42 DIRECT-CURRENT DYNAMOS AND MOTORS. 
35. Example.—The armature of a bipolar ring dynamo is 
required to generate 10 kilowatts at 120 volis when 
running at 350 revolutions per minute, the total 
number of armature inductors beng 400. What 
should be the total flux through the armature ? 
Solution.—In bipolar dynamos, according to Par. 20, 
there are 2 paths for the current in the armature; hence, 
N», the number of pairs of parallel branches, is 7, and we 
have from (21): 
6 X 10° X 120 
d= ae ee 5,140,000 lines of force. 
By (6) the given armature has a diameter of about 
2000 ‘ 
D, = 3.8 X —— a5 it ef ecuse 
Inserting this value into formula (13), we have 
21.7 = / £30k ae 
22 X (1—m')x(m+m’') 
In the present case W = 10,000 watts. Taking s =1 
(Par. 16), k = .12 (Table 3), and assuming m’ = .10: 
(Table 4), the above equation becomes: 
1X.12X10 000 
6.28 xX (1—.10) x (m+. 10) 
1200 
~ (6.28 X.90X m)+(6.28X.90X. 10)’ 
from which follows : 
__ 1200 — (21.7° K 6.28 x .90 x .10)_ 934 _ 
21.77 X 6.28 x .90 2660 
The approximate length of the armature is, therefore: 
DL, = .85 X 21.7 = 7.6 inches. 
From Table 16, the proper value of the field density, for 
cast steel poles, is found to be B, = 28,000 lines per sq. 
in.; hence by (22): 
D=1.3 X 28,000 X 21.7 x 7.6=6,000,000 lines of force, 
which shows that the result obtained by (18) is practical. 
21.7%°= 
. 35. 
