PROBLEM SOLUTIONS: Chapter 1 Solutions Manual for Fitzgerald and Kingsleys Electric Machinery 7th Edition by Umans

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Problem 1-1

Part (a):

Rc = lc = lc = 0 A/Wb

µAc

µ r µ 0A c

Rg = g

= 5.457 × 106

A/Wb

µ A

0 c

Part (b):

Φ = N I

= 2.437 × 10−5

Wb

Rc + Rg

Part (c):

λ = N Φ = 2.315 × 10−3

Wb

Part (d):

L = λ

= 1.654 mH I

Problem 1-2

Part (a): lc lc 5

Rc =

=

= 2.419 × 10 A/Wb µAc µ r µ 0A c

Rg = g

= 5.457 × 106

A/Wb

µ A

0 c

2

Part (b):

Φ =

N I

= 2.334 × 1 0−5

Wb Rc + Rg

Part (c):

λ = N Φ = 2.217 × 10−3

Wb

Part (d):

L = λ

= 1.584 mH I

Problem 1-3

Part (a):

s

= 287 turns N = Lg

µ0 Ac

Part (b):

I = Bcore

= 7.68 A

µ0 N/g

Problem 1-4

Part (a):

N = s L(g + lcµ0 /µ) =

s L(g + lcµ0 /(µr µ0 )) = 129 turns

µ0 Acµ0 Ac

Part (b):

I =

Bcore

= 20.78 A µ0 N/(g + lc µ0 /µ)

3

Problem 1-5

Part (a):

Part (b):

Bg = Bm = 2.1 T

For Bm = 2.1 T, µr = 37.88 and thus

I = Bm g +

lc = 158 A µ N µ

0 r

Part (c):

Problem 1-6

Part (a):

B = µ0 N I

g 2g

4

Bc = Bg A g = µ0 N I 1 x

2g X0 Ac

Part (b): Will assume lc is “large” and lp is relatively “small”. Thus, Bg

Ag = BpAg = Bc Ac

We can also write

2gHg + Hp lp + Hc lc = N I ;

and

Bg = µ0 Hg ; Bp = µHp Bc = µHc

These equations can be combined to give

Bg = µ0 N I

= µ0 N I

µ0

µ0

Ag

µ0

µ0 1 −

x

2g + lp +

l c 2g + lp + l c

µ

µ

Ac

µ

X0 µ

and

Bc = 1 x Bg

X0

Problem 1-7

From Problem 1-6, the inductance can be found as

L =

N AcBc =

µ0 N 2 Ac

I

2g + µ0

(lp + (1 − x/X0 ) lc)

µ

from which we can solve for µr

5

µ = µ = L lp + (1 − x/X0 ) lc

= 88.5

µ

µ N 2 A

− 2gL r 0

c

0

Problem 1-8

Part (a):

L = µ0 (2N )2

Ac 2g

and thus

s

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):

B = B = 2µ0 N I

c g 2g

and thus

I = Bc

g

= 37.1 A

µ0 N

Problem 1-9

Part (a):

L = µ0 N 2 Ac

2g

6

and thus

s

2gL

N =

= 77.6 Ac

which rounds to N = 78 turns for which L = 12.33 mH.

Part (b): g = 0.121 cm

Part(c):

B = B = µ0 (2N )(I /2) c g

and thus

I = 2Bc

g

= 37.1 A

µ0N

Problem 1-10

Part (a):

L = µ0 (2N )

2 Ac

2(g + ( µ0 )lc ) µ

and thus

u

2(g + ( µ0

)lc)L

u

N = 0.5 µ = 38.8

Ac

which rounds to N = 39 turns for which L = 12.33 mH.

Part (b): g = 0.121 cm

7

Part(c):

Bc = Bg = 2µ0 N I

2(g + µ

l c0)

0

µ

and thus

Bc(g + µ

0 lc)

I = µ

= 40.9 A µ0 N

Problem 1-11

Part (a): From the solution to Problem 1-6 with x = 0

Bg 2g + 2

µ0 (lp + lc)

I = µ

= 1.44 A µ0 N

Part (b): For Bm = 1.25 T, µr = 941 and thus I = 2.43 A Part

(c):

Problem 1-12

µ0 N 2 Ac µ0

! −4

g =

−

lc = 7.8 × 1 0 m L µ

8

Problem 1-13

Part (a):

lc = 2π Ri + Ro − g = 22.8 cm

2

Ac

Part (b):

Rg

Part (c):

L =

Part (d):

I =

Part (e):

= h(Ro − Ri ) = 1.62 cm2

Rc = lc = 0

µAc

= g

= 7.37 × 106 H−1

µ A

0 c

N 2

= 7.04 × 1 0−4

H Rc + Rg

Bg Ac (Rc + Rg ) = 20.7 A N

λ = LI = 1.46 × 1 0−2

Wb

9

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):

L = 5.94 × 10−4

H

Part (d):

I = 24.6 A

Part (e):

