Electric Machinery 6Ed Fitzgerald, Kingsley, Uman Solution Manual

Electric Machinery

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1 PROBLEM SOLUTIONS: Chapter 1 Problem 1.1 Part (a): R c = l c µA c = l c µ r µ 0 A c = 0 A/Wb R g = g µ 0 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 0 A c = 1 . 591 × 10 5 A/Wb R g = g µ 0 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
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2 Problem 1.3 part (a): N = Lg µ 0 A c = 110 turns part (b): I = B core µ 0 N/g = 16 . 6 A Problem 1.4 part (a): N = L ( g + l c µ 0 ) µ 0 A c = L ( g + l c µ 0 / ( µ r µ 0 )) µ 0 A c = 121 turns part (b): I = B core µ 0 N/ ( g + l c µ 0 ) = 18 . 2 A Problem 1.5 part (a): part (b): µ r = 1 + 3499 1 + 0 . 047(2 . 2) 7 . 8 = 730 I = B g + µ 0 l c µ 0 N = 65 . 8 A
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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 0 part (b): Equations 2 gH g + H c l c = NI ; B g A g = B c A c and B g = µ 0 H g ; B c = µH c can be combined to give B g = NI 2 g + µ 0 µ A g A c ( l c + l p ) = NI 2 g + µ 0 µ 1 x X 0 ( l c + l p ) Problem 1.7 part (a): I = B g + µ 0 µ ( l c + l p ) µ 0 N = 2 . 15 A part (b): µ = µ 0 1 + 1199 1 + 0 . 05 B 8 = 1012 µ 0 I = B g + µ 0 µ ( l c + l p ) µ 0 N = 3 . 02 A
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4 part (c): Problem 1.8 g = µ 0 N 2 A c L µ 0 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 0 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 0 A c = 1 . 33 × 10 7 A/Wb; R c = l c µA c = 3 . 16 × 10 5 A/Wb
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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 0 N 2 A c g + l c r Problem 1.13 L = µ 0 N 2 A c g + l c 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
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6 Problem 1.15 part (a): R 3 = R 2 1 + R 2 2 = 4 . 27 cm part (b): L = µ 0 A g N 2 g + µ 0 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):
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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 0 NI B sat µ 0 l c µ = 0 . 36 mm part (b): From Eq.3.21 W gap = A c gB 2 sat 2 µ 0 = 0 . 207 J; W core = A c l c B 2 sat 2 µ = 0 . 045 J Thus W tot = W gap + W core = 0 . 252 J. From Eq. 1.47, (1 / 2) LI 2 = 0 . 252 J. Q.E.D. Problem 1.18 part (a): Minimum inductance = 4 mH, for which g = 0.0627 mm, N = 20 turns and V rms = 6.78 V part (b): Maximum inductance = 144 mH, for which g = 4.99 mm, N = 1078 turns and V rms = 224 V Problem 1.19 part (a): L = µ 0 πa 2 N 2 2 πr = 56 . 0 mH part (b): Core volume V core (2 πr ) πa 2 = 40 . 0 m 3 . Thus W = V core B 2 2 µ 0 = 4 . 87 J part (c): For T = 30 sec, di dt = (2 πrB ) / ( µ 0 N ) T = 2 . 92 × 10 3 A / sec v = L di dt = 163 V Problem 1.20 part (a): A cu = f w ab ; Vol cu = 2 ab ( w + h + 2 a ) part (b):
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8 B = µ 0 J cu A cu g part (c): J cu = NI A cu part (d): P diss = Vol cu ( ρJ 2 cu ) part (e): W mag = Vol gap B 2 2 µ 0 = gwh B 2 2 µ 0 part (f): L R = ( 1 2 ) LI 2 ( 1 2 ) RI 2 = W mag ( 1 2 ) P diss = 2 W mag P diss = µ 0 whA 2 cu ρg Vol cu Problem 1.21 Using the equations of Problem 1.20
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