which is the actual load, and from which

actual load 41,403
apparent load = —————— = ———— = 51,754
power factor  .8

The current therefore at 440 volts is

apparent load51,754
——————— =————— =117.6 amperes
volts440

EXAMPLE.—A 50 horse power single phase 440 volt motor, having a full load efficiency of .92 and power factor of .8, is to be operated at a distance of 1,000 feet from the alternator. The wires are to be spaced 6 inches apart and the frequency is 60, and % loss 5. Determine: A, electrical horse power; B, watts; C, apparent load; D, current; E, size of wires; F, drop; G, voltage at the alternator.

A. Electrical horse power

brake horse power  50
E. H. P. = ————————— × —— = 54.3
efficiency .92

or,

54.3 × 746 = 40,508 watts

0000 2 No. 0 4 No. 3 8 No. 6 16 No. 9 32 No. 12 64 No. 15
000 2 " 14 " 48 " 7 16 " 10 32 " 1364 " 16
00 2 " 24 " 58 " 816 " 1132 " 1464 " 17
0 2 " 34 " 68 " 916 " 1232 " 1564 " 18
1 2 " 44 " 78 " 1016 " 1332 " 1664 " 19
2 2 " 54 " 88 " 1116 " 1432 " 1764 " 20
3 2 " 64 " 98 " 1216 " 1532 " 1864 " 21
4 2 " 74 " 108 " 1316 " 1632 " 1964 " 22
5 2 " 84 " 118 " 1416 " 1732 " 2064 " 23
6 2 " 94 " 128 " 1516 " 1832 " 2164 " 24
7 2 " 104 " 138 " 1616 " 1932 " 2264 " 25
8 2 " 114 " 148 " 1716 " 2032 " 2364 " 26
9 2 " 124 " 158 " 1816 " 2132 " 2464 " 27
10 2 " 134 " 168 " 1916 " 2232 " 2564 " 28
11 2 " 144 " 178 " 2016 " 2332 " 2664 " 29
12 2 " 154 " 188 " 2116 " 2432 " 2764 " 30
13 2 " 164 " 198 " 2216 " 2532 " 28
14 2 " 174 " 208 " 2316 " 2632 " 29
15 2 " 184 " 218 " 2416 " 2732 " 30
16 2 " 194 " 228 " 2516 " 28
17 2 " 204 " 238 " 2616 " 29
18 2 " 214 " 248 " 2716 " 30
19 2 " 224 " 258 " 28
20 2 " 234 " 268 " 29
21 2 " 244 " 278 " 30