Introduction to Three-phase Circuits. Balanced 3-phase systems Unbalanced 3-phase systems

Transcription

1 Intrductin t Three-hase Circuits Balanced 3-hase systems Unbalanced 3-hase systems 1

2 Intrductin t 3-hase systems Single-hase tw-wire system: Single surce cnnected t a lad using tw-wire system Single-hase three-wire system: Tw surces cnnected t tw lads using three-wire system Surces have EQUAL magnitude and are IN PHASE 2

3 Circuit r system in which AC surces erate at the same frequency but different hases are knwn as lyhase. Balanced Tw-hase three-wire system: Tw surces cnnected t tw lads using three-wire system Surces have EQUAL frequency but DIFFFERENT hases Tw Phase System: A generatr cnsists f tw cils laced erendicular t each ther The vltage generated by ne lags the ther by 90. 3

4 Balanced Three-hase fur-wire system: Three surces cnnected t 3 lads using fur-wire system Surces have EQUAL frequency but DIFFFERENT hases Three Phase System: A generatr cnsists f three cils laced 120 aart. The vltage generated are equal in magnitude but, ut f hase by 120. Three hase is the mst ecnmical lyhase system. 4

5 AC Generatin Three things must be resent in rder t rduce electrical current: a) Magnetic field b) Cnductr c) Relative mtin Cnductr cuts lines f magnetic flux, a vltage is induced in the cnductr Directin and Seed are imrtant

6 GENERATING A SINGLE PHASE S N Mtin is arallel t the flux. N vltage is induced.

7 GENERATING A SINGLE PHASE S N Mtin is 45 t flux. Induced vltage is f maximum.

8 GENERATING A SINGLE PHASE S x N Mtin is erendicular t flux. Induced vltage is maximum.

9 GENERATING A SINGLE PHASE S N Mtin is 45 t flux. Induced vltage is f maximum.

10 GENERATING A SINGLE PHASE S N Mtin is arallel t flux. N vltage is induced.

11 GENERATING A SINGLE PHASE S N Ntice current in the cnductr has reversed. Mtin is 45 t flux. Induced vltage is f maximum.

12 GENERATING A SINGLE PHASE S N Mtin is erendicular t flux. Induced vltage is maximum.

13 GENERATING A SINGLE PHASE S N Mtin is 45 t flux. Induced vltage is f maximum.

14 GENERATING A SINGLE PHASE S N Mtin is arallel t flux. N vltage is induced. Ready t rduce anther cycle.

15 GENERATION OF THREE-PHASE AC S Three ltages will be induced acrss the cils with 120 hase difference x x N

16 Practical THREE PHASE GENERATOR The generatr cnsists f a rtating magnet (rtr) surrunded by a statinary winding (statr). Three searate windings r cils with terminals a-a, b-b, and c-c are hysically laced 120 aart arund the statr. As the rtr rtates, its magnetic field cuts the flux frm the three cils and induces vltages in the cils. The induced vltage have equal magnitude but ut f hase by 120.

17 THREE-PHASE WAEFORM Phase 1 Phase 2 Phase Phase 2 lags hase 1 by 120. Phase 2 leads hase 3 by 120. Phase 3 lags hase 1 by 240. Phase 1 leads hase 3 by 240.

18 WHY WE STUDY 3 PHASE SYSTEM? ALL electric wer system in the wrld used 3-hase system t GENERATE, TRANSMIT and DISTRIBUTE One hase, tw hase, r three hase ican be taken frm three hase system rather than generated indeendently. Instantaneus wer is cnstant (nt ulsating). thus smther rtatin f electrical machines High wer mtrs refer a steady trque Mre ecnmical than single hase less wire fr the same wer transfer The amunt f wire required fr a three hase system is less than required fr an equivalent single hase system. 18

19 3-hase systems Generatin, Transmissin and Distributin 19

20 3-hase systems Generatin, Transmissin and Distributin 20

21 Y and cnnectins Balanced 3-hase systems can be cnsidered as 3 equal single hase vltage surces cnnected either as Y r Delta () t 3 single three lads cnnected as either Y r SOURCE CONNECTIONS Y cnnected surce LOAD CONNECTIONS Y cnnected lad cnnected surce cnnected lad Y-Y Y- -Y - 21

