1. 1 Chapter 1. Three-Phase System

2. 1.1: Review of Single-Phase System The Sinusoidal voltage v 1 (t) = V m sin  t i v1v1 Load AC generator v2v2 2

3. 3 1.1: Review of Single-Phase System The Sinusoidal voltage v(t) = V m sin  t where V m = the amplitude of the sinusoid  = the angular frequency in radian/s t = time

4. 4 The angular frequency in radians per second

5. 5 A more general expression for the sinusoid (as shown in the figure): v 2 (t) = Vm sin (  t +  ) where  is the phase

6. 6 A sinusoid can be expressed in either sine or cosine form. When comparing two sinusoids, it is expedient to express both as either sine or cosine with positive amplitudes. We can transform a sinusoid from sine to cosine form or vice versa using this relationship: sin (ωt ± 180 o) = - sin ωt cos (ωt ± 180 o ) = - cos ωt sin (ωt ± 90 o ) = ± cos ωt cos (ωt ± 90 o ) = + sin ωt

7. 7 Sinusoids are easily expressed in terms of phasors. A phasor is a complex number that represents the amplitude and phase of a sinusoid. v(t) = V m cos (ωt + θ) Time domainPhasor domain Time domain Phasor domain

8. 8 Time domain Phasor domain θ V1V1 V2V2 v 2 (t) = Vm sin (  t +  ) v 1 (t) = Vm sin  t or

9. 9 1.1.1: Instantaneous and Average Power The instantaneous power is the power at any instant of time. p(t) = v(t) i(t) Where v(t) = V m cos (  t +  v ) i(t) = I m cos (  t +  i ) Using the trigonometric identity, gives

10. 10 The average power is the average of the instantaneous power over one period.

11. 11 The effective value is the root mean square (rms) of the periodic signal. The average power in terms of the rms values is Where

12. 12 1.1.2: Apparent Power, Reactive Power and Power Factor The apparent power is the product of the rms values of voltage and current. The reactive power is a measure of the energy exchange between the source and the load reactive part.

13. 13 The power factor is the cosine of the phase difference between voltage and current. The complex power:

14. 14 1.2: Three-Phase System In a three phase system the source consists of three sinusoidal voltages. For a balanced source, the three sources have equal magnitudes and are phase displaced from one another by 120 electrical degrees. A three-phase system is superior economically and advantage, and for an operating of view, to a single- phase system. In a balanced three phase system the power delivered to the load is constant at all times, whereas in a single-phase system the power pulsates with time.

15. 15 1.3: Generation of Three-Phase Three separate windings or coils with terminals R-R’, Y-Y’ and B-B’ are physically placed 120 o apart around the stator.

16. 16 V or v is generally represented a voltage, but to differentiate the emf voltage of generator from voltage drop in a circuit, it is convenient to use e or E for induced (emf) voltage.

17. 17 The instantaneous e.m.f. generated in phase R, Y and B: e R = E mR sin  t e Y = E mY sin (  t -120 o ) e B = E mB sin (  t -240 o ) = E mB sin (  t +120 o )

18. ERER Three-phase Load Three-phase AC generator VYVY IRIR VRVR EYEY EBEB ZRZR IYIY IBIB ZBZB ZYZY VBVB ININ 18

19. The instantaneous e.m.f. generated in phase R, Y and B: e R = E mR sin ωt e Y = E mY sin (ωt -120 o ) e B = E mB sin (ωt -240 o ) = E mB sin (ωt +120 o ) In phasor domain: E R = E Rrms 0o0o E Y = E Yrms -120 o E B = E Brms 120 o Phase voltage 120 o -120 o 0o0o E Rrms = E Yrms = E Brms = E p Magnitude of phase voltage 19

20. ERER Three-phase Load Three-phase AC generator VYVY IRIR VRVR EYEY EBEB ZRZR IYIY IBIB ZBZB ZYZY VBVB ININ Line voltage E RY E RY = E R - E Y

21. 21 Line voltage E RY = E R - E Y 120 o -120 o 0o0o -E Y E RY = E p 0o0o - E p -120 o = 1.732 E p E RY 30 o = √3 E p = E L 30 o

