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Exercicios resolvidos-Maquinas Elétricas -FitzgeraldMáquinas Elétricas Universidade Federal do Paraná (UFPR) 152 pag.Document shared on docsity1PROBLEM SOLUTIONS: Chapter 1Problem 1. Part (a): Rc= lc μAc= lc μrμ 0 Ac=0 A/WbRg=g μ 0 Ac=1. 017 × 106 A/Wbpart (b):Φ=NIRc+Rg=1. 224 × 10− 4 Wbpart (c):λ=N...

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Exercicios resolvidosMaquinas Elétricas Fitzgerald Máquinas Elétricas Universidade Federal do Paraná (UFPR) 152 pag.

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1

PROBLEM SOLUTIONS: Chapter 1

Problem 1.1 Part (a): Rc =

lc lc = = 0 A/Wb µr µ0 Ac µAc

Rg =

g = 1.017 × 106 µ0 Ac

A/Wb

part (b): Φ=

NI = 1.224 × 10−4 Rc + Rg

Wb

part (c): λ = NΦ = 1.016 × 10−2

Wb

part (d): L= Problem 1.2 part (a): Rc =

λ = 6.775 mH I

lc lc = = 1.591 × 105 µAc µrµ0 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 = NΦ = 8.787 × 10−3

Wb

part (d): L=

λ = 5.858 mH I

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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 /(µrµ0 )) = 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

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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 = NI;

Bg Ag = BcAc

and Bg = µ0 Hg ;

Bc = µHc

can be combined to give     NI NI  =       Bg =  Ag 2g + µµ0 2g + µµ0 1 − Xx0 (lc + lp ) Ac (lc + lp )

Problem 1.7 part (a):



part (b):

g+

I =B

  µ0 µ

(lc + lp )

µ0 N



 = 2.15 A

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

I =B

g+

  µ0 µ

(lc + lp )

µ0 N



 = 3.02 A

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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 cm 2

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 = NBg 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

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A/Wb

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5 part (c): L=

N2 = 0.311 mH Rg + Rg

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 =

ωNAc Bpeak √ = 19.2 V rms 2

part (b): Irms =

Vrms = 1.67 ωL

A rms;

√ Wpeak = 0.5L( 2 Irms)2 = 8.50 mJ

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

 R21 + R22 = 4.27 cm

µ0 Ag N 2   = 251 mH g + µµ0 lc

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

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

Problem 1.16 part (a):

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

LI = 99 turns; Ac Bsat

g=

µ0 NI µ0 lc − = 0.36 mm Bsat µ

part (b): From Eq.3.21 Wgap =

2 Ac gBsat = 0.207 J; 2µ0

Wcore =

Ac lcB 2sat = 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 m 3 . Thus  2  B W = Vcore = 4.87 J 2µ0 part (c): For T = 30 sec, di (2πrB)/(µ0 N) = = 2.92 × 103 T dt v=L

A/sec

di = 163 V dt

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

Vol cu = 2ab(w + h + 2a)

part (b):

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8

B = µ0



Jcu Acu g



part (c): Jcu =

NI Acu

part (d):   P diss = Volcu ρJ 2cu

part (e):

Wmag = Volgap



B2 2µ0



= gwh



B2 2µ0



part (f):  1 2 LI µ0 whA2cu 2Wmag Wmag L = = =  21 2 =  1  R RI ρgVolcu P diss P diss 2 2

Problem 1.21 Using the equations of Problem 1.20 P diss = 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

  A2 A1 I1 + λ1 = N1 (A1 B1 + A2 B2 ) = µ0 N 21 g2 g1   A2 I1 λ2 = N2 A2 B2 = µ0 N1 N2 g2

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9 part (b):

(i)

B1 = 0;

B2 =

µ0 N2 I2 g2

  A2 λ1 = N1 A2 B2 = µ0 N1 N2 I2 g2   A2 I2 λ2 = N2 A2 B2 = µ0 N 22 g2

(ii) (iii) part (c):

