Bài giảng Electric circuit theory - Chapter XI: Three-phase circuits - Nguyễn Công Phương

Nguy ễn Công Ph ươ ng Electric Circuit Theory Three-phase Circuits Contents I. Basic Elements Of Electrical Circuits II. Basic Laws III. Electrical Circuit Analysis IV. Circuit Theorems V. Active Circuits VI. Capacitor And Inductor VII. First Order Circuits VIII.Second Order Circuits IX. Sinusoidal Steady State Analysis X. AC Power Analysis XI. Three-phase Circuits XII. Magnetically Coupled Circuits XIII.Frequency Response XIV.The Laplace Transform XV. Two-port Networks Three-phase Circuits - sites.google.com/site/ncpdhbkhn 2 Three-phase Circuits 1. Introduction 2. Three-phase Source 3. Three-phase Load 4. Three-phase Circuit Analysis 5. Power in Three-phase Circuits Three-phase Circuits - sites.google.com/site/ncpdhbkhn 3 Introduction (1) • Polyphase: – Phase: branch, circuit or winding – Poly: many • Three-phase: three phases • Advantages: – Machine: less space & less cost – Transmission & distribution: less conducting material – Power delivered to a three-phase load is always constant – Single phase source from three-phase source Three-phase Circuits - sites.google.com/site/ncpdhbkhn 4 Three-phase Source (1) A C’ vAA’ vBB’ vCC’ B’ B t S C Stator A’ = ω vAA' V m sin t =ω − o vBB' V m sin( t 120 ) =ω + o vCC' V m sin( t 120 ) Three-phase Circuits - sites.google.com/site/ncpdhbkhn 5 Three-phase Source (2) = ω vAA' V m sin t vAA’ vBB’ vCC’ =ω − o vBB' V m sin( t 120 ) =ω + o t vCC' V m sin( t 120 ) + + = vAA' v BB ' v CC ' =+ωωo − ω o + ω o + ω o Vttm (sin sin cos120 cos t sin120 sin t cos120 cos t sin120 ) =V(sinω t + 2sin ω t cos120o ) m V ω −  CC ' =ω + ω 1  = 120 o Vm sin t 2sin t    0 2   120 o VAA ' v+ v + v = 0 120 o AA' BB ' CC ' VBB ' Three-phase Circuits - sites.google.com/site/ncpdhbkhn 6 Three-phase Source (3) = ω vAA' V m sin t vAA’ vBB’ vCC’ =ω − o vBB' V m sin( t 120 ) =ω + o t vCC' V m sin( t 120 ) + + = vAA' v BB ' v CC ' 0 Symmetrical (balanced) Three-phase Source: V ω CC ' - Same magnitude 120 o 120 o - Same frequency VAA ' o 120 o - Displaced from each other by 120 VBB ' Three-phase Circuits - sites.google.com/site/ncpdhbkhn 7 Three-phase Source (4) C A B’ C’ N A’ Load C C Star connection A’ A Load A B C’ B’ B’ Load B C B Delta connection B C’ A’ A Three-phase Circuits - sites.google.com/site/ncpdhbkhn 8 Three-phase Source (5) B’ C C A B’ C’ N A’ B C’ A’ A B A A B B C C N Three-phase Circuits - sites.google.com/site/ncpdhbkhn 9 Three-phase Source (6) iA A B’ i i vAB vCA vA C C BC iB B v C v v B BC vB i C’ i i B B CA C C vC A’ iA A vA iAB N vAB , vBC , vCA : line voltages vA, vB, vC: line voltages/phase voltages vA, vB, vC: phase voltages iAB , iBC , iCA : line currents iA, iB, iC: line currents/phase currents iA, iB, iC: phase currents Three-phase Circuits - sites.google.com/site/ncpdhbkhn 10 Three-phase Source (7) iA A V = V 0o A vAB vCA vA i = − o B B VB V 120 v BC vB o = iC VC V 120 C vC N = + = − VVVAB AN NB VVAN BN 1 3  = V 0o −V −120 o =V1 + + j  = 3V 30 o 2 2  = − o VBC 3V 90 VV= 3 30 o = − o line phase VCA 3V 210 Three-phase Circuits - sites.google.com/site/ncpdhbkhn 11 Three-phase Source (8) = o iA VA V 0 A