Solutions Manual for Power System Analysis and Design 6th Edition by Glover IBSN 9781305632134 Full Download: http://downloadlink.org/product/solutions-manual-for-power-system-analysis-and-design-6th-edition-by-glover-ibs Chapter 2 Fundamentals ANSWERS TO MULTIPLE-CHOICE TYPE QUESTIONS 2.1 b 2.19 a 2.2 a 2.20 A. c 2.3 c B. a 2.4 a C. b 2.5 b 2.21 a 2.6 c 2.22 a 2.7 a 2.23 b 2.8 c 2.24 a 2.9 a 2.25 a 2.10 c 2.26 b 2.11 a 2.27 a 2.12 b 2.28 b 2.13 b 2.29 a 2.14 c 2.30 (i) c 2.15 a (ii) b 2.16 b (iii) a 2.17 A. a (iv) d B. b 2.31 a C. a 2.32 a 2.18 c 1 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Full all chapters instant download please go to Solutions Manual, Test Bank site: downloadlink.org 2.1 (a) A1 = 6∠30° = 6 [ cos30° + j sin 30°] = 5.20 + j 3 5 = 6.40 ∠128.66° = 6.40e j128.66° −4 (c) A3 = ( 5.20 + j 3) + ( −4 + j 5 ) = 1.20 + j 8 = 8.01∠81.50° (b) A2 = −4 + j 5 = 16 + 25 ∠ tan −1 (d) A4 = ( 6∠30° )( 6.40 ∠128.66° ) = 38.414∠158.658° = −35.78 + j13.98 (e) A5 = ( 6∠30° ) / ( 6.40∠ − 128.66° ) = 0.94 ∠158.66° = 0.94e j158.66° 2.2 (a) I = 500∠ − 30° = 433.01 − j 250 (b) i(t ) = 4sin (ω t + 30° ) = 4cos (ω t + 30° − 90° ) = 4cos ( ω t − 60° ) I = ( 4 ) ∠ − 60° = 2.83∠ − 60° = 1.42 − j 2.45 ( ) (c) I = 5 / 2 ∠ − 15° + 4∠ − 60° = ( 3.42 − j 0.92 ) + ( 2 − j 3.46 ) = 5.42 − j 4.38 = 6.964∠ − 38.94° 2.3 (a) Vmax = 400 V; I max = 100 A (b) V = 400 2 = 282.84 V; I = 100 2 = 70.71A (c) V = 282.84∠30° V; I = 70.71∠ − 80° A 2.4 (a) I1 = 10∠0° − j6 6∠ − 90° = 10 = 7.5∠ − 90° A 8 + j6 − j6 8 I 2 = I − I1 = 10∠0° − 7.3∠ − 90° = 10 + j 7.5 = 12.5∠36.87° A V = I 2 ( − j 6 ) = (12.5∠36.87° ) ( 6∠ − 90° ) = 75∠ − 53.13° V (b) 2.5 (a) υ (t ) = 277 2 cos (ω t + 30° ) = 391.7cos (ω t + 30° ) V (b) I = V / 20 = 13.85∠30° A i(t ) = 19.58cos (ω t + 30° ) A (c) Z = jω L = j ( 2π 60 ) (10 × 10 −3 ) = 3.771∠90° Ω I = V Z = ( 277 ∠30° ) ( 3.771 ∠90° ) = 73.46 ∠ − 60° A i(t ) = 73.46 2 cos (ω t − 60° ) = 103.9cos (ω t − 60° ) A 2 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. (d) Z = − j 25 Ω I = V Z = ( 277∠30° ) ( 25∠ − 90° ) = 11.08∠120° A i(t ) = 11.08 2 cos (ω t + 120° ) = 15.67cos (ω t + 120° ) A 2.6 ( (a) V = 75 ) 2 ∠ − 15° = 53.03∠ − 15° ; ω does not appear in the answer. (b) υ (t ) = 50 2 cos (ω t + 10° ) ; with ω = 377, υ (t ) = 70.71cos ( 377t + 10° ) (c) A = A∠α ; B = B∠β ; C = A + B c(t ) = a(t ) + b(t ) = 2 Re Ce jωt The resultant has the same frequency ω. 2.7 (a) The circuit diagram is shown below: (b) Z = 3 + j8 − j 4 = 3 + j 4 = 5∠53.1° Ω (c) I = (100∠0° ) ( 5∠53.1° ) = 20∠ − 53.1° A The current lags the source voltage by 53.1° Power Factor = cos53.1° = 0.6 Lagging 2.8 Z LT = j ( 377 ) ( 30.6 × 10 −6 ) = j11.536 m Ω Z LL = j ( 377 ) ( 5 × 10 −3 ) = j1.885 Ω ZC = − j V= 1 = − j 2.88 Ω ( 377 ) ( 921 × 10−6 ) 120 2 2 ∠ − 30° V 3 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. The circuit transformed to phasor domain is shown below: 2.9 KVL : 120∠0° = ( 60∠0° )( 0.1 + j 0.5 ) + VLOAD ∴ VLOAD = 120∠0° − ( 60∠0° )( 0.1 + j 0.5 ) = 114.1 − j 30.0 = 117.9∠ − 14.7° V ← 2.10 (a) p(t ) = υ (t )i (t ) = 400cos (ω t + 30° ) 100cos (ω t − 80° ) 1 ( 400 )(100 ) cos110° + cos ( 2ω t − 50°) 2 = −6840.4 + 2 × 104 cos ( 2ω t − 50° ) W = (b) P = VI cos ( δ − β ) = ( 282.84 )( 70.71) cos ( 30° + 80° ) = −6840 W Absorbed = +6840 W Delivered (c) Q = VI sin (δ − β ) = ( 282.84 )( 70.71) sin110° = 18.79 kVAR Absorbed (d) The phasor current ( − I ) = 70.71∠ − 80° + 180° = 70.71 ∠100° A leaves the positive terminal of the generator. The generator power factor is then cos ( 30° − 100° ) = 0.3420 leading 2.11 (a) p(t ) = υ (t )i(t ) = 391.7 × 19.58cos2 (ω t + 30° ) 1 = 0.7669 × 10 4 1 + cos ( 2ω t + 60° ) 2 = 3.834 × 103 + 3.834 × 103 cos ( 2ω t + 60° ) W P = VI cos (δ − β ) = 277 × 13.85cos0° = 3.836 kW Q = VI sin (δ − β ) = 0 VAR Source Power Factor = cos (δ − β ) = cos ( 30° − 30° ) = 1.0 (b) p(t ) = υ (t )i(t ) = 391.7 × 103.9cos (ω t + 30° ) cos (ω t − 60° ) 1 = 4.07 × 10 4 cos90° + cos ( 2ω t − 30° ) 2 4 = 2.035 × 10 cos ( 2ω t − 30° ) W P = VI cos (δ − β ) = 277 × 73.46 cos ( 30° + 60° ) = 0 W Q = VI sin (δ − β ) = 277 × 73.46 sin 90° = 20.35 kVAR pf = cos (δ − β ) = 0 Lagging 4 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. (c) p(t ) = υ (t )i(t ) = 391.7 × 15.67 cos (ω t + 30° ) cos (ω t + 120° ) 1 = 6.138 × 103 cos ( −90° ) + cos ( 2ω t + 150° ) = 3.069 × 103 cos ( 2ω t + 150° ) W 2 P = VI cos (δ − β ) = 277 × 11.08cos ( 30° − 120° ) = 0 W Q = VI sin (δ − β ) = 277 × 11.08sin ( −90° ) = −3.069 kVAR Absorbed = +3.069 kVAR Delivered pf = cos (δ − β ) = cos ( −90° ) = 0 Leading 2.12 (a) pR (t ) = ( 359.3cos ω t )( 35.93cos ω t ) = 6455 + 6455cos2ω t W (b) px (t ) = ( 359.3cos ω t ) 14.37cos (ω t + 90° ) = 2582 cos ( 2 cot + 90° ) = −2582sin 2ω t W 2) ( X = ( 359.3 2 ) 2 (c) P = V 2 R = 359.3 (d) Q = V 2 2 10 = 6455 W Absorbed 25 = 2582 VAR S Delivered (e) ( β − δ ) = tan −1 ( Q / P ) = tan −1 ( 2582 6455 ) = 21.8° Power factor = cos (δ − β ) = cos ( 21.8° ) = 0.9285 Leading 2.13 Z = R − jxc = 10 − j 25 = 26.93 ∠ − 68.2° Ω i(t ) = ( 359.3 / 26.93 ) cos (ω t + 68.2° ) = 13.34 cos (ω t + 68.2° ) A (a) pR (t ) = 13.34 cos (ω t + 68.2° ) 133.4 cos (ω t + 68.2° ) = 889.8 + 889.8cos 2 (ω t + 68.2° ) W (b) px (t ) = 13.34 cos (ω t + 68.2° ) 333.5cos (ω t + 68.2° − 90° ) = 2224sin 2 (ω t + 68.2° ) W 2 ) 10 = 889.8 W ( (d) Q = I X = (13.34 2 ) 25 = 2224 VAR S (c) P = I 2 R = 13.34 2 2 2 (e) pf = cos tan −1 ( Q / P ) = cos tan −1 (2224 / 889.8) = 0.3714 Leading 5 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 2.14 (a) I = 2∠0° kA V = Z I = ( 3∠ − 45° )( 2∠0° ) = 6∠ − 45° kV υ (t ) = 6 2 cos ( ω t − 45° ) kV p(t ) = υ (t )i (t ) = 6 2 cos (ω t − 45° ) 2 2 cos ω t 1 = 24 cos ( −45° ) + cos ( 2ω t − 45° ) 2 = 8.49 + 12cos ( 2ω t − 45° ) MW (b) P = VI cos ( δ − β ) = 6 × 2cos ( −45° − 0° ) = 8.49 MW Delivered (c) Q = VI sin (δ − β ) = 6 × 2sin ( −45° − 0° ) = −8.49 MVAR Delivered = + 8.49 MVAR Absorbed (d) pf = cos (δ − β ) = cos ( −45° − 0° ) = 0.707 Leading ( 2.15 (a) I = 4 ) 2 ∠60° ( 2∠30°) = 2 ∠30° A i(t ) = 2 cos (ω t + 30° ) A with ω = 377 rad/s p(t ) = υ (t )i(t ) = 4 cos30° + cos ( 2ω t + 90° ) = 3.46 + 4 cos ( 2ω t + 90° ) W (b) υ(t), i(t), and p(t) are plotted below. (c) The instantaneous power has an average value of 3.46 W, and the frequency is twice that of the voltage or current. 