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