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Solutions Manual] Electric Machinery 6Ed Fitzgerald, Exercises of Electric Machines

Bogazici University Electric Machines

Solutions Manual] Electric Machinery 6Ed Fitzgerald

Typology: Exercises

2018/2019

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PROBLEM SOLUTIONS: Chapter 1

Problem 1.1

Part (a):

Rc=lc

µAc

=lc

µrµ0Ac

=0 A/Wb

Rg=g

µ0Ac

=1.017 ×106A/Wb

part (b):

Φ= NI

Rc+Rg

=1.224 ×10−4Wb

part (c):

λ=NΦ=1.016 ×10−2Wb

part (d):

L=λ

I=6.775 mH

Problem 1.2

part (a):

Rc=lc

µAc

=lc

µrµ0Ac

=1.591 ×105A/Wb

Rg=g

µ0Ac

=1.017 ×106A/Wb

part (b):

Φ= NI

Rc+Rg

=1.059 ×10−4Wb

part (c):

λ=NΦ=8.787 ×10−3Wb

part (d):

L=λ

I=5.858 mH

Partial preview of the text

Download Solutions Manual] Electric Machinery 6Ed Fitzgerald and more Exercises Electric Machines in PDF only on Docsity!

PROBLEM SOLUTIONS: Chapter 1

Problem 1. Part (a): Rc =

lc μAc

lc μrμ 0 Ac

= 0 A/Wb

Rg =

g μ 0 Ac = 1. 017 × 106 A/Wb

part (b):

N I

Rc + Rg

= 1. 224 × 10 −^4 Wb

part (c):

λ = N Φ = 1. 016 × 10 −^2 Wb

part (d):

L =

λ I

= 6. 775 mH

Problem 1. part (a): Rc =

lc μAc

lc μrμ 0 Ac

= 1. 591 × 105 A/Wb

Rg =

g μ 0 Ac

= 1. 017 × 106 A/Wb

part (b):

N I

Rc + Rg

= 1. 059 × 10 −^4 Wb

part (c):

λ = N Φ = 8. 787 × 10 −^3 Wb

part (d):

L =

λ I = 5. 858 mH

Problem 1. part (a):

N =

Lg μ 0 Ac

= 110 turns

part (b):

I =

Bcore μ 0 N/g

= 16. 6 A

Problem 1. part (a):

N =

L(g + lcμ 0 /μ) μ 0 Ac

L(g + lcμ 0 /(μrμ 0 )) μ 0 Ac = 121 turns

part (b):

I =

Bcore μ 0 N/(g + lcμ 0 /μ)

= 18. 2 A

Problem 1. part (a):

part (b):

μr = 1 +

1 + 0.047(2.2)^7.^8

I = B

g + μ 0 lc/μ μ 0 N

= 65. 8 A

part (c):

Problem 1.

g =

μ 0 N 2 Ac L

μ 0 μ

lc = 0. 353 mm

Problem 1. part (a):

lc = 2π(Ro − Ri) − g = 3. 57 cm; Ac = (Ro − Ri)h = 1. 2 cm^2

part (b):

Rg = g μ 0 Ac

= 1. 33 × 107 A/Wb; Rc = 0 A/Wb;

part (c):

L =

N 2

Rg + Rg

= 0. 319 mH

part (d):

I =

Bg(Rc + Rg )Ac N

= 3 3. 1 A

part (e):

λ = N BgAc = 10. 5 mWb

Problem 1. part (a): Same as Problem 1. part (b):

Rg =

g μ 0 Ac

= 1. 33 × 107 A/Wb; Rc =

lc μAc

= 3. 16 × 105 A/Wb

part (c):

L =

N 2

Rg + Rg

= 0. 311 mH

part (d):

I =

Bg(Rc + Rg)Ac N

= 3 3. 8 A

part (e): Same as Problem 1.9.

Problem 1.

Minimum μr = 340.

Problem 1.

L =

μ 0 N 2 Ac g + lc/μr

Problem 1.

