Máquinas elétricas fitzgerald resoluções

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 + µ Ac 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 = 10.5 mWb 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 √ =