PROBLEM SOLUTION KEY
Answers to selected problems (e.g., does not include graphs, diagrams, and proofs).
CHAPTER 1: ELECTRIC POWER SYSTEM
1.1. (a) Short circuit solely on transmission line T2 opens CBA 2 and 3. Supply S feeds
T1 through CBA 4, Bus 1 and CBA 1, and serves T3 directly through CBA 5. (b)
Short circuit solely on transmission line T3 activates CBA 6 and 5. Supply S feeds
T1 through CBA 4, Bus 1 and CBA 1, and serves T2 through CB4, Bus 1, and
CBA 1 and 2
1.2. (a) CBA 4 fails open. T3 is supplied directly through CBA 5; T2 is supplied through
CBA 5, 6, and 3; and T1 is supplied through CBA 5, 6, 3, and 2. It is evident that
the circuit breakers should be sized to pass all three load currents. In this case, the
full supply current passes through CBA 5. (b) CBA 4 fails closed. If CBA 4 cannot
be opened, then a short circuit on Bus 1 cannot be locally isolated since the faulty
CBA 4 directly connects the source to the shorted bus.
1.3. CBA 1, 2, 4, and 5 closed; CBA 3 and 6 open.
CHAPTER 2: ELECTRIC GENERATING STATIONS
2.1. 543 days
2.2. (a) CF1 = 70.8%, CF2 = 87.5%; (b) O&M2 = 1.2 ¢/kWh
2.3. Qcoal = 1125 MWt, Qnuclear = 2030 MWt
2.4. $20.5 million
2.5. ηorig = 66.3%, ηnew = 92.2%
2.6. Vout = 18 V, Iout = 3 A, Pout = 54 W
2.7. two derivations
2.8. (a) plot; (b) Pmax = 3.6 W, VPmax = 0.5 V
Electrical Energy Conversion and Transport: An Interactive Computer-Based Approach, Second Edition.
George G. Karady and Keith E. Holbert.
© 2013 Institute of Electrical and Electronics Engineers, Inc. Published 2013 by John Wiley & Sons, Inc.
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PROBLEM SOLUTION KEY
2.9. ηtrough = 40.2%, ηptower = 75.4%
2.10. η = 47.5%
2.11. (a) nC = 1.53 × 1018 atoms; (b) mC = 30.5 μg; (c) mC/mU-235 = 2.5 × 106
2.12. A = 0.028 km2
2.13. mSO2 = 4.21 × 107 kg/year, mash = 5.26 × 107 kg/year
2.14. P = 3.82 MW
2.15. (a) Ein = 1.6 × 105 MWh, Eout = 1.22 × 105 MWh; (b) ηta = 76%
CHAPTER 3: SINGLE-PHASE CIRCUITS
3.1. (a) T50 = 20 ms; (b) T60 = 16.7 ms
3.2. (a) VM = 169.7 V; (b) Vrms: {114–126 Vrms}
3.3. proof
3.4. phasor diagram
3.5. (a) Z50 = 50 + j29.7 Ω; (b) Z60 = 50 + j82.4 Ω
3.6. (a) Zeq = 32.6 + j38.2 Ω; (b) Yeq = 0.013 − j0.015 S
3.7. Ccap = 82.1 μF
3.8. V1 (P1 = −1.7 MW) operates as generator and V2 (P2 = 1.5 MW) serves as load
