ALTERNATING GENERATORS AND SYNCHRONOUS MOTORS. 205 
Oe ae K-40 8 
— X 2 == 
amounts to 0.01057 x io" 24 1,200 watts, the 
normal current being 24 amperes and the length of mean 
armature turn 40 inches. 
397. Armature Iron Losses.—For the calculation of iron 
losses the method applied in the previous paragraphs, such 
as $871 and 372, will give satisfactory results, when al- 
lowing for one-half of the hysteresis losses, which would 
result for a complete change of magnetization and taking 
the full value of eddy current loss, provided that the total 
height of armature lamination is about one-half of the 
pole are length. 
398. Iron Loss in Teeth.—The weight of the armature 
toothed parts is equal to 2 X 24 X 2.25 x 1.75 x 6 x 0.9 
Xx 0.28 = 280 pounds. The saturation is 80,000 lines per 
square inch and the specific losses amount to 5.7 watts. 
Therefore. the total loss in the teeth is equal to 280 X 5.7 
= 1,600 watts. 
399. Iron Loss in Core.—The weight of the laminated arma- 
| ture core is equal to 3.14 x 29.75 x 325x2x6x09 
x .28 = 920 pounds, and the saturation being 46,000 lines 
per square inch, the total losses will be equal to 920 
(1.4+ 0.8) = 2,020 watts. 
400. Temperature Rise.—tThe total losses in the armature 
are, therefore, 1,200 + 1,600 + 2,020 = 4,820 watts, and 
the cooling surface is composed of the 
Inner cylinders of............. 880 square inches 
external cylinders of........... 2,200. ** ES 
and the side faces of........... 3,800)... * fa 
MOMING HtOPAl OL). oii eek. 4,980 ‘** aS 
The temperature rise will be approximately 7’, = 75 X 
sa" F: 
