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### Transcript of Fitzgerald machine solution

• 1. 1 PROBLEM SOLUTIONS: Chapter 1 Problem 1.1 Part (a): Rc = lc Ac = lc r0Ac = 0 A/Wb Rg = g 0Ac = 1.017 106 A/Wb part (b): = NI Rc + Rg = 1.224 104 Wb part (c): = N = 1.016 102 Wb part (d): L = I = 6.775 mH Problem 1.2 part (a): Rc = lc Ac = lc r0Ac = 1.591 105 A/Wb Rg = g 0Ac = 1.017 106 A/Wb part (b): = NI Rc + Rg = 1.059 104 Wb part (c): = N = 8.787 103 Wb part (d): L = I = 5.858 mH www.mohandesyar.com www.mohandesyar.com

2. 2 Problem 1.3 part (a): N = Lg 0Ac = 110 turns part (b): I = Bcore 0N/g = 16.6 A Problem 1.4 part (a): N = L(g + lc0/) 0Ac = L(g + lc0/(r0)) 0Ac = 121 turns part (b): I = Bcore 0N/(g + lc0/) = 18.2 A Problem 1.5 part (a): part (b): r = 1 + 3499 1 + 0.047(2.2)7.8 = 730 I = B g + 0lc/ 0N = 65.8 A www.mohandesyar.com www.mohandesyar.com 3. 3 part (c): Problem 1.6 part (a): Hg = NI 2g ; Bc = Ag Ac Bg = Bg 1 x X0 part (b): Equations 2gHg + Hclc = NI; BgAg = BcAc and Bg = 0Hg; Bc = Hc can be combined to give Bg = NI 2g + 0 Ag Ac (lc + lp) = NI 2g + 0 1 x X0 (lc + lp) Problem 1.7 part (a): I = B g + 0 (lc + lp) 0N = 2.15 A part (b): = 0 1 + 1199 1 + 0.05B8 = 1012 0 I = B g + 0 (lc + lp) 0N = 3.02 A www.mohandesyar.com www.mohandesyar.com 4. 4 part (c): Problem 1.8 g = 0N2 Ac L 0 lc = 0.353 mm Problem 1.9 part (a): lc = 2(Ro Ri) g = 3.57 cm; Ac = (Ro Ri)h = 1.2 cm2 part (b): Rg = g 0Ac = 1.33 107 A/Wb; Rc = 0 A/Wb; part (c): L = N2 Rg + Rg = 0.319 mH part (d): I = Bg(Rc + Rg)Ac N = 33.1 A part (e): = NBgAc = 10.5 mWb Problem 1.10 part (a): Same as Problem 1.9 part (b): Rg = g 0Ac = 1.33 107 A/Wb; Rc = lc Ac = 3.16 105 A/Wb www.mohandesyar.com www.mohandesyar.com 5. 5 part (c): L = N2 Rg + Rg = 0.311 mH part (d): I = Bg(Rc + Rg)Ac N = 33.8 A part (e): Same as Problem 1.9. Problem 1.11 Minimum r = 340. Problem 1.12 L = 0N2 Ac g + lc/r Problem 1.13 L = 0N2 Ac g + lc/r = 30.5 mH Problem 1.14 part (a): Vrms = NAcBpeak 2 = 19.2 V rms part (b): Irms = Vrms L = 1.67 A rms; Wpeak = 0.5L( 2 Irms)2 = 8.50 mJ www.mohandesyar.com www.mohandesyar.com 6. 6 Problem 1.15 part (a): R3 = R2 1 + R2 2 = 4.27 cm part (b): L = 0AgN2 g + 0 lc = 251 mH part (c): For = 260 rad/sec and peak = NAgBpeak = 0.452 Wb: (i) Vrms = peak = 171 V rms (ii) Irms = Vrms L = 1.81 A rms (iii) Wpeak = 0.5L( 2Irms)2 = 0.817 J part (d): For = 250 rad/sec and peak = NAgBpeak = 0.452 Wb: (i) Vrms = peak = 142 V rms (ii) Irms = Vrms L = 1.81 A rms (iii) Wpeak = 0.5L( 2Irms)2 = 0.817 J Problem 1.16 part (a): www.mohandesyar.com www.mohandesyar.com 7. 