Electrical Circuits II (ECE233b)

Transcription

1 Eletril Ciruits (ECE2) Polyhse Ciruits Anestis Dounis The Uniersity of Western Ontrio Fulty of Engineering Siene

2 ThreePhse Ciruits Blned three hse iruit: ontins three oltge soures tht re equl in mgnitude nd re onethird of yle rt in time (i.e. oltges differ in hse y 12 ) n = rms n = 12 rms n = 24 rms = 12 rms n = rms n = 12 rms n = 24 rms n

3 ThreePhse Ciruits n = rms n = 12 rms n = 24 rms n n n ( t) m os t ( t) m os t 12 t t ( ) os 24 m where m f the lod is lned, the urrents rodued y the soures re: i ( t) m ost A i t t ( ) m os 12 A i ( t) os t 24 A The instntneous ower of the system is ( t) ( t) ( t) ( t) m ( t) ( t) i ( t) ( t) i ( t) ( t) i ( t) n n n 2 m m ( t) os W rmsrms os W 2 Note: the instntneous ower is onstnt where rms

4 Threehse Connetions Blned three hse ower soure n n n Note: only 4 wires re required n n = rms n = 12 rms n = 24 rms n An imortnt roerty of lned oltge is n n n = n n ddition, if lods re lned n n n = n This mens tht the totl urrent of the neutrl line is zero

5 Threehse Connetions n n = rms n = 12 rms n = 12 rms n The hse sequene is sid to e in sequene or in ositie hse sequene, mening tht n lgs n y 12. n n n = rms n = 12 rms n = 12 rms n The hse sequene is sid to e in sequene or in negtie hse sequene, mening tht n leds n y 12. n

6 onnetion hs no neutrl line Current in neutrl line is zero nd n e omitted during nlysis Threehse Connetions There re two ossile onfigurtions for soures nd lods to otin lned three hse networks. n Z Z or n Wye (Y)onneted lods Z or Delt ()onneted lods Z Z Z

7 Soure/od Connetions Sine the soure nd the lod n e onneted in either Y or, there re 4 ossile tye of onnetions Four ossiilities: Soures od Y Y Y Y Blned YY onnetion n Assuming: sequene n n = rms n n n = 12 rms n = 12 rms

8 YY Connetion n ineneutrl oltge n, n nd n Phse oltge n n n Phse oltges re defined s ineline oltge, nd ine oltge How re the hse nd line oltges relted? n = rms n = 12 rms n = 12 rms ine oltges = n n = 12 n = n 6 n Similrly for the other 2 hses: n n 9 n n 21

9 YY Connetion et the mgnitude of the line oltge e denoted y Phse oltges n = rms n = 12 rms n = 24 rms n ine oltges 9 21 where n For sequene: n ine oltge = Phse oltge

10 YY Connetion The oltge for the hse of negtiesequene lned Yonneted soure is n = rms. Find ll the hse nd line oltges nd drw the hsor digrm Phse oltges ine oltges n = rms n n n = 12 rms n n 9 n = 12 rms n n 15 Definition of negtie sequene sequene n For sequene: ine oltge = Phse oltge n n

11 YY Connetion ( sequene) n n n n n Z Z Z Wht out the line urrents, nd? Z Z Z n n n Z 12 Z 12 Z This mens tht the neutrl urrent n is n = = Note: Currents he the sme mgnitude nd nd lg y 12 nd 24, resetiely Sine there is no urrent in the neutrl line, this line n e reled with ny imedne or e n oen nd short iruit

12 YY Connetion n ineneutrl oltge n, n nd n Phse oltge n ineline oltge, nd ine oltge n n n For YY onnetion the line urrent ( ) onneting the soure to the lod is equl to the hse urrent ( Y ) flowing through the imedne. = Y For lned three hse systems we only need to nlyse one hse nd use the hse sequene to otin the oltges nd urrents of the other hses.

13 Exmle 1 The oltge for the hse of n sequene lned wyeonneted soure is n =12 9 rms. Determine the line oltges for this soure Exmle 2 A threehse Yonneted lod is sulied y n sequene lned hse Yonneted soure through trnsmission line with n imedne of 1j1 er hse. The lod imedne is 8j er hse. f the lod oltge for the hse is rms (i.e. = 14.2 rms t the lod end), determine the hse oltges of the soure.

14 Conneted Soures soures n e reled y equilent Y soures For sequene: ine oltge = Phse oltge or 1 Phse oltge = ine oltge soures (sequene) = rms = 12 rms = 12 rms n n 1 n 9 n Equilent Y soures re n n n 5

15 Solution Strtegy For Blned hse Ciruits 1. f soures nd/or lods re onneted in, then onert to equilent Yonfigurtion.. Anlyze single hse nd use the hse sequene to otin the oltges nd urrents in the other hses. 5. Conert to equilent lues in the originl system.

