Bài giải phần giải mạch P12

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Bài giải phần giải mạch P12

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Chapter 12, Solution 1. (a) If Vab = 400 , then Van = 400 3 ∠ - 30° = 231∠ - 30° V Vbn = 231∠ - 150° V Vcn = 231∠ - 270° V (b) For the acb sequence, Vab = Van − Vbn = Vp ∠0° − Vp ∠120°  1 3 Vab = Vp 1 + − j  = Vp 3∠ - 30°   2   2 i.e. in the acb sequence, Vab lags Van by 30°. Hence, if Vab = 400 , then Van = 400 3 ∠30° = 231∠30° V Vbn = 231∠150° V Vcn =...

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  1. Chapter 12, Solution 1. (a) If Vab = 400 , then 400 Van = ∠ - 30° = 231∠ - 30° V 3 Vbn = 231∠ - 150° V Vcn = 231∠ - 270° V (b) For the acb sequence, Vab = Van − Vbn = Vp ∠0° − Vp ∠120°  1 3 Vab = Vp 1 + − j  = Vp 3∠ - 30°    2 2  i.e. in the acb sequence, Vab lags Van by 30°. Hence, if Vab = 400 , then 400 Van = ∠30° = 231∠30° V 3 Vbn = 231∠150° V Vcn = 231∠ - 90° V Chapter 12, Solution 2. Since phase c lags phase a by 120°, this is an acb sequence. Vbn = 160∠(30° + 120°) = 160∠150° V Chapter 12, Solution 3. Since Vbn leads Vcn by 120°, this is an abc sequence. Van = 208∠(130° + 120°) = 208∠ 250° V
  2. Chapter 12, Solution 4. Vbc = Vca ∠120° = 208∠140° V Vab = Vbc ∠120° = 208∠260° V Vab 208∠260° Van = = = 120∠230° V 3 ∠30° 3 ∠30° Vbn = Van ∠ - 120° = 120∠110° V Chapter 12, Solution 5. This is an abc phase sequence. Vab = Van 3 ∠30° Vab 420∠0° or Van = = = 242.5∠ - 30° V 3 ∠30° 3 ∠30° Vbn = Van ∠ - 120° = 242.5∠ - 150° V Vcn = Van ∠120° = 242.5∠90° V Chapter 12, Solution 6. Z Y = 10 + j5 = 11.18∠26.56° The line currents are Van 220 ∠0° Ia = = = 19.68∠ - 26.56° A Z Y 11.18∠26.56° I b = I a ∠ - 120° = 19.68∠ - 146.56° A I c = I a ∠120° = 19.68∠93.44° A
  3. The line voltages are Vab = 200 3 ∠30° = 381∠30° V Vbc = 381∠ - 90° V Vca = 381∠ - 210° V The load voltages are VAN = I a Z Y = Van = 220∠0° V VBN = Vbn = 220∠ - 120° V VCN = Vcn = 220∠120° V Chapter 12, Solution 7. This is a balanced Y-Y system. + 440∠0° V ZY = 6 − j8 Ω − Using the per-phase circuit shown above, 440∠0° Ia = = 44∠53.13° A 6 − j8 I b = I a ∠ - 120° = 44∠ - 66.87° A I c = I a ∠120° = 44∠173.13° A Chapter 12, Solution 8. VL = 220 V , Z Y = 16 + j9 Ω Vp VL 220 I an = = = = 6.918∠ - 29.36° ZY 3 ZY 3 (16 + j9) I L = 6.918 A
