FROM -THE- LI BRARY- OF 
 WILL1AM-A HILLEBRAND 
 
PROBLEMS 
 
 IN 
 
 ALTERNATING CURRENT MACHINERY 
 
McGraw-Hill BookCompariy 
 
 PwSGsfiers 
 
 Ele c trie al Wor Id The Engineering and Mining Journal 
 En5ineering Record Engineering News 
 
 Railway Age Gazette American Machinist 
 
 Signal E,ngin<?9r American Engineer 
 
 Electric Railway Journal Coal Age 
 
 Metallurgical and Chemical Engineering P o we r 
 
PROBLEMS IN ALTERNATING 
 CURRENT MACHINERY 
 
 BY 
 WALDO V. LYON 
 
 INSTRUCTOR IN ELECTRICAL ENGINEERING, MASSACHUSETTS 
 INSTITUTE OF TECHNOLOGY. 
 
 FIRST EDITION 
 
 McGRAW-HILL BOOK COMPANY, INC. 
 239 WEST 39TH STREET, NEW YORK 
 
 6 BOUVERIE STREET, LONDON, E. C. 
 1914 
 
COPYRIGHT, 1914, BY THE 
 MCGRAW-HILL BOOK COMPANY, INC. 
 
 THE. MAPLE- PRESS. YORK- PA 
 
PREFACE 
 
 These problems chiefly concern the theory of the operation of 
 alternating current machinery, and are such as we give to the 
 fourth year students in electrical engineering in this subject. 
 
 In each chapter the problems of a similar nature are grouped 
 together, and those of each group are then arranged in the 
 approximate order of their difficulty. The groups in each chapter 
 follow each other in as logical an order as possible, both from point 
 of difficulty and the presentation of the subject. 
 
 In order that this collection of problems may be useful among 
 different classes of students a large variety has been included, 
 ranging from the very simple to those of considerable difficulty. 
 Wherever it is essential the data have been taken from actual 
 apparatus. This was possible through the courtesy of two of 
 the large manufacturing companies. In some of the problems 
 so few data are given that approximate methods of solution 
 must be used, but care has been taken to so state them that 
 the errors thus introduced need not be large. This lack of data 
 is frequently met in practice. 
 
 It has been thought best not to give introductory paragraphs 
 for each chapter as was done in the preceding volume of prob- 
 lems inasmuch as they would have to be of considerable length 
 to be of much value. 
 
 The answers to the problems will probably be ready for 
 publication in the fall of 1914. They will be available to all 
 but undergraduate students at a nominal price. Undergraduate 
 students can obtain them only on the recommendation of their 
 instructors. 
 
 WALDO V. LYON. 
 
 MASSACHUSETTS INSTITUTE OF TECHNOLOGY, 
 December, 1913. 
 
 995890 
 
CONTENTS 
 
 CHAPTER PAGE 
 
 I. TRANSFORMERS (98 PROBLEMS) 1 
 
 Iron loss, magnetizing current, number of turns, effective resist- 
 ance and reactance, copper loss, regulation, efficiency, auto- 
 transformer, induction regulator, parallel operation. 
 
 II. SYNCHRONOUS GENERATORS (99 PROBLEMS) 24 
 
 Generated electromotive force, form factor, harmonics, concen- 
 trated and distributed windings, leakage reactance, armature 
 reaction, synchronous reactance voltage, regulation-comparison 
 of methods, effect of power factor, efficiency, parallel operation, 
 hunting. 
 
 III. SYNCHRONOUS MOTORS (50 PROBLEMS) 52 
 
 Excitation, efficiency, power factor, maximum current, break- 
 down, effect of added resistance and reactance, transmission line 
 regulation, power factor compensation. 
 
 IV. INDUCTION MOTORS (80 PROBLEMS) 68 
 
 Flux distribution, exciting current, iron loss, winding pitch, 
 starting current, torque, speed regulation, breakdown torque, 
 effect of added resistance and reactance, Heyland diagram, 
 calculation of characteristics, concatenation, induction gen- 
 erator, necessary synchronous apparatus, calculation of char- 
 acteristics. 
 
 V. ROTARY CONVERTERS (61 PROBLEMS) 93 
 
 Voltage ratio, effect of flux distribution, relative outputs, trans- 
 former capacities, armature reaction, excitation, series field 
 turns, efficiency. 
 
 VI. POLYPHASE CIRCUITS (75 PROBLEMS) 105 
 
 Delta and Y connection, balanced and unbalanced loads, 
 neutral current, power measurement with unbalanced loads, 
 transformer connections and relative capacities, auto- trans- 
 formers, transmission line regulation and efficiency, combined 
 efficiency and regulation of generator, line and transformers, 
 induction and synchronous motor loads. 
 
 VII. NON-SINUSOIDAL WAVES (44 PROBLEMS) 126 
 
 Phase and line voltages, two-phase to three-phase transforma- 
 tion, neutral current, power measurement, effect of inductance 
 and capacity, cyclic order of voltages, unbalanced loads, trans- 
 former connections. 
 
 Vll 
 
INTRODUCTION 
 
 The great importance of problem work in the training of stu- 
 dents of engineering is now generally recognized. No other work 
 so efficiently develops analytical power and clear, logical think- 
 ing, so necessary to success in the engineering profession. Yet 
 notwithstanding the importance of problem work in general, the 
 first consistently developed book of electrical engineering prob- 
 lems was that prepared by Mr. Lyon in 1908, its wide use being 
 conclusive evidence of the needs that were felt among both 
 teachers and students for such a work, and of the appreciation of 
 the importance of the training which it exemplifies. 
 
 It requires a special gift to originate and develop problems 
 which shall give sound training in the fundamentals of engineering 
 and whose solution shall not only interest the student, but develop 
 his intellectual power as well. In general the problems must be 
 closely related to engineering practice, graded as to difficulty, 
 and must carefully avoid being mere mathematical puzzles. Mr. 
 Lyon, as his earlier book also shows, has a most unusual ability in 
 the preparation of problems for the electrical engineering field. 
 
 The present collection of problems relating to electrical 
 machinery, more particularly in the field of alternating current 
 engineering, has been prepared with the same point of view as 
 was Mr. Lyon's earlier work, and should likewise prove most 
 useful to both instructor and student. 
 
 H. E. CLIFFORD. 
 
 HARVARD UNIVERSITY, 
 December, 1913. 
 
 IX 
 
PROBLEMS IN ALTERNATING 
 CURRENT MACHINERY 
 
 CHAPTER I 
 TRANSFORMERS 
 
 1. The iron loss in a reactor is 240 watts of which 48 watts is 
 due to eddy currents. If the amount of iron in the magnetic 
 circuit were doubled by doubling the cross-section of the core, 
 what would be the iron loss for the same impressed voltage and 
 frequency? Neglect the resistance drop. 
 
 2. The iron loss in a reactor is 312 watts of which 86 watts is 
 due to eddy currents. If a similar reactor were constructed in 
 which laminations of twice the thickness were used, what would 
 be the iron loss for the same impressed voltage and frequency? 
 Neglect the resistance drop. 
 
 3. When 110 volts at 30 cycles is impressed on a reactor the 
 iron loss is 276 watts of which 204 watts is due to hysteresis. If 
 the impressed voltage and frequency are both doubled what will 
 be the iron loss? Neglect the resistance drop. 
 
 4. When 110 volts at 60 cycles is impressed on a reactor, the 
 iron loss is 248 watts of which 25 per cent, is due to eddy currents, 
 (a) What will be the iron loss when 220 volts at 60 cycles is 
 impressed on the reactor? (b) What will be the iron loss when 
 110 volts at 30 cycles is impressed on the reactor? Neglect the 
 resistance drop. 
 
 5. A reactor has two electric circuits having the same number 
 of turns which may be connected in series or in parallel. When 
 they are connected in series across 220-volt, 60-cycle mains the 
 iron loss is 326 watts, of which 89 watts is due to eddy currents. 
 What will be the iron loss if the coils are connected in parallel 
 across the same mains? Neglect the resistance drop. 
 
 6. With 425 volts at 25 cycles impressed on the low-tension 
 winding of a transformer 2500 watts is supplied at no load. If 
 the frequency of this impressed voltage is increased to 40 cycles 
 
 1 
 
2 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 without changing its root-mean-square value or form factor, 2020 
 watts is supplied at no load. What is the division between the 
 eddy current and hysteresis losses at 25 cycles? 
 
 7. With 440 volts at 50 cycles impressed on the low-tension 
 winding of a transformer 641 watts is supplied at no load. If 
 the frequency and voltage are each reduced 50 per cent, without 
 changing the form factor 278 watts is supplied at no load. 
 What is the division between the-eddy current and hysteresis 
 losses at 50 cycles? 
 
 8. With 440 volts at 60 cycles impressed on the low-tension 
 winding of a transformer 371 watts is supplied at no load. If the 
 voltage is reduced' 50 per cent, without changing the frequency or 
 the form factor 114 watts is supplied at no load. What is the di- 
 vision between the eddy current and hysteresis losses at 440 volts? 
 
 9. If a simple harmonic electromotive force of 2200 volts at 
 60 cycles is impressed on the low-tension winding of a transformer 
 the core loss is 940 watts, 23 per cent, of which is due to eddy 
 currents, (a) What will be the core loss if an electromotive 
 force of the same effective value and frequency but with a form 
 factor of 1.05 1 is impressed on the low-tension winding? (b) 
 What will be the core loss if an electromotive force of the same 
 effective value and frequency but with a form factor of 1.2 2 is 
 impressed on the low-tension winding? 
 
 10. The magnetic circuit of a transformer has a mean length 
 of 77.3 in. and an average cross-section of 34.4 sq. in. The 
 low-tension winding has 399 turns. Find the core loss, the no- 
 load current, and the power factor at no load when 2300 volts at 
 60 cycles is impressed on the low-tension winding. The curve of 
 core loss at 60 cycles and flux density, and the B-H curve are 
 given by the data on the following page : 
 
 11. A magnetic circuit has a mean length of 69.5 in. and an 
 average cross-section of 30.0 sq. in. It is wound with a coil of 
 94 turns. A constant voltage of 440 volts at 50 cycles is im- 
 pressed on this coil. How long an air-gap would it be necessary 
 to cut in the magnetic circuit in order that the coil would take a 
 current 20 times as great? What is the power factor before and 
 after cutting the air-gap? Neglect the resistance of the coil and 
 use the magnetic data given in problem 10. Assume that the core 
 loss at 50 cycles is 80 per cent, of the loss at 60 cycles. 
 
 1 Flat topped. 
 
 2 Peaked. 
 
TRANSFORMERS 
 
 Ordinates 
 
 Abscissae 
 
 Flux density 
 Kiloiines per sq. in. 
 
 Magnetizing force 
 4*- NI 
 
 H ~w l^- 
 
 Core loss at 60 cycles 
 Watts per cu. in. 
 
 40.0 
 
 2.0 
 
 0.132 
 
 53.0 
 
 3.0 
 
 0.206 
 
 61.5 
 
 4.0 
 
 0.262 
 
 67.0 
 
 5.0 
 
 0.300 
 
 73.6 
 
 7.0 
 
 0.352 
 
 80.0 
 
 10.0 
 
 0.406 
 
 86.0 
 
 15.0 
 
 0.457 
 
 89.6 
 
 20.0 
 
 0.490 
 
 This data is for a good grade of sheet steel. 
 
 For all of the problems to which this data applies assume that at the 
 working flux density the joints in the magnetic circuit require 75 additional 
 ampere turns. 
 
 12. The magnetic circuit of a 5000-kv.-a. transformer has a 
 mean length of 140 in. and an average cross-section, of 613 sq. 
 in. The number of turns in the high-tension winding is such that 
 with the rated voltage of 52,000 volts at 50 cycles impressed on 
 this winding, 17,500 watts is supplied at no load. How many 
 turns has the high-tension winding? What are the no-load cur- 
 rent and power factor? Use the magnetic data given in problem 
 10. Assume that the core loss at 50 cycles is 80 per cent, of the 
 loss at 60 cycles. 
 
 13. If the transformer described in problem 12 is connected to 
 a 66,000-volt, 50-cycle circuit, what will be the no-load current 
 and power? 
 
 14. The magnetic circuit of a lOOO-kv.-a. transformer has a 
 mean length of 95 in. and an average cross-section of 278 sq. in. 
 This transformer is designed to operate from a 66,000-volt, 
 60-cycle circuit. If the maximum flux density is to be 70,000 
 lines per square inch how many turns should the high-tension 
 winding have? What are the no-load current and power? Use 
 the magnetic data given in problem 10. 
 
 15. 1 At no load a 500-kv.-a. transformer takes a current of 
 
 1 Assume that the original flux density is 70 kilolines per sq. in., and that 
 the relation between the flux density and the permeability for this quality 
 of iron is expressed by: 
 
 5=102-0.017^ 
 
 between B equals 45 kilolines and B equals 95 kilolines per sq. in. In prob- 
 lems where this applies the effect of the joints in the magnetic circuit will 
 be neglected. 
 
4 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 3.35 amperes and a power of 2960 watts, when the voltage 
 impressed on the high-tension side is 11,000 volts at 60 cycles. 
 What current and power will this transformer take at no load if 
 12,700 volts at 60 cycles is impressed on the high-tension winding? 
 Assume that the core loss varies as the 1.7 power of the flux 
 density. 
 
 16. ' A 50-kv.-a. transformer takes a current of 0.1 ampere and 
 a power of 641 watts at no load when 30,000 volts at 50 cycles is 
 impressed on the high-tension winding. What current and power 
 will this transformer take at no load if 30,000 volts at 60 cycles is 
 impressed on the high-tension winding? Assume that the core 
 loss varies as the 1.3 power of the frequency and the 1.7 power of 
 the flux density. 
 
 17. The resistance to direct current of a reactor which has a 
 laminated magnetic circuit is 0.2 ohm. When 110 volts at 60 
 cycles is impressed on this reactor the current is 10 amperes and 
 the power is 550 watts. What are the apparent resistance and 
 reactance of the reactor? What is the actual inductance of the 
 reactor at this voltage and frequency? 
 
 18. l When 110 volts at 60 cycles is impressed on a reactor 
 which has a laminated magnetic circuit it absorbs 500 watts. At 
 this time the inductance of the reactor is 25 mil-henrys and the 
 resistance is negligibly small, (a) What are the current and 
 power factor? What are the apparent resistance and reactance 
 of the reactor? (b) If 146 volts at 60 cycles is impressed on 
 this reactor what will be the current and power factor? What 
 will be the apparent resistance and reactance of the reactor? 
 Assume that the iron loss varies as the 1.7 power of the flux 
 density. 
 
 19. A reactor with a laminated magnetic circuit has a negligibly 
 small resistance to direct current. When 110 volts at 25 cycles 
 is impressed on this reactor the current is 20 amperes and the 
 power is 1000 watts. If a very small air-gap is cut in the mag- 
 
 1 Assume that the orignal flux density is 70 kilolines per sq. in., and that 
 the relation between the flux density and the permeability for this quality 
 of iron is expressed by : 
 
 5=102-0.017^ 
 
 between B equals 45 kilolines and B equals 95 kilolines per sq. in. In prob- 
 lems where this applies the effect of the joints in the magnetic circuit will 
 be neglected. 
 
TRANSFORMERS 5 
 
 netic circuit the current becomes 100 amperes. By what amount 
 are the apparent resistance and reactance of the reactor changed 
 by the introduction of the air-gap? 
 
 20. A reactor with a laminated magnetic circuit has a resistance 
 to direct current of 0.5 ohm. When 110 volts at 60 cycles is 
 impressed on this reactor the apparent resistance and reactance 
 are 2 . ohms and 4.0 ohms respectively. What is the actual in- 
 ductance of the reactor at this voltage? 
 
 21. At no load with 220 volts at 60 cycles impressed on a 
 transformer the ratio of eddy current loss to hysteresis loss is 
 1:3. What is this ratio if 220 volts at 50 cycles is impressed on 
 the transformer? 
 
 22. At no load with 220 volts at 60 cycles impressed on a 
 transformer the ratio of eddy current loss to hysteresis loss is 1:3. 
 What is this ratio if 110 volts at 60 cycles is impressed on the 
 transformer? 
 
 23. At no load with 110 volts at 60 cycles impressed on a trans- 
 former the current is 1.3 amperes and the power is 52 watts. 
 What will be the no-load current and power taken by a trans- 
 former similar to this in every particular, except that the cross- 
 section of the magnetic circuit is double, when 220 volts at 60 
 cycles is impressed on it? 
 
 24. At no load with 110 volts at 30 cycles impressed on a trans- 
 former the current is 1.2 amperes and the power is 53 watts, of 
 .which 14 watts is dissipated in eddy currents. What will be the 
 
 current and power at no load if 220 volts at 60 cycles is impressed 
 on this transformer? 
 
 25. The high-tension winding of a transformer consists of two 
 coils which may be connected in series or in parallel. When 
 these coils are connected in series across 2200 volts at 60 cycles 
 the current is 0.30 amperes and the power is 140 watts on open 
 circuit. What is the no-load current and power if the coils are 
 connected in parallel across 1100 volts at 60 cycles? 
 
 26. 1 In problem 25 what is the no-load current and power when 
 the coils are connected in parallel across 1550 volts at 60 cycles? 
 
 1 Assume that the original flux density is 70 kilolines per sq. in., and that 
 the relation between the flux density and the permeability for this quality 
 of iron is expressed by: 
 
 5=102-0.017^ 
 
 between B equals 45 kilolines and B equals 95 kilolines per sq. in. In prob- 
 lems where this applies the effect of the joints in the magnetic circuit will 
 be neglected. 
 
6 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 Assume that the core losses vary as the 1.7 power of the flux 
 density. 
 
 27. The high-tension winding of a transformer consists of 
 two coils which may be connected in series or in parallel. When 
 these coils are connected in parallel across 22,000 volts at 60 
 cycles the no-load current is 0.077 amperes and the power is 371 
 watts, 28 per cent, of which is eddy current loss. What are the 
 no-load current and power when the coils are connected in series 
 across 22,000 volts at 30 cycles? 
 
 28. At no load, with 460 volts impressed on the primary of a 
 transformer, the current is 0.94 ampere and the power is 122 
 watts. Another transformer is similar to this one except that 
 the primary winding has twice as many turns and the cross-sec- 
 tion of the magnetic circuit is one-half as great. What no-load 
 current and power will this transformer take with 460 volts im- 
 pressed on its primary winding? 
 
 29. In a transformer having a ratio of transformation of NI: N 2 
 the windings are so designed that the current densities in the 
 primary and secondary are the same. What is the ratio of pri- 
 mary to secondary resistance if both windings have the same 
 mean length per turn? What is this ratio if the mean length 
 per turn of the primary is 15 per cent, greater than that of the 
 secondary? (b) If the current density in the primary had been 
 15 per cent, greater than in the secondary what would have 
 been this ratio of the resistances if the windings had the same 
 mean length per turn? 
 
 30. In a transformer having a ratio of transformation of NI : N 2 
 the windings are so designed that the heating losses in the pri- 
 mary and secondary are the same. What is the ratio of primary 
 to secondary resistance if both windings have the same mean 
 length per turn? What is this ratio if the mean length per turn 
 of the primary winding is 15 per cent, greater than that of the 
 secondary? (b) If the heating loss in the primary had been 15 
 per cent, greater than in the secondary what would have been 
 this ratio if both windings had the same mean length per turn? 
 
 31. What effect will be produced on the iron losses of a trans- 
 former and on the kilowatt rating, on the basis of the same total 
 losses, by doubling the number of turns in the primary and secon- 
 dary windings and halving the cross-section of the wires: first, 
 when the impressed voltage is doubled; and second, when the 
 impressed voltage is unchanged and the cross-section of the mag- 
 
TRANSFORMERS 7 
 
 netic circuit is halved? Assume that originally the iron and 
 copper losses are equal. 
 
 32. In a given transformer the ratio of the iron losses to the 
 copper losses at full load is 0.7. Another transformer with the 
 same rating has a magnetic circuit of the same length but of 25 
 per cent, greater cross-section, due to which the mean length of 
 one turn of the winding is 10 per cent, greater. Both transformers 
 are wound with the same size wire. The second transformer 
 is designed so as to have the same maximum flux density as the 
 first. What is the ratio of the total losses at full load, and what 
 is the ratio of the iron losses to the copper losses in the second 
 transformer at full load? Compare the amounts of copper in 
 the two transformers. Both transformers are designed for the 
 same voltage. 
 
 33. Two transformers have identical magnetic cores and the 
 same rated output. Both are wound for the same voltage but in 
 the first the iron losses are 25 per cent, greater than in the second 
 due to a higher flux density. In the first transformer the ratio of 
 the iron losses to the copper losses at full load is 0.7. Both 
 transformers have the same size of wire in their windings, but the 
 mean length of one turn is 5 per cent, greater in the second than in 
 the first. What is the ratio of the iron losses to the copper losses 
 at full load in the second transformer? What is the ratio of the 
 total losses at full load in the two transformers? Compare the 
 amounts of copper in the two transformers. Assume that the iron 
 losses vary as the 1.7 power of the flux density. 
 
 34. Two transformers have identical magnetic cores and the 
 same rated output. Both are wound for the same voltage but the 
 first has 15 per cent, more turns in all of its coils than the second, 
 due to which the mean length of one turn is 5 per cent, greater in 
 the first. The windings of each are of the same size wire. Assume 
 that the iron losses vary as the 1.7 power of the flux density. 
 The ratio of the iron losses to the copper losses at full load is 0.7 
 in the first transformer. What is the ratio of the iron losses to 
 the copper losses at full load in the second transformer? What is 
 the ratio of the total losses at full load in the two transformers? 
 
 35. In a given transformer the ratio of the iron losses to the 
 copper losses at full load is 0.75. Another transformer with the 
 same radiating capacity has a magnetic core of the same length 
 but with a cross-section 15 per cent, larger. The winding of the 
 second transformer is designed so that the iron losses are one- 
 
8 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 third greater than the iron losses of the first. The same size of 
 wire is used in winding both transformers but the mean length 
 of one turn is 5 per cent, greater in the second on account of the 
 larger magnetic core. Assume that the iron losses vary as the 
 1.7 power of the flux density. Both transformers have the same 
 voltage rating. The first transformer has a full-load capacity of 
 100 kw. and a full-load efficiency of 97.9 per cent, at unit power 
 factor. On the basis of the same total losses what should be the 
 full-load rating of the second transformer? What is the ratio 
 of the iron losses to the copper losses at full load in the second 
 transformer? 
 
 36. Three equal 10 to 1 transformers, each of 100 kv.-a. 
 capacity, are arranged with both the primaries and secondaries 
 in Y to receive power from a 3-phase, 2200-volt circuit. The 
 losses in each transformer at full load are 1960 watts of which 940 
 are core losses. If these transformers are connected in delta to 
 this circuit, what power will they deliver without exceeding their 
 full-load heating losses? Assume that the iron losses vary as 1.7 
 power of the flux density. 
 
 37. The name plate of a transformer gives the following 
 data: 100 kv.-a., 6600 : 220 volts, 25 cycles. The percentage 
 distribution of the losses at full load is: copper loss, 1.0 per cent. ; 
 eddy current loss, 0.3 per cent.; hysteresis loss 0.8 per cent. If 
 the insulation of the transformer will safely stand double volt- 
 age what should be its rating on the basis of the same total 
 heating losses when taking power from an 11,000-volt, 50-cycle 
 circuit? 
 
 38. The name plate of a transformer gives the following data: 
 500 kv.-a., 13,200 : 400 volts, 50 cycles. At full load the copper 
 loss is 4600 watts and the iron loss is 2800 watts. Of the latter 22 
 per cent, is due to eddy currents. If this transformer is to receive 
 power from an 11,000-volt, 60-cycle circuit what should be its 
 full-load rating on the basis of the same total heating losses? 
 
 39. A 10-kw. lighting transformer takes 122 watts on open 
 circuit and 178 watts on short circuit with full-load current in the 
 windings. This transformer is connected to the supply circuit 
 continuously. It delivers its rated load, however, during but 6 
 hours each day and is idle the remaining 18 hours. What is the 
 all-day efficiency? 
 
 40. A lOO-kv.-a. transformer supplies a lighting and power 
 load. The iron loss is 1010 watts and the copper loss is 1004 watts 
 
TRANSFORMERS 9 
 
 with full-load current. This transformer is connected to the 
 supply circuit continuously. During 7 hours the load is 60 kw- 
 at 0.75 power factor, and for the remaining 3 hours of the work- 
 ing day the load is 80 kw. at 0.83 power factor. The rest of the 
 time no load is supplied. What is the all-day efficiency? 
 
 41. A oOO-kv.-a. power transformer is connected to the supply 
 circuit for 12 hours each day. The iron loss is 3330 watts and 
 the copper loss is 4680 watts with full-load current. During 5 
 hours it delivers 400 kw. at 0.80 power factor and during 4 hours, 
 250 kw. at 0.75 power factor. The remaining hours it is idle. 
 What is the all-day efficiency? 
 
 42. l The magnetic core of a transformer has a mean length of 
 77.3 in. and a cross-section of 34.4 sq. in. How many turns 
 should there be in the high-tension winding if the applied voltage 
 is 22,500 volts at 60 cycles? The maximum flux density to be 
 used is 63,000 lines per square inch, at which the permeability of 
 the iron is 1600. The core loss per cubic inch of iron at this flux 
 density and frequency is 0.364 watts. What are the no-load 
 high-tension current and power factor? 
 
 43. l The magnetic core of a 25-kv.-a. transformer has a cross- 
 section of 17.5 sq. in. and an average length of 57.5 in. The 
 high-tension winding is designed for an impressed voltage of 
 22,000 volts at 60 cycles and for a maximum flux density in the 
 magnetic circuit of 70,000 lines per square inch. At this flux 
 density the permeability is 1860 and the iron loss is 0.37 watts 
 per cubic inch. How many turns should there be in the high- 
 tension winding? What are the no-load current and power 
 factor? 
 
 44. l A small experimental transformer has a magnetic circuit 
 with a net cross-section of 3.4 sq. in. and a mean length of 22.5 
 in. Assume that the permeability of the iron is 1800. There are 
 four coils, each of which has 232 turns, and a resistance of 0.23 
 ohm. (a) If all of the coils are connected for a 2 :1 ratio of trans- 
 formation what is the magnetizing component of the no-load 
 current when 220 volts at 60 cycles is impressed on the primary? 
 (b) If the eddy current and hysteresis losses are 50 watts what 
 is the no-load current? (c) What is the efficiency of this trans- 
 former when the secondary delivers 20 amperes at 110 volts and 
 100 per cent, power factor? 
 
 1 Assume that at this flux density 75 additional ampere turns are required 
 because of the joints in the magnetic circuit. 
 
10 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 TRANSFORMER DATA. 
 
 No. 
 
 Kv.-a. 
 
 Type 
 
 Cycles 
 
 Voltage 
 
 Turns 
 
 Magnetic circuit 
 
 H.T. 
 
 JL.T. 
 
 H.T. 
 
 L.T. 
 
 Area 
 sq. in. 
 
 Av. length 
 in. 
 
 A 
 B 
 C 
 D 
 
 E 
 F 
 
 50 
 100 
 500 
 500 
 1,000 
 3,333 
 
 Core 
 Shell 
 Core 
 Shell 
 Shell 
 Shell 
 
 60 
 60 
 60 
 25 
 60 
 60 
 
 22,500 
 11,000 
 11,000 
 13,200 
 32,400 
 55,000 
 
 2,300 
 2,200 
 2,300 
 425 
 11,500 
 6,600 
 
 900 
 
 180 
 
 34.4 
 66 
 
 88.25 
 276 
 372 
 348 
 
 77.31 
 47 
 131 
 68 
 86.8 
 92.8 
 
 558 
 
 18 
 
 733 
 
 88 
 
 45. 1 (a) How many turns should there be in the high- and 
 low-tension windings in order that the maximum flux density 
 shall be 69,000 lines per square inch in transformer No. A? What 
 is the core loss? What is the ratio of the no-load current to the 
 full-load current? Use the magnetic data given in problem 10. 
 (b) How many turns should there be in the high- and low-tension 
 windings in order that the maximum flux density shall be 69,000 
 lines per square inch in transformer No. C? What is the core loss? 
 What is the ratio of the no-load current to the full-load current? 
 Use the magnetic data given in problem 10. (c) How many 
 turns should there be in the high- and low-tension windings in 
 order that the maximum flux density shall be 69,000 lines per 
 square inch in transformer No. E? What is the core loss? What 
 is the ratio of the no-load current to the full-load current? Use 
 the magnetic data given in problem 10. 
 
 46. 1 (a) What are the no-load current and core loss in trans- 
 former No. B? Use the magnetic data given in problem 10. (b) 
 What are the no-load current and core loss in transformer No. D? 
 Use the magnetic data given in problem 10. (c) What are the 
 no-load current and core loss in transformer No. F.? Use the 
 magnetic data given in problem 10. 
 
 TRANSFORMER DATA 
 
 No. 
 
 Kv.-a. 
 
 Type 
 
 Cycles 
 
 Voltage 
 
 Turns 
 
 Magnetic circuit | Open circuit 
 
 H.T. 
 
 L.T. 
 
 H.T. 
 
 L.T. Area 
 
 j sq. in. 
 
 Av. 
 
 length 
 in. 
 
 Watts 
 
 ^Volt- 
 amps. 
 
 G 
 H 
 I 
 
 500 
 1,000 
 5,000 
 
 Shell 
 Shell 
 Shell 
 
 60 
 60 
 50 
 
 12,700 
 66,000 
 52,000 
 
 2,300 
 6,600 
 33,000 
 
 376 
 1,240 
 529 
 
 68 
 124 
 336 
 
 170 
 278 
 613 
 
 66.4 
 95 
 140 
 
 3,330 
 9,300 
 17,500 
 
 25,800 
 60,000 
 146,000 
 
 47. l (a) What are the maximum flux density, the permeability, 
 and the iron loss in watts per cubic inch for transformer No. G? 
 (b) What are the maximum flux density, the permeability, and 
 
 1 Assume that at this flux density 75 additional ampere turns are required 
 because of the joints in the magnetic circuit. 
 
TRANSFORMERS 11 
 
 the iron loss in watts per cubic inch for transformer No. H? 
 (c) What are the maximum flux density, the permeability, and 
 the iron loss in watts per cubic inch for transformer No. I? 
 
 48. A 100-kv.-a., 11,000 :2200-volt, 60-cycle transformer has 
 primary and secondary resistances of 6.0 ohms and 0.24 ohm 
 respectively, and primary and secondary leakage reactances of 
 16 ohms and 0.64 ohm respectively. What is the maximum per- 
 centage change that can occur in the mutual flux from no-load to 
 full-load current? At what power factor would this occur? 
 Neglect the no-load current. 
 
 49. A 500-kv.-a., 11,000 : 2300-volt, 60-cycle transformer has 
 primary and secondary resistances of 0.81 ohm and 0.0358 ohm 
 respectively, and primary and secondary leakage reactances of 
 3.7 ohm and 0.162 ohm respectively. With the low-tension wind- 
 ing short circuited what voltage should be impressed on the high- 
 tension winding so that the current will have its full-load value? 
 Neglect the no-load current. 
 
 50. A 7.5-kv.-a., 2080 : 208-volt, 60-cycle transformer has the 
 following constants: 
 
 ri = 7.53 ohms #1 = 14. 2 ohms 
 
 r 2 = 0.0662 ohm z 2 = 0.128 ohm 
 
 The core loss with 208 volts impressed on the secondary is 200 
 watts. Neglect the exciting current, (a) What is the regula- 
 tion of this transformer for full-load current at 0.8 power factor? 
 (b) What is the efficiency of the transformer for this load? 
 
 51. A 50-kv.-a., 30,000 : 440-volt, 50-cycle transformer has the 
 following constants: 
 
 ri = 120 ohms #1 = 428 ohms 
 
 r 2 = 0.0258 ohm x 2 = 0.092 ohm 
 
 The iron loss at the rated voltage is 641 watts. Neglect the 
 exciting current, (a) What is the regulation of this trans- 
 former for a load of 40 kw., at 0.75 power factor? (b) What is 
 the efficiency of the transformer for this load? 
 
 52. A 500-kv.-a., 13,200 : 425-volt, 25-cycle transformer has 
 primary and secondary resistances of 1.6 ohms and 0.00166 ohm 
 respectively, and primary and secondary leakage reactances of 
 14.5 ohms and 0.0151 ohm respectively. At no load with 425 
 volts impressed on the low-tension winding the power is 2.5 
 kw. and the kilovolt-amperes are 16.0. Neglect any change in 
 
12 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 the exciting current with change in the load. Calculate (a) the 
 regulation and (b) the efficiency of this transformer when 400 
 kw. is supplied to the high-tension winding at 13,200 volts and 
 0.75 power factor. Use the complete transformer diagram in the 
 solution of this problem, (c) Calculate the regulation and effi- 
 ciency of this transformer for the same load on the assumption 
 that the no-load current is negligible and that the core losses are 
 constant. 
 
 53. The following data are given concerning a 1000-kv.-a., 
 66,000 : 6600-volt, 60-cycle trafisf ormer : 
 
 
 No-load 
 
 Short-circuit 
 
 Current 
 Voltage 
 Power 
 
 9.1 
 6600 
 9300 
 
 15. 15 1 
 3240 
 7490 
 
 Assume that the primary and secondary resistances are equal 
 when reduced to the same side and that the primary and secon- 
 dary leakage reactances are similarly equal. Assume that the 
 magnetizing current varies as the generated voltage and that the 
 core loss varies as the square of this voltage. 
 
 Calculate (a) the regulation and (b) the efficiency when the 
 secondary delivers full-load current at 0.80 power factor and a 
 terminal voltage of 6680 volts, (c) Calculate the regulation and 
 efficiency of this transformer for the same load on the assumption 
 that the no-load current is negligible and that the core losses are 
 constant. 
 
 TRANSFORMER DATA 
 
 No. 
 
 Kv.-a. 
 
 Type 
 
 Cycles 
 
 Voltage 
 
 Open circuit 
 
 Short circuit 
 
 H.T. 
 
 L.T. 
 
 Watts 
 
 Volt- 
 amperes 
 
 Volts 
 
 Amperes 1 ! Watts 
 
 J 
 
 25 
 
 Core 
 
 60 
 
 22,000 
 
 440 
 
 371 
 
 1,700 
 
 1,020 
 
 1.136 
 
 351 
 
 K 
 
 50 
 
 Core 
 
 50 
 
 30,000 
 
 440 
 
 641 
 
 3,000 
 
 1,360 
 
 1.665 
 
 665 
 
 B 
 
 100 
 
 Shell 
 
 60 
 
 11,000 
 
 2,200 
 
 940 
 
 3,750 
 
 310 
 
 9.1 
 
 1,000 
 
 C 
 
 500 
 
 Core 
 
 60 
 
 11,000 
 
 2,300 
 
 2,960 
 
 7,710 
 
 345 
 
 45.5 
 
 3,375 
 
 D 
 
 500 
 
 Shell 
 
 25 
 
 13,200 
 
 425 
 
 2,500 
 
 16,000 
 
 1,100 
 
 37.9 
 
 4,600 
 
 H 
 
 1,000 
 
 Shell 
 
 60 
 
 66,000 
 
 6,600 
 
 9,300 
 
 60,000 
 
 3,240 
 
 15.15 
 
 7,490 
 
 F 
 
 3,333 
 
 Shell 
 
 60 
 
 55,000 
 
 6,600 
 
 13,940 
 
 161,000 
 
 2,350 
 
 60.6 
 
 18,850 
 
 I 
 
 5,000 
 
 Shell 
 
 50 
 
 52,000 
 
 33,000 
 
 17,500 
 
 146,000 
 
 1,800 
 
 151.5 
 
 27,000 
 
 54. Calculate (a) the regulation 2 and (b) the efficiency of the 
 transformer No. J for a load of 22 kw. at 0.85 power factor and 
 rated voltage. 
 
 1 Full-load current. 
 
 2 Neglect the exciting current. 
 
TRANSFORMERS 13 
 
 65. Calculate (a) the regulation 1 and (b) the efficiency of the 
 transformer No. K for a load of 40 kw at 0.78 power factor 
 and rated voltage. 
 
 56. Calculate (a) the regulation 1 and (b) the efficiency of the 
 transformer No. B for a load of 100 kw. at 0.90 power factor and 
 rated voltage. 
 
 57. Calculate (a) the regulation 1 and (b) the efficiency of the 
 transformer No. C for a load of 460 kw. at 0.91 power factor 
 and rated voltage. 
 
 58. Calculate (a) the regulation 1 and (b) the efficiency of the 
 transformer No. D for a load of 500 kw. at 0.93 power factor 
 and rated voltage. 
 
 59. Calculate (a) the regulation 1 and (b) the efficiency of the 
 transformer No. H for a load of 1100 kw. at unit power factor and 
 rated voltage. 
 
 60. Calculate (a) the regulation 1 and (b) the efficiency of the 
 transformer No. F for a load of 3500 kw. at unit power factor 
 and rated voltage. 
 
 61. Calculate (a) the regulation 1 and (b) the efficiency of the 
 transformer No. I for a load of 5100 kw. at 0.92 
 
 power factor and rated voltage. 
 
 62. The resistance of the high-tension wind- 
 ing (a-c) of a 2:1 auto transformer is 0.072 ohms 
 and that of the low-tension winding (b-c) is 0.035 
 ohms. The leakage reactance of the low-tension 
 winding is 0.106 ohm and that of the remaining 
 portion of the winding (a-b) is 0.108 ohm. If FlG 
 an electromotive force of 12 volts at the rated 
 
 frequency is impressed on the high-tension winding and the low- 
 tension winding is short-circuited what will be the current and 
 power taken from the line? 
 
 63. The resistance of the high-tension winding (a-c) see 
 Fig. 1 of a 25-kv.-a., 550 :110-volt, auto-transformer is 0.0682 
 ohm and that of the low-tension winding (b-c) is 0.0042 ohm. 
 The leakage reactance of the winding (a-b) is 0.214 ohm and that 
 of the low-tension winding is 0.0132 ohm. If the winding (a-b) 
 is short-circuited what percentage of the rated voltage should 
 be applied to the low-tension winding (b-c) in order that full- 
 load current should exist in the windings? What power would be 
 supplied? 
 
