Electrical Engineering Problem Solutions: Transformers and Motors, Exercises of Electrical and Electronics Engineering

Download Electrical Engineering Problem Solutions: Transformers and Motors and more Exercises Electrical and Electronics Engineering in PDF only on Docsity! SOLUTION MANUAL Fitzgerald & Kingsley’s Electric Machinery [7th Edition] Máquinas elétricas de Fitzgeral e Kingsley [7th edição] 1 PROBLEM SOLUTIONS: Chapter 1 Problem 1-1 Part (a): Rc = lc µAc = lc µrµ0Ac = 0 A/Wb Rg = g µ0Ac = 5.457 × 106 A/Wb Part (b): Φ = NI Rc + Rg = 2.437 × 10−5 Wb Part (c): λ = NΦ = 2.315 × 10−3 Wb Part (d): L = λ I = 1.654 mH Problem 1-2 Part (a): Rc = lc µAc = lc µrµ0Ac = 2.419 × 105 A/Wb Rg = g µ0Ac = 5.457 × 106 A/Wb c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 4 Bc = Bg ( Ag Ac ) = ( µ0NI 2g )( 1 − x X0 ) Part (b): Will assume lc is “large” and lp is relatively “small”. Thus, BgAg = BpAg = BcAc We can also write 2gHg + Hplp + Hclc = NI ; and Bg = µ0Hg; Bp = µHp Bc = µHc These equations can be combined to give Bg =   µ0NI 2g + ( µ0 µ ) lp + ( µ0 µ ) ( Ag Ac ) lc   =   µ0NI 2g + ( µ0 µ ) lp + ( µ0 µ ) ( 1 − x X0 ) lc   and Bc = ( 1 − x X0 ) Bg Problem 1-7 From Problem 1-6, the inductance can be found as L = NAcBc I = µ0N 2Ac 2g + µ0 µ (lp + (1 − x/X0) lc) from which we can solve for µr c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 5 µr = µ µ0 = L ( lp + (1 − x/X0) lc ) µ0N2Ac − 2gL = 88.5 Problem 1-8 Part (a): L = µ0(2N)2Ac 2g and thus N = 0.5 √ 2gL Ac = 38.8 which rounds to N = 39 turns for which L = 12.33 mH. Part (b): g = 0.121 cm Part(c): Bc = Bg = 2µ0NI 2g and thus I = Bcg µ0N = 37.1 A Problem 1-9 Part (a): L = µ0N 2Ac 2g c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 6 and thus N = √ 2gL Ac = 77.6 which rounds to N = 78 turns for which L = 12.33 mH. Part (b): g = 0.121 cm Part(c): Bc = Bg = µ0(2N)(I/2) 2g and thus I = 2Bcg µ0N = 37.1 A Problem 1-10 Part (a): L = µ0(2N)2Ac 2(g + (µ0 µ )lc) and thus N = 0.5 √ √ √ √ 2(g + (µ0 µ )lc)L Ac = 38.8 which rounds to N = 39 turns for which L = 12.33 mH. Part (b): g = 0.121 cm c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 9 Problem 1-14 See solution to Problem 1-13 Part (a): lc = 22.8 cm Ac = 1.62 cm2 Part (b): Rc = 1.37 × 106 H−1 Rg = 7.37 × 106 H−1 Part (c): L = 5.94 × 10−4 H Part (d): I = 24.6 A Part (e): λ = 1.46 × 10−2 Wb c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 10 Problem 1-15 µr must be greater than 2886. Problem 1-16 L = µ0N 2Ac g + lc/µr Problem 1-17 Part (a): L = µ0N 2Ac g + lc/µr = 36.6 mH Part (b): B = µ0N 2 g + lc/µr I = 0.77 T λ = LI = 4.40 × 10−2 Wb Problem 1-18 Part (a): With ω = 120π c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 11 Vrms = ωNAcBpeak√ 2 = 20.8 V Part (b): Using L from the solution to Problem 1-17 Ipeak = √ 2Vrms ωL = 1.66 A Wpeak = LI2 peak 2 = 9.13 × 10−2 J Problem 1-19 B = 0.81 T and λ = 46.5 mWb Problem 1-20 Part (a): R3 = √ (R2 1 + R2 2) = 4.49 cm Part (b): For lc = 4l + R2 + R3 − 2h; and Ag = πR2 1 L = µ0AgN 2 g + (µ0/µ)lc = 61.8 mH Part (c): For Bpeak = 0.6 T and ω = 2π60 c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 14 Problem 1-23 For part (b), Ipeak = 11.9 A. For part (c), Ipeak = 27.2 A. Problem 1-24 L = µ0AcN 2 g + (µ0/µ)lc Bc = µ0NI g + (µ0/µ)lc Part (a): For I = 10 A, L = 23 mH and Bc = 1.7 T N = LI AcBc = 225 turns g = µ0NI Bc − µ0lc µ = 1.56 mm Part (b): For I = 10 A and Bc = Bg = 1.7 T, from Eq. 3.21 Wg = ( B2 g 2µ0 ) Vg = 1.08 J c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 15 Wcore = ( B2 c 2µ ) Vg = 0.072 J based upon Vcore = Aclc Vg = Acg Part (c): Wtot = Wg + Wcore = 1.15 J = 1 2 LI2 Problem 1-25 Lmin = 3.6 mH Lmax = 86.0 mH Problem 1-26 Part (a): N = LI BAc = 167 g = µ0NI 2BAc = 0.52 mm Part (b): N = LI 2BAc = 84 g = µ0NI BAc = 0.52 mm c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 16 Problem 1-27 Part (a): N = LI BAc = 167 g = µ0NI 2BAc − (µ0/µ)lc = 0.39 mm Part (b): N = LI 2BAc = 84 g = µ0NI BAc − (µ0/µ)lc = 0.39 mm Problem 1-28 Part (a): N = 450 and g = 2.2 mm Part (b): N = 225 and g = 2.2 mm Problem 1-29 Part (a): L = µ0N 2A l = 11.3 H where A = πa2 l = 2πr c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 19 λ2 = N2A2Bg2 = µ0N1N2 A2 g2 I1 Part (b) (i): Bg1 = 0 Bg2 = µ0N2 g2 I2 (ii) λ1 = N1(A1Bg1 + A2Bg2) = µ0N1N2 A2 g2 I2 λ2 = N2A2Bg2 = µ0N 2 2 A2 g2 I2 Part (c) (i): Bg1 = µ0N1 g1 I1 Bg2 = µ0N1 g2 I1 + µ0N2 g2 I2 (ii) λ1 = N1(A1Bg1 + A2Bg2) = µ0N 2 1 ( A1 g1 + A2 g2 ) I1 + µ0N1N2 A2 g2 I2 λ2 = N2A2Bg2 = µ0N1N2 A2 g2 I1 + µ0N 2 2 A2 g2 I2 Part (d): L1 = µ0N 2 1 ( A1 g1 + A2 g2 ) L2µ0N 2 2 A2 g2 c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 20 L12 = µ0N1N2 A2 g2 Problem 1-33 Rg = g µ0Ac R1 = l1 µAc R2 = l2 µAc RA = lA µAc Part (a): L1 = N2 1 Rg + R1 + R2 + RA/2 LA = LB = N2 R where R = RA + RA(Rg + R1 + R2) RA + Rg + R1 + R2 Part (b): L1B = −L1A = N1N 2(Rg + R1 + R2 + RA/2) L12 = N2(Rg + R1 + R2) 2RA(Rg + R1 + R2) + R2 A c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 21 Part (c): v1(t) = L1A diA dt + L1B diB dt = L1A d(iA − iB) dt Problem 1-34 Part (a): L12 = µ0N1N2D(w − x) 2g Part (b): v2(t) = −ωIo ( µ0N1N2Dwε 4g ) cos ωt Problem 1-35 Part (a): H = 2N1i1 (Ro + Ri) Part (b): v2(t) = N2w(n∆) dB(t) dt Part (c): v0(t) = GN2w(n∆)B(t) c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 24 For f = 60 Hz, ρ = 7.65 × 103 kg/m3, Core loss = 1.50 W/kg Problem 1-40 Brms = 1.1 T and f = 60 Hz, Vrms = ωNAcBrms = 86.7 V Core volume = Aclc = 1.54× 10−3 m3. Mass density = 7.65× 103 kg/m3. Thus, the core mass = (1.54 × 10−3)(7.65 × 103) = 11.8 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× 11.8 = 15.3 W. The total exciting VA for the core is 2.0× 11.8 = 23.6 VA. Thus, its reactive component is given by √ 23.62 − 15.32 = 17.9 VAR. The rms energy storage in the air gap is 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 VArms = √ 15.32 + (1917 + 17.9)2 = 1935 VA and the rms current is Irms = VArms/Vrms = 22.3 A. Problem 1-41 Part(a): Area increases by a factor of 4. Thus the voltage increases by a factor of 4 to e = 1096 cos(377t). c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 25 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 1-42 From Fig. 1.19, the maximum energy product for samarium-cobalt occurs at (approxi- mately) 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 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 1-43 From Fig. 1.19, the maximum energy product for neodymium-iron-boron occurs at (ap- proximately) 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 c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 26 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 1-44 From Fig. 1.19, the maximum energy product for samarium-cobalt occurs at (approxi- mately) B = 0.47 T and H = -360 kA/m. Thus the maximum energy product is 1.69 × 105 J/m3. Thus, we want Bg = 1.3 T, Bm = 0.47 T and Hm = −360 kA/m. hm = −g ( Hg Hm ) = −g ( Bg µ0Hm ) = 2.87 mm Am = Ag ( Bg Bm ) = 2πRh ( Bg Bm ) = 45.1 cm2 Rm = √ Am π = 3.66 cm Problem 1-45 From Fig. 1.19, the maximum energy product for samarium-cobalt occurs at (approxi- mately) B = 0.63 T and H = -482 kA/m. Thus the maximum energy product is 3.03 × 105 J/m3. Thus, we want Bg = 1.3 T, Bm = 0.47 T and Hm = −360 kA/m. hm = −g ( Hg Hm ) = −g ( Bg µ0Hm ) = 2.15 mm Am = Ag ( Bg Bm ) = 2πRh ( Bg Bm ) = 31.3 cm2 c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 29 AmBm = AgAg = µ0HgAg gives Bg = ( µ0dAm µ0dAg + µRgAm ) Br = 0.95 T c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 30 PROBLEM SOLUTIONS: Chapter 2 Problem 2-1 At 60 Hz, ω = 120π. primary: (Vrms)max = N1ωAc(Brms)max = 3520 V, rms secondary: (Vrms)max = N2ωAc(Brms)max = 245 V, rms At 50 Hz, ω = 100π. Primary voltage is 2934 V, rms and secondary voltage is 204 V, rms. Problem 2-2 N = √ 2Vrms ωAcBpeak = 147 turns Problem 2-3 N = √ 300 75 = 2 turns Problem 2-4 Part (a): R1 = ( N1 N2 )2 R2 = 9.38 Ω I1 = V1 I1 = 1.28 A V2 = ( N2 N1 ) V1 = 48 V P2 = V 2 2 R2 = 14.6 W c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 31 Part (b): For ω = 2πf = 6.28 × 103 Ω X1 = ωL = 2.14 Ω I1 = ∣ ∣ ∣ ∣ V1 R1 + jX1 ∣ ∣ ∣ ∣ = 1.25 A I2 = ( N1 N2 ) I1 = 0.312 A V2 = I2R2 = 46.8 V P2 = V2I2 = 14.6 W Problem 2-5 For ω = 100]π, ZL = RL + jωL = 5.0 + j0.79 Ω VL = 110 ( 20 120 ) = 18.3 V IL = ∣ ∣ ∣ ∣ VL IL ∣ ∣ ∣ ∣ = 3.6 A and IH = IL ( 120 20 ) = 604 mA 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 (1.5 