LOSSES IN ARMATURES 



111 



Example. In order to explain the application of the formulas 

 for armature heating, numerical values will be assumed for the 

 dimensions in Fig. 35. 

 Let D = 32 in., 

 d = 23 in., 

 i a = 15 in., 



n = 5 (the number of air ducts in the armature core), 

 N = 400 revolutions per minute, 



whence 



v = 400 X ~ 



3,360 



and 



v d = 1,120. 



FIG. 35. Section through armature core. 



Let us further assume that the total armature losses, con- 

 sisting of hysteresis and eddy-current losses in teeth and core, 

 together with the PR losses in the portion of the armature wind- 

 ing that is buried in the slots, amount to 7 kw. 



The cooling surfaces to be considered are: 



1. The outside cylindrical surface of area AI = TT X 32 X 

 15 = 1,510 sq. in. 



2. The inside cylindrical surface of area A 2 = TT X 23 X 

 15 = 1,085 sq. in. 



