60 Hz 120 900 rmin 8 se m f n P At full load the power supplied by this machine is out 400 hp 746 Whp 298 kW P If the output power is 298 kW the output torque is out load m 298000 W 3162 N m 2 rad 1 min 900 rmin 1 r 60 s P 142 b The input power is OUT IN 298 kW 331 kW 090 P P c The mechanical speed is 900 rmin m n d The armature current is IN 331 kW 978 A 3 PF 3 2300 V 085 A L T P I I V A 978 318 A I The phase voltage is 2300 V 3 0 1328 0 V V S A Therefore A is E A A A R jX E V I I 1328 0 V 08 978 318 A 11 978 318 A A j E 2063 276 V A E e The magnitude of the armature current is 978 A f The power converted from electrical to mechanical form is given by the equation conv IN CU P P P 2 2 CU 3 3 978 A 08 23 kW A A P I R conv IN CU 331 kW 23 kW 308 kW P P P g The mechanical core and stray losses are given by the equation mech core stray conv OUT 308 kW 298 kW 10 kW P P P P P 515 The Yconnected synchronous motor whose nameplate is shown in Figure 521 has a perunit synchronous reactance of 070 and a perunit resistance of 002 a What is the rated input power of this motor b What is the magnitude of E at rated conditions A c If the input power of this motor is 12 MW what is the maximum reactive power the motor can simultaneously supply Is it the armature current or the field current that limits the reactive power output d How much power does the field circuit consume at the rated conditions e What is the efficiency of this motor at full load f What is the output torque of the motor at the rated conditions Express the answer both in newtonmeters and in poundfeet 143 SOLUTION The base quantities for this motor are base 6600 V VT base 6600 V 3811 V 3 V base base 1404 A A L I I base rated 3 PF 3 6600 V 1404 A 10 1605 MW T L S P V I a The rated input power of this motor is IN 3 PF 3 6600 V 1404 A 10 1605 MW T L P V I b At rated conditions 10 0 pu and V 10 0 pu I so A is given in perunit quantities as E A A A R jX E V I SI A 1 0 002 10 0 070 1 0 A j E 120 355 pu A E The base phase voltage of this motor is base 3811 V V so A is E 120 355 3811 V 4573 355 V A E c From the capability diagram we know that there are two possible constraints on the maximum reactive powerthe maximum stator current and the maximum rotor current We will have to check each one separately and limit the reactive power to the lesser of the two limits The stator apparent power limit defines a maximum safe stator current This limit is the same as the rated input power for this motor since the motor is rated at unity power factor Therefore the stator apparent power limit is 1605 MVA If the input power is 12 MW then the maximum reactive power load that still protects the stator current is 2 2 2 2 1605 MVA 12 MW 107 MVAR Q S P Now we must determine the rotor current limit The perunit power supplied to the motor is 12 MW 1605 MW 0748 The maximum is 4573 V or 120 pu so with set to maximum and the motor consuming 12 MW the torque angle ignoring armature resistance is EA A E 144 1 1 070 0748 sin sin 84 3 3 10 120 S A X P V E The negative sign on comes from the fact that EA lags V in a motor At rated voltage and 12 MW of power supplied the armature current will be 1 0 120 84 0366 469 pu 070 A A A S R jX j V E I In actual amps this current is 1404 A 0396 469 556 469 A A I The reactive power supplied at the conditions of maximum A and 12 MW power is E 3 sin 3 3811 V 556 A sin 469 46 MVAR A Q V I Therefore the field current limit occurs before the stator current limit for these conditions and the maximum reactive power that the motor can supply is 46 MVAR under these conditions d At rated conditions the field circuit consumes field 125 V 52 A 650 W F F P V I e The efficiency of this motor at full load is OUT IN 21000 hp 746 Whp 100 100 976 1605 MW P P f The output torque in SI and English units is OUT load 21000 hp 746 Whp 124700 N m 1 min 2 rad 1200 rmin 60 s 1 r m P load 5252 21000 hp 5252 91910 lb ft 1200 rmin m P n 516 A 480V 500kVA 08PFlagging Yconnected synchronous generator has a synchronous reactance of 04 and a negligible armature resistance This generator is supplying power to a 480V 80kW 08 PFleading Yconnected synchronous motor with a synchronous reactance of 20 and a negligible armature resistance The synchronous generator is adjusted to have a terminal voltage of 480 V when the motor is drawing the rated power at unity power factor a Calculate the magnitudes and angles of for both machines E A b If the flux of the motor is increased by 10 percent what happens to the terminal voltage of the power system What is its new value c What is the power factor of the motor after the increase in motor flux SOLUTION a The motor is Yconnected and is operating at rated voltage 480 V rated power 80 kW and unity power factor The current flowing in the motor is 145 m m 80 kW 962 A 3 PF 3 480 V 10 A L T P I I V so This machine is Yconnected so the phase voltage is V m 962 0 A A I 480 3 277 V The internal generated voltage of the motor is m m m A jXS E V IA m 277 0 V 20 962 0 A A j E m 337 348 V A E The current supplied to the motor comes from the generator so the internal generated voltage of the generator is g g g A jXS E V IA g 277 0 V 04 962 0 A A j E g A 280 79 V E EAg j04 Vm IAg IAm EAm j20 Vg jX I A Sm V I A EAm jX SgI A V I A EAg Generator Motor b The power supplied by the generator to the motor will be constant as the field current of the motor is varied because the power is a function of the load on the motor and that load has not changed The 10 increase in flux will raise the internal generated voltage of the motor to 11337 V 371 V To make finding the new conditions easier we will make the angle of the phasor the reference phase angle during the following calculations The resulting phasor diagram is shown below Note that and are at the same angle so they just add together A g E S g A jX I S m A jX I 146 I A jX Sm V I A EAm jX SgI A EAg m g Then by Kirchhoffs Voltage Law A g A m S g S m A j X X E E I or A g A m A S g S m j X X E E I Note that this combined phasor diagram looks just like the diagram of a synchronous motor so we can apply the power equation for synchronous motors to this system 3 sin A g A m S g S m E E P X X where g m From this equation 1 1 24 80 kW sin sin 380 3 3 280V 371 V S g S m A g A m X X P E E Therefore 280 0 V 371 380 V 953 31 A 24 A g A m A S g S m j X X j E E I The phase voltage of the system would be 280 0 V 04 953 31 A 285 77 V A g S g A jX j V E I The phase voltage of the system can also be calculated as 371 38 V 2 953 31 A 2846 77 V A g S m A jX j V E I These two calculations agree as we would expect If we now make V the reference as we usually do all of the phases will shift by 77 and these voltages and currents become A g 280 77 V E 285 0 V V 371 303 A m E V A 953 108 A I The new terminal voltage is 3 285 V 494 V T V so the system voltage has increased c The impedance angle is 108 the opposite of the current angle The power factor of the motor is now 0982 lead PF cos 108 ing 147 Note The reactive power in the motor is now motor 3 sin 3 280 V 953 A sin 108 15 kVAR A Q V I The motor is now supplying 15 kVAR to the system Note that an increase in machine flux has increased the reactive power supplied by the motor and also raised the terminal voltage of the system This is consistent with what we learned about reactive power sharing in Chapter 4 517 A 440V 60 Hz threephase Yconnected synchronous motor has a synchronous reactance of 15 per phase The field current has been adjusted so that the torque angle is 25 when the power supplied by the generator is 80 kW a What is the magnitude of the internal generated voltage in this machine E A b What are the magnitude and angle of the armature current in the machine What is the motors power factor c If the field current remains constant what is the absolute maximum power this motor could supply SOLUTION a The power supplied to the motor is 80 kW This power is given by the equation 3 sin A S V E P X The phase voltage of the generator is 440 3 254 V so the magnitude of A is E 15 80 kW 373 V 3 sin 3 254 V sin 25 S A X P E V b The armature current in this machine is given by 254 0 V 373 25 119 28 A 15 A A jXS j V E I The power factor of the motor is PF cos 28º 0883 leading c The maximum power that the motor could supply at this field current is max 3 3 254 V 373 V 1894 kW 15 A S V E P X 518 A 460V 200kVA 085PFleading 400Hz fourpole Yconnected synchronous motor has negligible armature resistance and a synchronous reactance of 090 per unit Ignore all losses a What is the speed of rotation of this motor b What is the output torque of this motor at the rated conditions c What is the internal generated voltage of this motor at the rated conditions d With the field current remaining at the value present in the motor in part c what is the maximum possible output power from the machine SOLUTION a The speed of rotation of this motor is sync 120 400 Hz 120 12000 rmin 4 fse n P 148 b Since all losses are ignored rated rated rated PF 200 kVA 085 170 kW IN OUT P P S The output torque of this motor is OUT load 170 kW 135 N m 1 min 2 rad 12000 rmin 60 s 1 r m P c The phase voltage of this motor is 460 V 3 266 V The rated armature current of this motor is 170 kW 251 A 3 PF 3 460 V 085 A L T P I I V Therefore A 251 318 A The base impedance of this motor is I 2 2 base base base 3 3 266 V 106 200000 VA V Z S so the actual synchronous reactance is 090 pu 106 0954 S X The internal generated voltage of this machine at rated conditions is given by A jXS E V IA 266 0 V 0954 251 318 A 441 274 V A j E d The maximum power that the motor could supply at these conditions is MAX 3 3 266 V 441 V 369 kW 0954 A S V E P X 519 A 100hp 440V 08PFleading connected synchronous motor has an armature resistance of 03 and a synchronous reactance of 40 Its efficiency at full load is 96 percent a What is the input power to the motor at rated conditions b What is the line current of the motor at rated conditions What is the phase current of the motor at rated conditions c What is the reactive power consumed by or supplied by the motor at rated conditions d What is the internal generated voltage E of this motor at rated conditions A e What are the stator copper losses in the motor at rated conditions f What is P at rated conditions conv g If E is decreased by 10 percent how much reactive power will be consumed by or supplied by the motor A SOLUTION a The input power to the motor at rated conditions is OUT IN 100 hp 746 Whp 777 kW 096 P P b The line current to the motor at rated conditions is 777 kW 120 A 3 PF 3 440 V 085 L T P I V 149 The phase current to the motor at rated conditions is 120 A 693 A 3 3 LI I c The reactive power supplied by this motor to the power system at rated conditions is rated 3 sin 3 440 V 693 A sin3687 549 kVAR A Q V I d The internal generated voltage at rated conditions is A A A R jX E V I SI A 440 0 V 03 693 3687 A 40 693 3687 A A j E 634 217 V A E e The stator copper losses at rated conditions are 2 2 CU 3 3 693 A 03 43 kW A A P I R f conv at rated conditions is P conv IN CU 777 kW 43 kW 734 kW P P P g If A E is decreased by 10 the new value if A E 09634 V 571 V To simplify this part of the problem we will ignore A R Then the quantity EA sin will be constant as A E changes Therefore 1 1 1 2 1 2 634 V sin sin sin sin 217 242 571 V A A E E Therefore 440 0 V 571 242 619 19 A 40 A A jXS j V E I and the reactive power supplied by the motor to the power system will be 3 sin 3 440 V 619 A sin 19 266 kVAR A Q V I 520 Answer the following questions about the machine of Problem 519 a If E 43015 V and 4400 V is this machine consuming real power from or supplying real power to the power system Is it consuming reactive power from or supplying reactive power to the power system A V b Calculate the real power P and reactive power Q supplied or consumed by the machine under the conditions in part a Is the machine operating within its ratings under these circumstances c If E 47020 V and 4400 V is this machine consuming real power from or supplying real power to the power system Is it consuming reactive power from or supplying reactive power to the power system A V d Calculate the real power P and reactive power Q supplied or consumed by the machine under the conditions in part c Is the machine operating within its ratings under these circumstances SOLUTION 150 151 a This machine is a generator supplying real power to the power system because EA is ahead of V It is consuming reactive power because EA cos V b This machine is acting as a generator and the current flow in these conditions is 430 15 440 0 V 284 168 A 03 40 A A A S R jX j E V I The real power supplied by this machine is 3 cos 3 440 V 284 A cos 168 359 kW A P V I The reactive power supplied by this machine is 3 sin 3 440 V 284 A sin 168 108 kVAR A Q V I In other words this machine is consuming reactive power c This machine is a motor consuming real power from the power system because EA is behind V It is supplying reactive power because EA cos V d This machine is acting as a motor and the current flow in these conditions is 440 0 V 470 20 401 49 A 03 40 A A A S R jX j V E I The real power consumed by this machine is 3 cos 3 440 V 401 A cos 49 527 kW A P V I The reactive power supplied by this machine is 3 sin 3 440 V 401 A sin 49 45 kVAR A Q V I In other words this machine is supplying reactive power It is operating within its rating limits Chapter 6 Induction Motors 61 A 220V threephase sixpole 50Hz induction motor is running at a slip of 35 percent Find a The speed of the magnetic fields in revolutions per minute b The speed of the rotor in revolutions per minute c The slip speed of the rotor d The rotor frequency in hertz SOLUTION a The speed of the magnetic fields is 2 sync 120 50 Hz 120 1000 rmin 6 f e n P b The speed of the rotor is sync 1 1 0035 1000 rmin 965 rmin nm s n c The slip speed of the rotor is slip sync 0035 1000 rmin 35 rmin n sn d The rotor frequency is slip 35 rmin 6 175 Hz 120 120 re n P f 62 Answer the questions in Problem 61 for a 480V threephase twopole 60Hz induction motor running at a slip of 0025 SOLUTION a The speed of the magnetic fields is sync 120 60 Hz 120 3600 rmin 2 fse n P b The speed of the rotor is sync 1 1 0025 3600 rmin 3510 rmin nm s n c The slip speed of the rotor is slip sync 0025 3600 rmin 90 rmin n sn d The rotor frequency is slip 90 rmin 2 15 Hz 120 120 re n P f 63 A threephase 60Hz induction motor runs at 715 rmin at no load and at 670 rmin at full load a How many poles does this motor have b What is the slip at rated load c What is the speed at onequarter of the rated load 152 d What is the rotors electrical frequency at onequarter of the rated load SOLUTION a This machine has 10 poles which produces a synchronous speed of sync 120 60 Hz 120 720 rmin 10 ef n P b The slip at rated load is sync sync 720 670 100 100 694 720 m n n s n c The motor is operating in the linear region of its torquespeed curve so the slip at ¼ load will be 02500694 00171 s The resulting speed is sync 1 1 00171 720 rmin 708 rmin nm s n d The electrical frequency at ¼ load is 00171 60 Hz 103 Hz r e f sf 64 A 50kW 460V 50Hz twopole induction motor has a slip of 5 percent when operating a fullload conditions At fullload conditions the friction and windage losses are 700 W and the core losses are 600 W Find the following values for fullload conditions a The shaft speed m n b The output power in watts c The load torque load in newtonmeters d The induced torque ind in newtonmeters e The rotor frequency in hertz SOLUTION a The synchronous speed of this machine is sync 120 50 Hz 120 3000 rmin 2 fse n P Therefore the shaft speed is sync 1 1 005 3000 rmin 2850 rmin nm s n b The output power in watts is 50 kW stated in the problem c The load torque is OUT load 50 kW 1675 N m 2 rad 1 min 2850 rmin 1 r 60 s m P d The induced torque can be found as follows conv OUT FW core misc 50 kW 700 W 600 W 0 W 513 kW P P P P P 153 conv ind 513 kW 1719 N m 2 rad 1 min 2850 rmin 1 r 60 s m P e The rotor frequency is 005 50 Hz 25 Hz r e f sf 65 A 208V fourpole 60Hz Yconnected woundrotor induction motor is rated at 30 hp Its equivalent circuit components are R 0100 0070 100 1 R2 X M X 0210 0210 1 X2 P 500 W 0 400 W mech Pmisc Pcore For a slip of 005 find a The line current IL b The stator copper losses PSCL c The airgap power PAG d The power converted from electrical to mechanical form Pconv e The induced torque ind f The load torque load g The overall machine efficiency h The motor speed in revolutions per minute and radians per second SOLUTION The equivalent circuit of this induction motor is shown below 010 j021 V IA R1 jX1 R2 s s R 1 2 jX2 jXM 007 j021 j10 133 a The easiest way to find the line current or armature current is to get the equivalent impedance of the rotor circuit in parallel with F Z jXM and then calculate the current as the phase voltage divided by the sum of the series impedances as shown below 154 010 j021 V IA R1 jX1 RF jXF The equivalent impedance of the rotor circuit in parallel with jXM is 2 1 1 1318 0386 1374 163 1 1 1 1 10 140 021 F M Z j jX Z j j The phase voltage is 208 3 120 V so line current is LI 1 1 120 0 V 010 021 1318 0386 L A F F V I I R jX R jX j j 780 228 A L A I I b The stator copper losses are 2 2 SCL 1 3 3 780 A 010 1825 W A P I R c The air gap power is 2 2 2 AG 3 2 3 A F R P I I s R Note that is equal to 2 3 A F I R 2 2 3 2 R I s since the only resistance in the original rotor circuit was R2 s and the resistance in the Thevenin equivalent circuit is F R The power consumed by the Thevenin equivalent circuit must be the same as the power consumed by the original circuit 2 2 2 2 AG 3 2 3 3 780 A 1318 240 kW A F R P I I R s d The power converted from electrical to mechanical form is conv AG 1 1 005 240 kW 228 kW P s P e The induced torque in the motor is AG ind sync 240 kW 1274 N m 2 rad 1 min 1800 rmin 1 r 60 s P f The output power of this motor is OUT conv mech core misc 228 kW 500 W 400 W 0 W 219 kW P P P P P The output speed is sync 1 1 005 1800 rmin 1710 rmin nm s n Therefore the load torque is 155 OUT load 219 kW 1223 N m 2 rad 1 min 1710 rmin 1 r 60 s m P g The overall efficiency is OUT OUT IN 100 100 3 cos A P P P V I 219 kW 100 846 3 120 V 780 A cos228 h The motor speed in revolutions per minute is 1710 rmin The motor speed in radians per second is 2 rad 1 min 1710 rmin 179 rads 1 r 60 s m 66 For the motor in Problem 65 what is the slip at the pullout torque What is the pullout torque of this motor SOLUTION The slip at pullout torque is found by calculating the Thevenin equivalent of the input circuit from the rotor back to the power supply and then using that with the rotor circuit model 1 1 TH 1 1 10 010 021 00959 02066 02278 651 010 021 10 M M jX R jX j j Z j R j X X j 156 TH 1 1 10 120 0 V 1175 06 V 01 023 10 M M j jX R j X X j V V The slip at pullout torque is 2 max 2 2 TH TH 2 R s R X X max 2 2 0070 0164 00959 02066 0210 s The pullout torque of the motor is 2 TH max 2 2 sync TH TH TH 2 3V R R X X 2 max 2 2 3 1175 V 1885 rads 00959 00959 02066 0210 max 210 N m 67 a Calculate and plot the torquespeed characteristic of the motor in Problem 65 b Calculate and plot the output power versus speed curve of the motor in Problem 65 SOLUTION a A MATLAB program to calculate the torquespeed characteristic is shown below Mfile prob67am Mfile create a plot of the torquespeed curve of the induction motor of Problem 65 First initialize the values needed in this program r1 0100 Stator resistance x1 0210 Stator reactance r2 0070 Rotor resistance x2 0210 Rotor reactance xm 100 Magnetization branch reactance vphase 208 sqrt3 Phase voltage nsync 1800 Synchronous speed rmin wsync 1885 Synchronous speed rads Calculate the Thevenin voltage and impedance from Equations 641a and 643 vth vphase xm sqrtr12 x1 xm2 zth jxm r1 jx1 r1 jx1 xm rth realzth xth imagzth Now calculate the torquespeed characteristic for many slips between 0 and 1 Note that the first slip value is set to 0001 instead of exactly 0 to avoid divide byzero problems s 0150 50 Slip s1 0001 157 nm 1 s nsync Mechanical speed Calculate torque versus speed for ii 151 tindii 3 vth2 r2 sii wsync rth r2sii2 xth x22 end Plot the torquespeed curve figure1 plotnmtindkLineWidth20 xlabelbfitnm ylabelbf auind title bfInduction Motor TorqueSpeed Characteristic grid on The resulting plot is shown below b A MATLAB program to calculate the outputpower versus speed curve is shown below Mfile prob67bm Mfile create a plot of the output pwer versus speed curve of the induction motor of Problem 65 First initialize the values needed in this program r1 0100 Stator resistance x1 0210 Stator reactance r2 0070 Rotor resistance x2 0210 Rotor reactance xm 100 Magnetization branch reactance vphase 208 sqrt3 Phase voltage nsync 1800 Synchronous speed rmin wsync 1885 Synchronous speed rads Calculate the Thevenin voltage and impedance from Equations 641a and 643 158 vth vphase xm sqrtr12 x1 xm2 zth jxm r1 jx1 r1 jx1 xm rth realzth xth imagzth Now calculate the torquespeed characteristic for many slips between 0 and 1 Note that the first slip value is set to 0001 instead of exactly 0 to avoid divide byzero problems s 0150 50 Slip s1 0001 nm 1 s nsync Mechanical speed rmin wm 1 s wsync Mechanical speed rads Calculate torque and output power versus speed for ii 151 tindii 3 vth2 r2 sii wsync rth r2sii2 xth x22 poutii tindii wmii end Plot the torquespeed curve figure1 plotnmpout1000kLineWidth20 xlabelbfitnm rmbfrmin ylabelbfitPOUT rmbfkW title bfInduction Motor Ouput Power versus Speed grid on The resulting plot is shown below 68 For the motor of Problem 65 how much additional resistance referred to the stator circuit would it be necessary to add to the rotor circuit to make the maximum torque occur at starting conditions when the shaft is not moving Plot the torquespeed characteristic of this motor with the additional resistance inserted 159 SOLUTION To get the maximum torque at starting the max must be 100 Therefore s 2 max 2 2 TH TH 2 R s R X X 2 2 2 100 00959 02066 0210 R 2 R 0428 Since the existing resistance is 0070 an additional 0358 must be added to the rotor circuit The resulting torquespeed characteristic is 69 If the motor in Problem 65 is to be operated on a 50Hz power system what must be done to its supply voltage Why What will the equivalent circuit component values be at 50 Hz Answer the questions in Problem 65 for operation at 50 Hz with a slip of 005 and the proper voltage for this machine SOLUTION If the input frequency is decreased to 50 Hz then the applied voltage must be decreased by 56 also If this were not done the flux in the motor would go into saturation since T dt v N 1 and the period T would be increased At 50 Hz the resistances will be unchanged but the reactances will be reduced to 56 of their previous values The equivalent circuit of the induction motor at 50 Hz is shown below 160 010 j0175 V IA R1 jX1 R2 s s R 1 2 jX2 jXM 007 j0175 j833 133 a The easiest way to find the line current or armature current is to get the equivalent impedance of the rotor circuit in parallel with and then calculate the current as the phase voltage divided by the sum of the series impedances as shown below F Z M jX 010 j0175 V IA R1 jX1 RF jXF The equivalent impedance of the rotor circuit in parallel with M is jX 2 1 1 1308 0386 136 165 1 1 1 1 833 140 0175 F M Z j jX Z j j The line voltage must be derated by 56 so the new line voltage is 1733 V T V The phase voltage is 1733 3 100 V so line current is LI 1 1 100 0 V 010 0175 1308 0386 L A F F V I I R jX R jX j j 660 217 A L A I I b The stator copper losses are 2 2 SCL 1 3 3 66 A 010 1307 W A P I R c The air gap power is 2 2 2 AG 3 2 3 A F R P I I s R Note that is equal to 2 3 A F I R 2 2 3 2 R I s since the only resistance in the original rotor circuit was R2 s and the resistance in the Thevenin equivalent circuit is F R The power consumed by the Thevenin equivalent circuit must be the same as the power consumed by the original circuit 2 2 2 2 AG 3 2 3 3 66 A 1308 171 kW A F R P I I R s 161 d The power converted from electrical to mechanical form is conv AG 1 1 005 171 kW 1625 kW P s P e The induced torque in the motor is AG ind sync 171 kW 1089 N m 2 rad 1 min 1500 rmin 1 r 60 s P f In the absence of better information we will treat the mechanical and core losses as constant despite the change in speed This is not true but we dont have reason for a better guess Therefore the output power of this motor is OUT conv mech core misc 1625 kW 500 W 400 W 0 W 1535 kW P P P P P The output speed is sync 1 1 005 1500 rmin 1425 rmin nm s n Therefore the load torque is OUT load 1535 kW 1029 N m 2 rad 1 min 1425 rmin 1 r 60 s m P g The overall efficiency is OUT OUT IN 100 100 3 cos A P P P V I 1535 kW 100 834 3 100 V 660 A cos217 h The motor speed in revolutions per minute is 1425 rmin The motor speed in radians per second is 2 rad 1 min 1425 rmin 1492 rads 1 r 60 s m 610 A threephase 60Hz twopole induction motor runs at a noload speed of 3580 rmin and a fullload speed of 3440 rmin Calculate the slip and the electrical frequency of the rotor at noload and fullload conditions What is the speed regulation of this motor Equation 368 SOLUTION The synchronous speed of this machine is 3600 rmin The slip and electrical frequency at no load conditions is sync nl nl sync 3600 3580 100 100 056 3600 n n s n nl 00056 60 Hz 033 Hz r e f sf The slip and electrical frequency at full load conditions is sync nl fl sync 3600 3440 100 100 444 3600 n n s n fl 00444 60 Hz 267 Hz r e f sf The speed regulation is 162 nl fl fl 3580 3440 SR 100 100 407 3440 n n n 611 The input power to the rotor circuit of a sixpole 60 Hz induction motor running at 1100 rmin is 5 kW What is the rotor copper loss in this motor SOLUTION This synchronous speed of this motor is sync 120 60 Hz 120 1200 rmin 6 fse n P The slip of the rotor is sync nl nl sync 1200 1100 100 100 833 1200 n n s n The air gap power is the input power to the rotor so AG 5 kW P The power converted from electrical to mechanical form is conv AG 1 1 00833 5 kW 4584 W P s P The rotor copper losses are the difference between the air gap power and the power converted to mechanical form so RCL AG conv 5000 W 4584 W 416 W P P P 612 The power crossing the air gap of a 60 Hz fourpole induction motor is 25 kW and the power converted from electrical to mechanical form in the motor is 232 kW a What is the slip of the motor at this time b What is the induced torque in this motor c Assuming that the mechanical losses are 300 W at this slip what is the load torque of this motor SOLUTION a The synchronous speed of this motor is sync 120 60 Hz 120 1800 rmin 4 fse n P The power converted from electrical to mechanical form is conv AG 1 P s P so conv AG 234 kW 1 1 0064 25 kW P s P or 64 b The speed of the motor is sync 1 1 0064 1800 rmin 1685 rmin nm s n The induced torque of the motor is 163 conv ind m 234 kW 1326 N m 2 rad 1 min 1685 rmin 1 r 60 s P Alternately the induced torque can be found as AG ind sync 250 kW 1326 N m 2 rad 1 min 1800 rmin 1 r 60 s P c The output power of this motor is out conv mech 23400 W 300 W 23100 W P P P out load m 231 kW 1309 N m 2 rad 1 min 1685 rmin 1 r 60 s P 613 Figure 618a shows a simple circuit consisting of a voltage source a resistor and two reactances Find the Thevenin equivalent voltage and impedance of this circuit at the terminals Then derive the expressions for the magnitude of TH and for V RTH given in Equations 641b and 644 SOLUTION The Thevenin voltage of this circuit is TH 1 1 M M jX R j X X V V The magnitude of this voltage is TH 2 2 1 1 M M X V V R X X If 1 XM X then 2 2 2 1 1 1 M M R X X X X so TH 1 M M X V V X X The Thevenin impedance of this circuit is 1 1 TH 1 1 M M jX R jX Z R j X X 164 1 1 1 1 TH 1 1 1 1 M M M M jX R jX R j X X Z R j X X R j X X 2 2 2 1 1 1 1 1 1 1 1 TH 2 2 1 1 M M M M M M R X X R X X R X j R X X X X X Z R X X 2 M 2 2 2 1 1 1 TH TH TH 2 2 2 2 1 1 1 1 M M M M M R X R X X X X X Z R jX j R X X R X X 2 1 M The Thevenin resistance is 2 1 TH 2 2 1 1 M M R X R R X X If 1 XM R then 2 2 2 1 1 1 M M R X X X X so 2 TH 1 1 M M X R R X X The Thevenin reactance is 2 2 1 1 1 TH 2 2 1 1 M M M M 2 R X X X X X X R X X If 1 XM R and 1 XM X then 2 2 2 1 1 1 M M M X X R X X X and 2 2 2 1 1 M M X X X R so 2 1 TH 1 2 M M X X X X X 614 Figure P61 shows a simple circuit consisting of a