  4. Chapter 12, Solution 9. Van 120 ∠0° Ia = = = 4.8∠ - 36.87° A Z L + Z Y 20 + j15 I b = I a ∠ - 120° = 4.8∠ - 156.87° A I c = I a ∠120° = 4.8∠83.13° A As a balanced system, I n = 0 A Chapter 12, Solution 10. Since the neutral line is present, we can solve this problem on a per-phase basis. For phase a, Van 220 ∠0° Ia = = = 6.55∠36.53° Z A + 2 27 − j20 For phase b, Vbn 220 ∠ - 120° Ib = = = 10 ∠ - 120° ZB + 2 22 For phase c, Vcn 220 ∠120° Ic = = = 16.92 ∠97.38° ZC + 2 12 + j5 The current in the neutral line is I n = -(I a + I b + I c ) or - In = Ia + Ib + Ic - I n = (5.263 + j3.9) + (-5 − j8.66) + (-2.173 + j16.78) I n = 1.91 − j12.02 = 12.17 ∠ - 81° A
  5. Chapter 12, Solution 11. Vbc VBC 220∠10° Van = = = 3 ∠ - 90° 3 ∠ - 90° 3 ∠ - 90° Van = 127 ∠100° V VAB = VBC ∠120° = 220∠130° V VAC = VBC ∠ - 120° = 220∠ - 110° V If I bB = 30 ∠60° , then I aA = 30∠180° , I cC = 30 ∠ - 60° I aA 30∠180° I AB = = = 17.32∠210° 3 ∠ - 30° 3 ∠ - 30° I BC = 17.32∠90° , I CA = 17.32 ∠ - 30° I AC = -I CA = 17.32∠150° A I BC Z = VBC VBC 220 ∠0° Z= = = 12.7 ∠ - 80° Ω I BC 17.32 ∠90° Chapter 12, Solution 12. Convert the delta-load to a wye-load and apply per-phase analysis. Ia + 110∠0° V ZY − Z∆ ZY = = 20 ∠45° Ω 3
  6. 110∠0° Ia = = 5.5∠ - 45° A 20∠45° I b = I a ∠ - 120° = 5.5∠ - 165° A I c = I a ∠120° = 5.5∠75° A Chapter 12, Solution 13. First we calculate the wye equivalent of the balanced load. ZY = (1/3)Z∆ = 6+j5 Now we only need to calculate the line currents using the wye-wye circuits. 110 Ia = = 6.471∠ − 61.93° A 2 + j10 + 6 + j5 110∠ − 120° Ib = = 6.471∠178.07° A 8 + j15 110∠120° Ic = = 6.471∠58.07° A 8 + j15
  7. Chapter 12, Solution 14. We apply mesh analysis. 1 + j 2Ω A a + ZL 100∠0 o V ZL - I3 n I1 B C - - 100∠120 o V 100∠120 o V Z L = 12 + j12Ω + - + c I2 b 1 + j 2Ω 1 + j 2Ω For mesh 1, − 100 + 100∠120 o + I 1 (14 + j16) − (1 + j 2) I 2 − (12 + j12) I 3 = 0 or (14 + j16) I 1 − (1 + j 2) I 2 − (12 + j12) I 3 = 100 + 50 − j86.6 = 150 − j86.6 (1) For mesh 2, 100∠120 o − 100∠ − 120 o − I 1 (1 + j 2) − (12 + j12) I 3 + (14 + j16) I 2 = 0 or − (1 + j 2) I 1 + (14 + j16) I 2 − (12 + j12) I 3 = −50 − j86.6 + 50 − j86.6 = − j173.2 (2) For mesh 3, − (12 + j12) I 1 − (12 + j12) I 2 + (36 + j 36) I 3 = 0 (3) Solving (1) to (3) gives I 1 = −3.161 − j19.3, I 2 = −10.098 − j16.749, I 3 = −4.4197 − j12.016 I aA = I 1 = 19.58∠ − 99.3 A o I bB = I 2 − I 1 = 7.392∠159.8 o A I cC = − I 2 = 19.56∠58.91o A
