11.1 Balanced Three Phase Voltage Sources

Transcription

1 BAANCED THREE- PHASE CIRCUITS C.T. Pn 1 CONTENT 11.1 Blnced Thee-Phse Voltge Souces 11.2 Blnced Thee-Phse ods 11.3 Anlysis of the Wye-WyeCicuits 11.4 Anlysis of the Wye-Delt Cicuits 11.5 Powe Clcultions in Blnced Thee-Phse Cicuits 11.6 Mesuement of Avege Powe in Thee-Phse Cicuits C.T. Pn 2

2 11.1 Blnced Thee Phse Voltge Souces A sketch of thee-hse geneto 2008 Peson Eduction Intentionl Pictue fom EECTRIC CIRCUITS 8th EDITION, by Nilsson Riedel C.T. Pn Blnced Thee Phse Voltge Souces () Positive sequence, o ositive hse sequence ( b c) ( 正序, 正相序 ) hse voltge in time domin v () t = V cosωt n ( ω o ) ( ω o ) ( ωt o ) v () t = V cos t 120 bn v () t = V cos t 240 cn = V cos v n v cn n v bn b c C.T. Pn 4

3 11.1 Blnced Thee Phse Voltge Souces Phse voltge in mlitude hso domin u V u V u V n bn cn = V 0 o = V 120 = V 120 o o V n V cn line cuent = hse cuent V bn C.T. Pn Blnced Thee Phse Voltge Souces u u u V n + V bn + V cn = 0 V u cn V u n V u n V u cn V u bn V u bn C.T. Pn 6

4 11.1 Blnced Thee Phse Voltge Souces -connection b V u n V u cn c c Ciculting cuent Seldom used V u bn b line to line voltge = hse voltge u u u u u u V = V, V = V, V = V b n bc bn c cn C.T. Pn Blnced Thee Phse Voltge Souces (b) Negtive sequence o negtive hse sequence ( c b) ( 負序, 負相序 ) hse voltge in time domin v () t = V cosωt n ( ω o ) ( ω o ) v () t = V cos t+ 120 bn v () t = V cos t 120 cn v n v cn n v bn b c C.T. Pn 8

5 11.1 Blnced Thee Phse Voltge Souces Phse voltge in mlitude hso domin u V u V u V n bn cn = V 0 o = V 120 o = V 120 o V n V cn n V bn b c C.T. Pn Blnced Thee Phse Voltge Souces V bn V n V bn V cn V cn u u u V n + V bn + V cn = 0 V n Although -connection is ossible but is seldom used due to little unblnce cticlly C.T. Pn 10

6 11.2 Blnced Thee-Phse ods + I Y-connected od b V b + I b Z b I b Z u I I c Z =Z b =Z c =Z Y line cuent = hse cuent u I = I c V bc I c Z c C.T. Pn 11 b c Blnced Thee-Phse ods V b + V bc line to line voltge I I b I c Z b I b Z u I I c Z c Vb = Vn Vbn = 3V 30 Vbc = Vbn Vcn = 3V 90 V = V V = 3V 150 c cn n C.T. Pn 12

7 11.2 Blnced Thee-Phse ods Vn = V 0 Vbn = V 120 Vcn = V 120 V = 3V 30 b V c V bn V cn V n V b V bc C.T. Pn Blnced Thee-Phse ods A B C IA IB IC c I b Z A b I bc Z B I c Z C -connected od Z A =Z B =Z C =Z hse voltge = line voltge u V = V AB b C.T. Pn 14

8 A B C 11.2 Blnced Thee-Phse ods IA IB IC c I b Z A b I bc Z B I c Z C Given Vb = V 0 Vbc = V 120 V = V 120 Then hse cuent Vb Ib = = I θ Z Vbc Ibc = = I θ 120 Z V I = = I θ c c Z C.T. Pn 15 c 11.2 Blnced Thee-Phse ods A B C IA IB IC line cuent IA = Ib + Ic = I I b c = 3I θ 30 Z A Z B Similly IB = 3I θ 150 I = 3I θ + 90 c I b b I bc Z C I c θ is negtive I B C.T. Pn 16 C V u c I bc I c V u bc I C I b V u b IA

