ENGIN. LIBRARY UC-NRLF Meeh, dept. PRACTICAL CALCULATION OF TRANSMISSION LINES PEACTICAL CALCULATION OF TBANSMISSION LINES FOR DISTRIBUTION OF DIRECT AND ALTERNATING CURRENTS BY MEANS OF OVERHEAD, UNDER- GROUND, AND INTERIOR WIRES FOR PURPOSES OF LIGHT, POWER, AND TRACTION BY L. W. ROSENTHAL, E.E. o ASSOCIATE MEMBER, A.I.E.E. NEW YORK MCGRAW PUBLISHING COMPANY 239 WEST 39TH STREET 1909 Engineering library COPYRIGHT, 1909, BY THE McGKAW PUBLISHING COMPANY NEW YORK Stanbopc Jpresa F. H.G1LSON COMPANY BOSTON. U.S.A. PREFACE. THIS little book is offered to the engineering profession with the hope that it may be of practical help in the rapid and accurate calculation of transmission lines. Its existence is the outcome of the belief that this field is in part barren. Its chief mission is to substitute a direct solution for the trial method which was formerly a necessary evil. The arrangement of the formulas, tables and text has been dictated solely by the needs of the rapid worker. All sections except the last include the important effects of temperature and specific conductivity,, The section relating to direct-current railways is novel in the form of its tables, and the methods outlined in it have been found rapid and comprehensive., The alternating-current division presents a new and original method for the solution of these problems. It is the only method known to the author which determines the size of wire directly from the volt loss in the line, and it also possesses unique features of scope, accu- racy and simplicity. The chapter on single-phase railways is in accord with most of the consistent data published on the subject, although further accurate investigations of installed lines may modify to some extent the present accepted values. The author desires to call particular attention to the fallacies of some familiar methods of calculating alternating-current transmis- sion lines which heretofore have been in common use. It will be evident from Tables 11, 28, and 36 that their results are wholly erroneous under certain practical conditions, and indicate wires which may be entirely at variance with the specified requirements. The scope of this book has been confined to methods of calcula- tion. Hence, the most desirable limits of line losses are not discussed, but the tables are sufficiently extended to cover all cases likely to arise in practice. There is no discussion of the character- istics of alternating-current transmission lines, although the tables of wire factors render apparent many of their important features. Furthermore, the book does not include either the determination of iii 254556 iv PREFACE the size of conductors for conditions of maximum economy or the consideration of alternating-current circuits in series. The preparation of the tables has involved thousands of calcula- tions, but thorough checks by the methods of differences and curve plotting have probably eliminated almost all errors of material influence. However, some discrepancies may have crept in, and the author would be glad to learn of them. L. W. R. NEW YORK CITY. December, 1908. CONTENTS. CHAPTER I. DIRECT-CURRENT DISTRIBUTION FOR LIGHT AND POWER. PAR. PAGES 1. INTRODUCTION 6^ 2. PROPERTIES OF CONDUCTORS : 5 3. CURRENT-CARRYING CAPACITY 6 4. PARALLEL RESISTANCE OF WIRES 9 5. GIVEN ITEMS 10 6. FORMULAS 10 7. AMPERE-FEET 10 8. EXAMPLES. . 11 CHAPTER II. DISTRIBUTION FOR DIRECT-CURRENT RAILWAYS. 9. INTRODUCTION 16 10. RESISTANCE OF RAILS 16 11. PARALLEL RESISTANCE OF RAILS AND FEEDERS 17 12. NEGATIVE CONDUCTORS 18 13. POSITIVE CONDUCTORS 18 14. RESISTANCE OF CIRCUIT 19 15. GIVEN ITEMS 19 16. EXAMPLES . . 20 CHAPTER III. ALTERNATING-CURRENT TRANSMISSION BY OVERHEAD WIRES. 17. INTRODUCTION 29 18. OUTLINE OF METHOD 30 19. RANGE OF APPLICATION 31 20. MAXIMUM ERROR 31 21. TRANSMISSION SYSTEMS 32 22. BALANCED LOADS 33 23. TEMPERATURE 33 24. SPECIFIC CONDUCTIVITY 33 25. SOLID AND STRANDED CONDUCTORS 33 26. SKIN EFFECT 33 27. WIRE SPACING 34 v vi CONTENTS PAR. PAGE 28. ARRANGEMENT OF WIRES 34 29. FREQUENCY 34 30. MULTIPLE CIRCUITS 34 31. CURRENT-CARRYING CAPACITY 34 32. TRANSMISSION VOLTAGE 35 33. VOLT LOSS 35 34. POWER TRANSMITTED 35 35. POWER LOSS 36 36. POWER-FACTOR 36 37. WIRE FACTOR 36 38. GIVEN ITEMS 37 39. SIZE OF WIRE 37 40. PER CENT VOLT LOSS 37 41. CHARGING CURRENT 38 42. CAPACITY EFFECTS 38 43. EXAMPLES . . 39 CHAPTER IV. ALTERNATING-CURRENT TRANSMISSION BY UNDERGROUND CABLES. 44. INTRODUCTION 63 45. MAXIMUM ERROR 63 46. TEMPERATURE 64 47. PROPERTIES OF CONDUCTORS 64 48. THICKNESS OF INSULATION 64 49. CURRENT-CARRYING CAPACITY 64 50. CAPACITY EFFECTS 65 51 . EXAMPLES 65 CHAPTER V. INTERIOR WIRES FOR ALTERNATING-CURRENT DISTRIBUTION. 52. INTRODUCTION 75 53. PROPERTIES OF CONDUCTORS 75 54. SPACING OF WIRES 75 55. AMPERE-FEET 76 56. EXAMPLES 76 CHAPTER VI. DISTRIBUTION FOR SINGLE PHASE RAILWAYS. 57. INTRODUCTION 85 58. METHOD OF CALCULATION 85 59. IMPEDANCE OF RAIL 85 60. PERMEABILITY OF RAIL 86 61. IMPEDANCE AND WEIGHT OF RAIL 86 62. IMPEDANCE OF RAIL AND FREQUENCY 86 CONTENTS vii PAR. PAGE 63. FORMULA FOR RAIL IMPEDANCE 86 64. POWER-FACTOR OF TRACK, , 87 65. HEIGHT OF TROLLEY 87 66. EFFECT OF CATENARY CONSTRUCTION , . . 87 67. IMPEDANCE OF COMPLETE CIRCUIT 87 68. MULTIPLE TRACKS .. = .,.. , 88 69. EXAMPLES . . ,.....-..* _ 88 TABLES. CHAPTER I. DIRECT-CURRENT DISTRIBUTION FOR LIGHT AND POWER. No. PAGE 1. Values of T for copper and aluminum 6 2. Values of F for copper and aluminum . 6 3. Properties of copper and aluminum , 7 4. Ampere-feet per volt drop and current-carrying capacity 8 5. Values of H for copper or aluminum. , 8 Formulas for direct-current wiring 14 6. Values of a for copper and aluminum. 15 CHAPTER II. DISTRIBUTION FOR DIRECT CURRENT RAILWAYS. 7. Values of T l for steel 17 8. Equivalents of copper of 100 per cent conductivity. . , 20 9. Resistance to direct current of one steel rail 24 Formulas for direct-current railway circuits , 25 10. Values of A for wires and rails 26 CHAPTER III. ALTERNATING-CURRENT TRANSMISSION BY OVERHEAD WIRES. 11. Error in per cent of volt loss , 31 12. Maximum error in per cent of true values at. 20 cent 32 13. Values of c for overhead wires 39 14. Reactance factors ,...., 39 Formulas for a. c. transmission by overhead wires 50 15. Values of volt loss factors . . 51 16. Values of A for balanced loads 52 17. Values of B for balanced loads. , 52 18. Values of M for overhead copper wires at 15 cycles per second 53 19. Values of M for overhead copper wires at 25 cycles per second 54 20. Values of M for overhead copper wires at 40 cycles per second 55 21. Values of M for overhead copper wires at 60 cycles per second. ..... 56 22. Values of M for overhead copper wires at 125 cycles per second 57 23. Values of M for overhead aluminum wires at 15 cycles per second. . . 58 24. Values of M for overhead aluminum wires at 25 cycles per second. . . 59 25. Values of M for overhead aluminum wires at 40 cycles per second. . . 60 26. Values of M for overhead aluminum wires at 60 cycles per second. . . 61 27. Values of M for overhead aluminum wires at 125 cycles per second. . 62 ix X TABLES CHAPTER IV. ALTERNATING-CURRENT TRANSMISSION BY UNDERGROUND CABLES. No. PAGE 28. Error in per cent of true volt loss ' 63 29. Maximum error in per cent of true values at 20 cent 64 30. Values of c for underground cables 65 Formulas for a. c. transmission by underground cables 71 31. Values of V Q '" = F (1 - 0.01 F ) 72 32. Values of A for balanced loads 72 33. Values of B for balanced loads 72 34. Values of M for multiple conductor copper cables 73 35. Values of M for multiple conductor copper cables 74 CHAPTER V. INTERIOR WIRES FOR ALTERNATING-CURRENT DISTRIBUTION. 36. Error in per cent of true volt loss 76 Formulas for a. c. interior wiring 81 37. Values of a and b for balanced loads 82 38. Values of B for balanced loads 82 39. Values of M for copper wires in interior conduits 83 40. Values of M for copper wires in molding or on cleats ... 84 CHAPTER VI. DISTRIBUTION FOR SINGLE-PHASE RAILWAYS. 41. Test and calculated values of impedance per mile 88 Formulas for single-phase railway circuits 92 42. Values of M for single-phase railway circuits 93 PART I. DIRECT-CURRENT DISTRIBUTION BY MEANS OF OVERHEAD, UNDERGROUND AND INTERIOR WIRES FOR PURPOSES OF LIGHT, POWER AND TRACTION DIRECT-CURRENT DISTRIBUTION CHAPTER I. DIRECT-CURRENT DISTRIBUTION FOR LIGHT AND POWER. 1. Introduction. Problems in direct-current transmission and distribution are relatively simple. The same formula covers all conditions of installation and operation whether by overhead, underground or interior wires. The formulas give accurate results, and all items of influence are easily included. The tables are concise but comprehensive and will cover almost all the usual and unusual requirements of varied practice. 2. Properties of Conductors. The resistance of stranded and solid conductors of the same cross section and length is practically the same. Table 3, page 8, gives the properties of wires at 20 cent, or 68 fahr. for copper of 100 per cent and aluminum of 62 per cent conductivity in Matthiessen's standard scale. The resistance at any other temperature and conductivity may be found for copper or aluminum from formulas (1) and (2). v Ohms resistance per 1000 feet = > * . . S Ohms resistance per mile = 54/700 X T ^ ^ o S = Cross section of metal in circular mils. T = Temperature factor. Table 1, page 6. Thus, at 40 cent, the resistance per mile of No. 1 copper wire of 98 per cent conductivity is 6 TRANSMISSION CALCULATIONS Table i. Values of T for Copper and Aluminum. Temperature Copper. Alumi- num in degrees. Conductivity in Matthiessen's standard scale. Cent. Fahr. 1.00 0.99 0.98 0.97 0.96 0.62 32 0.926 0.936 0.946 0.955 0.965 .495 10 50 0.963 0.973 0.983 0.993 1.003 .553 20 68 1.000 1.010 1.020 1.031 1.042 .613 30 86 1.038 1.048 1.059 1.069 1.081 .674 40 104 1.077 1.088 1.099 1.110 1.122 .738 50 122 1.117 1.128 1.140 1.150 1.163 .802 | For copper or aluminum ot other conductivity divide value under 1.00 by given con- ductivity. The weight of a unit length of copper or aluminum wire may be calculated from the following: Pounds per unit length = FS (3) F = Weight factor. Table 2, page 6. S = Cross section of metal in circular mils. Table 2. Values of JFfor Copper and Aluminum. Unit Length. Copper Wires. Aluminum Wires Solid. Stranded. Solid. Stranded. 1000 feet Mile 0.00303 0.0160 0.00305 0.0161 0.00091 0.00482 0.00092 0.00486 3. Current-carrying Capacity. Table 4, page 8, gives the current-carrying capacity of wires under various conditions. Overhead wires larger than No. have been considered stranded, while those smaller than No. have been taken solid. The values for interior wires have been taken from the National Electric Code. The results for bare and insulated overhead wires are calculated from formula (4) for copper of 100 per cent conduc- tivity and a temperature rise of 40 cent, or 72 fahr. The DIRECT-CURRENT DISTRIBUTION Table 3. Properties of Copper and Aluminum at 20 Cent, or 68 Fahr. Conductivity in Matthiessen's Standard Scale ; Copper 100 Per Cent, Aluminum 62 Per Cent. Copper. Aluminum. B.&S. Area, Diame- f cr Ohms Pounds Ohms Pounds A.W.G. Cir. Mils. tcr, Inches. Resistance. Weight. Resistance. Weight. 1000 Ft. Mile. 1000 Ft. Mile. 1000 Ft. Mile. 1000 Ft. Mile. Bare Stranded Wires. 1,000,000 1.152 0.0104 0.0547 3050 16100 0.0167 0.0882 920 4860 800,000 1.035 0.0129 0.0683 2440 12880 0.0209 0.110 736 3890 700,000 0.963 0.0148 0.0781 2140 11270 0.0239 0.126 644 3400 600,000 0.891 0.0173 0.0911 1830 9660 0.0278 0.147 552 2920 500,000 0.819 0.0207 0.109 1530 8050 0.0334 0.176 460 2430 400,000 0.728 0.0259 0.137 1220 6440 0.0417 0.220 368 1940 300,000 0.630 0.0345 0.182 915 4830 0.0557 0.294 276 1460 250,000 0.590 0.0414 0.219 762 4030 0.0668 0.353 230 1220 0000 211,600 0.530 0.0489 0.258 645 3410 0.0790 0.417 195 1030 000 167,800 0.470 0.0617 0.326 513 2710 0.0997 0.526 154 816 00 133,100 0.420 0.0778 0.411 406 2140 0.126 0.664 122 647 105,500 0.375 0.0981 0.518 322 1700 0.158 0.837 97.1 513 1 83,690 0.330 0.124 0.653 255 1350 0.200 1.06 77.0 407 2 66,370 0.291 0.156 0.824 203 1070 0.252 1.33 61.0 323 3 52,630 0.261 0.197 1.04 160 845 0.318 1.68 48.5 256 4 41,740 0.231 0.248 1.31 127 671 0.401 2.12 38.5 203 Bare Solid Wires. 