22 Balance Three-Phase Surces Tw ssible cnfiguratins: Three-hase vltage surces: (a) Y-cnnected ; (b) Δ-cnnected 22

23 Balanced 3-hase systems LOAD CONNECTIONS Y cnnectin cnnectin a a n Z 1 Z c Z b b c Z 2 Z 3 Z a c b Balanced lad: Z 1 = Z 2 = Z 3 = Z Y Z a = Z b = Z c = Z Y Z Z 3 Unbalanced lad: each hase lad may nt be the same. 23

24 Phase Sequence The hase sequence is the time rder in which the vltages ass thrugh their resective maximum values. cn an + bn a n b c an 0 bn 120 cn 120 RMS hasrs! v v v an bn cn (t) (t) (t) cs( t) cs( t 120 cs( t 120 ) ) an bn cn 24

25 Phase Sequence a cn 120 cn an + n an 0 bn b c 120 bn 120 Phase sequence : an leads bn by 120 and bn leads cn by 120 This is a knwn as abc sequence r sitive sequence 25

26 Phase Sequence What if different hase sequence? bn 120 v v cn an ( t) 2 cs( t) (t) 2 cs( t 120 ) an 0 cn an 0 v bn (t) 2 cs( t 120 ) bn cn 120 Phase sequence : an leads cn by 120 and cn leads bn by 120 This is a knwn as acb sequence r negative sequence RMS hasrs! an cn bn 26

27 Examle Determine the hase sequence f the set f vltages. Slutin: v v v an bn cn 200 cs( t 10) 200 cs( t 230) 200 cs( t 110) The vltages can be exressed in hasr frm as an bn cn (200 / (200 / (200 / 2) 10 2) 230 2) 110 We ntice that an leads cn by 120 and cn in turn leads bn by 120. Hence, we have an acb sequence.

28 c cn I I I a b c an I a a n + 0 Z Y Z Y bn I b ZY I c I n b 0 I n I b I c I a Balanced 3-hase Y-Y Line current = hase current A line currents B ab N Z Y a The wire cnnecting n and N can be remved! an 3 Z Y b nb Z Y 30 bc ca a C bn b 0 nc an 0 bn 120 n 3 90 cn na cn 120 n 60 Phase vltages measured between the neutral and any line (line t neutral vltage) line-line vltages OR Line vltages 28

29 Balanced 3-hase systems Balanced Y-Y Cnnectin ab an 0 nb

30 Balanced 3-hase systems Balanced Y-Y Cnnectin ab an nb 0 60 cn 3 30 an bn 30

31 Balanced 3-hase systems Balanced Y-Y Cnnectin ab an nb an nb 31

32 Balanced 3-hase systems Balanced Y-Y Cnnectin ab an 0 nb an nb bn bc bn nc 3 90 nc 32

33 Balanced 3-hase systems Balanced Y-Y Cnnectin na ab an 0 nb cn an nb bn bc bn nc 3 90 nc ca cn na

34 Balanced 3-hase systems Balanced Y-Y Cnnectin ab an 0 nb ca 30 cn 30 ab bn an bc bn nc ca cn na bc L 3 where L ab bc ca and an bn cn Line vltage LEADS hase vltage by 30 34

35 Balanced 3-hase systems Balanced Y-Y Cnnectin Fr a balanced Y-Y cnnectin, analysis can be erfrmed using an equivalent er-hase circuit: e.g. fr hase A: a I a A an n + I n =0 N Z Y cn c bn b I b I c B Z Y Z Y C 35

36 Balanced 3-hase systems Balanced Y-Y Cnnectin Fr a balanced Y-Y cnnectin, analysis can be erfrmed using an equivalent er-hase circuit: e.g. fr hase A: an a n + I a A N Z Y I a Z an Y Based n the sequence, the ther line currents can be btained frm: I b Ia 120 I c Ia120 36

37 Balanced 3-hase systems Balanced Y- Cnnectin c cn an a n + bn b I b I c I a B Z I AB A Z I BC Z C I CA an 0 bn 120 cn 120 ab 330 AB bc 3 90 BC ca 3150 CA I I I AB CA BC AB Z BC Z CA Z Using KCL, Phase currents I a I I b I I I AB AB AB I I I BC BC BC I CA ( I AB ( c ICA ) )