22. ERER Three-phase Load Three-phase AC generator VYVY IRIR VRVR EYEY EBEB ZRZR IYIY IBIB ZBZB ZYZY VBVB ININ Line voltage E YB E YB = E Y - E B

23. Line voltage E YB = E Y - E B 120 o -120 o 0o0o -E B E YB = E p -120 o - E p 120 o = 1.732 E p E YB -90 o = √3 E p = E L -90 o

24. ERER Three-phase Load Three-phase AC generator VYVY IRIR VRVR EYEY EBEB ZRZR IYIY IBIB ZBZB ZYZY VBVB ININ Line voltage E BR E BR = E B - E R

25. 25 Line voltage E BR = E B - E R 120 o -120 o 0o0o -E R E BR = E p 120 o - E p 0o0o = 1.732 E p E BR 150 o = √3 E p = E L 150 o For star connected supply, E L = √3 E p

26. 26 120 o -120 o 0o0o Phase voltages E R = E p 0o0o E Y = E p -120 o E B = E p 120 o Line voltages E RY = E L 30 o E YB = E L -90 o E BR = E L 150 o It can be seen that the phase voltage E R is reference.

27. 27 Phase voltages E R = E p -30 o E Y = E p -150 o E B = E p 90 o Line voltages E RY = E L 0o0o E YB = E L -120 o E BR = E L 120 o Or we can take the line voltage E RY as reference.

28. ERER Three-phase Load Three-phase AC generator VYVY IRIR VRVR EYEY EBEB ZRZR IYIY IBIB ZBZB ZYZY VBVB E RY Delta connected Three-Phase supply E RY = E R = E p 0o0o

29. ERER Three-phase Load Three-phase AC generator VYVY IRIR VRVR EYEY EBEB ZRZR IYIY IBIB ZBZB ZYZY VBVB E YB E BR For delta connected supply, E L = E p Delta connected Three-Phase supply

30. 30 Connection in Three Phase System 4-wire system (neutral line with impedance) 3-wire system (no neutral line ) 4-wire system (neutral line without impedance) Star-Connected Balanced Loads a) 4-wire system b) 3-wire system 3-wire system (no neutral line ), delta connected load Delta-Connected Balanced Loads a) 3-wire system

31. ERER Three-phase Load Three-phase AC generator VYVY IRIR VRVR EYEY EBEB ZRZR IYIY IBIB ZBZB ZYZY VBVB ININ ZNZN VNVN 4-wire system (neutral line with impedance) V N = I N Z N Voltage drop across neutral impedance: 1.1

32. ERER Three-phase Load Three-phase AC generator VYVY IRIR VRVR EYEY EBEB ZRZR IYIY IBIB ZBZB ZYZY VBVB ININ ZNZN VNVN 4-wire system (neutral line with impedance) I R + I Y + I B = I N Applying KCL at star point 1.2

33. ERER Three-phase Load Three-phase AC generator VYVY IRIR VRVR EYEY EBEB ZRZR IYIY IBIB ZBZB ZYZY VBVB ININ ZNZN VNVN 4-wire system (neutral line with impedance) Applying KVL on R-phase loop 33

34. ERER Three-phase Load Three-phase AC generator IRIR VRVR ZRZR ININ ZNZN VNVN Applying KVL on R-phase loop E R – V R – V N = 0 E R – I R Z R – V N = 0 I R = Thus E R – V N ZRZR 1.3 4-wire system (neutral line with impedance) 34

35. 35 ERER Three-phase Load Three-phase AC generator VYVY IRIR VRVR EYEY EBEB ZRZR IYIY IBIB ZBZB ZYZY VBVB ININ ZNZN VNVN 4-wire system (neutral line with impedance) Applying KVL on Y-phase loop

36. 36 Three-phase Load Three-phase AC generator VYVY EYEY IYIY ZYZY ININ ZNZN VNVN Applying KVL on Y-phase loop 4-wire system (neutral line with impedance) E Y – V Y – V N = 0 E Y – I Y Z Y – V N = 0 I Y = Thus E Y – V N ZYZY 1.4