0000 211,600 0.460 0.0489 0.258 641 3380 0.0790 0.417 193 1020 000 167,800 0.410 0.0617 0.326 508 2680 0.0997 0.526 153 809 00 133,100 0.365 0.0778 0.411 403 2130 0.126 0.664 121 640 105,500 0.325 0.0981 0.518 320 1690 0.158 0.837 96.1 508 1 83,690 0.289 0.124 0.653 253 1340 0.200 1.06 76.3 403 2 66,370 0.258 0.156 0.824 201 1060 0.252 1.33 60.4 320 3 52,630 0.229 0.197 1.04 159 841 0.318 1.68 48.0 253 4 41,740 0.204 0.248 1.31 126 667 0.401 2.12 38.1 201 5 33,100 0.182 0.313 1.65 100 529 0.505 2.67 30.2 159 6 26,250 0.162 0.394 2.08 79.5 420 0.637 3.37 23.9 127 8 16,510 0.129 0.627 3.31 50.0 264 1.01 5.35 15.1 79.5 10 10,380 0.102 0.997 5.27 31.4 166 1.61 8.51 9.5 50.0 12 6,530 0.0808 1.59 8.37 19.8 104 2.56 13.5 6.0 31.8 14 4,107 0.0641 2.52 13.3 12.4 65.6 4.07 21.5 3.7 19.8 16 2,583 0.0508 4.01 21.2 7.8 41.3 6.48 34.2 2.4 12.4 18 1,624 0.0403 6.37 33.7 4.9 26.0 10.3 54.4 1.5 7.8 8 TRANSMISSION CALCULATIONS Table 4. Ampere-Feet per Volt Drop and Current-Carrying Capacity. Ampere-Feet Current-Carrying Capacity in Amperes per Wire. Cir.Mils or A.W.G. per Volt Drop at 20 Cent* Interior Wires. Nat'l Elec. Code.t Overhead Bare Wires. Overhead Wires, Rubber Insulation. Underground Copper Cables. Copper. Alumi- num. Rub- ber. W'th'r- proof. Copper. Alumi- num. Copper. Alumi- num. 1-Condr 3-Condr 1000000 48,100 29,900 650 1000 2460 1950 1090 870 690 800000 38,700 23,900 550 840 2080 1670 930 730 610 700000 33,800 20,900 500 760 1880 1500 840 660 560 600000 28,900 18,000 450 680 1670 1330 750 590 500 500000 24,100 15,000 390 590 1480 1170 660 520 450 400000 19,300 12,000 330 500 1240 980 550 440 390 300000 14,500 9,000 270 400 990 790 440 350 320 250 250000 12,100 7490 240 350 900 720 400 320 290 220 0000 10,200 6330 210 312 760 610 340 270 260 190 000 8110 5020 177 262 630 500 280 230 220 170 00 6430 3970 150 220 540 430 240 190 190 150 5100 3170 127 185 460 370 200 160 170 130 4030 2500 107 156 350 280 150 120 150 110 3210 1980 90 130 300 250 120 100 120 94 2540 1570 76 110 250 210 100 84 110 81 2010 1250 65 92 210 170 88 69 91 68 1600 990 54 77 170 140 74 59 76 57 1270 785 46 65 150 120 63 50 64 48 8 798 495 33 46 100 82 44 35 45 10 502 311 24 32 12 314 195 17 23 14 198 123 12 16 16 125 77 6 8 18 78 49 3 5 * Product of feet of one wire and amperes for total drop of one volt in both wires, t These values are for copper wires; for aluminum wires, multiply by 0.84. Table 5. Values of H for Copper or Aluminum. Type of Installation. Solid Wires. Stranded Wires. 389 342 205 190 Overhead wires rubber insulation 165 152 DIRECT-CURRENT DISTRIBUTION 9 current per wire for lead-covered underground cables is based on tests for a temperature rise from 70 fahr. initial to 150 fahr. final.* Formula (4) may be solved for C to find the temperature rise under given conditions. Amperes per wire = H \~r ..... (4)f d = Diameter of wire in inches. C = Temperature rise in degrees centigrade. H= Heat factor. Table 5, page 8. T = Temperature factor at final temperature. Table 1, page 6. The size of wires is sometimes determined by their current- carrying capacity, especially in interior work where the runs are short. For longer lines it is usually advisable to calculate the loss and then note that the wire has sufficient carrying capacity, while for shorter stretches it is often better to select a wire of proper current capacity and then find by calculation whether the loss is within the specified limit. 4. Parallel Resistance of Wires. The parallel resistance at any temperature for a number of wires of any conductivity is given by the following formulas: i n Ohms resistance per 1000 feet = - u> + 1 + T T 12 * Ohms resistance per mile = - "/ (5) . . . T T 1 i * 2 Where all the wires have the same temperature and conductivity, formulas (5) reduce to those below. Ohms resistance per 1000 feet - 10 ' 350 X T . S, + S 2 + - Ohms resistance per mile = 54 r 700 X T ^ S l + S 2 4- In formulas (5) and (6) Sit S 2 = Circular mils in respective wires. T, T l} T 2 = Temperature factors of respective wires. Table 1, page 6. * Standard Underground Cable Co.'s Handbook No. XVII, page 192. t Based on formulas in Foster's " Electrical Engineers' Pocketbook," 1908, page 208. 10 TRANSMISSION CALCULATIONS Thus, the resistance per mile at 30 cent, of one No. 00 trolley wire of 97 per cent conductivity in parallel with one 1,000,000- cir. mil aluminum feeder of 62 per cent conductivity is 54,700 133,100 1,000,000 1.069 1.674 = 0.0758 ohm. Similarly, the parallel resistance per 1000 feet at 40 cent, of one No. and one No. 4 copper wires of 97 per cent conductivity 18 10,350 X 1.110 = Q 078Q ohm 105,500 + 41,740 5. Given Items. The problem may require the determina- tion of the size of wire from the volt loss, or the volt loss from the given size of wire. In either case the other required items are conductor metal and temperature, current and distance of transmission. If power in watts is specified, then the voltage at load or source must also be given. The term " source " is used in this book to designate the point from which the circuit, or the part of circuit under consideration, starts. Thus in direct-current distribution it may signify the generator, rotary converter, storage battery, connection to main, to feeder or to sub-feeder, or simply a certain point of the circuit. 6. Formulas. Formulas for the complete solution of direct- current problems are given on page 14, and the method of pro- cedure will be clear from the arrangement of the table. The size of wire may be determined from the volt drop, or from the per cent volt drop, which in direct-current systems is always equal to the per cent power loss. The value of a is found in Table 6, page 15, and the size of each wire is noted from Table 3, page 7, corresponding to the required circular mils. The per cent volt drop is expressed in terms of power as well as current, so that problems involving voltage at source and watts at load may be solved without preliminary approximation. It should be carefully noted that percent loss, V or V , is expressed as a whole number, and that the length of transmission, I, is the dis- tance from source to load, which is the same as the length of one wire. In formula (14), r is the resistance per foot of one wire, equal to the values in Table 3, columns 4 or 8, divided by 1000. 7. Ampere-Feet. The size of wires may be readily deter- mined from the ampere-feet per volt drop as given by formula (7). DIRECT-CURRENT DISTRIBUTION 11 The wire having a corresponding value is then noted from Table 4, page 8. Ampere-feet per volt drop = (7) / = Amperes. I = Distance from source to load, in feet. v = Drop in volts. The values in Table 4 are for copper and aluminum of 100 per cent and 62 per cent conductivity, respectively, at 20 cent. For other conductivities or temperatures, formula (8), page 14, is more convenient. In case of distributed loads, as in interior lighting work, II is the sum of the products given by each load, /, multiplied by its respective distance, I. Thus if 6 amperes are to be trans- mitted 25 ft., 3 amperes 50 ft., and 2 amperes 100 ft., II = 6 X 25 + 3 X 50 + 2 X 100 = 500 ampere-feet. 8. Examples. Examples in practice may take innumerable forms, but the method of procedure in any case will be clear from the table of formulas on page 14. The following problems are typical and serve to illustrate the simplicity of the calcu- lations for direct-current distribution. . Example i. A copper circuit of 97 per cent conductivity is to deliver 200 amperes for a distance of 1000 feet with a loss of 10 volts at 40 cent. GIVEN ITEMS. I = 200 amperes; v = 10 volts; I = 1000 feet. From Table 6, page 15, a = 23.0. REQUIRED ITEMS. Size of each wire. From (8), = 23 X 200 X 1000 = 46Q 000 ci] 10 Use 500,000 cir. mil wires. Volt drop. From (9), _ 23 X 200 X 1000 500,000 = 9.2 volts. 12 TRANSMISSION CALCULATIONS Watt loss. From Tables 1 and 3, T = 1.110, and r = 0.0000207 ohm. From (14), p = 2 X 1.11 X 0.0000207 X (200) 2 X 1000 = 1840 watts. By assuming the voltage at the load at any value, say 100, the per cent drop and per cent power loss are found to be the same, 9.2 per cent. Example 2. A motor is to take 25 kw. at a distance of 240 feet with a loss of 5 per cent of the 110 volts generated, while the circuit is to consist of copper wires of 98 per cent conductivity with a temperature of 30 cent. GIVEN ITEMS. w= 25,000 watts; e = 110 volts; 7 =5per cent; Z = 240 feet. From (19), v= 0.05 X 110= 5.5 volts. From Table 6, page 15, a = 21.9. REQUIRED ITEMS. Size of each wire. From (8), s = 21.9 X 239 X 240 = 5.5 Use No. 0000 wires, for which S= 211,600 cir. mils. Per cent volt drop. From (11), V "' = 2L9 X 25,000 X 240 = . - 0.01 X 211,600 X (HO) 2 From Table 31, page 72, 7 = 5.4 per cent. Volt drop. From (19), v = 0.054 X 110 = 5.9 volts. Volts at load. From (13), e= 110 - 5.9 = 104.1 volts. Wattloss. From (14), p = - 054 X 25 ' QOQ = 1430 watts. 1 ~~ U.Uo4 Watts at generator. From (15), w = 25,000 + 1430= 26,430 watts. From Table 4, page 8, it is seen that the wire should have weatherproof insulation or else be increased to 250,000 cir, mils. DIRECT-CURRENT DISTRIBUTION 13 Example 3. The load on a feeder is to consist of 20 amperes at 50 feet, 25 amperes at 100 feet, and 40 amperes at 150 feet, from the source. Calculate the required size of a uniform circuit of 100 per cent conductivity for a total loss of 2 volts at 20 cent. From (7), Ampere-feet per volt drop 20 X 50 + 25 X 100 X 40 X 150 2 From Table 4, page 8, the required size of each wire is No. for copper and No. 000 for aluminum. Example 4. Copper mains of 98 per cent conductivity are to deliver 500 amperes to a point 550 feet from a rotary converter with a loss of 3 per cent of the voltage at load. Calculate the size of wire of 98 per cent conductivity and at 50 cent, if 220 volts are generated. GIVEN ITEMS. 7 = 500 amperes; e =220 volts; V = 3 per cent; Z = 550 feet. From (18), v = ^ 03 X - 6.4 volts. From Table 6, page 15, a = 23.6. REQUIRED ITEMS. Size of each wire. From (8), a = 23.6 X 500 X 550 _ 1;010>000 cir Use 1,000,000 cir. mils. Volt drop. - From (9), t; = 23 ' 6 X ^ * 55 = 6.5 volts. 1,UUU,UUU Per cent volt drop. From (19), 7 = -5A = 2.95 per cent. 2^ ,2\j Volts at load. From (13), e = 220 - 6:5= 213.5 volts. Watts at load. From (16), w = 213.5 X 500 = 106,750 watts. Example 5. Find the combined resistance of 1500 feet of 500,000 cir. mils of 98 per cent copper wire at 20 cent, in parallel with the same length of 1,000,000 cir. mils of aluminum wire of 62 per cent conductivity at 30 cent. From (5) and Table 1, 1.02 1.674 TRANSMISSION CALCULATIONS Formulas For Direct-Current Wiring. Required Items. For Voltage Given at Load. For Voltage Given at Source. Circular mils in each wire. a from Table 6, page 15. 0- ^ . . (8) V Then find wire from Table 3, page 7. Volt drop. V = an 9 s Per cent volt drop or Per cent power loss. V a11 v (10} Y f " awl (in 1 0.01 Se* 0.01 Se 0.01 e' v F from Table 31, page 72. Volts at source or load. e =e + v = e(l + 0.01 F) (12) e =e -u=e (l-0.01F ) (13) Watt loss. T from Table 1, page 6. r from Table 3, page 7. Trin OOlFw) 0.01 F n w ,-.., l-0.01Fo ' Watts at source. MJ O = W + p = IV (1 + 0. f)1F^ w tlfo l-0.01Fo * Amperes. I w (16) 7 w (17) j. . . ^iw; e (l -0.01F ) ' When given V, find v from v = 0.01 Ve = 0.01 Fe /(I + 0.01 F). . . (18) When given F , find v from v = 0.01 F e = 0.01 F e/ (1 - 0.01 F ) . . (19) When given to, find I from (16) or (17) above. NOTATION. a = Resistance factor. Table 6, page 15. e = Volts at load. e = Volts at source. / = Amperes from source. I = Distance from source to load, in feet. p = Total power loss in watts, r = Resistance per foot of one wire, in ohms. Tablet, page 7. S = Circular mils in each wire. T = Temperature factor. Table 1, page 6. xVolt drop in per cent of volts at v= \ load. f Power loss in per cent of power at V load. (Volt drop in per cent of volts at source. ~ f Power loss in per cent of power at ^ source. O"' = Volt loss factor. Table 31, page 72. v Volt drop. w = Watts at load. ? = Watts at source. DIRECT-CURRENT DISTRIBUTION 15 Table 6. Values of a For Copper and Aluminum. Temperature in degrees. Copper. Alumi- num. Conductivity in Matthiessen's standard scale. Cent. Fahr. 1.00 0.99 0.98 0.97 0.96 0.62 10 20 32 50 68 19.2 19.9 20.7 19.4 20.1 20.9 19.6 20.3 21. 1 19.8 20.5 21.3 20.0 20.8 21.6 30.9 32.2 33.4 30 40 50 86 104 122 21.5 22.3 23.1 21.7 22.5 23.3 21.9 22.7 23.6 22.2 23.0 23.8 22.4 23.2 24.1 34.7 36.0 37.3 For copper or aluminum of other conductivity, divide value under 1.00 by given conduc- tivity. CHAPTER II. DISTRIBUTION FOR DIRECT-CURRENT RAILWAYS. 