Solutions manual for fitzgerald and kingsleys electric machinery 7th edition by umans

1
Solutions Manual for Fitzgerald and Kingsleys Electric Machinery
7th Edition by Umans
Full clear download (no error formatting) at :
https://downloadlink.org/p/solutions-manual-for-fitzgerald-and-
kingsleys-electric-machinery-7th-edition-by-umans/
PROBLEM SOLUTIONS: Chapter 1
Problem 1-1
Part (a):
lc lc
Rc = = = 0 A/Wb
µAc µrµ0Ac
g 6
Rg = = 5.457 × 10 A/Wb µ0Ac
Part (b):
NI −5
Φ = RR = 2.437 × 10 Wb c + g
Part (c):
λ = NΦ = 2.315 × 10−3 Wb
Part (d):

2
λ
L = = 1.654 mH
I
Problem 1-2
Part (a):
lc lc 5
Rc = = = 2.419 × 10 A/Wb µAc µrµ0Ac
g 6
Rg = = 5.457 × 10 A/Wb µ0Ac
Part (b):
NI −5
Φ = R R = 2.334 × 10 Wb c + g
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
µ0Ac
Part (b):

3
Bcore
I = = 7.68 A
µ0N/g
Problem 1-4
Part (a):
N = L(g +
lcµ0/µ) = sL(g + lcµ0/(µrµ0)) = 129 turns µ0Ac µ0Ac
Part (b):
Bcore
I = = 20.78 A µ0N/(g +
lcµ0/µ)
c 2014 by McGraw-Hill Education. This is proprietary material solely for authorized
Problem 1-5
Part (a):
Part (b):
Bg = Bm = 2.1 T
For Bm = 2.1 T, µr = 37.88 and thus

4
Bm g +
lc =I =
158 A µ0N µr
Part (c):
Problem 1-6
Part (a):
Bg = µ0NI
2g
Ag µ0NI x
Bc = Bg c = 1 − X0
A 2g
Part (b): Will assume lc is “large” and lp is relatively “small”. Thus,
BgAg = BpAg = BcAc
We can also write
2gHg + Hplp + Hclc = NI;
and
Bg = µ0Hg; Bp = µHp Bc = µHc

5
These equations can be combined to give Bg = µ0 0
µ0 Ag = µ0 0 µ0 x µ NI µ NI
2g + µ lp + µ Ac lc 2g + µ lp + µ 1 − X0lc
and
x
Bc = 1 − X0 Bg
Problem 1-7
From Problem 1-6, the inductance can be found as
NAcBc µ0N2Ac
L = =
I x/X0)lc)
from which we can solve for µr c 2014 by McGraw-Hill Education. This is proprietary
material solely for authorized µ Llp + (1 − x/X0)lc
µr = µ0 = µ0N2Ac − 2gL = 88.5
Problem 1-8
Part (a):
µ0(2N)2Ac
L =
2g
and thus

6
2gL
N = 0.5 = 38.8
Ac
which rounds to N = 39 turns for which L = 12.33 mH.
Part (b): g = 0.121 cm
Part(c):
Bc = Bg = 2µ0NI
2g
and thus
I = Bcg
= 37.1 A
µ0N
Problem 1-9
Part (a):
µ0N2Ac
L =
2g

7
and thus
2gL
N = = 77.6
Ac
Part (b): g = 0.121 cm
Part(c):
Bc = Bg = µ0(2N)(I/2)
2g
and thus
I = 2Bcg
= 37.1 A
µ0N
Problem 1-10
Part (a):
µ0(2N)2Ac
L =
µ
and thus
µ0
2(g + ( µ )lc)L
u

8
N = 0.5u
t = 38.8
Ac
Part (b): g = 0.121 cm
Part(c):
Bc = Bg = 2µ0NI
µ
and thus
Bc(g + µµ0lc)
I = = 40.9 A µ0N
Problem 1-11
Part (a): From the solution to Problem 1-6 with x = 0
Bg 2g + 2µ
µ0(lp + lc)
I == 1.44 A
µ0N
Part (b): For Bm = 1.25 T, µr = 941 and thus I = 2.43 A
Part (c):

