98 THE MAGNETIC CIRCUIT [ART. 37 



Referring to Art. 36, eq. (50) may be interpreted as follows: 

 Let k a ' be the air-gap factor for the slotted stator and a smooth- 

 body rotor; let k a " be the same factor for the slotted rotor and a 

 smooth-body stator. Then the air-gap factor of the actual 

 machine 



k a =k a 'Xk a " (51) 



In an induction motor the magnetic flux is distributed in the 

 air-gap approximately according to the sine law, due to the dis- 

 tributed polyphase windings. Therefore the value of M deter- 

 mined from eq. (44) gives only the average value of the m.m.f. 

 required for the air-gap. With a sine-wave distribution of the 

 flux the maximum m.m.f. is n/2 times larger than the average 

 value. 



Prob. 20. What is the permeance of the air-gap of a 16-pole direct- 

 current machine, the armature of which has a diameter of 250 cm. and 

 is provided with 324 slots, 12 by 15 mm. ? The gross length of the armature 

 is 23 cm., and it is provided with three ventilating ducts, 10 mm. wide 

 each. The axial length of the poles is 21.5 cm. The pole shoes cover 

 05 per cent of the periphery, and are not chamfered. The length of the 

 air-gap is 10 mm. Ans. About 900 perrn. 



Prob. 21. The machine mentioned in problem 14 has 120 slots, 3 by 

 6.5 cm. The pole-shoes are shaped according to the arc of a circle of a 

 radius equal to 90 cm. and subtending 36 degrees ; the pole-tips are formed 

 by quadrants of a radius equal to 2.5 cm. Check the value of the field 

 current (52 amp.) given in problem 15, by the method of equivalent 

 permeances. 



Prob. 22. What is the maximum m.m.f. across the air-gap of an 

 induction motor, if the gross average flux density in the air-gap (total 

 flux divided by the gross area of the air-gap, not including the vents) 

 is 3 kl./sq.cm., and the clearance is 1.2 rnm.? The bore is 64 cm.; the 

 stator is provided with 48 open slots, 22 by 43 mm. The rotor has 91 

 half-closed slots, the slot opening being 3 mm. The machine has a vent 

 7 mm. wide for every 9 cm. of the laminations. 



Ans. 820 amp. turns. 



Prob. 23. Show that Jt/a = 1.2 +2.93 log (s/2a), if the fringing 

 lines of force are assumed to be concentric quadrants (Fig. 27, to the left) 

 with the points c as the center; the average length of path in the part 

 bcc* is estimated to be equal to 1.2a, and the average width 0.72a. Hint: 

 The permeance of an infinitesimal tube of force of a radius x and of a 

 width dx is p'dx/(fax). Integrate this expression between the limits 



stromes in Induktionsmotoren, Elektrotechnik und Maschinenbau, Vol. 28 

 (1910), p. 743. 