9. Introduction. Electric railways almost always use the track rails for the return of current. Besides this, other differ- ences between circuits for railways and those for power and light- ing purposes are the higher voltage employed on the trolley, the greater per cent loss allowed in the line and the moving and variable nature of the loads. Due to the track rails the calculations of railway feeders and working conductors are somewhat complex and uncertain. However, the recent character of bonding has made the calcula- tions more reliable by giving a higher and more permanent value to the conductivity of the grounded side. In electric railways such as the open conduit and the double trolley systems, the track rails are not used for the return current, and the calcula- tions for their circuits are therefore simpler and more definite. Where the feeders and working conductors form a complete copper circuit, the formulas on page 14 may be used. 10. Resistance of Rails. Table 9, page 24, gives the equiva- lent copper section and the resistance of third rails or track rails at 20 cent., for various values of relative resistance of steel to copper. The corresponding value for any number of rails is found by multiplying the equivalent copper section, or by divid- ing the resistance, by the number. The resistance at any other temperature is found by multiplying the value in the table by the temperature factor of resistance for steel, T 1} Table 7, page 17, while the equivalent section of copper is equal to the tabular value divided by 7\; or, the values may be found for one rail for any other condition of relative resistance and temperature by means of the following formulas: Equivalent cir. mils of 100 per cent copper = 125,000 X Pounds per yard T l X Relative resistance 10 DISTRIBUTION FOR DIRECT-CURRENT RAILWAYS 17 Ohms resistance per 1000 feet T t X Relative resistance 12.1 X Pounds per yard Ohms resistance per mile T \ X Relative resistance 2.28 X Pounds per yard (21) (22) T l = Temperature factor of steel. Table 7, page 17. As an example, the total resistance per mile at 30 cent, of two 65-lb. track rails having a relative resistance of 13.2 is 1.05 X 13.2 2.28 X 65 X 2 = 0.0468 ohm. Table 7. Values of T for Steel. Temperature, deg. cent 10 20 30 40 50 Temperature, deg. fahr 32 50 68 86 104 122 Values of T, 0.91 0.96 1.00 1.05 1.09 1.14 ii. Parallel Resistance of Rails and Feeders. The parallel resistance at any temperature of copper and aluminum feeders of any conductivity with rails of any relative resistance is easily calculated by the following method: 10,350 Ohms resistance per 1000 feet = Ohms resistance per mile T, T, 54,700 (23) (24) + . r, S l Total equivalent cir. mils of copper in rails. Table 9, page 24. *S 2 = Total cir. mils in copper feeders. $ 3 = Total cir. mils in aluminum feeders. 7\ = Temperature factor of steel. Table 7, page 17. T 2 = Temperature factor of copper. Table 1, page 6. T 3 = Temperature factor of aluminum. Table 1, page 6. Thus the resistance per mile at 30 cent, of two 80-lb. rails of relative resistance of 12 in parallel with one 500,000-cir. mil 18 TRANSMISSION CALCULATIONS copper feeder of 97 per cent conductivity and one 500,000-cir. mil aluminum feeder of 62 per cent conductivity is 833^000 X 2 500,000 500,000 = 1.05 1.069 1.674 12. Negative Conductors. The track rails and negative feeders carry the return current. The size of rails is fixed by conditions other than electrical conductivity and usually give a total resistance much below that of the positive side. Electro- lytic conditions may require that negative feeders be connected to the rails at certain points, but otherwise the rails generally have ample conductivity without any copper reinforcement. The section of negative conductors may be based on a maxi- mum allowable drop in the return. The size of the negative feeder is found by subtracting the equivalent copper section of the rails from the total circular mils required. The additional resistance of bonds rnay be included by increasing the true relative resistance of the rail to an apparent value. As an example of the determination of the size of negative feeder, suppose that S in formula (25) on page 26 should come out 2,000,000 cir. mils for the negative side of a circuit for which the track is to consist of two 70-lb. rails of an apparent relative resistance of 14 (including bonding). The required feeders in parallel with the track should have 2,000,000 625,000 X 2, or 750,000 cir. mils of copper of 100 per cent con- ductivity. Based on Table 8, page 20, this is equivalent to 773,000 cir. mils of copper of 97 per cent conductivity, or 1,210,000 cir. mils of aluminum of 62 per cent conductivity. 13. Positive Conductors. The positive conductors consist of auxiliary feeders in parallel with trolleys or third rails. Trolley wires vary in size from Nos. to 0000, while third rails usually range from 60 to 100 pounds per yard. The drop in the positive conductors is found by subtracting the calculated negative drop from the total that is allowed. Then for a third-rail system, the method is exactly like the determination of negative feeders in paragraph 12; while for systems using trolley wires, the total section of positive conductors is calculated from formula (25), page 25. The size of the auxiliary feeder is given by the difference between the total circular mils required and that of the trolley wire. DISTRIBUTION FOR DIRECT-CURRENT RAILWAYS 19 The temperature is included in the calculation of the circular mils by means of A in Table 10, page 26. For conductivities other than 100 per cent, the value of S in formula (25) is divided by the given conductivity. Thus if S comes out 1,000,000 cir. mils of 100 per cent copper, the required section of copper of 97 per 1 ooo 000 cent conductivity is ' 1 , or 1,030,000 cir. mils, as may be 0.97 noted from Table 8, page 20. 14. Resistance of Circuit. The total resistance of the circuit is obtained by adding the resistance of the grounded side and the overhead. Where there are no negative feeders, the resist- ance of the rails may be taken directly from Table 9, page 24, and if no positive feeders are used, the resistance of the trolleys is given in Table 3, page 7. Thus the resistance per mile at 20 cent, of a single-track road having two 60-lb. rails of an apparent relative resistance of 14 (including bonding) and a No. 00 trolley of 100 per cent conductivity is 2^2 + o 411 = o.462 ohm. 15. Given Items. The determination of the proper loading on which to base the feeder systems is usually more difficult than the calculation of the feeders. In general the requirement will be for a maximum drop with a very severe condition of loading or for a much smaller drop with a distributed loading of an average value. In either case the loads and their positions are first settled and then the formulas are applied successively to the separate loads; or, where the feeder system is uniform through- out, one determination is made, using the total of the loads at their center of gravity. In case the feeder system is known the total loss is easily calculated by considering the loads separately, or by considering their combined effect where the feeder system is uniform throughout the area of loading. For convenience in calculation the formulas have been ex- pressed in terms of, current. The per cent volt loss has also been given in terms of power, so that no preliminary approxi- mation is necessary when the voltage is known only at the source. 16. Examples. Although^ in practice, these problems occur in a great many different forms, the application of the formulas .will be clear from the following typical examples. 20 TRANSMISSION CALCULA TIONS Table 8. Equivalents of Copper of 100 Per Cent Conductivity. Conductivity in Matthiessen's Standard Scale. 1.00 0.99 98 0.97 0.96 0.62 1,000,000 900,000 800,000 700,000 1,010,000 909,000 808,000 707,000 1,020,000 918,000 816,000 714,000 1,030,000 928,000 825,000 722,000 1,040,000 938,000 834,000 730,000 1,640,000 1,450,000 1,290,000 1,130,000 600,000 500,000 400,000 300,000 606,000 505,000 404,000 303,000 612,000 510,000 408,000 306,000 618,000 515,000 412,000 309,000 625 : 000 521,000 417,000 312,000 984,000 820,000 656,000 492,000 For other sections, divide given circular mils of 100 per cent copper by given conductivity. For other conductivity, divide circular mils under 1 .00 by given conductivity. To find equivalent circular mils of 100 per cent copper in a wire, multiply given circular mils by given conductivity. Example 6. Find the negative drop per 100 amperes at 30 cent, in a mile of track consisting of two 90-lb. rails having a relative resistance of 12.5 (including bonding) in parallel with 97 per cent copper feeders of 0, 500,000, 750,000, and 1,000,000 cir. mils, respectively. From Table 9, page 24, equivalent cir. mils of 100 per cent copper in two rails at 20 cent. = 1,800,000. From Table 7, page 17, temperature factor for steel at 30 cent., T l = 1.05. From Table 1, page 6, temperature factor for 97 per cent copper at 30 cent., T= 1.069. Then from (24) and (28), the following values are derived: Rails. . Copper Feeder. Resistance per Mile. Drop per Mile per 100 Amperes. Lb. per Yd. Cir. Mils. Ohm. Volts. 2-90 0.0319 3.19 2-90 1-500,000 0.0251 2.51 2-90 1-750,000 0.0226 2.26 2-90 1-1,000,000 0.0206 2.06 Owing to the comparatively large section of the rails, the combined resistance is reduced but 18 per cent for an increase DISTRIBUTION FOR DIRECT-CURRENT RAILWAYS 21 in the negative feeder from 500,000 to 1,000,000 cir. mils, equal to an increase of 100 per cent. Example 7. For a temperature of 40 cent., determine the line voltage at successive cars which take 100 amperes each at respective locations of 500, 2000, 3000, and 5000 feet from a power station generating 550 volts, if the circuit consists of four 80-lb. track rails having a relative resistance of 14 (including bonding), and two No. 00 trolleys of 97 per cent conductivity in parallel with one 500,000 cir. mil aluminum feeder of 62 per cent conductivity. From Tables 7 and 9, the resistance of the tracks is, Resistance per 1000 feet = - 0145 X i- 09 = 0.00395 ohm. From (23) and Tables 1 and 3, the resistance of the overhead is, Resistance per 1000 feet 10350 133,100 X 2 500,000 1.11 1.738 = Q 01962 ohm< Hence the total resistance per 1000 feet of road is 0.00395 + 0.01962 = 0.02357 ohm. 1st Car Total drop = 0.02357 X 400 X0.5 = 4.7 volts. Line volts = 550 - 4.7 = 545.3. 2d Car Additional drop = 0.02357 X 300 X 1 .500 = 10.6 volts. Line volts - 545.3 - 10.6 - 534.7. 3d Car Additional drop = 0.02357 X 200 X 1 .000 = 4.7 volts. Line volts = 534.7 - 4.7 = 530. 4th Car Additional drop = 0.02357 X 100X2.000 = 4.7 volts. Line volts = 530 - 4.7 = 525.3. In the above typical problem, the resistance of the grounded side is but 20 per cent of the resistance of the overhead. Example 8. A single-track road with two 75-lb. rails having a relative resistance of 13 plus 10 per cent for bonding (apparent relative resistance = 14.3), is to supply four cars with 150 amperes each when located at 0.8, 1.0, 1.5 and 2.5 miles from a substation generating 550 volts. If the minimum line e.m.f. is to be 400 volts at 50 cent., what size copper feeder of 97 per cent conductivity should be in parallel with a No. 00 trolley of 96 per cent conductivity? 22 TRANSMISSION CALCULATIONS From (22) and Table 7, page 17, Resistance per mile of two rails = 1<14 X 14 ' 3 = 0.0477 ohm. 2.28 X 75 X 2 Total drop in track = 0.0477 X 150 (0.8 + 1.0 + 1.5 + 2.5) - 41.5 volts. Allowable drop in overhead = 150 - 41.5= 108.5 volts. From (25), in which IL is the sum of products of each load multiplied by its distance from source, and from Table 10, page 26, s = 61,100 X 150(0.8 + 1.0 + 1.5 +W = 490>000 ci , mils lOo.O of 100 per cent copper. Based on Table 8, page 20, the equivalent section of 100 per cent copper in trolley is 133,100 X 0.96= 128,000 cir. mils. Required section of 100 per cent copper in feeder - 490,000 - 128,000 = 362,000 cir. mils. Required section of 97 per cent copper in feeder = 5^2,220 , 373,000 cir. mils. 0.97 Example 9. Find the size of a third rail of relative resistance 7.5 (including bonding) required to start two cars taking 1000 amperes at a point midway between substations 8 miles apart, if the drop at 20 cent, is to be 25 per cent of the 600 volts at rotaries. Each track rail is to weigh 80 Ib. per yard and to have a relative resistance of 13 (including bonding). Distance from either substation to cars = f = 4 miles. Current from each substation = ^-^ = 500 amperes. Total allowable drop = 600 X 0.25 = 150 volts. From Table 9, page 24, resistance per mile of two-track rails = 0.0713 = 003565 ohm Drop in track = 0.03565 X 500 X 4 = 71.3 volts. Allowable drop in third rail = 150 - 71.3= 78.7 volts. From (25), in which A from Table 10, page 26 = 54,700, s = 54,700 x 500 XJ = DISTRIBUTION FOR DIRECT-CURRENT RAILWAYS 23 From Table 9, page 24, this section corresponds to an 85-lb. rail. Then from above and Table 9, Total resistance per mile= R = 0.03565 + 0.0387 = 0.07435 ohm. From (28), total drop = 0.07435 X 500 X 4 = 148.7 volts. From (37), voltage at cars = 600 - 148.7= 451.3 volts. From (40), drop in per cent of volts at load is V = 148.7^-0.01 X 451.3 = 33.0 per cent. From (41), drop in per cent of volts at substation is 7 = 148.7-r-0.01 X 600 = 24.8 per cent. Watt loss. From above, R = 0.07435 ohm. From (35), p = 0.07435 X (500) 2 X 8 = 148,700 watts. Example 10. Calculate the size of positive feeders at 30 cent, required for a uniformly distributed load of 600 amperes per mile between two substations 6 miles apart with an average loss of 5 per cent of the 650 volts generated, if there are to be eight 100-lb. track rails and four 70-lb. third rails having relative resistances (including bonding) of 12.5 and 8.0, respectively. Since the average loss is to be 5 per cent, the maximum is 10 per cent or 65 volts. The result is equivalent to the total load concentrated at a point one-quarter the distance between sub- stations from either one. fiOO V fi Total load per substation = ^ = 1800 amperes. Center of gravity of load = f = 2 miles from either sub- station. From Tables 7 and 9, resistance per mile of four tracks = 0.0548 X 1.05-5-8 =0.00719 ohm. Drop in track = 0.00719 X 1800 X 2 = 25.9 volts. Allowable positive drop = 65 - 25.9= 39.1 volts. From (25), 5 = 57,400 X 1800 X 2-4-39.1 = 5,280,000 cir. mils of 100 per cent copper. From Tables 7 and 9, third-rail section =1,090,000 X 4-*- 1.05 = 4,150,000 cir. mils of 100 per cent copper. Required section of 100 per cent copper feeder = 5,280,000 - 4,150,000 = 1,130,000 cir. mils. Required section of 97 per cent copper feeder = 1,130,000-4-0.97 = 1,165,000 cir. mils. Required section of 62 per cent aluminum feeder = l,130,000-s-0.62 = 1,823,000 cir. mils. TRANSMISSION CALCULATIONS o I s o o 2 1 I I a I HIS o o o o o o o o o o' o' o o' o o o' o o o o o o o o d o' o' o' o' o' o o' o o o o o o o o o' o' o o' o' o' o' o' 22 = 2 o o o o o o o o oooo' oooo o o o o o o o o o' o' o' o' odd o' o o o o o o o o o' o' o' o" d o' o' d o o o o o o o o oooo' oooo . r^ C >O m n coo's oooo o'o'o'd o'o'oo' o o o o o o o o d d o' d d d d o' iT\rr\ oooo oooo oo oo o m r^ Ills ills o' o d o" d o' o' o' p o o o o o o o oo rs \oo 111! llli )' o' o' o' o' d o' d ! .S8. g . 