37. Three-phase Load Three-phase AC generator EBEB IBIB ZBZB VBVB ININ ZNZN VNVN 4-wire system (neutral line with impedance) Applying KVL on B-phase loop E B – V B – V N = 0 E B – I B Z B – V N = 0 I B = Thus E B – V N ZBZB 1.5 37

38. 38 4-wire system (neutral line with impedance) I R + I Y + I B = I N Substitute Eq. 1.2, Eq.1.3, Eq. 1.4 and Eq. 1.5 into Eq. 1.1: = E B – V N ZBZB + E Y – V N ZYZY E R – V N ZRZR + VNVN ZNZN E Y – V N + + E B – V N = VNVN ZNZN ZRZR ZRZR ZYZY ZYZY ZBZB ZBZB ERER ZRZR + EYEY ZYZY + EBEB ZBZB = 1 ZNZN + 1 ZRZR + 1 ZYZY VNVN + 1 ZBZB

39. 39 4-wire system (neutral line with impedance) ERER ZRZR + EYEY ZYZY + EBEB ZBZB = 1 ZNZN + 1 ZRZR + 1 ZYZY VNVN + 1 ZBZB VNVN = ERER ZRZR + EYEY ZYZY + EBEB ZBZB 1 ZNZN + 1 ZRZR + 1 ZYZY + 1 ZBZB 1.6

40. 4-wire system (neutral line with impedance) VNVN = ERER ZRZR + EYEY ZYZY + EBEB ZBZB 1 ZNZN + 1 ZRZR + 1 ZYZY + 1 ZBZB 1.6 V N is the voltage drop across neutral line impedance or the potential different between load star point and supply star point of three-phase system. We have to determine the value of V N in order to find the values of currents and voltages of star connected loads of three-phase system. 40

41. Example ERER Three-phase Load Z Y = 2 Ω IRIR VRVR EYEY EBEB Z R = 5 Ω IYIY IBIB Z B = 10 Ω VBVB ININ Z N =10 Ω VNVN Find the line currents I R,I Y and I B. Also find the neutral current I N. E L = 415 volt

42. ERER Three-phase Load Three-phase AC generator VYVY IRIR VRVR EYEY EBEB ZRZR IYIY IBIB ZBZB ZYZY VBVB ININ ZNZN VNVN 3-wire system (no neutral line )

43. ERER Three-phase Load Three-phase AC generator VYVY IRIR VRVR EYEY EBEB ZRZR IYIY IBIB ZBZB ZYZY VBVB VNVN 3-wire system (no neutral line ) No neutral line = open circuit, Z N = ∞ 43

44. 44 VNVN = ERER ZRZR + EYEY ZYZY + EBEB ZBZB 1 ZNZN + 1 ZRZR + 1 ZYZY + 1 ZBZB 1.6 3-wire system (no neutral line ) ∞= ZNZN ∞ 1 ∞ =0

45. 45 VNVN = ERER ZRZR + EYEY ZYZY + EBEB ZBZB 1 ZRZR + 1 ZYZY + 1 ZBZB 1.7 3-wire system (no neutral line )

46. Example ERER Three-phase Load Z Y = 2 Ω IRIR VRVR EYEY EBEB Z R = 5 Ω IYIY IBIB Z B = 10 Ω VBVB VNVN E L = 415 volt Find the line currents I R,I Y and I B. Also find the voltages V R, V Y and V B.