3.9. (a) P1 = 1.1 kW, P2 = 789 W, P3 = 302 W, P4 = 139 W; (b) Q1 = −1.76 kVA,
Q2 = 394 VA, Q3 = −302 VA, Q4 = 93 VA
3.10. (a) R = 64.3 Ω, X = 76.6 Ω; (b) P = 311 W, Q = 371 VA; (c) pf = 0.643 leading
3.11. (a) P1 = 512 W (absorbed), P2 = −1.45 kW (supplied); (b) Q1 = 1.76 kVA
(absorbed), Q2 = −1.65 kVA (supplied); (c) PZ1 = 53 W (absorbed), QZ2 = 26.5 VA
(absorbed), QZ3 = −132 VA (supplied), PZ4 = 883 W (absorbed); (d) Pfeeder
(180°) = 17.1 kW
3.12. (a) Qcap = −332 kVA; (b) pf = 0.908 (lagging)
3.13. (a) diagram; (b) Qcap = −46.4 kVA, Ccap = 2543 μF; (c) graph; (d) Cpf1 = 2137 μF
CHAPTER 4: THREE-PHASE CIRCUITS
4.1. proof
4.2. (a) IY = 12.9 A; (b) IΔphase = 7.47 A, IΔline = 12.9 A
4.3. Ia = 75.1 A ∠−35.5°, Ib = 82.8 A ∠−177°, Ic = 52 A ∠65.7°
4.4. Ia = 7.2 A ∠−11.9°, Ib = 7.6 A ∠−149°, Ic = 5.5 A ∠95.5°
4.5. (a) Z1Y = 5.33 − j2.58 Ω, Z2Δ = 19.6 − j12.1 Ω; (b) Iload = 72.8 A; (c) VLL_bus
= 597 V; (d) Pbus = 70 kW, Qbus = 27.5 kVA
4.6. (a) Iline = 7.11 A ∠−40.7°, Vload_ll = 205 V ∠29.3°; (b) Vload_ll = 215 V ∠26.3°
4.7. (a) diagram; (b) Zload1 = 25 + j23.4 Ω, Zload2 = 21.9 + j7.8 Ω; (c) Igen = 45.3 A = 97.99%
PROBLEM SOLUTION KEY
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4.8. (a) Iab = 33.3 A, Ibc = 48 A, Ica = 80 A, Ia = 84.6 A, Ib = 23.4, Ic = 69.7 A;
(b) Phases a and c are overloaded
4.9. (a) P3 = 13.3 kW, |S3| = 15.1 kVA, Q3 = −7.29 kVA; (b) pf3 = 0.876 (leading);
(c) R3 = 0.905 Ω
4.10. proof
CHAPTER 5: TRANSMISSION LINES AND CABLES
5.1. L = 0.487 mH/mile, X = 0.184 Ω/mile
5.2. C = 24.6 nF/mile, which is negligible
5.3. R = 0.119 Ω/mile, L = 2.14 mH/mile, C = 13.9 nF/mile
5.4. R = 0.0342 Ω/km, L = 0.927 mH/mile, C = 12.2 nF/km
5.5. (a) γline = 2.4 × 10−5 + 1.19 × 10−3/km, Zsurge = 316 − 6.3j Ω; (b) Zser = 4.3 + 110.7j
Ω, Zpar = 0.75 − 1759j Ω
5.6. Vsup = 14.8 kV, Reg = 7.0%
5.7. Vload = 12.9 kV, Reg = 7.4%
5.8. δ = 43.1°
5.9. (a) R = 0.0113 Ω/km, XL = 0.324 Ω/km; (b) C = 13.2 μF/km, YC = 4.96 μS/km;
(c) diagram; (d) Iload = 563 A, Vload = 164 kV
5.10. (a) Zline = 10.6 + j20.7 Ω; (b) (1) |VLN | = 13.1 kV, |I| = 257 A, P = 7.11 MW,
Q = 7.21 MVAR (ind); (2) |VLN | = 9.66 kV, |I| = 240 A, P = 6.84 MW,
Q = 1.31 MVAR (ind); (3) |VLN| = 10.93 kV, |I| = 219 A, P = 6.53 MW,
Q = 2.97 MVAR (ind); (c) plot
