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PROBLEM SOLUTIONS: Chapter 1
Solutions Manual for Fitzgerald and Kingsleys Electric Machinery 7th Edition by Umans Link download full:
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Problem 1-1 Part (a):
R c =
lc µA c
R g =
g µ Ac
=
lc µr µ 0A c
=
0
A/ W b
5.457 × 106
A/ W b
=
Part ( b):
Φ=
N I R c + R g
=
2.437 × 10−5
W b
Part (c): λ = N Φ = 2.315 × 10−3
W b
Part (d):
L
=
λ
I
=
1.654
mH
Problem 1-2 Part (a):
R c =
lc µA c
R g =
=
lc µr µ0 A c
g µ Ac
=
=
2.419 × 105
5.457 × 106
A/ W b
A/ W b
2
Part ( b):
Φ=
N I R + R
=
2.334 × 10−5
Wb
2
Part ( b):
Φ=
N I R c + R g
=
2.334 × 10−5
Wb
Part (c):
λ = N Φ = 2.217 × 10−3
W b
Part (d):
L
=
λ
I
=
1.584
mH
Problem 1-3 Part (a): s
=
Lg µ0 Ac
=
287 tur ns
I =
Bco r e µ0 N/ g
=
7.68
N
Part ( b):
A
Problem 1-4
Part (a): s
N
=
s
L(g
+ lc µ0 / µ)
µ0 Ac
=
L(g
+ lc µ0 / (µr µ0 ))
µ0 Ac
=
Part ( b): I =
Bco r e µ0 N/ (g + lc µ0 / µ)
=
20.78
A
129 tur ns
3
Problem 1-5 Part (a):
Part ( b):
Bg For Bm
=
2.1 T, µr
=
=
Bm
=
2.1 T
hus 37.88 and thus I =
Bm µ0 N
g+
lc µr
Part (c):
Problem 1-6 Part (a):
Bg
=
µ0 N I 2g
=
158 A
4
Ag Ac
Bc = B g
=
µ0 N I 2g
1−
x X0
Part (b): Will assume lc is “large” hus, large” and l p is r el el atively el y “small” small”. Thus
Bg Ag
=
B p Ag
=
Bc Ac
We can also wr ite 2gHg
+
H p l p
+
Hc lc
= N I ;
B p
=
µH p
Bc
and Bg
=
µ0 Hg ;
=
µHc
These equa tions can be combined to give Bg
µ0 N I
=
2g +
µ0 µ
l p
+
µ0 µ
Ag Ac
µ0 N I
=
lc
2g +
µ0 µ
l p
+
µ0 µ
and Bc
=
1−
x Bg X0
Problem 1-7 From Problem 1-6, the indu ctance ce can be found as N Ac Bc L= I from which we can solve for µr
=
2g +
µ0 µ
µ0 N 2 Ac (l p + (1 − x/ X0 ) lc )
1−
x X0
lc
5
µr
=
µ µ0
L l p =
+
(1 − x/ X0 ) lc
µ0 N 2 Ac − 2gL
=
Problem 1-8 Part (a):
L
µ0 (2N )2 Ac 2g
=
hus and thus s
N which rounds to N
Part (b):
=
2gL Ac
0.5
=
39 turns for which L
g
=
=
=
38.8
12.33 mH.
0.121 cm
Par t(c):
Bc
=
Bg
=
2µ0 N I 2g
hus and thus I =
Bc g µ0 N
=
37.1 A
Problem 1-9 Part (a):
L
=
µ0 N 2 Ac 2g
88.5
6
hus and thus s
N
which rounds to N
=
Part (b):
2gL Ac
=
78 turns for which L
g
=
=
=
77.6
12.33 mH.
0.121 cm
Par t(c):
Bc
=
Bg
=
)(I / 2) µ0 (2N )(I 2g
hus and thus I =
2Bc g µ0 N
=
37.1 A
Problem 1-10
Part (a):
L
=
µ0 (2N )2 Ac 2(g + ( µµ0 )lc )
hus and thus u u
N
which rounds to N
Part (b):
=
=
0.5
2(g
( µµ0 )lc )L
+
Ac
39 turns for which L
g
=
0.121 cm
=
=
38.8
12.33 mH.