6 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 2.16 (a) Z = 10 + j 120 π × 0.04 = 10 + j15.1 = 18.1∠56.4° Ω pf = cos56.4° = 0.553 Lagging (b) V = 120 ∠0° V The current supplied by the source is I = (120 ∠0° ) (18.1∠56.4° ) = 6.63∠ − 56.4° A The real power absorbed by the load is given by P = 120 × 6.63 × cos56.4° = 440 W which can be checked by I 2 R = ( 6.63 ) 10 = 440 W 2 The reactive power absorbed by the load is Q = 120 × 6.63 × sin 36.4° = 663VAR (c) Peak Magnetic Energy = W = LI 2 = 0.04 ( 6.63 ) = 1.76 J 2 Q = ωW = 377 × 1.76 = 663VAR is satisfied. 2.17 (a) S = V I * = Z I I * = Z I 2 = jω LI 2 Q = Im[ S ] = ω LI 2 ← di = − 2ω L I sin (ω t + θ ) dt p(t ) = υ (t ) ⋅ i(t ) = −2ω L I 2 sin (ω t + θ ) cos (ω t + θ ) (b) υ (t ) = L = −ω L I 2 sin 2 (ω t + θ ) ← = − Q sin 2 (ω t + θ ) ← Average real power P supplied to the inductor = 0 ← Instantaneous power supplied (to sustain the changing energy in the magnetic field) has a maximum value of Q. ← 2.18 (a) S = V I * = Z I I * = Re Z I 2 + j Im Z I 2 = P + jQ ∴P = Z I 2 cos ∠Z ; Q = Z I 2 sin ∠Z ← (b) Choosing i(t ) = 2 I cos ω t , Then υ (t ) = 2 Z I cos (ω t + ∠Z ) ∴ p(t ) = υ (t ) ⋅ i(t ) = Z I 2 cos (ω t + ∠Z ) ⋅ cos ω t = Z I 2 cos ∠Z + cos ( 2ω t + ∠Z ) = Z I 2 [ cos ∠Z + cos2ω t cos ∠Z − sin 2ω t sin ∠Z ] = P (1 + cos2ω t ) − Q sin 2ω t ← 7 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 1 jωC (c) Z = R + jω L + From part (a), P = RI 2 and Q = QL + QC 1 2 I ωC which are the reactive powers into L and C, respectively. where QL = ω LI 2 and QC = − Thus p(t ) = P (1 + cos2ω t ) − QL sin 2ω t − QC sin 2ω t ← If ω 2 LC = 1, QL + QC = Q = 0 ← p(t ) = P (1 + cos2ω t ) Then * 150 5 2.19 (a) S = V I * = ∠10° ∠ − 50° = 375 ∠60° 2 2 = 187.5 + j 324.8 P = Re S = 187.5 W Absorbed Q = Im S = 324.8 VAR SAbsorbed (b) pf = cos ( 60° ) = 0.5 Lagging (c) QS = P tan QS = 187.5 tan cos−1 0.9 = 90.81VAR S QC = QL − QS = 324.8 − 90.81 = 234 VAR S 2.20 Y1 = 1 1 = = 0.05∠ − 30° = ( 0.0433 − j 0.025 ) S = G1 − jB1 Z1 20∠30° Y2 = 1 1 = = 0.04∠ − 60° = ( 0.02 − j 0.03464 ) S = G2 + jB2 Z 2 25∠60° P1 = V 2 G1 = (100 ) 0.0433 = 433 W Absorbed 2 Q1 = V 2 B1 = (100 ) 0.025 = 250 VAR S Absorbed 2 P2 = V 2 G2 = (100 ) 0.02 = 200 W Absorbed 2 Q2 = V 2 B2 = (100 ) 0.03464 = 346.4 VAR SAbsorbed 2 8 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 2.21 (a) φL = cos−1 0.6 = 53.13° QL = P tan φL = 500 tan 53.13° = 666.7 kVAR φS = cos−1 0.9 = 25.84° QS = P tan φS = 500 tan 25.84° = 242.2 kVAR QC = QL − QS = 666.7 − 242.2 = 424.5 kVAR SC = QC = 424.5 kVA (b) The Synchronous motor absorbs Pm = ( 500 ) 0.746 = 414.4 kW and Q 0.9 m = 0 kVAR Source PF = cos tan −1 ( 666.7 914.4 ) = 0.808 Lagging 2.22 (a) Y1 = 1 1 1 = = = 0.16∠ − 51.34° Z1 ( 4 + j 5 ) 6.4∠51.34° = ( 0.1 − j 0.12 ) S Y2 = 1 1 = = 0.1S Z 2 10 P = V 2 ( G1 + G2 ) ⇒ V = P = G1 + G2 1000 = 70.71 V ( 0.1 + 0.1) P1 = V 2 G1 = ( 70.71) 0.1 = 500 W 2 P2 = V 2 G2 = ( 70.71) 0.1 = 500 W 2 (b) Yeq = Y1 + Y2 = ( 0.1 − j 0.12 ) + 0.1 = 0.2 − j 0.12 = 0.233∠ − 30.96° S I S = V Yeq = 70.71( 0.233) = 16.48A 9 