 1 Neglect the exciting current. 
 
14 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 64. The resistance of the high-tension winding (a-c) see Fig. 
 1 of a 50-kv.-a., 440 : 110-volt auto-transformer is 0.02156 ohm 
 and that of low-tension winding (6-c) is 0.00216 ohm. The 
 leakage reactance of the winding (a-b) is 0.068 and that of the 
 low-tension winding is 0.00756 ohm. If the high-tension winding 
 is short-circuited, what voltage should be impressed on the low- 
 tension winding in order that there will be full-load current in 
 the windings? What power would be supplied? 
 
 65. Calculate the regulation of the auto-transformer described 
 in problem 63 for a load of 22 kw. at 0.80 power factor on the 
 low- tension side. If the core loss at the rated voltage is 288 
 watts what is the efficiency of the transformer at this load? 
 
 66. Calculate the regulation of the auto-transformer described 
 in problem 64 for a load of 46 kw. at 0.87 power factor on the 
 low-tension side. If the core loss is 512 watts at the rated 
 voltage what is the efficiency of the transformer at this load? 
 
 AUTO-TRANSFORMER DATA 
 
 
 Kv.-a. 
 
 Voltage 
 
 Short circuit 
 
 Core loss at 
 rated veltage 
 
 Watts 
 
 Current* 
 
 Voltsi 
 
 A 1 10 
 
 155:110 
 
 168 
 
 64.5 
 
 7.2 
 
 122 
 
 B 
 
 20 
 
 440-220 
 
 320 
 
 45.5 
 
 20.2 
 
 264 
 
 C 
 
 25 
 
 660:220 
 
 340 
 
 37.9 
 
 29.0 
 
 371 
 
 D 
 
 50 
 
 550:110 
 
 620 
 
 91.0 25.0 
 
 664 
 
 E 
 
 100 
 
 6,600:440 
 
 940 
 
 15.15 275.0 
 
 980 
 
 F 
 
 500 
 
 11,000:330 
 
 3,850 
 
 45.5 460.0 
 
 2,980 
 
 67. Calculate the regulation of the auto-transformer No. A for 
 a load of 10 kw. at unit power factor on the low-tension side. 
 What is the efficiency of the transformer at this load? 
 
 68. Calculate the regulation of the auto -transformer No. B 
 for a load of 16 kw. at 0.78 power factor on the high-tension side. 
 What is the efficiency of the transformer at this load? 
 
 69. Calculate the regulation of the auto-transformer No. C 
 for a load of 22 kw. at 0.80 power factor on the high-tension 
 side. What is the efficiency of the transformer at this load? 
 
 70. Calculate the regulation of the auto-transformer No. D. 
 for a load of 55 kw. at unit power factor on the low-tension side. 
 What is the efficiency of the transformer at this load? 
 
 71. Calculate the regulation of the auto-transformer No. E 
 for a load of 93 kw. at 0.87 power factor on the low-tension side. 
 What is the efficiency of the transformer at this load? 
 
 1 The low-tension winding is short-circuited and voltage is applied to the 
 high-tension winding so that there is full-load current in the windings. 
 
TRANSFORMERS 15 
 
 72. Calculate the regulation of the auto-transformer No. F for 
 a load of 510 kw. at 0.90 power factor on the low-tension side. 
 What is the efficiency of the transformer at this load? 
 
 73. If the high-tension winding of the auto-transformer No. E is 
 short-circuited what voltage should be applied to the low-tension 
 winding so that there will be full-load current in the windings? 
 
 74. If the high-tension winding of the auto-transformer No. F 
 is short-circuited what voltage should be applied to the low-ten- 
 sion winding so that there will be full-load current in the windings? 
 
 75. An induction regulator is connected in a 2300-volt cir- 
 cuit as shown in Fig. 2. With no load on the line and with 2300 
 volts impressed on the circuit as shown in the ft 
 
 figure the voltage measured from a to c is 
 
 2550 volts. If the winding b-c is short-cir- 230QV 
 
 cuited and a reduced voltage of 210 volts is 
 
 impressed on the winding a-b, the line current - a 
 
 in (o-a) is 130 amperes and the power supplied F IG . 2. 
 
 is 2.2 kw. (a) When the station supplies 250 
 
 kw. at 0.78 power factor and a voltage of 2300 volts at a-b what is 
 
 the line voltage at a-cf (b) This regulator is adjusted to reduce 
 
 the voltage so that with no load on the line and with 2300 volts 
 
 impressed on the circuit (at a-b) the voltage at a-c is 2050. 
 
 The short-circuit data are unchanged. What is the voltage at a-c 
 
 when the station supplies 300 kw. at a leading power factor of 
 
 0.92 and a line voltage of 2300 volts at a-b? 
 
 76. An induction regulator is connected in a 6600-volt circuit 
 as shown in Fig. 2. With no load on the line and with 6600 volts 
 impressed on the circuit (at a-b) the voltage measured from a to c 
 is 7370 volts. If the winding, b-c, is short-circuited, and a reduced 
 voltage of 600 volts is impressed on the winding, a-b, the line 
 current (in o-a) is 76 amperes and power supplied is 4.2 kw. 
 
 (a) When the station supplies 430 kw. at 0.80 power factor and 
 a voltage of 6600 volts at a-b what is the line voltage at a-cf 
 
 (b) This regulator is adjusted to reduce the voltage so that 
 with no load on the line and with 6600 volts impressed on the 
 circuit (at a-b) the voltage at a-c is 5850 volts. The short- 
 circuit data are unchanged. What is the voltage at a-c when the 
 station supplies a line current (in o-a) of 81 amperes at a leading 
 power factor of 0.83 and a line voltage of 6600 volts at a-b? 
 
 77. A small experimental transformer has four equal coils 
 which may be inter-connected in different ways. In the first case 
 
16 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 all of the coils are connected to form a regular transformer with a 
 ratio of transformation of 2:1, and in the second case they are 
 connected to form an auto-transformer with the same ratio of 
 transformation. When connected as a regular transformer with 
 the low-tension winding short circuited the current, voltage and 
 power, measured on the high-tension side are 8 amperes, 52 volts, 
 and 59 watts respectively. On open circuit with 220 volts 
 impressed on the high-tension side the core loss, is 26 watts. 
 Assume that the paths of the leakage flux are the same for each 
 connection. Compare (a) the regulations and (b) the effi- 
 ciencies of this transformer for the two cases given above for a 
 load of 1.5 kw. at 0.9 power factor and 110 volts on the low-tension 
 side. 
 
 78. Two transformers, one connected as a regular and the 
 other as an auto-transformer, have identical magnetic cores and 
 the same amount of copper in their windings. Each of the trans- 
 formers gives a ratio of transformation of 330:220 volts, and the 
 windings are so designed that the core losses in each are 122 watts 
 when 330 volts is impressed on their primaries. The primary and 
 secondary resistances of the regular transformer are 0.097 ohm 
 and 0.0431 ohm respectively. The resistance of the auto- 
 transformer measured on the low-tension side is also 0.0431 ohm. 
 The leakage reactance of the primary winding of the regular 
 transformer is 0.37 ohm. Assume that all of the windings have 
 the same mean length per turn, and that the leakage reactance of 
 the coils vary as the square of the number of turns, (a) On the 
 basis of the same total heating losses, what is the ratio of the out- 
 puts of these two transformers? (b) What is the regulation and 
 efficiency of each transformer for a load of 10 kw. at 0.87 power 
 factor and 220 volts? 
 
 79. Two transformers, one connected as a regular and the 
 other as an auto-transformer, have identical magnetic cores and 
 the same amount of copper in their windings. Each of these 
 transformers gives a ratio of transformation of 330:110 volts and 
 the windings are so designed that the core losses of each are 371 
 watts when 330 volts is impressed on their primaries. The pri- 
 mary and secondary resistances of the regular transformer are 
 0.0306 ohm and 0.0034 ohm respectively. The resistances of the 
 auto-transformer measured on the high- and low-tension sides are 
 0.017 and 0.0034 ohm respectively. The equivalent leakage 
 reactance of the regular transformer referred to the high-tension 
 
TRANSFORMERS 17 
 
 side is 0.202 ohm. Assume that the leakage reactances of the 
 windings vary as the square of the number of turns in them. 
 
 (a) If the high-tension winding of the regular transformer has 
 186 turns how many turns are there in each of the other windings 
 of the regular and auto-transformer? 
 
 (b) The regular transformer is rated to deliver 25 kw. at 110 
 volts. If the losses at full load are the same for both transformers 
 what is the kilovolt-ampere rating of the auto-transformer? 
 
 (c) What is the regulation and efficiency of each transformer 
 when it is delivering its rated full-load current at 0.83 power fac- 
 tor and its rated voltage? 
 
 80. Two transformers, one connected as a regular, and the 
 other as an auto-transformer, have identical magnetic cores and 
 the same amount of copper in their windings. Each of these 
 transformers gives a ratio of transformation of 550:220 volts and 
 the windings are so designed that the volts per turn are the same 
 for each. With 25 volts impressed on the high-tension winding 
 of the regular transformer and with the low-tension winding 
 short circuited the current and power are 91 amperes and 665 
 watts. At the rated voltage the core loss in the regular trans- 
 former is 641 watts. Assume that the resistances and leakage 
 reactances of all of the windings vary as the square of the number 
 of turns in them. 
 
 (a) The regular transformer is rated to deliver 50 kv.-a. at 
 220 volts. If the losses at full load are the same for both 
 transformers, what is the kilovolt-ampere rating of the auto- 
 transformer? 
 
 (b) What are the regulation and efficiency of each transformer 
 when it is delivering its rated full-load power at 0.9 power factor? 
 
 81. Two transformers, one connected as a regular and the 
 other as an auto-transformer, have identical magnetic cores and 
 the same amount of copper in their windings. Each of these 
 transformers gives a ratio of transformation of 440:110 volts, 
 and the windings are so designed that the volts per turn are the 
 same for each. With 14 volts impressed on the high-tension 
 winding of the regular transformer and with the low-tension wind- 
 ing short circuited the current is 227 amperes and the power is 
 1004 watts. The core loss of this transformer at rated impressed 
 voltage is 1010 watts. Assume that the resistances and leakage 
 reactances of all of the windings vary as the square of the number 
 of turns in them. 
 
18 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 (a) The regular transformer is rated to deliver 100 kw.-a. at 
 110 volts. If the losses at full load are the same for both 
 transformers, what is the kilovolt-ampere rating of the auto- 
 transformer? 
 
 (b) What voltage should be impressed on the high-tension 
 winding of the auto-transformer so that there will be full-load 
 current in the short-circuited low-tension winding? 
 
 (c) What are the regulation and efficiency of each transformer 
 for full-load current at 0.8 power factor? 
 
 82. A lO-kv.-a. and a 25-kv.-a. transformer, each of which has 
 a ratio of transformation of 5:1, have their primaries connected in 
 parallel across an 1100-volt circuit. Their secondaries are 
 also connected in parallel and supply 152 amperes at 0.84 power 
 factor to an induction motor. Referred to the secondary sides 
 the equivalent resistances are 0.0865 ohm and 0.0272 ohm 
 respectively and the equivalent reactances are 0.143 ohm and 
 0.0856 ohm respectively. 
 
 What current does each transformer take from the circuit? 
 Compare these currents with their rated full-load values. 
 
 83. The following short-circuit data are given on two 100- 
 kv.-a., ll,000:460-volt transformers: 
 
 Type Amperes 1 Voltage Watts 
 
 Core 
 Shell 
 
 9.1 
 9.1 
 
 265 
 310 
 
 1004 
 1000 
 
 These transformers are connected in parallel on both the high- 
 and low-tension sides and supply a load of 186 kw. at 0.93 power 
 factor. What percentage of its full-load current does each trans- 
 former supply? 
 
 84. The following short-circuit data are given on two trans- 
 formers which have a ratio of transformation of 11,000:2300 
 volts: 
 
 Kv.-a. 
 
 Type 
 
 Amperes 1 
 
 Volts 
 
 Watts 
 
 100 
 500 
 
 Core 
 Core 
 
 9.1 
 45.5 
 
 265 
 345 
 
 1004 
 i 3375 
 
 These transformers are connected in parallel on both the high- 
 and low-tension sides and supply a total current of 293 amperes 
 at 0.92 power factor on the low-tension side. What is the current 
 
 1 Full-load current. 
 
TRANSFORMERS 
 
 19 
 
 in each transformer? Compare the copper losses with their full- 
 load values. 
 
 85. -The following short-circuit data are given on two 500- 
 kv.-a., ll,000:2300-volt transformers: 
 
 Type 
 
 Amperes x 
 
 Voltage 
 
 Watts 
 
 Core 
 Shell 
 
 45.5 
 45.5 
 
 345 
 345 
 
 3375 
 4680 
 
 These transformers are connected in parallel on both the high- 
 and low-tension sides and supply a total current of 452 amperes at 
 0.95 power factor on the low-tension side. What current does 
 each transformer supply? 
 
 86. The following short-circuit data are given on three 
 66,000 :6600-volt transformers: 
 
 No. 
 
 Kv.-a. 
 
 Type 
 
 Amperes 1 
 
 Volts | Watts 
 
 A 
 B 
 C 
 
 1000 
 3333 
 5000 
 
 Shell 
 Shell 
 Shell 
 
 15.15 
 50.5 
 
 75.8 
 
 3240 
 
 2820 
 3600 
 
 7,490 
 18,850 
 27,000 
 
 These transformers are connected in parallel on both their 
 high- and low-tension sides, and supply a load of 9200 kw. at 
 a power factor of 0.93 and their rated voltage. Compare the 
 division of the total current between the transformers with their 
 ratings. 
 
 87. The following short-circuit data are given on three 
 ll,000:460-volt transformers: 
 
 Kv.-a. 
 
 Type 
 
 Amperes l 
 
 Volts 
 
 Watts 
 
 100 
 500 
 500 
 
 Core 
 Shell 
 Core 
 
 - 9.1 
 45.5 
 45.5 
 
 265 
 917 
 345 
 
 1004 
 4600 
 3375 
 
 These transformers are connected in parallel on both the high- 
 and low-tension sides, and supply a load of 754 kw. at 0.88 power 
 factor and their rated voltages. What current does each trans- 
 former supply? Compare the copper loss in each transformer 
 with the copper loss at full load. 
 
 88. Consider the transformers, Nos. B and C, described in 
 problem 86. The high-tension windings receive power from the 
 same 66,000-volt circuit, while the low-tension windings deliver 
 power to independent circuits. Each transformer delivers its 
 
 1 Full-load current. 
 
20 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 rated full-load current, the first at unit power factor and the sec- 
 ond at 0.88 power factor. If the low-tension circuits are now 
 connected in parallel what is the current in each transformer? 
 Compare these currents with their full-load values. 
 
 Assume that the currents taken by the low-tension circuits and 
 the power factors at which they operate remain unchanged. 
 
 89. Two 50-kw. transformers are connected in parallel on 
 both the high- and low-tension sides. Their constants are given 
 in the following table: 
 
 Open-circuit voltage 
 
 High tension 
 
 Low tension 
 
 High tension 
 
 Low tension 
 
 1*1 
 
 xi r 2 
 
 X 2 
 
 22,500 
 22,400 
 
 2310 
 2320 
 
 61.6 
 61.6 
 
 110. 0.661 
 110. 0.661 
 
 1.16 
 1.16 
 
 They are alike except for the difference in their ratios of trans- 
 formation. These transformers supply a combined load of 93 
 kw. at 0.89 power factor on the low-tension side at a terminal 
 voltage of 2300 volts. What is the current in each transformer? 
 
 90. The following data are given on two transformers which 
 are operating with both their high- and low-tension windings 
 in parallel: 
 
 Kv.-a. 
 
 Type 
 
 Rated open-circuit voltage 
 
 Short circuit 
 
 High tension 
 
 Low tension 
 
 Amperes 1 
 
 Volts 
 
 Watts 
 
 100 
 500 
 
 Core 
 Shell 
 
 11,000 
 13,200 
 
 440 
 520 
 
 9.1 
 37.9 
 
 265 
 1,100 
 
 1,004 
 4,600 
 
 The ratios of transformation are slightly different. These trans- 
 formers take a combined power of 460 kw. at 0.90 power factor 
 from an 11,000-volt circuit. 
 
 (a) What is the current in each transformer? 
 
 (b) If the ratios of transformation are made equal by the 
 removal of a few turns from the high-tension winding of the 500- 
 kw. transformer, what is the current taken by each transformer 
 for this load? Assume that the short circuit data would be 
 unchanged. 
 
 91. The following data are given on two transformers which 
 are operating in parallel on both the high- and low-tension sides : 
 
 Kv.-a. 
 
 Type 
 
 Rated open-circuit voltage 
 
 Short circuit 
 
 High tension | Low tension 
 
 Amperes 1 | Volts 
 
 Watts 
 
 500 
 500 
 
 Core 
 Shell 
 
 13,200 
 12,700 
 
 2,340 
 2,170 
 
 37.9 
 39.4 
 
 413 
 332 
 
 3,375 
 4,680 
 
 1 Full-load current at rated voltages. 
 
TRANSFORMERS 21 
 
 The ratios of transformation are slightly different. These 
 transformers take a combined power of 960 kw. at 0.92 power 
 factor from a 12,800-volt circuit. 
 
 (a) What is the current in each transformer? 
 
 (b) If the ratios of transformation are made equal by removing 
 a few turns from the high-tension winding of the shell-type trans- 
 former what is the current taken by each transformer for this 
 load? Assume that the short-circuit data would be unchanged. 
 
 92. Consider the transformers described in problem 82. 
 They are operating in parallel on both the high- and low-tension 
 sides. What are the least values of resistance and reactance that 
 should be added to each transformer on the low-tension side in 
 order that the currents supplied by the transformers shall be in 
 phase and in proportion to their capacities? 
 
 (b) For a given load on the transformers compare the total 
 heating in the transformers and reactors with that in the trans- 
 formers alone before the addition of the reactors. 
 
 93. Consider the transformers described in problem 83. 
 They are operating in parallel on both the high- and low-tension 
 sides. 
 
 (a) Compare the sum of their rated outputs with the kilovolt- 
 ampere load they can deliver without overloading either of them. 
 
 (b) A reactor of negligible resistance is added on the low- 
 tension side of one transformer of such a value that the resultant 
 impedance volts of the transformer and reactor on short circuit 
 with full-load current is the same as that of the other transformer 
 alone when it also carries full-load current. What is the imped- 
 ance of this reactor? By what amount can the combined output 
 of the transformers be increased by the addition of this reactor 
 without overloading either transformer? 
 
 94. Consider the transformers described in problem 84. 
 They are operating in parallel on both the high- and low-tension 
 sides. 
 
 (a) Compare the sum of their rated outputs with the greatest 
 kilovolt-ampere load they can deliver without overloading either 
 of them. 
 
 (b) A reactor of negligible resistance is added on the low-ten- 
 sion side of one transformer of such a value that the resultant 
 impedance volts of the transformer and reactor on short circuit 
 with full-load current is the same as that of the other transformer 
 alone when it also carries full-load current. What is the imped- 
 
22 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 ance of this reactor? By what amount can the combined output 
 of the transformers be increased by the addition of this reactor 
 without overloading either transformer? 
 
 95. The following data are given on two transformers which 
 are operating in parallel on both their high- and low-tension 
 sides : 
 
 Kv.-a. 
 
 Type 
 
 Voltage 
 
 Short circuit 
 
 High tension Low tension 
 
 Volts 
 
 Amperes* | Watts 
 
 1,000 
 3,333 
 
 Shell 
 Shell 
 
 66,000 
 66,000 
 
 6,600 
 6,600 
 
 3,360 
 2,780 
 
 15.15 
 50.5 
 
 7,490 
 18,850 
 
 These transformers are delivering a combined load of 4333 
 kv.-a. at 6600 volts. A reactor of negligible resistance is added 
 in series with one of these transformers on the low-tension side 
 so that they divide this load in proportion to their ratings. As- 
 sume that this does not effect the low-tension voltage. 
 
 (a) What per cent, of its rated capacity is the load on each 
 transformer before the reactor is added? 
 
 (b) What is the ractance of this reactor? By what amount is 
 the total copper loss in the transformers reduced? 
 
 96. Two similar 5-kw. lighting transformers which give a 
 ratio of transformation of 1100:110 volts are connected in series 
 on both the high- and low-tension sides. With both low-tension 
 windings short circuited and with 326 volts impressed on the high- 
 tension windings connected in series the current is 4.55 amperes 
 and the power supplied is 101 watts. On the low-tension side 
 these transformers supply power to a three-wire system. The 
 resistances of the lamp loads on the two sides of the neutral are 
 2.8 ohms and 2.1 ohms. The high-tension line voltage is 2200 
 volts. What is the current in the neutral conductor? What is 
 the voltage across each lamp load? If each lamp takes approxi- 
 mately 50 watts how many should be turned off on one side in 
 order that all of the lamps shall burn with the same brilliancy? 
 
 97. The transformers described in problem 96 are connected 
 in series on both the high- and low-tension sides, and are deliver- 
 ing power to a three-wire circuit on each side of which there are 
 two equal lamp loads taking 32 amperes at 110 volts. 
 
 What is the high-tension line voltage? If one of these loads is 
 short circuited what will be the voltage across the other on the 
 assumption that its resistance and the high-tension line voltage 
 are both unchanged? 
 
 1 Full-load current. 
 
TRANSFORMERS 23 
 
 98. The transformers described in problem 96 are connected 
 in parallel on the high-tension and in series on the low-tension 
 sides and are delivering power to a three-wire lighting circuit. 
 The resistances of the lamp loads on the two sides of the system 
 are 2.8 and 2.1 ohms. What is the current in the neutral conduc- 
 tor? If the neutral conductor is disconnected from the trans- 
 formers what is the voltage across each lamp load? 
 
CHAPTER II 
 SYNCHRONOUS GENERATORS 
 
 1. With normal excitation the resultant air-gap flux in a 60- 
 cycle alternator is 10 6 lines per pole. The flux density is con- 
 stant under the pole face and is zero between the poles. 
 
 If the ratio of pole arc to pole pitch is 0.75 and the coil pitch 
 is 1.0, what is the generated armature voltage per turn? What 
 is the form factor of this e.m.f.? Sketch the graphs of the 
 flux density and the e.m.f. 
 
 2. In problem 1 if the ratio of pole arc to pole pitch is 0.75 and 
 the coil pitch 0.67, what is the generated armature voltage per 
 turn? What is the form factor of this e.m.f.? Sketch the 
 graphs of the flux density and the e.m.f. 
 
 3. In problem 1 if the ratio of pole arc to pole pitch is 0.5 and 
 the coil pitch is 1.0, what is the generated armature voltage per 
 turn? What, is the form factor of this e.m.f.? Sketch the 
 graphs of the flux density and the e.m.f. 
 
 4. With normal excitation the resultant air-gap in a 60-cycle 
 alternator is 10 6 lines per pole. The flux density is constant under 
 the pole face and decreases uniformly to zero at points midway 
 between the poles. 
 
 If the ratio of pole arc to pole pitch is 0.75 and the coil pitch 
 is 1.0, what is the generated armature voltage per turn? What 
 is the form factor of this e.m.f.? Sketch the graphs of the flux 
 density and the e.m.f. 
 
 5. In problem 4 if the ratio of pole arc to pole pitch is 0.75 
 and the coil pitch is 0.67, what is the generated armature vol- 
 tage per turn? What is the form factor of this e.m.f.? Sketch 
 the graphs of the flux density and the e.m.f. 
 
 6. In problem 4 if the ratio of pole arc to pole pitch is 0.5 and the 
 coil pitch is 1.0, what is the generated armature voltage per turn? 
 What is the form factor of this e.m.f.? Sketch the graphs of the 
 flux density and the e.m.f. 
 
 7. With normal excitation the resultant air-gap flux in a 60- 
 cycle alternator is 10 6 lines per pole. This flux is sinusoidally 
 distributed along the air-gap. 
 
 24 
 
S YNCHRONO US GENERA TORS 25 
 
 If the coil pitch is 1.0, what is the generated armature voltage 
 per turn? What is the form factor of this e.m.f.? 
 
 8. In problem 7 if the coil pitch is 0.8, what is the generated 
 armature voltage per turn? What is the form factor of this 
 e.m.f.? 
 
 9. In problem 7 if the coil pitch is 0.67, what is the generated 
 armature voltage per turn? What is the form factor of this 
 e.m.f.? 
 
 10. With normal excitation the resultant air-gap flux in a 
 60-cycle alternator is 10 6 lines per pole. The equation of the 
 curve which represents the flux density in the air-gap is: 
 
 B = Bi sin x+Ba sin 3x. 
 
 x is the electrical angle measured from the point midway between 
 the poles. Take B 3 = 0.3#i. 
 
 If the coil pitch is 1.0, what is the generated armature voltage 
 per turn? What is the form factor of this e.m.f.? 
 
 11. In problem 10 if the coil pitch is 0.8, what is the generated 
 armature voltage per turn? What is the form factor of this 
 e.m.f.? 
 
 12. In problem 10 if the coil pitch is 0.67 what is the generated 
 armature voltage per turn? What is the form factor of this 
 e.m.f.? 
 
 13. With normal excitation the resultant air-gap flux in a 
 60-cycle alternator is 10 6 lines per pole. The equation of the 
 curve which represents the flux density in the air gap is: 
 
 B = B l si 
 
 x is the electrical angle measured from the point midway between 
 the poles. Take B 3 = 0.35i. 
 
 If the coil pitch is 1.0, what is the generated armature voltage 
 per turn? What is the form factor of this e.m.f.? 
 
 14. With normal excitation the resultant air-gap flux in a 60- 
 cycle alternator is 10 6 lines per pole. The equation of the curve 
 which represents the flux density in the air-gap is: 
 
 B = Bi sin x-\-Bs sin (3# +TT) 
 
 x is the electrical angle measured from the point midway between 
 the poles. Take B 3 = 0.3#i. 
 
26 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 If the coil pitch is 1.0 what is the generated armature voltage 
 per turn? What is the form factor of this e.m.f.? 
 
 15. With normal" excitation the resultant air-gap flux in a 60- 
 cycle alternator is 10 6 lines per pole. The equation of the curve 
 which represents the flux density in the air-gap is: 
 
 B Bi sin x-\-Bs, sin 5x 
 
 x is the electrical angle measured from the point midway between 
 the poles. Take B b = 0.2#i. 
 
 If the coil pitch is 1.0, what is the generated armature voltage 
 per turn? What is the form factor of this e.m.f.? 
 
 16. In problem 15 if the coil pitch is 0.8, what is the generated 
 armature voltage per turn? What is the form factor of this 
 e.m.f.? 
 
 17. In problem 15 if the coil pitch is 0.67 what is the generated 
 armature voltage per turn? What is the form factor of this 
 e.m.f.? 
 
 18. With normal excitation the resultant air-gap flux in, a 
 60-cycle alternator is 10 6 lines per pole. The equation of the 
 curve which represents the flux density in the air-gap is: 
 
 B = Bi sin xB$ sin 5x 
 
 x is the electrical angle measured from the point midway between 
 the poles. 
 
 If the coil pitch is 1.0, what is the generated armature voltage 
 per turn? What is the form factor of this e.m.f.? 
 
 19. With normal excitation the resultant air-gap flux in a 60- 
 cycle alternator is 10 6 lines per pole. The flux density is con- 
 stant under the pole face and is zero between the poles. The 
 armature has 8 equally spaced slots per pole. The ratio of pole 
 arc to pole pitch is 0.67 and the coil pitch is 1.0. 
 
 (a) If the armature winding consists of 2 inductors in series 
 per pole placed in adjacent slots, what is the generated armature 
 voltage per pole? What is the form factor of this e.m.f. ? Sketch 
 the graphs of the flux density and the e.m.f. 
 
 (b) Compare this e.m.f. and form factor with what they 
 would have been had the winding been concentrated i.e., with 
 2 inductors in series per pole, placed in the same slot. 
 
 20. In problem 19 if the armature winding consists of 4 induc- 
 tors in series per pole placed in adjacent slots, what is the gene- 
 
SYNCHRONOUS GENERATORS 27 
 
 rated armature voltage per pole? What is the form factor of this 
 e.m.f.? Sketch the graphs of the flux density and the e.m.f. 
 
 (b) Compare this e.m.f. and form factor with what they would 
 have been had the winding been concentrated i.e., with 4 induc- 
 tors in series per pole placed in the same slot. 
 
 21. In problem 19 if the armature winding consists of 4 induc- 
 tors in series per pole placed in alternate slots, what is the gene- 
 rated armature voltage per pole? What is the form factor of 
 this e.m.f.? Sketch the graphs of the flux density and the e.m.f. 
 
 (b) Compare this e.m.f. and form factor with what they would 
 have been had the winding been concentrated i.e., with 4 induc- 
 tors in series per pole placed in the same slot. 
 
 22. With normal excitation the resultant air-gap flux in a 60- 
 cycle alternator is 10 6 lines per pole. The flux density is con- 
 stant under the pole face and decreases uniformly to zero at points 
 midway between the poles. The armature has 8 equally spaced 
 slots per pole. The ratio of pole arc to pole pitch is 0.67 and the 
 coil pitch is 1.0. 
 
 If the armature winding consists of 2 inductors in series per 
 pole placed in adjacent slots, what is the generated armature 
 voltage per pole? What is the form factor of this e.m.f.? 
 Sketch the graphs of the flux density and the e.m.f. 
 
 23. With normal excitation the resultant air-gap flux in a 
 60-cycle alternator is 10 6 lines per pole and is sinusoidally dis- 
 tributed along the air-gap. The armature has 12 equally spaced 
 slots per pole. The coil pitch is 1.0. 
 
 (a) If the armature winding consists of 6 inductors in series 
 per pole placed in alternate slots, what is the generated armature 
 voltage per pole? What is the form factor of this e.m.f.? 
 
 (b) Compare this e.m.f. with what it would have been had the 
 6 inductors per pole been concentrated in one slot. 
 
 24. In problem 23 if the armature winding consists of 12 
 inductors in series per pole, one in each slot, what is the generated 
 armature voltage per pole? What is the form factor of this 
 e.m.f.? 
 
 (b) Compare this e.m.f. with what it would have been had the 
 12 inductors per pole been concentrated in one slot. 
 
 25. In problem 23 if the armature winding consists of 8 
 inductors in series per pole placed in adjacent slots, what is the 
 generated armature voltage per pole? What is the form factor 
 of this e.m.f.? 
 
28 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 (b) Compare this e.m.f. with what it would have been had the 
 8 inductors per pole been concentrated in one slot. 
 
 26. With normal excitation the resultant air-gap flux in a 60- 
 cycle alternator is 10 6 lines per pole. The equation of the curve 
 which represents the flux density in the air-gap is 
 
 = i sn x--z sn x 
 
 x is the electrical angle measured from a point midway between 
 the poles. Take B 3 = 0.3#i. 
 
 (a) If the armature winding consists of 3 equally spaced 
 inductors per pole connected so as to give a winding which has a 
 spread of one-third the pole pitch and a coil pitch of 1.0, what is 
 the generated armature voltage per pole? What is the form 
 factor of this e.m.f.? (b) Compare this e.m.f. and form factor 
 with what they would have been had the winding been concen- 
 trated in one slot per pole. 
 
 27. A single-phase turbo-alternator is rated to deliver 1000 
 kv.-a. "at 5200 volts when driven at 1500 rev. per min. The 
 armature has 48 slots, 8 of which, or 2 per pole, carry no inductors, 
 so that the spread of the winding is 0.83. In each of the other 
 40 slots there are 4 inductors in series. What is the no-load 
 terminal voltage when the air-gap flux is 40 megalines per pole 
 and the speed is 1510 rev. per min.? 
 
 28. A single-phase alternator is rated to deliver 250 kv.-a. 
 at 2200 volts when driven at 375 rev. per min. The field struc- 
 ture has 24 poles and the armature, 240 slots. The spread of the 
 armature winding is 0.6 and all of the inductors are connected in 
 series. The coil pitch is one. In any belt each of the four cen- 
 tral slots contains 3 inductors while each of the outer slots con- 
 tains 2 inductors. What is the no-load terminal voltage when the 
 air-gap flux is 3.9 megalines per pole and the speed is 370 rev. per 
 min.? 
 
 29. A two-phase water turbine driven generator is rated to 
 deliver 900 kv.-a. at 5000 volts. The field structure has 46 poles 
 and the armature has 2 slots per pole per phase. There are 11 
 inductors in series per slot. The coil pitch is one. What is 
 the no-load terminal voltage when the air-gap flux is 5.4 megalines 
 per pole and the generator is driven at its normal speed of 120 
 rev. per min.? 
 
 30. A 2-phase, 60-cycle, engine driven alternator is rated to 
 deliver 3500 kv.-a. at 11,500 volts. The armature has 576 
 
S YNCHRONO US GENERA TORS 29 
 
 slots with 5 inductors in series per slot. The coil pitch is one. 
 The field structure has 96 poles. What is the generated armature 
 voltage when the resultant air-gap flux is 6.3 megalines per pole 
 and the speed is 74 rev. per min.? 
 
 31. The armature of a 3-phase, 50-cycle, 12-pole alternating 
 current generator has 54 slots with 2 inductors per slot. The 
 inductors are arranged symmetrically in the following order, which 
 repeats itself for every pair of poles, i.e., for every 9 slots. In the 
 first, third, fourth, sixth, seventh and ninth slots both inductors 
 are in phases 1, 3, 2, 1, 3 and 2 respectively. In the second, 
 fifth and eighth slots the top inductors are in phases 3, 1, and 2 
 respectively, and the bottom inductors are in phases 1, 2 and 
 3 respectively. The inductors in each phase are connected in 
 series and the phases are connected in Y. 
 
 (a) If the resultant no-load air-gap flux is sinusoidally distrib- 
 uted and has a value of 8X 10 5 lines per pole what is the terminal 
 voltage? 
 
 (b) What are the reduction factors for the fifth and seventh 
 harmonics in the phase voltage? 
 
 (c) For what harmonics are the reduction factors zero? 
 
 32. The armature core of a 3-phase alternating-current gener- 
 ator has 12 slots per pole. Each slot contains 2 coil sides, so that 
 while there are 8 inductors per slot each coil has 4 turns. The 
 coil pitch is 10 slots. 
 
 What are the total reduction factors for the fundamental and 
 the third, fifth and seventh harmonics in the phase voltage? 
 
 33. A 3-phase, Y-connected, 25-cycle, alternating-current 
 generator is rated to deliver 7500 kv.-a at a terminal potential 
 difference of 12,000 volts. The field structure has 12 poles and 
 the armature has 180 slots. There are 4 inductors in series per 
 slot. What is the no-load terminal voltage when the air-gap 
 flux is 52 megalines per pole and the generator is driven at its 
 rated speed? 
 
 34. (a) In problem 27, what is the armature reaction in 
 ampere terms per pole when the alternator is delivering its rated 
 load at unit power factor? 
 
 (b) What is the leakage reactance voltage at tliis load? The 
 armature inductors are 41 in. long. Assume that the leakage 
 flux is 7 lines per ampere per inch of inductor. 
 
 (c) If the armature winding had been equally distributed in 
 5 adjacent slots instead of in 10 what would have been the leakage 
 
30 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 reactance voltage at this load? Assume 7 leakage lines as 
 in(b). 
 
 35. (a) In problem 28 what is the armature reaction in ampere 
 turns per pair of poles when the alternator is delivering 200 kw. 
 at 0.8 power factor, and its rated voltage? 
 
 (b) What is the leakage reactance voltage at this load? The 
 armature inductors are 23 in. long. Assume that the leakage 
 flux is 8 lines per ampere per inch of inductor. 
 
 (c) If the armature winding had been equally distributed in 
 the four central slots, i.e., with 4 inductors per slot, what would 
 have been the leakage reactance voltage at this load? Assume 
 8 leakage lines as in (b) . 
 
 36. (a) In problem 29 what is the armature reaction in ampere 
 turns per pair of poles when the alternator is delivering full-load 
 kilo volt-amperes ? 
 
 (b) What is the leakage reactance voltage per phase at this 
 load? Assume that the leakage flux is 96 lines per ampere per 
 inductor. 
 
 (c) If the armature winding had been concentrated, i.e., with 
 one slot per pole per phase, holding 22 inductors, what would 
 have been the leakage reactance voltage? Assume the same 
 leakage flux per inductor as in (b). 
 
 37. (a) In problem 33 what is the armature reaction in ampere 
 turns per pole when the alternator is delivering 7650 kw. at 0.9 
 power factor and its rated voltage? 
 
 (b) What is the leakage reactance voltage per phase at this 
 load? The armature inductors are 48 in. long. Assume that 
 the leakage flux is 6.5 lines per ampere per inch of inductor. 
 
 38. A three-phase, Y-connected alternating-current generator 
 which is rated to deliver 760 kv.-a. at 2200 volts has an armature 
 with 6 slots per pole. There are 3 inductors in series per slot. 
 
 What are the cross-magnetizing and demagnetizing ampere 
 turns per pole due to armature reaction when the alternator 
 delivers its full-load current (a) if the power factor is such that 
 the current in any phase reaches its maximum value 60 degrees 
 after the inductors of that phase pass the center of the pole? 
 (b) if the power factor is such that the current in any phase 
 reaches its maximum value of 30 degrees before the inductors of 
 that phase pass the center of the pole? 
 
 39. A three-phase, 1500-kv.-a., 5500-volt alternating-current 
 generator has a field structure with 72 poles and an armature 
 
SYNCHRONOUS GENERATORS 
 
 31 
 
 with 216 slots. There are 12 inductors in series per slot. The 
 phase windings are connected in star. 
 
 (a) What are the cross-magnetizing and demagnetizing 
 ampere turns per pole due to armature reaction when the 
 alternator delivers its full-load current at 0.85 power factor (in- 
 ductive load)? Assume that for this power factor the current 
 in any phase is 0.7 of its maximum value at the time the in- 
 ductors of that phase are under the center of the pole. 
 
 (b) What are the cross-magnetizing and demagnetizing ampere 
 turns per pole due to armature reaction when the alternator 
 delivers its full-load current at 0.85 power factor (condensive 
 load)? Assume that for this power factor the current in any 
 phase is 0.2 of its maximum value at the time the inductors of 
 that phase are midway between the poles. 
 
 DATA ON ALTERNATING-CURRENT GENERATORS 
 
 
 A 
 
 B 
 
 Rated output (kv.-a.) 
 
 100 
 
 5000 
 
 Phases . .... 
 
 2 
 
 3, Y 
 
 Frequency 
 
 60 
 
 60 
 
 Line voltage 
 
 480 
 
 6600 
 
 Armature iron: 
 Effective length of core 
 
 5 75 in. 
 