7
Par t(c):
Bc
=
Bg
2µ0 N I µ 2(g + µ0 lc 0)
=
hus and thus I =
Bc (g
µ
+ µ0 lc )
µ0 N
=
40.9 A
Problem 1-11 Part (a): From the solution to Problem 1-6 with x I =
Part (b): For Bm
=
Bg 2g + 2
µ0 µ
(l p + lc )
µ0 N
1.25 T, µr
=
=
0
=
1.44 A
941 and thus I = 2.43 A
Part (c):
Problem 1-12
g
=
µ0 N 2 Ac L
!
−
µ0 µ
lc
=
7.8 × 10−4 m
8
Problem 1-13 Part (a):
lc
=
2π
Ac
=
R i + R o − g = 22.8 cm 2
h(R o − R i )
=
1.62 cm2
Part ( b):
R c =
R g =
g µ0 Ac
lc µA c
=
=
0
1 7.37 × 106 H−
Part (c):
L
=
N 2 R c + R g
=
7.04 × 10−4 H
Part (d):
I =
Bg Ac (R c + R g ) N
=
20.7 A
Part (e):
λ = LI = 1.46 × 10−2 W b
9
Problem 1-14
See solution to Problem 1-13 Part (a):
lc
=
22.8 cm
Ac
=
1.62 cm2
Part ( b):
R c = 1.37 × 106 H−1
R g = 7.37 × 106 H−1 Part (c):
L
=
5.94 × 10−4 H
Part (d):
I =
24.6 A
Part (e):
λ = 1.46 × 10−2 W b
10
Problem 1-15
µr must be b e gr eater than 2886. Problem 1-16
L
=
µ0 N 2 Ac g + lc / µr
Problem 1-17
Part (a):
L
=
µ0 N 2 Ac = 36.6 mH g + lc / µr
Part ( b):
B
=
µ0 N 2 I g + lc / µr
=
0.77 T
λ = LI = 4.40 × 10−2 W b Problem 1-18
Part (a): With ω = 120π
11
Vr m s =
N Ac B peak √ = 20.8 V 2
ω
Part (b): Using L from the solution to Problem 1-17 I peak
W peak
=
=
√ 2Vr ms ωL
2 LI peak 2
=
=
1.66 A
9.13
10−2 J
Problem 1-19
B
=
0.81 T and λ = 46.5 mW b
Problem 1-20 Part (a): q
R 3
=
(R 12
+
R 22 )
=
4.49 cm
Part (b): For
lc
=
4l + R 2
+ R 3
− 2h;
and Ag
L
=
=
π
R 2
µ0 Ag N 2 g + (µ0 / µ)lc
Part (c): For B pea k = 0.6 T and ω = 2π 60
=
61.8 mH
12
λ pea k =
Vr ms =
Irms
W peak
=
Ag N B peak
ω λ p eak
=
√ 2
Vr ms
ωL
=
=
23.2 V
0.99 A
1 2 1 √ LI pea k = L( 2Ir ms ) 2 2 2
Part (d): For ω = 2π 50
Vr ms =
19.3 V
I rms = 0.99 A
W peak Problem 1-21 Part (a);
=
61.0 mJ
=
61.0 mJ
13
Part ( b):
Emax = 4f N Ac B pea k = 118 V
part par t (c): For µ = 1000µ0 I peak
Problem 1-22 Part (a);
Part (b): I pe ak = 0.6 A
Part (c): I pea k = 4.0 A
=
lc B pea k = 0.46 A µ N
14
Problem 1-23
par t (b), I pea k = 11.9 A. For part par t (c), I pe ak = 27.2 A. For part
Problem 1-24
Part (a): For
I =
10 A, L
=
g
I =
=
Bc
=
µ0 N I g + (µ / µ)lc
23 mH and Bc N
Part (b): For
µ0 Ac N 2 g + (µ0 / µ)lc
L
=
10 A and Bc
=
LI Ac Bc
=
=
225 tur ns
µ0 lc µ0 N I − Bc µ =
Wg
Bg =
=
1.7 T
=
1.56 mm
1.7 T, from Eq. 3.21