kΩ). Thus the transformer turns ratio N to give maximum power must be N = √ Rs Rload = 4.47 Under these conditions, the source voltage will see a total resistance of Rtot = 3 kΩ and the source current will thus equal I = Vs/Rtot = 4 mA. Thus, the power delivered to the load will equal Pload = I2(N2Rload) = 24 mW c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 34 Problem 2-11 Part (a): At 60 Hz, all reactances increase by a factor of 1.2 over their 50-Hz values. Thus Xm = 55.4 Ω Xl,1 = 33.4 mΩ Xl,2 = 30.4 Ω Part (b): For V1 = 240 V I1 = V 1 X1 + Xm = 4.33 A V2 = NV1 ( Xm Xm + Xl,1 ) = 6883 V Problem 2-12 The load voltage as referred to the high-voltage side is V ′ L = 447 × 2400/460 = 2332 V. Thus the load current as referred to the high voltage side is I ′ L = PL V ′ L = 18.0 A and the voltage at the high voltage terminals is VH = |V ′ L + jXl,1I ′ L| = 2347 V and the power factor is pf = PL VHI ′ L = 0.957 lagging here we know that it is lagging because the transformer is inductive. c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 35 Problem 2-13 At 50 Hz, Xl = 39.3 × (5/6) = 32.8 Ω. The load voltage as referred to the high-voltage side is V ′ L = 362× 2400/460 = 1889 V. Thus the load current as referred to the high voltage side is I ′ L = PL V ′ L = 18.3 A and the voltage at the high voltage terminals is VH = |V ′ L + jXl,1I ′ L| = 1981 V Problem 2-14 Part (a): Part (b): Îload = 40 kW 240 V ejφ = 166.7 ejφ A where φ is the power-factor angle. Referred to the high voltage side, ÎH = 5.02 ejφ A. c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 36 V̂H = ZHÎH Thus, (i) for a power factor of 0.87 lagging, VH = 7820 V and (ii) for a power factor of 0.87 leading, VH = 7392 V. part (c): Problem 2-15 Part (a): c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 39 Problem 2-18 Following the methodology of Example 2.6, efficiency = 98.4 percent and regulation = 1.25 percent. Problem 2-19 Part (a): The core cross-sectional area increases by a factor of two thus the primary voltage must double to 22 kV to produce the same core flux density. Part (b): The core volume increases by a factor of 2 √ 2 and thus the excitation kVA must increase by the same factor which means that the current must increase by a factor of √ 2 to 0.47 A and the power must increase by a factor of 2 √ 2 to 7.64 kW. Problem 2-20 Part (a): |Zeq,H| = Vsc,H Isc,H = 14.1 Ω Req,H = Psc,H I2 sc,H = 752 mΩ Xeq,H = √ |Zeq,H|2 −R2 eq,H = 14.1 Ω and thus Zeq,H = 0.75 + j14.1 Ω Part (b): With N = 78/8 = 9.75 Req,L = Req,H N2 = 7.91 mΩ c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 40 Xeq,L = Xeq,H N2 = 148 mΩ and thus Zeq,L = 7.9 + j148 mΩ Part (c): From the open-circuit test, the core-loss resistance and the magnetizing reac- tance as referred to the low-voltage side can be found: Rc,L = V 2 oc,L Poc,L = 742 Ω Soc,L = Voc,LIoc,L = 317 kVA; Qoc,L = √ S2 oc,L − P 2 oc,L = 305 kVAR and thus Xm,L = V 2 oc,L Qoc,L = 210 Ω The equivalent-T circuit for the transformer from the low-voltage side is thus: c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 41 Problem 2-21 Part (a): |Zeq,H| = Vsc,H Isc,H = 14.1 Ω Req,H = Psc,H I2 sc,H = 752 mΩ Xeq,H = √ |Zeq,H|2 −R2 eq,H = 14.1 Ω and thus Zeq,H = 0.75 + j14.1 Ω Part (b): With N = 78/8 = 9.75 Req,L = Req,H N2 = 7.91 mΩ Xeq,L = Xeq,H N2 = 148 mΩ and thus Zeq,L = 7.9 + j148 mΩ Part (c): From the open-circuit test, the core-loss resistance and the magnetizing reac- tance as referred to the low-voltage side can be found: Rc,L = V 2 oc,L Poc,L = 742 Ω c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 44 Part (c): Pdiss = Poc,L + Psc,H = 18.4 kW Problem 2-24 Solution the same as Problem 2-22 Problem 2-25 Part (a): 7.69 kV:79.6 kV, 10 MVA Part (b): 17.3 A, 48.0 kW Part (c): Since the number of turns on the high-voltage side have doubled, this will occur at a voltage equal to twice that of the original transformer, i.e. 3.84 kV. Part (d): The equivalent-circuit parameters referred to the low-voltage side will be un- changed from those of Problem 2-22. Those referred to the high-voltage side will have 4 times the values of Problem 2-22. Rc,L = 1.32 kΩ Rc,H = 132 kΩ Xm,L = 491 Ω Xm,H = 49.1 kΩ RL = 38.0 mΩ RH = 3.80 Ω XL = 302 mΩ XH = 30.2 Ω Problem 2-26 Part (a): Under this condition, the total transformer power dissipation is 163.7 kW. Thus the efficiency is c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 45 η = 100 × 25 MW 25 MW + 163.7 kW = 99.4% From Problem 2-20, the transformer equivalent series impedance from the low voltage side is Zeq,L = 7.91 + j148 mΩ. The transformer rated current is Irated = 3125 A and thus under load the transformer high-side voltage (neglecting the effects of magnetizing current) referred to the primary is |V ′ H| = |VL − IratedZeq,L| = 7.989 kV and thus the voltage regulation is 100 × (7.989 − 8.00)/7.989 = 0.14%. Part (b): Same methodology as part (a) except that the load is 22.5 MW and the current is Î = Irated 6 φ where φ = cos−1 (0.9) = 25.8◦. In this case, the efficiency is 99.3% and the regulation is 1.94%. Problem 2-27 Part (a): c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 46 Part (b); Problem 2-28 Part (a): The transformer loss will be equal to the sum of the open-circuit and short- circuit losses, i.e. 313 W. With a load of 0.85 × 25 = 21.25 kW, the efficiency is equal to η = 21.25 21.25 + 0.313 = 0.9855 = 98.55% Part (b): The transformer equivalent-circuit parameters are found as is shown in the solution to Problem 2-23. Rc,L = 414 Ω Rc,H = 41.4 kΩ Xm,L = 193 Ω Xm,H = 19.3 kΩ RL = 17.1 mΩ RH = 1.71 Ω XL = 64.9 mΩ XH = 6.49 Ω The desired solution is 0.963 leading power factor, based upon a MATLAB search for the load power factor that corresponds to rated voltage at both the low- and high-voltage terminals. c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 49 Problem 2-36 Part (a): (i) 480 V:13.8 kV, 675 kVA (ii) Zeq = 0.0031 + j0.0215 Ω (iii) Zeq = 2.57 + j17.8 Ω Part (b): (i) 480 V:7.97 kV, 675 MVA (ii) Zeq = 0.0031 + j0.0215 Ω (iii) Zeq = 0.86 + j5.93 Ω Problem 2-37 Following the methodology of Example 2.8, Vload = 236 V, line-to-line. Problem 2-38 Part (a): The rated current on the high-voltage side of the transformer is Irated,H = 25 MVA√ 3 × 68 kV = 209 A The equivalent series impedance reflected to the high-voltage side is Zeq,H = N2Zeq,L = 1.55 + j9.70 Ω and the corresponding line-neutral voltage magnitude is VH = Irated,H |Zeq,H| = 2.05 kV corresponding to a line-line voltage of 3.56 kV. c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 50 Part (b): The apparent power at the high-voltage winding is S = 18/.75 = 24 MVA and the corresponding current is Iload = 24 MVA√ 3 × 68 kV = 209 A The power factor angle θ = − cos−1 (0.75) = −41.4◦ and thus Îload = 209 6 −41.4◦ With a high-side line-neutral voltage VH = 69 kV/ √ 3 = 39.8 kV, referred to the high- voltage side, the line-neutral load voltage referred to the high-voltage side is thus V ′ load = |VH − ÎloadZeq,H| = 38.7 kV Referred to the low-voltage winding, the line-neutral load voltage is Vload = ( 13.8 69 ) V ′ load = 7.68 kV corresponding to a line-line voltage of 13.3 kV. Problem 2-39 Part (a): The line-neutral load voltage Vload = 24 kV/ √ 3 = 13.85 kV and the load current is Îload = ( 375 MVA√ 3 24 kV ) ejφ = 9.02ejφ kA where φ = cos−1 0.89 = 27.1◦. The transformer turns ratio N = 9.37 and thus referred to the high voltage side, V ′ load = NVload = 129.9 kV and Î ′ load = Îload/N = 962ejφ A. Thus, the transformer high-side line- neutral terminal voltage is VH = |V ′ L + jXtÎ ′ load| = 127.3 kV c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 51 corresponding to a line-line voltage of 220.6 kV. Part (b): In a similar fashion, the line-neutral voltage at the source end of the feeder is given by Vs = |V ′ L + (Zf + jXt)Î ′ load| = 126.6 kV corresponding to a line-line voltage of 219.3 kV. Problem 2-40 Problem 2-41 Part (a): For a single transformer Req,H = Psc I2 sc = 342 mΩ Ssc = VscIsc = 8.188 kVA Qsc = √ S2 sc − P 2 sc = 8.079 kVAR and thus Xeq,H = Qsc I2 sc = 2.07 Ω c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 54 Î2 = I1 N ( jXm R′ 2 + j(Xm + X ′ 2) ) = 4.995 6 0.01◦ A Part (b): With Rb = 0.1 mΩ and R′ b = N2Rb = 90 mΩ Î2 = I1 N ( jXm R′ 2 + R′ b + j(Xm + X ′ 2) ) = 4.988 6 2.99◦ A Problem 2-46 This solution uses the methodology of Problem 2-45. Part (a): Part (b): Problem 2-47 The base impedance on the high-voltage side of the transformer is Zbase,H = V 2 rated,H Prated = 136.1 Ω c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 55 Thus, in Ohms referred to the high-voltage side, the primary and secondary impedances are Z = (0.0029 + j0.023)Zbase,H = 0.29 + j23.0 mΩ and the magnetizing reactance is similarly found to be Xm = 172 Ω. Problem 2-48 From the solution to Problem 2-20, as referred to the low voltage side, the total series impedance of the transformer is 7.92 + j148.2 mΩ, the magnetizing reactance is 210 Ω and the core-loss resistance is 742 Ω. The low-voltage base impedance of this transformer is Zbase,L = (8 × 103)2 25 × 106 = 2.56 Ω and thus the per-unit series impedance is 0.0031 + j0.0579, the per-unit magnetizing reac- tance is 82.0 and the per-unit core-loss resistance is 289.8. Problem 2-49 From the solution to Problem 2-23, as referred to the low voltage side, the total series impedance of the transformer is 20.6+ j259 mΩ, the magnetizing reactance is 395 Ω and the core-loss resistance is 1780 Ω. The low-voltage base impedance of this transformer is Zbase,L = (3.81 × 103)2 2.5 × 106 = 5.81 Ω and thus the per-unit series impedance is 0.0035 + j0.0446, the per-unit magnetizing reac- tance is 68.0 and the per-unit core-loss resistance is 306.6. Problem 2-50 Part (a): (i) The high-voltage base impedance of the transformer is Zbase,H = (7.97 × 103)2 2.5 × 103 = 2.54 kΩ c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 56 and thus the series reactance referred high-voltage terminal is XH = 0.075Zbase,H = 191 Ω (ii) The low-voltage base impedance is 2.83 Ω and thus the series reactance referred to the low-voltage terminal is 212 mΩ. Part (b): (i) Power rating: 3 × 25 kVA = 75 kVA Voltage rating: √ 3 × 7.97 kV : √ 3 × 266 V = 13.8 kV : 460 V (ii) The per-unit impedance remains 0.075 per-unit (iii) Referred to the high-voltage terminal, XH = 191 Ω (iv) Referred to the low-voltage terminal, XL = 212 mΩ Part (c): (i) Power rating: 3 × 25 kVA = 75 kVA Voltage rating: √ 3 × 7.97 kV : 266 V = 13.8 kV : 266 V (ii) The per-unit impedance remains 0.075 per-unit (iii) Referred to the high-voltage terminal, XH = 191 Ω (iv) Referred to the low-voltage terminal, the base impedance is now Zbase,L = 2662/(75 × 103) = 0.943 Ω and thus XL = 0.943 × 0.075 = 70.8mΩ Problem 2-51 Part (a): 500 V at the high-voltage terminals is equal to 500/13.8 × 103 = 0.0362 per unit. Thus the per-unit short-circuit current will be Isc = 0.0363 0.075 = 0.48 perunit (i) The base current on the high-voltage side is Ibase,L = 75 × 103 √ 3 × 13.8 × 103 = 3.14 A c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 59 pf = P |S| = 0.92 and it is leading. c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 60 PROBLEM SOLUTIONS: Chapter 3 Problem 3-1 By analogy to Example 3.1, T = −2B0Rl [I1 sin α + I2 cosα] = −7.24 × 10−2 [I1 sinα + I2 cos α] N·m Thus Part (a): T = −0.362 cos α N·m Part (b): T = −0.362 sin α N·m Part (c): T = −0.579 (I1 sinα + I2 cos α) N·m Problem 3-2 T = −0.579 N · m Problem 3-3 Part (a): From the example, the magnetic flux density in each air gap is Bg = 0.65 T. Since the iron permeability is assumed to be infinite, all of the stored energy is in the air gap. The total air gap volume is 2gAg and thus the stored energy is found from Eq. 3.21 as Wfld = 2gAg ( B2 g 2µ0 ) = 67.2 J Part (b): (i) The winding inductance is given by L = µ0N 2Ag 2g = 1.26 H c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 61 (ii) λ = LI = 12.6 Wb (iii) From Eq. 3.19 Wfld = λ2 2L = 63 J Problem 3-4 Wfld = L(x)i2 2 Part (a): For i = 7.0 A and x = 1.30 mm, Wfld = 1.646 J. Part (b): For i = 7.0 A and x = 2.5 mm, Wfld = 1.112 J. Thus ∆Wfld = −0.534 J. Problem 3-5 Part (a): For x = x0, L = L0. Wfld = ( L0 2 ) I2 0 sin2 (ωt) and < Wfld >= L0I 2 0 4 = 0.858 J Part (b): < Pdiss >= RwI2 0 2 = 3.31 W c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 64 Part (c): Pdiss = i2L(t)R = I2 0R e−2(R/L)t Wdiss = ∫ ∞ 0 Pdissdt = 1 2 LI2 0 Problem 3-10 Part (a): The power dissipated is equal to Pdiss = I2R = 1.3 × 106 and the stored energy is W = 1 2 LI2 Given that τ = L/R = 4.8, we can find