 17 5 in. 
 
 Diameter of core at air-gap 
 Slots (open) 
 
 26. 4 in. 
 96 
 
 148 in. 
 360 
 
 Depth of slot 
 
 1 488 in. 
 
 2 2 in 
 
 Width of slot 
 
 37 in. 
 
 58 in. 
 
 Field magnets: 
 Poles 
 
 8 
 
 30 
 
 Pole arc: Pole pitch 
 
 70 
 
 768 
 
 Armature copper: 
 Inductors per slot 
 
 8 
 
 4 
 
 Size of inductor 
 
 (2 in parallel) 
 No. 8 wire 
 
 (2 in parallel) 
 ^X g-in. strap 
 
 Thickness of insulation between 
 inductor and sides of slot. 
 Mean length oer turn. . 
 
 0.042 in. 
 44. 4 in. 
 
 0.114 in. 
 107. 26 in. 
 
 40. Calculate the cross-magnetizing and demagnetizing ampere 
 turns in alternator A for full-load current and such a power 
 factor that the current is lagging and has a value of 0.70 of its 
 maximum value in any phase when that phase is opposite the 
 center of the poles. 
 
 41. Calculate the cross-magnetizing and demagnetizing ampere 
 turns in alternator B for full-load current and such a power factor 
 
32 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 that the current is lagging and has a value of 0.80 of its maximum 
 value in any phase when that phase is opposite the center of 
 the poles. 
 
 42. A 25-cycle alternating-current generator has an armature 
 core with slots that are 1 .35 in. wide and 2.625 in. deep. There are 
 four inductors in series per slot, (a) Calculate the slot reactance 
 per inch of slot on the assumptions that the path of the leakage 
 flux across the inductors is parallel to the bottom of the slot, 
 that the permeability of the iron is great and that the inductors 
 completely fill the slot. 
 
 (b) If the slot reactance is calculated on the assumption that 
 all of the leakage flux per slot is linked with all of the inductors 
 in that slot, what value of leakage flux per ampere per inch of 
 inductor will give the same slot reactance as calculated by the 
 preceding method? 
 
 43. A 125-cycle alternating-current generator has an armature 
 core with slots that are 1.0 in. wide and 3.5 in. deep. Each slot 
 holds 12 inductors in series which completely fill it. (a) Calculate 
 the leakage reactance per inch of slot on the assumption that the 
 path of the leakage flux across the inductors is parallel to the 
 bottom of the slot, and that the permeability of the iron is great. 
 
 (b) If the 12 inductors fill the slot to a depth of but 2.5 in. 
 calculate the leakage reactance per inch of slot on the same 
 assumptions. 
 
 44. A 60-cycle alternating-current generator has an armature 
 core with slots that are 1.1 in. wide and 2.75 in. deep and 15 in. 
 long. There are 6 inductors in series per slot which fill it to a 
 depth of 2.0 in. (a) Calculate the slot reactance on the assump- 
 tion that the path of the leakage flux between the sides of the 
 slot is parallel to the bottom of the slot, and that the permea- 
 bility of the iron is great. 
 
 (b) If the slot reactance is calculated on the assumption that 
 all of the leakage flux per slot is linked with all of the inductors 
 in that slot, what value of leakage flux per ampere per inch of 
 inductor will give the same slot reactance as calculated by the 
 preceding method? 
 
 45. A 3-phase, 25-cycle altenator is rated to deliver 850 
 kv.-a. at 5000 volts. The slots in the armature core are 1.31 
 in. wide and 3.625 in. deep. The gross length of the armature 
 core is 14.5 in. and the effective length, deducting for the ven- 
 tilating ducts and the insulation on the laminations, is 10.75 
 
S YNCHRONO US GENERA TORS 33 
 
 in. There are 14 inductors in series per slot which occupy a 
 space 0.75 in. by 2.75 in. The thickness of the insulation 
 between the inductors and the armature core is the same at the 
 sides and at the bottom of the slot. The mean length of one 
 turn is 93 in. Each coil consists of 14 turns. 
 
 (a) Calculate the leakage reactance per coil on the following 
 assumptions. Where the inductors are embedded in iron the 
 path of the leakage flux between the sides of the slot is parallel to 
 the bottom of the slot. For the portions of the coil that are 
 not embedded in the armature core the leakage flux links with 
 all of the turns in that coil and has a value of 0.8 line per ampere 
 per inch of turn. 
 
 (b) If the leakage reactance is calculated on the assumption 
 that all of the leakage flux per coil links with all of the turns in 
 the coil, and that the length of the coil is but twice the length of 
 the inductor, i.e., twice the gross length of the armature core, 
 what value of leakage flux per ampere per inch of inductor will 
 give the same value of leakage reactance as calculated by the 
 preceding method? 
 
 46. (a) Calculate the slot and coil end leakage reactance for 
 alternator A (see problem 40). Use the most exact method at 
 your command for the data given. 
 
 (b) Calculate the slot and coil end leakage reactance for 
 alternator B (see problem 41). Use the most exact method at 
 your command for the data given. 
 
 47. A special 60-cycle generator has a field structure with 6 
 poles and an armature with 72 slots. There are 9 inductors in 
 series per slot. The coils in adjacent slots are connected in 
 series by pairs so that each of the 6 independent armature wind- 
 ings thus formed has a spread of one-sixth and a pitch of 1.0. 
 When the 6 windings are arranged to form a 3-phase, Y-con- 
 nected armature winding, the synchronous reactance voltage 
 is 46 volts for an armature current of 30 amperes. Assume 
 that the armature leakage flux is 72 lines per inductor per am- 
 pere, and that the approximate formula for armature reac- 
 
 77 
 tion, 0.707-p-' holds exactly in every case. 
 
 (a) What is the synchronous reactance voltage with an arma- 
 ture current of 30 amperes when the 6 armature windings are 
 connected in series to form a single-phase, open-circuit armature 
 winding? 
 
34 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 (b) The synchronous reactance voltage is 24 per cent, of the 
 open-circuit phase voltage when the armature winding is con- 
 nected in Y. With the same field current what per cent, of the 
 open-circuit armature voltage will the synchronous reactance 
 voltage be when the windings are connected to form a single- 
 phase, open-circuit armature winding? 
 
 48. In problem 47 what is the synchronous reactance voltage 
 with an armature current of 60 amperes i.e., 30 amperes per 
 inductor when the 6 windings are connected to form a single- 
 phase, 2-circuit armature winding? 
 
 (b) With the same field current as in 47 (b) what per cent, of 
 the open-circuit armature voltage will the synchronous react- 
 ance voltage be when the windings are connected to form a 
 single-phase, 2-circuit armature winding? 
 
 49. In problem 47 what is the synchronous reactance voltage 
 with an armature current of 30 amperes when the 6 windings 
 are connected to form a 2-phase armature winding? 
 
 (b) With the same field current as in 47 (b) what per cent, of 
 the open-circuit armature voltage will the synchronous reactance 
 voltage be when the windings are connected to form a 2-phase 
 armature winding? 
 
 60. Two three-phase, 60-cycle alternating-current generators 
 have the same current and voltage rating. The first has 12 
 slots per pole and the second has 9 slots per pole. Each alter- 
 nator has the same number of turns per phase. The first has 
 4 poles and the second 6 but the dimensions of the magnetic 
 circuit are such that the same number of ampere turns on the 
 field produces the same flux per pole in each. The shape of the 
 armature slots is such that the leakage flux per ampere per 
 inductor is also the same for each. 
 
 The armature leakage reactance and the synchronous react- 
 ance of the first alternator are respectively 0.54 ohms and 2.13 
 ohms. What are the corresponding constants of the second 
 alternator? 
 
 51. A 3-phase, delta-connected alternating-current generator 
 is rated to deliver 15 kv.-a. at 230 volts when running at a speed 
 of 1200 rev. per min. There are 6 field poles each of which is 
 wound with 398 turns. The armature has 72 slots with 8-induc- 
 tors in series per slot. Each inductor is 5 in. long. The hot 
 resistance of the armature measured between any two terminals 
 is 0.126 ohm. The effective resistance is 1.6 times the ohmic 
 
SYNCHRONOUS GENERATORS 
 
 35 
 
 resistance. In calculating the leakage reactance assume 8 
 leakage lines per ampere per inch of inductor. The open- and 
 short-circuit characteristics are given by the following data: 
 
 Field current 
 
 Open circuit 
 
 Short circuit 
 
 Terminal voltage 
 
 Armature current 
 
 2.0 
 
 104 
 
 22 
 
 4.0 
 
 199 
 
 44 
 
 5.0 
 
 240 
 
 55 
 
 6.0 
 
 275 
 
 
 7.5 
 
 ' 315 
 
 
 (a) Calculate the regulation of this generator by the general 
 method for an inductive load of 15 kw. at 0.8 power factor. 
 What is the field current calculated by this method? 
 
 (b) Calculate the regulation of this generator for the specified 
 load by the synchronous impedance method. What is the field 
 current calculated by this method? 
 
 (c) Calculate the regulation of this generator for the specified 
 load by the magnetomotive force method. What is the field 
 current calculated by this method? 
 
 52. A 760-kv.-a, 2200-volt alternating-current generator 
 delivers energy directly to a 3-phase, 50-cycle system. The 
 neutral pf the generator is grounded. The field structure 
 consists of 64 poles each of which is wound with 50 turns. The 
 armature core has 384 slots with 3 inductors in series per slot. 
 The length of the armature core parallel to the shaft is 10 in. 
 The effective resistance of the armature winding is 0.172 ohm per 
 phase. In calculating the leakage reactance assume 6.5 leakage 
 lines per ampere per inch of slot per inductor. The open- and 
 short-circuit characteristics are given by the following data: 
 
 
 Open circuit 
 
 Short circuit 
 
 
 Terminal voltage 
 
 Armature current 
 
 50 
 
 80 
 100 
 
 1060 
 1650 
 1950 
 
 280 
 450 
 
 120 
 
 2160 
 
 
 150 
 
 2420 
 
 
 200 
 
 2650 
 
 
 (a) Calculate the regulation of this generator for 25 per cent, 
 overload current at unit power factor by the general method. 
 What is the field current calculated by this method? 
 
36 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 (b) Calculate the regulation of this generator for the specified 
 load by the synchronous impedance method. What is the field 
 current calculated by this method? 
 
 (c) Calculate the regulation of this generator for the specified 
 load by the magnetomotive force method. What is the field 
 current calculated by this method? 
 
 53. A 3-phase, Y-connected, 1500-kv.-a., 5500-volt alternating- 
 current generator is driven by a 2000-h.p. reciprocating engine 
 that runs at 83 rev. per min. at full load. The field structure 
 has 72 poles each of which is wound with 35 turns. The armature 
 core has 1 slot per pole per phase with 12 inductors in series per 
 slot. Each inductor is 9 in. long. The armature winding has 
 an effective resistance of 0.36 ohm per phase. In calculating 
 the leakage reactance assume 7.5 equivalent leakage lines per 
 ampere per inch of inductor. The data for the open- and short- 
 circuit characteristics are: 
 
 Open circuit Short circuit 
 
 rituu current 
 Terminal voltage Armature current 
 
 150 
 200 
 250 
 300 
 350 
 
 5100 
 5900 
 6500 
 6800 
 7100 
 
 300 
 400 
 
 
 (a) Calculate the regulation of this generator by the general 
 method for 20 per cent, overload current at 0.85 power factor 
 (induction load). What is the field current calculated by this 
 method? 
 
 (b) Calculate the regulation of this generator for the specified 
 load by the synchronous impedance method. What is the field 
 current calculated by this method? 
 
 (c) Calculate the regulation of this generator for the specified 
 load by the magnetomotive force method. What is the field 
 current calculated by this method? 
 
 54. A 3750-kv.-a. alternating-current generator delivers energy 
 at 2200 volts to a 2-phase, 30-cycle system. The field structure 
 has 20 poles, each of which is wound with 60 turns. The arma- 
 ture has 360 slots with one inductor per slot. Each inductor is 
 20.5 in. long. The armature resistance by direct-current measure- 
 ment is 0.0196 ohm per phase at 25 C. Assume that the ratio 
 of effective resistance to ohmic resistance is 1.3 at 25 C. In 
 
SYNCHRONOUS GENERATORS 
 
 37 
 
 calculating the leakage reactance assume that the leakage flux 
 is 6.5 lines per ampere per inch of inductor. The data for the 
 open- and short-circuit characteristics are: 
 
 Field ampere turns 
 
 Open circuit 
 
 Short circuit 
 
 per pole 
 
 Terminal voltage 
 
 Armature current 
 
 4,000 
 
 8,000 
 12,000 
 16,000 
 
 680 
 1360 
 2000 
 2480 
 
 606 
 1210 
 
 18.000 
 
 2660 
 
 
 (a) Calculate the regulation of this generator for a condensive 
 load of 3500 kw. at 0.92 power factor by the general method. 
 What is the field current calculated by this method? Assume 
 that the temperature of the armature windings is 70 C. 
 
 (b) Calculate the regulation of this generator for the specified 
 load by the synchronous impedance method. What is the field 
 current calculated by this method? 
 
 (c) Calculate the regulation of this generator for the specified 
 load by the magneto-motive force method. What is the field 
 current calculated by this method? 
 
 55. A 3-phase water-wheel generator whose armature winding 
 is Y-connected is rated to deliver 5000 kv.-a. at 6600 volts. 
 Normal speed is 240 rev. per min. The field structure has 30 
 poles with 67.5 turns per pole. The armature core has 360 
 slots with 2 inductors in series per slot. The length of the 
 armature core is 21.5 in. The measured resistance of the arma- 
 ture between any two terminals is 0.0836 ohm at 25 C. The 
 ratio of effective resistance to ohmic resistance is 1.65 at 25 C. 
 In calculating the leakage reactance assume that the leakage flux 
 is 6.5 lines per ampere per inch of slot per inductor. 
 
 The open- and short-circuit characteristics are given by the 
 following data: 
 
 . 
 
 Open circuit 
 
 Short circuit 
 
 
 Terminal voltage 
 
 Armature current 
 
 100 
 150 
 200 
 
 4800 
 6500 
 7400 
 
 680 
 1020 
 
 250 
 
 7900 
 
 
 (a) Calculate the regulation of this generator by the general 
 method for full-load kilovolt-amperes at 0.8 power factor (indue- 
 
38 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 tive load). What is the field current calculated by this method? 
 Assume that the temperature of the armature windings is 
 70 C. 
 
 (b) Calculate the regulation of this generator for the specified 
 load by the synchronous impedance method. What is the field 
 current calculated by this method? 
 
 (c) Calculate the regulation of this generator for the specified 
 load by the magnetomotive-force method. What is the field 
 current calculated by this method? 
 
 56. A 2-phase, 60-cycle alternating-current generator is rated 
 to deliver 100 kv.-a. at 480 volts. The armature has an effective 
 resistance of 0.138 ohm and a leakage reactance of 0.159 ohm per 
 phase. When the power factor of the load is 0.8 the armature 
 demagnetizing ampere turns are 11.8 and the cross-magnetizing 
 ampere turns are 15.2 per pole per ampere. The field poles 
 are each wound with 265 turns. The data for the open circuit 
 characteristic are: 
 
 Field current 
 
 Open-circuit voltage 
 
 10 
 15 
 20 
 25 
 
 400 
 500 
 560 
 
 ! 598 
 
 What is the regulation of this generator when delivering full- 
 load current at 0.8 power factor (inductive load) ? 
 
 (a) Assume that the cross-magnetizing and demagnetizing 
 ampere turns act on magnetic circuits of the same reluctance as 
 that of the resultant field. This is a modification of the general 
 method. 
 
 (b) Assume that the cross-magnetizing ampere turns act on a 
 magnetic circuit whose reluctance is determined by the lower 
 part of the saturation curve, and that the demagnetizing ampere 
 turns act on the same magnetic circuit as do the impressed field 
 ampere turns. This is the Blondel method. 
 
 57. A 3-phase, 25-cycle alternating-current generator is rated 
 to deliver 850 kv.-a. at 5000 volts. The armature windings are 
 connected in Y, and have an effective resistance of 0.398 ohm and 
 a leakage reactance of 1.46 ohms per phase. When the power 
 factor of the load is unity the armature demagnetizing ampere 
 turns are 5.8 and the cross-magnetizing ampere turns are 31 per 
 
SYNCHRONOUS GENERATORS 
 
 39 
 
 pole per ampere. The field poles are each wound with 
 turns. The data for the open-circuit characteristic are: 
 
 79.5 
 
 Field ampere turns per pole Open-circuit terminal voltage 
 
 6,000 
 10,000 
 12,000 
 14,000 
 
 4330 
 5460 
 5800 
 6060 
 
 What is the regulation of this generator when delivering 30 
 per cent, overload current at unit power factor? 
 
 (a) Assume that the cross-magnetizing and demagnetizing 
 ampere turns act on magnetic circuits of the same reluctance as 
 that of the resultant field. This is a modification of the general 
 method. 
 
 (b) Assume that the cross-magnetizing ampere turns act on a 
 magnetic circuit whose reluctance is determined by the lower 
 part of the saturation curve, and that the demagnetizing ampere 
 turns act on the same magnetic circuit as do the impressed field 
 ampere turns. This is the Blondel method. 
 
 58. A 3-phase, Y-connected alternating-current generator is 
 rated to deliver 1000 kv.-a. at 13,800 volts. The armature 
 has an effective resistance of 2.18 ohms per phase. The data for 
 the open-circuit characteristic and the full-load current satura- 
 tion curve at zero power factor are: 
 
 Field current 
 
 Open-circuit terminal 
 voltage 
 
 Saturation curve 
 42 amp., at zero P.F. 
 
 41 2 
 
 
 
 
 50 
 
 8,800 
 
 
 110 
 140 
 180 
 
 15,600 
 17,250 
 18,900 
 
 10,750 
 13,250 
 15,600 
 
 What is the regulation for an inductive load of 50 amperes 
 at 0.85 power factor? 
 
 59. A 3-phase, water-wheel generator is rated to deliver 5000 
 kv.-a. at 6600 volts. The armature winding is Y-connected 
 and has an effective resistance of 0.081 ohm per phase. The data 
 for the open-circuit characteristic and the full-load current satura- 
 tion curve at zero power factor are: 
 
 4 
 
40 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 Field current 
 
 Open-circuit terminal 
 voltage 
 
 Saturation curve 
 438 amp. at zero P.F. 
 
 64 
 
 3050 
 
 
 
 100 
 
 4800 
 
 1800 
 
 150 
 
 6500 
 
 4200 
 
 200 
 
 7400 
 
 5700 
 
 275 
 
 8150 
 
 6900 
 
 What is the regulation for an inductive load of 500 amperes 
 at 0.8 power factor? 
 
 60. A 3-phase, Y-connected alternating-current generator is 
 rated to deliver 1640 kv.-a. at 13,500 volts. The armature has 
 an effective resistance of 1.52 ohms per phase and a synchronous 
 reactance of 37.4 ohms per phase. 
 
 (a) What is the regulation of this alternator on an inductive 
 load taking 1500 kw. at 0.85 power factor? 
 
 (b) What is the regulation on a condensive load taking 1500 
 kw. at 0.85 power factor? 
 
 61. A 2-phase alternating-current generator is rated to deliver 
 3500 kw. at 10,000 volts. The armature has an effective resist- 
 ance of 0.64 ohm per phase and a synchronous reactance of 23.7 
 ohms per phase. 
 
 (a) What is the regulation of this alternator on a non-induc- 
 tive load taking the rated kv. a.? 
 
 (b) What is the regulation on an inductive load taking the 
 rated kv.-a. at zero power factor? 
 
 (c) What is the regulation on a condensive load taking the 
 rated kv.-a. at zero power factor? 
 
 62. A 760-kv.-a, 2200-volt, 3-phase alternating-current gen- 
 erator has an effective armature resistance of 0.17 ohm per phase. 
 The armature winding is connected in Y. With the armature 
 short-circuited the armature current is 338 amperes when the 
 field current is 60 amperes. The open-circuit characteristic 
 data are: 
 
 Field current 
 
 Open-circuit terminal voltage 
 
 50 
 80 
 100 
 120 
 150 
 200 
 
 1060 
 1650 
 1950 
 2160 
 2420 
 2650 
 
S YNCHRONO US GENERA TORS 4 1 
 
 (a) What is the regulation of this generator, calculated by the 
 magnetomotive-force method, for an inductive load which re- 
 quires full-load current at 0.5 power factor? 
 
 (b) What is the regulation for a condensive load which re- 
 quires full-load current at 0.5 power factor? 
 
 63. A 3-phase, A-connected alternating-current generator 
 has a full-load capacity of 15 kv.-a. at 230 volts. The effective 
 armature resistance and synchronous reactance are respectively 
 0.302 and 4.36 ohms per phase. Three reactors each of which 
 has an effective resistance of 2.5 ohms and a reactance of 10 
 ohms are connected in A across the terminals of the generator. 
 If the terminal voltage is adjusted to its rated value to what will 
 it rise when the coils are removed? 
 
 (b) If the coils are connected in Y and the terminal voltage 
 adjusted as before to what will it rise when the coils are removed? 
 
 (c) What is the power output of the generator in each case? 
 
 64. In problem 63 if the open-circuit voltage of the generator 
 is adjusted to 300 volts to what will it fall when the coils are 
 connected in A across the terminals? What is the power out- 
 put of the generator? 
 
 65. A 3-phase, 1500-kv.-a., 5500-volt alternating-current 
 generator delivers full-load current to an inductive load at 0.85 
 power factor. The effective resistance and synchronous react- 
 ance of this alternator are respectively 0.36 and 8.5 ohms per 
 phase. The armature winding is Y-connected. With the field 
 excitation unchanged what will be the terminal voltage if the 
 alternator delivers its rated current to a condensive load at 0.85 
 power factor? 
 
 66. A special 6-pole, 60-cycle alternating-current generator 
 has- six similar and independent armature windings. The 
 windings are equally spaced so that their voltages differ by 30 
 degrees. 
 
 On the basis of equal armature and field heating losses and of 
 equal frequencies compare the rated outputs kilovolt-amperes 
 and terminal voltage of this alternator when the windings are 
 connected (1) for an open-coil single-phase winding and (2) for a 
 2-phase winding. 
 
 67. (a) In problem 66 compare the rated outputs on the same 
 basis when the windings are connected (1) for an open-coil single- 
 phase winding and (2) for a 3-phase mesh winding. 
 
 (b) Compare the rated outputs on the same basis when the 
 
42 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 windings are connected (1) for an open-coil single-phase winding 
 and (2) for a 3-phase star winding. 
 
 68. In problem 66 compare the rated outputs on the same 
 basis when the windings are connected (1) for a 2-phase winding 
 and (2) for a 3-phase star winding. 
 
 69. Concerning the alternator described in problem 51 the 
 following additional data are given. The field current is sup- 
 plied at 110 volts. The friction and windage loss is 310 watts at 
 normal speed, and the core loss due to rotation is 480 watts for 
 an armature generated voltage of 240. Assume that the core 
 loss is constant. What is the efficiency of this alternator at the 
 load described in problem 51? (1) Calculate the field current 
 by the synchronous impedance method, (2) by the magneto- 
 motive-force method. 
 
 70. Concerning the alternator described in problem 52 the 
 following additional data are given. The resistance of the field 
 circuit is 0.516 ohm. The friction and windage loss is 6.2 kw. 
 at normal speed. The core loss due to rotation is 11.0 kw. at 
 2200 volts and may be assumed to vary as the square of the 
 generated armature voltage. 
 
 (a) What is the efficiency of this generator at the load de- 
 scribed in problem 52? 
 
 (b) What is the efficiency of this generator when delivering 
 the same current at 0.80 power factor? 
 
 Calculate the field current (1) by the general method and (2) 
 by the magnetomotive-force method. 
 
 71. Concerning the alternator described in problem 53 the 
 following additional data are given. The resistance of the 
 field circuit is 0.376 ohm. The friction and windage loss is 8.4 
 kw. The core loss due to rotation is 20.2 kw. at 5500 volts, and 
 should be assumed to vary as the square of the generated arma- 
 ture voltage. 
 
 (a) What is the efficiency of the generator at the load de- 
 scribed in problem 53? 
 
 (b) What is the efficiency of this generator when delivering 
 the same power at unit power factor? 
 
 Calculate the field current (1) by the synchronous impedance 
 method, and (2) by the magnetomotive-force method. 
 
 72. The full-load capacity of a 3-phase, 25-cycle alternating- 
 current generator is 850 kv.-a. at 5000 volts. The armature 
 windings are connected in Y and have an effective resistance of 
 
S YNCHRONO US GENERA TORS 43 
 
 0.398 ohm per phase. The synchronous reactance is 15.4 ohms 
 per phase. With the armature short-circuited the armature 
 current is 108 amperes when the field current is 50 amperes. 
 The resistance of the field circuit is 0.82 ohm. The friction and 
 windage is 7.6 kw. The core loss due to rotation is 20.2 kw. 
 at 5200 volts and may be assumed to be constant. The open- 
 circuit characteristic is given by the following data: 
 
 Field current Open-circuit terminal voltage 
 
 125 
 150 
 175 
 
 5460 
 5800 
 6060 
 
 What is the efficiency of this generator when delivering 10 per 
 cent, overload current at 0.80 power factor (inductive)? 
 
 Calculate the field current (1) by the synchronous impedance 
 method, and (2) by the magentomotive force method. 
 
 73. Concerning the alternator described in problem 54 the 
 following additional data are given. The resistance of the field 
 circuit is 0.37 ohm. The friction and windage is 19 kw. The 
 core loss is given by: 
 
 Terminal voltage on open circuit Core loss 
 
 1360 
 2000 
 2480 
 
 18.0 kw. 
 36.5 kw. 
 60.0 kw. 
 
 What is the efficiency of this generator for the load described 
 in problem 54? Calculate the field current (1) by the general 
 method, (2) by the synchronous impedance method, and (3) by 
 the magnetomotive-force method. 
 
 74. Concerning the alternator described in problem 58 the 
 following additional data are given. The resistance of the field 
 circuit is 0.541 ohm. The friction and windage loss is 9.2 kw. 
 The core loss is given by: 
 
 Terminal voltage on open circuit | Core loss 
 
 8,800 
 13,000 
 15,600 
 17,250 
 
 7.5 kw. 
 16.6 kw. 
 
 25.4 kw. 
 
 33.5 kw. 
 
 What is the efficiency of this generator for the load described 
 in problem 58? 
 
44 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 75. Concerning the alternator described in problem 59 the 
 following additional data are given. The resistance of the field 
 circuit is 0.549 ohm at 25 C. The temperature of the field 
 under load conditions is 68 C. The friction and windage loss 
 is 38 kw. The core loss is given by: 
 
 Terminal voltage on open circuit 
 
 Core loss 
 
 4800 
 6000 
 6600 
 7500 
 
 45 kw. 
 73 kw. 
 90 kw. 
 123 kw. 
 
 What is the efficiency of this generator for the load described in 
 problem 59? 
 
 76. Two alternators of the same design are operating in 
 parallel. The first delivers 980 kw. at 0.95 power factor, and 
 the second 720 kw. at 0.73 power factor. What adjustments 
 should be made to have these alternators operate under the best 
 conditions? When these have been made what power will each 
 deliver and at what power factor will it operate? 
 
 77. Two 3-phase alternators connected in parallel are driven 
 by shunt motors whose speed load characteristics for particular 
 field excitations are given by the following data: The speed of 
 the first motor falls uniformly from 600 rev. per min. at no load 
 to 530 rev. per min. at full load of 100 kw. on the alternator. 
 The speed of the second motor falls uniformly from 590 rev. per 
 min. at no load to 550 rev. per min. at full load of 100 kw. on the 
 alternator. 
 
 (a) For what load will the alternators divide the load equally? 
 
 (b) What will be the load on each alternator when their com- 
 bined load is 200 kw.? 
 
 (c) What is the greatest load that can be delivered without 
 overloading either alternator? 
 
 78. A 3-phase, 2200-volt alternator which is rated to deliver 
 760 kv.-a. is connected in parallel through transformers with a 
 3-phase, 5500-volt alternator which is rated to deliver 1500 
 kv.-a. The first alternator has 64 poles and is driven by an 
 engine whose speed falls from 94 rev. per min. at no load to 
 91 rev. per min. at full load on the alternator. The second al- 
 ternator has 72 poles and is driven by an engine whose speed falls 
 from 83 rev. per min. at no load to 79 rev. per min. at full load 
 on the alternator. 
 
SYNCHRONOUS GENERATORS 45 
 
 (a) What is the greatest combined load that the alternators 
 can deliver without overloading either by more than 25 per 
 cent.? 
 
 (b) What is the load on each alternator when the first is run- 
 ning at 91.6 rev. per min.? 
 
 (c) What is the frequency when they are delivering a com- 
 bined load of 2000 kw.? 
 
 79. Two 3-phase, 60-cycle alternating-current generators are 
 operating in parallel. The first has a capacity of 1000 and the 
 second a capacity of 1500 kv.-a. The first is driven by a prime 
 mover so adjusted that the frequency falls from 61 cycles at no 
 load to 59.6 cycles at full load. The second has a different speed- 
 load characteristic, the frequency falling from 61.4 cycles at no 
 load to 59.2 cycles at full load. 
 
 When these alternators are jointly delivering 2000 kw. what is 
 the load on each? What is the frequency? If the speed-load 
 characteristic of the second is shifted parallel to itself until the 
 alternators divide this load properly, what is the new value of the 
 no-load frequency of this alternator? At what frequency will 
 they now operate when delivering 2000 kw.? 
 
 80. Two 3-phase, 11,000-volt, 60-cycle alternating-current 
 generators, operating in parallel, are driven by prime movers 
 which have the same speed-load characteristic. The armature 
 windings of the alternators are Y-connected and have an effective 
 resistance of 0.94 ohm and a synchronous reactance of 56 ohms per 
 phase. The total load supplied is 1700 kw. at 0.83 power factor. 
 The excitations are adjusted so that the terminal voltage is 
 11,000 volts and one of them is operating at unit power factor. 
 What are the excitation voltages and the phase angle between 
 them? 
 
 81. Two identical 3-phase, Y-connected alternators operating 
 in parallel are driven by prime movers that have such dissimilar 
 speed-load characteristics that when the first alternator is 
 delivering 400 kw. at 0.8 power factor the second is running at 
 no load, (a) If the excitations of the alternators are adjusted so 
 that the terminal voltage is 5000 volts and the armature current 
 of the second alternator is zero, what are the excitation voltages 
 and their phase displacement ? (b) If the excitations are adj usted 
 so that the terminal voltage is 5000 volts and the total armature 
 copper loss is reduced to its least value what current will each 
 alternator deliver? The effective armature resistance and the 
 
46 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 synchronous reactance of each alternator are respectively 0.42 
 and 15.4 ohms per phase. 
 
 82. Two identical 3-phase alternators, connected in parallel, 
 are driven by prime movers that have dissimilar speed-load 
 characteristics. When the excitations of the alternators are 
 equal the first delivers 100 amperes at 0.9 power factor (lagging) 
 and the second, 75 amperes at 0.7 power factor (lagging). 
 
 (a) What per cent, of the total load does each alternator 
 deliver? 
 
 (b) What is the power factor of the load? 
 
 (c) If the field excitations are adjusted so that both alter- 
 nators operate at the same power factor what current will each 
 deliver? 
 
 (d) If the field excitations are adjusted so that the total 
 armature copper loss is reduced to its least value at what power 
 factor will each alternator operate? 
 
 83. Two identical 2-phase alternators, connected in parallel, 
 are driven by prime movers which have somewhat dissimilar 
 speed-load characteristics. The first alternator delivers 3200 
 kw. at 2210 volts and has an armature current of 760 amperes, 
 while the second delivers 3700 kw. at 2210 volts and has an 
 armature current of 1100 amperes. The effective armature 
 resistance and the synchronous reactance of each alternator are 
 respectively 0.0256 and 1.12 ohms per phase. The excitations 
 of the alternators are now adjusted so as to reduce the total 
 armature copper loss to its least value, but the terminal voltage 
 is maintained at 2210 volts. 
 
 (a) At what power factor will each alternator be operating? 
 
 (b) What is the reduction in the total armature copper loss? 
 
 (c) What is the change in the copper loss of each alternator? 
 
 84. Two identical 3-phase alternators, operating in parallel 
 on a balanced load, are driven by prime movers with different 
 speed-load characteristics. The power output of each alter- 
 nator is measured by two wattmeters. Show that when the 
 differences between the wattmeter readings for each alternator 
 are the same the total armature copper loss is reduced to its least 
 value for the given load. 
 
 85. Two dissimilar Y-connected alternators, operating in 
 parallel, supply a load of 1500 kw. at 0.85 power factor and a 
 terminal potential difference of 5000 volts. The alternators 
 being of the same capacity are adjusted to deliver equal loads. 
 
S YNCHRONO US GENERA TORS 47 
 
 The effective armature resistances are respectively 0.40 and 0.52 
 ohm per phase. 
 
 What should be the armature current of each alternator in 
 order that their combined armature copper loss will be reduced 
 to its least value? 
 
 86. An alternator which has a capacity of 1650 kv.-a. is 
 operated in parallel with one which has a capacity of 1000 kv.-a. 
 Their armature resistances are 1.56 ohms and 2.08 ohms re- 
 spectively. For a combined load of 2200 kw. at 0.83 power 
 factor what load should each deliver and at what power factor 
 should it operate if the current and power outputs are propor- 
 tional to their ratings? With this division of the load the excita- 
 tions are adjusted so that the combined armature copper loss 
 is reduced to a minimum. At what power factor should each 
 alternator operate for this latter condition? 
 
 87. Two identical 3-phase, Y-connected alternators are ope- 
 rating in parallel with a common potential difference of 2200 
 volts. The effective armature resistance and the synchronous 
 reactance of these alternators are respectively 0.158 and 2.12 ohms 
 per phase. The characteristics of the prime movers are so dis- 
 similar that of the total load of 1500 kw. at 0.85 power factor 
 the first alternator supplies 840 kw. The excitations of the 
 alternators are adjusted so that both are operating at the same 
 power factor. What are their excitation voltages and the phase 
 displacement between them? 
 
 88. Two identical 3-phase, Y-connected alternators, operat- 
 ing in parallel, are driven by prime movers which have the same 
 speed-load characteristic. The effective armature resistance 
 and the synchronous reactance of the alternators are respec- 
 tively 2.18 ohms and 92 ohms per phase. 
 
 The alternators supply 1830 kw. at 13,800 volts to an induction 
 motor load that is operating at 0.83 power factor. The excita- 
 tions of the alternators are adjusted so that the first supplies a 
 current of 40 amperes. 
 
 (a) What current does the second alternator supply? 
 
 (b) What are the excitation voltages of the alternators and 
 their phase displacement? 
 
 89. Two identical 3-phase, Y-connected alternators, operating 
 in parallel, are driven by prime movers that have the same speed- 
 load characteristic. The alternators are rated to deliver 1000 
 kv.-a. at 2400 volts. The effective armature resistance and the 
 
48 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 synchronous reactance are respectively 0.067 and 2.64 ohms per 
 phase. Assume that the rotational losses, both core loss and 
 friction, are constant and equal. 
 
 (a) If these alternators are delivering 2000 kw. at unit power 
 factor and their rated voltage, by what amount can the load be 
 shifted from one to the other if their field excitations are adjusted 
 so that there is an interchange current equal to the full-load 
 current, viz., 240 amperes? 
 
 (b) If these alternators are delivering 1500 kw. at 0.75 power 
 factor and their rated voltage, by what amount can the load be 
 shifted from one to the other if their field excitations are adjusted 
 so that there is an interchange current equal to the full-load cur- 
 rent, viz., 240 amperes. 
 
 90. The alternators described in problem 89 are operating in 
 parallel with non-inductive resistances of 0.8 ohm inserted in 
 each phase of each alternator. 
 
 (a) If the alternators are jointly delivering 2000 kw. at unit 
 power factor and their rated voltage to a load by what amount 
 can this load be shifted from one to the other if the excitations 
 are adjusted so that there is an interchange current equal to the 
 full-load current, viz., 240 amperes? 
 
 (b) If the alternators are jointly delivering 1500 kw. at 0.75 
 power factor and their rated voltage to a load, by what amount 
 can this load be shifted from one to the other if the excitations 
 are adjusted so that there is an interchange current equal to 
 the full-load current, viz., 240 amperes? 
 
 91. Two identical 3-phase alternators, rigidly coupled together 
 so that their excitation voltages are in phase, are driven by 
 a shunt motor. The effective armature resistance and the 
 synchronous reactance of each alternator are respectively 0.302 
 and 4.36 ohms per phase. The terminals of the alternators are 
 electrically connected as they would be for parallel operation but 
 no external load is supplied. When the field excitations are ad- 
 justed so that the excitation voltages are respectively 200 and 
 300 volts per phase, what is the armature current? If the 
 armature windings are connected in delta what is the terminal 
 voltage? What is the electrical output of the alternator 
 which is acting as a generator? If the rotational losses are 
 supplied by the shunt motor what is the mechanical output of 
 the alternator which is acting as a motor? If the rotational 
 losses are 1620 watts what power does the shunt motor supply? 
 
SYNCHRONOUS GENERATORS 49 
 
 92. If the alternators described in problem 91 are mechanically 
 coupled together so that their excitation voltages differ in phase 
 by 30 degrees what will be the armature current when the 
 excitation voltages are each 300 volts? What is their terminal 
 voltage? What is the electrical output of the alternator which is 
 acting as a generator? If the rotational losses are supplied by 
 the shunt motor what is the mechanical output of the alternator 
 which is acting as a motor? What power does the shunt motor 
 supply if the rotational losses are 1620 watts? 
 
 93. Two identical, Y-connected, 60-cycle alternators are 
 rigidly coupled together and are driven at their rated speed of 
 1200 rev. per min. The alternators have revolving fields and 
 the coupling is so made that the north poles of the first are 10 
 degrees (mechanical) ahead, i.e., in the direction of rotation, of 
 the corresponding north poles of the second. The corresponding 
 terminals are connected through non-inductive resistances of 
 1.5 ohms each. The effective armature resistance and the 
 synchronous reactance of each alternator are respectively 0.302 
 and 4.36 ohms per phase. The field currents are adjusted so 
 that the excitation voltages are respectively 200 and 300 volts 
 per phase. What is the current? What is the electrical out- 
 put of the alternator which is acting as a generator? If the 
 rotational losses are supplied by the driving motor what is the 
 mechanical output of the alternator which is acting as a motor? 
 If the rotational losses are 1620 watts what is the output of the 
 driving motor? 
 
