Solutions Manual for Fitzgerald and Kingsleys Electric Machinery 7th ...

May 13, 2018 | Author: Anonymous | Category: Documents
Share Embed


Short Description

Solutions Manual for Fitzgerald and Kingsleys Electric Machinery 7th Edition Umans Download - Free download as PDF File ...

Description

PROBLEM SOLUTIONS: Chapter 1

Solutions Manual for Fitzgerald and Kingsleys Electric Machinery 7th Edition by Umans Link download full:

https://testbankservice.com/download/solutions-manual-for-fitzge https://testbankservice.com /download/solutions-manual-for-fitzgerald-and-kingsleys-electric-ma rald-and-kingsleys-electric-machinery-7th-edition-by-umans/  chinery-7th-edition-by-umans/ 

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

=



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

View more...

Comments

Copyright © 2017 DOCIT Inc.