168 ELECTRICAL ENGINEERING 



145,200 lines per square inch is found to be 30; the permeance of 

 the path through the teeth is 



_A lj ui_43.4X(2.54) 2 X30_ 

 "IT 3.5 X 2.54 



and the permeance of the air path is 



_ A 2 _ 58 X (2.54)** _ 

 2 ~1T =: 3.5X2.54 : 



the flux of 6,300,000 crossing the gap divides between the two 

 paths in proportion to their permeances, and therefore the actual 

 flux passing through the teeth is 



<$ 040 



6,300,000 X rV = 6,300,000 X . = 6,040,000; 



6 040 000 

 and the corrected density in the teeth is - - = 139,300; 



4o.4 



the number of ampere turns per inch corresponding to this is 1200 

 and the number per pair of poles is 1200 X 3.5 = 4200. The 

 error introduced in this case by considering that the teeth carried 

 the whole flux would be 5250 - 4200 = 1050 ampere turns which 

 is an error of 25 per cent. 



The last section is the armature core below the slots; its section 

 is the product of the net length of iron in the armature by the 

 depth of iron below the slots; it is (8 - 3 X T 5 ?r) X 0.9 X 5 = 31.5 

 sq. ins.; the length of the path through the armature is estimated 

 from the drawing to be 12.3 ins. 



The flux carried by the armature section is 



3> 6,300,000 



= -- -- = 3,150,000 lines, 



and the flux density is 



3,150,000 

 31.5 



100,000 lines per square inch. 



The number of ampere turns per inch required is 75 and the num- 

 ber per pair of poles is 12.3 X 75 = 860. 



The results of the calculations above are tabulated in Fig. 133; 

 the total number of ampere turns required per pair of poles to pro- 

 duce a flux across the air gap of 6,300,000 lines and a voltage of 

 150 volts is 16,548. 



