EXAMPLE.—What is the proper size of wire for a 10 H.P. motor, run at 220 volts, allowable drop 2 per cent. on 200 foot circuit.
Substituting the given values in the formula on page 758:
10 × 746 × 200 × 21.6 Circular mils = ————————————————————- = 36,991. 220 × 4.4 × .9
The nearest larger value to this result, in the table of carrying capacities of copper wire (page 731), is 41,740, corresponding to No. 4 wire, B. & S. gauge.
_In all cases the size of the wire thus formed should be compared with that allowed by the underwriters for full load current of motor, plus 25 per cent. of that current_, and if the size calculated happen to be smaller than the allowable size, it should be increased to the latter, otherwise it will not pass inspection.
——————————————————————————————————————————————————————————————————————————————- H.P. Per Cent. Eff. Watts Input 50 Volts 100 Volts 220 Volts 500 Volts ¾ 70 800 16 7 4 2 1½ 70 1600 32 15 7 3 3 75 2980 60 27 14 6 5 80 4660 93 42 21 9 7½ 85 6580 132 60 30 13 10 85 8780 176 80 40 18 15 85 13200 264 120 60 26 20 85 17600 352 160 80 35
25 85 21900 438 199 100 44 30 90 24900 498 226 113 50 40 90 33200 664 301 151 66 50 90 41400 828 376 188 83
60 90 49700 994 452 226 99 70 90 58000 1160 527 264 116 80 90 66300 1330 608 302 133 90 90 74600 1490 678 339 149
100 90 82900 1660 755 377 166 120 90 99500 1990 905 453 199 150 90 124000 2480 1130 564 248 ——————————————————————————————————————————————————————————————————————————————-
——————————————————————————————————————————————————————————————————————————————- Appx. Appx. K.W. 125 Vs. 250 Vs. 500 Vs. H.P. K.W. 125 Vs. 250 Vs. 500 Vs. H.P. 1. 8 4 2 1.3 30. 240 120 60 40. 2. 16 8 4 2.7 37.5 300 150 75 50. 3. 24 12 6 4.0 40. 320 160 80 53. 5. 40 20 10 6.7 50. 400 200 100 67. 7.5 60 30 15 10. 60. 480 240 120 80. 10. 80 40 20 13. 75. 600 300 150 100. 12.5 100 50 25 17. 100. 800 400 200 134. 15. 120 60 30 20. 125. 1000 500 250 167. 20. 160 80 40 27. 150. 1200 600 300 201. 25. 200 100 50 34. 200. 1600 800 400 268. ——————————————————————————————————————————————————————————————————————————————-
To determine the current required for a motor, as for instance, the 10 H. P. motor under consideration, _multiply the horsepower by 746, and divide the product by the voltage of the motor multiplied by its efficiency as follows_: (10 × 746) ÷ (220 × .90) = 37.6 amperes.