500 ELEMENTS OF ELECTRICAL ENGINEERING. 



Note The magnetic reluctance of the air between two similar iron spheres, r 

 centimeters -in radius and ^centimeters apart, center to center, is equal to (d r)/27rr/r, 

 when </is large compared with r. In this case r may be dropped from the numerator 

 as negligible and we have i/27rr as the magnetic reluctance of the air between two iron 

 spheres, each r centimeters in radius. 



In the above problem the poles of the iron rod may be taken to be two iron spheres 

 2 inches in diameter at a distance apart which is large compared with 2 inches. The 

 magnetic reluctance of the air return path is therefore approximately equal to 0.0627 

 oersted ( = i/27rr). The magnetic reluctance of the iron part of the magnetic circuit 

 in this problem is about 0.002 oersted and it is entirely negligible, inasmuch as the 

 larger reluctance of the air part of the circuit is calculated by an approximate formula. 



195. The magnetic reluctance of the two air gaps of a large 

 bipolar generator is 0.0006 oersted, the reluctance of the arma- 

 ture core is 0.00005 oersted, the reluctance of the leakage paths 

 through the air from pole-piece to pole-piece is 0.0022 oersted, 

 and the useful flux through the armature is 6,000,000 maxwells. 

 Find : (a) the magnetomotive force between the pole-pieces in 

 ampere -turns ; (b) the magnetomotive force in ampere-turns 

 across the two air gaps ; (c) the magnetomotive force in ampere- 

 turns across the armature core ; (d) the leakage flux ; (e) the 

 total field flux ; (/) the magnetic leakage coefficient of the 

 machine ; and (g) the total magnetic reluctance from pole-piece 

 to pole-piece. Ans. (a) 3,110 ampere-turns; (b) 2,870.8 am- 

 pere-turns ; (c) 239.2 ampere-turns ; (d) 1,773,000 maxwells ; 

 (e) 7,773,000 maxwells; (/) 1.295 ; (g) 0.000502 oersted. 



196. A 6-pole generator with a full load rating of 275 kilo- 

 watts at 550 volts has a simplex wave armature winding (2 paths 

 through the armature); the armature has 1,200 conductors and 

 the commutator has 300 segments. At full load the brushes 

 have a forward lead of four commutator segments. Find the 

 number of series turns required on each field core C (when the 

 field windings are arranged like Fig. 33, Chap. II.) to counter- 

 act the demagnetizing action of the armature at full load. Ans. 

 8 turns. 



Note. The number of ampere-turns required on the field per magnetic circuit to 

 balance D demagnetizing ampere-turns on the armature per magnetic circuit is equal to 



field reluctance \ 

 leakage reluctance / 