 94. Two identical, 3-phase, Y-connected alternators, rigidly 
 coupled to the same prime mover, are operating in parallel and 
 supply 1500 kw. at 0.83 power factor and a terminal potential 
 difference of 5000 volts. The effective armature resistance and 
 the synchronous reactance of each alternator are respectively 
 0.42 and 15.4 ohms per phase. The mechanical coupling is so 
 made that the excitation voltages are in phase, and the field 
 currents are adjusted so that one of these voltages is 50 per cent, 
 greater than the other. What is the output of each alternator? 
 
 95. Two 3-phase, Y-connected alternators, each of which is 
 rated to deliver 760 kv.-a. at 2200 volts, are rigidly coupled to the 
 same prime mover. Each of these alternators has 64 field poles, 
 an effective armature resistance of 0.172 ohm and a synchronous 
 reactance of 2.12 ohms per phase. With equal field excitations 
 they operate at full load with undue heating and on examination 
 
50 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 it is found that their corresponding field poles are displaced by 
 an angle of 1 degree and 6 minutes. 
 
 (a) When they deliver 1500 kw. at 0.87 power factor and 
 their rated voltage, what is the output of each alternator if the 
 field excitations are equal? 
 
 (b) If this displacement of the field poles is reduced to zero 
 what will be the reduction in the armature copper loss of each 
 alternator? The load delivered is still 1500 kw. at 0.87 power 
 factor and their rated voltage. 
 
 96. Two identical, 2-phase, 3750-kv.-a. water-wheel genera- 
 tors, operating in parallel, are driven by prime movers which have 
 the same speed-load characteristic. The effective armature resis- 
 tance and the synchronous reactance of these alternators are 
 respectively 0.0254 and 1.12 ohms per phase. The alternators 
 are jointly delivering 7500 kw. at unit power factor and their 
 rated terminal potential difference of 2200 volts. The excita- 
 tions are equal. 
 
 Instantaneous records show that due to hunting the maximum 
 displacement of their excitation voltages is 30 degrees. What is 
 the synchronizing power at the time of maximum displace- 
 ment? What is the maximum value of the effective armature 
 current? 
 
 97. If the alternators described in problem 96 are jointly 
 delivering 5250 kw. at 0.7 power factor and their rated voltage 
 to an inductive load, what is the synchronizing power at the 
 time the displacement of their excitation voltages is 30 degrees? 
 What are the effective armature currents at this time? 
 
 98. If the alternators described in problem 96 are jointly 
 delivering 5250 kw. at 0.7 power factor and their rated voltage 
 to a condensive load, what is the synchronizing power at the 
 time the displacement of their excitation voltages is 30 degrees? 
 What are the effective armature currents at this time? 
 
 99. Two identical 3-phase, Y-connected alternators, each of 
 which is rated to deliver 1640 kv.-a. at 13,500 volts are operating 
 in parallel. The speed-load characteristics of the prime movers 
 are the same and the load requires 70 amperes from each alter- 
 nator at 0.83 power factor and the rated voltage. The effective 
 armature resistance and the synchronous reactance of the alter- 
 nators are respectively 1.52 ohms and 37.4 ohms per phase. 
 Oscillograph records show that due to hunting the greatest 
 value of the effective armature current is 87 amperes. 
 
SYNCHRONOUS GENERATORS 51 
 
 What is the maximum displacement between the excitation 
 voltages? 
 
 What is the synchronizing power at this maximum displace- 
 ment? 
 
CHAPTER III 
 SYNCHRONOUS MOTORS 
 
 1. A 3-phase, 60-cycle, A-connected alternating-current 
 generator is rated to deliver 15 kv.-a. at 230 volts. The field 
 structure has 6 poles each of which is wound with 398 turns. 
 The armature core has 4 slots per pole per phase with 8 inductors 
 in series per slot. Each inductor is 5 in. long. The hot resist- 
 ance of the armature measured between any two terminals is 
 0.126 ohm. The effective resistance is 1.6 times the ohmic 
 resistance. In calculating the leakage reactance assume 8 
 leakage lines per ampere per inch of inductor. The open- and 
 short-circuit characteristics are given by the following data: 
 
 
 Open circuit 
 
 Short circuit 
 
 
 Terminal voltage 
 
 Armature current 
 
 2.0 
 4.0 
 5.0 
 6 
 
 104 
 199 
 240 
 275 
 
 22 
 
 44 
 55 
 
 7.5 
 
 315 
 
 
 This generator is running as an overexcited synchronous motor 
 and receives 15 kw. at its rated voltage and 0.8 power factor. 
 
 (a) Calculate the field current by the general method. 
 
 (b) Calculate the field current by the synchronous-impedance 
 method. 
 
 (c) Calculate the field current by the magnetomotive-force 
 method. 
 
 2. A 3-phase, 5000-volt, 25-cycle synchronous motor has a 
 full-load capacity of 1100 h.p. The field structure consists of 32 
 poles each of which is wound with 79.5 turns. The armature 
 core has 192 slots with 14 inductors in series per slot. The 
 armature winding is Y-connected. The effective resistance of 
 the armature winding is 0.38 ohm per phase. In calculating the 
 leakage reactance, assume 75 leakage lines per ampere per 
 inductor. The open- and short-circuit characteristics are given 
 by the following data: 
 
 52 
 
SYNCHRONOUS MOTORS 
 
 53 
 
 Field ampere- 
 
 Open circuit 
 
 Short circuit 
 
 turns per pole 
 
 Terminal voltage 
 
 Armature current 
 
 4,000 
 6,000 
 10,000 
 
 3200 
 4330 
 5460 
 
 108 
 162 
 
 12,000 
 
 5800 
 
 
 14,000 
 
 6060 
 
 
 This motor receives 760 kw. at its rated voltage and 0.83 power 
 factor. The excitation is less than normal. 
 
 (a) Calculate the field current by the general method. 
 
 (b) Calculate the field current by the synchronous-impedance 
 method. 
 
 (c) Calculate the field current by the magnetomotive-force 
 method. 
 
 3. A 3-phase, 5500-volt, 50-cycle synchronous motor has a 
 full-load capacity of 2000 h.p. The field structure has 72 
 poles each of which is wound with 35 turns. The armature core 
 has one slot per pole per phase with 12 inductors per slot. The 
 armature winding is connected in Y. The armature resistance 
 measured between any two terminals is 0.536 ohm, and the ratio 
 of effective to ohmic resistance is 1.35. In calculating the 
 leakage reactance assume 67 leakage lines per inductor per 
 ampere. The data for the open- and short-circuit characteristics 
 are: 
 
 
 Open circuit 
 
 Short circuit 
 
 Field current 
 
 Terminal voltage 
 
 Armature current 
 
 150 
 
 200 
 250 
 
 5100 
 5900 
 6500 
 
 300 
 
 400 
 
 300 
 
 6800 
 
 
 350 
 
 7100 
 
 
 This motor receives 960 kw. at its rated voltage and 0.68 power 
 factor. The excitation is greater than normal. 
 
 (a) Calculate the field current by the general method. 
 
 (b) Calculate the field current by the synchronous-impedance 
 method. 
 
 (c) Calculate the field current by the magnetomotive-force 
 method. 
 
 4. A 3-phase, 2200-volt, 50-cycle synchronous motor is rated 
 to deliver 1000 h.p. when operating at unit power factor. The 
 
54 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 field structure has 64 poles each of which is wound with 50 turns. 
 The resistance of the field circuit is 0.516 ohm. The armature 
 core has 384 slots with 3 inductors in series per slot. Each in- 
 ductor is 10 in. long. The hot resistance of the armature meas- 
 ured between terminals is 0.264 ohm. The ratio of effective 
 to ohmic resistance is 1.3. In calculating the leakage reactance 
 assume 6.5 lines per ampere per inch of inductor. The armature 
 windings are connected in Y. The open- and short-circuit 
 characteristics are given by the following data: 
 
 
 Open circuit 
 
 Short circuit 
 
 
 Terminal voltage 
 
 Armature current 
 
 40 
 50 
 60 
 
 1450 
 1840 
 2220 
 
 225 
 
 280 
 
 80 
 
 2860 
 
 
 100 
 
 3380 
 
 
 The rotational losses at normal voltage are 17.2 kw. 
 
 This motor deliveres 1000 h.p. and the excitation greater than 
 normal and is adjusted so that it is operating from a 2200-volt 
 circuit at a power factor of 0.93. 
 
 (a) Calculate the voltage impressed on the field circuit Use 
 the general method. 
 
 (b) Calculate the voltage impressed on the field circuit. Use 
 the synchronous-impedance method. 
 
 (c) Calculate the voltage impressed on the field circuit. Use 
 the magnetomotive-force method. 
 
 5. A 2-phase, 10,000-volt synchronous motor is rated to deliver 
 4500 h.p. when operating at unit power factor. The armature 
 has an effective resistance of 0.64 ohm per phase. The open- 
 and short-circuit characteristics are given by the following data: 
 
 Open circuit 
 
 Short circuit 
 
 rieia current 
 
 Terminal voltage 
 
 Armature current 
 
 100 
 200 
 250 
 
 7,700 
 10,200 
 10,900 
 
 220 
 430 
 
 300 
 
 11,500 
 
 
 375 
 
 12,200 
 
 
 This motor is delivering 3760 h.p. with an efficiency, exclusive 
 of field loss, of 95.6 per cent., and the field current is adjusted 
 so that it takes a leading current at 0.91 power factor. 
 
SYNCHRONOUS MOTORS 55 
 
 (a) Calculate the field current by the synchronous-impedance 
 method. 
 
 (b) Calculate the field current by the magnetomotive-force 
 method. 
 
 6. The full-load capacity of a 3-phase, 13,500-volt, synchro- 
 nous motor is 2200 h.p. The armature windings, which are 
 connected in Y, have an effective resistance of 1.52 ohm per 
 phase. The open- and short-circuit characteristics are given by 
 the following data: 
 
 
 Open circuit 
 
 Short circuit 
 
 Field current 
 
 Terminal voltage 
 
 Armature current 
 
 50 
 100 
 150 
 200 
 
 7,500 
 10,100 
 14,700 
 15,800 
 
 75 
 155 
 227 
 
 250 
 
 16,700 
 
 
 Calculate the field current by the synchronous impedance 
 method when this motor receives 1640 kv.-a. at 0.75 power 
 factor, (a) with excitation greater than normal, (b) with excita- 
 tion less than normal. 
 
 7. The synchronous motor described in problem 6 receives 
 1250 kw. at 0.75 power factor. 
 
 Calculate the field current by the magnetomotive-force 
 method (a) with excitation greater than normal, (b) with excita- 
 tion less than normal. 
 
 8. Concerning the alternator described in problem 1 the fol- 
 lowing additional data are given: The field current is supplied 
 at 110 volts. The friction and windage loss is 310 watts at 
 normal speed. The core loss is 280 watts, 480 watts, and 610 
 watts for generated armature voltages of 199 volts, 240 volts, 
 and 275 volts respectively. 
 
 What is the efficiency of this generator when it is running as 
 an overexcited synchronous motor and receives 15 kw. at 0.8 
 power factor? 
 
 Calculate the field current (a) by the general method, (b) by 
 the synchronous-impedance method, and (c) by the magneto- 
 motive-force method. 
 
 9. Concerning the synchronous motor described in problem 2 
 the following additional data are given: The resistance of the 
 field circuit is 0.82 ohm. The friction and windage loss at 
 
56 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 normal speed is 7.6 kw. The core loss due to rotation is 22.3 
 kw. at 5460 volts and may be assumed to vary as the square of 
 the generated armature voltage. What is the efficiency of 
 this motor under the conditions described in problem 2? 
 
 Calculate the field current (a) by the general method, (b) by 
 the synchronous-impedance method, and (c) by the magneto- 
 motive-force method. 
 
 10. Concerning the synchronous motor described in problem 
 3 the following additional data are given. The field circuit has 
 a resistance of 0.376 ohm. The friction and windage loss is 8.4 
 kw. The core loss due to rotation is 20.2 kw. at 5500 volts and 
 may be assumed to vary as the square of the generated armature 
 voltage. What is the efficiency of this motor under the conditions 
 described in problem 3 ? 
 
 Calculate the field current (a) by general method, (b) by the 
 synchronous-impedance method, and (c) by the magnetomotive- 
 force method. 
 
 11. A 3-phase, 2200-volt synchronous motor is rated to 
 deliver 1000 h.p. when operating at unit power factor. The 
 armature windings, which are connected in Y, have an effective 
 resistance of 0.172 ohm per phase. The resistance of the field 
 circuit is 0.576 ohm. The friction, windage and core losses are 
 17.2 kw. and may be assumed constant. The open- and short- 
 circuit characteristics are given by the following data: 
 
 . 
 
 Open circuit 
 
 Short circuit 
 
 
 Terminal voltage 
 
 Armature current 
 
 40 
 50 
 60 
 
 80 
 
 1450 
 1840 
 2220 
 2860 
 
 225 
 
 280 
 
 100 
 
 3380 
 
 
 What is the efficiency of this motor when it delivers 960 h.p. and 
 is operating at 0.83 power factor from a 2200-volt circuit (a) if 
 the excitation is greater than normal, (b) if the excitation is less 
 than normal? 
 
 Calculate the field current by the synchronous impedance 
 method. 
 
 12. A 2-phase, 2200-volt synchronous motor has a full-load 
 capacity of 5000 h.p. when operating at unit power factor. The 
 armature has an effective resistance of 0.0255 ohm per phase. 
 
SYNCHRONOUS MOTORS 
 
 57 
 
 The resistance of the field circuit is 0.37 ohm. The friction and 
 windage loss is 19.0 kw. The core loss due to rotation is 45.2 
 kw. at 2200 volts and may be assumed constant. The open- 
 and short-circuit characteristics are given by the following data: 
 
 Field current 
 
 Open circuit 
 
 Short circuit 
 
 
 Terminal voltage 
 
 Armature current 
 
 66.7 
 133.0 
 200 
 
 680 
 1360 
 2000 
 
 606 
 1210 
 
 267 
 
 2480 
 
 
 300.0 
 
 2660 
 
 
 What is the efficiency of this motor when it receives 3200 kw. 
 at 0.80 power factor (a) if the excitation is greater than normal, 
 (b) if the excitation is less than normal? 
 
 Calculate the field current by the magnetomotive-force 
 method. 
 
 13. The following test data are given on a 1340-h.p., 11,000- 
 volt, 3-phase synchronous motor. The armature effective resist- 
 ance is 0.94 ohm per phase, and the resistance of the field winding 
 is 3.11 ohms. The armature windings are connected in Y. 
 
 Field 
 current 
 
 Open circuit ter- 
 minal voltage ( V) 
 
 Core loss 
 and V 
 
 Terminal voltage 
 for/ = 60, PF = Q 
 
 20 
 
 8,400 
 
 11,600 
 
 
 30 
 40 
 50 
 
 11,000 
 12,700 
 13 800 
 
 19,400 
 25,600 
 
 9 800 
 
 60 
 70 
 
 14,700 
 15.500 
 
 
 11,100 
 12.100 
 
 The friction and windage loss is 12.1 h.p. 
 
 What is the efficiency of this motor when it receives 68 amperes 
 per terminal at 11,550 volts if the excitation is greater than 
 normal and is adjusted so that the power factor is 78.4 per cent.? 
 
 14. The following test data are given on a 6500-h.p., 6600- 
 volt, 3-phase synchronous motor. The effective resistance of 
 the armature winding is 0.081 ohm per phase, and the resistance 
 of the field winding is 0.549 ohm. The armature windings are 
 connected in Y. 
 
58 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 Field 
 current 
 
 Open circuit ter- 
 minal voltage (V) 
 
 Core loss 
 and V 
 
 Terminal voltage 
 7a = 438, PF=0 
 
 100 
 150 
 
 200 
 250 
 
 4800 
 6470 
 7400 
 7930 
 
 45 kw. 
 85 kw. 
 120 kw. 
 
 1750 
 4200 
 5700 
 6600 
 
 275 
 
 8150 
 
 
 6900 
 
 The friction and windage loss is 62.8 h.p. 
 
 What is the efficiency of this motor when it receives 450 
 amperes at unit power factor, and its rated voltage? 
 
 15. The following test data are given on a 1340-h.p., 13,800- 
 volt, 3-phase synchronous motor. The armature windings, 
 which are connected in Y, have an effective resistance of 2.18 
 ohms per phase. The resistance of the field circuit is 0.541 ohm. 
 
 Field 
 current 
 
 Open circuit ter- 
 minal voltage (V) 
 
 Terminal voltage 
 la = 42, PF = 
 
 Core loss 
 and V 
 
 50 
 
 8,800 
 
 
 7 5 
 
 80 
 110 
 140 
 180 
 
 13,000 
 15,600 
 17,250 
 
 18.900 
 
 . 
 
 10,750 
 13,250 
 15.600 
 
 16.6 
 25.4 
 33.5 
 
 The friction and windage loss is 11.2 h.p. 
 
 What is the efficiency of this motor when it delivers 1200 h.p. 
 if the excitation is greater than normal and is adj usted so that 
 the motor takes an armature current of 47.6 amperes at a 
 terminal voltage of 13,500 volts? 
 
 16. A synchronous motor, whose armature windings are 
 connected in Y, has an effective resistance of 0.94 ohm and a 
 synchronous reactance of 56 ohms per phase. This motor 
 receives a line current of 60 amperes at a terminal potential 
 difference of 11,000 volts and the excitation is adjusted so that 
 the power factor is 0.85. 
 
 What power does the motor receive? What is the exci- 
 tation voltage (a) if the excitation is greater than normal, 
 (b) if the excitation is less than normal? 
 
 17. A 2-phase synchronous motor has an effective armature re- 
 sistance of 0.64 ohm and a synchronous reactance of 23.7 ohms 
 per phase. The motor receives 3000 kw. at a line potential 
 difference of 10,000 volts and the excitation adjusted so that 
 the line current is 200 amperes. 
 
SYNCHRONOUS MOTORS 59* 
 
 At what power factor is this motor operating? What is the 
 excitation voltage (a) if the excitation is greater than normal, 
 (b) if the excitation is less than normal? 
 
 18. A 5000-h.p. synchronous motor is operated from a 2200- 
 volt, 2-phase circuit. The effective resistance of the armature 
 is 0.0254 ohm per phase. The field winding has 60 turns per pole 
 and a total resistance of 0.37 ohm. 
 
 Ampere turns per 
 pole 
 
 Open circuit terminal 
 voltage 
 
 Short circuit armature 
 current 
 
 4,000 
 8,000 
 12,000 
 
 680 
 1360 
 2000 
 
 606 
 1210 
 
 16,000 
 18,000 
 
 2480 
 2660 
 
 
 At no load this motor takes 67.5 kw. at 2200 volts when the 
 excitation is adjusted so that it operates at 0.86 power factor. 
 How much must the excitation be increased in order that the 
 motor will deliver its rated load and operate at this same power 
 factor? Use the magnetomotive-force method for calculating 
 the field current. 
 
 19. A 3-phase synchronous motor, whose armature windings 
 are connected in A, has an effective resistance of 0.302 ohm and 
 a synchronous reactance of 4.36 ohms per phase. This motor 
 receives a line current of 40 amperes at 230 volts and the field 
 current is adjusted so that the excitation voltage is 310 volts. 
 
 What power does the motor receive? At what power factor 
 is it operating? 
 
 20. A 2400-volt, 3-phase, synchronous motor has a full-load 
 capacity of 1340 h.p. The effective resistance of the armature 
 is 0.067 ohm per phase and the synchronous reactance is 2.64 
 ohms per phase. The armature windings are connected in Y. 
 The resistance of the field circuit is 0.427 ohm. The rotational 
 losses at normal voltage are 26.6 kw. 
 
 This motor is operated from a 2400-volt circuit at 0.82 power 
 factor and is delivering 1220 h.p. What is the line current, 
 and what is the necessary excitation voltage? Calculate the 
 excitation voltage for both under- and overexcitation. 
 
 21. A 135-h.p. synchronous motor is operated from a 480-volt, 
 2-phase circuit. The effective resistance of the armature is 
 0.135 ohm per phase at normal running temperature. The 
 rotational losses at normal voltage are 5.7 kw. 
 
60 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 Field current Terminal voltage on open ; Armature current on 
 circuit short circuit 
 
 10 
 17 
 
 397 
 527 
 
 158 
 
 25 
 
 600 
 
 
 35 
 
 655 
 
 
 With the greatest allowable excitation the motor takes a 
 current of 135 amperes when delivering its rated load. At what 
 power factor does it then operate? What is the field current? 
 Use what you consider the best method for calculating the field 
 current. 
 
 22. A 3-phase synchronous motor, whose armature windings are 
 connected in Y, has an effective resistance of 0.172 ohm and a 
 synchronous reactance of 2.12 ohms per phase. This motor 
 receives 750 kw. at a line potential difference of 2200 volts, 
 and the field current is adjusted so that the excitation voltage 
 is 2800 volts. 
 
 What is the line current? At what power factor is the motor 
 operating? 
 
 23. A 2200-volt, 2-phase synchronous motor has a full-load 
 capacity of 5000 h.p. The full-load efficiency of the armature, 
 with the field current adjusted for unit power factor, is 96.1 per 
 cent. The armature winding has an effective resistance of 
 0.025 ohm and a synchronous reactance of 1.12 ohms per phase. 
 The motor receives a constant power of 3000 kw. at 2200 volts. 
 If the current is limited to 130 per cent, of its full-load value, 
 what is the greatest allowable excitation voltage? At what 
 power factor would the motor be operating? 
 
 24. A 3-phase, 13,500-volt synchronous motor is rated to 
 deliver 2200 h.p. when operating at unit power factor. The 
 armature has an effective resistance of 1.52 ohms per phase. 
 The armature windings are connected in Y. 
 
 . . . Open circuit terminal Short circuit armature 
 
 Field current 
 
 voltage current 
 
 50 
 
 7,500 
 
 75 
 
 100 
 
 10,100 
 
 155 
 
 150 
 
 14,700 
 
 227 
 
 The rotational losses at normal voltage are 68 h.p. 
 
 If the maximum allowable current is 125 per cent, of the full- 
 
SYNCHRONOUS MOTORS 61 
 
 load current over what range should it be possible to vary the 
 field current when the motor is delivering a constant load of 
 2000 h.p.? Use the magnetomotive-force method for calculat- 
 ing the field current. 
 
 25. A 3-phase, 230-volt, synchronous motor, whose armature 
 windings are connected in A, has an effective resistance of 0.302 
 ohm and a synchronous reactance of 4.36 ohms per phase. The 
 rotational losses are 750 watts and may be assumed constant. 
 The motor delivers a constant load of 20 h.p. What is the 
 least excitation voltage with which the motor will run? What 
 is the armature current at the instant of breakdown? 
 
 26. A 1340-h.p., 3-phase synchronous motor receives a con- 
 stant power of 860 kw. from an 11,000-volt circuit. The arma- 
 ture windings, which are connected in Y, have an effective 
 resistance of 0.94 ohm and a synchronous reactance of 56 ohms 
 per phase. 
 
 Over what range can the excitation voltage be varied so that 
 the current will not exceed 135 per cent, of its full-load value, 
 which is 62.5 amperes? 
 
 27. A 3-phase, 5000-volt synchronous motor, whose armature 
 windings are connected in Y, has an effective resistance of 0.40 
 ohm and a synchronous reactance of 15.4 ohms per phase. The 
 rotational losses are 30 kw. and may be assumed constant. The 
 greatest excitation voltage that can be obtained is 3520 volts per 
 phase. 
 
 When the motor is delivering its full load of 1100 h.p. over 
 what range can the power factor be varied? What is the arma- 
 ture current at each of the limiting conditions? Compare these 
 with the full-load armature current. 
 
 28. Neglecting the field copper loss, the efficiency of a 2200- 
 h.p., 3-phase synchronous motor is 96.0 per cent, at full load 
 when the motor is operating at unit power factor. Assume that 
 the rotational losses are constant. The armature windings, 
 which are connected in Y, have an effective resistance of 1.52 
 ohms and a synchronous reactance of 37.4 ohms per phase. 
 With the motor delivering 2000 h.p. the field current is adjusted 
 so that the motor takes a leading current of 85 amperes from a 
 13,500-volt constant potential circuit. 
 
 (a) At what power factor is the motor operating? What is 
 the excitation voltage? 
 
 (b) If this load is thrown off what current will the motor take, 
 
62 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 and at what power factor will it be operating? The excitation 
 is unchanged. 
 
 29. The rotational losses of a 1340-h.p., 2400-volt, 3-phase 
 synchronous motor are 43 h.p., and may be assumed constant. 
 The armature windings, which are connected in Y, have an 
 effective resistance of 0.067 ohm and a synchronous reactance 
 of 2.64 ohms per phase. 
 
 (a) What is the least power factor at which the motor can be 
 operated at no load so that the current will not exceed 135 per 
 cent, of its full-load value? If the motor is overexcited what is 
 the excitation voltage? 
 
 30. When operating at unit power factor the full-load losses 
 of a 1100-h.p., 5000-volt, 3-phase synchronous motor are: 
 Armature copper loss=13.9 kw: Field copper loss=18.6 kw: 
 Rotational losses = 27. 8 kw. Assume that the rotational losses 
 are constant. The armature windings, which are connected in 
 Y, have a synchronous impedance of 15.4 ohms per phase. 
 
 If the current is limited to 130 per cent, of its full-load value 
 what is the least power factor at which this motor can be oper- 
 ated when it is delivering full load? What is the necessary 
 excitation voltage if the motor is overexcited? 
 
 31. The synchronous motor described in problem 14 is rated 
 to deliver 6500 h.p. when operating at unit power factor. If 
 the maximum allowable current is 130 per cent, of its full-load 
 value what is the least power factor at which it can operate when 
 delivering its rated load? What is the greatest allowable value 
 of the field current at full load under this condition? 
 
 32. A 20-h.p., 230-volt, 3-phase synchronous motor has an 
 effective resistance of 0.302 ohm and a synchronous reactance of 
 4.36 ohms per phase. The armature windings are connected 
 in A. The rotational losses are 750 watts and may be assumed 
 constant. 
 
 With the maximum excitation voltage of 315 volts at what 
 load will this motor break down? Compare the current at 
 breakdown with the full-load current for normal excitation. 
 
 33. If the ratio of resistance to synchronous reactance is 
 increased to 0.30 by inserting equal resistances in series with 
 each phase of the synchronous motor what will be the results 
 called for (a) in problem 25; (b) in problem 26; (c) in problem 
 27; (d) in problem 29; (e) in problem 30; (f) in problem 32. 
 
 34. A 3-phase synchronous motor receives line currents of 
 
SYNCHRONOUS MOTORS 63 
 
 226 amperes at a terminal potential difference of 2180 volts and 
 delivers 812 h.p. The no-load rotational losses are 23 h.p., and 
 may be assumed constant. The armature windings, which are 
 Y-connected, have an effective resistance of 0.172 ohm and a 
 synchronous reactance of 2.12 ohms per phase. 
 
 At what power factor is the motor operating? Neglecting the 
 field copper losses, what is the efficiency? 
 
 35. At no load a 3-phase synchronous motor takes line currents 
 of 1.52 amperes at 11,000 volts if the field current is adjusted for 
 a minimum armature current. The armature windings, which 
 are connected in Y, have an effective resistance of 0.94 ohm and 
 a synchronous reactance of 56 ohms. 
 
 When the motor receives 960 kw. at 11,000 volts and the 
 excitation is adjusted so that the line current is 72 amperes, what 
 is the output? If the excitation is greater than normal, what is 
 the excitation voltage? 
 
 36. The no-load rotational losses of a 1340-h.p., 3-phase syn- 
 chronous motor are 34.6 kw. The armature windings are con- 
 nected in Y and have an effective resistance of 0.067 ohm and a 
 synchronous reactance of 2.64 ohms per phase. This motor is 
 operated from a 2400-volt circuit and when delivering its rated 
 output the excitation is adjusted so that the motor takes a leading 
 current of 0.83 power factor. 
 
 What is the power input? What is the excitation voltage? 
 
 37. At no load a 150-h.p., 2-phase synchronous motor takes 
 7.6 kw. at 480 volts when the excitation is adjusted for a minimum 
 armature current. Under load this motor takes 108 amperes at 
 485 volts when the excitation is adjusted so that the power factor 
 is 0.78. The effective resistance of the armature is 0.118 ohm 
 and the synchronous reactance is 2.13 ohms. 
 
 What is the output? What is the excitation voltage (a) if 
 the field current is greater than normal? (b) if the field current 
 is less than normal? 
 
 38. A 3-phase, 6600-volt synchronous motor is operated at 
 its rated voltage and delivers its rated output of 6500 h.p. 
 In order that the motor shall take a large leading current the 
 field current is adjusted for an excitation voltage of 8100 volts. 
 The no-load rotational losses are 138 kw. and may be assumed 
 constant. The armature windings are connected in Y and have 
 an effective resistance of 0.081 ohm and a synchronous reactance 
 of 2.8 ohms. 
 
64 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 What is the power factor? What current does the motor 
 take? 
 
 39. A 2-phase, 10,000-volt synchronous motor has a rated 
 capacity of 4500 h.p. The armature has an effective resistance 
 of 0.64 ohm and a synchronous reactance of 23.7 ohms per phase. 
 At no load with an excitation greater than normal the motor 
 takes 74 kw. at 10,000 volts and the line current is 41 amperes. 
 
 With the field excitation unchanged what current will the 
 motor take when it is delivering its rated output? At what 
 power factor will the motor be operating? 
 
 40. At the end of a 3-phase transmission line are induction 
 motors which take a total load of 1200 kw. at 0.87 power factor. 
 An additional load of 600 h.p. should be provided for. Find the 
 kilovolt-ampere capacity of a synchronous motor that will 
 supply this load and will at the same time make it possible to 
 adjust the power factor of the entire load to unity. Assume that 
 the efficiency of the synchronous motor is 92 per cent. 
 
 41. An induction motor load at the end of a 3-phase trans- 
 mission line takes 6000 kw. at 0.90 power factor. A synchronous 
 motor is operating in parallel with the induction motors in order 
 to improve the power factor. The motor has a full-load capacity 
 of 6500 h.p. when operating at unit power factor from a 6600-volt 
 circuit. The armature windings, which are connected in Y, have 
 an effective resistance of 0.081 ohm per phase. The synchronous 
 impedance is 3.57 ohms per phase. The friction and windage 
 losses are 62.8 h.p. and the core losses at normal voltage are 86 
 kw. The latter may be assumed to be constant. The synchro- 
 nous motor is operated so that the resultant power factor of the 
 load is unity, and the line voltage is 6600 volts. If the current 
 is limited to 125 per cent, of its full-load value what is the 
 greatest power that the synchronous motor can supply? What 
 is the necessary excitation voltage of the synchronous motor at 
 this time? 
 
 42. Induction motors at the end of a 3-phase transmission 
 line deliver a total power of 3500 h.p. and operate at a resultant 
 efficiency and power factor of 91.6 per cent, and 0.903 respectively. 
 The line voltage is 13,200 volts. A 1640-kv.-a. synchronous 
 motor is operated in parallel with these motors to improve the 
 power factor of the load, and to supply an additional load of 
 1000 h.p. The armature windings of this synchronous motor 
 are connected in Y and have an effective resistance of 1.52 ohms 
 
SYNCHRONOUS MOTORS 65 
 
 per phase. The rotational losses at normal voltage are 57 h.p. 
 and may be assumed to be constant. 
 
 Field current 
 
 Open circuit terminal 
 voltage 
 
 Short circuit armature 
 current 
 
 50 
 100 
 150 
 200 
 
 7,500 
 10,100 
 14,700 
 15,800 
 
 75 
 155 
 227 
 
 250 
 
 16,700 
 
 
 At what per cent, of its rated capacity must the synchronous 
 motor be operated so that the power factor of the entire load shall 
 be unity and the line voltage 13,200 volts? Calculate the neces- 
 sary field current of the motor for this condition by the magneto- 
 motive-force method. 
 
 43. A 1000-h.p. synchronous motor is operating at the end 
 of a 3-phase transmission line which has a resistance of 0.58 ohm 
 and a reactance of 0.64 ohm per conductor. The motor is over- 
 excited so that it takes a line current of 216 amperes at a power 
 factor of 0.82, and with a line potential difference of 2210 volts. 
 
 What is the line voltage at the generating station? What is 
 the efficiency of transmission? 
 
 44. A synchronous motor operating at the end of a 3-phase" 
 transmission line takes a constant power of 1500 kw. The 
 resistance and reactance of the line are respectively 2.2 ohms and 
 2.6 ohms per conductor. The line voltage at the generating 
 station is maintained at 5740 volts. The excitation of the 
 motor is adjusted so that the line loss has its least possible value. 
 
 What is the terminal voltage at the motor? What is the 
 efficiency of transmission? 
 
 45. A synchronous motor operating at the end of a 3-phase 
 transmission line which has a resistance of 3.6 ohm and a re- 
 actance of 4.1 ohm per conductor takes a constant power of 
 720 kw. from the line and the excitation is adjusted so that the 
 line voltage is 5000 volts. The line voltage at the generating 
 station is also maintained at 5000 volts. 
 
 (a) What current does the motor take? At what power 
 factor is it operating? 
 
 (b) If the motor is Y-connected and has an effective resist- 
 ance of 0.40 ohm and a synchronous reactance of 15.4 ohms per 
 phase, what is the necessary excitation voltage? 
 
66 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 46. A 1200-h.p. synchronous motor is operated at the end of 
 a 3-phase transmission line which has a resistance of 3.3 ohms 
 and a reactance of 3.7 ohms per conductor. The line voltage 
 at the generating station is 5000 volts, and the motor delivers 
 energy to a constant load of 1000 h.p. The rotational losses of 
 the motor are 29.8 kw. and may be assumed constant. The 
 armature windings are connected in Y and have a resistance of 
 0.40 ohm and a synchronous reactance of 15.4 ohms per phase. 
 
 If the maximum allowable line current is 135 amperes wnat is 
 the greatest possible terminal voltage at the motor? What is 
 the necessary excitation voltage of the motor? 
 
 47. At the end of a transmission line which has a resistance of 
 12.2 ohms and a reactance of 17.4 ohms per conductor there is a 
 synchronous motor which delivers a constant load of 1300 h.p. 
 The armature windings of the motor, which are connected in Y, 
 have an effective resistance of 0.94 ohm and a synchronous 
 reactance of 56 ohms per phase. At no load with an impressed 
 voltage of 11,000 volts the motor takes a current of 1.52 amperes 
 when the field current is adjusted for unit power factor. The 
 line voltage at the generating station is maintained constant at 
 11,500 volts. 
 
 If the line current is limited to 120 per cent, of its full-load 
 value, which is 62.5 amperes, what is the least additional react- 
 ance that should be inserted in each line so that a terminal poten- 
 tial difference of 12,000 volts may be obtained at the motor? 
 What is the necessary excitation voltage of the motor in this 
 case? 
 
 48. A 13,800-volt, 3-phase synchronous motor has a full-load 
 capacity of 1350 h.p. The armature windings are connected 
 in Y and have an effective resistance of 2.18 ohms and a syn- 
 chronous reactance of 92 ohms per phase. At no load with an 
 impressed voltage of 13,800 volts the motor takes 28.2 kw. at 
 unit power factor when the excitation is normal. This motor 
 is operated at the end of a transmission line which has a resist- 
 ance of 22 ohms and a reactance of 28 ohms per conductor. 
 The line voltage at the generating station is maintained con- 
 stant at 14,000 volts. 
 
 When the motor is delivering 1200 h.p. what is the greatest 
 potential difference at the motor if the armature current is 
 limited to 125 per cent, of its full-load value? What is the neces- 
 sary excitation voltage of the motor? 
 
SYNCHRONOUS MOTORS 67 
 
 49. At the end of a 3-phase transmission line which has a 
 resistance of 0.62 ohm and a reactance of 0.64 ohm per conductor 
 there is a synchronous motor whose armature windings are con- 
 nected in Y and have an effective resistance of 0.172 ohm and a 
 synchronous reactance of 2.12 ohms per phase. The rotational 
 losses of the motor are 17.2 kw. and may be assumed constant. 
 
 When instruments in the generating station indicate that the 
 transmission line is receiving 780 kw. at 2250 volts and a power 
 factor of 0.88 (leading), what is the output of the motor? What 
 is the line voltage at the motor? 
 
 50. A 3-phase transmission line has a resistance of 12.6 ohms 
 and a reactance of 16.4 ohms per conductor. The generating 
 station which delivers energy to this line maintains a constant 
 line potential difference of 14,000 volts. At the end of the line 
 there is a synchronous motor whose armature windings are 
 connected in Y and have an effective resistance of 1.52 ohms 
 and a synchronous reactance of 37.4 ohms per phase. The 
 rotational losses of this motor are 36.8 kw. and may be assumed 
 constant. When the motor is delivering 2000 h.p. and the 
 excitation voltage is adjusted to its greatest value of 16,700 
 volts what is the terminal voltage of the motor? At what 
 power factor is the motor operating? 
 
CHAPTER IV 
 INDUCTION MOTORS 
 
 1. A 2-phase induction motor has a stator with 2 slots per 
 pole per phase. There are 4 inductors in series per slot. The 
 coil pitch is 4 slots. The effective value of the current in each 
 phase of the stator winding is 7.07 amperes, and the current 
 in phase two lags behind the current in phase one by 90 electrical 
 degrees. Place the first slot in phase one at the extreme left 
 of the paper and draw the zero lines to allow for a maximum 
 ordinate of 2 in. Use the following scales in these plots. 
 
 Abscissae Pole pitch = 3 in. 
 
 Ordinates 40 ampere turns = 1 in. 
 
 (a) Plot the distribution of the magnetomotive force in the 
 air-gap due to the stator currents at the time that the current in 
 phase one is a maximum. 
 
 (b) Plot the distribution of magnetomotive force in the air- 
 gap at one-eighth of a period later than in (a) . 
 
 (c) Plot the distribution of the magnetomotive force in the 
 air-gap at one-quarter of a period later than in (a). 
 
 2. A 2-phase induction motor has a stator with 4 slots per 
 pole per phase. There are 4 inductors in series per slot. The 
 coil pitch is 8 slots. The effective value of the current in each 
 phase of the stator winding is 7.07 amperes, and the current in 
 phase two lags behind the current in phase one by 90 electrical 
 degrees. Place the first slot in phase one at the extreme left of 
 the paper and draw the zero lines to allow for a maximum or- 
 dinate of 3 in. Use the following scales in this plots : 
 
 Abscissae Pole pitch = 3 in. 
 
 Ordinates 40 ampere turns = 1 in. 
 
 (a) Plot the distribution of the magnetomotive force in the 
 air-gap due to the stator currents at the time that the current in 
 phase one is a maximum. 
 
 Note. Unless otherwise stated assume that the ratio of the turns in the 
 stator and rator windings is the same as the ratio of transformation. 
 
 68 
 
INDUCTION MOTORS 69 
 
 (b) Plot the distribution of magnetomotive force in the air-gap 
 at one-eighth of a period later than in (a). 
 
 (c) Plot the distribution of the magnetomotive force in the 
 air-gap at one-quarter of a period later than in (a). 
 
 3. A 3-phase induction motor has a stator with 2 slots per 
 pole per phase. There are 4 inductors in series per slot. The 
 coil pitch is 6 slots. The effective value of the current in each 
 phase of the stator winding is 7.07 amperes, and the current 
 in phase two lags behind the current in phase one by 120 elec- 
 trical degrees and leads the current in phase three by the same 
 amount. Place the first slot in phase one at the extreme left 
 of the paper and draw the zero lines to allow for a maximum 
 ordinate of 2 in. Use the following scales in these plots: 
 
 Abscissae Pole pitch = 3 in. 
 
 Ordinates 40 ampere turns = 1 in. 
 
 (a) Plot the distribution of the magnetomotive force in the 
 air-gap due to the stator currents at the time that the current in 
 phase one is a maximum. 
 
 (b) Plot the distribution of the magnetomotive force in the 
 air-gap at one-eighth of a period later than in (a). 
 
 (c) Plot the distribution of the magnetomotive force in the 
 air-gap at one-quarter of a period later than in (a). 
 
 4. A 3-phase induction motor has a stator with 4 slots per pole 
 per phase. There are 4 inductors in series per slot. The coil 
 pitch is 12 slots. The effective value of the current in each 
 phase of the stator winding is 7.07 amperes, and the current in 
 phase two lags behind the current in phase one by 120 electrical 
 degrees and leads the current in phase three by the same amount. 
 Place the first slot in phase one at the extreme left of the paper 
 and draw the zero lines to allow for a maximum ordinate of 
 4 in. Use the following scales in these plots: 
 
 Abscissas Pole pitch = 3 in. 
 
 Ordinates 40 ampere turns = 1 in. 
 
 (a) Plot the distribution of the magnetomotive force in the air- 
 gap due to the stator currents at the time that the current in phase 
 one is a maximum. 
 
 (b) Plot the distribution of the magnetomotive force in the 
 air-gap at one-eighth of a period later than in (a). 
 
70 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 (c) Plot the distribution of the magnetomotive force in the 
 air-gap at one-quarter of a period later than in (a). 
 
 6. A 3-phase induction motor has a stator with 4 slots per pole 
 per phase. There are 4 inductors per slot, two of which are in 
 one coil and two in another. The coil pitch is 10 slots. There are 
 two coil sides in each slot. The effective value of the current in 
 each phase of the stator winding is 7.07 amperes, and the current 
 in phase two lags behind the current in phase one by 120 electrical 
 degrees and leads the current in phase three by the same amount. 
 Place the first slot in phase one at the extreme left of the paper and 
 draw the zero lines to allow for a maximum ordinate of 4 in 
 Use the following scales in these plots: 
 
 Abscissae Pole pitch = 3 in. 
 
 Ordinates 40 ampere turns = 1 in. 
 
 (a) Plot the distribution of the magnetomotive force in the air- 
 gap due to the stator currents at the time that the current in phase 
 one is a maximum. 
 
 (b) Plot the distribution of the magnetomotive force in the 
 air-gap at one-eighth of a period later than in (a). 
 
 (c) Plot the distribution of the magnetomotive force in the 
 air-gap at one-quarter of a period later than in (a). 
 
 6. Assume that the stator winding of a 3-phase induction motor 
 is uniformly distributed. The coil pitch is unity and the phase 
 spread is one-third the polar pitch. The current in phase two 
 lags behind the current in phase one by 120 degrees and leads 
 the current in phase three by the same amount. Place the be- 
 ginning of phase one at the extreme left of the paper and draw 
 the zero lines to allow for a maximum ordinate of 3 in. As 
 ordinates let 1 in. equal the maximum ampere turns per pole 
 per phase, and as abscissae let 3 in. equal the pole pitch. 
 
 (a) Plot the distribution of the magnetomotive force in the 
 air-gap at the time that the current in phase one is a maximum. 
 
 (b) Plot the distribution of the magnetomotive force in the 
 air-gap at one-eight of a period later than in (a) . 
 
 (c) Plot the distribution of the magnetomotive force in the 
 air-gap at a time one-quarter of a period later than in (a). 
 
 7. Assume that the stator winding of a 3-phase induction 
 motor is uniformly distributed. The coil pitch is unity and 
 the phase spread is two-thirds the polar pitch. The current in 
 phase J;wo lags behind the current in phase one by 120 degrees 
 
INDUCTION MOTORS 71 
 
 and leads the current in phase three by the same amount. Place 
 the beginning of phase one at the extreme left of the paper and 
 draw the zero lines to allow for a maximum ordinate of 3 in. As 
 ordinates let 1 in. equal the maximum ampere turns per pole per 
 phase, and as abscissae let 3 in. equal the pole pitch. 
 
 (a) Plot the distribution of the magnetomotive force in the 
 air-gap at the time that the current in phase one is a maximum. 
 
 (b) Plot the distribution of the magnetomotive force in the 
 air-gap at one-eighth of a period later than in (a). 
 
 (c) Plot the distribution of the magnetomotive force in the 
 air-gap at a time one-quarter of a period later than in (a) . 
 
 8. At no load a 25 h.p. induction motor takes 865 watts at 
 0.18 power factor from a 250- volt, 3-phase circuit. The effective 
 resistance of the stator winding is 0.15 ohms per phase. Neg- 
 lect the leakage reactance. What would have been the no-load 
 power and power" factor if the motor had been designed with 
 an air-gap which would have made the reluctance of the magnetic 
 circuit 25 per cent, greater? 
 
 9. A 200-h.p. induction motor is designed to operate from a 
 3-phase, 980-volt, 20-cycle circuit. Under this condition the 
 no-load power and power factor are 6.62 kw. and 0.086. The 
 friction and windage loss is 4.0 kw. The stator winding is delta- 
 connected and has an effective resistance of 0.387 ohms per 
 phase. What will be the no-load power and power factor if 
 this motor is operated from a 3-phase, 1100-volt, 25-cycle cir- 
 cuit? Neglect the leakage reactance and assume that the 
 core losses vary as B 1 ' 7 / 1 ' 3 
 
 10. At no load a 570-h.p., 3-phase induction motor takes 
 15.1 kw. at 0.081 power factor. The friction and windage 
 loss is 12.0 kw. The effective resistance of the stator winding 
 is 0.757 ohm per phase. The windings are connected in delta. 
 There are 8 inductors in series per slot. If the motor is rewound 
 with the same size wire but with 6 inductors in series per 
 slot, what no-load current and power will it take from the same 
 circuit? Neglect the leakage reactance and assume that the core 
 losses vary as B 1 ' 7 / 1-3 . 
 
 11. Both the stators and rotors of two 150-h.p., 3-phase in- 
 duction motors are of the same design except that one is wound 
 for 6 poles and the other for 8 poles. Each stator has the same 
 number of slots with the same number of inductors per slot. 
 The six-pole motor takes 5.7 kw. from a 500 volt, 38-cycle circuit 
 
72 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 at no load and the line current is 46 amperes. The friction and 
 windage loss is 3.9 kw. The effective resistance of the stator 
 winding is 0.082 ohm per phase. The effective resistance of the 
 stator winding of the 8-pole motor is 11 per cent, less than this. 
 The windings are connected in Y. What current and power 
 will the 8-pole motor take at no load from the same circuit? 
 Assume that the friction and windage loss varies as the speed 
 and neglect the leakage reactance. Where approximations are 
 made state what they are. 
 
 12. At no load a 335-h.p., 3-phase induction motor takes a 
 line current of 15.5 amperes and absorbs 10.1 kw. from a 2000- 
 volt, 50-cycle circuit. The effective resistance and the leakage 
 reactance of the stator winding are respectively, 0.248 and 0.76 
 ohm per phase. The windings are connected in Y. The friction 
 and windage loss is 2.5 kw. What current and power will this 
 motor take at no load from a 2200-volt, 60-cycle circuit? As- 
 sume that the core losses vary as B 1 ' 1 } 1-3 and that the generated 
 voltage due to the air-gap flux is equal to the impressed voltage 
 less the magnetizing current multiplied by the leakage reactance. 
 
 13. At no load a 150-h.p., 3-phase induction motor takes a 
 line current of 46 amperes and absorbs 5.65 kw. from a 500- volt, 
 38-cycle circuit. The friction and windage loss is 4.1 kw. The 
 stator has 12 slots per pole and the winding pitch is 12 slots. 
 If this motor is rewound with the same number of turns using 
 a pitch of 10 slots, what current and power will it take at no 
 load from the same circuit? Neglect the resistance and the 
 leakage reactance, and assume that the core losses vary as B 1 ' 7 / 1- 3 
 
 14. The stator of a 150-h.p., 3-phase, 500-volt induction 
 motor has 108 slots, and in each of the following cases it is wound 
 for 6 poles with 4 inductors per slot. With a winding pitch 
 of 18 slots the core losses are 3.9 kw. and the magnetizing cur- 
 rent is 45 amperes. If a winding pitch of 14 slots is used what 
 will be the core losses and the magnetizing current at the same 
 voltage and frequency? 
 
 15. The stator of a 25-h.p., 3-phase, 250-volt induction 
 motor has 72 slots, and in each of the following cases it is wound 
 with 8 inductors per slot. If this motor is wound for 8 poles 
 with a winding pitch of 8 slots the core losses are 590 watts, 
 the friction and windage loss is 220 watts, and the magnetizing 
 current is 11 amperes. What current and power will this motor 
 take from the same circuit at no load if it is wound for 6 poles 
 
INDUCTION MOTORS 73 
 
 with a winding pitch of 10 slots? Neglect the resistance and 
 leakage reactance and assume that the friction and windage loss 
 varies as the speed and that the core losses vary as B i- 7 / 1 - 3 . 
 
 16. The stator of a 25-h.p., 3-phase, 50-cycle induction motor 
 is wound for 250 volts. With an impressed voltage of 110 volts 
 the starting current is 61 amperes at 0.336 power factor. The ef- 
 fective resistance and leakage reactance of the stator winding are 
 respectively 0.159 ohm and 0.46 ohm per phase. The stator 
 has 72 slots with 9 inductors per slot and the rotor has 120 slots 
 with 2 inducters per slot. Both the stator and rotor windings 
 connected in Y. The ohmic resistance of the rotor winding is 
 0.015 ohm per phase. 
 
 What is the leakage inductance of the rotor winding per phase? 
 What is the ratio of effective to ohmic resistance for the rotor 
 winding at 50 cycles? 
 
 If the stator is rewound for 500 volts by using twice as 
 many turns of wire of one-half the size, what voltage should 
 be impressed to have a starting current of 30 amperes? 
 
 17. The full-load line current taken by a 200-h.p., 3-phase, 980- 
 volt induction motor is 101 amperes. The stator has 216 slots 
 with 5 inductors per slot, and the ohmic resistance of the winding 
 is 0.246 ohm per phase. The rotor has 288 slots with 1 inductor 
 per slot and the ohmic resistance of the winding is 0.010 ohm per 
 phase. Both the stator and rotor windings are connected in delta. 
 The leakage inductances of the stator and rotor windings are 
 respectively 3.9 mil-henrys and 0.19 mil-henry per phase. At 
 the rated frequency of 50 cycles the ratios of effective to ohmic 
 resistance are respectively 1.6 and 1.9 for the stator and rotor. 
 
 (a) What voltage should be impressed on this motor in order 
 that the starting current will be twice the full-load current? 
 
 (b) If the stator winding is reconnected in Y what voltage 
 should be impressed on the motor in order that the starting cur- 
 rent will be twice the full-load current ? In this case the rated volt- 
 age becomes 1700 volts. 
 
 18. At full load the stator and rotor copper losses of a 500-h.p., 
 3-phase, 2000-volt induction motor are respectively 4.14 per 
 cent, and 1.82 per cent. The core loss is 2470 watts and the fric- 
 tion and windage loss is 11.0 kw. The magnetizing component 
 of the line current is 23 amperes. At the rated frequency of 60 
 cycles the ratio of effective resistance to ohmic resistance is 1.55 
 for the stator winding and 1.85 for the rotor winding. At the 
 
74 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 rated frequency the leakage reactances of the windings are 3.2 
 times their effective resistances. What line voltage should be 
 impressed on this motor when starting to give a line current of 200 
 amperes? 
 
 19. The ohmic resistances of the stator and rotor windings of 
 an indication motor are respectively 0.04 ohm and 0.01 ohm. 
 At normal freqency the ratios of effective resistance to ohmic 
 resistance are respectively 1.5 and 1.8 for the stator and rotor 
 windings and the leakage reactances are 3.1 times their effective 
 resistances. The stator winding has 108 inductors per phase and 
 the rotor winding has 54 inductors per phase. Assume that the 
 core loss due to leakage flux and the increase in the resistance 
 due to a non-uniform distribution of current over the cross- 
 section of the inductors both vary directly as the frequency. 
 
 (a) If the voltage impressed on an induction motor when 
 starting is increased 30 per cent., how much is the starting current 
 increased? How much is the starting torque increased? 
 
 (b) If the frequency of the voltage impressed on an induction 
 motor when starting is increased 10 per cent., how much is the 
 starting current decreased? How much is the starting torque 
 decreased. 
 
 20. A 570-h.p. induction motor is designed to receive power 
 from a 3-phase, 1900-volt, 22.5-cycle circuit. The ohmic re- 
 sistances of the stator and rotor windings are 0.488 ohm and 
 0.0138 ohm per phase, and at the rated frequency the effective 
 resistances are 1.3 and 1.5 times as great. The leakage reactances 
 at the rated frequency are three times the effective resistances. 
 The stator has 864 inductors per phase and the rotor 144 in- 
 ductors per phase. The motor is wound for 36 poles, and both of 
 the windings are connected in delta. The friction and windage 
 loss is 12 kw. 
 
 What is the slip at full load if the generated voltage in the stator 
 winding due to the air-gap flux is 93 per cent, of the impressed 
 voltage? What is the starting torque with full voltage impressed 
 on the stator windings? 
 
 21. The effective resistances of the stator and rotor windings of 
 a 3-phase inducton motor are 0.071 ohm and 0.0218 ohm per 
 phase, and the leakage reactances at normal frequency are 0.22 
 ohm and 0.066 ohm per phase. The stator has 108 slots with 5 
 inductors per slot, and the rotor has 126 slots with 2 inductors 
 per slot. Both of the windings are connected in Y. At starting, 
 
INDUCTION MOTORS 75 
 
 with full impressed voltage, the current is 3.6 times the full-load 
 value and the torque is 0.62 of the full-load torque. What will 
 be the starting current and the starting torque with full impressed 
 voltage if resistances of 0.2 ohm are inserted in each phase of the 
 rotor winding? 
 
 22. With full-load current the slip and brake torque of a 500- 
 h.p., 3-phase, 60-cycle induction motor are 1.82 per cent, and 
 16,300 pound-feet. The ohmic resistance and leakage inductance 
 of the rotor winding are 0.24 ohm and 3.1 mil-henrys per phase. 
 What will be the slip, the brake torque and the output when the 
 current has its full-load value if resistances of 2 ohms are inserted 
 in each phase of the rotor winding? Estimate the copper loss 
 in each of these additional resistance units. 
 
 23. The resistance and leakage inductance of the rotor wind- 
 ings of a 3-phase, 38-cycle induction motor are 0.013 ohm and 
 0.27 mil-henry per phase. The full-load torque and slip are 
 2,190 Ib.-ft. and 2.6 per cent. 
 
 (a) To what per cent, of its normal full-load value should the 
 air-gap flux be reduced so that the motor will deliver its full- 
 load torque at one-half the full-load speed? 
 
 (b) What resistance should be inserted in each phase of the 
 rotor so that the motor will deliver its full-load torque at one- 
 half the full-load speed? Assume that the air-gap flux has 
 its normal value. 
 
 (c) Compare the rotor currents in (a) and (b). 
 
 24. The full-load distribution of losses in a 10-h.p., 220- volt, 
 3-phase induction motor is: 
 
 Stator copper loss 3.9 per cent. 
 
 Rotor copper loss 4.8 per cent. 
 
 Core loss 3.1 per cent. 
 
 Friction and windage loss 4.0 per cent. 
 
 (a) What are the slip and efficiency at full load? 
 
 (b) What are the slip and efficiency at one-half of full load? 
 
 25. At full-load the slip of a 335-h.p., 2000-volt, 50-cycle, 6-pole 
 induction motor is 1.8 per cent. The ohmic resistances of the 
 stator and rotor windings are 0.165 ohm and 0.0127 ohm per 
 phase, and the leakage inductances are 2.4 mil-henrys and 0.22 
 mil-henry per phase. Both of the windings are connected in Y. 
 The stator has 72 slots with 10 inductors per slot, and the rotor 
 has 90 slots with 2 inductors per slot. The ratios of effective 
 
76 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 resistance to ohmic resistance at 50 cycles are 1.45 and 1.75 for 
 the stator and rotor windings. At full load the voltage gener- 
 ated in the stator winding by the mutual flux is 93 per cent, of 
 the impressed voltage. 
 
 To what per cent, of its full-load value should the air-gap 
 flux be reduced so that the starting current will be 175 amperes? 
 What is the starting torque for this condition? What are the 
 full-load torque and rotor current? 
 
 26. In problem 25 what resistance should be inserted in each 
 phase of the rotor winding so that when starting with full im- 
 pressed voltage the rotor current will be twice its full-load 
 value? What is the starting torque for this condition? What 
 is the running torque with full-load current? 
 
 27. In problem 25 what reactance should be inserted in each 
 phase of the rotor winding so that when starting with full im- 
 pressed voltage the rotor current will be twice its full-load 
 value? What is the starting torque for this condition? What 
 is the running torque with full-load current? 
 
 28. At the instant of starting on a reduced voltage of 500 volts 
 a 500-h.p., 3-phase, 60-cycle induction motor takes a line current 
 of 152 amperes at 0.31 power factor. The starting torque is 
 790 pound-feet. The motor is wound for 44 poles. The 
 ohmic resistance of the rotor winding is 0.0306 ohm measured 
 between terminals when the winding is not short-circuited. 
 The effective resistance is 1.7 times as great at the rated fre- 
 quency. The stator winding has 704 inductors per phase 
 and the rotor winding has 220 inductors per phase. The fric- 
 tion and windage loss is 11. kw. With an impressed voltage 
 of 2000 volts what brake torque would be delivered when the 
 slip is 1.8 per cent? Assume that the voltage generated in the 
 stator winding by the air-gap flux is 94 per cent, of the im- 
 pressed voltage. 
 
 29. A 570-h.p., 3-phase induction motor is arranged so that 
 it may be connected to the line through a compensator at start- 
 ing. The motor is wound for 36 poles and a line voltage of 1900 
 volts at 22.5 cycles. When the compensator reduces the im- 
 pressed to 600 volts the line current is 200 amperes and the power 
 is 58.8 kw. at the instant of starting. The stator and rotor ohmic 
 resistances are equal when reduced to the same side and the ef- 
 fective resistances are 1.5 times as great as the ohmic. The 
 friction and windage loss at full load is 12. kw. Assume that 
 
INDUCTION MOTORS 77 
 
 the voltage generated in the stator winding at full load is 94 
 per cent, of the impressed voltage. What is the slip when the 
 motor delivers its rated load? What is the starting torque with 
 600 volts impressed on the motor? 
 
 30. At full load the slip of a 500-h.p., 3-phase induction motor 
 is 1.80 per cent. The motor is wound for 12 poles and an im- 
 pressed voltage of 2200 volts at 25 cycles. When the rotor wind- 
 ing is not short circuited the hot resistance referred to the stator 
 measured between terminals is 0.431 ohm, and the effective 
 resistance is 1.65 times as much. What is the starting torque 
 when the impressed voltage is adjusted so that the rotor cur- 
 rent is twice its full-load value? 
 
 31. At full load the slip of a 1000-h.p., 3-phase induction motor 
 is 1.72 per cent. The motor is wound for 12 poles and an im- 
 pressed voltage of 2200 volts at 25 cycles. When the rotor wind- 
 ing is not short circuited the hot resistance measured between 
 terminals is 0.0895 ohm, and the effective resistance is 1.6 times 
 as great. The ratio of transformation from stator to rotor is 
 2200 to 1500. The effective resistance and leakage reactance of 
 the stator winding between terminals are 0.195 ohm and 0.59 
 ohm. 
 
 What voltage should be impressed on this motor so that the 
 starting torque will be the same as the full-load torque? What 
 is the starting current for this condition and how does it compare 
 with the full-load current? 
 
 32. In problem 31 what resistance should be inserted in each 
 phase of the rotor winding in order that the starting torque 
 for the rated voltage will be equal to the full-load torque? 
 What is the starting current for this condition and how does 
 it compare with the full-load current? 
 
 33. The ohmic resistances of the stator and rotor windings of a 
 500-h.p., 3-phase induction motor are 0.24 ohm and 0.0153 ohm 
 per phase, and the effective resistances at the rated frequency are 
 respectively 1.5 and 1.6 times as great. The leakage reactances 
 at the rated frequency are respectively 1.1 ohms and 0.075 ohm 
 per phase. The motor is wound for 44 poles and an impressed 
 voltage of 2000 volts at 60 cycles. The stator has 2112 inductors 
 and the rator 660 inductors. Both windings are connected in Y. 
 At full load the voltage generated in the stator winding by the 
 air-gap flux is 94 per cent, of the impressed voltage. 
 
 What resistance should be inserted in each phase of the rotor 
 
78 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 in order that the starting torque with full impressed voltage may 
 have its maximum value? What is this torque? What is the 
 starting current for this condition and how does it compare with 
 the full-load current? 
 
 34. The ohmic resistances of the stator and rotor windings 
 of a 150-h.p., 3-phase railway induction motor are 0.054 ohm and 
 0.014 ohm per phase, and the effective resistances at the rated 
 frequency are respectively 1.45 and 1.65 times as great. The 
 leakage reactances at the rated frequency are respectively 0.24 
 ohm and 0.071 ohm per phase. The motor is wound for 6 poles 
 and an impressed voltage of 500 volts at 38 cycles. The ratio of 
 transformation from stator to rotor winding is 15 to 7, and 
 both windings are connected in Y. With full impressed voltage 
 the measured slip for a brake torque of 1470 pound-feet is 2.26 
 per cent. 
 
 What resistance should be inserted in each phase of the rotor 
 winding so that the starting torque with full impressed voltage 
 will have its maximum value? What is this torque? 
 
 INDUCTION MOTOR DATA 
 
 
 A 
 
 B 
 
 C 
 
 D 
 
 E 
 
 Horse-powc 
 Line voltag 
 Type of wii 
 Frequency . 
 Poles 
 
 >r 
 
 25 
 250 
 Y 
 50 
 8 
 0.096 
 
 120 
 500 
 Y 
 38 
 6 
 0.04 
 0.01 
 2 : 1 
 
 200 
 980 
 A 
 20 
 24 
 0.246 
 
 335 
 2000 
 Y 
 50 
 6 
 0.165 
 0.0127 
 4 : 1 
 
 570 
 
 1900 
 A 
 22.5 
 36 
 0.488 
 0.0138 
 6 : 1 
 
 e 
 
 iding 
 
 
 
 Ohmic resig 
 per phase. 
 Ratio of in 
 Magnetizin 
 Core loss 
 
 tance ( Stator. . 
 \ Rotor. . 
 insformation 
 
 g current (line).. 
 
 11 
 590 
 220 
 
 39 
 1120 
 4000 
 
 
 
 
 Friction an 
 No load 
 
 Blocked.. - 
 
 d windage loss . . . 
 Line voltage . . . 
 Line current. . . 
 Power 
 
 
 
 
 500 
 34.5 
 
 4400 
 500 
 800 
 220,000 
 
 2000 
 15.3 
 10,100 
 440 
 170 
 40,500 
 
 1900 
 57.6 
 17,200 
 600 
 200 
 65,000 
 
 
 
 220 
 173 
 
 20,400 
 
 Line voltage. . . 
 Line current . . . 
 Power 
 
 100 
 57 
 3,100 
 
 35. Draw the Heyland diagram for motor No. A. (a) What 
 are the full-load power factor, slip and efficiency, (b) What is 
 
INDUCTION MOTORS 79 
 
 the break-down torque? (c) At what load is the power factor a 
 maximum? 
 
 36. Draw the Heyland diagram for motor No. B. (a) What 
 are the power factor, the slip, and the efficiency at full load? 
 (b) at one-half of full load? (c) What is the break-down torque? 
 (d) At what load is the power factor a maximum? 
 
 37. Draw the Heyland diagram for motor No. C. (a) What 
 are the power factor, the slip and the efficiency at full load? 
 (b) What is the break-down torque? fc) At what load is the 
 power factor a maximum? 
 
 38. Draw the Heyland diagram for motor No. D. (a) What 
 are the power factor, the slip and the efficiency at full load? (b) 
 at one-half full load? (c) What is the break-down torque? 
 
 39. Draw the Heyland diagram for motor No. E. (a) What 
 are the power factor, the slip and the efficiency at full load? 
 (b) What is the break-down torque? (c) At what load is the 
 power factor a maximum? 
 
 40. l The following data are given on a 5-h.p., 110- volt, 8- 
 pole, 60-cycle, 3-phase induction motor. The stator and rotor 
 windings are both connected in delta, and the former has an ef- 
 fective resistance of 0.24 ohm and a leakage reactance of 0.70 ohm 
 per phase. The ohmic resistance of the rotor winding referred 
 to the stator is 0.36 ohm per phase. At no load the motor takes 
 300 watts at 110 volts and a power factor of 0.15. 
 
 When this motor takes 5100 watts at 110 volts and a power 
 factor of 0.835 what are the slip, the power output and the 
 torque? 
 
 41. l A 3-phase, 1000-h.p. induction motor is taking 916 kw. 
 from a 2200-volt circuit at a power factor of 0.914. At no load 
 with an impressed voltage of 2200 volts the line current is 75 .1 
 amperes and the power is 15.2 kw. The effective resistance and 
 the leakage reactance of the stator winding, which is Y-connected, 
 are 0.118 ohm and 0.32 ohm per phase. The ohmic resistance of 
 the rotor winding referred to the stator is 0.10 ohm per phase. 
 What is the output for the specified load? What is the slip? 
 
 42. l A 3-phase, 2600-h.p. induction motor is taking 2272 kw. 
 from a 6470-volt circuit, and the line current is 245 amperes. At 
 no load with an impressed voltage of 6400 volts the line current 
 is 89.5 amperes and the power is 47.5 kw. The ohmic resistance of 
 
 1 Use either the transformer diagram or the equivalent circuit in the 
 solution of this problem. 
 
80 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 the stator winding between terminals is 0.561 ohm at 25 C. The 
 ohmic resistance of the open-circuited rotor winding between 
 terminals is 0.0766 ohm at 25 C. The leakage reactance of the 
 stator winding is 3.64 ohms between terminals. With the rotor 
 blocked 375 kw. is supplied when the line current is 369 amperes. 
 The temperature of the windings at this time is 25 C. Assume 
 that the ratios of effective resistance to ohmic resistance are the 
 same for the stator and rotor windings at 25 C. 
 
 What is the output for the specified load? What is the slip? 
 What is the distribution of the losses? The temperature at this 
 time is 73 C. 
 
 43. l A 3-phase, 2000-volt, 60-cycle induction motor has a full- 
 load capacity of 500 h.p. The stator winding has an effective 
 resistance of 0.36 ohm and a leakage reactance of 1.1 ohms per 
 phase. The rotor winding has an ohmic resistance of 0.157 ohm 
 and a leakage inductance of 1.96 mil-henry s per phase referred 
 to the stator. Both of the windings are connected in Y. The 
 magnetizing current is 23 amperes, the core loss is 2470 watts, 
 and the friction and windage is 11 kw. 
 
 What are the slip, the power factor, and the efficiency when the 
 motor delivers its rated output? The impressed voltage has 
 such a value that the voltage generated in the stator winding by 
 the air-gap flux is 1850 volts. 
 
 44. l A 3-phase, 1900-volt, 22.5 cycle, 36-pole induction motor 
 has a full-load capacity of 570 h.p. The ohmic resistances of 
 the stator and rotor windings are 0.488 ohm and 0.0138 ohm per 
 phase, and their effective resistances at the rated frequency are 
 respectively 1.4 and 1.6 times as great. Both of the windings 
 are connected in delta, and the ratio of transformation of stator 
 to rotor is 6 to 1. The magnetizing component of the line cur- 
 rent is 57.2 amperes, the core loss is 2.8 kw., and the friction 
 and windage loss is 12 kw. With the rotor blocked the line 
 current is 200 amperes when the impressed voltage is 600 volts. 
 Assume that the ratios of leakage reactance to effective resistance 
 at the rated frequency are the same for both stator and rotor 
 windings. 
 
 What are the slip, the power factor, and the efficiency when the 
 motor delivers its rated output? Assume that the voltage 
 generated in the stator winding by the air-gap flux is 1790 volts. 
 
 ^se either the transformer diagram or the equivalent circuit in the 
 solution of this problem. 
 
INDUCTION MOTORS 81 
 
 45. ! A 3-phase, 500-volt, 38-cycle, 6-pole railway induction 
 motor has a full-load capactiy of 120 h.p. The ohmic resistances 
 of the stator and rotor windings are 0.04 and 0.01 ohm per phase, 
 and their effective resistances at the rated frequency are respec- 
 tively 1.45 and 1.60 times as great. Both of the windings are 
 connected in Y, and their ratio of transformation is 2 to 1. The 
 magnetizing component of the line current is 34 amperes, and 
 the total no-load losses are 4400 watts. With the rotor blocked 
 the line current is 800 amperes when the impressed voltage is 500 
 volts. Assume that the leakage reactances of the stator and rotor 
 windings at the rated frequency are equal when reduced to the 
 same side. 
 
 What are the slip, the power factor and the efficiency when 
 the motor delivers a torque of 1200 pound-feet? Assume that 
 the voltage generated in the stator winding by the air-gap flux is 
 475 volts. 
 
 46. 1 The full-load capacity of a 3-phase, 2000-volt, 50-cycle 
 induction motor is 335 h.p. The motor is wound for 6 poles, 
 and both the stator and rotor windings are connected in Y. 
 The effective resistance of the stator winding is 0.23 ohm and the 
 ohmic resistance of the rotor winding referred to the stator is 
 0.203 ohm per phase. At 50 cycles the leakage reactances of the 
 windings, referred to the stator, are respectively 0.70 ohm and 
 0.83 ohm per phase. At no load the motor takes 15.2 amperes 
 from a 2000-volt circuit at a power factor of 0.187. The core loss 
 is 7400 watts at this time. The motor is operating with a slip of 
 1.43 per cent., and the terminal voltage has such a value that 
 the voltage generated in the stator winding is the same as at 
 no load. What is the output of the motor? What is the power 
 factor? What is the efficiency? 
 
 47. l A 3-phase, 500-volt, 38-cycle railway induction motor has 
 a full-load capacity of 150 h.p. The ohmic resistances of the 
 stator and rotor windings are 0.052 ohm and 0.013 ohm per phase, 
 and the effective resistances are respectively 1.40 and 1.55 times 
 as great at the rated frequency. The motor is wound for 8 poles. 
 Both stator and rotor windings are connected in Y and have a 
 ratio of transformation of 9 to 5. At no load with an impressed 
 voltage of 500 volts the motor takes a current of 46.6 amperes at 
 a power factor of 0.141. With the rotor blocked the motor takes 
 
 1 Use either the transformer diagram or the equivalent circuit in the 
 solution of this problem. 
 
82 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 a current of 610 amperes when the impressed voltage is 500 volts. 
 Assume that the leakage reactances are in the same ratio as the 
 effective resistances at the rated frequency. 
 
 What are the slip, the power factor and the efficiency when the 
 motor delivers a torque of 1950 pound-feet? Assume that the 
 voltage generated in the stator winding by the air-gap flux is the 
 same as at no load. 
 
 48. 1 At no load when the line voltage has its rated value of 1900 
 volts, a 3-phase, 570-h.p. induction motor takes a line current of 
 57.2 ampers at a power factor of 0.091. The speed of the rotor is 
 74.6 rev. per min., and the slip is one revolution in 23 minutes. 
 The ohmic resistances of the stator and rotor winding are 0.488 
 ohm and 0.0138 ohm per phase. The motor is wound for 36 
 poles and both the stator and the rotor windings are connected in 
 delta, and have a ratio of transformation of 6 to 1. With the 
 rotor blocked the line current is 200 amperes and the power sup- 
 plied is 59.2 kw. when the impressed voltage is 600 volts. Assume 
 that the ratios of effective to ohmic resistance are the same 
 for the stator and rotor, and that the ratios of leakage reactance 
 to effective resistance are also the same for each winding. 
 
 What is the friction and windage loss? What is the core loss? 
 
 49. l A 3-phase, 2000- volt, 60-cycle induction motor is rated 
 to deliver 500 h.p. at full load. The ohmic resistances of the 
 stator and rotor windings are respectively 0.24 ohm and 0.0153 
 ohm per phase, and the ratios of effective to ohmic resistance are 
 respectively 1.5 and 1.6. With the rotor blocked the power 
 factor is 0.29. Assume that the ratios of leakage reactance to 
 effective resistance at the rated frequency are the same for the 
 stator and rotor. The motor is wound for 44 poles, and both 
 the stator and rotor windings are connected in Y and have a 
 ratio of transformation of 16 to 5. The friction and windage loss 
 is 11.0 kw. and the core loss is 2470 watts at the rated voltage. 
 
 With the impressed voltage adjusted so that the voltage gen- 
 erated in the stator winding by the air gap flux is 1120 volts per 
 phase, what is the slip in revolutions per minute at no load? 
 What are the impressed voltage, the current and the power for 
 this condition? 
 
 50. 1 The full-load capacity of a 3-phase, 980-volt, 20-cycle 
 induction motor is 200 h.p. The effective resistance and leakage 
 
 ^se either the transformer diagram or the equivalent circuit in the 
 solution of this problem. 
 
INDUCTION MOTORS 
 
 83 
 
 reactance of the stator winding are 0.342 ohm and 1.08 ohms re- 
 spectively. The ohmic resistance and leakage inductance of the 
 rotor winding referred to the stator are 0.141 ohm and 5.1 mil- 
 henrys. The motor is wound for 24 poles and both the stator and 
 rotor windings are connected in delta. At no load the motor 
 takes a line current of 39.4 amperes and absorbs 6680 watts when 
 the impressed voltage has its rated value. The friction and wind- 
 age loss is 4.0 kw. 
 
 If the air-gap flux is assumed to be constant the maximum 
 torque occurs when the slip equals the ratio of the resistance of the 
 rotor winding to its leakage reactance at the rated frequency. 
 In this case how much must the impressed voltage be increased 
 in order that the air-gap flux will have its no-load value when the 
 torque is a maximum? 
 
 
 F 
 
 G 
 
 H 
 
 ' 
 
 J 
 
 Horse-power 
 
 
 50 
 
 150 
 
 500 
 
 1000 
 
 2600 
 
 Line voltage 
 
 
 2200 
 
 500 
 
 2200 
 
 2200 
 
 6400 
 
 Frequency 
 
 
 60 
 
 60 
 
 25 
 
 25 
 
 25 
 
 
 
 Y 
 
 A 
 
 
 
 
 Poles 
 
 
 
 
 12 
 
 12 
 
 36 
 
 Ohmic resistance between / Stator. . . 
 
 6.307 
 
 0.050 
 
 0.357 
 
 0.130 
 
 0.561 
 
 
 I Rotor 
 
 
 
 0286 
 
 0772 
 
 0766 
 
 
 
 hot 
 
 hot 
 
 25 C 
 
 25 C 
 
 25 C. 
 
 Ratio of transformation .... 
 
 
 
 2200 610 
 
 2200 1500 
 
 6400 2076 
 
 No load, temp. = 
 
 Voltage (line) . 
 Current (line) 
 
 2200 
 4.14 
 
 500 
 36.7 
 
 2200 
 39.9 
 
 2200 
 75.1 
 
 6400 
 89.5 
 
 25 C. 
 
 Power 
 
 1620 
 
 7550 
 
 13,600 
 
 15,200 
 
 47,500 
 
 
 Voltage (line) . 
 
 620 
 
 80 
 
 750 
 
 400 
 
 2200 
 
 Blockedtemp. = 25C. 
 
 Current (line) 
 
 25 
 
 245 
 
 288 
 
 346.5 
 
 369 
 
 
 Power 
 
 10,600 
 
 12,500 
 
 116,000 
 
 68,000 
 
 375,000 
 
 Blocked with full im- 
 
 [Current (line) . 
 
 
 935 
 
 1960 
 
 1210 
 
 pressed voltage. 
 
 Power, (kw.) 
 
 1050 
 
 . 1960 
 
 2950 
 
 51. 1 What are the power factor, the output, and the efficiency 
 of the induction motor No. F when the slip is 1.55 per cent. 
 and the voltage generated by the air-gap flux in the stator wind- 
 ing is 2060 volts? Assume that the ratios of effective to ohmic 
 resistance are respectively 1.1 and 1.20 for the stator and rotor 
 windings and that the leakage reactances are equal when reduced 
 to the same side. 
 
 52. l What are the slip, the power factor, and the efficiency 
 of the induction motor No. G when it delivers full load and the 
 
 Use the transformer diagram in the solution of this problem. 
 
84 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 voltage generated in the stator winding by the air-gap flux is 
 the same as at no load? Assume that the ratios of effective to 
 ohmic resistance are respectively 1.4 and 1.6 for the stator and 
 rotor windings, and that the ratio of the leakage reactances 
 at the rated frequency is equal to the ratio of the effective 
 resistances. 
 