5 . >o o oooo OOjQ O ill! tn ITS O < oooo oooo oooo oooo oo r^ T r^ oooo oooo OO (S t-s slis o o o o 8888 o o o o' oooo oooo o m M rq His SSS8 o o o o' ISIi 888S o' o' d d ST T * o o o o' d o' o' fs,- rr, o^ "^ ^" O rr, O O O O O O OOOO o' d o' o' ills o o d d sss; OO O OO liSs oo'oo 0000 oooo r^^^t'ir* 0000 oooo tr\ o r-% ITS 888 i o' o' d o' 88 IS o o o o DISTRIBUTION FOR DIRECT-CURRENT RAILWAYS 25 Formulas for D. C. Railway Circuits. Required Items. Positive or Negative Alone. Positive and Negative Total. Circular mils or Total resistance. A from Table 10, page 26. .-*... S = . . (25) v Volt drop. .-^..() See Par. 14 , page 19. V = EIL. . . (28) Per cent volt drop or Per cent power loss from amperes. y AIL ,.> 9) y RIL (30) OOlSe 0.01 Se y AIL , 31) y RIL (32) 0.01 Se 0.01 Se ' Per cent volt drop from watts at load. y , AwL , 33 , y,n RwL (34) O.OlSe* 0.01 Sej' TO from Table 31, page 72. Watt loss. T from Table 1 or 7. R from Table 3, page 7. p = TRI Z L = 0.01 Vw = - 01T ^ . . * (35) Volts at source. e = e + v = e(l +0 01 V} e (36) l) 1-0.01F ' Volts at load. .-^-.-n^-^-o-oi^cw) Watts at source. 7/1 7/J 4 T) M(\ \ OIF 1 ) W (38) W 1-0.01F/ Amperes. z w w w n (39) e e (l-0.01F ) e When given F, find v from v = 0.01 Fe = 0.01 Ve / (1 + 0.01 F) . . . When given F , find v from v = 0.01F e = 0.01F e / (1 - 0.01F ) . . When given w, find I from (39) above. (40) (41) NOTATION. ^ = Resistance factor. Table 10, page 26. S = Equivalent cir. mils of copper of 100 per e = Volts at load. cent conductivity. e = Volts at source. T = Temperature factor. Table 1, page 6, or / = Total current at source, in amperes. Table 7, page 17. L Distance from source to load, in miles or V = Volt loss in per cent of volts at load. 1000 feet. ^0 = Volt loss in per cent of volts at source. p = Power loss in watts. V n '" = Volt loss factor. Table 31, page 72. / Ohms resistance per mile for L in miles. v = Volt drop. R = ] Ohms resistance per 1000 feet for L in w = Watts at load. ' 1000 feet. w'o = Watts at source. 26 TRANSMISSION CALCULATIONS Table 10. Values of A for Wires and Rails. Temperature in degrees. Wires. Rails. With or without feeders. Cent Fahr. For L in. Miles. ForL in. 1000 Ft For L in. Miles. For I, in. 1000 Ft. 32 50,600 9,580 49,800 9.420 10 50 52,700 9,970 52,500 9,950 20 68 54,700 10,350 54,700 10,350 30 86 56,800 10,740 57,400 10,870 40 104 58,900 11,140 59,600 11,280 50 122 61,100 11,560 62,400 11,800 The above values for wires are for copper of 100 per cent conductivity in Matthiessen's standard scale. The above values for rails are for equivalent circular mils of copper of 100 per cent conductivity in Matthiessen's standard scale. PART II. ALTERNATING-CURRENT TRANSMISSION BY MEANS OF OVERHEAD, UNDERGROUND AND INTERIOR WIRES FOR PURPOSES OF LIGHT, POWER AND TRACTION 27 ALTERNATING-CURRENT TRANSMISSION CHAPTER III. ALTERNATING-CURRENT TRANSMISSION BY OVERHEAD WIRES 17. Introduction. The methods in common use for calcu- lating alternating-current transmission lines from the volt loss are either indirect or inaccurate. This condition results from the fact that it is impossible to devise an exact formula for any alternating-current system which will directly indicate the size of wire required for the transmission of a given amount of power with a given volt loss. Methods of the indirect class require that the size wire be known, and hence are trial methods. Approximations of the second class may be sufficiently close under certain conditions but give wholly erroneous results for other practical cases. The essential feature^pLthe desirable method is that it should directly determine for any system and frequency, the size of wire, power loss and power-factor at generator when given the volt loss, length of line, power received, load power-factor, voltage at generator or load and distance between wires. The following method accomplishes this result with commercial accuracy over a sufficiently wide range of conditions to cover almost air practical cases. One familiar method of calculating alternating-current lines assumes that the impedance volts equals the line loss, or in Figs. 1 and 2 AB = AC. Several other methods in common use assume that the line loss equals the projection of the impedance volts on the vector of delivered voltage. The errors of these approximations in terms of the actual volt loss are shown in Table 11, for copper wires on 36-inch centers and a true volt loss of 10 per cent of the generated volts. The errors for leading power-factors, larger wires or greater spacings exceed those shown in Table 11, and it is evident then, that both methods may lead to erroneous results. The 29 30 TRANSMISSION CALCULATIONS first assumption gives wires which are too large, while the other assumption gives wires too small. By stating the errors of Table 11 in terms of the calculated values, the percentages shown would be greatly decreased for the first assumption and similarly increased in the second case. 18. Outline of Method. All items which depend on the frequency, size of wire, load power-factor and wire spacing were grouped into one term, which is denoted by M and called the wire factor. All other variables that determine the size of wire Fig. 2 were grouped into the second member of the equation of M. Then after introducing the proper transmission factors, the equation took the form shown in the table of formulas opposite the size of wire. The remaining formulas are simple derivations either from the equation of M or from well known relations. The vector diagram from which the fundamental equation was derived is shown in Fig. 1 for lagging current and in Fig. 2 for leading current, wherein, ALTERNATING-CURRENT TRANSMISSION 31 01 = Current at load. OA = Volts at load. AD= Resistance volts. DB = Reactance volts. AB= Impedance volts. OB Volts at source. AC= OB - OA = Volt loss.

Q = Power-factor angle at source. 19. Range of Application. The formulas apply for all volt losses less than 20 per cent of the generated voltage or 25 per cent of the voltage at load, and for a sufficient range of wire sizes to cover the usual and unusual cases in practice. At high standard frequencies and for power-factors near 100 per cent the values of M for the largest wires were omitted in order to confine the maximum possible error within the limits prescribed in Table 12, page 32. Table n. Error in Per Cent of True Volt Loss. Size of Wire. A.W.G. 25 Cycles per Sec. 60 Cycles per Sec. 125 Cycles per Sec. Lagging Power Factor of Load in Per Cent. 100 90 80 100 90 80 100 90 80 Assuming Impedance Volts = Volt Loss. 0000 4 36 11 1 4 1 7 110 53 12 29 10 13 2 200 125 44 55 33 7 29 16 1 Assuming Projection of Impedance Volts = Volt Loss. 0000 4 5 1 19 8 1 4 1 1 43 22 6 8 4 4 1 20. Maximum Error. As previously stated, any direct method must be approximate. However, all errors introduced by the following method are easily within conservative limits and usually negligible. In determining the size of wire from the volt loss the maximum error is about 10 per cent of the wire factor M, which is usually less than the difference of M for successive wires in commercial sizes. The maximum error occurs near per cent and 20 per cent volt losses but gradually 32 TRANSMISSION CALCULATIONS reduces to a zero error near 10 per cent loss. The point where the formula for M should be correct was arbitrarily fixed at 10 per cent volt loss, since this is about the mean value in practice, After the wire is determined, the actual per cent drop is calcu- lated. In order to minimize its error and that of the remaining items, each column in the tables of M has been divided into a maximum of three sections. It will be observed that any value of M is found between the section letters a and b, b and c, or c and d, depending upon the size of wire, frequency, spac- ing and power-factor. The maximum errors in the calculations are shown in Table 12 below. These errors occur near per cent and 20 per cent volt losses but gradually reduce to zero near 10 per cent loss. In practical problems the calculated value of M is most often between the section letters c and d, and therefore the errors of the remaining calculations are negligible. It should be observed that the values in Table 12 are expressed in per cent of the true result. Thus for an error of 2 per cent with a true volt loss of 5 per cent, the calculated value would be no greater than 5.1 and no less than 4.9 per cent. Table 12. Maximum Errors in Per Cent of True Values at 20 Cent. Calculated Items. For M in Section a--b. b-c. c d. Volt loss or per cent volt loss 6.0 1.5 0.0 3.0 0.8 0.0 1.5 1.3 2.3 4.0 1.0 0.0 2.0 0.5 0.0 !.o 1.0 1.5 2.0 0.5 0.0 1.0 0.3 0.0 0.5 0.5 0.8 Xilovolts at load or source Power loss or kilowatts at source, when given E . . Power loss or per cent power loss, when given Z? Kilowatts at source when given E G . ... . . Amperes per wire, when given E Amperes per wire, when given En . . ... Power-factor at source, when given E Power-factor at source, when given E 21. Transmission Systems. Systems for alternating-current transmission are either one, two or three-phase. These results are applicable to one and three-phase systems, and to the two- phase system with either three or four wires of equal size. ALTERNATING-CURRENT TRANSMISSION 33 Accurate formulas cannot be given for ,the two-phase three-wire system where the size of the common wire is different from the others. However, the result may be .approximated by calcu- lating for three equal wires and then making the proper allow- ance for the larger cross section of the common lead. (See Example 16, page 44.) In the three-phase four- wire system the neutral wire carries no current when the system is balanced, and hence the results are exactly similar to those for the three- phase system with three wires. 22. Balanced Loads. A balanced load is assumed in all the formulas. In practice such is often not the case, although the variation from a balanced condition is usually small. The effect of unbalancing is to alter the voltage in proportion to the amount of unbalancing. The other items are also affected, but seldom is the discrepancy of any practical importance. 23. Temperature. All results have been calculated for a temperature of 20 cent, or 68 fahr. The value of M varies directly with the temperature, but it also depends to a lesser degree on other conditions. However, as the error of commission is much less than the error of omission, the values of A in Table 16, page 52, have been calculated for various temperatures. 24. Specific Conductivity. The calculations have been made for a specific conductivity of 100 per cent for copper and 62 per cent for aluminum. The value of M varies inversely with the specific conductivity but also depends somewhat upon other conditions. However, sufficiently accurate results are obtained by including the proper conductivity of the conductor by the method shown in Table 16. 25. Solid and Stranded Conductors. - Wires smaller than No. have been considered solid, while the larger sizes have been taken stranded. However, since the inductance of a stranded wire is between that of a solid wire of the same cross section of metal and one of the same diameter, the discrepancy in the final results for any ordinary variation from the assumed conditions is not appreciable. 