Electric Machinery

Electric Machinery 6Ed Fitzgerald, Kingsley, Uman Solution Manual
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Unformatted text preview: 1 PROBLEM SOLUTIONS: Chapter 1 Problem 1.1 Part (a): R c = l c µA c = l c µ r µ A c = 0 A/Wb R g = g µ A c = 1 . 017 × 10 6 A/Wb part (b): Φ = NI R c + R g = 1 . 224 × 10- 4 Wb part (c): λ = N Φ = 1 . 016 × 10- 2 Wb part (d): L = λ I = 6 . 775 mH Problem 1.2 part (a): R c = l c µA c = l c µ r µ A c = 1 . 591 × 10 5 A/Wb R g = g µ A c = 1 . 017 × 10 6 A/Wb part (b): Φ = NI R c + R g = 1 . 059 × 10- 4 Wb part (c): λ = N Φ = 8 . 787 × 10- 3 Wb part (d): L = λ I = 5 . 858 mH 2 Problem 1.3 part (a): N = s Lg µ A c = 110 turns part (b): I = B core µ N/g = 16 . 6 A Problem 1.4 part (a): N = s L ( g + l c µ /µ ) µ A c = s L ( g + l c µ / ( µ r µ )) µ A c = 121 turns part (b): I = B core µ N/ ( g + l c µ /µ ) = 18 . 2 A Problem 1.5 part (a): part (b): µ r = 1 + 3499 p 1 + 0 . 047(2 . 2) 7 . 8 = 730 I = B g + µ l c /µ µ N = 65 . 8 A 3 part (c): Problem 1.6 part (a): H g = NI 2 g ; B c = A g A c B g = B g 1 − x X part (b): Equations 2 gH g + H c l c = NI ; B g A g = B c A c and B g = µ H g ; B c = µH c can be combined to give B g =   NI 2 g + µ µ A g A c ( l c + l p )   =   NI 2 g + µ µ 1 − x X ( l c + l p )   Problem 1.7 part (a): I = B   g + µ µ ( l c + l p ) µ N   = 2 . 15 A part (b): µ = µ 1 + 1199 √ 1 + 0 . 05 B 8 = 1012 µ I = B   g + µ µ ( l c + l p ) µ N   = 3 . 02 A 4 part (c): Problem 1.8 g = µ N 2 A c L − µ µ l c = 0 . 353 mm Problem 1.9 part (a): l c = 2 π ( R o − R i ) − g = 3 . 57 cm; A c = ( R o − R i ) h = 1 . 2 cm 2 part (b): R g = g µ A c = 1 . 33 × 10 7 A/Wb; R c = 0 A/Wb; part (c): L = N 2 R g + R g = 0 . 319 mH part (d): I = B g ( R c + R g ) A c N = 33 . 1 A part (e): λ = NB g A c = 10 . 5 mWb Problem 1.10 part (a): Same as Problem 1.9 part (b): R g = g µ A c = 1 . 33 × 10 7 A/Wb; R c = l c µA c = 3 . 16 × 10 5 A/Wb 5 part (c): L = N 2 R g + R g = 0 . 311 mH part (d): I = B g ( R c + R g ) A c N = 33 . 8 A part (e): Same as Problem 1.9. Problem 1.11 Minimum µ r = 340. Problem 1.12 L = µ N 2 A c g + l c /µ r Problem 1.13 L = µ N 2 A c g + l c /µ r = 30 . 5 mH Problem 1.14 part (a): V rms = ωNA c B peak √ 2 = 19 . 2 V rms part (b): I rms = V rms ωL = 1 . 67 A rms; W peak = 0 . 5 L ( √ 2 I rms ) 2 = 8 . 50 mJ 6 Problem 1.15 part (a): R 3 = q R 2 1 + R 2 2 = 4 . 27 cm part (b): L = µ A g N 2 g + µ µ l c = 251 mH part (c): For ω = 2 π 60 rad/sec and λ peak = NA g B peak = 0 . 452 Wb: (i) V rms = ωλ peak = 171 V rms (ii) I rms = V rms ωL = 1 . 81 A rms (iii) W peak = 0 . 5 L ( √ 2 I rms ) 2 = 0 . 817 J part (d): For ω = 2 π 50 rad/sec and λ peak = NA g B peak = 0 . 452 Wb: (i) V rms = ωλ peak = 142 V rms (ii) I rms = V rms ωL = 1 . 81 A rms (iii) W peak = 0 . 5 L ( √ 2 I rms ) 2 = 0 . 817 J Problem 1.16 part (a): 7 part (b): E max = 4 fNA c B peak = 345 V Problem 1.17 part (a): N = LI A c B sat = 99 turns; g = µ NI B sat − µ l c µ = 0 . 36 mm part (b): From Eq.3.21 W gap = A c gB 2 sat...
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