7 part (b): Emax = 4fNAcBpeak = 345 V Problem 1.17 part (a): N = LI AcBsat = 99 turns; g = 0NI Bsat 0lc = 0.36 mm part (b): From Eq.3.21 Wgap = AcgB2 sat 20 = 0.207 J; Wcore = AclcB2 sat 2 = 0.045 J Thus Wtot = Wgap + Wcore = 0.252 J. From Eq. 1.47, (1/2)LI2 = 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 = 0a2 N2 2r = 56.0 mH part (b): Core volume Vcore (2r)a2 = 40.0 m3 . Thus W = Vcore B2 20 = 4.87 J part (c): For T = 30 sec, di dt = (2rB)/(0N) T = 2.92 103 A/sec v = L di dt = 163 V Problem 1.20 part (a): Acu = fwab; Volcu = 2ab(w + h + 2a) part (b): www.mohandesyar.com www.mohandesyar.com 8. 8 B = 0 JcuAcu g part (c): Jcu = NI Acu part (d): Pdiss = Volcu J2 cu part (e): Wmag = Volgap B2 20 = gwh B2 20 part (f): L R = 1 2 LI2 1 2 RI2 = Wmag 1 2 Pdiss = 2Wmag Pdiss = 0whA2 cu gVolcu Problem 1.21 Using the equations of Problem 1.20 Pdiss = 115 W I = 3.24 A N = 687 turns R = 10.8 = 6.18 msec Wire size = 23 AWG Problem 1.22 part (a): (i) B1 = 0N1I1 g1 ; B2 = 0N1I1 g2 (ii) 1 = N1(A1B1 + A2B2) = 0N2 1 A1 g1 + A2 g2 I1 (iii) 2 = N2A2B2 = 0N1N2 A2 g2 I1 www.mohandesyar.com www.mohandesyar.com 9. 9 part (b): (i) B1 = 0; B2 = 0N2I2 g2 (ii) 1 = N1A2B2 = 0N1N2 A2 g2 I2 (iii) 2 = N2A2B2 = 0N2 2 A2 g2 I2 part (c): (i) B1 = 0N1I1 g1 ; B2 = 0N1I1 g2 + 0N2I2 g2 (ii) 1 = N1(A1B1 + A2B2) = 0N2 1 A1 g1 + A2 g2 I1 + 0N1N2 A2 g2 I2 (iii) 2 = N2A2B2 = 0N1N2 A2 g2 I1 + 0N2 2 A2 g2 I2 part (d): L11 = N2 1 A1 g1 + A2 g2 ; L22 = 0N2 2 A2 g2 ; L12 = 0N1N2 A2 g2 Problem 1.23 RA = lA Ac ; R1 = l1 Ac ; R2 = l2 Ac ; Rg = g 0Ac part (a): L11 = N2 1 R1 + R2 + Rg + RA/2 = N2 1 Ac l1 + l2 + lA/2 + g (/0) www.mohandesyar.com www.mohandesyar.com 10. 10 LAA = LBB = N2 RA + RA||(R1 + R2 + Rg) = N2 Ac lA lA + l1 + l2 + g (/0) lA + 2(l1 + l2 + g (/0)) part (b): LAB = LBA = N2 (R1 + R2 + Rg) RA(RA + 2(R1 + R2 + Rg)) = N2 Ac lA l1 + l2 + g (/0) lA + 2(l1 + l2 + g (/0)) LA1 = L1A = LB1 = L1B = NN1 RA + 2(R1 + R2 + Rg) = NN1Ac lA + 2(l1 + l2 + g (/0)) part (c): v1 = d dt [LA1iA + LB1iB] = LA1 d dt [iA iB] Q.E.D. Problem 1.24 part (a): L12 = 0N1N2 2g [D(w x)] part (b): v2 = d2 dt = I0 dL12 dt = N1N20D 2g dx dt = N1N20D 2g w 2 cos t Problem 1.25 part (a): H = N1i1 2(Ro + Ri)/2 = N1i1 (Ro + Ri) part (b): v2 = d dt [N2(tn)B] = N2tn dB dt part (c): vo = G v2 dt = GN2tnB www.mohandesyar.com www.mohandesyar.com 11. 11 Problem 1.26 Rg = g 0Ag = 4.42 105 A/Wb; Rc = lc Ag = 333 A/Wb Want Rg 0.05Rc 1.2 104 0. By inspection of Fig. 1.10, this will be true for B 1.66 T (approximate since the curve isnt that detailed). Problem 1.27 part (a): N1 = Vpeak t(Ro Ri)Bpeak = 57 turns part (b): (i) Bpeak = Vo,peak GN2t(Ro Ri) = 0.833 T (ii) V1 = N1t(Ro Ri)Bpeak = 6.25 V, peak Problem 1.28 part (a): From the M-5 magnetization curve, for B = 1.2 T, Hm = 14 A/m. Similarly, Hg = B/0 = 9.54 105 A/m. Thus, with I1 = I2 = I I = Hm(lA + lC g) + Hgg N1 = 38.2 A part (b): Wgap = gAgapB2 20 = 3.21 Joules part (c): = 2N1AAB = 0.168 Wb; L = I = 4.39 mH Problem 1.29 part (a): www.mohandesyar.com www.mohandesyar.com 12. 