26. Skin Effect. Owing to a decreasing current density toward the center of wires carrying alternating currents, the effective resistance is increased in direct proportion to the product of its cross section and the frequency. In the usual sizes of transmission wire the effect is negligible, and even in the 34 TRANSMISSION CALCULATIONS largest sizes shown for. the higher frequencies the maximum additional resistance is less than 5 per cent, which value decreases very rapidly with the size of wire. 27. Wire Spacing. - - Inductance increases directly with the distance from center to center of the wires. However, the effect on the impedance of the line for large variations in the spacing is not great, even at the higher commercial frequencies. In order to cover all cases in practice, the wire factor has been calculated for three different spacings at each frequency. In general the results for wires on 18 and 60-inch centers will be sufficiently accurate for spacings less than 27 inches and greater than 48 inches, respectively, while the values for 36 inches will cover spacings from 27 to 48 inches. But, where greater accuracy is desired for spacings other than shown, the values of M may be readily interpolated from the tables. 28. Arrangement of Wires. It is assumed in the two and three-phase systems with three wires that the conductors are placed at the three corners of an equilateral triangle. For other arrangements, with the wires properly transposed, little error is introduced by taking the distance from center to center of the wires as the average of the distances between wires 1 and 2, 2 and 3, and 1 and 3. In the two-phase four-wire system the distance between wires of the same phase should be used. 29. Frequency. The tables have been calculated for all standard frequencies. The values for 15 cycles per second are necessary for the design of single-phase railway systems, while 25, 40, 60, and 125 cycles per second cover the remaining systems of transmission for purposes of light, power, and traction. How- ever, where an odd frequency is to be employed, the required value of M may be interpolated from the tables. 30. Multiple Circuits. Where circuits are in parallel from the source to receiver, the load should be proportioned between them and each line calculated separately. 31. Current-carrying Capacity. The current-carrying capa- city in amperes per wire is shown in Table 4, page 8, for both copper and aluminum. The values in the table are based on a temperature rise of 40 cent, or 72 fahr., but the current- carrying capacity for any temperature elevation may be found from formula (4), page 9. In general, overhead transmission lines are of sufficient length ALTERNATING-CURRENT TRANSMISSION 35 to insure the necessary current-carrying capacity when the conditions of volt loss are met. However, it is desirable to note from Table 4 that such is the case, after each determination of wire. 32. Transmission Voltage. The voltage is taken between wires, either at the load or source. The term "source" designates the generator, the secondary terminals of step-up or step-down transformers, or a certain point of the circuit from which the calculation is made. In general the voltage at the source is given, although in some cases of transmission to a single center of distribution, the voltage at the load is specified. The formu- las have been stated in terms of both voltages in order to cover all cases without any preliminary approximation. For convenience the voltage has been expressed in kilovolts, thousands of volts. In a two-phase four-wire system the voltage between the wires of the same phase is specified. With lagging current the voltage at the load is always less than the voltage at the generator. With leading current the voltage at load is less when the sum of the power-factor angles of load and line is less than 90 degrees, but it gradually becomes greater than the voltage at the generator as their sum increases above 90 degrees. 33. Volt Loss. In an alternating-current system the volt loss in the line is the difference between the voltages at the generator and at the load. The line loss is always less than the impedance volts and almost always greater than the projection of the impedance volts on the vector of delivered volts. It may be greater or less than the resistance volts. (See Figs. 1 and 2, page 30.) The per cent volt loss may be expressed in terms of the volts at load or source. However, either may be obtained from the other by means of the simple equations shown below the table of formulas on page 50. For convenience in calculations the per cent loss is expressed as a whole number. 34. Power Transmitted. The power transmitted is always specified at the load, and is usually expressed as true power in kilowatts or as apparent power in kilo volt-amperes. Where the load is given in amperes the equivalent value of kilowatts may be obtained from equations (59) or (60) solved for W. The values of B in Table 17, page 52, serve to make the formula 36 TRANSMISSION CALCULATIONS for amperes per wire applicable to all systems of trans- mission. For balanced loads in one-phase, two-phase four-wire or three- phase circuits the current is the same in each wire, while in the two-phase three-wire system, the current in the common lead is 1.41 times that in each of the other wires. In the two-phase three-wire system / represents the amperes in each of the outer wires, this being the same as the current per phase. 35. Power Loss. The power loss in any system depends only upon the current and resistance. The per cent power loss may be greater or less than the per cent volt loss. It is less than the per cent volt loss, V, when the power-factor at the source is less than the power-factor of the load, and is greater when the reverse is true. An exception to this statement may occur when capacity effects in the line are included. (See Example 20, page 48, and Example 25, page 69.) The power at the source is the sum of the power at load and the power loss in line. 36. Power-Factor. The power-factor of the load is always given. The values of M have been calculated from 95 per cent lead to 75 per cent lag, for all frequencies except 125 cycles per second, giving a range sufficient to cover almost all practical cases of transmission. The power-factor at the load should be known accurately, as the value of M varies greatly with it. The power-factor at the source depends upon the power-factor at the load, the per cent power loss and the per cent volt loss. For current leading at the load, the power-factor is less leading at the source than at the load, and even may become lagging at the source. (See Example 24, page 68.) Lagging current at the load is always accompanied by lagging current at the source. The power-factor at 'the source is lower than the power- factor of the load when the volt loss, V, is a larger percentage than the power loss, P, and it is higher at the source when the reverse is true. This statement may be modified by capacity effects in the line. (See Example 20, page 48, and Example 25, page 69.) 37. Wire Factor. The wire factor, M, depends upon the impedance of the wires, and on the power-factor angles of the load and line. The values have been calculated for copper and aluminum at all standard frequencies, for a range of sizes and load power-factors sufficient to cover all practical cases likely ALTERNATING-CURRENT TRANSMISSION 37 to arise in transmission design. Tables 18 to 22 give the results for copper, and Tables 23 to 27 for aluminum. Those values which, under certain conditions, might lead to errors greater than previously specified have been omitted. 38. Given Items. The given problem may be of two kinds; either the line loss is specified and the size of wire is required, or the size of wire is specified and the line loss is required. In both cases the additional items to be given are system of transmission, conductor metal, temperature, frequency, wire spacing, distance of transmission, power delivered, power-factor of load, and the voltage between wires at the source of load. 39. Size of Wire. The size of wire is generally determined from the per cent volt loss, although in some instances it is fixed by the condition of per cent power loss. Formulas are given for both methods of procedure. When the per cent volt loss is given, the value of A is taken from Table 16, page 52, for the given system of transmission, temperature and specific conductivity. From Table 15, page 51, V or VQ is found, depending upon whether the voltage is given at the load or source. The value of M is then calculated from formula (46) or (47), and the size of wire is noted from Tables 18 to 27, opposite the required value of M. When given the per cent power loss, R is calculated from for- mula (48) or (49), and the size of wire is found in Table 3, page 7, opposite the required resistance per mile. The wire found by either method is the required size of each conductor. 40. Per Cent Volt Loss. It is desirable to calculate the per cent volt loss after the size of wire is determined. The value of M whether between the section letters a and b, b and c, or c and d, is noted from the proper table. The value of V" or F " is calculated from formula (50) or (51), and then located in Table 15, page 51, in the column headed by the same section letters as noted above for M. The corresponding value of per cent volt loss in terms of the volts at load or source, respectively, is then obtained in the column headed by V or V . Where the calcu- lated value does not appear in the table, the corresponding value of V or V is easily found by interpolation. The remaining calculations are clearly outlined in the table of formulas on page 50. When the voltages at the load and source become known, either formula for any required item may be used. 38 TRANSMISSION CALCULATIONS 41. Charging Current. The total charging current for a single-phase circuit or for each phase of a two-phase four-wire circuit is 0.000122 EJL 2 D log,, -j- The charging current per wire of a three-phase three-wire circuit is 0.000141 EJL E = Kilovolts between wires at source. / = Frequency in cycles per second. L = Distance from source to load, in miles. D = Distance between center of wires. d = Diameter of wire in same unit as D. 42. Capacity Effects. Capacity influences the voltage loss, the power loss and the power-factor at generator. In all systems except those of unusual length or high commercial frequency, capacity may be entirely neglected without any disturbing error. However, the effect of capacity may be easily included in the calculations by assuming that the same result is produced by substituting for the distributed capacity of the line one-half its total at each end.* The true power-factor of the load is then replaced by an apparent value determined as follows: . .'.... (44) After if has been calculated, the corresponding apparent power-factor is taken from Table 14, page 39-, and this value is used in the table of M for the determination of the size of wire, volt loss and power loss. The true power-factor at the source is obtained by the follow- ing method: Formula (61) is solved for K ' ', using the apparent power-factor of load for K. The value of t ' corresponding to KQ is noted from Table 14, page 39, and substituted in formula (45) for the reactance factor. The true power-factor at source is then obtained from Table 14, by finding K corresponding to t . (See Example 20, page 48, and Example 25, page 69.) (45) W Q * Method of H. Fender, Proceedings of A.I.E.E., June, 1908, page 772. ALTERNATING-CURRENT TRANSMISSION 39 In formulas (44) and (45) above, c = Capacity factor. Table 13, page 39. E = Kilovolts between wires at load. E Q =s Kilovolts between wires at source. / = Frequency in cycles .per second. L = Distance from source to load, in miles. = Reactance factor for true power-factor of load. Table 14. = Reactance factor for apparent power factor of load. = Reactance factor for true power-factor at source. = Reactance factor for apparent power-factor at source. W = Kilowatts at load. W = Kilowatts at source. Table 13. Values of c for Overhead Wires. Cir. Mils or A.W.G. 1-Phase System, or 2- Phase 4-Wire System 3-Phase 3-Wire System. Distance Between Center of Wires, in Inches. 18 36 60 18 36 60 1,000,000 to 500, 000 500,000 to 0000 0000 to 8 0.00004 0.00004 0.00003 0.00003 0.00003 0.00003 0.00003 0.00003 0.00002 0.00008 0.00007 0.00006 0.00007 0.00006 0.00005 0.00006 0.00006 0.00005 Table 14. Reactance Factors. Power-factors in per cent. 95 Lead. 98 Lead. 100 98 Lag. 95 Lag. 90 Lag. 85 Lag. 80 Lag 75 Lag. Reactance factors -0.33 -0.20 0.20 0.33 0.49 0.62 0.75 0.88 43. Examples. Examples illustrating the application of the formulas on page 50 have been chosen at random and are solved below in detail. The directness, simplicity, range and accuracy of the method will at once be apparent. As a comparison some of the examples are based on problems already in print, although in most cases direct solutions are impossible by other methods. In order to prove the accuracy of the method, the exact result 40 TRANSMISSION CALCULATIONS has been inclosed in brackets after many of the calculated values. Example n. A transformer substation of an electric railway is to be supplied with 3200 kw. at 15 cycles per second and 80 per cent lagging power-factor over 40 miles of single-phase copper line with its wires on 60-inch centers. The loss is to be 15 per cent of the 33,000 volts generated, the temperature 20 cent., and the wires are to be of 100 per cent conductivity. GIVEN ITEMS. W = 3200 kw.; E Q = 33 kv.; 7 = 15 per cent; L = 40 miles; K = 0.80 lag. From Table 16, page 52, A = 0.200. From Table 15, page 51, V ' = 10.9. REQUIRED ITEMS. Size of each wire. From (47), M= (33) 2 X10.9 . 