9 11.2 Blnced Thee-Phse ods Similly fo negtive sequence ( c b) Y-connected cse V line voltge mgnitude = = 3 V hse voltge mgnitude RV RV = 30 u line cuent I = hse cuent I C.T. Pn Blnced Thee-Phse ods -connected cse I line cuent mgnitude = = I hse cuent mgnitude RI RI = 30 line voltge = hse voltge 3 C.T. Pn 18

10 11.3 Anlysis of the Wye-Wye Cicuits Blnced voltge souce lus blnced lod A thee-hse Y-Y blnced system : A Z S n V n V bn Z S I O b Z o I B B Z Z N V cn Z S c Zl I C C Z C.T. Pn Anlysis of the Wye-Wye Cicuits uv uv v V n V Nn = Zs + Zl + Z uv uv v V bn V Nn I B = Zs + Zl + Z uv uv v V cn V Nn IC = Zs + Zl + Z uv v V Nn Io = Zo uuv uuv uuv uv Fom KC: IA + IB + IC Io = 0 uv uv uv V n + V bn + V cn uv 3 1 V Nn( + ) = 0 ZT ZT Zo uuuv V = 0 Nn C.T. Pn 20 The neutl wie cn be oen o shot cicuited.

11 11.3 Anlysis of the Wye-Wye Cicuits The thee-hse cicuit cn be decomosed into thee single-hse equivlent cicuits. The -hse cicuit is nomlly chosen fo clculting the line cuent nd the hse voltge in Y-Y stuctue. C.T. Pn Anlysis of the Wye-Wye Cicuits A single-hse equivlent cicuit Z s I A V n Z Once, e obtined, othe hse quntities I A cn be obtined t once fom the hse sequence infomtion. V AN C.T. Pn 22

12 11.4 Anlysis of the Wye-Delt Cicuits Blnced thee-hse Y-connected voltge lus blnced thee-hse -connected lod. Fo blnced thee-hse lod, fom Y tnsfomtion one cn obtin ZY = 1 3 Z C.T. Pn Anlysis of the Wye-Delt Cicuits A V n V bn b B Z => Z Y n Z Z Y V cn Z Z Y c C C.T. Pn 24

13 11.4 Anlysis of the Wye-Delt Cicuits The hse cuent cn be clculted fom comlex ohm s lw. A single-hse equivlent cicuit cn be dwn to find the line cuent. Z Y I = A Vn Z + Z l Y C.T. Pn Anlysis of the Wye-Delt Cicuits Othe hse, line quntities cn be obtined diectly fom the infomtion of hse-sequence. I B u I CA 30 u I BC 30 I C 30 ositive sequence u B I C u I BC 30 u I CA 30 I B 30 u B negtive sequence C.T. Pn 26

14 11.5 Powe Clcultion in Blnced Thee-Phse Cicuit (i) Use oot men squed hsos (ii) The thee-hse owe = sum of thee single hse owe C.T. Pn Powe Clcultion in Blnced Thee-Phse Cicuit Fo Y-connected Y lod with Z Y =Z θ nd ositive sequence v = 2V cos wt n P v = 2V cos( wt 120 ) bn P v = 2V cos( wt ) cn P VP in = 2IPcos( wt θ ), IP = Z i = 2I cos( wt θ 120 ) bn P i = 2I cos( wt θ ) cn P C.T. Pn 28

15 11.5 Powe Clcultion in Blnced Thee-Phse Cicuit instntneous owe t () = + + A B C = v i + v i + v i n n bn bn cn cn 1 QcosAcos B= [cos( A+ B) + cos( A B)] 2 t () = 3VI cosθ => constn t P P C.T. Pn Powe Clcultion in Blnced Thee-Phse Cicuit Comlex owe fo ech hse u * S = P + jq = V * S = P 0 ( I + jq = V 0( θ) = VI IP θ) = VI P P θ P P θ u Sb = Pb + jqb = V 120 P θ = VI P P θ Sb = P b + jqb = V Sc = I Pc + jqc = V 120 IP P θ + = VI θ 120 = VI P P θ P P θ u S = P + jq = V 120 I θ 120 = VI θ c c c P P P Comlex owe fo the thee hse u u u u = + + = θ S 3φ S S b S c 3V I P P = 3V I cosθ 3φ Q 3φ = 3V I P sin θ u S = 3V I θ P C.T. Pn 3φ 30

