Hawkins Electrical Guide v. 04 (of 10) Questions, Answers, & Illustrations, A progressive course of study for engineers, electricians, students and those desiring to acquire a working knowledge of electricity and its applications

LAMP TABLE FOR RUBBER COVERED WIRES Fig . 829.—Wiring for lights requiring unusually long feeders. The first step is to find an expression for the resistance of the wire which may be later substituted in Ohm's law formula. Accordingly, the resistance of any conductor is equal to its length in feet multiplied by its resistance per mil foot and the product divided by its area in circular mils , thus: resistance in ohms = length in feet × resistance per mil foot circular mils or ohms = feet × 10.8 circular mils . . . . (1) (calling the resistance per mil foot 10.8 instead of 10.79 to facilitate calculation). LAMP TABLE FOR RUBBER COVERED WIRES Showing the maximum number of 16 candle power 110 to 240 volt lamps in parallel that may be carried by the various sizes of wire without violating the underwriters' rules. Wire size B. & S. gauge Amperes. 3.1.watt lamps. 3.5.watt lamps. 4.watt lamps. At 110 volts. 220 V. At 110 volts. 220 V. 220 V. 230 V. 240 V. 0000 210 462 924 412 825 722 754 787 000 177 389 778 347 695 608 636 663 00 150 330 660 294 589 515 539 562 0 127 279 558 249 499 436 456 476 1 107 235 470 210 420 367 384 401 2 90 197 396 176 353 309 323 337 3 76 167 334 149 298 261 273 285 4 65 143 286 127 255 223 233 243 5 54 118 237 106 212 185 194 202 6 46 101 202 90 180 158 165 172 8 33 72 145 64 129 113 118 123 10 24 52 105 47 94 82 86 90 12 17 37 74 33 66 58 61 63 14 12 26 52 23½ 47 41 43 45 16 3 6 13 .. 11 .. 20 21 22 Now, according to Ohm's law, volts = amperes × ohms . . . . (2) hence, substituting in (2) the value for the resistance in ohms, as obtained in (1): volts = amperes × feet × 10.8 circular mils or using the usual symbols E = I × feet × 10.8 circular mils . . . . (3) or expressed in words, formula (3) means that the volts lost or drop between the beginning and end of a circuit is equal to the current flowing through the circuit multiplied by the product of the conductors' length in feet multiplied by the resistance of one mil foot of wire, divided by the area of the conductor in circular mils. LAMP TABLE FOR WEATHER PROOF WIRES Showing the maximum number of 16 candle power 120 to 240 volt lamps in parallel that may be carried by various sizes of weather proof wire without violating the underwriters' rules. Wire size B. & S. gauge Amperes. 3.1.watt lamps. 3.5.watt lamps. 4.watt lamps. 110 V. 220 V. 110 V. 220 V. 220 V. 230 V. 240 V. 0000 312 686 1372 612 1225 1072 1121 1170 000 262 576 1152 514 1029 900 941 982 00 220 484 968 432 864 756 790 825 0 185 407 814 363 726 636 665 693 1 156 343 686 306 612 536 560 585 2 131 288 576 257 514 450 470 491 3 220 242 484 216 432 378 395 412 4 92 202 404 180 361 316 330 345 5 77 169 338 151 302 264 276 288 6 65 143 286 127 255 223 233 243 8 46 101 202 90 180 158 165 172 10 32 70 140 62 125 110 115 120 12 23 50 101 45 90 79 82 86 14 16 35 70 31 62 55 57 60 Now, since the length of the circuit is given as the "run" or distance one way, that is, one half the total length of wire in the circuit, formula (3) must be multiplied by 2 to get the total drop, that is: E = I × feet × 10.8 X 2 circular mills = I × feet × 21.6 circular mills . . . . (4) Solving the last equation for the unknown quantity, the following equation is obtained for size of wire: Figs . 