9
Problem 1-12
µ0N2Ac µ0 −4
g = − !lc = 7.8 × 10 m
L µ
Problem 1-13
Part (a):
Ri + Ro − g = 22.8 cmlc = 2π
2
Ac = h(Ro − Ri) = 1.62 cm2
Part (b):
lc
Rc = = 0 µAc
g 6 −1
Rg = = 7.37 × 10 H µ0Ac

10
Part (c):
N2 −4
L = R R = 7.04 × 10
H c + g
Part (d):
I = BgAc(Rc + Rg) = 20.7 A
N
Part (e):
λ = LI = 1.46 × 10−2 Wb
Problem 1-14
See solution to Problem 1-13
Part (a): lc = 22.8 cm
Ac = 1.62 cm2
Part (b):
Rc = 1.37 × 106 H−1
Rg = 7.37 × 106 H−1
Part (c):

11
L = 5.94 × 10−4 H
Part (d):
I = 24.6 A
Part (e):
λ = 1.46 × 10−2 Wb
Problem 1-15
µr must be greater than 2886.
Problem 1-16
µ0N2Ac
L =
g + lc/µr
Problem 1-17
Part (a):
µ0N2Ac

12
L = = 36.6 mH g +
lc/µr
Part (b):
µ0N2
B = I = 0.77 T g +
lc/µr
λ = LI = 4.40 × 10−2 Wb
Problem 1-18
Part (a): With ω = 120π
Vrms ωNAcBpeak
8 V
Part (b): Using L from the solution to Problem 1-17
Vrms
Ipeak = = 1.66 A ωL
LIpeak2 −2
Wpeak = = 9.13 × 10 J
2
Problem 1-19
B = 0.81 T and λ = 46.5 mWb
Problem 1-20

13
Part (a):
R3 = q(R2
1 + R2
2) = 4.49 cm
Part (b): For
lc = 4l + R2 + R3 − 2h;
and
Ag = πR21
µ0AgN2
L = = 61.8 mH g +
(µ0/µ)lc
Part (c): For Bpeak = 0.6 T and ω = 2π60
λpeak = AgNBpeak
ωλpeak
Vrms 2 V
Vrms
Irms = = 0.99 A ωL
Wpeak 0 mJ
Part (d): For ω = 2π50

14
Vrms = 19.3 V
Irms = 0.99 A
Wpeak = 61.0 mJ
Problem 1-21
Part (a);
Part (b):
Emax = 4fNAcBpeak = 118 V
part (c): For µ = 1000µ0
Ipeak = lcBpeak = 0.46 A
µN
Problem 1-22
Part (a);

15
Part (b): Ipeak = 0.6 A
Part (c): Ipeak = 4.0 A
Problem 1-23
For part (b), Ipeak = 11.9 A. For part (c), Ipeak = 27.2 A.
Problem 1-24

16
µ0AcN2
L =
g + (µ0/µ)lc
Bc = µ0NI
g + (µ0/µ)lc
Part (a): For I = 10 A, L = 23 mH and Bc = 1.7 T
LI
N = = 225 turns
AcBc
g = µ0NI − µ0lc = 1.56 mm
Bc µ
Part (b): For I = 10 A and Bc = Bg = 1.7 T, from Eq. 3.21
Bg2
Wg = Vg = 1.08
J
2µ0
Bc2
Wcore = Vg =
0.072 J
2µ
based upon
Vcore = Aclc Vg = Acg

17
Part (c):
1 2
Wtot = Wg + Wcore = 1.15 J = LI
2
Problem 1-25
Lmin = 3.6 mH Lmax = 86.0 mH
Problem 1-26
Part (a):
LI
N = = 167
BAc
g = µ0NI
= 0.52 mm
2BAc
Part (b):
LI
N = = 84
2BAc
g = µ0NI
= 0.52 mm
BAc
Problem 1-27

18
Part (a):
LI
N = = 167
BAc
g = µ0NI
− (µ0/µ)lc = 0.39 mm 2BAc
Part (b):
LI
N = = 84
2BAc
g = µ0NI
− (µ0/µ)lc = 0.39 mm
BAc 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):
µ0N2A
L = = 11.3 H
l
where