(i) (ii) (iii)

µ0 N2 I2 µ0 N1 I1 + g2 g2     A2 A2 A1 + I1 + µ0 N1 N2 I2 λ1 = N1 (A1 B1 + A2 B2 ) = µ0 N 21 g1 g2 g2     A2 A2 λ2 = N2 A2 B2 = µ0 N1 N2 I1 + µ0 N 22 I2 g2 g2 B1 =

µ0 N1I1 ; g1

B2 =

part (d): L11 = N 12



A2 A1 + g1 g2



;

L22 = µ0 N 22



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 =

N 21 µAc N 21 = l1 + l2 + lA /2 + g (µ/µ0 ) R1 + R2 + Rg + RA /2

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A2 g2



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10

LAA = LBB =

N 2 µAc N2 = RA + RA ||(R1 + R2 + Rg ) lA



lA + l1 + l2 + g (µ/µ0 ) lA + 2(l1 + l2 + g (µ/µ0 ))



part (b): LAB = LBA =

N 2 (R1 + R2 + Rg ) N 2 µAc = lA RA (RA + 2(R1 + R2 + Rg ))

LA1 = L1A = −LB1 = −L1B =



l1 + l2 + g (µ/µ0 ) lA + 2(l1 + l2 + g (µ/µ0 ))

−NN1µAc −NN1 = lA + 2(l1 + l2 + g (µ/µ0 )) RA + 2(R1 + R2 + Rg )

part (c): v1 =

d d [iA − iB ] [LA1iA + LB1 iB ] = LA1 dt dt

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

µ0 N1 N2 [D(w − x)] 2g

part (b):

v2



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

Problem 1.25 part (a): H=

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

v2 =

d dB [N2(tn∆)B] = N2 tn∆ dt dt

part (b):

part (c): vo = G



v2 dt = GN2 tn∆B

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11 Problem 1.26 Rg =

g = 4.42 × 105 µ0 Ag

A/Wb;

Rc =

333 lc A/Wb = µAg µ

Want Rg ≤ 0.05Rc ⇒ µ ≥ 1.2 × 104µ0 . 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 ωt(Ro − Ri )Bpeak

turns

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;

L=

λ = 4.39 mH I

Problem 1.29 part (a):

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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 = ωNAcBrms = 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 gAcBrms = 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 = VA rms/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 P c = 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 2 cm 2 = 3.40 cm2 Am = 0.47 and

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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 2 cm2 = 2.54 cm2 Am = 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.     Bg Hg = −g = 2.65 mm hm = −g Hm µ0 Hm Am = Ag



Bg Bm



Rm =

= 2πRh





Bg Bm



= 26.0 cm 2

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

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14

Hm d + Hg g = Ni;

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=

  g + Bg µdA R Am

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.

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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 N=

√

2Vrms = 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

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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 N= = 6.32 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:

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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 V2 = 7960 V (i) V1 = N Xm + Xl2 and (ii)

Isc =

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

Problem 2.9 part (a): I1 =

V1 = 3.47 A; Xl1 + Xm

V2 = NV1



Xm Xl1 + Xm



= 2398 V

part (b): Let X ′l2 = Xl2 /N 2 and Xsc = Xl1 + Xm ||(Xm + X ′l2 ). For Irated = 50 kVA/120 V = 417 A V1 = IratedXsc = 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 = NVL + jXH IH = 2381 9.6◦

V

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

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18 Problem 2.11 part (a):

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

A

where φ is the power-factor angle. Referred to the high voltage side, ˆIH = 9.38 ejφA. VˆH = ZHIˆH 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):

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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( ˆVL + Zt IL ) which gives VH = 33.7 kV. part (b): ˆVsend = N(VˆL + (Zt + Zf )IL )

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

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 = and thus

 2 = 107.7 mΩ |Zeq,L |2 − Req,L 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; a...