16 11.5 Powe Clcultion in Blnced Thee-Phse Cicuit Fo blnced Y-connected cicuits linecuent I = hsecuent = I linevoltgemgnitudev P = Q 3φ 3φ = 3VI cosθ 3VI sinθ = 3V P P C.T. Pn Powe Clcultion in Blnced Thee-Phse Cicuit Fo blnced -connected cicuits linevlotgev 3φ 3VI cosθ 3I Q3 φ = 3VI sinθ u S = 3 VI θ foy & lods f = cosθ = hsevoltgev linecuent mgnitude I P = = P P C.T. Pn 32

17 11.5 Powe Clcultion in Blnced Thee-Phse Cicuit Exmle 1 V n V cn V bn Z G Z G Z G IA IB IC u B Z u I BC Z Z u I CA C.T. Pn Powe Clcultion in Blnced Thee-Phse Cicuit V n V cn V bn Z G Z G Z G IA IB IC u B Z u I BC Z Z u I CA uv V Z Z Z n G l = V, ositviesequence = 0.2+ j0.5ω Find( ) linecuents uv uv uv ( b) V, V, V AB BC CA = 0.3+ j0.9ω () c hsecuents = j85.8ω C.T. Pn 34

18 11.5 Powe Clcultion in Blnced Thee-Phse Cicuit Solution: ()The -hse equivlent cicuit n Z G IA A N 1 Z = Z Y 3 1 Zy = ( j 85.8) = j 28.6 Ω = = = A Z + Z + Z 40+ j30 G l y Fom ositivesequencecondition v I B = = A v I = = A C.T. Pn C Powe Clcultion in Blnced Thee-Phse Cicuit u u ( b) V AB = V AN ( 3 30 ) v = Z ( 3 30 ) Y = (39.5+ j28.6) 3 30 = V uv uv uv uv V BC = V AB 120, V CA = V AB 120 C.T. Pn 36

19 11.5 Powe Clcultion in Blnced Thee-Phse Cicuit u I BC u I CA 30 u B () c hsecuents Fomtheeltionbetweenlinend hse cuent of connected lod 1 v B = 30 = A 3 I BC = B 120 ICA = B 120 C.T. Pn Powe Clcultion in Blnced Thee-Phse Cicuit Exmle 2 I A V AB = 600 V ( ms) I B I C S S = P + jq P = 480kW f = 0.8 lgging Z l = j0.025ω C.T. Pn 38

20 11.5 Powe Clcultion in Blnced Thee-Phse Cicuit I A I B I C S Find: ( ) line cuent mgnitude IA ( b) Vb u () c Comlex owe Ssulied by the voltge souce. Solution : ( ) Choose V AB s the efeence ie.., V AB = V( ms) C.T. Pn Powe Clcultion in Blnced Thee-Phse Cicuit The -hse equivlent cicuit Z Y 480kW PΦ = = 160kW 3 f = 0.8 lgging 1 cos 0.8= C.T. Pn 40

21 11.5 Powe Clcultion in Blnced Thee-Phse Cicuit Fom single hse comlex owe * 600* 600 VI = I = IA kW = IA cos IA = A I = A A o fom the fomul I 3VI cosθ = P T I cos(cos 0.8) = W = Ams ( ) C.T. Pn Powe Clcultion in Blnced Thee-Phse Cicuit (b) Fom the single hse cicuit Z Y u u V = V + ZI n AN l A 600 = 0 + ( j 0.025) = V V = V = = V( ms) C.T. Pn 42 b

