Energy Conversion - Electric Machinery book Fitzgerald solution PDF

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1 PROBLEM SOLUTIONS: Chapter 1 Problem 1.1 Part (a): lc lc Rc = = =0 A/Wb µAc µr µ0 Ac g Rg = = 1.017 × 106 A/Wb µ0 Ac part (b): NI Φ= = 1.224 × 10−4 Wb Rc + Rg part (c): λ = N Φ = 1.016 × 10−2 Wb part (d): λ L= = 6.775 mH I Problem 1.2 part (a): lc lc Rc = = = 1.591 × 105 A/Wb µAc µr µ0 Ac g Rg = =...

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1

PROBLEM SOLUTIONS: Chapter 1

Problem 1.1 Part (a): Rc =

lc lc = =0 µAc µr µ0 Ac

A/Wb

Rg =

g = 1.017 × 106 µ0 Ac

A/Wb

part (b): Φ=

NI = 1.224 × 10−4 Rc + Rg

Wb

part (c): λ = N Φ = 1.016 × 10−2

Wb

part (d): L=

λ = 6.775 mH I

Problem 1.2 part (a): Rc =

lc lc = = 1.591 × 105 µAc µr µ0 Ac

Rg =

g = 1.017 × 106 µ0 Ac

A/Wb

A/Wb

part (b): Φ=

NI = 1.059 × 10−4 Rc + Rg

Wb

part (c): λ = N Φ = 8.787 × 10−3 part (d): L=

λ = 5.858 mH I

Wb

2 Problem 1.3 part (a): N=




Lg = 110 turns µ0 Ac

part (b): I=

Bcore = 16.6 A µ0 N/g

Problem 1.4 part (a): N=




L(g + lc µ0 /µ) = µ0 Ac




L(g + lc µ0 /(µr µ0 )) = 121 turns µ0 Ac

part (b): I=

Bcore = 18.2 A µ0 N/(g + lc µ0 /µ)

Problem 1.5 part (a):

part (b): 3499 µr = 1 +  = 730 1 + 0.047(2.2)7.8 I=B



g + µ0 lc /µ µ0 N



= 65.8 A

3 part (c):

Problem 1.6 part (a): Hg =

NI ; 2g

Bc =



Ag Ac



  x Bg = Bg 1 − X0

part (b): Equations 2gHg + Hc lc = N I;

Bg Ag = Bc Ac

and Bg = µ0 Hg ;

Bc = µHc

can be combined to give     N I N I =        Bg =  Ag µ0 x (l (l 1 − + l ) + l ) 2g + µµ0 2g + c p c p Ac µ X0 Problem 1.7 part (a):



part (b):

g+

I = B



µ0 µ



(lc + lp )

µ0 N



 = 2.15 A

  1199 µ = µ0 1 + √ = 1012 µ0 1 + 0.05B 8 

I = B

g+



µ0 µ



(lc + lp )

µ0 N



 = 3.02 A

4 part (c):

Problem 1.8 g=



µ0 N 2 Ac L



−



µ0 µ



lc = 0.353 mm

Problem 1.9 part (a): lc = 2π(Ro − Ri ) − g = 3.57 cm;

Ac = (Ro − Ri )h = 1.2 cm2

part (b): Rg =

g = 1.33 × 107 µ0 Ac

A/Wb;

Rc = 0

A/Wb;

part (c): L=

N2 = 0.319 mH Rg + Rg

part (d): I=

Bg (Rc + Rg )Ac = 33.1 A N

part (e): λ = N Bg Ac = 10.5 mWb Problem 1.10 part (a): Same as Problem 1.9 part (b): Rg =

g = 1.33 × 107 µ0 Ac

A/Wb;

Rc =

lc = 3.16 × 105 µAc

A/Wb

5 part (c): N2 = 0.311 mH Rg + Rg

L= part (d): I=

Bg (Rc + Rg )Ac = 33.8 A N

part (e): Same as Problem 1.9. Problem 1.11

Minimum µr = 340. Problem 1.12 L=

µ0 N 2 Ac g + lc /µr

Problem 1.13 L=

µ0 N 2 Ac = 30.5 mH g + lc /µr

Problem 1.14 part (a): Vrms =

ωN Ac Bpeak √ = 19.2 V rms 2

part (b): Irms =

Vrms = 1.67 A rms; ωL

√ Wpeak = 0.5L( 2 Irms )2 = 8.50 mJ

6 Problem 1.15 part (a): R3 = part (b): L=

 R12 + R22 = 4.27 cm

µ0 Ag N 2   = 251 mH g + µµ0 lc

part (c): For ω = 2π60 rad/sec and λpeak = N Ag Bpeak = 0.452 Wb: (i) (ii) (iii)

