2. 11.1 basic concepts ： 3-phase source 11.2 key point and mast :the balanced 3-phase circuit 11.3 mast: the unbalanced 3-phase circuit 11.4 mast ： power of 3-phase circuit nodus Chapter 11 Three-Phase Circuit

3. Supplements ： the merits of Three-phase Alternating Current :  With same size ， 3-phase generator 、 Electric Motor have higher in output power,simplier in construction,and better capacity than single phase  transmission electricity in three phase economically  Single phase instantaneous power is time-variant while total instantaneous power of three phase is time-invariant 。  After the rectify of three phase,the output waveforms is smooth 。

4. A X Y C B Z N S rotor  Stator include three loops ： A  X B  Y C  Z top end Three loops are displaced by 120 o rotor with magnetic pole and rotates in speed of  。 Three sets of coils are used to produce three balanced voltages 。 S N 11.1 Generation of Three-Phase Alternating electro-motive Force stator

5. 2. waveform  t t 0 uAuA uBuB uCuC B + – Y uBuB A + – X uAuA C + – Z uCuC 1. Instantaneous expression

6. 120° 4. The character of Balanced three-phase voltage  t t 0 uAuA uBuB u uCuC 3. Phase expression

7. it refers to the order in which three-phase voltage generated Positive phase sequence ： A—B—C—A A B C  Negative phase sequence ： A—C—B—A A B C  If you reverse any two phase voltages,the sequence will reverse 5. Phase Sequence link electric motor

8. A X Y C B Z N 一、 the source is defined as a Y

9. N A B C B C A + – X + – + – Y Z (6) neutral line (3) line current (4) phase current (5) line voltage phase voltage (2) phase line Introduce The notions of the balance phase voltage as a Wye: Three-wire systems Four-wire systems ？ ＝ line current

10. A B C A + – X + – + – B C Y Z Introduction of notion ： (2) Port line A 、 B 、 C (4) Line current (5) Phase current (3) Line voltage (1) Phase voltage Line current ≠ phase current Phase voltage =line voltage 二、  connection of source ？

11. A + – X + – + – B C Y Z A B C N Line current ＝ phase current 30 o 11.2 Relation Between Line Voltage (Current) and Phase Voltage (Current) Relation Between Line Voltage (Current) and Phase Voltage (Current) of symmetric Y connection 3-phase source

12. Line voltage is symmetric( the size is equation ， Phase difference 120 o ) Generally ： (2) Phase voltage is symmetric ， and line voltage is also symmetric (4) Line voltage’s phase lead phase voltage 30 o 。 (1) Line current is equal to phase current Conclusion ： for the symmetric 3-phase circuit of Y connection

13. Symmetric 3-phase source of ∆ connection ， the relation between line current and phase current Given a b c Z Z Z 30 o Line current is symmetric b a c Line current ：

14. (2) Phase current is symmetry,line current is also symmetry 。 (4) Line current’s phase lags the corresponding phase current’s phase by 30 o 。 (1) Line voltage is identical to the corresponding phase voltage 。 Reference direction Conclusion ： for a  (3) The magnitude of line current is identical to that of phase current pl II3 

15. usual connections of 3-phase circuit ZLZL ZLZL + + + _ _ _ ZLZL Z4Z4 Z4Z4 Z4Z4 A B C + _ + Nn Z1Z1 Z2Z2 - - ZNZN ZLZL ZLZL ZLZL Z3Z3 Y - Y Y - ∆

16. A B C N Right figure show “the Y connection of the loads of lamp 、 electric motor” 。 the connection methods of 3-phase circuit loads ——Y and △

17. 11.3 compute of symmetric 3-phase circuit + _ + Nn Z Z Z - - Z1Z1 ZNZN Z1Z1 Z1Z1 + Non-neural line is no affection on circuit 一、 symmetric 3-pjase four- wise 3-lphase 3-wise

18. Z L =Z+Z 1 ZLZL ZLZL + - + - + - n N ZLZL ZNZN

19. 3. It is no affection to circuit to have the neutral conductor or not 。 none the neutral wire (Y–Y connection ， three-wire wye-wye circuit) 4. Under balanced conditions ， each phase voltage and current is symmetric.Just need to compute one phase voltage,current ;then you can write other phase voltage,current by the symmetric relation 1. U nN =0 ， electric potential of the center point of source is equal to the center point of loads 。 2. The neutral conductor carries no current 。 5. The compute of Each phase is independence ， each own current is determined by its own phase voltage and impedance.it has no relation to other two phase 。 6. Draw compute circuit of single phase,compute symmetric 3- phase can conclude to compute single phase 。 Conclusion : symmetric 3-phase circuit （ Y-Y)