830 and 831.—Symmetrical and unsymmetrical distribution. When a main is supplied by a feeder, the junction of the two, if practicable, is located at the electrical center of the main, as indicated in fig. 830 , so that the distribution is symmetrical, that is, the ampere feet each way from the junction are the same. This is nearly always practicable in surface wiring, and when it is practiced it is only necessary to calculate the wire size for one-half of the main, as the other half is identical. In fig. 830 there are four lamps on each side of the junction, J; the center of each group is at a distance, M, so that the lamp feet in each half of the main are 5 × M. The lamp feet of the feeder would be 10 × N, N being the distance from the feeder fuse block to the junction, J. In concealed work, however, it does not always happen that a feeder can be made to join a main at its electrical center; when this is not practicable, each end of the main should be figured separately. In fig. 831 , for instance, the main has five lamps on one side and two on the other, and the distances from the junction to the centers of the two groups are at unequal distances S and S'. If the distance S be 14 feet, and the lamps, 16 c. p., the lamp feet in the left hand main equals 5 × 14 = 70, while in the main to the right, taking S' at 10 feet, there are only 2 × 10 = 20 lamp feet. Hence what appears to be one continuous main in this case would have to be treated as two mains, and each part figured separately. circular mils = I × feet × 21.6 E = amperes × feet × 21.6 "drop" . . . . (5) The following practical example is given to illustrate the application of the formula just obtained: EXAMPLE.—What size wire should be used on a 250 volt circuit to transmit a current of 200 amperes a distance of 350 feet to a center of distribution with a loss of three per cent. under full load? The volts lost or drop is equal to 250 × .03 = 7.5 volts. PROPERTIES OF COPPER WIRE Number of gauge B. & S. Diameter in mils Area in circular mils Weight in pounds Resistance at 68° Fahr. 1,000 feet mile Feet per pound 1,000 Feet mile 0000 460 211,600 640.5 3,381 1.561 .04893 .2583 000 409.6 167,800 508 2,682 1.969 .06170 .3258 00 364.8 133,100 402.8 2,127 2.482 .07780 .4108 0 324.9 105,500 319.5 1,687 3.130 .09811 .5180 1 289.3 83,690 253.3 1,337 3.947 .12370 .6531 2 257.6 66,370 200.9 1,062 4.977 .1560 .8237 3 229.4 52,630 159.3 841.1 6.276 .1967 1.0386 4 204.3 41,740 126.4 667.4 7.914 .2480 1.3094 5 181.9 33,100 100.2 529.0 9.980 .3128 1.6516 6 162.0 26,250 79.46 419.5 12.580 .3944 2.0824 7 144.3 20,820 63.02 332.7 15.87 .4973 2.6257 8 128.5 16,510 49.98 263.9 20.01 .6271 3.3111 9 114.4 13,090 39.63 209.2 25.23 .7908 4.1754 10 101.9 10,380 31.13 166.0 31.82 .9972 5.2652 11 90.74 8,234 24.93 131.6 40.12 1.257 6.6370 12 80.81 6,530 19.77 104.4 50.59 1.586 8.374 13 71.96 5,178 15.68 82.79 63.79 2.000 10.560 14 64.08 4,107 12.43 65.63 80.44 2.521 13.311 15 57.07 3,257 9.858 52.05 101.4 3.179 16.785 16 50.82 2,583 7.818 41.28 127.9 4.009 21.168 17 45.26 2,048 6.200 32.74 161.3 5.055 26.690 18 40.30 1,624 4.917 25.96 203.4 6.374 33.655 19 35.89 1,288 3.899 20.59 256.5 8.038 42.440 20 31.96 1,022 3.092 16.33 323.4 10.14 15.340 Substituting the given value in formula (5) circular mils = 350 × 200 × 21.6 7.5 = 201,600. Diameter = 2 √ 201,600 = 449 circular mils or .449 in. From the table (on page 731 or on page 742 ) the nearest ( larger ) size of wire is 0000 B. & S. gauge. 