 53. * What are the power factor, the torque and the efficiency 
 of the induction motor No. H when the slip is 1.86 per cent., and 
 the voltage generated by the air-gap flux in the stator winding 
 is 93 per cent, of the rated impressed voltage? Assume that 
 the ratios of effective to ohmic resistance are the same for the 
 stator and rotor windings at 25 C., and that the ratio of the 
 ohmic resistances is equal to the ratio of the leakage reactances 
 of the two windings at the rated frequency. The temperature 
 under the running condition is 65 C. 
 
 54. 1 What are the power factor, the slip, and the efficiency 
 of the induction motor No. I when it delivers a torque of 22 , 000 
 pound-feet, and the voltage generated by the air-gap flux in 
 the stator winding is 2040 volts? Assume that the ratios of 
 the ohmic resistance, the effective resistances, and the leakage 
 reactances of the stator and rotor windings are equal at the rated 
 frequency and a temperature of 25 C. The running tempera- 
 ture is 70 C. 
 
 55. 1 What are the power factor, the slip, and the efficiency 
 when the induction motor No. J delivers 2750 h.p. and the 
 voltage generated by the air-gap flux in the stator winding is 
 92 per cent, of the rated impressed voltage? Assume that the 
 ratio of the effective to the ohmic resistance of the rotor is 
 20 per cent, greater than for the stator, and that the ratios 
 of the leakage reactances and of the effective resistances are 
 equal at the rated frequency and a temperature of 25 C. The 
 running temperature is 70 C. 
 
 56. What are the power factor, the torque and the efficiency 
 for the induction motor No. I when the slip is 1.6 per cent, 
 and the impressed voltage has its rated value? Make the same 
 assumptions in regard to the resistances and reactances as were 
 made in problem 54. 
 
 57. What are the power factor, the slip and the efficiency 
 when the induction motor No. J delivers a torque of 164,000 
 pound-feet and the impressed voltage is 6400 volts? Make 
 
 1 Use the transformer diagram in the solution of this problem. 
 
INDUCTION MOTORS 85 
 
 the same assumptions in regard to the resistances and reactances 
 as were made in problem 55. 
 
 58. What are the power factor, the slip and the efficiency when 
 the induction motor No. H delivers 560 h.p. and the impressed 
 voltage has its rated value ? Make the same assumption in regard 
 to the resistances and reactances as were made in problem 53. 
 
 59. Two 3-phase, 220-volt. 60-cycle induction motors are con- 
 nected in concatenation across a 220-volt circuit. Each is rated 
 to deliver 10 h.p. and is wound for 6 poles. What is the no-load 
 speed? When they deliver 10 h.p. what is the torque developed 
 by each motor? 
 
 60. Two 3-phase, 220-volt. 60-cycle induction motors are con- 
 nected in concatenation across a 220-volt circuit. Each is rated 
 to deliver 10 h.p. but one iswound for 6 poles and the other for 
 8 poles. What is the no-load speed? When they deliver 10 h.p. 
 what is the torque developed by each motor? What per cent, 
 of its full-load current does each motor carry? 
 
 61. Two 3-phase, 220-volt, 60-cycle induction motors are con- 
 nected in concatenation across a 220-volt circuit. One is rated 
 to deliver 10 h.p. and the other 15 h.p., but both are wound for 
 6 poles. What is the no-load speed? What load is delivered 
 when the 15-h.p. motor takes its full-load current? What per 
 cent, of its full-load value is the current in the 10-h.p. motor? 
 What is the torque developed by each? 
 
 62. Two 3-phase, 220 volt 60-cycle induction motors are con- 
 nected in concatenation across a 220-volt circuit. One is rated 
 to deliver 10 h.p. and is wound for 6 poles, and the other is rated 
 to deliver 15 h.p. and is wound for 4 poles. What is the no-load 
 speed? What is the greatest load that can be delivered and 
 have neither motor take more than its full-load current? When 
 they deliver 15 h.p. what torque does each develop? 
 
 63. The two induction motors, M and P, are connected in 
 concatenation. The stator of the first receives power at its 
 rated voltage and frequency, and the stator of the second is 
 short-circuited. Neglect the no-load component of the current 
 and the core loss due to the leakage flux. 
 
 (a) What is the total output when the slip of the first motor 
 is 51.3 per cent.? 
 
 (b) What is the power developed by each motor? 
 
 (c) What are the copper losses in the stator and rotor of each 
 motor? 
 
86 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 INDUCTION MOTOR DATA 
 
 Number 
 
 K 
 
 L 
 
 M 
 
 N 
 
 P 
 
 Horse-power 
 
 500 
 
 570 
 
 150 
 
 150 
 
 120 
 
 Line voltage 
 
 2000 
 
 1900 
 
 500 
 
 500 
 
 500 
 
 Connection (both stator and rotor) 
 
 Y 
 
 A 
 
 Y 
 
 Y 
 
 Y 
 
 Frequency 
 
 60 
 
 22.5 
 
 38 
 
 38 
 
 38 
 
 Poles 
 
 44 
 
 36 
 
 6 
 
 8 
 
 6 
 
 Magnetizing current (per phasa.) ' 
 
 23 
 
 33 
 
 45 
 
 46 
 
 34 
 
 Core loss at no load 1 ' 2 
 
 2470 
 
 2800 
 
 1300 
 
 1360 
 
 1200 
 
 / Stator 
 Ohmic resistance per phase < R 
 
 0.24 
 0.0153 
 
 0.488 
 0.0138 
 
 0.054 
 0.014 
 
 0.052 
 0.013 
 
 0.04 
 0.01 
 
 Ratio of effective resistance f Stator 
 
 1.3 
 
 1.35 
 
 1.25 
 
 1.25 
 
 1.2 
 
 to ohmic resistance at the -j Rotor 
 
 1.4 
 
 1.55 
 
 1.4 
 
 1.35 
 
 1.4 
 
 rated frequency. 
 
 
 
 
 
 
 Leakage inductance per / Stator 
 
 2.6 
 
 13. 
 
 0.091 
 
 0.086 
 
 0.068 
 
 phase (mil-henry). \ Rotor 
 
 0.17 
 
 0.37 
 
 0.024 
 
 0.022 
 
 0.018 
 
 Ratio of transformation 
 
 16 to 5 
 
 6 tol 
 
 15 to 7 
 
 9 to 5 
 
 2 to 1 
 
 64. The two induction motors, N and P, are connected in 
 concatenation. The stator of the first receives power at its 
 rated voltage and frequency, and the stator of the second is 
 short-circuited. Neglect the no-load component of the current 
 and the core loss due to the leakage flux. Assume that the core 
 loss due to the mutual flux varies as the product of the frequency 
 and the square of the flux density. 
 
 (a) What is the total output when the slip of the second motor 
 is 4.0 per cent.? 
 
 (b) What is the power developed by each motor? 
 
 (c) What are the copper losses in the stator and rotor of each 
 motor? 
 
 (d) What are the core losses in the stator and rotor of each 
 motor? 
 
 65. The two induction motors, M and N, are connected in con- 
 catenation. The stator of the first receives power at its rated 
 voltage and frequency, and the stator of the second is short- 
 circuited. Neglect the no-load component of the current. As- 
 sume that the loss caused by the leakage flux varies as the fre- 
 quency and the square of the current. Assume that the core 
 loss due to the mutual flux varies as the product of the frequency 
 and the square of the flux density. 
 
 (a) What is the total output when the speed is 228 rev. per min. ? 
 
 1 At the rated voltage. 
 
 2 Assume that the ratio of the core loss in the stator to that in the rotor 
 for the same mean flux density and frequency is 4 to 3. 
 
INDUCTION MOTORS 87 
 
 (b) What is the power developed by each motor? 
 
 (c) What are the effective resistance losses in the stator and 
 rotor of each motor? 
 
 (d) What are the core losses in the stator and rotor of each 
 motor? 
 
 66. The two induction motors, K and L, are connected in 
 concatenation. The stator of the first receives power at its 
 rated voltage and frequency, and the stator of the second is 
 short-circuited. Neglect the no-load component of the current. 
 Assume that the loss caused by the leakage flux is proportional 
 to the frequency and the square of the current. Assume that 
 the core loss due to the mutual flux varies as the frequency and 
 the square of the flux density. 
 
 (a) What is the total output when the slip of the second motor 
 is 1.55 per cent.? 
 
 (b) What is the power developed by each motor? 
 
 (c) What are the effective resistance losses in the stator and 
 rotor of each motor? 
 
 (d) What are the core losses in the stator and rotor of each 
 motor? 
 
 67. The two induction motors, M and P, are connected in 
 concatenation. The stator of the first receives power at its 
 rated frequency, and the stator of the second is short-circuited. 
 Do not neglect the no-load component of the current, but assume 
 that it is proportional and wattless with respect to the generated 
 voltage. Assume that the loss caused by the leakage flux varies 
 as the frequency and the square of the current. Assume that 
 the core loss due to the mutual flux varies as the product of the 
 frequency and the square of the flux density. 
 
 (a) What is the total output when the speed is 268 rev. per 
 min. and the voltage generated by the air-gap flux in the rotor 
 winding of the second motor is 68 volts per phase? 
 
 (b) What power do they receive from the line and at what 
 power factor do they operate? 
 
 (c) What is the power developed by each motor? 
 
 68. The two induction motors, K and L, are connected in 
 concatenation. The stator of the first receives power at its 
 rated frequency, and the stator of the second is short-circuited. 
 Do not neglect the no-load component of the currrent, but assume 
 that it is proportional and wattless with respect to the generated 
 voltage. Assume that the loss caused by the leakage flux varies 
 
88 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 as the frequency and the square of the current. Assume that 
 the core loss due to the mutual flux varies as the product of the 
 frequency and the square of the flux density. 
 
 (a) What is the total output when the slip of the second is 
 1.55 per cent, and the voltage generated by the air-gap flux in 
 the rotor winding of the second motor is 430 volts per phase? 
 
 (b) What power do they receive from the line and at what 
 power factor do they operate? 
 
 (c) What is the power developed by each motor? 
 
 69. A 1000-h.p. induction motor is operated as an induction 
 generator in parallel with a synchronous generator having a 
 full-load capacity of 1000 kv.-a. The induction machine, which 
 is wound for 12 poles, is driven at a constant speed of 250.5 rev. 
 per min. The speed of the synchronous generator falls uni- 
 formly from 1530 rev. per min. at no load to 1500 rev. per min. 
 at full load, when the frequency is 25 cycles. The load deliv- 
 ered by the induction generator is proportional to the slip which 
 at full load is 1.7 per cent. 
 
 When the total load supplied is 1500 kw. what is the load 
 delivered by each? At what speed should the induction gener- 
 ator be driven so that both will deliver their rated loads at the 
 same time? 
 
 70. An induction generator and a synchronous generator, each 
 rated to deliver 2500 kv.-a., are operated in parallel. The speed 
 of the induction generator falls from 1520 at no load to 1498 at 
 full load, and the speed of the synchronous generator falls 
 from 1525 at no load to 1490 at full load. The load on the 
 induction generator is proportional to the slip which at full 
 load is 1.8 per cent. 
 
 What is the greatest load that .can be delivered without over 
 loading either generator? To what value should the full-load 
 speed of the synchronous generator be adjusted so that both gen- 
 erators will deliver their rated loads at the same time? 
 
 71. A 4-pole induction generator and a 2-pole synchronous 
 generator are operating in parallel. The induction generator is 
 driven at a constant speed, but the speed of the synchronous 
 generator falls from 3660 rev. per min. at no load to 3590 at 
 full load of 2000 kw. The resistance of the rotor windings 
 between terminals refered to the stator is 0.73 ohm. Neglect 
 the^stator resistance and reactance, the rotor reactance, and the 
 losses. The excitation of the synchronous generator is adjusted 
 
INDUCTION MOTORS 89 
 
 so that the terminal voltage is 6400 volts at all loads. The 
 speed of the induction generator is adjusted so that when the 
 synchronous machine is delivering no power the load on the 
 induction generator is 500 kw. What is the division of the load 
 when 3500 kw. is required? What is the frequency at this 
 time? At what speed should the induction generator be driven 
 so that both will deliver their rated loads, viz., 2000 kw., at the 
 same time? 
 
 72. A 500-kw., 3-phase induction generator is operated with 
 a synchronous motor floated across its terminals. At no load, 
 when running as an induction motor, it takes a line current of 31 
 amperes at 2000 volts. The resistance of the rotor winding be- 
 tween terminals is 0.0228 ohm, and the ratio of transformation 
 from stator to rotor is 16 to 5. Neglect the resistance and 
 reactance of the stator and the reactance of the rotor windings 
 of the induction generator, and all of the losses in both machines. 
 The excitation of the synchronous motor is adjusted so that the 
 terminal voltage is 2000 volts. The load supplied by these 
 machines is 450 kw. at a power factor of 85 per cent. What is 
 the line current supplied by the induction generator? What is 
 the frequency of this system if the induction generator is 
 wound for 44 poles and is driven at 165 rev. per min.? If the 
 synchronous motor has a synchronous reactance of 6.2 ohms 
 between terminals what is its necessary excitation voltage? 
 
 73. An induction generator supplies power to a load and to 
 an over-excited synchronous motor. At no load as an induction 
 motor it takes a line current of 98 amperes at 2200 volts. The 
 resistance of the rotor winding between terminals is 0.058 ohm 
 and the ratio of transformation from stator to rotor is 22 to 15. 
 Neglect the resistance and reactance of the stator winding and 
 the reactance of the rotor winding of the induction generator 
 and all of its losses. The rotational losses of the synchronous 
 motor are 23.3 kw. and the resistance and synchronous reactance 
 are respectively 0.26 ohm and 3.04 ohm between terminals. 
 The induction generator delivers 1150 kw. of which the syn- 
 chronous motor receives 450 kw. The excitation of the latter 
 is adjusted so that the terminal voltage is 2200 volts. The power 
 factor of the load exclusive of the synchronous motor is 0.83. 
 The induction generator is wound for 12 poles and is driven at 
 a speed of 254 rev. per min. 
 
 What is the frequency of the system. At what power factor 
 
90 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 is the synchronous motor operating? What is the excitation 
 voltage of the synchronous motor? What is the power output 
 of the synchronous^ motor? 
 
 74. A 2000-kw. induction generator is operated in parallel 
 with a synchronous generator of the same capacity. At no load, 
 when running as an induction motor, it takes a line current of 
 90 amperes at 6400 volts. The resistance of the rotor winding 
 between terminals is 0.0766 ohm and the ratio of transformation 
 from stator to rotor is 6400 to 2076. Neglect the resistance 
 and reactance of the stator and the reactance of the rotor winding 
 and all of its losses. The resistance and synchronous reactance 
 of the synchronous generator between terminals are respectively 
 0.65 ohm and 14.2 ohms. The induction generator delivers 
 1800 kw. and the synchronous generator 1200 kw. The excita- 
 tion of the latter is adjusted so that the terminal voltage is 6400 
 volts, and the power factor of the load is 0.85. The induction 
 generator is wound for 36 poles and is driven at a speed of 82.4 
 rev. per min. 
 
 What is the frequency of the system? At what power factor 
 does the synchronous generator operate? What is the excita- 
 tion voltage of the synchronous generator? 
 
 75. An induction generator and a synchronous generator are 
 operated in parallel and supply a load of 700 kw. at 0.86 power 
 factor. Data concerning the induction generator are: 370 kw., 
 2000 volts, 44 poles, Y wound. The rotor resistance is 0.166 
 ohm per phase refered to the stator. The magnetizing current 
 is 23 amperes at the rated voltage, and the core loss is 2.47 kw. 
 Neglect the resistance and reactance of the stator winding. 
 Data concerning the synchronous generator are: 500 kw., 2200 
 volts, 64 poles, Y wound. The effective resistance and synchron- 
 ous reactance of the armature winding are respectively 0.224 
 ohm and 2.76 ohms per phase. The induction generator is 
 driven at 166.4 rev. per min. and the synchronous generator 
 at 113.0 rev. per min. What is the division of the load if the 
 excitation of the synchronous machine is adjusted so that the 
 terminal voltage is 2200 volts? What is the necessary excitation 
 voltage of the synchronous generator? 
 
 76. An induction generator and a synchronous generator are 
 operated in parallel and supply a load of 950 kw. at 0.83 power 
 factor. Data concerning the induction generator are: 450 kw., 
 2000 volts, 36 poles, A wound. The rotor resistance is 0.50 ohm 
 
INDUCTION MOTORS 91 
 
 per phase refered to the stator. The magnetizing current is 33 
 amperes per phase and the core loss is 2.8 kw. Neglect the 
 resistance and reactance of the stator winding. Data concern- 
 ing the synchronous generator are: 850 kw., 2000 volts, 32 poles, 
 Y wound. The effective resistance and synchronous reactance 
 are respectively 0.064 ohm and 2.40 ohms per phase. The 
 induction generator is driven at 85.0 rev. per min. and the syn- 
 chronous generator, at 94.0 rev. per min. What is the division 
 of the load if the excitation of the synchronous generator is 
 adjusted so that the terminal voltage is 2000 volts? What is 
 the necessary excitation voltage of the synchronous generator? 
 
 77. l The induction motor No. G is operated as a generator on a 
 60-cycle circuit in parallel with synchronous apparatus. What 
 are the slip, the power factor and the efficiency when it re- 
 ceives 150 h.p. from the prime mover and the voltage generated 
 in the stator winding by the air-gap flux is 460 volts? Assume 
 that the ratios of effective to ohmic resistance are respectively 
 1.4 and 1.6 for the stator and rotor windings, and that the 
 ratio of the leakage reactances at the rated frequency is the same 
 as ratio of the effective resistances. 
 
 78. 1 The induction motor No. H is operated as a generator on 
 a 2200-volt, 25-cycle circuit in paralled with synchronous appa- 
 ratus. What are the output, the power factor and the efficiency 
 when the slip is 1.86 per cent, and the voltage generated in the 
 stator winding by the air-gap flux is assumed to be 93 per cent, 
 of the terminal voltage? Assume that the ratios of effective to 
 ohmic resistance are the same at 25 Centigrade, and that the 
 ratio of the ohmic resistances is equal to the ratio of the leakage 
 reactances of the two windings at the rated frequency. The 
 temperature under the running condition is 65 C. 
 
 79. 1 The induction motor No. I is operated as a generator 
 on a 25-cycle circuit in parallel with syncrohnous apparatus. 
 What are the slip, the power factor and the efficiency when it re- 
 ceives 1000 h.p. from the prime mover and the voltage generated 
 in the stator winding by the air-gap flux is 2040 volts? Assume 
 that the ratios of the ohmic resistances, the effective resistances 
 and the leakage reactances of the stator and rotor windings are 
 equal at the rated frequency and a temperature of 25 C. 
 The running temperature is 70 C. 
 
 80. 1 The induction motor No. / is operated as a generator 
 
 *See page 83. 
 
92 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 on a 6400-volt, 25-cycle circuit in parallel with synchronous 
 apparatus. What are the output, the power factor, and the 
 efficiency when the slip is 1.9 per cent, and the voltage gener- 
 ated in the stator winding by the air-gap flux is (assumed to be) 
 92 per cent, of the terminal voltage? Assume that the ratio of 
 the effective to the ohmic resistance for the rotor is 20 per cent, 
 greater than for the stator, and that the ratios of the leakage 
 reactances and of the effective resistances are equal at the rated 
 frequency and a temperature of 25 C. The running temperature 
 is 70 C. 
 
CHAPTER V 
 ROTARY CONVERTERS 
 
 1. Assume that the graph representing the flux density in the 
 air-gap of a rotary converter is rectangular and is constant over 
 the entire pole pitch. Also assume that the armature winding 
 is uniformly distributed. Calculate the ratio of the single- 
 phase alternating-current voltage to the direct-current voltage. 
 Assume that the coil pitch and phase spread are each unity. 
 
 2. In problem 1 calculate the ratio of the four-phase alternat- 
 ing-current voltage to the direct-current voltage. Assume that 
 the coil pitch is one and that the phase spread is one-half. 
 
 3. In problem 1 calculate the ratio of the three-phase alternat- 
 ing-current voltage to the direct-current voltage. Assume that 
 the coil pitch is one and that the phase spread is two-thirds. 
 
 4. In problem 1 calculate the ratio of the six-phase alternating- 
 current voltage to the direct-current voltage. Assume that the 
 coil pitch is one and that the phase spread is one- third. 
 
 5. In problem 1 calculate the ratio of the twelve-phase alternat- 
 ing-current voltage to the direct-current voltage. Assume that 
 the coil pitch is one and that the phase spread is one-sixth. 
 
 6. Assume that the air-gap flux density in a rotary converter 
 is constant under the poles and is zero between them. The 
 ratio of pole arc to pole pitch is two-thirds. Also assume that 
 the armature winding is uniformly distributed. Calculate the 
 ratio of the single-phase alternating-current voltage to the 
 direct-current voltage. Assume that the coil pitch and the phase 
 spread are each unity. 
 
 7. In problem 6 calculate the ratio of the four-phase alternat- 
 ing-current voltage to the direct-current voltage. Assume that 
 the coil pitch is one and that the phase spread is one-half. 
 
 8. In problem 6 calculate the ratio of the three-phase alternat- 
 ing-current voltage to the direct-current voltage. Assume that 
 the coil pitch is one and that the phase spread is two-thirds. 
 
 9. In problem 6 calculate the ratio of the six-phase alternating- 
 
94 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 current voltage to the direct-current voltage. Assume that the 
 coil pitch is one and that the phase spread is one-third. 
 
 10. In problem 6 calculate the ratio of the twelve-phase alter- 
 nating-current voltage to the direct-current voltage. Assume 
 that the coil pitch is one and that the phase spread is one-sixth. 
 
 11. The graph representing the flux density in the air-gap of 
 a rotary converter is a simple harmonic function. Assume that 
 the armature winding is uniformly distributed. Calculate the 
 ratio of the single-phase alternating-current voltage to the 
 direct-current voltage. Assume that the coil pitch and the 
 phase spread are each unity. 
 
 12. In problem 11 calculate the ratio of the four-phase alter- 
 nating-current voltage to the direct-current voltage. Assume 
 that the coil pitch is one and that the phase spread is one-half. 
 
 13. In problem 11 calculate the ratio of the three-phase alter- 
 nating-current voltage to the direct-current voltage. Assume 
 that the coil pitch is one and that the phase spread is two-thirds. 
 
 14. In problem 11 calculate the ratio of the six-phase alter- 
 nating-current voltage to the direct-current voltage. Assume 
 that the coil pitch is one and that the phase spread is one-third. 
 
 15. In problem 1 1 calculate the ratio of the twelve-phase alter- 
 nating-current voltage to the direct-current voltage. Assume 
 that the coil pitch is one and that the phase spread is one-sixth. 
 
 16. The graph representing the flux distribution in the air- 
 gap of a rotary converter is B = Bisin.x + B 3 sin 3x. x is the 
 angular displacement measured from the neutral point. Take 
 the third harmonic component of the flux density as 0.3 of the 
 fundamental. Assume that the armature winding is uniformly 
 distributed. Calculate the ratio of the single-phase alternating- 
 current voltage to the direct-current voltage. Assume that the 
 coil pitch and the phase spread are each unity. 
 
 17. In problem 16 calculate the ratio of the four-phase alter- 
 nating-current voltage to the direct-current voltage. Assume 
 that the coil pitch is one and that the phase spread is one-half. 
 
 18. In problem 16 calculate the ratio of the three-phase alter- 
 nating-current voltage to the direct-current voltage. Assume 
 that the coil pitch is one and that the phase spread is two-thirds. 
 
 19. In problem 16 calculate the ratio of the six-phase alter- 
 nating-current voltage to the direct-current voltage. Assume 
 that the coil pitch is one and that the phase spread is one-third. 
 
 20. In problem 16 calculate the ratio of the twelve-phase 
 
ROTARY CONVERTERS 95 
 
 alternating-current voltage to the direct-current voltage. As- 
 sume that the coil pitch is one and that the phase spread is 
 one-sixth. 
 
 21. The graph representing the flux distribution in the air- 
 gap of a rotary converter is B = Bisin.x .Bssin 3x. x is the 
 angular displacement measured from the neutral point. Take 
 the third harmonic component of the flux density as 0.3 of the 
 fundamental. Assume that the armature winding is uniformly 
 distributed. Calculate the ratio of the single-phase alternating- 
 current voltage to the direct-current voltage. Assume that the 
 coil pitch and the phase spread are each unity. 
 
 22. In problem 21 calculate the ratio of the four-phase alter- 
 nating-current voltage to the direct-current voltage. Assume 
 that the coil pitch is one and that the phase spread is one-half. 
 
 23. In problem 21 calculate the ratio of the three-phase 
 alternating-current voltage to the direct-current voltage. As- 
 sume that the coil pitch is one and that the phase spread is 
 two-thirds. 
 
 24. In problem 21 calculate the ratio of the six-phase alter- 
 nating-current voltage to the direct-current voltage. Assume 
 that the coil pitch is one and that the phase spread is one-third. 
 
 25. In problem 21 calculate the ratio of the twelve-phase 
 alternating-current voltage to the direct-current voltage. As- 
 sume that the coil pitch is one and that the phase spread is one- 
 sixth. 
 
 26. In a single-phase rotary converter, assume that the cur- 
 rents on the direct- and alternating- current sides are respectively 
 steady and sinusoidal, and neglect all of the losses in calculating 
 their relative values, (a) Calculate the ratio of the average, 
 heating in a conductor at one of the alternating-current taps to 
 that in a conductor midway between the taps when the rotary 
 is operating at unit power factor, (b) Calculate this ratio 
 when the rotary is operating at 0.7 power factor. 
 
 27. In a four-phase rotary converter assume that the currents 
 on the direct- and alternating-current sides are respectively 
 steady and sinusoidal, and neglect all of the losses in calculating 
 their relative values, (a) Calculate the ratio of the average 
 heating in a conductor at one of the alternating-current taps to 
 that in a conductor midway between the taps when the rotary 
 is operating at unit power factor, (b) Calculate this ratio 
 when the rotary is operating at 0.7 power factor. 
 
96 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 28. In a three-phase rotary converter assume that the cur- 
 rents on the direct- and alternating-current sides are respec- 
 tively steady and sinusoidal, and neglect all of the losses in 
 calculating their relative values, (a) Calculate the ratio of the 
 average heating in a conductor at one of the alternating-current 
 taps to that in a conductor midway between the taps when the 
 rotary is operating at unit power factor, (b) Calculate this 
 ratio when the rotary is operating at 0.7 power factor. 
 
 29. In a six-phase rotary converter assume that the currents 
 on the direct- and alternating-current sides are respectively 
 steady and sinusoidal, and neglect all of the losses in calculating 
 their relative values, (a) Calculate the ratio of the average 
 heating in a conductor at one of the alternating-current taps to 
 that in a conductor midway between the taps when the rotary 
 is operating at unit power factor, (b) Calculate this ratio 
 when the rotary is operating at 0.7 power factor. 
 
 30. In a twelve-phase rotary converter assume that the cur- 
 rents on the direct- and alternating-current sides are respec- 
 tively steady and sinusoidal, and neglect all of the losses in 
 calculating their .relative values, (a) Calculate the ratio of the 
 average heating in a conductor at one of the alternating-current 
 taps to that in a conductor midway between the taps when the 
 rotary is operating at unit power factor, (b) Calculate this 
 ratio when the rotary is operating at 0.7 power factor. 
 
 31. In a single-phase rotary converter assume that the cur- 
 rents on the direct- and alternating-current sides are respec- 
 tively steady and sinusoidal and neglect all of the losses in cal- 
 culating their relative values, (a) Calculate the relative out- 
 puts when operating as a rotary converter at unit power factor 
 and as a direct-current generator on the basis of the same arma- 
 ture copper loss, (b) Calculate the relative outputs when the 
 rotary is operating at 0.7 power factor, (c) Calculate the rela- 
 tive outputs when operating as a rotary converter and as a 
 synchronous generator at unit power factor on the basis of the 
 same armature copper loss, (d) Calculate the relative outputs 
 when both are operating at 0.7 power factor. 
 
 32. In a four-phase rotary converter assume that the currents 
 on the direct- and alternating-current sides are respectively steady 
 and sinusoidal, and neglect all of the losses in calculating their 
 relative values, (a) Calculate the relative outputs when operat- 
 ing as a rotary converter at unit power factor and as a direct- 
 
ROTARY CONVERTERS 97 
 
 current generator on the basis of the same armature copper loss. 
 (b) Calculate the relative outputs when the rotary is operating 
 at 0.7 power factor, (c) Calculate the relative outputs when 
 operating as a rotary converter and as a synchronous generator 
 at unit power factor on the basis of the same armature copper 
 loss, (d) Calculate the relative outputs when both are operat- 
 ing at 0.7 power factor. 
 
 33. In a three-phase rotary converter assume that the cur- 
 rents on the direct- and alternating-current sides are respectively 
 steady and sinusoidal, and neglect all of the losses in calculating 
 their relative values, (a) Calculate the relative outputs when 
 operating as a rotary converter at unit power factor and as a 
 direct-current generator on the basis of the same armature 
 copper loss, (b) Calculate the relative outputs when the 
 rotary is operating at 0.7 power factor, (c) Calculate the rela- 
 tive outputs when operating as a rotary converter and as a syn- 
 chronous generator at unit power factor on the basis of the 
 same armature copper loss, (d) Calculate the relative outputs 
 when both are operating at 0.7 power factor. 
 
 34. In a six-phase rotary converter assume that the currents 
 on the direct- and alternating-current sides are respectively 
 steady and sinusoidal, and neglect all of the losses in calculating 
 their relative values, (a) Calculate the relative outputs when 
 operating as a rotary converter at unit power factor and as a 
 direct-current generator on the basis of the same armature copper 
 loss, (b) Calculate the relative outputs when the rotary is 
 operating at 0.7 power factor, (c) Calculate the relative out- 
 puts when operating as a rotary converter and as a synchronous 
 generator at unit power factor on the basis of the same armature 
 copper loss, (d) Calculate the relative outputs when both are 
 operating at 0.7 power factor. 
 
 35. In a twelve-phase rotary converter assume that the 
 currents on the direct- and alternating-current sides are respec- 
 tively steady and sinusoidal and neglect all of the losses in cal- 
 culating their relative values, (a) Calculate the relative out- 
 puts when operating as a rotary converter at unit power factor 
 and as a direct-current generator on the basis of the same armature 
 copper loss, (b) Calculate the relative outputs when the rotary 
 is operating at 0.7 power factor, (c) Calculate the relative out- 
 puts when operating as a rotary converter and as a synchronous 
 generator at unit power factor on the basis of the same armature 
 
98 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 copper loss, (d) Calculate the relative outputs when both are 
 operating at 0.7 power factor. 
 
 36. A 5-kilowatt single-phase rotary converter supplies power 
 on the direct-current side at 110 volts. It receives energy on the 
 alternating-current side through a transformer from a 2200-volt 
 circuit. What should be the transformer's ratio of transforma- 
 tion? If the rotary has an efficiency of 85 per cent, what should 
 be the current rating of the high-tension winding of the trans- 
 former? 
 
 37. A 100-kilowatt, 4-phase rotary converter supplies power 
 on the direct-current side at 230 volts. It receives energy on 
 the alternating-current side through two single-phase trans- 
 formers from a 2-phase circuit. The voltage between adjacent 
 high-tension conductors is 1555 volts, and between alternate 
 conductors is 2200 volts. The high-tension windings of the 
 transformers are connected across the 2200-volt lines and the 
 low-tension windings are connected in star with the neutral 
 point grounded. The efficiency of the rotary at full load and 
 unit power factor is 94 per cent. What should be the full-load 
 current and voltage ratings of the high- and low-tension windings 
 of the transformers? 
 
 38. A 4-phase, 50-kilowatt rotary converter supplies power on 
 the direct-current side at 220 volts. It receives energy on the 
 alternating-current side from a 2-phase line through two 
 single-phase transformers which have double primary and 
 secondary windings. The voltage between the adjacent high- 
 tension conductors is 1625 volts and between alternate conduc- 
 tors is 2300 volts. The high-tension windings of the transform- 
 ers are connected in star and the low-tension windings in mesh. 
 The efficiency of the rotary at full-load output and 0.90 power 
 factor is 90.2 per cent. What are the primary and secondary 
 currents and voltages when the rotary delivers its rated load at 
 0.90 power factor? 
 
 39. A 4-phase 100-kilowatt rotary converter delivers power 
 on the direct-current side at 220 volts. It receives energy from 
 a three-phase, 11,000-volt circuit through Scott-connected trans- 
 formers. The low-tension windings are connected in star. The 
 efficiency of the rotary at full load and unit power factor is 92.5 
 per cent. What are the ratios of transformation of each of the 
 transformers? What are the full-load current ratings of the 
 primary and secondary windings of each transformer? 
 
ROTARY CONVERTERS 99 
 
 40. A 4-phase, 100-kilowatt, 230-volt rotary converter receives 
 its power from a three-phase, 6600-volt circuit through Scott-con- 
 nected transformers. The low-tension windings are double and 
 are connected in mesh. The efficiency of the rotary at full load 
 and 0.90 power factor is 91.8 per cent. What are the high- and 
 low-tension currents and voltages of the transformers when the 
 rotary delivers its rated load and operates at 0.90 power factor? 
 
 41. A three-phase, 200-kilowatt rotary converter supplies 
 direct current at 110 volts. It receives power from a three-phase, 
 6600-volt circuit through three single-phase transformers which 
 have their high-tension windings connected in delta and their low- 
 tension windings connected in Y. At full load and unit power 
 factor the rotary has an efficiency of 93 per cent. What should 
 be the full-load current and voltage ratings of the high -and low- 
 tension windings of the transformers? 
 
 42. A three-phase, 250-kilowatt, 220-volt rotary converter 
 receives power through three single-phase transformers from a 
 3-phase, 2200-volt circuit. The high-tension windings of the 
 transformers are connected in Y and the low-tension windings in 
 delta. The rotary has an overload current capacity of 15 per 
 cent, when operating at 0.9 power factor and at this time the 
 efficiency is 91.8 per cent. What should be the overload current 
 and voltage ratings of the high- and low-tension windings of the 
 transformers? 
 
 43. A three-phase, 300-kilowatt, 500-volt rotary converter 
 receives power through three single-phase transformers from a 
 three-phase, 11,000-volt circuit. Both the high- and low-tension 
 windings of the transformers are connected in Y. When deliver- 
 ing its rated load and operating at 0.85 power factor the rotary 
 has an efficiency of 91.6 per cent. What is the ratio of trans- 
 formation of the transformers? What should be the full-load 
 current ratings of the high- and low-tension windings of the 
 transformers? 
 
 44. A three-phase, 150-kilowatt, 230-volt rotary converter 
 receives power through two single-phase transformers from a 
 three-phase, 2200-volt circuit. Both the primary and secondary 
 windings of the. transformers are connected in V. Neglecting 
 the losses what should be the current and voltage ratings of the 
 high- and low- tension windings of the transformers? 
 
 45. A 6-phase, 1500-kilowatt, 650-volt rotary converter de- 
 livers power to a railway system. It receives power through 
 
100 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 three single-phase transformers from a three-phase, 13,200- 
 volt transmission line. The transformers are connected in Y on 
 the high-tension side and diametrically on the low-tension side. 
 In order to maintain a constant line voltage of 13,200 volts at the 
 substation the rotary is over-compounded so that when delivering 
 15 per cent, overload current at 650 volts it operates at 0.85 power 
 factor with an efficiency of 92.7 per cent. What is the ratio of 
 transformation for each transformer? What should be the 
 overload current ratings of the high- and low-tension windings of 
 the transformers? 
 
 46. A six-phase, 1000-kw., rotary converter receives power 
 through three transformers from a three-phase, 11,000-volt circuit. 
 The rotary delivers power at 650 volts to a railway system. The 
 transformers are connected in Y on the high-tension and in double 
 delta on the low-tension side. What should be the current and 
 voltage ratings and the ratio of transformation for each of these 
 transformers? In determining the rating neglect the losses. 
 
 47. A 6-phase, 2000-kw., rotary converter receives power from 
 a 3-phase, 13,000-volt transmission line. The transformers are 
 connected in delta on the high-tension and in double Y on the 
 low-tension side. The rotary delivers power at 650 volts to a 
 railway system. It has an overload current capacity of 20 per 
 cent, and an efficiency of 92.8 per cent, when delivering this load 
 and operating at 0.85 power factor. What should be the over- 
 load current and voltage ratings of the high- and low-tension 
 windings of the transformers? What should be their ratio of 
 transformation? 
 
 48. A 12-phase, 3000-kw., 650-volt rotating converter receives 
 power from a three-phase transmission line through three single- 
 phase transformers, whose high-tension windings are connected 
 in Y and whose low-tension windings are connected in double 
 chord. The high-tension line voltage is 22,000 volts. Neglect- 
 ing the losses what should be the current and voltage ratings of 
 the high- and low-tension windings of the transformers? 
 
 49. l A 6-phase, 25-cycle, 600-volt rotary converter has an 
 efficiency of 92.5 per cent, when delivering 750 kw. and operating 
 at 0.90 power factor. The armature winding has 4 inductors 
 
 1 In calculating the armature reaction do not neglect the distribution of 
 the winding, and assume that the constant usually given as 0.707 is 0.60. 
 This makes an approximate correction for the effect of the ratio of pole arc 
 to pole pitch. 
 
ROTARY CONVERTERS 101 
 
 in series per slot and 24 slots per pole. Neglect the resistance 
 and the leakage reactance of the armature winding. What are 
 the demagnetizing ampere turns per pole for the Speerfibd' 
 load? 
 
 50. 1 A 3-phase, 25-cycle rotary converter has a /^idl-lp'acV 
 capacity of 300 kw. at 600 volts. The armature has 96 slots 
 with 6 inductors in series per slot. The field structure has 4 
 poles. Neglect the resistance and leakage reactance of the arma- 
 ture winding. When this rotary receives a line current of 540 
 amperes at 367 volts and is overexcited so that it operates at a 
 power factor of 0.90 what are the demagnetizing ampere turns 
 per pole? 
 
 51. l A 6-phase, 60-cycle, 600-volt rotary converter has a full- 
 load capacity of 1000 kw. when operating at unit power factor. 
 The armature has 180 slots with 6 inductors in series per slot. 
 The field structure has 12 poles each of which is wound with 864 
 turns. When this rotary is overexcited so that it operates at 
 0.95 power factor and is delivering its rated output on the direct- 
 current side the efficiency is 92.9 per cent. How much greater 
 is the field current than its normal value, i.e., if the power factor 
 were unity? 
 