λ = 1.46 × 10−2

Wb

10

Problem 1-15

µr must be greater 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 = 0.77 T

g + lc/µr

λ = LI = 4.40 × 1 0−2

Wb

Problem 1-18

Part (a): With ω = 120π

11

ωN Ac Bpeak

Vrms = √2 = 20.8 V

Part (b): Using L from the solution to Problem 1-17

Ipeak = √2Vrms

= 1.66 A

ωL

LI 2

W = peak

= 9.13 10−2 J

peak 2

Problem 1-19

B = 0.81 T and λ = 46.5 mWb

Problem 1-20

Part (a):

q

R3 = (R2

1 + R2

2 ) = 4.49 cm

Part (b): For

lc = 4l + R2 + R3 − 2h;

and

Ag = πR21

L =

µ0 AgN 2

= 61.8 mH g + (µ0 /µ)lc

Part (c): For Bpeak = 0.6 T and ω = 2π60

12

λpeak = Ag N Bpeak

ωλpeak

Vrms = √2 = 23.2 V

Irms =

Vrms

= 0.99 A

ωL

Wpeak = 1

LIpeak2

=

1

L( √

2Irms ) 2 = 61.0 mJ 2 2

Part (d): For ω = 2π50

Vrms = 19.3 V

Irms = 0.99 A

Wpeak = 61.0 mJ

Problem 1-21

Part (a);

13

Part (b):

Emax = 4f N Ac Bpeak = 118 V

part (c): For µ = 1000µ0

I = lc

Bpeak

= 0.46 A

peak µN

Problem 1-22

Part (a);

Part (b): Ipeak = 0.6 A

Part (c): Ipeak = 4.0 A

14

Problem 1-23

For part (b), Ipeak = 11.9 A. For part (c), Ipeak = 27.2 A.

Problem 1-24

L =

µ0 AcN 2

g + (µ0 /µ)lc

Bc =

µ0 N I

g + (µ 0/µ)l c

Part (a): For I = 10 A, L = 23 mH and Bc = 1.7 T

N = LI

= 225 turns A B

c c

g = µ0 N I

− µ0 lc

= 1.56 mm

Bc µ

Part (b): For I = 10 A and Bc = Bg = 1.7 T, from Eq. 3.21

B2

Wg = g

Vg = 1.08 J 2µ0

15

B2

Wcore = c

Vg = 0.072 J 2µ

based upon

Vcore = Aclc Vg = Acg

Part (c):

Wtot = 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):

LI

N = BA c = 167

g = µ0

N I

= 0.52 mm

2BAc

Part (b):

N = 2BALI

c = 84

g = µ0

N I

= 0.52 mm

BAc

16

Problem 1-27

Part (a):

LI

N = BA c = 167

g = µ0

N I

− (µ /µ)l = 0.39 mm

2BAc 0 c

Part (b):

N = 2BALI

c = 84

g = µ0

N I

− (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):

L = µ0 N 2A

= 11.3

H l

where

A = πa2 l = 2πr

17

Part (b):

W =

B2

× Volume = 6.93 × 107

J 2µ0

where

Volume = (πa2)(2πr)

Part (c): For a flux density of 1.80 T,

I = lB = 2πrB = 6.75 kA

µ0 N

µ0 N and

∆I −3 6.75 × 103

V = L

= 113 × 10

= 1.90 kV ∆t 40

Problem 1-30

Part (a):

Copper cross − sectional area ≡ Acu = fw ab

Copper volume = Volcu = fw b (a + w )(h + w ) wh

2 2

Part (b):

B = µ0 JcuAcu

g

Part (c):

N I

Jcu = A cu

18

Part (d):

P = ρJ 2 Vol

diss cu cu

Part (e):

B2 µ J

2 A

2 wh

Wstored =

× gap volume = 0 cu cu

2µ0

2g

Part (f ): 1 I 2L

Wstore

= 2

Pdiss

I 2R

and thus

L

= 2

W stored

=

µ 0Acu2wh

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

Part (a) (i):

Bg1 = µ0 N1 I1 Bg2 = µ0 N1 I 1

(ii)

g1 g2

λ = N (A B + A B ) = µ N 2 A1 + A2 I

1 1 1 g1 2 g2 0

1 g1 g2 1

19

λ2 = N2 A2Bg2 = µ0 N1 N2 A2

I1 g2

Part (b) (i):

B = 0 B = µ0

N2

I

g1 g2 g2 2

(ii)

λ 1 = N 1(A 1Bg1 + A 2Bg2 ) = µ0N 1 N2 A2

I 2

g2

λ2 = N2 A2Bg2 = µ0 N 2 A2

I2

2 g2

Part (c) (i):

Bg1 = µ0 N1 I1 Bg2 = µ0 N1 I1 + µ0 N2 I 2

g1 g2 g2

(ii)

λ1 = N1 (A1Bg1 + A2 Bg2) = µ0 N 2 A1 + A2 I1 + µ0 N1 N2

A2 I2

1 g1 g2

g2

λ 2 = N 2A 2Bg2 = µ0 N 1 N2 A2

I 1 + µ 0N 22 A2

I 2

g2g2

Part (d):

L1 = µ0 N 2 A1 +

A2 L2 µ0 N 2 A2

1 g g 2

g

1 2 2

20

L = µ N

N A2

12 0 1 2 g2

Problem 1-33

Rg = g R1 = l1

µ A c

µA

0 c

R2 = l2 RA = lA

µAc µAc

Part (a):

N2

L1 =

1

Rg + R1 + R2 + RA/2

N 2

LA = LB = R

where

R = RA + RA(Rg + R1 + R2 )

RA + Rg + R1 + R2

Part (b):

L1B = −L1A =

N1 N

2( Rg + R1 + R2 + RA/2)

L12 = N

2 (Rg + R1 + R2 )

2RA (Rg + R1 + R2) + 2

A