20
Part (b):
B2 7
W = × Volume = 6.93 × 10 J
2µ0
where
Volume = (πa2)(2πr)
Part (c): For a flux density of 1.80 T,
lB 2πrB
I = = = 6.75 kA µ0N
µ0N
and
∆I 6.75
103
V = L = 113 × 10−3 × = 1.90 kV
∆t 40
Problem 1-30
Part (a):
Copper cross − sectional area ≡ Acu = fwab
w w
Copper volume = Volcu = fwb (a + )(h + ) − wh
2 2
Part (b):
µ0JcuAcu

21
B =
g
Part (c):
NI
Jcu = Acu
c 2014 by McGraw-Hill Education. This is proprietary material solely for authorized Part (d):
Pdiss = ρJcu2 Volcu
Part (e):
B2 µ0Jcu2 A2cuwh
Wstored = × gap volume =
2µ0 2g
Part (f):
WPstorediss = 12II22RL
and thus
L Wstored µ0A2cuwh = 2 =
R Pdiss gρVolcu
Problem 1-31
Pdiss = 6.20 W I = 155 mA N = 12,019 turns
R = 258 Ω L = 32 H τ = 126 msec Wire size = 34 AWG
Problem 1-32

22
Part (a) (i):
Bg1 = µ0N1 I1 Bg2 = µ0N1
I1 g1 g2
(ii)
λ1 = N1(A1Bg1 + A2Bg2) = µ0N12A1 + A2I1
g1 g2
A
λ2 = N2A2Bg2 = µ0N1N2 2I1
g2
Part (b) (i):
Bg1 = 0 Bg2 = µ0N2 I2
g2
(ii)
A2
λ1 = N1(A1Bg1 + A2Bg2) = µ0N1N2 I2 g2
2 A2 λ2
= N2A2Bg2 = µ0N2 I2
g2
Part (c) (i):
Bg1 = µ0N1 I1 Bg2 = µ0N1 I1 + µ0N2 I2 g1
g2 g2
(ii)

23
λ1 = N1(A1Bg1 + A2Bg2) = µ0N12 A1 + A2I1 + µ0N1N2 A2 I2
g1 g2 g2
A2 2 A2
λ2 = N2A2Bg2 = µ0N1N2 I1 + µ0N2 I2
g2 g2
Part (d):
L1 = µ0N12 A1 + A2 L2µ0N22 A2 g1 g2
g2
A2
L12 = µ0N1N2
g2
Problem 1-33
g l1
Rg = R1 =
µ0Ac µAc
l2 lA R2 = RA =
µAc µAc
Part (a):
N2
L1 = 1
Rg + R1 + R2 + RA/2

24
N2 LA
= LB = R
where
R = RA + RRA(RRg + RR1 + RR2)
A + g + 1 + 2
Part (b):
L1B = −L1A = R R N1N
RR
2( g + 1 + 2 + A/2)
N2(Rg + R1 + R2)
L12 = 2RA(Rg + R1 + R2) + R2A
Part (c):
v1(t) = L1A diA + L1B diB = L1A d(iA − iB) dt
dt dt
Problem 1-34
Part (a):
L12 = µ0N1N2D(w − x)
2g
Part (b):
v2(t) = −ωIo µ0N1N2Dwcosωt

25
4g
Problem 1-35
Part (a):
H = 2N1i1
(Ro + Ri)
Part (b):
dB(t) v2(t)
= N2w(n∆)
dt
Part (c):
v0(t) = GN2w(n∆)B(t)
Problem 1-36
Must have
µ0 µ0 B lc
lc < 0.05g + lc ⇒µ = > 19µ0 µ µ H g
For g = 0.05 cm and lc = 30 cm, must have µ > 0.014. This is satisfied over the approximate
range 0.65 T ≤ B ≤ 1.65 T.
Problem 1-37
Part (a): See Problem 1-35. For the given dimensions, Vpeak = 20 V, Bpeak = 1 T and ω = 100π
rad/sec

26
Vpeak
N1 = = 79 turns ω(Ro −
Ri)(n∆)
Part (b): (i)
V0,peak
Bpeak = = 0.83 T
GN2(R0 − Ri)(n∆)
(ii)
Vpeak = ωN1(Ro − Ri)(n∆)Bpeak = 9.26 V
Problem 1-38
Part (a): From the M-5 dc-magnetization characteristic, Hc = 19 A-turns/m at Bc =
Bg = 1.3 T. For Hg = 1.3 T/µ0 = 1.03 × 106 A-turns/m
I = Hc(lA + lC − g) +
Hgg = 30.2 A N1
Part(b):
Bg2
Wgap = gAC = 3.77 J
2µ0
For µ = Bc/Hc = 0.0684 H/m
Bc2 −3
Wc = (lAAA + lBAB + (lC − g)AC) = 4.37 × 10 J
2µ