47. 3-wire system (no neutral line ),delta connected load ERER Three-phase Load Three-phase AC generator VYVY IRIR VRVR EYEY EBEB ZRZR IYIY IBIB ZBZB ZYZY VBVB

48. 3-wire system (no neutral line ),delta connected load ERER Three-phase Load Three-phase AC generator IRIR EYEY EBEB IYIY IBIB V RY Z RY Z BR Z YB V YB V BR IrIr IbIb IyIy

49. 3-wire system (no neutral line ),delta connected load ERER Three-phase Load Three-phase AC generator IRIR EYEY EBEB IYIY IBIB V RY Z RY Z BR Z YB V YB V BR IrIr IbIb IyIy E RY =V RY E YB =V YB E BR =V BR

50. 50 3-wire system (no neutral line ),delta connected load Phase currents 30 o I r = V RY Z RY = E RY Z RY = ELEL -90 o I y = V YB Z YB = E YB Z YB = ELEL 150 o I b = V BR Z BR = E BR Z BR = ELEL

51. 3-wire system (no neutral line ),delta connected load ERER Three-phase Load Three-phase AC generator IRIR EYEY EBEB IYIY IBIB V RY Z RY Z BR Z YB V YB V BR IrIr IbIb IyIy E RY =V RY E YB =V YB E BR =V BR Line currents I R = IrIr IbIb - = ELEL Z RY 30 o - 150 o ELEL Z BR I Y = IyIy IrIr - = ELEL Z YB -90 o - 30 o ELEL Z RY

52. 3-wire system (no neutral line ),delta connected load ERER Three-phase Load Three-phase AC generator IRIR EYEY EBEB IYIY IBIB V RY Z RY Z BR Z YB V YB V BR IrIr IbIb IyIy E RY =V RY E YB =V YB E BR =V BR Line currents I B = IbIb IyIy - = ELEL Z BR 150 o - -90 o ELEL Z YB

53. Z RY Z BR Z YB ZRZR ZBZB ZYZY Star to delta conversionZ RY = ZRZYZRZY + ZYZBZYZB + ZBZRZBZR ZBZB Z YB = ZRZYZRZY + ZYZBZYZB + ZBZRZBZR ZRZR Z BR = ZRZYZRZY + ZYZBZYZB + ZBZRZBZR ZYZY

54. Example ERER Three-phase Load Z Y = 2 Ω IRIR VRVR EYEY EBEB Z R = 5 Ω IYIY IBIB Z B = 10 Ω VBVB VNVN Find the line currents I R,I Y and I B. E L = 415 volt Use star-delta conversion.

55. ERER Three-phase Load Three-phase AC generator VYVY IRIR VRVR EYEY EBEB ZRZR IRIR IBIB ZBZB ZYZY VBVB ININ ZNZN VNVN 4-wire system (neutral line without impedance) = 0 Ω V N = I N Z N = I N (0) = 0 volt 55

56. 56 4-wire system (neutral line without impedance) For 4-wire three-phase system, V N is equal to 0, therefore Eq. 1.3, Eq. 1.4, and Eq. 1.5 become, I B = EBEB ZBZB 1.5 E B – V N I Y = EYEY ZYZY 1.4 E Y – V N I R = ERER ZRZR 1.3 E R – V N

57. Example ERER Three-phase Load Z Y = 2 Ω IRIR VRVR EYEY EBEB Z R = 5 Ω IYIY IBIB Z B = 10 Ω VBVB ININ VNVN Find the line currents I R,I Y and I B. Also find the neutral current I N. E L = 415 volt

58. 58 The instantaneous e.m.f. generated in phase R, Y and B: e R = E mR sin  t e Y = E mY sin (  t -120 o ) e B = E mB sin (  t -240 o ) = E mB sin (  t +120 o )

59. 59 1.4: Phase sequences RYB and RBY V R leads V Y, which in turn leads V B. This sequence is produced when the rotor rotates in the counterclockwise direction. (a) RYB or positive sequence

60. 60 (b) RBY or negative sequence V R leads V B, which in turn leads V Y. This sequence is produced when the rotor rotates in the clockwise direction.