voltage source two resistors and two reactances in parallel with each other If the resistor is allowed to vary but all the other components are constant at what value of will the maximum possible power be supplied to it Prove your answer Hint Derive an expression for load power in terms of V and and take the partial derivative of that expression with respect to Use this result to derive the expression for the pullout torque Equation 654 RL RL RS X S RL X L RL SOLUTION The current flowing in this circuit is given by the equation L S S L L R jX R jX V I 2 2 L S L S L V I R R X X The power supplied to the load is 165 166 2 2 2 2 L L L S L S L V R P I R R R X X 2 2 2 2 2 2 2 2 S L S L L S L L S L S L R R X X V V R R R P R R R X X To find the point of maximum power supplied to the load set L P R 0 and solve for L R 2 2 2 2 2 0 S L S L L S L R R X X V V R R R 2 2 2 S L S L L S L R R X X R R R 2 2 2 2 2 S S L L S L S L 2 2 L R R R R X X R R R 2 2 2 2 S L S L L2 R R X X R 2 2 2 S S L L R X X R Therefore for maximum power transfer the load resistor should be 2 2 L S S L R R X X 615 A 460V 60Hz fourpole Yconnected induction motor is rated at 25 hp The equivalent circuit parameters are R 015 0154 20 1 R2 X M X 0852 1066 1 X2 P 400 W 150 W 400 W FW Pmisc Pcore For a slip of 002 find a The line current IL b The stator power factor c The rotor power factor d The rotor frequency e The stator copper losses SCL P f The airgap power PAG g The power converted from electrical to mechanical form Pconv h The induced torque ind i The load torque load j The overall machine efficiency k The motor speed in revolutions per minute and radians per second l What is the starting code letter for this motor SOLUTION The equivalent circuit of this induction motor is shown below IA 015 j0852 V R1 jX1 R2 167 s s R 1 2 jX2 jXM 0154 j1066 j20 7546 a The easiest way to find the line current or armature current is to get the equivalent impedance F Z by of the rotor circuit in parallel with jX M and then calculate the current as the phase voltage divided the sum of the series impedances as shown below IA 015 j0852 V R1 jX1 RF jXF The equivalent impedance of the rotor circuit in parallel with jXM is 2 1 1 6123 325 6932 280 1 1 1 1 20 770 1066 F M Z j jX Z j j The phase voltage is 460 3 266 V so line current is LI 1 1 266 0 V 015 0852 6123 325 L A F F V I I R jX R jX j j stator power factor is 355 332 A L A I I b The PF cos 332 083 7 lagging c To find the rotor power factor we m st find the im u pedance angle of the rotor 1 1 2 2 1066 X tan tan 788 770 R R s d The rotor frequency is 002 60 H f sf z 12 Hz r s Therefore the rotor power factor is PF cos788 0991 lagging R e The stator copper losses are 2 2 SCL 1 3 3 355 A 015 567 W A P I R f The air gap power is 2 2 2 AG 3 2 3 A F R P I I s R Note that is equal to 2 3 A F I R 2 2 3 2 R I s since the only resistance in the original rotor circuit was R2 s and the resistance in the Thevenin equivalent circuit is F R The power consumed by the Thevenin equivalent circuit must be the same as the power consumed by the original circuit 2 2 2 2 AG 3 2 3 3 355 A 6123 2315 kW A F R P I I R s g The power converted from electrical to mechanical form is conv AG 1 1 002 2315 kW 2269 kW P s P h The synchronous speed of this motor is sync 120 60 Hz 120 1800 rmin 4 fse n P sync 2 rad 1 min 1800 rmin 1885 rads 1 r 60 s Therefore the induced torque in the motor is AG ind sync 2315 kW 1228 N m 2 rad 1 min 1800 rmin 1 r 60 s P i The output power of this motor is OUT conv mech core misc 2269 kW 400 W 400 W 150 W 2174 kW P P P P P The output speed is sync 1 1 002 1800 rmin 1764 rmin nm s n Therefore the load torque is OUT load 2174 kW 1177 N m 2 rad 1 min 1764 rmin 1 r 60 s m P j The overall efficiency is OUT OUT IN 100 100 3 cos A P P P V I 2174 kW 100 917 3 266 V 355 A cos 332 k The motor speed in revolutions per minute is 1764 rmin The motor speed in radians per second is 2 rad 1 min 1764 rmin 1847 rads 1 r 60 s m 168 l The equivalent circuit of this induction motor at starting conditions is shown below 015 j0852 V IA R1 jX1 R2 jX2 jXM 0154 j1066 j20 The easiest way to find the line current or armature current is to get the equivalent impedance of the rotor circuit in parallel with and then calculate the current as the phase voltage divided by the sum of the series impedances as shown below F Z M jX 015 j0852 V IA R1 jX1 RF jXF The equivalent impedance of the rotor circuit in parallel with jXM is 2 1 1 0139 1013 1023 822 1 1 1 1 20 0154 1066 F M Z j jX Z j j The phase voltage is 460 3 266 V so line current is LI 1 1 266 0 V 015 0852 0139 1023 L A F F V I I R jX R jX j j 1402 812 A L A I I The starting kVA of the motor is start 3 3 266 V 140 A 1117 kVA A S V I The locked rotor kVAhp is 1117 kVA kVAhp 447 25 hp Therefore this motor is Starting Code Letter D 616 For the motor in Problem 615 what is the pullout torque What is the slip at the pullout torque What is the rotor speed at the pullout torque SOLUTION The slip at pullout torque is found by calculating the Thevenin equivalent of the input circuit from the rotor back to the power supply and then using that with the rotor circuit model 169 1 1 TH 1 1 20 015 0852 0138 08172 0830 804 015 0852 20 M M jX R jX j j Z j R j X X j TH 1 1 20 266 0 V 255 041 V 015 0852 20 M M j jX R j X X j V V The slip at pullout torque is 2 max 2 2 TH TH 2 R s R X X max 2 2 0154 00815 0138 08172 1066 s The synchronous speed of this motor is sync 120 60 Hz 120 1800 rmin 4 ef n P sync 2 rad 1 min 1800 rmin 1885 rads 1 r 60 s This corresponds to a rotor speed of max max sync 1 1 00815 1800 rmin 1653 rmin n s n The pullout torque of the motor is 2 2 TH 2 TH TH sync 2 TH max 3 X X R R V 2 max 2 2 3 255 V 1885 rads 0138 0138 08182 1066 max 2721 N m 617 If the motor in Problem 615 is to be driven from a 460V 50Hz power supply what will the pullout torque be What will the slip be at pullout SOLUTION If this motor is driven from a 50 Hz source the resistances will be unchanged and the reactances will be increased by a ratio of 56 The resulting equivalent circuit is shown below 015 j0710 V IA R1 jX1 R2 s s R 1 2 jX2 jXM 0154 j0890 j1667 7546 V 2125 V The phase voltage must be derated by 56 so 170 171 ting the Thevenin equivalent of the input circuit from the rotor back to the power supply and then using that with the rotor circuit model The slip at pullout torque is found by calcula 1 1 TH 1667 015 0710 0138 06822 jXM R jX j j Z j 1 1 0696 786 015 0710 1667 M R j X X j TH 1 1 1667 2125 0 V 203 049 V 015 0710 1667 M M j jX R j X X j V V The slip at pullout torque is 2 max R s 2 2 TH TH 2 R X X max 2 2 0154 00976 0138 06822 0890 s The synchronous speed of this motor is sync 120 50 Hz 120 150 ef n 0 rmin 4 P sync 2 rad 1 min 1500 rmin 1571 rads 1 r 60 s This corre spon ds to a rotor speed of 976 1500 rmin 1354 rmin max max sync 1 1 00 n s n The pullout torque of the motor is 2 TH max 3V 2 2 sync TH TH TH 2 R R X X 2 max 2 2 3 203 V 1571 rads 0138 0138 06822 0890 max 2293 N m 618 Plot the follo wing quantities for the motor in Problem 615 as slip varies from 0 to 10 a ind b SOLU it program is shown below It follows the calculations perform for Problem values Mfile create a plot of the induced torque power ut and efficiency of the induction Pconv c Pout d Efficiency At what slip does Pout equal the rated power of the machine TION This problem is ideally suited to solution w h a MATLAB program An appropriate ed 618 but repeats them at many of slip and then plots the results Note that it plots all the specified values versus m n which varies from 1620 to 1800 rmin corresponding to a range of 0 to 10 slip Mfile prob618m converted power o motor of Problem 615 as a function of slip First initialize the values needed in this program r1 0015 Stator resistance 172 ch reactance ed rads tions 641a and 643 1 jx1 r1 jx1 xm torquespeed characteristic for many slips between 0 and 01 Note that the first slip value ical speed 60 Mechanical speed versus speed tindii 3 vth2 r2 sii rth r2sii2 xth x22 tindii wmii outii pconvii pmech pcore pmisc 1 1jxm 1r2siijx2 r1 jx1 zf imagiarealia fii poutii pinii 100 t the torquespeed curve gure1 20 itnm rmbfrmin x1 0852 Stator reactance r2 0154 Rotor resistance x2 1066 Rotor reactance xm 20 Magnetization bran vphase 460 sqrt3 Phase voltage nsync 1800 Synchronous speed rmin wsync 1885 Synchronous spe pmech 400 Mechanical losses W pcore 400 Core losses W pmisc 150 Miscellaneous losses W Calculate the Thevenin voltage and impedance from Equa vth vphase xm sqrtr12 x1 xm2 zth jxm r rth realzth xth imagzth Now calculate the is set to 0001 instead of exactly 0 to avoid divide byzero problems s 0000101 Slip s1 0001 nm 1 s nsync Mechan wm nm 2pi Calculate torque Pconv Pout and efficiency for ii 1lengths Induced torque wsync Power converted pconvii Power output p Power input zf ia vphase pinii 3 vphase absia cosatan Efficiency ef end Plo fi plotnmtindbLineWidth xlabelbf ylabelbf auind rmbfNm 173 er converted versus speed gure2 20 itnm rmbfrmin put power versus speed gure3 20 itnm rmbfrmin efficiency gure4 idth20 itnm rmbfrmin title bfInduced Torque versus Speed grid on Plot pow fi plotnmpconv1000bLineWidth xlabelbf ylabelbfitPrmbfconv kW title bfPower Converted versus Speed grid on Plot out fi plotnmpout1000bLineWidth xlabelbf ylabelbfitPrmbfout kW title bfOutput Power versus Speed axis1620 1800 0 50 grid on Plot the fi plotnmeffbLineW xlabelbf ylabelbfeta title bfEfficiency versus Speed grid on The four plots are shown below This machine is rated at 75 kW It produces an output power of 75 kW at 31 slip or a speed of 2907 rmin 619 A dc test is performed on a 460V connected 100hp induction motor If V 21 V and 72 A what is the stator resistance Why is this so DC IDC R1 SOLUTION If this motors armature is connected in delta then there will be two phases in parallel with one phase between the lines tested R1 R1 R1 VDC Therefore the stator resistance 1 will be R 1 1 1 DC 1 DC 1 1 1 2 3 R R R V R I R R R DC 1 DC 3 3 21 V 0438 2 2 72 A V R I 620 A 208V sixpole Yconnected 25hp design class B induction motor is tested in the laboratory with the following results No load 208 V 240 A 1400 W 60 Hz Locked rotor 246 V 645 A 2200 W 15 Hz Dc test 135 V 64 A Find the equivalent circuit of this motor and plot its torquespeed characteristic curve 175 SOLUTION From the DC test 1 135 V 2 R 64 A 1 R 0105 R1 R1 R1 VDC IDC In the noload test the line voltage is 208 V so the phase voltage is 120 V Therefore 1 nl 120 V 500 60 Hz 240 A M A V X X I In the lockedrotor test the line voltage is 246 V so the phase voltage is 142 V From the lockedrotor test at 15 Hz LR LR LR LR 142 V 0220 645 A A V Z R jX I 1 1 LR LR LR 2200 W cos cos 3682 3 246 V 645 A P S Therefore LR LR cos LR 0220 cos 3682 0176 R Z 1 2 0176 R R 2 R 0071 LR LR in LR 02202 sin 3682 0132 X Z s At a frequency of 60 Hz LR 60 Hz LR 0528 15 Hz X X For a Design Class B motor the split is 1 X 0211 and 2 X 0317 Therefore 5000 0211 4789 XM The resulting equivalent circuit is shown below 176 0105 j0211 V IA R1 jX1 R2 s s R 1 2 jX2 jXM 0071 j0317 j4789 I2 A MATLAB program to calculate the torquespeed characteristic of this motor is shown below Mfile prob620m Mfile create a plot of the torquespeed curve of the induction motor of Problem 620 First initialize the values needed in this program r1 0105 Stator resistance x1 0211 Stator reactance r2 0071 Rotor resistance x2 0317 Rotor reactance xm 4789 Magnetization branch reactance vphase 208 sqrt3 Phase voltage nsync 1200 Synchronous speed rmin wsync 1257 Synchronous speed rads Calculate the Thevenin voltage and impedance from Equations 641a and 643 vth vphase xm sqrtr12 x1 xm2 zth jxm r1 jx1 r1 jx1 xm rth realzth xth imagzth Now calculate the torquespeed characteristic for many slips between 0 and 1 Note that the first slip value is set to 0001 instead of exactly 0 to avoid divide byzero problems s 0150 50 Slip s1 0001 nm 1 s nsync Mechanical speed Calculate torque versus speed for ii 151 tindii 3 vth2 r2 sii wsync rth r2sii2 xth x22 end Plot the torquespeed curve figure1 plotnmtindbLineWidth20 xlabelbfitnm ylabelbf auind title bfInduction Motor TorqueSpeed Characteristic 177 grid on The resulting plot is shown below 621 A 460V 10 hp fourpole Yconnected Insulation class F Service Factor 115 induction motor has the following parameters R 054 0488 5112 1 R2 X M X 2093 3209 1 X2 P 150 W 50 W 150 kW FW Pmisc Pcore For a slip of 002 find a The line current IL b The stator power factor c The rotor power factor d The rotor frequency e The stator copper losses SCL P f The airgap power PAG g The power converted from electrical to mechanical form Pconv h The induced torque ind i The load torque load j The overall machine efficiency k The motor speed in revolutions per minute and radians per second 178 l Sketch the power flow diagram for this motor 179 ise in this motor given its insulation class OL shown below m What is the starting code letter for this motor n What is the maximum acceptable temperature r o What does the service factor of this motor mean S UTION The equivalent circuit of this induction motor is IA 054 j2093 V R1 jX1 R2 s s R 1 2 jX2 jXM 0488 j3209 j5112 23912 a The easiest way to find the line current or armature current is to get the equivalent impedance F Z by of the rotor circuit in parallel with jX M and then calculate the current as the phase voltage divided the sum of the series impedances as shown below 054 j2093 V IA R jX 1 1 RF jXF The equivalent impedance of the rotor circuit in parallel with jXM is 2 1 1 1798 110 ZF 1 2113 317 1 1 1 1 5112 2441 3209 M j jX Z j j The phase voltage is 460 3 266 V so line current is LI 1 1 266 L A I I 0 V 054 2093 1798 1110 F F V R jX R jX j j stator power factor is 1170 355 A L A I I b The PF cos 355 081 4 lagging c To find the rotor power factor we must find the im pedance angle of the rotor 1 1 2 3209 tan tan 750 X 2 244 R R s d The rotor frequency is 002 60 H r s f sf 180 the rotor power factor is st tor copper losses are f The air gap power is z 12 Hz Therefore PF cos750 0991 lagging R e The a 2 2 SCL 1 3 3 117 A P I R 054 222 W A 2 2 2 AG 3 2 3 A F R P I I s R Note that 2 3 is equal to A F I R 2 2 3 2 R I s since the only resistance in the original rotor circuit was R2 s and the resistance in the Thevenin equivalent circuit is F R The power consumed by the Thevenin equivalent circuit must be the same as the power consumed the original circuit by 2 2 2 2 3 3 3 117 A 1798 738 kW R P I I R AG 2 A F s electrical to mechanical form is g The power converted from conv AG 1 1 002 738 kW 723 kW P s P h The synchronous speed of this motor is sync 1800 rmin 4 n P 120 60 Hz 120 fse sync 2 rad 1 min 1800 rmin 1885 rads 1 r 60 s Therefore the induced torque in the motor is AG 738 kW P ind sync 392 N m 2 rad 1 min 1800 rmin 1 r 60 s i The output power of this motor is ut speed is OUT conv mech core misc P P P P P 723 kW 150 W 150 W 50 W 688 kW The outp sync 1 002 1800 rmin 1764 rmin 1 nm s n Therefore the load torque is OUT P load 688 kW 372 N m 2 rad 1 min 1764 rmin 1 r 60 s m j The overall efficiency is OUT 100 P P OUT IN 100 3 A cos P V I 688 kW 100 905 3 266 V 117 A cos 355 181 otor speed in revolutions per minute is 1764 rmin The motor speed in radians per second is k The m 2 rad 1 min 1764 rmin 1847 rads 1 r 60 s m l The power flow diagram for this motor is 688 kW 222 W 738 kW 760 kW 150 W 150 W 50 W 723 kW 150 W m The equivalent circuit of this induction motor at starting conditions is shown below 054 j2093 V IA R1 1 2 2 jX R jX jXM 0488 j3209 j5112 The easiest way to find the line current or armature current is to get the equivalent impedance of the rotor circuit in parallel with and then calculate the current as the phase voltage divided by the sum F Z M jX of the series impedances as shown below 054 j2093 V IA R1 jX1 RF jXF The equivalent impedance of the rotor circuit in parallel with jXM is 2 1 1 0435 2864 290 813 F M Z j 1 1 1 1 5112 0488 3209 jX Z j j 182 e voltage is 460 3 The phas 266 V so line current is LI 1 1 266 0 V 054 2093 0435 290 L A F F V I I R jX R jX j j ing kVA of the motor is 523 A L A I I 79 The start start 3 3 266 V 523 A 417 kVA A S V I The locked rotor kVAhp is 417 kVA kVAhp 417 10 hp Therefore this motor is Starting Code Letter D torquespeed characteristic of the motor in Problem 621 What is the starting torque of this torquespeed characteristic of this motor is shown below e create a plot of the torquespeed curve of the e values needed in this program istance ance rmin rads any lue stead of exactly 0 to avoid divide x22 622 Plot the motor SOLUTION A MATLAB program to calculate the Mfile prob622m Mfil induction motor of Problem 621 First initialize th r1 054 Stator resistance x1 2093 Stator reactance r2 0488 Rotor res x2 3209 Rotor reactance xm 5112 Magnetization branch react vphase 460 sqrt3 Phase voltage nsync 1800 Synchronous speed wsync 1885 Synchronous speed Calculate the Thevenin voltage and impedance from Equations 641a and 643 vth vphase xm sqrtr12 x1 xm2 zth jxm r1 jx1 r1 jx1 xm rth realzth xth imagzth Now calculate the torquespeed characteristic for m slips between 0 and 1 Note that the first slip va is set to 0001 in byzero problems s 00150 50 Slip s1 0001 nm 1 s nsync Mechanical speed Calculate torque versus speed for ii 1lengths tindii 3 vth2 r2 sii wsync rth r2sii2 xth end Plot the torquespeed curve 183 nmtindbLineWidth20 abelbfitnm nduction Motor TorqueSpeed Characteristic figure1 plot xl ylabelbf auind title bfI grid on The resulting plot is shown below The starting torque of this motor is about 623 A 460V fourpole 75hp 60Hz Yconnected threephase induction motor develops its fullload induced torque at 35 percent slip when operating at 60 Hz and 460 V The perphase circuit model impedances of a losses ma lected in this problem a Find the value of the rotor resistance 179 N m the motor are R1 0058 X M 18 X1 032 X 0386 2 Mechanic l core and stray y be neg R2 smax and the rotor speed at maximum torque for this m b Find max otor SOL c Find the starting torque of this motor d What code letter factor should be assigned to this motor UTION The equivalent circuit for this motor is 058 j032 V IA R1 jX1 R2 s s R 1 2 jX2 jXM j0386 j18 I2 The Thevenin equivalent of the input circuit is 1 1 TH 1 1 18 058 032 0559 0332 0651 307 058 032 18 M M jX R jX j j Z j R j X X j TH 1 1 18 266 0 V 261 18 V 058 032 18 M M j jX R j X X j V V a If losses are neglected the induced torque in a motor is equal to its load torque At full load the output power of this motor is 75 hp and its slip is 12 so the induced torque is 1 0035 1800 rmin 1737 rmin m n ind load 75 hp 746 Whp 3076 N m 2 rad 1min 1737 rmin 1 r 60 s The induced torque is given by the equation 2 TH 2 ind 2 2 sync TH 2 TH 2 3 V R s R R s X X Substituting known values and solving for R2 s yields 2 TH 2 ind 2 2 sync TH 2 TH 2 3 V R s R R s X X 2 2 2 2 2 3 261 V 3076 N m 1885 rads 0559 0332 0386 R s R s 2 2 2 205932 57983 0559 0516 R s R s 2 2 2 0559 0516 3552 R s R s 2 2 2 2 03125 1118 0516 3552 R s R s R s 184 2 2 2 2434 0516 0 R R s s 2 02346 2199 R s 2 R 00082 0077 These two solutions represent two situations in which the torquespeed curve would go through this specific torquespeed point The two curves are plotted below As you can see only the 0077 solution is realistic since the 00082 solution passes through this torquespeed point at an unstable location on the back side of the torquespeed curve b The slip at pullout torque can be found by calculating the Thevenin equivalent of the input circuit from the rotor back to the power supply and then using that with the rotor circuit model The Thevenin equivalent of the input circuit was calculated in part a The slip at pullout torque is 2 max 2 2 TH TH 2 R s R X X max 2 2 0077 00846 0559 0332 0386 s The rotor speed a maximum torque is pullout sync 1 1 00846 1800 rmin 1648 rmin n s n and the pullout torque of the motor is 2 TH max 2 2 sync TH TH TH 2 3V R R X X 185 186 2 max 2 2 3 261 V 1885 rads 0559 0559 0332 0368 max 373 N m c The starting torque of this motor is the torque at slip s 1 It is 2 TH 2 ind 2 2 sync TH 2 TH 2 3 V R s R R s X X 2 ind 2 2 3 261 V 0077 933 N m 1885 rads 0559 0077 0332 0368 d To determine the starting code letter we must find the lockedrotor kVA per horsepower which is equivalent to finding the starting kVA per horsepower The easiest way to find the line current or armature current at starting is to get the equivalent impedance F Z of the rotor circuit in parallel with jXM at starting conditions and then calculate the starting current as the phase voltage divided by the sum of the series impedances as shown below 033 j042 V IAstart R1 jX1 RF jXF The equivalent impedance of the rotor circuit in parallel with jXM at starting conditions s 10 is start 2 1 1 00838 03782 0385 790 1 1 1 1 18 0077 0386 F M Z j jX Z j j The phase voltage is 460 3 266 V so line current start is LI start 1 1 266 0 V 0058 032 00838 03782 L A F F R jX R jX j j V I I start 373 785 A L A I I Therefore the lockedrotor kVA of this motor is rated 3 3 460 V 373 A 297 kVA T L S V I and the kVA per horsepower is 297 kVA kVAhp 396 kVAhp 75 hp This motor would have starting code letter C since letter C covers the range 355 400 624 Answer the following questions about the motor in Problem 621 a If this motor is started from a 460V infinite bus how much current will flow in the motor at starting b If transmission line with an impedance of 050 j035 per phase is used to connect the induction motor to the infinite bus what will the starting current of the motor be What will the motors terminal voltage be on starting c If an ideal 141 stepdown autotransformer is connected between the transmission line and the motor what will the current be in the transmission line during starting What will the voltage be at the motor end of the transmission line during starting SOLUTION a The equivalent circuit of this induction motor is shown below 054 j2093 V IA R1 jX1 R2 s s R 1 2 jX2 jXM 0488 j3209 j5112 23912 The easiest way to find the line current or armature current is to get the equivalent impedance of the rotor circuit in parallel with and then calculate the current as the phase voltage divided by the sum of the series impedances as shown below F Z M jX 054 j2093 V IA R1 jX1 RF jXF The equivalent impedance of the rotor circuit in parallel with jXM is 2 1 1 0435 2864 290 813 1 1 1 1 5112 0488 3209 F M Z j jX Z j j The phase voltage is 460 3 266 V so line current is LI 1 1 266 0 V 054 2093 0435 290 L A F F R jX R jX j j V I I 523 79 A L A I I 187 b If a transmission line with an impedance of 050 j035 per phase is used to connect the induction motor to the infinite bus its impedance will be in series with the motors impedances and the starting current will be bus line line 1 1 L A F F R jX R jX R jX V I I 266 0 V 050 035 054 2093 0435 290 L A j j j I I 480 746 A L A I I The voltage at the terminals of the motor will be 1 1 A F F R jX R jX V I 480 746 A 054 2093 0435 290 j j V 2442 43 V V Therefore the terminal voltage will be 3 2442 V 423 V Note that the terminal voltage sagged by about 92 during motor starting c If an ideal 141 stepdown autotransformer is connected between the transmission line and the motor the motors impedances will be referred across the transformer by the square of the turns ratio a 14 The referred impedances are 2 1 1 196 0058 01137 R a R 2 1 1 196 032 0627 X a X 2 196 0435 0853 F F R a R 2 196 290 5684 F F X a X Therefore the starting current referred to the primary side of the transformer will be bus line line 1 1 L A F F R jX R jX R jX V I I 266 0 V 050 035 01137 0627 0853 5684 L A j j j I I 390 776 A L A I I The voltage at the motor end of the transmission line would be the same as the referred voltage at the terminals of the motor 1 1 A F F R jX R jX V I 390 776 A 01137 0627 0853 5684 j j V 249 37 V V Therefore the line voltage at the motor end of the transmission line will be 3 249 V 431 V Note that this voltage sagged by 63 during motor starting which is less than the 92 sag with case of acrosstheline starting 625 In this chapter we learned that a stepdown autotransformer could be used to reduce the starting current drawn by an induction motor While this technique works an autotransformer is relatively expensive A much less expensive way to reduce the starting current is to use a device called Y starter described earlier in this chapter If an induction motor is normally connected it is possible to reduce its phase 188 voltage and hence its starting current by simply reconnecting the stator windings in Y during starting and then restoring the connections to when the motor comes up to speed Answer the following questions about this type of starter V a How would the phase voltage at starting compare with the phase voltage under normal running conditions b How would the starting current of the Yconnected motor compare to the starting current if the motor remained in a connection during starting SOLUTION a The phase voltage at starting would be 1 3 577 of the phase voltage under normal running conditions b Since the phase voltage decreases to 1 3 577 of the normal voltage the starting phase current will also decrease to 577 of the normal starting current However since the line current for the original delta connection was 3 times the phase current while the line current for the Y starter connection is equal to its phase current the line current is reduced by a factor of 3 in a Y starter For the connection 3 LI I For the Yconnection Y Y LI I But 3 I I Y Y so 3 L L I I 626 A 460V 50hp sixpole connected 60Hz threephase induction motor has a fullload slip of 4 percent an efficiency of 91 percent and a power factor of 087 lagging At startup the motor develops 175 times the fullload torque but draws 7 times the rated current at the rated voltage This motor is to be started with an autotransformer reduced voltage starter a What should the output voltage of the starter circuit be to reduce the starting torque until it equals the rated torque of the motor b What will the motor starting current and the current drawn from the supply be at this voltage SOLUTION a The starting torque of an induction motor is proportional to the square of TH V 2 2 start2 TH2 2 start1 TH1 1 T T V V V V If a torque of 175 rated is produced by a voltage of 460 V then a torque of 100 rated would be produced by a voltage of 2 rated 2 rated 100 175 460 V VT 2 2 460 V 348 V 175 VT b The motor starting current is directly proportional to the starting voltage so 2 1 1 rated rated 348 V 0757 0757 7 530 460 V L L L I I I I I The input power to this motor is 189 OUT IN 50 hp 746 Whp 410 kW 091 P P The rated current is equal to IN rated 410 kW 5915 A 3 PF 3 460 V 087 T P I V Therefore the motor starting current is 2 5445 rated 530 5915 A 3135 A LI I The turns ratio of the autotransformer that produces this starting voltage is 460 V 1322 348 V SE C C N N N so the current drawn from the supply will be start line 3135 A 237 A 1377 1322 I I 627 A woundrotor induction motor is operating at rated voltage and frequency with its slip rings shorted and with a load of about 25 percent of the rated value for the machine If the rotor resistance of this machine is doubled by inserting external resistors into the rotor circuit explain what happens to the following a Slip s b Motor speed n m c The induced voltage in the rotor d The rotor current e ind f Pout g PRCL h Overall efficiency SOLUTION a The slip s will increase b The motor speed m will decrease n c The induced voltage in the rotor will increase d The rotor current will increase e The induced torque will adjust to supply the loads torque requirements at the new speed This will depend on the shape of the loads torquespeed characteristic For most loads the induced torque will decrease 190 f The output power will generally decrease OUT ind m P g The rotor copper losses including the external resistor will increase h The overall efficiency will decrease 628 A 460V 75 hp fourpole Yconnected induction motor has the following parameters R 0058 0037 924 1 R2 X M X 0320 0386 1 X2 P 650 W 150 W 600 kW FW Pmisc Pcore For a slip of 001 find a The line current IL b The stator power factor c The rotor power factor d The rotor frequency e The stator copper losses SCL P f The airgap power PAG g The power converted from electrical to mechanical form Pconv