Vrms = ωλpeak = 171 V rms Vrms = 1.81 A rms Irms = ωL √ Wpeak = 0.5L( 2Irms )2 = 0.817 J

part (d): For ω = 2π50 rad/sec and λpeak = N Ag Bpeak = 0.452 Wb: (i) (ii) (iii) Problem 1.16 part (a):

Vrms = ωλpeak = 142 V rms Vrms Irms = = 1.81 A rms ωL √ Wpeak = 0.5L( 2Irms )2 = 0.817 J

7 part (b): Emax = 4f N Ac Bpeak = 345 V Problem 1.17 part (a): N=

LI = 99 turns; Ac Bsat

g=

µ0 N I µ0 lc = 0.36 mm − Bsat µ

part (b): From Eq.3.21 Wgap =

2 Ac gBsat = 0.207 J; 2µ0

Wcore =

2 Ac lc Bsat = 0.045 J 2µ

Thus Wtot = Wgap + Wcore = 0.252 J. From Eq. 1.47, (1/2)LI 2 = 0.252 J. Q.E.D. Problem 1.18 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 1.19 part (a): L=

µ0 πa2 N 2 = 56.0 mH 2πr

part (b): Core volume Vcore ≈ (2πr)πa2 = 40.0 m3 . Thus  2 B W = Vcore = 4.87 J 2µ0 part (c): For T = 30 sec, (2πrB)/(µ0 N ) di = = 2.92 × 103 dt T v=L

A/sec

di = 163 V dt

Problem 1.20 part (a): Acu = fw ab; part (b):

Volcu = 2ab(w + h + 2a)

8

B = µ0



Jcu Acu g



part (c): Jcu =

NI Acu

part (d): 2 Pdiss = Volcu ρJcu

part (e):

Wmag = Volgap



B2 2µ0



= gwh



B2 2µ0



part (f): 1 2 LI 2Wmag Wmag L µ0 whA2cu = 12 2 = 1 = = R Pdiss ρgVolcu 2 RI 2 Pdiss

Problem 1.21 Using the equations of Problem 1.20 Pdiss = 115 W I = 3.24 A N = 687 turns R = 10.8 Ω τ = 6.18 msec Wire size = 23 AWG Problem 1.22 part (a):

(i) (ii) (iii)

B1 =

µ0 N1 I1 ; g1

B2 =

µ0 N1 I1 g2 µ0 N12



λ1 = N1 (A1 B1 + A2 B2 ) =   A2 λ2 = N2 A2 B2 = µ0 N1 N2 I1 g2

A2 A1 + g1 g2



I1

9 part (b):

(i) (ii) (iii)

B1 = 0;

B2 =

µ0 N2 I2 g2

 A2 λ1 = N1 A2 B2 = µ0 N1 N2 I2 g2   A2 λ2 = N2 A2 B2 = µ0 N22 I2 g2 

part (c):

(i) (ii) (iii)

µ0 N1 I1 µ0 N2 I2 + g2 g  2    A A2 A2 1 λ1 = N1 (A1 B1 + A2 B2 ) = µ0 N12 I1 + µ0 N1 N2 I2 + g1 g2 g2     A2 A2 λ2 = N2 A2 B2 = µ0 N1 N2 I1 + µ0 N22 I2 g2 g2 B1 =

µ0 N1 I1 ; g1

B2 =

part (d): L11 = N12



A1 A2 + g1 g2



;

L22 = µ0 N22



A2 g2



;

L12 = µ0 N1 N2



Problem 1.23

RA =

lA ; µAc

R1 =

l1 ; µAc

R2 =

l2 ; µAc

Rg =

g µ0 Ac

part (a): L11 =

N12 µAc N12 = R1 + R2 + Rg + RA /2 l1 + l2 + lA /2 + g (µ/µ0 )