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 = NAcBrms = 46.7 V Core volume = Aclc = 1.05 103 m3 . Mass density = 7.65 103 kg/m3 . Thus, the core mass = (1.05 103 )(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 = gAcB2 rms 0 = 3.61 Joules corresponding to an rms reactive power of VARgap = Wgap = 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 I,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, Am = 0.8 0.47 2 cm2 = 3.40 cm2 and www.mohandesyar.com www.mohandesyar.com 13. 13 lm = 0.2 cm 0.8 0(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, Am = 0.8 0.63 2 cm2 = 2.54 cm2 and lm = 0.2 cm 0.8 0(4.70 105) = 0.27 cm Thus the volume is 2.540.25 = 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. hm = g Hg Hm = g Bg 0Hm = 2.65 mm Am = Ag Bg Bm = 2Rh Bg Bm = 26.0 cm2 Rm = Am = 2.87 cm Problem 1.35 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 = RHm + Br where Br = 1.26 T and R = 1.0670. The magnetic circuit must satisfy www.mohandesyar.com www.mohandesyar.com 14. 14 Hmd + Hgg = Ni; BmAm = BgAg 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 cm2 d = Hg Hm g = 1.69 mm part (b): i = Bg dAg RAm + g 0 Brd R 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 ux variation will correspond to a sinusoidal current variation. www.mohandesyar.com www.mohandesyar.com 15. 15 PROBLEM SOLUTIONS: Chapter 2 Problem 2.1 At 60 Hz, = 120. primary: (Vrms)max = N1Ac(Brms)max = 2755 V, rms secondary: (Vrms)max = N2Ac(Brms)max = 172 V, rms At 50 Hz, = 100. Primary voltage is 2295 V, rms and secondary voltage is 143 V, rms. Problem 2.2 N = 2Vrms AcBpeak = 167 turns Problem 2.3 N = 75 8 = 3 turns Problem 2.4 Resistance seen at primary is R1 = (N1/N2)2 R2 = 6.25. Thus I1 = V1 R1 = 1.6 A and V2 = N2 N1 V1 = 40 V Problem 2.5 The maximum power will be supplied to the load resistor when its im- pedance, as reected to the primary of the ideal transformer, equals that of the source (2 k). Thus the transformer turns ratio N to give maximum power must be N = Rs Rload = 6.32 Under these conditions, the source voltage will see a total resistance of Rtot = 4 k and the current will thus equal I = Vs/Rtot = 2 mA. Thus, the power delivered to the load will equal Pload = I2 (N2 Rload) = 8 mW www.mohandesyar.com www.mohandesyar.com 16. 16 Here is the desired