h The induced torque ind i The load torque load j The overall machine efficiency k The motor speed in revolutions per minute and radians per second l Sketch the power flow diagram for this motor m What is the starting code letter for this motor 191 SOLUTION The equivalent circuit of this induction motor is shown below 0058 j0320 V IA R1 jX1 R2 s s R 1 2 jX2 jXM 0037 j0386 j924 3663 a The easiest way to find the line current or armature current is to get the equivalent impedance of the rotor circuit in parallel with and then calculate the current as the phase voltage divided by the sum of the series impedances as shown below F Z M jX 0058 j0320 V IA R1 jX1 RF jXF The equivalent impedance of the rotor circuit in parallel with jXM is 2 1 1 2970 1512 3333 270 1 1 1 1 924 370 0386 F M Z j jX Z j j The phase voltage is 460 3 266 V so line current is LI 1 1 266 0 V 0058 032 297 1512 L A F F R jX R jX j j V I I 752 312 A L A I I b The stator power factor is PF cos 312 0855 lagging c To find the rotor power factor we must find the impedance angle of the rotor 1 1 2 2 0386 tan tan 596 0037 001 R X R s d The rotor frequency is 001 60 Hz 06 Hz r s f sf Therefore the rotor power factor is PF cos596 0995 lagging R 192 e The stator copper losses are 2 2 SCL 1 3 3 752 A 0058 984 W A P I R f The air gap power is 2 2 2 AG 3 2 3 A F R P I I s R Note that is equal to 2 3 A F I R 2 2 3 2 R I s since the only resistance in the original rotor circuit was R2 s and the resistance in the Thevenin equivalent circuit is F R The power consumed by the Thevenin equivalent circuit must be the same as the power consumed by the original circuit 2 2 2 2 AG 3 2 3 3 752 A 297 504 kW A F R P I I R s g The power converted from electrical to mechanical form is conv AG 1 1 001 504 kW 499 kW P s P h The synchronous speed of this motor is sync 120 60 Hz 120 1800 rmin 4 fse n P sync 2 rad 1 min 1800 rmin 1885 rads 1 r 60 s Therefore the induced torque in the motor is AG ind sync 504 kW 2674 N m 2 rad 1 min 1800 rmin 1 r 60 s P i The output power of this motor is OUT conv mech core misc 499 kW 650 W 600 W 150 W 485 kW P P P P P The output speed is sync 1 1 001 1800 rmin 1782 rmin nm s n Therefore the load torque is OUT load 485 kW 260 N m 2 rad 1 min 1782 rmin 1 r 60 s m P j The overall efficiency is OUT OUT IN 100 100 3 cos A P P P V I 485 kW 100 945 3 266 V 752 A cos 312 k The motor speed in revolutions per minute is 1782 rmin The motor speed in radians per second is 2 rad 1 min 1782 rmin 1866 rads 1 r 60 s m 193 l The power flow diagram for this motor is 485 kW 984 W 504 kW 513 kW 600 W 650 W 150 W 499 kW 500 W m The equivalent circuit of this induction motor at starting conditions is shown below 0058 j0320 V IA R1 jX1 R2 jX2 jXM 0037 j0386 j924 The easiest way to find the line current or armature current is to get the equivalent impedance of the rotor circuit in parallel with and then calculate the current as the phase voltage divided by the sum of the series impedances as shown below F Z M jX 0058 j0320 V IA R1 jX1 RF jXF The equivalent impedance of the rotor circuit in parallel with jXM is 2 1 1 00341 03707 0372 847 1 1 1 1 924 0037 0386 F M Z j jX Z j j The phase voltage is 460 3 266 V so line current is LI 1 1 266 0 V 0058 032 00341 03707 L A F F R jX R jX j j V I I 194 195 3817 824 A L A I I The starting kVA of the motor is start 3 3 266 V 3817 A 3046 kVA A S V I The locked rotor kVAhp is 3046 kVA kVAhp 406 75 hp Therefore this motor is Starting Code Letter D 629 Plot the torquespeed characteristic of the motor in Problem 628 What is the starting torque of this motor SOLUTION A MATLAB program to calculate the torquespeed characteristic of this motor is shown below Mfile prob629m Mfile create a plot of the torquespeed curve of the induction motor of Problem 628 First initialize the values needed in this program r1 0058 Stator resistance x1 0320 Stator reactance r2 0037 Rotor resistance x2 0386 Rotor reactance xm 924 Magnetization branch reactance vphase 460 sqrt3 Phase voltage nsync 1800 Synchronous speed rmin wsync 1885 Synchronous speed rads Calculate the Thevenin voltage and impedance from Equations 641a and 643 vth vphase xm sqrtr12 x1 xm2 zth jxm r1 jx1 r1 jx1 xm rth realzth xth imagzth Now calculate the torquespeed characteristic for many slips between 0 and 1 Note that the first slip value is set to 0001 instead of exactly 0 to avoid divide byzero problems s 000550 50 Slip s1 0001 nm 1 s nsync Mechanical speed Calculate torque versus speed for ii 1lengths tindii 3 vth2 r2 sii wsync rth r2sii2 xth x22 end Plot the torquespeed curve figure1 plotnmtindbLineWidth20 xlabelbfitnm ylabelbf auind title bfInduction Motor TorqueSpeed Characteristic grid on The resulting plot is shown below 630 Answer the following questions about a 460V connected twopole 100hp 60Hz starting code letter F induction motor a What is the maximum current starting current that this machines controller must be designed to handle b If the controller is designed to switch the stator windings from a connection to a Y connection during starting what is the maximum starting current that the controller must be designed to handle c If a 1251 stepdown autotransformer starter is used during starting what is the maximum starting current that it must be designed to handle SOLUTION a Starting code letter F corresponds to a 500 560 kVAhp so the maximum starting kVA of this motor is start 100 hp 560 560 kVA S Therefore start 560 kVA 703 A 3 3 460 V T S I V b The line voltage will still be 460 V when the motor is switched to the Yconnection but now the phase voltage will be 460 3 266 V Before in 196 TH 2 TH 2 TH 2 TH 2 460 V V I R R j X X R R j X X But the line current in a connection is 3 times the phase current so TH 2 TH 2 TH 2 TH 3 797 V 3 L V I I 2 R R j X X R R j X X After in Y Y Y Y TH 2 TH 2 TH 2 TH 2 266 V L V I I R R j X X R R j X X Therefore the line current will decrease by a factor of 3 when using this starter The starting current with a Y starter is start 703 A 234 A 3 I c A 1251 stepdown autotransformer reduces the phase voltage on the motor by a factor 08 This reduces the phase current and line current in the motor and on the secondary side of the transformer by a factor of 08 However the current on the primary of the autotransformer will be reduced by another factor of 08 so the total starting current drawn from the line will be 64 of its original value Therefore the maximum starting current drawn from the line will be start 064 703 A 450 A I 631 When it is necessary to stop an induction motor very rapidly many induction motor controllers reverse the direction of rotation of the magnetic fields by switching any two stator leads When the direction of rotation of the magnetic fields is reversed the motor develops an induced torque opposite to the current direction of rotation so it quickly stops and tries to start turning in the opposite direction If power is removed from the stator circuit at the moment when the rotor speed goes through zero then the motor has been stopped very rapidly This technique for rapidly stopping an induction motor is called plugging The motor of Problem 621 is running at rated conditions and is to be stopped by plugging a What is the slip s before plugging b What is the frequency of the rotor before plugging c What is the induced torque before plugging ind d What is the slip s immediately after switching the stator leads e What is the frequency of the rotor immediately after switching the stator leads f What is the induced torque immediately after switching the stator leads ind SOLUTION a The slip before plugging is 002 see Problem 621 b The frequency of the rotor before plugging is 002 60 Hz 12 Hz r e f sf c The induced torque before plugging is 392 Nm in the direction of motion see Problem 621 d After switching stator leads the synchronous speed becomes 1800 rmin while the mechanical speed initially remains 1764 rmin Therefore the slip becomes 197 sync sync 1800 1764 198 1800 m n n s n e The frequency of the rotor after plugging is 198 60 Hz 1188 Hz r e f sf f The equivalent circuit for this motor is 054 j2093 V IA R1 jX1 R2 s s R 1 2 jX2 jXM 0488 j3209 j5112 The Thevenin equivalent of the input circuit is 1 1 TH 1 1 5112 054 2093 04983 2016 2076 761 054 2093 5112 M M jX R jX j j Z j R j X X j TH 1 1 512 266 0 V 256 058 V 054 2093 512 M M j jX R j X X j V V The induced torque immediately after switching the stator leads is 2 TH 2 ind 2 2 sync TH 2 TH 2 3 V R s R R s X X 2 ind 2 2 3 256 V 0488 1962 1885 rads 04983 0488 1962 2016 3209 2 ind 2 2 3 256 V 02487 1885 rads 04983 02487 2016 3209 ind 93 N m opposite the direction of motion 632 A 460V 10 hp twopole Yconnected induction motor has the following parameters R 054 2093 5112 1 X1 X M P 150 W 50 W 150 kW FW Pmisc Pcore The rotor is a dualcage design with a tightlycoupled high resistance outer bar and a looselycoupled low resistance inner bar see Figure 625c The parameters of the outer bar are 198 R2o X2o 480 375 The resistance is high due to the lower cross sectional area and the reactance is relatively low due to the tight coupling between the rotor and stator The parameters of the inner bar are R2i X2i 0573 465 The resistance is low due to the high cross sectional area but the reactance is relatively high due to the quite loose coupling between the rotor and stator Calculate the torquespeed characteristic for this induction motor and compare it to the torque speed characteristic for the singlecage design in Problem 621 How do the curves differ Explain the differences SOLUTION The dualcage rotor has two current paths in parallel the inner cage and the outer cage As a result the impedance of the rotor is calculated as the parallel combination of these two current paths For any given slip the impedance of the rotor can be calculated as 1 1 1 R i i o Z o R jX R jX where iR is the resistance of the inner rotor cage i X is the reactance of the inner rotor cage and so forth Also recall that rotor reactance varies with rotor frequency The rotor reactance is given by the equation o X sX where s is the slip and o X is the rotor reactance at lockedrotor conditions The rotor impedance and any slip can thus be calculated as 1 1 1 R i oi o Z oo R jsX R jsX where oi X is the reactance of the inner rotor cage at lockedrotor conditions and Xoo is the reactance of the outer rotor cage at lockedrotor conditions We must apply this equation to calculate the rotor impedance at any slip and then divide the resulting reactance by the slip to get to the equivalent impedance at lockedrotor conditions the reactance at lockedrotor conditions is the term that goes into the torque equation A MATLAB program to calculate the torquespeed characteristic of this motor is shown below Mfile prob632m Mfile create a plot of the torquespeed curve of the induction motor of Problem 632 First initialize the values needed in this program r1 054 Stator resistance x1 2093 Stator reactance Resistances and reactances of the dualcage rotor r2a 48 Outer bar rotor resistance x2a 375 Outer bar rotor reactance r2b 0573 Inner bar rotor resistance x2b 465 Inner bar rotor reactance Resistance and reactance of the singlecage rotor 621 r2 0488 x2 3209 xm 5112 Magnetization branch reactance vphase 460 sqrt3 Phase voltage nsync 1800 Synchronous speed rmin wsync 1885 Synchronous speed rads 199 Calculate the Thevenin voltage and impedance from Equations 200 xm sqrtr12 x1 xm2 Now calculate the torquespeed characteristic for many Slip nsync Mechanical speed Calculate torque for the singlecage rotor th2 r2 sii h x22 Calculate resistance and reactance of the doublecage ix2a 1r2b jsiix2b Convert the reactace back t Xo the reactance at Effective rotor reactance Calculate induced torque for doublecage rotor h x2eff2 Plot the torquespeed curve orquespeed curves 20 ind2kLineWidth20 TorqueSpeed Characteristic 641a and 643 vth vphase zth jxm r1 jx1 r1 jx1 xm rth realzth xth imagzth slips between 0 and 1 Note that the first slip value is set to 0001 instead of exactly 0 to avoid divide byzero problems s 00150 50 s1 0001 nm 1 s for ii 1 lengths tind1ii 3 v wsync rth r2sii2 xt end rotor at this slip and then use those values to calculate the induced torque for ii 1 lengths yr 1r2a jsi zr 1yr Effective rotor impedance r2eff realzr Effective rotor resistance x2eff imagzr Effective rotor reactance synchronous speed x2eff x2effsii tind2ii 3 vth2 r2eff sii wsync rth r2effsii2 xt end figure1 Plot the t plotnmtind1bLineWidth hold on plotnmt xlabelbfitnm ylabelbf auind title bfInduction Motor legend Singlecage designDoublecage design grid on hold off 201 The resulting plot is shown below Chapter 7 DC Machinery Fundamentals 71 The following information is given about the simple rotating loop shown in Figure 76 B 0 4 T VB 48 V m 50 l R 0 4 r 0 25 m 500 rads a Is this machine operating as a motor or a generator Explain b What is the current i flowing into or out of the machine What is the power flowing into or out of the machine c If the speed of the rotor were changed to 550 rads what would happen to the current flow into or out of the machine d If the speed of the rotor were changed to 450 rads what would happen to the current flow into or out of the machine 202 SOLUTION a If the speed of rotation of the shaft is 500 rads then the voltage induced in the rotating loop will be ind 2 e rlB ind 2 025 m 05 m 04 T 500 rads 50 V e Since the external battery voltage is only 48 V this machine is operating as a generator charging the battery b The current flowing out of the machine is approximately ind 50 V 48 V 50 A 04 B e V i R Note that this value is the current flowing while the loop is under the pole faces When the loop goes beyond the pole faces will momentarily fall to 0 V and the current flow will momentarily reverse Therefore the average current flow over a complete cycle will be somewhat less than 50 A ind e 203 c If the speed of the rotor were increased to 550 rads the induced voltage of the loop would increase to ind 2 e rlB ind 2 025 m 05 m 04 T 550 rads 55 V e and the current flow out of the machine will increase to ind 55 V 48 V 175 A 04 B e V i R d If the speed of the rotor were decreased to 450 rads the induced voltage of the loop would fall to ind 2 e rlB ind 2 025 m 05 m 04 T 450 rads 45 V e Here eind is less than B V so current flows into the loop and the machine is acting as a motor The current flow into the machine would be ind 48 V 45 V 75 A 04 VB e i R 72 Refer to the simple twopole eightcoil machine shown in Figure P71 The following information is given about this machine Figure P71 B 10 T in air gap l 03 m length of coil sides radius of coils 0 10 m r CCW 1800 rmin n The resistance of each rotor coil is 004 a Is the armature winding shown a progressive or retrogressive winding b How many current paths are there through the armature of this machine c What are the magnitude and the polarity of the voltage at the brushes in this machine d What is the armature resistance of this machine RA e If a 1 resistor is connected to the terminals of this machine how much current flows in the machine Consider the internal resistance of the machine in determining the current flow f What are the magnitude and the direction of the resulting induced torque g Assuming that the speed of rotation and magnetic flux density are constant plot the terminal voltage of this machine as a function of the current drawn from it 204 SOLUTION a This winding is progressive since the ends of each coil are connected to the commutator segments ahead of the segments that the beginnings of the coils are connected to b There are two current paths in parallel through the armature of this machine this is a simplex lap winding c The voltage is positive at brush x with respect to brush y since the voltage in the conductors is positive out of the page under the North pole face and positive into the page under the South pole face The voltage produced by this machine can be found from Equations 732 and 733 A ZvBl Zr Bl E a a where Z is the number of conductors under the pole faces since the ones between the poles have no voltage in them There are 16 conductors in this machine and about 12 of them are under the pole faces at any given time 2 rad 1 min 1800 rmin 1885 rads 1 r 60 s 12 cond 010 m 1885 rads 10 T 03 m 339 V 2 current paths A Zr Bl E a 205 d There are 8 coils on this machine in two parallel paths with each coil having a resistance of 004 Therefore the total resistance A R is 004 004 004 004 004 004 004 004 004 004 004 004 004 004 004 004 RA 008 A R e The voltage produced by this machine is 339 V as found in part c Therefore the current flowing in the machine will be load 339 V 314 A 008 10 A A A E I R R f The induced torque is given by Equation 746 ind 12 cond 010 m 03 m 10 T 339 A 2 current paths ZrlBIA a ind 412 N m CW opposite to the direction of rotation e The terminal voltage of this machine is given by T A A V E I R A Mfile prob72m Mfile to create a plot of the terminal voltage vs load current for the machine of problem 72 First initialize the values needed in this program Ea 314 Stator resistance Ra 008 Stator reactance Get the line currents to calculate the anser for Ia 0135 Calculate the terminal voltage Vt Ea Ia Ra Plot the terminal voltage figure1 plotIaVtbLineWidth20 xlabelbfCurrent A ylabelbfitVT rmbfV title bfTerminal Voltage vs Load axis 0 35 0 32 grid on hold off The resulting plot is shown below 206 73 Prove that the equation for the induced voltage of a single simple rotating loop ind 2 e 76 is just a special case of the general equation for induced voltage in a dc machine EA K 738 SOLUTION From Equation 738 EA K where 2 ZP K a For the simple rotation loop Z 2 There are 2 conductors P 2 There are 2 poles a 1 There is one current path through the machine Therefore 2 2 2 2 2 1 ZP K a and Equation 738 reduces to Equation 76 74 A dc machine has 8 poles and a rated current of 120 A How much current will flow in each path at rated conditions if the armature is a simplex lapwound b duplex lapwound c simplex wavewound SOLUTION a Simplex lapwound 1 8 8 paths a mP 207 Therefore the current per path is 120 A 15 A 8 AI I a b Duplex lapwound 2 8 16 paths a mP Therefore the current per path is 120 A 75 A 16 AI I a c Simplex wavewound 2 2 1 2 path a m s Therefore the current per path is 120 A 60 A 2 AI I a 75 How many parallel current paths will there be in the armature of an 20pole machine if the armature is a simplex lapwound b duplex wavewound c triplex lapwound d quadruplex wavewound SOLUTION a Simplex lapwound 120 20 paths a mP b Duplex wavewound 2 22 4 path a m s c Triplex lapwound 320 60 paths a mP d Quadruplex wavewound 2 24 8 path a m s 76 The power converted from one form to another within a dc motor was given by conv ind A A m P E I Use the equations for EA and ind Equations 738 and 749 to prove that E I A A ind m that is prove that the electric power disappearing at the point of power conversion is exactly equal to the mechanical power appearing at that point SOLUTION conv A A P E I Substituting Equation 838 for A E conv A P K I conv A P K I But from Equation 749 ind K A I so 208 conv ind P 77 An eightpole 25kW 120V dc generator has a duplex lapwound armature which has 64 coils with 10 turns per coil Its rated speed is 3600 rmin a How much flux per pole is required to produce the rated voltage in this generator at noload conditions b What is the current per path in the armature of this generator at the rated load c What is the induced torque in this machine at the rated load d How many brushes must this motor have How wide must each one be e If the resistance of this winding is 0011 per turn what is the armature resistance of this machine RA SOLUTION a 2 A ZP E K a In this machine the number of current paths is 2 8 16 a mP The number of conductor is 64 coils 10 turnscoil 2 conductorsturn 1200 Z The equation for induced voltage is 2 A ZP E a so the required flux is 1200 cond 8 poles 2 rad 1 min 120 V 3600 rmin 2 16 paths 1 r 60 s 120 V 36000 000333 Wb b At rated load the current flow in the generator would be 25 kW 208 A 120 V A I There are a m P 28 16 parallel current paths through the machine so the current per path is 208 A 13 A 16 IA I a c The induced torque in this machine at rated load is ind 2 A ZP a I ind 1200 cond 8 poles 000333 Wb 208 A 2 16 paths ind 661 N m 209 d This motor must have 8 brushes since it is lapwound and has 8 poles Since it is duplexwound each brush must be wide enough to stretch across 2 complete commutator segments e There are a total of 64 turns on the armature of this machine so the number of turns per path is 10 640 650 turns 40 turnspath 16 paths P N The total resistance per path is 40 0011 044 P R Since there are 16 parallel paths through the machine the armature resistance of the generator is 044 00275 16 paths RA 78 Figure P72 shows a small twopole dc motor with eight rotor coils and 10 turns per coil The flux per pole in this machine is 0006 Wb a If this motor is connected to a 12V dc car battery what will the noload speed of the motor be b If the positive terminal of the battery is connected to the rightmost brush on the motor which way will it rotate c If this motor is loaded down so that it consumes 600 W from the battery what will the induced torque of the motor be Ignore any internal resistance in the motor SOLUTION a At no load T A V E K If K is known then the speed of the motor can be found The constant K is given by 2 ZP K a On the average about 6 of the 8 coils are under the pole faces at any given time so the average number of active conductors is Z 6 coils4 turnscoil2 conductorsturn 48 conductors There are two poles and two current paths so 48 cond 2 poles 764 2 2 2 paths ZP K a 210 The speed is given by 12 V 262 rads 764 0006 Wb EA K 1 r 60 s 262 rads 2500 rmin 2 rad 1 min nm b If the positive terminal of the battery is connected to the rightmost brush current will flow into the page under the South pole face producing a CW torque CW rotation c If the motor consumes 600 W from the battery the current flow is 600 W 50 A 12 V B P I V Therefore the induced torque will be ind 764 0006 Wb 50 A 229 N m CW K IA 79 Refer to the machine winding shown in Figure P73 211 a How many parallel current paths are there through this armature winding b Where should the brushes be located on this machine for proper commutation How wide should they be c What is the plex of this machine d If the voltage on any single conductor under the pole faces in this machine is e what is the voltage at the terminals of this machine SOLUTION a This is a duplex twopole lap winding so there are 4 parallel current paths through the rotor b The brushes should be shorting out those windings lying between the two poles At the time shown those windings are 1 2 9 and 10 Therefore the brushes should be connected to short out commutator 212 213 segments bcd and jkl at the instant shown in the figure Each brush should be two commutator segments wide since this is a duplex winding c Duplex see above d There are 16 coils on the armature of this machine Of that number an average of 14 of them would be under the pole faces at any one time Therefore there are 28 conductors divided among 4 parallel paths which produces 7 conductors per path Therefore T A V e E 7 for noload conditions 710 Describe in detail the winding of the machine shown in Figure P74 If a positive voltage is applied to the brush under the north pole face which way will this motor rotate SOLUTION This is a 2pole retrogressive lap winding If a positive voltage is applied to the brush under the North pole face the rotor will rotate in a counterclockwise direction Chapter 8 DC Motors and Generators Problems 81 to 812 refer to the following dc motor 30 hp 110 A Prated ILrated V 240 V 2700 turns per pole T N F 1800 rmin 14 turns per pole nrated NSE 019 75 RA RF 002 100 to 400 RS Radj Rotational losses 3550 W at full load Magnetization curve as shown in Figure P81 Note An electronic version of this magnetization curve can be found in file p81magdat which can be used with MATLAB programs Column 1 contains field current in amps and column 2 contains the internal generated voltage EA in volts In Problems 81 through 87 assume that the motor described above can be connected in shunt The equivalent circuit of the shunt motor is shown in Figure P82 214 81 If the resistor is adjusted to 175 what is the rotational speed of the motor at noload conditions Radj SOLUTION At noload conditions 240 V A T E V The field current is given by adj 240 V 240 V 0960 A 175 75 250 T F F V I R R From Figure P91 this field current would produce an internal generated voltage of 241 V at a speed of 1200 rmin Therefore the speed n with a voltage Ao E on A E of 240 V would be A Ao o E n E n 240 V 1200 rmin 1195 rmin 241 V A o Ao E n n E 82 Assuming no armature reaction what is the speed of the motor at full load What is the speed regulation of the motor SOLUTION At full load the armature current is adj 240 V 110 A 109 A 250 T A L F L F V I I I I R R The internal generated voltage A E is 240 V 109 A 019 2193 V A T A A E V I R The field current is the same as before and there is no armature reaction so EAo is still 241 V at a speed of 1200 rmin Therefore on 2193 V 1200 rmin 1092 rmin 241 V A o Ao E n n E The speed regulation is nl fl fl 1195 rmin 1092 rmin SR 100 100 94 1092 rmin n n n 215 83 If the motor is operating at full load and if its variable resistance is increased to 250 what is the new speed of the motor Compare the fullload speed of the motor with 175 to the fullload speed with 250 Assume no armature reaction as in the previous problem Radj Radj Radj SOLUTION If Radj is set to 250 the field current is now adj 240 V 240 V 0739 A 250 75 325 T F F V I R R Since the motor is still at full load A E is still 2183 V From the magnetization curve Figure P81 the new field current FI would produce a voltage EAo of 212 V at a speed of 1200 rmin Therefore on 2183 V 1200 rmin 1236 rmin 212 V A o Ao E n n E Note that Radj has increased and as a result the speed of the motor n increased 84 Assume that the motor is operating at full load and that the variable resistor is again 175 If the armature reaction is 1000 Aturns at full load what is the speed of the motor How does it compare to the result for Problem 82 Radj SOLUTION The field current is again 096 A and the motor is again at full load conditions However this time there is an armature reaction of 1200 Aturns and the effective field current is AR 1000 A turns 096 A 059 A 2700 turns F F F I I N From Figure P91 this field current would produce an internal generated voltage EAo of 185 V at a speed of 1200 rmin The actual internal generated voltage at these conditions is on A E 240 V 109 A 019 2193 V A T A A E V I R Therefore the speed n with a voltage of 240 V would be 2193 V 1200 rmin 1423 rmin 185 V A o Ao E n n E If all other conditions are the same the motor with armature reaction runs at a higher speed than the motor without armature reaction 85 If can be adjusted from 100 to 400 what are the maximum and minimum noload speeds possible with this motor Radj SOLUTION The minimum speed will occur when Radj 100 and the maximum speed will occur when Radj 400 The field current when Radj 100 is adj 240 V 240 V 137 A 100 75 175 T F F V I R R From Figure P91 this field current would produce an internal generated voltage EAo of 2715 V at a speed of 1200 rmin Therefore the speed n with a voltage of 240 V would be on 216 A Ao o E n E n 240 V 1200 rmin 1061 rmin 2715 V A o Ao E n n E The field current when Radj 400 is adj 240 V 240 V 0505 A 400 75 500 T F F V I R R From Figure P81 this field current would produce an internal generated voltage EAo of 167 V at a speed of 1200 rmin Therefore the speed n with a voltage of 240 V would be on A Ao o E n E n 240 V 1200 rmin 1725 rmin 167 V A o Ao E n n E 86 What is the starting current of this machine if it is started by connecting it directly to the power supply How does this starting current compare to the fullload current of the motor VT SOLUTION The starting current of this machine ignoring the small field current is start 240 V 1260 A 019 T L A V I R The rated current is 110 A so the starting current is 115 times greater than the fullload current This much current is extremely likely to damage the motor 87 Plot the torquespeed