L =

μ 0 N 2 Ac g + lc/μr

= 3 0. 5 mH

Problem 1. part (a): Vrms = ωN AcBpeak √ 2

= 19. 2 V rms

part (b):

Irms =

Vrms ωL

= 1. 67 A rms; Wpeak = 0. 5 L(

2 Irms)^2 = 8. 50 mJ

part (b):

Emax = 4f N AcBpeak = 3 45 V

Problem 1. part (a):

N =

LI

AcBsat

= 99 turns; g =

μ 0 N I Bsat

μ 0 lc μ

= 0. 3 6 mm

part (b): From Eq.3.

Wgap =

AcgB^2 sat 2 μ 0

= 0. 207 J; Wcore =

AclcB^2 sat 2 μ

= 0. 045 J

Thus Wtot = Wgap + Wcore = 0.252 J. From Eq. 1.47, (1/2)LI^2 = 0.252 J. Q.E.D.

Problem 1. part (a): Minimum inductance = 4 mH, for which g = 0.0627 mm, N = 20 turns and Vrms = 6.78 V part (b): Maximum inductance = 144 mH, for which g = 4.99 mm, N = 1078 turns and Vrms = 224 V

Problem 1. part (a):

L =

μ 0 πa^2 N 2 2 πr

= 56. 0 mH

part (b): Core volume Vcore ≈ (2πr)πa^2 = 40.0 m^3. Thus

W = Vcore

B^2

2 μ 0

= 4. 87 J

part (c): For T = 30 sec,

di dt

(2πrB)/(μ 0 N ) T

= 2. 92 × 103 A/sec

v = L di dt

= 163V

Problem 1. part (a):

Acu = fwab; Volcu = 2ab(w + h + 2a)

part (b):

B = μ 0

JcuAcu g

part (c):

Jcu =

N I

Acu

part (d):

Pdiss = Volcu

ρJ^2 cu

part (e):

Wmag = Volgap

B^2

2 μ 0

= gwh

B^2

2 μ 0

part (f):

L R

LI^2

RI^2

Wmag ( (^1) 2

Pdiss

2 Wmag Pdiss

μ 0 whA^2 cu ρgVolcu

Problem 1. Using the equations of Problem 1.

Pdiss = 115 W I = 3 .24 A N = 687 turns R = 10.8 Ω τ = 6.18 msec Wire size = 23AWG

Problem 1. part (a):

(i) B 1 =^

μ 0 N 1 I 1 g 1

; B 2 =

μ 0 N 1 I 1 g 2

(ii) λ 1 = N 1 (A 1 B 1 + A 2 B 2 ) = μ 0 N 12

A 1

g 1

A 2

g 2

I 1

(iii) λ 2 = N 2 A 2 B 2 = μ 0 N 1 N 2

A 2

g 2

I 1

LAA = LBB =

N 2

RA + RA||(R 1 + R 2 + Rg )

N 2 μAc lA

[

lA + l 1 + l 2 + g (μ/μ 0 ) lA + 2(l 1 + l 2 + g (μ/μ 0 ))

]

part (b):

LAB = LBA =

N 2 (R 1 + R 2 + Rg ) RA(RA + 2(R 1 + R 2 + Rg ))

N 2 μAc lA

[

l 1 + l 2 + g (μ/μ 0 ) lA + 2(l 1 + l 2 + g (μ/μ 0 ))

]

LA1 = L1A = −LB1 = −L1B =

−N N 1

RA + 2(R 1 + R 2 + Rg )

−N N 1 μAc lA + 2(l 1 + l 2 + g (μ/μ 0 ))

part (c):

v 1 = d dt

[LA1iA + LB1iB] = LA d dt

[iA − iB]

Q.E.D.

Problem 1. part (a):

L 12 =

μ 0 N 1 N 2 2 g

[D(w − x)]

part (b):

v 2 = dλ 2 dt

= I 0

dL 12 dt

N 1 N 2 μ 0 D 2 g

dx dt

= −

N 1 N 2 μ 0 D 2 g

% ωw 2

cos ωt

Problem 1. part (a):

H =

N 1 i 1 2 π(Ro + Ri)/ 2

N 1 i 1 π(Ro + Ri)

part (b):

v 2 =

d dt [N 2 (tn∆)B] = N 2 tn∆

dB dt

part (c):

vo = G

v 2 dt = GN 2 tn∆B

Problem 1.

