102 ELECTRICAL MACHINERY 



lines per sq.cm. and 20% of them lie in the pole fringe 

 where the average flux density is 5000 lines per sq.cm. 

 Find the same quantities as in previous problem. 



To calculate the generated e.m.f. the method of precedure 

 is to find: (1) the length of inductor in the denser field 

 and calculate the e.m.f. generated by it (A) ; (2) the length 

 of inductor in the weak field and calculate the e.m.f. gener- 

 ated by it (B). The sum of (A) and (B) gives the total 

 generated voltage. 



As the machine is lap wound and is 12 pole, there must 

 be 20 coils per path. 



The length of inductor per path 



-20X4X20X2.54 = 4070 cm.; 



The length of inductor in the dense field 



= 4070X50% = 2035 cm. (A); 



The length of inductor in the pole fringe 



= 4070X20% = 814 c.m (B); 



4X200X2.54X12 

 I he peripheral speed = - - = 1275 cm./sec. 



E.m.f. generated by inductors (A) 



= 1275 X 2035 X 9000 X 10 ~ 8 = 233 volts 

 E.m.f. generated by inductors (B) 



= 1275 X 814 X 5000 X 10 ~ 8 = 52 volts 



Total e.rn.f. per path =285 volts 



The conductor of which the winding is formed is a double 

 No. 8 and so has a cross-section of 33,020 circular mils. 



Allowing 600 circular mils per ampere gives a capacity 

 per path of 55 amperes. As there are 12 paths in parallel 

 and each can carry 55 amperes the whole armature has a 