characteristic of this motor assuming no armature reaction and again assuming a fullload armature reaction of 1200 Aturns Assume that the armature reaction increases linearly with increases in armature current SOLUTION This problem is best solved with MATLAB since it involves calculating the torquespeed values at many points A MATLAB program to calculate and display both torquespeed characteristics is shown below Mfile prob87m Mfile to create a plot of the torquespeed curve of the the shunt dc motor with and without armature reaction Get the magnetization curve Note that this curve is defined for a speed of 1200 rmin load p81magdat ifvalues p81mag1 eavalues p81mag2 n0 1200 First initialize the values needed in this program vt 240 Terminal voltage V rf 75 Field resistance ohms radj 175 Adjustable resistance ohms ra 019 Armature resistance ohms il 01110 Line currents A 217 218 nf 2700 Number of turns on field far0 1000 Armature reaction 110 A Atm Calculate the armature current for each load ia il vt rf radj Now calculate the internal generated voltage for each armature current ea vt ia ra Calculate the armature reaction MMF for each armature current far ia 55 far0 Calculate the effective field current with and without armature reaction Ther term ifar is the field current with armature reaction and the term ifnoar is the field current without armature reaction ifar vt rf radj far nf ifnoar vt rf radj Calculate the resulting internal generated voltage at 1200 rmin by interpolating the motors magnetization curve ea0ar interp1ifvalueseavaluesifar ea0noar interp1ifvalueseavaluesifnoar Calculate the resulting speed from Equation 913 nar ea ea0ar n0 nnoar ea ea0noar n0 Calculate the induced torque corresponding to each speed from Equations 855 and 856 tindar ea ia nar 2 pi 60 tindnoar ea ia nnoar 2 pi 60 Plot the torquespeed curves figure1 plottindnoarnnoarbLineWidth20 hold on plottindarnarkLineWidth20 xlabelbf auind Nm ylabelbfitnm rmbfrmin title bfShunt DC Motor TorqueSpeed Characteristic legendNo armature reactionWith armature reaction axis 0 125 800 1600 grid on hold off The resulting plot is shown below For Problems 88 and 89 the shunt dc motor is reconnected separately excited as shown in Figure P83 It has a fixed field voltage V of 240 V and an armature voltage V that can be varied from 120 to 240 V F A 88 What is the noload speed of this separately excited motor when 175 and a V 120 V b 180 V c V 240 V Radj A VA A SOLUTION At noload conditions A A E V The field current is given by adj 240 V 240 V 096 A 175 76 250 F F F V I R R From Figure P91 this field current would produce an internal generated voltage EAo of 241 V at a speed of 1200 rmin Therefore the speed n with a voltage of 240 V would be on 219 A Ao o E n E n A o Ao E n n E a If VA 120 V then A E 120 V and 120 V 1200 rmin 598 rmin 241 V n b If VA 180 V then A E 180 V and 180 V 1200 rmin 986 rmin 241 V n c If VA 240 V then A E 240 V and 240 V 1200 rmin 1195 rmin 241 V n 89 For the separately excited motor of Problem 88 a What is the maximum noload speed attainable by varying both V and A Radj b What is the minimum noload speed attainable by varying both V and A Radj c What is the motors efficiency at rated conditions Note Assume that 1 the brush voltage drop is 2 V 2 the core loss is to e determined at an armature voltage equal to the armature voltage under full load and 3 stray load losses are 1 percent of full load SOLUTION a The maximum speed will occur with the maximum VA and the maximum Radj The field current when Radj 400 is adj 240 V 240 V 0505 A 400 75 475 T F F V I R R From Figure P81 this field current would produce an internal generated voltage EAo of 167 V at a speed of 1200 rmin At noload conditions the maximum internal generated voltage on A A E V 240 V Therefore the speed n with a voltage of 240 V would be A Ao o E n E n 240 V 1200 rmin 1725 rmin 167 V A o Ao E n n E b The minimum speed will occur with the minimum VA and the minimum Radj The field current when Radj 100 is adj 240 V 240 V 137 A 100 75 175 T F F V I R R 220 From Figure P81 this field current would produce an internal generated voltage EAo of 271 V at a speed of 1200 rmin At noload conditions the minimum internal generated voltage on A A E V 120 V Therefore the speed n with a voltage of 120 V would be A Ao o E n E n 120 V 1200 rmin 531 rmin 271 V A o Ao E n n E For Problems 810 to 811 the motor is connected cumulatively compounded as shown in Figure P84 810 If the motor is connected cumulatively compounded with 175 Radj a What is the noload speed of the motor b What is the fullload speed of the motor c What is its speed regulation d Calculate and plot the torquespeed characteristic for this motor Neglect armature effects in this problem SOLUTION At noload conditions 240 V A T E V The field current is given by adj 240 V 240 V 0960 A 175 75 250 T F F V I R R From Figure P81 this field current would produce an internal generated voltage EAo of 241 V at a speed of 1200 rmin Therefore the speed n with a voltage of 240 V would be on A Ao o E n E n 240 V 1200 rmin 1195 rmin 241 V A o Ao E n n E At full load conditions the armature current is adj 110 A 096 A 109 A T A L F L F V I I I I R R EA The internal generated voltage is 221 0 V 109 A 021 2171 V S 24 A T A A E V I R R The equivalent field current is 222 SE 0 F F A N I I I 14 turns 96 A 109 A 153 A 2700 turns NF ould produce an internal generated voltage EAo From Figure P81 this field current w of 279 V at a speed on of 1200 rmin Therefore 2171 A o E n n V 1200 rmin 934 rmin 279 V EAo The speed regulation is nl fl SR 10 n n fl 1195 rmin 934 rmin 0 100 279 934 rmin n ue ed characteristic can best be plotted with a MATLAB program An appropriate program is shown below rob810m plot of the torquespeed curve of the curve Note that this curve is ialize the values needed in this program hms ohms ated voltage for e field current for each armature age at The torq spe Mfile p Mfile to create a a cumulatively compounded dc motor without armature reaction Get the magnetization defined for a speed of 1200 rmin load p81magdat ifvalues p81mag1 eavalues p81mag2 n0 1200 First init vt 240 Terminal voltage V rf 75 Field resistance ohms radj 175 Adjustable resistance o ra 021 Armature series resistance il 01110 Line currents A nf 2700 Number of turns on shunt field nse 14 Number of turns on series field Calculate the armature current for each load ia il vt rf radj Now calculate the internal gener each armature current ea vt ia ra Calculate the effectiv current if vt rf radj nse nf ia Calculate the resulting internal generated volt 1200 rmin by interpolating the motors magnetization curve 223 erp1ifvalueseavaluesif Calculate the resulting speed from Equation 913 Calculate the induced torque corresponding to each Plot the torquespeed curves bLineWidth20 Motor TorqueSpeed 00 plot is shown below ea0 int n ea ea0 n0 speed from Equations 855 and 856 tind ea ia n 2 pi 60 figure1 plottindn xlabelbf auind Nm ylabelbfitnm rmbfrmin title bfCumulativelyCompounded DC Characteristic axis 0 125 800 12 grid on The resulting Compare this torquespeed curve to that of the shunt motor in Problem 87 Both curves are plotted on 811 pounded and is operating at full load What will the new speed the same scale to facilitate comparison The motor is connected cumulatively com of the motor be if Radj is increased to 250 How does the new speed compared to the fullload speed calculated in Proble 10 SOLUTION If m 8 Radj is increased to 250 the field current is given by adj 240 V 240 V 0739 A 250 75 250 T F F V I R R At full load conditions the armature current is 110 A 0739 A 1092 A A L F I I I The internal generated voltage A E is 240 V 1092 A 021 2171 V A T A A S E V I R R The equivalent field current is SE 14 turns 0739 A 1092 A 1305 A 2700 turns F F A F N I I I N From Figure P91 this field current would produce an internal generated voltage EAo of 268 V at a speed of 1200 rmin Therefore on 2171 V 1200 rmin 972 rmin 268 V A o Ao E n n E The new fullload speed is higher than the fullload speed in Problem 810 For Problem 812 the motor is now connected differentially compounded as shown in Figure P84 812 The motor is now connected differentially compounded a If R 175 what is the noload speed of the motor adj b What is the motors speed when the armature current reaches 20 A 40 A 60 A c Calculate and plot the torquespeed characteristic curve of this motor SOLUTION a At noload conditions 240 V The field current is given by A T E V adj 240 V 240 V 096 A 175 75 275 F F F V I R R From Figure P81 this field current would produce an internal generated voltage EAo of 241 V at a speed of 1200 rmin Therefore the speed n with a voltage of 240 V would be on A Ao o E n E n 240 V 1200 rmin 1195 rmin 241 V A o Ao E n n E b At IA 20A the internal generated voltage A E is 240 V 20 A 021 2358 V A T A A S E V I R R The equivalent field current is SE 14 turns 096 A 20 A 0856 A 2700 turns F F A F N I I I N 224 From Figure P81 this field current would produce an internal generated voltage EAo of 229 V at a speed of 1200 rmin Therefore on 2358 V 1200 rmin 1236 rmin 229 V A o Ao E n n E At IA 40A the internal generated voltage A E is 240 V 40 A 021 2316 V A T A A S E V I R R The equivalent field current is SE 14 turns 096 A 40 A 0753 A 2700 turns F F A F N I I I N From Figure P81 this field current would produce an internal generated voltage EAo of 214 V at a speed of 1200 rmin Therefore on 2316 V 1200 rmin 1299 rmin 214 V A o Ao E n n E At IA 60A the internal generated voltage A E is 240 V 60 A 021 2274 V A T A A S E V I R R The equivalent field current is SE 14 turns 096 A 60 A 0649 A 2700 turns F F A F N I I I N From Figure P81 this field current would produce an internal generated voltage EAo of 196 V at a speed of 1200 rmin Therefore on 2274 V 1200 rmin 1392 rmin 196 V A o Ao E n n E c The torquespeed characteristic can best be plotted with a MATLAB program An appropriate program is shown below Mfile prob812m Mfile to create a plot of the torquespeed curve of the a cumulatively compounded dc motor without armature reaction Get the magnetization curve Note that this curve is defined for a speed of 1200 rmin load p81magdat ifvalues p81mag1 eavalues p81mag2 n0 1200 First initialize the values needed in this program vt 240 Terminal voltage V rf 75 Field resistance ohms radj 175 Adjustable resistance ohms 225 ra 021 Armature series resistance ohms il 01110 Line currents A 226 shunt field Calculate the armature current for each load Now calculate the internal generated voltage for Calculate the effective field current for each armature rf radj nse nf ia Calculate the resulting internal generated voltage at erp1ifvalueseavaluesif Calculate the resulting speed from Equation 913 Calculate the induced torque corresponding to each Plot the torquespeed curves bLineWidth20 C Motor TorqueSpeed 00 nf 2700 Number of turns on nse 14 Number of turns on series field ia il vt rf radj each armature current ea vt ia ra current if vt 1200 rmin by interpolating the motors magnetization curve ea0 int n ea ea0 n0 speed from Equations 855 and 856 tind ea ia n 2 pi 60 figure1 plottindn xlabelbf auind Nm ylabelbfitnm rmbfrmin title bfDifferentiallyCompounded D Characteristic axis 0 250 800 16 grid on The resulting plot is shown below Compare this torquespeed curve to that of the shunt motor in Problem 87 and the cumulatively compounded motor in Problem 810 813 A 75hp 120V series dc motor has an armature resistance of 01 and a series field resistance of 008 At full load the current input is 56 A and the rated speed is 1050 rmin Its magnetization curve is shown in Figure P85 The core losses are 220 W and the mechanical losses are 230 W at full load Assume that the mechanical losses vary as the cube of the speed of the motor and that the core losses are constant 227 a What is the efficiency of the motor at full load b What are the speed and efficiency of the motor if it is operating at an armature current of 40 A c Plot the torquespeed characteristic for this motor Note An electronic version of this magnetization curve can be found in file p85magdat which can be used with MATLAB programs Column 1 contains field current in amps and column 2 contains the internal generated voltage EA in volts SOLUTION a The output power of this motor at full load is OUT 75 hp 746 Whp 5595 W P The input power is IN 120 V 56 A 6720 W T L P V I Therefore the efficiency is 228 OUT IN 5595 W 100 100 833 6720 W P P b If the armature current is 40 A then the input power to the motor will be IN 120 V 40 A 4800 W T L P V I The internal generated voltage at this condition is 2 120 V 40 A 010 008 1128 V A T A A S E V I R R and the internal generated voltage at rated conditions is 2 120 V 56 A 010 008 1099 V A T A A S E V I R R The final speed is given by the equation 2 2 2 2 2 1 2 2 1 Ao A A A 1 o E n E K E K E n since the ratio 2 1 Ao Ao E E is the same as the ratio 2 1 Therefore the final speed is 1 2 2 1 1 2 Ao A A Ao E E n n E E From Figure P85 the internal generated voltage EAo2 for a current of 40 A and a speed of 1200 rmin is on EAo2 120 V and the internal generated voltage EAo1 for a current of 56 A and a speed of 1200 rmin is on EAo1 133 V 1 2 2 1 1 2 1128 V 133 V 1050 rmin 1195 rmin 1099 V 120 V Ao A A Ao E E n n E E The power converted from electrical to mechanical form is conv 1128 V 40 A 4512 W A A P E I The core losses in the motor are 220 W and the mechanical losses in the motor are 230 W at a speed of 1050 rmin The mechanical losses in the motor scale proportionally to the cube of the rotational speedm so the mechanical losses at 1326 rmin are 3 3 2 mech 1 1195 rmin 230 W 230 W 339 W 1050 rmin n P n Therefore the output power is OUT conv mech core 4512 W 339 W 220 W 3953 W P P P P and the efficiency is OUT IN 3953 W 100 100 824 4800 W P P c A MATLAB program to plot the torquespeed characteristic of this motor is shown below Mfile prob913m Mfile to create a plot of the torquespeed curve of the the series dc motor in Problem 913 229 230 Get the magnetization curve Note that this curve is defined for a speed of 1200 rmin load p85magdat ifvalues p85mag1 eavalues p85mag2 n0 1200 First initialize the values needed in this program vt 120 Terminal voltage V ra 036 Armature field resistance ohms ia 9158 Armature line currents A Calculate the internal generate voltage ea ea vt ia ra Calculate the resulting internal generated voltage at 1200 rmin by interpolating the motors magnetization curve Note that the field current is the same as the armature current for this motor ea0 interp1ifvalueseavaluesiaspline Calculate the motors speed using the known fact that the motor runs at 1050 rmin at a current of 58 A We know that Ea2 K phi2 n2 Eao2 n2 Ea1 K phi1 n1 Eao1 n1 Ea2 Eao1 n2 n1 Ea1 Eao2 where Ea0 is the internal generated voltage at 1200 rmin for a given field current Speed will be calculated by reference to full load speed and current n1 1050 1050 rmin at full load Eao1 interp1ifvalueseavalues58spline Ea1 vt 58 ra Get speed Eao2 interp1ifvalueseavaluesiaspline n eaEa1 Eao1 Eao2 n1 Calculate the induced torque corresponding to each speed from Equations 855 and 856 tind ea ia n 2 pi 60 Plot the torquespeed curve figure1 plottindnbLineWidth20 hold on xlabelbf auind Nm ylabelbfitnm rmbfrmin title bfSeries DC Motor TorqueSpeed Characteristic grid on hold off The resulting torquespeed characteristic is shown below 814 A 20hp 240V 76A 900 rmin series motor has a field winding of 33 turns per pole Its armature resistance is 009 and its field resistance is 006 The magnetization curve expressed in terms of magnetomotive force versus EA at 900 rmin is given by the following table EA V 95 150 188 212 229 243 F A turns 500 1000 1500 2000 2500 3000 Armature reaction is negligible in this machine a Compute the motors torque speed and output power at 33 67 100 and 133 percent of fullload armature current Neglect rotational losses b Plot the terminal characteristic of this machine Note An electronic version of this magnetization curve can be found in file prob814magdat which can be used with MATLAB programs Column 1 contains magnetomotive force in ampereturns and column 2 contains the internal generated voltage EA in volts SOLUTION Note that this magnetization curve has been stored in a file called prob814magdat The first column of the file is an array of mmfvalues and the second column is an array of eavalues These values are valid at a speed 900 rmin Because the data in the file is relatively sparse it is important that interpolation be done using smooth curves so be sure to specify the spline option in the MATLAB interp1 function on load prob814magdat 231 mmfvalues prob814mag1 eavalues prob814mag2 Eao interp1mmfvalueseavaluesmmfspline a Since full load corresponds to 76 A this calculation must be performed for armature currents of 253 A 507 A 76 A and 1013 A If A I 233 A then 240 V 253 A 009 006 2362 V A T A A S E V I R R The magnetomotive force is 33 turns 253 A 835 A turns NIA F which produces a voltage EAo of 134 V at 900 rmin Therefore the speed of the motor at these conditions is on 2362 V 900 rmin 1586 rmin 134 V A o Ao E n n E The power converted from electrical to mechanical form is conv 2362 V 253 A 5976 W A A P E I Since the rotational losses are ignored this is also the output power of the motor The induced torque is conv ind 5976 W 36 N m 2 rad 1 min 1586 rmin 1 r 60 s m P If A 507 A then I 240 V 507 A 009 006 2324 V A T A A S E V I R R The magnetomotive force is 33 turns 507 A 1672 A turns NIA F which produces a voltage EAo of 197 V at 900 rmin Therefore the speed of the motor at these conditions is on 2324 V 900 rmin 1062 rmin 197 V A o Ao E n n E The power converted from electrical to mechanical form is conv 2324 V 507 A 11780 W A A P E I Since the rotational losses are ignored this is also the output power of the motor The induced torque is conv ind 11780 W 106 N m 2 rad 1 min 1062 rmin 1 r 60 s m P If A 76 A then I 240 V 76 A 009 006 2286 V A T A A S E V I R R The magnetomotive force is 33 turns 76 A 2508 A turns NIA F which produces a voltage EAo of 229 V at 900 rmin Therefore the speed of the motor at these conditions is on 2286 V 900 rmin 899 rmin 229 V A o Ao E n n E 232 233 electrical to mechanical form is The power converted from conv 2286 V 76 A 17370 W A A P E I Since the rotational losses are ignored this is also the output power of the motor The induced torque is conv 17370 W 185 N m 1 r 60 s P 1 A then ind 2 rad 1 min 899 rmin m If I 10 3 A 240 V 1013 A 009 006 2248 V A T A A S E V I R R 33 turns 1013 A 3343 A turns NIA F which produces a voltage The magnetomotive force is EAo of 252 V at 900 rm on in Therefore the speed of the motor at these conditions is 2248 V 900 rmin 803 rmin 252 V A Ao E n n E o electrical to mechanical form is The power converted from conv 2248 V 1013 A 22770 W A A P E I Since the rotational losses are ignored this is also the output power of the motor The induced torque is conv 22770 W 271 N m 1 r 60 s P A LAB program to plot the torquespeed characteristic of this motor is shown below Mfile seriestscurvem the series dc motor in Problem 814 eded in this program al voltage V Armature field resistance ohms d oltage ea tage at zation ind 2 rad 1 min 803 rmin m b A M T Mfile to create a plot of the torquespeed curve of the Get the magnetization curve Note that this curve is defined for a speed of 900 rmin load prob814magdat mmfvalues prob814mag1 eavalues prob814mag2 n0 900 First initialize the values ne vt 240 Termin ra 015 ia 15176 Armature line currents A ns 33 Number of series turns on fiel Calculate the MMF for each load f ns ia Calculate the internal generate v ea vt ia ra Calculate the resulting internal generated vol 900 rmin by interpolating the motors magneti 234 curve Specify cubic spline interpolation to provide 55 and 856 ind ea ia n 2 pi 60 ottindnbLineWidth20 tauind Nm fSeries DC Motor TorqueSpeed Characteristic istic is shown below good results with this sparse magnetization curve ea0 interp1mmfvalueseavaluesfspline Calculate the motors speed from Equation 913 n ea ea0 n0 Calculate the induced torque corresponding to each speed from Equations 8 t Plot the torquespeed curve figure1 pl hold on xlabelbf ylabelbfitnm rmbfrmin title b axis 0 700 0 5000 grid on hold off The resulting torquespeed character 815 A 300hp 440V 560A 863 rmin shunt dc motor has been tested and the following data were taken Blockedrotor test 149 V exclusive of brushes A V VF 440 V I 500 A 752 A FI A Noload operation 440 V includi A V 235 ng brushes 750 A FI I A 231 A n 863 rmin What i s this motors efficiency at the rated conditions Note Assume that 1 the brush voltage drop is 2 V 2 the core loss is to be determined at an armature voltage equal to the armature voltage under full load and 3 stray load losses are 1 percent of full load SOLUTION The armature resistance of this motor is br 149 V 00298 R VA br 500 A A AI Under noload conditions the core and mechanical losses taken together that is the rotational losses of this motor are equal to the product of the internal generated voltage A E and the armature current A I since this is no output power from the motor at noload conditions refore the rotational losses t rated speed can be found as The a 440 V 2 V 231 A 00298 4373 V A brush A A A E V V I R rot conv 4373 V 231 A 101 kW A A P P E I The input power to the motor at full load is IN 440 V 560 A 2464 T L P V I kW ut power from the motor at full load is er losses are h losses are r at full load is The outp OUT IN CU rot brush stray P P P P P P The copp 2 CU P I R V 560 A 2 00298 440 V 752 A 1265 kW A A F FI The brus brush brush P V I 2 V 560 A 112 kW A Therefore OUT IN P P CU rot brush stray P P P P OUT 2464 kW 141 kW 1265 kW 101 kW 112 kW 246 kW 206 kW P The moto s efficiency OUT 206 kW 100 P IN 100 836 2464 kW P roblems 816 to 819 refer to a 240V 100A dc motor which has both shunt and series windings Its set to 1 om he magnetization curve for this motor at 3000 rmin is P characteristics are RA 014 N 1500 turns F 005 15 turns RS NSE 200 3000 rmin RF nm 0 to 300 currently 20 Radj This motor has c pensating windings and interpoles T shown in Figure P86 Note An electronic version of this magnetization curve can be found in file p86magdat which can be used with MATLAB programs Column 1 contains field current in amps and column 2 contains the internal generated voltage EA in volts 236 816 The motor des 120 ristic for this motor eeds can be achieved by adjusting OL column of the file is an array of iavalues and the second column is an array of eavalues These cribed above is connected in shunt a What is the noload speed of this motor when Radj b What is its fullload speed c What is its speed regulation d Plot the torquespeed characte e Under noload conditions what range of possible sp Radj S UTION Note that this magnetization curve has been stored in a file called p86magdat The first values are valid at a speed on 3000 rmin These values can be used with the MATLAB interp1 function to look up an internal generated voltage as follows load p86magdat ifvalues p86mag1 values p86mag2 ea 237 valuesifspline Ea interp1ifvaluesea a If Radj 120 the total field resistance is 320 and the resulting field current is adj 200 120 F RF R 240 V 075 A VT I EAo This field current would produce a voltage of 245 V at a speed of 3000 rmin The actual s e actual speed will be on A E is 240 V o th 240 V 3000 rmin 9 rmin 245 V A o Ao E n n E 293 b At full load 100 A 075 A 9925 A A L F I I I and 240 V 9925 A 014 2261 V A T A A E V I R The fore the spee re d at full load will be 2261 V 3000 rmin 2769 rmin 245 V Ao E A o E c The speed regulation of this motor is n n nl fl fl 2769 r n 2939 rmin 2769 rmin SR 100 100 616 min n n d A MATLAB program to calculate the torquespeed characteristic is shown below Mfile prob816m ues needed in this program erminal voltage V Field resistance ohms r Mfile to create a plot of the torquespeed curve of the the shunt dc motor with and without armature reaction Get the magnetization curve Note that this curve is defined for a speed of 3000 rmin load p86magdat ifvalues p86mag1 eavalues p86mag2 n0 3000 First initialize the val vt 240 T rf 200 radj 120 Adjustable resistance ohms ra 014 Armature resistance ohms il 01100 Line currents A nf 1500 Number of turns on shunt field Calculate the armature current for each load ia il vt rf radj Now calculate the internal generated voltage fo each armature current ea vt ia ra 238 Calculate the resulting internal generated voltage at magnetization a0 interp1ifvalueseavaluesif ea0 n0 Calculate the induced torque corresponding to each pi 60 abelbf auind Nm hunt DC Motor TorqueSpeed Characteristic Calculate the effective field current if vt rf radj 1200 rmin by interpolating the motors curve e Calculate the resulting speed from Equation 913 n ea speed from Equations 855 and 856 tind ea ia n 2 Plot the torquespeed curves figure1 plottindnbLineWidth20 xl ylabelbfitnm rmbfrmin title bfS axis 0 80 2700 3000 grid on The resulting torquespeed curve is shown below e If Radj is maximum at noload conditions the total resistance is 500 and adj 240 V 048 A 200 300 T F F V I R R This field current would produce a voltage EAo of 195 V at a speed of on 3000 rmin The actual EA is 240 V so the actual speed will be 240 V 3000 rmin 3692 rmin 195 V A o E Ao If Radj is minimum at noload cond n n E itions the total resistance is 200 and adj 240 V 12 A 200 0 T F F V I R R This field current would produce a voltage EAo of 282 V at a speed of on 3000 rmin The actual EA is 240 V so the actual speed will be 240 V 3000 rmin 2553 rmin 282 V A o Ao E n n E 817 This machine is now connected as a R cumulatively compounded dc motor with 120 a What is the noload speed of this motor motor OL adj b What is its fullload speed c What is its speed regulation d Plot the torquespeed characteristic for this S UTION a The field current will be adj 240 V 075 A T F V I 200 120 RF R and the effective field current will be At no load AI 0 A and A 240 V T A A S E V I R R 075 A I I SE 15 turns 0 A 075 A 1500 turns F F A F N I N tage of 245 V at a speed of 3000 rmin The actual is 240 V so the actual speed at full load will be EAo on A E This field current would produce a vol 240 V 3000 rmin 2939 rmin 245 V A o Ao E n n E b The field current will be adj 240 V 075 A 200 120 T F F V I R R At full load A L F I I I 100 A 075 A 9925 A and 240 V 9925 A 014 005 2211 V A T A A S E V I R R 239 and the effective field current will be 240 SE 15 turns 075 A 9925 A 174 A 1500 turns F F A F N I I I N This field current would produce a voltage of 292 V at a speed of 3000 rmin The actual is 240 V so the actual speed at full load will be EAo on A E 2211 V 3000 rmin 2272 rmin 292 V A o Ao E n n E c The speed regulation of this motor is nl fl 2939 rmin 2272 rmin SR 100 2272 rmi n n n fl 100 294 n ATLAB program to calculate the torquespeed characteristic is shown below mu taively compounded dc mtor Get the magnetization curve Note that this curve is ad p86magdat 1 tance ohms djustable resistance ohms rmature resistance ohms Line currents A Number of turns on shunt field d armature nf ia e at ting the motors magnetization 13 d A M Mfile prob817m Mfile to create a plot of the torquespeed curve of a cu l defined for a speed of 3000 rmin lo ifvalues p86mag eavalues p86mag2 n0 3000 First initialize the values needed in this program vt 240 Terminal voltage V rf 200 Field resis radj 120 A ra 014 A il 01100 nf 1500 nse 15 Number of turns on series fiel Calculate the armature current for each load ia il vt rf radj Now calculate the internal generated voltage for each armature current ea vt ia ra Calculate the effective field current for each current if vt rf radj nse Calculate the resulting internal generated voltag 1200 rmin by interpola curve ea0 interp1ifvalueseavaluesif Calculate the resulting speed from Equation 9 n ea ea0 n0 241 om Equations 855 and 856 ottindnbLineWidth20 r TorqueSpeed aracteristic ote that is curve is plotted on the same scale as the Calculate the induced torque corresponding to each speed fr tind ea ia n 2 pi 60 Plot the torquespeed curves figure1 pl xlabelbf auind Nm ylabelbfitnm rmbfrmin title bfCumulatively Compounded DC Moto Ch axis 0 80 2300 3000 grid on The resulting torquespeed curve is shown below N shunt motor in Problem 816 