20. + _ + + Nn Z Z Z - - Given in symmetric 3-phase source, line voltage is 380V ， symmetric load Z ＝ 100  30  Solve line current solve ： connection neural Nn ， choose A phase to compute - + N n Z From symmetric ， get Example ： Y - Y

21. B C A Electric motor applied example of △ connection of 3- phase loads

22. + – + – + – A B C N Z Z Z a b c n solution ： connect the neutral wire Nn ， choose A phase as example to compute - + N n Z/3 Base on symmetric Solution 二 : Transfer to Y - Y

23. 二、 example ： Z Z Z A B C a b c + – + – + – Given in symmetric 3- phase circuit,line voltage 380V ， Symmetric load Z ＝ 100  30  Solve line current 。 solution 一 From symmetric,get choose A phase to compute - + B b Z Aa ∆ - ∆ connection

24. + _ + _ _ + N Z Z Z A B C a b c A B C + – + – + – n Z Z Z Y - ∆ ∆ - Y

25. (2) Phase current is symmetric,so line current is symmetric 。 (4) Line current’s phase lags phase current 30 o 。 (1)Line voltage is equal to phase voltage 。 Reference direction Conclusion ： for  connection

26. In symmetric 3-phase circuit ， line voltage of source380V ， |Z 1 |=10  ， cos  1 =0.6(lag) ， Z 2 = –j50  ， Z N =1+ j2  。 solve ： line current 、 phase current ， and qualitatively draw phase diagram (A phase as example) 。 + _ _ _ + + N A C B Z1Z1 Z2Z2 ZNZN N' solve ： + _ Z1Z1 example ：

27. + _ Z1Z1 According to symmetric ， get B 、 C phase line current,phase current ：

28. 30 o –53.1 o –18.4 o

29. + + + _ _ _ Z1Z1 Z1Z1 Z1Z1 Z2Z2 Z2Z2 ZnZn Z4Z4 Z4Z4 Z4Z4 Z3Z3 Z3Z3 Z3Z3 A B C Z2Z2 example ：

30. Z 4 /3 n" Z 4 /3 Z 2 /3 n' + + + _ _ _ Z1Z1 Z1Z1 Z1Z1 Z2Z2 ZnZn Z4Z4 Z3Z3 Z3Z3 Z3Z3 A B C N solution ： firstly transfer  —Y ， the n choose A phase to compute ： Symmetric circuit connect N, n’, n” Neutral wire impedance Z n is short circuit

31. + _ Z1Z1 Z3Z3 Z 2 /3 Z 4 /3 + + + _ _ _ Z1Z1 Z1Z1 Z1Z1 Z2Z2 ZnZn Z4Z4 Z3Z3 Z3Z3 Z3Z3 A B C N

32. (1)If ∆ connection ， can transfer 3-phase source,loads to equivalence of Y,also can compute directly ； (2) In Y-Y connection ， the resistance of neutral line can be neglect ； (3)Draw single phase circuit ， solve voltage and current of one phase ： (4) According to the relation between line phase and phase  connection 、 Y connection ， solve current and voltage of circuit 。 (5)Form symmetric ， get to the voltage and current of two phase 。 The conclusion of compute method of symmetric 3-phase circuit

33. Source unbalanced  degree small (assure by system ) 。 Circuit Reference (loads) unbalanced  many situations 。 Discuss ： source is symmetric ， loads are unbalanced ( (Low Voltage Power Line)) 。 Analysis method ： unbalanced Analysis method of Complex AC current 。 Can not only compute 。 11.4 Definition of None-symmetric Three-phase Circuit 1. Have neutral line 2. Non-neutral line nodus

34. + _ + _ _ + N N' ZaZa ZbZb ZcZc 1. Have neutral line

35. A N' N C B R R R R N A B C Equivalent circuit

36. About unbalanced 3-phase loads △ connection Balanced source with Phase voltage U p =220V 的； each resistance respectively R A =5Ω, R B =10Ω, R C =20Ω 。 Solve: under △ connection load phase voltage 、 load current and neutral wire current 。 solution RARA RBRB RCRC i N i A i B i C u A u B u C A C B Through loads are unbalanced ， because of existence of neutral wire ， load phase voltage is equal to source phase voltage and the effective value is 220V 。 Use complex number to express: example

37. RARA RBRB RCRC i N i A i B i C u A u B u C A C B

38. Main zero line can not have fuse and switch A B C N... 一 floor 二 floor 三 floor general drawing of Illumination circuit instance

39. N ZbZb ZcZc + - + - + - N'N' ZaZa 2. No neutral wire ABC Neutral point displacement

40. Every phase voltage of load ： 2. No neutral wire Phase voltage is unbalanced Line (phase) current is unbalanced + _ + _ _ + N N'N' ZaZa ZbZb ZcZc A B C

