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

Part b: lc doubles therefore so does the current.

PROBLEM SOLUTIONS: Chapter 1

Part (b):

Φ = NI

Rc+ Rg = 2.334 × 10−5 WbPart (c):

λ = NΦ = 2.217 × 10−3 WbPart (d):

L = λ

I = 1.584 mHProblem 1-3

Problem 1-5Part (a):

Part (b):

Bg = Bm= 2.1 TFor Bm = 2.1 T, µr = 37.88 and thus

Bg = µ0NI

2g

µr = µ

µ0

=L



lp+ (1 − x/X0) lc



µ0N2Ac− 2gL = 88.5Problem 1-8

Part (a):

L = µ0(2N)

2Ac

2gand thus

I = Bcg

µ0N = 37.1 AProblem 1-9

Part (a):

L = µ0N

2Ac

2g

I = 2Bcg

µ0N = 37.1 AProblem 1-10

Bc = Bg = 2µ0NI

2(g + µ0

µlc0)and thus

I = Bc(g +

µ 0

µlc)

µ0N = 40.9 AProblem 1-11

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

Vrms = ωNA√cBpeak

2 = 20.8 VPart (b): Using L from the solution to Problem 1-17

Ipeak=

√2Vrms

ωL = 1.66 A

Wpeak = LI

2 peak

2 = 9.13 × 10−2 JProblem 1-19

√2Irms)2 = 61.0 mJ

Part (a);

Part (b): Ipeak = 0.6 A

Part (c): Ipeak = 4.0 A

Wg =

B2 g

2µ0



Vg = 1.08 J

Wcore =

B2 c

= 0.52 mm

Part (b):

N = LI2BAc

= 84

g = µ0NI

BAc

= 0.52 mm

N = LI2BAc

= 84

g = µ0NI

BAc − (µ0/µ)lc = 0.39 mmProblem 1-28

Part (a): N = 450 and g = 2.2 mm

A = πa2 l = 2πr

Part (b):

W = B

2

2µ0 × Volume = 6.93 × 107 Jwhere

Volume = (πa2)(2πr)Part (c): For a flux density of 1.80 T,

I = lB

µ0N =

2πrB

µ0N = 6.75 kAand

Copper cross − sectional area ≡ Acu= fwab

Copper volume = Volcu = fwb

Jcu = NI

Acu

Rg + R1 + R2+ RA/2

LA = LB = N

2

Rwhere

A

Part (a):

L12 = µ0N1N2D(w − x)

2gPart (b):

Part (b):

v2(t) = N2w(n∆)dB(t)

dtPart (c):

v0(t) = GN2w(n∆)B(t)

Wgap = gAC

B2 g

L = 2Wgap

I2 = 8.25 mHProblem 1-39

For f = 60 Hz, ρ = 7.65 × 103 kg/m3, Core loss = 1.50 W/kg

Wgap = gAcB

2 rms

µ0

= 5.08 J

corresponding to an rms reactive power of

VARgap= ωWgap = 1917 VAR

Thus, the total rms exciting VA for the magnetic circuit is

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

Am =

 0.80.47

(ap-Thus,

Am =

 0.80.63



2 cm2 = 2.54 cm2

and