818 The motor is reconnected differentially compounded with 120 Derive the shape of its torque speed characteristic SOLUTION A MATLAB program to calculate the torquespeed characteristic of this motor is shown below Mfile prob818m Mfile to create a plot of the torquespeed curve of a tization curve Note that this curve is ad p86magdat 1 Radj differentially compounded dc mtor Get the magne defined for a speed of 3000 rmin lo ifvalues p86mag eavalues p86mag2 n0 3000 242 tage V Field resistance ohms djustable resistance ohms rmature resistance ohms Line currents A f 1500 Number of turns on shunt field d e for armature f vt rf radj nse nf ia internal generated voltage at ating the motors magnetization curve 13 ea ea0 n0 om Equations 855 and 856 ottindnbLineWidth20 tor TorqueSpeed aracteristic First initialize the values needed in this program vt 240 Terminal vol rf 200 radj 120 A ra 014 A il 01100 n nse 15 Number of turns on series fiel Calculate the armature current for each load ia il vt rf radj Now calculate the internal generated voltag each armature current ea vt ia ra Calculate the effective field current for each current i Calculate the resulting 1200 rmin by interpol ea0 interp1ifvalueseavaluesif Calculate the resulting speed from Equation 9 n Calculate the induced torque corresponding to each speed fr tind ea ia n 2 pi 60 Plot the torquespeed curves figure1 pl xlabelbf auind Nm ylabelbfitnm rmbfrmin title bfDifferentially Compounded DC Mo Ch axis 0 80 2900 3500 grid on The resulting torquespeed curve is shown below 819 A series motor is now constructed from this machine by leaving the shunt field out entirely Derive the torquespeed characteristic of the resulting motor SOLUTION This motor will have extremely high speeds since there are only a few series turns and the flux in the motor will be very small A MATLAB program to calculate the torquespeed characteristic of this motor is shown below Mfile prob819m Mfile to create a plot of the torquespeed curve of the a series dc motor This motor was formed by removing the shunt field from the cumulativelycompounded machine of Problem 817 Get the magnetization curve Note that this curve is defined for a speed of 3000 rmin load p86magdat ifvalues p86mag1 eavalues p86mag2 n0 3000 First initialize the values needed in this program vt 240 Terminal voltage V rf 200 Field resistance ohms radj 120 Adjustable resistance ohms ra 014 Armature resistance ohms il 201100 Line currents A nf 1500 Number of turns on shunt field nse 15 Number of turns on series field Calculate the armature current for each load ia il vt rf radj 243 Now calculate the internal generated voltage for each armature current ea vt ia ra Calculate the effective field current for each armature current if nse nf ia Calculate the resulting internal generated voltage at 1200 rmin by interpolating the motors magnetization curve ea0 interp1ifvalueseavaluesif Calculate the resulting speed from Equation 913 n ea ea0 n0 Calculate the induced torque corresponding to each speed from Equations 855 and 856 tind ea ia n 2 pi 60 Plot the torquespeed curves figure1 plottindnbLineWidth20 xlabelbf auind Nm ylabelbfitnm rmbfrmin title bfSeries DC Motor TorqueSpeed Characteristic grid on The resulting torquespeed characteristic is shown below The extreme speeds in this characteristic are due to the very light flux in the machine To make a practical series motor out of this machine it would be necessary to include 20 to 30 series turns instead of 15 244 820 An automatic starter circuit is to be designed for a shunt motor rated at 20 hp 240 V and 75 A The armature resistance of the motor is 012 and the shunt field resistance is 40 The motor is to start with no more than 250 percent of its rated armature current and as soon as the current falls to rated value a starting resistor stage is to be cut out How many stages of starting resistance are needed and how big should each one be 245 F SOLUTION The rated line current of this motor is 75 A and the rated armature current is 75 A 6 A 69 A The maximum desired starting current is 2569 A 1725 A Therefore the total initial starting resistance must be A L I I I start1 240 V 1391 1725 A RA R start1 1391 012 1271 R The current will fall to rated value when A E rises to 240 V 1391 69 A 144 V A E At that time we want to cut out enough resistance to get the current back up to 1725 A Therefore start2 240 V 144 V 0557 1725 A RA R start2 0557 012 0437 R With this resistance in the circuit the current will fall to rated value when A rises to E 240 V 0557 69 A 2016 V A E At that time we want to cut out enough resistance to get the current back up to 1725 A Therefore start3 240 V 2016 V 0223 1725 A RA R start3 0223 012 0103 R With this resistance in the circuit the current will fall to rated value when A rises to E 240 V 0223 69 A 2246 V A E If the resistance is cut out when A E reaches 2246 V the resulting current is 240 V 2246 V 128 A 1725 A 012 AI so there are only three stages of starting resistance The three stages of starting resistance can be found from the resistance in the circuit at each state during starting start1 1 2 3 1217 R R R R start2 2 3 0437 R R R start3 3 0103 R R Therefore the starting resistances are 1 R 0780 2 R 0334 3 R 0103 821 A 10hp 120V 1000 rmin shunt dc motor has a fullload armature current of 70 A when operating at rated conditions The armature resistance of the motor is 012 and the field resistance is 40 RA RF The adjustable resistance in the field circuit may be varied over the range from 0 to 200 and is currently set to 100 Armature reaction may be ignored in this machine The magnetization curve for this motor taken at a speed of 1000 rmin is given in tabular form below Radj EA V 5 78 95 112 118 126 130 IF A 000 080 100 128 144 288 400 a What is the speed of this motor when it is running at the rated conditions specified above b The output power from the motor is 10 hp at rated conditions What is the output torque of the motor c What are the copper losses and rotational losses in the motor at full load ignore stray losses d What is the efficiency of the motor at full load e If the motor is now unloaded with no changes in terminal voltage or what is the noload speed of the motor Radj f Suppose that the motor is running at the noload conditions described in part e What would happen to the motor if its field circuit were to open Ignoring armature reaction what would the final steadystate speed of the motor be under those conditions g What range of noload speeds is possible in this motor given the range of field resistance adjustments available with Radj Note An electronic version of this magnetization curve can be found in file prob821magdat which can be used with MATLAB programs Column 1 contains field current in amps and column 2 contains the internal generated voltage EA in volts SOLUTION a If Radj 100 the total field resistance is 140 and the resulting field current is adj 120 V 0857 A 100 40 T F F V I R R This field current would produce a voltage EAo of 828 V at a speed of 1000 rmin The actual on A E is 120 V 70 A 012 1116 V A T A A E V I R so the actual speed will be 1116 V 1000 rmin 1348 rmin 828 V A o Ao E n n E b The output power is 10 hp and the output speed is 1000 rmin at rated conditions therefore the torque is out out 10 hp 746 Whp 712 N m 2 rad 1 min 1000 rmin 1 r 60 s m P c The copper losses are 246 2 2 CU 70 A 012 120 V 0857 A 691 W A A F F P I R V I The power converted from electrical to mechanical form is conv 1116 V 70 A 7812 W A A P E I The output power is OUT 10 hp 746 Whp 7460 W P Therefore the rotational losses are rot conv OUT 7812 W 7460 W 352 W P P P d The input power to this motor is IN 120 V 70 A 0857 A 8503 W T A F P V I I Therefore the efficiency is OUT IN 7460 W 100 100 877 8503 W P P e The noload A E will be 120 V so the noload speed will be 120 V 1000 rmin 1450 rmin 828 V A o Ao E n n E f If the field circuit opens the field current would go to zero drops to res A E AI ind n to a very high speed If FI 0 A EAo 85 V at 1800 rmin so 230 V 1000 rmin 46000 rmin 5 V A o Ao E n n E In reality the motor speed would be limited by rotational losses or else the motor will destroy itself first g The maximum value of Radj 200 so adj 120 V 0500 A 200 40 T F F V I R R This field current would produce a voltage of 506 V at a speed of 1000 rmin The actual EAo on A E is 120 V so the actual speed will be 120 V 1000 rmin 2372 rmin 506 V A o Ao E n n E The minimum value of Radj 0 so adj 120 V 30 A 0 40 T F F V I R R This field current would produce a voltage of about 1264 V at a speed of 1000 rmin The actual EAo on A E is 120 V so the actual speed will be 247 120 V 1000 rmin 949 rmin 1264 V A o Ao E n n E 822 The magnetization curve for a separately excited dc generator is shown in Figure P87 The generator is rated at 6 kW 120 V 50 A and 1800 rmin and is shown in Figure P88 Its field circuit is rated at 5A The following data are known about the machine 0 18 RA VF 120 V adj 0 to 40 R 20 F R N F 1000 turns per pole Answer the following questions about this generator assuming no armature reaction a If this generator is operating at no load what is the range of voltage adjustments that can be achieved by changing Radj b If the field rheostat is allowed to vary from 0 to 30 and the generators speed is allowed to vary from 1500 to 2000 rmin what are the maximum and minimum noload voltages in the generator 248 Note An electronic version of this magnetization curve can be found in file p87magdat which can be used with MATLAB programs Column 1 contains field current in amps and column 2 contains the internal generated voltage EA in volts SOLUTION a If the generator is operating with no load at 1800 rmin then the terminal voltage will equal the internal generated voltage The maximum possible field current occurs when 0 The current is EA adj R max adj 120 V 6 A 20 0 F F F V I R R From the magnetization curve the voltage EAo at 1800 rmin is 135 V Since the actual speed is 1800 rmin the maximum noload voltage is 135 V The minimum possible field current occurs when Radj 40 The current is max adj 120 V 20 A 20 40 F F F V I R R From the magnetization curve the voltage EAo at 1800 rmin is 795 V Since the actual speed is 1800 rmin the minimum noload voltage is 795 V b The maximum voltage will occur at the highest current and speed and the minimum voltage will occur at the lowest current and speed The maximum possible field current occurs when Radj 0 The current is max adj 120 V 6 A 20 0 F F F V I R R From the magnetization curve the voltage EAo at 1800 rmin is 135 V Since the actual speed is 2000 rmin the maximum noload voltage is A Ao o E n E n 2000 rmin 135 V 150 V 1800 rmin A Ao o n E n E 250 251 um possible field current occurs and minimum speed and field current The maximum e resistance is The minim adjustabl Radj 30 The current is max ad 120 V 24 A 20 30 F F F V I R R j netization curve the voltage EAo From the mag at 1800 rmin is 931 V Since the actual speed is 1500 aximum noload voltage is rmin the m A Ao o E n E n 1500 rmin 931 V 776 V 1800 rmin A A E o o n E n ature current of the generator in Problem 822 is 50 A the speed of the generator is 1700 rmin rminal voltage is 106 V how much field current must be flowing in the generator 823 If the arm and the te SOLUTION The internal generated voltage of this generator is 106 V 50 A 018 115 V A T A A E V I R at a speed of 1700 rmin This corresponds to an EAo at 1800 rmin of A E Ao o n E n 1800 rmin 115 V 1218 V 1700 rmin o Ao A n E E n From the magnetization curve this value of EAo requires a field current of 42 A 824 Assuming that the generator in Problem 822 has an armature reaction at full load equivalent to 400 when 5 A Aturns of magnetomotive force what will the terminal voltage of the generator be IF nm 1700 rmin and A 50 A SOLUTION When I FI is 5 A and the armature current is 50 A the magnetomotive force in the generator is net AR 000 turns 5 A 400 A turns 4600 A turns F 1 NI F F or 4600 A turns 1000 turns 46 A I F net F F N EAo The equivalent internal generated voltage of the generator at 1800 rmin would be 126 V The actual t 1700 rmin would be voltage a 1700 rmin 126 V 19 V 1800 rmin A Ao o n E n E 1 Therefore the terminal voltage would be 119 V 50 A 018 110 V T A A A V E I R 8 The machine in Problem 822 is reconnecte 25 d as a shunt generator and is shown in Figure P89 The shunt stor Radj is adjusted to 10 and the generators speed is 1800 rmin field resi a What is the noload terminal voltage of the generator b Assuming no armature reaction what is the terminal voltage of the generator with an armature current of 20 A 40 A c Assuming an armature reaction equal to 300 Aturns at full load what is the terminal voltage of the generator with an armature current of 20 A 40 A d Calculate and plot the terminal characteristics of this generator with and without armature reaction SOLUTION a The total field resistance of this generator is 30 and the noload terminal voltage can be found from the intersection of the resistance line with the magnetization curve for this generator The magnetization curve and the field resistance line are plotted below As you can see they intersect at a terminal voltage of 121 V 252 b At an armature current of 20 A the internal voltage drop in the armature resistance is As shown in the figure below there is a difference of 36 V between and at a terminal voltage of about 116 V V 63 20 A 018 T V A E A MATLAB program to locate the position where there is exactly 36 V between the and lines is shown below This program created the plot shown above Note that there are actually two places where the difference between the and lines is 36 volts but the lowvoltage one of them is unstable The code shown in bold face below prevents the program from reporting that first unstable point EA T V EA T V Mfile prob825b1m Mfile to create a plot of the magnetization curve and the field current curve of a shunt dc generator determining the point where the difference between them is 36 V Get the magnetization curve This file contains the three variables ifvalues eavalues and n0 clear all load p87magdat ifvalues p87mag1 eavalues p87mag2 n0 1800 First initialize the values needed in this program rf 20 Field resistance ohms radj 10 Adjustable resistance ohms ra 018 Armature series resistance ohms if 00056 Field current A n 1800 Generator speed rmin Calculate Ea versus If Ea interp1ifvalueseavaluesif Calculate Vt versus If Vt rf radj if Find the point where the difference between the two lines is 36 V This will be the point where the line line Ea Vt 36 goes negative That will be a close enough estimate of Vt diff Ea Vt 36 This code prevents us from reporting the first unstable location satisfying the criterion waspos 0 for ii 1lengthif if diffii 0 waspos 1 end if diffii 0 waspos 1 break end end We have the intersection Tell user 253 disp Ea num2strEaii V disp Vt num2strVtii V disp If num2strifii A Plot the curves figure1 plotifEabLineWidth20 hold on plotifVtkLineWidth20 Plot intersections plotifii ifii 0 Eaii k plot0 ifii Vtii Vtiik plot0 ifii Eaii Eaiik xlabelbfitIF rmbfA ylabelbfitEA rmbf or itVT title bfPlot of itEA rmbf and itVT rmbf vs field current axis 0 5 0 150 setgcaYTick0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 setgcaXTick0 05 10 15 20 25 30 35 40 45 50 legend Ea lineVt line4 hold off grid on At an armature current of 40 A the internal voltage drop in the armature resistance is As shown in the figure below there is a difference of 72 V between and at a terminal voltage of about 110 V The program to create this plot is identical to the one shown above except that the gap between and is 72 V The resulting terminal voltage is about 110 V 40 A 018 72 V T V A E EA T V 254 c The rated current of this generator is 50 A so 20 A is 40 of full load If the full load armature reaction is 300 Aturns and if the armature reaction is assumed to change linearly with armature current then the armature reaction will be 120 Aturns The demagnetizing effect of armature reaction is equivalent to a reduction in field current of 120 A t 012 A 1000 t FI The figure below shows that a triangle consisting of 36 V and 120 Aturns1000 turns 012 A fits exactly between the and lines at a terminal voltage of 114 V EA T V 255 The rated current of this generated is 50 A so 40 A is 80 of full load If the full load armature reaction is 300 Aturns and if the armature reaction is assumed to change linearly with armature current then the armature reaction will be 240 Aturns The demagnetizing effect of armature reaction is equivalent to a reduction in field current of 240 A t 024 A 1000 t FI The figure below shows that a triangle consisting of 72 V and 240 Aturns1000 turns 024 A fits exactly between the and lines at a terminal voltage of 105 V EA T V 256 d A MATLAB program to calculate the terminal characteristic of this generator without armature reaction is shown below Mfile prob825d1m Mfile to calculate the terminal characteristic of a shunt dc generator without armature reaction Get the magnetization curve This file contains the three variables ifvalues eavalues and n0 clear all load p87magdat ifvalues p87mag1 eavalues p87mag2 n0 1800 First initialize the values needed in this program rf 20 Field resistance ohms radj 10 Adjustable resistance ohms ra 018 Armature series resistance ohms if 000056 Field current A n 1800 Generator speed rmin Calculate Ea versus If Ea interp1ifvalueseavaluesif Calculate Vt versus If Vt rf radj if Find the point where the difference between the two lines is exactly equal to iara This will be the point where the line line Ea Vt iara goes negative 257 258 ia 0155 for jj 1lengthia Get the voltage difference diff Ea Vt iajjra This code prevents us from reporting the first unstable location satisfying the criterion waspos 0 for ii 1lengthif if diffii 0 waspos 1 end if diffii 0 waspos 1 break end end Save terminal voltage at this point vtjj Vtii iljj iajj vtjj rf radj end Plot the terminal characteristic figure1 plotilvtbLineWidth20 xlabelbfitIL rmbfA ylabelbfitVT rmbfV title bfTerminal Characteristic of a Shunt DC Generator hold off axis 0 50 0 120 grid on The resulting terminal characteristic is shown below A MATLAB program to calculate the terminal characteristic of this generator with armature reaction is shown below Mfile prob825d2m Mfile to calculate the terminal characteristic of a shunt dc generator with armature reaction Get the magnetization curve This file contains the three variables ifvalues eavalues and n0 clear all load p87magdat ifvalues p87mag1 eavalues p87mag2 n0 1800 First initialize the values needed in this program rf 20 Field resistance ohms radj 10 Adjustable resistance ohms ra 018 Armature series resistance ohms if 000056 Field current A n 1800 Generator speed rmin nf 1000 Number of field turns Calculate Ea versus If Ea interp1ifvalueseavaluesif Calculate Vt versus If Vt rf radj if Find the point where the difference between the Ea 259 armature reaction line and the Vt line is exactly 260 equal to iara This will be the point where the line Eaar Vt iara goes negative ia 0155 for jj 1lengthia Calculate the equivalent field current due to armature reaction iar iajj 50 300 nf Calculate the Ea values modified by armature reaction Eaar interp1ifvalueseavaluesif iar Get the voltage difference diff Eaar Vt iajjra This code prevents us from reporting the first unstable location satisfying the criterion waspos 0 for ii 1lengthif if diffii 0 waspos 1 end if diffii 0 waspos 1 break end end Save terminal voltage at this point vtjj Vtii iljj iajj vtjj rf radj end Plot the terminal characteristic figure1 plotilvtbLineWidth20 xlabelbfitIL rmbfA ylabelbfitVT rmbfV title bfTerminal Characteristic of a Shunt DC Generator wAR hold off axis 0 50 0 120 grid on The resulting terminal characteristic is shown below Note that the armature reaction reduces the terminal voltage for any given load current relative to a generator without armature reaction 826 If the machine in Problem 825 is running at 1800 rmin with a field resistance 10 and an armature current of 25 A what will the resulting terminal voltage be If the field resistor decreases to 5 while the armature current remains 25 A what will the new terminal voltage be Assume no armature reaction Radj SOLUTION If AI 25 A then 25 A 018 A A I R 45 V The point where the distance between the and curves is exactly 45 V corresponds to a terminal voltage of 114 V as shown below EA T V 261 If R decreases to 5 the total field resistance becomes 29 and the terminal voltage line gets shallower The new point where the distance between the and curves is exactly 45 V corresponds to a terminal voltage of 125 V as shown below adj EA T V Note that decreasing the field resistance of the shunt generator increases the terminal voltage 827 A 120V 50A cumulatively compounded dc generator has the following characteristics R R A S 0 21 N F 1000 turns 262 RF 20 SE 25 turns N Radj to 30 set to 10 0 nm 1800 r min The machine has the magnetization curve shown in Figure P87 Its equivalent circuit is shown in Figure P810 Answer the following questions about this machine assuming no armature reaction a If the generator is operating at no load what is its terminal voltage b If the generator has an armature current of 20 A what is its terminal voltage c If the generator has an armature current of 40 A what is its terminal voltage d Calculate and plot the terminal characteristic of this machine SOLUTION a The total field resistance of this generator is 30 and the noload terminal voltage can be found from the intersection of the resistance line with the magnetization curve for this generator The magnetization curve and the field resistance line are plotted below As you can see they intersect at a terminal voltage of 121 V 263 b If the armature current is 20 A then the effective field current contribution from the armature current SE 25 20 A 05 A 1000 A F N I N and the A A S I R R voltage drop is 20 A 021 42 V A A S I R R The location where the triangle formed by SE A F N N I and I R exactly fits between the and lines corresponds to a terminal voltage of 121 V as shown below A A EA T V 264 c If the armature current is 40 A then the effective field current contribution from the armature current SE 25 40 A 10 A 1000 A F N I N and the S voltage drop is A A R R I 40 A 021 84 V A A S I R R The location where the triangle formed by A F I N NSE and exactly fits between the and lines corresponds to a terminal voltage of 121 V as shown below ARA EA VT I 265 A MATLAB program to locate the position where the triangle exactly fits between the and lines is shown below This program created the plot shown above EA T V Mfile prob827bm Mfile to create a plot of the magnetization curve and the Mfile to create a plot of the magnetization curve and the field current curve of a cumulativelycompounded dc generator when the armature current is 20 A Get the magnetization curve This file contains the three variables ifvalues eavalues and n0 clear all load p87magdat ifvalues p87mag1 eavalues p87mag2 n0 1800 First initialize the values needed in this program rf 20 Field resistance ohms radj 10 Adjustable resistance ohms ra 021 Armature series resistance ohms if 00026 Field current A n 1800 Generator speed rmin nf 1000 Shunt field turns nse 25 Series field turns Calculate Ea versus If Ea interp1ifvalueseavaluesif Calculate Vt versus If Vt rf radj if 266 267 alues modified by mmf due to the current a interp1ifvalueseavaluesif ia nsenf t where the line Eaa Vt 42 goes ff Eaa Vt 42 the first unstable isfying the criterion f spos 1 i 0 waspos 1 ak d sp Ifa num2strifii ia nsenf A urves abLineWidth20 otifVtkLineWidth20 ot0 ifiiiansenf Eaaii Eaaiik ifiiiansenf ifiiiansenfVtii Eaaiib lot of itEA rmbf and itVT rmbf vs field YTick0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 0 25 30 35 40 45 50 Calculate the Ea v armature ia 20 Ea Find the point where the difference between the enhanced Ea line and the Vt line is 42 V This will be the poin negative di This code prevents us from reporting location sat waspos 0 for ii 1lengthi if diffii 0 wa end if diffi bre end en We have the intersection Tell user disp Eaa num2strEaaii V disp Ea num2strEaii V disp Vt num2strVtii V disp If num2strifii A di Plot the c figure1 plotifE hold on pl Plot intersections plotifii ifii 0 Vtii k plot0 ifii Vtii Vtiik pl Plot compounding triangle plotifii ifiiiansenfVtii Vtiib plotifii ifiiiansenfVtii Eaaiib plot xlabelbfitIF rmbfA ylabelbfitEA rmbf or itEA rmbfV title bfP current axis 0 5 0 150 setgca 150 setgcaXTick0 05 10 15 2 268 lineVt line4 MATLAB program to calculate and plot the terminal characteristic of this generator is shown low legend Ea hold off grid on d A be Mfile prob827dm Mfile to calculate the terminal characteristic of a cumulatively compounded dc generator without armature reaction Get the magnetization curve This file contains the three variables ifvalues eavalues and n0 clear all load p87magdat ifvalues p87mag1 eavalues p87mag2 n0 1800 First initialize the values needed in this program rf 20 Field resistance ohms radj 10 Adjustable resistance ohms ra 021 Armature series resistance ohms if 00026 Field current A n 1800 Generator speed rmin nf 1000 Shunt field turns nse 25 Series field turns Calculate Ea versus If Ea interp1ifvalueseavaluesif Calculate Vt versus If Vt rf radj if Find the point where the difference between the two lines is exactly equal to iara This will be the point where the line line Ea Vt iara goes negative ia 0155 for jj 1lengthia Calculate the Ea values modified by mmf due to the armature current Eaa interp1ifvalueseavaluesif iajjnsenf Get the voltage difference diff Eaa Vt iajjra This code prevents us from reporting the first unstable location satisfying the criterion waspos 0 for ii 1lengthif if diffii 0 waspos 1 end 269 i 0 waspos 1 ak d tage at this point ljj iajj vtjj rf radj d erminal characteristic a Cumulatively nded DC Generator ng 0 0 130 own below Compare it to the terminal characteristics of the shunt dc generators in Problem 825 d if diffi bre end en Save terminal vol vtjj Vtii i en Plot the t figure1 plotilvtbLineWidth20 xlabelbfitIL rmbfA ylabelbfitVT rmbfV string bfTerminal Characteristic of Compou title stri hold off axis 0 5 grid on The resulting terminal characteristic is sh If the machine described in Problem 827 is reconnected as a differentially compounded dc g 828 enerator ATLAB program to calculate and plot the terminal characteristic of this generator is own below what will its terminal characteristic look like Derive it in the same fashion as in Problem 827 SOLUTION A M sh Mfile prob828m Mfile to calculate the terminal characteristic of a 270 ly compounded dc generator without armature reaction the iables ifvalues eavalues and n0 87mag2 0 1800 am esistance ohms min se 25 Series field turns interp1ifvalueseavaluesif rf radj if the line line Ea Vt iara goes r jj 1lengthia alues modified by mmf due to the Eaa interp1ifvalueseavaluesif iajjnsenf diff Eaa Vt iajjra the first unstable isfying the criterion f spos 1 i 0 waspos 1 ak d tage at this point ljj iajj vtjj rf radj differential Get the magnetization curve This file contains three var clear all load p87magdat ifvalues p87mag1 eavalues p n First initialize the values needed in this progr rf 20 Field resistance ohms radj 10 Adjustable resistance ohms ra 021 Armature series r if 00026 Field current A n 1800 Generator speed r nf 1000 Shunt field turns n Calculate Ea versus If Ea Calculate Vt versus If Vt Find the point where the difference between the two lines is exactly equal to iara This will be the point where negative ia 0121 fo Calculate the Ea v armature current Get the voltage difference This code prevents us from reporting location sat waspos 0 for ii 1lengthi if diffii 0 wa end if diffi bre end en Save terminal vol vtjj Vtii i 271 d erminal characteristic a Differentially nded DC Generator ng 0 0 120 cumulatively compounded dc generator in Problem 928 and the shunt dc generators in Problem 925 d en Plot the t figure1 plotilvtbLineWidth20 xlabelbfitIL rmbfA ylabelbfitVT rmbfV string bfTerminal Characteristic of Compou title stri hold off axis 0 5 grid on The resulting terminal characteristic is shown below Compare it to the terminal characteristics of the A cumulatively compounded dc generator