41. N Center of loads N‘ do not overlap center of source N ， it is called center displacement 。 A CB N' voltage of center displace ment A C N'N' N + _ + _ _ + ZaZa ZbZb ZcZc B The function of neutral :make phase voltage of Y unbalanced load is symmetric 。

42. (1) in common case ， 3-phase 4- wise ， neutral line impedance is zero 。 A C B N N' N A C B N 380 220 Additional Examples ： illumination circuit (2) Assume neutral line break ， A phase lamp do not connect Lamp is dim 。 A C B N N' N A C B 380 N 190

43. Lamp voltage exceed the rating work voltage ， and break 。 A C B N N' N A C B N 380 Notice ： in order to avoid neutral wire breaking ， the main neutral wire of illuminating circuit can not fix up fuse or on-off switch (3)Neutral wire break (3-wire wye-wye circuit), A phase arise short circuit

44. Analysis ： show in figure given first floor lamps turn off ， second, third floor lamp go through,but the quantity is not equal 。（ given the quantity of second floor lamps is 1/4 that of third floor ） how is the result ？ Result ： voltage of two floor lamps is exceed to the rating voltage ， lamp burn out ； three flour lamp is dim 。 Analyze A C B R2R2 R3R3

45. when loads are unbalanced and no neutral line,every phase voltage is not equal.sometimes it exceeds rating voltage, ， sometimes can not reach the rating voltage ;under those conditions,load can not work normally 。 For example ， in illuminating circuit,every phase loads can not be balanced so we can not use three-wire wye-wye to supply electricity and make sure the neutral line can count on 。 the function of the existence of neutral line ， make Y connection unbalanced loads get equal phase voltage 。 in order to avoid neutral wire breaking ， the main neutral wire of illuminating circuit can not fix up fuse or on-off switch. Conclusion On Neutral Line

46. solve ： A C B N' R C R Phase Sequencer circuit 。 Given 1/(  C)=R 3-phase source are balanced 。 solve ： the tolerant voltage of lamp 。

47. N A C B N' If choose the phase that connect with capacitance as A phase ， so B phase voltage is higher than C phase voltage 。 C phase is dim (positive sequence) 。 So measure the phase sequence 。 B phase lamp is lighter ，

48. Z Z Z A1A1 A2A2 A3A3 S circuit show in diagram ， 3- phase source is symmetric 。 When on-off S close ， current meter reading is5A 。 solve ： after S open the reading of each current meter Solution: after S opening ， the current of current A 2 is equal to that of symmetric loads 。 While the current of A 1 、 A 3 is equal to phase current of symmetric load Reading of Current meter A 2 =5A Reading of current meterA 1 、 A 3 = example

49. 3 、 conclude to compute single phase ， other two phases voltage and current can solve by symmetric relation directly 。 1. electric potential of the neutral point is equal to that of loads 。 2 、 neutral line carries no current 。∴ so no neutral line  Balanced Y-Y circuit  Unbalanced 3-phase circuit （ four-wire Y-Y) Under unbalanced 3-phase circuit,every phase voltage and current are unbalanced ， Need to compute respectively  Unbalanced 3-phase circuit （ three-wire Y-Y) electric potential of the neutral point is not equal to that of loads 。 First Must compute U N' Y-Y conclusion ： nodus

50. 11.5 Power of Three-phase Circuit 1. Instantaneous Power of 3-phase load ： p p  t 0 U p I p cos 

51. Single Phase ： Instantaneous Power Impulse tt p 0 3U p I p cos  p  t 0 U p I p cos  the sum of three phase instantaneous power is invariant namely the average power P Three phase instantaneous power is invariant ， torque m  p can get balanced mechanical torque 。

52. Balanced three-phase load Z=|Z|  P p =U p I p cos  Total power of three phase P=3P p =3U p I p cos  Z Single phase load power the Average Power of balanced three-phase circuit ： units W The angle between U and I P 2. the Average Power of balanced three-phase circuit : P Balanced three- phase load

53. (1)  is the phase difference angle between phase voltage and phase current,not line voltage and line current (2) cos  is every phase power factor,in balanced 3-phase circuit it is 3-phase power factor ： cos  A = cos  B = cos  C = cos  。 3.Reactive power of balanced 3-phase circuit Q=Q A +Q B +Q C = 3Q p average power balanced 3-phase ：  Is decided by the character of loads units ： Var (Volt Ampere Reactive ) note

54. Generally speaking ， P 、 Q 、 S is the sum of the phase 。 Power factor id defined by ： cos  =P/S (unbalanced  is none meaning 4. The VA of Balanced three-phase circuit