38 Balanced 3-hase systems Balanced Y- Cnnectin I c I CA 30 I b IL 3I 30 I BC I a b I 30 I I I AB AB AB I BC BC I CA ( I IBC 3 30 where IL Ia Ib Ic and I IAB IBC ICA Ic ICA 3 30 Phase current LEADS line current by 30 I I AB AB I a ( ) ) 38

39 ca a Balanced 3-hase - ab I a Z A Z Line-line vltage is the same as hase vltage in - ab 0 bc 120 c + bc b I b I c B I AB Z I BC C I CA cn 120 ab AB bc BC I I AB BC AB Z BC Z Using KCL, Phase currents I a I b I I I AB AB AB I I BC BC I CA ( I AB ( ) ) line currents ca CA I CA CA Z I I BC 3 30 c ICA

40 Balanced 3-hase systems Balanced - Cnnectin ca a ab I a Z A Z ab 0 bc 120 c + bc b I b I c B I AB Z I BC C I CA cn 120 Alternatively, by transfrming the cnnectins t the equivalent Y cnnectins er hase equivalent circuit analysis can be erfrmed. 40

41 Balanced 3-hase systems Balanced -Y Cnnectin a I a A ab 0 ca ab L1 N Z Y bc 120 ca 120 c + bc b I b I c B Z Y Z Y C Hw t find I a? L1 - ab Z Y I a Z Y I b 0 I a I b Z ab Y Since circuit is balanced, I b = I a -120 I I I (1 1 ( 120 )) a b a Therefre I a 30 Z Y 3 Ia

42 Balanced 3-hase systems Balanced -Y Cnnectin a I a A ab 0 ca ab N Z Y bc 120 ca 120 c + bc b I b I c B Z Y Z Y C Hw t find I a? (Alternative) Transfrm the delta surce cnnectin t an equivalent Y and then erfrm the er hase circuit analysis 42

43 A balanced Y-Y system, shwing the surce, line and lad imedances. Line Imedance Surce Imedance Lad Imedance Equivalent Circuit 43

44 Three-hase Circuits Unbalanced 3-hase systems Pwer in 3-hase system 44

45 UNBALANCED DELTA-CONNECTED LOAD The line currents will nt be equal nr will they have a 120 hase difference as was the case with balanced lads. 45

46 UNBALANCED FOUR-WIRE, WYE-CONNECTED LOAD On a fur-wire system the neutral cnductr will carry a current when the lad is unbalanced The vltage acrss each f the lad imedances remains fixed with the same magnitude as the line t neutral vltage. The line currents are unequal and d nt have a 120 hase difference. 46

47 UNBALANCED THREE-WIRE, WYE-CONNECTED LOAD The cmmn int f the three lad imedances is nt at the tential f the neutral and is marked "O" instead f N. The vltages acrss the three imedances can vary cnsiderably frm line t neutral magnitude, as shwn by the vltage triangle which relates all f the vltages in the circuit. Draw the circuit diagram and select mesh currents as shwn in Fig. Write the crresnding matrix equatins (Crammer Rule) 47

48 UNBALANCED THREE-WIRE, WYE-CONNECTED LOAD Nw the vltages acrss the three imedances are given by the rducts f the line currents and the crresnding imedances. 48

49 POWER IN BALANCED THREE-PHASE LOADS Since the hase imedances f balanced wye r delta lads cntain equal currents, the hase wer is ne-third f the ttal wer. The vltage acrss is line vltage The current is hase current. The angle between & I is the angle n the imedance. Phase wer Ttal wer Fr a balanced Δ-cnnected lads: The vltage acrss is hase vltage The current is line current. The angle between & I is the angle n the imedance. Phase wer Ttal wer Fr a balanced Y-cnnected lads: 49

50 POWER IN BALANCED THREE-PHASE LOADS 50

51 Remember: INSTANTANEOUS THREE-PHASE POWER The instantaneus Single-hase wer 51

52 INSTANTANEOUS THREE-PHASE POWER The instantaneus 3-hase wer t = AN Ia + BN Ib + CN IC 52

53 3 = 3 (l S 3 ) = 3 (S 1 /2) 1 2 (l S 1 ) 2 (S 1 ) =