ALTERNATING GENERATORS AND SYNCHRON OUS MOTORS. 57 
F 300,000 
will be 73 x 1,900 
equal to 3 X 0.0163 = 0.049 square inch. Therefore the. 
= 92 amps. The conductor section is 
ao = 1880 amps. per square inch, 
(Ans.), which is a good figure. 
Length of mean turn: The next step is to find the length 
of mean turn. 
The bore being 51 inches, the circumference is equal to 
51 X 3.14 = 160.14 inches and with 14 poles the pole pitch 
amounts to 11.14 inches. Further, the width of the arma- 
ture core equals 10.5 inches. 
From formula 12, the mean length of a turn will therefore 
be equal to 2(10.5 + 2 x 11.44) = 66.8 inches, Ans. 
It is of interest to state that in the machine in question, the 
length has been found by actual measurement to be 69 © 
inches. 
Copper loss; With the mean length of, say, 67 inches, 
the total length of the armature winding will be 462 xX 67 
= 28,954 inches, and, having a sectional area of 3 x 0.0163 
square inches, the weight of the armature copper will 
amount to 28,954 x 0.049 x 0.319 = 450 Ibs. Ans. 
From the curve in Fig. 31 it is found, that with a current | 
density of 1,880, the energy loss in 450 lbs. of copper is 
equal to 450 x 9 = 4,050 watts. Ans. The same result 
would, of course, be obtained by calculating the armature 
resistance in ohms and multiplying the result by the square 
of the current in amperes, that is, C?R. | 
Armature resistance: Supposing that the temperature of > 
the air is 60 degrees Fahrenheit, the temperature rise 
in the armature 75 degrees, and the specific resistance of 
current density is 
0.8 pid, 
copper i08 ohms per cubie inch. Then the resistance of . 
28,954 X08 _ 
0.049 x 10° 
the whole armature winding is equal to 
9.48 ohm. Ans. 