v = − o vAB vCA A VBN V 120 i B B o = v VCN V 120 BC vB i V C C V AB vC C N V CA = o VAB 3V 30 VA = − o VBC 3V 90 VB = − o VCA 3V 210 VBC Three-phase Circuits - sites.google.com/site/ncpdhbkhn 12 Three-phase Circuits 1. Introduction 2. Three-phase Source 3. Three-phase Load 4. Three-phase Circuit Analysis 5. Power in Three-phase Circuits Three-phase Circuits - sites.google.com/site/ncpdhbkhn 13 Three-phase Load Z ZA B B A B A ZAB ZCA ZBC ZC C C Z Z ZZ+ ZZ + ZZ Z = CA AB Z = AB BC CA A + + AB ZAB Z BC Z CA ZC Z Z ZZ+ ZZ + ZZ Z = AB BC Z = AB BC CA B + + BC ZAB Z BC Z CA ZA Z Z ZZ+ ZZ + ZZ Z = BC CA Z = AB BC CA C + + CA ZAB Z BC Z CA ZB Three-phase Circuits - sites.google.com/site/ncpdhbkhn 14 Three-phase Circuits 1. Introduction 2. Three-phase Source 3. Three-phase Load 4. Three-phase Circuit Analysis 5. Power in Three-phase Circuits Three-phase Circuits - sites.google.com/site/ncpdhbkhn 15 Three-phase Circuit Analysis (1) • Y–Y, Y–∆, ∆–∆, ∆–Y • 2 kinds of three-phase circuit: balanced & unbalanced • Balanced three-phase circuit: – Balanced three-phase source and balanced three-phase load – Balanced three-phase source : same magnitude, same frequency, displaced from each other by 120 o – Balanced three-phase load : three identical loads • Unbalanced three-phase circuit: – Unbalanced three-phase source and/or unbalanced three-phase load • To solve a balanced one: – Exploit the symmetry of a balanced three-phase circuit, or – Treat it like a normal three-source circuit • To solve an unbalanced one: – Treat it like a normal three-source circuit Three-phase Circuits - sites.google.com/site/ncpdhbkhn 16 Three-phase Circuit Analysis (2) IC I Suppose VN = 0 A V c 1 1 1 1  V V V C VA ZY +++ =++A B C Z   Vn A Y ZZZZYYYN  ZZZ YYY C a N n + + = IN VA V B V C 0 ZN – V + B ZY →V = 0 →V = 0 IB n Nn B b − V 0o − 0 V 0o → =VA V n = = I A ZY ZY ZY o + + = V V ×1 −120 IA I B I C 0 I =B = A =I × 1 −120 o B A I = 0 ZY ZY N V V ×1 120 o =C = A = × o IC I A 1 120 ZY ZY Three-phase Circuits - sites.google.com/site/ncpdhbkhn 17 Three-phase Circuit Analysis (3) I V 0o C IA =VA = I A V V c Z Z C A ZY Y Y ZY C A I= I × 1 −120 o a B A N I n ZN N I= I × 1 120 o – V C A + B I ZY B B b For a balanced Y–Y system: 1. Draw a single phase equivalent circuit VA (a-phase) + I – A 2. Find the current in the A-phase ZY 3. Write down the currents in the two N n other phases Three-phase Circuits - sites.google.com/site/ncpdhbkhn 18 Ex Three-phase Circuit Analysis (4) IC IA ZY = 3 + j4 Ω; find phase currents? 220 0o V c Z For a balanced Y–Y system: Z Y A Y 1.P Draw a single phase equivalent C a o circuit (a-phase) 220 120 V N I n ZN N 2. Find the current in the A-phase – P − o 3. Write down the currents in the two 220 120 V + I ZY P B B other phases b o o 220 0 220 0 o I = = = 44 − 53.1o A 220 0 V A + ZY 3j 4 + I – A I= I × 1 −120o = 44 −53.10 − 120 o = 44 −173.1o A B A ZY = × o = −o + o = o N n IC I A 1 120 44 53.1 120 44 66.9 A Three-phase Circuits - sites.google.com/site/ncpdhbkhn 19 Three-phase Circuit Analysis • Y–Y, Y–∆, ∆–∆, ∆–Y • 2 kinds of three-phase circuit: balanced & unbalanced • Balanced three-phase circuit : – Balanced three-phase source and balanced