is operating properly as a fla 829 tcompounded dc generator The a the same direction as before will an output voltage be built up at its c in which a voltage builds up will the generator be cumulatively or compounded n the opposite e voltage to produce a field current that increases the residual flux starting a positive feedback chain machine is then shut down and its shunt field connections are reversed If this generator is turned in terminals Why or why not b Will the voltage build up for rotation in the opposite direction Why or why not For the direction of rotation differentially SOLUTION a The output voltage will not build up because the residual flux now induces a voltage i direction which causes a field current to flow that tends to further reduce the residual flux b If the motor rotates in the opposite direction the voltage will build up because the reversal in voltage due to the change in direction of rotation causes th c The generator will now be differentially compounded 830 A threephase synchronous machine is mechanically connected to a shunt dc machine forming a motor generator set as shown in Figure P811 The dc machine is connected to a dc power system supplying a constant 240 V and the ac machine is connected to a 480V 60Hz infinite bus The dc machine has four poles and is rated at 50 kW and 240 V It has a perunit armature resistance of 003 The ac machine has four poles and is Yconnected It is rated at 50 kVA 480 V and 08 PF and its saturated synchronous reactance is 30 per phase All losses except the dc machines armature resistance may be neglected in this problem Assume that the magnetization curves of both machines are linear a Initially the ac machine is supplying 50 kVA at 08 PF lagging to the ac power system 1 How much power is being supplied to the dc motor from the dc power system 2 How large is the internal generated voltage A E of the dc machine 3 How large is the internal generated voltage A of the ac machine E b The field current in the ac machine is now increased by 5 percent What effect does this change have on the real power supplied by the motorgenerator set On the reactive power supplied by the motor generator set Calculate the real and reactive power supplied or consumed by the ac machine under these conditions Sketch the ac machines phasor diagram before and after the change in field current c Starting from the conditions in part b the field current in the dc machine is now decreased by 1 percent What effect does this change have on the real power supplied by the motorgenerator set On the reactive power supplied by the motorgenerator set Calculate the real and reactive power supplied or consumed by the ac machine under these conditions Sketch the ac machines phasor diagram before and after the change in the dc machines field current d From the above results answer the following questions 1 How can the real power flow through an acdc motorgenerator set be controlled 2 How can the reactive power supplied or consumed by the ac machine be controlled without affecting the real power flow SOLUTION a The power supplied by the ac machine to the ac power system is AC cos 50 kVA 08 40 kW P S 272 and the reactive power supplied by the ac machine to the ac power system is 1 AC sin 50 kVA sin cos 08 30 kvar Q S The power out of the dc motor is thus 40 kW This is also the power converted from electrical to mechanical form in the dc machine since all other losses are neglected Therefore conv 40 kW A A T A A A P E I V I R I 2 40 kW 0 T A A A V I I R The base resistance of the dc machine is 2 2 base basedc base 230 V 1058 50 kW VT R P Therefore the actual armature resistance is 003 1058 00317 A R Continuing to solve the equation for conv we get P 2 40 kW 0 T A A A V I I R Multiplying by 1 and rearranging terms produces 2 40 kW 0 A A T A I R V I 2 00317 240 40000 0 A A I I 2 7571 1261800 0 A A I I 17 05 A AI and 240 V 1705 A 00317 2346 V A T A A E V I R Therefore the power into the dc machine is 4092 kW T A V I while the power converted from electrical to mechanical form which is equal to the output power is 2364 V 1705 A 40 kW A A E I The internal generated voltage A E of the dc machine is 2346 V The armature current in the ac machine is 50 kVA 601 A 3 3 480 V A S I V 601 3687 A A I A Therefore the internal generated voltage A of the ac machine is E A jXS E V I 277 0 V 30 601 3687 A 411 205 V A j E b When the field current of the ac machine is increased by 5 it has no effect on the real power supplied by the motorgenerator set This fact is true because P and the speed is constant since the MG set is tied to an infinite bus With the speed unchanged the dc machines torque is unchanged so the total power supplied to the ac machines shaft is unchanged If the field current is increased by 5 and the OCC of the ac machine is linear A E increases to 273 105 411 V 432 V A E The new torque angle can be found from the fact that since the terminal voltage and power of the ac machine are constant the quantity EA sin must be constant sin sin A A E E 1 1 411 V sin sin sin sin 205 184 432 V A A E E Therefore the armature current will be 432 184 V 277 0 V 635 442 A 30 A A jXS j E V I The resulting reactive power is 3 sin 3 480 V 635 A sin 442 368 kvar T L Q V I The reactive power supplied to the ac power system will be 368 kvar compared to 30 kvar before the ac machine field current was increased The phasor diagram illustrating this change is shown below V EA1 jX SI A EA2 IA2 IA1 c If the dc field current is decreased by 1 the dc machines flux will decrease by 1 The internal generated voltage in the dc machine is given by the equation EA K and is held constant by the infinite bus attached to the ac machine Therefore A E on the dc machine will decrease to 0992346 V 23225 V The resulting armature current is dc 240 V 23225 V 2445 A 00317 T A A A V E I R The power into the dc motor is now 240 V2445 A 587 kW and the power converted from electrical to mechanical form in the dc machine is 2325 V2445 A 568 kW This is also the output power of the dc machine the input power of the ac machine and the output power of the ac machine since losses are being neglected The torque angle of the ac machine now can be found from the equation ac 3 sin A S V E P X 1 1 ac 568 kW 30 sin sin 299 3 3 277 V 411 V S A P X V E The new EA of this machine is thus 411 299 V and the resulting armature current is 411 299 V 277 0 V 732 212 A 30 A A jXS j E V I The real and reactive powers are now 274 275 3 cos 3 480 V 732 A cos212 567 kW T L P V I 3 sin 3 480 V 732 A sin 212 22 kvar T L Q V I The phasor diagram of the ac machine before and after the change in dc machine field current is shown below V EA1 jX SI A EA2 I A2 IA1 d The real power flow through an acdc motorgenerator set can be controlled by adjusting the field current of the dc machine Note that changes in power flow also have some effect on the reactive power of the ac machine in this problem Q dropped from 368 kvar to 22 kvar when the real power flow was adjusted The reactive power flow in the ac machine of the MG set can be adjusted by adjusting the ac machines field current This adjustment has basically no effect on the real power flow through the MG set Chapter 9 SinglePhase and SpecialPurpose Motors 91 A 120V 14hp 60Hz fourpole splitphase induction motor has the following impedances 200 256 605 R1 X1 X M 280 256 R2 X 2 At a slip of 005 the motors rotational losses are 51 W The rotational losses may be assumed constant over the normal operating range of the motor If the slip is 005 find the following quantities for this motor a Input power b Airgap power c Pconv d Pout e ind f load g Overall motor efficiency h Stator power factor SOLUTION The equivalent circuit of the motor is shown below 20 j256 V I1 R1 jX1 s R2 50 j05X2 j05XM j128 j3025 s R 2 50 2 jX2 j128 j05XM Forward Reverse 05ZB 05ZF j3025 The impedances F Z and B Z are 2 2 2 2 M F M R s jX jX Z R s jX jX 276 277 The slip s 005 so 2 280 005 56 R s 56 256 605 2882 2805 56 256 605 F j j Z j j j 2 2 2 2 2 2 M B M R s jX jX Z R s jX jX The slip s 005 so 2 2 280 2 005 1436 R s 1436 256 605 1321 2486 1436 256 605 B j j Z j j j a The input current is 1 1 1 I 05 05 F B R jX Z Z V 1 120 0 V I 4862 462 A 200 256 05 2882 2805 05 1321 2486 j j j IN cos 120 V 4862 A cos 462 4038 W P VI b The airgap power is 2 2 AG 1 05 4862 A 1441 3406 W F F P I R 2 2 AG 1 05 4862 A 0661 156 W B B P I R AG AG AG 3406 W 156 W 325 W F B P P P c The power converted from electrical to mechanical form is conv AG 1 1 005 3406 W 323 W F F P s P conv AG 1 1 005 156 W 148 W B B P s P conv conv conv 323 W 148 W 308 W F B P P P d The output power is OUT conv rot 308 W 51 W 257 W P P P e The induced torque is AG ind sync 325 W 172 N m 2 rad 1 min 1800 rmin 1 r 60 s P f The load torque is OUT load 257 W 144 N m 2 rad 1 min 095 1800 rmin 1 r 60 s m P g The overall efficiency is OUT IN 257 W 100 100 636 4038 W P P h The stator power factor is PF cos 462 0692 lagging 92 Repeat Problem 91 for a rotor slip of 0025 The impedances F Z and B Z are 2 2 2 2 M F M R s jX jX Z R s jX jX The slip s 0025 so 2 280 0025 112 R s 112 256 605 2481 4653 112 256 605 F j j Z j j j 2 2 2 2 2 2 M B M R s jX jX Z R s jX jX The slip s 0025 so 2 2 280 2 0025 1418 R s 1418 256 605 1305 2485 1418 256 605 B j j Z j j j a The input current is 1 1 1 I 05 05 F B R jX Z Z V 1 120 0 V I 3874 609 A 200 256 05 2481 4653 05 1305 2485 j j j IN cos 120 V 3874 A cos 609 2261 W P VI b The airgap power is 2 2 AG 1 05 3874 A 1241 1862 W F F P I R 2 2 AG 1 05 3874 A 0653 98 W B B P I R AG AG AG 1862 W 98 W 1764 W F B P P P c The power converted from electrical to mechanical form is conv AG 1 1 0025 1862 W 1815 W F F P s P conv AG 1 1 0025 98 W 96 W B B P s P conv conv conv 1815 W 96 W 1719 W F B P P P d The output power is OUT conv rot 1719 W 51 W 1209 W P P P e The induced torque is AG ind sync 1764 W 0936 N m 2 rad 1 min 1800 rmin 1 r 60 s P f The load torque is 278 OUT load 1209 W 0658 N m 2 rad 1 min 0975 1800 rmin 1 r 60 s m P g The overall efficiency is OUT IN 1209 W 100 100 535 2261 W P P h The stator power factor is PF cos 609 0486 lagging 93 Suppose that the motor in Problem 91 is started and the auxiliary winding fails open while the rotor is accelerating through 400 rmin How much induced torque will the motor be able to produce on its main winding alone Assuming that the rotational losses are still 51 W will this motor continue accelerating or will it slow down again Prove your answer SOLUTION At a speed of 400 rmin the slip is 1800 rmin 400 rmin 0778 1800 rmin s The impedances F Z and B Z are 2 2 2 2 M F M R s jX jX Z R s jX jX The slip s 0778 so 2 280 0778 360 R s 360 256 605 3303 2645 360 256 605 F j j Z j j j 2 2 2 2 2 2 M B M R s jX jX Z R s jX jX The slip s 0778 so 2 2 280 2 0778 2291 R s 2291 256 605 2106 2533 2291 256 605 B j j Z j j j The input current is 1 1 1 I 05 05 F B R jX Z Z V 1 120 0 V I 1721 476 A 200 256 05 3303 2645 05 2106 2533 j j j The airgap power is 2 2 AG 1 05 1721 A 3303 9783 W F F P I R 2 2 AG 1 05 1721 A 2106 6238 W B B P I R AG AG AG 9783 W 6238 W 3545 W F B P P P The power converted from electrical to mechanical form is 279 conv AG 1 1 0778 9783 W 2172 W F F P s P conv AG 1 1 0778 6238 W 1385 W B B P s P conv conv conv 2172 W 1385 W 787 W F B P P P The output power is OUT conv rot 787 W 51 W 277 W P P P The induced torque is AG ind sync 3545 W 188 N m 2 rad 1 min 1800 rmin 1 r 60 s P Assuming that the rotational losses are still 51 W this motor will still be able to speed up because is 787 W while the rotational losses are 51 W so there is more power than it required to cover the rotational losses The motor will continue to speed up conv P 94 Use MATLAB to calculate and plot the torquespeed characteristic of the motor in Problem 91 ignoring the starting winding SOLUTION This problem is best solved with MATLAB since it involves calculating the torquespeed values at many points A MATLAB program to calculate and display both torquespeed characteristics is shown below Note that this program shows the torquespeed curve for both positive and negative directions of rotation Also note that we had to avoid calculating the slip at exactly 0 or 2 since those numbers would produce dividebyzero errors in F Z and B Z respectively Mfile prob94m Mfile create a plot of the torquespeed curve of the singlephase induction motor of Problem 94 First initialize the values needed in this program r1 200 Stator resistance x1 256 Stator reactance r2 280 Rotor resistance x2 256 Rotor reactance xm 605 Magnetization branch reactance v 120 SinglePhase voltage nsync 1800 Synchronous speed rmin wsync 1885 Synchronous speed rads Specify slip ranges to plot s 000120 Offset slips at 0 and 2 slightly to avoid divide by zero errors s1 00001 s201 19999 Get the corresponding speeds in rpm nm 1 s nsync Caclulate Zf and Zb as a function of slip zf r2 s jx2 jxm r2 s jx2 jxm zb r2 2s jx2 jxm r2 2s jx2 jxm 280 Calculate the current flowing at each slip i1 v r1 jx1 05zf 05zb Calculate the airgap power pagf absi12 05 realzf pagb absi12 05 realzb pag pagf pagb Calculate torque in Nm tind pag wsync Plot the torquespeed curve figure1 plotnmtindColorbLineWidth20 xlabelitnm rmrmin ylabel auind rmNm title Single Phase Induction motor torquespeed characteristicFontSize12 grid on hold off The resulting torquespeed characteristic is shown below 95 A 220V 15hp 50Hz sixpole capacitorstart induction motor has the following mainwinding impedances 130 201 105 R1 X1 X M 173 201 R2 X 2 At a slip of 005 the motors rotational losses are 291 W The rotational losses may be assumed constant over the normal operating range of the motor Find the following quantities for this motor at 5 percent slip a Stator current b Stator power factor 281 c Input power d AG P e Pconv f out P g ind h load i Efficiency SOLUTION The equivalent circuit of the motor is shown below 130 j201 V 2200 V I1 R1 jX1 s R2 50 j05X2 j05XM j201 j105 s R 2 50 2 jX2 j201 j05XM j105 Forward Reverse 05ZB 05ZF The impedances F Z and B Z are 2 2 2 2 M F M R s jX jX Z R s jX jX The slip s 005 so 2 173 005 206 R s 206 201 105 1912 5654 206 201 105 F j j Z j j j 2 2 2 2 2 2 M B M R s jX jX Z R s jX jX The slip s 005 so 2 2 173 2 005 0887 R s 0887 201 105 0854 1979 0887 201 105 B j j Z j j j 282 a The input stator current is 1 1 1 I R 05 05 F B jX Z Z V 1 220 0 V I 1732 273 A 130 201 05 1912 5654 05 0854 1979 j j j b The stator power factor is i put power is PF cos 273 0889 l agging c The n IN cos 2 P VI 20 V 1732 A cos 273 3386 W d The airgap power is power converted from electrical to mechanical form is 2 AG 1 05 P I R 1732 A 2 956 2868 W F F 2 2 AG 1 05 1732 A 0427 128 W B B P I R AG AG AG 2868 W 128 W 2740 W F B P P P e The conv AG 1 1 005 2868 W 2725 W F F P s P conv AG 1 1 005 128 W 122 W B B P s P conv conv conv 2725 W 122 W 2603 W F B P P P f The output power is synchronous speed for a 6 pole 50 Hz machine is 1000 rmin so induced torque is OUT conv rot P P P 2603 W 291 W 2312 W g The AG 2740 W 2617 N m P ind sync 2 rad 1 min 1000 rmin 1 r 60 s h The load torque is OUT P load 2312 W 2324 N m 2 rad 1 min 095 1000 rmin 1 r 60 s m i The overall efficiency is OUT 2312 W 100 P IN 100 683 3386 W P 96 Find the induced torque in the motor in Problem 95 if it is operating at 5 percent slip and its terminal voltage is a 190 V b 208 V c 230 V The impedances F Z and B Z are 283 2 2 2 2 M F M R jX jX s Z R s jX jX so The slip s 005 2 173 005 206 R s 206 j j Z j 201 105 1912 5654 206 201 105 F j j 2 2 2 2 2 2 M B M R s jX jX Z R s jX jX The slip s 005 so 2 2 173 2 005 0887 R s 0887 Z j 201 105 0854 1979 0887 201 105 B j j j j a If T 1900 V V 1 1 1 I 05 05 F B R jX Z Z V 1 190 0 V I 1496 273 A 130 201 05 1912 5654 05 0854 1979 j j j 2 2 AG 1 05 1496 A 956 2140 W F F P I R 2 2 AG 1 05 1496 A 0427 956 W B B P I R AG AG AG 2140 W 956 W 2044 W F B P P P AG ind sync 2044 W 1952 N m 2 rad 1 min 1000 rmin 1 r 60 s P b If T 2080 V V 1 1 1 I 05 05 F B R jX Z Z V 1 208 0 V I 1637 273 A 130 201 05 1912 5654 05 0854 1979 j j j 2 2 AG 1 05 1637 A 956 2562 W F F P I R 2 2 AG 1 05 1637 A 0427 114 W B B P I R AG AG AG 2562 W 114 W 2448 W F B P P P AG ind sync 2448 W 2338 N m 2 rad 1 min 1000 rmin 1 r 60 s P c If T 2300 V V 1 1 1 I 05 05 F B R jX Z Z V 1 230 0 V I 1810 273 A 130 201 05 1912 5654 05 0854 1979 j j j 2 2 AG 1 05 1810 A 956 3132 W F F P I R 284 2 2 AG 1 05 1732 A 0427 140 W B B P I R AG AG AG 3132 W 140 W 2992 W F B P P P 285 AG ind sync 2992 W 2857 N m 2 rad 1 min 1000 rmin 1 r 60 s P Note that the induced torque is proportional to the square of the terminal voltage 7 mpressor er fan in sal motorfor its high torque nd runFor its high starting torque and relatively constant re lowstartingtorque applications and a splitphase motor is appropriate tor would also do iversal Motorfor easy speed control with solidstate drives plus high torque under loaded 98 lar application a threephase stepper motor must be capable of stepping in 10 increments 8 the relationship between mechanical angle and electrical angle in a 9 What type of motor would you select to perform each of the following jobs Why a Vacuum cleaner b Refrigerator c Air conditioner co d Air condition e Variablespeed sewing mach e f Clock g Electric drill SOLUTION a Univer b Capacitor start or Capacitor start a speed at a wide variety of loads c Same as b above d SplitphaseFans a e Universal MotorDirection and speed are easy to control with solidstate drives f Hysteresis motorfor its easy starting and operation at nsync A reluctance mo nicely g Un conditions For a particu How many poles must it have SOLUTION From Equation 91 threephase stepper motor is 2 m P e so 60 2 2 12 pol 10 e m P es 99 How many pulses per second must be supplied to the control unit of the motor in Problem 98 to achieve a rotational speed of 600 rmin SOLUTION From Equation 920 pulses 1 n n 3 m P t a table showing step size versus number of poles for threephase and fourphase stepper motors so pulses 3 3 12 poles 600 rmin 21600 pulsesmin 360 pulsess m n P n 910 Construc 286 SOLUTION For 3phase stepper motors e 60 and for 4phase stepper motors e 45 Therefore Number of poles Mechanical Step Size 3phase 60 e 4phase 45 e 2 45 60 4 30 2 25 6 20 15 8 15 1 125 10 12 9 12 10 75 Appendix A Review of ThreePhase Circuits A1 Three impedances of 4 j3 are connected and tied to a threephase 208V power line Find P Q S and the power factor of this load I IL SOLUTION Z Z Z 240 V IL I 4 3 j Z Here and so 208 V VL V 4 3 5 3687 Z j 208 V 416 A 5 V I Z 3 3 416 A 7205 A LI I 2 2 208 V 3 cos 3 cos 3687 2077 kW 5 V P Z 2 2 208 V 3 sin 3 sin 3687 1558 kvar 5 V Q Z 2 2 2596 kVA S P Q PF cos 08 lagging A2 Figure PA1 shows a threephase power system with two loads The connected generator is producing a line voltage of 480 V and the line impedance is 009 j016 Load 1 is Yconnected with a phase impedance of 253687 and load 2 is connected with a phase impedance of 520 287 a What is the line voltage of the two loads b What is the voltage drop on the transmission lines c Find the real and reactive powers supplied to each load d Find the real and reactive power losses in the transmission line e Find the real power reactive power and power factor supplied by the generator SOLUTION To solve this problem first convert the deltaconnected load 2 to an equivalent wye by dividing the impedance by 3 and get the perphase equivalent circuit 2770 V Line 0090 j016 Z1 2 Z 3687 52 1 Z 20 1 67 2 Z load V a The phase voltage of the equivalent Yloads can be found by nodal analysis load load load 277 0 V 0 009 016 25 3687 167 20 j V V V load load load 5443 606 277 0 V 04 3687 06 20 0 V V V load 5955 5334 1508 606 V load 2532 73 V V 288 Therefore the line voltage at the loads is 3 439 V VL V b The voltage drop in the transmission lines is line gen load 277 0 V 2532 73 413 52 V V V V c The real and reactive power of each load is 2 2 1 2532 V 3 cos 3 cos 3687 616 kW 25 V P Z 2 2 1 2532 V 3 sin 3 sin 3687 462 kvar 25 V Q Z 2 2 2 2532 V 3 cos 3 cos 20 1084 kW 167 V P Z 2 2 2 2532 V 3 sin 3 sin 20 395 kvar 167 V Q Z d The line current is line line line 413 52 V 225 86 A 009 016 Z j V I Therefore the loses in the transmission line are 2 2 line line line 3 3 225 A 009 137 kW P I R 2 2 line line line 3 3 225 A 016 243 kvar Q I X e The real and reactive power supplied by the generator is gen line 1 2 137 kW 616 kW 1084 kW 1837 kW P P P P gen line 1 2 243 kvar 462 kvar 395 kvar 31 kvar Q Q Q Q The power factor of the generator is gen 1 1 gen 31 kvar PF cos tan cos tan 0986 lagging 1837 kW Q P A3 Figure PA2 shows a oneline diagram of a simple power system containing a single 480 V generator and three loads Assume that the transmission lines in this power system are lossless and answer the following questions a Assume that Load 1 is Yconnected What are the phase voltage and currents in that load b Assume that Load 2 is connected What are the phase voltage and currents in that load c What real reactive and apparent power does the generator supply when the switch is open d What is the total line current when the switch is open LI e What real reactive and apparent power does the generator supply when the switch is closed f What is the total line current when the switch is closed LI g How does the total line current compare to the sum of the three individual currents LI 1 2 I I I3 If they are not equal why not 289 SOLUTION Since the transmission lines are lossless in this power system the full voltage generated by will be present at each of the loads 1 G a Since this load is Yconnected the phase voltage is 1 480 V 277 V 3 V The phase current can be derived from the equation 3 cos P V I as follows 1 100 kW 1337 A 3 cos 3 277 V 09 P I V b Since this load is connected the phase voltage is 2 48 0 V V The phase current can be derived from the equation 3 S V I as follows 2 80 kVA 5556 A 3 3 480 V S I V c The real and reactive power supplied by the generator when the switch is open is just the sum of the real and reactive powers of Loads 1 and 2 1 100 kW P 1 1 tan tan cos PF 100 kW tan2584 484 kvar Q P P 2 cos 80 kVA 08 64 kW P S 2 sin 80 kVA 06 48 kvar Q S 1 2 100 kW 64 kW 164 kW G P P P 1 2 484 kvar 48 kvar 964 kvar QG Q Q d The line current when the switch is open is given by 3 cos L L P I V where tan 1 G G Q P 1 1 964 kvar tan tan 3045 164 kW G G Q P 164 kW 2288 A 3 cos 3 480 V cos 3045 L L P I V 290 e The real and reactive power supplied by the generator when the switch is closed is just the sum of the real and reactive powers of Loads 1 2 and 3 The powers of Loads 1 and 2 have already been calculated The real and reactive power of Load 3 are 3 80 kW P 1 3 tan tan cos PF 80 kW tan 3179 496 kvar Q P P 1 2 3 100 kW 64 kW 80 kW 244 kW G P P P P 1 2 3 484 kvar 48 kvar 496 kvar 468 kvar QG Q Q Q f The line current when the switch is closed is given by 3 cos L L P I V where tan 1 G G Q P 1 1 468 kvar tan tan 1086 244 kW G G Q P 244 kW 2988 A 3 cos 3 480 V cos 1086 L L P I V g The total line current from the generator is 2988 A The line currents to each individual load are 1 1 1 100 kW 1336 A 3 cos 3 480 V 09 L L P I V 2 2 80 kVA 962 A 3 3 480 V L L S I V 3 3 3 80 kW 1132 A 3 cos 3 480 V 085 L L P I V The sum of the three individual line currents is 343 A while the current supplied by the generator is 2988 A These values are not the same because the three loads have different impedance angles Essentially Load 3 is supplying some of the reactive power being consumed by Loads 1 and 2 so that it does not have to come from the generator A4 Prove that the line voltage of a Yconnected generator with an acb phase sequence lags the corresponding phase voltage by 30 Draw a phasor diagram showing the phase and line voltages for this generator SOLUTION If the generator has an acb phase sequence then the three phase voltages will be 0 an V V 240 bn V V 120 cn V V The relationship between line voltage and phase voltage is derived below By Kirchhoffs voltage law the linetoline voltage ab is given by V ab a b V V V 0 240 ab V V V 1 3 3 3 2 2 2 2 ab V V j V V j V V 3 1 3 2 2 ab V j V 3 30 ab V V 291 Thus the line voltage lags the corresponding phase voltage by 30 The phasor diagram for this connection is shown below Van Vbn Vcn Vab Vbc A5 Find the magnitudes and angles of each line and phase voltage and current on the load shown in Figure P23 SOLUTION Note that because this load is connected the line and phase voltages are identical 120 0 V 120 120 V 208 30 V ab an bn V V V 120 120 V 120 240 V 208 90 V bc bn cn V V V 120 240 V 120 0 V 208 150 V ca cn an V V V 292 208 30 V 208 10 A 10 20 ab ab Z V I 208 90 V 208 110 A 10 20 bc bc Z V I 208 150 V 208 130 A 10 20 ca ca Z V I 208 10 A 208 130 A 36 20 A a ab ca I I I 208 110 A 208 10 A 36 140 A b bc ab I I I 208 130 A 208 110 A 36 100 A c ca bc I I I A6 Figure PA4 shows a small 480V distribution system Assume that the lines in the system have zero impedance a If the switch shown is open find the real reactive and apparent powers in the system Find the total current supplied to the distribution system by the utility b Repeat part a with the switch closed What happened to the total current supplied Why SOLUTION a With the switch open the power supplied to each load is 5 9 86 kW cos 30 10 3 480 V cos 3 2 2 1 Z V P 2 2 1 480 V 3 sin 3 sin 30 3456 kvar 10 V Q Z 4604 kW cos 3687 4 3 277 V cos 3 2 2 2 Z V P 2 2 2 277 V 3 sin 3 sin 3687 3453 kvar 4 V Q Z 5986 kW 4604 kW 1059 kW 2 1 TOT P P P TOT 1 2 3456 kvar 3453 kvar 6909 kvar Q Q Q The apparent power supplied by the utility is 2 2 TOT TOT TOT 1264 kVA S P Q The power factor supplied by the utility is 293 1 1 TOT TOT 6909 kvar PF cos tan cos tan 0838 lagging 1059 kW Q P The current supplied by the utility is TOT 1059 kW 152 A 3 PF 3 480 V 0838 L T P I V b With the switch closed is added to the circuit The real and reactive power of is 3P 3P 0 kW cos 90 5 3 277 V cos 3 2 2 3 Z V P 2 2 3 277 V 3 sin 3 sin 90 4606 kvar 5 V P Z TOT 1 2 3 5986 kW 4604 kW 0 kW 1059 kW P P P P TOT 1 2 3 3456 kvar 3453 kvar 4606 kvar 2303 kvar Q Q Q Q The apparent power supplied by the utility is 2 2 TOT TOT TOT 1084 kVA S P Q The power factor supplied by the utility is 1 1 TOT TOT 2303 kVAR PF cos tan cos tan 0977 lagging 1059 kW Q P The current supplied by the utility is TOT 1059 kW 1304 A 3 PF 3 480 V 0977 L T P I V c The total current supplied by the power system drops when the switch is closed because the capacitor bank is supplying some of the reactive power being consumed by loads 1 and 2 294 Appendix B Coil Pitch and Distributed Windings B1 A 2slot threephase stator armature is wound for twopole operation If fractionalpitch windings are to be used what is the best possible choice for winding pitch if it is desired to eliminate the fifthharmonic component of voltage SOLUTION The pitch factor of a winding is given by Equation B19 2 kp sin To eliminate the fifth harmonic we want to select so that 0 2 sin 5 This implies that 180 n 2 5 where n 0 1 2 3 or 72 144 5 2 180 n These are acceptable pitches to eliminate the fifth harmonic Expressed as fractions of full pitch these pitches are 25 45 65 etc Since the desire is to have the maximum possible fundamental voltage the best choice for coil pitch would be 45 or 65 The closest that we can approach to a 45 pitch in a 24slot winding is 1012 pitch so that is the pitch that we would use At 1012 pitch 0 966 2 sin 150 kp for the fundamental frequency 0 259 2 sin 5 150 kp for the fifth harmonic 295 B2 Derive the relationship for the winding distribution factor in Equation B22 kd SOLUTION The above illustration shows the case of 5 slots per phase but the results are general If there are 5 slots per phase each with voltage where the phase angle of each voltage increases by from slot to slot then the total voltage in the phase will be Ai E An A A A A A A E E E E E E E 5 4 3 2 1 The resulting voltage can be found from geometrical considerations These n phases when drawn endtoend form equallyspaced chords on a circle of radius R If a line is drawn from the center of a chord to the origin of the circle it forma a right triangle with the radius at the end of the chord see voltage above The hypotenuse of this right triangle is R its opposite side is and its smaller angle is A E EA5 2 2 E Therefore R sin 2 E 2 2 sin 2 1 E R 1 The total voltage also forms a chord on the circle and dropping a line from the center of that chord