Rg = g μ 0 Ag

= 4. 42 × 105 A/Wb; Rc = lc μAg

A/Wb

Want Rg ≤ 0. 05 Rc ⇒ μ ≥ 1. 2 × 104 μ 0. By inspection of Fig. 1.10, this will be true for B ≤ 1 .66 T (approximate since the curve isn’t that detailed). Problem 1. part (a):

N 1 = Vpeak ωt(Ro − Ri)Bpeak

= 57 turns

part (b):

(i) Bpeak = Vo,peak GN 2 t(Ro − Ri)

= 0. 833 T

(ii) V 1 = N 1 t(Ro − Ri)ωBpeak = 6. 25 V, peak

Problem 1. part (a): From the M-5 magnetization curve, for B = 1.2 T, Hm = 14 A/m. Similarly, Hg = B/μ 0 = 9. 54 × 105 A/m. Thus, with I 1 = I 2 = I

I = Hm(lA + lC − g) + Hgg N 1

= 3 8. 2 A

part (b):

Wgap =

gAgapB^2 2 μ 0

= 3. 21 Joules

part (c):

λ = 2N 1 AAB = 0. 168 Wb; L =

λ I = 4. 3 9 mH

Problem 1. part (a):

lm = − 0 .2 cm

μ 0 (− 3. 60 × 105 )

= 0.3 5 cm

Thus the volume is 3. 40 × 0 .3 5 = 1.20 cm^3 , which is a reduction by a factor of 5.09/1.21 = 4.9.

Problem 1. 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/m^3. Thus,

Am =

2 cm^2 = 2.54 cm^2

and

lm =^ −^0 .2 cm

μ 0 (− 4. 70 × 105 )

= 0.27 cm

Thus the volume is 2. 54 × 0 .25 = 0.688 cm^3 , which is a reduction by a factor of 5.09/0.688 = 7.4.

Problem 1. 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/m^3. Thus, we want Bg = 1.2 T, Bm = 0.47 T and Hm = −360 kA/m.

hm = −g

Hg Hm

= −g

Bg μ 0 Hm

= 2. 65 mm

Am = Ag

Bg Bm

= 2πRh

Bg Bm

= 26. 0 cm^2

Rm =

Am π

= 2. 87 cm

Problem 1. From Fig. 1.19, the maximum energy product for neodymium-iron-boron oc- curs at (approximately) Bm = 0.63 T and Hm = -470 kA/m. The magnetization curve for neodymium-iron-boron can be represented as

Bm = μRHm + Br

where Br = 1.26 T and μR = 1. 067 μ 0. The magnetic circuit must satisfy

Hmd + Hg g = N i; BmAm = Bg Ag

part (a): For i = 0 and Bg = 0.5 T, the minimum magnet volume will occur when the magnet is operating at the maximum energy point.

Am =

Bg Bm

Ag = 4. 76 cm^2

d = −

Hg Hm

g = 1. 69 mm

part (b):

i =

[

dAg μR Am +^

g μ 0

− B μrR^ d

]

N

For Bg = 0.75, i = 17.9 A. For Bg = 0.25, i = 6.0 A.

Because the neodymium-iron-boron magnet is essentially linear over the op- erating range of this problem, the system is linear and hence a sinusoidal flux variation will correspond to a sinusoidal current variation.

Here is the desired MATLAB plot:

Problem 2. The maximum power will be supplied to the load resistor when its im- pedance, as reflected to the primary of the idealtransformer, equals that of the source (2 kΩ). Thus the transformer turns ratio N to give maximum power must be

N =

Rs Rload

Under these conditions, the source voltage will see a total impedance of Ztot = 2 + j2 kΩ whose magnitude is 2

2 kΩ. The current will thus equal I = Vs/|Ztot| = 2

2 mA. Thus, the power delivered to the load will equal

Pload = I^2 (N 2 Rload) = 16 mW

Here is the desired MATLAB plot:

Problem 2.