27
L = 2(WgapI2+ Wc) = 8.26 mH
Part (c):
2Wgap
L = I2 = 8.25 mH
Problem 1-39
Part (a):
Part (b): Loop area = 191 J/m3
Part (c):
f × Loop area Core
loss =
ρ
For f = 60 Hz, ρ = 7.65 × 103 kg/m3, Core loss = 1.50 W/kg
Problem 1-40
Brms = 1.1 T and f = 60 Hz,
Vrms = ωNAcBrms = 86.7 V

28
Core volume = Aclc = 1.54×10−3 m3. Mass density = 7.65×103 kg/m3. Thus, the core mass = (1.54
× 10−3)(7.65 × 103) = 11.8 kg.
At B = 1.1 T rms = 1.56 T peak, core loss density = 1.3 W/kg and rms VA density is
2.0 VA/kg. Thus, the core loss = 1.3 ×11.8 = 15.3 W. The total exciting VA for the core is
2.0 × 11.8 = 23.6 VA. Thus, its reactive component is given by The
rms energy storage in the air gap is
gAcBrms2
Wgap = = 5.08 J
µ0
corresponding to an rms reactive power of
VARgap = ωWgap = 1917 VAR
Thus, the total rms exciting VA for the magnetic circuit is
VArms = q15.32 + (1917 + 17.9)2 = 1935 VA
and the rms current is Irms = VArms/Vrms = 22.3 A.
Problem 1-41
Part(a): Area increases by a factor of 4. Thus the voltage increases by a factor of 4 to e = 1096
cos(377t).

29
Part (b): lc doubles therefore so does the current. Thus I = 0.26 A.
Part (c): Volume increases by a factor of 8 and voltage increases by a factor of 4. There Iφ,rms
doubles to 0.20 A.
Part (d): Volume increases by a factor of 8 as does the core loss. Thus Pc = 128 W.
Problem 1-42
From Fig. 1.19, the maximum energy product for samarium-cobalt occurs at (approximately)
B = 0.47 T and H = -360 kA/m. Thus the maximum energy product is 1.69 × 105 J/m3.
Thus,
0.8 2 2
Am = 2 cm = 3.40 cm
0.47
and
0.8
lm = 0.2 cm µ (−3.60 × 105)! = 0.35 cm
0
Thus the volume is 3.40 × 0.35 = 1.20 cm3, which is a reduction by a factor of 5.09/1.21 = 4.9.
Problem 1-43
From Fig. 1.19, the maximum energy product for neodymium-iron-boron occurs at
(approximately) B = 0.63 T and H = -470 kA/m. Thus the maximum energy product is 2.90 × 105
J/m3.
Thus,
0.8 2
2
Am = 2 cm = 2.54 cm

30
0.63
and c 2014 by McGraw-Hill Education. This is proprietary material solely for authorized
0.8
lm = −0.2 cm µ0(−4.70 × 105)! = 0.27 cm
Thus the volume is 2.54×0.25 = 0.688 cm3, which is a reduction by a factor of 5.09/0.688 =
7.4.
Problem 1-44
From Fig. 1.19, the maximum energy product for samarium-cobalt occurs at (approximately)3
B = 0.47 T and H = -360 kA/m. Thus the maximum energy product is 1.69 × 105 J/m . Thus, we
want Bg = 1.3 T, Bm = 0.47 T and Hm = −360 kA/m.
H B
hm = −g Hmg = −g µ0H
gm ! = 2.87 mm
Bg Bg
Am = Ag = 2πRh = 45.1 cm2
Bm Bm
Am
Rm = = 3.66 cm
π
Problem 1-45
s

31
Solutions Manual for Fitzgerald and Kingsleys Electric Machinery
7th Edition by Umans
Full clear download (no error formatting) at :
https://downloadlink.org/p/solutions-manual-for-fitzgerald-and-
kingsleys-electric-machinery-7th-edition-by-umans/
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