61. 61 1.5: Connection in Three Phase System 1.5.1: Star Connection a) Three wire system

62. 62 Star Connection b) Four wire system

63. 63 Wye connection of Load

64. 64 1.5.2: Delta Connection

65. 65 Delta connection of load

66. 66 1.6: Balanced Load Connection in 3-Phase System

67. 67 Example ERER Three-phase Load Z Y = 20 Ω IRIR VRVR EYEY EBEB Z R = 20 Ω IYIY IBIB Z B = 20 Ω VBVB VNVN E L = 415 volt Find the line currents I R,I Y and I B. Also find the voltages V R, V Y and V B. Wye-Connected Balanced Loads b) Three wire system

68. 68 Wye-Connected Balanced Loads b) Three wire system V N = = 0 volt V R = E R V Y = E Y V B = E B

69. 69 Example ERER Three-phase Load Z Y = 20 Ω IRIR VRVR EYEY EBEB Z R = 20 Ω IYIY IBIB Z B = 20 Ω VBVB ININ VNVN Find the line currents I R,I Y and I B. Also find the neutral current I N. E L = 415 volt 1.6.1: Wye-Connected Balanced Loads a) Four wire system

70. 70 For balanced load system, I N = 0 and Z 1 = Z 2 = Z 3 1.6.1: Wye-Connected Balanced Loads a) Four wire system

71. 71 Wye-Connected Balanced Loads b) Three wire system

72. 72 1.6.2: Delta-Connected Balanced Loads Phase currents: Line currents:

73. 73 1.7: Unbalanced Loads

74. 74 1.7.1: Wye-Connected Unbalanced Loads Four wire system For unbalanced load system, I N  0 and Z 1  Z 2  Z 3

75. 75 1.7.2: Delta-Connected Unbalanced Loads Phase currents: Line currents:

76. 76 1.8 Power in a Three Phase System

77. 77 Power Calculation The three phase power is equal the sum of the phase powers P = P R + P Y + P B If the load is balanced: P = 3 P phase = 3 V phase I phase cos θ

78. 78 1.8.1: Wye connection system: I phase = I L and Real Power, P = 3 V phase I phase cos θ Reactive power, Q = 3 V phase I phase sin θ Apparent power, S = 3 V phase I phase or S = P + jQ

79. 79 1.8.2: Delta connection system V LL = V phase P = 3 V phase I phase cos θ

80. 80 1.9: Three phase power measurement

81. 81 Power measurement In a four-wire system (3 phases and a neutral) the real power is measured using three single-phase watt-meters.

82. 82 Three Phase Circuit Four wire system, Each phase measured separately

83. 83 watt-meter connection Current coil (low impedance) voltage coil (high impedance) W

84. Example ERER Three-phase Load Ω IRIR VRVR EYEY EBEB Z R = 5 IYIY IBIB Z B = 20 VBVB ININ VNVN Find the three-phase total power, P T. E L = 415 volt a) Four wire system 30 o Z Y = 1090 o 45 o Ω Ω WRWR WBWB WYWY

85. Example ERER Three-phase Load Ω IRIR VRVR EYEY EBEB Z R = 5 IYIY IBIB Z B = 20 VNVN Find the three-phase total power, P T. E L = 415 volt b) Three wire system 30 o Z Y = 1090 o 45 o Ω Ω WRWR WBWB WYWY

86. Example ERER Three-phase Load Ω IRIR VRVR EYEY EBEB Z R = 5 IYIY IBIB Z B = 20 VNVN Find the three-phase total power, P T. E L = 415 volt b) Three wire system 30 o Z Y = 1090 o 45 o Ω Ω WRWR WBWB WYWY VBVB

87. 87 Three Phase Circuit Three wire system, The three phase power is the sum of the two watt- meters reading

88. 88 The three phase power (3-wire system) is the sum of the two watt-meters reading Proving: Instantaneous power: p A = v A i A p B = v B i B p C = v C i C p T = p A + p B + p C = v A i A + v B i B +v C i C = v A i A + v B i B +v C i C = v A i A + v B (-i A -i C ) +v C i C

89. 89 The three phase power (3-wire system) is the sum of the two watt-meters reading Proving: Instantaneous power: p T = p AB + p CB p T = v A i A + v B (-i A –i C ) +v C i C = (v A – v B )i A + (v C – v B )i C = v AB i A + v CB i C

90. 90 Power measurement In a four-wire system (3 phases and a neutral) the real power is measured using three single-phase watt-meters. In a three-wire system (three phases without neutral) the power is measured using only two single phase watt-meters. The watt-meters are supplied by the line current and the line-to-line voltage.