A2 g2



10

LAA = LBB =

  lA + l1 + l2 + g (µ/µ0 ) N 2 µAc N2 = RA + RA ||(R1 + R2 + Rg ) lA lA + 2(l1 + l2 + g (µ/µ0 ))

part (b): LAB = LBA =

  N 2 µAc N 2 (R1 + R2 + Rg ) l1 + l2 + g (µ/µ0 ) = RA (RA + 2(R1 + R2 + Rg )) lA lA + 2(l1 + l2 + g (µ/µ0 ))

LA1 = L1A = −LB1 = −L1B =

−N N1 µAc −N N1 = RA + 2(R1 + R2 + Rg ) lA + 2(l1 + l2 + g (µ/µ0 ))

part (c): v1 =

d d [LA1 iA + LB1 iB ] = LA1 [iA − iB ] dt dt

Q.E.D. Problem 1.24 part (a): L12 =

µ0 N1 N2 [D(w − x)] 2g

part (b):

v2

  N1 N2 µ0 D dx dλ2 dL12 = I0 =− dt dt 2g dt    N1 N2 µ0 D ǫ ωw = − cos ωt 2g 2

=

Problem 1.25 part (a): H=

N1 i1 N1 i1 = 2π(Ro + Ri )/2 π(Ro + Ri )

v2 =

dB d [N2 (tn∆)B] = N2 tn∆ dt dt

part (b):

part (c): vo = G



v2 dt = GN2 tn∆B

11 Problem 1.26 Rg =

g = 4.42 × 105 µ0 Ag

A/Wb;

Rc =

lc 333 A/Wb = µAg µ

Want Rg ≤ 0.05Rc ⇒ µ ≥ 1.2 × 104 µ0 . By inspection of Fig. 1.10, this will be true for B ≤ 1.66 T (approximate since the curve isn’t that detailed). Problem 1.27 part (a): N1 =

Vpeak = 57 turns ωt(Ro − Ri )Bpeak

part (b): (i) (ii)

Vo,peak = 0.833 T GN2 t(Ro − Ri ) V1 = N1 t(Ro − Ri )ωBpeak = 6.25 V, peak

Bpeak =

Problem 1.28 part (a): From the M-5 magnetization curve, for B = 1.2 T, Hm = 14 A/m. Similarly, Hg = B/µ0 = 9.54 × 105 A/m. Thus, with I1 = I2 = I I=

Hm (lA + lC − g) + Hg g = 38.2 A N1

part (b): Wgap =

gAgap B 2 = 3.21 Joules 2µ0

part (c): λ = 2N1 AA B = 0.168 Wb; Problem 1.29 part (a):

L=

λ = 4.39 mH I

12 part (b): Area = 191 Joules part (c): Core loss = 1.50 W/kg. Problem 1.30 Brms = 1.1 T and f = 60 Hz, Vrms = ωN Ac Brms = 46.7 V Core volume = Ac lc = 1.05 × 10−3 m3 . Mass density = 7.65 × 103 kg/m3 . Thus, the core mass = (1.05 × 10−3 )(7.65 × 103 ) = 8.03 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 × 8.03 = 10.4 W. The total exciting VA√for the core is 2.0 × 8.03 = 16.0 VA. Thus, its reactive component is given by 16.02 − 10.42 = 12.2 VAR. The rms energy storage in the air gap is Wgap =

2 gAc Brms = 3.61 Joules µ0

corresponding to an rms reactive power of VARgap = ωWgap = 1361 Joules Thus, the total rms exciting VA for the magnetic circuit is VArms = sqrt10.42 + (1361 + 12.2)2 = 1373 VA and the rms current is Irms = VArms /Vrms = 29.4 A. Problem 1.31 part(a): Area increases by a factor of 4. Thus the voltage increases by a factor of 4 to e = 1096cos377t. 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 1.32 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 Am = 2 cm2 = 3.40 cm2 0.47 and

13

lm = −0.2 cm



0.8 µ0 (−3.60 × 105 )



= 0.35 cm

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 1.33 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 Am = 2 cm2 = 2.54 cm2 0.63 and lm = −0.2 cm



0.8 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 1.34 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, we want Bg = 1.2 T, Bm = 0.47 T and Hm = −360 kA/m.     Hg Bg hm = −g = −g = 2.65 mm Hm µ0 Hm Am = Ag



