. A text-book of electrical engineering;. 0 12000 14400 /X (y j from the curve for stampings = o8 23 4 9 ^*=^*(j)^ = 4(f)^= 3-2 9-2 16 36 3. the air-gap, we have the fundamental equation (43) on p. 61, •b ■-9 7 or X, = :?^ = o-85,.i,. ^ o 477 We have, therefore, O = 0-2 0-4 0-5 o-6. 10^ Bj = ^- = ^- = 2400 4820 6020 7220^!7 03Xg = o-& = o-Z2Bg= 770 1540 1930 2310 4. Pole-Cores. In calculating the ampere-turns for the poles and yoke we must takeinto account the fact that the flux in them is larger than the armature flux,on account of leakage. Assuming that the leakage coefficie
. A text-book of electrical engineering;. 0 12000 14400 /X (y j from the curve for stampings = o8 23 4 9 ^*=^*(j)^ = 4(f)^= 3-2 9-2 16 36 3. the air-gap, we have the fundamental equation (43) on p. 61, •b ■-9 7 or X, = :?^ = o-85,.i,. ^ o 477 We have, therefore, O = 0-2 0-4 0-5 o-6. 10^ Bj = ^- = ^- = 2400 4820 6020 7220^!7 03Xg = o-& = o-Z2Bg= 770 1540 1930 2310 4. Pole-Cores. In calculating the ampere-turns for the poles and yoke we must takeinto account the fact that the flux in them is larger than the armature flux,on account of leakage. Assuming that the leakage coefficient O™ _ _ V =we have V~ --T2. 0^ = i-2 43 11200 14000 16800 lyo Electrical Engineering (y] from the curve for wrought iron = i6 ^.=.(?).-==(t).= - B,. 5. ^ = i-2«D = A, 74y j from the curve for cast-iron x. = (f) = -(f).- 0-2024 3250 4190 ■5 62-5 0-40-48 6500 19 418 14 100 175 1250 0-5 06 . 10*06 072. 10* 8100 531160 9720 107 2350. 1000 aooo 6000 X woo 3000 woo Ampere turns Fig- 153 The total ampere-turns can now be found by adding up the five com-ponents, thus - ^ fi^ $ = 0-2 0-4 0-5 * Xa= 3^5 77 P 12^6 Xi = 3-2 9-2 16 36 Xg = 770 1540 1930 2310 Xj,= 20 62-5 175 1250 Xy= 90 , 418 II6O 2350 XX = Xa + Xt + Xg + X„ + Xy= 890 2040 3290 5960 57- Characteristics of Separately-exdted Dynamo 171 In Fig. 153 the values of armature flux O are plotted as ordinates, andthe values of Xj,, Xy, Xg, and HX as abscissae. The curve Xg is naturallya straight line. The values of X^ and Xf are so small that they have beenomitted in the figure. It is easy to see from the figure how the total ampere-turns HiX are used up in the various parts of the machine. To produce theflux OA, for example, we have to employ an excitation AE. Of this, a partAB is required for the poles, a part AC for the yoke, and the larger partAD for the air-gap, and AE = AB + AC + AD. Near the origin the curve for HX coincides very nearly with the straig
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