55. i1i1 i2i2 u loads 一、 measurement of single phase power— electromotor wattmeter In AC circuit ， is the angle between Pointer Deflection Angel) Current loop Voltage loop Measurement of The Power of Three-phase Circuit

56. A C B N The total power is equal to the algebra sum of three wattmeter 3-phase four-wise connection （ loads is Y connection ）， use three Power Meter to measure ： at this time each voltage loop measure the phase voltage of each phase 二、 Measurement of The Power of Three-phase Circuit loads

57. The total power is the algebraic sum of the meter readings 。 A C B For a three-wire system,only two wattmeters are needed ：

58. Conclusion ： three-phase total power is equal to the sum of two meter readings 。 Just single meter reading is no meaning. the Principal of measure power using the two-wattmeter method ：

59. Given ： electromotor power is 2.5 kW, 。 Solve ： wattmeter W 1 、 W 2 readings 。 electric -motor A C B ( given electromotor is Y connection ， N is middle point 。 ) Electric source line voltage is380V ， each phase are balanced 。 supplement : example

60. （ 1 ） solve the current A C B ZBZB ZAZA ZCZC N

61. A C B ZBZB ZAZA ZCZC N If ： （ 2 ） solve the relation of phase between voltage and current so

62. A C B ZBZB ZAZA ZCZC N Phase difference If electromotor 

63. from phasor diagram to get Phase difference W 1 show number is 833.3 w W 2 show number is 1666.7w

64. 1. Under the condition i A +i B +i C =0 ， only use two wattmeter methods 。 Two wattmeter methods can not be used in unbalanced 3-phase four-wise circuit 。 3. If meter connect right,one’s reading is negative ， the index of power meter is reversion ， if reverse current loop connection ， index points to positive number, but the reading is negative 。 2. The algebra sum of reading of two wattmeter is the total power of 3-phasor,single wattmeter reading is no meaning 。 4. There are three connection ways of using two wattmeter method to measure 3-phase power ， notice the same polarity port of power meter 。 Note ：

65. Load  connection method Balanced loads ： Each phase voltage and current should be computed respectively 。 loads are unbalanced ： Loads are balanced ： voltage and current are balanced ， just need compute one phase 。 solve the electric -instrument showing number,just need to compute the effective value and no need to compute phase 。 Load Y connection method Electric source are balanced ： The conclusion of three –phase AC circuit -----three- phase circuit computer

66. Three-phase total power ： When Loads are balanced ： Non-relative to connective methods The conclusion of three –phase AC circuit - ----three-phase power computer

67. solve ： (1) the total power of releasing by line current and source ； (2) use two wattmeter method to measure power of electric motor, draw connection draw,solve two wattmeter reading 。 solution ： (1) D A B C Z1Z1 Electric motor ZDZD One phase circuit ： Example :U l =380V, Z 1 =30+j40 , electric motor P D =1700W, cos Φ =0.8(lagged) 。

68. Electromotor loads ： total current Also ：

69. D A B C Z1Z1 electromotor (2) Two instrument ‘s connection methods shown in right figure 。 W1W1 * * * * W2W2

70. Instrument W 1 show number P 1 ： P 1 =U AC I A2 cos  1 = 380  3.23cos(– 30  + 36.9  ) = 380  3.23cos(6.9  ) =1219W Instrument W 2 的 show number P 2 ： P 2 =U BC I B2 cos  2 = 380  3.23cos(–90  +156.9 º ) =380  3.23cos(66.9 º ) =481.6W A B C 66.9  N 6.9 

71. Analyze ： in balanced loads ， what is each phase load’s impendence angle ， power instrument show number is negative number ？ Assume loads is Y connection and inductive ， phase current (namely line current )lag phase voltage  U AN ， U BN ， U CN is phase voltage 。 P=P 1 +P 2 =U AC I A cos  1 +U BC I B cos  2 C B A N  1 =  –30  ，  2 =  + 30 

72. P 1 =U AC I A cos  1 =U AC I A cos(  –30  ) P 2 =U BC I B cos  2 =U BC I B cos(  +30  )  >60 o negative numberPositive number  < –60 o negative number Positive number capacitive zero  =60 o P1P1 P2P2 P =P 1 +P 2  = –60 o zero  =0 Resistance ind ucti ve

73. If loads is  connection (balanced) ： phase voltage =line voltage 。 P =P 1 +P 2 =U AC I A cos  1 +U BC I B cos  2 A C B  1 =  –30   2 =  +30    30   1 1  2 2 Assignment ： 11---1. 2. 8. 12. 15.

74. Some questions in assignment ： The phase relation of three-phase voltage 、 current ， must be clarify The analysis of unbalanced loads ， the displacement of middle point Link Thesis