  8. Chapter 12, Solution 15. Convert the delta load, Z ∆ , to its equivalent wye load. Z∆ Z Ye = = 8 − j10 3 (12 + j5)(8 − j10) Z p = Z Y || Z Ye = = 8.076∠ - 14.68° 20 − j5 Z p = 7.812 − j2.047 Z T = Z p + Z L = 8.812 − j1.047 Z T = 8.874 ∠ - 6.78° We now use the per-phase equivalent circuit. Vp 210 Ia = , where Vp = Zp + ZL 3 210 Ia = = 13.66 ∠6.78° 3 (8.874 ∠ - 6.78°) I L = I a = 13.66 A Chapter 12, Solution 16. (a) I CA = - I AC = 10∠(-30° + 180°) = 10∠150° This implies that I AB = 10 ∠30° I BC = 10∠ - 90° I a = I AB 3 ∠ - 30° = 17.32∠0° A I b = 17.32∠ - 120° A I c = 17.32∠120° A VAB 110 ∠0° (b) Z∆ = = = 11∠ - 30° Ω I AB 10 ∠30°
  9. Chapter 12, Solution 17. Convert the ∆-connected load to a Y-connected load and use per-phase analysis. ZL Ia + Van ZY − Z∆ ZY = = 3 + j4 3 Van 120 ∠0° Ia = = = 19.931∠ - 48.37° Z Y + Z L (3 + j4) + (1 + j0.5) But I a = I AB 3 ∠ - 30° 19.931∠ - 48.37° I AB = = 11.51∠ - 18.37° A 3 ∠ - 30° I BC = 11.51∠ - 138.4° A I CA = 11.51∠101.6° A VAB = I AB Z ∆ = (11.51∠ - 18.37°)(15∠53.13°) VAB = 172.6∠34.76° V VBC = 172.6∠ - 85.24° V VCA = 172.6∠154.8° V Chapter 12, Solution 18. VAB = Van 3 ∠30° = (440 ∠60°)( 3 ∠30°) = 762.1∠90° Z ∆ = 12 + j9 = 15∠36.87°
  10. VAB 762.1∠90° I AB = = = 50.81∠53.13° A Z ∆ 15∠36.87° I BC = I AB ∠ - 120° = 50.81∠ - 66.87° A I CA = I AB ∠120° = 50.81∠173.13° A Chapter 12, Solution 19. Z ∆ = 30 + j10 = 31.62 ∠18.43° The phase currents are Vab 173∠0° I AB = = = 5.47 ∠ - 18.43° A Z ∆ 31.62 ∠18.43° I BC = I AB ∠ - 120° = 5.47 ∠ - 138.43° A I CA = I AB ∠120° = 5.47 ∠101.57° A The line currents are I a = I AB − I CA = I AB 3 ∠ - 30° I a = 5.47 3 ∠ - 48.43° = 9.474∠ - 48.43° A I b = I a ∠ - 120° = 9.474∠ - 168.43° A I c = I a ∠120° = 9.474∠71.57° A Chapter 12, Solution 20. Z ∆ = 12 + j9 = 15∠36.87° The phase currents are 210∠0° I AB = = 14∠ - 36.87° A 15∠36.87° I BC = I AB ∠ - 120° = 14∠ - 156.87° A I CA = I AB ∠120° = 14∠83.13° A The line currents are I a = I AB 3 ∠ - 30° = 24.25∠ - 66.87° A I b = I a ∠ - 120° = 24.25∠ - 186.87° A I c = I a ∠120° = 24.25∠53.13° A
  11. Chapter 12, Solution 21. − 230∠120° − 230∠120° (a) I AC = = = 17.96∠ − 98.66° A(rms) 10 + j8 12.806∠38.66° 230∠ − 120 230∠0° I bB = I BC + I BA = I BC − I AB = − 10 + j8 10 + j8 (b) = 17.96∠ − 158.66° − 17.96∠ − 38.66° = −16.729 − j6.536 − 14.024 + j11.220 = −30.75 + j4.684 = 31.10∠171.34° A Chapter 12, Solution 22. Convert the ∆-connected source to a Y-connected source. Vp 208 Van = ∠ - 30° = ∠ - 30° = 120 ∠ - 30° 3 3 Convert the ∆-connected load to a Y-connected load. Z (4 + j6)(4 − j5) Z = Z Y || ∆ = (4 + j6) || (4 − j5) = 3 8+ j Z = 5.723 − j0.2153 ZL Ia + Van Z − Van 120 ∠30° Ia = = = 15.53∠ - 28.4° A Z L + Z 7.723 − j0.2153 I b = I a ∠ - 120° = 15.53∠ - 148.4° A I c = I a ∠120° = 15.53∠91.6° A