22 11.5 Powe Clcultion in Blnced Thee-Phse Cicuit (c) V n 1.57 o u S = 3VI o I = = kva C.T. Pn Mesuing Avege Powe in Thee- Phse Cicuits The totl owe of genel cicuit with n teminls i 1 i 2 + v 1 _ + v 2 _ + v 3 _ i 3 n = vi + vi + + v i n 1 n 1 C.T. Pn 44

23 11.6 Mesuing Avege Powe in Thee- Phse Cicuits Fo thee-hse thee wie system, whethe blnced o not, we need only two wttmetes to mesue the totl owe. The two-wttmete wttmete method educes to detemining the mgnitude nd lgebic sign of the vege owe indicted by ech wttmete. C.T. Pn Mesuing Avege Powe in Thee- Phse Cicuits Conside the Y-connected Y blnced lod with e hse imednce Zφ = Z θ s n exmle. I B Z φ = Z θ I C C.T. Pn 46

24 11.6 Mesuing Avege Powe in Thee- Phse Cicuits I B Z φ = Z θ I C Assume idel wttmetes nd blnced thee-hse lod u * W1 : mesue Re VAB I A wtt u * W2 : mesue Re VCB I C wtt C.T. Pn Mesuing Avege Powe in Thee- Phse Cicuits u u u Assume VAN, VBN, VCN with ositive sequence I B V u CN u u u VAB = VAN VBN u u u VCB = VCN VBN u The ngle between VAB nd IA is 30 + θ u The ngle between V nd I is 30 -θ C.T. BN Pn I C 48 V u θ I C θ V u CB θ V u AN V u AB I B CB C Z φ = Z θ

25 11.6 Mesuing Avege Powe in Thee- Phse Cicuits 1 2 [ θ] [ θ] W =Re VI 30 + = VI cos(30 + θ) W =Re VI 30 = VI cos(30 θ) P = W + W = VI[cos30 cosθ sin30 sinθ i 1 2 = + cos30 cosθ + sin30 sin θ] 3VI cosθ I B Z φ = Z θ C.T. Pn I C Mesuing Avege Powe in Thee- Phse Cicuits If f = cosθ > 0.5, ie.., θ < 60 then W & W e ositive. 1 2 If f = 0.5, ie.., θ = 60 then W = 0, W = P. 1 2 If f < 0.5, ie.., θ >60 then W < 0, W > 0, P = W + W. T 1 2 T 1 2 Revese the hse sequence, the edings of W 1 nd W 2 will be intechnged. I B Z φ = Z θ C.T. Pn I C 50

26 11.6 Mesuing Avege Powe in Thee- Phse Cicuits Exmle 3 : Assume the following blnced thee-hse (Pob ) Y-connected lod is fed with ositive sequence souce nd the wttmete is idel. Find the eding of this wttmete. c.c. V u CN b c.c. Z I B I C = Z θ A B C φ C.T. Pn BC 51 N V u BN V u θ V u AN 11.6 Mesuing Avege Powe in Thee- Phse Cicuits Z φ I B I C = Z θ V u CN V u BN u The ngle between V 90 - BC nd IA is θ u W = Re[ V I ] = VI cos(90 θ) BC A Since Q = 3VI sinθ T = VI sinθ V u BC θ V u AN C.T. Pn 52

27 11.6 Mesuing Avege Powe in Thee- Phse Cicuits Z φ I B I C = Z θ V u CN V u BN V u BC θ V u AN This wttmete connection cn be used to mesue the ective owe of blnced thee-hse lod with scling fcto of 3. C.T. Pn 53 SUMMARY Objective 1 : Know the definition of blnced thee- hse cicuits s well s thei secil fetues. Objective 2 : Know the shotcut of nlyzing blnced, thee-hse wye-wye connected cicuit by solving only one single equivlent cicuit. C.T. Pn 54

28 SUMMARY Objective 3 : Know the shotcut of nlyzing blnced, thee-hse wye-delt connected cicuit. Objective 4 : Be ble to clculte the owe in ny thee-hse cicuit. Objective 5 : Be ble to mesue the vege owe in thee-hse cicuits. C.T. Pn 55 SUMMARY Chte oblems : C.T. Pn 56