Solutions Manual For Power System Analysis And Design 6th Edition By .
Solutions Manual for Power System Analysis and Design 6th Edition by Glover IBSN 9781305632134 Full Download: -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: -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
Solutions Manual for Power System Analysis and Design 6th Edition by Glover IBSN 9781305632134 Full Download: -glover-ibsn-9781305632134/ Full all chapters instant download please go to Solutions Manual, Test Bank site: downloadlink.org
Bruksanvisning för bilstereo . Bruksanvisning for bilstereo . Instrukcja obsługi samochodowego odtwarzacza stereo . Operating Instructions for Car Stereo . 610-104 . SV . Bruksanvisning i original
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