  12. Chapter 12, Solution 23. V AB 208 (a) I AB = = Z∆ 25∠60 o 208 3∠ − 30 o I a = I AB 3∠ − 30 = o = 14.411∠ − 90 o 25∠60 o I L =| I a |= 14.41 A  208 3  (b) P = P1 + P2 = 3VL I L cosθ = 3 (208)   25  cos 60 = 2.596 kW o   Chapter 12, Solution 24. Convert both the source and the load to their wye equivalents. Z∆ ZY = = 20 ∠30° = 17.32 + j10 3 Vab Van = ∠ - 30° = 240.2∠0° 3 We now use per-phase analysis. 1+jΩ Ia + Van 20∠30° Ω − Van 240.2 Ia = = = 11.24∠ - 31° A (1 + j) + (17.32 + j10) 21.37 ∠31° I b = I a ∠ - 120° = 11.24∠ - 151° A I c = I a ∠120° = 11.24∠89° A
  13. But I a = I AB 3 ∠ - 30° 11.24 ∠ - 31° I AB = = 6.489∠ - 1° A 3 ∠ - 30° I BC = I AB ∠ - 120° = 6.489∠ - 121° A I CA = I AB ∠120° = 6.489∠119° A Chapter 12, Solution 25. Convert the delta-connected source to an equivalent wye-connected source and consider the single-phase equivalent. 440 ∠(10° − 30°) Ia = 3 ZY where Z Y = 3 + j2 + 10 − j8 = 13 − j6 = 14.32 ∠ - 24°.78° 440 ∠ - 20° Ia = = 17.74∠4.78° A 3 (14.32 ∠ - 24.78°) I b = I a ∠ - 120° = 17.74∠ - 115.22° A I c = I a ∠120° = 17.74∠124.78° A Chapter 12, Solution 26. Transform the source to its wye equivalent. Vp Van = ∠ - 30° = 72.17 ∠ - 30° 3 Now, use the per-phase equivalent circuit. Van I aA = , Z = 24 − j15 = 28.3∠ - 32° Z
  14. 72.17 ∠ - 30° I aA = = 2.55∠ 2° A 28.3∠ - 32° I bB = I aA ∠ - 120° = 2.55∠ - 118° A I cC = I aA ∠120° = 2.55∠122° A Chapter 12, Solution 27. Vab ∠ - 30° 220∠ - 10° Ia = = 3 ZY 3 (20 + j15) I a = 5.081∠ - 46.87° A I b = I a ∠ - 120° = 5.081∠ - 166.87° A I c = I a ∠120° = 5.081∠73.13° A Chapter 12, Solution 28. Let Vab = 400∠0° Van ∠ - 30° 400∠ - 30° Ia = = = 7.7 ∠30° 3 ZY 3 (30 ∠ - 60°) I L = I a = 7.7 A Van VAN = I a Z Y = ∠ - 30° = 230.94∠ - 30° 3 Vp = VAN = 230.9 V
  15. Chapter 12, Solution 29. VL P = 3Vp I p cos θ , Vp = , IL = Ip 3 P = 3 VL I L cos θ P 5000 IL = = = 20.05 = I p 3 VL cos θ 240 3 (0.6) Vp VL 240 ZY = = = = 6.911 Ip 3 IL 3 (20.05) cos θ = 0.6  → θ = 53.13° Z Y = 6.911∠ - 53.13° (leading) Z Y = 4.15 − j5.53 Ω P 5000 S= = = 8333 pf 0.6 Q = S sin θ = 6667 S = 5000 − j6667 VA Chapter 12, Solution 30. Since this a balanced system, we can replace it by a per-phase equivalent, as shown below. + ZL Vp -
  16. 3V 2 p VL S = 3S p = , Vp = Z*p 3 V 2L (208) 2 S= = = 1.4421∠45 o kVA Z * p 30∠ − 45 o P = S cosθ = 1.02 kW Chapter 12, Solution 31. PP (a) Pp = 6,000, cosθ = 0.8, Sp = = 6 / 0.8 = 7.5 kVA cos θ Q p = S P sin θ = 4.5 kVAR S = 3S p = 3(6 + j 4.5) = 18 + j13.5 kVA For delta-connected load, Vp = VL= 240 (rms). But 3V 2 p 3V 2 p 3(240) 2 S=  → Z*p = = , Z P = 6.144 + j 4.608Ω Z*p S (18 + j13.5) x10 3 6000 (b) Pp = 3VL I L cosθ  → IL = = 18.04 A 3 x 240 x0.8 (c ) We find C to bring the power factor to unity Qc 4500 Qc = Q p = 4.5 kVA  → C= = = 207.2 µF ωV rms 2πx60 x 240 2 2 Chapter 12, Solution 32. S = 3 VL I L ∠θ S = S = 3 VL I L = 50 × 10 3 5000 IL = = 65.61 A 3 (440)