the stored energy as W = 1 2 τ Pdiss = 0.5 × (1.3 × 106) × 4.8 = 3.12 MJ Part (b): i(t) = I0(0.7 + 0.3e−t/τ) where I0 is the field current prior to reducing the terminal voltage. Thus, the stored energy is W (t) = 1 2 Li2(t) = W0(0.49 + 0.21e−t/τ + 0.09e−2t/τ) where W0 = 3.12 MJ. c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 65 Problem 3-11 Part (a): Wfld(λ, x) = λ2 2L(x) = ( λ2 2L0X 2 0 ) x2 Part (b): ffld(λ, x) = −∂Wfld ∂x ∣ ∣ ∣ ∣ λ = − ( λ2 L0X2 0 ) x Part (c): For i = I0, λ = L(x)I0 = L0I0 (x/X0)2 and thus from Part (b) ffld(x) = −L0I 2 0X2 0 x3 The force acts to decrease x. Problem 3-12 Part (a): Four poles Part (b): Tfld = I2 0 2 ( dL(θm) dθm ) = −L2I 2 0 sin 2θm c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 66 Problem 3-13 Part (a): Tfld = I2 0 2 ( dL(θm) dθm ) = −3L6I 2 0 sin 6θm Part (b): λ(t) = I0(L0 + L6 sin (6θm)) and thus v(t) = dλ(t) dt = 6I0L6 cos 6θm The power which must be supplied to the coil is p(t) = I0v(t) = 6I2 0L6 cos 6θm Problem 3-14 The coil inductance is equal to L = µ0N 2Ac/(2g) and hence the lifting force is equal to ffld = i2 2 dL dg = − ( µ0N 2Ac 4g2 ) i2 where the minus sign simply indicates that the force acts in the direction to reduce the gap (and hence lift the mass). The required force is equal to 118 N (the mass of the slab times the acceleration due to gravity, 9.8 m/sec2). Hence, setting g = gmim and solving for i gives imin = ( 2gmin N ) √ ffld µ0Ac = 223 mA and vmin = iminR = 0.51 V. c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 69 Problem 3-17 Part (a): πR2Bδ = 2πRhBg πR2Hδ = 2πRhHg δHδ + gHg = Ni Bδ = µ0Hδ = µ0N (δ + (R/(2h))g) λ = πR2NBδ = ( µ0πR2N2 (δ + (R/(2h))g) ) i L = λ i = µ0πR2N2 (δ + (R/(2h))g) Part (b): ffld = −∂Wfld ∂δ ∣ ∣ ∣ ∣ λ = ( λ2 2L2 ) dL dδ = ( i2 2 ) dL dδ Thus (i) ffld = − ( 1 2µ0πR2N2 ) λ2 and (ii) ffld = − ( µ0πR2N2 2(δ + (R/(2h))g))2 ) i2 Part (c): The net force includes both ffld and the spring force. c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 70 fnet = ffld(δ, I0) + K(δ0 − δ) Part (c): Stable equilibrium will occur at the position where fnet = 0 and dfnet dδ < 0; in this case at x = 4.25 mm. Problem 3-18 Part (a): L = µ0Ac(2N)2 2g = 2µ0AgN 2 g Part (b): ffld = i2 2 dL dg = −2.22 × 103 N P = |ffld| 2Ac = 1.15 × 106 N/m2 c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 71 Problem 3-19 Part (a): Part (b): From 2Ni = 2Hclc + 2Hgg Bc = Bg = B = µ0Hg we can solve for i as a function of Bc i = Hc(Bc)lc + Bcg/µ0 N where we have recognized that Hc can be expressed as a function of Bc via a spline fit with MATLAB. For an infinitely-permeable core B = µ0Ni g c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 74 This force is positive, acting to increase x and hence force the coil further into the slot. Part (d): ffld = 10.2 N/m. Problem 3-22 W ′ fld = ( µ0H 2 2 ) × coil volume = ( µ0πr2 0N 2 2h ) i2 Thus f = dW ′ fld dr0 = ( µ0πr0N 2 h ) I2 0 and hence the pressure is P = f 2πr0h = ( µ0N 2 2h2 ) I2 0 The pressure is positive and hence acts in such a direction as to increase the coil radius r0. Problem 3-23 Part (a): Wfld(q, x) = ∫ q 0 v(q′, x)dq′ Part (b): ffld = − ∂Wfld ∂x ∣ ∣ ∣ ∣ ∣ q c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 75 Part (c): W ′ fld = vq − dWfld ⇒ dW ′ fld = qdv + fflddx Thus W ′ fld = ∫ v 0 q(v′, x)dv′; ffld = ∂W ′ fld ∂x ∣ ∣ ∣ ∣ ∣ v Problem 3-24 Part (a): Wfld = ∫ q 0 v(q′, x)dq′ = q2 2C = xq2 2ε0A W ′ fld = ∫ v 0 q(v′, x)dv′ = Cv2 2 = ε0Av2 2x Part (b): ffld = ∂W ′ fld ∂x ∣ ∣ ∣ ∣ ∣ v = Cv2 2 = ε0Av2 2x2 and thus ffld(V0, δ) = ε0AV 2 0 2δ2 Problem 3-25 Part (a): Tfld = ( V 2 dc 2 ) dC dθ = ( Rd 2g ) V 2 dc c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 76 Part (b): In equilibrium, Tfld + Tspring = 0 and thus θ = θ0 + ( Rd 2gK ) V 2 dc Part (c): Problem 3-26 Part (a): L11 = µ0N 2 1 A 2g0 L22 = µ0N 2 2 A 2g0 Part (b): L12 = µ0N1N2A 2g0 Part (c): W ′ fld = 1 2 L11i 2 1 + 1 2 L22i 2 2 + L12i1i2 = µ0A 4g0 (N1i1 + N2i2) 2 Part (d): ffld = ∂W ′ fld ∂g0 ∣ ∣ ∣ ∣ ∣ i1,i2 = −µ0A 4g2 0 (N1i1 + N2i2) 2 c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 79 Problem 3-29 Part (a): Winding 1 produces a radial magnetic which, under the assumption that g << r0, Br,1 = µ0N1 g i1 The