V 2 = V 1

Xm Xl 1 + Xm

= 266 V

Problem 2. part (a): Referred to the secondary

Lm, 2 =

Lm, 1 N 2 = 150 mH

part(b): Referred to the secondary, Xm = ωLm, 2 = 56.7 Ω, Xl 2 = 84.8 mΩ and Xl 1 = 69.3 mΩ. Thus,

(i) V 1 = N

Xm Xm + Xl 2

V 2 = 7960 V

and

(ii) Isc =

V 2

Xsc

V 2

Xl 2 + Xm||Xl 1

= 1730 A

Problem 2. part (a):

I 1 =

V 1

Xl 1 + Xm

= 3. 47 A; V 2 = N V 1

Xm Xl 1 + Xm

= 2398 V

part (b): Let X l′ 2 = Xl 2 /N 2 and Xsc = Xl 1 + Xm||(Xm + X′ l 2 ). For Irated = 50 kVA/120 V = 417 A

V 1 = IratedXsc = 23. 1 V

I 2 =

N

Xm Xm + Xl 2

Irated = 15. 7 A

Problem 2.

IL = Pload VL

= 55. 5 A

and thus

IH =

IL

N

= 10. 6 A; VH = N VL + jXHIH = 2381 9. 6 ◦^ V

The power factor is cos (9. 6 ◦) = 0.986 lagging.

Problem 2. part (a):

part (b): Following methodology of Problem 2.11, (i) for a power factor of 0.85 lagging, VH = 4956 V and (ii) for a power factor of 0.85 leading, VH = 4000 V. part (c):

Problem 2. part (a): Iload = 160 kW/2340 V = 68.4 A at = cos−^1 (0.89) = 27. 1 ◦

V^ ˆt,H = N ( VˆL + ZtIL)

which gives VH = 33.7 kV. part (b):

Vˆsend = N ( VˆL + (Zt + Zf )IL)

which gives Vsend = 33.4 kV. part (c):

Ssend = Psend + jQsend = Vˆsend Iˆ send∗ = 164 kW − j 64 .5 kVAR

Thus Psend = 164 kW and Qsend = − 64 .5 kVAR. Problem 2. Following the methodology of Example 2.6, efficiency = 98.4 percent and regulation = 1.25 percent.

Problem 2. part (a):

|Zeq,L| =

Vsc,L Isc,L = 107.8 mΩ

Req,L = Psc,L I^2 sc,L

= 4.78 mΩ

Xeq,L =

|Zeq,L|^2 − R^2 eq,L = 107.7 mΩ

and thus

Zeq,L = 4.8 + j108 mΩ

part (b):

Req,H = N 2 Req,L = 0.455 Ω

Xeq,H = N 2 Xeq,L = 10.24 Ω

Zeq,H = 10.3 + j 0 .46 mΩ

part (c): From the open-circuit test, the core-loss resistance and the magne- tizing reactance as referred to the low-voltage side can be found:

Rc,L =

V (^) oc^2 ,L Poc,L

Soc,L = Voc,LIoc,L = 497 kVA; Qoc,L =

Soc^2 ,L − P (^) oc^2 ,L = 45.2 kVAR

and thus

Solutions Manual] Electric Machinery 6Ed Fitzgerald, Exercises of Electric Machines

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Partial preview of the text

Download Solutions Manual] Electric Machinery 6Ed Fitzgerald and more Exercises Electric Machines in PDF only on Docsity!

PROBLEM SOLUTIONS: Chapter 1

N I

L =

N I

L =

N =

I =

= 16. 6 A

N =

I =

= 18. 2 A

1 + 0.047(2.2)^7.^8

I = B

= 65. 8 A

L =

N 2

I =

= 3 3. 1 A

L =

N 2

I =

= 3 3. 8 A

L =

L =

N =

LI

= 0. 045 J

L =

B^2

= 4. 87 J

= 163V

N I

B^2

B^2

LI^2

RI^2

; B 2 =

A 1

A 2

I 1

A 2

I 1

LAA = LBB =

N 2

[

]

LAB = LBA =

[

]

LA1 = L1A = −LB1 = −L1B =

−N N 1

Q.E.D.

L 12 =

= I 0

H =

= 0. 833 T

= 3 8. 2 A

[

]

N

N =

V 2 = V 1

= 266 V

V 2 = 7960 V

V 2

V 2

= 1730 A

I 1 =

V 1

= 3. 47 A; V 2 = N V 1

= 2398 V

I 2 =

N

= 55. 5 A

IH =

IL