to the origin forms a right triangle For this triangle the hypotenuse is R the opposite side is and the angle is A E 2 A 2 E n Therefore R A sin 2 E n 2 2 sin 2 1 n E R A 2 Combining 1 and 2 yields 296 2 sin 2 1 2 sin 2 1 n E E A 2 sin 2 sin n E EA Finally sin 2 2 sin n n nE E k A d since is defined as the ratio of the total voltage produced to the sum of the magnitudes of each component voltage dk B3 A threephase fourpole synchronous machine has 96 stator slots The slots contain a doublelayer winding two coils per slot with four turns per coil The coil pitch is 1924 a Find the slot and coil pitch in electrical degrees b Find the pitch distribution and winding factors for this machine c How well will this winding suppress third fifth seventh ninth and eleventh harmonics Be sure to consider the effects of both coil pitch and winding distribution in your answer SOLUTION a The coil pitch is 1924 or 1425 Note that these are electrical degrees Since this is a 4pole machine the coil pitch would be 7125 mechanical degrees There are 96 slots on this stator so the slot pitch is 36096 375 mechanical degrees or 75 electrical degrees b The pitch factor of this winding is 0 947 2 sin 142 5 sin 2 kp The distribution factor is sin 2 2 sin n n kd The electrical angle between slots is 75 and each phase group occupies 8 adjacent slots Therefore the distribution factor is 956 0 2 sin15 8 2 8 15 sin sin 2 sin 2 n n kd 297 The winding factor is 0 905 0 947 0956 d p w k k k B4 A threephase fourpole winding of the doublelayer type is to be installed on a 48slot stator The pitch of the stator windings is 56 and there are 10 turns per coil in the windings All coils in each phase are connected in series and the three phases are connected in The flux per pole in the machine is 0054 Wb and the speed of rotation of the magnetic field is 1800 rmin a What is the pitch factor of this winding b What is the distribution factor of this winding c What is the frequency of the voltage produced in this winding d What are the resulting phase and terminal voltages of this stator SOLUTION a The pitch factor of this winding is 0 966 2 sin150 sin 2 kp b The coils in each phase group of this machine cover 4 slots and the slot pitch is 36048 75 mechanical degrees or 15 electrical degrees Therefore the distribution factor is 958 0 2 sin15 4 2 4 15 sin sin 2 sin 2 n n kd c The frequency of the voltage produces by this winding is 1800 rmin 4 poles 60 Hz 120 120 m se n P f d There are 48 slots on this stator with two coils sides in each slot Therefore there are 48 coils on the machine They are divided into 12 phase groups so there are 4 coils per phase There are 10 turns per coil so there are 40 turns per phase group The voltage in one phase group is 533 V 40 turns 0966 0958 0054 Wb 60 Hz 2 2 e d p P G f N k k E There are two phase groups per phase connected in series this is a 4pole machine so the total phase voltage is 1066 V 2 EG V Since the machine is connected 1066 V V VT B5 A threephase Yconnected sixpole synchronous generator has six slots per pole on its stator winding The winding itself is a chorded fractionalpitch doublelayer winding with eight turns per coil The distribution factor 0956 and the pitch factor 0981 The flux in the generator is 002 Wb per pole and the speed of rotation is 1200 rmin What is the line voltage produced by this generator at these conditions kd k p SOLUTION There are 6 slots per pole 6 poles 36 slots on the stator of this machine Therefore there are 36 coils on the machine or 12 coils per phase The electrical frequency produced by this winding is 298 1200 rmin 6 poles 60 Hz 120 120 m se n P f The phase voltage is 480 V 96 turns 0981 0956 002 Wb 60 Hz 2 2 e d P p f N k k V Therefore the line voltage is 831 V 3 V VL B6 A threephase Yconnected 50Hz twopole synchronous machine has a stator with 18 slots Its coils form a doublelayer chorded winding two coils per slot and each coil has 60 turns The pitch of the stator coils is 89 a What rotor flux would be required to produce a terminal linetoline voltage of 6 kV b How effective are coils of this pitch at reducing the fifthharmonic component of voltage The seventh harmonic component of voltage SOLUTION a The pitch of this winding is 89 160 so the pitch factor is 0 985 2 sin 160 kp The phase groups in this machine cover three slots each and the slot pitch is 20 mechanical or 20 electrical degrees Thus the distribution factor is 960 0 2 sin 20 3 2 3 20 sin sin 2 sin 2 n n kd The phase voltage of this machine will be 6 coils 60 turnscoi l 0985 0960 50 Hz 2 2 e d P p f N k k V V 75621 The desired phase voltage is 6 kV 3 3464 V so 0 046 Wb 75621 3464 V b The fifth harmonic 0 643 2 sin 5 160 kp The seventh harmonic 0 342 2 sin 7 160 kp Since the fundamental voltage is reduced by 0985 the fifth and seventh harmonics are suppressed relative to the fundamental by the fractions 5th 0 653 0 985 0 643 299 7th 0 347 0 985 0 342 In other words the 5th harmonic is suppressed by 347 relative to the fundamental and the 7th harmonic is suppressed by 653 relative to the fundamental frequency B7 What coil pitch could be used to completely eliminate the seventhharmonic component of voltage in ac machine armature stator What is the minimum number of slots needed on an eightpole winding to exactly achieve this pitch What would this pitch do to the fifthharmonic component of voltage SOLUTION To totally eliminate the seventh harmonic of voltage in an ac machine armature the pitch factor for that harmonic must be zero 2 sin 7 0 kp 180 n 2 7 n 0 1 2 7 2 180 n In order to maximize the fundamental voltage while canceling out the seventh harmonic we pick the value of n that makes as nearly 180 as possible If n 3 then 1543 and the pitch factor for the fundamental frequency would be 0 975 2 sin 154 3 kp This pitch corresponds to a ratio of 67 For a twopole machine a ratio of 67 could be implemented with a total of 14 slots If that ratio is desired in an 8pole machine then 56 slots would be needed The fifth harmonic would be suppressed by this winding as follows 0 434 2 sin 5 154 3 kp B8 A 138kV Yconnected 60Hz 12pole threephase synchronous generator has 180 stator slots with a doublelayer winding and eight turns per coil The coil pitch on the stator is 12 slots The conductors from all phase belts or groups in a given phase are connected in series a What flux per pole would be required to give a noload terminal line voltage of 138 kV b What is this machines winding factor kw SOLUTION a The stator pitch is 1215 45 so 144 and 0 951 2 sin144 kp Each phase belt consists of 180 slots12 poles6 25 slots per phase group The slot pitch is 2 mechanical degrees or 24 electrical degrees The corresponding distribution factor is 962 0 2 5 sin 24 2 2 24 52 sin sin 2 sin 2 n n kd 300 Since there are 60 coils in each phase and 8 turns per coil all connected in series there are 480 turns per phase The resulting voltage is 480 turns 0951 0962 60 Hz 2 2 e d P p f N k k V 117061 V The phase voltage of this generator must be 7967 V 13 8 kV 3 so the flux must be 0 068 Wb 117061 7967 V b The machines winding factor is 0 915 0951 0962 d p w k k k 301 Appendix C Salient Pole Theory of Synchronous Machines C1 A 138kV 50MVA 09powerfactorlagging 60Hz fourpole Yconnected synchronous generator has a directaxis reactance of 25 a quadratureaxis reactance of 18 and an armature resistance of 02 Friction windage and stray losses may be assumed negligible The generators opencircuit characteristic is given by Figure P51 Figure P41 a How much field current is required to make the terminal voltage V or line voltage equal to 138 kV when the generator is running at no load T L V b What is the internal generated voltage of this machine when it is operating at rated conditions How does this value of compare to that of Problem 42b E A c What fraction of this generators fullload power is due to the reluctance torque of the rotor SOLUTION a If the noload terminal voltage is 138 kV the required field current can be read directly from the opencircuit characteristic It is 350 A b This generator is Yconnected so A L I I At rated conditions the line and phase current in this generator is 50 MVA 2092 A 3 3 13800 V A L L P I I V at an angle of 258 302 The phase voltage of this machine is 3 7967 V T V V The internal generated voltage of the machine is E V I I A A A R jX q A 7967 0 020 2092 258 A 18 2092 258 A A j E A 10485 178 V E Therefore the torque angle is 178 The directaxis current is 90 sin A d I I 2092 A sin 436 722 d I 1443 722 A d I The quadratureaxis current is A cos q I I 2092 A cos 436 178 q I q 1515 178 A I Therefore the internal generated voltage of the machine is q q d d A A A jX jX R I I I V E 7967 0 020 2092 258 25 1443 722 18 1515 178 A j j E A 11496 178 V E EA is approximately the same magnitude here as in Problem 42b but the angle is quite different c The power supplied by this machine is given by the equation 2 3 3 sin sin 2 2 d q A d d q X X V E V P X X X 2 3 7967 11496 3 7967 25 18 sin 178 sin 356 25 2 25 18 P 303 304 al rotor term is 336 MW and the reluctance term is 86 MW so the reluctance torque of the power in this generator c owerversustorqueangle curve for this generator At what angle is the power of the a 336 MW 86 MW 422 MW P The cylindric accounts for about 20 C2 A 14pole Yconnected threephase waterturbinedriven generator is rated at 120 MVA 132 kV 08 PF lagging and 60 Hz Its directaxis reactance is 062 and its quadrature axis reactance is 040 All rotational losses may be neglected a What internal generated voltage would be required for this generator to operate at the rated conditions b What is the voltage regulation of this generator at the rated conditions Sketch the p generator maximum d How does the maximum power out of this generator compare to the maximum power available if it were of cylindrical rotor construction SOLUTION At rated conditions the line and phase current in this generator is 5249 A 120 MVA P I at an angle of 3687 q A I 3 13 2 kV 3 L L A V E V I I A A A R jX 3687 A 5249 0 40 0 7621 0 j EA V 9038 10 7 EA Therefore th e torque angle is 107 The directaxis current is 90 sin A d I I 79 3 5249 A sin 4757 dI A 79 3 3874 dI The quad at re r u axis current is A cos q I I 5249 A cos 47 qI 57 10 7 A 3541 10 7 qI Therefore t e h internal generated voltage of the machine is q q d d A A A jX jX R I I I V E 79 3 0 62 3874 0 7621 0 j EA 0 40 3541 10 7 j V 9890 10 7 EA b The vo ltage regulation of this generator is 29 8 100 7621 fl V 7621 9890 100 fl nl V c The power supplied by this machine is given by the equation V P V E X V X X X X A d d q d q 3 3 2 2 sin sin 2 0 62 0 40 sin 2 2 sin 0 62 P 0 40 0 62 3 7621 3 7621 9890 2 MW 77 3 sin 2 364 7 sin P A plot of power supplied as a function of torque angle is shown below The peak power occurs at an angle of 706 and the maximum power that the generator can supply is 3924 MW d If this generator were nonsalient would occur when 90 and would be 3647 MW Therefore the salientpole generator has a higher maximum power than an equivalent nonsalint C3 achine is to be used as a motor pole motor is given by MAX P MAX P pole generator Suppose that a salientpole m a Sketch the phasor diagram of a salient synchronous machine used as a motor b Write the equations describing the voltages and currents in this motor c Prove that the torque angle between E A and V on this 1 cos sin A q A A I X I R tan sin cos A q A A V I X I R SOLUTION 305 307 C4 If the machine in Problem C1 is running as a motor at the rated conditions what is the maximum torque that can be drawn from its shaft without it slipping poles when the field current is zero SOLUTION When the field current is zero A 0 so E P V E X V X X X X A d d q d q 3 3 2 2 sin sin 2 3 7621 2 062 040 sin 2 148sin 2 MW 2 062 040 P At 45 148 MW can be drawn from the motor Chapter S1 Introduction to Power Electronics S11 Calculate the ripple factor of a threephase halfwave rectifier circuit both analytically and using MATLAB SOLUTION A threephase halfwave rectifier and its output voltage are shown below 6 56 23 sin A M v t V t sin 2 3 B M v t V t sin 2 3 C M v t V t SOLUTION If we find the average and rms values over the interval from 6 to 56 one period these values will be the same as the average and rms values of the entire waveform and they can be used to calculate the ripple factor The average voltage is 5 6 6 1 3 sin 2 DC M V v t dt V t d T t 5 6 6 3 3 3 3 3 3 cos 08270 2 2 2 2 2 M M DC M M V V V t V V The rms voltage is 5 6 2 2 2 rms 6 1 3 sin 2 M V v t dt V t d T t 5 6 2 rms 6 3 1 1 sin2 2 2 4 VM V t t 308 2 rms 3 1 5 1 sin 5 sin 2 2 6 6 4 3 3 VM V 2 2 rms 3 1 5 3 1 3 sin sin 2 3 4 3 3 2 3 4 2 2 M M V V V 3 2 2 rms 3 1 3 3 3 3 08407 2 3 4 2 2 2 3 4 M M M V V V V The resulting ripple factor is 2 2 rms DC 08407 1 100 1 100 183 08270 M M V V r V V The ripple can be calculated with MATLAB using the ripple function developed in the text We must right a new function halfwave3 to simulate the output of a threephase halfwave rectifier This output is just the largest voltage of vA t vB t and vC t at any particular time The function is shown below function volts halfwave3wt Function to simulate the output of a threephase halfwave rectifier wt Phase in radians omega x time Convert input to the range 0 wt 2pi while wt 2pi wt wt 2pi end while wt 0 wt wt 2pi end Simulate the output of the rectifier a sinwt b sinwt 2pi3 c sinwt 2pi3 volts max a b c The function ripple is reproduced below It is identical to the one in the textbook function r ripplewaveform Function to calculate the ripple on an input waveform Calculate the average value of the waveform nvals sizewaveform2 temp 0 for ii 1nvals temp temp waveformii end average tempnvals 309 Calculate rms value of waveform temp 0 for ii 1nvals temp temp waveformii2 end rms sqrttempnvals Calculate ripple factor r sqrtrms average2 1 100 Finally the test driver program is shown below Mfile testhalfwave3m Mfile to calculate the ripple on the output of a three phase halfwave rectifier First generate the output of a threephase halfwave rectifier waveform zeros1128 for ii 1128 waveformii halfwave3iipi64 end Now calculate the ripple factor r ripplewaveform Print out the result string The ripple is num2strr dispstring When this program is executed the results are testhalfwave3 The ripple is 182759 This answer agrees with the analytical solution above S12 Calculate the ripple factor of a threephase fullwave rectifier circuit both analytically and using MATLAB SOLUTION A threephase halfwave rectifier and its output voltage are shown below 310 T12 sin A M v t V t sin 2 3 B M v t V t sin 2 3 C M v t V t SOLUTION By symmetry the rms voltage over the interval from 0 to T12 will be the same as the rms voltage over the whole interval Over that interval the output voltage is 2 2 sin sin 3 3 C B M M v t v t v t V t V t 2 2 2 2 sin cos cos sin sin cos cos sin 3 3 3 3 M M v t V t t V t t 2 2cos sin 3cos 3 M v t V t t Note that the period of the waveform is 2 T so T12 is 6 The average voltage over the interval from 0 to T12 is 6 6 0 0 1 6 6 3 3 cos sin DC M M V v t dt V t dt V t T 3 3 16540 DC M M V V V The rms voltage is 6 2 2 rms 0 1 6 3 M cos V v t dt V T 2 t dt 311 6 2 rms 0 18 1 1 sin2 2 4 M V V t t rms 18 1 3 9 3 sin 16554 12 4 3 2 4 M M M V V V V The resulting ripple factor is 2 2 rms DC 16554 1 100 1 100 42 16540 M M V V r V V The ripple can be calculated with MATLAB using the ripple function developed in the text We must right a new function fullwave3 to simulate the output of a threephase halfwave rectifier This output is just the largest voltage of vA t vB t and vC t at any particular time The function is shown below function volts fullwave3wt Function to simulate the output of a threephase fullwave rectifier wt Phase in radians omega x time Convert input to the range 0 wt 2pi while wt 2pi wt wt 2pi end while wt 0 wt wt 2pi end Simulate the output of the rectifier a sinwt b sinwt 2pi3 c sinwt 2pi3 volts max a b c min a b c The test driver program is shown below Mfile testfullwave3m Mfile to calculate the ripple on the output of a three phase fullwave rectifier First generate the output of a threephase fullwave rectifier waveform zeros1128 for ii 1128 waveformii fullwave3iipi64 end Now calculate the ripple factor r ripplewaveform Print out the result 312 string The ripple is num2strr dispstring When this program is executed the results are testfullwave3 The ripple is 42017 This answer agrees with the analytical solution above S13 Explain the operation of the circuit shown in Figure S11 What would happen in this circuit if switch S1 were closed SOLUTION Diode D1 and D2 together with the transformer form a fullwave rectifier Therefore a voltage oriented positivetonegative as shown will be applied to the SCR and the control circuit on each half cycle 1 Initially the SCR is an open circuit since v1 VBO for the SCR Therefore no current flows to the load and vLOAD 0 2 Voltage v1 is applied to the control circuit charging capacitor C1 with time constant RC1 3 When vC VBO for the DIAC it conducts supplying a gate current to the SCR 4 The gate current in the SCR lowers its breakover voltage and the SCR fires When the SCR fires current flows through the SCR and the load 5 The current flow continues until iD falls below IH for the SCR at the end of the half cycle The process starts over in the next half cycle 313 If switch S1 is shut the charging time constant is increased and the DIAC fires later in each half cycle Therefore less power is supplied to the load S14 What would the rms voltage on the load in the circuit in Figure S11 be if the firing angle of the SCR were a 0 b 30 c 90 SOLUTION The input voltage to the circuit of Figure S11 is t t v 339sin ac where 377 rads Therefore the voltage on the secondary of the transformer will be t t v 169 5 sin ac a The average voltage applied to the load will be the integral over the conducting portion of the half cycle divided by the period of a half cycle For a firing angle of 0 the average voltage will be ave 0 0 0 1 1 sin cos T M M V v t dt V t dt V t T ave 1 2 1 1 0637 1695 V 108 V M M V V V b For a firing angle of 30 the average voltage will be ave 6 6 6 1 1 sin cos T M M V v t dt V t dt V T t ave 1 3 2 3 1 0594 1695 V 101 V 2 2 M M V V V c For a firing angle of 90 the average voltage will be ave 2 2 2 1 1 sin cos T M M V v t dt V t dt V T t 314 ave 1 1 1 0318 1695 V 54 V M M V V V S15 For the circuit in Figure S11 assume that V for the DIAC is 30 V C1 is 1 F R is adjustable in the range 120 k and that switch S1 is open What is the firing angle of the circuit when R is 10 k What is the rms voltage on the load under these conditions BO Note Problem 35 is significantly harder for many students since it involves solving a differential equation with a forcing function This problem should only be assigned if the class has the mathematical sophistication to handle it SOLUTION At the beginning of each half cycle the voltages across the DIAC and the SCR will both be smaller then their respective breakover voltages so no current will flow to the load except for the very tiny current charging capacitor C and vloadt will be 0 volts However capacitor C charges up through resistor R and when the voltage vCt builds up to the breakover voltage of D1 the DIAC will start to conduct This current flows through the gate of SCR1 turning the SCR ON When it turns ON the voltage across the SCR will drop to 0 and the full source voltage vSt will be applied to the load producing a current flow through the load The SCR continues to conduct until the current through it falls below IH which happens at the very end of the half cycle Note that after D1 turns on capacitor C discharges through it and the gate of the SCR At the end of the half cycle the voltage on the capacitor is again essentially 0 volts and the whole process is ready to start over again at the beginning of the next half cycle To determine when the DIAC and the SCR fire in this circuit we must determine when vCt exceeds VBO for D1 This calculation is much harder than in the examples in the book because in the previous problems the source was a simple DC voltage source while here the voltage source is sinusoidal However the principles are identical a To determine when the SCR will turn ON we must calculate the voltage vCt and then solve for the time at which vCt exceeds VBO for D1 At the beginning of the half cycle D1 and SCR1 are OFF and the voltage across the load is essentially 0 so the entire source voltage vSt is applied to the series RC circuit To determine the voltage vCt on the capacitor we can write a Kirchhoffs Current Law equation at the node above the capacitor and solve the resulting equation for vCt since the DIAC is an open circuit at this time 0 2 1 i i 0 1 C C dt v C d R v v 315 1 1 1 RC v RC v dt v d C C t RC V RC v dt v d M C C sin 1 The solution can be divided into two parts a natural response and a forced response The natural response is the solution to the differential equation 1 0 C C d v v dt RC The solution to the natural response differential equation is e t RC vC n t A where the constant A must be determined from the initial conditions in the system The forced response is the steadystate solution to the equation 1 M sin C C d V v v dt RC RC t It must have a form similar to the forcing function so the solution will be of the form 1 2 sin cos vC f t B t B t where the constants 1 B and 2 B must be determined by substitution into the original equation Solving for 1 B and 2 B yields 1 2 1 2 1 sin cos sin cos M sin d V B t B t B t B t t dt RC RC 1 2 1 2 1 cos in sin cos VM sin B t B s t B t B t RC RC t cosine equation 1 2 1 0 B RC B 2 1 B RC B sine equation 2 1 1 VM B RC B R C 2 1 1 1 VM RC B RC B R C 2 1 1 VM RC B RC R C 2 2 2 1 1 M R C V B RC R C Finally 316 1 2 2 2 1 VM B R C and 2 2 2 2 1 RC VM B R C Therefore the forced solution to the equation is 2 2 2 2 2 2 sin cos 1 1 M M C f V RC V v t t R C R C t and the total solution is C C n C f v t v t v t 2 2 2 2 2 2 sin cos 1 1 t M M RC C V RC V v t Ae t t R C R C The initial condition for this problem is that the voltage on the capacitor is zero at the beginning of the halfcycle 0 2 2 2 2 2 2 0 sin 0 1 1 M M RC C V RC V v Ae R C R C cos 0 0 2 2 2 0 1 RC VM A R C 2 2 2 1 RC VM A R C Therefore the voltage across the capacitor as a function of time before the DIAC fires is 2 2 2 2 2 2 2 2 2 sin cos 1 1 1 t M M M RC C RC V V RC V v t e t t R C R C R C If we substitute the known values for R C and VM this equation becomes 100 42 1114 sin 42 cos t vC t e t t This equation is plotted below It reaches a voltage of 30 V at a time of 350 ms Since the frequency of the waveform is 60 Hz the waveform there are 360 in 160 s and the firing angle is 360 350 ms 756 160 s or 1319 radians 317 Note This problem could also have been solved using Laplace Transforms if desired b The rms voltage applied to the load is 2 2 2 rms 1 M sin V v t dt V T t dt 2 rms 1 1 sin2 2 4 VM V t t 2 rms 1 1 sin2 sin2 2 4 VM V rms 03284 0573 971 V M M V V V S16 One problem with the circuit shown in Figure S11 is that it is very sensitive to variations in the input voltage For example suppose the peak value of the input voltage were to decrease Then the time that it takes capacitor C1 to charge up to the breakover voltage of the DIAC will increase and the SCR will be triggered later in each half cycle Therefore the rms voltage supplied to the load will be reduced both by the lower peak voltage and by the later firing This same effect happens in the opposite direction if increases How could this circuit be modified to reduce its sensitivity to variations in input voltage v t ac vac t SOLUTION If the voltage charging the capacitor could be made constant or nearly so then the feedback effect would be stopped and the circuit would be less sensitive to voltage variations A common way to do this is to use a zener diode that fires at a voltage greater than for the DIAC across the RC charging circuit This diode holds the voltage across the RC circuit constant so that the capacitor charging time is not much affected by changes in the power supply voltage BO V vC R S17 Explain the operation of the circuit shown in Figure S12 and sketch the output voltage from the circuit 318 SOLUTION This circuit is a singlephase voltage source inverter 1 Initially suppose that both SCRs are OFF Then the voltage on the transformer T3 will be 0 and voltage VDC will be dropped across SCR1 and SCR2 as shown 2 Now apply a pulse to transformer T1 that turns on SCR1 When that happens the circuit looks like Since the top of the transformer is now grounded a voltage VDC appears across the upper winding as shown This voltage induces a corresponding voltage on the lower half of the winding charging capacitor C1 up to a voltage of 2VDC as shown Now suppose that a pulse is applied to transformer T2 When that occurs SCR2 becomes a short circuit and SCR1 is turned OFF by the reverse voltage applied to it by capacitor C1 forced commutation At that time the circuit looks like Now the voltages on the transformer are reversed charging capacitor C1 up to a voltage of 2VDC in the opposite direction When SCR1 is triggered again the voltage on C1 will turn SCR2 OFF The output voltage from this circuit would be roughly a square wave except that capacitor C2 filters it somewhat Note The above discussion assumes that transformer T3 is never in either state long enough for it to saturate S18 Figure S13 shows a relaxation oscillator with the following parameters R1 variable 1500 R2 10 C F VDC 100 V BO 30 V V 05 mA IH a Sketch the voltages v and for this circuit t C v t D v t o b If R1 is currently set to 500 k calculate the period of this relaxation oscillator 319 SOLUTION a The voltages vCt vDt and vot are shown below Note that vCt and vDt look the same during the rising portion of the cycle After the PNPN Diode triggers the voltage across the capacitor decays with time constant 2 R1R2 R1 R2 C while the voltage across the diode drops immediately b When voltage is first applied to the circuit the capacitor C charges with a time constant 1 R1 C 500 k100 F 050 s The equation for the voltage on the capacitor as a function of time during the charging portion of the cycle is 320 1 t R C vC t A B e where A and B are constants depending upon the initial conditions in the circuit Since vC0 0 V and vC 100 V it is possible to solve for A and B A vC 100 V A B vC0 0 V B 100 V Therefore 050 100 100 V t vC t e The time at which the capacitor will reach breakover voltage is found by setting vCt VBO and solving for time t1 1 100 V 30 V 050 ln 178 ms 100 V t Once the PNPN Diode fires the capacitor discharges through the parallel combination of R1 and R2 so the time constant for the discharge is 1 2 2 1 2 500 k 15 k 10 F 00015 s 500 k 15 k R R C R R The equation for the voltage on the capacitor during the discharge portion of the cycle is 2 t vC t A B e 2 BO t vC t V e The current through the PNPN diode is given by 2 BO 2 t D V i t e R If we ignore the continuing trickle of current from R1 the time at which iDt reaches IH is 2 2 2 BO 00005 A 1500 ln 00015 ln 55 ms 30 V H I R t R C V Therefore the period of the relaxation oscillator is T 178 ms 55 ms 1835 ms and the frequency of the relaxation oscillator is f 1T 545 Hz S19 In the circuit in Figure S14 T1 is an autotransformer with the tap exactly in the center of its winding Explain the operation of this circuit Assuming that the load is inductive sketch the voltage and current applied to the load What is the purpose of SCR What is the purpose of D2 This chopper circuit arrangement is known as a Jones circuit 2 321 SOLUTION First assume that SCR1 is triggered When that happens current will flow from the power supply through SCR1 and the bottom portion of transformer T1 to the load At that time a voltage will be applied to the bottom part of the transformer which is positive at the top of the winding with respect to the bottom of the winding This voltage will induce an equal voltage in the upper part of the autotransformer winding forward biasing diode D1 and causing the current to flow up through capacitor C This current causes C to be charged with a voltage that is positive at its bottom with respect to its top This condition is shown in the figure above Now assume that SCR2 is triggered When SCR2 turns ON capacitor C