z-directed Lorentz force acting on coil 2 will be equal to the current in coil 2 multiplied by the radial field Br,1 and the length of coil 2. fz = 2πr0N2Br,1i2 = 2πr0µ0N1N2 g i1i2 Part (b): The self inductance of winding 1 can be easily written based upon the winding-1 flux density found in part (a) L11 = 2πr0lµ0N 2 1 g The radial magnetic flux produced by winding 2 can be found using Ampere’s law and is a function of z. Bz =       0 0 ≤ z ≤ x −µ0N2i2(z−x) gh x ≤ z ≤ x + h −µ0N2i2 g x + h ≤ z ≤ l Based upon this flux distribution, one can show that the self inductance of coil 2 is L22 = 2πr0µ0N 2 2 g ( l − x − 2h 3 ) Part (c): Based upon the flux distribution found in part (b), the mutual inductance can be shown to be L12 = 2πr0µ0N1N2 g ( x + h 2 − l ) c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 80 Part (d): ffld = d dx [ 1 2 L11i 2 1 + 1 2 L22i 2 2 + L12i1i2 ] = −πr0µ0N 2 2 g i22 + 2πr0µ0N1N2 g i1i2 Note that this force expression includes the Lorentz force of part (a) as well as a reluctance force due to the fact that the self inductance of coil 2 varies with position x. Substituting the given expressions for the coil currents gives: ffld = −πr0µ0N 2 2 g I2 2 cos2 ωt + 2πr0µ0N1N2 g I1I2 cos ωt Problem 3-30 The solution follows that of Example 3.8 with the exception of the magnet properties of samarium-cobalt replaced by those of neodymium-boron-iron for which µR = 1.06µ0, H ′ c = −940 kA/m and Br = 1.25 T. The result is ffld = { -203 N at x = 0 cm -151 N at x = 0.5 cm Problem 3-31 Part (a): Because there is a winding, we don’t need to employ a “fictitious” winding. Solving Hmd + Hgg0 = Ni; BmwD = Bg(h − x)D in combination with the constitutive laws Bm = µR(Hm −Hc); Bg = µ0Hg gives Bm = µ0(Ni + Hcd) dµ0 µR + wg0 (h−x) c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 81 Note that the flux in the magnetic circuit will be zero when the winding current is equal to I0 = −Hcd/N . Hence the coenergy can be found from integrating the flux linkage of the winding from an initial state where it is zero (i.e. with i = I0) to a final state where the current is equal to i. The flux linkages are given by λ = NwDBm and hence W ′ fld(i, x) = ∫ i I0 λ(i′, x)di′ = µ0wDN dµ0 µR + wg0 (h−x) [ Ni2 2 + Hc ( i + Hcd 2N )] The force is then ffld = dW ′ fld dx = −µ0w 2DNg0 (µ0d(h−x) µR + wg0)2 [ Ni2 2 + Hc ( i + Hcd 2N )] (i) for i = 0, ffld = dW ′ fld dx = −µ0w 2Dg0(Hcd)2 2(µ0d(h−x) µR + wg0)2 where the minus sign indicates that the force is acting upwards to support the mass against gravity. (ii) The maximum force occurs when x = h fmax = −µ0wD(Hcd)2 2 = −Mmaxa where a is the acceleration due to gravity. Thus Mmax = µ0wD(Hcd)2 2a Part (b): Want f(Imin,x=h = −a Mmax 2 = −µ0wD(Hcd)2 4 Substitution into the force expression of part (a) gives c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 84 where µR = 1.05µ0 and H ′ c = −712 kA/m. Solving gives Bg =    µ0R1(−Hctm) 2hx + gR1 + 2µ0R2 1 htm µR(R2 3 −R2 2 )    = 0.624 T and Bx = ( 2h R1 ) Bg = 0.595 T Part (b): We can replace the magnet by an equivalent winding of Ni = −Hctm. The flux linkages of this equivalent winding can then be found to be λ = N(2πR1h)Bg =    2πµ0hR2 1N 2 2hx + gR1 + 2µ0R2 1 htm µR(R2 3 −R2 2 )    i = Li The force can then be found as ffld = i2 2 dL dx = −2πµ0(hR1) 2(Ni)2 ( 2hx + gR1 + 2µ0R2 1 htm µR(R2 3 −R2 2 ) )2 = −2πµ0(hR1) 2(−Hctm)2 ( 2hx + gR1 + 2µ0R2 1 htm µR(R2 3 −R2 2 ) )2 = −195 N Part (c): c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 85 Part (d): The spring force will be of the form fspring = −K(x− X0) − ffld(X0) to guarantee that there will be equilibrium at x = X0. It also must be larger than the magnitude of the magnetic force at x = 0 (ffld(0) = −246 N). Thus, we must have K > ffld(X0) + |ffld(0)| X0 = −195 + 246 5 × 10−4 = 103 N/mm Problem 3-34 Part (a): If the plunger is stationary at x = 0.5X0, the inductance will be constant at L(0.5X0) = 0.5L0. Thus i(t) = V0 Rc (1 − e−t/τ) where τ = L(0.5X0)/Rc. The magnetic force will thus be ffld = i2 2 dL dx = − L0 2X0 ( V0 Rc )2 (1 − e−t/τ)2 A force of this magnitude must therefore be applied to maintain the plunger at this position. Part (b): The steady-state current will be