Solution manual Fitzgerald Kingsley Electric Machinery

277 Pages·2014·2.766 MB·English

by  Stephen D. Umans

SOLUTION MANUAL

Fitzgerald & Kingsley’s Electric Machinery [7th Edition]
Máquinas elétricas de Fitzgeral e Kingsley [7th edição]
1
PROBLEM SOLUTIONS: Chapter 1
Problem 1-1
Part (a):
l l
c c
= = = 0 A/Wb
c
R µA µ µ A
c r 0 c
g
= = 5.457 106 A/Wb
g
R µ A ×
0 c
Part (b):
NI
Φ = = 2.437 10−5 Wb
+ ×
c g
R R
Part (c):
λ = NΦ = 2.315 10−3 Wb
×
Part (d):
λ
L = = 1.654 mH
I
Problem 1-2
Part (a):
l l
= c = c = 2.419 105 A/Wb
c
R µA µ µ A ×
c r 0 c
g
= = 5.457 106 A/Wb
g
R µ A ×
0 c
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instructor use. Not authorized for sale or distribution in any manner. This document may
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2
Part (b):
NI
Φ = = 2.334 10−5 Wb
+ ×
c g
R R
Part (c):
λ = NΦ = 2.217 10−3 Wb
×
Part (d):
λ
L = = 1.584 mH
I
Problem 1-3
Part (a):
Lg
N = = 287 turns
sµ0Ac
Part (b):
B
core
I = = 7.68 A
µ N/g
0
Problem 1-4
Part (a):
L(g +l µ /µ) L(g +l µ /(µ µ ))
c 0 c 0 r 0
N = = = 129 turns
s µ0Ac s µ0Ac
Part (b):
B
core
I = = 20.78 A
µ N/(g +l µ /µ)
0 c 0
c 2014 by McGraw-Hill Education. This is proprietary material solely for authorized
(cid:13)
instructor use. Not authorized for sale or distribution in any manner. This document may
not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in
whole or part.
3
Problem 1-5
Part (a):
Part (b):
B = B = 2.1 T
g m
For B = 2.1 T, µ = 37.88 and thus
m r
B l
m c
I = g + = 158 A
µ N µ
(cid:18) 0 (cid:19)(cid:18) r(cid:19)
Part (c):
Problem 1-6
Part (a):
µ NI
0
B =
g
2g
c 2014 by McGraw-Hill Education. This is proprietary material solely for authorized
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instructor use. Not authorized for sale or distribution in any manner. This document may
not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in
whole or part.
4
A µ NI x
g 0
B = B = 1
c g
A 2g − X
(cid:18) c(cid:19) (cid:18) (cid:19)(cid:18) 0(cid:19)
Part (b): Will assume l is “large” and l is relatively “small”. Thus,
c p
B A = B A = B A
g g p g c c
We can also write
2gH +H l +H l = NI;
g p p c c
and
B = µ H ; B = µH B = µH
g 0 g p p c c
These equations can be combined to give
µ NI µ NI
0 0
B = =
g 2g + µ0 l + µ0 Ag l  2g + µ0 l + µ0 1 x l 
µ p µ Ac c µ p µ − X0 c
 (cid:16) (cid:17) (cid:16) (cid:17)(cid:16) (cid:17)   (cid:16) (cid:17) (cid:16) (cid:17)(cid:16) (cid:17) 
and
x
B = 1 B
c g
− X
(cid:18) 0(cid:19)
Problem 1-7
From Problem 1-6, the inductance can be found as
NA B µ N2A
c c 0 c
L = =
I 2g + µ0 (l +(1 x/X )l )
µ p − 0 c
from which we can solve for µ
r
c 2014 by McGraw-Hill Education. This is proprietary material solely for authorized
(cid:13)
instructor use. Not authorized for sale or distribution in any manner. This document may
not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in
whole or part.
5
L l +(1 x/X )l
µ p 0 c
−
µ = = (cid:18) (cid:19) = 88.5
r
µ µ N2A 2gL
0 0 c
−
Problem 1-8
Part (a):
µ (2N)2A
0 c
L =
2g
and thus
2gL
N = 0.5 = 38.8
s Ac
which rounds to N = 39 turns for which L = 12.33 mH.
Part (b): g = 0.121 cm
Part(c):
2µ NI
0
B = B =
c g
2g
and thus
B g
c
I = = 37.1 A
µ N
0
Problem 1-9
Part (a):
µ N2A
0 c
L =
2g
c 2014 by McGraw-Hill Education. This is proprietary material solely for authorized
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instructor use. Not authorized for sale or distribution in any manner. This document may
not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in
whole or part.
6
and thus
2gL
N = = 77.6
s Ac
which rounds to N = 78 turns for which L = 12.33 mH.
Part (b): g = 0.121 cm
Part(c):
µ (2N)(I/2)
0
B = B =
c g
2g
and thus
2B g
c
I = = 37.1 A
µ N
0
Problem 1-10
Part (a):
µ (2N)2A
0 c
L =
2(g +(µ0)l )
µ c
and thus
2(g +(µ0)l )L
µ c
N = 0.5 = 38.8
vu Ac
u
t
which rounds to N = 39 turns for which L = 12.33 mH.
Part (b): g = 0.121 cm
c 2014 by McGraw-Hill Education. This is proprietary material solely for authorized
(cid:13)
instructor use. Not authorized for sale or distribution in any manner. This document may
not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in
whole or part.
7
Part(c):
2µ NI
0
B = B =
c g 2(g + µ0l 0)
µ c
and thus
B (g + µ0l )
c µ c
I = = 40.9 A
µ N
0
Problem 1-11
Part (a): From the solution to Problem 1-6 with x = 0
B 2g +2 µ0 (l +l )
g µ p c
I = = 1.44 A
(cid:16) µ(cid:16)N(cid:17) (cid:17)
0
Part (b): For B = 1.25 T, µ = 941 and thus I = 2.43 A
m r
Part (c):
Problem 1-12
µ N2A µ
g = 0 c 0 l = 7.8 10−4 m
c
L − µ ! ×
c 2014 by McGraw-Hill Education. This is proprietary material solely for authorized
(cid:13)
instructor use. Not authorized for sale or distribution in any manner. This document may
not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in
whole or part.
8
Problem 1-13
Part (a):
R +R
i o
l = 2π g = 22.8 cm
c
2 −
(cid:18) (cid:19)
A = h(R R) = 1.62 cm2
c o i
−
Part (b):
l
c
= = 0
c
R µA
c
g
= = 7.37 106 H−1
g
R µ A ×
0 c
Part (c):
N2
L = = 7.04 10−4 H
+ ×
c g
R R
Part (d):
B A ( + )
g c c g
I = R R = 20.7 A
N
Part (e):
λ = LI = 1.46 10−2 Wb
×
c 2014 by McGraw-Hill Education. This is proprietary material solely for authorized
(cid:13)
instructor use. Not authorized for sale or distribution in any manner. This document may
not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in
whole or part.
9
Problem 1-14
See solution to Problem 1-13
Part (a):
l = 22.8 cm
c
A = 1.62 cm2
c
Part (b):
= 1.37 106 H−1
c
R ×
= 7.37 106 H−1
g
R ×
Part (c):
L = 5.94 10−4 H
×
Part (d):
I = 24.6 A
Part (e):
λ = 1.46 10−2 Wb
×
c 2014 by McGraw-Hill Education. This is proprietary material solely for authorized
(cid:13)
instructor use. Not authorized for sale or distribution in any manner. This document may
not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in
whole or part.