Bg 2 Vg 2µ0
=
1.08 J
15
Wcore
Bc 2 Vg 2µ
=
=
0.072 J
based up on Vco r e =
Ac l c
Vg
=
Ac g
Part (c):
Wt o t
=
Wg + Wcore
=
1.15
J=
1 2
LI 2
Problem 1-25
Lmin
=
3.6 mH
Lmax = 86.0 mH
Problem 1-26 Part (a):
N
g
=
LI B Ac
µ0 N I 2B Ac
=
=
=
167
0.52 mm
Part ( b):
N
g
=
=
LI 2B Ac
µ0 N I B Ac
=
=
84
0.52 mm
16
Problem 1-27 Part (a):
N
g
=
=
LI B Ac
167
=
µ0 N I − (µ0 / µ)lc 2BA c
=
0.39 mm
Part ( b):
N
g
=
=
LI 2B Ac
=
µ0 N I − (µ0 / µ)lc B Ac
84
=
0.39 mm
Problem 1-28
Part (a): N
=
450 and g = 2.2 mm
Part (b): N
=
225 and g = 2.2 mm
Problem 1-29
Part (a):
L
=
µ0 N 2 A l
=
11.3 H
wher e A = π a2
l
=
2π r
17
Part ( b):
W
=
B2 2µ0
×
Volume
=
6.93 × 107 J
wher e Volume
=
(π a2 )( 2π r )
Part (c): For a flux density of 1.80 T, I =
lB µ0 N
=
2π r B µ0 N
=
6.75 kA
and V =
L
∆I ∆t
3 3 6.75 × 10 − = 113 × 10 40
=
1.90 kV
Problem 1-30 Part (a):
Copper cross − sec tiona l area ≡ Acu
Copper volume = Volcu
= f w b
(a
+
Part ( b):
B
=
µ0 Jcu Acu g
Part (c):
Jcu =
N I Acu
w )(h 2
= f w a b
+
w ) − wh 2
18
Part (d):
Pdiss
=
ρJc2u Volcu
Part (e):
Wstor ed
=
B2 2µ
×
2 µ0 Jcu A2cuwh 2g
gap volume =
Part (f ):
Wstor e Pdiss
= 2
I L
I 2 R
hus and thus L R
=
2
Wstor ed Pdiss
2
=
µ0 Acu w h gρVolcu
Problem 1-31
Pdiss
R
=
=
6.20 W
258 Ω L
=
I =
32 H
τ
155 mA
=
N
126 msec
=
12, 019 tur ns
Wire size
=
Problem 1-32
Part (a) (i):
Bg1
=
µ0 N1 I1 g1
Bg2
=
µ0 N1 I g2
(ii) λ 1
= N1 (A1 Bg1 +
A2Bg2 )
=
µ0 N 2
A1 g1
+
A2 I g2
34 AWG
19
λ 2
= N2 A2 Bg2 =
µ0 N1 N2
A2 I1 g2
Part (b) (i):
Bg1
=
0
Bg2
=
µ0 N2 I g2
(ii)
λ
= N
(A B g
λ 2
+
A Bg )
= N2 A2 Bg2 =
=
µ N N
µ0 N 2
A2 I g2
A2 I2 g2
Part (c) (i):
Bg1
=
µ0 N1 I1 g1
Bg2
=
µ0 N1 µ0 N2 I1 + I g2 g2
(ii) λ 1
= N1 (A1 Bg1 +
λ
A2 Bg2 )
=
= N
A Bg
=
L1
µ0 N 12
A1 g1
µ0 N 2
µ N N
A1 A2 A2 + I1 + µ0 N1 N2 I2 g2 g2 g1 A2 I g2
+
µ N 22
A2 I g2
Part (d): =
+
A2 g2
L2 µ0 N2
A2 g2
20
L12
=
µ0 N1 N2
A2 g2
Problem 1-33
R g =
g µ Ac
R 1 =
l1 µAc
R 2 =
l2 µAc
R A =
lA µAc
Part (a):
L1
=
N12 R g + R 1 + R 2 + R A / 2
LA
=
N 2 LB = R
wher e
R = R A +
R A (R g + R 1 + R 2 ) R A + R g + R 1 + R 2
Part ( b):
L1B = −L 1A =
L12
=
N1 N 2( R g + R 1 + R 2
+ R A / 2)
N 2 (R g + R 1 + R 2 ) 2R A (R g + R 1 + R 2 ) +
2 A
×
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