21.6 = ohms per foot run in circuit where wires are one mil in diameter;
E = voltage at the motor;
L = drop in percentage of the voltage at the motor;
K = efficiency of the motor expressed as a decimal.The average values for K are about as follows: 1 H.P., .75; 3 H.P., .80; 5 H.P., .80; 10 H.P. and over, 90 per cent.
Figs. 839 and 840.—Wrong and right methods of loop wiring. In general, when a large percentage of loss is allowed with lamps at short distances, the size of wire, calculated simply in accordance with the resistance rules, will be found too small to carry the current safely. This fact is often overlooked, and even though wires may have been correctly calculated for a uniform percentage of loss, they will become painfully hot simply because the table of carrying capacity was not consulted. The cross connection of mains wherever possible, for the purpose of equalizing the pressure, will also often reduce the heating effects of the current. An example of this is shown in the above figures. A circle of lights was wired as in [fig. 839], and after the current had been turned on, the wires of the circle became hot, and there was quite a perceptible difference of candle power between the lights near A and those near B. Investigation disclosed the fact that the loop, contrary to instructions, had been left open. A few inches of wire as shown in dotted lines remedied the fault. A better arrangement, however, is shown in [fig. 840].
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]:
Circular mils = 10 × 746 × 200 × 21.6 220 × 4.4 × .9 = 36,991.
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.
This value increased by 25 per cent. of itself (37.6 × .25 = 9.4 amp.) is equal to 37.6 plus 9.4 = 47 amperes. In the table of carrying capacities of copper wire ([page 731]), 46 amperes is given as the allowable carrying capacity of No. 6, B. & S. gauge, rubber covered wire; therefore No. 5 wire must be used.