three-phase load – Balanced three-phase source : same magnitude, same frequency, displaced from each other by 120 o – Balanced three-phase load : three identical loads • Unbalanced three-phase circuit: – Unbalanced three-phase source and/or unbalanced three-phase load • To solve a balanced one: – Exploit the symmetry of a balanced three-phase circuit, or – Treat it like a normal three-source circuit • To solve an unbalanced one: – Treat it like a normal three-source circuit Three-phase Circuits - sites.google.com/site/ncpdhbkhn 20 Three-phase Circuit Analysis (5) V =V 0o I A VC VA C A IA = − o ICA a VB V 120 C c Z I N ∆ AB = o Z∆ VC V 120 – V B I Z∆ + I BC = − B B b KVL for AabBNA: Z∆IVVAB A B − → =VVVA B = AB I AB Z∆ Z ∆ VV Line currents: =BC = AB × −o = × − o IBC 1 120I AB 1 120 - Same magnitude Z∆ Z ∆ - Same frequency VV =CA = AB × o = × o - Displaced from ICA 1 120I AB 1 120 each other by 120 o Z∆ Z ∆ Three-phase Circuits - sites.google.com/site/ncpdhbkhn 21 Three-phase Circuit Analysis (6) V o =AB = × V V IC IA IAB I AB 1 0 C A A Z∆ ICA a C c = × − o Z I IBC I AB 1 120 N ∆ AB Z∆ o – V I= I × 1 120 B I Z∆ CA AB + I BC B B b KCL for a: = − IA I AB I CA = − o Iphase I line 3 30 → = o − o IA I AB (1 0 1 120 ) IC = + − Phase currents: I AB (1 0,5j 0,866) - Same magnitude = − o I I AB 3 30 CA - Same frequency I AB - Displaced from I= I 3 −150 o B AB I each other by 120 o B = o IC I AB 3 90 I IBC A Three-phase Circuits - sites.google.com/site/ncpdhbkhn 22 Ex Three-phase Circuit Analysis (7) = o I I Z∆ = 30 + j40 Ω;V A 220 15 V . Find currents? VC VA C A A I = o CA a Method 1 Vline V phase 3 30 C c Z∆ I = × o N Z AB VAB3 V A 1 30 ∆ – V B I Z∆ =3 × 220 15o+ 30 o = 381 45o V + I BC B B b V V 381 45 o =ab = AB = = − o = − o I AB 7.62 8.1 A Iphase I line 3 30 Z∆ Z ∆ 30+ j 40 = −o − o = − o IBC 7.62 8.1 120 7.62 128.1 A = −o + o = o ICA 7.62 8.1 120 7.62 111.9 A = −o = −o × −o = − o IA I AB 3 30 7.62 8.1 3 30 13.20 38.1 A = −o = −o − o = − o IB I A 120 13.2 38.1 120 13.20 158.1 A = o = −o + o = o IC I A 120 13.2 38.1 120 13.20 81.9 A Three-phase Circuits - sites.google.com/site/ncpdhbkhn 23 Ex Three-phase Circuit Analysis (8) = o I I Z∆ = 30 + j40 Ω;V A 220 15 V . Find currents? VC VA C A A ICA a Method 2 C c Z I N ∆ AB Z∆ – V B I Z∆ + I BC B B b IC IA VC V c Z A Z Y A Y C a N I n ZN N – V + B I ZY B B b Three-phase Circuits - sites.google.com/site/ncpdhbkhn 24 Three-phase Circuit Analysis • Y–Y, Y–∆, ∆–∆, ∆–Y • 2 kinds of three-phase circuit: balanced & unbalanced • Balanced three-phase circuit : – Balanced three-phase source and balanced three-phase load – Balanced three-phase source : same magnitude, same frequency, displaced from each other by 120 o – Balanced three-phase load : three identical loads • Unbalanced three-phase circuit: – Unbalanced three-phase source and/or unbalanced three-phase load • To solve a balanced one: – Exploit the symmetry of a balanced three-phase circuit, or – Treat it like a normal three-source circuit • To solve an unbalanced one: – Treat it like a normal three-source circuit Three-phase Circuits - sites.google.com/site/ncpdhbkhn 25 Three-phase Circuit Analysis (9) = VAB I AB Z A I ICA a ∆ VCA A V Z =BC = × − o V Z∆ ∆ IBC I AB 1 120 AB Z∆ IAB I Z + ∆ C – B c BC b =VCA = × o ICA I AB 1 120 Z∆ VBC IC I = − B IA I AB I CA → = o − o = + − = − o IA I AB (1 0 1 120 ) Iab(1 0.50j 0.87) I ab 3 30 = − o IB I AB 3 150 = o IC I AB 3 90 Three-phase Circuits - sites.google.com/site/ncpdhbkhn 26 Three-phase Circuit Analysis (10) A I ICA a VCA A Method 2 Z∆ VAB Z∆ IAB I Z + ∆ C – B c BC b VBC I C IB IC IA VC V c Z A Z Y A Y C a N I n ZN N – V + B I ZY B B b Three-phase Circuits - sites.google.com/site/ncpdhbkhn 27 Three-phase Circuit Analysis • Y–Y, Y–∆, ∆–∆, ∆–Y • 2 kinds of three-phase circuit: balanced & unbalanced • Balanced three-phase circuit : – Balanced three-phase source and balanced three-phase load – Balanced three-phase source : same magnitude, same frequency, displaced from each other by 120 o – Balanced three-phase load : three identical loads • Unbalanced three-phase circuit: – Unbalanced three-phase source and/or unbalanced three-phase load • To solve a balanced one: – Exploit the symmetry of a balanced three-phase circuit, or – Treat it like a normal three-source circuit • To solve an unbalanced one: – Treat it like a normal three-source circuit Three-phase Circuits - sites.google.com/site/ncpdhbkhn 28 Three-phase Circuit Analysis (11) = o a VAB V 0 A VCA IA Z V = V −120 o Y BC VAB N = o ZY ZY VCA V 120 VBC c b IB + C – B KVL for AaNbBA: IC V V 0o − = →−=AB = ZYAIIV Z YB AB IA I B ZY ZY = × −o →−= − −o = o IB I A 1 120IA I B I A (1 1 120 )I A 3 30 V → = V − o = V − o = o I A 30 IB 150 IC 90 3ZY 3ZY 3ZY = × − o = × o I A 1 120 I A 1 120 Three-phase Circuits - sites.google.com/site/ncpdhbkhn 29 Three-phase Circuit Analysis (12) a A VCA IA ZY VAB Method 2 N ZY ZY VBC c b IB + C – B IC IC IA VC V c Z A Z Y A Y C a N I n ZN N – V + B I ZY B B b Three-phase Circuits - sites.google.com/site/ncpdhbkhn 30 Three-phase Circuit Analysis • Y–Y, Y–∆, ∆–∆, ∆–Y • 2 kinds of three-phase circuit: balanced & unbalanced • Balanced three-phase circuit: – Balanced three-phase source and balanced three-phase load – Balanced three-phase source : same magnitude, same frequency, displaced from each other by 120 o – Balanced three-phase load : three identical loads • Unbalanced three-phase circuit: – Unbalanced three-phase source and/or unbalanced three-phase load • To solve a balanced one: – Exploit the symmetry of a balanced three-phase circuit, or – Treat it like a normal three-source circuit • To solve an unbalanced one: – Treat it like a normal three-source circuit Three-phase Circuits - sites.google.com/site/ncpdhbkhn 31 Three-phase Circuit Analysis • Y–Y, Y–∆, ∆–∆, ∆–Y • 2 kinds of three-phase circuit: balanced & unbalanced • Balanced three-phase circuit: – Balanced three-phase source and balanced three-phase load – Balanced three-phase source : same magnitude, same frequency, displaced from each other by 120 o – Balanced three-phase load : three identical loads • Unbalanced three-phase circuit : – Unbalanced three-phase source and/or unbalanced three-phase load • To solve a balanced one: – Exploit the symmetry of a balanced three-phase circuit, or – Treat it like a normal three-source circuit • To solve an unbalanced one: – Treat it like