Bg Bm



Rm =

= 2πRh





Bg Bm



= 26.0 cm2

Am = 2.87 cm π

Problem 1.35 From Fig. 1.19, the maximum energy product for neodymium-iron-boron occurs at (approximately) Bm = 0.63 T and Hm = -470 kA/m. The magnetization curve for neodymium-iron-boron can be represented as Bm = µR Hm + Br where Br = 1.26 T and µR = 1.067µ0. The magnetic circuit must satisfy

14

Hm d + Hg g = N i;

Bm Am = 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.   Bg Ag = 4.76 cm2 Am = Bm d=−



Hg Hm



g = 1.69 mm

part (b):

i=

  dA Bg µR Agm +

N

g µ0



−

Br d µR



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 operating range of this problem, the system is linear and hence a sinusoidal flux variation will correspond to a sinusoidal current variation.

15

PROBLEM SOLUTIONS: Chapter 2

Problem 2.1 At 60 Hz, ω = 120π. primary: (Vrms )max = N1 ωAc (Brms )max = 2755 V, rms secondary: (Vrms )max = N2 ωAc (Brms )max = 172 V, rms At 50 Hz, ω = 100π. Primary voltage is 2295 V, rms and secondary voltage is 143 V, rms. Problem 2.2 √ 2Vrms N= = 167 turns ωAc Bpeak Problem 2.3 N=




75 =3 8

turns

Problem 2.4 Resistance seen at primary is R1 = (N1 /N2 )2 R2 = 6.25Ω. Thus I1 =

V1 = 1.6 A R1

and V2 =



N2 N1



V1 = 40 V

Problem 2.5 The maximum power will be supplied to the load resistor when its impedance, as reflected to the primary of the ideal transformer, equals that of the source (2 kΩ). Thus the transformer turns ratio N to give maximum power must be
Rs N= = 6.32 Rload Under these conditions, the source voltage will see a total resistance of Rtot = 4 kΩ and the current will thus equal I = Vs /Rtot = 2 mA. Thus, the power delivered to the load will equal Pload = I 2 (N 2 Rload ) = 8

mW

16 Here is the desired MATLAB plot:

Problem 2.6 The maximum power will be supplied to the load resistor when its impedance, as reflected to the primary of the ideal transformer, equals that of the source (2 kΩ). Thus the transformer turns ratio N to give maximum power must be
Rs = 6.32 N= Rload Under these conditions, the source √ voltage will see a total impedance of Ztot = 2 kΩ. The current will thus equal I = 2 + j2 kΩ whose magnitude is 2 √ 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:

17 Problem 2.7 V2 = V1



Xm Xl1 + Xm



= 266 V

Problem 2.8 part (a): Referred to the secondary Lm,2 =

Lm,1 = 150 mH N2

part(b): Referred to the secondary, Xm = ωLm,2 = 56.7 Ω, Xl2 = 84.8 mΩ and Xl1 = 69.3 mΩ. Thus,   Xm (i) V1 = N V2 = 7960 V Xm + Xl2 and (ii) Isc =

V2 V2 = = 1730 A Xsc Xl2 + Xm ||Xl1

Problem 2.9 part (a): V1 = 3.47 A; I1 = Xl1 + Xm

V2 = N V1



Xm Xl1 + Xm



= 2398 V

part (b): Let Xl′2 = Xl2 /N 2 and Xsc = Xl1 + Xm ||(Xm + Xl′2 ). For Irated = 50 kVA/120 V = 417 A V1 = Irated Xsc = 23.1 V

I2 =

1 N



Xm Xm + Xl2



Irated = 15.7 A

Problem 2.10 IL =

Pload = 55.5 A VL

and thus IH =

IL = 10.6 A; N

VH = N VL + jXH IH = 2381 9.6◦

The power factor is cos (9.6◦ ) = 0.986 lagging.

V

18 Problem 2.11 part (a):

part (b): 30 kW jφ Iˆload = e = 93.8 ejφ 230 V

A

where φ is the power-factor angle. Referred to the high voltage side, IˆH = 9.38 ejφ A. VˆH = ZH IˆH Thus, (i) for a power factor of 0.85 lagging, VH = 2413 V and (ii) for a power factor of 0.85 leading, VH = 2199 V. part (c):

19 Problem 2.12 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 2.13 part (a): Iload = 160 kW/2340 V = 68.4 A at 

= cos−1 (0.89) = 27.1◦

Vˆt,H = N (VˆL + Zt IL ) which gives VH = 33.7 kV. part (b): Vˆsend = N (VˆL + (Zt + Zf )IL )