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 2.23 S = V I * = (120∠0° )(15∠ − 30° ) = 1800∠ − 30° = 1558.85 − j 900 P = Re S = 1558.85 W Delivered Q = Im S = −900 VAR S Delivered = +900 VAR SAbsorbed 2.24 S1 = P1 + jQ1 = 10 + j 0; S2 = 10∠ cos−1 0.9 = 9 + j 4.359 10 × 0.746 ∠ − cos−1 0.95 = 9.238∠ − 18.19° = 8.776 − j 2.885 0.85 × 0.95 SS = S1 + S2 + S3 = 27.78 + j1.474 = 27.82 ∠3.04° S3 = PS = Re(SS ) = 27.78 kW QS = Im(SS ) = 1.474 kVAR SS = SS = 27.82 kVA 2.25 SR = VR I * = RI I * = I 2 R = (20)2 3 = 1200 + j 0 SL = VL I * = ( jX L I )I * = jX L I 2 = j8(20)2 = 0 + j 3200 SC = VC I * = (− jIXC )I * = − jX C I 2 = − j 4(20)2 = 0 − j1600 Complex power absorbed by the total load SLOAD = SR + SL + SC = 2000∠53.1° Power Triangle: Complex power delivered by the source is SSOURCE = V I * = (100 ∠0° )( 20∠ − 53.1° ) = 2000∠53.1° * The complex power delivered by the source is equal to the total complex power absorbed by the load. 2.26 (a) The problem is modeled as shown in figure below: PL = 120 kW pfL = 0.85 Lagging θ L = cos−1 0.85 = 31.79° Power triangle for the load: 10 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. QL = PL tan ( 31.79° ) SL = PL + jQL = 141.18∠31.79° kVA = 74.364 kVAR I = SL / V = 141,180 / 480 = 294.13A Real power loss in the line is zero. Reactive power loss in the line is QLINE = I 2 X LINE = ( 294.13 ) 1 2 = 86.512 kVAR ∴ SS = PS + jQS = 120 + j ( 74.364 + 86.512 ) = 200.7∠53.28° kVA The input voltage is given by VS = SS / I = 682.4 V (rms) The power factor at the input is cos53.28° = 0.6 Lagging (b) Applying KVL, VS = 480 ∠0° + j1.0 ( 294.13∠ − 31.79° ) = 635 + j 250 = 682.4∠21.5° V (rms) ( pf )S = cos ( 21.5° + 31.79° ) = 0.6 Lagging 2.27 The circuit diagram is shown below: Pold = 50 kW; cos−1 0.8 = 36.87° ; θOLD = 36.87°; Qold = Pold tan (θ old ) = 37.5 kVAR ∴ Sold = 50,000 + j 37,500 θ new = cos−1 0.95 = 18.19°; Snew = 50,000 + j 50,000 tan (18.19° ) = 50,000 + j16, 430 Hence Scap = Snew − Sold = − j 21,070 VA ∴C = 21,070 ( 377 )( 220 ) 2 = 1155µ F ← 11 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 2.28 S1 = 15 + j 6.667 S 2 = 3 ( 0.96 ) − j 3 sin ( cos −1 0.96 ) = 2.88 − j 0.84 S3 = 15 + j 0 STOTAL = S1 + S2 + S3 = ( 32.88 + j 5.827 ) kVA (i) Let Z be the impedance of a series combination of R and X * V V2 Since S = V I = V = * , it follows that Z Z * ( 240 ) V2 Z = = = (1.698 − j 0.301) Ω S ( 32.88 + j 5.827 )103 2 * ∴ Z = (1.698 + j 0.301) Ω ← (ii) Let Z be the impedance of a parallel combination of R and X ( 240 ) = 1.7518 Ω ( 32.88)103 2 240 ) ( X= = 9.885 Ω ( 5.827 )103 2 R= Then ∴ Z = (1.7518 j 9.885 ) Ω ← 2.29 Since complex powers satisfy KCL at each bus, it follows that S13 = (1 + j1) − (1 − j1) − ( 0.4 + j 0.2 ) = −0.4 + j1.8 ← S31 = − S13* = 0.4 + j1.8 ← Similarly, S23 = ( 0.5 + j 0.5 ) − (1 + j1) − ( −0.4 + j 0.2 ) = −0.1 − j 0.7 ← S32 = − S23* = 0.1 − j 0.7 ← At Bus 3, SG 3 = S31 + S32 = ( 0.4 + j1.8 ) + ( 0.1 − j 0.7 ) = 0.5 + j1.1 ← 2.30 (a) For load 1: θ1 = cos−1 (0.28) = 73.74° Lagging S1 = 125∠73.74° = 35 + j120 S2 = 10 − j 40 S3 = 15 + j 0 STOTAL = S1 + S2 + S3 = 60 + j80 = 100∠53.13° kVA = P + jQ ∴ PTOTAL = 60 kW; QTOTAL = 80 kVAR; kVA TOTAL = STOTAL = 100 kVA. ← Supply pf = cos ( 53.13° ) = 0.6 Lagging ← S * 100 × 103 ∠ − 53.13° = = 100∠ − 53.13° A V* 1000∠0° At the new pf of 0.8 lagging, PTOTAL of 60kW results in the new reactive power Q′ , such that (b) ITOTAL = 12 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. θ ′ = cos−1 ( 0.8 ) = 36.87° and Q′ = 60 tan ( 36.87° ) = 45 kVAR ∴ The required capacitor’s kVAR is QC = 80 − 45 = 35 kVAR ← V 2 (1000 ) It follows then XC = * = = − j 28.57 Ω SC j 35000 2 and C= 106 = 92.85µ F ← 2π ( 60 )( 28.57 ) S ′* 60,000 − j 45,000 = = 60 − j 45 = 75∠ − 36.87° A V* 1000∠0° The supply current, in magnitude, is reduced from 100A to 75A ← The new current is I ′ = 2.31 (a) I12 = V1∠δ1 − V2 ∠δ 2 V1 V = ∠δ1 − 90° − 2 ∠δ 2 − 90° X ∠90° X X V V Complex power S12 = V1 I12* = V1∠δ1 1 ∠90° − δ1 − 2 ∠90° − δ 2 X X V12 V1V2 = ∠90° − ∠90° + δ1 − δ 2 X X ∴ The real and reactive power at the sending end are P12 = Q12 = V12 VV cos90° − 1 2 cos ( 90° + δ1 − δ 2 ) X X V1V2 = sin (δ1 − δ 2 ) ← X V12 VV sin 90° − 1 2 sin ( 90° + δ1 − δ 2 ) X X V = 1 V1 − V2 cos (δ1 − δ 2 ) ← X Note: If V1 leads V2 , δ = δ1 − δ 2 is positive and the real power flows from node 1 to node 2. If V1 Lags V2 , δ is negative and power flows from node 2 to node 1. (b) Maximum power transfer occurs when δ = 90° = δ1 − δ 2 ← PMAX = V1V2 ← X 2.32 4 Mvar minimizes the real power line losses, while 4.5 Mvar minimizes the MVA power flow into the feeder. 13 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 2.33 Qcap Mvar Losses 0 0.42 0.84 0.5 0.4 0.8 1 0.383 0.766 1.5 0.369 0.738 2 0.357 0.714 2.5 0.348 0.696 3 0.341 0.682 3.5 0.337 0.675 4 0.336 0.672 4.5 0.337 0.675 5 0.341 0.682 5.5 0.348 0.696 6 0.357 0.714 6.5 0.369 0.738 7 0.383 0.766 7.5 0.4 0.801 8 0.42 0.84 8.5 0.442 0.885 9 0.467 0.934 9.5 0.495 0.99 0.525 1.05 10 2.34 MW Losses 7.5 Mvars 2.35 14 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. (.3846 + .4950 ) + j (10 − 1.923 − 4.950 ) − (.4950 − j 4.950 ) − (.4950 − j 4.950 ) V10 (.3846 + .4950 ) + j (10 − 1.923 − 4.95) V20 1.961∠ − 48.69° = 1.961∠ − 78.69° 0.8796 + j 3.127 −0.4950 + j 4.950 V10 1.961∠ − 48.69° = −0.4950 + j 4.950 −0.8796 + j 3.127 V20 1.961∠ − 78.69° 2.36 Note that there are two buses plus the reference bus and one line for this problem. After converting the voltage sources in Fig. 2.29 to current sources, the equivalent source impedances are: Z S1 = Z S 2 = ( 0.1 + j 0.5 ) // ( − j 0.1) = = ( 0.1 + j 0.5 )( − j 0.1) 0.1 + j 0.5 − j 0.1 ( 0.5099∠78.69°)( 0.1∠ − 90°) = 0.1237∠ − 87.27° 0.4123∠75.96° = 0.005882 − j 0.1235 Ω The rest is left as an exercise to the student. 