a normal three-source circuit Three-phase Circuits - sites.google.com/site/ncpdhbkhn 32 Ex. 1 Three-phase Circuit Analysis (13) Solve for currents. ICc IAa SupposeV = 0 o N C 220 0 V c 220 0o 220 −120o 220 120 o + + N A − a = 20 j10 j10 Vn o n 11+ + 1 + 1 220 120 V 1+ j2 Ω InN – o j10 Ω 20j 10− j 10 1 + j 2 220 −120 V + = 57.46 −122o V B IBb 220 0o − V 220 0o − 57.46 −122 o →I = n = = 12.76 11o A Aa 20 20 220 −120 o − V 220 −120o − 57.46 −122 o I = n = = 16.26 150.7o A Bb j10 j10 220 120o − V 220 120o − 57.46 −122 o I = n = = 25.21 −161.6o A Cc − j10 − j10 V 57.46 −122 o I = n = = 25,70 174,6o A nN 1+ j 2 1+ j 2 Three-phase Circuits - sites.google.com/site/ncpdhbkhn 33 Ex. 2 Three-phase Circuit Analysis (14) Solve for currents. ICc IAa − + += 220 0o V c j10Ipurple j 10( I purple I aqua ) C A = − 220 −120o + 220 120 o N a 220 120o V n + + = – o 20Iaquaj 10( I aqua I purple ) 220 −120 V j10 Ω + I = − 220 −120o + 220 0o B Bb = = IAa I aqua 38.11A I = −19.05 + j 43.21A purple I= I =−19.05 + j 43.21A = Cc purple Iaqua 38.11A =− − =− + IBb I purple I aqua 57.16j 43.21A Three-phase Circuits - sites.google.com/site/ncpdhbkhn 34 Three-phase Circuit Analysis (15) Ex. 3 V ZL A A – o o o + = = − = Z1 VA 220 0 V;VB 220 120 V;VC 220 120 V Z VB L Z=Ω5; Z =Ω 10; Z =j 20; Ω Z =− j 30; Ω N – + Z2 L 1 2 3 B Z find currents? VC L Z – + 3 C (2ZZIZIZIVV+ ) − −=−  L1 red 1 green L blue A B −+++ −=  ZI1red( ZZZI 123 ) green ZI 3 blue 0  − −++=−  ZIZIZZIVVL red3 green(2 L 3 ) blue B C  = + Ired 13.57j 0.48 A → =− − → Igreen 6.89j 12.59 A currents...  = − Iblue 2.06j 11.02 A Three-phase Circuits - sites.google.com/site/ncpdhbkhn 35 Three-phase Circuit Analysis (16) Ex. 4 = o = −o = o VB Z VA 220 0 V;VB 220 120 V;VC 220 120 V 1 =Ω = Ω =− Ω V Z110 ; Z 2j 20 ; Z 3 j 30 ;findcurrents? A Z2 + – VC Z3 + − =− (ZZIZIV1 2 ) blue 2 green B  − ++ =−  ZIZZIV2blue( 2 3 ) green C Three-phase Circuits - sites.google.com/site/ncpdhbkhn 36 Three-phase Circuits 1. Introduction 2. Three-phase Source 3. Three-phase Load 4. Three-phase Circuit Analysis 5. Power in Three-phase Circuits Three-phase Circuits - sites.google.com/site/ncpdhbkhn 37 Power in Three-phase Circuits (1) Z = Z φ Y Z ZY A Y B = ω =ω − φ N vAN V2 sin t iA I2 sin( t ) =ω − o =ω − φ − o vBN V2 sin( t 120 ) iB I2 sin( t 120 ) Z =ω + o =ω − φ + o Y vCN V2 sin( t 120 ) iC I2 sin( t 120 ) =++= + + C ptotal pppvivivi a b c ANA BNB CNC =2VI [sinωωφ t sin( t −+− ) sin( ω t 120)sin(o ωφ t −−+ 120) o +sin(ωt + 120o )sin( ω t −+ φ 120 o )] 1 sin Asin B = [cos( A − B) − cos( A + B)] 2 → = φ ptotal 3 VI cos Three-phase Circuits - sites.google.com/site/ncpdhbkhn 38 Power in Three-phase Circuits (2) = φ ptotal 3 VI cos = φ Pp VI cos = Sp VI = φ Qp VI sin S = + = ˆ pP p jQ p VI Three-phase Circuits - sites.google.com/site/ncpdhbkhn 39 Power in Three-phase Circuits (3) Ex. A three-phase balanced Y–Y system has a phase voltage of 220 V. The total real power absorbed by the load is 2400 W, the power factor angle of the load is 20 o. Find the line current? p 2400 P =total = = 800W =V Icos 2= 0o 220 I × 0.94 p 3 3 p p p 800 →I = = 3.87 A p 0.94× 220 → = = IL I p 3.87 A Three-phase Circuits - sites.google.com/site/ncpdhbkhn 40