 52. l At full load and when operating at unit power factor a 
 3-phase, 25-cycle rotary converter receives a line current of 499 
 amperes at 367 volts and delivers on the direct-current side 500 
 amperes at 600 volts. The armature has 96 slots with 6 inductors 
 in series per slot. The field structure has 4 poles with 2340 turns 
 in the shunt windings per pole. The resistance of this field 
 circuit with the regulating rheostat cut out is 120.5 ohms. When 
 this rotary receives its rated current what is the least power 
 factor at which it can be operated and still maintain its rated 
 voltage? Normal excitation, i.e., when operating at unit 
 power factor, is 3.25 amperes. 
 
 53. l At full load and when operating at unit power factor a 6- 
 phase, 25-cycle compound rotary converter receives a line current 
 of 840 amperes at a voltage of 212 volts between adjacent slip 
 rings and delivers on the direct-current side 1667 amperes at 600 
 volts. The armature has 168 slots with 6 inductors in series per 
 
 1 In calculating the armature reaction do not neglect the distribution of 
 the winding, and assume that the constant usually given as 0.707 is 0.60. 
 This makes an approximate correction for the effect of the ratio of pole arc 
 to pole pitch. 
 
102 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 slot. The field structure has 8 poles with 1501 turns in the shunt 
 winding and 2 turns in the series winding per pole. A current of 
 .56 amperes in the shunt field winding alone is the normal excita- 
 tion and produces a voltage of 600 volts on the direct-current 
 .side at no load when running as a generator. 
 
 Whep" this rotary delivers 1500 amperes at 600 volts what should 
 be the shunt field current in order that the current on the alter- 
 nating-current side shall be leading and not exceed its full-load 
 value and the power factor have its least value? What is this 
 power factor? Neglect any change in the efficiency. 
 
 54. l At full load and when operating at unit power factor a 
 6-phase, 60-cycle rotary converter receives a line current of 840 
 amperes at a voltage of 424 volts between diametrical points of 
 the armature and delivers on the direct-current side a current of 
 1667 amperes at 600 volts. The armature has 15 slots per pole 
 with 6 inductors in series per slot. On each field pole there are 
 864 turns in the shunt winding and 2 turns in the series winding. 
 A current of 9.3 amperes in the shunt field winding alone is the 
 normal excitation and produces a voltage of 600 volts on the 
 direct-current side at no load when running as a generator. 
 When this rotary delivers 1460 amperes at 600 volts what are 
 the limits of the shunt field current in order that the current 
 on the alternating-current side shall not exceed its full-load 
 value and the power factor have its least value? What are 
 these limiting power factors? Neglect any change in the 
 efficiency. 
 
 55. l At full load and when operating at unit power factor a 
 6-phase, 25-cycle rotary converter takes a line current of 840 
 amperes at a voltage of 424 between diametrical points and 
 delivers 1667 amperes at 600 volts. The armature has 21 slots 
 per pole with 6 inductors in series per slot. Each field pole is 
 wound with 1501 turns. The resistance of the field circuit with 
 the regulating rheostat cut out is 76.6 ohms. A current of 5.6 
 amperes in the field winding is the normal excitation: i.e., for 
 full load and unit power factor. With an output of 850 kw. at 
 600 volts what is the least power factor at which this rotary can 
 be operated when overexcited? What per cent, is the current 
 
 1 In calculating the armature reaction do not neglect the distribution of 
 the winding, and assume that the constant usually given as 0.707 is 0.60. 
 This makes an approximate correction for the effect of the ratio of pole arc 
 to pole pitch. 
 
ROTARY CONVERTERS 103 
 
 on the alternating-current side of its full-load value? Neglect 
 any change in the efficiency. 
 
 56. J At full load and when operating at unit power factor a 
 6-phase, 60-cycle rotary converter takes a line current of 840 
 amperes at 212 volts per phase and delivers 1000 kw. at 600 
 volts. The armature has 180 slots with 6 inductors in series 
 per slot. The field structure has 12 poles with 864 turns in the 
 shunt winding and 2 turns in the series winding per pole. The 
 resistance of the shunt field circuit is 39.65 ohms with the regu- 
 lating rheostat cut out. A current of 9.3 amperes in the shunt 
 field alone is the normal excitation, i.e., for full load and unit 
 power factor. With an output of 800 kw. at 600 volts what is 
 the least power factor at which the rotary can be operated when 
 overexcited? What per cent, is the current of its full-load value? 
 Neglect any change in this efficiency. 
 
 57. l A 3-phase, 25-cycle rotary converter has a full-load capa- 
 city of 300 kw. at 600 volts. The armature has 96 slots with 
 6 inductors in series per slot. The field structure has 4 poles 
 with 2340 turns in the shunt winding per pole. The resistance 
 of this field circuit with all of the regulating rheostat cut out is 
 120.5 ohms. A current of 3.25 amperes in the shunt field winding 
 alone gives the normal excitation, i.e., for full load at unit power 
 factor. In order to maintain the terminal voltage on the direct- 
 current side at 600 volts when full load is delivered, it is neces- 
 sary to overexcite the rotary so that it takes a line current of 
 600 amperes at a power factor of 86 per cent. 
 
 Calculate the least number of series field turns that must be 
 added in order that this excitation may be produced. What 
 should be the current in the shunt field winding? 
 
 58. l Calculate the efficiency of the rotary A when it receives 300 
 kw. at a power factor of 0.90 and with a line voltage of 380 
 volts. The excitation is greater than normal. Assume that 
 the running temperature is 70 C. 
 
 59. 1 Calculate the efficiency of the rotary A when it delivers 300 
 kw. at 625 volts and the excitation is greater than normal and is 
 adjusted so that the line current on the alternating-current side 
 is 520 amperes. Assume that the running temperature is 70 C. 
 
 1 In calculating the armature reaction do not neglect the distribution of 
 the winding, and assume that the constant usually given as . 707 is . 60. 
 This makes an approximate correction for the effect of the ratio of pole arc 
 to pole pitch. As a first approximation it may be assumed that the effi- 
 ciency is 95 per cent. 
 
 8 
 
104 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 DATA ON ROTARY CONVERTERS 
 
 
 A 
 
 B 
 
 C 
 
 Kilowatt 
 
 300 
 
 1000 
 
 1000 
 
 Voltage (D.C.) 
 
 600 
 
 600 
 
 600 
 
 Current (D.C.) 
 
 500 
 
 1667 
 
 1AA7 
 
 Phases 
 
 3 
 
 6 
 
 Q 
 
 Frequency 
 
 25 
 
 25 
 
 60 
 
 Poles 
 
 4 
 
 8 
 
 12 
 
 Armature slots 
 
 96 
 
 168 
 
 180 
 
 Inductors per slot 
 Turns per pole: 
 Shunt. 
 
 6 
 2340 
 
 6 
 1501 
 
 6 
 864 
 
 Series 
 
 9 
 
 2 
 
 2 
 
 Resistance at 25 C. : 
 Armature between D.C. brushes 
 Shunt field 2 (alone) 
 
 0.030 
 120.5 
 
 0.00688 
 76 6 
 
 0.00589 
 39 65 
 
 Series field 
 Friction and windage (kw.) 
 
 0.00248 
 3.2 
 
 0.00042 
 6.7 
 
 0.0006095 
 9.1 
 
 D. C. SATURATION CURVES AND CORE LOSSES 
 
 Ordinates 
 
 Abscissae 
 
 D.C. 
 
 voltage 
 
 
 Shunt 
 
 field currents 
 
 Core 
 
 losses 
 
 
 
 A 
 
 
 B 
 
 
 C 
 
 A 
 
 I 
 
 B \ C 
 
 
 550 
 600 
 650 
 700 
 
 2 
 
 3 
 4 
 5 
 
 .70 
 
 .25 
 .00 
 .00 
 
 4 
 5 
 6 
 
 7 
 
 .90 
 
 .60 
 .36 
 .44 
 
 7 
 9 
 11 
 14 
 
 9 
 3 
 2 
 
 
 1.96 
 2.52 
 3.25 
 4.23 
 
 4 
 5 
 6 
 
 7 
 
 .68 
 .50 
 .40 
 .70 
 
 12. 
 
 14. 
 17. 
 21. 
 
 4 
 
 7 
 7 
 6 
 
 60. l Calculate the efficiency of the rotary B when it delivers 
 1000 kw. at 650 volts and the shunt field rheostat is cut out. 
 Assume that the running temperature is 70 C. What are the 
 line current and power factor on the alternating-current side? 
 
 61. * Calculate the efficiency of the rotary C when it delivers 
 1000 kw. at 650 volts and the excitation is greater than normal 
 and is adjusted so that the power factor is 0.95. Assume that 
 the running temperature is 70 C. 
 
 1 In calculating the armature reaction do not neglect the distribution of 
 the winding, and assume that the constant usually given as 0.707 is 0.60. 
 This makes an approximate correction for the effect of the ratio of pole arc 
 to pole pitch. 
 
 2 The shunt field has a regulating rheostat in series with it, the loss in 
 whichjshould be included in calculating the efficiency. 
 
CHAPTER VI 
 POLYPHASE CIRCUITS 
 
 1. Three equal impedance units, each of which has an equiva- 
 lent resistance of 2.0 ohms and a reactance of 1.25 ohms are 
 connected in delta across a three-phase 220-volt circuit. What 
 current does each unit take? What is the line current? What is 
 the total power supplied? 
 
 2. The three impedance units described in problem 1 are con- 
 nected in Y across a three-phase, 220-volt circuit. What cur- 
 rent does each unit take? What is the total power supplied? 
 
 3. Six equal impedance units each of which has an equivalent 
 resistance of 2.5 ohms and a reactance of 1.5 ohms are connected 
 across a three-phase, 220-volt circuit three in delta and three in Y. 
 What is the line current? What is the total power supplied? 
 
 4. Three equal impedance units each of which has an equivalent 
 resistance of 2 ohms and a condensive reactance of 1 ohm are 
 connected in delta across a three-phase, 220-volt circuit. At the 
 same point three other equal impedance units, each of which has 
 an equivalent resistance of 1.5 ohms and an inductive reactance 
 of 1 ohm, are connected in Y across the circuit. What is the line 
 current? What is the total power supplied? 
 
 5. Three equal resistances are connected in delta across a three- 
 phase circuit. What should be the relative value of three other 
 equal resistances which will take the same power when connected 
 in Y across the circuit? 
 
 6. Three equal impedance units, each of which has an equiva- 
 lent resistance of 2.0 ohms and a reactance of 1.0 ohm are con- 
 nected in delta across a three-phase, 220-volt circuit. Three 
 other equal impedance units are connected in Y across the same 
 circuit. What should be their equivalent resistance and react- 
 ance in order that they will take the same line current and the 
 same total power? 
 
 7. Three equal impedance units each of which has an equiva- 
 lent resistance of 2.0 ohms and an inductive reactance of 1.0 
 ohm are connected in delta at the end of a transmission line, each 
 conductor of which has a resistance of 0.2 ohm and an inductive 
 
 105 
 
106 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 reactance of 0.3 ohm. If the line voltages at the generating 
 station are each 2200 volts what is the line current? (b) What 
 is the voltage at the load? (c) What is the efficiency of trans- 
 mission? 
 
 8. In problem 7 if the reactance of the impedance units had been 
 condensive instead of inductive what would have been (a) the 
 line current, (b) the voltage at the load, and (c) the efficiency of 
 transmission? 
 
 9. Two equal resistances of 100 ohms each are connected in 
 series across two mains of a three-phase 220-volt circuit and 
 from their junction a resistance of 50 ohms is connected to the 
 neutral conductor of the system. The line voltages are balanced 
 and the voltages from the lines to the neutral conductor are 
 equal, (a) What are the line currents? (b) What is the neutral 
 current? (c) What is the total power absorbed? 
 
 10. Two equal impedances, each of which has an equivalent 
 resistance of 2.0 ohms and an inductive reactance of 1.0 ohm 
 are connected in series across two mains of a three-phase, 220-volt 
 circuit, and from their junction another unit which has a resist- 
 ance of 1.0 ohm and a condensive reactance of 1.0 ohm is connected 
 to the neutral conductor of the system. The line voltages are 
 balanced and the voltages from the lines to the neutral are equal, 
 (a) What are the line currents? (b) What is the neutral current? 
 (c) What is the total power absorbed? 
 
 11. Three non-inductive resistances of 5, 10 and 15 ohms are 
 connected in delta across the lines of a three-phase, 220-volt 
 circuit, (a) What is the total power absorbed? (b) What are 
 the line currents? 
 
 12. Three impedance units which are represented by the 
 expressions, Zi = 5+./5, z 2 = 5+j!Q, z 3 = 5 zjlQ, are connected in 
 delta across the lines 1-2, 2-3, 3-1 respectively of a three-phase, 
 220-volt circuit. If Viz leads VM by 120 degrees (a) what is 
 the total power absorbed? (b) What are the line currents? 
 
 If Viz lags F 2 3 by 120 degrees, (c) what is the total power 
 absorbed? (d) What are the line currents? 
 
 13. Three non-inductive resistances of 5, 10 and 15 ohms are 
 connected in Y across the lines of a three-phase, 220-volt circuit. 
 (a) What is the total power absorbed? (b) What are the line 
 currents? 
 
 14. Three impedance units which are represented by the ex- 
 pressions, zi 5+J5, 22 = 5+jlO, 23 = 5 j'10, are connected in Y 
 
POLYPHASE CIRCUITS 107 
 
 from the mains 1, 2 and 3 respectively of a three-phase, 220-volt 
 circuit to a common point. If V i2 leads Vzs by 120 degrees, (a) 
 what are the line currents? (b) What is the total power absorbed? 
 (c) What is the voltage between the "common point" and the 
 true neutral of the system? 
 
 15. In problem 14 if F 2 3 leads Viz by 120 degrees a what are 
 the line currents? (b) What is the total power absorbed? (c) 
 What is the voltage between the " common point" and the true 
 neutral of the circuit? 
 
 16. Three impedance units which are represented by the ex- 
 pressions Zi=10-fjO, 2 2 = 0+jlO, 3 = jW, are connected in 
 Y from the mains 1, 2 and 3 respectively of a three-phase, 
 220-volt circuit to a common point. If Viz leads Vzs by 120 de- 
 grees (a) what are the line currents? (b) What is the total power 
 absorbed? (c) What is the voltage between the " common point" 
 and the true neutral of the system? 
 
 17. In problem 16 if Vzz leads Viz by 120 degrees (a) what are 
 the line currents? (b)What is the total power absorbed? (c) 
 What is the voltage between the "common point" and the true 
 neutral of the system? 
 
 18. Three voltmeters are connected in Y from the mains of a 
 three-phase, 220-volt circuit to a common point. The resistances 
 of the voltmeters are Ri ohms and Rz ohms and R 3 ohms respec- 
 tively. What does each instrument indicate? 
 
 19. Three unequal lamp loads are connected between the 
 mains and neutral conductor of a three-phase transmission line. 
 The mains and neutral conductor each have a resistance of 0.1 
 ohm and negligible reactance. The resistances of the lamp 
 loads are 1.0 ohm, 1.5 ohms and 2.0 ohms respectively. At 
 the generating station the line voltages are each 220 volts and the 
 voltages between the mains and neutral conductors are equal, 
 (a) What are the line and neutral currents? (b) What is the 
 voltage across each lamp load? (c) What is the efficiency of 
 transmission? 
 
 20. Three unequal lamp loads are connected in delta at the 
 end of a three-phase transmission line which has a resistance of 
 0.1 ohm per conductor and negligible reactance. The resistances 
 of the lamp loads are 1.0 ohm, 1.5 ohms and 2.0 ohms respectively. 
 If the line voltages at the generating station are each 220 volts, 
 (a) what are the line currents? (b) What are the line voltages at 
 the load? (c) What is the efficiency of transmission? 
 
108 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 21. Three impedance units whose values are represented by 
 the expressions, 2i = 2.0+./ 1.0, 22 = 1.5+.? 0.5, and z 3 = 2.5 + 
 j 1.0, are connected in delta at the end of a three-phase transmis- 
 sion line each conductor of which has a resistance of 0.1 ohm and 
 a reactance of 0.15 ohm. If the line voltages at the generating 
 station are each 11,000 volts, (a) what are the line currents? 
 (b) What are the line voltages at the load? (c) What is the effi- 
 ciency of transmission? 
 
 22. The power supplied to a three-phase induction motor is 
 measured by the two- wattmeter method. One wattmeter indi- 
 cates 5770 watts and the other 2930 watts. What is the power 
 supplied? At what power factor is the motor operating? 
 
 23. The only instrument available for measuring the power 
 taken by a three-phase, 230-volt induction motor is a wattmeter 
 of suitable range. Measurements are made as follows: The 
 current coil of the wattmeter is inserted in main 1 and the poten- 
 tial coil, first between mains 1 and 2 and then between mains 
 1 and 3. If the two wattmeter readings thus obtained are 5760 
 and 3380 watts respectively what is the power supplied to the 
 motor? If the line voltage is 230 volts what is the line current? 
 At what power factor is this motor operating? 
 
 24. The power taken by a balanced three-phase load is meas- 
 ured by two wattmeters. The current coils of the wattmeters are 
 connected to current transformers which are in lines 1 and 2 
 respectively. The potential coils are connected to potential 
 transformers which are across lines 2 and 3 and lines 1 and 3 
 respectively. The line voltages are each 230 volts and the line 
 currents are each 150 amperes. The wattmeters each indicate 19.6 
 kw. What is the power supplied? What is the power factor? 
 
 25. The power taken by an unbalanced three-phase load is 
 measured by two wattmeters. The current coils of the watt- 
 meters are connected to current transformers which are in lines 
 1 and 2 respectively, and the potential coils are connected to 
 potential transformers which are across lines 2 and 3 and lines 1 
 and 3 respectively. The line voltages are each 230 volts. The 
 currents in lines 1 and 2 are 150 amperes and 200 amperes re- 
 spectively. The first wattmeter indicates 21. 2 kilowatts and the 
 second indicates 18.1 kilowatts. What is the power supplied to 
 the load? 
 
 26. A 3-phase, 500-volt, Y-connected alternating-current 
 generator with equal line voltages and a grounded neutral supplies 
 
POLYPHASE CIRCUITS 109 
 
 energy to an unbalanced Y-connected load, the neutral of which is 
 not grounded. The line currents are, /i = 141.4 amperes, 7 2 = 100 
 amperes and 7 3 = 100 amperes. One wattmeter is used and it is 
 connected with its current coil in line 1 and its potential coil 
 across lines 1 and 2. If this wattmeter indicates 70.7 kw. what 
 is the total power supplied to the load? If the power factors 
 of each of the three phases are equal what is the voltage between 
 the neutral of the load and the neutral of the generator? What 
 is the power factor? 
 
 27. An unbalanced lamp load, consisting of 115-volt lamps, is 
 connected in Y across the lines of a balanced 3-phase, 200-volt 
 circuit. The line currents are 70.7 amperes, 50 amperes and 50 
 amperes respectively. What is the power supplied to this load? 
 What is the voltage across each phase of the load? 
 
 28. In problem 27 i the resistances of the lamp load are 
 assumed to be constant, what will be the currents in the lines and 
 neutral when the neutral point of the load is connected to the 
 neutral conductor of the circuit? The voltages between the 
 neutral conductor and the lines are equal. 
 
 29. An unbalanced lamp load is connected in delta across the 
 lines of a balanced 3-phase, 230-volt circuit. The resistances 
 of these loads between lines 1 and 2, 2 and 3, and 3 and 1 are 10 
 ohms, 8 ohms and 6 ohms respectively, (a) What are the line 
 currents? (b) If the power is measured by two wattmeters which 
 have their current coils in lines 1 and 2 what will each instrument 
 indicate? 
 
 30. An unbalanced lamp load, consisting of 115-volt lamps, is 
 connected between the lines and neutral conductor of a balanced 
 3-phase, 200-volt circuit. The resistances of the loads between 
 lines 1, 2, and 3 and the neutral conductor are 6, 8, and 10 ohms 
 respectively. The line voltages are equal and the voltages from 
 the lines to the neutral conductor are also equal. What would 
 two wattmeters indicate which have their current coils in lines 
 1 and 2 and their potential coils across lines 1 and 3 and lines 2 and 
 3 respectively? What is the total power supplied? What is the 
 current in the neutral conductor? 
 
 31. Three single-phase transformers each of which has a ratio 
 of transformation of 2.5 to 1 are connected in delta on the high- 
 tension side. The low-tension windings are not connected but 
 supply three separate single-phase loads. The first of these loads 
 is 90 kilowatts at unit power factor, the second is 60 kilowatts 
 
110 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 at 0.7 power factor, and the third is 30 kilowatts at unit power 
 factor. The high-tension line voltages are 600 volts. Neglect 
 the losses in the transformers. What are the high-tension line 
 currents? 
 
 32. Three unequal single-phase motor loads are connected 
 across the lines of a balanced 3-phase, 230-volt circuit. The first 
 takes 106 amperes at 0.78 power factor, the second takes 142 
 amperes at 0.82 power factor, and the third takes 28.4 kilowatts 
 at 0.77 power factor. What are the line currents? 
 
 33. Three unequal single-phase motor loads are connected 
 between the lines and neutral conductor of a balanced 3-phase, 
 350-volt circuit. The voltages from the lines to the neutral are 
 each 202 volts. The first load takes 20 kilowatts at 0.82 power 
 factor, the second takes 28 kilowatts at 0.75 power factor, and the 
 third takes 36 kilowatts at 0.79 power factor. What is the 
 current in the neutral conductor? 
 
 34. From the terminals of a 3-phase, 550-volt, 60-cycle 
 alternating-current generator runs an artificial transmission line 
 which has a resistance of 0.1 ohm and an inductance of 1.0 milli- 
 henry per conductor. At the end of this line is a balanced load of 
 three reactors connected in Y. The equivalent resistance and 
 reactance of these reactors should be assumed to be constant. 
 The power output of the generator is measured by the two-watt- 
 meter method with the current coils of the instruments in lines 1 
 and 2. The first wattmeter indicates 40 kw. and the second, 
 100 kw. 
 
 If line 3 is opened at the load what power will the instruments 
 indicate? The terminal voltage of the generator is constant. 
 
 35. In problem 34 if line 3 is opened between the terminal 
 of the generator and the potential coils of the wattmeter, what 
 power will the instruments indicate? 
 
 36. From the terminals of a 3-phase, 500-volt, 25-cycle alter- 
 nating-c n rrent generator runs an artificial transmission line which 
 has a resistance of 0.3 ohm and an inductance of 3 millihenrys 
 per conductor. At the end of this line is a balanced load of three 
 reactors connected in delta. The equivalent resistance and 
 reactance of these reactors should be assumed to be constant. 
 The power output of the generator is measured by the two- 
 wattmeter method with the current coils of the instruments in 
 lines 1 and 2. The first wattmeter indicates 50 kw. and the 
 second, 25 kw. 
 
POLYPHASE CIRCUITS 111 
 
 If line 3 is opened what power will the instruments indicate? 
 The terminal voltage of the generator is constant. 
 
 37. In problem 36 if line 3 is opened between the terminal of 
 the generator and the potential coils of the wattmeters what 
 power will each instrument indicate? The terminal voltage of 
 the generator is constant. 
 
 38. Two alternating-current generators, operating in parallel, 
 deliver power to a balanced, 3-phase load. The output of each 
 generator is measured by the two-wattmeter method. The 
 terminal voltage is 2210 volts. The wattmeter readings are: 
 
 First generator Second generator 
 
 TFi = 196kw. TFi = 172kw. 
 
 TF 2 = 312kw. TF 2 = 88kw. 
 
 The similarly numbered instruments are connected in the same 
 lines. What is the total power supplied? What is the power 
 factor of the load? 
 
 39. Two alternating-current generators, operating in parallel, 
 deliver power to a balanced 3-phase load. The output of each 
 generator is measured by the two-wattmeter method. The ter- 
 minal voltage is 6650 volts. The wattmeter readings are: 
 
 First generator Second generator 
 
 W i = 412 kw. Wi = 183 kw. 
 
 TF 2 = 626 kw. W 2 = 457 kw. 
 
 The similarly numbered instruments are connected in the same 
 lines. What is the total power supplied? What is the power 
 factor of the load? 
 
 40. It is desired to transform 200 kw. from 2-phase to 3-phase 
 by Scott-connected transformers. The two-phase line voltage 
 is 2200 volts and the three-phase line voltage is 230 volts. What 
 should be the current and voltage rating, and the ratio of trans- 
 formation of each transformer? 
 
 41. It is desired to transform 100 kw. from 3-phase at 6600 
 volts to 2-phase at 110 volts by the means of Scott-connected 
 transformers. What should be the current and voltage rating, 
 and the ratio of transformation of each transformer? 
 
 42. Three single-phase transformers are connected in delta on 
 both the primary and secondary sides. With a primary voltage 
 on open circuit of 22,000 volts the secondary voltage is 440 volts. 
 The short-circuit characteristic data of each transformer are: 
 
112 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 V = 1020 volts, /= 1.136 amperes, P = 351 watts. If the second- 
 ary terminal voltage is 440 volts when there is a single-phase 
 load taking 40 kw. at 0.8 power factor connected across one phase 
 of the secondary, what will be the percentage rise in this voltage 
 when the load is thrown off? 
 
 43. (a) On the basis of the same heating loss compare the 
 full-load kilowatt output at unit power factor of two single-phase 
 transformers connected in open delta with their name-plate 
 rating, (b) On the basis -of the same heating loss compare the 
 full-load kilowatt output at 0.87 power factor both lagging and 
 leading of two single-phase transformers connected in open delta 
 with 87 per cent, of their name-plate rating. In each case the load 
 is balanced. 
 
 44. (a) Three single-phase transformers with both primaries 
 and secondaries connected in delta supply a balanced load of 100 
 kw. at unit power factor. If it is necessary to remove one of these 
 transformers from the line by what per cent, will the copper 
 loss in each of the other two transformers be increased? (b) 
 If the load had been a balanced one taking 87 kw. at 0.87 power 
 factor both leading and lagging -what would have been the 
 per cent, increase in the copper loss in each of the other 
 transformers? 
 
 45. (a) On the basis of the same copper loss in each trans- 
 former compare the full-load kilowatt output at unit power 
 factor of two equal single-phase transformers connected in T 
 with their name-plate rating, (b) On the basis of the same copper 
 loss in each transformer compare the full-load kilowatt output 
 at 0.87 power factor of two equal single-phase transformers 
 connected in T with 87 per cent, of their name-plate rating. 
 
 46. On the basis of the same copper loss in each transformer 
 compare the full-load kilowatt output of two equal single-phase 
 transformers connected in open delta with the full-load output ol 
 the same transformers connected in T (a) at unit power factor, 
 (b) at 0.87 power factor both lagging and leading. 
 
 47. Compare the regulation of three lOO-kv.-a. transformers 
 whose primaries and secondaries are connected in delta with 
 that of two 100-kv-a. transformers whose primaries and second- 
 aries are connected in open delta when a single-phase load of 
 100 kw. at 0.8 power factor is delivered on the secondary side. 
 The secondary or low-tension voltage under load conditions is 2200 
 volts and the ratio of transformation is 5 to 1. The short-circuit 
 
POLYPHASE CIRCUITS 113 
 
 of each transformer are: F = 310 volts, 7 = 9.1 amperes (full-load) 
 P = 1000 watts. In the case of the open delta the load is connected 
 across one transformer. 
 
 48. In problem 47 compare the regulation in the two cases 
 when a single-phase load of 150 kw. at 0.8 power factor is sup- 
 plied, which in the case of the open delta is connected across the 
 terminals of the two transformers, i.e., across the open side. 
 
 49. Three transformers whose primaries are connected in Y 
 and whose secondaries are connected in delta are in parallel on the 
 primary side with three others whose primaries and secondaries 
 are both connected in Y. If one secondary terminal of the first 
 set is connected to one corresponding terminal of the second set 
 what are the greatest and least voltages that can exist between 
 the other secondary terminals of the two sets? The line voltages 
 on the secondary sides are 1 100 volts for the two sets of transformers. 
 
 50. Three transformers whose primaries and secondaries are 
 both connected in delta are in parallel on the primary side with 
 three others whose primaries are connected in delta and whose 
 secondaries are connected in Y. If one secondary terminal of 
 the first set is connected to a corresponding terminal of the 
 second set, what are the greatest and least voltages that can exist 
 between the other secondary terminals of the two sets. The line 
 voltages are 1100 volts for the two sets of transformers. 
 
 51. Three transformers whose primaries and secondaries are 
 both connected in delta are in parallel on the primary side with 
 three others whose primaries are connected in Y and whose second- 
 aries are connected in delta. If one secondary terminal of the first 
 set is connected to a corresponding terminal of the second set 
 what are the greatest and least voltages that can exist between 
 the other secondary terminals of the two sets. The line voltages 
 are 1100 volts for the two sets of transformers. 
 
 52. Three auto-transformers are connected as shown in Fig. 3 
 to receive power from a 3-phase, 11,000-volt circuit. The ratio 
 of transformation for each transformer from high tension to low 
 tension is 2 to 1. What is the secondary line voltage, and what is 
 the phase relation of the corresponding primary and secondary 
 line voltages on open circuit? 
 
 53. Three auto-transformers are connected as shown in Fig. 4 
 to receive power from a 3-phase, 11,000-volt circuit. The ratio 
 of transformation f ron high tension to low tension is 2 to 1 . What 
 is the secondary line voltage, and what is the phase relation of 
 
114 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 the corresponding primary and secondary line voltages on open 
 circuit? 
 
 54. Three auto-transformers are connected as shown in Fig. 5 
 to receive power from a 3-phase, 11,000-volt circuit. The ratio 
 of transformation for each transformer from high tension to low 
 tension is 2 to 1. What is the secondary line voltage, and what 
 
 L.T. 
 
 L.T. 
 
 H.T. 
 
 FIG. 4. 
 
 is the phase relation of the corresponding primary and secondary 
 line voltages on open circuit? 
 
 55. A 1500-kw., 5500-volt, 3-phase generator delivers power 
 to a transmission line through three single-phase transformers 
 which have their low-tension windings connected in delta and their 
 high-tension windings connected in Y. The following data on this 
 generator and the transformers are given: 
 
 GENERATOR 
 
 Field current 
 
 Open-circuit terminal 
 voltage 
 
 Short-circuit armature 
 current 
 
 150 
 200 
 250 
 
 5100 
 5900 
 6500 
 
 300 
 400 
 
 300 
 
 6800 
 
 
 350 
 
 7100 
 
 
 The core loss at the rated voltage is 20.2 kw. and the friction 
 and windage is 8.4 kw. Both of these losses may be assumed 
 constant. The effective resistance of the armature is 0.362 ohm 
 per phase. The resistance of the field winding is 0.376 ohm. 
 The armature windings are connected in Y. 
 
 TRANSFORMER 
 
 Voltage 
 
 Short circuit 
 
 -ivv.-a, 
 
 High tension 
 
 Low tension 
 
 Amperes 1 
 
 Volts 
 
 Watts 
 
 500 
 
 12,700 
 
 5500 
 
 39.4 
 
 332 
 
 4680 
 
 1 Full-load current. 
 
POLYPHASE CIRCUITS 
 
 115 
 
 The core loss at the rated voltage is 3330 watts and may be 
 assumed constant. 
 
 (a) When the power delivered to the transmission is 1280 kw. 
 at a power factor of 0.83 and a line voltage of 22,300 volts, what 
 is the combined efficiency of the generator and transformers? 
 Calculate the generator field current by the magnetomotive-force 
 method. 
 
 (b) If this load is removed from the line and the field excita- 
 tion of the generator is unchanged, what is the high-tension line 
 voltage? 
 
 56. A 1640-kw., 13,500-volt, 3-phase generator delivers 
 power to a transmission line through three single-phase trans- 
 formers which are connected in Y on both the high- and low- 
 tension sides. The following data are given on this generator 
 and transformers: 
 
 GENERATOR 
 
 Field current 
 
 Open-circuit terminal 
 voltage 
 
 Short-circuit armature 
 current 
 
 50 
 100 
 150 
 200 
 
 7,500 
 10,100 
 14,700 
 15 800 
 
 75 
 155 
 
 227 
 
 250 
 
 16.700 
 
 
 The core loss at the rated voltage is 21.3 kw., and the friction 
 and windage is 8.8 kw. Both of the losses may be assumed con- 
 stant. The effective resistance of the armature is 1.52 ohms per 
 phase. The resistance of the field circuit is 0.392 ohms. The 
 armature windings are connected in Y. 
 
 TRANSFORMERS 
 
 Kv.-a 
 
 Voltage 
 
 Short circuit 
 
 High tension 
 
 Low tension 
 
 Amperes 1 
 
 Volts 
 
 Watts 
 
 500 
 
 38,000 
 
 7800 
 
 13.15 
 
 1190 
 
 3375 
 
 The core loss at the rated voltage is 2960 watts, and may be 
 assumed constant. 
 
 (a) When the output of the generator is 1460 kw., at 0.92 
 power factor and a terminal voltage of 13,600 volts what is the 
 combined efficiency of the generator and transformers? Cal- 
 culate field current of the generator by the magnetomotive-force 
 method. 
 
 1 Full-load current. 
 
116 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 (b) If this load is removed from the line and the field excita- 
 tion of the generator is unchanged what is the high-tension line 
 voltage? 
 
 57. A 1000-kw., 2400-volt, 3-phase generator delivers power 
 to a transmission line through three single-phase transformers 
 which are connected in delta on the low-tension and in Y on the 
 high-tension side. Each transformer has a ratio of transforma- 
 tion of 5.29 to 1. The resistances of the high- and low-tension 
 windings are 2.02 ohms and 0.072 ohm respectively. The effective 
 resistance of the generator is 0.067 ohm per phase. The armature 
 windings are connected in Y. With the high-tension windings 
 of the transformers short circuited and with a field excitation 
 of 100 amperes for the generator the armature current is 458 
 amperes. With this same excitation the open-circuit terminal 
 voltage of the generator is 2220 volts. The rotation losses in the 
 generator are 31.2 kw. at normal voltage and the core losses 
 in each transformer are 1.8, kw. 
 
 (a) What is the combined efficiency of the generator armature 
 and the transformers when a balanced load of 954 kw. at 0.91 
 power factor is delivered on the high-tension side of the trans- 
 formers at a line voltage of 22,400 volts? 
 
 (b) What would be the high-tension line voltage if this load 
 were removed and the excitation of the generator unchanged? 
 
 58. A 760-kw., 2200-volt, 3-phase generator delivers power to 
 a transmission line through three single-phase transformers which 
 have both their high and low-tension windings connected in Y. 
 With the high-tension windings of the transformers short-cir- 
 cuited the output of the generator is 37.8 kilowatts, the armature 
 current is 450 amperes when the terminal voltage is 133 volts. 
 With the transformers on open circuit and with the same field 
 excitation, the terminal voltage of the generator is 1780 volts. 
 The effective resistance of the armature is 0.172 ohm per phase. 
 The armature windings are connected in Y. The rotation losses 
 of the generator are 17.2 kw. at normal voltage and the core loss 
 in each transformer is 1670 watts. The transformers have a 
 ratio of transformation of 10 to 1. 
 
 (a) What is the combined efficiency of the generator armature 
 and the transformers when a balanced load of 680 kw. at 0.9 
 power factor is delivered to the transmission line at 22,400 volts? 
 
 (b) What would be the high-tension line voltage if this load 
 were removed and the excitation of the generator unchanged? 
 
POLYPHASE CIRCUITS 117 
 
 59. A 1000-kw., 11,000-volt, 3-phase generator delivers power 
 to a transmission line at the end of which is a 1200-h.p. induction 
 motor. With the induction motor running at no load the exci- 
 tation of the generator is adjusted so that the terminal voltage, 
 the line current and the total power measured at the motor are 
 10,600 volts, 20.2 amperes and 20.4 kilowatts respectively. At 
 the same time the terminal voltage and the total power measured 
 at the generator are 10,980 volts and 29.8 kw. The field excita- 
 tion of the generator is adjusted so that, when the motor is deliver- 
 ing full load, its terminal voltage is 11,000 volts. The efficiency 
 and power factor of the motor at full load are 0.921 and 0.906 
 respectively. 
 
 What is the terminal voltage of the generator when the motor 
 delivers full load? What is the efficiency of transmission at this 
 time? 
 
 60. A 1000-kw., 13,800-volt, 3-phase generator delivers power 
 directly to a transmission line at the end of which is an induction 
 motor load. The resistance and reactance of the transmission 
 line are 16.8 ohms and 17. 2 ohms per conductor. The generator 
 has an effective armature reactance of 2.18 ohms per phase. The 
 field current is supplied at 120 volts. 
 
 Field 
 current 
 
 Open-circuit terminal 
 voltage 1 
 
 Terminal voltage with an armature 
 current of 42 amp. at zero 
 power factor 
 
 50 
 
 8,800 
 
 
 80 
 
 13,000 
 
 
 110 
 140 
 180 
 
 15,600 
 17,250 
 18,900 
 
 10,750 
 13,250 
 15,600 
 
 On open circuit the rotational losses are 16.6 kw. and 25.4 kw. 
 when the terminal voltages are 13,000 and 15,600 volts respec- 
 tively. 
 
 What should be the field excitation of the generator so 
 that the line voltage at the motor load will be 13,200 volts when 
 the motors take 926 kw. at 0.91 power factor? What is the effi- 
 ciency of the generator and the line? 
 
 61. At the end of a 3-phase transmission line is a motor load 
 requiring 3000 kw. The line voltage at the load should be 32,000 
 volts and the power factor of the load is 0.90. What should be 
 
 1 The generator is Y-connected. 
 
118 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 the resistance and reactance of the line per conductor so that 
 the efficiency of transmission will be 90 per cent, and the voltage 
 regulation, 12 per cent? 
 
 62. A 1640-kv.-a., 13,500-volt, 3-phase generator delivers 
 power directly to a transmission line which has a resistance of 
 30.2 ohms and a reactance of 24 ohms per conductor. With the 
 far end of the line short-circuited and with a field excitation of 150 
 amperes the line current is 138 amperes. With the same field 
 excitation the open-circuit terminal voltage of the generator is 
 14,700 volts. The effective resistance of the armature is 1.52 
 ohms per phase. The generator is Y-connected. 
 
