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Electric Machinery 6Ed Fitzgerald, Kingsley, Uman Solution Manual

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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 . 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 . 591 × 10 5 A/Wb R g = g µ 0 A c . 017 × 10 6 A/Wb part (b): R c + R g . 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 = s Lg µ 0 A c = 110 turns part (b): I = B core µ 0 N/g =16 . 6A Problem 1.4 part (a): N = s L ( g + l c µ 0 ) µ 0 A c = s 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 . 2A Problem 1.5 part (a): part (b): µ r =1+ 3499 p 1+0 . 047(2 . 2) 7 . 8 =730 I = B ± g + µ 0 l c µ 0 N ² =65 . 8A
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3 part (c): Problem 1.6 part (a): H g = NI 2 g ; B c = ± A g A c ² B g = B g ± 1 x X 0 ² part (b): Equations 2 gH g + H c l c = ; 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 = 2 g + ³ µ 0 µ ´³ A g A c ´ ( l c + l p ) = 2 g + ³ µ 0 µ 1 x X 0 ´ ( l c + l p ) Problem 1.7 part (a): I = B g + ³ 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 ² = 1012 µ 0 I = B g + ³ 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 ² ± µ 0 µ ² 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 . 2c m 2 part (b): R g = g µ 0 A c . 33 × 10 7 A/Wb; R c =0 A/Wb; part (c): L = N 2 R g + R g . 319 mH part (d): I = B g ( R c + R g ) A c N =33 . 1A part (e): λ = NB g A c =10 . 5mW b Problem 1.10 part (a): Same as Problem 1.9 part (b): R g = g µ 0 A c . 33 × 10 7 A/Wb; R c = l c µA c . 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 . 8A 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 . 5m H Problem 1.14 part (a): V rms = ωNA c B peak 2 =19 . 2V r m s part (b): I rms = V rms ωL =1 . 67 A rms; W peak . 5 L ( 2 I rms ) 2 =8 . 50 mJ
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6 Problem 1.15 part (a): R 3 = q R 2 1 + R 2 2 =4 . 27 cm part (b): L = µ 0 A g N 2 g + ± µ 0 µ ² 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 . 5 L ( 2 I rms ) 2 . 817 J part (d): For ω π 50 rad/sec and λ peak = g B peak . 452 Wb: (i) V rms = peak = 142 V rms (ii) I rms = V rms . 81 A rms (iii) W peak . 5 L ( 2 I rms ) 2 . 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 . 3 6m m part (b): From Eq.3.21 W gap = A c gB 2 sat 2 µ 0 . 207 J; W core = A c l c B 2 sat 2 µ . 045 J Thus W tot = W gap + W core . 252 J. From Eq. 1.47, (1 / 2) LI 2 . 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 .78V 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 . 0m H part (b): Core volume V core (2 ) 2 =40 . 3 .Thu s W = V core ± B 2 2 µ 0 ² . 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 ;V o l cu ab ( w + h +2 a ) part (b):
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8 B = µ 0 ± J cu A cu g ² part (c): J cu = NI A cu part (d): P diss =Vol cu ( ρJ 2 cu ) part (e): W mag gap ± B 2 2 µ 0 ² = gwh ± B 2 2 µ 0 ² 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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