4 WIRING TABLE FOR LIGHT AND POWER CIRCUITS VOLTS PERCENTAGE OF LOSS 2000 1.7 1.5 1.4 1.2 1.1 1.0 0.75 0.5 1000 3.4 2.9 2.7 2.4 2.2 2.0 1.5 1.0 500 6.5 5.7 6.2 4.8 4.3 3.9 2.9 2.0 220 13.7 12.0 11.0 10.3 9.3 8.3 6.5 4.4 110 — — 20.0 18.5 17.0 15.4 12.0 8.4 52 — — — — — — 22.4 16.1 VOLTS PERCENTAGE OF LOSS (continued) 2000 0.45 0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 1000 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 500 1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.2 220 3.9 3.5 3.1 2.7 2.2 1.8 1.4 0.9 0.45 110 7.6 6.8 6.0 5.2 4.4 3.5 2.7 1.8 0.9 52 14.7 13.3 11.8 10.3 8.8 7.1 5.5 3.7 1.9 ACTUAL VOLTS LOST Carrying Capacity Amperes. Size B. & S. 35 30 27.5 25 22.5 20 15 10 300 0000 345800 296400 271700 247000 222300 197600 148200 98800 245 000 274400 235200 215600 196000 176400 156800 117600 78400 215 00 217525 186450 170912 155375 139837 124300 93225 62150 190 0 172550 147900 135575 123250 110925 98600 73950 49300 160 1 136850 117300 107525 97750 87975 78200 58650 39100 135 2 108500 93000 85250 77500 69750 62000 46500 31000 115 3 86100 73800 67650 61500 55350 49200 36900 24600 100 4 68250 58500 53625 48750 43875 39000 29250 19500 90 5 54250 46500 42625 38750 34875 31000 23250 15500 80 6 43050 36900 33825 30750 27675 24600 18450 12300 60 8 26965 23130 21202 19275 17347 15420 11565 7710 40 10 16975 14550 13337 12125 10912 9700 7275 4850 30 12 10675 9150 8388 7625 6862 6100 4575 3050 22 14 6720 5760 5280 4800 4320 3840 2880 1920 5 5 16 4235 3630 3328 3025 2723 2420 1815 1210 ACTUAL VOLTS LOST (continued) Carrying Capacity Amperes. Size B. & S. 9 8 7 6 5 4 3 2 1 300 0000 88920 79040 69160 59280 49400 39520 29640 19760 9880 245 000 70560 62720 54880 47040 39200 31360 23520 15680 7840 215 00 55935 49720 43505 37290 31075 24860 18645 12430 6215 190 0 44370 39440 34510 29580 24650 19720 14790 9860 4930 160 1 35190 31280 27370 23460 19550 15640 11730 7820 3910 135 2 27900 24800 21700 18600 15500 12400 9300 6200 3100 115 3 22140 19680 17220 14760 12300 9840 7380 4920 2460 100 4 17550 15600 13650 11700 9750 7800 5850 3900 1950 90 5 13950 12400 10850 9300 7750 6200 4650 3100 1550 80 6 11070 9840 8610 7380 6150 4920 3690 2460 1230 60 8 6939 6168 5397 4626 3855 3084 2313 1542 771 40 10 4365 3880 3395 2910 2425 1940 1455 970 486 30 12 2745 2440 2135 1830 1525 1220 915 610 305 22 14 1728 1536 1344 1152 960 768 576 384 192 5 5 16 1089 968 847 726 605 484 363 242 121 RULE.— Multiply current in amperes by single distance and refer to the nearest corresponding number under column of actual volts lost, to find size of wire. It should also be noted that the underwriters prohibit the use of wire smaller than No. 14 B. & S. gauge, except as allowed for fixture work and pendant cord. TABLE OF WIRE EQUIVALENTS GAUGE B. & S. NUMBER OF WIRES 2 4 8 1 6 3 2 6 4 1 2 8 2 5 6 5 1 2 1 0 2 4 2 0 4 8 4 0 9 6 8 1 9 2 1 6 3 8 4 0000 0 3 6 9 12 15 18 21 24 27 30 33 36 39 000 1 4 7 10 13 16 19 22 25 28 31 34 37 40 00 2 5 8 11 14 17 20 23 26 29 32 35 38 0+6 0 3 6 9 12 15 18 21 24 27 30 33 36 39 1+7 1 4 7 10 13 16 19 22 25 28 31 34 37 40 2+8 2 5 8 11 14 17 20 23 26 29 32 35 38 3+9 4+6 3 6 9 12 15 18 21 24 27 30 33 36 39 4+10 5+7 4 7 10 13 16 19 22 25 28 31 34 37 40 5+11 6+8 5 8 11 14 17 20 23 26 29 32 35 38 6+12 7+9 6 9 12 15 18 21 24 27 30 33 36 39 7+13 8+10 7 10 13 16 19 22 25 28 31 34 37 40 8+14 9+11 8 11 14 17 20 23 