. A text-book of electrical engineering;. Fig. 151 If we make the simple but approximate assumption that the lines offorce cross the air-gap in the manner shown in Fig. 152 the cross-section ofthe gap Ag is the mean of the area of the pole-face and the total area ofthe tops of the teeth facing the pole. If the double air-gap has a length Ig = 0-4 cm,the area of the pole-face is {D + lg)TT . ^^— = 104 sq. cm. The tops of the teeth have an area 4 (DiT - 366) 62 sq. cm. 168 Electrical Engineering As the mean of these two values, we get . 104 + 62 oAg = —^ = 83 sq. cm. The cross-section of the rou
. A text-book of electrical engineering;. Fig. 151 If we make the simple but approximate assumption that the lines offorce cross the air-gap in the manner shown in Fig. 152 the cross-section ofthe gap Ag is the mean of the area of the pole-face and the total area ofthe tops of the teeth facing the pole. If the double air-gap has a length Ig = 0-4 cm,the area of the pole-face is {D + lg)TT . ^^— = 104 sq. cm. The tops of the teeth have an area 4 (DiT - 366) 62 sq. cm. 168 Electrical Engineering As the mean of these two values, we get . 104 + 62 oAg = —^ = 83 sq. cm. The cross-section of the round pole-cores can be seen from the drawing to be 77- ^j, = =43sq. cm. Since the flux of each pole divides into two equal parts in the yoke, the effective cross-section of the yoke is twice that of the actual magnet ring, that is, . ^„ = = 74sq. Fig. 152 The lengths of the paths in the different portions of the magnetic circuitcan be found from the drawing. It is important to notice that the air-pathIg is twice the gap between armature and pole, while the path in the teethis twice the depth of a tooth. Similarly l^, is twice the length of a pole,since the flux passes through both a north and a south pole. For ly, however,we put the simple length through the yoke from pole to pole. The pole-shoesneed not be separately considered. The values so obtained are collectedin the following table: Armature core(stampings) Armature teeth(stampings) Air-gap Pole-core(W. I.) Yoke(C. I.) la-7 At = 41-5 lt = A ^. = 83Ig = 0-4 ^i> = 43K = 12-5 ^.= 74 ly= 22 We have now to determine the number of ampere-turns required todrive the flux through each individual part of the magnetic circuit. Forthis purpose we shall assume four different values of the flux, and workthem out side by side. By dividing the flux by the cross-section of the partund
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