applies a reversebiased voltage to SCR1 shutting it off Current then flow through the capacitor SCR2 and the load as shown below This current charges C with a voltage of the opposite polarity as shown SCR2 will cut off when the capacitor is fully charged Alternately it will be cut off by the voltage across the capacitor if SCR1 is triggered before it would otherwise cut off In this circuit SCR1 controls the power supplied to the load while SCR2 controls when SCR1 will be turned off Diode D2 in this circuit is a freewheeling diode which allows the current in the load to continue flowing for a short time after SCR1 turns off 322 S110 A seriescapacitor forced commutation chopper circuit supplying a purely resistive load is shown in Figure S15 DC 120 V V R 20 k 8 m IH A LOAD 250 R BO 200 V V 150 F C a When SCR is turned on how long will it remain on What causes it to turn off 1 b When SCR turns off how long will it be until the SCR can be turned on again Assume that three time constants must pass before the capacitor is discharged 1 c What problem or problems do these calculations reveal about this simple seriescapacitor forced commutation chopper circuit d How can the problems described in part c be eliminated 323 Solution a When the SCR is turned on it will remain on until the current flowing through it drops below IH This happens when the capacitor charges up to a high enough voltage to decrease the current below IH If we ignore resistor R because it is so much larger than RLOAD the capacitor charges through resistor RLOAD with a time constant LOAD RLOADC 250 150 F 00375 s The equation for the voltage on the capacitor as a function of time during the charging portion of the cycle is LOAD t R C vC t A B e where A and B are constants depending upon the initial conditions in the circuit Since vC0 0 V and vC VDC it is possible to solve for A and B A vC VDC A B vC0 VDC B VDC Therefore LOAD DC DC V t R C vC t V V e The current through the capacitor is C C d i t C v t dt LOAD DC DC t R C C d i t C V V e dt LOAD DC LOAD A t R C C V i t e R Solving for time yields 2 2 LOAD DC DC ln 00375 ln C C i t R i t R t R C V V 324 The current through the SCR consists of the current through resistor R plus the current through the capacitor The current through resistor R is 120 V 20 k 6 mA and the holding current of the SCR is 8 mA so the SCR will turn off when the current through the capacitor drops to 2 mA This occurs at time 2 mA 250 00375 ln 0206 s 120 V t b The SCR can be turned on again once the capacitor has discharged The capacitor discharges through resistor R It can be considered to be completely discharged after three time constants Since RC 20 k150 F 3 s the SCR will be ready to fire again after 9 s c In this circuit the ON time of the SCR is much shorter than the reset time for the SCR so power can flow to the load only a very small fraction of the time This effect would be less exaggerated if the ratio of R to RLOAD were smaller d This problem can be eliminated by using one of the more complex series commutation circuits described in Section 35 These more complex circuits provide special paths to quickly discharge the capacitor so that the circuit can be fired again soon S111 A parallelcapacitor forced commutation chopper circuit supplying a purely resistive load is shown in Figure S16 DC 120 V V 1 R 20 k 5 m IH A load 250 R BO 250 V V 15 F C a When SCR is turned on how long will it remain on What causes it to turn off 1 b What is the earliest time that SCR can be turned off after it is turned on Assume that three time constants must pass before the capacitor is charged 1 c When SCR turns off how long will it be until the SCR can be turned on again 1 d What problem or problems do these calculations reveal about this simple parallelcapacitor forced commutation chopper circuit e How can the problems describe in part d be eliminated 325 SOLUTION a When SCR1 is turned on it will remain on indefinitely until it is forced to turn off When SCR1 is turned on capacitor C charges up to VDC volts with the polarity shown in the figure above Once it is charged SCR1 can be turned off at any time by triggering SCR2 When SCR2 is triggered the voltage across it drops instantaneously to about 0 V which forces the voltage at the anode of SCR1 to be VDC volts turning SCR1 off Note that SCR2 will spontaneously turn off after the capacitor discharges since VDC R1 IH for SCR2 b If we assume that the capacitor must be fully charged before SCR1 can be forced to turn off then the time required would be the time to charge the capacitor The capacitor charges through resistor R1 and the time constant for the charging is R1C 20 k15 F 03 s If we assume that it takes 3 time constants to fully charge the capacitor then the time until SCR1 can be turned off is 09 s Note that this is not a very realistic assumption In real life it is possible to turn off SCR1 with less than a full VDC volts across the capacitor c SCR1 can be turned on again after the capacitor charges up and SCR2 turns off The capacitor charges through RLOAD so the time constant for charging is RLOADC 250 15 F 000375 s and SCR2 will turn off in a few milliseconds d In this circuit once SCR1 fires a substantial period of time must pass before the power to the load can be turned off If the power to the load must be turned on and off rapidly this circuit could not do the job e This problem can be eliminated by using one of the more complex parallel commutation circuits described in Section 35 These more complex circuits provide special paths to quickly charge the capacitor so that the circuit can be turned off quickly after it is turned on S112 Figure S17 shows a singlephase rectifierinverter circuit Explain how this circuit functions What are the purposes of C1 and C2 What controls the output frequency of the inverter 326 SOLUTION The last element in the filter of this rectifier circuit is an inductor which keeps the current flow out of the rectifier almost constant Therefore this circuit is a current source inverter The rectifier and filter together produce an approximately constant dc voltage and current across the two SCRs and diodes at the right of the figure The applied voltage is positive at the top of the figure with respect to the bottom of the figure To understand the behavior of the inverter portion of this circuit we will step through its operation 1 First assume that SCR1 and SCR4 are triggered Then the voltage V will appear across the load positivetonegative as shown in Figure a At the same time capacitor C1 will charge to V volts through diode D3 and capacitor C2 will charge to V volts through diode D2 a 2 Now assume that SCR2 and SCR3 are triggered At the instant they are triggered the voltage across capacitors C1 and C2 will reverse bias SCR1 and SCR4 turning them OFF Then a voltage of V volts will appear across the load positivetonegative as shown in Figure b At the same time capacitor C1 will charge to V volts with the opposite polarity from before and capacitor C2 will charge to V volts with the opposite polarity from before 327 Figure b 3 If SCR1 and SCR4 are now triggered again the voltages across capacitors C1 and C2 will force SCR2 and SCR3 to turn OFF The cycle continues in this fashion Capacitors C1 and C2 are called commutating capacitors Their purpose is to force one set of SCRs to turn OFF when the other set turns ON The output frequency of this rectifierinverter circuit is controlled by the rates at which the SCRs are triggered The resulting voltage and current waveforms assuming a resistive load are shown below S113 A simple fullwave ac phase angle voltage controller is shown in Figure S18 The component values in this circuit are R 20 to 300 k currently set to 80 k C 015 F 328 BO 40 V for PNPN Diode D1 V BO 250 V for SCR1 V sin volts s M v t V t where M V 1697 V and 377 rads a At what phase angle do the PNPN diode and the SCR turn on b What is the rms voltage supplied to the load under these circumstances Note Problem S113 is significantly harder for many students since it involves solving a differential equation with a forcing function This problem should only be assigned if the class has the mathematical sophistication to handle it SOLUTION At the beginning of each half cycle the voltages across the PNPN diode and the SCR will both be smaller then their respective breakover voltages so no current will flow to the load except for the very tiny current charging capacitor C and vloadt will be 0 volts However capacitor C charges up through resistor R and when the voltage vCt builds up to the breakover voltage of D1 the PNPN diode will start to conduct This current flows through the gate of SCR1 turning the SCR ON When it turns ON the voltage across the SCR will drop to 0 and the full source voltage vSt will be applied to the load producing a current flow through the load The SCR continues to conduct until the current through it falls below IH which happens at the very end of the half cycle Note that after D1 turns on capacitor C discharges through it and the gate of the SCR At the end of the half cycle the voltage on the capacitor is again essentially 0 volts and the whole process is ready to start over again at the beginning of the next half cycle To determine when the PNPN diode and the SCR fire in this circuit we must determine when vCt exceeds VBO for D1 This calculation is much harder than in the examples in the book because in the previous problems the source was a simple DC voltage source while here the voltage source is sinusoidal However the principles are identical a To determine when the SCR will turn ON we must calculate the voltage vCt and then solve for the time at which vCt exceeds VBO for D1 At the beginning of the half cycle D1 and SCR1 are OFF and the voltage across the load is essentially 0 so the entire source voltage vSt is applied to the series RC circuit To determine the voltage vCt on the capacitor we can write a Kirchhoffs Current Law equation at the node above the capacitor and solve the resulting equation for vCt since the PNPN diode is an open circuit at this time 1 2 0 i i 329 1 0 C C v v C d v R dt 1 1 1 C C d v v dt RC RC v 1 M sin C C d V v v dt RC RC t The solution can be divided into two parts a natural response and a forced response The natural response is the solution to the equation 1 0 C C d v v dt RC The solution to the natural response equation is e t RC vC n t A where the constant A must be determined from the initial conditions in the system The forced response is the steadystate solution to the equation 1 M sin C C d V v v dt RC RC t It must have a form similar to the forcing function so the solution will be of the form 1 2 sin cos vC f t B t B t where the constants 1 B and 2 B must be determined by substitution into the original equation Solving for 1 B and 2 B yields 1 2 1 2 1 sin cos sin cos M sin d V B t B t B t B t t dt RC RC 1 2 1 2 1 cos in sin cos VM sin B t B s t B t B t RC RC t cosine equation 1 2 1 0 B RC B 2 1 B RC B sine equation 2 1 1 VM B RC B R C 2 1 1 1 VM RC B RC B R C 2 1 1 VM RC B RC R C 2 2 2 1 1 M R C V B RC R C Finally 330 1 2 2 2 1 VM B R C and 2 2 2 2 1 RC VM B R C Therefore the forced solution to the equation is 2 2 2 2 2 2 sin cos 1 1 M M C f V RC V v t t R C R C t and the total solution is C C n C f v t v t v t 2 2 2 2 2 2 sin cos 1 1 t M M RC C V RC V v t Ae t t R C R C The initial condition for this problem is that the voltage on the capacitor is zero at the beginning of the halfcycle 0 2 2 2 2 2 2 0 sin 0 1 1 M M RC C V RC V v Ae R C R C cos 0 0 2 2 2 0 1 RC VM A R C 2 2 2 1 RC VM A R C Therefore the voltage across the capacitor as a function of time before the PNPN diode fires is 2 2 2 2 2 2 2 2 2 sin cos 1 1 1 t M M M RC C RC V V RC V v t e t t R C R C R C If we substitute the known values for R C and VM this equation becomes 833 3576 791 sin 3576 cos t vC t e t t This equation is plotted below 331 It reaches a voltage of 40 V at a time of 48 ms Since the frequency of the waveform is 60 Hz the waveform there are 360 in 160 s and the firing angle is 360 48 ms 1037 160 s or 1810 radians Note This problem could also have been solved using Laplace Transforms if desired b The rms voltage applied to the load is 2 2 2 rms 1 M sin V v t dt V T t dt 2 rms 1 1 sin2 2 4 VM V t t sin 2 4 sin 2 1 2 1 2 rms VM V Since 1180 radians the rms voltage is rms 01753 0419 710 V M M V V V S114 Figure S19 shows a threephase fullwave rectifier circuit supplying power to a dc load The circuit uses SCRs instead of diodes as the rectifying elements a What will the load voltage and ripple be if each SCR is triggered as soon as it becomes forward biased At what phase angle should the SCRs be triggered in order to operate this way Sketch or plot the output voltage for this case b What will the rms load voltage and ripple be if each SCR is triggered at a phase angle of 90 that is half way through the halfcycle in which it is forward biased Sketch or plot the output voltage for this case 332 SOLUTION Assume that the three voltages applied to this circuit are sin A M v t V t sin 2 3 B M v t V t sin 2 3 C M v t V t The period of the input waveforms is T where 2 T For the purpose of the calculations in this problem we will assume that is 377 rads 60 Hz a The when the SCRs start to conduct as soon as they are forward biased this circuit is just a three phase fullwave bridge and the output voltage is identical to that in Problem 32 The sketch of output voltage is reproduced below and the ripple is 42 The following table shows which SCRs must conduct in what order to create the output voltage shown below The times are expressed as multiples of the period T of the input waveforms and the firing angle is in degrees relative to time zero Start Time t Stop Time t Positive Phase Negativ e Phase Conducting SCR Positive Conducting SCR Negative Triggered SCR Firing Angle T 12 T 12 c b SCR3 SCR5 SCR5 30 T 12 3T 12 a b SCR1 SCR5 SCR1 30 3T 12 5T 12 a c SCR1 SCR6 SCR6 90 5T 12 7T 12 b c SCR2 SCR6 SCR2 150 7T 12 9T 12 b a SCR2 SCR4 SCR4 210 9T 12 11T 12 c a SCR3 SCR4 SCR3 270 11T 12 T 12 c b SCR3 SCR5 SCR5 330 333 T12 b If each SCR is triggered halfway through the halfcycle during which it is forward biased the resulting phase a b and c voltages will be zero before the first half of each halfcycle and the full sinusoidal value for the second half of each halfcycle These waveforms are shown below These plots were created by the MATLAB program that appears later in this answer and the resulting output voltage will be 334 A MATLAB program to generate these waveforms and to calculate the ripple on the output waveform is shown below The first function biphasecontrollerm generates a switched ac waveform The inputs to this function are the current phase angle in degrees the offset angle of the waveform in degrees and the firing angle in degrees function volts biphasecontrollerwttheta0fire Function to simulate the output of an ac phase angle controller that operates symmetrically on positive and negative half cycles Assume a peak voltage VM 120 SQRT2 170 V for convenience wt Current phase in degrees theta0 Starting phase angle in degrees fire Firing angle in degrees Degrees to radians conversion factor deg2rad pi 180 Remove phase ambiguities 0 wt 360 deg ang wt theta0 while ang 360 ang ang 360 end while ang 0 ang ang 360 end Simulate the output of the phase angle controller if ang fire ang 180 volts 170 sinang deg2rad elseif ang fire180 ang 360 volts 170 sinang deg2rad else 335 336 volts 0 end The main program below creates and plots the threephase waveforms calculates and plots the output waveform and determines the ripple in the output waveform Mfile prob314bm Mfile to calculate and plot the three phase voltages when each SCR in a threephase fullwave rectifier triggers at a phase angle of 90 degrees Calculate the waveforms for times from 0 to 130 s t 0121600130 deg zerossizet rms zerossizet va zerossizet vb zerossizet vc zerossizet out zerossizet for ii 1lengtht Get equivalent angle in degrees Note that 160 s 360 degrees for a 60 Hz waveform theta 21600 tii Calculate the voltage in each phase at each angle vaii biphasecontrollertheta090 vbii biphasecontrollertheta12090 vcii biphasecontrollertheta12090 end Calculate the output voltage of the rectifier for ii 1lengtht vals vaii vbii vcii outii max vals min vals end Calculate and display the ripple disp The ripple is num2strrippleout Plot the voltages versus time figure1 plottvabLinewidth20 hold on plottvbrLinewidth20 plottvckLinewidth20 titlebfPhase Voltages xlabelbfTime s ylabelbfVoltage V grid on legendPhase aPhase bPhase c hold off Plot the output voltages versus time figure2 plottoutbLinewidth20 titlebfOutput Voltage xlabelbfTime s ylabelbfVoltage V axis 0 130 0 260 grid on hold off When this program is executed the results are prob314b The ripple is 309547 S115 Write a MATLAB program that imitates the operation of the PulseWidth Modulation circuit shown in Figure 355 and answer the following questions a Assume that the comparison voltages and have peak amplitudes of 10 V and a frequency of 500 Hz Plot the output voltage when the input voltage is v t x v t y v t ft in 10sin 2 V and f 50 Hz b What does the spectrum of the output voltage look like What could be done to reduce the harmonic content of the output voltage c Now assume that the frequency of the comparison voltages is increased to 1000 Hz Plot the output voltage when the input voltage is v t ft in 10sin 2 V and f 50 Hz d What does the spectrum of the output voltage in c look like e What is the advantage of using a higher comparison frequency and more rapid switching in a PWM modulator SOLUTION The PWM circuit is shown below 337 a To write a MATLAB simulator of this circuit note that if then and if then 0 Similarly if then 0 and if then The output voltage is then A MATLAB function that performs these calculations is shown below Note that this function arbitrarily assumes that 100 V It would be easy to modify the function to use any arbitrary dc voltage if desired in v in v xv yv uv vv DC V DC V in v xv uv vin yv vv DC V u v v v v out function voutvuvv voutvin vx vy Function to calculate the output voltage of a PWM modulator from the values of vin and the reference voltages vx and vy This function arbitrarily assumes that VDC 100 V vin Input voltage vx x reference vy y reference vout Ouput voltage vu vv Components of output voltage 338 fire Firing angle in degrees vu if vin vx vu 100 else vu 0 end vv if vin vy vv 0 else vv 100 end Caclulate vout vout vv vu Now we need a MATLAB program to generate the input voltage vin t and the reference voltages vx t and v y t After the voltages are generated function vout will be used to calculate vout t and the frequency spectrum of vout t Finally the program will plot vin t xv t and v y t vout t and the spectrum of Note that in order to have a valid spectrum we need to create several cycles of the 60 Hz output waveform and we need to sample the data at a fairly high frequency This problem creates 4 cycles of and samples all data at a 20000 Hz rate t vout t vout Mfile probs115am Mfile to calculate the output voltage from a PWM modulator with a 500 Hz reference frequency Note that the only change between this program and that of part b is the frequency of the reference fr Sample the data at 20000 Hz to get enough information for spectral analysis Declare arrays fs 20000 Sampling frequency Hz t 01fs415 Time in seconds vx zerossizet vx vy zerossizet vy vin zerossizet Driving signal vu zerossizet vx vv zerossizet vy vout zerossizet Output signal fr 500 Frequency of reference signal T 1fr Period of refernce signal Calculate vx at fr 500 Hz for ii 1lengtht vxii vreftiiT vyii vxii end 339 Calculate vin as a 50 Hz sine wave with a peak voltage of 340 10 V for ii 1lengtht vinii 10 sin2pi50tii end Now calculate vout for ii 1lengtht voutii vuii vvii voutvinii vxii vyii end Plot the reference voltages vs time figure1 plottvxbLinewidth10 hold on plottvykLinewidth10 titlebfReference Voltages for fr 500 Hz xlabelbfTime s ylabelbfVoltage V legendvxvy axis 0 130 10 10 hold off Plot the input voltage vs time figure2 plottvinbLinewidth10 titlebfInput Voltage xlabelbfTime s ylabelbfVoltage V axis 0 130 10 10 Plot the output voltages versus time figure3 plottvoutbLinewidth10 titlebfOutput Voltage for fr 500 Hz xlabelbfTime s ylabelbfVoltage V axis 0 130 120 120 Now calculate the spectrum of the output voltage spec fftvout Calculate sampling frequency labels len lengtht df fs len fstep zerossizet for ii 2len2 fstepii df ii1 fsteplenii2 fstepii end Plot the spectrum figure4 plotfftshiftfstepfftshiftabsspec titlebfSpectrum of Output Voltage for fr 500 Hz xlabelbfFrequency Hz ylabelbfAmplitude Plot a closeup of the near spectrum positive side only figure5 plotfftshiftfstepfftshiftabsspec titlebfSpectrum of Output Voltage for fr 500 Hz xlabelbfFrequency Hz ylabelbfAmplitude setgcaXlim0 1000 When this program is executed the input reference and output voltages are 341 b The output spectrum of this PWM modulator is shown below There are two plots here one showing the entire spectrum and the other one showing the closein frequencies those under 1000 Hz which will have the most effect on machinery Note that there is a sharp peak at 50 Hz which is there desired frequency but there are also strong contaminating signals at about 850 Hz and 950 Hz If necessary these components could be filtered out using a lowpass filter 342 c A version of the program with 1000 Hz reference functions is shown below Mfile probs115bm Mfile to calculate the output voltage from a PWM modulator with a 1000 Hz reference frequency Note that the only change between this program and that of part a is the frequency of the reference fr Sample the data at 20000 Hz to get enough information for spectral analysis Declare arrays fs 20000 Sampling frequency Hz t 01fs415 Time in seconds vx zerossizet vx vy zerossizet vy vin zerossizet Driving signal vu zerossizet vx vv zerossizet vy vout zerossizet Output signal fr 1000 Frequency of reference signal T 1fr Period of refernce signal Calculate vx at 1000 Hz for ii 1lengtht vxii vreftiiT vyii vxii end Calculate vin as a 50 Hz sine wave with a peak voltage of 10 V for ii 1lengtht vinii 10 sin2pi50tii end 343 344 Now calculate vout for ii 1lengtht voutii vuii vvii voutvinii vxii vyii end Plot the reference voltages vs time figure1 plottvxbLinewidth10 hold on plottvykLinewidth10 titlebfReference Voltages for fr 1000 Hz xlabelbfTime s ylabelbfVoltage V legendvxvy axis 0 130 10 10 hold off Plot the input voltage vs time figure2 plottvinbLinewidth10 titlebfInput Voltage xlabelbfTime s ylabelbfVoltage V axis 0 130 10 10 Plot the output voltages versus time figure3 plottvoutbLinewidth10 titlebfOutput Voltage for fr 1000 Hz xlabelbfTime s ylabelbfVoltage V axis 0 130 120 120 Now calculate the spectrum of the output voltage spec fftvout Calculate sampling frequency labels len lengtht df fs len fstep zerossizet for ii 2len2 fstepii df ii1 fsteplenii2 fstepii end Plot the spectrum figure4 plotfftshiftfstepfftshiftabsspec titlebfSpectrum of Output Voltage for fr 1000 Hz xlabelbfFrequency Hz ylabelbfAmplitude Plot a closeup of the near spectrum positive side only figure5 plotfftshiftfstepfftshiftabsspec titlebfSpectrum of Output Voltage for fr 1000 Hz xlabelbfFrequency Hz ylabelbfAmplitude setgcaXlim0 1000 When this program is executed the input reference and output voltages are 345 d The output spectrum of this PWM modulator is shown below 346 347 e Comparing the spectra in b and d we can see that the frequencies of the first large sidelobes doubled from about 900 Hz to about 1800 Hz when the reference frequency was doubled This increase in sidelobe frequency has two major advantages it makes the harmonics easier to filter and it also makes it less necessary to filter them at all Since large machines have their own internal inductances they form natural lowpass filters If the contaminating sidelobes are at high enough frequencies they will never affect the operation of the machine Thus it is a good idea to design PWM modulators with a high frequency reference signal and rapid switching Appendix E Errata for Electric Machinery Fundamentals 5e Current at 15 April 2011 Please note that some or all of the following errata will be corrected in future reprints of the book so they may not appear in your copy of the text PDF pages with these corrections are attached to this appendix please provide them to your students 1 Page 145 Problem 23 was printed incorrectly in the first printing of this text By accident a portion if Problem 24 was printed instead of the appropriate text The correct text is 23 Consider a simple power system consisting of an ideal voltage source an ideal stepup transformer a transmission line an ideal stepdown transformer and a load The voltage of the source is The impedance of the transmission line is and the impedance of the load is S 480 0 V V line 3 4 Z j load 30 40 Z j a Assume that the transformers are not present in the circuit What is the load voltage and efficiency of the system b Assume that transformer 1 is a 15 stepup transformer and transformer 2 is a 51 stepdown transformer What is the load voltage and efficiency of the system c What transformer turns ratio would be required to reduce the transmission line losses to 1 of the total power produced by the generator 2 Page 147 Problem 213 the transformer is Y connected 3 Page 264 Problem 46 the generator should be 2pole connected 60 Hz instead of Y connected 4 Page 269 Problem 425 the problem should say Make a plot of the terminal voltage versus the load impedance angle instead of Make a plot of the terminal voltage versus the load power factor 5 Page 301 Problem 54 the synchronous reactance should be 25 6 Page 304 Problem 512 parts b and i are incorrect The correct problems is given below with the changes in red 512 Figure P53 shows a small industrial plant supplied by an external 480 V threephase power supply The plant includes three main loads as shown in the figure Answer the following questions about the plant The synchronous motor is rated at 100 hp 460 V and 08PF leading The synchronous reactance is 11 pu and armature resistance is 001 pu The OCC for this motor is shown in Figure P54 a If the switch on the synchronous motor is open how much real reactive and apparent power is being supplied to the plant What is the current LI in the transmission line The switch is now closed and the synchronous motor is supplying rated power at rated power factor 348 b What is the field current in the motor c What is the torque angle of the motor c What is the power factor of the motor d How much real reactive and apparent power is being supplied to the plant now What is the current LI in the transmission line Now suppose that the field current is increased to 30 A e What is the real and reactive power supplied to the motor f What is the torque angle of the motor g What is the power factor of the motor h How much real reactive and apparent power is being supplied to the plant now What is the current LI in the transmission line i How does the line current in part d compare to the line current in part h Why 7 Page 305 Problem 517 the power supplied by the generator is 80 kW 8 Page 358 Figure 634 one of the numbers in the table of NEMA starting code letters are incorrect The correct table is given below with the corrected error in red Nominal code letter Locked rotor kVAhp Nominal code letter Locked rotor kVAhp A 0315 L 9001000 B 315355 M 10001120 C 355400 N 11201250 D 400450 P 12501400 E 450500 R 14001600 F 500560 S 16001800 G 560630 T 18002000 H 630710 U 20002240 J 710800 V 2240up K 800900 9 Page 400 Problem 623 the motor develops its fullload induced torque at 35 percent slip 10 Page 402 Problem 631 this problem refers to the motor of Problem 621 not the motor of problem 623 11 Page 402 Problem 632 the parameters of the outer bar are 349 R2o X2o 480 375 and the parameters of the inner bar are R2i X2i 0573 465 12 Page 553 Problem 84 the armature reaction is 1000 A turns at full load 350 13 Page 667 Problem C1 this problem should begin with the sentence A 138kV 50MVA 09 powerfactorlagging 60Hz fourpole Yconnected synchronous generator has a directaxis reactance of 25 a quadratureaxis reactance of 18 and an armature resistance of 02