5.11. (a) R = 7.15 Ω, XL = 96.6 Ω, YC = 0.698j mS; (b) diagram; (c) ZTh = 8.38 + j119 Ω,
VTh = 315 kV; (d) plot; (e) Ishort = 2.65 kA
5.12. (a) R = 2.97 Ω, XL = 63.8 Ω, YC = 0.988j mS, Xsupply1 = 4.41 Ω, Xsupply2 = 2.78 Ω;
(b) diagram; (c) PmaxXfr = 779 MW at δmax = 92.4°
5.13. (a) R = 4.63 Ω, L = 0.148 H, C = 1.61 μF; (b) diagram; (c) |VLL_source| = 437 kV,
Ssup = 888 + j827 MVA; (d) |S10%| = 433 MVA
5.14. Hmax = 2.62 A/m
CHAPTER 6: ELECTROMECHANICAL ENERGY CONVERSION
6.1. ΦIronError = 0.015%, BIronError = 0.014%, HIronError = 0.006%, HGapError = 0.014%
6.2. (a) Hcore ≅ 100 A/m, μr = 9950; (b) Φcore = 2.45 mWb, Icoil = 0.224 A;
(c) Lcoil = 3.06 H; (d) Icoil2 = 28.6 A
6.3. (a) Φgap = 1.68 mWb; (b) Hgap = 540 kA/m, Bgap = 0.679 T; (c) Lcoil = 112 mH;
(d) Icoil = 4.66 A; (e) Lb = 108 mH
6.4. (a) drawing; (b) Hdc = 9.55 kA/m, Bdc = 0.012 T; (c) Fdc = 540 N
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PROBLEM SOLUTION KEY
6.5. (a) drawing; (b) Fa_max = 58 N, Fb_max = 72 N, Fc_max = 58 N; (c) phase B in middle
is most stressed
6.6. Fmiddle,max = 315 N, Fside,max = 157.5 N
6.7. Relations/plots where B(gmin) = 0.335 T, B(gmax) = 0.077 T and F(gmin) = 112 N,
F(gmax) = 5.95 N
6.8. Vcoil = 67 V
6.9. (a) Bgap = 0.377 T; (b) plot
6.10. (a) I(gapmax) = 2.8 A; (b) Fmax = 78.5 N; (c) plot, iron saturation does not occur
6.11. (a) ℓPmag = 0.331 cm; (b) IVcoil = 1.105 A
6.12. Fmag = 1026 N
6.13. (a) Bgap = 0.77 T; (b) Ecoil = 1.16 kV
CHAPTER 7: TRANSFORMERS
7.1. Vsup = 7.64 V
7.2. Vload = 226 V
7.3. ISC = 2.88 kA ∠−90°
7.4. (a) IAB = 65.1 A ∠−36.9°, IBC = 20.8 A ∠−161.4°, ICA = 122.4 A ∠88.2°; (b)
Ia = 4.3 A ∠−36.9°, Ib = 1.4 A ∠−161.4°, Ic = 8.2 A ∠88.2°; (c) Va = 8.0 kV,
Vb = 7.3 kV, Vc = 4.9 kV, Vab = 1.5 kV, Vbc = 2.5 kV, Vca = 3.2 kV; (d) IGND = 5.6 A
∠64.7°
7.5. ISC = 681 A
7.6. Vsource = 230 V
7.7. (a) Zhigh = 242 Ω, Zlow = 4.5 Ω; (b) Iload = 250 A, Vsup = 110 kV
7.8. (a) Zin = 0.646 + j1.684 Ω; (b) IS = 66.5 A ∠−35.5°, Vsup = 2.9 kV ∠1.3°,
Pin = 154 kW, pfin = 0.8 (lagging); (c) P5% = 216 kW
7.9. (a) Reg = 9.9%; (b) plot; (c) εmax(pf = 0.8 lag) = 92.9%, εmax(pf = 0.9 lead) =
94.1%
7.10. (a) Re,s = 0.013 Ω, Xe,s = 0.087 Ω, Rc,p = 2.6 kΩ, Xm,p = 700 Ω; (b) and (c) plots;
(d) explanation