2 X 3200 X 40 From Table 18, page 53, use No. 00, for which M = 0.494 (o - d). Per cent volt loss. From (51), T/ __ 0.494 X 0.2 X 3200 X 40 = n 6 f c _^\ From Table 15, page 51, V = 15.3 per cent. [15.3 per cent.] Kttovolts at load. From (53), E = 33 (1 - 0.153) = 28.0 kv. [28.0 kv.] VoUloss. From (54), v = 1000 (33 - 28) - 5000 volts. [5000 volts.] Per cent power loss. From Table 3, page 7, R = 0.411 ohm. [21.0 per cent.] Kilowatts at source. From (57), W = 3200 (1 + 0.21) = 3870 kw. [3870 kw.] ALTERNATING-CURRENT TRANSMISSION 41 Watt loss. From (58), P= 1000 (3870-3200) = 670,000 watts. [670,000 watts.] Amperes per wire. From Table 17, page 52, B = 1.000. From (60), 33 (1 - From Table 4, page 8, No. 00 conductors will safely carry this current. Power-factor at source. From (61), = (1 + 0.21) (1 - 0.153) 0.80 = 0.820. [0.820.] Example 12. A lighting transformer 10,000 feet distant is to receive 50 kw. at 95 per cent lagging power-factor and 125 cycles per second with a loss of 5 per cent of the 2000 volts at load at a temperature of 30 cent. The line is to be single-phase and of copper conductors of 100 per cent conductivity on 18- inch centers. GIVEN ITEMS. T7=50kw.; #=2kv.; 7= 5 per cent; L = 10,000 feet; K = 0.95 lag. From Table 16, page 52, A = 0.0393. From Table 15, page 51, V = 4.8. REQUIRED ITEMS. Size of each wire. From (46), ^ (2) 2 X 4.8 0.0393 X 50 X 10 From Table 22, use No. 0, for which M = 0.905 (b - c). Per cent volt loss. From (50), v/ , = 0.905 X 0.0393 X 50 X 10 = 445/5 \ From Table 15, page 51, 7= 4.75 per cent. [4.75 per cent.] Kilovolts at source. From (52), E = 2 (1 + 0.0475) = 2.10 kv. [2.10 kv.] Per cent power loss. From Table 3, page 7, R = 0.518 ohm. 42 TRANSMISSION CALCULATIONS From (55), Kilowatts at source. From (57), W = 50 (1 + 0.0282) = 51.4 kw. [51.4 kw.] Amperes per wire. From Table 17, page 52, A = 1.000. From (59), / = ^_ = 26.3 amperes. [26.3 amperes.] 2 X 0.95 Power-factor at source. From (61), Example 13. (From Electric Journal, 1907, page 231, Ex. 2.) GIVEN ITEMS. TF=75kw.; #=2kv.; Wires = No. 4; L = 5000 feet; K = 0.95 lag. System: 1-phase, 60 cycles, 18-inch spacing. Assume wires of 100 per cent copper and temperature of 20 cent. From Table 16, page 52, A = 0.0379. From Table 21, page 56, M= 1.38 (c - d). REQUIRED ITEMS. Per cent volt loss. From (50), r/ , _ 1.38X0.0379 X 75 X 5_ 4 9(Hc _ ^ (2) 2 From Table 15, page 51, V = 5.40 per cent. [5.40 per cent.] Kilovolts at source. From (52), E= 2 (1 + 0.054) = 2.11 kv. [2.11 kv.] Volt loss. From (54), v= 1000 (2.11 - 2.00) = 110 volts. [110 volts.] Example 14. (From "Alternating Currents " by Franklin and Esty, 1907, page 321.) ALTERNATING-CURRENT TRANSMISSION 43 GIVEN ITEMS. TF=1000kw.; # =23kv.; E = 20 kv.; L = 30 miles; K = 0.85 lag. System: 1-phase, 60 cycles, 18-inch spacing. Assume temperature of 20 cent, and copper of 100 per cent conductivity. From Table 16, page 52, A = 0.200. 7= -^=15 per cent. From Table 15, page 51, V = 13.1. REQUIRED ITEMS Size of each wire. From (46), (20) 2 X13.1 0.2 X 1000 X 30 From Table 21, page 56, use No. 1, for which M= 0.936 (c - d}. Per cent volt loss. From (50), T/" _ 0.936 X 0.2 X 1000 X 30 __+ Anff ^ ~7207" From Table 15, page 51, V = 15.75 per cent. [15.5 per cent.] Example 15. A load of 750 kw. of 2-phase power at 25 cycles per second and 100 per cent power-factor is to be delivered over 5 miles of aluminum line of 4 wires with 36 inches between con- ductors of the same phase, with a loss of 7.5 per cent of the 6600 volts generated, at 30 cent. GIVEN ITEMS. W= 750 kw.; E = 6.6 kv.; 7 = 7.5 per cent; L = 5 miles; K= 1.00. From Table 16, page 52, A = 0.104. From Table 15, page 51, F '= 6.4. REQUIRED ITEMS. Size of each wire. From (47), M- ()' xa.4 -- 0716 . 0.104 X 750 X 5 44 TRANSMISSION CALCULATIONS From Table 23, page 58, use No. 0, for which M = 0.755 (c - d). Per cent volt loss. From (51), y 0.755X0.104X75X5 = 6 ?6 (c _ ^ (6.6) 2 From Table 15, page 51, F = 8.16 per cent. [8.10 per cent.] Kilovolts at load. From (53), E = 6.6 (1 - 0.0816) = 6.06 kv. [6.07 kv.] Per cent power loss. From Table 3, page 7, R = 0.837 ohm. From (55), p = 0.837 X 0.104 X 750 X 5 =88Qpercent [8 . 82 per cent>] (6.06) Kilowatts at source. From (57), W = 750 (1 + 0.0889) = 817 kw. [816 kw.] Amperes per wire. From Table 17, page 52, B = 0.500. From (59), 7 = ' 5 X 75Q = 61.9 amperes. [61.8 amperes.] 6.06 From Table 4, page 8, No. aluminum conductors will easily carry this current. Power-factor at source. From (61), K = (1 + 0.0889). (1 - 0.0816) = 1.00. [1.00.] Example 16. A two-phase three-wire line 30 miles long is to deliver 2000 kw. at 40 cycles per second and 98 per cent lag- ging power-factor with a loss of 10 per cent of the 20,000 volts at load. Calculate the size of copper wires of 98 per cent conduc- tivity on 36-inch centers for a temperature of 50 cent. GIVEN ITEMS. W= 2000 kw.; E= 20 kv.; V= 10 per cent; L= 30 miles; K = 0.98 lag. From Table 16, page 52, 'A = ^~ = 0.138. U. C/O From Table 15, page 51, V = 9.1. ALTERNATING-CURRENT TRANSMISSION 45 REQUIRED ITEMS. Size of each wire. From (46), M= 0.138 X 2000 X 30 From Table 20, page 55, use three No. 00 wires, or increase the common lead to No. 000. For No. 00, M = 0.466 (c - d). Per cent volt loss. From (50), T/// _ 0.466 X 0.138 X 2000 X 30 _ Q R4 / ^ (20) 2 From Table 15, page 51, 7= 10.84 per cent. Since the volt loss in the common lead is 1.41 times that in each of the outer wires, Per cent volt loss in outer wire = - : - = 4.50 per cent. With a common lead one number larger in the A. W. Gauge (26 per cent larger) than each outer wire, Approximate per cent volt loss in common lead = 10.84- 4.5 1.26 Therefore, in the above case with No. 00 outer wires and No. 000 for the common lead, Approximate total per cent volt loss = 4.50 + 5.03 = 9.53 per cent. Amperes per wire. From Table 17, page 52, B = 0.500. From (59), in each outer wire, In common wire, /= 51 X 1.41 = 71.9 amperes. Example 17. Two substations located respectively at 40 and 50 miles away are supplied from a No. 00, 25-cycle three- phase aluminum line with wires on 60-inch centers. The total load at the nearer substation (No. 1) is 2000 kw. at 95 per cent lagging power-factor, while the other (No. 2) takes 1000 kw. at 98 per cent lagging power-factor, the generator voltage being 33,000 and the temperature 30 cent. 46 TRANSMISSION CALCULATIONS GIVEN ITEMS. For No. 1 : W= 2000 kw. ; E = 33 kv. ; L = 40 miles ; K = 0.95 lag; M = 0.690 (c - d) . For No. 2: W= 1000 kw.; L = 10 miles; K = 0.98 lag; M= 0.655 (c - d). From Table 16, page 52, A = 0.104. REQUIRED ITEMS. Per cent volt drop to No. 1. From (51), T/ // _ 0.690 X 0.104 X 2000 X 40 _ - 9ft . A VQ : (33) 2 From Table 15, page 51, V = 6.26 per cent. KilovoltsatNo. 1. From (53), # = 33 (1 - 0.0626) = 30.9 kv. Per cent volt drop from No. 1 to No. 2. E =30.9 kv. from above. From (51), T/ _ 0.655 X 0.104 X 1000 X 10 _ n 71 / ^ VQ : (30.9) 2 From Table 15, page 51, V = 0.81 per cent. Kilovolts at No. 2. From (53), E = 30.9 (1 -0.0081) = 30.6 kv. Example 18.* A three-phase load of 5000 kw. and 25 cycles is to be delivered at 30,000 volts to a receiver having 95 per cent lagging power-factor, over 40 miles of No. 00 copper wires of 100 per cent conductivity on 48-inch centers, at a temperature of 20 cent. GIVEN ITEMS. W= 5000 kw.; E= 30 kv.; L= 40 miles; K= 0.95 lag. By interpolation from Table 19, page 54, M = 0.457 (c d). From Table 16, page 52, A = 0.100. REQUIRED ITEMS. Per cent volt loss. From (50), T/// _ 0.457 X 0.1 X 5000 X 40 _ n A n , ,. "" } * Problem of H. Fender, Proceedings of A.I.E.E., June, 1908, page 771. ALTERNATING-CURRENT TRANSMISSION 47 From Table 15, page 51, V = 11.3 per cent. [11.3 per cent.] Kilovolts at source. From (52), E Q = 30 (1 + 0.113) = 33.4 kv. [33.4 kv.] Volt loss. From (54), v = 1000 (33.4 - 30.0) = 3400 volts. [3400 volts.] Per cent power loss. From Table 3, page 7, R = 0.411 ohm. From (55), Kilowatts at source. From (57), W Q = 5000 (1 + 0.101) = 5505 kw. [5505 kw.] Amperes per wire. From Table 17, page 52, B = 0.578. From (59), 7 = ' 578 X 500 = 101.4 amperes. [101.4 amperes.] From Table 4, page 8, No. 00 wires have ample current- carrying capacity. Power-factor at source. From (61), = o.94 lag. [0.94 lag.] Example 19. A three-phase load of 10,000 kw. at 60 cycles per second and 98 per cent leading power-factor is to be delivered over 65 miles of three aluminum wires on 60-inch centers, with a loss of 15 per cent of the 44,000 volts generated; temperature 40 cent. GIVEN ITEMS. W = 10,000 kw. ; E = 44 kv. ; F = 15 per cent; L = 65 miles; K= 0.98 lead. From Table 16, page 52, A = 0.108. From Table 15, page 51, TV = 10.9. REQUIRED ITEMS. . Size of each wire. From (47), M== (44) 2 X 10.9 0.108 X 10,000 X 65 48 TRANSMISSION CALCULATIONS From Table 26, page 61, use No. 0000, for which M = 0.316 (a - 6). Per cent volt loss. From (51), T , _ 0.316 X 0.108 X 10,000 X 65 _ 1 1 A , M (44)* From Table 15, page 51, F = 17.3 per cent. [17.4 per cent.] Per cent power loss. From Table 3, page 7, R = 0.417 ohm. From (56), Power-factor at source. From (61), K = (1 + 0.23) (1 - 0.173) 0.98= 1.00. [0.994 lead.] Example 20.* A three-phase load of 7500 kw. at 90 per cent lagging power-factor and 60 cycles per second is to be delivered over 140 miles of three copper wires of 100 per cent conductivity on 96-inch centers, with a voltage loss of 18.7 per cent of the volts at load, the voltage at the source being 71,200. Include the effect of capacity and assume a temperature of 20 cent. GIVEN ITEMS. W= 7500 kw.; J = 71.2 kv.; V= 18.7 per cent; L= 140 miles; K= 0.90 lag. From Table 16, page 52, A - 0.100. From (63), V = |yj^ - 15-8 per cent. From Table 15, page 51, T V = 11.2. REQUIRED ITEMS. Apparent power-factor of load. Assuming that the required size of wire will be between Nos. 0000 and 8, c is approxi- mately 0.00005. (From Table 13, page 39.) From Table 14, page 39, the reactance factor corresponding to 90 per cent lagging power-factor = 0.49. * Based on Problem of H. Fender, Proceedings of A.I.E.E., June, 1908, page 774. ALTERNATING-CURRENT TRANSMISSION 49 From (44), page 38, , = Q 4Q _ Q.QOQQ5[71.2 (1 - 0.158)] 2 60 X 140 = Q 2g 7500 By interpolation in Table 14, page 39, the apparent power- factor of load, K', = 0.96 lag Size of each wire. From (47), 0.1 X 7500 X 140 From Table 21, page 56, use No. 00, for which M is found by interpolation to be 0.600 (c d). Per cent volt loss. From (51), T7 0.600 X 0.1 X 7500 X 140 _ 10 A ( ,, T : (71.2)' From Table 15, Page 51,F = 16.7 per cent. [17.2 per cent.] Kilovolts at load. From (53), E = 71.2 (1 - 0.167) = 59.3 kv. [59.0 kv.] Per cent power loss. From Table 3, page 7, R = 0.411 ohm. From (55), Kilowatts at source. From (57), W = 7500 (1 + 0.133) = 8500 kw. [8500 kw.] Power-factor at generator (not including capacity). From (61), K '= (1 + 0.133) (1 - 0.167) 0.96 = 0.91. From Table 14, page 39, ? ' m 0.46. From (45), page 38, 0.00005 X(71.2) 2 X 60 X 140 n 91 8500 From Table 14, page 39, the true power-factor at load, K (corresponding to Z ), = 0.98 lag. [0.98 lag.] 50 TRANSMISSION CALCULATIONS Formulas for A. C. Transmission by Overhead Wires. Required Items. For Voltage Given at Load. For Voltage Given at Source. Size of each wire from per cent volt loss. A from Table 16, page 52. V from Table 15, page 51. V Q ' from Table 15, page 51. M- E2V ' (16) Jf-Sffii' AWL AWL ;< Then find wire from Tables 18 to 27, pages 53 to 62. Size of each wire from per cent power loss. PE 2 K 2 E _P[E,(l-OMV n )KV R . (48) AWL AWL (49) Then find wire from Table 3, page 7. Per cent volt loss. V" MAWL ,r ft x T/// . MAWL - J2 ' ( &U J T ^ V from Table 15, page 51. FQ from Table 15, page 51. Kilovolts at source or load. E = E (1 + 0. OIF). (52) E = # (1-0.01F ). (53) Volt loss. v = 1000 (E - E) = 10 VE = 10 V E . . . (54) Per cent power loss. R from Table 3, page 7. p RA WL p _ RAWL FJT H 01 V \'K"\* E*K* ' l J2j O\ i U.U1K JJVJ3 (56) Kilowatts at source. W, = w (H - 0.01P) , . (57) Watt loss. j>= 1000(JF -TF) = 10 PW. . (58) Amperes per wire. (See par. 34, page 35.) B from Table 17, page 52. 