16 Exmle For the network shown elow omute the line oltges t the lod. od rms rms.1 j.1 28 rms.1 j.1.1 j j4 j4 j4

17 n Conneted od n n n Z Z Z The hse urrent t the lod Z Δ Z Δ θ 12 Z Δ 12 Z Δ Phse oltges ( sequene) n = rms n = 12 rms n = 24 rms ine oltges 9 where 21 where Z Δ The hse urrents he the sme mgnitude, howeer nd lg y 12 nd 24, resetiely.

18 n Conneted od n n n Z Z Z The hse urrent t the lod ine urrents (KC t node, & ) The mgnitude of the hse urrent in the lod nd the line urrent is For ositie hse sequene ine urrent = Phse urrent 1 or Phse urrent = ine urrent

19 Conneted od Conerting onneted lod to Yonneted lod Z Z Z f = Z / then oth the nd Y lods re equilent Proof: rwin ook (7 th edition 424) Alterntie roof: next slide

20 Conneted od n n Z Z 1 Z For sequene: Phse oltge = 1 ine oltge Phse urrent = 1 ine urrent

21 Conneted od n Z Y n Z Z φ n Y φ θ θ φ θ φ θ Z Z 1 Z Rtio of nd Z : Z Z Y Δ 1

22 Reltionshi etween Y nd ods ine oltge Phse oltge ine urrent Phse urrent n n Yod n Z od Z 1 1 Z od medne

23 Exmle 4 An sequene threehse oltges soure onneted in lned wye sulies ower to lned deltonneted lod. The line urrents for the hse is = 12 4 A rms. Find the hse urrents in the deltonneted lod.

24 Exmle 5 A lne hse deltonneted soure with linetoline oltges of 6 olts sulies ower through lines hing n imedne of.j.4 Ohms to hse deltonneted lod, Z = (j6) Ohms, s shown. Clulte 1. The line urrents 2. The hse urrents t the lod. The linetoline oltges 6 15 rms 6 rms. j.4. j.4 Z Z Z 6 9 rms. j.4

25 Power Reltionshi θ θ For oth Y or onneted lods, the rel nd retie ower t eh lod is: P * os(θ θi ) S P jq where Q sin(θ θi ) The totl rel nd retie ower for ll three hses is P T θ i n os(θ n θ i ) Q T θ 1 Z i Z sin(θ Z θ i ) i Phse nd mgnitude of totl ower S T i S P Q T 2 T 2 T

26 Power Reltionshi θ θ n n 1 θ i i Yonneted system onneted system For oth Y or onneted lods, the rel nd retie ower t eh lod is: 1 P os(θ θi ) os(θ θi ) 1 Q sin(θ θi ) sin(θ θi ) The totl ower for ll three hses is P os(θ θ ) os(θ θ ) S T i T Q T sin(θ θ ) i i sin(θ i θ i ) θ T Z Z Z i S

27 Exmle 6 For the rolem desried in exmle 5, lulte the ower roided to the lod nd the ower lost in the lines. Exmle 7 A threehse Yonneted lod is sulied y n sequene lned hse Yonneted soure through trnsmission line with n imedne of 1j1 er hse. The lod imedne is 8j er hse. f the lod oltge for the hse is rms (i.e. = 14.2 rms t the lod end), determine the rel nd retie ower nd the omlex ower t eh soure nd eh lod. n ddition, lulte the totl omlex ower t the soures nd t the lods.

28 Exmle 8 A 48 rms line feeds two lned hse lods. f the two lods re rted s follows: od 1: 5kA t.8 f lgging od 2: 1kA t.9 f lgging determine the mgnitude of the line urrent from the 48 rms soure.

29 Power Ftor Corretion n hse system, ower ftor orretion is erformed in similr mnner s for 1hse system Three itors re required to inrese the ower ftor Citors re onneted to lod either in or Y formtion

30 Exmle 9 n lned hse system shown in elow, the line oltge is 4.5 K rms t 6 Hz. Find the lues of the itors C suh tht the totl lod hs ower ftor of.9 lgging. Blned three hse soure n C C C Blned lod 24MA.78 f lgging

31 Exmle 1 An industril lnt is to e sulied hse, 28 olts, linetoline nd will he the following lods: od Rel Power (kw) Retie Power (Krs) 1 2 lgging lgging 7) Wht is the KA of the lnt nd the ower ftor? 8) Wht is the line urrent? 9) How mny KARS of itors er hse must e dded t the serie entrne to ring the ower ftor to.9 lg? 1) Wht is the line urrent with the lnt oerting t.9 lgging ower ftor nd no hnge in the rel ower requirement?