2.37 After converting impedance values in Figure 2.30 to admittance values, the bus admittance matrix is: Ybus −1 0 0 1 1 1 −1 1 + 1 + 1 + 1 − j1 − − j 1 − 2 3 4 3 4 = 1 1 1 1 1 0 − − j1 − j 1 + j + j − j 3 4 3 4 2 1 1 1 1 1 − − j + j 4 − j 3 0 4 4 4 Writing nodal equations by inspection: −1 0 1 V10 1∠0° 0 −0.25 −1 ( 2.083 − j1) ( −0.3333 + j1) V20 = 0 0 ( −0.3333 + j1) ( 0.3333 − j 0.25 ) V30 0 − j 0.25 − j 0.25 ( 0.25 − j 0.08333) V40 2∠30° 0 ( −0.25 ) 15 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 2.38 The admittance diagram for the system is shown below: YBUS where Y11 Y = 21 Y31 Y41 Y12 Y13 Y22 Y23 Y32 Y42 Y33 Y43 Y14 Y24 = Y34 Y44 0 −8.5 2.5 5.0 2.5 −8.75 5.0 0 j S 5.0 5.0 −22.5 12.5 0 12.5 −12.5 0 Y11 = y10 + y12 + y13 ; Y22 = y20 + y12 + y23 ; Y23 = y13 + y23 + y34 Y44 = y34 ; Y12 = Y21 = − y12 ; Y13 = Y31 = − y13 ; Y23 = Y32 = − y23 Y34 = Y43 = − y34 and 2.39 (a) Yc + Yd + Y f −Yd −Yc −Y f −Yd Yb + Yd + Ye −Yb −Ye V1 I1 = 0 −Ye V2 = I 2 = 0 V I 0 3 3 Ye + Y f + Yg V4 I 4 −Y f −Yc −Yb Ya + Yb + Yc 0 4 2.5 V1 0 −14.5 8 8 −17 4 5 V2 0 (b) j = 4 4 −8.8 0 V3 1∠ − 90° 0 −8.3 V4 0.62∠ − 135° 2.5 5 −1 −1 YBUS V = I ; YBUS YBUS V = YBUS I 16 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. −1 YBUS = Z BUS where 0.7187 0.6688 = j 0.6307 0.6194 0.6688 0.7045 0.7045 0.6307 0.6242 0.6840 0.6258 0.5660 0.6194 0.6258 Ω 0.5660 0.6840 −1 V = YBUS I V1 0 0 V V = 2 and I = V3 1∠ − 90° 0.62∠ − 135° V4 where Then solve for V1 , V2 , V3 , and V4 . 240 ∠0° = 138.56∠0° V (Assumed as Reference) 3 2.40 (a) VAN = VAB = 240∠30° V; VBC = 240∠ − 90° V; I A = 15∠ − 90° A VAN 138.56 ∠0° = = 9.24∠90° = ( 0 + j 9.24 ) Ω IA 15∠ − 90° ZY = (b) I AB = IA 3 Z∆ = ∠30° = 15 ∠ − 90° + 30° = 8.66∠ − 60°A 3 VAB 240∠30° = = 27.71∠90° = ( 0 + j 27.71) Ω I AB 8.66∠ − 60° Note: ZY = Z ∆ / 3 2.41 S3φ = 3VLL I L ∠ cos−1 ( pf ) = 3 ( 480 )( 20 ) ∠ cos−1 0.8 = 16.627 × 103 ∠36.87° = (13.3 × 103 ) + j (9.976 × 103 ) P3φ = Re S3φ = 13.3 kW Delivered Q3φ = I m S3φ = 9.976 kVAR Delivered 2.42 (a) With Vab as reference Van = Ia = 208 3 Z∆ = 4 + j 3 = 5∠36.87° Ω 3 ∠ − 30° Van 120.1∠ − 30° = = 24.02∠ − 66.87° A ( Z ∆ / 3) 5∠36.87° 17 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. S3φ = 3Van I a* = 3 (120.1∠ − 30° )( 24.02∠ + 66.87° ) = 8654∠36.87° = 6923 + j 5192 P3φ = 6923 W; Q3φ = 5192 VAR; both absorbed by the load pf = cos ( 36.87° ) = 0.8 Lagging; S3φ = S3φ = 8654 VA (b) Vab = 208∠0° V I a = 24.02∠ − 66.87° A 13.87∠ − 36.87° A 2.43 (a) Transforming the ∆-connected load into an equivalent Y, the impedance per phase of the equivalent Y is Z2 = 60 − j 45 = ( 20 − j15 ) Ω 3 With the phase voltage V1 = 1203 3 = 120 V taken as a reference, the per-phase equivalent circuit is shown below: Total impedance viewed from the input terminals is Z = 2 + j4 + I= ( 30 + j 40 )( 20 − j15) = 2 + j 4 + 22 − j 4 = 24 Ω ( 30 + j 40 ) + ( 20 − j15) V1 120∠0° = = 5∠0° A Z 24 The three-phase complex power supplied = S = 3V1 I * = 1800 W P = 1800 W and Q = 0 VAR delivered by the sending-end source 18 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. (b) Phase voltage at load terminals V2 = 120∠0° − ( 2 + j 4 )( 5 ∠0° ) = 110 − j 20 = 111.8∠ − 10.3° V The line voltage magnitude at the load terminal is (VLOAD )L -L = 3 111.8 = 193.64 V (c) The current per phase in the Y-connected load and in the equiv.Y of the ∆-load: I1 = V2 = 1 − j 2 = 2.236∠ − 63.4° A Z1 I2 = V2 = 4 + j 2 = 4.472 ∠26.56° A Z2 The phase current magnitude in the original ∆-connected load (I ) ph ∆ = I2 3 = 4.472 3 = 2.582 A (d) The three-phase complex power absorbed by each load is S1 = 3V2 