2 _ BW I- BW (60) EK ' ' ' ' ( ' ^ (l-0.01F )^' k Power-factor at source. K (l+0.01P)JK_ rl K ~ 1 + 0.017 o 01 PUI 001F^7^ (&\\ When given E and V When given J8 Q and V When given W, find find F from F- F /(l - find F froinF = F/(l W from TF - TF'JT - 0.01 F ) ... . (62) + 0.01F) (63) . (64r\ NOTATION. A = Transmission factor. Table 16, page 52. B = Transmission constant. Table 17, p. 52. E = Kilovolts (1000 volts) between wires at load. E = Kilovolts (1000 volts) between wires at source. See par. 32, p. 35. / = Amperes per wire. See par. 32, p. 35. K Power-factor of load, in decimals. K = Power-factor at source, in decimals. L = Distance from source to load, in miles or 1000 feet. M = Wire factor Tables 18 to 27, pp. 53 to 62. P = Power loss in per cent of power at load. = H = Power loss in watts. Resistance per mile of one wire, in ohms. Table 3, p. 7. = Volt loss in per cent of volts at load. = Volt loss in per cent of volts at source. *Y, F", F "= Volt loss factors. Table 15, p. 51 = Volt loss (volts at source volts at load). = Real kilowatts at load. = Real kilowatts at source. = Kilovolt-amperes (apparent power) at load. ALTERNATING-CURRENT TRANSMISSION 51 Table 15. Values of Volt Loss Factors. V V" V " y V" V " or V V o or V V ' Vo a-b b-c c-d a-6 b-c c-d V a-6 b-c c~d a-6 b-c c-d On 00 On 8 7 4 . U 0.2 .u 0.2 .V 0.2 0.2 0.2 0.2 0.2 0.2 0.2 8.2 7.6 7.7 7.6 7.3 6.9 7.0 6.9 6.8 0.4 0.4 0.4 0.4 0.4 0.4 0.4 0.4 0.4 8.4 7.8 7.9 7.8 7.5 7.0 7.1 7.1 6.9 0.6 0.6 0.6 0.6 0.5 0.6 0.6 0.6 0.6 8.6 7.9 8.0 8.0 7.7 7.2 7.3 7.2 7.1 0.8 0.8 0.8 0.8 0.7 0.8 0.8 0.8 0.7 8.8 8.1 8.2 8.1 7.9 7.3 7.4 7.4 7.2 .0 .0 .0 .0 0.9 .0 .0 .0 0.9 9.0 8.3 8.4 8.3 8.0 7.5 7.5 7.5 7.4 .2 .2 .2 .2 1.1 If .2 .2 .2 .1 9.2 8.4 8.5 8.5 8.2 7.6 7.6 7.6 7.5 7 .4 .6 .4 .6 .4 .6 .4 .5 . J 1.5 .6 .6 .5 .4 9.6 8.8 8.9 8.8 8.5 7.8 7.9 7.9 , j 7.8 .8 .8 .8 .7 1.7 .7 .8 .7 .6 9. 8.9 9.0 9.0 8.7 8.0 8.0 8.1 7.9 2.0 2.0 2.0 1.9 1.8 1.9 2.0 1.9 .8 10.0 9.1 9.2 9.1 8.9 8.1 8.1 8.2 8.1 2.2 2.2 2.2 2.1 2.0 2.1 2.2 2.1 .9 10.2 9.2 9.3 9.3 9.1 8.2 8.2 8.3 8.2 2.4 2.3 2.4 2.3 2.2 2.3 2.3 2.2 2.1 10.4 9.4 9.5 9.5 9.3 8.4 8.4 8.5 8.4 2.6 2.5 2.6 2.5 2.4 2.5 2.5 2.4 2.2 10.6 9.6 9.7 9.7 9.5 8.5 8.5 8.6 8.5 2.8 2.7 2.8 2.7 2.6 2.t 2.7 2.6 2.4 10.8 9.7 9.8 9.8 9.6 8.6 8.6 8.8 8.7 3.0 2.9 3.0 3") 2.8 3/\ 2.7 2/\ 2.8 3/t 2.9 2.8 3r\ 2.6 20 11.0 9.9 1 A 1 10.0 1 f\ 1 10.0 ml 9.8 8.7 8.7 8.9 8.9 3.2 3.4 3. 1 3.3 2 3.4 .0 3.2 .V 3.1 .U 3.2 3. 1 3.2 .U 3.1 .O 2.9 11.2 -11.4 Iv. I 10.2 IU. 1 10.3 . 1 10.3 10.2 9.0 8.9 9.1 9.1 3.6 3.5 3.6 3.4 3.3 3.3 3.4 3.3 3.1 3 a 11.6 10.4 | A 7 10.4 in ft 10.4 in o 10.3 in n 9.1 9-7 9.0 9' 9.3 9C 9.3 9e 3.8 4.0 3.8 4.0 3.8 3.6 3.7 3.8 3.6 .J 3.4 12. C 12.5 IU. / 11.1 IU .0 ii. i IU .C 11.3 IU. / 11.1 . J 9.6 ./ 9.5 9.8 9.8 4.2 4.0 4.2 4.0 3.8 3.8 3.9 3.8 3.6 13.0 11.5 11.5 11.7 11.6 9.8 9.7 10.1 10.2 4.4 4.2 4.4 4.2 4.0 4.0 4.1 4.0 3.8 13.5 11.9 11.9 12.1 12.0 10.1 9.9 10.4 10.5 4.6 4.4 4.5 4.3 4.2 4.2 4.3 4.1 3.9 14.0 12.3 12.2 12.6 12.5 10.4 10.2 10.7 10.8 4.8 4.6 4.7 4.5 4.4 4.3 4.4 4 1 4 1 14 5 1? 7 1? 6 13 1? 9 10 6 10 4 11 (I 11 1 5.0 4.8 4.9 4.7 4.5 4.5 4 6 4 4 4 3 15 13 1 n o n 4 13 4 10 9 10 6 11 ? 11 4 5.2 4.9 5.1 4.9 4./ 4.7 4.8 4.6 4.4 15.5 13.4 13.3 13.8 13.8 11.1 10.8 11.5 11.7 5.4 5.1 5.3 5.1 4.9 4.8 4.9 4.8 4.6 16.0 13.8 13.6 14.2 14.2 11.3 11.0 11.8 12.0 5.6 5.3 5.5 5.3 5.1 5.0 5.1 4.9 4.7 16.5 14.2 14.0 14.6 14.7 11.5 11.2 12.0 12.3 5.8 5.5 5.7 5.5 5.2 5.1 5.3 5.1 4.9 17.0 14.5 14.3 15.0 15.1 11.7 11.3 12.3 12.6 6.0 5.7 5.8 5.6 5.4 5 1 5 4 5 3 5 1 17 5 14 9 14 7 15 4 15 6 11 9 II 5 1? 5 12.9 6.2 5.8 6.0 5.8 5.6 5.4 5.6 5.4 5.2 18.0 15.3 15.0 15.8 16.0 12.1 11.6 12.7 13.2 6.4 6.0 6.2 6.0 5.8 5.6 5.7 5.6 5.4 18.5 15.6 15.3 16.2 16.5 12.3 11.8 13.0 13.4 6.6 6.2 6.4 6.2 5.9 5.8 5.9 5.7 5.5 19.0 16.0 15.6 16.6 16.9 12.5 11.9 13.2 13.7 6.8 6.4 6.5 6.3 6.1 5.9 6.0 5.9 5.7 20.0 16.7 16.2 17.4 17.8 12.8 12.2 13.6 14.2 7.0 6.5 6.7 6.5 6.3 6.1 6.2 6.0 5.8 21.0 17.4 16.8 18.2 18.7 7.2 6.7 6.9 6.7 6.5 6.2 6.3 6.2 6.0 22.0 18.0 17.4 19.0 19.6 7.4 6.9 7.0 6.9 6.6 6.3 6.4 6.3 6.2 23.0 18.7 18.0 19.7 20.5 .... .... 7.6 7.1 7.2 7.1 6.8 6.5 6.6 6.5 6.3 24.0 19.4 18 5 70 5 71 4 7.8 7.2 7.4 7.3 7.0 6.6 6.7 6.6 6.5 25.0 20.0 19.1 21 3 7? 3 52 TRANSMISSION CALCULATIONS Table 16 Values of A for Balanced Loads. 2- Phase Sys- Temperature In degrees. 1-Phase Sys- tem with 2 Equal Wires 2- Phase Sys- tem with 3 Equal Wires. tern with 4 Equal Wires, or 3-Phase Sys- tem with 3 Equal Wires. For L ForL ForL ForL ForL For L Cent. Fahr in in in in in in Miles. 1000 Ft. Miles. 1000 Ft. Miles 1000 Ft. 32 0.185 0.0351 0.112 0,0212 0926 0.0175 10 50 0.193 0.0365 0.116 0.0220 0.0963 0.0182 20 68 0.200 0.0379 0.121 0229 100 0.0190 30 86 0.208 0.0393 0.125 0237 0.104 0.0197 40 104 0.215 0.0408 0.130 0.0246 108 0204 50 122 223 0.0423 0.135 0,0255 0,112 0.0212 The above values are for copper wires of 100 per cent conductivity and aluminum wires of 62 per cent conductivity, in Matthiessen's standard scale, For copper of other conductivity, divide A by given conductivity For aluminum of other conductivity, divide A by given conductivity and multiply by 0.62 Table 17. Values of B for Balanced Loads. 1 -Phase System. 2-Phase System. 3- Phase System. 1.000 0.500 0.578 ALTERNATING-CURRENT TRANSMISSION 53 CO U O 10 8 ^ o 8 S ' - S S i - fSJ gl^^SS S K S =:? O X _ _ eg eg eg r^ TT m hM o- c* oo oo a S 1 i S Us S S S N S 5 iJ _ rg (N eg ~oo ?g ^ oo ^ r^ O O f^ I s ** eg eg rg o^^i * S - - - OOO H ts -= " CO Tfi 10 10 D- 00 . 3 54 TRANSMISSION CALCULATIONS I C?) s. I I 5 1 I S s 8 S 8 $ s SS.. . * S S 8 8 8 25 = a S S S S 3 i i SS.. * 1 o g g g S 2 S 8 S S 8: !* S 2.9 a S S S so co ro 0^ oo oo m ^ S ^ ^ S IS S N S S ^ S S-S g 8 S 8 S c^ ^ cA r>, 2 2 Ji - 8 - s a - w o *v s s s 5 g T> s - - 2 o s *5 O i-l CM CO "- . _ w (sj (N rr\ f^\ . ^. ^. rg rg mNO rOTrirCD - R 2 s a . -- (VI Pv] _ (VJ ~ ALTERNATING-CURRENT TRANSMISSION 55 t3 G & g, o % S i 5 S I & % 13 ,0 S S 1 S 5 1 1 S i S - S 5 1 S S 3 *.-.*. 5 5 S S 8 S _ _ (s) rs) 1 1 I S . R . 3-9 1-8 i S S 2 - ! 3 S-S S= **. J5 S S8 >o . >o >e> ON r^ u-i vooo s s K s . -- - iTsO^ OO ro r% .1-9 1 S i o t^ r>. oo in s o> t^ oo o <^ rN. CN g I g s s rs g a s . ~. 2- !2- _ ^ S ^ "'l-s^ SS K 00 - ~- *^ ^ ~ S I i . i S s s S cs * oo o "* o > oo t> Us T o o, I O o I S S 1 m \o < O 5 5 5 J i i I -. ^. R -. *. a . * -?s =~ in vO ' * ^ 2 o xr tn vo S ES rs M t' ^' o 10 00 ^ S - u IN P; S S S = 53^^,, j3 C s ti $ o o> ^,55 5 = S w S ^ S SJ a 3 K ^ g 5 K ^ 5 o - * $ o & ,00 ^ ^, I O O O O f^ ^.o^ ~ ^^^u^ O tNt^c^i^ O O O O o "'* ' f>l N ej 1 O fNt->.-rt^ t^rr^fvjfo, - B vO 1^. OO^jO fM iri^OvTO^ o o o ' ' ' pi X to t- o c^i o oi^-r^t^ OTC-INO i^mvooo a r~^i^ oo o^ O O <^r,r^o^ e m o f^o< & O O od tfl .S .a I CO I O (N GO _ (VI 5 ? 2 2 S g ^ | ggg SR88. 2 2 s g in gigs g^.^ss, S 8 a S S 5 S 31 SISi S S o g g g m .3 . 2 2 8 S s I l-i s 5.S s i * CM 00 ^ <>&CMr^ I r^ O O OO ^ ?J T^ O ^ t^ GO fM US gS 2 2 a 8 S SSS-K CM r^ . o o ALTERNATING-CURRENT TRANSMISSION 59 S I 10 (S E I I I j OJ 2 ctf H U^ NO O t*. S 8 .?. 8 8 1 5 1 5 1- 3 Ra.s.2. S S S S i -1 S -2 2 rN oo N tr\ t^ co vO O rA 21 s KJ S 5? . - 3 a S i 1 K s SI O - ^T s r^O^OO^ ** ^ oo - ^* oo a s t> ^* o 8 K = 2 * S 8 S S S S S 1 = 2-S S 5-SS a S S S S 1 S i . [i s 9 S ?5 S S ? s s s i it^o^K^^OOOiA^-mr-afA O I ^ oo o fs u*% o-* u^ ~ 1*% r^ o o X I S 2i S N N cvi f^ ^ ^- u^ rCoo s .3-2 B 2 s S i 5 s Sa B5!!-- -S 3 2-S 5 S IS ^ i 1 1 1 1 1 1 1 1 1 60 TRANSMISSION CALCULATIONS i 9 9 9 < S S S'S-i * S S 9 S 1 1 s-s s i i a S. CO 0> g - NO r* m r-^ O^ >O OO m - 5 s ? O O 1 * ^T (N ir% : S 2 I 5Q * gj O - * I < 1 a S i S E T OO O t>t ON CN! m rv m NO O -a ?5 a ?; mo*N~- * pg CN| S? S o S S SS -^ in NO t^ oo .. 81 ;.s 2 s g.g i S S-S S S I lolS in * O? r^ S ? O S ^ O O SinooS f u A to 3 $ K C* m co N n L> c 10 en rNOmi^3;Toog i c co en S i ^ S S -*^ ^ ^ S 3 K I 2 . ^ *. S . " 3 g ^SSJQS^SSorllinoJv. J 3 S 1 8 : : : i : ': ': S I-R S 1 f. ^ ^ * 5 IO en : : : : : : : : :g ?- S = . S. R. a Q . t- u-> m n m vo r>. oo oo O~ _ _ _ __ ^ ( ^- ) 1 S SooSSSoSr^o^ooS^^oo^? 5 1 CO * 1 5 8 S S 1 1 S S f. 5 f- 5 ^- ^ -fi I S ooo-^-oo^^^g r^^gjo ga -1- T . in M5 I-N 55 __ _ _ ^ 03 g : : : :2^ S^, SowSfiJJn^iB 3 s i g : : : : . ^ ? ^. T "^ ". ^ ". _ 3 ? 5 1 8 : : : : : : :*5S^ : ' lino'oo co -' a > > A S j ^ , S Sr^SN^o^SoJQSJs^S S?* ^ i* ^T n ui vO I-N oo _ _ __ __ N - ^^j- } H S 2S^S8S2K^2gP^S 5^^ * ^ 'T T * "^ "^ ^ **> ^ ^ _' _J _' ' (sj S 1 s Kfsssisssssassa?; ss* 8% * S! J^ T * ^ ** -0 t>. <^ __ _ _ _ N - 5 i-i o en .0 S ^ 1^0" ^ "" S5 ^ "* ^ c en M O S ^> "^^W OO>a "(N ^" O^ O .0 ON S ^ - - S * NO 00 ^ 00 13 o ' ' fs CN{ s 6 c 3 M S S{s ^*o^ ^o ^OO 00 O> ON O u rA n- sO ON rsvO'Q ' ' ' N \o NO t^^oo ON y,^ "^ OOCN!^ o o o o o o o ' ' eg' S : : : iSSSS^^^NC B iri NO r^ oo^o rNi^u-iON^:! o o o o . 10 t \O ON U^ r^ilO r^. ONOOOO ^ N CN) i * m r> oo om NO "ti ' ' ' ' CN| Cs| S (X "E V O | s s 2 ?; a ? s K 5 s s 5 ^ ~ ' ' _ ' ' ' ' CVJ CNj CN| I PH .S 8 ON ^* I^N ^TITNONO r^(Airsr> OO ON O^O O r^r^^j O O ' PN| CN{ CN| 8 d s *o bi J o 05 ., 1 5 S I 8 tt 2 S 5 S S R ^ ' ' ' M PNJ I ^actor 10 o> ^S NO^f^ K 00 S O^S ^ S S f 1) fS S 8 ,1 S S l.l S 8 ,S ^. 82^ g o o o o o o o s PM ^- r^ o A .' ?r 2R t^ ^ S o "^ ^ . "^ ^ ' ' N vN S S S S i 5s -^ s: s s^^ * O' - -' - - - $ S^^S S2^r N^^US 9 9 ^f S S ^ S oo w o M -r r^ - -a 000 AWL' ' * Then find wire from Table 34 or 35, page 73 or 74. Size of each wire from per cent power loss. PWK* R P[E a (l - 0.01 F )lTp AWL ''' AWL (68) Then find wire from Table 3, page 7. per cent volt loss. v MA WL , 69) V ,_XAWL (70) E 2 V from Table 31, page 72. Kilovolts at source or load. E =E(l + 0.01F). (71) * = * (1-0.017 ).(72) Volt loss. v = 1000 (E - E) = 10 VE = 10 FO^ O . . . (73) per cent power loss. R from Table 3, page 7. p RA WL ,-. p RAWL (76) Kilowatts at source. W Q =W(l + 0.01P) .... (76) Watt loss. p = 1000(TF -TF) = 10 PW . . .(77) Amperes per wire. (See par. 34, page 35.) B from Table 33, page 72. I- BW (78) I BW (70) 1 EK ' ' ' ' ( Power-factor at source. K _ (1 + O.OIP)K _ (l f-0.01P)(l-0.01F )lT.(80) 1+0.01F When given E and F , find V from V = V / (1 - When given E and V, find V from V Q = V / (1 When given W', find W from W = W'K . . . - 0.01 F ) . . . (81) + 0.01 F) (82) (83) NOTATION. A = Transmission factor. Table 32, page 72. P = B = Transmission constant. Table 33, p. 72. p = E = Kilovolts (1000 volts) between wires at R = load. See par. 32, p. 35. E = Kilovolts (1000 volts) between wires at V = source. See par. 32, p. 35. F = / = Amperes per wire. See par. 34, p. 35. V '" = K = Power-factor of load, in decimals. v = K = Power-factor at source, in decimals. L = Distance from source to load, in miles W = or 1000 ft. Wg = M = Wire factor, Tables 34 and 35, pp. 73 W* = and 74. Power loss in per cent of power at load. Power loss in watts. Resistance per mile of one wire, in ohms. Table 3, p. 7. Volt loss in per cent of volts at load. Volt loss in per cent of volts at source, Volt loss factor. Table 31, p. 72. Volt loss (volts at source volts a load). Real kilowatts at load. Real kilowatts at source. Kilovolt-amperes (apparent power) at load. 72 TRANSMISSION CALCULATIONS Table 31. Values of F '" = F (i-o.oi F ). FO FO'" Fo FO'" FO FO'" FO FO'" FO FO'" FO FO'" V FO'" 0.0 0.0 2.0 2.0 4.0 3.8 6.0 5.6 8.0 7.4 10.0 9.0 15.0 12.8 0.2 0.4 0.2 2.2 2.1 4.2 4.0 6.2 5.8 8.2 7.5 10.5 11.0 9.4 9.8 15.5 16 13.1 13.4 0.6 8 0.6 8 2.6 2 8 2.5 2 7 4.6 4 8 4 6 6 8 6 3 8 8 8 11.5 12 10.2 10 6 16.5 17 13.8 14 1 1.0 .0 3.0 2.9 5.0 4.8 7.0 6.5 9.0 8.2 12.5 10.9 17.5 14.4 1.2 .2 3.2 3.1 5.2 4.9 7.2 6.7 9.2 8.4 13.0 11.3 18.0 14.8 Table 32. Values of A for Balanced Loads. 2- Phase Sys- Temperature in Degrees. 1-Phase Sys- tem with 2 Equal Wires. 2-Phase Sys- tem with 3 Equal Wires. tem with 4 Equal Wires, or 3- Phase Sys- tem with 3 Equal Wires. Cent. Fa.hr. For Lin Miles. For Lin 1000ft. ForLin Miles. ForLin 1000ft. ForLin Miles. ForLin 1000ft. 32 0.185 0.0351 0.112 0.0212 0.0926 0.0175 10 50 0.193 0.0365 0.116 0.0220 0.0963 0.0182 20 68 0.200 0.0379 0.121 0.0229 0.100 0.0190 30 86 0.208 0.0393 0.125 0.0237 0.104 0.0197 40 104 0.215 0.0408 0.130 0.0246 0.108 0.0204 50 122 0.223 0.0423 0.135 0.0255 0.112 0.0212 The above values are for copper wires of 100 per cent conductivity and aluminum wires of 62 per cent conductivity, in Matthiessen's standard scale. For copper of other conductivity, divide A by given conductivity. For aluminum by other conductivity, divide A by given conductivity and multiply by 0.62. Table 33. Values of B for Balanced Loads. 1-Phase System. 2- Phase System. 3- Phase System. 1.000 0.500 0.578 TRANSMISSION BY UNDERGROUND CABLES 73 g o o' o' o' o o' o' o' ' o 00 o^ o o' o o o o o o ' CM m' m' IO 00 3a Ssil ggsv KRI o' o' o' CM in 1 8 ' CM tX in I 1 10 p o o' o' o' o o o o CM' >o r^ 1 ^ mr^ oor^oo cMomcM OOCMOM o' o o o o o o" o ' CM n-i in 12 9i ISI1 liss R^9^ PM . oo oooo oo CM n m i 8 V 3 8 CMCM cMn-^-m voooom f^ ?> o' ' CM tX in I 8 O o' O O O O O O CM ' m' a g2 StM-2 SCM-T- moo-rx iH o' o o o o o o o ' CM tn in i 8 o o o o o o o o ' CM' cX in 10 en ^rr\ CMO^f^O inm 2 s s % s s o a s cS a s- o o oooo o o ' * CM' tn m JJ 6 A.W.G. <=> -* ooo II <3 I 74 TRANSMISSION CALCULATIONS S s O OO O <^ fN ^- TJ- u*i O O^ fN fN ^ f^O^tfN Jnir\ in NO r^ ao o ^r r>. m ao o ' ' ' ' C^' M tN UN S r> "T fs o o o^ IT* $ $ Sv I f . R S S S S C> ' ' ' i "^" *r\ o o "^roo^o ^r> 0* 0* 0* ' ' ' ' s o> ZS S|^^ gSoos _^o 5 f^fP\ tA^-invO r^ ON 'Sr OO(Ni^-* O o' OOOO O O *- S \ f\ u-i s :1 Slii S|2^ R29S o do' o* o' o o ' -0 OOO^fNU^ OOroSvO O O O O O O O O R ?5 S $ *f K ?: 8 2 S R 2 ? S d o o* o* o* o' d o' ' s rX ^' 3 S s 00 o> SS &- vO T tr\ ^2oom Or^t^T tNM OOOOf^ t^ t<> (^ OO* OOOO O O 1 s : : ^ SSSi Sg.sa ssaa d do d d o' d cq tn >X I : S 5 1 E g a s s s s odd o o ri tn iri 1 s 5 6 pt 3 3 I d in CO I CHAPTER V. INTERIOR WIRES FOR ALTERNATING-CURRENT DISTRIBUTION. 