7.11. (a) Re = 0.26 Ω, Xe = 0.83 Ω, Rc = 220 Ω, Xm = 22.4 Ω; (b) η0.6lag,max = 98.8%,
η0.8lag,max = 99.1%, η0.9lag,max = 99.2%, η0.6lead,max = 98.9%, η0.8lead,max = 99.2%,
η0.9lead,max = 99.3%; (c) Vlag = 219 V, Vlead = 221 V
7.12. (a) diagram with wye-connected primary and delta-connected secondary;
(b) Ipri,ph,a = 196 A ∠−31.8°, Isec,ph,ab = 1.63 kA ∠−31.8°, Isec,ll,a = 2.83 kA ∠−1.8°;
(c) Vpri,ph,a = 20 kV ∠0°, Vpri,ll,ab = 34.6 kV ∠30°, Vsec,ll,ab = 2.4 kV ∠0°
7.13. (a) η = 97.82%; (b) η = 97.81% at 85.7% load
7.14. (a) diagram; (b) Xtr = 41 Ω, Rct = 6.85 Ω, Xmt = 5.1 Ω; (c) Imag = 17.6 A ∠−53.1°;
(d) Iload = 126 A ∠−33.3°, Vload = 120 V ∠−3.6°
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PROBLEM SOLUTION KEY
9.13. (a) diagrams; (b) s = 6.7%, pf = 0.922 (lagging); (c) Tout = 118 N·m; (d) ε = 81.9%;
(e) Istart = 126 A, Tstart = 76 N·m; (f) Tmax = 175 N·m, sTmax = 18.5%
9.14. (a) Rsta = 0.182 Ω, Xsta = 1.008 Ω = Xrot_t, Rc = 271 Ω, Xm = 75.7 Ω, Rrot_t = 0.274 Ω;
(b) Imot(nm) = 28.2 A, Irot(nm) = 26.7 A; (c) Tout(nm) = 112 N·m; (d) nop = 1681 rpm
9.15. (a) Im_rated = 5.18 A; (b) nrated = 1105 rpm; (c) npf0.6 = 1138 rpm; (d) nm0.5 = 1141 rpm;
(e) Tmax = 3.8 N·m, nTmax = 899 rpm
9.16. (a) Ista = 5.45 A, Ifor = 1.62 A, Irev = 5.08 A; (b) Pout = 107 W, Pin = 474 W;
(c) ε = 22.5%
CHAPTER 10: DC MACHINES
10.1. proof
10.2. Pout = 3.88 hp
10.3. Pout = 1.64 hp
10.4. Km = 0.633 V·s/A
10.5. Pin = 0.986 hp
10.6. nmot = 649 rpm
10.7. nmot = 86 rpm
10.8. Vf = 258 V
10.9. Vsup = 108 V, Tstart = 1126 N·m; motor can start the pump
10.10. (a) Km = 0.837 V·s/A; (b) nnl = 1367 rpm; (c) I99% = 24.7 A
10.11. (a) Km = 0.06 V·s/A; (b) n = 858 rpm, T = 96.5 N·m; (c) I = 18.2 A, n = 1938 rpm
10.12. (a) diagram; (b) Km = 0.651 V·s/A; (c) Tstart = 3175 rpm, n = 1241 rpm
10.13. (a) If = 2.14 A, Ia = 347 A, Iload = 345 A; (b) Vt(0) = 277 V, Vt(P) = 246 V;
(c) Reg = 12.6%; (d) P5% = 40.6 kW
10.14. (a) Km = 7.86 V·s/A; (b) Iload = 149 A, Ea = 393 V, n = 672 rpm, (c) Pin = 56.8 kW,
Pout = 75.8 hp, ε = 99.6%
10.15. derivation
CHAPTER 11: INTRODUCTION TO POWER ELECTRONICS AND
MOTOR CONTROL
11.1. (a) and (b) a1 = VM, otherwise an = 0 bn; (c) a1 (cosine) = b1 (sine)
11.2. Ifw(0) = 21.6 A, Ifw(1) = 0, Ifw(2) = 14.4 A, Ifw(3) = 0, Ifw(4) = 2.88 A, Ifw(5) = 0,
Ifw(6) = 14.4 A, . . .