I1* = 430 W + j 600 VAR S2 = 3V2 I 2* = 1200 W − j 900 VAR The three-phase complex power absorbed by the line is SL = 3 ( RL + jX L ) I 2 = 3 ( 2 + j 4 ) (5)2 = 150 W + j300 VAR The sum of load powers and line losses is equal to the power delivered from the supply: S1 + S2 + SL = ( 450 + j600 ) + (1200 − j 900 ) + (150 + j 300 ) = 1800 W + j 0 VAR 2.44 (a) The per-phase equivalent circuit for the problem is shown below: Phase voltage at the load terminals is V2 = 2200 3 = 2200 V taken as Ref. 3 Total complex power at the load end or receiving end is SR( 3φ ) = 560.1( 0.707 + j 0.707 ) + 132 = 528 + j 396 = 660∠36.87° kVA 19 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. With phase voltage V2 as reference, I = SR*( 3φ ) * 2 3V = 660,000∠ − 36.87° = 100∠ − 36.87° A 3 ( 2200∠0° ) Phase voltage at sending end is given by V1 = 2200∠0° + ( 0.4 + j 2.7 )(100∠ − 36.87° ) = 2401.7∠4.58° V The magnitude of the line to line voltage at the sending end of the line is (V1 ) L -L = 3V1 = 3 ( 2401.7 ) = 4160 V (b) The three-phase complex-power loss in the line is given by SL (3φ ) = 3 RI 2 + j 3 × I 2 = 3 ( 0.4 ) (100 2 ) + j 3 ( 2.7 )(100 ) 2 = 12 kW + j81kVAR (c) The three-phase sending power is SS (3φ ) = 3V1 I * = 3 ( 2401.7∠4.58° )(100∠36.87° ) = 540 kW + j 477 kVAR Note that SS (3φ ) = SR(3φ ) + SL ( 3φ ) 2.45 (a) IS = SS 3VLL = 25.001 × 103 3 ( 480 ) = 30.07 A (b) The ammeter reads zero, because in a balanced three-phase system, there is no neutral current. 2.46 (a) 20 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Using voltage division: VAN = Van = = 208 3 ∠0° Z∆ / 3 ( Z ∆ / 3) + Z LINE 10∠30° 10∠30° + ( 0.8 + j 0.6 ) (120.09 )(10∠30°) = 9.46 + j 5.6 = 109.3∠ − 0.62° V 1200.9∠30° 10.99∠30.62° Load voltage = VAB = 3 (109.3 ) = 189.3V Line-to-Line (b) Z eq = 10 ∠30 ° || (− j 20) = 11.547 ∠0 °Ω VAN = Van Z eq Z eq + Z LINE ( = 208 = 3 ) (11.54711.547 + 0.8 + j 0.6 ) 1386.7 = 112.2∠ − 2.78° V 12.362∠2.78° Load voltage Line-to-Line VAB = 3 (112.2 ) = 194.3 V 2.47 (a) I G1 = 15 × 103 ∠ − cos−1 0.8 = 23.53∠ − 36.87° A 8 ( 460 )( 0.8 ) 460 ∠0° − (1.4 + j1.6 )( 23.53∠ − 36.87° ) 3 = 216.9∠ − 2.73° V Line to Neutral VL = VG1− Z LINE1 I G1 = Load Voltage VL = 3 216.9 = 375.7 V Line to line 21 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 30 × 103 (b) I L = 3 ( 375.7 )( 0.8 ) ∠ − 2.73° − cos−1 0.8 = 57.63∠ − 39.6° A I G 2 = I L − I G1 = 57.63 ∠ − 39.6° − 23.53∠ − 36.87° = 34.14∠ − 41.49° A VG 2 = VL + Z LINE 2 I G 2 = 216.9∠ − 2.73° + ( 0.8 + j1)( 34.14∠ − 41.49° ) = 259.7∠ − 0.63° V Generator 2 line-to-line voltage VG 2 = 3 ( 259.7 ) = 449.8 V (c) SG 2 = 3VG 2 I G* = 3 ( 259.7∠ − 0.63° )( 34.14∠41.49° ) 2 = 20.12 × 103 + j17.4 × 103 PG 2 = 20.12 kW; QG 2 = 17.4 kVAR; Both delivered 2.48 (a) (b) pf = cos31.32° = 0.854 Lagging SL (c) I L = 3VLL = 26.93 × 103 3 ( 480 ) = 32.39 A (d) QC = QL = 14 × 103 VAR = 3 (VLL ) / X ∆ 2 X∆ = (e) 3 ( 480 ) 2 = 49.37 Ω 14 × 103 I C = VLL / X ∆ = 480 / 49.37 = 9.72 A I LINE = PL 3 VLL = 23 × 103 3 480 = 27.66 A 22 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 2.49 (a) Let ZY = Z A = Z B = ZC for a balanced Y-load Z ∆ = Z AB = Z BC = Z CA Using equations in Fig. 2.27 ZY2 + ZY2 + ZY2 = 3 ZY ZY Z∆ = and ZY = (b) Z A = ZB = ( j10 )( − j 25) j10 + j 20 − j 25 Z ∆2 Z = ∆ 3 Z∆ + Z∆ + Z∆ = − j 50 Ω ( j10 )( j 20 ) = j 40 Ω; Z j5 C = ( j 20 )( − j 25) = − j 100 Ω j5 2.50 Replace delta by the equivalent WYE: ZY = − j 2 Ω 3 Per-phase equivalent circuit is shown below: 2 Noting that j 1.0 − j = − j 2 , by voltage-divider law, 3 V1 = − j2 (100∠0°) = 105∠0° − j 2 + j 0.1 ∴υ1 (t ) = 105 2 cos (ω t + 0°) = 148.5cos ω t V ← In order to find i2 (t ) in the original circuit, let us calculate VA′B′ VA′B′ = VA′N ′ − VB′N ′ = 3 e j 30°VA′N ′ = 173.2∠30° Then I A′B′ = 173.2∠30° = 86.6∠120° − j2 ∴ i2 (t ) = 86.6 2 cos (ω t + 120° ) = 122.5cos (ω t + 120° ) A ← 23 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 2.51 On a per-phase basis S1 = ∴ I1 = 1 (150 + j120 ) = ( 50 + j 40 ) kVA 3 ( 50 − j 40 )103 = 2000 ( 25 − j 20 ) A Note: PF Lagging Load 2: Convert ∆ into an equivalent Y 1 (150 − j 48 ) = ( 50 − j16 ) Ω 3 2000∠0° ∴ I2 = = 38.1∠17.74° 50 − j16 Z 2Y = = ( 36.29 + j 11.61) A Note: PF Leading 1 S3 per phase = (120 × 0.6 ) − j 120 sin( cos−1 0.6 ) = ( 24 − j 32 ) kVA 3 ∴ I3 = ( 24 + j32 )103 = 2000 (12 + j 16 ) A Note:PF Leading Total current drawn by the three parallel loads IT = I1 + I 2 + I 3 ITOTAL = ( 73.29 + j 7.61) A Note:PF Leading Voltage at the sending end: VAN = 2000∠0° + ( 73.29 + j 7.61)( 0.2 + j 1.0 ) = 2007.05 + j 74.81 = 2008.44∠2.13° V Line-to-line voltage magnitude at the sending end = 2.52 (a) Let VAN be the reference: VAN = 2160 3 3( 2008.44 ) = 3478.62 V ← ∠0° ≃ 2400∠0° V Total impedance per phase Z = ( 4.7 + j 9 ) + ( 0.3 + j1) = ( 5 + j10 ) Ω ∴ Line Current = 2400∠0° = 214.7∠ − 63.4° A = I A ← 5 + j10 With positive A-B-C phase sequence, I B = 214.7∠ − 183.4° A; I C = 214.7∠ − 303.4° = 214.7∠56.6° A ← 24 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. (b) (VA′N ) LOAD = 2400∠0° − ( 214.7∠ − 63.4° )( 0.3 + j1) = 2400∠0° − 224.15∠9.9° = 2179.2 − j 38.54 = 2179.5∠ − 1.01° V ← (V ) B′N LOAD = 2179.5∠ − 121.01° V □ ; (VC ′N ) LOAD = 2179.5∠ − 241.01° V □ (c) S / Phase = (VA′N ) LOAD I A = ( 2179.5 )( 214.7 ) = 467.94 kVA ← Total apparent power dissipated in all three phases in the load S3φ = 3 ( 467.94 ) = 1403.82 kVA ← LOAD Active power dissipated per phase in load = ( P1φ ) LOAD = ( 2179.5 )( 214.7 ) cos ( 62.39° ) = 216.87 kW ← ∴ P3φ LOAD = 3 ( 216.87 ) = 650.61kW ← Reactive power dissipated per phase in load = ( Q1φ ) LOAD = ( 2179.5 )( 214.7 ) sin ( 62.39° ) = 414.65 kVAR ← ∴ Q3φ LOAD = 3 ( 414.65 ) = 1243.95 kVAR ← (d) Line losses per phase ( P1φ ) Total line loss ( P3φ ) LOSS = ( 214.7 ) 0.3 = 13.83 kW ← 2 LOSS = 13.83 × 3 = 41.49 kW ← 25 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Solutions Manual for Power System Analysis and Design 6th Edition by Glover IBSN 9781305632134 Full Download: http://downloadlink.org/product/solutions-manual-for-power-system-analysis-and-design-6th-edition-by-glover-ibs 26 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Full all chapters instant download please go to Solutions Manual, Test Bank site: downloadlink.org