52. Introduction. Interior wiring involves short runs of con- ductors; therefore the conductors are often determined by their current-carrying capacity rather than by conditions of maximum drop. It is generally advisable to note the required size of wire for both conditions. Table 4, page 8, gives the National Electric Code Standard for current-carrying capacity of interior wires. The formulas for interior wiring calculations are similar to the preceding ones for alternating-current transmission. The units of power, voltage and distance have been changed in order to facilitate calculation. All the required items are expressed in terms of the current per wire and the per cent volt loss is also given in terms of the power at load so that problems involv- ing watts at load and voltage at source may be solved without preliminary approximation. The ordinary error of the calculation is considerably less than stated in paragraph 45 for underground conductors, on account of the smaller wires usually employed. The error at 20 cent, of the common assumption that the resistance drop equals the volt loss is indicated in Table 36, page 76, for wires in conduits and on 3-inch centers. It is apparent that results based on the assumptions of Table 36 may be much greater or much less than the true values, even at power-factors near unity. The error is due to the power-factor angles of the load and line. (See Figs. 1 and 2, page 30.) 53. Properties of Conductors. The values of M have been calculated for wires of 100 per cent Matthiessen's Standard at a temperature of 20 cent, or 68 fahr. However, the effects of temperature and conductivity are introduced by means of a and b in Table 37, page 82. 54. Spacing of Wires. Table 39, page 83, gives the values of M for wires with a thickness of insulation of inch, while 75 76 TRANSMISSION CALCULATIONS Table 40, page 84, is calculated for conductors on 3-inch centers, Any slight variation from either of these spacings will not appre- ciably alter the results, even for the largest wires at the highest commercial frequency. Table 39 is to be used for wires in conduit, duplex cables, multi-conductor cables or twisted wires. Table 40 is for wires in molding, or for open work such as wires on cleats or knobs. 55. Ampere-Feet. A simple method of dealing with dis- tributed lamps is by use of formula (84) on page 8L The term II is the ampere-feet and is equal to the sum of the products given by each load, / multiplied by its distance I. Table 36. Error in Per Cent of True Volt Loss, Assuming Resistance Drop = Volt Loss. Size of Wire. A.W.G. 60 Cycles per Sec. 125 Cycles per Sec. Lagging Power- Factor of Load in Per Cent. 100 90 80 100 90 80 Wires in Conduit. 0000 4 14 2 -15 + 2 + 8 -15 + 8 + 18 -8 -35 .5 + 7 -39 2 + 16 f Wires on 3-Inch Centers. 0000 4 14 -7 -34 -4 + 8 -37 - 1 + 17 2 54 16 + 6 59 16 + 14 56. Examples. The application of formulas on page 81 are illustrated in the following examples. It should be carefully noted that the per cent volt and power losses are expressed as whole numbers and that I is the distance of transmission from source to load, which is the same as the length of one wire. In formulas (90) and (91), r is the resistance per foot of one wire, equal to the values in Table 3, column 4, divided by 1000. The meaning of the term source is stated in paragraph 32, page 35. WIRES FOR ALTERNATING-CURRENT DISTRIBUTION 77 Example 26. A group of lamps having a power-factor of 95 per cent lag is to be supplied with 20 amperes at 60 cycles per second over a single-phase copper cable 150 feet long. Cal- culate the size of wire for 2 volts drop at 30 cent, for copper of 97 per cent conductivity. GIVEN ITEMS. / = 20 amperes; v = 2 volts; I = 150 feet; K = 0.95 lag. From Table 37, page 82, a = 963 X 0.97 = 934. REQUIRED ITEMS. Size of each wire, From (84), 934 x 2 20 X 150 0.623. From Table 39, page 83, use No. 5, for which M = 0.623. Example 27. At 40 cent, a single-phase cable with No. 8 conductors of 97 per cent conductivity and 200 feet long, delivers 25 amperes at 98 per cent lagging power-factor from a trans- former which gives 100 volts at 125 cycles per second. GIVEN ITEMS. / = 25 amperes; e = 100 volts; I = 200 feet; K = 0.98 lag. From Table 39, page 83, M = 1.27, From Table 37, page 82, a = 928 X 0.97 = 900; b = -^ = 222. \j * y / REQUIRED ITEMS. Volt loss. From (85), 1.27 xsx 200 = 70voltg Volts at load. From (89), e = 100 - 7 = 93 volts. Per cent volt loss. From (86), Per cent power loss. From Table 3, page 7, r = 0.000627. 78 TRANSMISSION CALCULATIONS From (90), p _ 0.000627 X 222 X^25 X 200 = ? Watts at load. From Table 38, page 82, B = 1.000. From (94), w = 25 X 93 X ' 98 - 2280 watts. 1 .00 Watt loss. From (92), p= 0.076 X 2280== 173 watts. Power-factor at source. From (96), Example 28. A 60-cycle step-down transformer with 120 volts at its secondary terminals is connected to a load of 2000 watts at 90 per cent lagging power-factor over 200 feet of single- phase circuit on cleats. Determine the size of wire for 5 per cent drop at 20 cent, and copper of 98 per cent conductivity. GIVEN ITEMS. w = 2000 watts; e = 120 volts; 7 = 5 per cent; 1= 200 feet; K = 0.90 lag. From Table 31, page 72, 7 ///= 4 -8- From (98), v= 0.05 X 120 = 6 volts. From Table 37, page 82, a = 1000 X 0.98 - 980. From (95), (in which B from Table 38, page 82, = 1.000), 2000 120 (1 - 0.05) 0.90 REQUIRED ITEMS. Size of each wire. From (84), M= 98QX6 =1.51. 19.5 X 200 From Table 40, page 84, use No. 8, for which M= 1.20. Per cent volt loss. From (87), v '" = L2Q X 200 X 20 m 3 40 0.01 X 980 X (120) 2 WIRES FOR ALTERNATING-CURRENT DISTRIBUTION 79 From Table 31, page 72, V = 3.50 per cent. Volts at load. From (89), e= 120 (1 - 0.035) = 116 volts. Amperes per wire. From (94), = 19.2 amperes. 116 X 0.90 From Table 4, page 8, it is seen that the current will be safely carried by No. 8 wires. Example 29. A load of 1500 watts at 110 volts, 85 per cent lagging power-factor and 60 cycles per second is to be delivered 250 feet over a two-phase four- wire circuit on cleats, with a loss of 2 per cent of the generated volts. Calculate the size of wire of 98 per cent conductivity for a temperature of 30 cent. GIVEN ITEMS. W = 1500 watts; e = 110 volts; V = 2 per cent; 1= 250 feet; K=0.85 lag. From (98), v = - 02 X 11Q - 2.25 volts. JL ~ U.U^ ^ From Table 37, page 82, a = 963 X 0.98 = 944. From (94) (in which B from Table 38, page 82, = 0.500), REQUIRED ITEMS. Size of each wire. From (84), M = 944 X 2.25 8 X 250 From Table 40, page 84, use No. 8, for which M = 1.18. Per cent volt loss. From (86), Volts at source. From (88), e = 110 (1 + 0.0227) = 113 volts. 80 TRANSMISSION CALCULATIONS Example 30. A 125 volt, 125 cycle generator is to supply a three-phase load of 100 amperes per wire at 100 per cent power- factor over 225 feet of cable, with a loss of 3 per cent of the gen- erated volts. The wires are to have a conductivity of 99 per cent and a temperature of 30 cent. GIVEN ITEMS. 7 = 100 amperes; e = 125 volts; F = 3 per cent ; /= 225 feet; K= LOO. From (98), v=0.03 X 125=3.75 volts. From Table 37, page 82, a = 1110 X 0.99= 1100; & = -=181. REQUIRED ITEMS. Size of each wire. From (84), ,.1100 X 3.75_ ft100 100 X 225 ' From Table 39, page 77, use No. 00, for which M = 0.163. Volt toss. From (85), Per cent volt loss. From (98), Volts at toad From (89), e= 125 - 3.3= 121.7 volts. WIRES FOR ALTERNATING-CURRENT DISTRIBUTION 81 Formulas for A.C. Interior Wiring. Required Items. For Voltage Given at Load. For Voltage Given at Source. Size of each wire. a from Table 37, page 82. *- au (84) II ' Then find wire from Table 39 or 40, page 83 or 84. Volt loss. v = **!! a (85) Per cent volt loss. Mil y , 3fW>Z 0.01 ae 2 ' ' * (87) 0.01 ae O.Ole* F from Table 31, page 72. Volts at source or load. e Q = e + v = e(l + 0.01F) . (88) e !e(l-0.01F ) . (89) Per cent power loss. r from Table 3, page 7-6 from Table 37, page 82. *-!' <* p rbll (91) Watt loss. p 0.01 Pw (92) Watts at source. w o = w + p = w (1 + 0.01 P) . (93) Amperes per wire. (See Par. 34, page 35.) B from Table 38, page 82. j Bw Bw f (94) I _ Bw (95) Bw' eK e e (l-0.01F )' ' Power-factor at source. K (i+o.oip)*: /nooipvi OOIFW (96) 1+0.01F When given F, find v from v = 0.01 Fe = 0.01 Fe /(I + 0.01 F) . . . When given F , find v from v = 0.01 F () e = 0.01 F e / (1 - 0.01 F ) . . When given w or w', find I from (94) or (95) above. (97) (98) NOTATION. a = Transmission factor. Table 37, page 82. P = R = Transmission constant. Table 38, p. 82. p = 6 = Transmission factor. Table 37, p. 82. r = e = Volts between wires at load. See par. 32, p. 35. F = c = Volts between wires at source. See par. F = 32, p. 35. FO'" = 7 = Amperes per wire. See par. 34, p. 35. v = K = Power-factor of load, in decimals. K = Power-factor at source, in decimals. w = I = Distance from source to load, in feet. W Q = M= Wire factor. Tables 39 and 40, pp. w' = 83 and 84. Power loss in percent of power at load. Total power loss in watts. Resistance per foot of one wire. Table 3, p. 7. Volt loss in per cent of volts at load. Volt loss in per cent of volts at source. Volt loss factor. Table 31, p. 72. Volt loss (volts at source - volts at load). Real watts at load. Real watts at source. = Volt-amperes (apparent power) at load. 82 TRANSMISSION CALCULATIONS Table 37. Values of a and b for Balanced Loads. Temperature in Degrees. 1 -Phase Sys- tem with 2 Equal Wires. 2-Phase Sys- tem with 3 Equal Wires. 2-Phase Sys- tem with 4 Equal Wires. 3-Phase Sys- tem with 3 Equal Wires. Cent. Fahr. a 6 a b a 6 a b 10 20 32 50 68 1080 1040 1000 185 193 200 895 860 828 185 193 200 1080 1040 1000 185 193 200 1250 1200 1160 160 167 173 30 40 50 86 104 T22 MI 928 896 208 215 223 798 769 742 208 215 223 963 928 896 208 215 223 1110 1070 1030 179 186 193 The above values are for copper of 100 per cent conductivity. For other conductivity, multiply a and divide 6 by given conductivity. Table 38. Values of B for Balanced Loads. 1 -Phase System. 2-Phase System. 3-Phase System. 1.000 0.500 0.578 WIRES FOR ALTERNATING-CURRENT DISTRIBUTION 83 1 .s CA S S S 3 9 - - _ N V o 2 08 5r ** S - S 288: cs to I 3 < So fn ?R ? S o 3 !n S "" t- . o :<=> S8S5 SS.R i^2 SSSi 58 5.8. 2 5-2-2 2^E 8 5 s 2 .s ---- u-v O as i !z 2 5 s i S 1 s 1 . S i orsu^^ TOOOOO oor^o^ SS3 fsAmT -o ix M ONOr^ 3S l T*^GO s ssss 84 TRANSMISSION CALCULATIONS a IT 1 o fei s so I I 4 s s tApr, o^CMf^cs oo- OO s S 5? S r^O^ O^r^\t> S ". . -. . ". S S S -- -- \ v MIL E ( E o ' ' ' ( ' y ,,, MWL E*K ' V (} from Table 31, page 72. Kilovolts at source or load. E = E(l + 0.01 F) . (105) E=E (1 -001F ) . (106) Volt loss. v = 1000(# - E) = IOVE = WV E = 10 MIL. (107) Per cent power loss. R from Table 42, page 93. p RIL (-\OR\ P RIL 10J (1 - 0.01F ).K" (109 Watt loss. p = WPW=RPL (110) Kilowatts at source. W Q = W + 0.001 p = W(l + 0.01 P) (Ill) Amperes. '-- - (m) I w ~w> (113) Power-factor at source. TT (1+001P)JK" ,j 1 A oi P) n Q1F N 7T (114) 1 -f 0.01 F When given E and F , find F from F = F / (1 - .01F ) (115) When given E and F, find F f rorn F = F/(l + .OIF) (116) When given W or W, find I from (112) or (113) above. NOTATION. E = Kilovolts between trolley and rail, at load. E = Kilovolts between trolley and rail, at source. / = Total amperes from source. K = Power-factor of load, in decimals. /To = Power-factor at source, in decimals. L Distance from source to load, in miles. M = Wire factor. Table 42, page 93. p = Power loss in per cent of power at load. Power loss in watts. Resistance per mile of circuit. Table 42, page 93. Volt loss in per cent of volts at load. V = Volt loss in per cent of volts at source. /Y" = Volt loss factor. Table 31, page 72. v = Total loss in volts (volts at source - volts at load). W = Real kilowatts at load. W n = Real kilowatts at source. W = Kilovolt-amperes (apparent power) at load. I : v = DISTRIBUTION FOR SINGLE-PHASE RAILWAYS i s a I 8 Ji Ill g jliil ssss siss ssss ssss iss *\ u"\ O r\ SSSS N" 00 CN sss sill ssss i 1 i i 1 i c<^ NO oo ^" r*. m O so O* u"N ssss ssss ssss s fN 9- tr\ 1^ m si lisl III ii SS2 sii 00 s I i 8 lifili il BBi \O t^ CS if> r^ cs rA o Siil issl sisi O^ ao r^ ^ >r\ m Oj H Z2 I s s ss s s o 'T m tr -^j- IA m 000 000 ro O *0 SSSS ssss ss Us g ssss 00 s s III ssss ssss glssss ssss ssss 000 0000 r^ OO ss s s sss 00 xO .T|- ss .ssss ssss s ssss sss F: S 2^SS8 ^- c^c^tA-^- f^r^-^--^- S ? S s s S S 000 i.lsi sSii 111 |i|g 0000 0000 sis r^rj- ^-^ro^so f^\\o oo sooiri \pt^ \ovOvOt~ so\ot>.i^ sot^r^oo OO 0000 0000 OOOO s s I s s s ss^^ s 11^ UNIVERSITY OF CALIFORNIA LIBRARY BERKELEY Return to desk from which borrowed. This book is DUE n-rfielMt data Damed below. 1 ? 19497 I LD 21-100m-9,'48(B399sl6)476 tea- Engineering library 254556 tJL