. A text-book of electrical engineering;. f armatureAngle subtended by poleLength of each air-gap ...Cross-section of poles and of path in poles and yoke The armature is built up of laminations stamped out of wrought ironsheet and separated from each other by thin paper, the space lost in this way being 15 per cent, of thewhole. We shall assume that thewhole magnet is of cast iron. Asindicated in Fig. 60, a part of themagnetic flux produced in thefield magnets is lost by shall assume that a sixth partof the flux produced is lost inthis way, so that, if O be the fluxin th
. A text-book of electrical engineering;. f armatureAngle subtended by poleLength of each air-gap ...Cross-section of poles and of path in poles and yoke The armature is built up of laminations stamped out of wrought ironsheet and separated from each other by thin paper, the space lost in this way being 15 per cent, of thewhole. We shall assume that thewhole magnet is of cast iron. Asindicated in Fig. 60, a part of themagnetic flux produced in thefield magnets is lost by shall assume that a sixth partof the flux produced is lost inthis way, so that, if O be the fluxin the armature and <l)„j that inthe magnet, „= We will now determine theampere-turns required to producea flux O in the armature of2-5. 10* lines. We must first find the lengthand cross-section of each part ofthe magnetic circuit. The cross-section of the armature per-pendicular to the lines of force is found by multiplying the difference D — d,by the axial length L, and also by 0-85 to allow for the paper Fig. 60 Hence, Aa= L {D — dg).o-85 = 290 sq. cm. The mean value of l^ is evidently about 20 cm. 29. The Magnetic Circuit 69 To find the cross-section of the path in the air-gaps we must reduce thecylindrical surface of the armature in the ratio /S: 360. We have then oAg=D .IT .L . -2-=420 sq. cm. To find the length of the path in air we must multiply the length of thegap between armature and pole by 2, since the lines have to cross the gapboth on entering and on leaving the armature. We have, therefore, ^j = = 0-8 cm. For the field magnet we havs given that Aj^ = 400 sq. cm, Ijn = no know also that 0„ = 1-2.$ = ^ = ^ Bringing all the above results together, we have $ = 2-5. 10^ 0 = ^ Aa= 2()0, Ag = 420, A^ = 400, la = 20, Ig = 0-8, l^ = no. From these figures it follows that -B„ = 1 = 8600, £, = £=5960, S^ = 1^ = 7500. From the magnetisation curve for armature stampings in Fig. 58 onpage 64 we see that the ampere-t
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