# Maquinas eletricas

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1

PROBLEM SOLUTIONS: Chapter 1
Problem 1.1 Part (a): Rc = Rg = part (b): Φ= part (c): λ = N Φ = 1.016 × 10−2 part (d): L= Problem 1.2 part (a): Rc = lc lc = = 1.591 × 105 µAc µr µ0 Ac g = 1.017 × 106 µ0 Ac A/Wb λ = 6.775 mH I Wb NI = 1.224 × 10−4 Rc + Rg Wb lc lc = =0 µAc µr µ0 Ac g = 1.017 × 106 µ0 Ac A/Wb

A/Wb

Rg = part (b): Φ= part (c):A/Wb

NI = 1.059 × 10−4 Rc + Rg

Wb

λ = N Φ = 8.787 × 10−3 part (d): L= λ = 5.858 mH I

Wb

2 Problem 1.3 part (a): N= part (b): I= Problem 1.4 part (a): N= part (b): I= Problem 1.5 part (a): Bcore = 18.2 A µ0 N/(g + lc µ0 /µ) L(g + lc µ0 /µ) = µ0 Ac L(g + lc µ0 /(µr µ0 )) = 121 turns µ0 Ac Bcore = 16.6 A µ0 N/g Lg = 110 turns µ0 Ac

part(b): µr = 1 + 3499 1 + 0.047(2.2)7.8 g + µ0 lc /µ µ0 N = 730

I=B

= 65.8 A

3 part (c):

Problem 1.6 part (a): Hg = NI ; 2g Bc = Ag Ac Bg = Bg 1 − x X0

part (b): Equations 2gHg + Hc lc = N I; and Bg = µ0 Hg ; Bc = µHc  NI
µ0 µ

Bg Ag = Bc Ac

can be combined to give    NI = Bg =  Ag (lc + lp ) 2g + µ0 2g + Ac µ Problem 1.7 part (a): I = B part (b): 1199 µ = µ0 1 + √ 1 + 0.05B 8  I = B g+
µ0 µ

1−

x X0

(lc + lp )

g+

µ0 µ

(lc + lp )

  = 2.15 A

µ0 N

= 1012 µ0   = 3.02 A

(lc + lp )

µ0 N

4 part (c):

Problem 1.8 g= Problem 1.9 part (a): lc = 2π(Ro − Ri ) − g = 3.57 cm; part (b): Rg = part (c): L= part (d): I= part (e): λ = N Bg Ac = 10.5mWb Problem 1.10 part (a): Same as Problem 1.9 part (b): Rg = g = 1.33 × 107 µ0 Ac A/Wb; Rc = lc = 3.16 × 105 µAc A/Wb Bg (Rc + Rg )Ac = 33.1 A N N2 = 0.319 mH Rg + Rg g = 1.33 × 107 µ0 Ac A/Wb; Rc = 0 A/Wb; Ac = (Ro − Ri )h = 1.2 cm2 µ0 N 2 Ac L − µ0 µ lc = 0.353 mm

5 part (c): L= part (d): I= Bg (Rc + Rg )Ac = 33.8 A N N2 = 0.311 mH Rg + Rg

part(e): Same as Problem 1.9. Problem 1.11

Minimum µr = 340. Problem 1.12 L= Problem 1.13 L= Problem 1.14 part (a): Vrms = part (b): Irms = Vrms = 1.67 A rms; ωL √ Wpeak = 0.5L( 2 Irms )2 = 8.50 mJ ωN Ac Bpeak √ = 19.2 V rms 2 µ0 N 2 Ac = 30.5 mH g + lc /µr µ0 N 2 Ac g + lc /µr

6 Problem 1.15 part (a): R3 = part (b): L= µ0 Ag N 2 g+
µ0 µ 2 2 R1 + R2 =4.27 cm

= 251 mH

lc

part (c): For ω = 2π60 rad/sec and λpeak = N Ag Bpeak = 0.452 Wb: (i) (ii) (iii) Vrms = ωλpeak = 171 V rms Vrms = 1.81 A rms Irms = ωL √ Wpeak = 0.5L( 2Irms )2 = 0.817 J

part (d): For ω = 2π50 rad/sec and λpeak = N Ag Bpeak = 0.452 Wb: (i) (ii) (iii) Problem 1.16 part (a): Vrms = ωλpeak = 142 V rms Vrms Irms = = 1.81 A rms ωL √ Wpeak = 0.5L( 2Irms )2 = 0.817 J WWW.COMUNIDADEELETRICIDADE.BLOGSPOT.COM

7 part (b): Emax = 4f N Ac Bpeak = 345 V Problem 1.17 part (a): N= LI = 99 turns; Ac Bsat g= µ0 N I µ0 lc = 0.36 mm − Bsat µ

part (b): From Eq.3.21 Wgap =
2 Ac gBsat = 0.207 J; 2µ0

Wcore =

2 Ac lc Bsat = 0.045 J 2µ

Thus Wtot = Wgap + Wcore = 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 Vrms = 6.78 V part (b): Maximum inductance = 144 mH, for which g = 4.99 mm, N = 1078 turns and Vrms = 224 V Problem 1.19 part (a): L= µ0 πa2 N 2 = 56.0 mH 2πr

part (b): Core volume Vcore ≈ (2πr)πa2 = 40.0 m3 . Thus W = Vcore part (c): For T = 30 sec, di (2πrB)/(µ0 N ) = = 2.92 × 103 dt T di = 163 V dt A/sec B2 2µ0 = 4.87 J

v=L Problem 1.20 part (a):Acu = fw ab; part (b):

Volcu = 2ab(w + h + 2a)

8

B = µ0 part (c): Jcu = part (d):

Jcu Acu g

NI Acu

2 Pdiss = Volcu ρJcu

part (e): Wmag = Volgap part (f): L = R
1 2 1 2

B2 2µ0

= gwh

B2 2µ0

LI 2 = RI 2

Wmag 2Wmag µ0 whA2 cu = = 1 ρgVolcu Pdiss 2 Pdiss

Problem 1.21 Using the equations of Problem 1.20 Pdiss = 115 W I...