20 which gives Vsend = 33.4 kV. part (c): ∗ Ssend = Psend + jQsend = Vˆsend Iˆsend = 164 kW − j64.5 kVAR

Thus Psend = 164 kW and Qsend = −64.5 kVAR. Problem 2.14 Following the methodology of Example 2.6, efficiency = 98.4 percent and regulation = 1.25 percent. Problem 2.15 part (a): |Zeq,L | =

Vsc,L = 107.8 mΩ Isc,L

Req,L =

Psc,L = 4.78 mΩ 2 Isc,L

Xeq,L =

 2 |Zeq,L |2 − Req,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 + j0.46 mΩ part (c): From the open-circuit test, the core-loss resistance and the magnetizing reactance as referred to the low-voltage side can be found: Rc,L =

Soc,L = Voc,L Ioc,L = 497 kVA; and thus

2 Voc,L = 311 Ω Poc,L

Qoc,L =

 2 2 Soc,L − Poc,L = 45.2 kVAR

21

Xm,L =

2 Voc,L = 141 Ω Qoc,L

The equivalent-T circuit for the transformer from the low-voltage side is thus:

part (d): We will solve this problem with the load connected to the highvoltage side but referred to the low-voltage side. The rated low-voltage current is IL = 50 MVA/8 kV = 6.25 kA. Assume the load is at rated voltage. Thus the low-voltage terminal voltage is VL = |Vload + Zeq,L IL | = 8.058 kV and thus the regulation is given by (8.053-8)/8 = 0.0072 = 0.72 percent. The total loss is approximately equal to the sum of the open-circuit loss and the short-circuit loss (393 kW). Thus the efficiency is given by η=

50.0 Pload = = 0.992 = 99.2 percent Pin 50.39

part (e): We will again solve this problem with the load connected to the high-voltage side but referred to the low-voltage side. Now, IˆL = 6.25 25.8◦ kA. Assume the load is at rated voltage. Thus the low-voltage terminal voltage is VL = |Vload + Zeq,L IˆL | = 7.758 kV and thus the regulation is given by (7.758-8)/8 = -0.0302 = -3.02 percent. The efficiency is the same as that found in part (d), η = 99.2 percent. Problem 2.16 √ The core length of the second transformer is is 2 times that of the first, its core √ area of the second transformer is twice that of the first, and its volume is 2 2 times that of the first. Since the voltage applied to the second transformer is twice that of the first, the flux densitities will be the same. Hence, the core loss will be proportional to the volume and √ Coreloss = 2 23420 = 9.67 kW

22 The magnetizing inductance is proportional to the area and inversely pro√ portional to the core length√and hence is 2 times larger. Thus the no-load magnetizing current will be 2 times larger in the second transformer or √ Ino−load = 2 4.93 = 6.97 A Problem 2.17 part (a): Rated current at the high-voltage side is 20 kVA/2.4 kV = 8.33 A. Thus the total loss will be Ploss = 122 + 257 = 379 W. The load power is equal to 0.8 × 20 = 16 kW. Thus the efficiency is η=

16 = 0.977 = 97.7 percent 16.379

part (b): First calculate the series impedance (Zeq,H = Req,H + jXeq,H ) of the transformer from the short-circuit test data. Req,H =

Psc,H = 3.69 Ω 2 Isc,H

Ssc,H = Vsc,H Isc,H = 61.3 × 8.33 = 511 kV A Thus Qsc,H =

 2 2 Ssc,H − Psc,H = 442 VAR and hence Xeq,H =

Qsc,H = 6.35 Ω 2 Isc,H

The regulation will be greatest when the primary and secondary voltages of the transformer are in phase as shown in the following phasor diagram

Thus the voltage drop across the transformer will be equal to ∆V = |Iload ||Zeq,H | = 61.2 V and the regulation will equal 61.2 V/2.4 kV = 0.026 = 2.6 percent. Problem 2.18 For a power factor of 0.87 leading, the efficiency is 98.4 percent and the regulation will equal -3.48 percent. Problem 2.19 part (a): The voltage rating is 2400 V:2640 V. part (b): The rated current of the high voltage terminal is equal to that of the 240-V winding, Irated = 30 × 103/240 = 125 A. Hence the kVA rating of the transformer is 2640 × 125 = 330 kVA.

23 P...