11.3. PSpice solution
11.4. α70% = 98.7°
11.5. (a) R = 7.2 Ω; (b) plots; (c) THD = 63.6%
PROBLEM SOLUTION KEY
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CHAPTER 8: SYNCHRONOUS MACHINES
8.1. 1200 rpm
8.2. 750 rpm
8.3. proof that Σ Sk = 0 at each node
8.4. Egen = 23.4 + j14.1 kV, Sgen = 125 + j268 MVA
8.5. Isc = 5.87 kA
8.6. Egen = 35 kV
8.7. δ = 41.9°
8.8. Pnet = 373 MVA, Qnet = −1.49 MVA
8.9. (a) pf = 0.814; (b) ELL = 19 kV
8.10. (a) Erload = 51.9 kV, Esc = 1.27 kV; (b) Isc(0 MW) = 97.7 A, Isc(Pload) = 4.0 kA
8.11. (a) ns = 3600 rpm, ωm = 377 r/s; (b) Φdc = 2.5 Wb, Bdc = 0.056 T, Hdc = 44.5 kA/m;
(c) Idc = 16.3 A; (d) Xsyn = 2.6 kΩ; (e) diagram
8.12. (a) Zline = 3.15 + j22.5 Ω, Xg = 1.53 Ω, Xxfmr = 36.1 Ω; (b) Egen_ll = 41.6 kV ∠63.9°;
(c) δ350 = 50.3°
8.13. (a) Xg = 0.722 Ω, Xxfmr = 9.68 Ω, Zline = 8.4 + j60 Ω, Xc = 1.77 kΩ; (b)
Eg_ph = 24 kV ∠33.9°; (c) C950 = 206 μF
8.14. For 0°, TD = 1/60 s and Ipeak ≅ 15 kA
8.15. For PW = 40 ms, Ishort ≅ 1.7 kA and VCB,peak ≅ 1.7 MV
CHAPTER 9: INDUCTION MACHINES
9.1. (a) 16 slots/phase; (b) 30° between phases
9.2. Smotor = 19.4 + j17.7 kVA, Imotor = 31.6 A
9.3. Ccap = 155 μF
9.4. IMotorStart = 356 A
9.5. IMotorIn = 167 A
9.6. Vsup = 141 V, Imotor = 163 A
9.7. (a) ns = 3600 rpm; (b) two poles; (c) frotor = 1.5 Hz; (d) Irate = 314 A
9.8. (a) diagram; (b) nm = 878 rpm; (c) Iin(2.5%) = 18.7 A ∠−31.1°, pfin(2.5%) = 0.856
(lagging); (d) nT150 = 827 rpm; (e) ηmax = 81.8%, nηmax = 867 rpm
9.9. (a) nm = 1692 rpm; (b) PAirGap = 40.1 kW, Tdev = 213 N·m, Tload = 210 N·m
9.10. (a) Rsta = 1.4 Ω, Xsta = 131 Ω = Xrot_t, Rc = 1225 Ω, Xm = 247 Ω, Rrot_t = 1.865 Ω;
(b) diagram; (c) Tout(10%) = 3.93 N·m, Pout(10%) = 666 W, ε(10%) = 46.3%
9.11. (a) Pout = 6 hp; (b) Pag = 5.37 kW; (c) n = 3265 rpm; (d) Tout = 13.1 N·m
9.12. (a) Im = 33.6 A, pf = 0.784 (lagging), Vt = 434 V; (b) Pin = 20.3 kW, Qin = 16.1 kVA;
(c) Pag = 18.7 kW; (d) Pdev = 17.8 kW, Tdev = 99.3 N·m; (e) Pout = 23/65 hp,
Tout = 98.5 N·m; (f) ε = 87%
PROBLEM SOLUTION KEY
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11.6. (a) THD increases with delay angle; (b) Irms decreases with α
11.7. vC(t) = 5.4 ± 0.14 V, iL(t) = 250 ± 42 mA
11.8. (a) Ton = 5 μs; (b) L = 0.2 mH, C = 1.67 μF; (c) PSpice plots
11.9. (a) delay angle control is feasible; (b) α = 32.2°; (c) Pac = 3.66 kW, pfac = 0.97
11.10. (a) Vdc = 350 V, Pdc = 15.7 kW; (b) Vac_rms = 480 V if αr = 36.0°; (c) THD = 47.3%;
Ih1 = 40.5 A
11.11. (a) α = 178.5°; (b) derivation and plot; (c) Pxfr = 64.8 kW, pf = 0.90
11.12. (a) Km = 0.393 V·s/A; (b) Vmot = 376 V; (c) α = 24.9°
11.13. (a) finv = 36.7 Hz; (b) Vmot = 269 V, (c) plot; (d) nmot = 988 rpm
11.14. (a) Iavg,load ≅ 4 A for Lsm = 10 μH; (b) Iavg,load ≅ 3 A for Lsm = 200 mH
11.15. (a) Iavg,load ≅ 2.25 A; (b) Iavg,load ≅ 3.4 A for Csm = 700 μF