  17. For a Y-connected load, VL 440 I p = I L = 65.61 , Vp = = = 254.03 3 3 Vp 254.03 Z = = = 3.872 Ip 65.61 Z = Z ∠θ , θ = cos -1 (0.6) = 53.13° Z = (3.872)(cos θ + j sin θ) Z = (3.872)(0.6 + j0.8) Z = 2.323 + j3.098 Ω Chapter 12, Solution 33. S = 3 VL I L ∠θ S = S = 3 VL I L For a Y-connected load, IL = Ip , VL = 3 Vp S = 3 Vp I p S 4800 IL = Ip = = = 7.69 A 3 Vp (3)(208) VL = 3 Vp = 3 × 208 = 360.3 V
  18. Chapter 12, Solution 34. VL 220 Vp = = 3 3 Vp 200 Ia = = = 6.73∠58° ZY 3 (10 − j16) I L = I p = 6.73 A S = 3 VL I L ∠θ = 3 × 220 × 6.73∠ - 58° S = 1359 − j2174.8 VA Chapter 12, Solution 35. (a) This is a balanced three-phase system and we can use per phase equivalent circuit. The delta-connected load is converted to its wye-connected equivalent 1 Z '' y = Z ∆ = (60 + j 30) / 3 = 20 + j10 3 IL + Z’y Z’’y 230 V - Z y = Z ' y // Z '' y = (40 + j10) //( 20 + j10) = 13.5 + j 5.5 230 IL = = 14.61 − j 5.953 A 13.5 + j 5.5 (b) S = Vs I * L = 3.361 + j1.368 kVA (c ) pf = P/S = 0.9261
  19. Chapter 12, Solution 36. (a) S = 1 [0.75 + sin(cos-10.75) ] =0.75 + 0.6614 MVA S (0.75 + j 0.6614) x10 6 (b) S = 3V p I * p  → I*p = = = 59.52 + j 52.49 3V p 3x 4200 PL =| I p | 2 Rl = (79.36) 2 (4) = 25.19 kW (c) Vs = VL + I p (4 + j ) = 4.4381 − j 0.21 kV = 4.443∠ - 2.709 o kV Chapter 12, Solution 37. P 12 S= = = 20 pf 0.6 S = S∠θ = 20∠θ = 12 − j16 kVA But S = 3 VL I L ∠θ 20 × 10 3 IL = = 55.51 A 3 × 208 2 S = 3 Ip Zp For a Y-connected load, I L = I p . S (12 − j16) × 10 3 Zp = 2 = 3 IL (3)(55.51) 2 Z p = 1.298 − j1.731 Ω
  20. Chapter 12, Solution 38. As a balanced three-phase system, we can use the per-phase equivalent shown below. 110∠0° 110∠0° Ia = = (1 + j2) + (9 + j12) 10 + j14 1 2 1 (110) 2 Sp = I ZY = ⋅ ⋅ (9 + j12) 2 a 2 (10 2 + 14 2 ) The complex power is 3 (110) 2 S = 3S p = ⋅ ⋅ (9 + j12) 2 296 S = 551.86 + j735.81 VA Chapter 12, Solution 39. Consider the system shown below. a 5Ω A -j6 Ω 4Ω 100∠120° − + 100∠0° I1 + − 8Ω I3 j3 Ω 5Ω c − + C b B 10 Ω 100∠-120° I2 5Ω For mesh 1, 100 = (18 − j6) I 1 − 5 I 2 − (8 − j6) I 3 (1) For mesh 2, 100 ∠ - 120° = 20 I 2 − 5 I 1 − 10 I 3 20∠ - 120° = - I 1 + 4 I 2 − 2 I 3 (2)
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