equal to I0 = V0/Rc and since the force is independent of x we can write that at equilibrium fnet = 0 = K0(X0 − x0) − L0 2X0 ( V0 Rc )2 and thus x0 = 0.5X0 − L0 2K0X0 ( V0 Rc )2 c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 86 Problem 3-35 Part (a): Following the derivation of Example 3.1, for a rotor current of 8 A, the torque will be give by T = T0 sinα where T0 = −0.0048 N·m. The stable equilibrium position will be at α = 0. Part (b): J d2α dt2 = T0 sinα Part (c): The incremental equation of motion is J d2α dt2 = T0α and the natural frequency is ω = √ T0 J = 0.62 rad/sec corresponding to a frequency of 0.099 Hz. Problem 3-36 As long as the plunger remains within the core, the inductance is equal to L = µ0dπN2 ag ( ( a 2 )2 − x2 ) where x is the distance from the center of the solenoid to the center of the core. Hence the force is equal to ffld = i2 2 dL dx = −µ0dπN2i2x ag c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 89 Problem 3-38 c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 90 PROBLEM SOLUTIONS: Chapter 4 Problem 4-1 Part (a): ωm = rpm × π 30 = 125.7 rad/sec Part (b): f = rpm × poles 120 = 60 Hz (= 120π rad/sec) Part (c): The frequency will be 50 Hz if the speed is (5/6) × 1200 = 1000 rpm. Problem 4-2 (i) vb(t) = √ 2Va cos (ωt − 2π/3) vc(t) = √ 2Va cos (ωt + 2π/3) (ii) vab(t) = va(t) − vb(t) = √ 6Va cos (ωt + π/6) Problem 4-3 (i) With the wind turbine rotating at 0.5 r/sec = 30 rpm, the generator speed will be 300 rpm which will produce a line-line voltage of 480 × 300/900 = 160 V at a frequency of f = 300 × 8/120 = 20 Hz. (ii) At 1.75 r/sec = 105 rpm, the generator voltage will be 560 V line-line at a frequency of 70 Hz. c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 91 Problem 4-4 Part (a): Induction motor Part (b): 6 poles Problem 4-5 rpm = f × 120 poles = 6000 Problem 4-6 Part (a): Part (b): c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 94 Part (c): From Eq. 4.47 ΦP = ( 2 3 ) lRBag1,peak = 0.713 Wb Problem 4-12 From the solution to Problem 4-11, ΦP = 0.713 Wb. Vrms = ωNΦP√ 2 = 7.60 kV Problem 4-13 From the solution to Problem 4.9, ΦP = 0.713 Wb. Vrms = ωkwNaΦ√ 2 = 7.99 kV Problem 4-14 Part (a): Because this generator is ∆-connected, the rms phase voltage Vrms and line-line voltages are both equal to 575 V. Thus from Eq. 4.52 ΦP = √ 2 Vrms ωekwNph = 0.191 Wb and thus from Eq. 4-45 Bpeak = ( poles 2 ) ΦP 2lr = 1.39 T Part (b): From Eq. 4.45 If = πg × poles 4µ0kfNf = 20.1 A c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 95 Part (c): Operating at 50 Hz with the same air-gap flux (corresponding to a field current of 20.1 A), the open-circuit voltage will be (5/6) × 575 = 479 V. The armature turns must be increase from 12 to 18 turns/phase, which would produce a terminal voltage of 718 V. If the field current is then reduced to 20.1 × (690/718) = 19.3 A, the desired open-circuit voltage of 690 V will be achieved. Problem 4-15 Part (a): The air-gap flux density is plotted below and is of amplitude Bgap = µ0NfIf 2g = 1.16 T The peak fundamental amplitude is Bpeak,1 = ( 4 π ) Bgap = 1.47 T Part (b): In terms of the stator inner radius r, axial-length l and turns Nph, the peak flux linkage of the stator winding is equal to λpeak = πrlBgapNph and it varies as a saw-tooth function of time, giving rise to a square-wave of voltage, shown in the plot below, of magnitude Vpeak = 2λpeak T/2 = 476.7 V where T = 1/60 sec. The rms-fundamental component of this voltage is c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 96 V1,rms = 4 π Vpeak√ 2 = 477 V Problem 4-16 Fa = ia[A1 cos θa + A3 cos 3θa + A5 cos 5θa] = Ia cos ωt[A1 cos θa + A3 cos 3θa + A5 cos 5θa] Similarly, we can write Fb = ib[A1 cos (θa − 120◦) + A3 cos 3(θa − 120◦) + A5 cos 5(θa − 120◦)] = Ia cos (ωt − 120◦)[A1 cos (θa − 120◦) + A3 cos 3θa + A5 cos (5θa + 120◦)] and Fc = ic[A1 cos (θa + 120◦) + A3 cos 3(θa + 120◦) + A5 cos 5(θa + 120◦)] = Ia cos (ωt + 120◦)[A1 cos (θa + 120◦) + A3 cos 3θa + A5 cos (5θa − 120◦)] The total mmf will be Ftot = Fa + Fb + Fc = 3 2 Ia[A1 cos (θa − ωt)A5 cos (5θa + ωt)] = 3 2 Ia[A1 cos (θa − ωt)A5 cos 5 ( θa + ( ωt 5 ) ) ] c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.