 What is the combined electrical efficiency of the armature of 
 the generator and the transmission line for a load of 1500 kw. 
 at 0.90 power factor if the line voltage at the load is 13,200 volts? 
 To what value will the line voltage rise if this load is removed and 
 the field excitation of the generator is unchanged? 
 
 63. A 1000-kv.-a., 11,000-volt, 3-phase generator delivers 
 power directly to a transmission line which has a resistance of 
 8.42 ohms and a reactance of 6.8 ohms per conductor. 
 
 Field current 
 
 Open-circuit terminal voltage 
 
 Rotational losses 
 
 20 
 
 30 
 40 
 50 
 
 8,400 
 11,000 
 12,700 
 13,800 
 
 11,600 
 19,400 
 25,600 
 
 60 
 
 14700 
 
 
 70 
 
 15.500 
 
 
 The armature has an effective resistance of 0.94 ohm per phase, 
 and the windings are connected in Y. The field current is supplied 
 at 260 volts. With the far end of the line short-circuited 
 and with a field current of 40 amperes the line current is 115 
 amperes . 
 
 If the load at the end of the line requires 940 kw. at 0.92 power 
 factor what must be the excitation of the generator in order that 
 the line voltage at the load shall be 11,000 volts? 
 
 64. A 1000-kw., 13,800-volt, 3-phase alternating-current 
 generator^ delivers power over a transmission line to a synchronous 
 motor load. The resistance and reactance of the line are 16.5 
 and 17.1 ohms per conductor respectively. The armature of 
 the generator has an effective resistance of 2.18 ohms per phase. 
 The field current is supplied with 120 volts. 
 
POLYPHASE CIRCUITS 
 
 119 
 
 Field current 
 
 Open-circuit 
 terminal voltage 1 
 
 Core loss at 
 open circuit 
 
 50 
 
 80 
 110 
 140 
 180 
 
 8,800 
 13,000 
 15,600 
 17,250 
 18,900 
 
 7.5 
 
 16.6 
 25.4 
 33.5 
 
 With the synchronous motor running at no load and with under 
 excitation the line current has its full-load value of 42 amperes, 
 the line voltage at the motor is 13,200 volts and the power 
 supplied to the motor is 28.6 kw. At this time the field excita- 
 tion of the generator is 161 amperes. 
 
 When the motor load requires 1000 kw. what must be the 
 excitation of the generator in order that it shall operate at unit 
 power factor 2 and the line voltage at the load shall be 13,200 
 volts? What is the power factor of the load? 
 
 65. A 3500-kw., 10,000-volt, 2-phase alternating-current 
 generator delivers power through Scott transformers to a three- 
 phase transmission line. With the far end of this line short- 
 circuited the output of the generator is 490 kw. and the armature 
 current and terminal voltage are 220 amperes and 1630 volts. 
 On open circuit the high-tension line voltage is 33,000 volts when 
 the terminal voltage of the generator is 10,000 volts. The arma- 
 ture has an effective resistance of 0.64 ohm per phase The open- 
 and short-circuit characteristics are: 
 
 Field current 
 
 Open-circuit 
 terminal voltage 
 
 Short-circuit 
 armature current 
 
 100 
 200 
 250 
 
 7,700 
 10,200 
 10900 
 
 220 
 430 
 
 300 
 
 11,500 
 
 
 375 
 
 12.200 
 
 
 If there is a load requiring 3200 kw. at 0.91 power factor at the 
 end of the line, what must be the excitation of generator in order 
 that the line voltage at the load shall be 30,000 volts? (Use the 
 magnetomotive-force method when dealing with the generator.) 
 
 66. A 1000-kw., 13,800-volt, 3-phase alternating-current 
 generator delivers power over a transmission line which has a 
 
 1 The armature windings are connected in Y. 
 
 2 The excitation of the motor must also be properly adjusted for this to 
 occur. 
 
120 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 resistance of 16.4 ohms and a reactance of 17.2 ohms per con- 
 ductor to a 1340-h.p. synchronous motor. Both generator and 
 motor have their armatures connected in Y. This motor is 
 rated for 11,000 volts and has an effective armature resistance of 
 0.94 ohms and a synchronous reactance of 46 ohms per phase. 
 The effective resistance and synchronous reactance of the 
 generator are respectively 2.18 ohms, and 82 ohms per phase. 
 The rotation losses of the motor are 22 kw. and may be assumed 
 constant. For certain excitations of generator and motor the 
 line current and power factor at the motor are respectively 48 
 amperes and 0.92 (lagging) when the motor delivers 1140 h.p. 
 
 If the excitations are unchanged what will be the line current 
 and the terminal voltages of the generator and motor if the load 
 on the motor is thrown off? 
 
 67. A 1500-kv.-a., 5500-volt alternator delivers energy to 
 a high-tension transmission line through step-up transformers, 
 the low-tension windings of which are connected in delta and the 
 high-tension windings in Y. The neutrals of the generator and of 
 the high-tension transformer windings are grounded. The 
 transformers have a ratio of transformation of 6.3 to 1.0. 
 
 GENERATOR CHARACTERISTICS 
 
 Field current 
 
 Terminal voltage 
 on open circuit 
 
 Armature current 
 on short circuit 
 
 100 
 150 
 200 
 250 
 
 3500 
 5100 
 5900 
 6500 
 
 200 
 300 
 400 
 
 300 
 
 6800 
 
 
 350 
 
 7100 
 
 
 The effective resistance of the armature is 0.36 ohm per phase. 
 The resistance of the field circuit is 0.386 ohm. The power 
 required to drive the generator on open circuit with a terminal 
 voltage of 5500 volts is 28.6 kw. 
 
 TRANSFORMER CHARACTERISTICS 
 
 With the high-tension winding short-circuited and 170 volts 
 impressed on the low-tension winding the current supplied to a 
 transformer is 91 amperes and the power is 4.26 kw. At no load 
 and with 5500 volts impressed on the low-tension winding the 
 power is 3.12 kw. 
 
 With a balanced load of 1360 kw. at 0.83 power factor delivered 
 
POLYPHASE CIRCUITS 
 
 121 
 
 to the high-tension line at 60,000 volts what must be the gen- 
 erator's excitation? What is the combined efficiency of the gen- 
 erator and the transformers at this load? If this load were 
 removed what would be the high-tension line voltage if the 
 excitation of the generator were unchanged? Use (a) the syn- 
 chronous impedance method and (b) the magnetomotive-force 
 method when calculating the field current of the generator. 
 
 68. A 1000-kv.-a., 2400-volt alternator delivers energy to a 
 high-tension transmission line through step-up transformers both 
 the low- and high-tension windings of which are connected in Y. 
 The armature windings of the generator which are connected in 
 delta have an effective resistance of 0.20 ohm per phase. The 
 resistance of the field circuit is 0.427 ohm. 
 
 GENERATOR CHARACTERISTICS 
 
 Field 
 current 
 
 Open-circuit 
 terminal voltage 
 
 Terminal voltage 
 la = 139. P.F.=Q 
 
 Core loss on 
 open circuit 
 
 60 
 
 1470 
 
 
 7 6 
 
 100 
 
 2220 
 
 
 17 9 
 
 140 
 180 
 220 
 
 2700 
 2980 
 3180 
 
 1920 
 2370 
 2630 
 
 28.5 
 37.0 
 
 The friction and windage loss is 8.2 kw. 
 
 TRANSFORMER CHARACTERISTICS 
 
 The transformers have a ratio of transformation of 1390:12,700 
 volts. With the low-tension winding short-circuited and with 
 332 volts impressed on the high-tension winding the current is 
 26.2 amperes and the power is 3120 watts. The core loss at the 
 rated voltage is 2220 watts. 
 
 The heaviest load that the high-tension line requires is 1000 kw. 
 at 0.88 power factor and the necessary line voltage is 24,500 volts. 
 What must be the terminal voltage of the exciter for this load? 
 What is the combined efficiency of the generator and the trans- 
 formers at this load? 
 
 69. A 1500-kv.-a., 5500-volt alternator delivers energy to a 
 high-tension transmission line through three step-up transformers, 
 the low-tension windings of which are connected in delta and the 
 high-tension windings in Y. The neutrals of the generator and 
 of the high-tension transformer windings are grounded. The 
 transformers have a ratio of transformation of 3.2 to 1.0. The 
 calculated resistance and reactance of the high-tension line are 
 
122 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 48.6 ohms and 59.4 ohms per conductor respectively. The gen- 
 erator has an effective armature resistance of 0.36 ohm per phase. 
 The open- and short-circuit characteristic data for the generator 
 
 are: 
 
 Field 
 current 
 
 Terminal voltage 
 on open circuit 
 
 Armature current 
 on short circuit 
 
 100 
 150 
 200 
 
 3500 
 5100 
 5900 
 
 200 
 300 
 
 250 
 300 
 350 
 
 6500 
 6800 
 7100 
 
 
 With the far end of the transmission line short-circuited the 
 generator supplies a current of 165 amperes at a terminal potential 
 of 860 volts, and the power delivered to the transformers and line 
 is 146 kw. 
 
 If, when the far end of this transmission line delivers a balanced 
 load of 1450 kw. at a power factor of 0.93, the line potential differ- 
 ence at the load is 30,000 volts, to what value would this voltage 
 rise if the load were removed? Use what you consider the most 
 exact method of calculation. 
 
 70. A 1000-kv.-a., 13,800-volt generator delivers power over 
 a transmission line and through step-down transformers which are 
 connected in delta on both the high- and low-tension sides. The 
 armature windings of the generator, which are connected in Y, 
 have an effective resistance of 2.18 ohms per phase. The resist- 
 ance of the field circuit is 0.541 ohm. 
 
 GENERATOR CHARACTERISTICS 
 
 Field 
 current 
 
 Terminal voltage 
 on open circuit 
 
 Terminal voltage 
 la = 42 P.F.=Q 
 
 Core loss on 
 open circuit 
 
 50 
 
 8,800 
 
 
 7 5 
 
 80 
 
 13,000 
 
 
 16 6 
 
 110 
 140 
 180 
 
 15,600 
 17,250 
 18,900 
 
 10,750 
 13,250 
 15,600 
 
 25.4 
 33.5 
 
 The friction and windage loss is 7.9 kw. 
 
 TRANSFORMER CHARACTERISTICS 
 
 Each transformer has a ratio of transformation of 425:13,200 
 volts. With the low-tension winding short-circuited and with 
 1100 volts impressed on the high-tension winding the current 
 
POLYPHASE CIRCUITS 
 
 123 
 
 is 25.2 amperes and the power is 3070 watts. The core loss 
 at the rated voltage is 2130 watts. 
 
 The calculated resistance and reactance of the transmission 
 line are 15.7 ohms and 16.2 ohms per conductor. 
 
 With the low-tension windings of the transformers short-cir- 
 cuited what must be the terminal voltage of the generator in 
 order that it will deliver its full-load current? When the trans- 
 formers deliver a load of 1000 kv.-a. at 0.90 power factor and a 
 line voltage of 420 volts what is the necessary excitation of the 
 generator, and what is the combined efficiency of the generator, 
 line and transformers? 
 
 71. An 850-kv.-a., 11,000-volt, Y-connected generator delivers 
 energy over a transmission line and through transformers, that 
 are connected in Y on the high-tension and in delta on the low- 
 tension sides, to a load of induction motors and a synchronous 
 motor. The resistance and reactance of the line are 11.4 ohms 
 and 10.6 ohms per conductor respectively. The transformers are 
 each rated to deliver 300 kw. with a ratio of transformation of 
 6600:660 volts. 
 
 GENERATOR CHARACTERISTICS 
 
 Field 
 current 
 
 Open-circuit 
 phase voltage 
 
 Short-circuit 
 armature current 
 
 25 
 50 
 
 2200 
 4070 
 
 72 . 
 
 75 
 100 
 
 5500 
 6270 
 
 
 125 
 150 
 175 
 
 6930 
 7370 
 7700 
 
 
 The ohmic resistance of the armature between terminals is 
 3.32 ohms at the running temperature. The effective resis- 
 tance is 1.45 times the ohmic resistance. The field resistance is 
 0.723 ohm. The core and friction losses are 26 kw. at the rated 
 voltage. 
 
 TRANSFORMER CHARACTERISTICS 
 
 With the low-tension winding short-circuited and with 240 
 volts impressed on the high-tension winding the current is 47 
 amperes and the power, 2.8 kw. The core loss at the rated vol- 
 tage is 2.6 kw. 
 
 The induction motors take a constant load of 220 kw. at 0.83 
 
124 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 power factor. The synchronous motor takes a constant Joad of 
 460 kw., and, with the greatest allowable field excitation, its line 
 current is 610 amperes and the terminal voltage is boosted to 
 650 volts. 
 
 Calculate the terminal voltage of the generator's exciter and 
 the combined efficiency of the generator, line and transformers. 
 
 Use what you consider the most exact method. 
 
 72. A constant induction motor load taking 2500 kw. at 
 0.91 power factor and with a line voltage of 13,200 volts is at 
 the end of a short transmission line. For this load the efficiency 
 of transmission is 91.3 per cent, and the voltage regulation of the 
 line is 11.2 per cent. A synchronous motor of suitable capacity 
 is added at the load so that when running light with full-load 
 current the resultant power factor at the load is increased to 
 unity. Assume that the efficiency of this synchronous motor 
 at full load and unit power factor is 0.93. The voltage at the 
 load is maintained constant. 
 
 What is the necessary capacity of the synchronous motor? 
 What are the efficiency of transmission and the voltage regula- 
 tion of the line after the synchronous motor is added? 
 
 73. An induction motor load at the end of a three-phase 25-cycle 
 transmission line takes 7600 kw. at 0.912 power factor and 
 with a line voltage of 11,000 volts. The resistance and induc- 
 tance of the line are respectively 1.4 ohms and 6.5 millihenrys per 
 conductor. A synchronous motor, running light and taking 
 full-load current, is added at the generating station to improve 
 the power factor and thus increase the capacity of the line at the 
 load. Assume that the full-load efficiency of this motor is 94 
 per cent, when operating at unit power factor. The voltage 
 at the load is maintained constant. Induction motors, operating 
 at the same power factor as do the others, are added at the load 
 and the synchronous motor is adjusted so that the generating 
 station operates at unit power factor. For the same line cur- 
 rent in the station as was required before the addition of the 
 synchronous motor calculate the permissible increase in the 
 induction motor- load. What is the kilovolt-ampere capacity of 
 the necessary synchronous motor? What is the line voltage at 
 the generating station before and after the synchronous motor 
 is added? 
 
 74. An induction motor load taking 5400 kw. at 0.914 power 
 factor and with a line voltage of 13,200 volts is at the end of a 
 
POLYPHASE CIRCUITS 125 
 
 three-phase transmission line, which has a resistance of 2.38 ohms 
 and an inductance of 7.6 millihenrys per conductor. The full- 
 load capacity of the generating station is 320 amperes per line. 
 A synchronous motor is to be added at the load both to compensate 
 for power factor and to supply additional power. Assume that 
 the efficiency of this synchronous motor and its exciter is 92 per 
 cent, at full load and with unit power factor. The voltage at 
 the load is maintained constant. The frequency is 60 cycles. 
 
 Calculate the kilovolt-ampere capacity of the synchronous motor 
 so that the generating station can deliver its full-load current at 
 unit power factor. What additional power can the synchronous 
 motor supply? At what power factor does the synchronous 
 motor operate? What is the necessary line voltage at the 
 generating station before and after the synchronous motor 
 is added? 
 
 75. An induction motor load taking 26,000 kw. at 0.906 
 power factor and with a line voltage of 6600 volts is operated at 
 the end of a high-tension transmission line. At the ends of the line 
 there are step-up and step-down transformers, which have the 
 same ratio of transformation. The total equivalent resistance and 
 reactance of the line and transformers referred to the low-tension 
 sides are 0.431 ohm and 0.986 ohm at 25 cycles respectively. Syn- 
 chronous motors, running light, but taking their full-load current, 
 are added at the load to improve the power factor and thus 
 increase the capacity of the generating station. Assume that the 
 full-load efficiency of the synchronous motors and their exciters 
 is 0.92 at unit power factor. The voltage at the load is main- 
 tained constant. Induction motors operating at the same power 
 factor as do the others are added at the load and the synchronous 
 motors are adjusted so that the resultant power factor of the load 
 is increased to unity. 
 
 For the same line current as required before the addition of 
 the synchronous motors calculate the permissible increase in the 
 induction motor load. What is the necessary kilovolt-ampere 
 capacity of the synchronous motors? At what power factor was 
 the generating station operating before and after the synchronous 
 motors were added? What was the line voltage at the generating 
 station before and after the synchronous motors were added? 
 
CHAPTER VII 
 NON-SINUSOIDAL WAVES 
 
 1. The equations for the open-circuit phase voltages of a 
 three-phase, Y-connected ; alternating-current generator are 
 
 61 = 180 sin 0^+60 sin 3 cut 
 
 2n 2r 
 
 e 2 =180 sin (cut ^~)+60 sin 
 
 -x 
 e 3 =180 sin (art -) 60 sin 
 
 What is the equation of the line voltage, 612? What would a 
 voltmeter indicate when connected across one phase? When 
 connected across the line terminals? 
 
 2. The equations for the open-circuit phase voltages of a three- 
 phase, Y-connected, alternating-current generator are : 
 
 ei = 5300 sin ^+1200 sin 3 cut 
 
 2x 
 62 = 5300 sin (orf -) + 1200 sin 
 
 63 = 5300 sin (orf-y)+1200 sin 3(oj<-y). 
 
 What is the equation for the line voltage 6 i2 ? What would a 
 voltmeter indicate when connected across one phase? When 
 connected across the line terminals? 
 
 3. The equation for the voltage between line and neutral of 
 a four-phase generator is: 
 
 61 = 1600 sin 6^+500 sin 3 wt. 
 
 What is the equation for the voltage between adjacent line 
 terminals? What would a voltmeter indicate when connected 
 between line and neutral? Between adjacent lines? 
 
 4. The equation for the voltage between line and neutral of a 
 four-phase generator is: 
 
 61 = 1600 sin itrt+400 sin 5 wt. 
 
 What is the equation for the voltage between adjacent line ter- 
 
 126 
 
NON-SINUSOIDAL WAVES 127 
 
 minals? What would a voltmeter indicate when connected 
 between line and neutral? Between adjacent lines? 
 
 5. The phase voltage of a three-phase, Y-connected, alternat- 
 ing-current generator is 3750 volts. This consists of a funda- 
 mental and a third harmonic which is 30 per cent, of the funda- 
 mental. What is the line voltage? 
 
 6. The phase voltage of a three-phase, Y-connected, alternat- 
 ing-current generator is 2900 volts. This consists of a funda- 
 mental and a fifth harmonic which is 25 per cent, of the funda- 
 mental. What is the line voltage? 
 
 7. The phase voltage of a 3-phase, Y-connected, alternating- 
 current generator is 3980 volts, while the line voltage is 6600 
 volts at the same time. If it is assumed that the phase voltage con- 
 tains no harmonic higher than the seventh, what is the magnitude of 
 the third harmonic in the phase voltage? 
 
 8. The phase voltage of a three-phase, Y-connected, alternat- 
 ing-current generator is 3010 volts while the line voltage (at the 
 same time) is 5000 volts. What is the greatest value that a third 
 harmonic component of the phase voltage can have? 
 
 9. The phase voltage of a three-phase, Y-connected, alter- 
 nating-current generator is 138 volts while at the same time the 
 line voltage is 230 volts. If the phases of this generator were con- 
 nected in delta, what would be the unbalanced voltage tending to 
 set up a circulating current in the windings? 
 
 10. Two transformers are arranged after the Scott method 
 of connection to transform power from two-phase to three-phase. 
 Their ratios of transformation are 1:10 and 1:8.66. 
 
 If there are impressed on the two-phase side electromotive 
 forces whose equations are 
 
 6i = 1550 sin a)t+ 500 sin 3 cot, 
 
 and 6 2 = 1550 sin (cot ^+500 sin 3(cut ^), 
 
 what are the equations for the line voltages on the three-phase 
 side? 
 
 11. In problem 10 if there are impressed on the two-phase side 
 electromotive forces whose equations are, 
 
 61 = 1550 sin &>+300 sin 5 cot, 
 and e 2 = 1550 sin (cut ^+300 sin 5(tot ^) 
 
128 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 what are the equations for the line voltages on the three-phase 
 side? 
 
 12. In problem 10 there are impressed on the two-phase side 
 equal electromotive forces which contain a third and a fifth har- 
 monic. The effective value of these electromotive forces is 
 1100 volts and the harmonics are respectively 0.3 and 0.2 of the 
 fundamental. What are the effective values of the three-phase 
 line voltages? What per cents, of the fundamental components 
 are each of the harmonics in these line voltages? 
 
 13. The line voltage and the voltage to neutral of a balanced 
 three-phase circuit are respectively 230 volts and 139 volts. The 
 voltage to neutral contains a fundamental and a third harmonic 
 only. Three equal resistance units of 10 ohms each are connected 
 in Y across the lines of this circuit. 
 
 (a) What is the line current? What are the line and neutral 
 currents when the neutral point of the resistance units is connected 
 to the neutral of the circuit. 
 
 (b) If the power is measured by the two-wattmeter method 
 what would be the indicated power before and after the connection 
 to the neutral is made? If a wattmeter were connected in the 
 circuit with its current coil in the neutral and its potential coil 
 between the neutral and one of the lines, what would this watt- 
 meter indicate in the second case? 
 
 14. The voltage to neutral of a balanced three-phase circuit 
 is 134 volts, and it contains a fundamental and a fifth harmonic 
 which is 0.2 of the fundamental. Three equal resistance units of 
 10 ohms each are connected in Y across the lines of this circuit. 
 What is the line current? If the neutral point of the resistance 
 units and the neutral of the circuit are connected what is the cur- 
 rent in the neutral conductor? What is the total power supplied 
 in each case? 
 
 15. The line voltage and the voltage to neutral of a balanced 
 three-phase, 60-cycle circuit are respectively 230 volts and 139 
 volts. The voltage to neutral contains a fundamental and a 
 third harmonic only. Three equal air-core reactors, each having a 
 resistance of 5 ohms and an inductance of 0.015 henry, are con- 
 nected in Y across the lines of this circuit. 
 
 (a) What is the line current? What are the line and neutral 
 currents when the neutral point of the reactors is connected to 
 the neutral of the circuit? 
 
 (b) If the power is measured by the two-wattmeter method 
 
NON-SINUSOIDAL WAVES 129 
 
 what is the indicated power before and after the connection to the 
 neutral is made? If a wattmeter is connected in the circuit with 
 its current coil in the neutral and its potential coil between the 
 neutral and one of the lines, what would this wattmeter indicate 
 in the second case? 
 
 16. The voltage to neutral of a balanced, three-phase, 60- 
 cycle circuit is 134 volts, and it contains a fundamental and a fifth 
 harmonic which is 0.2 of the fundamental component. Three 
 equal air-core reactors, each having a resistance of 5.0 ohms and 
 an inductance of 0.015 henry, are connected in Y across the lines 
 of this circuit. What is the line current? If the neutral point of 
 these reactors and the neutral of the circuit are connected what is 
 the current in the neutral conductor? What is the total power 
 supplied in each case? 
 
 17. The line voltage and the voltage to neutral of a balanced 
 three-phase, 60-cycle circuit are 230 volts and 139 volts respect- 
 ively. The voltage to neutral contains a fundamental and a 
 third harmonic only. Three equal impedance units, each con- 
 sisting of a resistance of 5 ohms in series with a capacity of 25 
 microfarads are connected in Y across the lines of this circuit. 
 
 (a) What is the line current? What are the line and neutral 
 currents when the neutral point of this load is connected to the 
 neutral point of the circuit? 
 
 (b) If the power is measured by the two-wattmeter method 
 what is the indicated power before and after the connection to the 
 neutral is made? If a wattmeter is connected in the circuit with 
 its current coil in the neutral and its potential coil between the 
 neutral and one of the lines, what will it indicate in the second 
 case? 
 
 18. The voltage to neutral of a balanced three-phase, 60-cycle 
 circuit is 134 volts, and it contains a fundamental and a fifth 
 harmonic which is 0.2 of the fundamental component. Three 
 equal impedance units, each consisting of a resistance of 5 
 ohms in series with a capacity of 25 microfarads, are connected in 
 Y across the lines by this circuit. What is the line current? If 
 the neutral point of this load is connected to the neutral point of 
 the circuit what is the current in the neutral conductor? What 
 is the total power supplied in each case? 
 
 19. Three unequal lamp loads are connected between the 
 lines and neutral of a balanced three-phase circuit. The line 
 voltage is 230 volts and the voltage to neutral is known to contain 
 
130 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 a third harmonic which is 0.3 of the fundamental component. The 
 resistances of the lamp loads are 5, 8, and 10 ohms. What are 
 the line currents, the neutral current, and the total power 
 supplied? 
 
 20. Three equal lamp loads are connected in Y across the lines 
 of a balanced three-phase circuit. The line voltages are 230 
 volts and the voltage from line to neutral is 139 volts. The latter 
 voltage is known to contain harmonics. The resistance of each of 
 the lamp loads is 5.0 ohms. What will be the effect on the line 
 current if the neutral point of the load is connected through 
 a resistance of 2 ohms to the neutral of the circuit? By what 
 amount is the power supplied to the lamps increased? What is 
 the loss in the 2-ohm resistance? 
 
 21. The voltages from the lines to the neutral conductor of a 
 balanced three-phase circuit are each 140 volts. Three equal 
 resistance units of 10 ohms each are connected in Y across this 
 circuit. A voltmeter connected between the neutral point of 
 this load and the neutral conductor indicates 40 volts. What 
 power will these resistance units take if the neutral point of the 
 load is directly connected to the neutral conductor? What will 
 be the current in the neutral conductor? 
 
 22. The line voltages of a three-phase circuit are equal and 
 maintained constant. The voltages from the lines to the neutral 
 conductor, which are known to contain third harmonics, are also 
 maintained constant and are each equal to 140 volts. Three 
 equal resistance units of 10 ohms each are connected in Y across 
 this circuit. The measured voltage from the neutral of this 
 load to the neutral conductor is 40 volts. What current will 
 exist in the neutral conductor if a resistance unit of 5 ohms is 
 connected between these two points? What is the effective 
 voltage across each of the equal resistance units before and after 
 this additional resistance is inserted in the circuit? 
 
 23. Three hypothetical impedance units, a resistance, a re- 
 actor, and a condenser are so constructed that at 60 cycles the 
 values of their impedances may be represented in the complex 
 notation by: 2i = 10+jQ, z 2 = 0+jlO and z 2 = 0-j!0. These 
 impedance units are connected between the mains and neutral 
 of a balanced three phase, 60-cycle circuit The line voltages are 
 each 230 volts, and the voltages between the lines and the neutral 
 are equal and consist of a fundamental and a third harmonic 
 which is 0.3 of the fundamental component. The first impedance 
 
NON-SINUSOIDAL WAVES 131 
 
 unit is connected from line (1), the second, from line (2), and the 
 third, from line (3) to the neutral. 
 
 If the cyclic order of the line voltages is such that the funda- 
 mental component of Viz leads Vzs by 120 degrees what are the 
 line currents? What is the neutral current? What is the total 
 power supplied? 
 
 24. In problem 23 if the cyclic order of the line voltages is such 
 that the fundamental component of Viz lags Vzs by 120 degrees 
 what are the line currents? What is the neutral current? What 
 is the total power supplied? 
 
 25. Three hypothetical impedance units, a resistance, a react- 
 ance, and a condenser are so constructed that at 60 cycles their 
 values may be presented in the complex notation by: z\ 10+ JO, 
 22 = 0+jlO, 23 = jlO. These impedance units are connected 
 between the mains and neutral of a balanced three-phase, 60- 
 cycle circuit. The line voltages are each 230 volts, and the vol- 
 tages between the lines and the neutral are equal and consist of a 
 fundamental and a fifth harmonic which is 0.25 of the 
 fundamental component. The first impedance is connected 
 from line (1), the second, from line (2), and the third, from line 
 (3) to the neutral. If the cyclic order of the line voltages is 
 such that the fundamental component of Viz leads Vzs by 120 
 degrees what are the line currents? What is the neutral current? 
 What is the total power supplied? 
 
 26. In problem 25 if the cyclic order of the line voltages is 
 such that the fundamental component of Viz lags Vzs by 120 
 degrees what are the line currents? What is the neutral current? 
 What is the total power supplied? 
 
 27. Three hypothetical impedance units, a resistance, a 
 reactor, and a condensor are so constructed that at 60 cycles 
 their values may be represented in the complex notation by: 
 
 zi = 10-f-jO, 2 2 = 0+jlO, z 3 = jlO. These impedance units are 
 connected in delta across the lines of a balanced three-phase, 
 60-cycle circuit. The line voltages are each 230 volts and con- 
 sist of a fundamental and a fifth harmonic which is 0.25 of the 
 fundamental component. The first impedance is connected 
 between lines 1-2, the second, between lines 2-3, and the third, 
 between lines 3-1. 
 
 If the cyclic order of the line voltages is such that the funda- 
 mental component of Viz leads Vzs by 120 degrees what are the 
 line currents? 
 
132 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 28. In problem 27 if the cyclic order of the line voltages is such 
 that the fundamental component of Viz lags V%$ by 120 degrees 
 what are the line currents? 
 
 29. Three equal resistance units of 50 ohms each are connected in 
 Y across the mains of a balanced three-phase, 230-volt circuit. An 
 air-core reactor which has a resistance of 0.5 ohm and an induc- 
 tance of 5 millihenrys is connected from neutral point of this 
 load through a switch to the neutral conductor of the circuit. The 
 voltage across the open switch is 40 volts. There is no higher 
 harmonic than a third present. 
 
 (a) With the switch open what is the line current? What is 
 the total power supplied? 
 
 (b) With the switch closed what is the line current? What is 
 the neutral current? What is the total power supplied? What 
 are the voltages across the resistance units before and after the 
 switch is closed? What is the voltage across the reactor after 
 the switch is closed? 
 
 30. In problem 29 if the air-core reactor is replaced by a con- 
 denser which has a capacity of 50 microfarads what are the line 
 currents before and after the switch is closed? What is the 
 neutral current after the switch is closed? What is the total 
 power supplied with the switch open? With the switch closed? 
 What are the voltages across the resistance units before and after 
 the switch is closed? What is the voltage across the condenser 
 after the switch is closed? 
 
 31. The voltages between the lines and neutral conductor of a 
 balanced three-phase, 60-cycle circuit are each 140 volts. Three 
 equal condensers each of 5 microfarads' capacity are connected 
 in Y across the lines of this circuit, and from their common junc- 
 tion a non-inductive resistance unit of 10 ohms' resistance is con- 
 nected through a switch to the neutral conductor. The voltage 
 across the open switch is 45 volts. There is no higher harmonic 
 than a third present, (a) With the switch open what is the line 
 current? What is the total power supplied? (b) With the switch 
 closed what is the line current? What is the neutral current? 
 What is the total power supplied? 
 
 32. In problem 31 if the non-inductive resistance unit is 
 replaced by an air-core reactor which has a resistance of 0.5 
 ohm and an inductance of 5 millihenrys what are the line cur- 
 rents before and after the switch is closed? What is the neutral 
 current after the switch is closed? What is the total power sup- 
 
NON-SINUSOIDAL WAVES 133 
 
 plied with the switch closed? What are the voltages across the 
 condensers before and after the switch is closed? What is the 
 voltage across the reactor after the switch is closed? 
 
 33. Three equal lamp loads each of which has a resistance of 
 5 ohms are connected in Y across the lines of a balanced 230-volt, 
 3-phase circuit. Due to harmonics the voltages from the lines to 
 the neutral conductor of the circuit are each 140 volts. The 
 power supplied is measured by the two-wattmeter method. 
 
 (a) What are the line currents and the wattmeter readings? 
 
 (b) If the neutral point of the load is connected to the neutral 
 conductor what will the wattmeters read? What is the power 
 supplied? Compare the true power factor of this load with that 
 calculated from the wattmeter, voltmeter, and ammeter readings. 
 
 34. Three unequal lamp loads are connected between the lines 
 and the neutral conductor of a balanced 230-volt, 3-phase circuit 
 Due to third and fifth harmonics, which are respectively 0.3 and. 
 0.25 of the fundamental, the voltages from the lines to the neutral 
 conductor are each 140 volts. The effective line currents are 
 20 amperes, 30 amperes and 40 amperes respectively. What is 
 the current in the neutral conductor? 
 
 35. In problem 34 if fuses in the third line and in the neutral 
 conductor blow, what current will the lamp loads take, and what 
 will be the voltage across each of them? Assume that the resist- 
 ance of each lamp circuit is constant. 
 
 36. By mistake three equal air-core impedance units are con- 
 nected in delta across two of the mains and the neutral conductor 
 of a three-phase, 60-cycle circuit. Each of these impedance units 
 has a resistance of 5 ohms and reactance of 2 ohms at 60 cycles. 
 The voltages between the mains and neutral conductor are each 
 140 volts, and the voltage between any two of the mains is 230 
 volts. Assume that there are no harmonics higher than the third 
 present. What are the currents in each of the impedance units? 
 What are the currents in the mains and in the neutral conductor? 
 
 37. Two suitable transformers are arranged after the Scott 
 method of connection to transform from 2-phase to 3-phase. 
 Each of the two-phase line voltages is 2200 volts and consists of 
 a fundamental and a third harmonic which is 30 per cent, of the 
 fundamental. The 3-phase line voltages each have an effective 
 value of 230 volts. The frequency of the fundamental is 60 
 cycles. Neglect the resistance and the leakage reactance of the 
 transformers. 
 
134 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 (a) If three equal lamp loads, each of which has a resistance of 
 5 ohms, are connected in delta across the 3-phase lines what 
 current does each take? 
 
 (b) If three equal lamp loads, each of which has a resistance of 
 3 ohms, are connected in Y across the 3-phase lines what current 
 does each take? 
 
 38. In problem 37 three equal air-core reactors, each of which 
 has a resistance of 5 ohms and an inductance of 7.5 mil-henry s, 
 are connected in delta across the 3-phase lines. What current 
 does each take? What is the line current? 
 
 (b) If these reactors are connected in Y what current will 
 each take? 
 
 39. In problem 37 three equal impedance units, each con- 
 sisting of a resistance of 20 ohms in series with a condenser 
 of 50 microfarads' capacity, are connected in delta across the 3- 
 phase lines. What current does each unit take? What is the 
 line current? 
 
 (b) If these impedance units are connected in Y what current 
 will each take? 
 
 40. In problem 37 three hypothetical impedance units are 
 connected in delta across the 3-phase lines. The values of these 
 units at 60 cycles may be represented in the complex notation by : 
 *i = 10+jO, 2fe = 0+jlO, z 3 = 0-jlO. What current does each 
 unit take? What are the line currents? 
 
 41. In problem 37 the smaller transformer is tapped at a 
 point two-thirds from its line terminal to give a neutral on the 
 3-phase side. The three hypothetical impedance units described 
 in problem 40 are connected in Y between the lines and neutral. 
 What are the line currents? What is the neutral current? 
 
 42. A 3-phase, 60-cycle alternating-current generator supplies 
 power to a transmission line through step-up transformers whose 
 low tension windings are connected in Y and whose high-tension 
 windings are connected in delta. These transformers are rated at 
 3333 kv.-a. , and have a voltage ratio of 3,800 to 80,000 volts. With 
 the high-tension winding short-circuited and with 162 volts at 60 
 cycles impressed on the low-tension winding full-load current 
 exists in each winding and 18.85 kw. is supplied. The neutral 
 points of both the generator and the low-tension windings of the 
 transformers are connected to the same ground bus. When the 
 transformers are delivering no load on the high-tension side and 
 the excitation of the generator is adjusted so that its terminal 
 
NON-SINUSOIDAL WAVES 135 
 
 voltage is 6600 volts the voltage from line to neutral is found to 
 be 3850 volts. A third harmonic in the phase voltage is sus- 
 pected, and the oscillograph shows that one with a magnitude of 
 14 per cent, of the fundamental does exist. 
 
 What is the copper loss in each transformer when there is no 
 load delivered to the high-tension line. Compare this with the 
 full-load copper loss that would be produced in the transformers 
 if their neutral point was not grounded. 
 
 43. A 1000-kw., 6-phase, 25-cycle, 600-volt rotary converter is 
 supplied with power from a 3-phase, 13,200-volt transmission 
 line. The high-tension windings of the transformers are con- 
 nected in delta and the low-tension windings are connected to 
 diametrical points of the armature of the rotary. With the low- 
 tension winding of a transformer short-circuited and with 345 
 volts at 25 cycles impressed on the high-tension winding, full- 
 load current, or 25.2 amperes exists in the winding and 3070 
 watts is supplied. When the rotary is delivering about full load 
 and the excitation is adjusted for near unit power factor receive 
 the transformer 1120 kw. and the line voltage and current are 
 13,200 volts and 49.8 amperes. Oscillograph records show that 
 at this time there exists across diametrical points of the arma- 
 ture of the rotary a voltage which practically consists of a 
 fundamental and a third harmonic that is 8.5 per cent, of the 
 fundamental. 
 
 Calculate the total copper loss in eath transformer. If the 
 low-tension windings had been divided and connected in double Y 
 with no connection between their neutral points what would have 
 been the copper loss in each transformer under the same load 
 condition? 
 
 44. A3000-kw., 5000-volt, 60-cycle alternating-current generator 
 supplies power to a 3-phase transmission line through three 
 lOOO-kv.-a. transformers which are connected in Y on the low- 
 tension and in delta on the high-tension sides. With the low- 
 tension winding short-circuited and with 3240 volts at 60 cycles 
 impressed on the high-tension side full-load current exists in the 
 windings and 7490 watts is supplied. The ratio of transforma- 
 tion is 2890 to 66,000 volts. Both the neutral point of the arma- 
 ture winding and the neutral point of the low-tension windings 
 of the transformers are directly connected to the same bus. 
 When the transformers are delivering 2650 kw. at 0.875 power 
 factor and with a line voltage of 66,000 volts they become 
 
136 PROBLEMS IN ALTERNATING CURRENT MACHINERY 
 
 unduly hot. Oscillograph records show that at this time the 
 terminal voltage of each phase of the generator practically con- 
 sists of a fundamental and a third harmonic which is 12 per 
 cent, of the fundamental. 
 
 What is the copper loss in each transformer? If the neutral 
 point of the transformer windings is disconnected from the 
 ground bus what will be the copper loss in each transformer for 
 the same load condition? 
 
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