26 29 32 35 38 9+15 10+12 9 12 15 18 21 24 27 30 33 36 39 10+16 11+13 10 13 16 19 22 25 28 31 34 37 40 11+17 12+14 11 14 17 20 23 26 29 32 35 38 12+18 13+15 12 15 18 21 24 27 30 33 36 39 13+19 14+16 13 16 19 22 25 28 31 34 37 40 14+20 15+17 14 17 20 23 26 29 32 35 38 15+21 16+18 15 18 21 24 27 30 33 36 39 16+22 17+19 16 19 22 25 28 31 34 37 40 17+23 18+20 17 20 23 26 29 32 35 38 18+24 19+21 18 21 24 27 30 33 36 39 19+25 20+22 19 22 25 28 31 34 37 40 20+26 21+23 20 23 26 29 32 35 38 21+27 22+24 21 24 27 30 33 36 39 22+28 23+25 22 25 28 31 34 37 40 23+29 24+26 23 26 29 32 35 38 24+30 25+27 24 27 30 33 36 39 25+31 26+28 25 28 31 34 37 40 26+32 27+29 26 29 32 35 38 27+33 28+30 27 30 33 36 39 28+34 29+31 28 31 34 37 40 29+35 30+32 29 32 35 38 30+36 31+33 30 33 36 39 31+37 32+34 31 34 37 40 32+38 33+35 32 35 38 33+39 34+36 33 36 39 34+40 35+37 34 37 40 36+39 35 38 37+39 36 39 38+40 37 40 Ques. If the calculated size of wire be larger than any in the table how is the required area obtained? Ans. By using two or more smaller wires in parallel, whose combined area is equal to the required area. To facilitate finding the equivalent sizes the above table of wire equivalents has been prepared. Ques. How is the table of wire equivalents used? Ans. To use the table, find in the vertical column at left the size of conductor desired; then follow along horizontally until the size of wire that is desired to use for the strands, and the corresponding number at top of column will give the number of strands of that size wire required. Fig. 832.—Break down switch for use on three wire circuit, enabling it to be operated break down fashion with the two outers connected together and the neutral wire serving as one side of the resultant two wire circuit. Such circuits must be figured as two wire installations of one half the three wire voltage. The size of the neutral wire of a three wire circuit depends on conditions of operation. Three wire circuits for occasional two wire working, must have a neutral wire whose cross section is equal to the combined cross sections of the two outer wires. This plan is useful for buildings supplied from a central station, as it will be satisfactory for two wire operation in emergencies, and for three wire, two phase or three phase distribution should the central station ultimately be changed over to either of those alternating current systems. The expense for the extra copper in the beginning will not be nearly so great as that entailed by a change in the wiring later on should developments require it. It is permissible, however, to make the cross section of the neutral wire smaller than that of each outer wire, if one be reasonably sure that there will never be any changes such as those just mentioned, and if the drop in the two outer wires do not exceed 1½ per cent. Under such conditions, it will be found a very good rule to calculate the neutral wire of a principal feeder for a maximum unbalancing of 25 per cent, that is, a condition under which the current in one outer wire will be 75 per cent of the current in the other one, the current in the neutral being 25 per cent of that in the heavier loaded outer wire. Ques. What is the significance of the zig-zag line? Ans. The figures below this line give the gauge numbers of two wires which will have the same conductivity as the corresponding conductor in left hand column.

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